A graph is $k$-choosable (also known as $k$-list-colorable) if, for every function $f$ that maps vertices to sets of $k$ colors, there is a color assignment $c$ such that, for all vertices $v$, $c(v)\in f(v)$, and such that, for all edges $vw$, $c(v)\ne c(w)$.
Now suppose that a graph $G$ is not $k$-choosable. That is, there exists a function $f$ from vertices to $k$-tuples of colors that does not have a valid color assignment $c$. What I want to know is, how few colors in total are needed? How small can $\cup_{v\in G}f(v)$ be? Is there a number $N(k)$ (independent of $G$) such that we can be guaranteed to find an uncolorable $f$ that only uses $N(k)$ distinct colors?
The relevance to CS is that, if $N(k)$ exists, we can test $k$-choosability for constant $k$ in singly-exponential time (just try all $\binom{N(k)}{k}^n$ choices of $f$, and for each one check that it can be colored in time $k^n n^{O(1)}$) whereas otherwise something more quickly growing like $n^{kn}$ might be required.