Bisimplicial vertices in even-hole-free graphs

@article{AddarioBerry2008BisimplicialVI,
  title={Bisimplicial vertices in even-hole-free graphs},
  author={Louigi Addario-Berry and M. Chudnovsky and Fr{\'e}d{\'e}ric Havet and Bruce A. Reed and Paul D. Seymour},
  journal={J. Comb. Theory, Ser. B},
  year={2008},
  volume={98},
  pages={1119-1164}
}
Even-hole-free graphs still have bisimplicial vertices.
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
Coloring even-hole-free graphs with no star cutset
A hole is a chordless cycle of length at least 4. A graph is evenhole-free if it does not contain any hole of even length as an induced subgraph. In this paper, we study the class of even-hole-free
Coloring even-hole-free graphs with no star cutset
TLDR
The optimal upper bound for its chromatic number in terms of clique number and a polynomial-time algorithm to color any graph in this class of even-hole-free graphs with no star cutset is given.
Triangulated neighborhoods in even-hole-free graphs
On the Structure of (Even Hole, Kite)-Free Graphs
TLDR
It is proved that a connected ( even hole, kite)-free graph isDiamond-free, or the join of a clique and a diamond-free graph, or contains a cliques cutset, and the Vizing bound is established for (even hole), which is β-perfect in the sense of Markossian, Gasparian and Reed.
On the chromatic number of a family of odd hole free graphs
TLDR
It is proved that for (odd hole, full house)-free graph G, χ(G) ≤ ω(G)+ 1, and the equality holds if and only if ω (G) = 3 and G has H as an induced subgraph.
Even-hole-free graphs
TLDR
It is proved that every even-hole-free graphs has a node whose neighborhood is triangulated, which implies that in an even- hole-free graph, with n nodes and m edges, there are at most n+2m maximal cliques, and yields a fastest known algorithm for computing a maximum clique.
Even-hole-free graphs that do not contain diamonds: A structure theorem and its consequences
Coloring graphs with no even holes ≥ 6: the triangle-free case
TLDR
It is proved that the class of graphs with no triangle and no induced cycle of even length at least 6 has bounded chromatic number and the existence of C_4 is allowed.
Graphs without even holes or diamonds
An even hole is an induced chordless cycle of even length at least four. A diamond is an induced subgraph isomorphic to K_4-e. We show that graphs without even holes and without diamonds can be
...
...

References

SHOWING 1-10 OF 38 REFERENCES
Triangulated neighborhoods in even-hole-free graphs
Even-hole-free graphs
TLDR
It is proved that every even-hole-free graphs has a node whose neighborhood is triangulated, which implies that in an even- hole-free graph, with n nodes and m edges, there are at most n+2m maximal cliques, and yields a fastest known algorithm for computing a maximum clique.
Even‐hole‐free graphs part II: Recognition algorithm
TLDR
An algorithm that determines in polytime whether a graph contains an even hole is presented, based on a decomposition theorem for even‐hole‐free graphs obtained in Part I of this work.
ω-Perfect graphs
Abstractω-Perfect graph is defined and some classes of ω-perfect graphs are described, although the characterization of the complete class of ω-perfect graphs remains an open question. A bound on the
Even‐hole‐free graphs part I: Decomposition theorem
TLDR
A decomposition theorem for even-hole-free graphs is proved and this theorem is used in the second part of this paper to obtain a polytime recognition algorithm for even -hole- free graphs.
Detecting even holes
TLDR
This algorithm is simpler — it is able to search directly for even holes, while the algorithm of [4] made use of a structure theorem for even-hole-free graphs, proved in an earlier paper.
(b)
But (19) implies that u ∈ B 1 , a contradiction. This proves (36)
  • But (19) implies that u ∈ B 1 , a contradiction. This proves (36)
34) Some vertex of A 2 is complete to B 1,2
  • 34) Some vertex of A 2 is complete to B 1,2
1) holds. By (11) and since b is non-adjacent to w , it follows that M(w ) ⊆ M(b) and similarly M(w) ⊆ M(b ) Again by
  • M(b) ⊆ M(w) and M(b ) ⊆ M(w ), and therefore all the inclusions hold with equality, that is M(w) = M(w ) = M(b) = M(b ). Next we claim that either N(b) ∩ P ⊆ N(b ) ∩ P or N(b ) ∩ P ⊆ N(b) ∩ P. Suppose not
...
...