<Comment deleted by user>
@pkill yeah pretty much (though I thought g2 pawn moved one square and f2 pawn went on killing streak and h7 did hxg2 from h3)
edit: thanks
edit: thanks
<Comment deleted by user>
<Comment deleted by user>
I hope it is ok to write my solution here ... I did not read all the posts, not wanting to get spoiled. So I probably repeat only what others have written.
DON'T READ ON IF YOU WANT TO TRY YOURSELF!
In this position White gives check, so it must be Black's turn to move. How could white give check with the rook, that could not have been moved there? Some obstacle on c7 must have vanished. Only solution: A pawn on c7 captured on d8 with underpromotion to a rook. And the captured piece must be a knight, because the white king was not in check, so neither rook nor queen, and the black dark square bishop never moved.
So Black must have promoted to a knight, and the only pawn that could have done this was the h-pawn, capturing twice and promotion on f1. Now let's count black captures: a6, e6, d5, c4, and somewhere on g and f line. So all white pieces except one rook have been captured (Q R N N B B), because Black never captured a pawn: still 7 white pawns on the board, and one was used for promotion.
So now, no spare white pieces. Which means on h4 there must be a black piece. Queen or rook would give check, black square bishop never moved because it was blocked by g7 and e7, white square bishop can't be on a dark square anyway, two knights on the board, one just captured -> impossible. Where is the flaw???
Ok, there must be one remaining white piece. The h-pawn only captured once, promoting via h3, hxg2, g1=N. And now we see the solution: All squares where white pieces have been captured are light. White's dark square bishop must be on the board.
DON'T READ ON IF YOU WANT TO TRY YOURSELF!
In this position White gives check, so it must be Black's turn to move. How could white give check with the rook, that could not have been moved there? Some obstacle on c7 must have vanished. Only solution: A pawn on c7 captured on d8 with underpromotion to a rook. And the captured piece must be a knight, because the white king was not in check, so neither rook nor queen, and the black dark square bishop never moved.
So Black must have promoted to a knight, and the only pawn that could have done this was the h-pawn, capturing twice and promotion on f1. Now let's count black captures: a6, e6, d5, c4, and somewhere on g and f line. So all white pieces except one rook have been captured (Q R N N B B), because Black never captured a pawn: still 7 white pawns on the board, and one was used for promotion.
So now, no spare white pieces. Which means on h4 there must be a black piece. Queen or rook would give check, black square bishop never moved because it was blocked by g7 and e7, white square bishop can't be on a dark square anyway, two knights on the board, one just captured -> impossible. Where is the flaw???
Ok, there must be one remaining white piece. The h-pawn only captured once, promoting via h3, hxg2, g1=N. And now we see the solution: All squares where white pieces have been captured are light. White's dark square bishop must be on the board.
Happy retrograde puzzle world.
www.janko.at/Retros/
www.janko.at/Retros/
So I got it right. Nice.
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