Evenly Distributed Matches In Rounds w/Constraint

[cblaze22]

I have a situation where I can get the following to evenly distribute matches in such a way that the number of duplicates is at the min for each matchup.

Participates

1,2,3,4,5,6,7,8

1. Need to play 10 games each
2. Play each other at least once
3. 1 and 5 can't play each other

Is there a way to event distribute these participates in rounds and get the duplication of each matchup at a low, while not having 1 and 5 play each other

Round 1

1v2
3v4
5v6
7v8

[Ian Wakeling]

You can achieve this by modifying a standard schedule.  For example start with the round robin for 8 players that you can find by following the schedules link above.

modify round 4, so that instead of containing the games (1 5) and (4 6) it has the games (1 4) and (5 6).  To complete the 10 round schedule required, construct 3 new rounds with the following constraints:

(1) the game (4 6) appears once.
(2) the games (1 5), (1 4) and (5 6) do not appear.
(3) all the games in the 3 rounds are different.

It should be possible to do this by hand.

[cblaze22]

Can you explain the reasoning you picked 1/5 and 4/6 to do the switch?  Why not another combination, just curious.

[Ian Wakeling]

No particular reason, as you could choose to swap (1 5) with any of the three other games in round 4.

[cblaze22]

Ok so I did that and it works well.  However, when finding the other three rounds I am running into difficulty.  I add back the rounds that were skipped in the original processing, then start finding matchups that are least used.  Howver I end up having the last participants be 1 and 5 with 9 games and everyone else 10, and 1 and 5 can't play each other.  Is there some sort of algorithm I should use to find those matchups.  Keep in mind instead of 10 games it could be 20 so just trying to adhere to 10 games needs to also work for 20 games also.

[Ian Wakeling]

You could recycle the first two rounds and then add a 10th like this:

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

Which gives all games apart from (1 5) either once or twice.  But this trick will only work for adding a few rounds.  For a fully variable number of rounds, then I think you will need to invent some other way.  The more rounds there are,  the worse the balance must be.  By the time there are 19 rounds then it must be the case that some games are repeated twice, while others are repeated four times.