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my friends and I are doing beer Olympics. There are 12 teams with 6 games. Each game has its own “bracket”. Essentially there will be 6 winners, 1 from each bracket or game. Those teams will then compete in the playoffs. Yes I know a bracket with 12 teams doesn’t come out evenly, I have it figured out. My question to you is, how can schedule where each team plays one game for each game in the first round without playing another team twice. ALSO while having them go in a rotation.

To make this more clear, each “session” every team will he participating in 1 game. After all games finish, I would to be able to rotate these teams to a game they haven’t played yet, but also against a team they haven’t played yet as well. Ive tried trial and error but I’m currently having a problem with it. Can someone help me out?

Too add onto this, all 12 teams are playing simultaneously (6 games, 2 teams per). Once all the single games end, teams must rotate to another game all while playing a different team.

See if this helps.. First game of Event 1: T1 vs T2, 1st game of event 2: T3 vs T4, 1st game of event 3: T5 vs T6, 1st game of event 4: T7 vs T8, first game of event 5: T9 vs T10, first game of event 6: T11 vs T12. Now all the first games of each event are competed. Now for the 2nd games of each event. 2nd game of E1: T1 vs T12, 2nd game of E2: T2 vs T3, 2nd game of E3: T4 vs T5, 2nd game of E4: T6 vs T7, 2nd game of E5: T8 vs T9, 2nd game of E6: T10 vs T11. I need games 3-6 for each event to go in a rotation where each team plays a different game against a different team.

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  • $\begingroup$ How many players are there in a game? Are they all two player games? Are you trying to do a round robin in each bracket or a direct elimination or what? $\endgroup$
    – Dr Xorile
    Apr 11, 2019 at 23:03
  • $\begingroup$ I commented below and went more in depth... $\endgroup$
    – Eric
    Apr 11, 2019 at 23:27
  • $\begingroup$ Hi @Eric, welcome to Puzzling Stack Exchange! (Take the tour if you haven’t already!) This sounds more like a math problem than a puzzle. In general, math textbook style problems are considered off-topic here and may be better off posted elsewhere such as in MSE. Thanks! $\endgroup$
    – HTM
    Apr 11, 2019 at 23:30
  • $\begingroup$ I think I see the problem that's confusing everyone - you not only want a tournament, but you have different contests in the tournament, and don't want a single time to participate in any single contest more than once. I.e. if a team has already played and won beer pong, then they have to play beer chess or beer soccer (no, I'm not a drinker; how'd you tell?). $\endgroup$
    – Brandon_J
    Apr 11, 2019 at 23:33
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    $\begingroup$ @PiIsNot3 math puzzles are OK, and I don't think that this is a textbook-style problem. Not sure though; just my initial reaction. $\endgroup$
    – Brandon_J
    Apr 11, 2019 at 23:34

2 Answers 2

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Edit: okay, wow, I underestimated this problem! This is really tricky. I'm ran a search on my computer and found the below solution. I'd be interested to know if there's a simpler way to think about this.

The search is to find 6 rows of 12 elements where

  1. Each row has two 0s, two 1s, ..., Two 5s.
  2. Each column is a permutation of 0,1,2,3,4,5.
  3. If elements i and j are equal in one row then the ith and jth elements are not equal in any other row.

One of the commentators to the OPs problem referenced a slightly stricter problem involving mutually orthogonal Latin squares. This method (according to the link found in the answer to that problem) doesn't have a solution for $n=6$.

The above is a slightly less strict version. But it's a tricky puzzle! +1 for taking your beer games so seriously!

Here are 6 rows that satisfy the requirements:

 [0, 0, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1],
 [2, 1, 2, 1, 0, 5, 4, 3, 3, 4, 5, 0],
 [4, 2, 5, 4, 5, 2, 3, 0, 1, 1, 0, 3],
 [1, 3, 3, 5, 4, 0, 2, 1, 2, 0, 4, 5],
 [3, 5, 4, 0, 1, 1, 0, 4, 5, 2, 3, 2],
 [5, 4, 0, 3, 2, 3, 1, 2, 0, 5, 1, 4]

This corresponds to a match up of

 0 v  1   2 v 11   3 v 10   4 v  9   5 v  8   6 v  7  
 4 v 11   1 v  3   0 v  2   7 v  8   6 v  9   5 v 10  
 7 v 10   8 v  9   1 v  5   6 v 11   0 v  3   2 v  4  
 5 v  9   0 v  7   6 v  8   1 v  2   4 v 10   3 v 11  
 3 v  6   4 v  5   9 v 11   0 v 10   2 v  7   1 v  8  
 2 v  8   6 v 10   4 v  7   3 v  5   1 v 11   0 v  9

Note that you can run this in two ways. Either the columns are the rounds and the rows are the game or vice versa. And, of course, you can swap columns or rows to your heart's content.

Hope this helps

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  • $\begingroup$ okay let me try to explain more... so if you’ve ever seen a bracket, such as the March madness bracket. The first round or “round of 64” has 32 mini brackets or two lines joined together to represent the teams that are playing each other. For mine, I only have 12 teams, so the first round of “round of 12” has 6 mini brackets or 6 two lined objects joined together (it eventually minimizes to 6 teams then 3). There 6 individual brackets, 1 for each game. I need for every team to play the game once in the round of 12 and never play a team twice in that round of 12. BUTalso while having a rotation $\endgroup$
    – Eric
    Apr 11, 2019 at 23:25
  • $\begingroup$ A rotation so I can be able to call out each team and tell them where there next game is without having any problems. Is this making sense? $\endgroup$
    – Eric
    Apr 11, 2019 at 23:26
  • $\begingroup$ Okay. Then my answer will work. I'll try to make it clearer later but if you work through it it will give you the answer you're looking for $\endgroup$
    – Dr Xorile
    Apr 12, 2019 at 0:21
  • $\begingroup$ At least roughly. It's not possible to do it exactly. You'll have 11 rounds until everyone has met everyone else, so not quite enough for everyone to play every game twice $\endgroup$
    – Dr Xorile
    Apr 12, 2019 at 0:28
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    $\begingroup$ If I understand this correctly, in the second round, 1v3 2v4 5v12 6v11 7v10 8v9? Well done. $\endgroup$
    – LeppyR64
    Apr 12, 2019 at 11:17
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So maybe you have to change the separation. I think this is one sample that works:

1v2 3v4 5v6 7v8 9v10 11v12

1v3 2v4 5v7 6v8 9v11 10v12

1v4 2v5 3v6 7v10 8v11 9v12

1v5 2v9 3v10 4v11 6v7 8v12

1v6 2v7 3v8 4v9 5v12 10v11

1v7 2v8 3v9 4v10 5v11 6v12

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  • $\begingroup$ This doesn't work (if I understand correctly) because you can't have a number twice in the same column. Each player must play each game exactly once. $\endgroup$
    – Dr Xorile
    Apr 13, 2019 at 1:44

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