# Podcast Episode 235: Leon Festinger and the Alien Apocalypse

In 1955, aliens from the planet Clarion contacted a Chicago housewife to warn her that the end of the world was imminent. Psychologist Leon Festinger saw this as a unique opportunity to test a new theory about human cognition. In this week’s episode of the Futility Closet podcast we’ll follow him inside a UFO religion as it approaches the apocalypse.

We’ll also try to determine when exactly LBJ became president and puzzle over some wet streets.

See full show notes …

# Rubik’s Clock

Hungarian sculptor and architect Ernő Rubik presented this puzzle in 1988; it was originally created by Christopher C. Wiggs and Christopher J. Taylor. The puzzle has two sides, with nine clocks on each side, and the goal is to set all the clocks to 12 o’clock simultaneously.

There are two ways to adjust the clocks. Turning a wheel at any of the four corners will adjust the clock at that corner on both sides of the puzzle. And turning a wheel will also adjust the three clocks adjacent to that corner on one side of the puzzle or the other; which side is determined by the four buttons surrounding the central clock.

So, for example, pressing the northwest button “in” and then turning the northwest wheel will adjust the northwestern quartet of clocks and the corresponding corner clock on the other side of the puzzle. Pulling the northwest button “out” and turning the same wheel will adjust the northwestern clock on the front of the puzzle, its counterpart on the back, and the three clocks adjacent to it on that side.

This is more intuitive than it sounds. Here’s a simulator.

Since there are 14 independent clocks, with 12 settings each, there are a total of 1214 = 1,283,918,464,548,864 possible configurations. It turns out that no configuration requires more than 12 moves to solve; for comparison, in the “worst case” solving a Rubik’s cube can take 20 moves. The trouble, of course, is knowing how to go about it.

# Escalating Magic

Each number in this pandiagonal order-4 magic square is a three-digit prime:

277 197 631 431
661 401 307 167
137 337 491 571
461 601 107 367


Add 30 to each cell and you get a new magic square, also made up of 16 three-digit primes:

307 227 661 461
691 431 337 197
167 367 521 601
491 631 137 397


Add 1092 to each cell in that one and you get a magic square of four-digit primes:

1399 1319 1753 1553
1783 1523 1429 1289
1259 1459 1613 1693
1583 1723 1229 1489


(Allan William Johnson Jr., “Related Magic Squares,” Journal of Recreational Mathematics 20:1 [January 1988], 26-27, via Clifford A. Pickover, The Zen of Magic Squares, Circles, and Stars, 2011.)

# The Sea Island Problem

The Chinese mathematician Liu Hui offered this technique in a text composed about 500 years after Euclid. We’re on the mainland, and we want to find the height of a mountain on a distant island without crossing the sea.

Liu Hui showed that this can be accomplished by setting up two poles of a known height in a line with the mountain …

… and by appealing to a principle of complementary rectangles — here the red and the blue rectangles have the same area:

By using that principle it’s possible to recast the problem in terms of values that we can measure: the height of the poles (CD), the “offset” from which the top of the mountain can just be sighted from ground level over the top of each pole (DG and FH), and the distance between the poles (DF). Putting all that together we can find both the height of the mountain:

$\displaystyle \frac{CD \times DF}{FH - DG} + CD$

and the distance between the first pole and the mountain:

$\displaystyle \frac{DG \times DF}{FH - DG}$

without ever leaving the mainland. Penn State University mathematician Frank Swetz concluded that “in the endeavours of mathematical surveying, China’s accomplishments exceeded those realized in the West by about one thousand years.”

# Observation

“There is nothing at McDonald’s that makes it necessary to have teeth.” — Harvard nutritionist Jean Mayer, Time, 1973

# Fair Play

Any two lines drawn between opposite sides of a square and intersecting at right angles are equal.

# Seduction

In 1940 Bertrand Russell was invited to teach logic at the City College of New York.

A Mrs. Kay of Brooklyn opposed the appointment, citing Russell’s agnosticism and his alleged practice of sexual immorality.

In the lawsuit his works were described as “lecherous, libidinous, lustful, venerous, erotomaniac, aphrodisiac, irreverent, narrowminded, untruthful, and bereft of moral fiber.”

“Although he lost the case, the aging Russell was delighted to have been described as ‘aphrodisiac,'” writes Betsy Devine in Absolute Zero Gravity. “‘I cannot think of any predecessors,’ he claimed, ‘except Apuleius and Othello.'”

# Dance Lessons

The quicksort computer sorting algorithm demonstrated with Hungarian folk dance, from Romania’s Sapientia University.

Also:

The four queens puzzle solved using ballet.

Binary search through flamenco dance.

Merge sort via Transylvanian-Saxon folk dance.

Selection sort using Gypsy folk dance.

More.

(Via MetaFilter.)

01/19/2019 UPDATE: When Gavin Taylor showed these algorithms to his students at the United States Naval Academy, they asked whether they themselves could dance for extra credit. He said yes. So here are the U.S. Naval Academy midshipmen dancing the InsertionSort algorithm:

(Thanks, Gavin.)

# Morrie’s Law

$\displaystyle \cos \left ( 20^{\circ} \right ) \cdot \cos \left ( 40^{\circ} \right ) \cdot \cos \left ( 80^{\circ} \right ) = \frac{1}{8}$

Richard Feynman was so struck by this fact that he remembered ever afterward where he had learned it — from his childhood friend Morrie Jacobs as the two stood in Morrie’s father’s leather shop in Far Rockaway, Queens.

# Area Matters

If you know the vertices of a polygon, here’s an interesting way to find its area:

1. Arrange the vertices in a vertical list, repeating the first vertex at the end (see below).
2. Multiply diagonally downward both ways as shown.
3. Add the products on each side.
4. Find the difference of these sums.
5. Halve that difference to get the area.

This works for any polygon, no matter the number of points, so long as it doesn’t intersect itself. It’s a slight restatement of the shoelace formula.

(Thanks, Derek, Dan, and Kyle.)