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 Question from Don: 2 foursomes and 1 threesome for 6 rounds of golf

Don,

It is hard to find the "best" schedule for 11 players over 6 rounds. I have a program to do that and it is running, but it could take a long time as there are about $2000^5$ possibilities to check. As a first step, the program finds a schedule that is "close to" balanced. The best one it found is copied below.

Day 1 : (0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2)
Day 2 : (0, 0, 1, 2, 0, 1, 1, 2, 0, 1, 2)
Day 3 : (0, 1, 0, 2, 0, 1, 2, 1, 2, 1, 0)
Day 4 : (0, 1, 1, 2, 2, 0, 1, 1, 0, 2, 0)
Day 5 : (0, 1, 2, 0, 1, 1, 0, 2, 2, 0, 1)
Day 6 : (0, 1, 2, 2, 0, 2, 1, 0, 1, 0, 1)

The way to read it is that the players are the 11 positions in the sequence and the numbers indicate the groups. One Day 1 the groups are 1, 2, 3, 4; 5, 6, 7, 8; 9, 10, 11. On Day 2 group zero is 1, 2, 5, 8, and so on.

There are a couple of flaws. For one, player 1 is always in group zero. For another, some players are together 4 times. In a perfect world (which probably does not exist), no pair of players would be together more than two times. On the other hand, in a perfect world some pairs of players would play together only once, and I think that that this schedule has every pair of players together once (though I did not check carefully).

If an improvement is found, I will send a follow-up answer. A feasible alternative is to try generating a few million schedules at random and then sending the "best" one of those.

--Victoria

Don Responded

Victoria apologizing in advance but I don"t understand that (0,0,0,0,1,1,1,12,2,2) terminology it is okay for the fist round but I cannot follow after that. Could you just # players 1 to 11 for the 6 rounds.

Hi Don,

Rather than number the players from 1 to 11 I named them with letters, A, B, C, ... up to K. I then redid Victoria's table with the players names at the head of each column.

Each day there are three groups, group zero which is a foursome, group one which is a foursome and group two which is a threesome. To see which group a player is with on any day look down the column. For example player C is in group zero on day 1, group one on day 2, group zero on day 3, group one on day 4, and the threesomes on days 5 and 6.

A B C D E F G H I J K
Day 1 0 0 0 0 1 1 1 1 2 2 2
Day 2 0 0 1 2 0 1 1 2 0 1 2
Day 3 0 1 0 2 0 1 2 1 2 1 0
Day 4 0 1 1 2 2 0 1 1 0 2 0
Day 5 0 1 2 0 1 1 0 2 2 0 1
Day 6 0 1 2 2 0 2 1 0 1 0 1

Does this help?
Harley

Don wrote again

Question from Don:

Victoria answered my question about 2 foursomes and 1 threesome playing 6 rounds of golf. I was wondering if she could tweak it because player 4 plays in a threesome 4 times and player 9 3 times in a threesome whereas players 1and 2 never play in a threesome. I know I have been a bother but would appreciate any help you could give me.

Don,

It isn't so easy to make a few small changes and not spoil the balance completely. The way my program currently works is based on the false assumption that the groups are the same size. Since, in that case, the groups could be renumbered, it is ok to always put player 1 in group number 1, and of player 2 is not also in group 1, then s/he can go in group 2.

I've made a try at changing the schedule. But really I am just eyeballing it. You can probably do at least as well as me.

Day 1 : (2, 2, 0, 0, 1, 1, 1, 1, 0, 2, 2)
Day 2 : (0, 0, 1, 2, 0, 1, 1, 2, 0, 1, 2)
Day 3 : (0, 1, 0, 2, 0, 1, 2, 1, 2, 1, 0)
Day 4 : (2, 0, 1, 1, 2, 0, 1, 1, 0, 2, 0)
Day 5 : (0, 1, 2, 0, 1, 1, 0, 2, 2, 0, 1)
Day 6 : (0, 2, 2, 1, 0, 2, 1, 0, 1, 0, 1)

--Victoria

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