AMS 345/CSE 355 Computational Geometry

1. AMS 345/CSE 355 Computational Geometry Lecture: Polygons, Guarding Joe Mitchell

2. Do You Like Puzzles? Come to the new Stony Brook Puzzle Society Meets: Friday 1:05 - 2:30 pm at CSE 2120 Organizer:  PramodGanapathi pganapathi@cs.stonybrook.edu

3. Simple Polygons Definition: A simple polygon P is the (closed) region bounded by a “simple closed polygonal curve”.

4. Simple Polygon Definition in [O’Rourke]:

5. Simple Polygons Alternate Definition: P is a simple polygon if it is a simply connected (i.e., no “holes”) subset of the plane whose boundary consists of a connected finite union of straight line segments.

6. Simple Polygons Some definitions would allow this as a “degenerate” simple polygon

7. Definitions: Visibility, Diagonals p q For p,q in P, p is visible to q if segment pq lies within (closed) P

8. Definitions: Visibility, Diagonals p clearly sees q but does not clearly see q’ p sees q’ p q q’ For p,q in P, p is visible to q if segment pq lies within (closed) P p is clearly visible to q if p is visible to q AND the only points in common between pq and P are possibly p and q

9. Definitions: Visibility, Diagonals q vm vj p vi pq is a chord (not a diagonal) vivj is a diagonal vkvm is not a diagonal vk vivj is a diagonal if vi and vj are vertices that clearly see each other (versus: chordpq, with p and q on the boundary of P)

10. Diagonals

11. Triangulation Definition: A partition of P into triangles by a maximal set of noncrossing diagonals.

12. Triangulation: Existence Proof: Induction on n.

13. 3D: Polyhedra

14. Polyhedra: Tetrahedralization

15. Polyhedron with No Tetrahedralization

16. Combinatorics: Triangulations Proof: induction on n. True for n=3 (trivially). Assume true for n=k (the Induction Hypothesis). Take an n-gon, P, with n=k+1. We know it has a diagonal (since n>3). The diagonal partitions P into two polygons, of sizes n1 and n2 , with n1 +n2 =n+2 (since the endpoints of the diagonal are counted twice) By IH, the two subpolygons have triangulation with n1-2 and n2-2 triangles; glue together along the diagonal to get a total of (n1-2)+(n2-2)=n-2 triangles in the overall triangulation. Similar for number of diagonals.

17. Ears vi+1 vi-1 A diagonal of the form vi-1vi+1 is an ear diagonal; the triangle vi-1vivi+1 is an ear, and vi is the ear tip Note that there are at most n ears (and that a convex polygon has exactly n ears) vi

18. Ears Proof(1): There are n edges of P and n-2 triangles in any triangulation. Imagine dropping the n edges into the n-2 “pigeonholes” corresponding to the triangles: Each edge appears on boundary of some triangle. By pigeonhole principle, at least 2 triangles get 2 edges “dropped in their box”. (2) Consider the planar dual (excluding the face at infinity) of a triangulation of P. Claim: The dual graph for a triangulated simple polygon is a TREE. Any tree of 2 or more nodes has at least 2 nodes of degree 1.

19. Ear-Clipping Triangulation Ear-clipping applet Input: Simple polygon P vi+1 vi-1 vi pq is a diagonal, cutting off a single triangle (the “ear”) Naive: O(n3) Smarter: Keep track of “ear tip status” of each vi (initialize: O(n2) ) Each ear clip requires O(1) ear tip tests ( @ O(n) per test ) Thus, O(n2) total, worst-case

20. Number of Triangulations The diagonals, together with the edges of convex polygon P, form the complete graph, Kn

21. Number of Triangulations: Convex Polygons The first Catalan numbers, for n=1,2,3,… are given by: 1, 2, 5, 14, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452,…

22. Number of Triangulations: Convex Polygons

23. Number of Triangulations: Convex Polygons

24. Number of Triangulations: Simple Polygons Question: Is it possible to find a simple polygon P having exactly k triangulations, for every positive integer k?

25. Exercise Can you find a simple polygon that has exactly 3 different triangulations?

26. Number of Triangulations: Simple Polygons

27. Number of Triangulations: Simple Polygons Diagonals Set of all diagonals, together with the edges of P, forms the visibility graph of P Forced diagonals: Must be in ANY triangulation of P: Any diagonal that is not crossed by other diagonals. Example:

28. Number of Triangulations: Simple Polygons 5 1 2 Total number of different triangulations: 1 * 2 * C5-2 = 1 * 2 * 5 = 10 Example:

29. More Examples: HW1

30. Counting Triangulations: Example

31. Counting Triangulations: Example Step 1: Draw all diagonals Forced diagonals: Those not crossed by any other diagonals

32. Counting Triangulations: Example Step 2: Count number of triangulations in each separate subpolygon C4 = 14 4 By case analysis 3 Overall: 4*3*14

33. Counting Triangulations: Example How many triangulations of subpolygon? 5 Case analysis: (1) Use 03: one completion (since, once we use 03, diagonals 24, 14, and 14 are ruled out, leaving us with just 04 and 02; i.e., diagonal 03 splits polygon into two nonconvex quadrilaterals, each with a unique triangulation) 0 4 3 2 1

34. Counting Triangulations: Example How many triangulations? 5 • Case analysis: • Use 03: one completion • (2) Do NOT Use 03: Then, 24 is forced, and there are 3 triangulations possible (using 15 and 14, or 04 and 14, or 04 and 02). • Total for cases (1) and (2): 1+3=4 0 4 3 2 1

35. Counting Triangulations Efficient algorithm? Yes! (for simple polygons P, but not for polygons with holes) Idea: use “dynamic programming” (recursive solution)

36. Triangulation Simple polygon Planar Straight-Line Graph (PSLG) Polygon with holes Input: Set S of n points Input: Other shapes 3D: Surfaces and solids (tetrahedralization) Triangulation applet for simple polygons

37. Triangulation Theory in 2D Also with holes But, NOT true in 3D! Proofs: Induction on n. • Thm: A simple polygon has a triangulation. • Lem: An n-gon with n4 has a diagonal. • Thm: Any triangulation of a simple n-gon has n-3 diagonals, n-2 triangles. • Thm: The “dual” graph is a tree. • Thm: An n-gon with n4 has 2 “ears”. • Thm: The triangulation graph can be 3-colored.

38. Triangulating a Polygon fan Not practical! • Simple “ear-clipping” methods: O(n2 ) • Cases with simple O(n) algorithms: • Convex polygons (trivial!) • Monotone polygons, monotone mountains • General case (even with holes!): • Sweep algorithm to decompose into monotone mountains • O(n log n) • Best theoretical results: • Simple polygons: O(n) [Chazelle’90] • Polygons with h holes: O(n+h log1+ h), (n+h log h) [BC] • Good practical method: FIST [Held], based on clever methods of ear clipping (worst-case O(n2 ) )

39. TetrahedralizingPolyhedra

40. TetrahedralizingPolyhedra

41. Today, 9/5/13 Review triangulation Guarding problem Art Gallery Theorem Computing guard numbers by hand Examples Begin Convex Hulls

42. Guarding Polygons V(p) = visibility polygon of p inside P = set of all points q that p sees in P

43. Guarding Polygons Goal: Find a set of points (“guards”) within P so that their VP(p) sets cover P “Guard cover” “Point guards” versus “vertex guards” Regular visibility versus “clear visibility”

44. Min-Guard Coverage Problem Determine a small set of guards to see all of a given polygon P 5 guards suffice to cover P (what about 4 guards? 3?) Computing min # of guards, g(P), for n-gon P is NP-hard Challenge/open: Compute g(P) approximately

45. Art Gallery Theorem The Combinatorics of Guarding g(P) = min number of guards for P G(n) = max of g(P), over all n-gons P What is G(n)? Answer: G(n) = floor(n/3) In fact, floor(n/3) vertex guards suffice Answers a question of Victor Klee: How many guards are needed to see a simple n-gon? Proofs: Chvatal (induction); Fisk (simple coloring argument)

46. Chvatal Comb: Necessity of n/3 Guards in Some Cases w1 w2 Shows that some n-gons require at least n/3 guards, since we can place “independent witness points”, wi , near each tip, and must have a separate guard in each of their visibility regions (triangles) Can extend to cases where n is not a multiple of 3, showing lower bound of floor(n/3). Thus: G(n) ≥ floor(n/3)

47. Fisk Proof: Floor(n/3) Guards Suffice: G(n) ≤ floor(n/3) • Triangulate P (we know a triangulation exists) • 3-color the vertices (of triangulation graph) • Place guards at vertices in smallest color class (claim: every point of P is seen, since each triangle has a guard at a corner, and that guard sees all of the (convex) triangle)

48. Vertex Guarding a Simple Polygon • Vertex guarding applet 11 yellow vertices 11 blue vertices 16 white vertices Place guards at yellow (or blue) vertices: at most n/3 vertex guards (here, n=38)

49. Computing g(P) by Inspection • By inspection, find a large set of “visibility independent witness points” within P • If we find w indep witness points, then we know that g(P)≥w • By inspection, find a small set of m guards that see all of P: g(P)≤m • If we are lucky, m=w; otherwise, more arguments are needed!

50. Lower Bound on g(P) g(P)  4 Fact: If we can place w visibility independent witness points, then g(P) ≥ w.