Deductive Puzzle #12

From my last diagram it shows d4-pawn (final e6 pawn) cannot be f-pawn, as then white needs at least 6 captures - black is missing 5 pieces/pawns.

So final e6 pawn is a d-pawn

amiraz,

your diagram is missing the e6-pawn

put it in, then we need to do the piece-counts and capture counts in both our positions...

Yeah, but where's the e6 pawn?

Isn't black missing four pieces and two pawns?

 I am talking about the last diagram I posted, not the original one.

From MY last diagram, black's black-sq bishop is a promoted one (original dark-sq bishop was captured in situ), and must be promotion of a-pawn (current g4-pawn must be d-pawn, current d3-pawn must be c-pawn).

Therefore black's pawns must have made 6 captures, whereas in MY diagram only 5 white's pieces are missing ???

I must be missing something...

dsarkar - your position is not possible, since the black bishop at d8 has no way to reach that position; it would be blocked in by the pawns at e8 and g8.

I think amiraz has figured out the only possible final move sequence.

EDIT: Although he needs to add a pawn to his diagram . . .

 black's d8 bishop is not original - it was promoted from a-pawn - can you please explain why it is impossible?

Amiraz's position (your edited version) is good no doubt.

dsarkar -

You are correct, my apologies. So we have a promoted black bishop AND a promoted white Knight?

And that durn board editor messed up ;

its the d pawn

I think it's the h-pawn

g4 has to be the e-pawn, right?  Because if amiraz' diagram is correct only 4 white pieces were taken by pawns.  For g4 to be the d-pawn would require 3 captures to get to g4, plus 1 for the pawn to get to d3 and at least 1 for the a-pawn to promote.  And dsarkar's diagram shows the pawn on e7.  So either way it's the d-pawn and a promoted black bishop.

We have two options for what happened last White move; there could have been an en passant capture onto e6 or a third Knight could have moved to g6 from e5.  Note that White is missing 5 pieces, but one (the light-squared Bishop) had to have been captured on its home square by a non-pawn.

If there was an en passant capture as White's last move, then the Black Bishop on d8 is a promoted one because the original had to have been captured on its home square by a Knight.  This means that Black's pieces that were captured by White are 1 pawn, both Bishops, Queen, and both Knights.  This allows 5 captures by White's pawns (since one of the Bishops was captured by a Knight).

Consider the case of dxe6 e.p., then the Black d-pawn is currently on g5, having made 3 captures; the  Black c-pawn is on d3, having made one capture, and the Black a-pawn has made at least one capture in order to be promoted.  This is a total of 5 or more captures by pawns.  However, as I noted in the first paragraph, only 4 are available.  Thus, if an en passant capture was White's last move, then it was not made with the d-pawn.

Consider the case of fxe6 e.p..  Recall that one of Black's missing pawns was promoted to the dark-squared Bishop.  Also, the e-pawn was captured after the d-pawn made it to g5.  Looking at White's pawns, we note that the White pawns made captures on a3, g3, e6 (ep), b3 or b5, and a 4, 5, or 6.  What pieces could these be?  Black is missing 6 pieces, so it seems possible.  However, the dark-squared Bishop was captured on its home square and the a-pawn promoted and was not captured.  This leaves the two Knights, a Queen, and a light-squared Bishop to be captured on e6 (by the d-pawn pre- en passant, to get the Black d-pawn to g5, since the f-pawn is not available to be captured in this case), g3, a3, b-file, and a-file (each for the pawn currently on a6).  However, this is 4 pieces over 5 squares, and so is one capture too many.  Thus, an en passant capture was not made by White with the f-pawn as his last move.

This leaves N[e5]g6+ as the only possible last White move.  This means that at least 1 White pawn promoted to a Knight.  Whether it was the promoted Knight that was lost on g6 or another does not matter.  In order to minimize the number of captures, we require the 3 pawns on the a-file to come from a2, b2, and c2 originally.  Which pawn, then, is the one on e6?  It is either the d-pawn or the f-pawn.  Whichever pawn it is, the other one must have promoted.  In either case, the f-pawn has a specific role which does not include capturing and staying on g3.

Thus, the pawn currently sitting at g3 originally came from h2.

omnipaul, impeccable logic! Hats off to you!

Quite masterful :)