Black and White

blumenthal chess problem

By Oskar Blumenthal. White to mate in two moves.

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1.Qh1! and White will mate with 2. Qh8# or 2. c7#. From Schweizerische Schachzeitung, 1907.

April 12, 2016April 24, 2016 | Puzzles

The Barbershop Paradox

In 1894 Lewis Carroll published a conundrum that, he wrote, presents “a very real difficulty in the Theory of Hypotheticals.” Suppose that Allen, Brown, and Carr run a shop. At least one of them must always be present to mind the shop, and whenever Allen leaves he always takes Brown with him. Now, suppose that Carr is out. In that case then if Allen is out then Brown must be in, in order to tend the shop. But we know that this isn’t true — we’ve been told that whenever Allen is out then Brown is out.

Since the supposition that Carr is out leads to a falsehood, then it must itself be false. Confusingly, the laws of logic seem to require that Carr never leave the shop.

“I greatly hope that some of the readers of Mind who take an interest in logic will assist in clearing up these curious difficulties,” Carroll wrote. Modern logicians would say that this is a simple error in reasoning, rather than a logical disaster. But what is the error?

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Consider the troublesome statement “If Allen is out, then Brown is in” (supposing that Carr is out). We had evaluated this as false. But saying “If Allen is out, then Brown is in” is just the same as saying “Either Allen is in or Brown is in.” And this is true, not false — it admits the possibility that Allen remains in the shop, as of course he must. So the problem disappears. In simple terms, when Carr leaves the shop then Allen (and perhaps Brown) stays behind, as our common sense says he does. (Lewis Carroll, “A Logical Paradox,” Mind 3:11 [July 1894], 436-438.)

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April 7, 2016April 17, 2016 | Puzzles

The Falling Chain

Here are two identical rope ladders with slanting rungs. One falls to the floor, the other onto a table. The ladders are released at the same time and fall freely, but the one on the left falls faster, as if the table is “sucking” it downward. Why does this happen?

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Consider a single pencil falling at a shallow angle. When one end strikes the table, the opposite end jerks downward. In the ladder this action tugs downward on the string above, accelerating its descent. The other ladder doesn’t experience these tugs (yet), so it falls more slowly. (Anoop Grewal, Philip Johnson, and Andy Ruina, “A Chain That Accelerates, Rather Than Slows, Due to Collisions: How Compression Can Cause Tension,” American Journal of Physics 79:7 [July 2011], 723.)

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April 6, 2016April 17, 2016 | Puzzles · Science & Math

Podcast Episode 100: Lateral Thinking Puzzles

Here are five new lateral thinking puzzles to test your wits and stump your friends — play along with us as we try to untangle some perplexing situations using yes-or-no questions.

See full show notes …

April 4, 2016October 14, 2017 | Podcast · Puzzles

Perspective

perspective puzzle

AB and CD are consecutive ties across a pair of railroad tracks that appear to meet at O on the horizon, H. If the ties are parallel to the horizon and are equally spaced along the tracks, how can we draw the next tie in this perspective figure?

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In the figure, parallel lines meet at the horizon, which represents infinity. If all the ties are evenly spaced, then the railroad track is a series of identical rectangles, and the diagonals of these rectangles are parallel. If diagonal BC reaches the horizon at P, then the line through D that’s parallel to it is PD. That gives us E, the point where the next tie meets the left track, and all that remains is to draw EF parallel to the horizon. (From Ross Honsberger, More Mathematical Morsels, 1991.)

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April 1, 2016April 10, 2016 | Puzzles

Black and White

spencer chess problem

G.B. Spencer devised this ingenious diagram in 1906: It contains 16 separate chess problems, one on each rank and file. In each case, ignore the pieces not on that rank or file and find a way for White to mate in two moves.

For example, the solution to the problem on the first rank is 1. Bd4 Ke2 2. Ng3#. What are the other 15 solutions?

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Here’s the first move of each solution: Rank: 1: 1. Bd4 2: 1. Qb4 3: 1. Bgd6 4: 1. Qc8+ 5: 1. Ra3 6: 1. Kb7 7: 1. Qe5 8: 1. Rc7 File: a: 1. Bc2 b: 1. Qd4 c: 1. Qxc6 d: 1. Rxd4 e: 1. Qe4 f: 1. Bd5+ g: 1. Qd7 h: 1. Qe4 In general this type of problem is known as a “patchwork.”

March 26, 2016April 3, 2016 | Puzzles

Reading a Pitch

In 1972, Addison-Wesley published an ad for three textbooks on the back of The American Statistician magazine. The ad began with this line:

Y LUAEB H O DTYO AOOSGL

In Word Ways, David Silverman wrote, “Care to try and figure out the hidden message? Although the slogan doesn’t have the pizzazz of (say) ‘Let Esso put a tiger in your tank’, it does have the advantage of universality; it can equally well be applied to the sale of shoes, ships or sealing wax (or, for that matter, floor wax).”

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It’s a transposition cipher. Start with the letter S and skip backward three characters at a time through the string, counting spaces as characters and cycling back to the end as often as necessary: SAY HELLO TO A GOOD BUY

March 18, 2016March 27, 2016 | Puzzles

Dinner Wit

It’s said that when Frederick the Great hosted Voltaire at Sanssouci Palace, he sent him this puzzling note:

https://commons.wikimedia.org/wiki/File:Frederick-rebus.png — Image: Wikimedia Commons

It’s a rebus in French: deux mains sous Pé à cent sous scie? (“two hands under ‘p’ at hundred under saw”) means demain souper à Sanssouci? (“supper tomorrow at Sanssouci?”).

Voltaire replied “Ga!”: Gé grand, A petit! (“big ‘G’, small ‘a’!”) means j’ai grand appétit!, or “I am very hungry!”

March 13, 2016March 12, 2016 | Language · Puzzles

Town and Country

More chess masters reside in New York City than in the rest of the United States combined. We’re planning a chess tournament that all American masters are expected to attend, and we want to minimize the total intercity traveling done by the players. The New York players argue that, by this criterion, the tournament should be held in their city. The West Coast players argue that a city should be chosen near the center of the gravity of the players. Where should we hold the tournament?

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In New York. Label the New York players A₁, A₂, …, A_n, and label all the others B₁, B₂, …, B_m. Because most masters live in New York, n > m. Now let the straight-line distance between two given players, A_i and B_j, be denoted A_iB_j. If we hold the tournament in New York, the total distance traveled, T, will be A₁B₁ + A₂B₂ + … + A_mB_m. But if we hold the tournament elsewhere, then the total distance traveled by each pair A_i and B_i (i = 1, 2, … m) will be at least A_iB_i. Hence the total distance traveled by all the participants will be at least T plus the sum of the distances traveled by the unpaired New York masters A_m+1, A_m+2, …, A_n. (Proposed and solved by Chester McMaster in Pi Mu Epsilon Journal 1:8 [April 1953], 328.)

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