3/3/15CMPS 3130/6130 Computational Geometry1 CMPS 3130/6130 Computational Geometry Spring 2015 Delaunay Triangulations I Carola Wenk Based on: Computational.

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3/3/15CMPS 3130/6130 Computational Geometry1 CMPS 3130/6130 Computational Geometry Spring 2015 Delaunay Triangulations I Carola Wenk Based on: Computational Geometry: Algorithms and Applications Computational Geometry: Algorithms and Applications

3/3/15CMPS 3130/6130 Computational Geometry 2 Triangulation

3/3/15CMPS 3130/6130 Computational Geometry 3 Dual Graph

3/3/15CMPS 3130/6130 Computational Geometry 4 Delaunay Triangulation VD(P) P DT(P) Canonical straight-line embedding for DT(P): If P is in general position (no three points on a line, no four points on a circle) then every inner face of DT(P) is indeed a triangle. DT(P) can be stored as an abstract graph, without geometric information. (No such obvious storing scheme for VD(P).)

3/3/15CMPS 3130/6130 Computational Geometry 5 Straight-Line Embedding Lemma: DT(P) is a plane graph, i.e., the straight-line edges do not intersect. Proof: p p’ c pp’ is an edge of DT(P)  There is an empty closed disk D p with p and p’ on its boundary, and its center c on the bisector. Let qq’ be another Delaunay edge that intersects pp’  q and q’ lie outside of D p, therefore qq’ also intersects pc or p’c Similarly, pp’ also intersects qc’ or q’c’  (pc or p’c’) and (qc’ or q’c’) intersect  The edges are not in different Voronoi cells  Contradiction DpDp q q’ q c’ DpDp p p’

3/3/15CMPS 3130/6130 Computational Geometry 6 Characterization I of DT(P) Lemma: Let p,q,r  P let  be the triangle they define. Then the following statements are equivalent: a)  belongs to DT(P) b) The circumcenter of  is a vertex in VD(P) c) The circumcircle of  is empty (i.e., contains no other point of P) Characterization I: Let T be a triangulation of P. Then T=DT(P)  The circumcircle of any triangle in T is empty. pipi plpl pjpj pkpk non-empty circumcircle

3/3/15CMPS 3130/6130 Computational Geometry 7 Illegal Edges Definition: Let p i, p j, p k, p l  P. Then p i p j is an illegal edge  p l lies in the interior of the circle through p i, p j, p k. Lemma: Let p i, p j, p k, p l  P. Then p i p j is illegal  min  i < min  ’ i pipi plpl pjpj pkpk illegal edge 1≤i≤6 plpl pipi pjpj pkpk edge flip plpl pipi pjpj pkpk 11 22 33 44 55 66 ’1’1 ’2’2 ’3’3 ’4’4 ’5’5 ’6’6 Theorem (Thales): Let a, b, p, q be four points on a circle, and let r be inside and let s be outside of the circle, such that p,q,r,s lie on the same side of the line through a, b. Then  a,s,b <  a,q,b =  a,p,b <  a,r,b pq r s a b

3/3/15CMPS 3130/6130 Computational Geometry 8 Characterization II of DT(P) Definition: A triangulation is called legal if it does not contain any illegal edges. Characterization II: Let T be a triangulation of P. Then T=DT(P)  T is legal. Algorithm Legal_Triangulation(T): Input: A triangulation T of a point set P Output: A legal triangulation of P while T contains an illegal edge p i p j do //Flip p i p j Let p i, p j, p k, p l be the quadrilateral containing p i p j Remove p i p j and add p k p l return T Runtime analysis: –In every iteration of the loop the angle vector of T (all angles in T sorted by increasing value) increases –With this one can show that a flipped edge never appears again –There are O(n 2 ) edges, therefore the runtime is O(n 2 ) pipi pjpj pkpk edge flip plpl pipi pjpj pkpk plpl

3/3/15CMPS 3130/6130 Computational Geometry 9 Characterization III of DT(P) Definition: Let T be a triangulation of P and let  1,  2,…,  3m be the angles of the m triangles in T sorted by increasing value. Then A(T)=(  1,  2,…,  3m ) is called the angle vector of T. Definition: A triangulation T is called angle optimal  A(T) > A(T’) for any other triangulation of the same point set P. Let T’ be a triangulation that contains an illegal edge, and let T’’ be the resulting triangulation after flipping this edge. Then A(T’’) > A(T’). T is angle optimal  T is legal  T=DT(P) Characterization III: Let T be a triangulation of P. Then T=DT(P)  T is angle optimal. (If P is not in general position, then any triangulation obtained by triangulating the faces maximizes the minimum angle.)