8 players, 4 per game

[bussemac]

It's trivial to see that with 8 players (not in partnerships) and 4 per game, we need 3 games to play everyone at least once (and two people twice). What's the closest-to-even arrangement of players over 4 rounds (1 round = 2 simultaneous games). Each player will have 12 opponents, so obviously we can't visit everyone twice. If we go to 5 rounds, it seems like it might be possible since we have 15 opponents... if that is true, I can work with 5 rounds instead, and you'll just see one player an extra time. But I don't know if theory matches with reality here.

I'd love to know how to solve this problem in general for N players with 3 or 4 per game... we have a card game tournament where the number of players may be 6 (2x3), 8 (2x4), 12 (3x4), or 16 (4x4) based on how many players participated during the qualifying events, so knowing what the algorithm is would be extremely useful. If different numbers of players call for different numbers of rounds, we can be flexible with a range of 4 to 6 rounds.

Thanks!

[Ian Wakeling]

It's not as simple as you have been thinking and 4 rounds are needed in order to meet the everyone-at-least-once criterion.  The problem is that when you compare games in any two rounds, there must be some pairs of players who play together in both rounds. These repeated pairs limit the ability to cover all pairs at least once.  There are two 4 round options here in an old thread that you might find useful.

2x3 is easy as it is always possible to arrange the pairs into rounds. For example:
(1 2) (3 5) (6 4)
(5 6) (4 2) (3 1)
(4 3) (6 2) (5 1)
(1 6) (2 3) (5 4)
(2 5) (6 3) (4 1)

3x4 has similar problems to 2x4, the only balanced option is playing 6 rounds (see here), but then you have to divide the 12 players into 4 teams of 3 (see round 7), then each player gets to play  everyone who is not on their own team exactly twice.

4x4 is the only one of your problems with a perfect solution.  Use the 5 rounds corresponding to 'Part.0' that you can find here.

Hope that helps.

[bussemac]

Thanks, I see the solutions (with limitations) for 8 players and 12 players, and the perfect solution for 16 players. For 6 players, I would prefer to run 2 games of 3 players each (3 games of 2 players is possible but the 2 player game plays out very differently from a 3 or 4 which was used for qualifying purposes, and therefore I would rather stick to the the 3 player format).

Thanks again for the assistance!

[Ian Wakeling]

With 2 games of 3 players there is a schedule along the same lines as the first example for 2x4 (the first link in my message above).  This is:

(1 2 3) (4 5 6)
(1 5 3) (4 2 6)
(1 5 6) (4 2 3)
(1 2 6) (4 5 3)
(1 4) (2 5) (3 6)