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I have 6 golfers who are going to play 3 rounds and want to divide up into 6 3 somes in a manner that allows them to all be exposed to each other as many times as possible with a minimum of duplication.
If we asssign each a number form 1 to 6 how do we determine teh best way to accomplish this goal?
hi Dfudpucker
Here's one way. You'll see, many other combinations are possible.
123
145
146
234
256
356
I worked this like this:
Try to get 1 playing everyone. So 123 then 145 then 16+someone again , lets say 146.
Now try to get 4 playing everyone. I've already got 145 and 146 so let's add 234.
Now let's try to get 2 playing everyone. I've already got 123 and 234 so let's add 256.
Now I have one round left and the three who haven't played everyone yet are 3, 5 and 6, so put them together.
The diagram shows these in different colours.
Some duplication is inevitable as you only need 15 two-person-joins to connect everybody to everybody once and you will have 6 x 3 = 18 joins in 6 rounds of three people in each.
Bob
Last edited by Bob (2011-04-27 19:27:04)
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