The Clique-Transversal and the Clique-Independent Set Problems on Distance-Hereditary Graphs

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(1)Int. Computer Symposium, Dec. 15-17, 2004, Taipei, Taiwan.. The Clique-Transversal and the Clique-Independent Set Problems on Distance-Hereditary Graphs † Chuan-Min Lee and Maw-Shang Chang Department of Computer Science and Information Engineering National Chung Cheng University Ming-Hsiung, Chiayi 621, Taiwan, R.O.C (cmlee|mschang)@cs.ccu.edu.tw Abstract In this paper, we show that minimum clique-transversal and maximum clique-independent sets of a distance-hereditary graph have the same cardinality, and the clique-transversal set problem can be solved in O(n + m) time and the clique-independent set problem can be solved in O(n2 ) time for distancehereditary graphs.. k-fold clique-transversal problem [11], respectively. The clique-independent set problem is a special case of the clique r-packing problem [7]. Following the algorithm of [11] with k = 1, the clique-transversal set problem is polynomial-time solvable on balanced graphs. In [7], an efficient algorithm was proposed to solve the clique r-domination problem and the clique r-packing problem on dually chordal graphs. We can use this algorithm to solve the clique-transversal and the clique-independent set problems on a dually chordal graph, but the time complexity of this algorithm is proportional to the sum of the sizes of all maximal cliques of a dually chordal graph. Notice that a dually chordal graph may have a exponential number of maximal cliques.. Keywords: Algorithm, Clique-Transversal Set, Clique-Independent Set, Distance-Hereditary Graph.. 1. Introduction Let G = (V, E) be a finite, simple, undirected graph with |V | = n and |E| = m. 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. A clique-transversal set of G is a subset of vertices intersecting all maximal cliques of G. The clique-transversal set problem is to find a clique-transversal set of G of minimum cardinality. The cardinality τC (G) of a minimum clique-transversal set of G is called the cliquetransversal number of G. A clique-independent set of G is a collection of pairwise disjoint maximal cliques. The clique-independent set problem is to find a cliqueindependent set of G of maximum cardinality. The cardinality αC (G) of a maximum clique-independent set of G is called the clique-independence number of G. It is clear that the weak duality inequality αC (G) ≤ τC (G) holds for any graph G. The clique-transversal set problem is a special case of the generalized clique-transversal problem [9], the clique r-domination problem [7], and †. The clique-transversal set problem has been widely studied in [1, 2, 3, 14, 19, 21]. Eades et al. [13] showed that the problem of deciding whether a chordal graph has two disjoint minimum clique-transversal sets is NP-complete. Both the clique-transversal and the clique-independent set problems are NP-hard for cocomparability graphs, planar graphs, line graphs, total graphs, split graphs, undirected path graphs, and k-trees with unbounded k [8, 9, 15]. Furthermore, both problems are polynomial-time solvable for comparability graphs, strongly chordal graphs, and Helly circulararc graphs [4, 8, 9, 15]. In [20], Sheu extended the algorithm of [4] to solve the weighted version of the clique-transversal set√problem on weighted comparability graphs in O(m n + M (n)) time, where M (n) is the complexity of multiplying two n × n matrices. A graph G is clique-perfect if τC (F ) = αC (F ) for every induced subgraph F of G [15]. The following are examples of clique-perfect graph classes: chordal graphs without odd suns [18], strongly chordal graphs [8], and. This research was partially supported by National Science Council, ROC, under contract number NSC-90-2213-E-194034.. 1362.

(2) comparability graphs [4]. Dur´ an [12] et al. demonT S(G2 ) is a distance-hereditary graph, and the and αC (G) can15-17, be computed in Taiwan.twin set of G is T S(G1 ). G is said to be obstrated that τC (G) Int. Computer Symposium, Dec. 2004, Taipei, polynomial time for any clique-perfect graph G by ustained from G1 and G2 by a pendant vertex operation. (In the rest of the paper, we assume that ing integer linear programming. A graph G = (V, E) is T S(G) = T S(G1 ) whenever we say that G is obcalled distance-hereditary if every pair of vertices are tained from G1 and G2 by a pendant vertex opequidistant in every connected induced subgraph coneration.) taining them. For other features of distance-hereditary graphs, please refer to [5, 6, 16]. It has been shown [17] By Theorem 1, a distance-hereditary graph G has that finding a minimum-weighted clique-transversal its own twin set T S(G), the twin set T S(G) is a subset set on weighted distance-hereditary graphs can be of vertices of G, and it is defined recursively. The consolved in O(n + m) time and the clique-independence struction of G from disjoint distance-hereditary graphs number of a distance-hereditary graph can be comG1 and G2 as described in Theorem 1 involves only the puted in O(n3 ) time, but it remains open whether twin sets of G1 and G2 . distance-hereditary graphs are clique-perfect. In this Following Theorem 1, a binary ordered decomposipaper, we show that τC (G) = αC (G) for any distancetion tree can be obtained in linear-time [10]. In this dehereditary graph G, and the clique-transversal set composition tree, each leaf is a single vertex graph, and problem can be solved in O(n+m) time and the cliqueeach internal node represents one of the three operaindependent set problem can be solved in O(n2 ) time tions: pendant vertex operation (labelled by P), true for distance-hereditary graphs. Following the definition twin operation (labelled by T), and false twin operaof distance-hereditary graphs, every induced subgraph tion (labelled by F). This ordered decomposition tree of a distance-hereditary graph is distance-hereditary, is called a PTF-tree. It has 2n − 1 tree nodes. Figtoo. The equation τC (F ) = αC (F ) holds for every inure 1 illustrates an example of a PTF-tree. Hence, a duced subgraph F of G. Therefore, distance-hereditary PTF-tree of a distance-hereditary graph can be obgraphs are clique-perfect. tained in linear-time [10].. 2. Preliminaries. 3. Distance-Hereditary Clique-Perfect. The following theorem shows that distancehereditary graphs can be defined recursively.. Graphs. Are. In this section, we will prove that distancehereditary graphs are clique-perfect by induction. We observe that a graph G of a single vertex holds the duality equality αC (G) = τC (G). Suppose that G1 and G2 are two distance-hereditary graphs that hold the duality equality. We will show that a graph G obtained from G1 and G2 by any one of operations mentioned in Theorem 1 always holds the duality equality. It will reveal. Theorem 1. [10] Distance-hereditary graphs can be defined recursively as follows: 1. A graph consisting of only one vertex is distancehereditary, and the twin set is the vertex itself. 2. If G1 and G2 are disjoint distance-hereditary graphs with the twin sets T S(G1 ) and T S(G2 ), respectively, then the graph G = G1 ∪ G2 is a distance-hereditary graph and the twin set of G is T S(G1 ) ∪ T S(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 T S(G1 ) and T S(G2 ), respectively, then the graph G obtained by connecting every vertex of T S(G1 ) to all vertices of T S(G2 ) is a distance-hereditary graph, and the twin set of G is T S(G1 ) ∪ T S(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 T S(G1 ) and T S(G2 ), respectively, then the graph G obtained by connecting every vertex of T S(G1 ) to all vertices of. Figure 1. (a) A distance-hereditary graph G. (b) A PTF-tree of G.. 1363.

(3) that distance-hereditary graphs hold the duality equal(1) C(G[T S(G)]) = C12 (G), ity.Int. Since an induced subgraph of a distance-hereditary Computer Symposium, Dec. 15-17, 2004, Taipei, Taiwan. (2) C(G) = CT S (G1 ) ∪ CT S (G2 ) ∪ C12 (G), graph is also distance-hereditary, it will follow that (3) CT S (G) = C(G[T S(G)]), distance-hereditary graphs are clique-perfect. (4) CT S (G) = CT S (G1 ) ∪ CT S (G2 ), and Throughout this section, we assume that G = (V, E) is a distance-hereditary graph. For a subset V of V , (5) CE (G) = C(G). G[V ] is the subgraph induced by V . Before proving Proof. In the following, we just show the correctness that distance-hereditary graphs are clique-perfect, we of statement (1). The other statements of this lemma give some observations about maximal cliques based can be easily verified by definition. Notice that G is upon the recursive definition of distance-hereditary obtained by connecting every vertex of T S(G1 ) to all graphs. vertices of T S(G2 ), and T S(G) = T S(G1 ) ∪ T S(G2 ). Definition 1. We use C(G) to denote the collection Clearly, every clique in C12 (G) is also a clique of of all maximal cliques of G. Hence C(G[T S(G)]) is G[T S(G)]. Let c be a maximal clique in C(G[T S(G)]), the collection of all maximal cliques of G[T S(G)]. We c1 = c ∩ T S(G1 ), and c2 = c ∩ T S(G2 ). Suppose that use CT S (G) to denote the collection of all maximal c1 is not a maximal clique of G1 [T S(G1 )]. There excliques of G which are maximal cliques of G[T S(G)] ists a maximal clique c1 in C(G1 [T S(G1 )]) such that and use CT S (G) to denote the collection of all maxc1 ⊂ c1 . Then c1 ∪ c2 is a clique of G[T S(G)] and imal cliques of G which are not maximal cliques of c ⊂ (c1 ∪ c2 ), which contradicts that c is a maximal G[T S(G)]. Hence C(G) = CT S (G) ∪ CT S (G). Let clique of G[T S(G)]. Therefore, c1 ∈ C(G1 [T S(G1 )]. CE (G) = C(G) ∪ C(G[T S(G)]). CE (G) denotes the Similarly, we can prove that c2 ∈ C(G2 [T S(G2 )]). collection of all maximal cliques of G and all maximal Hence, c ∈ C12 (G). Conversely, let c1 be a maxicliques of G[T S(G)]. mal clique in C(G1 [T S(G1 )]) and c2 be a maximal clique in C(G2 [T S(G2 )]). Then c = c1 ∪ c2 is a Remark 1. Suppose that G is a graph of single vertex clique in C12 (G). Suppose that c is not a maximal and v is the vertex of G. Then, C(G) = C(G[T S(G)]) = clique of G[T S(G)]. There exists a maximal clique CT S (G) = CE (G) = {{v}} and CT S (G) = ∅. c ∈ C(G[T S(G)]) such that c ⊂ c . Then either Remark 2. A maximal clique of G[T S(G)] is not necc1 ⊂ (c ∩ T S(G1 )) or c2 ⊂ (c ∩ T S(G2 )). However, essarily a maximal clique of G. If all maximal cliques either of them contradicts that c1 and c2 are maxiof G[T S(G)] are maximal cliques of G, then C(G) = mal cliques of G[T S(G1 )] and G[T S(G2 )], respectively. CE (G). On the other hand, if all maximal cliques of Therefore, (c1 ∪ c2 ) ∈ C(G[T S(G)]). Following the disG[T S(G)] are not maximal cliques of G, then C(G) = 2 cussion above, C(G[T S(G)]) = C12 (G). CT S (G). Lemma 3. Suppose that G is a graph obtained from two Lemma 1. Suppose that G is a graph obtained from two disjoint distance-hereditary graphs G1 and G2 by a pendisjoint distance-hereditary graphs G1 and G2 by a false dant vertex operation. Then, we have twin operation. Then, we have (1) C(G[T S(G)]) = C(G1 [T S(G1 )]),. (1) C(G[T S(G)]) = C(G1 [T S(G1 )]) ∪ C(G2 [T S(G2 )]),. (2) C(G) = CT S (G1 ) ∪ CT S (G2 ) ∪ C12 (G),. (2) C(G) = C(G1 ) ∪ C(G2 ),. (3) CT S (G) = ∅,. (3) CT S (G) = CT S (G1 ) ∪ CT S (G2 ),. (4) CT S (G) = C(G),. (4) CT S (G) = CT S (G1 ) ∪ CT S (G2 ), and. (5) CE (G) = C(G) ∪ C(G1 [T S(G1 )]).. (5) CE (G) = CE (G1 ) ∪ CE (G2 ). Proof. By definition.. Proof. Notice that G is obtained by connecting every vertex of T S(G1 ) to all vertices of T S(G2 ), but T S(G) = T S(G1 ). By arguments similar to those for proving Lemma 2, this lemma can be easily proved. 2. 2. Definition 2. Suppose that G is a graph obtained from two disjoint distance-hereditary graphs G1 and G2 by a true twin operation or a pendant vertex operation. We use C12 (G) to denote {c1 ∪ c2 |c1 ∈ C(G1 [T S(G1 )]) and c2 ∈ C(G2 [T S(G2 )])}.. Lemma 4. Suppose that G is a graph obtained from two disjoint distance-hereditary graphs G1 and G2 by a true twin operation or a pendant vertex operation. If S is a clique-transversal set of G, then either S ∩ T S(G1 ) is a clique-transversal set of G1 [T S(G1 )] or S ∩ T S(G2 ) is a clique-transversal set of G2 [T S(G2 )].. Lemma 2. Suppose that G is a graph obtained from two disjoint distance-hereditary graphs G1 and G2 by a true twin operation. Then, we have. 1364.

(4) Proof. Assume for contrary that neither S ∩T S(G1 ) is Instead of proving that distance-hereditary graphs S(G115-17, )] nor S2004, ∩ T S(G a clique-transversal set of G1 [T Dec. hold the duality equality, we prove that they hold the Int. Computer Symposium, Taipei, Taiwan. 2) is a clique-transversal set of G2 [T S(G2 )]. There exstrong duality. We will show how to find a CT (G), a ist maximal cliques c1 and c2 of G1 [T S(G1 )] and CI(G), a W CT (G), a W CI(G), an SCT (G), and an G2 [T S(G2 )], respectively, such that S does not conECI(G) such that the four conditions of strong duality tain any vertex in them. By Lemma 2 and Lemma 3, are satisfied. c1 ∪ c2 is a maximal clique of G. However, S does not Lemma 5. Assume that G is a graph of single vertex contain any vertex in c1 ∪ c2 , which contradicts the asand v is the vertex of G. There exist the following sets: sumption that S is a clique-transversal set of G. 2 (1) CT (G) = {v}, To prove that distance-hereditary graphs are clique(2) SCT (G) = {v}, perfect, we introduce the following definitions. (3) W CT (G) = ∅, Definition 3. A strong clique-transversal set of G is a (4) W CI(G) = ∅,. subset of V that intersects all cliques in CE (G). We use SCT (G) to represent a strong clique-transversal set of G.. (5) ECI(G) = {{v}}, and (6) CI(G) = {{v}} such that G holds the strong duality.. Definition 4. A weak clique-transversal set of G is a subset of V that intersects all maximal cliques in CT S (G). We use W CT (G) to represent a weak cliquetransversal set of G.. Proof. The lemma can be easily verified by the definition. 2. Definition 5. A weak clique-independent set of G is a collection of pairwise disjoint cliques in CT S (G). We use W CI(G) to represent a weak clique-independent set of G.. Definition 8. Assume that S is a family of sets. Let min S denote a set of minimum cardinality in S. Lemma 6. Assume that G is formed from two disjoint distance-hereditary graphs G1 and G2 by a pendant vertex operation, and both G1 and G2 hold the strong duality. Suppose that XI(G1 ) = {c1 , . . . , ck1 }, XI(G2 ) = {d1 , . . . , dk2 }, and k = min{k1 , k2 }. Let ˜ = {ck+1 , · · · , ck } if X = {ci ∪ di |1 ≤ i ≤ k}. Let X 1 ˜ k1 > k and X = ∅ otherwise. There exist the following sets:. Definition 6. An expanded clique-independent set of G is a collection of pairwise disjoint cliques in CE (G). We use ECI(G) to represent an expanded cliqueindependent set of G. Definition 7. Let CT (G) and CI(G) denote a cliquetransversal set and a clique-independent set of G, respectively. We say that a distance-hereditary graph G holds the strong duality if there exist a CT (G), a CI(G), a W CT (G), a W CI(G), an SCT (G), and an ECI(G) such that the following four conditions are satisfied:. (1) CT (G) = min{SCT (G1 ) ∪ W CT (G2 ), SCT (G2 ) ∪ W CT (G1 )}, (2) SCT (G) = SCT (G1 ) ∪ W CT (G2 ), (3) W CT (G) = CT (G),. (1) |CT (G)| = |CI(G)|,. (4) W CI(G) = W CI(G1 ) ∪ W CI(G2 ) ∪ X,. (2) |W CT (G)| = |W CI(G)|,. ˜ and (5) ECI(G) = W CI(G1 ) ∪ W CI(G2 ) ∪ X ∪ X,. (3) |SCT (G)| = |ECI(G)|, and. (6) CI(G) = W CI(G) such that G holds the strong duality.. (4) W CI(G) ⊆ ECI(G). For simplicity, let XI(G) denote ECI(G) \ W CI(G).. Proof.. Remark 3. Suppose that G holds the strong duality. Since |W CT (G)| = |W CI(G)|, such a W CT (G) and a W CI(G) are a minimum weak clique-transversal set and a maximum weak clique-independent set of G, respectively. Hence XI(G) ⊆ C(G[T S(G)]).. (1) By (2) of Lemma 3, both SCT (G1 ) ∪ W CT (G2 ) and SCT (G2 ) ∪ W CT (G1 ) are clique-transversal sets of G. We let CT (G) = min{SCT (G1 ) ∪ W CT (G2 ), SCT (G2 ) ∪ W CT (G1 )}.. Remark 4. Since |CI(G)| ≤ αC (G) ≤ τC (G) ≤ |CT (G)|, G holds the duality equality if there exist a clique-transversal set CT (G) and a cliqueindependent set CI(G) satisfying the condition that |CT (G)| = |CI(G)|.. (2) Notice that T S(G) = T S(G1 ). Since SCT (G1 ) intersects all maximal cliques of C(G[T S(G1 )]), it intersects all maximal cliques of C(G[T S(G)]). Hence SCT (G1 )∪W CT (G2 ) is a strong clique-transversal set of G. We let SCT (G) = SCT (G1 ) ∪ W CT (G2 ).. 1365.

(5) (3) By (4) of Lemma 3, a weak clique-transversal (6) CI(G) = ECI(G) set Int. of G is also aSymposium, clique-transversal set of 2004, G. WeTaipei, let Taiwan. such that G holds the strong duality. Computer Dec. 15-17, W CT (G) = CT (G). Proof. By (2) and (5) of Lemma 2, we see that both Suppose that XI(G1 ) = {c1 , . . . , ck1 }, XI(G2 ) = SCT (G1 ) ∪ W CT (G2 ) and SCT (G2 ) ∪ W CT (G1 ) are {d1 , . . . , dk2 }, and k = min{k1 , k2 }. Let X = {ci ∪ not only clique-transversal sets of G but also strong ˜ di |1 ≤ i ≤ k} and X = {ck+1 , · · · , ck1 }. clique-transversal sets of G. Besides, X is a clique(4) By (2) and (4) of Lemma 3, X is a cliqueindependent set of G and W CI(G1 ) ∪ W CI(G2 ) ∪ X independent set of G and W CI(G1 ) ∪ W CI(G2 ) ∪ X is not only a clique-independent set of G but also is a weak clique-independent set of G. We let an expanded clique-independent set of G. FurtherW CI(G) = W CI(G1 ) ∪ W CI(G2 ) ∪ X. more, by (4) of Lemma 2, W CT (G1 ) ∪ W CT (G2 ) and W CI(G1 )∪W CI(G2 ) are a weak clique-transversal set (5) It is easy to verify that W CI(G1 ) ∪ W CI(G2 ) ∪ and a weak clique-independent set of G, respectively. ˜ is an expanded clique-independent set of G. We X ∪X Therefore we let ˜ let ECI(G) = W CI(G1 ) ∪ W CI(G2 ) ∪ X ∪ X. (1) CT (G) = min{SCT (G1 ) ∪ W CT (G2 ), SCT (G2 ) ∪ W CT (G1 )},. (6) By (4) of Lemma 3, a clique-independent set of G is also a weak clique-independent set of G. We let CI(G) = W CI(G). In the following, we show that CT (G), CI(G), W CT (G), W CI(G), SCT (G), and ECI(G) satisfy the conditions of strong duality. Since G1 and G2 hold the strong duality, by (2) and (3) of Definition 7, k1 = = |XI(G1 )| and |SCT (G1 )| − |W CT (G1 )| k2 = |SCT (G2 )| − |W CT (G2 )| = |XI(G2 )|. Clearly |CT (G)| = |W CT (G)| = |W CT (G1 )| + |W CT (G2 )| + k = |W CI(G1 )| + |W CI(G2 )| + k = |W CI(G)| = |CI(G)|. Next we verify that ˜ is |SCT (G)| = |ECI(G)|. If k = k1 , then X empty, |SCT (G)| = k1 + |W CT (G1 )| + |W CT (G2 )|, and |ECI(G)| = |W CI(G1 )| + |W CI(G2 )| + k1 . On the other hand, suppose that k = k2 . We have |SCT (G)| = |SCT (G1 )| + |W CT (G2 )| = k1 + |W CT (G1 )| + |W CT (G2 )| and |ECI(G)| = ˜ = |W CI(G1 )| + |W CI(G2 )| + |X| + |X| |W CI(G1 )| + |W CI(G2 )| + k1 . We can see that |SCT (G)| = |ECI(G)| in both cases. Finally, W CI(G) ⊆ ECI(G) is obvious. Thus G holds the strong duality. 2. (2) SCT (G) = CT (G), (3) W CT (G) = W CT (G1 ) ∪ W CT (G2 ), (4) W CI(G) = W CI(G1 ) ∪ W CI(G2 ), (5) ECI(G) = W CI(G1 ) ∪ W CI(G2 ) ∪ X, and (6) CI(G) = ECI(G). Since G1 and G2 hold the strong duality, by (2) and (3) of Definition 7, k1 = |SCT (G1 )| − |W CT (G1 )| = |XI(G1 )| and k2 = |SCT (G2 )| − |W CT (G2 )| = |XI(G2 )|. Hence |CT (G)| = |SCT (G)| = |ECI(G)| = |CI(G)| = k + |W CI(G1 )| + |W CI(G2 )|. Besides, |W CT (G)| = |W CI(G)|. Finally, W CI(G) ⊆ ECI(G) is obvious. Following the discussion above, G holds the strong duality. 2 Lemma 8. Assume that G is obtained from two disjoint distance-hereditary graphs G1 and G2 by a false twin operation, and both G1 and G2 hold the strong duality. There exist the following sets: (1) CT (G) = CT (G1 ) ∪ CT (G2 ), (2) SCT (G) = SCT (G1 ) ∪ SCT (G2 ), (3) W CT (G) = W CT (G1 ) ∪ W CT (G2 ),. Lemma 7. Assume that G is formed from two disjoint distance-hereditary graphs G1 and G2 by a true twin operation, and both G1 and G2 hold the strong duality. Suppose that XI(G1 ) = {c1 , . . . , ck1 }, XI(G2 ) = {d1 , . . . , dk2 }, and k = min{k1 , k2 }. Let X = {ci ∪di |1 ≤ i ≤ k}. There exist the following sets:. (4) W CI(G) = W CI(G1 ) ∪ W CI(G2 ), (5) ECI(G) = ECI(G1 ) ∪ ECI(G2 ), and (6) CI(G) = CI(G1 ) ∪ CI(G2 ) such that G holds the strong duality.. (1) CT (G) = min{SCT (G1 ) ∪ W CT (G2 ), SCT (G2 ) ∪ W CT (G1 )},. Proof. Following Lemma 1, CT (G1 ) ∪ CT (G2 ) is a clique-transversal set of G, and SCT (G1 ) ∪ SCT (G2 ) a strong clique-transversal set of G, . . . , etc. So we let. (2) SCT (G) = CT (G),. (1) CT (G) = CT (G1 ) ∪ CT (G2 ),. (3) W CT (G) = W CT (G1 ) ∪ W CT (G2 ),. (2) SCT (G) = SCT (G1 ) ∪ SCT (G2 ),. (4) W CI(G) = W CI(G1 ) ∪ W CI(G2 ),. (3) W CT (G) = W CT (G1 ) ∪ W CT (G2 ),. (5) ECI(G) = W CI(G1 ) ∪ W CI(G2 ) ∪ X, and. (4) W CI(G) = W CI(G1 ) ∪ W CI(G2 ),. 1366.

(6) (5) ECI(G) = ECI(G1 ) ∪ ECI(G2 ), and [5] H. J. Bandelt, and H. M. Mulder, Distance hereditary graphs, J. of Combin. Theory, Ser. B 41 (1986) 182–208. Int. Computer Symposium, Dec. 15-17, 2004, Taipei, Taiwan. (6) CI(G) = CI(G1 ) ∪ CI(G2 ). [6] A. Brandst¨ adt, V. B. Le, and J. P. Spinrad, Graph Since G1 and G2 hold the strong duality, it is easy to classes–A Survey, SIAM Monographs on Discrete Math. verify that all four conditions of the strong duality are and Applications (Philadelphia, 1999). satisfied. 2. [7] A. Brandst¨ adt, V. D. Chepoi, and F. F. Dragan, Clique r -domination and clique r -packing problems on dually chordal graphs, SIAM J. Discrete Math., 10 (1) (1997) 109–127. [8] G. J. Chang, M. Farber, and Zs. Tuza, Algorithmic aspects of neighborhood numbers, SIAM J. Discrete Math., 6 (1993) 24–29. [9] M. S. Chang, Y. H. Chen, G. J. Chang, and J. H. Yan, Algorithmic aspects of the generalised clique transversal problem on chordal graphs, Discrete Appl. Math. 66 (1996) 189–203. [10] M. S. Chang, S. Y. Hsieh, and G. H. Chen, Dynamic programming on distance-hereditary graphs, Lecture Notes in Computer Science, Vol. 1350 (1997) 344–353. [11] E. Dahlhaus, P. D. Manuel, M. Miller, Maximum hcolourable subgraph problem in balanced graphs, Inform. Process. Lett., 65 (1998) 301–303. [12] G. Dur´ an, M. C. Lin, and J. L. Szwarcfiter, On cliquetransversal and clique-independent sets, Ann. of Oper. Res. 116 (2002) 71–77. [13] P. Eades, M. Keil, P. D. Manuel, and M. Miller, Two minimum dominating sets with minimum intersection in chordal graphs, Nordic J. Comput., 3 (1996) 220–237. [14] P. Erd¨ os, T. Gallai and Zs. Tuza, Covering the cliques of a graph with vertices, Discrete Math., 108 (1992) 279– 289. [15] V. Guruswami, and C. P. Rangan, Algorithmic aspects of clique-transversal and clique-independent sets, Discrete Appl. Math. 100 (2000) 183–202. [16] P. L. Hammer, and F. Maffray, Completely separable graphs, Discrete Appl. Math. 27 (1990) 85–90. [17] C. M. Lee, M. S. Chang, and S. C. Sheu, The Clique Transversal and Clique Independence of Distance Hereditary Graphs, in: Proceedings of the 19th Workshop on Combinatorial Mathematics and Computation Theory, Taiwan, (2002) 64-69. [18] J. Lehel and Zs. Tuza, Neighborhood perfect graphs, Discrete Math., 61 (1986) 93–101. [19] Z. Lonc, I. Rival, Chains, antichains and fibres, J. Combin. Theory Ser. A 44 (1987) 207–228. [20] S. C. Sheu, The Weighted Clique Transversal Set Problem on Distance-Hereditary Graphs, Master Thesis, Department of Computer Science and Information Engineering, Chung Cheng University, Taiwan, 2001. [21] Zs. Tuza, Covering all cliques of a graph, Discrete Math., 86 (1990) 117–126.. Now we are ready to prove the main theorem. Theorem 2. Distance-hereditary graphs are cliqueperfect. Proof. We have explained the reasons that distancehereditary graphs are clique-perfect if they hold the strong duality. Based upon the recursive definition of distance-hereditary graphs, and Lemmas 5, 6, 7, and 8, we can prove that distance-hereditary graphs hold strong duality by induction. Hence, distance-hereditary graphs are clique-perfect. 2. 4. Conclusion We have shown distance-hereditary graphs are clique-perfect graphs. The proof is based upon the recursive definition of distance-hereditary graphs. Furthermore, as a byproduct of the inductive proof, we can design polynomial-time dynamic programming algorithms to find a minimum clique-transversal set and a maximum clique-independent set for a distancehereditary graph G. The constructive proofs of Lemma 6, 7, and 8 suggest that we build a CT (G) and a CI(G) bottom up according to the PTF-tree of G. If we store a clique-transversal set in a linked list and a clique-independent set in a linked list of cliques, respectively. Then, the union of two clique-transversal sets or two clique-independent sets can be done in constant time. We see that the clique-transversal set problem can be solved in O(n + m) and the cliqueindependent set problem can be solved in O(n2 ) time for distance-hereditary graphs. We conjecture that the algorithm for the clique-independent set problem on distance-hereditary graphs can be implemented in O(n + m) time.. References [1] T. Andreae, M. Schughart, and Zs. Tuza, Clique transversal sets of line graphs and complements of line graphs, Discrete Math., 88 (1991) 11–20. [2] T. Andreae, and C. Flotow, On covering all cliques of a chordal graph, Discrete Math., 149 (1996) 299–302. [3] T. Andreae, On the clique-transversal number of chordal graphs, Discrete Math., 191 (1998) 3–11. [4] V. Balachandran, P. Nagavamsi, and C. P. Rangan, Clique transversal and clique independence on comparability graphs, Inform. Process. Lett. 58 (1996) 181–184.. 1367.

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