To solve it, the first thing is to look at the two kings. At least we know that one of them must be black. Suppose it's the one on d6. Then what about the rook on d8 and queen on c6? They can't both be black, or the white king on c8 would be in an impossible double-check. And they can't both be white, or the black king would face a similar impossibility. Yet they can't be one black and one white without putting both kings in check.
So it must be the white king on d6 and the black on c8. Now, back to that queen and rook again. As before, they can't be different colours, and they can't both be black without another impossible double check. The sole remaining possibility is that they're both white, which leaves the black king in a double check that can only have been achieved if White's last move was a pawn capture from c7 to d8, promoting to a rook.
We can now deduce that the knight and the rook on f6 must both be white, or both sides would be in check. The pawn on b7 must be black (or it would be delivering a third check to the black king), which means that the bishop on a8 has to be white, and theresult of an earlier pawn promotion. And that pawn could only have promoted if the black pawn had first vacated a7. So the pawn now on a7 must be white.
There remains only the question of the last move. We know it was a capture, c7xd8, but what was taken? Whatever it was must have just moved there, since there is no way Black's king could have made the previous move, and a quick examination of the possibilities shows that only a knight could have moved to d8 without White's king having been left in check.
So that's the complete answer: the king on c8 and pawn on b7 are black, all the other pieces are white, and the last move was White's c7 (pawn) x d8 (knight).
William HartstonReuse content