Hi John,
It is not possible for each player to play with everyone else. There are 190 = (20 choose 2)
pairs of golfers. Each foursome accounts for 6 = (4 choose 2) of them. That's 30 pairs per
round, and 180 pairs in 6 rounds.
Here is a scheme that's as not too bad. I don't know if it is possible to do better. The construction
will seem a little bit strange. I'm doing it this way because it makes it easier to explain. In a 5x5
grid, write the numbers 1, 2, 3, ..., 25 as:
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 
21 
22 
23 
24 
25 
Now write down six sets of 5 groups of 5 numbers. The first one is the rows. The second one
is the columns. The third one starts with and entry in the first row and then includes the next
entry in the next row, wrapping around when needed. The first two groups of five are
1, 7, 13, 19, 25 and 2, 8, 14, 20, 21. Now do the same but advance by 2 in the next row
(e.g. 1, 8, 15, 17, 24), then advance by 3, and then by 4 to get the last two collections of five
groups of 5.
Now delete the numbers 21, 22, 23, 24, 25 where ever they appear. In all collections except the
first, you have 5 groups of 4 (and no pair is repeated). In the first collection you have 4 groups
of 5. Take 1 number from each group to form a group of 4.
Now you have 6 collections of 5 foursomes that cover 174 pairs of golfers (180, if it could be achieved,
would be best possible). The four golfers used to form the final foursome on the first day each play
with all but four players (the four in the group they came from), and play with each other twice
over the course of the six rounds.
I hope this helps. Have a great tour.
Victoria
