Illegal Position Contest!

It seems I have refuted my own reasoning. However, it cannot be done shorter than 75 moves after hxg4: the white bishop (b4) and king are optimally placed.

This seems illegal to me. To avoid a collision on the a1-h8 diagonal, the white b-pawn has to reach f7 (after four captures on dark squares) before the g7 pawn starts making its captures to reach b2. So the black f8 bishop and both black rooks will be trapped on the back rank, and Black only has three pieces for the white b-pawn to capture.

Actually both rooks can come out if bBf8 was captured at home, but there will be lack of captures for wPh5.

You're right, my proof was incomplete and should have noted that the captures by the white b-pawn and e-pawn account for all of White's captures, so no black piece can be captured at home.

I didn’t think about the rooks before. I thought bBf8 could not get captured by any of the wPs (as wPh5 captures on light squares), but the rooks also could not get captured unless we get rid of bBf8.

It could be a proof game problem if the white e-pawn only got to g4.

This position, made by me, requires a lot of analysis to prove its illegality.

Black has made three captures by the bfg pawns and one capture of the white a-rook on the first or second rank, for a total of four captures, but White is only missing three men.

Don't forget a common trick in retros - cross-captures.

Can you show me how cross-captures would keep the three remaining black pawns on the king side from needing to make at least two captures?

All bPs are on the right files: just bPh4 should be on h5, but that is illegal

You were probably thinking cross-captures on the kingside black pawns, but it is the queenside white pawns which cross-capture.

Now I understand the cross-capture you meant. Assuming the white rook couldn't get out invalidated my proof attempt.

Here's another try at an illegality proof.

If the white a-rook never leaves the first or second rank, Black lacks enough pawn captures. Otherwise, White must cross-capture to release the rook. Since no knights have been captured, the first black piece captured must be the f8 bishop. (The c7 pawn can't move to let the black queen out, since c6 will be needed for the b8 knight so it can return to b8 after the b7 pawn captures to let the a8 rook out.) So Black must play Pg7-g6 to release the f8 bishop, after which all three of Black's pawn captures will be on light squares, and the white queen bishop can't be captured but is no longer on the board, making the position illegal.

Well done for getting there in the end! I would have written much more analysis, but your analysis is sufficient. This position shows that pawns are very restrictive pieces and can freeze even knights in place.

@n9531l1 have you tried this one (at page 325)?

I think I might have glanced at it and thought it was too obvious, but that was probably wrong. Here's an attempt at an illegality proof.

An original white rook can't reach the 8th rank, so both white rooks promoted before the white king reached the 8th rank. If the white knight is original, it reached a8 before the b7 pawn moved, but then there is no square for a black rook where the white king can be effectively shielded from it on the way to the 8th rank, so the white knight is also promoted. The three promotions, one on a8, require five black pieces to be captured by white pawns. Those, along with the capture of the f8 bishop at home, leave Black with no knight left to make tempo moves until it can be captured at f8 on White's last move.

That is correct! A 5th capture is indeed necessary for the promotions because only two of the three promoted pieces can be the a or c pawn, where the third would be a different pawn and make unoptimal captures.