Hans-Jürgen Manthey: R: 1. Dg8xTg6+ Tg7-g6 2. Sg6-h8+ Th7-g7 3. Dh8-g8 b6-b5 4. Kg8xSf8 Te8-e7 5. a5-a6 Ta8-e8 6. a4-a5 Ta5-a8 7. c4-c5 Td5-a5 8. a3-a4 Td1-d5 9. a2-a3 Tb1-d1 10. c3-c4 b2-b1=T 11. c2-c3 a3xSb2 12. Sc4-b2 a4-a3 13. Sd6-c4 a5-a4 14. Se8-d6 a6-a5 15. e7-e8=S a7-a6 16. d6xDe7 Dd8-e7 17. d5-d6 e7-e6 18. Tc6-f6 Da8-d8 19. Tc3-c6 Dc6-a8 20. Dd4-h8 Se6-f8 21. Td3-c3 Sd8-e6 22. Kf8-g8 Df6-c6 23. Ke8-f8 Se6-d8 24. Kd8-e8 Sc5-e6+ 25. Kc8-d8 Se6-c5 26. Kb7-c8 Dh8-f6 27. Sf4-g6 Tg8-g5 28. Ka6-b7 Ta8-g8 29. Kb5-a6 Dd8-h8 30. Kc4-b5 Th8-h7 31. Td1-d3 Sc5-e6 32. Kc3-c4 Sa6-c5 33. Dd3-d4 Sb8-a6 34. Kd2-c3 Kg7-h6 35. Kc1-d2 Kf8-g7 36. Dd4-d3 Ke8-f8 37. O-O-O g5-g4 38. Dd1-d4 g6xLf5 39. d4-d5 h6XSg5 40. Le4-f5 h7-h6 41. Lg2-e4 Lf3-h5 42. Lf1-g2 Lb7-f3 43. Sd5-f4 Lc8-b7 44. g2-g3 b7-b6 45. Lf4-h2 Sg3-h1 46. Th1-h3 Lh2-g1 47. Lc1-f4 Sh5-g3 48. Sf3-g5 Le5-h2 49. Sc3-d5 Lg7-e5 50. Sb1-c3 Lf8-g7 51. d2-d4 Sf6-h5 52. Sg1-f3 Sg8-f6 53. b2-b3 g7-g6 54. h2-h4 (2021-07-18)
Henrik Juel: The animation engine does not catch the error of the two last single retractions both being black
The solution should end with 17. d5-d6 e7-e6 (2021-07-19, hidden) more ...
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A.Buchanan: There is a good try here. It seems that there are two possible pawn histories. In the one in the given solution, wBef cross-capture, while wBah are waylaid (= captured by officers on their own original files). In the other, sBef cross-capture, while wBah capture to promote. In the latter case, could sS or sT uncapture a white officer which, following various parity-fixing adventures, unpromotes on b8, and then gives White the necessary tempi? The Black cross-capture would have been wBe2-6 sBf7-f6xe5 wBf2-f7 sBe7xf6-f5. I could imagine a situation with sSd8, sTc8, wDb7. Then R: 1. Db8-b7 a7-a6 2. b7-b8=D Tb8-c8 3. a6xSb7 Sc5-b7 4. Kg8-f8 Sd3-c5 5. Kf8-g8 Sf4-d3 6. Kg8-f8 Sg6-Sf4 7. a5-a6 Sf8-g6 8. a4-a5 Te8-e7 retropat. In this line, wPh3 was not waylaid, so can't get that extra critical tempo. a3-a4 is not possible because then wTa1 could never have escaped its home. (2022-01-24)
Henrik Juel: This problem was awarded 5th Prize in the informal annual retro tourney (2022-01-23, hidden)
A.Buchanan: Looking quickly at this, it seems that there are two possible pawn histories. In the one in the given solution, wBef cross-capture, while wBah are waylaid (= captured by officers on their own original files). In the other, sBef cross-capture, while wBah capture to promote. In the latter case, could sS or sT uncapture a white officer which, following various parity-fixing adventures, unpromotes on b8, and then gives White the necessary tempi. I could imagine a situation with sSd8, sTc8, wDb7. Then R: 1. Db8-b7 a7-a6 2. b7-b8=D Tb8-c8 3. a6xSb7 Sc5-b7 4. Kg8-f8 Sd3-c5 5. Kf8-g8 Sf4-d3 6. Kg8-f8 Sg6-Sf4 7. a5-a6 Sf8-g6 8. a4-a5 Te8-e7 etc. The Black cross-capture would have been wBe2-6 sBf7-f6xe5 wBf2-f7 sBe7xf6-f5. What am I missing? (2022-01-24, hidden)
A.Buchanan: If this is a cook, then can it be fixed by sSa8 & wSh6? The free sS must be on d8 to prevent sTc8 from checking wK: the issue currently is wBa6xSb7 creates another sS which can go to f8. So a position which locks up the other sS rather than the harmless dark-squared sL may be a fix. Please confirm, refute etc (2022-01-24, hidden) more ...
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Michel Caillaud: cooked using Jacobi0.7.5 and constraints :
1.a4 e5 2.Ta3 d5 3.Sh3 Sd7 4.d3 Sb6 5.Lh6 gxh6(c1=L) 6.e3 Lg4 7.Sf4 Lxd1(Lg4) 8.Le2 Db8 9.Kf1 Lc8 10.Lh5 Lxa3(Lc3) 11.g4 L3xb2(Lb4) 12.Kg2 Kf8 13.Td1 Kg7 14.Kh3 Lf8 (2021-05-26)
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Henrik Juel: C+ Popeye 4.61 after analysis
Analysis
White is missing one man, so last move was not a pawn capture
White pawns captured all six missing black men, so last move was not d7-d5
Obviously last move was not made by Ke7 or Sd8f8, so last move was f7-f5 legitimizing the ep key (2020-11-02)
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Anton Baumann: C+ Gustav 4.1d (2021-10-22, hidden) more ...
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Anton Baumann: C+ Gustav 4.1d (2021-02-17, hidden) more ...
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SCHRECKE: C+, Gustav 4.1d, Brute Force (2020-09-20)
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