Is the converse true?












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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







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  • 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












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






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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














  • 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










1 Answer
1






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1












$begingroup$

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.






share|cite|improve this answer









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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.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 at 16:13









        Alex RavskyAlex Ravsky

        42.7k32483




        42.7k32483






























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