Math CentralQuandaries & Queries


Question from Mike:

Regarding arranging golf players so no person plays with anyone more than once. You have given examples for 16 and 24 players. If it can be done, i need a solution for 20 players, 4 players per team one round per day for 5 days
Thank you


For 20 players, there are 20 choose 2 = 190 pairings. Each foursome accounts for 4 choose 2 = 6 of them. Since 190 is not evenly divisible by 6, there is no collection of foursomes where each person plays with every other person exactly once.

If you have 5 foursomes per day, then you account for 30 pairings per day, and 150 pairings in all, so some people
will not play together.

You can construct a schedule as follows. First make a schedule for 25 golfers with 5 fivesomes per day for 6 days. This exists (it is an affine plane of order 5). You can make one in the same way as the 16 player example. It works because 5 is prime.

After making the schedule for 25 players, choose one of the fivesomes that play on day 6 -- say 21,22,23,24,25. Then, delete these players from every fivesome where they appear on days 1 through 5. This is your schedule. (That is, forget about day 6 -- these are the players that are not in your tournament, or the pairs who will will never play together.)

Hope this helps. If you need more info, write back.


About Math Central


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS