3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Do you know about resolvable 2-designs, or RBIBDs? $\endgroup$ – Théophile Jul 4 '17 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 '17 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 '17 at 20:07
2
$\begingroup$

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]]]
$\endgroup$
  • 1
    $\begingroup$ It seems to be correct. ¿How did you generate the table? $\endgroup$ – gonthalo Jul 4 '17 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 '17 at 19:45
  • 1
    $\begingroup$ @gonthalo Have fun with your card games. :) $\endgroup$ – Théophile Jul 4 '17 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.