In a Word

periergia
n. bombastic or laboured language

galimatias
n. confused language, meaningless talk, nonsense

taigle
v. to impede or hinder; hence, to fatigue; weary

obtrect
v. to disparage or decry

A paragraph from an unnamed “publication from a leading geographical society”:

The examples given suggest that the multiformity of environmental apprehension and the exclusivity of abstract semantic conceptions constitute a crucial distinction. Semantic responses to qualities, environmental or other, tend to abstract each individual quality as though it were to be considered in isolation, with nothing else impinging. But in actual environmental experience, our judgements of attributes are constantly affected by the entire milieu, and the connectivities such observations suggest reveal this multiform complexity. Semantic response is generally a consequence of reductive categorization, environmental response or synthesizing holism.

In The Jargon of the Professions, Kenneth Hudson suggests that the authors “should be locked up without food or water until they can produce an acceptable translation.” In Secret Language, Barry J. Blake adds, “I think the passage simply means that in experiencing the environment we need to look at it as a whole rather than at particular properties, though I am at a loss to decode the first sentence.”

March 21, 2017 | Language

Unquote

“The leading advocates of the need to subject everything to the competitive test of the market are tenured economists.” — Sheen Kassouf

March 20, 2017 | Quotations

Podcast Episode 146: Alone in the Wilderness

In 1913 outdoorsman Joseph Knowles pledged to spend two months in the woods of northern Maine, naked and alone, fending for himself “without the slightest communication or aid from the outside world.” In this week’s episode of the Futility Closet podcast we’ll follow Knowles’ adventures in the woods and the controversy that followed his return to civilization.

We’ll also consider the roots of nostalgia and puzzle over some busy brothers.

See full show notes …

A Mathless Math Puzzle

$hess bug puzzle$

Richard Hess posed this problem in the Spring 1980 issue of Pi Mu Epsilon Journal. At noon on Monday, a bug departs the upper left corner, X, of a p × q rectangle and crawls within the rectangle to the diagonally opposite corner, Y, arriving there at 6 p.m. He sleeps there until noon on Tuesday, when he sets out again for X, crawling along another path within the rectangle and reaching X at 6 p.m. Prove that at some time on Tuesday the bug was no farther than p from his location at the same time on Monday.

	Select Click for Answer
Well, suppose there are two bugs making their journeys simultaneously — one departs X as the other departs Y, each headed for the opposite corner, and they travel different paths within the rectangle, arriving at their destinations simultaneously. At some time during their journeys they’ll both be the same horizontal distance from the left side of the rectangle. Since both paths remain within the rectangle, at that moment the distance between them can’t be greater than p. The editors received this answer from six readers, including Hess. “Their solutions were characterized by the complete absence of mathematical symbols and mathematical jargon,” they noted. “While there is certainly no objection to mathematical solutions to mathematical problems, a simple word-solution intelligible to any layman is to be preferred. Some of the other submitted solutions were profuse with subscripts, coordinates, inequalities, vinculi, functional relations, intermediate value theorems, Greek symbols, graphs, continuous functions and derivatives — all reminiscent of the sledgehammer method of swatting a fly.” The original problem is here (PDF, page 135).

Click for Answer

March 19, 2017July 17, 2017 | Puzzles

The Greatest Show

Before making his name with mobile sculpture, Alexander Calder was captivated by the circus. On a visit to Ringling Brothers and Barnum & Bailey Circus in New York City at age 27, Calder traveled about the big top with a sketchpad, drawing tightrope walkers, horseback riders, and acrobats. Using a free pass, he returned to the circus every day for two weeks, and then set out to make a toy circus of his own.

He assembled it from wire, cloth, leather, corks, pipe cleaners, string, and wood. He worked on it for six years, until he had 55 performers, and then put on circus parties for friends, playing music and introducing a ringmaster who would direct each of the acts. When it became too fragile to handle, he gave the circus to the Whitney Museum of American Art in New York City, where it remains today.

“Sandy is evidently always happy, or perhaps up to some joke, for his face is always wrapped up in that same mischievous, juvenile grin,” his school yearbook description had read. “This is certainly the index to the man’s character in this case, for he is one of the best natured fellows there is.”

March 18, 2017March 18, 2017 | Art

Pentalpha

During a visit to Crete in 1938, Miss L.S. Sutherland described a game she saw played on a pentagram:

You have nine pebbles, and the aim is to get each on one of the ten spots. You put your pebble on any unoccupied spot, saying ‘one’, and then move it through another, ‘two’, whether this spot is occupied or not, to a third, ‘three’, which must be unoccupied when you reach it; these three spots must be in a straight line. If you know the trick, you can do this one-two-three trick, for each of your nine pebbles and find it a berth, and then you win your money. If you don’t know the trick, it’s extremely hard to do it.

To make this a bit clearer: The figure has 10 “spots,” the five points of the star and the five corners of the pentagon in the middle. A move consists of putting a pebble on any unoccupied spot, moving it through an adjacent spot (which may be occupied) and continuing in a straight line to the next adjacent spot, which must be unoccupied. You then leave the pebble there and start again with a new pebble, choosing any unoccupied spot to begin this next move. If you can fill 9 of the 10 spots in this way then you’ve won.

Can you find a solution?

	Select Click for Answer
Image: Wikimedia Commons One solution that’s easy to remember is to start your first move at a corner of the pentagon, arriving at a point of the star, and then contrive each subsequent move to end at the start of the preceding move (as above). Another solution:

March 17, 2017March 26, 2017 | Puzzles

Pandigital Pi

In the July/August 2006 issue of MIT Technology Review, Richard Hess noted that this expression:

$3 + \frac{16 - 8^{-5}}{97 + 2^{4}} \approx \pi - 3.3 \times 10^{-9}$

provides a good approximation to π using each of the digits 1-9 once. He challenged readers to do better, limiting themselves to the operators +, -, ×, ÷, exponents, decimal points, and parentheses.

The best solution received was from Joel Karnofsky:

$3.14 + \left ( 7^{-.9^{-6}} + 2/8 \right )^{5} \approx \pi - 9.3 \times 10^{-11}$

But Karnofsky noted that this is probably not the best possible. “Unfortunately, my estimate is that there are on the order of 1016 unique values that can be generated under the given conditions and I cannot see how to avoid checking essentially all of them to fnd a guaranteed best. With maybe a thousand computers I think this could be done in my lifetime.”

Indeed, eight months later Sergey Ioffe sent this solution, which he found “using a genetic-like algorithm applying mutations to a population of parse trees and keeping some number of best ones”:

$3 + 5^{-\left ( 7^{.1} \right )} + \frac{.49^{8}}{2^{6}} \approx \pi - 3.8 \times 10^{-13}$

Even that has now been surpassed — on the Contest Center’s ongoing pi approximation page, Oleg Vlasii offers this expression:

$\left ( \frac{2}{.98} - .3 \right ) \times \left (.4 + 5^{(7^{-.6}-.1)} \right ) \approx \pi - 4.1 \times 10^{-14}$

And it’s possible to do even better than this if zero is added as a tenth digit.

03/25/2017 UPDATE: Reader Danesh Forouhari wondered whether there’s a “unidigital” formula for pi. There is — Viète’s formula:

$\displaystyle \frac{2}{\pi } = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{{2 + \sqrt{2}}}}{2} \cdot \frac{\sqrt{2 + {\sqrt{{2 + \sqrt{2}}}}}}{2} \cdots$

March 16, 2017March 25, 2017 | Science & Math

Letter From Home

https://commons.wikimedia.org/wiki/File:German_POWs_lined_up_in_camp.png — Image: Wikimedia Commons

In Love Letters of the Great War (2014), Mandy Kirkby quotes this letter sent from Gleiwitz, Upper Silesia, in April 1917:

Dear Husband!

This is the last letter I am writing to you, because on the 24th I am going to marry another man. Then I don’t have to work any longer. I have already been working for three years as long as you are away from home. All the other men come home for leave, only you POWs never come. Nobody knows how long it will take until you come home. That’s why I am going to have a new husband. I will give the children to the orphanage. I don’t give a rat’s ass about a life like that! There is no way to survive with these few Pfennig benefits. At work they have a big mouth about the women. Now I don’t need to go to work, now the other man is going to work for me. All wives whose husbands are POWs will do the same thing and they will all get rid of the children. Three years at work are too much for the women and 20 Mark for benefit and 10 Mark child benefit are not enough. One cannot live on that. Everything is so expensive now. One pound of bacon costs 8 Mark, a shirt, 9 Mark.

Your wife

“We don’t know anything more about this unfortunate couple, but the strain of separation has brought the wife to breaking point,” Kirkby writes. “Whether she carried out her threat, we’ll never know.”

March 15, 2017 | History · Society

Black and White

By T.R. Dawson. “The problem is for each White man to capture the corresponding Black man in such a way that the routes traversed by each man never come in contact with one another.”

	Select Click for Answer
From Strand Magazine, October 1911.

March 14, 2017March 26, 2017 | Puzzles

A Guilty Key

In the contentious presidential election of 1876, the campaign of Democrat Samuel Tilden sent many enciphered messages to its agents in contested states. Two years after the election, the New York Daily Tribune published some of the deciphered telegrams, showing that Tilden’s campaign had tried to bribe election officials to win the race. Here’s one of the telegrams:

Since only 10 letters are used, it seems likely that the cipher refers to pairs of letters. So if each successive pair in the message is assigned to an arbitrary letter:

… then we have a simple cryptogram that can be solved to give the message:

Tilden’s campaign did the same thing with pairs of numbers. For example, this message:

… turns out to mean:

In 1879 the Tribune’s experts worked out the letter and number pairs that had corresponded to each letter of the alphabet:

But it wasn’t until 1952 that cryptographer William F. Friedman reconstructed the table that the agents had used to remember this system:

“It is amusing to note that the conspirators selected as their key a phrase quite in keeping with their attempted illegalities — HIS PAYMENT — for bribery seems to have played a considerable part in that campaign.”

(From Beaird Glover, Secret Ciphers of the 1876 Presidential Election, 1991.)

March 14, 2017March 13, 2017 | History

An idler's miscellany of compendious amusements

Author: Greg Ross