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Retroanalyse im Schach

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Kurze Beweispartien

von Mark Kirtley

11 - Valeri Surkov

Probleemblad, 1997

[r2qkbnr/p1pppppp/P7/8/8/8/1PPPPPPP/RNBQKBNR]

16+13. Kürzeste Beweispartie. a) Diagramm b) Ba6 nach a5

[r2qkbnr/p1pppppp/P7/8/8/8/1PPPPPPP/RNBQKBNR]

Lösung

a) Diagramm b) Ba6 nach a5
1. Ba2-a4  Bb7-b5
2. Ba4:b5  Sb8-a6
3. Ta1:a6  Lc8-b7
4. Ta6-a1  Lb7-a6
5. Bb5:a6
1. Ba2-a3  Bb7-b5
2. Ta1-a2  Bb5-b4
3. Ba3:b4  Lc8-a6
4. Ta2:a6  Sb8-c6
5. Ta6-a1  Sc6-a5
6. Bb4:a5

The slight twinning change creates a package of paradox: a tempo reversal between the two sides, reciprocal change between capturing and captured units, and a longer solution for the phase in which the white pawn is less advanced! Note that there is no change for the square on which the black Bishop is captured, but there is such a change for the black Knight. Is it possible to compose a two-phase problem that preserves such reciprocal change, but in which the square for neither captured unit changes?