Re: Ten chess-players...
- To: mathgroup at smc.vnet.net
- Subject: [mg104296] Re: [mg104219] Ten chess-players...
- From: John Fultz <jfultz at wolfram.com>
- Date: Sun, 25 Oct 2009 23:26:42 -0500 (EST)
- Reply-to: jfultz at wolfram.com
On Sat, 24 Oct 2009 02:38:47 -0400 (EDT), cmpbrn at gmail.com wrote: > 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 I've done quite the same thing for Scrabble competitions. Here's an= algorithm for round robin pairing for any number of players: http://en.wikipedia.org/wiki/Round-robin_tournament#Scheduling_algorithm And here's a simple function which implements the algorithm for a complete= round robin. For an odd number of players, just invent a "bye" player (bye's= opponent gets that round off) and pair as for even players. Pair[n_Integer?EvenQ] := Table[ Prepend[RotateRight[Range[n - 1] + 1, round - 1], 1][[{couple, -couple}]], {round, n - 1}, {couple, n/2}] An example output: In[28]:= Pair[8] // Grid Out[28]= {1,8}=09{2,7}=09{3,6}=09{4,5} {1,7}=09{8,6}=09{2,5}=09{3,4} {1,6}=09{7,5}=09{8,4}=09{2,3} {1,5}=09{6,4}=09{7,3}=09{8,2} {1,4}=09{5,3}=09{6,2}=09{7,8} {1,3}=09{4,2}=09{5,8}=09{6,7} {1,2}=09{3,8}=09{4,7}=09{5,6} Sincerely, John Fultz jfultz at wolfram.com User Interface Group Wolfram Research, Inc.