Richard,
At first glance it appears that it might be possible for no pair to play together more than twice, however for reasons similar to those that I have outlined in my
message of Nov 1st it is not possible. With 7 rounds you must have at least 3 pairs of players playing together three times.
But may be there is an alternative... If you played the schedule below then after 6 rounds everybody would have played exactly twice with 9 out of the other 11 players and not at all with the other 2 players.
Round
1 (A B C D) (E F G H) (I J K L)
2 (G H A J) (K L E B) (C D I F)
3 (A F C L) (E J G D) (I B K H)
4 (A B G L) (E F K D) (I J C H)
5 (K D A J) (C H E B) (G L I F)
6 (K H A F) (C L E J) (G D I B)
7 (A E I) (B F J) (C G K) (D H L)
If in the final round you played 4 threesomes, then this mops up all 12 of the pairs that have not yet occured together.
Alternatively you could use round 7 as the assignment of the 12 players to four teams of three. Then the first six rounds would constitute a tournament where each player played against all members of the other teams exactly twice. If you wanted a 7th round in this case, you could run a play-off between the four teams.
--
Monday, December 5 2005, 01:30 am --
Hope that helps,
Regards,
Ian.
PS send me and e-mail if you want the first schedule I discussed.
Thursday, December 1 2005, 01:33 pm
