Hard of Hearing

In 1979 Auberon Waugh was working as a columnist at Private Eye when his editor offered him a trip to Senegal to help celebrate the anniversary of the magazine’s sister publication. “All I would have to give in exchange was a short discourse in the French language on the subject of breast feeding.”

The assignment struck Waugh as strange but not unaccountable — he’d been writing a regular column in a medical magazine that had touched on that topic.

“So I composed a speech on this subject in French, with considerable labour, only to find when I landed in Dakar that the subject chosen was not breast-feeding but press freedom.” He’d misheard the editor.

“There was no way even to describe the misunderstanding, since la liberté de la Presse bears no resemblance to le nourrisson naturel des bébés.”

(From Waugh’s 1991 autobiography Will This Do?)

September 4, 2024September 3, 2024 | Humor · Language

The Right Track

Suppose you’re hiking in the woods and become lost. What’s the best path to follow to find the boundary? You know the forest’s shape and dimensions, but you don’t know where you are within it, nor which direction you’re facing.

This has remained an open problem ever since mathematician Richard Bellman posed it in the Bulletin of the American Mathematical Society in 1956. The best strategy will cover the shortest distance in the worst case; in forests of certain simple shapes this might be as straightforward as walking in a straight line or in a spiral, but other shapes are more troubling. We know how to escape squares and circles efficiently, but not equilateral triangles.

Mathematician Scott W. Williams classed this as a “million-buck problem” because solving it is expected to cultivate techniques of particular value to mathematics. It’s known as Bellman’s lost-in-a-forest problem.

September 3, 2024September 2, 2024 | Science & Math

The Territory

Much blood has … been spilled on the carpet in attempts to distinguish between science fiction and fantasy. I have suggested an operational definition: science fiction is something that could happen — but usually you wouldn’t want it to. Fantasy is something that couldn’t happen — though often you only wish that it could.

— Arthur C. Clarke, foreword, The Collected Stories of Arthur C. Clarke, 2000

September 3, 2024September 2, 2024 | Literature

Black and White

nabokov chess problem

“This seems to be my most ingenious two-mover,” wrote Vladimir Nabokov in 1970. He had composed it a year earlier in Montreux. White to mate in two moves.

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1. Qh5! From The Problemist, January 1970, via Nabokov’s Poems and Problems.

September 1, 2024August 31, 2024 | Puzzles

High and Dry

Since much of the Netherlands is below sea level, Dutch farmers needed a way to leap waterways to reach their various plots of land. Over time this evolved into a competitive sport, known as fierljeppen (“far leaping”) in which contestants sprint to the water, seize a 10-meter pole, and climb it as it lurches forward over the channel. The winner is the one who lands farthest from the starting point in the sand bed on the opposite side.

The current record holder is Jaco de Groot of Utrecht, who leapt, clambered, swayed, and fell 22.21 meters in 2017.

Below: In the Red Bull Stalen Ros in The Hague, two-person teams must navigate tandem bikes along a narrow 80-meter track. Participants are assessed on speed, design of bikes and attire, and creativity.

August 31, 2024August 30, 2024 | Entertainment

Board Walk

Al writes the numbers 1, 2, …, 2n on a blackboard, where n is an odd positive integer. He then picks any two numbers a and b, erases them, and writes instead |a – b|. He keeps doing this until one number remains. Prove that this number is odd.

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Denote by S the sum of all the numbers that remain on the blackboard. At the start this sum is S = 1 + 2 + … + 2n = n(2n + 1), which is odd. Each step reduces S by 2 min(a, b), which is an even number. Thus the parity of S never changes. Since it started odd, it will end odd. From Arthur Engel’s Problem-Solving Strategies, 2003. 09/01/2024 UPDATE: Readers Seth Cohen and Robert Filman point out a simpler solution. In Seth’s words: Call the number of odd numbers on the board n. At the start, n is odd. If a and b are even, so is a–b, so you don’t lose or gain any odd numbers, so n is still odd. If a and b are odd, a–b is even. So you lose two odd numbers and gain none, and n remains odd. If a and b are one odd and one even, a–b is odd. You lose an odd number and gain an odd number. Thus, n stays odd. No matter what you do, n is always odd, so once you’re down to one number, it’ll be odd. Thanks to both.

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August 30, 2024September 1, 2024 | Puzzles

Misc

To burn up is to burn down.
Litotes is an anagram of T.S. Eliot.
1012658227848 × 8 = 8101265822784
Three U.S. presidents died on July 4.
“Grasp the subject, the words will follow.” — Cato the Elder

August 30, 2024August 29, 2024 | Language · Quotations · Science & Math · Trivia

A Fistful of Scrawlers

From an 1897 Strand article: The “typewriter glove” was “a contrivance of wash-leather, upon which were embossed a set of rubber types. ‘Caps’ were on the left hand. Small letters on the right. The ink was supplied by a couple of pads, fixed to the palms of the gloves; and the alternate opening and shutting [of] the hands was supposed to bring it in contact with the type.

“Then, all that was necessary was for the operator to dab the impression of the particular letter he desired to use upon the paper in front of him. How the alignment was to be preserved, with even a tolerable degree of accuracy, the inventor did not deign to explain.”

(C.L. McCluer Stevens, “The Evolution of the Typewriter,” Strand 13:6 [June 1897], 649-656.)

August 29, 2024August 28, 2024 | Technology

Continuity

In 1590 Emperor Rudolf II commissioned Flemish painter Joris Hoefnagel to illuminate the Mira calligraphiae monumenta, an illustration of various scripts that had been begun 15 years earlier by court calligrapher Georg Bocskay.

The book contains a tiny demonstration of Hoefnagel’s skill in trompe-l’œil. On one page he painted a Maltese cross, a type of flower, depicting it as though the plant’s stem passes through a slit in the paper. On the overleaf he continues the idea — the foregoing text and images can be discerned through the page, and Hoefnagel has faithfully painted in the flower’s “stem” as if the insertion were real. It’s the only painted element on this side of the page.

August 29, 2024August 28, 2024 | Art

Sousselier’s Problem

https://commons.wikimedia.org/wiki/File:Petersen1_tiny.svg — Image: Wikimedia Commons

It appears that there was a club and the president decided that it would be nice to hold a dinner for all the members. In order not to give any one member prominence, the president felt that they should be seated at a round table. But at this stage he ran into some problems. It seems that the club was not all that amicable a little group. In fact each member only had a few friends within the club and positively detested all the rest. So the president thought it necessary to make sure that each member had a friend sitting on either side of him at the dinner. Unfortunately, try as he might, he could not come up with such an arrangement. In desperation he turned to a mathematician. Not long afterwards, the mathematician came back with the following reply. ‘It’s absolutely impossible! However, if one member of the club can be persuaded not to turn up, then everyone can be seated next to a friend.’ ‘Which member must I ask to stay away?’ the president queried. ‘It doesn’t matter,’ replied the mathematician. ‘Anyone will do.’

This problem, dubbed “Le Cercle Des Irascibles,” was posed by René Sousselier in Revue Française de Recherche Opérationelle in 1963. The remarkable solution was given the following year by J.C. Herz. In this figure, it’s possible to visit all 10 nodes while traveling on line segments alone, but there’s no way to close the loop and return to the starting node at the end of the trip (and thus to seat all the guests at a round table). But if we remove any node (and its associated segments), the task becomes possible. In the language of graph theory, the “Petersen graph” is the smallest hypohamiltonian graph — it has no Hamiltonian cycle, but deleting any vertex makes it Hamiltonian.

(Translation by D.A. Holton and J. Sheehan.)

August 28, 2024August 27, 2024 | Science & Math