Round Robin Tournament Scheduling

Volleyball Co-ed Rotating Pairs (12 M/12 F)

jla9182 · 6 · 6928

jla9182

  • Newbie
  • *
    • Posts: 3
on: August 07, 2010, 06:55:10 PM
I've seen similar requests posted, but can't find one exactly like this one.  We are planning a rotating pairs volleyball tournament with 12 men and 12 women.  Rules we would like to try to incorporate, in the order of importance (would like to do all of these, but haven't figured a way yet):

(1) all the men play with all of the women once
(2) no two men play against each other more than once
(3) no two women play against each other more than once
(4) no two man and woman play against each other more than once

If unable to accomodate all of these rules, the next preference is to reduce the times played against each other in each case (rules 2-4) to a minimum (preferably 2 or less).


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1140
Reply #1 on: August 09, 2010, 02:33:57 PM
I am not sure about the exact format you are intending to play.  Is each team comprised of three men and three women? And two simultaneous matches per round? Or something else?


jla9182

  • Newbie
  • *
    • Posts: 3
Reply #2 on: August 09, 2010, 04:41:09 PM
Sorry - left out the important part.  The tournament is coed doubles, so each team is comprised of one man and one woman.  Ultimately, everyone plays each round.  I'd like either 11 or 12 games, knowing that if you do 12, you will violate #2 & #3 and if you play 11, then you violate #1.  I guess 12 games would be preferrable.


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1140
Reply #3 on: August 10, 2010, 04:03:38 AM
How about the following schedule:

(M1 F9 v M10 F7) (M2 F11 v M5 F1) (M3 F10 v M8 F4) (M4 F12 v M11 F6) (M6 F3 v M9 F5) (M7 F2 v M12 F8)
(M1 F12 v M6 F2) (M2 F10 v M9 F8) (M3 F11 v M12 F5) (M4 F9 v M7 F3) (M5 F4 v M10 F6) (M8 F1 v M11 F7)
(M1 F11 v M7 F8) (M2 F1 v M12 F6) (M3 F4 v M9 F7) (M4 F10 v M6 F5) (M5 F3 v M11 F12) (M8 F2 v M10 F9)
(M1 F2 v M11 F5) (M2 F12 v M8 F7) (M3 F9 v M5 F6) (M4 F3 v M10 F8) (M6 F4 v M12 F11) (M7 F1 v M9 F10)
(M1 F4 v M8 F12) (M2 F3 v M7 F11) (M3 F2 v M6 F10) (M4 F1 v M5 F9) (M9 F6 v M12 F7) (M10 F5 v M11 F8)
(M1 F7 v M5 F10) (M2 F8 v M6 F9) (M3 F5 v M7 F12) (M4 F6 v M8 F11) (M9 F4 v M11 F2) (M10 F3 v M12 F1)
(M1 F10 v M4 F11) (M2 F9 v M3 F12) (M5 F8 v M12 F4) (M6 F7 v M11 F3) (M7 F6 v M10 F2) (M8 F5 v M9 F1)
(M1 F8 v M3 F6) (M2 F7 v M4 F5) (M5 F11 v M9 F2) (M6 F12 v M10 F1) (M7 F9 v M11 F4) (M8 F10 v M12 F3)
(M1 F5 v M12 F9) (M2 F6 v M11 F10) (M3 F7 v M10 F11) (M4 F8 v M9 F12) (M5 F2 v M8 F3) (M6 F1 v M7 F4)
(M1 F6 v M9 F3) (M2 F5 v M10 F4) (M3 F8 v M11 F1) (M4 F7 v M12 F2) (M5 F12 v M7 F10) (M6 F11 v M8 F9)
(M1 F3 v M2 F4) (M3 F1 v M4 F2) (M5 F7 v M6 F8) (M7 F5 v M8 F6) (M9 F11 v M10 F12) (M11 F9 v M12 F10)


That's 11 rounds and gives you  #2 and #3.  If you played any 12th round that involved forming teams from (M1 F1), (M2 F2), etc.. then this would give you #1 at the expense of violating #2 and #3.

Hope that helps.
« Last Edit: August 10, 2010, 04:07:00 AM by Ian »


jla9182

  • Newbie
  • *
    • Posts: 3
Reply #4 on: August 12, 2010, 06:29:29 PM
thanks - this is exactly what I needed!  How did you do it?


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1140
Reply #5 on: August 13, 2010, 03:29:29 AM
This may or may not help...  It came from this book on combinatorial designs.  The schedule is equivalent to two Orthogonal Latin Squares of a special type called a SOLSSOM.

There are some similar schedules on this message board for other number of players which you can find by following this link and the links that you will find there.
« Last Edit: August 13, 2010, 03:39:24 AM by Ian »