Problem: Polygon Triangulation. Trapezoidal Map. Solution via Trapezoidal Map. The trapezoidal map of non-crossing line segments

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1 Problem: Polygon Triangulation The trapezoidal map of non-crossing line segments Given a simple polygon P with n edges, compute a triangulation of its interior. P 1 2 Solution via Trapezoidal Map Given a set S of n nonintersecting segments in the plane, compute its trapezoidal map. Trapezoidal Map planar graph, vertices V, edges E, faces F V : endpoints, artificial vertices E: pieces of segments, vertical extensions F: set of trapezoids, each one incident to at most 4 segments (assuming no two endpoints have the same x-coordinate; not true in triangulation application, but can be achieved even there) 3 4

2 Randomized Incremental Construction 1. Compute trapezoidal map of {s 1 } T 1 From T r 1 to T r (I) Find s 1 s r 2. Insert segments s 2,..., s n in random order T n 5 6 From T r 1 to T r (II) Split From T r 1 to T r (III) s r 1. Find: Find the trapezoid containing the left endpoint of s r Merge 2. Split: Trace s r through T r 1 and split all the trapezoids intersected by s r 3. Merge: Remove parts of vertical extensions cut off by s r and merge the adjacent trapezoids s r 7 8

3 RIC Analysis (II) RIC Analysis (I) Cost of step T r 1 T r : Apply configuration spaces! Find: we ll care for that later... X: the set S of segments Π: set of all trapezoids defined by segments of S Split: constant time per traced ; is replaced by at most 4 new trapezoids. D( ): the (at most 4) segments incident to the trapezoid K( ): the set of segments intersecting O(number of removed trapezoids) = O(number of created trapezoids) Merge: O(number of trapezoids created in step Split) 9 10 Analysis of Update T r 1 T r (II) Analysis of Update T r 1 T r (I) Observation: The number of trapezoids created by Split is at most twice as large as the number of new trapezoids in T r. Configuration Spaces expected value of deg(s r, T r ) is 4 r E( T r ). T r 6r (each is incident to a segment endpoint, and each endpoint is charged by at most three segments). Proof: For every Merge operation above (below) s r, one new trapezoid below (above) s r survives. It follows that at most half of the previously created trapezoids are not in T r. Complexity of Split and Merge is O( { T r \ T r 1 } ) = O(deg(s r, T r )). Expected update cost T r 1 T r is O(1) Overall expected update cost is O(n) 11 12

4 Analysis of Find (I) Realization of Find Assume p r runs through a trapezoid. Then there is j r such that History approach: store all the trapezoids of T r, r = 1... n. T r 1 \ T r has pointers to all T r \ T r 1 with At most 4 pointers per Location of segment endpoint p r of s r : trace p r through the history graph T j \ T j 1 s r intersects length of history path to p r n 1 [s r K( )] j=0 T j \T j 1 expected time for history searches is proportional to the expected number n 1 r=0 K(r) of conflicts that appear during the algorithm Analysis of Find (II) Configuration spaces n 1 n 1 K(r) (k 1 (r) k 2 (r) + k 3 (r)) r=1 r=1 d(n 1)t 1 + n 1 t d(d 1)n r+1 r=1 r(r + 1) d 2 n 1 t r+1 r=1 r + 1 = O(nlog n), because Trapezoidal Map Conclusion Given a set S of n nonintersecting segments in the plane, its trapezoidal map T(S) can be computed in time O(nlog n). (The assumption that segment endpoints have different x-coordinates can be achieved by comparing them lexicographically.) t r+1 = E( T r ) = O(r + 1)

5 Special Case: S forms simple polygon P Trapezoidal Map Triangulation (I) Step 1: Within each trapezoid, connect the two polygon vertices Trapezoidal Map Triangulation (II) Step 2: Triangulate the resulting x-monotone polygons separately, in total time O(n) (Exercise) A fast method for the special case (I) Runtime will be O(nlog n). log (h) n := loglog...log n }{{} h times log n := max{h log (h) n 1} Example: log ( ) = 5 log n < 5 for all n. Definition: n N(h) := log (h) n, 0 h log n

6 A fast method for the special case (II) Generalized history management: keep several histories and for each p P a pointer to the history in charge. compute T 1 and initialize one history, in charge of all points FOR h = 1 TO log n DO FOR r = N(h 1) + 1 TO N(h) DO compute T r from T r 1 (* as usual *) (* Renew histories by tracing S through T r *) FOR ALL T r containing an endpoint DO make the root of a history in charge of all the points it contains FOR r = N(log n) + 1 TO n DO compute T r from T r 1 (* as usual *) 21 Analysis of the fast method (I) Split and Merge proceed as before in expected time O(n) Find will be faster on average, but we have log n additional Trace steps 22 Analysis of Find (I) In phase h, every trapezoid traced during the history search corresponds to a trapezoid that has been present in the beginning of phase h (root of a hierarchy) or was created during phase h Analysis of Find (II) For fixed X := S N(h), E(K h ) is the expected number of conflicts appearing in steps N(h 1) + 1 to N(h) when T(X) is computed. i := N(h 1) + 1, j := N(h) 1. Configuration spaces analysis is in conflict with a segment inserted in phase h expected cost of history search is at most proportional to n + K h, N(h) K h := K( ) S N(h). r=n(h 1)+1 T r \T r 1 E(K h ) j (k 1 k 2 + k 3 ) r=i d(j + 1 i) t i + i j t d(d 1)(j + 1) r+1 r=i r(r + 1) d 2 j t r+1 r=i r

7 Analysis of Find (III) Recall: t r+1 = O(r + 1). Analysis of Trace (I) Then E(K h ) = O(N(h) N(h 1)) + N(h) 1 1 O N(h) r r=n(h 1)+1 ( = O N(h) + N(h)log N(h) ) N(h 1) = O ( N(h) + N(h)log (h) n ) = O(n). (This also holds for a random set S N(h) and for the last insertion phase (i = N(log n)+1, j = n 1).) The total cost for Find over all h is then O(nlog n). 25 The expected cost T h of tracing S through T N(h) is at most proportional to the expected number of conflicts between trapezoids in T N(h) and segments in S, which is 1 ( ) n N(h) { T(R) y K( )}. R S, R =N(h) y S\R Up to a missing factor of d/n(h) this is exactly the bound for the expected number K N(h) of new conflicts when s N(h) is inserted that we derived from the configuration spaces. 26 Analysis of Trace (II) configuration spaces here k d 1 r (n r)t r (n r)t r k d 2 r (n r)t r+1 (n r)t r+1 k d 2 3 r(r+1) (n r)t d r+1 r+1 (n r)t r+1 Setting r = N(h), we obtain T h = k 1 k 2 + k 3 as T h (n N(h))t N(h) (n N(h))t N(h)+1 + d N(h) + 1 (n N(h))t N(h)+1 = O ( n(t N(h) t N(h)+1 ) + n ) = O(n), because t N(h) t N(h)+1. Fast Trapezoidal Map Conclusion Given a simple polygon P with n vertices in the plane, its trapezoidal map T(P) can be computed in time O(nlog n). (This is not optimal, because Chazelle has given a (rather complicated) O(n) algorithm for the problem.) The total cost for Trace over all h is then O(nlog n)