Meno’s Paradox

“[A] man cannot enquire either about that which he knows, or about that which he does not know; for if he knows, he has no need to enquire; and if not, he cannot; for he does not know the very subject about which he is to enquire.” — Socrates

August 31, 2021August 30, 2021 | Oddities

The Billups Neon Crossing Signal

After numerous accidents where the Illinois Central Railroad crossed Highway 7 near Grenada, Mississippi, in the 1930s, inventor Alonzo Billups came up with a one-of-a-kind solution. When a train approached the crossing, motorists were confronted with a lighted skull and crossbones, the glowing words “Stop-DEATH-Stop,” flashing neon arrows indicating the train’s direction, and an air raid siren.

The video here is a simulation; the actual gantry was removed due to a scarcity of neon in the war years. But two photographs survive.

In a Word

astriferous
adj. bearing or containing stars; starry

stelliferous
adj. filled with stars; starry

August 30, 2021August 28, 2021 | Art · Language

Podcast Episode 356: A Strawberry’s Journey

The modern strawberry has a surprisingly dramatic story, involving a French spy in Chile, a perilous ocean voyage, and the unlikely meeting of two botanical expatriates. In this week’s episode of the Futility Closet podcast we’ll describe the improbable origin of one of the world’s most popular fruits.

We’ll also discuss the answers to some of our queries and puzzle over a radioactive engineer.

See full show notes …

August 30, 2021August 29, 2021 | History · Podcast · Science & Math

Landscapes

Victoria Rose Richards embroiders moors.

The artist lives in Plymouth, England, and creates aerial views in needlework of the Devon countryside.

More at My Modern Met.

August 29, 2021 | Art

The Fortsas Hoax

In 1840, librarians and booksellers throughout Europe received a catalog describing a unique collection of books to be auctioned: Jean Nepomucene Auguste Pichauld, Comte de Fortsas, had collected 52 unique books, books of which only a single copy was known to exist. The count had died the preceding September, the message said, and as his heirs had no interest in books, the collection would be auctioned off.

Bibliophiles converged on Binche, Belgium, that August for the event, only to discover that the appointed address did not exist. Notices declared that the town’s library had acquired the books — but Binche had no library. In time it became clear that the Comte de Fortsas himself had never existed.

The whole thing had been an elaborate hoax put on by an antiquarian and retired military officer named Renier Hubert Ghislain Chalon. Ironically, the catalog of nonexistent books itself in time became a collectors’ item.

What will happen if someone now writes those books?

August 28, 2021 | Hoaxes · Literature

Slow Life

https://www.youtube.com/watch?v=Aw-E6LzJux8

This 2020 installation, by French artist Pierre Brault, is a reaction against accelerating lifestyles.

Fittingly, it takes a day to unfold.

August 27, 2021August 27, 2021 | Art

$50 for All

A puzzle by Ben H., a systems engineer at the National Security Agency, from the agency’s August 2016 Puzzle Periodical:

At a work picnic, Todd announces a challenge to his coworkers. Bruce and Ava are selected to play first. Todd places $100 on a table and explains the game. Bruce and Ava will each draw a random card from a standard 52-card deck. Each will hold that card to his/her forehead for the other person to see, but neither can see his/her own card. The players may not communicate in any way. Bruce and Ava will each write down a guess for the color of his/her own card, i.e. red or black. If either one of them guesses correctly, they both win $50. If they are both incorrect, they lose. He gives Bruce and Ava five minutes to devise a strategy beforehand by which they can guarantee that they each walk away with the $50.

Bruce and Ava complete their game and Todd announces the second level of the game. He places $200 on the table. He tells four of his coworkers — Emily, Charles, Doug and Fran — that they will play the same game, except this time guessing the suit of their own card, i.e. clubs, hearts, diamonds or spades. Again, Todd has the four players draw cards and place them on their foreheads so that each player can see the other three players’ cards, but not his/her own. Each player writes down a guess for the suit of his/her own card. If at least one of them guesses correctly, they each win $50. There is no communication while the game is in progress, but they have five minutes to devise a strategy beforehand by which they can be guaranteed to walk away with $50 each.

For each level of play — 2 players or 4 players — how can the players ensure that someone in the group always guesses correctly?

	Select Click for Answer
The 2-person case is simpler. Let r denote a red card and b denote a black card. Let’s list Bruce’s card’s color first. Then the 4 distinct possibilities are rr, bb, rb, and br. The following strategy is foolproof. Bruce guesses that his card is the same color as Ava’s (covering the cases rr and bb). Ava takes the opposite approach and assumes that her card’s color is different from Bruce’s (covering the cases rb and br). One of them will necessarily be correct. If Bruce and Ava properly employ these strategies, they cannot lose the game. Before diving into the 4-person case, let’s reexamine the 2-person case. We could interpret a red card as a 0 and a black card as a 1. Then Bruce’s strategy, as described above, handles the cases 00 and 11. His strategy could be interpreted as “the sum of our cards is even.” (0+0=0, 1+1=2) In this language, Ava’s strategy would be “the sum of our cards is odd.” Again, one of them must be right, no matter the cards. Note that Bruce and Ava have divided the sample space of all possible outcomes into 2 cases, based on the sum of the cards. The 4-person case is more general and depends upon the notion of modular arithmetic, and in particular remainders when dividing by 4. Rather than purely guessing “the suits are the same,” or “the suits are different,” the players must be more clever. Let’s map each suit to a number. Clubs are 0, diamonds are 1, hearts are 2, and spades are 3. So, if Emily looks out and sees hearts (Charles), hearts (Doug), and spades (Fran), she interprets this as 2+2+3=7. Suppose that Emily’s card’s suit is clubs, which she obviously doesn’t know. Consider the following strategies: Emily guesses her suit so that the total sum is a multiple of 4. She sees hearts, hearts, spades and guesses diamonds (2+2+3+1=8). Charles guesses his suit so that the total sum is 1 more than a multiple of 4. He sees clubs, hearts, spades and guesses clubs (0+2+3+0=5). Doug guesses his suit so that the total sum is 2 more than a multiple of 4. He sees clubs, hearts, spades and guesses diamonds (0+2+3+1=6). Fran guesses her suit so that the total sum is 3 more than a multiple of 4. She sees clubs, hearts, hearts and guesses spades (0+2+2+3=7). In this example, Fran guesses correctly because the total sum is 7, which is 3 more than a multiple of 4. For any combination of suits, one of them must be right. After all, the total sum must be a whole number. And all whole numbers have remainder 0, 1, 2, or 3 when divided by 4. As long as each of them sticks to his/her strategy, the group can ensure they won’t lose the game. Interestingly, it doesn’t matter which numbers they choose to use, as long as they all have different remainders when divided by 4. So, choosing any four consecutive whole numbers is a sufficient mapping.

Click for Answer

The 2-person case is simpler. Let r denote a red card and b denote a black card. Let’s list Bruce’s card’s color first. Then the 4 distinct possibilities are rr, bb, rb, and br. The following strategy is foolproof. Bruce guesses that his card is the same color as Ava’s (covering the cases rr and bb). Ava takes the opposite approach and assumes that her card’s color is different from Bruce’s (covering the cases rb and br). One of them will necessarily be correct. If Bruce and Ava properly employ these strategies, they cannot lose the game.

Before diving into the 4-person case, let’s reexamine the 2-person case. We could interpret a red card as a 0 and a black card as a 1. Then Bruce’s strategy, as described above, handles the cases 00 and 11. His strategy could be interpreted as “the sum of our cards is even.” (0+0=0, 1+1=2) In this language, Ava’s strategy would be “the sum of our cards is odd.” Again, one of them must be right, no matter the cards. Note that Bruce and Ava have divided the sample space of all possible outcomes into 2 cases, based on the sum of the cards.

The 4-person case is more general and depends upon the notion of modular arithmetic, and in particular remainders when dividing by 4. Rather than purely guessing “the suits are the same,” or “the suits are different,” the players must be more clever. Let’s map each suit to a number. Clubs are 0, diamonds are 1, hearts are 2, and spades are 3. So, if Emily looks out and sees hearts (Charles), hearts (Doug), and spades (Fran), she interprets this as 2+2+3=7. Suppose that Emily’s card’s suit is clubs, which she obviously doesn’t know. Consider the following strategies:

Emily guesses her suit so that the total sum is a multiple of 4. She sees hearts, hearts, spades and guesses diamonds (2+2+3+1=8).
Charles guesses his suit so that the total sum is 1 more than a multiple of 4. He sees clubs, hearts, spades and guesses clubs (0+2+3+0=5).
Doug guesses his suit so that the total sum is 2 more than a multiple of 4. He sees clubs, hearts, spades and guesses diamonds (0+2+3+1=6).
Fran guesses her suit so that the total sum is 3 more than a multiple of 4. She sees clubs, hearts, hearts and guesses spades (0+2+2+3=7).

In this example, Fran guesses correctly because the total sum is 7, which is 3 more than a multiple of 4. For any combination of suits, one of them must be right. After all, the total sum must be a whole number. And all whole numbers have remainder 0, 1, 2, or 3 when divided by 4. As long as each of them sticks to his/her strategy, the group can ensure they won’t lose the game. Interestingly, it doesn’t matter which numbers they choose to use, as long as they all have different remainders when divided by 4. So, choosing any four consecutive whole numbers is a sufficient mapping.

August 27, 2021August 26, 2021 | Puzzles

Dark Days

During a visit to the Auschwitz-Birkenau Museum in 2016, University of Michigan music theory professor Patricia Hall discovered handwritten manuscripts of popular songs of the day. This one, “The Most Beautiful Time of Life,” is a light foxtrot based on a song by Franz Grothe, a popular German film composer. The prisoners had arranged it for the available instruments and musicians and probably performed it before the commandant’s villa in Sunday concerts for the Auschwitz garrison in 1942 or 1943.

“This was for the SS personnel,” she explained. “It was about a three-hour concert that was broken up into stages, and at one point, they had a dance band so that soldiers could dance. Given the instrumentation of this foxtrot, I think that’s probably what it was used for.”

Two of the arrangers had signed their prison numbers to the manuscript, so Hall was able to identify them: Antoni Gargul, a Polish soldier and a violinist, and Maksymilian Piłat, a professional bassoonist. Both of them survived the war, and Hall believes the unidentified third author may have as well.

August 26, 2021August 25, 2021 | Art · History

Paint Job

chessboard

Given a standard chessboard, you can choose any rank or file and repaint each of its squares to the opposite color (white squares turn black, and black squares turn white). By doing this repeatedly, is it possible to produce a board with 63 white squares and one black square?

	Select Click for Answer
No. Repainting a rank or file in this way transforms its k black and (8 – k) white squares into (8 – k) black and k white squares. The number of black squares changes by (8 – k) – k = 8 – 2k. This is an even number, which means that the parity of the number of black squares is preserved. Since we start with an even number (32) of black squares, there is no way to reach the odd outcome 1. (From Viktor Prasolov’s Problems in Plane and Solid Geometry, 1988.)

Click for Answer

August 26, 2021August 25, 2021 | Puzzles