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ISRN Discrete Mathematics
Volume 2011 (2011), Article ID 540834, 20 pages
http://dx.doi.org/10.5402/2011/540834
Research Article

Weighted Maximum-Clique Transversal Sets of Graphs

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 𝐺 is a subset of vertices intersecting all maximum cliques of 𝐺. The maximum-clique transversal set problem is to find a maximum-clique transversal set of 𝐺 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 𝐺=(𝑉,𝐸) be a graph with vertex set 𝑉 and edge set 𝐸. Unless stated otherwise, it is understood that |𝑉|=𝑛 and |𝐸|=𝑚. For a graph 𝐺, we also use 𝑉(𝐺) and 𝐸(𝐺) to denote the vertex set and edge set of 𝐺, respectively. A graph 𝐺=(𝑉,𝐸) is a subgraph of 𝐺 if 𝑉𝑉 and 𝐸𝐸. We use 𝐺[𝑊] to denote a subgraph of 𝐺  induced by a subset 𝑊 of 𝑉, that is, 𝐺[𝑊] is the subgraph with vertex set 𝑊 in which two vertices are adjacent whenever they are adjacent in 𝐺. For any vertex 𝑣𝑉, the neighborhood of 𝑣 in 𝐺 is 𝑁𝐺(𝑣)={𝑢𝑉(𝑢,𝑣)𝐸} and the closed neighborhood of 𝑣 in 𝐺 is 𝑁𝐺[𝑣]=𝑁𝐺(𝑣){𝑣}. The degree of a vertex 𝑣 in 𝐺, denoted by deg𝐺(𝑣), is the number of edges incident with 𝑣. If deg𝐺(𝑣)=0, then 𝑣 is an isolated vertex of 𝐺. A clique is a subset of pairwise adjacent vertices of 𝑉. A maximal clique is a clique that is not a proper subset of any other clique. We use 𝐶(𝐺) to denote the collection of all maximal cliques of 𝐺. A clique is maximum if there is no clique of 𝐺 of larger cardinality. The clique number of 𝐺, denoted by 𝜔(𝐺), is the cardinality of a maximum clique of 𝐺. We use 𝑄(𝐺) to denote the collection of all maximum cliques of 𝐺.

A maximum-clique transversal set of a graph 𝐺=(𝑉,𝐸) is a subset of 𝑉 intersecting all maximum cliques of 𝐺. The maximum-clique transversal number of 𝐺, denoted by 𝜏𝑀(𝐺), is the minimum cardinality of a maximum-clique transversal set of 𝐺. The maximum-clique transversal set problem is to find a maximum-clique transversal set of 𝐺 of minimum cardinality. A maximum-clique independent set of 𝐺 is a collection of pairwise disjoint maximum cliques of 𝐺. The maximum-clique independence number, denoted by 𝛼𝑀(𝐺), is the maximum cardinality of a maximum-clique independent set of 𝐺. The maximum-clique independent set problem is to find a maximum-clique independent set of 𝐺 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 𝑘-trees, strongly chordal graphs, graphs with few 𝑃4s, 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 𝐺=(𝑉,𝐸) is a subset of 𝑉 intersecting all maximal cliques of 𝐺 and a clique independent set of 𝐺 is a collection of pairwise disjoint maximal cliques of 𝐺. The clique transversal number of 𝐺, denoted by 𝜏𝐶(𝐺), is the minimum cardinality of a clique transversal set of 𝐺. The clique independence number of 𝐺, denoted by 𝛼𝐶(𝐺), is the maximum cardinality of a clique independent set of 𝐺. The clique transversal (resp., independent) set problem is to find a clique transversal (resp., independent) set of 𝐺 of minimum (resp., maximum) cardinality. The clique transversal and clique independent set problems have been widely studied in [1, 320].

In this paper, we study the weighted version of the maximum-clique transversal set problem. Let 𝑤𝑉 be a function assigning to each vertex 𝑣 of 𝐺=(𝑉,𝐸) a weight 𝑤(𝑣) such that all arithmetic operations on vertex weights can be performed in time 𝑂(1). We call 𝑤 a vertex-weight function and call 𝐺=(𝑉,𝐸,𝑤) a weighted graph. We let 𝑤(𝑆)=𝑢𝑆𝑤(𝑢) for any subset 𝑆 of 𝑉 and let 𝑤(𝑆) be the weight of 𝑆. The weighted maximum-clique transversal set problem is to find a maximum-clique transversal set 𝑆 of a weighted graph 𝐺=(𝑉,𝐸,𝑤) such that 𝑤(𝑆) 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 𝐺=(𝐼𝑄,𝐸), where the vertices of 𝐺 can be partitioned into an independent set 𝐼 and a clique 𝑄.

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

alg1
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 𝐷 of a split graph 𝐺=(𝐼𝑄,𝐸,𝑤) of minimum weight in 𝑂(𝑛+𝑚) time.

Proof. The theorem holds trivially if 𝐺 has only one maximum clique. We may assume that 𝐺 has more than one maximum clique. We show the correctness of Algorithm 1 as follows.
Initially, 𝑆=𝐼 and 𝑤𝑁(𝑣)=𝑤(𝑁𝐺(𝑣)) for each vertex 𝑣𝑄. At each iteration of Steps (5)–(12), the algorithm removes from 𝑆 an element 𝑠 with deg𝐺(𝑠)<𝜔(𝐺)1, and 𝑤𝑁(𝑣) is decreased by the weight 𝑤(𝑠) if 𝑣 is adjacent to 𝑠. At the end of the last iteration of Steps (5)–(12), the set 𝑆 consists of all vertices 𝑠 with deg𝐺(𝑠)=𝜔(𝐺)1, and 𝑤𝑁(𝑣)=𝑤(𝑁𝐺(𝑣)𝑄)+𝑤(𝑁𝐺(𝑣)𝑆) for each vertex 𝑣𝑄.
For every maximum clique of 𝐺 other than 𝑄, it has exactly one vertex in 𝐼 and 𝜔(𝐺)1 vertices in 𝑄. For a vertex 𝑥𝐼 with deg𝐺(𝑥)<𝜔(𝐺)1, 𝑁𝐺[𝑥] is not a maximum clique of 𝐺. Therefore, 𝑄(𝐺)={𝑄}𝑠𝑆{𝑁𝐺[𝑠]}.
Assume that 𝐷 is a maximum-clique transversal set of 𝐺 of minimum weight. Let 𝑥1 and 𝑥2 be two vertices in 𝑄. For every maximum clique 𝑄 of 𝐺 other than 𝑄, it has exactly one vertex in 𝑆 and 𝜔(𝐺)1 vertices in 𝑄. It is clear that 𝑄 contains at least one vertex in {𝑥1,𝑥2} and thus 𝐷 contains at most two vertices in 𝑄.
Let 𝑣1 be a vertex in 𝑄 such that 𝑤(𝑣1)=min{𝑤(𝑣)𝑣𝑄}. Let 𝑣2 be a vertex in 𝑄{𝑣1} such that 𝑤(𝑣2)=min{𝑤(𝑣)𝑣𝑄{𝑣1}} and let 𝑣3 be a vertex in 𝑄 such that 𝑤(𝑆𝑄)𝑤𝑁(𝑣3)=min{𝑤(𝑆𝑄)𝑤𝑁(𝑣)𝑣𝑄}. We consider the following two cases.Case 1 (|𝐷𝑄|=2). It can be easily verified that the set {𝑣1,𝑣2} is a maximum-clique transversal set of 𝐺 of minimum weight, and thus 𝑤(𝐷)=𝑤(𝑣1)+𝑤(𝑣2).Case 2 (|𝐷𝑄|=1). Let 𝐷𝑄={𝑣} and let 𝐷={𝑣}(𝑆𝑁𝐺(𝑣)). If 𝑆𝑁𝐺(𝑣)=, then 𝑣 is adjacent to every vertex in 𝑆. We have 𝑤(𝐷)=𝑤(𝐷)=𝑤(𝑣). Suppose that 𝑆𝑁𝐺(𝑣). For any vertex 𝑥𝑆𝑁𝐺(𝑣), the maximum clique 𝑁𝐺[𝑥] does not contain the vertex 𝑣. Since 𝐷 does not contain any vertex in 𝑄{𝑣}, 𝑥 must be included in 𝐷. In other words, the set 𝑆𝑁𝐺(𝑣) is a subset of 𝐷. Then, the set 𝐷={𝑣}(𝑆𝑁𝐺(𝑣)) is a maximum-clique transversal set of 𝐺 and 𝑤(𝐷)=𝑤(𝐷). Note that 𝑆𝑄= and 𝑤𝑁(𝑣)=𝑤(𝑁𝐺(𝑣)𝑄)+𝑤(𝑁𝐺(𝑣)𝑆). We have 𝑤(𝑆𝑄)𝑤𝑁𝑣𝑤𝑁=𝑤(𝑆)+𝑤(𝑄)𝐺𝑣𝑁𝑄+𝑤𝐺𝑣=𝑁𝑆𝑤(𝑆)𝑤𝐺𝑣+𝑁𝑆𝑤(𝑄)𝑤𝐺𝑣𝑄=𝑤𝑆𝑁𝐺𝑣𝑣+𝑤𝐷=𝑤.(2.1) Since 𝑤(𝑆𝑄)𝑤𝑁(𝑣3)=min{𝑤(𝑆𝑄)𝑤𝑁(𝑣)𝑣𝑄}, the set 𝐷={𝑣3}(𝑆𝑁𝐺(𝑣3)) is a maximum-clique transversal set of 𝐺 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 𝑂(𝑛) time. Steps (2)–(4) and Steps (5)–(12) can be done in 𝑂(𝑣𝑉(𝐺)(deg𝐺(𝑣)+1))=𝑂(𝑛+𝑚) time. Hence, the running time of the algorithm is 𝑂(𝑛+𝑚) time.

3. Balanced Graphs

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

Let 𝐺 be a graph. Suppose that 𝑉(𝐺)={𝑣1,𝑣2,,𝑣𝑛}, 𝐶(𝐺)={𝐶1,𝐶2,,𝐶𝑝}, and 𝑄(𝐺)={𝑄1,𝑄2,,𝑄}. A clique matrix (resp., maximum-clique matrix) of 𝐺, is the (0,1)-matrix whose entry (𝑖,𝑗) is 1 if 𝑣𝑗𝐶𝑖 (resp., 𝑣𝑗𝑄𝑖), 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 𝐺 is balanced (resp., totally balanced) if a clique matrix of 𝐺 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 𝐺 is balanced (resp., totally balanced), then all clique matrices of 𝐺 are balanced (resp., totally balanced). Note that a maximum-clique matrix of a graph 𝐺 is a submatrix of some clique matrix of 𝐺. By definition, a submatrix of a balanced matrix is also balanced. We have the following lemma.

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

Let 1 (resp., 0) be a vector with 𝑛 1’s (resp., 0’s) and let 𝑥=(𝑥1,𝑥2,,𝑥𝑛) be a column vector. Let 𝐺=(𝑉,𝐸,𝑤) be a weighted graph with 𝑉={𝑣1,𝑣2,,𝑣𝑛}. Suppose that 𝑀 is a maximum-clique matrix of 𝐺. The weighted maximum-clique transversal set problem for 𝐺 can be formulated as the following integer linear programming problem:Minimize𝑛𝑖=1𝑤𝑣𝑖𝑥𝑖𝑥subjectto𝑀𝑥1𝑖=1or0for𝑖=1,2,,n.(3.1)

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

Theorem 3.2 (Fulkerson et al. [22]). If 𝐴 is a balanced matrix, then the polyhedra 𝑃1(𝐴)={𝑥𝐴𝑥1,𝑥0} and 𝑃2(𝐴)={𝑥𝐴𝑥1,𝑥0} have only integer extreme points.

Theorem 3.3. For any weighted balanced graph 𝐺, 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 𝑂(𝑚) 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 𝑀 of 𝐺 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 𝐺=(𝑉,𝐸) be a graph. A vertex 𝑣 is simplicial if all vertices of 𝑁𝐺[𝑣] form a clique. The ordering (𝑣1,𝑣2,,𝑣𝑛) of the vertices of 𝑉 is a perfect elimination ordering of 𝐺 if for all 𝑖{1,,𝑛}, 𝑣𝑖 is a simplicial vertex of the subgraph 𝐺𝑖 of 𝐺 induced by {𝑣𝑖,𝑣𝑖+1,,𝑣𝑛}. Let 𝑁𝑖[𝑣] denote the closed neighborhood of 𝑣 in 𝐺𝑖. 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 𝑖𝑗𝑘 if 𝑣𝑗 and 𝑣𝑘 belong to 𝑁𝑖[𝑣𝑖] in 𝐺𝑖, then 𝑁𝑖[𝑣𝑗]𝑁𝑖[𝑣𝑘].

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 𝑂(𝑚log𝑛) [27] or 𝑂(𝑛2) time [28].

Definition 4.1. Let 𝐺=(𝑉,𝐸) be a graph and 𝑄(𝐺)={𝑄1,𝑄2,,𝑄}. Let 𝑋(𝐺)=𝑖=1𝑄𝑖. The 𝒱𝒬-incidence graph of 𝐺, denoted by (𝐺), is defined as follows. The vertex set of (𝐺) is 𝑋(𝐺)𝑆(𝐺) where 𝑆(𝐺)={𝑠1,𝑠2,,𝑠}. In (𝐺), (1) 𝑆(𝐺) is an independent set, (2) two vertices of 𝑋(𝐺) are adjacent if they are adjacent in 𝐺, and (3) for 1𝑖, 𝑠𝑖𝑆(𝐺) is adjacent to 𝑣𝑗𝑋(𝐺) if 𝑣𝑗𝑄𝑖 in 𝐺.

Definition 4.2. A dominating set 𝑆 of a graph 𝐺 is a subset of 𝑉(𝐺) such that |𝑆𝑁𝐺[𝑣]|1 for every vertex 𝑣𝑉(𝐺). The domination number of 𝐺, denoted by 𝛾(𝐺), is the minimum cardinality of a dominating set of 𝐺. The weighted dominating set problem is to find a dominating set 𝑆 of a weighted graph 𝐺=(𝑉,𝐸,𝑤) such that 𝑤(𝑆) is minimized.

Lemma 4.3. Let 𝐺=(𝑉,𝐸,𝑤) be a weighted graph. Let 𝑤 be a vertex-weight function of (𝐺) defined by 𝑤(𝑣)= if 𝑣𝑆(𝐺) and 𝑤(𝑣)=𝑤(𝑣) if 𝑣𝑋(𝐺). A maximum-clique transversal set of 𝐺 of minimum weight is equivalent to a dominating set of (𝐺) of minimum weight.

Proof. Let ̃𝜏𝑀 be the minimum weight of a maximum-clique transversal set of 𝐺 and let ̃𝛾 be the minimum weight of a dominating set of (𝐺).
Suppose that 𝐷 is a maximum-clique transversal set of 𝐺 of minimum weight. Then every maximum clique of 𝐺 has at least one vertex in 𝐷. By the construction of (𝐺), a vertex 𝑠𝑆(𝐺) is adjacent to the vertices of a maximum clique of 𝐺. Therefore, 𝐷 is a dominating set of (𝐺). We have ̃𝛾̃𝜏𝑀.
Conversely, we let 𝐷 be a dominating set of (𝐺) of minimum weight. Clearly, 𝐷 does not contain any vertex in 𝑆(𝐺). For any vertex 𝑠𝑆(𝐺), the set 𝐷 has a vertex in 𝑁(𝐺)(𝑠). By the construction of (𝐺), 𝑁(𝐺)(𝑠) is a maximum clique of 𝐺. Therefore, 𝐷 is a maximum-clique transversal of 𝐺. We have ̃𝜏𝑀̃𝛾.
Following the discussion above, ̃𝜏𝑀=̃𝛾 and thus the lemma holds.

Lemma 4.4. Let 𝐺 be a strongly chordal graph. Then (𝐺) is a strongly chordal graph, and a strong elimination ordering of (𝐺) can be obtained from a strong elimination ordering of 𝐺 in 𝑂(𝑛+𝑚) time.

Proof. A strongly chordal graph is chordal. It has at most 𝑛 maximal cliques, and all of its maximal cliques can be enumerated in 𝑂(𝑛+𝑚) time [29]. Then, all maximum cliques 𝑄1,𝑄2,,𝑄 of a strongly chordal graph can be enumerated in 𝑂(𝑛+𝑚) time.
Let 𝐶(𝐺)={𝐶1,𝐶2,,𝐶𝑝} and 𝑆(𝐺)={𝑠1,𝑠2,,𝑠𝑝}. The vertex-clique incidence graph of 𝐺, denoted by (𝐺), is defined as follows. The vertex set of (𝐺) is 𝑉(𝐺)𝑆(𝐺). In (𝐺), (1) 𝑆(𝐺) is an independent set, (2) two vertices of 𝑉 are adjacent if they are adjacent in 𝐺, and (3) for 1𝑖𝑝, 𝑠𝑖𝑆(𝐺) is adjacent to 𝑣𝑗𝑉(𝐺) if 𝑣𝑗𝐶𝑖 in 𝐺. Let 𝑆 be a maximum subset of 𝑆(𝐺) such that 𝑁(𝐺)(𝑠) is a maximum clique of 𝐺 for each vertex 𝑠𝑆. Note that 𝑄(𝐺)𝐶(𝐺) and 𝑋(𝐺)𝑉(𝐺). Therefore, (𝐺) is isomorphic to the subgraph of (𝐺) induced by 𝑋(𝐺)𝑆.
Guruswami and Rangan [17] showed that (𝐺) is a strongly chordal graph with 𝑂(𝑛+𝑚) edges and 𝑂(𝑛) vertices, and that a strong elimination ordering of (𝐺) can be obtained from a given one for 𝐺 in 𝑂(𝑛+𝑚) time. Therefore, (𝐺) is also a strongly chordal graph and can be constructed from (𝐺) in 𝑂(𝑛+𝑚) time.
Suppose that 𝑛1=|𝑉((𝐺))| and 𝑛2=|𝑉((𝐺))|. Then, 𝑛2𝑛1. Let (𝑤1,𝑤2,,𝑤𝑛1) be a strong elimination ordering of (𝐺). Let (𝑤𝑥1,𝑤𝑥2,,𝑤𝑥𝑛2) be the ordering of the vertices in (𝐺) obtained by removing all the vertices in 𝑉((𝐺))𝑉((𝐺)) from the ordering (𝑤1,𝑤2,,𝑤𝑛1). It can be easily verified that the ordering (𝑤𝑥1,𝑤𝑥2,,𝑤𝑥𝑛2) is also a strong elimination ordering of (𝐺). Following the discussion above, the lemma holds.

Theorem 4.5. The weighted maximum-clique transversal set problem can be solved in 𝑂(𝑛+𝑚) time for a strongly chordal graph 𝐺 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 𝑂(𝑛+𝑚) 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 𝒮 be a set. Let 𝒫 be a set of some positive integers. A collection {𝑆𝑖}𝑖𝒫 of subsets of 𝒮 is said to satisfy the Helly property if 𝐽𝒫 and 𝑆𝑖𝑆𝑗 for all 𝑖,𝑗𝐽 implies 𝑗𝐽𝑆𝑗.

Let be a collection of nonempty sets. The intersection graph 𝐺 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 (𝒞,𝒜), where 𝒞 is a circle and 𝒜 is a collection of arcs of 𝒞. If a graph 𝐺 is the intersection graph of 𝒜, then 𝐺 is a circular-arc graph. If 𝒜 satisfies the Helly property, then 𝐺 is a Helly circular-arc graph and is called a Helly circular-arc model of 𝐺. For an arc 𝐴𝒜, let 𝑣(𝐴) be the vertex of 𝐺 corresponding to 𝐴. For 𝒜𝐴, let 𝑉(𝒜)=𝐴𝒜{𝑣(𝐴)}. Let 𝑝 be a point of 𝒞 and let 𝒜(𝑝) be the collection of arcs that contain 𝑝. If 𝑉(𝒜(𝑝)) is a maximal clique of 𝐺, then 𝑝 is called a clique point. Suppose that 𝑝1 and 𝑝2 are distinct points of 𝒞. If 𝒜(𝑝1)=𝒜(𝑝2), then 𝑝1 and 𝑝2 are equivalent. A clique point representation of is a maximum set of nonequivalent clique points of 𝒞.

Let =(𝒞,𝒜) be a circular-arc model of a Helly circular-arc graph 𝐺. Let 𝑋 be a maximal clique of 𝐺 and let 𝒲={𝑐1,𝑐2,,𝑐𝑝} be a clique point representation of . Due to Helly property, the arcs corresponding to vertices in 𝑋 have a point on the circle 𝒞 in common. It is clear that a point 𝑏 of 𝒞 is a clique point if and only if 𝑉(𝒜(𝑏)) is a maximal clique of 𝐺. Then, we have 𝐶(𝐺)={𝑉(𝒜(𝑐1)),𝑉(𝒜(𝑐2)),,𝑉(𝒜(𝑐𝑝))}.

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

Let 𝒜={𝐴1,𝐴2,,𝐴𝑛} and 𝐴𝑖=(𝑠𝑖,𝑡𝑖) for 1𝑖𝑛. An intersection segment (𝑠𝑖,𝑡𝑗) is the contiguous part of 𝒞 formed by two consecutive extremes 𝑠𝑖 and 𝑡𝑗, where 𝑠𝑖 is the starting point of some arc 𝐴𝑖𝒜 and 𝑡𝑗 is the ending point of an arc 𝐴𝑗𝒜 in clockwise direction. A point inside an intersection segment is called an intersection point. There are at most 𝑛 intersection segments and every clique point is an intersection point [13].

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

Theorem 5.2. Let =(𝒞,𝒜) be a Helly circular-arc model of a Helly circular-arc graph 𝐺. Let 𝒲={𝑐1,𝑐2,,𝑐𝑝} be a clique point representation of . Then, the clique number 𝜔(𝐺) and 𝒜(𝑐1),𝒜(𝑐2),,𝒜(𝑐𝑝) can be computed in linear time.

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

Lemma 5.3. Suppose that 𝐺 is a Helly circular-arc graph with a Helly circular-arc model =(𝒞,𝒜) and 𝜔(𝐺)>1. Let 𝑣 be a vertex of 𝐺. Then, there are at most deg𝐺(𝑣) maximal cliques of 𝐺 containing the vertex 𝑣.

Lemma 5.4. Suppose that 𝐺=(𝑉,𝐸) is a Helly circular-arc graph with a Helly circular-arc model (𝒞,𝒜).(1)The 𝒱𝒬-incidence graph (𝐺) has 𝑂(𝑛+𝑚) edges and 𝑂(𝑛) vertices.(2)The 𝒱𝒬-incidence graph (𝐺) is a Helly circular-arc graph and a Helly circular-arc model (𝒞,𝒜) of (𝐺) can be obtained from (𝒞,𝒜) in 𝑂(𝑛+𝑚) time.

Proof. (1) By Definition 4.1, 𝑉((𝐺))=𝑋(𝐺)𝑆(𝐺). Since 𝐺 is a Helly circular-arc graph, it has at most 𝑛 maximal cliques [17]. Then (𝐺) has 𝑂(𝑛) vertices. Without loss of generality, we assume that the clique number 𝜔(𝐺)>1. By Lemma 5.3 and the construction of (𝐺), every vertex 𝑣𝑋(𝐺) is adjacent to at most deg𝐺(𝑣) vertices in 𝑆(𝐺). Hence, the number of 𝐸((𝐺)) is 𝑂(𝑚+𝑣𝑋(𝐺)(deg𝐺(𝑣)+1))=𝑂(𝑛+𝑚).
(2) Let 𝒲={𝑐1,𝑐2,,𝑐𝑝} be a clique point representation of . The clique point representation 𝒬 can be constructed in linear time [31]. Following Theorem 5.2, the clique number 𝜔(𝐺) and the arc sets 𝒜(𝑐1),𝒜(𝑐2),,𝒜(𝑐𝑝) can be computed in linear time. Then 𝐶(𝐺)={𝑉(𝒜(𝑐1)),𝑉(𝒜(𝑐2)),,𝑉(𝒜(𝑐𝑝))}.
Let 𝑃={𝑝1,𝑝2,,𝑝} be a maximum subset of 𝒬 such that 𝑉(𝒜(𝑝𝑖)) is a maximum clique of 𝐺 for 1𝑖. The set 𝑃 and the arc sets 𝒜(𝑝1),𝒜(𝑝2),,𝒜(𝑝) can be computed in linear time. Then 𝑄(𝐺)={𝑉(𝒜(𝑝1)),𝑉(𝒜(𝑝2)),,𝑉(𝒜(𝑝))}.
Let 𝒜𝑃=𝑖=1𝒜(𝑝𝑖) and let 𝐴𝑊 be a set of arcs of 𝒞 such that each arc of 𝒜𝑊 contains exactly one clique point of 𝑃 and contains no extremes of arcs of 𝒜𝑃. It follows that (𝐺) is a Helly circular-arc graph and (𝒞,𝒜𝑃𝒜𝑊) is a Helly circular-arc graph model for (𝐺). Hence, (𝐺) is a Helly circular-arc graph, and a Helly circular-arc model (𝒞,𝒜) of (𝐺) can be obtained from (𝒞,𝒜) in 𝑂(𝑛+𝑚) time.

Theorem 5.5. The weighted maximum-clique transversal set problem can be solved in 𝑂(𝑛+𝑚) time for a Helly circular-arc graph 𝐺=(𝑉,𝐸,𝑤) if a Helly circular-arc model =(𝒞,𝒜) 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 𝑂(𝑛+𝑚) time for a circular-arc graph if a circular-arc model is given [32].

6. Comparability Graphs

A directed graph (or just digraph) 𝐷=(𝑉,𝐴) consists of a nonempty finite set 𝑉 of vertices and a finite set 𝐴 of ordered pairs of distinct vertices called arrows. We call 𝑉 the vertex set and 𝐴 the arrow set of 𝐷. We also use 𝑉(𝐷) and 𝐴(𝐷) to denote the vertex set and arrow set of 𝐷, respectively. For an arrow 𝑢,𝑣, the first vertex 𝑢 is its tail and the second vertex 𝑣 is its head. We also say that the arrow 𝑢,𝑣 leaves 𝑢 and enters 𝑣. For a vertex 𝑣 of 𝐷, we use the following notations:𝑁+𝐷(𝑣)={𝑤𝑉{𝑣}𝑣,𝑤𝐴},𝑁𝐷(𝑣)={𝑢𝑉{v}𝑢,𝑣𝐴}.(6.1) The sets 𝑁+𝐷(𝑣),𝑁𝐷(𝑣), and 𝑁𝐷(𝑣)=𝑁+𝐷(𝑣)𝑁𝐷(𝑣) are called the outneighborhood, inneighborhood, and neighborhood of 𝑣, respectively. We call the vertices in 𝑁+𝐷(𝑣),𝑁𝐷(𝑣), and 𝑁𝐷(𝑣) the outneighbors, inneighbors, and neighbors of 𝑣, respectively.

A directed walk in a digraph 𝐷=(𝑉,𝐴) from vertex 𝑢 to vertex 𝑣, or simply a directed (𝑢,𝑣)-walk is a sequence of vertices (𝑣0,𝑣1,,𝑣𝑛) such that 𝑢=𝑣0,𝑣=𝑣𝑛, and 𝑣𝑖1,𝑣𝑖 is an arrow in 𝐷 for 1𝑖𝑛, where 𝑛 is called the length of this walk. A directed path is a directed walk in which no vertex is repeated. A directed (𝑠,𝑡)-path is directed path starting at 𝑠 and ending at 𝑡. A directed Hamiltonian path is a directed path that visits each vertex of 𝐷 exactly once. A directed cycle is a directed (𝑣,𝑣)-walk in which no vertex is repeated except 𝑣. Arrow set 𝐴 is a transitive relation on 𝑉 if for all 𝑢,𝑣,𝑤𝑉, the following holds:If𝑢,𝑣𝐴and𝑣,𝑤𝐴,then𝑢,𝑤𝐴.(6.2)

Let 𝐺=(𝑉,𝐸) be an undirected graph. Then the directed graph 𝐷=(𝑉,𝐴) is an orientation of 𝐺 if for all (𝑥,𝑦)𝐸, either 𝑥,𝑦𝐴 or 𝑦,𝑥𝐴 and for all 𝑥,𝑦𝐴, (𝑥,𝑦)𝐸 holds. If 𝐴 is a transitive relation on 𝑉, then 𝐷 is a transitive orientation of 𝐺. If there are no directed cycles in 𝐷, then 𝐷 is an acyclic orientation of 𝐺. Assume that 𝐷=(𝑉,𝐴) is a directed graph. An undirected graph 𝐺=(𝑉,𝐸) is the underlying graph of 𝐷 if for all 𝑥,𝑦𝐴, (𝑥,𝑦)𝐸 and for all (𝑥,𝑦)𝐸, either 𝑥,𝑦𝐴 or 𝑦,𝑥𝐴.

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

fig1
Figure 1: (a) A comparability graph 𝐺. (b) A transitive orientation of 𝐺.

Given a comparability graph 𝐺=(𝑉,𝐸), a transitive orientation of 𝐺 can be found in linear time [33]. Chang et al. [1] solved the maximum-clique transversal set problem in 𝑂(𝑚𝑛) time for comparability graphs. In this section, we show how to use a transitive orientation of a comparability graph 𝐺=(𝑉,𝐸) to solve the weighted maximum-clique transversal set problem on 𝐺 in 𝑂(𝑛𝑚log(𝑛2/𝑚)) 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 𝐺 and the set of longest directed paths of a transitive orientation 𝐷 of 𝐺.

Proof. Let 𝐺=(𝑉,𝐸) be a comparability graph and let 𝐷=(𝑉,𝐴) be a transitive orientation of 𝐺. By the transitive relation on 𝑉, each directed path in 𝐷 corresponds to a clique of 𝐺. Let (𝑣1,𝑣2,,𝑣𝑘) be a longest directed path of 𝐷. If 𝑆={𝑣1,𝑣2,,𝑣𝑘} is not a maximum clique of 𝐺, then there exists a vertex 𝑣𝑉𝑆 such that 𝑆{𝑣} is a clique of 𝐺. With the help of Theorem 6.2, we know that there is a directed path of length 𝑘+1. However, it contradicts that the length of a longest directed path is 𝑘. Hence, each longest directed path of 𝐷 corresponds to a maximum clique of 𝐺.
Conversely, let 𝑆={𝑣1,𝑣2,,𝑣𝑘} be a maximum clique of 𝐺. With the help of Theorem 6.2, we know that there exists a directed path 𝑃 of length 𝑘. Assume that 𝑃 is not a longest directed path in 𝐷. Then there exists a path of length greater than 𝑘. By the transitive relation on 𝑉, there is a clique whose number of vertices is larger than 𝑘. However, it contradicts that 𝑆 is a maximum clique of 𝐺. Hence, each maximum clique of 𝐺 corresponds to a longest directed path in 𝐷. Following the discussion above, the lemma holds.

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

Lemma 6.4. Every longest directed path in a transitive orientation 𝐷 of a comparability graph 𝐺=(𝑉,𝐸) starts at a source vertex and ends at a sink vertex.

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

Lemma 6.5. Let 𝐺=(V,𝐸) be a comparability graph and let 𝐷 be a transitive orientation of 𝐺. Suppose that 𝐷 is the (𝑠,𝑡)-transitive orientation of 𝐺. Let 𝑃=(𝑠=𝑣0,𝑣1,𝑣2,,𝑣𝑖1,𝑣𝑖=𝑡) be an arbitrary directed (𝑠,𝑡)-path in 𝐷. Directed path 𝑃 has the longest length if and only if the path (𝑣1,𝑣2,,𝑣𝑖1) is a longest directed path in 𝐷.

Proof. It can be easily verified by Lemma 6.4.

Definition 6.6. Let 𝐺=(𝑉,𝐸) be an undirected graph. A vertex set 𝑆𝑉 is a vertex separator of 𝐺 if 𝐺[𝑉𝑆] is disconnected. A set 𝑆𝑉 is an (𝑎,𝑏)-vertex separator of 𝐺 if 𝑎 and 𝑏 are in distinct connected components of 𝐺[𝑉𝑆].

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

Theorem 6.7. Let 𝐺 be a comparability graph with a vertex-weight function 𝑤 and let 𝐷 be an (𝑠,𝑡)-longest-path digraph of 𝐺. Suppose that 𝐺 is the underlying graph of 𝐷. An (𝑠,𝑡)-vertex separator 𝑆 of 𝐺 of minimum weight 𝑤(𝑆) is equivalent to a maximum-clique transversal set of 𝐺 of minimum weight.

Definition 6.8. Let 𝐷=(𝑉,𝐴) be a flow network with a capacity function 𝑐. Let 𝑠 be the source vertex of the network, and let 𝑡 be the sink vertex. Let 𝑆 be a subset of vertices such that 𝑠𝑆 and 𝑡𝑉𝑆. Let 𝑆=𝑉𝑆. We use (𝑆;𝑆) to denote the set of arrows which leave from a vertex of 𝑆 and enter a vertex of 𝑆. The (𝑆;𝑆) is called a cut of 𝐷. Let 𝑐(𝑆)=𝑒(𝑆;𝑆)𝑐(𝑒) be the capacity of the cut determined by 𝑆. A minimum cut 𝑆 is a cut of 𝐷 such that 𝑐(𝑆) 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 𝐺 is a comparability graph with a vertex-weight function 𝑤 and 𝐷 is an (𝑠,𝑡)-longest-path digraph of 𝐺. We construct a flow network 𝐷=(𝑉,𝐴) from 𝐷 as follows:(1)𝑉={𝑠,𝑡}{𝑣1,𝑣2𝑣𝑉(𝐷){𝑠,𝑡}};(2)𝐴=𝐴1𝐴2𝐴3𝐴4, where 𝐴1={𝑠,𝑣1𝑠,𝑣𝐴(𝐷)}, 𝐴2={𝑣1,𝑣2𝑣𝑉(𝐷){𝑠,𝑡}}, 𝐴3={𝑢2,𝑣1𝑢,𝑣𝐴(𝐷) and 𝑢,𝑣𝑉(𝐷){𝑠,𝑡}}, and 𝐴4={𝑣2,𝑡𝑣,𝑡𝐴(𝐷)};(3)For each arrow 𝑣1,𝑣2𝐴2, let the weight 𝑤(𝑣) be its capacity. For each edge in 𝐴1𝐴3𝐴4, we assign the capacity to it.

Let (𝑆;𝑆) be a minimum cut of 𝐷. By the max-flow min-cut theorem, (𝑆;𝑆) does not contain any arrow in 𝐴1𝐴3𝐴4. The set (𝑆;𝑆) is a subset of 𝐴2. Suppose that (𝑆;𝑆)={𝑥11,𝑥21,𝑥12,𝑥22,,𝑥1𝑖,𝑥2𝑖}. Let 𝑆={𝑥1,𝑥2,,𝑥𝑖}. It can be easily verified that 𝑆 is an (𝑠,𝑡)-vertex separator of 𝐺 of minimum weight. Following Theorem 6.7, we know that the set 𝑆 is a maximal-clique transversal set of 𝐺 of minimum weight. Note that a minimum cut of a flow network can be computed in 𝑂(𝑛𝑚log(𝑛2/𝑚)) time [38]. Therefore, we have the following result.

Theorem 6.10. Given a comparability graph 𝐺 with a vertex-weight function 𝑤, the weighted maximum-clique transversal set problem can be solved in 𝑂(𝑛𝑚log(𝑛2/𝑚)) time.

7. Graphs of Bounded Treewidth 𝑘

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 𝑘 vertices is called a 𝑘-clique. A (𝑘+1)-clique is a 𝑘-tree. A 𝑘-tree with 𝑛+1 vertices can be obtained from a 𝑘-tree with 𝑛 vertices by making a new vertex adjacent to exactly all vertices of a 𝑘-clique. For a 𝑘-tree 𝐺, 𝜔(𝐺)=𝑘 if 𝐺 is a 𝑘-clique and 𝜔(𝐺)=𝑘+1 otherwise. For convenience, we define 𝑘-trees as having at least 𝑘+1 vertices. With this definition, the treewidth of a 𝑘-tree is 𝑘 and the clique number of a 𝑘-tree is 𝑘+1. Then, a 𝑘-clique is a (𝑘1)-tree and the treewidth of a 𝑘-clique is 𝑘1.

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

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 𝐺=(𝑉,𝐸) is a pair (𝑇,𝒮), where 𝑇 is a tree with nodes and 𝒮 is a collection of subsets 𝑆1,𝑆2,,𝑆 of 𝑉 such that a node 𝑖 in 𝑇 corresponds to the subset 𝑆𝑖𝒮 for 1𝑖 and the following three conditions are satisfied.(1)Every vertex 𝑥𝑉 appears in at least one subset 𝑆𝑖𝒮.(2)For every edge 𝑒𝐸, there is at least one subset 𝑆𝑖𝒮 containing both endpoints of 𝑒.(3)If a vertex 𝑥 appears in two subsets 𝑆𝑖,𝑆𝑗𝒮, then it appears in every subset 𝑆𝑘 for 𝑘 on the (unique) path from node 𝑖 to node 𝑗 in 𝑇.

Definition 7.2. The width of a tree decomposition (𝑇,𝒮) of a graph 𝐺 is the maximum cardinality minus one over all subsets of 𝒮. The treewidth of a graph 𝐺 is the minimum width over all tree decompositions of 𝐺.

Lemma 7.3 (Bodlaender [40]). If the treewidth of a graph 𝐺=(𝑉,𝐸) is at most 𝑘, then |𝐸|𝑘|V|(1/2)𝑘(𝑘+1).

By Lemma 7.3, 𝑂(|𝑉|+|𝐸|)=𝑂(𝑘𝑛) for a partial 𝑘-tree 𝐺=(𝑉,𝐸) with bounded 𝑘. It was shown in [40] that for each constant 𝑘 it can be determined in linear time whether a graph 𝐺 has treewidth at most 𝑘.

A tree decomposition (𝑇,𝒮) is rooted if the tree 𝑇 is equipped with some root node. A rooted tree decomposition is called nice if the following conditions are satisfied.(1)Every node of 𝑇 has at most two children.(2)If a node 𝑖 has two children 𝑗 and 𝑘 then 𝑆𝑖=𝑆𝑗=𝑆𝑘.(3)If a node 𝑖 has only one child 𝑗 then either |𝑆𝑖|=|𝑆𝑗|+1 and 𝑆𝑗𝑆𝑖 or |𝑆𝑖|=|𝑆𝑗|1 and 𝑆𝑖𝑆𝑗.

By [39], it is fairly easy to see that every graph with treewidth 𝑘 has a nice tree decomposition of width 𝑘 and that it can be obtained in linear time from an ordinary tree decomposition with the same width. Furthermore any graph on 𝑛 vertices has a nice tree decomposition with at most 4𝑛 nodes. In the following, we assume that a nice tree decomposition (𝑇,𝒮) of 𝐺 of width 𝑘 is part of the input.

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

Lemma 7.5. Given a nice tree decomposition (𝑇,𝒮) of a graph 𝐺 of bounded treewidth 𝑘, the maximum cliques of 𝐺=(𝑉,𝐸) can be enumerated in 𝑂(3k/3𝑛) time.

Proof. Note that a nice tree decomposition of a graph of bounded treewidth 𝑘 has at most 4𝑛 nodes. In other words, 𝒮 contains at most 4𝑛 subsets of 𝑉. Let 𝒮={𝑆1,𝑆2,,𝑆}. For 1𝑖, |𝑆𝑖|𝑘+1. By Lemma 7.4, a maximum clique is contained in a subset 𝑆𝑖𝒮. For each 𝑆𝑖𝒮, all maximal cliques of 𝐺[𝑆𝑖] can be enumerated in 𝑂(3𝑘/3) time by the algorithm in [42]. Therefore, all maximum cliques of 𝐺 can be extracted in 𝑂(3𝑘/3𝑛) time.

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

Proof. Clearly, 𝜔(𝐺)𝑘+1 and 𝐺 is also a partial (𝑘+1)-tree. By Definition 4.1, 𝑉((𝐺))=𝑋(𝐺)𝑆(𝐺), and each vertex in 𝑆(𝐺) is adjacent to all vertices of exactly one maximum clique of 𝐺. It can be easily verified that there exists a (𝑘+1)-tree 𝐻 such that 𝑉(𝐻)=𝑉(𝐺)𝑆(𝐺) and (𝐺) is a subgraph of 𝐻. Therefore, (𝐺) is a partial (𝑘+1)-tree. It was shown in [40] that for each constant 𝑘 there is a linear-time algorithm for finding a treewidth decomposition for a graph of bounded treewidth 𝑘. Hence, a tree decomposition of (𝐺) 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 𝑂(4𝑘𝑛) time for a graph 𝐺 of bounded treewidth 𝑘 with a tree decomposition (𝑇,𝒮).

Proof. Note that the weighted dominating set problem can be solved in 𝑂(4𝑘𝑛) time for a weighted graph 𝐺=(𝑉,𝐸,𝑤) of bounded treewidth 𝑘 if a tree decomposition of 𝐺 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 𝐺1 and 𝐺2 are disjoint distance-hereditary graphs with the twin sets 𝑇𝑆(𝐺1) and 𝑇𝑆(𝐺2), respectively, then the graph 𝐺=𝐺1𝐺2 is a distance-hereditary graph and the twin set of 𝐺 is 𝑇𝑆(𝐺1)𝑇𝑆(𝐺2). 𝐺 is said to be obtained from 𝐺1 and 𝐺2 by a false twin operation.(3)If 𝐺1 and 𝐺2 are disjoint distance-hereditary graphs with the twin sets 𝑇𝑆(𝐺1) and 𝑇𝑆(𝐺2), respectively, then the graph 𝐺 obtained by connecting every vertex of 𝑇𝑆(𝐺1) to all vertices of 𝑇𝑆(𝐺2) is a distance-hereditary graph, and the twin set of 𝐺 is 𝑇𝑆(𝐺1)𝑇𝑆(𝐺2). 𝐺 is said to be obtained from 𝐺1 and 𝐺2 by a true twin operation.(4)If 𝐺1 and 𝐺2 are disjoint distance-hereditary graphs with the twin sets 𝑇𝑆(𝐺1) and 𝑇𝑆(𝐺2), respectively, then the graph 𝐺 obtained by connecting every vertex of 𝑇𝑆(𝐺1) to all vertices of 𝑇𝑆(𝐺2) is a distance-hereditary graph, and the twin set of 𝐺 is 𝑇𝑆(𝐺1). 𝐺 is said to be obtained from 𝐺1 and 𝐺2 by a pendant vertex operation.

By Theorem 8.1, a distance-hereditary graph 𝐺 has its own twin set 𝑇𝑆(𝐺), the twin set 𝑇𝑆(𝐺) is a subset of vertices of 𝐺, and it is defined recursively. The construction of 𝐺 from disjoint distance-hereditary graphs 𝐺1 and 𝐺2 as described in Theorem 8.1 involves only the twin sets of 𝐺1 and 𝐺2.

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 2𝑛1 tree nodes and can be obtained in linear time [44].

Definition 8.2. Let 𝐺=(𝑉,𝐸) be a distance-hereditary graph and let 𝑇𝑆(𝐺) be the twin set of 𝐺. We use 𝑄(𝐺) to denote the collection of maximum cliques of 𝐺. Therefore 𝑄(𝐺[𝑇𝑆(𝐺)]) is the collection of all maximum cliques of 𝐺[𝑇𝑆(𝐺)]. Let 𝑄𝐸(𝐺)=𝑄(𝐺)𝑄(𝐺[𝑇𝑆(𝐺)]). The set 𝑄𝐸(𝐺) denotes the collection of all maximum cliques of 𝐺 and all maximum cliques of 𝐺[𝑇𝑆(𝐺)].

Definition 8.3. Suppose that 𝐺 is a distance-hereditary graph obtained from two disjoint distance-hereditary graphs 𝐺1 and 𝐺2 by a true twin operation or a pendant vertex operation. We use 𝑄𝑋12(𝐺) to denote the set {𝑄1𝑄2𝑄1𝑄(𝐺1[𝑇𝑆(𝐺1)]) and 𝑄2𝑄(𝐺2[𝑇𝑆(𝐺2)])}.

Lemma 8.4. Let 𝐺 be a distance-hereditary graph obtained from two disjoint distance-hereditary graphs 𝐺1 and 𝐺2 by a true twin operation or a pendant vertex operation. Then 𝑄𝑋12(𝐺)=𝑄(𝐺[𝑇𝑆(𝐺1)𝑇𝑆(𝐺2)]).

Proof. This lemma can be easily proved by contradiction.

Lemma 8.5. Suppose 𝐺 is a graph obtained from two disjoint distance-hereditary graphs 𝐺1 and 𝐺2 by a true twin operation or a pendant vertex operation. Let 𝒮 be a subset of 𝑉(𝐺) such that 𝒮 intersects with all maximum cliques of 𝐺[𝑇𝑆(𝐺1)𝑇𝑆(𝐺2)]. Then either 𝒮𝑇𝑆(𝐺1) is a maximum-clique transversal set of 𝐺1[T𝑆(𝐺1)] or 𝒮𝑇𝑆(𝐺2) is a maximum-clique transversal set of 𝐺2[𝑇𝑆(𝐺2)].

Proof. Assume for contrary that neither 𝒮𝑇𝑆(𝐺1) is a maximum-clique transversal set of 𝐺1[𝑇𝑆(𝐺1)] nor 𝒮𝑇𝑆(𝐺2) is a maximum-clique transversal set of 𝐺2[𝑇𝑆(𝐺2)]. There exist maximum cliques 𝑄1 and 𝑄2 of 𝐺1[𝑇𝑆(𝐺1)] and 𝐺2[𝑇𝑆(𝐺2)], respectively, such that 𝒮 does not contain any vertex in them. By Lemma 8.4, 𝑄1𝑄2 is a maximum clique of 𝐺[𝑇𝑆(𝐺1)𝑇𝑆(𝐺2)]. Then, 𝒮 does not contain any vertex in 𝑄1𝑄2, which contradicts the assumption that 𝒮 intersects all maximum cliques of 𝐺[𝑇𝑆(𝐺1)𝑇𝑆(𝐺2)]. Following the discussion above, the lemma holds.

Definition 8.6. Let 𝐺=(𝑉,𝐸,𝑤) be a weighted graph. Suppose that 𝑊={𝑉1,𝑉2,,𝑉}, where 𝑉𝑖 is a subset of 𝑉 for 1𝑖. Let min𝑊 be an element in 𝑊 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 𝐺=(𝑉,𝐸), we use 𝜔(𝐺) (resp., 𝜔𝑡(𝐺)) to denote the clique number of 𝐺 (resp., 𝐺[𝑇𝑆(𝐺)]) and use 𝒞𝒯(𝐺) (resp., 𝒞𝒯𝑡(𝐺)) to denote a maximum-clique transversal set of 𝐺 (resp., 𝐺[𝑇𝑆(𝐺)]) of minimum weight. A strong maximum-clique transversal set 𝑆 of a distance-hereditary graph 𝐺 is a subset of 𝑉(𝐺) such that 𝑆 intersects all cliques in 𝑄𝐸(𝐺). We use 𝒮𝒞𝒯(𝐺) to denote a strong maximum-clique transversal set of 𝐺 of minimum weight.

Lemma 8.7. Suppose that 𝐺 is a distance-hereditary graph of only one vertex 𝑣. Then, 𝜔(𝐺)=𝜔𝑡(𝐺)=1, and 𝒞𝒯(𝐺)=𝒞𝒯𝑡(𝐺)=𝒮𝒞𝒯(𝐺)={𝑣}.

Proof. It follows from the definitions.

Lemma 8.8. Suppose that 𝐺 is formed from two disjoint distance-hereditary graphs 𝐺1 and 𝐺2 by a “false twin” operation.(1)If 𝜔𝑡(𝐺1)=𝜔𝑡(𝐺2) and 𝜔(𝐺1)=𝜔(𝐺2), then 𝜔(𝐺)=𝜔(𝐺1)=𝜔(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1)=𝜔𝑡(𝐺2), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺1)𝒞𝒯(𝐺2), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1)𝒞𝒯𝑡(𝐺2), and 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺1)𝒮𝒞𝒯(𝐺2).(2)If 𝜔𝑡(𝐺1)=𝜔𝑡(𝐺2) and 𝜔(𝐺1)>𝜔(𝐺2), then 𝜔(𝐺)=𝜔(𝐺1), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1)=𝜔𝑡(𝐺2), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺1), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1)𝒞𝒯𝑡(𝐺2), and 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺1)𝒞𝒯𝑡(𝐺2).(3)If 𝜔𝑡(𝐺1)=𝜔𝑡(𝐺2) and 𝜔(𝐺2)>𝜔(𝐺1), then 𝜔(𝐺)=𝜔(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1)=𝜔𝑡(𝐺2), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺2), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺2)𝒞𝒯𝑡(𝐺1), and 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺2)𝒞𝒯𝑡(𝐺1).(4)If 𝜔𝑡(𝐺1)>𝜔𝑡(𝐺2) and 𝜔(𝐺1)=𝜔(𝐺2), then 𝜔(𝐺)=𝜔(𝐺1)=𝜔(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺1)𝒞𝒯(𝐺2), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1), and 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯𝑣(𝐺1)𝒞𝒯(𝐺2).(5)If 𝜔𝑡(𝐺2)>𝜔𝑡(𝐺1) and 𝜔(𝐺1)=𝜔(𝐺2), then 𝜔(𝐺)=𝜔(𝐺1)=𝜔(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺2), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺1)𝒞𝒯(𝐺2), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺2), and 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺2)𝒞𝒯(𝐺1).(6)If 𝜔𝑡(𝐺1)>𝜔𝑡(𝐺2) and 𝜔(𝐺1)>𝜔(𝐺2), then 𝜔(𝐺)=𝜔(𝐺1), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺1), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1), and 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺1).(7)If 𝜔𝑡(𝐺2)>𝜔𝑡(𝐺1) and 𝜔(𝐺2)>𝜔(𝐺1), then 𝜔(𝐺)=𝜔(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺2), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺2), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺2), and 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺2).(8)If 𝜔𝑡(𝐺1)>𝜔𝑡(𝐺2) and 𝜔(𝐺1)<𝜔(𝐺2), then 𝜔(𝐺)=𝜔(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺2), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1), and 𝒮𝒞𝒯(𝐺)=𝒞𝒯𝑡(𝐺1)𝒞𝒯(𝐺2).(9)If 𝜔𝑡(𝐺2)>𝜔𝑡(𝐺1) and 𝜔(𝐺2)<𝜔(𝐺1), then 𝜔(𝐺)=𝜔(𝐺1), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺2), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺1), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺2), and 𝒮𝒞𝒯(𝐺)=𝒞𝒯𝑡(𝐺2)𝒞𝒯(𝐺1).

Proof. In the following, we just show the correctness for Statement (4) since other statements can be verified in similar ways. Note that 𝐺=𝐺1𝐺2 and 𝑇𝑆(𝐺)=𝑇𝑆(𝐺1)𝑇𝑆(𝐺2). Since 𝜔𝑡(𝐺1)>𝜔𝑡(𝐺2) and 𝜔(𝐺1)=𝜔(𝐺2), 𝜔(𝐺)=𝜔(𝐺1)=𝜔(𝐺2), and 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1). Therefore, 𝑄(𝐺)=𝑄(𝐺1)𝑄(𝐺2) and 𝑄(𝐺[𝑇𝑆(𝐺)])=𝑄(𝐺1[𝑇𝑆(𝐺1)]). We have 𝒞𝒯(𝐺)=𝒞𝒯(𝐺1)𝒞𝒯(𝐺2) and 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1).
By definition, 𝑄𝐸(𝐺)=𝑄(𝐺)𝑄(𝐺[𝑇𝑆(𝐺)]) and thus 𝑄𝐸(𝐺)=𝑄(𝐺1)𝑄(𝐺2)𝑄(𝐺1[𝑇𝑆(𝐺1)]). Then 𝑄𝐸(𝐺)=𝑄𝐸(𝐺1)𝑄(𝐺2). Hence, 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺1)𝒞𝒯(𝐺2). Following the discussion above, the statement is true.

Lemma 8.9. Suppose that 𝐺 is formed from two disjoint distance-hereditary graphs 𝐺1 and 𝐺2 by a “pendant vertex” operation.(1)If 𝜔𝑡(𝐺1)+𝜔𝑡(𝐺2)>max{𝜔(𝐺1),𝜔(𝐺2)}, then 𝜔(𝐺)=𝜔𝑡(𝐺1)+𝜔𝑡(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1), 𝒞𝒯(𝐺)=min{𝒞𝒯𝑡(𝐺1),𝒞𝒯𝑡(𝐺2)}, 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1), and 𝒮𝒞𝒯(𝐺)=𝒞𝒯𝑡(𝐺1).(2)If 𝜔𝑡(𝐺1)+𝜔𝑡(𝐺2)=𝜔(𝐺1)>𝜔(𝐺2), then 𝜔(𝐺)=𝜔(𝐺1)=𝜔𝑡(𝐺1)+𝜔𝑡(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1), 𝒞𝒯(𝐺)=min{𝒮𝒞𝒯(𝐺1),𝒞𝒯𝑡(𝐺2)𝒞𝒯(𝐺1)}, 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1), and 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺1).(3)If 𝜔𝑡(𝐺1)+𝜔𝑡(𝐺2)=𝜔(𝐺2)>𝜔(𝐺1), then 𝜔(𝐺)=𝜔(𝐺2)=𝜔𝑡(𝐺1)+𝜔𝑡(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1), 𝒞𝒯(𝐺)=min{𝒮𝒞𝒯(𝐺2),𝒞𝒯𝑡(𝐺1)𝒞𝒯(𝐺2)}, 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1), and 𝒮𝒞𝒯(𝐺)=𝒞𝒯𝑡(𝐺1)𝒞𝒯(𝐺2).(4)If 𝜔𝑡(𝐺1)+𝜔𝑡(𝐺2)=𝜔(𝐺1)=𝜔(𝐺2), then 𝜔(𝐺)=𝜔(𝐺1)=𝜔(𝐺2)=𝜔𝑡(𝐺1)+𝜔𝑡(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1), 𝒞𝒯(𝐺)=min{𝒮𝒞𝒯(𝐺1)𝒞𝒯(𝐺2),𝒮𝒞𝒯(𝐺2)𝒞𝒯(𝐺1)}, 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1), and 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺1)𝒞𝒯(𝐺2).(5)If 𝜔𝑡(𝐺1)+𝜔t(𝐺2)<𝜔(𝐺1) and 𝜔(𝐺1)>𝜔(𝐺2), then 𝜔(𝐺)=𝜔(𝐺1), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺1), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1), and 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺1).(6)If 𝜔𝑡(𝐺1)+𝜔𝑡(𝐺2)<𝜔(𝐺2) and 𝜔(𝐺2)>𝜔(𝐺1), then 𝜔(𝐺)=𝜔(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺2), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1), and 𝒮𝒞𝒯(𝐺)=𝒞𝒯𝑡(𝐺1)𝒞𝒯(𝐺2).(7)If 𝜔𝑡(𝐺1)+𝜔𝑡(𝐺2)<𝜔(𝐺2) and 𝜔(𝐺2)=𝜔(𝐺1), then 𝜔(𝐺)=𝜔(𝐺1)=𝜔(𝐺2), 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1), 𝒞𝒯(𝐺)=𝒞𝒯(𝐺1)𝒞𝒯(𝐺2), 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1), and 𝒮𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺1)𝒞𝒯(𝐺2).

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, 𝑇𝑆(𝐺)=𝑇𝑆(𝐺1) and 𝐺 is obtained from 𝐺1 and 𝐺2 by connecting every vertex of 𝑇𝑆(𝐺1) to all vertices of 𝑇𝑆(𝐺2). Therefore, 𝜔𝑡(𝐺)=𝜔𝑡(𝐺1) and 𝒞𝒯𝑡(𝐺)=𝒞𝒯𝑡(𝐺1).
We now consider 𝜔(𝐺) and 𝒞𝒯(𝐺). In this case, 𝜔𝑡(𝐺1)+𝜔𝑡(𝐺2)=𝜔(𝐺2)>𝜔(𝐺1). Then, 𝜔(𝐺)=𝜔(𝐺2)=𝜔𝑡(𝐺1)+𝜔t(𝐺2) and thus 𝑄(𝐺)=𝑄(𝐺2)𝑄𝑋12(𝐺). Clearly, 𝒞𝒯(𝐺) is a subset of 𝑇𝑆(𝐺1)𝑉(𝐺2) and 𝒞𝒯(𝐺)𝑉(𝐺2) is a maximum-clique transversal set of 𝐺2. By Lemma 8.4, we know that 𝑄𝑋12(𝐺)=𝐺[𝑇𝑆(𝐺1)𝑇𝑆(𝐺2)]. Therefore, 𝒞𝒯(𝐺) intersects all maximum cliques of 𝐺[𝑇𝑆(𝐺1)𝑇𝑆(𝐺2)]. By Lemma 8.5, either 𝒞𝒯(𝐺)𝑇𝑆(𝐺1) is a maximum-clique transversal set of 𝐺1[𝑇𝑆(𝐺1)] or 𝒞𝒯(𝐺)𝑇𝑆(𝐺2) is a maximum-clique transversal set of 𝐺2[𝑇𝑆(𝐺2)]. If 𝒞𝒯(𝐺)𝑇𝑆(𝐺1) is a maximum-clique transversal set of 𝐺1[𝑇𝑆(𝐺1)], then we have 𝒞𝒯(𝐺)=𝒞𝒯(𝐺2)𝒞𝒯𝑡(𝐺1). If 𝒞𝒯(𝐺)𝑇𝑆(𝐺2) is a maximum-clique transversal set of 𝐺2[𝑇𝑆(𝐺2)], then 𝒞𝑇(𝐺)𝑉(𝐺2) is not only a maximum-clique transversal set of 𝐺 but also a strong maximum-clique transversal set of 𝐺2. We have 𝒞𝒯(𝐺)=𝒮𝒞𝒯(𝐺2). Hence, 𝒞𝒯(𝐺)=min{𝒞𝒯𝑡(𝐺1)𝒞𝒯(𝐺2),𝒮𝒞𝒯(𝐺2)}.
We now consider 𝒮𝒞𝒯(𝐺). By definition, 𝒮𝒞𝒯(𝐺) intersects all maximum cliques in 𝑄(𝐺[𝑇𝑆(𝐺)]) and 𝑄(𝐺), respectively. Recall that 𝑇𝑆(𝐺)=𝑇𝑆(𝐺1) and 𝑄(𝐺)=𝑄(𝐺2)𝑄𝑋12(𝐺). Therefore, 𝒮𝒞𝒯(𝐺)𝑇𝑆(𝐺1) is a maximum-clique transversal set of 𝐺[𝑇𝑆(𝐺1)] and it also intersects all maximum cliques in