Round Robin Tournament Scheduling

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Title: 2 teams of 6 golfers, playing foursomes where 2 must be from each team, x 3 days
Post by: senlac1066 on June 12, 2023, 03:52:47 PM

HI
Could someone help 
I have two groups of 6 players, we are playing three times each (once a day for three days !) , and i would like to create foursomes using 2 from each team but maximising the number of different partners over the 3 days.( a partner is anyone in their foursome)

What is the maximum number of  different partners for players.. my own trial and error show 10 people can play 8 people, 2 can play 7 at best.

Can all players play with 8 different partners..9 surely not !! ?

If that is possible what combination of players (9 fourballs made up of 4 players) would deliver that ?

thank you for your consideration, and if you have time for your solution
regards

Title: Re: 2 teams of 6 golfers, playing foursomes where 2 must be from each team, x 3 days
Post by: Ian Wakeling on June 13, 2023, 12:55:44 PM

Hi,

I definitely agree that all-plays-9 is impossible.  My experiments on the computer are suggesting that "10 people can play 8 people, 2 can play 7" is also not possible, so I am interested to see the schedule if you have achieved this.

The optimal schedule here is slightly counterintuitive, it has 6 people who play 7 others, and 6 people who play 9 others - it has slightly more unique pairs than you are claiming as 10*8 + 2*7 = 94, while 6*7 + 6*9 = 96.  I have given two examples below that you might like to use, where group A are player numbers 1 to 6, and group B are player numbers 7 to 12:

Schedule 1

 (1  2  7  8)  (3  4  9 10)  (5  6 11 12)
 (1  3  7 11)  (2  5  9 12)  (4  6  8 10)
 (1  6  7  9)  (2  4 10 11)  (3  5  8 12)

(1 7) (4 10) & (5 12) always in same foursome
(1 4 5 7 10 12) have 7 different partners
(2 3 6 8 9 11) have 9 different partners

Schedule 2

 (1  2  7  8)  (3  4  9 10)  (5  6 11 12)
 (1  2  9 11)  (3  4  7 12)  (5  6  8 10)
 (1  2 10 12)  (3  4  8 11)  (5  6  7  9)

(1 2) (3 4) & (5 6) always in same foursome
Team A have 7 different partners
Team B have 9 different partners

Title: Re: 2 teams of 6 golfers, playing foursomes where 2 must be from each team, x 3 days
Post by: senlac1066 on July 07, 2023, 04:27:56 AM

hi
 thank you for looking at this, i have taken time to reply as i couldnt work out where i was getting a better answer but found it..!!
i had a solution that worked as long as each foursome was on a different day, i hadnt got the matches on 3 days, with the condition that no player can play twice on that day... obvious once i found it but only after you did!!

i tried to solve it using chatgpt, each time you subtely ask the q you get a different answer but interesting, i couldnt rely on any answer but it did say 8 was possible for each player the 30th time i rephrased the question so maybe i will get someone to code all possible combinations and run it
i will let you know if anything interesting comes of it@!
many thanks
J