Golf Scheduling

[turney]

Problem:

12 players going on a golf vacation to play 9 rounds of golf, 4 golfers in each group.  Trying to come up with a schedule that evenly distributes how many times each player plays with each other.  The minimum number of times you play with any one player is 2 with a max of 4 or 5.  By hand I can get everyone with everyone at least once with one pair playing 5 times.  I would assume I could get a better distribution using an algorithm of some sort but have been unable to find anything useful.

Thanks

[Ian Wakeling]

Turney,

See my reply to Richard William's request for an interesting option if you only wanted 7 rounds.

For 9 rounds you can achieve best possible social balance. There are 12*11/2=66 possible pairings of players. In the schedule below 36 pairs of players occur together twice, while the remaining 30 pairs of players occur three times.

(F B C I)  (G D E A)  (K H J L)
(E G H I)  (C D L J)  (B A F K)
(A J B G)  (H D I K)  (L E C F)
(C B J K)  (L G E F)  (D A I H)
(A L K E)  (H C G B)  (J I D F)
(A H F J)  (G L D B)  (I E K C)
(B I L A)  (J F H E)  (D K G C)
(K F I G)  (L H A C)  (B E J D)
(E K B H)  (F C A D)  (I J G L)

The schedule has similar balance for position in the foursomes, each player appears either 2 or 3 times in each of the four positions in a foursome.  If you wanted you could use this to vary the tee-off order as fairly as possible.

Hope the helps,

Ian.

[turney]

That's great!

I figured there was some way to do it, thanks a lot.  What method/program did you use?

Thanks again

[turney]

How about 8 rounds instead of 9?  

[Ian Wakeling]

Turney,

Here is an 8 round schedule.  This time 54 pairs occur twice within a foursome, 12 pairs occur three times.  Players exactly twice in each position within a foursome.

(A B L I)  (K C J G)  (F E H D)
(C F I L)  (G K D B)  (H J A E)
(J H G I)  (B A F C)  (L E D K)
(G L B J)  (A K F D)  (H I E C)
(E J B F)  (D A I G)  (L C K H)
(K B H A)  (C G E F)  (I D L J)
(F L G H)  (B I K E)  (J D C A)
(E G A L)  (D H C B)  (I F J K)

Both designs were found with some statistical software for finding experimental designs.  This may or may not help, but to a mathematician both these schedules would be called resolvable partially balanced incomplete block designs. A fully balanced design, where all pairs occur 3 times is possible if you have 11 rounds.

Regards,

Ian.

[Linda_Ball]

Would you be able to list the 11 rounds with perfect balance please.  I have 12 ladies for golf and we have 22 weeks over the summer, so 11 rounds of 3 foursomes would be perfect.  Thanks for your help.
Linda

[Ian Wakeling]

Here is an example of an 11 round schedule:

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

You could also follow the schedules link near the top of the page and select '12 items' and 'Whist'.  If you ignore the pairs and treat each table as a foursome, then this will also work.  If you do this, then note that player 12 is always in the 1st foursome.

[Linda_Ball]

I wondered about that. But I didn't understand what it meant when it said it was considering "pairs of pairs".
Thank you. You will have helped me avoid a lot of chatter among the ladies about who plays with who.  
I had tried drawing out of a hat for fun and it was surprising how skewed it was. I also tried looking through some bridge movements but they are all pairs so nothing there worked either.

Linda

[Ian Wakeling]

Linda,

By "pairs of pairs", it just means that at each table 2 pairs of players play each other. Classic 'doubles' format if you like.

The footnote goes on to say "at the end of the tournament each player has been paired once with each player and opposed each player twice".  If you ignore 'pairs' and just concentrate on the 'tables' aspect, then this is equivalent to saying "at the end of the tournament each player has sat down at a table with every other player exactly 3 times".  Which is exactly what you want if tables are considered as foursomes.

Ian.

[Linda_Ball]

Thank you so much! I couldn't be happier.
My friend who is a math teacher in Taupo New Zealand found your site when I posed the problem to him. The internet is amazing. You can find information on pretty much everything.
Linda