Mate in 4 (Petrovic)

I am looking at d7 +, Kxp, d8 (kn) + the only two forcing moves but don't see a real continuation after that.

Originally posted by sonhouse
I am looking at d7 +, Kxp, d8 (kn) + the only two forcing moves but don't see a real continuation after that.

There obviously isn't one. After 2...Ka8, White can't even give check, even if he were allowed two moves in a row unanswered.

Originally posted by BigDoggProblem
There obviously isn't one. After 2...Ka8, White can't even give check, even if he were allowed two moves in a row unanswered.

Well thats what I am working on, I gave two checks in a row but thats half my allotment of moves! So it has to be mate in two from the point where I promoted to a knight. It looks like my own pieces get in the way.....

Originally posted by sonhouse
Well thats what I am working on, I gave two checks in a row but thats half my allotment of moves! So it has to be mate in two from the point where I promoted to a knight. It looks like my own pieces get in the way.....

I didn't find any either. Did you read the part about ep and o-o-o? I think that's the only way to mate in 4.

Well, if en passant is legal in the position, it's just a fairly simple mate in 2, so I'm going to assume that it is not. It would be a pretty nice looking mate in 2, though - after dxc! black has a large number of possible moves, but they all lose to d7++.

Originally posted by OrangeKing
Well, if en passant is legal in the position, it's just a fairly simple mate in 2

In chess problems, en passant is assumed to be illegal, unless it can be proven legal.

Castling is legal, unless it can be proven illegal.

Legality of castling takes precedence over illegality of en passant.

Given those conventions, your statement is false.

I was just wondering why David113's first solution wasnt right?

1. dc6 d3
2. O-O-O d2
3. Rd2 Bxe6
4.d7#

I dont see any way out of it. can someone tell me why this is wrong?

Originally posted by hypothetical
I was just wondering why David113's first solution wasnt right?

1. dc6 d3
2. O-O-O d2
3. Rd2 Bxe6
4.d7#

I dont see any way out of it. can someone tell me why this is wrong?

It's not wrong. However, a full solution requires a good explanation of the retro analysis.

Originally posted by BigDoggProblem
Given those conventions, your statement is false.

Actually, nothing you said makes my statement false. 🙂

I simply said that if en passant were legal in the position (which you have already stated it is not), this position has a rather elegant mate in two. Had I stated that this position is a mate in two, and that en passant is legal, you'd be absolutely correct that my statement is false, but I didn't.

Actually, all I was trying to say is that having the composition presented by a mate-in-four is, in and of itself, proof that en passant isn't legal here, even if you knew nothing of the conventions you've mentioned.

Edit: Providing, of course, that we assume the composition is presented correctly, which I do. 🙂

Originally posted by OrangeKing
Actually, nothing you said makes my statement false. 🙂

I simply said that if en passant were legal in the position (which you have already stated it is not), this position has a rather elegant mate in two. Had I stated that this position is a mate in two, and that en passant is legal, you'd be absolutely correct that my sta ...[text shortened]... Providing, of course, that we assume the composition is presented correctly, which I do. 🙂

Originally posted by OrangeKing
Actually, nothing you said makes my statement false. 🙂

Not true. To understand why, a careful review of the history of the position is required. What might the last few moves have been? How did the pawns form such a strange structure? Can White castle?

I simply said that if en passant were legal in the position (which you have already stated it is not)

I stated no such thing. I said that it is assumed illegal unless otherwise provable.

Actually, all I was trying to say is that having the composition presented by a mate-in-four is, in and of itself, proof that en passant isn't legal here, even if you knew nothing of the conventions you've mentioned.

That statement is also false. The solution claims the opposite of your proof.

Can someone please explain why this is wrong. I'm sure it is, but can't see why:

Assuming en-passant is legal:

1. d5c6xc5 ***
2. d7 Checkmate


(where *** denotes absolutely any legal move Black can make, and my apologies for not knowing the correct way to denote en-passant capture).

This then makes mate in two, so surely can't be right, so surely en-passant is not inolved?

Originally posted by Zeddicus
Can someone please explain why this is wrong. I'm sure it is, but can't see why:

Assuming en-passant is legal:

1. d5c6xc5 ***
2. d7 Checkmate


(where *** denotes absolutely any legal move Black can make, and my apologies for not knowing the correct way to denote en-passant capture).

This then makes mate in [b]two, so surely can't be right, so surely en-passant is not inolved?[/b]

You can't assume en passant is legal. You have to prove it.

Originally posted by BigDoggProblem
You can't assume en passant is legal. You have to prove it.

Ok, fair enough. But I'm still confused about the mate in 2 being possible when we're asked for a mate in 4. It's almost like a trick question, or is it standard for chess problems to actually mean "n moves or less"??

Originally posted by Zeddicus
Ok, fair enough. But I'm still confused about the mate in 2 being possible when we're asked for a mate in 4. It's almost like a trick question, or is it standard for chess problems to actually mean "n moves or less"??

That's the thing though. You can't correctly claim that a mate in 2 is possible. You can't even consider en passant as your first move until you first establish a way to prove that en passant is legal.

Mate in 4 means Mate in 4. There should be no shorter solutions.

Here is the retro-analysis of the position.

White is missing QN = 2 units
Black is missing QNNRR = 5 units

The White pawns must have captured 5 times to achieve their current structure. One possible scheme: d6-g6 could have come from the e-h files respectively, and b7 came from the c-file. This uses up all of White's possible captures.

Black has no doubled pawns, but we must explain how the White pawns got behind them. The trouble lies in the doubled White d-pawns. One of them must be the original d-pawn, because there are not enough possible captures to allow both of them to leave the file. Black will have to make two captures: one to clear the d-file (...dxc or ...dxe), and another to put a pawn on d4, behind the two wP's (...cxd4 or ...exd4). This uses up Black's two possible captures.

What were the last moves? Since White currently has the move, Black must have moved last. The only piece that can retract a move is the c-pawn (-1...Kc7?? and White can't retract the check by Pd6). Did it come from c6 or c7? If -1...c6-c5 was the last move, White could only have moved the King or Rook before that move - Castling is therefore illegal in the diagrammed position. BUT - if -1...c7-c5 was the last move, White could have played -2.c6xb7 before that, and castling is legal in the diagrammed position (we assume it is, because it can no longer be proven illegal). This brings us to the key realization of the problem:

En passant is only legal if White can castle.

The conventions say castling is assumed legal, and en passant is assumed illegal. So which shall it be? In this problem, the two assumptions contradict. One must take precedence....