Triangulate

# Triangulate - CSE 4101/5101 Prof. Andy Mirzaian Polygon...

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CSE 4101/5101 Polygon Triangulation Prof. Andy Mirzaian

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References: [M. de Berge et al ’00] chapter 3 [Preparata-Shamos’85] chapter 6 [O’Rourke’98] chapter 1 Applications : q Graphics: Ray Shooting q Robotics: Geodesic Shortest Paths inside polygon, visibility q GIS: Planar Point Location q GIS: Elevation estimate on polyhedral terrain: q GIS: Piece-wise linear interpolation of bi-variate function f(x,y). q . . . 2
Polygon Triangulation Guarding and Art Gallery 3

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Art Gallery Problem [Victor Klee 1973] How many camera guards do we need to guard a given gallery and how do we decide where to place them? It’s NP-hard to determine the MINIMUM number of camera guards for an arbitrary given simple polygon [Aggarwal 1984]. Let P be an n-vertex simple polygon. If P is convex, then a single guard anywhere inside P is sufficient. n guards for P are always sufficient; one guard at each vertex . [This does not work for 3D polytopes!] Can we use less than n guards? Yes. Use Triangulation of P. 4
A simple polygon P 5

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A triangulation of P 6
Dual Tree of the Triangulation 7

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Diagonal of a simple polygon P: Any line-segment between two non-adjacent vertices of P that is completely inside P. yes no Proof : Let x be any convex vertex of the polygon (e.g., an extreme vertex, say, the lowest-leftmost). (case a) yz is a diagonal (case b) xw is a diagonal x y z x y z w m ax [Shaded triangle does not contain any vertex of the polygon] LEMMA 1 Any simple n-gon with n>3 admits at least one diagonal. Such a diagonal can be found in O(n) time. 8
Proof : By induction on n. Basis (n=3): Obvious. Ind. Step (n>3): By previous Lemma, a diagonal d of P exists and can be found in O(n) time, and divides P into simple polygons P1 & P2 with, say, n1 & n2 vertices, where d is an edge of both. Note, n = n1+n2 -2. Triangulations T1 & T2 of P1 & P2 can be obtained recursively. Now set T = T1T2 with d as an extra diagonal. Total computation time: Time(n) = Time(n1) + Time(n2) + O(n) = O( n2). By induction hypothesis: T1 has n1–3 diagonals and n1–2 triangles, T2 has n2–3 diagonals and n2–2 triangles, These imply: T has n–3 diagonals and n–2 triangles. P1

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## This note was uploaded on 02/13/2012 for the course CSE 4101 taught by Professor Mirzaian during the Winter '12 term at York University.

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Triangulate - CSE 4101/5101 Prof. Andy Mirzaian Polygon...

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