Recomm. for undergrads: yes
Posted by: Jon Noel
on: April 12th, 2014
Question   Given $ a,b\geq2 $, what is the smallest integer $ t\geq0 $ such that $ \chi_\ell(K_{a,b}+K_t)= \chi(K_{a,b}+K_t) $?

The list chromatic number of a graph $ G $, denoted $ \chi_\ell(G) $, is the minimum $ k $ such that for every assignment of lists of size $ k $ to the vertices of $ G $ there is a proper colouring in which every vertex is mapped to a colour in its own list. For more background on the list chromatic number, see [3].

Given graphs $ G $ and $ H $, the join of $ G $ and $ H $, denoted $ G+H $, is obtained by taking disjoint copies of $ G $ and $ H $ and adding all edges between them. Ohba [1] proved that for every graph $ G $ there exists $ t\geq0 $ such that $ \chi_\ell(G+K_t)= \chi(G+K_t) $. The question above asks to determine the minimum value of $ t $ in the case that $ G $ is a complete bipartite graph. It seems that it was first studied in [4], although this is unclear; for the time being, we have chosen to attribute this problem to J. Allagan.

Define $ \phi(a,b) $ to be the minimum $ t $ such that $ \chi_\ell(K_{a,b}+K_t)= \chi(K_{a,b}+K_t) $. Note that, if $ G $ is a complete multipartite graph with at most one non-singleton part, then we see that $ \chi_\ell(G)=\chi(G) $ by colouring the vertices of the non-singleton part last. Thus, if $ a $ or $ b $ is equal to 1, then $ \phi(a,b)=0 $. As it turns out, $ \phi(2,2)=\phi(2,3)=0 $ and $ \phi(3,3)=\phi(2,4)=1 $. This can be deduced from the following result of [2] and the fact that $ \chi_\ell(K_{3,3})=\chi_\ell(K_{4,2})=3 $:

Theorem  (Noel, Reed, Wu (2012))   If $ |V(G)|\leq 2\chi(G)+1 $, then $ \chi_\ell(G)=\chi(G) $.

The above result of [2] implies that if $ a+b\geq 5 $, then $ \phi(a,b)\leq a+b-5 $. However it seems that, for most values of $ a,b $, this bound is far from tight.

A simple observation is that, since $ \chi_\ell(K_{a,b}+K_t)\geq \chi_\ell(K_{a,b}) $ for all $ t $, we must have \[\phi(a,b)\geq \chi_\ell(K_{a,b}) - \chi(K_{a,b}) = \chi_\ell(K_{a,b}) -2.\]

The following is a result of Allagan [4]:

Theorem  (Allagan (2009))   If $ a\geq5 $, then \[\lfloor \sqrt{a}\rfloor - 1 \leq \phi(a,2)\leq \left\lceil\frac{-7+\sqrt{8a+17}}{2}\right\rceil.\]

This implies that $ \phi(a,2)=1 $ for $ 4\leq a\leq 8 $ and that $ \phi(a,2)=2 $ for $ 9\leq a\leq 13 $.

Bibliography

[1] K. Ohba. On chromatic-choosable graphs, J. Graph Theory. 40 (2002) 130--135. MathSciNet.

[2] J. A. Noel, B. A. Reed, H. Wu. A Proof of a Conjecture of Ohba. Submitted. pdf.

[3] J. A. Noel. Choosability of Graphs with Bounded Order: Ohba's Conjecture and Beyond. Master's Thesis, McGill University. pdf.

[4] J. A. D. Allagan. Choice Numbers, Ohba Numbers and Hall Numbers of some complete $ k $-partite graphs. PhD Thesis. Auburn University. 2009.


* indicates original appearance(s) of problem.

Reply

Comments are limited to a maximum of 1000 characters.
More information about formatting options