Computing Maximally Separated Sets in the Plane and Independent Sets in the Intersection Graph of Unit Disks

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1 Computing Maximally Separated Sets in the Plane and Independent Sets in the Intersection Graph of Unit Disks Pankaj K. Agarwal Ý Mark Overmars Þ Micha Sharir Ü Abstract Let Ë be a set of Ò points in Ê. Given an integer Ò, we wish to find a maximally separated subset Ë of size ; this is a subset for which the minimum among the pairwise distances between its points is as large as possible. The decision problem associated with this problem is to determine whether there exists Á Ë, Á, so that all pairwise distances in Á are at least 2, say. This problem can also be formulated in terms of disk-intersection graphs: Let be the set of unit disks centered at the points of Ë. The disk-intersection graph of connects pairs of points by an edge if the disks centered at those points intersect. Á now forms an independent set in the graph. This problem is known to be NP-Complete if is part of the input. In this paper we first present a linear-time approximation algorithm for constant. Next we give Ç Ò ÓÐÝÐÓ Òµµ exact algorithms for the cases and. We also present a simpler Ç µ Ò - time algorithm (as compared with the recent algorithm in [5]) for arbitrary values of. 1 Introduction Let Ë be a set of Ò points in the plane. We are interested in finding a small subset Á of Ë such that the pairwise distances between points in Á are large. To be more precise, let Á be a subset of Ë of cardinality, for Ò. We define the separation distance Áµ to be the minimum among the pairwise distances between its points. We call Á Æ-separated if Áµ Æ. We call Á a maximally separated subset of Ë if Áµ Á ¼ µ for all subsets Á ¼ Ë of size. Note that a set can have Å Ò µ maximally separated -sets. Let ˵ ÑÜ ÁËÁ Áµ. In this paper we study algorithms for computing such maximally separated subsets. We mostly consider small (constant) values of, but we also address the general case. For the case the problem is equivalent to finding a diametral pair of Ë and thus can be solved in Ç Ò ÐÓ Òµ time [6]. For larger, the problem becomes much more complicated and is known to be NP-Complete if is part of the input [8]. Work by P.A. and M.S. was supported by a grant from the U.S.-Israeli Binational Science Foundation. Work by P.A. was also supported by NSF under grants CCR EIA , EIA , EIA , and CCR Work by M.S. was also supported by NSF Grants CCR and CCR , by a grant from the Israel Science Fund (for a Center of Excellence in Geometric Computing), and by the Hermann Minkowski MINERVA Center for Geometry at Tel Aviv University. Part of the research was done during the 2003 Bellairs workshop on computational geometry, organized by Godfried Toussaint, and the 2003 Dagstuhl Workshop on Computational Geometry. Ý Department of Computer Science, Duke University, Durham, NC , USA. pankaj@cs.duke.edu Þ Department of Computer Science, Utrecht University, the Netherlands. markov@cs.uu.nl Ü School of Computer Science, Tel Aviv University, Tel Aviv 69978, Israel; and Courant Institute of Mathematical Sciences, New York University, New York, NY 10012, USA. michas@post.tau.ac.il 1

2 INTRODUCTION 2 Finding small well-separated subsets is important in certain pattern-matching problems, where the points in the subset form a representation of the total set of points. For example, Vleugels and Veltkamp [18] described a method for fast indexing of multimedia databases using vantage objects. These vantage objects are points in the feature space for the matching problem. It has been observed that the chosen vantage objects best be well-separated. The decision problem associated with the problem of computing a maximally separated subset of size asks us to determine whether a Æ-separated subset Á of size exists for a given Æ ¼. This problem can also be formulated in terms of disk-intersection graphs: Let be the set of disks of radius Æ centered at the points of Ë. The disk-intersection graph of has the disks as nodes and two disks are connected by an edge if they intersect. Clearly, a Æ-separated subset Á is the set of centers of an independent set in. So the decision problem is equivalent to the problem of finding an independent set of size in the disk-intersection graph. Recently, the problem of computing the maximum independent set in intersection graphs have received much attention because of its application in geographic information systems (GIS); see [2, 9, 10] and references therein. Related work. The problem of computing an independent set in a graph is one of the earliest problems known to be NP-Complete [11]. In fact, for a general graph with Ò vertices, there cannot be a polynomialtime algorithm with approximation ratio better than Ò, for any ¼, unless ÆÈ È È [13]. The best known algorithm finds an independent set of size Å ÐÓ ÒµÒµ, where is the size of the maximum independent set in the graph [7]. However, better algorithms are known for intersection graphs of geometric objects. The maximum independent set in the intersection graph of intervals on a line can be computed in polynomial time, but the problem remains NP-Ccomplete for intersection graphs of orthogonal segments, unit disks, and unit squares [8]. For example, µ-approximation algorithms have been proposed for intersection graphs of unit disks, unit squares, and arbitrary disks [9, 15], and Ç ÐÓ Òµ-approximation algorithm is known for intersection graphs of rectangles [2]. Little is known about computing maximally spearable sets. Formann and Wagner [10] developed a -approximation algorithm under the Ä -metric. Alber and Fiala [5] present an algorithm that computes an independent set of cardinality in arbitrary disk-intersection graphs in time Ò Ç µ. Their algorithm, however, is rather complicated, and they do not consider cases involving small values of. Moreover, since they consider the entire graph, their algorithm takes Å Ò µ time even for small values of. Our results. In this paper we mostly focus on small values of and develop exact and approximation algorithms. The paper contains four main results: (i) For constant values of, we present a simple, linear-time algorithm that returns a subset Á of size such that Áµ µ ˵. Such an approximation algorithm is suitable for the patternmatching application mentioned above (Section 2). (ii) We present Ç Ò ÓÐÝÐÓ Òµµ algorithms for computing maximally separated subsets of size and (Sections 3 and 4). (iii) We also present a simpler Ò Ç µ -time algorithm (as compared with the algorithm in [5]) for arbitrary values of (Section 5). Our approximation algorithm relies on a standard bucketing technique but with an additional twist. Our exact algorithms for analyze the underlying geometric structure using results from the theory of ararngements and show that one can represent this structure implicitly, which is sufficient for our purpose.

3 AN -APPROXIMATION ALGORITHM 3 2 An -Approximation Algorithm In this section we show that for any constant and any ¼ we can find in linear time a subset Á of Ë of cardinality such that Áµ µ ˵. If, then we can compute an -approximation of a diametral pair in Ç Òµ time [1], so assume that. Using induction, we assume that for all ¼ an -approximation of ¼ ˵ can be computed in linear time. We compute the smallest axis-parallel bounding box of Ë. Let Û be the width of and the height of. Without loss of generality we may assume that Û. We first consider the case in which ˵ Û µ. We subdivide the box into vertical strips ¼, each of width Û µ and let Ë Ë. Any solution will use points from at most of these strips. For each strip, we compute an -approximation of a maximally separated set in Ë Ò Ë. The best among those solutions is the answer we are looking for. So let us assume the solution does not use strip. Let us first consider the case ¼. Let Ð be the point on the left border of the box. Let Á ¼ be an -approximate maximally separated set of size in Ë Ò Ë ¼. Then Á ¼ Ð is a solution because Ð lies at distance at least Û µ from all points in Á ¼. A similar procedure works for. Now consider a value of between 1 and. Let Ë Ð Ë Ë and Ë Ö Ë Ë. Since both Ë Ð and Ë Ö are nonempty and the distance between points of Ë Ð and Ë Ö is at least Û µ, there exists a solution that uses points from both Ë Ð and Ë Ö. Let us assume we use Ø points from Ë Ð and Ø points from Ë Ö. We compute an -approximate maximally separated set Á Ð of size Ø in Ë Ð and a set Á Ö of size Ø in Ë Ö. Á Ð Á Ö form a solution to the problem. We need to repeat this for every value of Ø between 1 and. So for each strip we must solve µ problems with a size smaller than. In total we need to solve Ç µ problems. Denoting by Ì Ò µ the maximum time needed to -approximate ˵ over sets Ë of Ò points, we thus obtain a total cost of Ç Ì Ò µµ. So we are left with the case in which the maximal separation distance ˵ is larger than Û µ. Û Let Æ µ. We split the bounding box of the set Ë into Ç µ grid cells of size at most Æ Æ. we choose an arbitrary point of Ë from each nonempty cell of the grid. Let be the resulting set of representative points; Ç µ. We compute a maximally separable set Á of size for. We claim that Áµ µ ˵. Indeed, let Ë be a maximally separated set of size. Assuming is small enough, these points will lie in different cells. Let ¼ be the representative point from the cell in which lies, and let Á ¼ ¼ ¼. It is easily seen that Á ¼ µ ˵ Æ Ëµ Û Now, as ˵ Û µ, it follows that Á ¼ µ µ ˵. Since we solve the problem exactly for, Áµ Á ¼ µ µ ˵. The running time bound Ì Ò µ thus satisfies the recurrence Ì Ò µ Ç Ì Ò µ µµ, where ѵ is the time needed to compute exactly a maximally separated subset of size in a set of Ñ points. Clearly, the solution of this recurrence is Ç Òµ, for any constant, where the constant of proportionality depends exponentially on. That is, we have: Theorem 2.1 For a set Ë of Ò points in the plane and any constants and we can compute in Ç Òµ time a subset Á Ë of size such that Áµ µ ˵.

4 COMPUTING A MAXIMALLY SEPARATED TRIPLE 4 3 Computing a Maximally Separated Triple Let Ë be a set of Ò points in Ê. We wish to compute a maximally separated triple in Ë. Our overall approach consists of three steps. First, we perform a binary search on the pairwise distances of Ë, and for each distance Æ we determine whether Ë contains a Æ-separated triple. Next, in order to compute a Æ-separated triple, we draw a sufficiently small grid on the bounding box of Ë so that each point of a Æ-separated triple of Ë lies in a distinct grid cell. We thus reduce the problem of computing a Æ-separated triple to a multi-colored variant of this problem. Finally, we compute a trichomatic Æ-separated triple in Ç Ò ÐÓ ¼ Òµ time. For simplicity, we describe these steps in the reverse order. That is, we first describe the algorithm for the multi-colored version, then we show how to reduce the original decision problem to the multi-colored problem, and finally we sketch the binary-search procedure. We need a few notations. For a point Ê, let µ denote the disk of unit radius centered at. For a set of points in Ê, let à µ Ì µ. à µ is a convex region bounded by circular arcs that lie on the boundaries of the disks µ, and each disk contributes at most one such arc to à µ. à µ can be constructed in time Ç ÐÓ µ. 3.1 Computing a trichromatic 1-separated triangle Let Ë, Ë, and Ë be three sets of points in Ê that satisfy the following property: () There is a constant Æ so that, for, Ë is contained in a disk Ç of radius Æ centered at a point and. Without loss of generality, we assume that ¼ ¼µ ¼µ, and µ. See Figure 1. Set Ò Ë, for. We wish to compute a -separated triple in Ë Ë Ë. (Clearly, no other triple of points in Ë Ë Ë can be 1-separated.) Ï Ï Ã µ à µ Ç Ç Ç Æ Ç Ç Æ Ã µ à µ (i) (ii) (iii) Figure 1: (i) An instance of three point sets (contained in the shaded disks) with property (). (ii) The annuli Ï Ï and their top intersection. (iii) à µ, à µ, and the edges of (drawn as thick lines). Let Ë Ë denote the bipartite graph Õµ Ë Õ Ë Õ Using the algorithm of Katz and Sharir [16], we compute in Ç Ò Ò Ò Ò µ ÐÓ Òµ time a family Ù Ù, which is a partition of into complete bipartite graphs, satisfying µ Ç Ò Ò Ò Ò µ ÐÓ Òµ

5 COMPUTING A MAXIMALLY SEPARATED TRIPLE 5 For each Ù, let Ê Ã µ à µ. Set Ê Ì Ù Ê. The following lemma is a straightforward reformulation of the original problem. Lemma 3.1 There exists a -separated triple in Ë Ë Ë if and only if Ë Ê. The following simple observation is crucial for our algorithm. Lemma 3.2 Let È be a set of points lying in a disk of radius Æ centered at a point. Then Ã È µ lies between two concentric circles of radius Æ and Æ centered at. Lemma 3.3 For each Ù, the upper (resp., lower) boundaries of à µ and à µ cross at exactly one point. Proof: Let Ï (resp., Ï ) denote the annulus bounded by the concentric circles of radii Æ and Æ centered at (resp., ). By Lemma 3.2, à µ (resp., à µ) is contained in Ï (resp., Ï ). Therefore à µ à µ Ï Ï. Since Æ and, the inner circles of Ï and Ï intersect and thus Ï Ï consists of two connected components, where lies above the Ü-axis and below the Ü-axis; see Figure 1(ii). Moreover, by the choice of Æ, lies fully to the right of Ç, to the left of Ç, and above both these disks. This implies that within, the boundary of each µ, for, is the graph of a strictly monotone decreasing function, and thus à µ is also the graph of a strictly decreasing function within. By a fully symmetric argument, à µ is the graph of a strictly monotone increasing function within. Moreover, à µ is contained in the upper boundary of à µ, and similarly for à µ, because lies above Ç and Ç. This is easily seen to imply the assertion of the lemma. Lemma 3.3 implies that Ê consists of a connected portion of à µ and a connected portion of à µ. The leftmost and the rightmost points of Ê partition Ê into two parts, which we refer to as its upper and lower boundaries. Let be the set of circular arcs forming the upper boundary of Ê ; we have. Set Ë Ù ; È Ù µ. Let Ä denote the lower envelope of. Lemma 3.4 A point Ë lies inside Ê if and only if lies below the lower envelope Ä. Proof: If Ê, then it lies below the upper boundary of each Ê, thereby implying that lies below Ä. Conversely, suppose that lies below Ä. Then lies below the upper boundary of every Ê. Let be the same as in the proof of Lemma 3.3. Since, Ç, and thus Ë is also contained in. The argument in the proof of Lemma 3.3 implies that lies above the lower boundaries of every à µ and of every à µ. Hence, lies in each Ê and thus also in Ê. In view of Lemma 3.4, we may proceed as follows. For each, we compute à µ, à µ, Ê, and. The total time spent in this step is Ç Ù µ ÐÓ Ò Ò µ Ç Ò Ò Ò Ò µ ÐÓ Ò Ò µ Since each arc in is a portion of the upper boundary of a unit-radius disk, two arcs of intersect in at most one point. Hence, we can compute the lower envelope Ä of in Ç ÐÓ Òµ time using the algorithm of Hershberger [14]. For each edge of Ä we store the index such that. Finally, for each point Ë we determine whether lies below or above Ä. If lies above Ä, then the test yields an arc of Ä that lies below. If this arc belongs to then we deduce that Ê (by Lemma 3.4). Then, scanning the points of in additional Ç µ time, we are certain to find a -separated triple µ Ë.

6 COMPUTING A MAXIMALLY SEPARATED TRIPLE 6 The total running time of the algorithm is Ç Ò Ò Ò Ò µ ÐÓ Ò Ò µ Ò ÐÓ Ò Ò µµ. Hence, we obtain the following result. Theorem 3.5 Let Ë, Ë, and Ë be three sets of points in Ê that satisfy property (), and put Ò Ë, for. Then one can construct, in Ç Ò Ò Ò Ò µ ÐÓ Ò Ò µ Ò ÐÓ Ò Ò µµ time, a -separated triple in Ë Ë Ë, if one exists, or determine that no such triple exists. 3.2 Reduction to -partite graphs and finding a maximally separated triple Let Ë be a set of Ò points in Ê. We wish to compute a -separated triple in Ë. We fix a small constant, and set. We draw a square grid of size in the plane. For, let denote the grid cell µµ µµ, and let Ë Ë. Let denote the set of nonempty grid cells (i.e., those with Ë ). We construct a graph µ where ¼ µ if ÑÒ ¼ ¼. Lemma 3.6 If is not connected or spans more than columns or rows, then a -separated triple in Ë can be computed in Ç Òµ time. Proof: Omitted. Informally, if is disconnected, then the problem reduces to computing the diameter of various pairwise-disjoint subsets of Ë. If spans more than rows or columns, then a -separated triple can easily be computed in Ç Òµ time. By the above lemma, it remains to consider the case where is connected and spans at most rows and columns. Clearly, in this case µ. We consider all triples and determine whether Ë Ë Ë contains a -separated triple, where Ë Ë, for. If the maximum distance between two of these three cells, say, and, is less than, then no -separated triple in Ë Ë Ë exists. Hence, we can assume that the maximum distance between every pair of is at least. There are four cases to consider, depending on the number of edges of between these three cells: (i) ¼; that is, is an independent set in. Then any triple in Ë Ë Ë is -separated and we return any of them. (ii) ; suppose, without loss of generality, that µ and µ µ. We compute the diametral pair Õµ of Ë Ë. If Õ, then we return Õ Öµ, where Ö is any point of Ë. If Õ, no triple in Ë Ë Ë is 1-separated. (iii) ; suppose, without loss of generality, that µ µ and µ. We compute Ã Ë µ and Ã Ë µ. If a point Ë lies neither in Ã Ë µ nor in Ã Ë µ, then there exists a pair Õ Öµ Ë Ë so that Õµ Öµ and thus Õ Öµ is -separated. If Ë Ã Ë µ Ã Ë µ then, arguing as in the proof of Lemma 3.1, no triple in Ë Ë Ë is 1-separated. (iv) ; that is, µ µ µ. In other words, for any pair we have ÑÒ ÜÝ Ü Ý ÑÜ ÜÝ Ü Ý By the triangle inequality, this implies that any Ü, Ý satisfy ÜÝ. We claim that our choice of implies that there exist points Ê so that for each pair of distinct points, and Ë is contained in the disk Ç of radius Æ centered at, for. To see this, let, be such that. Let Ê be a point such that is equilateral and lies on the same side of the line through and as (our choice of is easily seen to imply that does not intersect such a line). It can be shown that is fully contained in the disk of radius

7 COMPUTING A MAXIMALLY SEPARATED QUADRUPLE 7 centered at. We can therefore use Theorem 3.5 to compute a -separated triple in Ë Ë Ë, if one exists, or to determine that no such triple exists. The total running time of the algorithm is dominated by the overall cost of handling case (iv), and is thus, by Theorem 3.5, Ç Ò ÐÓ Òµ. We thus obtain the following main result of this section. Theorem 3.7 Let Ë be a set of Ò points in Ê. We can compute, in Ç Ò ÐÓ Òµ time, a -separated triple in Ë, if one exists, or determine that no such triple exists. Finally, we run a binary search on the Ò pairwise distances in Ë. The th smallest pairwise distance Æ in Ë, for any Ò, can be computed in time Ç Ò ÐÓ Òµ [16], and by Theorem 3.7, we can determine whether a Æ -separated triple exists in Ë within the same time bound. Hence, we obtain the following. Theorem 3.8 Let Ë be a set of Ò points in Ê. We can compute, in Ç Ò ÐÓ Òµ time, a maximally separated triple in Ë. 4 Computing a Maximally Separated Quadruple Our overall approach is the same as in Section 3. We first consider a multi-colored version of this problem, in which we are given four sets Ë Ë Ë, and Ë of points, placed reasonably far from each other, and we wish to compute a -separated quadruple in Ë Ë Ë Ë. We present an algorithm for a special configuration of these sets, which we refer to as a diamond configuration. We then sketch the overall algorithm, which, as above, reduces the general problem to a constant number of multi-colored instances. 4.1 The diamond configuration Let Ë, Ë, Ë, and Ë be four sets of points in Ê that satisfy the following property: ( ) There is a constant Æ so that each Ë is contained in a disk Ç of radius Æ centered at so that,, and Æ. Without loss of generality assume that ¼ ¼µ, lies on the Ü-axis to the right of, and (resp., ) lies below (resp., above) the Ü-axis (in symmetric positions). The conditions on the s imply that. See Figure 2 (i). Set Ò Ë, for. «Ç Ç Ç Ç Ã Ã µ µ µ µ Ã Ã µ µ µ (i) (ii) (iii) Figure 2: (i) A diamond configuration. (ii) Implicit representation of Ê ; µ does not appear on Ê. (iii) Implicit representation of Ê ; µ appears on Ê. For a point Ë, let Ë Õ Ë Õ and Ë µ Õ Ë Õ. (We ignore for the time being the issue of efficient construction of these sets; this will be addressed later on.) We remove

8 COMPUTING A MAXIMALLY SEPARATED QUADRUPLE 8 from Ë any point for which one of these sets is empty, because such a cannot be part of a 1-separated quadruple. Set à ÕË Õµ à µ ÕË µ The following lemma is fairly straightforward. Õµ Ê Ã Ã µ µ Ê Lemma 4.1 There exists a -separated quadruple in Ë Ë Ë Ë if and only if Ë Ê. The following is a variant of Lemma 3.3, proved in a similar manner. Lemma 4.2 (i) For any Ë, à and à µ lies on the upper boundaries of both regions. (ii) For any Ë, à and µ. Ë Ê intersect above the Ü-axis at exactly one point, which and µ intersect above the Ü-axis exactly once, and similarly for à µ Definition 4.3 For a point Ë, let denote, as in Lemma 4.2, the unique intersection point of the upper boundaries of à and à µ, and let «(resp., ) denote the intersection point of the upper boundary of à (resp., à µ ) with µ, if such a point exists. Consider the upper envelope of the upper boundaries of Ã, à µ, and µ. The preceding analysis implies that the envelope has one of the following two structures: (i) Either µ does not appear on the envelope, and then it consists of a connected portion of the upper boundary of à and a connected portion µ of the upper boundary of à µ, meeting at the point (see Figure 2 (ii)); or (ii) µ appears on the envelope, and then it consists of a connected portion of the upper boundary of Ã, a connected portion Æ of the upper boundary of µ, and a connected portion µ of the upper boundary of à µ, so that the first and second portions meet at «and the second and third portions meet at (see Figure 2 (iii)). Let Ë, Ä µ ) denote the lower envelope of µ µ Ë, and Æ Ë. Let Ä (resp., Ä µ, (resp., µ, ). The following lemma is a corollary of Lemma 3.4 and Lemma 4.2. Its proof uses the obvious observation that the lower envelope of Ä, Ä µ, and Ä µ is the same as the lower envelope of the upper boundaries of the regions Ê, for Ë. (We follow here the convention that if a lower envelope is undefined at some Ü, it is assumed to be there.) Corollary 4.4 A point Õ Ë lies in Ê if and only if Õ lies below each of Ä, Ä µ, and Ä µ. We thus compute Ä, Ä µ, and Ä µ separately, and determine whether any point of Ë lies above any of them. If the answer is yes, we can conclude that a -separated quadruple in Ë Ë Ë Ë exists, and we can compute it in additional linear time. Otherwise no such quadruple exists. Computing these envelopes explicitly is however expensive, so we represent them implicitly. We first describe the implicit representation of the envelopes and of its arcs, following a similar representation used by Agarwal et al. [4], and then present the algorithm for computing the envelopes. Implicit representation of à à µ, and of the lower envelopes. For a subset É Ë, let Ä É denote the lower envelope of arcs in the set É. We represent Ä É by the sequence of its breakpoints in increasing order of their Ü-coordinates. The breakpoints are defined so that each portion of Ä Ê between

9 COMPUTING A MAXIMALLY SEPARATED QUADRUPLE 9 two consecutive breakpoints is contained in a single (such a may overlap with many s, but there is (at least) one point Ë such that is fully contained in ). We maintain implicitly, by recording a point Ê that satisfies. Let Õµ Õ Ë. To represent each implicitly, we choose a parameter Ò Ò, compute a family Ì µ Ùµ È of canonical subsets of and associate with each µ its intersection region à µ µ Ù, such that µ Ç ÐÓ Ò µ, and such that for any Ë, à can be represented as the intersection of Ç Ò µ ÐÓ Ò µ of Ì these canonical regions à µ. Let  denote the set of indices of these canonical subsets (i.e., à  à µ ). Katz and Sharir [16] have shown that the construction of such a family of canonical sets, and of the corresponding sets of indices  Ë, can be accomplished in time Ç ÐÓ Ò µ. For each canonical subset µ, the construction of à µ can be performed in Ç µ ÐÓ µ µ time. We store the vertices of the upper boundary of à µ in a list, sorted in increasing order of their Ü-coordinates. For each such boundary vertex, we also store the disk whose boundary appears on à µ immediately to its right. The total time spent in computing this implicit representation of the à s is È Ù Ç µ ÐÓ Ò µ Ç ÐÓ Ò µ. As shown by Agarwal et al. [4], each of the following three operations on the arcs in and à can be performed in Ç Ò µ ÐÓ Ò µ time. (S1) Leftmost and rightmost points: Given a point Ë, compute the leftmost and the rightmost points of Ã. (S2) Intersection point(s) with a vertical line: Given a vertical line and a point Ë, determine the intersection point(s) of with or with Ã. (S3) Intersection points with a unit disk: Given a unit disk and a point Ë, determine the intersection point(s) of with or with Ã. (S4) Crossing point of two arcs: Given two points Õ Ë and an Ü-interval contained in the Ü-span of both and Õ, determine whether the top boundaries of à and Ã Õ cross in. If so, return their crossing point. If they weakly cross in, i.e., overlap over some subinterval  of and their vertical order to the right of  is the reverse of their vertical order to the left of Â, then return the leftmost endpoint of their common overlap in. In a fully analogous fashion, we choose a parameter Ò ¼ Ò and process Ë in Ç ¼ ÐÓ Ò µ time to compute an implicit representation of all the arcs µ. Each of the operations (S1) (S4) on µ can be performed in Ç Ò ¼ µ ÐÓ Ò µ time. Ë Computing Ä, Ä µ, and Ä µ. Using the subroutine (S1), we first compute the lefmost point of à and the rightmost point of à µ, for each Ë. Next, using (S3) and (S4), we compute the intersection point of the upper boundaries of à and à µ, the intersection point «of the upper boundaries of and µ. By comparing à and µ, and the intersection point of the upper boundaries of à µ the Ü-coordinates of these three points, we can determine whether the upper boundary of Ê is of the first by the by the interval kind (without µ) or of the second kind (with µ). In the first case, we represent the arc interval Ü µ Ü µ and by the set  of indices as described above, and the arc µ

10 COMPUTING A MAXIMALLY SEPARATED QUADRUPLE 10 Ü µ Ü µ and by the corresponding set  ¼ of indices. In the second case, we represent by the interval Ü µ Ü «µ, Æ by the interval Ü «µ Ü µ, and µ by the interval Ü µ Ü µ ; in addition, each arc is associated with its corresponding set  of indices. The Ü-coordinate of any point Ë is at most Æ and the Ü-coordinate of the rightmost point of Ê is at least Æ. This easily implies that lies below à (i.e., the vertical ray emanating upward from intersects à ). Since this holds for any point Ë, Theorem 2.8 of Agarwal et al. [4] (concerning the pseudo-diskness of the regions à ) implies that, for any Õ Ë, and Õ cross in at most one point. A similar argument proves the corresponding claim for µ. Hence, each of µ is a collection of pseudo-segments. We can therefore compute the lower envelopes Ä Ä µ using the divide-and-conquer algorithm of Hershberger [14], mentioned above. In the main step of this algorithm, we have available the envelopes Ä Ä of two subsets, and we need to merge these envelopes to compute Ä in a given Ü-interval. Using the subroutine (S4), we can perform this step in Ç Ò µ ÐÓ Ò µ time. Plugging this bound into Hershberger s algorithm, we can compute an implicit representation of Ä in overall time Ç Ò Ò µ ÐÓ Ò ÐÓ Ò ÐÓ Ò µ. Similarly, we can compute an implicit representation of Ä µ in time Ç Ò Ò ¼ µ ÐÓ Ò ÐÓ Ò ¼ ÐÓ Ò µ. Computing Ä µ is easier, since no implicit representation is needed here: We simply have to compute the lower envelope of Ò upper unit circular arcs, which behave as pseudo-segments, so their envelope can be computed in Ç Ò ÐÓ Ò µ time (as in [14]). Finally, for each point Ë, we determine in Ç Ò µ ÐÓ Ò Ò ¼ µ ÐÓ Ò µ time, using the subroutine (S2), whether lies above Ä or above Ä µ. Testing whether lies above Ä µ is easy to accomplish in Ç ÐÓ Òµ time. The total time spent is thus The only nontrivial part in the merge step is computing the crossing point of two arcs and Õ Ç Ò Ò µ Ò Ò ¼ In summary, we have shown: ÐÓ Ò Ò µ ÐÓ Ò Ò µ ¼ µ ÐÓ Ò Ò µ. (4.1) Lemma 4.5 Let Ë, Ë, Ë, and Ë be four sets of points in Ê that satisfy property ( ). Let Ò Ë, for, and let Ò Ò, Ò ¼ Ò, be two parameters. One can determine whether Ë Ë Ë Ë contains a -separated quadruple, and, if so, compute such a quadruple, within the time bound in (4.1). In particular, bounding Ò Ò Ò Ò by Ò and choosing ¼ Ò ÐÓ Ò, we obtain: Theorem 4.6 For a set Ë of Ò points in the plane of the form Ë Ë Ë Ë, where the Ë s satisfy property ( ), one can determine whether Ë contains a -separated quadruple, and, if so, compute such a quadruple, in Ç Ò ÐÓ ¼ Òµ time. 4.2 Reducing to -partite graphs and finding a maximally separated quadruple As in Section 3, we construct a square grid of size, for a sufficiently small constant parameter ¼. Let Ë be as before. Following the same argument used earlier, we can assume that is connected and spans at most rows and columns. We now try all quadruples and determine whether the corresponding quadruple Ë Ë Ë Ë contains a -separated quadruple. We can assume that the maximum distance between any pair of these cells is at least one and that the subgraph induced by these four cells is connected, because if the former assumption is violated then no -separated quadruple exists, and if the latter is violated then the problem can be reduced to finding a -separated triple (or two

11 AN EXACT ALGORITHM FOR AN ARBITRARY VALUE OF 11 pairs of diametral points). Following a case analysis similar to the previous section, but somewhat more involved, we obtain: Theorem 4.7 Let Ë be a set of Ò points in Ê. A -separated quadruple in Ë can be computed in time Ç Ò ÐÓ ¼ Òµ. Finally, by performing a binary search on the pairwise distances in Ë, as above, we obtain the following. Theorem 4.8 Let Ë be a set of Ò points in Ê. A maximally separated quadruple in Ë can be computed in Ç Ò ÐÓ Òµ time. 5 An Exact Algorithm for an Arbitrary Value of Let Ë be a set of Ò points in Ê, and let ¼ be an integer. We describe an Ò Ç µ -time algorithm for computing a maximally separated subset of Ë of size. Because of lack of space we only provide a sketch of the algorithm and omit the details. As in the previous sections, we focus on the decision problem: Given a set of Ò unit disks and an integer Ò, is there a subset Á of pairwise-disjoint disks? Suppose lies inside a strip Ï of width Û. By a sweep-line algorithm we can compute the largest subset of pairwise disjoint disks in Ç Ò Û µ time as follows. Our algorithm is similar to the one by Gonzalez [12] for computing a -center. For a subset and a vertical line, let µ be the set of disks that intersect. We sweep a vertical line from left to right, stopping at the leftmost and the rightmost point of each disk in. At any time, the algorithm maintains a family Á Á Ù of subsets of pairwise-disjoint disks that satisfies the following invariants: (I.1) For every Ù, no disk in Á is contained in the (closed) halfplane lying to the right of the sweep line. (I.2) For Ü Ý, Á Ü µ Á Ý µ. (I.3) If there is a subset of pairwise-disjoint disks so that no disk in lies completely to the right of, then there is a subset Á so that Á µ µ ( Á µ may be empty) and Á. Since at most Ç Ûµ pairwise-disjoint disks of can intersect, it can be argued that Ò Ç Ûµ. Whenever the sweep-line passes through the leftmost or the rightmost point of a disk in, the family can be updated in Ò Ç Ûµ time. We leave it to the reader to verify that the invariants (I.1) (I.3) ensure that the algorithm computes the largest subset of pairwise-disjoint disks in. Hence, if Û, we are done. So assume that Û. By invoking the above algorithm at most Ò times, we can determine in Ò Ç µ time whether there is a subset Á of pairwise-dsjoint disks that lie in a horizontal strip of width at most. Hence, we can assume that no subset of pairwise-disjoint disks lies in a horizontal strip of width. Using a straightforward packing lemma, we can prove the following. Lemma 5.1 Let be a set of Ò disks in Ê, and let Á be a subset of pairwise-disjoint disks that lie in a horizontal strip of width larger than. Then there exists a horizontal line tangent to one of the disks in that intersects Ç µ disks of Á. In view of this lemma, we guess the horizontal line and a subset of Ø Ç µ pairwise-disjoint disks that intersect. Let (resp., ) be the set of disks that lie completely above (resp., below) and do not intersect any disk in. We guess an integer ¼ Ð Ø. We recursively determine whether

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