Mate in 1? (hard)

A retro just in time for April Fool's, too. I'm always trying this sort of evil "joke".

(12+11) "White to move". "Mate in 1".

I hope it's more fun to solve than it was to compose. Really. Pawns hurt my brain.

castle  

...And giving just the answer "castle" with a smiley is WRONG. I hope.

Are you sure white can castle in this position? Prove it, by giving the score of a legal game that results in this position where white can still castle.

Mate for White or Black? Because white cannot avoid the double threat of Qxe2# and Qxf2#. I am not sure whether you want a mate in one turn or one move though...

Oh, and there are two additional mating lines for black: 1. Nf1 with either Bxf2# or Ng2# and 1. Rf1 with Ng2#.

Well, let's assume that there exists at least a game which results in this position with white still having the right to castle. If that's true, there's also at least where the same position is reached but white doesn't have the right to castle. So given just the end position, one can't discriminate between the two cases. It's not up to the solver to prove a castle case exists, the puzzle starting position should state which one it is. :P

And if the goal is mate for White, clearly you have a very clever April (Chess) Fool's joke haha.

Ultimately, I agree with Griever3216.

This is a retro problem. By convention, "Castling is permitted unless it is proved not permissible." (Codex of Chess Composition, article 16.1, Rio de Janeiro 2009).

As such, the burden of proof is on the solver to establish that white can possibly castle, by providing a game history where it is permitted (which by convention proves that white can castle, regardless of the others that disallow castling.)

And if it is not permissible, explain completely why there is no possible game leading to the position with white maintaining castling rights.

Ah, see, I didn't know what a retro problem was. I stand corrected.

I don’t think there is a mate in one for white because white can’t castle in this position. I probably missed something though.

Analysis:

To get to c1, the black king had to go through b1 or b2, so b3 had to be played already.  Before b3 was played, the two rooks, the knight, and the bishop already had to be in the a1 corner, which is impossible unless the black knight on a1 was promoted.

The pawn on b6 captured one white piece to get to c7- only two other white pieces are missing, but the knight needs three captures to get to the d file from the a file.

The only way to get another white piece to capture is to promote the h pawn.  This is possible if it captures the black g pawn, but then there are not enough captures for the a7 pawn to go from the d file to the a file.  It needs three captures, but there are only two from black’s a7 pawn and the original second knight. 

There is no way to promote another black piece to capture without losing castling rights, since the rook can’t move from h1 and there are no extra white pieces for the black h pawn to capture to move onto the g file.

Very well reasoned! What you wrote is thus far correct, but not the whole story. Two further important points to address:

- What if the black Pb6 came from a7, and the promoted knight was from the bPc7? (The answer is very similar to what you already wrote.)

- What if the black Bb1, instead of the black knight, was promoted? (Hint: this potentially involves the white a-pawn and b-pawn not being what they seem.)

You're almost there.

Thanks for the interesting problem by the way.

Here is the rest, as far as I can figure out:

If the black Pb6 came from a7, and the promoted knight was from the bPc7:

This requires one less capture for black to promote the knight, so there is no need to promote a white piece on the g file.  This allows a black piece promotion if the black h-pawn captures the white g pawn and promotes, then the white h pawn captures the black g pawn to reach g4.  Now there are two black pieces for the white d pawn to capture, but it still can’t reach a7 because there is no pawn on the a file. 

If the black Bb1, instead of the black knight, was promoted:

The only way for the Bb1 to be promoted is if the king entered before the rooks.  This means the b pawn must have captured on a3, the configuration in the bottom left was created, and then the a pawn captured on b2.  This bishop can be promoted because it only requires two captures from the d file, leaving the third missing white piece to be captured by the c7 pawn. 

There needs to be two black pieces for the a and b pawns to capture- one is the original LS bishop, one can be promoted on the g file (as discussed above).  This leaves no black pieces for the white d pawn to capture to reach the b file, from where it could capture the black a pawn.

This approach of promoting the black Bb1 still wouldn’t work if the Pb6 came from a7 and the promoted pawn came from the c file.  This case would require one less capture for black to promote the bishop, leaving an extra white piece for no purpose and giving the Pd2 one less available capture.

Well done. The reasoning is once again correct. Just about all the important branches have been covered, barring small changes to the capturing schemes that don't actually reduce the number of captures. Therefore, white cannot castle, and 1. 0-0#?? is illegal.

To put the cherry on top, black mates in 1 regardless of white's next move with 1...Qxe2#/Qxf2#, as joepi96 pointed out.

SOLVED

Very nice problem.

Please prove that this position is possible with a game where it happened

Looks like the site lagged out on me I wanted to post my attempt at this. I tried really hard.  

Who cares about where it happened? As long as the game is legal, no matter however unsound and with so many stupid mistakes from both sides, the position is still legal.

@justus_jep: Still far from the desired position. Looks like you attempt to keep White's castling right, but well, still a lot of things to shuffle in.

amazing/....