Minimal 4-regular planer graph

Question

I'm asked to draw a minimal 4-regular planer graph and to give number of its verteces and edges. I tried Octahedral graph (see picture) with 6 verteces and 12 edges, but it doesn't work. I'm not sure there is even smaller 4-regular planer graph.

Octahedral graph

Answer 1

For a 4-regular graph with $v$ vertices and $e$ edges, the handshaking lemma tells us that $v=2e$. In particular $v$ is even.

We have the trivial solution of the empty graph ($v=e=0$), which is vacuously planar and 4-regular. If that is considered cheating, let's exclude it and assume $v\ge 1$.

We also could have $v=2$ and $v=4$, but only if we allow multiple edges between vertices, Again, this is cheating if what we actually want is a simple non-empty 4-regular planar graph.

So let $A$ be one of the vertices of our graph (which is possible as $v\ge 1$). Then it has four pairwise distinct (it's a simple graph!) neighbours $B,C,D,E$, which are also all distinct from $A$ (no loops in simple graphs!). So that already gives us $v\ge 5$. As $v$ must be even, we must have $v\ge 6$ (and accordingly $e\ge 12$).

As we can actually exhibit a simple 4-regular plane graph with six vertices and twelve edges (the octahedral graph), that is indeed a minimal one.