I need to arrange 5 goups of 4 golfers to play over a 7 day period with minimum duplication of foursomes

Hi rccody,

Well each player has 19 potential partners. To play with a fresh set of partners each day you would need at least 21 potential partners, so there is going to be a small amount of duplication of partners. Each person will have to play with two of the others twice and the rest once. You have enough partners for 6 days of unique pairings. On the last day they will have to play with at least one player from a previous grouping. A strategy for rotating the players so that each player from a particular foursome goes to a different foursome the next day should be quite simple to come up with.

Regards
  Kelvin

If you number the foursomes from 1 to 5 then the following rotation guarantees unique partners for the first six days:

Player Day1 Day2 Day3 Day4 Day5 Day6
1      1    2    3    4    5    1
2      1    3    5    2    4    2
3      1    4    2    5    3    3
4      1    5    4    3    2    4
5      2    3    4    5    1    1
6      2    4    1    3    5    2
7      2    5    3    1    4    3
8      2    1    5    4    3    5
9      3    4    5    1    2    1
10     3    5    2    4    1    2
11     3    1    4    2    5    5
12     3    2    1    5    4    4
13     4    5    1    2    3    1
14     4    1    3    5    2    5
15     4    2    5    3    1    3
16     4    3    2    1    5    4
17     5    1    2    3    4    5
18     5    2    4    1    3    2
19     5    3    1    4    2    3
20     5    4    3    2    1    4

KelvinY:

Thanks for your reply to my question, however your reply/calculation does not seem to be accurate.

When I validate your reply, foursome number 1 on Day 2 contains the same players (player numbers 8, 11, 14, 1nd 17) as foursome number 5 on day 6 (players 8,11,14, and 17).

Do you have another suggestion?

Regards,

rccody

http://mathworld.wolfram.com/SocialGolferProblem.html

KelvinY:

Thanks for your reply to my question, however your reply/calculation does not seem to be accurate.

When I validate your reply, foursome number 1 on Day 2 contains the same players (player numbers 8, 11, 14, and 17) as foursome number 5 on day 6 (players 8,11,14, and 17).

Do you have another suggestion?

Regards,

rccody

http://www.cs.brown.edu/~sello/golf.html

rccody,

Yes. I noticed the problem shortly after I posted, but thought I would wait for a response. My strategy was a little too simple. It works up to day 5 then breaks down on day 6. It's possible to still have foursomes with all new partners on day, but it would probably be better to start introducing one previous player so that on day 7 you still have two new players available to each partner. I'll have a rethink and post another solution shortly.

KelvinY

Any luck in developing yet a revised formula/format?

Thanks!

rccody

as the links i've referenced indicate,
5 groups of 4 players can play for5 days with each player playing with 3 diffrerent players each day.

ozo

I appreciate the reply, but I am trying to get 5 groups of 4 to play for 6 days.  Is this possible?  Each person has the ability to play with 19 other golfers.

What do you mean when you state "as the links I've refernced indicate"?

Thanks!

rccody

KelvinY

ozo

Can anyone help me with a solution to my question?

Thanks!

rccody