# Constructing a 1-planar graph that has no rectilinear drawing

A 1-planar graph is a graph that can be drawn in the Euclidean plane in such a way that each edge has at most one crossing point, where it crosses a single additional edge. 1 planar graph

I read the following paper, which stated the following fact about the rectilinear drawing of a 1-plane graph.

Thomassen, C. (1988), Rectilinear drawings of graphs. J. Graph Theory, 12: 335-341.https://doi.org/10.1002/jgt.3190120306

Theorem 3.1. Let $$G$$ be a graph represented as a normal drawing of index $$0$$ or $$1$$. Then there exists a rectilinear drawing equivalent to $$G$$ if and only if $$G$$ contains no $$B-$$ or $$W-$$configuration.

From this theorem we know that a 1-planar graph satisfies any 1-plane drawing of it without $$B-$$ and $$W-$$configuration, then it has no rectilinear drawing.

But I haven't constructed a 1-planar graph without rectilinear drawing so far. It is even doubtful that it exists. This is one of my confusing points. Any advice would be helpful and thanks in advance!

• Every finite graph can be drawn in the Euclidean plane such that any two edges have at most one point in common (in particular, two edges with a common end have no other point in common), and such that every point that is not a vertex and that is on two edges is on precisely two edges and is a cross-point for these two edges. We call such a drawing normal.

• The maximum number of crosspoints on an edge is called the cross-index of the drawing.

The main result of the following paper states that a graph which has a 1-planar straight-line drawing has at most $$4n-9$$ edges, where $$n$$ is (as usual) the number of vertices:
On the other hand, it is not hard to come up with a 1-planar graph with $$4n-8$$ edges: take any drawing of a planar graph in which every face is a quadrilateral and add a pair of crossing edges to each face. For instance, applying this to the cube gives $$K_8$$ with a matching removed.