Approximating min-max k-clustering

Transcription

1 Aroximating min-max k-clustering Asaf Levin July 24, 2007 Abstract We consider the roblems of set artitioning into k clusters with minimum total cost and minimum of the maximum cost of a cluster. The cost function is given by an oracle, and we assume that it satisfies some natural structural constraints. That is, we assume that the cost function is monotone, the cost of a singleton is zero, and we assume that for all S S the following holds c(s) + c(s ) c(s S ). For this roblem we resent a (2k )-aroximation algorithm for k 3, a 2-aroximation algorithm for k = 2, and we also show a lower bound of k on the erformance guarantee of any olynomial-time algorithm. We then consider secial cases of this roblem arising in vehicle routing roblems, and resent imroved results. Introduction We are given a ground set E with a cost function, c, defined over all subsets of E. Assume that c satisfies the following roerties:. c({i}) = 0 for all i E. 2. c(s) c(s ) for all S S. That is, c is a monotone cost function. 3. If S S, then c(s) + c(s ) c(s S ). The k MIN-MAX roblem is to find a artition of E into k (disjoint) subsets S, S 2,... S k so that max i c(s i ) is minimized. The k MIN-SUM roblem is to find a artition of E into k (disjoint) subsets S, S 2,... S k so that k i= c(s i) is minimized. We assume that c is given as an oracle that evaluates the value of c for a given subset of E in O() time, and we are interested in olynomial-time (in size of E) algorithms. The sets S, S 2,..., S k (of a feasible solution) are called the blocks of the artition or clusters. For a artition P we denote its cost by c(p) (where the definition of the cost of the solution deends on the roblem we consider). For a artition of E into E, E 2,..., E k a refinement is a artition of E into k subsets E, E 2,..., E k such that for all i k there exists j(i) such that E i E j(i). In recent years the study of clustering roblems has increased dramatically. In most roblems that were investigated in the literature the goal is either to minimize the maximum cost of a cluster or to minimize the sum of costs of the clusters. This leaves the definition of a cost of a cluster different for different roblems. To obtain a olynomial reresentation of the roblem, the cost function is usually given in a comact form. In this aer we assume that the cost function satisfies Proerties, 2 and 3, and we investigate the otimization roblems that result from such cost structure. The assumtion that the cost function is given by an Deartment of Statistics, The Hebrew University, Jerusalem 9905, Israel. levinas@mscc.huji.ac.il Dagstuhl Seminar Proceedings 0726 Fair Division htt://dros.dagstuhl.de/ous/volltexte/2007/228

2 oracle relaces the comact form of the cost function (that is usually assumed) and is articularly interesting in cases in which the cost of a cluster is evaluated by solving an NP-hard roblem or erforming a simulation study of some engineering roblem (so in these cases it is unclear how to obtain a comact reresentation of the cost function). For examle, in the well-known vehicle routing roblem a cluster is a subset of the vertex set to be served by a common vehicle, and the cost of a cluster is the otimal solution of the traveling saleserson roblem over this vertex set. Given such a heavy comutational rocedure to evaluate the cost of a cluster, it is clear that we are interested in trying to minimize the number of such rocedure calls. The k CLUSTERING PROBLEM (kcluster) is defined as follows: Given a comlete undirected grah G = (V, A) where the edges are endowed with a metric length function l : A R +, the goal is to artition the vertex set into k sets V, V 2,..., V k so as to minimize the following objective function max i max u,v Vi l(u, v). We next note that the kcluster is a secial case of the k min-max roblem. To see this last claim note that we can define a cost function c(s), where S V, to be max u,v S l(v, u). To see that the resulting roblem is an instance of the k min-max roblem, we need to show that c satisfies the requested roerties. Proerties and 2 are trivially satisfied. To see that Proerty 3 is also satisfied, consider a air of non-disjoint sets S, S 2, and assume that v S S 2. Fix arbitrary vertices u, w S S 2. If u, w S or u, w S 2, then clearly l(u, w) c(s ) + c(s 2 ). If u S and w S 2, then by the triangle inequality l(u, w) l(u, v)+l(v, w) c(s )+c(s 2 ). Therefore, for all airs u, w S S 2 we conclude that l(u, w) c(s ) + c(s 2 ). Proerty 3 follows by taking the maximum over all airs u and w. Gonzales [2] and Hochbaum and Shmoys [5] showed that the kcluster roblem has a 2-aroximation algorithm. Hsu and Nemhauser [6] and Hochbaum [4] showed that kcluster cannot be aroximated within a factor better than 2 (assuming P NP ). Gonzales [2] showed that the same lower bound on the aroximability of the kcluster alies also to the secial case where the inut is restricted to oints in 3-dimensional Euclidean sace. Feder and Greene [] showed that the kcluster roblem where the inut is restricted to be a set of oints in 2-dimensional Euclidean sace is not aroximable within.969. As we noted above the k min-max roblem generalizes the kcluster, and therefore it is NP-hard for all k 3. We then consider secial cases of this roblem arising in vehicle routing roblems, and resent imroved results. In articular we consider the following roblem. The MIN-MAX RURAL POSTMEN COVER PROB- LEM is defined as follows. The inut is a comlete grah G = (N, E), a length function l : E R +, a subset E E, and an integer number k > 0. The outut is a set of k aths P,..., P k, such that E P i. The goal is to minimize the maximum length of a ath in the solution, that is, to minimize max i l(p i ). We note that this roblem is a secial case of the min-max k-clustering roblem. However, using the secial structure of this roblem, we are able to resent a 7-aroximation algorithm for this roblem (so the aroximation ratio is indeendent of k). Paer outline. In Section 2 we rovide an aroximation algorithm for the k min-max roblem whose aroximation ratio is 2 for k = 2 and ((k ) α + ) for all k 3 where α = 2 denotes the aroximation ratio of an aroximation algorithm for the kcluster roblem, and we conclude this section by showing that any olynomial-time algorithm cannot guarantee an aroximation ratio that is better than k for the k min-max roblem. 2 The k min-max roblem In this section, we first rovide an aroximation algorithm and analyze its erformance guarantee. Our aroximation algorithm is a 2-aroximation if k = 2, and a ((k )α + )-aroximation algorithm where α is the aroximation ratio of any aroximation algorithm for the kcluster roblem. Afterwards, we will show a lower bound of k on the erformance guarantee of any olynomial-time algorithm for the k min-max roblem. Recall that for k = 2 there is a olynomial-time algorithm that solves (otimally) the kcluster roblem. 2

3 This algorithm is based on guessing the otimal cost (that is one out of O( A ) values that are known in advance), and then defining an exensive edge to be an edge whose cost is greater than the current guess. The current guess is not smaller than the otimal cost if and only if the grah of the exensive edges is biartite. This results an O( A log A ) time algorithm for the 2Cluster roblem (by using a binary search on the otimal cost guess value). In this case where k = 2 we denote α =. For higher values of k, one can show [6, 4] a similar reduction from k-coloring of a simle grah that shows that aroximating the kcluster within a factor 2 ε is NP-hard (for all ε > 0). That is, given an inut grah G c = (V c, E c ) for the k-coloring roblem we let G = (V c, E) to be a comlete grah with edge cost c({i, j}) = if and only if (i, j) E c and otherwise c({i, j}) = 2. Then the kcluster instance has a solution with cost one if and only if G c is k-colorable, and otherwise the otimal solution to the kcluster instance costs two. For the kcluster roblem Gonzales [2] and Hochbaum and Shmoys [5] rovided 2-aroximation algorithms (for all values of k). Using the results of [2, 5], when k 3 we denote α = 2. So α is either or 2 deending on the value of k. Algorithm. Comute a cost function c defined as c (S) = max i,j S {c({i, j})}. 2. Using an α-aroximation algorithm for the kcluster roblem, find a artition of E into k subsets such that its cost with resect to c is minimized. Theorem 2 Algorithm is a ((k ) α + )-aroximation algorithm. Proof. Note that by monotonicity of the cost function, we conclude that for every S E, c (S) c(s). Moreover, for i, j and k we note that c ({i, j}) + c ({j, k}) = c({i, j}) + c({j, k}) c({i, j, k}) c({i, k}) = c ({i, k}) where the first inequality holds by roerty 3 and the second inequality holds by roerty 2, and therefore there is an α-aroximation algorithm for the kcluster instance that we create in ste 2 (i.e., the metric assumtion holds for c ). Denote by OPT the otimal solution with resect to c, and by APX the solution of Algorithm. Assume that APX OPT. Let G be a grah defined as follow: For every cluster in OPT there is a vertex in G and for every air I and J of clusters of OPT, there is an edge in G between their vertices if and only if there is a cluster C in APX such that both I C and J C. For every edge (i, j) in G reresenting the clusters I and J in OPT let v i I C and v j J C where C is a cluster in APX that causes this edge to exist. Then by Proerty 3 of c, the following is satisfied: c(i) + c(j) + c({v i, v j }) c(i) + c(j {v i, v j }) c(i J) Note that c({v i, v j }) c (APX), but APX is an α-aroximation algorithm with resect to c, and therefore c (APX) α c (OPT) α c(opt). For each connected comonent C of G, we union all the clusters that corresond to vertices in C. Denote by the number of connected comonents of G. Finally, we receive a artition of E into sets that costs at most c(opt)+(k ) α c(opt) ((k ) α + ) c(opt). But APX is a refinement of the resulting one, and therefore costs less. Therefore, c(apx) ((k ) α + ) c(opt). We next show that our algorithm is best ossible u to a constant factor. Theorem 3 No olynomial-time algorithm for the k min-max roblem has an aroximation ratio better than k. 3

4 Proof. Consider a ground set E of size n where n = k 2 and is an arbitrary large integer to be selected afterwards. Consider a artition of E into k equal size sets E, E 2,..., E k. For a set S E denote by n(s) = {i : E i S }. We define a cost function c as follows: { 0 if S = c(s) = { } min n(s), otherwise. We next show that c satisfies roerties, 2 and 3. For i E, c({i}) = 0 because {i} =, and therefore roerty holds. Let S S, then n(s ) n(s) as E i S imlies that E i S. Moreover, S, and therefore S. Therefore, for S S, c(s ) c(s), and hence roerty 2 holds. It remains to consider roerty 3. Let S, S E such that S S. If S = (and similarly if S = ), then the roerty clearly holds as both sides of the inequality equal to c(s ). It remains to consider the case where S 2 and S 2. Note that n(s S ) n(s) + n(s ) as if (S S ) E i, then either S E i or S E i (or both). Moreover, S S + S, and therefore S S + S. Therefore, c(s S ) c(s) + c(s ), and roerty 3 also holds. The otimal solution is the artition of E into E, E 2,..., E k. By definition n(e i ) = for all i, and therefore the otimal solution has a unit cost. In order to rove the theorem it suffices to show that any olynomial-time algorithm cannot guarantee a solution whose cost is at most k. Assume otherwise; then the algorithm must identify (in olynomial time) a set S such that c(s) k and S k (this is so as the set with the largest cardinality in the returned solution has at least k elements). Consider this set S; then since S k, we conclude that c(s) = n(s). Since the algorithm is not aware of the otimal artition, it must identify a set whose size is at least + and that intersects at most k blocks of the otimal artition. Now assume that the otimal artition is selected randomly. Then the robability that a given set S will intersect at most k blocks of the artition is at most k ( ). k When an algorithm queries a set S and finds out that n(s ) = k, it learns only that any set that contains S has a value of n that equals k. However, for all remaining sets the algorithm does not gain new information. Therefore, the exected number of stes until such S will be found is at least k ( k). Note that this last bound is also a lower bound on the number of stes that the assumed algorithm erforms. However, k ( k) e/k k where the last inequality holds because e for all k 2. Since can be arbitrary large (significantly larger than k), the last ( k) k lower bound on the number of stes that the algorithm erforms is exonential in the size of E, and this is a contradiction to the assumtion that the algorithm erforms only a olynomial number of stes. References S [] T. Feder and D.H. Greene, Otimal algorithms for aroximate clustering, Proceedings of the 20th Annual ACM Symosyum on Theory of Comuting (STOC 988), , 988. [2] T.F. Gonzalez, Clustering to minimize the maximum intercluster distance, Theoretical Comuter Science, 38, , 985. [3] P. Hansen and B. Jaumard, Cluster analysis and mathematical rogramming, Mathematical Programming, 79, 9-25, 997. [4] D.S. Hochbaum, When are NP-hard location roblems easy?, Annalls of Oerations Research,, 20-24,

5 [5] D.S. Hochbaum and D.B. Shmoys, A unified aroach to aroximation algorithms for bottleneck roblems, Journal of the ACM, 33, , 986. [6] W.L. Hsu, and G.L. Nemhauser, Easy and hard bottleneck location roblems, Discrete Alied Mathematics,, , 979. [7] C. L. Monma and S. Suri, Partitioning oints and grahs to minimize the maximum or the sum of diameters, in Grah Theory, Combinatorics and Alications, , Alavi et al. (editors), Wiley, 99. 5