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.
How many distinct colors are needed to lower-bound the choosability of a graph?
graph theory;co.combinatorics;graph colouring;exp time algorithms
Daniel Krl and Ji Sgall answered your question to the negative. From the abstract of their paper:A graph $G$ is said to be $(k,\ell)$-choosable if its vertices can be colored from any lists $L(v)$ with $|L(v)| \ge k$, for all $v\in V(G)$, and with $|\bigcup_{v\in V(G)} L(v)| \le \ell$. For each $3 \le k \le \ell$,  we construct a graph $G$ that is $(k,\ell)$-choosable but not $(k,\ell+1)$-choosable.So, $N(k)$ does not exist if $k\ge 3$. Krl and Sgall also show that $N(2)=4$. Of course, $N(1)=1$.Daniel Krl, Ji Sgall: Coloring graphs from lists with bounded size of their union. Journal of Graph Theory 49(3): 177-186 (2005)