in reply to Round robin tournament problem

I only see an easy solution for for 2n players. I have provided working brute-force code for those cases.
#!/usr/bin/perl -w use strict; my @name = qw(a b c d e f g h); die "You must have an even number of players" if @name % 2; my %tournament; $tournament{$_} = [] for @name; for my $round (1 .. $#name) { $round--; for my $player (@name) { next if $tournament{$player}->[$round]; for my $opponent (@name) { next if $opponent eq $player; next if $tournament{$opponent}->[$round]; next if @{$tournament{$player}} && grep /^$opponent$/ , @{ +$tournament{$player}}; $tournament{$player}->[$round] = $opponent; $tournament{$opponent}->[$round] = $player; last; } } } for my $round (1 .. $#name) { $round--; print "\n\nRound: $round\n"; my %seen; for my $player (@name) { next if $seen{$player}; print "$player plays $tournament{$player}->[$round]\n"; $seen{$tournament{$player}->[$round]} = 1; } }
Update: Originally, I stated that this was all that's possible, but as blokhead points out non 2n solutions with backtracking is possible.

Hope this gets you started - L~R

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Re: Re: Round robin tournament problem
by waswas-fng (Curate) on Oct 21, 2003 at 23:56 UTC
    It is possible just not a ballanced table, you will have a table like the following:
    Round 1 A-B C-D E Sits Out Round 2 B-C D-E A Sits Out ...