I was wondering if I could trouble you to assist me with the following: I am working on a schedule for 24 golfers. 6 groups of 4. I have 8 golf days (twice per week for a month).

Ideally, I would like to schedule all 24 golfers in 6 different groups of 4 on each day. Here is the catch.....no golfer in any group can be grouped togther more than once. Every group of 4 each day will have 4 new golfers that have never played together before. Is this possible?

Any assistance that you could provide will be greatly appreciated.

Thank you.

John

Dear John,

Such a tournament is not quite possible. Consider player 1, for example. For him to play with 3 different golfers each round for 8 rounds, he would need 24 other golfers; but, there are only 23 others available. It is, however, possible to come close with 24 golfers. We can arrange for each pair of the 24 golfers to play together exactly once, except for 12 disjoint pairs who would play together twice. Such an arrangement is described in the CRC HANDBOOK OF COMBINATORIAL DESIGNS, Edited by Charles J. Colbourn and Jeffrey H. Dinitz (CRC Press, 1996). See pages 264-5, Example 8.28 (the resolvable 2-(24, 4, 1) covering on 48 blocks). We have it listed below.

To answer the original question -- for a tournament having n players playing each day in foursomes, each player grouped with each other exactly once -- there are two necesary conditions:

1. Each player plays n-1 others, so that n-1 must be divisible by 3; and
   
   
2. Since they play in foursomes, n must be divisible by 4.

Put these conditions together you get that the number of golfers must be 12k + 4. The Handbook cited above says that these conditions can be satisfied for all k >= 0. Thus it is possible with n = 4, 16, 28, and so on. We provide below an example with n = 16 golfers.