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) non-self-intersecting 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, (pq) over bar boolean AND C = phi. We draw G(C)(P) straight-line and call this a visibility drawing. 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 when 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; outside obstacles lie in the unbounded component of the visibility drawing, whereas inside obstacles lie in the complement of 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 has an outside-obstacle representation, which does not hold for a specific graph with eight vertices. Finally, we show NP-hardness of the outside-obstacle graph sandwich problem: given graphs G and H on the same vertex set, is there a graph K such that G subset of K subset of H and K has an outside-obstacle representation. Our proof also shows that the simple-polygon visibility graph sandwich problem, the inside-obstacle graph sandwich problem, and the single-obstacle graph sandwich problem are all NP-hard.