579 problem(s) found in 5756 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT K='En passant als Schlüssel' AND K='Rochade'] [download as LaTeX]
1 - P0000016
Nikita M. Plaksin
Alexander Kislyak
(F) Die Schwalbe 98 04/1986
(11+9)
Welches war der erste Zug der beiden Könige?
Nikita M. Plaksin
Alexander Kislyak
(F) Die Schwalbe 98 04/1986
(11+9)
Welches war der erste Zug der beiden Könige?
R: 1. bxc3ep+ c2-c4 2. b5-b4+
Erster K-Zug: Weiß: 0-0; Schwarz: 0-0
Erster K-Zug: Weiß: 0-0; Schwarz: 0-0
Keywords: En passant, First Move?, Last Moves? (3), Castling (wksk), Castling in the retro play, En passant in the retro play, Volet Pawn, Type C
Genre: Retro
FEN: 7k/1p2pPpp/b3p3/8/8/2pK2P1/pP1PPPP1/2B3RN
Reprints: 566 Ukrainisches Album 1986-1990
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-20 more...
Genre: Retro
FEN: 7k/1p2pPpp/b3p3/8/8/2pK2P1/pP1PPPP1/2B3RN
Reprints: 566 Ukrainisches Album 1986-1990
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-20 more...
*) 1. ... 0-0-0 2. Txf2 Dxg1#
1) 1. cxd2+ Kxd2 2. Txf2 Dxg1#
R: 1. ... Da8-h8 2. d3xTe4 Da2-a8 3. c2xSd3 Sc1-d3+ 4. b4xLc5 Db1-a2 5. b3-b4 Ba2xSb1=D 6. Sa3-b1 und weiter z.B. (Beispielauflösung mri)
6. ... Te8-e4 7. Sc4-a3 Sg4-h2 8. Se5-c4 Th2-g2 9. Sf3-e5 g2-g1=L 10. Sg1-f3 Sh6-g4 11. Sf3xDg1 Th8-e8 12. Sd4-f3 f3xSg2 13. Se3-g2 f4-f3 14. Sc6-d4 Tg2-h2 15. Sa7-c6 Kh2-h1 16. Sc8-a7 Dh1-g1 17. Sb6xLc8 Tg1-g2 18. Sf5-e3 Dc6-h1 19. Sh4-f5 Kh1-h2 20. h2-h3 a3-a2 21. Sa4-b6 Sa2-c1 22. Sb6-a4 Sb4-a2 23. Sa4-b6 Sd5-b4 24. Sb6-a4 Se3-d5 25. Dh3-f1 Sf1-e3 26. Dg4-h3 f5-f4 27. Df4-g4 a4-a3 28. Db4-f4 a5-a4 29. Da3-b4 Kg2-h1 30. Sf3-h4 Kh3-g2 31. Da2-a3 g2-g1=T 32. Db1-a2 Kg4-h3 33. Dd1-b1 h3xTg2 34. Tg1-g2 Kg5-g4 35. Sh4-f3 Kf6-g5 36. Th1-g1 Se3-f1 37. Sf3-h4 h4-h3 38. Sg1-f3 e6xLf5 39. Lh3-f5 Ke7-f6 40. Lf1-h3 Sc4-e3 41. g2-g3 Ke8-e7 42. Sa4-b6 Sg8-h6 43. a2xTb3 Tb6-b3 44. Sh3-g1 Ta6-b6 45. Sb6-a4 Ta8-a6 46. Sg1-h3 a7-a5 47. Sh3-g1 Sa5-c4 48. Sg1-h3 Sb3-a5 49. Sc4-b6 Sc1-b3 50. Sa3-c4 Sb3xLc1 51. Sb1-a3 Sa5-b3 52. Sh3-g1 h5-h4 53. Sg1-h3 c4-c3 54. Sh3-g1 Lf8-c5 55. Sg1-h3 Dc7-c6 56. Sh3-g1 Dd8-c7 57. Sg1-h3 c5-c4 58. Sh3-g1 h7-h5 59. Sg1-h3 e7-e6 60. Sh3-g1 Sc6-a5 61. Sg1-h3 Sb8-c6 62. Sh3-g1 c7-c5 63. Sg1-h3
1) 1. cxd2+ Kxd2 2. Txf2 Dxg1#
R: 1. ... Da8-h8 2. d3xTe4 Da2-a8 3. c2xSd3 Sc1-d3+ 4. b4xLc5 Db1-a2 5. b3-b4 Ba2xSb1=D 6. Sa3-b1 und weiter z.B. (Beispielauflösung mri)
6. ... Te8-e4 7. Sc4-a3 Sg4-h2 8. Se5-c4 Th2-g2 9. Sf3-e5 g2-g1=L 10. Sg1-f3 Sh6-g4 11. Sf3xDg1 Th8-e8 12. Sd4-f3 f3xSg2 13. Se3-g2 f4-f3 14. Sc6-d4 Tg2-h2 15. Sa7-c6 Kh2-h1 16. Sc8-a7 Dh1-g1 17. Sb6xLc8 Tg1-g2 18. Sf5-e3 Dc6-h1 19. Sh4-f5 Kh1-h2 20. h2-h3 a3-a2 21. Sa4-b6 Sa2-c1 22. Sb6-a4 Sb4-a2 23. Sa4-b6 Sd5-b4 24. Sb6-a4 Se3-d5 25. Dh3-f1 Sf1-e3 26. Dg4-h3 f5-f4 27. Df4-g4 a4-a3 28. Db4-f4 a5-a4 29. Da3-b4 Kg2-h1 30. Sf3-h4 Kh3-g2 31. Da2-a3 g2-g1=T 32. Db1-a2 Kg4-h3 33. Dd1-b1 h3xTg2 34. Tg1-g2 Kg5-g4 35. Sh4-f3 Kf6-g5 36. Th1-g1 Se3-f1 37. Sf3-h4 h4-h3 38. Sg1-f3 e6xLf5 39. Lh3-f5 Ke7-f6 40. Lf1-h3 Sc4-e3 41. g2-g3 Ke8-e7 42. Sa4-b6 Sg8-h6 43. a2xTb3 Tb6-b3 44. Sh3-g1 Ta6-b6 45. Sb6-a4 Ta8-a6 46. Sg1-h3 a7-a5 47. Sh3-g1 Sa5-c4 48. Sg1-h3 Sb3-a5 49. Sc4-b6 Sc1-b3 50. Sa3-c4 Sb3xLc1 51. Sb1-a3 Sa5-b3 52. Sh3-g1 h5-h4 53. Sg1-h3 c4-c3 54. Sh3-g1 Lf8-c5 55. Sg1-h3 Dc7-c6 56. Sh3-g1 Dd8-c7 57. Sg1-h3 c5-c4 58. Sh3-g1 h7-h5 59. Sg1-h3 e7-e6 60. Sh3-g1 Sc6-a5 61. Sg1-h3 Sb8-c6 62. Sh3-g1 c7-c5 63. Sg1-h3
Anton Baumann: Auszeichnung Informalturnier 1986: 2.Preis
Preisbericht: 'Die Schwalbe' 06/2011 S.124 (2023-01-02)
Henrik Juel: How is the SE corner released, without ruining the castling? (2023-01-02)
Mario Richter: Good question, Henrik! I first thought that releasing the SE corner without ruining White's castling right is impossible, but the trick is to uncapture a black Queen in the SE corner at the right moment.
Perhaps Theodore Hwa can use ths problem as a test case for his latest improvement to Retractor 2 ... (2023-01-02)
Henrik Juel: Thanks, Mario
In view of the prize I suspected that the problem was correct, but I did not find the uncapture trick (2023-01-02)
Henrik Juel: C+ Popeye 4.61, because with Black to move White may not castle (2023-01-02)
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Preisbericht: 'Die Schwalbe' 06/2011 S.124 (2023-01-02)
Henrik Juel: How is the SE corner released, without ruining the castling? (2023-01-02)
Mario Richter: Good question, Henrik! I first thought that releasing the SE corner without ruining White's castling right is impossible, but the trick is to uncapture a black Queen in the SE corner at the right moment.
Perhaps Theodore Hwa can use ths problem as a test case for his latest improvement to Retractor 2 ... (2023-01-02)
Henrik Juel: Thanks, Mario
In view of the prize I suspected that the problem was correct, but I did not find the uncapture trick (2023-01-02)
Henrik Juel: C+ Popeye 4.61, because with Black to move White may not castle (2023-01-02)
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3 - P0000050
Andrey Frolkin
(T) Die Schwalbe 102 12/1986
(13+13)
Vor mindestens 50 Einzelzügen mußte rochiert werden!
Andrey Frolkin
(T) Die Schwalbe 102 12/1986
(13+13)
Vor mindestens 50 Einzelzügen mußte rochiert werden!
R: 1. ... La2-b1 2. Dh7-g8 Lb1-a2 3. Kg8-h8 La2-b1 4. Kf7-g8 Lb1-a2 5. Ke8-f7 La2-b1 6. Kd8-e8 Lb1-a2 7. Kc7-d8 La2-b1 8. Kb6-c7 Lb1-a2 9. Kc5-b6 La2-b1 10. Kd4-c5 Lb1-a2 11. Ke4-d4 La2-b1 12. Kf3-e4 Lb1-a2 13. Kg4-f3 La2-b1 14. Kh3-g4 Lb1-a2 15. Kh2-h3 La2-b1 16. Kg1-h2 Lb1-a2 17. h2xTg3 Th3-g3 18. Dg8-h7 Th8-h3 19. Sg3-h1 h7xTg6 20. Tg5-g6 La2-b1 21. Ta5-g5 Lb1-a2 22. Ta2-a5 f5-f4 23. Tb2-a2 La2-b1 24. Tb1-b2 f6-f5 25. Tf1-b1 Lb1-a2 26. 0-0
James Malcom: Lastly, here is a PG that may or may not be the shortest: 1. Nf3 c5 2. Ne5 Qb6 3. Nc3 Qb3 4. axb3 c4 5. Nd5 c3 6. Ra6 Nf6 7. Rd6 Ng4 8. Re6 Ne3 9. Nf4 Nxf1 10. Nh5 Ne3 11. Ng3 Nc4 12. bxc4 dxe6 13. Nf5 Bd7 14. Ng3 Ba4 15. Nf5 Bb3 16. Ng3 Ba2 17. Nf5 Nc6 18. Ng3 Na5 19. Nf5 Nb3 20. Ng3 Na1 21. b3 a6 22. Ba3 Ra7 23. Bc5 Kd8 24. Qc1 Kc8 25. Qa3 Kc7 26. Qa4 Bb1 27. Qe8 f6 28. Qf7 Kc8 29. Qg8 Kb8 30. Bd6+ Ka8 31. Bb8 Ba2 32. Nd7 Bb1 33. O-O Ba2 34. Rb1 f5 35. Rb2 Bb1 36. Ra2 f4 37. Ra5 Ba2 38. Rg5 Bb1 39. Rg6 hxg6 40. Nh1 Rh3 41. Qh7 Rg3 42. hxg3 Ba2 43. Kh2 Bb1 44. Kh3 Ba2 45. Kg4 Bb1 46. Kf3 Ba2 47. Ke4 Bb1 48. Kd4 Ba2 49. Kc5 Bb1 50. Kb6 Ba2 51. Kc7 Bb1 52. Kd8 Ba2 53. Ke8 Bb1 54. Kf7 Ba2 55. Kg8 Bb1 56. Kh8 Ba2 57. Qg8 Bb1 (2020-11-08)
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Keywords: Castling in the retro play
Genre: Retro
FEN: kB3bQK/rp1Np1p1/p3p1p1/8/2P2p2/1Pp3P1/2PPPPP1/nb5N
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-02-02 more...
Genre: Retro
FEN: kB3bQK/rp1Np1p1/p3p1p1/8/2P2p2/1Pp3P1/2PPPPP1/nb5N
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-02-02 more...
4 - P0000101
Leonid M. Borodatow
Evgeny V. Kharichev
6209v Die Schwalbe 110 04/1988
(12+10) cooked
h#3
b) sBh5 nach h6
Leonid M. Borodatow
Evgeny V. Kharichev
6209v Die Schwalbe 110 04/1988
(12+10) cooked
h#3
b) sBh5 nach h6
a) 1. 0-0-0 Sxc7 2. hxg3 Sd5 3. g2 Se7#
Beispielauflösung mri:
R: 1. Kf1-e1 Ta2-a1 2. Kg2-f1 Tb2-a2 3. La2-b1 Tb1-b2 4. Kf3-g2 Tf1-b1 5. Ke3-f3 g5xDh4 6. Lh2-g1 Tg1-f1 7. Dd4-h4 Tg2-g1 8. Lg1-h2 Th2-g2 9. Kd3-e3 Tg2-h2 10. Kc3-d3 Th2-g2 11. Kb2-c3 Tg2-h2 12. Kc1-b2 Th2-g2 13. Kd1-c1 Tg2-h2 14. Ke1-d1 Th2-g2 15. Da1-d4 Tg2-h2 16. Dd1-a1 d6xLc5 17. La3-c5 Th2-g2 18. Lc1-a3 Tg2-h2 19. b2-b3 Th2-g2 20. Ld5-a2 Tg2-h2 21. Sc5-a6 Th2-g2 22. Lg2-d5 h6-h5 23. Lf1-g2 Tg2-h2 24. Lh2-g1 Tg1-g2 25. Sa4-c5 g2-g1=T 26. Sb6-a4 f3xTg2 27. Tg1-g2 f4-f3 28. g2-g3 f5-f4 29. Le5-h2 h7-h6 30. Lg7-e5 f6-f5 31. Lf8-g7 f7-f6 32. e7xLf8=L Lg7-f8 33. e6-e7 Lf8-g7 34. d5xTe6 Tg6-e6 35. Sc8-b6 Tg8-g6 36. Sb6xLc8 g6-g5 37. c4xDd5 Dg5-d5 38. Sd5-b6 Dd8-g5 39. Sg3-h1 e7xTd6 40. Ta6-d6 Th8-g8 41. Ta1-a6 g7-g6 42. b3xSc4 Sa5-c4 43. a2xSb3 Sd4-b3 44. Sc3-d5 Sc6-a5 45. Sb1-c3 Sb8-c6 46. Th1-g1 Sf5-d4 47. Se4-g3 Sh6-f5 48. Sg5-e4 Sg8-h6 49. Sf3-g5 Sh6-g8 50. Sg1-f3 Sg8-h6 51. h2-h3
b) 1. Kf7 f4 2. Kg6 c3,c4+ 3. Kh5 g4#
in b) fehlt Schwarz ein Retrotempo zur Aufrechterhaltung des Rochaderechts, es kann maximal eine Stellung wie z.B. r3kB2/pppp1p1p/3p2p1/8/8/7P/1PPPPPP1/1NBQKBRN erreicht werden, in der entweder sTh8 oder sLf8 nicht als Schlagobjekt für die notwendigen weißen Bauernschläge zur Verfügung stand.
Der Dual 2. ... c3+,c4+ in b) war von den Autoren in Kauf genommen worden., die (schon bei der Lösungsveröffentlichung bekannten) NLs aber natürlich nicht. Ein Schwalbe-Löser ("PS") kritisierte den "völlig unnötigen Vorwärtsballast" und schlug implizit vor, das Vorwärtsspiel komplett wegzulassen und stattdessen direkt zu fragen: "Ist die 0-0-0 erlaubt?"
Cook: a) 1. hxg3 Sxg3 2. c6 Sf5 3. 0-0-0 Sd6#
b) 1. Kf7 f4 2. Kg6 c4+ 3. Kh5 g4#
Beispielauflösung mri:
R: 1. Kf1-e1 Ta2-a1 2. Kg2-f1 Tb2-a2 3. La2-b1 Tb1-b2 4. Kf3-g2 Tf1-b1 5. Ke3-f3 g5xDh4 6. Lh2-g1 Tg1-f1 7. Dd4-h4 Tg2-g1 8. Lg1-h2 Th2-g2 9. Kd3-e3 Tg2-h2 10. Kc3-d3 Th2-g2 11. Kb2-c3 Tg2-h2 12. Kc1-b2 Th2-g2 13. Kd1-c1 Tg2-h2 14. Ke1-d1 Th2-g2 15. Da1-d4 Tg2-h2 16. Dd1-a1 d6xLc5 17. La3-c5 Th2-g2 18. Lc1-a3 Tg2-h2 19. b2-b3 Th2-g2 20. Ld5-a2 Tg2-h2 21. Sc5-a6 Th2-g2 22. Lg2-d5 h6-h5 23. Lf1-g2 Tg2-h2 24. Lh2-g1 Tg1-g2 25. Sa4-c5 g2-g1=T 26. Sb6-a4 f3xTg2 27. Tg1-g2 f4-f3 28. g2-g3 f5-f4 29. Le5-h2 h7-h6 30. Lg7-e5 f6-f5 31. Lf8-g7 f7-f6 32. e7xLf8=L Lg7-f8 33. e6-e7 Lf8-g7 34. d5xTe6 Tg6-e6 35. Sc8-b6 Tg8-g6 36. Sb6xLc8 g6-g5 37. c4xDd5 Dg5-d5 38. Sd5-b6 Dd8-g5 39. Sg3-h1 e7xTd6 40. Ta6-d6 Th8-g8 41. Ta1-a6 g7-g6 42. b3xSc4 Sa5-c4 43. a2xSb3 Sd4-b3 44. Sc3-d5 Sc6-a5 45. Sb1-c3 Sb8-c6 46. Th1-g1 Sf5-d4 47. Se4-g3 Sh6-f5 48. Sg5-e4 Sg8-h6 49. Sf3-g5 Sh6-g8 50. Sg1-f3 Sg8-h6 51. h2-h3
b) 1. Kf7 f4 2. Kg6 c3,c4+ 3. Kh5 g4#
in b) fehlt Schwarz ein Retrotempo zur Aufrechterhaltung des Rochaderechts, es kann maximal eine Stellung wie z.B. r3kB2/pppp1p1p/3p2p1/8/8/7P/1PPPPPP1/1NBQKBRN erreicht werden, in der entweder sTh8 oder sLf8 nicht als Schlagobjekt für die notwendigen weißen Bauernschläge zur Verfügung stand.
Der Dual 2. ... c3+,c4+ in b) war von den Autoren in Kauf genommen worden., die (schon bei der Lösungsveröffentlichung bekannten) NLs aber natürlich nicht. Ein Schwalbe-Löser ("PS") kritisierte den "völlig unnötigen Vorwärtsballast" und schlug implizit vor, das Vorwärtsspiel komplett wegzulassen und stattdessen direkt zu fragen: "Ist die 0-0-0 erlaubt?"
Cook: a) 1. hxg3 Sxg3 2. c6 Sf5 3. 0-0-0 Sd6#
b) 1. Kf7 f4 2. Kg6 c4+ 3. Kh5 g4#
Neufassung 5949.
A.Buchanan: It would be great if someone can retrieve 6209v from Die Schwalbe, please, to confirm the diagram and intended solution. If castling right remains, then sBf must promote on g1=T. There is then some jiggling around because Tg1-h2 to unpin Lg1 and let it go b1. In (a) this is doable, but in (b), Black runs out of tempi however there is an alternative no-castling mate with sKh5. However that's just a rough sketch, and I don't want to spend more time until the actual published form is confirmed (2022-01-08)
Ladislav Packa: Is everything allright here? Because solution a) is not a solution - after 3 .... Se7 # ?? 4.Kc7 / b8! (2022-01-09)
Mario Richter: Indeed, the given solution only works with wLh2 instead of g1. I only have access to the solution, but not to original diagram. Perhaps someoe with access to it can give us the correct details? (2022-01-10)
A.Buchanan: Yes I didn't check the solution to a) matches one of the actual ones. With wLh2, it's still cooked, as is b) in any case. Possible to rearrange the free pieces to make b) sound, but I don't see how to fix a) at the same time. (2022-01-10)
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A.Buchanan: It would be great if someone can retrieve 6209v from Die Schwalbe, please, to confirm the diagram and intended solution. If castling right remains, then sBf must promote on g1=T. There is then some jiggling around because Tg1-h2 to unpin Lg1 and let it go b1. In (a) this is doable, but in (b), Black runs out of tempi however there is an alternative no-castling mate with sKh5. However that's just a rough sketch, and I don't want to spend more time until the actual published form is confirmed (2022-01-08)
Ladislav Packa: Is everything allright here? Because solution a) is not a solution - after 3 .... Se7 # ?? 4.Kc7 / b8! (2022-01-09)
Mario Richter: Indeed, the given solution only works with wLh2 instead of g1. I only have access to the solution, but not to original diagram. Perhaps someoe with access to it can give us the correct details? (2022-01-10)
A.Buchanan: Yes I didn't check the solution to a) matches one of the actual ones. With wLh2, it's still cooked, as is b) in any case. Possible to rearrange the free pieces to make b) sound, but I don't see how to fix a) at the same time. (2022-01-10)
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5 - P0000112
Dmitri W. Pronkin
Andrey Frolkin
6386v Die Schwalbe 113 10/1988
(14+14) cooked
BP in 45.0
Dmitri W. Pronkin
Andrey Frolkin
6386v Die Schwalbe 113 10/1988
(14+14) cooked
BP in 45.0
1. f4 g5 2. f5 g4 3. f6 g3 4. fxe7 gxh2 5. g4 d5 6. g5 d4 7. g6 d3 8. g7 dxc2 9. d4 f5 10. Lf4 c1=T 11. Lg3 Tc6 12. Dd3 Tg6 13. Da6 Sf6 14. g8=L c5 15. Lb3 c4 16. d5 Kf7 17. e8=T c3 18. Te4 c2 19. Ta4 c1=L 20. e4 Lf4 21. Sd2 h5 22. 0-0-0 h4 23. Te1 h3 24. Ld1 Th4 25. d6 L8h6 26. d7 Dh8 27. d8=T Le6 28. Tc8 S8d7 29. Tc2 Kg8 30. e5 Lf7 31. e6 Tf8 32. e7 Lb8 33. e8=S f4 34. Te7 f3 35. Se2 f2 36. Tg1 h1=S 37. b4 h2 38. Lh3 f1=S 39. b5 Sf2 40. b6 h1=L 41. bxa7 b6 42. a8=L Lb7 43. Th1 Lc8 44. Lag2 Se3 45. Lf1 S6e4
Cook: 1. b4 c5 2. b5 c4 3. b6 c3 4. bxa7 d5 5. e4 d4 6. f4 d3 7. f5 dxc2 8. d4 g5 9. Lf4 c1=L 10. d5 g4 11. f6 g3 12. fxe7 gxh2 13. g4 c2 14. Lg3 Lf4 15. g5 c1=T 16. g6 Tc6 17. Dd3 f5 18. Sd2 h5 19. g7 Tg6 20. 0-0-0 Sf6 21. Te1 h4 22. d6 h3 23. g8=L Th4 24. Lb3 L8h6 25. Ld1 Kf7 26. d7 Dh8 27. d8=T Le6 28. Td4 Sbd7 29. e5 Tf8 30. e8=T Kg8 31. Tc8 Lf7 32. e6 b6 33. e7 Lb8 34. e8=S f4 35. Te7 f3 36. Se2 f2 37. Tg1 h1=S 38. Da6 h2 39. Lh3 f1=S 40. Ta4 Sf2 41. Tc2 h1=L
Cook: 1. b4 c5 2. b5 c4 3. b6 c3 4. bxa7 d5 5. e4 d4 6. f4 d3 7. f5 dxc2 8. d4 g5 9. Lf4 c1=L 10. d5 g4 11. f6 g3 12. fxe7 gxh2 13. g4 c2 14. Lg3 Lf4 15. g5 c1=T 16. g6 Tc6 17. Dd3 f5 18. Sd2 h5 19. g7 Tg6 20. 0-0-0 Sf6 21. Te1 h4 22. d6 h3 23. g8=L Th4 24. Lb3 L8h6 25. Ld1 Kf7 26. d7 Dh8 27. d8=T Le6 28. Td4 Sbd7 29. e5 Tf8 30. e8=T Kg8 31. Tc8 Lf7 32. e6 b6 33. e7 Lb8 34. e8=S f4 35. Te7 f3 36. Se2 f2 37. Tg1 h1=S 38. Da6 h2 39. Lh3 f1=S 40. Ta4 Sf2 41. Tc2 h1=L
Michel Caillaud: cooked by Stelvio 0.93 :
1.b4 c5 2.b5 c4 3.b6 c3 4.bxa7 d5 5.e4 d4 6.f4 d3 7.f5 dxc2 8.d4 g5 9.Lf4 c1=L 10.d5 g4 11.f6 g3 12.fxe7 gxh2 13.g4 c2 14.Lg3 Lf4 15.g5 c1=T 16.g6 Tc6 17.Dd3 f5 18.Sd2 h5 19.g7 Tg6 20.0-0-0 Sf6 21.Te1 h4 22.d6 h3 23.g8=L Th4 24.Lb3 L8h6 25.Ld1 Kf7 26.d7 Dh8 27.d8=T Le6 28.Td4 Sbd7 29.e5 Tf8 30.e8=T Kg8 31.Tc8 Lf7 32.e6 b6 33.e7 Lb8 34.e8=S f4 35.Te7 f3 36.Se2 f2 37.Tg1 h1=S 38.Da6 h2 39.Lh3 f1=S 40.Ta4 Sf2 41.Tc2 h1=L... (2022-12-20)
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1.b4 c5 2.b5 c4 3.b6 c3 4.bxa7 d5 5.e4 d4 6.f4 d3 7.f5 dxc2 8.d4 g5 9.Lf4 c1=L 10.d5 g4 11.f6 g3 12.fxe7 gxh2 13.g4 c2 14.Lg3 Lf4 15.g5 c1=T 16.g6 Tc6 17.Dd3 f5 18.Sd2 h5 19.g7 Tg6 20.0-0-0 Sf6 21.Te1 h4 22.d6 h3 23.g8=L Th4 24.Lb3 L8h6 25.Ld1 Kf7 26.d7 Dh8 27.d8=T Le6 28.Td4 Sbd7 29.e5 Tf8 30.e8=T Kg8 31.Tc8 Lf7 32.e6 b6 33.e7 Lb8 34.e8=S f4 35.Te7 f3 36.Se2 f2 37.Tg1 h1=S 38.Da6 h2 39.Lh3 f1=S 40.Ta4 Sf2 41.Tc2 h1=L... (2022-12-20)
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Keywords: Unique Proof Game, Move Length Record, Non-standard material, Castling, Promotion (tLTlTSsslL)
Genre: Retro
FEN: 1bb1Nrkq/3nRb2/Qp4rb/8/R3n2r/4n1BB/P1RNNn2/2KB1B1R
Reprints: 583 Ukrainisches Album 1986-1990
80 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Silvio Baier, 2023-03-02 more...
Genre: Retro
FEN: 1bb1Nrkq/3nRb2/Qp4rb/8/R3n2r/4n1BB/P1RNNn2/2KB1B1R
Reprints: 583 Ukrainisches Album 1986-1990
80 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Silvio Baier, 2023-03-02 more...
6 - P0000136
Dmitri W. Pronkin
Andrey Frolkin
6631v Die Schwalbe 117 06/1989
Preis
(14+14)
BP in 57.5
Dmitri W. Pronkin
Andrey Frolkin
6631v Die Schwalbe 117 06/1989
Preis
(14+14)
BP in 57.5
1. a4 h5 2. a5 h4 3. a6 h3 4. axb7 hxg2 5. h4 d5 6. h5 d4 7. h6 d3 8. h7 dxc2 9. d4 a5 10. Lh6 c1=T 11. e4 Tc5 12. Se2 Th5 13. e5 c5 14. e6 Sc6 15. b8=T a4 16. Tb4 a3 17. Ta4 c4 18. b4 c3 19. b5 c2 20. b6 c1=T 21. b7 Tc4 22. b8=T Da5+ 23. Tbb4 Lb7 24. S1c3 0-0-0 25. exf7 e5 26. Tc1 Lc5 27. f8=T a2 28. Tf3 a1=T 29. Sa2 g1=T 30. Tfa3 Tg6 31. f4 Te6 32. f5 g5 33. f6 g4 34. f7 g3 35. f8=T g2 36. Tf5 g1=T 37. Lf8 Tg7 38. Sg3 e4 39. Ld3 e3 40. 0-0 e2 41. Tcc3 e1=T 42. Lc2 T1e3 43. d5 Tdd7 44. d6 Tdf7 45. d7+ Kb8 46. Dd6+ Ka8 47. Dc7 Sge7 48. d8=T+ Sc8 49. Tdd3 Thg8 50. h8=T Tae1 51. Th6 T1e2 52. T1f2 Tce4 53. Kf1 Ld4 54. Tfc5 Se5 55. Sf5 Sc4 56. Sd6 Sb2 57. Tbc4 Sb6 58. Db8+
Der absolute KBP-Längenrekord.
Der absolute KBP-Längenrekord.
See P1338946 cooked.
paul: Compare with P0002278 & P0002279 (2010-04-30)
Mu-Tsun Tsai: This one is by far the toughest retro I've ever solved. Very little certain information can be determined by structural consideration alone, even with long and complicated argument. It took me five days to complete solving this. (2012-07-22)
A.Buchanan: @Mu-Tsun: that's an interesting data point - thanks for posting. (2017-09-07)
Henrik Juel: The current record is 58.5 moves in a proof game problem by the authors + Keym, Die Schwalbe 2017 (2017-09-07)
Henrik Juel: I just learned that the 58.5 move proof game has been cooked... (2017-09-07)
A.Buchanan: In retrospect, my earlier comment about "interesting data point" is a bit weak. It's actually great that for such an extreme problem, someone took substantial time to independently validate it. It's like doing science: people want to do their own new stuff, and are unwilling to take the time to validate what's already been claimed. This one has survived 30+ years, and maybe the use of constraints e.g. in Jacobi can eventually allow it to be HC+. (2021-05-29)
Olaf Jenkner: This problem is the current record, because P1338946 (58.5 moves) has been cooked. (2021-11-25)
Reto: This is C+ up to 51.0 moves with Stelvio 2.0. This ties the record for partial testing of an SPG. Took 1200 CPU hours of strategy seeking (finding 378 0+0 strategies) and another 13h of strategy playing these strategies. If this can ever be completely solved, then it needs to be the case that all strategies have 0+0 free moves, otherwise playing is utterly hopeless.
@Andrew: There is absolutely no way a brute-force based program like Jacobi ever stands a chance at solving something like this, no matter how many conditions you add. (2023-12-14)
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paul: Compare with P0002278 & P0002279 (2010-04-30)
Mu-Tsun Tsai: This one is by far the toughest retro I've ever solved. Very little certain information can be determined by structural consideration alone, even with long and complicated argument. It took me five days to complete solving this. (2012-07-22)
A.Buchanan: @Mu-Tsun: that's an interesting data point - thanks for posting. (2017-09-07)
Henrik Juel: The current record is 58.5 moves in a proof game problem by the authors + Keym, Die Schwalbe 2017 (2017-09-07)
Henrik Juel: I just learned that the 58.5 move proof game has been cooked... (2017-09-07)
A.Buchanan: In retrospect, my earlier comment about "interesting data point" is a bit weak. It's actually great that for such an extreme problem, someone took substantial time to independently validate it. It's like doing science: people want to do their own new stuff, and are unwilling to take the time to validate what's already been claimed. This one has survived 30+ years, and maybe the use of constraints e.g. in Jacobi can eventually allow it to be HC+. (2021-05-29)
Olaf Jenkner: This problem is the current record, because P1338946 (58.5 moves) has been cooked. (2021-11-25)
Reto: This is C+ up to 51.0 moves with Stelvio 2.0. This ties the record for partial testing of an SPG. Took 1200 CPU hours of strategy seeking (finding 378 0+0 strategies) and another 13h of strategy playing these strategies. If this can ever be completely solved, then it needs to be the case that all strategies have 0+0 free moves, otherwise playing is utterly hopeless.
@Andrew: There is absolutely no way a brute-force based program like Jacobi ever stands a chance at solving something like this, no matter how many conditions you add. (2023-12-14)
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Keywords: Unique Proof Game, Move Length Record, Non-standard material (TTTTTTtttttt), Castling, Aristocrat, Superseded by (P1397486)
Genre: Retro
FEN: kQ3Br1/1b3rr1/1n1Nr2R/q1R4r/R1Rbr3/R1RRr3/NnB1rR2/5K2
Reprints: 584 Ukrainisches Album 1986-1990
86 Shortest Proof Games 11/1991
(6) diagrammes 103 10-12/1992
H18 FIDE Album 1989-1991 1997
feenschach 137, p. 368, 08-09/2000
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-12-25 more...
Genre: Retro
FEN: kQ3Br1/1b3rr1/1n1Nr2R/q1R4r/R1Rbr3/R1RRr3/NnB1rR2/5K2
Reprints: 584 Ukrainisches Album 1986-1990
86 Shortest Proof Games 11/1991
(6) diagrammes 103 10-12/1992
H18 FIDE Album 1989-1991 1997
feenschach 137, p. 368, 08-09/2000
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-12-25 more...
a) 1. 0-0? droht 2. Tf8+ Kd7 3. Lxe6#
1. ... Lxc5! 2. Sxc5 0-0-0!
1. Thf1! 0-0-0 2. Lxe6+ Td7 3. Tf8#
b) 1. Thf1? Lxc5 2. Sxc5 0-0-0!
1. 0-0! Lxc5 2. Sxc5 ... (0-0-0?) 3. Tf8#
a) zuletzt h6-h5 und beide Rochaden sind noch möglich. Schläge von Weiß: Bbxc, Bcxd, Bhxg, Bfxg-g8=L.
b) gegenseitiger Ausschluss der beiden Rochaden. 3 Fälle sind zu betrachten:
b1) zuletzt f7xDe6; Weiß muss 4x geschlagen haben: Bfxexdxc8=L, d.h. s0-0-0 ist unzulässig.
b2) zuletzt d7xDe6; zu den 4 Schlägen von a) wird auch der sBf gebraucht, der sich auf f1 umwandeln muss, also w0-0 unzulässig.
b3) zuletzt h6-h5: wBf zieht über f7 zum UW-Feld g8, g.h. s0-0-0 unzulässig.
1. ... Lxc5! 2. Sxc5 0-0-0!
1. Thf1! 0-0-0 2. Lxe6+ Td7 3. Tf8#
b) 1. Thf1? Lxc5 2. Sxc5 0-0-0!
1. 0-0! Lxc5 2. Sxc5 ... (0-0-0?) 3. Tf8#
a) zuletzt h6-h5 und beide Rochaden sind noch möglich. Schläge von Weiß: Bbxc, Bcxd, Bhxg, Bfxg-g8=L.
b) gegenseitiger Ausschluss der beiden Rochaden. 3 Fälle sind zu betrachten:
b1) zuletzt f7xDe6; Weiß muss 4x geschlagen haben: Bfxexdxc8=L, d.h. s0-0-0 ist unzulässig.
b2) zuletzt d7xDe6; zu den 4 Schlägen von a) wird auch der sBf gebraucht, der sich auf f1 umwandeln muss, also w0-0 unzulässig.
b3) zuletzt h6-h5: wBf zieht über f7 zum UW-Feld g8, g.h. s0-0-0 unzulässig.
Henrik Juel: C+ Popeye 4.61 after analysis (2020-10-30)
A.Buchanan: A very harmonious problem. Suppose that both sides retain castling rights. wLh3 is obtrusive, so wBf/h promoted via g7, and escaped via e6 so sBe6 came from d7 later. Wh has made at least 4 pcs, while Bl has made 2 (D & Lc) dxe6, gxh6. If last move was dxe6, then sD & sLc were captured by officers in cage. Thus sBf was captured by wB, and therefore promoted on f1, disrupting wK. Therefore the last move was h6-h5. In (a) there is no problem retaining both castling rights, but in (b) Sg8 must have reached that square via f6 *after* the promotion, disrupting bK. However in (b) the castling rights are incompatible. In (a) 1. 0-0? Lxc5! 1. Tf1!, while in (b) the reverse.
Neither twin is a PRA problem: in (a) there is no conflict between the castling, while in (b) RS applies: the pre-emptive Wh castling means we are in a reality where Bl cannot castle. (2020-10-31)
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A.Buchanan: A very harmonious problem. Suppose that both sides retain castling rights. wLh3 is obtrusive, so wBf/h promoted via g7, and escaped via e6 so sBe6 came from d7 later. Wh has made at least 4 pcs, while Bl has made 2 (D & Lc) dxe6, gxh6. If last move was dxe6, then sD & sLc were captured by officers in cage. Thus sBf was captured by wB, and therefore promoted on f1, disrupting wK. Therefore the last move was h6-h5. In (a) there is no problem retaining both castling rights, but in (b) Sg8 must have reached that square via f6 *after* the promotion, disrupting bK. However in (b) the castling rights are incompatible. In (a) 1. 0-0? Lxc5! 1. Tf1!, while in (b) the reverse.
Neither twin is a PRA problem: in (a) there is no conflict between the castling, while in (b) RS applies: the pre-emptive Wh castling means we are in a reality where Bl cannot castle. (2020-10-31)
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Keywords: Castling key (wksg), Obvious promotion (L), Retro Strategy (RS)
Genre: Retro, 3#
FEN: r3k3/ppp1p2p/Nb2p3/P1PP2Pp/4N3/7B/3PPRP1/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2022-01-07 more...
Genre: Retro, 3#
FEN: r3k3/ppp1p2p/Nb2p3/P1PP2Pp/4N3/7B/3PPRP1/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2022-01-07 more...
8 - P0000250
Nikita M. Plaksin
Valery Liskovets
7577v Die Schwalbe 132 12/1991
(5+14)
ser-h#5 (AP)
Typ Tauber/Caillaud
Nikita M. Plaksin
Valery Liskovets
7577v Die Schwalbe 132 12/1991
(5+14)
ser-h#5 (AP)
Typ Tauber/Caillaud
1. Kd4 2. bxc3ep (zuletzt nur K,T-Zug oder c2-c4) 3. e5! (nicht 4. Tb4? weil zuletzt K,T-Zug und 0-0-0 nicht mehr möglich) 4. Tb4 (zuletzt e7xLd8=S möglich) 5. Tc4 & 1. 0-0-0# nicht 1. Td1#? wegen fehlender AP-Legalisierung
"das zeigt die Paradoxie der AP-Bedingung!" (MS). Gleich 1. bxc3 ep.? und dann 2. Kd4 geht natürlich nicht, weil zuvor außer K,T-Zug und c2-c4 auch S-Züge und c3-c4 möglich waren. "Interessanter und genau abgestimmter Lösungsverlauf" (GW) "Sehr schöne und ökonomische Verbindung zwischen konsequentem sh# und AP!" (TB) Mit 'Typ Tauber/Caillaud' ist gemeint, daß die einzelnen Stellungen doch als retroanlytischer Gesamtkomplex betrachtet werden dürfen/sollen (sonst wäre keine AP-Legalisierung möglich), obwohl sie ja aufgrund der retroanalytischen Neubewertung nach jedem Rückzug voneinander unabhängig sind. 6L./5+5P.
A.Buchanan: "By 'type Tauber/Caillaud' it is meant that the individual positions may/should be regarded as a retroanalytic overall complex (otherwise no AP legalization would be possible), although they are independent of each other due to the retroanalytic reassessment after each retraction." (2023-06-29)
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A.Buchanan: "By 'type Tauber/Caillaud' it is meant that the individual positions may/should be regarded as a retroanalytic overall complex (otherwise no AP legalization would be possible), although they are independent of each other due to the retroanalytic reassessment after each retraction." (2023-06-29)
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Keywords: Castling (wl), a posteriori (AP), Promotion (S), Seriesmover, Consequent, En passant
Genre: Retro, Fairies
FEN: 1N6/1ppppppn/n7/1Pq1k3/1pP1r3/4p3/1r6/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-29 more...
Genre: Retro, Fairies
FEN: 1N6/1ppppppn/n7/1Pq1k3/1pP1r3/4p3/1r6/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-29 more...
9 - P0000254
Leonid M. Borodatow
7642 Die Schwalbe 133 02/1992
(16+10)
Welches waren die letzten 7 Einzelzüge, wenn dabei keine Zugwiederholungen vorkamen?
Leonid M. Borodatow
7642 Die Schwalbe 133 02/1992
(16+10)
Welches waren die letzten 7 Einzelzüge, wenn dabei keine Zugwiederholungen vorkamen?
R: 1. 0-0-0# Ke4-d4 2. e5xf6ep+ f7-f5 3. Tg6-b6+ Kf5-e4 4. c7-c8=L
Die von einigen Lösern angeführte Abweichung 2. f5-f6+ Kd4-e4 3. Lh6-g7+ (und mehrdeutig weiter) ließe sich durch die Erweiterung '... keine Zugwiederholungen und keine Pendelzüge ...' (mühsam) kitten. Beim Autor hieß es bei dieser ich-weiß-nicht-wie-vielten Fassung nur 'letzte 9 (!) Einzelzüge ohne Wiederholung).
HHS meint ohnehin, daß es das ganze auch ohne die einengende Zusatzbedingung schon gibt.
Das von einem Löser angegebene 1. Ld3-h7# Th1-h8 2. Lh8-g7 Tg1-h1 3. Se1-g3 g2-g1=T 4. Th7-h8=L scheitert allerdings an der Schlagbilanz.
Die von einigen Lösern angeführte Abweichung 2. f5-f6+ Kd4-e4 3. Lh6-g7+ (und mehrdeutig weiter) ließe sich durch die Erweiterung '... keine Zugwiederholungen und keine Pendelzüge ...' (mühsam) kitten. Beim Autor hieß es bei dieser ich-weiß-nicht-wie-vielten Fassung nur 'letzte 9 (!) Einzelzüge ohne Wiederholung).
HHS meint ohnehin, daß es das ganze auch ohne die einengende Zusatzbedingung schon gibt.
Das von einem Löser angegebene 1. Ld3-h7# Th1-h8 2. Lh8-g7 Tg1-h1 3. Se1-g3 g2-g1=T 4. Th7-h8=L scheitert allerdings an der Schlagbilanz.
Keywords: En passant, Last Moves?, Non-standard material, Castling (wl), Promotion (L), Valladao Task (WWW)
Genre: Retro
FEN: qrB2brr/Bp2p1BB/pR3P2/1Q6/2Pk1P2/B1p2R2/2P3N1/2KR1N2
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2019-08-11 more...
Genre: Retro
FEN: qrB2brr/Bp2p1BB/pR3P2/1Q6/2Pk1P2/B1p2R2/2P3N1/2KR1N2
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2019-08-11 more...
10 - P0000548
Alexander Kislyak
3463 Die Schwalbe 68 04/1981
Dr. L. Ceriani zum Gedenken
(10+13) cooked
BP in 46,0
Alexander Kislyak
3463 Die Schwalbe 68 04/1981
Dr. L. Ceriani zum Gedenken
(10+13) cooked
BP in 46,0
AL: 1. d4 a5 2. c4 a4 3. b4 a3 4. Lb2 axb2 5. a4 c5 6. a5 b5 7. a6 Lb7 8. a7 Da5 9. bxa5 Sc6 10. a6 0-0-0 11. a8=S Sa5 12. Sc7 b4 13. a7 b3 14. a8=S Sf6 15. Sb6 Kb8 16. Se6 dxe6 17. Sd7 Ka8 18. Sxf8 Td5 19. cxd5 c4 20. d6 c3 21. d7 c2 22. d5 c1=L 23. d6 Ld5 24. Ta4 Se8 25. Tf4 f6 26. Sc3 b1=L 27. d8=S b2 28. d7 Lg6 29. Sb7 b1=L 30. d8=S Lf7 31. Sg6 Lf5 32. Sd6 Lg4 33. Sf5 hxg6 34. Sb7 gxf5 35. Sc5 Th4 36. S5e4 fxe4 37. e3 Lf5 38. Ke2 Tg4 39. h4 Sb3 40. h5 Sa1 41. h6 Ld2 42. h7 Le1 43. h8=S Sc7 44. Sg6 Sb5 45. Sh4 Sa3 46. S4f3 exf3+
Korrektur siehe 3463v
James Malcom: How in the world can a non-unique PG be cooked? I'm unmarking it, for now. (2021-01-25)
A.Buchanan: Maybe the intended theme was not forced? Haven’t looked at this problem though (2021-01-26)
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James Malcom: How in the world can a non-unique PG be cooked? I'm unmarking it, for now. (2021-01-25)
A.Buchanan: Maybe the intended theme was not forced? Haven’t looked at this problem though (2021-01-26)
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Keywords: Non-Unique Proof Game, Ceriani-Frolkin Theme, Non-standard material (ll), Castling, konsekutive Umwandlungen 8, Promotion, Superseded by (P0000582)
Genre: Retro
FEN: k7/4pbp1/4pp2/3b1b2/5Rr1/n1N1Pp2/4KPP1/n2QbBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-01-26 more...
Genre: Retro
FEN: k7/4pbp1/4pp2/3b1b2/5Rr1/n1N1Pp2/4KPP1/n2QbBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-01-26 more...
11 - P0000582
Alexander Kislyak
3463v Die Schwalbe 72 12/1981
Dr. L. Ceriani zum Gedenken
(10+13) cooked
BP in 46,0
Alexander Kislyak
3463v Die Schwalbe 72 12/1981
Dr. L. Ceriani zum Gedenken
(10+13) cooked
BP in 46,0
AL: 1. b4 a5 2. c4 a4 3. d4 a3 4. Lb2 axb2 5. a4 b5 6. a5 c5 7. a6 Da5 8. a7 Sc6 9. bxa5 Lb7 10. a6 0-0-0 11. a8=S b4 12. Sc7 Sa7 13. Se6 Sb5 14. Sxf8 Sh6 15. a7 f6 16. a8=S Sf7 17. Sc7 b3 18. Sce6 dxe6 19. Ta4 Td5 20. cxd5 c4 21. d6 c3 22. d5 c2 23. Sc3 Sg5 24. d7+ Kb8 25. d6 c1=L 26. d8=S b1=L 27. d7 b2 28. Sf7 Lf5 29. Sd6 Lh3 30. d8=S b1=L 31. S8f7 Lg6 32. Sh6 Lf7 33. Sg6 hxg6 34. Shf5 Th4 35. Tf4 gxf5 36. Sde4 fxe4 37. e3 Lf5 38. Ke2 Tg4 39. h4 Sa3 40. h5 Sb1 41. h6 Ka8 42. h7 Ld5 43. h8=S Ld2 44. Sg6 Le1 45. Sh4 Sh3 46. Shf3 exf3+
Cook: NL: 1. b4 a5 2. c4 a4 3. d4 a3 4. Lb2 axb2 5. a4 b5 6. a5 c5 7. a6 Sc6 8. a7 Lb7 9. Ta6 Da5 10. bxa5 0-0-0 11. a8=S b4 12. Sc7 b3 13. Se6 Sb4 14. Sxf8 Sf6 15. Te6 dxe6 16. a6 Td5 17. cxd5 c4 18. a7 c3 19. a8=S c2 20. Sc3 c1=L 21. Sb6+ Kb8 22. Sc8 b1=L 23. d6 Lg6 24. d7 b2 25. d8=D b1=L 26. Da5 Lbf5 27. d5 Se4 28. d6 f6 29. d7 Lf7 30. Sg6 hxg6 31. d8=T Lh3 32. Td4 Th4 33. Df5 gxf5 34. Sd6 Sg5 35. Tf4 Ka8 36. S6e4 fxe4 37. e3 Lf5 38. Ke2 Tg4 39. h4 Sc2 40. h5 Sa3 41. h6 Sb1 42. h7 Ld5 43. h8=S Sh3 44. Sg6 Ld2 45. Sh4 Le1 46. Shf3 exf3+
Cook: NL: 1. b4 a5 2. c4 a4 3. d4 a3 4. Lb2 axb2 5. a4 b5 6. a5 c5 7. a6 Sc6 8. a7 Lb7 9. Ta6 Da5 10. bxa5 0-0-0 11. a8=S b4 12. Sc7 b3 13. Se6 Sb4 14. Sxf8 Sf6 15. Te6 dxe6 16. a6 Td5 17. cxd5 c4 18. a7 c3 19. a8=S c2 20. Sc3 c1=L 21. Sb6+ Kb8 22. Sc8 b1=L 23. d6 Lg6 24. d7 b2 25. d8=D b1=L 26. Da5 Lbf5 27. d5 Se4 28. d6 f6 29. d7 Lf7 30. Sg6 hxg6 31. d8=T Lh3 32. Td4 Th4 33. Df5 gxf5 34. Sd6 Sg5 35. Tf4 Ka8 36. S6e4 fxe4 37. e3 Lf5 38. Ke2 Tg4 39. h4 Sc2 40. h5 Sa3 41. h6 Sb1 42. h7 Ld5 43. h8=S Sh3 44. Sg6 Ld2 45. Sh4 Le1 46. Shf3 exf3+
James Malcom: Again, how is this cooked? (2021-01-25)
A.Buchanan: The AL has 5 Wh S promotions in 46.0 moves but the cook shows this is not necessary. (2021-01-26)
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A.Buchanan: The AL has 5 Wh S promotions in 46.0 moves but the cook shows this is not necessary. (2021-01-26)
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Keywords: Non-Unique Proof Game, Ceriani-Frolkin Theme (SSSSS), Non-standard material (ll), Castling, Promotion, konsekutive Umwandlungen 8
Genre: Retro
FEN: k7/4pbp1/4pp2/3b1b2/5Rr1/2N1Pp1n/4KPP1/1n1QbBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-01-26 more...
Genre: Retro
FEN: k7/4pbp1/4pp2/3b1b2/5Rr1/2N1Pp1n/4KPP1/1n1QbBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-01-26 more...
1. ... Kgxf4 2. Tf6 e6 3. Tf8 Sg7#
Cook: 1. ... exf6ep 2. 0-0-0 gxf4 3. Td7 a8=D
Failed try, as Bg4xBh5 allows bRh8 to have escaped the cage prior to ep.
Cook: 1. ... exf6ep 2. 0-0-0 gxf4 3. Td7 a8=D
Failed try, as Bg4xBh5 allows bRh8 to have escaped the cage prior to ep.
Anton Baumann: Sollte eine Verbesserung von P0000777 sein.
Autorabsicht: "letzter Zug f7-f5 sonst retropatt! Jetzt ist aber die Rochade illegal, da der sTh8 nur über e8 aus seiner Ecke kommt. Daher Verführung 1. ... exf6 e.p. 2.0-0-0? Lxf4 3.Td7 a8=D#. Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 sg7#"
Aber: Der s h-Bauer kann auch auf h5 geschlagen worden sein; so konnte der sT über h6 aus seiner Ecke entkommen. Die vermeintliche Verführung ist also NL! (2022-12-14)
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Autorabsicht: "letzter Zug f7-f5 sonst retropatt! Jetzt ist aber die Rochade illegal, da der sTh8 nur über e8 aus seiner Ecke kommt. Daher Verführung 1. ... exf6 e.p. 2.0-0-0? Lxf4 3.Td7 a8=D#. Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 sg7#"
Aber: Der s h-Bauer kann auch auf h5 geschlagen worden sein; so konnte der sT über h6 aus seiner Ecke entkommen. Die vermeintliche Verführung ist also NL! (2022-12-14)
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Keywords: Castling (sg), Superseded by (P0000058)
Genre: h#, Retro
FEN: r3k3/P7/p1r3pP/4PpBN/3ppnKn/6P1/1PP2PPq/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
Genre: h#, Retro
FEN: r3k3/P7/p1r3pP/4PpBN/3ppnKn/6P1/1PP2PPq/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
a) 1. ... Kxa1 2. 0-0#
b) 1. ... Kxh1 2. 0-0-0#
b) 1. ... Kxh1 2. 0-0-0#
A.Buchanan: The term "retro" is jungle not garden - that means we should not expect an axiomatic definition. The current problem is a case in point. Case law has established that neither simple employment of the castling convention nor existence of check are sufficient to make a problem "retro". But all this problem has is the quirky use of Codex Article 15 to force BTM. So I think this problem has to be retro. The key point is that nothing hinges on the retro-ness. If the problem included 50M or DP, then one would expect a more solid foundation. As it is, all we need is the free direct mate in 1 that comes as part of the retro paradigm. (2021-11-26)
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Keywords: Castling (wk), No legal last move for Black, Minimal, Miniature
Genre: Retro
Computer test: C+ Popeye v4.87
FEN: 8/8/8/8/8/4p3/4Q3/Rk2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-26 more...
Genre: Retro
Computer test: C+ Popeye v4.87
FEN: 8/8/8/8/8/4p3/4Q3/Rk2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-26 more...
14 - P0000775
Ferad Kakabadze
1422 Die Schwalbe 30 12/1974
(11+5) C+
#3
Welches waren die letzten 2 Einzelzüge?
Ferad Kakabadze
1422 Die Schwalbe 30 12/1974
(11+5) C+
#3
Welches waren die letzten 2 Einzelzüge?
1. e8=D! f5 2. exf6ep e5 3. Dxe5#
R: 1. b2-b1=L 0-0
R: 1. b2-b1=L 0-0
Henrik Juel: The forward play is C+ Popeye 4.61
1.e8=D thr. 2.Dc8 thr. 3.Dc3#
1... f5 2.exf6ep e5 3.Dxe5#
(1... f6 does not prevent the threat, so 2.exf6 is not considered dualistic) (2020-09-26)
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1.e8=D thr. 2.Dc8 thr. 3.Dc3#
1... f5 2.exf6ep e5 3.Dxe5#
(1... f6 does not prevent the threat, so 2.exf6 is not considered dualistic) (2020-09-26)
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Keywords: Last Moves?, Valladao Task, En passant, Promotion, Castling in the retro play, Obvious promotion
Genre: Retro, 3#
FEN: 8/3pPp2/3Pp3/4P3/1N3P2/B7/P1P2P2/kb3RK1
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2020-09-27 more...
Genre: Retro, 3#
FEN: 8/3pPp2/3Pp3/4P3/1N3P2/B7/P1P2P2/kb3RK1
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2020-09-27 more...
1. ... exf6ep 2. 0-0-0? Lxf4 3. Td7 a8=D# try
1. ... Kxf4 2. Tf6 e6 3. Tf8 Sg7# solution
Cook: 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
1. ... Kxf4 2. Tf6 e6 3. Tf8 Sg7# solution
Cook: 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
See P0000674
Henrik Juel: The stipulation should probably read h#2.5: 1.exf6ep 0-0-0 2.Bxf4 Rd7 3.a8Q# (2003-08-18)
Anton Baumann: Letzter Zug war f7-f5, sonst retropatt. Die weissen Bauern haben 6x geschlagen. Der s a-Bauer wurde so ebenfalls als Schlagopfer benötigt und musste 2x schlagen, um sich umwandeln zu können. Weiss hat also auf h6 den s Bauern geschlagen. Unabhängig davon ob der sTe6 ein Umwandlungsturm ist oder nicht, muss der sTh8 über e8 aus der Ecke gekommen sein. Also ist die Rochade illegal!
Autorabsicht: 1. ... exf6 e.p. 2.0-0-0 Lxf4 3.Td7 a8=D# ist die Verführung!
Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 Sg7#
leider geht auch: 1. ... Kxf4 2.Te7 e6 3.Tb8 axb8=D,T# (2022-12-13)
A.Buchanan: I agree Anton. I suggest as a fix -bBa6 +bDc2. I think retro, solution and try all work correctly now. Do you folks agree? (2022-12-13)
Anton Baumann: Der Vorschlag von Andrew ergibt eine korrekte Lösung, aber nicht im Sinne des Autors: ohne einen s Stein auf a6 ist f6-f5+ als letzter Zug möglich, d.h. kein Retropatt weil Weiss a6-a7 zurücknehmen kann! 1. ... exf6 e.p. ist deshalb nicht zulässig. LB wollte aber dass f7-f5 der letzte Zug war, damit aber die Rochade illegal ist! (2022-12-14)
Anton Baumann: Mögliche Korrektur: a6=sL, g2=sD (2022-12-15)
A.Buchanan: Yes I agree. I had overlooked the unmove of wPa7 (2022-12-15)
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Henrik Juel: The stipulation should probably read h#2.5: 1.exf6ep 0-0-0 2.Bxf4 Rd7 3.a8Q# (2003-08-18)
Anton Baumann: Letzter Zug war f7-f5, sonst retropatt. Die weissen Bauern haben 6x geschlagen. Der s a-Bauer wurde so ebenfalls als Schlagopfer benötigt und musste 2x schlagen, um sich umwandeln zu können. Weiss hat also auf h6 den s Bauern geschlagen. Unabhängig davon ob der sTe6 ein Umwandlungsturm ist oder nicht, muss der sTh8 über e8 aus der Ecke gekommen sein. Also ist die Rochade illegal!
Autorabsicht: 1. ... exf6 e.p. 2.0-0-0 Lxf4 3.Td7 a8=D# ist die Verführung!
Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 Sg7#
leider geht auch: 1. ... Kxf4 2.Te7 e6 3.Tb8 axb8=D,T# (2022-12-13)
A.Buchanan: I agree Anton. I suggest as a fix -bBa6 +bDc2. I think retro, solution and try all work correctly now. Do you folks agree? (2022-12-13)
Anton Baumann: Der Vorschlag von Andrew ergibt eine korrekte Lösung, aber nicht im Sinne des Autors: ohne einen s Stein auf a6 ist f6-f5+ als letzter Zug möglich, d.h. kein Retropatt weil Weiss a6-a7 zurücknehmen kann! 1. ... exf6 e.p. ist deshalb nicht zulässig. LB wollte aber dass f7-f5 der letzte Zug war, damit aber die Rochade illegal ist! (2022-12-14)
Anton Baumann: Mögliche Korrektur: a6=sL, g2=sD (2022-12-15)
A.Buchanan: Yes I agree. I had overlooked the unmove of wPa7 (2022-12-15)
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Keywords: Castling (sg), Valladao Task, Superseded by (P0000058)
Genre: h#, Retro
FEN: r3k3/P7/p3r1pP/4PpBN/3nnpKP/6PB/1P2PPb1/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
Genre: h#, Retro
FEN: r3k3/P7/p3r1pP/4PpBN/3nnpKP/6PB/1P2PPb1/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
16 - P0000792
Klaus Wenda
1557 Die Schwalbe 33 06/1975
2. Preis
(13+12)
#2 Längstzüger
b) sTa7 nach d7
Klaus Wenda
1557 Die Schwalbe 33 06/1975
2. Preis
(13+12)
#2 Längstzüger
b) sTa7 nach d7
Anton Baumann: Autorabsicht: Die weiss-schwarzen Rochaden schliessen sich gegenseitig aus.
a) 1.O-O? Tf8! daher: 1.Tf1! O-O 2.Sxe7#
b) 1.Tf1? O-O! daher: 1.O-O! Tf8 2.Sxg7#
Aber in der Urfassung (= nebenstehendes Diagramm) geht in a) und b) die NL:
1.Tg1 O-O 2.Txg7,Sf5xh6#
Korrektur in 'Schwalbe' 04/1976 S.464: sLb7 nach g6, sBc5 nach b7
Ausgezeichnet wurde die korrigierte Fassung 1557v (vergl. 'Die Schwalbe' 06/1977 S.82) (2022-12-09)
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a) 1.O-O? Tf8! daher: 1.Tf1! O-O 2.Sxe7#
b) 1.Tf1? O-O! daher: 1.O-O! Tf8 2.Sxg7#
Aber in der Urfassung (= nebenstehendes Diagramm) geht in a) und b) die NL:
1.Tg1 O-O 2.Txg7,Sf5xh6#
Korrektur in 'Schwalbe' 04/1976 S.464: sLb7 nach g6, sBc5 nach b7
Ausgezeichnet wurde die korrigierte Fassung 1557v (vergl. 'Die Schwalbe' 06/1977 S.82) (2022-12-09)
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Keywords: Maximummer, Castling (wksk)
Genre: Retro, Fairies
FEN: 4k2r/rb2pNbp/1P5p/p1pppN2/8/8/PPPPP2P/2BQK2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
Genre: Retro, Fairies
FEN: 4k2r/rb2pNbp/1P5p/p1pppN2/8/8/PPPPP2P/2BQK2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
Kees: +wKb5 1. Txc8 (2. Lxe7#)
0-0 is illegal for K or T must have made a move.
-1. a7-a5? Not possible with position of wL and bS (2022-11-23)
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0-0 is illegal for K or T must have made a move.
-1. a7-a5? Not possible with position of wL and bS (2022-11-23)
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Keywords: Castling (sk), Add pieces
Genre: Retro
FEN: 2nBk2r/3pp3/1p1p2P1/p4NN1/PP4p1/7b/PP2P1Pp/2R2B2
Input: Gerd Wilts, 1995-06-03
Genre: Retro
FEN: 2nBk2r/3pp3/1p1p2P1/p4NN1/PP4p1/7b/PP2P1Pp/2R2B2
Input: Gerd Wilts, 1995-06-03
a) 1. T2xh3 Txh3 2. 0-0 Th8#
b) 1. La4 0-0 2. Tf8 Te1#
a) White cannot cross-capture, so because of sTh2, w0-0 impossible.
wBe captured sD, either to be captured on f-file or to promote without disrupting sKe8.
b) sTh2 can return to a8 only via e8 as Black cannot cross-capture, so s0-0 is impossible.
wBe was waylaid or promoted after disrupting sKe8.
Cook: 1. T2xh3 Lb4 2. Txg3 Txh8#
1. Sd2 Lf6 2. Tf8 Sd6#
(cooked by MR)
b) 1. La4 0-0 2. Tf8 Te1#
a) White cannot cross-capture, so because of sTh2, w0-0 impossible.
wBe captured sD, either to be captured on f-file or to promote without disrupting sKe8.
b) sTh2 can return to a8 only via e8 as Black cannot cross-capture, so s0-0 is impossible.
wBe was waylaid or promoted after disrupting sKe8.
Cook: 1. T2xh3 Lb4 2. Txg3 Txh8#
1. Sd2 Lf6 2. Tf8 Sd6#
(cooked by MR)
See P0003736 a companion problem.
milan: [-bPa7b6wPg3bSa1+bPg3bBa5-b4 bPc6-d6] h#2 1.2...bRh2? illegal move? M.Frelih (2023-03-30)
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milan: [-bPa7b6wPg3bSa1+bPg3bBa5-b4 bPc6-d6] h#2 1.2...bRh2? illegal move? M.Frelih (2023-03-30)
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Keywords: Cant Castler, Castling (wksk), Cross-capture (s,w), Superseded by (P1399805)
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: B3k2r/pN1p1p2/1pp3p1/bb3p2/8/p1B3PP/5P1r/nn2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-16 more...
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: B3k2r/pN1p1p2/1pp3p1/bb3p2/8/p1B3PP/5P1r/nn2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-16 more...
Set Play
1. ... Kxh1 2. Kf2#
1. ... Kxh3 2. Sf4#
1. ... Kxf3 2. Lb7#
BTM
1. ... hxg5 2. Sc3
2. ... Kxh1 3. Kf2#
2. ... Kxh3 3. Lf1#
2. ... Kxf3 3. 0-0#
Cook: 1. ... hxg5 2. Kd2 Kxh3 3. Df1# 2. ... Kxf3 3. Lb7#
1. ... Kxh1 2. Kf2#
1. ... Kxh3 2. Sf4#
1. ... Kxf3 2. Lb7#
BTM
1. ... hxg5 2. Sc3
2. ... Kxh1 3. Kf2#
2. ... Kxh3 3. Lf1#
2. ... Kxf3 3. 0-0#
Cook: 1. ... hxg5 2. Kd2 Kxh3 3. Df1# 2. ... Kxf3 3. Lb7#
A.Buchanan: We can +wBd2, but then need to e.g. transpose wLg7 & wBg5, otherwise the wBB are impossibly concentrated in corner. Then I think it's sound. (2021-10-08)
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Keywords: Castling (wk), No legal last move for Black, Superseded by (P1394496)
Genre: Retro
Computer test: C- Popeye v4.87
FEN: 8/6BR/B6p/6PP/3P2Np/5P1P/4N1kP/Q3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-12 more...
Genre: Retro
Computer test: C- Popeye v4.87
FEN: 8/6BR/B6p/6PP/3P2Np/5P1P/4N1kP/Q3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-12 more...
a) 1. 0-0-0? Sg6 2. Lb8 Sxe7! illegal castling
1. Kf7! Sxh7 2. Ke6 Sg5#
b) 1. 0-0-0! Sg6 2. Lb8 Sxe7#
1. Kf7? Sh7 2. Ke6 Sg5+? 3. hxg5!
R: 1. Ke3-e4 Sd6-b7 2. h2xDg3 Dg2-g3+ 3. Kd2-e3 Db7-g2 4. Ke1-d2 Dc8-b7 5. e4-e5 Dd8-c8 6. f3xLe4 Lb7-e4 7. Kf1-e1 Lc8-b7 8. Ke1-f1 b7-b6 9. Kf1-e1 Lb6-a7 10. ... La5-b6 11. ... Ld2-a5 12. ... Lc1-d2 13. ... d2xTc1=L 14. ... e3xTd2 15. ... e4-e3 16. ... f5xLe4
Missing: Wh: QRRBB Bl: QRBB
Captures: Wh: gxf3xe (inc B), hxg + [Bf8] Bl: axbxc (for Ba7) fxexdxc/e (g1 not possible)
Assume Black can castle: so neither bK nor bRa have moved. Before bPb7-b6 (which releases QB) *all* Bl captures have been made. wPd can freely advance, all Wh units released except for rooks & Bf1. gxRf3 is forced, and now all Wh units are free and can be captured. To avoid deadlock, wB was captured on e4 not e6. Sequence must be f7-f5 Rh8-f8-f6-...-f3gxRf3.
In this position wLe4 must be played back to f1. With bPh7, wBe4 must retract either by stopping on f7, (disrupting bK) or via f5 (implying retraction of f6-f5, in which case bRa8 is an imposter).
In the alternative route via h7, wB crosses over f7 harmlessly. While wB is on g5 & h4, f7 must be occupied by a static shielding knight, but there is no tempo issue. After all this excitement, b7-b6 if followed by simple and non-unique play to reach the diagram.
(Gerd's earlier solution: Weiße Schläge: h2xg3, gxfxe, sLf8. Schwarze Schläge: a7xb6xc5; fxexdxc1=L In dieser Stellung muß der wLe4 nach f1 zurückgespielt werden. Mit sBh7 kann der wLe4 nur entweder über f7 nach f1 zurück, so daß der sK bereits gezogen haben muß, oder der wLe4 kann über f5 zurück, wozu aber der sTa8 nach h8 zurückgezogen werden müsste, um f6-f5 zurücknehmen zu können.)
1. Kf7! Sxh7 2. Ke6 Sg5#
b) 1. 0-0-0! Sg6 2. Lb8 Sxe7#
1. Kf7? Sh7 2. Ke6 Sg5+? 3. hxg5!
R: 1. Ke3-e4 Sd6-b7 2. h2xDg3 Dg2-g3+ 3. Kd2-e3 Db7-g2 4. Ke1-d2 Dc8-b7 5. e4-e5 Dd8-c8 6. f3xLe4 Lb7-e4 7. Kf1-e1 Lc8-b7 8. Ke1-f1 b7-b6 9. Kf1-e1 Lb6-a7 10. ... La5-b6 11. ... Ld2-a5 12. ... Lc1-d2 13. ... d2xTc1=L 14. ... e3xTd2 15. ... e4-e3 16. ... f5xLe4
Missing: Wh: QRRBB Bl: QRBB
Captures: Wh: gxf3xe (inc B), hxg + [Bf8] Bl: axbxc (for Ba7) fxexdxc/e (g1 not possible)
Assume Black can castle: so neither bK nor bRa have moved. Before bPb7-b6 (which releases QB) *all* Bl captures have been made. wPd can freely advance, all Wh units released except for rooks & Bf1. gxRf3 is forced, and now all Wh units are free and can be captured. To avoid deadlock, wB was captured on e4 not e6. Sequence must be f7-f5 Rh8-f8-f6-...-f3gxRf3.
In this position wLe4 must be played back to f1. With bPh7, wBe4 must retract either by stopping on f7, (disrupting bK) or via f5 (implying retraction of f6-f5, in which case bRa8 is an imposter).
In the alternative route via h7, wB crosses over f7 harmlessly. While wB is on g5 & h4, f7 must be occupied by a static shielding knight, but there is no tempo issue. After all this excitement, b7-b6 if followed by simple and non-unique play to reach the diagram.
(Gerd's earlier solution: Weiße Schläge: h2xg3, gxfxe, sLf8. Schwarze Schläge: a7xb6xc5; fxexdxc1=L In dieser Stellung muß der wLe4 nach f1 zurückgespielt werden. Mit sBh7 kann der wLe4 nur entweder über f7 nach f1 zurück, so daß der sK bereits gezogen haben muß, oder der wLe4 kann über f5 zurück, wozu aber der sTa8 nach h8 zurückgezogen werden müsste, um f6-f5 zurücknehmen zu können.)
Henrik Juel: To make the retroplay plausible one should uncapture bQ early on g3 and retract it to d8. The wB could also get back to f1 via f5, but this would require retracting bRa8 to h8 before retracting bPf5, so castling is still illegal in part a). (2003-04-10)
Gerd Wilts: Hello Henrik, thank you for pointing out the inaccuracy of the solution, I will make the solution more precise soon. And thank you for adding all the other solutions! (2003-04-11)
A.Buchanan: Have posted a solution based on GW&HJ ideas. More often a j’adoube of a rook pawn signals a tempo idea, but not here. The surprising motivation harmonizes with accurate and varied forward play. (2021-10-24)
A.Buchanan: This problem was featured in yesterday's Monthly International Zoom Call, and was a great success, with audible gasps of appreciation, as the significance of h7 was realized :) (2021-10-24)
A.Buchanan: Oh dear, Alfred Pfeiffer has silently reverted the German keyword to "mit Umwandlungsfigur". I'm not going to enter an "edit war" with him, but I would appreciate if he can explain his position here. This is a problematic concept to keyword, but to me "mit Umwandlungfigur" is weak and inaccurate. What is the intended distinction with the existing keywords "Umwandlung" or "Umwandlungen"? It's hopeless. We have in PDB very many poor promotion keywords, and I would like to clean up progressively. I don't know if a native German speaker would care to engage with Alfred on this point. (2021-10-30)
A.Buchanan: To be clearer: to me the German definition seems pretty good. I think the term itself should give more of a clue what's happening :-) (2021-10-30)
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Gerd Wilts: Hello Henrik, thank you for pointing out the inaccuracy of the solution, I will make the solution more precise soon. And thank you for adding all the other solutions! (2003-04-11)
A.Buchanan: Have posted a solution based on GW&HJ ideas. More often a j’adoube of a rook pawn signals a tempo idea, but not here. The surprising motivation harmonizes with accurate and varied forward play. (2021-10-24)
A.Buchanan: This problem was featured in yesterday's Monthly International Zoom Call, and was a great success, with audible gasps of appreciation, as the significance of h7 was realized :) (2021-10-24)
A.Buchanan: Oh dear, Alfred Pfeiffer has silently reverted the German keyword to "mit Umwandlungsfigur". I'm not going to enter an "edit war" with him, but I would appreciate if he can explain his position here. This is a problematic concept to keyword, but to me "mit Umwandlungfigur" is weak and inaccurate. What is the intended distinction with the existing keywords "Umwandlung" or "Umwandlungen"? It's hopeless. We have in PDB very many poor promotion keywords, and I would like to clean up progressively. I don't know if a native German speaker would care to engage with Alfred on this point. (2021-10-30)
A.Buchanan: To be clearer: to me the German definition seems pretty good. I think the term itself should give more of a clue what's happening :-) (2021-10-30)
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Keywords: Cant Castler, Castling (sg), Promotion (l), Obvious promotion (l), Corridor, Retro Shield
Genre: h#, Retro
Computer test: Forward: C+ Popeye V4.87 Retro: non-trivial reasoning
FEN: r3kN2/bnppp1pp/1p6/2p1P3/4K3/3P2P1/PPP1PP2/N6n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-30 more...
Genre: h#, Retro
Computer test: Forward: C+ Popeye V4.87 Retro: non-trivial reasoning
FEN: r3kN2/bnppp1pp/1p6/2p1P3/4K3/3P2P1/PPP1PP2/N6n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-30 more...
Schwarz hat bereits rochiert!
See P1398939
Henrik Juel: Black captured b7xLc6 and axbxc; White captured [Lf8], e2xd3, and fxexdxc6xb7-b8=D.
Ke8 to b7, Th8 to b8, Kb7 to g8, Tb8 to f8, Da4 to b8 for unpromotion, -22.c6xDb7 Da8 -23.d5xLc6 Dd8 -24.d4 Lb7 -25.e3xTd4 Lc8 -26.f2xSe3 c6 -27.h5 b7xLc6, Lc6 to f1, f2xSe3 etc.
Black may not castle, because he already did.
It is not possible to avoid the early castling:
-11... Ka7 -12.Db4 Ka6 -13.Db8 Ka7 -14.D=b7 Kb8 -15.c6xDb7 Kc8 -16.d5xLc6 Kd8 -17.d4 Ke8 -18.e3xTd4 Da8 -19.f2xSe3 Dd8 -20.h5 Lb7 -21.h4 Lc8 and two white pawn retractions are missing (2012-07-23)
Yoav Ben-Zvi: The solution does not require a full analysis since if Black King and Rook never moved then Black is almost immediately in retro-stalemate. An alternative stipulation is "First move of Black King?". (2014-06-03)
A.Buchanan: Maybe there is intentional irony? (2014-06-03)
Henrik Juel: Nowadays composers are not afraid to use the real stipulation in this type of resolution retro, like 'Release the position'.
Formerly a formal stipulation was popular, like 'Mate in 1', with the real stipulation understood but not mentioned.
Here, the stipulation question may well be ironic, but you will get no solution points, if you just answer 'No'; you still need to explain how the position arose (2014-06-03)
more ...
comment
Henrik Juel: Black captured b7xLc6 and axbxc; White captured [Lf8], e2xd3, and fxexdxc6xb7-b8=D.
Ke8 to b7, Th8 to b8, Kb7 to g8, Tb8 to f8, Da4 to b8 for unpromotion, -22.c6xDb7 Da8 -23.d5xLc6 Dd8 -24.d4 Lb7 -25.e3xTd4 Lc8 -26.f2xSe3 c6 -27.h5 b7xLc6, Lc6 to f1, f2xSe3 etc.
Black may not castle, because he already did.
It is not possible to avoid the early castling:
-11... Ka7 -12.Db4 Ka6 -13.Db8 Ka7 -14.D=b7 Kb8 -15.c6xDb7 Kc8 -16.d5xLc6 Kd8 -17.d4 Ke8 -18.e3xTd4 Da8 -19.f2xSe3 Dd8 -20.h5 Lb7 -21.h4 Lc8 and two white pawn retractions are missing (2012-07-23)
Yoav Ben-Zvi: The solution does not require a full analysis since if Black King and Rook never moved then Black is almost immediately in retro-stalemate. An alternative stipulation is "First move of Black King?". (2014-06-03)
A.Buchanan: Maybe there is intentional irony? (2014-06-03)
Henrik Juel: Nowadays composers are not afraid to use the real stipulation in this type of resolution retro, like 'Release the position'.
Formerly a formal stipulation was popular, like 'Mate in 1', with the real stipulation understood but not mentioned.
Here, the stipulation question may well be ironic, but you will get no solution points, if you just answer 'No'; you still need to explain how the position arose (2014-06-03)
more ...
comment
Keywords: Castling (sk), Castling Paradox (sk hidden)
Genre: Retro
FEN: 4k2r/2pppppp/7P/2p5/Q1P5/PPRP4/RBpP2P1/N1K5
Reprints: 71 32 personaggi e 1 autore 1955
12 Europe Echecs 12 08/1959
222 FIDE Album 1914-1944/III 1975
341 Eigenartige Schachprobleme , p. 110, 2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-23 more...
Genre: Retro
FEN: 4k2r/2pppppp/7P/2p5/Q1P5/PPRP4/RBpP2P1/N1K5
Reprints: 71 32 personaggi e 1 autore 1955
12 Europe Echecs 12 08/1959
222 FIDE Album 1914-1944/III 1975
341 Eigenartige Schachprobleme , p. 110, 2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-23 more...
a) 1. Kd8 Se5 2. Te8 Sxf7#
b) 1. 0-0 Sd5 2. Sh8 Sxe7#
b) 1. 0-0 Sd5 2. Sh8 Sxe7#
Originalquelle?
nicht sicher, ob Zwilling b) auch schon im Original.
Artikel von Petrovici zum "Thema Than" 03/2016 gibt als Originalquelle an: "Europe Echecs 10/1951"
hans: Counting problem
1. Kd8 Se5 2. Te8 Sxf7# !!
1. 0-0 Sd5 2. Sh8 Sxe7# ??
Black needs a tempo to get on move, and the only piece to do that is Th8, so 0-0 is illegal. (2010-06-20)
Mario Richter: The term usually used to describe this kind of problems is "Parity problem". The bRh8 might have been on h8 all the time, since the tempo move might also have been made by the black Queen, but in this case, the bK must have already moved, thereby losing the right to castle too.. (2010-06-21)
Ladislav Packa: You both are right. Black would be in this position made an uneven number moves.
This can be achieved by using Rook moves (eg Rg8-f8-h8), or by the Queen moves while she was still alive (eg Kf8-g8, then Qf8-e8-d8 and then the King back). But both possibilities preventing black castling. (2013-10-04)
A.Buchanan: Parity change could also be achieved without triangulation by e.g. SxDe8 after the black queen has moved once (2022-03-15)
comment
nicht sicher, ob Zwilling b) auch schon im Original.
Artikel von Petrovici zum "Thema Than" 03/2016 gibt als Originalquelle an: "Europe Echecs 10/1951"
hans: Counting problem
1. Kd8 Se5 2. Te8 Sxf7# !!
1. 0-0 Sd5 2. Sh8 Sxe7# ??
Black needs a tempo to get on move, and the only piece to do that is Th8, so 0-0 is illegal. (2010-06-20)
Mario Richter: The term usually used to describe this kind of problems is "Parity problem". The bRh8 might have been on h8 all the time, since the tempo move might also have been made by the black Queen, but in this case, the bK must have already moved, thereby losing the right to castle too.. (2010-06-21)
Ladislav Packa: You both are right. Black would be in this position made an uneven number moves.
This can be achieved by using Rook moves (eg Rg8-f8-h8), or by the Queen moves while she was still alive (eg Kf8-g8, then Qf8-e8-d8 and then the King back). But both possibilities preventing black castling. (2013-10-04)
A.Buchanan: Parity change could also be achieved without triangulation by e.g. SxDe8 after the black queen has moved once (2022-03-15)
comment
Keywords: Castling (sk), Parity Argument, Cant Castler, Than theme
Genre: h#, Retro
Computer test: rawbats
FEN: r1b1k2r/1ppppppp/p5n1/8/8/P1NN3P/1PPPPPPR/nRBK1B2
Reprints: 787 Themes-64 10-12/1961
(7) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2018-03-22 more...
Genre: h#, Retro
Computer test: rawbats
FEN: r1b1k2r/1ppppppp/p5n1/8/8/P1NN3P/1PPPPPPR/nRBK1B2
Reprints: 787 Themes-64 10-12/1961
(7) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2018-03-22 more...
1. Dxc2 Tc1 2. 0-0-0? Txc2#
1. Sd7 0-0-0 2. 0-0 Tg1#
Black qside castling is lost as wPa promotes but other rights can be maintained.
S: axb,cxb,dxcxb,fxexBd,gxfxe
W: a=X,g=X,fxe
1. Sd7 0-0-0 2. 0-0 Tg1#
Black qside castling is lost as wPa promotes but other rights can be maintained.
S: axb,cxb,dxcxb,fxexBd,gxfxe
W: a=X,g=X,fxe
1. bxa5 Sb6 2. axb4 Ta8#
Henrik Juel: analysis
Black captured a3xb2 (not a7xb6, as the SW corner then cannot be released)
The missing white man is [Pe2], which must have promoted on b8
Now the white captures can be seen: axb, exdxcxb, and [Lf8] on f8
The only free white officer is Dd8, so the retroplay must include unpromotion of it on b8
It is impossible to shield Ke8 from check by Db8, so Ke8 has moved and Black may not castle (2022-04-27)
Henrik Juel: solution
1.bxa5 Sb6 2.axb4 Ta8#
not 1.0-0? Lxb6 2.Txa8 Txa8
HC+ Popeye 4.61 (2022-04-27)
Henrik Juel: Andrew, I should have been more precise in recommending your suggestion to introduce the new keyword Organ Pipes
I was talking about the standard meaning of Organ Pipes, which dates back to Sam Loyd, 1859: black LTTL in a problem where each L interferes with each T and vice versa, forming four Grimshaws
You have probably used a pattern search to find all occurrences of black or white LTTL, and you have neglected the interferences
This problem has LTTL in the wrong color, and there are no interferences (2022-04-27)
James Malcom: I agree with Henrik, although White organ pipes still exist, few and far. (2022-04-28)
James Malcom: Until Andrew bumbles back, I've updated the English definition to "A problem *utilizing* the pattern bishop, rook, rook, bishop in a straight line to create multiple Grimshaws." This specifies the purpose and formation of the Organ Pipes. (2022-04-28)
Henrik Juel: Thanks, James (2022-04-28)
A.Buchanan: Hi Guys. I wondered about other uses of organ pipes. In a first pass population of the 500-odd records with the current PDB interface, there’s not much time for thinking. I have multiple tabs open to eliminate the wait time associated with PDB refresh. One could go back and eliminate those which are not grimshaws. Alternatively (and this would be my preference) accept that this is a visual pattern which may occur in non-Grimshaw context. Then use the keyword Grimshaw(4x) to indicate when it’s really Grimshaw. I think it can also occur with white pattern, as a problem by Pal Benko shows. So I blitzed through the d# records, if someone wants to complete the rest that’s good. I think the existence of a few false transient positives is an acceptable price even if we take the narrow definition of Organ Pipes. (2022-04-28)
A.Buchanan: By the way, these days Deepl is good enough to give us decent translation to Feench & German and as a matter of policy whenever I make a change to a definition I try to align the other two. Other users who do not maintain the glossary are encouraged to propose definitions where there is a gap (E.g. Grimshaw). There are a lot of undefined terms, many very recent. (2022-04-28)
A.Buchanan: Hi Henrik feel free to respond to my response to your message, say thanks for the tags I have added, or add tags yourself to complete the work. The ones I’ve added were mostly the d# and it was a deliberate decision to blast through as a first pass and just add them for now anyway, not “neglected”. Now, I really don’t feel like continuing (2022-04-28)
Henrik Juel: Andrew, I believe that most PDB users appreciate your contributions to the site
I certainly do, so please continue your good work (2022-04-28)
comment
Black captured a3xb2 (not a7xb6, as the SW corner then cannot be released)
The missing white man is [Pe2], which must have promoted on b8
Now the white captures can be seen: axb, exdxcxb, and [Lf8] on f8
The only free white officer is Dd8, so the retroplay must include unpromotion of it on b8
It is impossible to shield Ke8 from check by Db8, so Ke8 has moved and Black may not castle (2022-04-27)
Henrik Juel: solution
1.bxa5 Sb6 2.axb4 Ta8#
not 1.0-0? Lxb6 2.Txa8 Txa8
HC+ Popeye 4.61 (2022-04-27)
Henrik Juel: Andrew, I should have been more precise in recommending your suggestion to introduce the new keyword Organ Pipes
I was talking about the standard meaning of Organ Pipes, which dates back to Sam Loyd, 1859: black LTTL in a problem where each L interferes with each T and vice versa, forming four Grimshaws
You have probably used a pattern search to find all occurrences of black or white LTTL, and you have neglected the interferences
This problem has LTTL in the wrong color, and there are no interferences (2022-04-27)
James Malcom: I agree with Henrik, although White organ pipes still exist, few and far. (2022-04-28)
James Malcom: Until Andrew bumbles back, I've updated the English definition to "A problem *utilizing* the pattern bishop, rook, rook, bishop in a straight line to create multiple Grimshaws." This specifies the purpose and formation of the Organ Pipes. (2022-04-28)
Henrik Juel: Thanks, James (2022-04-28)
A.Buchanan: Hi Guys. I wondered about other uses of organ pipes. In a first pass population of the 500-odd records with the current PDB interface, there’s not much time for thinking. I have multiple tabs open to eliminate the wait time associated with PDB refresh. One could go back and eliminate those which are not grimshaws. Alternatively (and this would be my preference) accept that this is a visual pattern which may occur in non-Grimshaw context. Then use the keyword Grimshaw(4x) to indicate when it’s really Grimshaw. I think it can also occur with white pattern, as a problem by Pal Benko shows. So I blitzed through the d# records, if someone wants to complete the rest that’s good. I think the existence of a few false transient positives is an acceptable price even if we take the narrow definition of Organ Pipes. (2022-04-28)
A.Buchanan: By the way, these days Deepl is good enough to give us decent translation to Feench & German and as a matter of policy whenever I make a change to a definition I try to align the other two. Other users who do not maintain the glossary are encouraged to propose definitions where there is a gap (E.g. Grimshaw). There are a lot of undefined terms, many very recent. (2022-04-28)
A.Buchanan: Hi Henrik feel free to respond to my response to your message, say thanks for the tags I have added, or add tags yourself to complete the work. The ones I’ve added were mostly the d# and it was a deliberate decision to blast through as a first pass and just add them for now anyway, not “neglected”. Now, I really don’t feel like continuing (2022-04-28)
Henrik Juel: Andrew, I believe that most PDB users appreciate your contributions to the site
I certainly do, so please continue your good work (2022-04-28)
comment
1. e3 d5 2. Lc4 d4 3. Se2 d3 4. 0-0 dxc2 5. d4 Kd7 6. d5 Kd6 7. Dd4 Sd7 8. Ld2 c1=L 9. Lb4+ c5 10. dxc6ep+
Keywords: Unique Proof Game, En passant, Non-standard material (l), Castling (wk), Promotion (l), Valladao Task (sww)
Genre: Retro
Computer test: Ergänzung Stelvio 1.2 C+: Keine Lösung: BP 8.5, 9.0. 10. dxc6ep++ Doppelschach.
FEN: r1bq1bnr/pp1npppp/2Pk4/8/1BBQ4/4P3/PP2NPPP/RNb2RK1
Reprints: 92 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-25 more...
Genre: Retro
Computer test: Ergänzung Stelvio 1.2 C+: Keine Lösung: BP 8.5, 9.0. 10. dxc6ep++ Doppelschach.
FEN: r1bq1bnr/pp1npppp/2Pk4/8/1BBQ4/4P3/PP2NPPP/RNb2RK1
Reprints: 92 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-25 more...
26 - P0001273
Luigi Ceriani
145 Europe Echecs 84 01/1965
(6+7)
h#1 (wer? wie?)
Die F.P.I. wurde zweimal erreicht: das erstemal nach einer geraden, das zweitemal nach einer ungeraden Anzahl von Zügen.
Vertikaler Zylinder
Luigi Ceriani
145 Europe Echecs 84 01/1965
(6+7)
h#1 (wer? wie?)
Die F.P.I. wurde zweimal erreicht: das erstemal nach einer geraden, das zweitemal nach einer ungeraden Anzahl von Zügen.
Vertikaler Zylinder
Henrik Juel: The FPI (the initial array, but with Black to move) may be reached, e.g., by playing the white knights out, playing Ta1 to b1 and Th1 to g1, then correct the tempo by playing Tb1-a1-h1-b1, and finally moving rooks and knights back into the initial array.
The stipulation condition implies that all four castlings are illegal. Solution: 1.Txa7 Txh8#, not 1.0-0-0? Tc1 nor 0... 0-0-0? 1.Tc8. I do not see any tries involving king-side castling, although they would be nice to have also. (2010-12-08)
A.Buchanan: I am evidently being slow again: please explain. "The FPI was achieved twice: the first time after an even number of moves, the second time after an odd number of moves." To achieve FPI requires an odd number of single moves. Should the stip read "even number of *White* moves"? And then what does it mean to achieve FPI for the *second* time: does that mean game array with White to move again? I agree that the bottom line is to have eliminated all castling rights. (2020-12-25)
Henrik Juel: In P0001272 the FPI was reached once, implying that either White or Black has lost castling rights
I also do not understand this problem (2020-12-26)
Hans-Jürgen Manthey: da der lezte Zug b5x~a6 bzw b4x~a3 gewesen sein kann, sind alle Rochaden, sowie die Farbwahl zulässig.
Oder kann mir mal einer erklären was zum Teufel F.P.I. auf Deutsch heißt ?? (2020-12-26)
A.Buchanan: Siehe die Beschreibung des Schlüsselworts. Aber hier macht es nicht viel Sinn. Ich veröffentliche in ein paar Tagen einen Artikel in Problemas, der besser erklärt und einige neue Beispiele enthält. (2020-12-27)
Hans-Jürgen Manthey: Habe mich über F.P.I. schlau gemacht und FPI, Vertikale Zylinder sowie viele weitere Märchenschach-Bedingungen sind alsolut nicht mein Fall... (2020-12-28)
A.Buchanan: @H-JM: Ich bin kein großer Fan von Feenschach, aber es hat seinen Platz in der Welt der Komposition. Es vervielfacht den Gestaltungsraum für Schachkompositionen erheblich, und viele coole künstlerische Effekte können nur mit Feenschach erzielt werden. FPI ist bislang kein gutes Beispiel: Aber die zugrunde liegende Idee ist gut und reif für eine breitere Nutzung. (2020-12-28)
Henrik Juel: My goodness, Andrew, you are approaching Mario's command of all languages...
I believe, however, that Feenschach is an old-fashioned name for Märchenschach
It came about as a literal translation of Fairy Chess, but now lives on only in the magazine name feenschach
We did almost the same in Denmark: feskak in the 1930s, but now fantasiskak (2020-12-28)
A.Buchanan: Hi Henrik: thanks for your kind feedback. I try to draft in German, and then back translate in google translate to English to see what it thinks, and then go forwards again to correct dumb errors and give me better word choices. But there's still errors, which I am always keen to hear about, so thanks for reminding me about Märchenschach. I now understand that Märchen are "fairy tales" not "fairies" and that the term is used in English by folklorists. Encyclopedia Britannica characterizes them by: "Their usual theme is the triumph over difficulty, with or without supernatural aid, of the one least likely to succeed." I like the detail at the end, which suggests that the key move should be an unlikely one! (2020-12-29)
Henrik Juel: I had three years of german in middle school, but never used the language actively
I believe that most german problemists can read english as well as we can read german, so I do not make the extra effort to communicate in german; besides remembering to capitalize nouns, I would have to check masculinum, femininum, or neutrum (gender?), and check which prepositions entail accusative, dative, or both (kasus?) (2020-12-29)
Olaf Jenkner: (:-) (2020-12-29)
more ...
comment
The stipulation condition implies that all four castlings are illegal. Solution: 1.Txa7 Txh8#, not 1.0-0-0? Tc1 nor 0... 0-0-0? 1.Tc8. I do not see any tries involving king-side castling, although they would be nice to have also. (2010-12-08)
A.Buchanan: I am evidently being slow again: please explain. "The FPI was achieved twice: the first time after an even number of moves, the second time after an odd number of moves." To achieve FPI requires an odd number of single moves. Should the stip read "even number of *White* moves"? And then what does it mean to achieve FPI for the *second* time: does that mean game array with White to move again? I agree that the bottom line is to have eliminated all castling rights. (2020-12-25)
Henrik Juel: In P0001272 the FPI was reached once, implying that either White or Black has lost castling rights
I also do not understand this problem (2020-12-26)
Hans-Jürgen Manthey: da der lezte Zug b5x~a6 bzw b4x~a3 gewesen sein kann, sind alle Rochaden, sowie die Farbwahl zulässig.
Oder kann mir mal einer erklären was zum Teufel F.P.I. auf Deutsch heißt ?? (2020-12-26)
A.Buchanan: Siehe die Beschreibung des Schlüsselworts. Aber hier macht es nicht viel Sinn. Ich veröffentliche in ein paar Tagen einen Artikel in Problemas, der besser erklärt und einige neue Beispiele enthält. (2020-12-27)
Hans-Jürgen Manthey: Habe mich über F.P.I. schlau gemacht und FPI, Vertikale Zylinder sowie viele weitere Märchenschach-Bedingungen sind alsolut nicht mein Fall... (2020-12-28)
A.Buchanan: @H-JM: Ich bin kein großer Fan von Feenschach, aber es hat seinen Platz in der Welt der Komposition. Es vervielfacht den Gestaltungsraum für Schachkompositionen erheblich, und viele coole künstlerische Effekte können nur mit Feenschach erzielt werden. FPI ist bislang kein gutes Beispiel: Aber die zugrunde liegende Idee ist gut und reif für eine breitere Nutzung. (2020-12-28)
Henrik Juel: My goodness, Andrew, you are approaching Mario's command of all languages...
I believe, however, that Feenschach is an old-fashioned name for Märchenschach
It came about as a literal translation of Fairy Chess, but now lives on only in the magazine name feenschach
We did almost the same in Denmark: feskak in the 1930s, but now fantasiskak (2020-12-28)
A.Buchanan: Hi Henrik: thanks for your kind feedback. I try to draft in German, and then back translate in google translate to English to see what it thinks, and then go forwards again to correct dumb errors and give me better word choices. But there's still errors, which I am always keen to hear about, so thanks for reminding me about Märchenschach. I now understand that Märchen are "fairy tales" not "fairies" and that the term is used in English by folklorists. Encyclopedia Britannica characterizes them by: "Their usual theme is the triumph over difficulty, with or without supernatural aid, of the one least likely to succeed." I like the detail at the end, which suggests that the key move should be an unlikely one! (2020-12-29)
Henrik Juel: I had three years of german in middle school, but never used the language actively
I believe that most german problemists can read english as well as we can read german, so I do not make the extra effort to communicate in german; besides remembering to capitalize nouns, I would have to check masculinum, femininum, or neutrum (gender?), and check which prepositions entail accusative, dative, or both (kasus?) (2020-12-29)
Olaf Jenkner: (:-) (2020-12-29)
more ...
comment
Keywords: Cant Castler (wbsb), Fake game array, Castling (wbsb), Constrained problem, Vertical Cylinder (Vertikaler)
Genre: Retro, Fairies
FEN: r3k2r/P3p3/P3P3/8/8/p3p3/p7/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2020-12-26 more...
Genre: Retro, Fairies
FEN: r3k2r/P3p3/P3P3/8/8/p3p3/p7/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2020-12-26 more...
* 1. ... dxc5 2. Dxh3 0-0-0#
1. Db2 Le2+ 2. Kc2 Ld1#
1. Db2 Le2+ 2. Kc2 Ld1#
Henrik Juel: the five missing black men were captured by white pawns (exfxgxh, fxgxj), so with Black to move last move was with Ta1 or Ke1, and White may not castle
C+ Popeye 4.61 (2022-11-26)
A.Buchanan: Pleasant White tempo play in both phases (2022-11-27)
comment
C+ Popeye 4.61 (2022-11-26)
A.Buchanan: Pleasant White tempo play in both phases (2022-11-27)
comment
Keywords: Cant Castler (wl), Castling (wl)
Genre: Retro, h#
Computer test: HC+ Popeye 4.61 with simple retro logic
FEN: 8/7p/7P/2pr2pP/2bP2Pb/2pk1BRP/6qN/R3KNrn
Reprints: 553 FIDE Album 1959-1961 1966
157 Europe Echecs 90 07/1966
(10) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-27 more...
Genre: Retro, h#
Computer test: HC+ Popeye 4.61 with simple retro logic
FEN: 8/7p/7P/2pr2pP/2bP2Pb/2pk1BRP/6qN/R3KNrn
Reprints: 553 FIDE Album 1959-1961 1966
157 Europe Echecs 90 07/1966
(10) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-27 more...
Rosalie Fay: White has lost only the bishops. So the pawn on c5 is not [Pa7] (because that entails 2 White units captured on black squares). White has played axbxcxdxe7, dxe, fxe, hxg, gxfxe. Black has 7 units, so white pawns have captured all missing Black units, but none on the a or h files.
Black has 2 pawns on the c-file, so one has captured. Thus [bPa7] and [bPh7] pawns have collectively captured no more than once. So at least one of them must have promoted, in order to either get to a file where White made a capture, or replace a captured unit; it didn't capture en route to promotion, so it displaced a white rook and thus spoilt one White castling right.
White would mate by 1 Rd1 & 2 Rd6 or 1 Rf1 & 2 Rf6, except that Black threatens Qxe2+. So either 1 0-0 or 1 0-0-0, though it's impossible to say which is legal. (2022-11-24)
Henrik Juel: one solution, but in two parts
if Ta1 has moved, 1.0-0 thr. 2.Dc8,Tf6#
if Th1 has moved, 1.0-0-0 thr. 2.Dg8,Td6# (2022-11-25)
Hans-Jürgen Manthey: nach der möglichen Zugfolge: 1. Sb1-c3 c7-c6 2. Sc3-d5 Dd8-b6 3. Sg1-f3 Db6-b3 4. a2xDb3 a7-a5 5. Sd5-b4 a5-a4 6. Sb4-a2 e7-e5 7. Sf3-h4 Lf8-c5 8. Sh4-g6 f7-f5 9. Sg6-f4 g7-g5 10. Sf4-g6 Lc5-e3 11. d2xLe3 f5-f4 12. g2-g3 Ta8-a5 13. g3xf4 g5-g4 14. f4xe5 g4-g3 15. h2xg3 h7-h5 16. Lf1-g2 h5-h4 17. Lc1-d2 h4-h3 18. Ld2-b4 Th8-h4 19. c2-c3 Th4-c4 20. Sa2-c1 a4-a3 21. Lb4-c5 a3-a2 22. Dd1-d4 Sb8-a6 23. Dd4-h4 Sa6-c7 24. Dh4-d8+ Ke8-f7 25. b3xTc4 Sc7-d5 26. c4xSd5 Sg8-e7 27. d5-d6 Ta5-b5 28. d6xSe7 d7-d6 --- folgt nun
29. Lg2-e4 Lc8-e6 30. Le4-b1 a2xLb1D 31. Th1-g1 Db1-d3 32. Sc1-b3 Le6-c4 33. Th1-g1 h3-h2 34. Dd8-e8+ Kf7-e6 35. Tg1-h1 Tb5-b6 36. Th1-g1 Le6-c4 37. Tg1-h1 h3-h2 38. Th1-g1 h2-h1D 39. Sc1-b3 Dh1-e4 40. f2-f3 Lc4-a6 41. f3xDe4 d6xLc5 42. Dd8-e8+ Kf7-e6 43. Tg1-h1 Dd3-b5 matt in 2:
1. OOO droht 2. De8-g8/Td1-d6# - 1. ... Db5-d3 2. Sb3xc5# oder:
29. Sc1-b3 h3xLg2 30. Sb3-d2 g2-g1D+ 31. Sd2-f1 Dg1-g2 32. Sf1-d2 Dg2-e4 33. f2-f3 Tb5-b6 34. f3xDe4 d6xLc5 35. Sd2-b3 Lc8-e6 36. Ta1-b1 Le6-c4 37. Ta1-b1 Kf7-e6 38. Tb1-a1 Lc4-a6 39. Ta1-b1 a2-a1D 40. Dd8-e8 Da1-a4 41. Tb1-a1 Da4-b5 matt in 2: 1. OO bel. 2. De8-c8/Tf1-f6# (2023-02-22)
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Black has 2 pawns on the c-file, so one has captured. Thus [bPa7] and [bPh7] pawns have collectively captured no more than once. So at least one of them must have promoted, in order to either get to a file where White made a capture, or replace a captured unit; it didn't capture en route to promotion, so it displaced a white rook and thus spoilt one White castling right.
White would mate by 1 Rd1 & 2 Rd6 or 1 Rf1 & 2 Rf6, except that Black threatens Qxe2+. So either 1 0-0 or 1 0-0-0, though it's impossible to say which is legal. (2022-11-24)
Henrik Juel: one solution, but in two parts
if Ta1 has moved, 1.0-0 thr. 2.Dc8,Tf6#
if Th1 has moved, 1.0-0-0 thr. 2.Dg8,Td6# (2022-11-25)
Hans-Jürgen Manthey: nach der möglichen Zugfolge: 1. Sb1-c3 c7-c6 2. Sc3-d5 Dd8-b6 3. Sg1-f3 Db6-b3 4. a2xDb3 a7-a5 5. Sd5-b4 a5-a4 6. Sb4-a2 e7-e5 7. Sf3-h4 Lf8-c5 8. Sh4-g6 f7-f5 9. Sg6-f4 g7-g5 10. Sf4-g6 Lc5-e3 11. d2xLe3 f5-f4 12. g2-g3 Ta8-a5 13. g3xf4 g5-g4 14. f4xe5 g4-g3 15. h2xg3 h7-h5 16. Lf1-g2 h5-h4 17. Lc1-d2 h4-h3 18. Ld2-b4 Th8-h4 19. c2-c3 Th4-c4 20. Sa2-c1 a4-a3 21. Lb4-c5 a3-a2 22. Dd1-d4 Sb8-a6 23. Dd4-h4 Sa6-c7 24. Dh4-d8+ Ke8-f7 25. b3xTc4 Sc7-d5 26. c4xSd5 Sg8-e7 27. d5-d6 Ta5-b5 28. d6xSe7 d7-d6 --- folgt nun
29. Lg2-e4 Lc8-e6 30. Le4-b1 a2xLb1D 31. Th1-g1 Db1-d3 32. Sc1-b3 Le6-c4 33. Th1-g1 h3-h2 34. Dd8-e8+ Kf7-e6 35. Tg1-h1 Tb5-b6 36. Th1-g1 Le6-c4 37. Tg1-h1 h3-h2 38. Th1-g1 h2-h1D 39. Sc1-b3 Dh1-e4 40. f2-f3 Lc4-a6 41. f3xDe4 d6xLc5 42. Dd8-e8+ Kf7-e6 43. Tg1-h1 Dd3-b5 matt in 2:
1. OOO droht 2. De8-g8/Td1-d6# - 1. ... Db5-d3 2. Sb3xc5# oder:
29. Sc1-b3 h3xLg2 30. Sb3-d2 g2-g1D+ 31. Sd2-f1 Dg1-g2 32. Sf1-d2 Dg2-e4 33. f2-f3 Tb5-b6 34. f3xDe4 d6xLc5 35. Sd2-b3 Lc8-e6 36. Ta1-b1 Le6-c4 37. Ta1-b1 Kf7-e6 38. Tb1-a1 Lc4-a6 39. Ta1-b1 a2-a1D 40. Dd8-e8 Da1-a4 41. Tb1-a1 Da4-b5 matt in 2: 1. OO bel. 2. De8-c8/Tf1-f6# (2023-02-22)
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Keywords: Partial Retro Analysis (PRA), Castling (wb)
Genre: Retro, 2#
FEN: 4Q3/1p2P3/brp1k1N1/1qp1P3/4P3/1NP1P1P1/1P2P3/R3K2R
Reprints: 223 Europe Echecs 130 09/1969
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-26 more...
Genre: Retro, 2#
FEN: 4Q3/1p2P3/brp1k1N1/1qp1P3/4P3/1NP1P1P1/1P2P3/R3K2R
Reprints: 223 Europe Echecs 130 09/1969
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-26 more...
hans: 1. Th2xf2 De1xf2#!
1. Th2xh3 0-0#? (Castling illegal)
R: -1. Kf1xSe1 Sg2-e1 -2. Ke1-f1 Sf4-g2+ -3. Tg1-h1 Sd5-f4 -4. Tf1-g1 Se3-d5 -5. Tg1-f1 Sf1-e3 -6. Tg1-h1 Tg2-h2 -7. Th2-h1-g1 Tg1-g2 -8. Th1-h2 g2-g1=T -9 h2-h3 h3xSg2 (2010-06-26)
HENRI: Hans solution has a tipo. Their is no D (=Queen in german) mating. One should read the solution in german : 1. Th2xf2 Ke1xf2#! 1.Th2xh3 0-0#?
or in english : 1. Rh2xf2 Ke1xf2#! 1.Rh2xh3 0-0#? (Castling illegal) (2023-10-25)
comment
1. Th2xh3 0-0#? (Castling illegal)
R: -1. Kf1xSe1 Sg2-e1 -2. Ke1-f1 Sf4-g2+ -3. Tg1-h1 Sd5-f4 -4. Tf1-g1 Se3-d5 -5. Tg1-f1 Sf1-e3 -6. Tg1-h1 Tg2-h2 -7. Th2-h1-g1 Tg1-g2 -8. Th1-h2 g2-g1=T -9 h2-h3 h3xSg2 (2010-06-26)
HENRI: Hans solution has a tipo. Their is no D (=Queen in german) mating. One should read the solution in german : 1. Th2xf2 Ke1xf2#! 1.Th2xh3 0-0#?
or in english : 1. Rh2xf2 Ke1xf2#! 1.Rh2xh3 0-0#? (Castling illegal) (2023-10-25)
comment
Keywords: Cant Castler, Castling (wk)
Genre: h#, Retro
FEN: 8/8/8/8/8/PPP3PP/B1rPPP1r/BNk1K2R
Input: Gerd Wilts, 1995-06-03
Genre: h#, Retro
FEN: 8/8/8/8/8/PPP3PP/B1rPPP1r/BNk1K2R
Input: Gerd Wilts, 1995-06-03
R: 1. 0-0-0
Deemed stipulation: "Erster Zug des wTd1?"
Henrik Juel: White pawns captured all 13 missing black men
Retracting the castling is the only way to give Black a retraction, e.g. Kc2-b3 (2020-12-01)
comment
Henrik Juel: White pawns captured all 13 missing black men
Retracting the castling is the only way to give Black a retraction, e.g. Kc2-b3 (2020-12-01)
comment
Keywords: Type A, Last Move? (0-0-0), Castling (wl), Economy record (Last Move? Type A), First Move? (T0), Economy record (First move)
Genre: Retro
FEN: 8/P1p5/PN6/1P6/P1N5/Pk6/pP6/2KR4
Reprints: 342 Europe Echecs 241 01/1979
1.57A Eigenartige Schachprobleme , p. 194, 2010
1 Die Schwalbe 360-1, p. 737, 12/2020
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-28 more...
Genre: Retro
FEN: 8/P1p5/PN6/1P6/P1N5/Pk6/pP6/2KR4
Reprints: 342 Europe Echecs 241 01/1979
1.57A Eigenartige Schachprobleme , p. 194, 2010
1 Die Schwalbe 360-1, p. 737, 12/2020
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-28 more...
31 - P0001550
Michel Caillaud
421 Europe Echecs 295 07/1983
A. Hazebrouck gewidmet
1. Preis
(13+12) C+
BP in 35.5
Michel Caillaud
421 Europe Echecs 295 07/1983
A. Hazebrouck gewidmet
1. Preis
(13+12) C+
BP in 35.5
1. h4 c5 2. h5 c4 3. Th4 c3 4. Tc4 b5 5. g4 b4 6. Lg2 b3 7. Lc6 bxa2 8. b4 a5 9. b5 a4 10. b6 a3 11. La4 Sc6 12. b7 d5 13. b8=D d4 14. Dd6 d3 15. Dg6 dxc2 16. d3 fxg6 17. Ld2 c1=L 18. Db3 c2 19. La5 Lh6 20. Sd2 c1=L 21. Sf1 Lcg5 22. f4 Kf7 23. 0-0-0 a1=L 24. fxg5 a2 25. gxh6 Kf6 26. Sh2 Kg5 27. Tf1 Lf6 28. Tff4 a1=L 29. Kb1 Lae5 30. d4 Lb7 31. dxe5 Dd2 32. exf6 Te8 33. f7 Sf6 34. Dd3 Sd7 35. Ld1 Sdb8 36. Sgf3+
5 Frolkin-Ceriani-Umwandlungen: 4 schwarze Läufer und 1 weiße Dame! Eine der bahnbrechenden frühen KBPs.
5 Frolkin-Ceriani-Umwandlungen: 4 schwarze Läufer und 1 weiße Dame! Eine der bahnbrechenden frühen KBPs.
Silvio Baier: Der wesentliche thematische Inhalt ist bereits nach 31,5 Zügen erreicht. Bis dahin ist es C+ (Euclide 0.98). (2010-08-04)
James Malcom: Is this fully C+ then? (2021-01-27)
Henrik Juel: No, only the first 31.5 moves are tested OK (2021-01-27)
James Malcom: No Henrik, as in is the entire problem testable. (2021-01-27)
Henrik Juel: I guess that testing the entire problem would take an unreasonably long time (2021-04-06)
A.Buchanan: It might be possible these days: the motivation for stopping at 31.5 is that the promotion theme had been demonstrated by then. But there’s still e.g. bSb8 as random impostor (2021-04-07)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 50:24:34 Stunden.(hh:mm:ss)
Keine Lösung: BP 34.5, BP 35.0. (2023-05-08)
Henrik Juel: Thanks for your patience, Moldenhauer; more than two days... (2023-05-08)
comment
James Malcom: Is this fully C+ then? (2021-01-27)
Henrik Juel: No, only the first 31.5 moves are tested OK (2021-01-27)
James Malcom: No Henrik, as in is the entire problem testable. (2021-01-27)
Henrik Juel: I guess that testing the entire problem would take an unreasonably long time (2021-04-06)
A.Buchanan: It might be possible these days: the motivation for stopping at 31.5 is that the promotion theme had been demonstrated by then. But there’s still e.g. bSb8 as random impostor (2021-04-07)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 50:24:34 Stunden.(hh:mm:ss)
Keine Lösung: BP 34.5, BP 35.0. (2023-05-08)
Henrik Juel: Thanks for your patience, Moldenhauer; more than two days... (2023-05-08)
comment
Keywords: Ceriani-Frolkin Theme (llllD), Unique Proof Game, Castling, Promotion (llllD), Impostor (s)
Genre: Retro
Computer test: Computerprüfung: C+ Stelvio 1.11 50:24:34 Stunden.(hh:mm:ss) Keine Lösung: BP 34.5, BP 35.0.
FEN: 1n2rb1r/1b2pPpp/2n3pP/B5kP/2R2RP1/3Q1N2/3qP2N/1K1B4
Reprints: 25 Shortest Proof Games 11/1991
(1) diagrammes 103 10-12/1992
(A) Quartz 22 10-12/2002
(8-a) Die Schwalbe 250A 08/2011
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-04-07 more...
Genre: Retro
Computer test: Computerprüfung: C+ Stelvio 1.11 50:24:34 Stunden.(hh:mm:ss) Keine Lösung: BP 34.5, BP 35.0.
FEN: 1n2rb1r/1b2pPpp/2n3pP/B5kP/2R2RP1/3Q1N2/3qP2N/1K1B4
Reprints: 25 Shortest Proof Games 11/1991
(1) diagrammes 103 10-12/1992
(A) Quartz 22 10-12/2002
(8-a) Die Schwalbe 250A 08/2011
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-04-07 more...
1. d3 h6 2. Lg5 hxg5 3. h4 Txh4 4. a4 Txa4 5. Th6 Sxh6 6. g4 Sxg4 7. Lg2 Sxf2 8. Lc6 Sxc6 9. d4 Sxd4 10. c4 Sxe2 11. c5 Sxg1 12. c6 dxc6 13. Dd4 Dxd4 14. Ta3 Dxb2 15. Th3 Lxh3 16. Sc3 0-0-0 17. Sd5 Txd5= patt
Henrik Juel: Black makes 15 captures.
The first 13.0 moves are correct by Euclide 1.0, but it takes some 4 minutes, so a complete test would probably take more than a day (2014-12-09)
paul: Jacobi checked the last 16 moves (in about 20 min). (2018-05-10)
Mario Richter: I would like to label this problem as a "Massacre PG", but the current definition of that term only knows of two-sided MPGs. Perhaps we could make the same differentiation as for "Homebase"?!
Btw., wasn't "popeye" best for such massacres?
rawbats confirms complete correctness of the problem after approx. 25 minutes. (2018-05-11)
Henrik Juel: Yes, Mario, Popeye is fairly good at this type
C+ by Popeye 4.61 in 57 minutes (2018-05-11)
A.Buchanan: @Mario: yes I agree it would be good to make massacre more specific. There are 69 massacres currently, almost all 2-sided, but maybe also some series-movers (unique or non-unique) which are currently not counted? (2018-05-12)
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The first 13.0 moves are correct by Euclide 1.0, but it takes some 4 minutes, so a complete test would probably take more than a day (2014-12-09)
paul: Jacobi checked the last 16 moves (in about 20 min). (2018-05-10)
Mario Richter: I would like to label this problem as a "Massacre PG", but the current definition of that term only knows of two-sided MPGs. Perhaps we could make the same differentiation as for "Homebase"?!
Btw., wasn't "popeye" best for such massacres?
rawbats confirms complete correctness of the problem after approx. 25 minutes. (2018-05-11)
Henrik Juel: Yes, Mario, Popeye is fairly good at this type
C+ by Popeye 4.61 in 57 minutes (2018-05-11)
A.Buchanan: @Mario: yes I agree it would be good to make massacre more specific. There are 69 massacres currently, almost all 2-sided, but maybe also some series-movers (unique or non-unique) which are currently not counted? (2018-05-12)
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Keywords: Unique Proof Game, Castling, Homebase (w), Rex solus (w), Massacre PG
Genre: Retro
Computer test: C+ rawbats. C+ by Popeye 4.61 in 57 minutes. C+ Stelvio 1.2 2 Sekunden. Keine Lösung: BP 16.0, BP 16.5.
FEN: 2k2b2/ppp1ppp1/2p5/3r2p1/r7/7b/1q3n2/4K1n1
Reprints: 116 Shortest Proof Games 11/1991
20a 64 Proof Games 2012
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-21 more...
Genre: Retro
Computer test: C+ rawbats. C+ by Popeye 4.61 in 57 minutes. C+ Stelvio 1.2 2 Sekunden. Keine Lösung: BP 16.0, BP 16.5.
FEN: 2k2b2/ppp1ppp1/2p5/3r2p1/r7/7b/1q3n2/4K1n1
Reprints: 116 Shortest Proof Games 11/1991
20a 64 Proof Games 2012
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-21 more...
1. Lf1 2. Kh3 3. Kh4 4. Lh3 5. fxg3ep 6. Kh5 7. Kg6 8. Kf7 9. Ke8 10. 0-0-0 11. Td7 a8=D#
The cluster of tries: Kh2-g1-f1-e1-d2-...-e8 0-0-0 Td7 is 12 moves: just too long.
In the diagram, e.p. is not permitted because last move might be g3-g4. So first shift sK to h4. Following the e.p., check that wK does now have prior move: Kf4-f3. Now sK scampers to e8 where through consequent forgetfulness it recovers castling rights.
The cluster of tries: Kh2-g1-f1-e1-d2-...-e8 0-0-0 Td7 is 12 moves: just too long.
In the diagram, e.p. is not permitted because last move might be g3-g4. So first shift sK to h4. Following the e.p., check that wK does now have prior move: Kf4-f3. Now sK scampers to e8 where through consequent forgetfulness it recovers castling rights.
Keywords: En passant, Castling (sg), Seriesmover, Consequent, Valladao Task, Promotion in the mating move (D), Switchback (l), Promotion (D), Königswanderung
Genre: Retro, Fairies
FEN: r7/P1p5/pp6/5pp1/5pP1/5K1b/4rp1k/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
Genre: Retro, Fairies
FEN: r7/P1p5/pp6/5pp1/5pP1/5K1b/4rp1k/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
34 - P0001637
Andrej N. Kornilow
Andrey Frolkin
506v Europe Echecs 364 04/1989
(16+10) C+
BP in 13,5
2 solutions
Andrej N. Kornilow
Andrey Frolkin
506v Europe Echecs 364 04/1989
(16+10) C+
BP in 13,5
2 solutions
1. Sc3 c5 2. Sd5 c4 3. Sxe7 c3 4. Sxg8 Dh4 5. Sh6 g5 6. Sxf7 Lg7 7. Se5 0-0 8. Sxd7 Lh8 9. Sxb8 Ld7 10. Sc6 Tc8 11. Sxa7 La4 12. Sb5 Tc6 13. Sa3 b5 14. Sb1
und
1. Sf3 c5 2. Se5 c4 3. Sxd7 c3 4. Sxb8 Ld7 5. Sc6 Tc8 6. Sxa7 La4 7. Sc6 b5 8. Sxe7 Tc6 9. Sxg8 Dh4 10. Sh6 g5 11. Sxf7 Lg7 12. Se5 0-0 13. Sf3 Lh8 14. Sg1
und
1. Sf3 c5 2. Se5 c4 3. Sxd7 c3 4. Sxb8 Ld7 5. Sc6 Tc8 6. Sxa7 La4 7. Sc6 b5 8. Sxe7 Tc6 9. Sxg8 Dh4 10. Sh6 g5 11. Sxf7 Lg7 12. Se5 0-0 13. Sf3 Lh8 14. Sg1
Moldenhauer: Ergänzung: Stelvio 1.2. Keine Lösungen BP 12.5, BP 13.0.
Es ist wirklich eine Rarität dass es 2 komplette Lösungen gibt. (2023-05-20)
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Es ist wirklich eine Rarität dass es 2 komplette Lösungen gibt. (2023-05-20)
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Keywords: Unique Proof Game, Homebase (W), Initial Game Array (W), Castling
Genre: Retro
Computer test: Natch 2.2 Copyright (C) 1997,98,99,2001,2002,2003 Pascal Wassong Ergänzung: Stelvio 1.2. Keine Lösungen BP 12.5, BP 13.0.
FEN: 5rkb/7p/2r5/1p4p1/b6q/2p5/PPPPPPPP/RNBQKBNR
Reprints: 569 Ukrainisches Album 1986-1990
140 Shortest Proof Games 11/1991
(10) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-21 more...
Genre: Retro
Computer test: Natch 2.2 Copyright (C) 1997,98,99,2001,2002,2003 Pascal Wassong Ergänzung: Stelvio 1.2. Keine Lösungen BP 12.5, BP 13.0.
FEN: 5rkb/7p/2r5/1p4p1/b6q/2p5/PPPPPPPP/RNBQKBNR
Reprints: 569 Ukrainisches Album 1986-1990
140 Shortest Proof Games 11/1991
(10) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-21 more...
R: 1. Kg1xSh1 Sf2-h1+ 2. 0-0, dann 1. Th8#
R: 1. Kg1-h1 Lb8xSa7+ 2. Te7-f7, dann 1. Sc6#
R: 1. Kg1-h1 Sb6xDc8,Sb6xDd5,Se3xDd5+ 2. Dc5-c8,Dc5-d5,Dc5-d5+, dann 1. Df8#
R: 1. Kg1-h1 Lb8xSa7+ 2. Te7-f7, dann 1. Sc6#
R: 1. Kg1-h1 Sb6xDc8,Sb6xDd5,Se3xDd5+ 2. Dc5-c8,Dc5-d5,Dc5-d5+, dann 1. Df8#
Henrik Juel: This problem demonstrates an advantage of type Høeg over type Proca:
another way of generating variations
2.0-0 cannot be an uncapture, of course
2.Te7-f7 and 2.Dc5-c8 etc. could be uncaptures, but no matter what Black supplements, he is mated (2023-04-08)
A.Buchanan: Thanks Henrik: is an alternative for White R: 1. Kg1-h1 Lb8xDa7+ 2. Dc5-a7/Da3-a7, dann 1. Df8# (2023-04-08)
Henrik Juel: No Andrew, when White moves his uncaptured queen back to c5 or a3, Black supplements a queen or bishop on a7, preventing the mate on f8 (2023-04-08)
A.Buchanan: Thanks! (2023-04-08)
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another way of generating variations
2.0-0 cannot be an uncapture, of course
2.Te7-f7 and 2.Dc5-c8 etc. could be uncaptures, but no matter what Black supplements, he is mated (2023-04-08)
A.Buchanan: Thanks Henrik: is an alternative for White R: 1. Kg1-h1 Lb8xDa7+ 2. Dc5-a7/Da3-a7, dann 1. Df8# (2023-04-08)
Henrik Juel: No Andrew, when White moves his uncaptured queen back to c5 or a3, Black supplements a queen or bishop on a7, preventing the mate on f8 (2023-04-08)
A.Buchanan: Thanks! (2023-04-08)
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Keywords: Defensive Retractor, Type Høeg, Castling (wk)
Genre: Retro
FEN: 2nk4/b4Rp1/8/3n1q2/8/8/8/5R1K
Reprints: 729 FIDE Album 1962-1964 1968
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-07 more...
Genre: Retro
FEN: 2nk4/b4Rp1/8/3n1q2/8/8/8/5R1K
Reprints: 729 FIDE Album 1962-1964 1968
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-07 more...
1. dxc3ep 2. Kxb5 3. Kxc6 4. Kd7 5. Ke8 6. 0-0 Th8#
Aber es geht auch R: 1. Kb2-a1!?
Aber es geht auch R: 1. Kb2-a1!?
Henrik Juel: The intention may be 1.dxc3ep 2.Kxb5xc6-d7-e8 6.0-0 Rh8#, but -1.Kb2 seems possible. Should bBc1 be moved to a3? (2003-04-28)
GW: Yes, that's the solution, and the problem seems indeed to be cooked. I don't know if it has been corrected. (2003-04-28)
James Malcom: I believe that the intent was that the sLc1 must be the orginal one, and thus an impediment to wK foreplay, as the wBc6 "must" have captured the e7 Black pawn on its way from f2. The problem is that it can take the alternative path fxsLf3xDe3xTc5-c6, and there are enough White pieces to capture for the Black b pawn to make it d2 and for the d pawn to make it to c2. (2021-09-14)
James Malcom: It turns out this problem actually was corrected a long time later: P1012052 (2021-09-14)
comment
GW: Yes, that's the solution, and the problem seems indeed to be cooked. I don't know if it has been corrected. (2003-04-28)
James Malcom: I believe that the intent was that the sLc1 must be the orginal one, and thus an impediment to wK foreplay, as the wBc6 "must" have captured the e7 Black pawn on its way from f2. The problem is that it can take the alternative path fxsLf3xDe3xTc5-c6, and there are enough White pieces to capture for the Black b pawn to make it d2 and for the d pawn to make it to c2. (2021-09-14)
James Malcom: It turns out this problem actually was corrected a long time later: P1012052 (2021-09-14)
comment
Keywords: En passant, Castling (sk), Seriesmover, Consequent, Non-standard material (sLb1), Valladao Task, Promotion in the retro play (sLb1), Obvious promotion ((sLb1)), Superseded by (P1012052)
Genre: Retro, Fairies
FEN: 7r/5pnR/2P3pR/pPpn3p/1kPp4/3P4/P2pP3/Kbb5
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-09-14 more...
Genre: Retro, Fairies
FEN: 7r/5pnR/2P3pR/pPpn3p/1kPp4/3P4/P2pP3/Kbb5
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-09-14 more...
1. Sf3 f5 2. Sd4 Kf7 3. Sb5 Kg6 4. Sd6 exd6 5. e4 Le7 6. e5 Lh4 7. e6 Df6 8. e7 Dxb2 9. e8=S Se7 10. Lc4 Tf8 11. 0-0 Tf6 12. f3 Kh5 13. Te1 Th6 14. Lf7+ Sg6 15. Te7 Dxc1 16. De1 Dxd2 17. De6 dxe6 18. Td7 Dg5 19. Sxg7#
Cook: 1. d4 g5 2. Sf3 Lg7 3. e4 Lxd4 4. e5 Lxb2 5. Dd6 exd6 6. e6 f5 7. e7 Kf7 8. e8=S Se7 9. Sxg5+ Kg6 10. Lc4 Kh5 11. 0-0 Tg8 12. Te1 Tg6 13. f3 Th6 14. Lf7+ Sg6 15. Te7 Lf6 16. Se6 dxe6 17. Td7 Lh4 18. Lg5 Dxg5 19. Sg7#
Cook: 1. d4 g5 2. Sf3 Lg7 3. e4 Lxd4 4. e5 Lxb2 5. Dd6 exd6 6. e6 f5 7. e7 Kf7 8. e8=S Se7 9. Sxg5+ Kg6 10. Lc4 Kh5 11. 0-0 Tg8 12. Te1 Tg6 13. f3 Th6 14. Lf7+ Sg6 15. Te7 Lf6 16. Se6 dxe6 17. Td7 Lh4 18. Lg5 Dxg5 19. Sg7#
Kostas Prentos: A correction was published in Phenix, 2009 (No.186/Pg.7979)
Solution:
1. Sf3 f5 2. Sd4 Kf7 3. Sb5 Kg6 4. Sd6 exd6 5. e4 Le7 6. e5 Lh4 7. e6 Dg5 8. e7 Dg3 9. e8=T Se7 10. Lc4 Tf8 11. 0-0 Tf6 12. f3 Kh5 13. Te1 Th6 14. Lf7+ Sg6 15. T1e6 dxe6 16. Txc8 Sd7 17. Th8 Txh8 18. Le8 (2022-12-08)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.2 00:33:56 Minuten. (hh:mm:ss)
Keine Lösung: BP 17.5, BP 18.0.
Beispiel: 1.Sf3 f5 2.d4 Kf7 3.e4 g5 4.Lc4+ Kg6 5.e5 Lg7 6.e6 Lxd4
7.0–0 Lxb2 8.Dd6 exd6 9.e7 Kh5 10.e8S Se7 11.Te1 Tf8 12.Sxg5 Tf6
13.f3 Th6 14.Lf7+ Sg6 15.Te7 Lf6 16.Se6 dxe6 17.Td7 Lh4 18.Lg5 Dxg5
19.Sg7# (2023-05-30)
comment
Solution:
1. Sf3 f5 2. Sd4 Kf7 3. Sb5 Kg6 4. Sd6 exd6 5. e4 Le7 6. e5 Lh4 7. e6 Dg5 8. e7 Dg3 9. e8=T Se7 10. Lc4 Tf8 11. 0-0 Tf6 12. f3 Kh5 13. Te1 Th6 14. Lf7+ Sg6 15. T1e6 dxe6 16. Txc8 Sd7 17. Th8 Txh8 18. Le8 (2022-12-08)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.2 00:33:56 Minuten. (hh:mm:ss)
Keine Lösung: BP 17.5, BP 18.0.
Beispiel: 1.Sf3 f5 2.d4 Kf7 3.e4 g5 4.Lc4+ Kg6 5.e5 Lg7 6.e6 Lxd4
7.0–0 Lxb2 8.Dd6 exd6 9.e7 Kh5 10.e8S Se7 11.Te1 Tf8 12.Sxg5 Tf6
13.f3 Th6 14.Lf7+ Sg6 15.Te7 Lf6 16.Se6 dxe6 17.Td7 Lh4 18.Lg5 Dxg5
19.Sg7# (2023-05-30)
comment
Keywords: Unique Proof Game, Castling, Promotion
Genre: Retro
FEN: rnb5/pppR1BNp/3pp1nr/5pqk/7b/5P2/P1P3PP/RN4K1
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2006-10-15 more...
Genre: Retro
FEN: rnb5/pppR1BNp/3pp1nr/5pqk/7b/5P2/P1P3PP/RN4K1
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2006-10-15 more...
1. Sf3 e5 2. Sd4 Dg5 3. Sc6 De3 4. fxe3 h5 5. Kf2 h4 6. De1 h3 7. Kg3 Th4 8. Df2 Ta4 9. Df4 hxg2 10. h4 dxc6 11. h5 Sd7 12. h6 Sb6 13. h7 Ld7 14. h8=T 0-0-0 15. T8h6 Le8 16. Tf6 Tdd4 17. exd4 Lc5 18. dxc5 g1=L 19. cxb6 Lc5 20. bxa7 La3 21. bxa3 g5 22. Lb2 g4 23. Kh4 g3 24. Ld4 g2 25. Sc3 g1=S 26. Tb1 Sh3 27. Tb6 Sf2 28. Ta6 Sd1 29. Lf2 b6 30. a8=D+
Keywords: Ceriani-Frolkin Theme (Tls), Unique Proof Game, Non-standard material, Castling, Allumwandlung (DTls)
Genre: Retro
FEN: Q1k1b1n1/2p2p2/Rpp2R2/4p3/r4Q1K/P1N5/P1PPPB2/3n1B1R
Reprints: 136 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Kevin Begley, 2011-05-18 more...
Genre: Retro
FEN: Q1k1b1n1/2p2p2/Rpp2R2/4p3/r4Q1K/P1N5/P1PPPB2/3n1B1R
Reprints: 136 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Kevin Begley, 2011-05-18 more...
1. a4 c5 2. a5 Db6 3. Ta4 Sc6 4. Td4 Tb8 5. b4 cxd4 6. b5 e5 7. d3 La3 8. Le3 Lc1 9. bxc6 Db2 10. c7 b5 11. a6 Tb7 12. axb7 h5 13. b8=L Lb7 14. c8=D+ Ke7 15. Sf3 d5 16. Dh3 Th6 17. g4 Ta6 18. Dg3 h4 19. c4 hxg3 20. h4 Ta1 21. h5 Kf6 22. Sh2 dxe3 23. h6 Kg5 24. h7 Kh4 25. c5 a5 26. c6 a4 27. c7 b4 28. c8=T b3 29. h8=S Se7 30. Sg6 fxg6 31. Lh3 Sc6 32. 0-0 Sb4 33. Ld6 Sc2 34. Td8 Se1
"Promenades to Power"
Es läßt sich beweisen, daß die UWs in D,T,L,S zwingend erfolgt sind (a2-b8=L,h2-b8=S und entweder b2-c8=D/c2-c8=T oder b2-c8=T/c2-c8=D).
Erich Bartel: weitere Nachdrucke:
3) 160 Die Allumwandlung im Problemschach VIII 1966a---
4) Schach ohne Grenzen 1969.-- (2007-01-09)
Sally: Vier erzwungene Umwandlungen. (2012-02-21)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 da NUPG sonst cooked in 00:03:51 Minuten. (hh:mm:ss)
Keine Lösung: BP 33.0, BP 33.5.
Beispiel: 1.Sf3 Sf6 2.a4 Sd5 3.a5 Sf4 4.Ta4 c5 5.Td4 Db6 6.b4 Sc6 7.b5 Tb8 8.bxc6 cxd4 9.c7 e5 10.d3 La3 11.Le3 Lc1 12.c4 Db2 13.c5 b5 14.c6 Tb6 15.axb6 Ke7 16.b7 Kf6 17.b8L Lb7 18.c8D b4 19.c7 d5 20.Dh3 b3 21.c8T h5 22.g4 h4 23.Dg3 hxg3 24.h4 Th6 25.Lh3 dxe3 26.0–0 Sg2 27.Sh2 Se1 28.h5 Kg5 29.Td8 Ta6 30.Ld6 Kh4 31.h6 Ta1 32.h7 a5 33.h8S a4 34.Sg6+ fxg6
Umwandlungen: 17.b8L, 18.c8D, 21.c8T, 33.h8S. (2023-04-19)
comment
Es läßt sich beweisen, daß die UWs in D,T,L,S zwingend erfolgt sind (a2-b8=L,h2-b8=S und entweder b2-c8=D/c2-c8=T oder b2-c8=T/c2-c8=D).
Erich Bartel: weitere Nachdrucke:
3) 160 Die Allumwandlung im Problemschach VIII 1966a---
4) Schach ohne Grenzen 1969.-- (2007-01-09)
Sally: Vier erzwungene Umwandlungen. (2012-02-21)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 da NUPG sonst cooked in 00:03:51 Minuten. (hh:mm:ss)
Keine Lösung: BP 33.0, BP 33.5.
Beispiel: 1.Sf3 Sf6 2.a4 Sd5 3.a5 Sf4 4.Ta4 c5 5.Td4 Db6 6.b4 Sc6 7.b5 Tb8 8.bxc6 cxd4 9.c7 e5 10.d3 La3 11.Le3 Lc1 12.c4 Db2 13.c5 b5 14.c6 Tb6 15.axb6 Ke7 16.b7 Kf6 17.b8L Lb7 18.c8D b4 19.c7 d5 20.Dh3 b3 21.c8T h5 22.g4 h4 23.Dg3 hxg3 24.h4 Th6 25.Lh3 dxe3 26.0–0 Sg2 27.Sh2 Se1 28.h5 Kg5 29.Td8 Ta6 30.Ld6 Kh4 31.h6 Ta1 32.h7 a5 33.h8S a4 34.Sg6+ fxg6
Umwandlungen: 17.b8L, 18.c8D, 21.c8T, 33.h8S. (2023-04-19)
comment
Keywords: Allumwandlung, Castling (wk), Non-Unique Proof Game
Genre: Retro
FEN: 3R4/1b4p1/3B2p1/3pp3/p5Pk/1p1Pp1pB/1q2PP1N/rNbQnRK1
Reprints: 53 Caissa's Wild Roses 1935
Chess unlimited 1969
150 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Gunter Jordan, 2020-07-01 more...
Genre: Retro
FEN: 3R4/1b4p1/3B2p1/3pp3/p5Pk/1p1Pp1pB/1q2PP1N/rNbQnRK1
Reprints: 53 Caissa's Wild Roses 1935
Chess unlimited 1969
150 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Gunter Jordan, 2020-07-01 more...
a) 1. Ta1! 0-0? illegal 1. ... hxg6,~ 2. Ta8#
1. Tff7? hxg6!
b) 1. Tff7! h5,~ Tb8#
1. Ta1? 0-0!
In a), Schwarz muss zuletzt mit K oder T gezogen haben, deshalb s0-0 illegal.
1. Tff7? hxg6!
b) 1. Tff7! h5,~ Tb8#
1. Ta1? 0-0!
In a), Schwarz muss zuletzt mit K oder T gezogen haben, deshalb s0-0 illegal.
hans: 1. b4 a5 2. Bb2 Na6 3. Bd4 Nc5 4. Nc3 Na4 5. Bc5 Nb6 6. b5 Nh6 7. a4 Nf5 8.
Na2 Nh4 9. Nb4 axb4 10. a5 Rb8 11. a6 Na8 12. a7 Nb6 13. a8=Q Rg8 14. Qa2
Ra8 15. Qe6 Ra3 16. Qg6 hxg6 17. Qb1 Na8 18. Ba7 b6 19. Qb3 Bb7 20. Qd3 Bf3
21. Qf5 gxf5 22. Nh3 Bh5 23. Nf4 Bg6 24. e3 Rh8 25. Bc4 Rh6 26. f3 Bh7 27.
Ng6 Bg8 28. Nh8 Rd6 29. h3 g6 30. Rf1 Bh6 31. Rf2 Bf4 32. Rf1 Bh2 33. f4
Rdd3 34. Rf3 Rac3 35. Rg3 Kf8 36. Rg5 Kg7 37. Rh5 Kf6 38. Rh7 Qc8 39. Rg7
Bh7 40. Rg8 Qa6 41. Rb8 Qa3 42. Rb7 Qb2 43. Bb8 Rb3 44. Rba7 Qd4 45. R7a2
Ra3 46. Rb2 Ra7 47. Rba2 Rb7 48. Ba7 Rb8 49. Rb2 Rg8 50. Rba2 Rg7 51. Bb8
Bg8 52. Rb2 Rh7 53. Rba2 Rh5 54. Rb2 Rg5 55. Rba2 Rg3 56. Rb2 Rf3 57. Rba2
Rf2 58. Rb2 Ra3 59. Rba2 Ra7 60. Rb2 Rb7 61. Ba7 Rb8 62. Rba2 Bh7 63. Ra6
Rg8 64. Bb8 Rg7 65. Ra7 Bg8 66. Rb7 Rh7 67. Ba7 Rh5 68. Rb8 Bh7 69. Rg8 Rg5
70. Rg7 Bg8 71. Rh7 Rg3 72. Rh6 Rgf3 73. Bb3 Kg7 74. Ba4 Kf8 75. Rh7 Ke8
76. Rg7 Bh7 77. Bb8 Kd8 78. Bb3 Kc8 79. Rg8+ Kb7 80. Ba4 Qg7 81. Re8 Bg8
82. Rf8 Bg3 83. O-O-O {50 moves rule} 1/2-1/2 (2012-11-14)
Olaf Jenkner: Warum nicht z.B. 83. Te8 remis?
Was bedeutet das Schlüsselwort unused? (2012-11-14)
Henrik Juel: The castling shows that White never moved his king before; otherwise the position could be reached faster, and the 50 moves rule could not be applied.
The keyword unused seems non-sensical here and should be deleted (2012-11-14)
A.Buchanan: This problem is very interesting. It's orthodox 50M, so that castling does not reset the counter, and the timing works out nicely. But it makes me wonder... suppose we have a problem where it is B32 which was the last reset. Then W83 0-0-0 would prove that the position had already been at 50.0. If 50M convention (which needs to be rewritten because it's a mess) operated like 3Rep, then the game end would have been mandatory after B82. So castling W83 would be illegal. Is this how one would want 50M convention to work? Or should there be a carve-out to say: if you can definitely prove that no-one claimed, then no-one claimed. And should there be a similar carve-out for 3Rep rule? See https://www.thehoppermagazine.com/AA010 (2022-01-20)
A.Buchanan: Personally, I think no carve-outs. The 50M convention should be rewritten as: "A position is considered as a draw if it can be proved that the last 50.0 moves in the proof game combined with the solution did not contain a capture or pawn move. Unless expressly stipulated, this applies only to retro-problems." (2022-01-20)
Thomas Volet: What if the composer intends the retroplay to go beyond 50 non-P and non-capturing moves? (2022-01-20)
A.Buchanan: Hi Thomas - thanks for your question. I think it’s best if we take this offline. I will email you, if that’s ok (2022-01-21)
James Malcom: This is the Volet problem in question: P0008399 In hindsight, it is quite humorous that the 75 move rule was later introduced in the 21st century. (2022-01-21)
Thomas Volet: The question was not directed specifically to P0008399, which is just one of several compositions with the property at issue. (2022-01-22)
A.Buchanan: I've emailed Thomas. My earlier comment should have include checkmate as a third mechanism to zero the count. (2022-01-22)
James Malcom: You can always carry it on over to MatPlus. (2022-01-24)
more ...
comment
Na2 Nh4 9. Nb4 axb4 10. a5 Rb8 11. a6 Na8 12. a7 Nb6 13. a8=Q Rg8 14. Qa2
Ra8 15. Qe6 Ra3 16. Qg6 hxg6 17. Qb1 Na8 18. Ba7 b6 19. Qb3 Bb7 20. Qd3 Bf3
21. Qf5 gxf5 22. Nh3 Bh5 23. Nf4 Bg6 24. e3 Rh8 25. Bc4 Rh6 26. f3 Bh7 27.
Ng6 Bg8 28. Nh8 Rd6 29. h3 g6 30. Rf1 Bh6 31. Rf2 Bf4 32. Rf1 Bh2 33. f4
Rdd3 34. Rf3 Rac3 35. Rg3 Kf8 36. Rg5 Kg7 37. Rh5 Kf6 38. Rh7 Qc8 39. Rg7
Bh7 40. Rg8 Qa6 41. Rb8 Qa3 42. Rb7 Qb2 43. Bb8 Rb3 44. Rba7 Qd4 45. R7a2
Ra3 46. Rb2 Ra7 47. Rba2 Rb7 48. Ba7 Rb8 49. Rb2 Rg8 50. Rba2 Rg7 51. Bb8
Bg8 52. Rb2 Rh7 53. Rba2 Rh5 54. Rb2 Rg5 55. Rba2 Rg3 56. Rb2 Rf3 57. Rba2
Rf2 58. Rb2 Ra3 59. Rba2 Ra7 60. Rb2 Rb7 61. Ba7 Rb8 62. Rba2 Bh7 63. Ra6
Rg8 64. Bb8 Rg7 65. Ra7 Bg8 66. Rb7 Rh7 67. Ba7 Rh5 68. Rb8 Bh7 69. Rg8 Rg5
70. Rg7 Bg8 71. Rh7 Rg3 72. Rh6 Rgf3 73. Bb3 Kg7 74. Ba4 Kf8 75. Rh7 Ke8
76. Rg7 Bh7 77. Bb8 Kd8 78. Bb3 Kc8 79. Rg8+ Kb7 80. Ba4 Qg7 81. Re8 Bg8
82. Rf8 Bg3 83. O-O-O {50 moves rule} 1/2-1/2 (2012-11-14)
Olaf Jenkner: Warum nicht z.B. 83. Te8 remis?
Was bedeutet das Schlüsselwort unused? (2012-11-14)
Henrik Juel: The castling shows that White never moved his king before; otherwise the position could be reached faster, and the 50 moves rule could not be applied.
The keyword unused seems non-sensical here and should be deleted (2012-11-14)
A.Buchanan: This problem is very interesting. It's orthodox 50M, so that castling does not reset the counter, and the timing works out nicely. But it makes me wonder... suppose we have a problem where it is B32 which was the last reset. Then W83 0-0-0 would prove that the position had already been at 50.0. If 50M convention (which needs to be rewritten because it's a mess) operated like 3Rep, then the game end would have been mandatory after B82. So castling W83 would be illegal. Is this how one would want 50M convention to work? Or should there be a carve-out to say: if you can definitely prove that no-one claimed, then no-one claimed. And should there be a similar carve-out for 3Rep rule? See https://www.thehoppermagazine.com/AA010 (2022-01-20)
A.Buchanan: Personally, I think no carve-outs. The 50M convention should be rewritten as: "A position is considered as a draw if it can be proved that the last 50.0 moves in the proof game combined with the solution did not contain a capture or pawn move. Unless expressly stipulated, this applies only to retro-problems." (2022-01-20)
Thomas Volet: What if the composer intends the retroplay to go beyond 50 non-P and non-capturing moves? (2022-01-20)
A.Buchanan: Hi Thomas - thanks for your question. I think it’s best if we take this offline. I will email you, if that’s ok (2022-01-21)
James Malcom: This is the Volet problem in question: P0008399 In hindsight, it is quite humorous that the 75 move rule was later introduced in the 21st century. (2022-01-21)
Thomas Volet: The question was not directed specifically to P0008399, which is just one of several compositions with the property at issue. (2022-01-22)
A.Buchanan: I've emailed Thomas. My earlier comment should have include checkmate as a third mechanism to zero the count. (2022-01-22)
James Malcom: You can always carry it on over to MatPlus. (2022-01-24)
more ...
comment
Keywords: 50 move rule, Castling (wl)
Genre: Retro
FEN: nB3RbN/1kppppq1/1p4p1/1P3p2/Bp3P1n/4PrbP/2PP1rP1/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-25 more...
Genre: Retro
FEN: nB3RbN/1kppppq1/1p4p1/1P3p2/Bp3P1n/4PrbP/2PP1rP1/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-25 more...
42 - P0001967
Nenad Petrovic
628 Sahovski vjesnik 1950
Dr. Fabel und Dr. Ceriani gewidmet
2. Preis
(15+15)
Längste Beweispartie?
(AL: 1021,0)
Nenad Petrovic
628 Sahovski vjesnik 1950
Dr. Fabel und Dr. Ceriani gewidmet
2. Preis
(15+15)
Längste Beweispartie?
(AL: 1021,0)
1. S S 50. S b6 100. S h6 250. S h3 300. a3 S 450. a6 S 500. b3 S 550. g3 S 600. 0-0 S 649. S hxSg 699. h3 S 899. h7 S 949. axSb Ke8 950 Tf1 Kd8 951. Tg1 Ke8 952. Tf1 Dd8 1020. Kd1 Kd8 1021. De1 Ke8=
Henrik Juel: Note that in problems castling acts like capture and pawn move with respect to the 50 moves rule. After 949.a6xSb7 there are 4 moves left by [Pa7]; but each camp can shift only KDT, on d1-g1 and b8-e8, respectively, so the triple repetition rule now limits the length of the game. (2004-09-09)
A.Buchanan: In some problems it's certainly the case that the 50-move rule operates incorrectly in this way. Such problems are fine, but obviously wouldn't want to impose this as a standard. Different composers can make different assumptions here (2023-06-20)
comment
A.Buchanan: In some problems it's certainly the case that the 50-move rule operates incorrectly in this way. Such problems are fine, but obviously wouldn't want to impose this as a standard. Different composers can make different assumptions here (2023-06-20)
comment
Keywords: 50 move rule, Non-Unique Proof Game, Longest Proof Game, Castling
Genre: Retro
FEN: Nrq1kb2/1PpppppP/1p6/8/8/pP4P1/BRPPPPp1/BrnKQ1Rb
Reprints: (I) Problem 5-6 12/1951
Problem 7-9 03/1952
1439 FIDE Album 1945-1955 1964
(130) Problem 91-94 04/1964
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2012-04-08 more...
Genre: Retro
FEN: Nrq1kb2/1PpppppP/1p6/8/8/pP4P1/BRPPPPp1/BrnKQ1Rb
Reprints: (I) Problem 5-6 12/1951
Problem 7-9 03/1952
1439 FIDE Album 1945-1955 1964
(130) Problem 91-94 04/1964
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2012-04-08 more...
1. Da1! droht (unparierbar) 2. Dh8#
1. ... 0-0-0? ist illegal, da sK oder sTa8 schon gezogen haben muss
1. ... 0-0-0? ist illegal, da sK oder sTa8 schon gezogen haben muss
Keywords: Cant Castler, Castling (sg), Minimal, Miniature, Homebase
Genre: Retro, 2#
Computer test: C+ Popeye 4.61
FEN: r3k3/p1p5/Q3K3/8/8/8/8/8
Reprints: 304 Chess Strategy (Loyd) 1878
73 150 Schachkuriositäten 1910
63 Sam Loyd and his Chess Problems 1913
43 64 Schach-Scherze 1915
32 Retrograde Analysis 1915
168 Allgemeine Zeitung Chemnitz 27/11/1927
Arbeiter-Zeitung (Wien) 27/11/1932
8 Comoedia 09/07/1933
(II) Problem 37-40 09/1956
(D10) feenschach 27 04/1975
84 100 Classics of the Chessboard 1983
(7a) Die Schwalbe 145 08/1995
Thema Danicum 95 1999
52 Opfer-Opfer-Matt Gaudium 21 10/2000
Outrageous Chess Problems 2005
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-05-26 more...
Genre: Retro, 2#
Computer test: C+ Popeye 4.61
FEN: r3k3/p1p5/Q3K3/8/8/8/8/8
Reprints: 304 Chess Strategy (Loyd) 1878
73 150 Schachkuriositäten 1910
63 Sam Loyd and his Chess Problems 1913
43 64 Schach-Scherze 1915
32 Retrograde Analysis 1915
168 Allgemeine Zeitung Chemnitz 27/11/1927
Arbeiter-Zeitung (Wien) 27/11/1932
8 Comoedia 09/07/1933
(II) Problem 37-40 09/1956
(D10) feenschach 27 04/1975
84 100 Classics of the Chessboard 1983
(7a) Die Schwalbe 145 08/1995
Thema Danicum 95 1999
52 Opfer-Opfer-Matt Gaudium 21 10/2000
Outrageous Chess Problems 2005
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-05-26 more...
1. 0-0-0?? illegal
1. Td1! ... 2. Td3#
1. Td1! ... 2. Td3#
45 - P0002296
Andrey Frolkin
Dmitri W. Pronkin
2179 diagrammes 91 10-12/1989
Lob
(13+15) C+
BP in 25.0
Andrey Frolkin
Dmitri W. Pronkin
2179 diagrammes 91 10-12/1989
Lob
(13+15) C+
BP in 25.0
1. a4 h5 2. Ta3 h4 3. Tc3 h3 4. Tc6 bxc6 5. g3 La6 6. Lg2 Lc4 7. Le4 Sa6 8. f3 Db8 9. Kf2 Db4 10. Ke3 0-0-0 11. Kf4 Dxd2+ 12. Kf5 Dxc2 13. Dd5 Kb7 14. Dxf7 Kb6 15. a5+ Kc5 16. b4+ Kd4 17. Lb2+ Ke3 18. La1 Kf2 19. e3 d5 20. Se2 Td6 21. Td1 Tdh6 22. Td4 g6+ 23. Ke6 Ke1 24. Kd7 e6 25. Ke8 Sf6+
paul: Checked for the first 19 moves, by Euclide. (2011-07-01)
Moldenhauer: Computerprüfung: C+ Stelvio 1. Std. 16 Min.
Keine Lösung BP 24.0 (2023-02-06)
comment
Moldenhauer: Computerprüfung: C+ Stelvio 1. Std. 16 Min.
Keine Lösung BP 24.0 (2023-02-06)
comment
Keywords: Unique Proof Game, Promenade (K), Castling, Belfort (Kk)
Genre: Retro
Computer test: Stelvio 1.0
FEN: 4Kb1r/p1p2Q2/n1p1pnpr/P2p4/1PbRB3/4PPPp/2q1N2P/BN2k3
Reprints: 114 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Silvio Baier, 2023-03-02 more...
Genre: Retro
Computer test: Stelvio 1.0
FEN: 4Kb1r/p1p2Q2/n1p1pnpr/P2p4/1PbRB3/4PPPp/2q1N2P/BN2k3
Reprints: 114 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Silvio Baier, 2023-03-02 more...
46 - P0002316
Alexander Kislyak
3587 feenschach 60 05/1982
A. G. Kuznecow gewidmet
1. Preis
(15+13)
BP in 44.0
Alexander Kislyak
3587 feenschach 60 05/1982
A. G. Kuznecow gewidmet
1. Preis
(15+13)
BP in 44.0
1. a4 Sa6 2. a5 Sc5 3. a6 b6 4. Ta4 Sb7 5. axb7 c5 6. b8=L g6 7. Lg3 Dc7 8. Lh4 Dg3 9. hxg3 Lh6 10. Tg4 f5 11. Sf3 f4 12. Sd4 f3 13. Sc3 Lf4 14. gxf4 a5 15. Lg3 a4 16. Lh2 a3 17. g3 a2 18. Lg2 a1=D 19. 0-0 Da2 20. Kh1 Sf6 21. Lg1 Sd5 22. Kh2 Se3 23. Kh3 Sxd1 24. Kh4 Se3 25. Kg5 Tf8 26. Kh6 Tf5 27. Kg7 c4 28. Kh8 Kf7 29. Sd1 Kf6 30. Kg8 c3+ 31. Kf8 De6 32. Ke8 Kg7 33. Kd8 Kh6 34. Kc7 Ta7+ 35. Kb8 Tb7+ 36. Ka8 Tb8+ 37. Ka7 Tb7+ 38. Ka6 Tf7 39. Kb5 Kh5 40. Ka4 h6 41. Ka3 Sd5 42. Ka2 Sc7+ 43. Ka1 Ta7+ 44. Kb1 Da2
Paulo Roque: Is not possible: 38... Kh5 39.Kb5 (because black rook at square f5).
Probably the continuation of the author:
38...Tf7 39.Kb5 Kh5 40.Ka4 h6 41.Ka3 Sd5 42.Ka2 Sc7+ 43.Ka1 Ta7+ 44.Kb1 Da2# (2008-12-11)
James Malcom: Fixed. (2021-02-15)
comment
Probably the continuation of the author:
38...Tf7 39.Kb5 Kh5 40.Ka4 h6 41.Ka3 Sd5 42.Ka2 Sc7+ 43.Ka1 Ta7+ 44.Kb1 Da2# (2008-12-11)
James Malcom: Fixed. (2021-02-15)
comment
Keywords: Promotion (L), Promenade (K), Castling (wk), Non-standard material, Non-Unique Proof Game
Genre: Retro
FEN: 2b5/r1nppr2/1p4pp/7k/3N1PR1/2p2pP1/qPPPPPB1/1KBN1RB1
Reprints: 152 Shortest Proof Games 11/1991
(7) feenschach 103 01-09/1992
(8) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-02-15 more...
Genre: Retro
FEN: 2b5/r1nppr2/1p4pp/7k/3N1PR1/2p2pP1/qPPPPPB1/1KBN1RB1
Reprints: 152 Shortest Proof Games 11/1991
(7) feenschach 103 01-09/1992
(8) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-02-15 more...
1. Sf3 e6 2. Se5 Df6 3. Sxd7 Sxd7 4. a4 Sb6 5. a5 Ld7 6. axb6 0-0-0 7. Txa7 La4 8. Txa4 Td4 9. Txd4 Kb8 10. bxc7+ Ka8 11. Ta4#
Cook: 1. Sf3 e6 2. Se5 Df6 3. Sxd7 Kd8 4. Sxb8 Ld7 5. a4 Lb5 6. axb5 Kc8 7. Txa7 Kxb8 8. Ta3 c6 9. bxc6 Ta4 10. c7 Ka8 11. Txa4
Cook: 1. Sf3 e6 2. Se5 Df6 3. Sxd7 Kd8 4. Sxb8 Ld7 5. a4 Lb5 6. axb5 Kc8 7. Txa7 Kxb8 8. Ta3 c6 9. bxc6 Ta4 10. c7 Ka8 11. Txa4
Moldenhauer: Computerprüfung: Cooked Stellung Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 9.5, BP 10.0.
Eine Notation von Stelvio:
1.Sf3 e6 2.Se5 Df6 3.Sxd7 Kd8 4.Sxb8 Ld7 5.Sc6+ Kc8 6.Sxa7+ Kb8
7.a4 Lb5 8.axb5 Txa7 9.b6 Ka8 10.bxc7 Ta4 11.Txa4#
Schlüsselwort Rochade? (2023-05-02)
comment
Keine Lösung: BP 9.5, BP 10.0.
Eine Notation von Stelvio:
1.Sf3 e6 2.Se5 Df6 3.Sxd7 Kd8 4.Sxb8 Ld7 5.Sc6+ Kc8 6.Sxa7+ Kb8
7.a4 Lb5 8.axb5 Txa7 9.b6 Ka8 10.bxc7 Ta4 11.Txa4#
Schlüsselwort Rochade? (2023-05-02)
comment
Keywords: Unique Proof Game, Castling
Genre: Retro
FEN: k4bnr/1pP2ppp/4pq2/8/R7/8/1PPPPPPP/1NBQKB1R
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2006-10-15 more...
Genre: Retro
FEN: k4bnr/1pP2ppp/4pq2/8/R7/8/1PPPPPPP/1NBQKB1R
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2006-10-15 more...
BTM: 1. ... Lxb7+ 2. Ke3 0-0-0 3. Sb6#
WTM: 1. Sf6+! Kd8,Kf8 2. Dc7#,Ld6#
If WTM, b000 rights are already lost. Under Keym AP, Black attempts to steal the move. White disruption of castling now counts as win for White, so the only chance is 1. … BxQb7+ 2. Ke3! thr 3. Sf6+ disrupting castling but 2. … 0-0-0 3. Sb6#. 2. Kd3,Kf5? Be4+ 3. ~ 0-0-0! as b7 is no longer occupied, or 2. Kd3,Kd4,Kd5? 0-0-0! pinning wS or 2. Ke5,Kf4? 0-0-0! as wL is blocked. As usual when flip of player to move in d#n, Black gets an extra move rather than White losing one (c.f. Codex Article 15). Sublime miniature!
WTM: 1. Sf6+! Kd8,Kf8 2. Dc7#,Ld6#
If WTM, b000 rights are already lost. Under Keym AP, Black attempts to steal the move. White disruption of castling now counts as win for White, so the only chance is 1. … BxQb7+ 2. Ke3! thr 3. Sf6+ disrupting castling but 2. … 0-0-0 3. Sb6#. 2. Kd3,Kf5? Be4+ 3. ~ 0-0-0! as b7 is no longer occupied, or 2. Kd3,Kd4,Kd5? 0-0-0! pinning wS or 2. Ke5,Kf4? 0-0-0! as wL is blocked. As usual when flip of player to move in d#n, Black gets an extra move rather than White losing one (c.f. Codex Article 15). Sublime miniature!
VL: AP after Keym. Solution:
I: 1.Sf6+.
II: Bl's try to be on move. 0... Bxb7+! 1.Ke3! O-O-O (legalizing!) 2.Sb6#. (2007-01-26)
more ...
comment
I: 1.Sf6+.
II: Bl's try to be on move. 0... Bxb7+! 1.Ke3! O-O-O (legalizing!) 2.Sb6#. (2007-01-26)
more ...
comment
Keywords: Castling (sg), a posteriori (AP) (Type Keym), Homebase (s), Aristocrat, Miniature
Genre: Retro, 2#
FEN: r1b1k3/1Q1N4/8/8/4K3/8/7B/8
Reprints: Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
Genre: Retro, 2#
FEN: r1b1k3/1Q1N4/8/8/4K3/8/7B/8
Reprints: Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
49 - P0002665
Bernd Schwarzkopf
120v feenschach 2 03/1971
(1+1)
a) Diagramm
b) wKh8, sLa7
Ergänze zu einem IC!
a) sK, 2 sTT, 3 sBB
b) sK, 2 sTT, 4 sBB
Bernd Schwarzkopf
120v feenschach 2 03/1971
(1+1)
a) Diagramm
b) wKh8, sLa7
Ergänze zu einem IC!
a) sK, 2 sTT, 3 sBB
b) sK, 2 sTT, 4 sBB
a) sKg8, sTg7,f8, sBf7,g6,h6
b) sKc8 sTb7,d8, sBa6,b6,c7,d7
b) sKc8 sTb7,d8, sBa6,b6,c7,d7
Henrik Juel: In part b) the set of added men should be changed to sK, 2 sTT, 4 sBB
Both parts rely on the last move being a black castling with check (2021-10-31)
comment
Both parts rely on the last move being a black castling with check (2021-10-31)
comment
Keywords: Illegal cluster, Castling (sksg), Aristocrat, Miniature, Aristocrat, Miniature
Genre: Retro
FEN: K7/7b/8/8/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-11-01 more...
Genre: Retro
FEN: K7/7b/8/8/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-11-01 more...
1. c4 c5 2. Db3 Db6 3. Dh3 Dh6 4. b3 b6 5. La3 La6 6. Lb4 Lb5 7. Sa3 Sa6 8. 0-0-0 0-0-0 9. Kb1 Kb8 10. Tc1 Tc8 11. Tc3 Tc6 12. Tg3 Tg6 13. Tg4 Tg5 14. g3 g6 15. Lg2 Lg7 16. Ld5 Ld4 17. Sf3 Sf6 18. Tc1 Tc8 19. Tc3 Tc6 20. Te3 Te6 21. Te4 Te5 22. Tef4 Tef5 23. e4 e5
Keywords: Unique Proof Game, Castling (wgsg), Symmetrical position, Capture-free
Genre: Retro
Computer test: Mondenhauer C+ Stelvio 1.11 00:09:39 Minuten. (hh:mm:ss) Keine Lösung: BP 22.0, BP 22.5.
FEN: 1k6/p2p1p1p/np3npq/1bpBprr1/1BPbPRR1/NP3NPQ/P2P1P1P/1K6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-03-27 more...
Genre: Retro
Computer test: Mondenhauer C+ Stelvio 1.11 00:09:39 Minuten. (hh:mm:ss) Keine Lösung: BP 22.0, BP 22.5.
FEN: 1k6/p2p1p1p/np3npq/1bpBprr1/1BPbPRR1/NP3NPQ/P2P1P1P/1K6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-03-27 more...
51 - P0003009
Luigi Ceriani
117 La Genesi delle Posizioni 1961
(13+12) cooked
Welches war der erste Zug der sD und des sK?
Luigi Ceriani
117 La Genesi delle Posizioni 1961
(13+12) cooked
Welches war der erste Zug der sD und des sK?
AL in der Version von "hans" (PDB 2012-07-26):
R: 1. ... Ld8xLc7 2. Lb6-c7 Kc8-b8 3. Lc7-b6 Kb7-c8 4. Lb6-c7 Lc7-d8 5. Lg1-b6 Lb8-c7 6. Lb6-g1 Kc8-b7 7. Ld8-b6 Kb7-c8 8. d7-d8=L Lh7-g8 9. e6xSd7 Sc5-d7 10. Sc3-a4 Sa4-c5 11. Sd1-c3 d7-d6 12. Sc3-d1 Le5-b8 13. Se4-c3 Lg7-e5 14. Sf6-e4 c7-c6 15. Sg8-f6 Lh6-g7 16. g7-g8=S Kc8-b7 17. f6xTg7 Tg8-g7 18. e5-e6 Td8-g8 19. e4-e5 0-0-0 20. e3-e4 Lf8-h6 21. f5-f6 g7-g6 22. f4-f5 Le4-h7 23. f3-f4 Lb7-e4 24. f2-f3 Lc8-b7 25. Kb5-a5 b7xSa6 26. Kc4-b5 Sc5-a4 27. Kd3-c4 Se4-c5 28. Ke2-d3 Ta5-a3 29. Ta4-a2 Sf6-e4 30. La2-b1 Sg8-f6 31. Tb1-b2 Tf5-a5 32. Lb2-c1 Tf6-f5 33. Th1-b1 Tg6-f6 34. Ke1-e2 Th6-g6 35. Dd1-a1 Th7-h6 36. Lc1-b2 Th8-h7 37. b2-b3 Th7-h8 38. Lc4-a2 Th6-h7 39. Lf1-c4 Th7-h6 40. Ta1-a4 Th8-h7 41. Sc5-a6 h6xSg5 42. Se6-g5 h7-h6 43. a3xSb4 Sa6-b4 44. Sd8-e6 Sb8-a6 45. Se6xDd8 Sh6-g8 46. Sf4-e6 Sg8-h6 47. Sh3-f4 Sh6-g8 48. Sa4-c5 Sg8-h6 49. Sc3-a4 Sh6-g8 50. a2-a3 Sg8-h6 51. Sb1-c3 Sh6-g8 52. e2-e3 Sg8-h6 53. Sg1-h3
Cook: (Mario Richter, PDB 2023-06-30)
R: 1. ... Ld8xLc7 2. Lb6-c7 Kc8-b8 3. Lc7-b6 Lh7-g8 4. Lb6-c7 Lc7-d8 5. Ld4-b6 Lb8-c7 6. Lb6-d4 Kb7-c8 7. Ld8-b6 Kc7-b7 8. d7-d8=L Kd8-c7 9. e6xSd7 Sb6-d7 10. Sc5-a4 Sa4-b6 11. Se4-c5 d7-d6 12. Sf6-e4 Ke8-d8 13. Sg8-f6 Le5-b8 14. Sh6-g8 Lg7-e5 15. Sg8-h6 Lf8-g7 16. g7-g8=S c7-c6 17. f6xDg7 Dh6-g7 18. f5-f6 g7-g6 19. e5-e6 Dc6-h6 20. e4-e5 Da8-c6 21. e3-e4 Dd8-a8 22. f4-f5 Le4-h7 23. f3-f4 Lb7-e4 24. f2-f3 Lc8-b7 25. Kb5-a5 b7xSa6 26. Sb8-a6 Sc5-a4 27. Sc6xTb8 Se4-c5 28. Se5-c6 Sf6-e4 29. Kc4-b5 Ta6-a3 30. Kd3-c4 Sg8-f6 31. Ta4-a2 Tb6-a6 32. La2-b1 Tc6-b6 33. Tb1-b2 Te6-c6 34. Lb2-c1 Tf6-e6 35. Th1-b1 Tg6-f6 36. Dd1-a1 Th6-g6 37. Ke2-d3 Th5-h6 38. Ke1-e2 Th6-h5 39. Lc1-b2 Th7-h6 40. b2-b3 Th8-h7 41. Lc4-a2 Th7-h8 42. Lf1-c4 Th8-h7 43. Sc4-e5 Th7-h8 44. Ta1-a4 Th6-h7 45. Sa3-c4 Ta8-b8 46. Sb1-a3 Th8-h6 47. a3xSb4 h6xSg5 48. Sf3-g5 h7-h6 49. Sg1-f3 Sa6-b4 50. e2-e3 Sb8-a6 51. a2-a3
R: 1. ... Ld8xLc7 2. Lb6-c7 Kc8-b8 3. Lc7-b6 Kb7-c8 4. Lb6-c7 Lc7-d8 5. Lg1-b6 Lb8-c7 6. Lb6-g1 Kc8-b7 7. Ld8-b6 Kb7-c8 8. d7-d8=L Lh7-g8 9. e6xSd7 Sc5-d7 10. Sc3-a4 Sa4-c5 11. Sd1-c3 d7-d6 12. Sc3-d1 Le5-b8 13. Se4-c3 Lg7-e5 14. Sf6-e4 c7-c6 15. Sg8-f6 Lh6-g7 16. g7-g8=S Kc8-b7 17. f6xTg7 Tg8-g7 18. e5-e6 Td8-g8 19. e4-e5 0-0-0 20. e3-e4 Lf8-h6 21. f5-f6 g7-g6 22. f4-f5 Le4-h7 23. f3-f4 Lb7-e4 24. f2-f3 Lc8-b7 25. Kb5-a5 b7xSa6 26. Kc4-b5 Sc5-a4 27. Kd3-c4 Se4-c5 28. Ke2-d3 Ta5-a3 29. Ta4-a2 Sf6-e4 30. La2-b1 Sg8-f6 31. Tb1-b2 Tf5-a5 32. Lb2-c1 Tf6-f5 33. Th1-b1 Tg6-f6 34. Ke1-e2 Th6-g6 35. Dd1-a1 Th7-h6 36. Lc1-b2 Th8-h7 37. b2-b3 Th7-h8 38. Lc4-a2 Th6-h7 39. Lf1-c4 Th7-h6 40. Ta1-a4 Th8-h7 41. Sc5-a6 h6xSg5 42. Se6-g5 h7-h6 43. a3xSb4 Sa6-b4 44. Sd8-e6 Sb8-a6 45. Se6xDd8 Sh6-g8 46. Sf4-e6 Sg8-h6 47. Sh3-f4 Sh6-g8 48. Sa4-c5 Sg8-h6 49. Sc3-a4 Sh6-g8 50. a2-a3 Sg8-h6 51. Sb1-c3 Sh6-g8 52. e2-e3 Sg8-h6 53. Sg1-h3
Cook: (Mario Richter, PDB 2023-06-30)
R: 1. ... Ld8xLc7 2. Lb6-c7 Kc8-b8 3. Lc7-b6 Lh7-g8 4. Lb6-c7 Lc7-d8 5. Ld4-b6 Lb8-c7 6. Lb6-d4 Kb7-c8 7. Ld8-b6 Kc7-b7 8. d7-d8=L Kd8-c7 9. e6xSd7 Sb6-d7 10. Sc5-a4 Sa4-b6 11. Se4-c5 d7-d6 12. Sf6-e4 Ke8-d8 13. Sg8-f6 Le5-b8 14. Sh6-g8 Lg7-e5 15. Sg8-h6 Lf8-g7 16. g7-g8=S c7-c6 17. f6xDg7 Dh6-g7 18. f5-f6 g7-g6 19. e5-e6 Dc6-h6 20. e4-e5 Da8-c6 21. e3-e4 Dd8-a8 22. f4-f5 Le4-h7 23. f3-f4 Lb7-e4 24. f2-f3 Lc8-b7 25. Kb5-a5 b7xSa6 26. Sb8-a6 Sc5-a4 27. Sc6xTb8 Se4-c5 28. Se5-c6 Sf6-e4 29. Kc4-b5 Ta6-a3 30. Kd3-c4 Sg8-f6 31. Ta4-a2 Tb6-a6 32. La2-b1 Tc6-b6 33. Tb1-b2 Te6-c6 34. Lb2-c1 Tf6-e6 35. Th1-b1 Tg6-f6 36. Dd1-a1 Th6-g6 37. Ke2-d3 Th5-h6 38. Ke1-e2 Th6-h5 39. Lc1-b2 Th7-h6 40. b2-b3 Th8-h7 41. Lc4-a2 Th7-h8 42. Lf1-c4 Th8-h7 43. Sc4-e5 Th7-h8 44. Ta1-a4 Th6-h7 45. Sa3-c4 Ta8-b8 46. Sb1-a3 Th8-h6 47. a3xSb4 h6xSg5 48. Sf3-g5 h7-h6 49. Sg1-f3 Sa6-b4 50. e2-e3 Sb8-a6 51. a2-a3
s.a. 32Pe1A
Korrekturversuch zu P0005036
hans: Good motivation to capture Qd8 on the spot, to make long castling possible, which is needed to retrack the captured bRg7 just on time. Also a minor-promotion. I like this one, and I think the stipulation asks for which move black queen makes.
R: -1. …Ld8xLc7+ -2. Lb6c7+ Kc8b8 -3. Lc7b6 Kb7c8 -4. Lb6c7 Lc7d8 -5. Lg1b6 Lb8c7+ -6.Lb6g1 Kc8b7 -7. Ld8b6 Kb7c8 -8. d7d8=L Lh7g8 -9. e6xSd7 Sc5d7 -10. Sc3a4 Sa4c5+ -11.Sd1c3 d7d6 -12.Sc3d1 Le5b8 -13.Se4c3 Lg7e5 -14.Sf6e4 c7c6 -15.Sg8f6 Lh6g7 -16.g7g8=S Kc8b7 -17.f6xTg7 Tg8g7 -18.e5e6 Td8g8 -19.e4e5 0-0-0!! -20.e3e4 Lf8h6 -21.f6f5 g7g6 -21.f4f5 Le4h7 -22.f3f4 Lb7e4 -23.f2f3 Lc8b7 -24. Kb5a5 b7xSa6 and cage can be undone while black plays only with Ta3 and Sa4.
captures white axSb, SxDd8, exSd7, fxTg7, black hxSg, bxSa6, LxLc7 (2012-07-26)
Henrik Juel: Very similar to P0005036 and with identical stipulation question:
What was the first move by black queen and by black king.
Ceriani's abbreviations for move, queen, king, black, (and white) are t., D, R, n, (and b); in his ortho reconstruction problems the color abbreviations are capitalized, e.g. N=11 meaning 11 black moves (2012-07-26)
Thomas Volet: This composition appears on p.197 of Ceriani's 1961 book as his correction of P0005036 (which appeared in his earlier book). On that page, he discusses the clever cook in P0005036 ("ma questa bella posizione e demolita") with the WhP unpromoting at h8, uncapturing to the g file, and uncapturing back to the h file. (2012-08-02)
Mario Richter: I'm sorry to say this, but Ceriani's correction attempt still leaves room for cooks ... (2023-06-30)
comment
Korrekturversuch zu P0005036
hans: Good motivation to capture Qd8 on the spot, to make long castling possible, which is needed to retrack the captured bRg7 just on time. Also a minor-promotion. I like this one, and I think the stipulation asks for which move black queen makes.
R: -1. …Ld8xLc7+ -2. Lb6c7+ Kc8b8 -3. Lc7b6 Kb7c8 -4. Lb6c7 Lc7d8 -5. Lg1b6 Lb8c7+ -6.Lb6g1 Kc8b7 -7. Ld8b6 Kb7c8 -8. d7d8=L Lh7g8 -9. e6xSd7 Sc5d7 -10. Sc3a4 Sa4c5+ -11.Sd1c3 d7d6 -12.Sc3d1 Le5b8 -13.Se4c3 Lg7e5 -14.Sf6e4 c7c6 -15.Sg8f6 Lh6g7 -16.g7g8=S Kc8b7 -17.f6xTg7 Tg8g7 -18.e5e6 Td8g8 -19.e4e5 0-0-0!! -20.e3e4 Lf8h6 -21.f6f5 g7g6 -21.f4f5 Le4h7 -22.f3f4 Lb7e4 -23.f2f3 Lc8b7 -24. Kb5a5 b7xSa6 and cage can be undone while black plays only with Ta3 and Sa4.
captures white axSb, SxDd8, exSd7, fxTg7, black hxSg, bxSa6, LxLc7 (2012-07-26)
Henrik Juel: Very similar to P0005036 and with identical stipulation question:
What was the first move by black queen and by black king.
Ceriani's abbreviations for move, queen, king, black, (and white) are t., D, R, n, (and b); in his ortho reconstruction problems the color abbreviations are capitalized, e.g. N=11 meaning 11 black moves (2012-07-26)
Thomas Volet: This composition appears on p.197 of Ceriani's 1961 book as his correction of P0005036 (which appeared in his earlier book). On that page, he discusses the clever cook in P0005036 ("ma questa bella posizione e demolita") with the WhP unpromoting at h8, uncapturing to the g file, and uncapturing back to the h file. (2012-08-02)
Mario Richter: I'm sorry to say this, but Ceriani's correction attempt still leaves room for cooks ... (2023-06-30)
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Keywords: First Move? (kd), Castling in the retro play (sg)
Genre: Retro
FEN: 1k4b1/p1b1pp2/p1pp2p1/K5p1/NP6/rP6/RRPP2PP/QBB5
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-07-02 more...
Genre: Retro
FEN: 1k4b1/p1b1pp2/p1pp2p1/K5p1/NP6/rP6/RRPP2PP/QBB5
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-07-02 more...
1. 0-0-0 Ke2 2. Kb8 Td1 3. Ka8 Txd8#
1. 0-0-0 0-0? usw. ist nicht erlaubt, da Weiß im letzten Zug einen schwarzen Stein geschlagen haben muss, was nur mit K oder T ging.
1. 0-0-0 0-0? usw. ist nicht erlaubt, da Weiß im letzten Zug einen schwarzen Stein geschlagen haben muss, was nur mit K oder T ging.
A.Buchanan: Tried some experiments to see if other formats of position entry are possible, but having reverted those, the solution animation no longer works. The usual PDB problem of hidden & undocumented variables between stipulation, diagrams & solution animation, sigh. (2021-04-01)
A.Buchanan: I found a "fix" - I added a bogus empty diagram b! How random! (2021-04-02)
A.Buchanan: For h#, I run Popeye without opti noca, so can see all the retro tries explicitly. For d#, I would apply opti noca if relevant, else lines of play might be blocked (2021-04-04)
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A.Buchanan: I found a "fix" - I added a bogus empty diagram b! How random! (2021-04-02)
A.Buchanan: For h#, I run Popeye without opti noca, so can see all the retro tries explicitly. For d#, I would apply opti noca if relevant, else lines of play might be blocked (2021-04-04)
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a) 1. Df7! h4 2. 0-0 h5 3. Lh8 hxg6 4. Sg7 gxh7#
b) 1. De6! h4 2. 0-0 h5 3. Kh8 hxg6 4. Tg8 Txh7#
1. 0-0-0?? 0-0 2. Kb8 Txc1 3. Ka8 Txb1 4. Tb8 Ta1# (0-0-0 ist illegal)
1. 0-0-0? ist nicht gestattet, da wegen der schwarzen Bauernstellung der weiße Bauer a2 auf a8 umgewandelt wurde!
b) 1. De6! h4 2. 0-0 h5 3. Kh8 hxg6 4. Tg8 Txh7#
1. 0-0-0?? 0-0 2. Kb8 Txc1 3. Ka8 Txb1 4. Tb8 Ta1# (0-0-0 ist illegal)
1. 0-0-0? ist nicht gestattet, da wegen der schwarzen Bauernstellung der weiße Bauer a2 auf a8 umgewandelt wurde!
Marcin Banaszek: Die Urdruck-Stellung ist mit sSe6 (nicht sSd5) und über die Zwilling-Stellung ist dort keine Erwähnung. (2022-06-09)
A.Buchanan: Yes I don't know where the twin concept was introduced to this record (the visible "Last Update" history does not mention it). The twin adds very little: the retro try is identical, and the solution shares most of the White moves, ending in a duller mate. Did the 1983 reprint include it? (2022-06-10)
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A.Buchanan: Yes I don't know where the twin concept was introduced to this record (the visible "Last Update" history does not mention it). The twin adds very little: the retro try is identical, and the solution shares most of the White moves, ending in a duller mate. Did the 1983 reprint include it? (2022-06-10)
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set play
1. ... a3,a4? 2. cxb2 0-0?# castling illegal
1. ... Kf1! 2. cxb2 Kg2#
1. ... Kd1! 2. cxb2 Kc2#
actual play
1. c3xb2? 0-0#? castling illegal
1. Ka1! Kd1 2. c2+ Kxc2#
1. Kxb2! Kd1 2. Ka1 Kc2#
1. ... a3,a4? 2. cxb2 0-0?# castling illegal
1. ... Kf1! 2. cxb2 Kg2#
1. ... Kd1! 2. cxb2 Kc2#
actual play
1. c3xb2? 0-0#? castling illegal
1. Ka1! Kd1 2. c2+ Kxc2#
1. Kxb2! Kd1 2. Ka1 Kc2#
A.Buchanan: Not sure what is intended here: should one include set play? It's a bit unclear. (2021-10-22)
A.Buchanan: I have marked stip from "h#2" to "h#2* 2 solutions" (2021-10-22)
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A.Buchanan: I have marked stip from "h#2" to "h#2* 2 solutions" (2021-10-22)
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1. Kb3 Kd2 2. Ka4 Kc3 3. a2 Txa2#
Nicht 1. Kb3 0-0-0?? 2. Ka2 Td3 3. Ka1 Txa3#, da Weiß zuletzt mit K oder T gezogen hat
Nicht 1. Kb3 0-0-0?? 2. Ka2 Td3 3. Ka1 Txa3#, da Weiß zuletzt mit K oder T gezogen hat
* 1. ... Tg1 2. e4 g3#
1. Kg3 0-0? 2. h4 Tf3#
1. Kg3 Tg1! 2. Kh4 g3#
1. Kg3 0-0? 2. h4 Tf3#
1. Kg3 Tg1! 2. Kh4 g3#
A.Buchanan: I suppose set play is intended, because it's the only function of sBe5. But I think it's fine without. (2021-10-22)
Mario Richter: The set play is not mentioned in the "official solution" ('Arbejder-Skak' 08/1957 p.231) (2021-10-23)
A.Buchanan: Thanks. Odd! (2021-10-23)
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Mario Richter: The set play is not mentioned in the "official solution" ('Arbejder-Skak' 08/1957 p.231) (2021-10-23)
A.Buchanan: Thanks. Odd! (2021-10-23)
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Keywords: Cant Castler, Castling (wk)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic I suppose set play is intended, because it's the only function of sBe5. But I think it's fine without.
FEN: 8/8/8/4p1pp/6Pk/4p2P/4P1PR/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-24 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic I suppose set play is intended, because it's the only function of sBe5. But I think it's fine without.
FEN: 8/8/8/4p1pp/6Pk/4p2P/4P1PR/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-24 more...
1. Kxd8 Ta7 2. Te8 Td7#
1. 0-0 0-0-0? 2. Th8 Txg1# try
1. 0-0 0-0-0? 2. Th8 Txg1# try
White castling not possible as sBa= on a1. (Also sBgxh= or sBhxg=)
A.Buchanan: Retractor 2.1.1 does not notice that promotion happened on a1 (2021-11-25)
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A.Buchanan: Retractor 2.1.1 does not notice that promotion happened on a1 (2021-11-25)
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1. Kd8 Txh8 2. Kc8 Txg8#
Henrik Juel: The diagram pawns captured all missing men, so the missing pawns [Pa7,Ph2] promoted on a1,h8, and Ta1,Th8 have moved, ruling out the apparent solutions 1.Ke7 0-0-0 2.Kxe6 The1#, 1.Dg1+ Ke2 2.0-0 Taxg1#
C+ by Popeye 4.61 and analysis (2021-08-31)
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C+ by Popeye 4.61 and analysis (2021-08-31)
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1. ... Sf5? 2. 0-0-0?? Sd6# - aber die s0-0-0 ist illegal, den zuletzt muß Schwarz mit K oder T gezogen haben.
1. ... Sg8! 2. Td8 Sc7#
1. ... Sg8! 2. Td8 Sc7#
Mario Richter: Luboš Kekely (Slovakia) correctly points out that 1. ... Sf5! 2. 0-0-0 Sd6# is only a try and not a solution (last black move must have been by King or Rook, so the queenside castling is illegal).
I have changed the solution accordingly. (2023-03-10)
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I have changed the solution accordingly. (2023-03-10)
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not: 1. Sd7 0-0-0 2. 0-0 Tdg1#
but: 1. Lxe6 Th7 2. 0-0-0 Ta8#
and: 1. Ke7 0-0 2. Kxe6 Tae1#
but: 1. Lxe6 Th7 2. 0-0-0 Ta8#
and: 1. Ke7 0-0 2. Kxe6 Tae1#
A.Buchanan: Hard to see this as anything but a version of P0563989. Popeye reports again three apparent mates, with all 4 castlings. However pawn capture analysis indicates that Black qside castling is illegal, while the two kingside castlings are mutually exclusive. (2022-01-08)
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Keywords: Cant Castler (sg), Castling, Partial Retro Analysis (PRA)
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: r1b1kn1r/1n6/1pPpP3/1p1p4/1bp5/1qpp4/1PP5/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-08 more...
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: r1b1kn1r/1n6/1pPpP3/1p1p4/1bp5/1qpp4/1PP5/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-08 more...
s) 1. Kf8 g6 2. Se7 Th8#
w) 1. ... Kg1 Td8 2. Lh2 Td1#
w) 1. ... Kg1 Td8 2. Lh2 Td1#
Rochaderecht
Henrik Juel: C+ by Popeye 4.61 and analysis
If Black could castle, both parts would be cooked:
1.0-0-0?? Txg7 2.Sb8 Tc7#
1... Kg1 0-0-0?? 2.Lh2 Td1#
But he cannot, because explaining wTh7 by h7-h8=T would require nine white pawn captures, and only eight black men are missing; hxg8=T is even worse, requiring 10 captures (Th7 being original or promoted further towards the west obviously would require Ke8 to have moved) (2021-07-04)
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Henrik Juel: C+ by Popeye 4.61 and analysis
If Black could castle, both parts would be cooked:
1.0-0-0?? Txg7 2.Sb8 Tc7#
1... Kg1 0-0-0?? 2.Lh2 Td1#
But he cannot, because explaining wTh7 by h7-h8=T would require nine white pawn captures, and only eight black men are missing; hxg8=T is even worse, requiring 10 captures (Th7 being original or promoted further towards the west obviously would require Ke8 to have moved) (2021-07-04)
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*) 1. ... 0-0 2. Tf8 Tc1 3. Tf7 Tc8#
1) 1. 0-0 Tf1 2. Txf2 Txf2 3. Kh8 Tf8#
1) 1. 0-0 Tf1 2. Txf2 Txf2 3. Kh8 Tf8#
Henrik Juel: Ljubomir had an anonymous co-author?? (2021-10-31)
A.Buchanan: I've no idea. There were three users called "anonymous" of different kinds, so I merged them into one, that's all. (2021-10-31)
A.Buchanan: I think anticipated in 1948 by Peter Kniest P0003606 (2021-11-01)
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A.Buchanan: I've no idea. There were three users called "anonymous" of different kinds, so I merged them into one, that's all. (2021-10-31)
A.Buchanan: I think anticipated in 1948 by Peter Kniest P0003606 (2021-11-01)
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Keywords: Cant Castler, Castling (wksk), Homebase (w), anticipated (P0003606)
Genre: h#, Retro
FEN: 4k2r/3pp1pp/pp6/8/8/8/3PPP1P/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-01 more...
Genre: h#, Retro
FEN: 4k2r/3pp1pp/pp6/8/8/8/3PPP1P/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-01 more...
*) 1. ... 0-0 2. Dh4 Txf3#
1) 1. Kh4 Kf2 2. fxg2 Sf3#
1) 1. Kh4 Kf2 2. fxg2 Sf3#
SCHRECKE: NL: 1. Dg5,Kh4 gxf3 2. Kh4,Dg5 Sf1# (2023-09-13)
Ladislav Packa: Retro content is not needed, the solution is preserved even without it.
Pg4 Pg2 Sh2 Ke1 Rh1 (5)- Ph5 Kg3 (2) h#2* C+
1...0-0 2.h5-h4 Rf1-f3 #
1.Kg3-h4 Ke1-f2 2.h5*g4 Sh2-f3 # (2023-09-14)
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Ladislav Packa: Retro content is not needed, the solution is preserved even without it.
Pg4 Pg2 Sh2 Ke1 Rh1 (5)- Ph5 Kg3 (2) h#2* C+
1...0-0 2.h5-h4 Rf1-f3 #
1.Kg3-h4 Ke1-f2 2.h5*g4 Sh2-f3 # (2023-09-14)
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*) 1. ... 0-0#
1) 1. Dc5 Tf1+ 2. Df2 Txf2#
1) 1. Dc5 Tf1+ 2. Df2 Txf2#
a) 1. Txh3 0-0 2. Sd3 Txf4 3. Sd6 Ta4#
b) 1. Sd1 Txh4 2. Se3 Th8 3. Sd6 Ta8#
b) 1. Sd1 Txh4 2. Se3 Th8 3. Sd6 Ta8#
Adrian Storisteanu: Fix for P0534003. (2015-10-17)
Yuri Bilokin: version: -wPh3 8/3pq3/5r2/5p2/4nppp/5r2/kpP2n2/1b2K2R (3+13) h#3 2 Sol
1.Rh3 0-0 2.Sd3 Rxf4 3.Sd6 Ra4# (MM)
1.Sd1 Rxh4 2.Se3 Rh8 3.Sd6 Ra8# (MM) (2021-06-24)
Ladislav Packa: Yuri: Without wPh3, the castling is impossible (theme Cant Castler). But after wPh3-d5, two solutions are OK. (2021-06-24)
Yuri Bilokin: Ladislav, sorry, was wrong. Thank you for your vigilance. A slightly different edition, too, with two solutions: 8/2p1q3/4r3/5pPp/4npp1/5r2/kpP2n2/1b2K2R (4+13) h#3 2 Sol. (2021-06-25)
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Yuri Bilokin: version: -wPh3 8/3pq3/5r2/5p2/4nppp/5r2/kpP2n2/1b2K2R (3+13) h#3 2 Sol
1.Rh3 0-0 2.Sd3 Rxf4 3.Sd6 Ra4# (MM)
1.Sd1 Rxh4 2.Se3 Rh8 3.Sd6 Ra8# (MM) (2021-06-24)
Ladislav Packa: Yuri: Without wPh3, the castling is impossible (theme Cant Castler). But after wPh3-d5, two solutions are OK. (2021-06-24)
Yuri Bilokin: Ladislav, sorry, was wrong. Thank you for your vigilance. A slightly different edition, too, with two solutions: 8/2p1q3/4r3/5pPp/4npp1/5r2/kpP2n2/1b2K2R (4+13) h#3 2 Sol. (2021-06-25)
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66 - P0003659
Maurice Jago
8202v Die Schwalbe , p. 148, 10/1980
version of cooked 8202 Die Schwalbe 08/1977
(12+14) cooked
h#2
b) sBb4->b5
Maurice Jago
8202v Die Schwalbe , p. 148, 10/1980
version of cooked 8202 Die Schwalbe 08/1977
(12+14) cooked
h#2
b) sBb4->b5
a) 1. Sf2 Dxf2+ 2. Kh1 0-0-0#
b) 1. Sb1 Ta4 2. Sxg3 Th4#
Cook: 1. Txg3 0-0-0 2. La2/Lb3/Lc4 Dxh1#
1. La2/Lb3/Lc4 0-0-0 2. Txg3 Dxh1#
b) 1. Sb1 Ta4 2. Sxg3 Th4#
Cook: 1. Txg3 0-0-0 2. La2/Lb3/Lc4 Dxh1#
1. La2/Lb3/Lc4 0-0-0 2. Txg3 Dxh1#
See P0000642
Mario Richter: Hier stimmt was nicht: In der Schwalbe 08/1977 erschien als Nr. 2202 die folgende Aufgabe: 8/Pp2p1p1/Bq1p3p/4p3/1p6/n4pP1/1PPPPr1k/R3KQ1n/; h#2, b) sBb4 nach b5 und der AL a) 1.Tf2-g2 0-0-0 2.Tg2xg3 Df1xh1#; b) 1.Sa3-b1 Ta1-a4 2.Sh1xg3 Ta4-h4#, die aber in a) und b) durch 1.f3xe2 La6xb7 2.Tf2-g2 Df1xg2# nebenlösig war.
Ferner ist in nebenstehendem Diagramm sowohl in a) als auch in b) sehr wohl die w0-0-0 möglich
(Weiß hat als letzten Zug Sc7-a8, Schwarz entwandelt auf b1 und h1, Weiß auf f8); und selbst unter der Annahme, daß die w0-0-0 nur mit sBb4 möglich sei, wäre die Aufgabe mehrfach nebenlösig, z.B. durch 1.Ld5-c4 0-0-0 2.Bd6-d5 Df1xh1# (2010-05-29)
A.Buchanan: Here, the source was incorrectly given as "8202v Die Schwalbe 08/1977". No, that was where the original problem appeared, with a "v" added. So when, as here, we suspect the diagram has a typo, we would have to look through many back issues of Die Schwalbe! It turns out that this correction (itself cooked) was published (as text) in p.148 Die Schwalbe 65 10/1980. I have corrected the source. The reference to the earlier buggy problem can then cleanly be put in the "after" field. The actual typo was missing bPc7. (2024-03-06)
A.Buchanan: With the typo fixed, I think the retro-logic works ok, but there is a single cook family, fortunately repairable (2024-03-06)
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Mario Richter: Hier stimmt was nicht: In der Schwalbe 08/1977 erschien als Nr. 2202 die folgende Aufgabe: 8/Pp2p1p1/Bq1p3p/4p3/1p6/n4pP1/1PPPPr1k/R3KQ1n/; h#2, b) sBb4 nach b5 und der AL a) 1.Tf2-g2 0-0-0 2.Tg2xg3 Df1xh1#; b) 1.Sa3-b1 Ta1-a4 2.Sh1xg3 Ta4-h4#, die aber in a) und b) durch 1.f3xe2 La6xb7 2.Tf2-g2 Df1xg2# nebenlösig war.
Ferner ist in nebenstehendem Diagramm sowohl in a) als auch in b) sehr wohl die w0-0-0 möglich
(Weiß hat als letzten Zug Sc7-a8, Schwarz entwandelt auf b1 und h1, Weiß auf f8); und selbst unter der Annahme, daß die w0-0-0 nur mit sBb4 möglich sei, wäre die Aufgabe mehrfach nebenlösig, z.B. durch 1.Ld5-c4 0-0-0 2.Bd6-d5 Df1xh1# (2010-05-29)
A.Buchanan: Here, the source was incorrectly given as "8202v Die Schwalbe 08/1977". No, that was where the original problem appeared, with a "v" added. So when, as here, we suspect the diagram has a typo, we would have to look through many back issues of Die Schwalbe! It turns out that this correction (itself cooked) was published (as text) in p.148 Die Schwalbe 65 10/1980. I have corrected the source. The reference to the earlier buggy problem can then cleanly be put in the "after" field. The actual typo was missing bPc7. (2024-03-06)
A.Buchanan: With the typo fixed, I think the retro-logic works ok, but there is a single cook family, fortunately repairable (2024-03-06)
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Keywords: Cant Castler, Castling (wg), Superseded by (P1415606)
Genre: h#, Retro
Computer test: both twins cooked by Popeye v4.87
FEN: NBr5/N1p3p1/1qPpp3/2bb4/1p6/n1pP1rP1/1PP1P2k/R3KQ1n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-03-06 more...
Genre: h#, Retro
Computer test: both twins cooked by Popeye v4.87
FEN: NBr5/N1p3p1/1qPpp3/2bb4/1p6/n1pP1rP1/1PP1P2k/R3KQ1n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-03-06 more...
a) 1. Ta6! Lxc3 2. Ta2 0-0#
b) 1. Ta5! Kd1 2. Ta2 Kc2#
b) 1. Ta5! Kd1 2. Ta2 Kc2#
Henrik Juel: In b) White may not castle, because last move was done by Ke1 or Th1 (2023-12-02)
A.Buchanan: Cute. Do wPe2 & bPe3 serve any purpose, however? (2023-12-03)
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A.Buchanan: Cute. Do wPe2 & bPe3 serve any purpose, however? (2023-12-03)
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Keywords: Cant Castler, Castling (wk), Superseded by (P1413924)
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & simple retro logic
FEN: 5n1q/6B1/5r2/7r/5p2/2p1p3/1P2P3/k3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-05 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & simple retro logic
FEN: 5n1q/6B1/5r2/7r/5p2/2p1p3/1P2P3/k3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-05 more...
a) 1. 0-0 Tcg3 2. Sh8 Txg7#
b) 1. 0-0-0 Tgc4 2. Sb8 Txc7#
b) 1. 0-0-0 Tgc4 2. Sb8 Txc7#
Keywords: Castling, Cant Castler, Obvious promotion (T), Twin
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: r3k2r/1pp3pp/2n3n1/8/6R1/2R4P/4P1P1/5B1K
Input: Gerd Wilts, 1995-06-03
Last update: Gunter Jordan, 2022-12-01 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: r3k2r/1pp3pp/2n3n1/8/6R1/2R4P/4P1P1/5B1K
Input: Gerd Wilts, 1995-06-03
Last update: Gunter Jordan, 2022-12-01 more...
1. Th8 0-0-0 2. Txf8 The1#
1. Sa6 0-0 2. Txd8 Tae1#
1. Th7 Sxg6 2. Td7 Th8#
1. Sa6 0-0 2. Txd8 Tae1#
1. Th7 Sxg6 2. Td7 Th8#
wie lautet die Retroanalyse? Warum sollten die Rochaden nicht möglich sein?
Henrik Juel: White pawns captured all 8 missing black men, incl. [Pa7,e7,h7]
For both white castlings to be possible, 4 black pawn captures are needed (axb, dxc, e2xd1/f1, and hxg)
but only 3 white men are missing, so the the white castlings are mutually exclusive (2021-11-22)
Henrik Juel: HC+ Popeye 4.61, if '2 solutions' is added to the stipulation
The 2 solutions are 1.Th7... and the one with the permissible castling (2021-11-22)
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Henrik Juel: White pawns captured all 8 missing black men, incl. [Pa7,e7,h7]
For both white castlings to be possible, 4 black pawn captures are needed (axb, dxc, e2xd1/f1, and hxg)
but only 3 white men are missing, so the the white castlings are mutually exclusive (2021-11-22)
Henrik Juel: HC+ Popeye 4.61, if '2 solutions' is added to the stipulation
The 2 solutions are 1.Th7... and the one with the permissible castling (2021-11-22)
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1. 0-0 Tf1 (0-0?) 2. Kh7 Txf5 3. Tg8 Th5#
1. Kf7 0-0 2. Td8 Lxf5 3. Ke8 Lg6#
If both players retain castling rights, then Wh last move was R: 1. L~xLb1. This wL is not original, and all captures by Wh are balanced, so wBc must have promoted on c8. sBc & sBd cross-captured in any case to release sTa1, but the sequence of moves (in forward direction) can only have been sBc7xd6 wBc6-c7 sBd7xc6 sLc8~ wBc7-c8=L. But under this sequence sTa8 could still not have escaped the cage. Contradiction, so some castling right is lost
1. Kf7 0-0 2. Td8 Lxf5 3. Ke8 Lg6#
If both players retain castling rights, then Wh last move was R: 1. L~xLb1. This wL is not original, and all captures by Wh are balanced, so wBc must have promoted on c8. sBc & sBd cross-captured in any case to release sTa1, but the sequence of moves (in forward direction) can only have been sBc7xd6 wBc6-c7 sBd7xc6 sLc8~ wBc7-c8=L. But under this sequence sTa8 could still not have escaped the cage. Contradiction, so some castling right is lost
Henrik Juel: C+ Popeye 4.61
The castlings are mutually exclusive, so
1. 0-0 0-0? 2. Kf7 Txf5 3. Tg8 Th5# is not a solution,
and the stipulation should be just h#3 (without 2.1...),
as the problem has one solution in two parts, depending on which castling is legal (2020-12-21)
A.Buchanan: I agree and have edited the stipulation. The problem came from a time when RS was the default not PRA, so the stipulation would have had 2.1.... in it. We should have an originalforderung field for this kind of situation. Also typo 2.Kh7 not 2.KF7. Nice harmonious problem (2020-12-22)
Henrik Juel: Thanks Andrew
Still a tiny typo: if ..., last move was LxLb1, not Lb8
I often confuse files b and g, so confusing rows 1 and 8 is refreshing... (2020-12-22)
A.Buchanan: Thanks - I am completely dyslexic for chess notation up-down left-right. It was shifting from descriptive to algebraic notation as a boy that did it, I think (2020-12-22)
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The castlings are mutually exclusive, so
1. 0-0 0-0? 2. Kf7 Txf5 3. Tg8 Th5# is not a solution,
and the stipulation should be just h#3 (without 2.1...),
as the problem has one solution in two parts, depending on which castling is legal (2020-12-21)
A.Buchanan: I agree and have edited the stipulation. The problem came from a time when RS was the default not PRA, so the stipulation would have had 2.1.... in it. We should have an originalforderung field for this kind of situation. Also typo 2.Kh7 not 2.KF7. Nice harmonious problem (2020-12-22)
Henrik Juel: Thanks Andrew
Still a tiny typo: if ..., last move was LxLb1, not Lb8
I often confuse files b and g, so confusing rows 1 and 8 is refreshing... (2020-12-22)
A.Buchanan: Thanks - I am completely dyslexic for chess notation up-down left-right. It was shifting from descriptive to algebraic notation as a boy that did it, I think (2020-12-22)
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Keywords: Partial Retro Analysis (PRA), Castling, Obvious promotion (L)
Genre: h#, Retro
FEN: 4k2r/1p1np1b1/p1pp4/3p1p2/7P/6p1/P3P1PP/rB2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-12-23 more...
Genre: h#, Retro
FEN: 4k2r/1p1np1b1/p1pp4/3p1p2/7P/6p1/P3P1PP/rB2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-12-23 more...
a) 1. f5 Le5 2. 0-0 Th8#
b) 1. Kd8 0-0-0 2. Te8 Lf8#
Try to deduce the diagram error, as we do have the intended solutions!
a) wRh7 is original, but cannot have escaped from cage by white cross-capture as there are too many captures needed. On the other hand, Black only has to make 4 captures so the cross-capture is ok.
b) changing the colour of Sa5 shifts the capture balance: now White can cross-capture but not Black, so White but not Black can castle.
Thus the retro-logic works with the diagram as it is, but the forward logic does not. How can we repair the diagram?
b) 1. Kd8 0-0-0 2. Te8 Lf8#
Try to deduce the diagram error, as we do have the intended solutions!
a) wRh7 is original, but cannot have escaped from cage by white cross-capture as there are too many captures needed. On the other hand, Black only has to make 4 captures so the cross-capture is ok.
b) changing the colour of Sa5 shifts the capture balance: now White can cross-capture but not Black, so White but not Black can castle.
Thus the retro-logic works with the diagram as it is, but the forward logic does not. How can we repair the diagram?
See P0000899 a companion problem.
A.Buchanan: Something odd going on here. There are numerous h#2 in both a&b. PRA not normally found in a problem with single solution for each twin. White has 8 pawns so trivially wRh7 must be original. (2018-10-13)
A.Buchanan: There are three similar problems by Brogi which are all twinned RS rather than PRA, but don't have diagram error. P0003743, P0003746 & P0003747. (2018-10-13)
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
Ladislav Packa: Cooked a) and b):
1...b8S and 2...R:h8# (2021-02-10)
A.Buchanan: I think the composer simply forgot that wPb7 can promote. Most of the cooks come from S promotions, but it's also possible to have QR. I've fixed it, rejigging the captures to make them still add up. I've posted in Discord, as I did with the partner composition, and will create entries for them here. (2022-03-15)
milan: +sLb8 sBa7=sT M.Frelih (2023-12-02)
A.Buchanan: Hi Milan - I don't think your suggestion quite works for b). In a) there are 0+2 spare captures, so Black can certainly cross-capture. But in b) there is 1+1 so neither side can cross-capture, so there is no solution. Please compare with P1399806, in which there are 1+2 & 2+1 spare captures, so both twins are sound. (2023-12-03)
milan: Hi Andrew my correction works only with 2.1... solutions, black or white knights on a5. are not important. (2023-12-03)
A.Buchanan: Hi Milan not really clear what you are doing, but if as well as the piece changes you proposed, you also change the stipulation to 2.1... then there is still only one solution. Even if you remove Sa5 entirely as well, there is no White cross-capture possible. (2023-12-04)
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A.Buchanan: Something odd going on here. There are numerous h#2 in both a&b. PRA not normally found in a problem with single solution for each twin. White has 8 pawns so trivially wRh7 must be original. (2018-10-13)
A.Buchanan: There are three similar problems by Brogi which are all twinned RS rather than PRA, but don't have diagram error. P0003743, P0003746 & P0003747. (2018-10-13)
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
Ladislav Packa: Cooked a) and b):
1...b8S and 2...R:h8# (2021-02-10)
A.Buchanan: I think the composer simply forgot that wPb7 can promote. Most of the cooks come from S promotions, but it's also possible to have QR. I've fixed it, rejigging the captures to make them still add up. I've posted in Discord, as I did with the partner composition, and will create entries for them here. (2022-03-15)
milan: +sLb8 sBa7=sT M.Frelih (2023-12-02)
A.Buchanan: Hi Milan - I don't think your suggestion quite works for b). In a) there are 0+2 spare captures, so Black can certainly cross-capture. But in b) there is 1+1 so neither side can cross-capture, so there is no solution. Please compare with P1399806, in which there are 1+2 & 2+1 spare captures, so both twins are sound. (2023-12-03)
milan: Hi Andrew my correction works only with 2.1... solutions, black or white knights on a5. are not important. (2023-12-03)
A.Buchanan: Hi Milan not really clear what you are doing, but if as well as the piece changes you proposed, you also change the stipulation to 2.1... then there is still only one solution. Even if you remove Sa5 entirely as well, there is no White cross-capture possible. (2023-12-04)
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Keywords: Castling (wgsk), Cant Castler (wgsk), Cross-capture (s,w), Superseded by (P1399806)
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: 4k2r/pPp2p1R/n1pB1ppp/npP5/1P6/5PPP/2P2P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-03 more...
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: 4k2r/pPp2p1R/n1pB1ppp/npP5/1P6/5PPP/2P2P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-03 more...
a) 1. Dd7 Ta5 2. 0-0-0 Ta8#
b) 1. Kf8 0-0 2. Te8 Txf7#
Whether Sb1 is White or Black, there are only the same two forward solutions, one with White Castling the other with Black. So it must be retro concerns that eliminate one solution in each case. bRa1 is original in any case, as Black has 8 pawns.
a) Suppose White can castle, then bRa1 must have come via cross-capturing White pawns. But Black has only lost 2 units. So White can't castle.
b) Suppose Black can castle, then bRa1 must have escaped its home by cross-capturing. But White has lost only 5 units. So Black can't castle.
This is not PRA. Rather, it shows twinned RS.
b) 1. Kf8 0-0 2. Te8 Txf7#
Whether Sb1 is White or Black, there are only the same two forward solutions, one with White Castling the other with Black. So it must be retro concerns that eliminate one solution in each case. bRa1 is original in any case, as Black has 8 pawns.
a) Suppose White can castle, then bRa1 must have come via cross-capturing White pawns. But Black has only lost 2 units. So White can't castle.
b) Suppose Black can castle, then bRa1 must have escaped its home by cross-capturing. But White has lost only 5 units. So Black can't castle.
This is not PRA. Rather, it shows twinned RS.
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
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s) 1. 0-0 Td1 2. Txd8 Txd8#
w) 1. ... 0-0-0 Tf8 2. Txf1 Txf1#
a2xLb3, d2xD/Tc3, f2xT/De3, d7xT/Dc6, f7xD/Te6. wLc1 opened no doors.
If both sides can still castle, then dxc6 preceded all white captures, but relies on prior release of wD/T. This could not have happened, so castling mutex.
This composition was marked as PRA, but it is RS.
w) 1. ... 0-0-0 Tf8 2. Txf1 Txf1#
a2xLb3, d2xD/Tc3, f2xT/De3, d7xT/Dc6, f7xD/Te6. wLc1 opened no doors.
If both sides can still castle, then dxc6 preceded all white captures, but relies on prior release of wD/T. This could not have happened, so castling mutex.
This composition was marked as PRA, but it is RS.
Keywords: Retro Strategy (RS), Castling, mutual exclusive, Quasi-symmetrical position
Genre: h#, Retro
Computer test: C+ Popeye 4.61 and analysis
FEN: 3Nk2r/1pp1p1p1/p1p1p2p/6N1/1P2n3/2P1P2P/1PP1P1P1/R3Kn2
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-30 more...
Genre: h#, Retro
Computer test: C+ Popeye 4.61 and analysis
FEN: 3Nk2r/1pp1p1p1/p1p1p2p/6N1/1P2n3/2P1P2P/1PP1P1P1/R3Kn2
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-30 more...
a) 1. 0-0-0 Txa2 2. Td7 Ta8#
b) 1. Kf8 0-0-0 2. Te8 Tf1#
In a&b have the same two forward solutions, one with White Castling the other with Black. So it must be retro concerns that eliminate one solution in each case. bRc7 is original in any case, as Black has 7 pawns, and bBb4 is promoted.
a) Black can only promote on g1, because 8 captures total are insufficient to reach further west, and wK will be disturbed on the way out. So White can't castle. Black can then cross-capture c&d pawns, so can castle himself.
b) Black only has 7 captures, and this allows for promotion on g1 (3 captures) and escape via h2, but does not allow for cross-capture of c&d pawns.
This is not PRA but twinned RS.
b) 1. Kf8 0-0-0 2. Te8 Tf1#
In a&b have the same two forward solutions, one with White Castling the other with Black. So it must be retro concerns that eliminate one solution in each case. bRc7 is original in any case, as Black has 7 pawns, and bBb4 is promoted.
a) Black can only promote on g1, because 8 captures total are insufficient to reach further west, and wK will be disturbed on the way out. So White can't castle. Black can then cross-capture c&d pawns, so can castle himself.
b) Black only has 7 captures, and this allows for promotion on g1 (3 captures) and escape via h2, but does not allow for cross-capture of c&d pawns.
This is not PRA but twinned RS.
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
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a) 1. Kf8 0-0-0 2. Te8 Tf1#
b) 1. Sd2 Le5 2. 0-0-0 Tb8#
In a&b have the same two forward solutions, one with White Castling the other with Black. So it must be retro concerns that eliminate one solution in each case. bRg8 is original in any case, as Black has 7 pawns, and bBa5 is promoted.
a) bhP must promote on g1 This allows for 4 captures of bPb3, but there is no chance for cross-capturing of b b&c pawns. White g and h pawn cannot reach Black cage to promote inside. So Black castling cannot be preserved. But bB once promoted can easily escape, so White capturing is unimpeded.
b) bhP can make 1 capture to promote, but now cannot escape without kicking wK, so White can't capture. Cross-capturing is still not possible, but wbP can promote on c8 without disrupting bK.
This is not PRA but twinned RS.
b) 1. Sd2 Le5 2. 0-0-0 Tb8#
In a&b have the same two forward solutions, one with White Castling the other with Black. So it must be retro concerns that eliminate one solution in each case. bRg8 is original in any case, as Black has 7 pawns, and bBa5 is promoted.
a) bhP must promote on g1 This allows for 4 captures of bPb3, but there is no chance for cross-capturing of b b&c pawns. White g and h pawn cannot reach Black cage to promote inside. So Black castling cannot be preserved. But bB once promoted can easily escape, so White capturing is unimpeded.
b) bhP can make 1 capture to promote, but now cannot escape without kicking wK, so White can't capture. Cross-capturing is still not possible, but wbP can promote on c8 without disrupting bK.
This is not PRA but twinned RS.
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
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1. d3 ... 2. Tc2 ... 3. Tc8#
1. Kf8 2. Ke8 3. 0-0 4. Kh8 5. Tg8 Sf7#
Version Bernd Schwarzkopf
Henrik Juel: Not 1.Kf8 2.Tg8 3.Tg7 4.Kg8 5.Kh8 6.Tg8, one move too many (2021-01-18)
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Henrik Juel: Not 1.Kf8 2.Tg8 3.Tg7 4.Kg8 5.Kh8 6.Tg8, one move too many (2021-01-18)
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Keywords: Seriesmover, Consequent, Castling
Genre: Retro, Fairies
FEN: 6kr/7p/7K/6N1/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
Genre: Retro, Fairies
FEN: 6kr/7p/7K/6N1/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
1. a1=S 2. fxg3ep 3. Kxh5 4. Kxg6 5. Kf7 6. Ke8 7. 0-0-0 a8=D#
8. TT
Henrik Juel: Solution: 1.a1S 2.fxg3ep 3.Kxh5xg6-f7-e8 7.0-0-0 a8Q#. Nice problem with open and hidden ep capture (exf3ep). (2003-04-28)
James Malcom: How is the key here justified? (2021-01-18)
James Malcom: I am still pondering. (2021-09-14)
Henrik Juel: James, I have forgotten all about this weird stipulation during the time since my last comment, and I don't know what I meant so say...
The definition of shc in the PDB is not complete; here is the Schwalbe definition:
Konsequenter Serienzüger: Ein Serienzüger, bei dem nach jedem Zug die Retroanalyse der Stellung neu durchgeführt wird (also ohne Kenntnis früherer Züge und Analysen). Beispielsweise wird eine Rochade wieder möglich, wenn König und Rochadeturm auf ihre entsprechenden Felder ziehen (weil in der neuen Analyse "vergessen" ist, dass beide bereits gezogen haben).
In english, something like
After each move the analysis of the position is done afresh, without knowledge of previous moves;
for instance, castling is possible when king and rook reach their original squares (e8 and a8), because it is 'forgotten' that they have already moved (2021-09-14)
Henrik Juel: You ask for a justification of the key, James; there is no need for justification, Black can do whatever he likes
In the position after 1.a1=S last move must have been g2-g4, so the ep capture 2.fxg3ep is legitimized
In the position after 2.fxg3ep last move must have been Ke3xPf3, with the double check being explained by exf3ep++; this is the 'hidden ep capture' I mentioned in my youthful comment (2021-09-14)
James Malcom: Thanks Henri! While the definition is incomplete, the reset each time is deducible from looking over shc problems. The justification here, however, was a level above me somehow. (2021-09-14)
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Henrik Juel: Solution: 1.a1S 2.fxg3ep 3.Kxh5xg6-f7-e8 7.0-0-0 a8Q#. Nice problem with open and hidden ep capture (exf3ep). (2003-04-28)
James Malcom: How is the key here justified? (2021-01-18)
James Malcom: I am still pondering. (2021-09-14)
Henrik Juel: James, I have forgotten all about this weird stipulation during the time since my last comment, and I don't know what I meant so say...
The definition of shc in the PDB is not complete; here is the Schwalbe definition:
Konsequenter Serienzüger: Ein Serienzüger, bei dem nach jedem Zug die Retroanalyse der Stellung neu durchgeführt wird (also ohne Kenntnis früherer Züge und Analysen). Beispielsweise wird eine Rochade wieder möglich, wenn König und Rochadeturm auf ihre entsprechenden Felder ziehen (weil in der neuen Analyse "vergessen" ist, dass beide bereits gezogen haben).
In english, something like
After each move the analysis of the position is done afresh, without knowledge of previous moves;
for instance, castling is possible when king and rook reach their original squares (e8 and a8), because it is 'forgotten' that they have already moved (2021-09-14)
Henrik Juel: You ask for a justification of the key, James; there is no need for justification, Black can do whatever he likes
In the position after 1.a1=S last move must have been g2-g4, so the ep capture 2.fxg3ep is legitimized
In the position after 2.fxg3ep last move must have been Ke3xPf3, with the double check being explained by exf3ep++; this is the 'hidden ep capture' I mentioned in my youthful comment (2021-09-14)
James Malcom: Thanks Henri! While the definition is incomplete, the reset each time is deducible from looking over shc problems. The justification here, however, was a level above me somehow. (2021-09-14)
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Keywords: Seriesmover, Consequent, En passant, Promotion (sD x2), Valladao Task, Castling, Promotion in the mating move
Genre: Retro, Fairies
FEN: r7/P1pp4/pp3pP1/4rpbP/5pPk/3n1K1b/p6q/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
Genre: Retro, Fairies
FEN: r7/P1pp4/pp3pP1/4rpbP/5pPk/3n1K1b/p6q/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
1. ... ... 2. 0-0?
1. ... Kxg2 2. e5#
1. ... Kxf4 2. Df6#
1. ... Kxg2 2. e5#
1. ... Kxf4 2. Df6#
Keywords: Castling (wk), No legal last move for Black, Castling in the forward play, Fabel-Opus (82)
Genre: Retro
Computer test: HC+ Popeye 4.61 with analysis
FEN: 8/8/2Q5/8/4PBN1/3P1k2/6R1/4K2R
Reprints: (IV) Problem 17-18 08/1952
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-08 more...
Genre: Retro
Computer test: HC+ Popeye 4.61 with analysis
FEN: 8/8/2Q5/8/4PBN1/3P1k2/6R1/4K2R
Reprints: (IV) Problem 17-18 08/1952
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-08 more...
a) 1. 0-0-0#
b) 1. ... Df1#
b) 1. ... Df1#
siehe P0004305
Henrik Juel: in part a last move could be Kc2-d3 or Kc4-d3
in part b Black has no last move, so he has the move
C+ Popeye 4.61 (2021-03-08)
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Henrik Juel: in part a last move could be Kc2-d3 or Kc4-d3
in part b Black has no last move, so he has the move
C+ Popeye 4.61 (2021-03-08)
comment
Keywords: Castling (wg), Whose move?
Genre: Retro
FEN: 8/8/8/2B5/6N1/1P1k1B1b/4N1qr/R3K3
Reprints: (VII) Problem 17-18 08/1952
1166 Problemista 127-130 04-07/1972
Input: Gerd Wilts, 1995-06-03
Last update: Marcin Banaszek, 2021-03-08 more...
Genre: Retro
FEN: 8/8/8/2B5/6N1/1P1k1B1b/4N1qr/R3K3
Reprints: (VII) Problem 17-18 08/1952
1166 Problemista 127-130 04-07/1972
Input: Gerd Wilts, 1995-06-03
Last update: Marcin Banaszek, 2021-03-08 more...
R: 1. 0-0! 0-0-0 (erzwungen wegen Retropattvermeidung) 2. b4-b5!, dann 1. Th8#
Eine mögliche Auflösung:
R: 1. 0-0 0-0-0 2. b4-b5! (nur so wird der sD nicht der schnellste Weg nach d8 versperrt) Dd8-d4 3. d5xLc6 Ld7-c6 4. e4xSd5 Lc8-d7 5. f3xSe4 d7xTe6 6. Ta6-e6 Se3-d5 7. Ta1-a6 Sc4-e3 8. a3xLb4 Ld6-b4 9. a2-a3 Lf4-d6 10. Tb1-a1 Lh6-f4 11. Tc1-b1 Lf8-h6 12. Ta1-c1 g7xLf6 13. Lg5-f6 Sf6-e4 14. Lc1-g5 Sa5-c4 15. d2-d3 d3xSc2 16. Sa3-c2 e4xDd3 17. Dc2-d3 f5xTe4 18. Th4-e4 Sg8-f6 19. Th8-h4 Sc6-a5 20. h7-h8=T Sb8-c6 21. h6-h7 g6xLf5 22. Lh3-f5 Sf6-g8 23. Lf1-h3 Sg8-f6 24. g2xTf3 Tf4-f3 25. h5-h6 Tf5-f4 26. h4-h5 Th5-f5 27. Sb1-a3 Th8-h5 28. Dd1-c2 h7xSg6 29. Sf4-g6 Sf6-g8 30. c2-c3 Sg8-f6 31. Sh3-f4 Sf6-g8 32. Sg1-h3 Sg8-f6 33. h2-h4
Eine mögliche Auflösung:
R: 1. 0-0 0-0-0 2. b4-b5! (nur so wird der sD nicht der schnellste Weg nach d8 versperrt) Dd8-d4 3. d5xLc6 Ld7-c6 4. e4xSd5 Lc8-d7 5. f3xSe4 d7xTe6 6. Ta6-e6 Se3-d5 7. Ta1-a6 Sc4-e3 8. a3xLb4 Ld6-b4 9. a2-a3 Lf4-d6 10. Tb1-a1 Lh6-f4 11. Tc1-b1 Lf8-h6 12. Ta1-c1 g7xLf6 13. Lg5-f6 Sf6-e4 14. Lc1-g5 Sa5-c4 15. d2-d3 d3xSc2 16. Sa3-c2 e4xDd3 17. Dc2-d3 f5xTe4 18. Th4-e4 Sg8-f6 19. Th8-h4 Sc6-a5 20. h7-h8=T Sb8-c6 21. h6-h7 g6xLf5 22. Lh3-f5 Sf6-g8 23. Lf1-h3 Sg8-f6 24. g2xTf3 Tf4-f3 25. h5-h6 Tf5-f4 26. h4-h5 Th5-f5 27. Sb1-a3 Th8-h5 28. Dd1-c2 h7xSg6 29. Sf4-g6 Sf6-g8 30. c2-c3 Sg8-f6 31. Sh3-f4 Sf6-g8 32. Sg1-h3 Sg8-f6 33. h2-h4
Henrik Juel: The key R: 1.0-0 'threatens' with white retrostalemate, even though White seems to have many pawn retractions available
All missing men were captured by pawns (and White promoted on h8)
R: 1... Kd7-c8? 2.d5xLc6+ Tb8-d8 3.b4-b5 Ke8-d7 4.e4xd5 Ld7-c6 5.f3xe4 Lc8-d7 and now White is retrostalemate
not 6.g2xf3 because of [Lf1]
not 6.d2-d3 because of [Lc1]
not 6.a3xLb4 because of [Ta1]
and not 6.b3-b4 because [Lf8] was captured on a dark square
R: 1... 0-0-0 handles Td8, but it also fixes the black king, so Dd4 must retract to d8 before d7xTf6 can be retracted, but there is still just time enough:
R: 2.b4-b5 Dd8-d4 3.d5xLc6 Ld7-c6 4.e4xd5 Lc8-d7 5.f3xe4 d7xTe6
The rest is easy: Retract Te6 to a1, a3xLb4, Lb4 to f8, g7xLf6, Lf6 to c1, d2-d3, and now the road towards h7 is free for Pc2 (2023-04-08)
A.Buchanan: Great! So which Typ is this? (2023-04-08)
Henrik Juel: Any type, there are no uncaptures in the solution, so anything goes (2023-04-09)
A.Buchanan: Ok I see - the sequence of retro moves is not VRZ play, but history of the game. After the key, there is no choice for either player until wPe4xd5. But isn’t there some Typ where black can checkmate too? R: 1. 0-0 0-0-0 then c1=D#! (2023-04-09)
Henrik Juel: A black checkmate is a possibility in the tries of defensive retractors, regardless of type
When Black has completed a retraction, he has the right to mate White with a forward move, if this is possible
I should have added the testing of mating in my general story about Høeg retractors
1. White chooses a man and moves it back
2. Black chooses which man (if any) to supplement on the abandoned square
Now the white retraction is completed, and White may mate with a forward move, if this is possible
If so, a solution has been found
If not
3. Black chooses a man and moves it back
4. White chooses which man (if any) to supplement on the abandoned square
Now the black retraction is completed, and Black may mate with a forward move, if this is possible
If so, a try has been found
If not, go to step 1. (2023-04-09)
A.Buchanan: Thanks - so does that mean that the solution given here is just a try? (2023-04-09)
Henrik Juel: No, Andrew, in the solution given here White mates, so it is a solution
A try requires Black to mate (2023-04-09)
A.Buchanan: Sorry I am apparently being slow: isn't R: 1. 0-0 0-0-0, dann c1=D#! a mate for Black, so White never gets to retract further? (2023-04-09)
Henrik Juel: You are not slow, Andrew, but I never really saw the black mate in your first 04-09 comment, sorry
R: 1.0-0 0-0-0 2.b4-b5, then 1.Th8# is the intended solution (and not an intended try), but it fails because following R: 1.0-0 0-0-0, Black mates with 1.c1=D#, so you have cooked the problem
It is easily repaired by adding the condition 'Ohne Vorwärtsverteidigung' (without forward defense), but maybe the author implied this condition (or maybe he never saw your black mate) (2023-04-09)
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All missing men were captured by pawns (and White promoted on h8)
R: 1... Kd7-c8? 2.d5xLc6+ Tb8-d8 3.b4-b5 Ke8-d7 4.e4xd5 Ld7-c6 5.f3xe4 Lc8-d7 and now White is retrostalemate
not 6.g2xf3 because of [Lf1]
not 6.d2-d3 because of [Lc1]
not 6.a3xLb4 because of [Ta1]
and not 6.b3-b4 because [Lf8] was captured on a dark square
R: 1... 0-0-0 handles Td8, but it also fixes the black king, so Dd4 must retract to d8 before d7xTf6 can be retracted, but there is still just time enough:
R: 2.b4-b5 Dd8-d4 3.d5xLc6 Ld7-c6 4.e4xd5 Lc8-d7 5.f3xe4 d7xTe6
The rest is easy: Retract Te6 to a1, a3xLb4, Lb4 to f8, g7xLf6, Lf6 to c1, d2-d3, and now the road towards h7 is free for Pc2 (2023-04-08)
A.Buchanan: Great! So which Typ is this? (2023-04-08)
Henrik Juel: Any type, there are no uncaptures in the solution, so anything goes (2023-04-09)
A.Buchanan: Ok I see - the sequence of retro moves is not VRZ play, but history of the game. After the key, there is no choice for either player until wPe4xd5. But isn’t there some Typ where black can checkmate too? R: 1. 0-0 0-0-0 then c1=D#! (2023-04-09)
Henrik Juel: A black checkmate is a possibility in the tries of defensive retractors, regardless of type
When Black has completed a retraction, he has the right to mate White with a forward move, if this is possible
I should have added the testing of mating in my general story about Høeg retractors
1. White chooses a man and moves it back
2. Black chooses which man (if any) to supplement on the abandoned square
Now the white retraction is completed, and White may mate with a forward move, if this is possible
If so, a solution has been found
If not
3. Black chooses a man and moves it back
4. White chooses which man (if any) to supplement on the abandoned square
Now the black retraction is completed, and Black may mate with a forward move, if this is possible
If so, a try has been found
If not, go to step 1. (2023-04-09)
A.Buchanan: Thanks - so does that mean that the solution given here is just a try? (2023-04-09)
Henrik Juel: No, Andrew, in the solution given here White mates, so it is a solution
A try requires Black to mate (2023-04-09)
A.Buchanan: Sorry I am apparently being slow: isn't R: 1. 0-0 0-0-0, dann c1=D#! a mate for Black, so White never gets to retract further? (2023-04-09)
Henrik Juel: You are not slow, Andrew, but I never really saw the black mate in your first 04-09 comment, sorry
R: 1.0-0 0-0-0 2.b4-b5, then 1.Th8# is the intended solution (and not an intended try), but it fails because following R: 1.0-0 0-0-0, Black mates with 1.c1=D#, so you have cooked the problem
It is easily repaired by adding the condition 'Ohne Vorwärtsverteidigung' (without forward defense), but maybe the author implied this condition (or maybe he never saw your black mate) (2023-04-09)
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1. 0-0-0! ... 2. Txd7#
R: 1. ... Lh2-g1 2. Lg1-f2 Sg7-h5 3. Lf2-g1 Sf5-g7 4. Lg1-f2 Sd4-f5 5. Lf2-g1 Sb3-d4 6. Lg1-f2 Sc5-b3 7. Lf2-g1 Sa6-c5 8. Lg1-f2 Sc5xBa6 9. Lf2-g1 Se4-c5 10. Lg1-f2 Sf2-e4 11. f3-f4 Sh1-f2 12. Lf2-g1 Lg1-h2 13. a5-a6 h2-h1=S 14. a4-a5 h3-h2 15. h2xSg3 Sh5-g3 16. Lg3-f2 Sg7-h5 17. Le5-g3 Sh5-g7 18. Lc3-e5 Sg7-h5 19. Ld2-c3 Sh5-g7 20. Lc1-d2 Sg3-h5 21. d2xSe3 Sd5-e3 22. a3-a4 Le3-g1 23. a2-a3 Lh6-e3 24. c3-c4 Lf8-h6 25. c2-c3 g7xSf6 26. Sh5-f6 Sf4-d5 27. f6-f7 etc.
R: 1. ... Lh2-g1 2. Lg1-f2 Sg7-h5 3. Lf2-g1 Sf5-g7 4. Lg1-f2 Sd4-f5 5. Lf2-g1 Sb3-d4 6. Lg1-f2 Sc5-b3 7. Lf2-g1 Sa6-c5 8. Lg1-f2 Sc5xBa6 9. Lf2-g1 Se4-c5 10. Lg1-f2 Sf2-e4 11. f3-f4 Sh1-f2 12. Lf2-g1 Lg1-h2 13. a5-a6 h2-h1=S 14. a4-a5 h3-h2 15. h2xSg3 Sh5-g3 16. Lg3-f2 Sg7-h5 17. Le5-g3 Sh5-g7 18. Lc3-e5 Sg7-h5 19. Ld2-c3 Sh5-g7 20. Lc1-d2 Sg3-h5 21. d2xSe3 Sd5-e3 22. a3-a4 Le3-g1 23. a2-a3 Lh6-e3 24. c3-c4 Lf8-h6 25. c2-c3 g7xSf6 26. Sh5-f6 Sf4-d5 27. f6-f7 etc.
Henrik Juel: 1.0-0-0 (2.Txd7#). -1... Lh2 -2.Lg1, Sh5 uncaptures wPa6 and unpromotes on h1, then retract h2xg3, Lf2 via f4 to c1, d2xe3, Lg1 to f8, g7xf6 etc. (2004-01-12)
Henrik Juel: C+ Popeye 4.61 (2023-07-14)
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Henrik Juel: C+ Popeye 4.61 (2023-07-14)
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-sBg5 1. Lxb5,Lxb3 Ta8#
-sBg5 (=Vollendung e.p.- Schlag). Nicht Ta1-d1 (=Vollendung w0-0-0), denn Schwarz hätte keinen letzten Zug
-sBg5 (=Vollendung e.p.- Schlag). Nicht Ta1-d1 (=Vollendung w0-0-0), denn Schwarz hätte keinen letzten Zug
Henrik Juel: a slight flaw is that we cannot say whether the entire move was fxg6ep or hxg6ep (2022-07-05)
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Keywords: Castling (wg), Complete an unfinished move, En passant, Joke
Genre: Retro, h#
FEN: 7k/7P/2P1PPPP/1P4p1/b7/1P6/8/R1K5
Reprints: Problem 101-102 09/1966
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-07-10 more...
Genre: Retro, h#
FEN: 7k/7P/2P1PPPP/1P4p1/b7/1P6/8/R1K5
Reprints: Problem 101-102 09/1966
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-07-10 more...
84 - P0004454
Immo Fuß
Die Schwalbe , p. 463, 03/1939
Allen Teilnehmern herzlich gewidmet
Internationaler Lösungswettkampf 1938
(13+11)
#3
Immo Fuß
Die Schwalbe , p. 463, 03/1939
Allen Teilnehmern herzlich gewidmet
Internationaler Lösungswettkampf 1938
(13+11)
#3
1. Tf1? 0-0! Thematic retro try
1. Dxd7+,Sf6+,exd7+,exf7+? Kf8!
1. Dxa8+? Sd8!
1. Sh7? Txh7!
1. 0-0! (droht 2. Txf7,Dxa8+) 0-0? illegal
1. ... Lxb2,Lxb4,b5,c6 2. Dxa8+ Sd8 3. Sxc7#
1. ... Kf8 2. Txf7+
1. ... Tg8 2. Dxd7+,Dxa8+
Suppose both players can castle, and derive a contradiction.
White captures: a3xb4,fxe,g2xh3,Bc8 & QR in cage a8--d8.
Black captures: a7xb6,fxe/g,Bc1. bPg promoted. bPf promoted or was captured by wPfxe.
So all captures accounted for. Pieces captured by pawns were wRB & bQXX
b6 is dark, so light wBf1 was not captured there. By elimination, a7xRb6, which released bQ.
Was an original officer captured on b4 to release wR?
bQ not yet released
bRh never moved, bRa captured in cage
bB wrong shade, bBc8 captured in cage
bS couldn't escape g1, and two others on board.
So it must have been a promoted officer captured on b4 earlier.
What was captured on h3, to open the line for promotion on g1? Must be original as wPa & bPa have not yet captured. For the same reasons as axb4, we can eliminate all 4 possible officer types.
Contradiction! So at least one player cannot castle.
We apply the Retro Strategy (RS) convention, and White is permitted to castle, while the range of possible histories shrinks to exclude all cases where Black might have castled. This common special case of RS is also known as Mutual Exclusion.
1. Dxd7+,Sf6+,exd7+,exf7+? Kf8!
1. Dxa8+? Sd8!
1. Sh7? Txh7!
1. 0-0! (droht 2. Txf7,Dxa8+) 0-0? illegal
1. ... Lxb2,Lxb4,b5,c6 2. Dxa8+ Sd8 3. Sxc7#
1. ... Kf8 2. Txf7+
1. ... Tg8 2. Dxd7+,Dxa8+
Suppose both players can castle, and derive a contradiction.
White captures: a3xb4,fxe,g2xh3,Bc8 & QR in cage a8--d8.
Black captures: a7xb6,fxe/g,Bc1. bPg promoted. bPf promoted or was captured by wPfxe.
So all captures accounted for. Pieces captured by pawns were wRB & bQXX
b6 is dark, so light wBf1 was not captured there. By elimination, a7xRb6, which released bQ.
Was an original officer captured on b4 to release wR?
bQ not yet released
bRh never moved, bRa captured in cage
bB wrong shade, bBc8 captured in cage
bS couldn't escape g1, and two others on board.
So it must have been a promoted officer captured on b4 earlier.
What was captured on h3, to open the line for promotion on g1? Must be original as wPa & bPa have not yet captured. For the same reasons as axb4, we can eliminate all 4 possible officer types.
Contradiction! So at least one player cannot castle.
We apply the Retro Strategy (RS) convention, and White is permitted to castle, while the range of possible histories shrinks to exclude all cases where Black might have castled. This common special case of RS is also known as Mutual Exclusion.
Kees: Only one castling is legal With black castling there's no #3
1. 0-0!
1. ... Kf8 2.Txf7+ Kg8 3 Dxa8#
1. ... c6 2. Dxa8+ Pd8 3. Pc7#
(axNb3, and for f1=N or g1=N wK must move. so bD must pass bK)
Somebody can better explain than me. (2022-02-14)
A.Buchanan: Hi Kees thanks for the solution which grabs the essence - I have used more words, please point out any slips I might have made! :) (2022-02-15)
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1. 0-0!
1. ... Kf8 2.Txf7+ Kg8 3 Dxa8#
1. ... c6 2. Dxa8+ Pd8 3. Pc7#
(axNb3, and for f1=N or g1=N wK must move. so bD must pass bK)
Somebody can better explain than me. (2022-02-14)
A.Buchanan: Hi Kees thanks for the solution which grabs the essence - I have used more words, please point out any slips I might have made! :) (2022-02-15)
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Keywords: Retro Strategy (RS), Castling, mutual exclusive (wksk)
Genre: Retro, 3#
Computer test: Popeye v4.87 for forward play Non-trivial thinking for retro logic
FEN: n3k2r/1ppppn2/1p2P3/3N2Np/QP6/b6P/1PPPP2P/4K2R
Reprints: (14) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-15 more...
Genre: Retro, 3#
Computer test: Popeye v4.87 for forward play Non-trivial thinking for retro logic
FEN: n3k2r/1ppppn2/1p2P3/3N2Np/QP6/b6P/1PPPP2P/4K2R
Reprints: (14) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-15 more...
85 - P0004536
Hans Gruber
2286 feenschach 40 11-12/1977
(1+2)
Ergänze wK, wT, 4wBB und 2sBB zu einem IC!
Hans Gruber
2286 feenschach 40 11-12/1977
(1+2)
Ergänze wK, wT, 4wBB und 2sBB zu einem IC!
a) +wKc1 wTd1 wBb2d2e3h2 sBe7g7
Zuvor 0-0+ illegal, da sLc5=Umwandlungsläufer! Umwandlung auf g1=L, Lf2+ Schach
Zuvor 0-0+ illegal, da sLc5=Umwandlungsläufer! Umwandlung auf g1=L, Lf2+ Schach
Bernd Schwarzkopf: In der Lösung muss es +wBd2 (statt e2) heißen; auf e2 steht im Diagramm schon ein wB. (2020-12-16)
A.Buchanan: @Bernd: I made the change you suggested, but having seen the result I think the original diagram was wrong, not the solution. In the diagram with just wKh1, sLc5 & one wB , the wB should be on d2 (or f2) not e2, in order to eliminate a cook. Otherwise, the final pawn can be on d2 or f2: both are ICs. (2020-12-17)
Henrik Juel: Andrew, the final pawn must be on d2, not f2
Without wPd2 last move could be Td8-d1+, so this position is no IC (2020-12-17)
A.Buchanan: Oh yes of course thanks. So what cooks does showing wBe2 in the diagram prevent? (2020-12-17)
Henrik Juel: I was asking myself the same question, Andrew
But there are so many IC's around that it is unadvisable to move Pe2 from the diagram to the stipulation list
We should have asked Hans, some 40 years ago... (2020-12-17)
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A.Buchanan: @Bernd: I made the change you suggested, but having seen the result I think the original diagram was wrong, not the solution. In the diagram with just wKh1, sLc5 & one wB , the wB should be on d2 (or f2) not e2, in order to eliminate a cook. Otherwise, the final pawn can be on d2 or f2: both are ICs. (2020-12-17)
Henrik Juel: Andrew, the final pawn must be on d2, not f2
Without wPd2 last move could be Td8-d1+, so this position is no IC (2020-12-17)
A.Buchanan: Oh yes of course thanks. So what cooks does showing wBe2 in the diagram prevent? (2020-12-17)
Henrik Juel: I was asking myself the same question, Andrew
But there are so many IC's around that it is unadvisable to move Pe2 from the diagram to the stipulation list
We should have asked Hans, some 40 years ago... (2020-12-17)
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Keywords: Illegal cluster, Castling (wg), Promotion (l)
Genre: Retro
FEN: 8/8/8/2b5/8/8/4P3/7k
Reprints: feenschach 45 01-03/1979
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-12-17 more...
Genre: Retro
FEN: 8/8/8/2b5/8/8/4P3/7k
Reprints: feenschach 45 01-03/1979
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-12-17 more...
wZug:
1. ... Kxb7! Txd8 2. Ka8 Le4#
1. Txd8+? Kxd8 2. Kf8 Th8#
Wh cxb,dxc
Bl axb,dxc,gxf,hxgxh
All missing pieces accounted for. Easy to see White has no last move.
R: 1. S6-d8? 0-0 means wK stuck behind Black lines.
1. fxe6ep is not permitted, and leads to no mate anyway.
1. ... Kxb7! Txd8 2. Ka8 Le4#
1. Txd8+? Kxd8 2. Kf8 Th8#
Wh cxb,dxc
Bl axb,dxc,gxf,hxgxh
All missing pieces accounted for. Easy to see White has no last move.
R: 1. S6-d8? 0-0 means wK stuck behind Black lines.
1. fxe6ep is not permitted, and leads to no mate anyway.
Corrected suspected typo, by putting wBe6 on f5.
Mario Richter: As given, the position is illegal: 3 white pawn captures, but only 2 missing black men. (2010-01-23)
A.Buchanan: I think this is a typo: wBe6 should be on f5. Then the two candidate solutions are C+ 1. Txd8+ Kxd8 2. Kf8 Th8# & 1. Kxb7 Txd8 2. Ka8 Le4# WinChloe has the same diagram. Note the uncastling try. En passant for once is not part of the problem. If others here concur, I will correct this diagram in place. (2021-10-20)
Henrik Juel: I think you are right, Andrew (2021-10-20)
A.Buchanan: Thanks Henrik. I've fixed the diagram on the basis of this suspected typo. (2021-10-20)
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Mario Richter: As given, the position is illegal: 3 white pawn captures, but only 2 missing black men. (2010-01-23)
A.Buchanan: I think this is a typo: wBe6 should be on f5. Then the two candidate solutions are C+ 1. Txd8+ Kxd8 2. Kf8 Th8# & 1. Kxb7 Txd8 2. Ka8 Le4# WinChloe has the same diagram. Note the uncastling try. En passant for once is not part of the problem. If others here concur, I will correct this diagram in place. (2021-10-20)
Henrik Juel: I think you are right, Andrew (2021-10-20)
A.Buchanan: Thanks Henrik. I've fixed the diagram on the basis of this suspected typo. (2021-10-20)
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Keywords: Whose move?, Castling
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic
FEN: 1bKN1rk1/1ppn1r1R/5p1P/4pP1p/2p2p1q/1PP3P1/PP2P3/1b6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-20 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic
FEN: 1bKN1rk1/1ppn1r1R/5p1P/4pP1p/2p2p1q/1PP3P1/PP2P3/1b6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-20 more...
R: 1. 0-0?
See P1388763
James Malcom: Black has no last move, so they must have castled illegally out of checkmate. (2021-04-18)
A.Buchanan: “dann” is used in retractor animation when shifting from retro to forward moves (2021-04-19)
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James Malcom: Black has no last move, so they must have castled illegally out of checkmate. (2021-04-18)
A.Buchanan: “dann” is used in retractor animation when shifting from retro to forward moves (2021-04-19)
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Keywords: Joke, Retract illegal move, Castling
Genre: Retro
FEN: 3R1rk1/5N1n/4KB1P/8/8/8/8/8
Reprints: (XV) Die Schwalbe 8 04/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-19 more...
Genre: Retro
FEN: 3R1rk1/5N1n/4KB1P/8/8/8/8/8
Reprints: (XV) Die Schwalbe 8 04/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-19 more...
1. ... Tfxg8 2. Lg6! h5 3. Lxf7+
2. ... fxg6=?,hxg6=?,Txg6#?
2. ... Tf8,Tg7 3. Lxf7+ Txf7#
2. Lxe6? Tg6+! 3. Kf5 0-0!
2. ... dxe6? 3. d7+!
2. ... fxe6=?
2. Lxh7? Tg6+! 3. Lxg6 0-0!
Therefore it's WTM
1. Dg2 h5,~ 2. Da8#
Black cannot steal the move, as White can prevent the castling justification.
2. ... fxg6=?,hxg6=?,Txg6#?
2. ... Tf8,Tg7 3. Lxf7+ Txf7#
2. Lxe6? Tg6+! 3. Kf5 0-0!
2. ... dxe6? 3. d7+!
2. ... fxe6=?
2. Lxh7? Tg6+! 3. Lxg6 0-0!
Therefore it's WTM
1. Dg2 h5,~ 2. Da8#
Black cannot steal the move, as White can prevent the castling justification.
A.Buchanan: Ingenious play, but Black cannot execute the castling. Lines include a good try, checkmate by Black and pat by Black. (2022-04-16)
A.Buchanan: There are two kinds of directmate Type Keym. In one White successfully pushes the move to Black, in the other (as here) Black unsuccessfully pulls it. (2022-04-17)
Ladislav Packa: The logic of this problem is foreign to me, but it is incorrect: 2.Bxe6 Rg6+ 3.Kf5 0-0 (2022-04-18)
A.Buchanan: Hi Ladislav thanks for this. You're right. So 2. Lxe6? Tg6! 3. Kf5 0-0! is another try. The solution must be 2. Lg6! I'll post the solution above (2022-04-18)
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A.Buchanan: There are two kinds of directmate Type Keym. In one White successfully pushes the move to Black, in the other (as here) Black unsuccessfully pulls it. (2022-04-17)
Ladislav Packa: The logic of this problem is foreign to me, but it is incorrect: 2.Bxe6 Rg6+ 3.Kf5 0-0 (2022-04-18)
A.Buchanan: Hi Ladislav thanks for this. You're right. So 2. Lxe6? Tg6! 3. Kf5 0-0! is another try. The solution must be 2. Lg6! I'll post the solution above (2022-04-18)
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Keywords: Castling (sk), a posteriori (AP) (Type Keym)
Genre: Retro, 2#
FEN: 4krQr/3p1p1p/3PpK1p/4PB2/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
Genre: Retro, 2#
FEN: 4krQr/3p1p1p/3PpK1p/4PB2/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
BTM 1. ... Txh6? 2. Sd7! Txc6+ 3. Kb5! (Kb7?) Td6 4. Sf8 Tf6 5. Sh7! Th6 6. Sf8 no castling
5. a6? Tfxf8 6. a7 Tf5+ 7. Kc6,K~ 0-0!
1. ... 0-0? 2. Se7+! Kh8 3. S5g6#
WTM 1. Td6 droht 2. Td8#
White to move has #2 since Black has lost castling rights. So Black pulls the move, but must castle at some point. If Black castles right away, then White has a different #2, so Black must be more subtle. 0... g6/g5/gxh6 leads to castling disruption, e.g. 1.Txg6/Te6+/Sg6. So Black only has 0... Txh6. This pins wSc6 and threatens 0-0, so 1.Sd7! (1. Sg6? Txg6 2. ~ 0-0) etc.
5. a6? Tfxf8 6. a7 Tf5+ 7. Kc6,K~ 0-0!
1. ... 0-0? 2. Se7+! Kh8 3. S5g6#
WTM 1. Td6 droht 2. Td8#
White to move has #2 since Black has lost castling rights. So Black pulls the move, but must castle at some point. If Black castles right away, then White has a different #2, so Black must be more subtle. 0... g6/g5/gxh6 leads to castling disruption, e.g. 1.Txg6/Te6+/Sg6. So Black only has 0... Txh6. This pins wSc6 and threatens 0-0, so 1.Sd7! (1. Sg6? Txg6 2. ~ 0-0) etc.
A.Buchanan: A key feature of adversarial A Posteriori is that any castling must be forced in a finite number of moves (but not necessarily limited by the number of moves in the stipulation goal). If the other side can prevaricate indefinitely, then that is sufficient to defeat the A Posteriori "steal" (2022-02-16)
A.Buchanan: Why this would be "PRA"? Maybe the idea is that we don't know who is first to move, yet whoever it is, White wins. But that only applies to "pull" scenarios such as this, where Black snatches the move because otherwise the game is lost. In other situations where White to avoid loss must "push" the move, then there is no way this can be described as PRA. The fundamental push/pull thing has a unity, and I don't think it's helpful to use "PRA" which only describes half of this, and was really designed for a different context. Strategically, these push/pull adversarial battles are amongst the most interesting AP problems. (2023-07-22)
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A.Buchanan: Why this would be "PRA"? Maybe the idea is that we don't know who is first to move, yet whoever it is, White wins. But that only applies to "pull" scenarios such as this, where Black snatches the move because otherwise the game is lost. In other situations where White to avoid loss must "push" the move, then there is no way this can be described as PRA. The fundamental push/pull thing has a unity, and I don't think it's helpful to use "PRA" which only describes half of this, and was really designed for a different context. Strategically, these push/pull adversarial battles are amongst the most interesting AP problems. (2023-07-22)
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Keywords: a posteriori (AP) (Type Keym), Cant Castler, Castling
Genre: Retro, 2#
FEN: 4k2r/6pr/K1N4R/P3N3/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-22 more...
Genre: Retro, 2#
FEN: 4k2r/6pr/K1N4R/P3N3/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-22 more...
1. Sa3 Sa6 2. Sc4 Sc5 3. Se5 Se4 4. c4 c5 5. Da4 Da5 6. b3 b6 7. Lb2 Lb7 8. Tc1 Tc8 9. Tc3 Tc6 10. Th3 Th6 11. e3 e6 12. Le2 Le7 13. Lh5 Lh4 14. Sgf3 Sgf6 15. 0-0 0-0 16. Tc1 Tc8 17. Tc3 Tc6 18. Td3 Td6 19. Td4 Td5 20. d3 d6 21. Dd7 Dd2 22. De7 De2 23. Sd7 Sd2 24. Tg4 Tg5 25. Le5 Le4 26. Lf4 Lf5 27. Sfe5 Sfe4 28. f3 f6 29. Le8 Le1 30. Lg3 Lg6 31. Lf2 Lf7 32. Thg3 Thg6 33. h3 h6 34. Kh2 Kh7
Cook: 1. Sa3 Sa6 2. Sc4 Sc5 3. Se5 Se4 4. c4 c5 5. Da4 Da5 6. b3 b6 7. Lb2 Lb7 8. Tc1 Tc8 9. Tc3 Tc6 10. Th3 Th6 11. e3 e6 12. Le2 Le7 13. Lh5 Lh4 14. Sgf3 Sgf6 15. 0-0 0-0 16. Tc1 Tc8 17. Tc3 Tc6 18. Td3 Td6 19. Td4 Td5 20. d3 d6 21. Dd7 Dd2 22. De7 De2 23. Sd7 Sd2 24. Tg4 Tg5 25. Le5 Le4 26. Lf4 Lf5 27. Sfe5 Sfe4 28. f3 Le1 29. Lg3 f6 30. Le8 Lg6 31. Lf2 Lf7 32. Thg3 Thg6 33. h3 h6 34. Kh2 Kh7
Cook: 1. Sa3 Sa6 2. Sc4 Sc5 3. Se5 Se4 4. c4 c5 5. Da4 Da5 6. b3 b6 7. Lb2 Lb7 8. Tc1 Tc8 9. Tc3 Tc6 10. Th3 Th6 11. e3 e6 12. Le2 Le7 13. Lh5 Lh4 14. Sgf3 Sgf6 15. 0-0 0-0 16. Tc1 Tc8 17. Tc3 Tc6 18. Td3 Td6 19. Td4 Td5 20. d3 d6 21. Dd7 Dd2 22. De7 De2 23. Sd7 Sd2 24. Tg4 Tg5 25. Le5 Le4 26. Lf4 Lf5 27. Sfe5 Sfe4 28. f3 Le1 29. Lg3 f6 30. Le8 Lg6 31. Lf2 Lf7 32. Thg3 Thg6 33. h3 h6 34. Kh2 Kh7
Keywords: Unique Proof Game, Castling, Symmetrical position, Capture-free
Genre: Retro
FEN: 4B3/p2NQbpk/1p1ppprp/2p1N1r1/2P1n1R1/1P1PPPRP/P2nqBPK/4b3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-13 more...
Genre: Retro
FEN: 4B3/p2NQbpk/1p1ppprp/2p1N1r1/2P1n1R1/1P1PPPRP/P2nqBPK/4b3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-13 more...
91 - P0005327
Valerian Onitiu
8397 The Fairy Chess Review 9 28/11/1949
(14+16)
BP in 19,0
Die PAS wurde verändert, indem die Position der schwarzen und weißen Steine geändert wurde (Free Chess).
Valerian Onitiu
8397 The Fairy Chess Review 9 28/11/1949
(14+16)
BP in 19,0
Die PAS wurde verändert, indem die Position der schwarzen und weißen Steine geändert wurde (Free Chess).
PAS: ssltktdl, LDTKTLSS (nnbrkrqb/pppppppp/8/8/8/8/PPPPPPPP/BQRKRBNN)
1. b3 0-0-0 (Kf8, Te8) 2. Lf6 gxf6 3. Sg3 Lg7 4. Se4 Lh6 5. Sc5 Le3 6. Sa6 Lb6 7. e3 Dg3 8. Lc4 Dd6 9. Le6 fxe6 10. Sf3 Kf7 11. 0-0 (Kg1,Tef1) Tg8 12. Tce1 Tg3 13. Dd1 Th3 14. g4 Tg8 15. Kg2 Tg5 16. Th1 Tb5 17. Sg5 Kg7 18. Df3 La5 19. Te2 Tb6
1. b3 0-0-0 (Kf8, Te8) 2. Lf6 gxf6 3. Sg3 Lg7 4. Se4 Lh6 5. Sc5 Le3 6. Sa6 Lb6 7. e3 Dg3 8. Lc4 Dd6 9. Le6 fxe6 10. Sf3 Kf7 11. 0-0 (Kg1,Tef1) Tg8 12. Tce1 Tg3 13. Dd1 Th3 14. g4 Tg8 15. Kg2 Tg5 16. Th1 Tb5 17. Sg5 Kg7 18. Df3 La5 19. Te2 Tb6
Originalforderung: "In No. 8397 pieces began game on normal ranks, but not in normal array. The two arrays were however similar, e.g., similar pieces were on c1 and f8. Black has just played his 19th move. What was the game?"
Mario Richter: I do not understand, why 1. ... (Kf8, Te8) is noticed as "0-0-0" (does "Free Chess" have a "Free Castling Rule"?) (2020-09-20)
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Mario Richter: I do not understand, why 1. ... (Kf8, Te8) is noticed as "0-0-0" (does "Free Chess" have a "Free Castling Rule"?) (2020-09-20)
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Keywords: Unique Proof Game, Castling
Genre: Retro
FEN: nnb5/ppppp2p/Nr1qppk1/b5N1/6P1/1P2PQ1r/P1PPRPKP/7R
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2020-09-20 more...
Genre: Retro
FEN: nnb5/ppppp2p/Nr1qppk1/b5N1/6P1/1P2PQ1r/P1PPRPKP/7R
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2020-09-20 more...
92 - P0005358
Thomas R. Dawson
(4) The Chess Amateur 08/1914
(9+12) C+
Black, having just made hist 25th move, is it possible for White to mate in 2?
(BP in 24,5)
Thomas R. Dawson
(4) The Chess Amateur 08/1914
(9+12) C+
Black, having just made hist 25th move, is it possible for White to mate in 2?
(BP in 24,5)
1. Sc3 a5 2. Se4 Sa6 3. b4 a4 4. b5 a3 5. d4 b6 6. d5 Lb7 7. d6 exd6 8. Lf4 Dh4 9. bxa6 Dh3 10. c4 Sf6 11. Sxf6+ gxf6 12. c5 Lf3 13. exf3 Lh6 14. Lb5 c6 15. Kf1 cxb5 16. Se2 dxc5 17. Ld6 Ld2 18. Tg1 La5 19. gxh3 h6 20. Tg5 hxg5 21. Kg2 Th4 22. Db1 Ta4 23. Db4 cxb4 24. Sf4 gxf4 25. Tg1
Black can have wasted only one move and so cannot have moved K or KR, which would require a second waste move returning to diagram. Hence Black is legally entitled to castle. Therefore White mates in 2 by: 1. Kh1! 0-0-0 2. Tc1, else 2. Tg8
Black can have wasted only one move and so cannot have moved K or KR, which would require a second waste move returning to diagram. Hence Black is legally entitled to castle. Therefore White mates in 2 by: 1. Kh1! 0-0-0 2. Tc1, else 2. Tg8
Moldenhauer: Computerprüfung: C+ Stelvio NUPG cooked 1 Sekunde.
Keine Lösung: BP 23.5, BP 24.0.
Wenn Schwarz seinen 25. b3 spielt ist kein Matt in 2 möglich.
Wenn angenommen wird das Weiß aus dieser Stellung nochmals
ziehen darf dann Matt in 2 Zügen.
Notation: 1.Sc3 a5 2.Se4 Sa6 3.b4 a4 4.b5 a3 5.d4 b6 6.d5 Lb7 7.d6 exd6
8.Lf4 Dh4 9.bxa6 Dh3 10.c4 Sf6 11.Sxf6+ gxf6 12.c5 Lf3 13.exf3 Lh6
14.Lb5 c6 15.Kf1 cxb5 16.Se2 dxc5 17.Ld6 Ld2 18.Tg1 La5 19.gxh3 h6
20.Tg5 hxg5 21.Kg2 Th4 22.Db1 Ta4 23.Db4 cxb4 24.Sf4 gxf4 25.Tg1 (2023-04-06)
A.Buchanan: Should the stip be something like:
Position after White’s 25th move.
Can Black play to avoid being mated in 2?
But both b3 and Rc8 seem to let Black escape (2023-04-06)
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Keine Lösung: BP 23.5, BP 24.0.
Wenn Schwarz seinen 25. b3 spielt ist kein Matt in 2 möglich.
Wenn angenommen wird das Weiß aus dieser Stellung nochmals
ziehen darf dann Matt in 2 Zügen.
Notation: 1.Sc3 a5 2.Se4 Sa6 3.b4 a4 4.b5 a3 5.d4 b6 6.d5 Lb7 7.d6 exd6
8.Lf4 Dh4 9.bxa6 Dh3 10.c4 Sf6 11.Sxf6+ gxf6 12.c5 Lf3 13.exf3 Lh6
14.Lb5 c6 15.Kf1 cxb5 16.Se2 dxc5 17.Ld6 Ld2 18.Tg1 La5 19.gxh3 h6
20.Tg5 hxg5 21.Kg2 Th4 22.Db1 Ta4 23.Db4 cxb4 24.Sf4 gxf4 25.Tg1 (2023-04-06)
A.Buchanan: Should the stip be something like:
Position after White’s 25th move.
Can Black play to avoid being mated in 2?
But both b3 and Rc8 seem to let Black escape (2023-04-06)
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Keywords: Castling, Non-Unique Proof Game
Genre: Retro
Computer test: Computerprüfung: C+ Stelvio NUPG cooked 1 Sekunde. Keine Lösung: BP 23.5, BP 24.0.
FEN: r3k3/3p1p2/Pp1B1p2/bp6/rp3p2/p4P1P/P4PKP/6R1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-06 more...
Genre: Retro
Computer test: Computerprüfung: C+ Stelvio NUPG cooked 1 Sekunde. Keine Lösung: BP 23.5, BP 24.0.
FEN: r3k3/3p1p2/Pp1B1p2/bp6/rp3p2/p4P1P/P4PKP/6R1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-06 more...
WTM: ???
BTM: 1. ... Kxf3 2. 0-0+! Ke3 3. Td1 Kf3 4. Td3#
not 2. Kd2? Kg2 3. Sf4+ Kf2,Kf3 4. Tf1#
Cook: 1.Tf1,Se2+,Sf2 all lead to mate in 3, and none require castling
BTM: 1. ... Kxf3 2. 0-0+! Ke3 3. Td1 Kf3 4. Td3#
not 2. Kd2? Kg2 3. Sf4+ Kf2,Kf3 4. Tf1#
Cook: 1.Tf1,Se2+,Sf2 all lead to mate in 3, and none require castling
A.Buchanan: I don't get the regular part of this. White can mate in 3 in various ways: 1.Tf1,Se2+,Sf2. Black just played R: 1.Kh2g2 Td~h1+, so White can't castle but doesn't need to. What's going on? (2022-02-15)
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Keywords: a posteriori (AP) (Type Keym), Castling, Rex solus, Miniature
Genre: Retro, 3#
FEN: 8/8/6R1/8/8/5PNN/6k1/4K2R
Input: Gerd Wilts, 1995-06-06
Last update: A.Buchanan, 2023-07-30 more...
Genre: Retro, 3#
FEN: 8/8/6R1/8/8/5PNN/6k1/4K2R
Input: Gerd Wilts, 1995-06-06
Last update: A.Buchanan, 2023-07-30 more...
R: 0-0! Kb6xc7 1. Ld8xc7+ Lf4-d6 2. Sb1-a3 De5-c7 3. Lg5-d8+ Da1-e5 3. Sd2-b1 Dc3-a1+ 4. Sf1-d2 Dg3-c3+ 5. Lf6-g5 Ld6-f4 6. Lc3-f6 Lb4-d6 7. c7xDb8=T Dg1-g3 8. Sg3-f1 g2-g1=D+ 9. Se4-g3 g3-g2 10. d6xDc7 g4-g3 11. Sd2-e4 g5-g4 12. Sf1-d2 g7-g5 13. d5-d6 Dg3-c7 14. Ld2-c3 Dg1-g3 15. Se3-f1 g2-g1=D+ 16. Lc3-d2 g3-g2 17. Sf1-e3 g4-g3 18. Sg3-f1 g5-g4 19. Se4-g3 g6-g5 20. Sd2-e4 h7xSg6 21. c7xDb8=T Lb4-a3 22. d6xDc7 Dg8-d8 23. Sf1-d2 Dg1-g8 24. Sg3-f1 g2-g1=D+ 25. Se4-g3 g3-g2 26. Sd2-e4 g4-g3 27. d5-d6 Dg3-c7 28. Sf1-d2 Dg1-g3 29. Se3-f1 g2-g1=D+ 30. Sf5-e3 g5-g4 31. Sd4-f5 g3-g2 32. Sb3-d4 Kc5-b6 33. Sd2-b3+ g4-g3 34. Sf1-d2 Dg3-b8 35. Tb8-a8 g7-g5 36. c7xDb8=T g5-g4 37. b6xDc7 Dg1-g3 38. Se3-f1 g2-g1=D+ 39. Sc4-e3 g6-g5 40. Sa3-c4 g3-g2 41. Sb1-a3 g4-g3 42. Sd2-b1 Dg3-c7 43. Sf1-d2 Dg1-g3 44. Ld2-c3 g2-g1=D 45. Lc3-d2 f3xLg2 46. Se3-f1 e4xTf3 47. Sc4-e3 Kd6-c5 48. Sa3-c4+ h7xSg6 49. Sb1-a3 Ke7-d6 50. Sf4-g6+ Ke8-e7 51. Lf1-g2 f5xDg4 52. g2xTh3 Th6-h3 53. Sd2-b1 e5-e4 54. Lb2-c3 Tf6-h6 55. Tc3-f3 Tf8-f6 56. Sh3-f4 Th8-f8 57. Sg1-h3 f7-f5 58. Sc4-d2 Sc5-a4 59. Sa3-c4 Se4-c5 60. Td3-c3 Lf8-b4+ 61. Td1-d3 Sf6-e4 62. Ta1-d1 Sg8-f6 63. Sb1-a3 e7-e5 64. Lc1-b2 Dc7-b8 65. b5-b6 Dd8-c7 66. b4-b5 c7-c6 67. b2-b4 Sc6-a5 68. Dd4-g4 Sb8-c6 69. Dd1-d4 Ta8-a7 70. d4-d5 a7-a6 71. d2-d4
hans: 4 Q-promotions on g1 and captured on b8 and c7. Sf1 for shield, so 0-0 is legal. (2010-04-30)
Hans-Jürgen Manthey: R: 0. O-O ! Kb6xc7 1. Ld8xc7+ Lf4-d6 2. Sb1-a3 De5-c7 3. Lg5-d8+ Da1-e5 3. Sd2-b1 Dc3-a1+ 4. Sf1-d2 Dg3-c3+ 5. Lf6-g5 Ld6-f4 6. Lc3-f6 Lb4-d6 7. c7xDb8=T Dg1-g3 8. Sg3-f1 g2-g1=D+ 9. Se4-g3 g3-g2 10. d6xDc7 g4-g3 11. Sd2-e4 g5-g4 12. Sf1-d2 g7-g5 13. d5-d6 Dg3-c7 14. Ld2-c3 Dg1-g3 15. Se3-f1 g2-g1=D+ 16. Lc3-d2 g3-g2 17. Sf1-e3 g4-g3 18. Sg3-f1 g5-g4 19. Se4-g3 g6-g5 20. Sd2-e4 h7xSg6 21. c7xDb8=T Lb4-a3 22. d6xDc7 Dg8-d8 23. Sf1-d2 Dg1-g8 24. Sg3-f1 g2-g1=D+ 25. Se4-g3 g3-g2 26. Sd2-e4 g4-g3 27. d5-d6 Dg3-c7 28. Sf1-d2 Dg1-g3 29. Se3-f1 g2-g1=D+ 30. Sf5-e3 g5-g4 31. Sd4-f5 g3-g2 32. Sb3-d4 Kc5-b6 33. Sd2-b3+ g4-g3 34. Sf1-d2 Dg3-b8 35. Tb8-a8 g7-g5 36. c7xDb8=T g5-g4 37. b6xDc7 Dg1-g3 38. Se3-f1 g2-g1=D+ 39. Sc4-e3 g6-g5 40. Sa3-c4 g3-g2 41. Sb1-a3 g4-g3 42. Sd2-b1 Dg3-c7 43. Sf1-d2 Dg1-g3 44. Ld2-c3 g2-g1=D 45. Lc3-d2 f3xLg2 46. Se3-f1 e4xTf3 47. Sc4-e3 Kd6-c5 48. Sa3-c4+ h7xSg6 49. Sb1-a3 Ke7-d6 50. Sf4-g6+ Ke8-e7 51. Lf1-g2 f5xDg4 52. g2xTh3 Th6-h3 53. Sd2-b1 e5-e4 54. Lb2-c3 Tf6-h6 55. Tc3-f3 Tf8-f6 56. Sh3-f4 Th8-f8 57. Sg1-h3 f7-f5 58. Sc4-d2 Sc5-a4 59. Sa3-c4 Se4-c5 60. Td3-c3 Lf8-b4+ 61. Td1-d3 Sf6-e4 62. Ta1-d1 Sg8-f6 63. Sb1-a3 e7-e5 64. Lc1-b2 Dc7-b8 65. b5-b6 Dd8-c7 66. b4-b5 c7-c6 67. b2-b4 Sc6-a5 68. Dd4-g4 Sb8-c6 69. Dd1-d4 Ta8-a7 70. d4-d5 a7-a6 71. d2-d4 (2021-07-17)
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Hans-Jürgen Manthey: R: 0. O-O ! Kb6xc7 1. Ld8xc7+ Lf4-d6 2. Sb1-a3 De5-c7 3. Lg5-d8+ Da1-e5 3. Sd2-b1 Dc3-a1+ 4. Sf1-d2 Dg3-c3+ 5. Lf6-g5 Ld6-f4 6. Lc3-f6 Lb4-d6 7. c7xDb8=T Dg1-g3 8. Sg3-f1 g2-g1=D+ 9. Se4-g3 g3-g2 10. d6xDc7 g4-g3 11. Sd2-e4 g5-g4 12. Sf1-d2 g7-g5 13. d5-d6 Dg3-c7 14. Ld2-c3 Dg1-g3 15. Se3-f1 g2-g1=D+ 16. Lc3-d2 g3-g2 17. Sf1-e3 g4-g3 18. Sg3-f1 g5-g4 19. Se4-g3 g6-g5 20. Sd2-e4 h7xSg6 21. c7xDb8=T Lb4-a3 22. d6xDc7 Dg8-d8 23. Sf1-d2 Dg1-g8 24. Sg3-f1 g2-g1=D+ 25. Se4-g3 g3-g2 26. Sd2-e4 g4-g3 27. d5-d6 Dg3-c7 28. Sf1-d2 Dg1-g3 29. Se3-f1 g2-g1=D+ 30. Sf5-e3 g5-g4 31. Sd4-f5 g3-g2 32. Sb3-d4 Kc5-b6 33. Sd2-b3+ g4-g3 34. Sf1-d2 Dg3-b8 35. Tb8-a8 g7-g5 36. c7xDb8=T g5-g4 37. b6xDc7 Dg1-g3 38. Se3-f1 g2-g1=D+ 39. Sc4-e3 g6-g5 40. Sa3-c4 g3-g2 41. Sb1-a3 g4-g3 42. Sd2-b1 Dg3-c7 43. Sf1-d2 Dg1-g3 44. Ld2-c3 g2-g1=D 45. Lc3-d2 f3xLg2 46. Se3-f1 e4xTf3 47. Sc4-e3 Kd6-c5 48. Sa3-c4+ h7xSg6 49. Sb1-a3 Ke7-d6 50. Sf4-g6+ Ke8-e7 51. Lf1-g2 f5xDg4 52. g2xTh3 Th6-h3 53. Sd2-b1 e5-e4 54. Lb2-c3 Tf6-h6 55. Tc3-f3 Tf8-f6 56. Sh3-f4 Th8-f8 57. Sg1-h3 f7-f5 58. Sc4-d2 Sc5-a4 59. Sa3-c4 Se4-c5 60. Td3-c3 Lf8-b4+ 61. Td1-d3 Sf6-e4 62. Ta1-d1 Sg8-f6 63. Sb1-a3 e7-e5 64. Lc1-b2 Dc7-b8 65. b5-b6 Dd8-c7 66. b4-b5 c7-c6 67. b2-b4 Sc6-a5 68. Dd4-g4 Sb8-c6 69. Dd1-d4 Ta8-a7 70. d4-d5 a7-a6 71. d2-d4 (2021-07-17)
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Keywords: Ceriani-Frolkin Theme, Castling, Non-standard material
Genre: Retro
FEN: RRb5/rpBp4/pkpb4/n7/n7/N6P/P1P1PP1P/4K2R
Reprints: (17) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-06
Last update: James Malcom, 2021-07-17 more...
Genre: Retro
FEN: RRb5/rpBp4/pkpb4/n7/n7/N6P/P1P1PP1P/4K2R
Reprints: (17) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-06
Last update: James Malcom, 2021-07-17 more...
Henrik Juel: Black: Ke8, Rh8, Sh1, Pf7,g6,h6 (12+16). 1.000. bSh1 is promoted after 3 captures; wRh7 came from h1 via row 8, ruining Black's castling. (2003-11-17)
Henrik Juel: Fedir had an anonymous co-author?? (2021-10-31)
Henrik Juel: How are the castlings mutually exclusive? (2021-10-31)
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Henrik Juel: Fedir had an anonymous co-author?? (2021-10-31)
Henrik Juel: How are the castlings mutually exclusive? (2021-10-31)
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Keywords: Colouring problem, Castling, mutual exclusive
Genre: Retro
FEN: B3K2R/1N3P1R/1P3PPP/8/8/6P1/2P1PPP1/R3K2N
Reprints: (43) Die Schwalbe 144 12/1993
Input: Gerd Wilts, 1995-06-07
Last update: A.Buchanan, 2021-10-31 more...
Genre: Retro
FEN: B3K2R/1N3P1R/1P3PPP/8/8/6P1/2P1PPP1/R3K2N
Reprints: (43) Die Schwalbe 144 12/1993
Input: Gerd Wilts, 1995-06-07
Last update: A.Buchanan, 2021-10-31 more...
Keywords: Castling, Non-standard material
Genre: Retro
FEN: 8/2ppp2p/1p6/1p6/brPN4/rqP3PP/nbrPPP1r/RNk1K2R
Input: Gerd Wilts, 1995-08-14
Last update: Gerd Wilts, 2004-09-12 more...
Genre: Retro
FEN: 8/2ppp2p/1p6/1p6/brPN4/rqP3PP/nbrPPP1r/RNk1K2R
Input: Gerd Wilts, 1995-08-14
Last update: Gerd Wilts, 2004-09-12 more...
1. Ke6 Th7 2. Dg5 Kf8 3. Dd8+ Kg7 4. De7+ Kg8 5. De8+ Kg7 6. Df7+ Kh6 7. Df6+ Kh5 8. Kf5 Tf7 9. Dxf7+ Kh4 10. Db3 Kh5 11. Dh3#
1. ... 0-0?
Schwarz kann nicht rochieren, da entweder der sK oder der sT zuletzt gezogen haben müssen.
1. ... 0-0?
Schwarz kann nicht rochieren, da entweder der sK oder der sT zuletzt gezogen haben müssen.
Korrektur: Bernhard Rittmeier
Sally: 1. Ke6 - Kh7, 2. Dg5 - Kf8, 3. Dd6+ - Kg7, 4. De7+ - Kg8,
5. De5+- Kg7, 6. Df7+- Kh6, 7. Df6+ - Kh5, 8. Kf5 - Tf7,
9. Dxf5+-Kh4,10. Db3 - Kh5,11. Dh3#.S.131 Schwalbe 8.1995.
Nr. 199 Mattaufg. mit 3 + 4 Steinen (Speckmann 1979) Dort
mit wk. f5,daum Verb. s. 21 Errata Liste ( = ) (2011-03-04)
Anton Baumann: C+ Gustav 4.1d (2021-01-30)
comment
Sally: 1. Ke6 - Kh7, 2. Dg5 - Kf8, 3. Dd6+ - Kg7, 4. De7+ - Kg8,
5. De5+- Kg7, 6. Df7+- Kh6, 7. Df6+ - Kh5, 8. Kf5 - Tf7,
9. Dxf5+-Kh4,10. Db3 - Kh5,11. Dh3#.S.131 Schwalbe 8.1995.
Nr. 199 Mattaufg. mit 3 + 4 Steinen (Speckmann 1979) Dort
mit wk. f5,daum Verb. s. 21 Errata Liste ( = ) (2011-03-04)
Anton Baumann: C+ Gustav 4.1d (2021-01-30)
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Keywords: Castling in the forward play, Cant Castler
Genre: Retro
FEN: 4k2r/8/8/3K4/8/8/3Q4/8
Reprints: 1125 Zuglängenrekorde im Wenigsteiner 1986
(7b) Die Schwalbe 154 08/1995
Input: Gerd Wilts, 1995-08-23
Last update: Rainer Staudte, 2013-11-19 more...
Genre: Retro
FEN: 4k2r/8/8/3K4/8/8/3Q4/8
Reprints: 1125 Zuglängenrekorde im Wenigsteiner 1986
(7b) Die Schwalbe 154 08/1995
Input: Gerd Wilts, 1995-08-23
Last update: Rainer Staudte, 2013-11-19 more...
98 - P0006423
Andrey Frolkin
9128 Die Schwalbe 157, p. 283, 02/1996
Leonid Borodatow gewidmet
(9+4)
#3 (AP)
Andrey Frolkin
9128 Die Schwalbe 157, p. 283, 02/1996
Leonid Borodatow gewidmet
(9+4)
#3 (AP)
White pushes the move (Keym AP)
1. ... f5! (f6?,fxe6?,fxg6? 2.0-0! ~ 3.Tf3#) 2. gxf6ep! exf6 3. 0-0! f5 4. Tf3#
(2. 0-0=?)
Valladao Task via the try: 1. exf7? e5! 1. ... e6? 2. f8=D e5 3. Da3#
Another try shows fully differentiated black Albino:
1. Kf1? fxg6!
1. ... f5? 2. Lxf5! h3 3. Txh3#
1. ... fxe6? 2. Lxe6! h3 3. Txh3#
1. ... f6? 2. Lf5! h3,fxg5 3. Txh3#,Th3#
And retro tries:
1. 0-0?? f5,~ 2. Tf3# (short solution)
1. f5? f6,~ 2. 0-0?? fxg5,~ 3. Tf3#
Wenn Schwarz zuletzt gezogen hat, dann kann Weiß nicht mehr rochieren: R: 1. Kh2-g3 Tf1(g1)-h1+. Da aber Weiß in einem direkten Matt beginnt, ist die Rochade nicht mehr zulässig. Der Zusatz "AP" in der Forderung ist überflüssig: es könnte höchstens der Anzug a posteriori dem Schwarzen übertragen werden. Das geht aber nicht, da Weiß in einem direkten Matt beginnt.
1. ... f5! (f6?,fxe6?,fxg6? 2.0-0! ~ 3.Tf3#) 2. gxf6ep! exf6 3. 0-0! f5 4. Tf3#
(2. 0-0=?)
Valladao Task via the try: 1. exf7? e5! 1. ... e6? 2. f8=D e5 3. Da3#
Another try shows fully differentiated black Albino:
1. Kf1? fxg6!
1. ... f5? 2. Lxf5! h3 3. Txh3#
1. ... fxe6? 2. Lxe6! h3 3. Txh3#
1. ... f6? 2. Lf5! h3,fxg5 3. Txh3#,Th3#
And retro tries:
1. 0-0?? f5,~ 2. Tf3# (short solution)
1. f5? f6,~ 2. 0-0?? fxg5,~ 3. Tf3#
Wenn Schwarz zuletzt gezogen hat, dann kann Weiß nicht mehr rochieren: R: 1. Kh2-g3 Tf1(g1)-h1+. Da aber Weiß in einem direkten Matt beginnt, ist die Rochade nicht mehr zulässig. Der Zusatz "AP" in der Forderung ist überflüssig: es könnte höchstens der Anzug a posteriori dem Schwarzen übertragen werden. Das geht aber nicht, da Weiß in einem direkten Matt beginnt.
Guus Rol: This is apparently AP after Keym. The move goes to black:
0. ... f5! 1.gxf6ep exf6 2.0-0(justifies the handover) f5 3.Tf3 (2007-02-13)
A.Buchanan: There are two kinds of directmate Type Keym. In one Black unsuccessfully pulls the move, in the other (as here) White successfully pushes it to Black. (2022-04-16)
A.Buchanan: I disagree with the German comment in the solution text. I think including "AP" is advisable in the stipulation. Keym AP riffs off Codex Article 15, but it's not the default. (2022-04-17)
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0. ... f5! 1.gxf6ep exf6 2.0-0(justifies the handover) f5 3.Tf3 (2007-02-13)
A.Buchanan: There are two kinds of directmate Type Keym. In one Black unsuccessfully pulls the move, in the other (as here) White successfully pushes it to Black. (2022-04-16)
A.Buchanan: I disagree with the German comment in the solution text. I think including "AP" is advisable in the stipulation. Keym AP riffs off Codex Article 15, but it's not the default. (2022-04-17)
more ...
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Keywords: a posteriori (AP) (Type Keym), Castling, En passant, Promotion (D), Valladao Task
Genre: Retro, 3#
FEN: 8/4pp2/4P1N1/6PP/5P1p/6kB/6P1/4K2R
Input: Gerd Wilts, 1996-06-12
Last update: A.Buchanan, 2022-04-18 more...
Genre: Retro, 3#
FEN: 8/4pp2/4P1N1/6PP/5P1p/6kB/6P1/4K2R
Input: Gerd Wilts, 1996-06-12
Last update: A.Buchanan, 2022-04-18 more...
99 - P0006428
Andrey Frolkin
9133v Die Schwalbe 157, p. 283, 02/1996
(13+12) cooked
BP in 20.0
May White castle?
Andrey Frolkin
9133v Die Schwalbe 157, p. 283, 02/1996
(13+12) cooked
BP in 20.0
May White castle?
Weiß darf noch rochieren: 1. h4 Sa6 2. h5 Sc5 3. h6 a6 4. hxg7 h5 5. f4 h4 6. f5 h3 7. f6 h2 8. fxe7 Sf6 9. g8=L Lh6 10. Lxf7 Kxf7 11. e8=T Kg6 12. Te6 Dg8 13. Tc6 Db3 14. axb3 dxc6 15. Ta4 Lh3 16. Te4 Tag8 17. Te8 hxg1=D 18. e4 Dd4 19. Lc4 Tg7 20. Lg8 Dd8
Ohne die Bedingung, daß Weiß rochieren darf, gibt es andere BPs: 1. f4 Sa6 2. f5 Sc5 3. f6 a6 4. fxe7 f5 5. e4 f4 6. Lb5 f3 7. Lc6 f2+ 8. Ke2 fxg1=D 9. h4 dxc6 10. h5 Lf5 11. h6 Sf6 12. hxg7 Dd5 13. g8=L Db3 14. Lxh7 Kf7 15. axb3 Lh6 16. Ta4 Tag8 17. Td4 Dxd4 18. e8=T Tg7 19. Lg8+ Kg6 20. Ke1 Dd8
Ohne die Bedingung, daß Weiß rochieren darf, gibt es andere BPs: 1. f4 Sa6 2. f5 Sc5 3. f6 a6 4. fxe7 f5 5. e4 f4 6. Lb5 f3 7. Lc6 f2+ 8. Ke2 fxg1=D 9. h4 dxc6 10. h5 Lf5 11. h6 Sf6 12. hxg7 Dd5 13. g8=L Db3 14. Lxh7 Kf7 15. axb3 Lh6 16. Ta4 Tag8 17. Td4 Dxd4 18. e8=T Tg7 19. Lg8+ Kg6 20. Ke1 Dd8
SH: Cooked?:
1. e4 f5 2. Ne2 f4 3. h4 f3 4. h5 fxe2 5. f4 exf1B 6. f5 Bc4 7. f6 Bb3 8. axb3 Na6 9. fxe7 Nc5 10. Ra6 Nf6 11. Rc6 dxc6 12. h6 Bh3 13. hxg7 Qd7 14. g8B Bh6 15. Bxh7 Kf7 16. Qf3 Rag8 17. Qe2 Rg7 18. Bg8+ Kg6 19. Qd1 a6 20. e8R Qd8 (2005-06-20)
Henrik Juel: The stipulation question was probably intended as an implicit condition; when answered affirmatively, there should be a unique solution
But SH's solution shows that the problem is cooked (2018-12-09)
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1. e4 f5 2. Ne2 f4 3. h4 f3 4. h5 fxe2 5. f4 exf1B 6. f5 Bc4 7. f6 Bb3 8. axb3 Na6 9. fxe7 Nc5 10. Ra6 Nf6 11. Rc6 dxc6 12. h6 Bh3 13. hxg7 Qd7 14. g8B Bh6 15. Bxh7 Kf7 16. Qf3 Rag8 17. Qe2 Rg7 18. Bg8+ Kg6 19. Qd1 a6 20. e8R Qd8 (2005-06-20)
Henrik Juel: The stipulation question was probably intended as an implicit condition; when answered affirmatively, there should be a unique solution
But SH's solution shows that the problem is cooked (2018-12-09)
more ...
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Keywords: Anti-Pronkin, Constrained problem, Unique Proof Game, Castling
Genre: Retro
FEN: 3qR1Br/1pp3r1/p1p2nkb/2n5/4P3/1P5b/1PPP2P1/1NBQK2R
Input: Gerd Wilts, 1996-06-12
Last update: James Malcom, 2022-02-08 more...
Genre: Retro
FEN: 3qR1Br/1pp3r1/p1p2nkb/2n5/4P3/1P5b/1PPP2P1/1NBQK2R
Input: Gerd Wilts, 1996-06-12
Last update: James Malcom, 2022-02-08 more...
100 - P0006644
Chris Patzke
9260 Die Schwalbe 159, p. 391, 06/1996
2. Preis Abt. II
(13+14)
BP in 16.0
Zeroposition
a) sD nach f3
b) +wBe3
Chris Patzke
9260 Die Schwalbe 159, p. 391, 06/1996
2. Preis Abt. II
(13+14)
BP in 16.0
Zeroposition
a) sD nach f3
b) +wBe3
a) 1. d3 a5 2. Lg5 a4 3. Lxe7 Kxe7 4. Kd2 Kd6 5. Kc3 Se7 6. Kb4 Sec6+ 7. Ka3 Le7 8. Sd2 Te8 9. Sf3 Lf8 10. Dd2 Txe2 11. Td1 Te5 12. Le2 Tea5 13. Se5 Df6 14. Lg4 Df3 15. Lxd7 Kd5+ 16. b4 axb3ep+
b) 1. b3 a5 2. La3 a4 3. Lxe7 axb3 4. Lb4 Df6 5. La5 Ke7 6. e3 Kd6 7. Lb5 Se7 8. Lxd7 Sec6 9. d3 Le7 10. Sd2 Te8 11. Sf3 Lf8 12. Dd2 Te5 13. 0-0-0 Tc5 14. Se5 Kd5 15. Kb2 Dd6 16. Ka3 Tcxa5+
b) 1. b3 a5 2. La3 a4 3. Lxe7 axb3 4. Lb4 Df6 5. La5 Ke7 6. e3 Kd6 7. Lb5 Se7 8. Lxd7 Sec6 9. d3 Le7 10. Sd2 Te8 11. Sf3 Lf8 12. Dd2 Te5 13. 0-0-0 Tc5 14. Se5 Kd5 15. Kb2 Dd6 16. Ka3 Tcxa5+
Henrik Juel: should the diagram have sD on d6? (2021-01-24)
Henrik Juel: With this change part a) (i.e. the current diagram position) is C+ Euclide 1.01
For part b) I stopped the test without results at pos. 62 after many hours (2021-01-25)
Hans-Jürgen Manthey: auch hier klappt der Player nur mit Änderungen:
bei a) 9. Sdf3
bei b) 11. Sdf3 (2021-01-25)
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Henrik Juel: With this change part a) (i.e. the current diagram position) is C+ Euclide 1.01
For part b) I stopped the test without results at pos. 62 after many hours (2021-01-25)
Hans-Jürgen Manthey: auch hier klappt der Player nur mit Änderungen:
bei a) 9. Sdf3
bei b) 11. Sdf3 (2021-01-25)
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Keywords: Unique Proof Game, En passant, Castling
Genre: Retro
FEN: rnb2b2/1ppB1ppp/2n5/r2kN3/8/Kp1P1q2/P1PQ1PPP/3R2NR
Input: Gerd Wilts, 1996-07-11
Last update: James Malcom, 2021-01-24 more...
Genre: Retro
FEN: rnb2b2/1ppB1ppp/2n5/r2kN3/8/Kp1P1q2/P1PQ1PPP/3R2NR
Input: Gerd Wilts, 1996-07-11
Last update: James Malcom, 2021-01-24 more...
Show statistic for complete result. Show search result faster by using ids.
https://pdb.dieschwalbe.de/search.jsp?expression=COMMENTDATE%3E%3D20200919+AND+NOT+K%3D%27Hilfsr%C3%BCckz%C3%BCger%27+AND+NOT+K%3D%27En+passant+als+Schl%C3%BCssel%27+AND+K%3D%27Rochade%27
The problems of this query have been registered by the following contributors:
Gerd Wilts (100)
Henrik Juel: Good motivation for the black castling
The further retroplay includes retracting Tf1-g1, Kd3 to g1, h2xTg3, Tg3 to f8, La6 to c8, d7xTSe6, and e6xPf7 (2021-04-19)
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