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Determine all connected regular planar graphs G such that the number of regions in a planar embedding of G equals its order.

I am not sure how to approach this problem. I know the solution involves Euler's identity (n - m + r = 2), and I know it is only a property of connected graphs with planar embeddings. I also realize that n = r in this situation. Any ideas on how to approach this?

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closed as off-topic by Did, Claude Leibovici, Leucippus, Henrik, choco_addicted Sep 30 '17 at 12:19

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Let $G$ be a $r$-regular planar graph of order $v$ with $e\ge v-1$ edges and $f=v$ faces. Then $r\ge2$ and

$$v-e+f=2 \iff 2v-e=2 \iff 2v-\frac{vr}{2}=2 \iff v(4-r)=4.$$

Since $v$ is a positive integer, we require $4-r$ to be a positive integer as well. Thus the only possibilities are when $r=2,$ or $3.$

When $r=2$, then we have a cycle, which is planar with $2$ regions, but $K_3$ is the smallest cycle and has order $3$, so this is impossible.

The only remaining possibility is when $r=3$ and we observe that $K_4$ is the only $3$-regular graph of order $4$ that also has $4$ regions.

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