Christmas problem-solving competition

The festive season is upon us, so let’s celebrate with a fun problem-solving contest with prizes! It’s something of a Christmas tradition for chess problemists to send one another greeting cards diagrammed with their own compositions – often of the unconventional variety. In keeping with that custom, here are three problems I’ve devised that may involve some thinking outside the box. Solve at least one of these positions for the chance to win one of two Redbubble.com gift certificates; more details are found at the end of this blog.

We start with a quirky setup where a white piece and a black one are swapped in the initial array. Yes, it’s an illegal position but hey, it’s Christmas and anything is possible! Other than that, this solves as a standard helpmate-in-2 problem with the duplex condition, i.e. there’s a second solution in which White moves first and is mated by Black.

(Incidentally, this problem came about as a response to a construction challenge – create a sound helpmate-in-2 setting by making a single change to the initial array. The task wasn’t that simple, partly because of the need to eliminate the Fool’s mate, which is effectively the solution of an unsound helpmate-in-2 in the normal array. Hence in the diagram, for the Black-to-move part, 1.f6 e4 2.g5 Qh5 fails only because the black queen no longer blocks d8.)

In the popular Chess960 variant (aka Fischer random chess), the two players start the game with the main pieces randomly placed (and mirrored) on their first ranks. Suppose that the above position arose in such a game, where each player’s bishop-and-knight pairs were originally interchanged. Now apparently there is a mate-in-1 for Black, but is it actually a legal move? In other words, is it really possible that it’s Black to play in the diagram? To solve this one, put on your detective’s hat and work out what could have occurred in the game, regardless of how unlikely any moves may have been. Provide a short explanation or proof for your answer, whether it’s “yes” or “no”.

Joke problems usually employ some kind of unorthodox idea or rule, without specifying what that condition could be. Solving them thus requires some lateral thinking; familiarity with classic examples of the genre could also help as the same rule may be in effect! The only other hint for No.3 is that it’s trickier than it may appear at first. Again, you have to supply a short explanation of why your mate-in-1 solution works.

To enter this competition, send your solutions to peterwong@virtualpieces.net. A correct solution to each problem earns one point, and your total number of points determines how many times you go in the draw for one of the prizes. So cracking all three problems will give you more chances, but you can still win by solving any one of them. The prizes are two Redbubble.com gift certificates, each to the value of 45 AUD. All sorts of merchandise are available from that site, including some with chess designs made by yours truly.

Be sure to mention your full name in your entry and use an email address that would be appropriate for receiving an online gift certificate, should you win. To be fair to all entrants, I won’t comment on the correctness of your answers while the competition is still open. The closing date is Jan. 5, 2020, and good luck!

Christmas problem-solving competition – solutions and results

8 Jan. 2020