1509 problem(s) found in 3745 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT K='Eindeutige Beweispartie' AND COOKED ] [download as LaTeX]
1 - 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#
1. d4 e5 2. d5 f5 3. d6 Le7 4. dxe7 d5 5. e4 d4 6. Dg4 fxg4 7. c4 d3 8. c5 Dd6 9. b4 Kd7 10. cxd6 c5 11. b5 c4 12. b6 Sc6 13. e8=T d2+ 14. Ke2 c3 15. Kd3 d1=L 16. Sd2 Sd4 17. Tf8 Kc6 18. Tf4 exf4 19. e5 h5 20. e6 Lb3 21. e7 Lf7 22. bxa7 b5 23. e8=T b4 24. Te6 b3 25. Th6 gxh6 26. d7 b2 27. d8=T Lce6 28. Td5 Te8 29. Kc4 b1=L 30. a8=T Lh7 31. Ta3 c2 32. Tg3 Se7 33. a4 Thf8 34. a5 Sg6 35. a6 Sh8 36. a7 fxg3 37. a8=T gxh2 38. T8a3 Lhg8 39. Th3 gxh3 40. Se4 Lxd5+
Cook: 1. b4 e5 2. d4 Le7 3. d5 f5 4. d6 h5 5. dxe7 d5 6. b5 Dd6 7. b6 Kd7 8. bxa7 b5 9. c4 b4 10. c5 d4 11. cxd6 c5 12. e8=T Sc6 13. Tf8 c4 14. e4 d3 15. Dg4 d2+ 16. Ke2 c3 17. Kd3 d1=L 18. Sd2 fxg4 19. Tf4 exf4 20. e5 Sd4 21. e6+ Kc6 22. e7 b3 23. e8=T b2 24. Te6 Lb3 25. Th6 Lf7 26. d7+ gxh6 27. d8=T Lce6 28. Td5 Te8 29. Kc4 b1=L 30. a8=T Lh7 31. Ta3 c2 32. Tg3 Se7 33. a4 Thf8 34. a5 Sg6 35. a6 Sh8 36. a7 fxg3 37. a8=T gxh2
38. T8a3 Lhg8 39. Th3 gxh3 40. Se4 Lxd5+
Cook: 1. b4 e5 2. d4 Le7 3. d5 f5 4. d6 h5 5. dxe7 d5 6. b5 Dd6 7. b6 Kd7 8. bxa7 b5 9. c4 b4 10. c5 d4 11. cxd6 c5 12. e8=T Sc6 13. Tf8 c4 14. e4 d3 15. Dg4 d2+ 16. Ke2 c3 17. Kd3 d1=L 18. Sd2 fxg4 19. Tf4 exf4 20. e5 Sd4 21. e6+ Kc6 22. e7 b3 23. e8=T b2 24. Te6 Lb3 25. Th6 Lf7 26. d7+ gxh6 27. d8=T Lce6 28. Td5 Te8 29. Kc4 b1=L 30. a8=T Lh7 31. Ta3 c2 32. Tg3 Se7 33. a4 Thf8 34. a5 Sg6 35. a6 Sh8 36. a7 fxg3 37. a8=T gxh2
38. T8a3 Lhg8 39. Th3 gxh3 40. Se4 Lxd5+
Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 29:20:32 Stunden. (hh:mm:ss)
Da NUPG C+
Notation: 1.b4 e5 2.d4 Le7 3.d5 f5 4.d6 h5 5.dxe7 d5 6.b5 Dd6 7.b6 Kd7 8.bxa7 b5
9.c4 b4 10.c5 d4 11.cxd6 c5 12.e8=T Sc6 13.Tf8 c4 14.e4 d3 15.Dg4 d2+ 16.Ke2 c3
17.Kd3 d1=L 18.Sd2 fxg4 19.Tf4 exf4 20.e5 Sd4 21.e6+ Kc6 22.e7 b3 23.e8=T b2
24.Te6 Lb3 25.Th6 Lf7 26.d7+ gxh6 27.d8=T Lce6 28.Td5 Te8 29.Kc4 b1=L 30.a8=T Lh7
31.Ta3 c2 32.Tg3 Se7 33.a4 Thf8 34.a5 Sg6 35.a6 Sh8 36.a7 fxg3 37.a8=T gxh2
38.T8a3 Lhg8 39.Th3 gxh3 40.Se4 Lxd5+
3 schwarze Läufer stehen zum Schluss auf den weißen Felder d5, f7, g8. (2023-09-21)
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Da NUPG C+
Notation: 1.b4 e5 2.d4 Le7 3.d5 f5 4.d6 h5 5.dxe7 d5 6.b5 Dd6 7.b6 Kd7 8.bxa7 b5
9.c4 b4 10.c5 d4 11.cxd6 c5 12.e8=T Sc6 13.Tf8 c4 14.e4 d3 15.Dg4 d2+ 16.Ke2 c3
17.Kd3 d1=L 18.Sd2 fxg4 19.Tf4 exf4 20.e5 Sd4 21.e6+ Kc6 22.e7 b3 23.e8=T b2
24.Te6 Lb3 25.Th6 Lf7 26.d7+ gxh6 27.d8=T Lce6 28.Td5 Te8 29.Kc4 b1=L 30.a8=T Lh7
31.Ta3 c2 32.Tg3 Se7 33.a4 Thf8 34.a5 Sg6 35.a6 Sh8 36.a7 fxg3 37.a8=T gxh2
38.T8a3 Lhg8 39.Th3 gxh3 40.Se4 Lxd5+
3 schwarze Läufer stehen zum Schluss auf den weißen Felder d5, f7, g8. (2023-09-21)
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Keywords: Non-Unique Proof Game, Ceriani-Frolkin Theme (TTTTT), Non-standard material (ll), Promotion (TlTTlTT)
Genre: Retro
FEN: 4rrbn/5b2/2k4p/3b3p/2KnN3/7p/2p2PPp/R1B2BNR
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2023-09-22 more...
Genre: Retro
FEN: 4rrbn/5b2/2k4p/3b3p/2KnN3/7p/2p2PPp/R1B2BNR
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2023-09-22 more...
3 - 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)
more ...
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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...
4 - 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. ... Kg5 2. 0-0?? Se7#
1. ... Txh7 2. Kf8 Txh8#
1. cxb3ep Txh7 2. Kf8?? Txh8#
1. cxb3ep Kg5 2. 0-0! Se7#
Wh captured 10 by pawns (including hxPg-g8) but not on the last move. wPa3 came from a2: wPa5 captured bPa.
Bl took only b3xa2.
In set play, Bl has no last move except for K or R.
For the main play, the intention is that if Bl retains castling rights, last moves were R: 1. b2-b4 b3xa2 e.g. 2. Bg1-d4 b4-b3 3. d4xc5 etc. However there are retro cooks e.g. R: 1. Bg1-d4 b3xa2 2. d4xc5 etc, with wPa5 the original wPb. wPb4 & wPb5 come from further East.
Cook: Retro logic for main play is unsound.
1. ... Txh7 2. Kf8 Txh8#
1. cxb3ep Txh7 2. Kf8?? Txh8#
1. cxb3ep Kg5 2. 0-0! Se7#
Wh captured 10 by pawns (including hxPg-g8) but not on the last move. wPa3 came from a2: wPa5 captured bPa.
Bl took only b3xa2.
In set play, Bl has no last move except for K or R.
For the main play, the intention is that if Bl retains castling rights, last moves were R: 1. b2-b4 b3xa2 e.g. 2. Bg1-d4 b4-b3 3. d4xc5 etc. However there are retro cooks e.g. R: 1. Bg1-d4 b3xa2 2. d4xc5 etc, with wPa5 the original wPb. wPb4 & wPb5 come from further East.
Cook: Retro logic for main play is unsound.
A.Buchanan: Not trivial to fix as shifting wPa5 to b6 leads to promotion cooks. Adding black material must be carefully done to avoid disrupting uniqueness of e.p. as tempo. (2022-03-11)
A.Buchanan: OK just to say I have found a form of this idea that works. This one was the very devil to fix - the final form includes different mates and retro logic, so becomes an "after" rather than a "correction", and I will find somewhere else to publish first. (2022-03-19)
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A.Buchanan: OK just to say I have found a form of this idea that works. This one was the very devil to fix - the final form includes different mates and retro logic, so becomes an "after" rather than a "correction", and I will find somewhere else to publish first. (2022-03-19)
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Keywords: Castling (sk), a posteriori (AP) (Type Petrovic), En passant as key, Tempo Move, Superseded by (P1409841)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/5p1p/5K1R/PPP2N2/1PpB4/P1P5/p1RNB1P1/Q7
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-20 more...
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/5p1p/5K1R/PPP2N2/1PpB4/P1P5/p1RNB1P1/Q7
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-20 more...
Kees: possible fix: Lb1=Sb1 -De1 +Ld1
White begins: 1.Kxb7 Lxe2 2.Kc8 La6#
(1.Txd8+ Kxd8 2.Kf8 Th8# illegal for white has no last move) (2023-06-07)
A.Buchanan: Your fix is good, Kees. It removes the cook, and sLd1 denies R: 1. c2xb3 as well as sLb1 did. Note R: 1 Sc6-d8 0-0+? as black castling rights were lost to let wK enter the back rank. (2023-06-08)
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White begins: 1.Kxb7 Lxe2 2.Kc8 La6#
(1.Txd8+ Kxd8 2.Kf8 Th8# illegal for white has no last move) (2023-06-07)
A.Buchanan: Your fix is good, Kees. It removes the cook, and sLd1 denies R: 1. c2xb3 as well as sLb1 did. Note R: 1 Sc6-d8 0-0+? as black castling rights were lost to let wK enter the back rank. (2023-06-08)
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Genre: h#, Retro
FEN: 1bKN1rk1/1ppn1r1R/5p1P/4pP1p/3p1p2/1PP3P1/PP2P3/Rb2q3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-08 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...
NL: 1. d4 d6 2. Kd2 Le6 3. g4 f5 4. g5 Lf7 5. g6 Sd7 6. e4 Db8 7. gxf7+ Kd8 8. Kc3 g5 9. Kc4 g4 10. Kd5 Lg7 11. Ke6 Kc8 12. f8=L g3 13. Kf7 g2 14. Ke8 gxf1=L 15. e5 Lg2 16. Se2 d5 17. Lg5 Lh3 18. e6
Die Autorlösung wurde nie publiziert und ist unbekannt.
Die Autorlösung wurde nie publiziert und ist unbekannt.
Moldenhauer: Computerprüfung: C+ NUPG cooked Stelvio 1.11 in 1 Sekunde.
Keine Lösung: BP 17.0. BP 17.5 cooked.
Beispiel BP 18.0: 1.d4 Sa6 2.Kd2 Sb8 3.Kc3 d6 4.e4 Le6 5.Se2 Sd7 6.e5 Db8 7.g4 Kd8
8.g5 f5 9.g6 Lf7 10.gxf7 Kc8 11.Kc4 g5 12.Kd5 Lg7 13.Ke6 d5 14.f8L g4 15.Kf7 g3
16.Ke8 g2 17.Lg5 gxf1L 18.e6 Lh3
Beispiel BP 17.5: 1.d4 Sa6 2.Kd2 Sc5 3.Ke3 d5 4.Kf4 Le6 5.Ke5 Kd7 6.e4 Db8
7.g4 Kc8 8.g5 f5 9.g6 Lf7 10.gxf7 Sd7+ 11.Ke6 g5 12.Se2 Lg7 13.e5 g4 14.Lg5 g3
15.f8L g2 16.Kf7 gxf1L 17.Ke8 Lh3 18.e6 (2023-05-07)
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Keine Lösung: BP 17.0. BP 17.5 cooked.
Beispiel BP 18.0: 1.d4 Sa6 2.Kd2 Sb8 3.Kc3 d6 4.e4 Le6 5.Se2 Sd7 6.e5 Db8 7.g4 Kd8
8.g5 f5 9.g6 Lf7 10.gxf7 Kc8 11.Kc4 g5 12.Kd5 Lg7 13.Ke6 d5 14.f8L g4 15.Kf7 g3
16.Ke8 g2 17.Lg5 gxf1L 18.e6 Lh3
Beispiel BP 17.5: 1.d4 Sa6 2.Kd2 Sc5 3.Ke3 d5 4.Kf4 Le6 5.Ke5 Kd7 6.e4 Db8
7.g4 Kc8 8.g5 f5 9.g6 Lf7 10.gxf7 Sd7+ 11.Ke6 g5 12.Se2 Lg7 13.e5 g4 14.Lg5 g3
15.f8L g2 16.Kf7 gxf1L 17.Ke8 Lh3 18.e6 (2023-05-07)
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Keywords: Non-Unique Proof Game, Non-standard material
Genre: Retro
FEN: rqk1KBnr/pppnp1bp/4P3/3p1pB1/3P4/7b/PPP1NP1P/RN1Q3R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2014-05-20 more...
Genre: Retro
FEN: rqk1KBnr/pppnp1bp/4P3/3p1pB1/3P4/7b/PPP1NP1P/RN1Q3R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2014-05-20 more...
9 - P0000759
Luis Alberto Garaza
1034 Die Schwalbe 21 06/1973
(9+9) cooked
Schwarz am Zug, Weiß gewinnt
Luis Alberto Garaza
1034 Die Schwalbe 21 06/1973
(9+9) cooked
Schwarz am Zug, Weiß gewinnt
1. ... fxg3ep 2. hxg3+ Kh5 3. f4 Kh6 4. h8=T+! Kg7
Cook: Intended a Posteriori combo of castling and ep rights retro-locks wTh in cage, contradicting pawn capture balance
Cook: Intended a Posteriori combo of castling and ep rights retro-locks wTh in cage, contradicting pawn capture balance
hans: only black move is fxg3e.p.
1. fxg3+ Kh5 2. h8D/T#
Why the keyword 'castling'?
Castling is illegal, for there are 7 black captures, and because white's last move is g2-g4, needs the wK let pass Rh1 (2012-05-18)
Anton Baumann: g2-g4 war nicht zwingend der letzte Zug! Schwarz ist patt, also unlösbar!
Korrektur in 'die Schwalbe' 12/2074 S.264: "wTa1 nach h1. Nun gewinnt nach 1. - fxg3 Weiss mit 2.hxg3+ Kh5 3.f4 Kh6 4.h8=T+! Kg7 und der umgewandelte Turm holt die störenden sBB ab, sodass Weiss rochieren kann und beweist, dass Schwarz nicht patt war. Schwarz versucht die Rochade zu verhindern, ist aber machtlos." (2022-12-06)
A.Buchanan: The forward play is more complicated, as Black can play more aggressively than 3. … Kh6 (2022-12-07)
Anton Baumann: Wenn Schwarz nicht 3.-Kh6! zieht, holt sich Weiss auf h8 die Dame, mit einfacherer Fortsetzung.
3.-g4? 4.f5 Kg5 (4.-g5 5.f6 ~ 6.h8=D) 5.h8=D Kxf5 6.De8 +_
3.-Kg4? 4.h8=D Kf3 5.Dxh3 g4 6.Dh2 Ke4 7.Dg2+ +_ (2022-12-09)
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1. fxg3+ Kh5 2. h8D/T#
Why the keyword 'castling'?
Castling is illegal, for there are 7 black captures, and because white's last move is g2-g4, needs the wK let pass Rh1 (2012-05-18)
Anton Baumann: g2-g4 war nicht zwingend der letzte Zug! Schwarz ist patt, also unlösbar!
Korrektur in 'die Schwalbe' 12/2074 S.264: "wTa1 nach h1. Nun gewinnt nach 1. - fxg3 Weiss mit 2.hxg3+ Kh5 3.f4 Kh6 4.h8=T+! Kg7 und der umgewandelte Turm holt die störenden sBB ab, sodass Weiss rochieren kann und beweist, dass Schwarz nicht patt war. Schwarz versucht die Rochade zu verhindern, ist aber machtlos." (2022-12-06)
A.Buchanan: The forward play is more complicated, as Black can play more aggressively than 3. … Kh6 (2022-12-07)
Anton Baumann: Wenn Schwarz nicht 3.-Kh6! zieht, holt sich Weiss auf h8 die Dame, mit einfacherer Fortsetzung.
3.-g4? 4.f5 Kg5 (4.-g5 5.f6 ~ 6.h8=D) 5.h8=D Kxf5 6.De8 +_
3.-Kg4? 4.h8=D Kf3 5.Dxh3 g4 6.Dh2 Ke4 7.Dg2+ +_ (2022-12-09)
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Keywords: Castling (wl), a posteriori (AP) (Type Petrovic), En passant as key
Genre: Retro, Studies
FEN: 8/7P/6p1/2p3p1/2p2pPk/2Pp1P1p/3PpP1P/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-07 more...
Genre: Retro, Studies
FEN: 8/7P/6p1/2p3p1/2p2pPk/2Pp1P1p/3PpP1P/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-07 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...
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...
paul: Cooked in 4: 1.Ke5-e6! b2×Ba1=B+ -2.Kf4×Pe5 e6-e5+ -3.Rc2×Sg2 Se1-g2+ -4.Re2×Pc2 & 1.Rxe1# (2023-06-14)
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Keywords: En passant, Defensive Retractor, Type Proca
Genre: Retro
FEN: 8/4p3/3PK1P1/8/1B5p/1P1qPP1p/P5Rp/b2k4
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2004-08-28 more...
Genre: Retro
FEN: 8/4p3/3PK1P1/8/1B5p/1P1qPP1p/P5Rp/b2k4
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2004-08-28 more...
NL: 1. b4 a5 2. h4 d6 3. Th3 g6 4. Te3 Lg7 5. g3 Lc3 6. Lh3 Ta6 7. Kf1 Tc6 8. Kg2 b6 9. Kf3 Dd7 10. Ke4 Tc4+ 11. Kd5 Dxh3 12. Sf3 Lg4 13. Sd4 Sd7 14. Sb3 Td4+ 15. Kc6 Se5+ 16. Kb7 Kd7 17. Dh1 Ke6 18. Kc8 Lh5 19. Da8 Kf5 20. g4+ Kf4 21. f3 Kg3 22. Kd8 Kf2 23. Ke8 Ke1
AL wurde nie veröffentlicht.
AL wurde nie veröffentlicht.
Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 01:00:21 Stunde. (hh:mm:ss)
1.Sf3 a5 2.b4 g6 3.h4 Lg7 4.Th3 Lc3 5.Sd4 Ta6 6.Te3 Tc6 7.g3 Tc4 8.Lh3 d6 9.Kf1 Dd7
10.Kg2 Dxh3+ 11.Kf3 Lg4+ 12.Ke4 Sd7 13.Kd5 b6 14.Sb3 Td4+ 15.Kc6 Se5+ 16.Kb7 Kd7
17.Dh1 Ke6 18.Kc8 Kf5 19.Da8 Lh5 20.g4+ Kf4 21.f3 Kg3 22.Kd8 Kf2 23.Ke8 Ke1
Keine Lösung: BP 22.5.
Wenn die Forderung KBP 25.0 richtig ist dann Cooked. (2023-09-28)
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1.Sf3 a5 2.b4 g6 3.h4 Lg7 4.Th3 Lc3 5.Sd4 Ta6 6.Te3 Tc6 7.g3 Tc4 8.Lh3 d6 9.Kf1 Dd7
10.Kg2 Dxh3+ 11.Kf3 Lg4+ 12.Ke4 Sd7 13.Kd5 b6 14.Sb3 Td4+ 15.Kc6 Se5+ 16.Kb7 Kd7
17.Dh1 Ke6 18.Kc8 Kf5 19.Da8 Lh5 20.g4+ Kf4 21.f3 Kg3 22.Kd8 Kf2 23.Ke8 Ke1
Keine Lösung: BP 22.5.
Wenn die Forderung KBP 25.0 richtig ist dann Cooked. (2023-09-28)
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Keywords: Interchange, Non-Unique Proof Game, Non-Unique Proof Game
Genre: Retro
FEN: Q3K1nr/2p1pp1p/1p1p2p1/p3n2b/1P1r2PP/1Nb1RP1q/P1PPP3/RNB1k3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2014-05-21 more...
Genre: Retro
FEN: Q3K1nr/2p1pp1p/1p1p2p1/p3n2b/1P1r2PP/1Nb1RP1q/P1PPP3/RNB1k3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2014-05-21 more...
1. h4 d5 2. h5 d4 3. h6 d3 4. hxg7 h5 5. g4 h4 6. g5 h3 7. e4 h2 8. Sh3 Th6 9. Tg1 h1=D 10. e5 Dc6 11. Lg2 Dc3 12. dxc3 Tc6 13. e6 Sh6 14. g8=L Lg7 15. exf7 Kf8 16. Sd2 e5 17. Sf3 e4 18. Sh4 d2 19. Ke2 e3 20. Kf3 e2 21. g6 e1=D 22. Tb1 Dee7 23. De1 d1=T 24. Lg5 Da3 25. bxa3 Td4 26. Tb3 Tb4 27. cxb4 Lb2 28. Le7 Kg7 29. f8=S Sa6 30. Lc4 Sf7 31. Sd7 Kh6 32. Sb6 cxb6 33. g7 Sc7 34. La6 bxa6 35. g8=T Lb7 36. Tg5 Tc8 37. Ta5 bxa5
Cook: 1. h4 d5 2. h5 e5 3. h6 f5 4. hxg7 h5 5. Sa3 h4 6. Sc4 h3 7. Sb6 h2 8. e4 Th3 9. exf5 Ta3 10. Sh3 Sa6 11. Tg1 h1=D 12. g4 Df3 13. f6 Dc3 14. dxc3 cxb6 15. Tb1 Sc7 16. La6 bxa6 17. bxa3 Lb4 18. cxb4 Sh6 19. g8=L d4 20. Ld5 Kf8 21. Lg2 e4 22. Ke2 e3 23. Kf3 e2 24. f7 Kg7 25. f8=S e1=L 26. Sg6 Lc3 27. Sh4 d3 28. g5 d2 29. De1 d1=T 30. g6 Td6 31. Lg5 Tc6 32. Le7 Sf7 33. Tb3 Kh6 34. g7 Lb2 35. g8=T Lb7 36. Tg5 Tc8 37. Ta5 bxa5
Cook: 1. h4 d5 2. h5 e5 3. h6 f5 4. hxg7 h5 5. Sa3 h4 6. Sc4 h3 7. Sb6 h2 8. e4 Th3 9. exf5 Ta3 10. Sh3 Sa6 11. Tg1 h1=D 12. g4 Df3 13. f6 Dc3 14. dxc3 cxb6 15. Tb1 Sc7 16. La6 bxa6 17. bxa3 Lb4 18. cxb4 Sh6 19. g8=L d4 20. Ld5 Kf8 21. Lg2 e4 22. Ke2 e3 23. Kf3 e2 24. f7 Kg7 25. f8=S e1=L 26. Sg6 Lc3 27. Sh4 d3 28. g5 d2 29. De1 d1=T 30. g6 Td6 31. Lg5 Tc6 32. Le7 Sf7 33. Tb3 Kh6 34. g7 Lb2 35. g8=T Lb7 36. Tg5 Tc8 37. Ta5 bxa5
A.Buchanan: The unsoundness in this non-unique PG is that 6 promotions (apparently intended) are not forced (2023-05-11)
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Keywords: Ceriani-Frolkin Theme (dLdtST), Promotion, Non-Unique Proof Game, konsekutive Umwandlungen 6
Genre: Retro
Computer test: Computerprüfung: many solutions Stelvio 1.11 48 Sekunden. Keine Lösung: BP 36.0, BP 36.5.
FEN: 2rq4/pbn1Bn2/p1r4k/p7/1P5N/PR3K1N/PbP2PB1/4Q1R1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-11 more...
Genre: Retro
Computer test: Computerprüfung: many solutions Stelvio 1.11 48 Sekunden. Keine Lösung: BP 36.0, BP 36.5.
FEN: 2rq4/pbn1Bn2/p1r4k/p7/1P5N/PR3K1N/PbP2PB1/4Q1R1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-11 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)
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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) 1. cxd3ep Ta5 2. Sf5 0-0 3. Sd4 Txe5#
2) 1. gxf3ep Th5 2. Sd5 0-0-0 3. Sf4 Txe5#
Cook: 707 cooks which do not begin with e.p., e.g. 1.Sg3-e2 Ra1-a2 2.Ke4-d3 Rh1-h3 3.e5-e4 Ra2-d2 #
What is the obscure intent behind "Einheitslösung!"? (= "One size fits all solution"?)
2) 1. gxf3ep Th5 2. Sd5 0-0-0 3. Sf4 Txe5#
Cook: 707 cooks which do not begin with e.p., e.g. 1.Sg3-e2 Ra1-a2 2.Ke4-d3 Rh1-h3 3.e5-e4 Ra2-d2 #
What is the obscure intent behind "Einheitslösung!"? (= "One size fits all solution"?)
Mario Richter: Im Nachdruck in problem 144-147 steht nur auf e5 ein sB (also keiner auf e3). Allerdings gibt es in allen drei Varianten (sBe5+e3, nur sBe5, nur sBe3) jede Menge NL, die ohne Rochade und e.p.-Key auskommen - am wenigsten dann, wenn nur auf e3 ein sB steht. (2010-10-10)
VL: Cf. P0002475. (2012-08-26)
A.Buchanan: Possibly "Einheitlösung!" is a request to unify retro implications derived across separate parts of the PRA solution. But cooked chessically irrespective of these philosophical issues, which still wait to be resolved 60 years on. (2022-02-16)
comment
VL: Cf. P0002475. (2012-08-26)
A.Buchanan: Possibly "Einheitlösung!" is a request to unify retro implications derived across separate parts of the PRA solution. But cooked chessically irrespective of these philosophical issues, which still wait to be resolved 60 years on. (2022-02-16)
comment
Keywords: a posteriori (AP) (Type Petrovic - ccee), En passant as key (2), Castling (wb), Symmetrical position, Symmetrical solution, Superseded by (P1401449,P1399112)
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/8/8/4p3/2pPkPp1/2n1p1n1/8/R3K2R
Reprints: (66) Problem 144, p. 52, 12/1971
Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/8/8/4p3/2pPkPp1/2n1p1n1/8/R3K2R
Reprints: (66) Problem 144, p. 52, 12/1971
Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
18 - 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)
comment
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. bxc3ep Lxe2 2. Sa3 0-0-0#
Cook: 2. ... Td1#
Cook: 2. ... Td1#
Sally: Der letzte Zug war: Bc2 - c4!
Nr.138 200 Ausgewählte S .Probleme T. Kardos(W. Fentze1983) (2010-09-30)
A.Buchanan: All 9+1 captures are visible by pawns. Note R: 1. Kd1-e1 d3xe2 is illegal because it costs 2 more captures for Black. So e.p. is unconditionally legal, requiring no AP justification. I see no reason why White has lost castling rights. Indeed someone might make a demo game ending with c2-c4. So this one seems cooked. Neither WinChloe nor yacpdb contains this problem. (2022-01-08)
A.Buchanan: If this is AP, it will be rendered sound by removing wPa2. But is this the intention? Does anyone have access to the collection of selected Kardos problems mentioned? (2024-01-25)
more ...
comment
Nr.138 200 Ausgewählte S .Probleme T. Kardos(W. Fentze1983) (2010-09-30)
A.Buchanan: All 9+1 captures are visible by pawns. Note R: 1. Kd1-e1 d3xe2 is illegal because it costs 2 more captures for Black. So e.p. is unconditionally legal, requiring no AP justification. I see no reason why White has lost castling rights. Indeed someone might make a demo game ending with c2-c4. So this one seems cooked. Neither WinChloe nor yacpdb contains this problem. (2022-01-08)
A.Buchanan: If this is AP, it will be rendered sound by removing wPa2. But is this the intention? Does anyone have access to the collection of selected Kardos problems mentioned? (2024-01-25)
more ...
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1. dxe3ep+ Kxc4 2. Lc3 Sxc2 3. Da5 axb3#
See the correction P1396159
Cook: 1. dxe3ep+ Kxc4 2. Td5 Sxc2 3. Ta5 axb3#
See the correction P1396159
Cook: 1. dxe3ep+ Kxc4 2. Td5 Sxc2 3. Ta5 axb3#
A.Buchanan: Easy enough to eliminate what I suppose is the intended solution (sL unblocking sD and blocking sT) by e.g. just removing sDf5. However I don't see a way to say goodbye to the cook instead. (2021-11-23)
James Malcom: I believe I may have a fix Andrew. If we substitute the Black queen with a Black rook, the same theme is kept even is if considered watered down.
h#3 8/3p4/3p1p2/3Kbr2/k1ppPp2/1n1r3P/P1p4P/N3nq1b (2021-11-24)
A.Buchanan: Hi James, hurray cool I thought I couldn't get through that way: I was confused in my mind about wPe2 blocking that line, but with the freshness of a new day, I can look at it again. The precise diagram that you propose doesn't quite work because R: 1. e3-e4 Sf3*e1+,Sg2*e1+,g2xf1=Q+,g2xh1=Q+, but this is all fixable with +bPd2,-bSe1. Now Popeye 4.87 says the forward solution is works uniquely, and Retractor 2.0 says R: 1. e2-e4 Rf3-d3+ 2. Ke4-d5 are unique. So for full C+ we only need some demo game that results in the position prior to Ke4-d5.
I don't see any "watering down" here: I think we have faithfully represented Tivadar's idea. Let's post it here as TK, corr JM&AB? (2021-11-24)
A.Buchanan: I have made an unedifying demo game with lichess, so our correction is fully C+. (2021-11-24)
James Malcom: "TK, corr JM&AB" sounds good to me. (2021-11-24)
more ...
comment
James Malcom: I believe I may have a fix Andrew. If we substitute the Black queen with a Black rook, the same theme is kept even is if considered watered down.
h#3 8/3p4/3p1p2/3Kbr2/k1ppPp2/1n1r3P/P1p4P/N3nq1b (2021-11-24)
A.Buchanan: Hi James, hurray cool I thought I couldn't get through that way: I was confused in my mind about wPe2 blocking that line, but with the freshness of a new day, I can look at it again. The precise diagram that you propose doesn't quite work because R: 1. e3-e4 Sf3*e1+,Sg2*e1+,g2xf1=Q+,g2xh1=Q+, but this is all fixable with +bPd2,-bSe1. Now Popeye 4.87 says the forward solution is works uniquely, and Retractor 2.0 says R: 1. e2-e4 Rf3-d3+ 2. Ke4-d5 are unique. So for full C+ we only need some demo game that results in the position prior to Ke4-d5.
I don't see any "watering down" here: I think we have faithfully represented Tivadar's idea. Let's post it here as TK, corr JM&AB? (2021-11-24)
A.Buchanan: I have made an unedifying demo game with lichess, so our correction is fully C+. (2021-11-24)
James Malcom: "TK, corr JM&AB" sounds good to me. (2021-11-24)
more ...
comment
Keywords: En passant as key, Superseded by (P1396159)
Genre: h#, Retro
Computer test: Popeye C-Version 4.87 & trivial retro-logic
FEN: 8/3p4/3p1p2/3KbqP1/k1ppPpp1/1n1r3P/P1p4P/N6b
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-24 more...
Genre: h#, Retro
Computer test: Popeye C-Version 4.87 & trivial retro-logic
FEN: 8/3p4/3p1p2/3KbqP1/k1ppPpp1/1n1r3P/P1p4P/N6b
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-24 more...
1. exf3ep e3 2. 0-0-0 Tb4 3. Tg4 Tb8#
I think that this cooked problem is an early A osteriori. I can see ways that it might be made sound, but I would like to see the published version. San anyone help please? Thanks.
Cook: 17 Black pawns, retro cook and numerous forward cooks
I think that this cooked problem is an early A osteriori. I can see ways that it might be made sound, but I would like to see the published version. San anyone help please? Thanks.
Cook: 17 Black pawns, retro cook and numerous forward cooks
klären wK im Schach, vielleicht wBb2?
Alfred Pfeiffer: 9 schwarze Bauern! (2012-02-07)
Ladislav Packa: Auch mit wBb2 NL, z.B. 1.exf3 e.p. Txg5 2.0-0-0 Tb5 3.Tg~ Tb8# (2012-02-07)
A.Buchanan: And adding to earlier comments, if we do swap sBb2 for wB, the retraction is still not unique with R: 1. f3-f4 Lh6xg5+ (2021-11-26)
A.Buchanan: I think that this cooked problem is an early A Posteriori. I can see ways that it might be made sound, but I would like to see the published version. Can anyone help please? Thanks. (2022-04-25)
Gerald Ettl: Verbesserungsvorschlag: -sBa2, -sBb2, -sSg7, +sBg7
r3k3/3b2p1/5p2/6b1/4pPRp/2pq2rp/2p1P1pB/2K3n1/ (2022-04-25)
Gerald Ettl: und +sSg8 (wegen exf4 Vermeidung) (2022-04-25)
Gerald Ettl: +sSa4 nicht g8 (2022-04-25)
A.Buchanan: Hi Gerald. Thanks for this. Your final proposed version is r3k3/3b2p1/5p2/6b1/n3pPRp/2pq2rp/2p1P1pB/2K3n1, yes? There are 16 candidate solutions, all using castling. But wPa cannot leave the a-file, and is required for capture balance, so Black cannot castle. Five of the candidates begin with e.p.: play might have just been R: 1. Kb1-c1 b3xPc2+, so e.p. is not permitted in any case. What am I missing? (2022-04-25)
Gerald Ettl: Hi Andrew,
Those were 2 mistakes I made. I improved it again.
r3k3/3b4/4pp2/6b1/4pPRp/n1pq2rp/PPp1P1pB/2K3n1/
(den wBa2 habe ich auf das Brett gestellt, da sonst wieder Lh6xXg5 geht.) (2022-04-26)
A.Buchanan: Hi Gerald: Deine Retro-Logik ist gut. Aber es gibt 18 Lösungsvorschläge für die Zukunft. Ich habe heute im Discord meinen eigenen Vorschlag zur AP-Korrektur von Kardos veröffentlicht. Ich werde ihn hier im PDB hinzufügen. (2022-04-26)
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Alfred Pfeiffer: 9 schwarze Bauern! (2012-02-07)
Ladislav Packa: Auch mit wBb2 NL, z.B. 1.exf3 e.p. Txg5 2.0-0-0 Tb5 3.Tg~ Tb8# (2012-02-07)
A.Buchanan: And adding to earlier comments, if we do swap sBb2 for wB, the retraction is still not unique with R: 1. f3-f4 Lh6xg5+ (2021-11-26)
A.Buchanan: I think that this cooked problem is an early A Posteriori. I can see ways that it might be made sound, but I would like to see the published version. Can anyone help please? Thanks. (2022-04-25)
Gerald Ettl: Verbesserungsvorschlag: -sBa2, -sBb2, -sSg7, +sBg7
r3k3/3b2p1/5p2/6b1/4pPRp/2pq2rp/2p1P1pB/2K3n1/ (2022-04-25)
Gerald Ettl: und +sSg8 (wegen exf4 Vermeidung) (2022-04-25)
Gerald Ettl: +sSa4 nicht g8 (2022-04-25)
A.Buchanan: Hi Gerald. Thanks for this. Your final proposed version is r3k3/3b2p1/5p2/6b1/n3pPRp/2pq2rp/2p1P1pB/2K3n1, yes? There are 16 candidate solutions, all using castling. But wPa cannot leave the a-file, and is required for capture balance, so Black cannot castle. Five of the candidates begin with e.p.: play might have just been R: 1. Kb1-c1 b3xPc2+, so e.p. is not permitted in any case. What am I missing? (2022-04-25)
Gerald Ettl: Hi Andrew,
Those were 2 mistakes I made. I improved it again.
r3k3/3b4/4pp2/6b1/4pPRp/n1pq2rp/PPp1P1pB/2K3n1/
(den wBa2 habe ich auf das Brett gestellt, da sonst wieder Lh6xXg5 geht.) (2022-04-26)
A.Buchanan: Hi Gerald: Deine Retro-Logik ist gut. Aber es gibt 18 Lösungsvorschläge für die Zukunft. Ich habe heute im Discord meinen eigenen Vorschlag zur AP-Korrektur von Kardos veröffentlicht. Ich werde ihn hier im PDB hinzufügen. (2022-04-26)
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1. dxc3ep dxc3+ 2. Ka5 c4 3. d2 c5 4. d1=L c6 5. Lg4 hxg4 6. Kb6 g5 7. Ka7 g6 8. Kb8 g7 9. Kc8 g8=D/T#
Cook: 1. b2 c5 2. b1=S c6 3. Sc3 dxc3+ 4. Ka5 cxd4 5. Kb6 d5 6. Ka7 d6 7. Kb8 dxc7+ 8. Ka7 c8=D 9. a5 Db7#
Cook: 1. b2 c5 2. b1=S c6 3. Sc3 dxc3+ 4. Ka5 cxd4 5. Kb6 d5 6. Ka7 d6 7. Kb8 dxc7+ 8. Ka7 c8=D 9. a5 Db7#
paul: See P0003212 as version. (2011-08-06)
Anton Baumann: C+?? Die Aufgabe ist eindeutig inkorrekt! (2021-01-27)
James Malcom: Fixed. (2021-01-27)
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Anton Baumann: C+?? Die Aufgabe ist eindeutig inkorrekt! (2021-01-27)
James Malcom: Fixed. (2021-01-27)
comment
Keywords: En passant as key, Promotion (D), konsekutive Umwandlungen 2 (L, D/T), En passant, Kindergarten Problem
Genre: h#, Retro
FEN: 8/2p5/p7/8/1kPp3p/1p1p2pP/3P2Pp/7K
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-01-27 more...
Genre: h#, Retro
FEN: 8/2p5/p7/8/1kPp3p/1p1p2pP/3P2Pp/7K
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-01-27 more...
23 - P0003339
Tivadar Kardos
2519 Revista Romana de Sah 01/1948
2. Preis
(10+15) cooked
h# in wieviel?
Tivadar Kardos
2519 Revista Romana de Sah 01/1948
2. Preis
(10+15) cooked
h# in wieviel?
1. ... cxd6ep 2. Tc8 dxe7 3. Tf8 exf8=S#
AL ('Revista Romana de Sah', 06/1948, S.225-226):
Bekanntlich beginnt im Hilfsmatt Schwarz, also hat in der Diagrammstellung Weiß den letzten Zug gemacht. Man findet aber, wenn man die Position analysiert, daß Weiß den letzten Zug nicht gemacht haben kann.
Der weiße Doppelbauer auf der d-Linie läßt sich nur durch Schlagen des einzig fehlenden schwarzen Steines, des sBa7, erklären. Dieser konnte aber erst nach Umwandlung geschlagen werden. Dazu mußte er zunächst bis a3 vorrücken, dann nach b2 schlagen, und dann umwandeln.
Also kann Weiß nicht zuletzt b2-b3 gezogen haben.
Die anderen schwarzen Bauern haben mindestens 5 Schläge gemacht, um die Position im Diagramm zu erreichen (man beachte, daß der sBh7 zweimal schlagen mußte, um den wBh2 durchzulassen)
also kann auch g2-g3 nicht der letzte Zug gewesen sein, denn sonst hätte der wLf1 nicht von einem sB geschlagen worden sein können.
Für alle anderen weißen Steine ist leicht zu sehen, daß sie den letzten Zug nicht gemacht haben können.
Es stellt sich heraus, dass der letzte Zug von Schwarz gemacht wurde, und eine einfache Analyse zeigt, daß dies nur möglich ist durch d7-d5 (sonst wäre es unmöglich, daß die weißen Steine ihre Position im Diagramm erreichen könnten).
In diesem Fall Weiß kann 1. ... c5:d6(e.p.) spielen, gefolgt von 2. Tc8 d:e7 3. Tf8 e:f8C matt!
Das kürzeste Hilfsmatt wird also in 2,5 Zügen erreicht!
Cook: 1. ... Txg5+ 2. Kxg5 f3 3. Lg6 gxf4#
1. ... Kxd5 2. Kf6 e4 3. Tg6,Dg6 e5#
AL ('Revista Romana de Sah', 06/1948, S.225-226):
Bekanntlich beginnt im Hilfsmatt Schwarz, also hat in der Diagrammstellung Weiß den letzten Zug gemacht. Man findet aber, wenn man die Position analysiert, daß Weiß den letzten Zug nicht gemacht haben kann.
Der weiße Doppelbauer auf der d-Linie läßt sich nur durch Schlagen des einzig fehlenden schwarzen Steines, des sBa7, erklären. Dieser konnte aber erst nach Umwandlung geschlagen werden. Dazu mußte er zunächst bis a3 vorrücken, dann nach b2 schlagen, und dann umwandeln.
Also kann Weiß nicht zuletzt b2-b3 gezogen haben.
Die anderen schwarzen Bauern haben mindestens 5 Schläge gemacht, um die Position im Diagramm zu erreichen (man beachte, daß der sBh7 zweimal schlagen mußte, um den wBh2 durchzulassen)
also kann auch g2-g3 nicht der letzte Zug gewesen sein, denn sonst hätte der wLf1 nicht von einem sB geschlagen worden sein können.
Für alle anderen weißen Steine ist leicht zu sehen, daß sie den letzten Zug nicht gemacht haben können.
Es stellt sich heraus, dass der letzte Zug von Schwarz gemacht wurde, und eine einfache Analyse zeigt, daß dies nur möglich ist durch d7-d5 (sonst wäre es unmöglich, daß die weißen Steine ihre Position im Diagramm erreichen könnten).
In diesem Fall Weiß kann 1. ... c5:d6(e.p.) spielen, gefolgt von 2. Tc8 d:e7 3. Tf8 e:f8C matt!
Das kürzeste Hilfsmatt wird also in 2,5 Zügen erreicht!
Cook: 1. ... Txg5+ 2. Kxg5 f3 3. Lg6 gxf4#
1. ... Kxd5 2. Kf6 e4 3. Tg6,Dg6 e5#
A.Buchanan: Popeye v.487. WinChloe has the same diagram, and although it just claims 1 solution, it's C? Yet WinChloe engine gives the cooks right away (2021-10-27)
Mario Richter: Shouldn't this problem get the Label "No legal last move for White"?
(White pawn d2 has captured the missing black piece (Pdxe), so black pawn a7 had to promote to get to the capture square. This requires one capture (Pa3xb2), together with bxc,cxd,hxg,gxh,gxh this accounts for all missing white pieces.)
Btw. "for me, "C?" stands for "Not yet computer-tested or computer test didn't come to a conlusion" ... (2021-10-28)
A.Buchanan: Hi Mario, thank you.
(1) The stipulation is non-standard. I think this is why the WinChloe database doesn't include a solution, and just marks it as "C?" However if I run the WinChloe solving engine manually it (like Popeye) gives 121 h#3 solutions and 4 h#2.5 solutions. I think you, me and Christian would all agree on the meaning of "C?"
(2) The problem does not rely on Article 15 to eliminate the first single move, but implicitly requests the shortest. Currently the PDB usage of "No legal last move..." is to indicate that Codex Article 15 applies, so I did not add it. But I'm ok if you want me to add the keyword. In my correction, I shift the stipulation to h#3 so it's clear that Article 15 and not any desire for brevity which push the problem to h#2.5.
(3) However the PDB usage is confused. Ideallly think the two "No legal..." keywords should be replaced by "Retro-stalemate", while two new keywords "Article 15" & "Whose mate?" indicate the choice of implications of the retropat for forward play. "Board rotation" and other jokes are also implications of retropat (and of illegal diagram).
(4) But this is complicated to implement, requiring either database-level access or screen-scraping. The latter is quite practical for some future time when I have more leisure.
(5) https://www.stere.ro/biblioteca-digitala/reviste/revista-romana-de-sah/ is an excellent archive of Revista Romana de Sah. My sincere compliments and thanks to whoever prepared it. And the magazines themselves contain thorough annual indices. p225-6 of 6/1948, contains the solution to Kardos. I haven't translated it, nor have I found time to locate the prize announcement.
Comments welcome! (2021-10-29)
Mario Richter: I still believe, that the label "No legal last move for White" is justified here, because this is exactly what the author wanted to show.
(And this is too, what the reprint in 'Problem (Zagreb)' 37-40 09/1956 is all about - it appears there inside asn article
Btw., I wouldn't call the presentation of the solution in 'Revista Romana de Sah', p.225, 06/1948, a "reprint" - it lacks the presentation of the position ... (2021-10-30)
A.Buchanan: Added the keyword.
Yes I know that's the usage of "reprint", but then there should be a field for the location of the solution :-) It's idiotic that if I want to say where the solution is contained, I have to embed it in text like this: 2519 Revista Romana de Sah , p. 225-6, 06/1948 (2021-10-30)
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Mario Richter: Shouldn't this problem get the Label "No legal last move for White"?
(White pawn d2 has captured the missing black piece (Pdxe), so black pawn a7 had to promote to get to the capture square. This requires one capture (Pa3xb2), together with bxc,cxd,hxg,gxh,gxh this accounts for all missing white pieces.)
Btw. "for me, "C?" stands for "Not yet computer-tested or computer test didn't come to a conlusion" ... (2021-10-28)
A.Buchanan: Hi Mario, thank you.
(1) The stipulation is non-standard. I think this is why the WinChloe database doesn't include a solution, and just marks it as "C?" However if I run the WinChloe solving engine manually it (like Popeye) gives 121 h#3 solutions and 4 h#2.5 solutions. I think you, me and Christian would all agree on the meaning of "C?"
(2) The problem does not rely on Article 15 to eliminate the first single move, but implicitly requests the shortest. Currently the PDB usage of "No legal last move..." is to indicate that Codex Article 15 applies, so I did not add it. But I'm ok if you want me to add the keyword. In my correction, I shift the stipulation to h#3 so it's clear that Article 15 and not any desire for brevity which push the problem to h#2.5.
(3) However the PDB usage is confused. Ideallly think the two "No legal..." keywords should be replaced by "Retro-stalemate", while two new keywords "Article 15" & "Whose mate?" indicate the choice of implications of the retropat for forward play. "Board rotation" and other jokes are also implications of retropat (and of illegal diagram).
(4) But this is complicated to implement, requiring either database-level access or screen-scraping. The latter is quite practical for some future time when I have more leisure.
(5) https://www.stere.ro/biblioteca-digitala/reviste/revista-romana-de-sah/ is an excellent archive of Revista Romana de Sah. My sincere compliments and thanks to whoever prepared it. And the magazines themselves contain thorough annual indices. p225-6 of 6/1948, contains the solution to Kardos. I haven't translated it, nor have I found time to locate the prize announcement.
Comments welcome! (2021-10-29)
Mario Richter: I still believe, that the label "No legal last move for White" is justified here, because this is exactly what the author wanted to show.
(And this is too, what the reprint in 'Problem (Zagreb)' 37-40 09/1956 is all about - it appears there inside asn article
Btw., I wouldn't call the presentation of the solution in 'Revista Romana de Sah', p.225, 06/1948, a "reprint" - it lacks the presentation of the position ... (2021-10-30)
A.Buchanan: Added the keyword.
Yes I know that's the usage of "reprint", but then there should be a field for the location of the solution :-) It's idiotic that if I want to say where the solution is contained, I have to embed it in text like this: 2519 Revista Romana de Sah , p. 225-6, 06/1948 (2021-10-30)
more ...
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Keywords: En passant as key, Promotion (S), Superseded by (P1394978), No legal last move for White
Genre: h#, Retro
Computer test: Popeye v4.87 & WinChloe both say cooked
FEN: 7b/4p1nP/2r1P1kq/1nPpKbrp/2pp1pRp/1P4P1/P3PP2/8
Reprints: (III) Problem 37-40 09/1956
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-30 more...
Genre: h#, Retro
Computer test: Popeye v4.87 & WinChloe both say cooked
FEN: 7b/4p1nP/2r1P1kq/1nPpKbrp/2pp1pRp/1P4P1/P3PP2/8
Reprints: (III) Problem 37-40 09/1956
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-30 more...
1. hxg3ep Lxb1 2. Kf5 fxe4 3. Kg6 e5+ 4. f5 exf6ep#
Cook: NL:
1. hxg3ep fxe4 2. f6 e5 3. Kf5 exf6 4. Kg6 Lxb1# uvm
Cook: NL:
1. hxg3ep fxe4 2. f6 e5 3. Kf5 exf6 4. Kg6 Lxb1# uvm
YM: The correct mechanism is in P1109109 (2010-07-13)
YM: Correction option: P1229434 (2021-06-28)
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YM: Correction option: P1229434 (2021-06-28)
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Keywords: En passant as key, Superseded by (P1109109)
Genre: h#, Retro
FEN: q2r4/4Kp2/3r3b/3p2pp/4nkPp/1n3P2/B4p1p/1b6
Reprints: 805 FIDE Album 1980-1982 1988
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-28 more...
Genre: h#, Retro
FEN: q2r4/4Kp2/3r3b/3p2pp/4nkPp/1n3P2/B4p1p/1b6
Reprints: 805 FIDE Album 1980-1982 1988
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-28 more...
1) 1. axb3ep bxc6+ 2. b5 cxb6ep#
2) 1. exd3ep g8=D,L+ 2. d5 cxd6ep#
PRA: 1 solution with 2 parts
2) 1. exd3ep g8=D,L+ 2. d5 cxd6ep#
PRA: 1 solution with 2 parts
Henrik Juel: White captured [sLc8] on c8 and axb, so last move was either b2-b4 or d2-d4
C+ Popeye 4.61, except for the promotion dual (2020-08-03)
A.Buchanan: There's no keyword for an untolerated dual promotion! (2020-08-03)
Henrik Juel: Yes and no, Andrew
What about the cooked label? (2020-08-03)
A.Buchanan: Hi Henrik: this is obviously intentional. I tried to replace the pawn promotion with wDh8-h7+ but I ran out of pieces. I guess that Bebesi had the same difficulty. The word "cooked" seems too strong to me, but it's obviously a significant defect though (2020-08-04)
A.Buchanan: The bar is higher for helpmates than other problems, and this is unsound. However it’s not cooked but dualled. Yet all we have is “cooked” so I suppose it is in PDB-speak. I’ve added that and removed the 2.1… (2023-08-07)
Joost de Heer: Is there an 'unsound' keyword? (2023-08-07)
A.Buchanan: Hi Joost: no there isn't. It can be added, the challenge for any new common keyword would be populating it for more than a few values. (2023-08-07)
A.Buchanan: This time round, I've managed to find a fix. Bebesi's correction in WinChloe 66992 has non-standard wL. I have a feeling I've seen a similar one by Andrey Frolkin & someone else recently with a different matrix, and also non-standard wL. I've sent my effort to Joaquim Crusats at Problemas for his assessment (2023-08-08)
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C+ Popeye 4.61, except for the promotion dual (2020-08-03)
A.Buchanan: There's no keyword for an untolerated dual promotion! (2020-08-03)
Henrik Juel: Yes and no, Andrew
What about the cooked label? (2020-08-03)
A.Buchanan: Hi Henrik: this is obviously intentional. I tried to replace the pawn promotion with wDh8-h7+ but I ran out of pieces. I guess that Bebesi had the same difficulty. The word "cooked" seems too strong to me, but it's obviously a significant defect though (2020-08-04)
A.Buchanan: The bar is higher for helpmates than other problems, and this is unsound. However it’s not cooked but dualled. Yet all we have is “cooked” so I suppose it is in PDB-speak. I’ve added that and removed the 2.1… (2023-08-07)
Joost de Heer: Is there an 'unsound' keyword? (2023-08-07)
A.Buchanan: Hi Joost: no there isn't. It can be added, the challenge for any new common keyword would be populating it for more than a few values. (2023-08-07)
A.Buchanan: This time round, I've managed to find a fix. Bebesi's correction in WinChloe 66992 has non-standard wL. I have a feeling I've seen a similar one by Andrey Frolkin & someone else recently with a different matrix, and also non-standard wL. I've sent my effort to Joaquim Crusats at Problemas for his assessment (2023-08-08)
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Keywords: En passant as key (2), Partial Retro Analysis (PRA), En passant as mating move (2), Superseded by (P1411659, P1413906)
Genre: h#, Retro
FEN: 3qn3/1p1p2P1/B1r3p1/1PP1Kp2/pPkPp3/p1p4n/N7/2b5
Reprints: 233 Ungarische Schachproblem Anthologie , p. 188, 1983
1 Problemas 44, p. 1456, 10/2023
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-04 more...
Genre: h#, Retro
FEN: 3qn3/1p1p2P1/B1r3p1/1PP1Kp2/pPkPp3/p1p4n/N7/2b5
Reprints: 233 Ungarische Schachproblem Anthologie , p. 188, 1983
1 Problemas 44, p. 1456, 10/2023
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-04 more...
1. dxe3ep Td1? 2. Le4 Txd6 3. Sg4 Tf6# usual AP try
1. dxe3ep 0-0-0! 2. Le4 Txd6 3. Sg4 Tf6#
Cook: 1. dxe3ep 0-0-0 2. Lf5 Txd6 3. Sf3 Td4#
not
1. dxe3ep Td1? 2. Lf5 Txd6 3. Sf3+ ?? usual AP try and also check Bl 3rd move
1. dxe3ep 0-0-0! 2. Le4 Txd6 3. Sg4 Tf6#
Cook: 1. dxe3ep 0-0-0 2. Lf5 Txd6 3. Sf3 Td4#
not
1. dxe3ep Td1? 2. Lf5 Txd6 3. Sf3+ ?? usual AP try and also check Bl 3rd move
Henrik Juel: The missing black man is dark-squared, so last move was not fxe4. In the solution Td6 should read Txd6. There seems to be a 'variation': 0... dxe3ep 1.0-0-0 Le4 2.Txd6 Sg4 3.Tf6#, so maybe the stipulation should have 1.1;2.1;1.1 added. (2004-09-23)
A.Buchanan: Well spotted Henrik. Your kindness is legendary, but in the harsh world of helpmates, this counts as a cooked I believe. In fact, it's the 2.Lf5/3.Sf3 version which is the cook. Yours is the actual solution! This is because 3.Sf3 would check if White hadn't castled, and we want the castling to be because of the AP condition only, not for 2 reasons, which one might term a "logical dual". The problem can be trivially fixed by sticking sSe5 on f6. I have no idea whether it's a typo or a cook: WinChloe has the same diagram, and the same confusion between cook & intended solution. The fact that AP does not appear in the stipulation might be a factor. WinChloe as far as I can see has no concept of AP as a keyword, and just picks it up based on a substring of the stipulation. In other compositions, T.Petrovic wasn't shy of flashing the AP acronym in the stipulation, but as usual the detective work peters out in the absence of primary records. (2020-12-08)
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A.Buchanan: Well spotted Henrik. Your kindness is legendary, but in the harsh world of helpmates, this counts as a cooked I believe. In fact, it's the 2.Lf5/3.Sf3 version which is the cook. Yours is the actual solution! This is because 3.Sf3 would check if White hadn't castled, and we want the castling to be because of the AP condition only, not for 2 reasons, which one might term a "logical dual". The problem can be trivially fixed by sticking sSe5 on f6. I have no idea whether it's a typo or a cook: WinChloe has the same diagram, and the same confusion between cook & intended solution. The fact that AP does not appear in the stipulation might be a factor. WinChloe as far as I can see has no concept of AP as a keyword, and just picks it up based on a substring of the stipulation. In other compositions, T.Petrovic wasn't shy of flashing the AP acronym in the stipulation, but as usual the detective work peters out in the absence of primary records. (2020-12-08)
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Keywords: En passant as key, Castling (wg), a posteriori (AP) (Type Petrovic), Superseded by (P1382802)
Genre: h#, Retro
Computer test: Popeye v4.85 + retro thinking
FEN: 1n1r4/p1B4p/1p1p4/p3n1p1/3pPk2/3b4/pr5q/R3K3
Reprints: (60) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-16 more...
Genre: h#, Retro
Computer test: Popeye v4.85 + retro thinking
FEN: 1n1r4/p1B4p/1p1p4/p3n1p1/3pPk2/3b4/pr5q/R3K3
Reprints: (60) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-16 more...
1. cxb3ep+ c3 2. Ld3 Tg4+ 3. Kh1 0-0-0#
paul: White captures was axb and g2xh3, so the retro move c3xb4 is not possible. If b3-b4, the retro check is not justified. So last move was b2-b4 (preceded by Rc3-a3). (2011-08-06)
A.Buchanan: There's a lot to unpack here. Firstly, the intended solution has a definite transposition dual in the order of Black's 2nd & 3rd moves. Second, the pawn capture count is wrong. What happened to the original wBh, given sBh2 never left the h-file? Thirdly, and not a showstopper, White's only solution involves castling: so that's a logical dual. (2020-12-08)
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A.Buchanan: There's a lot to unpack here. Firstly, the intended solution has a definite transposition dual in the order of Black's 2nd & 3rd moves. Second, the pawn capture count is wrong. What happened to the original wBh, given sBh2 never left the h-file? Thirdly, and not a showstopper, White's only solution involves castling: so that's a logical dual. (2020-12-08)
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28 - P0003444
Janko Furman
Miroslav Stosic
7273 Schach-Echo 11/1972
(8+14) cooked
h#2
b) Gespiegelt (a1<->h1) & wKd1->e1
Janko Furman
Miroslav Stosic
7273 Schach-Echo 11/1972
(8+14) cooked
h#2
b) Gespiegelt (a1<->h1) & wKd1->e1
a) 1. dxe3ep 0-0 2. Te4 Txf3#
b) 1. exd3ep 0-0-0 2. dxe2 Ld5#
Cook: a) 1. Kg3 Kf1 2. Df4 Th3#
1. Kg3 0-0 2. Dh4 Txf3#
b) 1. exd3ep 0-0-0 2. dxe2 Ld5#
Cook: a) 1. Kg3 Kf1 2. Df4 Th3#
1. Kg3 0-0 2. Dh4 Txf3#
A.Buchanan: This is a very heavy position to prevent the possibility of R: 1. c2xb3. I'm not sure why this was done. A much lighter position 8/8/8/5np1/1r1pPkr1/2Bp1p2/1p1P2P1/4K2R with 7 less units achieves the mates soundly. Am I missing something? (2022-02-16)
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Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wk,wg), Superseded by (P1399967), Twinning by board reflection
Genre: h#, Retro
FEN: 8/1p1pp1p1/8/5nq1/1r1pPkr1/1PBp1p2/Pp1P2P1/2n1K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
Genre: h#, Retro
FEN: 8/1p1pp1p1/8/5nq1/1r1pPkr1/1PBp1p2/Pp1P2P1/2n1K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
1) 1. gxf3ep 0-0-0 2. Te7 Th4#
2) 1. cxd3ep 0-0 2. Da4 Txa4#
2) 1. cxd3ep 0-0 2. Da4 Txa4#
AB: I think this is cooked. I don't see how AP castling can justify earlier ep here.
1. Kf3 Bf5 2. Kg2 Be4#
1. Kf3 Bxd7 2. Kg2 Bc6#
1. cxd3 Rf1 2. Qa4 Rxa4#
1. gxf3ep Ra5 2. Kf4 Rh4#
1. gxf3ep Rd1 2. Re7 Rh4#
Also, second solution given has typos 1. *c*xd3ep & 2. Q*a*4. (2002-03-21)
V.Liskovets: Indeed this problem is cooked, and I
failed to correct it preserving symmetry.
Here is a possible correction:
W: Ke1 Ra1 Rh1 Be6 Pd4 Pf4
B: Ke4 Qd7 Rf7 Pc3 Pc4 Pe3 Pg3 Pg4 Bh7 Rb2 Pb3
Another story, justifying its contents (e.p.).
In my opinion, all published treatments are
insatisfactory (the same concerns P0004295
(corrected) by Werner Kuntsche as well).
There is a way to make it sound under the
sophisticated HYBRID stipulation 'AP, pRA':
2 partial solutions legalized JOINTLY by
both castlings (details to be published). (2002-04-02)
VL: Sorry, Andrew, I see no issues with my version (excepting the mentioned special retro-convention/genre for justifying its soundness, of course!). h#2 (pRA&AP). The full solution consists of 2 partial AP-based ones: I 1.gxf3 e.p.(!?) 0-0-0! 2.Te7 Th4#; II 1.cxd3 e.p.(!?) 0-0! 2.Da4 Txa4#.
C+(popeye): h#2 & two h#1.5 after the keys (added Tb2 & Lh7 are cookstoppers).
The main features of the corresponding suitable AP-genre ("consolidated")are rather clear. However, not all subtle aspects have been analyzed thoroughly yet. I have got only a draft manuscript with few examples, still. (2022-02-15)
A.Buchanan: Yes Valery I think your correction is technically sound. Theological debate is something else, heh. Have posted it as separate entry. Please tell me if it was original here, or appeared somewhere else (2022-05-29)
Henrik Juel: I looked in Stella Polaris 1971 p.202-203, where the problem is given without AP under the diagram
It turns out that both the author and the editor (Jan Knöppel) consider the problem correct
After explaining the AP convention the latter writes [my translation]:
The author has elegantly applied the AP conventionn in combination with partial analysis. If we assume that last move was not f2-f4 or T-h1, the possibilities d2-d4, K-e1, and T-a1 remain, but the two latter disappear by 1... 0-0. The other assumption (last move was not d2-d4 or T-a1) is analogous.
So something like Valery's HYBRID stipulation was the intention, and the problem was considered correct in 1971; later developments do not change this fact (2022-05-29)
Henrik Juel: Of course, the obvious cooks 1.Kf3 etc. remain (2022-05-29)
A.Buchanan: I agree. Basic AP Petrovic does not allow multiple ep, nor does RS, so some alignment of AP with PRA is required. I would like to see this codified though. (2022-05-29)
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1. Kf3 Bf5 2. Kg2 Be4#
1. Kf3 Bxd7 2. Kg2 Bc6#
1. cxd3 Rf1 2. Qa4 Rxa4#
1. gxf3ep Ra5 2. Kf4 Rh4#
1. gxf3ep Rd1 2. Re7 Rh4#
Also, second solution given has typos 1. *c*xd3ep & 2. Q*a*4. (2002-03-21)
V.Liskovets: Indeed this problem is cooked, and I
failed to correct it preserving symmetry.
Here is a possible correction:
W: Ke1 Ra1 Rh1 Be6 Pd4 Pf4
B: Ke4 Qd7 Rf7 Pc3 Pc4 Pe3 Pg3 Pg4 Bh7 Rb2 Pb3
Another story, justifying its contents (e.p.).
In my opinion, all published treatments are
insatisfactory (the same concerns P0004295
(corrected) by Werner Kuntsche as well).
There is a way to make it sound under the
sophisticated HYBRID stipulation 'AP, pRA':
2 partial solutions legalized JOINTLY by
both castlings (details to be published). (2002-04-02)
VL: Sorry, Andrew, I see no issues with my version (excepting the mentioned special retro-convention/genre for justifying its soundness, of course!). h#2 (pRA&AP). The full solution consists of 2 partial AP-based ones: I 1.gxf3 e.p.(!?) 0-0-0! 2.Te7 Th4#; II 1.cxd3 e.p.(!?) 0-0! 2.Da4 Txa4#.
C+(popeye): h#2 & two h#1.5 after the keys (added Tb2 & Lh7 are cookstoppers).
The main features of the corresponding suitable AP-genre ("consolidated")are rather clear. However, not all subtle aspects have been analyzed thoroughly yet. I have got only a draft manuscript with few examples, still. (2022-02-15)
A.Buchanan: Yes Valery I think your correction is technically sound. Theological debate is something else, heh. Have posted it as separate entry. Please tell me if it was original here, or appeared somewhere else (2022-05-29)
Henrik Juel: I looked in Stella Polaris 1971 p.202-203, where the problem is given without AP under the diagram
It turns out that both the author and the editor (Jan Knöppel) consider the problem correct
After explaining the AP convention the latter writes [my translation]:
The author has elegantly applied the AP conventionn in combination with partial analysis. If we assume that last move was not f2-f4 or T-h1, the possibilities d2-d4, K-e1, and T-a1 remain, but the two latter disappear by 1... 0-0. The other assumption (last move was not d2-d4 or T-a1) is analogous.
So something like Valery's HYBRID stipulation was the intention, and the problem was considered correct in 1971; later developments do not change this fact (2022-05-29)
Henrik Juel: Of course, the obvious cooks 1.Kf3 etc. remain (2022-05-29)
A.Buchanan: I agree. Basic AP Petrovic does not allow multiple ep, nor does RS, so some alignment of AP with PRA is required. I would like to see this codified though. (2022-05-29)
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Keywords: Castling (wb), En passant as key (2), a posteriori (AP) (Type Petrovic), Superseded by (P1401508, P1399178), Quasi-symmetrical position, Partial Retro Analysis (PRA)
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/3q1r2/4B3/8/2pPkPp1/2p1p1p1/8/R3K2R
Reprints: (65) Problem 144-147 12/1971
(8) StrateGems SG19, p. 156, 07/2002
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-29 more...
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/3q1r2/4B3/8/2pPkPp1/2p1p1p1/8/R3K2R
Reprints: (65) Problem 144-147 12/1971
(8) StrateGems SG19, p. 156, 07/2002
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-29 more...
30 - 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. 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...
1. ... b6 2. Le7 Lxg2+ 3. Kd6 Sxc4#
Cook: Popeye V4.87
1. ... Sxc4 2. Da1 Kxb3 3. Le5 Lxg2#
1. ... bxa5 2. Le5 Kb4 3. d6 Lxg2#
1. ... Sxd1,Sxd3 2. Le5 Sb2,Sf2 3. d6 Lxg2#
1. ... Sxc4 2. Le5 Sb2,Sd2 3. d6 Lxg2#
WinChloe has same diagram, also says is cooked.
Cook: Popeye V4.87
1. ... Sxc4 2. Da1 Kxb3 3. Le5 Lxg2#
1. ... bxa5 2. Le5 Kb4 3. d6 Lxg2#
1. ... Sxd1,Sxd3 2. Le5 Sb2,Sf2 3. d6 Lxg2#
1. ... Sxc4 2. Le5 Sb2,Sd2 3. d6 Lxg2#
WinChloe has same diagram, also says is cooked.
Henrik Juel: Black is missing [Lc8], which was captured on c8, and a rook
Last move was not c3xTb4, because sTc4's check is impossible, so White has the move (2020-10-01)
A.Buchanan: Shifting sLd6 to c5 removes the cooks. What's the artistic intent here though? (2021-10-16)
Henrik Juel: Maybe the unique h#2 try plays a role: 1.Le5 b6 2.d6 Lxg2# (2021-10-16)
A.Buchanan: In the h#2 try, b5-b6 is cool as the sole tempo move available to White. However there are plenty of *pairs* of White moves e.g. Sxd1-f2, which lose two tempi. These lead directly to h#2.5 cooks, and I see no way to eliminate these without losing the h#2 try. (2021-10-17)
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Last move was not c3xTb4, because sTc4's check is impossible, so White has the move (2020-10-01)
A.Buchanan: Shifting sLd6 to c5 removes the cooks. What's the artistic intent here though? (2021-10-16)
Henrik Juel: Maybe the unique h#2 try plays a role: 1.Le5 b6 2.d6 Lxg2# (2021-10-16)
A.Buchanan: In the h#2 try, b5-b6 is cool as the sole tempo move available to White. However there are plenty of *pairs* of White moves e.g. Sxd1-f2, which lose two tempi. These lead directly to h#2.5 cooks, and I see no way to eliminate these without losing the h#2 try. (2021-10-17)
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Keywords: No legal last move for White
Genre: h#, Retro
FEN: 8/1p1p4/3br3/pP1k4/KPbp4/pn1p4/qN4p1/3n3B
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-07 more...
Genre: h#, Retro
FEN: 8/1p1p4/3br3/pP1k4/KPbp4/pn1p4/qN4p1/3n3B
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-07 more...
1. Df2 2. Db6 3. Kd4 4. bxc3ep 5. Dg6 6. De4 Lb6
Cook: 1. Kd4 2. Kc3 3. Db2 4. d2 5. Ld3 6. d4 Lxb4#
Cook: 1. Kd4 2. Kc3 3. Db2 4. d2 5. Ld3 6. d4 Lxb4#
Keywords: Seriesmover, Consequent, En passant
Genre: Retro, Fairies
FEN: 8/8/p7/B2pp3/KpP1k3/Np1p4/q7/1b6
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
Genre: Retro, Fairies
FEN: 8/8/p7/B2pp3/KpP1k3/Np1p4/q7/1b6
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
1. axb6ep+ Kb5 2. bxc7+ Tb6 3. 0-0-0 ...
K. Fabel: "Von Interesse ist, dass diese strittige Idee [Beweis des e.p.-Schlagrechts durch spätere Ausführung der Rochade] auch im direkten Mattproblem dargestellt werden kann, vergl. das Diagramm [diese Aufgabe P0004199]. (Vielleicht ist diese Aufgabe noch nicht korrekt doch es wird nicht schwer sein, sie zu verbessern). Falls Schwarz zuletzt Bb6-b5, Tb6-c6 oder Lb6-c7 gezogen hat, ergibt sich, dass Weiss im letzten oder vorletzten Zug den K oder T bewegt haben muss. Falls jedoch K und T noch nicht gezogen haben, muss b7-b5 der letzte Zug gewesen sein. Weiss spielt daher 1. ab6e.p.+ Kb5 2. bc7+ Tb6. Jetzt könnte Weiss mit Dxb6 sofort mattsetzen, aber er muss ja den e.p.-Schlag noch legalisieren. Daher 3. 0-0-0! und Matt erst im 5. Zuge."
Cook: 1. Sxc7+!
1. Lxc7!
K. Fabel: "Von Interesse ist, dass diese strittige Idee [Beweis des e.p.-Schlagrechts durch spätere Ausführung der Rochade] auch im direkten Mattproblem dargestellt werden kann, vergl. das Diagramm [diese Aufgabe P0004199]. (Vielleicht ist diese Aufgabe noch nicht korrekt doch es wird nicht schwer sein, sie zu verbessern). Falls Schwarz zuletzt Bb6-b5, Tb6-c6 oder Lb6-c7 gezogen hat, ergibt sich, dass Weiss im letzten oder vorletzten Zug den K oder T bewegt haben muss. Falls jedoch K und T noch nicht gezogen haben, muss b7-b5 der letzte Zug gewesen sein. Weiss spielt daher 1. ab6e.p.+ Kb5 2. bc7+ Tb6. Jetzt könnte Weiss mit Dxb6 sofort mattsetzen, aber er muss ja den e.p.-Schlag noch legalisieren. Daher 3. 0-0-0! und Matt erst im 5. Zuge."
Cook: 1. Sxc7+!
1. Lxc7!
Innerhalb des 2. Teils einer Artikelserie "Die Konventionen im Problemschach" von Karl Fabel.
A.Buchanan: The retro logic is fine, but forward play is savagely cooked. I wonder about changing wSc8 to sS. The retro stuff still works OK, indeed sSc8 is a fourth thematic unit to retract in the try. Ignoring proof of ep legality, there is a unique #4 beginning with ep, and no #5. Promising: but when can we interpolate w000? Any solution must begin 1. axb6ep+ Kb5. There are now numerous #4, but none include w000, so I think we must stick with the still unique line beginning 2. Dxc8 thr 3. 0-0-0. There are 3 black defenses to refute this. 2. ... Txb6,axb3,d5. So maybe #6 needed for White to prevail? (2022-03-28)
A.Buchanan: But if we shift to #6 then 1. Sxc7!,Qxc8+! mate without ep. Can anyone retrieve the original diagram for this one, please? (2022-03-28)
Mario Richter: The position here is identical to the original diagram, but it should be regarded more as a schematic example than a "real problem". Fabel in his quote above: "... perhaps the problem is still not correct, but it should be easy to improve it ..." (2022-03-28)
Henrik Juel: unfortunately K. Fabel forgot to give the easy correction... (2022-03-28)
A.Buchanan: Thanks for this. Valery Liskovetz, an AP expert, was kind enough to send a pdf of the relevant Problem issue, so I can confirm. It's easy enough to remove the two cooks (e.g. sTc8) but the difficulty is in validation of the intended solution. Is there any "exact" option in Popeye that forces the solution to include a waiting move? The actual solution might be expected to be included in that set. (2022-03-29)
Henrik Juel: Yes, Andrew
'stip exact-#5' would disregard #4 as a solution, I believe (2022-03-29)
A.Buchanan: Even with the "exact", it's still very complicated to investigate. I think this kind of problem requires an engine feature to disallow any mate or pat if the specified castling has not been executed. Gazing at the 20,000 lines of output, I don't think sTc8 works as 1. axb6ep+ Kb5 2. Qb7 bxa3! threatens axb2+ after castling. I wonder if the board might be flipped left-right except for wK, but I haven't explored this. (2022-05-31)
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A.Buchanan: The retro logic is fine, but forward play is savagely cooked. I wonder about changing wSc8 to sS. The retro stuff still works OK, indeed sSc8 is a fourth thematic unit to retract in the try. Ignoring proof of ep legality, there is a unique #4 beginning with ep, and no #5. Promising: but when can we interpolate w000? Any solution must begin 1. axb6ep+ Kb5. There are now numerous #4, but none include w000, so I think we must stick with the still unique line beginning 2. Dxc8 thr 3. 0-0-0. There are 3 black defenses to refute this. 2. ... Txb6,axb3,d5. So maybe #6 needed for White to prevail? (2022-03-28)
A.Buchanan: But if we shift to #6 then 1. Sxc7!,Qxc8+! mate without ep. Can anyone retrieve the original diagram for this one, please? (2022-03-28)
Mario Richter: The position here is identical to the original diagram, but it should be regarded more as a schematic example than a "real problem". Fabel in his quote above: "... perhaps the problem is still not correct, but it should be easy to improve it ..." (2022-03-28)
Henrik Juel: unfortunately K. Fabel forgot to give the easy correction... (2022-03-28)
A.Buchanan: Thanks for this. Valery Liskovetz, an AP expert, was kind enough to send a pdf of the relevant Problem issue, so I can confirm. It's easy enough to remove the two cooks (e.g. sTc8) but the difficulty is in validation of the intended solution. Is there any "exact" option in Popeye that forces the solution to include a waiting move? The actual solution might be expected to be included in that set. (2022-03-29)
Henrik Juel: Yes, Andrew
'stip exact-#5' would disregard #4 as a solution, I believe (2022-03-29)
A.Buchanan: Even with the "exact", it's still very complicated to investigate. I think this kind of problem requires an engine feature to disallow any mate or pat if the specified castling has not been executed. Gazing at the 20,000 lines of output, I don't think sTc8 works as 1. axb6ep+ Kb5 2. Qb7 bxa3! threatens axb2+ after castling. I wonder if the board might be flipped left-right except for wK, but I haven't explored this. (2022-05-31)
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Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Non-standard material (TLL)
Genre: Retro, n#
FEN: NQNB4/B1brpp2/k1rp4/Ppp5/Rp6/BP6/RPP5/R3K3
Reprints: (8) Problem 101-102 09/1966
(52) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-29 more...
Genre: Retro, n#
FEN: NQNB4/B1brpp2/k1rp4/Ppp5/Rp6/BP6/RPP5/R3K3
Reprints: (8) Problem 101-102 09/1966
(52) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-29 more...
1. fxe3ep? Dxa8# No justification for the ep, so just retro try.
White pcs: dxexf5,hxBg. Black none. Assume that Black can still castle. White can't have just played h4xg5 as sBh2 would be blocked. White might apparently just have moved D,Se5,Td4,Tc3,g4-g5,a5-a6,e3-e4 or e2-e4 but what might Black have played before? Only the last allows a move sDf1-e1. So set up for AP Type Petrovic is OK.
Cook: 446 candidate h#2
However 27 have no ep, while 59 have both ep & castling.
So a total of 86 viable solutions, 85 of which must be cooks.
White pcs: dxexf5,hxBg. Black none. Assume that Black can still castle. White can't have just played h4xg5 as sBh2 would be blocked. White might apparently just have moved D,Se5,Td4,Tc3,g4-g5,a5-a6,e3-e4 or e2-e4 but what might Black have played before? Only the last allows a move sDf1-e1. So set up for AP Type Petrovic is OK.
Cook: 446 candidate h#2
However 27 have no ep, while 59 have both ep & castling.
So a total of 86 viable solutions, 85 of which must be cooks.
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004341)
Genre: h#, Retro
Computer test: Popeye v4.87 indicates cook
FEN: r3k3/p1ppp3/Pp6/4NPP1/2PRPp2/2RK2PN/1PBn1PQp/2Brq1b1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
Genre: h#, Retro
Computer test: Popeye v4.87 indicates cook
FEN: r3k3/p1ppp3/Pp6/4NPP1/2PRPp2/2RK2PN/1PBn1PQp/2Brq1b1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
1. fxe3ep Dxa8#? because no AP justification for ep
1. 0-0-0 g7 2. Tf8 gxf8=D#
62 apparent h#2, so seems cooked.
Cook: 415 candidate solutions for h#2. 401 begin with ep, of which 48 contain 0-0-0 to justify. There are also 14 solutions without ep, of which 13 begin with 0-0-0. The odd one is 1. Kd8 Sc6+ 2. Kc8 Sge7#
1. 0-0-0 g7 2. Tf8 gxf8=D#
62 apparent h#2, so seems cooked.
Cook: 415 candidate solutions for h#2. 401 begin with ep, of which 48 contain 0-0-0 to justify. There are also 14 solutions without ep, of which 13 begin with 0-0-0. The odd one is 1. Kd8 Sc6+ 2. Kc8 Sge7#
Version zu P0004296
Autor: "In Nr.9, I intended to do something similiar to No. 8 [P0004199], but with a different key: the solution 1. PxP "e.p.", Qxa8 checkmate will not go, for if blackside does not castle, the "en-passant" capture cannot be justified. Though, well considering, there is the demolition 1. PxP "e.p." -any 2. 0-0-0! Qb7 or a8 checkmate.
A.Buchanan: Thanks Mario for retrieving the author's intent with this. So he knew it was cooked! Assume s000 rights remain. R: 1. b5xa6? as sBb promoted on b1. R: 1. h4xg5? as sBh retro-blocked. R: 1. Ke3-d3? impossible check from sBf4. R. 1. K~-d3 d3-d2? illegal check. I like these! So by elimination, R: 1. e2-e4 e3xTd2 2. T~d2 L~e1. So must have sLe1. (2022-03-22)
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Autor: "In Nr.9, I intended to do something similiar to No. 8 [P0004199], but with a different key: the solution 1. PxP "e.p.", Qxa8 checkmate will not go, for if blackside does not castle, the "en-passant" capture cannot be justified. Though, well considering, there is the demolition 1. PxP "e.p." -any 2. 0-0-0! Qb7 or a8 checkmate.
A.Buchanan: Thanks Mario for retrieving the author's intent with this. So he knew it was cooked! Assume s000 rights remain. R: 1. b5xa6? as sBb promoted on b1. R: 1. h4xg5? as sBh retro-blocked. R: 1. Ke3-d3? impossible check from sBf4. R. 1. K~-d3 d3-d2? illegal check. I like these! So by elimination, R: 1. e2-e4 e3xTd2 2. T~d2 L~e1. So must have sLe1. (2022-03-22)
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Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004342)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k1N1/p1pp4/P5P1/4PPp1/2PNPprB/3K1QRp/P2p1PB1/4b2n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k1N1/p1pp4/P5P1/4PPp1/2PNPprB/3K1QRp/P2p1PB1/4b2n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
1. bxc3ep Dg8#? (thematic try)
1. bxc3ep Dd5 2. 0-0-0 Db7#
Cook: 1. bxc3ep Ld6 2. 0-0-0 Dc4#
1. bxc3ep Dd5 2. 0-0-0 Db7#
Cook: 1. bxc3ep Ld6 2. 0-0-0 Dc4#
Author: "From the former [d.h. P0004341], the No. 10 was born, something similar to No. 8 [P0004199] but with Black-castling for it is a help-mate. The mating move 1. ... Qg8? will not go."
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Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P1399966)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k3/p2p4/8/NP6/BpPPN1PP/B2K1Ppp/QP1p1pP1/Rrb5
Reprints: (55) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k3/p2p4/8/NP6/BpPPN1PP/B2K1Ppp/QP1p1pP1/Rrb5
Reprints: (55) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
38 - P0004481
Matjaz Zigman
Länderkampf Mazedonien-Slowenien 1969
7. Platz
(13+10) cooked
ser-h#6** AP
Matjaz Zigman
Länderkampf Mazedonien-Slowenien 1969
7. Platz
(13+10) cooked
ser-h#6** AP
* 1. ... Sb5#
* 1. ... dxc3#
1. fxe3ep 2. e2 3. exf1=S 4. Se3? 5. Sd5 Sb5# (must castle for AP)
1. fxe3ep 2. e2 3. exf1=S 4. Sxd2 5. Sxb1 6. Sxa3 0-0-0!#
Luckily, illegal diagram can be fixed by removal of bPa4.
Then if White castling rights remain, ep is on. One ser-h#5 & 18 ser-h#6 tries exist, in which White does not bother to castle.
We don't know if this was a transcription error or a counting mistake by the composer.
Cook: Too many pawn captures required. For retro problem, illegal position implies unsound
* 1. ... dxc3#
1. fxe3ep 2. e2 3. exf1=S 4. Se3? 5. Sd5 Sb5# (must castle for AP)
1. fxe3ep 2. e2 3. exf1=S 4. Sxd2 5. Sxb1 6. Sxa3 0-0-0!#
Luckily, illegal diagram can be fixed by removal of bPa4.
Then if White castling rights remain, ep is on. One ser-h#5 & 18 ser-h#6 tries exist, in which White does not bother to castle.
We don't know if this was a transcription error or a counting mistake by the composer.
Cook: Too many pawn captures required. For retro problem, illegal position implies unsound
Mike Neumeier: The solution appears to be 1.fxe4 e.p. 2.e2 3.exf1=L 4.Ld3 5.Le4 6.Ld5 dxc3#. Was that the intention? (2013-02-18)
Arno Tüngler: There would even be a solution in 5 moves by 3.exf1=S 5.Sd5 Sb5#
However, in order to justify the e.p.-key there must be an AP prove that e2-e4 (and not a move by the wK or wRa1) was played as White's last move. Thus the only solution giving this in 6 moves is 1.fxe3 e.p. 2.e2 3.exf1=S 4.Sxd2 5.Sxb1 6.Sxa3 0-0-0!# (2013-02-18)
Mike Neumeier: It struck me as odd there was no solution listed. (2013-02-18)
Mike Neumeier: And, if we let black be the idle side, there is 1.Txg1 2.Lxh1 3.Kd1 4.Kc1 hxg1T#. What does the asterisk(*) mean? (2013-02-18)
Henrik Juel: The asterisk means that there is also a white mate if he had the move, in this case two mates, so maybe the stipulation should be ser-h#6**:
1.dxc3,Sb5# (2013-02-18)
Ladislav Packa: It is clear that e.p. is possible only when the wK and wRa1 done neither move.
What convention is used here for the right to castling? (2013-02-18)
Henrik Juel: White may castle, unless you can show that he has lost the right to castle. In this problem there is a major difficulty, I think: the position seems illegal.
White pawns captured b2xc3xd4xe5xf6, g2xf3, and axb, promoting on b8; Black captured bxc, dxc, and g3xh2. We cannot explain the destiny of [Ph7].
Reversing wPh3 and bPh2 seems to handle the illegality, but then the problem can be solved in 5 moves (2013-02-18)
Mike Neumeier: Thanks, Henrik. Maybe the stipulation was a typo. Considering all comments together, perhaps the stip should have been ser-h#5**, with the one solution being the 5-mover Arno gave. I do not think it can be proven, except by a stipulation of ser-h#5 that en passant is possible. Which leads to the question of convention. Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible? Whether there is castling or not here seems immaterial. It is just another 6-move solution. There are 19 6-movers (Popeye) with one queenside castling among them. But only the one 5-mover. (2013-02-18)
Henrik Juel: By convention, an en passant capture as first move is not permitted, unless the pawn double step can be shown by some kind of retro analysis.
(Conversely, a castling is permitted, unless it can be shown by retro analysis that the right to castle has been lost, i.e., that king and/or rook has moved) (2013-02-18)
A.Buchanan: Suggested repair: Remove bPa4. Add AP to stipulation.
As Henrik points out, the diagram as it stands is illegal. We can't swap wPh3 & bPh2, because that allows 2 h#1 cooks.
Suppose we remove bPa4 instead. Then we still at least 5 White pawn captures: bxcxdxexf6 & gxf3. But wPh3 did still move from h2, so there are 3 Black pawn captures bxc, dxc & gxh2. So [bhP] died without capturing or promoting. [waP] must have promoted - so either [baP] was captured to clear the way, or [waP] captured to promote, and later [baP] promoted. Either way, all the numbers add up, and the position is legal. wPe4 cannot have just come from d3. So AP is triggered. Forward logic works just as before without bPa4. There is 1 5-move try, and 18 6-move tries, but AP eliminates them all because of need for castling to retrospectively justify the ep. (2013-02-23)
A.Buchanan: Mike asked: "Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible?"
Answer: No. The stipulation does imply a default player to move, but otherwise cannot be used as a premise to determine game state (castling, en passant). (2022-03-21)
A.Buchanan: WinChloe & Yacpdb have the same cooked diagram as PDB, so I think this cook was in the design. I would like to post a sound version with bPa4 removed, and I think it would just be “Zigman correction” with no one else’s name mentioned in the credits. Is everyone ok with this? (2022-05-30)
Mario Richter: 'Bilten 1970' (printed 1971) contains on pp. 62-81 a complete report on the "Problem Duel Slovenia - Macedonia 1969", the problem here can be found on page 79 and is printed there exactly as given here.
Andrew's handling of the suggested correction (remove black pawn a4) is o.k. for me. (2022-05-31)
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Arno Tüngler: There would even be a solution in 5 moves by 3.exf1=S 5.Sd5 Sb5#
However, in order to justify the e.p.-key there must be an AP prove that e2-e4 (and not a move by the wK or wRa1) was played as White's last move. Thus the only solution giving this in 6 moves is 1.fxe3 e.p. 2.e2 3.exf1=S 4.Sxd2 5.Sxb1 6.Sxa3 0-0-0!# (2013-02-18)
Mike Neumeier: It struck me as odd there was no solution listed. (2013-02-18)
Mike Neumeier: And, if we let black be the idle side, there is 1.Txg1 2.Lxh1 3.Kd1 4.Kc1 hxg1T#. What does the asterisk(*) mean? (2013-02-18)
Henrik Juel: The asterisk means that there is also a white mate if he had the move, in this case two mates, so maybe the stipulation should be ser-h#6**:
1.dxc3,Sb5# (2013-02-18)
Ladislav Packa: It is clear that e.p. is possible only when the wK and wRa1 done neither move.
What convention is used here for the right to castling? (2013-02-18)
Henrik Juel: White may castle, unless you can show that he has lost the right to castle. In this problem there is a major difficulty, I think: the position seems illegal.
White pawns captured b2xc3xd4xe5xf6, g2xf3, and axb, promoting on b8; Black captured bxc, dxc, and g3xh2. We cannot explain the destiny of [Ph7].
Reversing wPh3 and bPh2 seems to handle the illegality, but then the problem can be solved in 5 moves (2013-02-18)
Mike Neumeier: Thanks, Henrik. Maybe the stipulation was a typo. Considering all comments together, perhaps the stip should have been ser-h#5**, with the one solution being the 5-mover Arno gave. I do not think it can be proven, except by a stipulation of ser-h#5 that en passant is possible. Which leads to the question of convention. Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible? Whether there is castling or not here seems immaterial. It is just another 6-move solution. There are 19 6-movers (Popeye) with one queenside castling among them. But only the one 5-mover. (2013-02-18)
Henrik Juel: By convention, an en passant capture as first move is not permitted, unless the pawn double step can be shown by some kind of retro analysis.
(Conversely, a castling is permitted, unless it can be shown by retro analysis that the right to castle has been lost, i.e., that king and/or rook has moved) (2013-02-18)
A.Buchanan: Suggested repair: Remove bPa4. Add AP to stipulation.
As Henrik points out, the diagram as it stands is illegal. We can't swap wPh3 & bPh2, because that allows 2 h#1 cooks.
Suppose we remove bPa4 instead. Then we still at least 5 White pawn captures: bxcxdxexf6 & gxf3. But wPh3 did still move from h2, so there are 3 Black pawn captures bxc, dxc & gxh2. So [bhP] died without capturing or promoting. [waP] must have promoted - so either [baP] was captured to clear the way, or [waP] captured to promote, and later [baP] promoted. Either way, all the numbers add up, and the position is legal. wPe4 cannot have just come from d3. So AP is triggered. Forward logic works just as before without bPa4. There is 1 5-move try, and 18 6-move tries, but AP eliminates them all because of need for castling to retrospectively justify the ep. (2013-02-23)
A.Buchanan: Mike asked: "Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible?"
Answer: No. The stipulation does imply a default player to move, but otherwise cannot be used as a premise to determine game state (castling, en passant). (2022-03-21)
A.Buchanan: WinChloe & Yacpdb have the same cooked diagram as PDB, so I think this cook was in the design. I would like to post a sound version with bPa4 removed, and I think it would just be “Zigman correction” with no one else’s name mentioned in the credits. Is everyone ok with this? (2022-05-30)
Mario Richter: 'Bilten 1970' (printed 1971) contains on pp. 62-81 a complete report on the "Problem Duel Slovenia - Macedonia 1969", the problem here can be found on page 79 and is printed there exactly as given here.
Andrew's handling of the suggested correction (remove black pawn a4) is o.k. for me. (2022-05-31)
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Keywords: Castling (wg), Seriesmover, a posteriori (AP) (Type Petrovic), Illegal position, En passant as key, Promotion (s), Valladao Task, Superseded by (P1401546)
Genre: Retro, Fairies
Computer test: Popeye v4.87 but simple retro-logic shows illegality
FEN: 8/8/5P2/2p1p3/p1pkPp2/N1p2P1P/2PP1PBp/RN2KRbr
Reprints: 104 Bilten 1970 1971
(72) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-01 more...
Genre: Retro, Fairies
Computer test: Popeye v4.87 but simple retro-logic shows illegality
FEN: 8/8/5P2/2p1p3/p1pkPp2/N1p2P1P/2PP1PBp/RN2KRbr
Reprints: 104 Bilten 1970 1971
(72) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-01 more...
*) 1. ... Ta4 2. Kxg4 Txf4#
1. fxg3ep Ta4#? (no proof of ep right)
1. fxg3ep 0-0-0! 2. g2 Td4#
A short retro try h#1 and 16 h#2 tries. B2 is unique tempo move
Cook: 1. Kxg4 Ta5 2. h4 Tg5#
1. Kxg4 Ta4 2. Kh4 Txf4#
1. fxg3ep Ta4#? (no proof of ep right)
1. fxg3ep 0-0-0! 2. g2 Td4#
A short retro try h#1 and 16 h#2 tries. B2 is unique tempo move
Cook: 1. Kxg4 Ta5 2. h4 Tg5#
1. Kxg4 Ta4 2. Kh4 Txf4#
Keywords: Castling (wg), En passant as key, a posteriori (AP) (Type Petrovic), Tempo Move
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 8/6p1/6pB/7p/5pPk/5p1p/5P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-06 more...
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 8/6p1/6pB/7p/5pPk/5p1p/5P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-06 more...
40 - P0005036
Luigi Ceriani
36 32 personaggi e 1 autore 1955
(14+12) cooked
Welches war der erste Zug der sD und des sK?
Luigi Ceriani
36 32 personaggi e 1 autore 1955
(14+12) cooked
Welches war der erste Zug der sD und des sK?
R: 1. ... Ld8xLc7 2. Lb6-c7 Kb7-b8 3. Lc7-b6 Kc8-b7 4. Lb6-c7 Lc7-d8 5. Lc5-b6 Lb8-c7 6. Lb6-c5 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 Lh7-g8 13. Sg8-f6 Lf4-b8 14. Sf6-g8 Lh6-f4 15. Sg8-f6 c7-c6 16. g7-g8=S Kc8-d8 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
alternative Auflösung (Mu-Tsun Tsai, PDB 2012-07-22, leicht gekürzt)
R: 1. ... Ld8xSc7+ 2. Sd5-c7 Kc7-b8 3. Sb6-d5 Kb7-c7 4. f2xSg3 Lc7-d8 5. Sd5-b6 Lb8-c7 6. Sf4-d5 Se4-g3 7. Se6-f4 Sc5-e4 8. Sb6-a4 Sa4-c5 9. Sc4-b6 Kc8-b7 10. Sd8-e6 Kc7-c8 11. d7-d8=S Kd8-c7 12. e6xTd7 Tb7-d7 13. Se3-c4 d7-d6 14. Sf5-e3 Le5-b8 15. Sg3-f5 Lg7-e5 16. Sf5-g3 Lf8-g7 17. Sh4-f5 g7-g6 18. Sf3-h4 Sg6-h8 19. Sh4-f3 Se5-g6 20. Sg6-h4 Lh7-g8 21. Sh8-g6 Le4-h7 22. Sg6-h8 Ke8-d8 23. Sh8-g6 Sf3-e5 24. h7-h8=S Ld5-e4 25. g6xDh7 Dh2-h7 26. g5-g6 Db8-h2 27. g4-g5 c7-c6 28. g3-g4 Dd8-b8 29. e5-e6 Tb8-b7 30. e4-e5 Lb7-d5 31. e3-e4 Lc8-b7 32. Kb5-a5 b7xSa6 33. Kc4-b5 Sd4-f3 34. Kd3-c4 Sc6-d4 35. Sc5-a6 Ta8-b8 36. Se4-c5 Sb8-c6 37. Ke2-d3 Sc5-a4 38. Ke1-e2 Da6-a3 39. Sg5-e4 Dh6-a6 40. Ta3-b3 Se4-c5 41. Db3-b2 Sf6-e4 42. Tb2-c2 Sg8-f6 43. Dd1-b3 Sf6-g8 44. Sf3-g5 Sg8-f6 45. Sg1-f3 Dh1-h6 46. Ld3-b1 h2-h1=D 47. Ta4-a3 h3-h2 48. Tb1-b2 h4-h3 49. b2-b4 h5-h4 50. Th4-a4 Sf6-g8 51. Th1-h4 Sg8-f6 52. h2xTg3 Tg6-g3 53. Lf1-d3 Th6-g6 54. e2-e3 Th7-h6 55. Sc2-a1 Th6-h7 56. Sa3-c2 Th7-h6 57. Ta1-b1 Th8-h7 58. Sb1-a3 h7-h5 59. c2-c3
alternative Auflösung (Mu-Tsun Tsai, PDB 2012-07-22, leicht gekürzt)
R: 1. ... Ld8xSc7+ 2. Sd5-c7 Kc7-b8 3. Sb6-d5 Kb7-c7 4. f2xSg3 Lc7-d8 5. Sd5-b6 Lb8-c7 6. Sf4-d5 Se4-g3 7. Se6-f4 Sc5-e4 8. Sb6-a4 Sa4-c5 9. Sc4-b6 Kc8-b7 10. Sd8-e6 Kc7-c8 11. d7-d8=S Kd8-c7 12. e6xTd7 Tb7-d7 13. Se3-c4 d7-d6 14. Sf5-e3 Le5-b8 15. Sg3-f5 Lg7-e5 16. Sf5-g3 Lf8-g7 17. Sh4-f5 g7-g6 18. Sf3-h4 Sg6-h8 19. Sh4-f3 Se5-g6 20. Sg6-h4 Lh7-g8 21. Sh8-g6 Le4-h7 22. Sg6-h8 Ke8-d8 23. Sh8-g6 Sf3-e5 24. h7-h8=S Ld5-e4 25. g6xDh7 Dh2-h7 26. g5-g6 Db8-h2 27. g4-g5 c7-c6 28. g3-g4 Dd8-b8 29. e5-e6 Tb8-b7 30. e4-e5 Lb7-d5 31. e3-e4 Lc8-b7 32. Kb5-a5 b7xSa6 33. Kc4-b5 Sd4-f3 34. Kd3-c4 Sc6-d4 35. Sc5-a6 Ta8-b8 36. Se4-c5 Sb8-c6 37. Ke2-d3 Sc5-a4 38. Ke1-e2 Da6-a3 39. Sg5-e4 Dh6-a6 40. Ta3-b3 Se4-c5 41. Db3-b2 Sf6-e4 42. Tb2-c2 Sg8-f6 43. Dd1-b3 Sf6-g8 44. Sf3-g5 Sg8-f6 45. Sg1-f3 Dh1-h6 46. Ld3-b1 h2-h1=D 47. Ta4-a3 h3-h2 48. Tb1-b2 h4-h3 49. b2-b4 h5-h4 50. Th4-a4 Sf6-g8 51. Th1-h4 Sg8-f6 52. h2xTg3 Tg6-g3 53. Lf1-d3 Th6-g6 54. e2-e3 Th7-h6 55. Sc2-a1 Th6-h7 56. Sa3-c2 Th7-h6 57. Ta1-b1 Th8-h7 58. Sb1-a3 h7-h5 59. c2-c3
Korrekturversuch s. P0003009
Henrik Juel: If you want a great solving challenge, this is the retro for you.
If you need a hint:
[Dd8] never moved, and Black castled (as you may have guessed).
If you need another hint:
Last move was Ld8xLc7+.
I gave up, but Nikolai told me the solution:
-1... Ld8xLc7 -2.Lb6 Kb7 -3.Lc7 Kc8 -4.Lb6 Lc7 -5.Lc5 Lb8 -6.Lb6 Kb7 -7.Ld8 Kc7 -8.L=d7 Kd8 -9.e6xSd7 Sb6 -10.Sc5 Sa4 -11.Sd4 d7 -12.Sf6 Lh7 -13.Sg8 Lf4 -14.Sf6 Lh6 -15.Sg8 c7 -16.S=g7 Kc8 -17.f6xTg7 Tg8 -18.e5 Td8 -19.e4 0-0-0 -20.e3 Lf8 -21.f5 g7 -22.f4 Le4 -23.f3 Lb7 -24.f2 Lc8 -25.Kb5 b7xSa6 etc. (2012-07-22)
Mu-Tsun Tsai: Once I heard "great challenge" I started working. But I came to a complete different conclusion. Not only [Qd8] can move, but the first move of the black king need not be castling. Here's the proof game. After playing
1.c3 h5 2.b4 Rh6 3.Na3 Rg6 4.Nc2 Rg3 5.hxg3 Nf6 6.Rb1 Ng8 7.Na1 Nf6 8.e3 Ng8 9.Bd3 Nf6 10.Rb2 Ng8 11.Bb1 Nf6 12.Qb3 Ng8 13.Rc2 Nf6 14.Qb2 Ng8 15.Rh4 Nf6 16.Rd4 h4 17.Rd5 h3 18.Ra5 h2 19.Ra3 h1=Q 20.Rb3 Qh5 21.Nf3 Qa5 22.Ke2 Qa3 23.Kd3 Na6 24.Kc4 Nc5 25.Kb5 Na4 26.Ne5 Nh5 27.Nd3 Nf4 28.Nc5 Nd3 29.Na6 bxa6+ 30.Ka5 Bb7 31.e4 Qb8 32.e5 Bc8 33.e6 Qb7 34.g4 Qf3 35.g5 Qh3 36.g6 Qh7 37.gxh7 Nf4 38.h8=N Bb7 39.Ng6 Be4 40.Ne5 Bh7 41.Nc4 Bg8 42.Ne3 Ng6 43.Nc4 Nh8 44.Ne3 g6 45.Nc4 Bg7 46.Ne3 Be5 47.Nc4 c6 48.Ne3,
you could either play
48...O-O-O 49.Nc4 Bb8 50.Ne3 d6 51.Nc4 Rd7 52.exd7+ Kb7 53.d8=N+ Kc7 54.Ne6+ Kb7 55.Nb6 Nc5+ 56.Na4 Ne4 57.Nc5+ Kc7 58.Nd7 Kb7 59.Nb6 Ng3 60.Nd5 Bc7+ 61.Nb6 Bd8 62.fxg3 Kc7 63.Nd5+ Kb8+ 64.Nc7 Bxc7+,
which is castling version, or,
48...Kd8 49.Nc4 Kc7 50.Ne3 Rd8 51.Nc4 Kc8 52.Ne3 Bb8 53.Nc4 d6 54.Ne3 Rd7 55.exd7+ Kb7 56.d8=N+ Kc7 57.Ne6+ Kb7 58.Nc4 Bh7 59.Nb6 Nc5+ 60.Na4 Ne4 61.Nd4 Kc7 62.Ne6+ Kc8 63.Nd4 Kb7 64.Ne6 Bg8 65.Nc5+ Kc7 66.Nd7 Kb7 67.Nb6 Ng3 68.Nd5 Bc7+ 69.Nb6 Bd8 70.fxg3 Kc7 71.Nd5+ Kb8+ 72.Nc7 Bxc7+
which is none castling version. Both reach the diagram position.
Why am I feeling cooking Ceriani's problems too much lately? (2012-07-22)
Mu-Tsun Tsai: Also in your retraction, -11.Nd4 should be -11.Ne4 I believe? It seems like your retraction also works just fine, and looks like it should be the intended solution. (2012-07-22)
Henrik Juel: Yes, the intended solution should have -11.Se4, and it can be shortened a bit.
I am impressed by your cook, Mu-Tsun, with two white pawn captures on g3 and promotion on h8, but it is also a little sad that a seemingly fine problem has been rendered worthless.
Probably several more Ceriani problems will be cooked, because they were not scrutinized well enough by testers and solvers in the old days; now, when I finally have cracked a Ceriani nut, I have no energy left to search for errors (2012-07-23)
Mu-Tsun Tsai: I've been thinking about how this problem might be fixed, but unfortunately I cannot come up with anything other than adding extra assumptions in the stipulation, for example "g3 pawn came from h2". The structure of this one is good, and either method of releasing the position (mine or the intended one) is quite subtle, so I feel sad about have to cook this one as well. (2012-07-23)
Thomas Volet: In his 1961 book Ceriani discusses the cook with the WhP unpromoting on h8 and uncapturing to the g file and back to the h file, and gives P0003009 as the corrected diagram position. (2012-08-02)
Mu-Tsun Tsai: This is a really late comment, but I do think this will make a great problem by changing the stip to "You don't know the first move of the black queen nor the black king"! (2023-06-29)
comment
Henrik Juel: If you want a great solving challenge, this is the retro for you.
If you need a hint:
[Dd8] never moved, and Black castled (as you may have guessed).
If you need another hint:
Last move was Ld8xLc7+.
I gave up, but Nikolai told me the solution:
-1... Ld8xLc7 -2.Lb6 Kb7 -3.Lc7 Kc8 -4.Lb6 Lc7 -5.Lc5 Lb8 -6.Lb6 Kb7 -7.Ld8 Kc7 -8.L=d7 Kd8 -9.e6xSd7 Sb6 -10.Sc5 Sa4 -11.Sd4 d7 -12.Sf6 Lh7 -13.Sg8 Lf4 -14.Sf6 Lh6 -15.Sg8 c7 -16.S=g7 Kc8 -17.f6xTg7 Tg8 -18.e5 Td8 -19.e4 0-0-0 -20.e3 Lf8 -21.f5 g7 -22.f4 Le4 -23.f3 Lb7 -24.f2 Lc8 -25.Kb5 b7xSa6 etc. (2012-07-22)
Mu-Tsun Tsai: Once I heard "great challenge" I started working. But I came to a complete different conclusion. Not only [Qd8] can move, but the first move of the black king need not be castling. Here's the proof game. After playing
1.c3 h5 2.b4 Rh6 3.Na3 Rg6 4.Nc2 Rg3 5.hxg3 Nf6 6.Rb1 Ng8 7.Na1 Nf6 8.e3 Ng8 9.Bd3 Nf6 10.Rb2 Ng8 11.Bb1 Nf6 12.Qb3 Ng8 13.Rc2 Nf6 14.Qb2 Ng8 15.Rh4 Nf6 16.Rd4 h4 17.Rd5 h3 18.Ra5 h2 19.Ra3 h1=Q 20.Rb3 Qh5 21.Nf3 Qa5 22.Ke2 Qa3 23.Kd3 Na6 24.Kc4 Nc5 25.Kb5 Na4 26.Ne5 Nh5 27.Nd3 Nf4 28.Nc5 Nd3 29.Na6 bxa6+ 30.Ka5 Bb7 31.e4 Qb8 32.e5 Bc8 33.e6 Qb7 34.g4 Qf3 35.g5 Qh3 36.g6 Qh7 37.gxh7 Nf4 38.h8=N Bb7 39.Ng6 Be4 40.Ne5 Bh7 41.Nc4 Bg8 42.Ne3 Ng6 43.Nc4 Nh8 44.Ne3 g6 45.Nc4 Bg7 46.Ne3 Be5 47.Nc4 c6 48.Ne3,
you could either play
48...O-O-O 49.Nc4 Bb8 50.Ne3 d6 51.Nc4 Rd7 52.exd7+ Kb7 53.d8=N+ Kc7 54.Ne6+ Kb7 55.Nb6 Nc5+ 56.Na4 Ne4 57.Nc5+ Kc7 58.Nd7 Kb7 59.Nb6 Ng3 60.Nd5 Bc7+ 61.Nb6 Bd8 62.fxg3 Kc7 63.Nd5+ Kb8+ 64.Nc7 Bxc7+,
which is castling version, or,
48...Kd8 49.Nc4 Kc7 50.Ne3 Rd8 51.Nc4 Kc8 52.Ne3 Bb8 53.Nc4 d6 54.Ne3 Rd7 55.exd7+ Kb7 56.d8=N+ Kc7 57.Ne6+ Kb7 58.Nc4 Bh7 59.Nb6 Nc5+ 60.Na4 Ne4 61.Nd4 Kc7 62.Ne6+ Kc8 63.Nd4 Kb7 64.Ne6 Bg8 65.Nc5+ Kc7 66.Nd7 Kb7 67.Nb6 Ng3 68.Nd5 Bc7+ 69.Nb6 Bd8 70.fxg3 Kc7 71.Nd5+ Kb8+ 72.Nc7 Bxc7+
which is none castling version. Both reach the diagram position.
Why am I feeling cooking Ceriani's problems too much lately? (2012-07-22)
Mu-Tsun Tsai: Also in your retraction, -11.Nd4 should be -11.Ne4 I believe? It seems like your retraction also works just fine, and looks like it should be the intended solution. (2012-07-22)
Henrik Juel: Yes, the intended solution should have -11.Se4, and it can be shortened a bit.
I am impressed by your cook, Mu-Tsun, with two white pawn captures on g3 and promotion on h8, but it is also a little sad that a seemingly fine problem has been rendered worthless.
Probably several more Ceriani problems will be cooked, because they were not scrutinized well enough by testers and solvers in the old days; now, when I finally have cracked a Ceriani nut, I have no energy left to search for errors (2012-07-23)
Mu-Tsun Tsai: I've been thinking about how this problem might be fixed, but unfortunately I cannot come up with anything other than adding extra assumptions in the stipulation, for example "g3 pawn came from h2". The structure of this one is good, and either method of releasing the position (mine or the intended one) is quite subtle, so I feel sad about have to cook this one as well. (2012-07-23)
Thomas Volet: In his 1961 book Ceriani discusses the cook with the WhP unpromoting on h8 and uncapturing to the g file and back to the h file, and gives P0003009 as the corrected diagram position. (2012-08-02)
Mu-Tsun Tsai: This is a really late comment, but I do think this will make a great problem by changing the stip to "You don't know the first move of the black queen nor the black king"! (2023-06-29)
comment
Keywords: First Move? (kd), Superseded by (P0003009)
Genre: Retro
FEN: 1k4bn/p1b1pp2/p1pp2p1/K7/NP6/qRP3P1/PQRP2P1/NBB5
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-07-02 more...
Genre: Retro
FEN: 1k4bn/p1b1pp2/p1pp2p1/K7/NP6/qRP3P1/PQRP2P1/NBB5
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-07-02 more...
41 - P0005275
Otto Kerekes
Tivadar Kardos
10 L'Echiquier de France 11/1956
(11+15) cooked
h#2
b) sBg2 statt wBg2
Otto Kerekes
Tivadar Kardos
10 L'Echiquier de France 11/1956
(11+15) cooked
h#2
b) sBg2 statt wBg2
a) 1. cxd4 Lxc4 2. Dd5 Ld3#
b)
b)
Paulo Roque: Illegale Stellung. Diagrammfehler? (2009-11-17)
Alfred Pfeiffer: außerdem in b) 9 schwarze Bauern. (2010-01-13)
A.Buchanan: Twin a) is also illegal, as too many Bl captures (2021-11-24)
more ...
comment
Alfred Pfeiffer: außerdem in b) 9 schwarze Bauern. (2010-01-13)
A.Buchanan: Twin a) is also illegal, as too many Bl captures (2021-11-24)
more ...
comment
Keywords: En passant as key, Castling, Illegal position, Superseded by (P1400824)
Genre: h#, Retro
FEN: 8/p1p2p1b/1B3pp1/q1p1b3/r1pPk3/1Bp3P1/PPP2PPr/3n1RK1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-23 more...
Genre: h#, Retro
FEN: 8/p1p2p1b/1B3pp1/q1p1b3/r1pPk3/1Bp3P1/PPP2PPr/3n1RK1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-23 more...
1. g4 h5 2. gxh5 d5 3. h6 Dd6 4. hxg7 Dxh2 5. gxh8=T Dxh1 6. Txh1
Duale sind möglich
Duale sind möglich
"These promoted piece games are anticipated in general idea by CMF" (TRD)
Moldenhauer: Computerprüfung: C+ KBP 5.5 cooked in 1 Sekunde.
Keine Lösung: vor BP 5.5.
Beispiellösung:1.g4 d5 2.g5 h6 3.gxh6 Dd6 4.hxg7 Ddxh2
5.gxh8=T Dxh1 6.Txh1 (2023-03-22)
comment
Moldenhauer: Computerprüfung: C+ KBP 5.5 cooked in 1 Sekunde.
Keine Lösung: vor BP 5.5.
Beispiellösung:1.g4 d5 2.g5 h6 3.gxh6 Dd6 4.hxg7 Ddxh2
5.gxh8=T Dxh1 6.Txh1 (2023-03-22)
comment
Keywords: Non-Unique Proof Game, Pronkin Theme (T), Homebase (w)
Genre: Retro
Computer test: Computerprüfung: C+ KBP 5.5 cooked in 1 Sekunde. Keine Lösung: vor BP 5.5. Beispiellösung:1.g4 d5 2.g5 h6 3.gxh6 Dd6 4.hxg7 Ddxh2 5.gxh8=T Dxh1 6.Txh1
FEN: rnb1kbn1/ppp1pp2/8/3p4/8/8/PPPPPP2/RNBQKBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-03-06 more...
Genre: Retro
Computer test: Computerprüfung: C+ KBP 5.5 cooked in 1 Sekunde. Keine Lösung: vor BP 5.5. Beispiellösung:1.g4 d5 2.g5 h6 3.gxh6 Dd6 4.hxg7 Ddxh2 5.gxh8=T Dxh1 6.Txh1
FEN: rnb1kbn1/ppp1pp2/8/3p4/8/8/PPPPPP2/RNBQKBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-03-06 more...
43 - P0005581
Leonid M. Borodatow
5647v Die Schwalbe 105 06/1987
Günter Lauinger gewidmet
(11+13) cooked
Wie oft stand der wK mindestens auf c7?
Leonid M. Borodatow
5647v Die Schwalbe 105 06/1987
Günter Lauinger gewidmet
(11+13) cooked
Wie oft stand der wK mindestens auf c7?
R: 1. ... De5-b8+ 2. Kc7xSc8 Sd6-c8+ 3. Kc8-c7 Sc4-d6+ 4. Kc7xSc8 Sd6xLc4+
Cook: R: 1. ... Dd6:Sb8+ 2. Kc7-c8 ...
Cook: R: 1. ... Dd6:Sb8+ 2. Kc7-c8 ...
Nikolai Beluhov: Diagram is now correct (missing wRe8 and bBf8 restored). (2011-05-05)
Nikolai Beluhov: This problem seems to be cooked by 1. ... Qd6:Nb8+ 2. Kc7-c8 Q~d6+ ...
Fortunately, this flaw is very easy to fix: just relocate bQb8 to c7, as in
L. Borodatov (correction)
k1KRRb2/P1qpp1p1/rpp5/p7/5p2/3PP3/P1Pp1PP1/4Bb2
(http://www.janko.at/Retros/d.php?ff=k1KRRb2/P1qpp1p1/rpp5/p7/5p2/3PP3/P1Pp1PP1/4Bb2)
(11 + 13) How often did the wK visit c7 at least? (2011-05-06)
Henrik Juel: The wK visited c7 at least four times, twice as shown in the indicated retroplay, and twice to let wK pass the two white rooks en route to f7, with Le1 screening on b8. (2011-05-06)
Anton Baumann: Informalturnier 1986 (PB in 'Die Schwalbe' 06/2011 S.124)
Auszeichnung: 2.Lob; ausgezeichnete Fassung: mit sDc7 statt b8 (2023-01-04)
comment
Nikolai Beluhov: This problem seems to be cooked by 1. ... Qd6:Nb8+ 2. Kc7-c8 Q~d6+ ...
Fortunately, this flaw is very easy to fix: just relocate bQb8 to c7, as in
L. Borodatov (correction)
k1KRRb2/P1qpp1p1/rpp5/p7/5p2/3PP3/P1Pp1PP1/4Bb2
(http://www.janko.at/Retros/d.php?ff=k1KRRb2/P1qpp1p1/rpp5/p7/5p2/3PP3/P1Pp1PP1/4Bb2)
(11 + 13) How often did the wK visit c7 at least? (2011-05-06)
Henrik Juel: The wK visited c7 at least four times, twice as shown in the indicated retroplay, and twice to let wK pass the two white rooks en route to f7, with Le1 screening on b8. (2011-05-06)
Anton Baumann: Informalturnier 1986 (PB in 'Die Schwalbe' 06/2011 S.124)
Auszeichnung: 2.Lob; ausgezeichnete Fassung: mit sDc7 statt b8 (2023-01-04)
comment
Genre: Retro
FEN: kqKRRb2/P2pp1p1/rpp5/p7/5p2/3PP3/P1Pp1PP1/4Bb2
Input: Gerd Wilts, 1995-06-04
Last update: Nikolai Beluhov, 2011-05-06 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)
comment
comment
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...
45 - P0006953
Josef Haas
feenschach 21, p. 258, 04/1974
(6+1) cooked
Welches war der letzte Zug?
Circe
Josef Haas
feenschach 21, p. 258, 04/1974
(6+1) cooked
Welches war der letzte Zug?
Circe
R: 1. 0-0
Cook: R: 1. Kf2-g1 Kh2xBh3[+wBh2]
Cook: R: 1. Kf2-g1 Kh2xBh3[+wBh2]
HBae: cook siehe feenschach-25, 10/1974, S. 372 (2015-12-02)
A.Buchanan: Bh2->L (2015-12-03)
A.Buchanan: A number of these problems by Haas were cooked. HBae gives a teasing link to an article which possibly includes some corrections. Has anyone got access to this article, please? (2021-05-13)
HBae: Hello Andrew, link to feenschach 10/1974 (website Kotesovec):
http://problem64.beda.cz/silo/feenschach25_1974.pdf (2021-05-13)
A.Buchanan: Thanks HBae! (2021-05-13)
A.Buchanan: In this one B-=D seems to be Type C, not type A?
In WinChloe there are a batch more from a few years later 34 p.143, feenschach 47 (juil. 79). If we're uploading anything, should maybe skip to the end? (2021-05-13)
comment
A.Buchanan: Bh2->L (2015-12-03)
A.Buchanan: A number of these problems by Haas were cooked. HBae gives a teasing link to an article which possibly includes some corrections. Has anyone got access to this article, please? (2021-05-13)
HBae: Hello Andrew, link to feenschach 10/1974 (website Kotesovec):
http://problem64.beda.cz/silo/feenschach25_1974.pdf (2021-05-13)
A.Buchanan: Thanks HBae! (2021-05-13)
A.Buchanan: In this one B-=D seems to be Type C, not type A?
In WinChloe there are a batch more from a few years later 34 p.143, feenschach 47 (juil. 79). If we're uploading anything, should maybe skip to the end? (2021-05-13)
comment
1. Lxa2 Txa2 2. 0-0-0 Txa7 3. Td7 Ta8#
1. Lxa2 0-0-0? 2. Dxg2 Tg1 3. 0-0 Txg2# but White can't castle
Cook: 1. Lxa2 Lf6 2. Lb1 Txa7 3. Kf8 Txa8#
1. Lxa2 Lf6 2. Lxb3 Txa7 3. Kf8 Txa8#
1. Dxg2 Txb1 2. Dxf2+ Kxf2 3. 0-0 Tg1#
1. Dg3 Txb1 2. Dxb3 Txb3 3. 0-0 Tg3#
1. Lxa2 0-0-0? 2. Dxg2 Tg1 3. 0-0 Txg2# but White can't castle
Cook: 1. Lxa2 Lf6 2. Lb1 Txa7 3. Kf8 Txa8#
1. Lxa2 Lf6 2. Lxb3 Txa7 3. Kf8 Txa8#
1. Dxg2 Txb1 2. Dxf2+ Kxf2 3. 0-0 Tg1#
1. Dg3 Txb1 2. Dxb3 Txb3 3. 0-0 Tg3#
paul: White Rook h1 left the S-E camp, so the white castle is no more possible.
Intention: 1.L×a2 T×a2 2.0-0-0 T×a7 3.Td7 Ta8#
1.L×a2 0-0-0 2.D×g2 Tg1 3.0-0 T×g2#
Cooked by 1.L×a2 Lf6 2.Lb1 T×a7 3.Rf8 T×a8# or
1.D×g2 T×b1 2.D×f2+ K×f2 3.0-0 Tg1# (2011-09-15)
milan: wBe5-f4 M.Frelih (2021-12-01)
A.Buchanan: @Milan: that removes all 4 cooks, and alas the intended try also. (2021-12-01)
comment
Intention: 1.L×a2 T×a2 2.0-0-0 T×a7 3.Td7 Ta8#
1.L×a2 0-0-0 2.D×g2 Tg1 3.0-0 T×g2#
Cooked by 1.L×a2 Lf6 2.Lb1 T×a7 3.Rf8 T×a8# or
1.D×g2 T×b1 2.D×f2+ K×f2 3.0-0 Tg1# (2011-09-15)
milan: wBe5-f4 M.Frelih (2021-12-01)
A.Buchanan: @Milan: that removes all 4 cooks, and alas the intended try also. (2021-12-01)
comment
Keywords: Cant Castler, Obvious promotion (l), Superseded by (P1396304)
Genre: Retro
FEN: r3k1qr/pp3p1p/2p5/3pB2p/8/1P6/P1PPPPPP/Rb2K3
Input: Gerd Wilts, 1996-08-11
Last update: A.Buchanan, 2021-12-01 more...
Genre: Retro
FEN: r3k1qr/pp3p1p/2p5/3pB2p/8/1P6/P1PPPPPP/Rb2K3
Input: Gerd Wilts, 1996-08-11
Last update: A.Buchanan, 2021-12-01 more...
47 - P0007435
Adamas
(47) feenschach 40 11-12/1977
(1+3) cooked
Welches war der letzte Zug?
Längstzüger
Adamas
(47) feenschach 40 11-12/1977
(1+3) cooked
Welches war der letzte Zug?
Längstzüger
R: 1. Bd2xDe1=L+
Cook: R: 1. Bd2xDc1=L+,d2xT/L/Sc1=L+
Cook: R: 1. Bd2xDc1=L+,d2xT/L/Sc1=L+
Mario Richter: Position is as given in 'feenschach'. It seems that R: 1. d3xDc1+,d3xT/L/Sc1+ also works and that adding a black pawn on the 7th rank (e.g. h7) would prevent the cook.
Can somebody confirm my observation? (2022-01-10)
Mario Richter: small typo: It seems that R: 1. d2xDc1+,d2xT/L/Sc1+ also works (2022-01-10)
Henrik Juel: Your observation, with last moves d2xDTLDc1=L+, seems correct (2022-01-10)
comment
Can somebody confirm my observation? (2022-01-10)
Mario Richter: small typo: It seems that R: 1. d2xDc1+,d2xT/L/Sc1+ also works (2022-01-10)
Henrik Juel: Your observation, with last moves d2xDTLDc1=L+, seems correct (2022-01-10)
comment
Keywords: Maximummer, Last Move? (BxD=L), Aristocrat, Miniature
Genre: Retro, Fairies
FEN: 8/8/8/8/8/4K1k1/8/2b1b3
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-10 more...
Genre: Retro, Fairies
FEN: 8/8/8/8/8/4K1k1/8/2b1b3
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-10 more...
48 - P0007436
Adamas
(48) feenschach 40 11-12/1977
(1+3) cooked
Welches war der letzte Zug?
Längstzüger
Adamas
(48) feenschach 40 11-12/1977
(1+3) cooked
Welches war der letzte Zug?
Längstzüger
R: 1. Bg2xTh1=L+
Cook: R: 1. Bg2xDf1=S+,Bg2xTf1=S+,Bg2xLf1=S+,Bg2xSf1=S+,
Cook: R: 1. Bg2xDf1=S+,Bg2xTf1=S+,Bg2xLf1=S+,Bg2xSf1=S+,
Henrik Juel: What is the other last move, Andrew? (2022-01-08)
A.Buchanan: I didn’t analyse these problems, just corrected the genre when I saw that it was incomplete. Here how about e.g. R: 1. Bc6-h8+ Kb7-a8 2. Bh8xPc6 (2022-01-09)
Mario Richter: R: 1. ... Bc6-h1+? is illegal - it leaves the wK in check!
The cooks are the discovered checks by 1. ... g2xYf1+ (2022-01-09)
comment
A.Buchanan: I didn’t analyse these problems, just corrected the genre when I saw that it was incomplete. Here how about e.g. R: 1. Bc6-h8+ Kb7-a8 2. Bh8xPc6 (2022-01-09)
Mario Richter: R: 1. ... Bc6-h1+? is illegal - it leaves the wK in check!
The cooks are the discovered checks by 1. ... g2xYf1+ (2022-01-09)
comment
Keywords: Maximummer, Last Move? (BxT=L), Aristocrat, Miniature
Genre: Retro, Fairies
FEN: K7/8/8/8/8/8/7k/5n1b
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-09 more...
Genre: Retro, Fairies
FEN: K7/8/8/8/8/8/7k/5n1b
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-09 more...
49 - P0007437
Ewald Reichel
(49) feenschach 40 11-12/1977
(4+8) cooked
Welches war der letzte Zug?
Längstzüger
Ewald Reichel
(49) feenschach 40 11-12/1977
(4+8) cooked
Welches war der letzte Zug?
Längstzüger
R: 1. Ba2xLb1=L+
Cook: R: 1. a2xSb1=L+
Cook: R: 1. a2xSb1=L+
Henrik Juel: What is the other last move, Andrew? (2022-01-08)
Henrik Juel: Well, maybe 1.Pa2xSb1=L+ also works (2022-01-08)
comment
Henrik Juel: Well, maybe 1.Pa2xSb1=L+ also works (2022-01-08)
comment
Keywords: Maximummer, Last Move? (BxL=L)
Genre: Retro, Fairies
FEN: 8/8/8/3p4/p2p4/R1pK4/1k1PP3/rbb5
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-09 more...
Genre: Retro, Fairies
FEN: 8/8/8/3p4/p2p4/R1pK4/1k1PP3/rbb5
Input: Gerd Wilts, 1996-08-21
Last update: Mario Richter, 2022-01-09 more...
50 - P0007858
Bernd Schwarzkopf
(16) feenschach 56 10/1981
(13+12) cooked
Für 3 DD ist der erste Zug eindeutig (sie zogen nie).
Bernd Schwarzkopf
(16) feenschach 56 10/1981
(13+12) cooked
Für 3 DD ist der erste Zug eindeutig (sie zogen nie).
Cook: (alternative Auflösung mri)
R: 1. a5-a6 Ka8-b8 2. Ta6xTa7 Kb8-a8 3. Tb6-a6 Ta8-a7 4. Tb3-b6 Ka7-b8 5. Th3-b3 Kb6-a7 6. a4-a5 Kc6-b6 7. Dh5-e8 Kd6-c6 8. Df3-h5 Ke6-d6 9. Th1-h3 Kf7-e6 10. Dd3-f3 Ke8-f7 11. De4-d3 Kd8-e8 12. Dd3-e4 f7-f6 13. Sf6-g8 Dg8-h8 14. De3-d3 Th8-h7 15. Se4-f6 Dh7-g8 16. Sc3-e4 Ke8-d8 17. Sb1-c3 Dd3-h7 18. De4-e3 Tg8-h8 19. Dh7-e4 Da3-d3 20. Dh8-h7 Da1-a3 21. h7-h8=D Th8-g8 22. g6xSh7 a2-a1=D 23. g5-g6 b3xTa2 24. Ta1-a2 b4-b3 25. g4-g5 Sf6-h7 26. h3xSg4 a5xSb4 27. Sc6-b4 Se3-g4 28. Sd8-c6 Sd1-e3 29. Se6xDd8 Sc3xDd1 30. Sf4-e6 Sb5-c3 31. h2-h3 h7-h6 32. Sh3-f4 Sg8-f6 33. Sg1-h3 Sa7-b5 34. Ta2-a1 Sc6-a7 35. Ta1-a2 Sb8-c6 36. a3-a4 a7-a5 37. a2-a3
vgl. P0007857
Hans-Jürgen Manthey: die Vorgabe ist nicht nötig. 2 Damen zogen nie, die dritte konnte über a2 oder a3 usw. gezogen haben !
R: 1. f7xSe8=D Sd6-e8 g6xSf7 Se5-f7 h5xTg6 Tg5-g6 a5-a6 Sc6-e5 a4-a5 Se4-d6 Ta6-a7 Ta5-g5 Tb6-a6 Ta8-a5 Tb3-b6 Ka7-b8 Th3-b3 Kb6-a7 Th4-h3 Kc5-b6 Th3-h4 Kd6-c5 Th2-h3 Ke6-d6 Th1-h2 Kf7-e6 h4-h5 Ke8-f7 h3-h4 Kd8-e8 h2-h3 f7-f6 Sf6-g8 Sc3-e4 Se4-f6 Sd1-c3 Sc5-e4 Ke8-d8 Se6-c5 Sc3xDd1 Sd8-e6 Dg8-h8 Se6xDd8 Th8-h7 Sg5-e6 Dh7-g8 Sf3-g5 Dd3-h7 Sg1-f3 Sd5-c3 Sh3-g1 Da3-d3 Sg5-h3 Da1-a3 Sf3-g5 a2-a1=D Sg1-f3 b3xTa2 Ta3-a2 b4-b3 Ta2-a3 a5xSb4 Ta1-a2 Sf6-d5 Sd5-b4 h7-h6 Sc3-d5 Sg8-f6 a3-a4 a7-a5 a2-a3 Sb8-c6 Sb1-c3 (2021-07-26)
Mario Richter: Geplant war ein First-Move-Rekord, bei dem hier für 3 Damen der erste Zug eindeutig ist (wOriginal-Dd1, sOriginal-Dd8 und wUW-De8 zogen nie).
Leider läßt sich die Stellung auch beginnend mit R: 1. a5-a6 Ka8-b8 2. Ta6xTa7+ auflösen, so daß diese Aufgabe wohl "cooked" ist.
vgl. P0007857 (2021-07-27)
comment
Hans-Jürgen Manthey: die Vorgabe ist nicht nötig. 2 Damen zogen nie, die dritte konnte über a2 oder a3 usw. gezogen haben !
R: 1. f7xSe8=D Sd6-e8 g6xSf7 Se5-f7 h5xTg6 Tg5-g6 a5-a6 Sc6-e5 a4-a5 Se4-d6 Ta6-a7 Ta5-g5 Tb6-a6 Ta8-a5 Tb3-b6 Ka7-b8 Th3-b3 Kb6-a7 Th4-h3 Kc5-b6 Th3-h4 Kd6-c5 Th2-h3 Ke6-d6 Th1-h2 Kf7-e6 h4-h5 Ke8-f7 h3-h4 Kd8-e8 h2-h3 f7-f6 Sf6-g8 Sc3-e4 Se4-f6 Sd1-c3 Sc5-e4 Ke8-d8 Se6-c5 Sc3xDd1 Sd8-e6 Dg8-h8 Se6xDd8 Th8-h7 Sg5-e6 Dh7-g8 Sf3-g5 Dd3-h7 Sg1-f3 Sd5-c3 Sh3-g1 Da3-d3 Sg5-h3 Da1-a3 Sf3-g5 a2-a1=D Sg1-f3 b3xTa2 Ta3-a2 b4-b3 Ta2-a3 a5xSb4 Ta1-a2 Sf6-d5 Sd5-b4 h7-h6 Sc3-d5 Sg8-f6 a3-a4 a7-a5 a2-a3 Sb8-c6 Sb1-c3 (2021-07-26)
Mario Richter: Geplant war ein First-Move-Rekord, bei dem hier für 3 Damen der erste Zug eindeutig ist (wOriginal-Dd1, sOriginal-Dd8 und wUW-De8 zogen nie).
Leider läßt sich die Stellung auch beginnend mit R: 1. a5-a6 Ka8-b8 2. Ta6xTa7+ auflösen, so daß diese Aufgabe wohl "cooked" ist.
vgl. P0007857 (2021-07-27)
comment
Keywords: First Move?
Genre: Retro
FEN: 1kb1QbNq/Rpppp1pr/P4p1p/8/8/8/1PPPPPP1/2B1KB2
Input: Gerd Wilts, 1996-09-03
Last update: Mario Richter, 2021-07-28 more...
Genre: Retro
FEN: 1kb1QbNq/Rpppp1pr/P4p1p/8/8/8/1PPPPPP1/2B1KB2
Input: Gerd Wilts, 1996-09-03
Last update: Mario Richter, 2021-07-28 more...
51 - P0008465
Thomas Volet
9375 Die Schwalbe 161 10/1996
Nikita Plaksin gewidmet
(14+9) cooked
Die Partie ist beendet. Weshalb?
Thomas Volet
9375 Die Schwalbe 161 10/1996
Nikita Plaksin gewidmet
(14+9) cooked
Die Partie ist beendet. Weshalb?
Thomas Volet: Problem P0008465 is faulty. (2000-11-27)
James Malcom: What is the author's solution and the cook? (2022-01-21)
Mario Richter: I do not the "official cook", but the following should work:
R: 1. Th1-g1 Kf5-e4 2. Th4-h1 Kg6-f5 3. Td4-h4 Kh6-g6 4. Td1-d4 Kg6-h6 5. Ta1-d1 Kh6-g6 6. Ta3-a1 Kg6-h6 7. Tb3-a3 Kh6-g6 8. Tb5-b3 Kg6-h6 9. Ta5-b5 Kh6-g6 10. Lb8-a7 Kg6-h6 11. Ta7-a5 Kh6-g6 12. Tb7-a7 Kg6-h6 13. La7-b8 Kh6-g6 14. Kg8-f8 Kg6-h6 15. Tb8-b7 Kh6-g6 16. Tf8-b8 Te8-e7 17. Lb8-a7 Tc8-e8 18. La7-b8 Tb8-c8 19. Te8-f8 Tb7-b8 20. Lb8-a7 Ta7-b7 21. Kf8-g8 Ta5-a7 22. Ke7-f8 Tb5-a5 23. Kd8-e7 Tb4-b5 24. Kc8-d8 Th4-b4 25. Kb7-c8 Th5-h4 26. Ka6-b7 Tg5-h5 27. Kb5-a6 Tf5-g5 28. Kc4-b5 Te5-f5 29. Kb3-c4 Te4-e5 30. Ka2-b3 Tb4-e4 31. Kb1-a2 Tb5-b4 32. Te7-e8 Ta5-b5 33. Kc1-b1 Ta7-a5 34. Kd1-c1 Tb7-a7 35. La7-b8 Tb8-b7 36. Ke1-d1 Th8-b8 37. Te8-e7 Kg5-h6 38. Tb8-e8 Tg8-h8 39. Tb7-b8 Th8-g8 40. Lb8-a7 Tg8-h8 41. Ta7-b7 Th8-g8 42. Ta5-a7 Tg8-h8 43. Tb5-a5 Th8-g8 44. Tb4-b5 Tg8-h8 45. Th4-b4 Tf8-g8 46. Th1-h4 Th8-f8 47. h2xLg3 (2022-01-23)
Thomas Volet: Mario, I do appreciate your tactfully lengthy unwind, but this effort at economy is hugely flawed. (2022-01-23)
comment
James Malcom: What is the author's solution and the cook? (2022-01-21)
Mario Richter: I do not the "official cook", but the following should work:
R: 1. Th1-g1 Kf5-e4 2. Th4-h1 Kg6-f5 3. Td4-h4 Kh6-g6 4. Td1-d4 Kg6-h6 5. Ta1-d1 Kh6-g6 6. Ta3-a1 Kg6-h6 7. Tb3-a3 Kh6-g6 8. Tb5-b3 Kg6-h6 9. Ta5-b5 Kh6-g6 10. Lb8-a7 Kg6-h6 11. Ta7-a5 Kh6-g6 12. Tb7-a7 Kg6-h6 13. La7-b8 Kh6-g6 14. Kg8-f8 Kg6-h6 15. Tb8-b7 Kh6-g6 16. Tf8-b8 Te8-e7 17. Lb8-a7 Tc8-e8 18. La7-b8 Tb8-c8 19. Te8-f8 Tb7-b8 20. Lb8-a7 Ta7-b7 21. Kf8-g8 Ta5-a7 22. Ke7-f8 Tb5-a5 23. Kd8-e7 Tb4-b5 24. Kc8-d8 Th4-b4 25. Kb7-c8 Th5-h4 26. Ka6-b7 Tg5-h5 27. Kb5-a6 Tf5-g5 28. Kc4-b5 Te5-f5 29. Kb3-c4 Te4-e5 30. Ka2-b3 Tb4-e4 31. Kb1-a2 Tb5-b4 32. Te7-e8 Ta5-b5 33. Kc1-b1 Ta7-a5 34. Kd1-c1 Tb7-a7 35. La7-b8 Tb8-b7 36. Ke1-d1 Th8-b8 37. Te8-e7 Kg5-h6 38. Tb8-e8 Tg8-h8 39. Tb7-b8 Th8-g8 40. Lb8-a7 Tg8-h8 41. Ta7-b7 Th8-g8 42. Ta5-a7 Tg8-h8 43. Tb5-a5 Th8-g8 44. Tb4-b5 Tg8-h8 45. Th4-b4 Tf8-g8 46. Th1-h4 Th8-f8 47. h2xLg3 (2022-01-23)
Thomas Volet: Mario, I do appreciate your tactfully lengthy unwind, but this effort at economy is hugely flawed. (2022-01-23)
comment
Keywords: 50 move rule
Genre: Retro
FEN: N4K2/B1pprPpp/1p2p3/2p5/N3k3/2P3P1/1PP1PPP1/5BR1
Input: Gerd Wilts, 1996-10-26
Last update: A.Buchanan, 2017-03-22 more...
Genre: Retro
FEN: N4K2/B1pprPpp/1p2p3/2p5/N3k3/2P3P1/1PP1PPP1/5BR1
Input: Gerd Wilts, 1996-10-26
Last update: A.Buchanan, 2017-03-22 more...
52 - P0008879
Arpad Molnar
R259 The Problemist 01/1997
(10+9) cooked
Welches waren die letzten 16 Einzelzüge?
Arpad Molnar
R259 The Problemist 01/1997
(10+9) cooked
Welches waren die letzten 16 Einzelzüge?
Cook: R: 1. a4xSb3+ Lb8-c7 2. d6xSc5 Ld8-b6 3. b6xSa5 e7xTd8=L
Keywords: Last Moves? (16), Non-standard material
Genre: Retro
FEN: 8/p1Bp1ppp/RBR5/pBpP4/1kP5/Rp6/1PK5/8
Input: Gerd Wilts, 1998-06-26
Last update: Gerd Wilts, 2006-10-08 more...
Genre: Retro
FEN: 8/p1Bp1ppp/RBR5/pBpP4/1kP5/Rp6/1PK5/8
Input: Gerd Wilts, 1998-06-26
Last update: Gerd Wilts, 2006-10-08 more...
1. Sc8 2. Sxb6 3. Sxa4 4. Sb2 5. Sd3 6. Se1 7. Sf3 8. exd3ep 9. Kd4 10. Ke4 11. Sd4 12. gxf3ep 13. Kf4 14. Kg4 15. f4 16. Sf5 exf3#
NL. Verbesserung 1998 erschienen: P1000933
Cook: 1. Lf7 2. Txe5 3. Kd5 4. Ke6 5. Kf6 6. Sg7 7. Se6 8. Tg6 9. Dh7 10. Dg7 dxe5#
außerdem Dual in der AL
1. Sc8 2. Sxb6 3. Sxa4 4. Sc5
NL. Verbesserung 1998 erschienen: P1000933
Cook: 1. Lf7 2. Txe5 3. Kd5 4. Ke6 5. Kf6 6. Sg7 7. Se6 8. Tg6 9. Dh7 10. Dg7 dxe5#
außerdem Dual in der AL
1. Sc8 2. Sxb6 3. Sxa4 4. Sc5
Keywords: Seriesmover, Consequent, En passant (x2)
Genre: Retro, Fairies
FEN: 8/4n3/1P2b3/3rPprn/P1kPpPpq/2p1p1p1/4P1Pb/7K
Input: Gerd Wilts, 1998-06-26
Last update: Mario Richter, 2021-09-16 more...
Genre: Retro, Fairies
FEN: 8/4n3/1P2b3/3rPprn/P1kPpPpq/2p1p1p1/4P1Pb/7K
Input: Gerd Wilts, 1998-06-26
Last update: Mario Richter, 2021-09-16 more...
1. Lf2 e3 2. Kf3 Txf7 3. Kg3+ Tb7 4. Tf3 Tg7#
Cook: NL:
1. Kg4 Kg2 2. f5 Kf1 3. Kg3 Tb5 4. f4 Tg5#
1. Kf2 e4 2. Ke1 Kg2 3. Tf3 Kxf3 4. Kf1 Tb1#
Cook: NL:
1. Kg4 Kg2 2. f5 Kf1 3. Kg3 Tb5 4. f4 Tg5#
1. Kf2 e4 2. Ke1 Kg2 3. Tf3 Kxf3 4. Kf1 Tb1#
Korrektur Timo Koistinen: +sDa2, +sBh6h7, ST 2001, p.179
Yuri Bilokin: Correction: bBd4-g1, bPh3 b4r2/1R3p2/8/8/7p/6kp/4P2p/6bK (3+8) (2022-04-21)
comment
Yuri Bilokin: Correction: bBd4-g1, bPh3 b4r2/1R3p2/8/8/7p/6kp/4P2p/6bK (3+8) (2022-04-21)
comment
Genre: h#
FEN: b4r2/1R3p2/8/8/3b3p/6kr/4P2p/7K
Reprints: 584 FIDE Album 1965-1967
Input: Gerd Wilts, 1996-06-06
Last update: Marcin Banaszek, 2016-01-22 more...
55 - P0500421
Reto List
(5) Die Schwalbe 84 12/1983
3. ehrende Erwähnung
(4+12) cooked
h#2
b) wSg2 nach h6
Reto List
(5) Die Schwalbe 84 12/1983
3. ehrende Erwähnung
(4+12) cooked
h#2
b) wSg2 nach h6
a) 1. Sh5 Tf4 2. Tg3 Txh4#
b) 1. Sf5 Lf3 2. Lg3 Lg4#
Cook: NL a)
1. La1 Ka7 2. Lg4 Sf4#
1. Lg4+ Ka7 2. Lf4 Sxf4#
b) 1. Sf5 Lf3 2. Lg3 Lg4#
Cook: NL a)
1. La1 Ka7 2. Lg4 Sf4#
1. Lg4+ Ka7 2. Lf4 Sxf4#
Kees: possible fix: +wBa7 +sSa5 (2023-06-07)
A.Buchanan: This fix certainly removes the cooks, but the twinning in the original problem feels clumsy. (2023-06-08)
comment
A.Buchanan: This fix certainly removes the cooks, but the twinning in the original problem feels clumsy. (2023-06-08)
comment
Genre: h#
FEN: K1b4q/6pr/3p1pp1/4b3/7p/3r2nk/5RN1/7B
Input: Gerd Wilts, 1996-06-06
Last update: Alfred Pfeiffer, 2019-04-26 more...
56 - P0501007
Thomas Brand
Jörg A. R. Kuhlmann
Bo Lindgren
Arno Tüngler
Kjell Widlert
3707v feenschach 63 12/1982
(1+8) cooked
h#8
Circe (nur Offiziere) Rex inklusive
Thomas Brand
Jörg A. R. Kuhlmann
Bo Lindgren
Arno Tüngler
Kjell Widlert
3707v feenschach 63 12/1982
(1+8) cooked
h#8
Circe (nur Offiziere) Rex inklusive
1. b1=S Kb2 2. a1=T Kxa1[+sTh8] 3. Te8 Kxb1[+sSg8] 4. c1=L Kxc1[+sLf8] 5. Ke1 Kc2 6. Se7 Kd3 7. f1=D+ Ke3 8. De2+ Kxe2[+sDd8]#
Korrektur in feenschach 1988
Cook: Michel Caillaud (2022-07-19): cooked using Jacobi 0.7.5 :
1. a1=L Kb3 2. b1=T Ka2 3. c1=D Kxa1[+sLf8] 4. Sb6 Kxb1[+sTa8] 5. Te8 Kxc1[+sDd8]+ 6. Ke1 Kc2 7. Sd5 Kd3 8. Se7+ Ke2#
Korrektur in feenschach 1988
Cook: Michel Caillaud (2022-07-19): cooked using Jacobi 0.7.5 :
1. a1=L Kb3 2. b1=T Ka2 3. c1=D Kxa1[+sLf8] 4. Sb6 Kxb1[+sTa8] 5. Te8 Kxc1[+sDd8]+ 6. Ke1 Kc2 7. Sd5 Kd3 8. Se7+ Ke2#
Michel Caillaud: cooked using Jacobi 0.7.5 :
1.a1=L Kb3 2.b1=T Ka2 3.c1=D Kxa1(Lf8) 4.Sb6 Kxb1(Ta8) 5.Te8 Kxc1(Dd8)+ 6.Ke1 Kc2 7.Sd5 Kd3 8.Se7+ Ke2# (2022-07-19)
more ...
comment
1.a1=L Kb3 2.b1=T Ka2 3.c1=D Kxa1(Lf8) 4.Sb6 Kxb1(Ta8) 5.Te8 Kxc1(Dd8)+ 6.Ke1 Kc2 7.Sd5 Kd3 8.Se7+ Ke2# (2022-07-19)
more ...
comment
Keywords: Circe (Rex inklusive), Allumwandlung
Genre: Fairies
FEN: n7/8/4pp2/8/8/2K5/pppk1p2/8
Reprints: The Problemist , p. 91, 09/1985
Rex Multiplex 12/1988
969 Minimalkunst im Schach 2006
My problems [Arno Tüngler] , p. 82, 06/2009
Input: Gerd Wilts, 1996-06-06
Last update: Mario Richter, 2022-07-19 more...
Genre: Fairies
FEN: n7/8/4pp2/8/8/2K5/pppk1p2/8
Reprints: The Problemist , p. 91, 09/1985
Rex Multiplex 12/1988
969 Minimalkunst im Schach 2006
My problems [Arno Tüngler] , p. 82, 06/2009
Input: Gerd Wilts, 1996-06-06
Last update: Mario Richter, 2022-07-19 more...
57 - P0501008
Karl Pohlheim
5883 Schach 12/1968
Ein Sylversterscherz
(1+8) cooked
ser-h=16
b) sGh8 nach h2
sDU=Grashüpfer
Karl Pohlheim
5883 Schach 12/1968
Ein Sylversterscherz
(1+8) cooked
ser-h=16
b) sGh8 nach h2
sDU=Grashüpfer
a) 1. Kg3 2. Kh2 3. Gh1 4. Kg2 5. Gg1 6. Kf2 7. Gf1 8. Ke2 9. Ge1 10. Kd2 11. Gd1 12. Kc2 13. Gc1 14. Kb2 15. Gb1 16. Ka1 Ka3=
b) 1. Kg5 2. Kh6 3. Gh7 4. Kg7 5. Gg6 6. Kf6 7. Gf5 8. Ke5 9. Ke4 10. Kd4 11. Gd3 12. Kc3 13. Gc2 14. Kb2 15. Gb1 16. Ka1 Ka3=
Cook: in b) 1. Ke5 2. Ke6 3. Gd5 4. Kf5 5. Gg4 6. Ke4 7. Ge3 8. Kd3 9. Kc2 10. Kb2 11. Ga8 12. Gb1 13. Ka1 14. Ga3 Kxa3=
b) 1. Kg5 2. Kh6 3. Gh7 4. Kg7 5. Gg6 6. Kf6 7. Gf5 8. Ke5 9. Ke4 10. Kd4 11. Gd3 12. Kc3 13. Gc2 14. Kb2 15. Gb1 16. Ka1 Ka3=
Cook: in b) 1. Ke5 2. Ke6 3. Gd5 4. Kf5 5. Gg4 6. Ke4 7. Ge3 8. Kd3 9. Kc2 10. Kb2 11. Ga8 12. Gb1 13. Ka1 14. Ga3 Kxa3=
HBae: Korrekturvorschlag b) sKf4 nach h7
1.Kh6 2.Gh5 3.Gh7 4.Kg7 ... usw. Eine Prüfung mit Popeye habe ich nach nach 13 Std. abgebrochen. (2021-06-06)
HBae: Mit sKh7 auch NL: 1.Ga8 2.Ga3 3.Kg6 4.Gg5 5.Kf5 6.Ke4 7.Gh4 8.Gh3 9.Gd4 10.Ke3 11.Ge2 12.Kd3 13.Kc3 14.Kb2 15.Gb1 16.Ka1 Kxa3= (2021-06-06)
comment
1.Kh6 2.Gh5 3.Gh7 4.Kg7 ... usw. Eine Prüfung mit Popeye habe ich nach nach 13 Std. abgebrochen. (2021-06-06)
HBae: Mit sKh7 auch NL: 1.Ga8 2.Ga3 3.Kg6 4.Gg5 5.Kf5 6.Ke4 7.Gh4 8.Gh3 9.Gd4 10.Ke3 11.Ge2 12.Kd3 13.Kc3 14.Kb2 15.Gb1 16.Ka1 Kxa3= (2021-06-06)
comment
Keywords: Seriesmover
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 1*2q*2q*2q*2q*2q*2q*2q/8/8/8/K4k2/8/8/8
Input: Gerd Wilts, 1996-06-06
Last update: Vaclav Kotesovec, 2020-07-01 more...
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: 1*2q*2q*2q*2q*2q*2q*2q/8/8/8/K4k2/8/8/8
Input: Gerd Wilts, 1996-06-06
Last update: Vaclav Kotesovec, 2020-07-01 more...
1. b1=D 2. Db4 3. De1 4. b4 5. b3 6. b2 7. b1=D 8. Db6 9. Dbg1 10. b5 11. b4 12. b3 13. b2 14. b1=D 15. Db8 16. c1=D 17. Dcf4 18. Dff1 19. Tf2 20. f5 21. f4 22. f3 23. Sh1 24. Dbh2 25. Tg3 26. Lh3 27. Dfg2 28. Def1 29. Ke1 Kc2 =
Cook: NL (s.PK137):
von George P.~Sphicas wird folgende von
Mike Neumeier gefundene NL mitgeteilt: 1.Tf2 2.b1\prom{S} 3.Sd2
4.Sf3 5.Sg1 6.b4 9.b1\prom{L} 10.La2 11.Lad5 12.Lf1 13.L5g2 14.b5
18.b1\prom{D} 19.Db4 20.De1 21.f5 23.f3 24.Sh1 25.Tg3 26.Lh3
27.L1g2 28.Df1 29.Ke1 K×c2=
Cook: NL (s.PK137):
von George P.~Sphicas wird folgende von
Mike Neumeier gefundene NL mitgeteilt: 1.Tf2 2.b1\prom{S} 3.Sd2
4.Sf3 5.Sg1 6.b4 9.b1\prom{L} 10.La2 11.Lad5 12.Lf1 13.L5g2 14.b5
18.b1\prom{D} 19.Db4 20.De1 21.f5 23.f3 24.Sh1 25.Tg3 26.Lh3
27.L1g2 28.Df1 29.Ke1 K×c2=
Erich Bartel: Nachdrucke:
1) G29 Fide-Album 1989-1991
2) 975 Minimalkunst im Schach 2006 (2013-03-07)
paul: See P1330761 as correction. (2022-07-10)
comment
1) G29 Fide-Album 1989-1991
2) 975 Minimalkunst im Schach 2006 (2013-03-07)
paul: See P1330761 as correction. (2022-07-10)
comment
Keywords: Promotion (dddd), Rex solus (w), Seriesmover
Genre: Fairies
FEN: 8/1p3p2/5r2/1p6/6p1/3K2nr/1pp1p1b1/3k4
Input: Gerd Wilts, 1996-06-06
Last update: Alfred Pfeiffer, 2013-07-17 more...
Genre: Fairies
FEN: 8/1p3p2/5r2/1p6/6p1/3K2nr/1pp1p1b1/3k4
Input: Gerd Wilts, 1996-06-06
Last update: Alfred Pfeiffer, 2013-07-17 more...
1. Df8 Lf1 2. Le6 Se3 3. Dc5+ Sc4+ 4. Kd5 Lg2#
NL:
1. Df4 Lf1 2. Le6 Se1 3. Kd5 Sd3 4. Dd6 Lg2#
NL:
1. Df4 Lf1 2. Le6 Se1 3. Kd5 Sd3 4. Dd6 Lg2#
Yuri Bilokin: correction bQf1-c1, then a1=b1 8/8/4k3/8/3B3p/2bK4/7N/2q5 (3+4) h#4
1.Qg5 Bg1 2.Bf6 Sf3 3.Qd5+ Sd4+ 4.Ke5 Bh2# (MM) (2023-05-24)
comment
1.Qg5 Bg1 2.Bf6 Sf3 3.Qd5+ Sd4+ 4.Ke5 Bh2# (MM) (2023-05-24)
comment
Keywords: Interchange (LS (20))
Genre: h#
FEN: 8/8/8/4k3/2B3p1/1bK5/6N1/5q2
Input: hpr, 1996-06-14
Last update: hpr, 1999-05-22 more...
Genre: h#
FEN: 8/8/8/4k3/2B3p1/1bK5/6N1/5q2
Input: hpr, 1996-06-14
Last update: hpr, 1999-05-22 more...
*) 1. ... f4 2. Se7 f5 3. 0-0 f6 4. Kh8 f7#
1) 1. Se7 f4 2. 0-0 f5 3. Te8 f6 4. Kh8 f7#
NL:
*) 1. ... f4 2. Sh6 f5 3. 0-0 f6 4. Kh8 f7#
1) 1. Se7 f4 2. 0-0 f5 3. Te8 f6 4. Kh8 f7#
NL:
*) 1. ... f4 2. Sh6 f5 3. 0-0 f6 4. Kh8 f7#
Keywords: Interchange (kt (9)), Castling (sk)
Genre: h#
FEN: 4k1nr/K6p/8/8/8/8/1B3P2/8
Input: hpr, 1996-07-01
Last update: Alfred Pfeiffer, 2013-08-21 more...
Genre: h#
FEN: 4k1nr/K6p/8/8/8/8/1B3P2/8
Input: hpr, 1996-07-01
Last update: Alfred Pfeiffer, 2013-08-21 more...
1. f2 Tb1 2. Sf3 Tb2 3. Ke1 Txc2 4. Sd2 Tc1#
Cook: NL
1. Sg2 Th1 2. f2 Th8 3. Ke1 Th2 4. Kf1 Th1#
Cook: NL
1. Sg2 Th1 2. f2 Th8 3. Ke1 Th2 4. Kf1 Th1#
Adrian Storisteanu: Possible fix:
Kh1 Rb1 / Kc2 Sd1 ppb2 d2 e3 g3 h2 h3 (2+8) h#4
1.e2 Ra1 2.Se3 Ra2 3.Kd1 Rxb2 4.Sc2 Rb1# (2015-07-24)
milan: milan frelih: i've found key solution,your move? (2015-07-24)
milan: Milan Frelih:[Ka1 h1,+bBb1,wPe3=bPe4] (2015-08-31)
Yuri Bilokin: Correction: wKa1-h1, wRc1-b1, bKd2-c2, bPc2-b2, bNe1-d1, bPe2-d2, bPf3-e3, -bPb3, -bPd3, -wPe3, -bPe4, +bQg3 8/8/8/8/8/4p1q1/1pkp4/1R1n3K (2+6) h#4 1.e2 Ra1 2.Se3 Ra2 3.Kd1 Rxb2 4.Sc2 Rb1# (MM) (2022-04-21)
comment
Kh1 Rb1 / Kc2 Sd1 ppb2 d2 e3 g3 h2 h3 (2+8) h#4
1.e2 Ra1 2.Se3 Ra2 3.Kd1 Rxb2 4.Sc2 Rb1# (2015-07-24)
milan: milan frelih: i've found key solution,your move? (2015-07-24)
milan: Milan Frelih:[Ka1 h1,+bBb1,wPe3=bPe4] (2015-08-31)
Yuri Bilokin: Correction: wKa1-h1, wRc1-b1, bKd2-c2, bPc2-b2, bNe1-d1, bPe2-d2, bPf3-e3, -bPb3, -bPd3, -wPe3, -bPe4, +bQg3 8/8/8/8/8/4p1q1/1pkp4/1R1n3K (2+6) h#4 1.e2 Ra1 2.Se3 Ra2 3.Kd1 Rxb2 4.Sc2 Rb1# (MM) (2022-04-21)
comment
Keywords: Interchange (ks (2)), Pure Round Trip (T)
Genre: h#
FEN: 8/8/8/8/4p3/1pp1Pp2/2pkp3/K1R1n3
Input: hpr, 1996-07-01
Last update: Alfred Pfeiffer, 2015-07-24 more...
Genre: h#
FEN: 8/8/8/8/4p3/1pp1Pp2/2pkp3/K1R1n3
Input: hpr, 1996-07-01
Last update: Alfred Pfeiffer, 2015-07-24 more...
1. Db6 b5 2. Ld8 bxc6 3. Ke7 c7 4. Df6 c8=S#
NL:
1. Da5 b5 2. Ld8 bxc6 3. Ke7 c7 4. f6 c8=S# ua
NL:
1. Da5 b5 2. Ld8 bxc6 3. Ke7 c7 4. f6 c8=S# ua
Keywords: Interchange (dl (8))
Genre: h#
FEN: 3qb1K1/3n1p2/2pk1b2/5P2/1P6/8/8/8
Input: hpr, 1996-07-01
Last update: hpr, 1999-04-06 more...
Genre: h#
FEN: 3qb1K1/3n1p2/2pk1b2/5P2/1P6/8/8/8
Input: hpr, 1996-07-01
Last update: hpr, 1999-04-06 more...
63 - P0501534
Jean-Michel Trillon
feenschach 7, p. 194, 01/1972
7. ehrende Erwähnung
(6+11) cooked
h#5
Jean-Michel Trillon
feenschach 7, p. 194, 01/1972
7. ehrende Erwähnung
(6+11) cooked
h#5
1. Sf3+ Se1 2. Se5 f3 3. Lg1 d4 4. Sc4 Sd3 5. Sb6 Sc5#
NL:
1. Ld4 Se3 2. Lxb2 Kxb2 3. Kb6 Sc4+ 4. Kc5 Kc3 5. d5 d4#
1. Lc5 Sd4 2. La3 bxa3 3. b2 Kc2 4. b1=T Se6 5. Tb6 Sc5#
NL:
1. Ld4 Se3 2. Lxb2 Kxb2 3. Kb6 Sc4+ 4. Kc5 Kc3 5. d5 d4#
1. Lc5 Sd4 2. La3 bxa3 3. b2 Kc2 4. b1=T Se6 5. Tb6 Sc5#
Yuri Bilokin: Correction: wNc2-g2, bPb2, +bRa2, +bPa3 8/q2p4/kbp5/p7/P4p2/pp6/rp1P1PNp/1K4nr (5+14) (2020-11-29)
comment
comment
Keywords: interchange in mating position (ls (50)), Checking key, Line opening, Line closing
Genre: h#
FEN: 8/q2p4/kbp5/p7/P4p2/1p6/1PNP1P1p/1K4nr
Input: hpr, 1996-07-01
Last update: Gunter Jordan, 2021-05-18 more...
Genre: h#
FEN: 8/q2p4/kbp5/p7/P4p2/1p6/1PNP1P1p/1K4nr
Input: hpr, 1996-07-01
Last update: Gunter Jordan, 2021-05-18 more...
1. Sd7 Le5 2. Sf6 Lxg3 3. Sxg4 Le1 4. Lb8 La5 5. Sh2 Lxe6#
NL:
1. Kd8 Lf6 2. Ke8 Lxg5 3. Kf7 Lf4+ 4. Kf6 Ld6 5. Kg5 Lxe7#
uvm
NL:
1. Kd8 Lf6 2. Ke8 Lxg5 3. Kf7 Lf4+ 4. Kf6 Ld6 5. Kg5 Lxe7#
uvm
Yuri Bilokin: Correction: wKd1-h1 nnk5/1p2p3/4p2p/2p3pP/6P1/4p1pB/4P1Pb/B6K (7+12) (2020-11-28)
comment
comment
Keywords: Interchange (ls (72))
Genre: h#
FEN: nnk5/1p2p3/4p2p/2p3pP/4p1P1/4P1pB/6Pb/B2K4
Input: hpr, 1996-07-01
Last update: hpr, 1999-05-23 more...
Genre: h#
FEN: nnk5/1p2p3/4p2p/2p3pP/4p1P1/4P1pB/6Pb/B2K4
Input: hpr, 1996-07-01
Last update: hpr, 1999-05-23 more...
1. Df5 Td4 2. Se6 Lc4 3. Df4 Td5#
NL:
1. Td4 Le4 2. Td6 Lg6 3. Sd5 Te4#
1. Dg8 Txe4+ 2. Kf5 Te8 3. Dg5 Le4#
NL:
1. Td4 Le4 2. Td6 Lg6 3. Sd5 Te4#
1. Dg8 Txe4+ 2. Kf5 Te8 3. Dg5 Le4#
Yuri Bilokin: correction wKa4-b4, bPb7-f7, bPb6-e3, bPg4-f3, bPf4(-bSf4), -bBa7, then rotate 90 8/8/1pr1pp2/2qkrp2/3B4/4R3/4K3/8 (3+8) h#3
1.Qd6 Re4 2.Rc5 Be3 3.Qc6 Rd4# (MM)
Blocking piece replacement (bQ-bR)
Blocking piece replacement (bR-bQ)
Pelle move (white)
Place exchange (bQ/bR)
Place exchange (wR/wB)
Model mate × 1 (2023-05-19)
comment
1.Qd6 Re4 2.Rc5 Be3 3.Qc6 Rd4# (MM)
Blocking piece replacement (bQ-bR)
Blocking piece replacement (bR-bQ)
Pelle move (white)
Place exchange (bQ/bR)
Place exchange (wR/wB)
Model mate × 1 (2023-05-19)
comment
Keywords: Interchange (TL ds)
Genre: h#
FEN: 8/bp6/1p2qr2/3Bk3/K1R1rnp1/8/8/8
Input: hpr, 1996-07-12
Last update: hpr, 1999-06-04 more...
Genre: h#
FEN: 8/bp6/1p2qr2/3Bk3/K1R1rnp1/8/8/8
Input: hpr, 1996-07-12
Last update: hpr, 1999-06-04 more...
1. Sg3 Ld3 2. Kg2 Lf2 3. Tg1 Le1 4. Sf1 Le4#
Cook: NL
1. Tg2 Lc5 2. Sg3 Kd4 3. Kg1 Le2 4. Kf2 Kd3#
Cook: NL
1. Tg2 Lc5 2. Sg3 Kd4 3. Kg1 Le2 4. Kf2 Kd3#
Adrian Storisteanu: Possible fix: +bPd4. (2015-07-25)
Yuri Bilokin: Correction: -bPe6, +bQd7 8/3q4/8/8/2K1n3/7n/7r/4rBBk (3+6)
Consecutive Umnov (mixed, BrB, 2), Place exchange (wB/bR), Place exchange (wB/bS) (2021-11-20)
comment
Yuri Bilokin: Correction: -bPe6, +bQd7 8/3q4/8/8/2K1n3/7n/7r/4rBBk (3+6)
Consecutive Umnov (mixed, BrB, 2), Place exchange (wB/bR), Place exchange (wB/bS) (2021-11-20)
comment
Keywords: Interchange (Lt Ls)
Genre: h#
FEN: 8/8/4p3/8/2K1n3/7n/7r/4rBBk
Input: hpr, 1996-07-15
Last update: Alfred Pfeiffer, 2015-07-25 more...
Genre: h#
FEN: 8/8/4p3/8/2K1n3/7n/7r/4rBBk
Input: hpr, 1996-07-15
Last update: Alfred Pfeiffer, 2015-07-25 more...
1. Db8 c7 2. Sb4 cxb8=S 3. Sa6 Sc6#
NL:
1. Dd2 c7 2. Sd7+ Ka3 3. Sdb8 c8=S#
1. Tb8 c7 2. Le4 c8=D 3. Ka8 Dxa6#
NL:
1. Dd2 c7 2. Sd7+ Ka3 3. Sdb8 c8=S#
1. Tb8 c7 2. Le4 c8=D 3. Ka8 Dxa6#
Yuri Bilokin: correction bBa8, wKb3-e6, wPe2-c3, +bPc4 n7/kr6/npP1K3/8/2p2q2/2P5/8/8 (3+7) (2022-06-05)
Yuri Bilokin: sorry bNa8, bPb6 (2022-06-05)
comment
Yuri Bilokin: sorry bNa8, bPb6 (2022-06-05)
comment
Genre: h#
FEN: b7/kr6/nnP5/8/5q2/1K6/4P3/8
Input: hpr, 1996-07-23
Last update: A.Buchanan, 2022-06-06 more...
Autorabsicht: 1. Lh2 Th1 2. Lg1 Th8 3. Lh2 Th1 4. Lg1 T1h7#
Cook: Viele NL, zB: 1. Sh4 Txg1 2. Sf3 gxf3 3. Kh6 Txg3 4. Kh5 Th1#
Cook: Viele NL, zB: 1. Sh4 Txg1 2. Sf3 gxf3 3. Kh6 Txg3 4. Kh5 Th1#
klären Autorlösung
Anton Baumann: Autorabsicht: 1.Lg2 Th1 2.Lg1 Th8 3.Lh2 Th1 4.Lg1 Th7#
Korrektur Version A. Bulawka: -sSg6, sKg7 nach f7, +sBe6, +sBf5, +sBg6 (vergl. yacpdb 320494) (2020-07-15)
Ladislav Packa: The modified version has an illegal position. (2020-07-15)
A.Buchanan: A good, sound, economical version is 320495 in YACPDB (2020-07-15)
A.Buchanan: Bulawka's correction 320494 is sound and legal (6 sB captures + 2 wL died at home). But Smirnov made a much lighter sound version 320495. (2021-05-16)
more ...
comment
Anton Baumann: Autorabsicht: 1.Lg2 Th1 2.Lg1 Th8 3.Lh2 Th1 4.Lg1 Th7#
Korrektur Version A. Bulawka: -sSg6, sKg7 nach f7, +sBe6, +sBf5, +sBg6 (vergl. yacpdb 320494) (2020-07-15)
Ladislav Packa: The modified version has an illegal position. (2020-07-15)
A.Buchanan: A good, sound, economical version is 320495 in YACPDB (2020-07-15)
A.Buchanan: Bulawka's correction 320494 is sound and legal (6 sB captures + 2 wL died at home). But Smirnov made a much lighter sound version 320495. (2021-05-16)
more ...
comment
Keywords: Superseded by (P1389613)
Genre: h#
FEN: 8/6k1/5pn1/8/8/6p1/PPKPPpP1/RR4b1
Input: hpr, 1996-07-23
Last update: A.Buchanan, 2021-05-16 more...
Genre: h#
FEN: 8/6k1/5pn1/8/8/6p1/PPKPPpP1/RR4b1
Input: hpr, 1996-07-23
Last update: A.Buchanan, 2021-05-16 more...
69 - P0502695
John Niemann
192 FEENSCHACH 3 05/1971
(1+0) cooked
Ergänze wB und sK, so daß ein korrektes Hilfsmatt entsteht
John Niemann
192 FEENSCHACH 3 05/1971
(1+0) cooked
Ergänze wB und sK, so daß ein korrektes Hilfsmatt entsteht
+sKd8, +wBg7
1. Ke7 Kg2 2. Kf6 Kf3 3. Kg5 g8=D+ 4. Kh4 Dg4#
Cook: +wBg7 +sKe8 = h#4
+wBg7 +sKe7,Ske1,Skf2 = h#3.5
1. Ke7 Kg2 2. Kf6 Kf3 3. Kg5 g8=D+ 4. Kh4 Dg4#
Cook: +wBg7 +sKe8 = h#4
+wBg7 +sKe7,Ske1,Skf2 = h#3.5
Duplicate Diagram: P0530804
Keywords: Aristocrat, Miniature, Kindergarten Problem, Constrained problem
Genre: Fairies, h#
FEN: 8/8/8/8/8/8/8/7K
Input: hpr, 1996-07-23
Last update: A.Buchanan, 2021-05-24 more...
Genre: Fairies, h#
FEN: 8/8/8/8/8/8/8/7K
Input: hpr, 1996-07-23
Last update: A.Buchanan, 2021-05-24 more...
1) 1. Ld3 e3 2. Le2 e4#
2) 1. Se3 g3 2. Sg2 g4#
3) 1. Sg5 Lg8 2. Sf7 Lh7#
4) 1. Sf8 Txd6 2. Sg6 Lxe6#
NL:
1. Dg3 dxe5 2. Df4 Sg7#
1. Th1 dxe5 2. Sxe5 g4#
1. Ld3 exd3 2. De4 dxe4#
2) 1. Se3 g3 2. Sg2 g4#
3) 1. Sg5 Lg8 2. Sf7 Lh7#
4) 1. Sf8 Txd6 2. Sg6 Lxe6#
NL:
1. Dg3 dxe5 2. Df4 Sg7#
1. Th1 dxe5 2. Sxe5 g4#
1. Ld3 exd3 2. De4 dxe4#
Kees: I don't know how to send a diagram, but I fixed this one with a whole different position, but with the 4 thematic solutions I can send a FEN-notation with the 4 solutions
3K4/1pr1B1p1/s1pp2P1/2Rpk1S1/2Pb1sP1/3pPp2/4P3/3q1r2
1.Fc3 exd3 2.Fd2 d4‡
1.Ch3 exf3 2.Cf2 f4‡
1.Cxg6 Ff8 2.Ce7 Fxg7‡
1.Ch5 Txc6 2.Cf6 Fxd6‡ (2023-06-09)
comment
3K4/1pr1B1p1/s1pp2P1/2Rpk1S1/2Pb1sP1/3pPp2/4P3/3q1r2
1.Fc3 exd3 2.Fd2 d4‡
1.Ch3 exf3 2.Cf2 f4‡
1.Cxg6 Ff8 2.Ce7 Fxg7‡
1.Ch5 Txc6 2.Cf6 Fxd6‡ (2023-06-09)
comment
Keywords: Interchange (Bl,Bs,Ls)
Genre: h#
FEN: 4K3/2pr1B1n/3pp2p/3Rbk1N/3Pb1nP/5P2/4P1P1/4q1r1
Input: hpr, 1996-07-28
Last update: hpr, 1999-06-05 more...
Genre: h#
FEN: 4K3/2pr1B1n/3pp2p/3Rbk1N/3Pb1nP/5P2/4P1P1/4q1r1
Input: hpr, 1996-07-28
Last update: hpr, 1999-06-05 more...
1. Kb5 Kf5 2. Kb6 Kg6 3. Kc7 Kf7 4. Kd6 Lh2#
Cook: NL
1. Ta6 Sg8 2. Td6 Kf4 3. Kd5 Kf5 4. c4 Se7#
Cook: NL
1. Ta6 Sg8 2. Td6 Kf4 3. Kd5 Kf5 4. c4 Se7#
(KSp h)
Adrian Storisteanu: See P0502965 (failed correction attempt / self-anticipation). (2015-07-25)
milan: milan frelih:[Ke5 to h4 +bPh6] 1.Kb5 Kh5 2.Kb6 Kg6 3.Kc7 Kf7 4.Kd6 Bh2# (2015-07-26)
Yuri Bilokin: correction b1=a1, then wKd5-h4, -bPf7 8/8/1r2N3/1p6/1k5K/8/8/5B2 (3+3) h#4
1.Ka5 Kg5 2.Ka6 Kf6 3.Kb7 Ke7 4.Kc6 Bg2# (MM) (2023-05-24)
comment
Adrian Storisteanu: See P0502965 (failed correction attempt / self-anticipation). (2015-07-25)
milan: milan frelih:[Ke5 to h4 +bPh6] 1.Kb5 Kh5 2.Kb6 Kg6 3.Kc7 Kf7 4.Kd6 Bh2# (2015-07-26)
Yuri Bilokin: correction b1=a1, then wKd5-h4, -bPf7 8/8/1r2N3/1p6/1k5K/8/8/5B2 (3+3) h#4
1.Ka5 Kg5 2.Ka6 Kf6 3.Kb7 Ke7 4.Kc6 Bg2# (MM) (2023-05-24)
comment
Keywords: Figurenspiel
Genre: h#
FEN: 8/6p1/2r2N2/2p1K3/2k5/8/8/6B1
Input: hpr, 1996-08-30
Last update: Alfred Pfeiffer, 2015-07-25 more...
Genre: h#
FEN: 8/6p1/2r2N2/2p1K3/2k5/8/8/6B1
Input: hpr, 1996-08-30
Last update: Alfred Pfeiffer, 2015-07-25 more...
72 - P0503022
Claude Goumondy
Turnier des Tschechischen Sportbundes 1981-1982
1. Lob
(6+6) cooked
h#2
b) wTe1 nach a4
c) sDe5 tauschen mit sSe4
d) sSg6 nach h4
Claude Goumondy
Turnier des Tschechischen Sportbundes 1981-1982
1. Lob
(6+6) cooked
h#2
b) wTe1 nach a4
c) sDe5 tauschen mit sSe4
d) sSg6 nach h4
a) 1. Ke6 Lxe4 2. Df6 Ld5#
b) 1. Kc5 Lxe5 2. Sd6 Ld4#
c) 1. Kxc6 Txe4 2. Sd7 Te6#
d) 1. Kd6 Txe5 2. Sc5 Te6#
NL:
d) 1. Ke6 Tf5 2. Sd6 Txe5#
b) 1. Kc5 Lxe5 2. Sd6 Ld4#
c) 1. Kxc6 Txe4 2. Sd7 Te6#
d) 1. Kd6 Txe5 2. Sc5 Te6#
NL:
d) 1. Ke6 Tf5 2. Sd6 Txe5#
Keywords: Figurenspiel
Genre: h#
FEN: 8/8/2P3nK/3kq2R/4n1r1/5B1r/7B/4R3
Input: hpr, 1996-09-03
Last update: hpr, 1999-06-06 more...
Genre: h#
FEN: 8/8/2P3nK/3kq2R/4n1r1/5B1r/7B/4R3
Input: hpr, 1996-09-03
Last update: hpr, 1999-06-06 more...
1. Th1 Sa1 2. Th4 Sh1 3. Tf4 Sc2#
Cook: NL
1. Tf1 Kh8 2. Tf2 Sa1 3. Df4 Sc2#
1. Df4 Sxa1 2. e1=L h6 3. Lf2 Sc2#
Cook: NL
1. Tf1 Kh8 2. Tf2 Sa1 3. Df4 Sc2#
1. Df4 Sxa1 2. e1=L h6 3. Lf2 Sc2#
teilweiser Rundlauf t
Adrian Storisteanu: Possible fix: bPe4 (instead of the bQ), +wPf5, +bPf6 (6+9). (2015-08-15)
Yuri Bilokin: correction bRd3(-bNd3), bBe4(-bQe4), bPf3(-bBf3) 6K1/8/8/7P/3Pb3/pN1rkpN1/3pp3/r7 (5+8) (2022-12-19)
A.Buchanan: This is very clean Yuri well done. I would shift bPa3 to a5. wK must be exactly on g8, else the problem is cooked (2022-12-20)
comment
Adrian Storisteanu: Possible fix: bPe4 (instead of the bQ), +wPf5, +bPf6 (6+9). (2015-08-15)
Yuri Bilokin: correction bRd3(-bNd3), bBe4(-bQe4), bPf3(-bBf3) 6K1/8/8/7P/3Pb3/pN1rkpN1/3pp3/r7 (5+8) (2022-12-19)
A.Buchanan: This is very clean Yuri well done. I would shift bPa3 to a5. wK must be exactly on g8, else the problem is cooked (2022-12-20)
comment
Keywords: Pure Round Trip
Genre: h#
FEN: 6K1/8/8/7P/3Pq3/pN1nkbN1/3pp3/r7
Input: hpr, 1996-09-05
Last update: Alfred Pfeiffer, 2015-08-15 more...
Genre: h#
FEN: 6K1/8/8/7P/3Pq3/pN1nkbN1/3pp3/r7
Input: hpr, 1996-09-05
Last update: Alfred Pfeiffer, 2015-08-15 more...
74 - P0503402
Laszlo Barna
3. Makuc-Moder-Gedenkturnier 03/1970
2. Preis
(4+9) cooked
h#2
b) wSc8 nach c6
c) wSc8 nach e6
d) wSc8 nach e4
Laszlo Barna
3. Makuc-Moder-Gedenkturnier 03/1970
2. Preis
(4+9) cooked
h#2
b) wSc8 nach c6
c) wSc8 nach e6
d) wSc8 nach e4
a) 1. Tf6 g8=D 2. Kc6 De8#
b) 1. Tc3 g8=L 2. Kc8 Le6#
c) 1. Te5 g8=S 2. Ke8 Sf6#
d) 1. Tc3 g8=T 2. Ke6 Te8#
NL:
1. Dh1 Sf6+ 2. Kd8 g8=D#
uvm
b) 1. Tc3 g8=L 2. Kc8 Le6#
c) 1. Te5 g8=S 2. Ke8 Sf6#
d) 1. Tc3 g8=T 2. Ke6 Te8#
NL:
1. Dh1 Sf6+ 2. Kd8 g8=D#
uvm
KSp Königsstern
t/Allumwandlung, u/61104
Daniel Novomesky: Add bPh5. C+ (2008-03-03)
Yuri Bilokin: correction +bPh5 2N4K/3k2PR/6pq/2rp1rpp/8/8/b7/b7 (4+10)
Allumwandlung (white)
JT Onkoud 50 theme (double)
Promotion (QRBS, 4)
Star (bK)
Mates on the same square × 2 (2023-06-26)
comment
t/Allumwandlung, u/61104
Daniel Novomesky: Add bPh5. C+ (2008-03-03)
Yuri Bilokin: correction +bPh5 2N4K/3k2PR/6pq/2rp1rpp/8/8/b7/b7 (4+10)
Allumwandlung (white)
JT Onkoud 50 theme (double)
Promotion (QRBS, 4)
Star (bK)
Mates on the same square × 2 (2023-06-26)
comment
Keywords: Allumwandlung, Figurenspiel
Genre: h#
FEN: 2N4K/3k2PR/6pq/2rp1rp1/8/8/b7/b7
Reprints: feenschach , p. 58, 04/1975
Input: Erich Bartel, 1996-09-17
Last update: hpr, 1999-06-12 more...
Genre: h#
FEN: 2N4K/3k2PR/6pq/2rp1rp1/8/8/b7/b7
Reprints: feenschach , p. 58, 04/1975
Input: Erich Bartel, 1996-09-17
Last update: hpr, 1999-06-12 more...
1. Kc3 De6 2. Kxd4 Dxd6+ 3. Ke4 De5#
NL
1. Kc3 Df5 2. Tf8 Kxd6 3. Kxd4 Dd3#
NL
1. Kc3 Df5 2. Tf8 Kxd6 3. Kxd4 Dd3#
Korrektur Hilmar Ebert: +sSe8 (erhält Quasi-Symmetrie)
Yuri Bilokin: Not all pieces in quasi-symmetry are necessary: the white pawn on h2 is superfluous. It is clear that more than 100 years ago there was such a direction in art, although some authors indulge in technical white pawns in our time. (2022-12-19)
A.Buchanan: The technical term for such a position is "symmetric constellation". "Quasi-symmetry" is where the position is nearly symmetric but not quite (e.g. game array except wPa2 is missing). In a symmetric constellation any redundant piece is a definite defect. A quasi-symmetric position is not an ornament, more of a consequence of the content. For example, a number of last move type DD problems are quasi-symmetric (2022-12-20)
Yuri Bilokin: Thanks for the interpretation (2022-12-20)
comment
Yuri Bilokin: Not all pieces in quasi-symmetry are necessary: the white pawn on h2 is superfluous. It is clear that more than 100 years ago there was such a direction in art, although some authors indulge in technical white pawns in our time. (2022-12-19)
A.Buchanan: The technical term for such a position is "symmetric constellation". "Quasi-symmetry" is where the position is nearly symmetric but not quite (e.g. game array except wPa2 is missing). In a symmetric constellation any redundant piece is a definite defect. A quasi-symmetric position is not an ornament, more of a consequence of the content. For example, a number of last move type DD problems are quasi-symmetric (2022-12-20)
Yuri Bilokin: Thanks for the interpretation (2022-12-20)
comment
Keywords: Pure Round Trip (D)
Genre: h#
FEN: 8/4K3/3p1r2/4Q3/3P1P2/8/1k2P2P/8
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 2001-12-21 more...
Genre: h#
FEN: 8/4K3/3p1r2/4Q3/3P1P2/8/1k2P2P/8
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 2001-12-21 more...
a) 1. Dd6 Lxb2 2. e6 Lc1 3. Ke5 Lxh6 4. f5 Lg7#
b) 1. Dd4 Le5 2. f6 Lxh2 3. e6 g3 4. Ke5 g4#
NL:
a) 1. Se6 Kxh7 2. Dc3 Kxh6 3. Lc1+ Kh5 4. Lf4 g4#
b) 1. Ke5 g3 2. Sf5 Kg8 3. e6 Lh6 4. Ld4 Lf4#
b) 1. Dd4 Le5 2. f6 Lxh2 3. e6 g3 4. Ke5 g4#
NL:
a) 1. Se6 Kxh7 2. Dc3 Kxh6 3. Lc1+ Kh5 4. Lf4 g4#
b) 1. Ke5 g3 2. Sf5 Kg8 3. e6 Lh6 4. Ld4 Lf4#
Korrekturvorschlag HJS Co+:wKh8, wLg7, wBg3, sKf5, sDf6, sTa2a8, sLb2g8, sBd5e4e7f7h4
Yuri Bilokin: Good fix, can be more economical: r5bK/4ppB1/5q2/3p1k2/4p3/1rb3P1/8/8 (3+10) h#3 b) wPg3-g2
a) 1.Qd6 Bxc3 2.e6 Bd2 3.Ke5 Bh6 4.f5 Bg7# (MM)
b) 1.Qd4 Be5 2.f6 Bh2 3.e6 g3 4.Ke5 g4# (MM) (2021-06-14)
comment
Yuri Bilokin: Good fix, can be more economical: r5bK/4ppB1/5q2/3p1k2/4p3/1rb3P1/8/8 (3+10) h#3 b) wPg3-g2
a) 1.Qd6 Bxc3 2.e6 Bd2 3.Ke5 Bh6 4.f5 Bg7# (MM)
b) 1.Qd4 Be5 2.f6 Bh2 3.e6 g3 4.Ke5 g4# (MM) (2021-06-14)
comment
Keywords: Hilfsmatt-Inder (LB), Pure Round Trip (L)
Genre: h#
FEN: 5n1K/4ppBp/5q1n/3p1k2/4p2p/6P1/rb5r/8
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-13 more...
Genre: h#
FEN: 5n1K/4ppBp/5q1n/3p1k2/4p2p/6P1/rb5r/8
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-13 more...
1. Dxe5+ Lf5 2. Dxd4 Lc2 3. De4 Lb3#
NL:
1. d1=S Le7 2. Sxb2 Sb3 3. f1=S Sc1#
NL:
1. d1=S Le7 2. Sxb2 Sb3 3. f1=S Sc1#
Korrekturvorschlag HJS Co+:wKg5, wLc2f6, wSc5, wBb2d4e5g4, sKa2, sDe4, sTa1h1, sLb1d8, sBb4d5e6f2f7
milan: +wPd6 M.Frelih (2021-12-01)
A.Buchanan: 3b4/5p2/4pB2/2NpP1K1/1p1Pq1P1/8/kPB2p2/rb5r as specified in Korrekturvorschlag HJS is C+ with 2 fewer units (2021-12-02)
comment
milan: +wPd6 M.Frelih (2021-12-01)
A.Buchanan: 3b4/5p2/4pB2/2NpP1K1/1p1Pq1P1/8/kPB2p2/rb5r as specified in Korrekturvorschlag HJS is C+ with 2 fewer units (2021-12-02)
comment
Keywords: Pure Round Trip (d)
Genre: h#
FEN: 3b2n1/5p2/4pB2/1pNpP1K1/3Pq1P1/8/kPBp1p1r/rb6
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-19 more...
Genre: h#
FEN: 3b2n1/5p2/4pB2/1pNpP1K1/3Pq1P1/8/kPBp1p1r/rb6
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-19 more...
1) 1. ... Txb8 2. Kc5 Txe8 3. Lc3 Te5+ 4. Kb4 Tb5#
2) 1. ... Te5 2. Dd6 Txe8 3. Kc5 Tb8 4. Ld4 Tb5#
NL:
1. ... Tb6 2. Kc3 Sb5+ 3. Kb4 Sd6+ 4. Ka5 Sc4#
2) 1. ... Te5 2. Dd6 Txe8 3. Kc5 Tb8 4. Ld4 Tb5#
NL:
1. ... Tb6 2. Kc3 Sb5+ 3. Kb4 Sd6+ 4. Ka5 Sc4#
Yuri Bilokin: correction +bNd2 1q2b2b/N7/8/1R6/3k4/pP6/3n4/7K (4+6)
AntiZielElement (W1, line obstruction)
Areal cycle (wR, with captures, 4) × 2
Exchange of moves (W1/W3)
JT Onkoud 50 theme
Long-trip (wR, 4) × 2. Many-ways (wR, 2) × 4
Zalokotsky theme (wR/wR, 3)
Model mate × 2.Mates on the same square × 2 (2023-05-08)
comment
AntiZielElement (W1, line obstruction)
Areal cycle (wR, with captures, 4) × 2
Exchange of moves (W1/W3)
JT Onkoud 50 theme
Long-trip (wR, 4) × 2. Many-ways (wR, 2) × 4
Zalokotsky theme (wR/wR, 3)
Model mate × 2.Mates on the same square × 2 (2023-05-08)
comment
Keywords: Pure Round Trip
Genre: h#
FEN: 1q2b2b/N7/8/1R6/3k4/pP6/8/7K
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
Genre: h#
FEN: 1q2b2b/N7/8/1R6/3k4/pP6/8/7K
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
1) 1. ... Lxb7 2. Le5 Lc8 3. Kd5 Lxd7 4. Ke4 Lc6#
2) 1. ... Th3 2. Ld2 Txh4 3. Kd3 Tf4 4. Ke3 Tf3#
NL:
1. ... Tf1 2. Kd3 Tf2 3. Sf3 Lb5+ 4. Ke3 Txe2#
2) 1. ... Th3 2. Ld2 Txh4 3. Kd3 Tf4 4. Ke3 Tf3#
NL:
1. ... Tf1 2. Kd3 Tf2 3. Sf3 Lb5+ 4. Ke3 Txe2#
Yuri Bilokin: correction +bQg4, +bPa6, -bNh4, -bPh2 8/1b1p4/p1Bn4/8/2kp1bq1/5R2/4p3/7K (3+9) h#4 0.2.1…
1...Rg3 2.Bd2 Rxg4 3.Kd3 Rf4 4.Ke3 Rf3# (MM)
1...Bxb7 2.Be5 Bc8 3.Kd5 Bxd7 4.Ke4 Bc6# (MM)
Areal cycle (wB, with captures, 4)
Areal cycle (wR, with captures, 4)
Exchange of functions (wRf3/wBc6, Mate / Passive guard)
Long-trip (wB, 4). Long-trip (wR, 4)
Model mate × 2 (2023-05-08)
comment
1...Rg3 2.Bd2 Rxg4 3.Kd3 Rf4 4.Ke3 Rf3# (MM)
1...Bxb7 2.Be5 Bc8 3.Kd5 Bxd7 4.Ke4 Bc6# (MM)
Areal cycle (wB, with captures, 4)
Areal cycle (wR, with captures, 4)
Exchange of functions (wRf3/wBc6, Mate / Passive guard)
Long-trip (wB, 4). Long-trip (wR, 4)
Model mate × 2 (2023-05-08)
comment
Keywords: Pure Round Trip
Genre: h#
FEN: 8/1b1p4/2Bn4/8/2kp1b1n/5R2/4p2p/7K
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
Genre: h#
FEN: 8/1b1p4/2Bn4/8/2kp1b1n/5R2/4p2p/7K
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
80 - P0503977
Leon Loewenton
H241 FIDE Olympia Turnier 14 Leipzig 1960
5. -9. Lob
(6+5) cooked
h#2
5.1...
Leon Loewenton
H241 FIDE Olympia Turnier 14 Leipzig 1960
5. -9. Lob
(6+5) cooked
h#2
5.1...
1) 1. f2 e8=S 2. Ld5 Txf2#
2) 1. Kd6 Tc2 2. Kc7 Le5#
3) 1. Kf6 e8=D 2. Kg7 Th2#
4) 1. Kf4 Th2 2. Kg3 Le5#
5) 1. Kd4 Le6 2. Kc3 Tb3#
NL:
1. Kxf5 e8=D 2. Kf6 Tb5#
2) 1. Kd6 Tc2 2. Kc7 Le5#
3) 1. Kf6 e8=D 2. Kg7 Th2#
4) 1. Kf4 Th2 2. Kg3 Le5#
5) 1. Kd4 Le6 2. Kc3 Tb3#
NL:
1. Kxf5 e8=D 2. Kf6 Tb5#
KSp Königsstern
Kees: Possible fix: +wBb4 -sLh1 (No 5.1... but 4.1... with only KSp Königsstern) (2023-06-05)
A.Buchanan: I agree with +wBb4 for soundness but let's keep sLh1. What's not to like about 5 well differentiated solutions? WinChloe has the same cooked diagram as we do here. (2023-06-06)
Yuri Bilokin: The fifth solution is not necessary, the author has kept it, so after adding the white pawn to the b4 square and without removing the black bishop, we fully preserve the author's intention.
8/4P3/1pP5/4kB2/1P6/4pp2/1R6/B2K3b (7+5) h#2 5.1...
Active sacrifice (black, delayed). BK moves only × 4. Delayed Umnov (bK-wB) × 2
Extended star (bK)
Promotion (QS, 2)
Battery mate × 3. Double-check mate. Mirror mate
Mate from initial bK square × 2 (2023-06-10)
more ...
comment
Kees: Possible fix: +wBb4 -sLh1 (No 5.1... but 4.1... with only KSp Königsstern) (2023-06-05)
A.Buchanan: I agree with +wBb4 for soundness but let's keep sLh1. What's not to like about 5 well differentiated solutions? WinChloe has the same cooked diagram as we do here. (2023-06-06)
Yuri Bilokin: The fifth solution is not necessary, the author has kept it, so after adding the white pawn to the b4 square and without removing the black bishop, we fully preserve the author's intention.
8/4P3/1pP5/4kB2/1P6/4pp2/1R6/B2K3b (7+5) h#2 5.1...
Active sacrifice (black, delayed). BK moves only × 4. Delayed Umnov (bK-wB) × 2
Extended star (bK)
Promotion (QS, 2)
Battery mate × 3. Double-check mate. Mirror mate
Mate from initial bK square × 2 (2023-06-10)
more ...
comment
Keywords: Figurenspiel
Genre: h#
FEN: 8/4P3/1pP5/4kB2/8/4pp2/1R6/B2K3b
Reprints: H241 Schach 19, p. 301, 10/1960
Input: hpr, 1996-10-03
Last update: A.Buchanan, 2023-06-06 more...
Genre: h#
FEN: 8/4P3/1pP5/4kB2/8/4pp2/1R6/B2K3b
Reprints: H241 Schach 19, p. 301, 10/1960
Input: hpr, 1996-10-03
Last update: A.Buchanan, 2023-06-06 more...
81 - P0504564
Leonid Makaronez
1634 Gazeta Czestochowska 22/01/1974
(4+7) cooked
h#2
b) wTc1 nach b1
c) wTc1 nach a5
Leonid Makaronez
1634 Gazeta Czestochowska 22/01/1974
(4+7) cooked
h#2
b) wTc1 nach b1
c) wTc1 nach a5
a) 1. Lg6 Tc7 2. e5 Le7#
b) 1. e5 Lf8 2. g5 Tb6#
c) 1. g5 Ld6 2. Lg6 Le5#
NL:
a) 1. e5 Le1 2. Lg6 Lh4#
ua
b) 1. e5 Lf8 2. g5 Tb6#
c) 1. g5 Ld6 2. Lg6 Le5#
NL:
a) 1. e5 Le1 2. Lg6 Lh4#
ua
SB- SB- SL
klären: fehlen irgendwelche Figuren?
Daniel Novomesky: Add black Pawns b6,d4. C+ (2007-11-04)
Yuri Bilokin: Currection: bPe2-f2, +bPb6, +bPd2, +bPh6 2B5/4ppp1/Kp3k1p/8/1B6/4p3/2bp1p2/2R5 (4+10) (2021-05-20)
Ladislav Packa: Correction: wKa6?f1, -bPe2, +bPa6, b6, g3 (4+9) C+ (2021-05-20)
Yuri Bilokin: Correction: The author's moves are quiet, without taking. Taking it can be more economical. "The captured girl is good for everyone." wKa6-f1, wBb4-a3, bPe2-b6, +bPa6 8/4ppp1/pp3k2/8/8/B3p2B/2b5/2R2K2 (4+8)
a) 1.Bg6 Rc7 2.e5 Be7# (MM)
b) 1.e5 Bf8 2.g5 Rxb6#
c) 1.g5 Bd6 2.Bg6 Be5# (MM) (2021-05-20)
Yuri Bilokin: It is possible without taking r1B5/1K2ppp1/5k2/8/8/B3p3/2bp4/2R5 (4+8) (2021-05-21)
comment
klären: fehlen irgendwelche Figuren?
Daniel Novomesky: Add black Pawns b6,d4. C+ (2007-11-04)
Yuri Bilokin: Currection: bPe2-f2, +bPb6, +bPd2, +bPh6 2B5/4ppp1/Kp3k1p/8/1B6/4p3/2bp1p2/2R5 (4+10) (2021-05-20)
Ladislav Packa: Correction: wKa6?f1, -bPe2, +bPa6, b6, g3 (4+9) C+ (2021-05-20)
Yuri Bilokin: Correction: The author's moves are quiet, without taking. Taking it can be more economical. "The captured girl is good for everyone." wKa6-f1, wBb4-a3, bPe2-b6, +bPa6 8/4ppp1/pp3k2/8/8/B3p2B/2b5/2R2K2 (4+8)
a) 1.Bg6 Rc7 2.e5 Be7# (MM)
b) 1.e5 Bf8 2.g5 Rxb6#
c) 1.g5 Bd6 2.Bg6 Be5# (MM) (2021-05-20)
Yuri Bilokin: It is possible without taking r1B5/1K2ppp1/5k2/8/8/B3p3/2bp4/2R5 (4+8) (2021-05-21)
comment
Keywords: Cycle
Genre: h#
FEN: 8/4ppp1/K4k2/8/1B6/4p2B/2b1p3/2R5
Input: Torsten Linss, 1996-10-09
Last update: hpr, 1999-07-09 more...
Genre: h#
FEN: 8/4ppp1/K4k2/8/1B6/4p2B/2b1p3/2R5
Input: Torsten Linss, 1996-10-09
Last update: hpr, 1999-07-09 more...
1.Dxa3 Tg1 2. Da8 Ta7 3. Dxg2 Txh7 4. Da8 Txg8#
NL:
1. Kb7 Td1 2. Dc5 dxc5 3. Ka6 Td7 4. Ka5 Ta7#
NL:
1. Kb7 Td1 2. Dc5 dxc5 3. Ka6 Td7 4. Ka5 Ta7#
Keywords: Figurenspiel
Genre: h#
FEN: 1k3qn1/7n/8/1p6/3P4/PpP1P2P/1P3PPp/R3K2R
Input: hpr, 1996-12-25
Last update: hpr, 1999-07-10 more...
Genre: h#
FEN: 1k3qn1/7n/8/1p6/3P4/PpP1P2P/1P3PPp/R3K2R
Input: hpr, 1996-12-25
Last update: hpr, 1999-07-10 more...
1. f4 h5 2. f3 h6 3. f2 h7 4. f1=T h8=L 5. Tg1 Le5#
Cook: 1. Kg3 Kb5 2. Kf2 Kc4 3. f4 Sd3+ 4. Ke2 Kc3 5. Kd1 Lf3#
Cook: 1. Kg3 Kb5 2. Kf2 Kc4 3. f4 Sd3+ 4. Ke2 Kc3 5. Kd1 Lf3#
Henrik Juel: Here's a correction: move the pawns one step ahead, then h#4. (2003-10-16)
Henrik Juel: A stronger correction is needed:
move Ph4 to h6, Pf5 to f2, and change stip to h#2.5
Then C+ Popeye 4.61 with solution
1... h7 2.f1=T h8=L 3.Tg1 Le5# (2021-05-23)
A.Buchanan: Hi Henrik: -wSe1 +wLh1 cheekily repairs the original h#5. Indeed you can push the pawns back to make a h#6, still sound 8/5p2/K7/8/8/7P/6Bk/7B You're welcome! :D (2021-05-23)
Henrik Juel: Back in the old days two light-squared white bishops were unthinkable
And some people considered the knight more economical than the bishop
The problem is from a construction contest: h# with white and black promotion (2021-05-24)
A.Buchanan: In some ways, Henrik, composition is about trying to find the best natural form that a problem is trying to take. I wasn't trying to think the unthinkable, but when I finally thought it as the only way to salvage and indeed amplify your idea here, I started looking forward to reading your response :) If you don't want it, that's absolutely fine - I will claim it as an "after HJ", and try to find some "Salon Des Refuses" in which to exhibit it (2021-05-24)
Adrian Storisteanu: Or you could send it to the "Salon des Fous":
Kd1 Bg2 Bh1 ph3 / Kh2 pa5 (4+2)
1.a4 h4 2.a3 h5 3.a2 h6 4.a1B h7 5.Bd4 h8B 6.Bg1 Be5# (2021-05-25)
A.Buchanan: :-) Adrian, your bishop line still works with wKa6: why change that? (2021-05-25)
Adrian Storisteanu: Old habit of putting the wK teasingly close to the main stage, while still not causing unnecessary cooks... (2021-05-25)
Henrik Juel: Your suggestion is fine, Andrew
Composing exercise 50 was announced in Thema Danicum no. 72, where the editor Holger Helledie asked for help mate problems with 1 white and 1 black promotion, as long as possible
Theodor Steudel pointed out that problems as short as possible were also relevant, and the submitted problems along with many old ones were shown in TD no. 73, where HH expanded the criteria to
1. As few men as possible (4 was achieved in 1927: Kh8 Pf7 - Ka1 Pb3, h#2)
2. Shortest or longest possible (from h#1.5 to h#10, the latter from 1956: Kc1 Pc2 - Kh7 Pc3e4)
3. Economy of the men
4. Publication date
Four more problems appeared in TD no. 74, where HH announced his resignation as editor
I considered my contributions as one of the many low points in my composing career, such as it is, and initially excluded them from the compilation, I am working on; but now I have changed my mind... (2021-05-25)
milan: wPh4-f4 wKa6-h4 wBg2-h1 bPf5-b5 1.b4 f5 2.b3 f6 3.b2 f7 4.b1=R f8=Q 5.R×h1 Qf2#
or wSe1-e3 bPf5-b5 ...4.b1=R.Q Dual 5.bRg1 Be5#i couldn't find the solution without dual.
M.Frelih (2021-05-25)
comment
Henrik Juel: A stronger correction is needed:
move Ph4 to h6, Pf5 to f2, and change stip to h#2.5
Then C+ Popeye 4.61 with solution
1... h7 2.f1=T h8=L 3.Tg1 Le5# (2021-05-23)
A.Buchanan: Hi Henrik: -wSe1 +wLh1 cheekily repairs the original h#5. Indeed you can push the pawns back to make a h#6, still sound 8/5p2/K7/8/8/7P/6Bk/7B You're welcome! :D (2021-05-23)
Henrik Juel: Back in the old days two light-squared white bishops were unthinkable
And some people considered the knight more economical than the bishop
The problem is from a construction contest: h# with white and black promotion (2021-05-24)
A.Buchanan: In some ways, Henrik, composition is about trying to find the best natural form that a problem is trying to take. I wasn't trying to think the unthinkable, but when I finally thought it as the only way to salvage and indeed amplify your idea here, I started looking forward to reading your response :) If you don't want it, that's absolutely fine - I will claim it as an "after HJ", and try to find some "Salon Des Refuses" in which to exhibit it (2021-05-24)
Adrian Storisteanu: Or you could send it to the "Salon des Fous":
Kd1 Bg2 Bh1 ph3 / Kh2 pa5 (4+2)
1.a4 h4 2.a3 h5 3.a2 h6 4.a1B h7 5.Bd4 h8B 6.Bg1 Be5# (2021-05-25)
A.Buchanan: :-) Adrian, your bishop line still works with wKa6: why change that? (2021-05-25)
Adrian Storisteanu: Old habit of putting the wK teasingly close to the main stage, while still not causing unnecessary cooks... (2021-05-25)
Henrik Juel: Your suggestion is fine, Andrew
Composing exercise 50 was announced in Thema Danicum no. 72, where the editor Holger Helledie asked for help mate problems with 1 white and 1 black promotion, as long as possible
Theodor Steudel pointed out that problems as short as possible were also relevant, and the submitted problems along with many old ones were shown in TD no. 73, where HH expanded the criteria to
1. As few men as possible (4 was achieved in 1927: Kh8 Pf7 - Ka1 Pb3, h#2)
2. Shortest or longest possible (from h#1.5 to h#10, the latter from 1956: Kc1 Pc2 - Kh7 Pc3e4)
3. Economy of the men
4. Publication date
Four more problems appeared in TD no. 74, where HH announced his resignation as editor
I considered my contributions as one of the many low points in my composing career, such as it is, and initially excluded them from the compilation, I am working on; but now I have changed my mind... (2021-05-25)
milan: wPh4-f4 wKa6-h4 wBg2-h1 bPf5-b5 1.b4 f5 2.b3 f6 3.b2 f7 4.b1=R f8=Q 5.R×h1 Qf2#
or wSe1-e3 bPf5-b5 ...4.b1=R.Q Dual 5.bRg1 Be5#i couldn't find the solution without dual.
M.Frelih (2021-05-25)
comment
1. Tc7 a4 2. Sb7 a5 3. Kd5 a6 4. Kc6 a7 5. Dd5 Ta6#
NL:
1. Ke3 h5 2. Kf3 h6 3. Ke2 h7 4. Kf1 h8=D 5. De2 Dh1#
NL:
1. Ke3 h5 2. Kf3 h6 3. Ke2 h7 4. Kf1 h8=D 5. De2 Dh1#
Yuri Bilokin: Correction: -bRf2, -wPh4, bBd7, bNe7-c1, bBe5-f4, bPf6-c3 8/r2b4/8/npp5/4kb2/2p5/Pp1q4/RKn5 (3+11) (2020-11-28)
comment
comment
1. Dh8 Le6 2. Lg7 f5#
Cook: NL 1. b5 Lxb2 2. h6 Lb1#
Cook: NL 1. b5 Lxb2 2. h6 Lb1#
Yuri Bilokin: Correction: wKe7-a7, wBa2-b3, wPf2-g5, bSc7-e2, -bPb6, -bPd7 8/K6p/6k1/6Pp/5P2/1B6/1q2n3/b1B5 (5+6) (2021-05-10)
comment
comment
1. Da1 exf5 2. Lb2 e4 3. c3 Ld4#
NL:
1. Ta1 exf5 2. Dd4 exd4 3. Lxd4 Lxd4#
NL:
1. Ta1 exf5 2. Dd4 exd4 3. Lxd4 Lxd4#
Korrekturvorschlag HJS Co+: -sBc4, +sTc5; ersetzt man einfach sBc4 durch sTc4 dann 2. Lösung 1. Dh4+ gxh4 2. Lb2 Lg3 3. Sd4 Le5#
Yuri Bilokin: Correction: bPc4-d6, bPg4-g5, +bSd5, -wPg3 6bk/6bp/3p1q2/3n1np1/4P3/4P3/r4B1K/8 (4+10) 1.Qa1 exf5 2.Bb2 e4 3.Sc3 Bd4# (MM) (2021-05-14)
comment
Yuri Bilokin: Correction: bPc4-d6, bPg4-g5, +bSd5, -wPg3 6bk/6bp/3p1q2/3n1np1/4P3/4P3/r4B1K/8 (4+10) 1.Qa1 exf5 2.Bb2 e4 3.Sc3 Bd4# (MM) (2021-05-14)
comment
1. Lg1 Th8 2. Df2+ Kh1 3. Te3 Txa8#
NL:
1. Lg1 Th8 2. De3 Txe8 3. c5 Txa8#
NL:
1. Lg1 Th8 2. De3 Txe8 3. c5 Txa8#
Yuri Bilokin: Correction: bPc7-d5, bBb5-c6, +bPb5, +bRc7 n3r3/qnr5/kbb5/Pp1p4/8/8/6K1/7R (3+10) (2021-06-24)
comment
comment
1. Kc3 b4 2. Kc4 b5 3. Kc5 bxa6 4. Kc6 Lg6 5. Tc5 Le8#
NL:
1. Kc3 Kd5 2. Kb4 Lc2 3. Ka5 Kc4 4. Ta1 b4#
NL:
1. Kc3 Kd5 2. Kb4 Lc2 3. Ka5 Kc4 4. Ta1 b4#
Yuri Bilokin: Correction: +bPe5, +bPe6, +bRg4, +bQh4 8/2p4B/pp1pp3/4p3/4K3/1P4rq/2k5/2r5 (3+10) (2021-05-15)
comment
comment
1. Tc1 Lg8 2. Lf7 Ke3 3. Kd5 d4 4. Tc6 Lxf7#
NL:
1. Ta1 La6 2. Le6 d4+ 3. Kc6 Kd3 4. Kd5 Lb7#
NL:
1. Ta1 La6 2. Le6 d4+ 3. Kc6 Kd3 4. Kd5 Lb7#
Yuri Bilokin: bRd1-d2, +bSa1, +bPa4, +bPg4 8/8/3p4/2k5/p3K1p1/1B1P4/b2r4/n7 (3+7) 1.Rc2 Bg8 2.Bf7 Ke3 3.Kd5 d4 4.Rc6 Bxf7# (MM) (2021-06-11)
comment
comment
1. d1=T Th1 2. Tg1+ Kf6 3. Tg4 Tf1#
Cook: NL
1. Kg5 Txe3 2. Kh5 Kf6 3. Kh6 Th3#
Cook: NL
1. Kg5 Txe3 2. Kh5 Kf6 3. Kh6 Th3#
versetzt man wBh2 nach h3, HJS gibt's ne neue eindeutige Lösung: 1. d1=T Tg1 2. Tf1 Kf6 3. Tf3 Tg4#,
Yuri Bilokin: Correction: +bbe8 4b3/6K1/8/8/4pk2/4p3/3p3P/4R3 (3+5) (2019-03-22)
Yuri Bilokin: version wKg7-c6, bPe4-g5 8/8/2K5/6p1/5k2/4p3/3p3P/4R3 (3+4)
1.d1=R Rh1 2.Rg1 Kd5 3.Rg4 Rf1# (IM) (2022-05-31)
comment
Yuri Bilokin: Correction: +bbe8 4b3/6K1/8/8/4pk2/4p3/3p3P/4R3 (3+5) (2019-03-22)
Yuri Bilokin: version wKg7-c6, bPe4-g5 8/8/2K5/6p1/5k2/4p3/3p3P/4R3 (3+4)
1.d1=R Rh1 2.Rg1 Kd5 3.Rg4 Rf1# (IM) (2022-05-31)
comment
Keywords: Bahnung, Promotion key, under-promotion (t)
Genre: h#
FEN: 8/6K1/8/8/4pk2/4p3/3p3P/4R3
Input: hpr, 1996-12-27
Last update: Alfred Pfeiffer, 2019-03-23 more...
Genre: h#
FEN: 8/6K1/8/8/4pk2/4p3/3p3P/4R3
Input: hpr, 1996-12-27
Last update: Alfred Pfeiffer, 2019-03-23 more...
1. Dh8 Lg7 2. Tc2 Lxb2 3. Dxb2 Kxd7 4. c3 Lxf7#
NL:
1. f5 Lxb4 2. Dg7 Lf8 3. Lc5 Lxc4+ 4. Ka3 Lxc5#
NL:
1. f5 Lxb4 2. Dg7 Lf8 3. Lc5 Lxc4+ 4. Ka3 Lxc5#
Yuri Bilokin: Correction: 8/5BB1/1p1pK3/4p3/8/pppq4/bkrn4/1n6 (3+12)
1.Qh7 Bg6 2.Rc1 Bxb1 3.Qxb1 Kxd6 4.c2 Bxe5# (MM) (2020-11-25)
comment
1.Qh7 Bg6 2.Rc1 Bxb1 3.Qxb1 Kxd6 4.c2 Bxe5# (MM) (2020-11-25)
comment
92 - P0505381
Lajos Riczu
Magyar Sakkelet 11/1970
1. Preis
Turnier des Ungarischen Schachbundes
(3+6) cooked
h#4
Lajos Riczu
Magyar Sakkelet 11/1970
1. Preis
Turnier des Ungarischen Schachbundes
(3+6) cooked
h#4
1. De3 Te1 2. De8 e4 3. Le7 e5 4. Tf8 e6#
NL:
1. Dg5 e4 2. Df5+ exf5 3. g5 Te1 4. Tg6 fxg6#
NL:
1. Dg5 e4 2. Df5+ exf5 3. g5 Te1 4. Tg6 fxg6#
Yuri Bilokin: Correction: bPg7-g5, +bNb7, +bPg3, +bPg4, +bPh2, +bRh3, +bNh4 5br1/1n3k1K/5p2/6p1/6pn/6pr/4P2p/6qR (3+12) h#4
1.Qe3 Re1 2.Qe8 e4 3.Be7 e5 4.Rf8 e6# (MM) (2022-04-20)
comment
1.Qe3 Re1 2.Qe8 e4 3.Be7 e5 4.Rf8 e6# (MM) (2022-04-20)
comment
1. Kd7 g4 2. Lh1 Lg2 3. Ke6 Lh3 4. Ld5 g5+ 5. f5 gxf6ep#
NL:
1. Td5 Lxb7 2. Sf6 g4 3. Sg8 Kxf8 4. Kd7 g5 5. Ke6 Lc8#
NL:
1. Td5 Lxb7 2. Sf6 g4 3. Sg8 Kxf8 4. Kd7 g5 5. Ke6 Lc8#
Korrekturvorschlag HJS Co+: wKg7, wLa8, wBg2, sKc7, sDd6, sTf8h8, sLb7c1, sSa5, sBb2b5c4e2e5f7h6h7
Yuri Bilokin: Correction: bQb1-a4, bPe3-a6, -bRd3 B4n1r/1bk2pKn/p2p2p1/4p3/qp6/8/6P1/8 (3+12) (2021-03-03)
A.Buchanan: @Yuri: no improvement appears in WinChloe. The "Korrekturvorschlag HJS Co+" B4r1r/1bk2pKp/3q3p/np2p3/2p5/8/1p2p1P1/2b5 is sound, but at 3+15 much more expensive than your economical 3+12. To my surprise, I found I can improve your version still further as B4n2/1bk2pKn/3q4/4p3/8/8/6P1/8 3+7! bQd6 is a very effective cook-stopper. (2021-03-04)
Viktoras Paliulionis: Nice improvement. Compare with P1235831. (2021-03-05)
more ...
comment
Yuri Bilokin: Correction: bQb1-a4, bPe3-a6, -bRd3 B4n1r/1bk2pKn/p2p2p1/4p3/qp6/8/6P1/8 (3+12) (2021-03-03)
A.Buchanan: @Yuri: no improvement appears in WinChloe. The "Korrekturvorschlag HJS Co+" B4r1r/1bk2pKp/3q3p/np2p3/2p5/8/1p2p1P1/2b5 is sound, but at 3+15 much more expensive than your economical 3+12. To my surprise, I found I can improve your version still further as B4n2/1bk2pKn/3q4/4p3/8/8/6P1/8 3+7! bQd6 is a very effective cook-stopper. (2021-03-04)
Viktoras Paliulionis: Nice improvement. Compare with P1235831. (2021-03-05)
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comment
Keywords: Bahnung, En passant as mating move, Model mate, Superseded by (P1388090)
Genre: h#
FEN: B4n1r/1bk2pKn/3p2p1/4p3/1p6/3rp3/6P1/1q6
Input: hpr, 1996-12-27
Last update: A.Buchanan, 2021-04-03 more...
Genre: h#
FEN: B4n1r/1bk2pKn/3p2p1/4p3/1p6/3rp3/6P1/1q6
Input: hpr, 1996-12-27
Last update: A.Buchanan, 2021-04-03 more...
1. Dc7 La6 2. Tc6 Db5#
NL:
1. Dc7 De1 2. Tc6 Db4#
NL:
1. Dc7 De1 2. Tc6 Db4#
1. Tg5 Lg8 2. De5 Df7#
NL:
1. Kg7 Dd2 2. Kh8 Dh6#
NL:
1. Kg7 Dd2 2. Kh8 Dh6#
1. Tc1 Tc2 2. Kd6 Tc7 3. Tc6 Td7#
Cook: NL 1. Kc6 Th2 2. Kb7 Th8 3. Ka8 Kxc7#
Cook: NL 1. Kc6 Th2 2. Kb7 Th8 3. Ka8 Kxc7#
Korrektur HJS Co+: sTc7 nach c3
Adrian Storisteanu: Another fix possibility: shift all the pieces one file right (Ke8/Ke5) -- 1.Rd1 Rd2 2.Ke6 Rd7 3.Rd6 Re7#. (2015-08-23)
Yuri Bilokin: bPa6-b7 (2021-06-23)
comment
Adrian Storisteanu: Another fix possibility: shift all the pieces one file right (Ke8/Ke5) -- 1.Rd1 Rd2 2.Ke6 Rd7 3.Rd6 Re7#. (2015-08-23)
Yuri Bilokin: bPa6-b7 (2021-06-23)
comment
1) 1. Da5 Tb5 2. Sd7 Tb6 3. Df5 Ld5#
2) 1. Da4 Lb5 2. Ld6 Lc4 3. Dd7 Te5#
NL:
1. Sa6 Txb5 2. Sc7 Tb6 3. Kd6 La4#
2) 1. Da4 Lb5 2. Ld6 Lc4 3. Dd7 Te5#
NL:
1. Sa6 Txb5 2. Sc7 Tb6 3. Kd6 La4#
Yuri Bilokin: Yuri Bilokin & Wenelin Alaikow correction a4=a1(-bPd), then +bPa4, +bPe5, +bPe6, bPg3-h2 8/8/4p3/4p3/pp2b3/2B1kn2/1q1R3p/1n1K4 (3+10) h#3 2.1...
1.Qa1 Bb2 2.Bd3 Bc1 3.Qd4 Re2#
1.Qa2 Rb2 2.Sd4 Rb3 3.Qf2 Bd2# (MM)
Ambush (wB). Ambush (wR). Anticipatory unpin
AntiZielElement (W1, line obstruction) × 2
Exchange of functions (wRd2/wBc3, Mating battery firing / Rear piece of mating battery + Line opening)
Play on the same square (W1, 2). Reciprocal batteries. Wigwag (bQ) × 2
Model mate × 1. Battery mate × 2. Double-check mate × 2 (2023-05-20)
comment
1.Qa1 Bb2 2.Bd3 Bc1 3.Qd4 Re2#
1.Qa2 Rb2 2.Sd4 Rb3 3.Qf2 Bd2# (MM)
Ambush (wB). Ambush (wR). Anticipatory unpin
AntiZielElement (W1, line obstruction) × 2
Exchange of functions (wRd2/wBc3, Mating battery firing / Rear piece of mating battery + Line opening)
Play on the same square (W1, 2). Reciprocal batteries. Wigwag (bQ) × 2
Model mate × 1. Battery mate × 2. Double-check mate × 2 (2023-05-20)
comment
Genre: h#
FEN: 8/1p2b3/2B1knp1/1q1R4/1n1K4/3p4/8/8
Input: Markus Manhart, 1997-01-06
Last update: hpr, 1999-08-29 more...
a) 1. Sb7+ Dc5 2. Sed6 Dxd5#
b) 1. De6 Df2 2. Sc3 d4#
Cook: NL
a) 1. f5 Da7 2. Ld4 Dxe7#
b) 1. De6 Df2 2. Sc3 d4#
Cook: NL
a) 1. f5 Da7 2. Ld4 Dxe7#
Version einer Aufgabe von Laszlo Apro, Scacco 1988
Adrian Storisteanu: See P0560003 for the inspiration?! (2015-11-07)
Adrian Storisteanu: Possible fix: +bPc7 (4+12). (2015-11-07)
A.Buchanan: Garai, one of my favourites composers, has 1372 helpmates in PDB. (Only Jonsson & Feather have more.) Only 29 are not C+, of which 24 are marked cooked. Therefore the chances are high that the bug here is a typo too. (2015-11-07)
Yuri Bilokin: Correction: bRd8-e8 b3r3/4q3/3n1p2/3pk3/4np2/K2PQ2r/8/b6B (4+11) (2020-11-24)
Gerald Ettl: Yuri Bilokin von Deiner Correction die Versetzung des sT ist in beiden Fällen immer noch NL. (2020-11-25)
A.Buchanan: I think Yuri's correction works in both twins, and is more economical than Adrian's. However in a) the line d8-d5 must be closed for the mate to work, so sTd8 is thematic. This consideration is not actually necessary for soundness though, which is why Yuri's sTe8 is sound! So sTd8 is a total weasel in both twins! (2020-11-25)
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Adrian Storisteanu: See P0560003 for the inspiration?! (2015-11-07)
Adrian Storisteanu: Possible fix: +bPc7 (4+12). (2015-11-07)
A.Buchanan: Garai, one of my favourites composers, has 1372 helpmates in PDB. (Only Jonsson & Feather have more.) Only 29 are not C+, of which 24 are marked cooked. Therefore the chances are high that the bug here is a typo too. (2015-11-07)
Yuri Bilokin: Correction: bRd8-e8 b3r3/4q3/3n1p2/3pk3/4np2/K2PQ2r/8/b6B (4+11) (2020-11-24)
Gerald Ettl: Yuri Bilokin von Deiner Correction die Versetzung des sT ist in beiden Fällen immer noch NL. (2020-11-25)
A.Buchanan: I think Yuri's correction works in both twins, and is more economical than Adrian's. However in a) the line d8-d5 must be closed for the mate to work, so sTd8 is thematic. This consideration is not actually necessary for soundness though, which is why Yuri's sTe8 is sound! So sTd8 is a total weasel in both twins! (2020-11-25)
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Genre: h#
FEN: b2r4/4q3/3n1p2/3pk3/4np2/K2PQ2r/8/b6B
Input: Markus Manhart, 1997-01-06
Last update: A.Buchanan, 2020-11-25 more...
1. Sb6 Le1 2. Da6 Kxc3 3. Kb5 a4+ 4. Ka5 Kb3#
Cook: NL 1. Sb6 Lf6 2. Da6 Kb3 3. Kb5 a4+ 4. Ka5 Lxc3#
Cook: NL 1. Sb6 Lf6 2. Da6 Kb3 3. Kb5 a4+ 4. Ka5 Lxc3#
Adrian Storisteanu: Possible fix: wKc2->d3 (3+4) [2.Qa6+]. (2015-11-07)
Yuri Bilokin: Correction: -bPc3, wKc2-d4, wBh4-f2 8/8/2k5/8/n2K4/r7/P4B2/8 (3+3) 1.Sb6 Be1 2.Ra6 Kc3 3.Kb5 a4+ 4.Ka5 Kb3# (2021-05-14)
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Yuri Bilokin: Correction: -bPc3, wKc2-d4, wBh4-f2 8/8/2k5/8/n2K4/r7/P4B2/8 (3+3) 1.Sb6 Be1 2.Ra6 Kc3 3.Kb5 a4+ 4.Ka5 Kb3# (2021-05-14)
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Genre: h#
FEN: 8/8/2k5/8/n6B/q1p5/P1K5/8
Input: Markus Manhart, 1997-01-06
Last update: Alfred Pfeiffer, 2015-11-07 more...
a) 1. Tb8 Th1 2. Tc8 Dh2#
b) 1. Tc8 Lb1 2. Tb8 Dc2#
NL:
a) 1. Kd6 Txf6+ 2. Ke7 Dxg7#
b) 1. Tc8 Lb1 2. Tb8 Dc2#
NL:
a) 1. Kd6 Txf6+ 2. Ke7 Dxg7#
Kees: Possible fix:Dd8-d7 Ld7-d8 +sSh8 -sLc1 -wBb2 h#1,5 ( a) 1. ...Th1 2.Tc8 Dh2# b) 1. ... Lb1 2. Tb8 2. Dc2#) (2023-06-05)
Gerald Ettl: Kees, das ist eine schlechte Verbesserung. Denn das Tempoverlustmanoever des sT ist hier eine wichtige Absicht des Autors, die beibehalten werden soll. (2023-06-05)
Ladislav Packa: White Rf7 Pg7 Qh7 Bg6 Pg5 Rh5 Pg4 Pb3 Pg2 Kg1 (10)
Black Ra8 Bd8 Bg8 Pb7 Kc7 Qd7 Pb6 Pc6 Pb4 (9)
b) bPc6-d6
C+ by Popeye v4.63 The author's idea is completely preserved. (2023-06-05)
A.Buchanan: I like Ladislav's fix a lot. I humbly suggest r2b2b1/1pkq1RPQ/1p1p2B1/6PR/3p4/8/4K3/6N1 saving two units. Some might say wSg1 is too passive for a White officer, but the way two units (bP & wK) block three rows is sufficiently amusing to offset this, imho. (2023-06-06)
A.Buchanan: The authors fixed it very well with P0523513 so any later fix is pointless. (2023-06-13)
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Gerald Ettl: Kees, das ist eine schlechte Verbesserung. Denn das Tempoverlustmanoever des sT ist hier eine wichtige Absicht des Autors, die beibehalten werden soll. (2023-06-05)
Ladislav Packa: White Rf7 Pg7 Qh7 Bg6 Pg5 Rh5 Pg4 Pb3 Pg2 Kg1 (10)
Black Ra8 Bd8 Bg8 Pb7 Kc7 Qd7 Pb6 Pc6 Pb4 (9)
b) bPc6-d6
C+ by Popeye v4.63 The author's idea is completely preserved. (2023-06-05)
A.Buchanan: I like Ladislav's fix a lot. I humbly suggest r2b2b1/1pkq1RPQ/1p1p2B1/6PR/3p4/8/4K3/6N1 saving two units. Some might say wSg1 is too passive for a White officer, but the way two units (bP & wK) block three rows is sufficiently amusing to offset this, imho. (2023-06-06)
A.Buchanan: The authors fixed it very well with P0523513 so any later fix is pointless. (2023-06-13)
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Keywords: Bahnung, Tempo Move (t t)
Genre: h#
FEN: r2q4/1pkb1RpQ/1p3nB1/6KR/1N4P1/5P2/1P1p4/2b5
Input: hpr, 1997-01-22
Last update: A.Buchanan, 2023-06-13 more...
Genre: h#
FEN: r2q4/1pkb1RpQ/1p3nB1/6KR/1N4P1/5P2/1P1p4/2b5
Input: hpr, 1997-01-22
Last update: A.Buchanan, 2023-06-13 more...
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https://pdb.dieschwalbe.de/search.jsp?expression=COMMENTDATE%3E%3D20200919+AND+NOT+K%3D%27Hilfsr%C3%BCckz%C3%BCger%27+AND+NOT+K%3D%27Eindeutige+Beweispartie%27+AND+COOKED+
The problems of this query have been registered by the following contributors:
Gerd Wilts (58)hpr (33)
Erich Bartel (1)
Hans-Jürgen Schäfer (3)
Torsten Linss (1)
Markus Manhart (4)
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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