Deductive puzzle

White to move. This position came from a standard chess game with the board properly oriented.

a. Can it be proved that White has a mate in one?

b. Is it possible for White to have mate in one, depending on what has happened in the game?

c. What is White's fastest mate if he does not have mate in one?

This is my composition, so I can't guarantee that it's correct, but I hope it is.

Here's what I see:

Assuming that black's last move was f7-f5, then white has a mate by en passant - pf6#. It does appear that black's last move was to f5, but I don't think it is absolutely necessary that it came from f7; that pawn could have been on f6, the white bishop then moved to h7+, followed by pf6-f7. So my initial answers to questions #1 and #2 are "No" and "Yes".

Question #3 is a bit more challenging.  It looks like there ought to be a King move (such as Kf1)that leads to mate in 2, but that pesky bishop at a1 can capture at b2 and stall things to three moves.

a. No

b. Yes, if white hasn't moved his king or h1-rook then castling gives checkmate in one.

c. If white cannot castle he has mate in three: 1.Kf1 Bxb2 2.Kg2+ Bc1 3.Qxc1#.

Can white castle do deliver mate?

I'll just tell you now that Question 3 was a throwaway. Kf1 is mate in three, and it's fastest if there's no mate in one.

Questions 1 and 2 are the important part, and if either of you got them both right (which I'm not guaranteeing) you're missing some analysis. You haven't explained a whole lot of things that need to be explained.

I'm no good at these

I'll put my reply in white:

0-0 is# is not possible. Why? White is moving down the board, else there is no way for the Black Bishop to get to the bottom left corner.

So, I see at least a few mates in one, incl e8(R) or (Q)#, Rf8#, Bxc4#

Is that it?

Remember, I said in my first post that the board is properly oriented. Your point about the Bishop is important in the final solution, though.

Oh, ok, I missed that : (.

So the bishop is a promoted piece.

It had to come from a2. There is another pawn on a2.

I dont see how either pawn can get there without at least 2 captures each, 4 in total, but only 3Whit pieces missing ? Hmmmm....

I think it can be proven that White does not have a mate in 1. First of all, O-O# is impossible because White's king must have moved to allow the Black king into White's camp via the kingside. Secondly, the Black f5 pawn must have come from f6, so exf6# is impossible. If the pawn had been on f7, there is no way that the White bishop could have reached e8.

Now we're getting somewhere. You still have a ways to go, though. Cobra91, why can't White's g-pawn capture over to f6 and e7 and promote to a Bishop? Also, do you know the answer to the problem rooperi has run into?

If black's a2-pawn and a1-bishop were originally his a- and c-pawns then only two captures are necessary to get them both on the a-file as long as white's a-pawn moves out of the way. That means that white's b2-pawn must have been his a-pawn and his a4-pawn must have been his c-pawn, requiring three captures.

I assume you mean that the b3-pawn must be the a-pawn, not the b2-pawn. I'll accept that... for now, although we may want to revisit it later. The biggest thing that needs explaining right now is why the e8-Bishop can't have been the result of an underpromotion of a pawn that captures over.

Sorry, b3, yes.

Now, if the e8-bishop were a promoted pawn then it would have to have been promoted on e8 or g8.

If e8, then it and white's e5-pawn would have required three captures at least.

If g8, then black's f-pawn would have had to move to let it past and therefore the en-passant capture would not now be possible.

Is it really impossible to promote on e8? Black only has nine pieces on the board, leaving seven off...

Hmmm... true, that only makes six captures so far.

Going back to white's a- and c-pawns, if the a-pawn captured to b3 first to allow black's a-pawn past then there would be no way for the c-pawn to get to a4 and if the c-pawn captured first the a-pawn would not have been able to get out of the way. Therefore, the pawn on a4 must be white's a-pawn, having captured first to b3 to let black's pawns past and then back to a4. Then the c-pawn captures to b3.

 There is only one way to account for Black's a1 bishop. Three captures were made by the White pawns: axb3 (after which Black's a-pawn dashed to underpromotion), followed by bxa4 and cxb3. Black's c-pawn would then need two captures to reach the a-file. However, it could not have captured White's dark-square bishop (since the bishop couldn't have escaped from its home square), so that just leaves the White knight and a promoted White pawn. Promotion of White's h-pawn would require another capture, and White's f-pawn would need yet another capture to get to the e-file. That's five total captures, and only seven Black pieces are missing. Since Black's light-square bishop was captured on its home square (it could not have escaped from c8), that leaves only one capture for the White g-pawn. It cannot reach e7 with only one capture, so this refutes the possibility of Black's f-pawn moving directly from f7 to f5. Instead, the g-pawn advanced to g6 (before the Black g-pawn had moved), then captured on h7 and promoted to a queen on h8. White's h-pawn promoted to a light-square bishop on g8 afterwards. Therefore, it's a mate in 3, not mate in 1.

Exactly! Congratulations, everybody. You've found the answers:

a. No

b. No

c. Mate in three with Kf1

You have way too much spare time, lol

Ah, a proof game! Nice. I thought about trying to make one to assure myself the puzzle was correct, but I never got around to it. I was getting sick of this position anyway by the time I actually got it to work.