Polygon Simplification by Minimizing Convex Corners

Transcription

1 Polygon Simplification by Minimizing Convex Corners Yeganeh Bahoo 1, Stephane Durocher 1, J. Mark Keil 2, Saee Mehrabi 3, Sahar Mehrpour 1, an Debajyoti Monal 1 1 Department of Computer Science, University of Manitoba, Winnipeg, Canaa 2 Department of Computer Science, University of Saskatchewan, Saskatoon, Canaa 3 Cheriton School of Computer Science, University of Waterloo, Waterloo, Canaa {bahoo,urocher,mehrpour,jyoti}@cs.umanitoba.ca, keil@cs.usask.ca, smehrabi@uwaterloo.ca Abstract. Let P be a polygon with r > 0 reflex vertices an possibly with holes. A subsuming polygon of P is a polygon P such that P P, each connecte component R of P subsumes a istinct component R of P, i.e., R R, an the reflex corners of R coincie with the reflex corners of R. A subsuming chain of P is a minimal path on the bounary of P whose two en eges coincie with two eges of P. Aichholzer et al. prove that every polygon P has a subsuming polygon with O(r) vertices. Let A e(p ) (resp., A v(p )) be the arrangement of lines etermine by the eges (resp., pairs of vertices) of P. Aichholzer et al. observe that a challenge of computing an optimal subsuming polygon P min, i.e., a subsuming polygon with minimum number of convex vertices, is that it may not always lie on A e(p ). We prove that in some settings, one can fin an optimal subsuming polygon for a given simple polygon in polynomial time, i.e., when A e(p min) = A e(p ) an the subsuming chains are of constant length. In contrast, we prove the problem to be NP-har for polygons with holes, even if there exists some P min with A e(p min) = A e(p ) an subsuming chains are of length three. Both results exten to the scenario when A v(p min) = A v(p ). 1 Introuction Polygon simplification is well stuie in computational geometry, with numerous applications in cartographic visualization, computer graphics an ata compression [8, 9]. Techniques for simplifying polygons an polylines have appeare in the literature in various forms. Common goals of these simplification algorithms inclue to preserve the shape of the polygon, to reuce the number of vertices, to reuce the space requirements, an to remove noise (extraneous bens) from the polygon bounary (e.g., [2, 4, 5]). In this paper we consier a specific version of polygon simplification introuce by Aichholzer et al. [1], which keeps reflex corners intact, but minimizes the number of convex corners. Aichholzer et al. showe that such a simplification can help achieve faster solutions for many geometric problems such as answering shortest path queries, computing Voronoi iagrams, an so on. Let P be a polygon with r reflex vertices an possibly with holes. A reflex corner of P consists of three consecutive vertices u, v, w on the bounary of P such that the Work of the author is supporte in part by the Natural Sciences an Engineering Research Council of Canaa (NSERC).

2 (a) (b) (c) () Fig. 1. (a) A polygon P, where the polygon is fille an the holes are empty regions. (b) A subsuming polygon P, where P is the union of the fille regions. A subsuming chain is shown in bol. (c) A min-convex subsuming polygon P min, where A e(p min) = A e(p ). () A polygon P such that for any min-convex subsuming polygon P min, A e(p ) A e(p min). angle uvw insie P is more than 180. We refer the vertex v as a reflex vertex of P. The vertices of P that are not reflex are calle convex vertices. By a component of P, we refer to a connecte region of P. A polygon P subsumes P if P P, each component R of P subsumes a istinct component R of P, i.e., R R, an the reflex corners of R coincie with the reflex corners of R. A k-convex subsuming polygon P contains at most k convex vertices. A min-convex subsuming polygon is a subsuming polygon that minimizes the number of convex vertices. Figure 1(a) illustrates a polygon P, an Figures 1(b) an (c) illustrate a subsuming polygon an a min-convex subsuming polygon of P, respectively. A subsuming chain of P is a minimal path on the bounary of P whose en eges coincie with a pair of eges of P, as shown in Figure 1(b). Aichholzer et al. [1] showe that for every polygon P with n vertices, r > 0 of which are reflex, one can compute in linear time a subsuming polygon P with at most O(r) vertices. Note that although a subsuming polygon with O(r) vertices always exists, no polynomial-time algorithm is known for computing a min-convex subsuming polygon. Fining an optimal subsuming polygon seems challenging since it oes not always lie on the arrangement of lines A e (P ) (resp., A v (P )) etermine by the eges (resp., pairs of vertices) of the input polygon. Figure 1(c) illustrates an optimal polygon P min for the polygon P of Figure 1(a), where A e(p min ) = A e(p ). On the other han, Figure 1() shows that a min-convex subsuming polygon may not always lie on A e (P ) or A v (P ). Note that the input polygon of Figure 1() is a simple polygon, i.e., it oes not contain any hole. Hence etermining min-convex subsuming polygons seems challenging even for simple polygons. In fact, Aichholzer et al. [1] pose an open question that asks to etermine the complexity of computing min-convex subsuming polygons, where the input is restricte to simple polygons. Let P be a simple polygon. In this paper we show that if there exists a min-convex subsuming polygon P min such that A e(p min ) = A e(p ) an the subsuming chains of P min are of constant length, then one can compute such an optimal subsuming polygon in polynomial time. In contrast, if P contains holes, then we prove the problem to be NP-har. The harness result hols even when the min-convex subsuming polygon P min lies on the arrangement A e(p ), an the length of every subsuming chain of P min is three. Both results exten to the scenario when A v (P min ) = A v(p ). 2

3 The rest of the paper is organise as follows. In Section 2 we escribe the techniques for computing subsuming polygons. Section 3 inclues the NP-harness result. Finally, Section 4 conclues the paper iscussing irections to future research. 2 Computing Subsuming Polygons In this section we show that for any simple polygon P, if there exists a min-convex subsuming polygon P min such that A e (P ) = A e (P min ) an the subsuming chains are of length at most t, then one can compute an optimal polygon in O(t O(1) n f(t) ) time. Therefore, if t = O(1), then the time complexity of our algorithm is polynomial in n. We first present efinitions an preliminary results on outerstring graphs, which will be an important tool for computing subsuming polygons. 2.1 Inepenent Set in Outerstring Graphs A graph G is a string graph if it is an intersection graph of a set of simple curves in the plane, i.e., each vertex of G is a mappe to a curve (string), an two vertices are ajacent in G if an only if the corresponing curves intersect. G is an outerstring graph if the unerlying curves lie interior to a simple cycle C, where each curve intersects C at one of its enpoints. Figure 2(a) illustrates an outerstring graph an the corresponing arrangement of curves. Later in our algorithm, the polygon will correspon to the cycle of an outerstring graph, an some polygonal chains attache to the bounary of the polygon will correspon to the strings of that outerstring graph. A set of strings is calle inepenent if no two strings in the set intersect, the corresponing vertices in G are calle an inepenent set of vertices. Let G be a weighte outerstring graph with a set T of weighte strings. A maximum weight inepenent set MWIS(T ) (resp., MWIS(G)) is a set of inepenent strings T T (resp., vertices) that maximizes the sum of the weights of the strings in T. Observe that MWIS(T ) is also a maximum weight inepenent set MWIS(G) of G. By MWIS(G) we enote the weight of MWIS(G). Let Γ (G) be the arrangement of curves that correspons to G, e.g., see Figure 2(a). Let R be a geometric representation of Γ (G), where C is represente as a simple polygon P, an each curve is represente as a simple polygonal chain insie P such that one of its enpoints coincies with a istinct vertex of P. Keil et al. [6] showe that given a geometric representation R of G, one can compute a maximum weight inepenent set of G in O(s 3 ) time, where s is the number of line segments in R. Theorem 1 (Keil et al. [6]). Given the geometric representation R of a weighte outerstring graph G, there exists a ynamic programming algorithm that computes a maximum weight inepenent set of G in O(s 3 ) time, where s is the number of straight line segments in R. Figure 2(b) illustrates a geometric representation R of some G, where each string is represente with at most 4 segments. Keil et al. [6] observe that any maximum weight inepenent set of strings can be triangulate to create a triangulation P t of P, as shown in Figure 2(c). Let T be the strings in R. Then the problem of fining MWIS(T ) can be 3

4 a b e c a (a) e c b w e = 2 e a w b = 7 w = 3 c w c = 5 b w b = 3 (b) e a c w w b v 1 v 2 (c) () (e) v 1 v 2 Fig. 2. (a) Illustration for G an Γ (G). (b) A geometric representation R of G. (c) A triangulate polygon obtaine from an inepenent set of G. () (e) Dynamic programming to fin maximum weight inepenent set. solve by iviing the problem into subproblems, each escribe using only two points of R. We illustrate how the subproblems are compute very briefly using Figure 2(). Let P (v 1, v 2 ) be the problem of fining MWIS(T v1,v 2 ), where T v1,v 2 consists of the strings that lie to the left of v 1 v 2. Let wv 1 v 2 be a triangle in P t, where w is a point on some string insie P (v 1, v 2 ); see Figure 2(). Since P t is a triangulation of the maximum weight string set, must be a string in the optimal solution. Hence P (v 1, v 2 ) can be compute from the solution to the subproblems P (v 1, w) an P (w, v 2 ), as shown in Figure 2(e). Keil et al. [6] showe that there are only a few ifferent cases epening on whether the points escribing the subproblems belong to the polygon or the strings. We will use this iea of computing MWIS(T ) to compute subsuming polygons. 2.2 Subsuming Polygons via Outerstring Graphs Let P = (v 0, v 1,..., v n 1 ) be a simple polygon with n vertices, r > 0 of which are reflex vertices. A convex chain of P is a path C ij = (v i, v i+1,..., v j 1, v j ) of strictly convex vertices, where the inices are consiere moulo n. Let P = (w 0, w 1,..., w m 1 ) be a subsuming polygon of P, where A e (P ) = A e (P ), an the subsuming chains are of length at most t. Let C qr = (w q,..., w r ) be a subsuming chain of P. Then by efinition, there is a corresponing convex chain C ij in P such that the eges (v i, v i+1 ) an (v j 1, v j ) coincie with the eges (w q, w q+1 ) an (w r 1, w r ). We call the vertex v i the left support of C qr. Since A e (P ) = A e (P ), the 4

5 z 8 z 15 z 7 z 5 v j z 11 v j 2 z 14 z 13 z 12 z 10 z 9 z 4 z 3 z 2 z 1 v j (a) v v j+1 j 3 (b) z 8 z 15 z 7 z 5 a z 14 z 13 z 12 z 11 z 10 z 9 z 4 z 3 z 2 z 1 v j e v j v j e b c (c) () Fig. 3. (a) Illustration for the polygon P (in bol), A e(p ) (in gray), an Q (in ashe lines). (b) Chains of v j. (c) Attaching the strings to Q. () Dynamic programming insie the gray region. chain C qr must lie on A e (P ). Moreover, since P is a min-convex subsuming polygon, the number of vertices in C qr woul be at most the number of vertices in C ij. We claim that the number of paths in A e (P ) from v i to v j is O(n t ). Since t is an upper boun on the length of the subsuming chains, any subsuming chain can have at most (t 1) line segments. Since there are only O(n) straight lines in the arrangement A e (P ), there can be at most n j paths of j eges, where 1 j t 1. Consequently, the number of caniate chains that can subsume C ij is O(n t ). Lemma 1. Given a simple polygon P with n vertices, every convex chain C of P has at most O(n t ) caniate subsuming chains in A e (P ), each of length at most t. In the following we construct an outerstring graph using these caniate subsuming chains. We first compute a simple polygon Q interior to P such that for each ege e in P, there exists a corresponing ege e in Q which is parallel to e an the perpenicular istance between e an e is ɛ, as shown in ashe line in Figure 3(a). We choose ɛ sufficiently small 4 such that for each component w of P, Q contains exactly one component insie w. We now construct the strings. Let v j be a convex corner of P. Let S j 4 Choose ɛ = δ/3, where δ is the istance between the closest visible pair of bounary points. 5

6 be the set of caniate subsuming chains such that for each chain in S j, the left support of the chain appears before v j while traversing the unboune face of P in clockwise orer. For example, the subsuming chains that correspon to v j are (v j 2, z 1, v j+1 ), (v j 3, z 13, z 2, v j+1 ), (v j 3, z 14, z 3, v j+1 ), (v j 3, z 11, z 4, v j+1 ), (v j 3, z 15, z 5, v j+1 ), (v j 3, z 8, z 5, v j+1 ), (v j 3, z 7, v j+1 ), as shown in Figure 3(b). For each of these chains, we create a unique enpoint on the ege e of Q, where e correspons to the ege v j v j+1 in P, as shown in Figure 3(c). We then attach these chains to Q by aing a segment from v j to its unique enpoint on Q. We attach the chains for all the convex vertices of P to Q. Later we will use these chains as the strings of an outerstring graph. We then assign each chain a weight, which is the number of convex vertices of P it can reuce. For example in Figure 3(b), the weight of the chain (v j 3, z 8, z 5, v j+1 ) is one. Although the strings are outsie of the simple cycle, it is straightforwar to construct a representation with all the strings insie a simple cycle Q: Consier placing a ummy vertex at the intersection points of the arrangement, an then fin a straight-line embeing of the resulting planar graph such that the bounary of Q correspons to the outerface of the embeing. Consequently, Q an its associate strings correspon to an outerstring graph representation R. Let G be the unerlying outerstring graph. We now claim that any MWIS(G) correspons to a min-convex subsuming polygon of P. Lemma 2. Let P be a simple polygon, where there exists a min-convex subsuming polygon that lies on A e (P ), an let G be the corresponing outerstring graph. Any maximum weight inepenent set of G yiels a min-convex subsuming polygon of P. Proof. Let T be a set of strings that correspon to a maximum weight inepenent set of G. Since T is an inepenent set, the corresponing subsuming chains o not create ege crossings. Moreover, since each subsuming chain is weighte by the number of convex corners it can remove, the subsuming chains corresponing to T can remove MWIS(G) convex corners in total. Assume now that there exists a min-convex subsuming polygon that can remove at least k convex corners. The corresponing subsuming chains woul correspon to an inepenent set T of strings in G. Since each string is weighte by the number of convex corners the corresponing subsuming chain can remove, the weight of T woul be at least k. 2.3 Time Complexity To construct G, we first place a ummy vertex at the intersection points of the chains, an then compute a straight-line embeing of the resulting planar graph such that all the vertices of Q are on the outerface. Therefore, the geometric representation use at most nt eges to represent each string. Since each convex vertex of P is associate with at most O(n t ) strings, there are at most n O(n t ) strings in G. Consequently, the total number of segments use in the geometric representation is O(tn 2+t ). A subtle point here is that the strings in our representation may partially overlap, an more than three strings may intersect at one point. Removing such egeneracy oes not increase the asymptotic size of the representation. Finally, by Theorem 1, one can compute the optimal subsuming polygon in O(t 3 n 6+3t ) time. 6

7 The complexity can be improve further to as follows. Let abc be a rectangle that contains all the intersection points of A e (P ). Then every optimal solution can be extene to a triangulation of the close region between abc an Q. Figure 3() illustrates this region in gray. We now can apply a ynamic programming similar to Section 2.1 to compute the maximum weight inepenent string set, where each subproblem fins a maximum weight set insie some subpolygon. Each such subpolygon can be escribe using two points v 1, v 2, each lying either on Q or on some string, an a subset of {a, b, c, } that helps enclosing the subpolygon. Since there are n O(n t ) strings, each containing at most t points, the number of vertices in the geometric representation is O(tn 1+t ). Therefore, the size of the ynamic programming table is O(tn 1+t ) O(tn 1+t ) O(1). Since there can be at most O(tn 1+t ) caniate triangles v 1 v 2 w, we take O(tn 1+t ) time to fill an entry of the table. Hence the ynamic program takes at most O(t 3 n 3+3t ) time in total. Theorem 2. Given a simple polygon P with n vertices such that there exists a minconvex subsuming polygon that lie on A e (P ) an the subsuming chains are of length at most t, one can compute such a min-convex subsuming polygon in O(t 3 n 3+3t ) time. 2.4 Generalizations We can generalize the results for any given line arrangements. However, such a generalization may increase the time complexity. For example, consier the case when the given line arrangement is A v (P ), which is etermine by the pairs of vertices of P. Since we now have O(n 2 ) lines in the arrangement A v (P ), the time complexity increases to O(t 3 (n 2 ) 3+3t ), i.e., O(t 3 n 6+6t ). 3 NP-harness of Min-Convex Subsuming Polygon In this section we prove that it is NP-har to fin a subsuming polygon with minimum number of convex vertices. We enote the problem by MIN-CONVEX-SUBSUMING- POLYGON. We reuce the NP-complete problem monotone planar 3-SAT [3], which is a variation of the 3-SAT problem as follows: Every clause in a monotone planar 3-SAT consists of either three negate variables (negative clause) or three non-negate variables (positive clause). Furthermore, the bipartite graph constructe from the variableclause inciences, amits a planar rawing such that all the vertices corresponing to the variables lie along a horizontal straight line l, an all the vertices corresponing to the positive (respectively, negative) clauses lie above (respectively, below) l. The problem remains NP-har even when each variable appears in at most four clauses [7]. The iea of the reuction is as follows. Given an instance of a monotone planar 3-SAT I with variable set X an clause set C, we create a corresponing instance P I of MIN-CONVEX-SUBSUMING-POLYGON. Let λ be the number of convex vertices in P I. The reuction ensures that if there exists a satisfying truth assignment of I, then P I can be subsume by a polygon with at most λ X C 2 3 C convex vertices, an vice versa. Given an instance I of monotone planar 3-SAT, we first construct an orthogonal polygon P o with holes. We enote each clause an variable using a istinct axis-aligne 7

8 c 1 = (x 1 x 3 x 4 ) c 2 =(x 1 x 2 x 3 ) c 1 = (x 1 x 3 x 4 ) c 2 =(x 1 x 2 x 3 ) x 1 x 2 x 3 x 4 c 3 =( x 2 x 3 x 4 ) c 4 = ( x 1 x 4 ) (a) c 3 =( x 2 x 3 x 4 ) (b) c 4 = ( x 1 x 4 ) l ab a b c (c) e l ab c 3 a b () (e) (f) 3 Fig. 4. (a) An instance I of monotone planar 3-SAT. (b) The orthogonal polygon P o corresponing to I. (c) (f) Illustration for the variable gaget. rectangle, which we refer to as the c-rectangle an v-rectangle, respectively. Each ege connecting a clause an a variable is represente as a thin vertical strip, which we call an ege tunnel. Figures 4(a) an (b) illustrate an instance of monotone planar 3- SAT an the corresponing orthogonal polygon, respectively. While aing the ege tunnels, we ensure for each v-rectangle that the tunnels coming from top lie to the left of all the tunnels coming from the bottom. Figure 4 (b) marks the top an bottom ege tunnels by upwar an ownwar rays, respectively. The v-rectangles, c-rectangles an the ege tunnels may form one or more holes, whereas the polygon is shown in iagonal line pattern. We now transform P o to an instance P I of MIN-CONVEX-SUBSUMING- POLYGON. We first introuce a few notations. Let abc be a convex quarangle an let l ab be an infinite line that passes through a an b. Assume also that l bc an l a intersect at some point e, an c,, e all lie on the same sie of l ab, as shown in Figures 4(c) (). Then we call the quarangle abc a tip on l, an the triangle ce a cap of abc. 3.1 Variable Gaget We construct variable gagets from the v-rectangles. We a some top-right (an the same number of top-left) tips at the bottom sie of the v-rectangle, as show in Figure 4(e). There are three top-right an top-left tips in the figure. For convenience we show only one top-left an one top-right tip in the schematic representation, as shown 8

9 in Figure 4(f). However, we assign weight to these tips to enote how many tips there shoul be in the exact construction. We will ensure a few more properties: (I) The caps o not intersect the bounary of the v-rectangle, (II) no two top-left caps (or, top-right caps) intersect, an (III) every top-left (resp., top-right) cap intersects all the top-right (resp., top-left) caps. Observe that each top-left tip contributes to two convex vertices such that covering them with a cap reuces the number of convex vertices by 1. The peak of the cap reaches very close to the top-left corner of the v-rectangle, which will later interfere with the clause gaget. Specifically, this cap will intersect any ownwar cap of the clause gaget coming through the top ege tunnels. Similarly, each top-right tip contributes to two convex vertices, an the corresponing cap intersects any upwar cap coming through the bottom ege tunnels. Note that the optimal subsuming polygon P cannot contain the caps from both the top-left an top-right tips. We assign the tips with a weight of C 2. In the harness proof this will ensure that either the caps of top-right tips or the caps of top-left tips must exist in P, which will correspon to the true an false configurations, respectively. 3.2 Clause Gaget Without loss of generality assume that each clause is incient to three ege tunnels, otherwise, we can create necessary multi-eges to satisfy this constraint. Figure 5(a) illustrates the transformation for a c-rectangle. Here we escribe the gaget for the positive clauses, an the construction for negative clauses is symmetric. We a three ownwar tips incient to the top sie of the c-rectangle, along its three ege tunnels. Each of these ownwar tip contributes to two convex vertices such that covering the tip with a cap reuces the number of convex vertices by 1. Besies, the corresponing caps reach almost to the bottom sie of the v-rectangles, i.e., they woul intersect the top-left caps of the v-rectangles. Let these tips be t 1, t 2, t 3 from left to right, an let γ 1, γ 2, γ 3 be the corresponing caps. We then a a own-left an a own-right tip at the top sie of the c-rectangle between t i an t i+1, where 1 i 2, as shown in Figure 5(a). Let the tips be t 1,..., t 4 from left to right, an let the corresponing caps be γ 1,..., γ 4. Note that the caps corresponing to t j an t j+1, where 1 j 4, intersect each other. Therefore, at most two of these four caps can exist at the same time in the solution polygon. Observe also that the caps corresponing to t 1, t 2, t 3 intersect the caps corresponing to {t 2}, {t 1, t 4}, {t 3}, respectively. Consequently, any optimal solution polygon containing none of {γ 1, γ 2, γ 3 } have at least 12 convex vertices along the top bounary of the c-rectangle, as shown in Figure 5(b). We now show that any optimal solution polygon P containing at least α > 0 caps from Γ = {γ 1, γ 2, γ 3 } have exactly 11 convex vertices along the top bounary of the c-rectangle. We consier the following three cases: Case 1 (α = 1): If γ 1 (resp., γ 3 ) is in P, then P must contain {γ 1, γ 3} (resp., {γ 2, γ 4}). Figure 5(c) illustrates the case when P contains γ 1. If γ 2 is in P, then P must contain {γ 2, γ 3}. In all the above scenario the number of convex vertices along the top bounary of the c-rectangle is 11. 9

10 t 1 t 1 t 2 t 2 t 3 t 4 t 3 (a) (b) (c) () (e) (f) Fig. 5. Illustration for the clause gaget. Case 2 (α = 2): If P contains {γ 1, γ 3 }, then either γ 1 or γ 4 must be in P. Otherwise, P contains either {γ 1, γ 2 } or {γ 2, γ 3 }. If that P contains {γ 1, γ 2 }, as in Figure 5(), then γ 3 must lie in P. In the remaining case, γ 2 must lie in P. Therefore, also in this case the number of convex vertices along the top bounary of the c-rectangle is 11. Case 3 (α = 3): In this scenario P cannot contain any of γ 1,..., γ 4. Therefore, as shown in Figure 5(e), the number of convex vertices along the top bounary of the c-rectangle is 11. As a consequence we obtain the following lemma. Lemma 3. If a clause is satisfie, then any optimal subsuming polygon reuces exactly three convex vertex from the corresponing c-rectangle. 3.3 Reuction Although we have alreay escribe the variable an clause gagets, the optimal subsuming polygon still may come up with some unexpecte optimization that interferes with the convex corner count in our harness proof. Figure 6(left) illustrates one such example. Therefore, we replace each convex corner that oes not correspon to the tips by a small polyline with alternating convex an reflex corners, as shown Figure 6(right). We now prove the NP-harness of computing optimal subsuming polygon. Theorem 3. Fining an optimal subsuming polygon is NP-har. Proof. Let I = (X, C) be an instance of the monotone planar 3-SAT an let P I be the corresponing instance of MIN-CONVEX-SUBSUMING-POLYGON. Let λ be the number of convex vertices in P I. We now show that I amits a satisfying truth assignment if an only if P I can be subsume using a polygon having at most λ X C 2 3 C convex vertices. 10

11 Fig. 6. Refinement of P I. First assume that I amits a satisfying truth assignment. For each variable x, we choose either the top-right caps or the top-left caps epening on whether x is assigne true or false. Consequently, we save at least X C 2 convex vertices. Consier any clause c C. Since c is satisfie, one or more of its variables are assigne true. Therefore, for each positive (resp., negative) clause, we can have one or more ownwar (resp., upwar) caps that enter into the v-rectangles. By Lemma 3, we can save at least three convex vertices from each c-rectangle. Therefore, we can fin a subsuming polygon with at most λ X C 2 3 C convex vertices. Assume now that some polygon P with at most λ X C 2 3 C convex vertices can subsume P I. We now fin a satisfying truth assignment for I. Note that the maximum number of convex vertices that can be reuce from the c-rectangles is at most 3 C. Therefore, P must reuce at least C 2 convex vertices from each v-rectangle. Recall that in each v-rectangle, either the top-right or the top-left caps can be chosen in the solution, but not both. Therefore, the v-rectangles cannot help reucing more than X C 2 convex vertices. If P contains the top-right caps of the v-rectangle, then we set the corresponing variable to true, otherwise, we set it to false. Since P has at most λ X C 2 3 C convex vertices, an each c-rectangle can help to reuce at most 3 convex vertices (Lemma 3), P must have at least one cap from γ 1, γ 2, γ 3 at each c-rectangle. Therefore, each clause must be satisfie. Recall that the ownwar (resp., upwar) caps coming from ege tunnels are esigne carefully to have conflict with the top-left (resp., top-right) caps of v-variables. Since top-left an top-right caps of v-variables are conflicting, the truth assignment of each variable is consistent in all the clauses that contains it. 4 Conclusion In this paper we have evelope a polynomial-time algorithm that can compute optimal subsuming polygons for a given simple polygon in restricte settings. On the other han, if the polygon contains holes, then we show the problem of computing an optimal subsuming polygon is NP-har. Therefore, the question of whether the problem is polynomial-time solvable for simple polygons, remains open. Our algorithm can fin an optimal solution if the optimal subsuming polygon lies on some prescribe arrangement of lines, e.g., A e (P ) or A v (P ). The running time of our algorithm epens on the length of the subsuming chains, i.e., the running time is polynomial if the subsuming chains are of constant length. However, there exist polygons 11

12 Fig. 7. Illustration for the case when the optimal subsuming polygon contains a subsuming chain of length Ω(n). The subsuming chain is shown in bol. whose optimal subsuming polygons contain subsuming chains of length Ω(n). Figure 7 illustrates such an example optimal solution that is lying on A e (P ). Therefore, it woul be interesting to fin algorithms whose running time is polynomial in the size of A e (P ) or A v (P ). Another interesting research irection woul be to examine whether there exists a goo approximation algorithm for the problem. References 1. O. Aichholzer, T. Hackl, M. Korman, A. Pilz, an B. Vogtenhuber. Geoesic-preserving polygon simplification. Int. J. Comp. Geom. & Appl., 24(4): , L. Arge, L. Deleuran, T. Mølhave, M. Revsbæk, an J. Truelsen. Simplifying massive contour maps. In Proc. ESA, volume 7501 of LNCS, pages Springer, M. e Berg an A. Khosravi. Optimal binary space partitions for segments in the plane. Int. J. Comp. Geom. & Appl., 22(3): , D. H. Douglas an T. K. Peucker. Algorithm for the reuction of the number of points require to represent a line or its caricature. The Canaian Cartographer, 10(2): , L. J. Guibas, J. Hershberger, J. S. B. Mitchell, an J. Snoeyink. Approximating polygons an subivisions with minimum link paths. Int. J. Comp. Geom. & Appl., 3(4): , J. M. Keil, J. S. B. Mitchell, D. Prahan, an M. Vatshelle. An algorithm for the maximum weight inepenent set problem on outerstring graphs. In Proc. CCCG, pages 2 7, D. Kempe. On the complexity of the reflections game, kempe/publications/reflections.pf. 8. W. A. Mackaness, A. Ruas, an L. T. Sarjakoski. Generalisation of Geographic Information: Cartographic Moelling an Applications. Elsevier, H. Ratschek an J. Rokne. Geometric Computations with Interval an New Robust Methos: Applications in Computer Graphics, GIS an Computational Geometry. Horwoo Publishing,