Waste Not …

A puzzle from Polish mathematician Paul Vaderlind:

John is swimming upstream in a river when he loses his goggles. He lets them go and continues upstream for 10 minutes, then decides to turn around and retrieve them. He catches up with them at a point one half mile from the point where he lost them. Is the river flowing faster than 1 mile per hour? (Assume that John swims at the same strength throughout.)

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Imagine an observer in a boat near the goggles. From her point of view, the goggles aren’t moving — John simply swims away from them for 10 minutes and will swim back in another 10 minutes (since he’s swimming at a constant speed). If the goggles have traveled half a mile in these 20 minutes, then the river is flowing at 1.5 miles per hour.

September 14, 2012February 6, 2013 | Puzzles

Twice Blessed

When Linus Pauling won his second Nobel Prize in 1962, he joked that receiving his second Nobel was less remarkable than receiving his first: The chance of anyone receiving his first Nobel Prize is one in several billion (the population of the world), while the chance of receiving his second is one in several hundred (the number of living people who have received one prize).

What’s wrong with this argument?

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Nobel prize winners are chosen independently of past awards, so if Pauling’s chance of winning one prize was (say) 1 in 3,000,000,000, then he’d have the same chance of winning the second, and his chance of winning two would be 1 in 3,000,000,000². (Actually they’re considerably better than that — not everyone on Earth is equally deserving of the Nobel prize.) If we knew that the Nobel committee would award the 1962 prize only to an existing laureate, then Pauling’s argument would hold. (From The Ivan Morris Puzzle Book, 1972.)

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September 13, 2012February 6, 2013 | Puzzles

Side Issue

A conundrum by Russian puzzle maven Boris Kordemsky:

A work train composed of a locomotive and five cars has just stopped at a railway station when word comes that a passenger train is approaching. The smaller train must make way for it to pass through, but the station has only one siding, and this will accommodate only three cars (or an engine and two cars). How can it arrange to let the passenger train through?

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The work train backs into the siding and uncouples its three rearmost cars. Then it moves forward a sufficient distance. The passenger train passes entirely through the station, backs into the siding, and couples these three cars. Then it moves forward again into the main railway, backs entirely through the station, and uncouples the three cars. The work train backs into the siding, which will now accommodate it, and the passenger train can move forward on its way. Now the work train can emerge from the siding and recouple the three waiting cars.

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September 12, 2012February 6, 2013 | Puzzles

Flying Fur

loyd cat and dog puzzle

Sam Loyd devised this puzzle for P.T. Barnum:

A trained cat and dog run a race, one hundred feet straight away and return. The dog leaps three feet at each bound and the cat but two, but then she makes three leaps to his two. Now, under those circumstances, what are the possible outcomes of the race?

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As we might have expected, it’s a trick question. It appears at first that the race will end in a tie, as the two animals are covering ground at the same rate. The trouble occurs at the far end of the course. The cat, which covers 2 feet at a bound, can turn around neatly at the halfway point. The dog, which covers 3 feet, cannot: It will arrive at the 99-foot mark and then must make an additional leap to get past the same point. “In all, the dog must make 68 leaps to go the distance. But it jumps only two-thirds as quickly as the cat, so that while the cat is making 100 leaps the dog cannot make quite 67.”

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September 11, 2012February 6, 2013 | Puzzles

Lattice Work

lattice work grid

Plot five points at random at the intersections of a coordinate grid. Between each pair of points a line segment can be drawn. Prove that the midpoint of at least one of these segments occurs at an intersection of grid lines.

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The coordinates of each point can be categorized by their parity — for example, (1, 1) is odd-odd and (4, 17) is even-odd. The midpoint of a line segment between two points will occur at a lattice point only if the two endpoints fall into the same category; for example, the two points above won’t produce such a midpoint, but (6, 8) and (14, 2) will because both are even-even. There are only four possible categories: even-even, odd-odd, even-odd, and odd-even. And because we are plotting five points, at least two points must fall into the same category. The midpoint between these two points will occur at a lattice point. From L.C. Larsen, Problem Solving Through Problems, 1983.

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September 10, 2012February 6, 2013 | Puzzles

Black and White

crane chess problem

By William Crane Jr., from the Sydney Town and Country Journal, 1877. White to mate in two moves.

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After the surprising 1. Bd3!, Black will be mated by 2. Qd1, 2. Bf6, or 2. Qxe4.

September 7, 2012February 6, 2013 | Puzzles

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