In Diagram D I give another familiar puzzle that first appeared in a book published in Brussels in 1789, Les Petites Aventures de Jerome Sharp.  Place seven counters on seven of the eight points in the following manner.  You must always touch a point that is vacant with a counter, and then move it along a straight line leading from that point to the next vacant point (in either direction), where you deposit the counter.  You proceed in the same way until all the counters are placed.  Remember you always touch a vacant place and slide the counter from it to the next place, which must be also vacant.  Now, by the “buttons and string” method of simplification we can transform the diagram into E. Then the solution becomes obvious.  “Always move to the point that you last moved from.”  This is not, of course, the only way of placing the counters, but it is the simplest solution to carry in the mind.

There are several puzzles in this book that the reader will find lend themselves readily to this method.

342.—­THE MANDARIN’S PUZZLE.

The rather perplexing point that the solver has to decide for himself in attacking this puzzle is whether the shaded numbers (those that are shown in their right places) are mere dummies or not.  Ninety-nine persons out of a hundred might form the opinion that there can be no advantage in moving any of them, but if so they would be wrong.

The shortest solution without moving any shaded number is in thirty-two moves.  But the puzzle can be solved in thirty moves.  The trick lies in moving the 6, or the 15, on the second move and replacing it on the nineteenth move.  Here is the solution:  2, 6, 13, 4, 1, 21, 4, 1, 10, 2, 21, 10, 2, 5, 22, 16, 1, 13, 6, 19, 11, 2, 5, 22, 16, 5, 13, 4, 10, 21.  Thirty moves.

343.—­EXERCISE FOR PRISONERS.

There are eighty different arrangements of the numbers in the form of a perfect knight’s path, but only forty of these can be reached without two men ever being in a cell at the same time.  Two is the greatest number of men that can be given a complete rest, and though the knight’s path can be arranged so as to leave either 7 and 13, 8 and 13, 5 and 7, or 5 and 13 in their original positions, the following four arrangements, in which 7 and 13 are unmoved, are the only ones that can be reached under the moving conditions.  It therefore resolves itself into finding the fewest possible moves that will lead up to one of these positions.  This is certainly no easy matter, and no rigid rules can be laid down for arriving at the correct answer.  It is largely a matter for individual judgment, patient experiment, and a sharp eye for revolutions and position.

          A
    +—­+—­+—­+—­+
    | 6| 1|10|15|
    +—­+—­+—­+—­+
    | 9|12| 7| 4|
    +—­+—­+—­+—­+
    | 2| 5|14|11|
    +—­+—­+—­+—­+
    |13| 8| 3|**|
    +—­+—­+—­+—­+

          B
    +—­+—­+—­+—­+
    | 6| 1|10|15|
    +—­+—­+—­+—­+
    |11|14| 7| 4|
    +—­+—­+—­+—­+
    | 2| 5|12| 9|
    +—­+—­+—­+—­+
    |13| 8| 3|**|
    +—­+—­+—­+—­+