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Dennis, 12 players and 7 rounds is a bit beyond what I can do by computer. Instead, I can make a mathematical suggestion. I'm hopeful that it will work out as you like, but I have not checked. There are a few steps. There is a schedule whereby 16 players can play in foursomes so that every pair is together exactly once over 5 days. You can find the archives. Delete from this schedule the players 13, 14, 15, and 16. This will leave one day with three foursomes, and four days with four threesomes. In each of these four days, break up one of the threesomes and use it to make up three foursomes. At this point every two players have been together exactly once except for the pairs in the threesomes that were broken up. There are 12 pairs of players who have not been together at this point. Use the remaining two days to get these players together, if possible. --Victoria | ||||||||||||
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Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |