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1258 cards
nid:1772046331702
IO r2
A&W
[Image Occlusion region 2]
12
lapses
3/4
users
192%
ease
nid:1772046331702
Cloze c2
Q: {{c3::image-occlusion:rect:left=.1591:top=.8923:width=.7185:height=.0742}}{{c2::image-occlusion:rect:left=.3252:top=.7428:width=.5272:height=.0923}}{{c1::image-occlusion:rect:left=.0549:top=.1782:width=.9041:height=.1203}}{{c4::image-occlusion:rect:left=.1645:top=.4824:width=.1234:height
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772046331702 |
7 | 130% | 28d | 33 |
| niklas | cid:1772209100380 |
4 | 215% | 12d | 14 |
| tomas | cid:1772090857647 |
1 | 230% | 1d | 6 |
nid:1766314094848
c1
DiskMat
A cyclic group of order \(n\) {{c1::is isomorphic to \(\lan...
9
lapses
3/4
users
210%
ease
nid:1766314094848
Cloze c1
Q: A cyclic group of order \(n\) {{c1::is isomorphic to \(\langle \mathbb{Z}_n,\oplus)\), and hence commutative.::has which useful property?}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094921 |
4 | 170% | 3d | 19 |
| lorenz | cid:1764867990841 |
4 | 170% | 1254d | 19 |
| niklas | cid:1762856074705 |
1 | 290% | 63d | 11 |
nid:1772327995541
Analysis
Wie lautet die Bernoulli Ungleichung?
9
lapses
2/4
users
192%
ease
nid:1772548090724
c1
A&W
ein perfektes Matching
8
lapses
2/4
users
162%
ease
nid:1772548090724
Cloze c1
Cloze answer: ein perfektes Matching
Q: Für alle \( k \) gilt: jeder \( k \)-reguläre bipartite Graph enthält {{c1::ein perfektes Matching}}.Theorem-name included
A: (Frobenius, 1917)Es gilt sogar: Der Graph ist die Vereinigung von perfekten Matchings.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772548090724 |
5 | 150% | 62d | 26 |
| niklas | cid:1772569386218 |
3 | 175% | 2d | 15 |
nid:1771973928570
c1
Analysis
e^a \cdot (\cos(b) + i \sin(b))
7
lapses
3/4
users
218%
ease
nid:1771973928570
Cloze c1
Cloze answer: e^a \cdot (\cos(b) + i \sin(b))
Q: Addition von komplexen Zahlen in Polarform: \(e^z = e^{a + ib} = {{c1:: e^a \cdot (\cos(b) + i \sin(b)) }}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771973928571 |
4 | 170% | 118d | 24 |
| niklas | cid:1771970299379 |
2 | 255% | 112d | 10 |
| tomas | cid:1772003104447 |
1 | 230% | 1d | 4 |
nid:1771365476565
c1
PProg
CPU state (registers, program counter)
7
lapses
3/4
users
208%
ease
nid:1771365476565
Cloze c1
Cloze answer: CPU state (registers, program counter)
Q: Process context includes:{{c1::CPU state (registers, program counter)}}{{c2::program state (stack, heap, resource handles)}}{{c3::additional management information}}.
A: A thread also has a context, but it is typically much smaller.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955105 |
4 | 170% | 13d | 17 |
| niklas | cid:1771364277648 |
2 | 225% | 13d | 10 |
| lorenz | cid:1771365476572 |
1 | 230% | 141d | 12 |
nid:1766314094781
c2
DiskMat
least upper bound
7
lapses
2/4
users
195%
ease
nid:1766314094781
Cloze c2
Cloze answer: least upper bound
Q: Consider the poset \((A;\preceq)\). If \(\{a,b\}\) has a {{c2::least upper bound}}, then it is called the {{c1::join of \(a\) and \(b\) (also denoted \(a \lor b\)).}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094832 |
4 | 170% | 9d | 14 |
| niklas | cid:1762856073631 |
3 | 220% | 16d | 11 |
nid:1774631277043
c1
A&W
Sei \(X\) eine Zufallsvariable mit Wertebereich \(W_X\subset...
7
lapses
1/4
users
130%
ease
nid:1774631277043
Cloze c1
Q: Sei \(X\) eine Zufallsvariable mit Wertebereich \(W_X\subseteq\mathbb{N}_0\).Dann gilt:\[ \mathbb{E}[X] = {{c1::\sum_{i=1}^{\infty}\Pr[X\ge i] :: \text{Schrankenform} }} \]Proof Included
A: Proof:\[ \mathbb{E}[X]=\sum_{i=0}^{\infty}i\cdot\Pr[X=i]=\sum_{i=0}^{\infty}\sum_{j=1}^{i}\Pr[X=i]=\sum_{j=1}^{\infty}\sum_{i=j}^{\infty}\Pr[X=i]=\sum_{j=1}^{\infty}\Pr[X\ge j].\quad\square \](Der Schlüsselschritt ist das Vertauschen der Summationsreihenfolge: Statt über \(i\) zu summieren und für jedes \(j\le i\) eine 1 zu zählen, wird über \(j\) summiert und alle \(i\ge j\) gezählt.)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774631277044 |
7 | 130% | 43d | 32 |
nid:1771973928521
c2
Analysis
Youngsche UngleichungFür jedes \(x, y \in \mathbb{R}\), \(\e...
7
lapses
1/4
users
130%
ease
nid:1771973928521
Cloze c2
Q: Youngsche UngleichungFür jedes \(x, y \in \mathbb{R}\), \(\epsilon > 0\) gilt: \[ {{c1:: 2|xy| }} \leq {{c2:: \epsilon x^2 + \frac{1}{\epsilon} y^2 }}\]Proof Included
A: Proof: Setze \(\gamma = \sqrt{\epsilon} > 0\). OBDA gelte \(x \cdot y \geq 0\). \[ 0 \leq (\gamma x - \frac{y}{\gamma})^2 = \gamma^2 x^2 - 2x\cdot y + \frac{1}{\gamma^2}y^2 \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771973928521 |
7 | 130% | 35d | 31 |
nid:1766314094913
c1
DiskMat
a^0
6
lapses
3/4
users
215%
ease
nid:1766314095056
DiskMat
What is the number of subgroups of \(\mathbb{Z}_n\)?
6
lapses
3/4
users
220%
ease
nid:1766314095056
Q: What is the number of subgroups of \(\mathbb{Z}_n\)?
A: The number of divisors of \(n\) (as the order of each subgroup divides the group order (which is n here) by Lagrange). If \(n\) is written \(n = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}\) then it is \(\prod_{i=1}^k (e_i+1)\).Note: This only holds because \(\mathbb{Z}_n\) is cyclic and therefore the subgroups are unique.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095224 |
3 | 190% | 4d | 13 |
| lorenz | cid:1766229407421 |
2 | 210% | 1112d | 12 |
| niklas | cid:1766000828774 |
1 | 260% | 15d | 10 |
nid:1771973928505
Analysis
Dreiecksungleichung (Subtraktion)
6
lapses
3/4
users
205%
ease
nid:1766314094777
c1
DiskMat
lower (upper) bound of \(S\)
6
lapses
2/4
users
198%
ease
nid:1766314094777
Cloze c1
Cloze answer: lower (upper) bound of \(S\)
Q: Consider the poset \((A; \preceq)\) and \( S \subseteq A\).\(a \in A\) is a {{c1::lower (upper) bound of \(S\)}} if {{c2::\(a \preceq b\) (\(a \succeq b) \) for all \(b \in S\)}}
A: Note that a is not necessarily in the subset S (difference to the least and greatest elements).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073624 |
5 | 180% | 11d | 20 |
| jonas | cid:1766314094825 |
1 | 215% | 24d | 13 |
nid:1766314094806
c2
DiskMat
commutative ring without zerodivisors (\( \forall a \ \foral...
6
lapses
2/4
users
190%
ease
nid:1766314094806
Cloze c2
Cloze answer: commutative ring without zerodivisors (\( \forall a \ \forall b \quad ab = 0 \rightarrow a = 0 \lor b = 0\) ).
Q: An {{c1::integral domain}} is a {{c2::commutative ring without zerodivisors (\( \forall a \ \forall b \quad ab = 0 \rightarrow a = 0 \lor b = 0\) ).}}
A: A domain of elements behaving like integers.Examples: \(\mathbb{Z}, \mathbb{R}\)Counterexample: \(\mathbb{Z}_m, m\) not prime
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073681 |
4 | 170% | 34d | 19 |
| jonas | cid:1766314094870 |
2 | 210% | 16d | 12 |
nid:1766314094941
DiskMat
State Corollary 5.11 about groups of prime order (what prope...
6
lapses
2/4
users
182%
ease
nid:1766314094941
Q: State Corollary 5.11 about groups of prime order (what property, what does each element satisfy). (Proof Included)
A: Corollary 5.11: Every group of prime order is cyclic, and in such a group every element except the neutral element is a generator.
Proof: Only \(1 \mid p\) and \(p \mid p\) for \(p\) prime. So for \(a \in G\), either \(\text{ord}(a) = 1\) (meaning \(a = e\)) or \(\text{ord}(a) = p\) (meaning \(a\) generates the whole group; Lagrange).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095075 |
3 | 175% | 8d | 11 |
| niklas | cid:1764859231418 |
3 | 190% | 12d | 18 |
nid:1766314094961
DiskMat
State Lemma 5.18 about the units of a ring and the property ...
6
lapses
2/4
users
190%
ease
nid:1766314094961
Q: State Lemma 5.18 about the units of a ring and the property their set satisfies? (Proof included)
A: Lemma 5.18: For a ring \(R\), \(R^*\) is a group (the multiplicative group of units of \(R\)).
Proof idea: Every element of \(R^*\) has an inverse by definition, so axiom G3 holds. The other group axioms (associativity, neutral element) are inherited from the ring.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991217 |
5 | 150% | 1569d | 23 |
| jonas | cid:1766314095101 |
1 | 230% | 4d | 7 |
nid:1772045507878
c1
A&W
Minimalgrad \(\delta(G) \geq |V|/2\)
6
lapses
2/4
users
198%
ease
nid:1772045507878
Cloze c1
Cloze answer: Minimalgrad \(\delta(G) \geq |V|/2\)
Q: Jeder Graph \(G = (V, E)\) mit \(|V| \geq 3\) und {{c1::Minimalgrad \(\delta(G) \geq |V|/2\)}} enthält {{c2::einen Hamiltonkreis}}.
A: Satz von Dirac
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772045507878 |
4 | 170% | 33d | 25 |
| niklas | cid:1772209100360 |
2 | 225% | 22d | 10 |
nid:1772046468683
c1
A&W
Das Problem „Gegeben ein Graph \(G = (V, E)\), enthält \(G\)...
6
lapses
2/4
users
205%
ease
nid:1772046468683
Cloze c1
Q: Das Problem „Gegeben ein Graph \(G = (V, E)\), enthält \(G\) einen Hamiltonkreis?" kann man in Zeit {{c1::\(O(|V|^2 \cdot 2^{|V|})\) entscheiden und, falls ja, einen solchen finden}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772046468684 |
4 | 170% | 69d | 23 |
| niklas | cid:1772209100367 |
2 | 240% | 30d | 9 |
nid:1773311192739
c1
A&W
In jedem Subgraphen gibt es einen Knoten mit Grad \(\leq k\)...
6
lapses
2/4
users
190%
ease
nid:1773311192739
Cloze c1
Cloze answer: In jedem Subgraphen gibt es einen Knoten mit Grad \(\leq k\)
Q: Heuristik:\(v_n\) := Knoten vom kleinsten Grad. Lösche \(v_n\).\(v_{n-1}\) := Knoten vom kleinsten Grad im Restgraph. Lösche \(v_{n-1}\). Iteriere.Falls \(G=(V,E)\) erfüllt:{{c1::In jedem Subgraphen gibt es ein
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773311192740 |
5 | 150% | 72d | 27 |
| niklas | cid:1773420068144 |
1 | 230% | 4d | 5 |
nid:1777984580493
c1
A&W
\(t(2n - 2) + O(n^2)\)
6
lapses
1/4
users
130%
ease
nid:1777984580493
Cloze c1
Cloze answer: \(t(2n - 2) + O(n^2)\)
Q: Reduktion Hamiltonkreis \(\to\) Long-PathFalls Long-Path für Graphen mit \(n\) Knoten in \(t(n)\) Zeit entscheidbar ist, dann lässt sich in {{c1::\(t(2n - 2) + O(n^2)\)}} Zeit entscheiden, ob ein Graph mit \(n\) Knoten einen Hamiltonkreis hat.
A: Konsequenz: Long-Path ist mindestens so schwer wie Hamiltonkreis, also ebenfalls NP-schwer.Idee: Konstruiere in \(O(n^2)\) Zeit einen Graphen \(G'\) mit \(n' \leq 2n - 2\) Knoten, sodass \(G\) einen Hamiltonkreis hat \(\Leftrightarrow\) \(G'\) einen Pfad der Länge \(n\) hat.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984580493 |
6 | 130% | 5d | 23 |
nid:1779193767077
c1
A&W
Sei \(G = (V, E)\), \(n := |V|\). Wird \(e\) gleichverteilt ...
6
lapses
1/4
users
130%
ease
nid:1779193767077
Cloze c1
Q: Sei \(G = (V, E)\), \(n := |V|\). Wird \(e\) gleichverteilt zufällig aus \(E\) gezogen, so gilt\[\Pr[\mu(G) = \mu(G/e)] \;\geq\; {{c1::1 - \tfrac{2}{n} }}.\]
A: Beweis-Skizze: Sei \(C\) min-Schnitt, \(k := |C| = \mu(G)\). Aus \(\deg(v) \geq k\) für alle \(v\) folgt \(|E| \geq kn/2\), also \[\Pr[e \notin C] = 1 - \tfrac{|C|}{|E|} \geq 1 - \tfrac{k}{kn/2} = 1 - \tfrac{2}{n}.\] Wegen \(e \notin C \Rightarrow \mu(G/e) = \mu(G)\) folgt die Behauptung.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779193767077 |
6 | 130% | 8d | 26 |
nid:1776332605880
c1
A&W
Seien \(\delta, \varepsilon > 0\). Falls \({{c1::N \geq 3\,\...
6
lapses
1/4
users
130%
ease
nid:1776332605880
Cloze c1
Q: Seien \(\delta, \varepsilon > 0\). Falls \({{c1::N \geq 3\,\frac{|U|}{|S|} \cdot \frac{1}{\varepsilon^2} \cdot \ln(\tfrac{2}{\delta})}}\), ist die Ausgabe \(Y\) von Target-Shooting mit Wahrscheinlichkeit mindestens \(1 - \delta\) im Intervall \[{
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1776332605882 |
6 | 130% | 30d | 28 |
nid:1772547552647
c2
A&W
State of the Art Matching:\( O({{c1::|E|^{1+o(1)} }}) \) für...
6
lapses
1/4
users
130%
ease
nid:1772547552647
Cloze c2
Q: State of the Art Matching:\( O({{c1::|E|^{1+o(1)} }}) \) für bipartite Graphen \( O({{c2::|V|^{1/2} \cdot |E|}}) \) für generelle Graphen (Hopcroft-Karp)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772547552649 |
6 | 130% | 67d | 28 |
nid:1774917595188
c1
Analysis
Unstetigkeitsstelle
6
lapses
1/4
users
130%
ease
nid:1774917595188
Cloze c1
Cloze answer: Unstetigkeitsstelle
Q: Dieser Graph hat eine {{c1::Unstetigkeitsstelle}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774917595188 |
6 | 130% | 4d | 29 |
nid:1774487166317
c1
Analysis
Die Riemansche-Zeta Funktion Reihe \(\displaystyle\zeta(s) =...
6
lapses
1/4
users
130%
ease
nid:1774487166317
Cloze c1
Q: Die Riemansche-Zeta Funktion Reihe \(\displaystyle\zeta(s) = {{c1:: \sum_{n=1}^\infty \frac{1}{n^s} }}\) konvergiert für {{c2::\(s > 1\)}} und divergiert für {{c2::\(s\leq1\)}}.
A: Oft als Referenzreihe im Vergleichssatz nützlich (wenn Wurzel/Quotient versagen).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487166318 |
6 | 130% | 8d | 30 |
nid:1766314094728
DiskMat
Why is Bézout's identity useful for finding modular inverses...
5
lapses
3/4
users
227%
ease
nid:1766314094728
Q: Why is Bézout's identity useful for finding modular inverses?
A: If \(\text{gcd}(a, m) = 1\), then \(ua + vm = 1\) for some \(u, v\). This means \(ua = 1 - vm\), so \(ua \equiv_m 1\), making \(u\) the multiplicative inverse of \(a\) modulo \(m\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990469 |
3 | 190% | 1180d | 19 |
| jonas | cid:1766314094750 |
1 | 230% | 20d | 8 |
| niklas | cid:1762106939359 |
1 | 260% | 50d | 8 |
nid:1766314094729
DiskMat
Why does \(ax \equiv_m 1\) have no solution when \(\text{gcd...
5
lapses
3/4
users
227%
ease
nid:1766314094729
Q: Why does \(ax \equiv_m 1\) have no solution when \(\text{gcd}(a, m) = d > 1\)?
A: We can rewrite \(ax \equiv_m 1\) as \(ax - 1 = km \Leftrightarrow ax - km = 1\). Now since, \(d \mid a\) and \(d \mid m\), then \(d \mid ax\) and \(d \mid km\) for any \(x\).Thus \(d \mid (ax - km)\), and \(ax - km = 1\).But \(d \nmid 1 \implies d \nmid (ax - km)\), which is a contradiction. Thus \(ax\) can never be congruent to \(1\) modulo \(m\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094751 |
2 | 195% | 8d | 12 |
| niklas | cid:1762106939361 |
2 | 255% | 6d | 11 |
| lorenz | cid:1764867990472 |
1 | 230% | 1512d | 11 |
nid:1766314094970
DiskMat
State Lemma 5.20 about division in integral domains: (The qu...
5
lapses
3/4
users
232%
ease
nid:1766314094970
Q: State Lemma 5.20 about division in integral domains: (The quotient has what property?)
A: Lemma 5.20: In an integral domain, if \(a \mid b\) (i.e., \(b = ac\) for some \(c\)), then \(c\) is unique and is denoted by \(c = b/a\) (the quotient).
Explanation: If \(b = ac_1\) and \(b = ac_2\), then \(a(c_1 - c_2) = 0\). Since \(a \neq 0\) in an integral domain, we must have \(c_1 - c_2 = 0\)\(\implies c_1 = c_2\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095113 |
2 | 210% | 3d | 10 |
| niklas | cid:1764859231486 |
2 | 255% | 4d | 14 |
| lorenz | cid:1764867991246 |
1 | 230% | 1601d | 12 |
nid:1771526451947
c1
A&W
Formale Definition der low-Werte:\(low[v] = {{c1::\min \left...
5
lapses
3/4
users
242%
ease
nid:1771526451947
Cloze c1
Q: Formale Definition der low-Werte:\(low[v] = {{c1::\min \left( dfs[v], \min_{(v,w) \in E} \begin{cases} dfs[w], & \text{falls } (v,w) \text{ Restkante} \\ low[w], & \text{falls } (v,w) \text{ Baumkante} \end{cases} \right)}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771535790938 |
3 | 265% | 11d | 15 |
| lorenz | cid:1771526451947 |
1 | 230% | 72d | 9 |
| tomas | cid:1771530245016 |
1 | 230% | 2d | 8 |
nid:1766314095018
DiskMat
What property does every finite field \(\text{GF}(q)\) have ...
5
lapses
2/4
users
178%
ease
nid:1766314095018
Q: What property does every finite field \(\text{GF}(q)\) have (and what does \(q\) satisfy)?
A: Theorem 5.40: The multiplicative group of every finite field \(\text{GF}(q)\) is cyclic (as \(q\) is a power of a prime, if \(\text{GF}(q)\) is cyclic).
This group has order \(q - 1\) and \(\varphi(q-1)\) generators.Note that even though q is not prime thus not every integer is coprime, GF(q) is not Z_q.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095172 |
3 | 145% | 4d | 17 |
| lorenz | cid:1764867991401 |
2 | 210% | 1009d | 17 |
nid:1766940295685
c2
DiskMat
all the atomic formulas
5
lapses
2/4
users
215%
ease
nid:1767089600366
DiskMat
Number of subgroups of \(\langle \mathbb{Z}_m \times \mathbb...
5
lapses
2/4
users
200%
ease
nid:1764867989741
c2
A&D
equivalence class of the relation defined as follows: \(u = ...
5
lapses
2/4
users
200%
ease
nid:1764867989741
Cloze c2
Cloze answer: equivalence class of the relation defined as follows: \(u = v\) if \(u\) reaches \(v\)
Q: A {{c1::connected component}} of \(G\) is a {{c2::equivalence class of the relation defined as follows: \(u = v\) if \(u\) reaches \(v\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1763363268619 |
3 | 190% | 2d | 14 |
| lorenz | cid:1764867989742 |
2 | 210% | 1049d | 12 |
nid:1764867990714
c2
DiskMat
A {{c1::field (Körper)}} is {{c2::a nontrivial commutative r...
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Cloze c2
Q: A {{c1::field (Körper)}} is {{c2::a nontrivial commutative ring \(F\) in which every nonzero element is a unit, so \(F^* = F \backslash \{0\}\)}}
A: Example: \(\mathbb{R}\), but not \(\mathbb{Z}\)Non-trivial: {0} is not a field. In particular, 1 = 0 (neutral element of mult. = neutral element of add.) causes trouble.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073684 |
4 | 185% | 3d | 17 |
| lorenz | cid:1764867990715 |
1 | 230% | 1907d | 9 |
nid:1771361604906
c1
A&W
Jeder \(u\)-\(v\)-Knotenseparator hat Grösse mindestens \(k ...
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nid:1771361604906
Cloze c1
Cloze answer: Jeder \(u\)-\(v\)-Knotenseparator hat Grösse mindestens \(k \); Jeder \(u\)-\(v\)-Kantenseparator hat Grösse mindestens \(k\)
Q: Sei \(G = (V, E)\) ein Graph und \(u, v \in V, u \neq v\). Dann gilt:
{{c1::Jeder \(u\)-\(v\)-Knotenseparator hat Grösse mindestens \(k \)}}\(\iff\){{c2::Es gibt mindestens \(k\) intern-knotendisjunkte \(u\)-\(v\)-Pfade.}}{{c1::Jeder \(u\)-\(v\)-Kantenseparator hat Grösse mindes
A: Satz von Karl Menger (Sohn vom sehr baseden Carl Menger)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771366536202 |
3 | 250% | 16d | 22 |
| lorenz | cid:1771361604907 |
2 | 210% | 70d | 16 |
nid:1771973928585
Analysis
Archimedisches Prinzip
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c1
Analysis
\[ \cos\!\left(\frac{\pi}{3}\right) = {{c1::\frac{1}{2} }} \...
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nid:1766580142830
A&D
Explain how union works in the optimised Union-Find:
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Q: Explain how union works in the optimised Union-Find:
A: Arrays:rep, where rep[v] gives the representative of \(v\).members, where members[rep[v]] which contains all members of the ZHK of \(v\)rank, where rank[rep[v]] contains the size of the ZHK of \(v\).We always merge the smaller ZHK into the bigger to minimise updates.We update the reps, then the member
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|---|---|---|---|---|---|
| lorenz | cid:1766580142830 |
5 | 150% | 685d | 19 |
nid:1777540083514
c2
A&W
Im Beweis von \(\mathbb{E}[T_{1,n}] \leq 2(n+1) \ln n + O(n)...
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nid:1777540083514
Cloze c2
Q: Im Beweis von \(\mathbb{E}[T_{1,n}] \leq 2(n+1) \ln n + O(n)\) für QuickSort beobachtet man, dass \(\mathbb{E}[T_{\ell, r}]\) nur von {{c1::\(r - \ell + 1\)}} abhängt, also nur von der Anzahl zu sortierender Elemente.Dies motiviert die rekursive Definition von Zahlen \(t_n\) m
A: Herleitung: \(\mathbb{E}[T_{\ell, r}] = \sum_{i=\ell}^{r} \Pr[t = i] \cdot (r - \ell + \mathbb{E}[T_{\ell, i-1}] + \mathbb{E}[T_{i+1, r}])\) (Linearität des Erwartungswerts + Satz über bedingten Erwartungswert), wobei \(t\) auf \(\{\ell, \ldots, r\}\) gleichverteilt ist (paarweise verschiedene Elemente). Daraus folgt durch Induktion über \(r - \ell\), dass \(\mathbb{E}[T_{\ell, r}] = t_{r - \ell + 1}\).
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|---|---|---|---|---|---|
| lorenz | cid:1777540083514 |
5 | 150% | 16d | 22 |
nid:1777383738516
c1
Analysis
Taylorreihe des natürlichen Logarithmus (konvergiert nur für...
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Cloze c1
Q: Taylorreihe des natürlichen Logarithmus (konvergiert nur für \(-1 < x \leq 1\)):\[ \ln(1 + x) = {{c1::\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} = x - \tfrac{1}{2} x^2 + \tfrac{1}{3} x^3 - \tfrac{1}{4} x^4 + \dots }}\]
A: Bei \(x = 1\) erhält man die alternierende harmonische Reihe mit Wert \(\ln 2\); bei \(x = -1\) divergiert die Reihe (harmonische Reihe).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777383738516 |
5 | 150% | 40d | 19 |
nid:1772928333395
c1
Analysis
\[ \cos\!\left(\frac{7\pi}{4}\right) = {{c1::\frac{\sqrt{2} ...
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nid:1772928333395
Cloze c1
Q: \[ \cos\!\left(\frac{7\pi}{4}\right) = {{c1::\frac{\sqrt{2} }{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333395 |
5 | 150% | 111d | 23 |
nid:1764859231444
DiskMat
State Fermat's Little Theorem (Corollary 5.14) (both totient...
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nid:1764859231444
Q: State Fermat's Little Theorem (Corollary 5.14) (both totient and prime):
A: Corollary 5.14 (Fermat's Little Theorem): For all \(m \geq 2\) and all \(a\) with \(\gcd(a, m) = 1\): \[a^{\varphi(m)} \equiv_m 1\]
In particular, for every prime \(p\) and every \(a\) not divisible by \(p\): \[a^{p-1} \equiv_p 1\]
Proof: This follows from Corollary 5.10 (\(a^{|G|} = e\)). Since \(\gcd(a, m)=1\), it is an element of \(\mathbb{Z}_m^*\) and thus an element of a group. \(\langle a \rangle\) there
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764859231445 |
5 | 165% | 4d | 17 |
nid:1771366536192
c2
A&W
\(|V| \geq k + 1\) und für alle Teilmengen \(X \subseteq V\)...
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nid:1771366536192
Cloze c2
Cloze answer: \(|V| \geq k + 1\) und für alle Teilmengen \(X \subseteq V\) mit \(|X| < k\) gilt: Der Graph \(G[V \setminus X]\) ist zusammenhängend
Q: Ein Graph \(G = (V, E)\) heisst {{c1::\(k\)-zusammenhängend}}, falls {{c2::\(|V| \geq k + 1\) und für alle Teilmengen \(X \subseteq V\) mit \(|X| < k\) gilt: Der Graph \(G[V \setminus X]\) ist zusammenhängend}}.
A: Man muss mindestens \(k\)-Knoten (und die inzidenten Kanten) löschen, um den Zusammenhang zu zerstören.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771366536201 |
5 | 210% | 13d | 23 |
nid:1766314094712
DiskMat
State Bézout's identity (Corollary 4.5).
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nid:1766314094712
Q: State Bézout's identity (Corollary 4.5).
A: For \(a, b \in \mathbb{Z}\) (not both 0), there exist \(u, v \in \mathbb{Z}\) such that:
\[\text{gcd}(a, b) = ua + vb\]
The GCD can be expressed as an integer linear combination.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990414 |
2 | 210% | 1958d | 12 |
| jonas | cid:1766314094733 |
1 | 230% | 14d | 8 |
| niklas | cid:1762106939325 |
1 | 290% | 119d | 10 |
nid:1766314094927
DiskMat
Which elements generate \(\mathbb{Z}_n\)? How can this be pr...
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nid:1766314094927
Q: Which elements generate \(\mathbb{Z}_n\)? How can this be proven?
A: \(\mathbb{Z}_n\) is generated by all \(a \in \mathbb{Z}_n\) for which \(\gcd(a, n) = 1\) (all elements coprime to \(n\)).
Proof:\(a\) generator \(\implies\)\(\gcd(a, n) = 1\)\(\mathbb{Z}_n = \langle a \rangle\)\(\implies\)\(1 \in \langle a \rangle\)\(\implies\)\(a^u = au \equiv_n 1\) for some \(u\)\(\implies\)\(\gcd(a, n) = 1\) (\(\gcd\) must divide both \(au-qn\) and 1).\(\gcd(a, n) = 1 \implies
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| niklas | cid:1764859231381 |
2 | 255% | 4d | 14 |
| jonas | cid:1766314095054 |
1 | 215% | 7d | 10 |
| lorenz | cid:1764867991106 |
1 | 230% | 2117d | 12 |
nid:1771973928567
c1
Analysis
Eulersche Formel:\[ \cos(t) = {{c1:: \frac{e^{it} + e^{-it} ...
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nid:1766314094853
c1
DiskMat
the order of \(1\) in the additive group if it is finite, an...
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nid:1766314094853
Cloze c1
Cloze answer: the order of \(1\) in the additive group if it is finite, and 0 if it is infinite.
Q: The characteristic of a ring is {{c1::the order of \(1\) in the additive group if it is finite, and 0 if it is infinite.}}
A: Example: the characteristic of \(\langle \mathbb{Z}_m;\oplus,\ominus,0,\odot,1\rangle\)is \(m\).
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|---|---|---|---|---|---|
| jonas | cid:1766314094926 |
3 | 190% | 8d | 13 |
| niklas | cid:1762856074719 |
1 | 275% | 10d | 10 |
nid:1767089604935
c1
LinAlg
\(x = 0\) is the only vector for which \(Ax = 0\)
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nid:1767089604935
Cloze c1
Cloze answer: \(x = 0\) is the only vector for which \(Ax = 0\)
Q: The columns of \(A\) are independent if and only if {{c1::\(x = 0\) is the only vector for which \(Ax = 0\)::Linear combination view}}.
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|---|---|---|---|---|---|
| jonas | cid:1767089604936 |
2 | 210% | 4d | 10 |
| lorenz | cid:1767105283315 |
2 | 210% | 1726d | 16 |
nid:1765372936281
c2
A&D
{{c1:: \(\sum_{i = 1}^{n} i^3\)::Sum}} \(=\) {{c2::\(\frac{...
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nid:1766580143526
A&D
Kruskal's Algorithm
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nid:1766580143526
Q: Kruskal's Algorithm
A: \(O(|E| \log |E| + |V| \log |V|)\)Outer loop: Iterate \(|E|\) times at most:Inner loop: find and union take \(O(\log |V|)\) per call amortised, thus \(O(|V| \log |V|)\) total.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766580143526 |
3 | 190% | 1130d | 16 |
| niklas | cid:1766568909602 |
1 | 245% | 19d | 5 |
nid:1764867991445
DiskMat
What is the minimum distance of two codewords in a polynomia...
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nid:1766448533056
c1
DiskMat
a field.
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DiskMat
What is the modus ponens logical rule?
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nid:1764867990087
DiskMat
When is a relation \(\rho\) on set \(A\) irreflexive?
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nid:1764867990087
Q: When is a relation \(\rho\) on set \(A\) irreflexive?
A: When \(a \ \not\rho \ a\) is true for all \(a \in A\), i.e., \(\rho \cap \text{id} = \emptyset\).Note that irreflexive is NOT the negation of reflexive!
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477382 |
3 | 190% | 23d | 17 |
| lorenz | cid:1764867990087 |
1 | 230% | 1377d | 9 |
nid:1765553400194
LinAlg
What is the 1-norm of a vector?
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nid:1768182518186
c1
LinAlg
at the same indices; rank
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nid:1768182518186
Cloze c1
Cloze answer: at the same indices; rank
Q: For \(A\) and \(MA\) (\(M\) invertible) they have:the independent columns {{c1:: at the same indices}}the same {{c1::rank}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182518186 |
3 | 190% | 1601d | 15 |
| niklas | cid:1768139535123 |
1 | 245% | 21d | 7 |
nid:1768182518580
LinAlg
Prove that the row space of \(A\) and \(MA\) is the same for...
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nid:1768182518580
Q: Prove that the row space of \(A\) and \(MA\) is the same for \(M\) invertible!
A: \(\textbf{R}(A) = \textbf{C}(A^\top) \overset{!}{=} \textbf{C}(A^\top M^\top) = \textbf{C}((MA)^\top) = \textbf{R}(MA)\)where ! holds because:
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182518580 |
3 | 190% | 1633d | 17 |
| niklas | cid:1768148472221 |
1 | 230% | 2d | 4 |
nid:1772547451587
c1
A&W
|V| \cdot |E|
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nid:1772046331702
IO r3
A&W
[Image Occlusion region 3]
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nid:1772046331702
Cloze c3
Q: {{c3::image-occlusion:rect:left=.1591:top=.8923:width=.7185:height=.0742}}{{c2::image-occlusion:rect:left=.3252:top=.7428:width=.5272:height=.0923}}{{c1::image-occlusion:rect:left=.0549:top=.1782:width=.9041:height=.1203}}{{c4::image-occlusion:rect:left=.1645:top=.4824:width=.1234:height
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772046331703 |
2 | 210% | 63d | 15 |
| niklas | cid:1772209100381 |
2 | 210% | 1d | 9 |
nid:1772046170351
A&W
Was ist die Laufzeit von Hamiltonkreise mit DP?
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nid:1772546471834
c1
A&W
gibt es mindestens einen Pfad mit Start- und Endkante in \( ...
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nid:1772546471834
Cloze c1
Cloze answer: gibt es mindestens einen Pfad mit Start- und Endkante in \( M' \)
Q: Seien \( M \), \( M' \) beliebige Matchings.Betrachte den Teilgraphen mit Kantenmenge \( M \oplus M' \).Falls \( |M| < |M'| \), so {{c1::gibt es mindestens einen Pfad mit Start- und Endkante in \( M' \)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772546471834 |
3 | 190% | 32d | 21 |
| niklas | cid:1772569386186 |
1 | 260% | 20d | 8 |
nid:1772545721385
c1
A&W
Für jede Kante in \( M_{\text{max}} \) gilt: {{c1::Mindesten...
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nid:1772545721385
Cloze c1
Q: Für jede Kante in \( M_{\text{max}} \) gilt: {{c1::Mindestens einer der beiden Endpunkte wird von einer Kante aus \( M_{\text{Greedy} } \) überdeckt}}
A: (Denn sonst könnten wir die Kante zu \( M_{\text{Greedy}} \) hinzufügen.)
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|---|---|---|---|---|---|
| niklas | cid:1772569386195 |
3 | 235% | 5d | 14 |
| lorenz | cid:1772545721385 |
1 | 230% | 72d | 9 |
nid:1772045795752
A&W
Wie kann man mit der Siebformel die Zahl der Hamiltonkreise ...
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nid:1773310950541
c1
A&W
|V|/2
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nid:1773310950541
Cloze c1
Cloze answer: |V|/2
Q: Es gibt bipartite Graphen und eine Reihenfolge \(V = \{v_1, \ldots, v_n\}\) der Knoten, für die der Greedy-Algorithmus \({{c1::|V|/2}}\) viele Farben benötigt.
A: Vollständig bipartiter Graph ohne ein perfektes Matching
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|---|---|---|---|---|---|
| lorenz | cid:1773310950541 |
3 | 190% | 82d | 19 |
| niklas | cid:1773420068142 |
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nid:1771973928521
c1
Analysis
2|xy|
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nid:1771973928521
Cloze c1
Cloze answer: 2|xy|
Q: Youngsche UngleichungFür jedes \(x, y \in \mathbb{R}\), \(\epsilon > 0\) gilt: \[ {{c1:: 2|xy| }} \leq {{c2:: \epsilon x^2 + \frac{1}{\epsilon} y^2 }}\]Proof Included
A: Proof: Setze \(\gamma = \sqrt{\epsilon} > 0\). OBDA gelte \(x \cdot y \geq 0\). \[ 0 \leq (\gamma x - \frac{y}{\gamma})^2 = \gamma^2 x^2 - 2x\cdot y + \frac{1}{\gamma^2}y^2 \]
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|---|---|---|---|---|---|
| lorenz | cid:1771973928522 |
3 | 190% | 17d | 20 |
| niklas | cid:1771969257001 |
1 | 275% | 140d | 9 |
nid:1772496585520
c1
Analysis
Es sei \((a_n)_{n \in \mathbb{N}_0}\) eine Folge in \(\mathb...
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users
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nid:1772496585520
Cloze c1
Q: Es sei \((a_n)_{n \in \mathbb{N}_0}\) eine Folge in \(\mathbb{R}\).Eine Teilfolge ist eine Folge der Form \(({a_n}_k)_{k \in \mathbb{N}_0}\) wobei \((n_k)_{k \in \mathbb{N}_0}\) eine {{c1:: Folge nicht-negativer ganze
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|---|---|---|---|---|---|
| lorenz | cid:1772496585520 |
2 | 210% | 82d | 13 |
| niklas | cid:1772520282865 |
2 | 270% | 205d | 13 |
nid:1772928333487
c1
Analysis
\[ \tan\!\left(\frac{\pi}{3}\right) = {{c1::\sqrt{3} }} \]
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nid:1771973928588
c3
Analysis
Beweis: Für alle \(a < b\) in \(\mathbb{R}\) existiert ein \...
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nid:1771973928588
Cloze c3
Q: Beweis: Für alle \(a < b\) in \(\mathbb{R}\) existiert ein \(\mathbb{Q}\) mit \(a < q < b\){{c1:: Wähle nach Archimedischem Prinzip \(n \in \mathbb{N}\) so dass \(\frac{1}{n} < b - a\).}}{{c2:: \(\frac{m}{n} \mid m \in \mathbb{Z}\) diese
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|---|---|---|---|---|---|
| niklas | cid:1771969342906 |
3 | 220% | 69d | 10 |
| lorenz | cid:1771973928589 |
1 | 230% | 104d | 8 |
nid:1774138447333
c2
Analysis
Eine Folge {{c1::konvergiert}} \(\Longleftrightarrow\) Sie i...
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nid:1774138447333
Cloze c2
Q: Eine Folge {{c1::konvergiert}} \(\Longleftrightarrow\) Sie ist {{c2:: eine Cauchy-Folge (für Folgen in \(\mathbb{R}\) und \(\mathbb{C}\))}}.
A: Dies gilt nicht für Folgen in \(\mathbb{Q}\), da sie zum Beispiel auf \(\sqrt{2}\) konvergieren können, was jedoch nicht in \(\mathbb{Q}\) liegt -> ergo konvergiert nie.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1774006743987 |
3 | 190% | 2d | 13 |
| lorenz | cid:1774138447335 |
1 | 230% | 67d | 9 |
nid:1762856073654
c1
A&D
closed walk without repeated vertices; at least three vertic...
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nid:1762856073654
Cloze c1
Cloze answer: closed walk without repeated vertices; at least three vertices
Q: In graph theory, a {{c2::cycle (Kreis)}} is a {{c1::closed walk without repeated vertices}} and {{c1::at least three vertices}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073667 |
2 | 210% | 121d | 16 |
| tomas | cid:1765551666552 |
2 | 210% | 50d | 12 |
nid:1766580144345
c1
A&D
the values of the vertices in the priority queue (see line d...
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users
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nid:1766580144345
Cloze c1
Cloze answer: the values of the vertices in the priority queue (see line decrease_key(H, v, d[v]))
Q: Prim's Algorithm Invariants:The distances "d[.] = " in the distance array are {{c1::the values of the vertices in the priority queue (see line decrease_key(H, v, d[v]))}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766580144345 |
4 | 170% | 701d | 18 |
nid:1764867991302
DiskMat
How do you perform polynomial division when the divisor is n...
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nid:1764867991302
Q: How do you perform polynomial division when the divisor is not monic (e.g., in \(\text{GF}(7)[x]\))?
A: If dividing by a non-monic polynomial like \(4x + 2\) in \(\text{GF}(7)[x]\):
Find the multiplicative inverse of the leading coefficient in the field
For \(4\) in \(\text{GF}(7)\): \(4 \cdot 2 \equiv_7 1\), so \(4^{-1} = 2\)
Multiply the polynomial by this inverse to make it monic
\(2 \cdot (4x + 2) = 8x + 4 \equiv_7 x + 4\)
Now divide by the monic polynomial
Example: \(3x^2 + 6x + 5\) divided by \(4x + 2\) become
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991302 |
4 | 170% | 1361d | 17 |
nid:1771526288993
c3
A&W
ggf update, wenn Algorithmus während des backtracks zum Knot...
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Cloze c3
Cloze answer: ggf update, wenn Algorithmus während des backtracks zum Knoten zurückkehrt
Q: Berechnung der low-Werte:{{c1::Initialisierung mit dfs-Wert}}{{c2::ggf update, wenn Restkanten gefunden werden}}{{c3::ggf update, wenn Algorithmus während des backtracks zum Knoten zurückkehrt}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771526288995 |
4 | 170% | 20d | 22 |
nid:1774487164708
c1
A&W
Die Anzahl der geordneten Auswahlen von \(k\) aus \(n\) Obje...
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Cloze c1
Q: Die Anzahl der geordneten Auswahlen von \(k\) aus \(n\) Objektenohne Zurücklegen (Reihenfolge wichtig) ist:\[P(n, k) = {{c1::\frac{n!}{(n-k)!} = n \cdot (n-1) \cdots (n-k+1) }}\]
A: Beispiel: Wie viele 3-stellige PINs aus den Ziffern 0–9 ohne Wiederholung?\(P(10,3) = 10 \cdot 9 \cdot 8 = 720\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
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| lorenz | cid:1774487164709 |
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nid:1777984580504
A&W
Reduktion Hamiltonkreis \(\to\) Long-Path: KonstruktionWie k...
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Q: Reduktion Hamiltonkreis \(\to\) Long-Path: KonstruktionWie konstruiert man aus \(G = (V, E)\) den Graphen \(G'\), sodass \(G\) genau dann einen Hamiltonkreis hat, wenn \(G'\) einen Pfad der Länge \(n\) besitzt?
A: Sei \(v^* \in V\) ein beliebig gewählter Knoten.Konstruiere \(G'\) wie folgt:Ersetze \(v^*\) durch zwei Kopien \(v_1^*\) und \(v_2^*\); jede Kopie erhält dieselben Nachbarn wie \(v^*\) in \(G\).Hänge an \(v_1^*\) und \(v_2^*\) je einen neuen Knoten \(w_1\) bzw. \(w_2\) an (jeweils Grad 1).Dann hat \(G'\) genau \(|V| + 1 = n + 1 \leq 2n - 2\) Knoten (für \(n \geq 3\)) und kann in \(O(n^2)\) Zeit gebaut werden.\(\Rightarrow\): Aus einem Hamiltonkreis i
| User | Card ID | Lapses | Ease | Interval | Reviews |
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| lorenz | cid:1777984580505 |
4 | 170% | 3d | 18 |
nid:1779193767139
A&W
Schreibe den Karger-Stein-Algorithmus als rekursiven Pseudoc...
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Q: Schreibe den Karger-Stein-Algorithmus als rekursiven Pseudocode.
A: Die Wahl \(t \approx n/\sqrt{2}\) garantiert, dass die Erfolgswkt. eines einzelnen Asts (\(n \to t\) Knoten) konstant ist; durch zwei unabhängige rekursive Aufrufe wird die Misserfolgswkt. quadriert, was den Gesamtfehler beherrschbar hält. Resultierende Laufzeit: \(\mathcal{O}(n^2 \log n)\), Erfolgswkt. \(\Omega(1/\log n)\); nach \(\Theta(\log^2 n)\) Wiederholungen Fehlerwkt. \(\leq 1/n\).
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| lorenz | cid:1779193767139 |
4 | 170% | 3d | 20 |
nid:1774487164866
c1
A&W
\Pr[A] \cdot \Pr[B \mid A] = \Pr[B] \cdot \Pr[A \mid B]
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Cloze c1
Cloze answer: \Pr[A] \cdot \Pr[B \mid A] = \Pr[B] \cdot \Pr[A \mid B]
Q: Für Ereignisse \(A, B\) mit \(\Pr[A], \Pr[B] > 0\) gilt:\[\Pr[A \cap B] = {{c1::\Pr[A] \cdot \Pr[B \mid A] = \Pr[B] \cdot \Pr[A \mid B]}}\]
A: Umgestellt ergibt sich direkt der Satz von Bayes.Beide Seiten sind gleich, weil \(\Pr[A \cap B]\) symmetrisch in \(A\) und \(B\) ist.
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| lorenz | cid:1774487164866 |
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nid:1780223730631
c3
A&W
\(O((n \log n + m)\log(n/\delta))\)
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Cloze c3
Cloze answer: \(O((n \log n + m)\log(n/\delta))\)
Q: ReachabilityCounting (Faktor-20-Approximation)Wähle {{c1::\(\ell = \lceil 2 \log_2(2n/\delta) \rceil\)}} Läufe. In jedem Lauf \(i\): ziehe \(r_{i,v} \sim \mathrm{Uniform}([0,1])\) und berechne \(x_{i,v} = \min_{u\in R(v)} r_{i,u}\) mit der Subroutine.Setze \(x_v = \mathrm{Median}(
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| lorenz | cid:1780223730633 |
4 | 170% | 5d | 18 |
nid:1777984580562
c1
A&W
ZufallsfärbungenSei \(G\) ein Graph mit einem Pfad \(P\) der...
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Cloze c1
Q: ZufallsfärbungenSei \(G\) ein Graph mit einem Pfad \(P\) der Länge \(k-1\).Eine zufällige Färbung mit \(k\) Farben erzeugt einen bunten Pfad der Länge \(k-1\) mit Wahrscheinlichkeit \(p \geq {{c1::e^{-k} }}\).Bei wiederholten Färbungen mit \(k\) Farben ist der Erwartu
A: Beweis von (1): \(\Pr[P \text{ wird bunt}] = \tfrac{k!}{k^k} \geq e^{-k}\) per Stirling.Beweis von (2): geometrisch verteilte Wartezeit mit Erfolgswahrscheinlichkeit \(p\), Erwartungswert \(1/p\).
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| lorenz | cid:1777984580562 |
4 | 170% | 5d | 18 |
nid:1777923968738
c1
A&W
Erwartete Anzahl KollisionenFür eine Hashfunktion \(h : U \t...
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Cloze c1
Q: Erwartete Anzahl KollisionenFür eine Hashfunktion \(h : U \to [m]\) mit \(\Pr[h(u) = i] = \tfrac{1}{m}\) gilt:\[\mathbb{E}[\#\text{Kollisionen}] \leq {{c1::\binom{n}{2} \cdot \tfrac{1}{m} }}.\]Insbesondere folgt mit der Wahl \(m = n^2\): \(\mathbb{E}[\#\text{Kollisionen}] {{c2::< 1}}\)
A: Beweisidee: Für jedes feste Paar \((i, j)\) mit \(s_i \neq s_j\) gilt \(\Pr[h(s_i) = h(s_j)] = \tfrac{1}{m}\) (wegen Unabhängigkeit / Zufallsfunktion). Es gibt höchstens \(\binom{n}{2}\) solche Paare, und Linearität des Erwartungswerts liefert die Schranke.
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|---|---|---|---|---|---|
| lorenz | cid:1777923968739 |
4 | 170% | 9d | 21 |
nid:1774487164478
c1
A&W
Für \(x, y \in \mathbb{R}\) und \(n \in \mathbb{N}_0\) gilt:...
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Cloze c1
Q: Für \(x, y \in \mathbb{R}\) und \(n \in \mathbb{N}_0\) gilt:\[(x + y)^n = {{c1::\sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k} }}\]
A: Speziell:\((1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k\)\((1-1)^n = 0 = \sum_{k=0}^n (-1)^k \binom{n}{k}\) (genutzt im Siebformel-Beweis!)
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| lorenz | cid:1774487164478 |
4 | 170% | 12d | 24 |
nid:1777538021737
c1
A&W
(\log nm)^3
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Cloze c1
Cloze answer: (\log nm)^3
Q: Designziel für Primzahltests: Laufzeit polynomiell in \(\log n\) (Darstellungsgrösse). Naives Trial-Division bis \(\sqrt{n}\) ist zu langsam für \(n \approx 2^{1000}\).Der \(\mathrm{ggT}\) zweier Zahlen \(m, n\) lässt sich mit dem Euklid-Algorithmus in \(O({{c1::(\log nm)^3}})\) berechn
A: Trivialerweise gilt: \(\mathrm{ggT}(a, n) > 1\) für ein \(a \in [n-1]\) \(\Rightarrow\) \(n\) nicht prim. Der Test sucht also einen kleinen gemeinsamen Faktor mit zufälligem \(a\).Problem: für \(n = p^2\) ist die Fehlerrate \(\approx 1 - 1/\sqrt{n}\), also fast \(1\). Der Test ist deshalb in der Praxis nutzlos.
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| lorenz | cid:1777538021737 |
4 | 170% | 9d | 22 |
nid:1774631277234
A&W
Zeige, dass die bedingten Wahrscheinlichkeiten \(\Pr[\cdot|B...
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Q: Zeige, dass die bedingten Wahrscheinlichkeiten \(\Pr[\cdot|B]\) für ein festes Ereignis \(B\) mit \(\Pr[B]>0\) einen gültigen Wahrscheinlichkeitsraum auf \(\Omega\) definieren.
A: Zu zeigen: \(\sum_{\omega\in\Omega}\Pr[\omega|B]=1\):\[ \sum_{\omega\in\Omega}\Pr[\omega|B] = \sum_{\omega\in\Omega}\frac{\Pr[\omega\cap B]}{\Pr[B]} = \sum_{\omega\in B}\frac{\Pr[\omega]}{\Pr[B]} = \frac{\Pr[B]}{\Pr[B]} = 1. \]Intuition: Bedingen setzt \(\Pr[\omega|B]=0\) für alle \(\omega\notin B\) und reskaliert die verbleibenden Wahrscheinlichkeiten mit \(1/\Pr[B]\), damit sie sich zu 1 summieren.Konsequenz: Alle Wahrscheinlichkeitsregeln (Komplem
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| lorenz | cid:1774631277234 |
4 | 170% | 29d | 23 |
nid:1774917593197
A&W
Sei \(X=\) Anzahl Würfe bis zum ersten Kopf mit \(\Pr[\text{...
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nid:1774917593197
Q: Sei \(X=\) Anzahl Würfe bis zum ersten Kopf mit \(\Pr[\text{Kopf}] = p\). Welche Methode verwenden wir bei einem gedächtnislosen Problem wie diesem?
A: Definiere \(K_1\) = "erster Wurf ist Kopf." Wende totale Erwartung bedingt auf \(K_1\) an:\(\mathbb{E}[X \mid K_1] = 1\) (sofort fertig)\(\mathbb{E}[X \mid \overline{K}_1] = 1 + \mathbb{E}[X]\) (gedächtnislos: nach Zahl startet der Prozess identisch neu, plus der eine verbrauchte Wurf)Einsetzen in \(\mathbb{E}[X] = 1 \cdot p + (1 + \mathbb{E}[X])(1-p)\) und Auflösen ergibt \(\mathbb{E}[X] = 1/p\). Vermeidet die direkte Berechnung von \(\sum k \cdot (1-p)^{k-1} p\).
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| lorenz | cid:1774917593197 |
4 | 170% | 32d | 25 |
nid:1772046206585
A&W
Was ist der Speicherbedarf von Hamiltonkreise mit DP?
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Q: Was ist der Speicherbedarf von Hamiltonkreise mit DP?
A: \(n\cdot2^n\)
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| lorenz | cid:1772046206585 |
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nid:1774631277033
A&W
In einer Gruppe von \(m\) Personen (mit \(n=365\) Tagen), wi...
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Q: In einer Gruppe von \(m\) Personen (mit \(n=365\) Tagen), wie gross ist die Wahrscheinlichkeit, dass alle Geburtstage verschieden sind? Leite die Formel her.Proof Included
A: (Geburtstagsproblem) Modell: Werfe \(m\) Bälle gleichverteilt in \(n\) Urnen. Sei \(A_j\) = "Ball \(j\) landet in einer leeren Urne."Mit dem Multiplikationssatz:\[ \Pr\!\left[\bigcap_{j=1}^m A_j\right] = \prod_{j=2}^{m} \frac{n-(j-1)}{n} = \prod_{j=2}^{m}\!\left(1-\frac{j-1}{n}\right). \]Obere Schranke mit \(1-x \le e^{-x}\):\[ \Pr[\text{alle verschieden}] \le \prod_{j=2}^{m} e^{-(j-1)/n} = e^{-m(m-1)/(2n)}. \]Also ist die Wahrscheinlichkeit für mindes
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| lorenz | cid:1774631277034 |
4 | 170% | 58d | 22 |
nid:1774487165282
Analysis
Wie beweist man \(\exp(z+w) = \exp(z) \cdot \exp(w)\)?
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Q: Wie beweist man \(\exp(z+w) = \exp(z) \cdot \exp(w)\)?
A: Beide Reihen \(\sum z^n/n!\) und \(\sum w^n/n!\) sind absolut konvergent.Das Cauchy-Produkt liefert:\[c_n = \sum_{k=0}^n \frac{z^{n-k}}{(n-k)!} \cdot \frac{w^k}{k!} = \frac{1}{n!}\sum_{k=0}^n \binom{n}{k} z^{n-k} w^k = \frac{(z+w)^n}{n!}\]Also \(\exp(z)\exp(w) = \sum c_n = \sum \frac{(z+w)^n}{n!} = \exp(z+w)\).
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| lorenz | cid:1774487165282 |
4 | 170% | 35d | 24 |
nid:1774917594967
c2
Analysis
injektiv
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Cloze c2
Cloze answer: injektiv
Q: Jede {{c1::streng monotone::Adjektiv}} Funktion ist {{c2::injektiv::Funktionseigenschaft}}.Proof Included
A: Proof: Nehme an wir haben eine streng monotone Funktion \(f\) die nicht injektiv ist.Dann gilt \(\exists x_1, x_2 \in \mathbb{D}\) sodass \(f(x_1) = f(x_2)\) weil nicht injektiv.Aber oBdA \(x_1 < x_2 \implies f(x_1) < f(x_2)\) was ein Widerspruch ist.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774917594967 |
4 | 170% | 27d | 21 |
nid:1771973928592
c1
Analysis
Eulersche Formel:\[ \sin(t) = {{c1:: \frac{e^{it} - e^{-it} ...
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Cloze c1
Q: Eulersche Formel:\[ \sin(t) = {{c1:: \frac{e^{it} - e^{-it} }{2i} ::\text{Exponentialform} }}\]
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| lorenz | cid:1771973928592 |
4 | 170% | 32d | 25 |
nid:1774487165343
c1
Analysis
Cauchy-Kriterium:\(\sum a_n\) konvergiert \(\iff\) für jedes...
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Cloze c1
Q: Cauchy-Kriterium:\(\sum a_n\) konvergiert \(\iff\) für jedes \(\varepsilon > 0\) existiert ein \(N\), so dass für alle \(n > m \geq N\) gilt:
\[{{c1::\left|\sum_{k=m+1}^n a_k\right| = |S_n - S_m| < \varepsilon}}\]
Proof Included
A: Direktes Cauchy-Kriterium auf die Partialsummenfolge.Man kann \(\sum_{k = m+1}^n a_k \) auch als \(S_n - S_{m} \) schreiben. Und für die Folge \(S_n\) gilt dann der Cauchy Satz. Falls also \(\exists N \in \mathbb{N}_0\) sodass \(\forall n > m > N gilt |S_n - S_m| < \epsilon\), konvergiert die Folge \(S_n\).
Die gilt per Annahme und deswegen konvergiert \(S_n\). Da die Folge der Partialsummen konvergiert, konvergiert die Reihe.
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| lorenz | cid:1774487165343 |
4 | 170% | 15d | 23 |
nid:1774917595550
c1
Analysis
0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon
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Cloze c1
Cloze answer: 0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon
Q: Die alternative Grenzwertdefinition schliesst \(x_0\) selbst aus:\[\begin{gathered}\forall \varepsilon > 0\;\exists \delta > 0 \text{ so dass für alle } x \in \mathbb{D}(f) \\ {{c1:: 0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon}}\end{gathered}\]
A: Durch \(0 < |x - x_0|\) ist der Grenzwert unabhängig vom Funktionswert bei \(x_0\) - selbst wenn \(f(x_0)\) nicht definiert ist.Da gilt \(0 < |x - x_0|\) kann \(x\) nicht den Wert \(x_0\) annehmen.
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| lorenz | cid:1774917595550 |
4 | 170% | 45d | 24 |
nid:1777383738534
c1
Analysis
Binomialreihe für beliebigen Exponenten \(p \in \mathbb{R}\)...
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Cloze c1
Q: Binomialreihe für beliebigen Exponenten \(p \in \mathbb{R}\) (konvergiert für \(-1 < x < 1\)):\[ (1 + x)^p = {{c1::\sum_{n=0}^{\infty} \binom{p}{n} x^n = 1 + p x + \frac{p(p-1)}{2!} x^2 + \dots }}\]
A: Verallgemeinerter Binomialkoeffizient: \(\binom{p}{n} = \dfrac{p(p-1)\cdots(p-n+1)}{n!}\). Für \(p \in \mathbb{N}_0\) bricht die Reihe nach endlich vielen Termen ab und ergibt den klassischen binomischen Lehrsatz.
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| lorenz | cid:1777383738534 |
4 | 170% | 33d | 18 |
nid:1774917595090
c1
Analysis
Es gelte \(\mathbb{D}(f) \cap [x_0,\, x_0 + \delta) \neq \em...
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Cloze c1
Q: Es gelte \(\mathbb{D}(f) \cap [x_0,\, x_0 + \delta) \neq \emptyset \;\forall \delta > 0\).Falls gilt \(\forall \varepsilon > 0 \;\exists \delta > 0\): \[{{c1::x \in \mathbb{D}(f) \cap [x_0,\, x_0 + \delta) \;\Rightarrow\; |f(x) - L| < \varepsilon }},\] hat \(f\) in \(x_
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| lorenz | cid:1774917595090 |
4 | 170% | 54d | 22 |
nid:1774487165385
c1
Analysis
Für die geometrische Reihe \(\sum_{n=0}^\infty q^n\) gilt \(...
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Cloze c1
Q: Für die geometrische Reihe \(\sum_{n=0}^\infty q^n\) gilt \(S_n = {{c1:: \frac{1 - q^{n + 1} }{1 - q} }}\)
A: \[ \begin{align} qS_n &= q + q^2 + \dots + q^{n + 1} \\ S_n - qS_n &= 1 - q^{n + 1} \\ S_n &= \frac{1 - q^{n + 1}}{1 - q} \end{align} \]
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| lorenz | cid:1774487165385 |
4 | 170% | 59d | 23 |
nid:1771973928491
c1
Analysis
Sei \(n \in \mathbb{N}\), \(n \ge 1\). Dann hat die Gleichun...
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Cloze c1
Q: Sei \(n \in \mathbb{N}\), \(n \ge 1\). Dann hat die Gleichung \(z^n = 1\) genau \(n\) Lösungen in \(\mathbb{C}\): \(z_1, z_2, \dots, z_n\) wobei: \[ z_j = {{c1:: \cos \frac{2\pi j}{n} + i \cdot \sin \frac{2 \pi j}{n} }}, \quad 1 \le j \le n \]
A: Die Lösungen liegen alle auf einem Kreis mit Radius 1 und sind gleichmässig verteilt (formen ein n-Eck). Beispiel: Für \(w = R \cdot e^{i \varphi}\) sind die Lösungen von \(z^n = w\) gleich der \(n\) komplexen Zahlen mit Betrag \(\sqrt[^n]{R}\) und Winkeln \(\varphi_k = \frac{\varphi}{n} + k \cdot \frac{2 \pi}{n}\) für \(k = 0, \dots, n - 1\).
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| lorenz | cid:1771973928491 |
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nid:1772928333323
c1
Analysis
streng monoton steigend
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Cloze c1
Cloze answer: streng monoton steigend
Q: Der Wertebereich von \(\arctan\) ist \({{c1::\left(-\frac{\pi}{2},\, \frac{\pi}{2}\right)}}\), und die Funktion ist {{c1::streng monoton steigend::Wachstumsverhalten}}.
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| lorenz | cid:1772928333323 |
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nid:1774138448037
Analysis
Trick: FixpunktSei eine Folge rekursiv definiert durch \(a_1...
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nid:1774138448037
Q: Trick: FixpunktSei eine Folge rekursiv definiert durch \(a_1 = c\) und \(a_{n+1} = f(a_n)\).
A: Falls \((a_n)\) konvergiert (z.B. nach Weierstrass), setzt man \(l = \lim_{n \to \infty} a_n \) \(= \lim_{n \to \infty} a_{n+1}\) und erhält die Fixpunktgleichung: \(l = f(l)\) Man löst diese Gleichung nach \(l\) auf und schliesst anhand der Eigenschaften der Folge (Vorzeichen, Monotonie, Beschränktheit) aus, welcher Kandidat der tatsächliche Grenzwert ist.
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| lorenz | cid:1774138448037 |
4 | 170% | 56d | 26 |
nid:1772928333414
c1
Analysis
\[ \sin\!\left(\frac{\pi}{4}\right) = {{c1::\frac{\sqrt{2} }...
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nid:1772928333414
Cloze c1
Q: \[ \sin\!\left(\frac{\pi}{4}\right) = {{c1::\frac{\sqrt{2} }{2} }} \]
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| lorenz | cid:1772928333414 |
4 | 170% | 95d | 19 |
nid:1761028602734
LinAlg
An linear combination of \(\lambda_1\textbf{v}_1 + \lambda_...
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users
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ease
nid:1761028602734
Q: An linear combination of \(\lambda_1\textbf{v}_1 + \lambda_2\textbf{v}_2 + \dots + \lambda_n\textbf{v}_n\) is convex if
A: it is both affine and conic
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761028602734 |
4 | 215% | 89d | 15 |
nid:1761491477391
DiskMat
How does antisymmetry appear in graph representation?
4
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users
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ease
nid:1761491477391
Q: How does antisymmetry appear in graph representation?
A: There is not a single cycle of length 2 (no edge from \(a\) to \(b\) AND from \(b\) to \(a\)).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477392 |
4 | 215% | 11d | 17 |
nid:1761491477501
DiskMat
What is the set \(\{0, 1\}^{\infty}\)?
4
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1/4
users
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ease
nid:1761491477501
Q: What is the set \(\{0, 1\}^{\infty}\)?
A: The set of semi-infinite binary sequences, or equivalently, the set of functions \(\mathbb{N} \to \{0,1\}\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477502 |
4 | 245% | 32d | 21 |
nid:1765296057127
c1
A&D
O(n)
4
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1/4
users
200%
ease
nid:1765296057127
Cloze c1
Cloze answer: O(n)
Q: Choose a tight bound!\({{c1::O(n)}} \leq {{c2::O(\log(n!))}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765296057127 |
4 | 200% | 8d | 12 |
nid:1772520447039
c2
A&W
\(\forall u, v \in V, u \neq v\) gibt mindestens \(k\) inter...
4
lapses
1/4
users
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nid:1772520447039
Cloze c2
Cloze answer: \(\forall u, v \in V, u \neq v\) gibt mindestens \(k\) intern-knotendisjunkte \(u\)-\(v\)-Pfade.; \(\forall u, v \in V, u \neq v\) gibt mindestens \(k\) kantendisjunkte \(u\)-\(v\)-Pfade.
Q: Sei \(G = (V, E)\) ein Graph. Dann gilt:
{{c1::\(G\) is \(k\)-knoten-zusammenhängend}}\(\iff\){{c2::\(\forall u, v \in V, u \neq v\) gibt mindestens \(k\) intern-knotendisjunkte \(u\)-\(v\)-Pfade.}}{{c1::\(G\) ist \(k\)-kanten-zusammenhängend}}\(\if
A: Satz von Karl Menger V2 (Sohn vom sehr baseden Carl Menger)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772520447040 |
4 | 200% | 9d | 14 |
nid:1766314094576
DiskMat
How are ordered pairs \((a, b)\) formally defined in set the...
3
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users
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ease
nid:1766314094825
c1
DiskMat
subset; \(A\times B\).; a relation on \(A\).
3
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users
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nid:1766314094825
Cloze c1
Cloze answer: subset; \(A\times B\).; a relation on \(A\).
Q: A relation \(\rho\) from a set \(A\) to a set \(B\) (also called an \((A,B)\)-relation) is a {{c1::subset}} of {{c1::\(A\times B\).}} If \(A = B\), then \(\rho\) is called {{c1::a relation on \(A\).}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094897 |
1 | 230% | 4d | 5 |
| lorenz | cid:1764867990770 |
1 | 230% | 2071d | 13 |
| niklas | cid:1762856074679 |
1 | 245% | 36d | 7 |
nid:1766940295781
DiskMat
In a finite group of order \(|G|\), for \(x^e = y\), \(d\) i...
3
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users
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ease
nid:1766940295781
Q: In a finite group of order \(|G|\), for \(x^e = y\), \(d\) is the inverse such that \(y^d = x\) iff: (Proof included)
A: \(ed \equiv_{|G|} 1\), i.e. \(d\) is the multiplicative inverse of \(e\) modulo \(|G|\).Proof\(ed = k \cdot |G| + 1\) (multiplicative inverse)\((x^e)^d = x^{ed} = x^{k\cdot |G| + 1}\)\((x^{|G|})^k \cdot x = 1^k \cdot x = x\)Thus this returns \(x\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295939 |
1 | 230% | 4d | 6 |
| lorenz | cid:1766448532968 |
1 | 230% | 1255d | 11 |
| niklas | cid:1766318730351 |
1 | 245% | 10d | 5 |
nid:1764867991514
LinAlg
The scalar product of \(\textbf{v} \cdot \textbf{v}\) is \(\...
3
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3/4
users
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nid:1764867991514
Q: The scalar product of \(\textbf{v} \cdot \textbf{v}\) is \(\leq or \geq\) to what?
A: \(\textbf{v} \cdot \textbf{v} \geq 0\) with equality exactly if \(\textbf{v} = \textbf{0}\).This is because we essentially square the entries and thus can't get negatives.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991514 |
1 | 230% | 1026d | 11 |
| niklas | cid:1761028945284 |
1 | 245% | 43d | 8 |
| tomas | cid:1765551644273 |
1 | 245% | 11d | 4 |
nid:1766314094559
c1
DiskMat
\(F \rightarrow G\)
3
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nid:1766314094620
DiskMat
What important property do equivalence classes have?
3
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nid:1766314094620
Q: What important property do equivalence classes have?
A: The set \(A / \theta\) of equivalence classes of an equivalence relation \(\theta\) on \(A\) is a partition of \(A\).
(Equivalence classes are disjoint and cover the entire set)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094628 |
2 | 180% | 9d | 12 |
| niklas | cid:1761491477412 |
1 | 260% | 40d | 9 |
nid:1766314094637
DiskMat
What is the meet of elements \(a\) and \(b\) in a poset?
3
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nid:1766314094711
DiskMat
How is the GCD related to ideals? (Lemma 4.4)
3
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nid:1766314094711
Q: How is the GCD related to ideals? (Lemma 4.4)
A: Let \(a, b \in \mathbb{Z}\) (not both 0). If \((a, b) = (d)\), then \(d\) is a greatest common divisor of \(a\) and \(b\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990411 |
2 | 210% | 1675d | 18 |
| jonas | cid:1766314094732 |
1 | 230% | 5d | 8 |
nid:1766314094928
c1
DiskMat
cyclic
3
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2/4
users
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ease
nid:1766314094985
DiskMat
If \(b(x)\) divides \(a(x)\), then so does:
3
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nid:1766314094985
Q: If \(b(x)\) divides \(a(x)\), then so does:
A: \(v \cdot b(x)\) for any nonzero \(v \in F\).
This holds because if \(a(x) = b(x) \cdot c(x)\), then \(a(x) = vb(x) \cdot (v^{-1} c(x))\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991292 |
2 | 210% | 991d | 12 |
| jonas | cid:1766314095135 |
1 | 230% | 4d | 8 |
nid:1766314094989
DiskMat
What does polynomial evaluation preserve?
3
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nid:1766314094989
Q: What does polynomial evaluation preserve?
A: Lemma 5.28: Polynomial evaluation is compatible with the ring operations:
- If \(c(x) = a(x) + b(x)\) then \(c(\alpha) = a(\alpha) + b(\alpha)\) for any \(\alpha\)
- If \(c(x) = a(x) \cdot b(x)\) then \(c(\alpha) = a(\alpha) \cdot b(\alpha)\) for any \(\alpha\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991309 |
2 | 210% | 1266d | 12 |
| jonas | cid:1766314095139 |
1 | 185% | 8d | 10 |
nid:1766314094994
DiskMat
If we want to use roots to check that a polynomial is irredu...
3
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2/4
users
228%
ease
nid:1766314094994
Q: If we want to use roots to check that a polynomial is irreducible, it has to have?
A: Degree \(2\) or \(3\).
Important: This doesn't work for polynomials of higher degrees! A degree \(4\) polynomial might be the product of two irreducible degree \(2\) polynomials, each with no roots.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764859231542 |
2 | 240% | 18d | 8 |
| jonas | cid:1766314095144 |
1 | 215% | 8d | 9 |
nid:1766314095000
c1
DiskMat
A ring \(R\) is a field if and only if {{c1:: \(\langle R \s...
3
lapses
2/4
users
235%
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nid:1766314095016
DiskMat
When does an irreducible polynomial exist in \(\text{GF}(p)[...
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nid:1766314095016
Q: When does an irreducible polynomial exist in \(\text{GF}(p)[x]\)?
A: For every prime \(p\) and every \(d > 1\), there exists an irreducible polynomial of degree \(d\) in \(\text{GF}(p)[x]\).
Result: we can construct a finite field with \(p^d\) elements by using an irreducible polynomial of degree \(d\) to cap the number of coefficients at \(d\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764859231591 |
2 | 210% | 4d | 14 |
| jonas | cid:1766314095170 |
1 | 230% | 9d | 9 |
nid:1766314111354
LinAlg
A linear combination of \(\lambda_1\textbf{v}_1 + \lambda_2...
3
lapses
2/4
users
220%
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nid:1766314111354
Q: A linear combination of \(\lambda_1\textbf{v}_1 + \lambda_2\textbf{v}_2 + \dots + \lambda_n\textbf{v}_n\) is conic if
A: \(\lambda_j \geq 0\) for \(j = 1, 2, \dots, n\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314111354 |
2 | 210% | 3d | 7 |
| lorenz | cid:1764867991507 |
1 | 230% | 1340d | 12 |
nid:1766314111376
LinAlg
Was ist der rank einer full rank matrix \(A \in \mathbb{R}^{...
3
lapses
2/4
users
235%
ease
nid:1767089600236
c1
DiskMat
factored uniquely into irreducible elements (up to associate...
3
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2/4
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228%
ease
nid:1767089600236
Cloze c1
Cloze answer: factored uniquely into irreducible elements (up to associates)
Q: In a Euclidean domain every element can be {{c1:: factored uniquely into irreducible elements (up to associates)}}
A: \(a, b\) associates (\(a \sim b\)) if \(a = ub\) for some unit \(u\).Proof sketch:
Consider a nonzero, nonunit \(a \in R\).
If a is irreducible, we are done.
Otherwise, \(a = bc\) with both \(b,c\) nonunits.
By the Euclidean property, we may assume \(\delta(b), \delta(c) < \delta(a)\).
If either factor is reducible, factor it
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1767089600236 |
2 | 210% | 1d | 9 |
| niklas | cid:1767082205233 |
1 | 245% | 5d | 7 |
nid:1768160640380
LinAlg
How do we find a basis for the row space \(R(A) = C(A^\top)\...
3
lapses
2/4
users
220%
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nid:1768160640380
Q: How do we find a basis for the row space \(R(A) = C(A^\top)\)?
A: The first \(r\) columns of \(R^\top\) where \(R\) is the RREF of \(A\) form a basis of the row space (the non-zero rows). In particular \(\dim(\textbf{R}(A)) = r\)This works because as noted before, multiplying by an invertible matrix \(M\) does not change the row-space of \(MA\) on the left.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182518113 |
2 | 210% | 768d | 11 |
| jonas | cid:1768160640388 |
1 | 230% | 1d | 4 |
nid:1766580143735
c1
A&D
\(O(|V| \log |V|)\)
3
lapses
2/4
users
228%
ease
nid:1766580143735
Cloze c1
Cloze answer: \(O(|V| \log |V|)\)
Q: The amortised runtime of union in the Union-Find datastructure is {{c1:: \(O(|V| \log |V|)\)}}.
A: Union takes \(\Theta(\min \{ |ZHK(u)| , |ZHK(v)| \}\). In the worst case, the minimum is \(|V| / 2\) as both have the same size.Therefore over all loops, this would take \(O(|V| \log |V|)\) time, as on average we only take \(O(\log |V|)\) time.The graph stays worst case, this is the average of the calls in the worst case.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766580143735 |
2 | 210% | 662d | 13 |
| niklas | cid:1766569467577 |
1 | 245% | 25d | 6 |
nid:1768344740183
c1
A&D
we know the graph is connected, i.e. \(m \geq n - 1\)
3
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2/4
users
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ease
nid:1768344740183
Cloze c1
Cloze answer: we know the graph is connected, i.e. \(m \geq n - 1\)
Q: We can run DFS in \(O(m)\) if {{c1:: we know the graph is connected, i.e. \(m \geq n - 1\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768344740183 |
2 | 210% | 717d | 12 |
| tomas | cid:1768391364319 |
1 | 230% | 7d | 5 |
nid:1766531635421
A&D
What is the tree condition for 2-3 Trees implementing a dict...
3
lapses
2/4
users
228%
ease
nid:1765372936179
A&D
When \(f = \Theta(g)\), this means?
3
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2/4
users
242%
ease
nid:1765372936179
Q: When \(f = \Theta(g)\), this means?
A: \(\exists C_1,C_2 \ge 0 \quad \forall n \in \mathbb{N}\) \(C_1 \cdot g(n) \leq f(n) \leq C_2 \cdot g(n)\)\(f\) grows asymptotically the same as \(g\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936179 |
2 | 210% | 977d | 13 |
| niklas | cid:1765295553120 |
1 | 275% | 20d | 13 |
nid:1766531635530
c1
A&D
it does not contain any cycles of odd length
3
lapses
2/4
users
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nid:1765372936170
A&D
What is l'Hôpital's Rule?
3
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nid:1765372936170
Q: What is l'Hôpital's Rule?
A: If \(\lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) = \infty\) (or both \(=0\)), and \(\lim_{x \to \infty} \frac{f'(x)}{g'(x)}\) exists (or is \(\pm\infty\)), then:
\(\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936170 |
2 | 210% | 1073d | 15 |
| niklas | cid:1765295341110 |
1 | 245% | 17d | 13 |
nid:1765372936194
c2
A&D
O(\log(n!))
3
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2/4
users
220%
ease
nid:1766531635629
A&D
Bellman-Ford
3
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nid:1766531635629
Q: Bellman-Ford
A: \(O(|V| \cdot |E|)\) (uses DP)We iterate over all edges in the "relaxation" thus the time complexity of that step is \(O(m)\) (the actual check is \(O(1)\)).As we relax \(n - 1\) (or \(n\) for negative cycle check) times, the total runtime is \(O(n \cdot m)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635631 |
2 | 210% | 1449d | 14 |
| tomas | cid:1766576739763 |
1 | 230% | 11d | 5 |
nid:1766580143624
A&D
Boruvka's Algorithm
3
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2/4
users
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nid:1766580143624
Q: Boruvka's Algorithm
A: \(O((|V| + |E|) \cdot \log |V|)\)During each iteration, we examine all edges to find the cheapest one: \(O(|V| + |E|)\):Run DFS to find the connected components: \(O(|V| + |E|)\)Find the cheapest one \(O(|E|)\)We iterate a total of \(\log_2 |V|\) times as each iteration halves the number of connected components.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766580143624 |
2 | 210% | 1513d | 14 |
| niklas | cid:1766567785295 |
1 | 245% | 16d | 6 |
nid:1766531635563
c1
A&D
\(\exists\) back edge
3
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2/4
users
242%
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nid:1765372936212
c1
A&D
O(\log(n!))
3
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2/4
users
228%
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nid:1764867990542
c1
DiskMat
Find a suitable statement \( T\).
3
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users
235%
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nid:1764867990542
Cloze c1
Cloze answer: Find a suitable statement \( T\).
Q: Proof method: Proof by Contradiction1. {{c1:: Find a suitable statement \( T\).}}2. {{c2:: Prove that \( T \) is false.}}3. {{c3:: Assume that \( S \) is false and prove that \( T \) is true (-> contradiction).}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073567 |
2 | 240% | 43d | 11 |
| lorenz | cid:1764867990542 |
1 | 230% | 1263d | 9 |
nid:1764867991070
c2
DiskMat
If {{c2:: the order \(\text{ord}(a)\) of \(a \in G\) is \(|G...
3
lapses
2/4
users
235%
ease
nid:1764867991498
c1
DiskMat
4
3
lapses
2/4
users
235%
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nid:1767918757756
IO r5
EProg
[Image Occlusion region 5]
3
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2/4
users
235%
ease
nid:1767918757756
Cloze c5
Q: {{c1::image-occlusion:rect:left=.2281:top=.3427:width=.0814:height=.2045:oi=1}}{{c2::image-occlusion:rect:left=.3053:top=.345:width=.1142:height=.2067:oi=1}}{{c3::image-occlusion:rect:left=.1625:top=.5221:width=.0693:height=.2181:oi=1}}{{c4::image-occlusion:rect:left=.1625:top=.713:width
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1767888505029 |
2 | 240% | 13d | 13 |
| lorenz | cid:1767918757756 |
1 | 230% | 476d | 10 |
nid:1768182517405
c1
LinAlg
unique
3
lapses
2/4
users
228%
ease
nid:1768344745873
c1
LinAlg
\(C(A^\top)\)
3
lapses
2/4
users
220%
ease
nid:1768344745873
Cloze c1
Cloze answer: \(C(A^\top)\)
Q: For a full row rank matrix \(A\), the unique solution to\[{{c1:: \min_{x \in \mathbb{R}^n} ||x||^2 \text{ s.t. } Ax = b}}\] is given by the vector \(\hat{x} = A^\dagger b\). This \(\hat{x}\) is in {{c1:: \(C(A^\top)\)}}. Proof Included
A: Proof By Lemma 6.4.5 we only need to show that \(\hat{x} = A^\dagger b\) satisfies \(A \hat{x} = b\) and that \(\hat{x} \in C(A^\top)\).\(A\hat{x} = AA^\dagger b = AA^\top (AA^\top)^{-1}b = b\) \(\hat{x} = A^\dagger b = A^\top ((AA^\top)^{-1} b) = A^\top y\) for some \(y\) thus \(x \in C(A^\top)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768344745873 |
2 | 210% | 1209d | 14 |
| niklas | cid:1768302903149 |
1 | 230% | 2d | 6 |
nid:1768263610799
c3
LinAlg
Assume \(Q\) is orthogonal and square. Then:{{c1::\(QQ^\top ...
3
lapses
2/4
users
220%
ease
nid:1768263610799
Cloze c3
Q: Assume \(Q\) is orthogonal and square. Then:{{c1::\(QQ^\top = I\)}}{{c2::\(Q^{-1} = Q^\top\)}}{{c3::The columns form an orthonormal basis for \(\mathbb{R}^n\).}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768214114072 |
2 | 210% | 2d | 6 |
| lorenz | cid:1768263610801 |
1 | 230% | 1639d | 12 |
nid:1768263610411
c1
LinAlg
Let \(A \in \mathbb{R}^{m \times n}\). Then \(N(A) = {{c1::N...
3
lapses
2/4
users
228%
ease
nid:1768263610411
Cloze c1
Q: Let \(A \in \mathbb{R}^{m \times n}\). Then \(N(A) = {{c1::N(A^\top A)::\text{another nullspace} }}\). Proof Included
A: \(N(A) = N(A^\top A)\) holds because:if \(x \in N(A)\) then \(Ax = 0 \implies A^\top Ax = A \cdot 0 \implies A^\top A x = 0\).if \(x \in N(A^\top A)\) then \(A^\top A x = 0\), which means \[ 0 = x^\top 0 = x^\top A^\top Ax = (Ax)^\top(Ax) = ||Ax||^2 \implies Ax = 0 \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768210666216 |
2 | 225% | 2d | 9 |
| lorenz | cid:1768263610411 |
1 | 230% | 1763d | 12 |
nid:1771362440456
c1
A&W
einen Knoten mit Grad < \(k\)
3
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2/4
users
250%
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nid:1773311287370
c1
A&W
2
3
lapses
2/4
users
220%
ease
nid:1773311287370
Cloze c1
Cloze answer: 2
Q: Heuristik:\(v_n\) := Knoten vom kleinsten Grad. Lösche \(v_n\).\(v_{n-1}\) := Knoten vom kleinsten Grad im Restgraph. Lösche \(v_{n-1}\). Iteriere.Die Heuristik findet immer eine Färbung mit {{c1::2}} Farben für Bäume.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773311287370 |
2 | 210% | 38d | 14 |
| niklas | cid:1773420068155 |
1 | 230% | 2d | 3 |
nid:1773307908373
IO r1
A&W
[Image Occlusion region 1]
3
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2/4
users
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nid:1773307908373
Cloze c1
Q: {{c1::image-occlusion:rect:left=.1376:top=.5345:width=.6408:height=.0783}}{{c2::image-occlusion:rect:left=.0886:top=.6098:width=.903:height=.2198}}{{c3::image-occlusion:rect:left=.2343:top=.9079:width=.0768:height=.0783}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773307908375 |
2 | 210% | 67d | 15 |
| niklas | cid:1773420068090 |
1 | 245% | 5d | 6 |
nid:1771526674685
c1
A&W
\(v \neq root\), und \(v\) hat ein Kind \(u\) im DFS-Baum mi...
3
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2/4
users
242%
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nid:1771526674685
Cloze c1
Cloze answer: \(v \neq root\), und \(v\) hat ein Kind \(u\) im DFS-Baum mit \(low[u] \geq dfs[v]\)
Q: \(v\) ist genau dann Artikulationsknoten, wenn:{{c1::\(v \neq root\), und \(v\) hat ein Kind \(u\) im DFS-Baum mit \(low[u] \geq dfs[v]\)}} oder {{c2::\(v = root\), und \(v\) hat mindestens zwei Kinder im DFS-Baum.}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771535790933 |
2 | 255% | 13d | 14 |
| lorenz | cid:1771526674686 |
1 | 230% | 72d | 13 |
nid:1771973928588
c1
Analysis
Beweis: Für alle \(a < b\) in \(\mathbb{R}\) existiert ein \...
3
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users
242%
ease
nid:1771973928588
Cloze c1
Q: Beweis: Für alle \(a < b\) in \(\mathbb{R}\) existiert ein \(\mathbb{Q}\) mit \(a < q < b\){{c1:: Wähle nach Archimedischem Prinzip \(n \in \mathbb{N}\) so dass \(\frac{1}{n} < b - a\).}}{{c2:: \(\frac{m}{n} \mid m \in \mathbb{Z}\) diese
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771973928591 |
2 | 210% | 40d | 14 |
| niklas | cid:1771969342907 |
1 | 275% | 42d | 9 |
nid:1772928333578
c1
Analysis
\sin x \cos y \pm \cos x \sin y
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users
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nid:1771973928518
c1
Analysis
Für \(z \in \mathbb{C}\) gilt: \(z + \bar{z} = {{c1:: 2 \te...
3
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users
242%
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nid:1771973928498
c1
Analysis
Der Abstand zwischen zwei komplexen Zahlen \(z_1, z_2\) ist ...
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users
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nid:1771973928498
Cloze c1
Q: Der Abstand zwischen zwei komplexen Zahlen \(z_1, z_2\) ist \( d = {{c1:: |z_2 - z_1 | = |z_1 - z_2| ::\text{Beide Formen} }}\).
A: Hier gilt wieder die Dreiecksungleichung: \(|z + w| \leq |z| + |w|\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771973928498 |
2 | 210% | 67d | 18 |
| niklas | cid:1771970006360 |
1 | 245% | 45d | 5 |
nid:1771973928515
c1
Analysis
Division im Komplexen:\[ \frac{z}{w} = {{c1:: \frac{z \cdot ...
3
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users
228%
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nid:1772496585510
Analysis
Was ist ein Häufungspunkt?
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nid:1772496585510
Q: Was ist ein Häufungspunkt?
A: Grenzwert einer Teilfolge (Punkt, an den eine Folge immer wieder beliebig nahe herankommt)\[\forall \varepsilon > 0 \forall N \in \mathbb{N}_0 \exists n \geq N \text{ so dass } | a_n - A | < \varepsilon\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772520282861 |
2 | 225% | 90d | 15 |
| lorenz | cid:1772496585510 |
1 | 230% | 99d | 9 |
nid:1774138446805
c1
Analysis
beschränkte Folge reeller Zahlen
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users
228%
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nid:1774138446805
Cloze c1
Cloze answer: beschränkte Folge reeller Zahlen
Q: Jede {{c1::beschränkte Folge reeller Zahlen}} hat {{c2::einen Häufungspunkt}} und {{c2::eine konvergente Teilfolge}}.Proof Idea Included
A: (Bolzano-Weierstrass)Beachte: Dies gilt nur für die 1-norm!Proof Idea: Nested Intervals. Always bisect the interval. Since the sequence is infinite, at least one of the intervals must contain an infinite amount of terms.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1774006423271 |
2 | 225% | 1d | 11 |
| lorenz | cid:1774138446805 |
1 | 230% | 90d | 12 |
nid:1771973928615
c1
Analysis
Ordnungsvollständigkeit:Seien \(A, B \subseteq \mathbb{R}\),...
3
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users
228%
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nid:1771973928615
Cloze c1
Q: Ordnungsvollständigkeit:Seien \(A, B \subseteq \mathbb{R}\), sodass
{{c2:: \(A \neq \emptyset\), \(B \neq \emptyset\)}}
{{c2:: \(\forall a \in A \ \forall b \in B \ : \ a \leq b\)}}
Dann {{c1:: gibt es ein \(c \in \mathbb{R}\), sodass \[ \foral
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771974617624 |
2 | 225% | 22d | 8 |
| lorenz | cid:1772327995619 |
1 | 230% | 117d | 13 |
nid:1771841911706
c2
PProg
instruction stream (independent execution units within a pro...
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users
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ease
nid:1771841911706
Cloze c2
Cloze answer: instruction stream (independent execution units within a process)
Q: Each thread has its own {{c1::execution stack (method calls, local variables)}} and {{c2::instruction stream (independent execution units within a process)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771841911707 |
2 | 210% | 130d | 12 |
| niklas | cid:1771872607386 |
1 | 260% | 13d | 8 |
nid:1766498257927
c1
A&D
it must lie on a cycle
3
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2/4
users
212%
ease
nid:1766314094859
DiskMat
For what \(m\) is \(\mathbb{Z}^*_m\) cyclic? (Theorem 5.15)
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users
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nid:1766314094859
Q: For what \(m\) is \(\mathbb{Z}^*_m\) cyclic? (Theorem 5.15)
A: The group \(\mathbb{Z}^*_m\) is cyclic if and only if:• \(m = 2\)• \(m = 4\)• \(m = p^e\) (where p is an odd prime and \(e ≥ 1\))• \(m = 2p^e\) (where p is an odd prime and \(e ≥ 1\)) Example: Is \(\mathbb{Z}^*_{19}\) cyclic? What is a generator? Yes, \(\mathbb{Z}^*_{19}\) is cyclic (since \(19\) is an odd prime). 2 is a generator.Powers of 2: 2, 4, 8, 16, 13, 7, 14
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094933 |
3 | 175% | 8d | 18 |
nid:1766940295689
c1
DiskMat
empty clause \(\emptyset\) (formula with no literals)
3
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1/4
users
175%
ease
nid:1766940295689
Cloze c1
Cloze answer: empty clause \(\emptyset\) (formula with no literals)
Q: The {{c1::empty clause \(\emptyset\) (formula with no literals)}} corresponds to an {{c2::unsatisfiable formula}}.
A: A disjunction with no disjuncts is false.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295781 |
3 | 175% | 3d | 11 |
nid:1765372936266
c1
A&D
{{c1:: \(\sum_{i = 1}^{n} \sum_{k = 1}^{\textbf{i} } 1\)::Su...
3
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1/4
users
190%
ease
nid:1765372936266
Cloze c1
Q: {{c1:: \(\sum_{i = 1}^{n} \sum_{k = 1}^{\textbf{i} } 1\)::Sum}} \(=\) {{c2:: \(\sum_{i = 1}^n i = \frac{n(n + 1)}{2}\)}}
A: inner loop depends on outer
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936267 |
3 | 190% | 791d | 16 |
nid:1769211470058
A&D
How can you find the upper bound of a geometric series like ...
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1/4
users
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nid:1769211470058
Q: How can you find the upper bound of a geometric series like \(T = 7^1, 7^2, \ldots, 7^n\)?
A: Use the multiply-subract trick.Mutliply the series by its base: \(7T\)Subtract: \(7T - T = 7^{n+1} - 7^1\) (middle terms cancel)Factor: \(T(7-1) = 7^{n+1} - 7^1\)Divide: \(T = \frac{7^{n+1} - 7^1}{6}\)This trick works even if every term has a constant coefficient.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1769211470058 |
3 | 190% | 841d | 15 |
nid:1765372936203
c1
A&D
O(k^n)
3
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1/4
users
190%
ease
nid:1765372936203
Cloze c1
Cloze answer: O(k^n)
Q: Choose a tight bound!\({{c1::O(k^n)}} \leq {{c2::O(n!)}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936203 |
3 | 190% | 934d | 17 |
nid:1766531635539
c1
A&D
\(\exists\) toposort
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users
190%
ease
nid:1766531635539
Cloze c1
Cloze answer: \(\exists\) toposort
Q: {{c1:: \(\exists\) toposort}} \(\Longleftrightarrow\) {{c2:: \(\lnot \exists\) directed closed walk}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635540 |
3 | 190% | 1448d | 15 |
nid:1765372936281
c1
A&D
{{c1:: \(\sum_{i = 1}^{n} i^3\)::Sum}} \(=\) {{c2::\(\frac{...
3
lapses
1/4
users
190%
ease
nid:1765372936281
Cloze c1
Q: {{c1:: \(\sum_{i = 1}^{n} i^3\)::Sum}} \(=\) {{c2::\(\frac{n^2(n + 1)^2}{4}\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936281 |
3 | 190% | 1878d | 18 |
nid:1767105269557
DiskMat
What's the definition of an Euclidean domain?
3
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users
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nid:1767105269557
Q: What's the definition of an Euclidean domain?
A: A euclidean domain is an integral domain \(D\) together with a degree function \(d: D \setminus {0} \rightarrow \mathbb{N}\) such that:For every \(a\) and \(b \neq 0\) in \(D\) there exist \(q\) and \(r\) such that \(a = bq + r\) and \(d(r) < d(b)\) or \(r = 0\)For all nonzero \(a\) and \(b\) in \(D\), \(d(a) \leq d(ab)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1767105269557 |
3 | 190% | 1232d | 18 |
nid:1766448532960
c2
DiskMat
\(e\) coprime to \(|G|\)
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lapses
1/4
users
190%
ease
nid:1766448532960
Cloze c2
Cloze answer: \(e\) coprime to \(|G|\)
Q: In a finite group the function \(x \rightarrow x^e\) is {{c1:: a bijection}} if {{c2::\(e\) coprime to \(|G|\)}}.For \(x^e = y\), the inverse of \(y\) is {{c3:: the unique \(e\)-th root \(x = y^d\), with \(de \equiv_{|G|} 1\)}}.
A: Proof:We have \(ed = k \cdot |G| + 1\) for some \(k\). Thus, for any \(x \in G\) we have\[(x^e)^d = x^{ed} = x^{k \cdot |G| + 1} = \underbrace{(x^{|G|})^k}_{=1} \cdot x = x\]which means that the function \(y \mapsto y^d\) is the inverse function of the function \(x \mapsto x^e\) (which is hence a bijection). The under-braced term is equal to 1 because the order of \(x\) must divide the order of \(G\) (Lagrange).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448532960 |
3 | 190% | 1716d | 18 |
nid:1764867990499
DiskMat
How can we use the CRT to decompose remainders like \(R_{77}...
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users
190%
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nid:1764867990499
Q: How can we use the CRT to decompose remainders like \(R_{77}(n)\)?
A: We can decompose \(77 = 11 \cdot 7\) and then calculate:\(R_7(n) = 3\)\(R_{11}(n) = 5\)Then to find the result mod 77, we use the CRT.Find \(11^{-1} \pmod{7} = 2\) (since \(11 \cdot 2 = 22 \equiv 1 \pmod{7}\))Find \(7^{-1} \pmod{11} = 8\) (since \(7 \cdot 8 = 56 \equiv 1 \pmod{11}\))Calculate: \(x = 3 \cdot 11 \cdot 2 + 5 \cdot 7 \cdot 8 = 66 + 280 = 346 \equiv 38 \pmod{77}\)Therefo
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990499 |
3 | 190% | 1993d | 19 |
nid:1768182517987
LinAlg
Express \(\text{Sol}(A, b)\) in standard form:
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users
190%
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nid:1768182517987
Q: Express \(\text{Sol}(A, b)\) in standard form:
A: \(\textbf{Sol}(A, 0) = \textbf{N}(A)\) as we search for the zeros. We thus first find the nullspace, and then shift it by an arbitrary solution of \(Ax = b\).Let \(s\) be some solution of \(Ax = b\). Then \[ \textbf{Sol}(A, b) = \{s + x : x \in \textbf{N}(A)\} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182517988 |
3 | 190% | 722d | 16 |
nid:1768182518324
c1
LinAlg
unique
3
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1/4
users
190%
ease
nid:1768182518324
Cloze c1
Cloze answer: unique
Q: For \(A\) written in CR-Decomposition \(A = CR'\), \(R'\) is {{c1:: unique::property? and why proof?}}.
A: \(R'\) is unique because the \(C\) is linearly independent and there's only one way to write a vector (the columns of \(A\)) as the linear combination of independent vectors.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182518324 |
3 | 190% | 1187d | 18 |
nid:1768182517631
c3
LinAlg
one unique inverse \(-v\) for all \(v\)
3
lapses
1/4
users
190%
ease
nid:1768182517631
Cloze c3
Cloze answer: one unique inverse \(-v\) for all \(v\)
Q: In a vector space \(V\) three important properties hold:{{c1::\(0v = 0\) for all \(v\)}}{{c2:: there is only one \(0\)}}{{c3:: one unique inverse \(-v\) for all \(v\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182517632 |
3 | 190% | 1363d | 17 |
nid:1768182517848
c2
LinAlg
\(b \neq 0\)
3
lapses
1/4
users
190%
ease
nid:1768182517848
Cloze c2
Cloze answer: \(b \neq 0\)
Q: If {{c2::\(b \neq 0\)}}, \(\textbf{Sol}(A, b)\) is {{c1::not a subspace of \(\mathbb{R}^n\)}}.
A: Because it doesn't contain the zero vector!If \(b \neq 0\), the the solution space is "shifted" off the origin:
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182517849 |
3 | 190% | 1663d | 18 |
nid:1768182518360
LinAlg
How do we find a basis for the nullspace of \(A\)?
3
lapses
1/4
users
190%
ease
nid:1768182518360
Q: How do we find a basis for the nullspace of \(A\)?
A: Compute the RREF form \(R\) of \(A\) (\(MA\) has the same nullspace as \(A\): \(\textbf{N}(A) = \textbf{N}(MA)\))Remove any zero rows (because \(0^\top x = 0\) regardless of \(x\))Solve for \(Rx = 0\):We seperate the matrix into the identity and the "rest". Note that for this we take colu
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182518361 |
3 | 190% | 1767d | 17 |
nid:1765553400173
LinAlg
What is a property that always holds for linear transformati...
3
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1/4
users
190%
ease
nid:1765553400173
Q: What is a property that always holds for linear transformations?
A: \(T(0) = 0\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765553400173 |
3 | 190% | 1885d | 15 |
nid:1772549069397
c1
A&W
Big \(O\) von Matching-Algorithmen:Für bipartite Graphen
...
3
lapses
1/4
users
190%
ease
nid:1772549069397
Cloze c1
Q: Big \(O\) von Matching-Algorithmen:Für bipartite Graphen
\( O(|V|^{1/2} \cdot |E|) \) Hopcroft-Karp (ungewichtet)
\( O(|E|^{1+o(1)}) \) (mit polynominellen Gewichte)
Für allgemeine Graphen (mit polynominellen Gewichten)
\( O({{c1::|V|^{1/2
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772549069398 |
3 | 190% | 5d | 17 |
nid:1779487730642
A&W
Clarkson-Algorithmus.Welche erwartete Laufzeit hat der Algor...
3
lapses
1/4
users
190%
ease
nid:1779487730642
Q: Clarkson-Algorithmus.Welche erwartete Laufzeit hat der Algorithmus, und wie folgt sie aus \(\Pr[T \geq k] \leq 0.993^k n\)?
A: Der Algorithmus berechnet \(C(P)\) in erwartet \(O(n \log n)\) Zeit.Die Laufzeit ist \(O(nT)\). Setze \(k_0 := \lfloor -\log_{0.993} n \rfloor = O(\log n)\). Dann\[\begin{gathered}\mathbb{E}[T] = \sum_{k \geq 1} \Pr[T \geq k] \;\leq\; \sum_{k=1}^{k_0} 1 \;+\; \sum_{k > k_0} 0.993^{\,k} n \\= k_0 \;+\; \underbrace{\sum_{k' \geq 0} 0.993^{\,k'} \cdot 0.993^{\,k_0 + 1} n}_{\leq\, O(1)} \;=\; k_0 + O(1) = O(\log n).\end{gathered}\]Also \(\mathbb{E}[\text{Laufzeit}] = O(n \cdot \mathbb{E}[
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487730643 |
3 | 190% | 3d | 17 |
nid:1779798950990
c2
A&W
JarvisWrap: erwartete Anzahl Hüllen-Ecken \(h\) bei zufällig...
3
lapses
1/4
users
190%
ease
nid:1779798950990
Cloze c2
Q: JarvisWrap: erwartete Anzahl Hüllen-Ecken \(h\) bei zufälligen Punkten.Punkte zufällig in einem Quadrat: \(\mathbb{E}[h] = {{c1::O(\log n)}}\).Punkte zufällig in einem Kreis: \(\mathbb{E}[h] = {{c2::O(\sqrt[3]{n})}}\).
A: In diesen Verteilungen ist JarvisWrap also im Erwartungswert deutlich schneller als \(O(n^2)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779798950990 |
3 | 190% | 2d | 15 |
nid:1780223730656
c2
A&W
Analyse Teil 2: Schranke für \(\tilde n_v \geq 20 n_v\)Defin...
3
lapses
1/4
users
190%
ease
nid:1780223730656
Cloze c2
Q: Analyse Teil 2: Schranke für \(\tilde n_v \geq 20 n_v\)Definiere \(Y'_{i,v}=1\) falls \(x_{i,v} \leq \tfrac{1}{20 n_v}\), sonst \(0\). Mit dem Union-Bound-Fakt \(1-(1-x)^n \leq nx\):\[\Pr[Y'_{i,v}=1] = 1-\left(1-\tfrac{1}{20n_v}\right)^{n_v} \leq {{c1::\tfrac{1}{20} }}\]Aus \(\tilde n
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780223730657 |
3 | 190% | 4d | 16 |
nid:1773311325220
c1
A&W
\leq 5
3
lapses
1/4
users
190%
ease
nid:1773311325220
Cloze c1
Cloze answer: \leq 5
Q: Ist ein Graph planar (kann überkreuzungsfrei in der Ebene gezeichnet werden), so gibt es immer einen Knoten vom Grad \({{c1::\leq 5}}\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773311325220 |
3 | 190% | 4d | 16 |
nid:1776171249095
c1
A&W
Eine Zufallsvariable \(X\) mit Dichte\[f_X(i) = \begin{cases...
3
lapses
1/4
users
190%
ease
nid:1776171249095
Cloze c1
Q: Eine Zufallsvariable \(X\) mit Dichte\[f_X(i) = \begin{cases}{{c1:: \frac{e^{-\lambda} \lambda^i}{i!} }} & \text{für } i \in \mathbb{N}_0 \\ 0 & \text{sonst} \end{cases}\]heisst {{c2::poisson-verteilt}} mit Parameter \(\lambda\).Man schreibt das auch als \({{c2::X \sim \text{Po}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1776171249097 |
3 | 190% | 32d | 19 |
nid:1779193767097
c2
A&W
Wiederholt man \(\text{Cut}(G)\) \(\alpha \binom{n}{2}\)-mal...
3
lapses
1/4
users
190%
ease
nid:1779193767097
Cloze c2
Q: Wiederholt man \(\text{Cut}(G)\) \(\alpha \binom{n}{2}\)-mal und gibt den kleinsten erhaltenen Wert aus, so gilt:Laufzeit \({{c1::\mathcal{O}(\alpha n^4)}}\).Der ausgegebene Wert ist mit Wahrscheinlichkeit mindestens \({{c2::1 - e^{-\alpha} }}\) gleich \(\mu(G)\).
A: Begründung Erfolgswahrscheinlichkeit: jede Einzelausführung scheitert mit Wkt. \(\leq 1 - 1/\binom{n}{2}\). Bei \(\alpha\binom{n}{2}\) unabhängigen Wiederholungen ist die Misserfolgswkt. höchstens \((1 - 1/\binom{n}{2})^{\alpha\binom{n}{2}} \leq e^{-\alpha}\). Mit \(\alpha := \ln n\) erhält man Zeit \(\mathcal{O}(n^4 \log n)\) bei Fehlerwkt. \(\leq 1/n\); aber diese Laufzeit hatten wir bereits deterministisch.
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nid:1779193767128
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Bootstrapping. Hat man bereits einen \(\mathcal{O}(n^c)\)-Al...
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Q: Bootstrapping. Hat man bereits einen \(\mathcal{O}(n^c)\)-Algorithmus mit Erfolgswkt. \(\geq 1 - e^{-1}\), so liefert die gleiche Konstruktion einen Algorithmus mit Erfolgswkt. \(\geq 1 - e^{-\alpha}\) und Laufzeit\[{{c1::\mathcal{O}\!\left(\alpha\!\left(\tfrac{n^4}{t^2} + n^2 t^{c-2}\rig
A: Die Folge der Exponenten ist \(4 \to 3 \to 8/3 \approx 2.666 \to 5/2 = 2.5 \to 12/5 = 2.4 \to 7/3 \approx 2.333 \to \ldots\); sie konvergiert gegen \(2\). Den polylog-Faktor bringt erst die rekursive Verzweigung (siehe KargerStein-Pseudocode).
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nid:1774917594265
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Für eine beliebige Zufallsvariable \(X\) gilt: \[\operatorna...
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Cloze c1
Q: Für eine beliebige Zufallsvariable \(X\) gilt: \[\operatorname{Var}[X] = {{c1::\mathbb{E}[X^2] - \mathbb{E}[X]^2::\text{Linearität} }}\]Proof Included
A: Sei \(\mu := \mathbb{E}[X]\).Nach Definition gilt \(\operatorname{Var}[X] = \mathbb{E}[(X - \mu)^2] = \mathbb{E}[X^2 - 2\mu \cdot X + \mu^2]\) Aus der Linearität des Erwartungswertes (Satz 2.33) folgt \[\mathbb{E}[X^2 - 2\mu \cdot X + \mu^2] = \mathbb{E}[X^2] - 2\mu \cdot \mathbb{E}[X] + \mu^2\]Damit erhalten wir \[\operatorname{Var}[X] = \mathbb{E}[X^2] - 2\mu \cdot \mathbb{E}[X] + \mu^2 = \mathbb{E}[X^2] - \mathbb{E}[X]^2\]
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nid:1779487730649
c2
A&W
Linearzeit
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Cloze answer: Linearzeit
Q: Anmerkungen zum Clarkson-Algorithmus.Der Algorithmus funktioniert auch für die kleinste umschliessende Kugel (oder Ellipse) in {{c1::allen Dimensionen}}, mit anderen Konstanten statt der \(11\). Bei {{c1::fixer Dimension}} bleibt die Laufzeit \(O(n \log n)\).Es gibt auch
A: Idee von [Clarkson '95].
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1 - (1-x)^n \leq n x
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Cloze answer: 1 - (1-x)^n \leq n x
Q: Union-Bound-FaktFür alle \(x \in [0,1]\) und \(n \in \mathbb{N}\) gilt\[{{c1::1 - (1-x)^n \leq n x}}\]
A: Beweisidee: Für unabhängige \(X_1,\ldots,X_n \sim \mathrm{Ber}(x)\) ist \(\Pr[X_1=1 \vee \ldots \vee X_n=1] = 1-(1-x)^n\); die Union Bound liefert andererseits \(\Pr[X_1=1 \vee \ldots \vee X_n=1] \leq \sum_i \Pr[X_i=1] = n x\).Spezialfall der Booleschen Ungleichung, im Beweis von ReachabilityCounting (Teil 2) verwendet.
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nid:1779487730629
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A&W
Wachstum von \(P'\) pro Runde (\(r = 11\)).Nach Verdopplung ...
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Q: Wachstum von \(P'\) pro Runde (\(r = 11\)).Nach Verdopplung der Punkte in \(P' \setminus C^{\bullet}(R)\) ist die erwartete neue Punktzahl\[\mathbb{E}[|P'_{\text{neu}}|] \;\leq\; N + 3\frac{N}{r+1} \;=\; {{c1::\left(1 + \tfrac{3}{12}\right) N = \tfrac54 N}}.\]Per Linearität und Induktion
A: Setzt man \(r = 11\), wird \(\tfrac{3}{r+1} = \tfrac{3}{12} = \tfrac14\): genau der Wert, der das Wachstum klein genug hält.
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nid:1776174687031
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\(N\) und \(X\) seien zwei unabhängige Zufallsvariablen mit ...
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Cloze c1
Q: \(N\) und \(X\) seien zwei unabhängige Zufallsvariablen mit \(W_N \subseteq \mathbb{N}\). Weiter sei\[Z := \sum_{i=1}^{N} X_i,\]wobei \(X_1, X_2, \ldots\) unabhängige Kopien von \(X\) sind.Dann gilt:\[\mathbb{E}[Z] = {{c1::\mathbb{E}[N] \cdot \mathbb{E}[X]}}.\]
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Monte-Carlo Algorithmus für Long-Path: AnalyseWiederhole \(\...
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Cloze c3
Q: Monte-Carlo Algorithmus für Long-Path: AnalyseWiederhole \(\lceil \varepsilon \cdot e^k \rceil\) Mal: zufällig färben, dann \(\text{Bunt}\) ausführen.Laufzeit: \(O\!\big({{c1::\varepsilon \cdot (2e)^k \cdot k \cdot m}}\big)\), wobei \(m = |E|\).Antwortet der Algorithm
A: Faktor \((2e)^k\) statt \(e^k\): einmalige bunte Färbung kostet \(O(2^k \cdot k \cdot m)\) (Mengen \(S \subseteq [k]\) im DP), und es werden \(\Theta(\varepsilon \cdot e^k)\) Versuche gemacht.Bemerkung: \((2e)^k\) ist konstant in \(n\). Damit wird Long-Path für festes \(k\) in Polynomialzeit lösbar (FPT-Resultat von Alon, Yuster, Zwick 1995).
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nid:1773310695996
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chromatischer Zahl \(\geq r\)
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Cloze answer: chromatischer Zahl \(\geq r\)
Q: \(\forall k \in \mathbb{N},\ \forall r \in \mathbb{N}\): Es gibt Graphen ohne einen Kreis mit Länge \(\leq k\), aber mit {{c1::chromatischer Zahl \(\geq r\)}}.
A: Lokal sieht der Graph aus wie ein Baum (alle Knoten, die man von einem \(v\) aus in \(k/2\) Schritten erreichen kann).
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Seien \(X_1, \ldots, X_n\) unabhängige Bernoulli-verteilte Z...
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Cloze c3
Q: Seien \(X_1, \ldots, X_n\) unabhängige Bernoulli-verteilte Zufallsvariablen mit \(\Pr[X_i = 1] = p_i\) und \(\Pr[X_i = 0] = 1 - p_i\).
Dann gilt für \(X = \sum_{i=1}^{n} X_i\)\(\Pr[X \geq (1+\delta)\,\mathbb{E}[X]] \;\leq\; {{c1::e^{-\frac{1}{3}\delta^2\,\mathbb{E}[X]} }}\) für alle
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|E|
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Cloze answer: |E|
Q: Jeder Graph kann in Zeit \(O({{c1::|E|}})\) mit \(\Delta(G)+1\) Farben gefärbt werden.
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Seien \(A_1,\ldots,A_n\) paarweise disjunkt, \(B\subseteq\bi...
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Cloze c1
Q: Seien \(A_1,\ldots,A_n\) paarweise disjunkt, \(B\subseteq\bigcup A_i\), \(\Pr[B]>0\). Dann gilt für jedes \(i\):\[ \Pr[A_i|B] = {{c1::\frac{\Pr[B|A_i]\cdot\Pr[A_i]}{\sum_{j=1}^{n}\Pr[B|A_j]\cdot\Pr[A_j]} }}. \]Proof Included
A: (Satz von Bayes)Proof: Nach Definition gilt \(\Pr[A_i|B]=\Pr[A_i\cap B]/\Pr[B]\). Zähler: \(\Pr[A_i\cap B]=\Pr[B|A_i]\cdot\Pr[A_i]\). Nenner: \(\Pr[B]=\sum_j\Pr[B|A_j]\Pr[A_j]\) (totale Wahrscheinlichkeit). \(\square\)Zentrale Anwendung: Die Konditionierungsrichtung "umkehren" - von \(\Pr[B|A_i]\) (leicht zu messen) zu \(\Pr[A_i|B]\) (was wir wissen wollen).
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nid:1774631277135
A&W
Falls \(A\) und \(B\) unabhängig sind, beweise, dass \(\bar{...
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Q: Falls \(A\) und \(B\) unabhängig sind, beweise, dass \(\bar{A}\) und \(B\) ebenfalls unabhängig sind. Proof Included
A: Zu zeigen: \(\Pr[\bar{A}\cap B]=\Pr[\bar{A}]\cdot\Pr[B]\).\[\begin{gathered}\Pr[\bar{A}\cap B] = \Pr[B] - \Pr[A\cap B] \\ = \Pr[B] - \Pr[A]\Pr[B] \\ = (1-\Pr[A])\Pr[B] = \Pr[\bar{A}]\Pr[B]. \quad\square\end{gathered}\]Folgerung: Falls \(A_1,\ldots,A_n\) gemeinsam unabhängig sind, so auch jede Familie, die durch Ersetzen einiger \(A_i\) durch \(\bar{A}_i\) entsteht (Lemma 2.23).
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2(n+1) \ln n + O(n)
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Cloze c2
Cloze answer: 2(n+1) \ln n + O(n)
Q: Ein klassisches Beispiel für einen Las-Vegas-Algorithmus ist QuickSort: der Algorithmus sortiert {{c1::immer richtig}}, aber die konkrete Laufzeit hängt von der zufälligen Wahl der Pivot-Elemente ab.Mit \(\mathrm{Partition}(A, \ell, r, p)\) sei eine Prozedur bezeichnet, die mit
A: Im Worst Case ist die Laufzeit \(\Theta(n^2)\) (z.B. wenn das Pivot stets minimal oder maximal ist). Erwartet aber nur \(O(n \log n)\), und das unabhängig von der Eingabe, weil das Pivot zufällig gewählt wird.Annahme im Beweis: alle Elemente von \(A\) sind paarweise verschieden, sodass \(t\) eindeutig ist.
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Für eine beliebige Zufallsvariable \(X\) und \(a, b \in \mat...
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Cloze c1
Q: Für eine beliebige Zufallsvariable \(X\) und \(a, b \in \mathbb{R}\) gilt: \[\operatorname{Var}[a \cdot X + b] = {{c1::a^2 \cdot \operatorname{Var}[X]}}\]Proof Included
A: Beweis:\(\operatorname{Var}[X + b] = \mathbb{E}[(X + b - \mathbb{E}[X + b])^2]\) \(= \mathbb{E}[(X - \mathbb{E}[X])^2]\) \(= \operatorname{Var}[X]\) Mit Hilfe von \(\text{Var}[X] = \mathbb{E}[X^2] - \mathbb{E}[X]^2\) erhalten wir \(\operatorname{Var}[a \cdot X] = \mathbb{E}[(aX)^2] - \mathbb{E}[aX]^2\) \(= a^2 \mathbb{E}[X^2] - (a\mathbb{E}[X])^2 = a^2 \cdot \operatorname{Var}[X]\)
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A&W
Eine Zufallsvariable \(X\) mit Dichte\[f_X(i) = \begin{cases...
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Cloze c1
Q: Eine Zufallsvariable \(X\) mit Dichte\[f_X(i) = \begin{cases} {{c1::p \cdot (1 - p)^{i-1} }} & \text{für } i \in \mathbb{N} \\ 0 & \text{sonst} \end{cases}\]heisst {{c2::geometrisch verteilt}} mit Erfolgswahrscheinlichkeit \(p\).Man schreibt das auch als \({{c2::X \sim \t
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nid:1774631269375
c3
A&W
Die Rekursionsformel des Pascalschen Dreiecks lautet: \[\bin...
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Q: Die Rekursionsformel des Pascalschen Dreiecks lautet: \[\binom{n}{k} = {{c3::\binom{n-1}{k-1} + \binom{n-1}{k} }}\]
A: Intuition: Fixiere Element \(x\).\(x\) dabei → noch \(k-1\) aus \(n-1\) wählen\(x\) nicht dabei → alle \(k\) aus \(n-1\) wählenPascalsches Dreieck (Eintrag in Zeile \(n\), Position \(k\) ist \(\binom{n}{k}\)):\[\begin{array}{ccccccccc} & & & & 1 \\ & & & 1 & & 1 \\ & & 1 & & 2 & & 1 \\ &
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nid:1774358596854
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Seien \(A_1, \ldots, A_n\) paarweise disjunkte Ereignisse un...
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Cloze c1
Q: Seien \(A_1, \ldots, A_n\) paarweise disjunkte Ereignisse und sei \(B \subseteq A_1 \cup \cdots \cup A_n\).
Dann gilt: \[\Pr[B] = {{c1::\sum_{i=1}^{n} \Pr[B\mid A_i] \cdot \Pr[A_i]}}.\]
A: Satz von der totalen WahrscheinlichkeitBeispiel: Ziegenproblem
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nid:1774487164704
c1
A&W
Summe von Indikatoren
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Cloze answer: Summe von Indikatoren
Q: Um \(\mathbb{E}[X]\) zu berechnen, schreibe \(X\) als {{c1::Summe von Indikatoren}}:\[X = {{c1::X_{A_1} + X_{A_2} + \cdots + X_{A_n} }},\]dann gilt per Linearität:\[\mathbb{E}[X] = {{c2::\Pr[A_1] + \Pr[A_2] + \cdots + \Pr[A_n] }} \]
A: Unabhängigkeit nicht nötig!Beispiel: Erwartete Anzahl Fixpunkte einer zufälligen Permutation von \([n]\)?\(X_i = [i \text{ ist Fixpunkt}]\), \(\Pr[X_i = 1] = \frac{1}{n}\), also \(\mathbb{E}[X] = n \cdot \frac{1}{n} = 1\).
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nid:1774487165116
c2
A&W
minimales (gewichtsminimales) perfektes Matching
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Cloze c2
Cloze answer: minimales (gewichtsminimales) perfektes Matching
Q: Für \(n\) gerade und \(\ell : \binom{[n]}{2} \to \mathbb{N}_0\) kann man in Zeit \(O({{c1::n^3}})\) ein {{c2::minimales (gewichtsminimales) perfektes Matching}} in \(K_n\) finden.
A: Das ist der Blossom-Algorithmus.Dies wird im Christofides-Algorithmus für das metrische TSP benötigt.
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nid:1774487165205
Analysis
Was gilt auf dem Rand des Konvergenzkreises \(|x - a| = R\) ...
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Q: Was gilt auf dem Rand des Konvergenzkreises \(|x - a| = R\) einer Potenzreihe?
A: Keine allgemeine Aussage - kommt auf den Einzelfall an:\(\sum \frac{x^n}{n}\): divergiert für \(x = 1\), konvergiert für \(x = -1\) (Leibniz)\(\sum \frac{x^n}{n^2}\): konvergiert für alle \(|x| = 1\) (absolut)\(\sum x^n\): divergiert für alle \(|x| = 1\)
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c1
Analysis
Geometrische Reihe als Potenzreihendarstellung (konvergiert ...
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Cloze c1
Q: Geometrische Reihe als Potenzreihendarstellung (konvergiert für \(-1 < x < 1\)):\[ \frac{1}{1 - x} = {{c1::\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \dots }}\]
A: Werkzeug: Durch Substitution erhält man viele weitere Reihen, etwa \(\tfrac{1}{1+x} = \sum (-1)^n x^n\) oder \(\tfrac{1}{1-x^2} = \sum x^{2n}\).
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nid:1779798962573
c1
Analysis
homogen in den Variablen
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Cloze answer: homogen in den Variablen
Q: Eine DGl der Form \(y' = g\!\left(\dfrac{y}{x}\right)\) heisst {{c1::homogen in den Variablen}} und lässt sich durch die Substitution\[ {{c2::u = \frac{y}{x} }} \]in eine DGl mit {{c3::trennbaren Variablen}} überführen.
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c1
Analysis
Reihen mit nicht-negativen Gliedern (ab einem Index \(N\)): ...
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Cloze c1
Cloze answer: Reihen mit nicht-negativen Gliedern (ab einem Index \(N\)): \(0 \leq a_n \leq b_n\) für alle \(n \geq N\)
Q: Das Majoranten-/Minorantenkriterium gilt nur für {{c1::Reihen mit nicht-negativen Gliedern (ab einem Index \(N\)): \(0 \leq a_n \leq b_n\) für alle \(n \geq N\)}}.
A: Für alternierende Reihen ist es nicht direkt anwendbar, erst Absolutkonvergenz mit \(\sum |a_n|\) zeigen, dann folgt Konvergenz.Häufiger Fehler: Vergleich von \((-1)^n/n\) mit \(1/n\) über Majorante. Das scheitert, da \((-1)^n/n \not\geq 0\).
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nid:1774487165306
c1
Analysis
der erste weggelassene Term
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Cloze c1
Cloze answer: der erste weggelassene Term
Q: Beim Leibniz-Kriterium gilt die Fehlerabschätzung:\[|S - S_n| \leq {{c1::a_{n+1} }}\]D.h. der Fehler ist höchstens so gross wie {{c1::der erste weggelassene Term}}.
A: Nützlich zur numerischen Approximation alternierender Reihen.
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c1
Analysis
Es sei \(f : \mathbb{D}(f) \to \mathbb{R}\), es sei \(x_0 \i...
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Cloze c1
Q: Es sei \(f : \mathbb{D}(f) \to \mathbb{R}\), es sei \(x_0 \in \mathbb{R}\) und es gelte \[{{c1::\mathbb{D}(f) \cap (x_0 - \delta,\, x_0 + \delta) \neq \emptyset \quad \forall \delta > 0}}\]Dann ist \(L \in \mathbb{R}\) der Grenzwert/Limes von \(f(x)\) an der Stelle \(x_0\), falls gilt \[{{c2::\be
A: Beachte, dass die Funktion nicht unbedingt an der Stelle \(x_0\) des Grenzwerts definiert sein muss (siehe Sprungstelle, Definitionslücke).
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3 | 190% | 51d | 20 |
nid:1774487166307
c1
Analysis
Exponentialreihe:\[\exp(z) = {{c1:: \sum_{n=0}^\infty \frac{...
3
lapses
1/4
users
190%
ease
nid:1774487166307
Cloze c1
Q: Exponentialreihe:\[\exp(z) = {{c1:: \sum_{n=0}^\infty \frac{z^n}{n!} }}\]Diese Reihe konvergiert {{c2::absolut für alle \(z \in \mathbb{C}\)::Konvergenztyp}}.
A: (Konvergenzradius \(R = \infty\))
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487166308 |
3 | 190% | 64d | 19 |
nid:1774917594689
c2
Analysis
Es sei \(f : \mathbb{D}(f) \to \mathbb{R}\), es sei \(x_0 \i...
3
lapses
1/4
users
190%
ease
nid:1774917594689
Cloze c2
Q: Es sei \(f : \mathbb{D}(f) \to \mathbb{R}\), es sei \(x_0 \in \mathbb{R}\) und es gelte \[{{c1::\mathbb{D}(f) \cap (x_0 - \delta,\, x_0 + \delta) \neq \emptyset \quad \forall \delta > 0}}\]Dann ist \(L \in \mathbb{R}\) der Grenzwert/Limes von \(f(x)\) an der Stelle \(x_0\), falls gilt \[{{c2::\be
A: Beachte, dass die Funktion nicht unbedingt an der Stelle \(x_0\) des Grenzwerts definiert sein muss (siehe Sprungstelle, Definitionslücke).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774917594690 |
3 | 190% | 69d | 19 |
nid:1774487165589
c5
Analysis
Eine Potenzreihe hat die Form \({{c5:: \displaystyle\sum_{k=...
3
lapses
1/4
users
190%
ease
nid:1774487165589
Cloze c5
Q: Eine Potenzreihe hat die Form \({{c5:: \displaystyle\sum_{k=0}^\infty c_k (x - a)^k }}\), wobei:\(a\) ist {{c1::der Entwicklungspunkt (Zentrum)}}\(c_0, c_1, \ldots\) sind {{c2::die Koeffizienten}}\(x\) ist {{c3::das Argument}}\((a - R,\, a + R)\) ist {{c
A: Spezialfall \(a = 0\): \(\sum c_k x^k\) - Entwicklungspunkt im Ursprung.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487165590 |
3 | 190% | 57d | 18 |
nid:1772928333399
c1
Analysis
\[ \cos\!\left(\frac{11\pi}{6}\right) = {{c1::\frac{\sqrt{3}...
3
lapses
1/4
users
190%
ease
nid:1772928333399
Cloze c1
Q: \[ \cos\!\left(\frac{11\pi}{6}\right) = {{c1::\frac{\sqrt{3} }{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333399 |
3 | 190% | 106d | 20 |
nid:1771973928588
c2
Analysis
Beweis: Für alle \(a < b\) in \(\mathbb{R}\) existiert ein \...
3
lapses
1/4
users
190%
ease
nid:1771973928588
Cloze c2
Q: Beweis: Für alle \(a < b\) in \(\mathbb{R}\) existiert ein \(\mathbb{Q}\) mit \(a < q < b\){{c1:: Wähle nach Archimedischem Prinzip \(n \in \mathbb{N}\) so dass \(\frac{1}{n} < b - a\).}}{{c2:: \(\frac{m}{n} \mid m \in \mathbb{Z}\) diese
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771973928590 |
3 | 190% | 118d | 18 |
nid:1774487165742
c1
Analysis
konvergiert, aber die Reihe der Beträge \(\sum |a_k|\) diver...
3
lapses
1/4
users
190%
ease
nid:1774487165742
Cloze c1
Cloze answer: konvergiert, aber die Reihe der Beträge \(\sum |a_k|\) divergiert
Q: Eine Reihe heisst bedingt konvergent, wenn sie {{c1::konvergiert, aber die Reihe der Beträge \(\sum |a_k|\) divergiert}}.Example Included
A: (D.h. nicht absolut konvergiert..)Beispiel: \(\sum \frac{(-1)^n}{n}\) ist bedingt konvergent.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487165742 |
3 | 190% | 79d | 17 |
nid:1778588922431
c1
PProg
the lock-free version can be slower than the blocking one; a...
3
lapses
1/4
users
190%
ease
nid:1778588922431
Cloze c1
Cloze answer: the lock-free version can be slower than the blocking one; atomic operations are expensive and the CAS retry loop wastes work under contention
Q: Empirical comparison of a synchronised stack vs. a CAS-based lock-free stack under high contention shows that {{c1::the lock-free version can be slower than the blocking one}}, because {{c1::atomic operations are expensive and the CAS retry loop wastes work under contention}}. Addin
A: Moral: lock-freedom is a progress property, not an automatic performance property. Contention management (backoff, elimination, combining) is still needed to make lock-free structures fast.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922431 |
3 | 190% | 8d | 14 |
nid:1777538021707
c4
PProg
actions that start a thread
3
lapses
1/4
users
190%
ease
nid:1777538021707
Cloze c4
Cloze answer: actions that start a thread
Q: Synchronization actions (SA) in the JMM are: {{c1::read/write of a volatile variable}}; {{c2::lock and unlock of a monitor}}; {{c3::the first and last action of a thread (synthetic)}}; {{c4::actions that start a thread}};
A: SA are the building blocks of the synchronization order (SO). Anything else is an "ordinary" action.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777751412394 |
3 | 190% | 165d | 16 |
nid:1773754343154
c1
PProg
Overhead and synchronization barriers
3
lapses
1/4
users
190%
ease
nid:1773754343154
Cloze c1
Cloze answer: Overhead and synchronization barriers
Q: What factors limit scalability?Sequential part of the program (Amdahl's law)Data structures and algorithmsWork distribution strategyWork scheduling strategy{{c1::Overhead and synchronization barriers}}Memory access and caches
A: How much time is spent on synchronization, locking, context switching?Frequent context switches introduce delays that degrade parallel performance.High contention for shared resources or excessive synchronization barriers create bottlenecks that limit parallel efficiency.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773754343155 |
3 | 190% | 241d | 20 |
nid:1761491477383
DiskMat
When is a relation \(\rho\) on set \(A\) symmetric?
3
lapses
1/4
users
235%
ease
nid:1761491477383
Q: When is a relation \(\rho\) on set \(A\) symmetric?
A: When \(a \ \rho \ b \Longleftrightarrow b \ \rho \ a\) for all \(a, b \in A\), i.e., \(\rho = \hat{\rho}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477384 |
3 | 235% | 65d | 19 |
nid:1762856073615
c1
DiskMat
least (greatest) element of \(A\)
3
lapses
1/4
users
220%
ease
nid:1762856073615
Cloze c1
Cloze answer: least (greatest) element of \(A\)
Q: Consider the poset \((A; \preceq)\).\(a \in A\) is the {{c1::least (greatest) element of \(A\)}} if {{c2::\(a \preceq b\) (\(a \succeq b) \) for all \(b \in A\)}}
A: Note that a least or a greatest element need not exist. However, there can be at most one least element, as suggested by the word “the” in the definition. This follows directly from the antisymmetry of \(\preceq\). If there were two least elements, they would be mutually comparable, and hence must be equal.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073623 |
3 | 220% | 4d | 11 |
nid:1765198200589
A&D
How can one get a lower bound for the function \(n!\) ?
3
lapses
1/4
users
250%
ease
nid:1765198200589
Q: How can one get a lower bound for the function \(n!\) ?
A: One could simply take only the largest 90% of elements: \(n! \geq 1 \cdot 2 \cdot ... \cdot n \geq n/10 \cdot ... \cdot n\)\(\geq (n/10)^{0.9n}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765198200589 |
3 | 250% | 27d | 16 |
nid:1765296364773
c1
A&D
O(\log(n))
3
lapses
1/4
users
235%
ease
nid:1765296364773
Cloze c1
Cloze answer: O(\log(n))
Q: Choose a tight bound!\({{c1::O(\log(n))}}\leq {{c2::O(n)}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765296364773 |
3 | 235% | 10d | 17 |
nid:1766000828772
DiskMat
What is the number of generators of \(\mathbb{Z}_n^*\)?
3
lapses
1/4
users
205%
ease
nid:1766000828772
Q: What is the number of generators of \(\mathbb{Z}_n^*\)?
A: 1. Verify that \(\mathbb{Z}_n^*\)is cyclic (iff n = 2, 4, \(p^e\), \(2p^e\), with \(e \ge 1\) and \(p\) is an odd prime)2. If \(\mathbb{Z}_n^*\) is cyclic then it is isomorphic to \(\mathbb{Z}_{\varphi(n)}^+\) (by Lemma) 3. The number of generators of \(\mathbb{Z}_{\varphi(n)}^+\) is \(\varphi(\varphi(n))\) as it is the number of elements coprime to the group order.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766000828772 |
3 | 205% | 3d | 9 |
nid:1766488260288
A&D
Jump Game
3
lapses
1/4
users
205%
ease
nid:1766488260288
Q: Jump Game
A: \(O(n)\) (hyper-optimised version)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766488260289 |
3 | 205% | 1d | 12 |
nid:1766522811173
c1
A&D
a path
3
lapses
1/4
users
205%
ease
nid:1766522811173
Cloze c1
Cloze answer: a path
Q: The shortest walk is always {{c1::a path}}.
A: This is due to the triangle inequality, given that no negative cycles exist.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766522811173 |
3 | 205% | 1d | 9 |
nid:1766580201542
A&D
Cut and Paste Proof of Cut-Property:
3
lapses
1/4
users
190%
ease
nid:1766580201542
Q: Cut and Paste Proof of Cut-Property:
A: Let \((S, V \setminus S)\) be any cut of a graph \(G\).Let \(e = (u,v)\) be the minimal edge crossing this cut. We want to show that \(e \in T\). Assume \(e \not \in T\) for contradiction.Since \(T\) is a spanning tree, \(T \cup {u}\) contains a cycle, crossing the cut at least twice (once via \(e\) and once via another edge \(e’\).)W
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766580201542 |
3 | 190% | 1d | 11 |
nid:1769377096401
c1
EProg
(Husky) dog; Casting further down than dynamic type
3
lapses
1/4
users
190%
ease
nid:1769377096401
Cloze c1
Cloze answer: (Husky) dog; Casting further down than dynamic type
Q: Runtime Errors for Casting:
{{c1::(Husky) dog; Casting further down than dynamic type}}
{{c2:: (Cat) dog; Casting into sibling type}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1769377096402 |
3 | 190% | 2d | 9 |
nid:1771364277486
c2
PProg
during any possible execution, a memory location could be wr...
3
lapses
1/4
users
205%
ease
nid:1771364277486
Cloze c2
Cloze answer: during any possible execution, a memory location could be written from one thread, while concurrently being read or written from another thread.
Q: A program has a {{c1::data race}} if, {{c2::during any possible execution, a memory location could be written from one thread, while concurrently being read or written from another thread.}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771364277571 |
3 | 205% | 22d | 12 |
nid:1771366536198
c2
A&W
für alle Teilmengen \(X \subseteq E\) mit \(|X| < k\) gilt: ...
3
lapses
1/4
users
250%
ease
nid:1771366536198
Cloze c2
Cloze answer: für alle Teilmengen \(X \subseteq E\) mit \(|X| < k\) gilt: Der Graph \((V, E \setminus X)\) ist zusammenhängend
Q: Ein Graph \(G = (V, E)\) heisst {{c1::\(k\)-kanten-zusammenhängend}}, falls {{c2::für alle Teilmengen \(X \subseteq E\) mit \(|X| < k\) gilt: Der Graph \((V, E \setminus X)\) ist zusammenhängend}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771366536213 |
3 | 250% | 34d | 13 |
nid:1772569386221
c1
A&W
|E|
3
lapses
1/4
users
235%
ease
nid:1772569386221
Cloze c1
Cloze answer: |E|
Q: In \( k \)-regulären bipartiten Graphen kann man in Zeit \( O({{c1::|E|}}) \) ein perfektes Matching bestimmen.
A: Perfektes Matching in \(k\)-regulären bipartiten GraphenDas Skript erwähnt, dass es einen Algorithmus gibt, der in Zeit \(O(|E|)\) ein perfektes Matching in \(k\)-regulären bipartiten Graphen findet, sagt aber explizit: „Der allgemeine Fall ist deutlich schwieriger."Bewiesen wird im Skript nur der Spezialfall \(k = 2^k\) (Satz 1.54).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772569386221 |
3 | 235% | 4d | 11 |
nid:1772788241867
c1
Analysis
\[ \tan\!\left(\frac{5\pi}{3}\right) = {{c1::-\sqrt{3} }} \]
3
lapses
1/4
users
175%
ease
nid:1772788241867
Cloze c1
Q: \[ \tan\!\left(\frac{5\pi}{3}\right) = {{c1::-\sqrt{3} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772788241867 |
3 | 175% | 7d | 13 |
nid:1766314077300
c2
A&D
Eulerian walk (Eulerweg)
2
lapses
2/4
users
260%
ease
nid:1766314077300
Cloze c2
Cloze answer: Eulerian walk (Eulerweg)
Q: In graph theory, an {{c2::Eulerian walk (Eulerweg)}} is a {{c1::walk that contains every edge of the graph exactly once}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314077304 |
1 | 245% | 14d | 8 |
| niklas | cid:1762856073668 |
1 | 275% | 39d | 7 |
nid:1766314077314
c1
A&D
take the first element from the unsorted input and place it ...
2
lapses
2/4
users
238%
ease
nid:1766314077314
Cloze c1
Cloze answer: take the first element from the unsorted input and place it correctly in the sorted output
Q: In every iteration of insertion sort, we {{c1::take the first element from the unsorted input and place it correctly in the sorted output}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314077321 |
1 | 245% | 22d | 6 |
| lorenz | cid:1764867989687 |
1 | 230% | 823d | 11 |
nid:1766314094565
DiskMat
What is the Pigeonhole Principle?
2
lapses
2/4
users
230%
ease
nid:1766314094565
Q: What is the Pigeonhole Principle?
A: If a set of \(n\) objects is partitioned into \(k < n\) sets, then at least one of those sets contains at least \(\lceil \frac{n}{k} \rceil\) objects.
(If you have more pigeons than holes, at least one hole must contain multiple pigeons)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094572 |
1 | 230% | 16d | 8 |
| lorenz | cid:1764867989956 |
1 | 230% | 1316d | 9 |
nid:1766314094621
DiskMat
How are the rational numbers \(\mathbb{Q}\) defined using eq...
2
lapses
2/4
users
230%
ease
nid:1766314094621
Q: How are the rational numbers \(\mathbb{Q}\) defined using equivalence relations?
A: Let \(A = \mathbb{Z} \times (\mathbb{Z} \setminus \{0\})\) and \((a, b) \sim (c,d) \overset{\text{def}}{\Longleftrightarrow} ad = bc\)
Then: \(\mathbb{Q} \overset{\text{def}}{=} (\mathbb{Z} \times (\mathbb{Z} \setminus \{0\})) / \sim\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094629 |
1 | 230% | 23d | 7 |
| lorenz | cid:1764867990134 |
1 | 230% | 1082d | 8 |
nid:1766314094625
DiskMat
When is a poset \((A; \preceq)\) totally ordered (linearly o...
2
lapses
2/4
users
230%
ease
nid:1766314094635
DiskMat
When is a poset \((A; \preceq)\) well-ordered?
2
lapses
2/4
users
215%
ease
nid:1766314094664
c1
DiskMat
\(A^n\) (\(n\)-tuples) is countable
2
lapses
2/4
users
230%
ease
nid:1766314094664
Cloze c1
Cloze answer: \(A^n\) (\(n\)-tuples) is countable
Q: Which operations preserve countability?Let \(A\) and \(A_i\) for \(i \in \mathbb{N}\) be countable sets. Then: {{c1::\(A^n\) (\(n\)-tuples) is countable }}{{c2::\(\bigcup_{i\in \mathbb{N} } A_i\) (countable union) is countabl
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094677 |
1 | 230% | 16d | 9 |
| lorenz | cid:1764867990269 |
1 | 230% | 1819d | 10 |
nid:1766314094710
DiskMat
What important property do ideals in \(\mathbb{Z}\) have? (L...
2
lapses
2/4
users
230%
ease
nid:1766314094710
Q: What important property do ideals in \(\mathbb{Z}\) have? (Lemma 4.3)
A: For \(a, b \in \mathbb{Z}\), there exists \(d \in \mathbb{Z}\) such that \((a, b) = (d)\).
Every ideal can be generated by a single integer.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094731 |
1 | 230% | 15d | 9 |
| lorenz | cid:1764867990409 |
1 | 230% | 916d | 9 |
nid:1766314094737
DiskMat
Does \( p \mid a \land q \mid a \land \gcd(p, q) = 1 \implie...
2
lapses
2/4
users
222%
ease
nid:1766314094737
Q: Does \( p \mid a \land q \mid a \land \gcd(p, q) = 1 \implies pq \mid a \) hold? (Proof included)
A: Yes, but this has to be reproven before using.The proof technique is important. Replacing a neutral element by something it's equal to often is a smart move.
Proof: This is an important result for the exam:
\[p \mid a \land q \mid a \land \gcd(p, q) = 1 \implies pq \mid a\]
Which is the same as saying \(\exists k \in \mathbb{Z}\) such that \(a = pq \cdot k\).
Since \(p \mid a\) and \(q \mid a\), we have:
\[\exists k, k' \in \mathbb{Z} \text{ such
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094761 |
1 | 185% | 9d | 9 |
| niklas | cid:1762453251142 |
1 | 260% | 37d | 11 |
nid:1766314094897
c1
DiskMat
neutral to neutral: \(\psi(e_G) = e_h\)
2
lapses
2/4
users
230%
ease
nid:1766314094897
Cloze c1
Cloze answer: neutral to neutral: \(\psi(e_G) = e_h\)
Q: Lemma 5.5(ii): A group homomorphism \(\psi: G \rightarrow H\) maps {{c1::inverses to inverses: \(\psi(\widehat{a}) = \widetilde{\psi(a)}\)}} for all \(a\).{{c1::neutral to neutral: \(\psi(e_G) = e_h\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095008 |
1 | 230% | 9d | 8 |
| lorenz | cid:1764867991002 |
1 | 230% | 1107d | 10 |
nid:1766314094909
c1
DiskMat
For \(H\) to be a subgroup, it must have closure under {{c1:...
2
lapses
2/4
users
230%
ease
nid:1766314094983
DiskMat
How can you check if a polynomial of degree \(d\) is irreduc...
2
lapses
2/4
users
245%
ease
nid:1766314094983
Q: How can you check if a polynomial of degree \(d\) is irreducible?
A: To check if a polynomial of degree \(d\) is irreducible, check all monic irreducible polynomials of degree \(\leq d/2\) as possible divisors.
Why \(d/2\)? If \(a(x) = b(x) \cdot c(x)\) where \(b\) and \(c\) are non-constant, then \(\deg(b) + \deg(c) = \deg(a) = d\). So at least one of \(b\) or \(c\) has degree \(\leq d/2\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095133 |
1 | 230% | 3d | 8 |
| niklas | cid:1764859231520 |
1 | 260% | 21d | 11 |
nid:1766314095043
c1
DiskMat
\(a \ | \ bc\)
2
lapses
2/4
users
245%
ease
nid:1766940295685
c1
DiskMat
free symbols of a formula
2
lapses
2/4
users
230%
ease
nid:1766940295760
DiskMat
What does the semantics of a logic define?
2
lapses
2/4
users
222%
ease
nid:1766940295760
Q: What does the semantics of a logic define?
A: The semantics defines:1. A function \(free\) that assigns to each formula which symbols occur free2. A function \(\sigma\) that assigns truth values to formulas under interpretations3. The meaning and behavior of logical operators
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295901 |
1 | 200% | 3d | 8 |
| niklas | cid:1766418002707 |
1 | 245% | 9d | 10 |
nid:1766940295779
c1
DiskMat
restricted to a certain type of mathematical statement
2
lapses
2/4
users
222%
ease
nid:1766940295779
Cloze c1
Cloze answer: restricted to a certain type of mathematical statement
Q: A proof system is always {{c1::restricted to a certain type of mathematical statement}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295936 |
1 | 215% | 6d | 9 |
| lorenz | cid:1766448533176 |
1 | 230% | 1566d | 9 |
nid:1767089604933
LinAlg
Let \(T : \mathbb{R}^n \rightarrow \mathbb{R}^m\) be a linea...
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2/4
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nid:1767089604933
Q: Let \(T : \mathbb{R}^n \rightarrow \mathbb{R}^m\) be a linear transformation. There is a?
A: There is a unique \(m \times n\) matrix A such that \(T = T_A\) meaning that \(T(x) = T_A(x) = Ax\) for all \(x \in \mathbb{R}^n\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1767089604934 |
1 | 230% | 3d | 8 |
| lorenz | cid:1767105283299 |
1 | 230% | 1587d | 9 |
nid:1766531635612
A&D
What is the optimal substructure property of shortest paths?
2
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ease
nid:1766531635612
Q: What is the optimal substructure property of shortest paths?
A: Any subpath of a shortest path is itself the shortest path between its endpoints (requires no negative cycles).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635612 |
1 | 230% | 688d | 10 |
| niklas | cid:1766523798264 |
1 | 245% | 40d | 6 |
nid:1765198542527
A&D
Runtime to determine whether an Eulerian walk exists?
2
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users
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ease
nid:1766531635566
c1
A&D
\(\geq\)
2
lapses
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users
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nid:1765372936263
c1
A&D
{{c1:: \(\sum_{j = 1}^{n} \sum_{k = \textbf{j} }^{n} 1\)::Su...
2
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2/4
users
245%
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nid:1765372936263
Cloze c1
Q: {{c1:: \(\sum_{j = 1}^{n} \sum_{k = \textbf{j} }^{n} 1\)::Sum}} \(=\) {{c2:: \(\sum_{j = 1}^n (n - j + 1) = \frac{n(n + 1)}{2}\) }}
A: inner loop depends on outer
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936263 |
1 | 230% | 1172d | 12 |
| niklas | cid:1765297991538 |
1 | 260% | 68d | 9 |
nid:1766531635569
A&D
How do we get a topological sorting from DFS?
2
lapses
2/4
users
230%
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nid:1765372936167
c2
A&D
What are the prerequisites for \(f\) and \(g\) to apply l'Hô...
2
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2/4
users
230%
ease
nid:1765372936167
Cloze c2
Q: What are the prerequisites for \(f\) and \(g\) to apply l'Hôpital's?{{c1::\(f, g\) are differentiable (for sufficiently large \(x\))}}{{c2::\(\lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) = \infty\) (or both \(= 0\))}}{{c3::\(g'(x
A: Then: \(\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766535228866 |
1 | 230% | 1391d | 9 |
| niklas | cid:1766567318280 |
1 | 230% | 3d | 2 |
nid:1766580142755
A&D
Johnson's Algorithm
2
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2/4
users
238%
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nid:1766580142755
A&D
Johnson's Algorithm
2
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2/4
users
238%
ease
nid:1764867990975
DiskMat
Give an example of a direct product of groups and explain it...
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nid:1764867990975
Q: Give an example of a direct product of groups and explain its structure.
A: The group \(\langle \mathbb{Z}_5; \oplus \rangle \times \langle \mathbb{Z}_7; \oplus \rangle\):
- Carrier: \(\mathbb{Z}_5 \times \mathbb{Z}_7\)
- Neutral element: \((0, 0)\)
- Operation is component-wise: \((a, b) \star (c, d) = (a \oplus_5 c, b \oplus_7 d)\)
By the Chinese Remainder Theorem, this group is isomorphic to \(\langle \mathbb{Z}_{35}; \oplus \rangle\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990975 |
1 | 230% | 833d | 8 |
| niklas | cid:1764859231274 |
1 | 245% | 23d | 9 |
nid:1764867990386
DiskMat
Give the formal definition of the least common multiple \(\t...
2
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users
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nid:1764867990386
Q: Give the formal definition of the least common multiple \(\text{lcm}(a, b)\).
A: \[a \mid l \land b \mid l \land \forall m \ ((a \mid m \land b \mid m) \rightarrow l \mid m)\]
\(l\) is a common multiple of \(a\) and \(b\) which divides every common multiple of \(a\) and \(b\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990386 |
1 | 230% | 1020d | 8 |
| niklas | cid:1762106939307 |
1 | 260% | 30d | 7 |
nid:1764867989897
DiskMat
What's the difference between \(\equiv\), \(\leftrightarrow\...
2
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2/4
users
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nid:1764867989897
Q: What's the difference between \(\equiv\), \(\leftrightarrow\), and \(\Leftrightarrow\)?
A: \(\equiv\): links formulas to statements (not part of PL itself)
\(\leftrightarrow\): formula → formula (part of PL)
\(\Leftrightarrow\): statement → statement
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867989897 |
1 | 230% | 1031d | 9 |
| niklas | cid:1761491477262 |
1 | 260% | 38d | 7 |
nid:1764867990481
DiskMat
Why does the Chinese Remainder Theorem require \(m_1, \dots,...
2
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users
238%
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nid:1764867990481
Q: Why does the Chinese Remainder Theorem require \(m_1, \dots, m_r\) to be pairwise relatively prime?
A: If \(\text{gcd}(m_i, m_j) = d > 1\), then the system could be inconsistent (e.g., \(x \equiv 0 \pmod{6}\) and \(x \equiv 1 \pmod{4}\) has no solution) or have multiple solutions (destroying uniqueness).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990481 |
1 | 230% | 1054d | 11 |
| niklas | cid:1762106939367 |
1 | 245% | 26d | 9 |
nid:1764867991256
c1
DiskMat
\(0\) (all \(a_i\) are \(0\))
2
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users
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ease
nid:1764867990060
DiskMat
How many distinct relations are possible on a finite set \(A...
2
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users
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nid:1764867990060
Q: How many distinct relations are possible on a finite set \(A\) with \(|A|\) elements?
A: \(2^{|A \times A|} = 2^{|A|^2}\) (because \(\rho \subseteq A \times A\))
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990060 |
1 | 230% | 1399d | 9 |
| niklas | cid:1761491477366 |
1 | 260% | 67d | 10 |
nid:1764867991083
DiskMat
What property do the orders of elements in finite groups hav...
2
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nid:1764867991083
Q: What property do the orders of elements in finite groups have?
A: Lemma 5.6: In a finite group \(G\), every element has a finite order.
(This doesn't hold for infinite groups - elements can have infinite order.)Proof: Since the order is finite, elements must repeat. That means, there exist \(m > n \geq 0\) s.t. \(g^m = g^n\)\(\implies g^{m-n} = e\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991083 |
1 | 230% | 1449d | 9 |
| niklas | cid:1764859231366 |
1 | 275% | 24d | 9 |
nid:1764867990681
c1
DiskMat
The order of an element \(a\) in a group (denoted \(\text{or...
2
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users
222%
ease
nid:1764867990681
Cloze c1
Q: The order of an element \(a\) in a group (denoted \(\text{ord}(a)\)) is {{c1::the smallest \(m \ge 1\) such that \(a^m = e\). If such an \(m\) does not exist, \(\text{ord}(a) = \infty\)}}
A: \(\text{ord}(e) = 1\) in any group
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990681 |
1 | 230% | 1571d | 9 |
| niklas | cid:1762856073654 |
1 | 215% | 27d | 9 |
nid:1764867990108
DiskMat
How can we test whether a relation is transitive using compo...
2
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users
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ease
nid:1764867990108
Q: How can we test whether a relation is transitive using composition?
A: A relation \(\rho\) is transitive if and only if \(\rho^2 \subseteq \rho\).
(If all two-step paths are already direct edges, the relation is transitive)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990108 |
1 | 230% | 1678d | 11 |
| niklas | cid:1761491477396 |
1 | 230% | 30d | 9 |
nid:1766448532960
c1
DiskMat
a bijection
2
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users
238%
ease
nid:1766448532960
Cloze c1
Cloze answer: a bijection
Q: In a finite group the function \(x \rightarrow x^e\) is {{c1:: a bijection}} if {{c2::\(e\) coprime to \(|G|\)}}.For \(x^e = y\), the inverse of \(y\) is {{c3:: the unique \(e\)-th root \(x = y^d\), with \(de \equiv_{|G|} 1\)}}.
A: Proof:We have \(ed = k \cdot |G| + 1\) for some \(k\). Thus, for any \(x \in G\) we have\[(x^e)^d = x^{ed} = x^{k \cdot |G| + 1} = \underbrace{(x^{|G|})^k}_{=1} \cdot x = x\]which means that the function \(y \mapsto y^d\) is the inverse function of the function \(x \mapsto x^e\) (which is hence a bijection). The under-braced term is equal to 1 because the order of \(x\) must divide the order of \(G\) (Lagrange).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448532962 |
1 | 230% | 1880d | 9 |
| niklas | cid:1766318243105 |
1 | 245% | 6d | 6 |
nid:1764867991649
c1
LinAlg
row vector; tuple
2
lapses
2/4
users
245%
ease
nid:1768182518428
c2
LinAlg
Let \(V\) be a finitely generated vector space, \(F \subsete...
2
lapses
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users
230%
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nid:1768182518428
Cloze c2
Q: Let \(V\) be a finitely generated vector space, \(F \subseteq V\) a finite set of linearly independent vectors (note that \(F\) does not need to span \(V\)) and \(G \subseteq V\) a finite set of vectors with \(\textbf{Span}(G) = V\) (but they don't all need to be independent). Then the followin
A: We can use the lemma to argue that there can't be more than \(n\) independent vectors in a space of dimension \(n\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182518428 |
1 | 230% | 791d | 11 |
| niklas | cid:1768146856907 |
1 | 230% | 20d | 5 |
nid:1768344745614
LinAlg
Rewrite \(A^\dagger = R^\dagger C^\dagger\) in terms of \(A\...
2
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users
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nid:1768344745614
Q: Rewrite \(A^\dagger = R^\dagger C^\dagger\) in terms of \(A\), \(R\), \(C\):
A: \(\begin{aligned}
A^\dagger &= R^\top (RR^\top)^{-1} (C^\top C)^{-1} C^\top \\
&= R^\top (C^\top C R R^\top)^{-1} C^\top \\
&= R^\top (C^\top A R^\top)^{-1} C^\top
\end{aligned}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768344745614 |
1 | 230% | 1086d | 13 |
| niklas | cid:1768303035591 |
1 | 230% | 3d | 5 |
nid:1768344745392
c1
LinAlg
a right inverse; A A^\dagger = I
2
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230%
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nid:1768344745392
Cloze c1
Cloze answer: a right inverse; A A^\dagger = I
Q: For \(A \in \mathbb{R}^{m \times n}\) with \(\text{rank}(A) = m\), the pseudo-inverse \(A^\dagger \in \mathbb{R}^{n \times m}\) is {{c1::a right inverse}} of \(A\): \[ {{c1:: A A^\dagger = I }}\]Proof Included
A: Proof Since \(A^\top\) has full column rank, \(((A^\top)^\top A^\top) = AA^\top\) is invertible: \(AA^\dagger = AA^\top(A A^\top)^{-1} = I\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768344745392 |
1 | 230% | 1325d | 11 |
| niklas | cid:1768302430259 |
1 | 230% | 2d | 6 |
nid:1773311114136
c1
A&W
k+1
2
lapses
2/4
users
230%
ease
nid:1773311114136
Cloze c1
Cloze answer: k+1
Q: Gilt für die (gewählte) Reihenfolge \(|N(v_i) \cap \{v_1, \ldots, v_{i-1}\}| \leq k\) \(\forall\, 2 \leq i \leq n\), dann benötigt der Greedy-Algorithmus höchstens \({{c1::k+1}}\) viele Farben.
A: Heuristik:\(v_n\) := Knoten vom kleinsten Grad. Lösche \(v_n\).\(v_{n-1}\) := Knoten vom kleinsten Grad im Restgraph. Lösche \(v_{n-1}\). Iteriere.Falls \(G=(V,E)\) erfüllt:In jedem Subgraphen gibt es einen Knoten mit Grad \(\leq k\)\(\Rightarrow\) Heuristik liefert Reihenfolge \(v_1,\ldots,v_n\) für die der Greedy-Algorithmus höchstens \(k+1\) Farben benötigt
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773311114136 |
1 | 230% | 27d | 9 |
| niklas | cid:1773420068136 |
1 | 230% | 2d | 7 |
nid:1772547552647
c1
A&W
State of the Art Matching:\( O({{c1::|E|^{1+o(1)} }}) \) für...
2
lapses
2/4
users
230%
ease
nid:1772547552647
Cloze c1
Q: State of the Art Matching:\( O({{c1::|E|^{1+o(1)} }}) \) für bipartite Graphen \( O({{c2::|V|^{1/2} \cdot |E|}}) \) für generelle Graphen (Hopcroft-Karp)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772547552648 |
1 | 230% | 70d | 12 |
| niklas | cid:1772569386229 |
1 | 230% | 1d | 5 |
nid:1772702804038
A&W
Wahr oder falsch?Jede Brücke in einem Graphen ist zu mindest...
2
lapses
2/4
users
238%
ease
nid:1772702804038
Q: Wahr oder falsch?Jede Brücke in einem Graphen ist zu mindestens einem Artikulationspunkt inzident.
A: Falsch.Gegenbeispiel: der \(K_2\) (zwei Knoten, eine Kante). Die Kante ist eine Brücke, aber keiner der Endknoten ist ein Artikulationspunkt, denn das Entfernen eines Grad-1-Knotens lässt den Rest zusammenhängend.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772702804038 |
1 | 230% | 73d | 9 |
| niklas | cid:1772783275472 |
1 | 245% | 23d | 5 |
nid:1772046117792
c1
A&W
Hamiltonkreise mit DPFür alle \(S \subseteq [n]\) mit \(1 \i...
2
lapses
2/4
users
230%
ease
nid:1772046117792
Cloze c1
Q: Hamiltonkreise mit DPFür alle \(S \subseteq [n]\) mit \(1 \in S\) und alle \(x \in S\) mit \(x \neq 1\):
\[P_{S,x} := {{c1::\begin{aligned} &\begin{cases} 1, & \text{es gibt einen 1-x-Pfad, der genau die Knoten aus } S \text{ enthält} \\ 0, &
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772046117792 |
1 | 230% | 95d | 13 |
| niklas | cid:1772209100374 |
1 | 230% | 19d | 7 |
nid:1771973928629
c1
Analysis
Um zu beweisen, dass eine komplexe Zahl \(z\) pur imaginär i...
2
lapses
2/4
users
245%
ease
nid:1774138446805
c2
Analysis
einen Häufungspunkt; eine konvergente Teilfolge
2
lapses
2/4
users
230%
ease
nid:1774138446805
Cloze c2
Cloze answer: einen Häufungspunkt; eine konvergente Teilfolge
Q: Jede {{c1::beschränkte Folge reeller Zahlen}} hat {{c2::einen Häufungspunkt}} und {{c2::eine konvergente Teilfolge}}.Proof Idea Included
A: (Bolzano-Weierstrass)Beachte: Dies gilt nur für die 1-norm!Proof Idea: Nested Intervals. Always bisect the interval. Since the sequence is infinite, at least one of the intervals must contain an infinite amount of terms.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774138446806 |
1 | 230% | 68d | 9 |
| niklas | cid:1774006423272 |
1 | 230% | 1d | 7 |
nid:1772928333503
c1
Analysis
\[ \tan\!\left(\frac{5\pi}{6}\right) = {{c1::-\frac{\sqrt{3}...
2
lapses
2/4
users
238%
ease
nid:1772928333368
c1
Analysis
\[ \cos\!\left(\frac{5\pi}{6}\right) = {{c1::-\frac{\sqrt{3}...
2
lapses
2/4
users
238%
ease
nid:1771839292091
IO r3
PProg
[Image Occlusion region 3]
2
lapses
2/4
users
230%
ease
nid:1771839292091
Cloze c3
Q: {{c1::image-occlusion:rect:left=.5516:top=.2782:width=.1174:height=.0851:oi=1}}{{c2::image-occlusion:rect:left=.3149:top=.504:width=.1095:height=.0818:oi=1}}{{c2::image-occlusion:rect:left=.2425:top=.7396:width=.2562:height=.0785:oi=1}}{{c3::image-occlusion:rect:left=.7726:top=.504:width
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771839292092 |
1 | 230% | 109d | 12 |
| niklas | cid:1771872607479 |
1 | 230% | 2d | 4 |
nid:1771365476419
c2
PProg
some form of orchestration via threads
2
lapses
2/4
users
238%
ease
nid:1771365476419
Cloze c2
Cloze answer: some form of orchestration via threads
Q: {{c1::Synchronisation}} is {{c2::some form of orchestration via threads}}.
A: Typically used to prevent bad interleavings.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771365476422 |
1 | 230% | 183d | 9 |
| niklas | cid:1771364277507 |
1 | 245% | 30d | 4 |
nid:1761491477251
c1
DiskMat
F and G are equivalent
2
lapses
2/4
users
238%
ease
nid:1761491477251
Cloze c1
Cloze answer: F and G are equivalent
Q: {{c2::\(F \equiv G\)}} means {{c1::F and G are equivalent}}, i.e., {{c3:: their truth values are equal for all truth assignments to the propositional symbols appearing in \(F\) or \(G\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477252 |
1 | 245% | 14d | 6 |
| tomas | cid:1765551656856 |
1 | 230% | 7d | 4 |
nid:1762856074477
c1
A&D
Eulerian walk (Eulerweg) that ends at the start vertex
2
lapses
2/4
users
260%
ease
nid:1762856074477
Cloze c1
Cloze answer: Eulerian walk (Eulerweg) that ends at the start vertex
Q: In graph theory, a {{c2::closed Eulerian walk (Eulerzyklus)}} is an {{c1::Eulerian walk (Eulerweg) that ends at the start vertex}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856074511 |
1 | 275% | 154d | 9 |
| tomas | cid:1765551666572 |
1 | 245% | 23d | 5 |
nid:1764745041020
A&D
What is the Cut-Property (Schnittprinzip)?
2
lapses
2/4
users
238%
ease
nid:1764745041020
Q: What is the Cut-Property (Schnittprinzip)?
A: To join a set of disjoint connected components, we need to use an edge to join two of their vertices. The idea is that the cheapest such edge is always a safe edge.This is true only for distinct edge weights!
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764745041020 |
1 | 245% | 67d | 13 |
| tomas | cid:1765551666639 |
1 | 230% | 19d | 6 |
nid:1766573228813
A&D
Johnson's Algorithm
2
lapses
2/4
users
238%
ease
nid:1769446026075
c1
A&D
\(O(n!)\)
2
lapses
2/4
users
238%
ease
nid:1771363637254
c1
Analysis
obere Schranke
2
lapses
2/4
users
238%
ease
nid:1771363637254
Cloze c1
Cloze answer: obere Schranke
Q: Eine {{c1::obere Schranke}} einer Teilmenge \(X \subset \mathbb{R}\) ist ein Element \(y \in \mathbb{R}\) mit der folgenden Eigenschaft: {{c2::\(\forall x \in X\) \(x \leq y\)}}.
A: Eine untere Schranke ist entsprechend mit \(\geq\) definiert.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771363637255 |
1 | 245% | 145d | 10 |
| tomas | cid:1771364083972 |
1 | 230% | 9d | 10 |
nid:1771364277466
c2
PProg
an independently running instance of a program/application, ...
2
lapses
2/4
users
238%
ease
nid:1771364277466
Cloze c2
Cloze answer: an independently running instance of a program/application, typically on the operating system level
Q: A {{c1::process}} is {{c2::an independently running instance of a program/application, typically on the operating system level}}.
A: Similar to a thread, but usually more heavy-weight (since a whole program) and encapsulated in memory.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771364277500 |
1 | 245% | 18d | 7 |
| tomas | cid:1771363955099 |
1 | 230% | 4d | 5 |
nid:1771364277472
c2
PProg
circular waiting/blocking (no instructions are executed/CPU ...
2
lapses
2/4
users
238%
ease
nid:1771364277472
Cloze c2
Cloze answer: circular waiting/blocking (no instructions are executed/CPU time is used) between threads, so that the system (union of all threads) cannot make any progress anymore
Q: {{c1::Deadlock}} is {{c2::circular waiting/blocking (no instructions are executed/CPU time is used) between threads, so that the system (union of all threads) cannot make any progress anymore}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771364277521 |
1 | 245% | 19d | 7 |
| tomas | cid:1771363955017 |
1 | 230% | 10d | 15 |
nid:1771970403211
c3
Analysis
Argument ausrechnen:
\(\varphi = {{c1:: \arctan(\frac{y}{x})...
2
lapses
2/4
users
238%
ease
nid:1771970403211
Cloze c3
Q: Argument ausrechnen:
\(\varphi = {{c1:: \arctan(\frac{y}{x}) }}\) falls \(x > 0\).
\(\varphi = {{c1:: \arctan(\frac{y}{x}) + \pi }}\) falls \(x < 0\) und \(y \ge 0\)
\(\varphi = {{c1:: \arctan(\frac{y}{x}) - \pi }}\) falls \(x < 0\) und \(y < 0\).
A: Achtung: Bei der Umrechnung von Normal- in Polarform ist der Fall \(x=y=0\) ausgeschlossen.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771970403213 |
1 | 245% | 32d | 7 |
| tomas | cid:1772003104419 |
1 | 230% | 4d | 5 |
nid:1766314094616
c1
DiskMat
Complete relation \(A \times A\) → single equivalence class...
2
lapses
1/4
users
225%
ease
nid:1766314094616
Cloze c1
Cloze answer: Complete relation \(A \times A\) → single equivalence class \(A\)
Q: What are the two trivial equivalence relations on a set \(A\)?{{c1:: Complete relation \(A \times A\) → single equivalence class \(A\)}}{{c2:: Identity relation → equivalence classes are all singletons \(\{a\}\
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094623 |
2 | 225% | 7d | 10 |
nid:1766314094664
c2
DiskMat
Which operations preserve countability?Let \(A\) and \(A_i\)...
2
lapses
1/4
users
180%
ease
nid:1766314094664
Cloze c2
Q: Which operations preserve countability?Let \(A\) and \(A_i\) for \(i \in \mathbb{N}\) be countable sets. Then: {{c1::\(A^n\) (\(n\)-tuples) is countable }}{{c2::\(\bigcup_{i\in \mathbb{N} } A_i\) (countable union) is countabl
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094676 |
2 | 180% | 8d | 9 |
nid:1766314094748
DiskMat
Proof method: "Indirect Proof of an Implication"
2
lapses
1/4
users
195%
ease
nid:1766314094748
Q: Proof method: "Indirect Proof of an Implication"
A: Indirect proof of \( S \implies T \): Assume T is false, prove that S is false.Follows from \( (\neg B \to \neg A) \models (A \to B) \)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094774 |
2 | 195% | 8d | 11 |
nid:1766314094775
c2
DiskMat
there exists no \(b \in A\) with \(b \prec a\) (\(b \succ a ...
2
lapses
1/4
users
210%
ease
nid:1766314094775
Cloze c2
Cloze answer: there exists no \(b \in A\) with \(b \prec a\) (\(b \succ a \) )
Q: Consider the poset \((A; \preceq)\) and \( S \subseteq A\).\(a \in A\) is a {{c1::minimal (maximal) element of \(A\)}} if {{c2::there exists no \(b \in A\) with \(b \prec a\) (\(b \succ a \) )}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094821 |
2 | 210% | 3d | 10 |
nid:1766314094778
c2
DiskMat
\(a\) is the greatest (least) element of the set of all lowe...
2
lapses
1/4
users
165%
ease
nid:1766314094778
Cloze c2
Cloze answer: \(a\) is the greatest (least) element of the set of all lower (upper) bounds of \(S\).
Q: Consider the poset \((A; \preceq)\) and \( S \subseteq A\).\(a \in A\) is the {{c1::greatest lower (least upper) bound of \(S\)}} if {{c2::\(a\) is the greatest (least) element of the set of all lower (upper) bounds of \(S\). }}
A: Note that greatest (least) refers to the operation \(\preceq\) and not to order by \(>\) or \(<\) (smaller, bigger).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094827 |
2 | 165% | 9d | 13 |
nid:1766314094948
DiskMat
\(\mathbb{Z}_m^*\) is defined as?
2
lapses
1/4
users
195%
ease
nid:1766314094948
Q: \(\mathbb{Z}_m^*\) is defined as?
A: \[ \overset{\text{def}}{=} \ \{a \in \mathbb{Z}_m \ | \ \gcd(a, m) = 1\} \]This is the set of all elements in \(\mathbb{Z}_m\) that are coprime to \(m\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095082 |
2 | 195% | 1d | 11 |
nid:1766940295803
DiskMat
\(F[x]_{m(x)}^*\) is defined as:
2
lapses
1/4
users
195%
ease
nid:1766940295803
Q: \(F[x]_{m(x)}^*\) is defined as:
A: \[\{ a(x) \in F[x]_{m(x)} \ | \ \gcd(a(x), m(x)) = 1 \}\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295973 |
2 | 195% | 2d | 8 |
nid:1765372936263
c2
A&D
{{c1:: \(\sum_{j = 1}^{n} \sum_{k = \textbf{j} }^{n} 1\)::Su...
2
lapses
1/4
users
210%
ease
nid:1765372936263
Cloze c2
Q: {{c1:: \(\sum_{j = 1}^{n} \sum_{k = \textbf{j} }^{n} 1\)::Sum}} \(=\) {{c2:: \(\sum_{j = 1}^n (n - j + 1) = \frac{n(n + 1)}{2}\) }}
A: inner loop depends on outer
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936264 |
2 | 210% | 841d | 12 |
nid:1765198542546
A&D
Let \(W\) be a walk and let \(u\) be a vertex, what is \(\te...
2
lapses
1/4
users
210%
ease
nid:1765198542546
Q: Let \(W\) be a walk and let \(u\) be a vertex, what is \(\text{deg}_W(u)\)? (generally)
A: The number of edges incident to \(u\) which are part of \(W\) but repetitions are included, therefore it is possible that \(\text{deg}(u) < \text{deg}_W(u)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765198542546 |
2 | 210% | 925d | 11 |
nid:1766531635590
A&D
What is the handshake lemma in directed graphs?
2
lapses
1/4
users
210%
ease
nid:1766531635590
Q: What is the handshake lemma in directed graphs?
A: \[ \sum_{v \in V} \deg_{out}(v) = \sum_{v \in V} \deg_{in}(v) = |E| \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635590 |
2 | 210% | 930d | 14 |
nid:1764867989852
c1
A&D
Dijkstra's
2
lapses
1/4
users
210%
ease
nid:1764867989852
Cloze c1
Cloze answer: Dijkstra's
Q: Prim's Algorithm is similar to {{c1:: Dijkstra's}} with the difference that {{c1:: \(d[v] = \min \{d[v], w(v*, v)\}\) instead of \(d[v^*] + w(v^*, v)\) }}.
A: Dijkstra's find the shortest distance to each vertex, thus it tracks the total distance.Prim's needs to build the MST. Since we add vertex \(v\) to the MST in the loop, we now want to know the new least distance to the MST for each node.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867989852 |
2 | 210% | 956d | 14 |
nid:1766531635457
c3
A&D
Order of calculation (what depends on what entries, what var...
2
lapses
1/4
users
210%
ease
nid:1766531635457
Cloze c3
Cloze answer: Order of calculation (what depends on what entries, what variable incremented first)
Q: Steps of giving a DP solution:{{c1::Define the DP table (dimensions, index, range; meaning of entry): ex: DP[1..n+1][1..k+1]}}{{c2::Computation of entries (Base case, recursive formula, pay attention to bounds!)}}{{c3::Order of calculation (what depends on w
A: SMIROST (Size, Meaning, Initialisation, Recursive Relation, Order, Solution, Time)Smiling Monkey In Red Overall S
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635458 |
2 | 210% | 980d | 11 |
nid:1766580143889
A&D
Prim's Algorithm
2
lapses
1/4
users
210%
ease
nid:1766580143889
Q: Prim's Algorithm
A: \(O((|V| + |E|) \log |V|)\) (Adjacency List, otherwise \(\Theta(|V|^2)\) like Dijkstra's)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766580143891 |
2 | 210% | 1081d | 14 |
nid:1764867989799
c1
A&D
complete
2
lapses
1/4
users
210%
ease
nid:1764867989799
Cloze c1
Cloze answer: complete
Q: A graph \(G\) is {{c1::complete}} when it's set of edges is {{c2::\(\{\{u, v\} \ | \ u, v \in V, u \neq v\}\) }}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867989799 |
2 | 210% | 1132d | 12 |
nid:1766531635467
A&D
Maximum Subarray Sum
2
lapses
1/4
users
210%
ease
nid:1766531635467
Q: Maximum Subarray Sum
A: \(\Theta(n)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635468 |
2 | 210% | 1239d | 15 |
nid:1765372936269
c1
A&D
{{c1:: \(\sum_{i = 1}^{n} i\)::Sum}} \(=\) {{c2::\(\frac{n(...
2
lapses
1/4
users
210%
ease
nid:1765372936269
Cloze c1
Q: {{c1:: \(\sum_{i = 1}^{n} i\)::Sum}} \(=\) {{c2::\(\frac{n(n + 1)}{2}\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936270 |
2 | 210% | 1346d | 13 |
nid:1765372936275
c2
A&D
{{c1:: \(\sum_{i = 1}^{n} i^2\)::Sum}} \(=\) {{c2::\(\frac{...
2
lapses
1/4
users
210%
ease
nid:1765372936275
Cloze c2
Q: {{c1:: \(\sum_{i = 1}^{n} i^2\)::Sum}} \(=\) {{c2::\(\frac{n(n + 1)(2n + 1)}{6}\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936275 |
2 | 210% | 1727d | 16 |
nid:1764867991211
DiskMat
If \(uv = vu = 1\) for some \(v \in R\) (we write \(v = u^{-...
2
lapses
1/4
users
210%
ease
nid:1764867991211
Q: If \(uv = vu = 1\) for some \(v \in R\) (we write \(v = u^{-1}\)), then \(u\) is a?
A: Unit.
Example The units of \(\mathbb{Z}\) are \(-1\) and \(1\). Therefore \(\mathbb{Z}^* = \{-1, 1\}\). In contrast, \(\mathbb{R}^* = \mathbb{R} \backslash \{0\}\), as we can divide any two numbers.
The set of units of \(R\) is denoted by \(R^*\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991211 |
2 | 210% | 991d | 14 |
nid:1764867990128
DiskMat
What is the quotient set \(A / \theta\)?
2
lapses
1/4
users
210%
ease
nid:1764867990128
Q: What is the quotient set \(A / \theta\)?
A: \[A / \theta \overset{\text{def}}{=} \{[a]_{\theta} \ | \ a \in A\}\]
The set of all equivalence classes of \(\theta\) on \(A\) (also called "\(A\) modulo \(\theta\)" or "\(A\) mod \(\theta\)").
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990128 |
2 | 210% | 1026d | 13 |
nid:1766448532935
c2
DiskMat
a basis \(g\), which is then exponentiated
2
lapses
1/4
users
210%
ease
nid:1766448532935
Cloze c2
Cloze answer: a basis \(g\), which is then exponentiated
Q: The Diffie-Hellman Key-Agreement selects two public values:{{c1:: a large prime \(p\)}}{{c2:: a basis \(g\), which is then exponentiated}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448532936 |
2 | 210% | 1069d | 13 |
nid:1764867990878
c2
DiskMat
\(\gcd(a, n) = 1\), i.e. \(a\) and \(n\) are coprime
2
lapses
1/4
users
210%
ease
nid:1764867990878
Cloze c2
Cloze answer: \(\gcd(a, n) = 1\), i.e. \(a\) and \(n\) are coprime
Q: We can reduce the exponent \(a^m\) modulo \(n\) by {{c1::the \(\text{ord}(a)\)}} iff. {{c2::\(\gcd(a, n) = 1\), i.e. \(a\) and \(n\) are coprime}}.
A: \((a^{\operatorname{ord}(a)})^q \cdot a^r \equiv_n a^r\)This is because if \(\gcd(a, n) = 1\) then there exists an \(m\) for which \(a^m = e\) (same as for the mult. inverse since \(a^{m-1}\) is the inverse).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990879 |
2 | 210% | 1164d | 15 |
nid:1764867991346
DiskMat
In a field, you can:
2
lapses
1/4
users
210%
ease
nid:1764867991346
Q: In a field, you can:
A: add
subtract
multiply
divide by any nonzero element.
You can divide, because in a field the multiplicative monoid is also a group (without \(0\), thus \(0\) cannot be divided by - no inverse).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991346 |
2 | 210% | 1217d | 12 |
nid:1764867991333
DiskMat
How is Lagrange interpolation for polynomials in a field def...
2
lapses
1/4
users
210%
ease
nid:1764867991333
Q: How is Lagrange interpolation for polynomials in a field defined?
A: Let \(\beta_i = a(\alpha_i)\) for \(i = 1, \dots, d+1\) where \(\alpha_i\) distinct for all \(i.\)\(a(x)\) is given by Lagrange's Interpolation formula: \[a(x) = \sum_{i=1}^{d+1} \beta_i u_i(x)\] where the polynomial \(u_i(x)\) is: \[u_i(x) = \frac{(x - \alpha_1) \cdots (x - \alpha_{i-1})(x - \alpha_{i+1}) \cdots (x - \alpha_{d+1})}{(\alpha_i - \alpha_1) \cdots (\alpha_i - \alpha_{i-1})(\alpha_i - \alpha_{i+1}) \cdots (\alpha_i - \alpha_{d+1})}\]
Note tha
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765655178920 |
2 | 210% | 1373d | 15 |
nid:1764867991070
c1
DiskMat
it has "volle Ordung"
2
lapses
1/4
users
210%
ease
nid:1764867991070
Cloze c1
Cloze answer: it has "volle Ordung"
Q: If {{c2:: the order \(\text{ord}(a)\) of \(a \in G\) is \(|G|\)}}, {{c1:: it has "volle Ordung"}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991071 |
2 | 210% | 1388d | 13 |
nid:1764867990871
DiskMat
How does one show the injectivity of a function?
2
lapses
1/4
users
210%
ease
nid:1764867990871
Q: How does one show the injectivity of a function?
A: Assume \(a \not= b\) and show that\(f(a) \neq f(b)\). Equivalently (by contrapositive), assume \(f(a) = f(b)\) and show that \(a = b\).Example: \(f(x) = 2x\), if \(f(a) = f(b)\), then \(2a = 2b\), which implies \(a = b\). Hence \(f\) is injective.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990871 |
2 | 210% | 1436d | 15 |
nid:1764867991398
DiskMat
When is there a finite field with \(q\) elements?
2
lapses
1/4
users
210%
ease
nid:1764867991398
Q: When is there a finite field with \(q\) elements?
A: \(\text{GF}(q)\) is only finite if and only if \(q\) is a power of a prime, i.e. \(q = p^k\) for \(p\) prime.
Any two fields of the same size \(q\) are isomorphic.Why: to construct an extension field, use \(\mathbb{Z}_p\) for coefficients. To be a field, \(p\) must be prime. In a polynomial with degree \(k-1\), each coefficient can take any of the \(p\) values from the coefficient field.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991398 |
2 | 210% | 1480d | 14 |
nid:1764867991265
c2
DiskMat
integral domain
2
lapses
1/4
users
210%
ease
nid:1764867991265
Cloze c2
Cloze answer: integral domain
Q: The degree of the product of two polynomials is {{c1::equal to the sum of their degrees}} if \(R\) is an {{c2::integral domain}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991265 |
2 | 210% | 1524d | 13 |
nid:1764867991385
c2
DiskMat
\(F[x]_{m(x)} =\) {{c2::\(F[x]_{m(x)}^* \cup \{0\}\)}} iff {...
2
lapses
1/4
users
210%
ease
nid:1764867991385
Cloze c2
Q: \(F[x]_{m(x)} =\) {{c2::\(F[x]_{m(x)}^* \cup \{0\}\)}} iff {{c1:: \(m(x)\) is irreducible}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991386 |
2 | 210% | 1655d | 13 |
nid:1764867990613
c2
DiskMat
\(a \preceq b\) (\(a \succeq b) \) for all \(b \in A\)
2
lapses
1/4
users
210%
ease
nid:1764867990613
Cloze c2
Cloze answer: \(a \preceq b\) (\(a \succeq b) \) for all \(b \in A\)
Q: Consider the poset \((A; \preceq)\).\(a \in A\) is the {{c1::least (greatest) element of \(A\)}} if {{c2::\(a \preceq b\) (\(a \succeq b) \) for all \(b \in A\)}}
A: Note that a least or a greatest element need not exist. However, there can be at most one least element, as suggested by the word “the” in the definition. This follows directly from the antisymmetry of \(\preceq\). If there were two least elements, they would be mutually comparable, and hence must be equal.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990614 |
2 | 210% | 2256d | 13 |
nid:1769307700918
c1
EProg
false since it implies everything
2
lapses
1/4
users
210%
ease
nid:1769307700918
Cloze c1
Cloze answer: false since it implies everything
Q: The strongest precondition is {{c1::false since it implies everything}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1769307700918 |
2 | 210% | 337d | 12 |
nid:1768182517514
c2
LinAlg
A vector space \(V\) is called {{c1::finitely generated}} if...
2
lapses
1/4
users
210%
ease
nid:1768182517514
Cloze c2
Q: A vector space \(V\) is called {{c1::finitely generated}} if {{c2::there exists a finite subset \(G \subseteq V\) with \(\textbf{Span}(G) = V\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182517515 |
2 | 210% | 700d | 11 |
nid:1768182517485
c1
LinAlg
singular
2
lapses
1/4
users
210%
ease
nid:1768182517485
Cloze c1
Cloze answer: singular
Q: A matrix \(A\) that is not invertible is called {{c1:: singular}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182517485 |
2 | 210% | 713d | 14 |
nid:1768870077025
c1
LinAlg
a complete set of real eigenvectors if and only if \(B\) doe...
2
lapses
1/4
users
210%
ease
nid:1768870077025
Cloze c1
Cloze answer: a complete set of real eigenvectors if and only if \(B\) does
Q: \(A \in \mathbb{R}^{n \times n}\) and \(B \in \mathbb{R}^{n \times n}\) are similar matrices. The matrix \(A\) has {{c1::a complete set of real eigenvectors if and only if \(B\) does :: EVs}}. Proof Included
A: Proof \(\lambda, v\) EW, EV pair for matrix \(A\) iff \(Av = \lambda v \Leftrightarrow \lambda S^{-1}v = S^{-1}Av = S^{-1}ASS^{-1}v = B(S^{-1}v)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768870077026 |
2 | 210% | 765d | 13 |
nid:1768182517624
c1
LinAlg
Three equivalent statements:{{c1::\(T_A : \mathbb{R}^m \righ...
2
lapses
1/4
users
210%
ease
nid:1768182517624
Cloze c1
Q: Three equivalent statements:{{c1::\(T_A : \mathbb{R}^m \rightarrow \mathbb{R}^m\) is bijective.::Transformation}}{{c2::There is an \(m \times m\) matrix \(B\) such that \(BA = I\).}}{{c3::The columns of \(A\) are linearly independent.}}
A: The third one can be derived from the fact that if \(BA = I\), there is only a single \(x \in \mathbb{R}^m\) such that \(A \textbf{x} = 0\).It is also intuitively clear that if not all columns were linearly independent, we'd actually have a tall linear transformation and would be losing information.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182517626 |
2 | 210% | 882d | 14 |
nid:1768608740503
c1
LinAlg
imaginary (or zero) eigenvalues
2
lapses
1/4
users
210%
ease
nid:1768608740503
Cloze c1
Cloze answer: imaginary (or zero) eigenvalues
Q: Real antisymmetric matrices always have {{c1::imaginary (or zero) eigenvalues}}.
A: Antisymmetric means \(A^T=-A\).
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nid:1768608741704
c2
LinAlg
\(\det(A - \lambda I) = 0\)
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Cloze answer: \(\det(A - \lambda I) = 0\)
Q: Let \(A \in \mathbb{R}^{n \times n}\).\(\lambda \in \mathbb{R}\) is a {{c1::real eigenvalue}} of \(A\) if and only if {{c2::\(\det(A - \lambda I) = 0\)}}.
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nid:1768263610432
c1
LinAlg
making sure that the sum of all the \(t_k = 0\), which can b...
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Cloze answer: making sure that the sum of all the \(t_k = 0\), which can be achieved by shifting the graph on the x-axis
Q: If the columns of \(A\) are pairwise orthogonal, we get \(A^\top A\) a diagonal matrix which is very easy to invert, i.e. makes Least Squares easier.We can convert any \(A\) to have orthogonal columns by {{c1:: making sure that the sum of all the \(t_k = 0\), which can
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nid:1768608739788
LinAlg
What is special about the characteristic polynomial?
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Q: What is special about the characteristic polynomial?
A: The characteristic polynomial is always monic.The polynomial \(\det(A - zI)\) has a leading \((-1)\) if the degree is odd. Therefore working with the characteristic one is easier.
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nid:1768263611378
LinAlg
Intuition on where the normal equations \(A^\top A\hat{x} = ...
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Q: Intuition on where the normal equations \(A^\top A\hat{x} = A^\top b\) come from:
A: In the previous case, we had \(\mathbf{e} = (\mathbf{b} - proj_S(\mathbf{b})) \ \bot \ \mathbf{a}\). Here, the same orthogonality condition holds for all columns of \(A\) (that we are projecting on).This is the same as stating \(A^\top (\mathbf{b} - proj_S(\mathbf{b})) = 0\) which by substituting \(proj_S(b) = \mathbf{p} = A \mathbf{\hat{x}}\) gives \(A^\top \mathbf{b} - A^\top A\mathbf{\hat{x}} = 0\) which we can restate as \(A^\top A \mathbf{\hat{x}} = A^\top \ma
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nid:1768263611201
c1
LinAlg
\(Ax\) to be the projection of \(b\) onto \(C(A)\)
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Cloze answer: \(Ax\) to be the projection of \(b\) onto \(C(A)\)
Q: When solving Least Squares (asking for a minimiser of \(||Ax - b||^2\)) we are asking {{c1::\(Ax\) to be the projection of \(b\) onto \(C(A)\)}}.
A: Least Squares is basically projection without multiplying by \(A\) at the end.It's also basically the Pseudoinverse.
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nid:1768944601191
c1
LinAlg
diagonalisable; the EW \(1\) has algebraic multiplicity 2 bu...
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Cloze answer: diagonalisable; the EW \(1\) has algebraic multiplicity 2 but geometric multiplicity 1
Q: \(A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) is invertible but not {{c1::diagonalisable}} since {{c1::the EW \(1\) has algebraic multiplicity 2 but geometric multiplicity 1}}.
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For \(A \in \mathbb{R}^{m \times n}\) with \(\text{rank}(A) ...
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Q: For \(A \in \mathbb{R}^{m \times n}\) with \(\text{rank}(A) = n\), we define the pseudo-inverse \(A^\dagger \in \mathbb{R}^{n \times m}\) as \[ A^\dagger = {{c1::(A^\top A)^{-1} A^\top }}\]
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nid:1768182518208
LinAlg
Give an example of a non-finitely generated vector space:
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Q: Give an example of a non-finitely generated vector space:
A: \(\mathbb{R}[x]\) is not finitely generated for example.
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nid:1764867991551
LinAlg
What special conditions (other than the 3 basic conditions) ...
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Q: What special conditions (other than the 3 basic conditions) make a set of vectors linearly dependent?
A: If:one of the vectors is 0one vector \(\textbf{v}\) is contained twice
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nid:1774631277127
A&W
Bei \(m\) fairen Münzwürfen sei \(X\) = Anzahl (möglicherwei...
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Q: Bei \(m\) fairen Münzwürfen sei \(X\) = Anzahl (möglicherweise überlappender) Vorkommen von "KKK" (drei aufeinanderfolgende Köpfe). Bestimme \(\mathbb{E}[X]\).
A: Ansatz: "KKK" kann an Positionen \(i=1,\ldots,m-2\) beginnen. Definiere:\[X_i = \begin{cases}1 & \text{Würfe } i,i+1,i+2 \text{ sind alle Kopf}\\ 0 & \text{sonst}\end{cases}.\]Dann ist \(X=X_1+\cdots+X_{m-2}\).Jeder Term: \(\mathbb{E}[X_i]=\Pr[X_i=1]=(1/2)^3=1/8\).Ergebnis: \(\mathbb{E}[X]=(m-2)\cdot\tfrac{1}{8}=\dfrac{m-2}{8}\).(Die Überlappungen spielen keine Rolle, Linearität des Erwartungswerts erledigt das automatisch.)
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nid:1777984580570
c1
A&W
\varepsilon \cdot (2e)^k \cdot k \cdot m
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Cloze c1
Cloze answer: \varepsilon \cdot (2e)^k \cdot k \cdot m
Q: Monte-Carlo Algorithmus für Long-Path: AnalyseWiederhole \(\lceil \varepsilon \cdot e^k \rceil\) Mal: zufällig färben, dann \(\text{Bunt}\) ausführen.Laufzeit: \(O\!\big({{c1::\varepsilon \cdot (2e)^k \cdot k \cdot m}}\big)\), wobei \(m = |E|\).Antwortet der Algorithm
A: Faktor \((2e)^k\) statt \(e^k\): einmalige bunte Färbung kostet \(O(2^k \cdot k \cdot m)\) (Mengen \(S \subseteq [k]\) im DP), und es werden \(\Theta(\varepsilon \cdot e^k)\) Versuche gemacht.Bemerkung: \((2e)^k\) ist konstant in \(n\). Damit wird Long-Path für festes \(k\) in Polynomialzeit lösbar (FPT-Resultat von Alon, Yuster, Zwick 1995).
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nid:1779193767087
c2
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Sei \(\hat{p}(G)\) die Wahrscheinlichkeit, dass \(\text{Cut}...
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Q: Sei \(\hat{p}(G)\) die Wahrscheinlichkeit, dass \(\text{Cut}(G)\) den Wert \(\mu(G)\) ausgibt, und \(\hat{p}(n) := \inf_{|V(G)| = n} \hat{p}(G)\). Dann gilt für \(n \geq 3\)\[\hat{p}(n) \;\geq\; {{c1::\left(1 - \tfrac{2}{n}\right) \cdot \hat{p}(n - 1)}},\]und durch Auflösung der Rekursion (mit \(\ha
A: Beweis der Rekursion: Mit \(E_1 := \{\mu(G) = \mu(G/e)\}\) und \(E_2 := \{\text{Cut}(G/e) \text{ liefert } \mu(G/e)\}\) gilt\[\hat{p}(G) = \Pr[E_1 \cap E_2] = \Pr[E_1] \cdot \Pr[E_2 \mid E_1] \geq (1 - 2/n) \cdot \hat{p}(n-1).\] Daraus folgt: erwartete Wiederholungen bis zum ersten Treffer von \(\mu(G)\) sind höchstens \(\binom{n}{2}\).
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nid:1779487730636
c1
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Schranke für \(\Pr[T \geq k]\).Sei \(T\) die Rundenzahl und ...
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Q: Schranke für \(\Pr[T \geq k]\).Sei \(T\) die Rundenzahl und \(X_k = |P'|\) nach \(\min\{T, k\}\) Runden. Aus \(X_k \geq 2^{k/3}\) (falls \(T \geq k\)) und \(\mathbb{E}[X_k] \leq (\tfrac54)^k n\) folgt\[2^{k/3} \Pr[T \geq k] \;\leq\; \mathbb{E}[X_k] \;\leq\; \left(\tfrac54\right)^k n,\]als
A: Der entscheidende Schritt: \(\mathbb{E}[X_k] \geq \mathbb{E}[X_k \mid T \geq k]\cdot \Pr[T \geq k] \geq 2^{k/3}\Pr[T \geq k]\). Für \(k \geq -\log_{0.993} n\) wird die Schranke \(\leq 1\).
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nid:1779798951084
A&W
Satz + Optimalität von LocalRepair.Was besagt der Laufzeit-S...
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Q: Satz + Optimalität von LocalRepair.Was besagt der Laufzeit-Satz für bereits \(x\)-sortierte Punkte, und warum ist der Algorithmus optimal?
A: Für nach \(x\)-Koordinate sortierte Punkte in allgemeiner Lage berechnet LocalRepair die konvexe Hülle von \(\{p_1, \ldots, p_n\}\) in Zeit \(O(n)\).Da Sortieren (\(\Omega(n \log n)\)) auf ConvexHull reduzierbar ist, ist LocalRepair (inkl. Sortieren) optimal.
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A&W
robuster als JarvisWrap: kann nie in eine unendliche Schleif...
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Cloze c1
Cloze answer: robuster als JarvisWrap: kann nie in eine unendliche Schleife laufen
Q: Anmerkungen zu LocalRepair.Degeneriertheiten sind einfach einzubeziehen (lexikographisch sortieren, Duplikate danach entfernen, Test adaptieren).Numerisch {{c1::robuster als JarvisWrap: kann nie in eine unendliche Schleife laufen}}.Liefert nebenbei {{c2::eine Tri
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nid:1777984580546
A&W
Algorithmus \(\text{Bunt}(G, i)\)Wie berechnet man \(P_i(v)\...
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Q: Algorithmus \(\text{Bunt}(G, i)\)Wie berechnet man \(P_i(v)\) für alle \(v \in V\), gegeben \(P_{i-1}(u)\) für alle \(u \in V\)?
A: Initialisierung: \(P_0(v) = \{\{\gamma(v)\}\}\) für alle \(v \in V\).Antwort JA \(\Leftrightarrow\) nach \(k-1\) Iterationen ist \(P_{k-1}(v) \neq \emptyset\) für irgendein \(v\).
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nid:1779193766994
c1
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0
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Cloze answer: 0
Q: Für einen nicht zusammenhängenden Multigraphen \(G\) gilt \(\mu(G) = {{c1::0}}\).
A: Begründung: man muss gar keine Kante entfernen, der Graph zerfällt schon.
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c2
A&W
Liniensegment und Konvexität.Für \(v_0, v_1 \in \mathbb{R}^d...
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Q: Liniensegment und Konvexität.Für \(v_0, v_1 \in \mathbb{R}^d\) ist das verbindende Liniensegment \(\overline{v_0 v_1} := {{c1::\{(1-\lambda)v_0 + \lambda v_1 \mid \lambda \in \mathbb{R},\ 0 \leq \lambda \leq 1\} }}\).Eine Menge \(C \subseteq \mathbb{R}^d\) heisst konvex, fa
A: In Worten: mit je zwei Punkten enthält eine konvexe Menge auch die ganze Verbindungsstrecke.
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c1
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Es gibt einen ganzzahligen maximalen Fluss
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Cloze c1
Cloze answer: Es gibt einen ganzzahligen maximalen Fluss
Q: Ford-Fulkerson mit ganzzahligen Kapazitäten Sei \(N = (V, A, c, s, t)\) ein Netzwerk mit \(c : A \to \mathbb{N}_0^{\leq U}\) (für \(U \in \mathbb{N}\)), ohne entgegen gerichtete Kanten. Dann:{{c1::Es gibt einen ganzzahligen maximalen Fluss}}.Er kann in Zeit {{c2::\(\
A: Begründung der Laufzeit: Höchstens \((n-1)U = \mathcal{O}(nU)\) Augmentierungsschritte, jeder Schritt (Pfadsuche per BFS/DFS in \(N_f\), Augmentierung, Update) braucht \(\mathcal{O}(m)\) Zeit. Die Ganzzahligkeit folgt induktiv: Start mit \(f \equiv 0\), und jeder Schritt erhält die Ganzzahligkeit, da \(\varepsilon = \min_i \varepsilon_i\) ganzzahlig ist.
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Sei \(\hat{p}(G)\) die Wahrscheinlichkeit, dass \(\text{Cut}...
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Q: Sei \(\hat{p}(G)\) die Wahrscheinlichkeit, dass \(\text{Cut}(G)\) den Wert \(\mu(G)\) ausgibt, und \(\hat{p}(n) := \inf_{|V(G)| = n} \hat{p}(G)\). Dann gilt für \(n \geq 3\)\[\hat{p}(n) \;\geq\; {{c1::\left(1 - \tfrac{2}{n}\right) \cdot \hat{p}(n - 1)}},\]und durch Auflösung der Rekursion (mit \(\ha
A: Beweis der Rekursion: Mit \(E_1 := \{\mu(G) = \mu(G/e)\}\) und \(E_2 := \{\text{Cut}(G/e) \text{ liefert } \mu(G/e)\}\) gilt\[\hat{p}(G) = \Pr[E_1 \cap E_2] = \Pr[E_1] \cdot \Pr[E_2 \mid E_1] \geq (1 - 2/n) \cdot \hat{p}(n-1).\] Daraus folgt: erwartete Wiederholungen bis zum ersten Treffer von \(\mu(G)\) sind höchstens \(\binom{n}{2}\).
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nid:1779193767128
c1
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Bootstrapping. Hat man bereits einen \(\mathcal{O}(n^c)\)-Al...
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Q: Bootstrapping. Hat man bereits einen \(\mathcal{O}(n^c)\)-Algorithmus mit Erfolgswkt. \(\geq 1 - e^{-1}\), so liefert die gleiche Konstruktion einen Algorithmus mit Erfolgswkt. \(\geq 1 - e^{-\alpha}\) und Laufzeit\[{{c1::\mathcal{O}\!\left(\alpha\!\left(\tfrac{n^4}{t^2} + n^2 t^{c-2}\rig
A: Die Folge der Exponenten ist \(4 \to 3 \to 8/3 \approx 2.666 \to 5/2 = 2.5 \to 12/5 = 2.4 \to 7/3 \approx 2.333 \to \ldots\); sie konvergiert gegen \(2\). Den polylog-Faktor bringt erst die rekursive Verzweigung (siehe KargerStein-Pseudocode).
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c3
A&W
Indikatorvariablen im Beweis des Sampling Lemmas.Für \(s \in...
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Cloze c3
Q: Indikatorvariablen im Beweis des Sampling Lemmas.Für \(s \in P'\) und \(R, Q \subseteq P'\):\(\operatorname{out}(s, R) := 1\) genau dann, wenn {{c1::\(s \notin C^{\bullet}(R)\) (\(s\) liegt ausserhalb von \(C(R)\))}}.\(\operatorname{ess}(s, Q) := 1\) genau dann,
A: Ausserdem: \(\sum_{s \in P' \setminus R} \operatorname{out}(s, R) = |P' \setminus C^{\bullet}(R)|\) und \(\sum_{s \in Q} \operatorname{ess}(s, Q) \leq 3\) (höchstens 3 essentielle Punkte, vgl. kleine bestimmende Menge).
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A&W
Wahr oder falsch?Wenn \(X\) eine nicht-negative Zufallsvaria...
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Q: Wahr oder falsch?Wenn \(X\) eine nicht-negative Zufallsvariable ist mit \(\mathbb{E}[X] > 100\), dann ist \(\Pr[X > 10] \geq 1/2\).
A: Falsch.Markov liefert obere, keine unteren Schranken. Gegenbeispiel: \(X = 10000\) mit Wahrscheinlichkeit \(1/50\), sonst \(0\). Dann ist \(\mathbb{E}[X] = 200 > 100\), aber \(\Pr[X > 10] = 1/50 < 1/2\).
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A&W
Runtime Brute Force ob ein Hamiltonkreis existiert?
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Q: Runtime Brute Force ob ein Hamiltonkreis existiert?
A: \(O(n!)\)
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c1
A&W
Sei \(PB_n := \{a \in [n-1] \mid a^{n-1} \equiv_n 1\} = \{a ...
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Cloze c1
Q: Sei \(PB_n := \{a \in [n-1] \mid a^{n-1} \equiv_n 1\} = \{a \in \mathbb{Z}_n^* \mid a^{n-1} \equiv_n 1\}\) die Menge der Pseudoprimzahlbasen von \(n\).\(n\) heisst Carmichael-Zahl, falls {{c1::\(n\) nicht prim ist und \(PB_n = \mathbb{Z}_n^*\)}}.Kleinste Beispiele: \({{
A: Auf Carmichael-Zahlen versagt der Fermat-Test komplett: Jede Pseudoprimzahlbasis täuscht "Primzahl" vor, und \(PB_n\) ist die ganze multiplikative Gruppe \(\mathbb{Z}_n^*\). Die Fehlerwahrscheinlichkeit \(|PB_n|/(n-1) = \varphi(n)/(n-1)\) ist nahe \(1\).Genau hier setzt der Miller-Rabin-Test an.
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c1
A&W
Monte-Carlo, Fehlerwahrscheinlichkeit \(< \tfrac{1}{2}\) Sei...
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Cloze c1
Q: Monte-Carlo, Fehlerwahrscheinlichkeit \(< \tfrac{1}{2}\) Sei \(\varepsilon > 0\) und \(A\) ein randomisierter Algorithmus, der immer JA oder NEIN ausgibt, mit\[\Pr[A(I) \text{ korrekt}] \geq \tfrac{1}{2} + \varepsilon.\]Sei \(A_\delta\) der Algorithmus, der\[N = {{c1::\left\lce
A: Achtung: \(N\) skaliert hier quadratisch in \(\varepsilon^{-1}\) (statt linear wie bei einseitigem Fehler oder Las-Vegas) und nutzt nicht die erste richtige Antwort, sondern den Mehrheitsentscheid.Beweis nutzt die Chernoff-Schranke \(\Pr[X \leq (1-\eta)\mathbb{E}[X]] \leq e^{-\eta^2 \mathbb{E}[X]/2}\) auf die Anzahl korrekter Aufrufe.
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c2
A&W
Das Array \(a\) darf nicht verändert werden (Zugriffe \(a[i]...
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Cloze c2
Cloze answer: Das Array \(a\) darf nicht verändert werden (Zugriffe \(a[i]\) sind beliebig oft erlaubt)
Q: Floyd's Cycle Finding (Aufgabenstellung)Gegeben ein Array \(a[1, \ldots, n]\) mit \(a[i] \in \{1, \ldots, n-1\}\). Finde zwei Indizes \(i \neq j\) mit \(a[i] = a[j]\), unter folgenden Einschränkungen:Laufzeit \(O({{c1::n}})\){{c2::Das Array \(a\) darf nicht verändert werd
A: Ein Duplikat existiert sicher, denn \(a\) hat \(n\) Einträge mit Werten aus einer Menge der Grösse \(n-1\) (Schubfachprinzip).Diese Variante hat keine direkte praktische Bedeutung, zeigt aber, was mit guter algorithmischer Herangehensweise möglich ist.
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| lorenz | cid:1777923968742 |
2 | 210% | 6d | 15 |
nid:1777984580514
c1
A&W
Durchprobieren aller möglichen Pfade liefert für das Long-Pa...
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Cloze c1
Q: Durchprobieren aller möglichen Pfade liefert für das Long-Path Problem eine Laufzeit von {{c1::\(O(n^{B+2})\)}}.
A: Das ist polynomiell in \(n\), aber exponentiell in \(B\): unbrauchbar, sobald \(B\) mit \(n\) wächst.Ziel der nachfolgenden Konstruktion (Color-Coding): Laufzeit \(O(c^B)\) für eine Konstante \(c\), also unabhängig von \(n\) im Exponenten.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984580514 |
2 | 210% | 25d | 13 |
nid:1779487730539
c2
A&W
die geschlossene, von \(C\) berandete Kreisscheibe
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nid:1779487730539
Cloze c2
Cloze answer: die geschlossene, von \(C\) berandete Kreisscheibe
Q: SmallEnclDisk-Problem.Gegeben eine endliche Punktemenge \(P \subseteq \mathbb{R}^2\), bestimme {{c1::den Kreis kleinsten Radius, der \(P\) umschliesst}}.Für einen Kreis \(C\) bezeichnet \(C^{\bullet}\) {{c2::die geschlossene, von \(C\) berandete Kreisscheibe}}. Wir sagen \(C\) ums
A: Punkte in \(P\) dürfen also auf dem Rand \(C\) liegen.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487730539 |
2 | 210% | 15d | 11 |
nid:1779798951092
c2
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eine Triangulierung der Punkte (lokale Verbesserung dient au...
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nid:1779798951092
Cloze c2
Cloze answer: eine Triangulierung der Punkte (lokale Verbesserung dient auch zur Berechnung guter, etwa Delaunay-, Triangulierungen)
Q: Anmerkungen zu LocalRepair.Degeneriertheiten sind einfach einzubeziehen (lexikographisch sortieren, Duplikate danach entfernen, Test adaptieren).Numerisch {{c1::robuster als JarvisWrap: kann nie in eine unendliche Schleife laufen}}.Liefert nebenbei {{c2::eine Tri
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779798951092 |
2 | 210% | 8d | 10 |
nid:1780223730623
c1
A&W
Minimum von \(n_v\) Zufallszahlen in \([0,1]\); 1/n_v; \(1/x...
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Cloze c1
Cloze answer: Minimum von \(n_v\) Zufallszahlen in \([0,1]\); 1/n_v; \(1/x_v\)
Q: Idee hinter der Erreichbarkeits-ApproximationWähle \(r_v \leftarrow \mathrm{Uniform}([0,1])\) und setze \(x_v = \min_{u \in R(v)} r_u\).Dann ist \(x_v\) das {{c1::Minimum von \(n_v\) Zufallszahlen in \([0,1]\)}}, also grob \(x_v \approx {{c1::1/n_v}}\).Folglich ist {{c1::\(1/x
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780223730623 |
2 | 210% | 8d | 10 |
nid:1778164855869
c1
A&W
(n - 1)\,U; \((n - 1)\,U\)
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Cloze c1
Cloze answer: (n - 1)\,U; \((n - 1)\,U\)
Q: Sei \(N = (V, A, c, s, t)\) ein Netzwerk mit \(c : A \to \mathbb{N}_0\), \(n := |V|\), \(m := |A|\), \(U := \max_{e \in A} c(e)\). Dann gilt\[\operatorname{val}(f) \;\leq\; \operatorname{cap}(\{s\}, V \setminus \{s\}) \;\leq\; {{c1::(n - 1)\,U}},\]und der Ford-Fulkerson-Algorithmus benötigt höchsten
A: Der triviale Schnitt \((\{s\}, V \setminus \{s\})\) hat höchstens \(n - 1\) ausgehende Kanten, jede mit Kapazität \(\leq U\). Da jeder Augmentierungsschritt den Fluss um mindestens \(1\) erhöht, ist \((n - 1)\,U\) auch eine obere Schranke für die Anzahl der Schritte.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778164855869 |
2 | 210% | 10d | 14 |
nid:1778588912019
c2
A&W
nm
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Cloze c2
Cloze answer: nm
Q: Das Maximum Bipartite Matching Problem kann durch {{c1::Anwendung von Ford-Fulkerson auf \(N_G\)}} in Zeit \(O({{c2::nm}})\) gelöst werden.
A: In \(N_G\) ist \(U = 1\), also liefert Ford-Fulkerson mit Laufzeit \(O(mnU)\) hier \(O(mn)\). Besser geht es mit Hopcroft-Karp in \(O((m + n)\sqrt{n})\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588912019 |
2 | 210% | 10d | 11 |
nid:1779487730574
A&W
CompleteEnumeration(\(P\)).Beschreibe das Vorgehen und die L...
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Q: CompleteEnumeration(\(P\)).Beschreibe das Vorgehen und die Laufzeit des naiven Enumerationsalgorithmus für \(C(P)\).
A: Man durchläuft \(\binom{n}{3}\) Mengen \(Q\), berechnet jedes \(C(Q)\) in \(O(1)\) und prüft \(P \subseteq C^{\bullet}(Q)\) in \(O(n)\).Laufzeit: \(O(n^4)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487730574 |
2 | 210% | 7d | 14 |
nid:1779798951092
c3
A&W
\(O(n \log h)\); \(O(n)\)
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Cloze c3
Cloze answer: \(O(n \log h)\); \(O(n)\)
Q: Anmerkungen zu LocalRepair.Degeneriertheiten sind einfach einzubeziehen (lexikographisch sortieren, Duplikate danach entfernen, Test adaptieren).Numerisch {{c1::robuster als JarvisWrap: kann nie in eine unendliche Schleife laufen}}.Liefert nebenbei {{c2::eine Tri
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779798951094 |
2 | 210% | 9d | 11 |
nid:1780223730648
c1
A&W
Analyse Teil 1: Schranke für \(\tilde n_v \leq n_v/20\)Defin...
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Cloze c1
Q: Analyse Teil 1: Schranke für \(\tilde n_v \leq n_v/20\)Definiere \(Y_{i,v} = 1\) falls \(x_{i,v} \geq 20/n_v\), sonst \(0\). Da \(x_{i,v}\) das Minimum von \(n_v\) Uniformen ist:\[\Pr[Y_{i,v}=1] = {{c1::\left(1 - \tfrac{20}{n_v}\right)^{n_v} \leq e^{-20} }}\]Aus \(\tilde n_v \leq n_v/
A: Es gilt \(\mathbb{E}\!\left[\sum_i Y_{i,v}\right] \leq \ell\, e^{-20} \leq \tfrac{\ell}{2\cdot 2e}\), weshalb die Chernoff-Voraussetzung \(t = \ell/2 \geq 2e\,\mathbb{E}[\sum_i Y_{i,v}]\) erfüllt ist.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780223730648 |
2 | 210% | 6d | 13 |
nid:1778164855828
c1
A&W
Beweis von „kein Pfad in \(N_f\) \(\Rightarrow\) \(\exists\)...
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Cloze c1
Q: Beweis von „kein Pfad in \(N_f\) \(\Rightarrow\) \(\exists\) Schnitt \((S, T)\) mit \(\operatorname{cap}(S, T) = \operatorname{val}(f)\)“. Definiere\[\begin{gathered}S := {{c1::\{v \in V : v \text{ ist von } s \text{ in } N_f \text{ erreichbar}\} }}, \\ T := V \setminus S.\end{gathered}\]Da k
A: Die zwei Bedingungen geben gleichzeitig die untere und die obere Schranke aus dem schwachen Dualitätslemma scharf: alle vorwärtsführenden Kanten sind saturiert, alle rückwärtsführenden Kanten tragen keinen Fluss.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778164855831 |
2 | 210% | 16d | 14 |
nid:1776175078408
c1
A&W
nicht-negative
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Cloze c1
Cloze answer: nicht-negative
Q: Für jede {{c1::nicht-negative}} Zufallsvariable \(X\) und alle \(t > 0\), gilt\[\Pr\left[X \geq t\right] \leq {{c2::\frac{\mathbb{E}[X]}{t} }}.\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1776175078408 |
2 | 210% | 38d | 12 |
nid:1776175111067
c2
A&W
Für eine {{c1::beliebige}} Zufallsvariable \(X\) und alle \(...
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nid:1776175111067
Cloze c2
Q: Für eine {{c1::beliebige}} Zufallsvariable \(X\) und alle \(t > 0\), gilt\[\Pr\left[|X - \mathbb{E}[X]| \geq t\right] \leq {{c2::\frac{\text{Var}[X]}{t^2} }}.\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1776175111068 |
2 | 210% | 9d | 12 |
nid:1779798950874
c2
A&W
der Schnitt aller Halbebenen, die \(S\) enthalten
2
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nid:1779798950874
Cloze c2
Cloze answer: der Schnitt aller Halbebenen, die \(S\) enthalten
Q: Konvexe Hülle \(\operatorname{conv}(S)\).Die konvexe Hülle einer Menge \(S \subseteq \mathbb{R}^d\) ist {{c1::der Schnitt aller konvexen Mengen, die \(S\) enthalten}}:\[\operatorname{conv}(S) := {{c1::\bigcap_{S \subseteq C \subseteq \mathbb{R}^d,\ C \text{ konvex} } C}}.\]Äquivalent:
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779798950875 |
2 | 210% | 9d | 11 |
nid:1780223730681
c2
A&W
\(\tilde n_v = -1/\log_2(1-x_v)\)
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nid:1780223730681
Cloze c2
Cloze answer: \(\tilde n_v = -1/\log_2(1-x_v)\)
Q: Bessere Approximation (Faktor \(1\pm\varepsilon\))Mit {{c1::\(\ell = \lceil \tfrac{100}{\varepsilon^2}\log(2n/\delta)\rceil\)}} Läufen und Rückgabe {{c2::\(\tilde n_v = -1/\log_2(1-x_v)\)}} statt \(1/x_v\) erhält man\[(1-\varepsilon)n_v \leq \tilde n_v \leq (1+\varepsilon)n_v\]mit Wahrsch
A: Hintergrund: \(-1/\log_2(1-x) = \ln(2)/x \pm O(1)\). Der Faktor \(1/\varepsilon^2\) ist typisch für solche Konzentrations-Argumente.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780223730682 |
2 | 210% | 10d | 13 |
nid:1777538021800
c2
A&W
Korrektheit:Falls \(n\) prim: {{c1::Ausgabe immer korrekt}}....
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nid:1777538021800
Cloze c2
Q: Korrektheit:Falls \(n\) prim: {{c1::Ausgabe immer korrekt}}.Falls \(n\) nicht prim: Falsche Ausgabe 'Primzahl' mit Wahrscheinlichkeit {{c2::\(\leq \tfrac{1}{2}\), tatsächlich sogar \(\leq \tfrac{1}{4}\)}}.
A: Im Gegensatz zum Fermat-Test funktioniert Miller-Rabin auch für Carmichael-Zahlen.Beispiel \(n = 561\) mit \(a = 2\):\(n - 1 = 560 = 2^4 \cdot 35\), und \(2^{280} \equiv_{561} 1\), aber \(2^{140} \equiv_{561} 67 \notin \{1, 560\}\). Also ist \(2\) ein Miller-Rabin-Zertifikat dafür, dass \(561\) nicht prim ist.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777538021802 |
2 | 210% | 19d | 12 |
nid:1776171659227
c1
A&W
Für \(n \geq 2\) heisst eine Zufallsvariable \(X\) mit Dicht...
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Cloze c1
Q: Für \(n \geq 2\) heisst eine Zufallsvariable \(X\) mit Dichte\[f_X(k) = \begin{cases} {{c1::\binom{k-1}{n-1} \cdot p^n \cdot (1 - p)^{k-n} }} & \text{für } k = 1, 2, \ldots \\ 0 & \text{sonst} \end{cases}\]{{c2::negativ binomialverteilt}} mit {{c3::Ordnung}} \(n\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1776171659230 |
2 | 210% | 27d | 16 |
nid:1777540083545
A&W
Skizziere den Beweis von \(\mathbb{E}[T] \leq 8n\) für Quick...
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nid:1777540083545
Q: Skizziere den Beweis von \(\mathbb{E}[T] \leq 8n\) für QuickSelect.Wie wird \(T\) zerlegt, und wieso gilt \(\mathbb{E}[N_j] \leq 2\)?
A: Zerlegung: Definiere \(N_j\) als Anzahl Aufrufe von QuickSelect mit\[(3/4)^j n < r_i - \ell_i + 1 \leq (3/4)^{j-1} n.\]Da jeder Partition-Aufruf \(r_i - \ell_i\) Vergleiche braucht:\[T \leq \sum_{j=1}^{\infty} N_j \cdot (3/4)^{j-1} n.\]Schlüsselbeobachtung für \(\mathbb{E}[N_j] \leq 2\): Wählt man als Pivot eines der mittleren \(\tfrac{1}{2}(r_i - \ell_i + 1)\) Elemente, so wird die Intervallgrösse mindestens um Faktor \(3/4\) reduziert. Die Wahrscheinlichkeit dafür ist \
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777540083546 |
2 | 210% | 15d | 14 |
nid:1778164855799
c1
A&W
im Restnetzwerk \(N_f\) gibt es keinen gerichteten s-t-Pfad
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nid:1778164855799
Cloze c1
Cloze answer: im Restnetzwerk \(N_f\) gibt es keinen gerichteten s-t-Pfad
Q: Sei \(N\) ein Netzwerk ohne entgegen gerichtete Kanten. Dann gilt:\(f\) ist maximaler Fluss \(\iff\) {{c1::im Restnetzwerk \(N_f\) gibt es keinen gerichteten s-t-Pfad}}.Für jeden maximalen Fluss \(f\) gibt es einen s-t-Schnitt \((S, T)\) mit {{c2::\(\operatorname{val}(f) = \operator
A: Der zweite Punkt liefert konstruktiv das Maxflow-Mincut-Theorem.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778164855800 |
2 | 210% | 24d | 13 |
nid:1773753822869
c1
A&W
isomorph
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nid:1773753822869
Cloze c1
Cloze answer: isomorph
Q: Es ist kein polynomieller Algorithmus bekannt, um zu entscheiden, ob zwei Graphen {{c1::isomorph}} sind.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773753822869 |
2 | 210% | 14d | 14 |
nid:1773307783473
IO r4
A&W
[Image Occlusion region 4]
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nid:1773307783473
Cloze c4
Q: {{c1::image-occlusion:polygon:left=.011:top=.2474:points=.0836,.2506 .4728,.2474 .4728,.3534 .011,.3566 .011,.3052 .0836,.3052}}{{c2::image-occlusion:rect:left=.0572:top=.4433:width=.1363:height=.045}}{{c2::image-occlusion:rect:left=.0924:top=.5815:width=.1869:height=.0514}}{{c3::image-o
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773307783483 |
2 | 210% | 61d | 16 |
nid:1777923968748
c1
A&W
\(k + r = s\ell\)
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nid:1777923968748
Cloze c1
Cloze answer: \(k + r = s\ell\)
Q: Floyd's Cycle Finding (Schlüsseleigenschaft)Definiere die Folge \(x_0 := n,\; x_i := a[x_{i-1}]\) für \(i \geq 1\). Sei der Pfad \(k \geq 1\) Kanten lang und der Kreis \(\ell \geq 3\) Kanten lang.Wähle \(0 \leq r < \ell\) und \(s \geq 1\) mit {{c1::\(k + r = s\ell\)}}. Dann gilt:\[
A: Anschauung: Sobald die Folge \(x_0, x_1, \ldots\) den Kreis erreicht hat (nach \(k\) Schritten), läuft sie unendlich im Kreis. Wir suchen einen Schritt-Index \(i\), bei dem ein einfach-laufender und ein doppelt-laufender Pointer den gleichen Knoten betreten. Dafür muss \(2i - i = i\) ein Vielfaches der Kreislänge \(\ell\) sein und \(i \geq k\) gelten: also \(i = k + r\) mit \(k + r \equiv 0 \pmod{\ell}\).Wahl: \(s = 1\) für \(k \leq \ell\), und \(s = \lceil k/\ell \rceil\) für \(k > \
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777923968750 |
2 | 210% | 28d | 13 |
nid:1773914249130
c1
A&W
Für Ereignisse \(A_1, \ldots, A_n\) (mit \(n \geq 2\)) gilt\...
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nid:1773914249130
Cloze c1
Q: Für Ereignisse \(A_1, \ldots, A_n\) (mit \(n \geq 2\)) gilt\[\Pr\left[\bigcup_{i=1}^{n} A_i\right] = {{c1::\sum_{\ell=1}^{n} (-1)^{\ell+1} \cdot \sum_{1 \leq i_1 < \cdots < i_\ell \leq n} \Pr[A_{i_1} \cap \cdots \cap A_{i_\ell}]}}\]
A: \[= \sum_{i=1}^{n} \Pr[A_i] - \sum_{1 \leq i_1 < i_2 \leq n} \Pr[A_{i_1} \cap A_{i_2}] + \ldots - \ldots + \ldots + (-1)^{n+1} \cdot \Pr[A_1 \cap \cdots \cap A_n].\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773914249130 |
2 | 210% | 43d | 17 |
nid:1776174099848
c1
A&W
Für zwei unabhängige Zufallsvariablen \(X\) und \(Y\) sei \(...
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nid:1776174099848
Cloze c1
Q: Für zwei unabhängige Zufallsvariablen \(X\) und \(Y\) sei \(Z := X + Y\). Es gilt:\[f_Z(\alpha) = {{c1::\sum_{\beta \in W_X} f_X(\beta) \cdot f_Y(\alpha - \beta)}}.\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1776174099848 |
2 | 210% | 29d | 15 |
nid:1774487164608
c1
A&W
Die Anzahl der Möglichkeiten, \(k\) Objekte aus \(n\) Sorten...
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users
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nid:1774487164608
Cloze c1
Q: Die Anzahl der Möglichkeiten, \(k\) Objekte aus \(n\) Sorten mit Zurücklegen zu wählen (Reihenfolge egal, Multiset) ist:\[{{c2::\binom{n + k - 1}{k} }} = {{c1::\frac{(n+k-1)!}{k!\,(n-1)!} }} \]
A: Auch bekannt als „Sterne und Striche“ (Stars and Bars).Beispiel: Wie viele Möglichkeiten, 3 Kugeln aus {rot, blau, grün} mit Zurücklegen zu ziehen?\(\binom{3+3-1}{3} = \binom{5}{3} = 10\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487164608 |
2 | 210% | 19d | 17 |
nid:1776333574545
A&W
Wahr oder falsch?Wenn \(A\) und \(B\) unabhängige Ereignisse...
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nid:1776333574545
Q: Wahr oder falsch?Wenn \(A\) und \(B\) unabhängige Ereignisse sind, dann ist \(\Pr[A \cup B] = \Pr[A] + \Pr[B]\).
A: Falsch.Verwechselt Unabhängigkeit mit Disjunktheit. Für unabhängige Ereignisse gilt \(\Pr[A \cup B] = \Pr[A] + \Pr[B] - \Pr[A]\Pr[B]\). Die Formel \(\Pr[A \cup B] = \Pr[A] + \Pr[B]\) gilt nur für disjunkte Ereignisse.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1776333574545 |
2 | 210% | 34d | 13 |
nid:1777538021737
c3
A&W
Designziel für Primzahltests: Laufzeit polynomiell in \(\log...
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nid:1777538021737
Cloze c3
Q: Designziel für Primzahltests: Laufzeit polynomiell in \(\log n\) (Darstellungsgrösse). Naives Trial-Division bis \(\sqrt{n}\) ist zu langsam für \(n \approx 2^{1000}\).Der \(\mathrm{ggT}\) zweier Zahlen \(m, n\) lässt sich mit dem Euklid-Algorithmus in \(O({{c1::(\log nm)^3}})\) berechn
A: Trivialerweise gilt: \(\mathrm{ggT}(a, n) > 1\) für ein \(a \in [n-1]\) \(\Rightarrow\) \(n\) nicht prim. Der Test sucht also einen kleinen gemeinsamen Faktor mit zufälligem \(a\).Problem: für \(n = p^2\) ist die Fehlerrate \(\approx 1 - 1/\sqrt{n}\), also fast \(1\). Der Test ist deshalb in der Praxis nutzlos.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777538021739 |
2 | 210% | 40d | 14 |
nid:1772496585226
IO r2
A&W
[Image Occlusion region 2]
2
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nid:1772496585226
Cloze c2
Q: {{c1::image-occlusion:rect:left=.186:top=.2984:width=.5344:height=.2754}}{{c2::image-occlusion:rect:left=.183:top=.5891:width=.8119:height=.3672}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772496585228 |
2 | 210% | 76d | 17 |
nid:1772545892871
c2
A&W
abwechselnd Kanten aus \( M \) und nicht aus \( M \) enthält...
2
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210%
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nid:1772545892871
Cloze c2
Cloze answer: abwechselnd Kanten aus \( M \) und nicht aus \( M \) enthält und der in von \( M \) nicht überdeckten Knoten beginnt und endet
Q: Ein {{c1::M-augmentierender Pfad}} ist ein Pfad, der {{c2::abwechselnd Kanten aus \( M \) und nicht aus \( M \) enthält und der in von \( M \) nicht überdeckten Knoten beginnt und endet}}.
A: \( \Rightarrow \) durch Tauschen entlang \( M \) können wir das Matching vergrössern
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772545892872 |
2 | 210% | 68d | 14 |
nid:1774631277262
c1
A&W
\Pr[A]
2
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nid:1774631277262
Cloze c1
Cloze answer: \Pr[A]
Q: Für die Indikatorvariable \(X_A\) eines Ereignisses \(A\) gilt:\[ \mathbb{E}[X_A] = {{c1::\Pr[A]}}. \]Proof Included
A: Proof: \(\mathbb{E}[X_A]=1\cdot\Pr[X_A=1]+0\cdot\Pr[X_A=0]=\Pr[A].\quad\square\)Das ist die Brücke zwischen Ereignissen (Wahrscheinlichkeit) und Zufallsvariablen (Erwartungswert): Die Wahrscheinlichkeit eines Ereignisses entspricht dem Erwartungswert seiner Indikatorvariable.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774631277262 |
2 | 210% | 59d | 13 |
nid:1777381484956
c1
Analysis
Eine Funktion der Form \[ z = f(t) = {{c1::z_0 \cdot a^t = z...
2
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210%
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nid:1777381484956
Cloze c1
Q: Eine Funktion der Form \[ z = f(t) = {{c1::z_0 \cdot a^t = z_0 \cdot e^{kt} }} \]mit \(z_0 > 0\) heisst (verallgemeinerte) Exponentialfunktion mit {{c2::Anfangswert \(z_0 = z(0)\)}} und {{c3::Wachstumsfaktor / Basis \(a\)}}.
A: Wichtigste Eigenschaft: Die relative Zunahme ist konstant: \[ \frac{z(t+s)}{z(t)} = \frac{z_0 a^{t+s} }{z_0 a^t} = a^s \](unabhängig von \(t\), nur abhängig von \(s\)).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777381484957 |
2 | 210% | 24d | 11 |
nid:1777383153571
c1
Analysis
\ln(a) \cdot a^x
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Cloze c1
Cloze answer: \ln(a) \cdot a^x
Q: Ableitung der Exponentialfunktion \(f(x) = a^x\) (mit \(a > 0\)):\[ f'(x) = {{c1::\ln(a) \cdot a^x}} \]
A: Spezialfall \(a = e\): \((e^x)' = e^x\), denn \(\ln(e) = 1\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777383153571 |
2 | 210% | 4d | 13 |
nid:1777383153590
c1
Analysis
Definitionen der hyperbolischen Funktionen:\[ \sinh(x) = {{c...
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nid:1777383153590
Cloze c1
Q: Definitionen der hyperbolischen Funktionen:\[ \sinh(x) = {{c1::\tfrac{1}{2}(e^x - e^{-x})}}, \quad \cosh(x) = {{c1::\tfrac{1}{2}(e^x + e^{-x})}} \]
A: Es gilt die Identität \(\cosh^2(x) - \sinh^2(x) = 1\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777383153591 |
2 | 210% | 23d | 13 |
nid:1777383153581
c3
Analysis
Ableitungen der trigonometrischen Funktionen (im Bogenmass):...
2
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210%
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nid:1777383153581
Cloze c3
Q: Ableitungen der trigonometrischen Funktionen (im Bogenmass):\(\sin'(x) = {{c1::\cos(x)}}\)\(\cos'(x) = {{c2::-\sin(x)}}\)\(\tan'(x) = {{c3::\dfrac{1}{\cos^2(x)} = 1 + \tan^2(x)}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777383153582 |
2 | 210% | 7d | 11 |
nid:1771973928582
Analysis
Archimedisches Prinzip (Epsilon Variante)
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nid:1771973928582
Q: Archimedisches Prinzip (Epsilon Variante)
A: Für jedes \(\epsilon > 0\) existiert \(n \in \mathbb{N}\) mit \(\frac{1}{n} < \epsilon\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771973928582 |
2 | 210% | 33d | 17 |
nid:1774487165263
Analysis
Wie lautet die Cauchy-Schwarz Ungleichung im euklidischen Ra...
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nid:1774487165263
Q: Wie lautet die Cauchy-Schwarz Ungleichung im euklidischen Raum?
A: Für alle \(x, y \in \mathbb{R}^n\) gilt:\[|x \cdot y| \leq \|x\| \cdot \|y\|\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487165263 |
2 | 210% | 36d | 13 |
nid:1780146657604
Analysis
Ist die folgende Aussage wahr? Sei \(f : [a,b] \to \mathbb{R...
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nid:1780146657604
Q: Ist die folgende Aussage wahr? Sei \(f : [a,b] \to \mathbb{R}\) eine Funktion. Dann gibt es \(c \in [a,b]\) mit \(\int_a^b f(x)\,dx = f(c)(b-a)\).Ja.Nein.
A: (b) Nein.Der Mittelwertsatz der Integralrechnung verlangt, dass \(f\) stetig ist; ohne diese Voraussetzung ist die Aussage falsch. Gegenbeispiel auf \([0,1]\): \(f(x) = 0\) für \(x \in [0, \tfrac{1}{2})\) und \(f(x) = 1\) für \(x \in [\tfrac{1}{2}, 1]\). Dann ist \(\int_0^1 f(x)\,dx = \tfrac{1}{2}\), aber \(f\) nimmt den Wert \(\tfrac{1}{2}\) nie an, es gibt also kein \(c\) mit \(f(c) = \tfrac{1}{2}\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780146657604 |
2 | 210% | 6d | 10 |
nid:1774917594762
c1
Analysis
Es sei \(f : \mathbb{D}(f) \to \mathbb{R}\). Dann gilt:\[ \l...
2
lapses
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users
210%
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nid:1774917594762
Cloze c1
Q: Es sei \(f : \mathbb{D}(f) \to \mathbb{R}\). Dann gilt:\[ \lim_{x \to x_o} f(x) = L \] genau dann, wenn {{c1::für jede konvergente Folge \((x_n)_{n \in \mathbb{N}_0}\), welche gegen \(x_0\) konvergiert, gilt: \[ \lim_{n \to \infty} f(x_n) = L \]}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774917594762 |
2 | 210% | 35d | 16 |
nid:1778839549927
c2
Analysis
\sup L(f) = \inf U(f); \int_a^b f\, dx := \sup L(f) = \inf ...
2
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users
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nid:1778839549927
Cloze c2
Cloze answer: \sup L(f) = \inf U(f); \int_a^b f\, dx := \sup L(f) = \inf U(f)
Q: Eine beschränkte Funktion \(f : [a,b] \to \mathbb{R}\) heisst {{c1::Riemann-integrierbar}}, falls \[ {{c2::\sup L(f) = \inf U(f)}}. \] Der gemeinsame Wert heisst {{c1::Riemann-Integral}} von \(f\) und wird geschrieben als \[{{c2:: \int_a^b f\, dx := \sup L(f) = \inf U(f)}}. \]
A: \(a\) heisst untere, \(b\) obere Integrationsgrenze, \(f\) der Integrand.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778839549927 |
2 | 210% | 8d | 11 |
nid:1773149513661
c1
Analysis
\(A \in \mathbb{R}\) ist ein Häufungspunkt einer Folge genau...
2
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nid:1773149513661
Cloze c1
Q: \(A \in \mathbb{R}\) ist ein Häufungspunkt einer Folge genau dann, wenn {{c1::eine konvergente Teilfolge \(({a_n}_k)_{k \in \mathbb{N_0} }\) existiert mit \[ \lim_{k \rightarrow \infty} {a_n}_k = A \]::Teilfolge}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773149513661 |
2 | 210% | 20d | 12 |
nid:1774487165476
c2
Analysis
Sei \(\rho = {{c1:: \limsup_{n\to\infty} |c_n|^{1/n} }}\). D...
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nid:1774487165476
Cloze c2
Q: Sei \(\rho = {{c1:: \limsup_{n\to\infty} |c_n|^{1/n} }}\). Dann:\[R = {{c2:: \begin{cases} 0 & \rho = \infty\\ \infty & \rho = 0 \\ \rho^{-1} & 0 < \rho < \infty \end{cases} }}\]Proof Included
A: (Formel von Hadamard)Proof (Die Hadamard-Formel ist einfach das Wurzelkriterium umgestellt für |z|)Potenzreihe \(\sum c_k z^k = \sum a_k\) für \(a_k = c_k z^k\)Wurzelkriterium: \(|a_k|^{1/k} = |(c_k)^{1/k}| \ |z|\)\(\limsup |(c_k)^{1/k}| \ |z| = |z| \limsup |(c_k)^{1/k}| < 1\)\(\implies |z| < \frac{1}{\limsup |c_k|^{1/k}}\)
Konvergiert also genau wenn \(|z| < \rho\) also innerhalb des Kreises mit Radius \(\rho\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774703261736 |
2 | 210% | 23d | 12 |
nid:1777924043303
c1
Analysis
Zum Lösen einer linearen, homogenen DGl mit konstanten Koeff...
2
lapses
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users
210%
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nid:1777924043303
Cloze c1
Q: Zum Lösen einer linearen, homogenen DGl mit konstanten Koeffizienten\[ a_n u^{(n)}(x) + a_{n-1} u^{(n-1)}(x) + \dots + a_1 u'(x) + a_0 u(x) = 0 \]verwendet man den Eulerschen Ansatz:\[ {{c1::u(x) = e^{\lambda x} }} \]Einsetzen liefert die charakteristische Gleichung {{c2::
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924043303 |
2 | 210% | 29d | 12 |
nid:1774487165756
c1
Analysis
0; - also konvergiert die Reihe nur für \(z = 0\); \infty;...
2
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nid:1774487165756
Cloze c1
Cloze answer: 0; - also konvergiert die Reihe nur für \(z = 0\); \infty; - die Reihe konvergiert für alle \(z\)
Q: Wurzelkriterium:wenn \((c_k)^{1/k}\) nicht beschränkt ist, setzen wir \(\rho = {{c1::0}}\){{c1:: - also konvergiert die Reihe nur für \(z = 0\)}} wenn \((c_k)^{1/k}\) beschränkt ist und \(\limsup (c_k)^{1/k} = 0\), setzen wir \(\rho ={{c1:: \infty}}\){{c1::&nbs
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487165757 |
2 | 210% | 44d | 16 |
nid:1774310308141
c1
Analysis
monoton fallend
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nid:1774310308141
Cloze c1
Cloze answer: monoton fallend
Q: Die Folge \(\sup \{ a_k \mid k \ge n \}\) ist {{c1::monoton fallend::property}}.
A: Für das Infinum: monoton steigend.Dies gilt, da \(n = 2\) weniger Terme als \(n = 1\) vergleicht (d.h. \(\{a_k \mid k \ge n + 1\} \subseteq \{a_k \mid k \ge n\}\)), weswegen es nur kleiner sein kann.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774310308142 |
2 | 210% | 15d | 12 |
nid:1772928333200
c1
Analysis
\[\tan(x \pm y) = {{c1:: \frac{\tan x \pm \tan y}{1 \mp \tan...
2
lapses
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users
210%
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nid:1772928333200
Cloze c1
Q: \[\tan(x \pm y) = {{c1:: \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y} }}\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333200 |
2 | 210% | 62d | 17 |
nid:1779798962597
c1
Analysis
in der allgemeinen Lösung der dazugehörigen homogenen DGl di...
2
lapses
1/4
users
210%
ease
nid:1779798962597
Cloze c1
Cloze answer: in der allgemeinen Lösung der dazugehörigen homogenen DGl die auftretende Konstante durch eine Funktion in der unabhängigen Variablen ersetzt
Q: Variation der Konstanten (DGl erster Ordnung)Einen Ansatz für die partikuläre Lösung einer inhomogenen, linearen DGl erhält man, indem man {{c1::in der allgemeinen Lösung der dazugehörigen homogenen DGl die auftretende Konstante durch eine Funktion in der unabhängigen Variablen ersetzt}}.
A: Dieser Ansatz heisst Variation der Konstanten.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779798962597 |
2 | 210% | 29d | 12 |
nid:1774631277856
c1
Analysis
Eine Teleskopreihe \(\sum_{k=1}^\infty (b_k - b_{k-1})\) kon...
2
lapses
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users
210%
ease
nid:1774631277856
Cloze c1
Q: Eine Teleskopreihe \(\sum_{k=1}^\infty (b_k - b_{k-1})\) konvergiert genau dann, wenn {{c1::\(\lim_{n\to\infty} b_n\) existiert}}.
A: In diesem Fall gilt \(\sum_{k=1}^\infty (b_k - b_{k-1}) = \lim_{n\to\infty} b_n - b_0\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774631277856 |
2 | 210% | 62d | 16 |
nid:1779798962641
c2
Analysis
von \(x_0\) und \(y_0\) abhängen kann
2
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users
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nid:1779798962641
Cloze c2
Cloze answer: von \(x_0\) und \(y_0\) abhängen kann
Q: Satz von Picard-Lindelöf-Peano (erster Ordnung)Das Anfangswertproblem \(y' = f(x, y)\), \(y(x_0) = y_0\) besitzt für "vernünftige" Funktionen \(f\) {{c1::eine eindeutige Lösung}} \(x \mapsto y(x)\), \(x \in I\),wobei das Intervall \(I\) {{c
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779798962642 |
2 | 210% | 32d | 10 |
nid:1774138446782
c2
Analysis
sie beschränkt ist
2
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Cloze c2
Cloze answer: sie beschränkt ist
Q: Für eine {{c1:: monotone Folge reeller Zahlen \((a_n)_{n \in \mathbb{N}_0}\)}} gilt: Sie konvergiert genau dann, wenn {{c2::sie beschränkt ist}}.
A: (Weierstrass)Falls die Folge monoton wachsend ist, gilt: \[ \lim_{n \rightarrow \infty} a_n = \sup \{a_n \mid n \in \mathbb{N}_0\} \]Falls die Folge monoton fallend ist, gilt:\[\lim_{n \rightarrow \infty} a_n = \inf \{ a_n \mid n \in \mathbb{N}_0\}\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774138446782 |
2 | 210% | 65d | 13 |
nid:1772928333435
c1
Analysis
\[ \sin\!\left(\frac{5\pi}{6}\right) = {{c1::\frac{1}{2} }} ...
2
lapses
1/4
users
210%
ease
nid:1772928333435
Cloze c1
Q: \[ \sin\!\left(\frac{5\pi}{6}\right) = {{c1::\frac{1}{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333435 |
2 | 210% | 88d | 15 |
nid:1777383153677
c1
Analysis
\(x_0\) ist ein Endpunkt des Intervalls \(I\)
2
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1/4
users
210%
ease
nid:1777383153677
Cloze c1
Cloze answer: \(x_0\) ist ein Endpunkt des Intervalls \(I\)
Q: Sei \(I \subset \mathbb{R}\) ein Intervall, \(f : I \to \mathbb{R}\), und \(x_0\) eine lokale Extremalstelle von \(f\). Dann ist mindestens eine der folgenden Aussagen wahr:{{c1::\(x_0\) ist ein Endpunkt des Intervalls \(I\)}}{{c2::\(f\) ist an \(x_0\) nicht differenzierbar}}
A: Bei Extremwertaufgaben muss man also alle drei Typen von Kandidaten separat prüfen: Randpunkte, Nicht-Differenzierbarkeitsstellen und kritische Punkte.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777383153679 |
2 | 210% | 48d | 13 |
nid:1777383738477
c1
Analysis
Satz von Taylor (Lagrange-Restglied): Sei \(f\) auf einem of...
2
lapses
1/4
users
210%
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nid:1777383738477
Cloze c1
Q: Satz von Taylor (Lagrange-Restglied): Sei \(f\) auf einem offenen Intervall \(I\) (mit \(x_0 \in I\)) beliebig oft differenzierbar. Dann gilt für jedes \(x \in I\)\[ f(x) = P_n(x) + R_n(x) \]mit dem Restglied\[ R_n(x) = {{c1::\frac{1}{(n+1)!}\, f^{(n+1)}(c)\, (x - x_0)^{n+1} }} \]
A: Das Restglied quantifiziert den Approximationsfehler. Geht \(R_n(x) \to 0\) für \(n \to \infty\), so konvergiert die Taylorreihe gegen \(f(x)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777383738478 |
2 | 210% | 52d | 13 |
nid:1777383738469
c1
Analysis
Das \(n\)-te Taylorpolynom von \(f\) an der Entwicklungsstel...
2
lapses
1/4
users
210%
ease
nid:1777383738469
Cloze c1
Q: Das \(n\)-te Taylorpolynom von \(f\) an der Entwicklungsstelle \(x_0\) ist\[ P_n(x) = {{c1::\sum_{k=0}^{n} \frac{f^{(k)}(x_0)}{k!} (x - x_0)^k }} \]
A: Voraussetzung: \(f\) ist auf einem offenen Intervall \(I\) mit \(x_0 \in I\) genügend oft differenzierbar.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777383738469 |
2 | 210% | 53d | 13 |
nid:1772928333383
c1
Analysis
\[ \cos\!\left(\frac{4\pi}{3}\right) = {{c1::-\frac{1}{2} }}...
2
lapses
1/4
users
210%
ease
nid:1772928333383
Cloze c1
Q: \[ \cos\!\left(\frac{4\pi}{3}\right) = {{c1::-\frac{1}{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333383 |
2 | 210% | 91d | 17 |
nid:1777383738497
c1
Analysis
Taylorreihe des Sinus (konvergiert für alle \(x \in \mathbb{...
2
lapses
1/4
users
210%
ease
nid:1777383738497
Cloze c1
Q: Taylorreihe des Sinus (konvergiert für alle \(x \in \mathbb{R}\)):\[ \sin(x) = {{c1::\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1} }{(2n+1)!} = x - \tfrac{1}{3!} x^3 + \tfrac{1}{5!} x^5 - \dots}} \]
A: Nur ungerade Potenzen, weil \(\sin\) eine ungerade Funktion ist. Entwicklungsstelle \(a = 0\) (Maclaurin-Reihe).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777383738497 |
2 | 210% | 54d | 13 |
nid:1777381484934
c1
Analysis
Die Exponentialfunktion \(\exp : \mathbb{R} \to \mathbb{R}^+...
2
lapses
1/4
users
210%
ease
nid:1777381484934
Cloze c1
Q: Die Exponentialfunktion \(\exp : \mathbb{R} \to \mathbb{R}^+\) ist gegeben durch \[ \exp(x) = e^x = {{c1::\lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n}} \]
A: Alternative Darstellung als Reihe: \(\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}\) (siehe Standardreihen).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777381484935 |
2 | 210% | 60d | 12 |
nid:1772928333345
c1
Analysis
\[ \cos\!\left(\frac{\pi}{6}\right) = {{c1::\frac{\sqrt{3} }...
2
lapses
1/4
users
210%
ease
nid:1772928333345
Cloze c1
Q: \[ \cos\!\left(\frac{\pi}{6}\right) = {{c1::\frac{\sqrt{3} }{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333346 |
2 | 210% | 99d | 15 |
nid:1774487165212
c5
Analysis
Form
Strategie
2
lapses
1/4
users
210%
ease
nid:1774487165212
Cloze c5
Q: Form
Strategie
A: (\(0\) und \(\infty\) sind hier Kurzschreibweisen für das Verhalten im Grenzwert: \(0\) steht für „geht gegen \(0\)" und \(\infty\) für „geht gegen \(\infty\)".)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1775072804338 |
2 | 210% | 53d | 13 |
nid:1772928333334
c1
Analysis
\cos\theta
2
lapses
1/4
users
210%
ease
nid:1772928333334
Cloze c1
Cloze answer: \cos\theta
Q: \[ \cos(-\theta) = {{c1::\cos\theta}} \]
A: \(\cos\) ist eine gerade Funktion.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333334 |
2 | 210% | 92d | 14 |
nid:1774917595110
c1
Analysis
Falls gilt \[{{c1:: \forall N > 0 \ \exists \delta > 0 \text...
2
lapses
1/4
users
210%
ease
nid:1774917595110
Cloze c1
Q: Falls gilt \[{{c1:: \forall N > 0 \ \exists \delta > 0 \text{ s.d. } \ \forall x \in C \ (0 < |x - c| < \delta \implies f(x) > N) }}\] hat \(f\) in \(c\) {{c2::den uneigentlichen Grenzwert \(\infty\) d.h. \(\lim_{x \to c} f(x) = \infty\)}}.
A: Das gleiche kann auch \(f(x) < -N\) für \(-\infty\) gelten.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774917595110 |
2 | 210% | 65d | 13 |
nid:1772928333361
c1
Analysis
\[ \cos\!\left(\frac{2\pi}{3}\right) = {{c1::-\frac{1}{2} }}...
2
lapses
1/4
users
210%
ease
nid:1772928333361
Cloze c1
Q: \[ \cos\!\left(\frac{2\pi}{3}\right) = {{c1::-\frac{1}{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333361 |
2 | 210% | 104d | 13 |
nid:1772928333456
c1
Analysis
-1
2
lapses
1/4
users
210%
ease
nid:1772928333456
Cloze c1
Cloze answer: -1
Q: \[ \sin\!\left(\frac{3\pi}{2}\right) = {{c1::-1}} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333456 |
2 | 210% | 103d | 15 |
nid:1772496585317
Analysis
Wann konvergiert eine Folge \((a_n)_{n \in \mathbb{N_0}}\)?
2
lapses
1/4
users
210%
ease
nid:1772496585317
Q: Wann konvergiert eine Folge \((a_n)_{n \in \mathbb{N_0}}\)?
A: \[\text{Wenn }\forall \varepsilon > 0 \; \exists N > 0 \text{, so dass } \forall n > N : |a_n - L| < \varepsilon\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772496585317 |
2 | 210% | 99d | 16 |
nid:1772928333357
c1
Analysis
0
2
lapses
1/4
users
210%
ease
nid:1772928333357
Cloze c1
Cloze answer: 0
Q: \[ \cos\!\left(\frac{\pi}{2}\right) = {{c1::0}} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333357 |
2 | 210% | 106d | 13 |
nid:1772928333443
c1
Analysis
\[ \sin\!\left(\frac{7\pi}{6}\right) = {{c1::-\frac{1}{2} }}...
2
lapses
1/4
users
210%
ease
nid:1772928333443
Cloze c1
Q: \[ \sin\!\left(\frac{7\pi}{6}\right) = {{c1::-\frac{1}{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333443 |
2 | 210% | 111d | 13 |
nid:1771973928646
Analysis
Wie lautet \(re^{i\varphi}\) ausgeschrieben mit \(\cos\) und...
2
lapses
1/4
users
210%
ease
nid:1771973928646
Q: Wie lautet \(re^{i\varphi}\) ausgeschrieben mit \(\cos\) und \(\sin\)?
A: \(re^{i \varphi} = r (\cos(\varphi) + i \sin(\varphi))\)Herleitung:\[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \dots = \sum_{k = 0}^\infty \frac{1}{k!}x^k \]Setzen wir in diese formel \(x = it\) ein, so erhalten wir \(e^{it} = \cos(t) + i \sin(t)\), \(t \in \mathbb{R}\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771973928646 |
2 | 210% | 106d | 13 |
nid:1774487165288
c1
Analysis
Quotient- und Wurzelkriterium versagen bei Reihen vom Typ {{...
2
lapses
1/4
users
210%
ease
nid:1774487165288
Cloze c1
Q: Quotient- und Wurzelkriterium versagen bei Reihen vom Typ {{c1::\(\sum \frac{1}{n^s}\) (p-Reihen)}}, da aufeinanderfolgende Terme asymptotisch gleich schnell wachsen (\(\rho = 1\)).
A: In diesem Fall: Verdichtungssatz oder Grenzwertkriterium verwenden.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487165288 |
2 | 210% | 78d | 17 |
nid:1772928333387
c1
Analysis
0
2
lapses
1/4
users
210%
ease
nid:1772928333387
Cloze c1
Cloze answer: 0
Q: \[ \cos\!\left(\frac{3\pi}{2}\right) = {{c1::0}} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333387 |
2 | 210% | 112d | 13 |
nid:1772928333452
c1
Analysis
\[ \sin\!\left(\frac{4\pi}{3}\right) = {{c1::-\frac{\sqrt{3}...
2
lapses
1/4
users
210%
ease
nid:1772928333452
Cloze c1
Q: \[ \sin\!\left(\frac{4\pi}{3}\right) = {{c1::-\frac{\sqrt{3} }{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333452 |
2 | 210% | 119d | 16 |
nid:1779487674922
c1
PProg
instantaneously between invocation and response; composabili...
2
lapses
1/4
users
210%
ease
nid:1779487674922
Cloze c1
Cloze answer: instantaneously between invocation and response; composability; high-level objects (software)
Q: Linearizability vs. sequential consistency. Linearizability: an operation takes effect {{c1::instantaneously between invocation and response}}; uses a sequential specification, locality implies {{c1::composability}}; good for {{c1::high-level objects (software)}
A: Real hardware is even weaker than sequentially consistent (but you can buy SC back at a cost, e.g. memory barriers / volatile). For high-level software, linearizability is the more appropriate concept.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487674923 |
2 | 210% | 1d | 7 |
nid:1778588922485
c1
PProg
to atomically swing a next-pointer and update the deletion m...
2
lapses
1/4
users
210%
ease
nid:1778588922485
Cloze c1
Cloze answer: to atomically swing a next-pointer and update the deletion mark in a single CAS over the (ref, mark) pair
Q: Lock-free list-set with AtomicMarkableReference: the central idea is {{c1::to atomically swing a next-pointer and update the deletion mark in a single CAS over the (ref, mark) pair}}. remove is split into two steps: {{c2::first set the mark bit on the victi
A: After the first step, the victim is observably deleted (any contains will return false) even if the second step is delayed or performed by a different thread.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922486 |
2 | 210% | 6d | 9 |
nid:1778588922352
c2
PProg
their wakeup latency is higher than a spinlock's (a scheduli...
2
lapses
1/4
users
210%
ease
nid:1778588922352
Cloze c2
Cloze answer: their wakeup latency is higher than a spinlock's (a scheduling round-trip is involved)
Q: Properties of scheduled (waiting) locks: {{c1::they require support from the runtime system / OS scheduler}}, {{c2::their wakeup latency is higher than a spinlock's (a scheduling round-trip is involved)}}, and {{c3::their internal queues need their own protection (spinl
A: Competitive spinning gets the best of both worlds for short critical sections: low latency when the lock is released quickly, but no CPU burned when the wait turns out to be long.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922355 |
2 | 210% | 7d | 9 |
nid:1777984596431
c5
PProg
contains() still has to acquire locks
2
lapses
1/4
users
210%
ease
nid:1777984596431
Cloze c5
Cloze answer: contains() still has to acquire locks
Q: Trade-offs of the optimistic list. Good: {{c1::no contention during traversal}}, {{c2::traversals are wait-free}}, {{c3::fewer lock acquisitions than hand-over-hand}}. Bad: {{c4::the list has to be traversed twice (search
A: Wait-free: every call finishes in a finite number of steps regardless of what other threads do (in particular it never waits for other threads).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596431 |
2 | 210% | 10d | 9 |
nid:1779664155915
PProg
What does the bank account look like in ScalaSTM (on Java), ...
2
lapses
1/4
users
210%
ease
nid:1779664155915
Q: What does the bank account look like in ScalaSTM (on Java), and why use new Runnable() instead of just atomic?
A: The mutable balance lives in a Ref:Ideal world with an atomic keyword:Real ScalaSTM version (Java 7 has no lambdas), each transaction is a Runnable:There is also no compiler support enforcing that Refs are only accessed inside a transaction.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779664155915 |
2 | 210% | 13d | 11 |
nid:1771365476487
c1
PProg
Cilk-style programming
2
lapses
1/4
users
210%
ease
nid:1771365476487
Cloze c1
Cloze answer: Cilk-style programming
Q: {{c1::Cilk-style programming}} is a parallel programming idiom: To compute a program, {{c2::execute code and spawn new tasks if required}}. Before returning, {{c3::wait for all spawned tasks to complete}}.
A: The system manages the eventual execution of the spawned tasks potentially in parallel.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771365476492 |
2 | 210% | 48d | 12 |
nid:1777562257030
c2
PProg
Correctness relies on no memory reordering, which requires e...
2
lapses
1/4
users
210%
ease
nid:1777562257030
Cloze c2
Cloze answer: Correctness relies on no memory reordering, which requires expensive memory barriers in real hardware
Q: Mutual exclusion built only from atomic registers (Filter, Bakery, Peterson) is not used in practice for four reasons: {{c1::Space lower bound is linear in the maximum number of threads}}.{{c2::Correctness relies on no memory reordering, which requires expensive memory barriers
A: The way out: extend the model. Modern multiprocessor architectures provide special instructions for atomically reading and writing at once (TAS, CAS, LL/SC), enabling \(O(1)\) space mutexes with practical performance.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778616609722 |
2 | 210% | 51d | 11 |
nid:1774487168256
c1
PProg
Mutual exclusion: at least one resource is held in non-share...
2
lapses
1/4
users
210%
ease
nid:1774487168256
Cloze c1
Cloze answer: Mutual exclusion: at least one resource is held in non-shareable mode
Q: What are the four necessary conditions for deadlock (Coffman conditions)?{{c1::Mutual exclusion: at least one resource is held in non-shareable mode}}{{c2::Hold and wait: a thread holds at least one resource while waiting for another}}{{c
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487168256 |
2 | 210% | 59d | 15 |
nid:1774487168266
c1
PProg
Bandwidth
2
lapses
1/4
users
210%
ease
nid:1774487168266
Cloze c1
Cloze answer: Bandwidth
Q: {{c1::Bandwidth}} of a pipeline is {{c2::the amount of work being processed in parallel at any given time}}.
A: Distinct from throughput (items/time) and latency (time/item). Bandwidth captures how many elements are simultaneously in-flight across all stages.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487168266 |
2 | 210% | 62d | 15 |
nid:1777538021707
c2
PProg
lock and unlock of a monitor
2
lapses
1/4
users
210%
ease
nid:1777538021707
Cloze c2
Cloze answer: lock and unlock of a monitor
Q: Synchronization actions (SA) in the JMM are: {{c1::read/write of a volatile variable}}; {{c2::lock and unlock of a monitor}}; {{c3::the first and last action of a thread (synthetic)}}; {{c4::actions that start a thread}};
A: SA are the building blocks of the synchronization order (SO). Anything else is an "ordinary" action.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777751408437 |
2 | 210% | 81d | 11 |
nid:1774487167493
c2
PProg
atomicity of compound operations (e.g. i++)
2
lapses
1/4
users
210%
ease
nid:1774487167493
Cloze c2
Cloze answer: atomicity of compound operations (e.g. i++)
Q: The Java volatile keyword guarantees {{c1::visibility, every read of a volatile field sees the most recent write by any thread}}, but does not guarantee {{c2::atomicity of compound operations (e.g. i++)}}.
A: Use volatile for simple flags (e.g. volatile boolean running). For compound operations, use synchronized or AtomicInteger.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487167493 |
2 | 210% | 106d | 12 |
nid:1777538021611
c1
PProg
Number of interleavings for 2 threads with \(k\) statements ...
2
lapses
1/4
users
210%
ease
nid:1777538021611
Cloze c1
Q: Number of interleavings for 2 threads with \(k\) statements each: \[{{c1::\binom{2k}{k} = O\!\left(\dfrac{4^k}{\sqrt{2k} }\right) }}\]
A: Derivation: the merged trace has length \(2k\). Once we fix which \(k\) of the \(2k\) positions belong to thread 1, the interleaving is determined (sampling without replacement).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777538021611 |
2 | 210% | 115d | 13 |
nid:1777560365972
c3
PProg
threads that enter a critical section eventually leave it
2
lapses
1/4
users
210%
ease
nid:1777560365972
Cloze c3
Cloze answer: threads that enter a critical section eventually leave it
Q: When constructing mutex algorithms from atomic registers, three assumptions are made: {{c1::atomic reads and writes of primitive-type variables}}{{c2::no reorderings of read/write sequences (not true in practice; assumed here for simplicity)}} {{c3::threads that en
A: Threads may therefore stall, die, or pause arbitrarily long outside a CS. This matters for arguments like starvation freedom: an algorithm must not rely on the "other" thread continuing to make progress.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777814143287 |
2 | 210% | 163d | 11 |
nid:1774359509898
c1
PProg
decreasing span without increasing work too much
2
lapses
1/4
users
210%
ease
nid:1774359509898
Cloze c1
Cloze answer: decreasing span without increasing work too much
Q: Designing parallel algorithms is about {{c1::decreasing span without increasing work too much}}.
A: Amdahl's Law describes the limit of speedup due to sequential parts of a program
\(T_\infty\) (span) in the DAG is the practical representation of the "sequential fraction" in Amdahl's Law
\(T_\infty\) is the fundamental cause of the speedup limit - it represents the longest sequential dependency
If we reduce \(T_\infty\), we get closer to ideal speedup
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774359509898 |
2 | 210% | 183d | 13 |
nid:1777538021707
c5
PProg
actions that determine whether a thread has terminated
2
lapses
1/4
users
210%
ease
nid:1777538021707
Cloze c5
Cloze answer: actions that determine whether a thread has terminated
Q: Synchronization actions (SA) in the JMM are: {{c1::read/write of a volatile variable}}; {{c2::lock and unlock of a monitor}}; {{c3::the first and last action of a thread (synthetic)}}; {{c4::actions that start a thread}};
A: SA are the building blocks of the synchronization order (SO). Anything else is an "ordinary" action.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777751414358 |
2 | 210% | 176d | 11 |
nid:1774487167488
c3
PProg
Elastic; creates threads on demand, reuses idle ones.
2
lapses
1/4
users
210%
ease
nid:1774487167488
Cloze c3
Cloze answer: Elastic; creates threads on demand, reuses idle ones.
Q: The four standard ExecutorService pool types:newFixedThreadPool(n) - {{c1::Fixed n threads; excess tasks are queued.}}newSingleThreadExecutor() - {{c2::Exactly 1 thread; tasks execute sequentially.}}new
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774631279571 |
2 | 210% | 207d | 13 |
nid:1780255536530
DDCA
State the duality principle.
2
lapses
1/4
users
210%
ease
nid:1780255536530
Q: State the duality principle.
A: A valid Boolean equation stays valid if you swap every AND with OR and every 0 with 1, leaving variables and their complements unchanged.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780255536530 |
2 | 210% | 14d | 13 |
nid:1761491477299
DiskMat
What is the logical rule for proof by contradiction?
2
lapses
1/4
users
240%
ease
nid:1761491477299
Q: What is the logical rule for proof by contradiction?
A: \((\lnot A \rightarrow B) \land \lnot B \models A\)
Alternative: \((A \lor B) \land \lnot B \models A\)
(If assuming \(\lnot A\) leads to something false, then \(A\) must be true)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477300 |
2 | 240% | 14d | 16 |
nid:1761491477305
DiskMat
How does an indirect proof of \(S \Rightarrow T\) work?
2
lapses
1/4
users
240%
ease
nid:1761491477305
Q: How does an indirect proof of \(S \Rightarrow T\) work?
A: An indirect proof assumes that \(T\) is false and proves that \(S\) is false under this assumption. This works because \(\lnot B \rightarrow \lnot A \models A \rightarrow B\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477306 |
2 | 240% | 15d | 16 |
nid:1761491477387
DiskMat
When is a relation \(\rho\) on set \(A\) antisymmetric?
2
lapses
1/4
users
240%
ease
nid:1761491477387
Q: When is a relation \(\rho\) on set \(A\) antisymmetric?
A: When \(a \ \rho \ b \land b \ \rho \ a \Longleftrightarrow a = b\) for all \(a, b \in A\), i.e., \(\rho \cap \hat{\rho} \subseteq \text{id}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477388 |
2 | 240% | 7d | 8 |
nid:1761491477545
DiskMat
List all types of symbols meaning equivalence:
2
lapses
1/4
users
270%
ease
nid:1761491477545
Q: List all types of symbols meaning equivalence:
A: Equivalences\(\equiv\) (formula→statement)\(\leftrightarrow\) (formula→formula)\(\Leftrightarrow\) (statement→statement)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477546 |
2 | 270% | 9d | 16 |
nid:1762856073655
c1
A&D
walk that contains every edge of the graph exactly once
2
lapses
1/4
users
240%
ease
nid:1762856073655
Cloze c1
Cloze answer: walk that contains every edge of the graph exactly once
Q: In graph theory, an {{c2::Eulerian walk (Eulerweg)}} is a {{c1::walk that contains every edge of the graph exactly once}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073669 |
2 | 240% | 90d | 11 |
nid:1762856074658
DiskMat
Sketch step-by-step how Cantor's diagonalization argument ca...
2
lapses
1/4
users
255%
ease
nid:1762856074658
Q: Sketch step-by-step how Cantor's diagonalization argument can be used to prove that the set \(\{0,1\}^\infty\) is uncountable.
A: Proof by contradiction: Assume a bijection to \(\mathbb{N}\) exists.That means there exists for each \(n\in \mathbb{N}\) a corresponding sequence of 0 and 1s, and vice-versa.We now construct a new sequence \(\alpha\) of 0s and 1s, by always taking the \(i\)-th bit from the \(i\)-th sequence, and inverting it.This new sequence does not agree with every existing sequence in at least one place.However, there is no&n
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856074690 |
2 | 255% | 32d | 13 |
nid:1763364140092
c2
A&D
the subgraph obtained after removing it and all it's inciden...
2
lapses
1/4
users
255%
ease
nid:1763364140092
Cloze c2
Cloze answer: the subgraph obtained after removing it and all it's incident edges is disconnected
Q: A vertex in a connected graph is a {{c1::cut vertex}} if {{c2::the subgraph obtained after removing it and all it's incident edges is disconnected}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1763364140092 |
2 | 255% | 70d | 9 |
nid:1764859231476
c2
DiskMat
(nontrivial, \(0 \neq 1\)) commutative ring
2
lapses
1/4
users
240%
ease
nid:1764859231476
Cloze c2
Cloze answer: (nontrivial, \(0 \neq 1\)) commutative ring
Q: An {{c1::integral domain \(D\)}} is a {{c2::(nontrivial, \(0 \neq 1\)) commutative ring}} without {{c3::zerodivisors (\(ab = 0 \implies a = 0 \lor b = 0\))}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764859231478 |
2 | 240% | 13d | 7 |
nid:1764859231569
DiskMat
What is \(F[x]_{m(x)}\)?
2
lapses
1/4
users
240%
ease
nid:1764859231569
Q: What is \(F[x]_{m(x)}\)?
A: Let \(m(x)\) be a polynomial of degree \(d\) over \(F\). Then: \[F[x]_{m(x)} \ \overset{\text{def}}{=} \ \{a(x) \in F[x] \ | \ \deg(a(x)) < d\}\]
This is the set of all polynomials over \(F\) with degree strictly less than \(d\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764859231570 |
2 | 240% | 6d | 14 |
nid:1764859231577
DiskMat
What is the cardinality of \(F[x]_{m(x)}\)?
2
lapses
1/4
users
255%
ease
nid:1764859231577
Q: What is the cardinality of \(F[x]_{m(x)}\)?
A: Lemma 5.34: Let \(F\) be a finite field with \(q\) elements and let \(m(x)\) be a polynomial of degree \(d\) over \(F\). Then: \[|F[x]_{m(x)}| = q^d\]
Explanation: Each polynomial of \(\deg d - 1\) has \(d\) coefficients (from \(0, \dots, d - 1\)), and each coefficient can be any of \(q\) elements from \(F\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764859231578 |
2 | 255% | 13d | 12 |
nid:1764860842101
c1
DiskMat
\(a \ | \ (b + c)\)
2
lapses
1/4
users
225%
ease
nid:1764860842101
Cloze c1
Cloze answer: \(a \ | \ (b + c)\)
Q: In any commutative ring, if \(a \ | \ b\) and \(a \ | \ c\), then {{c1:: \(a \ | \ (b + c)\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764860842101 |
2 | 225% | 4d | 6 |
nid:1765298196426
c2
A&D
{{c1:: \(\sum_{i = 1}^{n} i\log(i)\)}} \(\leq\) {{c2::\(\su...
2
lapses
1/4
users
240%
ease
nid:1765298196426
Cloze c2
Q: {{c1:: \(\sum_{i = 1}^{n} i\log(i)\)}} \(\leq\) {{c2::\(\sum_{i = 1}^n n \log(n) = n^2 \log n\)::Sum}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765298196426 |
2 | 240% | 2d | 11 |
nid:1765300949586
A&D
Selection Sort
2
lapses
1/4
users
270%
ease
nid:1765300949586
Q: Selection Sort
A: Best Case: \(O(n^2)\)Worst Case: \(O(n^2)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765388611001 |
2 | 270% | 49d | 12 |
nid:1765301119701
A&D
Insertion Sort
2
lapses
1/4
users
225%
ease
nid:1765301119701
Q: Insertion Sort
A: Best Case: \(O(n \log n)\)Worst Case: \(O(n^2)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765388611006 |
2 | 225% | 2d | 5 |
nid:1766314818238
c2
DiskMat
sent over the network to their partner
2
lapses
1/4
users
240%
ease
nid:1766314818238
Cloze c2
Cloze answer: sent over the network to their partner
Q: For Diffie-Hellman key agreement, both Alice and Bob {{c1:: choose \(x_A, x_B\) (their private keys) at random}}.They then compute {{c2:: \(y_A := R_p(g^{x_A})\) and \(y_B\) analogously, which are their public keys}} which is {{c2:: sent over the network to thei
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766314818239 |
2 | 240% | 6d | 9 |
nid:1766485563842
c1
A&D
\(\leq \log_2(n)\)
2
lapses
1/4
users
225%
ease
nid:1766485563842
Cloze c1
Cloze answer: \(\leq \log_2(n)\)
Q: The height of a 2-3 Tree for \(n\) keys is {{c1::\(\leq \log_2(n)\)}} thus \(h={{c2::O(\log(n))::\textbf{O-notation} }}\).
A: Note that for the case \(n = 1\) the root has one leaf with the key.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766485563842 |
2 | 225% | 1d | 8 |
nid:1766488312297
A&D
Longest Common Subsequence
2
lapses
1/4
users
195%
ease
nid:1766488312297
Q: Longest Common Subsequence
A: \(\Theta(n \cdot m)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766488406684 |
2 | 195% | 8d | 12 |
nid:1766488967649
A&D
Edit Distance
2
lapses
1/4
users
210%
ease
nid:1766488967649
Q: Edit Distance
A: \(\Theta(n \cdot m)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766488967650 |
2 | 210% | 9d | 14 |
nid:1766499628105
c2
A&D
\(\exists\) directed closed walk
2
lapses
1/4
users
210%
ease
nid:1766499628105
Cloze c2
Cloze answer: \(\exists\) directed closed walk
Q: {{c1::\(\exists\) back edge}} \(\Longleftrightarrow\){{c2::\(\exists\) directed closed walk}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766499628106 |
2 | 210% | 6d | 9 |
nid:1768239387172
c2
EProg
Generics - type erasure means List<String> becomes just List...
2
lapses
1/4
users
210%
ease
nid:1768239387172
Cloze c2
Cloze answer: Generics - type erasure means List<String> becomes just List at runtime, so the check is impossible t instanceof List<String>
Q: The cases where instanceof causes a compile error:{{c1::Primitives - instanceof only works with reference types}}{{c2::Generics - type erasure means List<String> becomes just List at runtime, so the check is impossible t instanceof List<Str
A: However:Animal a = getanimal() could get a Dog which might implement List thus a instanceof List is not a compile error.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768239387173 |
2 | 210% | 4d | 10 |
nid:1768301518838
c1
A&D
a for loop over all unmarked nodes
2
lapses
1/4
users
225%
ease
nid:1768301518838
Cloze c1
Cloze answer: a for loop over all unmarked nodes
Q: DFS Pseudocode needs to include {{c1:: a for loop over all unmarked nodes}}, when we're not sure whether the graph is connected.
A: Otherwise we aren't visiting ZHKs that aren't connected to our chosen first node.DFS(g):
all vertices unmarked
for u unmarked:
visit(u)
visit(u):
mark u
for v adjacent to u:
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768301518838 |
2 | 225% | 12d | 10 |
nid:1769377807780
c1
EProg
attributes inside a subclass; shadowed
2
lapses
1/4
users
210%
ease
nid:1769377807780
Cloze c1
Cloze answer: attributes inside a subclass; shadowed
Q: We cannot override {{c1::attributes inside a subclass}}, they are {{c1::shadowed}}.
A: class Animal {
String name = "Animal";
String getName() { return "Animal"; }
}
class Dog extends Animal {
String name = "Dog"; // Shadows Animal.name (doesn't override it)
@Override
String getName() { return Dog"; } // Overrides Animal.getName()
}Animal a = new Dog();
System.out.println(a.name);
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1769377807780 |
2 | 210% | 1d | 8 |
nid:1771364277474
c2
PProg
it can never enter a/any critical section
2
lapses
1/4
users
240%
ease
nid:1771364277474
Cloze c2
Cloze answer: it can never enter a/any critical section
Q: A thread {{c1::starves}} if {{c2::it can never enter a/any critical section}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771364277526 |
2 | 240% | 14d | 7 |
nid:1771364277497
c1
PProg
Concurrency
2
lapses
1/4
users
255%
ease
nid:1771364277497
Cloze c1
Cloze answer: Concurrency
Q: {{c1::Concurrency}} means {{c2::dealing with multiple things at the same time}}.
A: (As opposed to parallelism: doing multiple things at the same time.)Involves managing shared resources and their interactions. Often used interchangeably with parallelism.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771364277596 |
2 | 255% | 39d | 9 |
nid:1771366536188
c1
A&W
(Knoten-)Zusammenhang
2
lapses
1/4
users
225%
ease
nid:1771366536188
Cloze c1
Cloze answer: (Knoten-)Zusammenhang
Q: Es gilt immer:{{c1::(Knoten-)Zusammenhang}} ≤ {{c2::Kanten-Zusammenhang}} ≤ {{c3::minimaler Grad}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771366536191 |
2 | 225% | 6d | 10 |
nid:1771535790927
c2
A&W
low[w] > dfs[v]
2
lapses
1/4
users
240%
ease
nid:1771535790927
Cloze c2
Cloze answer: low[w] > dfs[v]
Q: Eine Baumkante \(e = (v,w)\) (\(v\) Elternknoten, \(w\) Kindknoten) ist genau dann {{c1::eine Brücke}}, wenn \({{c2::low[w] > dfs[v]}}\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771535790930 |
2 | 240% | 33d | 10 |
nid:1771535790935
c1
A&W
|V| + |E|
2
lapses
1/4
users
255%
ease
nid:1771535790935
Cloze c1
Cloze answer: |V| + |E|
Q: Alle low-Werte sind in Zeit \(O({{c1::|V| + |E|}})\) berechenbar.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771535790939 |
2 | 255% | 24d | 9 |
nid:1771969133128
c3
Analysis
Properties Absolutbetrag:
{{c1::\(|x| \geq 0\) für alle \(x\...
2
lapses
1/4
users
240%
ease
nid:1771969133128
Cloze c3
Q: Properties Absolutbetrag:
{{c1::\(|x| \geq 0\) für alle \(x\).::PSD}}
{{c2:: \(x \leq |x|, \forall x \in X\)::Vergleich}}
{{c3:: \(|xy| = |x||y| \forall x, y \in \mathbb{R}\).::Multiplikation}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771969133128 |
2 | 240% | 4d | 8 |
nid:1771872607447
c2
PProg
during any possible execution with the same inputs, its obse...
2
lapses
1/4
users
210%
ease
nid:1771872607447
Cloze c2
Cloze answer: during any possible execution with the same inputs, its observable behaviour (results, output, ...) may change if events happen in different order
Q: A program has a {{c1::race condition}} if, {{c2::during any possible execution with the same inputs, its observable behaviour (results, output, ...) may change if events happen in different order}}.
A: E.g. scheduler interactions causing different interleavings, changing network latency
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772532891703 |
2 | 210% | 2d | 5 |
nid:1772569386185
c1
A&W
wechselt ab zwischen Kanten aus \( M \) und \( M' \)
2
lapses
1/4
users
240%
ease
nid:1772569386185
Cloze c1
Cloze answer: wechselt ab zwischen Kanten aus \( M \) und \( M' \)
Q: Seien \( M \), \( M' \) beliebige Matchings.Betrachte den Teilgraphen mit Kantenmenge \( M \oplus M' \).Jeder Knoten hat Grad \( \leq 2 \) \( \Rightarrow \) Kollektion von Pfaden und Kreisen.Jeder
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772569386185 |
2 | 240% | 27d | 9 |
nid:1772788241826
c1
Analysis
\[ \sin\!\left(\frac{5\pi}{4}\right) = {{c1::-\frac{\sqrt{2}...
2
lapses
1/4
users
240%
ease
nid:1772788241826
Cloze c1
Q: \[ \sin\!\left(\frac{5\pi}{4}\right) = {{c1::-\frac{\sqrt{2} }{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772788241826 |
2 | 240% | 92d | 7 |
nid:1772788241864
c1
Analysis
1
2
lapses
1/4
users
240%
ease
nid:1772788241864
Cloze c1
Cloze answer: 1
Q: \[ \tan\!\left(\frac{5\pi}{4}\right) = {{c1::1}} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772788241864 |
2 | 240% | 2d | 11 |
nid:1773420068088
IO r2
A&W
[Image Occlusion region 2]
2
lapses
1/4
users
225%
ease
nid:1773420068088
Cloze c2
Q: {{c1::image-occlusion:rect:left=.1376:top=.5345:width=.6408:height=.0783}}{{c2::image-occlusion:rect:left=.0886:top=.6098:width=.903:height=.2198}}{{c3::image-occlusion:rect:left=.2343:top=.9079:width=.0768:height=.0783}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1773420068088 |
2 | 225% | 1d | 6 |
nid:1773420068121
A&W
Wahr oder falsch?Jeder \(k\)-reguläre bipartite Graph \(G = ...
2
lapses
1/4
users
210%
ease
nid:1773420068121
Q: Wahr oder falsch?Jeder \(k\)-reguläre bipartite Graph \(G = (A \cup B, E)\) für \(k \geq 1\) hat ein Matching der Größe \(|A|\).
A: Wahr.Hall-Satz: Da \(G\) \(k\)-regulär ist, hat jeder Knoten in \(X\) Grad \(k\), jeder in \(N(X)\) Grad \(\leq k\). Weil in bipartiten Graphen die Gradsumme links gleich der Gradsumme rechts ist, folgt \(|N(X)| \geq |X|\). Damit ist Halls Bedingung erfüllt und ein Matching der Größe \(|A|\) existiert. Es gilt sogar: \(E\) lässt sich in \(k\) disjunkte perfekte Matchings partitionieren (iteratives Entfernen eines perfekten Matchings liefert jeweils einen \((k-1)\)-reg
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1773420068122 |
2 | 210% | 1d | 8 |
nid:1765551656886
DiskMat
Describe the three steps of a proof by contradiction of stat...
2
lapses
1/4
users
210%
ease
nid:1765551656886
Q: Describe the three steps of a proof by contradiction of statement \(S\).
A: 1. Find a suitable statement \(T\)
2. Prove that \(T\) is false
3. Assume \(S\) is false and prove that \(T\) is true (a contradiction)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1765551656886 |
2 | 210% | 1d | 6 |
nid:1766410023689
A&D
Runtime of operations in an adjacency matrix?
2
lapses
1/4
users
210%
ease
nid:1766410023689
Q: Runtime of operations in an adjacency matrix?
A: 1. Check if \(uv \in E\): \(O(1)\)2. Vertex \(u\) , find all adjacent vertices in: \(O(n)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1766410023689 |
2 | 210% | 30d | 12 |
nid:1771363954980
c3
PProg
{{c1::Divide and conquer style parallelism (also called recu...
2
lapses
1/4
users
210%
ease
nid:1771363954980
Cloze c3
Q: {{c1::Divide and conquer style parallelism (also called recursive splitting)}} means: solve a problem by {{c2::recursively solving smaller sub-problems and combining their results}}.
A: Solve the sub-problems in separate threads to gain a speedup.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955012 |
2 | 210% | 4d | 7 |
nid:1771363955001
c1
PProg
parallelism
2
lapses
1/4
users
210%
ease
nid:1771363955001
Cloze c1
Cloze answer: parallelism
Q: The maximum possible speedup ({{c1::parallelism}}) is {{c2::\(\frac{T_1}{T_\infty} \)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955096 |
2 | 210% | 12d | 11 |
nid:1771363955014
c1
PProg
Scheduling overhead
2
lapses
1/4
users
210%
ease
nid:1771363955014
Cloze c1
Cloze answer: Scheduling overhead
Q: {{c1::Scheduling overhead}} is the {{c2::extra time spent by the system or the algorithm}} to distribute work on {{c3::multiple threads/tasks}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955146 |
2 | 210% | 9d | 10 |
nid:1771363955028
c1
PProg
Work partitioning
2
lapses
1/4
users
210%
ease
nid:1771363955028
Cloze c1
Cloze answer: Work partitioning
Q: {{c1::Work partitioning}} is the {{c2::split-up of a program}} into smaller tasks that can be executed in {{c3::parallel}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955196 |
2 | 210% | 2d | 8 |
nid:1771578182870
c3
PProg
idle time due to task dependencies or waiting for data excha...
2
lapses
1/4
users
210%
ease
nid:1771578182870
Cloze c3
Cloze answer: idle time due to task dependencies or waiting for data exchange
Q: Parallel execution can introduce inefficiencies such as {{c1::communication overhead}}, {{c2::load imbalance}}, and {{c3::idle time due to task dependencies or waiting for data exchange}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771578182872 |
2 | 210% | 1d | 8 |
nid:1771616439344
c4
Advanced Finance
taking too many or too few risks
2
lapses
1/4
users
210%
ease
nid:1771616439344
Cloze c4
Cloze answer: taking too many or too few risks
Q: Agency problems include a manager:{{c1:: not putting in sufficient effort}}{{c2:: wasting money on personal benefits}}{{c3:: overinvesting in search of power or prestige}}{{c4:: taking too many or too few risks}}{{c5:: focusing on short-term results at
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771616439348 |
2 | 210% | 2d | 11 |
nid:1766314077328
c1
A&D
\(u\) reaches \(v\) (erreicht)
1
lapses
1/4
users
245%
ease
nid:1766314077328
Cloze c1
Cloze answer: \(u\) reaches \(v\) (erreicht)
Q: For \(u, v \in V\) we say that {{c1::\(u\) reaches \(v\) (erreicht)}} if {{c2::there is a walk with endpoints \(u\) and \(v\) (or a path)}}.
A: Reachability is an equivalence relation.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314077343 |
1 | 245% | 17d | 7 |
nid:1766314077330
c1
A&D
connected component
1
lapses
1/4
users
215%
ease
nid:1766314077330
Cloze c1
Cloze answer: connected component
Q: A {{c1::connected component}} of \(G\) is a {{c2::equivalence class of the relation defined as follows: \(u = v\) if \(u\) reaches \(v\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314077346 |
1 | 215% | 4d | 7 |
nid:1766314077346
c1
A&D
cut edge
1
lapses
1/4
users
230%
ease
nid:1766314077346
Cloze c1
Cloze answer: cut edge
Q: An edge in a connected graph is a {{c1::cut edge}} if {{c2::the subgraph obtained after removing it (keeping the vertices) is disconnected}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314077373 |
1 | 230% | 10d | 7 |
nid:1766314077348
c2
A&D
A graph \(G\) is {{c1::complete}} when it's set of edges is ...
1
lapses
1/4
users
215%
ease
nid:1766314077348
Cloze c2
Q: A graph \(G\) is {{c1::complete}} when it's set of edges is {{c2::\(\{\{u, v\} \ | \ u, v \in V, u \neq v\}\) }}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314077377 |
1 | 215% | 17d | 11 |
nid:1766314077369
c1
A&D
<
1
lapses
1/4
users
230%
ease
nid:1766314077369
Cloze c1
Cloze answer: <
Q: In BFS enter/leave ordering for all \(v\), enter[v] {{c1:: <}} leave[v].
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314077403 |
1 | 230% | 27d | 7 |
nid:1766314094599
DiskMat
How does the inverse of a relation appear in matrix and grap...
1
lapses
1/4
users
230%
ease
nid:1766314094599
Q: How does the inverse of a relation appear in matrix and graph representations?
A: Matrix: The transpose of the matrix
Graph: Reversing the direction of all edges
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094606 |
1 | 230% | 29d | 6 |
nid:1766314094602
DiskMat
Is composition of relations associative?
1
lapses
1/4
users
230%
ease
nid:1766314094602
Q: Is composition of relations associative?
A: Yes: \(\rho \circ (\sigma \circ \phi) = (\rho \circ \sigma) \circ \phi\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094609 |
1 | 230% | 28d | 6 |
nid:1766314094636
DiskMat
For what types of posets is well-ordering primarily of inter...
1
lapses
1/4
users
230%
ease
nid:1766314094636
Q: For what types of posets is well-ordering primarily of interest?
A: Infinite posets. (Every totally ordered finite poset is automatically well-ordered)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094646 |
1 | 230% | 9d | 9 |
nid:1766314094645
DiskMat
What is the preimage \(f^{-1}(T)\) of a subset \(T \subseteq...
1
lapses
1/4
users
230%
ease
nid:1766314094645
Q: What is the preimage \(f^{-1}(T)\) of a subset \(T \subseteq B\)?
A: \[f^{-1}(T) \overset{\text{def}}{=} \{a \in A \ | \ f(a) \in T\}\]
The set of values in \(A\) that map into \(T\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094656 |
1 | 230% | 14d | 8 |
nid:1766314094698
DiskMat
What is a composite number?
1
lapses
1/4
users
215%
ease
nid:1766314094698
Q: What is a composite number?
A: An integer greater than 1 that is not prime (i.e., it has divisors other than 1 and itself).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094719 |
1 | 215% | 16d | 10 |
nid:1766314094752
c3
DiskMat
Assume that \( S \) is false and prove that \( T \) is true...
1
lapses
1/4
users
230%
ease
nid:1766314094752
Cloze c3
Cloze answer: Assume that \( S \) is false and prove that \( T \) is true (-> contradiction).
Q: Proof method: Proof by Contradiction1. {{c1:: Find a suitable statement \( T\).}}2. {{c2:: Prove that \( T \) is false.}}3. {{c3:: Assume that \( S \) is false and prove that \( T \) is true (-> contradiction).}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094781 |
1 | 230% | 13d | 10 |
nid:1766314094759
c2
DiskMat
\((a \ \rho \ b \wedge b \ \rho \ c) \implies a \ \rho \ c \...
1
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1/4
users
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nid:1766314094759
Cloze c2
Cloze answer: \((a \ \rho \ b \wedge b \ \rho \ c) \implies a \ \rho \ c \) is true.
Q: A relation is {{c1::transitive}} if {{c2::\((a \ \rho \ b \wedge b \ \rho \ c) \implies a \ \rho \ c \) is true.}}
A: Examples: \( \le, \ge, <, |, \equiv_m\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094793 |
1 | 230% | 19d | 7 |
nid:1766314094762
c2
DiskMat
{{c1::Ein Körper}} ist eine Menge {{c1::\( \mathbb{K}\) mit ...
1
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users
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nid:1766314094762
Cloze c2
Q: {{c1::Ein Körper}} ist eine Menge {{c1::\( \mathbb{K}\) mit Operationen \(+ , *\)}} mit folgenden Eigenschaften:{{c2::- \( (\mathbb{K}, +)\) ist eine abelsche Gruppe- \( (\mathbb{K} \backslash \{0\}, *)\) ist eine abelsche Gruppe- Distributivität:&
A: Beispiel: \( \mathbb{Q}, \mathbb{R}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094798 |
1 | 230% | 13d | 10 |
nid:1766314094763
c1
DiskMat
injective (or one-to-one)
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nid:1766314094763
Cloze c1
Cloze answer: injective (or one-to-one)
Q: A function is {{c1::injective (or one-to-one)}} if {{c2::for \(a \ne b\) we have \(f(a) \ne f(b)\), i.e. no "collisions"}}.
A: Example: \(f(x) = x\), counterexample: \(f(x) = x^2, x \in \mathbb{R}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094801 |
1 | 230% | 12d | 7 |
nid:1766314094774
c2
DiskMat
totally ordered (also: linearly ordered) by \(\preceq\)
1
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users
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nid:1766314094774
Cloze c2
Cloze answer: totally ordered (also: linearly ordered) by \(\preceq\)
Q: A poset \((A; \preceq)\) is called {{c2::totally ordered (also: linearly ordered) by \(\preceq\)}} if {{c1::any two elements of the poset are comparable.}}
A: Example: \((\mathbb{Z}; \ge)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094820 |
1 | 230% | 16d | 8 |
nid:1766314094788
DiskMat
Is the set \(\{0,1\}^*\) (finite binary sequences) countable...
1
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users
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nid:1766314094788
Q: Is the set \(\{0,1\}^*\) (finite binary sequences) countable?
A: Yes. A possible injection to \(\mathbb{N}\) is to add a "1" at the beginning of each sequence and interpret it in binary.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094845 |
1 | 215% | 9d | 9 |
nid:1766314094807
c1
DiskMat
field (Körper)
1
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users
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nid:1766314094807
Cloze c1
Cloze answer: field (Körper)
Q: A {{c1::field (Körper)}} is {{c2::a nontrivial commutative ring \(F\) in which every nonzero element is a unit, so \(F^* = F \backslash \{0\}\)}}
A: Example: \(\mathbb{R}\), but not \(\mathbb{Z}\)Non-trivial: {0} is not a field. In particular, 1 = 0 (neutral element of mult. = neutral element of add.) causes trouble.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094872 |
1 | 230% | 9d | 8 |
nid:1766314094832
c1
DiskMat
A partial function \(A \to B\) is a relation from \(A\) to \...
1
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users
215%
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nid:1766314094832
Cloze c1
Q: A partial function \(A \to B\) is a relation from \(A\) to \(B\) such that {{c1::\(\forall a \in A \; \forall b,b' \in B \; (a \mathop{f} b \land a\mathop{f} b' \to b = b')\) (well-defined).}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094904 |
1 | 215% | 5d | 8 |
nid:1766314094873
c1
DiskMat
right inverse element
1
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users
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nid:1766314094873
Cloze c1
Cloze answer: right inverse element
Q: A {{c1::right inverse element}} of \(a\) in \(\langle S; *, e \rangle\) is {{c2::an element \(b\) such that \(a * b = e\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094952 |
1 | 230% | 14d | 6 |
nid:1766314094875
DiskMat
Lemma about uniqueness of the inverse:
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users
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nid:1766314094875
Q: Lemma about uniqueness of the inverse:
A: Lemma 5.2: In a monoid \(\langle M; *, e \rangle\), if \(a \in M\) has a left inverse and a right inverse, then they are equal. In particular, \(a\) has at most one inverse.Proof: \(L\) left inverse, \(R\) right inverse.\(L = L * e = L * (a * R) \) \(= (L * a) * R = e * R = R\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094959 |
1 | 230% | 23d | 8 |
nid:1766314094876
c3
DiskMat
G1 (associativity)
1
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users
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nid:1766314094876
Cloze c3
Cloze answer: G1 (associativity)
Q: A {{c1::group}} is an algebra \(\langle G; *, \widehat{\ \ }, e \rangle\) satisfying {{c2::three}} axioms: {{c3::G1 (associativity)}}, {{c4::G2 (neutral element)}}, and {{c5::G3 (inverse elements)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094962 |
1 | 230% | 10d | 5 |
nid:1766314094890
c1
DiskMat
right cancellation
1
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users
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ease
nid:1766314094890
Cloze c1
Cloze answer: right cancellation
Q: In a group, the {{c1::right cancellation}} law states: \(a = b\) {{c2::\(\Leftrightarrow\)}} {{c3::\(ac = bc\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314094992 |
1 | 260% | 18d | 6 |
nid:1766314094924
DiskMat
What is the group generated by a, denoted \(\langle a \rangl...
1
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users
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nid:1766314094924
Q: What is the group generated by a, denoted \(\langle a \rangle\) defined as?
A: For a group \(G\) and \(a \in G\), the group generated by \(a\), denoted \(\langle a \rangle\), is defined as: \[\langle a \rangle \ \overset{\text{def}}{=} \ \{a^n \ | \ n \in \mathbb{Z}\}\]
This is a group, the smallest subgroup of \(G\) containing the element \(a\).
For finite groups: \(\langle a \rangle = \{e, a, a^2, \dots, a^{\text{ord}(a)-1}\}\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095049 |
1 | 230% | 2d | 9 |
nid:1766314094925
c2
DiskMat
smallest
1
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users
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nid:1766314094925
Cloze c2
Cloze answer: smallest
Q: The {{c2:: smallest}} subgroup of a group \(G\) containing \(a \in G\) is {{c1:: the group generated by \(a\), \(\langle a \rangle\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095051 |
1 | 230% | 17d | 9 |
nid:1766314094942
DiskMat
For which order is every group cyclic?
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nid:1766314094942
Q: For which order is every group cyclic?
A: If the order of the group is prime, it is cyclic!Every element has order 1 or \(|G|\) (Lagrange). Therefore, it is either the neutral element or a generator of the entire group.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095076 |
1 | 230% | 14d | 10 |
nid:1766314094972
c3
DiskMat
greatest \(i\) for which \(a_i \neq 0\)
1
lapses
1/4
users
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ease
nid:1766314094972
Cloze c3
Cloze answer: greatest \(i\) for which \(a_i \neq 0\)
Q: The {{c1::degree of \(a(x)\), denoted \(\deg(a(x))\)}}, is the {{c3::greatest \(i\) for which \(a_i \neq 0\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095116 |
1 | 230% | 14d | 7 |
nid:1766314094978
c2
DiskMat
also is an integral domain
1
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users
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nid:1766314094978
Cloze c2
Cloze answer: also is an integral domain
Q: If \(D\) is an {{c1::integral domain}}, then \(D[x]\) {{c2::also is an integral domain}}.
A: Lemma 5.22(1)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095125 |
1 | 200% | 3d | 10 |
nid:1766314094988
DiskMat
How do you find the GCD of two polynomials?
1
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users
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nid:1766314094988
Q: How do you find the GCD of two polynomials?
A: To find \(\gcd(a(x), b(x))\):
Find a common factor \((x - \alpha)\) using the roots method:
Try all possible elements of the field to find roots
If \(\alpha\) is a root of both, then \((x - \alpha)\) is a common factor
Use division with remainder to reduce to smaller polynomials
Repeat the process on the smaller polynomialsAfter they have no roots anymore, try all monic polynomials up to degree d/2 to
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095138 |
1 | 230% | 18d | 7 |
nid:1766314094992
c1
DiskMat
no roots
1
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users
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nid:1766314094992
Cloze c1
Cloze answer: no roots
Q: An irreducible polynomial of degree \(\geq 2\) has {{c1:: no roots}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095142 |
1 | 230% | 7d | 10 |
nid:1766314094995
DiskMat
State Theorem 5.31 about the number of roots a polynomial ca...
1
lapses
1/4
users
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nid:1766314094995
Q: State Theorem 5.31 about the number of roots a polynomial can have.
A: Theorem 5.31: For a field \(F\), a nonzero polynomial \(a(x) \in F[x]\) of degree \(d\) has at most \(d\) roots.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095145 |
1 | 230% | 3d | 7 |
nid:1766314095059
c2
DiskMat
\(a*e = a\) (\(e*a = a\))
1
lapses
1/4
users
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ease
nid:1766314095059
Cloze c2
Cloze answer: \(a*e = a\) (\(e*a = a\))
Q: A {{c1::right (left) neutral element}} is an element such that for all \(a \in G\), {{c2:: \(a*e = a\) (\(e*a = a\))}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314095227 |
1 | 215% | 3d | 7 |
nid:1766314111367
LinAlg
Is the empty set of vectors linearly dependent or independen...
1
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users
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nid:1766314111367
Q: Is the empty set of vectors linearly dependent or independent?
A: It is linearly independent by definition, since there is no vector it could be a combination of.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314111367 |
1 | 230% | 4d | 7 |
nid:1766314111406
LinAlg
What does \(N(A) = \mathbb{R}^n\) mean?
1
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users
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nid:1766314111406
Q: What does \(N(A) = \mathbb{R}^n\) mean?
A: it means \(A = \boldsymbol{0}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766314111411 |
1 | 230% | 3d | 5 |
nid:1766940295633
c1
DiskMat
For any \(i\) and \(k\), if \(t_1, \dots, t_k\) are terms, t...
1
lapses
1/4
users
200%
ease
nid:1766940295633
Cloze c1
Q: For any \(i\) and \(k\), if \(t_1, \dots, t_k\) are terms, then {{c1::\(P_i^{(k)}(t_1, \dots, t_k)\) is a formula}}, called an {{c2::atomic formula}}.
A: A formula in 1st order logic with no logical connectives (like \(\lnot, \land, \lor, \rightarrow \)) and no quantifiers (\(\forall, \exists\))
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295669 |
1 | 200% | 6d | 8 |
nid:1766940295636
c1
DiskMat
The {{c1::set of statements \(\mathcal{S}\)}} is {{c2:: a s...
1
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1/4
users
230%
ease
nid:1766940295636
Cloze c1
Q: The {{c1::set of statements \(\mathcal{S}\)}} is {{c2:: a subset of the finite bit strings \(\Sigma^*\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295678 |
1 | 230% | 5d | 8 |
nid:1766940295674
c2
DiskMat
assigns to each formula \(F = (f_1, f_2, \dots, f_k) \in \La...
1
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1/4
users
200%
ease
nid:1766940295674
Cloze c2
Cloze answer: assigns to each formula \(F = (f_1, f_2, \dots, f_k) \in \Lambda^*\) a subset \({{c1
Q: The {{c3::semantics}} of a logic defines a function {{c1::\(free\)}} which {{c2::assigns to each formula \(F = (f_1, f_2, \dots, f_k) \in \Lambda^*\) a subset \({{c1::free}}(F) \subseteq \{1, \dots, k\}\) of the indices}}.
A: If \(i \in free(F)\), then the symbol is said to occur free in \(F\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295752 |
1 | 200% | 10d | 8 |
nid:1766940295686
c1
DiskMat
closed
1
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1/4
users
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nid:1766940295686
Cloze c1
Cloze answer: closed
Q: A formula is {{c1::closed}} if it {{c2::contains no free variables}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295776 |
1 | 230% | 10d | 9 |
nid:1766940295694
c1
DiskMat
formal language
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users
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nid:1766940295694
Cloze c1
Cloze answer: formal language
Q: \( L = \{s \ | \ \tau(s) = 1\} \) is a set of strings called a {{c1:: formal language}}. It defines a {{c2:: predicate \(\tau\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295788 |
1 | 230% | 8d | 10 |
nid:1766940295695
c1
DiskMat
For a set \(Z\) of atomic formulas, a {{c1::truth assignment...
1
lapses
1/4
users
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nid:1766940295695
Cloze c1
Q: For a set \(Z\) of atomic formulas, a {{c1::truth assignment \(\mathcal{A}\)}} is {{c2::a function \(\mathcal{A}: Z \rightarrow \{0, 1\}\)}}.
A: A truth assignment \(\mathcal{A}\) is suitable for a formula \(F\) if it contains all atomic formulas appearing in \(F\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295790 |
1 | 230% | 9d | 9 |
nid:1766940295714
c1
DiskMat
syntax
1
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users
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nid:1766940295714
Cloze c1
Cloze answer: syntax
Q: The {{c1::syntax}} of a logic defines {{c2::an alphabet \(\Lambda\) (of allowed symbols)}} and specifies {{c2::which strings in \(\Lambda^*\) are formulas (i.e. syntactically correct)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295825 |
1 | 230% | 7d | 7 |
nid:1766940295715
DiskMat
For \(F \vdash_K G\), what is \(F\) called in a calculus?
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nid:1766940295715
Q: For \(F \vdash_K G\), what is \(F\) called in a calculus?
A: The premises or preconditions.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295826 |
1 | 215% | 6d | 9 |
nid:1766940295739
c2
DiskMat
A set \(M\) of formulas is {{c1::unsatisfiable}} if and only...
1
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1/4
users
200%
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nid:1766940295739
Cloze c2
Q: A set \(M\) of formulas is {{c1::unsatisfiable}} if and only if {{c2::\(\mathcal{K}(M) \vdash_{Res} \emptyset\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295870 |
1 | 200% | 4d | 8 |
nid:1766940295754
DiskMat
Propositional logic is (in relation to predicate logic):
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users
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nid:1766940295754
Q: Propositional logic is (in relation to predicate logic):
A: embedded into predicate logic as a special case. We extend it by the concept of predicates.Predicates of the form \(P()\) act as propositional symbols.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295892 |
1 | 230% | 11d | 5 |
nid:1766940295762
c1
DiskMat
function symbol
1
lapses
1/4
users
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nid:1766940295762
Cloze c1
Cloze answer: function symbol
Q: A {{c1::function symbol}} is of the form {{c2::\(f_i^{(k)}\) with \(i, k \in \mathbb{N}\)}}, where {{c2::\(k\) denotes the number of arguments (the arity) of the function}}.
A: Function symbols for \(k = 0\) are called constants.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295904 |
1 | 230% | 6d | 7 |
nid:1766940295780
c1
DiskMat
every true statement has a proof: \(\phi(s, p) = 1 \Longleft...
1
lapses
1/4
users
230%
ease
nid:1766940295780
Cloze c1
Cloze answer: every true statement has a proof: \(\phi(s, p) = 1 \Longleftarrow \tau(s) = 1\)
Q: A proof system is {{c2::complete}} if {{c1:: every true statement has a proof: \(\phi(s, p) = 1 \Longleftarrow \tau(s) = 1\)}}.
A: Note that the use of \(\Longleftarrow\) is not the correct formalism.For all \(s \in \mathcal{S}\) with \(\tau(s) = 1\) there exists a \(p \in \mathcal{P}\) such that \(\phi(s, p) = 1\), is the correct formal definition.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295938 |
1 | 230% | 7d | 6 |
nid:1766940295793
c1
DiskMat
they are of the same type
1
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users
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nid:1766940295793
Cloze c1
Cloze answer: they are of the same type
Q: We are allowed to swap quantifier order in a formula if:{{c1:: they are of the same type}}{{c2:: the variables never appear in the same predicate}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1766940295958 |
1 | 230% | 5d | 6 |
nid:1767734963666
DiskMat
What is really important for the prenex form due to the bind...
1
lapses
1/4
users
215%
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nid:1767734963666
Q: What is really important for the prenex form due to the binding of quantifiers?
A: We need to wrap the entire expression in parentheses \(\forall \exists (...)\) otherwise, it's not prenex!
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| jonas | cid:1767734963666 |
1 | 215% | 5d | 6 |
nid:1766531635603
A&D
BFS (Breadth First Search)
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users
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nid:1766531635603
Q: BFS (Breadth First Search)
A: \(O(|V|+|E|)\) (Adjacency List)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635603 |
1 | 230% | 502d | 8 |
nid:1764867989867
A&D
Pre- and Postordering in BFS:
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users
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nid:1764867989867
Q: Pre- and Postordering in BFS:
A: Same as with pre-/postordering, we can use enter-/leave-ordering here: enter step at which vertex \(v\) is first encountered.leave step at which vertex \(v\) is dequeued
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867989867 |
1 | 230% | 562d | 8 |
nid:1765372936200
c2
A&D
O(k^n)
1
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users
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nid:1765372936200
Cloze c2
Cloze answer: O(k^n)
Q: Choose a tight bound!\({{c1::O(n^k)}} \leq {{c2::O(k^n)}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936201 |
1 | 230% | 682d | 9 |
nid:1766271258634
A&D
In what situation is the array the correct underlying datast...
1
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users
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nid:1766271258634
Q: In what situation is the array the correct underlying datastructure for a list?
A: When we have a fixed upper bound for the size of the list.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766271258635 |
1 | 230% | 694d | 8 |
nid:1766531635503
A&D
How can we make Knapsack polynomial using approximation?
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1/4
users
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ease
nid:1766531635503
Q: How can we make Knapsack polynomial using approximation?
A: Round the profits and solve the Knapsack problem for those rounded profits:\(\overline{p_i} := K \cdot \lfloor \frac{p_i}{K} \rfloor\). We then only have to compute every K'th entry of the DP-table.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635503 |
1 | 230% | 731d | 10 |
nid:1766531635474
A&D
Subsequence
1
lapses
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users
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ease
nid:1766531635474
Q: Subsequence
A: Teilfolge
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635474 |
1 | 230% | 740d | 8 |
nid:1764867989708
c2
A&D
incident (inzident oder anliegend)
1
lapses
1/4
users
230%
ease
nid:1764867989708
Cloze c2
Cloze answer: incident (inzident oder anliegend)
Q: In an edge \(e = \{u, v\}\), we call \(u\) {{c1::adjacent (adjazent oder benachbart)}} to \(v\) (and the other way around) and \(e\) {{c2::incident (inzident oder anliegend)}} to \(u, v\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867989709 |
1 | 230% | 752d | 8 |
nid:1765372936339
A&D
How does extract_max work for a maxHeap?
1
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1/4
users
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ease
nid:1765372936339
Q: How does extract_max work for a maxHeap?
A: The extract max operation works by taking the root node, the biggest element in the heap by it’s definition and restoring the heap condition.We remove the root and replace it by the element that is most to the right (last element in the array storing the heap).Then we "versickern" this small element, until the heap condition is restored. We swap it with the larger of the child nodes, until it's bigger than both of it's children.&nb
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936339 |
1 | 230% | 762d | 11 |
nid:1766580144028
c1
A&D
\(n-x\) components (different values)
1
lapses
1/4
users
230%
ease
nid:1766580144028
Cloze c1
Cloze answer: \(n-x\) components (different values)
Q: After adding \(x\) edges to the Union-Find datastructure, the repr array contains {{c1::\(n-x\) components (different values)}}.
A: Each added edge removes one unconnected component.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766580144028 |
1 | 230% | 748d | 11 |
nid:1765372936234
c2
A&D
Master Theorem: If {{c1:: \(b = \log_2(a)\)}} then {{c2:: \(...
1
lapses
1/4
users
230%
ease
nid:1765372936234
Cloze c2
Q: Master Theorem: If {{c1:: \(b = \log_2(a)\)}} then {{c2:: \(T(n) \leq O(n^{\log_2 a} \cdot \log n)\)}}.
A: The recursive and non-recursive work is balanced.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936234 |
1 | 230% | 800d | 12 |
nid:1765198542351
A&D
What is the sum of all natural numbers between 1 and \(n\)?
1
lapses
1/4
users
230%
ease
nid:1765198542351
Q: What is the sum of all natural numbers between 1 and \(n\)?
A: \(= \frac{n(n+1)}{2}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765198542351 |
1 | 230% | 827d | 9 |
nid:1764867989631
c2
A&D
cycle (Kreis)
1
lapses
1/4
users
230%
ease
nid:1764867989631
Cloze c2
Cloze answer: cycle (Kreis)
Q: In graph theory, a {{c2::cycle (Kreis)}} is a {{c1::closed walk without repeated vertices}} and {{c1::at least three vertices}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867989631 |
1 | 230% | 835d | 8 |
nid:1766531635499
A&D
What is pseudo-polynomial time?
1
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1/4
users
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ease
nid:1766531635499
Q: What is pseudo-polynomial time?
A: Runtime dependent on a number \(W\) (like in knapsack) which is not correlated polynomially to input length but exponentially.The DP-table get's 10x for \(W = 10 \rightarrow 100\) but the input size (binary) only grows from \(\log_2(10) \approx 3 \rightarrow \approx 6\) so x2.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635500 |
1 | 230% | 840d | 11 |
nid:1765372936324
A&D
Merge Sort
1
lapses
1/4
users
230%
ease
nid:1765372936324
Q: Merge Sort
A: Best Case: \(O(n \log n)\)Worst Case: \(O(n \log n)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765383739470 |
1 | 230% | 861d | 10 |
nid:1765372936167
c1
A&D
\(f, g\) are differentiable (for sufficiently large \(x\))
1
lapses
1/4
users
230%
ease
nid:1765372936167
Cloze c1
Cloze answer: \(f, g\) are differentiable (for sufficiently large \(x\))
Q: What are the prerequisites for \(f\) and \(g\) to apply l'Hôpital's?{{c1::\(f, g\) are differentiable (for sufficiently large \(x\))}}{{c2::\(\lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) = \infty\) (or both \(= 0\))}}{{c3::\(g'(x
A: Then: \(\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936167 |
1 | 230% | 840d | 12 |
nid:1765372936330
A&D
Heapsort
1
lapses
1/4
users
230%
ease
nid:1765372936330
Q: Heapsort
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765383739478 |
1 | 230% | 864d | 11 |
nid:1766580157417
A&D
Can Kruskal's Algorithm be executed in \(O(|E| + |V|\log|V|)...
1
lapses
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users
230%
ease
nid:1766580157417
Q: Can Kruskal's Algorithm be executed in \(O(|E| + |V|\log|V|)\) time?
A: No, we need to sort the edges which takes at least \(|E| \log |E|\) time.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766580157417 |
1 | 230% | 855d | 11 |
nid:1766580143726
A&D
Floyd-Warshall
1
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users
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nid:1766580143726
Q: Floyd-Warshall
A: \(O(|V|^3)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766580143728 |
1 | 230% | 863d | 11 |
nid:1765372936146
A&D
Simplify \(\frac{a^{kn}}{b^{k'n}} =\)
1
lapses
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users
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ease
nid:1765372936146
Q: Simplify \(\frac{a^{kn}}{b^{k'n}} =\)
A: \(\frac{e^{\ln(a^{kn})}}{e^{\ln(b^{k'n})}} = e^{kn \cdot \ln(a) - k'n \cdot ln(b)}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936146 |
1 | 230% | 923d | 12 |
nid:1765198542383
A&D
Which functions \(f(n)\) have \(\lim_{n\rightarrow \infty} f...
1
lapses
1/4
users
230%
ease
nid:1765198542383
Q: Which functions \(f(n)\) have \(\lim_{n\rightarrow \infty} f(n)\) undefined?
A: Typically functions that oscilate as they approach infinity such as \(f(n) = \sin n\) or \(f(n) = (-1)^n\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765198542383 |
1 | 230% | 1011d | 9 |
nid:1765372936324
A&D
Merge Sort
1
lapses
1/4
users
230%
ease
nid:1765372936324
Q: Merge Sort
A: Best Case: \(O(n \log n)\)Worst Case: \(O(n \log n)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765383739472 |
1 | 230% | 1018d | 11 |
nid:1765372936300
A&D
What do we have to pay attention to in the I.H. and the I.S....
1
lapses
1/4
users
230%
ease
nid:1765372936300
Q: What do we have to pay attention to in the I.H. and the I.S. in an induction proof?
A: We should change the variable name from \(n\) to \(k\) (for example) as not to confuse it.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936300 |
1 | 230% | 1086d | 11 |
nid:1766531635431
c2
A&D
The height of a 2-3 Tree for \(n\) keys is {{c1::\(\leq \log...
1
lapses
1/4
users
230%
ease
nid:1766531635431
Cloze c2
Q: The height of a 2-3 Tree for \(n\) keys is {{c1::\(\leq \log_2(n)\)}} thus \(h={{c2::O(\log(n))::\textbf{O-notation} }}\).
A: Note that for the case \(n = 1\) the root has one leaf with the key.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635433 |
1 | 230% | 1109d | 9 |
nid:1765372936291
c1
A&D
{{c1:: \(\sum_{i = 1}^{n} i\log(i)\)::Sum}} \(\leq\) {{c2::...
1
lapses
1/4
users
230%
ease
nid:1765372936291
Cloze c1
Q: {{c1:: \(\sum_{i = 1}^{n} i\log(i)\)::Sum}} \(\leq\) {{c2::\(O(n \log(n))\)::O-notation}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936292 |
1 | 230% | 1185d | 9 |
nid:1766531635639
A&D
Bellman-Ford optimisation in a DAG?
1
lapses
1/4
users
230%
ease
nid:1766531635639
Q: Bellman-Ford optimisation in a DAG?
A: In an acyclic graph, topological sorting is already an algorithm that gives us the most-efficient order to calculate the cost in.Because we can be sure that any predecessors already have the correct \(l\)-good bound distance (guaranteed by topo-sort, no backedges), we can simply relax once.Thus we can compute the correct cheapest path in one "relaxation": \(O(|E|)\).Therefore with toposort: \(O(|V| + |E|)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635639 |
1 | 230% | 1221d | 11 |
nid:1765372936324
A&D
Merge Sort
1
lapses
1/4
users
230%
ease
nid:1765372936324
Q: Merge Sort
A: Best Case: \(O(n \log n)\)Worst Case: \(O(n \log n)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765383739471 |
1 | 230% | 1302d | 9 |
nid:1765372936244
A&D
If \(T(n) = aT(n/ 2) + Cn^b\), then we get which type of O-...
1
lapses
1/4
users
230%
ease
nid:1765372936244
Q: If \(T(n) = aT(n/ 2) + Cn^b\), then we get which type of O-Notation?
A: \(T(n) = \Theta(...)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936244 |
1 | 230% | 1411d | 11 |
nid:1765372936324
A&D
Merge Sort
1
lapses
1/4
users
230%
ease
nid:1765372936324
Q: Merge Sort
A: Best Case: \(O(n \log n)\)Worst Case: \(O(n \log n)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765383739469 |
1 | 230% | 1440d | 9 |
nid:1765372936222
A&D
What is the form of the recursive equations solved by the Ma...
1
lapses
1/4
users
230%
ease
nid:1765372936222
Q: What is the form of the recursive equations solved by the Master Theorem?
A: \(T(n) \leq aT(n/2) + Cn^b\)where \(a\), \(C > 0\) and \(b \geq 0\) are constants.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936222 |
1 | 230% | 1460d | 11 |
nid:1765372936231
c2
A&D
\(T(n) \leq O(n^b)\)
1
lapses
1/4
users
230%
ease
nid:1765372936231
Cloze c2
Cloze answer: \(T(n) \leq O(n^b)\)
Q: Master Theorem: If {{c1:: \(b > \log_2(a)\)}} then {{c2:: \(T(n) \leq O(n^b)\)}}.
A: This is the case for which the work outside the recursion dominates.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936232 |
1 | 230% | 1480d | 11 |
nid:1765372936182
c2
A&D
\(g \geq \Omega(f)\)
1
lapses
1/4
users
230%
ease
nid:1765372936182
Cloze c2
Cloze answer: \(g \geq \Omega(f)\)
Q: {{c2::\(g \geq \Omega(f)\)}} \( \Leftrightarrow\) {{c1::\( f \leq O(g)\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936183 |
1 | 230% | 1488d | 9 |
nid:1766580161426
A&D
In every connected graph \(G\), when executing Kruskal using...
1
lapses
1/4
users
230%
ease
nid:1766580161426
Q: In every connected graph \(G\), when executing Kruskal using Union-Find, the representative repr[u] changes \(O(\dots)\) times:
A: \(O(\log_2 |V|)\), as we always at least double the size of the representative (we merge smaller into bigger, and repr[u] changes if it's the smaller one).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766580161426 |
1 | 230% | 1501d | 11 |
nid:1765372936286
c2
A&D
\(\log(n!)\)
1
lapses
1/4
users
230%
ease
nid:1765372936286
Cloze c2
Cloze answer: \(\log(n!)\)
Q: {{c1:: \(\sum_{i = 1}^{n} \log(i)\)::Sum}} \(=\) {{c2::\(\log(n!)\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765372936286 |
1 | 230% | 1588d | 11 |
nid:1766531635418
A&D
Worst case for search in a binary tree?
1
lapses
1/4
users
230%
ease
nid:1766531635418
Q: Worst case for search in a binary tree?
A: Binary trees are not necessarily balanced, hence it is possible that \(h >> \log_2 n\).Worst case example if inserted in ascending order:
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635418 |
1 | 230% | 1632d | 9 |
nid:1765372936327
A&D
Quicksort
1
lapses
1/4
users
230%
ease
nid:1765372936327
Q: Quicksort
A: Best Case: \(O(n \log n)\)Worst Case: \(O(n^2)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765383739474 |
1 | 230% | 1699d | 9 |
nid:1764867989717
c2
A&D
Hamiltonian cycle (Hamiltonkreis)
1
lapses
1/4
users
230%
ease
nid:1764867989717
Cloze c2
Cloze answer: Hamiltonian cycle (Hamiltonkreis)
Q: In graph theory, a {{c2::Hamiltonian cycle (Hamiltonkreis)}} is a {{c1::cycle that contains every vertex}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867989719 |
1 | 230% | 1764d | 9 |
nid:1765655148922
A&D
Runtime Determine if Hamiltonian path exists?
1
lapses
1/4
users
230%
ease
nid:1765655148922
Q: Runtime Determine if Hamiltonian path exists?
A: Hamiltonian walk - exponential, we have to brute-force
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765655148922 |
1 | 230% | 1764d | 9 |
nid:1766531635615
c1
A&D
triangle inequality
1
lapses
1/4
users
230%
ease
nid:1766531635615
Cloze c1
Cloze answer: triangle inequality
Q: The {{c1::triangle inequality}} in a weighted graph is {{c2::\(d(u, v) \leq d(u, w) + d(w, v)\)}}.
A: This holds, since if the path through \(w\) was actually cheaper, then \(d(u, v)\) would be wrong.Does not hold in graphs with negative cycles.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766531635615 |
1 | 230% | 1834d | 11 |
nid:1766271258597
c1
A&D
datastructure
1
lapses
1/4
users
230%
ease
nid:1766271258597
Cloze c1
Cloze answer: datastructure
Q: A {{c1:: datastructure}} is the implementation of the wishlist of operations defined in our ADT.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766271258597 |
1 | 230% | 1874d | 9 |
nid:1764867991079
c2
DiskMat
order of \(G\)
1
lapses
1/4
users
230%
ease
nid:1764867991079
Cloze c2
Cloze answer: order of \(G\)
Q: For a finite group \(G\), {{c1::\(|G|\)}} is called the {{c2::order of \(G\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991079 |
1 | 230% | 810d | 8 |
nid:1764867991067
c2
DiskMat
If {{c2:: the order \(\text{ord}(a)\) of \(a \in G\) is 2}},...
1
lapses
1/4
users
230%
ease
nid:1764867991067
Cloze c2
Q: If {{c2:: the order \(\text{ord}(a)\) of \(a \in G\) is 2}}, {{c1:: a is it's own self-inverse}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991068 |
1 | 230% | 852d | 8 |
nid:1767535579762
c1
DiskMat
We can solve \(R_a(b^c)\) by using the fact that {{c1:: \(R_...
1
lapses
1/4
users
230%
ease
nid:1767535579762
Cloze c1
Q: We can solve \(R_a(b^c)\) by using the fact that {{c1:: \(R_a(b^c) = R_a(b^{R_{\varphi(a)}(c)})\)}} if \(a, b\) coprime.
A: Note that we can't simply reduce by \(a\)!
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1767535579762 |
1 | 230% | 856d | 8 |
nid:1764867991005
DiskMat
Give an example of a group homomorphism involving the logari...
1
lapses
1/4
users
230%
ease
nid:1764867991005
Q: Give an example of a group homomorphism involving the logarithm function.
A: The logarithm function is a group homomorphism from \(\langle \mathbb{R}^{>0}; \cdot \rangle\) to \(\langle \mathbb{R}; + \rangle\) because: \[\log(a \cdot b) = \log a + \log b\]
It's also an isomorphism because the logarithm is bijective on positive reals.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991005 |
1 | 230% | 856d | 8 |
nid:1764867991186
DiskMat
Why do we need \(\mathbb{Z}_m^*\) for multiplication, rather...
1
lapses
1/4
users
230%
ease
nid:1764867991186
Q: Why do we need \(\mathbb{Z}_m^*\) for multiplication, rather than just using \(\mathbb{Z}_m\)?
A: \(\mathbb{Z}_m\) (with \(\oplus\)) is not a group with respect to multiplication modulo \(m\) because elements that are not coprime to \(m\) don't have a multiplicative inverse.
For example, in \(\mathbb{Z}_6\), the element \(2\) has no multiplicative inverse because \(\gcd(2, 6) = 2 \neq 1\).
Thus we need \(\mathbb{Z}_m^*\) (elements coprime to \(m\)) to form a group with \(\odot\) (multiplication mod \(m\)).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991186 |
1 | 230% | 860d | 8 |
nid:1765655179118
c2
DiskMat
cyclic for every \(n\)
1
lapses
1/4
users
230%
ease
nid:1765655179118
Cloze c2
Cloze answer: cyclic for every \(n\)
Q: The group \(\langle \mathbb{Z}_n; \oplus \rangle\) is {{c2::cyclic for every \(n\)}}, where {{c3:: 1}} is always a generator.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765655179118 |
1 | 230% | 902d | 8 |
nid:1764867990443
c2
DiskMat
\(a \equiv_m b \Longleftrightarrow R_m(a) = R_m(b)\) (congr...
1
lapses
1/4
users
230%
ease
nid:1764867990443
Cloze c2
Cloze answer: \(a \equiv_m b \Longleftrightarrow R_m(a) = R_m(b)\) (congruence iff same remainder)
Q: What are the two key properties of the remainder function \(R_m\)? (Lemma 4.16)(i) {{c1:: \(a \equiv_m R_m(a)\) (the remainder represents the equivalence class)}}(ii) {{c2:: \(a \equiv_m b \Longleftrightarrow R_m(a) = R_m(b)\) (congru
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766229398882 |
1 | 230% | 908d | 8 |
nid:1766448533677
c1
DiskMat
The semantics of propositional logic are defined as:{{c1::\(...
1
lapses
1/4
users
230%
ease
nid:1766448533677
Cloze c1
Q: The semantics of propositional logic are defined as:{{c1::\(\mathcal{A}(F) = \mathcal{A}(A_i)\) for any atomic formula \(A_i\)}}for \(\land, \lor, \lnot\) the semantics are identical to before.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448533677 |
1 | 230% | 965d | 11 |
nid:1764867990475
DiskMat
What does "unique up to order" mean in the Fundamental Theor...
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nid:1764867990475
Q: What does "unique up to order" mean in the Fundamental Theorem of Arithmetic?
A: Every integer has exactly one prime factorization if we don't care about the order of factors. For example, \(12 = 2^2 \cdot 3 = 3 \cdot 2 \cdot 2 = 2 \cdot 3 \cdot 2\) are all the same factorization, just written differently.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990475 |
1 | 230% | 963d | 9 |
nid:1766448533765
DiskMat
For CNF construction, how do you form literals from a row in...
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nid:1766448533765
Q: For CNF construction, how do you form literals from a row in the truth table?
A: - If \(A_i = 0\) in the row, take \(A_i\)- If \(A_i = 1\) in the row, take \(\lnot A_i\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448533765 |
1 | 230% | 976d | 11 |
nid:1764867990795
c1
DiskMat
A function \(f:\mathbb{N}\to\{0,1\}\) is called computable i...
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nid:1764867990795
Cloze c1
Q: A function \(f:\mathbb{N}\to\{0,1\}\) is called computable if {{c1::there is a computer program that, for every \(n\in\mathbb{N}\), when given \(n\) as input, outputs \(f(n)\).}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990795 |
1 | 230% | 1005d | 11 |
nid:1764867990348
DiskMat
Which of the following are countable: \(\mathbb{N}\), \(\mat...
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nid:1764867990348
Q: Which of the following are countable: \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\), \(\{0,1\}^*\), \(\{0,1\}^{\infty}\)?
A: Countable: \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\), \(\{0,1\}^*\)
Uncountable: \(\mathbb{R}\), \(\{0,1\}^{\infty}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990348 |
1 | 230% | 987d | 9 |
nid:1764867991454
DiskMat
What is the left cancellation law in a group?
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nid:1764867991454
Q: What is the left cancellation law in a group?
A: Left cancellation law: \(a * b = a * c \ \implies \ b = c\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991454 |
1 | 230% | 1000d | 12 |
nid:1764867990296
DiskMat
Why is \((\mathbb{N}; |)\) NOT totally ordered?
1
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nid:1764867990296
Q: Why is \((\mathbb{N}; |)\) NOT totally ordered?
A: Because \(2 \nmid 3\) and \(3 \nmid 2\) (they are incomparable).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990296 |
1 | 230% | 1002d | 9 |
nid:1764867990269
c3
DiskMat
\(A^*\) (finite sequences) is countable
1
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nid:1764867990269
Cloze c3
Cloze answer: \(A^*\) (finite sequences) is countable
Q: Which operations preserve countability?Let \(A\) and \(A_i\) for \(i \in \mathbb{N}\) be countable sets. Then: {{c1::\(A^n\) (\(n\)-tuples) is countable }}{{c2::\(\bigcup_{i\in \mathbb{N} } A_i\) (countable union) is countabl
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766229398360 |
1 | 230% | 1016d | 11 |
nid:1764867990164
DiskMat
When is the lexicographic order on \(A \times B\) totally or...
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nid:1764867990164
Q: When is the lexicographic order on \(A \times B\) totally ordered?
A: When both \((A; \preceq)\) and \((B; \sqsubseteq)\) are totally ordered.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990164 |
1 | 230% | 1020d | 9 |
nid:1764867990755
c1
DiskMat
the truth value depends on the interpretation of the symbols
1
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nid:1764867990755
Cloze c1
Cloze answer: the truth value depends on the interpretation of the symbols
Q: A logical formula is generally not a mathematical statement, because {{c1::the truth value depends on the interpretation of the symbols}}.
A: (so we can't prove/disprove it)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990755 |
1 | 230% | 1063d | 9 |
nid:1764867990887
DiskMat
How many divisors does \(n\) expressed as a factor of prime ...
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users
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nid:1764867990887
Q: How many divisors does \(n\) expressed as a factor of prime numbers \(n = \prod_{i = 1}^m p_i^{e_i}\) have?
A: \(n\) has \(\# _ {\text{div}(n)} = \prod_{i = 1}^m (e_i + 1)\) divisors.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990887 |
1 | 230% | 1056d | 12 |
nid:1766448533830
c1
DiskMat
empty set \(\emptyset\)
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nid:1766448533830
Cloze c1
Cloze answer: empty set \(\emptyset\)
Q: The {{c1::empty set \(\emptyset\)}} is a {{c2::clause}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448533831 |
1 | 230% | 1061d | 11 |
nid:1764867991020
c2
DiskMat
neutral element; nullspace
1
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nid:1764867991020
Cloze c2
Cloze answer: neutral element; nullspace
Q: For a homomorphism \(h: G \rightarrow H\), the {{c1::kernel \(\ker(h)\)}} is the set of all elements mapped to the {{c2::neutral element}} (essentially the {{c2::nullspace}}).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991020 |
1 | 230% | 1083d | 11 |
nid:1764867990259
DiskMat
Is \(\mathbb{N} \times \mathbb{N}\) countable?
1
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users
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nid:1764867990259
Q: Is \(\mathbb{N} \times \mathbb{N}\) countable?
A: Yes, the set \(\mathbb{N} \times \mathbb{N}\) (= \(\mathbb{N}^2\)) of ordered pairs of natural numbers is countable.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990259 |
1 | 230% | 1113d | 9 |
nid:1764867989950
DiskMat
What is the logical principle behind case distinction?
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nid:1764867989950
Q: What is the logical principle behind case distinction?
A: For every \(k\) we have:
\[(A_1 \lor \dots \lor A_k) \land (A_1 \rightarrow B) \land \dots \land (A_k \rightarrow B) \models B\]
(If at least one case occurs, and all cases imply \(B\), then \(B\) holds)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867989950 |
1 | 230% | 1133d | 9 |
nid:1764867990542
c2
DiskMat
Prove that \( T \) is false.
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nid:1764867990542
Cloze c2
Cloze answer: Prove that \( T \) is false.
Q: Proof method: Proof by Contradiction1. {{c1:: Find a suitable statement \( T\).}}2. {{c2:: Prove that \( T \) is false.}}3. {{c3:: Assume that \( S \) is false and prove that \( T \) is true (-> contradiction).}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766229399206 |
1 | 230% | 1147d | 9 |
nid:1767291036660
c2
DiskMat
Associativity
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nid:1767291036660
Cloze c2
Cloze answer: Associativity
Q: An abelian group has the following properties:{{c1::Closure}}{{c2::Associativity}}{{c3::Identity}}{{c4::Inverse}}{{c5::Commutativity}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1767291036662 |
1 | 230% | 1185d | 12 |
nid:1766920111886
c1
DiskMat
\(|a|\)
1
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nid:1766920111886
Cloze c1
Cloze answer: \(|a|\)
Q: \(\gcd(a, 0) = \) {{c1::\(|a|\)}}
A: This is why \(0\) isn't in \(Z_m^* \) and \(F[x]^*_{m(x)}\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766920111887 |
1 | 230% | 1199d | 12 |
nid:1764867989903
DiskMat
What is \(\lnot \forall x P(x)\) equivalent to?
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nid:1764867989903
Q: What is \(\lnot \forall x P(x)\) equivalent to?
A: \(\lnot \forall x P(x) \equiv \exists x \lnot P(x)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867989903 |
1 | 230% | 1295d | 9 |
nid:1767534763076
DiskMat
Uncountability Proof by Complement (with example \([0,1] \se...
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nid:1767534763076
Q: Uncountability Proof by Complement (with example \([0,1] \setminus \mathbb{Q}\)):
A: Find \(B\) uncountable such that \(A \subseteq B\).
Show that \(B \backslash A\) countable which proves that \(A\) uncountable.
You have to prove this implication in the exam:
Assume \(A\) is countable towards contradiction.
We have shown that \(B \ \backslash \ A\) is countable.
Thus \(A \cup (B \ \backslash \ A)\) also countable (Theorem 3.22: Union of countable is countable).
But \(A \cup
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1767534763076 |
1 | 230% | 1278d | 12 |
nid:1764867991416
c1
DiskMat
The set \(\mathcal{C} = {{c1::\text{Im}(E)}}\) is called the...
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nid:1764867991416
Cloze c1
Q: The set \(\mathcal{C} = {{c1::\text{Im}(E)}}\) is called the {{c2::set of codewords}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991417 |
1 | 230% | 1280d | 9 |
nid:1764867990962
c1
DiskMat
left inverse
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nid:1764867990962
Cloze c1
Cloze answer: left inverse
Q: A function \(f: A \rightarrow B\) has a {{c1::left inverse}} if and only if \(f\) is {{c2::injective}} (not in script).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990962 |
1 | 230% | 1303d | 10 |
nid:1764867991229
DiskMat
What is the characteristic of \(\mathbb{Z}_m\)?
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nid:1764867991229
Q: What is the characteristic of \(\mathbb{Z}_m\)?
A: The characteristic of \(\mathbb{Z}_m\) is \(m\).
Explanation: The characteristic is the order of \(1\) in the additive group. In \(\mathbb{Z}_m\), adding \(1\) to itself \(m\) times gives: \[\underbrace{1 + 1 + \cdots + 1}_{m \text{ times}} = m \equiv_m 0\]
So \(\text{ord}(1) = m\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991229 |
1 | 230% | 1316d | 9 |
nid:1764867991382
DiskMat
When does an element of \(F[x]_{m(x)}\) have an inverse?
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nid:1764867991382
Q: When does an element of \(F[x]_{m(x)}\) have an inverse?
A: Lemma 5.36: The congruence equation \[a(x)b(x) \equiv_{m(x)} 1\] for a given \(a(x)\) has a solution \(b(x) \in F[x]_{m(x)}\) if and only if \(\gcd(a(x), m(x)) = 1\). The solution is unique.
In other words: \[ F[x]_{m(x)}^* = \{a(x) \in F[x]_{m(x)} \ | \ \gcd(a(x), m(x)) = 1\} \]
This is analogous to \(\mathbb{Z}_m^*\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991382 |
1 | 230% | 1316d | 9 |
nid:1764867991373
DiskMat
What are the equivalence classes modulo \(m(x)\) in a polyno...
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nid:1764867991373
Q: What are the equivalence classes modulo \(m(x)\) in a polynomial field?
A: Lemma 5.33: Congruence modulo \(m(x)\) is an equivalence relation on \(F[x]\), and each equivalence class has a unique representation of degree less than \(\deg(m(x))\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991373 |
1 | 230% | 1379d | 9 |
nid:1768521665087
c1
DiskMat
not
1
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nid:1768521665087
Cloze c1
Cloze answer: not
Q: \(0\) is {{c1::not}} in \(A^*\) where {{c2::\(A\) is a multiplicative algebra like \(\mathbb{Z}_{25}\)}}. Justification Included
A: \(\gcd(0, n) = n\) and not \(1\)!
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768521665088 |
1 | 230% | 1398d | 12 |
nid:1764867991413
c1
DiskMat
codeword
1
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nid:1764867991413
Cloze c1
Cloze answer: codeword
Q: The {{c2::output \((c_0, \dots, c_{n-1})\)}} of an encoding function is called a {{c1::codeword}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991414 |
1 | 230% | 1409d | 9 |
nid:1764867991315
DiskMat
\(\alpha \in F\) is a root of \(a(x)\) if and only if:
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nid:1764867991315
Q: \(\alpha \in F\) is a root of \(a(x)\) if and only if:
A: \((x - \alpha)\) divides \(a(x)\).
Corollary: An irreducible polynomial of degree \(\geq 2\) has no roots.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991315 |
1 | 230% | 1433d | 9 |
nid:1766448533127
c1
DiskMat
A proof system \(\Pi\) is {{c1:: a quadruple \(\Pi = (\mathc...
1
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users
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nid:1766448533127
Cloze c1
Q: A proof system \(\Pi\) is {{c1:: a quadruple \(\Pi = (\mathcal{S, P}, \tau, \phi)\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448533127 |
1 | 230% | 1451d | 9 |
nid:1766448532960
c3
DiskMat
In a finite group the function \(x \rightarrow x^e\) is {{c1...
1
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nid:1766448532960
Cloze c3
Q: In a finite group the function \(x \rightarrow x^e\) is {{c1:: a bijection}} if {{c2::\(e\) coprime to \(|G|\)}}.For \(x^e = y\), the inverse of \(y\) is {{c3:: the unique \(e\)-th root \(x = y^d\), with \(de \equiv_{|G|} 1\)}}.
A: Proof:We have \(ed = k \cdot |G| + 1\) for some \(k\). Thus, for any \(x \in G\) we have\[(x^e)^d = x^{ed} = x^{k \cdot |G| + 1} = \underbrace{(x^{|G|})^k}_{=1} \cdot x = x\]which means that the function \(y \mapsto y^d\) is the inverse function of the function \(x \mapsto x^e\) (which is hence a bijection). The under-braced term is equal to 1 because the order of \(x\) must divide the order of \(G\) (Lagrange).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448532961 |
1 | 230% | 1477d | 9 |
nid:1764867990569
c2
DiskMat
an inverse relation
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nid:1764867990569
Cloze c2
Cloze answer: an inverse relation
Q: The definition of {{c2::an inverse relation}} is \( a \ \rho \ b \iff{{c1:: b \ \hat{\rho} \ a}}\).
A: Example: Inverse of parent relation is childhood relation. Also written as \( \rho^{-1}\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990569 |
1 | 230% | 1548d | 9 |
nid:1766448533150
c1
DiskMat
efficiently computable
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nid:1766448533150
Cloze c1
Cloze answer: efficiently computable
Q: We require that the proof verification function \(\phi\) is {{c1::efficiently computable}}, otherwise the proof system is not useful.
A: A proof system is useless if verification is infeasible.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448533150 |
1 | 230% | 1518d | 9 |
nid:1766448533306
c1
DiskMat
The notation {{c1::\(\mathcal{A} \models F\)}} means that {{...
1
lapses
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nid:1766448533306
Cloze c1
Q: The notation {{c1::\(\mathcal{A} \models F\)}} means that {{c2::\(\mathcal{A}\) is a model for \(F\)}}.
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|---|---|---|---|---|---|
| lorenz | cid:1766448533306 |
1 | 230% | 1538d | 9 |
nid:1764867991099
c1
DiskMat
the group generated by \(a\), \(\langle a \rangle\)
1
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nid:1764867991099
Cloze c1
Cloze answer: the group generated by \(a\), \(\langle a \rangle\)
Q: The {{c2:: smallest}} subgroup of a group \(G\) containing \(a \in G\) is {{c1:: the group generated by \(a\), \(\langle a \rangle\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991099 |
1 | 230% | 1556d | 9 |
nid:1764867991253
c1
DiskMat
degree of \(a(x)\), denoted \(\deg(a(x))\)
1
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nid:1764867991253
Cloze c1
Cloze answer: degree of \(a(x)\), denoted \(\deg(a(x))\)
Q: The {{c1::degree of \(a(x)\), denoted \(\deg(a(x))\)}}, is the {{c3::greatest \(i\) for which \(a_i \neq 0\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991253 |
1 | 230% | 1561d | 9 |
nid:1764867990311
c1
DiskMat
For \(f: \mathbb{R} \to \mathbb{R}\) with \(f(x) = x^2\), wh...
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1/4
users
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nid:1764867990311
Cloze c1
Q: For \(f: \mathbb{R} \to \mathbb{R}\) with \(f(x) = x^2\), what are:
1. Range: {{c1::\(\mathbb{R}^{\geq 0}\) (non-negative reals)}}2. Preimage of \([4, 9]\): {{c2::\([-3, -2] \cup [2, 3]\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990311 |
1 | 230% | 1591d | 9 |
nid:1764867990323
DiskMat
What is the principle behind the proof step of composing imp...
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users
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nid:1764867990323
Q: What is the principle behind the proof step of composing implications?
A: If \(S \Rightarrow T\) and \(T \Rightarrow U\) are both true, then \(S \Rightarrow U\) is also true (transitivity of implication).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990323 |
1 | 230% | 1641d | 9 |
nid:1764867991337
DiskMat
Why is a polynomial of degree \(d\) uniquely determined by \...
1
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users
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nid:1764867991337
Q: Why is a polynomial of degree \(d\) uniquely determined by \(d + 1\) values of \(a(x)\)?
A: This \(a(x)\) is unique since if there was another \(a'(x)\) then \(a(x) - a'(x)\) would have at most degree \(d\) and thus at most \(d\) roots. But since \(a(x) - a'(x)\) has the same \(d + 1\) roots, it's \(0 \implies a(x) = a'(x)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991337 |
1 | 230% | 1645d | 9 |
nid:1766448533167
c2
DiskMat
predicate \(\tau\)
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lapses
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nid:1766448533167
Cloze c2
Cloze answer: predicate \(\tau\)
Q: \( L = \{s \ | \ \tau(s) = 1\} \) is a set of strings called a {{c1:: formal language}}. It defines a {{c2:: predicate \(\tau\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448533168 |
1 | 230% | 1677d | 12 |
nid:1766448533885
c2
DiskMat
there is a literal \(L\) such that \(L \in K_1\), \(\lnot L ...
1
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users
230%
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nid:1766448533885
Cloze c2
Cloze answer: there is a literal \(L\) such that \(L \in K_1\), \(\lnot L \in K_2\)
Q: A clause \(K\) is {{c1::resolvent}} of clauses \(K_1\) and \(K_2\) if {{c2::there is a literal \(L\) such that \(L \in K_1\), \(\lnot L \in K_2\)}}.
A: \[K = (K_1 \setminus \{L\}) \cup (K_2 \setminus \{\lnot L\})\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448533886 |
1 | 230% | 1681d | 12 |
nid:1767648242888
DiskMat
How do we construct a field \(GF(p^q)\)?
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Q: How do we construct a field \(GF(p^q)\)?
A: We take the field \(GF(p)[x]_{m(x)}\) where \(m(x)\) is an irreducible polynomial of degree \(q\).Then \(GF(p)[x]_{m(x)}\) has \({|F|}^q\) polynomials in it, as all of degree less than \(q\) are coprime to \(m(x)\), by definition of irreducible. And this field is isomorphic to \(GF(p^q)\).
Example: The field \(GF(2)[x]\) \({x^2 + x + 1}\) is isomorphic to \(GF(2^2 = 4)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1767648242888 |
1 | 230% | 1744d | 12 |
nid:1764867991055
c1
DiskMat
In a group, for \(n \geq 1\), the positive power is defined ...
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nid:1764867991055
Cloze c1
Q: In a group, for \(n \geq 1\), the positive power is defined recursively: {{c1::\(a^n = a \cdot a^{n-1}\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991055 |
1 | 230% | 1763d | 9 |
nid:1764867990250
DiskMat
If two sets each dominate the other, what can we conclude?
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nid:1764867990250
Q: If two sets each dominate the other, what can we conclude?
A: For sets \(A\) and \(B\):
\[A \preceq B \land B \preceq A \quad \Rightarrow \quad A \sim B\]
If there's an injection \(f: A \to B\) and an injection \(g: B \to A\), then there's a bijection between \(A\) and \(B\).Bernstein-Schröder Theorem
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990250 |
1 | 230% | 1772d | 10 |
nid:1764867990892
DiskMat
What exponentiation operation is valid in modular arithmetic...
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nid:1764867990892
Q: What exponentiation operation is valid in modular arithmetic?
A: This is allowed:\(a \equiv_n b\) and then \(a^x \equiv_n b^x\)But this on the other hand is illegal:\(a \equiv_n b\) and \(c \equiv_n d\) and then doing \(a^c \equiv_n b^d\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990892 |
1 | 230% | 1786d | 13 |
nid:1764867991429
c2
DiskMat
number of positions at which the string is non-zero
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nid:1764867991429
Cloze c2
Cloze answer: number of positions at which the string is non-zero
Q: The {{c1::Hamming weight}} of a string in a finite alphabet \(\mathcal{A}\) is the {{c2::number of positions at which the string is non-zero}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991430 |
1 | 230% | 1798d | 9 |
nid:1765193120869
DiskMat
What is a zerodivisor and in which structure do they exist?
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nid:1765193120869
Q: What is a zerodivisor and in which structure do they exist?
A: A zerodivisor is an element \(a \neq 0\) in a commutative ring for which there exists a \(b \neq 0\) such that \(ab = 0\).This is commonly encountered for the polynomial rings formed over \(\text{GF}[x]_{m(x)}\) with \(m(x)\) not irreducible (i.e. it's not a field).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765193120869 |
1 | 230% | 1797d | 9 |
nid:1766448533482
c1
DiskMat
axiom \(A\); these axioms
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nid:1766448533482
Cloze c1
Cloze answer: axiom \(A\); these axioms
Q: An {{c1::axiom \(A\)}} is a {{c2::statement taken as true in a theory}}. {{c3::Theorems}} are the statements which {{c4::follow from {{c1::these axioms}} (\(A \models T\))}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448533483 |
1 | 230% | 1805d | 9 |
nid:1764867991290
DiskMat
What is the GCD in a polynomial field?
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nid:1764867991290
Q: What is the GCD in a polynomial field?
A: The monic polynomial \(g(x)\) of largest degree such that \(g(x) \ | \ a(x)\) and \(g(x) \ | \ b(x)\) is called the greatest common divisor of \(a(x)\) and \(b(x)\), denoted \(\gcd(a(x), b(x))\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991290 |
1 | 230% | 1849d | 9 |
nid:1764867990867
c1
DiskMat
\(-\infty\)
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nid:1764867990867
Cloze c1
Cloze answer: \(-\infty\)
Q: The degree of the polynomial \(0\) is defined as {{c1::\(-\infty\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990867 |
1 | 230% | 1885d | 9 |
nid:1766448533987
c2
DiskMat
atomic formula
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nid:1766448533987
Cloze c2
Cloze answer: atomic formula
Q: For any \(i\) and \(k\), if \(t_1, \dots, t_k\) are terms, then {{c1::\(P_i^{(k)}(t_1, \dots, t_k)\) is a formula}}, called an {{c2::atomic formula}}.
A: A formula in 1st order logic with no logical connectives (like \(\lnot, \land, \lor, \rightarrow \)) and no quantifiers (\(\forall, \exists\))
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448533988 |
1 | 230% | 1916d | 12 |
nid:1766448533015
DiskMat
A ring has the following properties:
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nid:1766448533015
Q: A ring has the following properties:
A: Additive Group:closureassociativityidentityinversecommutativeMultiplicative group:closureassociativityidentitydistributivity
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1766448533015 |
1 | 230% | 1930d | 11 |
nid:1764867990075
DiskMat
How is composition of relations represented in matrix and gr...
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nid:1764867990075
Q: How is composition of relations represented in matrix and graph form?
A: Matrix: Matrix multiplication
Graph: Natural composition - there's a path from \(a\) to \(c\) if there's a path \(a \to b\) in graph 1 and \(b \to c\) in graph 2
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990075 |
1 | 230% | 1961d | 13 |
nid:1764867991451
c2
DiskMat
at least \(n - k + 1\) positions
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nid:1764867991451
Cloze c2
Cloze answer: at least \(n - k + 1\) positions
Q: Two codewords in a polynomial code with degree \(k-1\) cannot agree at {{c1:: \(k\) positions (else they'd be equal)}}, so they disagree in {{c2:: at least \(n - k + 1\) positions}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867991452 |
1 | 230% | 1963d | 9 |
nid:1764867990874
DiskMat
Reduce \(R_{11}(9^{2024})\)
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nid:1764867990874
Q: Reduce \(R_{11}(9^{2024})\)
A: As \(9^{10} \equiv_{11} 1\) (see Fermat little theorem and 11 prime), we can reduce the exponent modulo 10 (see Lagrange's theorem in chapter 5). Thus \(R_{11}(9^{2024}) = R_{11}(9^{4}) = R_{11}(-2^{4}) = 5\).For this to work however, we need the number and the order of the group (modulo remainder) to be coprime, i.e. \(\gcd(9, 11) = 1\).If the modulus itself is prime then it always works and the order of the element can be used to reduce the exponent
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1764867990874 |
1 | 230% | 1980d | 9 |
nid:1767918757948
EProg
What does 5 % 0 produce in Java?
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nid:1767918757948
Q: What does 5 % 0 produce in Java?
A: Runtime error, division by 0
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1767918757949 |
1 | 230% | 377d | 11 |
nid:1769307700300
c1
EProg
false
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nid:1769307700300
Cloze c1
Cloze answer: false
Q: The weakest precondition for an empty program with postcondition false is {{c1::false}}.
A: As only false implies false.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1769307700300 |
1 | 230% | 396d | 10 |
nid:1768263609443
EProg
instanceof can result in a Compile-/Runtime-/No error?
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nid:1768263609443
Q: instanceof can result in a Compile-/Runtime-/No error?
A: instanceof never throws an exception, just compile errors.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768263609443 |
1 | 230% | 412d | 11 |
nid:1765655188156
EProg
Which of the following is (or are) NOT a Java keyword? - vol...
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nid:1765655188156
Q: Which of the following is (or are) NOT a Java keyword? - volatile- mod- strictfp- loop- transient- do- use
A: loop, use and mod
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765655188156 |
1 | 230% | 509d | 12 |
nid:1765655188143
c2
EProg
repetition (Wiederholung)
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nid:1765655188143
Cloze c2
Cloze answer: repetition (Wiederholung)
Q: Not every EBNF language (Sprache) can be described just with{{c2:: repetition (Wiederholung)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765655188144 |
1 | 230% | 515d | 9 |
nid:1765655188125
c1
EProg
last
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nid:1765655188125
Cloze c1
Cloze answer: last
Q: The convention for EBNF is that the rule being considered is written {{c1::last}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765655188125 |
1 | 230% | 550d | 11 |
nid:1767918757856
EProg
5 == 5 || String.yourStupidAss() evaluates to ???
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nid:1767918757856
Q: 5 == 5 || String.yourStupidAss() evaluates to ???
A: Compile Error, even if it shortcircuits.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1767918757857 |
1 | 230% | 579d | 8 |
nid:1765655188137
c1
EProg
ihre Sprachen gleich sind.
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nid:1765655188137
Cloze c1
Cloze answer: ihre Sprachen gleich sind.
Q: Zwei EBNF-Beschreibungen sind äquivalent falls {{c1:: ihre Sprachen gleich sind.}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765655188137 |
1 | 230% | 657d | 9 |
nid:1768263609578
c1
EProg
Primitives - instanceof only works with reference types
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nid:1768263609578
Cloze c1
Cloze answer: Primitives - instanceof only works with reference types
Q: The cases where instanceof causes a compile error:{{c1::Primitives - instanceof only works with reference types}}{{c2::Generics - type erasure means List<String> becomes just List at runtime, so the check is impossible t instanceof List<Str
A: However:Animal a = getanimal() could get a Dog which might implement List thus a instanceof List is not a compile error.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768263609578 |
1 | 230% | 715d | 11 |
nid:1765655188119
c1
EProg
Terminal; Literal
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nid:1765655188119
Cloze c1
Cloze answer: Terminal; Literal
Q: Ein Symbol (auf der RHS) wie z.B. 1, a, A in EBNF wird {{c1::Terminal}} oder auch {{c1::Literal}} gennant.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1765655188119 |
1 | 230% | 962d | 13 |
nid:1768944601791
c1
LinAlg
\(n - \dim(N(A))\) so it's \(n\) minus the geometric multip...
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users
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nid:1768944601791
Cloze c1
Cloze answer: \(n - \dim(N(A))\) so it's \(n\) minus the geometric multiplicity of \(\lambda = 0\)
Q: \(A \in \mathbb{R}^{n \times n}\) arbitrary non-symmetric has rank {{c1:: \(n - \dim(N(A))\) so it's \(n\) minus the geometric multiplicity of \(\lambda = 0\) ::in terms of multiplicities}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768944601791 |
1 | 230% | 682d | 10 |
nid:1768608740128
c1
LinAlg
\(A,B\) share an EV
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nid:1768608740128
Cloze c1
Cloze answer: \(A,B\) share an EV
Q: If \(AB = BA\), then {{c1::\(A,B\) share an EV::EVs of A, B}}.
A: Assume \(AB = BA\).If \(\lambda, v\) an EW-EV pair of \(A\) then \(A(Bv) = (AB)v = B(Av) = \lambda Bv\) thus \(Bv\) is an eigenvector of \(A\).Then \(Bv\) is a multiple of some \(v\) of that EW \(\lambda\) (easiest to see for \(A\) complete set of real EVs) \(\implies\) \(Bv = \lambda'v\) thus that \(v\) is also an EV of \(B\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768608740128 |
1 | 230% | 689d | 10 |
nid:1768263610700
c1
LinAlg
x_1 + N(A) ; \(x_1 \in R(A)\) is unique such that \(Ax_1 = ...
1
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users
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nid:1768263610700
Cloze c1
Cloze answer: x_1 + N(A) ; \(x_1 \in R(A)\) is unique such that \(Ax_1 = b\)
Q: Suppose that \(\{x \in \mathbb{R}^n \ | \ Ax = b \} \not = \emptyset\). Then \[ \{x \in \mathbb{R}^n \ | \ Ax = b \} = {{c1::x_1 + N(A) }}\] where {{c1:: \(x_1 \in R(A)\) is unique such that \(Ax_1 = b\)}}.
A: This means that if there's more than one solution to the system (i.e. the nullspace is not \(= \{0\}\)), then the set of all solutions is a specific solution + the entire nullspace.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768263610700 |
1 | 230% | 688d | 8 |
nid:1768944602558
c1
LinAlg
all its eigenvalues are real and the geometric multiplicitie...
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nid:1768944602558
Cloze c1
Cloze answer: all its eigenvalues are real and the geometric multiplicities are the same as the algebraic multiplicities of all it's eigenvalues
Q: A matrix has a complete set of real eigenvectors if {{c1::all its eigenvalues are real and the geometric multiplicities are the same as the algebraic multiplicities of all it's eigenvalues::in terms of multiplicities}}.
A: Example \(I\) has eigenvalue \(1\) with geometric multiplicity \(n\) (\(\dim(N(I - 1 \cdot I)) = n\)) and algebraic multiplicity \(n\) (As the characteristic polynomial of \(I\), \(P(z) = (z - 1)(z - 1) \dots (z - 1)\) with that repeated \(n\) times).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768944602558 |
1 | 230% | 725d | 10 |
nid:1768944603219
c1
LinAlg
a real eigenvalue \(\lambda\)
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nid:1768944603219
Cloze c1
Cloze answer: a real eigenvalue \(\lambda\)
Q: Every symmetric matrix \(A \in \mathbb{R}^{n \times n}\) has {{c1::a real eigenvalue \(\lambda\)::existence}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768944603219 |
1 | 230% | 738d | 10 |
nid:1768944603346
c2
LinAlg
\(v_1, \dots, v_n\) are an orthonormal basis of eigenvector...
1
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nid:1768944603346
Cloze c2
Cloze answer: \(v_1, \dots, v_n\) are an orthonormal basis of eigenvectors (the \(V\) in diagonalisation) and \(\lambda_1, \dots, \lambda_n\) the associated eigenvectors
Q: We can write \(A\) as the sum of {{c1::rank \(1\) matrices}}: \[A = {{c2::\sum_{k = 1}^n \lambda_i v_i v_i^\top}}\]where {{c2:: \(v_1, \dots, v_n\) are an orthonormal basis of eigenvectors (the \(V\) in diagonalisation) and \(\lambda_1, \dots, \lambd
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768944603347 |
1 | 230% | 741d | 12 |
nid:1768182517631
c2
LinAlg
there is only one \(0\)
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nid:1768182517631
Cloze c2
Cloze answer: there is only one \(0\)
Q: In a vector space \(V\) three important properties hold:{{c1::\(0v = 0\) for all \(v\)}}{{c2:: there is only one \(0\)}}{{c3:: one unique inverse \(-v\) for all \(v\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182517633 |
1 | 230% | 756d | 8 |
nid:1768608741058
c1
LinAlg
\(\frac{1}{z} = {{c1:: \frac{\overline{z} }{|z|^2} :: \text{...
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users
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nid:1768608741058
Cloze c1
Q: \(\frac{1}{z} = {{c1:: \frac{\overline{z} }{|z|^2} :: \text{in terms of z} }}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768608741058 |
1 | 230% | 790d | 13 |
nid:1768263611504
c1
LinAlg
\(A\) has linearly independent columns
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users
230%
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nid:1768263611504
Cloze c1
Cloze answer: \(A\) has linearly independent columns
Q: \(A^\top A\) is invertible if and only if {{c1::\(A\) has linearly independent columns}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768263611504 |
1 | 230% | 809d | 8 |
nid:1768944603363
c2
LinAlg
Given \(n\) vectors \(v_1, \dots, v_n \in \mathbb{R}^n\) we ...
1
lapses
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users
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nid:1768944603363
Cloze c2
Q: Given \(n\) vectors \(v_1, \dots, v_n \in \mathbb{R}^n\) we call their {{c1::Gram matrix}} the {{c2::\(n \times n\) matrix of inner products \(G_{ij} = v_i^\top v_j\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768944603364 |
1 | 230% | 807d | 10 |
nid:1768263609972
c1
LinAlg
norm; inner product
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nid:1768263609972
Cloze c1
Cloze answer: norm; inner product
Q: Orthogonal matrices preserve the {{c1::norm}} and {{c1::inner product}} of vectors.
A: In other words, if \(Q \in \mathbb{R}^{n \times n}\) is orthogonal, then, for all \(x, y \in \mathbb{R}^n\):\[ ||Qx|| = ||x|| \text{ and } (Qx)^\top(Qy) = x^\top y \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768263609972 |
1 | 230% | 824d | 8 |
nid:1768263611647
c1
LinAlg
\(A\) to have independent columns, i.e. they form a basis fo...
1
lapses
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users
230%
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nid:1768263611647
Cloze c1
Cloze answer: \(A\) to have independent columns, i.e. they form a basis for \(C(A)\)
Q: For a projection to exist using our formula \(P = A (A^\top A)^{-1} A^\top\) we need {{c1:: \(A\) to have independent columns, i.e. they form a basis for \(C(A)\)}}.
A: Otherwise the projection is not unique.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768263611648 |
1 | 230% | 824d | 8 |
nid:1768182518317
c1
LinAlg
linearly independent columns; \(MA\) has linearly independen...
1
lapses
1/4
users
230%
ease
nid:1768182518317
Cloze c1
Cloze answer: linearly independent columns; \(MA\) has linearly independent colums
Q: For \(A\) a matrix and \(M\) an invertible matrix:\(A\) has {{c1::linearly independent columns}} if and only if {{c1::\(MA\) has linearly independent colums}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768182518317 |
1 | 230% | 837d | 10 |
nid:1768944601064
LinAlg
Proof that the Rayleigh Quotient has it's maximum and minimu...
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nid:1768944601064
Q: Proof that the Rayleigh Quotient has it's maximum and minimum at the largest/smallest EWs?
A: It is easy to see that \(R(v_{\max}) = \lambda_{\max}\) and \(R(v_{\min}) = \lambda_{\min}\). See: \(R(v_{\text{max}}) = \frac{v_{\text{max}}^\top A v_{\text{max}}}{v_{\text{max}}^\top v_{\text{max}}} = \frac{v_{\text{max}}^\top (\lambda_{\text{max}} v_{\text{max}})}{v_{\text{max}}^\top v_{\text{max}}} = \lambda_{\text{max}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1768944601064 |
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nid:1768608739773
c1
LinAlg
not correlated
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nid:1768608739773
Cloze c1
Cloze answer: not correlated
Q: The eigenvalues of \(AB\) and \(BA\) are {{c1::not correlated}}.
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| lorenz | cid:1768608739773 |
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nid:1764867991528
LinAlg
The Cauchy-Schwarz Inequality tells us that for \(\textbf{v}...
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nid:1764867991528
Q: The Cauchy-Schwarz Inequality tells us that for \(\textbf{v}, \textbf{w} \in \mathbb{R}^m\)
A: \(|\textbf{v} \cdot \textbf{w}| \leq ||\textbf{v}|| \ ||\textbf{w}||\).This equality holds exactly if one vector is the scalar multiple of the other.This essentially means that: the length of the projecton of v onto w is smaller than the both of their lengths multiplied.This explains the equality part: if they are already aligned, their projection doesn't lose any length...
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| lorenz | cid:1764867991528 |
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nid:1768182517842
c1
LinAlg
\(R = I\)
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nid:1768182517842
Cloze c1
Cloze answer: \(R = I\)
Q: \(A\) is invertible if and only if for \(\text{RREF}(A,I) = (R, M)\) we have {{c1::\(R = I\)}}.
A: Since we have \(R = MA\), \(M\) is the inverse of \(A\).
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| lorenz | cid:1768182517842 |
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nid:1768425680760
LinAlg
How can we use Gauss-Jordan to simplify the determinant calc...
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nid:1768425680760
Q: How can we use Gauss-Jordan to simplify the determinant calculations?
A: We can use Gauss-Jordan to make any matrix upper triangular (then the determinant is the product of the diagonals).We are allowed to use:Row addition / substractionExchanging rows (change sign)Multiply rows (multiply the determinant at the end)
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| lorenz | cid:1768425680760 |
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nid:1768425682505
c1
LinAlg
the parity of the number of row swaps necessary to get back ...
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nid:1768425682505
Cloze c1
Cloze answer: the parity of the number of row swaps necessary to get back to the identity
Q: The \(\text{sgn}(\sigma)\) where \(\sigma\) is a permutation is {{c1:: the parity of the number of row swaps necessary to get back to the identity ::swaps}}.
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|---|---|---|---|---|---|
| lorenz | cid:1768425682505 |
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nid:1768263611355
c2
LinAlg
\(z = 0\); \(z^\top b = 0 \neq 1\)
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nid:1768263611355
Cloze c2
Cloze answer: \(z = 0\); \(z^\top b = 0 \neq 1\)
Q: Applications of the certificate of no solutions:Assume \(A \in \mathbb{R}^{m \times n}\) has linearly independent rows.Since {{c1::the rows are linearly independent}}, the only solution to \(z^\top A = 0\) is {{c2::\(z = 0\)}}. Hence {{c2::\(z^\top b = 0 \neq 1\)}}.
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| lorenz | cid:1768527254333 |
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nid:1768344745450
LinAlg
Why is the pseudoinverse (for \(A\) with full row-rank) \(A^...
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nid:1768344745450
Q: Why is the pseudoinverse (for \(A\) with full row-rank) \(A^\top (AA^\top)^{-1}\)?
A: It uses the multiplication by \(A^\top\) to choose an \(\hat{x}\) that lies in the row-space, thus minimising the norm.
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|---|---|---|---|---|---|
| lorenz | cid:1768344745450 |
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nid:1768608739855
c1
LinAlg
Every polynomial \(P(z) = a_n z^n + a_{n - 1} z^{n - 1} + \d...
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nid:1768608739855
Cloze c1
Q: Every polynomial \(P(z) = a_n z^n + a_{n - 1} z^{n - 1} + \dots + a_1 z + a_0\) with \(a_n \neq 0\) has {{c1:: a zero \(\lambda \in \mathbb{C} \)}}.
A: Fundamental theorem of algebra
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| lorenz | cid:1768608739855 |
1 | 230% | 945d | 11 |
nid:1768521670852
c1
LinAlg
The determinant expressed in terms of co-factors is: \[\det(...
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nid:1768521670852
Cloze c1
Q: The determinant expressed in terms of co-factors is: \[\det(A) = {{c1:: \sum_{j = 1}^n A_{ij}C_{ij} }}\]
A: in which we multiply the cofactor of every element by the element itself, as is clear in the example for a 3x3.
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| lorenz | cid:1768521670852 |
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nid:1768263611327
c1
LinAlg
\(I - P\)
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Cloze c1
Cloze answer: \(I - P\)
Q: Let \(S^\perp\) be the orthogonal complement of \(S\) and \(P\) the projection matrix onto \(S\).Then {{c1::\(I - P\)}} is the projection matrix that maps {{c2::\(b \in \mathbb{R}^m\) to \(\text{proj}_{S^\perp}(b)\)}}.Proof Included
A: Since \(b = e + \text{proj}_S(b) = e + Pb\) with \(e \in S^\perp\) Thus \[ (I - P)b = b - Pb = e = \text{proj}_{S^\perp}(b) \]This is true, since it holds that indeed \(I - P\) is also idempotent: \((I - P)^2 = I - 2P + P^2 = I -P - P + P= I - P\)
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| lorenz | cid:1768263611327 |
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nid:1768425681409
c1
LinAlg
Multilinearity of the determinant:\[ \begin{vmatrix} ta & tb...
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nid:1768425681409
Cloze c1
Q: Multilinearity of the determinant:\[ \begin{vmatrix} ta & tb \\ c & d \end{vmatrix} = {{c1:: t \cdot \begin{vmatrix} a & b \\ c & d \end{vmatrix} }}\]
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| lorenz | cid:1768425681409 |
1 | 230% | 998d | 8 |
nid:1768521672527
c1
LinAlg
Given a permutation matrix \(P \in \mathbb{R}^{n \times n}\)...
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nid:1768521672527
Cloze c1
Q: Given a permutation matrix \(P \in \mathbb{R}^{n \times n}\) corresponding to a permutation \(\sigma\), then \(\det(P) = {{c1::\text{sgn}(\sigma)}}\)
A: (this is as \(P\) is also an orthogonal matrix, see 3.). We sometimes write \(\text{sgn}(P)\).For the permutation matrix, each row contains only one entry: a \(1\). Thus the only permutation \(\sigma\) in the product that doesn't have a \(0\) factor is the permutation corresponding to the matrix \(P\) itself. The product is \(1 \cdot 1 \dots \cdot 1\) thus we get \(\text{sgn}(\sigma) = \text{sgn}(P)\).
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|---|---|---|---|---|---|
| lorenz | cid:1768521672527 |
1 | 230% | 1007d | 8 |
nid:1768263610888
c1
LinAlg
A; \(QQ^\top \) is the projection onto \(A\), and \(C(Q) = C...
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nid:1768263610888
Cloze c1
Cloze answer: A; \(QQ^\top \) is the projection onto \(A\), and \(C(Q) = C(A)\)
Q: \(QQ^\top A = {{c1::A}}\) because {{c1::\(QQ^\top \) is the projection onto \(A\), and \(C(Q) = C(A)\)}}.
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| lorenz | cid:1768263610888 |
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nid:1768608739736
LinAlg
Prove some \(x \in \mathbb{C}\) is actually in \(\mathbb{R}\...
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nid:1768608739736
Q: Prove some \(x \in \mathbb{C}\) is actually in \(\mathbb{R}\)?
A: Show that \(x = \overline{x} \implies x \in \mathbb{R}\)
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| lorenz | cid:1768608739736 |
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nid:1768944602222
c1
LinAlg
\(n\) real eigenvalues
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Cloze c1
Cloze answer: \(n\) real eigenvalues
Q: Spectral Theorem: Any symmetric matrix \(A \in \mathbb{R}^{n \times n}\) has {{c1::\(n\) real eigenvalues::EW}} and {{c1::an orthonormal basis of \(\mathbb{R}^{n \times n}\) consisting of it's eigenvectors::EV}}.
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| lorenz | cid:1768944602223 |
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nid:1764867991521
LinAlg
The euclidian norm of \(\textbf{v}\) is defined as?
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nid:1764867991521
Q: The euclidian norm of \(\textbf{v}\) is defined as?
A: \(|| \textbf{v} || := \sqrt{\textbf{v} \cdot \textbf{v}}\)This is also called the 2-norm.
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| lorenz | cid:1764867991521 |
1 | 230% | 1125d | 11 |
nid:1768263610707
LinAlg
Why does \(QR\) give \(A\)?
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nid:1768263610707
Q: Why does \(QR\) give \(A\)?
A: \(QQ^\top\) is the projection on the span of the \(q_i\)'s and thus also on the \(a_i\)'s (\(C(Q) = C(A)\)).Thus \(QQ^\top A = A\) and therefore \(QR = QQ^\top A = A\).
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| lorenz | cid:1768263610707 |
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nid:1768263608594
c1
LinAlg
are orthogonal
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Cloze c1
Cloze answer: are orthogonal
Q: Let \(A \in \mathbb{R}^{n \times n}\) be a symmetric matrix and \(\lambda_1 {{c2::\neq}} \lambda_2 \in \mathbb{R}\) two {{c2::distinct}} eigenvalues of \(A\) with corresponding eigenvectors \(v_1, v_2\).Then \(v_1\) and \(v_2\) {{c1::are orthogonal}}. Proof Included
A: \(\lambda_1 v_1 ^\top v_2 = (Av_1)^\top v_2\) \( = v_1^\top A ^\top v_2 = \) \(v_1^\top (Av_2)\) \( = \lambda_2 v_1^\top v_2\)
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|---|---|---|---|---|---|
| lorenz | cid:1768263608595 |
1 | 230% | 1117d | 11 |
nid:1768608740846
LinAlg
What is the fundamental theorem of algebra?
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nid:1768608740846
Q: What is the fundamental theorem of algebra?
A: Any degree \(n\) polynomial \(P(z) = a_n z^n + a_{n-1} z^{n-1} + \dots + a_1 z + a_0\) (with \(n \geq 1\) and \(a_n \neq 0\)) has at least one zero \(\lambda \in \mathbb{C}\) such that \(P(\lambda) = 0\).
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| lorenz | cid:1768608740846 |
1 | 230% | 1139d | 11 |
nid:1768608742500
c1
LinAlg
For a complex vector \(v\) we have \(||v|| =\) {{c1:: \(v^*v...
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nid:1768608742500
Cloze c1
Q: For a complex vector \(v\) we have \(||v|| =\) {{c1:: \(v^*v = \overline{v}^\top v = \sum_{i = 1}^n \overline{v_i}v_i = \sum_{i = 1}^n |v_i|^2\)}}.
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|---|---|---|---|---|---|
| lorenz | cid:1768608742500 |
1 | 230% | 1180d | 11 |
nid:1768344745223
c1
LinAlg
\(C(A^\top)\)
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Cloze c1
Cloze answer: \(C(A^\top)\)
Q: \(A^\dagger A\) is the projection matrix onto {{c1::\(C(A^\top)\)}}.
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|---|---|---|---|---|---|
| lorenz | cid:1768344745223 |
1 | 230% | 1208d | 11 |
nid:1768944602019
c2
LinAlg
\(A^\top A\); \(AA^\top\)
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nid:1768944602019
Cloze c2
Cloze answer: \(A^\top A\); \(AA^\top\)
Q: Given a real matrix \(A \in \mathbb{R}^{n \times n}\), the {{c1::non-zero eigenvalues}} of {{c2::\(A^\top A\)}} are the same ones as of {{c2::\(AA^\top\)}}. Proof Included
A: Shared EWs: For \((A^\top A)v_k = \lambda_k v_k\) we get \(AA^\top A v_k = \lambda_k Av_k\) and thus \(Av_k\) EV and \(\lambda_k\) is an EW of \(AA^\top\).Orthogonality: For \(j \neq k\) we have \((Av_j)^\top (Av_k) = v_j^\top A^\top Av_k = v_j^\top \lambda_k v_k = \lambda_k v_j^\top v_k = 0\)
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|---|---|---|---|---|---|
| lorenz | cid:1768944602020 |
1 | 230% | 1225d | 12 |
nid:1768608742013
c1
LinAlg
possibly with repetitions
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nid:1768608742013
Cloze c1
Cloze answer: possibly with repetitions
Q: Any degree \(n\) polynomial \(P(z)\) (with \(n \geq 1\)) has {{c1::\(n\) zeros \(\lambda_1, \dots, \lambda_n \in \mathbb{C}\)}}, {{c1::possibly with repetitions}}, such that \[P(z) = a_n (z-\lambda_1)(z - \lambda_2) \cdots (z - \lambda_n)\]
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| lorenz | cid:1768608742013 |
1 | 230% | 1234d | 11 |
nid:1768344745894
LinAlg
What is the pseudoinverse in the case where \(A \in \mathbb{...
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nid:1768344745894
Q: What is the pseudoinverse in the case where \(A \in \mathbb{R}^{n \times m}\) has independent columns?
A: Because \(rank(A) = r = n\) and thus \(m \geq n\)\(R(A)\) spans \(\mathbb{R}^n\)(rows span the space)\(C(A) \subseteq\) \(\mathbb{R}^m\) (as \(A\) is not necessarily square)We therefore first project \(b\) into \(C(A)\) and then invert, which is Least Squares.
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| lorenz | cid:1768344745895 |
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nid:1764867991504
LinAlg
A linear combination of \(\lambda_1\textbf{v}_1 + \lambda_2...
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nid:1764867991504
Q: A linear combination of \(\lambda_1\textbf{v}_1 + \lambda_2\textbf{v}_2 + \dots + \lambda_n\textbf{v}_n\) is affine if
A: \(\lambda_1 + \lambda_2 + \dots + \lambda_n = 1\)
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| lorenz | cid:1764867991504 |
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nid:1768263611621
LinAlg
In QR decomposition \(R\) is invertible because?
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nid:1768263611621
Q: In QR decomposition \(R\) is invertible because?
A: \(N(A) = \{0\}\) since \(A\) has independent columns and thus \(N(R) = \{0\}\):\(Rx = 0\) then \(Ax = QRx = 0\) thus \(Q\cdot 0 = 0\)Thus \(x \in N(A) \implies x = 0\)Thus \(R \in \mathbb{R}^{n \times n}\) (square) must be invertible.
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| lorenz | cid:1768263611621 |
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nid:1764867991560
c2
LinAlg
Name the three definitions for linear independence:{{c1::Non...
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nid:1764867991560
Cloze c2
Q: Name the three definitions for linear independence:{{c1::None of the vectors is a linear combination of the other ones.}}{{c2::There are no scalars \(\lambda_1, ..., \lambda_n\) besides 0, 0, ..., 0 such that \(\sum_{i = 1}^n \lambda_i v_i = \mathbf{0}\). (\
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| lorenz | cid:1766491319679 |
1 | 230% | 1292d | 9 |
nid:1768263610994
LinAlg
Certificate of no solutions:Given \(P = \{x \in \mathbb{R}^n...
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nid:1768263610994
Q: Certificate of no solutions:Given \(P = \{x \in \mathbb{R}^n \mid Ax = b \}\) we have: \(P = \left\{ x \in \mathbb{R}^3 \;\middle|\; \begin{aligned} x_1 + 2x_2 - x_3 &= 1 \\ 2x_1 + 4x_2 - 2x_3 &= 0 \end{aligned} \right\}\)Provide the system
A: The system \(D = \{ z \in \mathbb{R}^m | A^\top z = 0, b^\top z = 1 \}\) then is: \[D = \left\{ z \in \mathbb{R}^2 \;\middle|\; \begin{aligned} z_1 + 2z_2 &= 0 \\ 2z_1 + 4z_2 &= 0 \\ -z_1 - 2z_2 &= 0 \\ z_1 &= 1 \end{aligned} \right\}\]One equation per each column of \(A\).\(P = \emptyset\) and \(D \neq \emptyset\) because \(z = (1, -\frac{1}{2})^\top \in D\).
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nid:1768608742035
LinAlg
Does \(Av = v\) mean \(1\) is an eigenvalue of \(A\)?
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nid:1768608742035
Q: Does \(Av = v\) mean \(1\) is an eigenvalue of \(A\)?
A: No, we need to have \(v \neq 0\) to have that relationship hold!
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| lorenz | cid:1768608742035 |
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nid:1768608741184
c1
LinAlg
the vectors \(v \neq 0\), \(v \in N(A - \lambda_i I)\), in t...
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nid:1768608741184
Cloze c1
Cloze answer: the vectors \(v \neq 0\), \(v \in N(A - \lambda_i I)\), in the nullspace
Q: All the eigenvectors for \(\lambda_i\) are {{c1::the vectors \(v \neq 0\), \(v \in N(A - \lambda_i I)\), in the nullspace::subspace}}.
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|---|---|---|---|---|---|
| lorenz | cid:1768608741184 |
1 | 230% | 1340d | 11 |
nid:1768263610822
c1
LinAlg
symmetric
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Cloze c1
Cloze answer: symmetric
Q: A projection matrix is always {{c1:: symmetric ::property?}} (note that this needs to be reproven in the exam, proof included)
A: \(P^\top = (A(A^\top A)^{-1} A^\top)^\top =\) \((A^\top)^\top {(A^\top A)^{-1}}^\top A^\top = A(A^\top A)^{-1} A^\top = P\)We use the fact that for invertible matrices \({M^{-1}}^\top = {M^\top}^{-1}\).
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nid:1768425682248
c1
LinAlg
multiplicative
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nid:1768425682248
Cloze c1
Cloze answer: multiplicative
Q: The sign of a permutation is {{c1::multiplicative::property}}: \(\text{sgn}(\sigma \circ \lambda) = {{c1:: \text{sgn}(\sigma) \cdot \text{sgn}(\lambda)}}\).
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| lorenz | cid:1768425682248 |
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nid:1768182517703
LinAlg
How do we find the inverse of \(A\) using Gauss-Jordan?
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nid:1768182517703
Q: How do we find the inverse of \(A\) using Gauss-Jordan?
A: We do \(\text{RREF}(A, I)\) which gives us \((R, j_1, \dots, j_r, M)\) where in the case that \(A\) is invertible:\(R\) is \(I\) and \(r = n\)\(M = A^{-1}\)
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| lorenz | cid:1768182517703 |
1 | 230% | 1446d | 11 |
nid:1768608739489
c1
LinAlg
the same
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nid:1768608739489
Cloze c1
Cloze answer: the same
Q: The eigenvectors of \(A^{-1}\) are {{c1::the same}} as those of \(A\).
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nid:1768944603654
LinAlg
How to recover a matrix \(A\) from it's eigenvectors and eig...
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nid:1768944603654
Q: How to recover a matrix \(A\) from it's eigenvectors and eigenvalues (complete set)?
A: \(V\) the matrix with the eigenvectors of \(A\), orthogonal. Then we know \(AV = VD\) (\(Av_i = \lambda_i v_i\) in matrix form), with \(D = \Lambda\) the matrix with the eigenvalues on the diagonal.Thus \(AVV^\top = VDV^\top \implies A = VDV^\top\) .
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| lorenz | cid:1768944603654 |
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nid:1767105283735
LinAlg
What can we use to speed up long matrix multiplications, for...
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nid:1767105283735
Q: What can we use to speed up long matrix multiplications, for example \(w^\intercal (vw^\intercal) v\)?
A: We can use associativity: \(w^\intercal (vw^\intercal) v = (w^\intercal v)(w^\intercal v)\).
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nid:1768263610143
LinAlg
How do we get the \(QR\) decomposition for \(A\) with linear...
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Q: How do we get the \(QR\) decomposition for \(A\) with linearly independent columns?
A: \(Q\) is the result of Gram-Schmidt on \(A\)\(R = Q^\top A\)
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| lorenz | cid:1768263610143 |
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nid:1768182517756
c1
LinAlg
For \(A\) a matrix and \(M\) an invertible matrix:\(C(A) = \...
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Cloze c1
Q: For \(A\) a matrix and \(M\) an invertible matrix:\(C(A) = \) {{c1::Not equal to \(\textbf{C}(MA)\), the column space changes!}}
A: \(\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}\) after RREF is \(\begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}\) which spans a completely different line.
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| lorenz | cid:1768182517756 |
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nid:1773914070689
c1
A&W
Für Ereignisse \(A_1, \ldots, A_n\) gilt\[\Pr\left[\bigcup_{...
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Cloze c1
Q: Für Ereignisse \(A_1, \ldots, A_n\) gilt\[\Pr\left[\bigcup_{i=1}^{n} A_i\right] \leq {{c1::\sum_{i=1}^{n} \Pr[A_i]}}.\]
A: (Boolsche Ungleichung)
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| lorenz | cid:1773914070689 |
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nid:1774358417548
c1
A&W
Seien \(A\) und \(B\) Ereignisse mit \(\Pr[B] > 0\).
Die be...
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Cloze c1
Q: Seien \(A\) und \(B\) Ereignisse mit \(\Pr[B] > 0\).
Die bedingte Wahrscheinlichkeit \(\Pr[A|B]\) von \(A\) gegeben \(B\) ist definiert durch: \[\Pr[A|B] := {{c1::\frac{\Pr[A \cap B]}{\Pr[B]} }}\]
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| lorenz | cid:1774358417548 |
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nid:1774631269214
c1
A&W
\Pr[A]
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Cloze c1
Cloze answer: \Pr[A]
Q: Für \(X_A\) eine Indikator-Zufallsvariable gilt \(\mathbb{E}[X_a] = {{c1:: \Pr[A] }}\).
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| lorenz | cid:1774631269215 |
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nid:1774917593082
c1
A&W
Für eine Zufallsvariable \(X\) mit \(\mu = \mathbb{E}[X]\) d...
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Cloze c1
Q: Für eine Zufallsvariable \(X\) mit \(\mu = \mathbb{E}[X]\) definieren wir die Varianz \(\operatorname{Var}[X]\) durch: \[\operatorname{Var}[X] := {{c1::\mathbb{E}[(X - \mu)^2]}}\]
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| lorenz | cid:1774917593082 |
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nid:1776171659227
c3
A&W
Ordnung
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Cloze c3
Cloze answer: Ordnung
Q: Für \(n \geq 2\) heisst eine Zufallsvariable \(X\) mit Dichte\[f_X(k) = \begin{cases} {{c1::\binom{k-1}{n-1} \cdot p^n \cdot (1 - p)^{k-n} }} & \text{für } k = 1, 2, \ldots \\ 0 & \text{sonst} \end{cases}\]{{c2::negativ binomialverteilt}} mit {{c3::Ordnung}} \(n\).
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| lorenz | cid:1776171659229 |
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nid:1778588912048
c3
A&W
Entfernen von Kreisen
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Cloze c3
Cloze answer: Entfernen von Kreisen
Q: Fluss \(\to\) kantendisjunkte Pfade. Gegeben ein ganzzahliger maximaler Fluss \(f\) in \(N_G^{*}\):Beginnend bei \(u\), laufe entlang gerichteter, {{c1::ungebrauchter Kanten mit Fluss 1}} bis man bei \(v\) ankommt; durchlaufene Kanten werden als gebraucht markiert.Wiederh
A: Kreise können entstehen, wenn der Fluss interne Zyklen mit Fluss 1 enthält; diese sind für die Pfade irrelevant und werden weggelassen.
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| lorenz | cid:1778588912049 |
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nid:1779487730601
c2
A&W
Zwei gegeneinander wirkende Kräfte im Clarkson-Algorithmus.S...
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Cloze c2
Q: Zwei gegeneinander wirkende Kräfte im Clarkson-Algorithmus.Sei \(B^{*} \subseteq P\) ein Zertifikat mit \(C(B^{*}) = C(P)\), \(|B^{*}| \leq 3\). Betrachte \(P'\) nach \(k\) Runden:\(|P'|\) wächst stark: ein Punkt aus \(B^{*}\) wurde mindestens {{c1::\(k/3\)}}-mal ve
A: Warum \(k/3\)? Solange \(P \not\subseteq C^{\bullet}(Q)\), liegt mindestens ein Punkt aus \(B^{*}\) (höchstens 3 Stück) ausserhalb und wird verdoppelt.
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| lorenz | cid:1779487730603 |
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nid:1779798950934
c3
A&W
\(O(n^3)\)
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Cloze c3
Cloze answer: \(O(n^3)\)
Q: Erster naiver Ansatz für ConvexHull.Gehe durch jedes der {{c1::\(n(n-1)\)}} geordneten Paare \(qr\) und prüfe per Orientierungstest über alle \(n - 2\) übrigen Punkte, {{c2::ob \(qr\) eine Randkante ist}}.Laufzeit zum Finden aller Randkanten: {{c3::\(O(n^3)\)}} (danach nur noch ko
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| lorenz | cid:1779798950934 |
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nid:1779798951076
c1
A&W
\(2(n-1) - h = O(n)\); \(O(n)\)
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Cloze c1
Cloze answer: \(2(n-1) - h = O(n)\); \(O(n)\)
Q: LocalRepair: Laufzeitanalyse.Start mit \(2(n-1)\) Ecken, Ende mit \(h\) Ecken, also genau {{c1::\(2(n-1) - h = O(n)\)}} erfolgreiche (entfernende) Tests. Pro Punkt \(p_i\) gibt es zudem {{c2::zwei erfolglose Tests (einmal unten, einmal oben)}}.Nach dem anfänglichen Sortieren in \(
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| lorenz | cid:1779798951076 |
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nid:1780223730589
c2
A&W
O(n + m)
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Cloze c2
Cloze answer: O(n + m)
Q: Zählen erreichbarer Knoten (ungerichtet)Gegeben ein ungerichteter Graph \(G\), bestimme für jeden Knoten \(v\) die Anzahl erreichbarer Knoten.Idee: Berechne mit DFS die {{c1::Zusammenhangskomponenten}}; die Antwort für \(v\) ist dann die {{c1::Grösse der Komponente von \(v\)}}.
A: Konkret markiert ConnectedComponents jede Komponente mit einer eigenen Zahl, danach zählt man die Komponentengrössen \(\mathrm{cnt}[c]\) und setzt \(\mathrm{res}[v] = \mathrm{cnt}[\mathrm{comp}[v]]\).
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nid:1774631269382
c1
A&W
bipartiten
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Cloze c1
Cloze answer: bipartiten
Q: Hopcroft-Karp findet in einem {{c1::bipartiten}} Graphen in \(O({{c2::\sqrt{|V|} \cdot |E|}})\) ein {{c3::maximales Matching}}.
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| lorenz | cid:1774631269383 |
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nid:1777923968745
c2
A&W
Floyd's Cycle Finding (Graph-Reformulierung)Definiere den ge...
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Cloze c2
Q: Floyd's Cycle Finding (Graph-Reformulierung)Definiere den gerichteten Graphen \(D = (V, A)\) mit {{c1::\(V = [n]\)}} und {{c2::\(A = \{(i, a[i]) \mid 1 \leq i \leq n\}\)}}.Eigenschaften:Jeder Knoten hat genau {{c3::eine ausgehende Kante}}Knoten \(n\) hat {{c4::keine e
A: Notation: Pfad hat \(k \geq 1\) Kanten, Kreis hat \(\ell \geq 3\) Kanten, mit \(k + \ell \leq n\).
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| lorenz | cid:1777923968748 |
1 | 230% | 3d | 6 |
nid:1778588912041
c1
A&W
Sei \(f\) ein ganzzahliger maximaler Fluss in \(N_G^{*}\). D...
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Cloze c1
Q: Sei \(f\) ein ganzzahliger maximaler Fluss in \(N_G^{*}\). Dann gilt:Flusswerte: {{c1::\(f(e) \in \{0, 1\}\)}} für alle Kanten \(e\).Für jeden Knoten \(w \notin \{u, v\}\): {{c2::\(\operatorname{indeg}_f(w) = \operatorname{outdeg}_f(w)\) (Flusserhaltung in inneren Knoten)}}.
A: \(\operatorname{indeg}_f(w)\) und \(\operatorname{outdeg}_f(w)\) bezeichnen die Ein- bzw. Ausgrade bezüglich der Kanten mit Fluss 1.
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| lorenz | cid:1778588912043 |
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nid:1779487730608
c1
A&W
Sampling Lemma.Seien \(r, N \in \mathbb{N}\), \(r \leq N\), ...
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Cloze c1
Q: Sampling Lemma.Seien \(r, N \in \mathbb{N}\), \(r \leq N\), und \(P'\) eine Multimenge mit \(|P'| = N\). Für \(R\) zufällig gleichverteilt aus \(\binom{P'}{r}\) gilt\[\mathbb{E}\big[\,|P' \setminus C^{\bullet}(R)|\,\big] \;\leq\; {{c1::3\,\frac{N - r}{r + 1} }} \;\leq\; {{c1::3\,\frac
A: \(|P' \setminus C^{\bullet}(R)|\) ist die Anzahl der Punkte aus \(P'\), die ausserhalb von \(C(R)\) liegen. Diese Schranke kontrolliert das Wachstum von \(P'\) pro Runde.
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| lorenz | cid:1779487730608 |
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nid:1779798950926
c2
A&W
(r_x - q_x)(p_y - q_y) - (p_x - q_x)(r_y - q_y) > 0
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Cloze c2
Cloze answer: (r_x - q_x)(p_y - q_y) - (p_x - q_x)(r_y - q_y) > 0
Q: Orientierungstest für allgemeines \(q\) (nicht im Ursprung).\(p\) liegt links von \(q, r\) \(\iff\) {{c1::\(p - q\) liegt links von \(o,\ r - q\)}} \(\iff\)\[{{c2::(r_x - q_x)(p_y - q_y) - (p_x - q_x)(r_y - q_y) > 0}}.\]
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| lorenz | cid:1779798950927 |
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nid:1780223730608
c1
A&W
\(n_v = |R(v)|\)
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Cloze c1
Cloze answer: \(n_v = |R(v)|\)
Q: Approximation der Erreichbarkeit (gerichtet)Die Menge der von \(v\) erreichbaren Knoten ist \(R(v) = \{u \in V : u \text{ von } v \text{ erreichbar}\}\), und man schreibt {{c1::\(n_v = |R(v)|\)}} für ihre Grösse.Da exaktes Zählen vermutlich nicht nahe-linear geht, approximiert man
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nid:1773913363614
c1
A&W
Elementarereignissen
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Cloze c1
Cloze answer: Elementarereignissen
Q: Ein diskreter Wahrscheinlichkeitsraum ist bestimmt durch eine Ergebnismenge \(\Omega = \{\omega_1, \omega_2, \ldots\}\) von {{c1::Elementarereignissen}}.
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| lorenz | cid:1773913363617 |
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nid:1774487164722
c1
A&W
Eine Zufallsvariable auf \(\Omega\) ist {{c1::eine Funktion ...
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Cloze c1
Q: Eine Zufallsvariable auf \(\Omega\) ist {{c1::eine Funktion \(X\colon \Omega \to \mathbb{R}\)}}.\[\Pr[X = x] := {{c2::\Pr[\{\omega \in \Omega : X(\omega) = x\}]}}.\]
A: Zufallsvariablen abstrahieren Ergebnisse zu numerischen Werten.Beispiel: Bei 2 Würfelwürfen ist \(X =\) "Summe der Augenzahlen" eine Zufallsvariable.
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| lorenz | cid:1774487164722 |
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nid:1774631277253
c1
A&W
Für eine Zufallsvariable \(X:\Omega\to\mathbb{R}\) ist der W...
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Cloze c1
Q: Für eine Zufallsvariable \(X:\Omega\to\mathbb{R}\) ist der Wertebereich:\[ W_X := {{c1::X(\Omega) = \{x\in\mathbb{R}\mid\exists\omega\in\Omega:\, X(\omega)=x\} }}\]
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| lorenz | cid:1774631277254 |
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nid:1777538021777
c2
A&W
\(PB_n\) ist eine {{c1::Untergruppe von \(\mathbb{Z}_n^*\)}}...
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Cloze c2
Q: \(PB_n\) ist eine {{c1::Untergruppe von \(\mathbb{Z}_n^*\)}}, denn\[a^{n-1} \equiv_n 1 \;\wedge\; b^{n-1} \equiv_n 1 \;\Longrightarrow\; (ab)^{n-1} \equiv_n 1.\]Folglich: ist \(n\) nicht Carmichael, dann ist \(|PB_n|\) ein {{c2::echter Teiler von \(|\mathbb{Z}_n^*|\)}}, also\[|PB_n| \leq {{c3
A: Genau diese Schranke liefert die Fehlerwahrscheinlichkeit \(\leq \tfrac{1}{2}\) des Fermat-Tests für Nicht-Carmichael-Zahlen: \(\Pr[\text{Fehler}] = |PB_n|/(n-1) \leq \tfrac{1}{2}\).Der Beweis nutzt den Satz von Lagrange: jede echte Untergruppe einer endlichen Gruppe hat höchstens halb so viele Elemente wie die Gruppe selbst.
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| lorenz | cid:1777538021777 |
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nid:1778164855606
c1
A&W
der Nettoabfluss der Quelle
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Cloze c1
Cloze answer: der Nettoabfluss der Quelle
Q: Der Wert eines Flusses \(f\) ist definiert als {{c1::der Nettoabfluss der Quelle}}:\[\operatorname{val}(f) := {{c1::\operatorname{netoutflow}(s) := \sum_{u \in V : (s,u) \in A} f(s, u) \;-\; \sum_{u \in V : (u,s) \in A} f(u, s). }}\]
A: Eingehende Kanten an der Quelle werden abgezogen.
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nid:1779193767119
A&W
Schreibe den modifizierten Algorithmus \(\text{Cut}_t(G)\), ...
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Q: Schreibe den modifizierten Algorithmus \(\text{Cut}_t(G)\), der die Kantenkontraktion bei \(t\) Knoten abbricht und dann einen randomisierten \(\mathcal{O}(t^4)\)-Algorithmus mit Erfolgswkt. \(\geq 1 - e^{-1}\) verwendet. Gib die untere Schranke für \(\hat{p}_t(n)\) und die resultierende Lau
A: Untere Schranke für die Erfolgswkt. einer Einzelausführung:\[\hat{p}_t(n) \;\geq\; \tfrac{n-2}{n} \cdot \tfrac{n-3}{n-1} \cdots \tfrac{t}{t+2} \cdot \tfrac{t-1}{t+1} \cdot \hat{p}_t(t) \;\geq\; \tfrac{t(t-1)}{n(n-1)} \cdot \tfrac{e-1}{e}.\]Nach \(\alpha / p_t(n)\)-maligem Wiederholen: Fehlerwkt. \(\leq e^{-\alpha}\) und Laufzeit\[\mathcal{O}\!\left(\alpha\!\left(\tfrac{n^4}{t^2} + n^2 t^2\right)\right).\]Wahl \(t := \sqrt{n}\)
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| lorenz | cid:1779193767119 |
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nid:1779487730553
A&W
Eindeutigkeit von \(C(P)\): Beweisidee.Angenommen, es gäbe z...
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Q: Eindeutigkeit von \(C(P)\): Beweisidee.Angenommen, es gäbe zwei verschiedene kleinste umschliessende Kreise \(C_1, C_2\) mit gleichem Radius \(r\) und Mittelpunkten \(z_1 \neq z_2\). Welchen kleineren Kreis konstruiert man als Widerspruch, und welchen Radius \(\hat r\) hat er?
A: Da beide umschliessen, gilt \(P \subseteq C_1^{\bullet} \cap C_2^{\bullet}\).Sei \(C\) der Kreis mit Mittelpunkt \(z = \tfrac{1}{2}(z_1 + z_2)\) (Mitte der Strecke \(z_1 z_2\)), und \(\hat r\) der Abstand von \(z\) zu den beiden Schnittpunkten von \(C_1\) und \(C_2\). Dann\[P \subseteq C_1^{\bullet} \cap C_2^{\bullet} \subseteq C^{\bullet}, \qquad \hat r = \sqrt{\,r^2 - \left(\tfrac{|z_1 z_2|}{2}\right)^2\,}.\]Wegen \(\hat r < r\) waren \(C_1, C_2\) keine kleinsten umschliessenden Kreise.
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| lorenz | cid:1779487730554 |
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nid:1779798950982
c2
A&W
\(O(n)\)
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Cloze c2
Cloze answer: \(O(n)\)
Q: JarvisWrap: Laufzeit-Spezialfälle.Aus \(O(nh)\) folgt:Da \(h \leq n\), läuft JarvisWrap in {{c1::\(O(n^2)\) (statt \(O(n^3)\) beim naiven Ansatz)}}.Ist \(h = O(1)\) (z.B. \(\operatorname{conv}(P)\) ein Dreieck), so {{c2::\(O(n)\)}}.
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| lorenz | cid:1779798950983 |
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nid:1779798950998
c3
A&W
nicht einmal verschieden (z.B. in einem Feld gegeben)
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Cloze c3
Cloze answer: nicht einmal verschieden (z.B. in einem Feld gegeben)
Q: JarvisWrap: Umgang mit Degeneriertheiten (Kollinearitäten, gleiche \(x\)-Koordinaten, Duplikate).Startpunkt \(q_0\): nimm den Punkt mit {{c1::lexikographisch kleinster Koordinate (unter kleinster \(x\)-Koordinate den mit kleinster \(y\)-Koordinate)}}.Test "\(p\) rechts vo
A: Software ohne Berücksichtigung dieser Fälle hat geringen praktischen Nutzen.
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| lorenz | cid:1779798951001 |
1 | 230% | 8d | 7 |
nid:1774631276995
c1
A&W
Sei \(A_1,\ldots,A_n\) eine Partition von \(\Omega\) mit \(\...
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Cloze c1
Q: Sei \(A_1,\ldots,A_n\) eine Partition von \(\Omega\) mit \(\Pr[A_i]>0\) für alle \(i\). Dann:\[ \mathbb{E}[X] = {{c1::\sum_{i=1}^{n}\mathbb{E}[X\mid A_i]\cdot\Pr[A_i]}}. \]Proof Included
A: (Gesetz der totalen Erwartung, nicht im Skript!) Proof:\[\begin{align} \mathbb{E}[X] &=\sum_{x}x\cdot\Pr[X=x] \\ &\overset{\text{totale W'keit}}{=}\sum_x x\sum_i\Pr[X=x|A_i]\Pr[A_i] \\ &=\sum_i\Pr[A_i]\underbrace{\sum_x x\Pr[X=x|A_i]}_{=\mathbb{E}[X|A_i]} \end{align}\](Verwendet das Gesetz der totalen Wahrscheinlichkeit um \(\Pr[X=x]\) zu expandieren, dann wird die Summationsreihenfolge vertauscht.)
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| lorenz | cid:1774631276995 |
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nid:1777538021722
c1
A&W
Für die Anwendung in der RSA-Kryptographie möchte man zufäll...
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Cloze c1
Q: Für die Anwendung in der RSA-Kryptographie möchte man zufällige grosse Primzahlen erzeugen (z.B. mit \(1000\) Bits, also \(n \approx 2^{1000}\)). Man erzeugt zufällig Zahlen mit \(1000\) Bits und testet sie auf prim.Dass das funktioniert, garantiert der Primzahlsatz:\[\pi(x) := |\
A: Die RSA-Verschlüsselung beruht auf der Asymmetrie: Ob \(n\) prim ist, kann man schnell entscheiden (randomisiert), aber einen nicht-trivialen Teiler von \(n\) effizient zu finden ist nicht bekannt. Quantencomputer können das via Shor-Algorithmus.
| User | Card ID | Lapses | Ease | Interval | Reviews |
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| lorenz | cid:1777538021723 |
1 | 230% | 12d | 11 |
nid:1778164855856
c2
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der Algorithmus kann unendlich laufen
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Cloze c2
Cloze answer: der Algorithmus kann unendlich laufen
Q: Eigenschaften des Ford-Fulkerson-Algorithmus bezüglich Termination:Allgemein: {{c1::Terminierung ist nicht garantiert}}.Bei Kapazitäten in \(\mathbb{R}\): {{c2::der Algorithmus kann unendlich laufen}}.Bei Kapazitäten in \(\mathbb{N}_0\): {{c3::Flüsse und Restkapazitäten ble
A: Bei reellen (sogar irrationalen) Kapazitäten gibt es Beispiele, in denen Ford-Fulkerson zwar konvergiert, aber gegen einen falschen Wert oder gar nicht. Die Wahl des augmentierenden Pfads (z.B. via BFS bei Edmonds-Karp) liefert eine Polynomialzeitschranke unabhängig von \(U\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778164855856 |
1 | 230% | 7d | 10 |
nid:1779798950883
c3
A&W
Darstellung der konvexen Hülle (endliches \(P\) in der Ebene...
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Cloze c3
Q: Darstellung der konvexen Hülle (endliches \(P\) in der Ebene).Der Rand von \(\operatorname{conv}(P)\) ist {{c1::ein Polygon, dessen Ecken Punkte aus \(P\) sind}}. Die Berechnung von \(\operatorname{conv}(P)\) meint {{c2::die Bestimmung der Eckenfolge \((q_0, q_1, \ldots,
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|---|---|---|---|---|---|
| lorenz | cid:1779798950884 |
1 | 230% | 15d | 7 |
nid:1779798950926
c1
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\(p - q\) liegt links von \(o,\ r - q\)
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Cloze c1
Cloze answer: \(p - q\) liegt links von \(o,\ r - q\)
Q: Orientierungstest für allgemeines \(q\) (nicht im Ursprung).\(p\) liegt links von \(q, r\) \(\iff\) {{c1::\(p - q\) liegt links von \(o,\ r - q\)}} \(\iff\)\[{{c2::(r_x - q_x)(p_y - q_y) - (p_x - q_x)(r_y - q_y) > 0}}.\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779798950926 |
1 | 230% | 6d | 10 |
nid:1780223730589
c1
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Zusammenhangskomponenten; Grösse der Komponente von \(v\)
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nid:1780223730589
Cloze c1
Cloze answer: Zusammenhangskomponenten; Grösse der Komponente von \(v\)
Q: Zählen erreichbarer Knoten (ungerichtet)Gegeben ein ungerichteter Graph \(G\), bestimme für jeden Knoten \(v\) die Anzahl erreichbarer Knoten.Idee: Berechne mit DFS die {{c1::Zusammenhangskomponenten}}; die Antwort für \(v\) ist dann die {{c1::Grösse der Komponente von \(v\)}}.
A: Konkret markiert ConnectedComponents jede Komponente mit einer eigenen Zahl, danach zählt man die Komponentengrössen \(\mathrm{cnt}[c]\) und setzt \(\mathrm{res}[v] = \mathrm{cnt}[\mathrm{comp}[v]]\).
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|---|---|---|---|---|---|
| lorenz | cid:1780223730591 |
1 | 230% | 6d | 10 |
nid:1773310887041
c1
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\chi(G)
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Cloze c1
Cloze answer: \chi(G)
Q: Es gibt eine Reihenfolge \(V = \{v_1, \ldots, v_n\}\) der Knoten, für die der Greedy-Algorithmus nur \({{c1::\chi(G)}}\) viele Farben benötigt.
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|---|---|---|---|---|---|
| lorenz | cid:1773310887041 |
1 | 230% | 29d | 9 |
nid:1774487164563
c1
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Für den Binomialkoeffizienten gilt:\[\binom{n}{k} = {{c1::\b...
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Cloze c1
Q: Für den Binomialkoeffizienten gilt:\[\binom{n}{k} = {{c1::\binom{n}{n-k} :: \text{Symmetrie} }}\]
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|---|---|---|---|---|---|
| lorenz | cid:1774487164564 |
1 | 230% | 44d | 12 |
nid:1777538021793
c3
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Sei \(n > 2\) ungerade, schreibe \(n - 1 = 2^k d\) mit \(d\)...
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Cloze c3
Q: Sei \(n > 2\) ungerade, schreibe \(n - 1 = 2^k d\) mit \(d\) ungerade.Vorüberlegung: Für \(n\) prim ist \((\mathbb{Z}_n, +, \cdot)\) ein {{c1::Körper}}, und in einem {{c1::Körper}} hat \(x^2 = 1\) genau zwei Lösungen: \({{c2::x = 1\ \text{und}\ x = n - 1\ (= -1)}}\).Ist \(n\) prim
A: Idee: aus \(a^{n-1} \equiv_n 1\) wiederholt Quadratwurzeln ziehen, solange das Ergebnis noch \(1\) ist. Im Körper sind die einzigen Quadratwurzeln von \(1\) genau \(\pm 1\), daher muss man irgendwann auf \(n-1\) treffen oder bei \(a^d \equiv_n 1\) landen.Ein \(a\), das diese Bedingung verletzt, ist ein Miller-Rabin-Zertifikat dafür, dass \(n\) nicht prim ist.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777538021793 |
1 | 230% | 23d | 8 |
nid:1778588912048
c1
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ungebrauchter Kanten mit Fluss 1
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Cloze c1
Cloze answer: ungebrauchter Kanten mit Fluss 1
Q: Fluss \(\to\) kantendisjunkte Pfade. Gegeben ein ganzzahliger maximaler Fluss \(f\) in \(N_G^{*}\):Beginnend bei \(u\), laufe entlang gerichteter, {{c1::ungebrauchter Kanten mit Fluss 1}} bis man bei \(v\) ankommt; durchlaufene Kanten werden als gebraucht markiert.Wiederh
A: Kreise können entstehen, wenn der Fluss interne Zyklen mit Fluss 1 enthält; diese sind für die Pfade irrelevant und werden weggelassen.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588912050 |
1 | 230% | 7d | 10 |
nid:1780484891688
A&W
Wahr oder falsch?Die erwartete Laufzeit von Quickselect ist ...
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Q: Wahr oder falsch?Die erwartete Laufzeit von Quickselect ist \(O(n)\).
A: Wahr.Randomisiertes Quickselect (Suche nach dem \(k\)-kleinsten Element mit zufälligem Pivot) hat erwartete Laufzeit \(O(n)\), da die Teilprobleme im Mittel geometrisch schrumpfen.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780484891689 |
1 | 230% | 6d | 7 |
nid:1780484891802
A&W
Wahr oder falsch?Wenn \(e\) eine Kante eines Multigraphen \(...
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Q: Wahr oder falsch?Wenn \(e\) eine Kante eines Multigraphen \(G\) ist, dann gilt immer \(\mu(G/e) \leq \mu(G)\), wobei \(\mu(G)\) die Kardinalität eines minimalen Schnitts in \(G\) bezeichnet.
A: Falsch.Kontraktion kann den minimalen Schnitt nur vergrössern oder gleich lassen: \(\mu(G/e) \geq \mu(G)\). Durch das Verschmelzen der Endknoten von \(e\) fallen genau jene Schnitte weg, die diese beiden Knoten trennen, sodass nie ein kleinerer Schnitt entsteht. (Genau das macht Kargers Algorithmus korrekt.)
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| lorenz | cid:1780484891802 |
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nid:1773311608808
c1
A&W
auch \(G\) mit \(k\) Farben färben
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Cloze c1
Cloze answer: auch \(G\) mit \(k\) Farben färben
Q: Sei \(G\) ein Graph, in dem man jeden Block mit \(k\) Farben färben kann.Dann kann man {{c1::auch \(G\) mit \(k\) Farben färben}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773311608808 |
1 | 230% | 25d | 9 |
nid:1774487164349
c2
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3n - 6
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Cloze c2
Cloze answer: 3n - 6
Q: Für einen zusammenhängenden planaren Graphen \(G\) mit \(n\) Knoten, \(m\) Kanten und \(f\) Flächen gilt:\[{{c1::n - m + f}} = 2.\]Daraus folgt für \(n \geq 3\): \[m \leq {{c2::3n - 6}}.\]
A: (Euler-Formel)Beweis der Ungleichung: Jede Fläche wird von mind. 3 Kanten begrenzt; jede Kante grenzt an max. 2 Flächen.Also \(3f \leq 2m\), einsetzen in Euler-Formel gibt \(m \leq 3n - 6\).Korollar: In jedem planaren Graphen gibt es einen Knoten mit Grad \(\leq 5\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487164351 |
1 | 230% | 24d | 9 |
nid:1776175078408
c2
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Für jede {{c1::nicht-negative}} Zufallsvariable \(X\) und al...
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Cloze c2
Q: Für jede {{c1::nicht-negative}} Zufallsvariable \(X\) und alle \(t > 0\), gilt\[\Pr\left[X \geq t\right] \leq {{c2::\frac{\mathbb{E}[X]}{t} }}.\]
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|---|---|---|---|---|---|
| lorenz | cid:1776175078409 |
1 | 230% | 34d | 11 |
nid:1777923968745
c1
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\(V = [n]\)
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Cloze c1
Cloze answer: \(V = [n]\)
Q: Floyd's Cycle Finding (Graph-Reformulierung)Definiere den gerichteten Graphen \(D = (V, A)\) mit {{c1::\(V = [n]\)}} und {{c2::\(A = \{(i, a[i]) \mid 1 \leq i \leq n\}\)}}.Eigenschaften:Jeder Knoten hat genau {{c3::eine ausgehende Kante}}Knoten \(n\) hat {{c4::keine e
A: Notation: Pfad hat \(k \geq 1\) Kanten, Kreis hat \(\ell \geq 3\) Kanten, mit \(k + \ell \leq n\).
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| lorenz | cid:1777923968746 |
1 | 230% | 14d | 11 |
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c1
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eindeutigen kleinsten umschliessenden Kreis \(C(P)\)
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Cloze c1
Cloze answer: eindeutigen kleinsten umschliessenden Kreis \(C(P)\)
Q: Für jede endliche Punktemenge \(P \subseteq \mathbb{R}^2\) gibt es einen {{c1::eindeutigen kleinsten umschliessenden Kreis \(C(P)\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487730547 |
1 | 230% | 8d | 10 |
nid:1779487730587
c1
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Man trifft das richtige \(Q\) mit Wahrscheinlichkeit {{c1::\...
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Cloze c1
Q: Man trifft das richtige \(Q\) mit Wahrscheinlichkeit {{c1::\(\geq 1 / \binom{n}{3}\)}}, die erwartete Anzahl Versuche ist also {{c2::\(\leq \binom{n}{3}\)}} und damit die erwartete Laufzeit {{c3::\(O(n^4)\)}}.
A: Idee zur Verbesserung: Man lernt, welche Punkte ausserhalb von \(C(Q)\) liegen: diese sind wichtiger für \(C(P)\) als die Punkte innerhalb.
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|---|---|---|---|---|---|
| lorenz | cid:1779487730590 |
1 | 230% | 17d | 10 |
nid:1780223730600
c1
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O(n + m); O(n(n + m))
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Cloze c1
Cloze answer: O(n + m); O(n(n + m))
Q: Zählen erreichbarer Knoten (gerichtet): LaufzeitFür einen einzelnen Startknoten \(s\) bestimmt ein DFS/BFS die Anzahl erreichbarer Knoten in Zeit \({{c1::O(n + m)}}\).Für alle Knoten ergibt das \({{c1::O(n(n + m))}}\).Geht es schneller? {{c3::Vermutlich nicht: Ei
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|---|---|---|---|---|---|
| lorenz | cid:1780223730602 |
1 | 230% | 7d | 7 |
nid:1780484891727
A&W
Wahr oder falsch?Sei \(N = (V, A, c, s, t)\) ein Netzwerk. W...
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Q: Wahr oder falsch?Sei \(N = (V, A, c, s, t)\) ein Netzwerk. Wenn \(c\) nur ganzzahlige Werte annimmt, so ist jeder maximale Fluss in \(N\) ganzzahlig.
A: Falsch.Der Ganzzahligkeitssatz garantiert nur, dass ein ganzzahliger maximaler Fluss existiert, nicht dass jeder maximale Fluss ganzzahlig ist. Bei mehreren maximalen Wegen kann man den Flusswert auch gebrochen (etwa \(0.5\)) auf parallele Pfade aufteilen.
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|---|---|---|---|---|---|
| lorenz | cid:1780484891727 |
1 | 230% | 10d | 7 |
nid:1774487164440
c4
A&W
Sobald \(L_i\) einen unüberdeckten Knoten enthält
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Cloze c4
Cloze answer: Sobald \(L_i\) einen unüberdeckten Knoten enthält
Q: BFS für augmentierende Pfade in bipartiten Graphen \(G = (A \uplus B, E)\):\(L_0 := \) {{c1::unüberdeckte Knoten aus \(A\)}}Für ungerades \(i\): \(L_i := \) {{c2::unbesuchte Nachbarn von \(L_{i-1}\) via Kanten in \(E \setminus M\)}}Für gerades \(i\): \(L_i :=
A: Laufzeit: \(O(|V| + |E|)\) für einen augmentierenden Pfad.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487164441 |
1 | 230% | 25d | 9 |
nid:1778588912086
c2
A&W
zwei gerichtete Kanten \((p, p')\) und \((p', p)\), je mit K...
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Cloze c2
Cloze answer: zwei gerichtete Kanten \((p, p')\) und \((p', p)\), je mit Kapazität \(\gamma_e\)
Q: Konstruktion Bild \(\to\) Netzwerk \(N = (P \cup \{s, t\}, \vec{E}, c, s, t)\):Neue Knoten \(s\) (Quelle) und \(t\) (Senke).Für jedes \(p \in P\): gerichtete Kante \((s, p)\) mit Kapazität {{c1::\(\alpha_p\)}}.Für jedes \(p \in P\): gerichtete Kante \((p, t)\) mit Ka
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|---|---|---|---|---|---|
| lorenz | cid:1778588912086 |
1 | 230% | 7d | 10 |
nid:1779487730601
c1
A&W
\(k/3\)
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Cloze c1
Cloze answer: \(k/3\)
Q: Zwei gegeneinander wirkende Kräfte im Clarkson-Algorithmus.Sei \(B^{*} \subseteq P\) ein Zertifikat mit \(C(B^{*}) = C(P)\), \(|B^{*}| \leq 3\). Betrachte \(P'\) nach \(k\) Runden:\(|P'|\) wächst stark: ein Punkt aus \(B^{*}\) wurde mindestens {{c1::\(k/3\)}}-mal ve
A: Warum \(k/3\)? Solange \(P \not\subseteq C^{\bullet}(Q)\), liegt mindestens ein Punkt aus \(B^{*}\) (höchstens 3 Stück) ausserhalb und wird verdoppelt.
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|---|---|---|---|---|---|
| lorenz | cid:1779487730601 |
1 | 230% | 13d | 10 |
nid:1779487730622
A&W
Sampling Lemma: Beweis (Kernrechnung).Wie kommt man von \(\m...
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Q: Sampling Lemma: Beweis (Kernrechnung).Wie kommt man von \(\mathbb{E}[|P' \setminus C^{\bullet}(R)|]\) auf die Schranke \(3\,\frac{N-r}{r+1}\)? (Stichworte: out/ess, Wechsel \(R \to Q\).)
A: \[\begin{gathered}\mathbb{E}\big[|P' \setminus C^{\bullet}(R)|\big] = \frac{1}{\binom{N}{r}} \sum_{R \in \binom{P'}{r}} \sum_{s \in P' \setminus R} \operatorname{out}(s, R) \\= \frac{1}{\binom{N}{r}} \sum_{R \in \binom{P'}{r}} \sum_{s \in P' \setminus R} \operatorname{ess}(s, R \cup \{s\}) \\= \frac{1}{\binom{N}{r}} \sum_{Q \in \binom{P'}{r+1}} \underbrace{\sum_{s \in Q} \operatorname{ess}(s, Q)}_{\leq 3} \;\leq\; \frac{3 \binom{N}{r+1}}{\binom{N}{r}} = 3\,\frac{N - r}{r + 1}.\end{gathered}\]Der
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487730622 |
1 | 230% | 7d | 10 |
nid:1779798950966
c1
A&W
\(q_h = q_0\) (die Hülle ist geschlossen)
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Cloze c1
Cloze answer: \(q_h = q_0\) (die Hülle ist geschlossen)
Q: Jarvis Wrap (Einwickeln).Startend bei \(q_0\) (kleinste \(x\)-Koordinate) hängt man wiederholt {{c1::\(q_h \leftarrow \texttt{FindNext}(q_{h-1})\)}} an, bis {{c1::\(q_h = q_0\) (die Hülle ist geschlossen)}}:
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|---|---|---|---|---|---|
| lorenz | cid:1779798950966 |
1 | 230% | 9d | 7 |
nid:1774487164522
c1
A&W
Die Anzahl der Anordnungen von \(n\) Objekten, von denen\(n_...
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Cloze c1
Q: Die Anzahl der Anordnungen von \(n\) Objekten, von denen\(n_1\) vom Typ 1, …, \(n_r\) vom Typ \(r\) sind (\(n_1 + \cdots + n_r = n\)), ist:\[{{c1::\frac{n!}{n_1!\, n_2!\, \cdots\, n_r!} }} = \binom{n}{n_1, n_2, \ldots, n_r}\](Multinomialkoeffizient)
A: Speziell für \(r=2\): \(\frac{n!}{k!\,(n-k)!} = \binom{n}{k}\).Beispiel: Anordnungen von „MISSISSIPPI“: \(\frac{11!}{1!\cdot 4!\cdot 4!\cdot 2!} = 34{,}650\).
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|---|---|---|---|---|---|
| lorenz | cid:1774487164523 |
1 | 230% | 35d | 12 |
nid:1777540083522
A&W
Im Beweis der QuickSort-Schranke \(t_n \leq 2(n+1) \ln n + O...
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Q: Im Beweis der QuickSort-Schranke \(t_n \leq 2(n+1) \ln n + O(n)\) startet man mit der Rekursion \(n \cdot t_n = \sum_{i=0}^{n-1}(n - 1 + t_i + t_{n-i-1})\).Wie kommt man von dort zur einfachen Ungleichung \(t_n \leq \tfrac{n+1}{n} t_{n-1} + 2\)?
A: Trick: Schreibe dieselbe Rekursion für \(n - 1\) auf:\[(n-1) \cdot t_{n-1} = \sum_{i=0}^{n-2} (n - 2 + t_i + t_{n-i-2})\]und subtrahiere von der ursprünglichen:\[n t_n - (n-1) t_{n-1} = 2(n-1) + 2 t_{n-1}.\]Umstellen liefert exakt\[t_n = \tfrac{n+1}{n} t_{n-1} + \tfrac{2(n-1)}{n} \leq \tfrac{n+1}{n} t_{n-1} + 2.\]Per Induktion folgt \(t_n \leq 2 \sum_{i=3}^{n+1} \tfrac{n+1}{i}\), und mit der harmonischen Reihe \(H_n = \ln n + O(1)\) folgt das Resultat.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777540083522 |
1 | 230% | 13d | 11 |
nid:1778164855883
c2
A&W
Ford-Fulkerson mit ganzzahligen Kapazitäten Sei \(N = (V, A,...
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Cloze c2
Q: Ford-Fulkerson mit ganzzahligen Kapazitäten Sei \(N = (V, A, c, s, t)\) ein Netzwerk mit \(c : A \to \mathbb{N}_0^{\leq U}\) (für \(U \in \mathbb{N}\)), ohne entgegen gerichtete Kanten. Dann:{{c1::Es gibt einen ganzzahligen maximalen Fluss}}.Er kann in Zeit {{c2::\(\
A: Begründung der Laufzeit: Höchstens \((n-1)U = \mathcal{O}(nU)\) Augmentierungsschritte, jeder Schritt (Pfadsuche per BFS/DFS in \(N_f\), Augmentierung, Update) braucht \(\mathcal{O}(m)\) Zeit. Die Ganzzahligkeit folgt induktiv: Start mit \(f \equiv 0\), und jeder Schritt erhält die Ganzzahligkeit, da \(\varepsilon = \min_i \varepsilon_i\) ganzzahlig ist.
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| lorenz | cid:1778164855883 |
1 | 230% | 13d | 11 |
nid:1780223730600
c3
A&W
Zählen erreichbarer Knoten (gerichtet): LaufzeitFür einen ei...
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nid:1780223730600
Cloze c3
Q: Zählen erreichbarer Knoten (gerichtet): LaufzeitFür einen einzelnen Startknoten \(s\) bestimmt ein DFS/BFS die Anzahl erreichbarer Knoten in Zeit \({{c1::O(n + m)}}\).Für alle Knoten ergibt das \({{c1::O(n(n + m))}}\).Geht es schneller? {{c3::Vermutlich nicht: Ei
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| lorenz | cid:1780223730601 |
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Seien \(\delta, \varepsilon > 0\). Falls \({{c1::N \geq 3\,\...
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Cloze c2
Q: Seien \(\delta, \varepsilon > 0\). Falls \({{c1::N \geq 3\,\frac{|U|}{|S|} \cdot \frac{1}{\varepsilon^2} \cdot \ln(\tfrac{2}{\delta})}}\), ist die Ausgabe \(Y\) von Target-Shooting mit Wahrscheinlichkeit mindestens \(1 - \delta\) im Intervall \[{
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nid:1778588912093
c1
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Schlüsselidentität Bildsegmentierung. Sei \((S, T)\) ein \(s...
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nid:1778588912093
Cloze c1
Q: Schlüsselidentität Bildsegmentierung. Sei \((S, T)\) ein \(s\)-\(t\)-Schnitt im Segmentierungs-Netzwerk \(N\) und \(A := S \setminus \{s\}\), \(B := T \setminus \{t\}\). Dann gilt\[\operatorname{cap}(S, T) = {{c1::q'(A, B) = \sum_{p \in A} \beta_p + \sum_{p \in B} \alpha_p + \sum_{e \in E
A: Beitrag zum Schnitt: Kanten \((s, p)\) mit \(p \in B\): Beitrag \(\sum_{p \in B} \alpha_p\).Kanten \((p, t)\) mit \(p \in A\): Beitrag \(\sum_{p \in A} \beta_p\).Bildkanten \((p, p')\) mit \(p \in A, p' \in B\): Beitrag \(\sum_{\ldots} \gamma_{(p, p')}\).Dieser Lösungsansatz wird in der Praxis verwendet, auch für höherdimensionale Bilder (Voxel), z.B. Computertomographie.
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nid:1779798950874
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der Schnitt aller konvexen Mengen, die \(S\) enthalten
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Cloze c1
Cloze answer: der Schnitt aller konvexen Mengen, die \(S\) enthalten
Q: Konvexe Hülle \(\operatorname{conv}(S)\).Die konvexe Hülle einer Menge \(S \subseteq \mathbb{R}^d\) ist {{c1::der Schnitt aller konvexen Mengen, die \(S\) enthalten}}:\[\operatorname{conv}(S) := {{c1::\bigcap_{S \subseteq C \subseteq \mathbb{R}^d,\ C \text{ konvex} } C}}.\]Äquivalent:
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nid:1779798951076
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zwei erfolglose Tests (einmal unten, einmal oben)
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Cloze c2
Cloze answer: zwei erfolglose Tests (einmal unten, einmal oben)
Q: LocalRepair: Laufzeitanalyse.Start mit \(2(n-1)\) Ecken, Ende mit \(h\) Ecken, also genau {{c1::\(2(n-1) - h = O(n)\)}} erfolgreiche (entfernende) Tests. Pro Punkt \(p_i\) gibt es zudem {{c2::zwei erfolglose Tests (einmal unten, einmal oben)}}.Nach dem anfänglichen Sortieren in \(
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| lorenz | cid:1779798951078 |
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nid:1780484891645
A&W
Wahr oder falsch?Wir können jeden Las-Vegas-Algorithmus mit ...
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Q: Wahr oder falsch?Wir können jeden Las-Vegas-Algorithmus mit bekannter erwarteter Laufzeit höchstens \(T\) in einen randomisierten Algorithmus umwandeln, dessen Laufzeit höchstens \(10T\) ist und der mit Wahrscheinlichkeit mindestens \(0.9\) erfolgreich ist.
A: Wahr.Markov-Ungleichung: Bricht man den Las-Vegas-Algorithmus nach \(10T\) Schritten ab, ist \(\Pr[\text{Laufzeit} > 10T] \leq T/(10T) = 1/10\). Mit Wahrscheinlichkeit \(\geq 0.9\) terminiert er also korrekt innerhalb von \(10T\).
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| lorenz | cid:1780484891645 |
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nid:1780484891777
A&W
Wahr oder falsch?Sei \(f\) ein maximaler Fluss in einem Netz...
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Q: Wahr oder falsch?Sei \(f\) ein maximaler Fluss in einem Netzwerk \(N = (V, A, c, s, t)\) ohne entgegen gerichtete Kanten. Dann gibt es keine zwei Knoten \(u, v \in V\), sodass sowohl die Kante \((u, v)\) als auch die Kante \((v, u)\) im Restnetzwerk \(N_f\) vorkommt.
A: Falsch.Wird eine Kante \((u, v) \in A\) teilweise genutzt, also \(0 < f(u, v) < c(u, v)\), so enthält \(N_f\) sowohl die Vorwärtskante \((u, v)\) (Restkapazität \(c - f\)) als auch die Rückwärtskante \((v, u)\) (Restkapazität \(f\)). Das kann auch bei maximalem Fluss auftreten.
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| lorenz | cid:1780484891778 |
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nid:1773330177039
A&W
Wahr oder falsch?Es gibt einen polynomiellen Algorithmus, de...
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Q: Wahr oder falsch?Es gibt einen polynomiellen Algorithmus, der für jeden planaren Graphen eine geeignete Einfärbung mit 6 Farben findet.
A: Wahr.Jeder planare Graph hat einen Knoten vom Grad \(\leq 5\). Entfernt man ihn rekursiv und färbt rückwärts greedy, genügen 6 Farben, und das in Polynomzeit. (Der Vier-Farben-Satz ist stärker, aber 6 Farben sind viel einfacher zu garantieren.)
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nid:1777540083530
c1
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\(k\)-kleinsten Wert
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Cloze c1
Cloze answer: \(k\)-kleinsten Wert
Q: Selektionsproblem: Bestimme in einer Folge \((A[1], \ldots, A[n])\) paarweise verschiedener Zahlen den {{c1::\(k\)-kleinsten Wert}}.Naiver Ansatz: Sortieren + Index, Laufzeit \(O({{c2::n \log n}})\).Schneller geht es mit QuickSelect (erwartet \(O({{c3::n}})\)):
A: Im Gegensatz zu QuickSort rekursiert QuickSelect nur in eine der beiden Hälften (oder gibt direkt zurück, falls Pivot bereits an der gesuchten Position liegt). Das macht den Unterschied zwischen \(O(n \log n)\) und \(O(n)\) erwartet.Bei einem Aufruf von QuickSelect ist die Anzahl Vergleiche \(T = \sum_{i=1}^{N} (r_i - \ell_i)\), wobei \(N\) die Anzahl Partition-Aufrufe ist.
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| lorenz | cid:1777540083531 |
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nid:1778164855813
c1
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\min_i \varepsilon_i
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Cloze c1
Cloze answer: \min_i \varepsilon_i
Q: Beweis von „Pfad in \(N_f\) \(\Rightarrow\) \(f\) nicht maximal“. Gegeben ein gerichteter s-t-Pfad in \(N_f\) mit Restkapazitäten \(\varepsilon_1, \ldots, \varepsilon_k\).Setze \(\varepsilon := {{c1::\min_i \varepsilon_i}} > 0\).Augmentiere \(f\) entlang des Pf
A: Zulässigkeit: bei \(+\varepsilon\) bleibt man unter \(c\), weil \(\varepsilon \leq c - f\); bei \(-\varepsilon\) bleibt man über \(0\), weil \(\varepsilon \leq f\).
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| lorenz | cid:1778164855814 |
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nid:1779193767012
c1
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Wir können das MIN-CUT-Problem auf den s-t-Mincut zurückführ...
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nid:1779193767012
Cloze c1
Q: Wir können das MIN-CUT-Problem auf den s-t-Mincut zurückführen: fixiere ein \(s\) und betrachte alle \(t \in V \setminus \{s\}\); jeder Schnitt ist ein s-t-Schnitt für ein passendes \(t\). Bei \((n - 1)\)-maligem Aufruf eines s-t-Mincut-Algorithmus mit Laufzeit \(\mathcal{O}(mn \log n)\) erh
A: Es genügt, ein einziges \(s\) zu fixieren, weil jeder Schnitt mindestens einen Knoten \(t \neq s\) auf der anderen Seite hat. Die Schranke \(m = \mathcal{O}(n^2)\) gilt im Multigraph nach Reduktion auf Kantengewichte (sonst wäre \(m\) unbeschränkt).
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| lorenz | cid:1779193767013 |
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nid:1779798950998
c2
A&W
JarvisWrap: Umgang mit Degeneriertheiten (Kollinearitäten, g...
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nid:1779798950998
Cloze c2
Q: JarvisWrap: Umgang mit Degeneriertheiten (Kollinearitäten, gleiche \(x\)-Koordinaten, Duplikate).Startpunkt \(q_0\): nimm den Punkt mit {{c1::lexikographisch kleinster Koordinate (unter kleinster \(x\)-Koordinate den mit kleinster \(y\)-Koordinate)}}.Test "\(p\) rechts vo
A: Software ohne Berücksichtigung dieser Fälle hat geringen praktischen Nutzen.
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nid:1772545581602
c1
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|E|
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Cloze c1
Cloze answer: |E|
Q: Mit einem Greedy-Algorithmus kann man in Zeit \( O({{c1::|E|}}) \) ein {{c3::inklusionsmaximales}} Matching \( M_{\text{Greedy}} \) bestimmen mit\[{{c2:: |M_{\text{Greedy} }| \geq |M_{\text{max} }| / 2, }}\]wobei \( M_{\text{max}} \) ein kardinalitätsmaximales Matching ist.
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| lorenz | cid:1772545581604 |
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nid:1774487164717
c2
A&W
Der Satz von Bayes lässt sich als Update-Regel schreiben:\[{...
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Cloze c2
Q: Der Satz von Bayes lässt sich als Update-Regel schreiben:\[{{c1::\text{Posterior} }} \propto {{c2::\text{Likelihood} \times \text{Prior} }}\]
A: \(\Pr[H \mid E] \propto \Pr[E \mid H] \cdot \Pr[H]\).Das \(\propto\) (proportional zu) bedeutet, dass die Normierungskonstante \(\Pr[E]\) im Nenner fehlt.
Intuition: Der Prior \(\Pr[H]\) ist unsere Ausgangsmeinung über \(H\). Die Likelihood \(\Pr[E \mid H]\) gewichtet, wie stark die beobachtete Evidenz \(E\) diese Meinung in Richtung \(H\) verschiebt.
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| lorenz | cid:1774487164718 |
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nid:1777538021785
c3
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Welche Anteile von \(a \in [n-1]\) sind Zertifikate dafür, d...
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nid:1777538021785
Cloze c3
Q: Welche Anteile von \(a \in [n-1]\) sind Zertifikate dafür, dass \(n\) nicht prim ist?ZertifikatBedingung an \(a\)
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| lorenz | cid:1777539214862 |
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nid:1778164855662
c1
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Die Kapazität eines s-t-Schnitts \((S, T)\) ist\[\operatorna...
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Cloze c1
Q: Die Kapazität eines s-t-Schnitts \((S, T)\) ist\[\operatorname{cap}(S, T) := {{c1::\sum_{(u, w) \in (S \times T) \cap A} c(u, w)}}.\]Wichtig: Die Kapazität {{c2::ignoriert die Kanten von \(T\) nach \(S\)}}.
A: Nur Kanten, die von \(S\) nach \(T\) zeigen, zählen. Rückwärtskanten von \(T\) nach \(S\) tragen nicht zur Kapazität bei.
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| lorenz | cid:1778164855663 |
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nid:1779193767002
c3
A&W
In einem Multigraphen \(G = (V, E)\) ist der Grad \(\deg(v)\...
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nid:1779193767002
Cloze c3
Q: In einem Multigraphen \(G = (V, E)\) ist der Grad \(\deg(v)\) eines Knoten \(v\) {{c1::die Anzahl inzidenter Kanten (nicht die Anzahl Nachbarn)}}. Es gelten\[\begin{gathered}|E| = {{c2::\tfrac{1}{2} \sum_{v \in V} \deg(v)}}, \\\mu(G) \leq {{c3::\min_{v \in V} \deg(v)}}.\end{gath
A: Die Schranke \(\mu(G) \leq \min_v \deg(v)\) folgt, weil das Isolieren des Knotens \(v\) mit kleinstem Grad einen gültigen Schnitt der Grösse \(\deg(v)\) liefert. Sie ist im Allgemeinen nicht scharf: es gibt Graphen mit \(\min \deg = 4\) und \(\mu = 3\).
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| lorenz | cid:1779193767002 |
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nid:1779193767067
A&W
Schreibe den Karger-Algorithmus \(\text{Cut}(G)\) (random ed...
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Q: Schreibe den Karger-Algorithmus \(\text{Cut}(G)\) (random edge contraction) als Pseudocode und gib seine Laufzeit an, unter der Annahme, dass Kantenkontraktion und das Ziehen einer gleichverteilt zufälligen Kante je in \(\mathcal{O}(n)\) Zeit möglich sind.
A: Laufzeit: \(\mathcal{O}(n^2)\) (genau \(n - 2\) Kontraktionen, jede in \(\mathcal{O}(n)\)). Achtung: das Ziehen einer gleichverteilt zufälligen Kante in einem Multigraph erfordert die Darstellung der Mehrfachkanten durch Kantengewichte. Der zurückgegebene Wert ist nie kleiner als \(\mu(G)\), aber im Allgemeinen zu gross.
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| lorenz | cid:1779193767067 |
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nid:1779798950892
c1
A&W
keine 3 Punkte auf einer gemeinsamen Geraden liegen; keine 2...
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Cloze c1
Cloze answer: keine 3 Punkte auf einer gemeinsamen Geraden liegen; keine 2 Punkte dieselbe \(x\)-Koordinate haben
Q: Vereinfachende Annahme: Allgemeine Lage.Für die ConvexHull-Algorithmen nimmt man an, dass {{c1::keine 3 Punkte auf einer gemeinsamen Geraden liegen}} und {{c1::keine 2 Punkte dieselbe \(x\)-Koordinate haben}}.
A: Diese Annahme schliesst Kollinearitäten und Mehrdeutigkeiten aus: Degeneriertheiten werden separat behandelt.
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| lorenz | cid:1779798950892 |
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nid:1772046933705
c1
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\Pi \in P \Rightarrow P = NP
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Cloze c1
Cloze answer: \Pi \in P \Rightarrow P = NP
Q: Ein Problem \(\Pi\) aus NP heißt NP-vollständig, falls gilt:\[{{c1::\Pi \in P \Rightarrow P = NP}}\]
A: Es gibt sehr viele NP-vollständige Probleme:
Hamiltonkreis
Rucksackproblem
Clique: Gibt es in einem Graphen \(k\) paarweise benachbarte Knoten?
Nullstelle mod \(n\): Hat ein Polynom mod \(n\) eine Nullstelle?
Satisfiability: Hat eine logische Formel eine Lösung?
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| lorenz | cid:1772046933705 |
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nid:1778164855676
c1
A&W
Schwache DualitätIst \(f\) ein Fluss und \((S, T)\) ein s-t-...
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Cloze c1
Q: Schwache DualitätIst \(f\) ein Fluss und \((S, T)\) ein s-t-Schnitt in einem Netzwerk, so gilt\[{{c1::\operatorname{val}(f) \;\leq\; \operatorname{cap}(S, T)}}.\]
A: Konsequenz: Findet man zu einem Fluss \(f\) einen Schnitt \((S, T)\) mit \(\operatorname{val}(f) = \operatorname{cap}(S, T)\), so ist \(f\) maximal. Der Schnitt ist dann ein einfaches Zertifikat für die Maximalität.
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| lorenz | cid:1778164855676 |
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nid:1779487730615
c1
A&W
Indikatorvariablen im Beweis des Sampling Lemmas.Für \(s \in...
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Cloze c1
Q: Indikatorvariablen im Beweis des Sampling Lemmas.Für \(s \in P'\) und \(R, Q \subseteq P'\):\(\operatorname{out}(s, R) := 1\) genau dann, wenn {{c1::\(s \notin C^{\bullet}(R)\) (\(s\) liegt ausserhalb von \(C(R)\))}}.\(\operatorname{ess}(s, Q) := 1\) genau dann,
A: Ausserdem: \(\sum_{s \in P' \setminus R} \operatorname{out}(s, R) = |P' \setminus C^{\bullet}(R)|\) und \(\sum_{s \in Q} \operatorname{ess}(s, Q) \leq 3\) (höchstens 3 essentielle Punkte, vgl. kleine bestimmende Menge).
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| lorenz | cid:1779487730617 |
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nid:1774631277414
c1
A&W
Hamiltonkreis
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Cloze c1
Cloze answer: Hamiltonkreis
Q: Es existiert ein {{c1::Hamiltonkreis}} in einem Graphen \(G\) mit gerader Zahl Knoten \(\implies\)perfektes Matching existiert.
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nid:1777538021785
c4
A&W
Welche Anteile von \(a \in [n-1]\) sind Zertifikate dafür, d...
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Cloze c4
Q: Welche Anteile von \(a \in [n-1]\) sind Zertifikate dafür, dass \(n\) nicht prim ist?ZertifikatBedingung an \(a\)
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| lorenz | cid:1777539221478 |
1 | 230% | 19d | 11 |
nid:1774487164737
c2
A&W
Bernoulli-verteilt
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Cloze c2
Cloze answer: Bernoulli-verteilt
Q: Eine Zufallsvariable \(X\) mit Wertebereich \(W_X = \{0, 1\}\) und Dichte\[f_X(\alpha) = \begin{cases} p & {{c1::\text{für } \alpha = 1}} \\ 1 - p & {{c1::\text{für } \alpha = 0}} \\ 0 & {{c1::\text{sonst} }} \end{cases}\]heisst {{c2::Bernoulli-verteilt}} mit {{c3::Erfolgswa
A: Modelliert einen einzelnen Münzwurf (mit verzerrter Münze).Indikator-ZV \(X_A\) ist genau \(\text{Bernoulli}(\Pr[A])\).
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| lorenz | cid:1774487164738 |
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nid:1778164855717
c1
A&W
nur endlich viele s-t-Schnitte
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Cloze c1
Cloze answer: nur endlich viele s-t-Schnitte
Q: Da es {{c1::nur endlich viele s-t-Schnitte}} gibt, folgt:s-t-MinCut ist {{c2::ein endliches algorithmisches Problem}},{{c3::ein minimaler Schnitt existiert immer}}.
A: Im Gegensatz dazu ist die Menge der zulässigen Flüsse im Allgemeinen unendlich (reelle Kapazitäten), aber das Supremum wird ebenfalls angenommen (Maxflow-Mincut).
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| lorenz | cid:1778164855719 |
1 | 230% | 32d | 10 |
nid:1772544804526
c1
A&W
überdeckt
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Cloze c1
Cloze answer: überdeckt
Q: Ein Knoten \( v \) wird von \( M \) {{c1::überdeckt}}, falls {{c2::es eine Kante \( e \in M \) gibt, die \( v \) enthält}}.
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| lorenz | cid:1772544804526 |
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nid:1777538021744
c4
A&W
Die multiplikative Gruppe modulo \(n\) ist\[\mathbb{Z}_n^* :...
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Cloze c4
Q: Die multiplikative Gruppe modulo \(n\) ist\[\mathbb{Z}_n^* := {{c1::\{a \in [n-1] \mid \mathrm{ggT}(a, n) = 1\} }}\]mit Multiplikation mod \(n\). Sie ist eine Gruppe der Ordnung\[\varphi(n) := {{c2::|\mathbb{Z}_n^*|}}\quad\text{(eulersche Phi-Funktion)}.\]Spezialfälle:\(n\) prim: \(\m
A: Beispiele: \(\mathbb{Z}_9^* = \{1, 2, 4, 5, 7, 8\}\), \(\varphi(9) = 6\). \(\mathbb{Z}_7^* = \{1, 2, 3, 4, 5, 6\}\), \(\varphi(7) = 6\).Konsequenz für den \(p^2\)-Fall beim Euklid-Test: \(\frac{|\mathbb{Z}_n^*|}{n-1} \approx 1 - \frac{1}{\sqrt{n}}\), die Fehlerrate ist also fast \(1\).
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nid:1777984580531
c1
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Sei \(G = (V, E)\) mit Färbung \(\gamma : V \to [k]\). Für \...
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Cloze c1
Q: Sei \(G = (V, E)\) mit Färbung \(\gamma : V \to [k]\). Für \(v \in V\) und \(i \in \{0, \dots, k-1\}\) sei\[P_i(v) := \left\{ S \subseteq [k] \,:\, {{c1::\begin{gathered} |S| = i+1 \text{ und } \exists \text{ in } v \text{ endender,} \\ \text{genau mit } S \text{ gefärbter bunter Pfad} \end{
A: \(P_i(v)\) sammelt also alle möglichen Farbmengen, die ein bunter Pfad der Länge \(i\) ausschöpfen kann, wenn er bei \(v\) endet.Ziel des Algorithmus: Berechne \(P_i(v)\) für alle \(v \in V\) und alle \(i \in \{0, 1, \dots, k-1\}\) per dynamischer Programmierung.Es existiert ein bunter Pfad der Länge \(k-1\) genau dann, wenn \(P_{k-1}(v) \neq \emptyset\) für ein \(v \in V\).
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A&W
Wahr oder falsch?Jeder 2-zusammenhängende Graph hat einen Ha...
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Q: Wahr oder falsch?Jeder 2-zusammenhängende Graph hat einen Hamiltonkreis.
A: Falsch.Gegenbeispiel: der vollständige bipartite Graph \(K_{2,3}\) ist 2-zusammenhängend, hat aber keinen Hamiltonkreis (in einem bipartiten Graphen müssten dafür beide Seiten gleich gross sein).
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c3
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\(f(u, w) = c(u, w)\) (sonst wäre \(w\) erreichbar)
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Cloze c3
Cloze answer: \(f(u, w) = c(u, w)\) (sonst wäre \(w\) erreichbar)
Q: Beweis von „kein Pfad in \(N_f\) \(\Rightarrow\) \(\exists\) Schnitt \((S, T)\) mit \(\operatorname{cap}(S, T) = \operatorname{val}(f)\)“. Definiere\[\begin{gathered}S := {{c1::\{v \in V : v \text{ ist von } s \text{ in } N_f \text{ erreichbar}\} }}, \\ T := V \setminus S.\end{gathered}\]Da k
A: Die zwei Bedingungen geben gleichzeitig die untere und die obere Schranke aus dem schwachen Dualitätslemma scharf: alle vorwärtsführenden Kanten sind saturiert, alle rückwärtsführenden Kanten tragen keinen Fluss.
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c1
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Wie modelliert man einen d-dimensionalen Hyperwürfel \(H_d\)...
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Cloze c1
Q: Wie modelliert man einen d-dimensionalen Hyperwürfel \(H_d\)? Knotenmenge: \({{c1::\{0,1\}^d}}\)Kantenmenge: {{c2::alle Knotenpaare, die sich in genau einer Koordinate unterscheiden}}
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A&W
Was besagt der Vierfarbensatz?
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Q: Was besagt der Vierfarbensatz?
A: Jeder planare Graph (jede Landkarte) lässt sich mit \(\leq 4\) Farben färben.Formal: Für jeden planaren Graphen \(G\) gilt \(\chi(G) \leq 4\).(Appel & Haken, 1976 - erster computergestützter Beweis)
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c5
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einem Pfad gefolgt von einem Kreis (\(\rho\)-Form)
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Cloze c5
Cloze answer: einem Pfad gefolgt von einem Kreis (\(\rho\)-Form)
Q: Floyd's Cycle Finding (Graph-Reformulierung)Definiere den gerichteten Graphen \(D = (V, A)\) mit {{c1::\(V = [n]\)}} und {{c2::\(A = \{(i, a[i]) \mid 1 \leq i \leq n\}\)}}.Eigenschaften:Jeder Knoten hat genau {{c3::eine ausgehende Kante}}Knoten \(n\) hat {{c4::keine e
A: Notation: Pfad hat \(k \geq 1\) Kanten, Kreis hat \(\ell \geq 3\) Kanten, mit \(k + \ell \leq n\).
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| lorenz | cid:1777923968749 |
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c1
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effizient entscheidbare Probleme; (einseitig) effizient veri...
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Cloze c1
Cloze answer: effizient entscheidbare Probleme; (einseitig) effizient verifizierbare Probleme
Q: \(P\) = {{c1::effizient entscheidbare Probleme}}
\(NP\) = {{c1::(einseitig) effizient verifizierbare Probleme}}
A: P = polynomiellNP = nichtdeterministisch polynomiell
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Welche drei Bestandteile bestimmen einen diskreten Wahrschei...
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Cloze c3
Q: Welche drei Bestandteile bestimmen einen diskreten Wahrscheinlichkeitsraum?{{c1::Eine Ergebnismenge \(\Omega = \{\omega_1, \omega_2, \ldots\}\) von Elementarereignissen.}}{{c2::Eine Wahrscheinlichkeitszuweisung \(\Pr[\omega_i] \in [0,1]\) für je
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c1
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Falls \(A_1, \ldots, A_n\) paarweise disjunkt sind, gilt\[\P...
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Cloze c1
Q: Falls \(A_1, \ldots, A_n\) paarweise disjunkt sind, gilt\[\Pr\!\left[\bigcup_{i=1}^{n} A_i\right] = {{c1::\sum_{i=1}^{n} \Pr[A_i]}}.\]Proof Included
A: (Additionssatz)Warum das aus Definition 2.1 folgt: \(\Pr[\bigcup A_i] = \sum_{\omega \in \bigcup A_i}\Pr[\omega]\). Da die \(A_i\) disjunkt sind, kommt jedes \(\omega\) in genau einem \(A_i\) vor, also ist dies gleich \(\sum_i\sum_{\omega\in A_i}\Pr[\omega]=\sum_i\Pr[A_i]\).Gegenbeispiel ohne Disjunktheit: Würfel, \(A=\{1,3,5\}\), \(B=\{5,6\}\). \(\Pr[A\cup B]=4/6 \ne 5/6 = \Pr[A]+\Pr[B]\).
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c2
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\(t \in T\)
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Cloze c2
Cloze answer: \(t \in T\)
Q: Beweis von „kein Pfad in \(N_f\) \(\Rightarrow\) \(\exists\) Schnitt \((S, T)\) mit \(\operatorname{cap}(S, T) = \operatorname{val}(f)\)“. Definiere\[\begin{gathered}S := {{c1::\{v \in V : v \text{ ist von } s \text{ in } N_f \text{ erreichbar}\} }}, \\ T := V \setminus S.\end{gathered}\]Da k
A: Die zwei Bedingungen geben gleichzeitig die untere und die obere Schranke aus dem schwachen Dualitätslemma scharf: alle vorwärtsführenden Kanten sind saturiert, alle rückwärtsführenden Kanten tragen keinen Fluss.
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| lorenz | cid:1778164855828 |
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nid:1778588912041
c3
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Sei \(f\) ein ganzzahliger maximaler Fluss in \(N_G^{*}\). D...
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Cloze c3
Q: Sei \(f\) ein ganzzahliger maximaler Fluss in \(N_G^{*}\). Dann gilt:Flusswerte: {{c1::\(f(e) \in \{0, 1\}\)}} für alle Kanten \(e\).Für jeden Knoten \(w \notin \{u, v\}\): {{c2::\(\operatorname{indeg}_f(w) = \operatorname{outdeg}_f(w)\) (Flusserhaltung in inneren Knoten)}}.
A: \(\operatorname{indeg}_f(w)\) und \(\operatorname{outdeg}_f(w)\) bezeichnen die Ein- bzw. Ausgrade bezüglich der Kanten mit Fluss 1.
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c2
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< 1
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Cloze c2
Cloze answer: < 1
Q: Erwartete Anzahl KollisionenFür eine Hashfunktion \(h : U \to [m]\) mit \(\Pr[h(u) = i] = \tfrac{1}{m}\) gilt:\[\mathbb{E}[\#\text{Kollisionen}] \leq {{c1::\binom{n}{2} \cdot \tfrac{1}{m} }}.\]Insbesondere folgt mit der Wahl \(m = n^2\): \(\mathbb{E}[\#\text{Kollisionen}] {{c2::< 1}}\)
A: Beweisidee: Für jedes feste Paar \((i, j)\) mit \(s_i \neq s_j\) gilt \(\Pr[h(s_i) = h(s_j)] = \tfrac{1}{m}\) (wegen Unabhängigkeit / Zufallsfunktion). Es gibt höchstens \(\binom{n}{2}\) solche Paare, und Linearität des Erwartungswerts liefert die Schranke.
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c1
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Zu einer Zufallsvariablen \(X\) mit Wertebereich \(W_X\) def...
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Cloze c1
Q: Zu einer Zufallsvariablen \(X\) mit Wertebereich \(W_X\) definieren wir {{c2::den Erwartungswert \(\mathbb{E}[X]\)}} durch\[{{c2::\mathbb{E}[X]}} := {{c1::\sum_{\alpha \in W_X} \alpha \cdot \Pr[X = \alpha]}},\]sofern die Summe absolut konvergiert.
A: Ansonsten sagen wir, dass der Erwartungswert undefiniert ist.Intuition: Gewichteter Durchschnitt aller möglichen Werte.
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c1
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n^3
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Cloze c1
Cloze answer: n^3
Q: Für \(n\) gerade und \(\ell : \binom{[n]}{2} \to \mathbb{N}_0\) kann man in Zeit \(O({{c1::n^3}})\) ein {{c2::minimales (gewichtsminimales) perfektes Matching}} in \(K_n\) finden.
A: Das ist der Blossom-Algorithmus.Dies wird im Christofides-Algorithmus für das metrische TSP benötigt.
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c3
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\(PB_n\) ist eine {{c1::Untergruppe von \(\mathbb{Z}_n^*\)}}...
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Cloze c3
Q: \(PB_n\) ist eine {{c1::Untergruppe von \(\mathbb{Z}_n^*\)}}, denn\[a^{n-1} \equiv_n 1 \;\wedge\; b^{n-1} \equiv_n 1 \;\Longrightarrow\; (ab)^{n-1} \equiv_n 1.\]Folglich: ist \(n\) nicht Carmichael, dann ist \(|PB_n|\) ein {{c2::echter Teiler von \(|\mathbb{Z}_n^*|\)}}, also\[|PB_n| \leq {{c3
A: Genau diese Schranke liefert die Fehlerwahrscheinlichkeit \(\leq \tfrac{1}{2}\) des Fermat-Tests für Nicht-Carmichael-Zahlen: \(\Pr[\text{Fehler}] = |PB_n|/(n-1) \leq \tfrac{1}{2}\).Der Beweis nutzt den Satz von Lagrange: jede echte Untergruppe einer endlichen Gruppe hat höchstens halb so viele Elemente wie die Gruppe selbst.
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| lorenz | cid:1777538021778 |
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c2
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ignoriert die Kanten von \(T\) nach \(S\)
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Cloze c2
Cloze answer: ignoriert die Kanten von \(T\) nach \(S\)
Q: Die Kapazität eines s-t-Schnitts \((S, T)\) ist\[\operatorname{cap}(S, T) := {{c1::\sum_{(u, w) \in (S \times T) \cap A} c(u, w)}}.\]Wichtig: Die Kapazität {{c2::ignoriert die Kanten von \(T\) nach \(S\)}}.
A: Nur Kanten, die von \(S\) nach \(T\) zeigen, zählen. Rückwärtskanten von \(T\) nach \(S\) tragen nicht zur Kapazität bei.
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c1
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dreiecksfreien
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Cloze c1
Cloze answer: dreiecksfreien
Q: Für alle \(k \geq 2\) gibt es einen {{c1::dreiecksfreien}} Graphen \(G_k\) mit \(\chi(G_k) \geq k\).
A: (Mycielski-Konstruktion)Konstruktion: Aus \(G_k = (V_k, E_k)\) mit \(V_k = \{v_1,\ldots,v_n\}\) bilde \(G_{k+1}\):Füge Knoten \(w_1,\ldots,w_n, z\) hinzu. \(w_i\) ist mit allen Nachbarn von \(v_i\) verbunden (aber nicht mit \(v_i\) selbst). \(z\) ist mit allen \(w_i\) verbunden.Der neue Graph ist dreiecksfrei und braucht eine Farbe mehr als \(G_k\).
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nid:1773307783473
IO r1
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[Image Occlusion region 1]
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Cloze c1
Q: {{c1::image-occlusion:polygon:left=.011:top=.2474:points=.0836,.2506 .4728,.2474 .4728,.3534 .011,.3566 .011,.3052 .0836,.3052}}{{c2::image-occlusion:rect:left=.0572:top=.4433:width=.1363:height=.045}}{{c2::image-occlusion:rect:left=.0924:top=.5815:width=.1869:height=.0514}}{{c3::image-o
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A&W
Wann ist der Erwartungswert \(\mathbb{E}[X] = \sum_{x\in W_X...
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Q: Wann ist der Erwartungswert \(\mathbb{E}[X] = \sum_{x\in W_X} x\cdot\Pr[X=x]\) undefiniert?
A: Falls die Summe nicht absolut konvergiert (z.B. positiver und negativer Anteil beide divergieren).Bemerkung:In der Vorlesung betrachten wir nur Zufallsvariablen mit definiertem Erwartungswert.Der Erwartungswert ist nur definiert, wenn die Summe absolut konvergiert, d.h. \(\sum_{x\in W_X}|x|\cdot\Pr[X=x]<\infty\).In endlichen Wahrscheinlichkeitsräumen ist dies immer erfüllt (endlich viele Terme). Bei unendlichen Räumen muss man aufpassen.
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c2
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2^n
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Cloze c2
Cloze answer: 2^n
Q: Es gilt:\[{{c1::\sum_{k=0}^{n} \binom{n}{k}::\text{Binomialsatz} }} = {{c2::2^n}}\]
A: Beweis: Setze \(x = y = 1\) im Binomialsatz: \((1+1)^n = \sum_{k=0}^n \binom{n}{k}\).Interpretation: Anzahl aller Teilmengen einer \(n\)-elementigen Menge ist \(2^n\).
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| lorenz | cid:1774693607608 |
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nid:1777923968751
A&W
Floyd's Cycle Finding (Phase 1)Wie wird mit nur drei Variabl...
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Q: Floyd's Cycle Finding (Phase 1)Wie wird mit nur drei Variablen ein \(i \geq 1\) mit \(x_i = x_{2i}\) berechnet, und welche Bedeutung hat das gefundene \(i\)?
A: igel := a[n]
hase := a[a[n]]
i := 1
while (igel != hase):
igel := a[igel]
hase := a[a[hase]]
i := i + 1Der Igel macht pro Iteration einen Schritt, der Hase zwei. Treffen sie sich, gilt \(x_i = x_{2i}\) mit \(i = k + r\) (Pfadlänge \(k\) plus Rest \(r\) wie in der Schlüsseleigenschaft).Laufzeit: \(O(n)\), denn \(i \leq k + \ell \leq n\).
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| lorenz | cid:1777923968753 |
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c1
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Monte-Carlo, einseitiger FehlerSei \(A\) ein randomisierter ...
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Cloze c1
Q: Monte-Carlo, einseitiger FehlerSei \(A\) ein randomisierter Algorithmus, der immer JA oder NEIN ausgibt, mit\[\begin{gathered}\Pr[A(I) = \text{JA}] = 1, \text{ falls } I \text{ JA-Instanz},\\ \Pr[A(I) = \text{NEIN}] \geq \varepsilon, \text{ falls } I \text{ NEIN-Instanz}.\end{gathered}\]S
A: Selbe Iterationsschranke wie bei Las-Vegas: \(N = \lceil \varepsilon^{-1} \ln \delta^{-1} \rceil\). Tatsächlich ist das hier äquivalent zur Las-Vegas-Sicht: man kann den einseitig-fehlerhaften MC-Algorithmus als Las-Vegas-Algorithmus auffassen, der '???' (alias 'JA') ausgibt, wenn er nicht sicher ist.
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A&W
Wahr oder falsch?Wenn \(G\) ein zusammenhängender Graph mit ...
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Q: Wahr oder falsch?Wenn \(G\) ein zusammenhängender Graph mit einem maximalen Grad von 100 ist, dann hat \(G\) eine korrekte Färbung mit 100 Farben, es sei denn, \(G\) ist ein vollständiger Graph.
A: Wahr.Satz von Brooks: Für einen zusammenhängenden Graphen gilt \(\chi(G) \leq \Delta(G)\), ausser \(G\) ist ein vollständiger Graph oder ein ungerader Kreis. Bei \(\Delta = 100\) ist \(G\) sicher kein ungerader Kreis (dort ist \(\Delta = 2\)), also genügen 100 Farben, sofern \(G\) nicht vollständig ist.
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c1
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die Berechnung von \(low[]\)
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Cloze c1
Cloze answer: die Berechnung von \(low[]\)
Q: Die um {{c1::die Berechnung von \(low[]\)}} ergänzte {{c2::Tiefensuche}} berechnet in einem zusammenhängenden Graphen alle Artikulationsknoten und Brücken in Zeit \(O({{c3::m}})\).
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c1
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Sei \(G = (V, E)\) ein zusammenhängender Graph. Der {{c4::Bl...
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Cloze c1
Q: Sei \(G = (V, E)\) ein zusammenhängender Graph. Der {{c4::Block-Graph}} von \(G\) ist der bipartite Graph \(T = (A \uplus B, E_T)\) mit\(A = {{c1::\{\text{Artikulationsknoten von } G\} }}\).
\(B = {{c2::\{\text{Blöcke von } G\} }}\).
\(\forall a \in A, b \i
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c1
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\chi(G) \leq 2
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Cloze c1
Cloze answer: \chi(G) \leq 2
Q: Spezialfall: \({{c1::\chi(G) \leq 2}}\iff G\) bipartit
A: „Ist G bipartit?" kann man in Zeit \(O(|E|)\) mit Breiten- oder Tiefensuche beantworten.
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nid:1774631269382
c2
A&W
Hopcroft-Karp findet in einem {{c1::bipartiten}} Graphen in ...
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Cloze c2
Q: Hopcroft-Karp findet in einem {{c1::bipartiten}} Graphen in \(O({{c2::\sqrt{|V|} \cdot |E|}})\) ein {{c3::maximales Matching}}.
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| lorenz | cid:1774999768204 |
1 | 230% | 52d | 9 |
nid:1774917592774
c1
A&W
Für die Varianz gilt: \[\mathbb{E}[(X - \mu)^2] = {{c1::\sum...
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Cloze c1
Q: Für die Varianz gilt: \[\mathbb{E}[(X - \mu)^2] = {{c1::\sum_{x \in W_X} (x - \mu)^2 \cdot \Pr[X = x]::\text{Summe} }}\]
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|---|---|---|---|---|---|
| lorenz | cid:1774917592774 |
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nid:1774358482736
c1
A&W
Multiplikationssatz: Seien \(A_1, \ldots, A_n\) Ereignisse.
...
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Cloze c1
Q: Multiplikationssatz: Seien \(A_1, \ldots, A_n\) Ereignisse.
Falls \(\Pr[A_1 \cap \cdots \cap A_n] > 0\), so gilt: \[\begin{align} \Pr[A_1 \cap \cdots \cap A_n] =& {{c1::\Pr[A_1] \cdot \Pr[A_2|A_1] \\ &\cdot \Pr[A_3|A_1 \cap A_2] \cdots \\ &\Pr[A_n|A_1 \cap \c
A: Proof: Expand each conditional probability by definition:\[ \Pr[A_1]\cdot\frac{\Pr[A_1\cap A_2]}{\Pr[A_1]}\cdot\frac{\Pr[A_1\cap A_2\cap A_3]}{\Pr[A_1\cap A_2]}\cdots\frac{\Pr[A_1\cap\cdots\cap A_n]}{\Pr[A_1\cap\cdots\cap A_{n-1}]}. \]All intermediate terms cancel (telescoping product), leaving \(\Pr[A_1\cap\cdots\cap A_n]\). \(\square\)Note: All conditional probabilities are well-defined because \(\Pr[A_1]\ge\Pr[A_1\cap A_2]\ge\cdots>0\).
Multiplikationssatz
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|---|---|---|---|---|---|
| lorenz | cid:1774358482737 |
1 | 230% | 70d | 9 |
nid:1774917592720
c1
A&W
Die Grösse \(\sigma := {{c1::\sqrt{\operatorname{Var}[X]} }}...
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nid:1774917592720
Cloze c1
Q: Die Grösse \(\sigma := {{c1::\sqrt{\operatorname{Var}[X]} }}\) heisst {{c2::Standardabweichung von \(X\)}}.
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| lorenz | cid:1774917592720 |
1 | 230% | 58d | 9 |
nid:1774917594111
c2
A&W
\(k\)-te zentrale Moment
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Cloze c2
Cloze answer: \(k\)-te zentrale Moment
Q: Für eine Zufallsvariable \(X\) nennen wir \(\mathbb{E}[X^k]\) das {{c1::\(k\)-te Moment}} und \(\mathbb{E}[(X - \mathbb{E}[X])^k]\) das {{c2::\(k\)-te zentrale Moment}}.
A: Der Erwartungswert ist also das erste Moment.
| User | Card ID | Lapses | Ease | Interval | Reviews |
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| lorenz | cid:1774917594112 |
1 | 230% | 52d | 9 |
nid:1774631276956
c3
A&W
n
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nid:1774631276956
Cloze c3
Cloze answer: n
Q: Für den Binomialkoeffizienten gelten:\(\binom{n}{0} = {{c1::1}}\)\(\binom{n}{n} = {{c2::1}}\)\(\binom{n}{1} = {{c3::n}}\)
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| lorenz | cid:1775000929043 |
1 | 230% | 68d | 9 |
nid:1771360670876
c1
A&W
\(k\)-zusammenhängend
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Cloze c1
Cloze answer: \(k\)-zusammenhängend
Q: Ein Graph \(G = (V, E)\) heisst {{c1::\(k\)-zusammenhängend}}, falls {{c2::\(|V| \geq k + 1\) und für alle Teilmengen \(X \subseteq V\) mit \(|X| < k\) gilt: Der Graph \(G[V \setminus X]\) ist zusammenhängend}}.
A: Man muss mindestens \(k\)-Knoten (und die inzidenten Kanten) löschen, um den Zusammenhang zu zerstören.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771360670877 |
1 | 230% | 60d | 10 |
nid:1771363498414
c2
A&W
Blöcke
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nid:1771363498414
Cloze c2
Cloze answer: Blöcke
Q: Sei \(G = (V, E)\). Wir definieren eine {{c3::Äquivalenzrelation}} auf \(E\) durch
\[{{c1::e \sim f :\iff \begin{cases} e = f, & \text{oder} \\ \exists \text{ Kreis durch } e \text{ und } f \end{cases} }}\]
Die {{c3::Äquivalenzklassen}} nennen wir {{c2::Blöcke}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771363498415 |
1 | 230% | 62d | 10 |
nid:1774487164608
c2
A&W
Die Anzahl der Möglichkeiten, \(k\) Objekte aus \(n\) Sorten...
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nid:1774487164608
Cloze c2
Q: Die Anzahl der Möglichkeiten, \(k\) Objekte aus \(n\) Sorten mit Zurücklegen zu wählen (Reihenfolge egal, Multiset) ist:\[{{c2::\binom{n + k - 1}{k} }} = {{c1::\frac{(n+k-1)!}{k!\,(n-1)!} }} \]
A: Auch bekannt als „Sterne und Striche“ (Stars and Bars).Beispiel: Wie viele Möglichkeiten, 3 Kugeln aus {rot, blau, grün} mit Zurücklegen zu ziehen?\(\binom{3+3-1}{3} = \binom{5}{3} = 10\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774631269206 |
1 | 230% | 73d | 9 |
nid:1772545581602
c2
A&W
Mit einem Greedy-Algorithmus kann man in Zeit \( O({{c1::|E|...
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nid:1772545581602
Cloze c2
Q: Mit einem Greedy-Algorithmus kann man in Zeit \( O({{c1::|E|}}) \) ein {{c3::inklusionsmaximales}} Matching \( M_{\text{Greedy}} \) bestimmen mit\[{{c2:: |M_{\text{Greedy} }| \geq |M_{\text{max} }| / 2, }}\]wobei \( M_{\text{max}} \) ein kardinalitätsmaximales Matching ist.
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|---|---|---|---|---|---|
| lorenz | cid:1772545581602 |
1 | 230% | 86d | 12 |
nid:1774631269283
c1
A&W
vollkommen wurscht ob unabhängig, du dummbatzi
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Cloze c1
Cloze answer: vollkommen wurscht ob unabhängig, du dummbatzi
Q: Die Linearität der Erwartung hält, wenn \(X_1,\ldots,X_n\) {{c1::vollkommen wurscht ob unabhängig, du dummbatzi}} sind?
| User | Card ID | Lapses | Ease | Interval | Reviews |
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| lorenz | cid:1774631269283 |
1 | 230% | 73d | 9 |
nid:1776174922324
c1
A&W
Sei \(X\) eine Zufallsvariable, die nur nicht-negative Werte...
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nid:1776174922324
Cloze c1
Q: Sei \(X\) eine Zufallsvariable, die nur nicht-negative Werte annimmt. Dann gilt für alle \(t \in \mathbb{R}\) mit \(t > 0\), dass\[{{c1::\Pr\left[X \geq t\right] \leq \frac{\mathbb{E}[X]}{t}.}}\]Oder äquivalent dazu,\[{{c2::\Pr\left[X \geq t \cdot \mathbb{E}[X]\right] \leq \frac{1}{t}.}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1776174922324 |
1 | 230% | 69d | 9 |
nid:1772547951495
c1
A&W
\forall X \subseteq A : |X| \leq |N(X)|
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Cloze c1
Cloze answer: \forall X \subseteq A : |X| \leq |N(X)|
Q: Ein bipartiter Graph \( G = (A \uplus B, E) \) enthält ein Matching \( M \) der Kardinalität \({{c2:: |M| = |A|}} \iff {{c1::\forall X \subseteq A : |X| \leq |N(X)| }}\).
A: (Hall, Heiratssatz)
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|---|---|---|---|---|---|
| lorenz | cid:1772547951497 |
1 | 230% | 90d | 12 |
nid:1772496585226
IO r1
A&W
[Image Occlusion region 1]
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nid:1772496585226
Cloze c1
Q: {{c1::image-occlusion:rect:left=.186:top=.2984:width=.5344:height=.2754}}{{c2::image-occlusion:rect:left=.183:top=.5891:width=.8119:height=.3672}}
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|---|---|---|---|---|---|
| lorenz | cid:1772496585227 |
1 | 230% | 92d | 9 |
nid:1771526674685
c2
A&W
\(v = root\), und \(v\) hat mindestens zwei Kinder im DFS-Ba...
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Cloze c2
Cloze answer: \(v = root\), und \(v\) hat mindestens zwei Kinder im DFS-Baum.
Q: \(v\) ist genau dann Artikulationsknoten, wenn:{{c1::\(v \neq root\), und \(v\) hat ein Kind \(u\) im DFS-Baum mit \(low[u] \geq dfs[v]\)}} oder {{c2::\(v = root\), und \(v\) hat mindestens zwei Kinder im DFS-Baum.}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771526674685 |
1 | 230% | 101d | 10 |
nid:1778839549943
c1
Analysis
gleichmässig
1
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Cloze c1
Cloze answer: gleichmässig
Q: Die Folge \((f_n)_{n \in \mathbb{N}_0}\) mit \(f_n : D \to \mathbb{R}\) konvergiert {{c1::gleichmässig}} gegen \(f : D \to \mathbb{R}\), falls {{c2::für jedes \(\varepsilon > 0\) ein Index \(N\) existiert, sodass für alle \(n \geq N\) und alle \(x \in D\) gilt \[ |f_n(x) - f(x)| < \varepsilon.
A: Entscheidender Unterschied zur punktweisen Konvergenz: \(N\) hängt nicht von \(x\) ab, sondern gilt uniform für alle \(x \in D\) gleichzeitig.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778839549944 |
1 | 230% | 22d | 9 |
nid:1778839549949
c1
Analysis
gleichmässig
1
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users
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nid:1778839549949
Cloze c1
Cloze answer: gleichmässig
Q: Sei \((f_n)_{n \in \mathbb{N}_0}\) eine Folge integrierbarer Funktionen \(f_n : [a,b] \to \mathbb{R}\), welche {{c1::gleichmässig}} gegen \(f : [a,b] \to \mathbb{R}\) konvergiert. Dann ist auch \(f\) integrierbar und es gilt \[ \int_a^b f\, dx = {{c2::\lim_{n \to \infty} \int_a^b f_n\, dx}}. \]
A: Integral und Limes dürfen vertauscht werden, sofern die Konvergenz gleichmässig ist. Bei punktweiser Konvergenz ist diese Vertauschung im Allgemeinen falsch.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778839549949 |
1 | 230% | 23d | 7 |
nid:1772626846633
c1
Analysis
e^x
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Cloze c1
Cloze answer: e^x
Q: \(\forall x \in \mathbb{R}: \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^n ={{c1::e^x}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772626846633 |
1 | 230% | 30d | 12 |
nid:1774917595591
c1
Analysis
Es sei \(X \neq \emptyset\) eine beliebige Menge.Dann ist di...
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Cloze c1
Q: Es sei \(X \neq \emptyset\) eine beliebige Menge.Dann ist die {{c1::Identitätsfunktion \(\text{id}_X\) }} definiert als \[ {{c2:: \text{id}_X(x) = x \ \ \forall x \in X }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774917595592 |
1 | 230% | 6d | 7 |
nid:1779487906843
Analysis
Welche der folgenden Bedingungen impliziert nicht, dass \(f ...
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nid:1779487906843
Q: Welche der folgenden Bedingungen impliziert nicht, dass \(f : \mathbb{R} \rightarrow \mathbb{R}\) stetig ist?Es gibt \(C \geq 0\), so dass \(|f(x) - f(y)| \leq C|x - y|\) für alle \(x, y \in \mathbb{R}\).Es gibt \(C \geq 0\), so dass \(|f(x) - f(y)| \leq C|x - y|\) f
A: (b) Diese Bedingung impliziert keine Stetigkeit. Gegenbeispiel: \(C = 1\), \(f(x) = 0\) für \(x < 0\) und \(f(x) = 1\) für \(x \geq 0\). Diese Funktion ist in \(0\) unstetig, erfüllt aber \(|f(x) - f(y)| \leq 1 \leq |x - y|\) für alle \(x, y\) mit \(|x - y| \geq 1\).(a) ist eine Lipschitz-Bedingung und impliziert Stetigkeit (\(\delta = \varepsilon / C\)). (c) impliziert ebenfalls Stetigkeit (\(\delta = \min\{\varepsilon/C, 1\}\)).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487906843 |
1 | 230% | 13d | 7 |
nid:1778839549911
c2
Analysis
Unterteilungspunkte
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Cloze c2
Cloze answer: Unterteilungspunkte
Q: Sei \([a,b] \subset \mathbb{R}\) ein kompaktes Intervall. Eine {{c1::Zerlegung}} von \([a,b]\) ist eine endliche Menge von Punkten \[ a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_n = b, \quad n \in \mathbb{N}. \] Die \(x_i\) heissen {{c2::Unterteilungspunkte}}.Eine {{c1::Ze
A: Eine Zerlegung erzeugt automatisch eine Partition von \([a,b]\) in die einpunktigen Mengen \(\{x_k\}\) und die offenen Intervalle \((x_{k-1}, x_k)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778839549911 |
1 | 230% | 25d | 7 |
nid:1779267598929
Analysis
Was ist \((1 + i)^{2000}\)?\(\sqrt{2}\,e^{500\pi i}\)\(-2^{1...
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Q: Was ist \((1 + i)^{2000}\)?\(\sqrt{2}\,e^{500\pi i}\)\(-2^{1000}\)\((2i)^{1000}\)\(2^{1000} e^{\pi i/4}\)\(2^{2000}\)
A: (c) \((2i)^{1000}\).Es gilt \((1+i)^2 = 1 + 2i + i^2 = 2i\), also \((1+i)^{2000} = \big((1+i)^2\big)^{1000} = (2i)^{1000}\).Zudem \((2i)^{1000} = 2^{1000} i^{1000} = 2^{1000}(i^4)^{250} = 2^{1000}\). Der Betrag ist \(\sqrt{2}^{\,2000} = 2^{1000}\), was (a) und (e) ausschliesst.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779267598929 |
1 | 230% | 18d | 10 |
nid:1779487907080
Analysis
Wir betrachten die Differentialgleichung \(u''(x) - e^x + e^...
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Q: Wir betrachten die Differentialgleichung \(u''(x) - e^x + e^u = 0\). Welche der folgenden Charakterisierungen ist korrekt?Es liegt eine homogene Differentialgleichung vor.Die Differentialgleichung ist nicht-linear.Die gegebene Differentialgleichung ist linear.
A: (b) Der Term \(e^u\) enthält die gesuchte Funktion nicht als erste Potenz, also liegt ein nicht-linearer Term in \(u\) vor.(a) ist falsch (der Term \(e^x\) ist eine Inhomogenität). (c) ist falsch wegen \(e^u\). (d) ist falsch, da der Eulersche Ansatz lineare DGl mit konstanten Koeffizienten voraussetzt.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487907081 |
1 | 230% | 13d | 7 |
nid:1774138448160
c1
Analysis
es gibt unendlich viele Folgeglieder, welche im Intervall \(...
1
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nid:1774138448160
Cloze c1
Cloze answer: es gibt unendlich viele Folgeglieder, welche im Intervall \((A - \epsilon, A + \epsilon)\) liegen
Q: Sei \(A\) ein Häufungspunkt der Folge \((a_n)_{n \in \mathbb{N}_0}\). Für jedes \(\epsilon > 0\) gilt {{c1::es gibt unendlich viele Folgeglieder, welche im Intervall \((A - \epsilon, A + \epsilon)\) liegen}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774138448160 |
1 | 230% | 22d | 8 |
nid:1774487165442
Analysis
Warum darf man bei Doppelreihen nicht einfach die Summations...
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nid:1774487165442
Q: Warum darf man bei Doppelreihen nicht einfach die Summationsreihenfolge tauschen?
A: Ohne die Bedingung, dass \(\sum_{i=0}^m \sum_{j=0}^m |a_{ij}| \leq B\) für alle \(m\), kann die Reihenfolge das Ergebnis ändern. Das ist im Wesentlichen die Forderung, dass die Doppelreihe absolut konvergiert.Gegenbeispiel:\[a_{ij} = \begin{cases} 1 & j = i \\ -1 & j = i+1 \\ 0 & \text{sonst} \end{cases}\]Zeilensummen: jede Zeile \(= 0\) → Gesamt \(= 0\)Spaltensummen: erste Spalte \(= 1\), Rest \(= 0\) → Gesamt \(= 1\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487165442 |
1 | 230% | 25d | 15 |
nid:1777924043306
c1
Analysis
Allgemeine Lösung (nur reelle Nullstellen): Hat das charakte...
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nid:1777924043306
Cloze c1
Q: Allgemeine Lösung (nur reelle Nullstellen): Hat das charakteristische Polynom einer linearen, homogenen DGl mit konstanten Koeffizienten ausschliesslich reelle Nullstellen \(\lambda_i\) mit Multiplizität \(m_i\) (\(i = 1, \dots, r\)), so ist die allgemeine Lösung\[ {{c1::u(x) = \sum_{
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924043306 |
1 | 230% | 27d | 9 |
nid:1779798962604
c1
Analysis
K(x)\,e^x
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Cloze c1
Cloze answer: K(x)\,e^x
Q: Variation der Konstanten: Beispiel \(y' - y = x\)Die homogene DGl \(y' - y = 0\) hat die allgemeine Lösung \(y_h(x) = K e^x\).Der Ansatz für die partikuläre Lösung (Variation der Konstanten) lautet daher\[ y_p(x) = {{c1::K(x)\,e^x}} \]
A: Die Konstante \(K\) wird durch die Funktion \(K(x)\) ersetzt.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779798962604 |
1 | 230% | 19d | 9 |
nid:1774487165521
c1
Analysis
\(\sum a_n\) divergiert sicher, wenn {{c1::\(\lim_{n\to\inft...
1
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nid:1774487165521
Cloze c1
Q: \(\sum a_n\) divergiert sicher, wenn {{c1::\(\lim_{n\to\infty} a_n \neq 0\) (d.h. \((a_n)\) ist keine Nullfolge)}}.
A: Beachte: \(\lim a_n = 0\) ist notwendig für Konvergenz, aber nicht hinreichend.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487165521 |
1 | 230% | 19d | 8 |
nid:1779487906803
Analysis
Sei \(f : D \rightarrow \mathbb{R}\) mit \(D \subseteq \math...
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Q: Sei \(f : D \rightarrow \mathbb{R}\) mit \(D \subseteq \mathbb{R}\). Welche der folgenden Aussagen ist äquivalent zur Stetigkeit von \(f\)?Für alle \(x \in D\) und \(\varepsilon > 0\) existiert ein \(\delta > 0\), so dass für alle \(z \in D\) gilt: \(z \in (x - \delta, x +
A: (a) Das ist genau die Stetigkeitsdefinition aus der Vorlesung, nur in Intervallschreibweise.(b) ist falsch, weil die Quantoren in der falschen Reihenfolge stehen. (c) ist echt stärker als Stetigkeit (das \(\delta\) gilt gleichzeitig für alle \(x\)) und beschreibt die gleichmässige Stetigkeit.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487906803 |
1 | 230% | 25d | 7 |
nid:1774138448149
c1
Analysis
1
1
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nid:1774138448149
Cloze c1
Cloze answer: 1
Q: Für alle Polynome \(P(n)\) mit \(P(n) > 0\), gilt für grosse \(n\): \[ \lim_{n \rightarrow \infty} \sqrt[n]{P(n)} = {{c1:: 1}} \]
A: (Die Wurzel dämpft diese vollständig ab.)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774138448149 |
1 | 230% | 43d | 12 |
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c1
Analysis
Integration durch Substitution.Mit \(y = g(x)\) gilt:\[ \int...
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Cloze c1
Q: Integration durch Substitution.Mit \(y = g(x)\) gilt:\[ \int f'(g(x))\, g'(x)\, dx = {{c1::\int \frac{df}{dy}\, dy}} = {{c2::f(g(x)) + C}}. \]
A: Herleitung: Aus der Kettenregel folgt \(\tfrac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)\). Beidseitiges Integrieren und formale Substitution \(dy = g'(x)\, dx\) führen auf die Formel.Substitution ist die Umkehrung der Kettenregel.
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nid:1777381485059
c4
Analysis
Mittelwert
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Cloze c4
Cloze answer: Mittelwert
Q: Eine Sinusschwingung hat die allgemeine Form \[ f(t) = A \sin(B(t + h)) + C \quad \text{oder} \quad f(t) = A \cos(B(t + h)) + C \]Dabei sind:\(|A|\) die {{c1::Amplitude}}\(|B|^{-1} \cdot 2\pi\) die {{c2::Periode}}\(|h|\) die {{c3::Phasenverschiebung}}\(|C|\)
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nid:1779798962565
c2
Analysis
die unabhängige Variable \(x\) nicht mehr explizit vorkommt
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Cloze c2
Cloze answer: die unabhängige Variable \(x\) nicht mehr explizit vorkommt
Q: Nach der linearen Substitution \(u = ax + by + c\) erhält man aus \(y' = f(ax + by + c)\)\[ \frac{du}{dx} = {{c1::a + b\,f(u)}} \]Da hier {{c2::die unabhängige Variable \(x\) nicht mehr explizit vorkommt}}, nennt man eine solche DGl {{c3::autonom}}.
A: Wegen \(\frac{du}{dx} = a + b\frac{dy}{dx} = a + b\,f(u)\).
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nid:1774138447415
Analysis
Trick: Rationalisieren
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Q: Trick: Rationalisieren
A: Binomische Formel \(a^2 - b^2 = (a - b)(a + b)\). Multipliziere die Gleichung mit \(\dots \times 1 = \dots \times \frac{\sqrt{n} + \sqrt{n + 1}}{\sqrt{n} - \sqrt{n + 1}}\). Beispiel: \({\sqrt{n^2 + 3} - n} \cdot 1 = \sqrt{n^2 + 3} - n \cdot \frac{\sqrt{n^2 + 3} + n}{\sqrt{n^2 + 3} + n}\) und dann mit \(a^2 - b^2\) vereinfachen.
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nid:1776290100388
c1
Analysis
\(\mathbb{Q}\) ist dicht in \(\mathbb{R}\) also {{c1::existi...
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Cloze c1
Q: \(\mathbb{Q}\) ist dicht in \(\mathbb{R}\) also {{c1::existiert eine Folge \((a_n) \to x\), \((a_n) \subset \mathbb{Q}\) für alle \(x \in \mathbb{R}\)::Folge}}.Proof Included
A: Äquivalent: Für alle \(a, b \in \mathbb{R}\) mit \(a < b\) existiert ein \(q \in \mathbb{Q}\) mit \(a < q < b\).Beweis: Sei \(x \in \mathbb{R}\). Für jedes \(n \in \mathbb{N}\) wähle \(q_n \in \mathbb{Q}\) mit \[x < q_n < x + \frac{1}{n}\]was nach der archimedischen Eigenschaft und der Existenz rationaler Zahlen zwischen je zwei reellen Zahlen möglich ist. Dann gilt \(|q_n - x| < \frac{1}{n}\), also \(q_n \to x\).
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nid:1777383153522
c1
Analysis
rechtsseitige Ableitung; linksseitige Ableitung
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Cloze c1
Cloze answer: rechtsseitige Ableitung; linksseitige Ableitung
Q: Falls der Grenzwert\[ \lim_{\substack{h \to 0 \\ h > 0}} \frac{f(a+h) - f(a)}{h} \]existiert, nennt man diesen die {{c1::rechtsseitige Ableitung}} von \(f\) an der Stelle \(a\). Analog ist die {{c1::linksseitige Ableitung}} definiert (mit \(h < 0\)).
A: \(f\) ist an der Stelle \(a\) genau dann differenzierbar, wenn beide einseitigen Ableitungen existieren und übereinstimmen.
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nid:1776774733437
c1
Analysis
\(x_0 \in \mathbb{R}\) ist ein Häufungspunkt eines Intervall...
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Cloze c1
Q: \(x_0 \in \mathbb{R}\) ist ein Häufungspunkt eines Intervalls \(D\) falls gilt {{c1::\[ \forall \epsilon > 0 \quad ((x_0 - \epsilon, x_0 + \epsilon) \setminus \{x_0\}) \cap D \neq \emptyset \]}}
A: Jedes Intervall um \(x_0\) hat mindestens einen Punkt, der nicht \(x_0\) ist.
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nid:1779487907186
Analysis
Ein Körper der Masse \(m\) wird mit Anfangsgeschwindigkeit \...
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Q: Ein Körper der Masse \(m\) wird mit Anfangsgeschwindigkeit \(v_0\) senkrecht in die Höhe geworfen. Der Luftwiderstand sei proportional zur Geschwindigkeit (positive Konstante \(k\)).Wie lautet die Bewegungsgleichung? (Hinw
A: (d) Mit \(F = m \tfrac{dv}{dt}\) und \(F = -mg - kv\) (beide Kräfte zeigen nach unten) folgt \(-mg - kv = m \tfrac{dv}{dt}\), also \(\tfrac{dv}{dt} + \tfrac{k}{m} v = -g\).(a), (b), (c) haben falsche Vorzeichen bei den Kräfte-Richtungen.
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nid:1779798962565
c1
Analysis
a + b\,f(u)
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Cloze c1
Cloze answer: a + b\,f(u)
Q: Nach der linearen Substitution \(u = ax + by + c\) erhält man aus \(y' = f(ax + by + c)\)\[ \frac{du}{dx} = {{c1::a + b\,f(u)}} \]Da hier {{c2::die unabhängige Variable \(x\) nicht mehr explizit vorkommt}}, nennt man eine solche DGl {{c3::autonom}}.
A: Wegen \(\frac{du}{dx} = a + b\frac{dy}{dx} = a + b\,f(u)\).
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nid:1774917594698
c1
Analysis
x \mapsto f(x) \quad \forall x \in D'
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Cloze c1
Cloze answer: x \mapsto f(x) \quad \forall x \in D'
Q: Die Einschränkung (Restriktion) von \(f: \mathbb{D}(f) \to \mathbb{R}\) auf \(D' \subset \mathbb{D}(f)\) ist:\[ f\mid_{D'} : D' \to \mathbb{R}, \quad {{c1::x \mapsto f(x) \quad \forall x \in D'}}\]Gleiche Zuordnung, aber nur auf der Teilmenge \(D'\) definiert.
A: Man beachte, dass \(f\) und \(f\mid_{D'}\) a priori zwei verschiedene Funktionen sind.
Beispiel \(\overline{f} : \mathbb{R}^+_0 \rightarrow \mathbb{R}^+_0\) \(f(x) = x^2\) ist bijektiv.
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nid:1779267598950
Analysis
Theorie Grenzwerte IV (Nullfolge = Folge gegen \(0\)): Welch...
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Q: Theorie Grenzwerte IV (Nullfolge = Folge gegen \(0\)): Welche Aussagen sind wahr?Falls \((a_n)\) konvergiert, dann ist \((a_{n+1} - a_n)\) eine NullfolgeFalls \((a_{n+1} - a_n)\) eine Nullfolge ist, konvergiert \((a_n)\)Jede Nullfolge ist beschränkt
A: (a) und (c) sind wahr.(a): \(\lim a_{n+1} = \lim a_n\), also \(\lim(a_{n+1} - a_n) = 0\).(c): Jede konvergente Folge ist beschränkt, unabhängig vom Grenzwert.(b) ist falsch: \(a_n = \sum_{k=1}^n \tfrac{1}{k}\) erfüllt \(a_{n+1} - a_n \to 0\), divergiert aber.
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nid:1771973928557
c1
Analysis
Reflexiv; Transitiv; Antisymmetrisch; Total
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Cloze c1
Cloze answer: Reflexiv; Transitiv; Antisymmetrisch; Total
Q: Auf den reellen Zahlen ist die Ordnung \(\le\): {{c1:: Reflexiv}}{{c1:: Transitiv}}{{c1:: Antisymmetrisch}}{{c1:: Total}}
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nid:1779798962533
c1
Analysis
Trennung der Variablen; Substitution; Variation der Konstant...
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Cloze c1
Cloze answer: Trennung der Variablen; Substitution; Variation der Konstanten
Q: Lösungstechniken für DifferentialgleichungenDGl erster Ordnung:{{c1::Trennung der Variablen}}{{c1::Substitution}}{{c1::Variation der Konstanten}} (lineare, inhomogene DGl)Lineare, inhomogene DGl beliebiger Ordnung:{{c2::geeignete A
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1 | 230% | 27d | 7 |
nid:1778839549918
c1
Analysis
Sei \(f : [a,b] \to \mathbb{R}\) eine Treppenfunktion bezügl...
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Cloze c1
Q: Sei \(f : [a,b] \to \mathbb{R}\) eine Treppenfunktion bezüglich der Zerlegung \(a = x_0 < \cdots < x_n = b\), wobei \(c_k\) der Funktionswert von \(f\) auf \((x_{k-1}, x_k)\) sei. Dann ist das Integral der Treppenfunktion definiert als \[ \int_a^b f(x)\, dx := {{c1::\sum_{k=1}^{n} c
A: Anschaulich: Summe der vorzeichenbehafteten Rechteckflächen \(c_k \cdot \text{Breite}\) über jedem Teilintervall.
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| lorenz | cid:1778839549918 |
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nid:1777924043310
c1
Analysis
Allgemeine Lösung (mit komplexen Nullstellen): Hat das chara...
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Cloze c1
Q: Allgemeine Lösung (mit komplexen Nullstellen): Hat das charakteristische Polynom \(r\) reelle Nullstellen \(\lambda_i\) (Multiplizität \(m_i\)) und \(s\) Paare komplexer Nullstellen \(a_j \pm i b_j\) (Multiplizität \(m_j\)), so ist die allgemeine Lösung\[ \begin{gathered} u(x) = \sum_
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| lorenz | cid:1777924043310 |
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nid:1778839549949
c2
Analysis
Sei \((f_n)_{n \in \mathbb{N}_0}\) eine Folge integrierbarer...
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Cloze c2
Q: Sei \((f_n)_{n \in \mathbb{N}_0}\) eine Folge integrierbarer Funktionen \(f_n : [a,b] \to \mathbb{R}\), welche {{c1::gleichmässig}} gegen \(f : [a,b] \to \mathbb{R}\) konvergiert. Dann ist auch \(f\) integrierbar und es gilt \[ \int_a^b f\, dx = {{c2::\lim_{n \to \infty} \int_a^b f_n\, dx}}. \]
A: Integral und Limes dürfen vertauscht werden, sofern die Konvergenz gleichmässig ist. Bei punktweiser Konvergenz ist diese Vertauschung im Allgemeinen falsch.
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nid:1779267598971
Analysis
Sei \(\sum_{n=1}^{\infty} a_n\) eine Reihe. Welche Aussagen ...
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Q: Sei \(\sum_{n=1}^{\infty} a_n\) eine Reihe. Welche Aussagen sind richtig?\(\sum a_n\) konvergiert, falls \(\lim_{n\to\infty} a_n = 0\)\(\sum a_n\) konvergiert, falls die Folge \((S_m)\) der Partialsummen \(S_m = \sum_{n=1}^m a_n\) konvergiertFalls \(\sum b_n\) konv
A: (b) ist richtig - das ist gerade die Definition der Konvergenz einer Reihe.(a) ist falsch: \(\sum \tfrac{1}{n}\) (harmonische Reihe) divergiert trotz \(a_n \to 0\).(c) ist falsch: z.B. \(b_n = \tfrac{1}{2^n}\), \(a_n = 1\); dann \(0 \leq b_n \leq a_n\) und \(\sum b_n\) konvergiert, aber \(\sum a_n\) divergiert (die Abschätzung läuft in die falsche Richtung).
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nid:1774487165599
c1
Analysis
Sei \(\sum a_n\) {{c1::absolut konvergent und \(\phi: \mathb...
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Cloze c1
Q: Sei \(\sum a_n\) {{c1::absolut konvergent und \(\phi: \mathbb{N}_0 \to \mathbb{N}_0\) eine Bijektion}}.Dann {{c2::konvergiert \(\sum a_{\phi(n)}\) ebenfalls absolut und:\[\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty a_{\phi(n)}\]}}
A: Umordnungssatz für absolut konvergente Reihen (Dirichlet)Merke: Bei absolut konvergenten Reihen darf man frei umordnen.
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nid:1777383153607
c4
Analysis
Ableitungen der Umkehrfunktionen:\(\ln'(x) = {{c1::\dfrac{1}...
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Cloze c4
Q: Ableitungen der Umkehrfunktionen:\(\ln'(x) = {{c1::\dfrac{1}{x} }}\)\(\arcsin'(x) = {{c2::\dfrac{1}{\sqrt{1 - x^2} } }}\)\(\arccos'(x) = {{c3::\dfrac{-1}{\sqrt{1 - x^2} } }}\)\(\arctan'(x) = {{c4::\dfrac{1}{1 + x^2} }}\)
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nid:1780146657589
Analysis
Welchen Wert hat das Integral \(\int_{-1}^{1} |x|\,dx\)?\(0\...
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Q: Welchen Wert hat das Integral \(\int_{-1}^{1} |x|\,dx\)?\(0\).\(\tfrac{1}{2}\).\(1\).\(2\).
A: (c) Der Wert ist \(1\).\(\int_{-1}^{1} |x|\,dx = 2\int_{0}^{1} x\,dx = 2 \cdot \tfrac{1}{2} = 1\).
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nid:1779267598981
Analysis
Was ist das grösste Intervall, welches \(1\) enthält und auf...
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Q: Was ist das grösste Intervall, welches \(1\) enthält und auf welchem \(\sin x\) bijektiv auf das Bild ist?\([-\pi, \pi]\)\([0, \pi]\)\([0, \tfrac{\pi}{2}]\)\([-1, \tfrac{\pi}{2}]\)\([-\tfrac{\pi}{2}, \tfrac{\pi}{2}]\)
A: (e) \([-\tfrac{\pi}{2}, \tfrac{\pi}{2}]\).An der Stelle \(x = 1 < \tfrac{\pi}{2}\) ist \(\sin x\) strikt monoton steigend. Das grösste Intervall um \(x=1\), auf dem \(\sin x\) monoton (und damit injektiv) ist, ist \([-\tfrac{\pi}{2}, \tfrac{\pi}{2}]\): die Grenzen sind Minimal- bzw. Maximalstelle. Eine Funktion ist genau dann injektiv, wenn sie bijektiv auf ihr Bild abbildet.(a), (b) sind nicht injektiv; (c), (d) sind injektiv, aber nicht das grösste solche Intervall.
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nid:1774487165594
c1
Analysis
Seien \(a_n, b_n > 0\). Dann:{{c1::\(\lim \frac{a_n}{b_n} = ...
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Cloze c1
Q: Seien \(a_n, b_n > 0\). Dann:{{c1::\(\lim \frac{a_n}{b_n} = g\) mit \(0 < g < \infty\)}} \(\implies\) \(\sum a_n\) und \(\sum b_n\) haben dasselbe Konvergenzverhalten{{c2::\(\lim \frac{a_n}{b_n} = 0\) und&nbs
A: Grenzwertkriterium (Limitenvergleich)Beispiel:\[\sum \frac{1}{n^2+3n}\]Vergleich mit \(1/n^2\), Grenzwert \(= 1\) → konvergiert.Proof Sketch
Ist \(\lim_{n \to \infty} \frac{a_n}{b_n} = g\) mit \(0 < g < \infty\)
So gilt \(\frac{a_n}{b_n} \leq g + \varepsilon\) und daher \(a_n \leq (g + \varepsilon) \, b_n\) für ein geeignetes \(\varepsilon > 0\) und alle genügend grossen \(n\).
Nach dem Majorantenkri
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nid:1774487165301
c2
Analysis
Sei \(\sum a_n\) {{c1::bedingt konvergent und \(L \in \mathb...
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Cloze c2
Q: Sei \(\sum a_n\) {{c1::bedingt konvergent und \(L \in \mathbb{R} \cup \{+\infty, -\infty\}\)}}.Dann {{c2::gibt es eine Bijektion \(\phi\), so dass:\[\sum_{n=0}^\infty a_{\phi(n)} = L\]}}
A: (Riemannscher Umordnungssatz)Merke: Bedingt konvergente Reihen können durch Umordnung jeden Grenzwert annehmen!
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nid:1779267598965
Analysis
Sei \(\sum_{k\geq1} a_k\) absolut konvergent. Was gilt für \...
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Q: Sei \(\sum_{k\geq1} a_k\) absolut konvergent. Was gilt für \(\sum_{k\geq1} a_k^2\)?konvergiert nicht notwendigerweisekonvergiert immer, aber nicht notwendigerweise absolutkonvergiert immer absolutkeine der obigen Aussagen trifft zu
A: (c) konvergiert immer absolut.Da \(\sum a_k\) konvergiert, ist \((a_k)\) eine Nullfolge, also beschränkt: \(|a_k| \leq C\). Dann \(|a_k|^2 \leq C|a_k|\), und mit dem Vergleichssatz folgt aus der absoluten Konvergenz von \(\sum a_k\) die absolute Konvergenz von \(\sum a_k^2\).
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nid:1777381485126
c2
Analysis
[0,\, \pi]
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Cloze c2
Cloze answer: [0,\, \pi]
Q: Wertebereiche der inversen trigonometrischen Funktionen (Hauptzweige):\(\arcsin : [-1, 1] \to {{c1::[-\pi/2,\, \pi/2]}}\)\(\arccos : [-1, 1] \to {{c2::[0,\, \pi]}}\)\(\arctan : \mathbb{R} \to {{c3::(-\pi/2,\, \pi/2)}}\)
A: Die Einschränkung auf einen Hauptzweig ist nötig, weil \(\sin\), \(\cos\), \(\tan\) als periodische Funktionen nicht injektiv sind und daher auf ihrem ganzen Definitionsbereich keine Inversen besitzen.
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nid:1777383153551
c1
Analysis
Die Menge der glatten Funktionen auf \(D\) ist definiert als...
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Cloze c1
Q: Die Menge der glatten Funktionen auf \(D\) ist definiert als\[ C^\infty(D) := {{c1::\bigcap_{n=0}^{\infty} C^n(D) }} \]Eine Funktion \(f \in C^\infty(D)\) ist also {{c2::beliebig oft stetig differenzierbar}}.
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nid:1772626803535
c6
Analysis
1
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Cloze c6
Cloze answer: 1
Q: \(\lim_{n\to\infty} x^{1/n} = {{c6::1}},\quad x > 0\)
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| lorenz | cid:1772626803536 |
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nid:1777383153627
c1
Analysis
Leibniz-Regel (Produktregel höhere Ordnung): Sind \(f, g\) a...
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Cloze c1
Q: Leibniz-Regel (Produktregel höhere Ordnung): Sind \(f, g\) an \(x_0\) \(n\)-fach differenzierbar, so ist auch \(f \cdot g\) \(n\)-fach differenzierbar mit\[ (f \cdot g)^{(n)}(x_0) = {{c1::\sum_{k=0}^{n} \binom{n}{k} f^{(k)}(x_0)\, g^{(n-k)}(x_0) }} \]
A: Spezialfall \(n = 1\): klassische Produktregel \((fg)' = f'g + fg'\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777383153627 |
1 | 230% | 42d | 11 |
nid:1774138446824
Analysis
Wie kann man einen Ausdruck in die Form von \(e^x\) bringen?
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Q: Wie kann man einen Ausdruck in die Form von \(e^x\) bringen?
A: Via \(\lim_{n \rightarrow \infty} (1 + \frac{x}{n})^n = e^x\).
Beispiel: Zunächst formen wir um: \((\frac{n}{n + 1})^n = (\frac{n + 1}{n})^{-n}\).
Dann trennen wir \((1 + \frac{1}{n})^{-n}\) und extrahieren den Exponenten \(((1 + \frac{1}{n})^n)^{-1}\).
Schliesslich können wir den Limes berechnen und erhalten \(e^{-1}\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774138446825 |
1 | 230% | 74d | 12 |
nid:1774487165318
c1
Analysis
Cauchy-Verdichtungssatz: Sei \((a_n)\) monoton fallend, \(a_...
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Cloze c1
Q: Cauchy-Verdichtungssatz: Sei \((a_n)\) monoton fallend, \(a_n \geq 0\):\[\sum_{n=0}^\infty a_n \text{ conv.} \iff {{c1::\sum_{n=0}^\infty 2^n a_{2^n} \text{ conv.} }}\]Proof Included
A: Anwendung: \(\sum 1/n^s\) für \(s > 1\) konvergiert: \(\sum 2^n \cdot 2^{-ns} = \sum 2^{n(1-s)}\) geometrisch mit \(q = 2^{1-s} < 1\).Proof
Weil \(a_n\) monoton fällt gilt \(2^n a_{2^n} \ge a_{2^k + 1} + a_{2^k + 2} + \dots + a_{2^{k + 1} - 1}\).
Wir benutzen das Majorantenkriterium mit\[\begin{align} \sum^n_{k = 0} 2^k a_{2^k} &\ge \sum_{k = 0}^n (a_{2^k + 1} + a_{2^k + 2} + \dots + a_{2^{k + 1} - 1}) \\ &= \sum^
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|---|---|---|---|---|---|
| lorenz | cid:1774487165318 |
1 | 230% | 57d | 12 |
nid:1779267598947
Analysis
Theorie Grenzwerte III: Sei \((a_n)\) eine Folge in \(\mathb...
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Q: Theorie Grenzwerte III: Sei \((a_n)\) eine Folge in \(\mathbb{R}\). Welche Aussagen sind wahr?Falls \(\varepsilon > 0\) und \(a \in \mathbb{R}\) existieren mit \(|a_n - a| < \varepsilon\ \forall n \geq 1\), dann konvergiert \((a_n)\)Falls \((a_n)\) konvergiert, is
A: (b) und (d) sind wahr.(b): \(b_n \to 2a\) nach Rechenregeln.(d): Monotone und beschränkte Folgen konvergieren.(a) ist falsch: \(a = 0\), \(a_n = (-1)^n\), \(\varepsilon = 2\).(c) ist falsch: \(a_n = \sum_{k=1}^n \tfrac{1}{k}\) erfüllt \(b_n = \tfrac{1}{n+1} \to 0\), divergiert aber (harmonische Reihe).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779267598947 |
1 | 230% | 32d | 8 |
nid:1772928333495
c1
Analysis
\[ \tan\!\left(\frac{2\pi}{3}\right) = {{c1::-\sqrt{3} }} \]
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nid:1772928333495
Cloze c1
Q: \[ \tan\!\left(\frac{2\pi}{3}\right) = {{c1::-\sqrt{3} }} \]
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|---|---|---|---|---|---|
| lorenz | cid:1772928333496 |
1 | 230% | 83d | 11 |
nid:1778839549940
c2
Analysis
Seien \(D \subset \mathbb{R}\), \((f_n)_{n \in \mathbb{N}_0}...
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Cloze c2
Q: Seien \(D \subset \mathbb{R}\), \((f_n)_{n \in \mathbb{N}_0}\) eine Folge von Funktionen \(f_n : D \to \mathbb{R}\) und \(f : D \to \mathbb{R}\). Die Folge \((f_n)\) konvergiert {{c1::punktweise}} gegen \(f\), falls {{c2::für jedes \(x \in D\) die reelle Folge \((f_n(x))_{n \in \mathbb{N}_0}
A: \(f\) heisst dann punktweiser Grenzwert der Folge \((f_n)\). Quantorenreihenfolge: zuerst \(x\), dann \(N(\varepsilon, x)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778839549940 |
1 | 230% | 37d | 8 |
nid:1778839549952
c3
Analysis
Mittelwertsatz der Integralrechnung.Ist \(f : [a,b] \to \mat...
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Cloze c3
Q: Mittelwertsatz der Integralrechnung.Ist \(f : [a,b] \to \mathbb{R}\) {{c1::stetig}}, so existiert ein {{c2::\(c \in [a,b]\)}} mit \[ f(c) = {{c3::\frac{1}{b-a} \int_a^b f(x)\, dx}}. \]
A: Der Ausdruck \(\tfrac{1}{b-a} \int_a^b f(x)\, dx\) ist der Mittelwert von \(f\) über \([a,b]\). Beweis-Idee: Zwischenwertsatz angewandt auf \(f\) zwischen Minimum und Maximum auf \([a,b]\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778839549953 |
1 | 230% | 37d | 8 |
nid:1779267598968
Analysis
Sei \(\sum_{k\geq1} a_k\) absolut konvergent und \(\sum_{k\g...
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nid:1779267598968
Q: Sei \(\sum_{k\geq1} a_k\) absolut konvergent und \(\sum_{k\geq1} b_k\) konvergent. Was gilt für \(\sum_{k\geq1} a_k b_k\)?konvergiert nicht notwendigerweisekonvergiert immer, aber nicht notwendigerweise absolutkonvergiert immer absolut
A: (c) konvergiert immer absolut.\((b_k)\) ist als Nullfolge beschränkt: \(|b_k| \leq C\). Dann \(|a_k b_k| \leq C|a_k|\), und mit dem Vergleichssatz folgt aus der absoluten Konvergenz von \(\sum a_k\) die absolute Konvergenz von \(\sum a_k b_k\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779267598968 |
1 | 230% | 36d | 8 |
nid:1777924043306
c2
Analysis
Allgemeine Lösung (nur reelle Nullstellen): Hat das charakte...
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nid:1777924043306
Cloze c2
Q: Allgemeine Lösung (nur reelle Nullstellen): Hat das charakteristische Polynom einer linearen, homogenen DGl mit konstanten Koeffizienten ausschliesslich reelle Nullstellen \(\lambda_i\) mit Multiplizität \(m_i\) (\(i = 1, \dots, r\)), so ist die allgemeine Lösung\[ {{c1::u(x) = \sum_{
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924043307 |
1 | 230% | 36d | 8 |
nid:1777381484987
c1
Analysis
k \cdot x^p
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Cloze c1
Cloze answer: k \cdot x^p
Q: Eine Funktion der Form \[ y = f(x) = {{c1::k \cdot x^p}} \]mit \(k, p \in \mathbb{R}\) heisst Potenzfunktion.
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|---|---|---|---|---|---|
| lorenz | cid:1777381484987 |
1 | 230% | 48d | 11 |
nid:1771973928579
c1
Analysis
Definition Absolutbetrag: \[|x| := {{c1:: \max\{x, -x\} = \...
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nid:1771973928579
Cloze c1
Q: Definition Absolutbetrag: \[|x| := {{c1:: \max\{x, -x\} = \begin{cases} x, & x \geq 0 \\ -x, & \text{sonst} \end{cases} :: \text{Funktion und Cases} }}\]
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|---|---|---|---|---|---|
| lorenz | cid:1771973928579 |
1 | 230% | 58d | 9 |
nid:1777383738542
c1
Analysis
Multiplikation von Potenzreihen: Seien \(f(x) = \sum_{n=0}^{...
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nid:1777383738542
Cloze c1
Q: Multiplikation von Potenzreihen: Seien \(f(x) = \sum_{n=0}^{\infty} a_n x^n\) und \(g(x) = \sum_{n=0}^{\infty} b_n x^n\) zwei Potenzreihen und \(x\) im Konvergenzintervall beider Reihen. Dann gilt\[ f(x) \cdot g(x) = {{c1::\sum_{n=0}^{\infty} \left(\sum_{k=0}^{n} a_k b_{n-k}\right) x^
A: Anwendung des Cauchy-Produkts auf Potenzreihen. Der Koeffizient von \(x^n\) im Produkt ist die Faltung \(\sum_{k=0}^{n} a_k b_{n-k}\) der Koeffizientenfolgen.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777383738542 |
1 | 230% | 54d | 10 |
nid:1774917595832
c2
Analysis
R
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nid:1774917595832
Cloze c2
Cloze answer: R
Q: Eine Funktion \(f: {{c1::D}} \rightarrow {{c2::R}}\) hat {{c1::einen Definitionsbereich \(\text{domain}(f) = \mathbb{D}(f) = D\)}} und {{c2::einen Wertebereich \(\text{range/image}(f) = R\)}}.
A: Der Input heisst unabhängige Variable (Argument) und der Output abhängige Variable.
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|---|---|---|---|---|---|
| lorenz | cid:1774917595833 |
1 | 230% | 75d | 9 |
nid:1777383153541
c1
Analysis
Eine Funktion \(f : D \to \mathbb{R}\) heisst \(n\)-fach ste...
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nid:1777383153541
Cloze c1
Q: Eine Funktion \(f : D \to \mathbb{R}\) heisst \(n\)-fach stetig differenzierbar, falls {{c1::die \(n\)-ten Ableitungen \(f^{(n)}\) auch noch stetige Funktionen sind}}. Die Menge dieser Funktionen wird mit {{c2::\(C^n(D)\)}} bezeichnet.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777383153542 |
1 | 230% | 51d | 8 |
nid:1773149513656
c1
Analysis
Falls für eine Folge gilt:\[{{c1:: \forall M > 0 \ \exists N...
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nid:1773149513656
Cloze c1
Q: Falls für eine Folge gilt:\[{{c1:: \forall M > 0 \ \exists N > 0 \text{ sodass } \forall n > N \ : \ a_n > M }}\] sagen wir, dass die Folge gegen unendlich divergiert und schreiben \(\lim_{n \rightarrow \infty} a_n = \infty\).
A: Genauso kann die Folge auch gegen \(-\infty\) divergieren.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773149513656 |
1 | 230% | 93d | 13 |
nid:1774487165324
Analysis
Welches Konvergenzkriterium wähle ich wann?
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nid:1774487165324
Q: Welches Konvergenzkriterium wähle ich wann?
A: Notwendiges Kriterium zuerst: \(a_n \to 0\)? Falls nein → divergiert sofort.Geometrisch/direkter Vergleich: Vergleichbar mit \(q^n\) oder \(1/n^s\)?Quotientenkriterium: Terme mit \(n!\), \(a^n\) oder einfachen Quotienten?Wurzelkriterium: Terme der Form \((\cdot)^n\) — mindestens so gut wie Quotient.Leibniz: Alternierende Reihe mit monoton fallenden \(|a_n| \to 0\)?Grenzwertkriterium: Ähnelt asymptotisc
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|---|---|---|---|---|---|
| lorenz | cid:1774487165324 |
1 | 230% | 77d | 12 |
nid:1774487165276
Analysis
Welches Konvergenzkriterium ist stärker: das Wurzel- oder da...
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nid:1774487165276
Q: Welches Konvergenzkriterium ist stärker: das Wurzel- oder das Quotientenkriterium?
A: Das Wurzelkriterium. Liefert der Quotient ein Ergebnis, so auch die Wurzel - aber nicht umgekehrt.
In der Praxis ist das Quotientenkriterium oft bequemer, besonders bei \(n!\) oder Potenzen.
Beide versagen bei \(\rho = 1\), z.B. bei \(p\)-Reihen.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487165276 |
1 | 230% | 89d | 12 |
nid:1777924043253
c4
Analysis
5-Schritt-Methodik zum Aufstellen einer Differentialgleichun...
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nid:1777924043253
Cloze c4
Q: 5-Schritt-Methodik zum Aufstellen einer Differentialgleichung für eine Grösse \(y(t)\):Veränderung in einem kleinen Zeitintervall \(\Delta t\) bilanzieren: {{c1::\(y(t + \Delta t) = y(t) + \text{Zuwachs} - \text{Abnahme}\)}}Differenz bilden: {{c2::\(\Delta y = y(t + \Delta t)
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|---|---|---|---|---|---|
| lorenz | cid:1777924043255 |
1 | 230% | 54d | 11 |
nid:1772928333431
c1
Analysis
\[ \sin\!\left(\frac{3\pi}{4}\right) = {{c1::\frac{\sqrt{2} ...
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nid:1772928333431
Cloze c1
Q: \[ \sin\!\left(\frac{3\pi}{4}\right) = {{c1::\frac{\sqrt{2} }{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333431 |
1 | 230% | 101d | 9 |
nid:1774138446942
c1
Analysis
kleinste Häufungspunkt ; grösste Häufungspunkt
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Cloze c1
Cloze answer: kleinste Häufungspunkt ; grösste Häufungspunkt
Q: \(\liminf_{n \rightarrow \infty} a_n\) ist der {{c1:: kleinste Häufungspunkt }} von \((a_n)\).\(\limsup_{n \rightarrow \infty} a_n\) ist der {{c1:: grösste Häufungspunkt }} von \((a_n)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774138446942 |
1 | 230% | 90d | 9 |
nid:1772928333178
c1
Analysis
\[\tan(x) = {{c1:: \frac{\sin x}{ \cos x} :: \text{in terms...
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nid:1772928333178
Cloze c1
Q: \[\tan(x) = {{c1:: \frac{\sin x}{ \cos x} :: \text{in terms of sin and cos} }}\]
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|---|---|---|---|---|---|
| lorenz | cid:1772928333178 |
1 | 230% | 62d | 9 |
nid:1771973928588
c4
Analysis
Beweis: Für alle \(a < b\) in \(\mathbb{R}\) existiert ein \...
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Cloze c4
Q: Beweis: Für alle \(a < b\) in \(\mathbb{R}\) existiert ein \(\mathbb{Q}\) mit \(a < q < b\){{c1:: Wähle nach Archimedischem Prinzip \(n \in \mathbb{N}\) so dass \(\frac{1}{n} < b - a\).}}{{c2:: \(\frac{m}{n} \mid m \in \mathbb{Z}\) diese
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|---|---|---|---|---|---|
| lorenz | cid:1771973928588 |
1 | 230% | 106d | 8 |
nid:1772928333376
c1
Analysis
\[ \cos\!\left(\frac{7\pi}{6}\right) = {{c1::-\frac{\sqrt{3}...
1
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users
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nid:1772928333376
Cloze c1
Q: \[ \cos\!\left(\frac{7\pi}{6}\right) = {{c1::-\frac{\sqrt{3} }{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333376 |
1 | 230% | 109d | 9 |
nid:1772928333418
c1
Analysis
\[ \sin\!\left(\frac{\pi}{3}\right) = {{c1::\frac{\sqrt{3} }...
1
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users
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nid:1772928333418
Cloze c1
Q: \[ \sin\!\left(\frac{\pi}{3}\right) = {{c1::\frac{\sqrt{3} }{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333418 |
1 | 230% | 102d | 13 |
nid:1777924043303
c2
Analysis
Zum Lösen einer linearen, homogenen DGl mit konstanten Koeff...
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nid:1777924043303
Cloze c2
Q: Zum Lösen einer linearen, homogenen DGl mit konstanten Koeffizienten\[ a_n u^{(n)}(x) + a_{n-1} u^{(n-1)}(x) + \dots + a_1 u'(x) + a_0 u(x) = 0 \]verwendet man den Eulerschen Ansatz:\[ {{c1::u(x) = e^{\lambda x} }} \]Einsetzen liefert die charakteristische Gleichung {{c2::
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924043305 |
1 | 230% | 71d | 8 |
nid:1774917595832
c1
Analysis
D
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users
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Cloze c1
Cloze answer: D
Q: Eine Funktion \(f: {{c1::D}} \rightarrow {{c2::R}}\) hat {{c1::einen Definitionsbereich \(\text{domain}(f) = \mathbb{D}(f) = D\)}} und {{c2::einen Wertebereich \(\text{range/image}(f) = R\)}}.
A: Der Input heisst unabhängige Variable (Argument) und der Output abhängige Variable.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774917595832 |
1 | 230% | 90d | 9 |
nid:1772928333491
c1
Analysis
\[ \tan\!\left(\frac{\pi}{2}\right) = {{c1::\text{undefined}...
1
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users
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nid:1772928333491
Cloze c1
Q: \[ \tan\!\left(\frac{\pi}{2}\right) = {{c1::\text{undefined} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333491 |
1 | 230% | 110d | 12 |
nid:1774487165294
c4
Analysis
WurzelkriteriumSei \(\rho = {{c4:: \limsup_{n\to\infty} |a_n...
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users
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nid:1774487165294
Cloze c4
Q: WurzelkriteriumSei \(\rho = {{c4:: \limsup_{n\to\infty} |a_n|^{1/n} }}\). Dann:\(\rho < 1\) \(\implies\) {{c1::\(\sum a_n\) konvergiert absolut}}\(\rho > 1\) \(\implies\) {{c1::\(\sum a_n\) divergiert}}\(\rho = 1\) \(\implies\) {{c1::keine Au
A: Wenn Quotientenkriterium versagt (\(\rho=1\)), versagt auch das Wurzelkriterium — aber nicht umgekehrt.Proof:
Convergence \(L < 1\)
\(\sum a_n \geq 0\), \(\displaystyle L = \limsup_{n\to\infty} \left| {a_n}^{1/n} \right| < 1\).
Choose \(q\) with \(L < q < 1\). Since \(\limsup \left| {a_n}^{1/n} \right| = L\), there exists \(N\) such that for all \(n \geq N: \left| {a_n}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774917594525 |
1 | 230% | 102d | 9 |
nid:1772928333410
c1
Analysis
\[ \sin\!\left(\frac{\pi}{6}\right) = {{c1::\frac{1}{2} }} \...
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users
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nid:1772928333410
Cloze c1
Q: \[ \sin\!\left(\frac{\pi}{6}\right) = {{c1::\frac{1}{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333410 |
1 | 230% | 125d | 9 |
nid:1772928333327
c1
Analysis
\[ {{c1::\sin^2\theta + \cos^2\theta :: \text{Identity} }} =...
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users
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nid:1772928333327
Cloze c1
Q: \[ {{c1::\sin^2\theta + \cos^2\theta :: \text{Identity} }} = {{c2::1}} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333327 |
1 | 230% | 128d | 9 |
nid:1774487165225
c1
Analysis
konvergente Teilfolgen
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Cloze c1
Cloze answer: konvergente Teilfolgen
Q: Eine divergente Folge kann trotzdem {{c1::konvergente Teilfolgen}} besitzen.
A: Beispiel: \(a_n = (-1)^n\) divergiert, aber \(a_{2n} = 1\) und \(a_{2n+1} = -1\) konvergieren.Umkehrung: Eine Folge konvergiert gegen \(L\) genau dann, wenn jede Teilfolge gegen \(L\) konvergiert.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487165228 |
1 | 230% | 132d | 9 |
nid:1774487165212
c1
Analysis
L'Hôpital, kürzen, Taylor
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nid:1774487165212
Cloze c1
Cloze answer: L'Hôpital, kürzen, Taylor
Q: Form
Strategie
A: (\(0\) und \(\infty\) sind hier Kurzschreibweisen für das Verhalten im Grenzwert: \(0\) steht für „geht gegen \(0\)" und \(\infty\) für „geht gegen \(\infty\)".)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487165212 |
1 | 230% | 141d | 9 |
nid:1772928333518
c1
Analysis
\[ \tan\!\left(\frac{4\pi}{3}\right) = {{c1::\sqrt{3} }} \]
1
lapses
1/4
users
230%
ease
nid:1772928333518
Cloze c1
Q: \[ \tan\!\left(\frac{4\pi}{3}\right) = {{c1::\sqrt{3} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772928333518 |
1 | 230% | 163d | 9 |
nid:1774631279995
c1
PProg
\(T_P \ge T_1 / p\); \(T_P \ge T_\infty\)
1
lapses
1/4
users
230%
ease
nid:1774631279995
Cloze c1
Cloze answer: \(T_P \ge T_1 / p\); \(T_P \ge T_\infty\)
Q: The work law is {{c1::\(T_P \ge T_1 / p\)}} and the span law is {{c1::\(T_P \ge T_\infty\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774631279995 |
1 | 230% | 2d | 6 |
nid:1777924071047
c2
PProg
n++ followed by if (n <= 0) means "after my insert there is ...
1
lapses
1/4
users
230%
ease
nid:1777924071047
Cloze c2
Cloze answer: n++ followed by if (n <= 0) means "after my insert there is at least one consumer still waiting", so signal exactly then, saving the call when n > 0 (no one is waiting)
Q: Sleeping-Barber producer-consumer signalling logic: The decrement-first / check-negative pattern: {{c1::m-- followed by if (m < 0) reserves a slot and detects "buffer was full" before waiting}};
A: The asymmetry < 0 vs <= 0 on the wait side vs the signal side is intentional: at the moment of my own increment, my counter has just become +1 if I was the only waiter, so the condition for "someone else is still waiting" is <= 0.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924071048 |
1 | 230% | 33d | 6 |
nid:1779457826322
c1
PProg
Calls overlap; the object may never be between calls (except...
1
lapses
1/4
users
230%
ease
nid:1779457826322
Cloze c1
Cloze answer: Calls overlap; the object may never be between calls (except periods of quiescence)
Q: Sequential versus concurrent objects: SequentialConcurrentState is meaningful only between method calls.{{c1::Calls overlap; the object may never be between calls (except periods of quiescen
A: A period of quiescence is an interval with no pending calls; only then does a concurrent object have a well-defined sequential-style state.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826325 |
1 | 230% | 7d | 8 |
nid:1771365476397
c1
PProg
Lockout
1
lapses
1/4
users
230%
ease
nid:1771365476397
Cloze c1
Cloze answer: Lockout
Q: {{c1::Lockout}} means {{c2::needlessly preventing a thread from entering a critical section}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771365476398 |
1 | 230% | 33d | 9 |
nid:1777562257075
c2
PProg
synchronised stampede of all waiters trying to acquire simul...
1
lapses
1/4
users
230%
ease
nid:1777562257075
Cloze c2
Cloze answer: synchronised stampede of all waiters trying to acquire simultaneously the moment the lock is released
Q: Exponential-backoff lockWhen a TATAS acquire attempt fails, the thread {{c1::sleeps for a random duration drawn from \([0, \text{limit}]\), then doubles limit up to maxDelay}} before retrying. This removes the {{c2::synchronised stampede of all waiter
A: Empirically this gives near-flat scaling under high contention (vs. linear-or-worse for TAS / TTAS). Trade-offs: a thread may be sleeping when the lock becomes free, so latency under low contention is worse; tuning minDelay / maxDelay is workload-specific.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777562257075 |
1 | 230% | 38d | 6 |
nid:1778588922566
c1
PProg
CAS(last.next, null, new); CAS(tail, last, new)
1
lapses
1/4
users
230%
ease
nid:1778588922566
Cloze c1
Cloze answer: CAS(last.next, null, new); CAS(tail, last, new)
Q: Initial Michael-Scott protocol. Enqueuer: read tail into last; {{c1::CAS(last.next, null, new)}}; on success, without retry try {{c1::CAS(tail, last, new)}}. Dequeuer: read head into fir
A: The asymmetry, enqueuer retries on the first CAS but not the second, is intentional: a failed second CAS means some other thread already advanced tail, so the enqueue's logical effect is already established.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922568 |
1 | 230% | 9d | 8 |
nid:1779457826332
c2
PProg
different points depending on whether the queue is empty (th...
1
lapses
1/4
users
230%
ease
nid:1779457826332
Cloze c2
Cloze answer: different points depending on whether the queue is empty (the call may fail) or non-empty (it does not fail)
Q: Important subtlety about linearization points: they can often be named in the code, but they may depend on {{c1::the execution, not only on the source code}}. Example: a deq() on a bounded queue linearizes at {{c2::different points depending on whether the queue is empty (the ca
A: We reason about executions to abstract away from the implementation, but this abstraction has to be mentally undone when actually analysing a program.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826333 |
1 | 230% | 3d | 5 |
nid:1779457826339
c1
PProg
it has no matching response; it contains no pending invocati...
1
lapses
1/4
users
230%
ease
nid:1779457826339
Cloze c1
Cloze answer: it has no matching response; it contains no pending invocations
Q: Terminology on histories: An invocation is pending if {{c1::it has no matching response}}; A subhistory is complete when {{c1::it contains no pending invocations}}. The operation complete(H) yields {{c2::\(H\) with all its pending invocations removed}}.
A: Pending invocations are exactly the calls still in flight at the end of the recorded history.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826339 |
1 | 230% | 9d | 8 |
nid:1779664155711
c2
PProg
Convoying (a thread holding a resource is descheduled while ...
1
lapses
1/4
users
230%
ease
nid:1779664155711
Cloze c2
Cloze answer: Convoying (a thread holding a resource is descheduled while other threads queue up waiting for it)
Q: Four problems of locks (motivation for transactional memory): {{c1::Deadlocks (threads take shared or dependent locks in different orders)}}; {{c2::Convoying (a thread holding a resource is descheduled while other threads queue up waiting for it)}}; {{c3::Priority inver
A: These are properties of the lock mechanism itself, not of a specific bug. TM aims to remove them by moving synchronisation from the programmer into the system.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779664155712 |
1 | 230% | 4d | 5 |
nid:1779664155858
c1
PProg
can be fast but has bounded resources and often cannot handl...
1
lapses
1/4
users
230%
ease
nid:1779664155858
Cloze c1
Cloze answer: can be fast but has bounded resources and often cannot handle big transactions
Q: Hardware TM (HTM): {{c1::can be fast but has bounded resources and often cannot handle big transactions::Pros and Cons}}. The first widely available x86 implementation was {{c2::Intel's Haswell (RTM)}}, which was largely removed soon after.
A: The Haswell instructions are xbegin (begin transaction), xend (end), xabort (abort).Other HTM examples: Sun/Oracle Rock (never released), IBM Blue Gene/Q (long retired). Pattern: xbegin L0 / transaction code / xend; on abort, execution jumps to L0.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779664155860 |
1 | 230% | 3d | 5 |
nid:1779664155878
c1
PProg
strong isolation; weak isolation
1
lapses
1/4
users
230%
ease
nid:1779664155878
Cloze c1
Cloze answer: strong isolation; weak isolation
Q: Design choice strong vs. weak isolation, concerning shared state accessed by a transaction that is also accessed outside a transaction: With {{c1::strong isolation}} the transactional guarantees still hold. It is {{c2::easier for porting existing code, but difficult to implement and incu
A: Strong isolation: transactional and non-transactional accesses are isolated against each other. Weak isolation leaves consistency to the programmer but is cheaper.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779664155880 |
1 | 230% | 3d | 5 |
nid:1777924071035
c4
PProg
always needs to re-check the condition (use while, not if)
1
lapses
1/4
users
230%
ease
nid:1777924071035
Cloze c4
Cloze answer: always needs to re-check the condition (use while, not if)
Q: A Java Condition (obtained via lock.newCondition()) offers {{c1::await()}}: must be called with the lock held; {{c2::atomically releases the lock and waits until the thread is signalled}}; on return, the lock is {{c3::gua
A: Crucial difference from wait/notify on intrinsic locks: a single lock can have multiple conditions, so producers and consumers can wait on different ones (notFull, notEmpty) and the right kind of waiter can be woken directly without notifying all.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924071039 |
1 | 230% | 12d | 6 |
nid:1777984596181
c2
PProg
writers and readers exclude each other
1
lapses
1/4
users
230%
ease
nid:1777984596181
Cloze c2
Cloze answer: writers and readers exclude each other
Q: A reader/writer lock has three states: {{c1::not held}}, {{c1::held for writing by exactly one thread}}, and {{c1::held for reading by one or more threads}}. The associated invariants are \(0 \leq \texttt{writers} \leq 1\), \(0 \leq
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596181 |
1 | 230% | 13d | 6 |
nid:1777984596236
c2
PProg
writersWaiting: number of writers trying to enter the CS; re...
1
lapses
1/4
users
230%
ease
nid:1777984596236
Cloze c2
Cloze answer: writersWaiting: number of writers trying to enter the CS; readersWaiting: number of readers trying to enter the CS
Q: The FIFO-fair RW lock keeps five counters:{{c1::writers: number of writers in the CS (\(\leq 1\))}}; {{c1::readers: number of readers in the CS}}; {{c2::writersWaiting: number of writers trying to enter the CS}}; {{c2::
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596237 |
1 | 230% | 13d | 6 |
nid:1779457826309
c3
PProg
a hybrid: thread-local pool plus a global pool guarded by ha...
1
lapses
1/4
users
230%
ease
nid:1779457826309
Cloze c3
Cloze answer: a hybrid: thread-local pool plus a global pool guarded by hazard pointers
Q: The ABA problem also affects the node pool itself. Three remedies: {{c1::thread-local node pools: no protection needed, but they fail when push/pop are not well balanced (one thread leaks, another starves)}};{{c2::hazard pointers on the global pool: expensive on reuse, equivalent to
A: Thread-local storage avoids contention entirely but is unsafe under imbalance; the hybrid captures the best of both.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826310 |
1 | 230% | 4d | 5 |
nid:1779457826312
c2
PProg
thread-local storage; processor-local storage
1
lapses
1/4
users
230%
ease
nid:1779457826312
Cloze c2
Cloze answer: thread-local storage; processor-local storage
Q: Remarks on the hazard-pointer stack: The safe Java version does {{c1::not actually improve performance over plain allocation plus garbage collection, it merely demonstrates how to solve ABA in principle}}. The hazard pointers live in {{c2::thread-local storage}}, and the scheme
A: Höfler's reference to Florian Negele's PhD thesis (ETH Zürich, 2014) on combining lock-free programming with cooperative multitasking.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826312 |
1 | 230% | 4d | 5 |
nid:1779457826312
c1
PProg
not actually improve performance over plain allocation plus ...
1
lapses
1/4
users
230%
ease
nid:1779457826312
Cloze c1
Cloze answer: not actually improve performance over plain allocation plus garbage collection, it merely demonstrates how to solve ABA in principle
Q: Remarks on the hazard-pointer stack: The safe Java version does {{c1::not actually improve performance over plain allocation plus garbage collection, it merely demonstrates how to solve ABA in principle}}. The hazard pointers live in {{c2::thread-local storage}}, and the scheme
A: Höfler's reference to Florian Negele's PhD thesis (ETH Zürich, 2014) on combining lock-free programming with cooperative multitasking.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826314 |
1 | 230% | 8d | 8 |
nid:1779487674914
c1
PProg
take effect in their real-time order
1
lapses
1/4
users
230%
ease
nid:1779487674914
Cloze c1
Cloze answer: take effect in their real-time order
Q: Quiescent consistency:Method calls separated by a period of quiescence (an interval with no pending calls) must {{c1::take effect in their real-time order}}; overlapping calls or calls not separated by quiescence {{c2::may be reordered arbitrarily}}. Relative to sequential consistenc
A: An example can be sequentially consistent but not quiescently consistent.An example can be quiescently consistent but not sequentially consistent (without a quiescent period, even deq and enq may be swapped). Unlike SC, quiescent consistency does not require preserving the program order within a thread.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487674915 |
1 | 230% | 5d | 5 |
nid:1777984596418
c2
PProg
neither node can currently be in the process of being delete...
1
lapses
1/4
users
230%
ease
nid:1777984596418
Cloze c2
Cloze answer: neither node can currently be in the process of being deleted
Q: Correctness argument for optimistic remove(c): Given that b and c {{c1::are both locked}}, b {{c1::is still reachable from head}}, and c {{c1::is still b's successor}}, {{c2::neither nod
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596420 |
1 | 230% | 13d | 6 |
nid:1777984596431
c4
PProg
the list has to be traversed twice (search + validation)
1
lapses
1/4
users
230%
ease
nid:1777984596431
Cloze c4
Cloze answer: the list has to be traversed twice (search + validation)
Q: Trade-offs of the optimistic list. Good: {{c1::no contention during traversal}}, {{c2::traversals are wait-free}}, {{c3::fewer lock acquisitions than hand-over-hand}}. Bad: {{c4::the list has to be traversed twice (search
A: Wait-free: every call finishes in a finite number of steps regardless of what other threads do (in particular it never waits for other threads).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596433 |
1 | 230% | 12d | 6 |
nid:1778588922316
c3
PProg
validate (as in lazy synchronisation: predecessors unmarked ...
1
lapses
1/4
users
230%
ease
nid:1778588922316
Cloze c3
Cloze answer: validate (as in lazy synchronisation: predecessors unmarked and still linked to their successors)
Q: Six steps of add on the lazy skip list: {{c1::find predecessors (lock-free traversal)}}, {{c2::lock the predecessors on every level the new node will occupy}}, {{c3::validate (as in lazy synchronisation: predecessors unmarked and still linked to their succe
A: The fully linked flag plays the same role here that the unmarked-flag plays in the lazy list: it certifies that a concurrent contains may rely on this node.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922317 |
1 | 230% | 13d | 6 |
nid:1778588922382
IO r1
PProg
[Image Occlusion region 1]
1
lapses
1/4
users
230%
ease
nid:1778588922382
Cloze c1
Q: {{c1::image-occlusion:rect:left=.3506:top=.7002:width=.2981:height=.2443:oi=1}}{{c2::image-occlusion:rect:left=.6796:top=.707:width=.2981:height=.2443:oi=1}}{{c3::image-occlusion:rect:left=.679:top=.3892:width=.2981:height=.2443:oi=1}}{{c4::image-occlusion:rect:left=.3524:top=.3891:width
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922382 |
1 | 230% | 16d | 8 |
nid:1778588922614
c1
PProg
one activity fails to recognise that a single memory locatio...
1
lapses
1/4
users
230%
ease
nid:1778588922614
Cloze c1
Cloze answer: one activity fails to recognise that a single memory location was modified temporarily by another activity, and therefore erroneously assumes the overall state has not changed
Q: The ABA problem: {{c1::one activity fails to recognise that a single memory location was modified temporarily by another activity, and therefore erroneously assumes the overall state has not changed}}. It is the fundamental limitation of plain CAS as a concurrency check, and it surf
A: Standard defences: tag the reference with a version counter that increments on every write (e.g. AtomicStampedReference); use LL/SC instead of CAS, since the store-conditional fails on any intervening write, not just on values; hazard pointers or epoch-based reclamation to ensure a freed node cannot reappear while another thread holds a reference to it.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922614 |
1 | 230% | 11d | 8 |
nid:1779457826339
c2
PProg
\(H\) with all its pending invocations removed
1
lapses
1/4
users
230%
ease
nid:1779457826339
Cloze c2
Cloze answer: \(H\) with all its pending invocations removed
Q: Terminology on histories: An invocation is pending if {{c1::it has no matching response}}; A subhistory is complete when {{c1::it contains no pending invocations}}. The operation complete(H) yields {{c2::\(H\) with all its pending invocations removed}}.
A: Pending invocations are exactly the calls still in flight at the end of the recorded history.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826340 |
1 | 230% | 5d | 5 |
nid:1779487674954
c1
PProg
\(\infty\)
1
lapses
1/4
users
230%
ease
nid:1779487674954
Cloze c1
Cloze answer: \(\infty\)
Q: Compare-And-Swap has consensus number {{c1::\(\infty\)}}.
A: Writing into proposed before the CAS guarantees that the winner's value is visible to all.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487674954 |
1 | 230% | 6d | 8 |
nid:1779664155711
c3
PProg
Priority inversion (a low-priority thread holds a resource a...
1
lapses
1/4
users
230%
ease
nid:1779664155711
Cloze c3
Cloze answer: Priority inversion (a low-priority thread holds a resource a high-priority thread is waiting on)
Q: Four problems of locks (motivation for transactional memory): {{c1::Deadlocks (threads take shared or dependent locks in different orders)}}; {{c2::Convoying (a thread holding a resource is descheduled while other threads queue up waiting for it)}}; {{c3::Priority inver
A: These are properties of the lock mechanism itself, not of a specific bug. TM aims to remove them by moving synchronisation from the programmer into the system.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779664155714 |
1 | 230% | 5d | 5 |
nid:1777984596431
c6
PProg
the algorithm is not starvation-free (an operation may have ...
1
lapses
1/4
users
230%
ease
nid:1777984596431
Cloze c6
Cloze answer: the algorithm is not starvation-free (an operation may have to retry arbitrarily often under unfavourable concurrency)
Q: Trade-offs of the optimistic list. Good: {{c1::no contention during traversal}}, {{c2::traversals are wait-free}}, {{c3::fewer lock acquisitions than hand-over-hand}}. Bad: {{c4::the list has to be traversed twice (search
A: Wait-free: every call finishes in a finite number of steps regardless of what other threads do (in particular it never waits for other threads).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596432 |
1 | 230% | 14d | 6 |
nid:1778588922345
c2
PProg
with a spinlock (unless implemented lock-free)
1
lapses
1/4
users
230%
ease
nid:1778588922345
Cloze c2
Cloze answer: with a spinlock (unless implemented lock-free)
Q: Waiting (scheduled) locks suspend a blocked thread instead of spinning. Typical building blocks: {{c1::semaphores, mutexes, and monitors}}. A monitor has two queues: waiting entry queue for threads trying to enter the monitor, and a waiting condition queue for
A: So spinlocks do not disappear when one moves to scheduled locks: they are reused at a lower level to protect the scheduler's data structures.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922347 |
1 | 230% | 13d | 6 |
nid:1778588922396
c1
PProg
suggests (not guarantees) that no other thread has written b...
1
lapses
1/4
users
230%
ease
nid:1778588922396
Cloze c1
Cloze answer: suggests (not guarantees) that no other thread has written between (a) and (c)
Q: The standard CAS update pattern (here for an AtomicInteger counter):A successful CAS {{c1::suggests (not guarantees) that no other thread has written between (a) and (c)}}. If a thread dies in
A: This is the canonical lock-free read-modify-write pattern. The reason for "suggests, not guarantees" is the ABA problem: a value can be changed and changed back between (a) and (c), and CAS cannot tell the difference.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922399 |
1 | 230% | 16d | 8 |
nid:1778588922411
c1
PProg
AtomicReference<Node> top
1
lapses
1/4
users
230%
ease
nid:1778588922411
Cloze c1
Cloze answer: AtomicReference<Node> top
Q: A lock-free stack uses a single {{c1::AtomicReference<Node> top}} as its only shared state. Both push and pop work by reading top, preparing the new top locally, and committing with {{c2::top.compareAndSet(oldTop, n
A: No locks are involved, so the structure is deadlock-free by construction. top is the only point of contention, which is also why a lock-free stack scales poorly under high concurrency unless combined with backoff or an elimination array.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922411 |
1 | 230% | 10d | 8 |
nid:1779457826303
c1
PProg
no other thread is already past its CAS without having seen ...
1
lapses
1/4
users
230%
ease
nid:1779457826303
Cloze c1
Cloze answer: no other thread is already past its CAS without having seen our hazard pointer (i.e. top still equals the node we just flagged)
Q: Hazard-pointer pop protects head before using it. After reading head = top.get() and calling setHazardous(head), the inner loop re-checks the condition head ==
A: Order matters: publish the hazard pointer first, then validate that top is unchanged. Without the re-validation a reclaimer could have slipped between the read and the setHazardous.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826303 |
1 | 230% | 8d | 8 |
nid:1779487674944
c2
PProg
there is no wait-free implementation of n-thread consensus w...
1
lapses
1/4
users
230%
ease
nid:1779487674944
Cloze c2
Cloze answer: there is no wait-free implementation of n-thread consensus with \(n > 1\) from read-write registers
Q: Atomic registers have consensus number {{c1::1}}. Corollary: {{c2::there is no wait-free implementation of n-thread consensus with \(n > 1\) from read-write registers}}.
A: With plain atomic read/write registers, not even two threads can reach consensus wait-free. This is the starting point for impossibility proofs (e.g. wait-free FIFO queue from registers).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487674944 |
1 | 230% | 8d | 6 |
nid:1777984596290
c2
PProg
test that predicate both before calling wait and after retur...
1
lapses
1/4
users
230%
ease
nid:1777984596290
Cloze c2
Cloze answer: test that predicate both before calling wait and after returning from it
Q: Four rules for using condition waits: {{c1::always have a condition predicate}}; {{c2::test that predicate both before calling wait and after returning from it}}; {{c3::always call wait in a while loop, never in a
A: Spurious wakeups, signal-and-continue semantics, and competing signallers all force every woken thread to re-check the predicate itself.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596290 |
1 | 230% | 10d | 6 |
nid:1778588922282
c2
PProg
it is a global operation that may touch the entire tree
1
lapses
1/4
users
230%
ease
nid:1778588922282
Cloze c2
Cloze answer: it is a global operation that may touch the entire tree
Q: Balanced trees (AVL, red-black, treap, ...) are awkward for concurrent sets because {{c1::rebalancing after add/remove is expensive}} and, more importantly, {{c2::it is a global operation that may touch the entire tree}}, which makes lock-free implementations parti
A: "Las Vegas" means: always correct, but the runtime is a random variable. Performance bounds are with high probability rather than worst-case.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922283 |
1 | 230% | 14d | 6 |
nid:1779457826288
c3
PProg
pointer tagging: does not cure the problem, only makes it mu...
1
lapses
1/4
users
230%
ease
nid:1779457826288
Cloze c3
Cloze answer: pointer tagging: does not cure the problem, only makes it much less likely (delays it)
Q: Four ways to defend against the ABA problem on CAS: {{c1::DCAS (double compare-and-swap): not available on most platforms}};{{c2::garbage collection: a node is not freed/reused while a pointer to it still exists, but it is far too slow for the inner loop of a runtime kernel}};
A: GC is itself the reason a lock-free GC cannot rely on GC to dodge ABA. Tagging can be practical but must be used very carefully.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826288 |
1 | 230% | 9d | 6 |
nid:1779457826362
c1
PProg
appending zero or more responses to pending invocations that...
1
lapses
1/4
users
230%
ease
nid:1779457826362
Cloze c1
Cloze answer: appending zero or more responses to pending invocations that took effect, and; discarding zero or more pending invocations that did not take effect
Q: Formal definition: a history \(H\) is linearizable if it can be extended to a history \(G\) by {{c1::appending zero or more responses to pending invocations that took effect, and}}{{c1::discarding zero or more pending invocations that did not take effect}}, such that \
A: The condition \(\to_G \subseteq \to_S\) is the crucial one: \(S\) may order overlapping calls freely, but it must respect every real-time precedence already fixed in \(G\). That real-time constraint is what distinguishes linearizability from mere sequential consistency.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826362 |
1 | 230% | 9d | 8 |
nid:1779487674891
c1
PProg
for every object \(x\) the projection \(H\,|\,x\) is lineari...
1
lapses
1/4
users
230%
ease
nid:1779487674891
Cloze c1
Cloze answer: for every object \(x\) the projection \(H\,|\,x\) is linearizable
Q: Composability theorem (locality): A history \(H\) is linearizable if and only if {{c1::for every object \(x\) the projection \(H\,|\,x\) is linearizable}}. Consequence for modularity: {{c2::linearizability of individual objects can be proved in isolation}}, and {
A: Linearizability is a local property: the correctness of the whole system follows from the correctness of the individual objects. This is exactly the property that sequential consistency lacks.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487674891 |
1 | 230% | 10d | 8 |
nid:1778588922289
c2
PProg
A skip list is a {{c1::sorted multi-level linked list}}. Eac...
1
lapses
1/4
users
230%
ease
nid:1778588922289
Cloze c2
Q: A skip list is a {{c1::sorted multi-level linked list}}. Each node has a probabilistic height with {{c2::\(\mathbb{P}(\text{height} = n) = 0.5^n\) (geometric distribution)}}, and {{c3::no rebalancing is ever performed}}.
A: The geometric height distribution makes the expected number of nodes at level \(k\) decrease by a factor of two with each level, which is what gives the logarithmic search.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922291 |
1 | 230% | 10d | 6 |
nid:1778588922345
c1
PProg
semaphores, mutexes, and monitors
1
lapses
1/4
users
230%
ease
nid:1778588922345
Cloze c1
Cloze answer: semaphores, mutexes, and monitors
Q: Waiting (scheduled) locks suspend a blocked thread instead of spinning. Typical building blocks: {{c1::semaphores, mutexes, and monitors}}. A monitor has two queues: waiting entry queue for threads trying to enter the monitor, and a waiting condition queue for
A: So spinlocks do not disappear when one moves to scheduled locks: they are reused at a lower level to protect the scheduler's data structures.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922345 |
1 | 230% | 10d | 6 |
nid:1778588922477
c3
PProg
read both at once
1
lapses
1/4
users
230%
ease
nid:1778588922477
Cloze c3
Cloze answer: read both at once
Q: java.util.concurrent.atomic.AtomicMarkableReference<V> packages a reference and a boolean mark into one atomic cell. Key methods: compareAndSet(expRef, newRef, expMark, newMark): {{c1::atomic CAS over the (reference, mark) pair}}; at
A: The bit-stealing trick relies on object addresses being 8-byte aligned, so the bottom three bits are always zero in a real pointer. Using one of them as a mark costs nothing - the slide jokes that a fully addressable 64-bit space is \(2^{64}\,\text{B} = 5.6 \cdot 10^{14}\,\text{PB}\), so the lost bit is not missed.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922479 |
1 | 230% | 10d | 6 |
nid:1779457826288
c1
PProg
DCAS (double compare-and-swap): not available on most platfo...
1
lapses
1/4
users
230%
ease
nid:1779457826288
Cloze c1
Cloze answer: DCAS (double compare-and-swap): not available on most platforms
Q: Four ways to defend against the ABA problem on CAS: {{c1::DCAS (double compare-and-swap): not available on most platforms}};{{c2::garbage collection: a node is not freed/reused while a pointer to it still exists, but it is far too slow for the inner loop of a runtime kernel}};
A: GC is itself the reason a lock-free GC cannot rely on GC to dodge ABA. Tagging can be practical but must be used very carefully.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826289 |
1 | 230% | 12d | 6 |
nid:1779457826345
c1
PProg
method calls of different threads do not interleave (each in...
1
lapses
1/4
users
230%
ease
nid:1779457826345
Cloze c1
Cloze answer: method calls of different threads do not interleave (each invocation is immediately followed by its matching response)
Q: A history is sequential if {{c1::method calls of different threads do not interleave (each invocation is immediately followed by its matching response)}}. The one exception allowed is {{c2::a single final pending invocation}}.
A: Sequential here is about non-interleaving of events, independent of how many threads appear.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826345 |
1 | 230% | 9d | 6 |
nid:1779487674900
c1
PProg
respect program order; need not be preserved
1
lapses
1/4
users
230%
ease
nid:1779487674900
Cloze c1
Cloze answer: respect program order; need not be preserved
Q: Properties of sequential consistency:operations of a single thread must {{c1::respect program order}}; the real-time order, by contrast, {{c1::need not be preserved}}.Therefore operations {{c2::of the same thread cannot be reordered}}, but operations {{c2::of different
A: Mnemonic: program order within a thread stays fixed, everything in between may be swapped. This corresponds exactly to what hardware caches and write buffers allow in practice.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487674902 |
1 | 230% | 11d | 8 |
nid:1777984596374
c2
PProg
a single slow thread holding "early" nodes blocks every othe...
1
lapses
1/4
users
230%
ease
nid:1777984596374
Cloze c2
Cloze answer: a single slow thread holding "early" nodes blocks every other thread that wants to reach "later" nodes (no overtaking)
Q: Disadvantages of hand-over-hand locking on a list: {{c1::potentially long sequence of acquire/release before the actual operation can take place}}; {{c2::a single slow thread holding "early" nodes blocks every other thread that wants to reach "later" nodes (no overtaking)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596375 |
1 | 230% | 13d | 6 |
nid:1778588922367
c5
PProg
cannot be used inside interrupt handlers
1
lapses
1/4
users
230%
ease
nid:1778588922367
Cloze c5
Cloze answer: cannot be used inside interrupt handlers
Q: Reasons to look beyond locks. They are {{c1::pessimistic by design: mutual exclusion is enforced even when no real conflict would occur}}; they impose {{c2::overhead in every call, even uncontended ones, and degrade badly under contention (Amdahl)}}; and their blocking
A: These five points together motivate non-blocking (lock-free / wait-free) data structures: progress should not depend on any one thread continuing to execute.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922371 |
1 | 230% | 15d | 6 |
nid:1778588922492
c3
PProg
b is no longer c's predecessor; the physical unlinking can b...
1
lapses
1/4
users
230%
ease
nid:1778588922492
Cloze c3
Cloze answer: b is no longer c's predecessor; the physical unlinking can be left to a later traversal that encounters the marked node
Q: Both steps are "try to" because {{c1::either CAS may fail if a concurrent thread has changed the relevant field in the meantime}}. On failure of step 1, {{c2::another thread already marked or modified c; resta
A: This is the core lock-free pattern: each thread that walks the list and stumbles over a marked node helps unlink it ("helping"). That property is what gives the algorithm its lock-freedom guarantee.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922494 |
1 | 230% | 14d | 6 |
nid:1779664155793
c1
PProg
as if they had been executed sequentially, one after another...
1
lapses
1/4
users
230%
ease
nid:1779664155793
Cloze c1
Cloze answer: as if they had been executed sequentially, one after another, in some serial order
Q: Serializability: Concurrently executed transactions behave {{c1::as if they had been executed sequentially, one after another, in some serial order}}. The transactions {{c2::appear serialized}} without actually running serially.
A: Example: if TXA and TXB run concurrently, a valid outcome is one that would also arise from the serial order TXB then TXA (or TXA then TXB).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779664155793 |
1 | 230% | 9d | 6 |
nid:1777924071026
c4
PProg
a waiting entry queue (threads trying to acquire the monitor...
1
lapses
1/4
users
230%
ease
nid:1777924071026
Cloze c4
Cloze answer: a waiting entry queue (threads trying to acquire the monitor); a waiting condition queue (threads inside the monitor that have called wait)
Q: A monitor adds, on top of mutual exclusion, the following condition mechanism: if a condition does not hold, {{c1::release the monitor lock}}, {{c2::wait for the condition to become true}}, and use {{c3::a signalling mechanism (rather than a busy-loop) to be woken when
A: The two queues are conceptually distinct: a thread that is signalled is moved from the condition queue back to the entry queue (under signal-and-continue), it does not get the lock immediately.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924071026 |
1 | 230% | 16d | 6 |
nid:1777924071026
c3
PProg
a signalling mechanism (rather than a busy-loop) to be woken...
1
lapses
1/4
users
230%
ease
nid:1777924071026
Cloze c3
Cloze answer: a signalling mechanism (rather than a busy-loop) to be woken when it does
Q: A monitor adds, on top of mutual exclusion, the following condition mechanism: if a condition does not hold, {{c1::release the monitor lock}}, {{c2::wait for the condition to become true}}, and use {{c3::a signalling mechanism (rather than a busy-loop) to be woken when
A: The two queues are conceptually distinct: a thread that is signalled is moved from the condition queue back to the entry queue (under signal-and-continue), it does not get the lock immediately.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924071028 |
1 | 230% | 14d | 6 |
nid:1777984596290
c4
PProg
ensure the state involved is protected by the lock associate...
1
lapses
1/4
users
230%
ease
nid:1777984596290
Cloze c4
Cloze answer: ensure the state involved is protected by the lock associated with the condition
Q: Four rules for using condition waits: {{c1::always have a condition predicate}}; {{c2::test that predicate both before calling wait and after returning from it}}; {{c3::always call wait in a while loop, never in a
A: Spurious wakeups, signal-and-continue semantics, and competing signallers all force every woken thread to re-check the predicate itself.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596291 |
1 | 230% | 13d | 6 |
nid:1778588922477
c1
PProg
atomic CAS over the (reference, mark) pair
1
lapses
1/4
users
230%
ease
nid:1778588922477
Cloze c1
Cloze answer: atomic CAS over the (reference, mark) pair
Q: java.util.concurrent.atomic.AtomicMarkableReference<V> packages a reference and a boolean mark into one atomic cell. Key methods: compareAndSet(expRef, newRef, expMark, newMark): {{c1::atomic CAS over the (reference, mark) pair}}; at
A: The bit-stealing trick relies on object addresses being 8-byte aligned, so the bottom three bits are always zero in a real pointer. Using one of them as a mark costs nothing - the slide jokes that a fully addressable 64-bit space is \(2^{64}\,\text{B} = 5.6 \cdot 10^{14}\,\text{PB}\), so the lost bit is not missed.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922480 |
1 | 230% | 13d | 6 |
nid:1779457826352
c1
PProg
they have the same per-thread projections, i.e. \(H\,|\,T = ...
1
lapses
1/4
users
230%
ease
nid:1779457826352
Cloze c1
Cloze answer: they have the same per-thread projections, i.e. \(H\,|\,T = G\,|\,T\) for every thread \(T\)
Q: Two histories \(H\) and \(G\) are equivalent iff {{c1::they have the same per-thread projections, i.e. \(H\,|\,T = G\,|\,T\) for every thread \(T\)}}.
A: Equivalence ignores how calls of different threads are interleaved in time; it only cares that each thread sees the same local sequence.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826352 |
1 | 230% | 10d | 6 |
nid:1778588922309
c3
PProg
pre: predecessor on each level; succ: successor on each leve...
1
lapses
1/4
users
230%
ease
nid:1778588922309
Cloze c3
Cloze answer: pre: predecessor on each level; succ: successor on each level
Q: The lazy skip-list find(x, pre, succ) returns {{c1::-1}} if the value is not present, otherwise {{c1::the level at which x was found}}. It also fills the output arrays {{c3::pre: predecessor on each level}} and {{c3::succ: succ
A: The level/pre/succ information is what add and remove need afterwards to splice or unlink the node at every level it appears in.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922311 |
1 | 230% | 16d | 6 |
nid:1778588922580
c2
PProg
having any thread that observes tail.next != null attempt CA...
1
lapses
1/4
users
230%
ease
nid:1778588922580
Cloze c2
Cloze answer: having any thread that observes tail.next != null attempt CAS(tail, tail, tail.next) to drag tail forward
Q: A second transient inconsistency in the lock-free queue. Now {{c1::tail still points to the original sentinel, which has just been removed from the queue}}. Without helping this is unrecoverable; the f
A: This is why the final dequeue code, before doing the actual dequeue, checks if (first == last) and, if so, helps advance tail.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922580 |
1 | 230% | 15d | 6 |
nid:1779457826291
c3
PProg
\(x - (x \bmod 32)\)
1
lapses
1/4
users
230%
ease
nid:1779457826291
Cloze c3
Cloze answer: \(x - (x \bmod 32)\)
Q: Pointer tagging exploits that aligned addresses leave some low bits unused: a pointer aligned modulo \(32\) frees up {{c1::\(5\)}} bits for a tag (in addition to the unused high address bits). Each time a pointer is stored in the data structure, the tag is {{c2::incremented by one}}, and
A: With 5 tag bits there are 32 distinct versions of each pointer, so a recycled pointer almost always carries a different tag and the stale CAS fails. It only delays ABA (the tag eventually wraps around), it does not eliminate it.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826293 |
1 | 230% | 13d | 9 |
nid:1778588922316
c5
PProg
mark the node as fully linked
1
lapses
1/4
users
230%
ease
nid:1778588922316
Cloze c5
Cloze answer: mark the node as fully linked
Q: Six steps of add on the lazy skip list: {{c1::find predecessors (lock-free traversal)}}, {{c2::lock the predecessors on every level the new node will occupy}}, {{c3::validate (as in lazy synchronisation: predecessors unmarked and still linked to their succe
A: The fully linked flag plays the same role here that the unmarked-flag plays in the lazy list: it certifies that a concurrent contains may rely on this node.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922321 |
1 | 230% | 12d | 6 |
nid:1778588922507
c1
PProg
finish the job; CAS the predecessor's next field past the ma...
1
lapses
1/4
users
230%
ease
nid:1778588922507
Cloze c1
Cloze answer: finish the job; CAS the predecessor's next field past the marked node and continue (repeating as needed if more marked nodes follow)
Q: Lock-free list-set traversal policy when stumbling over a logically deleted (marked) node: {{c1::finish the job}}, i.e. {{c1::CAS the predecessor's next field past the marked node and continue (repeating as needed if more marked nodes follow)}}.
A: This is the canonical helping pattern: every traverser shares responsibility for completing the physical-delete step that another thread left half-done. Without it, a thread that died right after step 1 of remove would leave a marked node in the list forever, which would not violate correctness, but would degrade performance and is incompatible with practical use.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922508 |
1 | 230% | 15d | 9 |
nid:1779457826288
c2
PProg
garbage collection: a node is not freed/reused while a point...
1
lapses
1/4
users
230%
ease
nid:1779457826288
Cloze c2
Cloze answer: garbage collection: a node is not freed/reused while a pointer to it still exists, but it is far too slow for the inner loop of a runtime kernel
Q: Four ways to defend against the ABA problem on CAS: {{c1::DCAS (double compare-and-swap): not available on most platforms}};{{c2::garbage collection: a node is not freed/reused while a pointer to it still exists, but it is far too slow for the inner loop of a runtime kernel}};
A: GC is itself the reason a lock-free GC cannot rely on GC to dodge ABA. Tagging can be practical but must be used very carefully.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826290 |
1 | 230% | 14d | 6 |
nid:1779457826299
c1
PProg
writes into the calling thread's own slot: hazarduous.set(id...
1
lapses
1/4
users
230%
ease
nid:1779457826299
Cloze c1
Cloze answer: writes into the calling thread's own slot: hazarduous.set(id, node)
Q: Two helper methods backing the global hazard array AtomicReferenceArray<Node> hazarduous (sized nThreads): setHazardous(node) {{c1::writes into the calling thread's own slot: hazarduous.set(id, node)}};isHazardous(
A: set touches only the current thread's index id; the check is a full linear scan over all nThreads slots, hence O(nThreads) per reuse.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826299 |
1 | 230% | 15d | 9 |
nid:1778588922352
c3
PProg
their internal queues need their own protection (spinlock or...
1
lapses
1/4
users
230%
ease
nid:1778588922352
Cloze c3
Cloze answer: their internal queues need their own protection (spinlock or lock-free)
Q: Properties of scheduled (waiting) locks: {{c1::they require support from the runtime system / OS scheduler}}, {{c2::their wakeup latency is higher than a spinlock's (a scheduling round-trip is involved)}}, and {{c3::their internal queues need their own protection (spinl
A: Competitive spinning gets the best of both worlds for short critical sections: low latency when the lock is released quickly, but no CPU burned when the wait turns out to be long.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922352 |
1 | 230% | 14d | 6 |
nid:1779664155817
c1
PProg
A big lock around all atomic sections: gives (nearly) all de...
1
lapses
1/4
users
230%
ease
nid:1779664155817
Cloze c1
Cloze answer: A big lock around all atomic sections: gives (nearly) all desired properties but is not scalable and is not done in practice
Q: Implementing TM, two approaches: {{c1::A big lock around all atomic sections: gives (nearly) all desired properties but is not scalable and is not done in practice}}. {{c2::Keep track of the operations performed by each transaction (concurrency control): the system guarantees at
A: With the big lock, the missing property is mainly scalability (concurrent transactions cannot run in parallel).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779664155817 |
1 | 230% | 20d | 8 |
nid:1778588922540
c3
PProg
without garbage collection the kernel must reuse queue nodes...
1
lapses
1/4
users
230%
ease
nid:1778588922540
Cloze c3
Cloze answer: without garbage collection the kernel must reuse queue nodes, which opens the door to the ABA problem
Q: Motivation for a lock-free unbounded queue: The scheduling queues at the heart of an OS kernel (ready / waiting-for-x / waiting-for-y) are accessed concurrently by {{c1::threads and interrupt service routines on different cores}}, and using (spin)locks for protection creates the usual problems (
A: This is the canonical motivating story for the Michael-Scott queue and for ABA: the Linux Big Kernel Lock (BKL) was a single coarse-grained spinlock and was eventually removed precisely because of its scalability problems.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922541 |
1 | 230% | 14d | 6 |
nid:1779457826325
c3
PProg
the associated sequential behaviour is correct
1
lapses
1/4
users
230%
ease
nid:1779457826325
Cloze c3
Cloze answer: the associated sequential behaviour is correct
Q: Linearizability: Each method should {{c1::appear to take effect instantaneously (atomically) at some single point between its invocation and its response}}.An object is linearizable if this holds {{c2::for all of its possible executions}}, and it is then correct iff {{c3::the associa
A: The single instant where the effect happens is the linearization point. Linearizability reduces reasoning about a concurrent object to reasoning about its sequential specification.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779457826328 |
1 | 230% | 14d | 6 |
nid:1777984596249
c1
PProg
exactly the readers waiting now are allowed through before t...
1
lapses
1/4
users
230%
ease
nid:1777984596249
Cloze c1
Cloze answer: exactly the readers waiting now are allowed through before the next writer
Q: Core logic of the FIFO-fair RW lock.release_write sets writersWait = readersWaiting, i.e. {{c1::exactly the readers waiting now are allowed through before the next writer}}. acquire_read blocks while {{c2::writers &
A: The clause writersWait <= 0 reads as: "the quota of preferred readers is used up, newly arriving readers must wait". When the next writer finishes, the quota is freshly set from the readers that happen to be waiting then.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596250 |
1 | 230% | 17d | 6 |
nid:1778588922600
c1
PProg
node pool
1
lapses
1/4
users
230%
ease
nid:1778588922600
Cloze c1
Cloze answer: node pool
Q: In an unmanaged environment (kernel, no GC), per-operation allocation is unacceptable, so dequeued nodes are returned to a {{c1::node pool}} and later reused by enqueues. Two consequences: {{c2::a node can be present in at most one in-place structure at a time (so per-instance po
A: The pool itself is implemented as just another lock-free stack with the same get/put CAS loop. The hidden cost is exactly the ABA exposure that the rest of this section addresses.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922603 |
1 | 230% | 17d | 9 |
nid:1771843744684
c1
PProg
native threading
1
lapses
1/4
users
230%
ease
nid:1771843744684
Cloze c1
Cloze answer: native threading
Q: In {{c1::native threading}} (most common), each JVM thread is mapped to a {{c2::dedicated operating system thread}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771843744684 |
1 | 230% | 51d | 9 |
nid:1779487674936
c1
PProg
a consensus protocol exists that uses arbitrarily many objec...
1
lapses
1/4
users
230%
ease
nid:1779487674936
Cloze c1
Cloze answer: a consensus protocol exists that uses arbitrarily many objects of class \(C\) and arbitrarily many atomic registers
Q: A class \(C\) solves n-thread consensus if {{c1::a consensus protocol exists that uses arbitrarily many objects of class \(C\) and arbitrarily many atomic registers}}. The consensus number of \(C\) is {{c2::the largest \(n\) for which \(C\) solves n-thread consensus}}.
A: The consensus number places synchronisation primitives into a hierarchy (Herlihy). It measures how many threads can reach consensus wait-free using the primitive.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1779487674936 |
1 | 230% | 17d | 9 |
nid:1771843744230
c2
PProg
safety properties
1
lapses
1/4
users
230%
ease
nid:1771843744230
Cloze c2
Cloze answer: safety properties
Q: {{c1::Exceptions, absence of deadlocks, and mutual exclusion}} are typical {{c2::safety properties}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771843744232 |
1 | 230% | 55d | 8 |
nid:1777562257035
c1
PProg
CMPXCHG mem, reg
1
lapses
1/4
users
230%
ease
nid:1777562257035
Cloze c1
Cloze answer: CMPXCHG mem, reg
Q: On x86 the {{c1::CMPXCHG mem, reg}} instruction implements compare-and-swap: It compares register A with a memory location and, if equal, copies the second operand into the location and sets ZF. To make it atomic across cores, it must be combined with {{c2::the LOCK
A: Plain CMPXCHG is atomic only with respect to other observers on the same core. The LOCK prefix locks the bus / cache line, which is what gives the instruction its inter-processor atomicity guarantee - and also what makes RMW operations 10-100x slower than ordinary loads.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777562257036 |
1 | 230% | 52d | 9 |
nid:1777562257039
c1
PProg
Load-Linked / Store-Conditional
1
lapses
1/4
users
230%
ease
nid:1777562257039
Cloze c1
Cloze answer: Load-Linked / Store-Conditional
Q: ARM (and MIPS, POWER, RISC-V) implement atomic RMW via a {{c1::Load-Linked / Store-Conditional}} pair: LDREX loads from memory and {{c2::marks the address as exclusive for this core}}; the subsequent STREX performs {{c3::a conditional store that only succeeds if no
A: If STREX fails, software retries the LL/SC pair. Compared with x86's monolithic CMPXCHG, LL/SC is more flexible (you can compute the new value between LL and SC) but requires explicit retry logic in software.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777562257040 |
1 | 230% | 53d | 9 |
nid:1778588922254
c1
PProg
it is reachable from head; it is reachable from its predeces...
1
lapses
1/4
users
230%
ease
nid:1778588922254
Cloze c1
Cloze answer: it is reachable from head; it is reachable from its predecessor
Q: Key invariant of the lazy list: if a node is not marked, then {{c1::it is reachable from head}} and {{c1::it is reachable from its predecessor}}.
A: This is exactly what makes the per-node validation in lazy remove/add work without rescanning the list: locking the two-node window and reading both marked flags is enough to certify reachability.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1778588922254 |
1 | 230% | 22d | 9 |
nid:1777560365843
c2
PProg
ordinary fields; locks
1
lapses
1/4
users
230%
ease
nid:1777560365843
Cloze c2
Cloze answer: ordinary fields; locks
Q: Accesses to volatile fields in Java do not count as {{c1::a data race}}. In terms of performance, volatile is slower than {{c2::ordinary fields}}, but faster than {{c2::locks}}.
A: Recommendation: only for experts; otherwise use the standard library (java.util.concurrent, AtomicInteger, …).Caveat: volatile guarantees visibility, but not atomicity of compound operations like i++.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777560365843 |
1 | 230% | 52d | 7 |
nid:1771840320732
c1
PProg
efficient
1
lapses
1/4
users
230%
ease
nid:1771840320732
Cloze c1
Cloze answer: efficient
Q: Context switching between threads is {{c1::efficient::cost?}}.
A: No change of address spaceNo automatic schedulingNo saving / (re-)loading of PCB (OS process) state
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771840320732 |
1 | 230% | 63d | 8 |
nid:1773756234990
c1
PProg
f + P(1-f) = P - f(P-1)
1
lapses
1/4
users
230%
ease
nid:1773756234990
Cloze c1
Cloze answer: f + P(1-f) = P - f(P-1)
Q: Gustafson's law:\(S_p = {{c1::f + P(1-f) = P - f(P-1)}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773756234991 |
1 | 230% | 55d | 8 |
nid:1771924799414
c1
PProg
non-deterministic
1
lapses
1/4
users
230%
ease
nid:1771924799414
Cloze c1
Cloze answer: non-deterministic
Q: The execution order is {{c1::non-deterministic}}.
A: The OS-Scheduler depends on all currently running processes.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771924799414 |
1 | 230% | 71d | 8 |
nid:1771842362463
c1
PProg
main() returns
1
lapses
1/4
users
230%
ease
nid:1771842362463
Cloze c1
Cloze answer: main() returns
Q: Threads can continue to run even if {{c1::main() returns}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771842362464 |
1 | 230% | 76d | 8 |
nid:1773755999965
c1
PProg
If \(f\) is the non-parallelizable serial fraction of the to...
1
lapses
1/4
users
230%
ease
nid:1773755999965
Cloze c1
Q: If \(f\) is the non-parallelizable serial fraction of the total work, this gives:\(S_p \leq {{c1::\dfrac{1}{f + \dfrac{1-f}{P} } }}\)
A: Note that the following equalities hold:\(W_{ser} = fT_1\)\(W_{par} = (1-f)T_1\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1773755999965 |
1 | 230% | 69d | 8 |
nid:1771843680785
c2
PProg
load the local data, program pointer etc. of the next thread...
1
lapses
1/4
users
230%
ease
nid:1771843680785
Cloze c2
Cloze answer: load the local data, program pointer etc. of the next thread/process to execute
Q: In terms of context switch, CPU needs to {{c1::store/save the local data, program pointer etc. of the current thread/process}}, and {{c2::load the local data, program pointer etc. of the next thread/process to execute}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771843680785 |
1 | 230% | 75d | 7 |
nid:1777560365978
c1
PProg
Mutual exclusion (statements from CSes of different processe...
1
lapses
1/4
users
230%
ease
nid:1777560365978
Cloze c1
Cloze answer: Mutual exclusion (statements from CSes of different processes must not interleave)
Q: The three conditions for a correct critical-section solution according to Ben-Ari are:{{c1::Mutual exclusion (statements from CSes of different processes must not interleave)}}.{{c2::Freedom from deadlock (if some processes are trying to enter a CS, at least one must eventually
A: Hierarchy: starvation freedom \(\Rightarrow\) deadlock freedom (if every individual gets through, then certainly some individual gets through), but not the converse.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777560365979 |
1 | 230% | 83d | 9 |
nid:1777924071035
c2
PProg
atomically releases the lock and waits until the thread is s...
1
lapses
1/4
users
230%
ease
nid:1777924071035
Cloze c2
Cloze answer: atomically releases the lock and waits until the thread is signalled
Q: A Java Condition (obtained via lock.newCondition()) offers {{c1::await()}}: must be called with the lock held; {{c2::atomically releases the lock and waits until the thread is signalled}}; on return, the lock is {{c3::gua
A: Crucial difference from wait/notify on intrinsic locks: a single lock can have multiple conditions, so producers and consumers can wait on different ones (notFull, notEmpty) and the right kind of waiter can be woken directly without notifying all.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924071037 |
1 | 230% | 72d | 7 |
nid:1777924070971
c1
PProg
spinning (with a blocking queue \(Q_S\))
1
lapses
1/4
users
230%
ease
nid:1777924070971
Cloze c1
Cloze answer: spinning (with a blocking queue \(Q_S\))
Q: Implementation of a semaphore without {{c1::spinning (with a blocking queue \(Q_S\))}}: The condition test plus queue operation must be {{c2::atomic (e.g. under an internal lock)}}. release only increments
A: Advantage over busy-wait: waiting threads consume no CPU. The price is context-switch overhead, which only pays off once waits are long enough to dwarf it (short critical sections may still favour spinning).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924070972 |
1 | 230% | 75d | 9 |
nid:1777924070974
c1
PProg
threads in bulk mode (all moving in lock-step); Bulk-Synchro...
1
lapses
1/4
users
230%
ease
nid:1777924070974
Cloze c1
Cloze answer: threads in bulk mode (all moving in lock-step); Bulk-Synchronous Parallel (BSP)
Q: Scaling a dot product to 1 million entries on 10 000 threads is impractical with semaphores or locks. It calls for a higher-level abstraction supporting {{c1::threads in bulk mode (all moving in lock-step)}}. The corresponding programming model is {{c1::Bulk-Synchronous Parallel (BSP)}}.
A: The full BSP model (Valiant, 1990) is more general and supports distributed memory: computation proceeds in supersteps consisting of local computation, communication, and a global barrier. In shared-memory code the barrier is the central building block.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924070974 |
1 | 230% | 75d | 9 |
nid:1777560365843
c1
PProg
a data race
1
lapses
1/4
users
230%
ease
nid:1777560365843
Cloze c1
Cloze answer: a data race
Q: Accesses to volatile fields in Java do not count as {{c1::a data race}}. In terms of performance, volatile is slower than {{c2::ordinary fields}}, but faster than {{c2::locks}}.
A: Recommendation: only for experts; otherwise use the standard library (java.util.concurrent, AtomicInteger, …).Caveat: volatile guarantees visibility, but not atomicity of compound operations like i++.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777560365844 |
1 | 230% | 79d | 10 |
nid:1774487167533
c1
PProg
invokeAll(tasks); invokeAny(tasks)
1
lapses
1/4
users
230%
ease
nid:1774487167533
Cloze c1
Cloze answer: invokeAll(tasks); invokeAny(tasks)
Q: exec.{{c1::invokeAll(tasks)}} submits all tasks and {{c2::blocks until all complete, returning a list of Futures}}.exec.{{c1::invokeAny(tasks)}} returns {{c2::the result of the first task to complete successfully}}.
A: invokeAll → wait for all. invokeAny → wait for any one. Both submit all tasks to the pool regardless.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487167533 |
1 | 230% | 92d | 11 |
nid:1777984596249
c2
PProg
writers > 0 or (writersWaiting > 0 and writersWait <= 0); wr...
1
lapses
1/4
users
230%
ease
nid:1777984596249
Cloze c2
Cloze answer: writers > 0 or (writersWaiting > 0 and writersWait <= 0); writersWait
Q: Core logic of the FIFO-fair RW lock.release_write sets writersWait = readersWaiting, i.e. {{c1::exactly the readers waiting now are allowed through before the next writer}}. acquire_read blocks while {{c2::writers &
A: The clause writersWait <= 0 reads as: "the quota of preferred readers is used up, newly arriving readers must wait". When the next writer finishes, the quota is freshly set from the readers that happen to be waiting then.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596253 |
1 | 230% | 79d | 9 |
nid:1777562256951
c1
PProg
Dekker's algorithm
1
lapses
1/4
users
230%
ease
nid:1777562256951
Cloze c1
Cloze answer: Dekker's algorithm
Q: {{c1::Dekker's algorithm}} fixes the previous mutex tries by combining the {{c2::flag-based approach (try 2)}} with a {{c2::turn variable used only for conflict resolution (try 3)}}.
A: When both flags are set, the process whose turn it is not temporarily lowers its flag and waits, then tries again.Properties: mutual exclusion, deadlock freedom, starvation freedom. But it is verbose: the code needs an inner conditional and a second wait loop. Peterson's algorithm achieves the same with a much shorter pre-protocol.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777562256951 |
1 | 230% | 86d | 10 |
nid:1777562257044
c1
PProg
Test-And-Set (TAS)
1
lapses
1/4
users
230%
ease
nid:1777562257044
Cloze c1
Cloze answer: Test-And-Set (TAS)
Q: Three common families of hardware atomic instructions: {{c1::Test-And-Set (TAS)}} - e.g. m68k TSL{{c2::Compare-And-Swap (CAS)}} - e.g. x86 LOCK CMPXCHG, SPARC CASA{{c3::Load-Linked / Store-Conditional}} - e.g. ARM LDREX/STREX
A: These primitives are needed because plain atomic registers cannot break the \(\Omega(n)\) space lower bound for \(n\)-thread mutex (Burns-Lynch). They also enable lock-free data structures.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777562257044 |
1 | 230% | 85d | 9 |
nid:1777984596249
c3
PProg
writers > 0 or readers > 0 or writersWait > 0
1
lapses
1/4
users
230%
ease
nid:1777984596249
Cloze c3
Cloze answer: writers > 0 or readers > 0 or writersWait > 0
Q: Core logic of the FIFO-fair RW lock.release_write sets writersWait = readersWaiting, i.e. {{c1::exactly the readers waiting now are allowed through before the next writer}}. acquire_read blocks while {{c2::writers &
A: The clause writersWait <= 0 reads as: "the quota of preferred readers is used up, newly arriving readers must wait". When the next writer finishes, the quota is freshly set from the readers that happen to be waiting then.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596256 |
1 | 230% | 60d | 7 |
nid:1772531506601
c1
PProg
"not runnable" and is no longer eligible for execution
1
lapses
1/4
users
230%
ease
nid:1772531506601
Cloze c1
Cloze answer: "not runnable" and is no longer eligible for execution
Q: If a thread calls the wait method in an Object or calls the join method in another thread object, the thread becomes {{c1::"not runnable" and is no longer eligible for execution}}.
A: It becomes executable as a result of an associated notify method being called by another thread, or if the thread with which it has requested a join, becomes terminated.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772531506601 |
1 | 230% | 85d | 8 |
nid:1777984596276
c2
PProg
neither writer priority nor reader-to-writer upgrading
1
lapses
1/4
users
230%
ease
nid:1777984596276
Cloze c2
Cloze answer: neither writer priority nor reader-to-writer upgrading
Q: Java's synchronized statement does not support reader/writer semantics. The library provides java.util.concurrent.locks.ReentrantReadWriteLock instead. Its methods readLock() and writeLock() each return {{c1::a lock object with it
A: Re-entrancy is orthogonal to the reader/writer distinction; some libraries do support upgrading a held read lock to a write lock in the same thread.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777984596277 |
1 | 230% | 59d | 7 |
nid:1771365476514
c1
PProg
Complex Instruction Set Computer
1
lapses
1/4
users
230%
ease
nid:1771365476514
Cloze c1
Cloze answer: Complex Instruction Set Computer
Q: CISC stands for {{c1::Complex Instruction Set Computer}}.
A: A fundamental CPU architecture model with complex, feature-rich instructions.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771365476516 |
1 | 230% | 91d | 9 |
nid:1777560365962
c2
PProg
the transitive closure of PO and SW
1
lapses
1/4
users
230%
ease
nid:1777560365962
Cloze c2
Cloze answer: the transitive closure of PO and SW
Q: Synchronizes-With (SW) pairs the specific actions that "see" each other: a volatile-write to x SW {{c1::a subsequent read of x (subsequent in SO)}}. Happens-Before (HB) is {{c2::the transitive closure of PO and SW}}.
A: Unlike PO or SO, SW is not a generic ordering; it specifically pairs the actions between which a synchronisation effect takes place.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777560365963 |
1 | 230% | 94d | 7 |
nid:1777562257030
c1
PProg
Space lower bound is linear in the maximum number of threads
1
lapses
1/4
users
230%
ease
nid:1777562257030
Cloze c1
Cloze answer: Space lower bound is linear in the maximum number of threads
Q: Mutual exclusion built only from atomic registers (Filter, Bakery, Peterson) is not used in practice for four reasons: {{c1::Space lower bound is linear in the maximum number of threads}}.{{c2::Correctness relies on no memory reordering, which requires expensive memory barriers
A: The way out: extend the model. Modern multiprocessor architectures provide special instructions for atomically reading and writing at once (TAS, CAS, LL/SC), enabling \(O(1)\) space mutexes with practical performance.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777562257030 |
1 | 230% | 87d | 9 |
nid:1777924071026
c2
PProg
wait for the condition to become true
1
lapses
1/4
users
230%
ease
nid:1777924071026
Cloze c2
Cloze answer: wait for the condition to become true
Q: A monitor adds, on top of mutual exclusion, the following condition mechanism: if a condition does not hold, {{c1::release the monitor lock}}, {{c2::wait for the condition to become true}}, and use {{c3::a signalling mechanism (rather than a busy-loop) to be woken when
A: The two queues are conceptually distinct: a thread that is signalled is moved from the condition queue back to the entry queue (under signal-and-continue), it does not get the lock immediately.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924071027 |
1 | 230% | 63d | 7 |
nid:1777924070991
c1
PProg
"after all processes have passed, barrier = 0" is violated: ...
1
lapses
1/4
users
230%
ease
nid:1777924070991
Cloze c1
Cloze answer: "after all processes have passed, barrier = 0" is violated: release(barrier) can be executed multiple times across iterations before any thread reaches the resetting acquire(barrier), so barrier drifts to \(2, 3, \dots\)
Q: A naive reusable barrier extends the one-shot barrier with a symmetric "second half" after the turnstile, intended to reset count and barrier.It is broken because the invariant {{c1:
A: Concrete scheduling: thread A finishes phase 1 with count = 1; meanwhile B and C race into the next iteration, both increment count back up, hit count == n, and each calls release(barrier). The semaphore is no longer a clean 0/1 toggle. The fix is a proper two-phase split with separate semaphores for the two halves.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924070991 |
1 | 230% | 88d | 9 |
nid:1777538021700
c2
PProg
ordering guarantee for memory accesses; not total across thr...
1
lapses
1/4
users
230%
ease
nid:1777538021700
Cloze c2
Cloze answer: ordering guarantee for memory accesses; not total across threads
Q: Program Order (PO) in the JMM is a {{c1::total order over the actions of a single thread, defined on an execution trace}}. PO gives no {{c2::ordering guarantee for memory accesses}} and is {{c2::not total across threads}}.
A: Its only purpose is to link an execution back to the original program.Intra-thread consistency: per thread, PO is consistent with the thread's isolated execution. E.g. taking the else-branch of an if whose condition was true makes the execution invalid.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777538021700 |
1 | 230% | 102d | 7 |
nid:1777924070984
c4
PProg
Once all processes have reached the barrier, all waiting pro...
1
lapses
1/4
users
230%
ease
nid:1777924070984
Cloze c4
Cloze answer: Once all processes have reached the barrier, all waiting processes can continue
Q: Four invariants a correct barrier must satisfy: {{c1::Each of the processes eventually reaches the acquire statement}}.{{c2::The barrier opens iff all processes have reached the barrier}}. {{c3::count gives the number of processes that have pas
A: The naive 1st-try implementation violates the latter two: the race on count++ breaks (3), and the single release on a 0-initialised semaphore breaks (4).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777924070986 |
1 | 230% | 98d | 9 |
nid:1777538021644
c1
PProg
an infinite loop (jmp test without ever reloading x)
1
lapses
1/4
users
230%
ease
nid:1777538021644
Cloze c1
Cloze answer: an infinite loop (jmp test without ever reloading x)
Q: For the home-made rendezvous GCC with optimisations compiles the loop into {{c1::an infinite loop (jmp test without ever reloading x)}}.
A: Reason: the compiler hoists the read of x out of the loop (register hoisting) because, from its sequential view, x is not modified inside the loop. Writes from other threads are invisible to it.Fix: volatile or a lock.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777538021644 |
1 | 230% | 111d | 9 |
nid:1774487167528
c3
PProg
steals a task from the back of another thread's deque (the o...
1
lapses
1/4
users
230%
ease
nid:1774487167528
Cloze c3
Cloze answer: steals a task from the back of another thread's deque (the oldest = largest task)
Q: How does the Fork/Join work-stealing scheduler work?Each worker thread has its own {{c1::deque (double-ended queue) of tasks}}.It processes {{c2::its own tasks LIFO from the front}}.When a thread runs out of work, it {{c3::steals
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774631279612 |
1 | 230% | 123d | 8 |
nid:1777562257000
c1
PProg
Filter Lock
1
lapses
1/4
users
230%
ease
nid:1777562257000
Cloze c1
Cloze answer: Filter Lock
Q: The {{c1::Filter Lock}} extends Peterson's lock from 2 to \(n\) processes by introducing {{c2::\(n-1\) levels}}: Each thread climbs from level 1 up to level \(n-1\), and at each level \(\ell\) it sets \(\texttt{level[me]} = \ell\) and \(\texttt{victim}[\ell] = \texttt{me}\), then waits while
A: Intuition: at each level, Peterson's mechanism filters out at most one thread (the current victim, if anyone else is still around). After traversing all \(n-1\) levels, at most one thread can be left, which then enters the CS.unlock just resets level[me] = 0.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777562257001 |
1 | 230% | 101d | 7 |
nid:1777538021715
c3
PProg
PO
1
lapses
1/4
users
230%
ease
nid:1777538021715
Cloze c3
Cloze answer: PO
Q: The Synchronization Order (SO) in the JMM is {{c1::a total order over all synchronization actions}}. All threads see the SA in {{c2::the same}} order. The SA inside a thread occur in {{c3::PO}}. SO is {{c4::consistent: every read in SO sees the latest write in SO}}.
A: The globality of SO is the reason that volatile and locks can synchronise threads at all.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777751564276 |
1 | 230% | 112d | 9 |
nid:1771843744640
c1
PProg
sequentially consistent
1
lapses
1/4
users
230%
ease
nid:1771843744640
Cloze c1
Cloze answer: sequentially consistent
Q: A {{c1::sequentially consistent}} interleaving is one where {{c2::the relative order of statements from one thread is preserved}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771843744641 |
1 | 230% | 106d | 9 |
nid:1774487167931
PProg
Why is \(T_p \geq T_\infty\) a strict lower bound?
1
lapses
1/4
users
230%
ease
nid:1774487167931
Q: Why is \(T_p \geq T_\infty\) a strict lower bound?
A: \(T_\infty\) is the length of the critical path - a chain of nodes where each depends on the previous. Even with infinite processors, these nodes must execute sequentially. No amount of parallelism can compress a dependency chain.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487167931 |
1 | 230% | 120d | 11 |
nid:1774487168095
PProg
Why should you never use synchronized(someInteger) or synchr...
1
lapses
1/4
users
230%
ease
nid:1774487168095
Q: Why should you never use synchronized(someInteger) or synchronized(someString) as a lock?
A: Because of object identity issues:Integer: autoboxing caches values -128 to 127, so different variables may share the same object - unrelated code could accidentally synchronize on the same lock.String: the string pool interns literals, so "lock" in different classes may be the same object.In both cases, you lose control over who else might be synchronizing on the same object, leading to unexpected contention or
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487168095 |
1 | 230% | 121d | 11 |
nid:1774487167493
c1
PProg
visibility, every read of a volatile field sees the most rec...
1
lapses
1/4
users
230%
ease
nid:1774487167493
Cloze c1
Cloze answer: visibility, every read of a volatile field sees the most recent write by any thread
Q: The Java volatile keyword guarantees {{c1::visibility, every read of a volatile field sees the most recent write by any thread}}, but does not guarantee {{c2::atomicity of compound operations (e.g. i++)}}.
A: Use volatile for simple flags (e.g. volatile boolean running). For compound operations, use synchronized or AtomicInteger.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487167494 |
1 | 230% | 131d | 11 |
nid:1774917598731
c1
PProg
NEW; RUNNABLE; start()
1
lapses
1/4
users
230%
ease
nid:1774917598731
Cloze c1
Cloze answer: NEW; RUNNABLE; start()
Q: A newly created Java thread starts in the {{c1::NEW}} state and transitions to {{c1::RUNNABLE}} when {{c1::start()}} is called.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774917598731 |
1 | 230% | 148d | 8 |
nid:1777538021652
c1
PProg
Pipelining (processors execute parts of multiple instruction...
1
lapses
1/4
users
230%
ease
nid:1777538021652
Cloze c1
Cloze answer: Pipelining (processors execute parts of multiple instructions simultaneously and may reorder them internally)
Q: Modern multiprocessors do not enforce a global ordering of all instructions for two main reasons: {{c1::Pipelining (processors execute parts of multiple instructions simultaneously and may reorder them internally)}}.{{c2::Per-processor local caches (loads and stores become visib
A: What exactly is guaranteed varies a lot across architectures (x86 vs ARM vs POWER vs Alpha …).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1777538021653 |
1 | 230% | 148d | 10 |
nid:1774310311659
PProg
What speed-up bound does Gustafson's Law specify?
1
lapses
1/4
users
230%
ease
nid:1774310311659
Q: What speed-up bound does Gustafson's Law specify?
A: Consider an infinite number of processors. Additionally, we assume that \(f < 1\), which is the same as saying the program has a parallel part. It follows that \(1 - f > 0\). \[ \begin{aligned} \lim_{P \to \infty} S_P &= f + P \cdot (1 - f) \\ &= f + (1 - f) \cdot \lim_{P \to \infty} P \\ &= \infty \end{aligned} \] Since \(P\) grows infinitely large and \(1 - f > 0\), \(S_P\) does not converge, meaning the speedup is unlimited.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774310311659 |
1 | 230% | 171d | 8 |
nid:1772531045442
c1
PProg
a shared buffer
1
lapses
1/4
users
230%
ease
nid:1772531045442
Cloze c1
Cloze answer: a shared buffer
Q: Producer-ConsumerProducer puts items into {{c1::a shared buffer}}, consumer takes them out.
A: For simplicity, buffer is unbounded (has no capacity limit); producing is always possible. But consumption only possible if buffer isn't empty.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772531045442 |
1 | 230% | 151d | 9 |
nid:1771365476472
c1
PProg
Thread mapping
1
lapses
1/4
users
230%
ease
nid:1771365476472
Cloze c1
Cloze answer: Thread mapping
Q: {{c1::Thread mapping}} describes {{c2::how a Java/JVM thread is related to an operating system thread}}.
A: In native threading (most common), each JVM thread is mapped to a dedicated operating system thread. In green threading, the JVM maps several threads to a single operating system thread.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771365476475 |
1 | 230% | 164d | 9 |
nid:1774359475784
c1
PProg
T_1 / p + T_\infty
1
lapses
1/4
users
230%
ease
nid:1774359475784
Cloze c1
Cloze answer: T_1 / p + T_\infty
Q: FJ work stealing scheduler: \[T_p = O({{c1::T_1 / p + T_\infty}})\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774359475784 |
1 | 230% | 208d | 8 |
nid:1774487167488
c4
PProg
Supports delayed and periodic task execution.
1
lapses
1/4
users
230%
ease
nid:1774487167488
Cloze c4
Cloze answer: Supports delayed and periodic task execution.
Q: The four standard ExecutorService pool types:newFixedThreadPool(n) - {{c1::Fixed n threads; excess tasks are queued.}}newSingleThreadExecutor() - {{c2::Exactly 1 thread; tasks execute sequentially.}}new
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774631279572 |
1 | 230% | 213d | 8 |
nid:1774362467471
IO r2
PProg
[Image Occlusion region 2]
1
lapses
1/4
users
230%
ease
nid:1774362467471
Cloze c2
Q: {{c1::image-occlusion:rect:left=.057:top=.0000:width=.9345:height=.9956}}{{c3::image-occlusion:rect:left=.057:top=.515:width=.9323:height=.4768}}{{c2::image-occlusion:rect:left=.057:top=.309:width=.9345:height=.6905}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774362467471 |
1 | 230% | 228d | 8 |
nid:1774487167070
c1
PProg
the serial fraction \(f\) in Amdahl's Law
1
lapses
1/4
users
230%
ease
nid:1774487167070
Cloze c1
Cloze answer: the serial fraction \(f\) in Amdahl's Law
Q: The span {{c2::\(T_\infty\)}} in a DAG corresponds to {{c1::the serial fraction \(f\) in Amdahl's Law}}.
A: (The longest chain of sequential dependencies that no amount of additional parallelism can overcome.)Designing parallel algorithms means decreasing span without increasing work too much - directly equivalent to reducing \(f\) in Amdahl's Law.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487167070 |
1 | 230% | 211d | 12 |
nid:1774487167075
c1
PProg
Latency of the first element through a pipeline \(= {{c1::\s...
1
lapses
1/4
users
230%
ease
nid:1774487167075
Cloze c1
Q: Latency of the first element through a pipeline \(= {{c1::\sum_{i} \text{time}(\text{stage}_i)}}\)
A: "Latency" by default refers to the first element. For a balanced pipeline this equals num_stages × max(stage_time).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487167075 |
1 | 230% | 227d | 12 |
nid:1771365476583
c2
PProg
a single answer from a collection via an associative operato...
1
lapses
1/4
users
230%
ease
nid:1771365476583
Cloze c2
Cloze answer: a single answer from a collection via an associative operator
Q: {{c1::Reductions}} produce {{c2::a single answer from a collection via an associative operator}}.
A: Examples: max, count, rightmost, sum.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771365476593 |
1 | 230% | 271d | 8 |
nid:1774487167926
PProg
Compare Big \(O\) of work, span and parallelism for these pa...
1
lapses
1/4
users
230%
ease
nid:1774487167926
Q: Compare Big \(O\) of work, span and parallelism for these parallel quicksort strategies:Parallelize only the recursive callsAlso parallelize the partition step (via pack)
A: VariantWorkSpanParallelismParallel recursive calls only\(O(n \log n)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487167926 |
1 | 230% | 266d | 12 |
nid:1774487168261
PProg
A pipeline has 4 stages with times [2, 4, 2, 2].Is it balanc...
1
lapses
1/4
users
230%
ease
nid:1774487168261
Q: A pipeline has 4 stages with times [2, 4, 2, 2].Is it balanced? What is the throughput? What is the latency of the 1st element? Of the 3rd?
A: Balanced? No — stage 2 takes 4 units; others take 2. Bottleneck is stage 2.Throughput \(= 1/\max(2,4,2,2) = 1/4\) items per time unit.Latency (1st element) \(= 2+4+2+2 = 10\)Latency (3rd element) \(= 10 + (4-2)\cdot(3-1) = 10+4 = 14\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487168261 |
1 | 230% | 293d | 12 |
nid:1774487167626
PProg
What does an exclusive parallel prefix sum compute?
1
lapses
1/4
users
230%
ease
nid:1774487167626
Q: What does an exclusive parallel prefix sum compute?
A: For input array \(A[0..n-1]\), it produces output \(B\) where \(B[i] = \sum_{j=0}^{i-1} A[j]\) (sum of all elements before index \(i\)). So \(B[0] = 0\) always.Example: \(A = [3,1,4,1,5] → B = [0,3,4,8,9]\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487167626 |
1 | 230% | 298d | 9 |
nid:1774487167037
c1
PProg
Pipeline throughput bound \(= {{c1::\dfrac{1}{\max_i(\text{s...
1
lapses
1/4
users
230%
ease
nid:1774487167037
Cloze c1
Q: Pipeline throughput bound \(= {{c1::\dfrac{1}{\max_i(\text{stage_time}_i)} }}\)(infinite stream, one execution unit per stage)
A: Throughput is limited by the slowest (bottleneck) stage. For a balanced pipeline all stage times are equal, so throughput = 1/stage_time.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1774487167037 |
1 | 230% | 324d | 9 |
nid:1771365476510
c2
PProg
a property of a system: "nothing bad ever happens"
1
lapses
1/4
users
230%
ease
nid:1771365476510
Cloze c2
Cloze answer: a property of a system: "nothing bad ever happens"
Q: A {{c1::safety property}} is {{c2::a property of a system: "nothing bad ever happens"}}.
A: Can be violated in finite time.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771365476513 |
1 | 230% | 388d | 9 |
nid:1772531107039
c1
PProg
notify()
1
lapses
1/4
users
230%
ease
nid:1772531107039
Cloze c1
Cloze answer: notify()
Q: {{c1::notify()}} wakes the highest-priority thread closest to front of object's internal queue.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1772531107039 |
1 | 230% | 412d | 9 |
nid:1771365476475
c2
PProg
Locally reason about one thread at a time
1
lapses
1/4
users
230%
ease
nid:1771365476475
Cloze c2
Cloze answer: Locally reason about one thread at a time
Q: Locality has several meanings in parallel programming:{{c2::Locally reason about one thread at a time}} (thread modularity) - simplifies correctness arguments.{{c3::Data locality}}: related memory locations are accessed shortly after each other - improves cache usage{{c
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771365476481 |
1 | 230% | 440d | 13 |
nid:1771365476576
c1
PProg
livelock
1
lapses
1/4
users
230%
ease
nid:1771365476576
Cloze c1
Cloze answer: livelock
Q: A {{c1::livelock}} is a situation in which {{c2::all threads starve by infinitely often trying to enter a critical section, but never succeeding}}.
A: Similar to a deadlock, the system makes no real progress, although the threads execute statements/use CPU time.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771365476585 |
1 | 230% | 523d | 9 |
nid:1771365476415
c1
PProg
Mutual exclusion
1
lapses
1/4
users
230%
ease
nid:1771365476415
Cloze c1
Cloze answer: Mutual exclusion
Q: {{c1::Mutual exclusion}} means preventing {{c2::more than one thread from being in a critical section, i.e. to execute a piece of code, at a given moment in time}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771365476420 |
1 | 230% | 596d | 9 |
nid:1771365476472
c2
PProg
how a Java/JVM thread is related to an operating system thre...
1
lapses
1/4
users
230%
ease
nid:1771365476472
Cloze c2
Cloze answer: how a Java/JVM thread is related to an operating system thread
Q: {{c1::Thread mapping}} describes {{c2::how a Java/JVM thread is related to an operating system thread}}.
A: In native threading (most common), each JVM thread is mapped to a dedicated operating system thread. In green threading, the JVM maps several threads to a single operating system thread.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1771365476473 |
1 | 230% | 597d | 9 |
nid:1781194821527
c1
DDCA
assign
1
lapses
1/4
users
230%
ease
nid:1781194821527
Cloze c1
Cloze answer: assign
Q: Behavioral constructs: a continuous {{c1::assign}} drives combinational logic from an expression; an {{c2::always}} block re-runs per its sensitivity list and can model combinational or sequential logic; an {{c3::initial}} block runs once at simulation start (mainly for
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781194821529 |
1 | 230% | 5d | 8 |
nid:1781529977546
c3
DDCA
clock generation (an always block toggling the clock for syn...
1
lapses
1/4
users
230%
ease
nid:1781529977546
Cloze c3
Cloze answer: clock generation (an always block toggling the clock for synchronous DUTs)
Q: The four typical parts of a testbench: {{c1::DUT instantiation (an instance of the module under test)}}; {{c2::signal declarations (reg for signals driven into the DUT, wire for signals observed from it)}}; {{c3::clock generation (an alwa
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781529977546 |
1 | 230% | 1d | 4 |
nid:1781529977639
c3
DDCA
not used by synthesis tools to infer real hardware delays
1
lapses
1/4
users
230%
ease
nid:1781529977639
Cloze c3
Cloze answer: not used by synthesis tools to infer real hardware delays
Q: Behavioral delays written with {{c1::#<delay> (e.g. assign #5 y = a & b;)}} are useful for {{c2::modeling abstract delays in early simulation or controlling event timing}}, but are {{c3::not used by synthesis tools to infer real hardware delays}}, so relying
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781529977641 |
1 | 230% | 1d | 4 |
nid:1781529977814
c1
DDCA
operate (data-processing)
1
lapses
1/4
users
230%
ease
nid:1781529977814
Cloze c1
Cloze answer: operate (data-processing)
Q: The three fundamental instruction categories: {{c1::operate (data-processing)}} instructions - arithmetic and logical transformations; {{c2::data movement}} instructions - transfers between memory and registers; {{c3::control-flow}} instructions - branches, jumps, calls
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781529977815 |
1 | 230% | 3d | 7 |
nid:1781529977883
c1
DDCA
N, Z, P; negative, zero, positive
1
lapses
1/4
users
230%
ease
nid:1781529977883
Cloze c1
Cloze answer: N, Z, P; negative, zero, positive
Q: LC-3 condition codes: most instructions that write a register set the {{c1::N, Z, P}} flags ({{c1::negative, zero, positive}}), and conditional branches like {{c2::BRn / BRz / BRp (and combinations such as BRnz)}} test them.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781529977884 |
1 | 230% | 3d | 7 |
nid:1781878852941
c1
DDCA
Superscalar
1
lapses
1/4
users
230%
ease
nid:1781878852941
Cloze c1
Cloze answer: Superscalar
Q: {{c1::Superscalar}} processing is the ability to fetch, decode, issue, and execute {{c2::multiple instructions in a single clock cycle}}, achieved by providing {{c2::multiple parallel execution units}} (several ALUs, load/store units, FPUs) and the logic to manage them.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781878852942 |
1 | 230% | 1d | 6 |
nid:1781878852963
c1
DDCA
clock gating - disables the clock to idle modules to cut dyn...
1
lapses
1/4
users
230%
ease
nid:1781878852963
Cloze c1
Cloze answer: clock gating - disables the clock to idle modules to cut dynamic power
Q: Power, latency, and reliability techniques: {{c1::clock gating - disables the clock to idle modules to cut dynamic power}};{{c2::DVFS - dynamically adjusts voltage and frequency to match the workload}};{{c3::prefetching - fetches data/instructions into caches before the
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781878852966 |
1 | 230% | 1d | 6 |
nid:1780952357974
DDCA
When does a positive edge-triggered D flip-flop update its o...
1
lapses
1/4
users
230%
ease
nid:1780952357974
Q: When does a positive edge-triggered D flip-flop update its output?
A: Only at the instant CLK transitions LOW-HIGH it samples D and copies it to Q. Q then stays fixed regardless of changes on D until the next rising edge, avoiding the transparency of a latch.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780952357974 |
1 | 230% | 2d | 5 |
nid:1781194821612
c1
DDCA
Hold-time constraint: the fastest path must satisfy \[{{c1::...
1
lapses
1/4
users
230%
ease
nid:1781194821612
Cloze c1
Q: Hold-time constraint: the fastest path must satisfy \[{{c1::\begin{gathered} t_{ccq} + t_{cd} \ge t_{h} \end{gathered} }}\]so new data does not reach FF2 too early; it is {{c2::independent of the clock frequency}} and is usually fixed by {{c3::adding buffers (delay) to the short path}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781194821614 |
1 | 230% | 4d | 8 |
nid:1781529977506
c3
DDCA
SAT solvers
1
lapses
1/4
users
230%
ease
nid:1781529977506
Cloze c3
Cloze answer: SAT solvers
Q: Three formal-verification techniques: {{c1::model checking}} explores the design's state space to check properties (often in temporal logic); {{c2::equivalence checking}} compares two design versions (e.g. RTL vs synthesized netlist) for logical equivalence; {{c3::SAT s
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781529977508 |
1 | 230% | 4d | 8 |
nid:1781529977865
c2
DDCA
procedure calls
1
lapses
1/4
users
230%
ease
nid:1781529977865
Cloze c2
Cloze answer: procedure calls
Q: Control-flow instruction types: {{c1::unconditional jumps/branches}} always change the PC (MIPS j, jr); {{c1::conditional branches}} change it only if a condition holds (MIPS beq, bne); {{c2::procedure calls}} save the
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781529977865 |
1 | 230% | 4d | 7 |
nid:1781878852821
c4
DDCA
side-effect free (functionally pure)
1
lapses
1/4
users
230%
ease
nid:1781878852821
Cloze c4
Cloze answer: side-effect free (functionally pure)
Q: Core principles of the dataflow model: execution is {{c1::data-driven}} - a node runs when all its input operands are available, not when a program counter reaches it;programs are represented as {{c2::directed graphs}}, where nodes are {{c2::operations}} and arcs carry {{c2::dat
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781878852823 |
1 | 230% | 2d | 7 |
nid:1781878852849
c4
DDCA
dynamic resource allocation
1
lapses
1/4
users
230%
ease
nid:1781878852849
Cloze c4
Cloze answer: dynamic resource allocation
Q: Challenges that have hindered general-purpose dataflow machines: the complexity of {{c1::managing and matching data tokens}};{{c2::memory latency}} can stall large portions of the graph if data fetching is not managed well;difficulty {{c3::expressing complex conditional
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781878852850 |
1 | 230% | 3d | 7 |
nid:1780255536617
DDCA
What does a multiplexer (MUX) do?
1
lapses
1/4
users
230%
ease
nid:1780255536617
Q: What does a multiplexer (MUX) do?
A: Selects one of \(2^n\) data inputs and routes it to a single output; the choice is made by n select lines (a data selector).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780255536617 |
1 | 230% | 8d | 10 |
nid:1780952357871
c1
DDCA
SRAM
1
lapses
1/4
users
230%
ease
nid:1780952357871
Cloze c1
Cloze answer: SRAM
Q: {{c1::SRAM}} is used for CPU caches and register files: fast access, lower density / higher cost per bit, and needs {{c2::no refresh}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780952357872 |
1 | 230% | 5d | 6 |
nid:1780952357894
c1
DDCA
level-sensitive: its mode (transparent vs opaque) is set by ...
1
lapses
1/4
users
230%
ease
nid:1780952357894
Cloze c1
Cloze answer: level-sensitive: its mode (transparent vs opaque) is set by the level of the gate G, not by an edge
Q: The D latch is {{c1::level-sensitive: its mode (transparent vs opaque) is set by the level of the gate G, not by an edge}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780952357895 |
1 | 230% | 4d | 6 |
nid:1781194821496
c1
DDCA
Structural; Behavioral
1
lapses
1/4
users
230%
ease
nid:1781194821496
Cloze c1
Cloze answer: Structural; Behavioral
Q: {{c1::Structural}} modeling describes a module by instantiating components (primitive gates or sub-modules) and wiring them together.{{c1::Behavioral}} modeling describes what the module does at a higher level (via assign and always) and lets the synthesis tool infe
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781194821497 |
1 | 230% | 7d | 8 |
nid:1781194821565
c1
DDCA
For a combinational path the {{c1::propagation delay \(t_{pd...
1
lapses
1/4
users
230%
ease
nid:1781194821565
Cloze c1
Q: For a combinational path the {{c1::propagation delay \(t_{pd}\)}} is the MAXIMUM time for the output to settle to its final value (latest it becomes valid), while the {{c2::contamination delay \(t_{cd}\)}} is the MINIMUM time before the output starts to change (earliest it may become invalid).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781194821566 |
1 | 230% | 6d | 8 |
nid:1781529977513
c4
DDCA
not a substitute for static timing analysis
1
lapses
1/4
users
230%
ease
nid:1781529977513
Cloze c4
Cloze answer: not a substitute for static timing analysis
Q: HDL functional simulation is the {{c1::most common}} verification method: the design under test (DUT) is simulated in software and driven by a {{c2::testbench}}. It primarily verifies {{c3::logical correctness and functional behavior}}, and is {{c4::not a substitute for static timing analysi
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781529977513 |
1 | 230% | 4d | 5 |
nid:1781194821571
c1
DDCA
critical (longest) path; shortest path
1
lapses
1/4
users
230%
ease
nid:1781194821571
Cloze c1
Cloze answer: critical (longest) path; shortest path
Q: The {{c1::critical (longest) path}} has the maximum total \(t_{pd}\) and sets the maximum clock frequency; the {{c1::shortest path}} has the minimum total \(t_{cd}\) and is what matters for avoiding {{c2::hold-time violations}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781194821571 |
1 | 230% | 9d | 6 |
nid:1781529977661
c1
DDCA
SDF (Standard Delay Format)
1
lapses
1/4
users
230%
ease
nid:1781529977661
Cloze c1
Cloze answer: SDF (Standard Delay Format)
Q: Gate-level timing simulation back-annotates delay information (e.g. from an {{c1::SDF (Standard Delay Format)}} file) onto the gate-level netlist and simulates it. It is {{c2::more computationally intensive than STA}} but can catch {{c3::dynamic effects or complex clocking issues}} that STA
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1781529977663 |
1 | 230% | 6d | 8 |
nid:1780255536493
DDCA
Why does static CMOS have very low static power dissipation?
1
lapses
1/4
users
230%
ease
nid:1780255536493
Q: Why does static CMOS have very low static power dissipation?
A: The PUN and PDN are mutually exclusive, so for stable inputs there is never a direct conducting path from \(V_{dd}\) to GND.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780255536493 |
1 | 230% | 12d | 7 |
nid:1780255536502
c1
DDCA
4; 2; 6
1
lapses
1/4
users
230%
ease
nid:1780255536502
Cloze c1
Cloze answer: 4; 2; 6
Q: A 2-input CMOS NAND uses {{c1::4}} transistors and an inverter uses {{c1::2}}, so AND built as NAND + NOT uses {{c1::6}} transistors.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780255536502 |
1 | 230% | 12d | 7 |
nid:1780255536667
c3
DDCA
(signed overflow)
1
lapses
1/4
users
230%
ease
nid:1780255536667
Cloze c3
Cloze answer: (signed overflow)
Q: ALU status flags: Z {{c1::(result is zero)}}, C {{c2::(carry-out of the MSB)}}, V {{c3::(signed overflow)}}, and N {{c4::(negative/sign = MSB of the result)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780255536668 |
1 | 230% | 12d | 7 |
nid:1780255536439
DDCA
What three quantities does computer architecture seek to bal...
1
lapses
1/4
users
230%
ease
nid:1780255536439
Q: What three quantities does computer architecture seek to balance?
A: Performance, energy efficiency, and cost-effectiveness.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780255536439 |
1 | 230% | 19d | 7 |
nid:1780255536667
c1
DDCA
(result is zero)
1
lapses
1/4
users
230%
ease
nid:1780255536667
Cloze c1
Cloze answer: (result is zero)
Q: ALU status flags: Z {{c1::(result is zero)}}, C {{c2::(carry-out of the MSB)}}, V {{c3::(signed overflow)}}, and N {{c4::(negative/sign = MSB of the result)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780255536670 |
1 | 230% | 13d | 7 |
nid:1780255536667
c4
DDCA
(negative/sign = MSB of the result)
1
lapses
1/4
users
230%
ease
nid:1780255536667
Cloze c4
Cloze answer: (negative/sign = MSB of the result)
Q: ALU status flags: Z {{c1::(result is zero)}}, C {{c2::(carry-out of the MSB)}}, V {{c3::(signed overflow)}}, and N {{c4::(negative/sign = MSB of the result)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780255536667 |
1 | 230% | 13d | 7 |
nid:1780255536608
DDCA
What does a decoder do?
1
lapses
1/4
users
230%
ease
nid:1780255536608
Q: What does a decoder do?
A: Translates an n-bit input code into one of \(2^n\) outputs, asserting exactly one line ('one-hot'); output \(Y_i\) equals minterm \(m_i\) of the inputs.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| lorenz | cid:1780255536608 |
1 | 230% | 22d | 7 |
nid:1761029886806
LinAlg
If columns \(v_1, v_2, ..., v_n\) of \(A\) are linearly inde...
1
lapses
1/4
users
290%
ease
nid:1761029886806
Q: If columns \(v_1, v_2, ..., v_n\) of \(A\) are linearly independent and \(A\lambda = A\mu = x\) are two ways of writing vector x as a linear combination of the vectors v then:
A: \(\lambda \ \text{and} \ \mu\) are the exact same vector of coefficients.Linear combinations are unique if all vectors are independent.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761029886806 |
1 | 290% | 36d | 9 |
nid:1761491477291
DiskMat
If \(F \models G\) in predicate logic, what can we conclude ...
1
lapses
1/4
users
230%
ease
nid:1761491477291
Q: If \(F \models G\) in predicate logic, what can we conclude via validity?
A: If \(F\) is valid, then \(G\) is also valid. (Logical consequence preserves validity)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477292 |
1 | 230% | 3d | 5 |
nid:1761491477341
DiskMat
What is the cardinality of the power set of a finite set wit...
1
lapses
1/4
users
260%
ease
nid:1761491477341
Q: What is the cardinality of the power set of a finite set with cardinality \(k\)?
A: \(|\mathcal{P}(A)| = 2^k\) (hence the alternative notation \(2^A\))
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477342 |
1 | 260% | 77d | 6 |
nid:1761491477349
DiskMat
What are the idempotence laws for sets?
1
lapses
1/4
users
260%
ease
nid:1761491477349
Q: What are the idempotence laws for sets?
A: \(A \cap A = A\)
\(A \cup A = A\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477350 |
1 | 260% | 83d | 6 |
nid:1761491477351
DiskMat
What are the commutativity laws for sets?
1
lapses
1/4
users
260%
ease
nid:1761491477351
Q: What are the commutativity laws for sets?
A: \(A \cap B = B \cap A\)
\(A \cup B = B \cup A\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477352 |
1 | 260% | 29d | 5 |
nid:1761491477439
DiskMat
What is the greatest lower bound (glb) of a subset \(S\) in ...
1
lapses
1/4
users
230%
ease
nid:1761491477439
Q: What is the greatest lower bound (glb) of a subset \(S\) in a poset?
A: The greatest element (by the relation, not just integer ordering) of the set of all lower bounds of \(S\). Also called the infimum.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477440 |
1 | 230% | 43d | 9 |
nid:1761491477479
DiskMat
When does set \(B\) dominate set \(A\) (denoted \(A \preceq ...
1
lapses
1/4
users
245%
ease
nid:1761491477479
Q: When does set \(B\) dominate set \(A\) (denoted \(A \preceq B\))?
A: When \(A \sim C\) for some subset \(C \subseteq B\), or equivalently, when there exists an injection \(A \to B\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477480 |
1 | 245% | 70d | 11 |
nid:1761491477481
DiskMat
What does it mean for a set \(A\) to be countable?
1
lapses
1/4
users
260%
ease
nid:1761491477481
Q: What does it mean for a set \(A\) to be countable?
A: \(A \preceq \mathbb{N}\) (i.e., there exists an injection \(A \to \mathbb{N}\))
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477482 |
1 | 260% | 25d | 8 |
nid:1761491477489
DiskMat
What are the two types of countable sets?
1
lapses
1/4
users
260%
ease
nid:1761491477489
Q: What are the two types of countable sets?
A: \(A\) is countable if and only if \(A \sim \mathbb{N}\) or \(A \sim \mathbf{n}\) for some \(n \in \mathbb{N}\) (i.e., \(A\) is finite or equinumerous with \(\mathbb{N}\)).
Conclusion: No cardinality level exists between finite and countably infinite.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477490 |
1 | 260% | 49d | 11 |
nid:1761491477505
DiskMat
What is a computable function \(f: \mathbb{N} \to \{0, 1\}\)...
1
lapses
1/4
users
305%
ease
nid:1761491477505
Q: What is a computable function \(f: \mathbb{N} \to \{0, 1\}\)?
A: A function for which there exists a program that, for every \(n \in \mathbb{N}\), when given \(n\) as input, outputs \(f(n)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477506 |
1 | 305% | 32d | 10 |
nid:1761491477525
DiskMat
What fundamental property distinguishes finite from infinite...
1
lapses
1/4
users
275%
ease
nid:1761491477525
Q: What fundamental property distinguishes finite from infinite sets regarding proper subsets?
A: A finite set never has the same cardinality as one of its proper subsets. An infinite set can (e.g., \(\mathbb{N} \sim \mathbb{O}\) where \(\mathbb{O}\) is the set of odd numbers).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1761491477526 |
1 | 275% | 10d | 8 |
nid:1762106939300
DiskMat
What is \(\text{gcd}(a, b)\)?
1
lapses
1/4
users
245%
ease
nid:1762106939300
Q: What is \(\text{gcd}(a, b)\)?
A: The unique positive greatest common divisor of \(a\) and \(b\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762106939301 |
1 | 245% | 12d | 7 |
nid:1762106939342
c1
DiskMat
\(a \equiv_m R_m(a)\) (the remainder represents the equival...
1
lapses
1/4
users
245%
ease
nid:1762106939342
Cloze c1
Cloze answer: \(a \equiv_m R_m(a)\) (the remainder represents the equivalence class)
Q: What are the two key properties of the remainder function \(R_m\)? (Lemma 4.16)(i) {{c1:: \(a \equiv_m R_m(a)\) (the remainder represents the equivalence class)}}(ii) {{c2:: \(a \equiv_m b \Longleftrightarrow R_m(a) = R_m(b)\) (congru
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762106939343 |
1 | 245% | 33d | 9 |
nid:1762106939348
DiskMat
State the Chinese Remainder Theorem (Theorem 4.19).
1
lapses
1/4
users
260%
ease
nid:1762106939348
Q: State the Chinese Remainder Theorem (Theorem 4.19).
A: Let \(m_1, m_2, \dots, m_r\) be pairwise relatively prime integers and let \(M = \prod_{i=1}^{r} m_i\). For every list \(a_1, \dots, a_r\) with \(0 \leq a_i < m_i\), the system
\[\begin{align} x &\equiv_{m_1} a_1 \\ x &\equiv_{m_2} a_2 \\ &\vdots \\ x &\equiv_{m_r} a_r \end{align}\]
has a unique solution \(x\) satisfying \(0 \leq x < M\).Why unique: If there are two solutions, then, for all \(i\):\(x \equiv_{m_i} a_i\) and
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762106939349 |
1 | 260% | 88d | 6 |
nid:1762106939370
c1
DiskMat
\(a \equiv_m a\) since \(m \mid (a - a) = 0\) ✓
1
lapses
1/4
users
230%
ease
nid:1762106939370
Cloze c1
Cloze answer: \(a \equiv_m a\) since \(m \mid (a - a) = 0\) ✓
Q: Verify that \(\equiv_m\) is reflexive, symmetric, and transitive.Reflexive: {{c1:: \(a \equiv_m a\) since \(m \mid (a - a) = 0\) ✓}}Symmetric: {{c2:: \(a \equiv_m b \Rightarrow m \mid (a-b) \Rightarrow m \mid (b-a) \Righ
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762106939371 |
1 | 230% | 45d | 7 |
nid:1762856073563
LinAlg
Was ist eine konjugiert-transponierte (auch: Hermitesch-tran...
1
lapses
1/4
users
260%
ease
nid:1762856073563
Q: Was ist eine konjugiert-transponierte (auch: Hermitesch-transponierte) Matrix?
A: \( \mathbf{A}^* = (\overline{\mathbf{A}})^\top = \overline{\mathbf{A}^\top}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073563 |
1 | 260% | 25d | 10 |
nid:1762856073577
c1
DiskMat
symmetric
1
lapses
1/4
users
245%
ease
nid:1762856073577
Cloze c1
Cloze answer: symmetric
Q: A relation ρ on a set A is called {{c1::symmetric}} if {{c2::\( a \ \rho \ b \iff b \ \rho \ a\) is true, i.e. if \( \rho = \hat{\rho}\)}}
A: Examples: \( \equiv_m\), marriage
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073578 |
1 | 245% | 24d | 6 |
nid:1762856073621
c1
DiskMat
meet of \(a\) and \(b\) (also denoted \(a \land b\)).
1
lapses
1/4
users
260%
ease
nid:1762856073621
Cloze c1
Cloze answer: meet of \(a\) and \(b\) (also denoted \(a \land b\)).
Q: Consider the poset \((A;\preceq)\). If \(\{a,b\}\) have a {{c2::greatest lower bound}}, then it is called the {{c1::meet of \(a\) and \(b\) (also denoted \(a \land b\)).}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073629 |
1 | 260% | 71d | 6 |
nid:1762856073628
c1
DiskMat
dominates (denoted \(A \preceq B\))
1
lapses
1/4
users
245%
ease
nid:1762856073628
Cloze c1
Cloze answer: dominates (denoted \(A \preceq B\))
Q: The set \(B\) {{c1::dominates (denoted \(A \preceq B\))}} if {{c2::there exists an injective function \(A \rightarrow B\).}}
A: Example: \(f(x): \mathbb{N} \rightarrow \mathbb{R} = x\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073642 |
1 | 245% | 42d | 5 |
nid:1762856073660
c2
DiskMat
\(\langle R, +, -, 0 \rangle\) is a commutative group
1
lapses
1/4
users
215%
ease
nid:1762856073660
Cloze c2
Cloze answer: \(\langle R, +, -, 0 \rangle\) is a commutative group
Q: {{c1::A ring \(\langle R, +, -, 0, \cdot, 1 \rangle\)}} is an algebra with the properties that{{c2::\(\langle R, +, -, 0 \rangle\) is a commutative group}}{{c3::\(\langle R, \cdot, 1 \rangle\) is a monoid}}{{c4::\( a(b+c) = (ab) + (ac), (b+c)a = (ba)
A: Examples: \(\mathbb{Z}, \mathbb{R}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856073673 |
1 | 215% | 26d | 7 |
nid:1762856074477
c2
A&D
closed Eulerian walk (Eulerzyklus)
1
lapses
1/4
users
245%
ease
nid:1762856074477
Cloze c2
Cloze answer: closed Eulerian walk (Eulerzyklus)
Q: In graph theory, a {{c2::closed Eulerian walk (Eulerzyklus)}} is an {{c1::Eulerian walk (Eulerweg) that ends at the start vertex}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856074510 |
1 | 245% | 88d | 7 |
nid:1762856074631
c2
DiskMat
expression using the propositional symbols \(A, B, C, \dots\...
1
lapses
1/4
users
245%
ease
nid:1762856074631
Cloze c2
Cloze answer: expression using the propositional symbols \(A, B, C, \dots\) and logical operators \(\land, \lor, \lnot, \ldots\)
Q: An {{c2::expression using the propositional symbols \(A, B, C, \dots\) and logical operators \(\land, \lor, \lnot, \ldots\)}} is called a {{c1::formula (of propositional logic)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856074667 |
1 | 245% | 5d | 7 |
nid:1762856074659
c1
DiskMat
composite
1
lapses
1/4
users
245%
ease
nid:1762856074659
Cloze c1
Cloze answer: composite
Q: An integer greater than \(1\) that is not a prime is called {{c1::composite}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856074691 |
1 | 245% | 75d | 5 |
nid:1762856074680
c1
DiskMat
The Fermat-Euler theorem states that for all \(m\ge 2\) and ...
1
lapses
1/4
users
260%
ease
nid:1762856074680
Cloze c1
Q: The Fermat-Euler theorem states that for all \(m\ge 2\) and all \(a\) with \(\gcd(a,m) = 1\),{{c1:: \[a^{\varphi(m)} \equiv_m 1\]and so in particular, for every prime \(p\) and every \(a\) not divisible by \(p\): \(a^{p-1} \equiv_p 1\).}}
A: We know \(a^{\operatorname{order}(a)} \equiv_m 1\). Since \(\operatorname{order}(a)\) divides \(| \mathbb{Z}_m^* | = \varphi(m)\) (Lagrange's), \(a^{\varphi(m)} \equiv_m a^{k \cdot \operatorname{order}(a)} \equiv_m (a^{\operatorname{order}(a)})^k \equiv_m 1^k \equiv_m 1\)This theorem is used for RSA.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856074708 |
1 | 260% | 26d | 8 |
nid:1762856074690
c1
LinAlg
\(\det (A^{-1}) =\) {{c1::\((\det (A))^{-1}\)}}
1
lapses
1/4
users
275%
ease
nid:1762856074690
Cloze c1
Q: \(\det (A^{-1}) =\) {{c1::\((\det (A))^{-1}\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1762856074715 |
1 | 275% | 9d | 12 |
nid:1763362644469
c1
A&D
adjacent (adjazent oder benachbart)
1
lapses
1/4
users
260%
ease
nid:1763362644469
Cloze c1
Cloze answer: adjacent (adjazent oder benachbart)
Q: In an edge \(e = \{u, v\}\), we call \(u\) {{c1::adjacent (adjazent oder benachbart)}} to \(v\) (and the other way around) and \(e\) {{c2::incident (inzident oder anliegend)}} to \(u, v\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1763362644470 |
1 | 260% | 223d | 7 |
nid:1763363435750
c2
A&D
connected and has no cycles (Kreise)
1
lapses
1/4
users
200%
ease
nid:1763363435750
Cloze c2
Cloze answer: connected and has no cycles (Kreise)
Q: A graph \(G\) is a {{c1::tree}} if it is {{c2::connected and has no cycles (Kreise)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1763363435750 |
1 | 200% | 64d | 8 |
nid:1763364155947
c2
A&D
the subgraph obtained after removing it (keeping the vertice...
1
lapses
1/4
users
245%
ease
nid:1763364155947
Cloze c2
Cloze answer: the subgraph obtained after removing it (keeping the vertices) is disconnected
Q: An edge in a connected graph is a {{c1::cut edge}} if {{c2::the subgraph obtained after removing it (keeping the vertices) is disconnected}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1763364155947 |
1 | 245% | 36d | 8 |
nid:1763493474474
DiskMat
What do we need to state before using the decomposition of a...
1
lapses
1/4
users
260%
ease
nid:1763493474474
Q: What do we need to state before using the decomposition of an \(n \in \mathbb{Z}\) into prime factors?
A: That this is allowed by the fundamental theorem of arithmetic.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1763493474474 |
1 | 260% | 18d | 9 |
nid:1764746595604
c1
A&D
BFS
1
lapses
1/4
users
260%
ease
nid:1764746595604
Cloze c1
Cloze answer: BFS
Q: We find the shortest walk in a graph using {{c1:: BFS}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764746595604 |
1 | 260% | 140d | 5 |
nid:1764859231354
c1
DiskMat
1 by definition
1
lapses
1/4
users
275%
ease
nid:1764859231354
Cloze c1
Cloze answer: 1 by definition
Q: The order \(\text{ord}(e)\) of \(e \in G\) is {{c1:: 1 by definition}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764859231355 |
1 | 275% | 33d | 7 |
nid:1764859231539
DiskMat
When is a polynomial of degree \(2\) or \(3\) irreducible?
1
lapses
1/4
users
245%
ease
nid:1764859231539
Q: When is a polynomial of degree \(2\) or \(3\) irreducible?
A: Corollary 5.30: A polynomial \(a(x)\) of degree \(2\) or \(3\) over a field \(F\) is irreducible if and only if it has no root.
Important: This doesn't work for polynomials of higher degrees! A degree \(4\) polynomial might be the product of two irreducible degree \(2\) polynomials, each with no roots.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764859231540 |
1 | 245% | 8d | 4 |
nid:1764859231560
DiskMat
Which of the following are fields: \(\mathbb{Z}, \mathbb{Q},...
1
lapses
1/4
users
260%
ease
nid:1764859231560
Q: Which of the following are fields: \(\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}, \mathbb{Z}_5, \mathbb{Z}_6, R[x]\)?
A: Fields: \(\mathbb{Q}, \mathbb{R}, \mathbb{C}, \mathbb{Z}_5\) (where \(5\) is prime)
Not fields:
- \(\mathbb{Z}\) (not all nonzero elements have multiplicative inverse, e.g., \(2\))
- \(\mathbb{Z}_6\) (since \(6\) is not prime, e.g., \(2\) has no inverse)
- \(R[x]\) for any ring \(R\) (polynomials don't have multiplicative inverses)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764859231561 |
1 | 260% | 7d | 10 |
nid:1764859231579
DiskMat
Is \(F[x]_{m(x)}\) a monoid, group, ring, field?
1
lapses
1/4
users
260%
ease
nid:1764859231579
Q: Is \(F[x]_{m(x)}\) a monoid, group, ring, field?
A: Lemma 5.35: \(F[x]_{m(x)}\) is a commutative ring with respect to addition and multiplication modulo \(m(x)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764859231580 |
1 | 260% | 60d | 7 |
nid:1764859231602
c2
DiskMat
The {{c2::output \((c_0, \dots, c_{n-1})\)}} of an encoding ...
1
lapses
1/4
users
245%
ease
nid:1764859231602
Cloze c2
Q: The {{c2::output \((c_0, \dots, c_{n-1})\)}} of an encoding function is called a {{c1::codeword}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764859231604 |
1 | 245% | 17d | 5 |
nid:1764860289620
c1
DiskMat
\(0a = 0\)
1
lapses
1/4
users
245%
ease
nid:1764860289620
Cloze c1
Cloze answer: \(0a = 0\)
Q: In any ring \(\langle R; +, -, 0, \cdot, 1 \rangle\), and for all \(a, b \in R\) \(a0 =\) {{c1::\(0a = 0\)}}.
A: The zero (neutral of additive group) pulls all other elements to 0 by multiplication.\(0a=(0-0)a=0a-0a=0\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764860289620 |
1 | 245% | 21d | 6 |
nid:1764860422155
c1
DiskMat
\(-(ab)\)
1
lapses
1/4
users
275%
ease
nid:1764860422155
Cloze c1
Cloze answer: \(-(ab)\)
Q: In any ring \(\langle R; +, -, 0, \cdot, 1 \rangle\), and for all \(a, b \in R\) \((-a)b =\) {{c1::\(-(ab)\)}}. (Proof included)
A: Proof: \(ab+(−a)b=(a+(−a))b=0⋅b=0\)Since \((−a)b\) satisfies \(ab+(−a)b=0\), we have \((−a)b=−(ab\)).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764860422155 |
1 | 275% | 29d | 8 |
nid:1764860775647
c1
DiskMat
\(a \ | \ c\), i.e. the relation | is transitive
1
lapses
1/4
users
245%
ease
nid:1764860775647
Cloze c1
Cloze answer: \(a \ | \ c\), i.e. the relation | is transitive
Q: In any commutative ring: If \(a \ | \ b\) and \(b \ | \ c\) then {{c1:: \(a \ | \ c\), i.e. the relation | is transitive}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1764860775647 |
1 | 245% | 8d | 4 |
nid:1765194177649
LinAlg
What is the rank of a matrix?
1
lapses
1/4
users
260%
ease
nid:1765194177649
Q: What is the rank of a matrix?
A: it is the number of independent columns, where independence is defined such that given a column vector \(v_j\) then \(v_j\) is not a linear combination of \(v_1, v_2 ... v_{j-1}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765194177649 |
1 | 260% | 75d | 7 |
nid:1765198200601
A&D
Runtime: Operations in an Adjacency List:
1
lapses
1/4
users
245%
ease
nid:1765198200601
Q: Runtime: Operations in an Adjacency List:
A: 1. Check if \(uv \in E \): \(O(1 + \min\{\text{deg}(u), \text{deg}(v) \})\) (we have to check the smaller of the two adjacency lists2. Vertex \(u\), find all adjacent vertices: \(O(1+\text{deg}(u) )\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765198200601 |
1 | 245% | 26d | 10 |
nid:1765294753798
c2
A&D
\(f \leq O(g)\) and \(f \neq \Theta(g)\)
1
lapses
1/4
users
230%
ease
nid:1765294753798
Cloze c2
Cloze answer: \(f \leq O(g)\) and \(f \neq \Theta(g)\)
Q: If \(\frac{f(n)}{g(n)}\) tends to {{c1:: 0}}, then {{c2::\(f \leq O(g)\) and \(f \neq \Theta(g)\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765294753799 |
1 | 230% | 5d | 8 |
nid:1765294947576
c1
A&D
\leq
1
lapses
1/4
users
260%
ease
nid:1765294947576
Cloze c1
Cloze answer: \leq
Q: If \(f \leq O(h)\) and \(g \leq O(h)\), then \(f + g {{c1::\leq}} O(h)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765294947576 |
1 | 260% | 19d | 12 |
nid:1765295484756
A&D
When \(f \geq \Omega(g)\), this means what exactly?
1
lapses
1/4
users
260%
ease
nid:1765295484756
Q: When \(f \geq \Omega(g)\), this means what exactly?
A: \(\exists C \ge 0 \quad \forall n \in \mathbb{N} \quad f(n) \ge C\cdot g(n)\)\(f\) grows asymptotically faster than \(g\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765295484757 |
1 | 260% | 2d | 14 |
nid:1765296240804
c2
A&D
O(n!)
1
lapses
1/4
users
230%
ease
nid:1765296240804
Cloze c2
Cloze answer: O(n!)
Q: Choose a tight bound!\({{c1::O(k^n)}} \leq {{c2::O(n!)}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765296240805 |
1 | 230% | 17d | 4 |
nid:1765296364773
c2
A&D
O(n)
1
lapses
1/4
users
230%
ease
nid:1765296364773
Cloze c2
Cloze answer: O(n)
Q: Choose a tight bound!\({{c1::O(\log(n))}}\leq {{c2::O(n)}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765296364774 |
1 | 230% | 17d | 5 |
nid:1765297403833
c1
A&D
\(b = \log_2(a)\)
1
lapses
1/4
users
260%
ease
nid:1765297403833
Cloze c1
Cloze answer: \(b = \log_2(a)\)
Q: Master Theorem: If {{c1:: \(b = \log_2(a)\)}} then {{c2:: \(T(n) \leq O(n^{\log_2 a} \cdot \log n)\)}}.
A: The recursive and non-recursive work is balanced.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765297403833 |
1 | 260% | 51d | 7 |
nid:1765297729656
A&D
For \(T(n) = 4T(n/2) + n\), which Master Theorem case applie...
1
lapses
1/4
users
230%
ease
nid:1765297729656
Q: For \(T(n) = 4T(n/2) + n\), which Master Theorem case applies?
A: Because \(b = 1\) and \(\log_2(a) = \log_2 4 = 2 > b\), therefore \(T(n) = \Theta(n^2)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765297729656 |
1 | 230% | 4d | 5 |
nid:1765298206873
c2
A&D
\(O(n \log(n))\)
1
lapses
1/4
users
260%
ease
nid:1765298206873
Cloze c2
Cloze answer: \(O(n \log(n))\)
Q: {{c1:: \(\sum_{i = 1}^{n} i\log(i)\)::Sum}} \(\leq\) {{c2::\(O(n \log(n))\)::O-notation}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765298206875 |
1 | 260% | 97d | 6 |
nid:1765298610771
A&D
Provide the outline of an induction proof.
1
lapses
1/4
users
245%
ease
nid:1765298610771
Q: Provide the outline of an induction proof.
A: We want to prove that ... for \(n \geq 5\)Base Case: Let \(n = 5\) .... So the property holds for \(n = 5\).Induction Hypothesis: We assume the property is true for some \(k \geq 5\)Induction Step: We must show that the property holds for \(k + 1\).By the principle of mathematical induction ... is true for all \(n \geq 5\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765298610771 |
1 | 245% | 57d | 6 |
nid:1765301887927
A&D
How do we create a maxHeap?
1
lapses
1/4
users
245%
ease
nid:1765301887927
Q: How do we create a maxHeap?
A: Insert the node \(v\) at the next free space in the tree, i.e. first to the left, then right (to conserve the tree structure). Then we restore the heap condition by reverse-“versickern” the element until it’s restored.Swap it with it’s parent nodes until the condition is restored.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765301887927 |
1 | 245% | 3d | 5 |
nid:1765300723241
A&D
Bubble Sort
1
lapses
1/4
users
275%
ease
nid:1765300723241
Q: Bubble Sort
A: Best Case: \(O(n^2)\) (\(O(n)\) if checking for swaps and aborting early)Worst Case: \(O(n^2)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765388610996 |
1 | 275% | 47d | 11 |
nid:1765300723241
A&D
Bubble Sort
1
lapses
1/4
users
245%
ease
nid:1765300723241
Q: Bubble Sort
A: Best Case: \(O(n^2)\) (\(O(n)\) if checking for swaps and aborting early)Worst Case: \(O(n^2)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765388610998 |
1 | 245% | 41d | 7 |
nid:1765300949586
A&D
Selection Sort
1
lapses
1/4
users
245%
ease
nid:1765300949586
Q: Selection Sort
A: Best Case: \(O(n^2)\)Worst Case: \(O(n^2)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765388611000 |
1 | 245% | 45d | 9 |
nid:1765653532362
c2
EProg
char
1
lapses
1/4
users
245%
ease
nid:1765653532362
Cloze c2
Cloze answer: char
Q: The 8 primitve types of Java are:{{c1:: byte}}{{c2:: char}}{{c3:: short}}{{c4:: int}}{{c5:: long}}{{c6:: float}}{{c7:: double}}{{c8:: boolean}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765653532368 |
1 | 245% | 30d | 7 |
nid:1765653532374
c1
EProg
copied
1
lapses
1/4
users
245%
ease
nid:1765653532374
Cloze c1
Cloze answer: copied
Q: Values given to a method in Java are always {{c1::copied}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1765653532379 |
1 | 245% | 44d | 9 |
nid:1766000828773
DiskMat
What is the number of generators of \(\mathbb{Z}_{25}^* \)?
1
lapses
1/4
users
245%
ease
nid:1766000828773
Q: What is the number of generators of \(\mathbb{Z}_{25}^* \)?
A: \(\varphi(\varphi(25)) = |\mathbb{Z}_{\varphi(25)}| = |\mathbb{Z}_{20}| = 8\) ( 1, 3, 7, 9, 11, 13, 17, 19 )
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766000828773 |
1 | 245% | 10d | 7 |
nid:1766245701439
c2
A&D
\(O(1)\) as we know the offset for each key
1
lapses
1/4
users
245%
ease
nid:1766245701439
Cloze c2
Cloze answer: \(O(1)\) as we know the offset for each key
Q: In an array we can:Insert in {{c1:: \(O(1)\) as we know the first empty cell in the array and can just write the key there}}Get in {{c2::\(O(1)\) as we know the offset for each key}}InsertAfter in {{c3::\(\Theta(l)\), since we ha
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766245701441 |
1 | 245% | 50d | 4 |
nid:1766246034328
c1
A&D
previous and next element
1
lapses
1/4
users
260%
ease
nid:1766246034328
Cloze c1
Cloze answer: previous and next element
Q: In a doubly linked list, we store a pointer to the {{c1:: previous and next element}} for each key.This increases {{c2::memory usage}} as a trade-off for {{c2:: speed}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766246034328 |
1 | 260% | 29d | 5 |
nid:1766246342851
c3
A&D
\(O(1)\) if we get the memory address of the element to ins...
1
lapses
1/4
users
275%
ease
nid:1766246342851
Cloze c3
Cloze answer: \(O(1)\) if we get the memory address of the element to insert after.
Q: In a singly and doubly linked list, the operation:Insert is {{c1::\(\Theta(1)\) as we know the memory address of the final element in the list and just have to set the null pointer to the new keys address. Without this pointer it's \(\Th
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766246342851 |
1 | 275% | 37d | 9 |
nid:1766248090341
c1
A&D
LIFO
1
lapses
1/4
users
245%
ease
nid:1766248090341
Cloze c1
Cloze answer: LIFO
Q: A stack is also called a {{c1:: LIFO}} queue.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766248090341 |
1 | 245% | 21d | 4 |
nid:1766319025292
c3
DiskMat
Describe the RSA protocol:{{c1:: Alice generates primes \(p\...
1
lapses
1/4
users
275%
ease
nid:1766319025292
Cloze c3
Q: Describe the RSA protocol:{{c1:: Alice generates primes \(p\) and \(q\)}}{{c2:: Set \(n = pq\) and \(f = \varphi(n) = (p - 1)(q - 1)\) }}{{c3:: Select \(e\): \(d \equiv_f e^{-1}\) the modular inverse (decryption)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766319025293 |
1 | 275% | 15d | 9 |
nid:1766319025292
c4
DiskMat
Send \(n\) and \(e\) to Bob
1
lapses
1/4
users
245%
ease
nid:1766319025292
Cloze c4
Cloze answer: Send \(n\) and \(e\) to Bob
Q: Describe the RSA protocol:{{c1:: Alice generates primes \(p\) and \(q\)}}{{c2:: Set \(n = pq\) and \(f = \varphi(n) = (p - 1)(q - 1)\) }}{{c3:: Select \(e\): \(d \equiv_f e^{-1}\) the modular inverse (decryption)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766319025297 |
1 | 245% | 9d | 4 |
nid:1766319174572
c1
DiskMat
Closure
1
lapses
1/4
users
245%
ease
nid:1766319174572
Cloze c1
Cloze answer: Closure
Q: A monoid has the following properties:{{c1::Closure}}{{c2::Associativity}}{{c3::Identity}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766319174572 |
1 | 245% | 3d | 5 |
nid:1766319253408
DiskMat
An abelian group has the following properties:
1
lapses
1/4
users
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ease
nid:1766319253408
Q: An abelian group has the following properties:
A: closureassociativityidentityinversecommutative
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766319253408 |
1 | 245% | 11d | 4 |
nid:1766319397636
DiskMat
A field has the following properties:
1
lapses
1/4
users
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ease
nid:1766319397636
Q: A field has the following properties:
A: Additive Group:closureassociativityidentityinversecommutativeMultiplicative group:closureassociativitydistributivityidentityno zero-divisorinverse
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766319397636 |
1 | 245% | 7d | 6 |
nid:1766408177022
c1
DiskMat
meaning or semantics
1
lapses
1/4
users
275%
ease
nid:1766408177022
Cloze c1
Cloze answer: meaning or semantics
Q: The truth function \(\tau : \mathcal{S} \rightarrow \{0,1\}\) defines the {{c1:: meaning or semantics}} in \(\mathcal{S}\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766408177022 |
1 | 275% | 23d | 8 |
nid:1766418002697
c2
DiskMat
an alphabet \(\Lambda\) (of allowed symbols); which strings ...
1
lapses
1/4
users
275%
ease
nid:1766418002697
Cloze c2
Cloze answer: an alphabet \(\Lambda\) (of allowed symbols); which strings in \(\Lambda^*\) are formulas (i.e. syntactically correct)
Q: The {{c1::syntax}} of a logic defines {{c2::an alphabet \(\Lambda\) (of allowed symbols)}} and specifies {{c2::which strings in \(\Lambda^*\) are formulas (i.e. syntactically correct)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002700 |
1 | 275% | 19d | 7 |
nid:1766418002702
c2
DiskMat
An interpretation consists of {{c1::a set \(\mathcal{Z} \sub...
1
lapses
1/4
users
260%
ease
nid:1766418002702
Cloze c2
Q: An interpretation consists of {{c1::a set \(\mathcal{Z} \subseteq \Lambda\) of \(\Lambda\)}}, {{c2::a domain (a set of possible values) for each symbol in \(\mathcal{Z}\)}}, and {{c3::a function that assigns to each symbol in \(\mathcal{Z}\) a value in the a
A: A set of symbols \(\mathcal{Z} \subseteq \Lambda\)
\(\Lambda\) is the "alphabet" or collection of all available symbols
\(\mathcal{Z}\) is the subset of symbols we're actually interpreting
A domain for each symbol
For each symbol in \(\mathcal{Z}\), there's a set of possible values it could take
Often the domain is defined in terms of the universe \(U\) where a symbol can be a fu
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002710 |
1 | 260% | 16d | 7 |
nid:1766418002746
c2
DiskMat
\(F \lor G\); \(F \vdash G \lor F\)
1
lapses
1/4
users
215%
ease
nid:1766418002746
Cloze c2
Cloze answer: \(F \lor G\); \(F \vdash G \lor F\)
Q: {{c1::\(F\) }} \(\vdash\) {{c2::\(F \lor G\)}} and {{c2::\(F \vdash G \lor F\)}} are valid derivation rules.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002790 |
1 | 215% | 4d | 5 |
nid:1766418002749
c1
DiskMat
\(F \rightarrow G\) is a tautology and thus that \(F \models...
1
lapses
1/4
users
245%
ease
nid:1766418002749
Cloze c1
Cloze answer: \(F \rightarrow G\) is a tautology and thus that \(F \models G\)
Q: If in a sound calculus \(K\) one can derive \(G\) from the set of formulas \(F\) (\(F \vdash_K G\)), then one has proved that {{c1::\(F \rightarrow G\) is a tautology and thus that \(F \models G\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002795 |
1 | 245% | 13d | 7 |
nid:1766418002768
DiskMat
For DNF construction from truth table, which rows do you use...
1
lapses
1/4
users
245%
ease
nid:1766418002768
Q: For DNF construction from truth table, which rows do you use?
A: Rows evaluating to 1.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002825 |
1 | 245% | 11d | 4 |
nid:1766418002773
c2
DiskMat
clause
1
lapses
1/4
users
260%
ease
nid:1766418002773
Cloze c2
Cloze answer: clause
Q: The {{c1::empty set \(\emptyset\)}} is a {{c2::clause}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002831 |
1 | 260% | 13d | 5 |
nid:1766418002791
c2
DiskMat
\(k\) denotes the number of arguments of the predicate (the ...
1
lapses
1/4
users
260%
ease
nid:1766418002791
Cloze c2
Cloze answer: \(k\) denotes the number of arguments of the predicate (the arity)
Q: A {{c1::predicate symbol}} is of the form {{c2::\(P_i^{(k)}\) with \(i, k \in \mathbb{N}\)}}, where {{c2::\(k\) denotes the number of arguments of the predicate (the arity)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002863 |
1 | 260% | 9d | 7 |
nid:1766418002792
c1
DiskMat
A variable
1
lapses
1/4
users
230%
ease
nid:1766418002792
Cloze c1
Cloze answer: A variable
Q: A term is defined inductively: {{c1::A variable}} is a termif {{c2::\((t_1, \dots, t_k)\) are terms}}, then {{c3::\(f^{(k)}(t_1, \dots, t_k)\) is a term}}.
A: For \(k = 0\) one writes no parentheses (constants).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002866 |
1 | 230% | 4d | 8 |
nid:1766418002817
c1
DiskMat
no existence quantifiers
1
lapses
1/4
users
260%
ease
nid:1766418002817
Cloze c1
Cloze answer: no existence quantifiers
Q: Skolem normal form has {{c1::no existence quantifiers}}.It is {{c2::equisatisfiable (not equivalent!)}} to the original formula.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002920 |
1 | 260% | 13d | 7 |
nid:1766418002818
c1
DiskMat
replacing all variables bound to an \(\exists\) by a functio...
1
lapses
1/4
users
245%
ease
nid:1766418002818
Cloze c1
Cloze answer: replacing all variables bound to an \(\exists\) by a function
Q: The Skolem transformation works by {{c1::replacing all variables bound to an \(\exists\) by a function}} whose arguments are {{c2::the universally quantified variables that precede it}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002922 |
1 | 245% | 3d | 6 |
nid:1766418002830
c1
DiskMat
no variable occurs both as a bound and as a free variable
1
lapses
1/4
users
245%
ease
nid:1766418002830
Cloze c1
Cloze answer: no variable occurs both as a bound and as a free variable
Q: Rectified form:{{c1::no variable occurs both as a bound and as a free variable}}{{c2::all quantifiers use distinct variable names}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002938 |
1 | 245% | 6d | 5 |
nid:1766418002830
c2
DiskMat
all quantifiers use distinct variable names
1
lapses
1/4
users
245%
ease
nid:1766418002830
Cloze c2
Cloze answer: all quantifiers use distinct variable names
Q: Rectified form:{{c1::no variable occurs both as a bound and as a free variable}}{{c2::all quantifiers use distinct variable names}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418002939 |
1 | 245% | 4d | 8 |
nid:1766418355297
c2
DiskMat
the variables never appear in the same predicate
1
lapses
1/4
users
245%
ease
nid:1766418355297
Cloze c2
Cloze answer: the variables never appear in the same predicate
Q: We are allowed to swap quantifier order in a formula if:{{c1:: they are of the same type}}{{c2:: the variables never appear in the same predicate}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766418355297 |
1 | 245% | 5d | 5 |
nid:1766484505751
c2
A&D
insert(x, W) Insert the key x into W, as long as it’s not sa...
1
lapses
1/4
users
245%
ease
nid:1766484505751
Cloze c2
Cloze answer: insert(x, W) Insert the key x into W, as long as it’s not saved there yet
Q: The ADT Dictionary implements the following methods:{{c1::search(x, W) returns the position of the key x in memory}}{{c2::insert(x, W) Insert the key x into W, as long as it’s not saved there yet}}{{c3::delete(x, W) find and delete
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766484505753 |
1 | 245% | 33d | 4 |
nid:1766484756595
c4
A&D
\(O(\log n)\)
1
lapses
1/4
users
230%
ease
nid:1766484756595
Cloze c4
Cloze answer: \(O(\log n)\)
Q: Search
Insertion
Deletion
Non-sorted array
{{c1::\(O(n)\)}}
{{c2::\(O(1)\)}}
{
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766484756597 |
1 | 230% | 31d | 6 |
nid:1766484876704
c1
A&D
\(O(h)\), where \(h\) is the height
1
lapses
1/4
users
245%
ease
nid:1766484876704
Cloze c1
Cloze answer: \(O(h)\), where \(h\) is the height
Q: The runtime of search in a binary tree is {{c1::\(O(h)\), where \(h\) is the height}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766484876704 |
1 | 245% | 42d | 6 |
nid:1766495679168
A&D
Subset Sum (Teilsummenproblem)
1
lapses
1/4
users
260%
ease
nid:1766495679168
Q: Subset Sum (Teilsummenproblem)
A: \(\Theta(n \cdot b)\) (Pseudo-Polynomial)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766495679169 |
1 | 260% | 9d | 12 |
nid:1766496919198
A&D
Longest Ascending Subsequence (Längste Aufsteigende Teilfolg...
1
lapses
1/4
users
260%
ease
nid:1766496919198
Q: Longest Ascending Subsequence (Längste Aufsteigende Teilfolge)
A: \(\Theta(n \log n)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766496919198 |
1 | 260% | 20d | 7 |
nid:1766500164961
A&D
How can we find a cross edge via DFS?
1
lapses
1/4
users
245%
ease
nid:1766500164961
Q: How can we find a cross edge via DFS?
A: If we find vertex with both pre- and post-values set, there's a cross edge.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766500164961 |
1 | 245% | 24d | 6 |
nid:1766500713117
c1
A&D
an adjacency list is better; an adjacency matrix is better
1
lapses
1/4
users
245%
ease
nid:1766500713117
Cloze c1
Cloze answer: an adjacency list is better; an adjacency matrix is better
Q: Which datastructure is best for DFS?In a sparse graph {{c1:: an adjacency list is better}}, in a dense graph {{c1:: an adjacency matrix is better}}.
A: \(|E| \geq |V|^2 / 10\), then DFS has the same runtime in the worst-case using adjacency matrices or lists as \(|V| + |E| \leq |V| + |V|^2 \)which is \(O(n^2)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766500713117 |
1 | 245% | 34d | 8 |
nid:1766523328098
A&D
BFS (Breadth First Search)
1
lapses
1/4
users
260%
ease
nid:1766523328098
Q: BFS (Breadth First Search)
A: \(O(|V|+|E|)\) (Adjacency List)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766523328099 |
1 | 260% | 30d | 8 |
nid:1766524219271
A&D
Dijkstra's Algorithm
1
lapses
1/4
users
260%
ease
nid:1766524219271
Q: Dijkstra's Algorithm
A: \(O((|V| + |E|) \log |V|)\) (or \(O(|V|^2)\)The runtime is calculated from \(O(n + (\#\text{extract-min} + \#\text{decrease-key}) \cdot \log n)\) which gives \(O((n + m) \cdot \log n)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766524219271 |
1 | 260% | 19d | 8 |
nid:1766524219271
A&D
Dijkstra's Algorithm
1
lapses
1/4
users
230%
ease
nid:1766524219271
Q: Dijkstra's Algorithm
A: \(O((|V| + |E|) \log |V|)\) (or \(O(|V|^2)\)The runtime is calculated from \(O(n + (\#\text{extract-min} + \#\text{decrease-key}) \cdot \log n)\) which gives \(O((n + m) \cdot \log n)\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766524328968 |
1 | 230% | 12d | 7 |
nid:1766568238909
c1
A&D
never contains a vertex already in the MST
1
lapses
1/4
users
230%
ease
nid:1766568238909
Cloze c1
Cloze answer: never contains a vertex already in the MST
Q: Prim's Algorithm Invariants: The priority queue \(H = V \setminus S\) (\(V\) set of all vertices, \(S\) vertices currently in the MST) {{c1::never contains a vertex already in the MST}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766568238910 |
1 | 230% | 5d | 3 |
nid:1766568909602
A&D
Kruskal's Algorithm
1
lapses
1/4
users
230%
ease
nid:1766568909602
Q: Kruskal's Algorithm
A: \(O(|E| \log |E| + |V| \log |V|)\)Outer loop: Iterate \(|E|\) times at most:Inner loop: find and union take \(O(\log |V|)\) per call amortised, thus \(O(|V| \log |V|)\) total.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766568909604 |
1 | 230% | 5d | 3 |
nid:1766574057724
c1
A&D
always negative \(\leq 0\)
1
lapses
1/4
users
260%
ease
nid:1766574057724
Cloze c1
Cloze answer: always negative \(\leq 0\)
Q: The height \(h(v)\) in Johnson's Algorithm is {{c1::always negative \(\leq 0\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766574057725 |
1 | 260% | 36d | 6 |
nid:1766742464527
IO r1
A&D
[Image Occlusion region 1]
1
lapses
1/4
users
230%
ease
nid:1766742464527
Cloze c1
Q: {{c5::image-occlusion:rect:left=.592:top=.4403:width=.0786:height=.0963:oi=1}}{{c10::image-occlusion:rect:left=.5847:top=.571:width=.0859:height=.0963:oi=1}}{{c12::image-occlusion:rect:left=.444:top=.6983:width=.0786:height=.0963:oi=1}}{{c3::image-occlusion:rect:left=.7912:top=.313:width
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766742464536 |
1 | 230% | 3d | 2 |
nid:1764744892590
c2
A&D
spanning, it connects all vertices
1
lapses
1/4
users
230%
ease
nid:1764744892590
Cloze c2
Cloze answer: spanning, it connects all vertices
Q: A Minimum Spanning Tree is a subgraph of a {{c1:: connected, undirected, weighted}} graph that fullfills:{{c2:: spanning, it connects all vertices}}{{c3:: acylic, it's a tree}}{{c4:: minimal, the sum of all edge weights in the Tree is minimal}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1766992688141 |
1 | 230% | 3d | 2 |
nid:1767084587767
c2
LinAlg
\((\lambda v) \cdot w = \lambda (v \cdot w)\) (scalars move...
1
lapses
1/4
users
230%
ease
nid:1767084587767
Cloze c2
Cloze answer: \((\lambda v) \cdot w = \lambda (v \cdot w)\) (scalars move freely)
Q: Scalar product properties: \(u, v, w \in \mathbb{R}^m\) be vectors and \(\lambda \in \mathbb{R}\) a scalar.{{c1::\(v \cdot w = w \cdot v\) (symmetry / commutatitivity}}{{c2:: \((\lambda v) \cdot w = \lambda (v \cdot w)\) (scalars move freely)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1767084587769 |
1 | 230% | 10d | 4 |
nid:1767087495269
c1
LinAlg
The {{c2::independent}} columns of \(A\), {{c1::span the col...
1
lapses
1/4
users
245%
ease
nid:1767087495269
Cloze c1
Q: The {{c2::independent}} columns of \(A\), {{c1::span the column space \(\textbf{C}(A)\) of \(A\)}}.
A: Proven by induction, adding elements that are a linear combination of other ones doesn't change span, thus we can iteratively remove the dependent columns.Lemma 2.11
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1767087495271 |
1 | 245% | 19d | 7 |
nid:1767439652577
LinAlg
What is the inverse of \(A = \begin{bmatrix} a & b \\ c & d ...
1
lapses
1/4
users
245%
ease
nid:1767439652577
Q: What is the inverse of \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)?
A: \[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -c \\ -b & a \end{bmatrix}\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1767439652577 |
1 | 245% | 5d | 5 |
nid:1767888505024
IO r3
EProg
[Image Occlusion region 3]
1
lapses
1/4
users
260%
ease
nid:1767888505024
Cloze c3
Q: {{c1::image-occlusion:rect:left=.2281:top=.3427:width=.0814:height=.2045:oi=1}}{{c2::image-occlusion:rect:left=.3053:top=.345:width=.1142:height=.2067:oi=1}}{{c3::image-occlusion:rect:left=.1625:top=.5221:width=.0693:height=.2181:oi=1}}{{c4::image-occlusion:rect:left=.1625:top=.713:width
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1767888505024 |
1 | 260% | 22d | 10 |
nid:1767888505024
IO r6
EProg
[Image Occlusion region 6]
1
lapses
1/4
users
260%
ease
nid:1767888505024
Cloze c6
Q: {{c1::image-occlusion:rect:left=.2281:top=.3427:width=.0814:height=.2045:oi=1}}{{c2::image-occlusion:rect:left=.3053:top=.345:width=.1142:height=.2067:oi=1}}{{c3::image-occlusion:rect:left=.1625:top=.5221:width=.0693:height=.2181:oi=1}}{{c4::image-occlusion:rect:left=.1625:top=.713:width
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1767888505025 |
1 | 260% | 14d | 11 |
nid:1767888762979
c1
EProg
&& is false
1
lapses
1/4
users
245%
ease
nid:1767888762979
Cloze c1
Cloze answer: && is false
Q: Java has short circuiting for the && and || operators.This means that if the left of {{c1:: && is false}} then the right isn't even executed{{c2:: || is true}} then the right i
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1767888762980 |
1 | 245% | 21d | 6 |
nid:1768138841525
c2
LinAlg
There is an \(m \times m\) matrix \(B\) such that \(BA = I\)...
1
lapses
1/4
users
230%
ease
nid:1768138841525
Cloze c2
Cloze answer: There is an \(m \times m\) matrix \(B\) such that \(BA = I\).
Q: Three equivalent statements:{{c1::\(T_A : \mathbb{R}^m \rightarrow \mathbb{R}^m\) is bijective.::Transformation}}{{c2::There is an \(m \times m\) matrix \(B\) such that \(BA = I\).}}{{c3::The columns of \(A\) are linearly independent.}}
A: The third one can be derived from the fact that if \(BA = I\), there is only a single \(x \in \mathbb{R}^m\) such that \(A \textbf{x} = 0\).It is also intuitively clear that if not all columns were linearly independent, we'd actually have a tall linear transformation and would be losing information.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768138841525 |
1 | 230% | 4d | 4 |
nid:1768138841525
c3
LinAlg
The columns of \(A\) are linearly independent.
1
lapses
1/4
users
230%
ease
nid:1768138841525
Cloze c3
Cloze answer: The columns of \(A\) are linearly independent.
Q: Three equivalent statements:{{c1::\(T_A : \mathbb{R}^m \rightarrow \mathbb{R}^m\) is bijective.::Transformation}}{{c2::There is an \(m \times m\) matrix \(B\) such that \(BA = I\).}}{{c3::The columns of \(A\) are linearly independent.}}
A: The third one can be derived from the fact that if \(BA = I\), there is only a single \(x \in \mathbb{R}^m\) such that \(A \textbf{x} = 0\).It is also intuitively clear that if not all columns were linearly independent, we'd actually have a tall linear transformation and would be losing information.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768138841526 |
1 | 230% | 6d | 7 |
nid:1768140101247
c1
LinAlg
\(R = MA\); \(M\) invertible
1
lapses
1/4
users
245%
ease
nid:1768140101247
Cloze c1
Cloze answer: \(R = MA\); \(M\) invertible
Q: For RREF on \(A, I\) we get \(R, M\) with the property that {{c1::\(R = MA\)::equation}} and {{c1::\(M\) invertible:: property of M}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768140101247 |
1 | 245% | 16d | 6 |
nid:1768146369419
c2
LinAlg
The {{c1::set of independent columns of \(A\)}} is {{c2::a b...
1
lapses
1/4
users
245%
ease
nid:1768146369419
Cloze c2
Q: The {{c1::set of independent columns of \(A\)}} is {{c2::a basis of the column space \(\textbf{C}(A)\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768146369420 |
1 | 245% | 22d | 6 |
nid:1768146519592
c1
LinAlg
has a basis \(B \subseteq G\)
1
lapses
1/4
users
230%
ease
nid:1768146519592
Cloze c1
Cloze answer: has a basis \(B \subseteq G\)
Q: Let \(V\) be a finitely generated vector space and let \(G \subseteq V\) be a finite subset with \(\textbf{Span}(G) = V\). Then \(V\) {{c1::has a basis \(B \subseteq G\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768146519592 |
1 | 230% | 6d | 8 |
nid:1768210767870
c1
LinAlg
that is closest to \(b\)
1
lapses
1/4
users
230%
ease
nid:1768210767870
Cloze c1
Cloze answer: that is closest to \(b\)
Q: The projection of a vector \(b \in \mathbb{R}^m\) onto a subspace \(S\) (of \(\mathbb{R}^m\)) is the point in \(S\) {{c1::that is closest to \(b\)}}. In other words \[ \text{proj}_S(b) = {{c1:: \text{argmin}_{p \in S} ||b - p|| }}\]
A: Where \(b = p + e \implies b - p = e\), with \(e\) the error.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768210767870 |
1 | 230% | 1d | 8 |
nid:1768240573172
IO r6
A&D
[Image Occlusion region 6]
1
lapses
1/4
users
230%
ease
nid:1768240573172
Cloze c6
Q: {{c1::image-occlusion:rect:left=.264:top=.1517:width=.4676:height=.1291:oi=1}}{{c2::image-occlusion:rect:left=.264:top=.3156:width=.4709:height=.1018:oi=1}}{{c3::image-occlusion:rect:left=.264:top=.4472:width=.472:height=.1043:oi=1}}{{c4::image-occlusion:rect:left=.2662:top=.5764:width=.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768240573177 |
1 | 230% | 3d | 2 |
nid:1768302182238
LinAlg
What is the pseudoinverse in the case where \(A \in \mathbb{...
1
lapses
1/4
users
245%
ease
nid:1768302182238
Q: What is the pseudoinverse in the case where \(A \in \mathbb{R}^{n \times m}\) has independent rows?
A: Because \(rank(A) = r = m\) and thus \(n \geq m\)\(C(A)\) spans \(\mathbb{R}^m\) (columns span the space)\(R(A) \subseteq\) \(\mathbb{R}^n\)There could be multiple \(x \in \mathbb{R}^n\) that map to \(T_A(x) = b\). We pick the one with the smallest norm \(||x||^2\).We know \(x = x_r + x_n\) for \(x_r \in R(A)\) and \(x_n \in N(A)\) thus we pick \(x = x_r + 0\) to get
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768302182238 |
1 | 245% | 7d | 6 |
nid:1768302385713
c1
LinAlg
For \(A \in \mathbb{R}^{m \times n}\) with \(\text{rank}(A) ...
1
lapses
1/4
users
230%
ease
nid:1768302385713
Cloze c1
Q: For \(A \in \mathbb{R}^{m \times n}\) with \(\text{rank}(A) = m\), we define the pseudo-inverse \(A^\dagger \in \mathbb{R}^{n \times m}\) as:\[ A^\dagger = {{c1::A^\top (A A^\top)^{-1} }}\]
A: For an \(A\) with full column-rank, we basically define, \(A^\dagger\) as the transpose of the pseudoinverse of the transpose:
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768302385713 |
1 | 230% | 3d | 5 |
nid:1768303179258
c1
LinAlg
any full rank (not just CR)
1
lapses
1/4
users
230%
ease
nid:1768303179258
Cloze c1
Cloze answer: any full rank (not just CR)
Q: We can compute the pseudoinverse from the {{c1:: any full rank (not just CR)}} factorisation of \(A\).
A: Note to Lorenz: Leave the "the" in, it's for maximum confusion .
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1768303179258 |
1 | 230% | 2d | 4 |
nid:1769360147747
A&D
Extra memory requirements of Heapsort?
1
lapses
1/4
users
230%
ease
nid:1769360147747
Q: Extra memory requirements of Heapsort?
A: \(O(1)\) as we simply arrange the array into a heap.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1769360147747 |
1 | 230% | 5d | 5 |
nid:1769376963519
c1
EProg
always a subtype of the static type
1
lapses
1/4
users
245%
ease
nid:1769376963519
Cloze c1
Cloze answer: always a subtype of the static type
Q: The dynamic type is {{c1::always a subtype of the static type}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1769376963519 |
1 | 245% | 6d | 6 |
nid:1769377883253
c1
EProg
casting to the static type of the parent
1
lapses
1/4
users
230%
ease
nid:1769377883253
Cloze c1
Cloze answer: casting to the static type of the parent
Q: We can access the parent's attribute of a subclass by {{c1:: casting to the static type of the parent}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1769377883253 |
1 | 230% | 2d | 5 |
nid:1769445714054
A&D
The depth \(h\) of a seach tree of any comparison-based algo...
1
lapses
1/4
users
230%
ease
nid:1769445714054
Q: The depth \(h\) of a seach tree of any comparison-based algorithm satisfies which bound?
A: \(h \geq \Omega(\log n)\) this is information theoretically the least amount of comparisons necessary.Note that \(h \not \leq O(n)\) necessarily as we could have a really stupid algorithm that compares thrice for example.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1769445714055 |
1 | 230% | 2d | 4 |
nid:1769445882673
A&D
Can (g, h) ever be in an MST? Prove it:
1
lapses
1/4
users
230%
ease
nid:1769445882673
Q: Can (g, h) ever be in an MST? Prove it:
A: No, because it's the heaviest edge in the cycle.If there was an MST containing it, we could remove it and replace it by another edge in the cycle.Then we preserve the tree property yet it's weight is strictly lower.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1769445882674 |
1 | 230% | 2d | 4 |
nid:1771363788400
c1
Analysis
eindeutig bestimmte Kenngrössen
1
lapses
1/4
users
260%
ease
nid:1771363788400
Cloze c1
Cloze answer: eindeutig bestimmte Kenngrössen
Q: Maximum und Minimum sind {{c1::eindeutig bestimmte Kenngrössen}} einer Menge, sofern {{c2::sie existieren}}.
A: (Es gibt nur ein Maximum und ein Minimum)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771363788400 |
1 | 260% | 155d | 9 |
nid:1771364277451
c1
PProg
Scheduling overhead
1
lapses
1/4
users
245%
ease
nid:1771364277451
Cloze c1
Cloze answer: Scheduling overhead
Q: {{c1::Scheduling overhead}} is {{c2::the extra time spent by the system or the algorithm to distribute work on multiple threads/tasks}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771364277454 |
1 | 245% | 20d | 7 |
nid:1771364277503
c3
PProg
increasing utilisation of a CPU's functional units
1
lapses
1/4
users
245%
ease
nid:1771364277503
Cloze c3
Cloze answer: increasing utilisation of a CPU's functional units
Q: {{c1::Instruction level parallelism (ILP)}} is {{c2::CPU-internal parallelisation of independent instructions}}, with the goal of improving performance by {{c3::increasing utilisation of a CPU's functional units}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771364277613 |
1 | 245% | 31d | 4 |
nid:1771364277511
c2
PProg
any resource (memory location, input source, output sink) sh...
1
lapses
1/4
users
230%
ease
nid:1771364277511
Cloze c2
Cloze answer: any resource (memory location, input source, output sink) shared by more than one thread
Q: A {{c1::shared resource}} is {{c2::any resource (memory location, input source, output sink) shared by more than one thread}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771364277641 |
1 | 230% | 26d | 7 |
nid:1771364277512
c3
PProg
additional management information
1
lapses
1/4
users
230%
ease
nid:1771364277512
Cloze c3
Cloze answer: additional management information
Q: Process context includes:{{c1::CPU state (registers, program counter)}}{{c2::program state (stack, heap, resource handles)}}{{c3::additional management information}}.
A: A thread also has a context, but it is typically much smaller.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771364277644 |
1 | 230% | 11d | 5 |
nid:1771364277518
c2
PProg
a management process, e.g. on the operating system level, th...
1
lapses
1/4
users
245%
ease
nid:1771364277518
Cloze c2
Cloze answer: a management process, e.g. on the operating system level, that performs context switches
Q: A {{c1::scheduler}} is {{c2::a management process, e.g. on the operating system level, that performs context switches}}.
A: I.e. it interrupts/pauses/sends to sleep the currently running process (or thread), performs a context switch, and selects the next process (or thread) to run.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771364277668 |
1 | 245% | 36d | 6 |
nid:1771366536186
c1
A&W
eine Brücke
1
lapses
1/4
users
230%
ease
nid:1771366536186
Cloze c1
Cloze answer: eine Brücke
Q: Sei \(G = (V, E)\) ein zusammenhängender Graph. Ist \(\{x, y\} \in E\) {{c1::eine Brücke::Eigenschaft?}}, so gilt:
\({{c2::\deg(x) = 1}}\) oder {{c3::\(x\) ist Artikulationsknoten}}.
A: (und analog für \(y\))Aber die Umkehrung gilt nicht!
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771366536187 |
1 | 230% | 8d | 5 |
nid:1771366536198
c1
A&W
\(k\)-kanten-zusammenhängend
1
lapses
1/4
users
260%
ease
nid:1771366536198
Cloze c1
Cloze answer: \(k\)-kanten-zusammenhängend
Q: Ein Graph \(G = (V, E)\) heisst {{c1::\(k\)-kanten-zusammenhängend}}, falls {{c2::für alle Teilmengen \(X \subseteq E\) mit \(|X| < k\) gilt: Der Graph \((V, E \setminus X)\) ist zusammenhängend}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771366536214 |
1 | 260% | 78d | 6 |
nid:1771535790926
c3
A&W
m
1
lapses
1/4
users
260%
ease
nid:1771535790926
Cloze c3
Cloze answer: m
Q: Die um {{c1::die Berechnung von \(low[]\)}} ergänzte {{c2::Tiefensuche}} berechnet in einem zusammenhängenden Graphen alle Artikulationsknoten und Brücken in Zeit \(O({{c3::m}})\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771535790928 |
1 | 260% | 55d | 5 |
nid:1771872607339
c2
PProg
Kernel-level thread: Managed by the OS
1
lapses
1/4
users
245%
ease
nid:1771872607339
Cloze c2
Cloze answer: Kernel-level thread: Managed by the OS
Q: The three levels of threads:{{c1::User-level thread: Managed by the application using a thread library}}{{c2::Kernel-level thread: Managed by the OS}}{{c3::CPU-level thread}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771872607340 |
1 | 245% | 34d | 6 |
nid:1771872607379
c1
PProg
an actual execution thread
1
lapses
1/4
users
260%
ease
nid:1771872607379
Cloze c1
Cloze answer: an actual execution thread
Q: Each call to start() method of a Thread object creates {{c1::an actual execution thread}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771872607379 |
1 | 260% | 104d | 5 |
nid:1771872607479
IO r2
PProg
[Image Occlusion region 2]
1
lapses
1/4
users
230%
ease
nid:1771872607479
Cloze c2
Q: {{c1::image-occlusion:rect:left=.5516:top=.2782:width=.1174:height=.0851:oi=1}}{{c2::image-occlusion:rect:left=.3149:top=.504:width=.1095:height=.0818:oi=1}}{{c2::image-occlusion:rect:left=.2425:top=.7396:width=.2562:height=.0785:oi=1}}{{c3::image-occlusion:rect:left=.7726:top=.504:width
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771872607481 |
1 | 230% | 2d | 4 |
nid:1771969055150
c2
Analysis
\(A \neq \emptyset\), \(B \neq \emptyset\); \(\forall a \i...
1
lapses
1/4
users
260%
ease
nid:1771969055150
Cloze c2
Cloze answer: \(A \neq \emptyset\), \(B \neq \emptyset\); \(\forall a \in A \ \forall b \in B \ : \ a \leq b\)
Q: Ordnungsvollständigkeit:Seien \(A, B \subseteq \mathbb{R}\), sodass
{{c2:: \(A \neq \emptyset\), \(B \neq \emptyset\)}}
{{c2:: \(\forall a \in A \ \forall b \in B \ : \ a \leq b\)}}
Dann {{c1:: gibt es ein \(c \in \mathbb{R}\), sodass \[ \foral
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771969055150 |
1 | 260% | 153d | 7 |
nid:1771969381133
Analysis
Dreiecksungleichung (Vektoren)
1
lapses
1/4
users
230%
ease
nid:1771969381133
Q: Dreiecksungleichung (Vektoren)
A: Für alle \(x, y, z \in \mathbb{R}^n\) gilt: \[ ||x - z|| \leq ||x - y|| + ||y - z|| \]wo \(||x||\) die euklidische Norm von \(x\) ist.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771969381133 |
1 | 230% | 101d | 7 |
nid:1771969600985
c1
Analysis
|z|^2
1
lapses
1/4
users
245%
ease
nid:1771969600985
Cloze c1
Cloze answer: |z|^2
Q: Für \(z \in \mathbb{C}\) gilt: \(z \cdot \bar{z} = {{c1:: |z|^2 }}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771969600985 |
1 | 245% | 109d | 5 |
nid:1771969965872
c1
Analysis
\(r = |z| \ge 0\) und \(\varphi \in (-\pi, \pi]\) der Polar...
1
lapses
1/4
users
275%
ease
nid:1771969965872
Cloze c1
Cloze answer: \(r = |z| \ge 0\) und \(\varphi \in (-\pi, \pi]\) der Polarwinkel \(\arg(z)\) (Argument) ist
Q: In der Polarform wird \(z = a + ib\) als {{c1:: \(r \cdot e^{i \varphi}\)}} dargestellt wo {{c1:: \(r = |z| \ge 0\) und \(\varphi \in (-\pi, \pi]\) der Polarwinkel \(\arg(z)\) (Argument) ist::Def. r und Winkel}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1771969965872 |
1 | 275% | 45d | 8 |
nid:1772209100380
IO r1
A&W
[Image Occlusion region 1]
1
lapses
1/4
users
260%
ease
nid:1772209100380
Cloze c1
Q: {{c3::image-occlusion:rect:left=.1591:top=.8923:width=.7185:height=.0742}}{{c2::image-occlusion:rect:left=.3252:top=.7428:width=.5272:height=.0923}}{{c1::image-occlusion:rect:left=.0549:top=.1782:width=.9041:height=.1203}}{{c4::image-occlusion:rect:left=.1645:top=.4824:width=.1234:height
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772209100382 |
1 | 260% | 41d | 7 |
nid:1772569386183
c1
A&W
einen augmentierenden Pfad
1
lapses
1/4
users
260%
ease
nid:1772569386183
Cloze c1
Cloze answer: einen augmentierenden Pfad
Q: Jedes Matching, das nicht {{c2::(kardinalitäts-)maximal}} ist, besitzt {{c1::einen augmentierenden Pfad}}.Theorem name included
A: (Berge, 1957)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772569386183 |
1 | 260% | 21d | 8 |
nid:1772569386183
c2
A&W
(kardinalitäts-)maximal
1
lapses
1/4
users
260%
ease
nid:1772569386183
Cloze c2
Cloze answer: (kardinalitäts-)maximal
Q: Jedes Matching, das nicht {{c2::(kardinalitäts-)maximal}} ist, besitzt {{c1::einen augmentierenden Pfad}}.Theorem name included
A: (Berge, 1957)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772569386184 |
1 | 260% | 21d | 6 |
nid:1772569386187
A&W
Wie funktioniert der Algorithmus um ein maximales Matching z...
1
lapses
1/4
users
275%
ease
nid:1772569386187
Q: Wie funktioniert der Algorithmus um ein maximales Matching zu finden?
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772569386187 |
1 | 275% | 14d | 9 |
nid:1772569386190
c2
A&W
Ein Matching \( M \) heisst {{c1::perfektes Matching}}, wenn...
1
lapses
1/4
users
245%
ease
nid:1772569386190
Cloze c2
Q: Ein Matching \( M \) heisst {{c1::perfektes Matching}}, wenn {{c2::jeder Knoten durch genau eine Kante aus \( M \) überdeckt wird, oder, anders ausgedrückt, wenn \( |M| = \frac{|V|}{2}\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772569386191 |
1 | 245% | 48d | 4 |
nid:1772569386198
c1
A&W
zwei
1
lapses
1/4
users
245%
ease
nid:1772569386198
Cloze c1
Cloze answer: zwei
Q: Jede Kante in \( M_{\text{Greedy}} \) kann höchstens {{c1::zwei}} Kanten aus \( M_{\text{max} } \) überdecken.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772569386198 |
1 | 245% | 13d | 8 |
nid:1772569386201
A&W
Inklusionsmaximal? Kardinalitätsmaximal?
1
lapses
1/4
users
245%
ease
nid:1772569386201
Q: Inklusionsmaximal? Kardinalitätsmaximal?
A: Sowohl als auch.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772569386201 |
1 | 245% | 39d | 4 |
nid:1772569386222
c1
A&W
|V| \cdot |E|
1
lapses
1/4
users
230%
ease
nid:1772569386222
Cloze c1
Cloze answer: |V| \cdot |E|
Q: In bipartiten Graphen kann man in Zeit \( O({{c1::|V| \cdot |E|}}) \) ein perfektes Matching bestimmen. Ist dies optimal?
A: Note, es geht mit Hopcroft-Karp in \(O(\sqrt{|V|} \cdot |E|)\) schneller.Augmentierende-Pfade-AlgorithmusMan startet mit einem beliebigen Matching und sucht iterativ \(M\)-augmentierende PfadeDiese baut schichtweise einen Layer-Graphen auf: \(L_0\) sind die unüberdeckten Knoten in \(A\), ungerade Schichten erreicht man über Kanten in \(E \setminus M\), gerade über Kanten in \(M\). Findet man einen unüberdeckten Knoten in \(B\), liefert
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772569386222 |
1 | 230% | 5d | 11 |
nid:1772569386236
c1
A&W
|E|
1
lapses
1/4
users
230%
ease
nid:1772569386236
Cloze c1
Cloze answer: |E|
Q: In \( 2^k \)-regulären bipartiten Graphen kann man in Zeit \( O({{c1::|E|}}) \) ein perfektes Matching bestimmen.
A: Satz 1.54 - Eulertour-basierter Algorithmus\(2^k\)-regulärer bipartiter Graph ist eulersch (alle Knoten haben geraden Grad).In jeder Zusammenhangskomponente berechnet man eine Eulertour in \(O(|E|)\)Dann läuft man diese ab und entfernt jede zweite Kante. Der verbleibende Graph ist \(2^{k-1}\)-regulär. Nach \(k\) Iterationen ist der Graph \(2^0 = 1\)-regulär, also selbst ein perfektes Matching. Die Gesamtl
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772569386236 |
1 | 230% | 1d | 5 |
nid:1772569386178
c2
A&W
es kein Matching \( M' \) gibt mit \( M \subseteq M' \) und ...
1
lapses
1/4
users
245%
ease
nid:1772569386178
Cloze c2
Cloze answer: es kein Matching \( M' \) gibt mit \( M \subseteq M' \) und \( |M'| > |M| \)
Q: Ein Matching \( M \subseteq E \) ist ein {{c1::inklusionsmaximales Matching}}, wenn {{c2::es kein Matching \( M' \) gibt mit \( M \subseteq M' \) und \( |M'| > |M| \)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772570517431 |
1 | 245% | 43d | 5 |
nid:1772698768089
c1
Analysis
einen Häufungspunkt, der mit dem Grenzwert übereinstimmt
1
lapses
1/4
users
290%
ease
nid:1772698768089
Cloze c1
Cloze answer: einen Häufungspunkt, der mit dem Grenzwert übereinstimmt
Q: Jede konvergente Folge hat genau {{c1:: einen Häufungspunkt, der mit dem Grenzwert übereinstimmt}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772698768090 |
1 | 290% | 129d | 12 |
nid:1772783275475
A&W
Wahr oder falsch?Für zwei Knoten \( a, b \) eines Graphen se...
1
lapses
1/4
users
245%
ease
nid:1772783275475
Q: Wahr oder falsch?Für zwei Knoten \( a, b \) eines Graphen sei \( a \sim b \) genau dann, wenn \( a = b \) gilt oder wenn \( a \) und \( b \) auf einem gemeinsamen Kreis liegen. Dann ist \( \sim \) eine Äquivalenzrelation.
A: Falsch.Die Relation ist nicht transitiv: Berühren sich zwei Kreise nur in einem Knoten \(b\), so gilt \(a \sim b\) und \(b \sim c\), aber \(a\) und \(c\) liegen auf keinem gemeinsamen Kreis (\(b\) ist ein Artikulationspunkt). Also keine Äquivalenzrelation.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772783275475 |
1 | 245% | 20d | 4 |
nid:1772783275526
c1
Analysis
0
1
lapses
1/4
users
260%
ease
nid:1772783275526
Cloze c1
Cloze answer: 0
Q: \(\forall x \in \mathbb{R}: \lim_{n\to\infty} \frac{x^n}{n!} ={{c1::0}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772783275527 |
1 | 260% | 128d | 9 |
nid:1772783275528
c1
Analysis
1
1
lapses
1/4
users
275%
ease
nid:1772783275528
Cloze c1
Cloze answer: 1
Q: \(\lim_{n\to\infty} n^{1/n} ={{c1::1}}\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772783275528 |
1 | 275% | 154d | 9 |
nid:1772788241820
c1
Analysis
\[ \sin\!\left(\frac{2\pi}{3}\right) = {{c1::\frac{\sqrt{3} ...
1
lapses
1/4
users
260%
ease
nid:1772788241820
Cloze c1
Q: \[ \sin\!\left(\frac{2\pi}{3}\right) = {{c1::\frac{\sqrt{3} }{2} }} \]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1772788241821 |
1 | 260% | 47d | 6 |
nid:1773134608434
c1
Analysis
eindeutigen
1
lapses
1/4
users
245%
ease
nid:1773134608434
Cloze c1
Cloze answer: eindeutigen
Q: Eine konvergente Folge besitzt genau einen {{c1::eindeutigen}} Grenzwert.Proof Included
A: Proof For contradiction, assume there are \(A, B\) limits.Then there exists \(N_A \in \mathbb{N}\) such that \(\forall n > N_A \ : \ |a_n - A| < \frac{\epsilon}{2}\) There must also be \(N_B\) such that \(\forall n > N_B \ : \ |a_n - B| < \frac{\epsilon}{2}\)But then for \(N := \max \{N_A, N_B\}\) it holds that \(n > N\) \(|A - B| \le |A - a_n| + |a_n -B| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\). As this hold
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1773134608434 |
1 | 245% | 4d | 5 |
nid:1773420068085
c1
A&W
3/2-Approximation Metrisches TSP
Bestimme minimalen Spannb...
1
lapses
1/4
users
245%
ease
nid:1773420068085
Cloze c1
Q: 3/2-Approximation Metrisches TSP
Bestimme minimalen Spannbaum \(T\)es gilt: \( \ell(T) \leq \text{opt}(K_n, \ell) \)
' \(X:=\) Knoten mit ungeradem Grad in \(T\)Bestimme minimales Matching \(M\) für \(X\) es gilt: \
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1773420068085 |
1 | 245% | 10d | 8 |
nid:1773420068117
A&W
Wahr oder falsch?Jeder Graph ohne Dreieck hat eine chromatis...
1
lapses
1/4
users
245%
ease
nid:1773420068117
Q: Wahr oder falsch?Jeder Graph ohne Dreieck hat eine chromatische Zahl von höchstens 100.
A: Falsch.Siehe Mycielski-Konstruktion.Konstruktion:Aus \(G_k = (V_k, E_k)\) mit \(V_k = \{v_1,\ldots,v_n\}\) bilde \(G_{k+1}\):Füge Knoten \(w_1,\ldots,w_n, z\) hinzu. \(w_i\) ist mit allen Nachbarn von \(v_i\) verbunden (aber nicht mit \(v_i\) selbst). \(z\) ist mit allen \(w_i\) verbunden.Der neue Graph ist dreiecksfrei und braucht eine Farbe mehr als \(G_k\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1773420068117 |
1 | 245% | 3d | 8 |
nid:1773420068135
c1
A&W
3/2-Approximation Metrisches TSP
Bestimme minimalen Spannb...
1
lapses
1/4
users
245%
ease
nid:1773420068135
Cloze c1
Q: 3/2-Approximation Metrisches TSP
Bestimme minimalen Spannbaum \(T\)es gilt: \( \ell(T) \leq \text{opt}(K_n, \ell) \)
' {{c1::\(X:=\) Knoten mit ungeradem Grad in \(T\)Bestimme minimales Matching \(M\) für \(X\) es gilt:&
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1773420068138 |
1 | 245% | 23d | 6 |
nid:1773773841684
c1
A&W
|V| + |E|
1
lapses
1/4
users
230%
ease
nid:1773773841684
Cloze c1
Cloze answer: |V| + |E|
Q: Einen 3-färbbaren Graphen kann man in Zeit \(O({{c1::|V| + |E|}})\) mit \(O({{c2::\sqrt{|V|} }})\) Farben färben.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1773773841685 |
1 | 230% | 1d | 5 |
nid:1774005500952
c1
Analysis
Für eine {{c1:: monotone Folge reeller Zahlen \((a_n)_{n \in...
1
lapses
1/4
users
230%
ease
nid:1774005500952
Cloze c1
Q: Für eine {{c1:: monotone Folge reeller Zahlen \((a_n)_{n \in \mathbb{N}_0}\)}} gilt: Sie konvergiert genau dann, wenn {{c2::sie beschränkt ist}}.
A: (Weierstrass)Falls die Folge monoton wachsend ist, gilt: \[ \lim_{n \rightarrow \infty} a_n = \sup \{a_n \mid n \in \mathbb{N}_0\} \]Falls die Folge monoton fallend ist, gilt:\[\lim_{n \rightarrow \infty} a_n = \inf \{ a_n \mid n \in \mathbb{N}_0\}\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1774005500953 |
1 | 230% | 17d | 12 |
nid:1774005965819
c2
Analysis
\((a_n)_{n \in \mathbb{N}_0}\) {{c1::eine konvergente Folge:...
1
lapses
1/4
users
215%
ease
nid:1774005965819
Cloze c2
Q: \((a_n)_{n \in \mathbb{N}_0}\) {{c1::eine konvergente Folge::Property}} \(\Longleftrightarrow\) \[ \lim_{n \rightarrow \infty} a_n = {{c2:: \limsup_{n \rightarrow \infty} a_n = \liminf_{n \rightarrow \infty} a_n }}\]
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1774005965819 |
1 | 215% | 18d | 6 |
nid:1774006045853
c2
Analysis
für unendlich viele Elemente \(A - \epsilon < a_n < A + \eps...
1
lapses
1/4
users
215%
ease
nid:1774006045853
Cloze c2
Cloze answer: für unendlich viele Elemente \(A - \epsilon < a_n < A + \epsilon\) gilt.
Q: Sei \((a_n)_{n \in \mathbb{N}_0}\) eine beschränkte Folge mit \(A = \limsup_{n \rightarrow \infty} a_n\). Dann ist \(A\) ein Häufungspunkt und für alle \(\epsilon > 0\) gilt, dass:{{c1::es nur endlich viele Elemente \(a_n\) gibt, für welche \(a_n \ge A + \e
A: Eine analoge Aussage gilt auch für den Limes inferior.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1774006045855 |
1 | 215% | 19d | 6 |
nid:1774006491519
c1
Analysis
beschränkt
1
lapses
1/4
users
260%
ease
nid:1774006491519
Cloze c1
Cloze answer: beschränkt
Q: Jede Cauchy-Folge ist {{c1:: beschränkt}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1774006491519 |
1 | 260% | 48d | 7 |
nid:1774474839885
c1
Analysis
Sei \(\sum a_n\) {{c1::bedingt konvergent und \(L \in \mathb...
1
lapses
1/4
users
230%
ease
nid:1774474839885
Cloze c1
Q: Sei \(\sum a_n\) {{c1::bedingt konvergent und \(L \in \mathbb{R} \cup \{+\infty, -\infty\}\)}}.Dann {{c2::gibt es eine Bijektion \(\phi\), so dass:\[\sum_{n=0}^\infty a_{\phi(n)} = L\]}}
A: (Riemannscher Umordnungssatz)Merke: Bedingt konvergente Reihen können durch Umordnung jeden Grenzwert annehmen!
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1774474839895 |
1 | 230% | 1d | 5 |
nid:1774474839891
c2
Analysis
die Koeffizienten
1
lapses
1/4
users
230%
ease
nid:1774474839891
Cloze c2
Cloze answer: die Koeffizienten
Q: Eine Potenzreihe hat die Form \({{c5:: \displaystyle\sum_{k=0}^\infty c_k (x - a)^k }}\), wobei:\(a\) ist {{c1::der Entwicklungspunkt (Zentrum)}}\(c_0, c_1, \ldots\) sind {{c2::die Koeffizienten}}\(x\) ist {{c3::das Argument}}\((a - R,\, a + R)\) ist {{c
A: Spezialfall \(a = 0\): \(\sum c_k x^k\) - Entwicklungspunkt im Ursprung.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| niklas | cid:1774474839909 |
1 | 230% | 3d | 4 |
nid:1765551644290
c1
LinAlg
The span of m linearly independent vectors is {{c1::\(\mathb...
1
lapses
1/4
users
245%
ease
nid:1765551644290
Cloze c1
Q: The span of m linearly independent vectors is {{c1::\(\mathbb{R}^m\)}}.
A: This also means that a matrix in \(\mathbb{R}^{n \times n}\) with rank(A) = n spans the entire space.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1765551644290 |
1 | 245% | 14d | 4 |
nid:1765551666570
c2
A&D
incident
1
lapses
1/4
users
245%
ease
nid:1765551666570
Cloze c2
Cloze answer: incident
Q: The {{c1::degree (Knotengrad) \(\deg(v)\)}} of a vertex \(v\) is the number of edges that are {{c2::incident}} to \(v\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1765551666580 |
1 | 245% | 25d | 6 |
nid:1765551666576
A&D
Cycle
1
lapses
1/4
users
245%
ease
nid:1765551666576
Q: Cycle
A: Kreis
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1765551666591 |
1 | 245% | 65d | 6 |
nid:1765551666578
A&D
What is the length of a walk?
1
lapses
1/4
users
245%
ease
nid:1765551666578
Q: What is the length of a walk?
A: The length of a walk \((v_0, v_1, \dots, v_k)\) is \(k\), i.e. the number of vertices minus 1.A walk of length \(l\) connects \(l + 1\) vertices.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1765551666594 |
1 | 245% | 48d | 6 |
nid:1765551666580
c2
A&D
for every two vertices \(u, v \in V\) \(u\) reaches \(v\)
1
lapses
1/4
users
230%
ease
nid:1765551666580
Cloze c2
Cloze answer: for every two vertices \(u, v \in V\) \(u\) reaches \(v\)
Q: A graph \(G\) is {{c1::connected (Zusammenhängend)}} if {{c2::for every two vertices \(u, v \in V\) \(u\) reaches \(v\)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1765551666598 |
1 | 230% | 93d | 7 |
nid:1765551666585
c1
A&D
direct predecessor (Vorgänger); direct successor (Nachfolger
1
lapses
1/4
users
230%
ease
nid:1765551666585
Cloze c1
Cloze answer: direct predecessor (Vorgänger); direct successor (Nachfolger
Q: In a directed graph, for the edge \(e = (u, v)\), \(u\) is the {{c1::direct predecessor (Vorgänger)}} of \(v\) and \(v\) the {{c1::direct successor (Nachfolger}} of \(u\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1765551666606 |
1 | 230% | 17d | 8 |
nid:1765551666588
c1
A&D
The {{c1::out-degree \(\deg_{\text{out} }(v)\) (Ausgangsgrad...
1
lapses
1/4
users
230%
ease
nid:1765551666588
Cloze c1
Q: The {{c1::out-degree \(\deg_{\text{out} }(v)\) (Ausgangsgrad)}} of a vertex in a directed graph is the {{c2::number of edges that have \(v\) as the start-vertex}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1765551666610 |
1 | 230% | 27d | 5 |
nid:1765551666614
c1
A&D
shortest length of a walk from \(u\) to \(v\)
1
lapses
1/4
users
230%
ease
nid:1765551666614
Cloze c1
Cloze answer: shortest length of a walk from \(u\) to \(v\)
Q: The distance \(d(u, v)\) in a directed graph is defined as {{c1:: shortest length of a walk from \(u\) to \(v\)}}.
A: Keep in mind in a weighted graph, this might mean the cheapest, which refers to cost not length.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1765551666649 |
1 | 230% | 46d | 9 |
nid:1765551666619
c1
A&D
the enter order equals the leave order
1
lapses
1/4
users
230%
ease
nid:1765551666619
Cloze c1
Cloze answer: the enter order equals the leave order
Q: In BFS enter/leave ordering, the FIFO queue guarantees that {{c1:: the enter order equals the leave order}} within a given level.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1765551666654 |
1 | 230% | 57d | 9 |
nid:1765551656956
c2
DiskMat
Well-definedness: \(\forall a \in A \ \forall b, b' \in B :...
1
lapses
1/4
users
230%
ease
nid:1765551656956
Cloze c2
Cloze answer: Well-definedness: \(\forall a \in A \ \forall b, b' \in B : (a \ f \ b \land a \ f \ b' \rightarrow b = b')\)
Q: What two properties must a relation \(f: A \to B\) have to be a function?{{c1:: Total-definedness: \(\forall a \in A \ \exists b \in B : a \ f \ b\) }}{{c2:: Well-definedness: \(\forall a \in A \ \forall b, b' \in B : (a
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1766410039197 |
1 | 230% | 1d | 4 |
nid:1766501315026
A&D
Find Closed Eulerian Path
1
lapses
1/4
users
230%
ease
nid:1766501315026
Q: Find Closed Eulerian Path
A: \(O(n+m)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1766501315056 |
1 | 230% | 4d | 5 |
nid:1766501315033
c2
A&D
\(\lnot \exists\) directed closed walk
1
lapses
1/4
users
230%
ease
nid:1766501315033
Cloze c2
Cloze answer: \(\lnot \exists\) directed closed walk
Q: {{c1:: \(\exists\) toposort}} \(\Longleftrightarrow\) {{c2:: \(\lnot \exists\) directed closed walk}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1766501315063 |
1 | 230% | 19d | 6 |
nid:1766501315038
c1
A&D
Cross edge, \(u, v\) in different subtrees
1
lapses
1/4
users
230%
ease
nid:1766501315038
Cloze c1
Cloze answer: Cross edge, \(u, v\) in different subtrees
Q: Pre-/Post-Ordering Classification for an edge \((u, v)\):\(\text{pre}(v) < \text{post}(v) < \text{pre}(u) < \text{post}(u)\): {{c1:: Cross edge, \(u, v\) in different subtrees}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1766501315070 |
1 | 230% | 19d | 8 |
nid:1766576733264
c1
A&D
Prim's Algorithm Invariants:\(\forall v \not \in S, v \neq s...
1
lapses
1/4
users
230%
ease
nid:1766576733264
Cloze c1
Q: Prim's Algorithm Invariants:\(\forall v \not \in S, v \neq s\), \(d[v] = \) {{c1:: \(\min \{ w(u, v) \ | \ (u, v) \in E, u \in S \}\)(\(\infty\) if no such edge exists)}}.
A: The 3rd invariant \[d[v] = \begin{cases} 0, & \text{if } v = s \text{ (the starting vertex)} \\ \min_{(u,v) \in E : u \in S} {w(u,v)}, & \text{if } v \in V \setminus S \text{ and } \exists (u,v) \in E \text{ with } u \in S \\ \infty, & \text{if } v \in V \setminus S \text{ and } \nexists (u,v) \in E \text{ with } u \in S \end{cases}\]ensures that d[v] always reflects the minimum cost to reach vertex v from the current MST. We always want to add the
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1766576733264 |
1 | 230% | 37d | 12 |
nid:1766576733286
c2
A&D
same(u,v) test if \(u, v\) in the same component
1
lapses
1/4
users
230%
ease
nid:1766576733286
Cloze c2
Cloze answer: same(u,v) test if \(u, v\) in the same component
Q: Union-Find datastructure methods:{{c1::make(u, v) creates the DS for \(F = \emptyset\)}}{{c2::same(u,v) test if \(u, v\) in the same component}}{{c3::union(u,v) merge ZHKs of \(u, v\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1766576733293 |
1 | 230% | 6d | 7 |
nid:1766576733289
A&D
Floyd-Warshall
1
lapses
1/4
users
230%
ease
nid:1766576733289
Q: Floyd-Warshall
A: \(O(|V|^3)\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1766576733296 |
1 | 230% | 8d | 7 |
nid:1766576739753
A&D
Floyd-Warshall, when is there a negative cycle?
1
lapses
1/4
users
230%
ease
nid:1766576739753
Q: Floyd-Warshall, when is there a negative cycle?
A: There exists a negative cycle \(\Leftrightarrow \exists v \in V \ : \ d^n_{v \rightarrow v} < 0\) In words: If there exists a path from a vertex to itself with negative weight (passing through any other vertex, i.e. \(n\)th iteration of the outer loop), then there exists a negative cycle that contains this vertex.We can perform a negative cycle check at the end, by going over all diagonals.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1766576739753 |
1 | 230% | 16d | 8 |
nid:1766656891070
c1
A&D
a cycle; undirected
1
lapses
1/4
users
230%
ease
nid:1766656891070
Cloze c1
Cloze answer: a cycle; undirected
Q: A graph with more than \(n-1\) edges has {{c1::a cycle}} if it is {{c1::undirected}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1766656891070 |
1 | 230% | 17d | 8 |
nid:1767089638548
c4
LinAlg
\(v \cdot v \geq 0\) with equality if and only if \(v = 0\)...
1
lapses
1/4
users
230%
ease
nid:1767089638548
Cloze c4
Cloze answer: \(v \cdot v \geq 0\) with equality if and only if \(v = 0\) (positive definiteness
Q: Scalar product properties: \(u, v, w \in \mathbb{R}^m\) be vectors and \(\lambda \in \mathbb{R}\) a scalar.{{c1::\(v \cdot w = w \cdot v\) (symmetry / commutatitivity}}{{c2:: \((\lambda v) \cdot w = \lambda (v \cdot w)\) (scalars move freely)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1767089638552 |
1 | 230% | 1d | 3 |
nid:1771363954967
c1
PProg
Amdahl's Law
1
lapses
1/4
users
230%
ease
nid:1771363954967
Cloze c1
Cloze answer: Amdahl's Law
Q: {{c1::Amdahl's Law}} specifies {{c2::the maximum amount of speedup that can be achieved for a program with a given sequential part.}}
A: The pessimistic view on scalability.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363954970 |
1 | 230% | 11d | 9 |
nid:1771363954970
c1
PProg
Cache coherence protocols
1
lapses
1/4
users
230%
ease
nid:1771363954970
Cloze c1
Cloze answer: Cache coherence protocols
Q: {{c1::Cache coherence protocols}} are hardware protocols that {{c2::ensure consistency across caches}}, typically by {{c3::tracking which locations are cached, and synchronising them if necessary}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363954981 |
1 | 230% | 29d | 10 |
nid:1771363954971
c2
PProg
execute code and spawn new tasks if required
1
lapses
1/4
users
230%
ease
nid:1771363954971
Cloze c2
Cloze answer: execute code and spawn new tasks if required
Q: {{c1::Cilk-style programming}} is a parallel programming idiom: To compute a program, {{c2::execute code and spawn new tasks if required}}. Before returning, {{c3::wait for all spawned tasks to complete}}.
A: The system manages the eventual execution of the spawned tasks potentially in parallel.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363954985 |
1 | 230% | 17d | 7 |
nid:1771363954971
c1
PProg
Cilk-style programming
1
lapses
1/4
users
230%
ease
nid:1771363954971
Cloze c1
Cloze answer: Cilk-style programming
Q: {{c1::Cilk-style programming}} is a parallel programming idiom: To compute a program, {{c2::execute code and spawn new tasks if required}}. Before returning, {{c3::wait for all spawned tasks to complete}}.
A: The system manages the eventual execution of the spawned tasks potentially in parallel.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363954986 |
1 | 230% | 62d | 11 |
nid:1771363954976
c2
PProg
resources required to set up an operation
1
lapses
1/4
users
230%
ease
nid:1771363954976
Cloze c2
Cloze answer: resources required to set up an operation
Q: {{c1::Context switch overhead}} refers to {{c2::resources required to set up an operation}}.
A: In terms of context switch, CPU needs to store/save the local data, program pointer etc. of the current thread/process, and load the local data, program pointer etc. of the next thread/process to execute.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955002 |
1 | 230% | 10d | 11 |
nid:1771363954980
c2
PProg
recursively solving smaller sub-problems and combining their...
1
lapses
1/4
users
230%
ease
nid:1771363954980
Cloze c2
Cloze answer: recursively solving smaller sub-problems and combining their results
Q: {{c1::Divide and conquer style parallelism (also called recursive splitting)}} means: solve a problem by {{c2::recursively solving smaller sub-problems and combining their results}}.
A: Solve the sub-problems in separate threads to gain a speedup.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955014 |
1 | 230% | 21d | 8 |
nid:1771363954981
c1
PProg
Deadlock
1
lapses
1/4
users
230%
ease
nid:1771363954981
Cloze c1
Cloze answer: Deadlock
Q: {{c1::Deadlock}} is {{c2::circular waiting/blocking (no instructions are executed/CPU time is used) between threads, so that the system (union of all threads) cannot make any progress anymore}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955016 |
1 | 230% | 33d | 8 |
nid:1771363954983
c1
PProg
divide and conquer parallelism
1
lapses
1/4
users
230%
ease
nid:1771363954983
Cloze c1
Cloze answer: divide and conquer parallelism
Q: The ForkJoin framework embraces {{c1::divide and conquer parallelism}}.
A: Tasks can be spawned (forked) and joined by the framework.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955027 |
1 | 230% | 44d | 7 |
nid:1771363954984
c1
PProg
functional unit
1
lapses
1/4
users
230%
ease
nid:1771363954984
Cloze c1
Cloze answer: functional unit
Q: A {{c1::functional unit}} is a component of a CPU (or core) that {{c2::performs a certain task}}, an {{c3::execution unit}} is one such example.
A: performing a task - e.g. executing integer arithmetic operations
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955028 |
1 | 230% | 10d | 8 |
nid:1771363954984
c2
PProg
performs a certain task
1
lapses
1/4
users
230%
ease
nid:1771363954984
Cloze c2
Cloze answer: performs a certain task
Q: A {{c1::functional unit}} is a component of a CPU (or core) that {{c2::performs a certain task}}, an {{c3::execution unit}} is one such example.
A: performing a task - e.g. executing integer arithmetic operations
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955029 |
1 | 230% | 18d | 7 |
nid:1771363954987
c1
PProg
granularity
1
lapses
1/4
users
230%
ease
nid:1771363954987
Cloze c1
Cloze answer: granularity
Q: The trick with {{c1::granularity}} is to find a size that {{c2::minimizes overhead}} while {{c3::maximizing parallelism}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955040 |
1 | 230% | 10d | 8 |
nid:1771363954993
c2
PProg
a property of a system: "something good eventually happens"
1
lapses
1/4
users
230%
ease
nid:1771363954993
Cloze c2
Cloze answer: a property of a system: "something good eventually happens"
Q: A {{c1::liveness property}} is {{c2::a property of a system: "something good eventually happens"}}.
A: Can only be violated in infinite time. Infinite loops and starvation are typical liveness properties.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955058 |
1 | 230% | 32d | 6 |
nid:1771363955001
c2
PProg
The maximum possible speedup ({{c1::parallelism}}) is {{c2::...
1
lapses
1/4
users
230%
ease
nid:1771363955001
Cloze c2
Q: The maximum possible speedup ({{c1::parallelism}}) is {{c2::\(\frac{T_1}{T_\infty} \)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955092 |
1 | 230% | 5d | 5 |
nid:1771363955014
c2
PProg
extra time spent by the system or the algorithm
1
lapses
1/4
users
230%
ease
nid:1771363955014
Cloze c2
Cloze answer: extra time spent by the system or the algorithm
Q: {{c1::Scheduling overhead}} is the {{c2::extra time spent by the system or the algorithm}} to distribute work on {{c3::multiple threads/tasks}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955147 |
1 | 230% | 25d | 6 |
nid:1771363955022
c1
PProg
Span
1
lapses
1/4
users
230%
ease
nid:1771363955022
Cloze c1
Cloze answer: Span
Q: {{c1::Span}} is the {{c2::critical path (height)}} of the task graph. It corresponds to {{c3::T_∞}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955172 |
1 | 230% | 9d | 7 |
nid:1771363955022
c2
PProg
critical path (height)
1
lapses
1/4
users
230%
ease
nid:1771363955022
Cloze c2
Cloze answer: critical path (height)
Q: {{c1::Span}} is the {{c2::critical path (height)}} of the task graph. It corresponds to {{c3::T_∞}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771363955173 |
1 | 230% | 6d | 8 |
nid:1771364083961
c1
Analysis
halboffenes
1
lapses
1/4
users
230%
ease
nid:1771364083961
Cloze c1
Cloze answer: halboffenes
Q: Ein {{c1::halboffenes}} Intervall zwischen \(a\) und \(b\) wäre z.B.:\({{c2::[a, b)}}={{c3::\{x \in \mathbb{R} \mid a \leq x < b\} }}\).
A: Das Intervall kann selbstverständlich auch in die andere Richtung geöffnet sein:\((a, b]=\{x \in \mathbb{R} \mid a < x \leq b\}\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771364083966 |
1 | 230% | 2d | 5 |
nid:1771578182870
c2
PProg
load imbalance
1
lapses
1/4
users
230%
ease
nid:1771578182870
Cloze c2
Cloze answer: load imbalance
Q: Parallel execution can introduce inefficiencies such as {{c1::communication overhead}}, {{c2::load imbalance}}, and {{c3::idle time due to task dependencies or waiting for data exchange}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771578182870 |
1 | 230% | 3d | 5 |
nid:1771616145174
c1
Advanced Finance
reductions in a firm's value that arise from agency problems
1
lapses
1/4
users
230%
ease
nid:1771616145174
Cloze c1
Cloze answer: reductions in a firm's value that arise from agency problems
Q: Agency costs are {{c1::reductions in a firm's value that arise from agency problems}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771616145174 |
1 | 230% | 15d | 11 |
nid:1771616439344
c5
Advanced Finance
focusing on short-term results at the expense of long-term r...
1
lapses
1/4
users
230%
ease
nid:1771616439344
Cloze c5
Cloze answer: focusing on short-term results at the expense of long-term results
Q: Agency problems include a manager:{{c1:: not putting in sufficient effort}}{{c2:: wasting money on personal benefits}}{{c3:: overinvesting in search of power or prestige}}{{c4:: taking too many or too few risks}}{{c5:: focusing on short-term results at
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771616439347 |
1 | 230% | 17d | 9 |
nid:1771770315370
c1
Advanced Finance
incentive missalignment
1
lapses
1/4
users
230%
ease
nid:1771770315370
Cloze c1
Cloze answer: incentive missalignment
Q: Family controlled companies struggle less with {{c1::incentive missalignment}} because {{c2::the shareholders and management are one and the same}}, they may, however have problems with {{c3::exploitation of minority shareholders}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771770315370 |
1 | 230% | 19d | 7 |
nid:1771771254633
c2
Advanced Finance
smaller and more independent
1
lapses
1/4
users
230%
ease
nid:1771771254633
Cloze c2
Cloze answer: smaller and more independent
Q: Boards in {{c1::the U.S. and UK}} are typically {{c2:: smaller and more independent}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771771254634 |
1 | 230% | 14d | 7 |
nid:1771780392187
c1
DDCA
Unique mapping from input values to output values; The same ...
1
lapses
1/4
users
230%
ease
nid:1771780392187
Cloze c1
Cloze answer: Unique mapping from input values to output values; The same input values produce the same output value every time.; No memory (output does not depend on past input values)
Q: What does the "functional" in functional specification signify?{{c1::Unique mapping from input values to output values}}{{c1::The same input values produce the same output value every time.}}{{c1::No memory (output does not depend on past input values)}}
A: Example: Full 1-bit adder
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771780392187 |
1 | 230% | 3d | 12 |
nid:1771780392199
DDCA
What's the formula for energy consumption?
1
lapses
1/4
users
230%
ease
nid:1771780392199
Q: What's the formula for energy consumption?
A: Power * Time
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771780392202 |
1 | 230% | 3d | 8 |
nid:1771780392204
DDCA
What is an implicant?
1
lapses
1/4
users
230%
ease
nid:1771780392204
Q: What is an implicant?
A: A product (AND) of literals.\((A \cdot B \cdot \overline{C}) \text{ , } (\overline{A} \cdot C) \text{ , } (B \cdot \overline{C})\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771780392207 |
1 | 230% | 4d | 7 |
nid:1771780392210
c1
DDCA
CNF
1
lapses
1/4
users
230%
ease
nid:1771780392210
Cloze c1
Cloze answer: CNF
Q: Product of Sums is equivalent to {{c1::CNF}}.
A: This is also the DeMorgan of SOP of \(\overline F\).
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771780392213 |
1 | 230% | 2d | 5 |
nid:1771780392217
DDCA
How can we build NOR from NOT and AND?
1
lapses
1/4
users
230%
ease
nid:1771780392217
Q: How can we build NOR from NOT and AND?
A: NOR is equivalent to AND with inputs complemented.\(A=\overline{(X+Y)}=\overline X \space\overline Y\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771780392220 |
1 | 230% | 4d | 7 |
nid:1771780392218
c4
DDCA
Transistor (MOS)
1
lapses
1/4
users
230%
ease
nid:1771780392218
Cloze c4
Cloze answer: Transistor (MOS)
Q: By combining:
{{c1::Conductors (Metal)}}
{{c2::Insulators (Oxide)}}
{{c3::Semiconductors}}
We get a {{c4::Transistor (MOS)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771780392223 |
1 | 230% | 3d | 5 |
nid:1771780392220
c2
DDCA
broken (i.e., the circuit is open)
1
lapses
1/4
users
230%
ease
nid:1771780392220
Cloze c2
Cloze answer: broken (i.e., the circuit is open)
Q: If the gate of the n-type transistor is supplied with {{c1::zero}} voltage, the connection between the source and drain is {{c2::broken (i.e., the circuit is open)}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771780392227 |
1 | 230% | 3d | 5 |
nid:1771780392223
DDCA
How does a decoder work?
1
lapses
1/4
users
230%
ease
nid:1771780392223
Q: How does a decoder work?
A: \(n\) possible inputs and \(2^n\) outputsExactly one of the outputs is 1 and all the rest are 0sThe output that is logically 1 is the output corresponding to the input pattern that the logic circuit is expected to detectA decoder is an "input pattern detector".Example: 2-to-4 decoder
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771780392230 |
1 | 230% | 3d | 5 |
nid:1771780392226
DDCA
What's the formula for dynamic power consumption?
1
lapses
1/4
users
230%
ease
nid:1771780392226
Q: What's the formula for dynamic power consumption?
A: \(C\cdot V^2\cdot f\)\(C =\) capacitance of the circuit (wires and gates)\(V =\) supply voltage\(f =\) charging frequency of the capacitor
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771780392233 |
1 | 230% | 3d | 7 |
nid:1771780392231
c1
DDCA
the delay between inputs changing and outputs responding
1
lapses
1/4
users
230%
ease
nid:1771780392231
Cloze c1
Cloze answer: the delay between inputs changing and outputs responding
Q: Timing specification describes {{c1::the delay between inputs changing and outputs responding}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771780392238 |
1 | 230% | 2d | 6 |
nid:1771794049785
c1
Advanced Finance
opinion that the statement is representative and in-line wit...
1
lapses
1/4
users
230%
ease
nid:1771794049785
Cloze c1
Cloze answer: opinion that the statement is representative and in-line with GAAP
Q: If an auditor finds no problems in a firm's financial statement, he issues an {{c1::opinion that the statement is representative and in-line with GAAP}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771794049785 |
1 | 230% | 11d | 11 |
nid:1771794112422
c2
Advanced Finance
the accounts of the firm have not been represented accuratel...
1
lapses
1/4
users
230%
ease
nid:1771794112422
Cloze c2
Cloze answer: the accounts of the firm have not been represented accurately
Q: If an auditor finds problems they can issue a {{c1::qualified opinion}} which states that {{c2::the accounts of the firm have not been represented accurately}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771794112423 |
1 | 230% | 14d | 11 |
nid:1771795613218
c2
Advanced Finance
syndicate ownership
1
lapses
1/4
users
230%
ease
nid:1771795613218
Cloze c2
Cloze answer: syndicate ownership
Q: {{c1::Keiretsu}} is a Japanese system of {{c2::syndicate ownership}} which centers around a main {{c3::bank}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771795613220 |
1 | 230% | 6d | 8 |
nid:1771795784415
c1
Advanced Finance
gives companies more space when they get in financial proble...
1
lapses
1/4
users
230%
ease
nid:1771795784415
Cloze c1
Cloze answer: gives companies more space when they get in financial problems
Q: The keiretsu system has positives in that it {{c1::gives companies more space when they get in financial problems}}.
A: This is because the company's lender is most likely the main group bank.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771795784415 |
1 | 230% | 14d | 6 |
nid:1771836465439
c1
PProg
the sequential part of a program
1
lapses
1/4
users
230%
ease
nid:1771836465439
Cloze c1
Cloze answer: the sequential part of a program
Q: Efficiency is heavily limited by {{c1::the sequential part of a program}}.
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771836465439 |
1 | 230% | 1d | 3 |
nid:1771836518739
c1
PProg
Efficiency
1
lapses
1/4
users
230%
ease
nid:1771836518739
Cloze c1
Cloze answer: Efficiency
Q: {{c1::Efficiency}} = {{c2::\(\frac{S_p}{p}\)}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771836518739 |
1 | 230% | 2d | 5 |
nid:1771836628438
c2
PProg
enforce mutual exclusion
1
lapses
1/4
users
230%
ease
nid:1771836628438
Cloze c2
Cloze answer: enforce mutual exclusion
Q: Locks are typically used to {{c2::enforce mutual exclusion}} by {{c1::guarding/protecting a critical section.}}
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771836628438 |
1 | 230% | 2d | 5 |
nid:1771914065795
Analysis
Ist die Menge \(A \neq \emptyset\) nach oben/unten unbeschrä...
1
lapses
1/4
users
230%
ease
nid:1771914065795
Q: Ist die Menge \(A \neq \emptyset\) nach oben/unten unbeschränkt, so definieren wir Supremum/Infinum:
A: \(\sup(A) = \infty\)/\(\inf(A) = -\infty\)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1771914065795 |
1 | 230% | 5d | 6 |
nid:1772090857637
c1
A&W
NP-vollständig
1
lapses
1/4
users
230%
ease
nid:1772090857637
Cloze c1
Cloze answer: NP-vollständig
Q: Das Problem „Gegeben ein Graph \(G = (V, E)\), enthält \(G\) einen Hamiltonkreis?" ist {{c1::NP-vollständig}}.
A: Karp (1972)
| User | Card ID | Lapses | Ease | Interval | Reviews |
|---|---|---|---|---|---|
| tomas | cid:1772090857637 |
1 | 230% | 3d | 5 |