Andrelious fixed: Mate in one

Black to move and mate in one.

To move 1. ... xe3, white must have played e2-e4 last move.

Last move was -e4 because of the bishop that checks on c6, only e3-e4 and e2-e4 remain. Pawn on e3 is not possible because it would check the black king.

Originally posted by Thomaster
To move 1. ... xe3, white must have played e2-e4 last move.

Last move was -e4 because of the bishop that checks on c6, only e3-e4 and e2-e4 remain. Pawn on e3 is not possible because it would check the black king.

so its either BxP++
or PxP ep++ isn't it?

Qh4++

Originally posted by iamatiger
so its either BxP++
or PxP ep++ isn't it?

Ahh, good point!

Needs more fixing then. I'll get right on it.

Originally posted by SwissGambit
[fen]8/pP6/pnb5/p7/p2pPk2/7p/PPPP1p1P/4n1bK w - - 0 1[/fen]

Black to move and mate in one.

Black to mate in 1

Try this one. Should be sound now.

Now I'm not sure that we can prove white's last move was e2-e4
The other possibilities are: g3xh4; f3xe4; e2xd3; and d3

Can we rule them all out?

Originally posted by iamatiger
Now I'm not sure that we can prove white's last move was e2-e4
The other possibilities are: g3xh4; f3xe4; e2xd3; and d3

Can we rule them all out?

A pawn on g3 would have checked the black king, so it certainly wasn't there last move.
f3xe4 and e2xd3 are not possible either: what piece must the pawn have captured? (black's missing three pieces, a knight, a rook, and a queen. Two of those are needed in order to get the pawn on h4 and g5, which leaves only one piece to be captured.)

All seven white pieces have been taken to get the black pawns on their place (places?). -d3 has been played earlier in the game, because the white black bishop had to go out (as well as the white a-rook).

Originally posted by Thomaster
A pawn on g3 would have checked the black king, so it certainly wasn't there last move.
f3xe4 and e2xd3 are not possible either: what piece must the pawn have captured? (black's missing three pieces, a knight, a rook, and a queen. Two of those are needed in order to get the pawn on h4 and g5, which leaves only one piece to be captured.)

All seven whit ...[text shortened]... rlier in the game, because the white black bishop had to go out (as well as the white a-rook).

Right you are. ... good analysis.

Originally posted by Thomaster
A pawn on g3 would have checked the black king, so it certainly wasn't there last move.
f3xe4 and e2xd3 are not possible either: what piece must the pawn have captured? (black's missing three pieces, a knight, a rook, and a queen. Two of those are needed in order to get the pawn on h4 and g5, which leaves only one piece to be captured.)

All seven whit ...[text shortened]... rlier in the game, because the white black bishop had to go out (as well as the white a-rook).

Nice logic! I didn't see all that.

Thank you!

I think I'm getting better and better at these puzzles. 🙂

Originally posted by SwissGambit
[fen]8/pb4p1/p7/p5P1/p2pPkrP/3P3p/PPP2p1P/4n1bK b - - 0 1[/fen]
Black to mate in 1

Try this one. Should be sound now.

Lot of superfluous pieces on the board. You should pare it down to minimum needed for mate. Like the rook, not needed, nor the 4 pawns in a row.

Originally posted by sonhouse
Lot of superfluous pieces on the board. You should pare it down to minimum needed for mate. Like the rook, not needed, nor the 4 pawns in a row.

Without those pieces, you can't prove that en passant is legal and thus can't mate in one. They're not superfluous.