SOLUTIONS 301-430

Here the powers are not all even in number, but by taking 9 from the 11 heap you immediately restore your winning position, thus—

And so on to the end. This solution is quite general, and applies to any number of matches and any number of heaps. A correspondent informs me that this puzzle game was first propounded by Mr. W.M.F. Mellor, but when or where it was published I have not been able to ascertain.

397.—THE MONTENEGRIN DICE GAME.—solution

The players should select the pairs 5 and 9, and 13 and 15, if the chances of winning are to be quite equal. There are 216 different ways in which the three dice may fall. They may add up 5 in 6 different ways and 9 in 25 different ways, making 31 chances out of 216 for the player who selects these numbers. Also the dice may add up 13 in 21 different ways, and 15 in 10 different ways, thus giving the other player also 31 chances in 216.

398.—THE CIGAR PUZZLE.—solution

Not a single member of the club mastered this puzzle, and yet I shall show that it is so simple that the merest child can understand its solution—when it is pointed out to him! The large majority of my friends expressed their entire bewilderment. Many considered that "the theoretical result, in any case, is determined by the relationship between the table and the cigars;" others, regarding it as a problem in the theory of Probabilities, arrived at the conclusion that the chances are slightly in favour of the first or second player, as the case may be. One man took a table and a cigar of particular dimensions, divided the table into equal sections, and proceeded to make the two players fill up these sections so that the second player should win. But why should the first player be so accommodating? At any stage he has only to throw down a cigar obliquely across several of these sections entirely to upset Mr. 2's calculations! We have to assume that each player plays the best possible; not that one accommodates the other.

The theories of some other friends would be quite sound if the shape of the cigar were that of a torpedo—perfectly symmetrical and pointed at both ends.

I will show that the first player should infallibly win, if he always plays in the best possible manner. Examine carefully the following diagram, No. 1, and all will be clear.

The first player must place his first cigar on end in the exact centre of the table, as indicated by the little circle. Now, whatever the second player may do throughout, the first player must always repeat it in an exactly diametrically opposite position. Thus, if the second player places a cigar at A, I put one at AA; he places one at B, I put one at BB; he places one at C, I put one at CC; he places one at D, I put one at DD; he places one at E, I put one at EE; and so on until no more cigars can be placed without touching. As the cigars are supposed to be exactly alike in every respect, it is perfectly clear that for every move that the second player may choose to make, it is possible exactly to repeat it on a line drawn through the centre of the table. The second player can always duplicate the first player's move, no matter where he may place a cigar, or whether he places it on end or on its side. As the cigars are all alike in every respect, one will obviously balance over the edge of the table at precisely the same point as another. Of course, as each player is supposed to play in the best possible manner, it becomes a matter of theory. It is no valid objection to say that in actual practice one would not be sufficiently exact to be sure of winning. If as the first player you did not win, it would be in consequence of your not having played the best possible.

The second diagram will serve to show why the first cigar must be placed on end. (And here I will say that the first cigar that I selected from a box I was able so to stand on end, and I am allowed to assume that all the other cigars would do the same.) If the first cigar were placed on its side, as at F, then the second player could place a cigar as at G—as near as possible, but not actually touching F. Now, in this position you cannot repeat his play on the opposite side, because the two ends of the cigar are not alike. It will be seen that GG, when placed on the opposite side in the same relation to the centre, intersects, or lies on top of, F, whereas the cigars are not allowed to touch. You must therefore put the cigar farther away from the centre, which would result in your having insufficient room between the centre and the bottom left-hand corner to repeat everything that the other player would do between G and the top right-hand corner. Therefore the result would not be a certain win for the first player.

399.—THE TROUBLESOME EIGHT.—solution

The conditions were to place a different number in each of the nine cells so that the three rows, three columns, and two diagonals should each add up 15. Probably the reader at first set himself an impossible task through reading into these conditions something which is not there—a common error in puzzle-solving. If I had said "a different figure," instead of "a different number," it would have been quite impossible with the 8 placed anywhere but in a corner. And it would have been equally impossible if I had said "a different whole number." But a number may, of course, be fractional, and therein lies the secret of the puzzle. The arrangement shown in the figure will be found to comply exactly with the conditions: all the numbers are different, and the square adds up 15 in all the required eight ways.

400.—THE MAGIC STRIPS.—solution

There are of course six different places between the seven figures in which a cut may be made, and the secret lies in keeping one strip intact and cutting each of the other six in a different place. After the cuts have been made there are a large number of ways in which the thirteen pieces may be placed together so as to form a magic square. Here is one of them:—

The arrangement has some rather interesting features. It will be seen that the uncut strip is at the top, but it will be found that if the bottom row of figures be placed at the top the numbers will still form a magic square, and that every successive removal from the bottom to the top (carrying the uncut strip stage by stage to the bottom) will produce the same result. If we imagine the numbers to be on seven complete perpendicular strips, it will be found that these columns could also be moved in succession from left to right or from right to left, each time producing a magic square.

401.—EIGHT JOLLY GAOL BIRDS.—solution

There are eight ways of forming the magic square—all merely different aspects of one fundamental arrangement. Thus, if you give our first square a quarter turn you will get the second square; and as the four sides may be in turn brought to the top, there are four aspects. These four in turn reflected in a mirror produce the remaining four aspects. Now, of these eight arrangements only four can possibly be reached under the conditions, and only two of these four can be reached in the fewest possible moves, which is nineteen. These two arrangements are shown. Move the men in the following order: 5, 3, 2, 5, 7, 6, 4, 1, 5, 7, 6, 4, 1, 6, 4, 8, 3, 2, 7, and you get the first square. Move them thus: 4, 1, 2, 4, 1, 6, 7, 1, 5, 8, 1, 5, 6, 7, 5, 6, 4, 2, 7, and you have the arrangement in the second square. In the first case every man has moved, but in the second case the man numbered 3 has never left his cell. Therefore No. 3 must be the obstinate prisoner, and the second square must be the required arrangement.

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