Round Robin Tournament Scheduling
Schedules - You must register to Post and Download => Requests => Topic started by: kanon87 on December 05, 2021, 04:47:29 PM
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The Whist Tournament format is nice because you don't play with someone more than once, but you play against people more than once. For 8 rounds, when the round robin is more than 16 people, we prefer to not play someone more than once. I'm guessing that this has been asked for before, but in my search I have not found the answer. Thank you in advance for any help that comes my way.
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This is an interesting question, and I am not aware of any off-the-shelf solutions I can point you towards. There are two points to be made.
Firstly it can't work for 16 players, since there will be a total of 16 opponents in the 8 rounds to play, but as there are only 15 possible opponents for any one player, then at least one of them must be repeated twice.
If there were slightly more players, then you can find solutions relatively easily. For example if you look on Ed Pegg's golfer page (https://www.mathpuzzle.com/MAA/54-Golf%20Tournaments/mathgames_08_14_07.html) for "28 golfers can play in foursomes for 9 days", then this works for you. It doesn't matter how you assign each group of 4 players into 2 pairs, it will always be the case that a player's partners and opponents will all be different.
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Here is a possible schedule for 24 players:
(4 8 v 11 20) (14 24 v 5 15) (19 2 v 12 16) (10 7 v 6 1) (22 13 v 21 18) (17 23 v 9 3)
(1 5 v 12 17) (15 21 v 6 16) (20 3 v 9 13) (11 8 v 7 2) (23 14 v 22 19) (18 24 v 10 4)
(2 6 v 9 18) (16 22 v 7 13) (17 4 v 10 14) (12 5 v 8 3) (24 15 v 23 20) (19 21 v 11 1)
(3 7 v 10 19) (13 23 v 8 14) (18 1 v 11 15) ( 9 6 v 5 4) (21 16 v 24 17) (20 22 v 12 2)
(5 19 v 20 21) (24 12 v 7 9) (11 4 v 13 16) ( 3 14 v 6 18) (23 2 v 1 10) (17 8 v 15 22)
(6 20 v 17 22) (21 9 v 8 10) (12 1 v 14 13) ( 4 15 v 7 19) (24 3 v 2 11) (18 5 v 16 23)
(7 17 v 18 23) (22 10 v 5 11) ( 9 2 v 15 14) ( 1 16 v 8 20) (21 4 v 3 12) (19 6 v 13 24)
(8 18 v 19 24) (23 11 v 6 12) (10 3 v 16 15) ( 2 13 v 5 17) (22 1 v 4 9) (20 7 v 14 21)
A player's partners are all different, and a player's opponents are all different, however there is some overlap, as some players will see the same player once as a partner and once as an opponent. Would that work for you?
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And this for 20 players is very similar.
(20 15 v 6 3) (16 11 v 18 19) (12 8 v 17 5) (14 13 v 10 4) ( 2 7 v 9 1)
(17 16 v 7 4) (13 12 v 19 20) ( 9 5 v 18 6) (15 14 v 11 1) ( 3 8 v 10 2)
(18 13 v 8 1) (14 9 v 20 17) (10 6 v 19 7) (16 15 v 12 2) ( 4 5 v 11 3)
(19 14 v 5 2) (15 10 v 17 18) (11 7 v 20 8) (13 16 v 9 3) ( 1 6 v 12 4)
( 7 18 v 2 19) (17 3 v 6 11) ( 9 13 v 12 5) ( 4 14 v 8 15) (20 10 v 16 1)
( 8 19 v 3 20) (18 4 v 7 12) (10 14 v 9 6) ( 1 15 v 5 16) (17 11 v 13 2)
( 5 20 v 4 17) (19 1 v 8 9) (11 15 v 10 7) ( 2 16 v 6 13) (18 12 v 14 3)
( 6 17 v 1 18) (20 2 v 5 10) (12 16 v 11 8) ( 3 13 v 7 14) (19 9 v 15 4)
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Playing with and against someone would be ok. Thank you for your help!