Can there be a bipartite graph that has an Eulerian circuit, Hamiltonian cycle, but is nonplanar?












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So far I have that the graph's chromatic number is two (not sure how that would help) as well as that since it has an Eulerian circuit, all vertices must have even degree. I am thinking of using Kuratowski's theorem and showing it has no subgraphs homeomorphic to $K_5$ or $K_{3,3}$ but I am unsure how to do that.



Should I start by showing since it is bipartite all of its subgraphs must be bipartite, hence cannot be homeomorphic to $K_5$ or $K_{3,3}$? Or would that be an incorrect approach? Thanks!






graph-theory







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  • 1




    $begingroup$
    Have you made an attempt to find such a graph?
    $endgroup$
    – Misha Lavrov
    Feb 3 at 0:58










  • $begingroup$
    Yes, $K_{4,4}$.
    $endgroup$
    – W.R.P.S
    Feb 3 at 7:59
















0












$begingroup$


So far I have that the graph's chromatic number is two (not sure how that would help) as well as that since it has an Eulerian circuit, all vertices must have even degree. I am thinking of using Kuratowski's theorem and showing it has no subgraphs homeomorphic to $K_5$ or $K_{3,3}$ but I am unsure how to do that.



Should I start by showing since it is bipartite all of its subgraphs must be bipartite, hence cannot be homeomorphic to $K_5$ or $K_{3,3}$? Or would that be an incorrect approach? Thanks!






graph-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Have you made an attempt to find such a graph?
    $endgroup$
    – Misha Lavrov
    Feb 3 at 0:58










  • $begingroup$
    Yes, $K_{4,4}$.
    $endgroup$
    – W.R.P.S
    Feb 3 at 7:59














0












0








0





$begingroup$


So far I have that the graph's chromatic number is two (not sure how that would help) as well as that since it has an Eulerian circuit, all vertices must have even degree. I am thinking of using Kuratowski's theorem and showing it has no subgraphs homeomorphic to $K_5$ or $K_{3,3}$ but I am unsure how to do that.



Should I start by showing since it is bipartite all of its subgraphs must be bipartite, hence cannot be homeomorphic to $K_5$ or $K_{3,3}$? Or would that be an incorrect approach? Thanks!






graph-theory







share|cite|improve this question











$endgroup$




So far I have that the graph's chromatic number is two (not sure how that would help) as well as that since it has an Eulerian circuit, all vertices must have even degree. I am thinking of using Kuratowski's theorem and showing it has no subgraphs homeomorphic to $K_5$ or $K_{3,3}$ but I am unsure how to do that.



Should I start by showing since it is bipartite all of its subgraphs must be bipartite, hence cannot be homeomorphic to $K_5$ or $K_{3,3}$? Or would that be an incorrect approach? Thanks!






graph-theory




graph-theory






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share|cite|improve this question













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edited Feb 20 at 22:39









Vinyl_cape_jawa

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asked Feb 2 at 23:41









EJJJJEJJJJ

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  • 1




    $begingroup$
    Have you made an attempt to find such a graph?
    $endgroup$
    – Misha Lavrov
    Feb 3 at 0:58










  • $begingroup$
    Yes, $K_{4,4}$.
    $endgroup$
    – W.R.P.S
    Feb 3 at 7:59














  • 1




    $begingroup$
    Have you made an attempt to find such a graph?
    $endgroup$
    – Misha Lavrov
    Feb 3 at 0:58










  • $begingroup$
    Yes, $K_{4,4}$.
    $endgroup$
    – W.R.P.S
    Feb 3 at 7:59








1




1




$begingroup$
Have you made an attempt to find such a graph?
$endgroup$
– Misha Lavrov
Feb 3 at 0:58




$begingroup$
Have you made an attempt to find such a graph?
$endgroup$
– Misha Lavrov
Feb 3 at 0:58












$begingroup$
Yes, $K_{4,4}$.
$endgroup$
– W.R.P.S
Feb 3 at 7:59




$begingroup$
Yes, $K_{4,4}$.
$endgroup$
– W.R.P.S
Feb 3 at 7:59










1 Answer
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$begingroup$

Actually yes and there are quite a few of them. How about, for each integer $n$ at least 2, the complete bipartite graph $K_{2n,2n}$ with $2n$ vertices on each side.



Note that $K_{3,3}$ is a subgraph of $K_{2n,2n}$ for all $n ge 2$ and indeed $K_{3,3}$ is a subgraph of $K_{m,m}$ for all integers odd or even $m ge 3$ .






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    $begingroup$

    Actually yes and there are quite a few of them. How about, for each integer $n$ at least 2, the complete bipartite graph $K_{2n,2n}$ with $2n$ vertices on each side.



    Note that $K_{3,3}$ is a subgraph of $K_{2n,2n}$ for all $n ge 2$ and indeed $K_{3,3}$ is a subgraph of $K_{m,m}$ for all integers odd or even $m ge 3$ .






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Actually yes and there are quite a few of them. How about, for each integer $n$ at least 2, the complete bipartite graph $K_{2n,2n}$ with $2n$ vertices on each side.



      Note that $K_{3,3}$ is a subgraph of $K_{2n,2n}$ for all $n ge 2$ and indeed $K_{3,3}$ is a subgraph of $K_{m,m}$ for all integers odd or even $m ge 3$ .






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Actually yes and there are quite a few of them. How about, for each integer $n$ at least 2, the complete bipartite graph $K_{2n,2n}$ with $2n$ vertices on each side.



        Note that $K_{3,3}$ is a subgraph of $K_{2n,2n}$ for all $n ge 2$ and indeed $K_{3,3}$ is a subgraph of $K_{m,m}$ for all integers odd or even $m ge 3$ .






        share|cite|improve this answer









        $endgroup$



        Actually yes and there are quite a few of them. How about, for each integer $n$ at least 2, the complete bipartite graph $K_{2n,2n}$ with $2n$ vertices on each side.



        Note that $K_{3,3}$ is a subgraph of $K_{2n,2n}$ for all $n ge 2$ and indeed $K_{3,3}$ is a subgraph of $K_{m,m}$ for all integers odd or even $m ge 3$ .







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 3 at 18:21









        MikeMike

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