Lecture #11 discussion

In the lecture, I forgot to mention why we chose to talk about the classes of perfect graphs we did. We showed that bipartite graphs and line graphs of bipartite graphs (and their complements) are all perfect.Let’s call a graph basic if it belongs to one of these four classes: bip, co-bip, line(bip), or co-line(bip).

Recall a graph was called a Berge graph if it did not contain (as an induced subgraph) any odd hole or odd antihole. Clearly, all perfect graphs are Berge graphs, else they could contain such a offending induced subgraph, which would contradict their perfectness. The other direction is the tricky one: showing that all Berge graphs are perfect. Which is what Maria Chudnovsky, Neil Robertson, Paul Seymour, and Robin Thomas eventually did in 2002.

So how did they proceed? Here’s a mile-high view of their approach. They showed that every Berge graph G either is basic (i.e., belongs to those four classes bip, co-bip, line(bip), or co-line(bip)), or else it has some structural “faults” (G has a 2-join, or its complement has a 2-join, or G has an M-join, or a skew partition). For us, it does not matter what exactly these structural faults are: what matters is (and this is what CRST showed) that minimal imperfect graphs cannot have any of these faults. So, if Berge graphs could be imperfect, a minimal counterexample would be basic, which is a contradiction. More on this can be found, e.g., off the links on Vasek Chvatal’s Perfect Graph Theorem Page.

Re Srivatsan’s question: given the strong perfect graph theorem, the perfect graph theorem follows as a corollary. Indeed, suppose G was perfect but \overline{G} was not. Then \overline{G} must not be a Berge graph, and contains either an odd hole or odd antihole as an induced subgraph. But then G contains an odd antihole or odd hole as an induced subgraph, so could not have been perfect.


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