# Round robin schedule problem

I run two dart leagues. One with 8 teams and 4 boards and one with 6 teams and 3 boards. I have a perfect schedule for the 8 teams league but the 6 team one gives me a dilemma.

I want the teams to:

1. Play each other once (round robin) no problem here many variations of the 15 permutations of the schedule work

2. Have the most even distribution of games on each board as possible (2,2,1 - 2,1,2 - 1,2,2) - this is fine

3. Not play the same board two times in a row. This is the problem. I set up a spreadsheet and have tried all 120 permutations of the 5 week schedule and different schedules and 4 repeats is the best I can find without unbalancing the distribution.

Am I missing something or is this one version of the optimal schedule?

6 team schedule

Week    Board 1     Board 2     Board 3
1       3 vs 4      5 vs 6      1 vs 2
2       4 vs 6      1 vs 3      2 vs 5
3       2 vs 6      3 vs 5      1 vs 4
4       1 vs 5      2 vs 4      3 vs 6
5       2 vs 3      1 vs 6      4 vs 5


Distribution of games played on each board by team

Team  Board 1     Board 2     Board 3
1       1           2           2
2       2           1           2
3       2           2           1
4       2           1           2
5       1           2           2
6       2           2           1


Teams playing on same board 2 times in a row

Weeks 1 & 2       Teams 2 and 4
Weeks 2 & 3       Teams 3 and 6


Thanks

## 2 Answers

A schedule such that no team plays the same board twice in a row:

\begin{matrix} 1-2 & 3-4 & 5-6\\ 3-5 & 1-6 & 2-4\\ 2-6 & 4-5 & 1-3\\ 1-4 & 3-6 & 2-5\\ 2-3 & 1-5 & 4-6 \end{matrix}

• Arthur, The problem with this schedule is that team 2 plays on board 1 - 3 times but never plays on board 2. – Mike Rowan Oct 22 '13 at 16:12

You are right that there is no schedule matching all of your criteria. Which schedule is then optimal is based on which criteria you consider more important. @Arthur's solution is optimal if you consider condition 3 more important than condition 2. Your own solution is optimal if you consider condition 2 more important. If neither 2 nor 3 is more important, then yet another option is:

1-2   3-4   5-6
3-5   1-6   2-4
4-6   3-2   1-5
1-3   4-5   2-6
2-5   1-4   3-6


Here, teams 4 and 6 play at the same boards in rounds 5 and 6. Teams 4 and 6 also have bad distributions, but still play on all boards. (Note that swapping the matches 1-4 and 3-6 in the last round yields a solution similar to @Arthur's)