First of all I will ask the reader to compare the original square
diagram with the circular one shown in Figs. 1, 2, and 3 below. If for
the moment we ignore the shading (the purpose of which I shall proceed
to explain), we find that the circular diagram in each case is merely a
simplification of the original square one--that is, the roads from A
lead to B, E, and M in both cases, the roads from L (London) lead to I,
K, and S, and so on. The form below, being circular and symmetrical,
answers my purpose better in applying a mechanical solution, and I
therefore adopt it without altering in any way the conditions of the
puzzle. If such a question as distances from town to town came into the
problem, the new diagrams might require the addition of numbers to
indicate these distances, or they might conceivably not be at all
practicable.
[Illustration: Figs. 1, 2, and 3]
Now, I draw the three circular diagrams, as shown, on a sheet of paper
and then cut out three pieces of cardboard of the forms indicated by the
shaded parts of these diagrams. It can be shown that every route, if
marked out with a red pencil, will form one or other of the designs
indicated by the edges of the cards, or a reflection thereof. Let us
direct our attention to Fig. 1. Here the card is so placed that the star
is at the town T; it therefore gives us (by following the edge of the
card) one of the circular routes from London: L, S, R, T, M, A, E, P, O,
J, D, C, B, G, N, Q, K, H, F, I, L. If we went the other way, we should
get L, I, F, H, K, Q, etc., but these reverse routes were not to be
counted. When we have written out this first route we revolve the card
until the star is at M, when we get another different route, at A a
third route, at E a fourth route, and at P a fifth route. We have thus
obtained five different routes by revolving the card as it lies. But it
is evident that if we now take up the card and replace it with the other
side uppermost, we shall in the same manner get five other routes by
revolution.
We therefore see how, by using the revolving card in Fig. 1, we may,
without any difficulty, at once write out ten routes. And if we employ
the cards in Figs. 2 and 3, we similarly obtain in each case ten other
routes. These thirty routes are all that are possible. I do not give the
actual proof that the three cards exhaust all the possible cases, but
leave the reader to reason that out for himself. If he works out any
route at haphazard, he will certainly find that it falls into one or
other of the three categories.