The following puzzle has an added interest from the circumstance that a
correct solution of it secured for a certain young Chinaman the hand of
his charming bride. The wealthiest mandarin within a radius of a hundred
miles of Peking was Hi-Chum-Chop, and his beautiful daughter, Peeky-Bo,
had innumerable admirers. One of her most ardent lovers was Winky-Hi,
and when he asked the old mandarin for his consent to their marriage,
Hi-Chum-Chop presented him with the following puzzle and promised his
consent if the youth brought him the correct answer within a week.
Winky-Hi, following a habit which obtains among certain solvers to this
day, gave it to all his friends, and when he had compared their
solutions he handed in the best one as his own. Luckily it was quite
right. The mandarin thereupon fulfilled his promise. The fatted pup was
killed for the wedding feast, and when Hi-Chum-Chop passed Winky-Hi the
liver wing all present knew that it was a token of eternal goodwill, in
accordance with Chinese custom from time immemorial.

The mandarin had a table divided into twenty-five squares, as shown in
the diagram. On each of twenty-four of these squares was placed a
numbered counter, just as I have indicated. The puzzle is to get the
counters in numerical order by moving them one at a time in what we call
"knight's moves." Counter 1 should be where 16 is, 2 where 11 is, 4
where 13 now is, and so on. It will be seen that all the counters on
shaded squares are in their proper positions. Of course, two counters
may never be on a square at the same time. Can you perform the feat in
the fewest possible moves?

[Illustration]

In order to make the manner of moving perfectly clear I will point out