1. Ra1 Kxg6 2. Qb1#

On any given street my state can be described by the street name, the direction I’m traveling on it, and the direction of my last turn (left or right). That means that each step determines that one that precedes it: If I’m running west on Primrose Lane having just turned right from Willow Avenue, then on the preceding step I was running on Willow Avenue toward Primrose Lane having just turned left, and so on backward. Since the island contains a finite number of such triples, I must eventually hit a triple for the second time. In most cases this implies a contradiction: Primrose Lane can’t be the first repeated triple, because I could only reach that triple via Willow Avenue, which would make Willow the first repeated triple. The only triple that doesn’t produce such a contradiction is the very first one … where my house is.

It will fit diagonally in a cubical box 1 meter on a side.

1. Rg6 and the queen will mate.

Unless the uptown and downtown trains arrive at the station at equal intervals, the man will find himself favoring one or the other. For example, if the the uptown train arrives at 1:00, 1:10, 1:20, etc., and the downtown train arrives at 1:01, 1:11, 1:21, etc., then he’ll have a 9 times greater chance of taking an uptown train.

See The Elevator Paradox.

Consider the two people separated by the shortest distance. At the signal, these two will shoot one another. Now:

• If another player has also shot at one of these two, then we’ve established that one player was shot twice, and hence that some other player (in the whole group) lacked an antagonist and stayed dry.
• If no other player shot at these two, then we can remove them from consideration and apply the same reasoning to the remaining group: The two closest neighbors fired on one another; if another player fired on them then we have our proof; if not then remove the pair and continue.

Because we’re removing players in pairs, the group under consideration always contains an odd number of players. But it can’t drop all the way to a single player, as he would be firing on no one, and we’ve said that every player fires on someone. Hence at some point we must discover a player who is fired on twice, and thus prove that some other player was not fired on at all.

From the 1987 Canadian Mathematical Olympiad, based on a problem from the 6th All Soviet Union Mathematical Competition in Voronezh, 1966.

(a) implies that both (b) and (f) are right, which is a contradiction, so (a) can’t be right. And (c) claims that (a) is right, so (c) can’t be right either.

We’ve eliminated (a) and (c). That means that if (b) is a correct statement then (d) is accurate. But (b) contradicts (d), so (b) must be false. And since (a), (b), and (c) are all false, then (d) is false as well.

Now we’re down to (e) and (f). (f) can’t be true, because it implies simultaneously that statements (a) through (d) are false and that (e), which agrees with this, is also false, another contradiction.

That leaves (e), whose truth is confirmed by the falsity of statements (a) through (d).

1. Rd8+ and the queen will mate.

Eight. Every bishop move involves a change of rank, so painting each rank a different color will do the trick. Using fewer than eight colors won’t do, as each long diagonal covers eight ranks.

1. After the ridiculous-looking Qc6 Black must succumb to a check down the b-file or a mate at close quarters by the queen.