Retroanalyse im Schach

Meisterwerke

in der Retroanalyse

16a - W. Keym

Die Schwalbe, 1971

1./2. Preis im Ceriani Memorial Tournier

[5r1n/4N1P1/QP1PBb2/1p1p1Kp1/1Pk1P1P1/p1PbpPpq/5n2/N1R5]

14+13. Matt in weniger als einem Zug.

[5r1n/4N1P1/QP1PBb2/1p1p1Kp1/1Pk1P1P1/p1PbpPpq/5n2/N1R5]

Lösung

There are 8 possible en-passant captures half-finished.

e5xd6 ep: This requires as last black move d7-d5. But because the white pawns captured 3 times, the bishop can't have captured on e6, so the check is illegal.
c5xd6 ep: Same reasoning.
c5xb6 ep: The white pawns captured 3 times before this move. The black pawns on d5 and e3 captured twice. So the missing pawns, which were captured by the white pawns, didn't capture, so they couldn't have promoted. But none of these pawns could've reached the b- or g-line. So this ep-capture is illegal too.
a5xb6 ep: The pawns on the a- and b-line, and the pawns on the d- and e-line cross-captured, once by black and once by white. Together with the capture on the g-line this accounts for all captures. But then, like in the previous case, the missing black pawns couldn't have reached squares where they could get captured.
d4xe3 ep: White captured twice, on b6 and on the g-line. So the d-pawn didn't capture, which means that the two black pawns on the d-line come from the c- and e-pawn. But then the d-pawn and the f-pawn were captured on their own line, so neither of them reached the b- or g-line.
f4xe3 ep: Black captures are [c/e]xd5 and Bxd3. Both captures were on white squares, and white misses the black-squared bishop.
h4xg3 epWhite captures are, again, a5xb6 and hxg. The black pawns on the d- and e-line captured twice, so the missing two pawns couldn't have promoted and got captured on their own line.
f4xg3 epThis is the only legal option. The missing h-pawn promoted on h1, after hxg was played. The other pawn got captured on its own line. A possible move-sequence could be for instance: a5xRb6; a7-..-a3; h6xQg7; h7-..-h1=Q; e2-e4; d4xBe3; d2-..-d6; c6xRd5;

So the only legal en-passant capture was f4xg3. So Black mates in 1/2 by removing the white pawn on g4.

Solution by Philippe Schnoebelen

16b - W. Keym

Die Schwalbe, 1971

1./2. Preis im Ceriani Memorial Tournier

[r6r/P7/1P6/P1P3P1/p1P1q1p1/1P2p1N1/B3N3/R6R]

12+6. Füge beide Könige hinzu, dann Matt in weniger als einem Zug!

[r6r/P7/1P6/P1P3P1/p1P1q1p1/1P2p1N1/B3N3/R6R]

Lösung

The white pawns on the queenside captured nine times, these are all the missing black pieces. This also means that the g5 pawn is the original g-pawn. Black captures are b5xa4, [h/f]5xg4 and gx[h/f]. The captures on a4 and g4 are on white squares, so the missing white bishop was captured by the g-pawn. This also means that the white h-pawn promoted before it got captured. This means that the capture on g4 was done by the black h-pawn. So the black f-pawn promoted on f1, breaking both white's castlings. The promotion on h8 breaks black's kingside castling. So the only castling allowed in this position is the black queenside castling.

Add the white king on d6, and the black king on c8, and mate with 'finish 000'

Solution by Werner Keym