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

Thanks

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

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  • $\begingroup$ 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
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    $\begingroup$ @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

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