High Hopes

A worm crawls along an elastic band that’s 1 meter long. It starts at one end and covers 1 centimeter per minute. Unfortunately, at the end of each minute the band is instantly and uniformly stretched by an additional meter. Heroically, the worm keeps its grip and continues crawling. Will it ever reach the far end?

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Surprisingly, yes. At the end of the first minute the worm has crawled 1/100 of the way to the far end. At the end of the second minute, it’s crawled (1/100 + 1/200) of the way. So it will reach the far end in t minutes when 1/100 (1/1 + 1/2 + 1/3 + … + 1/t) equals or exceeds 1. The expression in parentheses is the harmonic series, which can be made as large as one desires, but it’ll take a while: In this example the worm will reach its goal in t = 1.509269 × 10⁴³ minutes, or about 286,961,000,000,000,000,000,000,000,000,000,000 centuries.

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September 1, 2012February 6, 2013 | Puzzles · Science & Math

Black and White

wheeler sui-mate

By C.H. Wheeler, from the Dubuque Chess Journal, December 1877. “White to play and compel self-mate in two moves” — that is, White must force Black to checkmate him in two moves.

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I don’t have Wheeler’s solution, but I believe this is right: As things stand, only Black’s a4 bishop can move. With 1. Rd7, White can limit that bishop to three squares, from each of which a queen sacrifice will force it to the mating diagonal b1-h7: 1. Rd7 Bb5 2. Qd3+ Bxd3# 1. Rd7 Bc6 2. Qe4+ Bxe4# 1. Rd7 Bxd7 2. Qf5+ Bxf5#

August 31, 2012February 6, 2013 | Puzzles

Heads and Tails

Let’s play a coin-flipping game. At stake is half the money in my pocket. If the coin comes up heads, you pay me that amount; if it comes up tails, I pay you.

Initially this looks like a bad deal for me. If the coin is fair, then on average we should expect equal numbers of heads and tails, and I’ll lose money steadily. Suppose I start with $100. If we flip heads and then tails, my bankroll will rise to $150 but then drop to $75. If we flip tails and then heads, then it will drop to $50 and then rise to $75. Either way, I’ve lost a quarter of my money after the first two flips.

Strangely, though, the game is fair: In the long run my winnings will exactly offset my losses. How can this be?

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It’s true that if the number of wins equals the number of losses, as expected, I’ll lose money. But the sequence of flips won’t always match our expectations so precisely — it may favor heads or tails. And when it favors heads, I’ll make a lot of money. Suppose we play four games in a row, with two flips per game, starting with $100: Here I lose fully 75 percent of the games we play — but my winnings are so great in the fourth case that they offset the whole foregoing series of (relatively small) losses. Such cases are rare, but they’re so lucrative that in the long run my bankroll is unchanged.

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August 28, 2012February 6, 2013 | Puzzles · Science & Math

Black and White

“The Forlorn Hope,” by J. Paul Taylor, 1878. White to mate in two moves.

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The White King looked around, and noted how Our ranks were thinned and broken, so that now Our small successes were not worth the cost; And said, “We must resign, the day is lost!” “But not the Knight!” so spake a horseman brave: “My steed says neigh, the day we yet may save.” Then turned the King, and, frowning darkly, said: “Such jesting, good Sir Knight, may cost your head. He who would make a pun at such a time Is capable of any other crime; But one more chance I give you — Forward ride, And lead a hope forlorn. If good betide, Perhaps I may forgive you for the pun, And if you fall but little harm is done!” Forth spurred the Knight, regardless of his life, And rode at once where thickest raged the strife. To such a spot he dashed, that whilst he braved Five sorts of slaughter, still the day he saved; For should the foeman dare to lay him low, He leaves his King a prey to vassal’s blow; And should the King himself the victim slay, A priest steps in with stroke that none can stay; Whilst if they should decide to spare the Knight, And try to shield their Monarch from his might, They yet would fail, for still his falchion bright Descends with fell effect from left or right.

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August 24, 2012February 6, 2013 | Puzzles

“Logic”

Cries logical Bobby to Ned, will you dare
A bet, which has most legs, a mare, or no mare.
A mare, to be sure, replied Ned, with a grin,
And fifty I’ll lay, for I’m certain to win.
Quoth Bob, you have lost, sure as you are alive,
A mare has but four legs, and no mare has five.

— The Panorama of Wit, 1809

August 20, 2012August 20, 2012 | Puzzles

Ball Juggling

http://commons.wikimedia.org/wiki/File:Geometry_of_Solar_eclipses_and_Lunar_eclipses.svg — Image: Wikimedia Commons

In a solar eclipse, the moon casts its shadow on Earth. In a lunar eclipse, Earth casts its shadow on the moon.

Solar eclipses are more common than lunar eclipses, but we tend to have the opposite impression. Why?

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Because the moon’s shadow is smaller than Earth’s, a total solar eclipse is visible only from within a relatively narrow latitude on Earth’s surface, while a lunar eclipse is visible from anywhere on the night side of the planet. For the same reason, a total solar eclipse lasts for only a few minutes at any given location, while a lunar eclipse lasts for several hours.

August 20, 2012May 6, 2013 | Puzzles

Black and White

By Philip Hamilton Williams, Evening News and Post, 1895. White to mate in two moves.

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1. Qb1 and every Black move loses.

August 17, 2012February 6, 2013 | Puzzles

Up From Down

Mount Everest rises 29,029 feet above sea level, and Ecuador’s volcano Chimborazo rises only 20,702 feet. But because Earth bulges at the equator, Chimborazo is actually farther from the center of the planet. If we could connect the two peaks with a water pipe, in which direction would the water flow?

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Sea level gives us a convenient equilibrium water level. Since Everest rises higher, the water would flow toward Chimborazo.

August 12, 2012February 6, 2013 | Puzzles

Black and White

pierce chess problem

By W. Timbrell Pierce, 1873. White to mate in two moves.

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1. Qh5 Ke4 2. Rxc4#. (If the knight moves then the queen can mate immediately from e5.)

August 10, 2012February 6, 2013 | Puzzles

Spin City

Imagine two concentric roulette wheels, each divided into 100 sectors. Choose 50 sectors at random on each wheel, paint them black, and paint the rest white. Prove that we can now position the wheels so that at least 50 of the aligned sectors match.

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Follow a sector on the inner wheel through a complete revolution: That sector will find exactly 50 matches. The same is true for each of the 100 sectors on the wheel; altogether, as the wheel turns through 100 positions, there will be 100 × 50 = 5,000 matches. This means that the average number of matches per position is 50, so a position must exist with at least 50 matches.

August 9, 2012February 6, 2013 | Puzzles