The Holdout

Reader Joe Antognini sent this in: Brazilian mathematician Inder Taneja has found a way to render every number from 1 to 11,111 by starting with either of these strings:

1 2 3 4 5 6 7 8 9

9 8 7 6 5 4 3 2 1

and applying any of the operations addition, subtraction, multiplication, division, and exponentiation. Brackets are permitted. For example:

6439 = 1 + 2 × (34 × 5 × 6 + 789)

and

6439 = 9 × (8 + 7 + 6) + 54 × (32 + 1)

Intriguingly, there’s one hole: There doesn’t seem to be a way to render 10958 from the increasing sequence.

Taneja’s paper is here. (Thanks, Joe.)

01/29/2017 UPDATE: Taneja tells me that, while it can’t be calculated using only basic operations, 10958 can be reached using factorials or square roots. Here are two factorials:

10958 = 1 + 2 + 3!! + (-4 + 5! + 6 – 7) × 89

10958 = 1 × 2 × (3!! – 4! × (5 + 6) + 7! – 8 – 9)

| Science & Math

The Full Story

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U.S. senator Alan Cranston once lost a copyright suit to Adolf Hitler. Cranston, who had begun his career in journalism, spotted an abridged translation of Mein Kampf in a New York bookstore in 1939. He had read the full text in German and was concerned that the English adaptation omitted Hitler’s anti-Semitism and ambitions to dominate Europe.

To publicize the truth, Cranston worked with a friend to publish an anti-Nazi version of the book. “I wrote this, dictated it [from Hitler’s German text] in about eight days, to a battery of secretaries in a loft in Manhattan,” Cranston told the Los Angeles Times in 1988. They produced a tabloid edition of 32 pages, reducing Hitler’s 270,000 words to 70,000 to yield a “Reader’s Digest-like version [showing] the worst of Hitler.”

At 10 cents apiece, Cranston’s version sold half a million copies in 10 days. But by that time the original was a best-seller in Germany, and the publishers sued Cranston for undercutting the market. In June the U.S. Circuit Court of Appeals in New York ordered the presses stopped. The truth had gotten out, Cranston said, but “we had to throw away half a million copies.”

| History

Child Protection

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Should parents be licensed? We ask teachers to study full-time for years and to pass qualifying exams before we let them educate children for six hours a day. And we carefully assess the suitability of adoptive and foster parents. But anyone has the right to become a biological parent without any training at all in child development.

Philosopher Peg Tittle writes, “How many children have been punished because they could not do what their parents mistakenly thought they should be able to do at a certain age — remember X, carry Y, say Z? How many have been disadvantaged because they grew up on junk food — for their bodies as well as their minds? How many have been neglected because their parents didn’t notice the seeds of some talent?”

Today’s children are tomorrow’s citizens, so the public has a legitimate concern in this. Psychiatrist Jack Westman writes, “The way children are parented plays a vital role in the quality of all our lives. We no longer can afford to avoid defining and confronting incompetent parenting.”

Psychologist Roger McIntire writes, “We already license pilots, salesmen, scuba divers, plumbers, electricians, teachers, veterinarians, cab drivers, soil testers, and television repairmen. … Are our TV sets and toilets more important to us than our children?”

(Peg Tittle, ed., Should Parents Be Licensed?, 2004.)

| Society

The Real World

I had a growing feeling in the later years of my work at the subject that a good mathematical theorem dealing with economic hypotheses was very unlikely to be good economics: and I went more and more on the rules — (1) Use mathematics as a shorthand language, rather than an engine of inquiry. (2) Keep to them until you have done. (3) Translate into English. (4) Then illustrate by examples that are important in real life. (5) Burn the mathematics. (6) If you can’t succeed in 4, burn 3. This last I did often.

— Alfred Marshall, in a letter to A.L. Bowley, Jan. 27, 1906

| Quotations · Science & Math

Podcast Episode 136: The Boston Molasses Disaster

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In 1919 a bizarre catastrophe struck Boston’s North End: A giant storage tank failed, releasing 2 million gallons of molasses into a crowded business district at the height of a January workday. In this week’s episode of the Futility Closet podcast we’ll tell the story of the Boston Molasses Disaster, which claimed 21 lives and inscribed a sticky page into the city’s history books.

We’ll also admire some Scandinavian statistics and puzzle over a provocative Facebook photo.

See full show notes …

| Oddities · Podcast

New Light

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Our legal system assumes that a defendant is innocent until proven guilty beyond a reasonable doubt. But what constitutes a reasonable doubt? Law professors Ariel Porat and Alon Harel suggest that an “aggregate probabilities principle” might help to determine whether an accused party is innocent or guilty.

Suppose we’ve decided that the evidence must indicate a probability of 95 percent guilt before we’re willing to declare a defendant guilty. Mr. Smith is accused of two separate crimes, with a 90 percent probability of guilt in each case. Under the 95 percent rule he’d be acquitted of both crimes. But Porat and Harel point out that there’s a 10 percent chance that Smith is innocent of each crime, and aggregating the probabilities gives a 0.10 × 0.10 = 0.01 chance that Smith is innocent of both — that is, there’s a 99 percent chance that he’s guilty of at least one of the offenses.

On the other hand, consider Miller, who is also accused of two different crimes. Suppose that the evidence gives a 95 percent probability that he committed each crime. Normally he’d be convicted of both offenses, but aggregating the probabilities gives a 0.95 × 0.95 = 0.9025 chance that he’s guilty of both offenses, and hence he’d be acquitted of one.

In A Mathematical Medley (2010), mathematician George Szpiro points out that this practice can produce some paradoxical outcomes. Peter and Paul are each accused of a crime, each with a 90 percent chance of being guilty. Normally both would be acquitted. But suppose that each was accused of a similar crime in the past, Peter with a 90 percent chance of guilt and Paul with a 95 percent chance. Accordingly Peter was acquitted and Paul went to prison. But historically Peter has now been accused of two crimes, with a 90 percent chance of guilt in each case; according to the reasoning above he ought to be convicted of one of the two crimes and hence ought to go to jail today. Paul has also been accused of two crimes, with a 0.95 × 0.90 = 0.855 chance that he’s guilty of both. He’s already served one prison term, so the judge ought to acquit him today.

Szpiro writes, “Thus we have the following scenario: in spite of the evidence being identical, the previously convicted Peter is acquitted, while Paul, with a clean record, is incarcerated.”

(Ariel Porat and Alon Harel, “Aggregating Probabilities Across Offences in Criminal Law,” Public Law Working Paper #204, University of Chicago, 2008; George Szpiro, A Mathematical Medley, 2010.)

| Crime · Society

Crying Wolf

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A puzzle from reader Paul Sophocleous:

Van Helsing, who is of course famous for his part in the destruction of Dracula, has had many other encounters with supernatural creatures. In the early hours of one morning, he was woken by a loud knock at the door. “Come quickly!” cried the chief of police. “There’s been a ghastly attack at the manor house on the hill!”

Van Helsing dressed hurriedly and followed the chief. A grisly sight met him when he arrived. The front door of the house was open, and the beam of light that came from within shone on the body of a young man lying on the path. His throat had been torn out viciously, as though he had been attacked by some kind of hideous wild beast. Van Helsing looked around, but the grounds were dark, since the moon had set some time before, and he could see nothing else.

He stepped inside and found that several officers of the local constabulary were comforting a woman who appeared to be the maid. “It was horrible!” she cried. “I came down here after hearing some racket outside, and I found the young master at the door. ‘There’s something out there,’ he told me, ‘some beast, and I mean to drive it off.’ And he had in his hand the poker from the fireplace as a weapon. But when he opened the door, it was on him in a flash, a great beast, all hairy and shaggy, bigger than a man it was!”

Van Helsing stepped forward. “What was it?” he demanded.

The maid let out a little scream and gasped, “It was a werewolf!” And with that she fainted dead away.

“Could it be, Van Helsing?” said the chief, sounding worried.

Van Helsing shook his head. “Not a chance.”

Why not?

Click for Answer
| Puzzles

Taste

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Cheering news from India. My press clipping agency has sent me a letter from the correspondence column of an Indian paper about a cow that came into the bungalow of a Mr. Verrier Elwyn, who lives at Patengarth, Mandla District, and ate his copy of Carry On, Jeeves, ‘selecting it from a shelf which contained, among other works, books by Shakespeare, Thomas Hardy and Henry Fielding.’ A pretty striking tribute I look on that as.

— P.G. Wodehouse to William Townend, Sept. 3, 1929

| Literature

Traffic Planning

traffic planning - cars

Towns A and B are connected by two roads. Suppose that two cars connected by a rope of length 2r can travel from A to B without breaking the rope. How can we prove that two circular wagons of radius r, moving along these roads in opposite directions, will necessarily collide?

traffic planning - wagons

This can be solved neatly by creating a configuration space. Map each road onto a unit segment, and set these up as two sides of a square. The northern car’s progress is reflected by a point moving up the left side of the square, and the southern car’s by a point moving from left to right along the bottom. Now the motion of the two cars from A to B is represented by a continuous curve connecting (0,0) and (1,1).

The wagons start from opposite towns, so their motion is represented by a curve from (0,1) to (1,0), and it’s immediately clear that the two curves must intersect. The intersection point corresponds to the collision of the wagons.

This example, by N. Konstantinov, is reportedly common in Russian mathematical folklore; I found it in Serge Tabachnikov’s 2005 book Geometry and Billiards (of all places).

| Science & Math

Rough

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Image: Wikimedia Commons

SIR, — While surveying in northern Labrador I had occasion to visit the island of Nukusustok, a few miles to seaward of the village of Nain. On the slope of a hill, and about 300ft. inland, I found a golf ball in good condition. How did the ball come to be there, and so far inland? It is possible that the ball was driven by a golfer from an Atlantic liner during practice, drifted northward past Greenland, and was finally carried ashore by the Labrador current which runs from north to south along the Labrador coast.

I have sent the ball to Dunlops, the makers, who suggest that it was probably carried so far inland by a sea bird. Perhaps some of your readers could help in explaining the mystery.

Yours faithfully,

Thos. O. Hampson

— The Field, June 29, 1935

| Oddities