Round Robin Tournament Scheduling

Single Court Tennis Contract for Doubles

brakeshoe · 4 · 5103

brakeshoe

  • Newbie
  • *
    • Posts: 0
on: August 13, 2019, 09:43:58 PM
Looking for a way to use this tool to set up a balanced schedule for a weekly tennis contract.  There are 6 players and each week 4 of them play doubles, the other two sit out.  The 4 players on a given week would be chosen in such a way that all the players played the same number of weeks and players were rotated evenly.  Of course based on the number of weeks chosen the rotation would not be perfect so some combinations could be used one more time than the others.  From what I can see the current tool only sets up ten rounds of matches and picks 3 courts so that all players are playing each week.  For a contract schedule we would need to schedule one match a week for X weeks.  Where X might be a value like 33 or 34.  If this scenario is possible I would then want to run it for a second group using 8 players, 1 court and X weeks.  Many thanks in advance.


brakeshoe

  • Newbie
  • *
    • Posts: 0
Reply #1 on: August 19, 2019, 09:21:51 AM
I have looked at the Whilst calculator and it only seems to produce values for 7 weeks.  Repeating the same values would not result in a balanced schedule.  There are more unique combinations of matches then weeks in the season so it seems possible to generate a schedule with unique combinations for each week.  I looked at the VB code and could not determine how to change the weeks from 7, but I am not a VB expert by any means.  Season is looming so any help would be appreciated.
Thanks,


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1140
Reply #2 on: August 19, 2019, 12:09:21 PM
I think I can offer some partial help.  The following 2 whist schedules for 8 playes are complementary in the sense that no two games involve the same 4 players, so if you split each round over 2 weeks this gives 28 weeks.

(8 1 v 2 6)  (3 4 v 5 7)
(8 2 v 3 7)  (4 5 v 6 1)
(8 3 v 4 1)  (5 6 v 7 2)
(8 4 v 5 2)  (6 7 v 1 3)
(8 5 v 6 3)  (7 1 v 2 4)
(8 6 v 7 4)  (1 2 v 3 5)
(8 7 v 1 5)  (2 3 v 4 6)

(8 1 v 3 7)  (2 4 v 5 6)
(8 2 v 4 1)  (3 5 v 6 7)
(8 3 v 5 2)  (4 6 v 7 1)
(8 4 v 6 3)  (5 7 v 1 2)
(8 5 v 7 4)  (6 1 v 2 3)
(8 6 v 1 5)  (7 2 v 3 4)
(8 7 v 2 6)  (1 3 v 4 5)

For six players I have come up with the following 30 games.

(6 2 v 1 4)
(1 3 v 2 5)
(2 4 v 3 6)
(3 5 v 4 1)
(4 6 v 5 2)
(5 1 v 6 3)
(3 5 v 2 1)
(4 6 v 3 2)
(5 1 v 4 3)
(6 2 v 5 4)
(1 3 v 6 5)
(2 4 v 1 6)
(2 1 v 3 6)
(3 2 v 4 1)
(4 3 v 5 2)
(5 4 v 6 3)
(6 5 v 1 4)
(1 6 v 2 5)
(4 5 v 3 2)
(5 6 v 4 3)
(6 1 v 5 4)
(1 2 v 6 5)
(2 3 v 1 6)
(3 4 v 2 1)
(5 1 v 4 2)
(6 2 v 5 3)
(1 3 v 6 4)
(2 4 v 1 5)
(3 5 v 2 6)
(4 6 v 3 1)

It's not fully checked, but I think this gives you all partner pairs 4 times and all opposition pairs 8 times and each player has 10 byes.

So starting with the above you could add extra games by hand to reach 33 or 34.  Hope that helps.


brakeshoe

  • Newbie
  • *
    • Posts: 0
Reply #3 on: August 20, 2019, 10:35:52 AM
Ian:
The 6  and 8 player solutions look very good and adding a few more weeks or just resuming at the top seems like a big win.  Thanks very much for your help.  This was really stressing me out.