Algorithmic Graph Theory - perfect matching in bipartite graph when det(A) is not 0 The Next CEO of Stack OverflowProving that the characteristic polynomial of a bipartite graph has alternating positive and negative coefficientsLargest Matching whose removal does not leave Eulerian componentsRandomized Algorithm for finding perfect matchingsbipartite graph has perfect matchingHow do you find a perfect bipartite graph using Hall's Theorem?Proving that a Bipartite graph with has a perfect matching$2$-edge-connected graph has perfect matchingMaximum “$2$-to-$1$” matching in a bipartite graphRank of adjacency matrix of twin-free bipartite graph and maximum matchingGraph Theory as an approach to Category Theory
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Algorithmic Graph Theory - perfect matching in bipartite graph when det(A) is not 0
The Next CEO of Stack OverflowProving that the characteristic polynomial of a bipartite graph has alternating positive and negative coefficientsLargest Matching whose removal does not leave Eulerian componentsRandomized Algorithm for finding perfect matchingsbipartite graph has perfect matchingHow do you find a perfect bipartite graph using Hall's Theorem?Proving that a Bipartite graph with has a perfect matching$2$-edge-connected graph has perfect matchingMaximum “$2$-to-$1$” matching in a bipartite graphRank of adjacency matrix of twin-free bipartite graph and maximum matchingGraph Theory as an approach to Category Theory
$begingroup$
While learning about graphs, I came across theorem that I don't quite understand, and can't find a proof.
If G is bipartite, and $det(A) neq 0,$ then G has a perfect matching. (Given that matrix representation of G is A)
Ps.:
I found bits of information, that proof may be connected with Edmonds Theorem, which I cannot find, since Edmonds-Karp Algorithm is way more popular in google ;)
graph-theory algebraic-graph-theory bipartite-graph
asked yesterday
YurkeeYurkee
1063
New contributor
Yurkee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
While learning about graphs, I came across theorem that I don't quite understand, and can't find a proof.
If G is bipartite, and $det(A) neq 0,$ then G has a perfect matching. (Given that matrix representation of G is A)
Ps.:
I found bits of information, that proof may be connected with Edmonds Theorem, which I cannot find, since Edmonds-Karp Algorithm is way more popular in google ;)
graph-theory algebraic-graph-theory bipartite-graph
asked yesterday
YurkeeYurkee
1063
New contributor
Yurkee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Do you mean bipartite by chance?
$endgroup$
– Don Thousand
yesterday
$begingroup$
Do you mean "bipartite?"
$endgroup$
– saulspatz
yesterday
$begingroup$
thats exactly what I meant, I corrected this typo ;)
$endgroup$
– Yurkee
yesterday
$begingroup$
It's the next-to-last letter we are asking about.
$endgroup$
– saulspatz
yesterday
$begingroup$
Yes, that's the one where vertices can be divided into two disjoint and independent sets.
$endgroup$
– Yurkee
yesterday
add a comment |
$begingroup$
While learning about graphs, I came across theorem that I don't quite understand, and can't find a proof.
If G is bipartite, and $det(A) neq 0,$ then G has a perfect matching. (Given that matrix representation of G is A)
Ps.:
I found bits of information, that proof may be connected with Edmonds Theorem, which I cannot find, since Edmonds-Karp Algorithm is way more popular in google ;)
graph-theory algebraic-graph-theory bipartite-graph
asked yesterday
YurkeeYurkee
1063
New contributor
Yurkee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
While learning about graphs, I came across theorem that I don't quite understand, and can't find a proof.
If G is bipartite, and $det(A) neq 0,$ then G has a perfect matching. (Given that matrix representation of G is A)
Ps.:
I found bits of information, that proof may be connected with Edmonds Theorem, which I cannot find, since Edmonds-Karp Algorithm is way more popular in google ;)
graph-theory algebraic-graph-theory bipartite-graph
graph-theory algebraic-graph-theory bipartite-graph
asked yesterday
YurkeeYurkee
1063
New contributor
Yurkee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
YurkeeYurkee
1063
New contributor
Yurkee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Yurkee
asked yesterday
YurkeeYurkee
1063
New contributor
Yurkee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
YurkeeYurkee
1063
asked yesterday
YurkeeYurkee
1063
1063
New contributor
Yurkee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yurkee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yurkee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Do you mean bipartite by chance?
$endgroup$
– Don Thousand
yesterday
$begingroup$
Do you mean "bipartite?"
$endgroup$
– saulspatz
yesterday
$begingroup$
thats exactly what I meant, I corrected this typo ;)
$endgroup$
– Yurkee
yesterday
$begingroup$
It's the next-to-last letter we are asking about.
$endgroup$
– saulspatz
yesterday
$begingroup$
Yes, that's the one where vertices can be divided into two disjoint and independent sets.
$endgroup$
– Yurkee
yesterday
add a comment |
$begingroup$
Do you mean bipartite by chance?
$endgroup$
– Don Thousand
yesterday
$begingroup$
Do you mean "bipartite?"
$endgroup$
– saulspatz
yesterday
$begingroup$
thats exactly what I meant, I corrected this typo ;)
$endgroup$
– Yurkee
yesterday
$begingroup$
It's the next-to-last letter we are asking about.
$endgroup$
– saulspatz
yesterday
$begingroup$
Yes, that's the one where vertices can be divided into two disjoint and independent sets.
$endgroup$
– Yurkee
yesterday
$begingroup$
Do you mean bipartite by chance?
$endgroup$
– Don Thousand
yesterday
$begingroup$
Do you mean bipartite by chance?
$endgroup$
– Don Thousand
yesterday
$begingroup$
Do you mean "bipartite?"
$endgroup$
– saulspatz
yesterday
$begingroup$
Do you mean "bipartite?"
$endgroup$
– saulspatz
yesterday
$begingroup$
thats exactly what I meant, I corrected this typo ;)
$endgroup$
– Yurkee
yesterday
$begingroup$
thats exactly what I meant, I corrected this typo ;)
$endgroup$
– Yurkee
yesterday
$begingroup$
It's the next-to-last letter we are asking about.
$endgroup$
– saulspatz
yesterday
$begingroup$
It's the next-to-last letter we are asking about.
$endgroup$
– saulspatz
yesterday
$begingroup$
Yes, that's the one where vertices can be divided into two disjoint and independent sets.
$endgroup$
– Yurkee
yesterday
$begingroup$
Yes, that's the one where vertices can be divided into two disjoint and independent sets.
$endgroup$
– Yurkee
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let the vertices be $[n]=1,dots,n.$ The theorem follows from the definition of determinant: $$detA=sum_sigmain S_n(-1)^sigma a_isigma(i)$$ where $S_n$ is the set of at permutations on $[n].$ Since $a_ij$ is $0$ or $1$, there must be a permutation $sigma$ such that $i$ and $sigma(i)$ are adjacent for $iin [n].$
Since $sigma$ is a bijection, the two bipartition sets have the same cardinality, and the restriction of $sigma$ to one of them gives a perfect matching.
answered yesterday
saulspatzsaulspatz
17.1k31435
$endgroup$
$begingroup$
Is this proof from the Edmonds Theorem?
$endgroup$
– Yurkee
yesterday
1
$begingroup$
Edmonds had a number of theorems. The one I've heard called "Edmonds Theorem" has to do with packing of arborecences, so nothing to do with this.
$endgroup$
– saulspatz
yesterday
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
Let the vertices be $[n]=1,dots,n.$ The theorem follows from the definition of determinant: $$detA=sum_sigmain S_n(-1)^sigma a_isigma(i)$$ where $S_n$ is the set of at permutations on $[n].$ Since $a_ij$ is $0$ or $1$, there must be a permutation $sigma$ such that $i$ and $sigma(i)$ are adjacent for $iin [n].$
Since $sigma$ is a bijection, the two bipartition sets have the same cardinality, and the restriction of $sigma$ to one of them gives a perfect matching.
answered yesterday
saulspatzsaulspatz
17.1k31435
$endgroup$
$begingroup$
Is this proof from the Edmonds Theorem?
$endgroup$
– Yurkee
yesterday
1
$begingroup$
Edmonds had a number of theorems. The one I've heard called "Edmonds Theorem" has to do with packing of arborecences, so nothing to do with this.
$endgroup$
– saulspatz
yesterday
add a comment |
$begingroup$
Let the vertices be $[n]=1,dots,n.$ The theorem follows from the definition of determinant: $$detA=sum_sigmain S_n(-1)^sigma a_isigma(i)$$ where $S_n$ is the set of at permutations on $[n].$ Since $a_ij$ is $0$ or $1$, there must be a permutation $sigma$ such that $i$ and $sigma(i)$ are adjacent for $iin [n].$
Since $sigma$ is a bijection, the two bipartition sets have the same cardinality, and the restriction of $sigma$ to one of them gives a perfect matching.
answered yesterday
saulspatzsaulspatz
17.1k31435
$endgroup$
$begingroup$
Is this proof from the Edmonds Theorem?
$endgroup$
– Yurkee
yesterday
1
$begingroup$
Edmonds had a number of theorems. The one I've heard called "Edmonds Theorem" has to do with packing of arborecences, so nothing to do with this.
$endgroup$
– saulspatz
yesterday
add a comment |
$begingroup$
Let the vertices be $[n]=1,dots,n.$ The theorem follows from the definition of determinant: $$detA=sum_sigmain S_n(-1)^sigma a_isigma(i)$$ where $S_n$ is the set of at permutations on $[n].$ Since $a_ij$ is $0$ or $1$, there must be a permutation $sigma$ such that $i$ and $sigma(i)$ are adjacent for $iin [n].$
Since $sigma$ is a bijection, the two bipartition sets have the same cardinality, and the restriction of $sigma$ to one of them gives a perfect matching.
answered yesterday
saulspatzsaulspatz
17.1k31435
$endgroup$
Let the vertices be $[n]=1,dots,n.$ The theorem follows from the definition of determinant: $$detA=sum_sigmain S_n(-1)^sigma a_isigma(i)$$ where $S_n$ is the set of at permutations on $[n].$ Since $a_ij$ is $0$ or $1$, there must be a permutation $sigma$ such that $i$ and $sigma(i)$ are adjacent for $iin [n].$
Since $sigma$ is a bijection, the two bipartition sets have the same cardinality, and the restriction of $sigma$ to one of them gives a perfect matching.
answered yesterday
saulspatzsaulspatz
17.1k31435
answered yesterday
saulspatzsaulspatz
17.1k31435
answered yesterday
saulspatzsaulspatz
17.1k31435
answered yesterday
saulspatzsaulspatz
17.1k31435
17.1k31435
$begingroup$
Is this proof from the Edmonds Theorem?
$endgroup$
– Yurkee
yesterday
1
$begingroup$
Edmonds had a number of theorems. The one I've heard called "Edmonds Theorem" has to do with packing of arborecences, so nothing to do with this.
$endgroup$
– saulspatz
yesterday
add a comment |
$begingroup$
Is this proof from the Edmonds Theorem?
$endgroup$
– Yurkee
yesterday
1
$begingroup$
Edmonds had a number of theorems. The one I've heard called "Edmonds Theorem" has to do with packing of arborecences, so nothing to do with this.
$endgroup$
– saulspatz
yesterday
$begingroup$
Is this proof from the Edmonds Theorem?
$endgroup$
– Yurkee
yesterday
$begingroup$
Is this proof from the Edmonds Theorem?
$endgroup$
– Yurkee
yesterday
1
1
$begingroup$
Edmonds had a number of theorems. The one I've heard called "Edmonds Theorem" has to do with packing of arborecences, so nothing to do with this.
$endgroup$
– saulspatz
yesterday
$begingroup$
Edmonds had a number of theorems. The one I've heard called "Edmonds Theorem" has to do with packing of arborecences, so nothing to do with this.
$endgroup$
– saulspatz
yesterday
add a comment |
Yurkee is a new contributor. Be nice, and check out our Code of Conduct.
Yurkee is a new contributor. Be nice, and check out our Code of Conduct.
Yurkee is a new contributor. Be nice, and check out our Code of Conduct.
Yurkee is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Do you mean bipartite by chance?
$endgroup$
– Don Thousand
yesterday
$begingroup$
Do you mean "bipartite?"
$endgroup$
– saulspatz
yesterday
$begingroup$
thats exactly what I meant, I corrected this typo ;)
$endgroup$
– Yurkee
yesterday
$begingroup$
It's the next-to-last letter we are asking about.
$endgroup$
– saulspatz
yesterday
$begingroup$
Yes, that's the one where vertices can be divided into two disjoint and independent sets.
$endgroup$
– Yurkee
yesterday