A B C D E F G
A C D B G E F
A D B C F G E
A G B F E C D
A F C E G D B
A E D G F B C
A C E B G F D
A D G C F E B
A B F D E G C
A E F D C G B
A G E B D F C
A F G C B E D
A E B F C D G
A G C E D B F
A F D G B C E
Of course, at a circular table, A will be next to the man at the end of
the line.
I first gave this problem for six persons on ten days, in the _Daily
Mail_ for the 13th and 16th October 1905, and it has since been discussed
in various periodicals by mathematicians. Of course, it is easily seen
that the maximum number of sittings for _n_ persons is (_n_ - 1)(_n_
-2)/2 ways. The comparatively easy method for solving all cases where
_n_ is a prime+1 was first discovered by Ernest Bergholt. I then pointed
out the form and construction of a solution that I had obtained for 10
persons, from which E. D. Bewley found a general method for all even
numbers. The odd numbers, however, are extremely difficult, and for a
long time no progress could be made with their solution, the only numbers
that could be worked being 7 (given above) and 5, 9, 17, and 33, these
last four being all powers of 2+1. At last, however (though not without
much difficulty), I discovered a subtle method for solving all cases, and
have written out schedules for every number up to 25 inclusive. The case
of 11 has been solved also by W. Nash. Perhaps the reader will like to
try his hand at 13. He will find it an extraordinarily hard nut.
The solutions for all cases up to 12 inclusive are given in _A. in M._,
pp. 205, 206.