Obstructing Visibilities with One Obstacle

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 GC(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, pq ∩ C = ∅. We draw GC(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 = GC(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 outsideobstacle graph sandwich problem: given graphs G and H on the same vertex set, is there a graph K such that G ⊆ K ⊆ 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.