Allocating games in tournaments - example

TL;DR

After looking briefly at a promising way to arrange tournaments for multi-player games, let’s look at a few practical examples.

By example here we will also consider something a bit more… theoretical. In particular, we will focus on games with more than two players inside, and see where it goes.

Arrangements for multi-player games: numbers

As already discussed in previous post Allocating games in tournaments, finite affine planes are a promising way to get some tournament scheduled. BIBDs based on affine planes have the following characteristics:

  • $n$ is the number of players in each match;
  • $n^2$ is the total number of players in the tournament;
  • $n + 1$ is the number of rounds played, corresponding to the number of games played by each participant;
  • $n\cdot(n+1$ is the overall number of games played in the whole tournament.

This leads us to the following table:

players/game rounds total players total games
$n$ $n + 1$ $n^2$ $n \cdot (n + 1)$
2 3 4 6
3 4 9 12
4 5 16 20
5 6 25 30
7 8 49 56
8 9 64 72
9 10 81 90

Rows from $n = 5$ on become increasingly trickier because of the quick growth in the number of participants needed to arrange the tournament, but are anyway left for completeness.

You surely noticed the lack of an alternative for $n = 6$. It so happens that there is no finite projective plane possible for it, so we’re just leaving it out.

Example: 3-players games

This is the arrangement of lines for order-3 projective plane:

  0. (1, 4, 7, 10)
  1. (0, 4, 5, 6)
  2. (3, 4, 9, 11)
  3. (2, 4, 8, 12)
  4. (0, 1, 2, 3)
  5. (1, 6, 9, 12)
  6. (1, 5, 8, 11)
  7. (0, 10, 11, 12)
  8. (3, 6, 8, 10)
  9. (2, 5, 9, 10)
 10. (0, 7, 8, 9)
 11. (2, 6, 7, 11)
 12. (3, 5, 7, 12)

As before, we get rid of one line and all points inside, let’s choose line 7 and points 0, 10, 11, and 12:

  0. (1, 4, 7, 10)   --> (1, 4, 7)
  1. (0, 4, 5, 6)    --> (4, 5, 6)
  2. (3, 4, 9, 11)   --> (3, 4, 9)
  3. (2, 4, 8, 12)   --> (2, 4, 8)
  4. (0, 1, 2, 3)    --> (1, 2, 3)
  5. (1, 6, 9, 12)   --> (1, 6, 9)
  6. (1, 5, 8, 11)   --> (1, 5, 8)
  7. (0, 10, 11, 12) -->
  8. (3, 6, 8, 10)   --> (3, 6, 8)
  9. (2, 5, 9, 10)   --> (2, 5, 9)
 10. (0, 7, 8, 9)    --> (7, 8, 9)
 11. (2, 6, 7, 11)   --> (2, 6, 7)
 12. (3, 5, 7, 12)   --> (3, 5, 7)

To get the different rounds easily, it’s sufficient to look for all groups in the left hand side that share the same removed point:

round 1: (1, 4, 7) (3, 6, 8) (2, 5, 9)
round 2: (4, 5, 6) (1, 2, 3) (7, 8, 9)
round 3: (3, 4, 9) (1, 5, 8) (2, 6, 7)
round 4: (2, 4, 8) (1, 6, 9) (3, 5, 7)

So long… for now!

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