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I'm currently trying to find an algorithm to place players during a Mahjong tournament.

Here are the requirements :

  • Number of players in the tournament : $n$ with $n \equiv 0 \pmod 4$
  • Number of tables in the tournament : $n/4$ (4 players per table)
  • A tournament has $x$ rounds (determinated by the algorithm according to $n$)
  • A player must never meet another player twice

The goal is to find how many rounds I can set for $n$ players.

I found some documentation about Swiss-system tournament, and I would like to know if an equivalent exists for quartet instead of pairs.

Thanks for your help.

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2 Answers

Each player meets 3 players each round, and each player has $n-1$ other possible players to meet, so there can't be more than $(n-1)/3$ rounds. Whether it is actually possible to schedule as many as $(n-1)/3$ rounds, I don't know. People who organize bridge tournaments have to know about this kind of thing, and there is a lot of on it. I'd search for "bridge movements" and/or for "whist movements". Also, the "social golfer problem" is a good search term, as golfers go out in parties of 4.

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Well this is a good start. I've already worked on a similar case for Bridge movements, but the problem is totally different, in the sens that Bridge players are always together (teams of 2), then it can be considered as a $1/1$ tournament, and not a $1/1/1/1$. Golfers's problem seems to be closer from what I'm searching for, I'll have a look. –  zessx Jul 29 '12 at 13:16
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You could look at the list of bridge movements under individual movements. Also at this page which has a lot of information on how to construct them.

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