Black and White

By Joseph A. Potter. White to mate in two moves.

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1. Nc8 Kxc8 2. Qa8#

January 21, 2012January 30, 2013 | Puzzles

Vision Thing

A puzzle from Lewis Carroll:

A king wishes to dismiss his wise men, but he must obey an old law that says there must always be:

Seven blind of both eyes,
Ten blind of one eye,
Five that see with both eyes,
Nine that see with one eye.

How many wise men must he keep?

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Carroll’s answer: Five seeing, and seven blind Give us twelve, in all, we find; But all of these, ’tis very plain, Come into account again. For take notice, it may be true, That those blind of one eye are blind for two; And consider contrariwise, That to see with your eye you may have your eyes; So setting one against the other — For a mathematician no great bother — And working the sum, you will understand That sixteen wise men now trouble the land.

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January 20, 2012January 30, 2013 | Puzzles

Black and White

amusing old timer chess problem

The British Chess Magazine reprinted this “amusing old timer” in 1892. “We suspect that the amusement will be caused by the rapidity of solving it, and the prettiness and surprise of the mating position.” White to mate in two moves.

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1. Qf3+!! Kxf3+ 2. Ne5#

January 14, 2012January 30, 2013 | Puzzles

Pool Party

A billiard ball is resting on a table that measures 10 feet by 5 feet. A player hits it with no “English” and it strikes four different cushions and returns to its starting point. University of Alberta mathematician Murray Klamkin asks: How far did it travel?

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The ball’s path can be drawn as a straight line (green) across a field of mirrored tables. It strikes four cushions and returns to its original position on the table. Its starting and ending positions (the endpoints of the green line) can be connected to a corner of the table by lines that are parallel and equal (yellow). If we complete the parallelogram (blue), it becomes clear that the ball’s path is equal in length to two of the table’s diagonals, or about 22.36 feet.

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January 8, 2012January 30, 2013 | Puzzles

Black and White

loyd retrograde puzzle

By Sam Loyd. White is kicking himself — he has just completed a move, and realizes that he could have given mate instead. What was his move, and what was the mate?

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White has just captured a black bishop on b8 with a pawn on a7, which he promoted to a bishop. If instead he had advanced the pawn to a8 and promoted it to a bishop or a queen, it would have given mate. From Willard Fiske, Chess Tales & Chess Miscellanies, 1912.

January 7, 2012January 30, 2013 | Puzzles

Signs and Portents

In the November 1971 issue of Word Ways, reader Solomon Golomb noted that the following placard appeared in the window of a business establishment “and was considered quite ordinary by all passers-by.” Why?

signs and portents puzzle - 1

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The business was a restaurant in rural Denmark. In Danish, the sign reads:

January 5, 2012January 30, 2013 | Puzzles

Treasure Island

Babylas, Hilary, and Sosthenes have escaped the tower and divided their treasure into three bags. But now they must cross a river, and the boat can accommodate only two men at a time, or one man and a bag. None will trust another with his bag on the shore, but they agree that a man in the boat can be trusted to drop or retrieve a bag at either shore, as he’ll be too busy to tamper with it. How can they cross the river?

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Sosthenes crosses with his bag, leaves it on the farther shore, and returns for Hilary’s bag. He deposits this with his own bag and returns. Now Hilary takes Sosthenes across, drops him on the farther shore, collects his own bag there, and returns. Hilary drops his bag on the near shore, then takes Babylas over. He returns to collect Babylas’ bag for him, and returns again to retrieve his own. By André Sainte-Laguë. Maurice Kraitchik writes, “Perhaps it is easier to be honest.”

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January 4, 2012January 30, 2013 | Puzzles

Down and Out

Three robbers, Babylas, Hilary, and Sosthenes, are stealing a treasure chest from the top of an old tower. Unfortunately, they’ve had to destroy their ladder to avoid pursuit, so they’ll have to descend using a crude tackle — a single pulley and a long rope with a basket at each end.

Babylas weighs 170 pounds, Hilary 100 pounds, Sosthenes 80 pounds, and the treasure 60 pounds. If the difference in weight between the two baskets is greater than 20 pounds then the heavier basket will descend too quickly and injure its occupant (though the treasure chest can withstand this). How can the three of them safely escape the tower with the treasure?

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They send the chest down in an empty basket. Then Sosthenes climbs into the other basket and rides it down, bringing the chest up. They replace the chest with Hilary, who rides down, bringing Sosthenes up. Both get out, and the box descends on its own. Hilary re-enters the basket with the chest and rides up as Babylas descends. Both men get out, and the chest descends on its own again. Sosthenes goes down, bringing up the chest again, and the chest is sent down a fourth time. Hilary descends, bringing up Sosthenes; Sosthenes descends, bringing up the chest; and when Sosthenes has joined his friends the chest descends on its own.

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January 3, 2012January 30, 2013 | Puzzles

Black and White

white good companions chess problem

By Alain White. White to mate in two moves.

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1. Qa1 and Black must succumb: 1. … Qxa1+ 2. c3# 1. … Qa2 2. Qxc1# 1. … Qb2+ 2. Bc3# 1. … Qxb3 2. Qxc1# 1. … Qxc2 2. Qxc1# 1. … Bb2+ 2. c3# 1. … Ba3 2. c3# 1. … Bxd2 2. Qxb1# From Good Companions, March 1920.

December 31, 2011January 30, 2013 | Puzzles

Cleaning House

Upend Congress and pour its members onto Constitution Avenue. Each member has up to three enemies. Prove that it’s possible to pack the 535 members back into the House and Senate in such a way that none of them has more than one enemy in his chamber. (Enmity is always mutual — I am my enemy’s enemy — and the chambers need not retain their former sizes.)

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Fill the House and Senate arbitrarily, and then consider each member in turn. If he has two or more enemies in his current chamber, then move him to the opposite one, where he’ll have no more than one enemy. If we repeat this operation then eventually every member will have at most one enemy in his chamber. Adapted from Svetoslav Savchev and Titu Andreescu, Mathematical Miniatures, 2003. 12/30/2011 UPDATE: Wait, in paraphrasing Savchev and Andreescu’s solution I oversimplified it. (Thanks, John.) Here’s their solution: “Form two chambers in an arbitrary way and denote by e the total number of pairs of enemies who are members of the same chamber. If there is a parliamentarian A who has at least two enemies in his chamber, send him to the other one, where he may have no more than one enemy. So at least two pairs of enemies disappear, while no more than one new pair can be formed. It follows that e decreases at least by 1 after this operation. Repeating the same several times, we will arrive at a situation when e cannot be decreased by more (because it is a nonnegative integer). This means that each parliamentarian will have at most one enemy in his or her chamber.”

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December 27, 2011January 30, 2013 | Puzzles