Planar graph edge bound ($m\le 3n-6$)

A finite simple graph that is a connected graph and a planar graph, with $n\ge 3$ vertices and $m$ edges, satisfies $m\le 3n-6$. Choose a plane graph: each facial boundary walk has length at least three, so $3f\le 2m$. Substitute $f=2-n+m$ from the Euler formula for a connected planar graph.^[ref]

The complete graph $K_4$ attains equality: $6=3\cdot 4-6$. The bound also holds for a disconnected finite simple graph that is a planar graph with $n\ge 3$: place its connected components of a graph in disjoint regions and join them with new bridges in a graph, then apply the connected bound.

If a finite simple graph is a connected graph, a bipartite graph, and a planar graph, with $n\ge 3$ vertices, then $m\le 2n-4$. Every facial boundary walk has even length and length at least three, hence at least four. Thus $4f\le 2m$; substitute the Euler formula for a connected planar graph.^[ref]

The square cycle graph attains equality: $4=2\cdot 4-4$. The three-vertex graph path also attains equality: $2=2\cdot 3-4$. Its repeated bridges in a graph illustrate why counting facial boundary walk length correctly matters. Neither bound applies unchanged to a multigraph: two vertices may support arbitrarily many parallel edges in a plane graph.

The complete graph $K_5$ has $n=5$ and $m=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$, exceeding the planar graph edge bound $3\cdot 5-6=9$. Therefore it is not a planar graph. The complete bipartite graph $K_{3,3}$ has $n=6$ and $m=3\cdot 3=9$. The general bound allows twelve edges, but the bipartite planar graph edge bound allows only eight, proving nonplanarity.

These are necessary bounds, not a complete planarity test. For an original counterexample, subdivide every edge of $K_{3,3}$ once. The resulting simple graph is a connected graph with $n=15$, $m=18$, and is bipartite: put all original vertices in one part and all nine new vertices in the other. Both inequalities hold, $18\le 39$ and $18\le 26$. Nevertheless, a crossing-free graph drawing would let us trace each two-edge graph path as one curve, producing an impossible embedding of $K_{3,3}$. The graph subdivision preserves this obstruction. Kuratowski's theorem supplies a stronger characterization than these counts.