This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: AMS 345/CSE 355 (Fall, 2010) Joe Mitchell COMPUTATIONAL GEOMETRY Homework Set # 6 Solution Notes (1). [30 points] Let S be a set of n points in the plane in general position (no three are collinear, no four are cocircular). Let h denote the number of points of S that are vertices of the convex hull, CH ( S ) . (a). Consider the Delaunay diagram, D ( S ) , of the set S ; since no four points are cocircular, we know that the Delaunay diagram is a triangulation, with each face (except the face at infinity) being a triangle (and each point of S being a vertex of some triangle). As a function of n and h how many triangles does D ( S ) have? How many Delaunay edges are there in D ( S ) ? Actually, the fact that we are dealing with a Delaunay triangulation is irrelevant for this question: the combi- natorics depend solely on the input, T , being a triangulation. Let t denote the number of triangles in T . There is one face at infinity, so, when we view T as a planar graph, we know it has t + 1 faces, t of which are 3-sided and one of which is h-sided. Thus, the sum of the degrees of the faces is 3 t + h , which we know is twice the number of edges; thus, 2 e = 3 t + h . We also know from Eulers formula that ( t + 1) e + n = 2. Putting e = (3 t + h ) / 2 into this, we get ( t + 1) (3 t + h ) / 2 + n = 2. We solve for t to get t = 2 n h 2 for the number of triangles. Then, e = (3 t + h ) / 2 = 3 n h 3 is the number of edges. (b). Now we are interested in decomposing the convex hull of S into pentagons (5-sided polygons, not necessarily convex), such that each point of S is a vertex of some pentagon. Such a decomposition is called a pentagonalization of S . (i). Give an example of a set S with | S | 5 such that S does not have a pentagonalization. Justify briefly your claim. See below, left. The example has h = 4 points on the hull, with 1 point interior to the hull. The one interior point must be joined to at least two hull vertices in any decomposition of the hull into polygons such that each point of S is a vertex of some polygon. If the interior point is joined to two hull vertices, the hull is split into either ais a vertex of some polygon....
View Full Document
This note was uploaded on 03/13/2011 for the course AMS 345 taught by Professor Mitchell,j during the Spring '08 term at SUNY Stony Brook.
- Spring '08