end, they may actually save a move and perform the feat in sixteen! The rick consists in placing the man in the centre on the back of one of the corner men, and then working the pair into the centre before their final separation.
Here are the moves for getting the men into one or other of the above two positions. The numbers are those of the men in the order in which they move into the cell that is for the time being vacant. The pair is shown in brackets:—
Place 5 on 1. Then, 6, 9, 8, 6, 4, (6), 2, 4, 9, 3, 4, 9, (6), 7, 6, 1.
Place 5 on 9. Then, 4, 1, 2, 4, 6, (14), 8, 6, 1, 7, 6, 1, (14), 3, 4, 9.
Place 5 on 3. Then, 6, (8), 2, 6, 4, 7, 8, 4, 7, 1, 6, 7, (8), 9, 4, 3.
Place 5 on 7. Then, 4, (12), 8, 4, 6, 3, 2, 6, 3, 9, 4, 3, (12), 1, 6, 7.
The first and second solutions produce Diagram A ; the second and third produce Diagram B. There are only sixteen moves in every case. Having found the fewest moves, we had to consider how we were to make the burdened man do as little work as possible. It will at once be seen that as the pair have to go into the centre before separating they must take at fewest two moves. The labour of the burdened man can only be reduced by adopting the other method of solution, which, however, forces us to take another move.
403.—THE SPANISH DUNGEON.
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{{smallcaps|This can best be solved by working backwards—that is to say, you must first catch your square, and then work back to the original position. We must first construct those squares which are found to require the least amount of readjustment of the numbers. Many of these we know cannot possibly be reached. When we have before us the most favomrable possible arrangements, it then becomes a question of careful analysis to discover which position can be reached in the fewest moves. I am afraid, however, it is only after considerable study and experience that the solver is able to get such a grasp of the various "areas of disturbance" and methods of circulation that his judgment is of much value to him.
The second diagram is a most favourable magic square position. It will be seen that prisoners 4, 8, 13, and 14 are left in their original cells. This position may be reached in as few as thirty-seven moves. Here are the moves : 15, 14, 10, 6, 7, 3, 2, 7, 6, 11, 3, 3, 7, 6, 11, 10, 14, 3, 2, 11, 10, 9, 5, I, 6, 10, 9, 5, 1, 6, 10, 9, 5, 2, 12, 15, 3. This short solution will probably surprise many readers who may not find a way under from sixty to a hundred moves. The clever prisoner was No. 6, who in the original illustration will be seen with his arms extended calling out the moves. He and No. 10 did most of the work, each changing his cell five times. No. 12, the man with the crooked leg, was lame, and therefore fortunately had only to pass from his cell into the next one when his time came round.
404.—THE SIBERIAN DUNGEONS.
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In attempting to solve this puzzle it is clearly necessary to seek such magic squares as seem the most favourable for our purpose, and then carefully examine and try them for "fewest moves." Of course it at once occurs to us that if we can adopt a square in which a certain number of men need not leave their original cells, we may save moves on the one hand, but we may obstruct our movements on the other. For example, a magic square may be formed with the 6, 7, 13, and 16 unmoved; but in such case it is obvious that a solution is impossible, since cells 14 and 15 can neither be left nor entered without breaking the condition of no two men ever being in the same cell together.
The following solution in fourteen moves was found by Mr. G. Wotherspoon : 8—17, 16—21, 6—16, 14—8, 5—18, 4—14, 3—24, 11—20, 10—19, 2—23, 13—22, 12—6, 1—5, 9—13. As this solution is in what I consider the theoretical minimum number of moves, I am confident that it cannot be improved upon, and on this point Mr. Wotherspoon is of the same opinion.
405.—CARD MAGIC SQUARES.
Arrange the cards as follows for the three new squares:—
| 3 | 2 | 4 | 6 | 5 | 7 | 9 | 8 | 10 |
| 4 | 3 | 2 | 7 | 6 | 5 | 10 | 9 | 8 |
| 2 | 4 | 3 | 5 | 7 | 6 | 8 | 10 | 9 |
Three aces and one ten are not used. The summations of the four squares are thus : 9, 15, 18, and 27 — all different, as required.
