Anki Deck Changes

Note 1: ETH::A&D

Deck: ETH::A&D
Note Type: Horvath Cloze
GUID: C1$vvwFoR0

modified

Before

Front

In graph theory, an closed Eulerian walk (Eulerzyklus) is an Eulerian walk (Eulerweg) that ends at the start vertex.

Back

In graph theory, an closed Eulerian walk (Eulerzyklus) is an Eulerian walk (Eulerweg) that ends at the start vertex.

After

Front

In graph theory, a closed Eulerian walk (Eulerzyklus) is an Eulerian walk (Eulerweg) that ends at the start vertex.

Back

In graph theory, a closed Eulerian walk (Eulerzyklus) is an Eulerian walk (Eulerweg) that ends at the start vertex.

Field-by-field Comparison

Field	Before	After
Text	In graph theory, an {{c2::closed Eulerian walk (Eulerzyklus)}} is an {{c1::Eulerian walk (Eulerweg) that ends at the start vertex}}.	In graph theory, a {{c2::closed Eulerian walk (Eulerzyklus)}} is an {{c1::Eulerian walk (Eulerweg) that ends at the start vertex}}.

Tags: ETH::1._Semester::A&D::07._Graphs::2._Closed_Eulerian_Walks

Note 2: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Cloze
GUID: Fz6Ww3Tn2D

modified

Before

Front

The set of clauses associated with a formula \[F = (L_{11} \lor \dots \lor L_{1m_1}) \land \dots \land (L_{n1} \lor \dots \lor L_{nm_n})\] in CNF, denoted as {{c1::$\mathcal{K}(F)$}}, is the set {{c2::\[\mathcal{K}(F) = \{\{L_{11}, \dots, L_{1m_1}\}, \dots, \{L_{n1}, \dots, L_{nm_n}\}\}\]}}

Back

The set of clauses associated with a formula \[F = (L_{11} \lor \dots \lor L_{1m_1}) \land \dots \land (L_{n1} \lor \dots \lor L_{nm_n})\] in CNF, denoted as {{c1::$\mathcal{K}(F)$}}, is the set {{c2::\[\mathcal{K}(F) = \{\{L_{11}, \dots, L_{1m_1}\}, \dots, \{L_{n1}, \dots, L_{nm_n}\}\}\]}}

After

Front

The set of clauses associated with a formula \[F = (L_{11} \lor \dots \lor L_{1m_1}) \land \dots \land (L_{n1} \lor \dots \lor L_{nm_n})\] in CNF, denoted as {{c1::$\mathcal{K}(F)$}}, is the set {{c2::\[{{c1::\mathcal{K}(F)}} = \{\{L_{11}, \dots, L_{1m_1}\}, \dots, \{L_{n1}, \dots, L_{nm_n}\}\}\]}}

Back

The set of clauses associated with a formula \[F = (L_{11} \lor \dots \lor L_{1m_1}) \land \dots \land (L_{n1} \lor \dots \lor L_{nm_n})\] in CNF, denoted as {{c1::$\mathcal{K}(F)$}}, is the set {{c2::\[{{c1::\mathcal{K}(F)}} = \{\{L_{11}, \dots, L_{1m_1}\}, \dots, \{L_{n1}, \dots, L_{nm_n}\}\}\]}}

Field-by-field Comparison

Field	Before	After
Text	The set of clauses associated with a formula \[F = (L_{11} \lor \dots \lor L_{1m_1}) \land \dots \land (L_{n1} \lor \dots \lor L_{nm_n})\] in CNF, denoted as {{c1::$\mathcal{K}(F)$}}, is the set {{c2::\[\mathcal{K}(F) = \{\{L_{11}, \dots, L_{1m_1}\}, \dots, \{L_{n1}, \dots, L_{nm_n}\}\}\]}}	The set of clauses associated with a formula \[F = (L_{11} \lor \dots \lor L_{1m_1}) \land \dots \land (L_{n1} \lor \dots \lor L_{nm_n})\] in CNF, denoted as {{c1::$\mathcal{K}(F)$}}, is the set {{c2::\[{{c1::\mathcal{K}(F)}} = \{\{L_{11}, \dots, L_{1m_1}\}, \dots, \{L_{n1}, \dots, L_{nm_n}\}\}\]}}

Tags: ETH::1._Semester::DiskMat::6._Logic::5._Propositional_Logic::5._The_Resolution_Calculus_for_Propositional_Logic

Note 3: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Classic
GUID: JOM4&m6Z&$

modified

Before

Front

What is a cyclic group of order $n$ isomorphic to?

Back

What is a cyclic group of order $n$ isomorphic to?

Theorem 5.7: A cyclic group of order $n$ is isomorphic to $\langle \mathbb{Z}_n; \oplus \rangle$ (and hence abelian).

This means all cyclic groups of the same order have the same structure.

After

Front

What is a cyclic group of order $n$ isomorphic to?

Back

What is a cyclic group of order $n$ isomorphic to?

Theorem 5.7: A cyclic group of order $n$ is isomorphic to $\langle \mathbb{Z}_n; \oplus \rangle$ (and hence abelian).

This means all cyclic groups of the same order have the same structure.

Explanation:
You can easily create an isomophism. For any$[a], [b] \in \mathbb{Z}_n$,

$\varphi([a] + [b]) = \varphi([a+b])$$= g^{a+b} = g^a g^b = \varphi([a]) \varphi([b]).$

Field-by-field Comparison

Field	Before	After
Back	<p><strong>Theorem 5.7</strong>: A cyclic group of order $n$ is <strong>isomorphic</strong> to $\langle \mathbb{Z}_n; \oplus \rangle$ (and hence <strong>abelian</strong>).</p> <p>This means all cyclic groups of the same order have the same structure.</p>	<p><strong>Theorem 5.7</strong>: A cyclic group of order $n$ is <strong>isomorphic</strong> to $\langle \mathbb{Z}_n; \oplus \rangle$ (and hence <strong>abelian</strong>).</p> <p>This means all cyclic groups of the same order have the same structure.</p><p><b>Explanation:</b> <br>You can easily create an isomophism. For any$[a], [b] \in \mathbb{Z}_n$,</p><p>$\varphi([a] + [b]) = \varphi([a+b])$$= g^{a+b} = g^a g^b = \varphi([a]) \varphi([b]).$</p>

Tags: ETH::1._Semester::DiskMat::5._Algebra::3._The_Structure_of_Groups::5._Cyclic_Groups