I'll have something more to say in a while, but for the time being, note that there is essentially only one round-robin for 6 players, up to permutation of the players, permutation of the rounds, permutation of the the three matches in a round and swapping the two players in a match. Clearly there's no loss in assuming AB CD EF now x is not D as AB CD EF AC Bx this would force AC BE DF AD By EF again so x = E AD BF CE AE Bz (or F). wlog x = E AE BD CF AF B Then y = F, z = D AF BC DE Now a machine could insert the men in a few minutes, and I'm fairly sure that you can't get 5 rounds -- praps not even 4. I'll copy to math-fun, cos there's several people who can do this. Not the prettiest arrangement, but again without loss we can assume that the first round is Aa.Bb Cc.Dd Ee.Ff However, I still think that for sufficiently large values of 16, 20, 24, ... one can find a complete set of rounds. R. On Mon, 26 Jan 2004, R. Hess wrote:
Dear Richard,
Thanks for your schedule for 8 players which is correct. I made a mistake earlier and this should help my thinking. Now is there a way to extend to 12 or 20 players?
All the best, Dick