$G=(V,E). |V| = 25; |E| = 50.$ For every vertex $v \in V$, the degree of that vertex $d(v)=4.$ I am given that the shortest cycles in $G$ are 4-cycles (i.e. with 4 vertices).

For a contradiction I assumed that G is planar. Hence G would have a number of regions/faces $|R|$ with boundary lengths $b(r)$. From this I got:

\begin{equation*} \sum_{r \in R} b(r) = \sum_{v \in V}d(v) = 2|E| = 100, \end{equation*}

and by Euler's theorem: \begin{equation*} |V| - |E| + |R| = |R| - 25 \geq 2 \implies |R| \geq 27. \end{equation*}

So a lower bound on the first summation would be:

\begin{equation*} \sum_{r \in R} b(r) \geq 27 b(r)_{min}. \end{equation*}

where $b(r)_{min}$ is just the length of the smallest boundary for a region. My guess here is to say that $b(r)_{min}$ = 4, given the information I made bold at the start. This would give me the $100 \geq 108$ contradiction I'm after, but this is just a guess and I don't know why/if this is true. I know that the boundaries in a planar drawing are cycles, and maybe these cycles don't change/get shorter when going from planar to non-planar, but again I'm guessing here. Could someone help me finish this proof and tell me if my intuition is right/wrong? Thanks!


Suppose it exist. So each face has at least 4 edges. By double counting between edges and faces we have$$2 |E| \geq 4|F|\implies |E|\geq 2|F|$$ By Euler theorem we have $$|V|-|E|+|F|= 2$$

so $$ |E| \geq 2(2+|E|-|V|)$$

and thus $$2|V|-|E|\geq 4 \implies 50-50\geq 4$$ A contradiction.


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