Math CentralQuandaries & Queries


Question from Paul:

I have 18 golfers playing four rounds. I want two 3 balls and three 4 balls each day. I don't want anyone to play in a 3 ball more than once and I don't want anyone to play with the same person twice. I realise everybody can't play with everybody, I just want the best you can do.


There are 2 threesomes per round, for a total of 6 players. That means 24 players have been in threesomes over 4 rounds, so at least 6 players must be in threesomes twice. I don’t know if it is possible to achieve that number.

Here is a possible solution that’s not terrible. Start with the schedule here for 20 golfers over 4 rounds, The teams that never play together are those in the same column of the array. So delete teams 16 and 20 from this schedule (and renumber) and you will get a schedule with two 3s and two 4s in which no pair or players is together more than once.

This schedule has 9 players in threesomes twice, and 3 players never in them. The number of players involved is too large to search all possibilities with a computer.


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