Mixing
Is the converse true?
$begingroup$
Let $G$ be a connected bipartite graph with partite sets $X$ and $Y$ such that cardinality of $X$ equals to cardinality of $Y$ $geq$ 2. If every two vertices of $X$ have distinct degrees in $G$ then show that $G$ has a perfect matching. Is the converse true? Justify your answer.
graph-theory bipartite-graph matching-theory
asked Jan 29 at 17:58
bandana pandeybandana pandey
63
$endgroup$
add a comment |
$begingroup$
Let $G$ be a connected bipartite graph with partite sets $X$ and $Y$ such that cardinality of $X$ equals to cardinality of $Y$ $geq$ 2. If every two vertices of $X$ have distinct degrees in $G$ then show that $G$ has a perfect matching. Is the converse true? Justify your answer.
graph-theory bipartite-graph matching-theory
asked Jan 29 at 17:58
bandana pandeybandana pandey
63
$endgroup$
1
$begingroup$
For the converse: Try all vertices with equal degrees, e.g. |X|=|Y|=3 and the degrees 2.
$endgroup$
– user376343
Jan 29 at 18:11
add a comment |
$begingroup$
Let $G$ be a connected bipartite graph with partite sets $X$ and $Y$ such that cardinality of $X$ equals to cardinality of $Y$ $geq$ 2. If every two vertices of $X$ have distinct degrees in $G$ then show that $G$ has a perfect matching. Is the converse true? Justify your answer.
graph-theory bipartite-graph matching-theory
asked Jan 29 at 17:58
bandana pandeybandana pandey
63
$endgroup$
Let $G$ be a connected bipartite graph with partite sets $X$ and $Y$ such that cardinality of $X$ equals to cardinality of $Y$ $geq$ 2. If every two vertices of $X$ have distinct degrees in $G$ then show that $G$ has a perfect matching. Is the converse true? Justify your answer.
graph-theory bipartite-graph matching-theory
graph-theory bipartite-graph matching-theory
asked Jan 29 at 17:58
bandana pandeybandana pandey
63
asked Jan 29 at 17:58
bandana pandeybandana pandey
63
asked Jan 29 at 17:58
bandana pandeybandana pandey
63
asked Jan 29 at 17:58
bandana pandeybandana pandey
63
asked Jan 29 at 17:58
bandana pandeybandana pandey
63
63
1
$begingroup$
For the converse: Try all vertices with equal degrees, e.g. |X|=|Y|=3 and the degrees 2.
$endgroup$
– user376343
Jan 29 at 18:11
add a comment |
1
$begingroup$
For the converse: Try all vertices with equal degrees, e.g. |X|=|Y|=3 and the degrees 2.
$endgroup$
– user376343
Jan 29 at 18:11
1
1
$begingroup$
For the converse: Try all vertices with equal degrees, e.g. |X|=|Y|=3 and the degrees 2.
$endgroup$
– user376343
Jan 29 at 18:11
$begingroup$
For the converse: Try all vertices with equal degrees, e.g. |X|=|Y|=3 and the degrees 2.
$endgroup$
– user376343
Jan 29 at 18:11
add a comment |
1 Answer
1
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$begingroup$
We have to check that $G$ satisfies the conditions of Hall’s Marriage Theorem. Indeed, let $W$ be an arbitrary non-empty subset of $X$. Since all vertices of $W$ have distinct degrees (which are positive, because $G$ is connected),
there exists a vertex $vin W$ with degree $deg vge |W|$. Then $|N_G(W)|ge deg vge |W|.$
The converse is not true. A simple example it a bipartite graph whose set of edges is a perfect matching.
answered Jan 30 at 16:13
Alex RavskyAlex Ravsky
42.7k32483
$endgroup$
add a comment |
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$begingroup$
We have to check that $G$ satisfies the conditions of Hall’s Marriage Theorem. Indeed, let $W$ be an arbitrary non-empty subset of $X$. Since all vertices of $W$ have distinct degrees (which are positive, because $G$ is connected),
there exists a vertex $vin W$ with degree $deg vge |W|$. Then $|N_G(W)|ge deg vge |W|.$
The converse is not true. A simple example it a bipartite graph whose set of edges is a perfect matching.
answered Jan 30 at 16:13
Alex RavskyAlex Ravsky
42.7k32483
$endgroup$
add a comment |
$begingroup$
We have to check that $G$ satisfies the conditions of Hall’s Marriage Theorem. Indeed, let $W$ be an arbitrary non-empty subset of $X$. Since all vertices of $W$ have distinct degrees (which are positive, because $G$ is connected),
there exists a vertex $vin W$ with degree $deg vge |W|$. Then $|N_G(W)|ge deg vge |W|.$
The converse is not true. A simple example it a bipartite graph whose set of edges is a perfect matching.
answered Jan 30 at 16:13
Alex RavskyAlex Ravsky
42.7k32483
$endgroup$
add a comment |
$begingroup$
We have to check that $G$ satisfies the conditions of Hall’s Marriage Theorem. Indeed, let $W$ be an arbitrary non-empty subset of $X$. Since all vertices of $W$ have distinct degrees (which are positive, because $G$ is connected),
there exists a vertex $vin W$ with degree $deg vge |W|$. Then $|N_G(W)|ge deg vge |W|.$
The converse is not true. A simple example it a bipartite graph whose set of edges is a perfect matching.
answered Jan 30 at 16:13
Alex RavskyAlex Ravsky
42.7k32483
$endgroup$
We have to check that $G$ satisfies the conditions of Hall’s Marriage Theorem. Indeed, let $W$ be an arbitrary non-empty subset of $X$. Since all vertices of $W$ have distinct degrees (which are positive, because $G$ is connected),
there exists a vertex $vin W$ with degree $deg vge |W|$. Then $|N_G(W)|ge deg vge |W|.$
The converse is not true. A simple example it a bipartite graph whose set of edges is a perfect matching.
answered Jan 30 at 16:13
Alex RavskyAlex Ravsky
42.7k32483
answered Jan 30 at 16:13
Alex RavskyAlex Ravsky
42.7k32483
answered Jan 30 at 16:13
Alex RavskyAlex Ravsky
42.7k32483
answered Jan 30 at 16:13
Alex RavskyAlex Ravsky
42.7k32483
42.7k32483
add a comment |
add a comment |
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$begingroup$
For the converse: Try all vertices with equal degrees, e.g. |X|=|Y|=3 and the degrees 2.
$endgroup$
– user376343
Jan 29 at 18:11