Let $P$ be a set of $n$ points in the plane in general position. We show that
at least $\lfloor n/3\rfloor$ plane spanning trees can be packed into the
complete geometric graph on $P$. This improves the previous best known lower
bound $\Omega\left(\sqrt{n}\right)$. Towards our proof of this lower bound we
show that the center of a set of points, in the $d$-dimensional space in
general position, is of dimension either $0$ or $d$

Biniaz, Ahmad

García, Alfredo

English

arXiv

Let P be a set of n points in the plane in general position. We show that at least ¿n/3¿ plane spanning trees can be packed into the complete geometric graph on P. This improves the previous best known lower bound O(n). Towards our proof of this lower bound we show that the center of a set of points, in the d-dimensional space in general position, is of dimension either 0 or d

Repositorio Universidad de Zaragoza

Packing Plane Spanning Trees into a Point Set1Ahmad Biniaz∗ Alfredo Garćıa†2Abstract3Let P be a set of n points in the plane in general position. We show that at least bn/3c4plane spanning trees can be packed into the complete geometric graph on P . This improves5the previous best known lower bound Ω (√n). Towards our proof of this lower bound we6show that the center of a set of points, in the d-dimensional space in general position, is of7dimension either 0 or d.81 Introduction9In the two-dimensional space, a geometric graph G is a graph whose vertices are points in the10plane and whose edges are straight-line segments connecting the points. A subgraph S of G is11plane if no pair of its edges cross each other. Two subgraphs S1 and S2 of G are edge-disjoint12if they do not share any edge.13Let P be a set of n points in the plane. The complete geometric graph K(P ) is the geometric14graph with vertex set P that has a straight-line edge between every pair of points in P . We say15that a sequence S1, S2, S3, . . . of subgraphs of K(P ) is packed into K(P ), if the subgraphs in16this sequence are pairwise edge-disjoint. In a packing problem, we ask for the largest number of17subgraphs of a given type that can be packed into K(P ). Among all subgraphs, plane spanning18trees, plane Hamiltonian paths, and plane perfect matchings are of interest. Since K(P ) has19n(n− 1)/2 edges, at most bn/2c spanning trees, at most bn/2c Hamiltonian paths, and at most20n− 1 perfect matchings can be packed into it.21A long-standing open question is to determine whether or not it is possible to pack bn/2c22plane spanning trees into K(P ). If P is in convex position, the answer in the affirmative follows23from the result of Bernhart and Kanien [3], and a characterization of such plane spanning trees24is given by Bose et al. [5]. In CCCG 2014, Aichholzer et al. [1] showed that if P is in general25position (no three points on a line), then Ω(√n) plane spanning trees can be packed into K(P );26this bound is obtained by a clever combination of crossing family (a set of pairwise crossing27edges) [2] and double-stars (trees with only two interior nodes) [5]. Schnider [12] showed that28it is not always possible to pack bn/2c plane spanning double stars into K(P ), and gave a29necessary and sufficient condition for the existence of such a packing. As for packing other30spanning structures into K(P ), Aichholzer et al. [1] and Biniaz et al. [4] showed a packing of 231plane Hamiltonian cycles and a packing of dlog2 ne − 2 plane perfect matchings, respectively.32The problem of packing spanning trees into (abstract) graphs is studied by Nash-Williams [11]33and Tutte [13] who independently obtained necessary and sufficient conditions to pack k span-34ning trees into a graph. Kundu [10] showed that at least d(k − 1)/2e spanning trees can be35packed into any k-edge-connected graph.36In this paper we show how to pack bn/3c plane spanning trees into K(P ) when P is in37general position. This improves the previous Ω(√n) lower bound.38∗University of Waterloo, Canada. Supported by NSERC Postdoctoral Fellowship. ahmad.biniaz@gmail.com†Universidad de Zaragoza, Spain. Partially supported by H2020-MSCA-RISE project 734922 - CONNECTand MINECO project MTM2015-63791-R. olaverri@unizar.es12 Packing Plane Spanning Trees39In this section we show how to pack bn/3c plane spanning tree into K(P ), where P is a set of40n > 3 points in the plane in general position (no three points on a line). If n ∈ {3, 4, 5} then41one can easily find a plane spanning tree on P . Thus, we may assume that n > 6.42The center of P is a subset C of the plane such that any closed halfplane intersecting C43contains at least dn/3e points of P . A centerpoint of P is a member of C, which does not44necessarily belong to P . Thus, any halfplane that contains a centerpoint, has at least dn/3e45points of P . It is well known that every point set in the plane has a centerpoint; see e.g. [7,46Chapter 4]. We use the following corollary and lemma in our proof of the bn/3c lower bound;47the corollary follows from Theorem 3 that we will prove later in Section 3.48Corollary 1. Let P be a set of n > 6 points in the plane in general position, and let C be the49center of P . Then, C is either 2-dimensional or 0-dimensional. If C is 0-dimensional, then it50consists of one point that belongs to P , moreover n is of the form 3k+1 for some integer k > 2.51Lemma 1. Let P be a set of n points in the plane in general position, and let c be a centerpoint52of P . Then, for every point p ∈ P , each of the two closed halfplanes, that are determined by53the line through c and p, contains at least dn/3e+ 1 points of P .54HcpProof. For the sake of contradiction assume that a closed halfplane H, that55is determined by the line through c and p, contains less than dn/3e+1 points56of P . By symmetry assume that H is to the left side of this line oriented from57c to p; see the figure to the right. Since c is a centerpoint and H contains58c, the definition of centerpoint implies that H contains exactly dn/3e points59of P (including p and any other point of P that may lie on the boundary60of H). By slightly rotating H counterclockwise around c, while keeping c61on the boundary of H, we obtain a new closed halfplane that contains c but62misses p. This new halfplane contains less than dn/3e points of P ; this contradicts c being a63centerpoint of P .64Now we proceed with our proof of the lower bound. We distinguish between two cases65depending on whether the center C of P is 2-dimensional or 0-dimensional. First suppose that66C is 2-dimensional. Then, C contains a centerpoint, say c, that does not belong to P . Let67p1, . . . , pn be a counter-clockwise radial ordering of points in P around c. For two points p and68q in the plane, we denote by −→pq, the ray emanating from p that passes through q.69Since every integer n > 3 has one of the forms 3k, 3k+1, and 3k+2, for some k > 1, we will70consider three cases. In each case, we show how to construct k plane spanning directed graphs71G1, . . . , Gk that are edge-disjoint. Then, for every i ∈ {1, . . . , k}, we obtain a plane spanning72tree Ti from Gi. First assume that n = 3k. To build Gi, connect pi by outgoing edges to73pi+1, pi+2, . . . , pi+k, then connect pi+k by outgoing edges to pi+k+1, pi+k+2, . . . , pi+2k, and then74connect pi+2k by outgoing edges to pi+2k+1, pi+2k+2, . . . , pi+3k, where all the indices are modulo75n, and thus pi+3k = pi. The graph Gi, that is obtained this way, has one cycle (pi, pi+k, pi+2k, pi);76see Figure 1. By Lemma 1, every closed halfplane, that is determined by the line through c and77a point of P , contains at least k + 1 points of P . Thus, all points pi, pi+1, . . . , pi+k lie in the78closed halfplane to the left of the line through c and pi that is oriented from c to pi. Similarly,79the points pi+k, . . . , pi+2k lie in the closed halfplane to the left of the oriented line from c to80pi+k, and the points pi+2k, . . . , pi+3k lie in the closed halfplane to the left of the oriented line81from c to pi+2k. Thus, all the k edges outgoing from pi are in the convex wedge bounded by the82rays −→cpi and −−−→cpi+k, all the edges outgoing from pi+k are in the convex wedge bounded by −−−→cpi+k83and −−−→ci+2k, and all the edges from pi+2k are in the convex wedge bounded by −−−→cpi+2k and −−−→ci+3k.842Therefore, the spanning directed graph Gi is plane. As depicted in Figure 1, by removing the85edge (pi+2k, pi) from Gi we obtain a plane spanning (directed) tree Ti. This is the end of our86construction of k plane spanning trees.87p1p1+kp1+2kcp2p3 p1p2+kp2+2kcp2p3p1+kp1+2kFigure 1: The plane spanning trees T1 (the left) and T2 (the right) are obtained by removingthe edges (p1+2k, p1) and (p2+2k, p2) from G1 and G2, respectively.To verify that the k spanning trees obtained above are edge-disjoint, we show that two88trees Ti and Tj , with i 6= j, do not share any edge. Notice that the tail of every edge in Ti89belongs to the set I = {pi, pi+k, pi+2k}, and the tail of every edge in Tj belongs to the set90J = {pj , pj+k, pj+2k}, and I ∩ J = ∅. For contrary, suppose that some edge (pr, ps) belongs to91both Ti and Tj , and without loss of generality assume that in Ti this edge is oriented from pr92to ps while in Tj it is oriented from ps to pr. Then pr ∈ I and ps ∈ J . Since (pr, ps) ∈ Ti and93the largest index of the head of every outgoing edge from pr is r + k, we have that s 6 (r + k)94mod n. Similarly, since (ps, pr) ∈ Tj and the largest index of the head of every outgoing edge95from ps is s+ k, we have that r 6 (s+ k) mod n. However, these two inequalities cannot hold96together; this contradicts our assumption that (pr, ps) belongs to both trees. Thus, our claim,97that T1, . . . , Tk are edge-disjoint, follows. This finishes our proof for the case where n = 3k.98If n = 3k + 1, then by Lemma 1, every closed halfplane that is determined by the line99through c and a point of P contains at least k + 2 points of P . In this case, we construct Gi100by connecting pi to its following k + 1 points, i.e., pi+1, . . . , pi+k+1, and then connecting each101of pi+k+1 and pi+2k+1 to their following k points. If n = 3k + 2, then we construct Gi by102connecting each of pi and pi+k+1 to their following k + 1 points, and then connecting pi+2k+2103to its following k points. This is the end of our proof for the case where C is 2-dimensional.104Now we consider the case where C is 0-dimensional. By Corollary 1, C consists of one point105that belongs to P , and moreover n = 3k + 1 for some k > 2. Let p ∈ P be the only point of106C, and let p1, . . . , pn−1 be a counter-clockwise radial ordering of points in P \ {p} around p. As107in our first case (where C was 2-dimensional, c was not in P , and n was of the form 3k) we108construct k edge-disjoint plane spanning trees T1, . . . , Tk on P \ {p} where p playing the role of109c. Then, for every i ∈ {1, . . . , k}, by connecting p to pi, we obtain a plane spanning tree for P .110These plane spanning trees are edge-disjoint. This is the end of our proof. In this section we111have proved the following theorem.112Theorem 1. Every complete geometric graph, on a set of n points in the plane in general113position, contains at least bn/3c edge-disjoint plane spanning trees.11433 The Dimension of the Center of a Point Set115The center of a set P of n > d + 1 points in Rd is a subset C of Rd such that any closed116halfspace intersecting C contains at least α = dn/(d+ 1)e points of P . Based on this definition,117one can characterize C as the intersection of all closed halfspaces such that their complementary118open halfspaces contain less than α points of P . More precisely (see [7, Chapter 4]) C is the119intersection of a finite set of closed halfspaces H1, H2, . . . ,Hm such that for each Hi1201. the boundary of Hi contains at least d affinely independent points of P , and1212. the complementary open halfspace Hi contains at most α− 1 points of P , and the closure122of Hi contains at least α points of P .123Being the intersection of closed halfspaces, C is a convex polyhedron. A centerpoint of P is124a member of C, which does not necessarily belong to P . It follows, from the definition of the125center, that any halfspace containing a centerpoint has at least α points of P . It is well known126that every point set in the plane has a centerpoint [7, Chapter 4]. In dimensions 2 and 3, a127centerpoint can be computed in O(n) time [9] and in O(n2) expected time [6], respectively.128A set of points in Rd, with d > 2, is said to be in general position if no k + 2 of them lie in129a k-dimensional flat for every k ∈ {1, . . . , d− 1}.1 Alternatively, for a set of points in Rd to be130in general position, it suffices that no d+ 1 of them lie on the same hyperplane. In this section131we prove that if a point set P in Rd is in general position, then the center of P is of dimension132either 0 or d. Our proof of this claim uses the following result of Grünbaum.133Theorem 2 (Grünbaum, 1962 [8]). Let F be a finite family of convex polyhedra in Rd, let I be134their intersection, and let s be an integer in {1, . . . , d}. If every intersection of s members of F135is of dimension d, but I is (d− s)-dimensional, then there exist s+ 1 members of F such that136their intersection is (d− s)-dimensional.137C C CFigure 2: The dimension of a point set in the plane, that is not in general position, can be anynumber in {0, 1, 2}.Before proceeding to our proof, we note that if P is not in general position, then the138dimension of C can be any number in {0, 1, . . . , d}; see e.g. Figure 2 for the case where d = 2.139Observation 1. For every k ∈ {1, . . . , d+1} the dimension of the intersection of every k closed140halfspaces in Rd is in the range [d− k + 1, d].141Theorem 3. Let P be a set of n > d + 1 points in Rd, and let C be the center of P . Then,142C is either d-dimensional, or contained in a (d − s)-dimensional polyhedron that has at least143n− (s+ 1)(α− 1) points of P for some s ∈ {1, . . . , d} and α = dn/(d+ 1)e. In the latter case144if P is in general position and n > d+ 3, then C consists of one point that belongs to P , and n145is of the form k(d+ 1) + 1 for some integer k > 2.1461A flat is a subset of d-dimensional space that is congruent to a Euclidean space of lower dimension. The flatsin 2-dimensional space are points and lines, which have dimensions 0 and 1.4Proof. The center C is a convex polyhedron that is the intersection of a finite family H of closed147halfspaces such that each of their complementary open halfspaces contains at most α−1 points148of P [7, Chapter 4]. Since C is a convex polyhedron in Rd, its dimension is in the range [0, d].149For the rest of the proof we consider the following two cases.150(a) The intersection of every d+ 1 members of H is of dimension d.151(b) The intersection of some d+ 1 members of H is of dimension less than d.152First assume that we are in case (a). We prove that C is d-dimensional. Our proof follows153from Theorem 2 and a contrary argument. Assume that C is not d-dimensional. Then, C is154(d − s)-dimensional for some s ∈ {1, . . . , d}. Since the intersection of every s members of H155is d-dimensional, by Theorem 2 there exist s + 1 members of H whose intersection is (d − s)-156dimensional. This contradicts the assumption of case (a) that the intersection of every d + 1157members of H is d-dimensional. Therefore, C is d-dimensional in this case.158Now assume that we are in case (b). Let s be the largest integer in {1, . . . , d} such that everyintersection of s members of H is d-dimensional; notice that such an integer exists because everysingle halfspace in H is d-dimensional. Our choice of s implies the existence of a subfamily H′of s+ 1 members of H whose intersection is d′-dimensional for some d′ < d. Let s′ be an integersuch that d′ = d−s′. By Observation 1, we have that d′ > d−s, and equivalently d−s′ > d−s;this implies s′ 6 s. To this end we have a family H′ with s + 1 members for which everyintersection of s′ members is d-dimensional (because s′ 6 s and H′ ⊆ H), but the intersectionof all members of H′ is (d− s′)-dimensional. Applying Theorem 2 on H′ implies the existenceof s′ + 1 members of H′ whose intersection is (d − s′)-dimensional. If s′ < s, then this impliesthe existence of s′ + 1 6 s members of H′ ⊆ H, whose intersection is of dimension d − s′ < d.This contradicts the fact that the intersection of every s members of H is d-dimensional. Thus,s′ = s, and consequently, d′ = d− s′ = d− s. Therefore C is contained in a (d− s)-dimensionalpolyhedron I which is the intersection of the s + 1 closed halfspaces of H′. Let H1, . . . ,Hs+1be the complementary open halfspaces of members of H′, and recall that each Hi contains atmost α− 1 points of P . Let I be the complement of I. Then,n = |I ∪ I| = |I ∪H1 ∪ · · · ∪Hs+1|6 |I|+ |H1|+ · · ·+ |Hs+1| 6 |I|+ (s+ 1)(α− 1),where we abuse the notations I, I, and Hi to refer to the subset of points of P that they contain.159This inequality implies that I contains at least n− (s+ 1)(α− 1) points of P . This finishes the160proof of the theorem except for the part that P is in general position.161Now, assume that P is in general position and n > d + 3. By the definition of generalposition, the number of points of P in a (d − s)-dimensional flat is not more than d − s + 1.Since I is (d− s)-dimensional, this implies thatn− (s+ 1)(α− 1) 6 d− s+ 1.Notice that n is of the form k(d+ 1) + i for some integer k > 1 and some i ∈ {0, 1, . . . , d}.162Moreover, if i is 0 or 1, then k > 2 because n > d + 3. Now we consider two cases depending163on whether or not i is 0. If i = 0, then α = k. In this case, the above inequality simplifies to164k(d − s) 6 d − 2s, which is not possible because k > 2 and d > s > 1. If i ∈ {1, . . . , d}, then165α = k + 1. In this case, the above inequality simplifies to (k − 1)(d − s) + i 6 1, which is not166possible unless d = s and i = 1. Thus, for the above inequality to hold we should have d = s167and i = 1. These two assertions imply that n = k(d + 1) + 1, and that I is 0-dimensional and168consists of one point of P . Since C ⊆ I and C is not empty, C also consists of one point of P .1691705References171[1] O. 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Packing Plane Spanning Trees into a Point Set

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