Round Robin Tournament Scheduling

4 sets of 4 teams.

Darnellmoo · 8 · 3125

Darnellmoo

  • Newbie
  • *
    • Posts: 4
on: June 18, 2014, 04:32:47 AM
4 sets of 4 teams. 8 games, 8 rounds.
All 16 teams must play every game only once and teams from the same set cannot play against each other. Teams cannot play against the same team more than once.

Possible? ;D


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1141
Reply #1 on: June 19, 2014, 04:11:30 AM
Like this may be?

(A2 B3) (A3 B1) (A4 B4) (A1 B2) (C2 D3) (C3 D1) (C4 D4) (C1 D2)
(A1 B1) (A4 B3) (A3 B2) (A2 B4) (C1 D1) (C4 D3) (C3 D2) (C2 D4)
(A4 B2) (A1 B4) (A2 B1) (A3 B3) (C4 D2) (C1 D4) (C2 D1) (C3 D3)
(A3 B4) (A2 B2) (A1 B3) (A4 B1) (C3 D4) (C2 D2) (C1 D3) (C4 D1)
(A2 C3) (A3 C1) (A4 C4) (A1 C2) (B2 D3) (B3 D1) (B4 D4) (B1 D2)
(A1 C1) (A4 C3) (A3 C2) (A2 C4) (B1 D1) (B4 D3) (B3 D2) (B2 D4)
(A4 C2) (A1 C4) (A2 C1) (A3 C3) (B4 D2) (B1 D4) (B2 D1) (B3 D3)
(A3 C4) (A2 C2) (A1 C3) (A4 C1) (B3 D4) (B2 D2) (B1 D3) (B4 D1)

Where rounds are the rows.  It's the only way I can think of making it, but A never plays D, and B never plays C.


Darnellmoo

  • Newbie
  • *
    • Posts: 4
Reply #2 on: June 19, 2014, 04:15:54 AM
Is there anyway the sets play each other?


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1141
Reply #3 on: June 19, 2014, 04:29:47 AM

You could use the above as a starting point and then make swaps, moving teams between one game from the first four, and one game from the second four:

For example, the first rounds is:

(A2 B3) (A3 B1) (A4 B4) (A1 B2) (C2 D3) (C3 D1) (C4 D4) (C1 D2)

but could be modified to:

(A2 D3) (A3 B1) (A4 B4) (A1 B2) (C2 B3) (C3 D1) (C4 D4) (C1 D2)

each time you do this, create two new games, one between A and D and one between B and C.   Continue making swaps like this, but just make sure all the created games are unique.


Darnellmoo

  • Newbie
  • *
    • Posts: 4
Reply #4 on: June 19, 2014, 04:42:29 AM
Means there has to be 10x10 games in total?


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1141
Reply #5 on: June 19, 2014, 04:48:16 AM
Each swap also destroys two games, so the schedule is always 8 x 8.  Above (A2 B3) and (C2 D3) are no longer played.


Darnellmoo

  • Newbie
  • *
    • Posts: 4
Reply #6 on: June 19, 2014, 05:21:27 AM
Sorry I don't really get it. Can you do the illustrations?


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1141
Reply #7 on: June 20, 2014, 01:54:57 AM
If you modify the first round of the original schedule as I have illustrated above, then the net result is:

1 game less between set A and set B.
1 game less between set C and set D.
1 game more between set A and set D.
1 game more between set B and set C.

You need to make more swaps like this, any B set team for any D set team, within any of the first 4 rounds, until the balance of games between the various sets becomes more even.  Within the last four rounds the swaps will be of a different type, swapping any C set team with any D set team which results in:

1 game less between set A and set C.
1 game less between set B and set D.
1 game more between set A and set D.
1 game more between set B and set C.

If you can make 5 swaps of the first kind and 5 swaps of the second kind, such that all the created games between AD and BC are different, then the schedule will have best possible balance for inter-set games.
« Last Edit: June 20, 2014, 03:03:18 AM by Ian »