ISRN Discrete MathematicsVolume 2011 (2011), Article ID 540834, 20 pagesdoi:10.5402/2011/540834

Research Article

Weighted Maximum-Clique Transversal Sets of Graphs

Chuan-Min Lee

Department of Computer and Communication Engineering, Ming Chuan University, 5 De Ming Road, Guishan District, Taoyuan County 333, Taiwan

Received 16 November 2011; Accepted 26 December 2011

Academic Editors: E. Cheng, A. Pêcher, and E. Tomita

Copyright © 2011 Chuan-Min Lee. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

A maximum-clique transversal set of a graph is a subset of vertices intersecting all maximum cliques of . The maximum-clique transversal set problem is to find a maximum-clique transversal set of of minimum cardinality. Motivated by the placement of transmitters for cellular telephones, Chang, Kloks, and Lee introduced the concept of maximum-clique transversal sets on graphs in 2001. In this paper, we study the weighted version of the maximum-clique transversal set problem for split graphs, balanced graphs, strongly chordal graph, Helly circular-arc graphs, comparability graphs, distance-hereditary graphs, and graphs of bounded treewidth.

1. Introduction

All graphs considered in this paper are undirected, finite, and simple. Let be a graph with vertex set and edge set . Unless stated otherwise, it is understood that and . For a graph , we also use and to denote the vertex set and edge set of , respectively. A graph is a subgraph of if and . We use to denote a subgraph of   induced by a subset of , that is, is the subgraph with vertex set in which two vertices are adjacent whenever they are adjacent in . For any vertex , the neighborhood of in is and the closed neighborhood of in is . The degree of a vertex in , denoted by , is the number of edges incident with . If , then is an isolated vertex of . A clique is a subset of pairwise adjacent vertices of . A maximal clique is a clique that is not a proper subset of any other clique. We use to denote the collection of all maximal cliques of . A clique is maximum if there is no clique of of larger cardinality. The clique number of , denoted by , is the cardinality of a maximum clique of . We use to denote the collection of all maximum cliques of .

A maximum-clique transversal set of a graph is a subset of intersecting all maximum cliques of . The maximum-clique transversal number of , denoted by , is the minimum cardinality of a maximum-clique transversal set of . The maximum-clique transversal set problem is to find a maximum-clique transversal set of of minimum cardinality. A maximum-clique independent set of is a collection of pairwise disjoint maximum cliques of . The maximum-clique independence number, denoted by , is the maximum cardinality of a maximum-clique independent set of . The maximum-clique independent set problem is to find a maximum-clique independent set of of maximum cardinality.

Maximum-clique transversal sets were introduced by Chang et al. in 2001 [1]. One of the main objectives for their research on maximum-clique transversal sets is the placement of transmitter towers for cellular telephones. Chang et al. stated a cellular telephone tower placement problem as the maximum-clique transversal set problem. They considered the problem and presented fixed parameter and approximation results for planar graphs. They also investigated the problem for some other graph classes such as -trees, strongly chordal graphs, graphs with few s, comparability graphs, and distance-hereditary graphs. Recently, Lee [2] introduced some variations of the maximum-clique transversal set problem and presented complexity results for them on some well-known classes of graphs.

Maximum-clique transversal and maximum-clique independent sets are closely related to clique transversal and clique independent sets on graphs. A clique transversal set of a graph is a subset of intersecting all maximal cliques of and a clique independent set of is a collection of pairwise disjoint maximal cliques of . The clique transversal number of , denoted by , is the minimum cardinality of a clique transversal set of . The clique independence number of , denoted by , is the maximum cardinality of a clique independent set of . The clique transversal (resp., independent) set problem is to find a clique transversal (resp., independent) set of of minimum (resp., maximum) cardinality. The clique transversal and clique independent set problems have been widely studied in [1, 3–20].

In this paper, we study the weighted version of the maximum-clique transversal set problem. Let be a function assigning to each vertex of a weight such that all arithmetic operations on vertex weights can be performed in time . We call a vertex-weight function and call a weighted graph. We let for any subset of and let be the weight of . The weighted maximum-clique transversal set problem is to find a maximum-clique transversal set of a weighted graph such that is minimized.

We present polynomial-time algorithms (most of them with linear running time) for the weighted maximum-clique transversal set problem on split graphs, balanced graphs, strongly chordal graphs, Helly circular-arc graphs, comparability graphs, distance-hereditary graphs, and graphs of bounded treewidth.

2. Split Graphs

In this section, we consider the weighted maximum-clique transversal set problem on split graphs.

Definition 2.1. A split graph is a graph , where the vertices of can be partitioned into an independent set and a clique .

Throughout this section, we use to denote a split graph with a vertex-weight function . Without loss of generality, we may assume that has no isolated vertices and that the vertices of have been partitioned into an independent set and a maximum clique . We give Algorithm 1 to solve the weighted maximum-clique transversal set problem for a split graph .

Input: A split graph .

Output: A maximum-clique transversal set of of minimum weight.

  (1) ;

  (2) for each vertex   do

  (3)    ;

  (4) end for

  (5) for each vertex   do

  (6)     if     then

  (7)         ;

  (8)         for each   do

  (9)             ;

(10)         end for

(11)     end if

(12) end for

(13) Find a vertex such that ;

(14) Find a vertex such that ;

(15) Find a vertex such that ;

(16) if   then

(17)     ;

(18) else

(19)     ;

(20) end if

(21) Output the set ;

Algorithm 1: Finding a maximum-clique transversal set of a split graph of minimum weight.

Theorem 2.2. Algorithm 1 finds a maximum-clique transversal set of a split graph of minimum weight in time.

Proof. The theorem holds trivially if has only one maximum clique. We may assume that has more than one maximum clique. We show the correctness of Algorithm 1 as follows.Initially, and for each vertex . At each iteration of Steps (5)–(12), the algorithm removes from an element with , and is decreased by the weight if is adjacent to . At the end of the last iteration of Steps (5)–(12), the set consists of all vertices with , and for each vertex .For every maximum clique of other than , it has exactly one vertex in and vertices in . For a vertex with , is not a maximum clique of . Therefore, .Assume that is a maximum-clique transversal set of of minimum weight. Let and be two vertices in . For every maximum clique of other than , it has exactly one vertex in and vertices in . It is clear that contains at least one vertex in and thus contains at most two vertices in .Let be a vertex in such that . Let be a vertex in such that and let be a vertex in such that . We consider the following two cases.Case 1 (). It can be easily verified that the set is a maximum-clique transversal set of of minimum weight, and thus .Case 2 (). Let and let . If , then is adjacent to every vertex in . We have . Suppose that . For any vertex , the maximum clique does not contain the vertex . Since does not contain any vertex in , must be included in . In other words, the set is a subset of . Then, the set is a maximum-clique transversal set of and . Note that and . We have Since , the set is a maximum-clique transversal set of of minimum weight. Following the discussion above, the algorithm is correct.Clearly, Step (1) and Steps (13)–(21) of Algorithm 1 can be done in time. Steps (2)–(4) and Steps (5)–(12) can be done in time. Hence, the running time of the algorithm is time.

3. Balanced Graphs

In this section, we consider the weighted maximum-clique transversal set problem on balanced graphs.

Let be a graph. Suppose that , , and . A clique matrix (resp., maximum-clique matrix) of , is the (0,1)-matrix whose entry is 1 if (resp., ), and 0 otherwise. A (0,1)-matrix is balanced if it does not contain the vertex-edge incidence matrix of an odd cycle as a submatrix, or equivalently, if it does not contain a square submatrix of odd order with exactly two ones per row and column. A (0, 1)-matrix is totally balanced if it does not contain the vertex-edge incidence matrix of a cycle as a submatrix [21]. A graph is balanced (resp., totally balanced) if a clique matrix of is balanced (resp., totally balanced).

Balanced graphs have been considered in [2, 8, 12, 15, 22]. It can be easily verified that if a clique matrix of a graph is balanced (resp., totally balanced), then all clique matrices of are balanced (resp., totally balanced). Note that a maximum-clique matrix of a graph is a submatrix of some clique matrix of . By definition, a submatrix of a balanced matrix is also balanced. We have the following lemma.

Lemma 3.1. If a graph is balanced, then any maximum-clique matrix of is balanced.

Let 1 (resp., 0) be a vector with 1’s (resp., 0’s) and let be a column vector. Let be a weighted graph with . Suppose that is a maximum-clique matrix of . The weighted maximum-clique transversal set problem for can be formulated as the following integer linear programming problem:

Fulkerson et al. proved the following important property of balanced matrices.

Theorem 3.2 (Fulkerson et al. [22]). If is a balanced matrix, then the polyhedra and have only integer extreme points.

Theorem 3.3. For any weighted balanced graph , the weighted maximum-clique transversal set problem can be solved in polynomial time.

Proof. Balanced graphs are a subclass of hereditary clique-Helly graphs [8]. Prisner [23] showed that no connected hereditary clique-Helly graphs has more maximal cliques than edges. Then, a connected balanced graph has maximal cliques. Since all maximal cliques of a hereditary clique-Helly graph can be enumerated in polynomial time by the algorithms in [24], all the maximum cliques can be extracted in polynomial time. Therefore, a maximum-clique matrix of can be computed in polynomial time.Note that if the extreme points of the polyhedra defined by the linear relaxation of an integer linear programming problem are integers, then the optimal solution of the integer linear programming problem is equal to the optimal solution of its linear relaxation. It is wellknown that linear programming problems can be solved in polynomial time. Following Lemma 3.1 and Theorem 3.2, the weighted maximum-clique transversal set problem is polynomial-time solvable for balanced graphs.

4. Strongly Chordal Graphs

In this section, we consider the weighted maximum-clique transversal set problem for strongly chordal graphs.

Let be a graph. A vertex is simplicial if all vertices of form a clique. The ordering of the vertices of is a perfect elimination ordering of if for all , is a simplicial vertex of the subgraph of induced by . Let denote the closed neighborhood of in . Rose [25] showed the characterization that a graph is chordal if and only if it has a perfect elimination ordering. A perfect elimination ordering is called a strong elimination ordering if the following condition is satisfied.

For if and belong to in , then .

Farber [26] showed that a graph is strongly chordal if and only if it admits a strong elimination ordering. So far, the fastest algorithm to recognize a strongly chordal graph and give a strong elimination ordering takes [27] or time [28].

Definition 4.1. Let be a graph and . Let . The -incidence graph of , denoted by , is defined as follows. The vertex set of is where . In , (1) is an independent set, (2) two vertices of are adjacent if they are adjacent in , and (3) for , is adjacent to if in .

Definition 4.2. A dominating set of a graph is a subset of such that for every vertex . The domination number of , denoted by , is the minimum cardinality of a dominating set of . The weighted dominating set problem is to find a dominating set of a weighted graph such that is minimized.

Lemma 4.3. Let be a weighted graph. Let be a vertex-weight function of defined by if and if . A maximum-clique transversal set of of minimum weight is equivalent to a dominating set of of minimum weight.

Proof. Let be the minimum weight of a maximum-clique transversal set of and let be the minimum weight of a dominating set of .Suppose that is a maximum-clique transversal set of of minimum weight. Then every maximum clique of has at least one vertex in . By the construction of , a vertex is adjacent to the vertices of a maximum clique of . Therefore, is a dominating set of . We have .Conversely, we let be a dominating set of of minimum weight. Clearly, does not contain any vertex in . For any vertex , the set has a vertex in . By the construction of , is a maximum clique of . Therefore, is a maximum-clique transversal of . We have .Following the discussion above, and thus the lemma holds.

Lemma 4.4. Let be a strongly chordal graph. Then is a strongly chordal graph, and a strong elimination ordering of can be obtained from a strong elimination ordering of in time.

Proof. A strongly chordal graph is chordal. It has at most maximal cliques, and all of its maximal cliques can be enumerated in time [29]. Then, all maximum cliques of a strongly chordal graph can be enumerated in time.Let and . The vertex-clique incidence graph of , denoted by , is defined as follows. The vertex set of is . In , (1) is an independent set, (2) two vertices of are adjacent if they are adjacent in , and (3) for , is adjacent to if in . Let be a maximum subset of such that is a maximum clique of for each vertex . Note that and . Therefore, is isomorphic to the subgraph of induced by .Guruswami and Rangan [17] showed that is a strongly chordal graph with edges and vertices, and that a strong elimination ordering of can be obtained from a given one for in time. Therefore, is also a strongly chordal graph and can be constructed from in time.Suppose that and . Then, . Let be a strong elimination ordering of . Let be the ordering of the vertices in obtained by removing all the vertices in from the ordering . It can be easily verified that the ordering is also a strong elimination ordering of . Following the discussion above, the lemma holds.

Theorem 4.5. The weighted maximum-clique transversal set problem can be solved in time for a strongly chordal graph if a strong elimination ordering is given.

Proof. It follows from Lemmas 4.3 and 4.4 and the result that the weighted dominating set problem can be solved in time for a strongly chordal graph if a strong elimination ordering is given [30].

5. Helly Circular-Arc Graphs

In the section, we consider the weighted maximum-clique transversal set problem for Helly circular-arc graphs.

Definition 5.1. Let be a set. Let be a set of some positive integers. A collection of subsets of is said to satisfy the Helly property if and for all implies .

Let be a collection of nonempty sets. The intersection graph of is obtained by representing each set in as a vertex and connecting two vertices with an edge if and only if their corresponding sets intersect. A circular-arc model is a pair , where is a circle and is a collection of arcs of . If a graph is the intersection graph of , then is a circular-arc graph. If satisfies the Helly property, then is a Helly circular-arc graph and is called a Helly circular-arc model of . For an arc , let be the vertex of corresponding to . For , let . Let be a point of and let be the collection of arcs that contain . If is a maximal clique of , then is called a clique point. Suppose that and are distinct points of . If , then and are equivalent. A clique point representation of is a maximum set of nonequivalent clique points of .

Let be a circular-arc model of a Helly circular-arc graph . Let be a maximal clique of and let be a clique point representation of . Due to Helly property, the arcs corresponding to vertices in have a point on the circle in common. It is clear that a point of is a clique point if and only if is a maximal clique of . Then, we have .

If and are points of , we use to denote an arc of starting at and ending at in clockwise direction. For each arc , the points are called the extremes of . We also use and to denote the starting point and ending point of the arc , respectively. Without loss of generality, we assume that (1) all arcs of are open arcs, (2) no single arc entirely covers , and (3) no two extremes of distinct arcs of coincide.

Let and for . An intersection segment is the contiguous part of formed by two consecutive extremes and , where is the starting point of some arc and is the ending point of an arc in clockwise direction. A point inside an intersection segment is called an intersection point. There are at most intersection segments and every clique point is an intersection point [13].

Let be a weighted Helly circular-arc graph with a Helly circular-arc model . Lin et al. [31] proposed a linear-time algorithm that finds a clique point representation of . Based upon their algorithm, we have the following theorem.

Theorem 5.2. Let be a Helly circular-arc model of a Helly circular-arc graph . Let be a clique point representation of . Then, the clique number and can be computed in linear time.

Guruswami and Rangan [17] showed that a Helly circular-arc graph has at most maximal cliques. The following lemma can be easily verified according to the Helly property.

Lemma 5.3. Suppose that is a Helly circular-arc graph with a Helly circular-arc model and . Let be a vertex of . Then, there are at most maximal cliques of containing the vertex .

Lemma 5.4. Suppose that is a Helly circular-arc graph with a Helly circular-arc model . (1)The -incidence graph has edges and vertices.(2)The -incidence graph is a Helly circular-arc graph and a Helly circular-arc model of can be obtained from in time.

Proof. (1) By Definition 4.1, . Since is a Helly circular-arc graph, it has at most maximal cliques [17]. Then has vertices. Without loss of generality, we assume that the clique number . By Lemma 5.3 and the construction of , every vertex is adjacent to at most vertices in . Hence, the number of is .(2) Let be a clique point representation of . The clique point representation can be constructed in linear time [31]. Following Theorem 5.2, the clique number and the arc sets can be computed in linear time. Then .Let be a maximum subset of such that is a maximum clique of for . The set and the arc sets can be computed in linear time. Then .Let and let be a set of arcs of such that each arc of contains exactly one clique point of and contains no extremes of arcs of . It follows that is a Helly circular-arc graph and is a Helly circular-arc graph model for . Hence, is a Helly circular-arc graph, and a Helly circular-arc model of can be obtained from in time.

Theorem 5.5. The weighted maximum-clique transversal set problem can be solved in time for a Helly circular-arc graph if a Helly circular-arc model is given.

Proof. It follows from Lemmas 4.3 and 5.4, and the result that the weighted dominating set problem can be solved in time for a circular-arc graph if a circular-arc model is given [32].

6. Comparability Graphs

A directed graph (or just digraph) consists of a nonempty finite set of vertices and a finite set of ordered pairs of distinct vertices called arrows. We call the vertex set and the arrow set of . We also use and to denote the vertex set and arrow set of , respectively. For an arrow , the first vertex is its tail and the second vertex is its head. We also say that the arrow leaves and enters . For a vertex of , we use the following notations: The sets , and are called the outneighborhood, inneighborhood, and neighborhood of , respectively. We call the vertices in , and the outneighbors, inneighbors, and neighbors of , respectively.

A directed walk in a digraph from vertex to vertex , or simply a directed -walk is a sequence of vertices such that , and is an arrow in for , where is called the length of this walk. A directed path is a directed walk in which no vertex is repeated. A directed -path is directed path starting at and ending at . A directed Hamiltonian path is a directed path that visits each vertex of exactly once. A directed cycle is a directed -walk in which no vertex is repeated except . Arrow set is a transitive relation on if for all , the following holds:     

Let be an undirected graph. Then the directed graph is an orientation of if for all , either or and for all , holds. If is a transitive relation on , then is a transitive orientation of . If there are no directed cycles in , then is an acyclic orientation of . Assume that is a directed graph. An undirected graph is the underlying graph of if for all , and for all , either or .

An undirected graph is a comparability graph if and only if it has a transitive orientation. Figure 1 shows a comparability graph and its transitive orientation.

(a)

(b)

Figure 1: (a) A comparability graph . (b) A transitive orientation of .

Given a comparability graph , a transitive orientation of can be found in linear time [33]. Chang et al. [1] solved the maximum-clique transversal set problem in time for comparability graphs. In this section, we show how to use a transitive orientation of a comparability graph to solve the weighted maximum-clique transversal set problem on in time.

Definition 6.1. A tournament is an orientation of a complete graph.

Theorem 6.2 (Rédei [34]). Every tournament contains a directed Hamiltonian path.

Lemma 6.3. There exists a one-to-one correspondence between the set of maximum cliques of a comparability graph and the set of longest directed paths of a transitive orientation of .

Proof. Let be a comparability graph and let be a transitive orientation of . By the transitive relation on , each directed path in corresponds to a clique of . Let be a longest directed path of . If is not a maximum clique of , then there exists a vertex such that is a clique of . With the help of Theorem 6.2, we know that there is a directed path of length . However, it contradicts that the length of a longest directed path is . Hence, each longest directed path of corresponds to a maximum clique of .Conversely, let be a maximum clique of . With the help of Theorem 6.2, we know that there exists a directed path of length . Assume that is not a longest directed path in . Then there exists a path of length greater than . By the transitive relation on , there is a clique whose number of vertices is larger than . However, it contradicts that is a maximum clique of . Hence, each maximum clique of corresponds to a longest directed path in . Following the discussion above, the lemma holds.

Suppose that is a comparability graph and is a transitive orientation of . By the transitive relation on , has no directed cycle. It is known that an acyclic digraph has a topological sort of , that is, a linear ordering of all vertices in such that if contains an arrow , then precedes in the ordering [35]. Clearly, there exists at least one vertex such that no vertex enters it. We call such vertices the source vertices. Similarly, there exists at least one vertex such that no vertex leaves it. Such vertices are called the sink vertices. We add a new vertex to and add arrows from to every source vertex in . Correspondingly we add another new vertex and arrows from every sink vertex in to . Let be the resulting digraph. The digraph is called an -transitive orientation of .

Lemma 6.4. Every longest directed path in a transitive orientation of a comparability graph starts at a source vertex and ends at a sink vertex.

Proof. It can be easily verified according to the transitive relation on and the definition of a longest directed path.

Lemma 6.5. Let be a comparability graph and let be a transitive orientation of . Suppose that is the -transitive orientation of . Let be an arbitrary directed -path in . Directed path has the longest length if and only if the path is a longest directed path in .

Proof. It can be easily verified by Lemma 6.4.

Definition 6.6. Let be an undirected graph. A vertex set is a vertex separator of if is disconnected. A set is an -vertex separator of if and are in distinct connected components of .

In [1], Chang et al. remove some vertices (except for and ) and arrows from an -transitive orientation of a comparability graph to obtain a new digraph such that (1) every directed -path in is a longest directed path in and (2) every longest directed path in is a directed -path in . The construction of can be done in linear time. We call the -longest-path digraph of . By Lemmas 6.3 and 6.5, a subset of is a maximum clique if and only if , , and all vertices in can form a directed -path in . Therefore, we have the following theorem.

Theorem 6.7. Let be a comparability graph with a vertex-weight function and let be an -longest-path digraph of . Suppose that is the underlying graph of . An -vertex separator of of minimum weight is equivalent to a maximum-clique transversal set of of minimum weight.

Definition 6.8. Let be a flow network with a capacity function . Let be the source vertex of the network, and let be the sink vertex. Let be a subset of vertices such that and . Let . We use to denote the set of arrows which leave from a vertex of and enter a vertex of . The is called a cut of . Let be the capacity of the cut determined by . A minimum cut is a cut of such that is minimized.

Theorem 6.9 (Max-flow min-cut theorem [36, 37]). In every network, the maximum total value of a flow equals the minimum capacity of a cut.

Suppose that is a comparability graph with a vertex-weight function and is an -longest-path digraph of . We construct a flow network from as follows:(1);(2), where , , and , and ;(3)For each arrow , let the weight be its capacity. For each edge in , we assign the capacity to it.

Let be a minimum cut of . By the max-flow min-cut theorem, does not contain any arrow in . The set is a subset of . Suppose that . Let . It can be easily verified that is an -vertex separator of of minimum weight. Following Theorem 6.7, we know that the set is a maximal-clique transversal set of of minimum weight. Note that a minimum cut of a flow network can be computed in time [38]. Therefore, we have the following result.

Theorem 6.10. Given a comparability graph with a vertex-weight function , the weighted maximum-clique transversal set problem can be solved in time.

7. Graphs of Bounded Treewidth

In this section we show that the weighted maximum-clique transversal set problem can be solved in linear time for graphs of bounded treewidth.

A clique with vertices is called a -clique. A -clique is a -tree. A -tree with vertices can be obtained from a -tree with vertices by making a new vertex adjacent to exactly all vertices of a -clique. For a -tree , if is a -clique and otherwise. For convenience, we define -trees as having at least vertices. With this definition, the treewidth of a -tree is and the clique number of a -tree is . Then, a -clique is a -tree and the treewidth of a -clique is .

Subgraphs of -trees are called partial -trees. If a partial -tree is a subgraph of a -tree , then we call a -tree embedding for . The smallest such that a graph is a partial -tree is called the treewidth of . It is clear that a graph of treewidth is also a partial -tree for every . The class of partial -trees is exactly the class of graphs of treewidth at most .

The treewidth of a graph can be defined by the concept of tree decompositions of a graph (see, e.g., [39]).

Definition 7.1. A tree decomposition of a graph is a pair , where is a tree with nodes and is a collection of subsets of such that a node in corresponds to the subset for and the following three conditions are satisfied.(1)Every vertex appears in at least one subset .(2)For every edge , there is at least one subset containing both endpoints of .(3)If a vertex appears in two subsets , then it appears in every subset for on the (unique) path from node to node in .

Definition 7.2. The width of a tree decomposition of a graph is the maximum cardinality minus one over all subsets of . The treewidth of a graph is the minimum width over all tree decompositions of .

Lemma 7.3 (Bodlaender [40]). If the treewidth of a graph is at most , then .

By Lemma 7.3, for a partial -tree with bounded . It was shown in [40] that for each constant it can be determined in linear time whether a graph has treewidth at most .

A tree decomposition is rooted if the tree is equipped with some root node. A rooted tree decomposition is called nice if the following conditions are satisfied.(1)Every node of has at most two children.(2)If a node has two children and then .(3)If a node has only one child then either and or and .

By [39], it is fairly easy to see that every graph with treewidth has a nice tree decomposition of width and that it can be obtained in linear time from an ordinary tree decomposition with the same width. Furthermore any graph on vertices has a nice tree decomposition with at most nodes. In the following, we assume that a nice tree decomposition of of width is part of the input.

Lemma 7.4 (Bodlaender and Möhring [41]). Let be a tree decomposition for a graph . For every clique of , there exists a subset such that is a subset of .

Lemma 7.5. Given a nice tree decomposition of a graph of bounded treewidth , the maximum cliques of can be enumerated in time.

Proof. Note that a nice tree decomposition of a graph of bounded treewidth has at most nodes. In other words, contains at most subsets of . Let . For , . By Lemma 7.4, a maximum clique is contained in a subset . For each , all maximal cliques of can be enumerated in time by the algorithm in [42]. Therefore, all maximum cliques of can be extracted in time.

Lemma 7.6. Suppose that is a graph of bounded treewidth with a nice tree decomposition . (1)The -incidence graph is a partial -tree.(2)A tree decomposition of the -incidence graph can be constructed in linear time.

Proof. Clearly, and is also a partial -tree. By Definition 4.1, , and each vertex in is adjacent to all vertices of exactly one maximum clique of . It can be easily verified that there exists a -tree such that and is a subgraph of . Therefore, is a partial -tree. It was shown in [40] that for each constant there is a linear-time algorithm for finding a treewidth decomposition for a graph of bounded treewidth . Hence, a tree decomposition of can be constructed in linear time. Following the discussion above, the lemma holds.

Theorem 7.7. The weighted maximum-clique transversal set problem can be solved in time for a graph of bounded treewidth with a tree decomposition .

Proof. Note that the weighted dominating set problem can be solved in time for a weighted graph of bounded treewidth if a tree decomposition of is given [43]. By Lemmas 4.3, 7.5, and 7.6, the theorem holds.

8. Distance-Hereditary Graphs

This section deals with the weighted maximum-clique transversal set problem on distance-hereditary graphs. We show that the problem is linear-time solvable for distance-hereditary graphs.

A graph is distance-hereditary if any two distinct vertices have the same distance in every connected induced subgraph containing them. Chang et al. [44] showed that distance-hereditary graphs can be defined, recursively.

Theorem 8.1 (Chang et al. [44]). Distance-hereditary graphs can be defined recursively as follows. (1)A graph consisting of only one vertex is distance-hereditary, and the twin set is the vertex itself.(2)If and are disjoint distance-hereditary graphs with the twin sets and , respectively, then the graph is a distance-hereditary graph and the twin set of is . is said to be obtained from and by a false twin operation.(3)If and are disjoint distance-hereditary graphs with the twin sets and , respectively, then the graph obtained by connecting every vertex of to all vertices of is a distance-hereditary graph, and the twin set of is . is said to be obtained from and by a true twin operation.(4)If and are disjoint distance-hereditary graphs with the twin sets and , respectively, then the graph obtained by connecting every vertex of to all vertices of is a distance-hereditary graph, and the twin set of is . is said to be obtained from and by a pendant vertex operation.

By Theorem 8.1, a distance-hereditary graph has its own twin set , the twin set is a subset of vertices of , and it is defined recursively. The construction of from disjoint distance-hereditary graphs and as described in Theorem 8.1 involves only the twin sets of and .

Following Theorem 8.1, we can construct a binary ordered decomposition tree. In this decomposition tree, each leaf is a single vertex graph, and each internal node represents one of the three operations: pendant vertex operation (labeled by P), true twin operation (labeled by T), and false twin operation (labeled by F). This binary ordered decomposition tree is called a PTF-tree. It has tree nodes and can be obtained in linear time [44].

Definition 8.2. Let be a distance-hereditary graph and let be the twin set of . We use to denote the collection of maximum cliques of . Therefore is the collection of all maximum cliques of . Let . The set denotes the collection of all maximum cliques of and all maximum cliques of .

Definition 8.3. Suppose that is a distance-hereditary graph obtained from two disjoint distance-hereditary graphs and by a true twin operation or a pendant vertex operation. We use to denote the set and .

Lemma 8.4. Let be a distance-hereditary graph obtained from two disjoint distance-hereditary graphs and by a true twin operation or a pendant vertex operation. Then .

Proof. This lemma can be easily proved by contradiction.

Lemma 8.5. Suppose is a graph obtained from two disjoint distance-hereditary graphs and by a true twin operation or a pendant vertex operation. Let be a subset of such that intersects with all maximum cliques of . Then either is a maximum-clique transversal set of or is a maximum-clique transversal set of .

Proof. Assume for contrary that neither is a maximum-clique transversal set of nor is a maximum-clique transversal set of . There exist maximum cliques and of and , respectively, such that does not contain any vertex in them. By Lemma 8.4, is a maximum clique of . Then, does not contain any vertex in , which contradicts the assumption that intersects all maximum cliques of . Following the discussion above, the lemma holds.

Definition 8.6. Let be a weighted graph. Suppose that , where is a subset of for . Let be an element in such that the weight of the element is minimum.

In the paper [1], Chang et al. presented a linear-time algorithm for the maximum-clique transversal set problem. Based upon their algorithm, we develop a linear-time algorithm to solve the weighted maximum-clique transversal set problem by the dynamic programming technique.

For a distance-hereditary graph , we use (resp., ) to denote the clique number of (resp., ) and use (resp., ) to denote a maximum-clique transversal set of (resp., ) of minimum weight. A strong maximum-clique transversal set of a distance-hereditary graph is a subset of such that intersects all cliques in . We use to denote a strong maximum-clique transversal set of of minimum weight.

Lemma 8.7. Suppose that is a distance-hereditary graph of only one vertex . Then, , and .

Proof. It follows from the definitions.

Lemma 8.8. Suppose that is formed from two disjoint distance-hereditary graphs and by a “false twin” operation. (1)If and , then , , , , and .(2)If and , then , , , , and .(3)If and , then , , , , and .(4)If and , then , , , , and .(5)If and , then , , , , and .(6)If and , then , , , , and .(7)If and , then , , , , and .(8)If and , then , , , , and .(9)If and , then , , , , and .

Proof. In the following, we just show the correctness for Statement (4) since other statements can be verified in similar ways. Note that and . Since and , , and . Therefore, and . We have and .By definition, and thus . Then . Hence, . Following the discussion above, the statement is true.

Lemma 8.9. Suppose that is formed from two disjoint distance-hereditary graphs and by a “pendant vertex” operation. (1)If , then , , , , and .(2)If , then , , , , and .(3)If , then , , , , and .(4)If , then , , , , and .(5)If and , then , , , , and .(6)If and , then , , , , and .(7)If and , then , , , , and .

Proof. In the following, we just show the correctness for Statement (3) since other statements can be verified in similar ways.By Theorem 8.1, and is obtained from and by connecting every vertex of to all vertices of . Therefore, and .We now consider and . In this case, . Then, and thus . Clearly, is a subset of and is a maximum-clique transversal set of . By Lemma 8.4, we know that . Therefore, intersects all maximum cliques of . By Lemma 8.5, either is a maximum-clique transversal set of or is a maximum-clique transversal set of . If is a maximum-clique transversal set of , then we have . If is a maximum-clique transversal set of , then is not only a maximum-clique transversal set of but also a strong maximum-clique transversal set of . We have . Hence, .We now consider . By definition, intersects all maximum cliques in and , respectively. Recall that and . Therefore, is a maximum-clique transversal set of and it also intersects all maximum cliques in . Besides, is a maximum-clique transversal set of . Hence, .Following the discussion above, the statement is true.

Lemma 8.10. Suppose that a distance-hereditary graph is formed from two disjoint distance-hereditary graphs and by a “true twin” operation. Let . (1)If , then , , , , and .(2)If , then , , , , and .(3)If , then , , , , and .(4)If , then , , , , and .(5)If and , then , , , , and .(6)If and , then , , , , and .(7)If and , then , , , , and .

Proof. In the following, we just show that the correctness for Statement (3) since other statements can be verified in similar ways.By Theorem 8.1, , and is obtained from and by connecting every vertex in to all vertices in . By Lemma 8.4, we have . By Lemma 8.5, either is a maximum-clique transversal set of or is a maximum-clique transversal set of . Hence, we have and .We now consider and . In this case, . Then, and thus . Clearly, is a subset of and is a maximum-clique transversal set of . Note that . By Lemma 8.5, either is a maximum-clique transversal set of or is a maximum-clique transversal set of . If is a maximum-clique transversal set of , then we have . If is a maximum-clique transversal set of , then is not only a maximum-clique transversal set of but also a strong maximum-clique transversal set of . We have . Hence, .We now consider . By definition, intersects all maximum cliques in and , respectively. Recall that and . Clearly is a maximum-clique transversal set of . By Lemma 8.5, either is a maximum-clique transversal set of or is a maximum-clique transversal set of . If is a maximum-clique transversal set of , then we have . If is a maximum-clique transversal set of , then is not only a strong maximum-clique transversal set of but also a strong maximum-clique transversal set of . We have . Hence, . Following the discussion above, the statement is true.

We can develop a dynamic programming algorithm to solve the weighted maximum-clique transversal set problem in linear time for distance-hereditary graphs as follows.

Without loss of generality, we assume that is a connected distance-hereditary graph with a vertex-weight function . Given a PTF-tree of rooted at node , our algorithm starts from the leaves of and works upward to the root. For each node of , let be the subtree of rooted at node , and let represent the subgraph of induced by the leaves of . A node represents either a single-vertex subgraph of or a subgraph of obtained by applying one of pendant vertex, true twin, or false twin operations to children of node . We use and to denote the left and right children of node , respectively, if node is a nonleaf node of . Therefore and is formed from and . For each node , our algorithm computes , , , , and , in time by Lemmas 8.7–8.10. Then is a maximum-clique transversal set of of minimum weight. Note that has tree nodes and can be constructed in linear time. Hence, we obtain the following result.

Theorem 8.11. If is a weighted distance-hereditary graph, then maximum-clique transversal set of of minimum weight can be computed in linear time.

9. Conclusions

In this paper, we have presented polynomial-time algorithms (most of them with linear running time) for the weighted maximum-clique transversal set problem on split graphs, balanced graphs, strongly chordal graphs, Helly circular-arc graphs, comparability graphs, distance-hereditary graphs, and graphs of bounded treewidth. For further study, it is a great challenge to work on the complexity of this problem for other classes of graphs.

Acknowledgments

The author would like to thank the anonymous referees for their insightful comments which led to considerable improvements in the presentation of the paper. This research was partially supported by National Science Council in Taiwan, under the grant number NSC-97-2218-E-130-002-MY2.

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