Here is a little puzzle of the moving counters class that my readers will
probably find entertaining. Make a diagram of any convenient size similar
to that shown in our illustration, and provide six counters--three marked
to represent foxes and three to represent geese. Place the geese on the
discs 1, 2, and 3, and the foxes on the discs numbered 10, 11, and 12.
Now the puzzle is this. By moving one at a time, fox and goose
alternately, along a straight line from one disc to the next one, try to
get the foxes on 1, 2, and 3, and the geese on 10, 11, and 12--that is,
make them exchange places--in the fewest possible moves.
[Illustration]
But you must be careful never to let a fox and goose get within reach of
each other, or there will be trouble. This rule, you will find, prevents
you moving the fox from 11 on the first move, as on either 4 or 6 he
would be within reach of a goose. It also prevents your moving a fox from
10 to 9, or from 12 to 7. If you play 10 to 5, then your next move may be
2 to 9 with a goose, which you could not have played if the fox had not
previously gone from 10. It is perhaps unnecessary to say that only one
fox or one goose can be on a disc at the same time. Now, what is the
smallest number of moves necessary to make the foxes and geese change
places?