Paking Plane Spanning Trees into a Point Set Ahmad Biniaz Alfredo Garía Abstrat Let P be a set of n points in the plane in general position. We show that at least n/3 plane spanning trees an be paked into the omplete geometri graph on P. This improves the previous best known lower bound Ω ( n). Towards our proof of this lower bound we show that the enter of a set of points, in the d-dimensional spae in general position, is of dimension either 0 or d. 1 Introdution In the two-dimensional spae, a geometri graph G is a graph whose verties are points in the plane and whose edges are straight-line segments onneting the points. A subgraph S of G is plane if no pair of its edges ross eah other. Two subgraphs S 1 and S 2 of G are edgedisjoint if they do not share any edge. Let P be a set of n points in the plane. The omplete geometri graph K(P ) is the geometri graph with vertex set P that has a straight-line edge between every pair of points in P. We say that a sequene S 1, S 2, S 3,... of subgraphs of K(P ) is paked into K(P ), if the subgraphs in this sequene are pairwise edge-disjoint. In a paking problem, we ask for the largest number of subgraphs of a given type that an be paked into K(P ). Among all subgraphs, plane spanning trees, plane Hamiltonian paths, and plane perfet mathings are of interest. Sine K(P ) has n(n 1)/2 edges, at most n/2 spanning trees, at most n/2 Hamiltonian paths, and at most n 1 perfet mathings an be paked into it. A long-standing open question is to determine whether or not it is possible to pak n/2 plane spanning trees into K(P ). If P is in onvex position, the answer in the affirmative follows from the result of Bernhart and Kanien [3], and a haraterization of suh plane spanning trees is given by Bose et al. [5]. In CCCG 2014, Aihholzer et al. [1] showed that if P is in general position (no three points on a line), then Ω( n) plane spanning trees an be paked into K(P ); this bound is obtained by a lever ombination of rossing family (a set of pairwise rossing edges) [2] and double-stars University of Waterloo, Canada. Supported by NSERC Postdotoral Fellowship. ahmad.biniaz@gmail.om Universidad de Zaragoza, Spain. Partially supported by H2020-MSCA-RISE projet 734922 - CONNECT and MINECO projet MTM2015-63791-R. olaverri@unizar.es (trees with only two interior nodes) [5]. Shnider [12] showed that it is not always possible to pak n/2 plane spanning double stars into K(P ), and gave a neessary and suffiient ondition for the existene of suh a paking. As for paking other spanning strutures into K(P ), Aihholzer et al. [1] and Biniaz et al. [4] showed a paking of 2 plane Hamiltonian yles and a paking of log 2 n 2 plane perfet mathings, respetively. The problem of paking spanning trees into (abstrat) graphs is studied by Nash-Williams [11] and Tutte [13] who independently obtained neessary and suffiient onditions to pak k spanning trees into a graph. Kundu [10] showed that at least (k 1)/2 spanning trees an be paked into any k-edge-onneted graph. In this paper we show how to pak n/3 plane spanning trees into K(P ) when P is in general position. This improves the previous Ω( n) lower bound. 2 Paking Plane Spanning Trees In this setion we show how to pak n/3 plane spanning tree into K(P ), where P is a set of n 3 points in the plane in general position (no three points on a line). If n {3, 4, 5} then one an easily find a plane spanning tree on P. Thus, we may assume that n 6. The enter of P is a subset C of the plane suh that any losed halfplane interseting C ontains at least n/3 points of P. A enterpoint of P is a member of C, whih does not neessarily belong to P. Thus, any halfplane that ontains a enterpoint, has at least n/3 points of P. It is well known that every point set in the plane has a enterpoint; see e.g. [7, Chapter 4]. We use the following orollary and lemma in our proof of the n/3 lower bound; the orollary follows from Theorem 4 that we will prove later in Setion 3. Corollary 1 Let P be a set of n 6 points in the plane in general position, and let C be the enter of P. Then, C is either 2-dimensional or 0-dimensional. If C is 0- dimensional, then it onsists of one point that belongs to P, moreover n is of the form 3k + 1 for some integer k 2. Lemma 1 Let P be a set of n points in the plane in general position, and let be a enterpoint of P. Then, for every point p P, eah of the two losed halfplanes, that are determined by the line through and p, ontains at least n/3 + 1 points of P.
30 th Canadian Conferene on Computational Geometry, 2018 H p Figure 1: Illustration of the proof of Lemma 1. from to p i+2k. Thus, all the k edges outgoing from p i are in the onvex wedge bounded by the rays p i and p i+k, all the edges outgoing from p i+k are in the onvex wedge bounded by p i+k and i+2k, and all the edges from p i+2k are in the onvex wedge bounded by p i+2k and i+3k. Therefore, the spanning direted graph G i is plane. As depited in Figure 2, by removing the edge (p i+2k, p i ) from G i we obtain a plane spanning (direted) tree T i. This is the end of our onstrution of k plane spanning trees. Proof. For the sake of ontradition assume that a losed halfplane H, that is determined by the line through and p, ontains less than n/3 + 1 points of P. By symmetry assume that H is to the left side of this line oriented from to p as depited in Figure 1. Sine is a enterpoint and H ontains, the definition of enterpoint implies that H ontains exatly n/3 points of P (inluding p and any other point of P that may lie on the boundary of H). By slightly rotating H ounterlokwise around, while keeping on the boundary of H, we obtain a new losed halfplane that ontains but misses p. This new halfplane ontains less than n/3 points of P ; this ontradits being a enterpoint of P. p 1+k p 3 p 3 p 2 p 1 p 1+2k p 1 Now we proeed with our proof of the lower bound. We distinguish between two ases depending on whether the enter C of P is 2-dimensional or 0-dimensional. First suppose that C is 2-dimensional. Then, C ontains a enterpoint, say, that does not belong to P. Let p 1,..., p n be a ounter-lokwise radial ordering of points in P around. For two points p and q in the plane, we denote by pq, the ray emanating from p that passes through q. Sine every integer n 3 has one of the forms 3k, 3k + 1, and 3k + 2, for some k 1, we will onsider three ases. In eah ase, we show how to onstrut k plane spanning direted graphs G 1,..., G k that are edge-disjoint. Then, for every i {1,..., k}, we obtain a plane spanning tree T i from G i. First assume that n = 3k. To build G i, onnet p i by outgoing edges to p i+1, p i+2,..., p i+k, then onnet p i+k by outgoing edges to p i+k+1, p i+k+2,..., p i+2k, and then onnet p i+2k by outgoing edges to p i+2k+1, p i+2k+2,..., p i+3k, where all the indies are modulo n, and thus p i+3k = p i. The graph G i, that is obtained this way, has one yle (p i, p i+k, p i+2k, p i ); see Figure 2. By Lemma 1, every losed halfplane, that is determined by the line through and a point of P, ontains at least k + 1 points of P. Thus, all points p i, p i+1,..., p i+k lie in the losed halfplane to the left of the line through and p i that is oriented from to p i. Similarly, the points p i+k,..., p i+2k lie in the losed halfplane to the left of the oriented line from to p i+k, and the points p i+2k,..., p i+3k lie in the losed halfplane to the left of the oriented line p 1+k p 2 p 2+k p 2+2k p 1+2k Figure 2: The plane spanning trees T 1 (the top) and T 2 (the bottom) are obtained by removing the edges (p 1+2k, p 1 ) and (p 2+2k, p 2 ) from G 1 and G 2, respetively. To verify that the k spanning trees obtained above are edge-disjoint, we show that two trees T i and T j, with i j, do not share any edge. Notie that the tail of every edge in T i belongs to the set I = {p i, p i+k, p i+2k }, and the tail of every edge in T j belongs to the set J = {p j, p j+k, p j+2k }, and I J =. For ontrary, suppose that some edge (p r, p s ) belongs to both T i and T j, and without loss of generality assume that in T i this edge is oriented from p r to p s while in T j it is oriented from p s to p r. Then p r I and p s J. Sine (p r, p s ) T i and the largest index of the head of every outgoing edge from p r is r + k, we have that s (r + k) mod n. Similarly, sine (p s, p r ) T j and the largest index of the head of every outgoing edge from p s is s + k, we have that r (s + k) mod n. However, these two inequalities annot hold together; this ontradits our assumption
C C C Figure 3: The dimension of a point set in the plane, that is not in general position, an be any number in {0, 1, 2}. that (p r, p s ) belongs to both trees. Thus, our laim, that T 1,..., T k are edge-disjoint, follows. This finishes our proof for the ase where n = 3k. If n = 3k+1, then by Lemma 1, every losed halfplane that is determined by the line through and a point of P ontains at least k + 2 points of P. In this ase, we onstrut G i by onneting p i to its following k + 1 points, i.e., p i+1,..., p i+k+1, and then onneting eah of p i+k+1 and p i+2k+1 to their following k points. If n = 3k + 2, then we onstrut G i by onneting eah of p i and p i+k+1 to their following k + 1 points, and then onneting p i+2k+2 to its following k points. This is the end of our proof for the ase where C is 2-dimensional. Now we onsider the ase where C is 0-dimensional. By Corollary 1, C onsists of one point that belongs to P, and moreover n = 3k + 1 for some k 2. Let p P be the only point of C, and let p 1,..., p n 1 be a ounterlokwise radial ordering of points in P \ {p} around p. As in our first ase (where C was 2-dimensional, was not in P, and n was of the form 3k) we onstrut k edgedisjoint plane spanning trees T 1,..., T k on P \{p} where p playing the role of. Then, for every i {1,..., k}, by onneting p to p i, we obtain a plane spanning tree for P. These plane spanning trees are edge-disjoint. This is the end of our proof. In this setion we have proved the following theorem. Theorem 2 Every omplete geometri graph, on a set of n points in the plane in general position, ontains at least n/3 edge-disjoint plane spanning trees. 3 The Dimension of the Center of a Point Set The enter of a set P of n d + 1 points in R d is a subset C of R d suh that any losed halfspae interseting C ontains at least α = n/(d + 1) points of P. Based on this definition, one an haraterize C as the intersetion of all losed halfspaes suh that their omplementary open halfspaes ontain less than α points of P. More preisely (see [7, Chapter 4]) C is the intersetion of a finite set of losed halfspaes H 1, H 2,..., H m suh that for eah H i 1. the boundary of H i ontains at least d affinely independent points of P, and 2. the omplementary open halfspae H i ontains at most α 1 points of P, and the losure of H i ontains at least α points of P. Being the intersetion of losed halfspaes, C is a onvex polyhedron. A enterpoint of P is a member of C, whih does not neessarily belong to P. It follows, from the definition of the enter, that any halfspae ontaining a enterpoint has at least α points of P. It is well known that every point set in the plane has a enterpoint [7, Chapter 4]. In dimensions 2 and 3, a enterpoint an be omputed in O(n) time [9] and in O(n 2 ) expeted time [6], respetively. A set of points in R d, with d 2, is said to be in general position if no k+2 of them lie in a k-dimensional flat for every k {1,..., d 1}. 1 Alternatively, for a set of points in R d to be in general position, it suffies that no d + 1 of them lie on the same hyperplane. In this setion we prove that if a point set P in R d is in general position, then the enter of P is of dimension either 0 or d. Our proof of this laim uses the following result of Grünbaum. Theorem 3 (Grünbaum, 1962 [8]) Let F be a finite family of onvex polyhedra in R d, let I be their intersetion, and let s be an integer in {1,..., d}. If every intersetion of s members of F is of dimension d, but I is (d s)-dimensional, then there exist s + 1 members of F suh that their intersetion is (d s)-dimensional. Before proeeding to our proof, we note that if P is not in general position, then the dimension of C an be any number in {0, 1,..., d}; see e.g. Figure 3 for the ase where d = 2. Observation 1 For every k {1,..., d+1} the dimension of the intersetion of every k losed halfspaes in R d is in the range [d k + 1, d]. Theorem 4 Let P be a set of n d + 1 points in R d, and let C be the enter of P. Then, C is either d-dimensional, or ontained in a (d s)-dimensional polyhedron that has at least n (s + 1)(α 1) points of P for some s {1,..., d} and α = n/(d + 1). In the latter ase if P is in general position and n d + 3, then C onsists of one point that belongs to P, and n is of the form k(d + 1) + 1 for some integer k 2. 1 A flat is a subset of d-dimensional spae that is ongruent to a Eulidean spae of lower dimension. The flats in 2-dimensional spae are points and lines, whih have dimensions 0 and 1.
30 th Canadian Conferene on Computational Geometry, 2018 Proof. The enter C is a onvex polyhedron that is the intersetion of a finite family H of losed halfspaes suh that eah of their omplementary open halfspaes ontains at most α 1 points of P [7, Chapter 4]. Sine C is a onvex polyhedron in R d, its dimension is in the range [0, d]. For the rest of the proof we onsider the following two ases. (a) The intersetion of every d + 1 members of H is of dimension d. (b) The intersetion of some d + 1 members of H is of dimension less than d. First assume that we are in ase (a). We prove that C is d-dimensional. Our proof follows from Theorem 3 and a ontrary argument. Assume that C is not d- dimensional. Then, C is (d s)-dimensional for some s {1,..., d}. Sine the intersetion of every s members of H is d-dimensional, by Theorem 3 there exist s + 1 members of H whose intersetion is (d s)- dimensional. This ontradits the assumption of ase (a) that the intersetion of every d + 1 members of H is d-dimensional. Therefore, C is d-dimensional in this ase. Now assume that we are in ase (b). Let s be the largest integer in {1,..., d} suh that every intersetion of s members of H is d-dimensional; notie that suh an integer exists beause every single halfspae in H is d-dimensional. Our hoie of s implies the existene of a subfamily H of s+1 members of H whose intersetion is d -dimensional for some d < d. Let s be an integer suh that d = d s. By Observation 1, we have that d d s, and equivalently d s d s; this implies s s. To this end we have a family H with s + 1 members for whih every intersetion of s members is d-dimensional (beause s s and H H), but the intersetion of all members of H is (d s )-dimensional. Applying Theorem 3 on H implies the existene of s +1 members of H whose intersetion is (d s )-dimensional. If s < s, then this implies the existene of s + 1 s members of H H, whose intersetion is of dimension d s < d. This ontradits the fat that the intersetion of every s members of H is d-dimensional. Thus, s = s, and onsequently, d = d s = d s. Therefore C is ontained in a (d s)-dimensional polyhedron I whih is the intersetion of the s + 1 losed halfspaes of H. Let H 1,..., H s+1 be the omplementary open halfspaes of members of H, and reall that eah H i ontains at most α 1 points of P. Let I be the omplement of I. Then, n = I I = I H 1 H s+1 I + H 1 + + H s+1 I + (s + 1)(α 1), where we abuse the notations I, I, and H i to refer to the subset of points of P that they ontain. This inequality implies that I ontains at least n (s + 1)(α 1) points of P. This finishes the proof of the theorem exept for the part that P is in general position. Now, assume that P is in general position and n d + 3. By the definition of general position, the number of points of P in a (d s)-dimensional flat is not more than d s+1. Sine I is (d s)-dimensional, this implies that n (s + 1)(α 1) d s + 1. Notie that n is of the form k(d + 1) + i for some integer k 1 and some i {0, 1,..., d}. Moreover, if i is 0 or 1, then k 2 beause n d + 3. Now we onsider two ases depending on whether or not i is 0. If i = 0, then α = k. In this ase, the above inequality simplifies to k(d s) d 2s, whih is not possible beause k 2 and d s 1. If i {1,..., d}, then α = k + 1. In this ase, the above inequality simplifies to (k 1)(d s) + i 1, whih is not possible unless d = s and i = 1. Thus, for the above inequality to hold we should have d = s and i = 1. These two assertions imply that n = k(d + 1) + 1, and that I is 0-dimensional and onsists of one point of P. Sine C I and C is not empty, C also onsists of one point of P. Referenes [1] O. Aihholzer, T. Hakl, M. Korman, M. J. van Kreveld, M. Löffler, A. Pilz, B. Spekmann, and E. Welzl. Paking plane spanning trees and paths in omplete geometri graphs. Information Proessing Letters, 124:35 41, 2017. Also in CCCG 14, pages 233 238. [2] B. Aronov, P. Erdös, W. Goddard, D. J. Kleitman, M. Klugerman, J. Pah, and L. J. Shulman. Crossing families. Combinatoria, 14(2):127 134, 1994. Also in SoCG 91, pages 351 356. [3] F. Bernhart and P. C. Kainen. The book thikness of a graph. Journal of Combinatorial Theory, Series B, 27(3):320 331, 1979. [4] A. Biniaz, P. Bose, A. Maheshwari, and M. H. M. Smid. Paking plane perfet mathings into a point set. Disrete Mathematis & Theoretial Computer Siene, 17(2):119 142, 2015. [5] P. Bose, F. Hurtado, E. Rivera-Campo, and D. R. Wood. Partitions of omplete geometri graphs into plane trees. Computational Geometry: Theory and Appliations, 34(2):116 125, 2006. [6] T. M. Chan. An optimal randomized algorithm for maximum tukey depth. In Proeedings of the 15th Annual ACM-SIAM Symposium on Disrete Algorithms, SODA, pages 430 436, 2004. [7] H. Edelsbrunner. Algorithms in Combinatorial Geometry. Springer, 1987. [8] B. Grünbaum. The dimension of intersetions of onvex sets. Paifi Journal of Mathematis, 12(1):197 202, 1962.
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