I was doing a graph theory text book where one of the problems asks:
Is this graph planar?
As this graph contains a triangle, the best bound for $e$ is $3v-6$ which this satisfies $(14<18)$
So then I tried to look for a subgraph that could be smoothed into $K_5$ but quickly realised it wouldn't be possible because in this graph, Maximum degree is only $4$.
So there can only be 2 ways to end this problem now, either draw a planar embedding or find a subgraph that contains $K_{3,3}$.
For the first one, the thing to notice is that the inner square is basically a $K_4$ and the only way I know how to draw it as a plane graph is kind of like a triangle and its centroid system, in that case it is quiet clear that a planar representation won't be possible (because for instance the edge that joins centroid and outer square vertex must cross triangle).
I also tried for long to find a subdivision on $K_{3,3}$ that is isomorphic to a subgraph of the graph but I could not do so...
I would really appreciate any help...
Thank you!
Reverse the diagram - put the "outside" square into the middle and put the "inside" square on the outside. Then treat the central square as a single vertex.
All of the four outside vertices are connected to each other, and all of them are connected to this central "vertex".
That's $K_5$ and so the graph is not planar.