Podcast Episode 114: The Desperation of Donald Crowhurst

donald crowhurst

In 1968 British engineer Donald Crowhurst entered a round-the-world yacht race, hoping to use the prize money to save his failing electronics business. Woefully unprepared and falling behind, he resorted to falsifying a journey around the world. In this week’s episode of the Futility Closet podcast we’ll describe the desperate measures that Crowhurst turned to as events spiraled out of his control.

We’ll also get some updates on Japanese fire balloons and puzzle over a computer that turns on the radio.

See full show notes …

July 18, 2016October 5, 2017 | Oddities · Podcast

Road Work

Fed up with endless traffic detours in 1830, London printer Charles Ingrey published a pointed puzzle, Labyrinthus Londoninensis, or The Equestrian Perplexed.

“The object is to find a way from the Strand [lower left] to St. Paul’s [center], without crossing any of the Bars in the Streets supposed to be under repair.”

Mending our Ways, our ways doth oft-times mar,
So thinks the Traveller by Horse or Car,
But he who scans with calm and patient skill
This ‘Labyrinthine Chart of London’, will
One Track discover, open and unbarred,
That leads at length to famed St. Pauls Church Yard.

The image above is a bit too small to navigate, but the British Library has an interactive zoomable version (requires Flash).

I don’t have the solution, but The Court Journal of Dec. 14, 1833, hints that “the farthest way round is the nearest way home.”

07/06/2022 UPDATE: A solution! (Thanks, Paul.)

July 16, 2016July 7, 2022 | Puzzles

Footwork

A poser from Penn State mathematician Mark Levi’s Why Cats Land on Their Feet (2012):

Using only a stopwatch and a sneaker, how can you find an approximate value for $\sqrt{2}$ ?

	Select Click for Answer
Suspend the sneaker by its lace to make a pendulum. Using the stopwatch, record the number of swings it makes in one minute, and call this n₁. Double the lace to halve its length and again record the number of swings in one minute. Call this n₂. Now $\sqrt{2}\approx n_{2}/n_{1}.$ For better precision, record the number of swings over a longer period of time. To estimate $\sqrt{3}$ , use a lace 1/3 as long, and so on.

Click for Answer

July 15, 2016July 24, 2016 | Puzzles

Figures

A foolish young man said to John Wilkes (1725-1797), “Isn’t it strange that I was born on the first of January?”

“Not strange at all,” said Wilkes. “You could only have been conceived on the first of April.”

July 15, 2016July 14, 2016 | Humor

A Square Triangle

https://commons.wikimedia.org/wiki/File:3-Pascal.png — Image: Wikimedia Commons

The product of the six numbers surrounding any interior number in Pascal’s triangle is a perfect square.

July 14, 2016 | Science & Math

“On the Question of Choice”

A leaf was riven from a tree,
“I mean to fall to earth,” said he.

The west wind, rising, made him veer.
“Eastward,” said he, “I now shall steer.”

The east wind rose with greater force.
Said he, “‘Twere wise to change my course.”

With equal power they contend.
He said, “My judgment I suspend.”

Down died the winds; the leaf, elate,
Cried: “I’ve decided to fall straight.”

“First thoughts are best?” That’s not the moral;
Just choose your own and we’ll not quarrel.

Howe’er your choice may chance to fall,
You’ll have no hand in it at all.

— Ambrose Bierce

July 14, 2016July 13, 2016 | Poems

Hat Check

A puzzle from MIT Technology Review, July/August 2008:

Each of three logicians, A, B, and C, wears a hat that displays a positive integer. The number on one of the hats is the sum of the numbers on the other two. They make the following statements:

A: “I don’t know my number.”

B: “My number is 15.”

What numbers appear on hats A and C?

	Select Click for Answer
A is 10 and C is 5. The situation above, in which A doesn’t know his own number but B knows his after learning this, occurs if and only if (1) A and C have the same number or (2) A = 2C and B = 3C. Here’s the proof. If A and C have the same number, x, then B’s number must be 2x. Knowing this, A can conclude only that his own number must be either x or 3x, so he’ll declare that he doesn’t know his own number. B will see that both A and C bear the number x and declare that his own number is 2x. That takes care of (1). For (2), suppose that C’s number is x, A’s number is 2x, and B’s number is 3x. A can conclude only that his own number is either 2x or 4x, so he declares that he doesn’t know his own number. B can see that his own number must be x or 3x, but he realizes that he can eliminate x since, if that were the case, A would have seen two x‘s and could have concluded that his own number was 2x. So that shows that in cases (1) and (2) A and B will make the declarations presented in the puzzle statement. Now we have to show that these are the only cases that produce these two responses. So consider the three possibilities where A, then C, then B is the sum of the other two. (i) Here suppose C’s number is x, A’s is x + y, and B’s is y. If y = x, then A will see this and can conclude that his own number is 2x. If y ≠ x then A can’t know his own number. B knows that his own number must be either y or 2x + y. Neither of these numbers equals x, so both possibilities are consistent with A’s response, and B can’t know his number. (ii) Suppose C’s number is x + y, A’s is x, and B’s is y. A can’t tell whether his own number is x or x + 2y and so responds negatively. B knows that his number is either y or 2x + y. Neither number equals x + y, so B can’t eliminate either possibility and can’t know his own number. (iii) Here C’s number is x, A’s is y, B’s is x + y, and we know that x ≠ y (the case where x = y is covered under (1) above). So there are two possibilities: — If x > y then A responds negatively and B concludes that his own number must be either x + x or x – y. Neither of these possibilities equals x, so B can’t eliminate either one. — If x < y then A responds negatively and B concludes that his own number must be either x + y or y – x. Since x + y ≠ x, he can’t eliminate x + y. But he can eliminate y – x when y – x = x; that is, when y = 2x. This is (2) above, proving the claim. In this particular case we know that (1) is impossible since B claimed the number 15, an odd number. So (2) must be the case, and 3x = 15.

Click for Answer

A is 10 and C is 5. The situation above, in which A doesn’t know his own number but B knows his after learning this, occurs if and only if (1) A and C have the same number or (2) A = 2C and B = 3C.

Here’s the proof. If A and C have the same number, x, then B’s number must be 2x. Knowing this, A can conclude only that his own number must be either x or 3x, so he’ll declare that he doesn’t know his own number. B will see that both A and C bear the number x and declare that his own number is 2x.

That takes care of (1). For (2), suppose that C’s number is x, A’s number is 2x, and B’s number is 3x. A can conclude only that his own number is either 2x or 4x, so he declares that he doesn’t know his own number. B can see that his own number must be x or 3x, but he realizes that he can eliminate x since, if that were the case, A would have seen two x‘s and could have concluded that his own number was 2x.

So that shows that in cases (1) and (2) A and B will make the declarations presented in the puzzle statement. Now we have to show that these are the only cases that produce these two responses. So consider the three possibilities where A, then C, then B is the sum of the other two.

(i) Here suppose C’s number is x, A’s is x + y, and B’s is y. If y = x, then A will see this and can conclude that his own number is 2x. If y ≠ x then A can’t know his own number. B knows that his own number must be either y or 2x + y. Neither of these numbers equals x, so both possibilities are consistent with A’s response, and B can’t know his number.

(ii) Suppose C’s number is x + y, A’s is x, and B’s is y. A can’t tell whether his own number is x or x + 2y and so responds negatively. B knows that his number is either y or 2x + y. Neither number equals x + y, so B can’t eliminate either possibility and can’t know his own number.

(iii) Here C’s number is x, A’s is y, B’s is x + y, and we know that x ≠ y (the case where x = y is covered under (1) above). So there are two possibilities:

— If x > y then A responds negatively and B concludes that his own number must be either x + x or x – y. Neither of these possibilities equals x, so B can’t eliminate either one.

— If x < y then A responds negatively and B concludes that his own number must be either x + y or y – x. Since x + y ≠ x, he can’t eliminate x + y. But he can eliminate y – x when y – x = x; that is, when y = 2x. This is (2) above, proving the claim.

In this particular case we know that (1) is impossible since B claimed the number 15, an odd number. So (2) must be the case, and 3x = 15.

July 13, 2016July 24, 2016 | Puzzles

Class Warfare

“There is, on the whole, nothing on earth intended for innocent people so horrible as a school.” — George Bernard Shaw

“I sometimes think it would be better to drown children than to lock them up in present-day schools.” — Marie Curie

“Nearly 12 years of school … form not only the least agreeable, but the only barren and unhappy period of my life. … It was an unending spell of worries that did not then seem petty, of toil uncheered by fruition; a time of discomfort, restriction and purposeless monotony. … I would far rather have been apprenticed as a bricklayer’s mate, or run errands as a messenger boy, or helped my father to dress the front windows of a grocer’s shop. It would have been real; it would have been natural; it would have taught me more; and I should have done it much better.” — Winston Churchill

“Not one of you sitting round this table could run a fish-and-chip shop.” — Howard Florey, 1945 Nobel laureate in medicine, to the governing body of Queen’s College, Oxford, of which he was provost

July 12, 2016July 12, 2016 | Society

Comet Vintages

In “The Adventure of the Stockbroker’s Clerk,” Dr. Watson describes Sherlock Holmes as being as pleased as “a connoisseur who has just taken his first sip of a comet vintage.”

That’s a reference to a strange tradition in winemaking: Years in which a comet appears prior to the harvest tend to produce successful vintages:

1826 — Biela’s Comet
1832 — Biela’s Comet
1839 — Biela’s Comet
1845 — Great June Comet of 1845
1846 — Biela’s Comet
1852 — Biela’s Comet
1858 — Comet Donati
1861 — Great Comet of 1861
1874 — Comet Coggia
1985 — Halley’s Comet
1989 — Comet Okazaki-Levy-Rudenko

“For some unexplained reason, or by some strange coincidence, comet years are famous among vine-growers,” noted the New York Times in 1872. “The last comet which was fairly visible to human eyes [and that] remained blazing in the horizon for many months, until it faded slowly away, was seen in 1858, a year dear to all lovers of claret; 1846, 1832 and 1811 were all comet years, and all years of excellent wine.”

No one has even proposed a mechanism to explain how this might be, but it’s widely noted in the wine world: Critic Robert Parker awarded a perfect 100-point rating to the 1811 Château d’Yquem, and cognac makers still put stars on their labels to commemorate that exceptional year.

July 12, 2016July 11, 2016 | Oddities

Unquote

“Everybody wants to be somebody; nobody wants to grow.” — Goethe

July 11, 2016 | Quotations