i mean say from my attempt, move the bishop to Bh2 or Bh4. ok. both places, black can't castle. but that doesn't mean Bh4 MUST be a certain win, or Bh2 MUST be a certain win, because we can arrive at that position differently where it DOES matter which place it is.
hehehe, imagine there was a scenario that it had to be Bh4, but then theres a scenario that it has to be Bh2. Then there is no "Must" win if you choose either Bh4 or Bh2, because both have a chance of not winning in 2 =D
So whtat's the answer?
no idea, i dun't wanna bother with this anymore =S
Ya
[i]Suppose Black can castle[/i]. Then the Black Queen's Rook never got out to be captured by a pawn. In particular, the piece captured on c3 was not the Black Queen's Rook. It also was not the Black pawn, because if the pawn were from e7 it would have made two captures and the pawns on f6 and g6 two more, which is two too many, and if it were from any other square even more captures would be involved. Therefore the missing Black pawn promoted.
Continuing from where I left off, proving what must be necessary if Black CAN castle:
The White pawn on e4 comes from e2, otherwise it and the pawns on c3 and g3 would have made two more captures than is possible, and this pawn has always been on the efile, for otherwise it would have made two captures and the pawn on c3 a third, which is one too many, considering that Black's queen rook could not have gotten out to be captures by either of them. Therefore if the promoted Black pawn came from e7 it would have to have made at least one captures and the pawns on f6 and g6 two more, which is one too many. therefore the Black pawn that promoted did not come from e7. So the pawn on f6 comes from e7 and the promoted pawn comes from g7 or h7. Now, the White pawn on g3 cannot come from h2, or it plus the pawn on c3 would have made two captures, both on black squares, which is not possible, since Black's queen's bishop (traveling on white squares) would have to be one of the captured pieces. So the pawn on g3 comes from g2. This means that the black promoted pawn couldn't have come from g7, since it would then have had to have made a captures, the pawn on g6 another (from h7), and the pawn on f6 a third. So the black promoting pawn comes from h7.
Ok, thats about 2/3 of the solution... I think you guys can go from there.
Still a bit confusing.
Yeah, this is a pretty hard one. I've been working on it on and off all weekend, with the help of derUbermensch's partial solutions. keeganomahoney, this is probably a bad puzzle of this type (often called Retrograde Analysis) to start out on. If you go back through derUbermensch's posts, there's several there that are a bit easier. Basically, if there's only one page of comments, it should be an easier one to get. After seeing some of those, you might better appreciate the subtlety that exists in this problem.
Ok, now, on to the solution (which I'm figuring out as I type this, so bear with me):
[i]Suppose Black can castle[/i]. Then the Black Queen's Rook never got out to be captured by a pawn. In particular, the piece captured on c3 was not the Black Queen's Rook. It also was not the Black pawn, because if the pawn were from e7 it would have made two captures and the pawns on f6 and g6 two more, which is two too many, and if it were from any other square even more captures would be involved. Therefore the missing Black pawn promoted.
...
The White pawn on e4 comes from e2, otherwise it and the pawns on c3 and g3 would have made two more captures than is possible, and this pawn has always been on the efile, for otherwise it would have made two captures and the pawn on c3 a third, which is one too many, considering that Black's queen rook could not have gotten out to be captures by either of them. Therefore if the promoted Black pawn came from e7 it would have to have made at least one captures and the pawns on f6 and g6 two more, which is one too many. therefore the Black pawn that promoted did not come from e7. So the pawn on f6 comes from e7 and the promoted pawn comes from g7 or h7. Now, the White pawn on g3 cannot come from h2, or it plus the pawn on c3 would have made two captures, both on black squares, which is not possible, since Black's queen's bishop (traveling on white squares) would have to be one of the captured pieces. So the pawn on g3 comes from g2. This means that the black promoted pawn couldn't have come from g7, since it would then have had to have made a captures, the pawn on g6 another (from h7), and the pawn on f6 a third. So the black promoting pawn comes from h7.
Ok, so to sum up, what we have shown so far is that IF Black can castle, THEN the h7 pawn must have promoted and then been captured on c3 (or replaced a piece that was captured on c3). Now, did this pawn promote on h1 or g1? Now, we know that Black has made at least one capture with a pawn, from e7 to f6, and no other pawn captures (except for possibly one with the h-pawn). White is missing exactly two pieces, the h2 pawn and the light-squared Bishop. Since the light-squared Bishop was not captured on f6 (a dark square), it must have been either the White h-pawn which was captured there or a piece it replaced. If it was the pawn, this requires at least 2 captures, which is one too many as noted above. Thus, the White h-pawn also promoted.
This means that the two h-pawns had to have found a way around each other. The two ways of accomplishing this are having the White pawn move up to h6 and then capturing onto g7 or having the Black pawn move down to h3 and then capturing onto g2. As noted above, having two pawn captures by White onto dark squares is impossible. Thus, Black must have captured onto g2 before the White h-pawn promoted. Any further captures by this pawn is too many, so the black h-pawn promoted onto g1. Since Black can still castle, the White h-pawn promoted by capturing a piece located on g8.
Since we now know that White's f and g pawns were fixed before the promotion of Black's pawn and that the h-pawn was fixed on h2 until Black captured onto g2 when there was no pawn on that square, there is no way that the White Bishop could have found its way to h2.
Thus, IF Black can castle, THEN the h2 Bishop is an illusion. Therefore, IF the h2 Bishop is REAL, then Black can NOT castle and White has a mate in two.
(Side note: If the h4 Bishop is real, this does not guarantee either way that Black can or can not castle. This is why Lord-Chaos was able to get a position (post 37) wherein Black can not castle and yet the Bishop could land on either square. The position where the Bishop can land on h4 can be gotten in two importantly different ways - one in which Black can castle, as proved above, and one in which he can't, as shown by Lord-Chaos. Note the wording in the original problem. It doesn't indicate that White can only have that mate in exactly one of the situations, which I think is where we were all getting tripped up. At least, I know I was.)
ooh i get the puzzle question now. Even though my attempt got to the position, there IS a scenario that say, for example, the bishop is on h4 will mean a win. because in THAT scenario, Bh2 would be a non win =D. So if i move Bh4 or Bh2 in my attempt, yes, White would win, but there still is a chance that we can arrive at that position (say Bh4) and black can castle.
SO, its logical to start from saying that IF black can castle... aah makes more sense now. thanks for the clue. So our scenario gets as close as it can to trying to make black castle.