On the number of internal and external visibility edges of polygons

In this paper we prove that for any simple polygon P with n vertices, the sum of the number of strictly internal edges and the number of strictly external visibility edges of P is at least b 3n−1 2 c − 4. The internal visibility graph of a simple polygon P is the graph with vertex set equal to the vertex set of P , in which two vertices are adjacent if the line segment connecting them does not intersect the exterior of P . The external visibility graph of P is defined in a similar way, except that the line segments that generate its edges are not allowed to intersect the interior of P . A visibility edge is called strictly internal (resp. strictly external) if it is not an edge of P . In this paper we prove the following conjecture of Bagga [1]: For any simple polygon P with n vertices, the number of strictly internal visibility edges plus the number of strictly external visibility edges is at least b 3n−1 2 c − 4. In Figure 1 we present a family of polygons that achieve this bound. They have exactly n− 3 strictly internal visibility edges, and n−3 2 strictly external visibility edges. ∗ Supported by NSERC of Canada Figure 1: A sequence of polygons for which the number of strictly internal plus strictly external visibility edges is exactly b 3n−1 2 c − 4. Let int(P ) and ext(P ) denote the number of strictly internal and external visibility edges of P . Some observations will be used to prove that for any polygon P with n vertices, int(P ) + ext(P ) ≥ b 3n−1 2 c − 4. A vertex v of P will be called internal if it is in the interior of the convex hull Conv(P ) of P . An external vertex is a vertex of the convex hull of P . The following result is easy to prove: Lemma 1 Let P be a simple polygon with n vertices, k of which are internal. Then ext(P ) is at least k.