How to find a minumum vertex cover from a maximum matching in a bipartite graph?
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Konig's theorem states that for a bipartite graph the number of vertices in the minimum vertex cover equals the number of edges in a maximum matching.
https://en.wikipedia.org/wiki/K%C5%91nig%27s_theorem_(graph_theory)
For example given a bipartite graph with vertices 1,2,...,10 (with 5 on the left and 5 the right) and edges 1-6, 1-8, 2-8, 3-7, 3-8, 3-9, 4-8, 5-7, 5-8, 5-10. A maximal matching is 1-6, 2-8, 3-7, 5-10.
But how do you find a minimum vertex cover given a set of edges for a maximum matching? Such as 1, 8, 3, 5 in the example above.
graph-theory algorithms
asked Dec 17 '18 at 18:13
tyebilliontyebillion
1034
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add a comment |
$begingroup$
Konig's theorem states that for a bipartite graph the number of vertices in the minimum vertex cover equals the number of edges in a maximum matching.
https://en.wikipedia.org/wiki/K%C5%91nig%27s_theorem_(graph_theory)
For example given a bipartite graph with vertices 1,2,...,10 (with 5 on the left and 5 the right) and edges 1-6, 1-8, 2-8, 3-7, 3-8, 3-9, 4-8, 5-7, 5-8, 5-10. A maximal matching is 1-6, 2-8, 3-7, 5-10.
But how do you find a minimum vertex cover given a set of edges for a maximum matching? Such as 1, 8, 3, 5 in the example above.
graph-theory algorithms
asked Dec 17 '18 at 18:13
tyebilliontyebillion
1034
$endgroup$
add a comment |
$begingroup$
Konig's theorem states that for a bipartite graph the number of vertices in the minimum vertex cover equals the number of edges in a maximum matching.
https://en.wikipedia.org/wiki/K%C5%91nig%27s_theorem_(graph_theory)
For example given a bipartite graph with vertices 1,2,...,10 (with 5 on the left and 5 the right) and edges 1-6, 1-8, 2-8, 3-7, 3-8, 3-9, 4-8, 5-7, 5-8, 5-10. A maximal matching is 1-6, 2-8, 3-7, 5-10.
But how do you find a minimum vertex cover given a set of edges for a maximum matching? Such as 1, 8, 3, 5 in the example above.
graph-theory algorithms
asked Dec 17 '18 at 18:13
tyebilliontyebillion
1034
$endgroup$
Konig's theorem states that for a bipartite graph the number of vertices in the minimum vertex cover equals the number of edges in a maximum matching.
https://en.wikipedia.org/wiki/K%C5%91nig%27s_theorem_(graph_theory)
For example given a bipartite graph with vertices 1,2,...,10 (with 5 on the left and 5 the right) and edges 1-6, 1-8, 2-8, 3-7, 3-8, 3-9, 4-8, 5-7, 5-8, 5-10. A maximal matching is 1-6, 2-8, 3-7, 5-10.
But how do you find a minimum vertex cover given a set of edges for a maximum matching? Such as 1, 8, 3, 5 in the example above.
graph-theory algorithms
graph-theory algorithms
asked Dec 17 '18 at 18:13
tyebilliontyebillion
1034
asked Dec 17 '18 at 18:13
tyebilliontyebillion
1034
asked Dec 17 '18 at 18:13
tyebilliontyebillion
1034
asked Dec 17 '18 at 18:13
tyebilliontyebillion
1034
asked Dec 17 '18 at 18:13
tyebilliontyebillion
1034
1034
add a comment |
add a comment |
1 Answer
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The proof in the article you linked gives such an explicit construction. In your example, if $L = {1,2,3,4,5}$ , $R = {6,7,8,9,10}$ are the partite sets, let $Z$ be the set of vertices that are either unmatched vertices of $L$
or connected to an unmatched vertex in $L$ by an alternating path.
$4$ is the only unmatched vertex in $L$ and the only vertices we can reach by alternating paths are $2$ and $8$, so $Z = {2,4,8}$. A minimum vertex cover is then given by $(Lsetminus Z) cup (Rcap Z) = {1,3,5,8}$.
answered Dec 17 '18 at 18:49
BertrandBertrand
685
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1 Answer
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$begingroup$
The proof in the article you linked gives such an explicit construction. In your example, if $L = {1,2,3,4,5}$ , $R = {6,7,8,9,10}$ are the partite sets, let $Z$ be the set of vertices that are either unmatched vertices of $L$
or connected to an unmatched vertex in $L$ by an alternating path.
$4$ is the only unmatched vertex in $L$ and the only vertices we can reach by alternating paths are $2$ and $8$, so $Z = {2,4,8}$. A minimum vertex cover is then given by $(Lsetminus Z) cup (Rcap Z) = {1,3,5,8}$.
answered Dec 17 '18 at 18:49
BertrandBertrand
685
$endgroup$
add a comment |
$begingroup$
The proof in the article you linked gives such an explicit construction. In your example, if $L = {1,2,3,4,5}$ , $R = {6,7,8,9,10}$ are the partite sets, let $Z$ be the set of vertices that are either unmatched vertices of $L$
or connected to an unmatched vertex in $L$ by an alternating path.
$4$ is the only unmatched vertex in $L$ and the only vertices we can reach by alternating paths are $2$ and $8$, so $Z = {2,4,8}$. A minimum vertex cover is then given by $(Lsetminus Z) cup (Rcap Z) = {1,3,5,8}$.
answered Dec 17 '18 at 18:49
BertrandBertrand
685
$endgroup$
add a comment |
$begingroup$
The proof in the article you linked gives such an explicit construction. In your example, if $L = {1,2,3,4,5}$ , $R = {6,7,8,9,10}$ are the partite sets, let $Z$ be the set of vertices that are either unmatched vertices of $L$
or connected to an unmatched vertex in $L$ by an alternating path.
$4$ is the only unmatched vertex in $L$ and the only vertices we can reach by alternating paths are $2$ and $8$, so $Z = {2,4,8}$. A minimum vertex cover is then given by $(Lsetminus Z) cup (Rcap Z) = {1,3,5,8}$.
answered Dec 17 '18 at 18:49
BertrandBertrand
685
$endgroup$
The proof in the article you linked gives such an explicit construction. In your example, if $L = {1,2,3,4,5}$ , $R = {6,7,8,9,10}$ are the partite sets, let $Z$ be the set of vertices that are either unmatched vertices of $L$
or connected to an unmatched vertex in $L$ by an alternating path.
$4$ is the only unmatched vertex in $L$ and the only vertices we can reach by alternating paths are $2$ and $8$, so $Z = {2,4,8}$. A minimum vertex cover is then given by $(Lsetminus Z) cup (Rcap Z) = {1,3,5,8}$.
answered Dec 17 '18 at 18:49
BertrandBertrand
685
edited Dec 17 '18 at 19:00
answered Dec 17 '18 at 18:49
BertrandBertrand
685
answered Dec 17 '18 at 18:49
BertrandBertrand
685
answered Dec 17 '18 at 18:49
BertrandBertrand
685
685
add a comment |
add a comment |
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