Obstacle Numbers of Planar Graphs

Given finitely many connected polygonal obstacles $O_1,\dots,O_k$ in the plane and a set $P$ of points in general position and not in any obstacle, the {\em visibility graph} of $P$ with obstacles $O_1,\dots,O_k$ is the (geometric) graph with vertex set $P$, where two vertices are adjacent if the straight line segment joining them intersects no obstacle. The obstacle number of a graph $G$ is the smallest integer $k$ such that $G$ is the visibility graph of a set of points with $k$ obstacles. If $G$ is planar, we define the planar obstacle number of $G$ by further requiring that the visibility graph has no crossing edges (hence that it is a planar geometric drawing of $G$). In this paper, we prove that the maximum planar obstacle number of a planar graph of order $n$ is $n-3$, the maximum being attained (in particular) by maximal bipartite planar graphs. This displays a significant difference with the standard obstacle number, as we prove that the obstacle number of every bipartite planar graph (and more generally in the class PURE-2-DIR of intersection graphs of straight line segments in two directions) of order at least $3$ is $1$.Comment: Appears in the Proceedings of the 25th International Symposium on Graph Drawing and Network Visualization (GD 2017

Combinatorial Properties and Recognition of Unit Square Visibility Graphs

Unit square visibility graphs (USV) are described by axis-parallel visibility between unit squares placed in the plane. If the squares are required to be placed on integer grid coordinates, then USV become unit square grid visibility graphs (USGV), an alternative characterisation of the well-known rectilinear graphs. We extend known combinatorial results for USGV and we show that, in the weak case (i.e., visibilities do not necessarily translate into edges of the represented combinatorial graph), the area minimisation variant of their recognition problem is NP-hard. We also provide combinatorial insights with respect to USV, and as our main result, we prove their recognition problem to be NP-hard, which settles an open question.Peer Reviewe