Circle Method - Part 4

Odd number of teams

For an odd number of teams there are a couple possible approaches. One team must always be left out - perhaps it will be playing interdivision or interleague games when it is the one left out. Ignoring for now what we doing with that left-out team, we can build the matchups the same sort of way. One approach is to add another slot to get to an even number of teams but treat that slot as the bye. This is the same as the 8-team case, but with that eighth team not actually there.

1 X    1 7    1 6    1 5    1 4    1 3    1 2
2 7    X 6    7 5    6 4    5 3    4 2    3 X
3 6    2 5    X 4    7 3    6 2    5 X    4 7
4 5    3 4    2 3    X 2    7 X    6 7    5 6

And turning that into matchup rows...

1 out   2 v 7   3 v 6   4 v 5
1 v 7   out 6   2 v 5   3 v 4
1 v 6   7 v 5   out 4   2 v 3
1 v 5   6 v 4   7 v 3   out 2
1 v 4   5 v 3   6 v 2   7 out
1 v 3   4 v 2   5 out   6 v 7
1 v 2   3 out   4 v 7   5 v 6

The other approach is to keep the even number but have a dangling team that sort of misshapes the cirle and represents the team out of the loop. In this case rotate all of the teams and do not lock any in place.

1 7    7 6    6 5    5 4    4 3    3 2    2 1
2 6    1 5    7 4    6 3    5 2    4 1    3 7
3 5    2 4    1 3    7 2    6 1    5 7    4 6
4      3      2      1      7      6      5

And turning that into matchup rows...

1 v 7   2 v 6   3 v 5   4 out
7 v 6   1 v 5   2 v 4   3 out
6 v 5   7 v 4   1 v 3   2 out
5 v 4   6 v 3   7 v 2   1 out
4 v 3   5 v 2   6 v 1   7 out
3 v 2   4 v 1   5 v 7   6 out
2 v 1   3 v 7   4 v 6   5 out

In the next part we will go beyond the single matchups between teams.