Proof that the maximum number of edges of a 1-planar graph is 4n - 8

Question

The Wikipedia article about 1-planar graphs says that the maximum number of edges in such graphs is $4n - 8$

Every 1-planar graph with n vertices has at most 4n − 8 edges.[4] More strongly, each 1-planar drawing has at most n − 2 crossings; removing one edge from each crossing pair of edges leaves a planar graph, which can have at most 3n − 6 edges, from which the 4n − 8 bound on the number of edges in the original 1-planar graph immediately follows

This statement used the following 2 facts:

1. planar graph has at most $3n - 6$ edges
2. $1$-planar graph has at most $n - 2$ crossings

How do we prove the second fact?

If there are other proofs for the $4n - 8$ upper bound, I would also be interested in knowing about them.

Thank you.

Answer 1

The missing ingredient is the formula for Euler characteristic or Eulers polyhedron formula. If you have a planar embedding of a planar graph, then we have the formula

$$ V - E + F = 2 $$

where $V$ is the number of vertices, $E$ the edges and $F$ the number of faces (including one face for the outside, or equivalently counting faces when drawing on a sphere).

If a graph is maximally planar, then every face is a triangle. If any face has more than three sides one could add a diagonal of that face and the graph would still be planar.

Now if you have a maximally planar graph, we have $E=3V-6$ and by the Euler formula $F=2n-4$.

Now if you want to create a maximal 1-planar graph from the planar graph you can add at most one edge per pair of faces. pick two triangular faces that share an edge and connect the two remaining faces across that edge. This gives a maximal number of crossings of $F/2=n-2$ and a total number of edges for the 1-planar graph of $4n-8$.