Christmas problem-solving competition – solutions and results

The Christmas competition announced in my previous blog, presenting three unconventional chess problems to be solved, is now completed. The number of entries was good, but more importantly, readers who took part generally expressed how much they enjoyed tackling the positions. So thanks for all the positive feedback! The three tasks themselves vary considerably in difficulty, with the joke problem being particularly tricky and it was cracked by a minority of entrants only. Perhaps some solvers were lulled by how easy the first problem was, though that was certainly not my intention!

As mentioned previously, two prizes of Redbubble.com gift certificates are offered, and your chances of winning increase with the number of correct solutions submitted. The prize winners drawn are:

Ralf Krätschmer
Dennis Hale

Congratulations!

White mates: 1.Sc6 Se6 2.Sd8 Sxc7. Black mates: 1.e4 Qxc2 2.Ke2 Qxe4. In the first part, the white knight on d8 gets replaced by a black one for a smothered mate, and in the second part, the white king surprisingly steps away from a confined spot. The position is probably a unique one that derives a sound duplex helpmate from the initial array via a single change.

I like this comment from solver Zubrrr: “First one is simple but scenario reminds me of a spy movie: a lone hero wins the battle in villain’s camp.”

In an orthodox game position where all 32 units are present and no pawn has moved, it’s impossible for either side to lose a tempo. That means in such a position, if every knight and rook is on a square of the same colour as it was on originally (or a piece has swapped square-colours with its counterpart), then it must be White’s turn to play. However, in a Chess960 position where the four bishop-knight pairs were initially interchanged, there is just enough room for either queen to triangulate and lose a tempo. For example, 1.Sb3 Se6 2.Sg3 Sf4 3.Kf1 Se6 4.Qe1 Sf4 5.Qc1 Se6 6.Qd1 Sf4 7.Ke1 Se6 8.Sf1 Sf4 9.Sc1. The diagram position is thereby reached with Black to play, so it’s legal for Black to mate with …Sxg2.

The title of this retro problem is a pun, as it refers to not only the randomised pieces of Chess960, but also the shuffling back-and-forth of four pieces as they manoeuvre to help lose a tempo. A pity that no solver mentioned this, but here’s another good one, from Bob Meadley: “In this array there would be too much traitorous talk by the rooks and bishops side by side.”

In the past, the pawn promotion rules apparently did not specify that the new piece must belong to the promoting side. Some well-known joke problems exploit this ambiguity and their solutions involve promoting to a piece of the “wrong” colour. Adopting this idea, the current setting seems to be solved by 1.axb8=black rook – the only way to mate in 1. But such a white move would not be allowed if retro-analysis proves that it’s Black to play in the diagram, and indeed none of the black units could have made the last move, because the king and the bishop are enclosed and the pawns are on their starting rank. So if White had moved last, then it’s Black’s turn and the solution is 1…c6 mate. However, it turns out that Black could have moved last, because the joke promotion rule is in effect and the last move was either …a1=white rook or …bxa1=white rook! Therefore the normal convention of White-to-play applies, and 1.axb8=black rook mate is correct after all. In short, this problem functions as a double-bluff!

A special mention goes to these participants who managed to solve all three problems:

Stefan Felber
Ralf Krätschmer
Andrew Buchanan
George Meldrum

Well done, all!

Christmas problem-solving competition

20 Dec. 2019