Constrained Empty-Rectangle Delaunay Graphs

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1 CCCG 2015, Kingston, Ontario, August 10 12, 2015 Constrained Emty-Rectangle Delaunay Grahs Prosenjit Bose Jean-Lou De Carufel André van Renssen Abstract Given an arbitrary convex shae C, a set P of oints in the lane and a set S of line segments whose endoints are in P, a constrained generalized Delaunay grah of P with resect to C denoted CDG C (P ) is constructed by adding an edge between two oints and if and only if there exists a homothet of C with and on its boundary and no oint of P in the interior visible to both and. We study the case where the emty convex shae is an arbitrary rectangle and show that the constrained generalized Delaunay grah has sanning ratio at most 2 (2l/s + 1), where l and s are the length of the long and short side of the rectangle. 1 Introduction A geometric grah G is a grah whose vertices are oints in the Euclidean lane and whose edges are line segments between airs of oints. Every edge is weighted by the Euclidean distance between its endoints. A geometric grah G is called lane if no two edges intersect roerly. The distance between two vertices u and v in G, denoted by δ G (u, v), is defined as the sum of the weights of the edges along the shortest ath between u and v in G. A subgrah H of G is a t-sanner of G (for t 1) if for each air of vertices u and v, δ H (u, v) t δ G (u, v). The smallest value t for which H is a t-sanner is the sanning ratio or stretch factor of H. The sanning roerties of various geometric grahs have been studied extensively in the literature (see [5, 9] for an overview of the toic). We study this roblem in the resence of line segment constraints. Secifically, let P be a set of oints in the lane and let S be a set of line segments with endoints in P, with no two line segments intersecting roerly. The line segments of S are called constraints. Two vertices u and v can see each other or are visible to each other if and only if either the line segment uv does not roerly intersect any constraint or uv is itself a constraint. If two vertices u and v can see each other, the line segment Research suorted in art by FQRNT, NSERC, and Carleton University s President s 2010 Doctoral Fellowshi. School of Comuter Science, Carleton University, Ottawa, Canada. jit@scs.carleton.ca, jdecaruf@cg.scs.carleton.ca National Institute of Informatics (NII), Tokyo, Jaan. andre@nii.ac.j JST, ERATO, Kawarabayashi Large Grah Project. uv is a visibility edge. The visibility grah of P with resect to a set of constraints S, denoted Vis(P, S), has P as vertex set and all visibility edges as edge set. In other words, it is the comlete grah on P minus all edges that roerly intersect one or more constraints. This setting has been studied extensively within the context of motion lanning amid obstacles. Clarkson [7] was one of the first to study this roblem and showed how to construct a linear-sized (1 + ɛ)-sanner of Vis(P, S). Subseuently, Das [8] showed how to construct a sanner of Vis(P, S) with constant sanning ratio and constant degree. Bose and Keil [4] showed that the Constrained Delaunay Triangulation is a 4π 3/ sanner of Vis(P, S). The constrained Delaunay grah where the emty convex shae is an euilateral triangle was shown to be a 2-sanner [3]. Recently, it was shown that regardless of the emty convex shae C used, the constrained generalized Delaunay grah is a lane sanner with constant sanning ratio, where the sanning ratio deends on the erimeter and the width of C [2]. In this aer, we imrove the sanning ratio for the case where the emty convex shae is a rectangle. In the unconstrained setting, Chew [6] showed that the sanning ratio for suares is at most This was later imroved by Bonichon et al. [1], who showed a tight sanning ratio of We show that in the constrained setting the sanning ratio is at most 2 (2l/s + 1), where l and s are the length of the long and short side of C. For suares (the rectangles that minimize l/s), this imlies a ratio of Preliminaries Throughout this aer, we fix a convex shae C. We assume without loss of generality that the origin lies in the interior of C. A homothet of C is obtained by scaling C with resect to the origin, followed by a translation. Thus, a homothet of C can be written as x + λc = {x + λz : z C}, for some scaling factor λ > 0 and some oint x in the interior of C after translation. We refer to x as the center of the homothet x + λc. For a given set of vertices P and a set of constraints S, we now define the constrained generalized Delaunay grah. Given any two visible vertices and, let C(, ) be any homothet of C with and on its boundary. The

2 27 th Canadian Conference on Comutational Geometry, 2015 constrained generalized Delaunay grah contains an edge between and if and only if there exists a C(, ) such that there are no vertices of P in the interior of C(, ) visible to both and. Note that this imlies that constraints are not necessarily edges of the constrained generalized Delaunay grah. We assume that no four oints lie on the boundary of any homothet of C. 2.1 Auxiliary Lemmas Next, we resent three auxiliary lemmas that are needed to rove our main results. First, we reformulate a lemma that aears in [10]. Lemma 1 Let C be a closed convex curve in the lane. The intersection of two distinct homothets of C is the union of two sets, each of which is either a segment, a single oint, or emty. We say that a region R contains a vertex v if v lies in the interior or on the boundary of R. We call a region emty if it does not contain any vertex of P. Though the following lemma was alied to constrained θ-grahs in [3], the roerty holds for any visibility grah. Lemma 2 Let u, v, and w be three arbitrary oints in the lane such that uw and vw are visibility edges and w is not the endoint of a constraint intersecting the interior of triangle uvw. Then there exists a convex chain of visibility edges from u to v in triangle uvw, such that the olygon defined by uw, wv and the convex chain is emty and does not contain any constraints. Finally, we re-introduce a definition and lemma from [2]. Let and be two vertices that can see each other and let C(, ) be a convex olygon with and on its boundary. We look at the constraints that have as an endoint and the edge(s) of C(, ) on which lies, and extend them to half-lines that have as an endoint (see Figure 1a). Given the cyclic order of these half-lines around and the line segment, we define the clockwise neighbor of to be the half-line that minimizes the strictly ositive clockwise angle with. Analogously, we define the counterclockwise neighbor of to be the half-line that minimizes the strictly ositive counterclockwise angle with. We define the cone C that contains to be the region between the clockwise and counterclockwise neighbor of. Finally, let C(, ), the region of C(, ) that contains with resect to, be the intersection of C(, ) and C (see Figure 1b). Lemma 3 Let and be two vertices that can see each other and let C(, ) be any convex olygon with and on its boundary. If there is a vertex x in C(, ) (other than and ) that is visible to, then there is a vertex y (other than and ) in C(, ) that is visible to both and and triangle y is emty. r s (a) C(, ) C(, ) C(, ) (b) Figure 1: Defining the region of C(, ) that contains with resect to : (a) The clockwise and counterclockwise neighbor of are the half-lines through r and s, (b) C(, ) is marked in gray. 3 The Constrained Emty-Rectangle Delaunay Grah We look at the case where the emty convex shae is an arbitrary rectangle. We assume without loss of generality that the rectangle is axis-aligned. We do not, however, assume anything about the ratio between the height and width of the rectangle. We first show that if two visible vertices cannot see any vertices in C(, ) on one side of, then no vertex in C(, ) on the oosite side of can see any vertices beyond either. Lemma 4 Let and be two vertices that can see each other, such that is not vertical, and let C(, ) be any convex olygon with and on its boundary. If the region of C(, ) below does not contain any vertices visible to and, then no oint x in C(, ) above can see any vertices in C(, ) below. Proof. We rove the lemma by contradiction, so assume that there exists a vertex y in C(, ) below that is visible to x, but not to and. Since C(, ) is a convex olygon and x and y lie on oosite sides of, the visibility edge xy intersects. Let z be this intersection (see Figure 2). y z x w C(, ) Figure 2: If x can see a vertex below, then so can. Hence, zy and z are visibility edges. Since z is not a vertex, it is not the endoint of any constraints in-

3 CCCG 2015, Kingston, Ontario, August 10 12, 2015 tersecting the interior of triangle yz. It follows from Lemma 2 that there exists a convex chain of visibility edges between y and and this chain is contained in yz. However, this imlies that w, the neighbor of along this chain, is visible to and lies in C(, ) below. Next, we aly Lemma 2 on triangle w and find that the neighbor of along the chain from to w is visible to both and and lies in C(, ) below, contradicting that this region does not contain any vertices visible to and. Next, we introduce some notation for the following lemma. Let and be two vertices of the constrained generalized Delaunay grah that can see each other. Let R be a rectangle with and on its West and East boundary and let a, b, and r be the Northwest, Northeast, and Southwest corner of R. Let m 1,..., m k 1 be any k 1 oints on in the order they are visited when walking from to (see Figure 3). Let m 0 = and m k =. Consider the homothets S i of R with m i and m i+1 on their resective boundaries, for 0 i < k, such that a / ra = m i a i / r i a i, where a i, b i, r i are the Northwest, Northeast, and Southwest corner of S i. a b a b a 1 b 1 a a 2 b 2 0 b0 m 2 m 1 r 2 r 0 r1 r Figure 3: The total length of the sides of the rectangles S i euals that of C(, ). Lemma 5 We have x = 0 a b m 0 Figure 4: An inductive ath from to. k 1 ( mi a i + a i b i + b i m i+1 ) = a + ab + b. i=0 Proof. Let c = ( a + ab + b )/. Since for every S i we have that a / ra = m i a i / r i a i, we have ( m i a i + a i b i + b i m i+1 )/ m i m i+1 = c, for 0 i < k. Hence, we get k 1 ( mi a i + a i b i + b i m i+1 ) k 1 ( = c mi m i+1 ) i=0 roving the lemma. i=0 = c = a + ab + b, Before we rove the bound on the sanning ratio of the constrained generalized Delaunay grah, we first bound the length of the sanning ath between vertices and for the case where the rectangle C(, ) is artially emty. We call a rectangle C(, ) half-emty when C(, ) contains no vertices in C(, ) below that are visible to and C(, ) contains no vertices in C(, ) below that are visible to. We denote the x- and y-coordinate of a oint by x and y. Lemma 6 Let and be two vertices that can see each other. Let C(, ) be a rectangle with and on its boundary, such that it is half-emty. Let a and b be the corners of C(, ) on the non-half-emty side. The constrained generalized Delaunay grah contains a ath between and of length at most a + ab + b. Proof. We rove the lemma by induction on the rank of C(x, y) when ordered by size, for any two visible vertices x and y, such that C(x, y) is half-emty. We assume without loss of generality that lies on the West boundary, lies on the East boundary and that C(, ) is half-emty below. This imlies that a and b are the Northwest and Northeast corner of C(, ). We also assume without loss of generality that the sloe of is non-negative, i.e. x < x and y y (see Figure 4). We note that the case where lies on the West boundary, lies on the North boundary and C(, ) is halfemty below can be viewed as a secial case of the one above: We shrink C(, ) until one of and lies in a corner. This oint can now be viewed as being on both sides defining the corner and hence and are on oosite sides. An analogous statement holds for the case where lies on the West boundary, lies on the North boundary and C(, ) is half-emty above. Let r be the Southwest corner of C(, ). Let R be a homothet of C(, ) that is contained in C(, ) and whose West boundary is intersected by. Let a, b, r be the Northwest, Northeast, and Southwest corner of R and let m be the intersection of a r and. We call homothet R similar to C(, ) if and only if a / ra = ma / r a. Base case: If C(, ) is a rectangle of smallest area, then C(, ) does not contain any vertices visible to both and : Assume this is not the case and grow a rectangle R similar to C(, ) from to. Let x be the first vertex hit by R that is visible to and lies in C(, ). Note that this imlies that R is contained in C(, ). Therefore, R is smaller than C(, ). Furthermore, R is half-emty: By Lemma 4, the art below the line through and does not contain any vertices visible to or x in C(, ), and the art between the line through and x and the line through and does not contain any vertices visible to or x since x is the first visible vertex hit while growing R. However, this contradicts that C(, ) is the smallest half-emty rectangle. Hence, C(, ) does not contain any vertices visible to both and, which imlies that is an edge of

4 27 th Canadian Conference on Comutational Geometry, 2015 the constrained generalized Delaunay grah. Therefore the length of the shortest ath from to is at most a + ab + b. Induction ste: We assume that for all half-emty rectangles C(x, y) smaller than C(, ) the lemma holds. If is an edge of the constrained generalized Delaunay grah, the length of the shortest ath from to is at most a + ab + b. If is not an edge of the constrained generalized Delaunay grah, there exists a vertex in C(, ) that is visible from both and. We grow a rectangle R similar to C(, ) from to. Let x be the first vertex hit by R that is visible to and lies in C(, ) and let a and b be the Northwest and Northeast corner of R (see Figure 4). Note that this imlies that R is contained in C(, ). We also note that x is not necessarily an edge in the constrained generalized Delaunay grah, since if it is a constraint, there can be vertices visible to both and x above x. However, since R is half-emty and smaller than C(, ), we can aly induction on it and we obtain that the ath from to x has length at most a + a b + b x when x lies on the East boundary of R, and that the ath from to x has length at most a + a x when x lies on the North boundary of R. Let m 0 be the rojection of x along the vertical axis onto. Since m 0 is contained in R, x can see m 0. Since xm 0 and m 0 are visibility edges and m 0 is not the endoint of a constraint intersecting the interior of triangle xm 0, we can aly Lemma 2 and obtain a convex chain x = 0, 1,..., k = of visibility edges (see Figure 4). For each of these visibility edges i i+1, there is a homothet R i of C(, ) that falls in one of the following three tyes (see Figure 5): (i) i lies on the North boundary and i+1 lies in the Southeast corner, (ii) i lies on the West boundary and i+1 lies on the East boundary and the sloe of i i+1 is negative, (iii) i lies on the West boundary and i+1 lies on the East boundary and the sloe of i i+1 is not negative. Let a i and b i be the Northwest and Northeast corner of R i. We note that by convexity, these three tyes occur in the order Tye (i), Tye (ii), and Tye (iii). Let m i be the rojection of i along the vertical axis onto, let C i be the homothet of C(, ) with m i and m i+1 on its boundary that is similar to C(, ), and let a i and b i be the Northwest and Northeast corner of C i. Using these C i, we shift Tye (ii) and Tye (iii) rectangles down as far as ossible: We shift R i down until either i or i+1 lies in one of the North corners or the South boundary corresonds to the South boundary of C i. In the latter case, R i and C i are the same rectangle. Since all rectangles R i are smaller than C(, ), we can aly induction, rovided that we can show that R i is half-emty. For Tye (i) visibility edges, the art of the rectangle that lies below the line through i and i+1 is contained in R, which does not contain any visible i R i b i a i R i b i a i R i b i i+1 i i+1 i i+1 (i) (ii) (iii) Figure 5: The three tyes of rectangles along the convex chain. vertices, and the region of C(, ) below the convex chain, which is emty. For Tye (ii) and Tye (iii) visibility edges, the art of the rectangle that lies below the line through i and i+1 is contained in the region of C(, ) below the convex chain, which is emty, and the region of C(, ) below the line through and, which does not contain any visible vertices by Lemma 4. Hence, all R i are half-emty and we obtain an inductive ath of length at most: (i) i b i + b i i+1, (ii) i a i + a i b i + b i i+1, (iii) i a i + a i b i + b i i+1. To bound the total ath length, we erform case distinction on the location of x on R and whether the convex ath from x to goes down: (a) x lies on the East boundary of R and the convex ath does not go down, (b) x lies on the East boundary of R and the convex ath goes down, (c) x lies on the North boundary of R and the convex ath does not go down, (d) x lies on the North boundary of R and the convex ath goes down. Case (a): The vertex x lies on the East boundary of R and the convex ath does not go down. Recall that the length of the ath from to x is at most a + a b + b x, which is at most a + a b + b m 0. Since the convex chain does not go down, it cannot contain any Tye (i) or Tye (ii) visibility edges. Furthermore, since x lies on the East boundary of R, R and all C i are disjoint. Thus, Lemma 5 imlies that the boundaries above of R and all C i sum u to a + ab + b. Hence, if we can show that, for all R i, i a i + a i b i + b i i+1 m i a i + a i b i + b i m i+1, the roof of this case is comlete. By convexity, the sloe of i i+1 is at most that of and m i m i+1. Hence, when i+1 lies in the Northeast corner of R i, we have i+1 = b i and i a i + a i i+1 m i a i + a i b i + b i m i+1. If i+1 does not lie in the Northeast corner, R i = C i. Hence, since i and i+1 lie above, we have that i a i + a i b i + b i i+1 m i a i + a i b i + b i m i+1. Case (b): The vertex x lies on the East boundary of R and the convex ath goes down. Recall that the length of the ath from to x is at most a + a b + b x. Let

5 CCCG 2015, Kingston, Ontario, August 10 12, 2015 j be the lowest vertex along the convex chain. Since j lies above and has non-negative sloe, the descent of the convex ath is at most xm 0. Hence, when we charge this to R, we used a + a b + b m 0 of its boundary (see Figure 6). a a b m 0 Figure 6: Going down along the convex chain (blue) is charged to R (orange). b a b a b a 1 b 1 a a 0 Figure 7: Charging the ath from to j to C(, j ). Like in the Case (a), since x lies on the East boundary of R, R and all C i are disjoint. Thus, Lemma 5 imlies that the boundaries above of R and all C i sum u to a + ab + b. Hence, if we can show that, for all R i, the inductive ath length is at most m i a i + a i b i + b i m i+1, the roof of this case is comlete. For Tye (i) visibility edges, we have already charged b i i+1 to R, so it remains to show that i b i m i a i + a i b i + b i m i+1. This follows, since m i and m i+1 are the vertical rojections of i and i+1, which imlies that i b i = a i b i. For Tye (ii) visibility edges, we already charged b i i+1 i a i to R, so we can consider i i+1 to be horizontal and it remains to charge the remaining 2 i a i + a i b i. If i lies in the Northwest corner of R i, it follows that i a i = 0 and we have that i b i = a i b i m ia i + a i b i + b i m i+1. If i does not lie in the Northwest corner, R i is the same as C i. Hence, since we can consider i i+1 to be horizontal and i and i+1 lie above, it follows that 2 i a i + a i b i m i a i + a i b i + b i m i+1. Finally, Tye (iii) visibility edges are charged as in Case (a), hence we have that i a i + a i b i + b i i+1 m i a i + a i b i + b i m i+1, comleting the roof of this case. Case (c): Vertex x lies on the North boundary of R and the convex ath does not go down. Recall that the length of the ath from to x is at most a + a x. Since the convex chain does not go down, it cannot contain any Tye (i) or Tye (ii) visibility edges. Let j be the first vertex along the chain, such that R j 1 is the same as C j 1. Since lies on the East boundary of C(, ), this condition is satisfied for the last visibility edge along the convex chain, hence j exists. Let C(, j ) be the homothet of C(, ) that has and j on its boundary and is similar C(, ). Let a and b be the Northwest and Northeast corners of C(, j ) (see Figure 7). Since j is first vertex along the convex chain that does not lie in the Northeast corner of R j 1, we have that along the ath from to j the rojections of a x, all a i i+1, and a j 1 b j 1 onto a b are disjoint and the rojections of a, all i a i, and j 1 a j 1 onto a are disjoint. Hence, their lengths sum u to at most a + a b. Finally, since b j 1 j b j, the total length of the ath from to j is at most a + a b + b j, which is at most a + a b + b m j. All Tye (iii) visibility edges following j are charged as in Case (a), hence we have that i a i + a i b i + b i i+1 m i a i + a i b i + b i m i+1. We now aly Lemma 5 to C(, j ) and all C i following j and obtain that the total length of the ath from to is at most a + ab + b. Case (d): Vertex x lies on the North boundary of R and the convex ath goes down. Recall that the length of the ath from to x is at most a + a x and that 1 is the neighbor of x along the convex chain. Let C(, 1 ) be the homothet of C(, ) that has and 1 on its boundary and is similar to C(, ). Let a and b be the Northwest and Northeast corners of C(, 1 ). Since 1 lies to the right of R and lower than x, it lies on the East boundary of C(, 1 ). We first show that the length of the ath from to 1 is at most a + a b + b 1. If x 1 is a Tye (i) visibility edge, the length of the ath from x to 1 is at most xb 0 + b 0 1. Hence we have a ath from to 1 of length at most a + a x + xb 0 + b 0 1 = a + a b + b 0 1. Since a a and b 0 1 b 1, this imlies that the ath has length at most a + a b + b 1. If x 1 is a Tye (ii) visibility edge and x lies in the Northwest corner an analogous argument shows that the ath from to 1 is at most a + a b + b 1. If x 1 is a Tye (ii) visibility edge and R 0 = C 0, we have that the rojections of a x and a 0 b 0 onto a b are disjoint and the rojections of a and xa 0 onto a are disjoint. Hence, their total lengths sum u to at most a + a b. Finally, since b 0 1 b 1, the total length of the ath from to 1 is at most a + a b + b 1. Next, we observe, like in Case (b), that starting from 1 the convex ath cannot go down more than 1 m 1. Hence, when we charge this to C(, 1 ), we used a + a b + b m 1 of its boundary. Finally, we use arguments analogous to the ones in Case (b) to show that each inductive ath after 1 has length at most m i a i + a i b i + b i m i+1. We now aly Lemma 5 to C(, 1 ) and all C i following 1 and obtain that the total length of the ath from to is at most a + ab + b. Lemma 7 Let and be two vertices that can see each other. Let C(, ) be the rectangle with and on its boundary, such that lies in a corner of C(, ).

6 27 th Canadian Conference on Comutational Geometry, 2015 Let l and s be the length of the long and short side of C(, ). The constrained generalized Delaunay grah contains ( a ath between and of length at most 2l s + 1) ( x x + y y ). Proof. We assume without loss of generality that lies on the Southwest corner and lies on the East boundary. Note that this imlies that the sloe of is non-negative, i.e. x < x and y y. We rove the lemma by induction on the rank of C(x, y) when ordered by size, for any two visible vertices x and y, such that x lies in a corner of C(x, y). In fact, we show that the constrained generalized Delaunay grah contains a ath between x and y of length at most c ( x x )+d ( y y ) and derive bounds on c and d. Base case: If C(, ) is the smallest rectangle with in a corner, then C(, ) does not contain any vertices visible to both and : Let u be a vertex in C(, ) that is visible to both and. Let C(, u) be the rectangle with in a corner and u on its boundary. Since u lies in C(, ), C(, u) is smaller than C(, ), contradicting that C(, ) is the smallest rectangle with in a corner. Hence, C(, ) does not contain any vertices visible to both and, which imlies that is an edge of the constrained generalized Delaunay grah. Hence, the constrained generalized Delaunay grah contains a ath between and of length at most ( x x ) + ( y y ) c ( x x ) + d ( y y ), rovided that c 1 and d 1. Induction ste: We assume that for all rectangles C(x, y), with x in some corner of C(, ), smaller than C(, ) the lemma holds. If is an edge of the constrained generalized Delaunay grah, by the triangle ineuality, the length of the shortest ath from to is at most x x + y y. If there is no edge between and, there exists a vertex u in C(, ) that is visible from both and. We first look at the case where u lies below. Let g be the intersection of the South boundary of C(, ) and the line though arallel to the diagonal of C(, ) through, and let h be the Southeast corner of C(, ) (see Figure 8). If u lies in triangle g, by induction we have that the ath from to u has length at most c (u x x ) + d (u y y ) and the ath from u to has length at most c ( x u x ) + d ( y u y ). Hence, there exists a ath from to via u of length at most c ( x x ) + d ( y y ). If u lies in triangle gh, by induction we have that the ath from to u has length at most c (u x x ) + d (u y y ) and the ath from to u has length at most d ( x u x ) + c ( y u y ). When we take c and d to be eual, this imlies that there exists a ath from to via u of length at most c ( x x ) + d ( y y ). If there does not exist a vertex below that is visible to both and, than Lemma 3 imlies that there are no vertices in C(, ) below that are visible to and g Figure 8: Rectangle C(, ) with oints g and h. that there are no vertices in C(, ) below that are visible to. Hence, we can aly Lemma 6 and obtain that there exists a ath between and of length at most a + ab + b, where a and b are the Northwest and Northeast corner of C(, ). Since ab is ( x x ) and b a l s ( x x ), we can uer bound a + ab + b by c ( x x ) when c is at least ( 2l s + 1). Hence, since c and d need to be eual, we obtain that all cases work out when c = d = ( 2l s + 1). Finally, since ( x x + y y )/ is at most 2, we obtain the following theorem. Theorem 8 The constrained generalized Delaunay grah using an emty rectangle as emty convex shae has sanning ratio at most 2 ( 2l s + 1). References [1] N. Bonichon, C. Gavoille, N. Hanusse, and L. Perković. The stretch factor of L 1- and L -Delaunay triangulations. In ESA, ages , [2] P. Bose, J.-L. De Carufel, and A. van Renssen. Constrained generalized Delaunay grahs are lane sanners. EuroCG, ages , [3] P. Bose, R. Fagerberg, A. van Renssen, and S. Verdonschot. On lane constrained bounded-degree sanners. In LATIN, ages 85 96, [4] P. Bose and J. M. Keil. On the stretch factor of the constrained Delaunay triangulation. In ISVD, ages 25 31, [5] P. Bose and M. Smid. On lane geometric sanners: A survey and oen roblems. CGTA, 46(7): , [6] L. P. Chew. There is a lanar grah almost as good as the comlete grah. In SoCG, ages , [7] K. Clarkson. Aroximation algorithms for shortest ath motion lanning. In STOC, ages 56 65, [8] G. Das. The visibility grah contains a bounded-degree sanner. In CCCG, ages 70 75, [9] G. Narasimhan and M. Smid. Geometric Sanner Networks. Cambridge University Press, [10] K. Swaneoel. Helly-tye theorems for homothets of lanar convex curves. AMS, 131(3): , h