The Strong Perfect Graph Theorem Maria Chudnovsky Princeton University, Princeton NJ 08544 Neil Robertson1 Ohio State University, Columbus, Ohio 43210 Paul Seymour2 Princeton University, Princeton NJ 08544 Robin Thomas3 Georgia Institute of Technology, Atlanta, GA 30332 June 20, 2002; revised February 1, 2008 arXiv:math/0212070v1 [math.CO] 4 Dec 2002 1Supported by ONR grant N00014-01-1-0608, NSF grant DMS-0071096, and AIM. 2Supported by ONR grants N00014-97-1-0512 and N00014-01-1-0608, and NSF grant DMS-0070912. 3Supported by ONR grant N00014-01-1-0608, NSF grants DMS-9970514 and DMS-0200595, and AIM. Abstract A graph G is perfect if for every induced subgraph H, the chromatic number of H equals the size of the largest complete subgraph of H, and G is Berge if no induced subgraph of G is an odd cycle of length at least 5 or the complement of one. The “strong perfect graph conjecture” (Berge, 1961) asserts that a graph is perfect if and only if it is Berge. A stronger conjecture was made recently by Conforti, Cornu´ejols and Vuˇskovi´c— that every Berge graph either falls into one of a few basic classes, or it has a kind of separation that cannot occur in a minimal imperfect graph. In this paper we prove both these conjectures. 1 Introduction We begin with definitions of some of the terms we use which may be nonstandard. All graphs in this paper are finite and simple. The complement G of a graph G has the same vertex set as G, and distinct vertices u, v are adjacent in G just when they are not adjacent in G. A hole of G is an induced subgraph of G which is a cycle of length > 3. An antihole of G is an induced subgraph of G whose complement is a hole in G. A graph G is Berge if every hole and antihole of G has even length. A clique in G is a subset X of V (G) so that every two members of X are adjacent. A graph G is perfect if for every induced subgraph H of G, the chromatic number of H equals the size of the largest clique of H. In 1961 Claude Berge [1] proposed the so-called Strong Perfect Graph Conjecture, the main theorem of this paper: 1.1 A graph is Berge if and only if it is perfect. It is easy to see that every perfect graph is Berge, and so to prove (1.1) it remains to prove the converse. This has received a great deal of attention over the past 40 years, but so far has resisted solution. Most of the previous approaches on (1.1) fall into two classes: proving that the theorem holds for graphs with some particular graph excluded as an induced subgraph (there are a number of these for different subgraphs, but such an approach obviously cannot do the whole thing), and using linear programming methods to investigate the structure of a minimal counterexample. (There are rich connections with linear and integer programming - see [10] for example.) Our approach is different. Recently, Conforti, Cornu´ejols and Vuˇskovi´c[5] conjectured that every Berge graph either falls into one of four well-understood classes, or it admits one of several kinds of decomposition. They pointed out that if this could be proved, and if also it could be shown that no minimal counterexample to (1.1) admits any such decomposition, then 1.1 would follow (for certainly no minimal counterexample to 1.1 can fall into the four basic classes). We have been able to prove both (except we need a fifth class). Before we can be more precise we need more definitions. If X ⊆ V (G) we denote the subgraph of G induced on X by G|X. The line graph L(G) of a graph G has vertex set the set E(G) of edges of G, and e, f ∈ E(G) are adjacent in L(G) if they share an end in G. We need one other class of graphs, defined as follows. Let m,n ≥ 2 be integers, and let {a1,...,am}, {b1,...,bm}, {c1,...,cn}, {d1,...,dn} be disjoint sets. Let G have vertex set their union, and edges as follows: • ai is adjacent to bi for 1 ≤ i ≤ m, and cj is nonadjacent to dj for 1 ≤ j ≤ n ′ • there are no edges between {ai, bi} and {ai′ , bi′ } for 1 ≤ i < i ≤ m, and all four edges between ′ {cj , dj} and {cj′ , dj′ } for 1 ≤ j < j ≤ n • there are exactly two edges between {ai, bi} and {cj, dj} for 1 ≤ i ≤ m and 1 ≤ j ≤ n, and these two edges are disjoint. We call such a graph G a bicograph. Let us say a graph G is basic if either G or G is bipartite or is the line graph of a bipartite graph, or is a bicograph. (Note that if G is a bicograph then so is G.) It is easy to see that all basic graphs are perfect. 1 Now we turn to the various kinds of decomposition that we need. First, a decomposition due to Cornu´ejols and Cunningham [8] — a 2-join in G is a partition (X1, X2) of V (G) so that there exist disjoint nonempty Ai,Bi ⊆ Xi(i = 1, 2) satisfying: • every vertex of A1 is adjacent to every vertex of A2, and every vertex of B1 is adjacent to every vertex of B2, • there are no other edges between X1 and X2, • for i = 1, 2, every component of G|Xi meets both Ai and Bi, and • for i = 1, 2, if |Ai| = |Bi| = 1 and G|Xi is a path joining the members of Ai and Bi, then it has length ≥ 3. If X,Y ⊆ V (G) are disjoint, we say X is complete to Y (or the pair (X,Y ) is complete) if every vertex in X is adjacent to every vertex in Y ; and we say X is anticomplete to Y if there are no edges between X and Y . Our second decomposition is a very slight variant on the “homogenous sets” due to Chv´atal and Sbihi [3] — an M-join in G is a partition of V (G) into six nonempty sets, (A,B,C,D,E,F ), so that: • every vertex in A has a neighbour in B and a nonneighbour in B, and vice versa • the pairs (C, A), (A, F ), (F,B), (B, D) are complete, and • the pairs (D, A), (A, E), (E,B), (B,C) are anticomplete. A path in G is an induced subgraph of G which is non-null, connected and in which every vertex has degree ≤ 2 (this definition is highly nonstandard, and we apologise, but it avoids writing “induced” about 600 times), and an antipath is an induced subgraph whose complement is a path. The length of a path is the number of edges in it (and the length of an antipath is the number of edges in its complement). We therefore recognize paths and antipaths of length 0. If P is a path, P ∗ denotes the set of internal vertices of P , called the interior of P ; and similarly for antipaths. Let A, B be disjoint subsets of V (G). We say the pair (A, B) is balanced if there is no odd path between nonadjacent vertices in B with interior in A, and there is no odd antipath between adjacent vertices in A with interior in B. A set X ⊆ V (G) is connected if G|X is connected (so ∅ is connected); and anticonnected if G|X is connected. The third kind of decomposition we use is due to Chv´atal [2] — a skew partition in G is a partition (A, B) of V (G) so that A is not connected and B is not anticonnected. Skew partitions pose a difficulty that the other two decompositions do not, for it has not been shown that a minimal counterexample to 1.1 cannot admit a skew partition; indeed, this is a well-known open question, first raised by Chv´atal [2], the so-called “skew partition conjecture”. We get around it by confining ourselves to balanced skew partitions, which do not present this difficulty. We shall prove the following, a form of which was conjectured in [5]: 1.2 For every Berge graph G, either G is basic, or one of G, G admits a 2-join, or G admits an M-join, or G admits a balanced skew partition. 2 Proof of 1.1, assuming 1.2 and 4.8. Suppose that 1.1 is false, and let G be a counterexample with |V (G)| as small as possible. Since every perfect graph is Berge, it follows that G is Berge and not perfect. Every basic graph is perfect, and so G is not basic. It is shown in [8] that G does not admit a 2-join. Since Lov´asz [9] showed that the complement of a perfect graph is perfect, it follows that G is also a counterexample to 1.1 of minimum size, and therefore G also does not admit a 2-join. It is shown in [3] that G does not admit an M-join, and we shall prove in 4.8 that G does not admit a balanced skew partition. It follows that G violates 1.2, and therefore there is no such graph G. This proves 1.1. All nontrivial bicographs admit skew partitions, so if we delete “balanced” from 1.2 then we no longer need to consider bicographs as basic — four basic classes suffice.