By Francis Healey. White to mate in two moves.
You’re standing in a room with an uneven floor. Before you is a square table with four legs. The table wobbles, but by turning it gradually you manage to find a position in which all four feet are supported, eliminating the wobble (though now the tabletop isn’t level).
You wonder: Is this always possible? Assuming that the four legs are of equal length and that the surface of the floor varies smoothly, is it always possible to position a four-legged table so that all four legs are supported?
Three positions, “left,” “middle,” and “right,” are marked on a table. Three cards, an ace, a king, and a queen, lie face up in some or all three of the positions. If more than one card occupies a given position then only the top card is visible, and a hidden card is completely hidden; that is, if only two cards are visible then you don’t know which of them conceals the missing card.
Your goal is to have the cards stacked in the left position with the ace on top, the king in the middle, and the queen on the bottom. To do this you can move one card at a time from the top of one stack to the top of another stack (which may be empty).
The problem is that you have no short-term memory, so you must design an algorithm that tells you what to do based only on what is currently visible. You can’t recall what you’ve done in the past, and you can’t count moves. An observer will tell you when you’ve succeeded. Can you devise a policy that will meet the goal in a bounded number of steps, regardless of the initial position?
“It’s tricky to design an algorithm that makes progress, avoids cycling, and doesn’t do something stupid when it’s about to win,” wrote Dartmouth mathematician Peter Winkler in sharing this puzzle in his book Mathematical Puzzles: A Connoisseur’s Collection (2003). It’s called “The Conway Immobilizer” because it originated with legendary Princeton mathematician John H. Conway and because it’s said to have immobilized one solver in his chair for six hours.
A man presents himself as the The Amazing Sand Counter. He claims that if you put some quantity of sand into a bucket, he will know at a glance how many grains there are, but he won’t tell you the number. Can you devise a test that can verify this ability without telling you anything that you don’t already know? You can ask the Sand Counter to leave the room or turn away, for example, and you can ask him questions. How can you convince yourself that he knows how many grains of sand are in the bucket when he won’t actually tell you the number?
A mother is 21 years older than her son. Six years from now, she will be five times his age. Where’s the father?
I won’t give the answer to this one — if you do the math, you’ll know precisely where he is.
From Stephen Barr’s Experiments in Topology (1989) via Miodrag Petkovic’s Mathematics and Chess (1997):
This apartment contains eight rooms, each measuring 9 square meters, except for the top one, which measures 18 square meters. You have enough red paint to cover 27 square meters, enough yellow paint to cover 27 square meters, enough green paint to cover 18 square meters, and enough blue paint to cover 9 square meters. Can you paint the eight floors in four colors so that each room neighbors rooms of the other three colors?
By William Anthony Shinkman. White to mate in two moves.
Can two dice be weighted so that the probability of each of the numbers 2, 3, …, 12 is the same?
A bisecting arc is one that bisects the area of a given region. “What is the shortest bisecting arc of a circle?” Murray Klamkin asked D.J. Newman. Newman supposed that it was a diameter. “What is the shortest bisecting arc of a square?” Newman answered that it was an altitude through the center. Finally Klamkin asked, “And what is the shortest bisecting arc of an equilateral triangle?”
“By this time, Newman had suspected that I was setting him up (and I was) and almost was going to say the angle bisector,” Klamkin writes. “But he hesitated and said let me consider a chord parallel to the base and since this turns out to be shorter than an angle bisector, he gave this as his answer.”
Was he right?