Yablo’s Paradox

All the statements below this one are false.
All the statements below this one are false.
All the statements below this one are false.
All the statements below this one are false.
All the statements below this one are false.

These statements can’t all be false, because that would make the first one true, a contradiction. But neither can any one of them be true, as a true statement would have to be followed by an infinity of false statements, and the falsity of any one of them implies the truth of some that follow. Thus there’s no consistent way to assign truth values to all the statements.

This is reminiscent of the well-known liar paradox (“This sentence is false”), except that none of the sentences above refers to itself. MIT philosopher Stephen Yablo uses it to show that circularity is not necessary to produce a paradox.

Finding Mates

A remarkable spelling trick by American magician Howard Adams:

From a deck of cards choose five cards and their mates. A card’s mate is the card of the same value and color; for example, the mate of the five of clubs is the five of spades.

Arrange the cards in the order ABCDEabcde, where ABCDE are the chosen cards and abcde are the mates. Cut this packet as many times as you like, then deal five cards onto the table, reversing their order. Place the remaining five cards beside them in a second pile.

Now spell the phrase LAST TWO CARDS MATCH. As you say “L,” choose either pile at random and transfer a card from the top to the bottom. Do the same for A, S, and T. Now remove the top card from each pile and set them aside as a pair.

Perform the same procedure as you spell TWO, CARDS, and MATCH. When you’re finished, two cards will remain on the table. Not only do these cards match, but so do each of the other pairs!

Well Done

http://commons.wikimedia.org/wiki/File:Filippo_Palazzi_-_Giovinetta_alla_sorgente_(Napoli).jpg

Using a 7-quart and a 3-quart jug, how can you obtain exactly 5 quarts of water from a well?

That’s a water-fetching puzzle, a familiar task in puzzle books. Most such problems can be solved fairly easily using intuition or trial and error, but in Scripta Mathematica, March 1948, H.D. Grossman describes an ingenious way to generate a solution geometrically.

Let a and b be the sizes of the jugs, in quarts, and c be the number of quarts that we’re seeking. Here, a = 7, b = 3, and c = 5. (a and b must be positive integers, relatively prime, where a is greater than b and their sum is greater than c; otherwise the problem is unsolvable, trivial, or can be reduced to smaller integers.)

Using a field of lattice points (or an actual pegboard), let O be the point (0, 0) and P be the point (b, a) (here, 3, 7). Connect these with OP. Then draw a zigzag line Z to the right of OP, connecting lattice points and staying as close as possible to OP. Now “It may be proved that the horizontal distances from OP to the lattice-points on Z (except O and P) are in some order without repetition 1, 2, 3, …, a + b – 1, if we count each horizontal lattice-unit as the distance a.” In this example, if we take the distance between any two neighboring lattice points as 7, then each of the points on the zigzag line Z will be some unique integer distance horizontally from the diagonal line OP. Find the one whose distance is c (here, 5), the number of quarts that we want to retrieve.

Now we have a map showing how to conduct our pourings. Starting from O and following the zigzag line to C:

  • Each horizontal unit means “Pour the contents of the a-quart jug, if any, into the b-quart jug; then fill the a-quart jug from the well.”
  • Each vertical unit means “Fill the b-quart jug from the a-quart jug; then empty the b-quart jug.”

So, in our example, the map instructs us to:

  • Fill the 7-quart jug.
  • Fill the 3-quart jug twice from the 7-quart jug, each time emptying its contents into the well. This leaves 1 quart in the 7-quart jug.
  • Pour this 1 quart into the 3-quart jug and fill the 7-quart jug again from the well.
  • Fill the remainder of the 3-quart jug (2 quarts) from the 7-quart jug and empty the 3-quart jug. This leaves 5 quarts in the 7-quart jug, which was our goal.

You can find an alternate solution by drawing a second zigzag line to the left of OP. In reading this solution, we swap the roles of a and b given above, so the map tells us to fill the 3-quart jug three times successively and empty it each time into the 7-quart jug (leaving 2 quarts in the 3-quart jug the final time), then empty the 7-quart jug, transfer the remaining 2 quarts to it, and add a final 3 quarts. “There are always exactly two solutions which are in a sense complementary to each other.”

Grossman gives a rigorous algebraic solution in “A Generalization of the Water-Fetching Puzzle,” American Mathematical Monthly 47:6 (June-July 1940), pp. 374-375.

A Magic Mystery

collison bimagic square

In 1991, David Collison sent this figure to Canadian magic-square expert John Henricks, with no explanation, and then died.

It’s believed to be the first odd-ordered bimagic square ever discovered. Each row, column, and diagonal produces a sum of 369. The square remains magic if each number is squared, with a magic sum of 20,049.

No one knows how Collison created it.

UPDATE: Wait, Collison’s wasn’t the first — G. Pfeffermann published a 9th-order bimagic square as a puzzle in Les Tablettes du Chercheur in 1891. (Thanks, Baz.)

Regrets

On June 28, 2009, Stephen Hawking hosted a party for time travelers, but he sent out the invitations only afterward.

No one turned up.

He offered this as experimental evidence that time travel is not possible.

“I sat there a long time,” he said, “but no one came.”

Misc

  • To frustrate eavesdroppers, Herbert Hoover and his wife used to converse in Chinese.
  • Asteroids 30439, 30440, 30441, and 30444 are named Moe, Larry, Curly, and Shemp.
  • COMMITTEES = COST ME TIME
  • 15618 = 1 + 56 – 1 × 8
  • How is it that time passes but space doesn’t?

Webster’s Third New International Dictionary gives no pronunciation for YHWH.

The Tippe Top

The tippe top is a round top that, when spun, tilts to one side and leaps up onto its stem. This is perplexing, as the toy appears to be gaining energy — its center of mass rises with the flip.

How is this possible? The geometrical center of the top is higher than its center of mass. As the toy begins to topple to one side, friction with the underlying surface produces a torque that kicks it up onto its stem. It does gain potential energy, but it loses kinetic energy — in fact, during the inversion it actually reverses its direction of rotation.

Entire treatises have been written on the underlying physics, and the toy has occupied at least two Nobel Prize winners — below, Wolfgang Pauli and Niels Bohr play with one at the inauguration of the Institute of Physics at Lund, Sweden, in July 1954.

http://en.wikiquote.org/wiki/File:Pauli_wolfgang_c4.jpg

Constellation

German astronomer Karl Reinmuth discovered and named more than 400 asteroids. Among them are these eight:

1227 Geranium
1228 Scabiosa
1229 Tilia
1230 Riceia
1231 Auricula
1232 Cortusa
1233 Kobresia
1234 Elyna

Their initials spell G. STRACKE, for Gustav Stracke, a fellow astronomer who had asked that no planet be named after him. In this way Reinmuth could honor his colleague without contradicting his wish.

Shorthand

University of Michigan mathematician Norman Anning offered this “non-commutative soliloquy of an introspective epistemologist” in Scripta Mathematica in 1948:

[(N + H)ow + (T + W)hat](I know).

Expand the expression and you get Now I know how I know that I know what I know.