On Plane Constrained Bounded-Degree Spanners

Transcription

1 On Plane Constrained Bonded-Degree Spanners Prosenjit Bose 1, Rolf Fagerberg 2, André an Renssen 1, Sander Verdonschot 1 1 School of Compter Science, Carleton Uniersity, Ottaa, Canada. jit@scs.carleton.ca, arensse@connect.carleton.ca, sander@cg.scs.carleton.ca. 2 Dept. of Mathematics and Compter Science, Uniersity of Sothern Denmark. rolf@imada.sd.dk Abstract. Let P be a set of points in the plane and S a set of noncrossing 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 (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 adjacent to a ertex. 1 Introdction A Eclidean geometric graph G is a graph hose ertices are points in the plane and hose edges are line segments beteen pairs of points. Edges are eighted by their Eclidean length. The distance beteen to ertices and in G, denoted by d G (, ) or simply d(, ), is defined as the length of 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 ale t is the spanning ratio or stretch factor. The graph G is referred to as the nderlying graph of the t-spanner H. The spanning properties of arios geometric graphs hae been stdied extensiely in the literatre (see [6] 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 constraints sch that each constraint is a line segment beteen to ertices in P. The set of constraints is planar, i.e. no to constraints intersect properly. 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 Research spported in part by NSERC and the Danish Concil for Independent Research. De to space constraints, some proofs are omitted and aailable in the fll ersion of this paper.

2 2 On Plane Constrained Bonded-Degree Spanners 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 [4] as one of the first to stdy this setting in this context and shoed ho to constrct a linear-sized (1 + ɛ)-spanner of Vis(P, S). Sbseqently, Das [5] shoed ho to constrct a spanner of Vis(P, S) ith constant spanning ratio and constant degree. Bose and Keil [3] 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). A difficlty in proing the latter stems from the fact that the constrained Delanay graph is not necessarily a trianglation. 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. 2 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 Fig. 1). 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. 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 negatie cones. We se C i and C j to denote cones C i and C j ith apex. Note that for any to ertices and, C i if and only if C i. C 2,0 C 1,0 C 0,1 C 2 C 1,1 C 0,0 C 0 C 1 C 1 C 2 C 0 Fig. 1. The cones haing apex C 0,0 C 1,1 C 1,0 C 2,0 Fig. 2. The sbcones haing apex. Constraints are shon as thick line segments. Let ertex be an endpoint of a constraint c and let the other endpoint lie in cone Ci. The lines throgh all sch constraints c split C i into seeral

3 On Plane Constrained Bonded-Degree Spanners 3 parts. We call these parts sbcones and denote the j-th sbcone of Ci by Ci,j, nmbered in conterclockise order. When a constraint c = (, ) splits a cone of into to sbcones, e define to lie in both of these sbcones. We call a sbcone of a positie cone a positie sbcone and a sbcone of a negatie cone a negatie sbcone. We consider a cone that is not split as its on 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). More formally, e add an edge beteen to ertices and if can see, Ci,j, and for all points C i,j that can see ( ),, here and denote the projection of and on the bisector of Ci, respectiely, and xy denotes the length of the line segment beteen to points 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. Gien a ertex in a positie cone Ci 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. 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 of P. 3 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. To do this, e first mention a property of isibility graphs. 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. Theorem 1. The constrained half-θ 6 -graph is a 2-spanner of the isibility graph. Proof. Gien to ertices and sch that is a isibility edge, e assme.l.o.g. that C0,j. We proe that δ(, ) 2, here δ(x, y) denotes the length of the shortest path from x to y inside T xy in the constrained half-θ 6 - graph. We proe this by indction on the area of T (formally, indction on the rank, hen ordered by area, of the triangles T xy for all pairs of ertices x and y that can see each other). Let a and b be the pper left and right corner of T, and let A and B be the triangles a and b, respectiely (see Fig. 3). Or indctie hypothesis 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 cos α+sin α),

4 4 On Plane Constrained Bonded-Degree Spanners here α is the nsigned angle beteen and the bisector of C0. This expression is increasing for α [0, π/6]. Inserting the extreme ale π/6 yields a spanning ratio of 2. 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 (, ) mst be 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 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, respectiely (see Fig. 3). 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.l.o.g. that 0 lies to the left of, hich means that A is not empty. Let x be the intersection of and a 0 b 0. By definition x can see and. Since 0 is the closest isible ertex to, 0 can see x as ell. Otherise Lemma 1 old gie s a conex chain of ertices connecting 0 to x, all of hich old be closer and able to see. By applying Lemma 1 to triangle 0 x, a conex chain 0,..., k = of isibility edges connecting 0 and exists (see Fig. 3). 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 i 1, (iii) i C i 1 0 and i lies a 2 1 a 0 0 x b 0 Fig. 3. A conex chain from 0 to 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: 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, respectiely. Triangle B i lies beteen the conex chain and, so it mst be empty. 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. 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, respectiely. 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 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, respectiely. Since i can see i 1 and T i 1 i b

5 On Plane Constrained Bonded-Degree Spanners 5 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. 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. We can no bond δ(, ) by sing these bonds (see Fig. 4): δ(, ) 0 + k i=1 δ( i 1, i ) a 0 + a k i=1 ( i 1a i + a i i ). We see that the latter is eqal to a + a as reqired. a i a i i Fig. 4. 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 B i are also empty. Using the compted bonds on the lengths of the paths beteen the points along the conex chain, e can bond δ(, ) as in the preios case. Case (c): If a > π/2 and B is not empty, the conex chain can contain Type (iii) configrations and 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 Fig. 5). δ(, ) 0 + k δ( i 1, i ) (1) i=1 a 0 + a j ( i 1 a i + a i i ) + i=1 k ( i 1 b i + b i i ) (2) i=j+1 = a + a j + j b + b (3) b + b j + j b + b (4) = b + b (5)

6 6 On Plane Constrained Bonded-Degree Spanners b j a j b b j b Fig. 5. Visalization of the paths (thick lines) in the ineqalities of case (c) 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 triangle is contained in the other triangle. Lemma 3. The constrained half-θ 6 -graph is plane. Proof. Assme that to edges and xy cross at a point p. Since the to edges are contained in their canonical triangles, these mst intersect. By Lemma 2 e kno that at least one of the corners of one triangle lies inside the other. Assme.l.o.g. that the pper right corner of T xy lies inside T. Since and xy cross, this also means that either x or y mst lie in T. Assme.l.o.g. that C0,j and y T. If y C0,j, e look at triangle py. Using that both and y can see p, e get by Lemma 1 that either can see y or py contains a ertex. In both cases, can see a ertex in 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.

7 4 Bonding the Maximm Degree On Plane Constrained Bonded-Degree Spanners 7 In this section, e sho ho to constrct a bonded degree sbgraph G 9 (P ) 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 conterclockise order of the ertices in this sbcone that are neighbors of in the constrained halfθ 6 -graph (see Fig. 6). 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 (P ), e start ith a graph ith Fig. 6. The edges that are added to G 9(P ) for a negatie sbcone of a ertex ith canonical seqence 1, 2, 3 and 4 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 along the bisector of the cone containing the sbcone. This constrction is similar to the constrction of the nconstrained degree-9 half-θ 6 -graph described by Bonichon et al. [2]. Note that since eery edge of the canonical path is part of the constrained half-θ 6 -graph, G 9 (P ) is a sbgraph of the constrained half-θ 6 - graph. We proceed to proe that G 9 (P ) is a spanning sbgraph of the constrained half-θ 6 -graph ith spanning ratio 3. Theorem 2. G 9 (P ) is a 3-spanner of the constrained half-θ 6 -graph. Proof. We proe the lemma by shoing that for each edge of the constrained half-θ 6 -graph H that is not part of G 9 (P ), d G9(P )(, ) 3 d H (, ). We assme.l.o.g. that C 0. Let 0 be the ertex closest to on the canonical path and let 0, 1,..., k = be the ertices along the canonical path from 0 to (see Fig. 7). Let l j and r j denote the rays defining the conterclockise and clockise bondaries of C j 0 for 0 j k and let r denote the ray defining the clockise bondary of C 0. 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 be the intersection of r and the horizontal line throgh and let be the intersection of l k and r. The length of the path beteen and in G 9 (P ) can no be bonded as follos: d G9(P )(, ) 0 + m 0 + k j 1 j (6) j=1 k j=0 k 1 m j j + j m j+1 (7) j=0 = m m 0 (8) + 2 (9)

8 8 On Plane Constrained Bonded-Degree Spanners Let α be and let x be the intersection of and the line throgh perpendiclar to. Using some basic trigonometry, e get = cos α + sin α/ 3 and = 2 sin α/ 3. Ths the spanning ratio can be expressed as: d G9(P )(, ) cos α + 5 sin α 3 (10) Since this is a non-decreasing fnction on 0 < α π/3, its maximm ale is obtained hen α = π/3, here the spanning ratio is 3. It follos from Theorems 1 and 2 that G 9 (P ) is a 6-spanner of the isibility graph. Corollary 1. G 9 (P ) is a 6-spanner of the isibility graph. No, e bond the maximm degree. Lemma 4. When a ertex has at least to constraints in the same positie cone Ci, the closest ertex beteen to consectie constraints has as the closest ertex in the sbcone of C i that contains and is the only ertex on the canonical path of this sbcone. m 0 0 m 1 1 m2 Fig. 7. Bonding the length of the canonical path Proof. Since is the closest ertex in this positie sbcone Ci,j, e kno C i Ci,j contains only ertices and. Hence is the only isible ertex in C i Ci,j and is the only ertex along the canonical path of C i Ci,j. Lemma 5. Eery ertex in G 9 (P ) has degree at most c() + 9. Proof. To bond the degree of a ertex, e se a charging scheme that charges the edges added dring the constrction to the (sb)cones of that ertex. We proe that 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, respectiely. 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() + 9. In fact, e ill sho that a positie cone is charged at most max{2, c i () + 1}. We look at the canonical path in C i,j, created by a ertex. We se to indicate an arbitrary ertex along the canonical path. Let be the closest ertex along the canonical path and let Ci,k be the cone of that contains. The edges of G 9 (P ) are charged as follos (see Fig. 8):

9 On Plane Constrained Bonded-Degree Spanners 9 The edge is charged to C i,j and to C i,k An edge of the canonical path that lies in C i±1 is charged to C i 1 An edge of the canonical path that lies in C i±1 is charged to C i,k Note that each edge is charged once to each of its endpoints. We first proe that each positie cone has charge at most max{2, c i () + 1}. If the positie cone does not contain any constraints, a positie cone of a ertex containing is charged by the edge in that cone if is the closest isible ertex to and it is charged by the to adjacent negatie cones if the edges of the canonical path lie in those cones. Note that since all charges are shifted one cone toards the positie cone containing, other canonical paths cannot charge this positie cone of. Also note that the positie cone is charged at most 2 if is not the closest ertex to. If is the closest ertex to, the negatie cones adjacent Fig. 8. To edges of a canonical path and the associated charges to this positie cone cannot contain any ertices of the canonical path, since these ertices old be closer to than is. Hence, if is the closest ertex to, the positie cone containing is charged 1. If the cone contains constraints, e se Lemma 4 to get a charge of at most c i () 1 in total for all sbcones except the first and last one. We proe that these sbcones can be charged at most 1 each. We look at the first one. The only ay to charge this sbcone 2 is if is the closest ertex to in this sbcone and the adjacent negatie cone contains an edge to a ertex that is part of the same canonical path. Bt if is the closest ertex to, the negatie cones adjacent to this positie cone cannot contain any ertices of the canonical path, since these old be closer to than is. Hence, if is the closest ertex to, the positie cone containing is charged 1. Therefore each positie cone has charge at most max{2, c i () + 1}. Next, e proe that each negatie cone has charge at most c i () + 1. 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 the edges of the canonical paths lie in those cones (see Fig. 9). Sppose that lies in a positie cone of and is part of the canonical path of. Then lies in a negatie cone of, hich means that lies in a positie cone of and cannot Fig. 9. If is be part of a canonical path for. Ths eery negatie cone can be charged by only one edge in an adjacent positie present, the negatie cone does not contain edges haing cone. as endpoint. If this negatie cone does not contain any constraints, it remains to sho that if one of and is present, the negatie cone does not hae an edge to the closest ertex in that cone. We assme.l.o.g. that is present, Ci C i, and C i 1. Since and are neighbors on the canonical path, e kno that the triangle is part of the constrained half-θ 6 -graph and it is empty. Frthermore, since the constrained half-θ 6 -graph

10 10 On Plane Constrained Bonded-Degree Spanners is plane and is an edge of the constrained half-θ 6 -graph, cannot hae an edge to the closest ertex beyond. Hence the negatie cone does not hae an edge to the closest ertex in that cone. Using a similar argment it can be shon that if one of and is present, the negatie cone does not contain any constraints. Ths the charge of a negatie cone is at most c i () + 1. Corollary 2. If a positie cone has charge c i () + 2, it is charged for to edges in the adjacent negatie cones and does not contain any constraints haing as an endpoint. 4.1 Bonding the Maximm Degree Frther Using Corollary 2, e kno that the only sitation e need to modify to get the degree bond don to c() + 6 is the case here a positie cone is charged for to edges in the adjacent negatie cones and does not contain any constraints (see Fig. 10). If neither x nor y is the ertex closest to in their respectie cone, e do the folloing transformation on G 9 (P ). First, e add an edge beteen x and y. Next, e x y Fig. 10. A positie cone haing charge 2 look at the seqence of ertices beteen and the closest ertex along the canonical path. If this seqence incldes x, e remoe y. Otherise e remoe x. y y x x Fig. 11. Constrcting G 6(P ) (right) from G 9(P ) (left) We assme.l.o.g. that y is remoed. We look at ertex, the neighbor of ertex x on the canonical path of ertex containing x. Since x is not the closest ertex to, this ertex mst exist. The edge x is remoed if lies in a negatie cone of x and is not the closest ertex in this cone. The reslting graph is G 6 (P ) (see Fig. 11). It can be shon that the nely added edges do not intersect each other, the constraints and the remaining edges of G 9 (P ), hich implies that G 6 (P ) is plane. 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.

11 On Plane Constrained Bonded-Degree Spanners 11 Lemma 6. Let x be an edge of G 9 (P ) and let x lie in a negatie cone C i of. If x is not the ertex closest to in C i, then the edge x is sed by at most one canonical path. Proof. We proe the lemma by contradiction. Gien that x is not the ertex closest to in C i, assme that edge x is part of to canonical paths. This means that there exist ertices Ci+1 Cx i+1 and C i 1 Cx i 1 sch that and x are neighbors on a canonical path of and. Ths ertices x and x form to triangles in the constrained half-θ 6 -graph. By planarity, this implies that x is the only edge of in C i (see Fig. 12). This implies that x is the ertex closest to in C i. x Fig. 12. Edge x is part of to canonical paths. Lemma 7. G 6 (P ) is a 3-spanner of the half-θ 6 -graph. Proof. Since G 9 (P ) is a 3-spanner of the constrained half-θ 6 -graph, e need to look only at the edges that ere remoed from this graph. Let be a ertex sch that positie cone Ci has charge 2, let be the ertex hose canonical path charged 2 to Ci, and let x C i 1 and y C i+1 be the neighbors of on this canonical path. We assme.l.o.g. that y is remoed. Since this remoal potentially affects the spanning ratio of any path sing y, e need to look at the spanning path beteen and y and the spanning path beteen and any ertex on the canonical path that ses y. Since y is not the closest ertex to, Lemma 6 tells s that no other canonical path is affected. Since y is not the closest ertex to, there exists a spanning path beteen and y that does not se y. Since xy x + y, the length of the spanning path beteen and any ertex on the canonical path that ses y is not increased. Ths remoing y does not affect the spanning ratio. Next, e look at the other type of edge that is remoed. Let be the neighbor of ertex x on the canonical path of ertex containing x. Edge x is remoed if lies in C x i and is not the closest ertex in C x i. Since x is the last ertex on the canonical path of, e need to look only at the spanning path beteen x and and the spanning path beteen and x. Since y is not the closest ertex to, Lemma 6 tells s that no other canonical path is affected. Since is not the closest ertex to x, there exists a spanning path beteen x and that does not se x. By Lemma 6, x is part of only one canonical path and hence it is present in G 6 (P ). Ths there exists a spanning path beteen x and and remoing x does not affect the spanning ratio. Lemma 8. Eery ertex in G 6 (P ) has degree at most c() + 6. Proof. To bond the degree of a ertex, e look at the charges of the ertices. We proe that after the transformation each positie cone has charge at most

12 12 On Plane Constrained Bonded-Degree Spanners 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 2. Let be a ertex sch that one of its positie cones Ci has charge 2, let be the ertex hose canonical path charged 2 to Ci, and let x C i 1 and y C i+1 be the neighbors of on this canonical path (see Fig. 10). If x or y is the ertex closest to in C i 1 or C i+1, this edge has been charged to both that negatie cone and Ci. Hence e can remoe the charge to C i hile maintaining that the charge is an pper bond on the degree of. If neither x nor y is the closest ertex in C i 1 or C i+1, edge xy is added. We assme.l.o.g. 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 i 1 and this charge is remoed, e charge edge xy to C y i 1. Ths the charge of y does not change. It remains to sho that e can charge xy to x. We look at ertex, the neighbor of x on the canonical path of in C i 1. Since x is not the closest ertex to in C i 1, the canonical path and ertex exist. Since ertices x form a triangle in the constrained half-θ 6 -graph, Ci 1 x has charge at most 1. 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 C x i. Ths the charge of Ci 1 x is 0 and e can charge xy to it. If C x i and is the closest ertex to x in C x i, x has been charged to both Ci 1 x and Cx i. We replace the charge of Ci 1 x by xy and the charge of Cx i 1 remains 1. If C x i and is not the closest ertex to x in C x i, x is remoed. Since this edge as charged to Ci 1 x, e can charge xy to Cx i 1 and the charge of Ci 1 x remains 1. References 1. N. Bonichon, C. Gaoille, N. Hansse, and D. Ilcinkas. Connections beteen thetagraphs, Delanay trianglations, and orthogonal srfaces. In Proceedings of the 36th International Workshop on Graph Theoretic Concepts in Compter Science, pages , N. Bonichon, C. Gaoille, N. Hansse, and L. Perkoic. Plane spanners of maximm degree six. In Proceedings of the 37th International Colloqim on Atomata, Langages and Programming, pages 19 30, P. Bose and J. M. Keil. On the stretch factor of the constrained Delanay trianglation. In Proceedings of the 3rd International Symposim on Voronoi Diagrams in Science and Engineering, pages 25 31, K. Clarkson. Approximation algorithms for shortest path motion planning. In Proceedings of the 19th Annal ACM Symposim on Theory of Compting, pages 56 65, G. Das. The isibility graph contains a bonded-degree spanner. In Proceedings of the 9th Canadian Conference on Comptational Geometry, pages 70 75, G. Narasimhan and M. Smid. Geometric Spanner Netorks. Cambridge Uniersity Press, 2007.