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Part 33

The Canterbury Puzzles, and Other Curious Problems · Henry Ernest Dudeney — chapter 33 of 49 · ~1,240 words · public domain

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Also in the case of the solution to No. 67, the Coinage Puzzle, the numbers are, in shillings:--

2 2-1/2 3 4-1/2 5 5-1/2 7 7-1/2 8

If there are to be nine different numbers, 0 may occur once (as in the solution to No. 22). Yet one might construct squares with negative numbers, as follows:--

-2 -1 0 5 6 7 12 13 14

69.--The Frogs and Tumblers.

It is perfectly true, as the Professor said, that there is only one solution (not counting a reversal) to this puzzle. The frogs that jump are George in the third horizontal row; Chang, the artful-looking batrachian at the end of the fourth row; and Wilhelmina, the fair creature in the seventh row. George jumps downwards to the second tumbler in the seventh row; Chang, who can only leap short distances in consequence of chronic rheumatism, removes somewhat unwillingly to the glass just above him--the eighth in the third row; while Wilhelmina, with all the sprightliness of her youth and sex, performs the very creditable saltatory feat of leaping to the fourth tumbler in the fourth row. In their new positions, as shown in the accompanying diagram, it will be found that of the eight frogs no two are in line vertically, horizontally, or diagonally.

70.--Romeo and Juliet.

This is rather a difficult puzzle, though, as the Professor remarked when Hawkhurst hit on the solution, it is "just one of those puzzles that a person might solve at a glance" by pure luck. Yet when the solution, with its pretty, symmetrical arrangement, is seen, it looks ridiculously simple.

It will be found that Romeo reaches Juliet's balcony after visiting every house once and only once, and making fourteen turnings, not counting the turn he makes at starting. These are the fewest turnings possible, and the problem can only be solved by the route shown or its reversal.

71.--Romeo's Second Journey.

In order to take his trip through all the white squares only with the fewest possible turnings, Romeo would do well to adopt the route I have shown, by means of which only sixteen turnings are required to perform the feat. The Professor informs me that the Helix Aspersa, or common or garden snail, has a peculiar aversion to making turnings--so much so that one specimen with which he made experiments went off in a straight line one night and has never come back since.

72.--The Frogs who would a-wooing go.

This is one of those puzzles in which a plurality of solutions is practically unavoidable. There are two or three positions into which four frogs may jump so as to form five rows with four in each row, but the case I have given is the most satisfactory arrangement.

The frogs that have jumped have left their astral bodies behind, in order to show the reader the positions which they originally occupied. Chang, the frog in the middle of the upper row, suffering from rheumatism, as explained above in the Frogs and Tumblers solution, makes the shortest jump of all--a little distance between the two rows; George and Wilhelmina leap from the ends of the lower row to some distance N. by N.W. and N. by N.E. respectively; while the frog in the middle of the lower row, whose name the Professor forgot to state, goes direct S.

73.--The Game of Kayles.

To win at this game you must, sooner or later, leave your opponent an even number of similar groups. Then whatever he does in one group you repeat in a similar group. Suppose, for example, that you leave him these groups: o.o.ooo.ooo. Now, if he knocks down a single, you knock down a single; if he knocks down two in one triplet, you knock down two in the other triplet; if he knocks down the central kayle in a triplet, you knock down the central one in the other triplet. In this way you must eventually win. As the game is started with the arrangement o.ooooooooooo, the first player can always win, but only by knocking down the sixth or tenth kayle (counting the one already fallen as the second), and this leaves in either case o.ooo.ooooooo, as the order of the groups is of no importance. Whatever the second player now does, this can always be resolved into an even number of equal groups. Let us suppose that he knocks down the single one; then we play to leave him oo.ooooooo. Now, whatever he does we can afterwards leave him either ooo.ooo or o.oo.ooo. We know why the former wins, and the latter wins also; because, however he may play, we can always leave him either o.o, or o.o.o.o, or oo.oo, as the case may be. The complete analysis I can now leave for the amusement of the reader.

74.--The Broken Chessboard.

The illustration will show how the thirteen pieces can be put together so as to construct the perfect board, and the reverse problem of cutting these particular pieces out will be found equally entertaining.

Compare with Nos. 293 and 294 in A. in M.

75.--The Spider and the Fly.

Though this problem was much discussed in the Daily Mail from 18th January to 7th February 1905, when it appeared to create great public interest, it was actually first propounded by me in the Weekly Dispatch of 14th June 1903.

Imagine the room to be a cardboard box. Then the box may be cut in various different ways, so that the cardboard may be laid flat on the table. I show four of these ways, and indicate in every case the relative positions of the spider and the fly, and the straightened course which the spider must take without going off the cardboard. These are the four most favourable cases, and it will be found that the shortest route is in No. 4, for it is only 40 feet in length (add the square of 32 to the square of 24 and extract the square root). It will be seen that the spider actually passes along five of the six sides of the room! Having marked the route, fold the box up (removing the side the spider does not use), and the appearance of the shortest course is rather surprising. If the spider had taken what most persons will consider obviously the shortest route (that shown in No. 1), he would have gone 42 feet! Route No. 2 is 43.174 feet in length, and Route No. 3 is 40.718 feet.

I will leave the reader to discover which are the shortest routes when the spider and the fly are 2, 3, 4, 5, and 6 feet from the ceiling and the floor respectively.

76.--The Perplexed Cellarman.

Brother John gave the first man three large bottles and one small bottleful of wine, and one large and three small empty bottles. To each of the other two men he gave two large and three small bottles of wine, and two large and one small empty bottle. Each of the three then receives the same quantity of wine, and the same number of each size of bottle.

77.--Making a Flag.

The diagram shows how the piece of bunting is to be cut into two pieces. Lower the piece on the right one "tooth," and they will form a perfect square, with the roses symmetrically placed.

It will be found interesting to compare this with No. 154 in A. in M.

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