Retrograde Analysis Problem Inspired by Various Chess Memes

Question

Standard Retrograde Analysis conventions are used. In addition, the following rules also apply:

• White must begin the game with 1. d4 and 2. Bf4

• If capturing en passant is legal then it is mandatory.

White to play and mate in 4.

Answer 1

Looks interesting, let's see..

Since the white king is in check, the previous move can only have been "black somehow moves a pawn to b5.

There are three available starting squares for the pawn move, b7, b6 and c6.

Out of these, "pawn from b6 to b5" is not possible, because there is no legal previous move for white: the black king would have been under check from the a4 bishop during white's turn, because the bishop cannot have been off the black king's diagonal, and with the black pawn at b6, there are no white pieces that could have moved off the bishop's diagonal to deliver a discovered check. (The white king cannot have been at b5, because on black's previous move the a5 rook would have been giving check to the king)

How about "pawn c6 takes b5" then? Let's do a quick tally: white is missing a bishop, a knight, a queen, and a pawn. Black still has all the pawns, and the b2, e4, f5 and g4 pawns are off their initial files in such a way, that four captures are required to reach those spots. For "pawn c6 takes b5" to be possible, black would have needed to capture two more pieces (one to get to the c file, and another to get back), but there aren't enough captured white pieces, so that move is also out.

So the previous move was definitely "pawn b7 to b5", which allows for a previous white move "rook c6 to a6, discovered check".

White's starting move is then forced by the rules of the puzzle: "pawn c5 takes b5 en passant, check".

This starting move has a huge threat behind it; unless black takes the bishop at a4 (other legal moves are "rook to b5", "pawn to c6", "king to d8" and "king to f8") white can always play 2. h4 to force black to capture en passant, after which rook to g8 is instant checkmate.

After black takes the bishop at a4, white can just take the rook with the b3 pawn, which stops any further checks, and adds another threat of "rook a6 takes a8, checkmate". Black has no move that would protect against that and h4 simultaneously, so black's strongest resistance is

 1. cxb5 e.p Rxa4+
 2. bxa4     Rb8
 3. h4       gxh4 e.p. (forced by the extra rules of the puzzle)
 4. Rg8#

But what about

2. - 0-0-0!?!?,

I hear you say? Well, let's take a look.

Black is missing two bishops and a knight, one of which was taken by the white pawn at h7, and another by the pawn at d3. (White's d-pawn begun the game with d4).

With only one further capture available, the only way for the white pawns to reach their current files is by the d-pawn still being on the board, at either c5 or e6, both options accounting for the final available capture.

This means that white's a-pawn was one of the four pieces that the black pawns took to reach their current files, but that capture cannot have occurred on the a file, because that wouldn't help the black pawns. Therefore, white must have promoted their a-pawn at a8 (with no further captures available, the pawn cannot have left its file before the promotion), so the rook at a8 has moved, and castling isn't allowed.

This one was pretty tricky, I hope I didn't miss anything very important. :-)

Answer 2

I had come up with half of Bass' solution but did miss the possibility that Black could have castled. However, I think that one does not have to prove that the previous move was b7-b5.

Indeed, if Black's previous move was something else than b7-b5, then Bxb5+ is an allowed winning move. Indeed,

1. Bxb5+ - Rxb5
2. Rxa8#

or

1. Bxb5+ - c6
2. h4...

or

1. Bxb5+ - Kf8
2. h4...

or

1. Bxb5+ - Kd8
2. h4 - gxh4
3. Rg8 - Qe8
4. Rxe8#

So, combining the part of Bass' solution that explain what to do if the previous move was b7-b5, an excluded middle argument, and the lemma I just proved, we get a proof that the problem has a solution.