47 problem(s) found in 2174 milliseconds (displaying 47 problem(s)). [COMMENTDATE>=20220810 AND G='h#' AND NOT S='Uralski Problemist' AND K='Rochade'] [download as LaTeX]
*) 1. ... 0-0-0 2. Txf2 Dxg1#
1) 1. cxd2+ Kxd2 2. Txf2 Dxg1#
R: 1. ... Da8-h8 2. d3xTe4 Da2-a8 3. c2xSd3 Sc1-d3+ 4. b4xLc5 Db1-a2 5. b3-b4 Ba2xSb1=D 6. Sa3-b1 und weiter z.B. (Beispielauflösung mri)
6. ... Te8-e4 7. Sc4-a3 Sg4-h2 8. Se5-c4 Th2-g2 9. Sf3-e5 g2-g1=L 10. Sg1-f3 Sh6-g4 11. Sf3xDg1 Th8-e8 12. Sd4-f3 f3xSg2 13. Se3-g2 f4-f3 14. Sc6-d4 Tg2-h2 15. Sa7-c6 Kh2-h1 16. Sc8-a7 Dh1-g1 17. Sb6xLc8 Tg1-g2 18. Sf5-e3 Dc6-h1 19. Sh4-f5 Kh1-h2 20. h2-h3 a3-a2 21. Sa4-b6 Sa2-c1 22. Sb6-a4 Sb4-a2 23. Sa4-b6 Sd5-b4 24. Sb6-a4 Se3-d5 25. Dh3-f1 Sf1-e3 26. Dg4-h3 f5-f4 27. Df4-g4 a4-a3 28. Db4-f4 a5-a4 29. Da3-b4 Kg2-h1 30. Sf3-h4 Kh3-g2 31. Da2-a3 g2-g1=T 32. Db1-a2 Kg4-h3 33. Dd1-b1 h3xTg2 34. Tg1-g2 Kg5-g4 35. Sh4-f3 Kf6-g5 36. Th1-g1 Se3-f1 37. Sf3-h4 h4-h3 38. Sg1-f3 e6xLf5 39. Lh3-f5 Ke7-f6 40. Lf1-h3 Sc4-e3 41. g2-g3 Ke8-e7 42. Sa4-b6 Sg8-h6 43. a2xTb3 Tb6-b3 44. Sh3-g1 Ta6-b6 45. Sb6-a4 Ta8-a6 46. Sg1-h3 a7-a5 47. Sh3-g1 Sa5-c4 48. Sg1-h3 Sb3-a5 49. Sc4-b6 Sc1-b3 50. Sa3-c4 Sb3xLc1 51. Sb1-a3 Sa5-b3 52. Sh3-g1 h5-h4 53. Sg1-h3 c4-c3 54. Sh3-g1 Lf8-c5 55. Sg1-h3 Dc7-c6 56. Sh3-g1 Dd8-c7 57. Sg1-h3 c5-c4 58. Sh3-g1 h7-h5 59. Sg1-h3 e7-e6 60. Sh3-g1 Sc6-a5 61. Sg1-h3 Sb8-c6 62. Sh3-g1 c7-c5 63. Sg1-h3
1) 1. cxd2+ Kxd2 2. Txf2 Dxg1#
R: 1. ... Da8-h8 2. d3xTe4 Da2-a8 3. c2xSd3 Sc1-d3+ 4. b4xLc5 Db1-a2 5. b3-b4 Ba2xSb1=D 6. Sa3-b1 und weiter z.B. (Beispielauflösung mri)
6. ... Te8-e4 7. Sc4-a3 Sg4-h2 8. Se5-c4 Th2-g2 9. Sf3-e5 g2-g1=L 10. Sg1-f3 Sh6-g4 11. Sf3xDg1 Th8-e8 12. Sd4-f3 f3xSg2 13. Se3-g2 f4-f3 14. Sc6-d4 Tg2-h2 15. Sa7-c6 Kh2-h1 16. Sc8-a7 Dh1-g1 17. Sb6xLc8 Tg1-g2 18. Sf5-e3 Dc6-h1 19. Sh4-f5 Kh1-h2 20. h2-h3 a3-a2 21. Sa4-b6 Sa2-c1 22. Sb6-a4 Sb4-a2 23. Sa4-b6 Sd5-b4 24. Sb6-a4 Se3-d5 25. Dh3-f1 Sf1-e3 26. Dg4-h3 f5-f4 27. Df4-g4 a4-a3 28. Db4-f4 a5-a4 29. Da3-b4 Kg2-h1 30. Sf3-h4 Kh3-g2 31. Da2-a3 g2-g1=T 32. Db1-a2 Kg4-h3 33. Dd1-b1 h3xTg2 34. Tg1-g2 Kg5-g4 35. Sh4-f3 Kf6-g5 36. Th1-g1 Se3-f1 37. Sf3-h4 h4-h3 38. Sg1-f3 e6xLf5 39. Lh3-f5 Ke7-f6 40. Lf1-h3 Sc4-e3 41. g2-g3 Ke8-e7 42. Sa4-b6 Sg8-h6 43. a2xTb3 Tb6-b3 44. Sh3-g1 Ta6-b6 45. Sb6-a4 Ta8-a6 46. Sg1-h3 a7-a5 47. Sh3-g1 Sa5-c4 48. Sg1-h3 Sb3-a5 49. Sc4-b6 Sc1-b3 50. Sa3-c4 Sb3xLc1 51. Sb1-a3 Sa5-b3 52. Sh3-g1 h5-h4 53. Sg1-h3 c4-c3 54. Sh3-g1 Lf8-c5 55. Sg1-h3 Dc7-c6 56. Sh3-g1 Dd8-c7 57. Sg1-h3 c5-c4 58. Sh3-g1 h7-h5 59. Sg1-h3 e7-e6 60. Sh3-g1 Sc6-a5 61. Sg1-h3 Sb8-c6 62. Sh3-g1 c7-c5 63. Sg1-h3
a) 1. ... exf6ep 2. 0-0-0 Lxf4 3. Td7 a8=D#
b) 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
b) 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
Anton Baumann: Mattdual in b): 3.Tb8 axb8=D,T# (2022-12-16)
A.Buchanan: So Borodatow got it all working! In a) Black might have captured hxgxf and axPb. So castling rights might still be maintained with the ep. In b) on the other hand, it must be axb, bxa, exf and wPgxh6, so there was the cage. Promotion to TD is tolerated in the final move, although it may not be puristic, without the convention, too many mates would be excluded (2022-12-16)
comment
A.Buchanan: So Borodatow got it all working! In a) Black might have captured hxgxf and axPb. So castling rights might still be maintained with the ep. In b) on the other hand, it must be axb, bxa, exf and wPgxh6, so there was the cage. Promotion to TD is tolerated in the final move, although it may not be puristic, without the convention, too many mates would be excluded (2022-12-16)
comment
Keywords: En passant as key, Castling (sg), Cant Castler, Valladao Task
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & basic retro thinking
FEN: r3k3/P7/b3r1pP/4PpBP/3nnpKR/5PRB/5PP1/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & basic retro thinking
FEN: r3k3/P7/b3r1pP/4PpBP/3nnpKR/5PRB/5PP1/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 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)
comment
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)
comment
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...
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)
more ...
comment
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)
more ...
comment
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)
more ...
comment
milan: [-bPa7b6wPg3bSa1+bPg3bBa5-b4 bPc6-d6] h#2 1.2...bRh2? illegal move? M.Frelih (2023-03-30)
more ...
comment
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...
* 1. ... dxc5 2. Dxh3 0-0-0#
1. Db2 Le2+ 2. Kc2 Ld1#
1. Db2 Le2+ 2. Kc2 Ld1#
Henrik Juel: the five missing black men were captured by white pawns (exfxgxh, fxgxj), so with Black to move last move was with Ta1 or Ke1, and White may not castle
C+ Popeye 4.61 (2022-11-26)
A.Buchanan: Pleasant White tempo play in both phases (2022-11-27)
comment
C+ Popeye 4.61 (2022-11-26)
A.Buchanan: Pleasant White tempo play in both phases (2022-11-27)
comment
Keywords: Cant Castler (wl), Castling (wl)
Genre: Retro, h#
Computer test: HC+ Popeye 4.61 with simple retro logic
FEN: 8/7p/7P/2pr2pP/2bP2Pb/2pk1BRP/6qN/R3KNrn
Reprints: 553 FIDE Album 1959-1961 1966
157 Europe Echecs 90 07/1966
(10) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-27 more...
Genre: Retro, h#
Computer test: HC+ Popeye 4.61 with simple retro logic
FEN: 8/7p/7P/2pr2pP/2bP2Pb/2pk1BRP/6qN/R3KNrn
Reprints: 553 FIDE Album 1959-1961 1966
157 Europe Echecs 90 07/1966
(10) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-27 more...
1. ... cxd6ep 2. 0-0-0 0-0-0 3. Kd7 Sa7#
1. De7 c6 2. Th8 c7 3. Tf8 Sd6#
White pawn caps: axb,dxe,gxf,hxg.
Black: fxg,bxc,cxb.
wPb4 came from b3 to release wBa3, so bPb3 captured to reach that square.
All pcs accounted for means bPd never captured.
In the set play, there are 13 retro tries in which one or both players do not castle. The intention is that both castling rights are needed in order to imply the pawn double hop.
1. De7 c6 2. Th8 c7 3. Tf8 Sd6#
White pawn caps: axb,dxe,gxf,hxg.
Black: fxg,bxc,cxb.
wPb4 came from b3 to release wBa3, so bPb3 captured to reach that square.
All pcs accounted for means bPd never captured.
In the set play, there are 13 retro tries in which one or both players do not castle. The intention is that both castling rights are needed in order to imply the pawn double hop.
A.Buchanan: White pawn caps: axb,dxe,gxf,hxg definite.
Black: fxg and two to resolve c-file. But that may be c&d cross-capture, so in set play last move might have been c6xd5. So I think this problem is cooked. What am I missing? (2022-03-21)
Mario Richter: If Black's last move was c6xd5, how do you get white Bishop a3 out of his cage? (In that case, black pawn b7 never left the b-file). (2022-03-21)
A.Buchanan: I agree Mario thanks (2022-03-21)
Hans-Jürgen Manthey: beide Rochaden und im Satz ep. ist möglich:
R.: 1. ... d7-d5 2. Sa7-b5 b5xTc4 3. Tg4-c4 Dd8-h4 4. c4-c5 Th8-h3 5. c2-c4 c4xDb3 6. Dd3-b3 h3-h2 7. Dd1-d3 h4-h3 8. Sc8-a7 h5-h4 9. Th4-g4 c5-c4 10. Th1-h4 h7-h5 11. h2xSg3 Sf5-g3 12. b3-b4 Sh6-f5 13. Lb4-a3 c7-c5 14. Ld2-b4 g3-g2 15. Lc1-d2 a3-a2 16. a2xSb3 Sc5-b3 17. Lb3-a4 Sa6-c5 18. Lc4-b3 Sb8-a6 19. d2xLe3 Lc5-e3 20. Ld3-c4 Lf8-c5 21. Lf5-d3 a4-a3 22. Lh3-f5 a5-a4 23. Lf1-h3 g4-g3 24. g2xLf3 Lb7-f3 25. Sb6-c8 Lc8-b7 26. Sa4-b6 a7-a5 27. Sc3-a4 b7-b5 28. Sb1-c3 f5xSg4 29. Se5-g4 Sg8-h6 30. Sf3-e5 e7-e6 31. Sg5-f3 f6-f5 32. Sh3-g5 f7-f6 33. Sg1-h3 (2023-02-23)
comment
Black: fxg and two to resolve c-file. But that may be c&d cross-capture, so in set play last move might have been c6xd5. So I think this problem is cooked. What am I missing? (2022-03-21)
Mario Richter: If Black's last move was c6xd5, how do you get white Bishop a3 out of his cage? (In that case, black pawn b7 never left the b-file). (2022-03-21)
A.Buchanan: I agree Mario thanks (2022-03-21)
Hans-Jürgen Manthey: beide Rochaden und im Satz ep. ist möglich:
R.: 1. ... d7-d5 2. Sa7-b5 b5xTc4 3. Tg4-c4 Dd8-h4 4. c4-c5 Th8-h3 5. c2-c4 c4xDb3 6. Dd3-b3 h3-h2 7. Dd1-d3 h4-h3 8. Sc8-a7 h5-h4 9. Th4-g4 c5-c4 10. Th1-h4 h7-h5 11. h2xSg3 Sf5-g3 12. b3-b4 Sh6-f5 13. Lb4-a3 c7-c5 14. Ld2-b4 g3-g2 15. Lc1-d2 a3-a2 16. a2xSb3 Sc5-b3 17. Lb3-a4 Sa6-c5 18. Lc4-b3 Sb8-a6 19. d2xLe3 Lc5-e3 20. Ld3-c4 Lf8-c5 21. Lf5-d3 a4-a3 22. Lh3-f5 a5-a4 23. Lf1-h3 g4-g3 24. g2xLf3 Lb7-f3 25. Sb6-c8 Lc8-b7 26. Sa4-b6 a7-a5 27. Sc3-a4 b7-b5 28. Sb1-c3 f5xSg4 29. Se5-g4 Sg8-h6 30. Sf3-e5 e7-e6 31. Sg5-f3 f6-f5 32. Sh3-g5 f7-f6 33. Sg1-h3 (2023-02-23)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wgsg)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro-logic.
FEN: r3k3/6p1/4p3/1NPp4/BPp4q/Bp2PPPr/pP2PPpp/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro-logic.
FEN: r3k3/6p1/4p3/1NPp4/BPp4q/Bp2PPPr/pP2PPpp/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
hans: 1. Th2xf2 De1xf2#!
1. Th2xh3 0-0#? (Castling illegal)
R: -1. Kf1xSe1 Sg2-e1 -2. Ke1-f1 Sf4-g2+ -3. Tg1-h1 Sd5-f4 -4. Tf1-g1 Se3-d5 -5. Tg1-f1 Sf1-e3 -6. Tg1-h1 Tg2-h2 -7. Th2-h1-g1 Tg1-g2 -8. Th1-h2 g2-g1=T -9 h2-h3 h3xSg2 (2010-06-26)
HENRI: Hans solution has a tipo. Their is no D (=Queen in german) mating. One should read the solution in german : 1. Th2xf2 Ke1xf2#! 1.Th2xh3 0-0#?
or in english : 1. Rh2xf2 Ke1xf2#! 1.Rh2xh3 0-0#? (Castling illegal) (2023-10-25)
comment
1. Th2xh3 0-0#? (Castling illegal)
R: -1. Kf1xSe1 Sg2-e1 -2. Ke1-f1 Sf4-g2+ -3. Tg1-h1 Sd5-f4 -4. Tf1-g1 Se3-d5 -5. Tg1-f1 Sf1-e3 -6. Tg1-h1 Tg2-h2 -7. Th2-h1-g1 Tg1-g2 -8. Th1-h2 g2-g1=T -9 h2-h3 h3xSg2 (2010-06-26)
HENRI: Hans solution has a tipo. Their is no D (=Queen in german) mating. One should read the solution in german : 1. Th2xf2 Ke1xf2#! 1.Th2xh3 0-0#?
or in english : 1. Rh2xf2 Ke1xf2#! 1.Rh2xh3 0-0#? (Castling illegal) (2023-10-25)
comment
Keywords: Cant Castler, Castling (wk)
Genre: h#, Retro
FEN: 8/8/8/8/8/PPP3PP/B1rPPP1r/BNk1K2R
Input: Gerd Wilts, 1995-06-03
Genre: h#, Retro
FEN: 8/8/8/8/8/PPP3PP/B1rPPP1r/BNk1K2R
Input: Gerd Wilts, 1995-06-03
1. ... axb6ep 2. 0-0-0 0-0-0 3. Td7 a8=D#
This problem can be solved presuming that both castlings are executable. It implies that the last halfmoves have been.-1.Bb6xa7 sBb7-b5.
Bb5 could not come from c6 (sBc6xFb5), because all of the missing white pieces(5) have been captured on white squares except wLc1,which must have been captured by Bc7(Bc7xwLb6).Further,it has not played immediately before -1.Bb6-b5 because this would prevent white castling(forcing wT or wK to move)
Here is the proof that 4 remaining white pcs have been captured on white squares:
black d-pawn must have been captured on d4 and h,g f pawns had to promote after Bh3xwFg2,then wBh2moved to h4, enabling Bg4x Fh3 ..h1T, and finnaly sBf7
after fxg2... g1S. sLf1 prevented the check from Th1, and later has been captured by a white officer.
Conclusion:
Retro:-1.wBb6xFa7 Bb7-b5
Forward(AP) 1....axb6 e. p.2.0-0-0 0-0-0 3.Td7 a8Q# (Author)
This problem can be solved presuming that both castlings are executable. It implies that the last halfmoves have been.-1.Bb6xa7 sBb7-b5.
Bb5 could not come from c6 (sBc6xFb5), because all of the missing white pieces(5) have been captured on white squares except wLc1,which must have been captured by Bc7(Bc7xwLb6).Further,it has not played immediately before -1.Bb6-b5 because this would prevent white castling(forcing wT or wK to move)
Here is the proof that 4 remaining white pcs have been captured on white squares:
black d-pawn must have been captured on d4 and h,g f pawns had to promote after Bh3xwFg2,then wBh2moved to h4, enabling Bg4x Fh3 ..h1T, and finnaly sBf7
after fxg2... g1S. sLf1 prevented the check from Th1, and later has been captured by a white officer.
Conclusion:
Retro:-1.wBb6xFa7 Bb7-b5
Forward(AP) 1....axb6 e. p.2.0-0-0 0-0-0 3.Td7 a8Q# (Author)
Branko Koludrovic: P.S.
The black pawn a4(on the diagramm)came from c7 after capturing the white bishop on b6, then moving to b5 and capturing a white officer on a4. (2010-09-28)
A.Buchanan: I think sBf must have captured on g2 & then promoted on f1 to S, otherwise it’s impossible to unlock the southeast cage (2023-07-11)
A.Buchanan: So that means it’s sound. However I don’t see that it needs to have been sL shielding on f1 (2023-07-11)
more ...
comment
The black pawn a4(on the diagramm)came from c7 after capturing the white bishop on b6, then moving to b5 and capturing a white officer on a4. (2010-09-28)
A.Buchanan: I think sBf must have captured on g2 & then promoted on f1 to S, otherwise it’s impossible to unlock the southeast cage (2023-07-11)
A.Buchanan: So that means it’s sound. However I don’t see that it needs to have been sL shielding on f1 (2023-07-11)
more ...
comment
Keywords: a posteriori (AP), En passant as key, Castling (sgsgwg), Promotion (D), Valladao Task
Genre: h#, Retro
FEN: r3k3/P3p3/p7/Pp6/p6P/2P3PR/2PPP2b/R3K1nr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-11 more...
Genre: h#, Retro
FEN: r3k3/P3p3/p7/Pp6/p6P/2P3PR/2PPP2b/R3K1nr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-11 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 ...
comment
1. Kb3 Kd2 2. Ka4 Kc3 3. a2 Txa2#
Nicht 1. Kb3 0-0-0?? 2. Ka2 Td3 3. Ka1 Txa3#, da Weiß zuletzt mit K oder T gezogen hat
Nicht 1. Kb3 0-0-0?? 2. Ka2 Td3 3. Ka1 Txa3#, da Weiß zuletzt mit K oder T gezogen hat
1. ... Sf5? 2. 0-0-0?? Sd6# - aber die s0-0-0 ist illegal, den zuletzt muß Schwarz mit K oder T gezogen haben.
1. ... Sg8! 2. Td8 Sc7#
1. ... Sg8! 2. Td8 Sc7#
Mario Richter: Luboš Kekely (Slovakia) correctly points out that 1. ... Sf5! 2. 0-0-0 Sd6# is only a try and not a solution (last black move must have been by King or Rook, so the queenside castling is illegal).
I have changed the solution accordingly. (2023-03-10)
comment
I have changed the solution accordingly. (2023-03-10)
comment
*) 1. ... 0-0 2. Dh4 Txf3#
1) 1. Kh4 Kf2 2. fxg2 Sf3#
1) 1. Kh4 Kf2 2. fxg2 Sf3#
SCHRECKE: NL: 1. Dg5,Kh4 gxf3 2. Kh4,Dg5 Sf1# (2023-09-13)
Ladislav Packa: Retro content is not needed, the solution is preserved even without it.
Pg4 Pg2 Sh2 Ke1 Rh1 (5)- Ph5 Kg3 (2) h#2* C+
1...0-0 2.h5-h4 Rf1-f3 #
1.Kg3-h4 Ke1-f2 2.h5*g4 Sh2-f3 # (2023-09-14)
comment
Ladislav Packa: Retro content is not needed, the solution is preserved even without it.
Pg4 Pg2 Sh2 Ke1 Rh1 (5)- Ph5 Kg3 (2) h#2* C+
1...0-0 2.h5-h4 Rf1-f3 #
1.Kg3-h4 Ke1-f2 2.h5*g4 Sh2-f3 # (2023-09-14)
comment
*) 1. ... 0-0#
1) 1. Dc5 Tf1+ 2. Df2 Txf2#
1) 1. Dc5 Tf1+ 2. Df2 Txf2#
15 - 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)
more ...
comment
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)
more ...
comment
Keywords: Cant Castler, Castling (wg), Superseded by (P1415606)
Genre: h#, Retro
Computer test: both twins cooked by Popeye v4.87
FEN: NBr5/N1p3p1/1qPpp3/2bb4/1p6/n1pP1rP1/1PP1P2k/R3KQ1n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-03-06 more...
Genre: h#, Retro
Computer test: both twins cooked by Popeye v4.87
FEN: NBr5/N1p3p1/1qPpp3/2bb4/1p6/n1pP1rP1/1PP1P2k/R3KQ1n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-03-06 more...
a) 1. Ta6! Lxc3 2. Ta2 0-0#
b) 1. Ta5! Kd1 2. Ta2 Kc2#
b) 1. Ta5! Kd1 2. Ta2 Kc2#
Henrik Juel: In b) White may not castle, because last move was done by Ke1 or Th1 (2023-12-02)
A.Buchanan: Cute. Do wPe2 & bPe3 serve any purpose, however? (2023-12-03)
more ...
comment
A.Buchanan: Cute. Do wPe2 & bPe3 serve any purpose, however? (2023-12-03)
more ...
comment
Keywords: Cant Castler, Castling (wk), Superseded by (P1413924)
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & simple retro logic
FEN: 5n1q/6B1/5r2/7r/5p2/2p1p3/1P2P3/k3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-05 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & simple retro logic
FEN: 5n1q/6B1/5r2/7r/5p2/2p1p3/1P2P3/k3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-05 more...
a) 1. 0-0 Tcg3 2. Sh8 Txg7#
b) 1. 0-0-0 Tgc4 2. Sb8 Txc7#
b) 1. 0-0-0 Tgc4 2. Sb8 Txc7#
Keywords: Castling, Cant Castler, Obvious promotion (T), Twin
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: r3k2r/1pp3pp/2n3n1/8/6R1/2R4P/4P1P1/5B1K
Input: Gerd Wilts, 1995-06-03
Last update: Gunter Jordan, 2022-12-01 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: r3k2r/1pp3pp/2n3n1/8/6R1/2R4P/4P1P1/5B1K
Input: Gerd Wilts, 1995-06-03
Last update: Gunter Jordan, 2022-12-01 more...
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)
more ...
comment
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)
more ...
comment
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. fxe3ep d8=S 2. Kxd5 0-0-0# (Td1#?)
White has made 6 pawn captures with one missing black piece unaccounted for, that by parity can only have been captured by an officer. White's last move cannot have been a pawn capture. If R: 1.f2-f3, then sLg1 was promoted, implying 7 captures by black pawns - one too many. So if White can prove they retain castling rights, then the ep is on. Hence AP Petrovic is valid.
White has made 6 pawn captures with one missing black piece unaccounted for, that by parity can only have been captured by an officer. White's last move cannot have been a pawn capture. If R: 1.f2-f3, then sLg1 was promoted, implying 7 captures by black pawns - one too many. So if White can prove they retain castling rights, then the ep is on. Hence AP Petrovic is valid.
Henrik Juel: 0... fxe3ep 1.d8S Kxd5 2.0-0-0#. Not -1.f2? and Lg1 is caught. (2004-09-16)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
Henrik Juel: In principle I agree, Vaclav
But PDB does not (yet) allow HC+, so I find it acceptable to use the C+ label, when you also tell the whole story after 'Computer test:' below (2023-08-03)
A.Buchanan: Hopefully Gerd will have more time at some point, and can expand the functionality in this and other areas. In the meantime, engine solving of conventional retros including AP, is in its infancy. Retractor 2 has some effectiveness, but is still basic. And there is nothing that yet grasps the intricacies of castling/ep etc. However AP problems do often contain considerable forward chess, and the C+ tag is very useful to filter out those that have already been solved forwardly, without pretending that these are in any sense fully solved (2023-08-04)
Ladislav Packa: I know the definition of AP, but I don't understand the logic behind it. The move 1.fxe3 e.p. proves that White CAN castling. But the solution (2. ...Rd1#?) claims that castling is MANDATORY. From my point of view, AP is correct when only castling is necessary for the solution and the Rook move would be a dual. (2023-08-04)
A.Buchanan: @Ladislav: I am not sure how to help you. Maybe you can read this page from Retro Corner: https://www.janko.at/Retros/Glossary/APosteriori.htm (2023-08-05)
Ladislav Packa: Andrew, what should the article help me with? I quote the final sentence:
Some people still oppose this rule and argue that it should certainly not be the default convention. (2023-08-05)
Henrik Juel: You could view it this way, Ladislav
h#2 means that it is Black to move, so White made last move
What was last move? A little analysis shows just three possibilities: f2-f4, move by Ta1, or move by Ke1
So normally we cannot assume that last move was f2-f4
But if White can castle, then the last move was f2-f4
So if we could start with 0... 0-0-0, then 1.fxe3ep would be legitimate
AP says that you are allowed to reverse the sequence of events; first do the ep capture, then later legitimize it by castling
Was this helpful? (2023-08-05)
Ladislav Packa: Henrik, you don't have to explain that to me. I've done a few AP issues myself, like P1348357. But that doesn't mean I agree with AP's logic. I already wrote it - the term "you can castle" is applied as "you must castle". But these are only problems where, in addition to castling, the Rook move can also be used, I consider that a dual.
From that point of view, the P1000662 issue is perfectly fine for me. (2023-08-05)
Joost de Heer: AP: By castling, you prove a posteriori that the ep-capture was not just a try but the actual solution. Without castling, the ep-solution just is that: a try.
So: Try 1. fe3 ep e8=S 2. Kd5 Rd1 - but ep capture not allowed, as there is no proof that f2-f4 must've been the last move.
Solution 1. fe3 ep e8=S 2. Kd5 OOO - Now the ep capture was justified because white castled, thereby proving that the last move before the diagram position indeed was f2-f4. (2023-08-06)
Joost de Heer: See e.g. P1052919 : The try is an ep capture which is unjustified. (2023-08-06)
A.Buchanan: OK Ladislav: I think I get your point. If one solution with castling justifies the e.p., then based on that certainty, why shouldn't an alternative solution with no castling *then* be allowed as well? There are problems in which one twin shows 0-0-0 and the other shows 0-0. Each is based on the other in a similar way, so the idea of dependency is not new. Why are we not allowed to add other "parasitic" solutions as well? Why can we only have the "paying" solutions? We can't say that we are restricted to one solution: that's not the way chess problems operate! And this is just in the help world - in the adversarial world it might get even more complicated. Is this your issue, Ladislav? (2023-08-07)
Ladislav Packa: I don't want to unnecessarily prolong this discussion. However, I will add one more note: in this position, white castling is also possible without e.p. in Black's 1st move. If B1 were an indifferent move, then white can 1...0-0-0! The Codex of Chess Composition writes about it in Article 16 (1):
Castling convention. Casting is permitted unless it can be proven that it is not permissible.
In our case, 0-0-0 is possible because White's last move exists - e2-e4! It does not matter if it is this move or some a2-a3, both moves are equivalent. EP does not prove the possibility of casting, it would be legal even without it. (2023-08-07)
A.Buchanan: Ladislav was what I wrote your issue pls? Y/N :-) (2023-08-07)
Ladislav Packa: I have no problem, I'm just expressing my own opinion about the AP convention. (2023-08-07)
A.Buchanan: OK cos I think the point I raised is a real one that should be addressed by theory some day. Clearly from the nice problem that you composed Ladislav you understand the mechanics very well. From a justification perspective it's all a bit iffy, but that's why it's controversial. Under RS it's really the only way one can end up actually eping, and it's proved compositionally fertile. So that's enough to justify (2023-08-08)
Ladislav Packa: No need to apologize. This is a normal discussion with different views on the issue. Maybe it will come to some conclusion.
I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p. (2023-08-08)
Joost de Heer: "I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p."
You misinterpret AP. The e.p. capture does not authorize castling, castling provides a justification later on (hence the 'a posteriori') for the legality of ep.
Usually, for ep justification you need to examine all game trees that lead to the diagram, and only if all game trees end with the double-step, then ep is allowed.
With AP, you examine all the game trees including the actual play. If all those game trees have as last move before the diagram position the double step, then ep is possible.
In this case, if white doesn't castle, then there are game trees which don't have as last move the double step, and therefore AP logic dictates that the ep capture was illegal. However, all game trees which lead to the diagram and which have castling in the actual play have as last move before the diagram position the double step, hence AP dictates that the ep capture is legal. (2023-08-09)
Ladislav Packa: Joost: A simple question - is white allowed to castle after any 1st move by black (except e.p.)? (2023-08-09)
Joost de Heer: Of course he is. AP only is used to combine the ep justification with castling, not the castling right per se. (2023-08-09)
more ...
comment
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
Henrik Juel: In principle I agree, Vaclav
But PDB does not (yet) allow HC+, so I find it acceptable to use the C+ label, when you also tell the whole story after 'Computer test:' below (2023-08-03)
A.Buchanan: Hopefully Gerd will have more time at some point, and can expand the functionality in this and other areas. In the meantime, engine solving of conventional retros including AP, is in its infancy. Retractor 2 has some effectiveness, but is still basic. And there is nothing that yet grasps the intricacies of castling/ep etc. However AP problems do often contain considerable forward chess, and the C+ tag is very useful to filter out those that have already been solved forwardly, without pretending that these are in any sense fully solved (2023-08-04)
Ladislav Packa: I know the definition of AP, but I don't understand the logic behind it. The move 1.fxe3 e.p. proves that White CAN castling. But the solution (2. ...Rd1#?) claims that castling is MANDATORY. From my point of view, AP is correct when only castling is necessary for the solution and the Rook move would be a dual. (2023-08-04)
A.Buchanan: @Ladislav: I am not sure how to help you. Maybe you can read this page from Retro Corner: https://www.janko.at/Retros/Glossary/APosteriori.htm (2023-08-05)
Ladislav Packa: Andrew, what should the article help me with? I quote the final sentence:
Some people still oppose this rule and argue that it should certainly not be the default convention. (2023-08-05)
Henrik Juel: You could view it this way, Ladislav
h#2 means that it is Black to move, so White made last move
What was last move? A little analysis shows just three possibilities: f2-f4, move by Ta1, or move by Ke1
So normally we cannot assume that last move was f2-f4
But if White can castle, then the last move was f2-f4
So if we could start with 0... 0-0-0, then 1.fxe3ep would be legitimate
AP says that you are allowed to reverse the sequence of events; first do the ep capture, then later legitimize it by castling
Was this helpful? (2023-08-05)
Ladislav Packa: Henrik, you don't have to explain that to me. I've done a few AP issues myself, like P1348357. But that doesn't mean I agree with AP's logic. I already wrote it - the term "you can castle" is applied as "you must castle". But these are only problems where, in addition to castling, the Rook move can also be used, I consider that a dual.
From that point of view, the P1000662 issue is perfectly fine for me. (2023-08-05)
Joost de Heer: AP: By castling, you prove a posteriori that the ep-capture was not just a try but the actual solution. Without castling, the ep-solution just is that: a try.
So: Try 1. fe3 ep e8=S 2. Kd5 Rd1 - but ep capture not allowed, as there is no proof that f2-f4 must've been the last move.
Solution 1. fe3 ep e8=S 2. Kd5 OOO - Now the ep capture was justified because white castled, thereby proving that the last move before the diagram position indeed was f2-f4. (2023-08-06)
Joost de Heer: See e.g. P1052919 : The try is an ep capture which is unjustified. (2023-08-06)
A.Buchanan: OK Ladislav: I think I get your point. If one solution with castling justifies the e.p., then based on that certainty, why shouldn't an alternative solution with no castling *then* be allowed as well? There are problems in which one twin shows 0-0-0 and the other shows 0-0. Each is based on the other in a similar way, so the idea of dependency is not new. Why are we not allowed to add other "parasitic" solutions as well? Why can we only have the "paying" solutions? We can't say that we are restricted to one solution: that's not the way chess problems operate! And this is just in the help world - in the adversarial world it might get even more complicated. Is this your issue, Ladislav? (2023-08-07)
Ladislav Packa: I don't want to unnecessarily prolong this discussion. However, I will add one more note: in this position, white castling is also possible without e.p. in Black's 1st move. If B1 were an indifferent move, then white can 1...0-0-0! The Codex of Chess Composition writes about it in Article 16 (1):
Castling convention. Casting is permitted unless it can be proven that it is not permissible.
In our case, 0-0-0 is possible because White's last move exists - e2-e4! It does not matter if it is this move or some a2-a3, both moves are equivalent. EP does not prove the possibility of casting, it would be legal even without it. (2023-08-07)
A.Buchanan: Ladislav was what I wrote your issue pls? Y/N :-) (2023-08-07)
Ladislav Packa: I have no problem, I'm just expressing my own opinion about the AP convention. (2023-08-07)
A.Buchanan: OK cos I think the point I raised is a real one that should be addressed by theory some day. Clearly from the nice problem that you composed Ladislav you understand the mechanics very well. From a justification perspective it's all a bit iffy, but that's why it's controversial. Under RS it's really the only way one can end up actually eping, and it's proved compositionally fertile. So that's enough to justify (2023-08-08)
Ladislav Packa: No need to apologize. This is a normal discussion with different views on the issue. Maybe it will come to some conclusion.
I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p. (2023-08-08)
Joost de Heer: "I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p."
You misinterpret AP. The e.p. capture does not authorize castling, castling provides a justification later on (hence the 'a posteriori') for the legality of ep.
Usually, for ep justification you need to examine all game trees that lead to the diagram, and only if all game trees end with the double-step, then ep is allowed.
With AP, you examine all the game trees including the actual play. If all those game trees have as last move before the diagram position the double step, then ep is possible.
In this case, if white doesn't castle, then there are game trees which don't have as last move the double step, and therefore AP logic dictates that the ep capture was illegal. However, all game trees which lead to the diagram and which have castling in the actual play have as last move before the diagram position the double step, hence AP dictates that the ep capture is legal. (2023-08-09)
Ladislav Packa: Joost: A simple question - is white allowed to castle after any 1st move by black (except e.p.)? (2023-08-09)
Joost de Heer: Of course he is. AP only is used to combine the ep justification with castling, not the castling right per se. (2023-08-09)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Promotion (S), Valladao Task
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 + simple retro-logic
FEN: 8/3P2p1/2PP4/1ppPp3/2pkPp2/5PP1/6Pp/R3K1b1
Input: Gerd Wilts, 1999-02-27
Last update: A.Buchanan, 2023-09-11 more...
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 + simple retro-logic
FEN: 8/3P2p1/2PP4/1ppPp3/2pkPp2/5PP1/6Pp/R3K1b1
Input: Gerd Wilts, 1999-02-27
Last update: A.Buchanan, 2023-09-11 more...
1. Sf7 Df3 2. 0-0-0 Db7# (0-0-0 + !)
UL!
UL!
Anton Baumann: mögliche Korrektur: wBb4 nach b3, +sBd5, -wBg2
1.Sf7 Db4 2.0-0-0 Db7# (C+) (2024-01-03)
comment
1.Sf7 Db4 2.0-0-0 Db7# (C+) (2024-01-03)
comment
1. 0-0-0+ Ld3 2. Ld7 La6#
NL:
1. Dg8/Db7/Dc7 Dxe6+ 2. Te7 Dg8#
1. Df8 Dd4/Dc7/Dd6/Dd5 2. Lf7 Dd7#
NL:
1. Dg8/Db7/Dc7 Dxe6+ 2. Te7 Dg8#
1. Df8 Dd4/Dc7/Dd6/Dd5 2. Lf7 Dd7#
Rochade
F135
Verst-4
Yuri Bilokin: correction bRh7-f7(-bQf7), bBa1-b6, -bPa7, +bSf4, +bSh6 r3k3/5r1B/1b2b2n/4Q3/5n2/8/8/3K4 (3+7) (2022-08-28)
comment
F135
Verst-4
Yuri Bilokin: correction bRh7-f7(-bQf7), bBa1-b6, -bPa7, +bSf4, +bSh6 r3k3/5r1B/1b2b2n/4Q3/5n2/8/8/3K4 (3+7) (2022-08-28)
comment
1. Dh8 0-0 2. Da1 Lxf7#
Eric Huber: Anticipated by 'P0534152' Erkki Heinonen, 3. Makuc-Moder-Gedenkturnier 1971-1973, 4. ehrende Erwähnung. (2008-10-01)
Viktoras Paliulionis: This problem shows paradoxical idea - two mutually exclusive tempo moves (B1 & B2). P0534152 has only one tempo move. (2023-10-09)
more ...
comment
Viktoras Paliulionis: This problem shows paradoxical idea - two mutually exclusive tempo moves (B1 & B2). P0534152 has only one tempo move. (2023-10-09)
more ...
comment
Keywords: Castling (wk), Tempo Move (dd)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 6B1/5n1n/8/8/N7/b7/kq5r/4K2R
Input: Markus Manhart, 1998-02-10
Last update: Gerd Wilts, 2008-10-02 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 6B1/5n1n/8/8/N7/b7/kq5r/4K2R
Input: Markus Manhart, 1998-02-10
Last update: Gerd Wilts, 2008-10-02 more...
1) 1. 0-0 Lf6 2. Tf7 Th8#
2) 1. Tf8 Td7 2. Tf7 Td8#
3) 1. Tg8 Kc7 2. Tf8 Te7#
2) 1. Tf8 Td7 2. Tf7 Td8#
3) 1. Tg8 Kc7 2. Tf8 Te7#
in Vratnica von Datiaschwili/Gatseciladse steingetreu nachempfunden
Yuri Bilokin: you can see *) 1...Kc7 2.Rf8 Re7# (2022-09-11)
comment
Yuri Bilokin: you can see *) 1...Kc7 2.Rf8 Re7# (2022-09-11)
comment
Keywords: Echo, Castling (sk), Aristocrat
Genre: h#
Computer test: (Popeye C-Version 3.47 (1024 KB))
FEN: 4k2r/7R/1K6/8/7B/8/8/8
Reprints: 87 Vratnica-64 3 07-09/2001
Input: Markus Manhart, 1998-02-16
Last update: Erich Bartel, 2022-09-12 more...
Genre: h#
Computer test: (Popeye C-Version 3.47 (1024 KB))
FEN: 4k2r/7R/1K6/8/7B/8/8/8
Reprints: 87 Vratnica-64 3 07-09/2001
Input: Markus Manhart, 1998-02-16
Last update: Erich Bartel, 2022-09-12 more...
1) 1. Sg1 0-0-0 2. Sf3 Sc2 3. Sxe5 Txd4#
2) 1. Sf4 Td6 2. Sd5 Sc6 3. Se3 Txd4#
2) 1. Sf4 Td6 2. Sd5 Sc6 3. Se3 Txd4#
Yuri Bilokin: more economical -wPf6, -wPg4, -bPa4, bBg6-f5, bPf7-d7 8/4p3/2R1p1p1/2P1Pb2/1Nppk3/2p3p1/rr2n1P1/R3K3 (7+12) h#3 2.1…
1.Sg1 0-0-0 2.Sf3 Sc2 3.Sxe5 Rxd4#
1.Sf4 Rd6 2.Sd5 Sc6 3.Se3 Rxd4#
AntiZielElement (B2, line obstruction)
Castling (white, long). Check prevention (W-W)
Exchange of functions (wRa1/wRc6, Mate / Passive)
Long-trip (bS, 3) × 2.Play on the same square (W3, 2)
Mates on the same square × 2 (2023-08-23)
comment
1.Sg1 0-0-0 2.Sf3 Sc2 3.Sxe5 Rxd4#
1.Sf4 Rd6 2.Sd5 Sc6 3.Se3 Rxd4#
AntiZielElement (B2, line obstruction)
Castling (white, long). Check prevention (W-W)
Exchange of functions (wRa1/wRc6, Mate / Passive)
Long-trip (bS, 3) × 2.Play on the same square (W3, 2)
Mates on the same square × 2 (2023-08-23)
comment
25 - P0533723
Rudolf A. Pieprzyk
223 Problemista 02/1964
(4. Platz wurde annulliert)
IV Schlesischer Problemistenkongress
Blitzthematurnier 17.11.1963
(3+4) cooked
h#3
Rudolf A. Pieprzyk
223 Problemista 02/1964
(4. Platz wurde annulliert)
IV Schlesischer Problemistenkongress
Blitzthematurnier 17.11.1963
(3+4) cooked
h#3
1. Kc4 0-0-0 2. Lb4 La6+ 3. Kc3 Txd3#
NL:
1. c5 0-0-0 2. Ka2 Txd3 3. Ka1 Ta3#
1. Kc3 La6 2. Lb4 0-0-0 3. c5 Txd3#
NL:
1. c5 0-0-0 2. Ka2 Txd3 3. Ka1 Ta3#
1. Kc3 La6 2. Lb4 0-0-0 3. c5 Txd3#
Auszeichnung wurde in Problemista 04/1964 annulliert.
Yuri Bilokin: correction wBc8-e8, bKb3-d4, bBf8-e3, bPc7-f7, +bQa6, -bPd3 4B3/5p2/q7/8/3k4/4b3/8/R3K3 (3+4) h#3
1.Kc3 Bb5 2.Bc5 0-0-0 3.Bb4 Rd3#
Castling (white, long) (2023-06-10)
comment
Yuri Bilokin: correction wBc8-e8, bKb3-d4, bBf8-e3, bPc7-f7, +bQa6, -bPd3 4B3/5p2/q7/8/3k4/4b3/8/R3K3 (3+4) h#3
1.Kc3 Bb5 2.Bc5 0-0-0 3.Bb4 Rd3#
Castling (white, long) (2023-06-10)
comment
a) 1. Kf8 0-0-0 2. Kg8 Tf1 3. Kh8 Tf5 4. Tg8 Txh5#
b) 1. 0-0-0 Ta3 2. Kb8 Tb3 3. Ka8 Tb5 4. Tb8 Txa5#
b) 1. 0-0-0 Ta3 2. Kb8 Tb3 3. Ka8 Tb5 4. Tb8 Txa5#
SCHRECKE: NL in b):
z.B.: 1. Ke7 Tb1 2. Ke6 d3 3. Kd5 dxe4+ 4. Kc4 b3# (2022-11-14)
HBae: Mögliche Korrektur sBd6 nach d3, ohne wBe2 (9+10 (2022-11-14)
HBae: Mit der (berichtigten) oben genannten Korrektur ist das Problem C+ gemäß Popeye v4.37. Die lange weiße Rochade ist nur in a) erlaubt. (2022-11-14)
more ...
comment
z.B.: 1. Ke7 Tb1 2. Ke6 d3 3. Kd5 dxe4+ 4. Kc4 b3# (2022-11-14)
HBae: Mögliche Korrektur sBd6 nach d3, ohne wBe2 (9+10 (2022-11-14)
HBae: Mit der (berichtigten) oben genannten Korrektur ist das Problem C+ gemäß Popeye v4.37. Die lange weiße Rochade ist nur in a) erlaubt. (2022-11-14)
more ...
comment
27 - P0534198
Herbert Hultberg
2565v Eskilstuna-Kuriren 22/09/1937
Frans Hansson gewidmet
(8+5) cooked
h#3*
2.1...
Herbert Hultberg
2565v Eskilstuna-Kuriren 22/09/1937
Frans Hansson gewidmet
(8+5) cooked
h#3*
2.1...
*) 1. ... 0-0-0 2. Kf6 Tf1+ 3. Ke7 Txf7#
1) 1. Tc7 0-0 2. Kd6 Tad1 3. Ke7 Tf7#
NL:
*) 1. ... g3 2. Tf5 0-0-0 3. Tf6 Te1#
1. Ke4 0-0-0 2. Ke3 Td2 3. Tf4 Te1#
1) 1. Tc7 0-0 2. Kd6 Tad1 3. Ke7 Tf7#
NL:
*) 1. ... g3 2. Tf5 0-0-0 3. Tf6 Te1#
1. Ke4 0-0-0 2. Ke3 Td2 3. Tf4 Te1#
Yuri Bilokin: possibly -wPa2, -wPb6, wPh2-h3, bR7-a3, +bQf7, +bPb5, +bPb7, +bPg3 4n3/1p3q2/4r1P1/1p2k3/b7/r5pP/6P1/R3K2R (6+9) h+3 2.1…
1.Rb3 0-0-0 2.Kf6 Rhf1+ 3.Ke7 Rxf7#
1.Qc7 0-0 2.Kd6 Rad1+ 3.Ke7 Rf7#
AntiZielElement (B1, guard). Castling (white, long). Castling (white, short)
Check prevention (W-W) × 2. Delayed Umnov (bQ-wR)
Hideaway (bQ). Many-ways (bK, 2)
Epaulette mate × 2 (2023-08-23)
comment
1.Rb3 0-0-0 2.Kf6 Rhf1+ 3.Ke7 Rxf7#
1.Qc7 0-0 2.Kd6 Rad1+ 3.Ke7 Rf7#
AntiZielElement (B1, guard). Castling (white, long). Castling (white, short)
Check prevention (W-W) × 2. Delayed Umnov (bQ-wR)
Hideaway (bQ). Many-ways (bK, 2)
Epaulette mate × 2 (2023-08-23)
comment
1) 1. Lc8 Tb7 2. Ld7 Lf4 3. 0-0-0 Tb8#
2) 1. Ld4 Lf4 2. 0-0-0 Txb7 3. Td7 Tb8#
3) 1. 0-0-0+ Td7 2. Sb5 Lxf6 3. Sc7 Txd8#
2) 1. Ld4 Lf4 2. 0-0-0 Txb7 3. Td7 Tb8#
3) 1. 0-0-0+ Td7 2. Sb5 Lxf6 3. Sc7 Txd8#
klären autor?
Yuri Bilokin: author Yuri Bilikin
Castling (black, long) × 3 (2024-03-18)
Yuri Bilokin: slip of the pen, author Yuri Bilokin (2024-03-21)
comment
Yuri Bilokin: author Yuri Bilikin
Castling (black, long) × 3 (2024-03-18)
Yuri Bilokin: slip of the pen, author Yuri Bilokin (2024-03-21)
comment
29 - P0534722
Thomas R. Dawson
2828v Stratford Express 15/12/1934
(5+7)
h#2
b) alles eine Reihe tiefer
Thomas R. Dawson
2828v Stratford Express 15/12/1934
(5+7)
h#2
b) alles eine Reihe tiefer
a) 1. g2 Ta6 2. g1=T Tf2#
b) 1. Ke5 0-0-0 2. Ke4/Lf6 Te1#
NL in a)
1. Kf8 Sf4 /Sg7 2. Sf7 Se6#
1. Kf8 Tf2+ 2. Lf7 Ta8#
b) 1. Ke5 0-0-0 2. Ke4/Lf6 Te1#
NL in a)
1. Kf8 Sf4 /Sg7 2. Sf7 Se6#
1. Kf8 Tf2+ 2. Lf7 Ta8#
Dual in b)
Yuri Bilokin: possibly bRe8(-bBe8), +bNc7, +bPg6 4r1rn/2n1pk2/6p1/7N/6P1/6pp/R3K2R/8 (5+9) h#2 b) alles eine Reihe tiefer
a) 1.g2 Ra6 2.g1=R Rf2#
b) 1.Ke5 0-0-0 2.Ke4 Rhe1# (MM) (2022-12-16)
comment
Yuri Bilokin: possibly bRe8(-bBe8), +bNc7, +bPg6 4r1rn/2n1pk2/6p1/7N/6P1/6pp/R3K2R/8 (5+9) h#2 b) alles eine Reihe tiefer
a) 1.g2 Ra6 2.g1=R Rf2#
b) 1.Ke5 0-0-0 2.Ke4 Rhe1# (MM) (2022-12-16)
comment
1) 1. Tb1+ Ke2 2. Kc1 Kd3 3. Kd1 Txb1#
2) 1. Tb1+ Kf2 2. Kd1 Ke3 3. Ke1 Txb1#
2) 1. Tb1+ Kf2 2. Kd1 Ke3 3. Ke1 Txb1#
klären: welches ist die NL?
Bernd Schwarzkopf: Ein Heft „Die Schwalbe 10-12/1968“ gibt es nicht.
Das Problem ist in „Die Schwalbe 10-11/1968“ oder „12/1968“ nicht zu finden.
Vielleicht war P0544925 gemeint. (2023-04-26)
Henrik Juel: The cook is easily removed by adding '2 solutions' to the stipulation
1.Kb3 0-0-0 etc. are tries, because White just moved Ta1 or Ke1, so he may not castle (2023-04-26)
comment
Bernd Schwarzkopf: Ein Heft „Die Schwalbe 10-12/1968“ gibt es nicht.
Das Problem ist in „Die Schwalbe 10-11/1968“ oder „12/1968“ nicht zu finden.
Vielleicht war P0544925 gemeint. (2023-04-26)
Henrik Juel: The cook is easily removed by adding '2 solutions' to the stipulation
1.Kb3 0-0-0 etc. are tries, because White just moved Ta1 or Ke1, so he may not castle (2023-04-26)
comment
1) 1. Tb2 0-0-0 2. Tb4 Th3#
2) 1. Kd3 Ta4 2. c3 Th3#
3) 1. Kb2 0-0 2. c3 Tfb1#
2) 1. Kd3 Ta4 2. c3 Th3#
3) 1. Kb2 0-0 2. c3 Tfb1#
VL: One solution consists of two partial ones, and there is another, castling independent, solution. A rare scenario; cf. P1091926. Presently, due to Art.16 (3) of the Codex, the mark "RV" may be deleted from the stipulation. (2017-11-25)
A.Buchanan: Methodologically, I think that one applies PRA first. So there are two parts, and each has two solutions. One solution is shared across both parts (2023-08-22)
more ...
comment
A.Buchanan: Methodologically, I think that one applies PRA first. So there are two parts, and each has two solutions. One solution is shared across both parts (2023-08-22)
more ...
comment
Keywords: Castling (wk,wg), Partial Retro Analysis (PRA), Miniature, Homebase (w)
Genre: h#, Retro
Computer test: Popeye WIN32-Version 3.56 (2048 KB)
FEN: 8/8/8/8/2p5/2k5/2r5/R3K2R
Input: Michal Dragoun, 2001-09-20
Last update: A.Buchanan, 2023-08-22 more...
Genre: h#, Retro
Computer test: Popeye WIN32-Version 3.56 (2048 KB)
FEN: 8/8/8/8/2p5/2k5/2r5/R3K2R
Input: Michal Dragoun, 2001-09-20
Last update: A.Buchanan, 2023-08-22 more...
a) 1. 0-0 h3 2. Lh1 h2#
b) 1. h3 Sf4 2. hxg2 Sh5#
b) 1. h3 Sf4 2. hxg2 Sh5#
Keywords: Miniature, Twin, Stipulation change, Castling (in a)
Genre: h#, s#
Computer test: (Popeye WINDOWS-32Bit V4.05 (1575896 KB))
FEN: 8/8/6N1/8/6pp/6k1/6B1/4K2R
Input: hpr, 2007-04-29
Last update: Gunter Jordan, 2023-03-13 more...
Genre: h#, s#
Computer test: (Popeye WINDOWS-32Bit V4.05 (1575896 KB))
FEN: 8/8/6N1/8/6pp/6k1/6B1/4K2R
Input: hpr, 2007-04-29
Last update: Gunter Jordan, 2023-03-13 more...
1. a1=D+ Kb4 2. Da6 Se5 3. Kd8 Kc5 4. Dc8+ Kd6 5. Te8 Sf7#
Keywords: Epaulettenmatt, Minimal, Castling position
Genre: h#
FEN: 4k2r/p7/2N5/8/8/K7/p7/8
Input: Klaus Funk, 2013-07-15
Last update: Klaus Funk, 2013-07-25 more...
Genre: h#
FEN: 4k2r/p7/2N5/8/8/K7/p7/8
Input: Klaus Funk, 2013-07-15
Last update: Klaus Funk, 2013-07-25 more...
1. g1=S Sg3 2. Se2 Sf5 3. Sxc3 Sxc3 4. Kh7 0-0-0 5. Dg8 Th1#
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-08-15)
Yuri Bilokin: more economical -wPb2, bPb3-e5 6kq/8/6p1/4p3/p1p5/P1Pp1p2/3P1Pp1/RN2K2N (8+9) (2022-08-17)
comment
Yuri Bilokin: more economical -wPb2, bPb3-e5 6kq/8/6p1/4p3/p1p5/P1Pp1p2/3P1Pp1/RN2K2N (8+9) (2022-08-17)
comment
1. Sd1 Se3 2. Lb7 Sd5 3. 0-0-0 Sxc7 4. d5 Sa8 5. Tc7 Sxb6#
Keywords: Unpinning, Line opening, Remote selfblock, Liniensperrung, Check Protection, Castling, Selfblock, Model mate
Genre: h#
FEN: r1b1kb2/B1p3p1/1prp4/1np5/8/6pp/pn6/q4N1K
Input: Felber, Volker, 2013-09-03
Last update: Gunter Jordan, 2023-07-22 more...
Genre: h#
FEN: r1b1kb2/B1p3p1/1prp4/1np5/8/6pp/pn6/q4N1K
Input: Felber, Volker, 2013-09-03
Last update: Gunter Jordan, 2023-07-22 more...
1. g4 0-0 2. Kxh3 Tf3#
veröffentlicht im Rahmen des Weihnachtslösungsturniers 1958
zunächst ohne wBh3 abgedruckt, dann aber jede Menge NL. Die Korrektur erschien am 04.01.1959
Yuri Bilokin: Possibly -wPh3. –bPg5, +bBf3 2B5/8/8/8/7p/5bk1/8/4K2R (3+3)
1.Bg4 0-0 2.Kh3 Rf3# (MM)
Anticipatory self-pin
Castling (white, short)
Delayed Umnov (bB-wR)
Model mate × 1
Pin-mate (2024-02-29)
more ...
comment
zunächst ohne wBh3 abgedruckt, dann aber jede Menge NL. Die Korrektur erschien am 04.01.1959
Yuri Bilokin: Possibly -wPh3. –bPg5, +bBf3 2B5/8/8/8/7p/5bk1/8/4K2R (3+3)
1.Bg4 0-0 2.Kh3 Rf3# (MM)
Anticipatory self-pin
Castling (white, short)
Delayed Umnov (bB-wR)
Model mate × 1
Pin-mate (2024-02-29)
more ...
comment
* 1. ... h4 2. Kd3 0-0-0#
1. Db3 Ta5 2. Dd3 Sf3#
Mit Schwarz am Zug kann Weiß zuletzt nur K oder T gezogen haben, die w0-0-0 ist also illegal.
1. Db3 Ta5 2. Dd3 Sf3#
Mit Schwarz am Zug kann Weiß zuletzt nur K oder T gezogen haben, die w0-0-0 ist also illegal.
38 - P1398939
Edgar Fielder
Andrew Buchanan
Discord Chess Problems & Studies Server 07/022022
EF, version AB
(13+10)
h#1
Edgar Fielder
Andrew Buchanan
Discord Chess Problems & Studies Server 07/022022
EF, version AB
(13+10)
h#1
R: 1. Dg4-g3 Kd8-e8 2. Dg5-g4 Kc8-d8 3. Dg6-g5 Kb7-c8 4. Df6-g6 Tb8-h8 5. De6-f6 Kc8-b7 6. Dd6-e6 Kd8-c8 7. Dc6-d6 Ke8-d8 8. Db6-c6 Kf8-e8 9. Da7-b6 Kg8-f8 10. Db7-a7 Te8-b8 11. Db8-b7 Tf8-e8 12. b7-b8=D 0-0 13. c6xDb7 Da8-b7 14. d5xLc6 Dd8-a8 15. d4-d5 Lb7-c6 16. e3xSd4 Lc8-b7 17. f2xSe3 c6-c5 18. h5-h6 b7xLc6 is one of many possible histories, but all involve hidden castling
Retro solution (which works for Fielder's original too).
Pawn captures: white missing LSBf, captured by axbxc,bxc. Therefore wBf captured at least 3 times fxexdxc, also exd & sLf8 died at home. So one missing black unit unaccounted for currently.
Cage can only be unlocked by exd3, but wLf must be returned home prior to this. Can only be uncaptured by sBb7xLc6-c5. If wPf did not promote, it was captured by sBa on b6 after 5 captures or on c5 after 4 captures. The latter is not possible, as wBc4 never left its file. The former is not possible, because sD would be unavailable for capture until b7xc6, which only happened after the cage was formed.
Therefore wBf promoted, and must have captured four times to reach b-file, accounting for the final capture. When it promoted on b8, sBb had already moved, capturing wLf and therefore the cage was already formed. So wDg3 is the promoted unit. If wBexd3 was capture of sLc, then there is deadlock between the two releases of bishops, so sLc was captured by wBf, hence path to promotion shifted to light squares (i.e. via c6 not b6) so sBc5 was already in place. When b7-b8=D, all the other White pieces (except possibly for wPh6, bK & bRh8) were already in place.
There is now a time-critical retraction, as White has at most 7 moves (not 9, because must release wRh before wLf is free) before sBb7xLc6, with sD & sLc sealed inside. So R: 1. b7-b8=D ~ 2. c6xDb7 zB Da8-b7 3. d5xLc6 Dd8-a8 4. d4-d5 Lb7-c6 5. e3xSd4 Lc8-b7 6. f2xSe3 c6-c5 7. h5-h6 b7xLc6. Where was bK while all this was happening. There is only one spare tempo (h4-h5) If bK was off the back rank, there was not enough time to retract back there. So the only possibility is that Black was already castled, and can uncastle when the black cage is closed. The missing move can be R: 1. ... 0-0.
Since castling is illegal in the forward play, h#1 is unique with Donati theme as wQ revisits the promotion square:
1. c6 Db8# not
1. 0-0? Dxg7#
Retro solution (which works for Fielder's original too).
Pawn captures: white missing LSBf, captured by axbxc,bxc. Therefore wBf captured at least 3 times fxexdxc, also exd & sLf8 died at home. So one missing black unit unaccounted for currently.
Cage can only be unlocked by exd3, but wLf must be returned home prior to this. Can only be uncaptured by sBb7xLc6-c5. If wPf did not promote, it was captured by sBa on b6 after 5 captures or on c5 after 4 captures. The latter is not possible, as wBc4 never left its file. The former is not possible, because sD would be unavailable for capture until b7xc6, which only happened after the cage was formed.
Therefore wBf promoted, and must have captured four times to reach b-file, accounting for the final capture. When it promoted on b8, sBb had already moved, capturing wLf and therefore the cage was already formed. So wDg3 is the promoted unit. If wBexd3 was capture of sLc, then there is deadlock between the two releases of bishops, so sLc was captured by wBf, hence path to promotion shifted to light squares (i.e. via c6 not b6) so sBc5 was already in place. When b7-b8=D, all the other White pieces (except possibly for wPh6, bK & bRh8) were already in place.
There is now a time-critical retraction, as White has at most 7 moves (not 9, because must release wRh before wLf is free) before sBb7xLc6, with sD & sLc sealed inside. So R: 1. b7-b8=D ~ 2. c6xDb7 zB Da8-b7 3. d5xLc6 Dd8-a8 4. d4-d5 Lb7-c6 5. e3xSd4 Lc8-b7 6. f2xSe3 c6-c5 7. h5-h6 b7xLc6. Where was bK while all this was happening. There is only one spare tempo (h4-h5) If bK was off the back rank, there was not enough time to retract back there. So the only possibility is that Black was already castled, and can uncastle when the black cage is closed. The missing move can be R: 1. ... 0-0.
Since castling is illegal in the forward play, h#1 is unique with Donati theme as wQ revisits the promotion square:
1. c6 Db8# not
1. 0-0? Dxg7#
cheeky retro-active version of P0001136
A.Buchanan: Looking at the original, I am wondering why wBh cannot be on h5. It clearly cannot retract to h3, but how many times is it needed to retract if White retracts from the critical position 1Q3rk1/2pppppp/8/2p4P/2P5/PPRP4/RBpP1PP1/N1K5 (2022-02-07)
A.Buchanan: Maybe the distinguished composer felt that the more spare White moves has, the more creditable is the fact that Black had no way to avoid prior castling. The problem is still sound either way. Of course, with the h#1 version, wBh6 is mandatory. wD location was pretty much arbitrary in the original, but now there is only one wD location for soundness: g3. (e5 & g4 are unsound). (2022-02-07)
A.Buchanan: The solution given in Keym’s wonderful “Eigenartige Schachprobleme” uses an unnecessary move by wPh. (2022-02-23)
A.Buchanan: In offering this version, I do not think that pure retros are in anyway inferior to forward problems, which was perhaps the motivation for forward stipulations in the days of yore. Rather, the aesthetic tension between past and future never tires for me. It is not an issue to me whether the balance of interest in a retro-active problem be equally divided between the two directions. Rather, it’s essential that the position have two readings: that the pieces have two purposes. This formal requirement of double meaning which is ubiquitous and quotidian in a cryptic crossword should also find a place near to the hearth in chess problems. So Morse might welcome Dexter. (2022-02-23)
more ...
comment
A.Buchanan: Looking at the original, I am wondering why wBh cannot be on h5. It clearly cannot retract to h3, but how many times is it needed to retract if White retracts from the critical position 1Q3rk1/2pppppp/8/2p4P/2P5/PPRP4/RBpP1PP1/N1K5 (2022-02-07)
A.Buchanan: Maybe the distinguished composer felt that the more spare White moves has, the more creditable is the fact that Black had no way to avoid prior castling. The problem is still sound either way. Of course, with the h#1 version, wBh6 is mandatory. wD location was pretty much arbitrary in the original, but now there is only one wD location for soundness: g3. (e5 & g4 are unsound). (2022-02-07)
A.Buchanan: The solution given in Keym’s wonderful “Eigenartige Schachprobleme” uses an unnecessary move by wPh. (2022-02-23)
A.Buchanan: In offering this version, I do not think that pure retros are in anyway inferior to forward problems, which was perhaps the motivation for forward stipulations in the days of yore. Rather, the aesthetic tension between past and future never tires for me. It is not an issue to me whether the balance of interest in a retro-active problem be equally divided between the two directions. Rather, it’s essential that the position have two readings: that the pieces have two purposes. This formal requirement of double meaning which is ubiquitous and quotidian in a cryptic crossword should also find a place near to the hearth in chess problems. So Morse might welcome Dexter. (2022-02-23)
more ...
comment
Keywords: Cant Castler, Castling Paradox (hidden sK), Donati Theme (wD), Castling in the retro play (wk)
Genre: h#, Retro
Computer test: Substantial retro thinking to show castling illegal + trivial v4.87 for forward play
FEN: 4k2r/2pppppp/7P/2p5/2P5/PPRP2Q1/RBpP2P1/N1K5
Input: A.Buchanan, 2022-02-07
Last update: A.Buchanan, 2023-08-15 more...
Genre: h#, Retro
Computer test: Substantial retro thinking to show castling illegal + trivial v4.87 for forward play
FEN: 4k2r/2pppppp/7P/2p5/2P5/PPRP2Q1/RBpP2P1/N1K5
Input: A.Buchanan, 2022-02-07
Last update: A.Buchanan, 2023-08-15 more...
39 - P1399805
Giuseppe Brogi
Andrew Buchanan
Discord Chess Problems & Studies Server 15/03/2022
GB, correction AB
(8+14)
h#2
b) sSa1 -> wS
Giuseppe Brogi
Andrew Buchanan
Discord Chess Problems & Studies Server 15/03/2022
GB, correction AB
(8+14)
h#2
b) sSa1 -> wS
a) 1. T2xh3 Txh3 2. 0-0 Th8#
b) 1. La4 0-0 2. Tf8 Te1#
Colour of Sa1 is irrelevant to forward play, except for castling rights. Ignoring these retro concerns, there are two candidate solutions.
Suppose that both castling rights remain. Then sTh2 is original sTa8, and black b&c & white g&h pawn pairs have both cross-captured. However White has made 1 visible pawn capture, Black 6. In (a) this leaves 2+1 unaccounted for; in (b) 1+2. So at most one side can have cross-captured, so in each twin just one side can retain castling rights, allowing just one candidate to work in each.
b) 1. La4 0-0 2. Tf8 Te1#
Colour of Sa1 is irrelevant to forward play, except for castling rights. Ignoring these retro concerns, there are two candidate solutions.
Suppose that both castling rights remain. Then sTh2 is original sTa8, and black b&c & white g&h pawn pairs have both cross-captured. However White has made 1 visible pawn capture, Black 6. In (a) this leaves 2+1 unaccounted for; in (b) 1+2. So at most one side can have cross-captured, so in each twin just one side can retain castling rights, allowing just one candidate to work in each.
Keywords: Cant Castler (wksk), Castling (wksk), Cross-capture (s,w)
Genre: h#, Retro
FEN: 4k2r/p2p1Np1/1pp2p2/1b3p2/8/p4PPP/B4P1r/nn2K2R
Input: A.Buchanan, 2022-03-15
Last update: A.Buchanan, 2023-04-01 more...
Genre: h#, Retro
FEN: 4k2r/p2p1Np1/1pp2p2/1b3p2/8/p4PPP/B4P1r/nn2K2R
Input: A.Buchanan, 2022-03-15
Last update: A.Buchanan, 2023-04-01 more...
1. Txf3 0-0 2. Tf8 Txf8#
Thematischer Fehlversuch:
1. 0-0-0? Txh3 2. Td1 Th8+ 3. Td8!
Illegal:
1. Txf3 Txh3 2. Tf8 Te3#
Autor: Rochaden schließen sich gegenseitig aus. Die schwarze wird durch das Retrospiel f2-f3 Tf3xf7 ausgeschlossen, die weiße durch KxLe1. ... 'a posteriori'-Legalisierung durch Ausführung der Rochade.
Thematischer Fehlversuch:
1. 0-0-0? Txh3 2. Td1 Th8+ 3. Td8!
Illegal:
1. Txf3 Txh3 2. Tf8 Te3#
Autor: Rochaden schließen sich gegenseitig aus. Die schwarze wird durch das Retrospiel f2-f3 Tf3xf7 ausgeschlossen, die weiße durch KxLe1. ... 'a posteriori'-Legalisierung durch Ausführung der Rochade.
Henrik Juel: I do not know what Maximum exact means; neither do Popeye 4.61 and Märchenschachlexikon
Popeye 4.61 with 'con max' or 'con max ultra' both produced no solutions (2022-05-26)
Mario Richter: KW "en Passant as key" deleted, since I do not see any epkey here.
Classification after Branko Pavlovic:
1. Classical Maximummer:
Black has to make the geometrical longest move with the following restrictions:
a) Black attacks the white king according to the orthodox rules
b) If the absolutely longest moves are not executable, Black makes the relatively longest moves, i.e. one of the executable ones with maximum length.
2. Ultra-Maximummer:
As the classical Maximummer, but without a)
3. Exact-Maximummer:
Only the longest moves are "active". (2022-06-11)
A.Buchanan: Thanks Mario. I would prefer if this kind of distinction between checking, control & move was defined abstractly, independent of any particular fairy condition, to which it can then easily be applied. Also, prefixes "ultra & exact" are misleading (2023-06-29)
comment
Popeye 4.61 with 'con max' or 'con max ultra' both produced no solutions (2022-05-26)
Mario Richter: KW "en Passant as key" deleted, since I do not see any epkey here.
Classification after Branko Pavlovic:
1. Classical Maximummer:
Black has to make the geometrical longest move with the following restrictions:
a) Black attacks the white king according to the orthodox rules
b) If the absolutely longest moves are not executable, Black makes the relatively longest moves, i.e. one of the executable ones with maximum length.
2. Ultra-Maximummer:
As the classical Maximummer, but without a)
3. Exact-Maximummer:
Only the longest moves are "active". (2022-06-11)
A.Buchanan: Thanks Mario. I would prefer if this kind of distinction between checking, control & move was defined abstractly, independent of any particular fairy condition, to which it can then easily be applied. Also, prefixes "ultra & exact" are misleading (2023-06-29)
comment
Keywords: a posteriori (AP) (Type Petrovic), Maximummer (exact), Castling (wk)
Genre: h#, Retro, Fairies
FEN: r3k3/p4r2/8/8/8/p4P1p/4p1P1/4K2R
Input: A.Buchanan, 2022-05-26
Last update: Mario Richter, 2022-06-11 more...
Genre: h#, Retro, Fairies
FEN: r3k3/p4r2/8/8/8/p4P1p/4p1P1/4K2R
Input: A.Buchanan, 2022-05-26
Last update: Mario Richter, 2022-06-11 more...
41 - P1401711
Andrew Buchanan
OP008 The Hopper Magazine I01 24/12/2021
"Ak ja, retten, retten; hvad hjælper det, at du har retten, når du ikke har nogen magt?"
(3+3) C+
h#2.5 (2 solutions)
Position after Black's 5,696th move
Andrew Buchanan
OP008 The Hopper Magazine I01 24/12/2021
"Ak ja, retten, retten; hvad hjælper det, at du har retten, når du ikke har nogen magt?"
(3+3) C+
h#2.5 (2 solutions)
Position after Black's 5,696th move
1. ... Se6 2. Th2 Ta8 3. Th7 Txe8#
1. ... Txa2 2. Lg6 Tg2 3. Lh7 Sf7#
not 1. ... 0-0-0 2. Lg6 Tg1 3. Lh7 Sf7#?? (game just ended by 50M)
A full solution is really too big for PDB, but it is available at https://www.thehoppermagazine.com/AA084
1. ... Txa2 2. Lg6 Tg2 3. Lh7 Sf7#
not 1. ... 0-0-0 2. Lg6 Tg1 3. Lh7 Sf7#?? (game just ended by 50M)
A full solution is really too big for PDB, but it is available at https://www.thehoppermagazine.com/AA084
Henrik Juel: solutions
1...Sg5-e6 2.Ra2-h2 Ra1-a8 3.Rh2-h7 Ra8*e8 #
1...Ra1*a2 2.Be8-g6 Ra2-g2 3.Bg6-h7 Sg5-f7 #
not 1...0-0-0? 2.Be8-g6 Rd1-g1 3.Bg6-h7 Sg5-f7 #
because White has lost his right to castle, as Andrew will explain, I hope... (2023-07-29)
A.Buchanan: White has not necessarily lost the right to castle, but if he can castle then the 50M rule triggers before the mate can be executed. It is pretty complicated, sorry. (2023-07-29)
A.Buchanan: We don't do mottoes much these days. Mrs Baird was a big fan. On page 301 of the July 1916 Chess Amateur P.H. Williams wrote:
"I think it was Mrs Baird who did more to search Shakespeare for accidental (or deliberate) reference to chess, since all her retractors had Shakespearean mottoes, and her knowledge of the Avonian bard was obviously extensive. Her example was followed by other composers of retractors, who considered such positions would be incomplete without some reference to the poet."
But there are other great playwrights, and Asteroid 5696 was named after one such, whose bitter quotation accurately states this problem's theme. (2023-07-31)
Henrik Juel: The citation is by danish philosopher Søren Kierkegaard (1813-1855) and means something like
Oh yes, to be right, to be right; it does not help much that you are right, when you have no power. (2023-07-31)
Henrik Juel: Asteroid 5696 is named Ibsen after the norwegian playwright Henrik Ibsen (1828-1906), who must have borrowed the citation from Søren Kierkegaard (2023-07-31)
A.Buchanan: Thanks for this Henrik - I resolved the question here: https://philosophy.stackexchange.com/questions/101229/did-ibsen-originate-this-statement
Beyond the castling try, the other innovation here is in the solution beginning 1...Se6. Castling convention and 50M I think are trying to combine by Retro Strategy (being optimistic about 50M rights like we are with castling, rather than pessimistic like we are with ep). (2023-08-04)
more ...
comment
1...Sg5-e6 2.Ra2-h2 Ra1-a8 3.Rh2-h7 Ra8*e8 #
1...Ra1*a2 2.Be8-g6 Ra2-g2 3.Bg6-h7 Sg5-f7 #
not 1...0-0-0? 2.Be8-g6 Rd1-g1 3.Bg6-h7 Sg5-f7 #
because White has lost his right to castle, as Andrew will explain, I hope... (2023-07-29)
A.Buchanan: White has not necessarily lost the right to castle, but if he can castle then the 50M rule triggers before the mate can be executed. It is pretty complicated, sorry. (2023-07-29)
A.Buchanan: We don't do mottoes much these days. Mrs Baird was a big fan. On page 301 of the July 1916 Chess Amateur P.H. Williams wrote:
"I think it was Mrs Baird who did more to search Shakespeare for accidental (or deliberate) reference to chess, since all her retractors had Shakespearean mottoes, and her knowledge of the Avonian bard was obviously extensive. Her example was followed by other composers of retractors, who considered such positions would be incomplete without some reference to the poet."
But there are other great playwrights, and Asteroid 5696 was named after one such, whose bitter quotation accurately states this problem's theme. (2023-07-31)
Henrik Juel: The citation is by danish philosopher Søren Kierkegaard (1813-1855) and means something like
Oh yes, to be right, to be right; it does not help much that you are right, when you have no power. (2023-07-31)
Henrik Juel: Asteroid 5696 is named Ibsen after the norwegian playwright Henrik Ibsen (1828-1906), who must have borrowed the citation from Søren Kierkegaard (2023-07-31)
A.Buchanan: Thanks for this Henrik - I resolved the question here: https://philosophy.stackexchange.com/questions/101229/did-ibsen-originate-this-statement
Beyond the castling try, the other innovation here is in the solution beginning 1...Se6. Castling convention and 50M I think are trying to combine by Retro Strategy (being optimistic about 50M rights like we are with castling, rather than pessimistic like we are with ep). (2023-08-04)
more ...
comment
Keywords: Aristocrat, Miniature, 50 move rule, Castling, Exchange of roles (T/S, Guard/Mate), Chumakov theme (l/t, simplified), Retro Strategy (RS), Model mate (2), Constrained problem
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 and analysis
FEN: 4b2k/8/8/6N1/8/8/r7/R3K3
Reprints: AA084 The Hopper Magazine I04 13/07/2023
Input: A.Buchanan, 2022-06-09
Last update: A.Buchanan, 2023-08-27 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 and analysis
FEN: 4b2k/8/8/6N1/8/8/r7/R3K3
Reprints: AA084 The Hopper Magazine I04 13/07/2023
Input: A.Buchanan, 2022-06-09
Last update: A.Buchanan, 2023-08-27 more...
42 - P1407858
Manfred Nieroba
19040 Die Schwalbe 316, p. 622, 08/2022
(3+6)
h#2
b) sSa2 statt sBa2
c) sLa4->a7
d) sKh1->h2
Manfred Nieroba
19040 Die Schwalbe 316, p. 622, 08/2022
(3+6)
h#2
b) sSa2 statt sBa2
c) sLa4->a7
d) sKh1->h2
a) 1. Tg2 0-0-0 2. Th2 Lh3#
b) 1. Sc1 Ta3 2. b2 Th3#
c) 1. Lb8 Kf2 2. Lh2 Lg2#
d) 1. Th1 Txa2+ 2. Kg1 Tg2#
19040 (Manfred Nieroba). a) 1.Tg2 0-0-0 2.Th2 Lh3# (MM), b) 1.Sc1 Ta3 2.b2 Th3# (MM), c) 1.Lb8 Kg2 2.Lh2 Lg2#, d) 1.Th1 T:a2+ 2.Kg1 Tg2#. „Zweimal h3, zweimal g2 als Mattfelder – ein locker leichtes Unterhaltungsstückchen.“ (VZ) Ähnlich JS: „Jeweils Matt auf h3 und g2 mit Läufer und Turm, locker hingestellt. Gefällt mir gut!“ „Mattbildspielerei mit zwei aus vier Mustermatts ohne einheitliche Strategie, wobei Turm und Läufer wechselweise in zwei Variantenpaaren auf g2 und h3 mattsetzen. War nicht so mein Ding . . . “ (FRd) Insgesamt waren aber die Löser zufrieden: „Zweimal tauschen der weiße Turm und der weiße Läufer ihre Rolle als Deckungs- und Mattstein, wobei der Mattzug jeweils auf dasselbe Feld (h3 bzw. g2) erfolgt – sehr schön!“ (HJä) „Erstaunlich, was ein erfahrener Hilfsmatt-Komponist aus einer doch eher einfachen Stellung an Thematik herausholt: Zwei Mustermatts, Boros-Thema.“ (KHS) „Die maskierte Halbbatterie auf der Grundreihe wurde in einen Vierling umgemünzt, in dessen erster Phase ihr Feuer durch die lange Rochade eingeleitet wird. Die anderen Phasen fügen unter anderem einen Boros und einen Blockwechsel hinzu. In dieser Gewichtsklasse kann man damit zufrieden sein.“ (MRit)
b) 1. Sc1 Ta3 2. b2 Th3#
c) 1. Lb8 Kf2 2. Lh2 Lg2#
d) 1. Th1 Txa2+ 2. Kg1 Tg2#
19040 (Manfred Nieroba). a) 1.Tg2 0-0-0 2.Th2 Lh3# (MM), b) 1.Sc1 Ta3 2.b2 Th3# (MM), c) 1.Lb8 Kg2 2.Lh2 Lg2#, d) 1.Th1 T:a2+ 2.Kg1 Tg2#. „Zweimal h3, zweimal g2 als Mattfelder – ein locker leichtes Unterhaltungsstückchen.“ (VZ) Ähnlich JS: „Jeweils Matt auf h3 und g2 mit Läufer und Turm, locker hingestellt. Gefällt mir gut!“ „Mattbildspielerei mit zwei aus vier Mustermatts ohne einheitliche Strategie, wobei Turm und Läufer wechselweise in zwei Variantenpaaren auf g2 und h3 mattsetzen. War nicht so mein Ding . . . “ (FRd) Insgesamt waren aber die Löser zufrieden: „Zweimal tauschen der weiße Turm und der weiße Läufer ihre Rolle als Deckungs- und Mattstein, wobei der Mattzug jeweils auf dasselbe Feld (h3 bzw. g2) erfolgt – sehr schön!“ (HJä) „Erstaunlich, was ein erfahrener Hilfsmatt-Komponist aus einer doch eher einfachen Stellung an Thematik herausholt: Zwei Mustermatts, Boros-Thema.“ (KHS) „Die maskierte Halbbatterie auf der Grundreihe wurde in einen Vierling umgemünzt, in dessen erster Phase ihr Feuer durch die lange Rochade eingeleitet wird. Die anderen Phasen fügen unter anderem einen Boros und einen Blockwechsel hinzu. In dieser Gewichtsklasse kann man damit zufrieden sein.“ (MRit)
43 - P1408618
Udo Degener
Mirko Degenkolbe
9376 Phénix 335 10/2022
(9+7) C+
h#5
0.1.1...
Nullstellung
a) +wSe3
b) +wLc1
c) +wTa1
Udo Degener
Mirko Degenkolbe
9376 Phénix 335 10/2022
(9+7) C+
h#5
0.1.1...
Nullstellung
a) +wSe3
b) +wLc1
c) +wTa1
a) 1. ... Sf5 2. h5 e3 3. h4 Sxh4 4. Kg8 Sxg6 5. Th7 Lc4#
b) 1. ... e3 2. Th5 Lc4 3. Tf5 exf5 4. e4 fxg6 5. Lxb2 Lxb2#
c) 1. ... 0-0-0 2. Txf1 d4 3. Tf7 dxe5 4. Lf6 exf6 5. Th7 Td8#
b) 1. ... e3 2. Th5 Lc4 3. Tf5 exf5 4. e4 fxg6 5. Lxb2 Lxb2#
c) 1. ... 0-0-0 2. Txf1 d4 3. Tf7 dxe5 4. Lf6 exf6 5. Th7 Td8#
Keywords: Model mate, Castling (wl in c), Line opening (in a), Selfblock (in a+c), Model mate, Unpinning, Corner mate (in b+c), Line opening
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 7k/6b1/6pp/4p3/4P3/6p1/PPPPP1P1/4KB1r
Input: Felber, Volker, 2023-03-31
Last update: Gunter Jordan, 2023-04-02 more...
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 7k/6b1/6pp/4p3/4P3/6p1/PPPPP1P1/4KB1r
Input: Felber, Volker, 2023-03-31
Last update: Gunter Jordan, 2023-04-02 more...
1. ... e6 2. 0-0? Lxh7# (castling rights lost)
1. ... Txh7 2. Tf8 Te7#
1. cxb3ep Txh7 2. Tf8? Te7# (ep needs AP justification)
1. cxb3ep e6 2. 0-0! Lxh7#
White pawns have captured 9 times right-to-left, accounting for all but one missing Black unit. So by parity, there were no other pawn captures by White. So wBb never left that file, and Black cannot have just played Bb3xa2 behind it. sBa was waylaid on a-file by an officer.
So if Black moved last, it must have been Ke8 or Th8, and Black has lost castling rights.
On the other hand, if White moved last, then Black needs a tempo move. The only possible one is ep capture. (1. Tg8+? is check.) But the ep capture is only legal if Black retains castling rights, so Black's second move must indeed be 0-0 for A Posteriori justification.
In this case, last moves were R: 1. b2-b4 b3xDa2 & e.g. 2. Le3-d4,~ b4-b3 3. d4xSc5,~ Sa6-c5,~ with many ways for White to release Black.
1. ... Txh7 2. Tf8 Te7#
1. cxb3ep Txh7 2. Tf8? Te7# (ep needs AP justification)
1. cxb3ep e6 2. 0-0! Lxh7#
White pawns have captured 9 times right-to-left, accounting for all but one missing Black unit. So by parity, there were no other pawn captures by White. So wBb never left that file, and Black cannot have just played Bb3xa2 behind it. sBa was waylaid on a-file by an officer.
So if Black moved last, it must have been Ke8 or Th8, and Black has lost castling rights.
On the other hand, if White moved last, then Black needs a tempo move. The only possible one is ep capture. (1. Tg8+? is check.) But the ep capture is only legal if Black retains castling rights, so Black's second move must indeed be 0-0 for A Posteriori justification.
In this case, last moves were R: 1. b2-b4 b3xDa2 & e.g. 2. Le3-d4,~ b4-b3 3. d4xSc5,~ Sa6-c5,~ with many ways for White to release Black.
Keywords: a posteriori (AP), RIFACE Retro Solving Tourney (2022), En passant as key, Castling (sk), Tempo Move, waylaid (sBa)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro thinking
FEN: 4k2r/1N1p3p/3P4/1PPPP3/1PpBBP2/P1N3K1/p7/R6R
Input: A.Buchanan, 2023-05-20
Last update: A.Buchanan, 2023-09-11 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro thinking
FEN: 4k2r/1N1p3p/3P4/1PPPP3/1PpBBP2/P1N3K1/p7/R6R
Input: A.Buchanan, 2023-05-20
Last update: A.Buchanan, 2023-09-11 more...
45 - P1413924
Johannes Jacob Burbach
Alain C. White
Andrew Buchanan
PDB Website 05/12/2023
JJB & ACW, version AB
(4+6) C+
h#2
b) -sBg6h7
Johannes Jacob Burbach
Alain C. White
Andrew Buchanan
PDB Website 05/12/2023
JJB & ACW, version AB
(4+6) C+
h#2
b) -sBg6h7
a) 1. Ta5 Kd1 2. Ta2 Kc2# (not 1.Ta7?etc Lxc3 because White may not castle)
b) 1. Ta7 Lxc3 2. Ta2 0-0# (last move could be 0.h7-h8=L)
b) 1. Ta7 Lxc3 2. Ta2 0-0# (last move could be 0.h7-h8=L)
See P0003666
A.Buchanan: Apart from having removed wBe2/sBe3, I don't think this is an improvement. In the original, sBh6 gave and took. The version is more economical but there's two pawns difference between the twins. (2023-12-06)
more ...
comment
A.Buchanan: Apart from having removed wBe2/sBe3, I don't think this is an improvement. In the original, sBh6 gave and took. The version is more economical but there's two pawns difference between the twins. (2023-12-06)
more ...
comment
Keywords: Cant Castler, Castling (wk), Remove pieces
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 + trivial retro logic
FEN: 7B/6rp/6p1/7r/8/2p5/1P6/k3K2R
Input: A.Buchanan, 2023-12-05
Last update: A.Buchanan, 2023-12-05 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 + trivial retro logic
FEN: 7B/6rp/6p1/7r/8/2p5/1P6/k3K2R
Input: A.Buchanan, 2023-12-05
Last update: A.Buchanan, 2023-12-05 more...
1) 1. Sd2 Txh2 2. Lf1 Tg2 3. Dh2 Txf2+ 4. Kg3 Txe2 5. Sf3+ Kxf1 6. Dg2+ Txg2#
2) 1. Txh1 a4 2. Lh2 a5 3. Sg1 a6 4. Dh3 a7 5. Kg3 a8=D 6. f3 Db8#
2) 1. Txh1 a4 2. Lh2 a5 3. Sg1 a6 4. Dh3 a7 5. Kg3 a8=D 6. f3 Db8#
Yuri Bilokin: you can see 1...Rxg1 2.Bh1 Rxf1 3.Kg2 Rxf2+ 4.Kg1 Kxe2 5.Rg2 Ke1 6.Qh2 Rf1#
1...Rxh2 2.Sd2 Rxg2 3.Qh2 Rxf2+ 4.Kg3 Rxe2 5.Sf3+ Kf1 6.Qg2+ Rxg2#
1...Rxh2 2.Sd2 Rxg2 3.Qh2 Rxf2+ 4.Kg3 Kxe2 5.Sf3 Kf1 6.Qg2+ Rxg2# (2024-01-30)
comment
1...Rxh2 2.Sd2 Rxg2 3.Qh2 Rxf2+ 4.Kg3 Rxe2 5.Sf3+ Kf1 6.Qg2+ Rxg2#
1...Rxh2 2.Sd2 Rxg2 3.Qh2 Rxf2+ 4.Kg3 Kxe2 5.Sf3 Kf1 6.Qg2+ Rxg2# (2024-01-30)
comment
Keywords: Excelsior white (in 2), Castling position, Selfblock, Umnov, Promotion (D in 2)
Genre: h#
FEN: 8/8/8/6p1/4pppp/P3pkqn/4prbr/4KnbR
Input: Gunter Jordan, 2024-01-20
Last update: Gunter Jordan, 2024-01-20 more...
Genre: h#
FEN: 8/8/8/6p1/4pppp/P3pkqn/4prbr/4KnbR
Input: Gunter Jordan, 2024-01-20
Last update: Gunter Jordan, 2024-01-20 more...
47 - P1415606
Maurice Jago
Andrew Buchanan
PDB Website 07/03/2024
MJ, correction AB
corrects 8202v Die Schwalbe, p.148, 10/1980
(8+15)
h#2
b) sBb4->b5
Maurice Jago
Andrew Buchanan
PDB Website 07/03/2024
MJ, correction AB
corrects 8202v Die Schwalbe, p.148, 10/1980
(8+15)
h#2
b) sBb4->b5
Supersedes P0003659.
"The noted Cornish problemist, Dr. Maurice Jago was most prolific during the war when he was a lieutenant in the Royal Army Medical Corps. He became increasingly interested in the more flamboyant and unusual kinds of positions and problems." (www.keverellchess.com, now closed)
Yuri Bilokin: perhaps such a version does not harm the author's intention –wNa7, -wNa8, -wBb8, -wPd3, -bRc8, -bPc7, -bPe7 8/8/1Pqp4/2bb4/1p6/n1p2rP1/1PP1P2k/R3KQ1n (8+12) h#2 b) bPb4-b5 (2024-03-10)
Yuri Bilokin: slip of the pen -wNa7, -wNa8, -wBb8, -wPd3, -bRc8, -bPc7, -bPe7, -bPa4, -bPg6 8/8/1Pqp4/2bb4/1p6/n1p2rP1/1PP1P2k/R3KQ1n (8+10) h#2 b) bPb4-b5 (2024-03-10)
A.Buchanan: Hi Yuri thanks yes it’s more than intention here. In b need to ensure that White can’t have just played axb6, fxg3 or g2-g3, in order to eliminate the castling try: the extra pieces are all to ensure this (2024-03-11)
Yuri Bilokin: Hi Andrew. I apologize, I didn't go there, although the white horse on the a8 square still bothers me. (2024-03-11)
A.Buchanan: WP cap: axb BP caps: fxexdxc. bPh must have promoted, so either bPh3xg2-g1= or wPh was waylaid. In any case, all captures are accounted for. wPf promoted. Easy to see that White cannot have just played g2-g3 or f2xg3 or d2-d3 or axb6. So must have bPb4 rather than b5, allowing for White to have just played Sb5-a7 or b5-b6. If any of the "retro-dressing" White pieces e.g. Sa8 are removed, then White might have just played g2-g3 or d2-d3. At least the retro-dressing units have no possible last move (& wSa8 wBb8 have not just promoted), so there is some craft to their deployment. (2024-03-11)
A.Buchanan: bPg6 blocks a cook so must stay there, but can shift bPe7 to d7 (allowing the removal of wSa8 or wBb8) or to b3/b7 (allowing the removal of all three White officers: wSa7a8 & wBb8.) Maybe that's worth doing? (2024-03-11)
Yuri Bilokin: Thank you for the detailed analysis. Perhaps my poor view will allow you to get an optimal starting position. (2024-03-11)
Henrik Juel: a) 1.Sh1-f2 Qf1*f2 + 2.Kh2-h1 0-0-0 #
b 1.Sa3-b1 Ra1*a4 2.Sh1*g3 Ra4-h4 # (2024-03-11)
Henrik Juel: C+ Popeye 4.61 and analysis (2024-03-11)
A.Buchanan: I had wanted to keep to the spirit of P0003659, in which may be the author wanted to have a bunch of White retro-dressing that could not retract. But going back to the first version P0003659, the White position is much cleaner. So following Yuri being bothered by a horse, I am going to change the diagram here from NBr5/N1p1p3/1Pqp2p1/2bb4/pp6/n1pP1rP1/1PP1P2k/R3KQ1n by removing 4 white units. Hope this is agreeable to all (2024-03-13)
more ...
comment
"The noted Cornish problemist, Dr. Maurice Jago was most prolific during the war when he was a lieutenant in the Royal Army Medical Corps. He became increasingly interested in the more flamboyant and unusual kinds of positions and problems." (www.keverellchess.com, now closed)
Yuri Bilokin: perhaps such a version does not harm the author's intention –wNa7, -wNa8, -wBb8, -wPd3, -bRc8, -bPc7, -bPe7 8/8/1Pqp4/2bb4/1p6/n1p2rP1/1PP1P2k/R3KQ1n (8+12) h#2 b) bPb4-b5 (2024-03-10)
Yuri Bilokin: slip of the pen -wNa7, -wNa8, -wBb8, -wPd3, -bRc8, -bPc7, -bPe7, -bPa4, -bPg6 8/8/1Pqp4/2bb4/1p6/n1p2rP1/1PP1P2k/R3KQ1n (8+10) h#2 b) bPb4-b5 (2024-03-10)
A.Buchanan: Hi Yuri thanks yes it’s more than intention here. In b need to ensure that White can’t have just played axb6, fxg3 or g2-g3, in order to eliminate the castling try: the extra pieces are all to ensure this (2024-03-11)
Yuri Bilokin: Hi Andrew. I apologize, I didn't go there, although the white horse on the a8 square still bothers me. (2024-03-11)
A.Buchanan: WP cap: axb BP caps: fxexdxc. bPh must have promoted, so either bPh3xg2-g1= or wPh was waylaid. In any case, all captures are accounted for. wPf promoted. Easy to see that White cannot have just played g2-g3 or f2xg3 or d2-d3 or axb6. So must have bPb4 rather than b5, allowing for White to have just played Sb5-a7 or b5-b6. If any of the "retro-dressing" White pieces e.g. Sa8 are removed, then White might have just played g2-g3 or d2-d3. At least the retro-dressing units have no possible last move (& wSa8 wBb8 have not just promoted), so there is some craft to their deployment. (2024-03-11)
A.Buchanan: bPg6 blocks a cook so must stay there, but can shift bPe7 to d7 (allowing the removal of wSa8 or wBb8) or to b3/b7 (allowing the removal of all three White officers: wSa7a8 & wBb8.) Maybe that's worth doing? (2024-03-11)
Yuri Bilokin: Thank you for the detailed analysis. Perhaps my poor view will allow you to get an optimal starting position. (2024-03-11)
Henrik Juel: a) 1.Sh1-f2 Qf1*f2 + 2.Kh2-h1 0-0-0 #
b 1.Sa3-b1 Ra1*a4 2.Sh1*g3 Ra4-h4 # (2024-03-11)
Henrik Juel: C+ Popeye 4.61 and analysis (2024-03-11)
A.Buchanan: I had wanted to keep to the spirit of P0003659, in which may be the author wanted to have a bunch of White retro-dressing that could not retract. But going back to the first version P0003659, the White position is much cleaner. So following Yuri being bothered by a horse, I am going to change the diagram here from NBr5/N1p1p3/1Pqp2p1/2bb4/pp6/n1pP1rP1/1PP1P2k/R3KQ1n by removing 4 white units. Hope this is agreeable to all (2024-03-13)
more ...
comment
Show statistic for complete result. Show search result faster by using ids.
https://pdb.dieschwalbe.de/search.jsp?expression=COMMENTDATE%3E%3D20220810+AND+G%3D%27h%23%27+AND+NOT+S%3D%27Uralski+Problemist%27+AND+K%3D%27Rochade%27
The problems of this query have been registered by the following contributors:
Gerd Wilts (19)hpr (3)
HBae (1)
Markus Manhart (3)
Andreas Mokosch (5)
Michal Dragoun (1)
Klaus Funk (2)
Felber, Volker (3)
Mario Richter (2)
A.Buchanan (7)
Gunter Jordan (1)
Preisbericht: 'Die Schwalbe' 06/2011 S.124 (2023-01-02)
Henrik Juel: How is the SE corner released, without ruining the castling? (2023-01-02)
Mario Richter: Good question, Henrik! I first thought that releasing the SE corner without ruining White's castling right is impossible, but the trick is to uncapture a black Queen in the SE corner at the right moment.
Perhaps Theodore Hwa can use ths problem as a test case for his latest improvement to Retractor 2 ... (2023-01-02)
Henrik Juel: Thanks, Mario
In view of the prize I suspected that the problem was correct, but I did not find the uncapture trick (2023-01-02)
Henrik Juel: C+ Popeye 4.61, because with Black to move White may not castle (2023-01-02)
comment