MINIMUM DIAMETER COVERING PROBLEMS Esther M. Arkin y and Refael Hassin z May 20, 1997 Abstract A set V and a collection of (possibly non-disjoint) subsets are given. Also given is a real matrix describing distances between elements of V. A cover is a subset of V containing at least one representative from each subset. The multiple-choice minimum diameter problem is to select a cover of minimum diameter. The diameter is dened as the maximum distance of any pair of elements in the cover. The multiple-choice dispersion problem, which is closely related, asks us to maximize the minimum distance between any pair of elements in the cover. The problems are NP-hard. We present polynomial time algorithms for approximating special cases and generalizations of these basic problems, and we prove in other cases that no such algorithms exist (assuming P 6= N P ). 1 Introduction This paper deals with a class of multiple-choice problems. We assume that a set V of elements is given, together with subsets S 1 ; :::; S m. A cover is a subset of V that contains at least one representative from each S j ; j = 1; :::; m. Our interest is in problems in which the objective is to select a cover whose elements are close to each other with respect to a given distance function. Our measure of proximity is the diameter of the set. The diameter is dened as the maximum distance of any pair of elements in the cover. Thus, the multiple-choice cover problem is to compute a cover of minimum diameter. An illustrative way to describe our basic problem is as follows: There are m rms (or government oces, or points in some statistical experiment), j = 1; :::; m and each can choose for its location a point from a given subset S j. The rms all wish to communicate with one another, and thus want to be close to each other. Specically, the objective is to minimize the maximum distance between rms. In a more general case, it is assumed that the cover can be further partitioned into a given number, k, of clusters, and our goal is to minimize the maximum diameter of a cluster. We call this problem the multiple-choice k-partition problem. When the cardinalities of the sets S j are bounded by a constant `, the problem becomes `-choice k-partition. In particular, the pair-choice problem, i.e., ` = 2, will be treated separately. The dispersion of a set is dened as the distance between the two closest elements in the set. We also consider the variation of the problem in which the goal is to maximize the dispersion of the cover. This may be the goal when the location of obnoxious facilities is considered (see for example, [20]). It turns out that these problems are much harder to approximate than the diamter problems. Several (single-choice) k-partition problems have been studied in the literature. The single-choice 2- partition problem, namely, nding a minimum diameter bipartition of a given set is polynomially solvable. It can be solved by searching on the distinct distance values, d, and for each test value considering only edges of length larger than d and testing for bipartition of the resulting graph (see [3], [16]). The 3-partition problem is NP-hard even in the plane [12] [5]. The one-dimensional k-partition problem, where the elements are points on a line, is however polynomially solvable [3]. Assuming that the distance matrix obeys the triangle inequality, a 2-approximation for the k-partition problem can be obtained in time O(jV jk) [5] (see [9] for a dierent method to obtain a 2-approximation). Assuming P 6= N P, a 2? approximation for the single-choice k-partition problem (when k is an input to the problem) cannot be obtained in polynomial time for any > 0 ([5], [9]). This is true even when the elements are points in the 3-dimensional Euclidean space [5]. In the plane, a better error bound may be possible, but in any case it cannot be better than p 3 [5]. y Department of Applied Mathematics and Statistics, SUNY Stony Brook, Stony Brook, NY 11794-3600. Email: estie@ams.sunysb.edu. Partially supported by NSF Grants ECSE-8857642 and CCR-9204585. z Department of Statistics and Operations Research, Tel-Aviv University, Tel-Aviv 69978, Israel. Email: hassin@math.tau.ac.il. 1
Summary of results: The multiple-choice cover problem 1. The pair-choice cover problem of nding a vertex cover of minimum diameter is polynomially solvable (Theorem 2.2). 2. The triple-choice cover problem is NP-complete even when the sets are disjoint and represented by points in the plane (Theorem 2.4). 3. The triple-choice problem cannot be approximated with a constant error guarantee, assuming P 6= N P (Corollary 2.5). 4. Even if we assume the triangle inequality, no approximation of error guarantee of less than 2 is possible for the triple-choice cover problem (Corollary 2.6). 5. If we assume the triangle inequality then the multiple-choice cover problem can be approximated with a bound of 2 (Algorithm Multi Choice in Figure 2). 6. In the plane the error of the multile-choice cover problem can be bounded by 2= p 3 (Algorithm Plane Multi Choice in Figure 3). The pair-choice 2-partition problem 1. The pair-choice 2-partition problem is NP-hard when we allow the sets S j to be non-disjoint. This is also true when we assume the triangle inequality (Corollary 3.4). 2. The pair-choice 2-partition problem in the plane is polynomially solvable (Algorithm Plane Test in Figure 4). 3. Assuming P 6= N P, the pair-choice 2-partition problem cannot be approximated with a constant error guarantee (Corollary 3.5). 4. Even if we assume the triangle inequality, no approximation of error guarantee of less than 2 is possible for the pair-choice 2-partition problem (Corollary 3.4). The multiple-choice k-partition problem 1. If we assume the triangle inequality then the `-choice k-partition problem can be approximated with a bound of 2, for any xed ` and k (Remarks 3.8 and 3.9). 2. In the plane the multiple-choice 2-partition problem can be approximated within an error bound of 2= p 3 (Remark 3.10). The multiple-choice maximum dispersion problem 1. The pair-choice maximum dispersion problem of nding a vertex cover of maximum dispersion is polynomially solvable (Theorem 4.1). 2. The triple-choice maximum dispersion problem is NP-complete even when the sets are disjoint and represented by points in the plane (Theorem 4.3). 3. The triple-choice maximum dispersion problem cannot be approximated with a constant error guarantee, assuming P 6= N P, even when the elements are represented by points in the plane (Corollary 4.4). 4. The `-choice k-partition dispersion problem cannot be solved or approximated within a constant factor in polynomial time, even if ` = k = 2. 2
Several other multiple-choice problems have been studied in the literature. Gabow, Maheshwari, and Osterweil [7] proved that the following pair-choice problem is NP-complete: Given a directed acyclic graph with two distinguished vertices s; t, and a set of k pairs of vertices, determine whether there exists a path from s to t that uses at most one vertex from each of the given pairs. With the obvious modications, this proof can be used also for an equivalent result when the graph is undirected. Although their proof does not apply when the given pairs are disjoint; it is not hard to see that the problem remains NP-complete when the given pairs are disjoint (see [6]). Itai, Rodeh, and Tanimoto [11] proved that the following problem is NP-complete: Given a bipartite graph with k pairs of edges, nd whether there exists a perfect matching that uses at most one edge from each of the given subsets. Their proof does not apply when the given pairs are disjoint, however, the problem remains NP-complete for disjoint pairs, [6]. An important multiple-choice problem is the multiple-choice knapsack problem: A set of \items" is given, and each has a \cost" and a \value". The items are partitioned into classes. The goal is to select a subset that maximizes the total value subject to a budget constraint and so that at most one item is selected from each of the given classes. The problem is NP-hard but can be solved in pseudopolynomial time. Sinha and Zoltners [19] developed a Branch and Bound algorithm for it. (See [18] for more recent developments on this approach.) Lawler [14] presented a fully polynomial approximation scheme. Pair-choice problems that resemble ours but with a dierent objective functions are discussed by Arkin, Halldorson and Hassin [1]. The tree and tour cover problems on an edge-weighted graph are to compute a vertex cover whose vertices can be connected by a tree or a closed walk, respectively, of minimum total length. Both problems are NP-hard and constant factor approximation algorithms are described in [1]. Multiple choice p-center problems were studied by [10] who showed they are NP-complete. [8] gave polynomial approximation algorithms. 2 Multiple-choice cover of minimum diameter 2.1 The problem Let V = f1; : : :; ng be a given set. Also given is a symmetric real matrix D = (d ij ) 2 IR nn + dening the distances between pairs of elements of V. Let S 1 ; :::; S m V be given. A cover is a subset C V such that C \ S i 6= ; for all S i. A d-cover is a cover whose diameter, d(c) = maxfd ij : i; j 2 Cg, is at most d. Unless explicitly stated, we do not assume that D satises the triangle inequality, and thus, the diameter may dier from the longest \shortest path" in the set. In this section we observe the existence of a polynomial time algorithm for nding the minimum d for which a d-cover exists in the case where each set S i has cardinality two. A reduction from 3-SAT shows that the problem is NP-hard when the cardinality of the sets S i is three or more, and the distance matrix corresponds to the Euclidean distances between points in the plane. However, if the distances obey the triangle inequality, it is possible to approximate the minimum d and bound the error ratio by a constant in polynomial time. 2.2 Polynomial solution for pair-choice We now consider the case when each S i has cardinality 2. In this case, we can represent the sets S i using a graph G = (V; E), where V = f1; :::; ng is the set of elements, and, for every subset S i = fu; wg, we place an edge between element u and element w. The symmetric matrix D = (d ij ; i; j 2 V ) describes intervertex distances. A vertex cover is a subset C V such that C \ e 6= ; for all e 2 E. Problem 2.1 (Pair-Choice Cover Problem) Find a vertex cover of a minimum diameter. Theorem 2.2 Problem 2.1 can be solved in time O(n 2 log n). Proof: Algorithm Test in Figure 1 tests whether a particular value d is a feasible diameter by stating the problem as a 2-SAT problem. 2-SAT is the problem of nding whether there exists a truth assignment to the variables, such that a formula in conjunctive normal form, with at most two literals per clause, is satised. 3
Test input 1. A graph G = (V; E) with weights d ij i; j 2 V. 2. A value d 2 IR. returns 1. \TRUE" if G has a d-cover. 2. \FALSE" if G has no d-cover. begin Let x i be a variable for every i 2 V: for every (i; j) 2 E Dene a clause (x i _ x j ). for every i; j 2 V such that d ij > d Dene a clause (x i _ x j ). if the formula is satisable return \TRUE". else return \FALSE". end Test Figure 1: Algorithm Test The algorithm denes a variable, x i, for each vertex i 2 V, that is true if the vertex i is chosen and false otherwise. We now use a linear time algorithm that solves the 2-SAT problem, such as the one described by Even, Itai and Shamir [4] (linear in the number of clauses, which is O(n 2 ) in our case). To nd the optimal d we perform binary search on the values of D to nd the smallest value for which the test reply is \true". 2.3 NP-completeness of triple-choice We saw above how the pair-choice cover problem is related to 2-SAT. We now show that a 3-SAT problem can be written as a triple-choice cover problem, thus showing the triple-choice problem to be hard. Let V = f1; :::; ng be given with a symmetric real matrix D 2 IR nn + dening the distances between pairs of elements of V. Let S 1 ; :::; S m V be given. Problem 2.3 (Multiple-Choice Cover Problem) Find a subset C V, such that jc \ S i j 1 for i = 1; :::; m, and the diameter of C, d(c) = maxfd kl : k; l 2 Cg, is minimized. Theorem 2.4 Problem 2.3 is NP-hard even when S i i = 1; :::; m are disjoint, js i j 3 i = 1; :::; m, the elements of V are points in the plane, and D is the Euclidean distance matrix. Proof: The decision problem (does there exist a d-cover, for some constant d) is obviously in NP. We prove the theorem by a reduction from 3-satisability. Given a set of clauses K j = x j _ y j _ z j (j = 1; :::; `) where fx j ; y j ; z j g fu 1 ; u 1 ; : : : ; u n ; u n g. we consider a circle of diameter d 0 : For each variable, we place one point corresponding to the variable and one point corresponding to its negation diameterically opposite on the circle. That is, the distance between these points is d 0. We choose d 0 > d suciently close to d so that the distance between any two points which correspond to dierent variables is less than d. We dene an instance of Problem 2.3 with sets S 1 ; :::; S n such that S i consists of the points associated with u i and u i, and sets S n+1 ; :::; S n+` such that S n+j consists of the triplet of points whose locations coincide with those of x j ; y j ; z j. A feasible solution with diameter at most d is any set of points such that: 1. From each pair of points corresponding to u j ; u j, exactly one is selected. 2. From each triplet of points corresponding to x j ; y j ; z j, at least one is selected. 4
Thus, if there exists a subset of diameter at most d then there exists a satisfying assignment for K 1 ^ ^ K`. On the other hand, a satisfying assignment clearly determines a feasible set of diameter at most d. Thus the proof is complete. Note that in our proof the sets S i are not disjoint. However, it is easy to modify our construction to obtain disjoint sets by distinguishing among the points corresponding to dierent sets even when the above construction puts them at the same location. This is done by generating points that are very close to each other instead of a single point that appears in several sets S i. Corollary 2.5 Assuming P 6= N P, no polynomial approximation to Problem 2.3 with a bounded error guarantee exists. Proof: Suppose that a polynomial algorithm that guarantees a solution of diameter at most r times the optimal diameter exists. Consider the instance used in the proof of Theorem 2.4, except that whenever the distance between two points was d 0 we change it to rd 0, and all other distances are d. Now, the solutions associated with a satisfying assignment are of distance exactly d, while all other feasible solutions to the choice problem have diameter rd 0 > rd. Thus, the algorithm nds a satisfying assignment if one exists, in polynomial time. Note that in the construction used in the proof of Corollary 2.5, the triangle inequality does not hold. In the next section we show how to nd approximations when the triangle inequality is known to hold. However we have the following: Corollary 2.6 Assuming P 6= N P, no polynomial approximation to Problem 2.3 with an error guarantee of less than two exists, even when the triangle inequality is known to hold. Proof: We can use the same proof as before but modify the distances: The distance between two points that were oppositely located on the circle is dened to be 2 while every other distance between any other pair of points is dened to be 1. The triangle inequality holds and there exists a satisable assignment if and only if the solution to this instance of Problem 2.3 is 1. Otherwise, the solution is 2. Hence, an approximation algorithm which is guaranteed to produced a solution at most times the optimal, for < 2 must output a solution of diameter 1 when one exists. Such an algorithm can be used to solve the 3-satisability problem. Hence, if P 6= N P, it cannot be polynomial. 2.4 Approximation assuming the triangle inequality We now present a bounded error approximation for Problem 2.3 under the assumption that the distances satisfy the triangle inequality. Let S q be a set of minimum cardinality. For each element of i 2 S q, we nd the closest element in every other set S p, p 6= q, and let this distance be d ip. We get an approximate solution containing i, and P points that are at distance at most d i = maxfd ip j p 6= qg. m Denote N = j=1 js jj. Theorem 2.7 Algorithm Multi Choice nds a cover of diameter at most twice the optimum. Its running time is O(jS q jn). Proof: Consider a choice of i 2 S q such that d = d i. The algorithm nds a cover containing elements, all of which are at distance at most d from i. Consider any two sets, and the elements in them minimizing the distance to i, say j and k. By the triangle inequality, the interdistances between j; k satisfy d j;k d i;j + d i;k 2d so we have a 2d-cover. Clearly a cover that contains i must have a diameter of at least d i. Since a cover must contain a representative from S q, the minimum possible diameter of a cover is at least d = minfd i j i 2 S q g. Therefore, the cover C of diameter at most 2d which was found by the algorithm is a 2-approximation. The algorithm executes js q j iterations and each iteration scans all the elements in V n S q. Hence the time complexity is O(jS q jn). In the rest of this section we describe another polynomial approximation algorithm, which applies to geometric problems in the plane. While this algorithm compares favorably with the previous one for this case, guaranteeing an error ratio of at most 2= p 3 ( approximately 1.15), its complexity is O(n 3 N). Theorem 2.8 Algorithm Plane Multi Choice computes a cover whose diameter is at most 2= p 3 times the minimum diameter of a feasible solution, and this bound is tight, for this algorithm. The algorithm can be executed in time O(n 3 N). 5
Multi Choice input 1. A set V = f1; : : :; ng and subsets S 1 ; :::; S m V. 2. A symmetric matrix D = (d ij ) 2 IR nn + satisfying the triangle inequality. returns 1. A cover C of S 1 ; :::; S m of diameter at most 2d. begin Find q such that js q j = minfjs j j : j = 1; : : : ; mg. for every i 2 S q ( Compute d i, a 2-approximation for a minimum diameter cover containing i). d ip := minfd ij j j 2 S p g for p 6= q d i := maxfd ip j 1 p m; p 6= q g Find i 2 S q such that d i = min i fd i g. return d := d i, C := fi g [ fj j d i j dg. end Multi Choice Figure 2: Algorithm Multi Choice Plane Multi Choice Compute a disk B of minimum diameter that contains at least one point from each subset S 1 ; :::; S m. return V \ B. end Plane Multi Choice Figure 3: Algorithm Plane Multi Choice Proof: The smallest disk covering all sets is determined by two or three points which lie on its boundary. In the case of two points the diameters of the disk and the points are equal. Consider the case of three points. In this case, the convex hull of these points contains the center of the disk. If the triangle they form is not equilateral then we can shorten its longest edge by moving one of these points on the boundary of the disk. This change preserves the diameter of the disk while decreasing the diameter of the triangle. Hence, this change increases the ratio of the diameter of the smallest disk to the diameter of a feasible solution. We conclude that this ratio is maximized on an equilateral triangle, which yields the claimed bound. To nd a disk of minimum diameter one may simply check the feasibility of each of the O(n 3 ) disks dened by pairs and triplets of points of V and then choose the feasible one with the minimum diameter. This can easily be done in O(n 3 N) time. We close this section by pointing to a dierence in diculty between two problems that seem similar on the surface, namely that of nding the diameter, or the minimum radius disk covering the sets. For the single-choice (i.e., no-choice) problem, the minimum diameter of a set of n points in the plane can be found in O(n log n) time, and in fact also requires time (n log n) [17]. In contrast, nding the smallest radius disk covering n points in the plane can be done in linear time by Megiddo's algorithm [15]. In the multiple-choice problems we are considering, the contrast is even greater. The diameter problem is NP-hard (Theorem 2.4), while the smallest disk problem is polynomial. This dierence can be attributed to the fact that a constant number of points (3) determine the smallest disk, and thus simply examining all triples yields a polynomial time algorithm. 3 Multiple-choice 2-partition 3.1 The problem Let V = f1; :::; ng be given with a symmetric real matrix D 2 IR nn + dening the distances between pairs of elements of V. Let S 1 ; :::; S m V be given. A 2-partition of a set C consists of two disjoint subsets 6
C 1 ; C 2 such that C 1 [ C 2 = C. Consider a 2-partition C 1 ; C 2 of a cover C. We dene its diameter to be max`=1;2 maxfd ij : i; j 2 C`g. The problem we consider in this section is: Find a cover C and a 2-partition of C of minimum diameter. Clearly, this problem is harder than the minimum diameter cover discussed in the previous section, and therefore, we conclude that the problem is NP-hard even when the sets S i are disjoint, js i j 3, the elements of V are points in the plane, and D is the Euclidean distance matrix. In the next section we show that even the cases that were easy for the minimum diameter problem, namely when the sets S i have cardinality at most two, are NP-hard (except for the pair-choice problem in the plane). Approximation algorithms are discussed in Section 3.4 (see also Remarks 3.8 and 3.10). 3.2 The pair-choice problem Given are a graph G = (V; E) with V = f1; : : :; ng and a symmetric matrix D = (d ij ; i; j 2 V ) describing distances. Problem 3.1 (pair-choice 2-partition problem) Find a vertex cover C and a 2-partition of C, so that the diameter of the partition is minimized. To solve Problem 3.1, one could search on the distinct values of the set fd ij : i; j 2 V g. For a given test value d let G d = (V; E d ) be a graph with the edge set E d = f(i; j) : d ij dg. The test amounts to solving the following problem, which we show to be NP-hard: Problem 3.2 (pair-choice induced subgraph problem) Given graphs G = (V; E) and G d = (V; E d ), nd whether there exists a vertex cover C V of G such that the subgraph of G d induced by C is bipartite. Problem 3.2 is (polynomially) equivalent to Problem 3.1. Clearly, an algorithm for Problem 3.1 can be used to solve Problem 3.2. Suppose that the answer to Problem 3.2 is positive. Let P 1 ; P 2 be the two parts of the induced subgraph. Then P 1 [ P 2 = C and d ij d for all i; j 2 P l l = 1; 2. Thus the outcome of the test is positive and the optimal solution value d to Problem 3.1 satises d d. On the other hand, if the answer is negative then d > d. However, we have the following negative result: Theorem 3.3 Problem 3.2 is NP-complete. Proof: The problem is obviously in NP. We prove the theorem by reduction from 3-satisability. Consider a given set of clauses K j = (x j _ y j _ z j ) where fx j ; y j ; z j g fu 1 ; u 1 ; ; u n ; u n g j = 1; : : : ; `. We dene G = (V; E) and G d = (V; E d ) as follows: V has 3` vertices corresponding to the negations of the literals of K 1 ; :::; K`. E d consists of the edges (x j ; y j ); (x j ; z j ); (y j ; z j ) j = 1; :::; `. We then dene edges fq; pg 2 E in G for each pair of vertices of V whenever p and q correspond to a variable and its negation. Clearly this construction takes polynomial time. We now show that the formula is satisfyable if and only if there exists a vertex cover of G such that the subgraph it induces in G d is bipartite. A vertex cover must include either all vertices corresponding to v i or all the vertices corresponding to v i. This is the case because if some vertex v i is not chosen, all vertices corresponding to v i must be chosen, since there are edges between v i and all these latter vertices which must be covered. We note that it may happen that the vertex cover will include, for example, all the vertices corresponding to i and some or all of the vertices corresponding to i. The latter may then be deleted to form a new vertex cover (which clearly induces a bipartite graph in G d ) corresponding to a satisfying assignment for the 3-satisability problem. Set a literal to be true if the corresponding vertex is chosen to be part of the vertex cover. For the chosen induced graph to be bipartite, at least one vertex from each triangle in G d must be left out of the vertex cover. In other words, each clause must be satised, since the negation of at least one of its literals is false. The way E is dened guarantees that for each variable i, either all the vertices corresponding to i or all the vertices corresponding to i will be chosen. This is the case since otherwise there will be an edge in E between a vertex i and a vertex i that is not covered if neither i nor i are in the vertex cover. Conversely, such a satisfying assignment can be used to form a vertex cover that induces a bipartite subgraph in G d. Corollary 3.4 Problem 3.1 is NP-hard even when the distances d ij satisfy the triangle inequality. Furthermore, assuming P 6= N P, no polynomial approximation to Problem 3.1 with an error guarantee of less than 2 exists even when the triangle inequality holds. 7
Proof: An instance of Problem 3.2 can be generated from instances of Problem 3.1 that satisfy the triangle inequality. To show this we simply set the lengths of edges of G d obtained in the proof above to be 1, and all other intervertex distances to be 2. Furthermore, for this instance, there is a satisable assignment to the formula if and only if the solution to Problem 3.1 is 1, otherwise the solution is 2. Thus a solution that approximates the optimal solution with error less than 2, must be optimal. Note that if the distances are obtained as the Euclidean distances of points in the plane, the above proof fails. In fact, we show in the next section, that the geometric version of the problem is polynomially solvable. Corollary 3.5 Assuming P 6= N P, no polynomial approximation to Problem 3.1 with a bounded error exists. Proof: Suppose such an approximation with a bound r on the error ratio exists. Given an instance of 3-satisability, we construct the (undirected) graph G corresponding to it as described above. We then dene a distance matrix on its vertices; d ij = 1 if (i; j) is in G d ; d ij = r + 1 otherwise. There is a satisable assignment if and only if the solution to Problem 3.2 on this graph is 1, otherwise the solution is r + 1. Thus the approximation solution to Problem 3.1 must be optimal. 3.3 Pair-Choice 2-partition in the plane The single-choice 2-partition problem asks for a partition of a set of points such that the maximum of the diameters of the two parts is minimized. When the points lie in the plane and d ij is the Euclidean distance between i and j, it is well known and easy to verify that there exists an optimal 2-partition that is separable, i.e., a partition such that the convex hulls of its parts are disjoint (see for example, [3], [16]). In other words, the two parts can be separated by a line. Thus only O(n 2 ) partitions (the ones generated by lines connecting pairs of points) must be considered. In the pair-choice problem, we make the following observation: The separability property holds with respect to the points chosen to participate in the cover. The reason is that once these points are selected, the best way to partition them denes a single-choice 2-partition problem. In some cases the number of partitions to be examined can be further reduced as shown in [2]. The idea there is that for any xed, the maximum of the two diameters associated with the partitions generated by the lines y = + x is unimodal in. Hence the optimal partition within this set of partitions can be computed by applying binary search. However, in the pair-choice version this unimodality property does not hold. The method above cannot be extended to partition a set of points in the plane into 3 sets, minimizing the maximum diameter, since it is no longer true that the convex hulls of the parts are necessarily disjoint. In fact, the problem of (single-choice) 3-partitioning of a set of points in the plane is NP-hard [12, page 185] (see also [5].) We use the separability property to solve Problem 3.1. In Algorithm Plane Test (Figure 4), we describe how to test a given value. The complete algorithm consists of a binary search on the distinct values of D. Theorem 3.6 The pair-choice 2-partition problem in the plane can be solved in time O(n 4 log n). Proof: Algorithm Plane Test tests a given d value by checking O(n 2 ) partitions, checking each partition in O(n 2 ) time. Using binary search on the values of D requires O(log n) tests. 3.4 Approximation assuming the triangle inequality We now show how to obtain a bounded error approximation for Problem 3.1 under the assumption that the distances satisfy the triangle inequality. Theorem 3.7 If D satises the triangle inequality then the pair-choice 2-partition problem can be approximated within a factor of 2 in time O(n 2 log n). Proof: Algorithm 2 Part Test tests for a given value d whether a 2-partition of diameter at most 2d exists, or no partition with diameter d or smaller exists. To do so we select an arbitrary edge e = (v; w) 2 E that must be covered, and check for both endpoints of e whether a cover containing the endpoint exists. Let u be 8
Plane Test input 1. A graph G = (V; E) where V is a set of points in the plane and d ij is the Euclidean distance between i; j 2 V. 2. A value d 2 IR. returns 1. \TRUE" if G has a d-cover. 2. \FALSE" if G has no d-cover. begin for every separable 2-partition (V 1 ; V 2 ) for every (i; j) 2 E Dene a clause (x i _ x j ). for every i; j 2 V 1 such that d ij > d Dene a clause (x i _ x j ). for every i; j 2 V 2 such that d ij > d Dene a clause (x i _ x j ). if the formula is satisfyable return \TRUE". return \FALSE". end Plane Test Figure 4: Algorithm Plane Test 2 Part Test begin Let (v; w) = e 2 E for every u 2 fv; wg (Suppose u 2 C) Let C 1 = fug [ fv 2 V jd uv dg. Let E 0 = f(i; j) 2 Eji; j =2 C 1 g: Let G 0 be the graph induced by E 0. A u :=Test(G 0 ; d). if either A v =\TRUE" or A w =\TRUE" return \TRUE". else return \FALSE". end 2 Part Test Figure 5: Algorithm 2 Part Test 9
one of the endpoints of e. Given d, we assume that u 2 C. Let C 1 be the set of all nodes at distance at most d from u. The diameter of C 1 is, by the triangle inequality, at most 2d. Every pair i; j such that (i; j) 2 E and both i and j are not in C 1 must be represented in C 2. Let V 0 be this set of nodes. Apply Algorithm Test (given in Figure 1) which tests a value d for a pair-choice cover to these pairs in V 0 (see Theorem 2.2). If diameter d suces to cover the remaining pairs, we clearly obtain a 2-partition with diameter at most 2d (C 1 has diameter at most 2d and C 2 has diameter at most d). Otherwise, we claim that no 2-partition cover of diameter d containing u can exist: Such a partition would contain u and a subset of C 1 and hence the other part of the cover would have to cover a set including all of V 0 with a diameter of d. This, however, is clearly impossible. The complexity of this algorithm is log n times the complexity of Test, O(n 2 log n). Remark 3.8 A similar procedure applies for the multiple-choice problem where js i j ` i = 1; : : :; m, that is, for the multiple-choice 2-partition problem. Instead of the algorithm that tests a value d for V 0 we use the approximate algorithm Multi Choice, which nds a cover for the subsets S i which are not covered by C 1 of diameter d apx, such that d apx is bounded by twice the minimum diameter of a cover of these subsets. If d apx 2d, a 2-partition cover of diameter 2d was found. On the other hand, if d apx > 2d we have that a minimum diameter of a cover for the subsets which were not covered by C 1 is strictly greater than d, and by a similar argument to the one in the previous theorem, a 2-partition cover of diameter d cannot exist. The same error bound of 2 is obtained and the complexity of Algorithm Multi Choice is multiplied by ` log n, for a total running time of O(`2N log n). Remark 3.9 For the multi-choice k-partition problem when the distances obey the triangle inequality, the following test procedure gives a 2-approximation: For every u 2 S 1 repeat the following steps: Suppose u is in the cover. Let C 1 = fug [ fv 2 V jd uv dg. Select an uncovered set and for each element, j, of this set, repeat the following steps. Set C 2 = fjg [ fv =2 C 1 jd jv dg. Continue the procedure in this way, forming C 3 ; : : : ; C k. If for any of the choices, C 1 ; : : : ; C k cover all of the subsets then this is a k-partition 2d-cover. Otherwise, no k-partition d-cover exists. The complexity will be O(`kN) per test value. The result is again a 2-approximation. Remark 3.10 In the plane we approximate the multi-choice problem as follows: We compute for each separable partition (V 1 ; V 2 ) and each pair of sets T V 1, or, T V 2 such that jt j 2 f2; 3g the smallest covering disk. Then, for each pair of such disks we check feasibility of the cover they dene and select the feasible solution with the smallest maximum diameter as our solution. The error ratio is bounded by 2= p 3 as in Algorithm Plane Multi Choice. The complexity of the algorithm is O(n 6 N) since we test pairs of disks from O(n 3 ) candidates. 4 Multiple-choice cover of maximum dispersion In this section we discuss problems in which the data is as before, but the objective function is reversed: Let V = f1; : : :; ng be a given set. Also given is a symmetric real matrix D = (d ij ) 2 IR nn + dening the distances between pairs of elements of V. Let S 1 ; :::; S m V be given. A d-max-cover is a cover whose dispersion, ^d(c) = minfd ij : i; j 2 Cg, is at least d. In this section we show the existence of a polynomial time algorithm for nding the maximum d for which a d-max-cover exists in the case where each set S i has cardinality two. A reduction from 3-SAT shows that the problem is NP-hard when the cardinality of the sets S i is larger than two, even when this cardinality is three, and the distance matrix corresponds to the Euclidean distances between points in the plane. Furthermore, (even when the distances obey the triangle inequality) it is not possible to approximate the maximum d and bound the error ratio by a constant in polynomial time, unless P = N P. As in the diameter problems, when the cardinalities of the sets S i are two, we represent them by a graph G = (V; E). Theorem 4.1 The problem of nding the maximum d for which a d-max-cover exists can be solved in time O(n 2 log n), if all sets S i have cardinality two. Proof: We modify algorithm Test in Figure 1 to test whether a particular value d is a feasible dispersion. As before, for any edge (i; j) 2 E we have a clause (x i _ x j ). The other clauses are dened dierently. For 10
any pair of vertices k and l whose distance from each other is smaller than d, we know that we can choose at most one of the vertices, so we have clauses of the form, (x k _ x l ). Next, we show that a 3-SAT problem can be written as a triple-choice cover problem, thus showing the triple-choice dispersion problem to be hard. Problem 4.2 (Multiple-Choice Dispersion Cover Problem) Find a subset C V, such that jc \ S i j 1 for i = 1; :::; m, and the dispersion of C, ^d(c) = minfd kl : k; l 2 Cg, is maximized. Theorem 4.3 Problem 4.2 is NP-hard even when S i i = 1; :::; m are disjoint, js i j 3 i = 1; :::; m, the elements of V are points in the plane, and D is the Euclidean distance matrix. Proof: The decision problem (does there exist a d-max-cover, for some constant d) is obviously in NP. We prove the theorem by a reduction from 3-satisability. Given a set of clauses K j = x j _ y j _ z j (j = 1; :::; `) where fx j ; y j ; z j g fu 1 ; u 1 ; ; u n ; u n g, we represent each variable u i by 2 points at distance d 0 < d from each other. Each such pair of points is placed at distance greater than d from other pairs. We dene an instance of Problem 4.2 with sets S 1 ; :::; S n such that S i consists of the points associated with u i and u i, and sets S n+1 ; :::; S n+` such that S n+j consists of the triplet of points whose location coincides with these of x j ; y j ; z j. A feasible solution with dispersion at least d is any set of points such that: 1. From each pair of points corresponding to u j ; u j, exactly one is selected. 2. From each triplet of points corresponding to x j ; y j ; z j, at least one is selected. Thus, if there exists a subset of dispersion at least d then there exists a satisfying assignment for K 1 ^ ^ K`. On the other hand, a satisfying assignment clearly determines a feasible set of dispersion at least d. Thus the proof is complete. Corollary 4.4 Assuming P 6= N P, no polynomial approximation to Problem 4.2 with a bounded error guarantee exists, even for instances of the problem corresponding to points in the plane. Proof: Suppose that a polynomial algorithm that guarantees a solution of diameter at most r times the optimal diameter exists. Consider the instance used in the proof of Theorem 4.3, except that whenever the distance between two points was d 0 we change it to (1=r)d 0, and all other distances are d or greater. Now, the solutions associated with a satisfying assignment are of dispersion d or greater, while all other feasible solutions to the choice problem have dispersion (1=r)d 0 < (1=r)d. Thus, the algorithm nds a satisfying assignment if one exists, in polynomial time. Finally, we mention that the corresponding partition problems of max-min dispersion cannot be solved or approximated in polynomial time. The proofs are similar to those of Theorems 4.3 and 3.2, and Corollary 4.4, and are omitted. The only problem solvable in polynomial time for the min-max diameter was Problem 3.1 (pair-choice 2-partition) in the plane. However, this relied on the property that such a partition is separable by a line, a property that is clearly not true for the dispersion problem. The complexity of the pair-choice 2-partition maximum dispersion in the plane remains open. 5 Open Problems An interesting open question is whether a more ecient algorithm exists for Problem 2.1 in the geometric case where the input of size O(n) consists of the locations of n points in the plane and D is the Euclidean distance matrix. Several of the other algorithms that are polynomial have rather high worst case running times. Can these be improved? In particular Algorithm Plane Test whose running time is O(n 4 log n) seems improvable. The proof of Theorem 3.3 denes a pair-choice problem where the pairs (edges of G) are not disjoint. The question of whether the restricted problem where the pairs are disjoint can be solved polynomially, is open. Finally, the complexity of the pair-choice 2-partition maximum dispersion in the plane is open. 11
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