## Martin Gardner: Mathematical Games : A Cipher that Defeated Poe

Peter Bakke : I solved this cipher in 1987.

“Ge Jeasgdxv,

Zij gl mw, laam. xzy zmlwhfzek

ejlvdxw kwke tx lbr atgh lbmx aanu

bai Vsmukkss pwn vlwk agh gnumk

wdlnzweg jnbxw oaeg enwb zwmgy

mo mlw wnbx mw al pnfdcfpkh wzkex

hssf xkiyahul. Mk num yexdm wbxy

sbc hv wyx Phwkgnamcuk?”

In 1839, in a regular column Edgar Allan Poe contributed to a Philadelphia periodical, Alexander’s Weekly Messenger, Poe challenged readers to send him {cryptograms (monoalphabetic substitution ciphers), asserting that he would solve them all “forthwith.” One G. W. Kulp submitted a ciphertext in longhand. It was printed as shown above in the issue of February 26, 1840. Poe “proved ” in a subsequent column that the cipher as a hoax—”a jargon of random characters having no meaning whatsoever.” In 1975 Brian J.Winkel, a mathematician at Albion College, and Mark Lyster, a chemistry major in Winkel’s cryptology class, cracked Kulp’s cipher. It is not a simple substitution — Poe was right — but neither is it nonsense. Poe can hardly be blamed for his opinion. In addition to a major error by Kulp there are 15 minor errors, probably printer’s mistakes in reading the longhand. Winkel is an editor of a new quarterly, Cryptologia, available from Albion College, Albion, Mich. 49224, at $16 per year. The magazine stresses the mathematical and computational aspects of cryptology. The first issue (January, 1977) tells the story of Kulp’s cipher and gives it as a challenge to readers. So far only three readers have broken it.

## Symphony of Science / A Glorious Dawn / Carl Sagan / Stephen Hawking

## Symphony of Science / Carl Sagan

## Famous computer science problems / algorithms

## 31 Phrases That Only People In The Military Will U…

31 Phrases That Only People In The Military Will Understand shar.es/187Wy9

## How to use SurveyMonkey.com as a simulation tool

You are an Ambassador from which country?

## The width of the Veil Nebula is 3 fingers

So how does one calculate the width (in degrees) of any distant object – from say the Veil Nebula 2100 light years away to the width (in degrees) of that mountain just 21 miles away? Simple. But you will need a scientific calculator. Here’s one: http://web2.0calc.com/

Using the information given to you, say in an article about the Veil Nebula , calculate a ratio of distance over width.

In this case, we are told the Veil Nebula is 2100 light years away and it is 110 light years in width. It is very convenient that they gave us distance and width in the same units (light years), else we’d have to do that ourselves.

So, 2100 / 110 = 19.1

Using your scientific calculator (see above), enter 19.1 and hit the COT function (COTANGENT).

The answer of 5.9 (degrees), or about the width of your three fingers extended at arms length towards the fabulous night sky – in this case in the direction of the Swan Constellation.

Happy Asronomying !

(Conversely, of course, if you estimate the distance of an object and also estimate its width (in degrees), then you can determine the approximate width of the object {in feet, miles, light years, etc.}).

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## The Veil Nebula. What’s a few trillion miles between friends?

Article Source: http://www.chron.com/news/nation-world/space/article/NASA-s-Hubble-images-of-Veil-Nebula-supernova-6530074.php

I believe someone’s math is incorrect in this article. I’m not spoiling the fun, just correcting the record. The image is truly breathtaking and wondrous! But Numeracy counts. :~)

63,000 just didn’t seem big enough, and I was right. Somebody’s calculation re: the width of the Veil Nebula is off by a factor of 100.

Given:

A light year is approx 6 trillion miles, or 6,000,000,000,000 miles.

At a width of 110 light years, the Veil Nebula is therefore about 660,000,000,000,000 miles across.

The sun is about 93 million miles away, or 93,000,000.

If you do the math (660 trillion / 93 million), 110 light years across is not about 63,000 times the distance between earth and the sun as claimed in the article, but instead is about 7,000,000 (7 million) times! An even more incomprehensible number than 63,000, I know.

But, what’s a few trillion miles between friends?

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## Cat’s meow

**Albert Einstein:**

Then again, Einstein was a bit of a wag. Consider his explanation of wireless communication: “The wireless telegraph is not difficult to understand. The ordinary telegraph is like a very long cat. You pull the tail in New York, and it meows in Los Angeles. The wireless is the same, only without the cat.” This quote reportedly kept **Schrödinger** awake well past his bedtime.