1 Interval Graphs 2 Perfect Graphs

ORCO – Graphs and Discrete Structures November 18, 2020 – Lecture 6

Recall that for any graph G, ω(G) ≤ χ(G). We have seen various examples of graphs showing the gap between ω and χ can be arbitrarily large. In this lecture, we are going to see natural graph classes for which equality holds. A graph G with vertices v1, . . . , vn is said to be an interval graph if there exist intervals I1,...,In, such that any two vertices vi and vj are adjacent in G if and only if the corresponding intervals Ii and Ij intersect. The interval I1,...,In are called the interval representation of G. In such a graph G, consider the interval Ii with the rightmost left end. Note that by deﬁnition, all the neighbours of the corresponding vertex vi also intersect this left end, and so vi, together with its neighbours, forms a clique. Observation 1. Any interval graph G contains a vertex v whose neighbour- hood is a clique. In particular, v has degree at most ω(G) − 1. Note that each induced subgraph of an interval graph is also an interval graph (starting with an interval representation of the graph and removing some intervals, we can get interval representations of any induced subgraph of G). Applying Observation 2 from Lecture 2, we obtain: Theorem 2. Any interval graph G has chromatic number ω(G).

2 Perfect graphs

A graph G is perfect if and only if every induced subgraph H of G satisﬁes χ(H) = ω(H). In the previous section, we have proved that interval graphs are perfect. Another simple example is provided by bipartite graphs (right?). Examples of graphs that are not perfect include odd cycles of length at least 5. In this section, we will prove the so-called perfect graph theorem, which states that a graph is perfect if and only if its complement is perfect. It was conjectured by Berge in the early 1960s and proved by Lovász in 1972. We ﬁrst need an auxiliary result. Replicating a vertex v ∈ V (G) means adding a copy v0 of v and making it adjacent to v and all its neighbours.

1 Lemma 3. If G is perfect and H is obtained from G by replicating a vertex, then H is also perfect. Proof. The proof is by induction on n, the number of vertices of G. The lemma is clearly true for n = 1: the only graph on 1 vertex is the (perfect) graph K1, and replicating the only vertex of K1, we obtain the (perfect) graph K2. Now let G be a perfect graph on at least 2 vertices, and let G0 be the graph obtained by replicating a vertex v of G. Observe that every proper induced subgraph G00 of G0 is either isomorphic to an induced subgraph of G, or is obtained from a proper induced subgraph of G by replicating v. In either case, the resulting graph is perfect, and can be coloured by ω(G00) colours. It remains to prove that χ(G0) ≤ ω(G0). If ω(G0) = ω(G) + 1, then χ(G0) ≤ χ(G) + 1 = ω(G) + 1 = ω(G0) and we are done. So we may assume that ω(G0) = ω(G). This means that v belongs to no clique of G of size ω(G). Fix any colouring of G with ω(G) colours, and let A be the colour class of v. Every maximum clique of G intersects A \{v}. Therefore, the graph H = G − (A \{v}) has clique number ω(H) ≤ ω(G) − 1, so by the perfectness of G, χ(H) ≤ ω(G) − 1. Since A is independent, the set (A \{v}) ∪ {v0} = V (G0) − V (H) is also independent. We can therefore extend the (ω(G) − 1)-colouring of H to an ω(G)-colouring of G0, as desired. Theorem 4 (The perfect graph theorem (Lovász 1972)). A graph G is perfect if and only if its complement G is perfect. Proof. By symmetry, it is enough to prove that if G is perfect, then G is perfect. The proof is by induction on |V (G)|. The theorem is trivially true for graphs with |V (G)| = 1. Now assume that G is a graph on at least 2 vertices. Let K denote the set of all vertex sets of cliques of G, and let A be the set of all independent sets in G of size α(G). Every proper induced subgraph of G is the complement of a proper induced subgraph of G, and is hence perfect by induction. Therefore, it suﬃces to prove that χ(G) ≤ ω(G) = α(G). We will prove the existence of a set K ∈ K intersecting all A ∈ A. Then ω(G − K) = α(G − K) < α(G) = ω(G),

2 so by the induction hypothesis, χ(G) ≤ χ(G − K) + 1 = ω(G − K) + 1 ≤ ω(G), as desired. Suppose for a contradiction that there is no such K, i.e., for every K ∈ K there exists AK ∈ A such that K ∩ AK = ∅. We replace every vertex v in G by a complete graph Gv on

k(v) = |{K ∈ K : v ∈ AK }| vertices, putting all the edges between Gu and Gv whenever uv ∈ E(G). Let 0 0 S G be the graph thus obtained: it has vertex set V (G ) = v∈V (G) V (Gv), and 0 two vertices w ∈ V (Gu) and x ∈ V (Gv) are adjacent in G if and only if u = v or uv ∈ E(G). Clearly, G0 can be obtained by replicating certain vertices in the graph G[{v ∈ V (G): k(v) > 0}], which is an induced subgraph of G, and so is perfect. By Lemma 3, it follows that G0 is perfect, and thus χ(G0) ≤ ω(G0). (1) Let us now determine the values of χ(G0) and ω(G0). Firstly, every maxi- 0 0 S mal clique of G has the form G [ v∈X Gv] for some X ∈ K. Therefore, there exists a set X ∈ K such that X ω(G0) = k(v) v∈X

= |{(v, K): v ∈ X,K ∈ K, v ∈ AK }| X = |X ∩ AK | K∈K ≤ |K| − 1.

(Observe that since AK is independent and G[X] is complete, |X ∩ AK | ≤ 1 for all K. Moreover, |X ∩ AX | = 0 by the choice of AX .) Secondly, we compute the value of |V (G0)|. X |V (G0)| = k(v) v∈V (G)

= |{(v, K): v ∈ V (G),K ∈ K, v ∈ AK }| X = |AK | K∈K = |K| · α(G).

3 We can now ﬁnd the desired contradiction: |V (G0)| |V (G0)| χ(G0) ≥ ≥ = |K|. α(G0) α(G)

Hence, χ(G0) > ω(G0); contradicting (1)!

3 The strong perfect graph theorem

The following theorem was also conjectured by Berge. Its proof is over a hundred pages long, so we shall not give it here!

Theorem 5 (Strong perfect graph theorem (Chudnovsky, Robertson, Sey- mour and Thomas 2002)). A graph G is perfect if and only if G and G contains no induced odd cycle of length at least 5.

The strong perfect graph theorem clearly implies the perfect graph theorem (right?).