Ten chess-players...
- To: mathgroup at smc.vnet.net
- Subject: [mg104219] Ten chess-players...
- From: cmpbrn at gmail.com
- Date: Sat, 24 Oct 2009 02:38:47 -0400 (EDT)
Given 10 (1 to 10) chess-players, in one day they play 5 games (1-2, 6-10, 5-7, 4-8, 3-9). Then they need 8 more days to complete the championship (one gamer must play one time against any other player): 1-3, 2-10, 6-7, 5-8, 4-9 1-4, 2-3, 7-10, 6-8, 5-9 1-5, 2-4, 3-10, 7-8, 6-9 1-6, 2-5, 3-4, 7-9, 8-10 1-7, 2-6, 3-5, 4-10, 8-9 1-8, 2-7, 3-6, 4-5, 9-10 1-9, 2-8, 3-7, 4-6, 5-10 1-10, 2-9, 3-8, 4-7, 5-6 How can I get the 10*(10-1)/2 = 45 pairs distributed in the 9x5 matrix? What's about any other even number of players? Bruno