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COSC 6114

COSC 6114. Prof. Andy Mirzaian. Polygon Triangulation. References :. [M. de Berge et al] chapter 3 [Preparata-Shamos’85] chapter 6 [O’Rourke’98] chapter 1 Applications : Graphics: Ray Shooting Robotics: Geodesic Shortest Paths inside polygon, visibility GIS: Planar Point Location

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COSC 6114

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  1. COSC 6114 Prof. Andy Mirzaian Polygon Triangulation

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

  3. Polygon Triangulation Guarding and Art Gallery

  4. 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.

  5. A simple polygon P

  6. A triangulation of P

  7. Dual Tree of the Triangulation

  8. 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 y y x x w z z max [Shaded triangle does not contain any vertex of the polygon] Diagonalof a simple polygon P: Any line-segment between two non-adjacent vertices of P that is completely inside P. yes no LEMMA 1Any simple n-gon with n>3 admits at least one diagonal. Such a diagonal can be found in O(n) time.

  9. 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 inO(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. d P2 P1 P THEOREM 2Any simple n-gon P admits at least one Triangulation. Such a triangulation T can be computed in O(n2) time. Any such triangulation has n-3 diagonals, and n-2 triangles.

  10. Proof: Necessity: B R=4B=3G=4 G R R G R B G B R G THEOREM 3[Chvătal 1975, Fisk 1978]n/3 guards are always sufficient and sometimes necessary to guard any simple n-gon. • Sufficiency: • T = a triangulation of the n-gon. • 3-colour vertices of T (so that the vertices of each triangle get 3 different colours).This can be done (implicitly) by a DFS traversal on the dual tree of T. • Choose a colour least often used (break ties arbitrarily). • Place a guard at the vertices of the chosen colour. (Each triangle has a guard.)

  11. Geodesic Shortest Paths inside polygon

  12. Geodesic Shortest Paths inside polygon A B A shortest path: Walk along the path of the dual tree from triangle of A to B.

  13. Geodesic Shortest Paths inside polygon source Single-source shortest paths: Do a DFS on dual tree of the triangulation and maintain the visibility funnel for frontier diagonals.

  14. Geodesic Shortest Paths inside polygon source Single-source shortest paths

  15. Geodesic Shortest Paths inside polygon source Single-source shortest paths

  16. Simple Polygon Triangulation Algorithms • O(n2) time See Theorem 2. Also by “ear removal”, Lennes 1911. • O(n log n) Garey-Johnson-Preparata-Tarjan (plane sweep) 1978. • O(n log log n) Tarjan-van Wyk (balanced-cut & Jordan-sort) 1986-88. • O(n log* n) randomized Clarkson-Tarjan-van Wyk 1989. • O(n) Chazelle 1991. [Complicated. Can it be simplified?] • O(n) randomized Amato-Goodrich-Ramos 2000. [See LN15]A possible generic candidate for simpler & efficient polygon triangulation algorithm: • via pseudo-triangulationsMirzaian 1988 [See LN14]

  17. Garey-Johnson-Preparata-Tarjan • FACT: P is y-monotone if and only if it does not have any cusps. • A monotone polygon can easily be triangulated in linear time. • Subdivide the simple polygon into monotone sub-polygons by adding diagonals to cusps. y y y x x x y-monotone Cusp: concave local y-min or y-max vertex. 3 y-monotone sub-polygons.

  18. Garey-Johnson-Preparata-Tarjan How to partition the polygon into monotone sub-polygons by adding suitable diagonals

  19. Garey-Johnson-Preparata-Tarjan How to partition the polygon into monotone sub-polygons by adding suitable diagonals • Trapzoidize using plane sweep in O(n log n) time.

  20. Garey-Johnson-Preparata-Tarjan How to partition the polygon into monotone sub-polygons by adding suitable diagonals • Trapzoidize using plane sweep in O(n log n) time. • Remove visibility chords outside polygon.

  21. Garey-Johnson-Preparata-Tarjan How to partition the polygon into monotone sub-polygons by adding suitable diagonals • Trapzoidize using plane sweep in O(n log n) time. • Remove visibility chords outside polygon. • Add one supporting diagonal (if any) per trapezoid.These diagonals eliminate cusps and subdividepolygon into y-monotone sub-polygons.

  22. Garey-Johnson-Preparata-Tarjan How to partition the polygon into monotone sub-polygons by adding suitable diagonals • Trapzoidize using plane sweep in O(n log n) time. • Remove visibility chords outside polygon. • Add one supporting diagonal (if any) per trapezoid.These diagonals eliminate cusps and subdividepolygon into y-monotone sub-polygons. • Ignore visibility chords.

  23. Garey-Johnson-Preparata-Tarjan How to partition the polygon into monotone sub-polygons by adding suitable diagonals • Trapzoidize using plane sweep in O(n log n) time. • Remove visibility chords outside polygon. • Add one supporting diagonal (if any) per trapezoid.These diagonals eliminate cusps and subdividepolygon into y-monotone sub-polygons. • Ignore visibility chords. • Triangulate each y-monotone sub-polygon intotal O(n) time. [See next slides.]

  24. 17 Triangulating a y-monotone polygon 16 • Merge y-sorted left & right boundary chains of the polygon to obtain the y-sorted vertex list. • Advance along y-sorted vertex-list: • An “uncapped” chord (edge or diagonal) is one that is considered but doesn’t yet have an incident triangle above. These chords form a concave chain. • For each vertex v in y-sorted order, add downward visible chords and triangles from v to uncapped visible diagonals, starting from most recent & backwards. (Use a stack. See next slide.) 15 14 13 12 11 10 9 8 7 6 5 4 3 y 2 1 x

  25. Algorithm Triangulate y-monotone polygon P • merge the vertices of the left and right chains of P into y-sorted order, say, u1, u2, … , un. • push u1 and u2 into an initially empty stack S. • for j 3 .. n-1 do • if uj & top(S) are on different chainsthen pop all vertices from S and add a diagonal between uj and each popped vertex except the last. push uj-1 and uj onto S.else pop(S) pop the other vertices from S while they are visible from uj, and add a diagonal between uj and each popped vertex. push last popped vertex back onto S. push uj onto S. • add diagonals from the last vertex un to all stack vertices except first and last. • end

  26. This approach can also triangulate a polygon with polygonal obstacles inside, in O(n log n) time (applications in robotics). • A straight-line planar subdivision with n vertices can be triangulated in O(n log n) time and O(n) space. • Use the same approach: plane-sweep; trapzoidize; monotonize; and triangulate the resulting monotone polygons.

  27. If you do a DFS of the non-tree dual graph, you get a DFS tree, plus h back-edges. How can you deal with these back-edges that would correspond to opening corridors? THEOREM 4[Art Gallery Theorem for polygons with holes] A polygon P with h polygonal holes and n vertices (including vertices of the holes)can be guarded with (n+2h)/3 guards. Proof: Triangulate P. Dual graph is not a tree.Use h of the diagonals to connect the holes and the outer boundary of P. Double-up these h diagonals to open up narrow corridors.We now have a simple polygon with n+2h vertices. Now apply Theorem 3.

  28. Jordan Sorting Suppose line L and an unknown Jordan curve J intersect at k points X = {x1 , x2 , … , xk }.Given X in sorted order along J, obtain X in sorted order along L.[Hoffman-Mehlhorn-Rosenstiehl-Tarjan 1986] show how to do this in O(k) time using an efficient finger-search tree data structure.(Complexity of the unknown J could be much larger than k.) J x8 x7 x6 x2 x3 x4 x5 x12 x10 x11 L x1 x9

  29. O(n log log n) time polygon Triangulation:Tarjan–van Wyk 1988 v L L

  30. Exercises

  31. Give a simple linear time algorithm to triangulate any given star polygon.Assume a kernel point of the polygon is available. • Guarding the walls: Construct a simple polygon P and a placement of guards such that every boundary point of P is seen by some guard, but there is at least one point interior to P not seen by any guard. • Visibility: Consider an arbitrary simple polygon P. Two points a and b in P are visible from each other if the line-segment ab does not intersect the exterior of P.(a)Prove or disprove: Let v be any point in P such that every vertex of P is visible from v. Then every point in P is also visible from v. (b) Prove or disprove: Let v and w be any pair of points in P such that every vertex of P is visible from v or w. Then every point in P is also visible from v or w. • Suppose that a simple polygon P with n vertices is given, together with a set of diagonals that partitions P into convex quadrilaterals. How many point guards are sufficient to guard the interior of P? Why doesn't this contradict the Art Gallery Theorem 3? • Balanced Split: Let P be any simple polygon with n vertices. A diagonal d of P is said to be a balanced diagonal, if it splits P into two simple polygons, each with at most 2n/3 + 1 vertices.(a) Show that any simple polygon has a balanced diagonal.(b) Show the bound 2n/3 + 1 is tight, i.e., for every n>2, there are simple n-gons that have no diagonal that splits the polygon with both sides containing less than 2n/3 + 1 vertices each. (c) Give an O(n log n) time algorithm to find a balanced diagonal of P. [Hint: Use the dual graph of a triangulation. This has applications in divide-&-conquer algorithms.]

  32. Polygon Area: Prove the following fact about the algebraic area of a simple polygon as stated in the introductory Lecture Slide 1: • Polygon Triangulation by Horn Cutting:We are given a simple n-gon P = [ p0 , p1, ..., pn-1]. We want to find a triangulation T(P) of P by a horn-cutting method. A horn of P is any vertex pi of P, such that (the interior of) P contains (the interior of) triangle (pi-1 , pi , pi+1) formed by pi and its two neighboring vertices pi-1 and pi+1 (index arithmetic is done mod n). We can cut this horn by removing vertex pi from the sequence of polygon vertices, resulting in a sub-polygon P' with n-1 vertices. The line-segment (pi-1 pi+1) now becomes an edge of the new polygon P'. We say we have cut the horn pi (or the horn-triangle (pi-1 , pi , pi+1) from the simple polygon P).(a) Argue that the sub-polygon P' that results from cutting a horn of P is itself a simple polygon (i.e., it has a non-self-crossing boundary).(b) Prove that any simple polygon P with n >3 vertices has at least two horns.(c) Algorithmically show how in O(n) time we can test whether a given vertex pi is a horn of P.(d) Design and analyze an O(n2) time algorithm to compute a triangulation T(P) of simple polygon P by an iterative horn-cutting process.

  33. a a flip D D’ c c D d d b b • Diagonal Flip in Polygon Triangulation: Let P be a simple polygon with n vertices and T a triangulation of P. Consider a diagonal D=(a,b), incident to two triangles (a,b,c) and (a,b,d) in T. The union of these two triangles is the quadrangle (a,c,b,d). If this quadrangle is convex, then we can flip D to obtain the new diagonal D’=(c,d). This creates a new triangulation T’ where the two triangles (a,b,c) and (a,b,d) are replaced by the two new triangles (c,d,a) and (c,d,b). (See the figure below.). Note that flipping is reversible, i.e., you can go from T’ back to T by flipping D’.Let T and T’ be two arbitrary triangulations of P. We say T is flip-connected to T’, if there is a sequence of zero or more diagonal flips that transforms T to T’. Let the flip-distance, denoted FD(T,T’), be the minimum number of flips required to transform T to T’. FD(T,T’) = if T and T’ are not flip-connected. Flip-connectivity is an equivalence relation on the set of triangulations of P. So, it partitions the set into equivalence classes. (Can there be more than one equivalence class?)(a) Show that if P is a convex polygon, then T,T’, FD(T,T’) = O(n).(b) What if P is a simple polygon? Can there be a pair T, T’ such that FD(T,T’) =? Why or why not? If not, then derive a tight asymptotic upper-bound on FD(T,T’) over all triangulation pairs T and T’ of P.

  34. a b • Geodesic in a Simple Polygon: The following problem has a number of applications including robot path planning.We are given a simple n-gon P and a triangulation T of P. The triangulation is given by its list of triangles, and their incidence relationship, where two triangles are incident iff they share a common diagonal. We are also given two points a and b inside P. The problem is to find the shortest path from a to b inside P. Such a path is called the geodesic between a and b inside P (see the figure below). Design and analyze an O(n)-time algorithm to solve this problem. [Hint: after locating the triangles that contain a and b, "walk" along the triangles from a to b (which way?), while maintaining the two shortest paths from a to the two ends of the next diagonal on the walk. These two shortest paths form a funnel that encloses a prefix of the final geodesic. Carefully analyze how you update the funnel as you advance from one diagonal to the next on your "walk".]

  35. Polygon Area Bisection: We are given a simple n-vertex polygon P in the plane. We already know how to compute the area of polygon P in O(n) time. (a) The problem here is to find a vertical line L such that the area of P in the left half- plane defined by L is equal to the area of P in the right half-plane defined by L (Fig. (a)). Describe an O(n log n) time algorithm to solve this problem by the plane sweep method.(b) Develop an O(n) time algorithm to solve the above problem provided that we are already given TZ(P), the trapezoidization of the interior of P (Fig. (b)).(c) Can you solve the problem in O(n) time without the use of TZ(P) and without using any preprocessing algorithm that constructs a triangulation or trapzoidization of P? [Hint: Let A(x) denote the area in P to the left of the vertical line with x-coordinate x. Study the function A(x) and its derivative, and use prune-&-search.] (d) Prove or disprove: Any simple polygon can be cut in two equal area sub- polygons by a single straight line-segment s whose end points are on the boundary of P but s does not intersect the exterior of P (Fig. (c)). L P P P s (a) (b) (c)

  36. END

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