The difficult case of 7 persons on 15 occasions is solved as follows, and was given by me in The Canterbury Puzzles:—
| 1 2 3 4 5 7 6 |
| 1 6 2 7 5 3 4 |
| 1 3 5 2 6 7 4 |
| 1 5 7 4 3 6 2 |
| 1 5 2 7 3 4 6 |
In this case the 1 is a repeater, and there are two separate cycles, 2, 3, 4, 2, and 5, 6, 7, 5. We thus get five groups of three lines each, for a fourth line in any group will merely repeat the first line.
A solution for 8 persons on 21 occasions is as follows:—
| 1 8 6 3 4 5 2 7 |
| 1 8 4 5 7 2 3 6 |
| 1 8 2 7 3 6 4 5 |
The 1 is here a repeater, and the cycle 2, 3, 4, 5, 6, 7, 8. Every one of the 3 groups will give 7 lines.
Here is my solution for 9 persons on 28 occasions:—
| 2 1 9 7 4 5 6 3 8 |
| 2 9 5 1 6 8 3 4 7 |
| 2 9 3 1 8 4 7 5 6 |
| 2 9 1 5 6 4 7 8 3 |
There are here two repeaters, 1 and 2, and the cycle is 3, 4, 5, 6, 7, 8, 9. We thus get 4 groups of 7 lines each.
The case of 10 persons on 36 occasions is solved as follows:—