Towards Tight Bounds on Theta-Graphs

Transcription

1 Toards Tight Bonds on Theta-Graphs arxiv:10.633v1 [cs.cg] Apr 01 Prosenjit Bose Jean-Lo De Carfel Pat Morin André van Renssen Sander Verdonschot Abstract We present improved pper and loer bonds on the spanning ratio of -graphs ith at least six cones. Given a set of points in the plane, a -graph partitions the plane arond each vertex into m disjoint cones, each having apertre = /m, and adds an edge to the closest vertex in each cone. We sho that for any integer k 1, -graphs ith k + cones have a spanning ratio of 1 + sin/ and e provide a matching loer bond, shoing that this spanning ratio tight. Next, e sho that for any integer k 1, -graphs ith k + cones have spanning ratio at most 1 + sin//cos/ sin/. We also sho that - graphs ith k +3 and k +5 cones have spanning ratio at most cos//cos/ sin3/. This is a significant improvement on all families of -graphs for hich exact bonds are not knon. For example, the spanning ratio of the -graph ith 7 cones is decreased from at most to at most These spanning proofs also imply improved pper bonds on the competitiveness of the -roting algorithm. In particlar, e sho that the -roting algorithm is 1 + sin//cos/ sin/-competitive on -graphs ith k + cones and that this ratio is tight. Finally, e present improved loer bonds on the spanning ratio of these graphs. Using these bonds, e provide a partial order on these families of -graphs. In particlar, e sho that -graphs ith k + cones have spanning ratio at least 1 + tan/ + tan /, here is /k +. This is somehat srprising since, for eqal vales of k, the spanning ratio of -graphs ith k + cones is greater than that of -graphs ith k + cones, shoing that increasing the nmber of cones can make the spanning ratio orse. Keyords Comptational geometry, Spanners, Theta-graphs, Spanning Ratio, Tight bonds School of Compter Science, Carleton University. Research spported in part by FQRNT, NSERC, and Carleton University s President s 010 Doctoral Felloship. jit@scs.carleton.ca, jdecarf@cg.scs.carleton.ca, morin@scs.carleton.ca, andre@cg.scs.carleton.ca, sander@cg.scs.carleton.ca. 1

2 1 Introdction A geometric graph G is a graph hose vertices are points in the plane and hose edges are line segments beteen pairs of points. A graph G is called plane if no to edges intersect properly. Every edge is eighted by the Eclidean distance beteen its endpoints. The distance beteen to vertices and v in G, denoted by δ G, v, or simply δ, v hen G is clear from the context, is defined as the sm of the eights of the edges along the shortest path beteen and v in G. A sbgraph H of G is a t-spanner of G for t 1 if for each pair of vertices and v, δ H, v t δ G, v. The smallest vale 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 varios geometric graphs have been stdied extensively in the literatre see [3, 7] for a comprehensive overvie of the topic. Given a spanner, hoever, it is important to be able to rote, i.e. find a short path, beteen any to vertices. A roting algorithm is said to be c-competitive ith respect to G if the length of the path retrned by the roting algorithm is not more than c times the length of the shortest path in G []. The smallest vale c for hich a roting algorithm is c-competitive ith respect to G is the roting ratio of that roting algorithm. In this paper, e consider the sitation here the nderlying graph G is a straightline embedding of the complete graph on a set of n points in the plane ith the eight of an edge, v being the Eclidean distance v beteen and v. A spanner of sch a graph is called a geometric spanner. We look at a specific type of geometric spanner: -graphs. Introdced independently by Clarkson [5] and Keil [6], -graphs are constrcted as follos a more precise definition follos in Section : for each vertex, e partition the plane into m disjoint cones ith apex, each having apertre = /m. When m cones are sed, e denote the reslting -graph by the m -graph. The -graph is constrcted by, for each cone ith apex, connecting to the vertex v hose projection onto the bisector of the cone is closest. Rppert and Seidel [8] shoed that the spanning ratio of these graphs is at most 1/1 sin/, hen < /3, i.e. there are at least seven cones. This proof also shoed that the -roting algorithm defined in Section is 1/1 sin/-competitive on these graphs. Recently, Bonichon et al. [1] shoed that the 6 -graph has spanning ratio. This as done by dividing the cones into to sets, positive and negative cones, sch that each positive cone is adjacent to to negative cones and vice versa. It as shon that hen edges are added only in the positive cones, in hich case the graph is called the half- 6 - graph, the reslting graph is eqivalent to the Delanay trianglation here the empty region is an eqilateral triangle. The spanning ratio of this graph is, as shon by Che []. An alternative, indctive proof of the spanning ratio of the half- 6 -graph as presented by Bose et al. [], along ith an optimal local competitive roting algorithm on the half- 6 -graph. Tight bonds on spanning ratios are notoriosly hard to obtain. The standard Delanay trianglation here the empty region is a circle is a good example. Its spanning ratio has been stdied for over 0 years and the pper and loer bonds still do not match. Also, even thogh it as introdced abot 5 years ago, the spanning ratio of the 6 -graph has

3 only recently been shon to be finite and tight, making it the first and, ntil no, only -graph for hich tight bonds are knon. In this paper, e improve on the existing pper bonds on the spanning ratio of all - graphs ith at least six cones. First, e generalize the spanning proof of the half- 6 -graph given by Bose et al. [] to a large family of -graphs: the k+ -graph, here k 1 is an integer. We sho that the k+ -graph has a tight spanning ratio of 1 + sin/ see Section.1. We contine by looking at pper bonds on the spanning ratio of the other three families of -graphs: the k+3 -graph, the k+ -graph, and the k+5 -graph, here k is an integer and at least 1. We sho that the k+ -graph has a spanning ratio of at most 1 + sin//cos/ sin/ see Section.3. We also sho that the k+3 - graph and the k+5 -graph have spanning ratio at most cos//cos/ sin3/ see Section.. As as the case for Rppert and Seidel, the strctre of these spanning proofs implies that the pper bonds also apply to the competitiveness of -roting on these graphs. These reslts are smmarized in Table 1. Finally, e present improved loer bonds on the spanning ratio of these graphs see Section 5 and e provide a partial order on these families see Section 6. In particlar, e sho that -graphs ith k+ cones have spanning ratio at least 1+ tan/+ tan /. This is somehat srprising since, for eqal vales of k, the spanning ratio of -graphs ith k + cones is greater than that of -graphs ith k + cones, shoing that increasing the nmber of cones can make the spanning ratio orse. Crrent Spanning Crrent Roting Previos Spanning & Roting k+ -graph 1 + sin 1 1 sin [8] 1 1 sin [8] k+3 -graph cos cos sin sin cos cos sin 1 1 sin [8] k+ -graph 1 + sin cos sin 1 + sin cos sin 1 1 sin [8] k+5 -graph cos cos sin sin cos cos sin 1 1 sin [8] Table 1: An overvie of crrent and previos spanning and roting ratios of -graphs Preliminaries Let a cone be the region in the plane beteen to rays originating from the same vertex referred to as the apex of the cone. When constrcting a m -graph, for each vertex consider the rays originating from ith the angle beteen consective rays being 3

4 = /m see Figre 1. Each pair of consective rays defines a cone. The cones are oriented sch that the bisector of some cone coincides ith the vertical halfline throgh that lies above. We refer to this cone as C0 and nmber the cones in clockise order arond. The cones arond the other vertices have the same orientation as the ones arond. If the apex is clear from the context, e rite C i to indicate the i-th cone. For ease of exposition, e only consider point sets in general position: no to vertices lie on a line parallel to one of the rays that define the cones, no to vertices lie on a line perpendiclar to the bisector of one of the cones, and no three points are collinear. C 0 C 5 C 1 C C C 3 Figre 1: The cones having apex in the 6 -graph The m -graph is constrcted as follos: for each cone Ci of each vertex, add an edge from to the closest vertex in that cone, here the distance is measred along the bisector of the cone see Figre. More formally, e add an edge beteen to vertices and v if v Ci, and for all vertices Ci, v, here v and denote the orthogonal projection of v and onto the bisector of C i. Note that or assmptions of general position imply that each vertex adds at most one edge per cone to the graph. v Figre : Three vertices are projected onto the bisector of a cone of. Vertex v is the closest vertex Using the strctre of the m -graph, -roting is defined as follos. Let t be the destination of the roting algorithm and let be the crrent vertex. If there exists a direct

5 edge to t, follo this edge. Otherise, follo the edge to the closest vertex in the cone of that contains t. Finally, given a vertex in cone C of a vertex, e define the canonical triangle T to be the triangle defined by the borders of C and the line throgh perpendiclar to the bisector of C. We se m to denote the midpoint of the side of T opposite and α to denote the smaller nsigned angle beteen and m see Figre 3. Note that for any pair of vertices and in the m -graph, there exist to canonical triangles: T and T. m α Figre 3: The canonical triangle T 3 Some Geometric Lemmas First, e prove a fe geometric lemmas that are sefl hen bonding the spanning ratios of the graphs. We start ith a nice geometric property of the k+ -graph. Lemma 1 In the k+ -graph, any line perpendiclar to the bisector of a cone is parallel to the bondary of some cone. Proof. The angle beteen the bisector of a cone and the bondary of that cone is /. In the k+ -graph, since = /k +, the angle beteen the bisector and the line perpendiclar to this bisector is / = k +/ = k +/. Ths the angle beteen the line perpendiclar to the bisector and the bondary of the cone is / / = k. Since a cone bondary is placed at every mltiple of, the line perpendiclar to the bisector is parallel to the bondary of some cone. This property helps hen bonding the spanning ratio of the k+ -graph. Hoever, before deriving this bond, e prove a fe other geometric lemmas. We se xyz to denote the smaller angle beteen line segments xy and yz. Lemma Let a, b, c, and d be for points on a circle sch that cad bad adc. It holds that ac + cd ab + bd and cd bd. 5

6 b c c a d Figre : Illstration of the proof of Lemma Proof. This sitation is illstrated in Figre. Withot loss of generality, e assme that ad = 1. Since b and c lie on the same circle and abd and acd are the angle opposite to the same chord ad, the inscribed angle theorem implies that abd = acd. Frthermore, since cad adc, c lies to the right of the perpendiclar bisector of ad. First, e sho that ac + cd ab + bd by shoing that ac + cd + ad ab + bd + ad. Let c be the point on the circle hen e mirror c along the perpendiclar bisector of ad. Points c and c partition the circle into to arcs. Since cad bad adc, b lies on the pper arc of the circle. We focs on triangle acd. The locs of the point c sch that the perimeter of acd is constant defines an ellipse. This ellipse has major axis ad and goes throgh c and c. Since this major axis is horizontal, the ellipse does not intersect the pper arc of the circle. Hence, since b lies on the pper arc of the circle, hich is otside of the ellipse, the perimeter of abd is greater than that of acd, completing the first half of the proof. Next, e sho that cd bd. Using the sine la, e have that cd = sin cad/ sin acd and bd = sin bad/ sin abd. Since cad bad adc cad, e have that sin cad sin bad. Hence, since abd = acd, e have that cd bd. Lemma 3 Let, v and be three vertices in the k+x -graph, here x {, 3,, 5}, sch that C0 and v T, to the left of. Let a be the intersection of the side of T opposite to ith the left bondary of C0 v. Let Ci v denote the cone of v that contains and let c and d be the pper and loer corner of T v. If 1 i k 1, or i = k and c d, then max { vc + c, vd + d } va + a and max { c, d } a. Proof. This sitation is illstrated in Figre 5. We perform case distinction on max{ c, d }. Case 1: If c > d see Figre 5a, e need to sho that hen 1 i k 1, e have that vc + c va + a and c a. Since angles va and vc are both 6

7 a c C v i d a d C v i c d v v a b Figre 5: The to cases for the sitation here e apply Lemma : a c > d, b c d angles beteen the bondary of a cone and the line perpendiclar to its bisector, e have that va = vc. Ths, c lies on the circle throgh a, v, and. Therefore, if e can sho that cv av vc, Lemma proves this case. We sho cv av vc in to steps. Since Ci v and i 1, e have that avc = i. Hence, since av = avc + cv, e have that cv av. It remains to sho that av vc. We note that av i+1 and / vc, since c > d. Using that = /k + x and x {, 3,, 5}, e have the folloing. i k 1 i k + x 3 i k + x 3 i 3 i + 1 av vc Case : If c d see Figre 5b, e need to sho that hen 1 i k, e have that vd + d va + a and d a. Since angles va and vd are both angles beteen the bondary of a cone and the line perpendiclar to its bisector, e have that va = vd. Ths, hen e reflect d in the line throgh v, the reslting point d lies on the circle throgh a, v, and. Therefore, if e can sho that d v av vd, Lemma proves this case. 7

8 We sho d v av vd in to steps. Since Ci v and i 1, e have that av avc = i. Hence, since d v, e have that d v av. It remains to sho that av vd. We note that vd = dv = / dv and av = avd dv = i+1 dv. Using that = /k +x and x {, 3,, 5}, e have the folloing. i k i k + x 1 i k + x 1 i 1 i + 1 dv + dv av vd Lemma Let, v and be three vertices in the k+x -graph, sch that C 0, v T to the left of, and C v 0. Let a be the intersection of the side of T opposite to ith the left bondary of C v 0. Let c and d be the corners of T v opposite to v. Let β = av and let γ be the nsigned angle beteen v and the bisector of T v. Let c be a positive constant. If then c cos γ sin β cos β sin 1 + γ, max { vc + c c, vd + c d } va + c a. Proof. This sitation is illstrated in Figre 6. Since the angle beteen the bisector of a cone and its bondary is /, by the sine la, e have the folloing. cos γ vc = vd = v cos max { c, d } = v sin γ + cos γ tan sin β va = v cos a = v cos β + sin β tan 8

9 a c β T v v γ d Figre 6: Finding a constant c sch that vd + c d va + c a To sho that holds, e first mltiply both sides by cos// v and rerite as follos. cos max { vc + c c, vd + c d } v = cos γ + c sin γ cos + cos γ sin = cos γ + c sin + γ cos va + c a = sin β + c v cos β cos + sin β sin = sin β + c cos β Therefore, to prove that 1 implies, e rerite 1 as follos. cos γ sin β c cos β sin + γ cos γ sin β c cos β sin sin β + c cos β cos γ + c sin + γ + γ It remains to sho that c > 0. Since C0 v, e have that β 0, /. Moreover, e have that γ [0, /, by definition. This implies that sin/ + γ > sin β, or eqivalently, cos γ sin β > 0. Ths, e need to sho that cos/ β sin/ + γ > 0, or eqivalently, sin/ + / β > sin/ + γ. It sffices to sho that / + γ < / + / β < / γ. This follos from β 0, /, γ [0, /, and the fact that /7. 9

10 Upper Bonds In this section, e provide improved pper bonds for the for families of -graphs: the k+ -graph, the k+3 -graph, the k+ -graph, and the k+5 -graph. We first prove that the k+ -graph has a tight spanning ratio of 1 + sin/. Next, e provide a generic frameork for the spanning proof for the three other families of -graphs. After providing this frameork, e fill in the blanks for the individal families..1 Optimal Bonds on the k+ -Graph We start by shoing that the k+ -graph has a spanning ratio of 1 + sin/. At the end of this section, e also provide a matching loer bond, proving that this spanning ratio is tight. Theorem 5 Let and be to vertices in the plane. Let m be the midpoint of the side of T opposite and let α be the nsigned angle beteen and m. There exists a path connecting and in the k+ -graph of length at most 1 + sin cos cos α + sin α. Proof. We assme ithot loss of generality that C0. We prove the theorem by indction on the area of T formally, indction on the rank, hen ordered by area, of the canonical triangles for all pairs of vertices. Let a and b be the pper left and right corners of T and let y and z be the left and right intersections of the left and right bondaries of T and the bondaries of Ck+1, the cone of that contains see Figre 7. Or indctive hypothesis is the folloing, here δ, denotes the length of the shortest path from to in the k+ -graph: If ay is empty, then δ, b + b. If bz is empty, then δ, a + a. If neither ay nor bz is empty, then δ, max{ a + a, b + b }. Note that if both ay and bz are empty, the indction hypothesis implies that δ, min{ a + a, b + b }. We first sho that this indction hypothesis implies the theorem. Basic trigonometry gives s the folloing eqalities: m = cos α, m = sin α, am = bm = cos α tan/, and a = b = cos α/ cos/. Ths, the indction hypothesis gives s that 1 + sin δ, a + am + m = cos 10 cos α + sin α.

11 Base case: T has rank 1. Since the triangle is a smallest triangle, is the closest vertex to in that cone. Hence, the edge, is part of the k+ -graph and δ, =. From the triangle ineqality, e have min{ a + a, b + b }, so the indction hypothesis holds. Indction step: We assme that the indction hypothesis holds for all pairs of vertices ith canonical triangles of rank p to j. Let T be a canonical triangle of rank j + 1. If, is an edge in the k+ -graph, the indction hypothesis follos from the same argment as in the base case. If there is no edge beteen and, let v be the vertex closest to in C0, and let a and b be the pper left and right corners of T v see Figre 7. By definition, δ, v + δv,, and by the triangle ineqality, v min{ a + a v, b + b v }. a c y d b z z a v b c a a b a a a a v d z v z b b z y y y a b c Figre 7: The three cases of the indction step based on the cone of v that contains, in this case for the 1 -graph Withot loss of generality, e assme that v lies to the left of. We perform a case analysis based on the cone of v that contains : a C v 0, b C v i here 1 i k 1, c C v k. Case a: Vertex lies in C v 0 see Figre 7a. Let c and d be the pper left and right corners of T v, and let y and z be the left and right intersections of T v and the bondaries of C k+1. Since T v has smaller area than T, e apply the indctive hypothesis to T v. We need to prove all three statements of the indctive hypothesis for T. 1. If ay is empty, then cy is also empty, so by indction δv, vd + d. Since v, d, b, and b form a parallelogram, e have: δ, v + δv, b + b v + vd + d = b + b, 11

12 hich proves the first statement of the indction hypothesis.. If bz is empty, an analogos argment proves the second statement of the indction hypothesis. 3. If neither ay nor bz is empty, by indction e have δv, max{ vc + c, vd + d }. Assme, ithot loss of generality, that the maximm of the right hand side is attained by its second argment vd + d the other case is similar. Since vertices v, d, b, and b form a parallelogram, e have that: δ, v + δv, b + b v + vd + d b + b max{ a + a, b + b }, hich proves the third statement of the indction hypothesis. Case b: Vertex lies in Ci v here 1 i k 1 see Figre 7b. In this case, v lies in ay. Therefore, the first statement of the indction hypothesis for T is vacosly tre. It remains to prove the second and third statement of the indction hypothesis. Let a be the intersection of the side of T opposite and the left bondary of C0 v. Since T v is smaller than T, by indction e have δv, max{ vc + c, vd + d }. Since Ci v here 1 i k 1, e can apply Lemma 3. Note that point a in Lemma 3 corresponds to point a in this proof. Hence, e get that max{ vc + c, vd + d } va + a. Since v a + a v and v, a, a, and a form a parallelogram, e have that δ, a + a, proving the indction hypothesis for T. Case c: Vertex lies in Ck v see Figre 7c. Since v lies in ay, the first statement of the indction hypothesis for T is vacosly tre. It remains to prove the second and third statement of the indction hypothesis. Let a and b be the pper and loer left corners of T v, and let z be the intersection of T v and the loer bondary of Ck v, i.e. the cone of v that contains. Note that z is also the right intersection of T v and T v. Since v is the closest vertex to, T v is empty. Hence, b z v is empty. Since T v is smaller than T, e can apply indction on it. As b z v is empty, the indction hypothesis for T v gives δv, va + a. Since v a + a v and v, a, a, and a form a parallelogram, e have that δ, a + a, proving the second and third statement of the indction hypothesis for T. Since 1 + sin// cos/ cos α + sin α is increasing for α [0, /], for /3, it is maximized hen α = /, and e obtain the folloing corollary: Corollary 6 The k+ -graph is a 1 + sin /-spanner. The pper bonds given in Theorem 5 and Corollary 6 are tight, as shon in Figre 8: e place a vertex v arbitrarily close to the pper corner of T that is frthest from. 1

13 Likeise, e place a vertex v arbitrarily close to the loer corner of T that is frthest from. Both shortest paths beteen and visit either v or v, so the path length is arbitrarily close to 1 + sin// cos/ cos α + sin α, shoing that the pper bonds are tight. v v Figre 8: The loer bond for the k+ -graph. Generic Frameork for the Spanning Proof In this section, e provide a generic frameork for the spanning proof for the three other families of -graphs: the k+3 -graph, the k+ -graph, and the k+5 -graph. This frameork contains those parts of the spanning proof that are identical for all three families. In the sbseqent sections, e handle the single case that depends on each specific family and determines their respective spanning ratios. Theorem 7 Let and be to vertices in the plane. Let m be the midpoint of the side of T opposite and let α be the nsigned angle beteen and m. There exists a path connecting and in the k+x -graph of length at most cos α cos + c cos α tan + sin α, here c 1 is a fnction that depends on x {3,, 5} and. For the k+ -graph, c eqals 1/cos/ sin/ and for the k+3 -graph and k+5 -graph, c eqals cos// cos/ sin3/. Proof. We assme ithot loss of generality that C 0. We prove the theorem by indction on the area of T formally, indction on the rank, hen ordered by area, of the canonical triangles for all pairs of vertices. Let a and b be the pper left and right corners of T. Or indctive hypothesis is the folloing, here δ, denotes the length of the shortest path from to in the k+x -graph: δ, max{ a +c a, b +c b }. 13

14 We first sho that this indction hypothesis implies the theorem. Basic trigonometry gives s the folloing eqalities: m = cos α, m = sin α, am = bm = cos α tan/, and a = b = cos α/ cos/. Ths the indction hypothesis gives that cos α δ, a + c am + m = cos + c cos α tan + sin α. Base case: T has rank 1. Since the triangle is a smallest triangle, is the closest vertex to in that cone. Hence, the edge, is part of the k+x -graph and δ, =. From the triangle ineqality and the fact that c 1, e have max{ a + c a, b + c b }, so the indction hypothesis holds. Indction step: We assme that the indction hypothesis holds for all pairs of vertices ith canonical triangles of rank p to j. Let T be a canonical triangle of rank j + 1. If, is an edge in the k+x -graph, the indction hypothesis follos from the same argment as in the base case. If there is no edge beteen and, let v be the vertex closest to in T, and let a and b be the pper left and right corners of T v see Figre 9. By definition, δ, v + δv,, and by the triangle ineqality, v min{ a + a v, b + b v }. a c c d b a a a v d b a a a v c d b a a a v c d b a v b a b c d Figre 9: The for cases of the indction step based on the cone of v that contains, in this case for the 1 -graph Withot loss of generality, e assme that v lies to the left of. We perform a case analysis based on the cone of v that contains, here c and d are the left and right corners of T v, opposite to v: a C0 v, b Ci v here 1 i k 1, or i = k and c d, c Ck v and c > d, d Cv k+1. Case a: Vertex lies in C0 v see Figre 9a. Since T v has smaller area than T, e apply the indctive hypothesis to T v. Hence e have δv, max{ vc + c c, vd + 1

15 c d }. Since v lies to the left of, the maximm of the right hand side is attained by its first argment, vc + c c. Since vertices v, c, a, and a form a parallelogram, and c 1, e have that δ, v + δv, a + a v + vc + c c a + c a max{ a + c a, b + c b }, hich proves the indction hypothesis. Case b: Vertex lies in Ci v, here 1 i k 1, or i = k and c d see Figre 9b. Let a be the intersection of the side of T opposite and the left bondary of C0 v. Since T v is smaller than T, by indction e have δv, max{ vc +c c, vd +c d }. Since Ci v here 1 i k 1, or i = k and c d, e can apply Lemma 3. Note that point a in Lemma 3 corresponds to point a in this proof. Hence, e get that max { vc + c, vd + d } va + a and max { c, d } a. Since c 1, this implies that max{ vc +c c, vd +c d } va +c a. Since v a + a v and v, a, a, and a form a parallelogram, e have that δ, a + c a, proving the indction hypothesis for T. Case c and d Vertex lies in Ck v and c > d, or lies in Cv k+1 see Figres 9c and d. Let a be the intersection of the side of T opposite and the left bondary of C0 v. Since T v is smaller than T, e can apply indction on it. The actal application of the indction hypothesis varies for the three families of -graphs and, sing Lemma, determines the vale of c. Hence, these cases are discssed in the spanning proofs of the three families..3 Upper Bond on the k+ -Graph In this section, e improve the pper bonds on the spanning ratio of the k+ -graph, for any integer k 1. Theorem 8 Let and be to vertices in the plane. Let m be the midpoint of the side of T opposite and let α be the nsigned angle beteen and m. There exists a path connecting and in the k+ -graph of length at most cos α cos + cos α tan + sin α cos sin. Proof. We apply Theorem 7 sing c = 1/cos/ sin/. It remains to handle Case c, here C v k and c > d, and Case d, here Cv k+1. Recall that c and d are the left and right corners of T v, opposite to v, and a is the intersection of the side of T opposite and the left bondary of C v 0. Let β be a v 15

16 and let γ be the angle beteen v and the bisector of T v. Since T v is smaller than T, the indction hypothesis gives an pper bond on δv,. Since v a + a v and v, a, a, and a form a parallelogram, e need to sho that δv, va + c a for both cases in order to complete the proof. c a a a v γ β d b a a a v β γ c d b a b Figre 10: The remaining cases of the indction step for the k+ -graph: a lies in C v k and c > d, b lies in Cv k+1 Case c: When lies in Ck v and c > d, the indction hypothesis for T v gives δv, vc + c c see Figre 10a. We note that γ = β. Hence, the ineqality follos from Lemma hen c cos β sin β/cos/ β sin3/ β. Since this fnction is decreasing in β for / β, it is maximized hen β eqals /. Hence, c needs to be at least cos/ sin//1 sin, hich can be reritten to 1/cos/ sin/. Case d: When lies in Ck+1 v, lies above the bisector of T v see Figre 10b and the indction hypothesis for T v gives δv, d +c dv. We note that γ = β. Hence, the ineqality follos from Lemma hen c cos β sin β/cos/ β sin/+β, hich is eqal to 1/cos/ sin/. Since cos α/ cos/ + cos α tan/ + sin α/cos/ sin/ is increasing for α [0, /], for /, it is maximized hen α = /, and e obtain the folloing corollary: Corollary 9 The k+ -graph is a 1 + sin -spanner. cos sin Frthermore, e observe that the proof of Theorem 8 follos the same path as the -roting algorithm follos: if the direct edge to the destination is part of the graph, it follos this edge, and if it is not, it follos the edge to the closest vertex in the cone that contains the destination. Corollary 10 The -roting algorithm is 1 + sin -competitive on the cos sin k+ -graph. 16

17 . Upper Bonds on the k+3 -Graph and k+5 -Graph In this section, e improve the pper bonds on the spanning ratio of the k+3 -graph and the k+5 -graph, for any integer k 1. Theorem 11 Let and be to vertices in the plane. Let m be the midpoint of the side of T opposite and let α be the nsigned angle beteen and m. There exists a path connecting and in the k+3 -graph of length at most + sin α cos cos α cos α tan cos + cos sin 3. Proof. We apply Theorem 7 sing c = cos//cos/ sin3/. It remains to handle Case c, here C v k and c > d, and Case d, here Cv k+1. Recall that c and d are the left and right corners of T v, opposite to v, and a is the intersection of the side of T opposite and the left bondary of C v 0. Let β be a v and let γ be the angle beteen v and the bisector of T v. Since T v is smaller than T, the indction hypothesis gives an pper bond on δv,. Since v a + a v and v, a, a, and a form a parallelogram, e need to sho that δv, va + c a for both cases in order to complete the proof. c a a a v γ β d b a a a v γ β c b a b d Figre 11: The remaining cases of the indction step for the k+3 -graph: a lies in C v k and c > d, b lies in Cv k+1 Case c: When lies in C v k and c > d, the indction hypothesis for T v gives δv, vc + c c see Figre 11a. We note that γ = 3/ β. Hence, the ineqality follos from Lemma hen c cos3/ β sin β/cos/ β sin5/ β. Since this fnction is decreasing in β for / β 3/, it is maximized hen β eqals /. Hence, c needs to be at least cos/ sin//cos/ sin, hich is eqal to cos//cos/ sin3/. Case d: When lies in C v k+1, lies above the bisector of T v see Figre 11b and the indction hypothesis for T v gives δv, d + c dv. We note that γ = / + β. 17

18 Hence, the ineqality follos from Lemma hen c cos/ + β sin β/cos/ β sin3/ + β, hich is eqal to cos//cos/ sin3/. Theorem 1 Let and be to vertices in the plane. Let m be the midpoint of the side of T opposite and let α be the nsigned angle beteen and m. There exists a path connecting and in the k+5 -graph of length at most + sin α cos cos α cos α tan cos + cos sin 3. Proof. We apply Theorem 7 sing c = cos//cos/ sin3/. It remains to handle Case c, here C v k and c > d, and Case d, here Cv k+1. Recall that c and d are the left and right corners of T v, opposite to v, and a is the intersection of the side of T opposite and the left bondary of C v 0. Let β be a v and let γ be the angle beteen v and the bisector of T v. Since T v is smaller than T, the indction hypothesis gives an pper bond on δv,. Since v a + a v and v, a, a, and a form a parallelogram, e need to sho that δv, va + c a for both cases in order to complete the proof. c c a a a v γ 3 β a d b a a a v γ β b c d b a a a v γ β c d b Figre 1: The remaining cases of the indction step for the k+5 -graph: a lies in Ck v and c > d, b lies in Cv k+1 and c < d, c lies in Cv k+1 and c d Case c: When lies in Ck v and c > d, the indction hypothesis for T v gives δv, vc + c c see Figre 1a. We note that γ = 5/ β. Hence, the ineqality follos from Lemma hen c cos5/ β sin β/cos/ β sin7/ β. Since this fnction is decreasing in β for 3/ β 5/, it is maximized hen β eqals 3/. Hence, c needs to be at least cos/ sin3//cos/ sin, hich is less than cos//cos/ sin3/. Case d: When lies in Ck+1 v, the indction hypothesis for T v gives δv, max{ vc +c c, vd +c d }. If c < d see Figre 1b, the indction hypothesis for T v gives δv, vd + c d. We note that γ = β /. Hence, the ineqality follos from Lemma hen c cosβ / sin β/cos/ β sin/+β, hich is eqal to cos//cos/ sin3/. 18

19 If c d, the indction hypothesis for T v gives δv, vc + c c see Figre 1c. We note that γ = / β. Hence, the ineqality follos from Lemma hen c cos/ β sin β/cos/ β sin3/ β. Since this fnction is decreasing in β for 0 β /, it is maximized hen β eqals 0. Hence, c needs to be at least cos//cos/ sin3/. By looking at to vertices and in the k+3 -graph and the k+5 -graph, e can see that hen the angle beteen and the bisector of T is α, the angle beteen and the bisector of T is / α. Hence the orst case spanning ratio corresponds to the minimm of the spanning ratio hen looking at T and the spanning ratio hen looking at T. Theorem 13 The k+3 -graph and k+5 -graph are cos cos sin 3 -spanners. Proof. The spanning ratio of the k+3 -graph and the k+5 -graph is at most cos α cos min + cos α tan +sin α cos, cos sin 3 cos α + cos α tan +sin α cos. cos cos sin 3 Since cos α/ cos/ + c cos α tan/ + sin α is increasing for α [0, /], for /7, the minimm of these to fnctions is maximized hen the to fnctions are eqal, i.e. hen α = /. Ths the k+3 -graph and the k+5 -graph have spanning ratio at most cos cos cos cos tan + sin + cos sin 3 = cos cos sin 3. Frthermore, e observe that the proofs of Theorem 11 and Theorem 1 follo the same path as the -roting algorithm follos. Since in the case of roting, e are forced to consider the canonical triangle ith the sorce as apex, the argments that decreased the spanning ratio cannot be applied. Hence, e obtain the folloing corollary. Corollary 1 The -roting algorithm is 1 + sin cos -competitive on the cos sin 3 k+3 - graph and the k+5 -graph. 5 Loer Bonds In this section, e provide loer bonds for the k+3 -graph, the k+ -graph, and the k+5 -graph. For each of the families, e constrct a loer bond example by extending 19

20 the shortest path beteen to vertices and. For brevity, e describe only ho to extend one of the shortest paths beteen these vertices. To extend all shortest paths beteen and, the same transformation is applied to all eqivalent paths or canonical triangles. For example, hen constrcting the loer bond for the k+3 -graph, or first step is to ensre that there is no edge beteen and. To this end, the proof of Theorem 15 states that e place a vertex v 1 in the corner of T that is frthest from. Placing only this single vertex, hoever, does not prevent the edge from being present, as is still the closest vertex in T. Hence, e also place a vertex in the corner of T that is frthest from. Since these to modifications are essentially the same, bt applied to different canonical triangles, e describe only the placement of one of these vertices. The fll reslt of each step is shon in the accompanying figres. 5.1 Loer Bonds on the k+3 -Graph In this section, e constrct a loer bond on the spanning ratio of the k+3 -graph, for any integer k 1. Theorem 15 The orst case spanning ratio of the k+3 -graph is at least 3 cos + cos 3 + sin 3 cos + cos 3 + sin + sin 3. Proof. We constrct the loer bond example by extending the shortest path beteen to vertices and in three steps. We describe only ho to extend one of the shortest paths beteen these vertices. To extend all shortest paths, the same modification is performed in each of the analogos cases, as shon in Figre 13. v v v 1 v 1 v 1 v 3 a b c Figre 13: The constrction of the loer bond for the k+3 -graph 0

21 First, e place sch that the angle beteen and the bisector of the cone of that contains is /. Next, e ensre that there is no edge beteen and by placing a vertex v 1 in the pper corner of T that is frthest from see Figre 13a. Next, e place a vertex v in the corner of T v1 that lies otside T see Figre 13b. Finally, to ensre that there is no edge beteen v and, e place a vertex v 3 in T v sch that T v and T v3 have the same orientation see Figre 13c. Note that e cannot place v 3 in the loer right corner of T v since this old case an edge beteen and v 3 to be added, creating a shortct to. One of the shortest paths in the reslting graph visits, v 1, v, v 3, and. Ths, to obtain a loer bond for the k+3 -graph, e compte the length of this path. v v 1 m v 3 v v 3 Figre 1: The loer bond for the k+3 -graph Let m be the midpoint of the side of T opposite. By constrction, e have that v 1 m = /, m = v v 1 = v 3 v = /, v 3 v = 3/, v 1 = v 1 v = / /, and v v 3 = see Figre 1. We can express the varios line segments as follos: v 1 = cos cos sin 3 v 1 = sin v 1 v = cos cos v 1 v = v v 3 = sin 3 = sin 3 cos sin sin v 1 = sin cos v 1 sin v = sin 3 sin v 1

22 v 3 = sin sin v = sin sin v Hence, the total length of the shortest path is v 1 + v 1 v + v v 3 + v 3, hich can be reritten to 3 cos + cos 3 + sin + sin + sin 3 3 cos + cos 3, proving the theorem. 5. Loer Bond on the k+ -Graph The k+ -graph has the nice property that any line perpendiclar to the bisector of a cone is parallel to the bondary of a cone Lemma 1. As a reslt of this, if, v, and are vertices ith v in one of the pper corners of T, then T v is completely contained in T. The k+ -graph does not have this property. In this section, e sho ho to exploit this to constrct a loer bond for the k+ -graph hose spanning ratio exceeds the orst case spanning ratio of the k+ -graph. Theorem 16 The orst case spanning ratio of the k+ -graph is at least 1 + tan + tan. Proof. We constrct the loer bond example by extending the shortest path beteen to vertices and in three steps. We describe only ho to extend one of the shortest paths beteen these vertices. To extend all shortest paths, the same modification is performed in each of the analogos cases, as shon in Figre 15. v v v 3 v 1 v 1 v 1 a b c Figre 15: The constrction of the loer bond for the k+ -graph

23 First, e place sch that the angle beteen and the bisector of the cone of that contains is /. Next, e ensre that there is no edge beteen and by placing a vertex v 1 in the pper corner of T that is frthest from see Figre 15a. Next, e place a vertex v in the corner of T v1 that lies in the same cone of as and v 1 see Figre 15b. Finally, e place a vertex v 3 in the intersection of the left bondary of T v and the right bondary of T v to ensre that there is no edge beteen v and see Figre 15c. Note that e cannot place v 3 in the loer right corner of T v since this old case an edge beteen and v 3 to be added, creating a shortct to. One of the shortest paths in the reslting graph visits, v 1, v, v 3, and. Ths, to obtain a loer bond for the k+ -graph, e compte the length of this path. v v 1 m v 3 Figre 16: The loer bond for the k+ -graph Let m be the midpoint of the side of T opposite. By constrction, e have that v 1 m = m = v v 1 = v 3 v = v 3 v = / see Figre 16. We can express the varios line segments as follos: v 1 = v 1 = sin v 1 v = v 1 cos = tan v = tan v 1 = sin tan 1 v v 3 = v 3 = v 1 cos = tan 3

24 Hence, the total length of the shortest path is v 1 + v 1 v + v v 3 + v 3, hich can be reritten to 1 + tan + tan. 5.3 Loer Bonds on the k+5 -Graph In this section, e give a loer bond on the spanning ratio of the k+5 -graph, for any integer k 1. Theorem 17 The orst case spanning ratio of the k+5 -graph is at least 1 sec + 7 sec + sec 3 + sec 8 cos + tan + 1 sec tan. Proof. We constrct the loer bond example by extending the shortest path beteen to vertices and in to steps. We describe only ho to extend one of the shortest paths beteen these vertices. To extend all shortest paths, the same modification is performed in each of the analogos cases, as shon in Figre 17. v v v 1 v 1 a b Figre 17: The constrction of the loer bond for the k+5 -graph

25 First, e place sch that the angle beteen and the bisector of the cone of that contains is /. Next, e ensre that there is no edge beteen and by placing a vertex v 1 in the pper corner of T that is frthest from see Figre 17a. Finally, e place a vertex v in the corner of T v1 that lies otside T. We also place a vertex v in the corner of T v1 that lies in the same cone of as and v 1 see Figre 17b. Note that placing v creates a shortct beteen and v, as is the closest vertex in one of the cones of v. One of the shortest paths in the reslting graph visits, v, and. Ths, to obtain a loer bond for the k+5 -graph, e compte the length of this path. v v 1 m Figre 18: The loer bond for the k+5 -graph Let m be the midpoint of the side of T opposite. By constrction, e have that v 1 m = /, m = /, v 1 v = 3/, and v 1 v = v 1 + v 1 v = / + / 3/ = 3/ see Figre 18. We can express the varios line segments as follos: v 1 = cos cos v = cos cos v 1 v = v = sin sin + cos tan + cos tan v 1 + v 1 v v 1 v 1 v cos 3 5

26 Hence, the total length of the shortest path is v + v, hich can be reritten to 1 sec times the length of. + 7 sec + sec 3 + sec 8 cos + tan + 1 sec tan 6 Comparison In this section e prove that the pper and loer bonds of the for families of -graphs admit a partial ordering. We need the folloing lemma that can be proved by elementary calcls. Lemma 18 Let x [ 0, ] be a real nmber. Then the folloing ineqalities hold: 1. sinx x ith eqality if and only if x = 0.. cosx 1 x 3. sinx x x3 6 ith eqality if and only if x = 0. ith eqality if and only if x = 0.. cosx 1 x + x ith eqality if and only if x = tanx x ith eqality if and only if x = tan x x ith eqality if and only if x = 0. Using the above properties, e proceed to prove a nmber of relations beteen the for families of -graphs. 6

27 Lemma 19 Let bm and lbm denote the pper and loer bond on the m -graph: 1 + sin k+ if m = k + k 1 bm = cos k+3 cos k+3 sin 3 k sin k+ cos k+ sin k+ cos k+5 cos k+5 sin 3 k sin k+ if m = k + 3 k 1 if m = k + k 1 if m = k + 5 k 1 if m = k + k cos k+3+cos k+3+sin k+3+sin k+3+sin k cos k+3+cos k+3 3 lbm = 1 + tan k+ + tan k+ if m = k + 3 k 1 if m = k + k 1 sec k+5+7 sec k+5+ sec 3 k+5+sec k+5 8 cos k+5 tan + tan k sec k+5 k+5 Then the folloing ineqalities hold here k is an integer. if m = k + 5 k 1 bk < lbk + k 1 a bk < lbk + 3 k 1 b bk < lbk + k 1 c bk < lbk + 5 k 1 d bk + < lbk + k 1 e bk < lbk + k 1 f bk < lbk + 3 k 1 g bk < lbk + 5 k 1 h bk + 5 < lbk + k i Proof. We se the same strategy for each ineqality. We se the definitions of b and lb in combination ith Lemma 18. Notice that the restriction on k in each of these ineqalities ensres that e can apply Lemma 18. We are then left ith an algebraic ineqality that can be translated into a polynomial ineqality, hich is easy to verify. 7

28 a bk = 1 + sin k < 1 + k < k + 6 k + < 1 + sin k + by the definition of b, by Lemma 18-1, see belo, 3 by Lemma 18-3, = lbk + by the definition of lb. We no explain hy 3 holds. The ineqality 1 + < 1 + k k k + can be simplified to 19k + 19 k > 0. The largest real root of the polynomial involved in is negative. Moreover, 3 holds for k = 1. Therefore, 3 holds for any k 1. b The proof is analogos to the one of a. c The proof is analogos to the one of a. d We let so that cos k+1+5 fk =, 3 cos sin k+1+5 k+1+5 rk = sec + 7 sec k + 5 k + 5 sec 8 cos, k + 5 k + 5 gk = tan + sec tan k + 5 k sec 3 + k + 5 k + 5 bk = fk, rk + gk lbk + 5 =., 8

29 Using a proof similar to the one of a, e can prove that fk gk < rk. Using a proof similar to the one of a, e can prove that fk gk > 0, for k 1, ths e can proceed as follos fk gk < rk rk + gk fk < bk < lbk + 5, for k 1. e The proof is analogos to the one of a. f The proof is analogos to the one of a. g The proof is analogos to the one of d. h The proof is analogos to the one of d. i The proof is analogos to the one of d. We note that ineqalities a, b, c, and d imply that the spanning ratio is monotonic ithin each of the for families. We also note that increasing the nmber of cones of a -graph by from k + to k + increases the orst case spanning ratio, ths shoing that adding cones can make the spanning ratio orse instead of better. Therefore, the spanning ratio is non-monotonic beteen families. Corollary 0 We have the folloing partial order on the spanning ratios of the for families see Figre Tight Roting Bonds While improving the pper bonds on the spanning ratio of the k+ -graph, e also improved the pper bond on the roting ratio of the -roting algorithm. In this section e sho that this bond of 1 + sin// cos/ sin/ and the crrent pper bond of 1/ 1 sin/ on the 10 -graph are tight, i.e. e provide matching loer bonds on the roting ratio of the -roting algorithm on these families of graphs. 9

30 Figre 19: Partial order on the spanning ratios of the for families 7.1 Tight Roting Bonds for the k+ -Graph In this section e sho that the pper bond of 1 + sin//cos/ sin/ on the roting ratio of the -roting algorithm for the k+ -graph is a tight bond. Theorem 1 The -roting algorithm is graph and this bond is tight. 1 + sin -competitive on the cos sin k+ - Proof. Corollary 10 shoed that the roting ratio is at most 1 + sin//cos/ sin/, hence it sffices to sho that this is also a loer bond. We constrct the loer bond example on the competitiveness of the -roting algorithm on the k+ -graph by repeatedly extending the roting path from sorce to destination. First, e place in the right corner of T. To ensre that the -roting algorithm does not follo the edge beteen and, e place a vertex v 1 in the left corner of T. Next, to ensre that the -roting algorithm does not follo the edge beteen v 1 and, e place a vertex v 1 in the left corner of T v1. We repeat this step ntil e have created a cycle arond see Figre 0a. To extend the roting path frther, e again place a vertex v in the corner of the crrent canonical triangle. To ensre that the roting algorithm still rotes to v 1 from, e place v slightly otside of T v1. Hoever, another problem arises: vertex v 1 is no longer the vertex closest to v 1 in T v1, as v is closer. To solve this problem, e also place a vertex x 1 in T v1 v sch that v 1 lies in T x1 see Figre 0b. By repeating this process for times, e create a second cycle arond. 30

31 v 1 v 1 x 1 v 1 v 1 v a b Figre 0: Constrcting a loer bond example for -roting on the k+ -graph: a after constrcting the first cycle, b after adding v, the first vertex of the second cycle, and x 1, the axiliary vertex needed to maintain the first cycle To add more cycles arond, e repeat the same process as described above: place a vertex in the corner of the crrent canonical triangle and place an axiliary vertex to ensre that the previos cycle stays intact. Note that hen placing x i, e also need to ensre that it does not lie in T xi 1, to prevent shortcts from being formed. A loer bond example consisting of to cycles is shon in Figre 1. This ay e need to add axiliary vertices only to the k 1-th cycle, hen adding the k-th cycle, hence e can add an additional cycle sing only a constant nmber of vertices. Since e can place the vertices arbitrarily close to the corners of the canonical triangles, e ensre that v 1 = and that the distance beteen consective vertices v i and v i is alays 1/ cos/ times v i. Hence, hen e take = 1 and let the nmber of vertices approach infinity, e get that the total length of the path is 1 + sin/ i=0 tani // cos/, hich can be reritten to 1+ sin//cos/ sin/. 31

32 v 1 x 1 v 1 v v x Figre 1: A loer bond example for -roting on the k+ -graph, consisting of to cycles: the first cycle is colored orange and the second cycle is colored ble 7. Tight Roting Bonds for the 10 -Graph In this section e sho that the pper bond of 1/1 sin/ on the roting ratio of the -roting algorithm for the 10 -graph is a tight bond. Theorem The -roting algorithm is 1/ 1 sin /-competitive on the 10 -graph and this bond is tight. Proof. Rppert and Seidel [8] shoed that the roting ratio is at most 1/1 sin/, hence it sffices to sho that this is also a loer bond. We constrct the loer bond example on the competitiveness of the -roting algorithm on the 10 -graph by repeatedly extending the roting path from sorce to destination. First, e place in the right corner of T. To ensre that the -roting algorithm does not follo the edge beteen and, e place a vertex v 1 in the left corner of T. Next, to ensre that the -roting algorithm does not follo the edge beteen v 1 and, e place a vertex v 1 in the left corner of T v1. We repeat this step ntil e have created a cycle arond see Figre. To extend the roting path frther, e again place a vertex v in the corner of the crrent canonical triangle. To ensre that the roting algorithm still rotes to v 1 from, e place v slightly otside of T v1. Hoever, another problem arises: vertex v 1 is no 3

33 v 1 v x 1 v v 1 Figre : A loer bond example for -roting on the 10 -graph, consisting of to cycles: the first cycle is colored orange and the second cycle is colored ble longer the vertex closest to v 1 in T v1, as v is closer. To solve this problem, e also place a vertex x 1 in T v1 v sch that v 1 lies in T x1 see Figre 3. By repeating this process for times, e create a second cycle arond. v 1 x 1 v x Figre 3: The placement of vertices sch that previos cycles stay intact hen adding a ne cycle To add more cycles arond, e repeat the same process as described above: place a vertex in the corner of the crrent canonical triangle and place an axiliary vertex to ensre that the previos cycle stays intact. Note that hen placing x i, e also need to ensre that it does not lie in T xi 1, to prevent shortcts from being formed see Figre 3. This means that in general x i does not lie arbitrarily close to the corner of T vi v i+1. 33

34 This ay e need to add axiliary vertices only to the k 1-th cycle, hen adding the k-th cycle, hence e can add an additional cycle sing only a constant nmber of vertices. Since e can place the vertices arbitrarily close to the corners of the canonical triangles, e ensre that the distance to is alays sin/ times the distance beteen and the previos vertex along the path. Hence, hen e take = 1 and let the nmber of vertices approach infinity, e get that the total length of the path is i=0 sin/i, hich can be reritten to 1/ 1 sin/. 8 Conclsion We shoed that the k+ -graph has a tight spanning ratio of 1 + sin/. This is the first time tight spanning ratios have been fond for a large family of -graphs. Previosly, the only -graph for hich tight bonds ere knon as the 6 -graph. We also gave improved pper bonds on the spanning ratio of the k+3 -graph, the k+ -graph, and the k+5 -graph. We also constrcted loer bonds for all for families of -graphs and provided a partial order on these families. In particlar, e shoed that the k+ -graph has a spanning ratio of at least 1 + tan/ + tan /. This reslt is somehat srprising since, for eqal vales of k, the orst case spanning ratio of the k+ -graph is greater than that of the k+ -graph, shoing that increasing the nmber of cones can make the spanning ratio orse. There remain a nmber of open problems, sch as finding tight spanning ratios for the k+3 -graph, the k+ -graph, and the k+5 -graph. Similarly, for the and 5 -graphs, thogh pper and loer bonds are knon, these are far from tight. It old also be nice if e cold improve the roting algorithms for -graphs. At the moment, -roting is the standard roting algorithm for general -graphs, bt it is nclear hether this is the best roting algorithm for general -graphs: thogh e shoed that the crrent bonds on the competitiveness of the -roting algorithm are tight in case of the k+ -graph, this does not imply that there exists no algorithm that can do better on these graphs. As a special case, e note that the -roting algorithm is not on-competitive on the 6 -graph, bt a better tight algorithm is knon to exist []. References [1] N. Bonichon, C. Gavoille, N. Hansse, and D. Ilcinkas. Connections beteen thetagraphs, Delanay trianglations, and orthogonal srfaces. In Proceedings of the 36th International Conference on Graph Theoretic Concepts in Compter Science WG 010, pages 66 78,