Fish Scales

http://commons.wikimedia.org/wiki/File:Diracb.jpg

While a student at Cambridge, Paul Dirac attended a mathematical congress that posed the following problem:

After a big day’s catch, three fisherman go to sleep next to their pile of fish. During the night, one fisherman decides to go home. He divides the fish in three and finds that this leaves one extra fish. He throws this into the water, takes one third of the remaining fish, and departs.

The second fisherman awakes. Not knowing that the first has left, he too divides the fish into three piles, finds one fish left over, discards it, and takes a third of the remainder. The third fisherman does the same. What is the least number of fish that the fishermen could have started with?

Dirac proposed that they had begun with -2 fish. The first fisherman threw one into the water, leaving -3, and took a third of this, leaving -2. The second and third fisherman followed suit.

This story was recalled by “a well-meaning experimenter” in the Russian miscellany Physicists Continue to Laugh (1968). “I could tell many other stories about theoreticians and their work,” he wrote, “but they have told me that one theoretician is writing a story under the title ‘How Experimental Physicists Work.’ That, of course, will be presented upside down.”

All Greek

In 1948, George Washington University doctoral student Ralph Alpher was working on a cosmology thesis under physicist George Gamow. As the paper took shape, “Gamow, with the usual twinkle in his eye, suggested that we add the name of Hans Bethe to an Alpher-Gamow letter to the editor of the Physical Review,” listing the authors as Alpher-Bethe-Gamow.

Bethe agreed to join, and the result, now known as the αβγ paper, was published on April 1, 1948 (“believe it or not, a date not of our asking”). “The response was fascinating,” Alpher later recalled, “ranging from feature articles, Sunday supplement stories, newspaper cartoons and voluminous mail from religious fundamentalists, to a packed audience of over 200, including members of the press, at the traditionally public (though usually not in this sense) ‘defence’ of the thesis.”

Gamow added, “There was, however, a rumor that later, when the alpha, beta, gamma theory went temporarily on the rocks, Dr. Bethe seriously considered changing his name to Zacharias.”

Intro Zoology

http://books.google.com/books?id=grpBAAAAIAAJ&source=gbs_navlinks_s

How to tell a parrot from a carrot, from American physicist Robert W. Wood’s extracurricular How to Tell the Birds From the Flowers: A Manual of Flornithology for Beginners (1907):

The Parrot and the Carrot we may easily confound,
They’re very much alike in looks and similar in sound.
We recognize the Parrot by his clear articulation,
For Carrots are unable to engage in conversation.

Below: A further distinction.

http://books.google.com/books?id=grpBAAAAIAAJ&source=gbs_navlinks_s

Specialist Units

“Standards for inconsequential trivia,” offered by Philip A. Simpson in the NBS Standard, Jan. 1, 1970:

10-15 bismols = 1 femto-bismol
10-12 boos = 1 picoboo
1 boo2 = 1 boo-boo
10-18 boys = 1 attoboy
1012 bulls = 1 terabull
101 cards = 1 decacards
10-9 goats = 1 nanogoat
2 gorics = 1 paregoric
10-3 ink machines = 1 millink machine
109 los = 1 gigalos
10-1 mate = 1 decimate
10-2 mentals = 1 centimental
10-2 pedes = 1 centipede
106 phones = 1 megaphone
10-6 phones = 1 microphone
1012 pins = 1 terapin

Speechless

A puzzle from the Middle Ages, adapted by A.N. Prior:

Four people, on a certain occasion, say one thing each.

A says that 1 + 1 = 2.

B says that 2 + 2 = 4.

C says that 2 + 2 = 5.

Can D now say that exactly as many truths as falsehoods are uttered on this occasion?

“If what D says is true,” Prior writes, “that makes 3 truths to 1 falsehood, so that it is false; while if it is false, that makes two truths and two falsehoods, and it is true.”

Transversal of Primes

Choose a prime number p, draw a p×p array, and fill it with integers like so:

transversal of primes

Now: Can we always find p cells that contain prime numbers such that no two occupy the same row or column? (This is somewhat like arranging rooks on a chessboard so that every rank and file is occupied but no rook attacks another.)

The example above shows one solution for p=11. Does a solution exist for every prime number? No one knows.

Moving Target

What is the smallest integer that’s not named on this blog?

Suppose that the smallest integer that’s not named (explicitly or by reference) elsewhere on the blog is 257. But now the phrase above refers to that number. And that instantly means that it doesn’t refer to 257, but presumably to 258.

But if it refers to 258 then actually it refers to 257 again. “If it ‘names’ 257 it doesn’t, so it doesn’t,” writes J.L. Mackie, “but if it doesn’t, then it does, so it does.”

(Adduced by Max Black of Cornell.)

Tableau

http://www.sxc.hu/photo/29820

Mr. X, who thinks Mr. Y a complete idiot, walks along a corridor with Mr. Y just before 6 p.m. on a certain evening, and they separate into two adjacent rooms. Mr. X thinks that Mr. Y has gone into Room 7 and himself into Room 8, but owing to some piece of absent-mindedness Mr. Y has in fact entered Room 6 and Mr. X Room 7. Alone in Room 7 just before 6, Mr. X thinks of Mr. Y in Room 7 and of Mr. Y‘s idiocy, and at precisely 6 o’clock reflects that nothing that is thought by anyone in Room 7 at 6 o’clock is actually the case. But it has been rigorously proved, using only the most general and certain principles of logic, that under the circumstances supposed Mr. X just cannot be thinking anything of the sort.

— A.N. Prior, “On a Family of Paradoxes,” Notre Dame Journal of Formal Logic, 1961

Manual Labor

Dick and Jane are playing a game. Each holds up one or two fingers. If the total number of fingers is odd, then Dick pays Jane that number of dollars. If it’s even, then Jane pays Dick:

manual labor

At first blush this looks fair, but in fact it’s distinctly favorable for Jane. Let p be the proportion of times that Jane holds up one finger. Her average winnings when Dick holds up one finger are -2p + 3(1 – p), and her average winnings when he holds up two fingers are 3p – 4(1 – p). If she sets those equal to one another she gets p = 7/12. This means that if she raises one finger with probability 7/12, then on average she’ll win -2(7/12) + 3(5/12) = 1/12 dollar every round, no matter what Dick does. Dick’s best strategy is also to raise one finger 7/12 of the time, but the best this can do is to restrict his loss to 1/12 dollar on average. It’s not a fair game.