I'm trying to find a graph that is non-planar, non-hamiltonian and eulerian but I can't find anyone. Is this possible?



Take two copies of $K_5$ and identify one of the nodes in one with one of the nodes in the other.

  • $\begingroup$ But K5 is 4-regular. Eulerian graphs has even degrees. If I join two vertex, this will have odd degree. $\endgroup$ – emee Dec 10 '18 at 0:55
  • 1
    $\begingroup$ @emee: i'm not saying to connect the two vertices with an edge. I'm saying to combine them into one vertex. The combined vertex will have degree 8. $\endgroup$ – Henning Makholm Dec 10 '18 at 0:57
  • $\begingroup$ Sorry but I don't undestand. d(v) + d(u) = 2.8 > 5 [Ore's theorem] $\endgroup$ – emee Dec 12 '18 at 22:24
  • $\begingroup$ @emee: What is $v$ and $u$? The condition in Ore's theorem talks about pairs of distinct vertices. The graph I describe has eight vertices of degree 4 and one vertex of degree 8, so how do you get $2\cdot 8$ for even one pair of vertices? $\endgroup$ – Henning Makholm Dec 12 '18 at 22:44
  • $\begingroup$ I get it. Thanks so much. I was drawing another graph $\endgroup$ – emee Dec 12 '18 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.