AMS 345/CSE 355 Computational Geometry

Transcription

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: Pramod Ganapathi

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 For p,q in P, p is visible to q if segment pq lies within (closed) P p q

8 Definitions: Visibility, Diagonals 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 p clearly sees q but does not clearly see q p sees q p q q

9 Definitions: Visibility, Diagonals v i v j is a diagonal if v i and v j are vertices that clearly see each other (versus: chord pq, with p and q on the boundary of P) q pq is a chord (not a diagonal) v i v j is a diagonal v k v m is not a diagonal p v j v k v i v m

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 n 1 and n 2, with n 1 +n 2 =n+2 (since the endpoints of the diagonal are counted twice) By IH, the two subpolygons have triangulation with n 1-2 and n 2-2 triangles; glue together along the diagonal to get a total of (n 1-2)+(n 2-2)=n-2 triangles in the overall triangulation. Similar for number of diagonals.

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

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 v i-1 v i+1 Input: Simple polygon P v i pq is a diagonal, cutting off a single triangle (the ear ) Naive: O(n 3 ) Smarter: Keep track of ear tip status of each v i (initialize: O(n 2 ) ) Each ear clip requires O(1) ear tip tests O(n) per test ) Thus, O(n 2 ) total, worst-case Ear-clipping applet

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

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, , , , , , , , , , , , , , ,

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 Example: 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.

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

29 More Examples: HW1

30 Counting Triangulations: Example

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

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

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

34 Counting Triangulations: How many triangulations? Example 0 5 Case analysis: (1)Use 03: one completion (2) Do NOT Use 03: Then, 24 is 4 forced, and there are 3 triangulations possible (using 15 and 14, or 04 and 3 14, or 04 and 02). 2 Total for cases (1) and (2): 1+3=4 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 Input: Set S of n points Input: Other shapes Simple polygon Polygon with holes 3D: Surfaces and solids (tetrahedralization) Planar Straight-Line Graph (PSLG) Triangulation applet for simple polygons

37 Triangulation Theory in 2D 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. Also with holes But, NOT true in 3D! Thm: The triangulation graph can be 3-colored. Proofs: Induction on n.

38 Triangulating a Polygon Simple ear-clipping methods: O(n 2 ) 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] Not practical! Polygons with h holes: O(n+h log 1+ h), (n+h log h) [BC] Good practical method: FIST [Held], based on clever methods of ear clipping (worst-case O(n 2 ) ) fan

39 Tetrahedralizing Polyhedra

40 Tetrahedralizing Polyhedra

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 Answers a question of Victor Klee: How many guards are needed to see a simple n-gon? Proofs: Chvatal (induction); Fisk (simple coloring argument) 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

46 Chvatal Comb: Necessity of n/3 Guards in Some Cases w 1 w 2 Shows that some n-gons require at least n/3 guards, since we can place independent witness points, w i, 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) 1. Triangulate P (we know a triangulation exists) 2. 3-color the vertices (of triangulation graph) 3. 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) Fact: If we can place w visibility independent witness points, then g(p) w. g(p) 4

51 Witness Number Let w(p) = max # of independent witness points possible in a set of visibility independent witness points for P Then, g(p) w(p) Note: It is hard to find g(p); it is poly-time to find w(p) Some polygons have g(p)=w(p); I call these perfect polygons they are very special; most polygons P have a gap : g(p)>w(p)

52 Witness Number: Vertex Guards We say that a set, W, of points inside P are independent with respect to vertex guards if, for any two points of W, the set of vertices of P they see are disjoint Let w V (P) = max # of witness points possible in a set of witness points for P that are indep wrt vertex guards Then, g V (P) w V (P) Note: It is hard to find g V (P); a recent (unpublished) algorithm computes w V (P) in polytime.

53 Useful Polygon Example Godfried s Favorite Polygon g(p)=2, but w(p)=1

54 Useful Polygon Examples Godfried s Favorite Polygon variations

55 Examples For each of these polygons P, find the point guard number, g(p), and the vertex guard number, g V (P). Also, find the witness numbers w(p) and w V (P)

56 Visibility Graphs VG of P: (V,E), where E = set of all pairs of vertices of P that see each other, V = vertices (clear-vg: demand clear visibility: i.e., edge set E is set of all diagonals)

57 Art Gallery Theorem: Orthogonal (Rectilinear) Polygons

58 Polygons with Holes Art Galley Theorem: floor( (n+h)/3 ) guards suffice and are sometimes necessary (easy: floor( (n+2h)/3 ) suffice do you see why?)

59 Exterior Guarding: Fortress Problem

60 Edge Guards

61 Mirrored Galleries

62 Mobile Guards Find shortest route (path or tour) for a mobile guard within P: Watchman route problem Efficient algorithms for simple polygons P NP-hard for polygons with holes (as hard as the TSP)

63 Mobile Robotic Guard Visit all visibility polygons Subject to: stay inside polygonal domain P Watchman Route Problem

64 Guarding Polyhedra Note: Guards at vertices are NOT enough!

65 Scissors Congruence in 2D Dissections

66 Scissors Congruence in 2D

67 Scissors Congruence in 2D

68 Scissors Congruence in 2D

69 Scissors Congruence: Rectangles

70 Scissors Congruence: Rectangles

71 Scissors Congruence: Polygons

72 Scissors Congruence: Polygons

73 Scissors Congruence: Polygons

74 Fair Partitions: Polygons

75 Scissors Congruence: 3D

76 Scissors Congruence: 3D