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I have to arrange a card game tournament, in which the players will play in tables of 4. I want to arrange as little games as possible, but to prevent biases I want every 2 players to have played by each other by the end, but only once. Also, as the game has a more or less constant duration, the games should take place in N rounds, and in each of those no player should be idle.

Therefore the number of rounds should be $(N-1)/3$ and the number of tables $N/4$

It's easy to check that all possible solutions for N are of the form $4 + 12·k$, where $k$ is an integer. I managed to convince myself that there is no solution for $N = 16$, but I don't want to check manually for $N = 28$ and above. Can you find the general solution?

Note: This is not a theoretical problem, I actually want to arrange a card game tournament.

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    $\begingroup$ Do you know about resolvable 2-designs, or RBIBDs? $\endgroup$
    – Théophile
    Jul 4, 2017 at 19:19
  • $\begingroup$ I read about it, with that formulation I ask for the existence of a steiner system $S(2, 4, 4 + 12·k)$ for each integer k. However this case is not listed in the wikipedia page. $\endgroup$
    – gonthalo
    Jul 4, 2017 at 19:59
  • $\begingroup$ It's not exactly the same as a Steiner system, because it requires resolvability as well. (See the History section of the Wikipedia page on Steiner systems.) $\endgroup$
    – Théophile
    Jul 4, 2017 at 20:07

1 Answer 1

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I believe this should do it! Your object is called a Resolvable Balanced Incomplete Block Design; it has parameters $(v,k,\lambda)=(28,4,1)$.

References:

I generated the table using Sage online (https://sagecell.sagemath.org/), using the following commands:

K = designs.resolvable_balanced_incomplete_block_design(28,4)
K.is_resolvable(1)[1]

[[[4, 8, 11, 15],
  [13, 17, 20, 24],
  [2, 6, 22, 26],
  [1, 5, 12, 16],
  [10, 14, 21, 25],
  [3, 7, 19, 23],
  [0, 9, 18, 27]],
 [[2, 7, 12, 13],
  [11, 16, 21, 22],
  [3, 4, 20, 25],
  [0, 5, 15, 17],
  [9, 14, 24, 26],
  [6, 8, 18, 23],
  [1, 10, 19, 27]],
 [[1, 7, 9, 15],
  [10, 16, 18, 24],
  [0, 6, 19, 25],
  [3, 8, 13, 14],
  [12, 17, 22, 23],
  [4, 5, 21, 26],
  [2, 11, 20, 27]],
 [[5, 6, 10, 13],
  [14, 15, 19, 22],
  [1, 4, 23, 24],
  [2, 8, 9, 16],
  [11, 17, 18, 25],
  [0, 7, 20, 26],
  [3, 12, 21, 27]],
 [[0, 8, 10, 12],
  [9, 17, 19, 21],
  [1, 3, 18, 26],
  [6, 7, 11, 14],
  [15, 16, 20, 23],
  [2, 5, 24, 25],
  [4, 13, 22, 27]],
 [[3, 6, 16, 17],
  [12, 15, 25, 26],
  [7, 8, 21, 24],
  [0, 1, 11, 13],
  [9, 10, 20, 22],
  [2, 4, 18, 19],
  [5, 14, 23, 27]],
 [[3, 5, 9, 11],
  [12, 14, 18, 20],
  [0, 2, 21, 23],
  [4, 7, 10, 17],
  [13, 16, 19, 26],
  [1, 8, 22, 25],
  [6, 15, 24, 27]],
 [[1, 2, 14, 17],
  [10, 11, 23, 26],
  [5, 8, 19, 20],
  [4, 6, 9, 12],
  [13, 15, 18, 21],
  [0, 3, 22, 24],
  [7, 16, 25, 27]],
 [[0, 4, 14, 16],
  [9, 13, 23, 25],
  [5, 7, 18, 22],
  [2, 3, 10, 15],
  [11, 12, 19, 24],
  [1, 6, 20, 21],
  [8, 17, 26, 27]]]
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    $\begingroup$ It seems to be correct. ¿How did you generate the table? $\endgroup$
    – gonthalo
    Jul 4, 2017 at 19:45
  • $\begingroup$ @gonthalo Thanks for checking. I have reason to believe this new version works! As you can see, the players are partitioned in 7 groups of 4; there are 9 such partitions. $\endgroup$
    – Théophile
    Jul 4, 2017 at 19:45
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    $\begingroup$ @gonthalo Have fun with your card games. :) $\endgroup$
    – Théophile
    Jul 4, 2017 at 19:58

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