The Set Cover with Pairs Problem

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1 The Set Cover with Pairs Problem Refael Hassin Danny Segev Abstract We consider a generalization of the set cover problem, in which elements are covered by pairs of objects, and we are required to find a minimum cost subset of objects that induces a collection of pairs covering all elements. Formally, let U be a ground set of elements and let S be a set of objects, where each object i has a non-negative cost w i. For every {i, j} S, let C(i, j) be the collection of elements in U covered by the pair {i, j}. The Set Cover with Pairs problem asks to find a subset A S such that {i,j} A C(i, j) = U and such that i A w i is minimized. In addition to studying this general problem, we are also concerned with developing polynomial time approximation algorithms for interesting special cases. The problems we consider in this framework arise in the context of domination in metric spaces and separation of point sets. 1 Introduction 1.1 The General Framework Given a ground set U and a collection S of subsets of U, where each subset is associated with a nonnegative cost, the Set Cover problem asks to find a minimum cost subcollection of S that covers all elements. An equivalent formulation is obtained by introducing a covering function C : S 2 U, that specifies for each member of S the subset of U it covers. Set cover now becomes the problem of finding a subset A S of minimum cost such that C(A) = i A C(i) = U. We consider a generalization of this problem, in which the covering function C is defined for pairs of members of S, rather than for single members. Formally, let U = {e 1,..., e n } be a ground set of elements and let S = {1,..., m} be a set of objects, where each object i S has a non-negative cost w i. For every {i, j} S, let C(i, j) be the collection of elements in U covered by the pair {i, j}. The objective of the Set Cover with Pairs problem (SCP) is to find a subset A S such that C(A) = {i,j} A C(i, j) = U and such that w(a) = i A w i is minimized. We refer to the special case in which each object has a unit weight as the Cardinality SCP problem. SCP is indeed a generalization of the set cover problem. A set cover instance with U = {e 1,..., e n } and S 1,..., S m U can be interpreted as an SCP instance by defining C(i, j) = S i S j for every i j. Therefore, hardness results regarding set cover extend to SCP, and in particular the latter problem cannot be approximated within a ratio of (1 ɛ) ln n for any ɛ > 0, unless NP T IME(n O(log log n) ) [4]. 1.2 Applications In addition to studying the SCP problem, we are concerned with developing polynomial time approximation algorithms for interesting special cases, that arise in the context of domination in metric spaces and separation of point sets. An extended abstract of this paper appeared in Proceedings of the 25th Annual Conference on Foundations of Software Technology and Theoretical Computer Science, pages , School of Mathematical Sciences, Raymond and Beverly Sackler Faculty of Exact Sciences, Tel-Aviv University, Tel-Aviv 69978, Israel. {hassin,segevd}@post.tau.ac.il 1

2 Remote Dominating Set. Let M = (V, d) be a finite metric space with V = {v 1,..., v n }, and let r 1 r 2 be two covering radii. We refer to the elements of V as points or vertices, and assume that each v V is associated with a non-negative cost c v. A subset of points S V is called a remote dominating set if for every v V there is a point u S within distance r 1 of v, or a pair of points u 1 u 2 S within distance r 2 of v each. The remote dominating set problem (RDS) asks to find a minimum cost remote dominating set in M. RDS can be interpreted as a special case of SCP: The set of elements to cover is V, which is also the set of covering objects, and the collection of points covered by u v V is C(u, v) = {w V : min {d(u, w), d(v, w)} r 1 or max {d(u, w), d(v, w)} r 2 }. When d(u, v) {1, 2} for every u v and r 1 = r 2 = 1, RDS reduces to the standard Dominating Set problem. Therefore, hardness results regarding set cover extend to RDS, as the dominating set problem is equivalent to set cover with regard to inapproximability. We also consider two special cases of this problem, for which significantly better approximation algorithms are possible. In the Cardinality RDS on a Tree problem, the metric d is generated by a tree T = (V, E) with unit length edges, and the covering radii are r 1 = 1 and r 2 = 2. In the Cardinality Euclidean RDS problem, V is a set of points in the plane, and d(u, v) = u v 2. Group Cut on a Path. Let P = (V, E) be a path, in which each edge e E has a non-negative cost c e, and let G 1,..., G k be k groups, where each group is a set of at least two vertices. A group G i is separated by the set of edges F E if there is a representative v i G i such that no vertex in G i \ {v i } belongs to the connected component of P F that contains v i. The objective of the group cut on a path problem (GCP) is to find a minimum cost set of edges that separates all groups. Given a GCP instance we may assume without loss of generality that any optimal solution contains at least two edges. This assumption implies that GCP is a special case of SCP: The elements to cover are the groups G 1,..., G k, and the covering objects are the edges. The groups covered by pairs of edges are defined as follows. Let v 1,..., v r be the left-to-right order of the vertices in G i, and let [v i, v j ] be the set of edges on the subpath connecting v i and v j. The group G i is covered by a pair of edges e e E if {e, e } ([v 1, v 2 ] [v r 1, v r ]) or if e [v t 1, v t ] and e [v t, v t+1 ] for some 2 t r Our Results In Section 2 we study a natural extension of the greedy set cover algorithm [1, 10, 12] to approximate SCP. We define a class of functions, called feasible maps, that assign the elements in U to pairs of objects in the optimal solution, and characterize them by max and mean properties. We then present a conditional analysis of the greedy algorithm, based on the existence of such maps. Specifically, we prove an approximation guarantee of αh n for the weighted and cardinality versions of the problem, given the existence of feasible maps whose max and mean are at most α, respectively. We also prove that the unconditional approximation ratio for cardinality SCP is O( n log n). We continue the discussion with indications for the hardness of SCP. First, although the set cover problem becomes trivial when each subset contains a single element, we show that the corresponding special case of SCP, where each pair of objects covers at most one element, is at least as hard to approximate as set cover. Second, the analysis of the greedy set cover algorithm in [12] shows that the integrality gap of the natural LP-relaxation of set cover is O(log n). However, we demonstrate that this property does not extend to SCP, for which the integrality gap is Ω(n). As a first attempt at attacking the RDS problem, one might consider using the greedy SCP algorithm. However, we show in Section 3 that the approximation guarantee of this algorithm is Ω( n), mainly due to the observation that there are instances of RDS in which non-trivial feasible maps do not exist. Nevertheless, we provide a 2H n -approximation algorithm that constructs a remote dominating set by approximating two dependent set cover problems. 2

3 In Section 4 we proceed to the cardinality RDS problem on a tree T = (V, E). Although this problem can be solved to optimality in O( V 3 ) time using dynamic programming techniques (see Appendix B), we demonstrate that it can be well approximated much faster. We first show how to map a subset of problematic vertices of T to a small collection of pairs in the optimal solution. We then exploit the special structure of this map to present a linear time 2-approximation algorithm, and illustrate that in general graphs this algorithm does not guarantee a non-trivial approximation ratio. In Section 5 we present a polynomial time approximation scheme for the Euclidean RDS problem. Although we follow the general framework of Hochbaum and Maass for covering and packing problems in Euclidean spaces [9], our analysis is more involved. This is due to the use of two covering radii and the restriction that the set of points we choose must be a subset of V, instead of any set of points in the plane. Finally, in Section 6 we discuss the hardness of approximating GCP, and in particular prove that this problem is as hard to approximate as set cover. Moreover, we identify the exact point at which GCP becomes NP-hard, by showing that this problem is polynomial time solvable when the cardinality of each group is at most 3, but as hard to approximate as vertex cover when the bound on cardinality is 4. On the positive side, we prove the existence of a feasible map whose max is at most 2. This result enables us to show that the approximation ratio of the greedy SCP algorithm for this special case is 2H k, where k is the number of groups to be separated. 2 Set Cover with Pairs In this section we suggest a natural extension of the greedy set cover algorithm to approximate SCP, and present a conditional analysis based on the existence of a mapping of the elements in U to pairs of objects in the optimal solution, that satisfies certain properties. We then make use of these results to prove an approximation ratio of O( n log n) for cardinality SCP. We also prove that the special case in which each pair of objects covers at most one element is at least as hard to approximate as set cover, and demonstrate that the integrality gap of the natural LP-relaxation of SCP is Ω(n). 2.1 A Greedy Algorithm The greedy SCP algorithm iteratively picks the most cost-effective object or pair of objects until all elements are covered, where cost-effectiveness is defined as the ratio between the objects costs and the number of newly covered elements. Let GR be the set of objects already picked when an iteration begins, where initially GR =. We define: 1. For every i S \ GR, the current covering ratio of i is w i C(GR {i}) C(GR). 2. For every i j S \ GR, the current covering ratio of {i, j} is w i + w j C(GR {i, j}) C(GR). In each iteration we augment GR by adding a single object i S \ GR or a pair of objects i j S \ GR, whichever attains the minimum covering ratio. The algorithm terminates when U is completely covered. 3

4 2.2 Conditional Analysis Let F S be a feasible solution, that is, a set of objects that covers the elements of U, and let P (F ) = {{i, j} F : i j}. A function M : U P (F ) is a feasible map with respect to F if the pair of objects M(e) covers e, for every e U. Given a feasible map M, for every {i, j} P (F ) we use I M (i, j) to indicate whether at least one element is mapped to {i, j}. We define: max(m, F ) = max i F j i I M (i, j), mean(m, F ) = 1 F I M (i, j). In other words, max(m, F ) α if each object i F belongs to at most α pairs to which elements are mapped. Similarly, mean(m, F ) α if the average number of pairs, to which elements are mapped, an object belongs to is at most α. Clearly, mean(m, F ) max(m, F ). In Lemma 1 we show that given the existence of a feasible map M with respect to an optimal solution OPT, for which max(m, OPT) α, the greedy SCP algorithm constructs a solution whose cost is within factor αh n of optimum. In Lemma 2 we show that to obtain an approximation guarantee of αh n for cardinality SCP, the weaker condition of mean(m, OPT) α is sufficient. Lemma 1. If there exists an optimal solution OPT and a feasible map M such that max(m, OPT) α, then w(gr) αh n w(opt). Proof. For every {i, j} P (OPT), let M 1 (i, j) = {e U : M(e) = {i, j}}. By definition of M, {M 1 (i, j) : {i, j} P (OPT)} is a partition of U. In each iteration of the algorithm, we distribute the cost of the newly picked object or pair of objects among the new covered elements: If x new elements are covered, each such element is charged w i x or w i+w j x, depending on whether a single object or a pair of objects are picked. Let M 1 (i, j) = {e 1,..., e k }, where the elements of M 1 (i, j) are indexed by the order they were first covered by the greedy algorithm, breaking ties arbitrarily. Consider the iteration in which e l was first covered. One possibility of the greedy algorithm was to pick {i, j} (or if one of i and j was already picked, then take the other one), covering the elements e l,..., e k, and possibly other elements as well. Therefore, each element that was covered in this iteration is charged at most w i+w j k l+1, and the total cost charged to the elements of M 1 (i, j) satisfies charge(m 1 (i, j)) = k charge(e l ) l=1 Since w(gr) is charged to e 1,..., e n, we have w(gr) = = n charge(e j ) j=1 {i,j} P (OPT) H n = H n αh n {i,j} P (OPT) i OPT i OPT k l=1 i F j i w i + w j k l + 1 (w i + w j )H n. charge(m 1 (i, j)) (w i + w j )I M (i, j) w i I M (i, j) w i j i = αh n w(opt), where the last inequality holds since j i I M(i, j) max(m, OPT) α for every i OPT. 4

5 Lemma 2. If there exists an optimal solution OPT for cardinality SCP and a feasible map M such that mean(m, OPT) α, then GR αh n OPT. Proof. Using a charging argument similar to that in the proof of Lemma 1, for every {i, j} P (OPT) such that I M (i, j) = 1 we have charge(m 1 (i, j)) 2H n. Therefore, GR = charge(m 1 (i, j)) {i,j} P (OPT) 2H n = 2H n 1 2 {i,j} P (OPT) i OPT j i αh n OPT, I M (i, j) I M (i, j) where the last inequality holds since we have i OPT j i I M(i, j) α OPT, as mean(m, OPT) α. 2.3 Approximation Ratio for Cardinality SCP The conditional analysis in Lemma 2 is based on the existence of a feasible map M with respect to the optimal solution with small mean(m, OPT). In Lemma 3 we demonstrate that there are instances of cardinality SCP in which a non-trivial map does not exist, and show that the approximation ratio of the greedy SCP algorithm might be Ω( n). However, in Theorem 4 we prove that the cardinality of the solution constructed by the greedy algorithm is within factor 2nH n of the minimum possible. Lemma 3. The approximation guarantee of the greedy algorithm for cardinality SCP is Ω( n). Proof. Consider an SCP instance with U = {e 1,..., e n } and S = {a 1,..., a k, b 1,..., b n }, where n = ( k 2). The covering function is defined by: C(ai, b j ) = ; C(b i, b j ) = {e i, e j }; we define C(a i, a j ) such that each pair {a i, a j } covers a distinct element. Clearly, the optimal solution is {a 1,..., a k }, with OPT = k = O( n). However, the greedy SCP algorithm might obtain the solution GR = {b 1,..., b n } by picking a pair of b s in each iteration. Theorem 4. GR 2nH n OPT. Proof. We first observe that GR 2n, since in each iteration of the algorithm at least one element is covered using at most two objects. In addition, any feasible map M with respect to OPT certainly satisfies mean(m, OPT) OPT, since mean(m, OPT) max(m, OPT) OPT. By Lemma 2 we have GR H n OPT 2, and it follows that GR min { 2n, H n OPT 2} ( 2n ) 1 2 ( H n OPT 2) 1 2 = 2nH n OPT. 2.4 The Hardness of SCP: Additional Indications The Case C(i, j) 1. The set cover problem becomes trivial when each subset contains a single element. However, in Theorem 5 we prove that SCP remains at least as hard to approximate as set cover when each pair of objects covers at most one element. We refer to this special case as SCP 1. 5

6 Theorem 5. For any fixed ɛ > 0, a polynomial time α-approximation algorithm for SCP 1 would imply a polynomial time (1 + ɛ)α-approximation algorithm for set cover. Proof. Given a set cover instance I, with a ground set U = {e 1,..., e n } and a collection S = {S 1,..., S m } of subsets of U, we construct an instance ρ(i) of SCP 1 as follows. 1. Let k = n ɛ. 2. The set of elements is k t=1 {et 1,..., et n}. 3. The set of objects is ( k t=1 {St 1,..., St m}) {y 1,..., y n }. 4. For t = 1,..., k, i = 1,..., m and j = 1,..., n, the pair {S t i, y j} covers e t j if e j S i. 5. Other pairs do not cover any element. Let S S be a minimum cardinality set cover in I. Given a polynomial time α-approximation algorithm for SCP 1, we show how to find in polynomial time a set cover with cardinality at most (1 + ɛ)α S, for any fixed ɛ > 0. The construction of ρ(i) guarantees that the collection of objects {Si t, y 1,..., y n } covers the set of elements {e t j : e j S i }, for every t = 1,..., k. Therefore, since S is a set cover in I, the objects ( k t=1 {St i : S i S }) {y 1,..., y n } cover all elements of ρ(i). It follows that OPT(ρ(I)) k S + n, and we can find in polynomial time a feasible solution S to ρ(i) such that S α(k S + n). Let t be the index t for which S {S1 t,..., St m} is minimized. Then S = {S i : Si t S {S1 t,..., St m}} is a set cover in I with cardinality S = min S {S1, t..., Sm} t S t=1,...,k k α(k S + n) (1 + ɛ)α S. k Integrality Gap of LP-Relaxation. In contrast with the set cover problem, for which the integrality gap of the natural LP-relaxation is O(log n) [12], we show in Theorem 6 that the integrality gap of the corresponding relaxation of SCP is Ω(n). SCP can be formulated as an integer program by: minimize w i x i subject to i S {i,j}:e C(i,j) y {i,j} 1 e U (2.1) y {i,j} x i i j S (2.2) x i, y {i,j} {0, 1} i j S (2.3) The variable x i indicates whether the object i is chosen for the cover, and the variable y {i,j} indicates whether both i and j are chosen. Constraint (2.1) ensures that for each element e U we pick at least one pair of objects that covers it. Constraint (2.2) ensures that a pair of objects cannot cover any element unless we indeed pick both objects. The LP-relaxation of this integer program, (LP ), is obtained by replacing the integrality constraint (2.3) with x i 0 and y {i,j} 0. Theorem 6. The integrality gap of (LP ) is Ω(n), even for cardinality SCP. Proof. Consider the instance of cardinality SCP with U = {e 1,..., e n } and S = {1,..., 2n}. The elements covered by pairs of objects in S are: 6

7 1. C(i, n + 1) = C(i, n + 2) = = C(i, 2n) = {e i }, i = 1,..., n. 2. Other pairs do not cover any element. Since any integral solution must pick the objects 1,..., n and at least one of the objects n + 1,..., 2n, OPT n + 1. We claim that the fractional solution x i = 1 n and y {i,j} = 1 n for every i j is feasible for (LP ). Clearly, this solution is non-negative and satisfies constraint (2.2). In addition, = 1 for every e U, since the left-hand-side contains exactly n summands, each {i,j}:e C(i,j) y {i,j} of value 1 n. It follows that the cost of an optimal fractional solution is at most 2, and the integrality gap of (LP ) is at least n Remote Dominating Set In the following we show that there are instances of the problem in which a non-trivial map does not exist, and demonstrate that the greedy algorithm might construct a solution for RDS whose cost is Ω( n) times the optimum. On the positive side, we provide a 2H n -approximation algorithm for RDS that constructs a remote dominating set by approximating two dependent set cover problems. 3.1 The Greedy SCP Algorithm for RDS According to our interpretation of the RDS problem as a special case of SCP, the greedy algorithm picks in each iteration a single point or a pair of points, whichever attains the minimum ratio of cost to number of newly covered points. By modifying the construction in Lemma 3, we prove in Lemma 7 that the approximation ratio of this algorithm is Ω( n). Lemma 7. The approximation guarantee of the greedy algorithm for RDS is Ω( n). Proof. We construct an instance of RDS as follows (see Fig. 1). The set of points is V = A B X Y, where A = {a 1,..., a k }, B = {b ij : 1 i j k}, X = {x ij : 1 i j k} and Y = {y ij : 1 i j k}. The cost of points in A B is 1, and the cost of points in X Y is. The metric d is given by: 1. For every 1 i j k, d(a i, y ij ) = d(a j, y ij ) = 1 + ɛ. 2. For every 1 i j k, d(y ij, x ij ) = 1 2ɛ. 3. For every 1 i j k, d(b ij, x ij ) = ɛ. 4. We complete d to a metric on V according to the triangle inequality. Finally, the covering radii are r 1 = 1 and r 2 = 2. a 1 a i a j 1 + ɛ 1 + ɛ y 1,2 y 1,3 x 1,2 x 1,3 1 2ɛ ɛ y ij x ij b 1,2 b 1,3 b ij a k Figure 1: The metric space M = (V, d) The construction of M guarantees that for every v V \ A there are two points in A within distance 2. Therefore, the set of points A is an RDS in M and the cost of an optimal solution is at most k. However, the greedy algorithm might pick in each iteration a single point in B, and eventually obtain a solution whose cost is at least ( k 2). By choosing k = n, the number of points in V is O(n) and the approximation ratio of the greedy algorithm is Ω( n). 7

8 3.2 A 2H n -Approximation Algorithm Despite these negative results regarding the performance of the greedy SCP algorithm for the RDS problem, we show that this problem can still be approximated to within a logarithmic factor. Our algorithm constructs a remote dominating set by approximating two dependent set cover problems, (SC 1 ) and (SC 2 ). For v V, let N v = {u V : d(v, u) r 2 }. Using the greedy set cover algorithm, we construct an RDS in two phases: 1. We first approximate (SC 1 ): The set of elements to cover is V ; The covering sets are S = {N v : v V }; The cost of N v is c v. Let S 1 be the cover we obtain. V can now be partitioned into two sets: V 1, points within distance r 1 of some point in S 1 or within distance r 2 of two points in S 1, and V 2 = V \ V We then approximate (SC 2 ): The set of elements to cover is V 2 ; The covering sets are S = {N v : v V \ S 1 }; The cost of N v is c v. Let S 2 be the cover we obtain. Theorem 8. Let OPT be a minimum cost RDS. Then 1. S 1 S 2 is an RDS. 2. c(s 1 S 2 ) 2H n c(opt). Proof. Clearly, OPT is a feasible solution to (SC 1 ), and c(s 1 ) H n c(opt). By definition of V 1 and V 2, in order to complete S 1 to an RDS, it is sufficient to find a subset of V \ S 1 that satisfies the following property: For every v V 2 there is a point u in this subset such that d(v, u) r 2. The construction of S 2 guarantees this property, and S 1 S 2 is indeed an RDS. In addition, we claim that OPT \ S 1 is a feasible solution to (SC 2 ), which implies c(s 2 ) H V2 c(opt \ S 1 ). It is sufficient to show that for every v V 2 there is a point u OPT \ S 1 such that d(v, u) r 2. There are two cases: 1. There is a point u OPT such that d(v, u) r 1. Since v / V 1, there is no point in S 1 within distance r 1 of v, and u / S 1. Therefore, u OPT \ S 1 is within distance r 1 r 2 of v. 2. There is a pair of points u 1 u 2 OPT such that d(v, u 1 ) r 2 and d(v, u 2 ) r 2. Since v / V 1, at least one of these points is not in S 1, and we assume without loss of generality that u 1 / S 1. Therefore, u 1 OPT \ S 1 is within distance r 2 of v. It follows that c(s 1 S 2 ) = c(s 1 ) + c(s 2 ) H n c(opt) + H V2 c(opt \ S 1 ) 2H n c(opt). 4 Cardinality RDS on a Tree In this section we consider the minimum cardinality RDS problem on a tree T = (V, E) with unit length edges and covering radii r 1 = 1 and r 2 = 2. It would be convenient to work directly with the tree representation of the problem, instead of working with the related metric space. We constructively show that a minimum cardinality dominating set in T is a 2-approximation, by exploiting special properties of a partial map we find. We also prove that this bound is best possible, and demonstrate that in general graphs a minimum cardinality dominating set does not guarantee a non-trivial approximation ratio. 8

9 4.1 The Existence of Acyclic Mapping Graphs Let S be an RDS that contains at least two vertices. We denote by L V the set of vertices that are not covered by a single vertex in S. In other words, v L if there is no vertex in S within distance 1 of v. Given a partial map M L : L P (S), its mapping graph G(M L ) is defined by: 1. The set of vertices of G(M L ) is S. 2. For u v S, (u, v) is an edge of G(M L ) if there is a vertex w L such that M L (w) = {u, v}. Lemma 9. There is a partial map M L : L P (S) whose mapping graph G(M L ) is acyclic. Proof. We attach a distinct number ψ(x) to every x S, and map each vertex v L to a pair {x, y} P (S) that minimizes the sum ψ(x) + ψ(y) over all pairs covering v, breaking ties arbitrarily. Let M L be the partial map we obtain. We claim that G(M L ) is acyclic. Suppose G(M L ) contains a cycle (x 1, x 2,..., x k, x 1 ). We assume without loss of generality that ψ(x 1 ) < ψ(x j ) for every j = 2,..., k. By definition of G(M L ), we can find k distinct vertices v 1,2,..., v k 1,k, v k,1 L such that v j,j+1 was mapped to {x j, x j+1 } for every j = 1,..., k 1, and such that v k,1 was mapped to {x k, x 1 }. Since v j,j+1 was mapped to {x j, x j+1 }, there are two possibilities for the relative location of these vertices in T, called Y-connection and V-connection (see Fig. 2). x j S x j+1 S x j S x j+1 S / S L / S L / S L v j,j+1 L Y-connection v j,j+1 L V-connection Figure 2: Type of connections to a covering pair We denote by W the closed directed walk x 1 x 2 x k x 1 in T, that uses the unique paths between these vertices (thick edges in Fig. 2). Note that W passes only once at the distinct vertices x 1, x k 1, v k 1,k, x k, v k,1, x 1, in this particular order. Therefore, the vertices v k 1,k and v k,1 are connected to {x k 1, x k } and {x k, x 1 }, respectively, using a Y-connection with a common middle vertex (see Fig. 3). We claim that v k 1,k could not have been assigned to {x k 1, x k }: Since ψ(x 1 ) < ψ(x j ) for every j = 2,..., k, ψ(x 1 ) + ψ(x k ) < ψ(x k 1 ) + ψ(x k ), and {x 1, x k } covers v k 1,k with smaller sum of ψ. v k 1,k v k,1 x k 1 x k x 1 Figure 3: Configuration of v k 1,k, v k,1, x k 1, x k and x 1 9

10 4.2 A 2-Approximation Algorithm Based on the existence of a partial map whose mapping graph is acyclic, in Lemma 10 we constructively show that for every remote dominating set S in T there is a dominating set of cardinality at most 2 S 1. Lemma 10. Let S be an RDS in T. Then there is a dominating set of cardinality at most 2 S 1. Proof. The claim clearly holds when S = 1, as S is a dominating set in this case, and we consider the case S 2. Let L V be the set of vertices that are not covered by a single vertex in S. Since S dominates the set of vertices V \ L, to complete the proof it remains to show that there is a set of at most S 1 vertices that dominates L. By Lemma 9, there is a partial map M L : L P (S) whose mapping graph G(M L ) is acyclic. Consider a pair {u, v} P (S) for which I ML (u, v) = 1, and let {w 1,..., w k } be the set of vertices in L that are mapped to {u, v}. There are two cases: 1. k = 1. In this case we pick w 1 for the dominating set. 2. k 2. It is straightforward to see that w 1,..., w k are connected to {u, v} using a Y-connection with a common middle vertex x. In this case we pick x for the dominating set. Since G(M L ) is acyclic, the number of pairs {u, v} for which I ML (u, v) = 1 is at most S 1, and we have identified a set of at most S 1 vertices that dominates L. A minimum cardinality dominating set D in T can be found in linear time [2], and in the special case we consider, this set is also an RDS. Lemma 10 proves, in particular, the existence of a dominating set whose cardinality is at most 2 OPT 1, where OPT is a minimum cardinality RDS in T. We have as a conclusion the following theorem. Theorem 11. D 2 OPT 1. In Lemma 12 we show that the bound given in Theorem 11 is tight, by providing an instance with D = 2 OPT 1. We also demonstrate that in general graphs a minimum cardinality dominating set does not guarantee a non-trivial approximation ratio. Lemma 12. There are instances in which D = 2 OPT 1. In addition, when the underlying graph is not restricted to be a tree, there are instances with OPT = O(1) and D = Ω(n). Proof. Consider the tree T = (V, E) in Fig. 4. For every i = 1,..., 2n+1, any dominating set contains either v i or u i, and therefore D 2n + 1. In addition, since {v 1,..., v 2n+1 } is a dominating set, D = 2n + 1. However, OPT = {v 1, v 3,..., v 2n+1 }, and we have D = 2 OPT 1. u 1 u 2 u 3 u 4 u 2n u 2n+1 v 1 v 2 v 3 v 4 v 2n v 2n+1 Figure 4: The tree T = (V, E) Now suppose we modify the tree T by adding the vertices x and y, and connecting each of them to v 1,..., v 2n+1. The claim D = 2n + 1 still holds, but now OPT = {x, y}. This example shows that when the underlying graph is not restricted to be a tree, there are instances with OPT = O(1) and D = Ω(n). 10

11 5 Euclidean RDS In this section we present a polynomial time approximation scheme for the Euclidean RDS problem, following the general framework suggested by Hochbaum and Maass for covering and packing problems in Euclidean spaces [9]. The unifying idea behind their shifting strategy is to repeatedly apply a simple divide-and-conquer approach and select the best solution we find. To simplify the presentation, we denote by P = {p 1,..., p n } the set of points to be covered, and let D = r 2. We also assume that P is bounded in a rectangle I, where the length of the long edge of I is nd. Otherwise, we can partition P into sets for which this property is satisfied, and separately use the algorithm for each set. The Vertical Partitions. We divide I into pairwise disjoint vertical strips of width D. Given a shifting parameter l, the partition V 0 of I consists of strips of width ld. For every i = 1,..., l 1, let V i be the partition of I obtained by shifting V 0 to the right over distance id. For each partition V i we define a set of points OPT(V i ) as follows. For every strip J in the partition V i, let OPT(V i, J) be a minimum cardinality set of points in P that covers the points P J, where P J is the set of points in P located in the strip J. Then OPT(V i ) = J V i OPT(V i, J). Clearly, OPT(V i ) is an RDS. Lemma 13. Let OPT be a minimum cardinality Euclidean RDS, then ( min OPT(V i) ) OPT. i=0,...,l 1 l Proof. For each strip J V i we define three sets of points (see Fig. 5): 1. OPT J, points in OPT located in the strip J. 2. OPT J+, points in OPT located in a strip of width D to the right of J. 3. OPT J, points in OPT located in a strip of width D to the left of J. J J J+ D ld D Figure 5: The regions J, J+ and J The sets OPT J, OPT J+ and OPT J are pairwise disjoint, and their union covers P J, since points in OPT that are not within distance D of J cannot contribute to cover any point in J. Therefore, It is easy to verify that OPT(V i, J) OPT J + OPT J+ + OPT J. l 1 i=0 J V i OPT J+ = l 1 i=0 J V i OPT J = OPT, l 1 i=0 J V i OPT J = l OPT, 11

12 and we have min i=0,...,l 1 OPT(V i) 1 l 1 l 1 l l 1 OPT(V i ) i=0 l 1 i=0 l 1 i=0 J V i OPT(V i, J) J V i ( OPT J + OPT J+ + OPT J ) = 1 (l OPT + 2 OPT ) ( l = ) OPT. l The Horizontal Partitions. We are now concerned with the problem of finding a small set of points in P that covers P J, for a given strip J. We divide J into pairwise disjoint horizontal strips of height D. The partition H 0 of J consists of strips of height ld. For every i = 1,..., l 1, let H i be the partition of J obtained by shifting H 0 up over distance id. For each partition H i we define a set of points OPT(H i ) as follows. For every strip R in the partition H i, let OPT(H i, R) be a minimum cardinality set of points in P that covers the points P R, where P R is the set of points in P located in the strip R. Then OPT(H i ) = R H i OPT(H i, R). Clearly, OPT(H i ) is a set of points in P that covers P J. The next lemma follows from arguments similar to those in the proof of Lemma 13. Lemma 14. Let OPT J be a minimum cardinality set of points in P that covers P J, then ( min OPT(H i) ) OPT J. i=0,...,l 1 l Optimal Solution in an ld ld Square. Lemmas 13 and 14 show that in order to obtain a polynomial time approximation scheme for Euclidean RDS, it is sufficient to optimally solve the following problem: Given R, an ld ld square in I, find a minimum cardinality set of points in P that covers P R. The next lemma allows us to perform an exhaustive search for an optimal solution to this problem in time O(n O(l2) ). Lemma 15. There is a set of points S P, S = O(l 2 ), that covers P R. Proof. As a first step, we show how to find a set S 1 P R, S 1 = O(l 2 ), that satisfies the following property: For every point p P R there is a point p S 1 such that p p D. We construct S 1 by choosing in each step a point p P R for which there is no point p S 1 such that p p D. Clearly, this process eventually ends with a set S 1 that satisfies the required property, and it remains to show that S 1 = O(l 2 ). Let p i be the point chosen in step i, i = 1,..., k, and let B(c, r) be the disk whose center and radius are c and r, respectively. Then ( k ) ( k ( Volume B(p i, D) Volume B p i, D ) ) 2 i=1 = i=1 k Volume i=1 = πkd2 4, ( B ( p i, D )) 2 12

13 where the first equality follows from the observation that the disks B ( p i, D ) 2, i = 1,..., k, are pairwise disjoint since for every i j, p i p j > D. In addition, ( k ) Volume B(p i, D) Volume B(p, D) ((l + 2)D) 2, i=1 p P R and therefore S 1 = k 4(l+2)2 π = O(l 2 ). Let A be the set of points p P R for which there is a point p S 1 such that p p r 1, or a pair of points p p S 1 such that p p D and p p D. In order to complete S 1 to a set of points that covers P R, it is sufficient to find S 2 P R \ S 1 such that for every point p P R \ A there is a point p S 2 for which p p D. We find S 2 by applying the same scheme for finding S 1 : We choose in each step a point p P R \A for which there is no point p S 2 such that p p D. Clearly, S 2 satisfies the required property, and using a similar bounded volume argument we have S 2 = O(l 2 ). Theorem 16. There is a polynomial time approximation scheme for the Euclidean RDS problem. 6 Group Cut on a Path In this section we first discuss the hardness of approximating GCP, and prove that this problem is as hard to approximate as set cover. We also identify the exact point at which GCP becomes NP-hard. We then present a simple proof for the existence of a feasible map M with respect to the optimal solution for which max(m, OPT) 2. This result, combined with Lemma 1, enables us to show that the approximation ratio of the greedy SCP algorithm for this special case is 2H k. 6.1 Hardness Results By describing an approximation preserving reduction, we prove in Theorem 17 that GCP is as hard to approximate as set cover. A special case of this reduction also shows that GCP is as hard to approximate as vertex cover even when the cardinality of each group is at most 4. In addition, we prove in Lemma 18 that when the bound on cardinality is 3, the problem is polynomial time solvable. Theorem 17. A polynomial time approximation algorithm for the GCP problem with factor α would imply a polynomial time approximation algorithm for the set cover problem with the same factor. Proof. Given a set cover instance I, with a ground set U = {x 1,..., x n } and a collection S = {S 1,..., S m } of subsets of U, we construct an instance ρ(i) of GCP as follows (see Fig. 6). For each set S i S there is a corresponding edge e i = (u i, v i ) with unit cost, where the edges e 1,..., e m are vertex disjoint. We connect these edges to a path using zero cost edges (v i, u i+1 ), i = 1,..., m 1. For each element x j U we define a group G j = i:x j S i {u i, v i }. u 1 1 v 1 0 u 2 1 v u m 1 v m e 1 e 2 e m Figure 6: The resulting instance ρ(i) of GCP To complete the proof, it is sufficient to show that any solution for the set cover instance I can be translated in polynomial time to a solution for the GCP instance ρ(i) with identical cost, and vice versa. Suppose that S S is a collection of sets such that S S S = U and S = k. If we define the set of edges F = {e i : S i S } {edges of cost 0}, 13

14 then the cost of F is exactly k. We claim that F separates all groups. Consider some group G j. Since S covers U, there is a set S i S such that x j S i. By definition of F, it follows that e i F. In addition, since x j S i, {u i, v i } G j, and therefore F separates G j (see Fig. 7) edges in F u i G j e i v i G j Figure 7: F separates G j Conversely, suppose that F is a set of edges that separates G 1,..., G n with cost k. We assume without loss of generality that every zero cost edge belongs to F. If we define the collection of sets S = {S i : e i F }, then S = k. We show that S covers U. Consider some element x j U. Since F separates G j, we can choose a representative u G j that is separated from G j \ {u}. However, by construction of ρ(i), the vertices in G j appear in pairs, and there is a vertex v G j that is connected to u by an edge e i with unit cost. It follows that x j S i and by definition of S, S i S. Note that vertex cover is a special case of set cover in which each element belongs to exactly two sets. Therefore, the proof of Theorem 17 can be modified to show that GCP is as hard to approximate as vertex cover even when the cardinality of each group is at most 4. Lemma 18. GCP is polynomial time solvable when G i 3 for every i = 1,..., k. Proof. We assume that the left-to-right order of the vertices is v 1,..., v n, and denote by L i and R i the leftmost and rightmost vertices in G i, respectively. Consider the following linear program: minimize c e x e (LP ) subject to e E e [L i,r i ] x e 0 x e 1 i = 1,..., k e E A set of edges F E separates a group G i of cardinality at most 3 if and only if F [L i, R i ]. Therefore, an optimal integral solution to (LP ) is a minimum cost set of edges that separates G 1,..., G k. However, the rows of the coefficient matrix of (LP ) have the interval property, that is, each row contains a single block of consecutive 1 s. Such a matrix is totally unimodular, and any basic optimal solution to (LP ) is integral. 6.2 A Feasible Map with Small Max Let F E be any feasible solution, with F 2. In Lemma 19 we prove the existence of a feasible map M : {G 1,..., G k } P (F ) for which max(m, F ) 2. Lemma 19. There is a feasible map M : {G 1,..., G k } P (F ) with max(m, F ) 2. Proof. Let F = {e 1,..., e t }, where the edges in F are indexed by their left-to-right order on the path P. We first claim that each group G i is separated by one of the pairs {e j, e j+1 }, j = 1,..., t 1. When G i can be separated by a single edge in F, the claim clearly holds. Otherwise, there are three consecutive vertices in G i such that the middle vertex v is separated from the other two by a pair of edges in F. Consider the set of edges in F to the left of v, and let e j be the rightmost edge in this set. Then the pair {e j, e j+1 } separates v from the other two. 14

15 Using this observation we define a feasible map M in the obvious way: Each group is mapped to one of the pairs {e j, e j+1 } that separates it. It follows that max(m, F ) 2, since for every e j F we now have I M (i, j) = I M (e j 1, e j ) + I M (e j, e j+1 ) 2. i j Let OPT be a minimum cost set of edges that separates G 1,..., G k, and without loss of generality OPT 2. The next theorem follows from Lemmas 1 and 19. Theorem 20. The greedy SCP algorithm constructs a solution whose cost is at most 2H k c(opt). 7 Concluding Remarks There is a huge gap between the upper bound for approximating the cardinality SCP problem, that was established in Theorem 4, and the logarithmic lower bound that follows from the observation that SCP contains set cover as a special case. The first, and probably the most challenging, open problem is to obtain either an improved hardness result or an improved approximation algorithm. Another open problem in this context is to provide a non-trivial algorithm for the general problem. In addition, it would be interesting to study the seemingly simple special case, in which each pair of objects covers at most one element. We proved that this problem is at least as hard to approximate as set cover, but we do not know how to significantly improve the approximation guarantee. Moreover, we consider this case to demonstrate the main difficulty in approximating SCP, as it shows that the objective is to choose a dense set of objects that covers all elements. We suggest for future research the Partial SCP problem, a variant of SCP in which we are given an additional parameter k, and the objective is to cover at least k elements with minimum cost. This problem is closely related to the Dense k-subgraph problem, that required to find in a given graph G = (V, E) a subset of k vertices whose induced subgraph has maximum number of edges. This problem is NP-hard, and the currently best approximation guarantee in general graphs is O(n δ ), for some constant δ < 1 3, due to Feige, Kortsarz and Peleg [5]. The next theorem relates these problems, and shows that the approximation guarantee of dense k-subgraph can be improved by developing an o(n δ/2 )-approximation algorithm for partial SCP. We provide its proof in Appendix A. Theorem 21. A polynomial time α(k)-approximation algorithm for partial SCP would imply a 1 randomized polynomial time -approximation algorithm for dense k-subgraph, for any ɛ > 0. α 2 (k 2 )(1+ɛ) References [1] V. Chvátal. A greedy heuristic for the set covering problem. Mathematics of Operations Research, 4: , [2] E. J. Cockayne, S. E. Goodman, and S. T. Hedetniemi. A linear algorithm for the domination number of a tree. Information Processing Letters, 4:41 44, [3] E. Dahlhaus, D. S. Johnson, C. H. Papadimitriou, P. D. Seymour, and M. Yannakakis. The complexity of multiterminal cuts. SIAM Journal on Computing, 23: , [4] U. Feige. A threshold of ln n for approximating set cover. Journal of the ACM, 45: , [5] U. Feige, G. Kortsarz, and D. Peleg. The dense k-subgraph problem. Algorithmica, 29: ,

16 [6] N. Garg, V. V. Vazirani, and M. Yannakakis. Primal-dual approximation algorithms for integral flow and multicut in trees. Algorithmica, 18:3 20, [7] S. Guha and S. Khuller. Approximation algorithms for connected dominating sets. Algorithmica, 20: , [8] R. Hassin and A. Tamir. Improved complexity bounds for location problems on the real line. Operations Research Letters, 10: , [9] D. S. Hochbaum and W. Maass. Approximation schemes for covering and packing problems in image processing and VLSI. Journal of the ACM, 32: , [10] D. S. Johnson. Approximation algorithms for combinatorial problems. Journal of Computer and System Sciences, 9: , [11] F. T. Leighton and S. Rao. Multicommodity max-flow min-cut theorems and their use in designing approximation algorithms. Journal of the ACM, 46: , [12] L. Lovász. On the ratio of optimal integral and fractional covers. Discrete Mathematics, 13: , [13] P. Slavík. Improved performance of the greedy algorithm for partial cover. Information Processing Letters, 64: , A Proof of Theorem 21 We begin with introducing the Dense k-edges problem, whose input is identical to that of dense k-subgraph, but now the objective is to find a minimum cardinality set of vertices A V whose induced subgraph contains at least k edges. This problem is a special case of partial SCP: The elements to partially cover are the edges of G, of which we are required to cover at least k, and the set of covering objects is V. A pair of vertices u v V covers the edge (u, v), if it exists. We now prove that a polynomial time α(k)-approximation algorithm for dense k-edges can be used 1 to obtain a randomized polynomial time -approximation algorithm for dense k-subgraph, α 2 (k 2 )(1+ɛ) for any ɛ > 0. Let r be the maximum number of edges in a subgraph of G induced by any k vertices. We show how to find a random set of at most k vertices that induce a subgraph whose expected number of edges is at least, for any ɛ > 0. We can assume without loss of generality that r α 2 (k 2 )(1+ɛ) 1. r is known, since we can test every r = 0,..., min{ E, ( k 2) }, and return the best solution we find. 2. k 1+ 1 ɛ, since otherwise we can enumerate in polynomial time all O( V k ) subsets of k vertices and find an optimal solution. By definition of r, the minimum number of vertices that induce a subgraph with at least r edges is at most k. We apply the α(r)-approximation algorithm for the dense r-edges problem and obtain a set of vertices A, A α(r)k, that induce at least r edges. We then randomly choose a subset of k vertices from A. The expected number of edges induced by the chosen vertices is at least r k A k 1 A 1 r k α(r)k k 1 α(r)k 1 r α 2 (r)(1 + ɛ) r α 2 (k 2 )(1 + ɛ), where the first inequality holds since A α(r)k, and the second since k ɛ, by our second assumption. 16

17 B A Polynomial Time Algorithm for RDS on a Tree We assume that the tree T = (V, E) is rooted at an arbitrary vertex r V. For each v V, we denote by T v the subtree of T rooted at v, and by C(v) the set of children of v. Clearly, every vertex in T v at distance 2 of v must be covered by vertices in T v, and we define a collection of states for T v : 1. v is chosen for the cover. 2. v is not chosen for the cover, v is covered by a vertex in C(v), at least two vertices in C(v) are chosen for the cover. 3. v is not chosen for the cover, v is covered by a vertex in C(v), exactly one vertex in C(v) is chosen for the cover, all vertices in C(v) are covered. 4. v is not chosen for the cover, v is covered by a vertex in C(v), exactly one vertex in C(v) is chosen for the cover, vertices in C(v) might require to be covered by an additional vertex at distance v is not chosen for the cover, v is covered by two vertices in T v at distance 2, no vertex in C(v) is chosen for the cover, all vertices in C(v) are covered. 6. v is not chosen for the cover, v is covered by two vertices in T v at distance 2, no vertex in C(v) is chosen for the cover, vertices in C(v) might require to be covered by an additional vertex at distance v is not chosen for the cover, v is partially covered by a single vertex in T v at distance 2, no vertex in C(v) is chosen for the cover, all vertices in C(v) are covered. 8. v is not chosen for the cover, v is partially covered by a single vertex in T v at distance 2, no vertex in C(v) is chosen for the cover, vertices in C(v) might require to be covered by an additional vertex at distance v is not chosen for the cover, v is not covered by a vertex in C(v) and not partially covered by a single vertex in T v at distance 2, no vertex in C(v) is chosen for the cover, all vertices in C(v) are covered. 10. v is not chosen for the cover, v is not covered by a vertex in C(v) and not partially covered by a single vertex in T v at distance 2, no vertex in C(v) is chosen for the cover, vertices in C(v) might require to be covered by an additional vertex at distance 2. Let {1,..., 10} be the set of states, and let F (v, s) be the minimum cardinality of a solution for the subtree T v that satisfied the conditions of state s. A minimum cardinality set of vertices that covers all vertices of T corresponds to min s {1,...,6} F (r, s). We show how to compute in O( V 3 ) time the functions F (v, s) for every v V and s {1,..., 10} using dynamic programming. Base Step: When v is a leaf of T, F (v, 1) = 1 and F (v, s) = 0 for s 1. General Step: When v is an interval vertex of T, we assume that F (u, s) was already computed for every u C(v) and s {1,..., 10}, and find F (v, s) as follows. 17

18 F (v, 1) = 1 + min s {1,...,10} u C(v) F (u, s) F (v, 2) = min F (u, 1) + F (w, 1) + min F (z, s) {u,w} C(v) s {1,2,3,5,7,9} z C(v)\{u,w} F (v, 3) = min F (u, 1) + min F (w, s) u C(v) s {2,3,5,7} w C(v)\{u} F (v, 4) = min F (u, 1) + min F (w, s) u C(v) s {2,3,5,7,9} w C(v)\{u} F (v, 5) = min min F (u, 2) + min F (w, s) u C(v) s {2,3,5}, w C(v)\{u} min F (u, 3) + F (w, 3) + min F (z, s) {u,w} C(v) s {3,5} z C(v)\{u,w} F (v, 6) = min min F (u, 2) + min F (w, s) u C(v) s {2,3,5,7}, w C(v)\{u} min F (u, 3) + F (w, 3) + min F (z, s) {u,w} C(v) s {3,5,7} z C(v)\{u,w} F (v, 7) = min F (u, 3) + F (w, 5) u C(v) w C(v)\{u} F (v, 8) = min F (u, 3) + min F (w, s) u C(v) s {5,7} w C(v)\{u} F (v, 9) = F (u, 5) F (v, 10) = u C(v) min s {5,7} u C(v) F (u, s) 18