Schedule for 39 people ,in threesome of 19 round?

[sezuh]

Hi IAN,
I am looking for the table that shows the threesome pairings for a total of 39 players, Each player plays with all 38 possible partners exactly once, in 19 round. Is that possible?
thanks and kind regards

[Ian Wakeling]

Yes it's possible -  I believe that I mentioned this in a thread that might be a year or more old now.  Here  is an example:

( 1  2  3) ( 4 10 28) ( 5 11 29) ( 6 12 30) ( 7 16 19) ( 8 17 20) ( 9 18 21) (13 32 39) (14 33 37) (15 31 38) (22 26 36) (23 27 34) (24 25 35)
( 4  5  6) ( 7 13 31) ( 8 14 32) ( 9 15 33) (10 19 22) (11 20 23) (12 21 24) (16 35  3) (17 36  1) (18 34  2) (25 29 39) (26 30 37) (27 28 38)
( 7  8  9) (10 16 34) (11 17 35) (12 18 36) (13 22 25) (14 23 26) (15 24 27) (19 38  6) (20 39  4) (21 37  5) (28 32  3) (29 33  1) (30 31  2)
(10 11 12) (13 19 37) (14 20 38) (15 21 39) (16 25 28) (17 26 29) (18 27 30) (22  2  9) (23  3  7) (24  1  8) (31 35  6) (32 36  4) (33 34  5)
(13 14 15) (16 22  1) (17 23  2) (18 24  3) (19 28 31) (20 29 32) (21 30 33) (25  5 12) (26  6 10) (27  4 11) (34 38  9) (35 39  7) (36 37  8)
(16 17 18) (19 25  4) (20 26  5) (21 27  6) (22 31 34) (23 32 35) (24 33 36) (28  8 15) (29  9 13) (30  7 14) (37  2 12) (38  3 10) (39  1 11)
(19 20 21) (22 28  7) (23 29  8) (24 30  9) (25 34 37) (26 35 38) (27 36 39) (31 11 18) (32 12 16) (33 10 17) ( 1  5 15) ( 2  6 13) ( 3  4 14)
(22 23 24) (25 31 10) (26 32 11) (27 33 12) (28 37  1) (29 38  2) (30 39  3) (34 14 21) (35 15 19) (36 13 20) ( 4  8 18) ( 5  9 16) ( 6  7 17)
(25 26 27) (28 34 13) (29 35 14) (30 36 15) (31  1  4) (32  2  5) (33  3  6) (37 17 24) (38 18 22) (39 16 23) ( 7 11 21) ( 8 12 19) ( 9 10 20)
(28 29 30) (31 37 16) (32 38 17) (33 39 18) (34  4  7) (35  5  8) (36  6  9) ( 1 20 27) ( 2 21 25) ( 3 19 26) (10 14 24) (11 15 22) (12 13 23)
(31 32 33) (34  1 19) (35  2 20) (36  3 21) (37  7 10) (38  8 11) (39  9 12) ( 4 23 30) ( 5 24 28) ( 6 22 29) (13 17 27) (14 18 25) (15 16 26)
(34 35 36) (37  4 22) (38  5 23) (39  6 24) ( 1 10 13) ( 2 11 14) ( 3 12 15) ( 7 26 33) ( 8 27 31) ( 9 25 32) (16 20 30) (17 21 28) (18 19 29)
(37 38 39) ( 1  7 25) ( 2  8 26) ( 3  9 27) ( 4 13 16) ( 5 14 17) ( 6 15 18) (10 29 36) (11 30 34) (12 28 35) (19 23 33) (20 24 31) (21 22 32)
(13 33 38) (16 36  2) (19 39  5) (22  3  8) (25  6 11) (28  9 14) (31 12 17) (34 15 20) (37 18 23) ( 1 21 26) ( 4 24 29) ( 7 27 32) (10 30 35)
(14 31 39) (17 34  3) (20 37  6) (23  1  9) (26  4 12) (29  7 15) (32 10 18) (35 13 21) (38 16 24) ( 2 19 27) ( 5 22 30) ( 8 25 33) (11 28 36)
(15 32 37) (18 35  1) (21 38  4) (24  2  7) (27  5 10) (30  8 13) (33 11 16) (36 14 19) (39 17 22) ( 3 20 25) ( 6 23 28) ( 9 26 31) (12 29 34)
(22 27 35) (25 30 38) (28 33  2) (31 36  5) (34 39  8) (37  3 11) ( 1  6 14) ( 4  9 17) ( 7 12 20) (10 15 23) (13 18 26) (16 21 29) (19 24 32)
(23 25 36) (26 28 39) (29 31  3) (32 34  6) (35 37  9) (38  1 12) ( 2  4 15) ( 5  7 18) ( 8 10 21) (11 13 24) (14 16 27) (17 19 30) (20 22 33)
(24 26 34) (27 29 37) (30 32  1) (33 35  4) (36 38  7) (39  2 10) ( 3  5 13) ( 6  8 16) ( 9 11 19) (12 14 22) (15 17 25) (18 20 28) (21 23 31)

[sezuh]

Hi Ian,
thanks ever so much for the wonderful solution..very ....very much appreciated.
I thought I saw it in your previous posts, but before I posted this tread, I looked in all the posts with no luck unless I failed to see it??  
Have a very happy new Year.
sezuh          

[Ian Wakeling]

Hi Sezuh,

Have a look again at this thread.  In particular the comments about having 6i+3 players/girls.  You question was about the possibility of a threesome schedule for 39 players. But the proof of the generalization of Kirkman's Schoolgirl problem, implies the existence of an infinite number of threesome schedules, wherever the number of players is of the form 6i + 3.  So simply setting i=6 answers your question! The wikipedia page on Kirkman's schoolgirls points towards the original literature on the mathematical proof, and more discussion of this can be found in many textbooks on combinatorics.

Regards,

Ian.