| 1 10 8 3 6 5 4 7 2 9 |
| 1 10 6 5 2 9 7 4 3 8 |
| 1 10 2 9 3 8 6 5 7 4 |
| 1 10 7 4 8 3 2 9 5 6 |
The repeater is 1, and the cycle, 2, 3, 4, 5, 6, 7, 8, 9, 10. We here have 4 groups of 9 lines each.
My solution for 11 persons on 45 occasions is as follows:—
| 2 | 11 | 9 | 4 | 7 | 6 | 5 | 1 | 8 | 3 | 10 |
| 2 | 1 | 11 | 7 | 6 | 3 | 10 | 8 | 5 | 4 | 9 |
| 2 | 11 | 10 | 3 | 9 | 4 | 8 | 5 | 1 | 7 | 6 |
| 2 | 11 | 5 | 8 | 1 | 3 | 10 | 6 | 7 | 9 | 4 |
| 2 | 11 | 1 | 10 | 3 | 4 | 9 | 6 | 7 | 5 | 8 |
There are two repeaters, 1 and 2, and the cycle is, 3, 4, 5,... 11. We thus get 5 groups of 9 lines each.
The case of 12 persons on 55 occasions is solved thus:—
| 1 | 2 | 3 | 12 | 4 | 11 | 5 | 10 | 6 | 9 | 7 | 8 |
| 1 | 2 | 4 | 11 | 6 | 9 | 8 | 7 | 10 | 5 | 12 | 3 |
| 1 | 2 | 5 | 10 | 8 | 7 | 11 | 4 | 3 | 12 | 6 | 9 |
| 1 | 2 | 6 | 9 | 10 | 5 | 3 | 12 | 7 | 8 | 11 | 4 |
| 1 | 2 | 7 | 8 | 12 | 3 | 6 | 9 | 11 | 4 | 5 | 10 |
Here 1 is a repeater, and the cycle is 2, 3, 4, 5,... 12. We thus get 5 groups of 11 lines each.