Chair-free Berge Graphs are Perfect - CiteSeerX

Chair-free Berge Graphs are Perfect Antonio Sassano

∗

Report 10.95, July 1995 (revised August 1996) e-mail address: [email protected]

This work was partially supported by MURST, Roma, Italy and Progetto Finalizzato Trasporti II, CNR, Italy. ∗ Universit`a di Roma “La Sapienza” - Dipartimento di Informatica e Sistemistica - via Buonarroti, 12 - 00185 Roma, Italy.

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Abstract A graph G is called Berge if neither G nor its complement contains a chordless cycle with an odd number of nodes. The famous Berge’s Strong Perfect Graph Conjecture asserts that every Berge graph is perfect. A chair is a graph with nodes {a, b, c, d, e} and edges {ab, bc, cd, eb}. We prove that a Berge graph with no induced chair (chair-free) is perfect or, equivalently, that the Strong Perfect Graph Conjecture is true for chair-free graphs.

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1

Introduction

We use the standard graph-theoretical terminology and notation as in [1] and [8]. The graphs we consider are connected, finite, undirected without loops and multiple edges. We denote a graph by G(V, E), where V is the node set and E the edge set of G. An edge e of G with endpoints u and v is also denoted by uv. For each subset W of V , we denote by G[W ] the graph induced in G by the set W and by NG (W ) the set of all nodes in V − W adjacent to some node in W . We simply write G − v instead of G[V − {v}] and NG (v) instead of NG ({v}). A path is a set of nodes {v1 , v2 , . . . , vn } of V such that vi vj ∈ E if and only if j = i + 1 and i = 1, . . . , n − 1. If n ≥ 4 and there exists the edge v1 vn , the set v1 , . . . , vn is said to be a hole. The length of a path (hole) is the number of its nodes. An odd hole is a hole of odd length p ≥ 5. We will call p-hole a hole of length p. A stable set in a graph is a set of pairwise nonadjacent nodes; a clique is a set of pairwise adjacent nodes. As usual, α(G) denotes the size of the largest stable set in a graph G and ω(G) denotes the size of the largest clique in G. We simply write ω(G)-clique and α(G)-stable to denote a maximum clique and a maximum stable set in G. We also denote by χ(G) the least number of stable sets which cover all the nodes of G (chromatic number) and by θ(G) the least number of cliques which cover all the nodes of G (clique cover number). Clearly, ω(G) ≤ χ(G) and α(G) ≤ θ(G). With each graph G(V, E) we associate a polytope ST AB(G) in IRV (the |V |dimensional real space). This polytope is defined as the convex hull of the incidence vectors of all stable sets in G (stable set polytope). It is well known that an inequality P a facet (respectively a ridge) of ST AB(G) if the v∈W av xv ≤ a0 with W ⊆ V defines P polytope F = {x ∈ ST AB(G) : v∈W av xv = a0 } has dimension |V | − 1 (respectively |V | − 2). A facet-defining inequality is said to be non-trivial if and only if the right hand side is non zero. C.Berge [1] proposed to call a graph G perfect if χ(H) = ω(H) for each induced subgraph H of G. He also formulated two conjectures: (C1) a graph is perfect if and only if none of its induced subgraphs is an odd hole or its complement (odd anti-hole); (C2) a graph is perfect if and only if its complement is perfect. The latter conjecture, known as the Weak Perfect Graph Conjecture, was proved by Lov´asz in 1971 [10] and is now referred to as the Perfect Graph Theorem. An immediate corollary of the Perfect Graph Theorem is that a graph G is perfect if and only if α(G′ ) = θ(G′ ) for every induced subgraph G′ of G. The first conjecture, also known as the Strong Perfect Graph Conjecture (SPGC), concerns the structure of those graphs that are not perfect but all of whose induced 3

x b

b

c

d x

a c

a

d

Figure 1: chair and anti-chair subgraphs are perfect (minimal imperfect graphs). After three decades this conjecture is still open and it is certainly one of the most challenging open problems in graph theory (see [2] for an extended survey). If we call Berge a graph which does not contain an induced odd hole or an induced odd anti-hole then the SPGC can be rephrased as follows: a graph is perfect if and only if it is Berge. A usual line of attack to the SPGC consists of showing that the conjecture holds for a restricted class of graphs whose members do not contain, as an induced subgraph, a given (usually of small size) graph F . Classes of graphs for which this approach has been successful are (to cite only the most important classes obtained by forbidding a single graph F ) the P4 -free graphs [22], the diamond-free graphs [24, 19], the K4 -free graphs [25], the claw-free graphs [18], the paw-free graphs [15], the pan-free graphs [16], the bull-free graphs [6] and the dart-free graphs [23] (and, of course, their complements). The main purpose of this paper is to add one more class of graphs to this list by showing that chair-free Berge graphs are perfect. A chair is the graph (fig. 1) with nodes {a, b, c, d, x} and edges {ab, bc, cd, xb}. The anti-chair is the complement of the chair (fig. 1). In the rest of the paper we will denote a chair (with nodes labelled as in figure 1) as (x : a, b, c, d). Analogously, we will denote an anti-chair labeled as that displayed in figure 1 as (x : a, b, c, d). The class of chair-free graphs properly contains the class of claw-free graphs and the class of P4 -free graphs. On the other hand, the class of anti-chair-free graphs properly contains the class of diamond-free graphs and the class of paw-free graphs. As a consequence, this paper provides a common generalization of the results of Tucker [25], Parthasarathy and Ravindra [18], Seinsche [22] and Olariu [15]. Interesting subclasses of chair-free graphs have been also studied by De Simone and Galluccio [7] who proved that a chair-free graph G with no induced gem (a graph with nodes {a, b, c, d, e} and edges {ab, bc, cd, ea, eb, ec, ed}) or with no induced net (a graph with nodes {a, b, c, d, e, f } and edges {ab, bc, ac, da, eb, f c}) are perfect if and only if they are Berge, and by Olariu [14] who proved that a chair-free graph with no induced 5-path (P5 -free) is perfect if and only if it is Berge.

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The proof of our main result relies on the synergy of two main theoretical tools: the structural properties of minimal imperfect graphs and the properties of the reduction operation defined by Lov´asz and Plummer in [12]. The properties of minimal imperfect graphs have been studied by several authors and their fundamental contributions revealed an highly symmetric structure of such graphs and of the associated stable set polytopes ST AB(G). In what follows we list only the main graphical and polyhedral properties related to perfection and minimal imperfection that will be used in the rest of the paper. (P1) A graph G is perfect if and only if every nontrivial facet of ST AB(G) is defined P by an inequality v∈K xv ≤ 1 (clique inequality) where K is a maximal clique in G (Chv´atal [4]). (P2) A minimal imperfect graph G = (V, E) has exactly |V | cliques of size ω(G) and |V | stable sets of size α(G). For each node v ∈ V the graph G − v can be partitioned into α(G) cliques of size ω(G) and ω(G) stable sets of size α(G). Moreover, for every clique C in G of size ω(G), there exists a unique stable set S in G of size P α(G) such that S ∩ C = ∅. Finally, the inequality v∈V xv ≤ α(G) defines a facet of ST AB(G). (Lov´asz and Padberg [11, 17]). (P3) A minimal imperfect graph G with α(G) = 3 is an odd hole (Tucker [24]). (P4) A minimal imperfect graph G does not contain a nonempty set of nodes C such that G[V − C] is disconnected and with the property that some vertex v in C is adjacent to all the remaining nodes of G (star-cutset with center v). (Chv´atal [5]). (P5) Two non-adjacent nodes u and v of a minimal imperfect graph G with the property that NG (u) ∩ NG (v) contains a clique of size ω(G) − 1 are called a co-critical pair. A minimal imperfect Berge graph does not contain a set of nodes {v1 , . . . , vq } with the property that v1 vq ∈ E and {vi , vi+1 } is a co-critical pair for i = 1, . . . , q − 1 (Seb˝o [21]). As we remarked above, the second basic ingredient of our approach is the concept of clique reduction. A graph G|C = (V |C, E|C) is called the reduction of a graph G with respect to a maximal clique C [12] if and only if: (i) V |C = V − C; (ii) E|C = (E − {uv ∈ E : u ∈ C}) ∪ {uv : {u, v} ⊆ V − C, NG ({u, v}) ⊇ C}.

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It follows that the edges of G|C are either edges of G or “new” edges. We shall say that every edge uv ∈ E|C − E is a false edge of G|C. In [12] it is observed that if α(NG (C)) ≤ 2 then α(G|C) = α(G) − 1. Lov`asz and Plummer called reducible a clique C with α(NG (C)) ≤ 2. We slightly extend this definition by calling reducible any clique with the property that α(G|C) = α(G) − 1. Definition 1.1 A maximal clique C is reducible in G if and only if α(G|C) = α(G)−1.

The properties of reducible cliques have been first exploited by Lov´asz and Plummer in their polynomial algorithm for computing the stability number of claw-free graphs [12]. Subsequently, this operation has been crucial for deriving the description of the rank facets of the polytope ST AB(G) associated with a claw-free graph G [9], for designing a polynomial algorithm that finds the stability number of a chair and bull-free graph [8] and for computing good upper bounds for α(G) in general graphs [13]. Also in the realm of perfect graphs the reduction operation has interesting consequences. In fact, in [20] it is proved that the reduction of a maximum reducible clique in a minimal imperfect graph produces an imperfect graph. As remarked above, building on this result De Simone and Galluccio [7] proved that for the class of chair-and-gemfree graphs the SPGC is true. In all the above mentioned papers the crucial result is that the reduction of a reducible clique in a minimal imperfect graph produces an imperfect graph. Since this result is crucial also in this paper we report here, for sake of completeness, both its statement and its proof. Theorem 1.2 ([20]) Let G = (V, E) be a minimal imperfect graph with α(G) ≥ 3. If C is a reducible clique of size ω(G) then G|C is an imperfect graph. Proof. Suppose, conversely, that G|C is perfect. Since G is minimally imperfect, (P2) implies that the inequality X

xv ≤ α(G)

(1)

v∈V

defines a facet of ST AB(G). Moreover, property (P2) also implies that there exist exactly n = |V | stable sets in G of size α(G) whose incidence vectors satisfy (1) as an equality. Let M be the matrix whose rows are such incidence vectors. Let M ′ be the submatrix of M whose columns are indexed by the nodes of V |C and whose rows are the incidence vectors of the stable sets in G|C of size α(G) − 1. Property (P2) implies that there exists a unique maximum stable set in G which has an empty intersection with the maximum clique C. Moreover, by definition of reduction, every maximum stable set S of G with a non-empty intersection with C has the property that S − C is a maximum stable set of G|C (of size α(G) − 1). It follows 6

that M ′ has n − 1 rows. Write n′ = |V |C|. Since M has full rank, we have that n′ ≥ rank(M ′ ) ≥ n′ − 1. As a consequence, the inequality X

xv ≤ α(G|C)

(2)

v∈V |C

defines either a facet or a ridge of ST AB(G|C). Since the graph G|C is perfect, property (P1) implies that every non-trivial facet of ST AB(G|C) is defined by a clique inequality. Since α(G|C) ≥ 2, it follows that (2) defines a ridge of ST AB(G|C), and so (2) is a linear combination of two clique inequalities. Let K1 and K2 be the cliques in G|C corresponding to such inequalities. Clearly K1 ∪ K2 ⊇ V |C, and so θ(G|C) ≤ 2. Now, since G|C is perfect, and since α(G|C) ≥ 2, it follows that α(G|C) = θ(G|C) = 2, and so α(G) = 3. Hence, property (P3) implies that G is isomorphic to a 7-hole and so C is a clique of size 2 in G. Then G|C is isomorphic to a 5-hole, contradicting the assumption. The paper is organized as follows. In Section 2 we prove some general properties of chair and anti-chair free graphs related to the reduction operation. In Section 3 we prove that the strong perfect graph conjecture holds for chair free graphs.

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Properties of Chair and Anti-Chair-free graphs

As observed in the introduction, the proof of our main result relies on the properties of the reduction operation when applied to minimal imperfect chair and anti-chair free graphs. In this section, we first review some useful results concerning the structure of graphs obtained from chair and anti-chair free graphs by means of the reduction operation. Then we describe two useful properties of minimal imperfect anti-chair free graphs. We first introduce a new reduction operation: the reduction of a graph G with respect to a maximal stable set S. This operation consists of taking the complement of the reduction of G with respect to the clique S in G. Hence, the reduction of G with respect to a maximal stable set produces the graph G|S = G|S with node-set V |S = V − S and edge-set: E|S = {uv ∈ E : {u, v, s} induces a triangle in G for some s ∈ S}. In other words, the graph G|S is obtained from G by taking the subgraph induced by V − S and removing all the edges that do not form a triangle with some node s ∈ S. The edges of G that are removed in this construction will be called vanishing edges with respect to S (or simply vanishing edges whenever the reference to S is not ambiguous). Note that uv is a vanishing edge of G with respect to S if and only if it is a false edge of G|S. 7

∉E

false

∈E

v

u

v

u

v

z

w

z

w

z

w=u

(a)

(b)

(c)

Figure 2: Forbidden configurations for the reduction of a chair-free graph. A stable set is said to be reducible in G if and only if ω(G|S) = ω(G) − 1 (i.e. if and only if S is a reducible clique in G). In what follows we will use the same notation G|T (V |T, E|T ) to denote both the reduction of a graph G with respect to a clique or a stable set. It will be the nature of the subgraph induced by T in G that will specify the operation. Lemma 2.1 Let G(V, E) be a chair-free graph and let C be a maximal clique in G. Then G|C does not contain two false edges uv and wz (possibly with u ≡ w) with the property that uz ∈ / E|C, wu ∈ / E|C and vz ∈ / E|C. Proof. Suppose, conversely, that G|C contains two false edges {uv, wz} ⊆ E|C − E with the above properties. If w ≡ u (fig. 2 (a)), let w¯ a node in C − NG (w) (w¯ exists since w ∈ / C and C is a maximum clique in G). Since wz and wv are false we have that wz ¯ ∈ E and wv ¯ ∈ E. Moreover, since vz ∈ / E|C we have that there exists a node x ∈ C − {w} ¯ such that xv ∈ / E and xz ∈ / E. Now, since wv and wz are false, we have that xw ∈ E, and so (z : v, w, ¯ x, w) is a chair in G, a contradiction. Suppose, conversely, that w 6= u (fig. 2 (b,c)). Since vw ∈ / E|C we have that there exists a node y ∈ C such that yv ∈ / E and yw ∈ / E and, since uv and wz are false, we have that yu ∈ E and yz ∈ E. Moreover, since uz ∈ / E|C we have that there exists a node x ∈ C such that xu ∈ / E and xz ∈ / E. It follows that x 6= y and, since uv and wz are false, that xv ∈ E and wx ∈ E. But then (v : w, x, y, z) is a chair in G, a contradiction. The above lemma can be equivalently phrased in terms of reduction of a stable set in an anti-chair-free graph. Since in what follows we shall also use such a rephrasing, we prefer to state it formally (see figure 3): 8

∉E

vanishing

∈E

v

u

v

u

v

z

w

z

w

z

w=u

(a)

(b)

(c)

Figure 3: Forbidden configurations for an anti-chair-free graph Lemma 2.2 Let G(V, E) be an anti-chair-free graph and let S be a maximal stable set in G. Then G does not contain four nodes {u, v, w, u} (possibly with w ≡ u) with the property that the edges uv and wz are vanishing with respect to S and the edges uz, wu (if w 6≡ u) and vz belong to E and are not vanishing with respect to S. Another useful lemma is the following: Lemma 2.3 Let G(V, E) be an anti-chair-free graph and let P = {v1 , . . . , vk } with k ≥ 4 be an induced path (or an induced hole with k ≥ 5) in G. If there exists a node w ∈ /P adjacent to three consecutive nodes of P then P ⊆ NG (w). Proof. Suppose that a node w with the above properties exists and let vi+2 , vi+1 and vi be three consecutive nodes of P that are adjacent to w. If P − NG (w) 6= ∅ we can assume, without loss of generality, that vi−1 w ∈ / E, but then (vi−1 : vi , vi+1 , w, vi+2 ) is an anti-chair in G, a contradiction. Lemma 2.4 A minimally imperfect anti-chair free graph G(V, E) does not contain an induced graph isomorphic to F1 (figure 4). Proof. Suppose, conversely, that there exists a node x ∈ V which coincides with the node u¯ of some induced subgraph of G isomorphic to F1 . Since G is minimally imperfect, it does not contain a star-cutset with center x. It follows that the graph G[V − NG (x)] is connected, and so, for each induced subgraph of G isomorphic to F1 and with x ≡ u¯, there exists a path connecting w to u. Let H(W, F ) be the graph isomorphic to F1 with x ≡ u¯ and let P = (w ≡ y0 , y1 , . . . , yt , yt+1 ≡ u) (with t ≥ 1) be a shortest path connecting w to u in G[V − NG (x)]. 9

w

u

_ z

v

_ v

_ u

z

F1

Figure 4: Consider first the node yt and suppose that it is adjacent to both v¯ and z¯. It follows that z¯yt−1 ∈ / E and v¯yt−1 ∈ / E. In fact, if v¯ (¯ z ) is adjacent to the three consecutive nodes yt+1 , yt , yt−1 of P then t ≥ 2 and, by Lemma 2.3, v¯ (¯ z ) is also adjacent to w, a contradiction. As a consequence, we have that (yt−1 : yt , v¯, z¯, u¯}) is an anti-chair in G, a contradiction. Hence, we have that yt cannot be adjacent to both v¯ and z¯. Assume, without loss of generality, that yt z¯ ∈ / E. It follows that yt v¯ ∈ E (else (yt : u, v¯, z¯, u¯) is an anti-chair in G) and, consequently, that yt v ∈ E (else (yt : v¯, u¯, z¯, v) is an anti-chair in G). Moreover, we have that yt z ∈ / E, else the set {yt , u, z¯, u¯, z} induces a 5-hole in G and contradicts the hypothesis that G is minimally imperfect. Consider now the node yt−1 . If t = 1 then yt−1 ≡ w. In this case (w : yt , u, v¯, z¯) is an anti-chair in G, a contradiction. Hence, t ≥ 2 and yt−1 6= w. If v¯yt−1 ∈ E then we have that v¯ is adjacent to three consecutive nodes of P , namely {u ≡ yt+1 , yt , yt−1 }. Hence, by Lemma 2.3, we have that v¯w ∈ E, a contradiction. It / E, and so z¯yt−1 ∈ E (else (yt−1 : yt , u, v¯, z¯) is an anti-chair in G). follows that v¯yt−1 ∈ Consequently, we have that zyt−1 ∈ E (else (yt−1 : z¯, u¯, v¯, z) is an anti-chair in G) Now, if yt−1 v ∈ / E we have that the set {yt−1 , yt , v, u¯, z} induces a 5-hole in G, contradicting the assumption that G is minimally imperfect. It follows that yt−1 v ∈ E. Finally, consider the node yt−2 . If yt−2 ≡ w then (w : yt−1 , z¯, v, u¯) is an anti-chair in G, a contradiction. Hence yt−2 6≡ w and t ≥ 3. We have that yt−2 v ∈ / E, otherwise the node v would be adjacent to three consecutive nodes of P , namely {yt , yt−1 , yt−2 }, and, by Lemma 2.3, we would have wv ∈ E, which would contradict our assumptions. It follows that yt−2 z¯ ∈ E (else (yt−2 : yt−1 , z¯, v, u¯) is an anti-chair in G). Moreover, we have that yt−2 v¯ ∈ E (else (yt−2 : z¯, u, v¯, yt ) is an antichair in G). But then the set {yt−2 , yt−1 , v, u¯, v¯} induces a 5-hole in G, a contradiction.

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3

The SPGC holds for Chair-free graphs

In this section we will prove the three basic results that will immediately imply our main theorem. Namely, we will first show that every minimal imperfect chair-free graph contains a reducible clique. Then we will prove that the reduction of a minimally imperfect chair-free graph which is Berge cannot produce a hole or an anti-hole. Finally, we will show that the property of being chair-free is “weakly” hereditary with respect to the reduction operation. In fact, we will not prove that G|C is chair-free if G has this property, but the weaker result that every minimally imperfect Berge subgraph of G|C is chair-free if G is minimally imperfect, chair-free and Berge. The three results, along with the result expressed by Theorem 1.2, will be sufficient to prove that the Strong Perfect Graph Conjecture holds for chair-free graphs. Theorem 3.1 If G(V, E) is a minimal imperfect chair-free graph then G contains a reducible clique of size ω. Proof. Suppose, conversely, that every ω-clique of G is not reducible. We have the following: Claim 1. Every maximum stable set is reducible in G. Let S be a maximum stable set of G. By (P2) we have that there exists a unique maximum clique C of G with the property that C ∩ S = ∅. Since, by assumption, C is not reducible, we have that α(G|C) = α(G), and so S must be a maximum stable set of G|C of cardinality α(G). It follows that for each pair {u, v} of S there exists a node z ∈ C which is not adjacent to both u and v. Suppose now that S is not reducible. It follows that ω(G|S) = ω(G) and, consequently, that every edge of G with both endnodes in C is not vanishing. Choose, among all the pairs of nodes of S, a pair {u, v} which maximizes the quantity |NG ({u, v}) ∩ C|. Since C is not reducible, we have that there exists a node z ∈ C − NG ({u, v}). In addition, the maximality of S implies that there exists a node w ∈ S − {u, v} which is adjacent to z. Now, since |NG ({u, v}) ∩ C| ≥ |NG ({u, w}) ∩ C| and z ∈ NG ({u, w}) − NG ({u, v}), we have that there exists a node y ∈ (NG (v) − NG ({u, w})) ∩ C. Since the edge zy is not vanishing with respect to S, we have that there exists a node w′ ∈ S − {u, v, w} which is adjacent to both z and y. Again, since |NG ({u, v}) ∩ C| ≥ |NG ({v, w′ }) ∩ C| and z ∈ NG ({v, w′ }) − NG ({u, v}), we have that there exists a node t ∈ (NG (u) − NG ({v, w′ })) ∩ C. But then (w′ : v, y, t, u) is a chair in G, a contradiction. (End of Claim 1). Claim 1 asserts that if G is a minimally imperfect chair-free graph and all of its ω(G)cliques are not reducible then every maximum clique of G (which is a minimally imperfect anti-chair-free graph) is reducible. In Claim 4 we will show that if G is a minimally imperfect chair-free graph and all of its ω(G)-cliques are not reducible then its complement contains at least one maximum clique which is not reducible. Hence, the theorem 11

will follow by contradiction. To prove Claim 4 we first need to prove two preliminary claims. Claim 2. Let C be a ω-clique in a minimally imperfect anti-chair free graph G(V, E) and let S be the unique α-stable set with the property that C ∩ S = ∅. If S is not reducible in G then each triple {u, v, w} ⊆ S, with the property that uv is a false edge of G|C, induces a connected subgraph in G|C. Suppose, conversely, that there exists a triple {u, v, w} ⊆ S such that uv is false, wu ∈ / E|C and wv ∈ / E|C. Since wu ∈ / E|C we have that there exists a node in C, say u¯, with the property that u¯ u∈ / E and w¯ u∈ / E. Analogously, since wv ∈ / E|C, there exists a node v¯ in C with the property that v¯ v∈ / E and w¯ v∈ / E. Moreover, since uv is false, we have that u¯ 6= v¯ and v¯ u ∈ E and u¯ v ∈ E. Now, since S is not reducible in G we have that ω(G|S) = ω(G), and so C is a maximum clique in G|S. It follows that u¯v¯ is not vanishing with respect to S and, consequently, that there exists a node z ∈ S which is adjacent to both u¯ and v¯ in G. As a consequence we have that z ∈ S − {u, v, w}. Moreover, since C is a maximum clique in G, we have that there exists a node z¯ ∈ C with the property that z z¯ ∈ / E. If u¯ z ∈ / E then v¯ z ∈ E (since uv is false); but then (u : v¯, u¯, z¯, v) is an anti-chair in G. It follows that u¯ z ∈ E and, consequently, that v¯ z ∈ E (else (v : u¯, v¯, z¯, u) is an anti-chair in G). Moreover, we have that w¯ z∈ / E (else (w : z¯, u¯, v¯, z) is an anti-chair in G). In other words, the node set H = {u, v, z, u¯, v¯, z¯ w} induces in G a graph isomorphic to F1 , contradicting Lemma 2.4. (End of Claim 2). Claim 3. Let C be a reducible clique in a minimally imperfect anti-chair free graph G(V, E). If uv is a false edge in G|C then there exists a (unique) node u¯ in C with the u∈ / E and uz ∈ E for each z ∈ C − {¯ u}. As a consequence, {u, u¯} is a property that u¯ co-critical pair. (see (P5)). Since C is a maximal clique in G we have that there exist a node u¯ in C with the property that u¯ u∈ / E. Moreover, since uv is a false edge in G|C, we have that v¯ u ∈ E. Symmetrically, we have that there exists a node v¯ ∈ C − {¯ u} with the property that v¯v ∈ / E and v¯ u ∈ E. Suppose now that there exists a node z ∈ C − {¯ u} with the property that zu ∈ / E. It follows that zv ∈ E (since uv is a false edge) and, consequently, that (u : v¯, u¯, z, v) is an anti-chair in G, a contradiction. (End of Claim 3). We are now ready to prove the result which contradicts Claim 1 and completes the proof. Claim 4. If G is a minimally imperfect chair-free graph and all of its ω(G)-cliques are not reducible then at least one clique of G is not reducible. Suppose, conversely, that every clique in G(V , E) is reducible. Let C be one of such cliques and let uv be a false edge in G|C. By Claim 3 we have that there exist two nodes in C, say u¯ and v¯, with the property that {u, u¯} and {v, v¯} are co-critical pairs in G. 12

According to (P2) the graph G−u can be partitioned into α(G) cliques of size ω(G). Let Cv be the clique in the partition which contains v. Then the node u belongs to the stable set Su with the property that Su ∩ Cv = ∅. Since Cv is reducible we have that G|Cv contains at least one false edge xy with {x, y} ⊆ Su . We claim that there exists at least one false edge of G|Cv with one endpoint in u. If u ≡ x or u ≡ y then we are done. Hence, suppose that u ∈ / {x, y}. Since Su induces in G a non-reducible clique, we have, by Claim 2, that at least one of the edges {yu, xu} is false in G|Cv . Hence, we can assume, without loss of generality, that uy is false in G|Cv . Then, ¯ and by Claim 3, the node v is the unique node of Cv with the property that uv ∈ / E, so the pair {u, v} is co-critical in G. But then, the set {¯ u, v, u, v¯} and Seb˝o’s theorem (property P5) imply that G is either an odd hole or an odd anti-hole, contradicting the assumption that all the ω(G)-cliques of G are not reducible. (End of Claim 4). In the next theorems we show that the reduction G|C of a minimal imperfect, Berge and chair-free graph with respect to a reducible clique G is Berge. In particular, in the case of odd holes we will be able to prove the stronger result that if G is a chair-free graph with no induced odd hole then G|C does not contain an odd hole. In the case of odd-anti-hole we cannot prove a similar result (which is indeed false) but we will be able to prove the weaker result that if G is a minimal imperfect chair-free Berge graph then G|C does not contain an induced odd anti-hole. Indeed, to simplify the technicalities of the proof, we will prove the equivalent result that if G is a minimal imperfect anti-chair-free Berge graph and S is a reducible stable set in G then G|S does not contain an induced odd hole. Theorem 3.2 Let G(V, E) be a chair-free graph with no induced odd-hole and let C be a reducible clique in G. Then G|C does not contain an induced odd-hole. Proof. If G|C contains an induced odd hole then there must exist an induced subgraph of G, say G′ , with the property that G′ |C is an odd hole. The graph G′ is chair-free and does not contain an induced odd hole, and so we can prove the theorem by showing that the reduction G|C of a chair-free graph G with no induced odd holes is not an odd hole. Suppose, conversely, that G|C is an odd hole (v1 , . . . , v2q+1 ) with q ≥ 2. Since G does not contain odd holes, we have that there exists at least one false edge in E|C. Assume, without loss of generality, that v1 v2 is such an edge. By Lemma 2.1 we have that the edge v1 v2 is the unique false edge of G|C. Now, since v2 v2q+1 ∈ / E|C, we have that there exists a node u ∈ C with the property that uv2 ∈ / E and uv2q+1 ∈ / E. Moreover, since v1 v2 is false, we have that uv1 ∈ E. Analogously, there exists a node w ∈ C − {u} with the property that wv1 ∈ / E, wv3 ∈ /E and wv2 ∈ E. It follows that the set T = {v1 , u, w, v2 , . . . , v2q+1 } induces an odd cycle in G. Since the set {v1 , . . . , v2q+1 } induces a path in G and the set T does not induce and odd hole 13

in G, we have that there exists an edge uvj with j ∈ {3, . . . , 2q} or an edge wvj with j ∈ {4, . . . , 2q + 1}. Suppose first that uvj ∈ E for some j ∈ {4, . . . , 2q}. It follows that wvj ∈ E (else (v1 : vj , u, w, v2 ) is a chair in G). Now, since v1 vj ∈ / E, we have that there exists a node z ∈ C with the property that zv1 ∈ / E and zvj ∈ / E. But then z ∈ C − {u, w} and, since v1 v2 is a false edge, we have that zv2 ∈ E. It follows that (vj : v1 , u, z, v2 ) is a chair in G, a contradiction. A symmetric argument shows that wvj ∈ / E for j ∈ {4, . . . , 2q}. Consequently, we have that uv3 ∈ E or wv2q+1 ∈ E. Indeed, both edges belong to E. In fact, if uv3 ∈ E and wv2q+1 ∈ / E then (v3 : w, u, v1 , v2q+1 ) is a chair in G while if wv2q+1 ∈ E and uv3 ∈ / E then (v2q+1 : v3 , v2 , w, u) is a chair in G. But then, since uvj ∈ / E for j ∈ {4, . . . , 2q} and wvj ∈ / E for j ∈ {4, . . . , 2q}, we have that the set {u, v3 , v4 , . . . , v2q+1 , w} induces an odd hole in G, a contradiction. In order to prove the next result we need a new definition and a recent result obtained by Conforti, Cornu´ejols, Kapoor and Vuskovic [3] which turned out to be crucial in our proof. Definition 3.3 ([3]) A Mickey Mouse with Long Ears is a cycle M M = {x1 , . . . , xq } having only two chords x1 xh and x1 xh+1 with 4 ≤ h ≤ q − 3. Theorem 3.4 ([3]) If a graph G(V, E) without odd holes contains a Mickey Mouse with Long Ears M M = {x1 , . . . , xq } having chords x1 xh and x1 xh+1 then the set: X = (NG (x1 ) − {x2 , xq }) ∪ (NG (xh ) ∩ NG (xh+1 )) is a cutset of G which separates {x2 , . . . , xh−1 } from {xh+2 , . . . , xq }. We are now ready to prove that the reduction of a stable set in a minimal imperfect graph G with no induced odd holes and no induced anti-chair cannot produce a graph with an odd hole. Indeed, in the statement of the theorem we could relax the assumption of minimal imperfection and replace it by the weaker requirement that G has no star cutset. Theorem 3.5 Let G(V, E) be a minimal imperfect anti-chair-free graph with no induced odd hole and let S be a reducible stable set in G. Then G|S does not contain an induced odd hole. Proof. Suppose, conversely, that G|S(V |S, E|S) contains an odd hole H = (W, F ) with W = {u1 , . . . , u2q+1 } and q ≥ 2. Since G has no induced odd hole, we have that ˆ = G[W ] contains at least one vanishing edge. More specifically, we have that H ˆ is H an odd cycle such that every chord (edge connecting two non-consecutive nodes) is a vanishing edge. Claim 1. The edge u1 u3 is vanishing. 14

uj

s

s' ui+2 =uk

ui+1 ui

ui+3 uh

vanishing

Figure 5: ˆ contains chord which connects two nodes at distance 2 on the cycle Evidently, if H (short chord) then we can assume, without loss of generality, that u1 u3 is vanishing and ˆ has no short chord. As a first consequence we are done. Suppose, conversely, that H we have that q ≥ 3, in fact every chord of a 5-hole is short. ˆ does not contain two chords uj uh and uj uh+1 with By Lemma 2.2 we have that H j + 3 ≤ h ≤ j − 3 (i.e. two chords with a common endpoint and with the other two ˆ has only vanishing endpoints adjacent in H). It follows that every triangle induced in H edges. ˆ is diamond-free. Claim 1.1. H ˆ contains a diamond D induced by the nodes {ui , uj , uh , uk } Suppose, conversely, that H ˆ has and with edge-set {ui uj , ui uh , uj uh , uj uk , uh uk }. Since every triangle induced in H only vanishing edges we have that all the edges of D are vanishing. As a consequence ˆ Moreover, since H ˆ has no short chord we have that all of them are chords of H. we have that ui+2 6= uj and ui+2 6= uh . In addition, by Lemma 2.2, we have that ui+1 uj , ui+1 uh ∈ / E. It follows that ui+1 uk ∈ E (else (ui+1 : ui , uj , uh , uk ) is an anti-chair in G). Now, let s be the node of S with the property that {s, ui , ui+1 } induces a triangle in G. We have that suj , suh ∈ / E (since ui uj and ui uh are vanishing edges). It follows that suk ∈ E (else (s : ui , uj , uh , uk ) is an anti-chair in G). But then, the edge ui+1 uk is not vanishing (since {s, ui+1 , uk } is a triangle in G) and, consequently, we have that ˆ has k = i + 2. Moreover, since ui+2 ui+3 is not vanishing and every triangle in H only vanishing edges, we have that ui+3 uj ∈ / E and that ui+3 uh ∈ / E. It follows that ui+3 ui ∈ E (else (ui+3 : ui+2 , uh , uj , ui ) is an anti-chair in G). Observe now that the edge ui+3 ui is vanishing, and so sui+3 ∈ / E. As a consequence, since ui+2 ui+3 is not vanishing, we have that there exists a node s′ ∈ S − {s} with the property that {s′ , ui+2 , ui+3 } induces a triangle in G. Now, since ui+2 ≡ uk and since uj uk and uh uk are vanishing edges, we have that s′ uj ∈ / E and s′ uh ∈ / E. Moreover, ′ ′ since ui+3 ui is vanishing, we have that s ui ∈ / E. Consequently, (s : uk , uj , uh , ui ) is an anti-chair in G, a contradiction. (End of Claim 1.1) 15

ˆ contains a Mickey Mouse with long ears {x, z1 , . . . , zs , yt , yt−1 , . . . , y1 , x} Claim 1.2. H with the additional property that the two chords xzs and xyt and the edge zs yt are vanishing. Let H ′ = {x1 , . . . , x2t+1 } be the smallest (minimum number of nodes) odd cycle induced ˆ with t ≥ 2 and with no short chord. Observe that H ′ exists since H ˆ has the above in H properties. Claim 1.2.1 There does not exist an index i with 1 ≤ i ≤ h − 2 (modulo 2t + 1) with ˆ the property that the set {xi , xi+1 , xh , xh+1 } induces a K4 in H. Suppose, conversely, that such a subset exists. We can immediately observe that h ≥ i + 4. In fact, if h = i + 2 then xi xh is a short chord while h = i + 3 implies that xi+1 xh is a short chord. Analogously, we have that i ≥ h + 4 (see figure 6 (a)). Let C1 = {xi+1 , . . . , xh } and C2 = {xh+1 , . . . , xi }. Since H ′ is an odd cycle we have that either C1 or C2 is an odd cycle. Suppose, without loss of generality that C1 is ˆ and that odd. We have that xi+2 xh ∈ / E (else (xh , xi , xi+1 , xi+2 ) is a diamond in H) ˆ xi+1 xh−1 ∈ / E (else (xi+1 , xh−1 , xh , xh+1 ) is a diamond in H). But then C1 is an odd cycle without short chords, contradicting the minimality of H ′ . (End of Claim 1.2.1). ˆ does not contain an induced odd hole we have that H ′ contains a triangle. Now, since H We claim that there exists a triangle in H ′ of the form {xi , xj , xj+1 } (short triangle). Suppose, conversely, that every triangle in H ′ is not short and let T = {xi , xj , xh } be a triangle in H ′ . The triangle T defines two cycles C1 = {xi , xi+1 , . . . , xj , xh , xh+1 , . . . , xi } and C2 = {xj , xj+1 , . . . , xh }. Since H ′ is odd we have that either C1 or C2 is odd (figure 6 (b)). Moreover, since H ′ has no short chord, we have that both C1 and C2 have length at least 4. Now, by the assumption that no triangle in H ′ is short, we have that the sets {xj−1 , xj , xh } and {xj , xh , xh+1 } do not induce triangles in H ′ , and so C1 has no short chord. It follows, by the minimality of H ′ , that C1 is an even cycle and, consequently, that C2 is an odd cycle. But then, since {xh−1 , xh , xj } and {xj+1 , xj , xh } do not induce triangles in H ′ , we have that C2 has no short chord, contradicting the minimality of H ′. It follows that there exists a short triangle T = {xi , xj , xj+1 } induced in H ′ . Without loss of generality we assume that i = 1. Moreover, by Claim 1.1 we have that ˆ and so x1 xj−1 ∈ (xj−1 , x1 , xj , xj+1 ) is not a diamond in H, / E. Analogously, since ˆ (xj+2 , x1 , xj , xj+1 ) is not a diamond in H, we have that x1 xj+2 ∈ / E. Since H ′ has no short chord we can assume, without loss of generality, that 4 ≤ j ≤ 2t − 2. Let P1 = {x1 , x2 , . . . , xj } and P2 = {xj+1 , . . . , x1 } and suppose that the set F of the chords of H ′ connecting nodes of P1 − {x1 } to nodes of P2 − {x1 } is not empty. Let xa be the node with highest index in P1 which is the endnode of a chord in F and let xb be the node with highest index in P2 − {x1 } ∩ NH ′ (xa ). Analogously, let xc be the node with lowest index in P2 which is the endnode of a chord in F and let xd be 16

xi

xi+1 xj+1

xi-1

xi+2

xj

xh

C1

C2

xj-1

xh-1 xh+2

x h-1

C2

x h+1

xh

∉Ε

(a)

C1

x h+1

xi

(b)

Figure 6: the node with lowest index in P1 − {x1 } ∩ NH ′ (xc ) (figure 7 (a)). Evidently, a ≥ d and b ≥ c. Let P3 = {xa , xa+1 , . . . , xj }, P4 = {xj+1 , . . . , . . . , xc }, P5 = {x1 , . . . , xd } and P6 = {xb , . . . , x1 }. (see figure 7 (a)) Since xb 6= x1 and xd 6= x1 we have that |P5 | ≥ 2 and |P6 | ≥ 2. If xa ≡ xj and xb ≡ ˆ contradicting x2t+1 then x2t+1 xj+1 ∈ E (else (x2t+1 , x1 , xj , xj+1 ) is a diamond in H), Claim 1.2.1. It follows that xj x2t+1 ∈ / E. A symmetric argument shows that xj+1 x2 ∈ / E. As a consequence, we have that |P3 ∪ P6 | ≥ 4 and that |P4 ∪ P5 | ≥ 4. Another consequence of what is proved above (see figure 7 (a)) is that xj x2 ∈ / E ˆ and xj+1 x2t+1 ∈ (else (x2 , x1 , xj , xj+1 ) is a diamond in H) / E (else (x1 , x2t+1 , xj+1 , xj ) is ˆ a diamond in H). It follows that P1 induces a cycle without short chords in G, and, by the minimality of H ′ , it must be an even cycle. Analogously, we have that P2 induces an even cycle without short chords in G. Claim 1.2.2. The sets {xa , x1 , x2t+1 } and {xc , x1 , x2 } do not induce triangles in G. Due to the symmetry of the two statements we will prove only the first one. To this purpose suppose, by contradiction, that xa x1 ∈ E and xa x2t+1 ∈ E. It follows that ˆ Moreover, we have that a 6= 2 xa 6≡ xj (else (xj , xj+1 , x1 , x2t+1 ) is a diamond in H). ′ (else x2t+1 x2 is a short chord of H ) and a 6= 3 (else x1 xa is a short chord of H ′ ). It follows that 4 ≤ a ≤ j − 1. Consider now the set C ′ = {x1 , . . . , xa } and suppose that xa−1 x1 ∈ E. It follows ˆ and, consequently, that that x2t+1 xa−1 ∈ E (else (x2t+1 , x1 , xa , xa−1 ) is a diamond in H) ˆ the set {xa−1 , xa , x2t+1 , x1 } induces a K4 in H, contradicting Claim 1.2.1. It follows that ˆ we xa−1 x1 ∈ / E. Moreover, since xa x2 ∈ / E (else (xa , x2t+1 , x1 , x2 ) is a diamond in H), ′ ˆ have that C induces a cycle without short chords in H. It follows, by the minimality ˆ of H ′ , that C ′ induces an even cycle in H. ′ Consider now the set P = P2 − {x1 } = {x2t+1 , x2t , . . . , xj+1 }. Since P2 is even, we have that P ′ is odd. As a consequence, the set C ′′ = C ′ ∪ P ′ induces an odd cycle in G 17

x j-1

xa xd

xj xj+1

P3

P4

P5

P6

(a)

x1

xj+2

xa

xc

x2

x j+1

xj

x j+2

xb

x 2t

xa-1

x2t+1

x2

x 2t+1 ∉Ε

(b)

x1

Figure 7: (see figure 7 (b)). Moreover, since H ′ has no short chord, we have that no short chord of C ′′ can have both endnodes in C ′ or P ′ . In addition, we have seen above that xj+1 x2 ∈ / E and xj+2 x1 ∈ / E. Finally, we have that xa−1 x2t+1 ∈ / E (else (xa−1 , xa , x2t+1 , x1 ) is a diamond ˆ and that xa x2t ∈ ˆ (see figure 7 (b)). in H) / E (else (xa , x2t , x2t+1 , x1 ) is a diamond in H) ′′ ˆ and, since xj ∈ It follows that C induces an odd cycle without short chords in H ′ ′ ′′ H − C , it contradicts the minimality of H . (End of Claim 1.2.2.) Claim 1.2.3. The sets P3 ∪ P6 and P4 ∪ P5 induce cycle without short chords in G. By symmetry we will only prove the first statement. Suppose that P3 ∪ P6 has a short chord xh xk . Since H ′ has no short chord, we have that no short chord of P3 ∪ P6 can have both endnodes in P3 or P6 . Moreover, by construction, we have that every edge with one endnode in the set P6 − {x1 } cannot have the other endnode in P3 . It follows that xh ≡ x1 . Moreover, since x1 xj−1 ∈ / E we have that xk ≡ xa and, consequently, that xb ≡ x2t+1 . But then the set {xa , x1 , x2t+1 } contradicts Claim 2.1.2. (End of Claim 2.1.3) Suppose now that xa ≡ xd . It follows that P1 = P3 ∪ (P5 − {xd }) and, consequently, that P3 and P5 must have different parities. Hence, by construction, we have that either P4 ∪ P3 or P4 ∪ P5 induces an odd cycle without short chords in G, a contradiction. It follows that a ≥ d + 1. By symmetry we also have that b ≥ c + 1. Now, by Claim 1.2.3, we have that the set P3 ∪ P6 induces a cycle without short chords in G. It follows, by the minimality of H ′ , that P3 ∪ P6 is an even cycle and, consequently, that P3 and P6 have the same parity. An analogous argument shows that P4 and P5 have the same parity. Consequently, the set C1 = P3 ∪ P4 ∪ P5 ∪ (P6 − {x1 }) induces an odd cycle in H ′ . Since P3 ∪ P6 and P4 ∪ P5 do not have short chords it is easy to check that C1 has no 18

x3

xi

x i-1

x2

x i+1

x i-2 yt

zs

yt zs-1=xp

x1 =yt-1 y1

(a)

x4

zs zs-1

x1

z1

y1

x

vanishing

z1 x

(b)

Figure 8: short chord. It follows, by the minimality of H ′ that C1 has the same number of nodes as H ′ and, consequently, that xa xd ∈ E and xc xb ∈ E (i.e. a = d + 1 and b = c + 1). It follows that P3 and P5 have the same parity (since P1 is an even cycle), and so P3 , P4 , P5 and P6 have all the same parity. This implies that C2 = {xd } ∪ P3 ∪ P4 induces ˆ Moreover, if xa xc ∈ and odd cycle in H. / E then the cycle C2 has no short chord, contradicting the minimality of H ′ . It follows that xa xc ∈ E and, consequently, that ˆ (see figure 7 (b)). But then, the set xb xd ∈ E (else (xa , xb , xc , xd ) is a diamond in H) ˆ contradicting Claim 1.2.1. {xd , xa , xc , xb } induces a K4 in H, It follows that H ′ does not have a chord with one extremum in P1 − {x1 } and the other in P2 − {x1 }. Now, let Pˆ1 = {x1 , xj1 , . . . , xjr , xj } be a shortest path connecting x1 to xj in the subgraph of H ′ induced by P1 minus the edge x1 xj and Pˆ2 = {xj+1 , xh1 , . . . , xhs , x1 } be a shortest path connecting xj+1 to x1 in the subgraph of H ′ induced by P2 minus the ˆ and edge x1 xj+1 . We have that r ≥ 2 (else (x1 , xj1 , xj , xj+1 ) induces a diamond in H) ˆ ˆ s ≥ 2 (else (xh1 , x1 , xj , xj+1 ) induces a diamond in H). It follows that the set P1 ∪ Pˆ2 ˆ with the required properties. (End of induces a Mickey Mouse with long ears in H Claim 1.2). Let U = {x, z1 , . . . , zs , yt , yt−1 , . . . , y1 , x} be the node set of a Mickey Mouse with ˆ By the previous claim we can assume that it exists and has Long Ears induced in H. ˆ does not vanishing chords xzs and xyt and the vanishing edge zs yt . Moreover, since H contain induced odd holes, we have that both t and s are odd numbers greater than or equal to 3. Now, since {x} ∪ NG (x) is not a star-cutset in G we have that there exists a path P = {yt−1 ≡ x1 , x2 , . . . , xp ≡ zs−1 } connecting zs−1 to yt−1 in G − ({x} ∪ NG (x)). Since 19

zs−1 and yt−1 are separated, according to Theorem 3.4, by an extended star-cutset X, we have that there exists at least one node of P which belongs to X − NG (x) − {x} and, consequently, that is adjacent to both zs and yt . Let xi be the node with the previous properties and having minimum index in P (see figure 8 (a)). Note that i ≥ 2 and, since ˆ is diamond-free, that xi ∈ H / U . Moreover, we have that xi yj ∈ / E for j ∈ {2, . . . , t − 2} (else (yj : xi , zs , yt , x) is an anti-chair in G). Claim 1.3. If xi−1 yt ∈ E then p = 3. Suppose, conversely, that xi−1 yt ∈ E and p ≥ 4. By the minimality of i we have that xi−1 zs ∈ / E. But then xi zs−1 ∈ E (else (zs−1 : zs , yt , xi , xi−1 ) is an anti-chair in G). It follows that i = p − 1 and, since p ≥ 4, that xi yt−1 ∈ / E. But then (yt−1 : yt , zs , xi , zs−1 ) is an anti-chair in G, a contradiction. (End of Claim 1.3). Suppose first that i ≥ 4 (see figure 8 (a)) and, consequently, that yt−1 xi ∈ / E and yt−1 xi−1 ∈ / E. Since p ≥ 4 we have, by Claim 1.3, that xi−1 yt ∈ / E and,consequently, that xi−1 zs ∈ E (else (xi−1 : xi , zs , yt , x) is an anti-chair in G). But then (yt−1 : yt , xi , zs , xi−1 ) is an anti-chair in G, a contradiction. It follows that i ≤ 3. Suppose first that i = 3 and, consequently, that xi−1 ≡ x2 and xi ≡ x3 . We have that x2 yt−1 ∈ E and, by Claim 1.3, that x2 yt ∈ / E. Consequently, we have that x2 zs ∈ E (else (x2 : x3 , yt , zs , x) is an anti-chair in G). Now, recall that x3 ∈ / {y1 , . . . , yt−1 } and that x3 yj ∈ / E for j ∈ {2, . . . , t − 2}. It follows that x2 ∈ / {y2 , . . . , yt−2 } and, since x2 6= y1 , we have that the set C ′ = {x2 , zs , x, y1 , . . . , yt−1 ≡ x1 } induces an odd cycle in G. Since C ′ is not an odd hole we have that there exists j ∈ {1, . . . , t − 2} such that yj x2 ∈ E. As a consequence, x3 yj ∈ E (else (yj : x2 , x3 , zs , yt ) is an anti-chair in G) and, since x3 yj ∈ / E for j ∈ {2, . . . , t − 2}, we have that yj ≡ y1 (see figure 8 (b)). But, then x3 zs−1 ∈ E (else (zs−1 : zs , x3 , x2 , y1 ) is an anti-chair in G) and, consequently, (yt−1 : yt , zs , x3 , zs−1 ) is an anti-chair in G, a contradiction. As a consequence we have that i = 2 and that the node xi ≡ x2 is adjacent to both yt and zs in G. In addition, we have that y1 x2 ∈ E (else (y1 : x, yt , zs , x2 ) is an anti-chair in G). Now, since x2 yj ∈ / E for j ∈ {y2 , . . . , yt−1 } we have that for t ≥ 5 the set {x2 , y1 , y2 , . . . , yt−1 } induces an odd hole in G, a contradiction. It follows that t = 3 and, by a similar argument, that s = 3. In other words, we have that the set {x, z1 , z2 , z3 , y3 , y2 , y1 , x} induces a Mickey Mouse with Long Ears with vanishing chords xz3 and xy3 and the vanishing edge z3 y3 . Moreover, we have that x2 y1 ∈ E and that x2 z1 ∈ E (else (z1 : x, zs , yt , x2 ) is an anti-chair in G) (see figure 9 (a)). Claim 1.4. x2 y3 and x2 z3 are not vanishing edges. Suppose, conversely, that one of the two edges is vanishing. By symmetry, we can assume that x2 y3 is a vanishing edge. Since y3 z3 is a vanishing edge, we have, by ˆ we have Lemma 2.2, that x2 z3 is also vanishing. Moreover, since y3 is a node of H, that there exists a node s ∈ S which is adjacent to y3 . Since the edges y3 z3 , x2 y3 and 20

x2

x2

s _

z y2

y3

z3

y1

z2

z1 (a)

x

y2

y3

z3

y1

vanishing

z2

z1 x

(b)

Figure 9: xy3 are vanishing, we have that sx2 ∈ / E, sz3 ∈ / E and sx ∈ / E. It follows that sz2 ∈ E (else (s : y3 , x2 , z3 , z2 ) is an anti-chair in G) and, consequently, that sz1 ∈ E (else {s, z2 , z1 , x, y3 } induces a 5-hole in G). It follows that sy2 ∈ E (else (y2 : x2 , z2 , z1 , s) is an anti-chair in G) and sy1 ∈ E (else (y1 : x2 , z2 , z1 , s) is an anti-chair in G). But then (z3 : y3 , s, y2 , y1 ) is an anti-chair in G, a contradiction. (End of Claim 1.4) Claim 1.5. x2 y2 and x2 z2 are not vanishing edges. Suppose, conversely, that one of the two edges is vanishing. By symmetry, we can assume that x2 z2 is a vanishing edge. Since, by Claim 1.4, x2 z3 is not vanishing we have, by Lemma 2.2, that z3 z2 is not vanishing. But then the four nodes {x2 , z2 , y3 , z3 } contradict Lemma 2.2. (End of Claim 1.5). Claim 1.6. z2 z3 and y2 y3 are vanishing edges. Suppose, conversely, that one of the two edges is not vanishing. By symmetry, we ˆ can assume that z3 z2 is not a vanishing edge. Since both the endnodes belong to H, we have that there exists a node s ∈ S with the property that {s, z2 , z3 } induces a triangle in G. Observe that sy3 ∈ / E and sx ∈ / E since both y3 z3 and xz3 are vanishing edges. Moreover, we have that sz1 ∈ / E (else (y3 : z3 , s, z2 , z1 ) is an anti-chair in G) and, consequently, that sx2 ∈ / E (else (x : z1 , z2 , x2 , s) is an anti-chair in G). It follows that sy1 ∈ / E (else {s, y1 , x, z1 , z2 } induces a 5-hole in G). But then (y1 : x2 , z2 , z3 , s) is an anti-chair in G, a contradiction. (End of Claim 1.6). Claim 1.7. z2 z1 and y2 y1 are vanishing edges. Suppose, conversely, that one of the two edges is not vanishing. By symmetry, we ˆ we can assume that z1 z2 is not a vanishing edge. Since both the endnodes belong to H, have that there exists a node s ∈ S with the property that {s, z1 , z2 } induces a triangle in G. Moreover, since z2 z3 is vanishing by Claim 1.6, we have that sz3 ∈ / E.

21

Suppose first that sx2 ∈ E. It follows that either sy3 ∈ E or sx ∈ E (else (s : x2 , y3 , z3 , x) is an anti-chair in G). Moreover, since xy3 is a vanishing edge, the node s cannot be adjacent to both y3 and x. If sx ∈ / E then (x : z3 , x2 , z2 , s) is anti-chair in G. It follows that sx ∈ E and sy3 ∈ / E. But then (y3 : x, z1 , s, z2 ) is an anti-chair in G, and we get again a contradiction. It follows that sx2 ∈ / E and, consequently, that sy3 ∈ E (else (y3 : x2 , z1 , z2 , s) is an anti-chair in G) and sy2 ∈ E (else (y2 : x2 , z1 , z2 , s) is an anti-chair in G). But then the triangle {s, y2 , y3 } implies that y2 y3 is not vanishing, contradicting Claim 1.6. (End of Claim 1.7). Now, since z1 z2 is vanishing and, by Claim 1.5, x2 z2 is not vanishing, we have, by Lemma 2.2, that x2 z1 is not vanishing. A symmetric argument shows that x2 y1 is not a vanishing edge. Let z¯ be one of the two nodes adjacent to z2 in H. It follows that the edge z¯z2 is not vanishing and hence that z¯ ∈ / U . Moreover, we have that z¯z1 ∈ / E and z¯z3 ∈ / E. In fact, if z¯z1 (¯ z z3 ) was an edge of H then the edge z1 z2 (z2 z3 ) would be a short chord of ˆ H, contradicting the assumptions. On the other hand, if z¯z1 (¯ z z3 ) was a vanishing edge then the vanishing edges z¯z1 (¯ z z3 ) and z1 z2 would contradict Lemma 2.2. It follows that z¯ 6= x2 . Claim 1.8. z¯x ∈ E. Suppose, conversely, that z¯x ∈ / E. It follows that z¯y3 ∈ / E (else {¯ z , z2 , z1 , x, y3 } z : z2 , z3 , x2 , y3 ) is an induces a 5-hole in G) and, consequently, that z¯x2 ∈ E (else (¯ anti-chair in G. But then (x : z3 , x2 , z2 , z¯) is an anti-chair in G, a contradiction. (End of Claim 1.8). By the previous claim, we have z¯y3 ∈ / E (else {¯ z , y3 , x, z3 } induces a diamond in ˆ H) and, consequently, that z¯x2 ∈ E (else (¯ z : x, y3 , z3 , x2 ) is an anti-chair in G). Now, since z¯z2 is an edge of H, it is not vanishing. It follows that there exists a node s ∈ S with the property that {s, z¯, z2 } induce a triangle in G. Moreover, since z2 z3 and z1 z2 are vanishing by Claims 1.6 and 1.7, we have that sz3 ∈ / E and sz1 ∈ / E. Suppose first that sx2 ∈ E. It follows that either sy3 ∈ E or sx ∈ E (else (s : x2 , y3 , z3 , x) is an anti-chair in G). Moreover, since xy3 is a vanishing edge, the node s cannot be adjacent to both y3 and x. Now, if sy3 ∈ E then sx ∈ / E and the set {s, y3 , x, z1 , z2 } induces a 5-hole in G, a contradiction. If, conversely, sx ∈ E then sy3 ∈ / E and (y3 : x, z¯, s, z2 ) is an anti-chair in G, and we get again a contradiction. It follows that sx2 ∈ / E and, consequently, that sy3 ∈ E (else (y3 : x2 , z¯, z2 , s) is an anti-chair in G) (figure 9 (b)). But, since xy3 is a vanishing edge, we have that sx ∈ /E and, consequently, that the set {s, z2 , z1 , x, y3 } induces a 5-hole in G, a contradiction. This concludes the proof of Claim 1. (End of Claim 1). ˆ moreover, since By the previous claim we have that the edge u1 u3 is a short chord of H, u1 u2 is not vanishing, we have that there exists a node w ∈ S with the property that u1 w ∈ E and u2 w ∈ E. Analogously, since u2 u3 is not vanishing, there exists a node 22

w u 2q+1

u1

w u2q+1

u2

u2q u3 u5

u2

u1

u2q

u

u3

u4

u5

(a)

u

u4 (b)

Figure 10: u ∈ S with the property that u2 u ∈ E and u3 u ∈ E. Finally, since u1 u3 is a vanishing edge, we have that u1 u ∈ / E, u3 w ∈ / E and, consequently, that w 6= u. Claim 2. uu4 ∈ E and wu2q+1 ∈ E. We prove only the first statement since the proof of the second is analogous. If there exists a node u′ ∈ S such that u′ u1 ∈ / E, u′ u2 ∈ E, u′ u3 ∈ E and u′ u4 ∈ E, ′ then we can replace u by u and conclude that uu4 ∈ E. Suppose, conversely, that uu4 ∈ / E and that there does not exist a node u′ ∈ S with the property that u′ u1 ∈ / E, u′ u2 ∈ E, u′ u3 ∈ E and u′ u4 ∈ E. Consequently, since u3 u4 is not vanishing, we have that there exists a node z ∈ S − {u, w} with the property that zu3 ∈ E and zu4 ∈ E. Moreover, since u1 u3 is / E and, consequently, that zu2 ∈ E (else (z : u3 , u1 , u2 , w) vanishing, we have that u1 z ∈ is an anti-chair in G). But then the node z contradicts the assumption that a node u′ with the above properties does not exist. (End of Claim 2). Now, since wu2q+1 ∈ E and wu2 ∈ E we have that u2q+1 u2 cannot be a vanishing edge, and so u2q+1 u2 ∈ / E. An analogous argument shows that u2 u4 ∈ / E. It follows that uu2q+1 ∈ E (else (u : u2 , w, u1 , u2q+1 ) is an anti-chair in G) and that wu4 ∈ E (else (w : u2 , u, u3 , u4 ) is an anti-chair in G). Now, since u3 and u2q+1 are both adjacent to u, we have that u3 u2q+1 is not vanishing. Consequently, since q ≥ 2, we have that u3 u2q+1 ∈ / E. Analogously, we have that u1 u4 ∈ / E. Finally, since u4 and u2q+1 are adjacent to w (and to u) we have that u2q+1 u4 is not vanishing. If u2q+1 u4 ∈ E then q = 2 and it is easy to check that the set {u, w, u1 , . . . , u5 } induces an anti-hole in G, contradicting our assumptions. It follows that u2q+1 u4 ∈ / E and, consequently, that q ≥ 3 (see figure 10 (a)). / E or uu5 ∈ / E. Claim 3. Either wu2q ∈ Suppose, conversely, that wu2q ∈ E and uu5 ∈ E. Since u1 w ∈ E and wu2q ∈ E we have that u1 u2q ∈ / E, and so u3 u2q is a vanishing edge of E (else (u3 : u1 , w, u2q+1 , u2q ) is an anti-chair in G). Consequently, since uu3 ∈ E, we have that uu2q ∈ / E. 23

w

w u2q+1

u1

u2q

u2q+1

u2 u3

u2q

u3

u u5

u5

u

u4

u4 (a)

u2

u1

(b)

Figure 11: An analogous argument shows that u1 u5 is vanishing and wu5 ∈ / E (see figure 10 (b)). Now, since u is adjacent to both u2 and u5 , we have that u2 u5 ∈ / E. Analogously, since w is adjacent to both u2 and u2q we have that u2 u2q ∈ / E. It follows that u5 u2q ∈ /E (else the set {u2q , u5 , u, u2 , w} induces a 5-hole in G). Finally, we have that u5 u2q+1 ∈ /E since both u5 and u2q+1 are adjacent to u. But then, (u5 : u1 , w, u2q+1 , u2q ) is an antichair in G. (End of Claim 3). Claim 3 implies that either wu2q ∈ / E or uu5 ∈ / E. By symmetry we can assume, without loss of generality, that uu5 ∈ / E. If u5 u2 ∈ E (see figure 11 (a)) then it must be a vanishing edge. Consequently, by Lemma 2.2, we have that u5 u3 ∈ / E and u5 u1 ∈ / E. Moreover, we have that wu5 ∈ / E (else the nodes u2 and u5 would form a triangle with w). It follows that u2q+1 u5 ∈ E (else (u5 : u2 , w, u1 , u2q+1 ) is an anti-chair in G). But then the set {u5 , u4 , u3 , u1 , u2q+1 } induces a 5-hole in G, a contradiction. It follows that u5 u2 ∈ / E and, consequently, that u5 u3 ∈ E (else (u5 : u4 , u3 , u, u2 ) is an anti-chair in G (see figure 11 (b)). But then, we have that u2q+1 u5 ∈ E (else (u2q+1 : u, u3 , u4 , u5 ) is an anti-chair in G). It follows that wu5 ∈ / E (since u2q+1 u5 is a vanishing edge) and, consequently, that the set {u2q+1 , w, u2 , u3 , u5 } induces an odd hole in G, a contradiction. We now prove that the property of being chair-free is ”weakly” hereditary with respect to clique reduction. Namely, we prove that if G is minimally imperfect, Berge and chair-free then every minimally imperfect Berge subgraph of G|C is chair-free. Theorem 3.6 Let H(V, E) be a minimal imperfect chair-free graph which is Berge and let C be a reducible clique in H. If H ′ is a minimal imperfect subgraph of H|C which is Berge then H ′ is chair-free. Proof. Since H ′ is a subgraph of H|C we have that there exists a subgraph G of H such that H ′ = G|C. In addition, since H is minimally imperfect and chair-free, we have, 24

x1

x1

u=x0 s1

u=x0

xj

false

y

v

s'3

s2

s2

w=xt

s4

s1

s3

∉Ε (a)

w=xt

v

y

(b)

Figure 12: by Lemma 2.4, that H and G do not contain an induced subgraph isomorphic to the complement of F1 (figure 4). Finally, since H is chair-free and Berge we have that G has the same properties. We first prove two technical claims. Claim 1. If G|C contains a claw with nodes {v, u, w, y} edges {vu, vw, vy} and with the property that vu is a false edge then there exists a node u1 ∈ V − {v, u, w, y} with the property that u1 u ∈ E, u1 w ∈ E and, vu1 ∈ / E. Note first that, by Lemma 2.1, the edges vw and vy are not false. Moreover, since G|C is minimally imperfect, it does not contain a star-cutset with center v. It follows that there exists a path P = {u ≡ x0 , x1 , . . . , xt ≡ w} in V |C − (NG|C (v) − {u, w}) with t ≥ 2. We have that t is an even number for, otherwise, the set {v, x0 , . . . , xt } induces an odd hole in G|C, contradicting the assumption that G|C is Berge. Moreover, since x0 v ≡ uv is a false edge and x1 v ∈ / E, we have, by Lemma 2.1, that x0 x1 ∈ E. Since x1 v ∈ / E|C we have that there exists a node s1 ∈ C with the property that s1 x 1 ∈ / E and s1 v ∈ / E. Moreover, since vx0 is a false edge, we have that s1 x0 ∈ E. Analogously, since x0 xt ∈ / E|C, we have that there exists a node s2 ∈ C with the property that s2 xt ∈ / E, s2 x0 ∈ / E and s2 v ∈ E. It follows that s2 6= s1 (see figure 12 (a)). Observe now that if s1 xt ∈ E then the node s1 is the node u1 with the required properties and the claim is proved. Consequently, we can assume that s1 xt ∈ / E. Let us first examine the case t = 2. Since s1 x2 ∈ / E and s1 x1 ∈ / E we have that x1 x2 is not false, but then the node x1 is the node u1 with the required properties and we are done. Hence we can assume that t ≥ 4 and, consequently, that x1 s2 ∈ / E (else (s1 : x1 , s2 , v, xt ) is a chair in G). 25

u

u

v'

u1

v'

y' y'

w

u2

w'

w'

v

y

w

F2

v

y

F3

Figure 13: Suppose now that ys1 ∈ E (see figure 12 (a)). If ys2 ∈ / E then x1 y ∈ E (else (s2 : y, s1 , x0 , x1 ) is a chair in G) and, consequently, (x1 : xt , v, y, s1 ) is a chair in G, a contradiction. It follows that ys2 ∈ E. Now, since x0 y ∈ / E|C, we have that there exists a node s3 ∈ C − {s1 , s2 } with the property that s3 x0 ∈ / E and s3 y ∈ / E. Moreover, since x0 v is a false edge we have that NG ({x0 , v}) ⊇ C, and so vs3 ∈ E. As a consequence we have that xt s3 ∈ / E (else (y : x0 , s1 , s3 , xt ) is a chair in G) and that x1 s3 ∈ / E (else (s1 : x1 , s3 , v, xt ) is a chair in G). It follows that x1 y ∈ E (else (y : s3 , s1 , x0 , x1 ) is a chair in G) and, finally, that (s3 : xt , v, y, x1 ) is a chair in G (see figure 12 (a)), a contradiction. / E and, consequently, that ys2 ∈ E (else (y : xt , v, s2 , s1 ) is It follows that ys1 ∈ a chair in G) (see figure 12 (b)). Since x0 y ∈ / E|C, we have that there exists a node ′ ′ s3 ∈ C − {s1 , s2 } with the property that s3 x0 ∈ / E and s′3 y ∈ / E. Since vx0 is a false ′ ′ edge, we have that vs3 ∈ E. Moreover, we have that xt s3 ∈ E (else (y : xt , v, s′3 , s1 ) is a chair in G). Now, suppose that xj s1 ∈ E for some j ∈ {2, . . . , t − 1}. Since xj v ∈ / E|C and {s1 , s2 , s′3 } ⊆ NG ({xj , v}), we have that there exists a node s4 ∈ C − {s1 , s2 , s′3 } with the property that s4 xj ∈ / E and s4 v ∈ / E. Moreover, since vx0 is false we have that ′ / E (else (xj : v, s′3 , s4 , x0 ) is a chair in G) and, x0 s4 ∈ E. It follows that xj s3 ∈ ′ consequently, that (xj : x0 , s1 , s3 , v) is a chair in G, a contradiction. It follows that xj s1 ∈ / E for j ∈ {1, . . . , t − 1}. Suppose now that xj s′3 ∈ E for some j ∈ {2, . . . , t − 1}. It follows that (xj : /E v, s′3 , s1 , x0 ) is a chair in G, a contradiction. As a consequence, we have that xj s′3 ∈ for j ∈ {2, . . . , t − 1} (see figure 12 (b)). Now, if x1 s′3 ∈ E then the set {x1 , . . . , xt , s′3 } induces an odd hole in G. It follows / E and, consequently, that the set {x0 , . . . , xt , s1 , s′3 } induces an odd hole in that x1 s′3 ∈ G, contradicting the assumption that G is Berge. (End of Claim 1). 26

v1 v3

u

v1 v4

v2

v2 v5

w z

v3

(a)

v4

w

v5 (b)

u

Figure 14: Claim 2. The graph G does not contain a subgraph isomorphic to F2 and with the property that the set {u, w, v, y} induces a claw in G|C. If G contains an induced subgraph isomorphic to F2 and with the above property then uv is a false edge in G|C. It follows, by Claim 1, that there exists a node u1 ∈ V − {u, w, v, y} with the property that {u, w} ⊆ NG (u1 ), v ∈ / NG (u1 ). By symmetry, Claim 1 also implies that there exists a node u2 ∈ V − {u, w, v, y} with the property that {u, y} ⊆ NG (u2 ), v ∈ / NG (u2 ). Suppose now that u1 v ′ ∈ E. It follows that u1 w′ ∈ E ( else the set {u1 , v ′ , w′ , v, w} induces a 5-hole in G). But then u1 y ∈ E (else (w : y, w′ , u1 , u) is a chair in G) and, consequently, u1 y ′ ∈ E (else (u : y, u1 , w, y ′ ) is a chair in G). As a consequence, the graph G contains a subgraph isomorphic to the complement of F1 (see figure 4) induced by the nodes {u1 , w, y, u, v ′ , y ′ , w′ }, contradicting our assumptions. It follows that u1 v ′ ∈ / E and, by symmetry, that u2 v ′ ∈ / E. This implies that u1 6= u2 ′ ′ (else (w : y, u1 , u, v ) is a chair in G), u1 y ∈ E (else {u, u1 , w, y ′ , v ′ } is a 5-hole in G) and that u2 w′ ∈ E (else {u, u2 , y, w′ , v ′ } is a 5-hole in G). Moreover, we have that u1 y ∈ / E (else (y : w, u1 , u, v ′ ) is a chair in G) and, consequently, that u1 w′ ∈ / E (else ′ ′ (u : w, u1 , w , y) is a chair in G). A symmetric argument shows that u2 w ∈ E, u2 y ′ ∈ /E and u2 w ∈ / E (see graph F3 in figure 13). It follows that u1 u2 ∈ E (else the set {u1 , u, u2 , y ′ , w′ } induces a 5-hole in G) and, consequently, that the set {u1 , u2 , w, v, y} induces a 5-hole in G, a contradiction. (End of Claim 2). Let (v1 : v2 , v3 , v4 , v5 ) be a chair in G|C. By Lemma 2.1 we have that every chair induced in G|C contains at most one false edge. In the following three claims we will show that no edge of (v1 : v2 , v3 , v4 , v5 ) is false, contradicting the hypothesis that G is chair free. Claim 3. The edges v2 v3 and v1 v3 are not false. Suppose the claim does not hold. By symmetry, we can assume that v2 v3 is false (see figure 14 (a)). Since v2 v4 ∈ / E|C we have that there exists a node u in C such that v2 u ∈ / E and v4 u ∈ / E. Moreover, since v3 v5 ∈ / E|C we have that there exists a node w 27

in C such that v3 w ∈ / E and v5 w ∈ / E. Since v2 v3 is a false edge we have that wv2 ∈ E and uv3 ∈ E. Now, If uv5 ∈ E then (v3 : v5 , u, w, v2 ) is a chair in G. It follows that uv5 ∈ / E and, consequently, that uv1 ∈ E (else (v1 : u, v3 , v4 , v5 ) is a chair in G). Since v1 v2 ∈ / E, we have that there exists a node z ∈ C − {u, v} with the property that zv1 ∈ / E and zv2 ∈ / E. Since v2 v3 is a false edge, we have that zv3 ∈ E. If zv4 ∈ /E we have that zv5 ∈ E (else (v1 : z, v3 , v4 , v5 ) is a chair in G). But then (v3 : v5 , z, w, v2 ) is a chair in G, a contradiction. It follows that zv4 ∈ E. Moreover, we have that v4 w ∈ / E (else (v3 : v5 , v4 , w, v2 ) is a chair in G) and that v1 w ∈ / E (else (v1 : v2 , w, z, v4 ) is a chair in G). It follows that the set {v1 , v3 , v4 , u, z, w, v2 } induces in G a subgraph isomorphic to F2 of figure 13 with u ≡ v2 and v ≡ v3 (see figure 14 (a)). But, since v2 v3 is a false edge, we contradict Claim 2. (End of Claim 3). Claim 4. The edge v4 v5 is not false. Suppose, conversely, that v4 v5 is a false edge (see figure 14 (b)). Since v1 v4 ∈ / E|C we have that there exists a node u in C such that v1 u ∈ / E and v4 u ∈ / E. Moreover, since v3 v 5 ∈ / E|C we have that there exists a node w in C such that v3 w ∈ / E and v5 w ∈ / E. Since v4 v5 is a false edge we have that wv4 ∈ E and uv5 ∈ E and, consequently, that u 6= w. We have that v1 w ∈ / E (else (v4 : v1 , w, u, v5 ) is a chair in G). It follows that v2 w ∈ E (else (v1 : v2 , v3 , v4 , w) is a chair in G) and, consequently, that v2 u ∈ E (else (v4 : v2 , w, u, v5 ) is a chair in G). But then v3 u ∈ E (else (v1 : v4 , v3 , v2 , u) is a chair in G) and, consequently (see figure 14 (b1)), (v4 : v1 , v3 , u, v5 ) is a chair in G, a contradiction. (End of Claim 4). Claim 5. The edge v3 v4 is not false. Suppose, conversely, that v3 v4 is a false edge. Since v2 v4 ∈ / E|C we have that there / E and v4 u ∈ exists a node u in C such that v2 u ∈ / E. Moreover, since v3 v5 ∈ / E|C we have that there exists a node w in C such that v3 w ∈ / E and v5 w ∈ / E. Since v3 v4 is a false edge we have that wv4 ∈ E and uv3 ∈ E. Suppose first that wv2 ∈ E. It follows that uv5 ∈ E (else (v2 : u, w, v4 , v5 ) is a chair in G) and, consequently, that wv1 ∈ / E (else (v1 : v2 , w, v4 , v5 ) is a chair in G). But then uv1 ∈ / E (else (v2 : v4 , w, u, v1 ) is a chair in G), and so (see figure 15 (a)) (v2 : v1 , v3 , u, v5 ) is a chair in G, a contradiction. It follows that wv2 ∈ / E. Suppose now that uv1 ∈ / E, it follows that wv1 ∈ E (else (v2 : v1 , v3 , u, w) is a chair in G). Consequently, we have that uv5 ∈ E otherwise (v1 : u, w, v4 , v5 ) is a chair in G. But then (see figure 15 (b)) (v1 : v2 , v3 , u, v5 ) is a chair in G, a contradiction. It follows that uv1 ∈ E and, since v1 v4 ∈ / E, we have that there exists a node z ∈ C such that zv1 ∈ / E and zv4 ∈ / E. Since uv1 ∈ E and wv4 ∈ E we have that z 6= u and z 6= w. Moreover, since v3 v4 is a false edge we have that zv3 ∈ E. Now, if wv1 ∈ E then zv5 ∈ E (else (v1 : z, w, v4 , v5 ) is a chair in G). It follows that v2 z ∈ E (else (v1 : v2 , v3 , z, v5 ) is a chair in G). But then, (v5 : v2 , z, w, v1 ) is a chair in 28

v1

v1 u

v3

w

v2

v4

v5

(a)

u

v3

v4

v2

w

v1

(b)

v5

v3

u

w

z

v4

v2

v5

(c)

Figure 15: G, a contradiction. It follows that wv1 ∈ / E and, consequently, that v2 z ∈ E, otherwise (v1 : v2 , v3 , z, w) is a chair in G. But then (see figure 15 (c)) the nodes {z, u, w, v1 , v2 , v3 , v4 } induce in G the graph F2 of figure 13 with u ≡ v4 and v ≡ v3 . But, since v3 v4 is a false edge, we contradict Claim 2 and the Theorem follows. Theorem 3.7 A chair-free graph is perfect if and only if it is Berge. Proof. Necessity is obvious. To prove that a chair-free Berge graph is perfect we show that there does not exist a minimal imperfect chair-free graph which is Berge. Suppose, conversely that such a graph exists and let H be a minimal imperfect chair-free graph which is Berge and has the minimum number of nodes among all graphs with these properties. By Theorem 3.1 we have that H contains a reducible clique C and by Theorem 1.2 we have that H|C is imperfect. It follows that H|C contains a minimally imperfect subgraph H ′ . Since H is Berge, we have, by Theorems 3.2 and 3.5, that H|C is Berge. As a consequence, we have that the graph H ′ is Berge and, by Theorem 3.6, that it is also chair-free. But then, H ′ is a minimal imperfect chair-free graph which is Berge and has less nodes than H, a contradiction.

Ackowledgment I am deeply indebted to Myriam Preissmann and Frederic Maffray for their constructive suggestions which helped me improve and simplify an earlier version of this paper. I would also like to thank Carlo Mannino and Kristina Vuˇskovi´c whose valuable suggestions and remarks have improved the presentation.

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