Whisp schedule for pickleball

[Jschramm]

I am in need of help for a whisp schedule for these multiples on 5 courts (some players would have a bye each round) for a long pickleball round robin
24, 23, 22, 20 (5 courts)
21, 20, 19, 18, (4 courts)
17, 16, 15, 14 (3 courts)

[Ian Wakeling]

With the exception of 20 players on 5 courts, no balanced whist schedule is possible.  Really you need to consider what your objectives are?  How many rounds are you looking to play - is there a maximum because of time constraints?  Do you want all (or most) pairs of players to partner each other once?  Do you want all (or most) pairs of players to play each other (either as partners or opponents) at least once.  Is it important that all players have exactly the same number of games?

There is an 18 player schedule here that might help.  Does that look useful, or is it too long perhaps?

[Jschramm]

Ian
the opponents do not have to be equal for each player, the only requirement is that each person plays with every other player. It is timed games at 7 minutes each and it takes roughly 4 hours to do the entire thing. 

[Ian Wakeling]

I think it is important to at least try to balance the opponents.  If the number of players corresponds to a whist schedule (4n or 4n+1), then the possibility exists of trying to rearrange things to reduce the number of courts.  So for example if you follow the link near the top of page "visit the pages that inspired the forum" and then find the whist table for 16 items - then that schedule can be rearranged by turning table 2 into 5 new rounds as follows:


(16  1 v  9 14)  ( 3 10 v 12 13)  ( 6 15 v  7 11)
(16  2 v 10 15)  ( 4 11 v 13 14)  ( 7  1 v  8 12)
(16  3 v 11  1)  ( 5 12 v 14 15)  ( 8  2 v  9 13)
(16  4 v 12  2)  ( 6 13 v 15  1)  ( 9  3 v 10 14)
(16  5 v 13  3)  ( 7 14 v  1  2)  (10  4 v 11 15)
(16  6 v 14  4)  ( 8 15 v  2  3)  (11  5 v 12  1)
(16  7 v 15  5)  ( 9  1 v  3  4)  (12  6 v 13  2)
(16  8 v  1  6)  (10  2 v  4  5)  (13  7 v 14  3)
(16  9 v  2  7)  (11  3 v  5  6)  (14  8 v 15  4)
(16 10 v  3  8)  (12  4 v  6  7)  (15  9 v  1  5)
(16 11 v  4  9)  (13  5 v  7  8)  ( 1 10 v  2  6)
(16 12 v  5 10)  (14  6 v  8  9)  ( 2 11 v  3  7)
(16 13 v  6 11)  (15  7 v  9 10)  ( 3 12 v  4  8)
(16 14 v  7 12)  ( 1  8 v 10 11)  ( 4 13 v  5  9)
(16 15 v  8 13)  ( 2  9 v 11 12)  ( 5 14 v  6 10)

( 2  4 v  5  8)  ( 7  9 v 10 13)  (12 14 v 15  3)
( 3  5 v  6  9)  ( 8 10 v 11 14)  (13 15 v  1  4)
( 4  6 v  7 10)  ( 9 11 v 12 15)  (14  1 v  2  5)
( 5  7 v  8 11)  (10 12 v 13  1)  (15  2 v  3  6)
( 6  8 v  9 12)  (11 13 v 14  2)  ( 1  3 v  4  7)

The order of the rounds should then be randomized so that player 16 does not have all their byes at the end.  I am sorry that I don't have off-the-shelf solutions for many of your requests, but I will try to add more schedules if time permits.

[Ian Wakeling]

I looked at just one of the irregular scenarios, that for 15 players on 3 courts, and I have shown below the smallest possible schedule where all pairs of players partner at least once.

      court 1        court 2         court 3           byes
  ( 4 14 v 15  1) ( 6  2 v 10  7) (13  3 v  5  8)   (12 11  9)
  ( 5 15 v 13  2) ( 4  3 v 11  8) (14  1 v  6  9)   (10 12  7)
  ( 6 13 v 14  3) ( 5  1 v 12  9) (15  2 v  4  7)   (11 10  8)
  ( 7  6 v 15  4) ( 3  2 v 14 12) (10  8 v  1 11)   ( 9 13  5)
  ( 8  4 v 13  5) ( 1  3 v 15 10) (11  9 v  2 12)   ( 7 14  6)
  ( 9  5 v 14  6) ( 2  1 v 13 11) (12  7 v  3 10)   ( 8 15  4)
  ( 1  4 v 12  3) (14 15 v 11  5) ( 8  6 v  2  7)   (13 10  9)
  ( 2  5 v 10  1) (15 13 v 12  6) ( 9  4 v  3  8)   (14 11  7)
  ( 3  6 v 11  2) (13 14 v 10  4) ( 7  5 v  1  9)   (15 12  8)
  ( 7  9 v 13 10) ( 2 14 v  1 15) ( 6  5 v 12  8)   ( 3 11  4)
  ( 8  7 v 14 11) ( 3 15 v  2 13) ( 4  6 v 10  9)   ( 1 12  5)
  ( 9  8 v 15 12) ( 1 13 v  3 14) ( 5  4 v 11  7)   ( 2 10  6)
  (10  6 v  5  3) ( 4 13 v  2  9) (12 11 v 14  7)   (15  8  1)
  (11  4 v  6  1) ( 5 14 v  3  7) (10 12 v 15  8)   (13  9  2)
  (12  5 v  4  2) ( 6 15 v  1  8) (11 10 v 13  9)   (14  7  3)
  (10  2 v  8 14) (13 12 v  6 11) ( 7 15 v  3  9)   ( 4  5  1)
  (11  3 v  9 15) (14 10 v  4 12) ( 8 13 v  1  7)   ( 5  6  2)
  (12  1 v  7 13) (15 11 v  5 10) ( 9 14 v  2  8)   ( 6  4  3)

Note that players 1,2,3,13,14 & 15 have 15 games while the other players have 14 games.

[Jschramm]

Thank you, Ian!