I have a population of 28 golfers going on a 4 day golf trip.
Other than brute force is there an easy way to line up these golfers in seven foursomes per day so that no player plays with any other player more than once.
It helps that 7 is prime…
Day 1
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
17 18 19 20
21 22 23 24
25 26 27 28
Day 2
1 6 11 16
5 10 15 20
9 14 19 24
13 18 23 28
17 22 27 4
21 26 3 8
25 2 7 12
Day 3
1 10 19 28
5 14 23 4
9 18 27 8
13 22 3 12
17 26 7 16
21 30 11 20
25 34 15 24
Day 4
1 14 27 12
5 18 3 16
9 22 7 20
13 26 11 24
17 2 15 28
21 6 19 4
25 10 23 8
I think it should be relatively straightforward what’s going on here, but I’ll answer any questions you have. You could even extend the trip by a few days.
Ok Thanks, Yes, I see that you are going diagonally on each subsequent day.
I like to think of each column as a wheel on a suit case style lock. The first wheel is fixed, the second wheel moves one spot each day, the third two spots, and the fourth three spots. Since 1, 2, 3 are all relatively prime to 7 we don’t run into any trouble. The same technique would not work without modification would not work for a a group of 24 golfers for this reason.
I may be going a little mad here, but on day 3 why have you suddenly lost players 2 and 6? and where have 30 and 34 come from? Are you playing ringers here?
Oops.
Those players are equivalent modulo 28.
Day 3 (corrected)
1 10 19 28
5 14 23 4
9 18 27 8
13 22 3 12
17 26 7 16
21 2 11 20
25 6 15 24
This is a resolvable balanced incomplete block design where v = 28, k = 4, and λ = 1: