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 9 golfers - 4 games (Read 11580 times)
 Lee Perriam YaBB Newbies Posts: 3 9 golfers - 4 games 10/05/06 at 17:48:51   We have a group of nine golfers playing four rounds of golf in 3 x 3 balls.   Is it possible to arrange the groups of three so that everyone plays with everyone else at least once and no one has to play with the same person more than twice? Back to top IP Logged
 Ian Wakeling YaBB Moderator Posts: 874 Gender: Re: 9 golfers - 4 games Reply #1 - 10/06/06 at 03:33:58   Lee,   I am surprised that scheduling golf threesomes has not been raised here before. Anyway, the answer to your question is yes. With four rounds of threesomes it's possible to arrange that everyone plays everyone else exactly once.   (G F A)  (B D C)  (E H I) (F C H)  (I G D)  (A E B) (B I F)  (D A H)  (E C G) (F D E)  (H B G)  (C I A)   The example above is the first in a whole series of threesome schedules.  If the number of players is three more than an a multiple of 6 then the all-plays-all-exactly-once property is always possible.  The next example is 15 players, which is famously known in mathematics as Kirkman's Schoolgirl problem (search the internet for more info on this, and you will find out how the schedule below can be constructed).   (K O H)  (L N J)  (E A G)  (F I C)  (D M B) (B H N)  (A K I)  (F L D)  (O G C)  (M J E) (M K L)  (O D J)  (G N I)  (E F H)  (B C A) (C M N)  (H J G)  (L O A)  (I D E)  (K B F) (L B G)  (N E K)  (O I M)  (H C D)  (J A F) (I L H)  (A D N)  (C J K)  (G F M)  (B E O) (M H A)  (E C L)  (J I B)  (D G K)  (N F O)     Below is the third in the series for 21 players.   (G F T)  (B I J)  (A U D)  (O S L)  (Q H K)  (P R M)  (N E C) (K I U)  (T M H)  (S C B)  (J R L)  (A Q E)  (F P N)  (D O G) (H I R)  (L B D)  (T O P)  (U G Q)  (K A C)  (S J N)  (E F M) (A N M)  (E L P)  (S T D)  (G C R)  (U J H)  (I F Q)  (B K O) (I P A)  (L U T)  (C O M)  (H B F)  (Q R S)  (J D E)  (N K G) (E H G)  (I D C)  (O N R)  (M S U)  (T A J)  (K F L)  (Q B P) (M L I)  (Q C T)  (F U O)  (D N H)  (R A B)  (P J G)  (K E S) (C L H)  (E U R)  (N T I)  (D P K)  (F A S)  (O Q J)  (M G B) (C P U)  (B T E)  (H O A)  (G S I)  (J M K)  (R D F)  (L Q N) (R K T)  (D M Q)  (C J F)  (L G A)  (P H S)  (I E O)  (U N B)     Hope that helps,   Ian. Back to top IP Logged
 Lee Perriam YaBB Newbies Posts: 3 Re: 9 golfers - 4 games Reply #2 - 10/06/06 at 09:32:05   That's very useful. Thank you very much Back to top IP Logged
 mike G YaBB Newbies Posts: 2 Re: 9 golfers - 4 games Reply #3 - 10/24/06 at 23:38:48   now lets say that there are 10 players (2 groups of 5).  What are the pairings for 5 and that all players play with the same number of rounds with each person.  How many rounds would we need to play if we were to play each person 4 times each and how many for 5 times each?   Let me know.  Thanks     Mike G Back to top IP Logged
 Ian Wakeling YaBB Moderator Posts: 874 Gender: Re: 9 golfers - 4 games Reply #4 - 10/25/06 at 05:22:44   If there are m groups of k players and you want them all to play each other at least t times each then you need at least   r = t*(mk-1) / (k-1)   rounds.  Note that I say at least.  Clearly if r is not a whole number then you will need to increase it to the next highest whole number or may be more.  Even if you find whole number solutions for all the unknowns in the equation above, then there is no guarantee that you will be able to construct a schedule, for some combinations it's just not possible to make things work.  Your values of m=2 and k=5 are a good example of this, while t=4 and r=9 fit the equation perfectly the 9 round schedule is actually impossible.   You would need double up to 18 rounds when you could get everyone playing each other 8 times.   Ian. Back to top IP Logged
 mike G YaBB Newbies Posts: 2 Re: 9 golfers - 4 games Reply #5 - 10/25/06 at 09:52:34   I appreciate it.  The only way to make that happen would to spend an extra week together and I don't think that the wife would like that very much.   Mike G Back to top IP Logged
 Ian Wakeling YaBB Moderator Posts: 874 Gender: Re: 9 golfers - 4 games Reply #6 - 10/26/06 at 04:34:05   Mike,   My comments above relate to the situation where everyone plays everyone else the same number of times.  However there are other things you could do.  How about dividing your 10 players into 5 teams of 2 players each.  If the players are A to J, then make the teams up like this   (A B) (C D) (E F) (G H) (I J)   and then play the schedule below.   (B D F H I)   (A C E G J) (B C E G I)   (A D F H J) (A D E G I)   (B C F H J) (A D E H J)   (B C F G I) (B D E H I)   (A C F G J) (B D F G J)   (A C E H I) (A C F H I)   (B D E G J) (A D F G I)   (B C E H J)   In the 8 rounds each player will play with all players from opposing teams exactly 4 times, and never play with their own team mate.   Ian. Back to top IP Logged
 BG YaBB Newbies Posts: 5 Re: Golf threesomes, social golfer Reply #7 - 04/14/11 at 17:24:18   I am looking for the table that shows the threesome pairings for a total of 33 players.  Does anyone have that table? Back to top IP Logged
 Ian Wakeling YaBB Moderator Posts: 874 Gender: Re: 9 golfers - 4 games Reply #8 - 04/15/11 at 03:18:52   Table is below, each of the 16 rows is a round, 11 threesomes per round.  Each player plays with all 32 possible partners exactly once.  It would be a good idea to randomize the tee-off order of the threesomes in each round to prevent player 1 always being the first on the course.   Hope that helps.   (1 2 18) (3 4 5)  (19 21 20) (6 10 14) (22 30 26) (7 12 17) (23 33 28) (15 8 13) (31 29 24) (11 16 9) (27 25 32) (1 3 19) (2 4 21)  (18 5 20) (7 11 15) (23 31 27) (6 16 29) (22 13 32) (14 12 25) (30 9 28) (10 8 33) (26 17 24) (1 4 20) (2 5 19)  (18 3 21) (8 12 16) (24 32 28) (10 9 31) (26 15 25) (6 17 27) (22 11 33) (14 13 23) (30 7 29) (1 5 21) (2 3 20)  (18 4 19) (9 13 17) (25 33 29) (14 11 24) (30 8 27) (10 7 32) (26 16 23) (6 15 28) (22 12 31) (1 6 22) (7 8 9)  (23 25 24) (2 10 30) (18 14 26) (3 16 13) (19 29 32) (11 4 17) (27 33 20) (15 12 5) (31 21 28) (1 7 23) (6 8 25)  (22 9 24) (3 11 31) (19 15 27) (2 12 33) (18 17 28) (10 16 21) (26 5 32) (14 4 29) (30 13 20) (1 8 24) (6 9 23)  (22 7 25) (4 12 32) (20 16 28) (14 5 27) (30 11 21) (2 13 31) (18 15 29) (10 17 19) (26 3 33) (1 9 25) (6 7 24)  (22 8 23) (5 13 33) (21 17 29) (10 15 20) (26 4 31) (14 3 28) (30 12 19) (2 11 32) (18 16 27) (1 10 26) (11 12 13)  (27 29 28) (2 14 22) (18 6 30) (15 4 9) (31 25 20) (7 16 5) (23 21 32) (3 8 17) (19 33 24) (1 11 27) (10 12 29)  (26 13 28) (3 15 23) (19 7 31) (14 8 21) (30 5 24) (6 4 33) (22 17 20) (2 16 25) (18 9 32) (1 12 28) (10 13 27)  (26 11 29) (4 16 24) (20 8 32) (2 17 23) (18 7 33) (14 9 19) (30 3 25) (6 5 31) (22 15 21) (1 13 29) (10 11 28)  (26 12 27) (5 17 25) (21 9 33) (6 3 32) (22 16 19) (2 15 24) (18 8 31) (14 7 20) (30 4 23) (1 14 30) (15 16 17)  (31 33 32) (2 6 26) (18 10 22) (11 8 5) (27 21 24) (3 12 9) (19 25 28) (7 4 13) (23 29 20) (1 15 31) (14 16 33)  (30 17 32) (3 7 27) (19 11 23) (10 4 25) (26 9 20) (2 8 29) (18 13 24) (6 12 21) (22 5 28) (1 16 32) (14 17 31)  (30 15 33) (4 8 28) (20 12 24) (6 13 19) (22 3 29) (10 5 23) (26 7 21) (2 9 27) (18 11 25) (1 17 33) (14 15 32)  (30 16 31) (5 9 29) (21 13 25) (2 7 28) (18 12 23) (6 11 20) (22 4 27) (10 3 24) (26 8 19) Back to top IP Logged