Throughput

A bedeviling little puzzle from Dickinson College mathematician David Richeson.

A topologist, it is said, is someone who can’t tell his donut from his coffee mug.

April 10, 2021 | Puzzles · Science & Math

Double Duty

On Nov. 5, 1996, Election Day in the United States, the New York Times crossword puzzle carried a surprising clue:

39. Lead story in tomorrow’s newspaper (!), with 43A

43 across turned out to be ELECTED, but 39 across might be either CLINTON or BOBDOLE — both possibilities had seven letters. Was the Times venturing to guess the outcome of the day’s election?

No. Composer Jeremiah Farrell had contrived each of the seven down clues to admit of two possible answers, so that no matter which candidate won, the newspaper might claim a “correct” result.

Crossword editor Will Shortz called Farrell’s ambiguous effort his favorite puzzle of all time.

(Thanks, Andrew.)

April 7, 2021April 6, 2021 | Puzzles

Moving Day

https://commons.wikimedia.org/wiki/File:All_5_free_tetrominoes.svg — Image: Wikimedia Commons

Will these 5 tetrominoes fit into a 5 × 4 rectangle?

	Select Click for Answer
No. When colored like a checkerboard, the rectangle has an equal number of light and dark squares. So does each of the tetrominoes … with one exception. So the task is impossible.

April 5, 2021April 4, 2021 | Puzzles

Black and White

baird chess problem

A problem “for juvenile solvers” by Edith Baird, from Womanhood, May 1899. White to mate in two moves.

	Select Click for Answer
1. Bg5! leads to two pretty mates: 1. … Kf5 2. Rc5# 2. … Kd6 2. Bf4#

March 30, 2021March 29, 2021 | Puzzles

A Confusing Country

Philosopher Nelson Goodman published this puzzle anonymously in the Boston Post in 1931, at age 24. He later called it “by far the most popular and widely circulated of all my writings.”

All the men of a certain country are either nobles or hunters, and no one is both a noble and a hunter. The male inhabitants are so nearly alike that it is difficult to tell them apart, but there is one difference: nobles never lie, and hunters never tell the truth.

Three of the men meet one day and Ahmed, the first, says something. He says either, ‘I am a noble’, or ‘I am a hunter.’ (We don’t know yet which he said.)

Ali, the second man, heard what Ahmed said, and in reply to a query, answered, ‘Ahmed said, “I am a hunter”.’ Then Ali went on to say, ‘Azab is a hunter.’

Azab was the third man. He said, ‘Ahmed is a noble.’

Now the problem is, which is each? How do you know?

	Select Click for Answer
No one in this country can ever claim to be a hunter — a noble who did so would be lying, and a hunter who did so would be telling the truth. This means that Ali must have been lying when he attributed that claim to Ahmed — which tells us that Ali is a hunter. And if Ali always lies, then his statement that Azab is a hunter is false, which tells us that Azab is a noble. That means that we can trust Azab when he tells us that Ahmed is a noble. (From Goodman’s Problems and Projects, 1972.)

Click for Answer

March 23, 2021March 22, 2021 | Puzzles

Bright Idea

A jailer will send each of a group of n prisoners alone into a certain room. Each prisoner will visit the room infinitely often, but the order of the visits will be determined arbitrarily by the jailer. The prisoners can confer in advance, but once the visits have commenced they can communicate with one another only by means of a light in the room, which they can turn on or off. How can they ensure that some prisoner will eventually be able to determine that everyone has visited the room?

	Select Click for Answer
The prisoners agree that one of them (say, Alice) will try repeatedly to turn the light on, and each of the others will turn it off twice. That is, if Alice finds the light off she’ll turn it on, and if she finds it on she’ll leave it on. Each of the others will turn it off the first two times he finds it on and otherwise leave its state unchanged. Alice counts the number of times she finds the room dark after her first visit. When she’s returned to find the room dark 2n – 3 times, she can announce that everyone has visited the room. Each time Alice returns to the room and finds it dark, that’s a sign that one of the other n – 1 prisoners has visited it. If a prisoner exists who still hasn’t visited the room, then the light can’t have been turned off more than 2(n – 2) = 2n – 4 times. So once Alice’s total exceeds that, reaching 2n – 3, she knows that everyone’s visited the room. (Note that the light’s initial state doesn’t matter: If the first visitor isn’t Alice and he finds the light on, he’ll turn it off, and that sign of his presence won’t be evident to her. But she’ll have to turn the light on again before he can perform his second off-turning, and that ensures that she’ll register his second visit.) (From Peter Winkler’s excellent Mathematical Puzzles: A Connoisseur’s Collection, 2003.)

Click for Answer

March 16, 2021March 15, 2021 | Puzzles

Black and White

steinwig chess problem

A simple but pretty “lightweight” problem by R. Steinweg. White to mate in three moves.

	Select Click for Answer
1. Kc6 Ka7 2. c8=R! Ka6 3. Ra8# From Arthur Ford Mackenzie, Chess Lyrics: A Collection of Chess Problems, 1905.

March 10, 2021March 9, 2021 | Puzzles

“Piccadilly Underground Station”

This unusual puzzle by G.A. Roberts appeared in the January 1941 issue of Eureka, the journal of recreational mathematics published at Cambridge University. It concerns the Piccadilly Circus station of the London Underground, which lies on the Piccadilly line between Green Park and Leicester Square and on the Bakerloo line between Charing Cross and Oxford Circus.

At a given time there are on the platform, escalators and subways, and in the trains, 128 people, all of whom travel by train, and none of whom return immediately by the way they have come.

Those who have come via Leicester Square are equal in number to those who are about to travel via Leicester Square.

The number of people who arrived by Bakerloo Line is equal to the number who intend to leave by the Piccadilly Line.

The number of people who are travelling from the street to stations on the Piccadilly Line is equal to six-thirteenths of the number who change from the Piccadilly Line to the Bakerloo.

The number who arrive from Green Park and then change to the Bakerloo is equal to the number who are about to travel via Green Park.

The number who are travelling from the street to the Bakerloo is equal to four times the number who arrive in Piccadilly trains but do not use the Bakerloo Line, and of these, twice as many come from Green Park as from Leicester Square.

By how many does the number of people who use the Bakerloo Line exceed that of those who do not?

	Select Click for Answer
The solution appears in the May 1941 issue: “Of the 128 people at Piccadilly, 104 either had used or were about to use the Bakerloo Line. Hence the answer required is 80.” (The Eureka archive is published under a Creative Commons license.)

February 26, 2021February 26, 2021 | Puzzles

Black and White

A chess problem from the Jamaica Gleaner. There’s no black king on this board. Find a place for him such that White can mate him in two moves.

	Select Click for Answer
Put him on h5 and White can mate in two starting with 1. Qb3. From Arthur Ford Mackenzie, Chess: Its Poetry and Its Prose, 1887.

February 20, 2021February 19, 2021 | Puzzles

Route Cause

A problem submitted by France and shortlisted for the 17th International Mathematical Olympiad, Burgas-Sofia, Bulgaria, 1975:

A lake has six ports. Is it possible to arrange a series of routes that satisfy the following conditions?

Each route must include exactly three ports.
No two routes may contain the same three ports.
Any tourist who wants to visit two different arbitrary ports has a choice of exactly two routes.

	Select Click for Answer
Yes. The thing to notice is that we can’t have three routes in the pattern (A,B,C), (A,B,D), (A,C,D), because in that case it’s impossible to fulfill the third condition: If the remaining ports are E and F, then we’ve left ourselves only one permissible route (not two) that visits A and E. That’s (A, E, F). Any other three-port route that includes A and E would “overload” another segment. This means that if two of the routes are (A,B,C) and (A,B,D), then the second one that covers A,C must be (without loss of generality) (A,C,E). With that in mind the other routes can be worked out uniquely: (A,D,F), (A,E,F), (B,D,E), (B,E,F), (B,C,F), (C,D,E), (C,D,F).

Click for Answer

February 11, 2021February 10, 2021 | Puzzles