Podcast Episode 221: The Mystery Man of Essex County

In 1882, a mysterious man using a false name married and murdered a well-to-do widow in Essex County, New York. While awaiting the gallows he composed poems, an autobiography, and six enigmatic cryptograms that have never been solved. In this week’s episode of the Futility Closet podcast we’ll examine the strange case of Henry Debosnys, whose true identity remains a mystery.

We’ll also consider children’s food choices and puzzle over a surprising footrace.

Intro:

In 1972 two Canadian scientists set out to figure the number of monsters in Loch Ness.

Winston Churchill’s country home must always maintain a marmalade cat named Jock.

Sources for our feature on Henry Debosnys:

Cheri L. Farnsworth, Adirondack Enigma, 2010.

Craig P. Bauer, Unsolved!, 2017.

George Levi Brown, Pleasant Valley: A History of Elizabethtown, Essex County, New York, 1905.

Caroline Halstead Barton Royce, Bessboro: A History of Westport, Essex Co., N.Y., 1902.

“Debosnys Ciphers,” The Cipher Foundation (accessed Oct. 7, 2018).

Craig P. Bauer, “When Killers Leave Ciphers,” history.com, Nov. 14, 2017.

Nick Pelling, “Henry Debosnys and the Cimbria … ?” Cipher Mysteries, Nov. 16, 2015.

Nick Pelling, “Thoughts on the Debosnys Ciphers …” Cipher Mysteries, Nov. 7, 2015.

Nick Pelling, “The Person Not on the S.S.Cimbria …” Cipher Mysteries, Nov. 17, 2015.

“Guilty of Wife Murder,” [Washington D.C.] National Republican, March 8, 1883.

“Hangman’s Day,” [Wilmington, Del.] Daily Republican, April 28, 1883.

“A Murderer’s Story,” Burlington [Vt.] Weekly Free Press, Nov. 24, 1882.

“A Wife’s Fearful Death,” New York Times, Aug. 6, 1882.

“A Remarkable Man Hanged,” New York Times, April 28, 1883.

The Troy Times of Nov. 23, 1882, had noted, “The prisoner spends his time writing verses, or what he thinks is poetry, and he has over a ream of foolscap paper closely written. Much of this doggerel is written in Latin, French, and an unknown cipher, which Debosnys says is used in Europe quite extensively.” These six cryptograms came to light in 1957 — none has been solved:

https://commons.wikimedia.org/wiki/File:Debosnys-Cryptogram-3.png

https://commons.wikimedia.org/wiki/File:Debosnys-Cryptogram-4a.png

https://commons.wikimedia.org/wiki/File:Debosnys-Cryptogram-2b.png

https://commons.wikimedia.org/wiki/File:Debosnys-Cryptogram-2a.png

https://commons.wikimedia.org/wiki/File:Debosnys-Cryptogram-4b.png

https://commons.wikimedia.org/wiki/File:Debosnys-Cryptogram-1.png

Listener mail:

August Skalweit, Die Deutsche Kriegsernährungswirtschaft, 1927.

Emma Beckett, “Food Fraud Affects Many Supermarket Staples, So How Do You Choose the Good Stuff?” ABC, Sept. 3, 2018.

Stephen Strauss, “Clara M. Davis and the Wisdom of Letting Children Choose Their Own Diets,” Canadian Medical Association Journal 175:10 (Nov. 7, 2006), 1199–1201.

Benjamin Scheindlin, “‘Take One More Bite for Me’: Clara Davis and the Feeding of Young Children,” Gastronomica 5:1 (Winter 2005), 65-69.

Clara M. Davis, “Results of the Self-Selection of Diets by Young Children,” Canadian Medical Association Journal 41:3 (September 1939), 257.

This week’s lateral thinking puzzle was inspired by an item on the podcast No Such Thing as a Fish. Here are two corroborating links (warning — these spoil the puzzle).

You can listen using the player above, download this episode directly, or subscribe on Google Podcasts, on Apple Podcasts, or via the RSS feed at https://futilitycloset.libsyn.com/rss.

Please consider becoming a patron of Futility Closet — you can choose the amount you want to pledge, and we’ve set up some rewards to help thank you for your support. You can also make a one-time donation on the Support Us page of the Futility Closet website.

Many thanks to Doug Ross for the music in this episode.

If you have any questions or comments you can reach us at podcast@futilitycloset.com. Thanks for listening!

Cache and Carry

USC mathematician Solomon W. Golomb offered this problem in the Pi Mu Epsilon Journal, Fall 1971 (page 241):

Ted: I have two numbers x and y, where x + y = z. The sum of the digits of x is 43 and the sum of the digits of y is 68. Can you tell me the sum of the digits of z?

Fred: I need more information. When you added x and y how many times did you have to carry?

Ted: Let’s see. … It was five times.

Fred: Then the sum of the digits of z is 66.

Ted: That’s right! How did you know?

	Select Click for Answer
Solution by Robert C. Gebhardt (from the Fall 1972 issue, page 349): If no carrying were involved, we could get the answer by just adding the two digit sums together, 43 + 68 = 111. But each carry increases the final digit sum by only 1 rather than the 10 that would otherwise result. So each carry reduces our expected final digit sum by 9. In this case there are five carries, so the digit sum of the total is 43 + 68 – (5 × 9) = 66.

Click for Answer

October 15, 2018October 14, 2018 | Puzzles

Black and White

By Rudolf L’Hermet. White to mate in two moves.

	Select Click for Answer
1. Qa6! and Black can choose among four different mates: If he moves his knight, then 2. Qe6#. If he takes the rook, then 2. Qc4#. If he advances his pawn one square, then 2. Qd3#. If he advances his pawn two squares, then 2. Nf6#. From Benjamin Glover Laws, The Two-Move Chess Problem, 1890.

October 11, 2018 | Puzzles

The Hexagonal Tortoise Problem

In the 17th century, Korean aristocrat Choi Seok-jeong proposed a puzzle inspired by the pattern on a tortoise shell: Can you assign the numbers 1 to 30 to the vertices in this diagram so that each hexagon bears the same sum?

https://commons.wikimedia.org/wiki/File:Hexagonal-tortoise-problem.png — Image: Wikimedia Commons

	Select Click for Answer
Image: Wikimedia Commons

October 10, 2018 | Puzzles

Common Sense

A quickie from Raymond Smullyan: On the Island of Knights and Knaves, knights always tell the truth and knaves always lie. Every inhabitant is either a knight or a knave. One day a visiting anthropologist comes across a native and recalls that his name is either Paul or Saul, but he can’t remember which. He asks him his name, and the native replies “Saul.”

From this we can’t know whether the native is a knight or a knave, but we can tell with high probability. How?

	Select Click for Answer
If the native’s name isn’t Saul, it’s unlikely that he’d happen to choose that name for a lie. So he’s probably telling the truth, and is likely a knight (named Saul). Obligatory John Finnemore sketch:

October 7, 2018 | Puzzles

The Red Ball

An urn contains k black balls and one red ball. Peter and Paula are going to take turns drawing balls from the urn (without replacement), and whoever draws the red ball wins. Peter offers Paula the option to draw first. Should she take it? There seem to be arguments either way. If she draws first she might get the red ball straightaway, and it seems a shame to give up that opportunity. On the other hand, if she doesn’t succeed immediately then she’s only increased Peter’s chances of drawing the red ball himself. What should she do?

	Select Click for Answer
Imagine that Peter and Paula simply take turns drawing balls, without bothering to inspect their color, until the urn is exhausted. Then afterward they look to see who has the red ball. The winner in this procedure will always be the same as in the original, so each player’s chance of winning will be the same as in the original game. If k is odd, then the total number of balls is even, and when the urn is empty each player will have the same number of balls, so each of them has a 1/2 chance of winning. But when k is even, then the total number of balls is odd, and the player who draws first will have an extra ball when the drawing is done. So, generally, Paula should accept Peter’s offer to draw first — it may help her, and at worst it leaves her chances unchanged. (From Wolfgang Schwarz, 40 Puzzles and Problems in Probability and Mathematical Statistics, 2008.)

Click for Answer

October 6, 2018October 9, 2018 | Puzzles

Line Limit

You own a goat and a meadow. The meadow is in the shape of an equilateral triangle each side of which is 100 meters long. The goat is tied to a post at one corner of the meadow. How long should you make the tether in order to give the goat access to exactly half the meadow?

	Select Click for Answer
Draw a hexagon whose sides are 100 meters long. Now draw a circle, with the same center, whose radius r is the desired length of the tether: The difference in area between the hexagon and the circle is 6 times half the area of the meadow. The area of the circle is πr². The area of an equilateral triangle with side length a is $\displaystyle \frac{a^{2}\sqrt{3}}{4}$ . The area of a hexagon with side length a is $\displaystyle \frac{6a^{2}\sqrt{3}}{4}$ . For a = 100 the area of the hexagon is $\displaystyle 15000\sqrt{3}$ . If we want a circle half that size, then $\displaystyle \pi r^{2} = 7500\sqrt{3}$ and $\displaystyle r = \sqrt{\frac{7500\sqrt{3}}{\pi }} \approx 64.3037.$ (From Ian Stewart’s Cabinet of Mathematical Curiosities, 2009.)

Click for Answer

September 23, 2018September 24, 2018 | Puzzles

Black and White

I just ran across this in Benjamin Glover Laws’ The Two-Move Chess Problem, from 1890. It’s by G. Chocholous. White is to mate in two moves.

	Select Click for Answer
If White withdraws his light-squared bishop along the b1-h7 diagonal, then Black is facing two mate threats: White can move his king off the a-file, discovering mate by the a8 rook, or he can move his dark-squared bishop off the first rank, discovering mate by the h1 queen. A few sample lines: 1. Bf5 Rxa8+ 2. Ba7# 1. Bf5 Rxh1 2. Kb3# 1. Bf5 Rh4 [ready to block a rook check or take the queen] 2. Bh2# 1. Bf5 Rd8 [ready to block a queen check or take the rook] Bd4# (If 1. Bf5 c2 then 1. Bd4#!) At first it looks as though Black is doomed, but he does have one resource. If after e.g. 1. Bf5 he plays 1. … Rg8 then White has no way to mate on the next move: If the white king discovers check, Black can simply take the rook, and if the g1 bishop discovers check by the queen, the rook can interpose. The answer is 1. Bg6!, the only initial bishop move that blocks the rook’s route to g1 and keeps both mate threats alive.

Click for Answer

September 20, 2018 | Puzzles

Cube Route

Created by Franz Armbruster in 1967, “Instant Insanity” was the Rubik’s Cube of its day, a simple configuration task with a dismaying number of combinations. You’re given four cubes whose faces are colored red, blue, green, and yellow:

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

The task is to arrange them into a stack so that each of the four colors appears on each side of the stack. This is difficult to achieve by trial and error, as the cubes can be arranged in 41,472 ways, and only 8 of these give a valid solution.

One approach is to use graph theory — draw points of the four face colors and connect them to show which pairs of colors fall on opposite faces of each cube:

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

Then, using certain criteria (explained here), we can derive two directed subgraphs that describe the solution:

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

The first graph shows which colors appear on the front and back of each cube, the second which colors appear on the left and right. Each arrow represents one of the four cubes and the position of each of the two colors it indicates. So, for example, the black arrow at the top of the first graph indicates that the first cube will have yellow on the front face and blue on the rear.

This solution isn’t unique, of course — once you’ve compiled a winning stack you can rotate it or rearrange the order of the cubes without affecting its validity. B.L. Schwartz gives an alternative method, through inspection of a table, as well as tips for solving by trial and error using physical cubes, in “An Improved Solution to ‘Instant Insanity,'” Mathematics Magazine 43:1 (January 1970), 20-23.

September 18, 2018September 17, 2018 | Puzzles · Science & Math

Self-Study

For a puzzlers’ party in 1993, University of Wisconsin mathematician Jim Propp devised a “self-referential aptitude test,” a multiple-choice test in which each question except the last refers to the test itself:

1. The first question whose answer is B is question

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

2. The only two consecutive questions with identical answers are questions

(A) 6 and 7
(B) 7 and 8
(C) 8 and 9
(D) 9 and 10
(E) 10 and 11

3. The number of questions with the answer E is

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

The full 20-question test is here, the solution is here, and an interesting collection of solving routes is here.

(Jim Propp, “Self-Referential Aptitude Test,” Math Horizons 12:3 [February 2005], 35.)

September 12, 2018September 11, 2018 | Puzzles