Phil,
With 9 weeks there are going to be more than a few duplicate opponents; each player plays 9x5=45 times, so the optimal solution would be that a player plays 13 of the 29 possible opponents just once but the other 16 opponents twice [ 45 = 13 + 2x16 ]. I have not been able to find the optimal solution, the problem being that I end up with three pairs of players who opppose each other three times, but otherwise all other pairs meet either once or twice. In the schedule below the pairs who meet three times are (28 8) (29 9) and (30 7).
(16 11 5 4 2 23) ( 6 17 25 29 20 27) (26 14 9 24 1 15) (12 18 22 19 13 21) ( 3 28 7 8 30 10)
(17 12 6 5 3 24) ( 4 18 26 30 21 25) (27 15 7 22 2 13) (10 16 23 20 14 19) ( 1 29 8 9 28 11)
(18 10 4 6 1 22) ( 5 16 27 28 19 26) (25 13 8 23 3 14) (11 17 24 21 15 20) ( 2 30 9 7 29 12)
( 7 13 20 4 1 5) (22 26 23 8 17 15) ( 3 25 19 18 2 11) (27 9 29 12 14 10) ( 6 30 24 21 28 16)
( 8 14 21 5 2 6) (23 27 24 9 18 13) ( 1 26 20 16 3 12) (25 7 30 10 15 11) ( 4 28 22 19 29 17)
( 9 15 19 6 3 4) (24 25 22 7 16 14) ( 2 27 21 17 1 10) (26 8 28 11 13 12) ( 5 29 23 20 30 18)
(21 3 9 20 22 11) (12 10 25 28 15 5) (14 30 1 13 17 19) ( 7 27 18 6 16 8) (29 4 23 26 2 24)
(19 1 7 21 23 12) (10 11 26 29 13 6) (15 28 2 14 18 20) ( 8 25 16 4 17 9) (30 5 24 27 3 22)
(20 2 8 19 24 10) (11 12 27 30 14 4) (13 29 3 15 16 21) ( 9 26 17 5 18 7) (28 6 22 25 1 23)
If you have already played the first week as you suggest above, then you will need to rearrange my schedule by reassignment of the players as follows:
player 16 becomes player 1
player 11 becomes player 2
player 5 becomes player 3 etc...
Hope that helps,
Ian.
