Hi Richard,
You may like to know that it is possible to make a similar 16 player, 35 round schedule to the one described in the article above, where the rounds are partitioned into 7 sets of 5 rounds each, such that each set of 5 rounds forms a social square (a 2-(16,4,1) design). Also it's possible to arrange it so that no pair of players who play together in the last round of one partition, play together again in the first round of the next partition. This would seem to be optimal from the point of view of playing golf foursomes with 16 players; because of the partitioning, play can stop at any point during the schedule and there will be best possible balance for the number of times pairs of players have occured together within the same foursome, and since the whole thing is a 3-design, no three players will have played together in the same foursome more than once.
Ian.
Construction of a resolvable 3-(16,4,1) design with 7 2-(16,4,1) partitions.
Table 1. Resolvable 3-(8,4,1) design
i
0 {0,4,6,7} {1,2,3,5}
1 {1,5,0,7} {2,3,4,6}
2 {2,6,1,7} {3,4,5,0}
3 {3,0,2,7} {4,5,6,1}
4 {4,1,3,7} {5,6,0,2}
5 {5,2,4,7} {6,0,1,3}
6 {6,3,5,7} {0,1,2,4}
Table 2. Assignment of 10 new blocks derived from each old block {a,b,c,d} of Table 1.
j New Blocks Partition
0 { a , b , c , d } {a+8, b+8, c+8, d+8} i
1 {a+8, b+8, c , d } { a , b , c+8, d+8} i+1
2 {a+8, b , c+8, d } { a , b+8, c , d+8} i+2
3 {a+8, b , c , d+8} { a , b+8, c+8, d } i+4
4 { a , b , a+8, b+8} { c , d , c+8, d+8} i+5
Row i from Table 1 is combined with a row j from Table 2 to form four new blocks that contain all the players 0 to 15 exactly once. These four new blocks are one resolution class of the final 3-(16,4,1) design. Each of the 35 resolution classes, formed from all possible combinations of i and j, is placed into one of 7 separate partitions. The partition assignments being the residues mod 7, which are indicated by the final column of Table 2. An assignment depends on i, the row number in Table 1 from which the old block {a,b,c,d} was derived.
For example, the combination of i=5 & j=3 generates the four blocks:
{13,2,4,15} {14,0,1,11} {5,10,12,7} {6,8,9,3}
which are assigned to partition 2 (since 5 + 4 ≡ 2 mod 7).
The full partitioned schedule is given below.
Part.
0 { 0 4 6 7} { 1 2 3 5} { 8 12 14 15} { 9 10 11 13}
0 {14 11 5 7} { 8 9 2 4} { 6 3 13 15} { 0 1 10 12}
0 {13 2 12 7} {14 0 9 3} { 5 10 4 15} { 6 8 1 11}
0 {11 0 2 15} {12 5 6 9} { 3 8 10 7} { 4 13 14 1}
0 { 2 6 10 14} { 1 7 9 15} { 3 4 11 12} { 5 0 13 8}
1 { 8 12 6 7} { 9 10 3 5} { 0 4 14 15} { 1 2 11 13}
1 { 1 5 0 7} { 2 3 4 6} { 9 13 8 15} {10 11 12 14}
1 {14 3 13 7} { 8 1 10 4} { 6 11 5 15} { 0 9 2 12}
1 {12 1 3 15} {13 6 0 10} { 4 9 11 7} { 5 14 8 2}
1 { 3 0 11 8} { 2 7 10 15} { 4 5 12 13} { 6 1 14 9}
2 { 9 13 0 7} {10 11 4 6} { 1 5 8 15} { 2 3 12 14}
2 { 2 6 1 7} { 3 4 5 0} {10 14 9 15} {11 12 13 8}
2 { 8 4 14 7} { 9 2 11 5} { 0 12 6 15} { 1 10 3 13}
2 { 4 1 12 9} { 3 7 11 15} { 5 6 13 14} { 0 2 8 10}
2 {13 2 4 15} {14 0 1 11} { 5 10 12 7} { 6 8 9 3}
3 { 3 0 2 7} { 4 5 6 1} {11 8 10 15} {12 13 14 9}
3 {10 14 1 7} {11 12 5 0} { 2 6 9 15} { 3 4 13 8}
3 {14 3 5 15} { 8 1 2 12} { 6 11 13 7} { 0 9 10 4}
3 { 5 2 13 10} { 4 7 12 15} { 6 0 14 8} { 1 3 9 11}
3 { 9 5 8 7} {10 3 12 6} { 1 13 0 15} { 2 11 4 14}
4 { 4 1 3 7} { 5 6 0 2} {12 9 11 15} {13 14 8 10}
4 {11 8 2 7} {12 13 6 1} { 3 0 10 15} { 4 5 14 9}
4 {10 6 9 7} {11 4 13 0} { 2 14 1 15} { 3 12 5 8}
4 { 8 4 6 15} { 9 2 3 13} { 0 12 14 7} { 1 10 11 5}
4 { 6 3 14 11} { 5 7 13 15} { 0 1 8 9} { 2 4 10 12}
5 {12 9 3 7} {13 14 0 2} { 4 1 11 15} { 5 6 8 10}
5 { 5 2 4 7} { 6 0 1 3} {13 10 12 15} {14 8 9 11}
5 {11 0 10 7} {12 5 14 1} { 3 8 2 15} { 4 13 6 9}
5 { 0 4 8 12} { 6 7 14 15} { 1 2 9 10} { 3 5 11 13}
5 { 9 5 0 15} {10 3 4 14} { 1 13 8 7} { 2 11 12 6}
6 { 6 3 5 7} { 0 1 2 4} {14 11 13 15} { 8 9 10 12}
6 {13 10 4 7} {14 8 1 3} { 5 2 12 15} { 6 0 9 11}
6 {12 1 11 7} {13 6 8 2} { 4 9 3 15} { 5 14 0 10}
6 {10 6 1 15} {11 4 5 8} { 2 14 9 7} { 3 12 13 0}
6 { 1 5 9 13} { 0 7 8 15} { 2 3 10 11} { 4 6 12 14}
