Yajilin

The goal of this logic puzzle is simple: to draw an orthogonally connected, non-intersecting loop that passes through every white square on the board. The trouble is that the board contains some number of black squares, and these are hidden. The only clues to their location are the numbers in the gray squares. In the diagram above, there are exactly 3 black squares in the third file north of the “3” indicator. And there are no black squares on the third rank anywhere east of the “0” indicator.

Gray squares can’t be black, no two black squares are orthogonally adjacent, and there may be some black squares that aren’t referred to by any of the indicators.

Knowing all this (and knowing that a solution is possible), can you determine the location of all the black squares and draw a loop that passes through all the white ones?

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Image: Wikimedia Commons

October 22, 2024October 21, 2024 | Puzzles

Black and White

angelini retrograde problem 1

A logic problem in the shape of a chess puzzle, by Éric Angelini. White has just moved. What was his move?

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At first this appears impossible — the king must just have moved, but from where? Every possibility seems to require an impossible double check. But there is a legal sequence involving a discovered check and an en passant pawn capture. Play must have started from this position: Now 1. … Ne5+ 2. g4 fxg3+ e.p. 3. Kxe5. From Europe Echecs, April 1995.

October 15, 2024October 14, 2024 | Puzzles

In the Fold

Just found this on Wikimedia Commons — “Where Is the Fifth Pig?”, an anonymous puzzle created in occupied Holland in 1940:

October 14, 2024October 13, 2024 | History · Puzzles

A Little Clue

https://commons.wikimedia.org/wiki/File:RectangleAndCircleTeaser.jpg — Image: Wikimedia Commons

A circle is inscribed in a square, with a rectangle drawn from a corner of the square to a point on the circle, as shown. If this rectangle measures 6 inches by 12 inches, what’s the radius of the circle?

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With the information given we can draw a right triangle: Image: Wikimedia Commons We know that x = 12 and y = 6, so the Pythagorean theorem can lead us to discover that r is 30 inches. From the Wikibooks puzzle collection.

October 13, 2024October 12, 2024 | Puzzles

Also-Rans

“The dogs are, by placing two lines upon them, to be suddenly aroused to life and made to run. Query, How and where should these lines be placed, and what should be the forms of them?”

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Sam Loyd made this principle famous with his “trick mules puzzle” of 1858, but this “dog puzzle” had appeared in an American magic book the previous year, and indeed the same idea had appeared in Illustrirte Zeitung in 1843 and in an Iranian drawing of the 17th century, now held at Boston’s Museum of Fine Arts.

October 12, 2024October 11, 2024 | Puzzles

Backward and Forward

John Grant McLoughlin offered this problem in Crux Mathematicorum in April 2008:

Every four-digit numerical palindrome (e.g., 2772) is a multiple of 11. Why is this?

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Any such number can be written in the form abba, where a and b are integers, with a ranging from 1 to 9 and b from 0 to 9. But abba = 1000a + 100b + 10b + a = 1001a + 110b = 11(91a + 10b). So 11 is always a factor.

October 11, 2024October 10, 2024 | Puzzles

Our Best Friend

A Victorian illusion from Richard Wiseman’s blog: What is this?

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It’s a dog!

October 10, 2024October 9, 2024 | Puzzles

Smart Money

Mr. Smith goes to Atlantic City to gamble for a weekend. To guard against bad luck, he sets a policy at the start: In every game he plays, he’ll bet exactly half the money he has at the time, and he’ll make all his bets at even odds, so he’ll have an equal chance of winning and of losing this amount. In the end he wins the same number of games that he loses. Does he break even?

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No. Every win multiplies Smith’s holdings by 1.5, and every loss multiplies them by 0.5. One win and one loss (the order is immaterial) leaves him with 0.75 times his initial holdings, and n wins and n losses leaves him with 0.75ⁿ his initial assets. Ten such blows will leave him with only about 5 percent of his initial stake. From David L. Book, Problems for Puzzlebusters, 1992.

October 4, 2024 | Puzzles

Four Glasses

Martin Gardner published this puzzle in his “Mathematical Games” column in Scientific American in February 1979. You’re blindfolded and sitting before a lazy susan. On each corner is a glass. Some are right side up and some upside down. On each turn you can inspect any two glasses and, if you choose, reverse the orientation of either or both of them. After each turn the lazy susan will be rotated through a random angle. When all four glasses have the same orientation, a bell will sound. How can you reach this goal in a finite number of turns?

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This algorithm will do it in five turns: Choose a pair of glasses that are diagonally opposite and set both right side up. Choose two adjacent glasses. At least one must now be right side up. If the other is upside down, turn it right side up as well. If the bell rings, you’re done; if not, then only one glass is still upside down. Choose a diagonally opposite pair. If one is upside down, turn it up and you’re done. If both are right side up, turn one upside down. Now two glasses are upside down, and these two must be adjacent. Choose two adjacent glasses and reverse the orientation of both. If the bell doesn’t sound then there are now two glasses upside down, and they’re diagonally opposite one another. Choose a diagonally opposite pair and reverse both. The bell will sound.

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October 2, 2024October 1, 2024 | Puzzles

“Through the Looking-Glass”

C.S. Kipping published this unusual problem in Chess Amateur in 1923. In each position, White is to mate in two moves.

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(a) White castles, threatening 2. Nc7#. Black can capture the knight, but then 2. Rb8 is mate. (b) After 1. Rxe3, Black can’t escape 2. Re8#. The point is that in (a) both sides can castle legally, and in (b) they can’t. (In chess problems castling is deemed legal if king and rook stand on their original squares and it can’t be proven that either must have moved in a hypothetical game.) In (a) White needs to castle because the mere 1. Rf1 allows 1. … d2+, and 1. Rxd3 also won’t work because then it’s Black who can castle to safety. In the reflected position both sides lose the castling expedient, so a new solution becomes possible.

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September 27, 2024September 26, 2024 | Puzzles