Parenthood

We want to line up our six children for a photograph, but we can’t put Sally and John next to one another because they fight. In how many ways can we arrange the photo with this constraint?

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Solving this directly would require finding all the arrangements in which Sally and John are not side by side — 10 different cases, as it turns out. But another way is to consider the opposite problem: What if Sally and John are required to stand side by side? That amounts to treating them as one child, though we must remember that there are two ways to “attach” them, Sally & John or John & Sally. Five children can be arranged in 5 × 4 × 3 × 2 × 1 = 5! = 120 ways, so there are 2 × 120 = 240 ways to arrange the six children so that Sally and John will be side by side. Going back to the original problem, there would be 6 × 5 × 4 × 3 × 2 × 1 = 720 ways to arrange the six children if there were no constraints at all, so there must be 720 – 240 = 480 ways to line them up if Sally and John are to be kept apart. (Shawn Godin, “Pólya’s Paragon: Au Contraire, Mon Ami,” Crux Mathematicorum 30:1 [February 2004], 13-15.)

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January 22, 2020 | Puzzles

Black and White

kossolapov chess problem

By N. Kossolapov. White to mate in two moves.

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1. Qh3! and the queen will mate. From Novosty, 1963.

January 10, 2020 | Puzzles

The Smithy Code

In deciding a plagiarism case against author Dan Brown in 2006, British justice Peter Smith handed down a peculiar judgment: Certain letters in the text had been italicized with no explanation. Apparently inspired by Brown’s book The Da Vinci Code, Smith had hidden a message in the text.

The judgment included the sentence “The key to solving the conundrum posed by this judgment is in reading HBHG and DVC.” In context, those abbreviations refer to The Holy Blood and the Holy Grail, the book that Brown had been accused of plagiarizing, and The Da Vinci Code.

“I can’t discuss the judgement, but I don’t see why a judgement should not be a matter of fun,” Smith had said in handing down the opinion, which found Brown not guilty. He promised to confirm any correct solution.

He offered enough hints to reporters that Guardian media journalist Daniel Tench eventually solved it: It was a polyalphabetic cipher using a keyword based on the Fibonacci sequence, yielding the plaintext “Jackie Fisher who are you? Dreadnought.” Jackie Fisher was a British admiral whom Smith admired. (The code is described here; Tench describes the solving here.)

The Court of Appeal later said that Smith “was prompted by the extensive use in [The Da Vinci Code] of codes, and no doubt by his own interest in such things, to incorporate a coded message in his judgment, on which nothing turns. The judgment is not easy to read or to understand. It might have been preferable for him to have allowed himself more time for the preparation, checking and revision of the judgment.”

January 7, 2020January 6, 2020 | Puzzles

Podcast Episode 278: Lateral Thinking Puzzles

Here are six new lateral thinking puzzles — play along with us as we try to untangle some perplexing situations using yes-or-no questions.

See full show notes …

December 30, 2019January 6, 2020 | Podcast · Puzzles

Knights and Knaves

A logic puzzle by MIT mathematician Tanya Khovanova: You’re visiting an island on which every resident is either a knight or a knave. Knights always tell the truth, and knaves always lie. All the islanders know one another. You meet three islanders, Alice, Bob, and Charlie, and ask each one, “Of the two other islanders here, how many are knights?” Alice says, “Zero.” Bob says, “One.” What will Charlie say?

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If Alice is speaking the truth, then Bob and Charlie are liars. But in that case Bob would be uttering a true statement, which is a contradiction. So Alice is a knave. If that’s the case, then either Bob or Charlie (or both) is a knight. If Bob is a knave, then he’s lying when he says that there’s exactly one truth-teller among the other two. But that’s another contradiction: In that circumstance his statement would be right, since Alice is a knave and Charlie would be a knight. So Bob must be a knight, which means that we can trust what he says and Charlie is a knight too. So Charlie says, “One.” (Khovanova’s students pointed out that the very question in the puzzle implies that Charlie makes a predictable answer, which wouldn’t be the case if he were a liar. The result follows from there.) Obligatory John Finnemore sketch:

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December 29, 2019 | Puzzles

That Time Again

King William’s College has released its annual General Knowledge Paper, “The World’s Most Difficult Quiz,” a school tradition since 1904. There are 18 sets of 10 questions, each set treating a particular theme; divining the themes is difficult and useful.

This year’s quiz bears the customary warning at the top: Scire ubi aliquid invenire possis ea demum maxima pars eruditionis est, “The greatest part of knowledge is knowing where to find something.” If past quizzes are any model, then search engines may lead you astray.

The answers will be on the school website at the end of January. Meanwhile MetaFilter is coordinating a spreadsheet of proposed answers (warning: spoilers).

December 23, 2019December 23, 2019 | Puzzles · Trivia

Black and White

bekkelund chess problem

A prizewinning problem by Paul Bekkelund from the Norwegian chess magazine Sjakk-Nytt, 1947. White to mate in two moves.

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With 1. Ne4 White threatens to mate with 2. Rb8. Black can avoid this by capturing the b4 knight, and he has four ways to do this, but each blocks the b4 flight square, permitting mate: 1. … axb4 2. Rc5# 2. … Rxb4 2. Nc3# 3. … Bxb4 2. Bc4# 4. … Nxb4 2. Nd6#

December 21, 2019 | Puzzles

Gregarious

A problem proposed by Ashay Burungale of Satara, Maharashtra, India, in the November 2008 issue of American Mathematical Monthly: In a certain town of population 2n + 1, all relations are reciprocal: If Person 1 knows Person 2, then Person 2 knows Person 1. For any set A that consists of n citizens, there’s some person among the remaining n + 1 who knows everyone in A. Prove that there’s some citizen of the town who knows all the others.

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Kenneth Schilling of the University of Michigan offered this solution: Call any set of citizens all of whom know one another a clique. Now if there’s a clique B that has fewer than n + 1 citizens, then by hypothesis some citizen who’s not in B knows everyone in B, and we can add that person to B to form a larger clique. That means that a clique exists of size n + 1. The set A of citizens who aren’t in B has size n. So by hypothesis there’s some person in B who knows every person in A, and hence that person knows everyone in the town. (Ashay Burungale and Kenneth Schilling, “Getting to Know You, Directly and Indirectly,” American Mathematical Monthly 115:9 [November 2008], 862.)

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December 18, 2019 | Puzzles

A Hat Puzzle

Each of three people is wearing either a red hat or a blue hat. Each can see the color of the others’ hats but not her own. Each is told to raise her hand if she sees a red hat on another player. The first to guess the color of her own hat correctly wins.

All three raise their hands. A few minutes pass in which no guesses are made, and then one player says “Red” and wins. How did she know the color of her hat?

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All three players raised their hands, so each can see at least one red hat. This means that at least two of the hats are red; if two or more were blue then there’d be at least one player who didn’t raise her hand. But any player who can see a blue hat can immediately infer that her own hat must be red, because she can see a red-wearing player whose hand is raised. In the puzzle this doesn’t happen: All three players raise their hands and yet none of them makes this inference. That allows one of the players to conclude that none of them is wearing a blue hat; all three hats must be red.

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