2 problem(s) found in 7799 milliseconds (displaying 2 problem(s)). [COMMENTDATE>=20220810 AND G='Retro' AND G='Fairies' AND S='diagrammes'] [download as LaTeX]
1 - P0008449
Jean-Michel Trillon
3574 diagrammes 118 07/1996
(6+4)
Orthorekonstruktion in 12.5
Echecs sentinelles
Weiß am Zug
Jean-Michel Trillon
3574 diagrammes 118 07/1996
(6+4)
Orthorekonstruktion in 12.5
Echecs sentinelles
Weiß am Zug
1. e5 Kb8 2. e6 Ka8 3. e7 Kb8 4. e8=T Ka8 5. Th8 Kb8 6. Th1 Ka8 7. Te1 Kb8 8. Te2 Ka8 9. Te1[+wBe2] Kb8 10. Tb1+ Ka8 11. Tb8+ Kxb8 12. e4 Ka8 13. e3
1. dxc3ep 2. Kxb5 3. Kc6 4. Kd7 5. Ke8 6. 0-0 Txg7#
Un des PNd a pris 1 pièce blanche. pour venir de e7. Donc les PB b, c et d ne peuvent se croiser. Le PBg6 ne peut venir de f5 car il faudrait 2 prises noires! Seule reste le dernier coup blanc 0... c2-c4 Ceci implique que le FNb1 est issue de promtion ! C'est le PNb7, il aurait pris 2 pièces blanches. Les PNg5 et h5 ont pris 2 pièces blanches.
Un des PNd a pris 1 pièce blanche. pour venir de e7. Donc les PB b, c et d ne peuvent se croiser. Le PBg6 ne peut venir de f5 car il faudrait 2 prises noires! Seule reste le dernier coup blanc 0... c2-c4 Ceci implique que le FNb1 est issue de promtion ! C'est le PNb7, il aurait pris 2 pièces blanches. Les PNg5 et h5 ont pris 2 pièces blanches.
No. 11661 HN
A.Buchanan: “Obvious promotion” is intended to refer to something visible in the initial diagram position, not what is reached during series play (2022-09-14)
A.Buchanan: Similarly “non-standard material (2022-09-14)
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A.Buchanan: “Obvious promotion” is intended to refer to something visible in the initial diagram position, not what is reached during series play (2022-09-14)
A.Buchanan: Similarly “non-standard material (2022-09-14)
more ...
comment
Keywords: Seriesmover, Consequent, En passant as key, Promotion in the retro play (sLb1), Castling, Valladao Task
Genre: Retro, Fairies
FEN: 7r/5pnR/5PnR/1Pp2ppq/1kPp4/p2P4/P2pP3/Kbb4r
Input: Henri Nouguier, 2004-01-11
Last update: Mario Richter, 2022-09-14 more...
Genre: Retro, Fairies
FEN: 7r/5pnR/5PnR/1Pp2ppq/1kPp4/p2P4/P2pP3/Kbb4r
Input: Henri Nouguier, 2004-01-11
Last update: Mario Richter, 2022-09-14 more...
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The problems of this query have been registered by the following contributors:
Gerd Wilts (1)Henri Nouguier (1)
Joost de Heer: No berolina pawns.
Solution: 1.é5 Rb8 2.é6 Ra8 3.é7 Rb8 4.é8=T Ra8 5.Th8 Rb8 6.Th1 Ra8 7.Té1 Rb8 8.Té2 Ra8 9.Té1(+é2) Rb8 10.Tb1+ Ra8 11.Tb8+ R×b8 12.é4 Ra8 13.é3 (2023-08-27)
A.Buchanan: Quite delightful. Can it be validated in Jacobi as an A-B PG? (2023-08-28)
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