Round Robin Tournament Scheduling

Round Robin for 20 players

Laurie · 8 · 8311

Laurie

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on: May 09, 2009, 05:48:36 AM
Hi there,
I am new to this sort of organisation.

I am hosting a round robin on 16th May for 20 players who will play as pairs over 5 rounds.
I would like them not to play with or against anyone more than once during the whole competition.

Is this possible please?

Your Social Square for 16 players is ideal...is there one like it for 20 players please?
« Last Edit: May 09, 2009, 11:58:25 AM by admin »


Ian Wakeling

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Reply #1 on: May 09, 2009, 11:35:17 AM
It is indeed possible to have a schedule for 5 rounds:

  (1 17 16 14) (4  8 13  3) (11 20  7 15) (12  6  5 18) (2  9 10 19)
  (2 18 17 15) (5  9 14  4) (12 16  8 11) (13  7  1 19) (3 10  6 20)
  (3 19 18 11) (1 10 15  5) (13 17  9 12) (14  8  2 20) (4  6  7 16)
  (4 20 19 12) (2  6 11  1) (14 18 10 13) (15  9  3 16) (5  7  8 17)
  (5 16 20 13) (3  7 12  2) (15 19  6 14) (11 10  4 17) (1  8  9 18)


Fortunately you did not ask for a 6 round schedule, on the face of it, this might seem to be possible - 6 partners and 12 opponents, but it has been proven by the mathematicians that it can't exist.


Laurie

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Reply #2 on: May 09, 2009, 12:36:52 PM
 :) Thank you very much indeed for your prompt reply...just what we need...thank you
« Last Edit: May 09, 2009, 01:49:36 PM by admin »


Laurie

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Reply #3 on: May 09, 2009, 05:19:20 PM
 :) thank you for your help...we play a lot of round robins at our crown green bowling club here in the UK...can you please give me a format for 12 players playing in pairs so that they do not play with or against another player more than once throughout the competition. I find your social squares good for this...have you got one suitable please?

 :) What about 24 players playing in pairs...is there one for this please playing over 4/5 rounds ? Certainly not 6 as our members would not last that long as they are "recycled teenagers"

cheers and regards
Laurie


Ian Wakeling

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Reply #4 on: May 11, 2009, 04:01:46 AM
Laurie,

If you only have 12 players then it is only possible to play a single round.  No matter how you arrange the second round there will be at least three pairs of players who meet for a second time.  With 24 players it is possible to play up to seven rounds and still not have any repeated pairings.  For example:

(16 20 21  9) (5  2  3 19) (11 13 10  1) (14 15  7 22) ( 8 17  4 23) (12  6 18 24)
(17 21 15 10) (6  3  4 20) (12 14 11  2) ( 8 16  1 22) ( 9 18  5 23) (13  7 19 24)
(18 15 16 11) (7  4  5 21) (13  8 12  3) ( 9 17  2 22) (10 19  6 23) (14  1 20 24)
(19 16 17 12) (1  5  6 15) (14  9 13  4) (10 18  3 22) (11 20  7 23) ( 8  2 21 24)
(20 17 18 13) (2  6  7 16) ( 8 10 14  5) (11 19  4 22) (12 21  1 23) ( 9  3 15 24)
(21 18 19 14) (3  7  1 17) ( 9 11  8  6) (12 20  5 22) (13 15  2 23) (10  4 16 24)
(15 19 20  8) (4  1  2 18) (10 12  9  7) (13 21  6 22) (14 16  3 23) (11  5 17 24)


So just take any five rounds from the above for your competition.


Laurie

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Reply #5 on: May 11, 2009, 06:02:52 AM
Thank you for your reply...the  round robin for 24 players is ideal for our needs
What is the best you can do with 12 players please so that they play with or against other players the least times
Hope you don't mind me keep asking these questions
kind regards
Laurie


Ian Wakeling

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Reply #6 on: May 11, 2009, 12:18:56 PM
Laurie,

The 12 player/5 round problem is one that has come up before.

Please have a look here.

Regards,

Ian.


Laurie

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Reply #7 on: May 13, 2009, 05:13:15 PM
 :)thank you for your help in this matter...kind regards Laurie (Grange Bowling Club) :)