• Corpus ID: 201694472

Even-hole-free graphs still have bisimplicial vertices.

  title={Even-hole-free graphs still have bisimplicial vertices.},
  author={M. Chudnovsky and Paul D. Seymour},
  journal={arXiv: Combinatorics},
A {\em hole} in a graph is an induced subgraph which is a cycle of length at least four. A hole is called {\em even} if it has an even number of vertices. An {\em even-hole-free} graph is a graph with no even holes. A vertex of a graph is {\em bisimplicial} if the set of its neighbours is the union of two cliques. In an earlier paper~\cite{bisimplicial}, Addario-Berry, Havet and Reed, with the authors, claimed to prove a conjecture of Reed, that every even-hole-free graph has a bisimplicial… 

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Thus z 1 = c, and hence V (H) ∩ Z 1 = ∅, and the other H-neighbour of c is some z 2 ∈ Z 2 . But then H \ {b, c} is an odd induced path of G between z 2 , B with interior in A ∪ C ∪ D, contrary to (2)
    Thus T (x) contains one of v 1 , v 5 . Hence |X| = 2, and so S is a path s 1 -· · · -s k say, where v 1 ∈ T (s 1 ) and v 5 ∈ T (S k )
    • Similarly T (x) = {v 4 }, and T (x) = {v 2 , v 4 } since T (x) is a clique
    There do not exist distinct y 1 , y 2 , y 3 ∈ N (a)
      Let G be an even-hole-free graph, such that 1.2 holds for all graphs with fewer vertices than G. Let K be a non-dominating clique in G with |K| ≤ 2, and let a ∈ V (G) \ N [K] be splendid
        We may choose k ∈ K adjacent or equal to v, and so k is not anticomplete
          If z ∈ Z 1 , every induced path between z and B with interior in Z 2 ∪ A ∪ C ∪ D is odd
            If v ∈ A ∪ C then u ∈ D from the definition of D; and if v ∈ D, let v ∈ V (F ) where F is a component of G \ V (S) such that F is anticomplete to a and not anticomplete to A∪ C
            • G belongs to V (S)