The Canterbury Puzzles Part 28

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.

[Ill.u.s.tration]

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 a.n.a.lysis I can now leave for the amus.e.m.e.nt of the reader.

74.--_The Broken Chessboard._

The ill.u.s.tration 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.

[Ill.u.s.tration]

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 pa.s.ses 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.

[Ill.u.s.tration]

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 quant.i.ty 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.

[Ill.u.s.tration]

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

78.--_Catching the Hogs._

A very short examination of this puzzle game should convince the reader that Hendrick can never catch the black hog, and that the white hog can never be caught by Katrun.

Each hog merely runs in and out of one of the nearest corners and can never be captured. The fact is, curious as it must at first sight appear, a Dutchman cannot catch a black hog, and a Dutchwoman can never capture a white one! But each can, without difficulty, catch one of the other colour.

So if the first player just determines that he will send Hendrick after the white porker and Katrun after the black one, he will have no difficulty whatever in securing both in a very few moves.

It is, in fact, so easy that there is no necessity whatever to give the line of play. We thus, by means of the game, solve the puzzle in real life, why the Dutchman and his wife could not catch their pigs: in their simplicity and ignorance of the peculiarities of Dutch hogs, each went after the wrong animal.

The little principle involved in this puzzle is that known to chess-players as "getting the opposition." The rule, in the case of my puzzle (where the moves resemble rook moves in chess, with the added condition that the rook may only move to an adjoining square), is simply this. Where the number of squares on the same row, between the man or woman and the hog, is odd, the hog can never be captured; where the number of squares is even, a capture is possible. The number of squares between Hendrick and the black hog, and between Katrun and the white hog, is 1 (an odd number), therefore these individuals cannot catch the animals they are facing. But the number between Hendrick and the white hog, and between Katrun and the black one, is 6 (an even number), therefore they may easily capture those behind them.

79.--_The Thirty-one Game._

By leading with a 5 the first player can always win. If your opponent plays another 5, you play a 2 and score 12. Then as often as he plays a 5 you play a 2, and if at any stage he drops out of the series, 3, 10, 17, 24, 31, you step in and win. If after your lead of 5 he plays anything but another 5, you make 10 or 17 and win. The first player may also win by leading a 1 or a 2, but the play is complicated. It is, however, well worth the reader's study.

80.--_The Chinese Railways._

This puzzle was artfully devised by the yellow man. It is not a matter for wonder that the representatives of the five countries interested were bewildered. It would have puzzled the engineers a good deal to construct those circuitous routes so that the various trains might run with safety.

Diagram 1 shows directions for the five systems of lines, so that no line shall ever cross another, and this appears to be the method that would require the shortest possible mileage.

[Ill.u.s.tration]

The reader may wish to know how many different solutions there are to the puzzle. To this I should answer that the number is indeterminate, and I will explain why. If we simply consider the case of line A alone, then one route would be Diagram 2, another 3, another 4, and another 5. If 3 is different from 2, as it undoubtedly is, then we must regard 5 as different from 4. But a glance at the four diagrams, 2, 3, 4, 5, in succession will show that we may continue this "winding up" process for ever; and as there will always be an un.o.bstructed way (however long and circuitous) from stations B and E to their respective main lines, it is evident that the number of routes for line A alone is infinite. Therefore the number of complete solutions must also be infinite, if railway lines, like other lines, have no breadth; and indeterminate, unless we are told the greatest number of parallel lines that it is possible to construct in certain places. If some clear condition, restricting these "windings up,"

were given, there would be no great difficulty in giving the number of solutions. With any reasonable limitation of the kind, the number would, I calculate, be little short of two thousand, surprising though it may appear.

81.--_The Eight Clowns._

This is a little novelty in magic squares. These squares may be formed with numbers that are in arithmetical progression, or that are not in such progression. If a square be formed of the former cla.s.s, one place may be left vacant, but only under particular conditions. In the case of our puzzle there would be no difficulty in making the magic square with 9 missing; but with 1 missing (that is, using 2, 3, 4, 5, 6, 7, 8, and 9) it is not possible. But a glance at the original ill.u.s.tration will show that the numbers we have to deal with are not actually those just mentioned. The clown that has a 9 on his body is portrayed just at the moment when two b.a.l.l.s which he is juggling are in mid-air. The positions of these b.a.l.l.s clearly convert his figure into the recurring decimal .[.9]. Now, since the recurring decimal .[.9] is equal to 9/9, and therefore to 1, it is evident that, although the clown who bears the figure 1 is absent, the man who bears the figure 9 by this simple artifice has for the occasion given his _figure_ the value of the _number_ 1. The troupe can consequently be grouped in the following manner:--