62 problem(s) found in 4837 milliseconds (displaying 62 problem(s)). [COMMENTDATE>=20220810 AND G='h#' AND NOT A='Hage, Poul' AND G='Retro'] [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)
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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)
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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...
R: 1. ... Kg8xBf7,(If5-e4) Kb4xSb3,(If6-f5), dann 1. Ka4,(Ie6) Sc5#,(If8)
Cook: NL
R: 1. Ke7xLf7,xSf7,xBf7,(Id4-e4) d6xDe5,(Ic5-d4), dann 1. Kc2,(Id4) Db2#,(Ia1)
R: 1. Kg6xTf7,Ke6xTf7,(If3-e4,Id3-e4) d6xDe5,(Ie4,Ic4), dann 1. Te7,Tg7,(Id4) Db2#,(Ia1)
Cook: NL
R: 1. Ke7xLf7,xSf7,xBf7,(Id4-e4) d6xDe5,(Ic5-d4), dann 1. Kc2,(Id4) Db2#,(Ia1)
R: 1. Kg6xTf7,Ke6xTf7,(If3-e4,Id3-e4) d6xDe5,(Ie4,Ic4), dann 1. Te7,Tg7,(Id4) Db2#,(Ia1)
Anton Baumann: 'Die Schwalbe' 08/1984 S.308: der Autor korrigiert: wBe5 statt sBe5 (2022-12-23)
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Keywords: Help retractor, Kindergarten Problem
Pieces: = Imitator (I)
Genre: Retro, Fairies, h#
FEN: 8/5K2/8/4p3/4-I3/1k6/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-12-13 more...
Pieces: = Imitator (I)
Genre: Retro, Fairies, h#
FEN: 8/5K2/8/4p3/4-I3/1k6/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-12-13 more...
Kees: possible fix: Lb1=Sb1 -De1 +Ld1
White begins: 1.Kxb7 Lxe2 2.Kc8 La6#
(1.Txd8+ Kxd8 2.Kf8 Th8# illegal for white has no last move) (2023-06-07)
A.Buchanan: Your fix is good, Kees. It removes the cook, and sLd1 denies R: 1. c2xb3 as well as sLb1 did. Note R: 1 Sc6-d8 0-0+? as black castling rights were lost to let wK enter the back rank. (2023-06-08)
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White begins: 1.Kxb7 Lxe2 2.Kc8 La6#
(1.Txd8+ Kxd8 2.Kf8 Th8# illegal for white has no last move) (2023-06-07)
A.Buchanan: Your fix is good, Kees. It removes the cook, and sLd1 denies R: 1. c2xb3 as well as sLb1 did. Note R: 1 Sc6-d8 0-0+? as black castling rights were lost to let wK enter the back rank. (2023-06-08)
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Genre: h#, Retro
FEN: 1bKN1rk1/1ppn1r1R/5p1P/4pP1p/3p1p2/1PP3P1/PP2P3/Rb2q3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-08 more...
1. ... Kgxf4 2. Tf6 e6 3. Tf8 Sg7#
Cook: 1. ... exf6ep 2. 0-0-0 gxf4 3. Td7 a8=D
Failed try, as Bg4xBh5 allows bRh8 to have escaped the cage prior to ep.
Cook: 1. ... exf6ep 2. 0-0-0 gxf4 3. Td7 a8=D
Failed try, as Bg4xBh5 allows bRh8 to have escaped the cage prior to ep.
Anton Baumann: Sollte eine Verbesserung von P0000777 sein.
Autorabsicht: "letzter Zug f7-f5 sonst retropatt! Jetzt ist aber die Rochade illegal, da der sTh8 nur über e8 aus seiner Ecke kommt. Daher Verführung 1. ... exf6 e.p. 2.0-0-0? Lxf4 3.Td7 a8=D#. Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 sg7#"
Aber: Der s h-Bauer kann auch auf h5 geschlagen worden sein; so konnte der sT über h6 aus seiner Ecke entkommen. Die vermeintliche Verführung ist also NL! (2022-12-14)
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Autorabsicht: "letzter Zug f7-f5 sonst retropatt! Jetzt ist aber die Rochade illegal, da der sTh8 nur über e8 aus seiner Ecke kommt. Daher Verführung 1. ... exf6 e.p. 2.0-0-0? Lxf4 3.Td7 a8=D#. Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 sg7#"
Aber: Der s h-Bauer kann auch auf h5 geschlagen worden sein; so konnte der sT über h6 aus seiner Ecke entkommen. Die vermeintliche Verführung ist also NL! (2022-12-14)
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Keywords: Castling (sg), Superseded by (P0000058)
Genre: h#, Retro
FEN: r3k3/P7/p1r3pP/4PpBN/3ppnKn/6P1/1PP2PPq/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
Genre: h#, Retro
FEN: r3k3/P7/p1r3pP/4PpBN/3ppnKn/6P1/1PP2PPq/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
1. ... exf6ep 2. 0-0-0? Lxf4 3. Td7 a8=D# try
1. ... Kxf4 2. Tf6 e6 3. Tf8 Sg7# solution
Cook: 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
1. ... Kxf4 2. Tf6 e6 3. Tf8 Sg7# solution
Cook: 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
See P0000674
Henrik Juel: The stipulation should probably read h#2.5: 1.exf6ep 0-0-0 2.Bxf4 Rd7 3.a8Q# (2003-08-18)
Anton Baumann: Letzter Zug war f7-f5, sonst retropatt. Die weissen Bauern haben 6x geschlagen. Der s a-Bauer wurde so ebenfalls als Schlagopfer benötigt und musste 2x schlagen, um sich umwandeln zu können. Weiss hat also auf h6 den s Bauern geschlagen. Unabhängig davon ob der sTe6 ein Umwandlungsturm ist oder nicht, muss der sTh8 über e8 aus der Ecke gekommen sein. Also ist die Rochade illegal!
Autorabsicht: 1. ... exf6 e.p. 2.0-0-0 Lxf4 3.Td7 a8=D# ist die Verführung!
Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 Sg7#
leider geht auch: 1. ... Kxf4 2.Te7 e6 3.Tb8 axb8=D,T# (2022-12-13)
A.Buchanan: I agree Anton. I suggest as a fix -bBa6 +bDc2. I think retro, solution and try all work correctly now. Do you folks agree? (2022-12-13)
Anton Baumann: Der Vorschlag von Andrew ergibt eine korrekte Lösung, aber nicht im Sinne des Autors: ohne einen s Stein auf a6 ist f6-f5+ als letzter Zug möglich, d.h. kein Retropatt weil Weiss a6-a7 zurücknehmen kann! 1. ... exf6 e.p. ist deshalb nicht zulässig. LB wollte aber dass f7-f5 der letzte Zug war, damit aber die Rochade illegal ist! (2022-12-14)
Anton Baumann: Mögliche Korrektur: a6=sL, g2=sD (2022-12-15)
A.Buchanan: Yes I agree. I had overlooked the unmove of wPa7 (2022-12-15)
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Henrik Juel: The stipulation should probably read h#2.5: 1.exf6ep 0-0-0 2.Bxf4 Rd7 3.a8Q# (2003-08-18)
Anton Baumann: Letzter Zug war f7-f5, sonst retropatt. Die weissen Bauern haben 6x geschlagen. Der s a-Bauer wurde so ebenfalls als Schlagopfer benötigt und musste 2x schlagen, um sich umwandeln zu können. Weiss hat also auf h6 den s Bauern geschlagen. Unabhängig davon ob der sTe6 ein Umwandlungsturm ist oder nicht, muss der sTh8 über e8 aus der Ecke gekommen sein. Also ist die Rochade illegal!
Autorabsicht: 1. ... exf6 e.p. 2.0-0-0 Lxf4 3.Td7 a8=D# ist die Verführung!
Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 Sg7#
leider geht auch: 1. ... Kxf4 2.Te7 e6 3.Tb8 axb8=D,T# (2022-12-13)
A.Buchanan: I agree Anton. I suggest as a fix -bBa6 +bDc2. I think retro, solution and try all work correctly now. Do you folks agree? (2022-12-13)
Anton Baumann: Der Vorschlag von Andrew ergibt eine korrekte Lösung, aber nicht im Sinne des Autors: ohne einen s Stein auf a6 ist f6-f5+ als letzter Zug möglich, d.h. kein Retropatt weil Weiss a6-a7 zurücknehmen kann! 1. ... exf6 e.p. ist deshalb nicht zulässig. LB wollte aber dass f7-f5 der letzte Zug war, damit aber die Rochade illegal ist! (2022-12-14)
Anton Baumann: Mögliche Korrektur: a6=sL, g2=sD (2022-12-15)
A.Buchanan: Yes I agree. I had overlooked the unmove of wPa7 (2022-12-15)
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Keywords: Castling (sg), Valladao Task, Superseded by (P0000058)
Genre: h#, Retro
FEN: r3k3/P7/p3r1pP/4PpBN/3nnpKP/6PB/1P2PPb1/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
Genre: h#, Retro
FEN: r3k3/P7/p3r1pP/4PpBN/3nnpKP/6PB/1P2PPb1/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
a) 1. T2xh3 Txh3 2. 0-0 Th8#
b) 1. La4 0-0 2. Tf8 Te1#
a) White cannot cross-capture, so because of sTh2, w0-0 impossible.
wBe captured sD, either to be captured on f-file or to promote without disrupting sKe8.
b) sTh2 can return to a8 only via e8 as Black cannot cross-capture, so s0-0 is impossible.
wBe was waylaid or promoted after disrupting sKe8.
Cook: 1. T2xh3 Lb4 2. Txg3 Txh8#
1. Sd2 Lf6 2. Tf8 Sd6#
(cooked by MR)
b) 1. La4 0-0 2. Tf8 Te1#
a) White cannot cross-capture, so because of sTh2, w0-0 impossible.
wBe captured sD, either to be captured on f-file or to promote without disrupting sKe8.
b) sTh2 can return to a8 only via e8 as Black cannot cross-capture, so s0-0 is impossible.
wBe was waylaid or promoted after disrupting sKe8.
Cook: 1. T2xh3 Lb4 2. Txg3 Txh8#
1. Sd2 Lf6 2. Tf8 Sd6#
(cooked by MR)
See P0003736 a companion problem.
milan: [-bPa7b6wPg3bSa1+bPg3bBa5-b4 bPc6-d6] h#2 1.2...bRh2? illegal move? M.Frelih (2023-03-30)
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milan: [-bPa7b6wPg3bSa1+bPg3bBa5-b4 bPc6-d6] h#2 1.2...bRh2? illegal move? M.Frelih (2023-03-30)
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Keywords: Cant Castler, Castling (wksk), Cross-capture (s,w), Superseded by (P1399805)
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: B3k2r/pN1p1p2/1pp3p1/bb3p2/8/p1B3PP/5P1r/nn2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-16 more...
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: B3k2r/pN1p1p2/1pp3p1/bb3p2/8/p1B3PP/5P1r/nn2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-16 more...
* 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)
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C+ Popeye 4.61 (2022-11-26)
A.Buchanan: Pleasant White tempo play in both phases (2022-11-27)
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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)
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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)
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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)
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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)
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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. exd3ep Lxg4 2. f3 Le6#
Klären: Quelle = Schachmatt? - Felber, Volker: SCHACH ist korrekt, 6/1969, Seite 191 (2010-10-09)
Eliminierung von 4 Steinen zwischen sK und wT im h#2 (mit Hilfe von ep-Schlag)!
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Eliminierung von 4 Steinen zwischen sK und wT im h#2 (mit Hilfe von ep-Schlag)!
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Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 8/8/8/1pn4b/2kPpppR/2q1rB1p/1p4r1/3K4
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-01-09 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 8/8/8/1pn4b/2kPpppR/2q1rB1p/1p4r1/3K4
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-01-09 more...
1. exf3ep Lc2#
R: 1. f2-f4 f5xTe4 2. Tg4-e4 & e.g. f6-f5 3. Tg1-g4 f7-f6 4. Le4-h7 h5-h4 5. Lg2-e4 h6-h5 6. Lf1-g2 h7-h6 7. g2xh3
1.exf3e.p. Bc2# is the only possible solution, but this necessitates R: 1.f2-f4. Can we prove this?
(13+13) with 1+2 pawn captures. Bf8 captured at home, so to satisfy White appetite, the missing Black pawn (a or b) must have promoted via c2 on c1. Two more White units must be captured to allow this.
The kings cage can only be unlocked by retracting WPc2. But the clock is ticking as there are only 6 black moves which can be retracted.
The promoted piece was captured on e3 or h3. If either capture is undone, then a White bishop square is cut off, so WB must be replaced prior to this.
Now the order of the early moves is: WdP moves, WQB & WQR escape, BP promotes on c1 to X (capturing WR at some point), X captured by WP.
So the first White capture must be dxNe3 and the second White capture releases gxXh3. The second White capture releases WKB & WKR. WKR captured by original BfP.
The clock starts ticking with gxh3. Black has 6 pawn moves. WKB has 3 moves to reach h7. WR has 3 if it goes via d file, or 2 if it starts on g1 (in which case WfP or WQB must also move once). So certainly at least 6 White moves. Last move was therefore White (even if the stipulation didn't tell us), and it can only have been WfP coming from f3 or f4. If it had been coming from f3 it would have blocked WKB in its progress, so the last White move was indeed R: 1.f2-f4.
WKR did therefore move from g1-g4-e4, and R: 1. ... fxRe5 2. Rg4-e4. Prior to that, move order not unique, but counting still exact.
Note that WN loitering on b4, pretending to be part of the cage, is present on the board just to make up the numbers.
R: 1. f2-f4 f5xTe4 2. Tg4-e4 & e.g. f6-f5 3. Tg1-g4 f7-f6 4. Le4-h7 h5-h4 5. Lg2-e4 h6-h5 6. Lf1-g2 h7-h6 7. g2xh3
1.exf3e.p. Bc2# is the only possible solution, but this necessitates R: 1.f2-f4. Can we prove this?
(13+13) with 1+2 pawn captures. Bf8 captured at home, so to satisfy White appetite, the missing Black pawn (a or b) must have promoted via c2 on c1. Two more White units must be captured to allow this.
The kings cage can only be unlocked by retracting WPc2. But the clock is ticking as there are only 6 black moves which can be retracted.
The promoted piece was captured on e3 or h3. If either capture is undone, then a White bishop square is cut off, so WB must be replaced prior to this.
Now the order of the early moves is: WdP moves, WQB & WQR escape, BP promotes on c1 to X (capturing WR at some point), X captured by WP.
So the first White capture must be dxNe3 and the second White capture releases gxXh3. The second White capture releases WKB & WKR. WKR captured by original BfP.
The clock starts ticking with gxh3. Black has 6 pawn moves. WKB has 3 moves to reach h7. WR has 3 if it goes via d file, or 2 if it starts on g1 (in which case WfP or WQB must also move once). So certainly at least 6 White moves. Last move was therefore White (even if the stipulation didn't tell us), and it can only have been WfP coming from f3 or f4. If it had been coming from f3 it would have blocked WKB in its progress, so the last White move was indeed R: 1.f2-f4.
WKR did therefore move from g1-g4-e4, and R: 1. ... fxRe5 2. Rg4-e4. Prior to that, move order not unique, but counting still exact.
Note that WN loitering on b4, pretending to be part of the cage, is present on the board just to make up the numbers.
Jeliss: "Obstruction of passage square f3 to Bishop of same colour."
"Version 'Pittsburgh Leader' 08.06.1913"
Yoav Ben-Zvi: Appears as the first problem (D445) in the booklet on Dawson's RA problems by G.P. Jellis. The obstruction that occurs in the Try -1.Pf3-f4?, by WP of WB, is described as "obstruction of passage square". It is not considered by Dawson and his disciples to be a Retro opposition. Dawson's conception of RO was quite broad, it included cases where the interference was not by occupation of the target square, so the only valid reason that I can see to exclude this case is that the 2 pieces involved are both of the same color. Fabel's definition explicitly excludes "Monochrome RO". I conclude that it would be preferrable to interpret RO as a bi-chromatic interference. The keyword Retro opposition should be removed. (2018-04-07)
A.Buchanan: To my mind, RO involves some kind of parity-tempo issue between the sides, not just some kind of race-tempo. If it was just about "bi-chromatic interference", one might say that bPe4 blocks wBh7 from an immediate retreat, so it has to be wPf4 that retreats first, legitimizing the ep key. So I agree this is not RO. (2024-01-06)
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"Version 'Pittsburgh Leader' 08.06.1913"
Yoav Ben-Zvi: Appears as the first problem (D445) in the booklet on Dawson's RA problems by G.P. Jellis. The obstruction that occurs in the Try -1.Pf3-f4?, by WP of WB, is described as "obstruction of passage square". It is not considered by Dawson and his disciples to be a Retro opposition. Dawson's conception of RO was quite broad, it included cases where the interference was not by occupation of the target square, so the only valid reason that I can see to exclude this case is that the 2 pieces involved are both of the same color. Fabel's definition explicitly excludes "Monochrome RO". I conclude that it would be preferrable to interpret RO as a bi-chromatic interference. The keyword Retro opposition should be removed. (2018-04-07)
A.Buchanan: To my mind, RO involves some kind of parity-tempo issue between the sides, not just some kind of race-tempo. If it was just about "bi-chromatic interference", one might say that bPe4 blocks wBh7 from an immediate retreat, so it has to be wPf4 that retreats first, legitimizing the ep key. So I agree this is not RO. (2024-01-06)
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Keywords: Last Moves?, En passant as key
Genre: h#, Retro
FEN: nqb5/1rrpp1pB/KRp5/1p4B1/kN2pP1p/2P1P2P/PP2P2P/8
Reprints: D445 Retro-Opposition & Other Retro-Analytical Chess Problems 1989
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-06 more...
Genre: h#, Retro
FEN: nqb5/1rrpp1pB/KRp5/1p4B1/kN2pP1p/2P1P2P/PP2P2P/8
Reprints: D445 Retro-Opposition & Other Retro-Analytical Chess Problems 1989
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-06 more...
a) 1. Sg3 Kxh4 2. Le4 hxg3#
b) 1. hxg3ep Kh4 2. Lf5 hxg3#
b) 1. hxg3ep Kh4 2. Lf5 hxg3#
Henrik Juel: in a) last move could be Kg2-h3
in b) last move must be g2-g4 (2024-01-14)
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in b) last move must be g2-g4 (2024-01-14)
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Keywords: En passant as key, Zeroposition
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & trivial retro-logic
FEN: 8/7b/8/4pn2/4pkPp/4pp1K/7P/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-15 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & trivial retro-logic
FEN: 8/7b/8/4pn2/4pkPp/4pp1K/7P/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-15 more...
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)
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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)
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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...
a) 1. Sxb4 Sxc3 2. Ka5 Sxc4#
b) 1. cxb3ep Sc4 2. Sb4 Sxc3#
b) 1. cxb3ep Sc4 2. Sb4 Sxc3#
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)
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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)
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1. cxd3ep? bxc3 2. Lxf4 Lxg2# Last move not d2-d4
1. gxf3ep! bxc3 2. Lf4 Lf5# Last move f2-f4
White's last move was certainly one of the two double pawn hops. The last move couldn't be d2-d4 because Black's pawns had already captured 10 of White's pieces, and playing d2-d4 would have resulted in wBc1 being White's 17th piece. So, the only solution is 1.gxf3 e.p.
1. gxf3ep! bxc3 2. Lf4 Lf5# Last move f2-f4
White's last move was certainly one of the two double pawn hops. The last move couldn't be d2-d4 because Black's pawns had already captured 10 of White's pieces, and playing d2-d4 would have resulted in wBc1 being White's 17th piece. So, the only solution is 1.gxf3 e.p.
Viktoras Paliulionis: The last move couldn't be d2-d4 because Black's pawns had already captured 10 of White's pieces, and playing d2-d4 would have resulted in wBc1 being White's 17th piece. So, the only solution is 1.gxf3 e.p. (2023-12-30)
A.Buchanan: Yes that's right! (2023-12-31)
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A.Buchanan: Yes that's right! (2023-12-31)
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Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 4.87 & trivial retro-logic
FEN: 2n1q3/5prp/5K1b/3b4/2pPkPp1/2p1p1pB/1P4pP/8
Reprints: 546 FIDE Album 1962-1964 1968
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-31 more...
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 4.87 & trivial retro-logic
FEN: 2n1q3/5prp/5K1b/3b4/2pPkPp1/2p1p1pB/1P4pP/8
Reprints: 546 FIDE Album 1962-1964 1968
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-31 more...
18 - P0003206
Hans Joachim Schlüter
4443v Schach , p. 349, 11/1963
(5+8)
h#2
b) wTf1 tauschen mit wLg1
Hans Joachim Schlüter
4443v Schach , p. 349, 11/1963
(5+8)
h#2
b) wTf1 tauschen mit wLg1
a) 1. Kxb4 Lb6 2. a4 Tb1#
b) 1. cxb3ep gxf3 2. Sc1 Tg4#
b) 1. cxb3ep gxf3 2. Sc1 Tg4#
Keywords: En passant as key
Genre: h#, Retro
FEN: 8/8/8/p7/kPp5/p1p2p2/4n1Pp/5RBK
Input: Gerd Wilts, 1995-06-03
Last update: Felber, Volker, 2022-11-24 more...
Genre: h#, Retro
FEN: 8/8/8/p7/kPp5/p1p2p2/4n1Pp/5RBK
Input: Gerd Wilts, 1995-06-03
Last update: Felber, Volker, 2022-11-24 more...
1. exf3ep c3 2. Df7 Ke4 3. Dh5 exf3#
A.Buchanan: Can shift bQd5 to c4, while replacing bRg5 with bP to reach Meredith status. bBh4 can also be downgraded to bP. But probably even more economy is possible by shifting pieces one file to the right, keeping the ep, White tempo & model mate, which seem to be the main features (2022-11-30)
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Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 8/8/8/3q2r1/2p1pPkb/4K1pp/2PrP3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-30 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 8/8/8/3q2r1/2p1pPkb/4K1pp/2PrP3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-30 more...
a) 1. f3 Lxa7 2. Lg5 Lxf2#
b) 1. fxg3ep Lxc7 2. Lh5 Ld8#
b) 1. fxg3ep Lxc7 2. Lh5 Ld8#
Henrik Juel: a) C+ Popeye 4.61
b) the ep capture is unjustified, last move could be Ke2-f1 or f3xg4 (2023-08-06)
A.Buchanan: There was an error in the stipulation: "b) sLe3 nach h2 (-wSh2)" is completely incorrect. However "b) sLe3 nach f3" leads to a very nice problem which matches the solution already present. This problem is not in WinChloe, but I am sure this is the resolution of the mystery (2023-08-07)
Henrik Juel: You are probably right, Andrew; maybe there was a misprint in the source
Now part b) is C+ Popeye 4.61 after very simple analysis (2023-08-07)
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b) the ep capture is unjustified, last move could be Ke2-f1 or f3xg4 (2023-08-06)
A.Buchanan: There was an error in the stipulation: "b) sLe3 nach h2 (-wSh2)" is completely incorrect. However "b) sLe3 nach f3" leads to a very nice problem which matches the solution already present. This problem is not in WinChloe, but I am sure this is the resolution of the mystery (2023-08-07)
Henrik Juel: You are probably right, Andrew; maybe there was a misprint in the source
Now part b) is C+ Popeye 4.61 after very simple analysis (2023-08-07)
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Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye v4.61 & simple retro-logic for b)
FEN: 1B6/npp5/8/8/5pPk/4b2p/3p1p1N/5Kn1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-08-07 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.61 & simple retro-logic for b)
FEN: 1B6/npp5/8/8/5pPk/4b2p/3p1p1N/5Kn1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-08-07 more...
1. ... exf6ep 2. Sf5 f7#
White captured fxg and promoted [Pd2] on d8
Black captured axbxa2-a1=Y, b7xc6, d5xe4, and gxf
White cannot have moved last, and in fact Black's last move can only have been f7-f5 in order to give White a prior move. R: 1... d5-e4 would have retro-blocked wPd marching to its promotion.
So the retroplay was R: 1... f7-f5 2.f5xYg6
Numerous retro tries but none distinct.
White captured fxg and promoted [Pd2] on d8
Black captured axbxa2-a1=Y, b7xc6, d5xe4, and gxf
White cannot have moved last, and in fact Black's last move can only have been f7-f5 in order to give White a prior move. R: 1... d5-e4 would have retro-blocked wPd marching to its promotion.
So the retroplay was R: 1... f7-f5 2.f5xYg6
Numerous retro tries but none distinct.
"Autor Ing. Rudolf Buljan, Zagreb"
AB: Have therefore marked him in the author table as Croatian. Yugoslavia has been removed from the PDB country list, I see.
A.Buchanan: Any ideas how to distinguish usage of “whose move?” and “no legal last move for white” (or black) keywords (2020-10-01)
Henrik Juel: This problem is clearly the latter
I would use 'Whose move?' in retro problems where a deeper analysis is required to determine the move, often with stipulations like '#1 (who?)' (2020-10-01)
A.Buchanan: I think "Whose move?" is where there is no forward stip, or where we have to determine if the forward stip should be interpreted as orthodox or as half-duplex. "no legal last move for..." is where the mater/stalemater/etc is the usual one, but we add or remove a single move at the beginning of the forward stip. I guess a third possibility is "last move?" where in a Type A position we also have to figure out who moved last.
With respect, I don't think the level of analysis can be relevant. (2023-12-06)
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AB: Have therefore marked him in the author table as Croatian. Yugoslavia has been removed from the PDB country list, I see.
A.Buchanan: Any ideas how to distinguish usage of “whose move?” and “no legal last move for white” (or black) keywords (2020-10-01)
Henrik Juel: This problem is clearly the latter
I would use 'Whose move?' in retro problems where a deeper analysis is required to determine the move, often with stipulations like '#1 (who?)' (2020-10-01)
A.Buchanan: I think "Whose move?" is where there is no forward stip, or where we have to determine if the forward stip should be interpreted as orthodox or as half-duplex. "no legal last move for..." is where the mater/stalemater/etc is the usual one, but we add or remove a single move at the beginning of the forward stip. I guess a third possibility is "last move?" where in a Type A position we also have to figure out who moved last.
With respect, I don't think the level of analysis can be relevant. (2023-12-06)
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Keywords: En passant as key, No legal last move for White
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: 5nkb/1qp1p1n1/2p3PQ/4PpKb/4prRp/P4prB/1PP3PP/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: 5nkb/1qp1p1n1/2p3PQ/4PpKb/4prRp/P4prB/1PP3PP/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
1) 1. axb3ep bxc6+ 2. b5 cxb6ep#
2) 1. exd3ep g8=D,L+ 2. d5 cxd6ep#
PRA: 1 solution with 2 parts
2) 1. exd3ep g8=D,L+ 2. d5 cxd6ep#
PRA: 1 solution with 2 parts
Henrik Juel: White captured [sLc8] on c8 and axb, so last move was either b2-b4 or d2-d4
C+ Popeye 4.61, except for the promotion dual (2020-08-03)
A.Buchanan: There's no keyword for an untolerated dual promotion! (2020-08-03)
Henrik Juel: Yes and no, Andrew
What about the cooked label? (2020-08-03)
A.Buchanan: Hi Henrik: this is obviously intentional. I tried to replace the pawn promotion with wDh8-h7+ but I ran out of pieces. I guess that Bebesi had the same difficulty. The word "cooked" seems too strong to me, but it's obviously a significant defect though (2020-08-04)
A.Buchanan: The bar is higher for helpmates than other problems, and this is unsound. However it’s not cooked but dualled. Yet all we have is “cooked” so I suppose it is in PDB-speak. I’ve added that and removed the 2.1… (2023-08-07)
Joost de Heer: Is there an 'unsound' keyword? (2023-08-07)
A.Buchanan: Hi Joost: no there isn't. It can be added, the challenge for any new common keyword would be populating it for more than a few values. (2023-08-07)
A.Buchanan: This time round, I've managed to find a fix. Bebesi's correction in WinChloe 66992 has non-standard wL. I have a feeling I've seen a similar one by Andrey Frolkin & someone else recently with a different matrix, and also non-standard wL. I've sent my effort to Joaquim Crusats at Problemas for his assessment (2023-08-08)
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C+ Popeye 4.61, except for the promotion dual (2020-08-03)
A.Buchanan: There's no keyword for an untolerated dual promotion! (2020-08-03)
Henrik Juel: Yes and no, Andrew
What about the cooked label? (2020-08-03)
A.Buchanan: Hi Henrik: this is obviously intentional. I tried to replace the pawn promotion with wDh8-h7+ but I ran out of pieces. I guess that Bebesi had the same difficulty. The word "cooked" seems too strong to me, but it's obviously a significant defect though (2020-08-04)
A.Buchanan: The bar is higher for helpmates than other problems, and this is unsound. However it’s not cooked but dualled. Yet all we have is “cooked” so I suppose it is in PDB-speak. I’ve added that and removed the 2.1… (2023-08-07)
Joost de Heer: Is there an 'unsound' keyword? (2023-08-07)
A.Buchanan: Hi Joost: no there isn't. It can be added, the challenge for any new common keyword would be populating it for more than a few values. (2023-08-07)
A.Buchanan: This time round, I've managed to find a fix. Bebesi's correction in WinChloe 66992 has non-standard wL. I have a feeling I've seen a similar one by Andrey Frolkin & someone else recently with a different matrix, and also non-standard wL. I've sent my effort to Joaquim Crusats at Problemas for his assessment (2023-08-08)
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Keywords: En passant as key (2), Partial Retro Analysis (PRA), En passant as mating move (2), Superseded by (P1411659, P1413906)
Genre: h#, Retro
FEN: 3qn3/1p1p2P1/B1r3p1/1PP1Kp2/pPkPp3/p1p4n/N7/2b5
Reprints: 233 Ungarische Schachproblem Anthologie , p. 188, 1983
1 Problemas 44, p. 1456, 10/2023
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-04 more...
Genre: h#, Retro
FEN: 3qn3/1p1p2P1/B1r3p1/1PP1Kp2/pPkPp3/p1p4n/N7/2b5
Reprints: 233 Ungarische Schachproblem Anthologie , p. 188, 1983
1 Problemas 44, p. 1456, 10/2023
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-04 more...
a) 1. Dc4 Lxb6+ 2. c5 dxc6ep#
b) 1. fxe3ep Lxb4 2. Sc4 bxc3#
Lindner in 'Mattbilder eines Lebens':
In a) ist die Lösung der stellung b) nicht möglich, weil das e.p.-Schlagen durch Schwarz nicht legal ist. Der letzte Zug von Weiß muß nicht unbedingt e2-e4 geweseb sein. Es kommt als letzter zug auch Kh3-g2 in Betracht, mit den vorherigen Zügen h4:g3 e.p.+ und g2-g4.
In b) demgegenüber sind Kh3-g2 und vorher f4:g3 e.p. illegal, weil die Rücknahme von g2-g4 unmöglich ist: der sB würde 7 Schlagfälle benötigen, und es fehlen nur 6 weiße Steine. Der letzte weiße Zug muß also e2-e4 gewesen sein.
b) 1. fxe3ep Lxb4 2. Sc4 bxc3#
Lindner in 'Mattbilder eines Lebens':
In a) ist die Lösung der stellung b) nicht möglich, weil das e.p.-Schlagen durch Schwarz nicht legal ist. Der letzte Zug von Weiß muß nicht unbedingt e2-e4 geweseb sein. Es kommt als letzter zug auch Kh3-g2 in Betracht, mit den vorherigen Zügen h4:g3 e.p.+ und g2-g4.
In b) demgegenüber sind Kh3-g2 und vorher f4:g3 e.p. illegal, weil die Rücknahme von g2-g4 unmöglich ist: der sB würde 7 Schlagfälle benötigen, und es fehlen nur 6 weiße Steine. Der letzte weiße Zug muß also e2-e4 gewesen sein.
In 'Mattbilder eines Lebens' abgedruckt mit sTh7 statt h8 und der Quellenangabe: Europe Echecs, 1964
AB: (1) Where is wK?
(2) Why is 1.fxe3ep legal in b) but not a)? (2002-01-31)
Henrik Juel: wK is probably on g2. In part a) last move could have been Kh3-g2, I think (2002-02-01)
A.Buchanan: Very convincing, Henrik. I've repaired the diagram accordingly. (2023-05-28)
comment
AB: (1) Where is wK?
(2) Why is 1.fxe3ep legal in b) but not a)? (2002-01-31)
Henrik Juel: wK is probably on g2. In part a) last move could have been Kh3-g2, I think (2002-02-01)
A.Buchanan: Very convincing, Henrik. I've repaired the diagram accordingly. (2023-05-28)
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Keywords: En passant as key, En passant in the retro play
Genre: h#, Retro
FEN: 7r/2pn4/1nqRb3/B2Pp3/pb1kPp2/2p2Pp1/1PP2pKp/7r
Reprints: 501 Mattbilder eines Lebens , p. 379, 1996
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-05-28 more...
Genre: h#, Retro
FEN: 7r/2pn4/1nqRb3/B2Pp3/pb1kPp2/2p2Pp1/1PP2pKp/7r
Reprints: 501 Mattbilder eines Lebens , p. 379, 1996
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-05-28 more...
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)
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I have changed the solution accordingly. (2023-03-10)
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*) 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#
28 - P0003659
Maurice Jago
8202v Die Schwalbe , p. 148, 10/1980
version of cooked 8202 Die Schwalbe 08/1977
(12+14) cooked
h#2
b) sBb4->b5
Maurice Jago
8202v Die Schwalbe , p. 148, 10/1980
version of cooked 8202 Die Schwalbe 08/1977
(12+14) cooked
h#2
b) sBb4->b5
a) 1. Sf2 Dxf2+ 2. Kh1 0-0-0#
b) 1. Sb1 Ta4 2. Sxg3 Th4#
Cook: 1. Txg3 0-0-0 2. La2/Lb3/Lc4 Dxh1#
1. La2/Lb3/Lc4 0-0-0 2. Txg3 Dxh1#
b) 1. Sb1 Ta4 2. Sxg3 Th4#
Cook: 1. Txg3 0-0-0 2. La2/Lb3/Lc4 Dxh1#
1. La2/Lb3/Lc4 0-0-0 2. Txg3 Dxh1#
See P0000642
Mario Richter: Hier stimmt was nicht: In der Schwalbe 08/1977 erschien als Nr. 2202 die folgende Aufgabe: 8/Pp2p1p1/Bq1p3p/4p3/1p6/n4pP1/1PPPPr1k/R3KQ1n/; h#2, b) sBb4 nach b5 und der AL a) 1.Tf2-g2 0-0-0 2.Tg2xg3 Df1xh1#; b) 1.Sa3-b1 Ta1-a4 2.Sh1xg3 Ta4-h4#, die aber in a) und b) durch 1.f3xe2 La6xb7 2.Tf2-g2 Df1xg2# nebenlösig war.
Ferner ist in nebenstehendem Diagramm sowohl in a) als auch in b) sehr wohl die w0-0-0 möglich
(Weiß hat als letzten Zug Sc7-a8, Schwarz entwandelt auf b1 und h1, Weiß auf f8); und selbst unter der Annahme, daß die w0-0-0 nur mit sBb4 möglich sei, wäre die Aufgabe mehrfach nebenlösig, z.B. durch 1.Ld5-c4 0-0-0 2.Bd6-d5 Df1xh1# (2010-05-29)
A.Buchanan: Here, the source was incorrectly given as "8202v Die Schwalbe 08/1977". No, that was where the original problem appeared, with a "v" added. So when, as here, we suspect the diagram has a typo, we would have to look through many back issues of Die Schwalbe! It turns out that this correction (itself cooked) was published (as text) in p.148 Die Schwalbe 65 10/1980. I have corrected the source. The reference to the earlier buggy problem can then cleanly be put in the "after" field. The actual typo was missing bPc7. (2024-03-06)
A.Buchanan: With the typo fixed, I think the retro-logic works ok, but there is a single cook family, fortunately repairable (2024-03-06)
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Mario Richter: Hier stimmt was nicht: In der Schwalbe 08/1977 erschien als Nr. 2202 die folgende Aufgabe: 8/Pp2p1p1/Bq1p3p/4p3/1p6/n4pP1/1PPPPr1k/R3KQ1n/; h#2, b) sBb4 nach b5 und der AL a) 1.Tf2-g2 0-0-0 2.Tg2xg3 Df1xh1#; b) 1.Sa3-b1 Ta1-a4 2.Sh1xg3 Ta4-h4#, die aber in a) und b) durch 1.f3xe2 La6xb7 2.Tf2-g2 Df1xg2# nebenlösig war.
Ferner ist in nebenstehendem Diagramm sowohl in a) als auch in b) sehr wohl die w0-0-0 möglich
(Weiß hat als letzten Zug Sc7-a8, Schwarz entwandelt auf b1 und h1, Weiß auf f8); und selbst unter der Annahme, daß die w0-0-0 nur mit sBb4 möglich sei, wäre die Aufgabe mehrfach nebenlösig, z.B. durch 1.Ld5-c4 0-0-0 2.Bd6-d5 Df1xh1# (2010-05-29)
A.Buchanan: Here, the source was incorrectly given as "8202v Die Schwalbe 08/1977". No, that was where the original problem appeared, with a "v" added. So when, as here, we suspect the diagram has a typo, we would have to look through many back issues of Die Schwalbe! It turns out that this correction (itself cooked) was published (as text) in p.148 Die Schwalbe 65 10/1980. I have corrected the source. The reference to the earlier buggy problem can then cleanly be put in the "after" field. The actual typo was missing bPc7. (2024-03-06)
A.Buchanan: With the typo fixed, I think the retro-logic works ok, but there is a single cook family, fortunately repairable (2024-03-06)
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Keywords: Cant Castler, Castling (wg), Superseded by (P1415606)
Genre: h#, Retro
Computer test: both twins cooked by Popeye v4.87
FEN: NBr5/N1p3p1/1qPpp3/2bb4/1p6/n1pP1rP1/1PP1P2k/R3KQ1n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-03-06 more...
Genre: h#, Retro
Computer test: both twins cooked by Popeye v4.87
FEN: NBr5/N1p3p1/1qPpp3/2bb4/1p6/n1pP1rP1/1PP1P2k/R3KQ1n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-03-06 more...
a) 1. Ta6! Lxc3 2. Ta2 0-0#
b) 1. Ta5! Kd1 2. Ta2 Kc2#
b) 1. Ta5! Kd1 2. Ta2 Kc2#
Henrik Juel: In b) White may not castle, because last move was done by Ke1 or Th1 (2023-12-02)
A.Buchanan: Cute. Do wPe2 & bPe3 serve any purpose, however? (2023-12-03)
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A.Buchanan: Cute. Do wPe2 & bPe3 serve any purpose, however? (2023-12-03)
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Keywords: Cant Castler, Castling (wk), Superseded by (P1413924)
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & simple retro logic
FEN: 5n1q/6B1/5r2/7r/5p2/2p1p3/1P2P3/k3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-05 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & simple retro logic
FEN: 5n1q/6B1/5r2/7r/5p2/2p1p3/1P2P3/k3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-05 more...
a) 1. 0-0 Tcg3 2. Sh8 Txg7#
b) 1. 0-0-0 Tgc4 2. Sb8 Txc7#
b) 1. 0-0-0 Tgc4 2. Sb8 Txc7#
Keywords: Castling, Cant Castler, Obvious promotion (T), Twin
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: r3k2r/1pp3pp/2n3n1/8/6R1/2R4P/4P1P1/5B1K
Input: Gerd Wilts, 1995-06-03
Last update: Gunter Jordan, 2022-12-01 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: r3k2r/1pp3pp/2n3n1/8/6R1/2R4P/4P1P1/5B1K
Input: Gerd Wilts, 1995-06-03
Last update: Gunter Jordan, 2022-12-01 more...
a) 1. f5 Le5 2. 0-0 Th8#
b) 1. Kd8 0-0-0 2. Te8 Lf8#
Try to deduce the diagram error, as we do have the intended solutions!
a) wRh7 is original, but cannot have escaped from cage by white cross-capture as there are too many captures needed. On the other hand, Black only has to make 4 captures so the cross-capture is ok.
b) changing the colour of Sa5 shifts the capture balance: now White can cross-capture but not Black, so White but not Black can castle.
Thus the retro-logic works with the diagram as it is, but the forward logic does not. How can we repair the diagram?
b) 1. Kd8 0-0-0 2. Te8 Lf8#
Try to deduce the diagram error, as we do have the intended solutions!
a) wRh7 is original, but cannot have escaped from cage by white cross-capture as there are too many captures needed. On the other hand, Black only has to make 4 captures so the cross-capture is ok.
b) changing the colour of Sa5 shifts the capture balance: now White can cross-capture but not Black, so White but not Black can castle.
Thus the retro-logic works with the diagram as it is, but the forward logic does not. How can we repair the diagram?
See P0000899 a companion problem.
A.Buchanan: Something odd going on here. There are numerous h#2 in both a&b. PRA not normally found in a problem with single solution for each twin. White has 8 pawns so trivially wRh7 must be original. (2018-10-13)
A.Buchanan: There are three similar problems by Brogi which are all twinned RS rather than PRA, but don't have diagram error. P0003743, P0003746 & P0003747. (2018-10-13)
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
Ladislav Packa: Cooked a) and b):
1...b8S and 2...R:h8# (2021-02-10)
A.Buchanan: I think the composer simply forgot that wPb7 can promote. Most of the cooks come from S promotions, but it's also possible to have QR. I've fixed it, rejigging the captures to make them still add up. I've posted in Discord, as I did with the partner composition, and will create entries for them here. (2022-03-15)
milan: +sLb8 sBa7=sT M.Frelih (2023-12-02)
A.Buchanan: Hi Milan - I don't think your suggestion quite works for b). In a) there are 0+2 spare captures, so Black can certainly cross-capture. But in b) there is 1+1 so neither side can cross-capture, so there is no solution. Please compare with P1399806, in which there are 1+2 & 2+1 spare captures, so both twins are sound. (2023-12-03)
milan: Hi Andrew my correction works only with 2.1... solutions, black or white knights on a5. are not important. (2023-12-03)
A.Buchanan: Hi Milan not really clear what you are doing, but if as well as the piece changes you proposed, you also change the stipulation to 2.1... then there is still only one solution. Even if you remove Sa5 entirely as well, there is no White cross-capture possible. (2023-12-04)
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A.Buchanan: Something odd going on here. There are numerous h#2 in both a&b. PRA not normally found in a problem with single solution for each twin. White has 8 pawns so trivially wRh7 must be original. (2018-10-13)
A.Buchanan: There are three similar problems by Brogi which are all twinned RS rather than PRA, but don't have diagram error. P0003743, P0003746 & P0003747. (2018-10-13)
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
Ladislav Packa: Cooked a) and b):
1...b8S and 2...R:h8# (2021-02-10)
A.Buchanan: I think the composer simply forgot that wPb7 can promote. Most of the cooks come from S promotions, but it's also possible to have QR. I've fixed it, rejigging the captures to make them still add up. I've posted in Discord, as I did with the partner composition, and will create entries for them here. (2022-03-15)
milan: +sLb8 sBa7=sT M.Frelih (2023-12-02)
A.Buchanan: Hi Milan - I don't think your suggestion quite works for b). In a) there are 0+2 spare captures, so Black can certainly cross-capture. But in b) there is 1+1 so neither side can cross-capture, so there is no solution. Please compare with P1399806, in which there are 1+2 & 2+1 spare captures, so both twins are sound. (2023-12-03)
milan: Hi Andrew my correction works only with 2.1... solutions, black or white knights on a5. are not important. (2023-12-03)
A.Buchanan: Hi Milan not really clear what you are doing, but if as well as the piece changes you proposed, you also change the stipulation to 2.1... then there is still only one solution. Even if you remove Sa5 entirely as well, there is no White cross-capture possible. (2023-12-04)
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Keywords: Castling (wgsk), Cant Castler (wgsk), Cross-capture (s,w), Superseded by (P1399806)
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: 4k2r/pPp2p1R/n1pB1ppp/npP5/1P6/5PPP/2P2P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-03 more...
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: 4k2r/pPp2p1R/n1pB1ppp/npP5/1P6/5PPP/2P2P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-03 more...
a) 1. fxg1=S Dxe3 2. Sd2 Df2#
b) 1. ... Dxe3 2. c1=S Dc3#
a) R: 1. d2xc3
b) WTM, so solution in a) can't work here.
b) 1. ... Dxe3 2. c1=S Dc3#
a) R: 1. d2xc3
b) WTM, so solution in a) can't work here.
Retro
b) Weiß am Zug
Adrian Storisteanu: Two knight promotions differently motivated.
The technical challenge was in averting the short second solution in a) as set play, in order to keep the two phases nicely apart, and also, of course, preventing black tempi and potential white retro moves. (2000-11-20)
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b) Weiß am Zug
Adrian Storisteanu: Two knight promotions differently motivated.
The technical challenge was in averting the short second solution in a) as set play, in order to keep the two phases nicely apart, and also, of course, preventing black tempi and potential white retro moves. (2000-11-20)
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Keywords: No legal last move for White, Promotion (s,s)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro logic
FEN: 8/8/8/8/8/2P1p1P1/ppp1ppRP/bqQbknRK
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro logic
FEN: 8/8/8/8/8/2P1p1P1/ppp1ppRP/bqQbknRK
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
33 - P0008219
Jozsef Korponai
6690 Probleemblad 09-10/1969
1. ehrende Erwähnung
(12+7)
a) h#2
b) Last two single moves?
Jozsef Korponai
6690 Probleemblad 09-10/1969
1. ehrende Erwähnung
(12+7)
a) h#2
b) Last two single moves?
a) 1. Kxc5 h8=D 2. a1=T De5#
b) R: 1. a7-a8=S+ h2-h1=L
b) R: 1. a7-a8=S+ h2-h1=L
Erich Bartel: Auszeichnung: 1.ehrende Erwähnung.--
Nachdruck: 1) Jaarboek 1970.-- (2006-11-26)
Henrik Juel: Something is wrong: De5 does not mate. And the intention certainly was not -1.S=a7 L=h2, 0... Kxc5 1.h8DL Kb6 2.DLd4#. (2006-11-27)
Erich Bartel: Ich habe die Aufgabe aus dem Jaarboek 1970 übernommen, wo
sie wie angegeben im Diagramm und Lösung so steht. Es
scheint also bereits hier ein Fehler vorzuliegen. Zur Klärung
ist es wohl notwendig in der Ur-Quelle Probleemblad -die ich
nicht besitze- nachzusehen. (2006-11-27)
Siegfried Hornecker: No matter how the position is changed, there can be no line where the king, coming from b6, is not allowed to return to b6 when the queen checks on e5. Also, there would be no reason to not play again 2.h2-h1~ as a waiting move.
The retroanalytics say that the last move must(!) have been 1.a7-a8S+ h2-h1L so I think the correct stipulation is:
a) Last move (w+s)
b) h#2
I have no source for this but if you think about it, the whole thing makes sense! The last move was 1.a8=S+ after 1.h1=L and from the diagram position the h#2 runs 2. Kxc5 h8=D 3. a1=T De5#. That's what I think. (2006-12-12)
Jan Hein Verduin: I got the Jaarboek too, and mr. Hornecker is right. The stipulation there is "h#2; what where the last w+b moves?" The way the solution then is given (the retro moves are given first) probably caused the confusion. (2007-01-26)
Adrian Storisteanu: Then, ceci n'est pas un ... retractor!? (2024-03-17)
Henrik Juel: Right, Adrian
The Help retractor keyword should be deleted (2024-03-17)
comment
Nachdruck: 1) Jaarboek 1970.-- (2006-11-26)
Henrik Juel: Something is wrong: De5 does not mate. And the intention certainly was not -1.S=a7 L=h2, 0... Kxc5 1.h8DL Kb6 2.DLd4#. (2006-11-27)
Erich Bartel: Ich habe die Aufgabe aus dem Jaarboek 1970 übernommen, wo
sie wie angegeben im Diagramm und Lösung so steht. Es
scheint also bereits hier ein Fehler vorzuliegen. Zur Klärung
ist es wohl notwendig in der Ur-Quelle Probleemblad -die ich
nicht besitze- nachzusehen. (2006-11-27)
Siegfried Hornecker: No matter how the position is changed, there can be no line where the king, coming from b6, is not allowed to return to b6 when the queen checks on e5. Also, there would be no reason to not play again 2.h2-h1~ as a waiting move.
The retroanalytics say that the last move must(!) have been 1.a7-a8S+ h2-h1L so I think the correct stipulation is:
a) Last move (w+s)
b) h#2
I have no source for this but if you think about it, the whole thing makes sense! The last move was 1.a8=S+ after 1.h1=L and from the diagram position the h#2 runs 2. Kxc5 h8=D 3. a1=T De5#. That's what I think. (2006-12-12)
Jan Hein Verduin: I got the Jaarboek too, and mr. Hornecker is right. The stipulation there is "h#2; what where the last w+b moves?" The way the solution then is given (the retro moves are given first) probably caused the confusion. (2007-01-26)
Adrian Storisteanu: Then, ceci n'est pas un ... retractor!? (2024-03-17)
Henrik Juel: Right, Adrian
The Help retractor keyword should be deleted (2024-03-17)
comment
Keywords: Allumwandlung, Last Moves? (2), Type C
Genre: Retro, h#
FEN: N7/1Bp4P/pkP5/p1R5/P7/P5P1/p1KNPPp1/7b
Input: Gerd Wilts, 1996-09-14
Last update: A.Buchanan, 2024-03-18 more...
Genre: Retro, h#
FEN: N7/1Bp4P/pkP5/p1R5/P7/P5P1/p1KNPPp1/7b
Input: Gerd Wilts, 1996-09-14
Last update: A.Buchanan, 2024-03-18 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)
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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) 1. ... g4 2. e6 g5 3. Ke7 g6 4. Td8 gxf7 5. Td7 f8=D#
2) 1. ... g3 2. Kd7 g4 3. Td8 g5 4. Ke8 g6 5. Sd7 gxf7#
Numerous retro tries but none distinct
2) 1. ... g3 2. Kd7 g4 3. Td8 g5 4. Ke8 g6 5. Sd7 gxf7#
Numerous retro tries but none distinct
Keywords: Promotion in the mating move, Excelsior white (auto key), No legal last move for White, Minimal
Genre: h#, Retro
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 4k2r/4ppK1/5n1p/8/8/8/6P1/8
Reprints: 365 Minimalkunst im Schach 2006
Input: Gerd Wilts, 1996-06-06
Last update: A.Buchanan, 2023-12-08 more...
Genre: h#, Retro
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 4k2r/4ppK1/5n1p/8/8/8/6P1/8
Reprints: 365 Minimalkunst im Schach 2006
Input: Gerd Wilts, 1996-06-06
Last update: A.Buchanan, 2023-12-08 more...
1) 1. exd3ep b8=D 2. Kd4 Df4#
2) 1. exd3ep a8=S 2. d4 Sxb6#
Cook: e.p. not justified as R: 0.e3xd4 possible
2) 1. exd3ep a8=S 2. d4 Sxb6#
Cook: e.p. not justified as R: 0.e3xd4 possible
VL: Unjustified ep-capture because of –e3xd4. For correction it suffices e.g. to add bSe3.
This is possibly a unique 2-move helpmate with ep-key and 2 solutions of the type 1.2.1.1 (at least so currently in the PDB).
Publi?ation details. H551 in the Helpmate section of the July issue, with no preview. Here is the text of the solution published in the Nov.-77 issue: “…White may have just played P(e3)xd4. Very few solvers have spotted this. Full marks for claim of no solution with correct reasoning but, in fairness, one mark for composer's intention. White cannot have just played K(a5)a6 in reply to Pxb6+ as too many captures by Black are required.” (2023-12-01)
A.Buchanan: Yes Valery you are right this must be cooked. However, I'm not a fan of +bSe3 because it spoils one of the model mates. Separately, I mildly prefer bPf7 to bPe6. (2023-12-03)
VL: I do prefer bPf7, as well. I don't see any way to save at least one of the properties Kindergarten Problem and Model mate (2) after fixing (what for these keywords have been added just now?). Besides, Andrew, please, refine the specification ((full) number and publication month 07-08). (2023-12-03)
A.Buchanan: Ok have added the publication details. The keywords are useful to indicate the aim, even though the problem is cooked. I guess there is a way to keep the model mates but it’s very clunky and probably +bSe3 is better (2023-12-03)
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comment
This is possibly a unique 2-move helpmate with ep-key and 2 solutions of the type 1.2.1.1 (at least so currently in the PDB).
Publi?ation details. H551 in the Helpmate section of the July issue, with no preview. Here is the text of the solution published in the Nov.-77 issue: “…White may have just played P(e3)xd4. Very few solvers have spotted this. Full marks for claim of no solution with correct reasoning but, in fairness, one mark for composer's intention. White cannot have just played K(a5)a6 in reply to Pxb6+ as too many captures by Black are required.” (2023-12-01)
A.Buchanan: Yes Valery you are right this must be cooked. However, I'm not a fan of +bSe3 because it spoils one of the model mates. Separately, I mildly prefer bPf7 to bPe6. (2023-12-03)
VL: I do prefer bPf7, as well. I don't see any way to save at least one of the properties Kindergarten Problem and Model mate (2) after fixing (what for these keywords have been added just now?). Besides, Andrew, please, refine the specification ((full) number and publication month 07-08). (2023-12-03)
A.Buchanan: Ok have added the publication details. The keywords are useful to indicate the aim, even though the problem is cooked. I guess there is a way to keep the model mates but it’s very clunky and probably +bSe3 is better (2023-12-03)
more ...
comment
Keywords: En passant as key, Kindergarten Problem, Model mate (2)
Genre: h#, Retro
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/PP6/Kpp1p3/2pp4/1pkPp3/2p5/P7/8
Input: Michal Dragoun, 1998-04-09
Last update: A.Buchanan, 2023-12-03 more...
Genre: h#, Retro
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/PP6/Kpp1p3/2pp4/1pkPp3/2p5/P7/8
Input: Michal Dragoun, 1998-04-09
Last update: A.Buchanan, 2023-12-03 more...
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. ... Lxd3 2. La6/Lb5/Ld5/Le6/Lf7/Lg8 Kb1 3. a2+ Kc1 4. a1=L Lxa6/Lxa5/Le4/Lf5/Lg6/Lh7 5. Ka2 Lc4#/Lc4#/Lxd5#/Lxe6#/Lxf7#/Lxg8#
Es war zum Abschlus noch ein ganz besonderer Gag - "zum wiederholten Male stellt der Autor die Löser mit einem Retro-trick auf die Probe".
"denn eine rückshauende Betrachung der Stellung ergibt daß Weiß nicht zuletzt gezogen haben kann, mithin also am Zuge ist."
"aber da gibt es wohl jede Menge Variationsmöglichkeiten bei den Läuferzugen.
In diesem speziellen Fall blieben sie unbewertet, zumal es ja auch eine spezielle Spielart, das sog. "Varianten Hilfsmatt" geben soll."
"Fazit: eine gute Mischung!"
"Durch die Hilsmatt-chwemme war es eine recht einfache, aber nicht uninteressante Angelenheir, vor allem die Nr. 13412 und natürlich der Retro-Witz haben mir gefallen."
Dennoch "Ein kleines Selbstmatt dazwischen (so in neun Zügen oder mehr) ware schon gang recht gewesen."
Some retro tries h#6 but not distinct.
Cook: The 6 helpmate variants seem to be intended.
Es war zum Abschlus noch ein ganz besonderer Gag - "zum wiederholten Male stellt der Autor die Löser mit einem Retro-trick auf die Probe".
"denn eine rückshauende Betrachung der Stellung ergibt daß Weiß nicht zuletzt gezogen haben kann, mithin also am Zuge ist."
"aber da gibt es wohl jede Menge Variationsmöglichkeiten bei den Läuferzugen.
In diesem speziellen Fall blieben sie unbewertet, zumal es ja auch eine spezielle Spielart, das sog. "Varianten Hilfsmatt" geben soll."
"Fazit: eine gute Mischung!"
"Durch die Hilsmatt-chwemme war es eine recht einfache, aber nicht uninteressante Angelenheir, vor allem die Nr. 13412 und natürlich der Retro-Witz haben mir gefallen."
Dennoch "Ein kleines Selbstmatt dazwischen (so in neun Zügen oder mehr) ware schon gang recht gewesen."
Some retro tries h#6 but not distinct.
Cook: The 6 helpmate variants seem to be intended.
Keywords: No legal last move for White, Miniature
Genre: h#, Retro
Computer test: C+ Helpmate Analyzer 6 variant solutions
FEN: 8/8/8/8/2b5/pk1p4/1P6/KB6
Input: Felber, Volker, 1999-12-27
Last update: A.Buchanan, 2023-12-07 more...
Genre: h#, Retro
Computer test: C+ Helpmate Analyzer 6 variant solutions
FEN: 8/8/8/8/2b5/pk1p4/1P6/KB6
Input: Felber, Volker, 1999-12-27
Last update: A.Buchanan, 2023-12-07 more...
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)
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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...
R: 1. f7xDe8=D+, dann 1. fxg8=S#
R: 1. f7xDe8=D+ Kf6-e7 g7xLh8=L+ ... Tb8-a8 ... c7xTb8=T
R: 1. f7xDe8=D+ Kf6-e7 g7xLh8=L+ ... Tb8-a8 ... c7xTb8=T
No. 1449 HN
AV: Je crois que c'est un Fou blanc en h8 et non un Cavalier, et les derniers coups sont gxFh8=F+ Rf6-e7 fxDe8=D+ (au lieu de fxCg8=C#). Avec l'antérieur cxTb8=T, cela donne une "Allumwandlung" avec une pièce noire prise de la même nature. (2005-06-09)
Henrik Juel: You must be right, AV, because with wSh8 the retroplay might be the bland -1.f7xDe8=D Kf6 etc. without unpromotion on h8. (2005-06-13)
Michel Caillaud: Indeed the diagram is wrong : there should be a white Bishop on h8
AUW spread over retro play (-1.fxDe8=D+ Kf6-e7 -2.gxLh8=L+ ...-n.cxTb8=T) and forward play 1.fxSg8=S#
Matching nature of promoted piece and piece captured by promotion. (2023-03-25)
comment
AV: Je crois que c'est un Fou blanc en h8 et non un Cavalier, et les derniers coups sont gxFh8=F+ Rf6-e7 fxDe8=D+ (au lieu de fxCg8=C#). Avec l'antérieur cxTb8=T, cela donne une "Allumwandlung" avec une pièce noire prise de la même nature. (2005-06-09)
Henrik Juel: You must be right, AV, because with wSh8 the retroplay might be the bland -1.f7xDe8=D Kf6 etc. without unpromotion on h8. (2005-06-13)
Michel Caillaud: Indeed the diagram is wrong : there should be a white Bishop on h8
AUW spread over retro play (-1.fxDe8=D+ Kf6-e7 -2.gxLh8=L+ ...-n.cxTb8=T) and forward play 1.fxSg8=S#
Matching nature of promoted piece and piece captured by promotion. (2023-03-25)
comment
Keywords: Help retractor, Promotion in the retro play, Promotion in forward play, Allumwandlung
Genre: Retro, h#
FEN: R1brQnnB/pp1pk2p/3pp1p1/6N1/6Kp/8/8/8
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2024-01-21 more...
Genre: Retro, h#
FEN: R1brQnnB/pp1pk2p/3pp1p1/6N1/6Kp/8/8/8
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2024-01-21 more...
1. ... a8=T! 2. Tb5 Tc8#
1. Kb5? a8=S 2. Lc6 Sc7#
but White has no last move, so White to move in diagram
1. Kb5? a8=S 2. Lc6 Sc7#
but White has no last move, so White to move in diagram
A.Buchanan: If the unconventional party must move first due to retro considerations, then according to Footnote 19 of the Codex, the stipulation should state the number of White moves. This means that direct mates & self mates become half a move longer, but help mates become half a move shorter.
So this becomes sound if we were to remove the "*" from the stipulation. The 2.0 solution becomes a retro try, and the 1.5 set play becomes the solution. (2013-07-20)
A.Buchanan: Nice problem. Can sTd6 be downgraded to sB? (2023-12-06)
Ladislav Packa: Yes, it is possible. I used the same mechanism in another problem:
StrateGems SG63 July-September 2013 C0490, dedicated to Valery Liskovets
White Pb7 Pb6 Pa2 Pd2 Kh1
Black Pe7 Ph7 Pc6 Qh6 Pa5 Pb5 Kd5 Sc4 Ph4 Pc3 Se3 Bc2 Rg2 Bh2
pser-h#4 (legal parry) 2 solutions
I. 1.Kc5 dxc3 2.Sd5 3.Qd2 (Bd6?) b8S 4.Bd6 Sd7#
II. 1.Kd6 d4 2.Bd3 (Rg6?) b8R 3.Rg6 4.Re6 Rd8# (2023-12-06)
comment
So this becomes sound if we were to remove the "*" from the stipulation. The 2.0 solution becomes a retro try, and the 1.5 set play becomes the solution. (2013-07-20)
A.Buchanan: Nice problem. Can sTd6 be downgraded to sB? (2023-12-06)
Ladislav Packa: Yes, it is possible. I used the same mechanism in another problem:
StrateGems SG63 July-September 2013 C0490, dedicated to Valery Liskovets
White Pb7 Pb6 Pa2 Pd2 Kh1
Black Pe7 Ph7 Pc6 Qh6 Pa5 Pb5 Kd5 Sc4 Ph4 Pc3 Se3 Bc2 Rg2 Bh2
pser-h#4 (legal parry) 2 solutions
I. 1.Kc5 dxc3 2.Sd5 3.Qd2 (Bd6?) b8S 4.Bd6 Sd7#
II. 1.Kd6 d4 2.Bd3 (Rg6?) b8R 3.Rg6 4.Re6 Rd8# (2023-12-06)
comment
Keywords: No legal last move for White
Genre: h#, Retro
Computer test: Trivial retro analysis + Popeye Windows-32Bit v4.51
FEN: 8/P2p4/Ppkr4/p1rb4/p1p5/K1p5/P1P5/b7
Input: hpr, 2009-07-11
Last update: A.Buchanan, 2018-01-28 more...
Genre: h#, Retro
Computer test: Trivial retro analysis + Popeye Windows-32Bit v4.51
FEN: 8/P2p4/Ppkr4/p1rb4/p1p5/K1p5/P1P5/b7
Input: hpr, 2009-07-11
Last update: A.Buchanan, 2018-01-28 more...
a) 1. Txg2 Ta4 2. Tc2 Td4#
b) 1. c5 Ld4 2. c4 0-0-0#
NL:
a) 1. e2 Td1+ 2. Ke3 Lc1#
b) 1. e2 Td1+ 2. Ke3 Lc1#
b) 1. c5 Ld4 2. c4 0-0-0#
NL:
a) 1. e2 Td1+ 2. Ke3 Lc1#
b) 1. e2 Td1+ 2. Ke3 Lc1#
Yuri Bilokin: correction bPg3-g4, bRh2-h6, -wPg2, -wPh3, -bPh4, -bPh4
8/8/2p1p2r/4P3/4P1p1/3kpB2/1B6/R3K3 (6+6) h#2 2.1…
1.c5 Bd4 2.c4 0-0-0#
1.Rh2 Ra4 2.Rc2 Rd4#
Castling (white, long)
Exchange of functions (bPc6/bRh6, Passive / Self-block)
Tempo move (wB, waiting, type 2)
Battery mate. Royal battery mate (2024-02-22)
A.Buchanan: The proposed correction is sound but it loses the retro logic under which castling is illegal in one twin. (2024-02-23)
A.Buchanan: 8/4p3/2p1p3/4P2p/5p1p/1p1k2NP/1B4Pr/R3KN2 is sound I think and carries the retro content (2024-02-23)
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8/8/2p1p2r/4P3/4P1p1/3kpB2/1B6/R3K3 (6+6) h#2 2.1…
1.c5 Bd4 2.c4 0-0-0#
1.Rh2 Ra4 2.Rc2 Rd4#
Castling (white, long)
Exchange of functions (bPc6/bRh6, Passive / Self-block)
Tempo move (wB, waiting, type 2)
Battery mate. Royal battery mate (2024-02-22)
A.Buchanan: The proposed correction is sound but it loses the retro logic under which castling is illegal in one twin. (2024-02-23)
A.Buchanan: 8/4p3/2p1p3/4P2p/5p1p/1p1k2NP/1B4Pr/R3KN2 is sound I think and carries the retro content (2024-02-23)
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Genre: h#, Retro
Computer test: Popeye Windows-32Bit v4.51 (100000 KB)
FEN: 8/8/2p1p3/4P2p/4P2p/3kpBpP/1B4Pr/R3K3
Input: hpr, 2012-05-28
Last update: A.Buchanan, 2024-02-23 more...
43 - P1339695
Kostas Prentos
09 Belgrade Sake Tourney 2016
2. ehrende Erwähnung
(6+0)
h#2
2 Lösungen
Colorless chess
Kostas Prentos
09 Belgrade Sake Tourney 2016
2. ehrende Erwähnung
(6+0)
h#2
2 Lösungen
Colorless chess
Kostas Prentos: Solution:
1. Le7(=b) Kh5(=w, bKa4, wTb5, wLe8) 2. La3 Sb6(=w)#
1. Lg3(=b) Se3(=w) 2. Kh4(=b, wKa4) Th5(=w, wLe8)# (2022-12-08)
comment
1. Le7(=b) Kh5(=w, bKa4, wTb5, wLe8) 2. La3 Sb6(=w)#
1. Lg3(=b) Se3(=w) 2. Kh4(=b, wKa4) Th5(=w, wLe8)# (2022-12-08)
comment
Keywords: Colorless chess, Aristocrat
Genre: h#, Retro, Fairies
FEN: 4B3/8/8/1R6/K1N3KB/8/8/8
Reprints: 09 Problem Paradise 74, p. 47, 04-06/2016
Input: A.Buchanan, 2017-09-18
Last update: Alfred Pfeiffer, 2017-09-19 more...
Genre: h#, Retro, Fairies
FEN: 4B3/8/8/1R6/K1N3KB/8/8/8
Reprints: 09 Problem Paradise 74, p. 47, 04-06/2016
Input: A.Buchanan, 2017-09-18
Last update: Alfred Pfeiffer, 2017-09-19 more...
44 - P1339698
Kostas Prentos
05 Belgrade Sake Tourney 2016
3. ehrende Erwähnung
(9+0)
h#2
b) La2 nach e2
Colorless Chess
Kostas Prentos
05 Belgrade Sake Tourney 2016
3. ehrende Erwähnung
(9+0)
h#2
b) La2 nach e2
Colorless Chess
Kostas Prentos: Solution:
a) 1. cxd3 e.p.(=b) Tf6(=w) 2. Ld5(=b) Tf4(bKe4, wKg5, wLc1, wSf3)#
b) 1. dxc3 e.p.(=b) Ta4(=w) 2. Kd3(=b; bLe2, wKg5, wLc1, wSf3) Se1# (2022-12-08)
comment
a) 1. cxd3 e.p.(=b) Tf6(=w) 2. Ld5(=b) Tf4(bKe4, wKg5, wLc1, wSf3)#
b) 1. dxc3 e.p.(=b) Ta4(=w) 2. Kd3(=b; bLe2, wKg5, wLc1, wSf3) Se1# (2022-12-08)
comment
Keywords: Colorless chess
Genre: h#, Retro, Fairies
FEN: 8/8/R7/2P3K1/2PPK3/5N2/B7/2B5
Reprints: 05 Problem Paradise 74, p. 47, 04-06/2016
Input: A.Buchanan, 2017-09-18
Last update: Alfred Pfeiffer, 2017-09-19 more...
Genre: h#, Retro, Fairies
FEN: 8/8/R7/2P3K1/2PPK3/5N2/B7/2B5
Reprints: 05 Problem Paradise 74, p. 47, 04-06/2016
Input: A.Buchanan, 2017-09-18
Last update: Alfred Pfeiffer, 2017-09-19 more...
45 - P1339707
Kostas Prentos
04 Belgrade Sake Tourney 2016
Lob
(10+0)
h#2
Colorless chess
b) Bd3 nach b3
Kostas Prentos
04 Belgrade Sake Tourney 2016
Lob
(10+0)
h#2
Colorless chess
b) Bd3 nach b3
Kostas Prentos: Solution:
a) 1. fxe3 e.p,(=b) 0-0(wKg1, wTf1, wTa1, wTf5, bKd4) 2. Ke4 (bPd3) Ta4#
b) 1. Ke5(=b, bTf5, bPf4, wKe1, wTa1, wTh1) 0-0-0(wLh5) 2. Kxe4 The1# (2022-12-08)
comment
a) 1. fxe3 e.p,(=b) 0-0(wKg1, wTf1, wTa1, wTf5, bKd4) 2. Ke4 (bPd3) Ta4#
b) 1. Ke5(=b, bTf5, bPf4, wKe1, wTa1, wTh1) 0-0-0(wLh5) 2. Kxe4 The1# (2022-12-08)
comment
Keywords: Colorless chess
Genre: h#, Retro, Fairies
FEN: 8/8/8/5RPB/3KPP2/3P4/8/R3K2R
Reprints: 04 Problem Paradise 74, p. 49, 04-06/2016
Input: A.Buchanan, 2017-09-18
Last update: Alfred Pfeiffer, 2019-06-03 more...
Genre: h#, Retro, Fairies
FEN: 8/8/8/5RPB/3KPP2/3P4/8/R3K2R
Reprints: 04 Problem Paradise 74, p. 49, 04-06/2016
Input: A.Buchanan, 2017-09-18
Last update: Alfred Pfeiffer, 2019-06-03 more...
1. ... gxh6ep[+sBh7]! 2. Dxg4[+wBg2] Sxg4[+sDd8]#
1. Sf3? gxh5[+sBh7] 2. Dg4 Sxg4[+sDd8]#
R: 1. h7-h5 2. Sf3xBh2[+sBh7]
Five retro tries h#2 but not distinct:
1.Sh4-f3 g4*h5[+bPh7] 2.Qh3-g4 Sh2*g4[+bQd8] #
1.Sh4*g6[+wPg2] g2*h3[+bQd8] 2.h5*g4[+wPg2] Sh2*g4[+bPg7] #
1.Sh4*g6[+wPg2] g4*h5[+bPh7] 2.Qh3-g4 Sh2*g4[+bQd8] #
1.h5*g4[+wPg2] g2*h3[+bQd8] 2.Sh4-f3 Sh2*g4[+bPg7] #
1.h5*g4[+wPg2] g2*h3[+bQd8] 2.Sh4*g6[+wPg2] Sh2*g4[+bPg7] #
1. Sf3? gxh5[+sBh7] 2. Dg4 Sxg4[+sDd8]#
R: 1. h7-h5 2. Sf3xBh2[+sBh7]
Five retro tries h#2 but not distinct:
1.Sh4-f3 g4*h5[+bPh7] 2.Qh3-g4 Sh2*g4[+bQd8] #
1.Sh4*g6[+wPg2] g2*h3[+bQd8] 2.h5*g4[+wPg2] Sh2*g4[+bPg7] #
1.Sh4*g6[+wPg2] g4*h5[+bPh7] 2.Qh3-g4 Sh2*g4[+bQd8] #
1.h5*g4[+wPg2] g2*h3[+bQd8] 2.Sh4-f3 Sh2*g4[+bPg7] #
1.h5*g4[+wPg2] g2*h3[+bQd8] 2.Sh4*g6[+wPg2] Sh2*g4[+bPg7] #
Klären ob Originalforderung h#2
Alfred Pfeiffer: Originalforderung war h#2. In der Lösungsbesprechung (fs-1, 01/1971, S.17) steht: "Korrekterweise sollte bei Aufgaben dieser Art ein Hinweis gegeben werden, daß sie Retroanalyse erfordern!".
Bei der Scheinlösung hat weiß keinen letzten Zug. (2017-11-23)
Henrik Juel: Giving a hint about retro analysis would spoil the fun (2017-11-23)
A.Buchanan: Quite right Henrik. Solvers should always be on their toes (2023-12-07)
comment
Alfred Pfeiffer: Originalforderung war h#2. In der Lösungsbesprechung (fs-1, 01/1971, S.17) steht: "Korrekterweise sollte bei Aufgaben dieser Art ein Hinweis gegeben werden, daß sie Retroanalyse erfordern!".
Bei der Scheinlösung hat weiß keinen letzten Zug. (2017-11-23)
Henrik Juel: Giving a hint about retro analysis would spoil the fun (2017-11-23)
A.Buchanan: Quite right Henrik. Solvers should always be on their toes (2023-12-07)
comment
Keywords: Circe, En passant as key, No legal last move for White
Genre: Fairies, Retro, h#
Computer test: laut WinChloe
FEN: 8/8/6P1/6Pp/6Pn/4P1pq/3prk1N/4bb1K
Input: HBae, 2017-11-23
Last update: A.Buchanan, 2023-12-07 more...
Genre: Fairies, Retro, h#
Computer test: laut WinChloe
FEN: 8/8/6P1/6Pp/6Pn/4P1pq/3prk1N/4bb1K
Input: HBae, 2017-11-23
Last update: A.Buchanan, 2023-12-07 more...
* 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.
48 - 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)
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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)
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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...
49 - 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...
1. a4xb3ep+ Sxa5+ 2. Kd5 0-0-0# (by AP not Td1#?)
R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7,Sd6xLc7
With the typo wPd5 corrected to e5, the two forward candidate solutions are correct. How about the retro? Assume that White castling rights remain. bBg5 is promoted, and did so on c1 or g1. In either case, there were 6 pawn caps by Black, but in the former case, the blockaded wRh was not consumable. So the promotion must have been on g8, and White can't retract h2-h3. The missing White unit is light B, so was not just captured on e5 or b4. wRf6 is promoted, say wPd, which would have involved no captures.
This leaves b2-b4, preceded by Kb4-c4+ and before that Sd6-b7. Prior to b2-b4, why couldn't bQ have just played e.g. Qa8-a5+ instead? Because there would still be no prior move: wSb7 can't have just come from h5 or e6. Similarly Rc3-a3 has no precedent.
We don't know exactly what happened to sLc, but that doesn’t affect the soundness. It might have enabled wPc or wPd to capture. wPd did promote, but maybe it was wPc that after capturing promoted to T. Alternatively, Sd6xLc7 was possible.
R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7,Sd6xLc7
With the typo wPd5 corrected to e5, the two forward candidate solutions are correct. How about the retro? Assume that White castling rights remain. bBg5 is promoted, and did so on c1 or g1. In either case, there were 6 pawn caps by Black, but in the former case, the blockaded wRh was not consumable. So the promotion must have been on g8, and White can't retract h2-h3. The missing White unit is light B, so was not just captured on e5 or b4. wRf6 is promoted, say wPd, which would have involved no captures.
This leaves b2-b4, preceded by Kb4-c4+ and before that Sd6-b7. Prior to b2-b4, why couldn't bQ have just played e.g. Qa8-a5+ instead? Because there would still be no prior move: wSb7 can't have just come from h5 or e6. Similarly Rc3-a3 has no precedent.
We don't know exactly what happened to sLc, but that doesn’t affect the soundness. It might have enabled wPc or wPd to capture. wPd did promote, but maybe it was wPc that after capturing promoted to T. Alternatively, Sd6xLc7 was possible.
Henrik Juel: Popeye 4.61 with 'opt enp b3' found no solution (2022-05-27)
Gerald Ettl: Schaut so aus, als ob die sDa5 nach e5 muss. 1.axb3 ep Sa5+ 2.Kxd5 0-0-0# (2022-05-27)
Gerald Ettl: aber auch 2.-Td1# ? (2022-05-27)
Henrik Juel: Andrew, can you throw any light on this problem? (2022-05-28)
Henrik Juel: Gerald, in AP problems like this one White must castle to legitimize the ep capture key (2022-05-28)
A.Buchanan: Hi Henrik, Gerald. I found this one in WinChloe with the current diagram. It gave the "solution" 1. axb3ep+ Sxa5+ 2. Kd5?? 0-0-0#. However when I tried running the WinChloe solver just now, it said there's no solution. Quite right: Rf5 controls d5. I don't think we can shift Qa5 to e5 though.
My first thought is that wPd5 should be on e5. Check my suggested solution in the solution text. However, as you will see, I think we need a second thought as well! Hope the finest retro minds can figure this out. Failing that, we could ask Igor. (2022-05-28)
Mario Richter: The diagram in the reprint in 'Uralsky Problemist' is exactly as given here. The solution given there without further explanation is: 1. ab+ (e.p.) Sxa5+ 2. Kd5 0-0-0#!
Notice that the second black move doesn't contain a capture sign!
Some typos in Andrew's analysis:
The black potential promotion squares should be c1 and g1.
White's last move couldn't have been b3-b4?? simply because that would give an illegal check. (2022-05-28)
A.Buchanan: Thanks Mario for the careful reading. I’ve fixed the errors. Pd5 may be a typo in the reprint then. It also creates a bogus retraction d4-d5, so I don’t think it can be right. Now for Igor. (2022-05-29)
A.Buchanan: Joaquim Crusats spotted R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7, which I'm sure is the intended retraction. So we diagnose a single typo that occurred in the Uralsky reprint diagram: wPe5 appeared on d5. I suggest correcting it here. (2022-05-30)
A.Buchanan: Author confirmed this was a typo (2022-11-23)
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Gerald Ettl: Schaut so aus, als ob die sDa5 nach e5 muss. 1.axb3 ep Sa5+ 2.Kxd5 0-0-0# (2022-05-27)
Gerald Ettl: aber auch 2.-Td1# ? (2022-05-27)
Henrik Juel: Andrew, can you throw any light on this problem? (2022-05-28)
Henrik Juel: Gerald, in AP problems like this one White must castle to legitimize the ep capture key (2022-05-28)
A.Buchanan: Hi Henrik, Gerald. I found this one in WinChloe with the current diagram. It gave the "solution" 1. axb3ep+ Sxa5+ 2. Kd5?? 0-0-0#. However when I tried running the WinChloe solver just now, it said there's no solution. Quite right: Rf5 controls d5. I don't think we can shift Qa5 to e5 though.
My first thought is that wPd5 should be on e5. Check my suggested solution in the solution text. However, as you will see, I think we need a second thought as well! Hope the finest retro minds can figure this out. Failing that, we could ask Igor. (2022-05-28)
Mario Richter: The diagram in the reprint in 'Uralsky Problemist' is exactly as given here. The solution given there without further explanation is: 1. ab+ (e.p.) Sxa5+ 2. Kd5 0-0-0#!
Notice that the second black move doesn't contain a capture sign!
Some typos in Andrew's analysis:
The black potential promotion squares should be c1 and g1.
White's last move couldn't have been b3-b4?? simply because that would give an illegal check. (2022-05-28)
A.Buchanan: Thanks Mario for the careful reading. I’ve fixed the errors. Pd5 may be a typo in the reprint then. It also creates a bogus retraction d4-d5, so I don’t think it can be right. Now for Igor. (2022-05-29)
A.Buchanan: Joaquim Crusats spotted R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7, which I'm sure is the intended retraction. So we diagnose a single typo that occurred in the Uralsky reprint diagram: wPe5 appeared on d5. I suggest correcting it here. (2022-05-30)
A.Buchanan: Author confirmed this was a typo (2022-11-23)
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Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Promotion (Tl), Valladao Task
Genre: h#, Retro
Computer test: Once the diagram typo was fixed, C+ Popeye v4.87 & basic but tricky retro logic
FEN: 3n4/1Np1p1p1/4pn2/q1p1PRb1/pPk1pr2/r6P/P4PP1/R3K3
Reprints: 18 The Ural's Problemist 13, p. 5, 02/1998
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2022-05-30 more...
Genre: h#, Retro
Computer test: Once the diagram typo was fixed, C+ Popeye v4.87 & basic but tricky retro logic
FEN: 3n4/1Np1p1p1/4pn2/q1p1PRb1/pPk1pr2/r6P/P4PP1/R3K3
Reprints: 18 The Ural's Problemist 13, p. 5, 02/1998
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2022-05-30 more...
52 - 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)
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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)
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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...
53 - P1406941
Marco Bonavoglia
Best Problems 105, p. 561, 01/2023
(4+4)
h#2 a Posteriori
Ultrapatrouilleschach
Marco Bonavoglia
Best Problems 105, p. 561, 01/2023
(4+4)
h#2 a Posteriori
Ultrapatrouilleschach
1. dxe3ep 0-0-0+! 2. Kf7 d8=S#
R: 1. e2-e4
Autor: "Le ultime mosse sono state del Re bianco (protetto dalla Ta1) o Be2-e4 protetto dal Ke1. Il nero prende en passant e il bianco deve giustificarlo a posteriori arroccando. Tema Valladao of course." [Zuletzt zog der weiße König (geschützt durch Ta1) oder Be2-e4 geschützt durch Ke1. [mri: oder Td1-a1] Schwarz schlägt en passant und Weiß muss es anschließend durch die Rochade rechtfertigen. Valladao-Thema natürlich.]
R: 1. e2-e4
Autor: "Le ultime mosse sono state del Re bianco (protetto dalla Ta1) o Be2-e4 protetto dal Ke1. Il nero prende en passant e il bianco deve giustificarlo a posteriori arroccando. Tema Valladao of course." [Zuletzt zog der weiße König (geschützt durch Ta1) oder Be2-e4 geschützt durch Ke1. [mri: oder Td1-a1] Schwarz schlägt en passant und Weiß muss es anschließend durch die Rochade rechtfertigen. Valladao-Thema natürlich.]
Posthum innerhalb eines Nachrufes auf Marco Bonavoglia veröffentlicht.
Antonio Garofalo: "Marco, mentre era in ospedale, mi ha inviato questi due problemi, uno augurale, l'altro per la competizione fairies. Sfortunatamente egli non può vedere i suoi lavori pubblicati, essendo deceduto nelle prime ore del 5 dicembre 2022..." [Marco schickte mir, während er im Krankenhaus war, diese beiden Probleme [dieses und P1406940], eines für seine [Neujahrs-]Wünsche (P1406940), das andere für den Wettbewerb, Abteilung Märchenschach. Leider kann er seine veröffentlichten Werke nicht mehr sehen, denn er verstarb in den frühen Morgenstunden des 5. Dezember 2022.]
Joost de Heer: C+ Popeye with 'opt enp e2e3e4'.
Too bad there isn't a try without castling. (2023-04-04)
comment
Antonio Garofalo: "Marco, mentre era in ospedale, mi ha inviato questi due problemi, uno augurale, l'altro per la competizione fairies. Sfortunatamente egli non può vedere i suoi lavori pubblicati, essendo deceduto nelle prime ore del 5 dicembre 2022..." [Marco schickte mir, während er im Krankenhaus war, diese beiden Probleme [dieses und P1406940], eines für seine [Neujahrs-]Wünsche (P1406940), das andere für den Wettbewerb, Abteilung Märchenschach. Leider kann er seine veröffentlichten Werke nicht mehr sehen, denn er verstarb in den frühen Morgenstunden des 5. Dezember 2022.]
Joost de Heer: C+ Popeye with 'opt enp e2e3e4'.
Too bad there isn't a try without castling. (2023-04-04)
comment
Keywords: En passant as key, Ultrapatrol, Valladao Task
Genre: h#, Retro, Fairies
Computer test: laut 'Best Problems' C+, aber Prüfprogramm nicht angegeben
FEN: 4k2r/3P4/2n5/8/3pP3/8/8/R3K3
Input: Mario Richter, 2022-12-30
Last update: Mario Richter, 2022-12-30 more...
Genre: h#, Retro, Fairies
Computer test: laut 'Best Problems' C+, aber Prüfprogramm nicht angegeben
FEN: 4k2r/3P4/2n5/8/3pP3/8/8/R3K3
Input: Mario Richter, 2022-12-30
Last update: Mario Richter, 2022-12-30 more...
54 - P1408845
Andrew Buchanan
Phénix 11/2021
after J-F. Baudoin & J-C. Gandy
(9+13)
-1s -1w -1s, dann h#1
Andrew Buchanan
Phénix 11/2021
after J-F. Baudoin & J-C. Gandy
(9+13)
-1s -1w -1s, dann h#1
R: 1. ... h2-h1=S+ 2. a7-a8=L Kc8-d8 dann 1. h1=T a8=D#
AB: Mario was kind enough to check this 2-Jul-2021. I hope it's ok if I quote his email:
The trick with the light-squared black Bishop is fine, but makes things too complicated for rawbats. It will happily retract e.g. R: 1. ... h2-h1=S+ 2. Lb7-a8 Ke7xDd8, dann 1. Kd6 Df6# without noticing that the retractions lead to a conflict on the g-file: either wPh2 captured the original Bishop c8 on the g-file, then bPg7 was not available a a capture object for the remaining white pawn captures, or wPh2 captured the original bPg7 on the g-file, then the original light-squared black bishop c8 was not available a capture object for the remaining white pawn captures f2xe3xd4 (both on dark squares).
So brain-checking your retractor and your logical argumentation I found no defects, and I think the problem is sound now
AB: Mario was kind enough to check this 2-Jul-2021. I hope it's ok if I quote his email:
The trick with the light-squared black Bishop is fine, but makes things too complicated for rawbats. It will happily retract e.g. R: 1. ... h2-h1=S+ 2. Lb7-a8 Ke7xDd8, dann 1. Kd6 Df6# without noticing that the retractions lead to a conflict on the g-file: either wPh2 captured the original Bishop c8 on the g-file, then bPg7 was not available a a capture object for the remaining white pawn captures, or wPh2 captured the original bPg7 on the g-file, then the original light-squared black bishop c8 was not available a capture object for the remaining white pawn captures f2xe3xd4 (both on dark squares).
So brain-checking your retractor and your logical argumentation I found no defects, and I think the problem is sound now
Keywords: Help retractor, Allumwandlung
Genre: Retro, h#
FEN: B2k4/2pp4/8/8/1PPP4/1rpP2PN/1pppPK2/nrbq3n
Input: A.Buchanan, 2023-04-13
Last update: A.Buchanan, 2024-01-21 more...
Genre: Retro, h#
FEN: B2k4/2pp4/8/8/1PPP4/1rpP2PN/1pppPK2/nrbq3n
Input: A.Buchanan, 2023-04-13
Last update: A.Buchanan, 2024-01-21 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...
56 - P1409850
Andrew Buchanan
H1327 Problem Paradise 97, p. 11, 01-03/2022
(3+13) C+
h#3
b) sSb5<->sBa7
Andrew Buchanan
H1327 Problem Paradise 97, p. 11, 01-03/2022
(3+13) C+
h#3
b) sSb5<->sBa7
a) 1. ... axb7 2. Kc7 Kd5 3. Kd8 b8=D#
b) 1. Tc7 e7 2. Kd7 Kd5 3. Kc8 e8=D#
Unique set play retro try in the retro twin s). 1.Kc7 axb7 2.Kd8 b8=D#. Zero set play in b).
All pieces are necessary for soundness in at least one twin except for sBa4 ("the manager").
b) 1. Tc7 e7 2. Kd7 Kd5 3. Kc8 e8=D#
Unique set play retro try in the retro twin s). 1.Kc7 axb7 2.Kd8 b8=D#. Zero set play in b).
All pieces are necessary for soundness in at least one twin except for sBa4 ("the manager").
First publication had typo in stipulation
Henrik Juel: If you disregard retro considerations and blindly test with Popeye 4.61, you get lots of cooks in part a) and 1.Re7-c7 e6-e7 2.Kc6-d7 Ke4-d5 3.Kd7-c8 e7-e8=Q #
in part b) (2023-05-20)
Henrik Juel: Part b) is OK, because last move could be Kd4-e4, and a normal h#3 ensues
But in part a) White (not Black as intimated by the Keyword) has no last move, and the stipulation changes to h#2.5 with the solution 1...a6*b7 2.Kc6-c7 Ke4-d5 3.Kc7-d8 b7-b8=Q #
Nice symbol problem (2023-05-20)
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Henrik Juel: If you disregard retro considerations and blindly test with Popeye 4.61, you get lots of cooks in part a) and 1.Re7-c7 e6-e7 2.Kc6-d7 Ke4-d5 3.Kd7-c8 e7-e8=Q #
in part b) (2023-05-20)
Henrik Juel: Part b) is OK, because last move could be Kd4-e4, and a normal h#3 ensues
But in part a) White (not Black as intimated by the Keyword) has no last move, and the stipulation changes to h#2.5 with the solution 1...a6*b7 2.Kc6-c7 Ke4-d5 3.Kc7-d8 b7-b8=Q #
Nice symbol problem (2023-05-20)
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Keywords: No legal last move for White, Character problem (PP)
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 and simple analysis
FEN: 8/pp2rp2/P1k1P1p1/pn2nr2/p3K3/q3b3/8/8
Reprints: OP015 The Hopper Magazine I02 25/06/2022
Input: A.Buchanan, 2023-05-20
Last update: A.Buchanan, 2023-12-06 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 and simple analysis
FEN: 8/pp2rp2/P1k1P1p1/pn2nr2/p3K3/q3b3/8/8
Reprints: OP015 The Hopper Magazine I02 25/06/2022
Input: A.Buchanan, 2023-05-20
Last update: A.Buchanan, 2023-12-06 more...
1. Dxa2 Th4 2. Da4 Txa4#
1. Dxb2 0-0?? 2. Da3 Lxa3# because Ke1 moved to let Ka1 in
1. Dxb2 0-0?? 2. Da3 Lxa3# because Ke1 moved to let Ka1 in
1. ... d8=S 2. a1=T gxh8=D#
R: 1. b2-b1=L
R: 1. b2-b1=L
Yuri Bilokin: more economical -wBg5, -wSh7, -wPa3, bQh8(-bRh8) 6kq/3P1NP1/8/8/8/1K6/p1P5/1b6 (5+4) H#1.5
1...d8=S 2.a1=R gxh8=Q# (MM)
Tempo promotion
Promotion (QSr, 2, 1)
Tempo move (bP, waiting, type 1)
Model mate × 1 (2023-08-04)
Henrik Juel: C+ Popeye 4.61 (2023-08-04)
A.Buchanan: heh Yuri the 4th promotion is in the retraction: R: 1. b2-b1=L. The other pieces serve to force retro-uniqueness. (2023-08-04)
Olaf Jenkner: Vergleiche mit P1190770 oder P1190771 (2023-08-04)
A.Buchanan: Very nice! These remind me of my own P1408845. (2023-08-05)
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1...d8=S 2.a1=R gxh8=Q# (MM)
Tempo promotion
Promotion (QSr, 2, 1)
Tempo move (bP, waiting, type 1)
Model mate × 1 (2023-08-04)
Henrik Juel: C+ Popeye 4.61 (2023-08-04)
A.Buchanan: heh Yuri the 4th promotion is in the retraction: R: 1. b2-b1=L. The other pieces serve to force retro-uniqueness. (2023-08-04)
Olaf Jenkner: Vergleiche mit P1190770 oder P1190771 (2023-08-04)
A.Buchanan: Very nice! These remind me of my own P1408845. (2023-08-05)
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Keywords: Allumwandlung, Last Move?
Genre: h#, Retro
Computer test: HC+ forward popeye v4.61 retro thinking
FEN: 6kr/3P1NPN/8/6B1/8/PK6/p1P5/1b6
Input: Miguel Ambrona, 2023-07-27
Last update: A.Buchanan, 2023-08-04 more...
Genre: h#, Retro
Computer test: HC+ forward popeye v4.61 retro thinking
FEN: 6kr/3P1NPN/8/6B1/8/PK6/p1P5/1b6
Input: Miguel Ambrona, 2023-07-27
Last update: A.Buchanan, 2023-08-04 more...
1. exd3ep Lg8+ 2. d5 cxd6ep#
1. axb3ep bxc6+ 2. b5 cxb6ep#
PRA means one problem two parts
Cook: R: 1. Kf6-e5 Ld6-f8+
1. axb3ep bxc6+ 2. b5 cxb6ep#
PRA means one problem two parts
Cook: R: 1. Kf6-e5 Ld6-f8+
see P0003365
Henrik Juel: Intention:
White captured axSb and [Lc8]
If last move was b2-b4: 1.axb3ep bxc6+ 2.b5 cxb6ep#
If last move was d2-d4: 1.exd3ep Lb8+ 2.d5 cxd6ep# (2023-08-08)
Henrik Juel: Cook: last move could also be Kf6-e5
correction, e.g., move Sh3 to h5 (2023-08-08)
A.Buchanan: Good catch Henrik - I should have looked at this one more carefully! Luckily it admits an easy fix, but there is still the non-standard material. (2023-08-08)
Henrik Juel: I should have thought of a better birthday greeting, Andrew, but there you are (2023-08-08)
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Henrik Juel: Intention:
White captured axSb and [Lc8]
If last move was b2-b4: 1.axb3ep bxc6+ 2.b5 cxb6ep#
If last move was d2-d4: 1.exd3ep Lb8+ 2.d5 cxd6ep# (2023-08-08)
Henrik Juel: Cook: last move could also be Kf6-e5
correction, e.g., move Sh3 to h5 (2023-08-08)
A.Buchanan: Good catch Henrik - I should have looked at this one more carefully! Luckily it admits an easy fix, but there is still the non-standard material. (2023-08-08)
Henrik Juel: I should have thought of a better birthday greeting, Andrew, but there you are (2023-08-08)
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Keywords: Partial Retro Analysis (PRA), En passant as key (2), En passant as mating move (2), Non-standard material (L), Superseded by (P1413906)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 3r1b2/1p1p3B/B1r3p1/1PP1Kp2/pPkPp3/p1p4n/N7/2q5
Reprints: Schach-Echo , p. 324, 08/1984
23 125 ausgewählte Schachprobleme , p. 10, 1985
2 Problemas 44, p. 1456, 10/2023
Input: A.Buchanan, 2023-08-08
Last update: A.Buchanan, 2023-12-04 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 3r1b2/1p1p3B/B1r3p1/1PP1Kp2/pPkPp3/p1p4n/N7/2q5
Reprints: Schach-Echo , p. 324, 08/1984
23 125 ausgewählte Schachprobleme , p. 10, 1985
2 Problemas 44, p. 1456, 10/2023
Input: A.Buchanan, 2023-08-08
Last update: A.Buchanan, 2023-12-04 more...
1. axb3ep bxc6+ 2. b5 cxb6ep#
1. exd3ep Dxg8+ 2. d5 cxd6ep#
Wh captures axb, XxLc1.
Bl captures cxb/dxc, hxg, gxh/f and either fxe or wBe is waylaid. So the position is legal, with Wh last move being pawn double hop. In particular wKf6-e5 was not the last move
PRA, last move was either b2-b4 or d2-d4. So there is "one solution with two parts".
1. exd3ep Dxg8+ 2. d5 cxd6ep#
Wh captures axb, XxLc1.
Bl captures cxb/dxc, hxg, gxh/f and either fxe or wBe is waylaid. So the position is legal, with Wh last move being pawn double hop. In particular wKf6-e5 was not the last move
PRA, last move was either b2-b4 or d2-d4. So there is "one solution with two parts".
See P0003365 (cooked) & P1411659 (fixably cooked but non-standard thematic material).
This new version has an obvious promoted unit (non-thematic), but Joaquim Crusats (a noted constructor) could not find any way to do better
Ladislav Packa: Mr. Bebesi's first name is Gyula (not Gyulia). It is the Hungarian version of the name Julius. (2023-12-05)
A.Buchanan: Thanks - edited (2023-12-06)
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This new version has an obvious promoted unit (non-thematic), but Joaquim Crusats (a noted constructor) could not find any way to do better
Ladislav Packa: Mr. Bebesi's first name is Gyula (not Gyulia). It is the Hungarian version of the name Julius. (2023-12-05)
A.Buchanan: Thanks - edited (2023-12-06)
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Keywords: Partial Retro Analysis (PRA), Obvious promotion (l), En passant as key (2), En passant as mating move
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 3rn1bQ/1p1p2PR/B1r3pP/1PP1K2n/pPkPp3/b1p5/N7/2q5
Reprints: SuperProblem (Website) 06/12/2023
Input: A.Buchanan, 2023-12-04
Last update: A.Buchanan, 2023-12-06 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 3rn1bQ/1p1p2PR/B1r3pP/1PP1K2n/pPkPp3/b1p5/N7/2q5
Reprints: SuperProblem (Website) 06/12/2023
Input: A.Buchanan, 2023-12-04
Last update: A.Buchanan, 2023-12-06 more...
61 - 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)
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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)
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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...
62 - 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)
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"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)
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https://pdb.dieschwalbe.de/search.jsp?expression=COMMENTDATE%3E%3D20220810+AND+G%3D%27h%23%27+AND+NOT+A%3D%27Hage%2C+Poul%27+AND+G%3D%27Retro%27
The problems of this query have been registered by the following contributors:
Gerd Wilts (35)Michal Dragoun (2)
hpr (3)
Felber, Volker (1)
Henri Nouguier (1)
A.Buchanan (15)
HBae (1)
Mario Richter (2)
Miguel Ambrona (2)
Preisbericht: 'Die Schwalbe' 06/2011 S.124 (2023-01-02)
Henrik Juel: How is the SE corner released, without ruining the castling? (2023-01-02)
Mario Richter: Good question, Henrik! I first thought that releasing the SE corner without ruining White's castling right is impossible, but the trick is to uncapture a black Queen in the SE corner at the right moment.
Perhaps Theodore Hwa can use ths problem as a test case for his latest improvement to Retractor 2 ... (2023-01-02)
Henrik Juel: Thanks, Mario
In view of the prize I suspected that the problem was correct, but I did not find the uncapture trick (2023-01-02)
Henrik Juel: C+ Popeye 4.61, because with Black to move White may not castle (2023-01-02)
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