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Retroanalyse im Schach

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Meisterwerke

in der Retroanalyse

26 - W. Frangen

Fairy Chess Review, 1953

[5Nnr/1pR2ppn/3k1P2/2NPpKp1/3bQ2B/2PRP1B1/3P1NP1/8]

15+10. Matt in einem Zug!

[5Nnr/1pR2ppn/3k1P2/2NPpKp1/3bQ2B/2PRP1B1/3P1NP1/8]

Lösung

Suppose black moved last. Then his last move must've been e7-e5. White's last move was either Bh2xg3 or Kf4-f5. If it was the latter, black's move before that must've been g6-g5. But that implies that the h-pawn promoted on g8 to a knight, not on h8. So in both cases, white captured a piece on the g-line. If black's last move was e7-e5, this also means that Bd4 is promoted. The only pawns that could promote on a black square without capturing are the a- and the c-pawn. If the a-pawn promoted, the white a-pawn must have captured 3 times on its way to b8 or d8 (where it promoted to a bishop). Together with the 2 captures on the c- and d-file, the capture on the g-file, and the bishop on f8, this is 7 captures. So black's a-pawn didn't promote. If the c-pawn promoted, white must've captured b2xc3 and cxd. Together with a7xb8, Bf8 and the promotion on g8 this equals 5 captures. But none of these captures happened on the a-file. So the c-pawn couldn't have promoted either. So black didn't move last. So #1 with 1. ... Nh6!#


Solution by Philippe Schnoebelen