Black and White

loyd chess problem

By Sam Loyd. White to mate in two moves.

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1. Qa1!, threatening 2. Kf7#. If 1. … Kg8 then 2. Qa8#, and if 1. … Bg8 then 2. Kxg6#. From the Detroit Free Press, 1877.

July 6, 2015July 19, 2015 | Puzzles

Cube Route

You’re planning to make the wire skeleton of a cube by arranging 12 equal lengths of wire as shown and soldering them at the corners.

It occurs to you that you might be able to simplify the job by using one or more longer lengths of wire and bending them into right angles at the cube’s corners.

If you adopt that plan, what’s the smallest number of corners where soldering will still be necessary?

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Eight. An odd number of wires meet at each corner, so every corner will still need soldering.

June 15, 2015July 19, 2015 | Puzzles

Weighing in Verse

How can you find one bad coin among 12 using only three weighings in a pan balance? We published the classic solution in November 2013, but a clever alternative appeared in Eureka in October 1950. Here the aspiring counterfeiter is named Felix Fiddlesticks, and he’s trying to find the coin for this mother:

F set the coins out in a row
And chalked on each a letter, so,
To form the words: “F AM NOT LICKED”
(An idea in his brain had clicked).

And now his mother he’ll enjoin:

weighing in verse table

This plan will reveal the bad coin no matter which it is. For instance, if the coin marked O is heavy, then in the first two weighings the left pan will drop, and in the third weighing it will rise; no other bad coin will produce this result. The plan will also reveal reliably whether the bad coin is heavy or light.

Such cases number twenty-five,
And by F’s scheme we so contrive
No two agree in their effect,
As is with pen and patience checked:
And so the dud is found. Be as it may
It only goes to show CRIME DOES NOT PAY.

(Cedric Smith, “The Twelve Coin Problem,” The Mathematical Gazette, Vol. 89, No. 515 [July 2005], pp. 280-281. A generalized algorithm for solving such a problem for any number of coins is given by Michael Weatherfield in the same issue, pp. 275-279.)

June 14, 2015June 13, 2015 | Puzzles

Riddles in the Dark

In 1904, a 12-year-old J.R.R. Tolkien sent this rebus to a family friend, Father Francis Morgan. What does it say?

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This is from Puzzling Stack Exchange, which gives the following solution (unfortunately the second page of Tolkien’s letter is missing): Tolkien’s address, in the upper right, is Woodside House, Rednal (wood, S, eye, 500=D, E, house, red [N, A, 50=L]) The rest of the page reads: My dear wise owl Fr. Francis, (1000=M, eye, deer, Y’s, owl, “Fr.”, France, hiss) You are too bad (U, R, 2, bee, A, 500=D) not to come in (knot, 2, sea, O, 1000=M, E, inn) spite of Fr. Dennis. (spit, E, “OF:, “Fr.”, 500=D, hen, eye, S) I am so sorry you (Hyam, sew, saw, rye, yew) did not like the (500=D, eye, D, knot, 50=L, “bike” – B, tea, HE) word “piano” in my l… (words, pea, “note” – T, inn, M, eye, 50=L) …ast letter. So I… (“mast” – M, letter, sow, eye) As some of the commenters mention, the “word piano” interpretation seems uncertain, but it’s hard to know what was intended given so little context. Any ideas? UPDATE: Many readers point out that the “note” is a rest, which suggests that the word may be something like letterpress (letter, pea, rest – t) or padres (pod, rest – t), the latter of which might indeed have drawn an objection from Father Morgan. This may be the rest of the letter. (Thanks, Bruce.)

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June 13, 2015June 28, 2015 | Puzzles

Black and White

florentine chess problem

From a Florentine manuscript, 1600. White to mate in two moves.

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1. Qd5! and the pawn will queen, giving mate.

June 7, 2015June 14, 2015 | Puzzles

Three Coins

Three coins are lying on a table: a quarter, a half dollar, and a silver dollar. You claim one coin, I’ll claim the other two, and then we’ll toss all three. A coin that lands tails counts zero, and a coin that lands heads wins its value (in cents, 25, 50, or 100) for its owner. Whichever of us has the larger score wins all three coins. If all three coins land tails then we toss again.

Which coin should you claim to make the game fair — that is, so that each of us has an expected win of zero?

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Surprisingly, the game is fair regardless of which coin you choose. Here’s a table of the possible payoffs: In each case the holder of the single coin wins just enough (when he does win) to lead to a zero expectation for each player. David L. Silverman, who discovered this, found that this game is fair with any set of coins, no matter how they’re divided, provided that their values show a successive doubling (e.g., 25, 50, 100). From Martin Gardner, “Charles Addams’ Skier and Other Problems,” in Wheels, Life and Other Mathematical Amusements, 1983. UPDATE: The mathematical principle here is sound, but the presentation muddies it a bit. The expected outcomes are equal after a coin has been chosen, but properly speaking choosing a coin is part of the game. If both players start with nothing, as described, then choosing the silver dollar gives the best chance of winning. (Thanks, Seth and Tim.)

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June 3, 2015June 14, 2015 | Puzzles · Science & Math

Digit Spans

A puzzle by Anatoly Savin: Using each of the digits 1, 2, 3, and 4 twice, write out an eight-digit number in which there is one digit between the 1s, two digits between the 2s, three digits between the 3s, and four digits between the 4s.

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Two numbers work: 41312432 and its reverse, 23421314.

May 31, 2015November 12, 2023 | Puzzles

The Jeweler’s Observation

Prove that every convex polyhedron has at least two faces with the same number of sides.

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Consider the face with the largest number of sides. If that face has m sides, then it’s surrounded by m faces. Any face must have at least 3 sides. So altogether in this group we have m + 1 faces, and each face must have between 3 and m sides. At least two faces must have the same number of sides. From Arthur Engel’s Problem-Solving Strategies, 2003.

May 25, 2015May 31, 2015 | Puzzles

Black and White

ebert chess problem

By Hilmar Ebert. White to mate in two moves.

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1.Bh1! Kxh1 2.Rxh3#

May 21, 2015May 24, 2015 | Puzzles

Ride Sharing

You and I have to travel from Startville to Endville, but we have only one bicycle between us. So we decide to leapfrog: We’ll leave Startville at the same time, you walking and I riding. I’ll ride for 1 mile, and then I’ll leave the bicycle at the side of the road and continue on foot. When you reach the bike you’ll ride it for 1 mile, passing me at some point, then leave the bike and continue walking. And so on — we’ll continue in this way until we’ve both reached the destination.

Will this save any time? You say yes: Each of us is riding for part of the distance, and riding is faster than walking, so using the bike must increase our average speed.

I say no: One or the other of us is always walking; ultimately every inch of the distance between Startville and Endville is traversed by someone on foot. So the total time is unchanged — leapfrogging with the bike is no better than walking the whole distance on foot.

Who’s right?

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You are. My argument would be sound if each of us simply stood by the bike after dismounting until the pedestrian caught up. But instead we’re investing that extra time in walking, which accounts for the faster progress. Suppose the total trip is 2 miles, and each of us can walk at 4 mph and ride at 12 mph. I ride 1 mile in 5 minutes and leave the bike for you, walking the remaining 1 mile in 15 minutes. You take 15 minutes to reach the bike and then ride to Endville in 5 minutes. We both arrive at the destination in 20 minutes, where it would have taken 30 minutes if we’d walked the whole way.

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May 20, 2015May 31, 2015 | Puzzles