Bounded-Degree Spanners in the Presence of Polygonal Obstacles
BBounded-Degree Spanners in the Presence of Polygonal Obstacles
Andr´e van Renssen ∗ and Gladys Wong † University of Sydney, Australia
Abstract
Let V be a finite set of vertices in the plane and S be a finite set of polygonal obstacles.We show how to construct a plane 2-spanner of the visibility graph of V with respect to S . As this graph can have unbounded degree, we modify it in three easy-to-follow steps, inorder to bound the degree to 7 at the cost of slightly increasing the spanning ratio to 6. A geometric graph G consists of a finite set of vertices V ∈ R and a finite set of edges ( p, q ) ∈ E such that the endpoints p, q ∈ V . Every edge in E is weighted according to the Euclideandistance, | pq | , between its endpoints. For any two vertices x and y in G , their distance, d G ( x, y )or d ( x, y ) if the graph G is clear from the context, is defined as the sum of the Euclidean distanceof each constituent edge in the shortest path between x and y . A t -spanner H of G is a subgraphof G where for all pairs of vertices in G , d H ( x, y ) ≤ t · d G ( x, y ). The smallest t ≥ stretch factor or spanning ratio of H . For a comprehensiveoverview on spanners, see Bose and Smid’s survey [11] and Narasimhan and Smid’s book [17].Since spanners are subgraphs where all original paths are preserved up to a factor of t , thesegraphs have applications in the context of geometric problems, including motion planning andoptimizing network costs and delays.Another important factor considered when designing spanners is its maximum degree. Ifa spanner has a low maximum degree, each node needs to store only a few edges, making thespanner better suited for practical purposes. The best degree bound for plane spanners is 4 byBonichon et al. [3], whose spanner had a spanning ratio of 156.82. This result was improvedby Kanj et al. [16], who reduced the spanning ratio to 20. Bose et al. [2] showed that degree3 can be achieved in two special cases. In terms of lower bounds, Dumitrescu and Ghosh [15]showed that there exist point sets that require a spanning ratio of at least 1.4308. They alsostrengthened this bound to 2.1755 for spanners of degree 4 and 2.7321 for spanners of degree 3.Most research has focussed on designing spanners of the complete graph. This implicitlyassumes that every edge can be used to construct the spanner. However, unfortunately, in manyapplications this is not the case. In motion planning we need to move around physical obstaclesand in network design some connections may not be useable due to an area of high interferencebetween the endpoints that corrupts the messages. This naturally gives rise to the concept of obstacles or constraints . Spanners have been studied for the case where these obstacles form aplane set of line segments. It was shown that a number of graphs that are spanners without ∗ Email: [email protected] † Email: [email protected] a r X i v : . [ c s . C G ] S e p bstacles remain spanners in this setting [4, 6, 8, 12]. In this paper we consider more complexobstacles, namely simple polygons.Let S be a finite set of simple polygonal obstacles where each corner of each obstacle is avertex in V , such that no two obstacles intersect. Throughout this paper, we assume that eachvertex is part of at most one polygonal obstacle and occurs at most once along its boundary,i.e., the obstacles are vertex-disjoint simple polygons. Note that V can also contain verticesthat do not lie on the corners of the obstacles. Two vertices are visible to each other if andonly if the line segment connecting them does not properly intersect any obstacles (i.e., theline segment is allowed to touch obstacles at vertices or coincide with its boundary, but it isnot allowed to intersect the interior). The line segment between two visible points is called a visibility edge . The visibility graph of a given point set V and a given set of polygonal obstacles S , denoted V is ( V, S ), is the complete graph on V excluding all the edges that properly intersectsome obstacle. It is a well-known fact that the visibility graph is connected.Clarkson [13] was one of the first to study this problem, showing how to construct a (1 + (cid:15) )-spanner of V is ( V, S ). Modifying this result, Das [14] showed that it is also possible to constructa spanner of constant spanning ratio and constant degree. Recently, Bose et al. [6] constructeda 6-spanner of degree 6 + c , where c is the number of line segment obstacles incident to avertex. In the process, they also show how to construct a 2-spanner of the visibility graph.We generalize these results and construct a 6-spanner of degree at most 7 in the presence ofpolygonal obstacles, simplifying some of the proofs in the process. Leading up to this mainresult, we first construct the polygon-constrained half-Θ -graph, denoted G ∞ , a 2-spanner ofthe visibility graph of unbounded degree. We modify this graph in a sequence of three steps,each giving a plane 6-spanner of the visibility graph to bound the degree to 15, 10, and finally 7.Each of these graphs may be of independent interest. Specifically, the graphs with degree 10and 15 are constructed by solely removing edges from G ∞ . Furthermore, in the graph of degree15 the presence of an edge is determined by the vertices at its endpoints, whereas for the othertwo this involves the neighbors of these vertices. Hence, depending on the network model andcommunication cost, one graph may be more easily applicable than another. Throughout this paper, we assume that each vertex is part of at most one polygonal obstacle andoccurs at most once along its boundary, i.e., we have vertex-disjoint simple polygons. We notethat we can always duplicate the vertex and possibly split the polygon, depending on whetherwe allow a path to go through this vertex in order to reach the opposite side of the polygon. Anexample of this is shown in Figure 1. This operation does not affect the spanning ratio of theresulting graph, and the effect on the degree bound is minor: using Lemma 5 from [6], it can beshown that the degree of a vertex v of the original graph is 6 plus half the number of polygonedges incident on v . (a) (b) (c) Figure 1: (a) Two polygons share a vertex. (b) Modified input when paths are allowed to passbetween the polygons. (c) Modified input when paths are not allowed to pass between thepolygons. 2ext, we describe how to construct the polygon-constrained half-Θ -graph, G ∞ . Before wecan construct this graph, we first partition the plane around for each vertex u ∈ V into sixcones, each with angle θ = π and u as the apex. For ease of exposition, we assume that thebisector of one cone is a vertical ray going up from u . We refer to this cone as C u or C whenthe apex u is clear from the context. The cones are then numbered in counter-clockwise order( C , C , C , C , C , C ) (see Figure 2). Cones of the form C i are called positive cones , whereascones of the form C ui are negative cones . For any two vertices u and v in V , we have the propertythat if v ∈ C ui then u ∈ C vi . C C C C C C u Figure 2: The cones around u . u p p Figure 3: The positive cone C ui is split intotwo subcones and an edge is added in each.We are now ready to construct G ∞ . For every vertex u ∈ V , we consider each of its positivecones and add a single edge in such a cone. More specifically, we consider all vertices visible to u that lie in this cone and add an undirected edge to the vertex whose projection on the bisectorof the cone is closest to u . More precisely, the edge ( u, v ) is part of G ∞ when v is visible to u and | uv (cid:48) | < | uw (cid:48) | for all v (cid:54) = w , where v (cid:48) and w (cid:48) are the projections of v and w on the bisectorof the currently considered positive cone of u . For ease of exposition, we assume that no twovertices lie on a line parallel to one of the cone boundaries and no three vertices are collinear.This ensures that each vertex lies in a unique cone of each other vertex and their projecteddistances are distinct. If a point set is not in general position, one can rotate it by a small anglesuch that the resulting point set is in general position.Since every vertex is part of at most one obstacle, obstacles can affect the construction inonly a limited number of ways. Cones that are split in two (i.e., there are visible vertices onboth sides of the obstacle) are considered to be two subcones of the original cone and we add anedge in each of the two subcones using the original bisector (see Figure 3). If a cone is not split,the obstacle only changes the region of the cone that is visible from u . Since we only considervisible vertices when adding edges, this is already handled by the construction method.We note that the construction described above is similar to that of the constrained half-Θ -graph as defined by Bose et al. [6] for line segment constraints. In their setting a cone canbe split into multiple subcones and in each of the positive subcones an edge is added. Thissimilarity will form a crucial part of the planarity proof in the next section.Before we prove that G ∞ is a plane spanner, however, we first prove a useful visibilityproperty that will form a building block for a number of the following proofs. We note that thisproperty holds for any three points, not just vertices of the input. Lemma 1.
Let u , v , and w be three points where ( w, u ) and ( u, v ) are both visibility edges and u is not a vertex of any polygonal obstacle P where the open polygon P (cid:48) intersects (cid:52) wuv . Thearea A , bounded by ( w, u ) , ( u, v ) , and a convex chain formed by visibility edges between w and v inside (cid:52) wuv , does not contain any vertices and is not intersected by any obstacles. roof. We first construct the convex chain between w and v . Let Q be the set of verticesinside (cid:52) wuv . If Q = ∅ , ( w, v ) is the convex chain. If Q (cid:54) = ∅ , consider the convex hull of Q ∪ { v, w } . After removing edge ( w, v ) we obtain a convex chain between w and v inside (cid:52) wuv .By construction, the area A does not contain any vertices. a b b (cid:48) a (cid:48) w vu A Figure 4: Triangle wuv and the convex chain between w and v with consecutive vertices a and b and their intersections with the triangle’s boundary at a (cid:48) and b (cid:48) . The gray region A is empty.Next, we show, by contradiction, that every edge between two consecutive vertices a and b on the convex chain is a visibility edge. Extend ( a, b ) to the boundaries of the triangle and letthe intersections be a (cid:48) and b (cid:48) (see Figure 4). Since u is not a vertex of any (open) polygonalobstacle intersecting (cid:52) wuv and ( w, u ) and ( u, v ) are visibility edges, any polygonal obstaclecrossing ( a, b ) will have at least one vertex inside (cid:52) a (cid:48) ub (cid:48) . However, since (cid:52) a (cid:48) ub (cid:48) is contained in A , this contradicts that A is empty. Therefore, the convex chain consists of visibility edges.Finally, since A does not contain any vertices and is bounded by visibility edges, the vertices ofany polygonal obstacle intersecting this area have to be contained in the set of vertices bounding A . To also intersect A , this implies that u is one of these vertices, which is contradicts that u isnot a vertex of any such polygon. Hence, A does not contain any polygonal obstacles.For ease of notation, we define the canonical triangle of two vertices u and v with v ∈ C ui ,denoted (cid:53) vu , to be the equilateral triangle defined by the boundaries of C ui and the line through v perpendicular to the bisector of C ui . -Graph In this section we show that graph G ∞ is a plane 2-spanner of the visibility graph. We firstprove it is plane. Lemma 2. G ∞ is a plane graph.Proof. We prove this lemma by proving that G ∞ is a subgraph of the constrained half-Θ -graphintroduced by Bose et al. [6]. Recall that in their graph the set of obstacles is a plane set of linesegments.Given a set of polygonal obstacles, we convert them into line segments as follows: Eachboundary edge on polygonal obstacle P ∈ S forms a line segment obstacle l i . Since these edgesmeet only at vertices and no two obstacles intersect, this gives a plane set of line segments.Recall that the constrained half-Θ -graph constructs subcones the same way as G ∞ does. Thismeans that when considering the plane minus the interior of the obstacles in S , the constrainedhalf-Θ -graphs adds the same edges as G ∞ , while inside P the constrained half-Θ -graph mayadd additional edges (see Figure 5). Hence, G ∞ is a subgraph of the constrained half-Θ -graph.Since Bose et al. showed that their graph is plane, it follows that G ∞ is plane as well.4 (a) G ∞ with one obstacle P . l l l l l l l (b) The constrained half-Θ -graph withconstraints l , l , ..., l . Figure 5: G ∞ is a subgraph of the constrained half-Θ -graph. Edges coinciding with obstacleboundaries are drawn as slight arcs for clarity.Next, we show that G ∞ is a 2-spanner of the visibility graph. Lemma 3.
Let u and v be vertices where ( u, v ) is a visibility edge and v lies in a positive cone of u .Let a and b be the two corners of (cid:53) vu opposite to u and let m be the midpoint of ( a, b ) . There existsa path from u to v in G ∞ such that the path is at most · | uv | ≥ ( √ · cos ∠ muv + sin ∠ muv ) · | uv | in length.Proof. The following proof assumes without loss of generality that v lies in C u (Figure 6a). Weprove the lemma by induction on the area of (cid:53) vu (or the ordering of all triangles (cid:53) vu to be moreprecise). Consider (cid:53) vu for every visibility edge ( u, v ) and let δ ( u, v ) denote the length of theshortest path from u to v in G ∞ that lies within (cid:53) vu . Let A be the triangle (cid:52) uav and B be (cid:52) ubv (see Figure 6b). We use the following induction hypothesis. Induction hypothesis: δ ( u, v ) ≤ | ub | + | bv | , if A = ∅| ua | + | av | , if B = ∅ max {| ua | + | av | , | ub | + | bv |} , if A (cid:54) = ∅ and B (cid:54) = ∅ Base case: If (cid:53) vu is the smallest canonical triangle, v is the closest visible vertex to u inthat positive subcone and thus ( u, v ) is an edge in G ∞ and δ ( u, v ) = | uv | . Therefore, by triangleinequality, the inductive hypothesis holds. Induction step:
Assume that the induction hypothesis holds for every visibility edge withcanonical triangle smaller than (cid:53) vu . If ( u, v ) is an edge in the polygon-constrained half-Θ -graph, the induction hypothesis holds with the same argument as in the base case. Otherwise,let v be the closest visible vertex of u in the subcone of C u that contains v . This definesa canonical triangle (cid:53) v u with a and b at its corners opposite to u (see Figure 6c). Thus, δ ( u, v ) ≤ | uv | + δ ( v , v ) and | uv | ≤ min {| ua | + | a v | , | ub | + | b v |} . Without loss of generality,assume that ( u, v ) lies to the left of ( u, v ) and hence A (cid:54) = ∅ .Applying Lemma 1 on triangle (cid:52) uvv with visibility edges ( u, v ) and ( u, v ), there exists aconvex chain, v , ..., v k , v of visibility edges in (cid:52) uvv between u to v . Note that since v lies inthe same subcone as v , u cannot be a vertex of an obstacle intersecting the triangle, as required5 va bm (a) (cid:53) vu with a , b , and m . u va bA B (b) Triangles A and B . u va bv a b v v (c) The first step of the in-ductive argument. Figure 6: The various points and region defined with respect to (cid:53) vu .for the application of Lemma 1. Since v is the closest visible vertex to u , the vertices v , ..., v k , v on the convex chain are all above the line ( a , b ).Let a i and b i be the corners of canonical triangle of v i − and v i . Let m i be the midpoint of( a i , b i ) and define A i = (cid:52) a i v i − v i and B i = (cid:52) b i v i − v i . There are three possible arrangementsfor each visibility edge ( v i − , v i ) on the convex chain: (a) v i − lies in the cone C v i , (b) v i lies incone C v i − to the right of or on m i , and (c) v i lies to the left of m i in cone C v i − (as shown inFigure 7). We proceed to bound the length of the induction path in each of these cases. v i − v i a i b i A i B i (a) v i − lies in the cone C v i . v i − v i a i b i m i A i B i (b) v i lies in cone C v i − tothe left of m i . v i − a i b i m i A i B i v i (c) v i lies in cone C v i − tothe right of m i . Figure 7: The three configurations of visibility edges on the convex chain.
Type (a):
Since B i is contained in the empty region of Lemma 1 (it lies between the convexchain and ( u, v )), it is empty. Since ( v i − , v i ) is a visibility edge and the area of (cid:53) v i − v i is smallerthan the area of (cid:53) vu , the induction hypothesis implies that δ ( v i − , v i ) ≤ | v i − a i | + | a i v i | . Type (b):
Since ( v i − , v i ) is a visibility edge and the area of (cid:53) v i v i − is smaller than the areaof (cid:53) vu , the induction hypothesis implies that δ ( v i − , v i ) ≤ max {| v i − a i | + | a i v i | , | v i − b i | + | b i v i |} .Since v i lies to the right of or on m i , | a i v i | ≥ | b i v i | and therefore | v i − a i | + | a i v i | ≥ | v i − b i | + | b i v i | and δ ( v i − , v i ) ≤ | v i − a i | + | a i v i | . Type (c):
Since ( v i − , v i ) is a visibility edge and the area of (cid:53) v i v i − is smaller than the areaof (cid:53) vu , we can apply the induction hypothesis. If B i = ∅ then δ ( v i − , v i ) ≤ | v i − a i | + | a i v i | .Otherwise, δ ( v i − , v i ) ≤ | v i − b i | + | b i v i | , since | a i v i | < | b i v i | .Note that the three types of visibility edges appear in order on the convex chain whendirected from u to v . To complete the proof, we consider the three possible convex chains,depending on the location of v with respect to m and whether B is empty. Case (a): If v lies on mb , the convex chain can only contain Type (a) and Type (b)configurations. We can bound the total length of the induction path by summing up the twocomponents of the inductive lengths of each visibility edge ( v i − , v i ) (see Figure 8a). The6 va b (a) v lies on segment mb . uva b (b) v lies on segment am and B = ∅ . uva ba j b j (c) v lies on segment am and B (cid:54) = ∅ . Bounding the two parts of theconvex chain. uva ba i b i (d) v lies on segment am and B (cid:54) = ∅ . Summing up to obtain thetotal path length. Figure 8: Bounding the total length of the three possible compositions of the convex chain.horizontal components sum up to | av | and the components parallel to ua sum up to | ua | . Hence,we can conclude that δ ( u, v ) ≤ | ua | + | av | . Case (b): If v lies on am and B = ∅ , the convex chain can contain all three types of visibilityedges. However, since B is empty and using the empty area implied by Lemma 1, the areabetween the convex chain and ( u, v ) is empty. Hence, B i is empty for each canonical trianglealong the convex chain. Therefore, for visibility edges of Type (c), δ ( v i − , v i ) ≤ | v i − a i | + | a i v i | .The total length of the path can now be bounded the same way as was done in the previouscase (see Figure 8b). Thus, δ ( u, v ) ≤ | ua | + | av | . Case (c): If v lies on am and B (cid:54) = ∅ , the convex chain can again contain all three typesof visibility edges. Since v lies in A , both A and B are not empty. Thus, since | av | < | vb | , itsuffices to show that δ ( u, v ) ≤ | ub | + | bv | . Let ( v j , v j +1 ) be the first Type (c) visibility edgeand let a j and b j be the two corners of (cid:53) v j u . We can sum up the length of the path up to v j the same way we did in case (a), since this part only contains Type (a) and Type (b) visibilityedges (see Figure 8c). This gives that this part of the path has length at most | ua j | + | a j v j | .Next, since v lies on am , | a j v j | < | b j v j | . Thus, the length of the first part of the path is at most | ub j | + | b j v j | . Adding to this the lengths of all Type (c) confirgurations, we obtain the followingbound: δ ( u, v ) ≤ | ub | + | bv | (see Figure 8d).Thus, in all cases the induction hypothesis is satisfied. Using basic trigonometry, we canexpress the maximum value of the induction hypothesis as √ · cos ∠ muv + sin ∠ muv , for ∠ muv ∈ [0 , π/ ∠ muv = π/
6, where it has value 2. Therefore,polygon-constrained half-Θ -graph is a spanner with spanning ratio 2.7hese two lemmas imply the following theorem. Theorem 4. G ∞ is a plane -spanner of the visibility graph. We note that while every vertex in G ∞ has at most one edge in every positive subcone, itcan have an unbounded number of edges in its negative subcones. In the following sections, weproceed to bound the degree of the spanner. In this section, we introduce G , a subgraph of the polygon-constrained half-Θ -graph ofmaximum degree 15. We obtain G from G ∞ by, for each vertex, removing all edges fromits negative subcones except for the leftmost (clockwise extreme from u ’s perspective) edge,rightmost (counterclockwise) edge, and the edge to the closest vertex in that cone (see Figure 9a).Note that the edge to the closest vertex may also be the leftmost and/or the rightmost edge. uv v v v v (a) In C u only the edges to v , v , and v are kept. uv v v v v (b) The canonical path in C u with canonical sequence( v , v , v , v , v ). Figure 9: Transforming G ∞ into G .By simply counting the number of subcones (three positive, three negative, and at most oneadditional subcone caused by an obstacle), we obtain the desired degree bound. Furthermore,since G is a subgraph of G ∞ , it is also plane. Lemma 5.
The degree of each vertex u in G is at most .Proof. By construction, there could only be one edge lying in each positive subcone of vertex u . During the transformation to G , we removed all edges except at most three edges in eachnegative subcone of u . Temporarily ignoring obstacles, there are three positive and three negativecones and therefore, the degree bound without obstacles is 12. uP u Figure 10: Negative cone split by obstacle P u has two subcones and at most 6 edges.8ecall that each vertex u ∈ V is part of at most one obstacle and denote this obstacle by P u . If P u lies across multiple cones, the corresponding cones will only shrink in size and thenumber of subcones of u will not increase. Otherwise, P u splits a cone into two subcones. If P u lies within C ui , this creates an extra positive subcone with one edge and the degree bound willincreases to 13. If P u lies within C ui , this creates an extra negative subcone with at most threeedges and the degree bound increases to 15 (see Figure 10), completing the proof. Lemma 6. G is plane.Proof. G is constructed by removing edges from the negative subcones of vertices in G ∞ .Therefore, G is a subgraph of G ∞ . Since by Lemma 2 G ∞ is a plane graph, G is also a planegraph.Finally, we look at the spanning ratio of G . We assume without loss of generality thatwe look at two vertices u and v such that v ∈ C u and u ∈ C v . Given vertex u and a negativesubcone, we define the canonical sequence of this subcone as the vertices adjacent to u in G ∞ that lie in the subcone in counterclockwise order (see Figure 9b). The canonical path of C u refers to the path that connects the consecutive vertices in the canonical sequence of C u . Thisdefinition is not dissimilar to that of Bose et al. [6]. Lemma 7. If v i and v i +1 are consecutive vertices in the canonical sequence of the negativesubcone C u , then (cid:52) uv i v i +1 is empty.Proof. By Lemma 2, G ∞ is a plane graph and by construction ( u, v i ) , ( u, v i +1 ) ∈ G ∞ , thereforeno edges or polygonal obstacles can intersect these sides of (cid:52) uv i v i +1 . Hence, any obstacle oredge intersecting (cid:52) uv i v i +1 has at least one vertex within the triangle.Let A be the intersection of (cid:52) uv i v i +1 with the positive subcones of v i and v i +1 that contain u (see Figure 11a). We prove by contradiction that A = ∅ . Let x be the highest vertex in C v i ∩ A .Since ( u, v i ) and ( u, v i +1 ) exist in G ∞ , ( u, v i ) is a visibility edge and any obstacles blocking thevisibility between u and x has to have at least a vertex above x in C v i ∩ A . Therefore, ( u, x )is a visibility edge. By applying Lemma 1 to (cid:52) uv i x , we obtain a convex chain from v i to x .Since the neighbor of v i along the convex chain lies in C v i and is visible to v i , u cannot be theclosest visibility vertex of v i , contradicting that ( u, v i ) ∈ G ∞ . An analogous argument showsthat C v i +1 ∩ A is empty and hence, A is empty. uv i v i +1 A (a) Area A is empty. uv i v i +1 B (b) Area B , (cid:52) uv i v i +1 \ A ,is empty. Figure 11: Area A and B are both empty, hence (cid:52) uv i v i +1 is empty.Triangle (cid:52) uv i v i +1 \ A is denoted as area B . We show by contradiction that B is alsoempty. Let y ∈ B be the highest vertex in B (see Figure 11b). Since A = ∅ and no edges orobstacles pass through ( u, v i ) and ( u, v i +1 ), ( u, y ) is a visibility edge that lies within C u . Since y ∈ B ⊆ (cid:52) uv i v i +1 , C y lies within ( C v i ∪ C v i +1 ∪ B ). As u is the closest visible vertex in both9 v i and C v i +1 , u is also the closest visible vertex in C y and therefore, ( u, y ) exists in G ∞ . Thisimplies that y is in the canonical sequence between v i and v i +1 , contradicting that v i and v i +1 are consecutive. Hence, B = ∅ .Since A = ∅ and B = ∅ , (cid:52) uv i v i +1 = A ∪ B = ∅ . Finally, since any obstacle being fullycontained in (cid:52) uv i v i +1 would imply that it contains some vertices as well, it is empty of bothvertices and obstacles. Lemma 8. G contains an edge between every pair of consecutive vertices on a canonical path.Proof. We first prove that the edges on the canonical path are in G ∞ . Let v and v be a pairof consecutive vertices in the canonical sequence in C u . Assume, without loss of generality,that v ∈ C v and v ∈ C v . Let A be the area bounded by the positive subcones C v and C v that contain u (see Figure 12a). Let A (cid:48) be the set of vertices visible to v or v in A . We firstshow by contradiction that A (cid:48) = ∅ . According to Lemma 7, (cid:52) uv v is empty and thus does notcontain any vertices. We focus on the part of A to the left of uv . Consider vertex x ∈ A to theleft of uv where x has the smallest interior angle ∠ xuv . Since ( u, v ) is a visibility edge, theedge ( u, x ) is a visibility edge, as any obstacle blocking it implies the existence of a vertex withsmaller angle. Applying Lemma 1 to xuv , there exists a convex chain from v to x in (cid:52) v xu .The neighbor of v in the convex chain is a closer visible vertex than u in C v , contradictingthat ( u, v ) ∈ G ∞ . An analogous argument for the region to the right of uv contradicts that( u, v ) ∈ G ∞ . Therefore, A does not contain any vertices visible to v or v . uv v A (a) Area A is bounded by C v and C v . uv v B (b) Triangle B is bounded by C v and C v . uv v C (c) Triangle C is (cid:53) v v . Figure 12: A pair of consecutive vertices v and v along a canonical path and their surroundingempty areas.Next, let B be the intersection of C v and C v (see Figure 12b). Since ( v , v ) is part ofthe canonical path of C u , by Lemma 7, (cid:52) uv v = ∅ . Since B is contained in (cid:52) uv v , B is alsoempty.Finally, let area C be the canonical triangle (cid:53) v v (see Figure 12c). Let C (cid:48) be the set of visiblevertices to v and v in C . We show that C (cid:48) = ∅ , by contradiction. Since both A and B areempty, the highest vertex x ∈ C will have u as the closest visible vertex in C x . This impliesthat x occurs between v and v in the canonical path, contradicting that they are consecutivevertices. Therefore, C (cid:48) = ∅ .Since A , B , and C are all empty, v is the closest vertex of v in the subcone of C v thatcontains it. By Lemma 7 it is visible to v and thus, ( v , v ) is an edge in G ∞ .It remains to show that ( v , v ) is preserved in G . Since (cid:52) uv v is empty and G isplane (by Lemma 6), v is the rightmost vertex in C v . Hence, it is not removed and thus, thecanonical path in C u exists in G .Now that we know that these canonical paths exist in G , we can proceed to prove that itis a spanner. 10 emma 9. G is a -spanner of G ∞ .Proof. Consider an edge ( u, v ) in G ∞ and assume, without loss of generality, that v ∈ C u andlet v be the closest vertex that u has an edge to in C u . Consider the part of the canonicalpath between v and v and denote its vertices by v , v , ..., v k = v (see Figure 13). The pathfrom u via the vertices of the canonical path to v is an upper bound of δ ( u, v ), the length of theshortest path from u to v in G . u vc abv v v v c (cid:48) w a (cid:48) Figure 13: Negative cone C u with canonical sequence v , v , ..., v k = v .Let c and c (cid:48) be the intersections of the cone boundaries of C u with the line perpendicularto the bisector of C u that intersects v . Let a be the point on uc such that av is parallel to uc (cid:48) and let b be the point on uc such that bv is parallel to uc (cid:48) . Consider the reflection a (cid:48) of a overthe line cv . We now show that no vertices on the canonical path between v and v lie outside (cid:52) a (cid:48) cv ∪ (cid:52) acv , since any vertex x , below (cid:52) a (cid:48) cv would have v or v ∈ C x , or occur before v orafter v on the canonical sequence. If x occurs before v or after v on the canonical sequence, itis not part of the canonical path between v and v . If v ∈ C x is visible to x , it is closer than u and ( u, x ) could not exist in G ∞ . If v is not visible to x , then there exists a vertex y of theobstacle P blocking v in C x (as ( u, x ) needs to be visible for x to lie on the canonical path of u , P cannot block the cone entirely). Hence, C x contains vertices that are closer than u and sincethe cone cannot be completely blocked, the vertex z that minimizes the angle ∠ uxz is visible to x . Consequently, ( u, x ) could not exist in G ∞ . An analogous argument shows that if v ∈ C x ,( u, x ) does not exist.It remains to bound the length of the canonical path. By triangle inequality, | uv | ≤ | ub | + | bv | .For every consecutive pair of vertices v i − , v i , let w i be the intersection of the right cone boundaryof C v i − and the left cone boundary of C v i (see Figure 13). By triangle inequality, we have that | v i − v i | ≤ | v i − w i | + | w i v i | . Since all line segments of the form v i − w i are parallel to ac andall line segments of the form w i v i are parallel to av , summing them up gives | ab | − | bv | + | av | .Since v lies in (cid:52) ucc (cid:48) , | ab | ≤ | uc | . By construction (cid:52) ucc (cid:48) is an equilateral triangle and since av is parallel to uc (cid:48) , (cid:52) acv is also an equilateral triangle and | ac | = | cv | = | av | .Adding the upper bound on | uv | to the length of the canonical path, we get δ ( u, v ) ≤| ub | + | bv | + | ab | + | av | − | bv | . Since | ab | ≤ | ac | , | ub | ≤ | uc | , and | ac | = | cv | = | av | , wecan rewrite this to δ ( u, v ) ≤ | uc | + 2 · | cv | . Using basic trigonometric functions, we get that | uc | = (cos ∠ vuc + sin ∠ vuc/ √ · | uv | and | cv | = 2 · sin ∠ vuc/ √ · | uv | . This implies that11 ( u, v ) ≤ (cos ∠ vuc + √ · sin ∠ vuc ) · | uv | where 0 < ∠ vuc ≤ π . As this is an increasing function,it is maximal at ∠ vuc = π , where it attains value 3. Therefore, G is a 3-spanner of G ∞ .Since by Lemma 3, G ∞ is a 2-spanner of the visibility graph and by Lemma 9, G is a3-spanner of G ∞ , we obtain the main result of this section. Theorem 10. G is a plane -spanner of the visibility graph of degree at most 15. Observing that we need to maintain only the canonical paths between vertices, along with theedge to the closest vertex in each negative subcone for the proof of Lemma 9 to hold, we reducethe degree as follows. Consider each negative subcone C ui of every vertex u and keep only theedge incident to the closest vertex (with respect to the projection onto the bisector of the cone)and the canonical path in C ui (see Figure 14). We show that the resulting graph, denoted G ,has degree at most 10. Recall that at most one cone per vertex is split into two subcones andthus this cone can have two disjoint canonical paths, as we argue in Lemma 13. uv v v v v Figure 14: G keeps only the canonical path and the edge to the closest adjacent vertex.As in the previous section, we proceed to prove that G is a plane 6-spanner of the visibilitygraph of degree 10. Since G is a subgraph of G , it is also plane. Furthermore, since thecanonical path is maintained, the spanning property does not change. Lemma 11. G is a plane graph.Proof. Lemma 8 shows that G contains an edge between consecutive vertices along eachcanonical path. Since G consists of these canonical paths and the edge to the closest vertex, allof its edges are part of G . By Lemma 6, G is a plane graph and thus G is also plane. Lemma 12. G is a -spanner of G ∞ .Proof. According to Lemma 9, the canonical paths in G are the 3-spanning paths for eachedge in G ∞ . Since G preserves these canonical paths from G , these same paths in G stillserve as 3-spanning paths for each edge in G ∞ . Thus, G is a 3-spanner of G ∞ .It remains to upper bound the maximum degree of G . We do this by charging the edgesincident to a vertex to its cones. The four part charging scheme is described below (see Figure 15).Scenarios A and B handle the edge to the closest vertex, while Scenarios C and D handle theedges along the canonical path. Hence, the total charge of a vertex is an upper bound on itsdegree. Scenario A:
Edge ( u, v i ) lies in C v i j and v i is the closest vertex of u in C uj . Then ( u, v i ) ischarged to C v i j . 12 v i (a) Scenario A. uv i (b) Scenario B. uv i v i +1 (c) Scenario C. uv i v i +1 (d) Scenario D. Figure 15: The four scenarios of the charging scheme. The arrows in Scenarios C and D indicatethe cone these edges are charged to.
Scenario B:
Edge ( u, v i ) lies in C uj and v i is the closest vertex of u in C uj . Then ( u, v i ) ischarged in C uj . Scenario C:
Edge ( v i , v i +1 ) lies on a canonical path of u , with u ∈ C v i j , and v i +1 ∈ C v i j +1 .Edge ( v i , v i +1 ) is charged to C v i j . Similarly, edge ( v i , v i − ) on a canonical path of u , where v i − ∈ C v i j − , is charged to C v i j . Scenario D:
Edge ( v i , v i +1 ) lies on a canonical path of u , with u ∈ C v i j , and v i +1 ∈ C v i j − .Edge ( v i , v i +1 ) is charged to C v i j +1 . Similarly, edge ( v i , v i − ) on a canonical path of u , where v i − ∈ C v i j +1 , is charged to C v i j − .We note that every edge in G is charged to both of its endpoints. Scenarios A and Bconsider an edge formed and preserved as the shortest edge, i.e., edge ( u, v ) in Figure 14.Scenarios C and D look at edges on canonical paths, where Scenario C handles edges in negativesubcones and Scenario D takes care of the positive ones. Lemma 13.
The degree of every vertex in G is at most 10.Proof. Negative subcones:
We first prove that there can be only one edge charged to eachnegative subcone. According to the charging scheme, only Scenario B and D charge edges to anegative subcone. We first consider Scenario B, where u is the closest vertex of v i in its negativesubcone C v i j . By definition of G , edge ( u, v i ) is the only edge preserved and charged C v i j by thisscenario. If v i is part of a polygonal obstacle P and P splits the negative cone (see Figure 16),the number of negative subcones increases. Both negative subcones are charged 1 if they bothhave an edge belonging to scenario B. v i u v (cid:48) i P Figure 16: Two negative subcones can both be charged once from Scenario B.For Scenario D, the edge ( v i , v i +1 ) lies in C v i j − and is charged to C v i j +1 . Since ( v i , v i +1 ) is anedge in the canonical path of C uj , there is only one edge from this canonical path in C v i j − with v i as an endpoint. Next, we show that C v i j +1 is not charged by Scenario B, implying that its totalcharge is 1. Consider triangle (cid:52) uv i v i +1 . Since this triangle was present in G ∞ , by Lemma 7 it13ontains no polygonal obstacles. We split C v i j +1 into two parts: A lies outside (cid:52) uv i v i +1 and B lies inside (cid:52) uv i v i +1 (see Figure 17). uv i v i +1 AB Figure 17: Area A is shaded in gray and area B is shaded in violet.Consider area A . Since edge ( v i +1 , u ) existed in G ∞ , which by Lemma 2 G ∞ is plane, and G is a subgraph of G ∞ , there can be no edges between v i and any vertex in A . Thus, v i cannotbe charged by Scenario B for any vertex in A .Next, consider area B . Since ( v i , v i +1 ) is on the canonical path of C uj , both ( u, v i ) and( u, v i +1 ) exist in G ∞ by Lemma 8. According to Lemma 7, (cid:52) uv i v i +1 is empty and thus B isempty as well. Thus, v i cannot be charged by Scenario B for any vertex in B either.Finally, since B is empty and no polygonal obstacle can cross any side of (cid:52) uv i v i +1 (becauseof the planarity of G ∞ ), no edges can be charged by Scenario D if v i is a vertex of a polygonalobstacle P with polygon boundary in C v i j +1 .Thus, if a negative subcone is charged with an edge by Scenario D, it cannot be charged withany edge by Scenario B. Conversely, if a negative subcone is charged with an edge by ScenarioB, then it cannot be charged with any Scenario D edges either. Since both scenario B and Donly charge one edge to the negative subcone and there are three negative cones (disregardingobstacles), there are at most three edge charged in negative subcones for each vertex. Takingpolygonal obstacles into account and using the assumption that a vertex is part of at mostone obstacle, there are at most four subcones in total and thus at most four edges charged bynegative subcones to a vertex. Positive subcones:
We shift our attention to the positive subcones and prove that eachpositive subcone, C uj , is charged for at most 2 edges. As indicated in the charging scheme, onlyScenarios A and C charge edges to positive subcones.Scenario A charges at most 1 charge to each positive subcone, since by construction, G ∞ only adds one edge in each positive subcone of v i and G does not add additional edges.Scenario C charges at most 2 edges to each positive subcone C v i j , one for each of itsneighbouring negative subcones C v i j − and C v i j +1 . By the definition of the canonical path, everyvertex v i along the path has at most one edge in C v i j +1 (and at most one in C v i j − ) that belongsto the canonical path of subcone C uj .The above two arguments show that every positive subcone has charge at most 3. Usinga more careful analysis, we can show that each positive cone is charged by Scenario A or byScenario C, but not by both. By definition, charges from Scenario A apply when ( u, v i ) is theshortest edge of u in C uj . Hence, if there exists a vertex x in C v i j − or C v i j +1 such that ( x, v i ) is anedge in the canonical path, v i cannot be the closest vertex in C uj as x lies above the horizontalline through v i . Thus, ( u, v i ) would not be preserved during the construction of G . Therefore,if a positive subcone is charged for edge by Scenario A, it cannot have any edges charged byScenario C. Conversely, if a positive subcone is charged by Scenario C, it cannot be chargedwith edges from Scenario A. 14ence, a cone that is not affected by obstacles has at most 2 charges. Next, we look at whatchanges when v i lies on an obstacle P . If P does not split a cone, the above arguments stillapply. Hence, we focus on the case where P splits a positive cone C v i j into two subcones. Let u be the closest vertex of v i in the right subcone of C v i and let u (cid:48) be the closest vertex in its leftsubcone (see Figure 18). uv i v i +1 Pu (cid:48) v i − p i +1 Figure 18: Two positive subcones of v i are charged for one edge by Scenario C.By construction, the maximum number of charges from edges fitting Scenario A stays thesame, 1 per subcone. However, for edges from Scenario C, we show that the maximum chargeper subcone reduces to 1. Let p i +1 be the vertex following v i on the boundary of P on the sideof u . Since ( u, v i ) is an edge of G ∞ , p i +1 is further from v i than u is (or equal if p i +1 = u ), asotherwise p i +1 would be a closer visible vertex than u , contradicting the existence of ( u, v i ).For any vertex v i − in C v i or C v i , u is not visible since ( v i , p i +1 ) blocks visibility and therefore( v i − , v i ) cannot be part of the canonical path in C u . We note that if u = p i +1 , v i − may bevisible to u , but in this case it lies in a different subcone of u and hence would also not bepart of this canonical path. Hence, v i as the leftmost vertex of the canonical path of C u , andusing an analogous argument, v i is the rightmost vertex of the canonical path of C u (cid:48) . Therefore,these positive subcones cannot be charged more than 1 each by Scenario C. Using the fact thatScenario A and C cannot occur at the same time, we conclude that each positive cone is chargedat most 2, regardless of whether an obstacle splits it.Thus, each positive cone is charged at most 2 and each negative subcone is charged at most1. Since there are 3 positive cones and at most 4 negative subcones, the total degree bound is10. Putting the results presented in this section together, we obtain the following theorem. Theorem 14. G is a plane -spanner of the visibility graph of degree at most 10. In order to reduce the degree bound from 10 to 7, we ensure that at most 1 edge is charged toeach subcone. According to Lemma 13, the negative subcones and the positive subcones createdby splitting a cone into two already have at most 1 charge. Hence, we only need to reduce themaximum charge of positive cones that are not split by an obstacle from 2 to 1. The two edgescharged to a positive cone C vi come from edges in the adjacent negative subcones C vi − and C vi +1 (Scenario C), whereas cone C vi itself does not contain any edges. Hence, in order to reduce thedegree from 10 to 7, we need to resolve this situation whenever it occurs in G . We do so byperforming a transformation described below when a positive cone is charged twice for ScenarioC. The graph resulting from applying this transformation to every applicable vertex is referredto as G . 15et u be a vertex in G and let v be a vertex on its canonical path (in C ui ) whose positivecone is charged twice. Let ( x, v ) and ( v, y ) be the edges charged to C ui and assume without lossof generality that x occurs on the canonical path from u to y (see Figure 19a). If neither x nor y is the closest vertex in the respective negative subcone of v that contains them, we remove( v, y ) from the graph and add ( x, y ) (see Figure 19b). This reduces the degree of v , but maycause a problem at x . In order to solve this, we consider the neighbor w of x on the canonicalpath of v . Since x is not the closest vertex in v ’s subcone, this neighbor exists. If w ∈ C xi and w is not the closest vertex of x in the subcone that contains it, we remove ( x, w ) (see Figure 19c). v yx uw (a) The situation beforethe transformation. v yx uw (b) After ( x, y ) is addedand ( v, y ) is removed. v yx uw (c) The situation after thetransformation. Figure 19: Transforming G into G .The reasons for these choices are as follows. If an edge ( a, b ) is part of a canonical path and b is the closest vertex to a in the negative subcone that contains it, then this edge is chargedtwice at a . It is charged once as the edge to the closest vertex (Scenario B) and once as part ofthe canonical path (Scenario C). Hence, we can reduce the charge to a by 1 without changingthe graph.In the case where we end up removing ( v, y ), we need to ensure that there still exists aspanning path between u and y (and any vertex following y on the canonical path). Therefore,we insert the edge ( x, y ). This comes at the price that we now need to ensure that the totalcharge of x does not increase. Fortunately, this is always possible. Lemma 15.
The degree of every vertex in G is at most .Proof. Since each vertex v is part of at most one obstacle, v can have at most 7 subcones. Similarto G , we prove the degree bound of G by demonstrating that the charge of each subcone afterthe transformation is at most 1. Since we showed in Lemma 13 that every negative subconealready has a charge of at most 1 in G , we only need to prove that the transformation toconstruct G reduces the maximum charge of each positive cone (not split by an obstacle) from2 to 1.Let u be a vertex where ( u, v ) ∈ G ∞ and C vi is a positive cone that has 2 edge charges in G . Let x ∈ C vi − and y ∈ C vi +1 where ( x, v ) and ( v, y ) are two edges in the canonical path of C ui (see Figure 19a). If x is the closest vertex of v in C vi − , the edge ( x, v ) is counted in both C vi − (Scenario B) and C vi (Scenario C). Then by simply removing the charge of ( x, v ) in C vi wereduce the charge of this positive cone to 1 whilst maintaining that the total charge of a vertexis an upper bound on the degree of that vertex. Analogously, if ( v, y ) is the closest vertex, it iscounted twice at v . If neither ( x, v ) nor ( v, y ) is doubly counted and assuming, without loss ofgenerality, that x occurs on the canonical path from u to y , we remove one of the two charges of C vi by removing the edge ( v, y ). Cone C vi now has charge 1 as required.If edge ( v, y ) was removed, we then add the edge ( x, y ) in the transformation. For vertex y , edge ( v, y ) is removed and edge ( x, y ) is added, thus the degree of y remains unchanged.16harging ( x, y ) to the subcone C yi +1 (the one that ( v, y ) used to be charged to) will maintainthe charges as well, ensuring that it is still an upper bound on the degree.Next, we consider vertex x whose degree increased when ( x, y ) was added. Let w be theneighbor of x on the canonical path of C vi − . If w is the closest vertex of x in subcone C xi , ( x, w )is counted twice: once in C xi − and once in C xi . Removing the charge of ( x, w ) from C xi − allowsus to now charge ( x, y ) to this cone, leading to the desired charges. If ( x, w ) is not the closestvertex of x in subcone C xi , however, the last step of the transformation is performed and ( x, w )is removed. For vertex x , this maintains the degree, as edge ( x, y ) is added and edge ( x, w ) isremoved. Edge ( x, y ) is charged to the subcone that used to be charged for the removed edge( u, w ) and thus the charging bound of C xi remains unchanged. Finally, vertex w only has anedge removed and thus the charge for ( x, w ) is removed, resulting in its degree and charge toboth be reduced by 1. This completes the proof of the lemma. Lemma 16. G is a -spanner of G ∞ .Proof. For this lemma to stand, we have to prove that for any edge ( a, b ) in G ∞ , there exists apath from a to b in G where the length of the path δ ( a, b ) ≤ · | ( a, b ) | . Since G is provento be a 3-spanner of G ∞ in Lemma 12, it follows that there exists a spanning path betweenany pair of vertices in G except between those affected during the removal of edges. Therefore,we focus only on the paths affected by the transformation. We will use the same notation asin the G construction method. Hence, there are two edges potentially removed during thetransformation from G to G : ( v, y ) and ( x, w ).The removal of ( v, y ) interrupts the path between x and y along the canonical path of C u and hence the paths from u to any vertices on the canonical path after y . The original spanningpath in G includes ( x, v ) and ( v, y ). When removing ( v, y ), we also add edge ( x, y ), so as thenew spanning path in G , we consider the same path as that in G where ( x, v ) and ( v, y ) arereplaced by ( x, y ). By triangle inequality, | xy | ≤ | xv | + | vy | , therefore the spanning ratio of thespanning path from u to y and any vertices after y is still at most 3.Removing edge ( v, y ) also affects the spanning ratio between v and y . By construction,however, ( v, y ) is only removed if y is not the closest visible vertex of v in C v . Hence, thereexists a canonical path between v and y in C v . According to Lemma 12, the length of thiscanonical path is at most 3 · | vy | . Thus, there still exists a spanning path between v and y .Next, we consider the effect of removing edge ( x, w ). By construction, if it is removed, w is not the closest vertex to x in C x . We first show by contradiction that x is the end of thecanonical path of C v and thus has only one neighbour, w , on this path. Let vertex z be theother neighbor of x along the canonical path of C v . Consider the edge ( z, v ) ∈ G ∞ . This edgeeither intersects the boundary of (cid:52) uxv or z lies in its interior. However, by Lemma 7, z cannotlie inside and by Lemma 2 G ∞ is plane. Hence, z cannot exist and thus x is the last vertex onthe canonical path of C v . Since there are no further edges on the canonical path of C v , when( x, w ) is removed, the only paths affected are the one between x and w and the canonical pathfrom v to x .Since w is not the closest vertex to x in C x , there exists a canonical path of C x from x to w . As stated in Lemma 12, this path is a 3-spanning path of ( x, w ). Therefore, there exists aspanning path between x and w .Once ( x, w ) is removed, the path from x to v via the canonical path of C v no longer exists.However, since the edge ( x, v ) is not removed in the transformation, there still exists a 1-spanningpath between x and v .Since for all affected paths, there is still a spanning path with length at most 3 times thelength of the original path in G ∞ , G is a 3-spanner of G ∞ .17t remains to show that G is a plane graph. Lemma 17. G is a plane graph.Proof. By construction, each transformation performed when converting G to G includes theaddition of at most one edge ( x, y ) and the removal of at most two edges ( v, y ) and ( x, w ). ByLemma 11, G is a plane graph and thus the removal of ( v, y ) and ( x, w ) does not affect theplanarity of G . It remains to consider the added edges. Specifically, we need to show that:( x, y ) cannot intersect any edge from G , ( x, y ) cannot intersect an obstacle, and ( x, y ) cannotintersect another edge added during the construction of G .We first prove that the edge ( x, y ) cannot intersect any edge from G . Since x , v , and y areconsecutive vertices in the canonical sequence of a negative subcone of u , according to Lemma 7, (cid:52) uxv and (cid:52) uvy are empty. Furthermore, since the boundary edges of these triangles are edgesin G ∞ , which is plane by Lemma 2, no edges or obstacles can intersect these edges. Hence, since( x, y ) lies inside the quadrilateral uxvy , edge ( u, v ) is the only edge that can intersect ( x, y ).However, since x and y lie in C vi − and C vi +1 , both x and y are closer to u than v in C ui , implyingthat ( u, v ) is not part of G . Therefore, ( x, y ) does not intersect any edge from G .Next, we show that ( x, y ) cannot intersect any obstacles. Since by Lemma 7 no obstacles,vertices, or edges can pass through or exist within (cid:52) uxv and (cid:52) uvy , the only obstacles we needto consider are (cid:52) uxv and (cid:52) uvy themselves and the line segment obstacle uv . We note that inall three cases the obstacle splits the negative cone C ui into two subcones, and x and y lie indifferent negative subcones and thus on different canonical paths (see Figure 20). This impliesthat C ui is not charged twice by the same canonical path and thus, the transformation to G would not add ( x, y ). Therefore, ( x, y ) cannot intersect any obstacles. v yx u Figure 20: The quadrilateral uxvy in G where (cid:52) uvy is an obstacle itself. C u is split into twosubcones and x and y lie on two different canonical paths.Finally, we show that the added edge ( x, y ) cannot intersect another edge added during thetransformation from G to G . We prove this by contradiction. Let e be an added edge thatintersects ( x, y ). Edge e cannot have x or y as an endpoint, as this would imply that it doesnot intersect ( x, y ). Furthermore, since by Lemma 7 (cid:52) uxv and (cid:52) uvy are empty and e cannotintersect edges ( x, v ) and ( v, y ), as shown above, v is one of the endpoints of e . By constructionof G , e is added because either x or y has a positive cone charged by two edges from its adjacentnegative subcones (one of which being the edge to v ). However, this contradicts the fact that v lies in their positive cones ( C xi − and C yi +1 ) implied by the edge ( x, y ) being added duringthe transformation. Therefore, ( x, y ) cannot intersect any edge added while constructing G ,completing the proof.We summarize our main result in the following theorem. Theorem 18. G is a plane -spanner of the visibility graph of degree at most 7. Conclusion
Our goal was to expand upon the constrained half-Θ -graph suggested by Bose et al. [6]supporting more complex obstacles in the form of convex polygons, as opposed to line segmentsobstacles, whilst bounding the degrees of the resulting spanners. We presented a plane spanninggraph G ∞ of the visibility graph with spanning ratio 2 that avoids polygonal obstacles. Inaddition, we presented three bounded-degree 6-spanners. G is constructed by removing edgesin the negative subcones and results in a graph of degree 15. When we allow vertices to requirethe presence of edges between their neighbors, we can build G , reducing the degree boundto 10. Finally, if we are also allowed to add edges that are not part of G ∞ , we showed how toconstruct G , a plane 6-spanner of the visibility graph of degree 7.With the construction of these spanners, future work includes developing routing algorithmsfor them. There has been extensive research into routing algorithms, including routing on thevisibility graph in the presence of line segment obstacles [5, 9, 10]. For polygonal obstacles,Banyassady et al. [1] developed a routing algorithm on the visibility graph with polygonalobstacles, though this work requires some additional information to be stored at the vertices.Another logical next step is to reduce the degree bound further. Without obstacles, Boni-chon et al. [3] constructed a plane 156.82-spanner of degree 4. Kanj et al. [16] improved on thisby constructing a plane 20-spanner with maximum degree 4. Extending this to the setting withobstacles would be an interesting improvement on this paper. Additionally, Biniaz et al. [2]showed how to construct plane spanners of degree 3 in two special cases.Finally, it would be interesting to see if different constructions can provide bounded-degreespanners with smaller spanning ratios, even at the cost of increasing the degree bound slightly.Recently, Bose et al. [7] showed how to construct an (approximately) 4 . References [1] Bahareh Banyassady, Man-Kwun Chiu, Matias Korman, Wolfgang Mulzer, Andr´e vanRenssen, Marcel Roeloffzen, Paul Seiferth, Yannik Stein, Birgit Vogtenhuber, and MaxWillert. Routing in polygonal domains. In
Proceedings of the 28th International Sympo-sium on Algorithms and Computation , volume 92 of
Leibniz International Proceedings inInformatics , pages 10:1–10:13, 2017.[2] Ahmad Biniaz, Prosenjit Bose, Jean-Lou De Carufel, Cyril Gavoille, Anil Maheshwari,and Michiel H. M. Smid. Towards plane spanners of degree 3.
Journal on ComputationalGeometry , 8(1):11–31, 2017.[3] Nicolas Bonichon, Iyad Kanj, Ljubomir Perkovi´c, and Ge Xia. There are plane spanners ofdegree 4 and moderate stretch factor.
Discrete & Computational Geometry , 53(3):514–546,2015.[4] Prosenjit Bose, Jean-Lou De Carufel, and Andr´e van Renssen. Constrained generalizedDelaunay graphs are plane spanners.
Computational Geometry: Theory and Applications ,74:50–65, 2018.[5] Prosenjit Bose, Rolf Fagerberg, Andr´e van Renssen, and Sander Verdonschot. Competitivelocal routing with constraints.
Journal of Computational Geometry , 8(1):125–152, 2017.[6] Prosenjit Bose, Rolf Fagerberg, Andr´e van Renssen, and Sander Verdonschot. On planeconstrained bounded-degree spanners.
Algorithmica , 81(4):1392–1415, 2019.197] Prosenjit Bose, Darryl Hill, and Michiel Smid. Improved spanning ratio for low degreeplane spanners.
Algorithmica , 80(3):935–976, 2018.[8] Prosenjit Bose and J. Mark Keil. On the stretch factor of the constrained Delaunaytriangulation. In
Proceedings of the 3rd International Symposium on Voronoi Diagrams inScience and Engineering , pages 25–31, 2006.[9] Prosenjit Bose, Matias Korman, Andr´e van Renssen, and Sander Verdonschot. Constrainedrouting between non-visible vertices. In
Proceedings of the 23rd Annual InternationalComputing and Combinatorics Conference , volume 10392 of
Lecture Notes in ComputerScience , pages 62–74, 2017.[10] Prosenjit Bose, Matias Korman, Andr´e van Renssen, and Sander Verdonschot. Routing onthe visibility graph.
Journal of Computational Geometry , 9(1):430453, 2018.[11] Prosenjit Bose and Michiel Smid. On plane geometric spanners: A survey and open problems.
Computational Geometry: Theory and Applications , 46(7):818–830, 2013.[12] Prosenjit Bose and Andr´e van Renssen. Spanning properties of Yao and θ -graphs in thepresence of constraints. International Journal of Computational Geometry & Applications ,29(02):95–120, 2019.[13] Ken Clarkson. Approximation algorithms for shortest path motion planning. In
Proceedingsof the 19th Annual ACM Symposium on Theory of Computing , pages 56–65, 1987.[14] Gautam Das. The visibility graph contains a bounded-degree spanner. In
Proceedings ofthe 9th Canadian Conference on Computational Geometry , pages 70–75, 1997.[15] Adrian Dumitrescu and Anirban Ghosh. Lower bounds on the dilation of plane spanners.
International Journal of Computational Geometry & Applications , 26(02):89–110, 2016.[16] Iyad A. Kanj, Ljubomir Perkovic, and Duru T¨urkoglu. Degree four plane spanners: Simplerand better.
Journal of Computational Geometry , 8(2):3–31, 2017.[17] Giri Narasimhan and Michiel Smid.