Obstructing Visibilities with One Obstacle

Abstract

Obstacle representations of graphs have been investigated quite intensely over the last few years. We focus on graphs that can be represented by a single obstacle. Given a (topologically open) polygon $C$ and a finite set $P$ of points in general position in the complement of $C$, the visibility graph $G_C(P)$ has a vertex for each point in $P$ and an edge $pq$ for any two points $p$ and $q$ in $P$ that can see each other, that is, $\overline{pq} \cap C=\emptyset$. We draw $G_C(P)$ straight-line. Given a graph $G$, we want to compute an obstacle representation of $G$, that is, an obstacle $C$ and a set of points $P$ such that $G=G_C(P)$. The complexity of this problem is open, even for the case that the points are exactly the vertices of a simple polygon and the obstacle is the complement of the polygon-the simple-polygon visibility graph problem. There are two types of obstacles; an inside obstacle lies in a bounded component of the complement of the visibility drawing, whereas an outside obstacle lies in the unbounded component. We show that the class of graphs with an inside-obstacle representation is incomparable with the class of graphs that have an outside-obstacle representation. We further show that any graph with at most seven vertices or circumference at most 6 has an outside-obstacle representation, which does not hold for a specific graph with 8 vertices and circumference 8. Finally, we consider the outside-obstacle graph sandwich problem: given graphs $G$ and $H$ on the same vertex set, is there a graph $K$ such that $G \subseteq K \subseteq H$ and $K$ has an outside-obstacle representation? We show that this problem is NP-hard even for co-bipartite graphs. With slight modifications, our proof also shows that the inside-obstacle graph sandwich problem, the single-obstacle graph sandwich problem, and the simple-polygon visibility graph sandwich problem are all NP-hard.