12 golfers playing in foursomes

[e45324]

i have 12 golfers and 4 rounds of golf.   what is the math formula to have the players play with all of the other players at least once over 4 rounds   can I do it in four rounds or must I use 5.

[Ian Wakeling]

Unfortunately there is no formula to solve this problem.  Please have a look at this thread which links to another that tries to explain why there is no solution.

In fact you need at least six rounds before the all-plays-with-all criterion can be met.  There is an example of a schedule here.

The best compromise is to play five rounds, in which case there will be 6 pairs of players who never play together in the same foursome.  Ignore the example that you can find by following the first link above, instead use the schedule here, just ignore the doubles layout and turn each court into a foursome - the pairs that never play together are: (1 7), (2 8), (3 9), (4 10), (5 6), (11 12).

Hope that helps.

[rcschmidt]

I have 8 golfers that will be playing 10-12 rounds with two foursomes for each round.  How do I balance the foursomes so that everyone plays with each other the same number of times or close to the same num ber of times?

[Ian Wakeling]

For the first 7 rounds use this schedule and everyone will play with each other three times.  Without knowing the exact number of rounds you are going to play, I suggest you make a new assignment of the 8 players to the numbers 1 to 8 and then re-use 3 to 5 rounds from the same schedule.