Originally posted by Mephisto2Easy. If White may play 0-0-0, then Rd4 isn't the original Rh1 (which must have died in the SE corner), but a promoted Rook. The promoted Rook can only exit the 8th rank via d8, f8 or h8, all of which displace Black's K or R.
I don't see how you can prove that black can't castle
I might not have the hang of this yet.
I get:
If black may not caslte and white can not castle, there is 1 solution. (Rad1)
If black may not castle but white may, there are 2 solutions. (Rad1 and 0-0-0)
If black may castle, there is no solution.
Assuming this is supposed to have 1 solution only, the solution must be Rad1.
Where is the flaw in this logic?
Originally posted by jfkjmhAlso - doesn't Ra7 followed by Ra8 do the job (assuming that black can't castle)? I still don't get how to deal with the threat of castling, though!
I might not have the hang of this yet.
I get:
If black may not caslte and white can not castle, there is 1 solution. (Rad1)
If black may not castle but white may, there are 2 solutions. (Rad1 and 0-0-0)
If black may castle, there is no solution.
Assuming this is supposed to have 1 solution only, the solution must be Rad1.
Where is the flaw in this logic?
Originally posted by jfkjmhYour problem is with the case:
I might not have the hang of this yet.
I get:
If black may not caslte and white can not castle, there is 1 solution. (Rad1)
If black may not castle but white may, there are 2 solutions. (Rad1 and 0-0-0)
If black may castle, there is no solution.
Assuming this is supposed to have 1 solution only, the solution must be Rad1.
Where is the flaw in this logic?
"Black may castle".
In this position, you can't disprove this case, unless you are allowed proof A Posteriori, or proof after the fact. From retrograde analysis, we know that only one side may castle. By castling first, White deprives Black of the right to do so.
Not everyone is comfortable with the idea of AP proof, but that is how the logic of this problem is meant to work.
1 edit
Originally posted by Diapason1.Rxa7? does not prove that Black can't castle. 1.0-0-0! does.
Also - doesn't Ra7 followed by Ra8 do the job (assuming that black can't castle)? I still don't get how to deal with the threat of castling, though!
Edit: Perhaps this will help people understand AP proof. Imagine that you are a third-party observer to the game. You just walked in the room, and the players were at the position in the diagram. You are allowed to assume the position is legal. Once White plays 1.0-0-0, you know for certain that he has a forced mate the next move. If he had played 1.Rad1, you still would not know for sure.
Originally posted by BigDoggProblemThank you, I understand now. Is AP common knowledge on "mate in ..."problems or should it be stated explicitly beforehand?
Your problem is with the case:
"Black may castle".
In this position, you can't disprove this case, unless you are allowed proof A Posteriori, or proof after the fact. From retrograde analysis, we know that only one side may castle. By castling first, White deprives Black of the right to do so.
Not everyone is comfortable with the idea of AP proof, but that is how the logic of this problem is meant to work.
1 edit
Originally posted by jfkjmhArticle 16, section 2 of the FIDE codex of composition states:
Thank you, I understand now. Is AP common knowledge on "mate in ..."problems or should it be stated explicitly beforehand?
"(2) In case of mutual dependency of castling rights of each party, the party exercising this right first is entitled to do so."
http://www.saunalahti.fi/~stniekat/pccc/codex.htm
Edit: But it also states:
"Other conventions (which also affect other rights to move) should be expressly stipulated, for example:
(a) If an en-passant capture in the course of the solution has to be legalized by a subsequent castling (for example AP)."
So, the answer is: sometimes it is understood, sometimes it should be expressly stipulated. 😛
Originally posted by jfkjmhno no, rxa7 alsoworks
I might not have the hang of this yet.
I get:
If black may not caslte and white can not castle, there is 1 solution. (Rad1)
If black may not castle but white may, there are 2 solutions. (Rad1 and 0-0-0)
If black may castle, there is no solution.
Assuming this is supposed to have 1 solution only, the solution must be Rad1.
Where is the flaw in this logic?