1 2 3 6 4 5
1 3 4 2 5 6
1 4 5 3 6 2
1 5 6 4 2 3
1 6 2 5 3 4
-----------
1 2 4 5 6 3
1 3 5 6 2 4
1 4 6 2 3 5
1 5 2 3 4 6
1 6 3 4 5 2
It will now no longer be necessary to give the solutions in full, for
reasons that I will explain. It will be seen in the examples above that
the 1 (and, in the case of 5 persons, also the 2) is repeated down the
column. Such a number I call a "repeater." The other numbers descend in
cyclical order. Thus, for 6 persons we get the cycle, 2, 3, 4, 5, 6, 2,
and so on, in every column. So it is only necessary to give the two
lines 1 2 3 6 4 5 and 1 2 4 5 6 3, and denote the cycle and repeaters,
to enable any one to write out the full solution straight away. The
reader may wonder why I do not start the last solution with the numbers
in their natural order, 1 2 3 4 5 6. If I did so the numbers in the
descending cycle would not be in their natural order, and it is more
convenient to have a regular cycle than to consider the order in the
first line.
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.