2-colouring a graph without a monochromatic maximum clique

Importance: Medium ✭✭
Recomm. for undergrads: no
Posted by: Jon Noel
on: August 25th, 2013
Conjecture   If $ G $ is a non-empty graph containing no induced odd cycle of length at least $ 5 $, then there is a $ 2 $-vertex colouring of $ G $ in which no maximum clique is monochromatic.

A $ 2 $-division of a graph $ G $ is a partitioning of $ G $ into two subgraphs, neither of which contains a maximum clique. It is known that every perfect graph admits a $ 2 $-division. Thus, by the Strong Perfect Graph Theorem [CRS], a graph which does not contain an induced copy of an odd cycle of length at least $ 5 $ or its complement has a $ 2 $-division. Hoàng and McDiarmid [HMcD] also prove that a claw-free graph admits a 2-division if and only if it does not contain an induced odd cycle of length at least $ 5 $. The conjecture says that this holds for all graphs.

This problem was featured as unsolved problem #49 in Bondy and Murty's book "Graph Theory" [BM].

See also a posting on the American Institute of Mathematics website, contributed by Bruce Reed.


[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: The strong perfect graph theorem, Ann. of Math. (2) 164 (2006), no. 1, 51--229. MathSciNet

[HMcD] C.T. Hoàng, C. McDiarmid, On the divisibility of graphs, Discrete Math. 242 (1–3) (2002) 145–156.

[BM] J. A. Bondy and U. S. R. Murty. Graph theory, volume 244 of Graduate Texts in Mathematics. Springer, New York, 2008.

* indicates original appearance(s) of problem.