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10 couples, four rounds, no spousing

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Tom Barry(Guest)

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on: December 30, 2005, 08:16:06 PM
At the last minute, we added 2 couples to our annual fall golf outing (making a total of 10 couples), and I've been scratching my head trying to come up with the best combinations to play 4 rounds of golf without anyone playing in the same foresome twice AND not playing with your spouse (the point of the weekend is to socialize with people you don't get to see all year).

I've got one that has worked well in the past for 8 couples, but am really struggling with this one. Any ideas?

Tuesday, September 13 2005, 01:22 pm
Livonia, MI

I forgot to mention that each foursome is made of 2 men and 2 women.
Tuesday, September 13 2005, 01:37 pm
« Last Edit: December 30, 2005, 08:20:03 PM by admin »

Ian Wakeling

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Reply #1 on: December 30, 2005, 08:32:13 PM
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

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