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 Question from JimB: I cannot seem to get this to work. Previous answers cover 2-balls, but we are 4-balls. What we really want is for each player to play with each other, at least once and no more than twice. Ideally, each would also play equally in the first and the second 4-ball of the day. What I can get to leaves every player playing twice with 6 of the others, but every player NOT playing with one other. abcd efhg gebd cafh agdh ecbf chde gbfa a does not play with e b does not play with h c does not play with g d does not play with f Am I asking too much, or can it be done?

Hi Jim,

Unfortunately, the schedule you want does not exist.

We wrote a computer program to search through all of the possible schedules for 8 players in two foursomes daily for 4 days. Here is the most balanced schedule according to the measure we used. Here it is, the number in position i is the group (0 or 1) that player i is in on that day

Day 1 : (0, 0, 0, 0, 1, 1, 1, 1)
Day 2 : (0, 0, 1, 1, 0, 0, 1, 1)
Day 3 : (0, 0, 1, 1, 1, 1, 0, 0)
Day 4 : (0, 1, 0, 1, 0, 1, 0, 1)

This corresponds to....
abcd efgh
abef cdgh
abgh cdef
aceg bdfh

The tee times can be assigned arbitrarily so that "a" is not always in the first group.

Everyone plays everyone else at least once, with some playing three times. It isn't possible to have every pair of players play either 0 or 1 times over the first 2 days. No matter how you do it, each of the foursomes on each day after the first will contain either 2 pairs of 2 players who played together the previous day, or three players who played together the previous day.

Happy golfing!
--Victoria

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