Futility Closet

A group of men, all versed in logic, sits around a table. Fifteen of them are wearing red hats, the rest blue. Each can see the others' hats but not his own. On the table is a clock which strikes once each hour. The men are given the following instructions: 'You are not allowed to discuss the color of your hats. However, should any of you find that he is wearing a red hat, he should leave the table on the clock-strike immediately following his discovery.' Now it is assumed that no one is initially aware of the color of his own hat. Furthermore, since the men cannot see the color of their own hats, nor discuss it with their colleagues, nothing happens for a while. Then a guest arrives. She looks at the hats around the table, and says, clearly: 'At least one man here is wearing a red hat!' What happened and why?

That's from Edward Barbeau's Mathematical Fallacies, Flaws, and Flimflam (2000). "An induction argument can be used to show that once the guest has made her announcement all individuals with red hats will eventually leave the table," Barbeau writes. "This contradicts one's feeling that nothing should continue to happen; after all, apparently the guest has not provided any new information."

What's the argument? Barbeau gives no answer, but that won't stop me from torturing you with the question.

An island is a body of land surrounded by water, and a lake is a body of water surrounded by land.

Now suppose the northern hemisphere were all land, and the southern hemisphere water. Is one an island, or is the other a lake?

Solution to The Pup Tent Problem:

The test makers were expecting the answer (C), but several students pointed out that in fact (A) is correct — the shapes fit together to produce a surprisingly simple object with only five faces.

This is easiest to see if you imagine two square pyramids side by side: The tetrahedron fits precisely between them to make a smooth-sided "pup tent":

If you have polyhedral dice, try fitting the face of a 4-sided to an 8-sided die; you'll find that the adjoining sides are coplanar.

In 1980 the Educational Testing Service offered this question on an aptitude test:

In pyramids ABCD and EFGHI shown above, all faces except base FGHI are equilateral triangles of equal size. If face ABC were placed on face EFG so that the vertices of the triangles coincide, how many exposed faces would the resulting solid have?

(A) Five (B) Six (C) Seven (D) Eight (E) Nine

Which is correct?

(Answer)

Solution to A Curious Conversation:

It's the ace of diamonds.

If Val couldn't name the card knowing only its value, then it must be an ace, a queen, a 5, or a 4, as these are the only values that appear more than once.

If Colin knew that Val couldn't name it, it must be red, as all the red cards have values (A, Q, 5, 4) that appear more than once.

When Val understands that the card must be red, she can exclude from her list of possibilities the queen, the 5, and the 4, as each of those cards appears in a black suit.

That leaves only the ace of diamonds.

Adapted from Julian Havil's Impossible?, 2008.

UPDATE: Wait, this is cooked. Havil gives the problem but not the solution, and my solution above isn't valid. Thanks, Ben and Andrew, for pointing it out. If anyone can find a valid solution, let me know and I'll post it here. I don't think one is possible.

FWIW, this is based on a more famous "impossible puzzle," which is soluble but spectacularly hard.

UPDATE 2: In Canada or England, does the word "colour" mean "suit"? If so, the problem is soluble. It seems to have appeared originally on page 14 of a paper [PDF] by University of Victoria philosopher Audrey Yap. The first two steps above are right — Val can deduce that the card is red and must be A, Q, 5, or 4. The fact that she's able to say "I know the card now" at this point means it can't be an ace (because that value appears in both red suits). And the fact that Colin can then say "I know it too" means that it must be the five of diamonds. So, from the top:

Val says, "I don't know what the card is." (She knows it's a 5, but there are two 5s.)

Colin says, "I knew that you didn't know." (He knows it's a diamond, and thus an A or 5; but as each of these values appears more than once in the deck, he knows that Val won't have enough information yet to identify the suit.)

Val says, "I know the card now." (Colin's remark has just told her that the card is red, because all the red cards in this deck have values that appear in multiple suits. And the only red 5 is 5♦.)

Colin says, "I know it too." (He knows that Val was looking for an A or a 5, and that learning the card's color was helpful. Thus she can't have been looking for an A, as both aces are red. That leaves 5♦.)

So, if "colour" means "suit," the answer is 5♦. If "colour" means "color," then I still think it's insoluble. Thanks to the many smart people who wrote in about this, especially Helge.

You're standing with your friends Val and Colin when a stranger approaches and shows you 16 cards:

A♥ Q♥ 4♥
J♠ 8♠ 7♠ 4♠ 3♠ 2♠
K♣ Q♣ 6♣ 5♣ 4♣
A♦ 5♦

He shuffles the cards, selects one, and tells Val the card's value and Colin the card's color. Then he asks, "Do you know which card I have?"

Val says, "I don't know what the card is."

Colin says, "I knew that you didn't know."

Val says, "I know the card now."

Colin says, "I know it too."

What is the card?

(Solution)

Solution to Don't Even Try:

White: To mate, 1. Rb3#. To self-mate, 1. bxc8, promoting to a black king, pinning the pieces on b8 and c7 and leaving Black with only 1. … Kxc3#.

Black: To mate, 1. … Kxc3#. To self-mate, 1. … Nxd5+, forcing 2. Nxd5#.

"Notwithstanding the memo, many young solvers (and old ones too) have had a hard wrestle with this production, but all to no purpose, it was the first of its kind published. The consequence was that many communications were received by the composer doubting his sanity, denouncing the problem as a fraud and threatening that if they caught him on the other side of the channel they would give him a taste of their wrath. The receiver of these communications thought that the channel was a mighty fine affair!"

From Thomas B. Rowland, Chess Fruits, 1884.

White or Black to play and mate or self-mate in one move. That is, you must find a total of four moves from this position: a White move that mates Black instantly, a White move that forces Black to mate White instantly, and equivalent moves for Black.

"Memo: The above puzzle depends on a literal interpretation of the rule which provides that a Pawn on reaching the eighth square may become any piece irrespective of colour."

WARNING: "This monstrosity is the production of an erratic solver who has been sorely tried, puzzled and perplexed all the year round by the many posers and problems which have appeared from time to time in the numerous Chess columns. His aesthetic patience, resignation, fortitude, culture and hope all at once breaking down, he set to work and with wrathful spirit, regardless of all problem construction, devised it more for the sake of retaliation and revenge than to give pleasure. To prove his spiteful character; when composing it, he was overheard repeating, 'Since I cannot prove a lover to entertain these fair spoken days, I am determined to prove a villain.' Consequently, gentle reader, we warn you not to attempt it, except indeed that you are the happy possessor of that knowledge wherein you are able to puzzle others. It may look beastly simple, but to any young solver who may be foolhardy enough to venture it we offer a few words of advice–carefully study the above memo and note that–but 'hold enough,' no more can we divulge, fearful of bringing the fiery wrath of the exasperated composer upon our devoted heads."

(Solution)

Solution to Hopping List:

Our life is but a pilgrimage,
From birth unto the bier,
We journey onward as the Knight,
Our pathway never clear.
O'er chequered course of sixty-four,
Through ways that lead astray,
Now bright, then dark, we travel on,
Till darkness veils our day.

This literary knight's tour appeared originally in the Sussex Chess Magazine.

Start on d4, "Our", and jump from square to square in the manner of a chess knight to assemble an eight-line verse. Like a chess knight's tour, the correct solution visits every square on the board.

(Answer)