We have two responses for you

Paul,

Can you give more information? From your message I get the following:

• 14 players
• foursomes consisting of two two-man teams
• 18 weeks
  

How many foursomes each week? Do some players sit out each week?

I can tell you that there are 91 different pairs of players among 14 golfers. Each foursome takes care of six pairs. If there are three foursomes each week, that takes care of 18 pairs. Over 18 weeks, that's 324 pairs. So the best you can do is that some pairs are in the same foursome at least four times. That means they must be partners at least twice. Doing that well might not be possible.

A decent schedule might still exist. It all depends on what you're looking for.

Victoria

I'm not sure I understand the problem.

If everybody is meant to play each week you can't even divide the 14 into foursomes.

If "alternating" means each player plays every other week then a similar problem arises.

If only three foursomes play each week, that is a total of 108 pairs; but 14 players only allow 13*14/2 = 91 distinct pairs and some pairs will have to play twice.

If only one foursome plays each week, there are 72 (player,day) pairs in total and each player plays 5 or 6 times. The problem is then pretty easy, but I don't think that's what you meant.

RD

Paul replied

Thanks for the response. I know my question was not too clear. we have 14 players in an 18 week golf league. each player plays every week so thats 3 foursomes and a twosome. each week there will be 7 two man teams. we want the teams to change every week and the foursomes to alternate every week so each player is playing with each other player the same number of times or as close as possible.

First, each player has at most 13 partners over 18 weeks so some repeats will have to happen. The best you can hope for is no threepeats. Similarly, in each of 18 weeks of play there will be 19 pairings (six in each foursome and the twosome) out of a total of 91,
and 3 < 18*19/91 < 4; so some pairs will be in the same twosome or foursome at least four times. You cannot do better than these numbers.

That said, I'm afraid I don't see a simple solution. The asymmetries of the problem make it a bit difficult. Sorry!

RD

After 13 weeks each player will have been teamed with each other player. Here is a schedule for 2 man teams for 13 weeks. After that, start over.

Number the players 0, 1, 2, ..., 12 and P.

The 2-man teams are

Week 1: P&0; 1&12; 2&11; 3&10; ... 6&7 (See the pattern?)
Week 2: P&1; 2&0; 3&12; 4&11; ... 7&8 (compared to the line above, just increase the number of every player except P by 1, and when you hit 12 "wrap around" to 0 -- like on an odometer)
Week 3: P&2; 3&1; 4&0; 5&12; ... 8&9 (do the same as above, and keep doing it to get the next week.)
and so on

Or use this to make the overall schedule and renumber the weeks randomly.

Now, there are 91 possible pairs of players from among 14 golfers. Three foursomes and a twosome cover 19 pairs, so over 18 weeks you cover 342 pairs. So some pairs wll be together at least four times.

I'll leave it to you to make the foursomes. My suggestion is to choose the twosome first. Over the first 7 weeks use 2-man pair number 1, 3, 5, 7, 2, 4, 6 and then repeat. Then make the foursomes by putting the remaining twosomes into three groups. One way to do this is to take twosomes from opposite ends of the list (omitting the twsome who are going out alone). I didn't carefully check how well this works, but expect it isn't bad. A different method is to choose randomly.

And last, assign the tee times in some way so that P does not usually get to go out first!

Victoria