Scheduling 12 golfers in foursomes seems to be a common question here, but unfortunately there are no good solutions when the number of rounds is small. An alternative approach is presented below.
This message describes the basic problem with regard to a four round schedule for 12 players in foursomes. Five rounds has similar problems, the best possible solution will have 6 pairs of players who never play together and 30 pairs of players who play together twice. It may be best to avoid this situation all together, perhaps by playing in threesomes. For example in the first four rounds in the schedule below, no pair of players plays together twice, or in other words every player plays with exactly 8 out of other 11 players in the group.
Round 1 (A E I) (B F J) (C G K) (D H L)
Round 2 (A F K) (B E L) (C H I) (D G J)
Round 3 (A H J) (D E K) (B G I) (C F L)
Round 4 (A G L) (C E J) (D F I) (B H K)
Round 5 (A B C D) (E F G H) (I J K L)
If the optional fifth round of three foursomes is played, then everyone would play with each other exactly once. Alternatively, if round 5 is not played, it can still be used to define teams of four among the 12 players. So (A B C D) are team 1, (E F G H) are team 2, and (I J K L) are team 3. The way the schedule is presented above it can easily be seen that the opponents of player A from team 1 (in rounds one to 4), are always a player from team 2 and a player from team 3. In fact the same property applies to all 12 players and everyone will play with all players from opposing teams and none from their own team.
When its possible to play more than five rounds with 12 players then workable solutions using foursomes do exist. For example see
here for 7 rounds or
here for 8 & 9 rounds.
Ian.
