Discrete Mathematics 223 (2000) 125–133
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On the powers of graphs with bounded asteroidal number Ting-Yem Hoa; c , Jou-Ming Changb; c , Yue-Li Wanga; ∗ a Department
of Information Management, National Taiwan University of Science and Technology, Taipei, Taiwan, ROC b Institute of Computer Science and Information Engineering, National Central University, Chung-Li, Taiwan, ROC c Department of Information Management, National Taipei College of Business, Taipei, Taiwan, ROC Received 25 August 1998; revised 16 July 1999; accepted 20 December 1999
Abstract For an undirected graph G = (V; E), the kth power G k is the graph with the same vertex set as G such that two vertices are adjacent in G k if and only if their distance in G is at most k. A set of vertices A ⊆ V is an asteroidal set if for every vertex a∈A, the set A\{a} is contained in one connected component of G − NG [a], where NG [a] is the closed neighborhood of a in G. The asteroidal number of a graph G is the maximum cardinality of an asteroidal set in G. The class of graphs with asteroidal number at most s is denoted by A(s). In this paper, we show that if G k ∈A(s) for s¿2, then so is G k+1 . This generalizes a previous result for the family of AT-free graphs. Moreover, we consider the forbidden con gurations for the powers of graphs with bounded asteroidal number. Based on these forbidden con gurations, we show that every c 2000 Elsevier Science B.V. All rights reserved. proper power of AT-free graphs is perfect. Keywords: Asteroidal triple; AT-free graphs; Powers of graphs; Strong perfect graph conjecture
1. Introduction Let G = (V; E) be a graph consisting of the vertex set V and the edge set E, respectively. For a positive integer k, the kth power G k is the graph with the same vertex set as G such that two vertices are adjacent in G k if and only if their distance in G is at most k. Up to now several results concerning the family of graphs closed under power operations have been investigated. One of the rst results in this eld is due to Duchet [14]: If G k is chordal (i.e., a graph with the property that every cycle of length greater than three has a chord), then so is G k+2 . Consequently, all odd powers of chordal graphs are chordal, whereas this is not true in general for even powers. ∗
Correspondence address: Department of Information Management, National Taiwan University of Science and Technology, 43, Section 4, Kee-Lung Road, Taipei 106, Taiwan. E-mail address:
[email protected] (Y.-L. Wang). c 2000 Elsevier Science B.V. All rights reserved. 0012-365X/00/$ - see front matter PII: S 0 0 1 2 - 3 6 5 X ( 0 0 ) 0 0 0 4 3 - 1
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Table 1 Three type assertions related to the power operations Classes of graphs
G k ∈ C ⇒ G k+2 ∈ C
Chordal Strongly chordal Circular arc (Unit) interval AT-free Cocomparability m-trapezoid Dually chordal HHD-free Weak bipolarizable
[14]
G ∈ C ⇒ Gk ∈ C
G k ∈ C ⇒ G k+1 ∈ C
[12,20] [23]
[23]
[13] [6] [6]
[4]
[22] [22] [15] [15]
In contrast, Bandelt et al. [1] showed that all even powers of a distance–hereditary graph (i.e., a graph G with the property that the distance of any two vertices in each connected-induced subgraph equals their distance in G) are chordal. Let C be a class of graphs and k a positive integer. The assertion ‘G k ∈ C implies k+2 ∈ C’ (denoted by G k ∈ C ⇒ G k+2 ∈ C) is therefore called the Duchet-type asG sertion. Besides, two analogies of Duchet-type assertion that a graph class is weakly closed under power operations (G ∈C ⇒ G k ∈C) or strongly closed under power operations (G k ∈C ⇒ G k+1 ∈C) are also concerned in other papers [4,6,12,13,15,20,22,23]. Table 1 summarizes the existing results for the above three type assertions in various classes of graphs. For an overview of these classes of graphs, refer to [5,16]. An independent set of a graph G = (V; E) is a set of pairwise nonadjacent vertices. The open neighborhood NG (u) of a vertex u∈V is the set {v∈V | (u; v)∈E}; and the closed neighborhood NG [u] is NG (u) ∪ {u}. For a subset W ⊂ V , we use G − W to denote the subgraph of G induced by the vertex set V \ W (i.e., {v | v∈V and v 6∈ W }). An asteroidal triple (AT) of a graph G is an independent set of three vertices such that every pair of vertices are joined by a path avoiding the closed neighborhood of the third. Graphs without asteroidal triples are called asteroidal triple-free graphs (AT-free graphs). Lekkerkerker and Boland [19] rst introduced the concept of asteroidal triples to characterize the interval graphs. A graph is an interval graph if and only if it is chordal and AT-free. Recently, Corneil et al. [10] obtained a collection of interesting structural and algorithmic properties for AT-free graphs. Note that the class of AT-free graphs properly contains well-known classes of graphs such as interval, permutation, trapezoid and cocomparability. For further results on AT-free graphs please refer to [7,9 –11,18,21]. Walter [24] generalized the concept of asteroidal triples to the so-called asteroidal sets, and used asteroidal sets to characterize certain subclasses of chordal graphs. A set of vertices A ⊆ V is an asteroidal set if for every vertex a ∈ A, the set A \{a} is contained in one connected component of G − NG [a]. The asteroidal number of a graph G is the maximum cardinality of an asteroidal set in G. It is easy to see that
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each asteroidal set is an independent set and the AT-free graphs are those graphs with asteroidal number at most two. Kloks et al. [17] showed that the asteroidal number can be computed by ecient algorithms for some classes of graphs such as claw-free, HHD-free, circular-arc and circular permutation graphs, while the corresponding decision problem is NP-complete even when restricted to triangle-free 3-connected 3-regular planar graphs. Let s¿2 be a positive integer. The class of graphs with asteroidal number at most s is denoted by A(s). Obviously, all the classes of graphs with bounded asteroidal number constitute a hierarchy of families by set inclusion, i.e., A(s) ⊂ A(s + 1). In this paper, we show that each graph class A(s) for s¿2 is strongly closed under power operations (i.e., G k ∈ A(s) ⇒ G k+1 ∈ A(s)). This covers the previous result of [22] for the family of AT-free graphs. Besides, we consider the certain forbidden con gurations for the powers of graphs G ∈ A(s). We show that every proper power (G k for k¿2) of a graph G ∈A(s) does not contain K1; s+2 , K2; s+1 and C2s+1 induced subgraphs. The rst two forbidden con gurations generalize a partial result proposed in [13] that every proper power of a cocomparability graph has no K1; 4 and K2; 3 induced subgraphs. Moreover, based on these forbidden con gurations, we show that all proper powers of AT-free graphs are perfect.
2. Graphs with bounded asteroidal number and their powers All graphs considered in this paper are undirected, simple (i.e., without loops and multiple edges) and connected. For a graph G = (V; E), the distance dG (u; v) of two vertices u; v ∈ V is the number of edges of a shortest path from u to v in G. For convenience, we write G k = (V; E k ) where E k = {(u; v) | u; v ∈ V and dG (u; v)6k}. In particular, we call G 2 the square of G, and G k for k¿2 the proper power of G. A path joining two vertices u and v is termed a u-v path. The union of two paths P and P 0 with a common endvertex is denoted by P ⊕ P 0 . A vertex u misses a path P if there are no vertices of P adjacent to u; otherwise, we say that u intercepts P. For any two vertices u; v ∈ V , we denote DG (u; v) as the set of vertices that intercept all u-v paths in G. Note that a vertex w 6∈ DG (u; v) if and only if there exists a u-v path P in G such that w misses P. For graph–theoretic terminologies and notations not mentioned here we refer to [16]. Kloks et al. [17] show that a set A with cardinality |A|¿3 is an asteroidal set if and only if every triple of A is an AT. Based on this result, we have the following. Lemma 1. A set A with |A|¿3 is an asteroidal set of a graph G if and only if every three vertices u; v; w ∈A; u 6∈ DG (v; w); v 6∈ DG (u; w) and w 6∈ DG (u; v). Lemma 2. For every two vertices x and y of a graph G = (V; E); DGk (x; y) ⊆ DGk+1 (x; y).
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Proof. Let w ∈ V \{x; y}. We will prove that if w 6∈ DGk+1 (x; y) then w 6∈ DGk (x; y). Suppose that w misses some x-y path P = (x = u1 ; u2 ; : : : ; un = y) in G k+1 . Without loss of generality we assume that P is as short as possible. Clearly, w 6∈ {ui | 16i6n} and dGk (w; ui )¿dGk+1 (w; ui )¿2
for i = 1; : : : ; n:
(1)
Let F be the set consisting of edges of P that belong to E k+1 \E k . If F is empty, then P is still an x-y path in G k . In this case, since w misses P in G k , no further proof is necessary. For the remainder of the proof, we assume that F is nonempty. For each edge (ui ; ui+1 ) ∈ F, let P(ui ; ui+1 ) be a shortest path with length k + 1 joining ui and ui+1 in G. Let zi be the vertex in P(ui ; ui+1 ) that is adjacent to ui in G. It is easy to verify that zi cannot appear in P and (ui ; zi ); (zi ; ui+1 )∈E k . We claim that for any two edges (ui ; ui+1 ); (uj ; uj+1 )∈F, the corresponding vertices zi and zj are not the same vertex of G. Suppose not, and without loss of generality assume i ¡ j. Since zi and zj are the same vertex of G dG (ui ; uj+1 )6dG (ui ; zi ) + dG (zi ; uj+1 ) = dG (ui ; zi ) + dG (zj ; uj+1 ) = 1 + k: This implies that (ui ; uj+1 )∈E k+1 . Thus, (x = u1 ; : : : ; ui ; uj+1 ; : : : ; un = y) forms another x-y path with length less than P in G k+1 such that w misses the path. This leads a contradiction to the assumption. Next, we claim that for each edge (ui ; ui+1 ) ∈ F, the corresponding vertex zi is not adjacent to w in G k . It can be derived from (1) by the following implications: dGk+1 (w; ui )¿2 ⇒ dG (w; ui ) ¿ k + 1 ⇒ dG (w; zi ) + dG (zi ; ui ) ¿ k + 1 ⇒ dG (w; zi ) ¿ k ⇒ dGk (w; zi )¿2:
(2)
Hence, for each edge (ui ; ui+1 )∈F in P, substitute the edges (ui ; zi ) and (zi ; ui+1 ). The replacements of all these edges yield another x-y path P 0 with length n − 1 + |F| in G k+1 . Since (ui ; zi ); (zi ; ui+1 )∈E k , P 0 remains in G k . We conclude by (1) and (2) that no vertices of P 0 are adjacent to w in G k . Thus, w 6∈ DGk (x; y). This completes the proof. Theorem 3. Every asteroidal set of G k+1 is also an asteroidal set in G k . Proof. Since any two nonadjacent vertices of G k+1 are also nonadjacent in G k , the result is trivial for every asteroidal set of size 2. Let A be any asteroidal set of G k+1 with |A|¿3 and assume that u; v; w ∈A. By Lemma 2, if u misses a v-w path in G k+1 then u also misses a v-w path in G k . Thus, A is an asteroidal set of G k . Corollary 4. If G k ∈ A(s) for s¿2 then G k+1 ∈ A(s). Speciÿcally; if G k is AT-free then so is G k+1 .
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3. Forbidden conÿgurations In this section, the certain forbidden con gurations for the proper powers of graphs G ∈A(s) are considered. Part of these results will be used in the following section to prove that all proper powers of AT-free graphs are perfect. The following lemma is helpful for testing that a vertex v misses a certain shortest path in a graph G. Lemma 5. Let G = (V; E) be a graph and u; v; w ∈V . For any positive integer k¿2; if dG (u; v)¿2 and dG (v; w) ¿ k¿dG (u; w); then v misses every shortest u-w path in G. Proof. The lemma is trivial for the case (u; w) ∈ E. We now consider dG (u; w)¿2. Let P(u; w) be a shortest path joining u and w in G. Assume that there is a vertex x ∈NG [v] which is contained in P(u; w). Since dG (u; v)¿2 and dG (v; w) ¿ k¿dG (u; w), x 6∈ {u; v; w} and dG (x; w) ¡ k. Thus, dG (v; w)6dG (v; x) + dG (x; w)61 + (k − 1) = k, a contradiction. A graph G is a cocomparability graph if its complement G is transitively orientable. Recall the previous result proposed by Damaschke [13] that every proper power of cocomparability graphs has no induced subgraphs isomorphic to K1; 4 or K2; 3 . Since the cocomparability graphs are properly contained in the class of AT-free graphs, the following two lemmas generalize the result of [13]. Lemma 6. Every proper power G k of a graph G ∈ A(s) for s¿2 does not contain K1; s+2 as an induced subgraph. Proof. Suppose that the lemma is false. Let W = {x; y1 ; y2 ; : : : ; ys+1 ; ys+2 } be a set of vertices that induces a K1; s+2 in G k where yi for i = 1; : : : ; s + 2 are independent vertices. Then, for i; j ∈ {1; : : : ; s + 2} with i 6= j, dG (yi ; yj ) ¿ k¿dG (yi ; x). Because the subgraph induced by W in G k certainly does not include any two adjacent edges belonging to E, assume without loss of generality that ys+2 is the only vertex which may be adjacent to x in G, and dG (yi ; x)¿2 for i=1; : : : ; s+1. Let P(yi ; x) be a shortest path joining yi and x in G and let Y = W \ {x; ys+2 }. We now show that Y forms an asteroidal set of s + 1 vertices in G. For any two vertices yi ; yj ∈Y , since dG (x; yi )¿2 and dG (yi ; yj ) ¿ k¿dG (x; yj ), it follows from Lemma 5 that yi misses P(x; yj ) in G. Thus, for every three vertices yh ; yi ; yj ∈ Y in G, we conclude that yi misses P(yh ; x) ⊕ P(x; yj ), yj misses P(yh ; x) ⊕ P(x; yi ), and yh misses P(yi ; x) ⊕ P(x; yj ). By Lemma 1, Y is an asteroidal set of s + 1 vertices in G. Thus G 6∈ A(s) and the lemma follows. Lemma 7. Every proper power G k of a graph G ∈ A(s) for s¿2 does not contain K2; s+1 as an induced subgraph.
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Proof. Suppose that the lemma is false. Let X ∪ Y be a set of vertices that induces a K2; s+1 in G k where X = {x1 ; x2 } and Y = {y1 ; y2 ; : : : ; ys+1 } are two independent sets. Then dG (xi ; yj )6k for i ∈ {1; 2} and j ∈ {1; : : : ; s + 1}, dG (x1 ; x2 ) ¿ k, and dG (yi ; yj ) ¿ k for i; j ∈ {1; : : : ; s + 1} with i 6= j. We rst consider the subgraph K1; s+1 induced by the vertex set {x1 } ∪ Y in G k . By the same argument as the proof of Lemma 6, we may assume that dG (yi ; x1 )¿2 for i = 1; : : : ; s. Let Y 0 = Y \ {ys+1 }. Consequently, Y 0 forms an asteroidal set of s vertices in G, and for every two vertices yi ; yj ∈ Y 0 with i 6= j, yi 6∈ DG (yj ; x1 ) and yj 6∈ DG (yi ; x1 ). The aim of the following proof is to show that {x1 } ∪ Y 0 is an asteroidal set of s + 1 vertices in G. Hence, we only need to show that x1 6∈ DG (yi ; yj ) for yi ; yj ∈Y 0 . Let P(yi ; x2 ) be a shortest path joining yi and x2 in G for 16i6s. For any vertex yi ∈ Y 0 , since dG (yi ; x1 )¿2 and dG (x1 ; x2 ) ¿ k¿dG (yi ; x2 ), it follows from Lemma 5 that x1 misses P(yi ; x2 ) in G. So that for any two vertices yi ; yj ∈ Y 0 , x1 misses P(yi ; x2 ) ⊕ P(x2 ; yj ) in G. Thus, x1 6∈ DG (yi ; yj ). Since {x1 } ∪ Y 0 forms an asteroidal set of s + 1 vertices in G, it contradicts that G ∈A(s) and the lemma follows. Note that if a graph G is Cn -free (no chordless cycle of length n), it does not imply that every proper power of G remains Cn -free. For example, the square of a chordal graph is not necessarily chordal. Since every cycle C2s+2 for s¿2 contains an asteroidal set of size s + 1, a graph G ∈ A(s) has no such a cycle. In the following, we will show that every proper power G k of a graph G ∈A(s) contains no chordless cycle of length 2s + 1. Lemma 8. Every proper power G k of a graph G ∈ A(s) for s¿2 does not contain C2s+1 as an induced subgraph. Proof. Suppose that the lemma is false. Let C = (u0 ; u1 ; : : : ; u2s ; u0 ) be a chordless cycle in G k for some k¿2. Clearly, dG (ui ; uj )6k for |i − j| = 1 and dG (ui ; uj ) ¿ k for |i − j|¿2 (where the indices of the terms ui and uj are always taken modulo 2s + 1). Since C does not include any two consecutive edges belonging to E, without loss of generality we assume that dG (u0 ; u1 )¿2. Let W = {u1 ; u3 ; u5 ; : : : ; u2s−1 ; u0 }. We will show that W forms an asteroidal set of s + 1 vertices in G. For any three distinct vertices uh ; ui ; uj ∈W , let PC (ui ; uj ; uh ) denote the path joining ui and uj and avoiding uh in C. Since C is chordless, it follows that for every three vertices uh ; ui ; uj ∈ W in G k with |h − i|¿2, |i − j|¿2, and |h − j|¿2, uh misses PC (ui ; uj ; uh ), ui misses PC (uj ; uh ; ui ), and uj misses PC (uh ; ui ; uj ). Thus, {uh ; ui ; uj } forms an AT in G k . By Theorem 3, {uh ; ui ; uj } is also an AT in G. To complete the proof, we only need to verify that for every vertex uh ∈W \{u0 ; u1 }, the set {u0 ; u1 ; uh } constitutes an AT in G. Let P(ui ; uj ) be a shortest path joining ui and uj in G for i 6= j. For every vertex uh ∈W \ {u0 ; u1 }, since dG (u0 ; uh ) ¿ k and dG (uh ; u1 ) ¿ k¿dG (u0 ; u1 ), it follows from Lemma 5 that uh misses P(u0 ; u1 ) in G. Using the same argument as above, a similar proof shows that u0 misses P(ui−1 ; ui ) for i = 3; : : : ; h, and u1 misses P(uj ; uj+1 ) for j = h; : : : ; 2s − 1, respectively, in G. In particular, since dG (u0 ; u1 )¿2, u0 misses
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P(u1 ; u2 ), and u1 misses P(u2s ; u0 ) in G. Therefore, u0 misses the path P(u1 ; u2 ) ⊕ P(u2 ; u3 ) ⊕ · · · ⊕ P(uh−1 ; uh ), and u1 misses the path P(uh ; uh+1 ) ⊕ P(uh+1 ; uh+2 ) ⊕ · · · ⊕ P(u2s ; u0 ). Consequently, {u0 ; u1 ; uh } forms an AT in G. This completes the proof. We summarize the above results as the following theorem. Theorem 9. Let G be a graph in the class A(s) for s¿2. Then; every proper power G k has no K1; s+2 ; K2; s+1 and C2s+1 induced subgraphs. 4. Powers of AT-free graphs A graph G is called perfect if for every induced subgraph H of G, the chromatic number of H (i.e., the minimum number of colours is necessary to colour the vertices of H such that any two adjacent vertices have dierent colours) equals the largest size of a clique of H . For background results on perfect graphs (see [3,16]). A hole is a chordless cycle of length at least four, while an antihole is the complement of a hole. Berge [2] conjectured that a graph is perfect if and only if it contains no odd hole and no odd antihole. The ‘only if ’ part of this conjecture is easy to check. However, the converse remains open today and is referred to as the Strong Perfect Graph Conjecture (abbreviated SPGC). In the past few years, the SPGC has been proved to be valid when restricted to certain special classes of graphs. Since the odd antihole with at least ve vertices is AT-free, the AT-free graphs need not be perfect. However, the SPGC is true for AT-free graphs (see [10]). In the remainder, we will show that every proper power of an AT-free graph has no induced odd antihole. Using this property together with the forbidden con gurations given in the previous section, we will show that every proper power of an AT-free graph is perfect. Lemma 10. Let G = (V; E) be an AT-free graph and n¿5 be an odd integer. Then; every proper power G k does not contain C n as an induced subgraph. Proof. For the case n = 5, since C5 is isomorphic to its complement C 5 , the result follows directly from Lemma 7 that every proper power of AT-free graphs does not contain C 5 as an induced subgraph. For odd integer n¿7, we suppose to the contrary. Let (u0 ; u1 ; : : : ; un−1 ; u0 ) be a chordless cycle in the complement of G k , and let H be the subgraph of G k induced by the vertices ui for i = 0; : : : ; n − 1. That is, H is isomorphic to C n in which any two vertices ui and uj have the distance dG (ui ; uj )6k for |i − j|¿2 or dG (ui ; uj ) ¿ k for |i − j| = 1 (where the indices of the terms ui and uj are always taken modulo n). Let P(ui ; uj ) denote a shortest path joining ui and uj in G for i 6= j. We consider the following two cases. Case 1: at least one of edges (u0 ; u2 ) and (u0 ; un−2 ) in H does not belong to E. Without loss of generality we assume that 26dG (u0 ; u2 )6k. Clearly, the set {u0 ; u1 ; u2 } is an independent set in G. Since dG (u0 ; u1 ) ¿ k and dG (u1 ; u2 ) ¿ k¿dG (u0 ; u2 ), it
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follows from Lemma 5 that u1 misses P(u0 ; u2 ) in G. Similarly, we can show that u2 misses P(u1 ; u3 ) in G. Also, since dG (u0 ; u2 )¿2 and dG (u2 ; u3 ) ¿ k¿dG (u0 ; u3 ), it follows from Lemma 5 that u2 misses P(u0 ; u3 ) in G. Thus, u2 misses P(u0 ; u3 ) ⊕ P(u3 ; u1 ) in G. By the symmetric property of H , u0 misses P(u2 ; un−1 ) ⊕ P(un−1 ; u1 ) in G. So {u0 ; u1 ; u2 } forms an AT in G, a contradiction. Case 2: both the edges (u0 ; u2 ) and (u0 ; un−2 ) in H belong to E. Since the degree of u0 in H is even and no two vertices ui and ui+1 (i = 2; : : : ; n − 3) are adjacent to u0 in G, an easy argument shows that there exist at least two vertices um and um+1 for some m = 3; : : : ; n − 4 such that (u0 ; um ); (u0 ; um+1 ) 6∈ E; i.e., 26dG (u0 ; um ); dG (u0 ; um+1 )6k. Clearly, the set {u0 ; um ; um+1 } is an independent set in G. Since dG (u0 ; um )¿2 and dG (um ; um+1 ) ¿ k¿dG (u0 ; um+1 ), it follows from Lemma 5 that um misses P(u0 ; um+1 ) in G. Similarly, we can show that um+1 misses P(u0 ; um ) in G. Also, since 36m6n−4, dG (um ; u1 )6k and dG (um+1 ; u1 )6k. Thus, for j ∈ {m; m + 1}, we have dG (uj ; u0 )¿2 and dG (u0 ; u1 ) ¿ k¿dG (uj ; u1 ). This follows from Lemma 5 that u0 misses P(um ; u1 ) ⊕ P(u1 ; um+1 ) in G. So {u0 ; um ; um+1 } forms an AT in G, a contradiction. Corollary 4 shows that any proper power of an AT-free graphs remains AT-free. Since every chordless n-cycle for n¿6 contains an AT, AT-free graphs have no such a cycle. Thus, the following theorem directly follows from Lemma 8, Lemma 10 and the fact that the SPGC is true for AT-free graphs. Theorem 11. Every proper power of an AT-free graph is perfect. Consequently, all proper powers of AT-free graphs are perfect AT-free graphs. Since K1; 4 and K2; 3 are perfect AT-free graphs and by Theorem 9 every proper power of an AT-free graph has no K1; 4 and K2; 3 induced subgraphs, we conclude that the class of all proper powers of AT-free graphs is strictly contained in the class of perfect AT-free graphs. A previous result proposed in [8] showed that the class of cocomparability graphs is also strictly contained in the class of perfect AT-free graphs. Since the class of cocomparability graphs is the largest well-known subclass of perfect AT-free graphs, a natural question to ask is whether every proper power of an AT-free graph is a cocomparability graph. Acknowledgements The authors wish to thank Dr. Dieter Kratsch and two anonymous referees for many constructive comments that led to a much improved version of this work. References [1] H.-J. Bandelt, A. Henkmann, F. Nicolai, Powers of distance–hereditary graphs, Discrete Math. 145 (1995) 37–60. [2] C. Berge, Les problemes de coloration en theeorie des graphes, Publ. Inst. Statist. Univ. Paris 9 (1960) 123–160.
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