Scheduling for a bridge group of 12 couples

[lorjohnson]

I would like to set up a rotation of the players for a group of 12 couples for 12 sessions over the winter.
We meet twice a month and we have three homes each time with the host and 3 other couples playing at each of the 3 homes. (8 people)  Ideally each couple should play with each other couple four times.  Is there such a perfect schedule?  Ideally the couples should not play three times in a row and not too many back to back games.   I looked on the web but found nothing close to what we need.  Manually, I get some couples playing 5 together during the season while others only play 2 times together. Any help would be greatly appreciated.     'Thanks

[lorjohnson]

I would like to set up a rotation of the players for a group of 12 couples for 12 sessions over the winter.
We meet twice a month and we have three homes each time with the host and 3 other couples playing at each of the 3 homes. (8 people)  Ideally each couple should play with each other couple four times.  Is there such a perfect schedule?  Ideally the couples should not play three times in a row and not too many back to back games.   I looked on the web but found nothing close to what we need.  Manually, I get some couples playing 5 together during the season while others only play 2 times together. Any help would be greatly appreciated.     'Thanks

I guess I didn't explain my problem clearly enough.....I know that there are lots of round robin schedule out there ...but that is one competitor to another competitor ...however the home has FOUR couples (8 competitors) I am trying to have the 12 couples play with each other less than five times and more than two times.   If you have 3 other couples play with you  -- then 3 times 12 games = 36 games.  That's why ideally you should each play with each other four times....  If you look at all the pairing you have given me  -- there are 6 lines and I nee 3 lines of numbers.   I hope I have explained myself better and I hope you can come up with something that works....Thanks..

[Ian Wakeling]

I think, whenever 4 couples meet at a house, that you would like all 4 to play each other once.  Is that right? In which case the following is the basis for a solution with 11 sessions.

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

Each couple numbered 1 to 12 should meet every other couple exactly three times.  Of course you still need to try to optimize the assignment of houses.

Does that help?