CiceroLove,
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;
    }
}
[download]

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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