Polygon Triangulation

1. Polygon Triangulation • Motivation: Guarding art galleries • Art gallery theorem for simple polygons • Partitioning of polygons into monotone pieces • Triangulation of y-monotone polygons

2. Guarding art galleries “Art Gallery“ Problem

3. Guarding art galleries • The gallery is represented by a simple polygon • A guard is represented by a point within the polygon • Guards have a viewport of 360° • A polygon is completely guarded, if every point within the polygon is guarded by at least one of the watchmen

4. Guarding art galleries Visibility polygon: The visibility polygon vis(p) of a polygon P is defined by the set of all points that are visible from the base point p. Demo

5. Guarding art galleries Even if two polygons have the same number of vertices, one may be easier to guard than the other. We will consider the lower bound for the worst case: We want to determine the number of guards that suffice for an arbitrary polygon with n vertices.

6. Guarding art galleries If the polygon is complex, it is not obvious to see how many gurads are needed. Idea: Divide the polygon into pieces that are easy to guard

7. Guarding a triangulated polygon

8. w v u Triangulation of simple polygons Does every simple polygon admit a triangulation? If yes, what is the number of triangles? Does any triangulation lead to the same number of triangles?

9. Theorem Theorem: Every simple polygon admits a triangulation, and anytriangulation of a simple polygon with n vertices consistsof exactly n-2 triangles. Proof: By induction on n. Let n>3, and assume theorem is truefor all m<n. Let P be polygon with n vertices. We first prove the existence of a diagonal in P. Let v be leftmostvertex of P. Let u and v be two neighboring vertices of v.If uw lies in the interior of P we have found a diagonal. Else, there are one or more vertices inside the triangle defined by u, v, and w, or the diagonal uw. Let v´ be the farthest vertex from uw. The segment connecting v´ to v cannot intersect an edge of p (contradicts the definition of v´). Hence vv´ is a diagonal.

10. Case 2: uw not completely in P w w v u v u Proof of the existence of diagonals in P Consider leftmost vertex v of P Case 1: uw completely in P

11. Continuation of proof So a diagonal exists. Any diagonal cuts P in two simple sub-polygons P1 and P2. Let m1 be the number of vertices of P1and m2 the number of vertices of P2. Both m1 and m2 mustbe smaller than n, so by induction P1 and P2 can be triangulatedso P can be triangulated as well. Now we have to prove any triangulation of P contains n-2 triangles. Consider an arbitrary diagonal in some triangulation Tp.This diagonal cuts P into 2 subpolygons with m1 and m2 vertices.Every vertex of P occurs in exactly one of 2 subpolygons. Hencem1+m2 = n+2. So by induction any triangulation of Pi containsmi-2 triangles  (m1-2) + (m2-2) = n-2 triangles.

12. Number of triangles in any triangulation of a simple polygon with n vertices. Case 1: n=3 Case 2: n>3

13. Case 2: uw not completely in P w w v u v u Proof of the existence of diagonals in P Consider leftmost vertex v of P Case 1: uw completely in P

14. Upper and lower bounds for the number of guards We know that for any simple polygon with n vertices (n-2)guards are always enough. But can we do better?

15. Upper and lower bounds for the number of guards Idea: Compute a 3-coloring of the vertices and place guards on a color. Result: n/3guards are sufficient.

16. Theorem Theorem: Each simple polygon is 3-colorable. Proof: Dual graph is a binary tree, this means that we can find a 3-coloring using a simple DFS. Corollary: n/3guards are always sufficient to guard a simple Polygon with n vertices.

17. Example

18. Art gallery theorem Theorem: For a simple polygon with n vertices, n/3 cameras are occasionally necessary and always sufficient to have every point in the polygon visible from at least one of the cameras. Worst-case example?

19. Art gallery theorem Worst-case example: Demo