Tom,
If you had wanted 5 rounds then that would be impossible since each player would play a total of 10 players of the opposite gender over the 5 rounds. At least two of these 10 players would have to be the same person since you have the spouse avoiding specification. 4 rounds might just be possible, I can't see an elegant solution so have used a computer search to come up with the schedule below.
The husbands are H1..H10 and the wives W1..W10.
Foursome 1 Foursome 2 Foursome 3 Foursome 4 Foursome 5
-------------- --------------- --------------- -------------- --------------
(W1 H9 W3 H2) ( H5 W6 W9 H8) (W10 H3 W8 H4) (H6 W7 W4 H10) (H1 H7 W5 W2)
(W7 H4 H8 W1) ( W6 H2 H1 W10) ( W2 W3 H6 H5) (H3 W5 H9 W4) (W9 H10 H7 W8)
(H7 W10 W1 H6) ( H2 W9 W7 H3) ( W4 W8 H5 H1) (H4 W2 W6 H9) (W5 H8 H10 W3)
(W8 H6 H2 W5) (H10 W1 H3 W6) ( H9 H5 W10 W7) (W3 H1 H4 W9) (H8 W4 W2 H7)
If you only want three rounds then use the first three rows above, and each player gets to play with exactly 3 different players of the same gender and exactly six different players of the opposite gender.
If you use the full schedule of 4 rounds then it is very slightly unbalanced as W2 & H7 will occur together twice in a foursome.
Excepting these 2 players the rest will all play with 4 different players of the same gender and 8 out of 9 players of the opposite gender.
Anyone else know if this 4 round problem has a fully balanced solution?
Have a great golf trip,
Ian.
Wednesday, September 14 2005, 12:10 pm