On Plane Constrained Bounded-Degree Spanners

Transcription

1 Algorithmica manscript No. (ill be inserted by the editor) 1 On Plane Constrained Bonded-Degree Spanners 2 3 Prosenjit Bose Rolf Fagerberg André an Renssen Sander Verdonschot 4 5 Receied: date / Accepted: date Abstract Let P be a finite set of points in the plane and S a set of non-crossing line segments ith endpoints in P. The isibility graph of P ith respect to S, denoted Vis(P,S), has ertex set P and an edge for each pair of ertices, in P for hich no line segment of S properly intersects. We sho that the constrained half-θ 6 -graph P. Bose School of Compter Science Carleton Uniersity 5302 Herzberg Laboratories 1125 Colonel By Drie, Ottaa, Ontario K1S 5B6, Canada Tel.: (613) ext Fax: (613) jit@scs.carleton.ca R. Fagerberg Department of Mathematics and Compter Science Uniersity of Sothern Denmark Campsej 55 DK-5230 Odense M Denmark Tel.: rolf@imada.sd.dk A. an Renssen JST, ERATO Kaarabayashi Large Graph Project National Institte of Informatics Hitotsbashi, Chiyoda-k, Tokyo , Japan andre@nii.ac.jp S. Verdonschot School of Compter Science Carleton Uniersity 5302 Herzberg Laboratories 1125 Colonel By Drie, Ottaa, Ontario K1S 5B6, Canada sander@cg.scs.carleton.ca

2 2 Prosenjit Bose et al (hich is identical to the constrained Delanay graph hose empty isible region is an eqilateral triangle) is a plane 2-spanner of Vis(P,S). We then sho ho to constrct a plane 6-spanner of Vis(P,S) ith maximm degree 6 + c, here c is the maximm nmber of segments of S incident to a ertex. Keyords Plane Spanners Bonded-Degree Constraints Visibility Graph Introdction A geometric graph G is a graph hose ertices are points in the plane and hose edges are line segments beteen pairs of ertices. A graph G is called plane if no to edges intersect properly. Eery edge is eighted by the Eclidean distance beteen its endpoints. The distance beteen to ertices and in G, denoted by d G (,) or simply d(,) hen G is clear from the context, is defined as the sm of the eights of the edges along the shortest path beteen and in G. A sbgraph H of G is a t-spanner of G (for t 1) if for each pair of ertices and, d H (,) t d G (,). The smallest ale t for hich H is a t-spanner is the spanning ratio or stretch factor of H. The graph G is referred to as the nderlying graph of H. The spanning properties of arios geometric graphs hae been stdied extensiely in the literatre (see [6, 9] for a comprehensie oerie of the topic). Hoeer, most of the research has focsed on constrcting spanners here the nderlying graph is the complete Eclidean geometric graph. We stdy this problem in a more general setting ith the introdction of line segment constraints. Specifically, let P be a set of points in the plane and let S be a set of line segments ith endpoints in P, ith no to line segments intersecting properly. The line segments of S are called constraints. To ertices and can see each other if and only if either the line segment does not properly intersect any constraint or is itself a constraint. If to ertices and can see each other, the line segment is a isibility edge. The isibility graph of P ith respect to a set of constraints S, denoted Vis(P,S), has P as ertex set and all isibility edges as edge set. In other ords, it is the complete graph on P mins all edges that properly intersect one or more constraints in S. This setting has been stdied extensiely ithin the context of motion planning amid obstacles. Clarkson [7] as one of the first to stdy this problem and shoed ho to constrct a linear-sized (1 + ε)-spanner of Vis(P, S). Sbseqently, Das [8] shoed ho to constrct a spanner of Vis(P,S) ith constant spanning ratio and constant degree. Bose and Keil [4] shoed that the Constrained Delanay Trianglation is a 2.42-spanner of Vis(P,S). In this article, e sho that the constrained half-θ 6 - graph (hich is identical to the constrained Delanay graph hose empty isible region is an eqilateral triangle) is a plane 2-spanner of Vis(P, S) by generalizing the approach sed by Bose et al. [3]. A key difficlty in proing the latter stems from the fact that the constrained Delanay graph is not necessarily a trianglation (see Figre 1). We then generalize the elegant constrction of Bonichon et al. [2] to sho ho to constrct a plane 6-spanner of Vis(P,S) ith maximm degree 6 + c, here c = max{c() P} and c() is the nmber of constraints incident to a ertex.

3 On Plane Constrained Bonded-Degree Spanners 3 Fig. 1 The constrained half-θ 6 -graph is not necessarily a trianglation. The thick line segment represents a constraint Preliminaries We define a cone C to be the region in the plane beteen to rays originating from a ertex referred to as the apex of the cone. We let six rays originate from each ertex, ith angles to the positie x-axis being mltiples of π/3 (see Figre 2). Each pair of consectie rays defines a cone. For ease of exposition, e only consider point sets in general position: no to points define a line parallel to one of the rays that define the cones and no three points are collinear. These assmptions imply that e can consider the cones to be open. If a set of points is not in general position, one can easily find a sitable rotation of the point set to pt it in general position. C 0,1 C 0,0 C 1,2 C 0 C 2 C 1 C 2,0 C 1,1 C 1,0 C 1,0 C 1 C 0 C 2 C 1,1 C 2,0 C 0,0 Fig. 2 The cones haing apex Fig. 3 The sbcones haing apex. Constraints are shon as thick line segments Let (C 1,C 0,C 2,C 1,C 0,C 2 ) be the seqence of cones in conterclockise order starting from the positie x-axis. The cones C 0, C 1, and C 2 are called positie cones and C 0, C 1, and C 2 are called negatie cones. By sing addition and sbtraction modlo 3 on the indices, positie cone C i has negatie cone C i+1 as clockise next cone and negatie cone C i 1 as conterclockise next cone. A similar statement holds for

4 4 Prosenjit Bose et al negatie cones. We se Ci and C j to denote cones C i and C j ith apex. Note that for any to ertices and, Ci if and only if C i. 68 Let ertex be an endpoint of a constraint c and let the other endpoint lie 69 in cone Ci. The lines throgh all sch constraints c split C i into seeral parts. We call these parts sbcones and denote the j-th sbcone of Ci by Ci, j, nmbered in 71 conterclockise order (see Figre 3). When a constraint c = (,) splits a cone of 72 into to sbcones, e define to lie in both of these sbcones. We call a sbcone 73 of a positie cone a positie sbcone and a sbcone of a negatie cone a negatie 74 sbcone. We consider a cone that is not split to be a single sbcone We no introdce the constrained half-θ 6 -graph, a generalized ersion of the half-θ 6 -graph as described by Bonichon et al. [1]: for each positie sbcone of each ertex, add an edge from to the closest ertex in that sbcone that can see, here distance is measred along the bisector of the original cone (not the sbcone) (see Figre 4). More formally, e add an edge beteen to ertices and if can see, Ci, j, and for all ertices C i, j that can see,, here and denote the projection of and on the bisector of Ci and xy denotes the length of the line segment beteen to ertices x and y. Note that or assmption of general position implies that each ertex adds at most one edge to the graph for each of its positie sbcones. a m b α Fig. 4 Three ertices are projected onto the bisector of a cone of. Vertex is the closest ertex in the left sbcone and is the closest ertex in the right sbcone Fig. 5 Canonical triangle T Gien a ertex in a positie cone C i of ertex, e define the canonical triangle T to be the triangle defined by the borders of Ci and the line throgh perpendiclar to the bisector of Ci (see Figre 5). Note that for each pair of ertices there exists a niqe canonical triangle. We say that a region is empty if it does not contain any ertices.

5 On Plane Constrained Bonded-Degree Spanners Spanning Ratio of the Constrained Half-θ 6 -Graph In this section e sho that the constrained half-θ 6 -graph is a plane 2-spanner of the isibility graph Vis(P, S). To do this, e first proe a property of isibility graphs. Recall that a region is empty if it does not contain any ertices. Lemma 1 Let,, and be three arbitrary points in the plane sch that and are isibility edges and is not the endpoint of a constraint intersecting the interior of triangle. Then there exists a conex chain of isibility edges from to in triangle, sch that the polygon defined by, and the conex chain is empty and does not contain any constraints. Proof Let Q be the set of ertices of Vis(P,S) inside triangle. If Q is empty, no constraint can cross, since one of its endpoints old hae to be inside, so or conex chain is simply. Otherise, e bild the conex hll of Q {,}. Note that is part of the conex hll since Q lies inside to one side of the line throgh. When e remoe this edge, e get a conex chain from to in triangle. By the definition of a conex hll, the polygon defined by, and the conex chain is empty. x y Fig. 6 A conex chain from to and intersections and of the triangle and the line throgh x and y Next, e sho that to consectie ertices x and y along the conex chain can see each other. Let be the intersection of and the line throgh x and y and let be the intersection of and the line throgh x and y (see Figre 6). Since is not the endpoint of a constraint intersecting the interior of triangle and, by constrction, both and can see, any constraint crossing xy old need to hae an endpoint inside. Bt the polygon defined by, and the conex chain is empty, so this is not possible. Therefore x can see y. Finally, since the polygon defined by, and the conex chain is empty and consists of isibility edges, any constraint intersecting its interior needs to hae as an endpoint, hich is not alloed. Hence, the polygon does not contain any constraints. Theorem 1 Let and be ertices, ith in a positie cone of, sch that is a isibility edge. Let m be the midpoint of the side of T opposing, and let α be the nsigned angle beteen the lines and m. There exists a path connecting and in the constrained half-θ 6 -graph of length at most ( 3 cosα + sinα) that lies inside T.

6 6 Prosenjit Bose et al. Proof We assme ithot loss of generality that C0, j. We proe the theorem 119 by indction on the area of T. Formally, e perform indction on the rank, hen 120 ordered by area, of the triangles T xy for all pairs of ertices x and y that can see each 121 other. Let δ(x,y) denote the length of the shortest path from x to y in the constrained 122 half-θ 6 -graph that lies inside T xy. Let a and b be the pper left and right corner of T, 123 and let A and B be the triangles a and b (see Figre 7). Or indctie hypothesis 124 is the folloing: If A is empty, then δ(, ) b + b. If B is empty, then δ(, ) a + a. If neither A nor B is empty, then δ(, ) max{ a + a, b + b }. We first note that this indction hypothesis implies the theorem: sing the side of T as the nit of length, e hae that δ(,) ( 3 cosα + sinα) (see Figre 8). a b a m b a b A B α 2 1 a 0 0 b 0 Fig. 7 Triangles A and B Fig. 8 Canonical triangle T Fig. 9 Conex chain from 0 to Base case: Triangle T has minimal area. Since the triangle is a smallest canonical triangle, is the closest ertex to in its positie sbcone. Hence the edge is in the constrained half-θ 6 -graph, and δ(,) =. From the triangle ineqality, e hae that min{ a + a, b + b }, so the indction hypothesis holds. Indction step: We assme that the indction hypothesis holds for all pairs of ertices that can see each other and hae a canonical triangle hose area is smaller than the area of T. If is an edge in the constrained half-θ 6 -graph, the indction hypothesis follos by the same argment as in the base case. If there is no edge beteen and, let 0 be the isible ertex closest to in the positie sbcone containing, and let a 0 and b 0 be the pper left and right corner of T 0 (see Figre 9). By definition, δ(,) 0 + δ( 0,), and by the triangle ineqality, 0 min{ a 0 + a 0 0, b 0 + b 0 0 }. We assme ithot loss of generality that 0 lies to the left of, hich means that A is not empty. Since and 0 are isibility edges, by applying Lemma 1 to triangle 0, a conex chain 0,..., k = of isibility edges connecting 0 and exists (see Figre 9). Note that, since 0 is the closest isible ertex to, eery ertex along the conex chain lies aboe the horizontal line throgh 0.

7 On Plane Constrained Bonded-Degree Spanners When looking at to consectie ertices i 1 and i along the conex chain, there are three types of configrations: (i) i 1 C i 1, (ii) i C i 1 0 and i lies to the right of or has the same x-coordinate as i 1, (iii) i C i 1 0 and i lies to the left of i 1. Let A i = i 1 a i i and B i = i 1 b i i, the ertices a i and b i ill be defined for each case. By conexity, the direction of i i+1 is rotating conterclockise for increasing i. Ths, these configrations occr in the order Type (i), Type (ii), and Type (iii) along the conex chain from 0 to. We bond δ( i 1, i ) as follos (see Figre 10): Type (i): If i 1 C i 1, let a i and b i be the pper left and loer corner of T i i 1. Triangle B i lies beteen the conex chain and, so it mst be empty by Lemma 1. Since i can see i 1 and T i i 1 has smaller area than T, the indction hypothesis gies that δ( i 1, i ) is at most i 1 a i + a i i. a i i a i i b i a i i b i A i i 1 A i B i A i B i B i b i (i) i 1 (ii) i 1 (iii) Fig. 10 Charging the three types of configrations Type (ii): If i C i 1 0, let a i and b i be the left and right corner of T i 1 i. Since i can see i 1 and T i 1 i has smaller area than T, the indction hypothesis applies. Whether A i and B i are empty or not, δ( i 1, i ) is at most max{ i 1 a i + a i i, i 1 b i + b i i }. Since i lies to the right of or has the same x-coordinate as i 1, e kno i 1 a i + a i i i 1 b i + b i i, so δ( i 1, i ) is at most i 1 a i + a i i. Type (iii): If i C i 1 0 and i lies to the left of i 1, let a i and b i be the left and right corner of T i 1 i. Since i can see i 1 and T i 1 i has smaller area than T, e can apply the indction hypothesis. Ths, if B i is empty, δ( i 1, i ) is at most i 1 a i + a i i and if B i is not empty, δ( i 1, i ) is at most i 1 b i + b i i. Recall that a and b are the pper left and right corner of T and that B is the triangle b (see Figre 7). To complete the proof, e consider three cases: (a) a π/2, (b) a > π/2 and B is empty, (c) a > π/2 and B is not empty. Case (a): If a π/2, the conex chain cannot contain any Type (iii) configrations: for Type (iii) configrations to occr, i needs to lie to the left of i 1. Hoeer, by constrction, i lies to the right of the line throgh i 1 and. Hence, since a i 1 < a π/2, i lies to the right of i 1. We can no bond δ(,) as follos sing the bonds on Type (i) and Type (ii) configrations otlined aboe

8 8 Prosenjit Bose et al (see Figre 11): δ(,) 0 + k i=1 a 0 + a δ( i 1, i ) k i=1 = a + a We see that the latter is eqal to a + a as reqired. ( i 1 a i + a i i ) a i a i i Fig. 11 Visalization of the paths (thick lines) in the ineqalities of case (a) Case (b): If a > π/2 and B is empty, the conex chain can contain Type (iii) configrations. Hoeer, since B is empty and the area beteen the conex chain and is empty (by Lemma 1), all triangles B i are also empty. Hence sing the indction hypothesis, δ( i 1, i ) is at most i 1 a i + a i i for all i. Using these bonds on the lengths of the paths beteen the ertices along the conex chain, e can bond δ(,) as in the preios case. Therefore, δ(,) a + a as reqired. Case (c): If a > π/2 and B is not empty, the conex chain can contain Type (iii) configrations. Since B is not empty, the triangles B i need not be empty. Recall that 0 lies in A, hence neither A nor B are empty. Therefore, it sffices to proe that δ(,) max{ a + a, b + b } = b + b. Let T j j+1 be the first Type (iii) configration along the conex chain (if it has any), let a and b be the pper left and right corner of T j, and let b be the pper right corner of T j (see Figre 12). Note that since a > π/2 and j lies to the left of, a j is smaller than b j. δ(,) 0 + k i=1 a 0 + a δ( i 1, i ) j i=1 ( i 1 a i + a i i ) + = a + a j + j b + b k i= j+1 ( i 1 b i + b i i ) b + b j + j b + b = b + b

9 On Plane Constrained Bonded-Degree Spanners 9 b j a j b b j b Fig. 12 Visalization of the paths (thick lines) in the ineqalities of case (c) Since the expression 3 cosα + sinα is increasing for α [0,π/6], the maximm ale is attained by inserting the extreme ale π/6. This leads to the folloing corollary. Corollary 1 The constrained half-θ 6 -graph is a 2-spanner of the isibility graph. Next, e proe that the constrained half-θ 6 -graph is plane. Lemma 2 Let,, x, and y be for distinct ertices sch that the to canonical triangles T and T xy intersect. Then at least one of the corners of one canonical triangle is contained in the other canonical triangle. Proof If one triangle contains the other triangle, it contains all of its corners. Therefore e focs on the case here neither triangle contains the other. By definition, the pper bondaries of T and T xy are parallel, the left bondaries of T and T xy are parallel, and the right bondaries of T and T xy are parallel. Becase e assme that no to ertices define a line parallel to one of the rays that define the cones, e assme, ithot loss of generality, that the pper bondary of T lies belo the pper bondary of T xy. The pper bondary of T mst lie aboe the loer corner of T xy, since otherise the triangles do not intersect. If the pper left (right) corner of T lies to the right (left) of the right (left) bondary of T xy, the triangles cannot intersect. Hence, either one of the pper corners of T is contained in T xy or the pper bondary of T intersects both the left and right bondary of T xy. In the latter case, the fact that the left bondaries of T and T xy are parallel and the right bondaries of T and T xy are parallel, implies that the loer corner of T xy is contained in T.

10 10 Prosenjit Bose et al Lemma 3 The constrained half-θ 6 -graph is plane. Proof We proe the lemma by contradiction. Assme that to edges and xy cross at a point p. Since the to edges are contained in their canonical triangles, these triangles mst intersect. By Lemma 2 e kno that at least one of the corners of one triangle lies inside the other. We focs on the case here the pper right corner of T xy lies inside T. The other cases are analogos. Since and xy cross, this also means that either x or y mst lie in T. y p x Fig. 13 Edges and xy intersect at point p Assme ithot loss of generality that C0, 210 j and y T (see Figre 13). If y C0, j, e look at triangle py. Since both and y can see p, e get by Lemma that either can see y or py contains a ertex. In both cases, can see a ertex in 213 this sbcone that is closer than, contradicting the existence of the edge. If y / C0, j, there exists a constraint z sch that lies to one side of the line throgh z and y lies on the other side. Since this constraint cannot cross yp, z lies inside py and is therefore closer to than. Since by definition z can see, this also contradicts the existence of Bonding the Maximm Degree In this section, e sho ho to constrct a bonded degree sbgraph G 9 of the constrained half-θ 6 -graph that is a 6-spanner of the isibility graph. Gien a ertex and one of its negatie sbcones, e define the canonical seqence of this sbcone as the ertices in this sbcone that are neighbors of in the constrained half-θ 6 -graph, in conterclockise order (see Figre 14). These ertices all hae as their closest isible ertex in a positie sbcone. The canonical path is defined by connecting consectie ertices in the canonical seqence. This definition differs slightly from the one sed by Bonichon et al. [2]. To constrct G 9, e start ith a graph ith ertex set P and no edges. Then for each negatie sbcone of each ertex P, e add the canonical path and an edge beteen and the closest ertex along this path, here distance is measred sing the projections of the ertices onto the bisector of the cone containing the sbcone. A gien edge may be added by seeral ertices, bt it appears only once in G 9.

11 On Plane Constrained Bonded-Degree Spanners Fig. 14 The edges that are added to G 9 for a negatie sbcone of a ertex ith canonical seqence 1, 2, 3 and This constrction is similar to the constrction of the nconstrained degree-9 halfθ 6 -graph described by Bonichon et al. [2]. We proceed to proe that G 9 is a spanning sbgraph of the constrained half-θ 6 -graph ith spanning ratio 3. Lemma 4 G 9 is a sbgraph of the constrained half-θ 6 -graph. 244 Proof Gien a ertex, e look at one of its negatie sbcones, say C 232 0, j. The edges 233 added to G 9 for this sbcone can be diided into to types: edges of the canonical 234 path, and the edge beteen and the closest ertex along the canonical path. Since 235 eery ertex along the canonical path is by definition connected to in the constrained 236 half-θ 6 -graph, it remains to sho that the edges of the canonical path are part of the 237 constrained half-θ 6 -graph. Let and be to consectie ertices in the canonical path of C 238 0, j, ith 239 before in conterclockise order. By applying Lemma 1 on the isibility edges 240 and, e get a conex chain = x 0,x 1,...,x k 1,x k = of k 1 isibility edges, 241 hich together ith and form a polygon P empty of ertices and constraints. 242 Since P is empty, is not the endpoint of a constraint lying beteen and x Hence, x 1 cannot be in cone C0, otherise x 1 old be closer to than in the sbcone of that contains. Similarly, x k 1 cannot lie in cone C0. By conexity of 245 the chain, this implies that no ertex on the chain can lie in cone C 0 of another ertex 246 on the chain. Hence, since P is empty, all ertices x i can see. 247 We first sho that k = 1, i.e. that the chain is jst the line. We proe this 248 by contradiction, so assme that k > 1. Hence, there is at least one ertex x i ith 0 < i < k. As sch a ertex is not part of the canonical path in C 249 0, j, it mst see a closest ertex y different from in the sbcone of C x i 0 that contains. As ertices on empty, y mst therefore lie strictly otside of P, and yx i mst properly intersect either 253 or. Bt this contradicts the planarity of the constrained half-θ 6 -graph, as yx i, 254, and old all be edges of this graph. Hence, k = 1 and the chain is a single 255 isibility edge. 256 It remains to sho that is an edge of the constrained half-θ 6 -graph. Assme ithot loss of generality that lies in C2 (the case that lies in C 1 is similar). We need to sho that is the closest isible ertex in sbcone C2, j. We proe this by contradiction, so assme another ertex x in C2, 259 j is the closest. Vertex x lies in T, 260 hich is partitioned into a part inside P, a part to the right of, and a part belo

12 12 Prosenjit Bose et al (see Figre 15). If x lies to the right of, e old hae intersecting edges x and, contradicting planarity of the constrained half-θ 6 -graph. As P is empty, x mst lie belo (see Figre 15). x Fig. 15 T is partitioned into a part inside P (light gray), a part to the right of (hite), and a part belo (dark gray) Applying Lemma 1 on the isibility edges x and, e get a conex chain x = x 0,x 1,...,x k 1,x k = of isibility edges and an empty polygon Q. Vertex x 1 cannot lie in C0 x, as this old contradict that x is the closest isible ertex to in. Hence, since P and Q are empty, x can see. Since and are to consectie C 2, j ertices in the canonical seqence of C 0, j, x is not part of this canonical seqence. So it mst see a closest ertex y different from in the sbcone of C0 x that contains. Neither nor the conex chain from x to lie in C0 x. As P and Q are empty, xy mst properly intersect either or, contradicting the planarity of the constrained half-θ 6 -graph For ftre reference, e note that dring the proof of Lemma 4 the folloing to properties ere shon. Corollary 2 Let,, and be three ertices sch that and are neighbors along a canonical path of in C i. Vertex cannot lie in Ci or C i. Corollary 3 Let,, and be three ertices sch that and are neighbors along a canonical path of in C i. Triangle is empty and does not contain any constraints. Theorem 2 G 9 is a 3-spanner of the constrained half-θ 6 -graph. Proof We proe the theorem by shoing that for eery edge in the constrained half-θ 6 -graph, here lies in a negatie cone of, G 9 contains a spanning path beteen and of length at most 3. This path ill consist of a part of the canonical path in the sbcone of that contains pls the edge beteen and the closest canonical ertex in that sbcone. We assme ithot loss of generality that C 0. Let 0 be the ertex closest to on the canonical path in the sbcone C 0, j that contains and let 0, 1,..., k = be the ertices along the canonical path from 0 to (see Figre 16). Let l j and r j denote the rays defining the left and right bondaries of C j 0 for 0 j k and let r denote

13 On Plane Constrained Bonded-Degree Spanners 13 b m 0 m 1 m 2 m 3 m4 a Fig. 16 Bonding the length of the canonical path the ray defining the right bondary of C 0 (as seen from ). Let m j be the intersection of l j and r j 1, for 1 j k, and let m 0 be the intersection of l 0 and r. Let a be the intersection of r and the horizontal line throgh and let b be the intersection of l k and r. The length of the path beteen and in G 9 can no be bonded as follos: d G9 (,) 0 + k j=1 j 1 j m 0 + m = m 0 + k j=0 k j=1 m j j + m j j + k j=1 k j=1 j 1 m j j 1 m j Since lies in C 0 of each of the ertices along the canonical path, all m j j project onto b and all j 1 m j project onto m 0 b, hen projecting along lines parallel to the bondaries of C 0 instead of sing orthogonal projections. By Corollary 2 no edge on the canonical path can lie in C 0 of one of its endpoints, hence the projections of m j j onto b do not oerlap. For the same reason, the projections of j 1 m j onto m 0 b do not oerlap. Hence, e hae that k j=0 m j j = b and k j=1 j 1m j = m 0 b. d G9 (,) = m 0 + k j=0 m j j + = m 0 + b + m 0 b a + 2 a k j=1 j 1 m j

14 14 Prosenjit Bose et al Let α be a. Using some basic trigonometry, e get a = cosα + sinα/ 3 and a = 2 sinα/ 3. Ths the spanning ratio can be expressed as: d G9 (,) cosα + 5 sinα 3 Since this is a non-decreasing fnction in α for 0 < α π/3, its maximm ale is obtained hen α = π/3, here the spanning ratio is It follos from Theorems 1 and 2 that G 9 is a 6-spanner of the isibility graph. Corollary 4 G 9 is a 6-spanner of the isibility graph. To bond the degrees of the ertices, e se a charging scheme that charges the edges of a ertex to its cones. Smming the charge for all cones of a ertex then bonds its degree. Recalling that the edges of G 9 are generated by canonical paths, consider a canonical path in C i, j, created by a ertex. We se to indicate an arbitrary ertex along the canonical path, and e let be the closest ertex to along the canonical path. The edges of G 9 generated by this canonical path are charged to cones as follos: The edge is charged to C 301 i and to C An edge of the canonical path that lies in C i+1 is charged to Ci. 302 An edge of the canonical path that lies in C i 1 is charged to Ci. 303 An edge of the canonical path that lies in Ci+1 is charged to 304 C i 1. An edge of the canonical path that lies in Ci 1 is charged to 305 C i+1. i Essentially, the edge beteen and is charged to the cones that contain it and edges along the canonical path are charged to the adjacent cone that is closer to the cone of that contains. In other ords, all charges are shifted one cone toards the positie cone containing (see Figre 17). Fig. 17 To edges of a canonical path and the associated charges By Corollary 2, no edge on the canonical path can lie in Ci or C i, so the charging scheme aboe is exhastie. Note that each edge is charged to both of its endpoints and therefore the charge on a ertex is an pper bond on its degree (only an pper bond, since an edge can be generated and charged by seeral canonical paths).

15 On Plane Constrained Bonded-Degree Spanners Lemma 5 Let be a ertex that is incident to at least to constraints in the same positie cone Ci. Let C i, j be a sbcone beteen to constraints and let be the closest isible ertex in this sbcone. Let C 316 i,k be the sbcone of that contains and (hen is a constraint) intersects Ci, 317 j. Then is the only ertex on the canonical path of C 318 i,k. Proof Let 1 and 2 be the to constraints beteen hich sbcone Ci, j lies. By applying Lemma 1 on these isibility edges, e get a conex chain 1 = x 0,x 1,...,x k = 2 hich together ith 1 and 2 form a polygon P Ci, j empty of ertices and constraints. Since is the closest ertex isible to inside Ci, j, mst be the ertex on this chain closest to. In particlar, it is at least as close to as 1 and 2. Since 1 and 2 are constraints and P is empty, there can be no ertex other than in C i,k from hich is isible. Hence, is the only ertex on the canonical path of C i,k Lemma 6 Each positie cone C i of a ertex has a charge of at most max{2,c i ()+ 1}, here c i () is the nmber of incident constraints in Ci. 321 Proof Let be a ertex sch that is part of the canonical path of. We first sho that if this canonical path charges Ci, then mst lie in 322 C i. Assme lies in C j, j i. Since all charges of this canonical path are shifted one cone toards C, a charge to 323 C i old hae to come from C 324 j. Hoeer, by Corollary 2, no edge on the canonical path of a ertex in C j can lie in 325 C j. Next, e obsere that there can be only one sch ertex for each sbcone of Ci. 327 This follos becase is only part of canonical paths of ertices of hich is an 328 edge in the constrained half-θ 6 -graph, and there is at most one edge for each positie 329 sbcone. 330 If Ci is a single sbcone and is not the closest ertex to on its canonical path, 331 Ci is charged for at most to edges along a single canonical path. Hence, its charge 332 is at most 2. If is the closest ertex to, the negatie cones adjacent to this positie 333 cone cannot contain any ertices of the canonical path. If they did, these ertices 334 old be closer to than is, as distance is measred sing the projection onto the 335 bisector of the cone of. Hence, if is the closest ertex to, the positie cone 336 containing is charged 1. Ths, hen the positie cone is a single sbcone, the cone 337 is charged 2 if it has an edge of the canonical path in each adjacent negatie cone, 338 and at most 1 otherise. 339 Next, e look at the case here Ci is not a single sbcone. For each sbcone, 340 except the first and last, the canonical path of the ertex from that sbcone consists 341 only of, by Lemma 5. Hence, e get a charge of 1 per sbcone and a charge of at 342 most c i () 1 in total for all sbcones except the first and last sbcone. We complete 343 the proof by shoing that the ertices of the first and the last sbcone can add a 344 charge of at most 1 each. Consider the first sbcone Ci,0. The argment for the last sbcone is symmetric. 346 If is the closest ertex to on its canonical path, the negatie cones adjacent to this 347 positie cone cannot contain any ertices of the canonical path, since these old be 348 closer to than is. Hence, the ertex of this sbcone adds a charge of 1. If is 349 not the closest ertex to, e arge that is the end of the canonical path of the j

16 16 Prosenjit Bose et al ertex of the sbcone, implying that can add a charge of at most 1: Let x be the other endpoint of the constraint that defines the sbcone. Since is the closest isible ertex in this sbcone of, it cannot lie frther from than x. If is x, constraint splits C i and only one of these to parts intersects the first sbcone of. Hence is the end of the canonical path of. If is not x, lies closer to than x. Any ertex y before (in conterclockise order) on the canonical path old hae to lie in Ci+1 or C i 1, since by Corollary 2, y cannot lie in Ci or C i. Since y mst also lie in C i to be on this canonical path, ertex is not isible from y de to the constraint x. Hence, no sch ertex y can exist on the canonical path, implying that is the end of the canonical path. Smming p all charges, each positie cone is charged at most c i ()+1 if c i () 1, and at most 2 otherise. Hence, a positie cone is charged at most max{2,c i () + 1}. Corollary 5 If the i-th positie cone of a ertex has a charge of c i () + 2, then c i () = 0, i.e. it does not contain any constraints haing as an endpoint in C i and is charged for to edges in the adjacent negatie cones. Lemma 7 Each negatie cone C i of a ertex has a charge of at most c i () + 1, here c i () is the nmber of incident constraints in C i. Proof A negatie cone of a ertex is charged by the edge to the closest ertex in each of its sbcones and it is charged by the to adjacent positie cones if edges of canonical paths lie in those cones (see Figre 18). We first sho that ertices that do not lie in the positie sbcones directly adjacent to C i cannot hae an edge inoling along their canonical paths. Let be a ertex that does not lie in a positie sbcone directly adjacent to C i and let x be the constraint closest to C i that defines the bondary of the sbcone of that contains. For to hae an edge along its canonical path that is charged to C i, it needs to lie frther from than x, since otherise no ertex creating sch an edge is isible to. Hoeer, this implies that old not connect to, ths it old not part of the canonical path of. Fig. 18 If is present, the negatie cone does not contain edges haing as endpoint As can only be part of the canonical path of a single ertex in each of its positie sbcones, e need to consider only the charges to C i from the canonical path created by the closest isible ertices in the to positie sbcones directly adjacent to C i. Let these ertices be and.

17 On Plane Constrained Bonded-Degree Spanners Next, e sho that eery negatie cone can be charged by at most one edge in 380 total from its adjacent positie cones. Sppose that lies in a positie cone of and 381 is part of the canonical path of. Then lies in a negatie cone of, hich 382 means that lies in a positie cone of and cannot be part of a canonical path for. It remains to sho that this negatie cone of cannot be charged by an edge 383 from a canonical path of a different ertex 384. Since forms a triangle in constrained half-θ 6 -graph and this graph is planar, no edge of 385 can cross any of the edges of. This implies that either and lie inside or and lie inside Hoeer, by Corollary 3, triangles xyz formed by a ertex x and to ertices y and z that are neighbors along the canonical path of x are empty. Therefore, and 388 cannot lie inside and and cannot lie inside 389. Ths eery negatie cone 390 charged by at most one edge in total from its adjacent positie cones. 391 Finally, e sho that if one of or is present, the negatie cone does not 392 hae an edge to the closest ertex in that cone and it contains no constraint that has 393 as an endpoint. We first sho that if one of or is present, the negatie cone 394 does not hae an edge to the closest ertex in that cone. We assme ithot loss of generality that is present, Ci 395 C i, and C i 1. Since and are neighbors 396 on the canonical path of, e kno that the triangle is part of the constrained 397 half-θ 6 -graph and, by Corollary 3, this triangle is empty. Frthermore, since is 398 an edge of the constrained half-θ 6 -graph and, by Lemma 3, the constrained half-θ graph is plane, cannot hae an edge to the closest ertex beyond. Hence the 400 negatie cone does not hae an edge to the closest ertex in that cone. By the same 401 argment, the negatie cone cannot contain a constraint that has as an endpoint. It follos that if this negatie cone contains no constraint that has as an endpoint, it is charged at most 1, by one of,, or the edge to the closest. Also, if this negatie cone does contain constraints that hae as an endpoint, it is not charged by edges in the adjacent positie cones and hence its charge is at most c i () + 1, one for the closest in each of its sbcones. 402 Theorem 3 Eery ertex in G 9 has degree at most c() + 9. Proof From Lemmas 6 and 7, each positie cone has charge at most c i () + 2 and each negatie cone has charge at most c i ()+1, here c i () and c i () are the nmber of constraints in the i-th positie and negatie cone. Since a ertex has three positie and three negatie cones and the c i () and c i () sm p to c(), this implies that the total degree of a ertex is at most c() Bonding the Maximm Degree Frther In this section, e sho ho to redce the maximm degree frther, reslting in a plane 6-spanner G 6 of the isibility graph in hich the degree of any node is bonded by c() + 6. By Lemmas 6 and 7 e see that if e can aoid the case here a positie cone gets a charge of c i () + 2, then eery cone is charged at most c i () + 1, for a total charge of c()+6. By Corollary 5, this case only happens hen a positie cone has c i () = 0 and is charged for to edges in the adjacent negatie cones. This sitation is depicted

18 18 Prosenjit Bose et al in Figre 19, here x,, and y are all on the canonical path of. We constrct G 6 by performing a transformation on G 9 for all positie cones in this sitation. x y x (a) y x (b) y Fig. 19 A positie cone haing charge 2 Fig. 20 Transforming G 9 (a) into G 6 (b) 413 We no describe the transformation. We assme ithot loss of generality that the positie cone in qestion is C0. We call a ertex the closest canonical ertex in 415 a negatie sbcone of hen, among the ertices of the canonical path of in that 416 sbcone, is closest to. 417 We first note that if x is the closest canonical ertex in one of the at most to sbcones of C 2 that contain it, the edge x is charged to C0, since x is an edge of the canonical path indced by, and it is also charged to cone C 419 2, since it is the 420 closest canonical ertex in one of its sbcones. Since e need to charge it only once to accont for the degree of, e can remoe the charge to C0, redcing its charge by as desired. Similarly, if y is the closest canonical ertex in one of the at most to sbcones of C 1 that contain it, it is charged to both C0 and 423 C 1, so e can redce the charge to C0 by 1. Therefore, e only perform a transformation if neither x nor y is 425 the closest canonical ertex in the sbcones of that contain them. 426 In that case, the transformation proceeds as follos. First, e add an edge be- 427 teen x and y. Next, e look at the seqence of ertices beteen and the closest 428 canonical ertex on the canonical path indced by. If this seqence incldes x, 429 e remoe y. Otherise e remoe x. Note that by Corollary 3, triangles x and 430 y are empty and do not contain any constraints and therefore the edge xy does not 431 intersect any constraints. 432 We assme ithot loss of generality that y is remoed. By remoing y and 433 adding xy, e redce the degree of at the cost of increasing the degree of x. Hence, e need to find a ay to balance the degree of x. Since x lies in C and the edge x is part of the constrained half-θ 6 -graph, x lies on a canonical path of in C and, since 436 x is not the closest canonical ertex to on this canonical path, x has a neighbor 437 along this canonical path. We claim that x is the last ertex along the canonical path of in C and ths is niqely defined. This follos becase for any ertex z later 439 than x along that canonical path, either z mst lie in triangle x, contradicting its 440 emptiness by Corollary 3, or the edges z and x of the constrained half-θ 6 -graph 441 mst intersect, contradicting its planarity by Lemma 3. To balance the degree of x, e remoe edge x, if lies in C x and is not the closest canonical ertex in a sbcone of C x that contains it. Otherise x is not remoed. The sitation before the

19 On Plane Constrained Bonded-Degree Spanners transformation is shon in Figre 20 (a) and the sitation after the transformation is shon in Figre 20 (b). A cred line segment denotes a part of a canonical path pls the edge from its closest canonical ertex. To constrct G 6, e apply this transformation on each positie cone matching the sitation aboe. Note that since edge is part of the constrained half-θ 6 -graph, hich is plane, and G 9 is a sbgraph of the constrained half-θ 6 -graph, the edges added by this transformation cannot be part of G 9 as they cross. Hence, since only edges of G 9 are remoed, there are no conflicts among the transformations of different cones, i.e. no cone ill add an edge that as remoed by another cone and ice ersa. Before e proe that this constrction yields a graph of maximm degree 6 + c, e first sho that the reslting graph is still a 3-spanner of the constrained half-θ 6 -graph. Lemma 8 Let x be an edge of G 9 and let x lie in a negatie cone C i of. If x is not the closest canonical ertex in either of the at most to sbcones of C i that contain it, then the edge x is sed by at most one canonical path. Proof We proe the lemma by contraposition. Assme that edge x is part of to canonical paths of to ertices and. For and x to be neighbors on a canonical path of and, these ertices need to lie in Ci Cx i+1 or C i 1 Cx i 1, by Corollary By Corollary 3 and planarity of the constrained half-θ 6 -graph, and cannot lie in the same region, hence one lies in Ci+1 Cx i+1 and one lies in C i 1 Cx i 1. We assme 463 ithot loss of generality that Ci+1 Cx i+1 and C i 1 Cx i 1 (see Figre 21). 464 Ths x and x form to disjoint triangles in the constrained half-θ 6 -graph and, by Corollary 3, both triangles are empty. Frthermore, since the constrained half-θ graph is plane, no edge from can cross x or x, making x the only edge of in C i. Therefore, x is the closest canonical ertex in any sbcone of C 468 i that contains it. x Fig. 21 If edge x is part of to canonical paths, x is the only neighbor of in the negatie cone of Lemma 9 For eery edge in the constrained half-θ 6 -graph, there exists a path in G 6 of length at most 3. Proof In the proof of Theorem 2 e shoed that for eery edge in the constrained half-θ 6 -graph, here lies in a negatie cone of, G 9 contains a spanning path beteen and of length at most 3, consisting of a part of the canonical path in the sbcone of that contains pls the edge beteen and the closest canonical ertex in that sbcone. We no sho that G 6 also contains a spanning path beteen and of length at most 3. Note that in the constrction, e neer remoe an edge x ith x being the closest canonical ertex in a negatie sbcone of. This means to things: 1) For any spanning path in G 9, its last edge still exists in G 6. 2) By Lemma 8, any remoed edge is

20 20 Prosenjit Bose et al part of a single canonical path, so e need to arge only abot this single canonical path and the spanning paths sing it. Dring the constrction of G 6, to types of edges are remoed: Type 1, represented by y in Figre 20, and Type 2, represented by x in Figre 20. We first sho that no edge remoal of either of these types remoes edge x in Figre 20. A Type 1 remoal that has as the middle ertex in the configration, as shon in Figre 20, is called centered at. A Type 1 remoal of y affects the single canonical path containing x and y (see Figre 20). We note that no Type 1 remoal inoling can be centered at x or y, since lies in a positie cone of both x and y and a Type 1 remoal reqires both neighbors of the center ertex to lie in negatie cones. This implies that Type 1 remoals are non-oerlapping (i.e. their configrations do not share edges) and, in particlar, it implies that edge x is not remoed by this type of remoal. A Type 2 remoal of x affects the canonical path that contains and x (see Figre 20). As arged dring the constrction of G 6, x is the last ertex along a canonical path of and the edge x is remoed if lies in a negatie cone of x and is not a closest canonical ertex to x. We no sho that edge x cannot be remoed by a Type 2 remoal: For it to be remoed, e need that either x lies in a negatie cone of and is the last ertex along this canonical path, or lies in a negatie cone of x and x is the last ertex along this canonical path. Hoeer, since is not the last ertex along the canonical path that contains and x (it is folloed by y) and does not lie in a negatie cone of x, neither condition is satisfied. No that e kno that for eery Type 1 remoal, edge x is still present in the final G 6, e look at the spanning paths in G 6. Eery spanning path present in G 9 can be affected by seeral non-oerlapping Type 1 remoals, as ell as by a Type 2 remoal at either end. By applying the triangle ineqality to Figre 20, it follos that xy x + y. Combined ith the fact that for eery Type 1 remoal, x is present in G 6, it follos that there still exists a spanning path beteen and any ertex along its canonical path, except possibly the last ertex x on either end, as the edge connecting x to its neighbor along the canonical path cold be remoed by a Type 2 remoal. Hoeer, e perform a Type 2 remoal only hen and x are part of a Type 1 configration centered at and x is the edge of this configration that as not remoed (see Figre 20, here acts as the node called in the Type 2 argment aboe). Frthermore, e shoed that in this case x is still present in G 6. Hence, there exists a spanning path of length at most 3 beteen and any ertex along its canonical path. Ths, e hae proen that for eery edge in the constrained half-θ 6 -graph, here lies in a negatie cone of, also G 6 contains a spanning path beteen and of length at most 3. Lemma 10 Eery ertex in G 6 has degree at most c() + 6. Proof To bond the degree, e look at the charges of the ertices. We proe that after the transformation each positie cone has charge at most c i () + 1 and each negatie cone has charge at most c i () + 1. This implies that the total degree of a ertex is at most c() + 6. Since the charge of the negatie cones is already at most c i () + 1, e focs on positie cones haing charge c i ()+2. By Corollary 5, this means that these cones hae charge 2 and c i () = 0.

21 On Plane Constrained Bonded-Degree Spanners 21 Let be a ertex sch that one of its positie cones Ci 522 has charge 2, let be the ertex hose canonical path charged 2 to Ci, and let x C i 1 and y C 523 i+1 be the 524 neighbors of on this canonical path (see Figre 19). If x or y is the closest canonical ertex in a sbcone of C i 1 or C 525 i+1, this edge has been charged to both that negatie cone and Ci 526. Hence e can remoe the charge to C i hile maintaining that the charge 527 is an pper bond on the degree of. If neither x nor y is the closest canonical ertex in a sbcone of C i 1 or C 528 i+1, edge 529 xy is added. We assme ithot loss of generality that edge y is remoed. Ths y need not be charged, decreasing the charge of Ci to 1. Since y as charged to C y 530 i 1 and this charge is remoed, e charge edge xy to C y 531 i 1. Ths the charge of y does not 532 change. 533 It remains to sho that e can charge xy to x. Recall that x lies on the canonical path of in C 534 i 1, is the last ertex on this canonical path, and has as neighbor on 535 this canonical path (see Figre 20). Since ertices x and x each form a trian- 536 gle of neighboring ertices on a canonical path in the constrained half-θ 6 -graph, by 537 Corollary 3 they are empty and do not contain any constraints. This implies that x is not the endpoint of any constraint in Ci 1 x 538. Hence, since x is the last ertex along the canonical path of, Ci 1 x 539 has charge at most 1 by Lemma 6 and Corollary 5. No, 540 consider the neighbor of x. Vertex can be in one of to cones ith respect to x: Ci+1 x and Cx i. If Ci+1 x, x is charged to Cx i. Ths the charge of Ci 1 x 541 is 0 and e 542 can charge xy to it. If C x i and is the closest canonical ertex to x in a sbcone of C x i, x has been charged to both Ci 1 x and Cx i. We can remoe that charge from Ci 1 x and instead charge xy to it, hile keeping the charge of Ci 1 x at 1. If Cx i and is not the closest canonical ertex to x in a sbcone of C x i that contains it, x as remoed dring the transformation. Since this edge as charged to Ci 1 x, e can no charge xy to Cx i 1, hile keeping the charge of Ci 1 x at Lemma 11 G 6 is a plane sbgraph of the isibility graph. Proof Since G 9 is a plane sbgraph of the isibility graph by Lemmas 3 and 4, only the edges added in the transformation from G 9 to G 6 can iolate the lemma. An added edge xy can potentially intersect edges of G 6 that are in the constrained half-θ 6 -graph, other edges that ere added (recall that added edges are not in the constrained halfθ 6 -graph, so these to cases are disjoint), and constraints. First, e consider intersections of xy ith edges of G 6 that are in the constrained half-θ 6 -graph. Since xy as added in the transformation, x, y, and are part of a canonical path of some ertex (see Figre 20). Ths, in the constrained half-θ 6 - graph x and y form to triangles, each containing a pair of neighboring ertices along the canonical path, hich are empty by Corollary 3. Since the constrained halfθ 6 -graph is plane and xy lies inside xy, the only edge of the constrained half-θ 6 - graph that can intersect xy is. We no arge that is not in G 6. By constrction, can only be part of G 9 if is the closest ertex to on this canonical path, or if are neighboring ertices along another canonical path of some ertex t. The former cannot be the case, by the conditions for adding xy in the transformation (see Figre 20). Assme the latter is the case. If Ci, then either t C i+1 C i+1 or