Black and White

healey chess problem

By Frank Healey. White to mate in two moves.

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With the quiet 1. Qa1! White threatens to push his pawn, giving mate. Black can stop this only with 1. … c4, and then the queen can mate from a5. From A Collection of Two Hundred Chess Problems, 1866.

March 26, 2020March 26, 2020 | Puzzles

Secret Message

Jonathan Swift’s Journal to Stella, a collection of 65 letters written to his friend Esther Johnson, contains some puzzling passages, such as this one:

“He gave me al bsadnuk lboinlpl dfaonr ufainfbtoy dpionufnad, which I sent him again by Mr. Lewis.”

How should the obscured phrase be read?

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“A bank bill for fifty pound.” It’s a null cipher in which alternate letters are ignored. (From David Kahn’s The Codebreakers.)

March 24, 2020March 23, 2020 | Literature · Puzzles

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March 16, 2020March 17, 2020 | Puzzles

Black and White

shinkman chess problem

By William Anthony Shinkman. White to mate in two moves.

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1. Ra2! wins. By withdrawing the rook along the second rank, White keeps the black queen pinned, so she must follow it along that rank. (Black has no other options — his only other legal move, 1. … h3, loses to 2. Nf3#.) No matter which square she chooses, 2. Bb8 is mate. Note that White must withdraw the rook all the way to the a-file in order for this to work. If it stops at any other square on the second rank, then the queen can capture it and find herself free to block (or capture) the checking bishop, escaping mate in two. From Cumming’s Souvenir Chess Board, 1886.

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March 9, 2020March 9, 2020 | Puzzles

Take 5

An interesting problem from Crux Mathematicorum, March 2004: The increasing sequence 1, 5, 6, 25, 26, 30, 31, 125, 126, … consists of positive integers that can be formed by adding distinct powers of 5. That is, 1 = 5⁰, 5 = 5¹, 6 = 5⁰ + 5¹, and so on. What’s the 75th integer in this sequence?

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Write the sequence in base 5: 1, 10, 11, 100, 101, 110, … Now if we consider these expressions as a sequence of binary numbers, then the number in position n is just n — if we convert the sequence from binary to base 10 it’s just 1, 2, 3, 4 … 75 in binary is 1001011, so the number we want is 5⁶ + 5³ + 5 + 1 = 15,756. (“Mathematical Mayhem,” Crux Mathematicorum 30:2 [March 2004], 72-76.)

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March 3, 2020March 2, 2020 | Puzzles

Black and White

From Theophilus A. Thompson’s Chess Problems, 1873. White to mate in two moves.

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1. Ba8! and the queen will mate.

February 19, 2020February 18, 2020 | Puzzles

The Grand Tour

yoshigahara grand tour 1

Puzzle maven Nob Yoshigahara offered this puzzle in the September-October 2000 issue of MIT Technology Review, attributing it to a Professor Kotani. In the 4 × 4 complex of rooms above, two of the rooms are closed. This leaves a single way to tour the remaining rooms in a series of orthogonal moves, visiting each room once and returning to the starting point.

The 12 × 12 complex below has a similarly unique solution. What is it?

yoshigahara grand tour 2

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John Goodman offered this solution in the March 2001 issue:

February 5, 2020 | Puzzles

Black and White

ceriani puzzle

Luigi Ceriani published this curiously ambivalent retrograde puzzle in Europe Echecs in 1960. White is to mate in two moves.

The answer turns on whether White can castle kingside and whether Black can castle queenside. Castling is always deemed to be legal unless it can be proven otherwise. The question of White’s castling comes down to the origin of the rook on a6. If that started on h1, then White may not castle because the rook that’s currently on h1 obviously must have moved in order to get there. If the rook on a6 started on a1, then again White may not castle, because in that case the rook must have got to its present position via e1, since the pawn originally on a2 can have reached its present position on b3 only after the knight on a1 had itself moved there from b3.

That means that if the rook on a6 is not a promoted piece, White cannot castle. If it is a promoted piece, then Black cannot castle, because the white pawn that was promoted must have passed over either d7 or f7 first, which would have forced the black king to move if it had been on its original square. Hence either castling by White or castling by Black is impossible.

Well, if castling by Black is impossible, then White can win by castling himself, since then Black has no escape from 2. Rf8#. And if castling by White is impossible, then White can capture the b5 pawn en passant (followed by mate with the queen), because if Black’s king and rook have not moved then Black’s last move must have been b7-b5. (His last move cannot have been b6-b5 because that would leave White no possible previous move. As it is, with Black’s last move being b7-b5, White’s preceding move must have been R[c6]xa6+.)

(Via John M. Rice, An ABC of Chess Problems, 1970.) (Maybe I’m being obtuse, but isn’t it possible that neither side can castle? In that case both proposed mates fall apart and I don’t see how White can succeed.)

02/03/2020 UPDATE: Ach, I’d just given the answer to my own question: Castling is deemed to be legal unless it can be proven otherwise, and it can’t be proven that neither side can castle. (Thanks, David and Daniel.)

January 30, 2020February 3, 2020 | Puzzles

Genealogy

Reader Jack McLachlan found this curious entry among the marriage notices in the Scots Magazine of January 1790:

At Newburn, near Newcastle, Mr William Dormand, to Miss Hannah Hoy, of that place. The ceremony was attended by the father, mother, brother, sister, aunt, nephew, two husbands, and two wives, and yet there were only four persons present at the marriage.

No explanation is given. How is such an arrangement possible?

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We don’t know for certain, but Jack suggests this: William marries his maternal uncle’s widow, Hannah. Hannah’s brother (call him Fred) marries William’s widowed mother (call her Sue). Husbands – William and Fred Wives – Hannah and Sue Brother/Sister – Hannah and Fred Aunt/Nephew – William and Hannah Mother/Father – Sue and Fred (if we say that Fred is the stepfather of William) (Thanks, Jack.)

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January 30, 2020 | Puzzles

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.) 02/09/2020 UPDATE: Remarkably, Marin Mersenne addressed the same question in his 1636 musical treatise Harmonie universelle. Reader Jon Wild explains that in a section on combinatorics, Mersenne asks how many ways there are of forming non-repeating melodies from the six notes of the hexachord ut-re-mi-fa-sol-la, then asks how many are “good,” which for him means they do not contain the interval (ascending or descending) of a major sixth, found between “ut” and “la.” He follows the same reasoning to the same answer, that 240 of the 720 melodies must be ruled out. He actually writes out all 720 permutations, then explains that some of them are bad because they contain “ut la” or “la ut” (which is disagreeable, hard to sing, and painful to the ear, and whose frequency ratios are objectionable on numerological grounds). 120 of the listed permutations include the major sixth ascending and 120 descending; subtracting these from the 720 listed permutations leaves 480 good melodies. (Marin Mersenne, Harmonie universelle, 1636, Proposition 9 from the “Deuxième livre des chants” [second book on Melody].) (Thanks, Jon.)

Click for Answer

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.)

02/09/2020 UPDATE: Remarkably, Marin Mersenne addressed the same question in his 1636 musical treatise Harmonie universelle. Reader Jon Wild explains that in a section on combinatorics, Mersenne asks how many ways there are of forming non-repeating melodies from the six notes of the hexachord ut-re-mi-fa-sol-la, then asks how many are “good,” which for him means they do not contain the interval (ascending or descending) of a major sixth, found between “ut” and “la.” He follows the same reasoning to the same answer, that 240 of the 720 melodies must be ruled out.

Mersenne Harmonie universelle

He actually writes out all 720 permutations, then explains that some of them are bad because they contain “ut la” or “la ut” (which is disagreeable, hard to sing, and painful to the ear, and whose frequency ratios are objectionable on numerological grounds). 120 of the listed permutations include the major sixth ascending and 120 descending; subtracting these from the 720 listed permutations leaves 480 good melodies.

(Marin Mersenne, Harmonie universelle, 1636, Proposition 9 from the “Deuxième livre des chants” [second book on Melody].) (Thanks, Jon.)

January 22, 2020February 9, 2020 | Puzzles