If n is even, Bob can win by imagining the board covered with dominos, each domino covering two adjacent squares. Now whenever Alice marks a square, he simply marks the square that shares its domino; this strategy cannot fail (even if Alice is free to choose any square on the board, not just orthogonally adjacent ones!).

If n is odd, then there must be one square on the board that’s not covered by a domino, and Alice can win by imagining a tiling in which that square falls in a corner and making her first move there.

If Alice must begin with a square adjacent to a corner on a board with odd n, then Bob can win. Imagine that the board has the standard checkerboard coloring and that the corners are black, so Alice is making her first move by choosing a white square. There’s a tiling in which the uncovered square is black, and Bob can now win by completing these dominos. Alice can never select the uncovered square herself because all the squares that she chooses are white.

(From the 12th All Soviet Union Mathematical Competition, Tashkent, 1978, via Peter Winkler’s Mathematical Puzzles: A Connoisseur’s Collection, 2003.)

“Remove the two cigarettes forming the letter L in the original arrangement, and replace them in the way shown in our illustration. We have the square root of 1 minus 1 (that is 1 less 1), which is clearly 0.”

If that’s too obscure, “we can remove the same two cigarettes and, by placing one against the V and the other against the second I, form the word NIL, or nothing.”

From his Modern Puzzles and How to Solve Them, 1926.

10/09/2023 UPDATE: Another solution, by reader Drake Thomas:

(Thanks, Drake!)

The key is that the problem contains “no statement superfluous to its solution.” The correct solution must make use of every statement given; any solution that doesn’t do this is wrong.

So, for example, Statement 1 suggests that either Y.Y. or ‘Critic’ has seen Caliban in a green tie. If this weren’t the case, then the statement would be useless and any resulting solution would be wrong. This also tells us that Low cannot choose third, because if he did then Statement 1 again would be needless.

From Statement 2 we can infer that Y.Y. was not in Oxford in 1920 — if he had been, then this statement would be useless. Similarly, if no one has lent Caliban an umbrella then the second statement is not needed, so we know that someone has lent Caliban an umbrella.

Very well, who lent it to him? If Low lent it, then Statement 2 informs us that Low will not choose first. Statement 1 has already told us that Low will not choose last, so if Low lent Caliban the umbrella then Low will choose second. But that can’t be — Statement 3 can be of value only if Y.Y. or ‘Critic’ chooses second. Therefore Low didn’t lend Caliban an umbrella.

Is it possible that both ‘Critic’ and Y.Y. have lent Caliban an umbrella? No: In that case Statement 2 tells us that Low must choose first, and Statement 3 tells us that ‘Critic’ chooses second and Low third, so Statement 1 is not needed. So that possibility can’t be right; Caliban has received an umbrella from either ‘Critic’ or Y.Y., but not both. Similarly, if both Y.Y. and ‘Critic’ have seen Caliban in a green tie, then Statement 1 tells us that Low will choose first, but that would make Statement 2 useless. So either ‘Critic’ or Y.Y. has seen Caliban in a green tie, but not both.

Okay, suppose that Y.Y. both saw Caliban in a green tie and lent him an umbrella. In that case Statement 1 would tell us that Y.Y. cannot choose first, and if that’s the case then Statement 2 is not needed. Hence if Y.Y. saw Caliban in a green tie then he can’t have lent him an umbrella, which would mean that ‘Critic’ lent Caliban an umbrella. By the same logic, if ‘Critic’ saw Caliban in a green tie, Y.Y. must have lent Caliban an umbrella.

In both of those latter cases, Low must choose first (Statement 1 tells us that Low doesn’t choose last, and Statement 3 is useful only if Y.Y. or ‘Critic’ chooses second). And if that’s the case, then Statement 3 tells us that ‘Critic’ must come second and Y.Y. third. So the proper order is Low, ‘Critic’, Y.Y.

(The fanciful names refer to colleagues of Phillips at New Statesman.)

(Via Alex Bellos, Can You Solve My Problems?, 2017.)

Here’s the answer given in the journal:

No explanation is given. I had imagined something like a curved disk, but that wouldn’t give a circular plan view.

The nearest similar puzzle I know is the “architect’s puzzle,” in which you’re asked to imagine a single object that can cast three different shadows: a circle, a square, and a triangle. That’s possible, but it’s a distinctly different task, and it’s hard to see how to adapt that object into the one we’re seeking here.

09/27/2023 UPDATE: I’ve received a ton of mail about this puzzle — a million thanks to everyone who’s contributed; I’m constantly impressed by your intelligence, imagination, and resourcefulness. The consensus is that the solution was published upside down! And the puzzle as presented is a bit ambiguous. Reader Catalin Voinescu explains:

Note there are no additional lines to the semicircle that is the front elevation: that’s all there is to see from that direction. The plan view could be filled or not, and we can’t tell. So the figure must be the intersection of the bottom half of a horizontal cylinder surface (like a half-round gutter) and either a vertical cylinder surface, like a gutter downpipe, or a solid vertical cylinder. If you think of what a T joint of two cylinders of the same diameter looks like from the side, the joints appear as straight lines that bisect the angle between the cylinders. (They must be straight lines bisecting the angle for reasons of symmetry.)

So the given answer is one of two possibilities.

When the plan view is hollow, the figure is a closed curve in three dimensions: draw an ellipse with a long diameter times the short diameter, and bend the paper at 90 degrees along the short diameter. The top view is a circle, the side view a semicircle, the other side view is the angle made by the paper.

However, if the plan view was solid, the figure would be a curved surface: what you need to cut out of the half-round gutter to get the vertical cylinder to go through it. Viewed from the side, it would look like a triangle — like the given solution, but with a line at the bottom.

That second solution looks like this:

Here’s an interactive plot of the first solution, and here’s the second. Click and drag these figures to examine them. In both cases the side elevation is an inverted V, so the published solution is upside down (and I’m not crazy!).

My thanks to Drake Thomas and to Catalin for their work on all this, and thanks again to everyone else who wrote in about it. I’ve actually gone through the subsequent issues of the journal to see whether an erratum notice was ever published, and I don’t find one. I wonder how much mail this generated in 1958!

UNDERGROUND.

Merkuns never speak first, so the first alien is either a Kween or an Ozdak. Suppose it’s a Kween. That means the first statement is true and the second alien is a Merkun, which means that the second statement is false and thus the third alien is a Merkun. But this would require the third statement to be true (since we’re supposing that the third alien is a Merkun and the second statement is false), and that can’t be the case, as we know that a Merkun won’t speak first in a group. So the first alien is an Ozdak, and thus the second is either a Kween or an Ozdak. If it’s a Kween then the third is an Ozdak (the third can’t be a Kween), and if it’s an Ozdak then the third is a Merkun. The latter can’t be the case, because a Merkun would speak the truth after a lie is uttered, and we know that the third statement can’t be true. So the first and third aliens are Ozdaks and the second is a Kween.

A cannibal and a photographer cross the river together. Then the photographer rows back alone.

Now two cannibals cross and one of them returns.

Next two photographers cross and one of them returns along with a cannibal.

Now two photographers cross and one cannibal returns.

Finally two cannibals cross, and one of them returns to pick up the last cannibal.

I think this was originally a Mensa puzzle. I found it in James F. Fixx’s Solve It!, from 1978.

List all the possible rectangles in the set in order of increasing area:

{1 × 1, 1 × 2, 1 × 3, 1 × 4, 2 × 2, 1 × 5, 1 × 6, 2 × 3, 1 × 7, 1 × 8, 2 × 4, …}

These have areas, respectively, of {1, 2, 3, 4, 4, 5, 6, 6, 7, 8, 8, …}. If no two of the 10 smaller rectangles are congruent, then their minimum total area would be 1 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + 7 + 8 = 46, which is greater than that of the larger rectangle in which they must fit (5 × 9 = 45). So some duplication is unavoidable.

From Quantum, via Ross Honsberger’s Mathematical Delights, 2019.