Swiss Tournament. Team confrontation multiple rounds without repeat and ordered - php

Question

I try to make a tournament based on Swiss method.
I have for example 12 teams, and in each round, every team has points (0 to 100) and wins, loses or draws.
I want to find which teams play against each other, with these conditions:
Ordered by wins, draws, and points.
Not to be played previously.
In each round, I got foreach team the possible teams to play against in an array like this: (The key indicates the team id, and the values indicate ll possible teams to play separated by ",")
[2] => 4,11,6,10,3,8,7,12,
[5] => 4,11,9,10,3,8,1,12,
[4] => 2,5,6,10,8,7,12,
[11] => 5,9,10,3,8,7,
[9] => 5,11,6,3,8,7,12,
[6] => 2,4,9,10,3,7,12,
[10] => 2,5,4,11,6,8,7,12,
[3] => 5,11,9,6,8,7,
[8] => 2,5,4,11,9,10,3,1,12,
[7] => 2,4,11,9,6,10,3,1,12,
[1] => 5,4,11,9,6,3,8,7,
[12] => 2,5,4,9,6,10,8,7,
First I have all teams played before in a array: (key indicates team id)
Array (
[1] => 2,10,12,
[2] => 1,9,5,
[3] => 4,12,10,
[4] => 3,11,9,
[5] => 6,7,2,
[6] => 5,8,11,
[7] => 8,5,8,
[8] => 7,6,7,
[9] => 10,2,4,
[10] => 9,1,3,
[11] => 12,4,6,
[12] => 11,3,1, )
Then, I get all teams ordered by wins, loses and points in a array: (Keys indicates also team id)
Array
(
[1] => 2
[2] => 5
[3] => 4
[4] => 11
[5] => 9
[6] => 6
[7] => 10
[8] => 3
[9] => 8
[10] => 7
[11] => 1
[12] => 12
)
Finaly I try to find the possible match against two teams.
$checks = array();
$pairs = array();
for ($i = 1; $i <= count($list); $i++) {
for ($j = 1; $j <= count($list); $j++) {
if(strpos($plays[$list[$i]], $list[$j]) !== false || $list[$i] == $list[$j] ) {
}else{
if(!in_array($list[$i],$checks) && !in_array($list[$j],$checks)){
$pairs[] = $list[$i].",".$list[$j];
$checks[] = $list[$i];
$checks[] = $list[$j];
}
}
}
}
And finaly I print the array "$pairs". It shows:
Array
(
[0] => 2,4
[1] => 5,11
[2] => 9,6
[3] => 10,8
[4] => 3,7
)
Thats not correct because the team_id 1 and team_id 12 can not play in this round because the played before:
I don't know how to solve this.
Thanks again!