A taste of perfect graphs (continued)

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1 A taste of perfect graphs (continued) Recall two theorems from last class characterizing perfect graphs (and that we observed that the α ω theorem implied the Perfect Graph Theorem). Perfect Graph Theorem. A graph G is perfect if and only if G is perfect. Theorem. A graph G is perfect if and only if every induced subgraph H of G satisfies the inequalilty V (H) α(h)ω(h). We also proved a lemma regarding minimally imperfect graphs that plays a key role in the proof of the α ω theorem. Lemma. Let G be a minimally imperfect graph. Then G contains α(g)ω(g)+1 independent sets S 0, S 1,..., S α(g)ω(g) and α(g)ω(g)+1 cliques C 0, C 1,..., C α(g)ω(g) such that (a) each vertex of G belongs to precisely α(g) of the independent sets S i, (b) each clique C i has ω(g) vertices, (c) C i S i = for 0 i α(g)ω(g), (d) and C i S j = 1 for 0 i < j α(g)ω(g). Math Prof. Kindred - Lecture 24 Page 1

2 Proof of α ω theorem. ( ) Suppose that G is perfect, and let H be an induced subgraph of G. We have ω(h) = χ(h) V (H) α(h) = V (H) α(h)ω(h). ( ) We prove this direction by showing that if G is minimally imperfect, then V (G) α(g)ω(g) + 1. Let n = V (G). Consider the families {S i : 0 i α(g)ω(g)} and {C i : 0 i α(g)ω(g)} of independent sets and cliques described in the previous lemma. Let M S and M C be the n (α(g)ω(g) + 1) incidence matrices of these families. It follows from the lemma that M T S M C = J I. Notice that J I is a nonsingular matrix (with inverse 1 α(g)ω(g) J I). Thus, it must be a full rank matrix, i.e., We have rank(j I) = α(g)ω(g) + 1. α(g)ω(g) + 1 = rank(j I) = rank(m T S M C ) min(rank(m S ), rank(m C )) rank(m S) or rank(m C ) n, where the last inequality follows from the fact that the two matrices have n rows. Thus, we have shown that n α(g)ω(g) + 1. If a graph is perfect, then so are all of its induced subgraphs. This means that we can characterize perfect graphs by describing all minimally imperfect graphs. Math Prof. Kindred - Lecture 24 Page 2

3 Berge (1961) conjectured that the odd cycles (of length at least 5) and their complements were the only minimally imperfect graphs. This conjecture was known as the Strong Perfect Graph Conjecture (and is printed as such in your text). But it was proven in 2002 by Chudnovsky and colleagues. Strong Perfect Graph Theorem. A graph is perfect if and only if it contains no odd cycle of length at least five or its complement as an induced subgraph. The proof of this theorem was a major achievement, as much effort had been expended over the years on attempts to settle the Strong Perfect Graph Conjecture. Furthermore, a polynomial-time recognition algorithm for perfect graphs was developed shortly thereafter by Chudnovsky, et al., (2005). Math Prof. Kindred - Lecture 24 Page 3

4 Chordal graphs Definition A chord of a cycle C is an edge not in C whose endpoints lie in C. A chordless cycle in a graph G is a cycle of length at least 4 in G that has no chord (that is, the cycle is an induced subgraph). G is a chordal graph if it is simple and has no chordless cycle. Being chordal is a hereditary property, inherited by all the induced subgraphs of a chordal graph. Recall the notion of an x, y-cut. Definition Let x, y be nonadjacent vertices of G. An x, y-cut, or x, y-separator, is a set of vertices S such that x and y are in different components of G S. A minimal x, y-cut is an x, y-cut S such that no proper subset of S is an x, y-cut. (Note: a minimal x, y-cut is not necessarily the smallest x, y-cut in G.) Theorem. Let a and b be nonadjacent vertices of a connected chordal graph G, and let S be a minimal a, b-cut. Then S is a clique. Proof. Let S be such a cut, and let G[A], G[B] be components of G S with a A and b B. Every vertex in v S must be adjacent to some vertex of A and to some vertex of B otherwise, S would not be minimal. If S only contains one vertex, then we are done (it is clearly a clique). So assume u, v S. We claim that u v. Suppose not. Then let P be a shortest u, v-path all of whose internal vertices are in A, so P = u a 1 a 2 a s v. Math Prof. Kindred - Lecture 24 Page 4

5 Likewise, let Q be a shortest u, v-path all of whose internal vertices are in B, so Q = u b 1 b 2 b t v. a 1 u b a s A v S b t B Then the union of P and Q is an induced cycle of length at least 4. Hence, it must be that u v for any pair of vertices u, v S, thereby implying that S is a clique. Note that trees are clearly chordal graphs since they are acyclic. It turns out that chordal graphs, in general, have a treelike structure composed of complete graphs (just as trees are composed of copies of K 2.) In this spirit, we generalize the notion of a leaf in a tree to that of a simplicial vertex. Definition A vertex of G is simplicial if its neighborhood in G is a clique. We prove two results for simplicial vertices and chordal graphs that are analogous to the following facts for trees and leaves: If T is a tree with at least 2 vertices, then T has a leaf. If v is a leaf of T, then T v is also a tree. Theorem. If G is a (nonempty) chordal graph, then G has a simplicial vertex. Furthermore, if G is not a complete graph, then G has two nonadjacent simplicial vertices. Math Prof. Kindred - Lecture 24 Page 5

Small Survey on Perfect Graphs

Small Survey on Perfect Graphs Michele Alberti ENS Lyon December 8, 2010 Abstract This is a small survey on the exciting world of Perfect Graphs. We will see when a graph is perfect and which are families