Deductive Puzzle #9

White misses (a rook), a (promoted) pawn and a darksq. bishop.

Black misses (Q rook), Queen and dark sq. bishop.

Blacks B-pawn captured a piece.

This means that g-pawn captured 1 piece to promote, because the pawn promoted on a white square.

That means that White's G and H pawn also captured a piece so Blacks G-pawn could go through the G or H file.

So if White's E-pawn promoted, it must have taken two black pieces to go through the G file. Which brings the total pieces white captured to four.

But black is only missing three or two pieces so the e-pawn didn't promote.

So the captured piece on a6 must be white's rook.

Therefore it must be blacks Q rook that's standing on a5!

Correct me if I'm wrong.

This one looks a bit easier.

Sequence, again is key.

Black is missing three pieces – one is a pawn that promoted to light-squared bishop. The other two are Queen and Rook.

White is missing three pieces – one is e-pawn, Rook, and dark-squared Bishop.

So what did the b-pawn take in moving to a6? Could not have been the e-pawn, so it must have been either the Rook, the dark-squared Bishop, or the e-pawn (promoted to something). Can’t have been the dark-squared Bishop – that’s needed to get the second black light-squared Bishop. That has to come from the g-pawn capturing onto h2. So that leaves the Rook or the promoted e-pawn.

So it appears that the a5 rook could be white, if we can promote the e-pawn. In order to do that, the e-pawn has to make two captures – which means the black Queen and the black dark-squared bishop are needed to promote the e-pawn. But the e-pawn promotes on either h8 or c8, and would have to promote to either N, B, or Q, and then get to a6 to be captured. Looks possible.

Can the black Rook be at a5? Yes, as MBickley has shown.

So – if the White rook can be there, the puzzle is “cooked” again . . . BUT – to promote, the e-pawn needs two captures – the black Queen and the black king-side bishop. But that means the black Queen has to get out, and that means pc6 must be played. Now the black queen-side bishop has no way to get to a2, and has to stay at b7. And since the white queenside Bishop has to move to h2 to be captured there, pb3 is necessary, and there’s no way to get the promoted bishop to a2.

So MBickley’s analysis of how it came about is correct, and we have excluded the possibility that it was a White Rook.

Daxelson, does that mean that I was wrong?

 Well, you've got the right answer. But I don't think you excluded the possibility that the white e-pawn promoted and travelled over to be captured on a6. And I don't think this assertion is true: "That means that White's G and H pawn also captured a piece so Blacks G-pawn could go through the G or H file." MBickley showed how the g-pawn could promote.

Basically, MBickley's analysis is correct - it shows the solution.  But it does not exclude other possibilities - which is what I was trying to do in my answer (#22 above).

Still not perfect, but I've brushed up the solution for blacks queen rook at a5 that I had before.  Too lazy to correct it any further.

And by the way, I have read the question correctly, a white Ra5 is impossible. Figuring out why was a real treat, and kudos for daxelson for being the first to prove that it can't be whites rook on a5.  I held off on it =), didn't want to spoil everything, but it's quite obviously impossible as soon as you attempt to promote the g pawn.  Castling as the last move had more of a point then the OP made obvious, sneaky...

You have have the answer, but still lacking in the complete proof behind it...

A hint, there may be more than one way to arrive at the position... concentrate on which Bishop was promoted!

derUbermensches, you're clueless.  I referanced another post for the proof, I never intended to prove why white can't put his rook on a5 himself.  I knew why, but I didn't want to spoil the problem.  But since daxelson *has* proved it, the cats out of the bag, and I will try to describe the proof to you.

A: to get the position with the pawn on a6 white needs to sac his rook (as in my solution), or a piece promoted on the g file

B: To Promote a piece on the g file, one must sacrifice two pieces to bring a pawn over there (you cannot use the g pawn, as it must promote to create the second bishop).  But since the queen can't get out on the kingside until the king castles, and since bxa6 is not possible yet since a piece has not yet been promoted on the g file/we are trying not to use the rook, it must get out through c6.  A few moves later, an extra piece has been promoted on the g file!

C: this piece MUST be used on a6 since white has no other pieces to sacrifice, so this may as well be done right away

D: now black must promote his pawn, using the trick shown in my above diagram.  Since white cannot use the rook (as it's theroretically going to end up on a5), he must use the bishop

E: we weave the bishop over to a2, the pawn captures it, makes a bishop... but what's this?  the bishop can't get to a2 now!  Therefore, it is impossible for whites rook to end up on a5 and the diagrammed position to be achieved.

No, it does make sense, I just didn't want to be too specific. I think your proof is valid, to be sure, it simply doesn't follow the path of the book, so I can't be 100% certain. Mainly due to E, I hesitate to say you have completely solved it, as you seem to be sure that the a2 bishop is black's promoted one, or at least I read it that way. 

i dunno