On Monday I left you with this four part problem, composed by Marjan Kovacevic (Problem 1981). 

White mates in two moves here by playing 1. Kg1!, wanting to avoid checks while not blocking the bishop on h2. Black can only prevent Rf8# by castling, allowing Qxc7#

For the second part you replace the knight on a3 with a black bishop. This time, for very much the same reasons, 1. Ke1! is the solution.

Now you replace the bishop on a3 with a black rook. Now 1. Ke3! is the only way to mate in 2: 1. Kg1? would be met by Ra1!, pinning the rook on f1.

Finally, you replace the rook on a3 with a black queen. This time Kg1, Ke1 and Ke3 all fail to queen checks. Running Stockfish in ChessBase tells me there’s no solution. But it’s wrong: we can perform some simple retro-analysis.

The black bishop on b3 must be a promoted piece because the b7 and d7 pawns haven’t moved. So the original c8 bishop must have been captured at home. (If O-O-O is legal, this must have been by a white knight.) Black has seven pawns on the board, along with a promoted bishop, so the queen on a3 can’t be a promoted piece. It could only have been developed horizontally by the black king moving out of the way. Therefore, as we can prove the black king has moved, O-O-O is illegal. So White can mate in 2 by playing 1. Kg3!. Black has no checks and cannot castle so has no way of preventing Rf8#.

You’ll also notice that the four White solutions: Kg1, Ke1, Ke3 and Kg3, form a rather pretty star. I’ve shown you a couple of star flight problems in the past. Here, it’s the four white solutions rather than four black defences that make the star.

I do hope you enjoyed this: thanks to all our WhatsApp group members who commented.