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How to check if the given graph is planar or not?

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I am not able to find a sub-graph homeomorphic to $K_5$ or $K_{3,3}$.

Can someone please help me to find one? I am completely stuck

Charlotte
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  • If you have found the subgraph containing $K_{3,3}$ , you can merge edges (for example $0-6$ and $6-5$ to $0-5$) to get the desired $K_{3,3}$ to which the subgraph is homeomorphic. – Peter Sep 25 '19 at 12:44
  • For one direction (see the answer below) we do not need the homeomorphic part. If we find a subgraph containing $K_5$ or $K_{3,3}$ we have already shown that the graph is not planar. So, if not explicitely demanded in the exercise, you can omit the homeomorphic part. – Peter Sep 25 '19 at 12:51

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Consider the vertices labelled 0,3,6 in one partition and 1,5,4. The subgraph induced by these 6 vertices contain a $K_{3,3}$. Even the subgraph is not a $K_{3,3}$(subgraph has $K_{3,3}$ as subgraph) , This is enough for not planar.

sabeelmsk
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