Mate in 2
Logic
Level
2
It is White's move. What must white play in order to mate in two?
Clarification: Black plays optimally and Black can't castle.
This problem is the first one of the set Mate .
Qxd7+
Ng7+
Rxf8+
Bg7
Qd6
Bxe7
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Proof that black can't castle:
Assume black is able to castle. Also a piece is accounted for if it on the board or we know what captured it.
White's double h pawn means that white's two pawns must've captured 4 pieces/pawns. None of these could have been black's pawns (try to prove this). Therefore four of blacks pieces were captured by these pawns. Now these cannot be the bishop on c8 so all of black's pieces not including this bishop were captured. Now the black pawn currently on g2 must've come from c7. Therefore it captured 4 white pieces, none of which were pawns (try to prove this). However, white already has 4 pieces - which means that one of them (out of the queen, rook, bishop, knight and the other captured piece) came from a promotion. This must've been either the a or b pawn. However all of blacks pieces are accounted for and the only unaccounted pawn is the a pawn. Therefore whites a pawn must've captured blacks promoted a pawn on b8. How did black promote? He must've captured whites b pawn with his a pawn before promoting. Therefore all of whites pieces are accounted for which means that black captured whites light-squared bishop with his pawn currently on g2.
What was blacks last move? He cannot have moved the g2 pawn otherwise it can't've captured white's light-squared bishop. If black moved his queen then it came from g7 so white's last move must've been capturing on g8 with the rook. However all of blacks pieces are accounted for so that's not possible. Therefore blacks last move was either with his king or rook, a contradiction.
(Proof that it is possible)
I would say this is more retrograde chess than mate. The real hard part is proving black can't castle, not finding the mate in two even though it's a surprising move.