Consider the complete
bipartite graph , having six vertices , , , , , such that and are each connected to all of , , , and , and no other vertices are connected. As a bipartite graph, has usual chromatic number 2: one may color and in one color and , , , in another and no two adjacent vertices will have the same color. On the other hand, has list-chromatic number larger than 2, as the following construction shows: assign to and the lists {red, blue} and {green, black}. Assign to the other four vertices the lists {red, green}, {red, black}, {blue, green}, and {blue, black}. No matter which choice one makes of a color from the list of and a color from the list of , there will be some other vertex such that both of its choices are already used to color its neighbors. Thus, is not 2-choosable. On the other hand, it is easy to see that is 3-choosable: picking arbitrary colors for the vertices and leaves at least one available color for each of the remaining vertices, and these colors may be chosen arbitrarily. with three colors per vertex. No matter which colors are chosen for the three central vertices, one of the outer 27 vertices will be uncolorable, showing that the list chromatic number of is at least four. More generally, let be a positive integer, and let be the
complete bipartite graph . Let the available colors be represented by the different two-digit numbers in
radix . On one side of the bipartition, let the vertices be given sets of colors {{math|{
i0,
i1,
i2, ...}}} in which the first digits are equal to each other, for each of the possible choices of the first digit . On the other side of the bipartition, let the vertices be given sets of colors {{math|{0
a, 1
b, 2
c, ...}}} in which the first digits are all distinct, for each of the possible choices of the -tuple The illustration shows a larger example of the same construction, with . Then, does not have a list coloring for : no matter what set of colors is chosen for the vertices on the small side of the bipartition, this choice will conflict with all of the colors for one of the vertices on the other side of the bipartition. For instance if the vertex with color set {00,01} is colored 01, and the vertex with color set {10,11} is colored 10, then the vertex with color set {01,10} cannot be colored. Therefore, the list chromatic number of is at least . Similarly, if n=\tbinom{2k-1}{k}, then the complete bipartite graph is not -choosable. For, suppose that colors are available in total, and that, on a single side of the bipartition, each vertex has available to it a different -tuple of these colors than each other vertex. Then, each side of the bipartition must use at least colors, because every set of colors will be disjoint from the list of one vertex. Since at least colors are used on one side and at least are used on the other, there must be one color which is used on both sides, but this implies that two adjacent vertices have the same color. In particular, the
utility graph has list-chromatic number at least three, and the graph has list-chromatic number at least four. ==Properties==