White mates in 3

Originally posted by BigDoggProblem
You are correct that black must be deprived of the right to castle, but incorrect about [b]why black cannot castle.

White could have played Nxf8 in the past. We can't prove that Black recaptured, so this won't help us prove Black can't castle.

However, white plays 1.0-0-0! and it's a different story. The rook on d4 could not come from h1 ...[text shortened]... castle.

So, as others have pointed out, white's 1.0-0-0! destroys black's right to castle.[/b]

Wow. Pretty clever.

The solution has already been divulged within the thread, but here is the full solution in one post, for formality:

Eduardo Vasta
StrateGems 1999

#3

Black's last move was either ...g6, or he moved his King and Rook. If it was ...g6, then the Bf8 could not have moved, and therefore Bb8 is promoted. The promoting pawn must have passed through d2 or f2, checking the White King. If Black's last move was ...g6, then White can't castle. However, the opposite is also true - if White can castle, then Black's last move was not ...g6, and he cannot castle.

1.0-0-0!
Destroying Black's right to castle, and threatening 2.axb8Q#.
1...Bxa7 2.Ng7+ Kf8 3.Rd8#

The FIDE Codex (an agreement of problemists to follow certain conditions) states that castling is assumed to be legal unless it can be proved illegal. It also states that in a case where only one side may castle (mutually exclusive castlings), the right to castle belongs to the first side to choose to do so.

What do you think of this article?

http://www.amherst.edu/~philo/dept/faculty_folders/AG/Retroanalysis.pdf

Originally posted by THUDandBLUNDER
What do you think of this article?

http://www.amherst.edu/~philo/dept/faculty_folders/AG/Retroanalysis.pdf

I have read it before (and talked to the author a bit). In my opinion, pRA makes more logical sense, but the Codex conventions allow for better problems to be composed. His article gives 10 or so problems that he deems 'flawed' by reliance on the CCC, but those problems are considered classics!

Originally posted by BigDoggProblem
I have read it before (and talked to the author a bit).

Then you have probably also read a resultant discussion at
http://www.janko.at/Retros/Mails/1995/0021.htm
Be that as it may, hopefully others might find it interesting, too.

Originally posted by THUDandBLUNDER
Then you have probably also read a resultant discussion at
http://www.janko.at/Retros/Mails/1995/0021.htm
Be that as it may, hopefully others might find it interesting, too.

Thanks for the link...I didn't know of that discussion.

It was pretty much as I thought it would go...pragmatism vs. logic, etc.

I really really disagree with the guy who said "To castle just to stop black from castling is not chess". Retrograde problems in general aren't concerned with scenarios that are likely in competitive games. Even for a 'forced' mate, the positions are usually quite different from anything you see in a game. This shouldn't be a consideration because chess problems at best are an art form that is not dependant on chess, the game for validation.

PxB=Q+ Kd7
Qd5 Kc8/Ke8
Ra8#

I know that is wrong.

Originally posted by Black Queen
PxB=Q+ Kd7
Qd5 Kc8/Ke8
Ra8#

I know that is wrong.

Queens don't move like that.

Can you do

Rd1 PxP
PxB#

or
Rd1 BxP
NxP
But I can't continue the second one and the first one would be played by an idiot.

Originally posted by skeeter
Hmmm......Clearly the only way white can mate in two is Rad1.....Rd8++. However this only succeeds if it can be proven that black is unable to castle.

We can see that blacks f-file bishop is off the board and given that the e and g pawns have not been moved it would follow that it had to be captured. But by what? This is pivital and the missing black ...[text shortened]... ing or rook and thereby removing the option of castling. I think thats got it.

skeeter

i dont understand why people are intent of proving on the board that black can't castle.

Surely if someone tells you the parameters or the problem and if there's only one possible solution regardless of other possible variables then that solution must be the correct one.

What i'm saying is that if in setting the problem Thud insists that there is a solution, and the only way that solution can be made is if Black has lost it's right to castle then that solution must be correct.

Rd4 then Rd8

So my solution is that it's not a chess problem, Thud is trying to see how long it is before someone decides to answer the question rather than prove the answer.

lol, sorry. Thats my view.

Originally posted by KingsBishop
i dont understand why people are intent of proving on the board that black can't castle.

Surely if someone tells you the parameters or the problem and if there's only one possible solution regardless of other possible variables then that solution must be the correct one.

What i'm saying is that if in setting the problem Thud insists that there is a ...[text shortened]... decides to answer the question rather than prove the answer.

lol, sorry. Thats my view.

Chess problems must start from positions from legal chess games. Therefore, the history of the game (moves before the diagram) matters. It is too simplistic to say "Black cannot castle because the problem has no solution if he can." In fact, this flawed line of thought actually ruins the problem posted by Thud, because if we simply assume Black can't castle, there are two solutions: 1.Rad1 and 1.0-0-0.

W.Langstaff, Chess Amateur 1922

White mates in 2

And how will such simplistic reasoning solve the problem above?

Ke6 Anything
Rd8#

Originally posted by Black Queen
Ke6 Anything
Rd8#

Now what was Black's last move? If it were 0...g7-g5 then 1. Ke6 is met by 1...0-0, and no mate. While 1.hxg6! 0-0 2.h7# does the trick. On the other hand, if the last move were 0...Rh8 or 0...Ke8, then black cannot castle and 1.Ke6! solves the problem.

Originally posted by Black Queen
Ke6 Anything
Rd8#

After Ke6, what if Black castles?

Originally posted by ilywrin
Now what was Black's last move? If it were 0...g7-g5 then 1. Ke6 is met by 1...0-0, and no mate. While 1.hxg6! 0-0 2.h7# does the trick. On the other hand, if the last move were 0...Rh8 or 0...Ke8, then black cannot castle and 1.Ke6! solves the problem.

Right. This problem is an example of partial retro analysis. We can show White has a mate in two, but can't be certain which of the two lines is valid.