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# Christmas problem-solving competition – solutions and results

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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. While the helpmate was fairly easy, the Chess960 problem was deceptive and its correct response rate was 70%. The proof game was more complex, but everyone seemed comfortable with the genre and the irregular moves it entailed.

As mentioned previously, three equal prizes are offered – the choice of a Chess.com Diamond membership or a Redbubble.com gift certificate – and your chances of winning increase with the number of correct solutions submitted. The prize winners drawn are:

1e41-O
beugi19
sakksok_k

Congratulations!

1. Miraculous setup

White mates: 1.Nc6 Ne6 2.Nd8 Nxc7. 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.”

2. Shuffling pieces

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.Nb3 Ne6 2.Ng3 Nf4 3.Kf1 Ne6 4.Qe1 Nf4 5.Qc1 Ne6 6.Qd1 Nf4 7.Ke1 Ne6 8.Nf1 Nf4 9.Nc1. The diagram position is thereby reached with Black to play, so it’s legal for Black to mate with …Nxg2.

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.”

3. Two impostors

Since White is missing only the d-pawn and it couldn’t have reached f6 directly to be captured, it must have promoted – either to get to f6 itself or to replace another piece that was sacrificed there. The former option would take too long as …gxf6 must be played early to let out Black’s king-side pieces in time. What was the sacrificed piece that the pawn replaced then? A queen cannot be captured on f6 until the 4th move, which is still too slow, and though the knight from b1 can reach f6 in 3 moves, a promoted knight cannot return to b1 quickly enough. The bishop from c1 is thus the only suitable piece. 1.d4 Nc6 2.Bg5 Rb8 3.Bf6 gxf6 4.d5 Bh6 5.d6 Bf4 6.dxc7 Nh6 7.cxb8=B Rg8 8.Bxa7 Rg5 9.Be3 Ra5 10.Bc1 Ra8. To facilitate the promotion, Black has to sacrifice the original queen-side rook on b8. Hence the two impostors in the diagram are the white bishop on c1 and the black rook on a8.

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

1e41-O
mkkuhner
vinniethepooh
Zubrrr
beugi19
sakksok_k
chessrush123
Morcanie12
anselan
supesusik
BigDoggProblem
n9531l1

Well done, all!

Christmas problem-solving competition

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Peter Wong
Sydney