These puzzles, known as the ‘Thirty-four Game’ and the ‘Fifteen Game,’ on their introduction amongst us some years ago, created general interest wherever tried. Both games are played on the same board. The ‘blocks’ may be readily made by any lad. They can be cut out of cork, wood, or other suitable material, and may be square or round in shape, the numerals being written on the upper side, thus—
Let us describe the ‘15’ game first. Place the blocks, numbered 1 to 15, on the board irregularly, and then, by sliding them about from square to square, but without removing them from the board, arrange them in the following order:—
This looks simple enough in theory, but to any uninitiated reader who may think it easy of accomplishment in practice, we need merely say, ‘Try it.’
The ‘34’ puzzle consists in arranging the blocks, numbered 1 to 16, so that they will add perpendicularly, horizontally, and from corner to corner; also in arranging the 4 in each corner, the four outside numbers of the two centre lines, and the four groups in each corner; in all, sixteen different ways, and each produce the same total—34.
The ‘15’ puzzle would appear to have been, on its coming to England a few years ago, strictly a new introduction, but the ‘34’ was not only tolerably well-known, but seems to have been familiar to the Arabs in the ninth century. The nature of the game we have already described.
Now a word on the ‘15’ game. A skilful writer has pointed out how, while with two figures only two different arrangements can be obtained, three figures can be placed in six different orders, viz.:—3 1 2—3 2 1; 1 3 2—2 3 1; 1 2 3—2 1 3. Then, while there are thus six different orders that can be produced with three figures, if a fourth figure be placed in every one of the four different positions with regard to these six orders, we shall have four times the different number of orders that can be had with three figures. In the same way by multiplying these twenty-four different orders by five we shall have the number to be obtained with five figures. The rule, therefore, to find the number of different orders that may be obtained with any number of figures is to multiply all the different numbers used by each other. Thus, with six figures there will be 720 different orders; with seven, 5,040; with eight, 40,320, and so forth until we come to fifteen, when it will be found that the enormous number of 1,307,674,368,000 represents the different orders in which the fifteen numbers can be placed. This, of course, is the number of arrangements in which the cubes in the game of ‘15’ can be placed in the box. Let us consider how long it would take to test the solution of the different arrangements possible. Let us suppose that the cubes were worked to their consecutive or 15—14 order at the rate of one in every five minutes, which would, working day and night, be at the rate of about 105,000 arrangements a year. In this case it would take not less than twelve million years to test all the different arrangements!
The question now arises, whether out of this enormous number of different arrangements there are certain of them that cannot be shifted into the required consecutive order. The answer is, that exactly half of these arrangements are soluble, and the other half insoluble. The whole question turns upon the fact that in the arrangement of a certain number of figures in a row, otherwise than in consecutive order, the arrangement is made up either by an odd or an even number of transpositions. In the total number of different rows in which any given number of figures can be arranged, half of them will consist of rows containing an odd number, and the other half an even number of transpositions, just in the same way that in the numbers one to one hundred there are fifty odd and fifty even numbers. Take the case of six figures, 1, 2, 3, 4, 5, 6. If the first pair of figures is transposed the row will read 2, 1, 3, 4, 5, 6. In this row there is an odd number, namely, 1, of transpositions. If from this order the second pair is transposed the result will read 2, 1, 4, 3, 5, 6, and in this row there is an even number (2) of transpositions made from the original consecutive order. Suppose the last pair is transposed, the order will be 2, 1, 4, 3, 6, 5, and an odd number (3) of transpositions has been made. This may be continued until all the different rows possible have been worked, when it is evident that the first, third, fifth, etc., rows will each consist of an odd number of transpositions, while the intermediate rows will each contain an even number of transpositions.
Suppose we take the third row given above, 2, 1, 4, 3, 6, 5, this contains an odd number of transpositions. It is therefore impossible, by making an even number of transpositions each time, to bring 2, 1, 4, 3, 6, 5 to the original position of 1, 2, 3, 4, 5, 6, because it contains an odd number of transpositions, one of which will always remain, and cannot be eliminated by any even number of transpositions. Just in the same way we cannot, by the addition of even numbers, to any quantity or amount, to an odd number, produce an even number as the result. Every arrangement of a certain number of figures in a row contains therefore an odd or even number of transpositions which must be made before the figures can be brought into consecutive order, but it is immaterial whether these transpositions are made in such order as will the soonest produce the consecutive order, so long as they are made, and the number, whether odd or even, is noted.