Dominating Set on Bipartite Graphs

Transcription

1 Dominating Set on Bipartite Graphs Mathieu Liedloff Abstract Finding a dominating set of minimum cardinality is an NP-hard graph problem, even when the graph is bipartite. In this paper we are interested in solving the problem on graphs having a large independent set. Given a graph G with an independent set of size z, we show that the problem can be solved in time O (2 n z ), where n is the number of vertices of G. As a consequence, our algorithm is able to solve the dominating set problem on bipartite graphs in time O (2 n/2 ). Another implication is an algorithm for general graphs whose running time is O( n ). 1 Introduction In the last years, we experience a growing interest in the design and analysis of exact exponential time algorithms to solve NP-hard problems. Several surveys have been published recently (see e.g. [3, 14, 15]). Known results. Given a graph G = (V, E), a subset D V is a dominating set of G if every vertex of V either belongs to D or has at least one neighbor in D. The minimum dominating set problem requires to find a dominating set of minimum cardinality of a given graph. This problem is well known to be NP-hard and several exact exponential time algorithms for solving it on general graphs, as well as on some graph classes, have been published during the last years. The first non trivial algorithms for the problem was proposed in In [6], Fomin et al. provided an O( n ) time algorithm for general graphs and algorithms for split graphs, bipartite graphs and graphs of maximum degree three. The running time of these algorithms are O( n ), O( n ) and O( n ) respectively. Independently and using different approaches, Laboratoire d Informatique Théorique et Appliquée, Université Paul Verlaine - Metz, Metz Cedex 01, France. liedloff (at) univ-metz.fr 1

2 Randerath and Schiermeyer published an O( n ) time algorithm [10], and Grandoni gave an O( n ) time algorithm [8, 9], both for general graphs. By refining the analysis of this latter algorithm, Fomin et al. [2] established an O( n ) time algorithm using polynomial space and an O( n ) time algorithm using exponential space. By now, the best known algorithm is due to van Rooij and Bodlaender [13]. They obtained an O( n ) time and polynomial space algorithm, and an O( n ) time algorithm that uses exponential space. We are insterested in exact algorithms to find a minimum dominating set on special graph classes. As already mentioned, Fomin et al. provide in [6] algorithms for split graphs, bipartite graphs and graphs of maximum degree three. In [5], Fomin and Høie give an O( n ) time algorithm for graphs of maximum degree three. In [1], Dorn shows that a minimum dominating set in a planar graph can be found in sub-exponential time giving an O( n ) time algorithm. In [7], Gaspers et al. provide algorithms for chordal graphs, circle graphs and dense graphs Concerning bipartite graphs, the only previously known algorithm is the one published in [6] having running time O( n ). Since by now the general algorithm of Fomin et al. [2] has worst case running time O( n ), it has been asked whether there is an algorithm finding a minimum dominating set on bipartite graphs in time O(α n ), with α < 1.5. Our results. In this paper, we provide an algorithm computing a minimum dominating set on graphs having a large independent set. Given a graph G with an independent set of size z, the worst case running time of our algorithm is O (2 n z ). Moreover, the algorithm leads to two interesting results. First, we show that a minimum dominating set of a bipartite graph can be computed in time O( n ), providing the best known algorithm for this graph class. The second consequence is a new algorithm for arbitrary graphs having running time O( n ). Besides the O( n ) time algorithm of Fomin et al. using the Branch & Reduce paradigm, our Dynamic Programming based algorithm is the best known for general graphs. 2 Preliminaries Let G = (V, E) be an undirected and simple graph. The number of its vertices is denoted by n. For a vertex v V, we denote by N(v) the neighborhood of v and by N[v] = N(v) {v} the closed neighborhood of v. Given a subset of vertices S V, N[S] denotes the set v S N[S] and N(S) denotes the set N[S] \ S. A subset D V such that every vertex of V \D has at least one neighbor 2

3 in D is a dominating set of G. The minimum cardinality of a dominating set of a graph G is denoted by γ(g). A subset of vertices S V is an independent set of a graph G = (V, E) if the vertices of S are pairwise non adjacent. The problem asking to find an independent set of maximum size is well known to be NP-hard. An independent set S of a graph G is maximal if S is not properly contained in an other independent set. It is well-known that each maximal independent set of a graph G is also a dominating set of G. A graph G = (V, E) is bipartite if the set of its vertices can be partitioned into two independent sets. Through out this paper, we use the (by now standard) O notation introduced by Woeginger [14]. We write f(n) = O (g(n)) if f(n) p(n) g(n) for some polynomial p(n). 3 Solving the dominating set problem on graphs with a large independent set Let G be a graph with an independent set of size z with 1 z n. In this section we show that a minimum dominating set of a such graph can be found in time O (2 n z ). Note that although the maximum independent set problem is NP-hard, there exists several exponential time algorithms for solving this problem exactly [12, 11, 4]. E.g. the one by Fomin et al. has running time O( n ) [4]. Theorem 1 Given a graph G and an independent set of size z of G, a minimum dominating set of G can be computed in time O (2 n z ). Proof. Let G = (V, E) be a graph and I = {v 1, v 2,..., v z } V an independent set of size z. We denote by J the set V \ I. To compute a dominating set of minimum size, the presented algorithm uses two stages. The first one, called preprocessing, computes for each subset X J a subset D X I of minimum cardinality such that X N[D X ]. The second stage, called Domination, computes for each set D J J a dominating set D of smallest size for G fulfilling the condition D J = D J. The calculations done during the preprocessing stage will be used to speed up the execution of the domination stage. In the next, we describe this two stages, prove their correctness and study their running times. Stage : Preprocessing For each subset X J and each integer k, 1 k z, we denote by Opt[X, k] a set D X,k of smallest size, if any, such that : 3

4 (i) D X,k {v 1, v 2,..., v k }, and (ii) X N[D X,k ]. The computation of Opt[X, k] is done by using a dynamic programming approach. The algorithm starts by initializing Opt so that for all k, 1 k z, we let and for all subsets X J we let : Opt[, k] = (1) { {v1 } if X N[v Opt[X, 1] = 1 ], { } otherwise. (2) The special set { } denotes a sentinel meaning that a subset fulfilling the requested properties does not exist. Moreover, we let the cardinality of { } be. The computation of Opt[X, k] is done by considering the sets X by increasing order of cardinality and the values of k by increasing order. Finally, for every non empty subset X J and every integer k 2, the set Opt[X, k] is obtained from the following formula: Opt[X, k] = { The set of minimum cardinality chosen amongst Opt[X, k 1] and {v k } Opt[X \ N[v k ], k 1]. } (3) It is not hard to show that this stage computes for each subset X J and each integer k, 1 k z, a set D X,k = Opt[X, k] of smallest size such that (i) D X,k {v 1, v 2,..., v k } and (ii) X N[D X,k ]. Consider the initializations (1) and (2). If the set X is empty then a set D X,k = is clearly a set of minimum cardinality fulfilling (i) and (ii) for every k. If k is equal to 1, for every subset X, the only way to have X N[D X,k ] is to take v 1 in D X,k. However in that case, either X N[v 1 ] and D X,k = {v 1 } is a solution of minimum size fulfilling (i) and (ii), or X N[v 1 ] and thus no set D X,k {v 1 } can fulfill the condition (ii). As previously explained, in the latter case we store the sentinel { }. We consider now the formula (3). When Opt[X, k] is computed, all the sets Opt[X, k ] with k < k and X J were previously computed. There are two possibilities: either v k belongs to Opt[X, k], or v k Opt[X, k]. If v k Opt[X, k] then it sufficient to know an optimal solution for the subproblem X = X \N[v k ] and k = k 1, which is given by Opt[X, k ]. If v k Opt[X, k] then we have Opt[X, k] = Opt[X, k 1]. Finally, we keep the set having the smallest cardinality amongst these two possibilities. 4

5 Given a set X and an integer k, the computation of Opt[X, k] require no more than polynomial time. Since the computation of Opt[X, k] is done for every X J and every integer k, 1 k z, the running time of the preprocessing stage is O (z2 J ) = O (2 n z ). We note that the storage of all the sets Opt[X, k] request space bounded by O (2 n z ). Stage: Domination Given a set D J J, in this stage we compute a set D = D J D I of minimum size such that D is a dominating set of G respecting the condition D J = D J. As we will show in the next, the set D I I is given by D I = DI 1 D2 I where D1 I = I \ N(D J) and DI 2 = Opt[J \ (N[D J] N(DI 1 )), z] (see Fig. 1). J I D J D I 2 D I 1 Figure 1: An example of a graph having an independent set I together with sets D J, DI 1 and D2 I. (Dashed lines represent neighborhoods of the sets.) Claim 2 Amongst all dominating sets S of G such that S J = D J, the set D = D J DI 1 D2 I is one of smallest cardinality. Proof. First, we show that D = D J DI 1 D2 I is a dominating set. Let v be a vertex in J. If v N[D J ] then v has at least one neighbor in D J or belongs to D J. If v N[D J ], we distinguish two cases. (1) If v N(DI 1) then v has at least one neighbor in DI 1. (2) If v N(D1 I ) then v belongs to J \ (N[D J ] N(DI 1)) and thus, by definition of Opt[J \ (N[D J] N(DI 1 )), z], v has at least one neighbor in DI 2. Consider now v being a vertex of I. Then either v N(D J ) and thus v has at least one neighbor in D J, or v (I \ N(D J )) and then v is in DI 1. Now we show that D is of minimum cardinality amongst all dominating sets S fulfilling S J = D J. Since D J = D J and I is an independent set of G, the vertices of I \ N(D J ) cannot have any neighbor in the dominating set and thus should belong to the dominating set. Consequently, for any dominating set S we have DI 1 S. Then, since D J = D J, the vertices of J \ (N[D J ] N(DI 1 )) can only be dominated by some vertices of I. A 5

6 subset DI 2 I of minimum size such that J \ (N[D J] N(DI 1)) N[D2 I ] was computed during the preprocessing stage. Namely, DI 2 is given by Opt[J \ (N[D J ] N(DI 1)), z]. Note that by the minimality of D2 I, it follows that D1 I and DI 2 are disjoint. Consequently, D = D J DI 1 D2 I is a dominating set of minimum size fulfulling D J = D J. Note that given a set D J J, and using the results coming from the preprocessing stage, the computation of D = D J DI 1 D2 I can be done in polynomial time. Finally, by enumerating all the 2 J = 2 n z possible subsets for D J = D J, the algorithm finds a minimum dominating set D of the graph G in time O (2 n z ). Since every bipartite graph has an independent set of size at least n/2, as an immediate consequence of Theorem 1, we obtain the following corollary. Corollary 3 Given a bipartite graph G, a minimum dominating set of G can be computed in time O (2 n/2 ) = O( n ). 4 Dominating set on general graphs The algorithm described in the previous section can be used to solve the minimum dominating set problem on general graphs. By using a simple greedy strategy, we start by computing in polynomial time a maximal independent set I of the given graph G. (E.g. choose a vertex v of G, add it to I, remove N[v] from G, and continue this process until G is empty.) Let z = n. If I z then we use the algorithm of Theorem 1 to compute a minimum dominating set in time O ( n ) = O( n ). If I < z then γ(g) < z since I is also a (not necessary minimum) dominating set of G. In that case, we enumerate all subsets of vertices of G of size at most z and keep the one being a dominating set of minimum size. By using the Stirling approximation and standard calculations, we state that the time needed for a such enumeration is bounded by z ( ) ( ) n n n! ( ( n ) z ( n ) n z ) z = z k z (z)!(n z)! = O z = O( n ). z n z k=0 Since given a set D, checking whether it is a dominating set require a polynomial time, the overall running time is O( n ) which prove the following Theorem. Theorem 4 Given a graph G, a minimum dominating set of G can be computed in time O( n ). 6

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8 [13] van Rooij, J. M. M., H. L. Bodlaender, Design by measure and conquer: a faster exact algorithm for dominating set, Proceedings of STACS 2008, to appear. [14] Woeginger, G. J., Exact algorithms for NP-hard problems: A survey, Combinatorial Optimization - Eureka, You Shrink!, Springer-Verlag, 2003, Berlin, LNCS 2570, pp [15] Woeginger, G. J., Space and time complexity of exact algorithms: Some open problems, Proceedings of IWPEC 2004, Springer-Verlag, 2004, Berlin, LNCS 3162, pp