Round Robin Tournament Scheduling
Schedules - You must register to Post and Download => Requests => Topic started by: tomc on November 01, 2007, 10:57:46 AM
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Golf,
Can anyone help me identify a table or schedule that would provide minimum overlap of playing partners? Or put another way... Our group of 10 golfers want to mix up the rounds; not playing with the same person more than 3-4 times? We will be playing in groups of 5 for 6 rounds total.
Thanks for the help!
tomc
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I believe the best solution you can have to this problem is the following:
(2 10 1 8 6) (5 4 9 7 3)
(7 1 2 4 5) (6 8 10 3 9)
(9 2 8 1 4) (3 7 5 6 10)
(1 9 3 5 8) (10 6 4 2 7)
(8 5 7 6 1) (4 3 2 9 10)
(8 4 6 9 7) (3 5 1 10 2)
30 of the 45 possible pairings occur within 2 of the fivesomes, the remaining 15 pairings occur within 4 of the fivesomes.
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IAN,
Thanks for the quick response! Now we have a road map to avoid all the confrontations about playing with one person to much!
Tom
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Ian,
I put together a quick "crude" spreadsheet showing all the combinations with totals. I came up with the following:
1-2 plays together 3 times (1-2;3) & 1-3 plays together 2 times (1-3;2) 1-4;3 1-5;4 1-8;4 2-4;4 2-10;4 3-5;4 3-9;4 3-10;4 4-7;4 4-9;3 4-10;3 5-7;4 6-7;4 6-8;4 6-10;4 8-9;4 all other combinations play together 2 times!
Can you give us a table if we play eight rounds instead of six?
Would adding these two rounds spread out the possible combinations better? Or will adding two more rounds bring a five times playing together into the mix? In a perfect world there wouldn't be anyone playing together more than 4 times!
Thanks for your brain power!
Tom
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Tom,
Have another look at the schedule above, as I think 1-2 does play 4 times, and 1-4 only twice.
The 8 round problem has an interesting solution that you might want to go for:
(1 3 6 7 10) (2 4 5 8 9)
(1 4 5 8 10) (2 3 6 7 9)
(1 3 5 8 10) (2 4 6 7 9)
(2 3 5 8 9) (1 4 6 7 10)
(1 4 6 8 9) (2 3 5 7 10)
(1 4 5 7 9) (2 3 6 8 10)
(2 4 6 8 10) (1 3 5 7 9)
(1 3 6 8 9) (2 4 5 7 10)
Here 40 of the 45 possible pairings occur in exactly 4 of the fivesomes. The remaining pairings 1-2, 3-4, 5-6, 7-8, 9-10 never occur. On the face of it, not so good, but you could assign the 10 players to 5 teams of 2 and then you have a round-robin type tournament where players play 4 times with all members of opposing teams. Is that of any use?
Ian.
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Ian,
We are really trying to just spread out the eight rounds of golf to maximize the number each person will play together...
Can you provide a table that provides the best scenario in a eight round, five-some event? Even if some players play together no more than five times; but at least all play at least three times together? Or is this asking to much for the numbers?
Thanks again!
Tom