Midnight Oil

In 1960, MIT mathematician George B. Thomas Jr. received a letter from a waterfowl farmer in Maine. The farmer thought he had discovered an error in a problem in Thomas’ influential textbook Calculus and Analytic Geometry. A little bewildered, Thomas looked into it and discovered that there was indeed an error. He thanked the writer and promised to correct the mistake in future editions.

The two corresponded intermittently thereafter, but four years went by before Thomas realized that the farmer was novelist Henry Roth, author of Call It Sleep. Suffering a disastrous case of writer’s block, Roth had turned to farming and tutoring to support his family, and he had worked his way through every problem in Thomas’ book, ninety per chapter, “often struggling long into the night before arriving at the solution,” according to biographer Steven Kellman.

A copy of the textbook, “inscribed with notes,” is listed among Roth’s papers. In the preface to the fourth edition, Thomas wrote, “One of the author’s friends, Mr. Henry Roth, wrote that he feared that the new edition would be ‘rife with set theory.’ I believe that he, and others who have used the third edition, will find that only modest additions of set theory have been made.”

Point to Point

point to point 1

Here’s a triangle, ABC, and an arbitrary point, D, in its interior. How can we prove that AD + DB < AC + CB?

The fact seems obvious, but when the problem is presented on its own, outside of a textbook or some course of study, we have no hint as to what technique to use to prove it. Construct an equation? Apply the Pythagorean theorem?

“The issue is more serious than it first appears,” write Zbigniew Michalewicz and David B. Fogel in How to Solve It (2000). “We have given this very problem to many people, including undergraduate and graduate students, and even full professors in mathematics, engineering, or computer science. Fewer than five percent of them solved this problem within an hour, many of them required several hours, and we witnessed some failures as well.”

Here’s a dismaying hint: Michalewicz and Fogel found the problem in a math text for fifth graders in the United States. What’s the answer?

Click for Answer

Shifting Areas

shifting areas - 1

This square of 8 × 8 = 64 square units can apparently be reassembled into a rectangle of 5 × 13 = 65 square units:

shifting areas - 2

This paradox is described in W.W. Rouse Ball’s 1892 Mathematical Recreations and Essays; it seems to have been published first in 1868 in Zeitschrift für Mathematik und Physik.

In 1938 the Rockefeller Foundation’s Warren Weaver discovered an old trove of papers from the 1890s in which Lewis Carroll puzzled out the dimensions of all possible squares in which this illusion is possible (the other sizes include squares of 21 and 55 units on a side).

Regardless of publication, it’s not clear who first came up with the idea. Sam Loyd claimed to have presented it to the American Chess Congress in 1858. That would be interesting, as it was his son who later discovered that the four pieces can be assembled into a figure of 63 squares:

shifting areas - 3

(Warren Weaver, “Lewis Carroll and a Geometrical Paradox,” American Mathematical Monthly 45:4 [April 1938], 234-236.)

Self-Descriptive Squares

Lee Sallows has been working on a new experiment in self-reference that he calls self-descriptive squares, arrays of numbers that inventory their own contents. Here’s an example of a 4×4 square:

self-descriptive square 1

The sums of the rows and columns are listed to the right and below the square. These sums also tally the number of times that each row’s rightmost entry, or each column’s lowermost entry, appears in the square. So, for example, the sum of the top row is 3, and that row’s rightmost entry is 1; correspondingly, the number 1 appears three times in the square. Likewise, the sum of the rightmost column is 2, and the lowermost entry in that column, 4, appears twice in the square.

In this example this property extends to the diagonals — and, pleasingly, each sum applies to both ends of its diagonal. The northwest-southeast diagonal totals 2, and both -2 and 4 appear twice in the square. And the southwest-northeast diagonal totals 3, and both 1 and 0 appear three times.

“Easy to understand, but not so easy to produce!” he writes. “I’m still in the throes of figuring out the surprisingly complicated theory of such squares. It turns out there are just two basic squares of 3×3. One of them can be found at the centre of this 5×5 example, which is therefore a concentric self-descriptive square:”

self-descriptive square 2

(Thanks, Lee.)

Surprise Appearance

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Image: Wikimedia Commons

“Eight complete perfect dovetail shuffles, breaking pack exactly in center; that is, cutting off just 26 cards each time and dropping cards from each half alternately, brings the pack to its original order.”

— T. Nelson Downs, in a letter to fellow magician Edward G. “Tex” McGuire, 1923

11/14/2016 UPDATE: Sid Hollander and Harold VanAken sent this demonstration:

eight shuffles

Here’s what it looks like in the hands of a skilled shuffler (thanks to reader Sascha Müller):

Harms and the Man

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The International Statistical Classification of Diseases and Related Health Problems (ICD) is a list of more than 10,000 diseases and maladies that patients might present. The medical community uses it for recordkeeping — for example, a patient admitted to the hospital with whooping cough would be logged in the database with code A37. Reader Will Beattie sent me a list of some of the stranger complaints on the list:

  • Urban rabies – A821
  • Lobster-claw hand, bilateral – Q7163
  • Fall into well – W170
  • Complete loss of teeth, unspecified cause – K0810
  • Pecked by turkey – W6143
  • O’nyong-nyong fever – A921
  • Hang glider explosion injuring occupant – V9615
  • Contact with hot toaster – X151
  • Major anomalies of jaw size – M260
  • Intrinsic sphincter deficiency (ISD) – N3642
  • Underdosing of cocaine – T405X6
  • Prolonged stay in weightless environment – X52

Will says his favorite so far is “Burn due to water skis on fire – V9107.” It’s a dangerous world,” he writes. “Be safe out there.”

Related: Each year the Occupational Safety and Health Administration publishes a list of workplace deaths, with a brief description of each incident:

  • Worker died when postal truck became partially submerged in lake.
  • Worker was caught between rotating drum and loading hopper of a ready-mix truck.
  • Worker fatally engulfed in dry cement when steel storage silo collapsed.
  • Worker on ladder struck and killed by lightning.
  • Worker was pulled into a tree chipper machine.
  • Worker was caught between two trucks and crushed.
  • Worker died when his head was impaled by metal from the drive section of a Ferris wheel. The employee slipped after acknowledging he was clear and the wheel began to turn, trapping his head.
  • Worker was draining a tank; one of the employees climbed to the top of the tank and lit a cigarette and waved it over the opening in the tank. The tank exploded, killing the worker.
  • Worker was kicked by an elephant.
  • Sheriff Deputy was walking through the woods, working a cold case, and fell 161 feet into a sink hole.

It’s hard to pick the worst one. “Worker was operating a skid-steer cleaning out a dairy cattle barn near an outdoor manure slurry pit. The skid-steer and the worker fell off the end of the push-off platform into the manure slurry pit, trapping the worker in the vehicle. Worker died of suffocation due to inhalation of manure.”

Practicalities

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If 6 cats can kill 6 rats in 6 minutes, how many will be needed to kill 100 rats in 50 minutes?

It’s easy enough to work out that the answer is 12, but consider what this means. “When we come to trace the history of this sanguinary scene through all its horrid details, we find that at the end of 48 minutes 96 rats are dead, and that there remain 4 live rats and 2 minutes to kill them in,” observed Lewis Carroll in the Monthly Packet in February 1880. “The question is, can this be done?”

Consider the original statement: 6 cats can kill 6 rats in 6 minutes. What can this actually mean? Carroll counts at least four possibilities:

A. “All 6 cats are needed to kill a rat; and this they do in one minute, the other rats standing meekly by, waiting for their turn.”
B. “3 cats are needed to kill a rat, and they do it in 2 minutes.”
C. “2 cats are needed, and they do it in 3 minutes.”
D. “Each cat kills a rat all by itself, and takes 6 minutes to do it.”

Now try to apply these to our conclusion that 12 cats can kill 100 rats in 50 minutes. Cases A and B work out, but Case C can work only if we understand that fractional deaths are possible: that 2 cats could kill two-thirds of a rat in 2 minutes. Similarly, Case D works only if a cat can kill one-third of a rat in 2 minutes.

The only way to resolve this absurdity, it seems, is to supply extra cats. “In case C less than 2 extra cats would be of no use. If 2 were supplied, and if they began killing their 4 rats at the beginning of the time, they would finish them in 12 minutes, and have 36 minutes to spare, during which they might weep, like Alexander, because there were not 12 more rats to kill. In case D, one extra cat would suffice; it would kill its 4 rats in 24 minutes, and have 24 minutes to spare, during which it could have killed another 4. But in neither case could any use be made of the last 2 minutes, except to half-kill rats — a barbarity we need not take into consideration.”

“To sum up our results: If the 6 cats kill the 6 rats by method A or B, the answer is ’12’; if by method C, ’14’; if by method D, ’13’.”

(Another problem: “If a cat can kill a rat in a minute, how long would it be killing 60,000 rats? Ah, how long, indeed! My private opinion is that the rats would kill the cat.”)

Asked and Answered

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Image: Wikimedia Commons

During World War II, Alan Turing enrolled in the infantry section of the Home Guard so that he could learn to shoot a rifle. After completing this section of his training he stopped attending parades, as he had no further use for the service. Summoned to account for this, he explained that he was now an excellent shot and this was why he had joined.

“But it is not up to you whether to attend parades or not,” said Colonel Fillingham. “When you are called on parade, it is your duty as a soldier to attend.”

“But I am not a soldier.”

“What do you mean, you are not a soldier! You are under military law!”

“You know, I rather thought this sort of situation could arise,” Turing said. “I don’t know I am under military law. If you look at my form you will see that I protected myself against this situation.”

It was true. On his application form Turing had encountered the question “Do you understand that by enrolling in the Home Guard you place yourself liable to military law?” He could see no advantage in answering yes, so he answered no, and the clerk had filed the form without looking at it.

“So all they could do was to declare that he was not a member of the Home Guard,” remembered Peter Hilton. “Of course that suited him perfectly. It was quite characteristic of him. And it was not being clever. It was just taking this form, taking it at its face value and deciding what was the optimal strategy if you had to complete a form of this kind. So much like the man all the way through.”

(From Andrew Hodges, Alan Turing: The Enigma, 1992.)

Climbing Chains

Princeton mathematician John Horton Conway investigated this curious permutation:

3n ↔ 2n

3n ± 1 ↔ 4n ± 1

It’s a simple set of rules for creating a sequence of numbers. In the words of University of Calgary mathematician Richard Guy, “Forwards: if it divides by 3, take off a third; if it doesn’t, add a third (to the nearest whole number). Backwards: if it’s even, add 50%; if it’s odd, take off a quarter.”

If we start with 1, we get a string of 1s: 1, 1, 1, 1, 1, …

If we start with 2 or 3 we get an alternating sequence: 2, 3, 2, 3, 2, 3, …

If we start with 4 we get a longer cycle that repeats: 4, 5, 7, 9, 6, 4, 5, 7, 9, 6, …

And if we start with 44 we get an even longer repeating cycle: 44, 59, 79, 105, 70, 93, 62, 83, 111, 74, 99, 66, 44, …

But, curiously, these four are the only loops that anyone has found — start with any other number and it appears you can build the sequence indefinitely in either direction without re-encountering the original number. Try starting with 8:

…, 72, 48, 32, 43, 57, 38, 51, 34, 45, 30, 20, 27, 18, 12, 8, 11, 15, 10, 13, 17, 23, 31, 41, 55, 73, 97, …

Paradoxically, the sequence climbs in both directions: Going forward we multiply by 2/3 a third of the time and by roughly 4/3 two-thirds of the time, so on average in three steps we’re multiplying by 32/27. Going backward we multiply by 3/2 half the time and by roughly 3/4 half the time, so on average in two steps we’re multiplying by 9/8. And every even number is preceded by a multiple of three — half the numbers are multiples of three!

What happens to these chains? Will the sequence above ever encounter another 8 and close up to form a loop? What about the sequences based on 14, 40, 64, 80, 82 … ? “Again,” writes Guy, “there are many more questions than answers.”

(Richard K. Guy, “What’s Left?”, Math Horizons 5:4 [April 1998], 5-7; and Richard K. Guy, Unsolved Problems in Number Theory, 2004.)