The smallest possible number of moves is twenty-two--that is, eleven for
the foxes and eleven for the geese. Here is one way of solving the
puzzle:
10--5 11--6 12--7 5--12 6--1 7--6
---- ---- ---- ---- ---- ----
1--8 2--9 3--4 8--3 9--10 4--9
12--7 1--8 6--1 7--2 8--3
---- ---- ---- ---- ----
3--4 10--5 9--10 4--11 5--12
Of course, the reader will play the first move in the top line, then the
first move in the second line, then the second move in the top line, and
so on alternately.
[Illustration]
In _A. in M._, p. 230, I have explained fully my "buttons and string"
method of solving puzzles on chequered boards. In Diagram A is shown the
puzzle in the form in which it may be presented on a portion of the
chessboard with six knights. A comparison with the illustration on page
141 will show that I have there dispensed with the necessity of
explaining the knight's move to the uninstructed reader by lines that
indicate those moves. The two puzzles are the same thing in different
dress. Now compare page 141 with Diagram B, and it will be seen that by
disentangling the strings I have obtained a simplified diagram without
altering the essential relations between the buttons or discs. The reader
will now satisfy himself without any difficulty that the puzzle requires
eleven moves for the foxes and eleven for the geese. He will see that a
goose on 1 or 3 must go to 8, to avoid being one move from a fox and to
enable the fox on 11 to come on to the ring. If we play 1--8, then it is
clearly best to play 10--5 and not 12--5 for the foxes. When they are all
on the circle, then they simply promenade round it in a clockwise
direction, taking care to reserve 8--3 and 5--12 for the final moves. It
is thus rendered ridiculously easy by this method. See also notes on
solutions to Nos. 13 and 85.