Round Robin Tournament Scheduling
Schedules - You must register to Post and Download => Requests => Topic started by: chrisvasey on March 11, 2020, 02:27:53 PM
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Good afternoon
I have two golf professionals and 11 members playing 4 rounds of golf.
- each round will be 1 4some & 3 3somes.
- perfect scenario is no one plays with another player more than once AND the two professionals CANNOT play with each other at all.
The end result (hopefully)…..each pro will play in a 4-some twice and a 3-some twice.....therefore, playing with 10 of the 11 members.
Hope this makes some sense......I have got the result close, but just missing something.
Thank you
Chris
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I don't believe that this is possible. If you meet the criteria regarding the pros (don't play together, and play 10 members each), then you would have to have pairs of members who play each other twice. I think there would be as many as 8 of these pairs. The best solutions will have some asymmetry regarding the pros and the distribution of players over the threesomes and foursomes. For example consider the following:
( 3 12 6) ( 9 8 11) (4 13 1) (5 2 7 10)
(10 1 6) ( 9 12 7) (4 8 2) (5 13 3 11)
(11 2 6) (13 9 10) (4 7 3) (5 8 1 12)
( 2 1 3) ( 7 13 8) (5 9 6) (4 10 11 12)
Where players 4 and 5 are the two pros. Pro 4 meets 9 different members, while Pro 5 meets all 11 members. No pair of players meeet twice. The members have different numbers of opponents too, 10, 11 and 12 have two foursomes, while 6 and 9 are never in a foursome. Hope that helps.
Ian