Circle Method - Part 5

Balancing home/away

So far there has been no consideration of the home/away nature of the matchups as the goal has been to simply get all the matchups defined. But you could consider each "A v B" as "A @ B", such that "A" is away and "B" is home. Where does that put us toward even home/away splits for teams?

With an even number of teams it is not possible to balance the home/away perfectly. E.g., with 8 teams that means 7 opponents, so the most balanced is either 3A/4H or 4A/3H. Maybe you achieve balance by duplicating all the matchups and flipping them. A league may involve a total of 4 series against each opponent in the circle, and that could involve making the original matchups set, having one exact copy, then having two flipped copies. E.g., for two teams 1 & 2, you could have 1 @ 2, 1 @ 2, 2 @ 1, and 2 @ 1 occurring in the schedule.

But if you want to have as much balance as possible in the original set, the circle method almost does that automatically. The one issue is the one locked team. Revisiting the 10-team case...

 1 10     1  9     1  8     1  7     1  6     1  5     1  4     1  3     1  2
 2  9    10  8     9  7     8  6     7  5     6  4     5  3     4  2     3 10
 3  8  >  2  7  > 10  6  >  9  5  >  8  4  >  7  3  >  6  2  >  5 10  >  4  9
 4  7     3  6     2  5    10  4     9  3     8  2     7 10     6  9     5  8
 5  6     4  5     3  4     2  3    10  2     9 10     8  9     7  8     6  7

...laying out the matchups into rows taking the left-side team as away and the right-side team as home in each matchup...

1 @10   2 @ 9   3 @ 8   4 @ 7   5 @ 6
1 @ 9  10 @ 8   2 @ 7   3 @ 6   4 @ 5
1 @ 8   9 @ 7  10 @ 6   2 @ 5   3 @ 4
1 @ 7   8 @ 6   9 @ 5  10 @ 4   2 @ 3
1 @ 6   7 @ 5   8 @ 4   9 @ 3  10 @ 2
1 @ 5   6 @ 4   7 @ 3   8 @ 2   9 @10
1 @ 4   5 @ 3   6 @ 2   7 @10   8 @ 9
1 @ 3   4 @ 2   5 @10   6 @ 9   7 @ 8
1 @ 2   3 @10   4 @ 9   5 @ 8   6 @ 7

It is obvious going down the left column that team 1 has 9A/0H. But every other team is 4A/5H because of how they rotated around the circle taking each of the 4 other slots on the left and each of the 5 slots on the right. So simply flip 4 or 5 of the matchups involving team 1, which will give team 1 the 4 or 5 home and change the 4 or 5 teams on the other side of the flipped matchups to 5A/4H. Below flips the matchups of 1 vs odd--numbered teams.

1 @10   2 @ 9   3 @ 8   4 @ 7   5 @ 6
9 @ 1  10 @ 8   2 @ 7   3 @ 6   4 @ 5
1 @ 8   9 @ 7  10 @ 6   2 @ 5   3 @ 4
7 @ 1   8 @ 6   9 @ 5  10 @ 4   2 @ 3
1 @ 6   7 @ 5   8 @ 4   9 @ 3  10 @ 2
5 @ 1   6 @ 4   7 @ 3   8 @ 2   9 @10
1 @ 4   5 @ 3   6 @ 2   7 @10   8 @ 9
3 @ 1   4 @ 2   5 @10   6 @ 9   7 @ 8
1 @ 2   3 @10   4 @ 9   5 @ 8   6 @ 7

That gives us as much balance as we can with every team going 4H/5A or 5H/4A across their 9 matchups against the other 9 teams. If these are division opponents in a schedule, it is likely that each matchup will occur multiple times. Perhaps each team should play each other team in two 3-game series. That is a very simple case in which you could copy every matchup but flip the home/away. In that case every team would end up with 9H/9A as they play a series against every team twice, once home and once away.

In other scenarios it can get much trickier to lay out multiple matchups against each team and get the required balanced. Complicating factors include an odd number of series matchups between opponents (e.g., want to have 3 series against each opponent) and the number of games is not perfectly balanced (e.g., 96 division games would not divide evenly across 9 opponents so you might play 6 teams 10 games each and 3 teams 12 games each).

In the next part we will use the circle method for making interdivision/interleague matchups.