Motivation: Art gallery problem. Polygon decomposition. Art gallery problem: upper bound. Art gallery problem: lower bound

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1 CG Lecture 3 Polygon decomposition 1. Polygon triangulation Triangulation theory Monotone polygon triangulation 2. Polygon decomposition into monotone pieces 3. Trapezoidal decomposition 4. Conex decomposition Motiation: Art gallery problem Definition: two points q and r in a simple polygon P can see each other if the open segment qr lies entirely within P. A point p guards a region R P if p sees all q R Problem: Gien a polygon P, what is the minimum number of guards required to guard P, and what are their locations? 5. Other results 1 2 p R r q Simple obserations Conex polygon: all points are isible from all other points only one guard in any location is necessary! Star-shaped polygon: all points are isible from any point in the kernel only one guard located in its kernel is necessary. conex star-shaped 3 Art gallery problem: upper bound Theorem: Eery simple planar polygon with n ertices has a triangulation of size n-2 (proof later). n-2 guards suffice for an n-gon: Subdiide the polygon into n 2 triangles (triangulation). Place one guard in each triangle. 4 Art gallery problem: lower bound There exists a polygon with n ertices, for which n/3 guards are necessary. Therefore, n/3 guards are needed in the worst case. Can we improe the upper bound? Yes! In fact, at most n/3 guards are necessary. Simple polygon triangulation Input: a polygon P described by an ordered sequence of ertices < 0, n 1 >. Output: a partition of P into n 2 non-oerlapping triangles and the adjacencies between them

2 Simple polygon triangulation: obserations The triangulation is not unique. One of them suffices. The triangulation is always possible. No new ertices are required. The triangulation adds new edges, called diagonals, between existing ertices. Ideas? Not all diagonals are alid! 7 8 Triangulation theory A ertex is conex if its interior angle < π, otherwise it is concae. A diagonal is a new edge between two polygon ertices that is entirely inside the polygon. Lemma 1: eery polygon has a conex ertex. Proof: the highest ertex (the one with the largest y coordinate) is conex. Lemma 2: Eery polygon with n>3 ertices has a diagonal. 9 Diagonals in polygons Proof: let be a conex ertex and let a and b its adjacent ertices. Since P is a simple polygon and n>3, there is no edge between a and b. Consider the following two cases: 1. the new edge ab is a diagonal 2. Otherwise, there exists a ertex x which is the closest to with respect to a line L parallel to ab which is a diagonal. L a a x Case 1 b Case 2 b 10 Triangulation theory Theorem: Eery simple polygon with n ertices has a triangulation with n 3 diagonals and n 2 triangles. Proof: By induction on n: Basis: A triangle (n=3) has a triangulation (itself) with no diagonals and one triangle. P 2 Induction on n: For an n-ertex polygon, construct a diagonal diiding the polygon into two polygons P 1 and P 2 with n 1 and n 2 ertices such that n 1 +n 2 2 = n. Diagonals: (n 1 3)+(n 2 3)+1 = (n 1 +n 2 2) 3 = n 3 Triangles: (n 1 2)+(n 2 2) = (n 1 +n 2 2) 2 = n 2 11 P 1 Triangulation dual Definition: The triangulation dual T of a triangulation of a simple polygon P is a graph whose nodes are triangles and whose edges are adjacencies between triangles sharing an edge. Property: the triangulation dual is a tree whose node degree is 3. Proof: Degree 3 by construction: no triangle has more than three neighbors. No cycles: by contradiction. If it has a cycle, it is a polygon with a hole (not a simple polygon). If fact, T is a binary tree with root degree one or two! 12 2

3 Simple triangulation algorithm (1) Idea: Reduce the polygon by clipping a triangle at each iteration. The clipped triangle will be formed by three consecutie ertices ( i, i+1, i+2 ). The diagonal is ( i, i+2 ). Test for alidity: 1. The diagonal does not intersect other polygon edges. 2. The diagonal must be inside the polygon (test that diagonal is inside normal cone). 13 Simple triangulation algorithm proc triangulate(p) if P 3 output(p) and return; i 0; while diagonal( i, i+2 ) is not legal i++; output( i, i+1, i+2 ); remoe i+1 from P; triangulate(p); Complexity: n times n diagonal tests that take each O(n) O(n 3 ). Sources of inefficiencies: repeated diagonal tests diagonals are not sorted or ordered Precompute diagonals in O(n 2 ) Triangulation algorithms What is the lower bound? O(n 2 ) by precomputing diagonals Improe to O(nlogn) by ordering them (see later). Is less than O(nlogn) possible? O(n log n)-time triangulation algorithm 1) Partition the polygon into y-monotone pieces.( חתיכות מונוט וניות ) 2) Triangulate each y-monotone piece separately Monotone polygons: definition A simple polygon is called monotone with respect to a line L if for any line L perpendicular to L the intersection of the polygon with L is connected. A polygon is called monotone if there exists any such line l. A polygon that is monotone with respect to the x/y-axis is called x/ymonotone. Question: How can we check in O(n) time whether a polygon is y-monotone? L L y-monotone but not x-monotone polygon 18 Property of monotone polygons Definition: a ertex is an interior cusp iff it is a concae ertex whose adjacent ertices are both at or aboe (at or below) line L. L Theorem: If a polygon P has no interior cusps with respect to a line L, then it is monotone with respect to L. Proof sketch: partition P into two chains connecting the top and bottom ertices. Assume one of them is not monotone with respect to L. Then P must contain an interior cusp aboe or below. 19 3

4 Triangulating a y-monotone y polygon Sweep the polygon from top to bottom. Greedily triangulate anything possible aboe the sweep line, and then forget about this region. When procesing a ertex, the unhandled region aboe it always has a simple structure: Two y-monotone (left and right) chains, each containing at least one edge. If a chain consists of two or more edges, it is reflex, and the other chain consists of a single edge whose bottom endpoint has not been handled yet. Each diagonal is added in O(1) time. 20 Triangulating a Y-monotone Y polygon Continue sweeping while one chain contains only one edge, left chain while the other edge is concae. When a conex edge appears in the concae chain (or a second edge appears in the other one), triangulate as much as possible using a fan. Time complexity: O(k), where k is the complexity of the polygon. bottom right chain top 21 Stack operations on chains Monotone polygon triangulation Y-monotone polygons Vertex classification Polygon ertices classification: A start (resp., end) ertex is a ertex whose interior angle is less than π and its two neighboring ertices both lie below (resp., aboe) it. A split (resp., merge) ertex is a ertex whose interior angle is greater than π and its two neighboring ertices both lie below (resp., aboe) it. All other ertices are regular. split start merge end regular

5 Y-monotone polygons: properties Theorem: A polygon without split start and merge ertices is y-monotone. Proof: Since there are only merge start/end/regular ertices, the polygon must consist of two y-monotone chains. split Alternatiely, do a case analysis. To partition a polygon to monotone pieces, eliminate split (merge) ertices by adding diagonals upward (downward) from the ertex. Naturally, the diagonals must not intersect! end regular Split and merge ertices Classify all ertices. Sweep the polygon from top to bottom. Maintain the edges intersected by the sweep line L sorted by x coordinate). Maintain ertex eents in an eent queue Q sorted by y coordinate. Eliminate split/merge ertices by connecting them to other ertices. For each edge e, define helper(e) as the lowest ertex (seen so far) aboe the sweep line isible to the right of the edge. helper(e) is initialized by the upper endpoint of e. helper(e j ) e j i e k e i-1 e i 28 (cont.) A split ertex may be connected to the helper ertex of the edge immediately to its left. Howeer, a merge ertex should be connected to a ertex which has not been processed yet! Cleer idea: Eery merge ertex is the helper of some edge, and will be handled when this edge terminates. e j helper(e j ) i e i i e j helper(e j ) e k e k m 29 algorithm Input: A counterclockwise ordered list of ertices. The edge e i immediately follows the ertex i. Construct Q on the ertices of P using y-coordinates. (when two or more ertices hae the same y-coordinates, the ertex with the smaller x-coordinate has priority.) Initialize L to be empty. While Q is not empty: Pop ertex ; Handle. Note: No new eents are generated during execution. Handling a start ertex i : Adde i to L helper(e i ) i e 1 e 3 3 e e 9 9 e 8 e e e No split/merge ertex remains unhandled

6 Handling an end ertex i : If helper(e i-1 ) is a merge ertex, then connect i to helper(e i-1 ) Remoe e i-1 from L e 3 3 e 2 e e 9 e 8 9 e e e Handling a split ertex i : Find in L the edge e j directly to the left of i Connect i to helper(e j ) helper(e j ) i Insert e i into L helper(e i ) i e 3 3 e 2 e e 9 9 e 8 e e 5 e Handling a merge ertex i : If helper(e i-1 ) is a merge ertex, then connect i to helper(e i-1 ) Remoe e i-1 from L Find in L the edge e j directly to the left of i If helper(e j ) is a merge ertex, then connect i to helper(e j ) helper(e j ) i e 3 3 e 2 e e 9 e8 e e 6 e Handling a regular ertex i : If the polygon s interior lies to the left of i then: Find in L the edge e j directly to the left of i If helper(e j ) is a merge ertex, then connect i to helper(e j ) helper(e j ) i Else: If helper(e i-1 ) is a merge ertex, then connect i to helper(e i-1 ) Remoe e i-1 from L Insert e i into L helper(e i ) i 2 7 e 6 35 Monotone polygon algorithm Eent handling (1)

7 Eent handling (2) Example Time complexity of polygon triangulation Partitioning the polygon into monotone pieces: O(n log n) Triangulating all the monotone pieces: O(n) Total: O(n log n) 40 Trapezoidal decomposition Decompose a polygon ito trapezoids with two edges perpendicular to a gie line. Trapezoids are triially triangulated BUT we need new intermediate ertices. Useful for point search and other tasks. Obtained directly from a sweep line algorithm: at each ertex, compute the new supporting ertices left and right. Close trapezoids according to neighborhood relations. 41 Conex partitioning Problem: partition a polygon P into a small (fewest) number of conex pieces. Possibilities: Use only ertices of P Use only ertices on edges of P Use new internal ertices (Steiner points) Theorem: The smallest number of conex pieces λ that is needed to partition P is: r/2 + 1 λ r + 1 where r is the number of concae ertices. 42 Hertl and Melhorn algorithm Idea: triangulate polygon, then remoe non-essential diagonals. A diagonal is nonessential if its remoal creates non-conexities. Note: only concae ertices hae essential diagonals! Complexity: O(n log n) for triangulation. Is this strategy optimal? NO! Theorem: the number of pieces is neer worse than four times the optimal. 43 7

8 Other results Optimal O(n) triangulation algorithm [Chazelle 1991] Finding the minimum number of guards for a simple polygon is NP-hard [Aggarwal, 1984]. 3D ersion: find a tetrahedrization of a simple polyhedra. Not always possible without introducing new ertices! The decision problem (are new ertices necessary?) Is NP-complete. Algorithm runs in O(nr + r 2 log r) time [Chazelle and Palios, 1990]. 44 8