Chessametics

keith chessametic

Mike Keith published this remarkable invention in the Journal of Recreational Mathematics in 1975. Suppose a chess game goes:

P-K4   P-K4 
B-B4   P-R4 
Q-B3   P-R4 
QxP

This is checkmate, so White says “I win!” Now if the game score is written out in one column, including White’s exclamation:

P-K4
P-K4
B-B4
P-R4
Q-B3
P-R4
 QxP
----
IWIN

… we get a solvable alphametic — replace each letter (and the symbols x and -) with a unique digit and you get a valid sum. (The digits already shown count as numbers, and those numbers also remain available to replace letters.)

This example isn’t quite perfect — the two moves P-R4 are ambiguous, as is the final pawn capture. Keith’s Alphametics Page has an even better specimen, a pretty variation from a real 1912 game by Siegbert Tarrasch. It’s 13 moves long and forms a perfect alphametic with a unique solution.

April 16, 2019April 19, 2019 | Puzzles

Black and White

dehler chess problem

By Otto Georg Edgar Dehler. White to mate in two moves.

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1. Rb5! and Black can’t escape mate. Note that only b5 will do — if White withdraws the rook to, say, b3, then Black can play 1. … Nb6 and 2. Ra3+ is met with 2. … Na4. From Neue Welt, 1919.

April 4, 2019April 4, 2019 | Puzzles

Poser

What’s the next letter in this sequence?

OTTFFSSEN

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T. They’re the initial letters of the English number names: ONE, TWO, THREE …

April 2, 2019April 1, 2019 | Puzzles

A Geography Quiz

Before 1997 there were exactly three countries in the world that were both geographically contiguous and alphabetically adjacent in a list of nations. (One country changed its name.) What were they?

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Zaire, Zambia, and Zimbabwe. Zaire is now Democratic Republic of the Congo. (T.A. Hall, “A Geography Quiz,” Word Ways 45:1 [February 2012], 57-58.)

March 27, 2019March 26, 2019 | Puzzles · Trivia

Slitherlink

https://commons.wikimedia.org/wiki/File:SurizaL%C3%B6sung.png — Image: Wikimedia Commons

In this original logic puzzle by the Japanese publisher Nikoli, the goal is to connect lattice points to draw a closed loop so that each number in the grid denotes the number of sides on which the finished loop bounds its cell, as above: Each cell bearing a “1” is bounded on 1 side, a “2” on 2 sides, and so on.

Here’s a moderately difficult puzzle. Can you solve it? (A loop that merely touches a cell’s corner point without passing along any side is not considered to bound it.)

https://commons.wikimedia.org/wiki/File:Slitherlink-example.png — Image: Wikimedia Commons

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Image: Wikimedia Commons

March 16, 2019 | Puzzles

Black and White

tuxen chess problem

By Harry Viggo Tuxen. White to mate in two moves.

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1. Qh1! and the queen will mate. From Hvar Dag, 1917.

March 15, 2019March 14, 2019 | Puzzles

The Roving Princess

A puzzle by University College London mathematician Matthew Scroggs: A princess lives in a row of 17 rooms. Each day she moves to a new room adjacent to the last one (e.g., if she sleeps in Room 5 on one night, then she’ll sleep in Room 4 or Room 6 the following night). You can open one door each night. If you find her you’ll become her prince. Can you find her in a finite number of moves?

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Here’s one way to do it. Imagine a chessboard with 17 files and a large number of ranks. Each file represents one room, and each rank represents one night. On each night (rank), mark the room (square) where the princess is sleeping. Because the princess moves to an adjacent room each day, the marker will always move “diagonally” on the board, and hence will always occupy squares of the same color. Check the 17th room on the first night, the 16th on the second night, and so on. Suppose the 17th square in the first row is white. If the princess started on a white square, this will find her, as the diagonal you’re drawing must eventually intercept her path. If you reach Room 1 without finding her, this means she’s been traversing black squares. Go immediately back to Room 17 (which will be represented by a black square on row 18), and work your way down as before. This time your paths must cross.

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March 14, 2019March 17, 2019 | Puzzles

Sanity Check

From Raymond Smullyan: Every inhabitant of this island either a knight or a knave. Knights always tell the truth, and knaves always lie. Further, every inhabitant is either mad or sane. Sane inhabitants always answer questions correctly, and mad ones always answer incorrectly. You meet an inhabitant of the island and want to know whether he’s sane or mad. How can you determine this with a single yes-or-no question?

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Ask, “Are you a knight?” A sane knight will say yes, a mad knight will say no, a sane knave will say yes, and a mad knave will say no. So a sane inhabitant will answer yes and a mad inhabitant will answer no. From Logical Labyrinths, 2009. Obligatory John Finnemore sketch:

March 5, 2019 | Puzzles

For Pi Day

From the prolifically interesting Catriona Shearer: The red line is perpendicular to the bases of the three semicircles. What’s the total area shaded in yellow?

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At first it appears there’s not enough information, but there is! No matter the relative sizes of the two smaller semicircles, the area of the shaded region is π. There are any number of ways of proving this; here’s a pretty “proof without words” that relies on the fact that the Pythagorean theorem works with similar figures of any shape (here, semicircles). (Thanks to Bill Bowser for the illustration.)

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March 2, 2019March 3, 2019 | Puzzles · Science & Math

E Pluribus Unum

Replace each * with a different digit 1-9 to make this equation true:

$\displaystyle \frac{*}{**} + \frac{*}{**} + \frac{*}{**} = 1$

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$\displaystyle \frac{9}{12} + \frac{5}{34} + \frac{7}{68} = 1$ It’s easy to solve this by writing a program to try all possibilities, but reasoning it out is tricky. Donald Aucamp gives one solution in MIT Technology Review, March 2002. 03/18/2019 UPDATE: I should have explained that the double asterisks in the denominators are two-digit numbers. Interestingly, there are two other interpretations that both yield answers! Reader Nick Berry sent this: $\displaystyle \frac{1}{6 \times 3} + \frac{7}{4 \times 2} + \frac{5}{9 \times 8} = 1$ And Paul Wagner sent: $\displaystyle \frac{6}{3 + 9} + \frac{5}{7 + 8} + \frac{1}{2 + 4} = 1$ Thanks to both of them.

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February 27, 2019March 18, 2019 | Puzzles