Let be a graph. If and are two integers, a -colouring of is a function from to such that for each edge . Given a list assignment of , i.e.~a mapping that assigns to every vertex a set of non-negative integers, an -colouring of is a mapping such that for every . A list assignment is a --list-assignment if and for each vertex . Given such a list assignment , the graph G is --colourable if there exists a --colouring , i.e. is both a -colouring and an -colouring. For any real number , the graph is --choosable if it is --colourable for every --list-assignment . Last, is circularly -choosable if it is --choosable for any , . The circular choosability (or circular list chromatic number or circular choice number) of G is

**Problem**What is the best upper bound on circular choosability for planar graphs?

The problem was first posed in 2003 by Mohar (Problem 4 of link*) who suggested the answer should be between 4 and 5.

Some time later, Havet, Kang, Müller, and Sereni [HKMS] showed that in fact the answer is somewhere between 6 and 8. The upper bound extends a celebrated planar choosability proof due to Thomassen [T]. The lower bound is by way of an elementary, though rather large, construction.

## Bibliography

[HKMS] F. Havet, R. J. Kang, T. Müller, and J.-S. Sereni. Circular choosability. J. Graph Theory 61 (2009), no. 4, 241--270.

[T] C. Thomassen. Every planar graph is 5-choosable. J. Combinatorial Theory B 62 (1994) 180--181

* indicates original appearance(s) of problem.