Re: Ten chess-players...

*To*: mathgroup at smc.vnet.net*Subject*: [mg99296] Re: [mg99269] Ten chess-players...*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Sat, 2 May 2009 05:58:12 -0400 (EDT)*References*: <36a5dba2-11b9-45df-84c3-9e712e47fec9@q33g2000pra.googlegroups.com> <49F94E90.2090003@gmail.com> <200905010924.FAA01163@smc.vnet.net>*Reply-to*: drmajorbob at bigfoot.com

Here's an online schedule generator: http://www.teamopolis.com/tools/round-robin-generator.aspx WikiPedia has an algorithm: http://en.wikipedia.org/wiki/Round-robin_tournament and here is (I think) an implementation: Clear[schedule] schedule[players_?EvenQ, round_] /; round < players := Transpose@{Take[#, players/2], Reverse@Drop[#, players/2]} &@ Flatten@{1, RotateRight[Range[2, players], round - 1]} Bobby On Fri, 01 May 2009 04:24:17 -0500, Dr. Bruno Campanini <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? > > I do know how to get the matrix with a simple Visual Basic routine > (some 30 lines of code). > I was wondering if there is a shorter Mathematica simple formula > for the same job. > > Bruno > -- DrMajorBob at bigfoot.com