Measuring, Weighing and Packing Puzzles

Pouring puzzles look like arithmetic and are really about moves. A jug without graduations measures only two amounts, nothing and full, so the only honest pouring is to pour until the source runs dry or the target is full. Every position of the liquid is then a list of whole numbers, one per vessel, and the puzzle asks for the shortest walk from the starting list to the goal. A shortest walk can always be found and proved shortest by a breadth-first search, which lists every position reachable in one pouring, then in two, and so on until the goal first appears. Where Dudeney claims a fewest number of pourings, that search is how we check him.

The measures are the old English ones: two pints make a quart and four quarts a gallon. Money is in shillings and pence, twelve pence to the shilling, so that 2s. $4\tfrac12$d. is $28\tfrac12$ pence. The mixing puzzles turn on keeping separate account of each ingredient, and the packing puzzles on how tightly round or flat things will sit in a box.

The chapter ends with Dudeney’s river-crossing puzzles, a family that goes back to Alcuin’s wolf, goat and cabbages. They are pouring puzzles in another dress: a position says who is on which bank, a move is one trip of the boat, and the same breadth-first search finds the fewest trips and proves that no shorter way exists.

No. 362. The Wassail Bowl

No. 363. The Doctor’s Query

No. 364. The Barrel Puzzle

No. 365. New Measuring Puzzle

No. 366. The Honest Dairyman

No. 367. Wine and Water

No. 368. The Keg of Wine

No. 369. Mixing the Tea

No. 370. A Packing Puzzle

No. 371. Gold Packing in Russia

No. 372. The Barrels of Honey

No. 373. Crossing the Stream

No. 374. Crossing the River Axe

No. 375. Five Jealous Husbands

No. 376. The Four Elopements

No. 377. Stealing the Castle Treasure

The Wassail Bowl

The jugs measure only 5, 3 and the odd amounts they leave in each other, so the trick is to find a moment when the barrel itself holds a share, four pints, and can be poured straight down one throat. Write a position as six numbers:

the pints in the barrel, the five-pint jug and the three-pint jug, and the pints drunk so far by the three men X, Y and Z. A pour between vessels stops when the source is empty or the target full. A pour down a man’s throat takes the whole of the vessel, because nothing marks a part of it, and no man may take more than his four pints. Here is Dudeney’s sequence: $$\begin{array}{rcccccc} & \text{barrel} & 5 & 3 & X & Y & Z\\ \text{start} & 12 & 0 & 0 & 0 & 0 & 0\\ 1 & 7 & 5 & 0 & 0 & 0 & 0\\ 2 & 7 & 2 & 3 & 0 & 0 & 0\\ 3 & 7 & 0 & 3 & 2 & 0 & 0\\ 4 & 7 & 3 & 0 & 2 & 0 & 0\\ 5 & 4 & 3 & 3 & 2 & 0 & 0\\ 6 & 0 & 3 & 3 & 2 & 4 & 0\\ 7 & 0 & 5 & 1 & 2 & 4 & 0\\ 8 & 0 & 5 & 0 & 2 & 4 & 1\\ 9 & 0 & 2 & 3 & 2 & 4 & 1\\ 10 & 0 & 0 & 3 & 4 & 4 & 1\\ 11 & 0 & 0 & 0 & 4 & 4 & 4 \end{array}$$ That moment comes at the fifth and sixth pourings, when the barrel is down to exactly four pints and goes straight into Y. The other two shares are made up from the odd pints the jugs can produce, 2 and 2 for X and 1 and 3 for Z.

Eleven cannot be beaten. The position space is small, and a breadth-first search over it shows that no sequence of ten or fewer manipulations leaves every man with four pints. Dudeney is right.

Answer 11 manipulations

The Doctor’s Query

The crux is that the bottles end as they began, ten ounces each. So whatever spirits are now missing from the spirits bottle have been replaced by exactly the same volume of water. The water bottle is short of water by exactly the same amount as it holds of spirits. The two bottles are mirror images.

The water bottle’s mixture was forty parts water to one of spirits after the first pouring, and pouring some of it away does not change its proportion. By the mirror argument the spirits bottle now holds forty parts spirits to one of water. In figures: the quarter-ounce poured back holds $\tfrac14\cdot\tfrac{1}{41}$ oz of spirits and $\tfrac14\cdot\tfrac{40}{41}=\tfrac{10}{41}$ oz of water, so the spirits bottle holds $9\tfrac34+\tfrac{1}{164}=\tfrac{400}{41}$ oz of spirits against $\tfrac{10}{41}$ oz of water, which is 40 to 1. Dudeney agrees.

Answer 40 parts spirits to 1 of water

The Barrel Puzzle

Tilt the barrel on the edge of its base until the water just reaches the lip of the open top.

If at that moment the water surface also just reaches the edge of the base on the opposite side, the barrel is exactly half full. If the water covers the whole base with some to spare, it is more than half full. If part of the base shows dry, it is less.

The crux is symmetry. The water surface is then a plane through a point of the rim and the diametrically opposite point of the base edge, and the midpoint of those two points is the centre of the barrel. A barrel is symmetric about its centre: turning every point through half a turn about that centre carries the barrel onto itself and carries each side of any plane through the centre onto the other side. So such a plane cuts the barrel into two halves of equal volume. The argument uses nothing about the barrel’s bulge, only that it looks the same from either end, which is Dudeney’s remark that the method works for any symmetrically made vessel. (The thickness of the base and its height above the ground, which he also mentions, only move the reference points a little.)

Answer tilt it until the water reaches the lip

New Measuring Puzzle

The second ten-quart vessel is what makes this easier than it looks:

the two big vessels act as reservoirs, so an odd quantity can be parked in one while the measures are refilled. Write a position as the quarts in the two ten-quart vessels, then the five-quart and the four-quart measure. Dudeney’s eleven pourings run $$\begin{array}{rcccc} & 10 & 10 & 5 & 4\\ \text{start} & 10 & 10 & 0 & 0\\ 1 & 5 & 10 & 5 & 0\\ 2 & 5 & 10 & 1 & 4\\ 3 & 9 & 10 & 1 & 0\\ 4 & 9 & 6 & 1 & 4\\ 5 & 9 & 7 & 0 & 4\\ 6 & 9 & 7 & 4 & 0\\ 7 & 9 & 3 & 4 & 4\\ 8 & 9 & 3 & 5 & 3\\ 9 & 9 & 8 & 0 & 3\\ 10 & 4 & 8 & 5 & 3\\ 11 & 4 & 10 & 3 & 3 \end{array}$$ The parked quantities are the 1 of step 2 and the 3 of step 7, each held in a tall vessel while the other measure is refilled. The last move is the neat one: the five-quart measure, full, tops up the second vessel, which lacks exactly two quarts, and is left holding three.

A breadth-first search over every position shows that no sequence of ten or fewer pourings puts three quarts in each measure, so eleven is the least. Dudeney is right.

Answer 11 manipulations

The Honest Dairyman

Let B start with $m$ of milk and A with $w$ of water.

The crux is to follow the two liquids separately.

• First A gives B an amount $m$. Now B holds $m$ milk and $m$ water, half and half, and A holds $w-m$ water.

• Then B gives A an amount $w-m$ of its half-and-half mixture, which is $\tfrac{w-m}{2}$ of milk and $\tfrac{w-m}{2}$ of water. A now holds $\tfrac{w-m}{2}$ milk and $(w-m)+\tfrac{w-m}{2}=\tfrac32(w-m)$ water.

• The last step only pours some of A away, which leaves its proportions alone.

So London gets three parts water to one of milk, whatever the starting amounts. Dudeney agrees.

He also states when the operations can be carried out: $w$ must exceed $m$, or A has nothing left to double, and $w$ must be at most $3m$, or B has too little to double A. His conditions miss a third one. After the second step A holds $2(w-m)$ and B holds $2m-(w-m)=3m-w$. To pour from A into B until they are equal, A must hold at least as much as B: $2(w-m)\ge 3m-w$, which is $w\ge\tfrac53 m$. So the three operations are possible exactly when $$\tfrac53\,m\le w\le 3m .$$ With three pints of milk and four of water, for instance, the first two steps leave A with 2 pints and B with 5, and the final pour cannot be made. At $w=\tfrac53 m$ the cans are already equal and the last pour is a pour of nothing. The answer three to one is safe; Dudeney’s range of starting amounts is too generous.

Answer 3 parts water to 1 of milk

Wine and Water

This is plain bookkeeping, and the only trap is adding fractions of different glasses as if they were fractions of the same thing. Take the small glass to hold 1 unit and the large glass 2. The small glass has $\tfrac12$ unit of wine and the large one $\tfrac13\times 2=\tfrac23$. The tumbler receives 3 units in all, of which $\tfrac12+\tfrac23=\tfrac76$ is wine. So the wine is $\tfrac{7}{6}\div 3=\tfrac{7}{18}$ of the mixture and the water $\tfrac{11}{18}$. Dudeney agrees.

Answer $\tfrac{7}{18}$ wine, $\tfrac{11}{18}$ water

The Keg of Wine

The crux is that each draw removes the same fraction of whatever wine is in the keg. A jug of $j$ gallons takes $\tfrac{j}{10}$ of the contents, so after each draw and refill the keg keeps $1-\tfrac{j}{10}$ of its wine. After two draws it keeps $\bigl(1-\tfrac{j}{10}\bigr)^2$ of the original ten gallons, and this must be one half: $$1-\frac{j}{10}=\frac{1}{\sqrt2},\qquad j=10-5\sqrt2\approx 2.929 .$$ The negative square root would give a jug larger than the keg. Dudeney’s 2.93 gallons is this value rounded; the exact capacity is irrational, so no decimal ends it.

Answer $10-5\sqrt2\approx2.93$ gallons

Mixing the Tea

In pence, the teas cost 30, 27 and 21 a pound, and twenty pounds at $28\tfrac12$d. cost 570d. With $a$, $b$ and $c$ pounds of the three, $$a+b+c=20,\qquad 30a+27b+21c=570 .$$ Take away 21 times the first equation from the second: $9a+6b=150$, or $3a+2b=50$. The crux is that this forces $a$ to be even and makes $b=25-\tfrac32 a$. Then $c=20-a-b=\tfrac a2-5$. The condition $c\ge0$ gives $a\ge10$, and $b\ge0$ gives $a\le16$. So the possible mixes are $$(a,b,c)=(10,10,0),\ (12,7,1),\ (14,4,2),\ (16,1,3).$$ The first uses no Indian tea at all, and the grocer said to mix all three. Among the other three, the least of the best tea is 12 pounds: 12 lb at 2s. 6d., 7 lb at 2s. 3d. and 1 lb at 1s. 9d. Dudeney gives the same three mixes and chooses the same one. It is worth noticing that the two-tea mix of ten and ten would use even less of the best tea, and only the grocer’s word “three” rules it out.

Answer 12 lb, 7 lb and 1 lb

A Packing Puzzle

Dudeney’s packing uses layers in which the balls sit as closely as they can, each layer resting in the hollows of the one below. Stand the box so that the 14-inch depth runs along the rows and the $24\tfrac{9}{10}$-inch length runs through the layers.

In one layer. A row of touching balls two inches across fits seven to fourteen inches. The next row sits in the dips between them and holds six, offset by one inch. Neighbouring rows are $\sqrt3$ inches apart, since the centres of two touching balls in one row and the ball resting on them make an equilateral triangle of side 2. Thirteen rows need $2+12\sqrt3\approx22.785$ inches, just inside the $22\tfrac45$ available, so the first layer takes seven rows of 7 and six rows of 6, which is 85 balls.

The next layer. Each ball of the second layer rests in a hollow formed by three balls of the first. Its rows are shifted by $\sqrt3/3$ across, and they alternate 6 and 7 in the other order. Only twelve rows fit this time, since a thirteenth would need $2+(12+\tfrac13)\sqrt3\approx23.36$ inches, so the second layer holds $6\times7+6\times6=78$.

Between layers. A ball resting in a hollow sits $2\sqrt{2/3}\approx1.633$ inches above the layer below, which is the height of a regular tetrahedron of side 2. Fifteen layers need $2+14\times2\sqrt{2/3}\approx24.86$ inches, just inside $24\tfrac{9}{10}$. Alternating 85 and 78, eight layers of 85 and seven of 78 give $$8\times85+7\times78=680+546=1226 .$$ Every one of Dudeney’s measurements has been chosen to clear these three limits by less than a tenth of an inch. The crux is that close packing gains a whole row across the width and three extra layers along the length over the plain square stacking, which gives only $12\times11\times7=924$.

Whether 1226 is truly the greatest is another matter, and Dudeney does not prove it. A program tries this kind of packing in all six ways of standing the box and with every way of stacking the layers, and none beats 1226.

But packings that do not line up with the walls are not covered by that search. The only general ceiling we know how to prove comes from the Kepler conjecture, proved by Thomas Hales in 2005 (Annals of Mathematics 162), which says no packing of equal balls fills more than $\pi/\sqrt{18}$ of space. Reflecting the box in its own walls again and again turns any packing of the box into a packing of all space with the same density, so the box can hold at most $$\frac{\pi}{\sqrt{18}}\cdot\frac{24.9\times22.8\times14}{\tfrac43\pi} =\frac{7948.08}{4\sqrt2}\approx1405$$ balls. So 1226 is the best packing of the natural kind, and the true maximum lies somewhere from 1226 to 1405. Nothing here proves that Dudeney’s number is the greatest.

Answer 1226 by Dudeney’s packing (maximum unproved)

Gold Packing in Russia

Dudeney’s box is 100 inches square and 11 deep. Lay slabs flat, 12½ inches along one side and 11 along the other: eight end to end make 100 inches, and nine such rows make 99. Seventy-two slabs cover the floor except for a strip 100 inches long and 1 inch wide. Eleven layers of 72 fill the box to its depth, 792 slabs, and leave a slot 100 inches long, 1 inch wide and 11 deep, which takes the remaining eight slabs standing on edge, end to end.

Dudeney does not say whether this box is the only one, and it is. The crux is to cut the box across. Any straight line through a full box, parallel to an edge, passes through slabs whose extents along it are 12½, 11 or 1 inches, so the side $L$ and the depth $D$ are multiples of half an inch; and the volume is $800 \times 137\tfrac12 = 110{,}000$ cubic inches, so $L^2 D = 110{,}000$. Writing $L = h/2$ and $D = k/2$ in whole numbers, $h^2 k = 880{,}000 = 2^7 \cdot 5^4 \cdot 11$, so $h$ divides $2^3 \cdot 5^2 = 200$, and a side of at least 12½ inches leaves $L = 12\tfrac12$, 20, 25, 50 or 100.

Now cut the box horizontally at a height that misses every slab’s top and bottom. The flat slabs show as rectangles $12\tfrac12 \times 11$, and a slab on edge as a strip $12\tfrac12 \times 1$ (on its long edge, 11 inches tall) or $11 \times 1$ (on end, 12½ tall). At most twelve slabs stand on edge, so there are at most twelve strips, and the areas must add up to $L^2$. For each side this fixes the numbers: a side of 20 needs 2 flat slabs and 10 strips in every section, 25 needs 4 and 6, 50 needs 18 and 2, and 100 needs 72 and 8; 12½ cannot be made up at all. For each side these are the only numbers that fit, so every section at every height shows the same number $b$ of strips, all from slabs on their long edge, each 11 inches tall. Stacked through the depth, they make $bD/11$ slabs on edge. For 20 that is $10 \times 275/11 = 250$, and for 25 it is $6 \times 176/11 = 96$, both far over twelve.

That leaves 50, with 18 flat slabs and 2 strips in a section 50 inches square. A line across the section passes pieces whose widths, 12½, 11 or 1, add up to 50, with at most two 1s. The only way is $4 \times 12\tfrac12$, since $2 \times 12\tfrac12 + 2 \times 11$ is 47 and needs three strips, and every other mixture fails worse. So every piece is 12½ wide in that direction, which makes it 11 or 1 in the other, and a line the other way needs $11j + k = 50$ with $k \le 2$, which is impossible. The section cannot be tiled. Only Dudeney’s box remains.

Answer 100 by 100 inches, 11 deep

The Barrels of Honey

Count in half-barrels of honey. Each son gets 7 barrels and 7 half-barrels of honey, so if he has $f$ full, $h$ half-full and $e$ empty barrels, $$f + h + e = 7, \qquad 2f + h = 7 .$$ Subtracting, $f = e$, and then $h = 7 - 2f$. So a son’s share is one of $(0, 7, 0)$, $(1, 5, 1)$, $(2, 3, 2)$ or $(3, 1, 3)$, in full, half-full and empty. The four-of-a-kind rule removes the first two. Three shares from $(2, 3, 2)$ and $(3, 1, 3)$ must use exactly seven full barrels: that forces one $(3, 1, 3)$ and two $(2, 3, 2)$, which also uses seven half-full and seven empty. So one son takes 3 full, 1 half-full and 3 empty, and each of the others 2 full, 3 half-full and 2 empty.

Without the four-of-a-kind rule there is exactly one more way, $(3, 1, 3)$ twice and $(1, 5, 1)$, as Dudeney notes. Either way two brothers receive theirs alike.

Answer 3 full, 1 half, 3 empty to one son; 2, 3, 2 to each of the others

Crossing the Stream

The two boys are the key, since they are the only pair who can share the boat. Each parent has to cross alone, and each time someone must bring the boat back, which only a boy can usefully do. So: the boys cross and one returns; the father crosses and the other boy returns. That moves one parent in four crossings. Repeat for the mother: four more. Now both boys are on the near side with the dog; they cross together, one comes back, and he brings the dog, which weighs no more than a boy. That is eleven crossings.

A breadth-first search over every position confirms that eleven is the fewest, taking the dog as heavier than nothing and no heavier than a boy, and unable to row.

Answer Eleven crossings

Crossing the River Axe

Everything hangs on the rule that a man alone, on a bank or in the boat, may have beside him no more money than his own share, which keeps the large sacks away from Timothy and makes the shuttling long. A breadth-first search over every placing of the three men and three sacks, with the boat carrying two men or one man and a sack, finds that thirteen crossings are needed and suffice, as Dudeney says. There are fifty different shortest ways. Here is one, with G, J, T for the men and 8, 5, 3 for the sacks, showing what goes over or back on each crossing:

At every moment each man who is alone somewhere, on a bank or in the boat, holds no more than his own share: Timothy, for instance, never sits alone with the £500 or £800 sack. Dudeney also points out that his route never leaves two men with more than their joint shares, which the puzzle did not require. Even with that stricter rule the fewest is thirteen, and eight of the fifty routes obey it.

Answer Thirteen crossings

Five Jealous Husbands

With no jealousy, ten people and a boat of three need nine crossings: each round trip lands a net two people, so four round trips bring eight over and a final crossing takes the last three. The jealousy costs two more. Dudeney’s eleven crossings, with the boat’s load on each:

A breadth-first search over every division of the ten people confirms that eleven is the fewest.

Dudeney’s warning about nine-crossing answers is worth taking seriously. The search agrees that nine are enough if a bank is inspected only after the boat has left it, so that a wife may land among other men provided she rows straight back. No jealous husband would accept that, and with every landing counted the answer is eleven. His further point, that two women rowing back together go faster than one, does not change the number of crossings.

Answer Eleven crossings

The Four Elopements

The island is essential. Without it, four couples cannot cross at all in a boat of two: a search of every position finds the far bank unreachable, whereas three couples can manage without an island in eleven crossings. With the island, the fewest is seventeen passages, as Dudeney says, against the twenty-four he reports French writers giving. The search counts every trip between any two of the three lands, checks the boat and every land after each landing, and on a trip straight across checks the boat’s party together with anyone on the island, since the boat might put in there.

Dudeney’s seventeen passages, with the boat’s load and where it goes:

In it the girls do most of the rowing, a man and a girl never share the boat, and no man ever sets foot on the island.

Answer Seventeen passages

Stealing the Castle Treasure

The treasure is the tool.

It may fall as fast as it likes, so it can always be sent down alone, and at 75 lb it is exactly 15 lb lighter than the boy, so the boy can ride down against it. The weights that can balance within 15 lb are few: the boy against the treasure (90 and 75), the youth against the boy (105 and 90), and the man against the youth and the treasure together (195 and 180). Every move of a person uses one of these three.

Dudeney’s eleven manipulations:

A breadth-first search confirms that eleven is the fewest and finds eight shortest sequences, of which this is one. The answer is remarkably robust: it stays at eleven whether or not the treasure may share a basket with two people, and whether or not the 15 lb limit also applies to a person going up.

Answer Eleven manipulations