A taste of perfect graphs Remark Determining the chromatic number of a graph is a hard problem, in general, and it is even hard to get good lower bounds on χ(g). An obvious lower bound we have seen before is ω(g), the clique number of a graph. There are some interesting graphs for which χ(g) = ω(g), including the following: Bipartite graphs χ(g) = ω(g) = 2 unless G is edgeless Complements of bipartite graphs prob 5.1.38 of text: prove that χ(g) = ω(g) when G is bipartite Line graphs of bipartite graphs χ(l(g)) = χ (G) = (G) = ω(l(g)) when G is bipartite where we recall that vtcs. form a clique in L(G) corresponding edges in G have a common endpoint (star) or form a triangle Complements of line graphs of bipartite graphs Claim: if G is bipartite, χ(l(g)) = ω(l(g)). Definition The clique cover number, denoted θ(g), of a graph G is the minimum # of cliques in G needed to cover V (G). Examples Math 104 - Prof. Kindred - Lecture 23 Page 1
Notice that θ(g) = χ(g) since covering vtcs. with cliques in G Assume G is bipartite. We have because covering vtcs. with independent sets in G. χ(l(g)) = θ(l(g)) = β(g) Furthermore, vtcs. form a clique in L(G) corresponding edges in G have a common endpoint (star). since ω(l(g)) = α(l(g)) = α (G) independent sets in L(G) matchings in G. By König Egerváry theorem, we know α (G) = β(g). Comparability graphs A comparability graph has a vertex for each element of a partial order and an edge between any two comparable elements. In this case, χ(g) = min size of a partition of the poset into chains = max size of a chain = ω(g) Complements of comparability graphs Dilworth s theorem can be used to show that χ(g) = ω(g) when G is a comparability graph Math 104 - Prof. Kindred - Lecture 23 Page 2
While the above classes of graphs are all interesting, not every graph satisfying χ(g) = ω(g) is... Take any graph with at most 20 vtcs. and consider the disjoint union of it and K 20. Clearly this graph satisfies χ(g) = ω(g), but it is certainly not interesting in general. Goal: define a class of graphs satisfying χ(g) = ω(g), containing the interesting graphs mentioned above but not uninteresting ones. Strategy: make the property hereditary. Definition A family of graphs G is hereditary if every induced subgraph of a graph in G is also a graph in G. Definition A graph G is perfect if for all induced subgraphs H of G, χ(h) = ω(h). Otherwise, we say the graph is imperfect. Claude Berge first formalized notion of perfect graphs in the early 1960 s. Families of graphs that are perfect graphs All of the classes from previous list Complete graphs K n Edgeless graphs K n Wheel graphs with an odd # of vtcs. Interval graphs Intervals on the real line and the corresponding interval graph. Chordal graphs (proof by induction 1 ) Cographs (proof by induction) Math 104 - Prof. Kindred - Lecture 23 Page 3
Examples of graphs that are not perfect C 5, C 7, C 9,... (sometimes termed odd holes) C 5, C 7, C 9,... (sometimes termed odd antiholes) Definition An imperfect graph is minimally imperfect if each of its proper induced subgraphs is perfect. Examples Minimally imperfect graphs C 7 and C 7 1 1 7 2 7 2 6 3 6 3 5 4 5 4 Berge observed that all perfect graphs in the classes previously mentioned also have perfect complements, which led to the Perfect Graph Conjecture (1961). This conjecture was proved by Lovász (1972), resulting in what is now known as the Perfect Graph Theorem. Perfect Graph Theorem. A graph G is perfect if and only if G is perfect. Not long after the Perfect Graph Theorem, the following beautiful characterization of perfect graphs was proposed by Hajnal and proven by Lovász. 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). 1 A proof is given on page 227 of Introduction to Graph Theory by Douglas West. Math 104 - Prof. Kindred - Lecture 23 Page 4
Remark have Note that given the above inequality, if we consider H, we V (H) = V (H) α(h)ω(h) = ω(h)α(h). Thus, the above theorem implies the Perfect Graph Theorem. For a proof of the α ω theorem, we first need the following property of minimally imperfect graphs. Proposition. Let S be an independent set in a minimally imperfect graph G. Then ω(g S) = ω(g). Proof. We have the following string of inequalities: ω(g S) ω(g) χ(g) 1 since G is imperfect χ(g S) since χ(g) 1 + χ(g S) as vtcs. in S can all be assigned one color = ω(g S) by def n of minimally imperfect. Since we begin and end with ω(g S), equality must hold throughout; in particular, ω(g S) = ω(g). We can now establish a result on the structure of 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, Math 104 - Prof. Kindred - Lecture 23 Page 5
(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). Proof. Assume G = (V, E) is a minimally imperfect graph. Let S 0 be a maximum independent set of G, and let v S 0. The graph G v is perfect because G is minimally imperfect. Thus, χ(g v) = ω(g v) ω(g). This implies that for any v S 0, the set V v can be partitioned into a collection S v of ω(g) independent sets (namely the color classes in a coloring of G v using ω(g) colors). Denoting { S v : v S 0 } by {S 1, S 2,..., S α(g)ωg }, it can be seen that {S 0, S 1, S 2,..., S α(g)ωg } contains α(g)ω(g) + 1 independent sets of G satisfying property (a). if v S 0, then v is in... S 0 one of S ω(g)+1,..., S 2ω(G) one of S 2ω(G)+1,..., S 3ω(G). one of S (α(g) 1)ω(G)+1,..., S α(g)ω(g) By previous proposition, we also know that if v S 0, then v is not in any of S 1,..., S ω(g) ω(g S i ) = ω(g) for any 0 i α(g)ω(g). Hence, there exists a maximum clique C i of G that is disjoint from S i. Because no two vertices of C i can belong to a common independent set, we know C i S j 1 for i < j. But, in addition, because each of the Math 104 - Prof. Kindred - Lecture 23 Page 6
ω(g) vertices in C i lies in α(g) of the independent sets S j (by property (a)), we must have C i S j = 1 for 0 i < j α(g)ω(g). Example α(c 7 ) = 2 and ω(c 7 ) = 3 1 max independent set S 0 = {1, 2} 7 2 color classes in 3-coloring of C 7 {1} : S 1 = {2, 3}, S 2 = {4, 5}, S 3 = {6, 7} 6 5 4 3 color classes in 3-coloring of C 7 {2} : S 4 = {3, 4}, S 5 = {5, 6}, S 6 = {1, 7} C 1 = {1, 4, 6}, C 2 = {1, 3, 6}, C 3 = {1, 3, 5}, C 4 = {2, 5, 7}, C 5 = {2, 4, 7}, C 6 = {2, 4, 6} 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 the harder direction next class. Math 104 - Prof. Kindred - Lecture 23 Page 7