Round Robin Tournament Scheduling
Schedules - You must register to Post and Download => Requests => Topic started by: JennyE on April 02, 2012, 03:49:44 PM
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Not sure if this has already been discussed in which case apologies & please point me to the thread.
We need to schedule matches between 2 teams of 8 with everyone playing all rounds. The play is in pairs across 5 possible courts. Each member of a team plays once with a colleague and twice against (almost all) members of opposing teams but never in the same combination. Equal use of each court would be good.
How good a solution is possible?
Are there standard algorithms for sorting this sort of stuff out? Or even standard approaches? My maths is rusty!
Any ideas on a solution for 2 teams of 10 players?
Thank you in advance. :)
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Are you playing a doubles format? The five courts is confusing me slightly as with doubles you only need 4 courts for 16 players. If I am right then the four rounds here would give you all A team members against all B team members exactly once.
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Thanks for your speedy reply. The 5 courts was probably not helpful! Playing on the same court just gives a slight advantage. (This is croquet b.t.w.) I checked the schedule you gave but it seems to have only half the matches I need. It is doubles play. I need everyone to play 7 rounds. All play is simultaneous. So player A1 plays with players A2-A8 as a partner each time, and meets pairs of Team B in combination such that no pair from Team B is the same from round to round and each player from Team B is met twice (or once for one of the players as there are only 7 rounds). The same format should apply to every player.
Is this possible?
If necessary we could introduce an eighth round and allow players to sit out if that allows a better solution.
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My gut feeling is that this may well be impossible. The closest thing I can think of is to modify a spouse avoiding mixed doubles tournament. So start with this schedule. Consider H1 to H8 to be team A, and W1 to W8 to be team B, and rearrange each match so that it is two 'H' players against two 'W' players. This will meet your criterion that all pairs of players from the same team partner each other once, however every team member will have one opposing team member whom they never play against, and will play against the other 7 opponents exactly twice. If you want to go with this, then then you will have to try to balance the matches over the 5 courts as best you can as I know of no way to balance it.
Two teams of 10 players is more problematic still as the spouse avoiding schedule does not exist (if you insist that all play is simultaneous).
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Thanks. This is really helpful and looks like the best solution (it certainly beats my empirical attempts). I had a feeling that the 10 player problem was a step too far. Ah well.