Hi Simon,
There is no nice solution to your 8 player problem as it is not possible to avoid having each player play 3 times with one out of the other 7 players.
In the plan below, the pairings GC, DH, EB and AF all occur three times within a foursome,
all other pairings occur either once or twice.
Foursome 1 Foursome 2
--------- ---------
(A B C D) (E F G H)
(H C D G) (F A E B)
(B D H E) (C G F A)
(D H A F) (G E B C)
The rounds are arranged above so that each player occurs exactly once in each of the four positions within a foursome. Pairs of foursomes from different rounds all share 2 players in common, however there are never 3 players in common (this gives you the best possible social mixing that Richard and others like to have).
Regards,
Ian.
Monday, September 19 2005, 02:24 am
