Black and White

The English occultist Aleister Crowley, “the wickedest man in the world,” was a skilled chess player. In 1894 he published several problems in the Eastbourne Gazette under the pseudonym Ta Dhuibh. This one appeared on Feb. 21. How can White mate in two moves?

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White wins with the craziest move on the board, 1. Qd4!, offering the queen. If Black captures her then 2. Nxd4 is mate, but other options are no better: If Black moves his knight to any other square, then 2. Qa1#. (If 1. … Nxd2 then 2. Nxd2#.) If he moves his bishop then the queen mates from b2. If the king captures the knight then 2. Qd3#. If the a-pawn captures the knight then 2. Qa1#. The following year Crowley began a three-year course at Cambridge, where he later claimed to have beaten the president of the university chess club in his first year. He considering a professional career in chess and practiced two hours a day, but he gave up the idea when occultism captured his interest.

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May 9, 2018May 8, 2018 | Puzzles

A Royal Tour

king's tour

A problem by Kagen Schaefer:

Suppose a king tours a chessboard, visiting each square once, never crossing his own path, and finishing where he starts. Inevitably he’ll have to make some horizontal and vertical moves; for example, in the tour above he makes 14 horizontal and 16 vertical moves.

Show that in any such tour of an 8 × 8 chessboard the sum of the horizontal and vertical moves must be at least 28.

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The king must visit the perimeter squares of the chessboard in order; if he skips any he’ll be unable to visit them and get home again without crossing his own path. And since the perimeter squares alternate colors, the path to each successive perimeter square must involve at least one horizontal or vertical move (diagonal moves alone won’t bring the king to a new color). There are 28 perimeter squares, so the task requires at least 28 horizontal or vertical moves. Here’s a tour that requires exactly 28: (From John J. Watkins, Across the Board: The Mathematics of Chessboard Problems, 2004.)

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April 25, 2018 | Puzzles

Black and White

erdey chess problem

By Aladár Erdey. White to mate in two moves.

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1. Qf8! and 2. b5 or 2. Qe8 will mate. From Magyar Sakkvilág, 1911.

April 18, 2018 | Puzzles

Black and White

ferber chess problem

By Eugène Ferber, 1899. White to mate in two moves.

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1. Qc6! threatens 2. Qh1#. If the bishop moves to d3 or e4 to prevent this, then 2. Qc1 is mate.

March 29, 2018 | Puzzles

Black and White

bilfinger chess problem

By Carl Bilfinger. White to mate in two moves.

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1. Rd3! The king is forced to take the rook, and 2. Qf3 is mate.

March 6, 2018 | Puzzles

The Three-Hat Problem

In the October 2003 issue of MIT Technology Review, Donald Aucamp offered this conundrum:

Three logicians, A, B, and C, are wearing hats. Each hat displays a positive integer, and each logician can see his companions’ numbers but not his own. All of them know that the numbers are positive integers and that one of the numbers is the sum of the other two. The three then take turns in a contest to see who can determine his number first. In the first round, all three pass, but in the second round A correctly states his number is 50. What are the other two numbers, and how did A know that his was 50?

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This solution is by Howard Haber, who writes, “The key observation is that if two of the three integers are equal, then the person who sees the equal integers knows that his number is twice the same integer.” Disregard the actual values for the moment and focus on their ratio instead. For A, B, C respectively, this ratio must be 5, 2, 3. To see why, consider the first round. A, who sees the numbers 2, 3, concludes that he himself must be wearing either 1 (1 + 2 = 3) or 5 (2 + 3 = 5). A can’t get any further than this in the first round and watches the play continue, adopting for the moment the hypothesis that A, B, C are 1, 2, 3. He’s not surprised when B passes his turn, then, because that’s consistent with his idea: B would have seen A wearing 1 and C wearing 3 and been unable to decide whether he himself were wearing 2 (1 + 2 = 3) or 4 (1 + 3 = 4). So he’d pass. So far, so good. Now play passes to C. If A, B, C were 1, 2, 3, then C would see A wearing 1 and B wearing 2. That would tell C that he himself must be wearing 3 (1 + 2 = 3) or 1 (1 + 1 = 2). But unlike the first two players, C would have a way to eliminate one of his candidates: He would know that he himself could not be wearing 1, because if he were, then B would have seen the numbers 1 and 1 on his turn, and thus would have realized that his own (B’s) number must have been 2 (according to the key observation above). With that insight B would have had enough information to win the game on his turn — he could have declared that A, B, C were 1, 2, 1. But B passed his turn, revealing to C that B was unable to reach that conclusion. And seeing that, C could then have concluded that the numbers could not have been 1, 2, 1 and so must have been 1, 2, 3. But C too passes his turn. That tells A (finally!) that his starting hypothesis was wrong: A, B, C can’t be 1, 2, 3. That leaves only the second possibility he’d identified, where A, B, C are 5, 2, 3, and he announces this. “For the problem as stated [i.e., using actual values rather than ratios], just multiply all numbers above by 10.”

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This solution is by Howard Haber, who writes, “The key observation is that if two of the three integers are equal, then the person who sees the equal integers knows that his number is twice the same integer.”

Disregard the actual values for the moment and focus on their ratio instead. For A, B, C respectively, this ratio must be 5, 2, 3. To see why, consider the first round. A, who sees the numbers 2, 3, concludes that he himself must be wearing either 1 (1 + 2 = 3) or 5 (2 + 3 = 5). A can’t get any further than this in the first round and watches the play continue, adopting for the moment the hypothesis that A, B, C are 1, 2, 3. He’s not surprised when B passes his turn, then, because that’s consistent with his idea: B would have seen A wearing 1 and C wearing 3 and been unable to decide whether he himself were wearing 2 (1 + 2 = 3) or 4 (1 + 3 = 4). So he’d pass. So far, so good. Now play passes to C. If A, B, C were 1, 2, 3, then C would see A wearing 1 and B wearing 2. That would tell C that he himself must be wearing 3 (1 + 2 = 3) or 1 (1 + 1 = 2). But unlike the first two players, C would have a way to eliminate one of his candidates: He would know that he himself could not be wearing 1, because if he were, then B would have seen the numbers 1 and 1 on his turn, and thus would have realized that his own (B’s) number must have been 2 (according to the key observation above). With that insight B would have had enough information to win the game on his turn — he could have declared that A, B, C were 1, 2, 1. But B passed his turn, revealing to C that B was unable to reach that conclusion. And seeing that, C could then have concluded that the numbers could not have been 1, 2, 1 and so must have been 1, 2, 3.

But C too passes his turn. That tells A (finally!) that his starting hypothesis was wrong: A, B, C can’t be 1, 2, 3. That leaves only the second possibility he’d identified, where A, B, C are 5, 2, 3, and he announces this.

“For the problem as stated [i.e., using actual values rather than ratios], just multiply all numbers above by 10.”

March 3, 2018March 3, 2018 | Puzzles

Black and White

thompson chess problem

By Theophilus Thompson. White to mate in two moves.

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1. Qh7! and the queen will mate. From Chess Problems, 1873.

February 14, 2018February 13, 2018 | Puzzles

A Clueless Crossword

From Lee Sallows:

sallows clueless crossword

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(Thanks, Lee!)

February 13, 2018 | Puzzles

Sold!

https://www.flickr.com/photos/carleasingmadesimpletm/10723206113 — Image: Flickr

An old “puzzler” from NPR’s Car Talk:

Thirty buyers attended an auction of dozens of cars. Ten of the buyers bought fewer than 6 cars; eight bought more than 7 cars; five bought more than 8 cars; and one bought more than 9 cars.

Of the 30 buyers, how many bought 6, 7, 8, or 9 cars?

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Ten of the buyers bought fewer than 6 cars, and 1 bought more than 9. Subtract those 11 from the 30 and 19 are left. (From the Nov. 10, 2008, show.)

February 7, 2018 | Puzzles

Running Cargo

This passage is from Rudyard Kipling’s 1910 story “Brother Square-Toes.” What’s notable about the bolded section?

‘I’ll have to bide ashore and grow cabbages for a while, after I’ve run this cargo; but I do wish’ — Dad says, going over the lugger’s side with our New Year presents under his arm and young L’Estrange holding up the lantern — ‘I just do wish that those folk which made war so easy had to run one cargo a month all this winter. It ‘ud show ’em what honest work means.’

‘Well, I’ve warned ye,’ says Uncle Aurette. ‘I’ll be slipping off now before your Revenue cutter comes. Give my love to sister and take care o’ the kegs. It’s thicking to southward.’

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It contains no repeated words. At 66 words, it’s the longest such passage discovered by mathematician Mike Keith in a 2002 search of Project Gutenberg. Remarkably, the same story is notable for a second passage — 15 three-letter words in a row: “‘What?’ says Toby, ‘I thought it was Gert Schwankfelder.’ He put down his fiddle and took a good look at me. ‘Himmel!’ he says. ‘I have hit the wrong boy. It is not the new boy. Why are you not the new boy? Why are you not Gert Schwankfelder?'” (Mike Keith, “Special Word-Unit Windows in Running Text,” Word Ways 35:4 [November 2002], 262-266.)

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February 1, 2018January 31, 2018 | Puzzles