Date: Mon, 06 Apr 1998 23:38:23 -0400
From: Alex Bogomolny
I believe nothing is known about the number of round robin tournaments. If you take one solution for 2n players, and the ordering within a round and the rounds themselves matter, then, by permuting things, there are n! arrangements for each round and (2n-1)! orderings of the rounds, leading to (2n-1)n!(2n-1)! tournaments based on just one tournament. The interesting, and, I think, unknown question, is the number of inequivalent tournaments relative to these permutations. There is certainly more than one, even in small cases. The following is a solution for n=3.
{1,2}{3,4}{5,6}
{1,3}{2,5}{4,6}
{1,4}{2,6}{3,5}
{1,5}{2,4}{3,6}
{1,6}{2,3}{4,5}
Now, if you interchange the symbols 1 and 2, you get a new solution
{2,1}{3,4}{5,6}
{2,3}{1,5}{4,6}
{2,4}{1,6}{3,5}
{2,5}{1,4}{3,6}
{2,6}{1,3}{4,5}
This solution is new because {2,3} and {1,5} appear in the same round but not in ANY round of the first solution. Hence this solution is not one of the 5(3!)5!=3600 permutations of the first solution; there are at least 3600 solutions more than the first 3600.
Ignoring permutations, there are at least two solutions for n=3, but I am sure you can see that there are more than two ignoring permutations. How many altogether? Who knows?
William
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