The Spider and the Fly

The last part of Dudeney’s book has no story to hold it together. He called it simply Miscellaneous Puzzles: forty-two of them, from Nos. 73 to 114, each with its own small scene, a medieval game in one place, a cellar or a railway carriage in another. They follow here in his order across this chapter and the next two, and each chapter takes its name from one of its puzzles. None of them carries a number forward, so none has a Reckoning.

No. 73. The Game of Kayles

No. 74. The Broken Chessboard

No. 75. The Spider and the Fly

No. 76. The Perplexed Cellarman

No. 77. Making a Flag

No. 78. Catching the Hogs

No. 79. The Thirty-one Game

No. 80. The Chinese Railways

No. 81. The Eight Clowns

No. 82. The Wizard’s Arithmetic

No. 83. The Ribbon Problem

No. 84. The Japanese Ladies and the Carpet

No. 85. Captain Longbow and the Bears

No. 86. The English Tour

The Game of Kayles

Dudeney’s key is a copying strategy. If you can leave your opponent the rows of counters in matching pairs (two rows of three, say, or a row of one and a row of one), you win: whatever he takes from one row, you take the same from its twin, and you are sure to take the last counter. In a single unbroken row the first player wins at once by taking the middle counter, or the middle two, and copying from then on. Taking away the second counter before play spoils that easy win.

The start, one counter and then a row of eleven, is not a matching position, and the first player wins by knocking down the sixth or the tenth counter, counting the missing one as the second. Either move leaves rows of one, three and seven. Dudeney shows how every reply can be met by play that eventually leaves matching pairs, and then leaves “the complete analysis” to the reader.

That analysis was completed long after Dudeney, and it shows why rows of one, three and seven belong together. Independently, R. P. Sprague in 1935 and P. M. Grundy in 1939 showed that in any game of this kind, where both players have the same moves and whoever moves last wins, every position behaves like a single pile in the game of Nim and can be given a number. R. K. Guy and C. A. B. Smith worked out these numbers for Kayles in 1956. For a single row of $n$ counters they begin

and the number of a row is the smallest whole number that is not the number of any position one move away. The number of several rows together is found by writing their numbers in binary and adding without carrying, so that $1$, $3$ and $2$, which are $01$, $11$ and $10$, give $00$. A position is lost for the player about to move exactly when that number is 0, because every move from such a position makes it non-zero, and from any non-zero position some move makes it 0 again. Rows of one, three and seven have numbers 1, 3 and 2, which add in this way to 0; the starting position, with numbers 1 and 6, does not. Knocking down the sixth or tenth counter are the only first moves that reach 0.

Answer first player; knock down the 6th or 10th

The Broken Chessboard

The twelve pieces of five squares are the twelve different shapes that five squares can make, and the thirteenth is a square of four. Figure 9.1 shows Dudeney’s arrangement. Nothing short of trial finds it, but the colours help: every piece must go where its dark and light squares match the board, which rules out most placements before they are tried.

It is not the only one. Keeping the pieces face up, there are four arrangements in all, and four more that are the same ones with the whole board turned half round.

Turning a board a quarter turn swaps its colours, so these cover both ways of chequering it.

Answer four ways (eight with half turns)

The Spider and the Fly

Imagine the room as a cardboard box, cut open along some of its edges and laid flat.

Any route over the surfaces becomes a line on the flat card, and the shortest route across a given set of faces becomes a straight line. So the problem is to choose the faces the spider crosses, unfold them, and measure.

The direct route runs straight down the middle: 1 foot up to the ceiling, 30 feet along the ceiling and 11 feet down the far wall to the fly, which is 42 feet. Better routes cut diagonally across the side walls. Dudeney’s four candidates are in Figure 9.2. The best of them goes from the end wall onto the ceiling, crosses one side wall slantwise, then the floor, and reaches the far end wall. Laid flat, the ceiling, the side wall and the floor make a strip 36 feet across and 30 long, with the two end walls hinged to its ends. The spider then sits 1 foot beyond one end of the strip and the fly 1 foot beyond the other, so they are $30+1+1=32$ feet apart along it. Across the strip the spider is level with the middle of the ceiling and the fly with the middle of the floor, $6+12+6=24$ feet apart. The route is $$\sqrt{32^2+24^2}=\sqrt{1600}=40\text{ feet},$$ two feet shorter than the direct one, and it crosses five of the room’s six faces. Dudeney’s other two candidates measure $\sqrt{1864}\approx 43.17$ and $\sqrt{1658}\approx 40.72$ feet.

Dudeney ended by asking the reader for the shortest routes when the spider and the fly are 2, 3, 4, 5 and 6 feet from the ceiling and the floor. They are $\sqrt{1700}\approx 41.23$ and $\sqrt{1746}\approx 41.79$ feet at 2 and 3 feet, both across four faces, and at 4 feet and beyond the direct 42-foot route down the middle is best.

Answer 40 feet

The Perplexed Cellarman

There are twelve large bottles and twelve small, so each man takes four of each.

Count wine in small bottlefuls: there are $7\times 2+7=21$, so each man’s share is 7. A man with $\ell$ full large bottles and $s$ full small ones has $2\ell+s=7$, with neither more than four, so either $\ell=3$ and $s=1$, or $\ell=2$ and $s=3$. The seven full large bottles must be shared as three numbers from $\{2,3\}$ that add up to 7, which can only be $3+2+2$.

So one man receives three full large bottles and one full small, with one large and three small empty. Each of the other two receives two full large and three full small, with two large and one small empty. That is the only way to do it.

Answer full: 3 large + 1 small, and twice 2 large + 3 small

Making a Flag

The cut is a staircase. Figure 9.3 shows it running across the bunting in a zigzag of equal teeth, clear of every rose. Slide the right-hand piece down by one tooth and the teeth fit back together, one step along, while the outline closes into a square with the roses symmetrically placed. Moving one piece a step along a staircase changes the proportions of the whole while wasting nothing, and here the steps can be placed to weave between the roses. Dudeney compares it with a stepped dissection in his Amusements in Mathematics.

Answer a staircase cut

Catching the Hogs

Colour the board like a chessboard. Every move, by a person or a hog, goes to a square of the other colour. A capture happens when a chaser moves onto a hog, so the chaser must start its move on a square next to the hog, which means a square of the other colour. Now look at one chaser and one hog at the moment the chaser is about to move. After the chaser moves and the hog replies, both have changed colour, so whether they stand on the same colour or on different colours never changes. If they start on the same colour, the chaser can never be next to the hog when it is its turn, and the hog is never caught.

Hendrick has one square between him and the black hog, so they stand on the same colour, and he can chase it for ever. So can Katrün with the white hog. But four squares separate Hendrick from the white hog, and four separate Katrün from the black one, so those pairs start on different colours. On a board this small the chaser can then drive the hog into a corner and catch it; either capture takes at most six moves.

Hendrick and Katrün failed because each went after the hog in front of them. They should have changed hogs.

This is the principle chess players call “the opposition”. Dudeney’s rule is the same one: an odd number of squares between chaser and hog means no capture, an even number means a capture. In his revised edition he gives the number of squares between each chaser and the hog behind as six, but on his own board it is four; the rule is untouched, since both are even.

Answer each must chase the hog behind them

The Thirty-one Game

The plan of making 3, 10, 17, 24 and 31 works in a game with unlimited cards, because any card from 1 to 6 can be answered by the card that makes seven. Here there are only four cards of each value, and the sharper wins by exhausting the ones you need. Every time you answer his 4 with a 3, or his 3 with a 4, you use up the pair together, and with the 3s and 4s gone the key numbers cannot be reached.

The first player can still win, by leading a 5. If the opponent then plays another 5, making 10, answer with a 2 to make 12, and go on answering every 5 with a 2. The first player has turned exhaustion against the sharper. After his third 5, which makes 24, you play 2 to make 26, and the last 5, the only card that would make 31, is already on the table. If at any point he plays anything other than a 5, you step in and make the next of 10, 17, 24 and 31, and keep doing so.

A lead of 1 or 2 also wins, though the play is more complicated; a lead of 3, 4 or 6 loses against correct play.

In the first edition Dudeney named only the lead of 1 besides the 5, and he added the 2 when he revised the book.

Answer play first and lead a 5

The Chinese Railways

The difficulty is that the five lines enter side by side on one wall, in the order A, B, C, D, E, while the stations are scattered in a quite different order round the city, so a line cannot simply head for its station without cutting across its neighbours. The way out is to keep the lines together as a bundle, like the approach tracks to a crowded terminus, and let each one leave the bundle only when it reaches its own station, curving round the stations of the others. Diagram 1 in Figure 9.4 is Dudeney’s layout, which he thought used the least track.

He also asked how many solutions there are, and gave an unexpected answer: infinitely many, if lines have no width. Diagrams 2 to 5 show line A alone, wound one more turn around stations B and E each time. Every winding still leaves the other companies an unobstructed way to their stations, and nothing stops the winding from going on for ever. He added that with some sensible limit on the windings, the number of solutions would be a little short of two thousand, but he did not say which limit, so that count cannot be checked.

Answer Diagram 1; infinitely many

The Eight Clowns

Taken plainly, it cannot be done. The eight numbers 2 to 9 add up to 44, and the three rows of a magic square together hold every number once, so each row would have to add up to a third of 44, which is not a whole number. With No. 9 missing instead there is no difficulty: 1 to 8 add up to 36, each line to 12, and the square is easy to make.

Dudeney’s way out is in the picture. Clown No. 9 is juggling, and the two balls hang in the air just above his figure, like the dots that mark a recurring decimal. He is showing $0.\dot 9$, which is $0.999\ldots$. To see why that equals 1, call it $x$: then $10x=9.999\ldots$, so subtraction gives $9x=9$ and $x=1$. With No. 9 standing in for No. 1, the troupe forms $$\begin{array}{|c|c|c|} \hline 7 & & 5\\ \hline 2 & 4 & 6\\ \hline 3 & 8 & 0.\dot 9\\ \hline \end{array}$$ and every row, column and diagonal adds up to 12.

It is a joke answer, and the arithmetic behind it is sound.

Answer 9 read as $0.\dot 9=1$

The Wizard’s Arithmetic

Write the multiplier as a two-figure number with tens figure $a$ and units figure $b$, and let the number on the shelf be $N$, with $d$ figures. Putting $b$ in front and $a$ at the end makes $b\cdot 10^{d+1}+10N+a$, so the trick works when $$(10a+b)\,N=b\cdot 10^{d+1}+10N+a,$$ that is, $$(10a+b-10)\,N=b\cdot 10^{d+1}+a .$$ For the wizard’s 83 this reads $73N=3\,000\,008$, and $N=41096$.

The equation also shows how to hunt for others: for each multiplier, find a power of 10 that makes the right-hand side divisible by $10a+b-10$, and check that the quotient has the right number of figures. Since powers of 10 repeat their remainders on division by any number below 90, only a limited range of lengths needs trying.

There are exactly two more, as Dudeney says, and a search over every multiplier and that range of lengths confirms it. The multiplier 86 works with $N=8$: $86\times 8=688$. The other is 71, and its smallest shelf number has fifty-two figures, $$\begin{gathered} 16393442622950819672131147540983\\ 60655737704918032787, \end{gathered}$$ which, multiplied by 71, gives the same figures with a 1 in front and a 7 behind. Every multiplier that works has infinitely many shelf numbers. For 83, putting the block 41095890 in front of 41096 any number of times still gives a number the wizard can multiply by 83 in his way.

Answer 86 and 71 (with 83)

The Ribbon Problem

The ribbon number is the repeating block of the decimal for $\tfrac1{17}$: $\tfrac1{17}=0.\overline{0588235294117647}$. Working out that decimal by long division, the remainders run through every number from 1 to 16 before repeating, so the decimals of $\tfrac2{17},\tfrac3{17},\ldots,\tfrac{16}{17}$ are the same sixteen figures started at different places. Multiplying the ribbon number by $k$ gives the block of $\tfrac{k}{17}$, which is why it survives multiplication by anything up to 16, and not merely up to 9. Dudeney’s ring figure (Figure 9.5) writes the remainders inside the circle of figures: the place where each remainder stands is the place where the corresponding product begins.

The same happens for any prime $p$

whose decimal $\tfrac1p$ takes the longest possible period, $p-1$ figures. A ribbon that starts with 0 needs $p$ larger than 10. To end in 7, notice that $p$ times the block is a row of nines, so the last figure of the block times the last figure of $p$ must end in 9; with a block ending in 7, $p$ must end in 7 as well. The primes above 10 that end in 7 are 17, 37, 47, and so on, but $\tfrac1{37}=0.\overline{027}$ repeats after only three figures. The next is 47, whose decimal has the full period of 46 figures: $$0212765957446808510638297872340425531914893617,$$ and this number, multiplied by anything from 2 to 46, gives the same circle of figures. Dudeney’s printing of this number, in the 1908 edition and in the revised one, has a 3 where the sixth figure from the end should be 8.

Dudeney thought it improbable that a shorter number would do, and he was right. Among all numbers of fewer than 46 figures, the only ones that behave this way under 2 to 9, start with 0 and end with 7, are the 1/17 block itself and that block written twice.

Answer the 46-figure block of 1/47

The Japanese Ladies and the Carpet

Take the three equal mats first and see how to join them into one square, then reverse the process. Three unit squares side by side make a rectangle 3 long and 1 high, and the square of the same area has side $\sqrt3$. Dudeney finds that side with compasses: extend the base of the rectangle by its height, draw a semicircle on the whole, and the perpendicular raised where the rectangle ends meets the semicircle at height $\sqrt{3\times 1}$, the mean proportional of the two sides.

The semicircle calculation in the Haberdasher’s solution gives the reason: height squared equals the product of the two base segments, here $3\cdot1$.

Now cut the rectangle along the slanting line from its top left corner to the point on the base at distance $\sqrt3$ from the left, and down from the top at distance $\sqrt3$ from the right until this vertical meets the slanting line. With the two edges between the squares, that makes six pieces (Figure 9.6). The remarkable thing is how they go together. Nothing is turned. Pieces 5 and 6, below the slanting line, stay where they are. Pieces 4 and 3 slide together, right by $2\sqrt3-3$ and down by $2-\sqrt3$, and sit against the slanting line further along. Pieces 2 and 1 slide together, left by $3-\sqrt3$ and up by $\sqrt3-1$, and fill the top of the square. The six pieces make a square of side $\sqrt3$ exactly.

For the puzzle as set, run this backwards: cut the carpet into the six pieces, and pieces 1, 4 with 5, and 2 with 3 and 6 make the three mats. Dudeney also shows how to find the mats’ side directly from the carpet. On the carpet $QNLO$, mark $M$ on the side $OL$ with $LM$ half the diagonal $ON$, draw $NM$, and drop a perpendicular $LP$ from $L$ onto it. Then $LP$ is the side of the three mats, the carpet’s side divided by $\sqrt3$.

Answer six pieces, as in the figure

Captain Longbow and the Bears

The story of the bear on the Pole is part of the puzzle.

Dudeney says that eleven bears cannot form seven rows of four on their own; that impossibility is not proved here. A catalogue of tree-planting arrangements records constructions with six rows for eleven points and seven for twelve. The twelfth point is the impaled bear. In Dudeney’s arrangement (Figure 9.7) three of the eleven stand on the line running due north, and that line, a hundred miles long, ends at the Pole, where the twelfth bear completes the seventh row.

The other six rows are ordinary. The arrangement is symmetric about the northward line, with a bear far out on each side, and each of those two starts three rows of four, fanning across the group. Whether the seventh row is a hundred feet long or a hundred miles makes no difference, so long as it is straight.

Answer the dead bear on the Pole makes the seventh row

The English Tour

By rail alone it cannot be done: no route from A calls at every town once and ends at Z. Dudeney expected the solver to find that out, and then to read the conditions again. The traveller may not cut across country by road, but nothing is said about going by sea. Two towns, and only two, stand on the coast, one in the north-west and one in Kent. Travelling by rail to one of them, he takes a coasting steamer round to the other and finishes the tour by rail (Figure 9.8). With that one voyage there are four routes in all.

Answer by rail, and once by sea

Sources. The puzzles are retold from Nos. 73 to 86, and the illustrations reproduced from, Henry Ernest Dudeney, The Canterbury Puzzles and Other Curious Problems (1907; revised edition 1919), in the text of Project Gutenberg eBook 27635; the map for No. 86 was read from a scan of the 1908 printing. On Kayles: R. P. Sprague, “Über mathematische Kampfspiele”, Tôhoku Mathematical Journal 41 (1935), pages 438 to 444; P. M. Grundy, “Mathematics and games”, Eureka 2 (1939), pages 6 to 8; R. K. Guy and C. A. B. Smith, “The G-values of various games”, Proceedings of the Cambridge Philosophical Society 52 (1956), pages 514 to 526. The numerical answers are found again by the programs in the verify folder accompanying this book.