I can give it a try if you can fill in a few blanks for me.
Maybe I am misunderstanding something. In eight (two-person) matches it is only possible to play eight (of the nine) opponents. Or is a player simultaneously playing everyone in the group? That's the only way threesomes make sense. What group sizes did you have in mind?
Is it a nine hole course? That's what your message seems to suggest.
I read through your solution one time with the 9 players and adding the 10th with the 3-3-3 solution and that one seem to work. Here is a further explanation of my challenge:
We have 10 men.
We are playing four different courses of 18 holes each.
I need to have each guy play all other nine golfers, but we can't switch after 9 holes only after we move to another course (after each 18).
Example: Group 1,2,3,4 could be in the first foursome having 1 play 2 on the front nine and then 1 play 3 on the back nine. On the front nine, 3 would play 4 and then on the back 2 would play 4.
Is it possible with 8 9 holes rounds available to have all 10 guys play against each other if we can't switch except after 18 (meaning only 4 18 hole rounds)?
This isn't of utmost importance as your original 9 player scenario worked out and made our 10 person have to pair up with another for three of the four rounds, but it will suffice if another solution isn't available.
Thanks Jason. I'm happy that the 9 player solution gives you something. Playing as you describe, it is not possible to play all 45 possible maches.
In an 18 hole round, a foursome accounts for four matches, and a twosome accounts for one. I'll guess that a threecome accounts for two matches.
Going 4-4-2 you play 9 matches each day, and get 36 played over 4 days. Going 4-3-3 you play 8 matches each day, and 32 over all. There may or may not be repeats in the schedule, so it may be that fewer matches are actually played.
If you do something strange and creative, then maybe it is possible to get a better outcome. In a threesome, is it possible to have 1 play 2 and 2 play 3 on the front, and then have 1 play 3 on the back? In a foursome, it is possible to do something similar so that all 6 possible matches get played? If you do that, then these groupings have every player together at least once (I think), and so all matches can be played.
1: 1 2 3 4; 5 6 7 8; 9 10
2: 1 5 9; 2 6 10; 3 7 4 8
3: 1 6 3 7; 2 5 8 9; 3 4 10
4: 1 4 6 8; 3 5 7 9; 2 10
A similar thing would happen if you took the 9 man schedule and always added player 10 to the group containing 1.
I hope this makes some sense. Enjoy the trip.