929 problem(s) found in 4498 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20220810 AND G='h#' AND NOT K='vorweggenommen' AND G='h#'] [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)
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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)
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Ladislav Packa: Retro content is not needed, the solution is preserved even without it.
Pg4 Pg2 Sh2 Ke1 Rh1 (5)- Ph5 Kg3 (2) h#2* C+
1...0-0 2.h5-h4 Rf1-f3 #
1.Kg3-h4 Ke1-f2 2.h5*g4 Sh2-f3 # (2023-09-14)
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*) 1. ... 0-0#
1) 1. Dc5 Tf1+ 2. Df2 Txf2#
1) 1. Dc5 Tf1+ 2. Df2 Txf2#
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...
1. ... e5 2. g4 Le4#
1. Lxf4 e5 2. Lh2 Le4#
1. Lxf4 e5 2. Lh2 Le4#
A.Buchanan: Setplay doesn't care whether the position is legal. (2023-12-06)
A.Buchanan: So is this a retro problem? I think not. (2023-12-06)
comment
A.Buchanan: So is this a retro problem? I think not. (2023-12-06)
comment
Keywords: No legal last move for Black (setplay doesn't care), Tempo Move
Genre: h#
Computer test: HC+ Popeye v4.87 & trivial retro-logic
FEN: 8/8/5PBP/6pP/4PP2/8/5K1b/7k
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
Genre: h#
Computer test: HC+ Popeye v4.87 & trivial retro-logic
FEN: 8/8/5PBP/6pP/4PP2/8/5K1b/7k
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
34 - 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)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Promotion (S), Valladao Task
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 + simple retro-logic
FEN: 8/3P2p1/2PP4/1ppPp3/2pkPp2/5PP1/6Pp/R3K1b1
Input: Gerd Wilts, 1999-02-27
Last update: A.Buchanan, 2023-09-11 more...
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 + simple retro-logic
FEN: 8/3P2p1/2PP4/1ppPp3/2pkPp2/5PP1/6Pp/R3K1b1
Input: Gerd Wilts, 1999-02-27
Last update: A.Buchanan, 2023-09-11 more...
36 - P0500385
Zdravko Maslar
3371 Die Schwalbe 67, p. 209, 02/1981
2. ehrende Erwähnung
(6+9) C+
h#3
Zdravko Maslar
3371 Die Schwalbe 67, p. 209, 02/1981
2. ehrende Erwähnung
(6+9) C+
h#3
1. Sf3 Th1 2. Dxh1+ Kxg3 3. Da1 d4#
zunächst für die 3. ehrende Erwähnung vorgesehen, wegen Disqualifikation von P0532932 hochgerutscht (vergl. 'Die Schwalbe' 08/1984 S.308)
Anton Baumann: Nachträglich zur 2. ehrenden Erwähnung aufgerückt (vergl. 'Die Schwalbe' 08/1984 S.308) (2022-12-23)
comment
Anton Baumann: Nachträglich zur 2. ehrenden Erwähnung aufgerückt (vergl. 'Die Schwalbe' 08/1984 S.308) (2022-12-23)
comment
Keywords: Sacrifice of white pieces
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/5p2/3P1kpp/6pq/p2P2p1/P1B3Kn/R7
Reprints: 773 FIDE Album 1980-1982 1988
Input: Gerd Wilts, 1996-06-06
Last update: Mario Richter, 2022-12-23 more...
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/5p2/3P1kpp/6pq/p2P2p1/P1B3Kn/R7
Reprints: 773 FIDE Album 1980-1982 1988
Input: Gerd Wilts, 1996-06-06
Last update: Mario Richter, 2022-12-23 more...
37 - P0500421
Reto List
(5) Die Schwalbe 84 12/1983
3. ehrende Erwähnung
(4+12) cooked
h#2
b) wSg2 nach h6
Reto List
(5) Die Schwalbe 84 12/1983
3. ehrende Erwähnung
(4+12) cooked
h#2
b) wSg2 nach h6
a) 1. Sh5 Tf4 2. Tg3 Txh4#
b) 1. Sf5 Lf3 2. Lg3 Lg4#
Cook: NL a)
1. La1 Ka7 2. Lg4 Sf4#
1. Lg4+ Ka7 2. Lf4 Sxf4#
b) 1. Sf5 Lf3 2. Lg3 Lg4#
Cook: NL a)
1. La1 Ka7 2. Lg4 Sf4#
1. Lg4+ Ka7 2. Lf4 Sxf4#
Kees: possible fix: +wBa7 +sSa5 (2023-06-07)
A.Buchanan: This fix certainly removes the cooks, but the twinning in the original problem feels clumsy. (2023-06-08)
comment
A.Buchanan: This fix certainly removes the cooks, but the twinning in the original problem feels clumsy. (2023-06-08)
comment
Genre: h#
FEN: K1b4q/6pr/3p1pp1/4b3/7p/3r2nk/5RN1/7B
Input: Gerd Wilts, 1996-06-06
Last update: Alfred Pfeiffer, 2019-04-26 more...
Weiter NL!
klären Lösung
Anton Baumann: mögliche Korrektur: wLg2 nach h1, -sLf1, -sSa1, sBc7 nach f7 (3+12)
1.Kb6 Tg3 2.Ka7 Tg8 3.Tcb6 Ta8#
1.Ta6 Txc3+ 2.Kb6 Tc7 3.Lc5 Tb7#
1.Tb5 Td3 2.Db4 Txd6 3.Ld4 Txc6#
1.Kd4 Te3 2.Tc4 Txe5 3.Le3 Td5#
C+ Gustav 4.2a (2023-12-30)
comment
Anton Baumann: mögliche Korrektur: wLg2 nach h1, -sLf1, -sSa1, sBc7 nach f7 (3+12)
1.Kb6 Tg3 2.Ka7 Tg8 3.Tcb6 Ta8#
1.Ta6 Txc3+ 2.Kb6 Tc7 3.Lc5 Tb7#
1.Tb5 Td3 2.Db4 Txd6 3.Ld4 Txc6#
1.Kd4 Te3 2.Tc4 Txe5 3.Le3 Td5#
C+ Gustav 4.2a (2023-12-30)
comment
Genre: h#
FEN: 8/2p5/2rp1p2/p1k1p3/5q2/2p2R1K/5bB1/nrn2b2
Input: Gerd Wilts, 1996-06-06
Last update: hpr, 1999-05-17 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. f1=S Kf4 2. Se3 dxe3 3. d2 e4 4. d1=S e5 5. Se3 e6 6. Sf5 e7 7. Sg7 exf8=S#
Unterverwandlung!Springer!3-fach
Yuri Bilokin: more economical bNh6-f6(-bPf6), bPh5-d5, then a1=b1 6n1/6b1/6nk/4p3/8/4p1K1/4P1p1/8 (2+7) h#7
1.g1=S Kg4 2.Sf3 exf3 3.e2 f4 4.e1=S f5 5.Sf3 f6 6.Sg5 f7 7.Sh7 fxg8=S# (MM)
Slow Excelsior. Active sacrifice (black). Annihilation.
Excelsior (white). Long-trip (bS, promoted, 3) Phoenix. Promotion (Sss, 1, 2)
Model mate × 1 (2023-05-10)
more ...
comment
Yuri Bilokin: more economical bNh6-f6(-bPf6), bPh5-d5, then a1=b1 6n1/6b1/6nk/4p3/8/4p1K1/4P1p1/8 (2+7) h#7
1.g1=S Kg4 2.Sf3 exf3 3.e2 f4 4.e1=S f5 5.Sf3 f6 6.Sg5 f7 7.Sh7 fxg8=S# (MM)
Slow Excelsior. Active sacrifice (black). Annihilation.
Excelsior (white). Long-trip (bS, promoted, 3) Phoenix. Promotion (Sss, 1, 2)
Model mate × 1 (2023-05-10)
more ...
comment
Keywords: Excelsior white (auto key), under-promotion
Genre: h#
Computer test: C+ PY V3.52.--
FEN: 5n2/5p2/5pkn/7p/8/3p1K2/3P1p2/8
Reprints: 198 Ideal-Mate Encyclopedia Vol.1 1999
Input: Gerd Wilts, 1996-06-06
Last update: Erich Bartel, 2011-07-06 more...
Genre: h#
Computer test: C+ PY V3.52.--
FEN: 5n2/5p2/5pkn/7p/8/3p1K2/3P1p2/8
Reprints: 198 Ideal-Mate Encyclopedia Vol.1 1999
Input: Gerd Wilts, 1996-06-06
Last update: Erich Bartel, 2011-07-06 more...
1) 1. Lc3 Df6 2. Td4 Df1 3. Dc4 Db1#
2) 1. Lf7+ Kh7 2. Dc3 Kh6 3. Lc4 Df5#
2) 1. Lf7+ Kh7 2. Dc3 Kh6 3. Lc4 Df5#
Grimshaw
Yuri Bilokin: more economical bPg2-f3, bRd8-g2, bBg8-h5(-bNh5), -bPb5 7K/2qr4/8/4Q2b/3b4/3kpp2/3pn1r1/8 (2+10) 2.1…
1.Bc3 Qf6 2.Rd4 Qb6 3.Qc4 Qb1# (MM)
1.Bf7 Kh7 2.Qc3 Kh6 3.Bc4 Qf5# (MM)
Amazon theme. Blocking piece replacement (bB-bR)
Check prevention (W-W). Hideaway maneuver (wK, 2)
Long-trip (wQ, 3). Pelle move (white)
Play on the same square (B3, 2)
Model mate × 2 (2024-02-09)
comment
Yuri Bilokin: more economical bPg2-f3, bRd8-g2, bBg8-h5(-bNh5), -bPb5 7K/2qr4/8/4Q2b/3b4/3kpp2/3pn1r1/8 (2+10) 2.1…
1.Bc3 Qf6 2.Rd4 Qb6 3.Qc4 Qb1# (MM)
1.Bf7 Kh7 2.Qc3 Kh6 3.Bc4 Qf5# (MM)
Amazon theme. Blocking piece replacement (bB-bR)
Check prevention (W-W). Hideaway maneuver (wK, 2)
Long-trip (wQ, 3). Pelle move (white)
Play on the same square (B3, 2)
Model mate × 2 (2024-02-09)
comment
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 3r2bK/2qr4/8/1p2Q2n/3b4/3kp3/3pn1p1/8
Input: Gerd Wilts, 1996-06-06
Last update: hpr, 1999-05-20 more...
42 - P0501066
Henry Olof Axel Forsberg
599-603 Revista Romana de Sah 01/1935
1. Preis
Pauly Gedenkturnier 1935
(3+2) C+
h#2
b) sTa6 statt sDa6
c) sLa6 statt sDa6
d) sSa6 statt sDa6
e) sBa6 statt sDa6
Henry Olof Axel Forsberg
599-603 Revista Romana de Sah 01/1935
1. Preis
Pauly Gedenkturnier 1935
(3+2) C+
h#2
b) sTa6 statt sDa6
c) sLa6 statt sDa6
d) sSa6 statt sDa6
e) sBa6 statt sDa6
a) 1. Df6 Sc5 2. Db2 Ta4#
b) 1. Tb6 Tb1 2. Tb3 Ta1#
c) 1. Lc4 Se1 2. La2 Sc2#
d) 1. Sc5 Sc1 2. Sa4 Tb3#
e) 1. a5 Tb3+ 2. Ka4 Sc5#
b) 1. Tb6 Tb1 2. Tb3 Ta1#
c) 1. Lc4 Se1 2. La2 Sc2#
d) 1. Sc5 Sc1 2. Sa4 Tb3#
e) 1. a5 Tb3+ 2. Ka4 Sc5#
Das namensgebende [Stamm-]Problem für dieses Thema (=Forsberg-Zwilling).
Erich Bartel: weitere Nachdrucke:
13) 197 moving on 2008.---
14) Naef:Endspielstudien Hilfmattprobleme und Märchenschachaufgaben 1998,S.72.--- (2008-10-01)
Jacques Rotenberg: 5982, CHM 1995, G. Lestriguel
p.1315, Le Guide des Echecs 1993
5 p.8, The Modern Helpmate in Two, Z. Janevski & N. Stolev 1989
XIII p.151, Thèmes-64 10 (avr. 58)
XII bis p.38, Rex Multiplex 2 (avr. 82)
285, Svenska Miniatyrer I, A.Uddgren & A. Hildebrand 1996
p.76, Im Wunderland des Schachproblems, E. Ramin 1958
E-1 p.236, StrateGems 12 (oct. 00)
I33 p.1333, Phénix 14 (sept. 91)
3 p.219, Le Genre Aidé Supplément diagrammes 15 (juin 75)
161 p.241, Echecs, N. Ramini 1985
E p.14, Best Problems 21 (jan. 02)
259 p.238, Chess Problems : Introduction to an Art, M. Lipton, R.C.O. Matthews, J. Rice 1963
139, Skakopgaven 1942
135 p.31, Album FIDE 1914-44/III
9 p.403, feenschach 26 (déc. 74)
40 p.61, Europe Echecs 330 (juin 86)
p.185, feenschach 47 (juil. 79)
in WinChloe, you find also :
A p.5, Springaren 17 (mars 84)
p.50, feenschach 73 (jan. 85)
M p.171, The Problemist (sept. 81)
p.415, feenschach 97 (nov. 90)
133 p.104, Solving in Style, J. Nunn 1985
E-I p.43, Probleemblad 58/1 (jan. 00)
E-1 p.102, Probleemblad 58/3 (mai 00)
p.136, Springaren 70 (août 97)
4 p.4424, diagrammes 159 (oct. 06)
p.37, Problem-Forum 36 (déc. 08) (2012-03-10)
A.Buchanan: The position is unique (except for left-right reflection). In particular, it cannot be shifted up or down without breaking (b) and only the current location of wK makes (b) sound and (a)&(d) solvable. (2017-01-27)
Anton Baumann: in Revista de Sah 1936 soll Henry Forsberg eine weitere Möglichkeit gezeigt haben:
f) wSa6 statt sDa6: 1.Ka2 Sac5 2.Ka3 Ta4# (C+)
Ferner mit Märchenschachfiguren: vergl. P1349128 (2022-12-28)
more ...
comment
Erich Bartel: weitere Nachdrucke:
13) 197 moving on 2008.---
14) Naef:Endspielstudien Hilfmattprobleme und Märchenschachaufgaben 1998,S.72.--- (2008-10-01)
Jacques Rotenberg: 5982, CHM 1995, G. Lestriguel
p.1315, Le Guide des Echecs 1993
5 p.8, The Modern Helpmate in Two, Z. Janevski & N. Stolev 1989
XIII p.151, Thèmes-64 10 (avr. 58)
XII bis p.38, Rex Multiplex 2 (avr. 82)
285, Svenska Miniatyrer I, A.Uddgren & A. Hildebrand 1996
p.76, Im Wunderland des Schachproblems, E. Ramin 1958
E-1 p.236, StrateGems 12 (oct. 00)
I33 p.1333, Phénix 14 (sept. 91)
3 p.219, Le Genre Aidé Supplément diagrammes 15 (juin 75)
161 p.241, Echecs, N. Ramini 1985
E p.14, Best Problems 21 (jan. 02)
259 p.238, Chess Problems : Introduction to an Art, M. Lipton, R.C.O. Matthews, J. Rice 1963
139, Skakopgaven 1942
135 p.31, Album FIDE 1914-44/III
9 p.403, feenschach 26 (déc. 74)
40 p.61, Europe Echecs 330 (juin 86)
p.185, feenschach 47 (juil. 79)
in WinChloe, you find also :
A p.5, Springaren 17 (mars 84)
p.50, feenschach 73 (jan. 85)
M p.171, The Problemist (sept. 81)
p.415, feenschach 97 (nov. 90)
133 p.104, Solving in Style, J. Nunn 1985
E-I p.43, Probleemblad 58/1 (jan. 00)
E-1 p.102, Probleemblad 58/3 (mai 00)
p.136, Springaren 70 (août 97)
4 p.4424, diagrammes 159 (oct. 06)
p.37, Problem-Forum 36 (déc. 08) (2012-03-10)
A.Buchanan: The position is unique (except for left-right reflection). In particular, it cannot be shifted up or down without breaking (b) and only the current location of wK makes (b) sound and (a)&(d) solvable. (2017-01-27)
Anton Baumann: in Revista de Sah 1936 soll Henry Forsberg eine weitere Möglichkeit gezeigt haben:
f) wSa6 statt sDa6: 1.Ka2 Sac5 2.Ka3 Ta4# (C+)
Ferner mit Märchenschachfiguren: vergl. P1349128 (2022-12-28)
more ...
comment
Keywords: Forsberg Twins, Aristocrat, Model mate
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/q7/8/1R4K1/k2N4/8/8
Reprints: 6 The Chess Review , p. 25, 12/1945
141 Zwillinge und Mehrlinge 1952
Schach-Echo Vol.11, p. 151, 1953
5 The Problemist 01/1962
761 Szachy 06/1965
Problemschach [Sidler] , p. 140, 1968
Problem 127-132, p. 95, 09/1969
9) feenschach 26, p. 403, 12/1974
135 FIDE Album 1914-1944/III 1975
83 Schach ohne Partner [Grasemann] 1977
feenschach 47, p. 185, 07-09/1979
XII Rex Multiplex 2, p. 38, 04-06/1982
67 100 Classics of the Chessboard 1983
Belfort , p. 75, 1994
Tim Krabbé's Website (57) 2000
E-I Probleemblad 01-02/2000
harmonie 63 09/2000
König & Turm 4, p. 13, 03/2001
986 Minimalkunst im Schach 2006
Problemkiste 163 02/2006
633 Encyclopedia of Chess Problems 2012
Input: Gerd Wilts, 1996-06-06
Last update: Dieter Berlin, 2022-02-09 more...
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/q7/8/1R4K1/k2N4/8/8
Reprints: 6 The Chess Review , p. 25, 12/1945
141 Zwillinge und Mehrlinge 1952
Schach-Echo Vol.11, p. 151, 1953
5 The Problemist 01/1962
761 Szachy 06/1965
Problemschach [Sidler] , p. 140, 1968
Problem 127-132, p. 95, 09/1969
9) feenschach 26, p. 403, 12/1974
135 FIDE Album 1914-1944/III 1975
83 Schach ohne Partner [Grasemann] 1977
feenschach 47, p. 185, 07-09/1979
XII Rex Multiplex 2, p. 38, 04-06/1982
67 100 Classics of the Chessboard 1983
Belfort , p. 75, 1994
Tim Krabbé's Website (57) 2000
E-I Probleemblad 01-02/2000
harmonie 63 09/2000
König & Turm 4, p. 13, 03/2001
986 Minimalkunst im Schach 2006
Problemkiste 163 02/2006
633 Encyclopedia of Chess Problems 2012
Input: Gerd Wilts, 1996-06-06
Last update: Dieter Berlin, 2022-02-09 more...
1) 1. Kc3+ Sc4 2. Sb4 Sxe3 3. Kb3 Sf1 4. Tc3 Sd2#
2) 1. Tc3 Sf1 2. Kc4 Sxe3+ 3. Kb3+ Sc4 4. Sb4 Sd2#
2) 1. Tc3 Sf1 2. Kc4 Sxe3+ 3. Kb3+ Sc4 4. Sb4 Sd2#
Keywords: Pure Round Trip (S,S), Tempo Move (k,k)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/2n5/2r5/rk4K1/p3p3/pppNp3/3b4
Reprints: E226 FIDE Album 1989-1991 1997
Input: Gerd Wilts, 1996-06-06
Last update: Marcin Banaszek, 2018-02-13 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/2n5/2r5/rk4K1/p3p3/pppNp3/3b4
Reprints: E226 FIDE Album 1989-1991 1997
Input: Gerd Wilts, 1996-06-06
Last update: Marcin Banaszek, 2018-02-13 more...
44 - P0501209
Hilmar Ebert
Hans Gruber
Denis Blondel
12173v British Chess Magazine 1985
1. ehrende Erwähnung
(2+13) C+
h#4
Hilmar Ebert
Hans Gruber
Denis Blondel
12173v British Chess Magazine 1985
1. ehrende Erwähnung
(2+13) C+
h#4
1. Tc4 Txa2 2. Tb4 Txb2 3. Ta2 Txb1 4. Ta1 Txa1#
Yuri Bilokin: you can see
1...Rxb1 2.a1=Q Rxa1 3.b1=Q+ Rxb1 4.Ra1 Rxa1#
1...Rxb1 2.a1=Q Rxa1 3.b1=R Rxb1 4.Ra1 Rxa1#
1...Rxb1 2.a1=Q Rxa1 3.b1=B+ Rxb1 4.Ra1 Rxa1#
1...Rxb1 2.a1=Q Rxa1 3.b1=S Rxb1 4.Ra1 Rxa1# (2023-05-08)
more ...
comment
1...Rxb1 2.a1=Q Rxa1 3.b1=Q+ Rxb1 4.Ra1 Rxa1#
1...Rxb1 2.a1=Q Rxa1 3.b1=R Rxb1 4.Ra1 Rxa1#
1...Rxb1 2.a1=Q Rxa1 3.b1=B+ Rxb1 4.Ra1 Rxa1#
1...Rxb1 2.a1=Q Rxa1 3.b1=S Rxb1 4.Ra1 Rxa1# (2023-05-08)
more ...
comment
Keywords: Pure Round Trip (T), Tempo Move (ttt), Switchback (t), Minimal (T)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: b7/1n6/8/pp2p3/kr1pK3/rp6/pp6/Rn6
Reprints: 12173v British Chess Magazine 01/1986
Input: Gerd Wilts, 1996-06-06
Last update: A.Buchanan, 2017-03-20 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: b7/1n6/8/pp2p3/kr1pK3/rp6/pp6/Rn6
Reprints: 12173v British Chess Magazine 01/1986
Input: Gerd Wilts, 1996-06-06
Last update: A.Buchanan, 2017-03-20 more...
1. e1=L Txh3 2. f1=T Txh2 3. Kf2 Txh1 4. g1=S Th2#
Unterverwandlung!Springer, Unterverwandlung!Läufer, Unterverwandlung!Turm, Umwandlungswechsel!3-fach
Adrian Storisteanu: See also slight variation P0577817. (2015-09-12)
Gunter Jordan: Dies ist kein Umwandlungswechsel, dazu müssten die UW auf dem selben Feld erfolgen. (2023-01-11)
A.Buchanan: Why would we talk of helpmate promotions (particularly black ones) as underpromotions. Is a knight ever an under-promotion? And we still with all this under-promotion effort don’t know what the pieces are: it might be rook or bishop. Just label them as promotion:lts and be done (2023-01-12)
more ...
comment
Adrian Storisteanu: See also slight variation P0577817. (2015-09-12)
Gunter Jordan: Dies ist kein Umwandlungswechsel, dazu müssten die UW auf dem selben Feld erfolgen. (2023-01-11)
A.Buchanan: Why would we talk of helpmate promotions (particularly black ones) as underpromotions. Is a knight ever an under-promotion? And we still with all this under-promotion effort don’t know what the pieces are: it might be rook or bishop. Just label them as promotion:lts and be done (2023-01-12)
more ...
comment
Keywords: under-promotion, Minimal
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/8/5p1R/4pp2/2ppbrnn/4pppr/1K4kb
Input: Gerd Wilts, 1996-06-06
Last update: SBD, 2012-12-26 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/8/5p1R/4pp2/2ppbrnn/4pppr/1K4kb
Input: Gerd Wilts, 1996-06-06
Last update: SBD, 2012-12-26 more...
46 - P0501246
Vladislav Bunka
Vaclav Kotesovec
6326 feenschach 104 10/1992
(2+11)
h#3
b) wLg7 statt wTg7
c) wSg7 statt wTg7
Vladislav Bunka
Vaclav Kotesovec
6326 feenschach 104 10/1992
(2+11)
h#3
b) wLg7 statt wTg7
c) wSg7 statt wTg7
a) 1. Tg5 Tf7 2. e5 Ke7 3. Kf5 Txf6#
b) 1. Th5 Kxe7 2. Kg5 Ke6 3. Lf4 Lxf6#
c) 1. Kg5 Se8 2. Kh5 Kxe7 3. Tg5 Sxf6#
b) 1. Th5 Kxe7 2. Kg5 Ke6 3. Lf4 Lxf6#
c) 1. Kg5 Se8 2. Kh5 Kxe7 3. Tg5 Sxf6#
There was an error in position, corrected.
Henrik Juel: C+ Popeye 4.61 also (2023-04-14)
more ...
comment
Henrik Juel: C+ Popeye 4.61 also (2023-04-14)
more ...
comment
Keywords: Forsberg Twins, Minimal
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 1b1K4/4p1R1/1n3pbq/2r5/4pkpn/8/8/8
Input: Gerd Wilts, 1996-06-06
Last update: Vaclav Kotesovec, 2023-04-14 more...
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 1b1K4/4p1R1/1n3pbq/2r5/4pkpn/8/8/8
Input: Gerd Wilts, 1996-06-06
Last update: Vaclav Kotesovec, 2023-04-14 more...
1. Df8 Lf1 2. Le6 Se3 3. Dc5+ Sc4+ 4. Kd5 Lg2#
NL:
1. Df4 Lf1 2. Le6 Se1 3. Kd5 Sd3 4. Dd6 Lg2#
NL:
1. Df4 Lf1 2. Le6 Se1 3. Kd5 Sd3 4. Dd6 Lg2#
Yuri Bilokin: correction bQf1-c1, then a1=b1 8/8/4k3/8/3B3p/2bK4/7N/2q5 (3+4) h#4
1.Qg5 Bg1 2.Bf6 Sf3 3.Qd5+ Sd4+ 4.Ke5 Bh2# (MM) (2023-05-24)
comment
1.Qg5 Bg1 2.Bf6 Sf3 3.Qd5+ Sd4+ 4.Ke5 Bh2# (MM) (2023-05-24)
comment
Keywords: Interchange (LS (20))
Genre: h#
FEN: 8/8/8/4k3/2B3p1/1bK5/6N1/5q2
Input: hpr, 1996-06-14
Last update: hpr, 1999-05-22 more...
Genre: h#
FEN: 8/8/8/4k3/2B3p1/1bK5/6N1/5q2
Input: hpr, 1996-06-14
Last update: hpr, 1999-05-22 more...
1. Df5 Td4 2. Se6 Lc4 3. Df4 Td5#
NL:
1. Td4 Le4 2. Td6 Lg6 3. Sd5 Te4#
1. Dg8 Txe4+ 2. Kf5 Te8 3. Dg5 Le4#
NL:
1. Td4 Le4 2. Td6 Lg6 3. Sd5 Te4#
1. Dg8 Txe4+ 2. Kf5 Te8 3. Dg5 Le4#
Yuri Bilokin: correction wKa4-b4, bPb7-f7, bPb6-e3, bPg4-f3, bPf4(-bSf4), -bBa7, then rotate 90 8/8/1pr1pp2/2qkrp2/3B4/4R3/4K3/8 (3+8) h#3
1.Qd6 Re4 2.Rc5 Be3 3.Qc6 Rd4# (MM)
Blocking piece replacement (bQ-bR)
Blocking piece replacement (bR-bQ)
Pelle move (white)
Place exchange (bQ/bR)
Place exchange (wR/wB)
Model mate × 1 (2023-05-19)
comment
1.Qd6 Re4 2.Rc5 Be3 3.Qc6 Rd4# (MM)
Blocking piece replacement (bQ-bR)
Blocking piece replacement (bR-bQ)
Pelle move (white)
Place exchange (bQ/bR)
Place exchange (wR/wB)
Model mate × 1 (2023-05-19)
comment
Keywords: Interchange (TL ds)
Genre: h#
FEN: 8/bp6/1p2qr2/3Bk3/K1R1rnp1/8/8/8
Input: hpr, 1996-07-12
Last update: hpr, 1999-06-04 more...
Genre: h#
FEN: 8/bp6/1p2qr2/3Bk3/K1R1rnp1/8/8/8
Input: hpr, 1996-07-12
Last update: hpr, 1999-06-04 more...
1. Kc2! Lh1 2. Kxd3 Kg2 3. Ke4 Sc5#
Keywords: Cheney-Loyd (LK)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/4N3/8/2P1BBP1/1k1P4/4N2K/8
Reprints: 4 Problemschach-Jahrbuch 1993 D , p. 10, 1993
(1) Die Schwalbe , p. 498, 08/1994
Input: hpr, 1996-07-14
Last update: Dieter Berlin, 2023-05-12 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/4N3/8/2P1BBP1/1k1P4/4N2K/8
Reprints: 4 Problemschach-Jahrbuch 1993 D , p. 10, 1993
(1) Die Schwalbe , p. 498, 08/1994
Input: hpr, 1996-07-14
Last update: Dieter Berlin, 2023-05-12 more...
1. Tg8 h8=D 2. Kg6 Df6+ 3. Kh7 Dxf8 4. Kh8 Dh6#
1. Sf7 Df3 2. 0-0-0 Db7# (0-0-0 + !)
UL!
UL!
Anton Baumann: mögliche Korrektur: wBb4 nach b3, +sBd5, -wBg2
1.Sf7 Db4 2.0-0-0 Db7# (C+) (2024-01-03)
comment
1.Sf7 Db4 2.0-0-0 Db7# (C+) (2024-01-03)
comment
{ NL}
klären AL
SCHRECKE: Geplant war wohl:
1. Kd6 Lh2+ 2. e5 dxe6 e.p. 3. Ke7 Sd5#
Nebenlösungen, z.B.:
1. Df6 Kb4 2. De6 fxe6 3. Kd6 Lh2#
1. Kd7 Lh2 2. Sc7 Sc4 3. De8 Sb6# (2023-09-20)
comment
klären AL
SCHRECKE: Geplant war wohl:
1. Kd6 Lh2+ 2. e5 dxe6 e.p. 3. Ke7 Sd5#
Nebenlösungen, z.B.:
1. Df6 Kb4 2. De6 fxe6 3. Kd6 Lh2#
1. Kd7 Lh2 2. Sc7 Sc4 3. De8 Sb6# (2023-09-20)
comment
Genre: h#
FEN: 3bnq2/2k1p3/8/3P1P2/8/K3N3/8/6B1
Input: hpr, 1996-07-23
Last update: hpr, 1999-06-05 more...
{ NL}
klären AL
SCHRECKE: Satz:
1. ... c8=T+ 2. Kd7 Le6#
Es lösen:
1) 1. Ld8 cxd8=T+ 2. Ke7 Lh4#
2) 1. Kd7 cxb8=D 2. Kc6 Dc7#
3) 1. Ld6 c8=D+ 2. Ke7 Lh4#
4) 1. Kf8 Lf4 2. De8 Lh6# (2023-09-20)
comment
klären AL
SCHRECKE: Satz:
1. ... c8=T+ 2. Kd7 Le6#
Es lösen:
1) 1. Ld8 cxd8=T+ 2. Ke7 Lh4#
2) 1. Kd7 cxb8=D 2. Kc6 Dc7#
3) 1. Ld6 c8=D+ 2. Ke7 Lh4#
4) 1. Kf8 Lf4 2. De8 Lh6# (2023-09-20)
comment
Genre: h#
FEN: rq2k3/p1P1b3/p7/Pp3P2/1p6/1B4B1/8/7K
Input: hpr, 1996-07-23
Last update: hpr, 1999-06-05 more...
54 - P0502798
Achim Schöneberg
3849 Die Schwalbe 74 04/1982
3. Preis
(3+12) C+
h#4
b) sSe4 tauschen mit sLh5
Achim Schöneberg
3849 Die Schwalbe 74 04/1982
3. Preis
(3+12) C+
h#4
b) sSe4 tauschen mit sLh5
a) 1. Sxc5 Sxf5 2. Te4 Sd4 3. Ke3 Kg3 4. Sd3 Sc2#
b) 1. Kd4 Sxh5 2. Ld3 Kg3 3. Te4 Sg7 4. Ke3 Sxf5#
b) 1. Kd4 Sxh5 2. Ld3 Kg3 3. Te4 Sg7 4. Ke3 Sxf5#
klären Datum
Erich Bartel: die erste Zeile stimmt.Dort steht die Aufgabe auf Seite 459
als Urdruck. (2007-10-12)
Achim Schöneberg: Siehe P0502915 (2022-10-05)
comment
Erich Bartel: die erste Zeile stimmt.Dort steht die Aufgabe auf Seite 459
als Urdruck. (2007-10-12)
Achim Schöneberg: Siehe P0502915 (2022-10-05)
comment
Keywords: Interchange (kts,kls)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/2p4q/r1P2p1b/4n3/3krpNK/3pp3/6b1
Reprints: 3849 Die Schwalbe 06/1982
Input: hpr, 1996-07-28
Last update: hpr, 1999-06-05 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/2p4q/r1P2p1b/4n3/3krpNK/3pp3/6b1
Reprints: 3849 Die Schwalbe 06/1982
Input: hpr, 1996-07-28
Last update: hpr, 1999-06-05 more...
1) 1. Ld3 e3 2. Le2 e4#
2) 1. Se3 g3 2. Sg2 g4#
3) 1. Sg5 Lg8 2. Sf7 Lh7#
4) 1. Sf8 Txd6 2. Sg6 Lxe6#
NL:
1. Dg3 dxe5 2. Df4 Sg7#
1. Th1 dxe5 2. Sxe5 g4#
1. Ld3 exd3 2. De4 dxe4#
2) 1. Se3 g3 2. Sg2 g4#
3) 1. Sg5 Lg8 2. Sf7 Lh7#
4) 1. Sf8 Txd6 2. Sg6 Lxe6#
NL:
1. Dg3 dxe5 2. Df4 Sg7#
1. Th1 dxe5 2. Sxe5 g4#
1. Ld3 exd3 2. De4 dxe4#
Kees: I don't know how to send a diagram, but I fixed this one with a whole different position, but with the 4 thematic solutions I can send a FEN-notation with the 4 solutions
3K4/1pr1B1p1/s1pp2P1/2Rpk1S1/2Pb1sP1/3pPp2/4P3/3q1r2
1.Fc3 exd3 2.Fd2 d4‡
1.Ch3 exf3 2.Cf2 f4‡
1.Cxg6 Ff8 2.Ce7 Fxg7‡
1.Ch5 Txc6 2.Cf6 Fxd6‡ (2023-06-09)
comment
3K4/1pr1B1p1/s1pp2P1/2Rpk1S1/2Pb1sP1/3pPp2/4P3/3q1r2
1.Fc3 exd3 2.Fd2 d4‡
1.Ch3 exf3 2.Cf2 f4‡
1.Cxg6 Ff8 2.Ce7 Fxg7‡
1.Ch5 Txc6 2.Cf6 Fxd6‡ (2023-06-09)
comment
Keywords: Interchange (Bl,Bs,Ls)
Genre: h#
FEN: 4K3/2pr1B1n/3pp2p/3Rbk1N/3Pb1nP/5P2/4P1P1/4q1r1
Input: hpr, 1996-07-28
Last update: hpr, 1999-06-05 more...
Genre: h#
FEN: 4K3/2pr1B1n/3pp2p/3Rbk1N/3Pb1nP/5P2/4P1P1/4q1r1
Input: hpr, 1996-07-28
Last update: hpr, 1999-06-05 more...
a) 1. Sb3 Se3 2. Td4 Sd5 3. Kc4 Kc2 4. Sc5 Sb6#
b) 1. Kd5 Se1 2. Lc5 Sg2 3. Td4 Kc2 4. Kc4 Se3#
b) 1. Kd5 Se1 2. Lc5 Sg2 3. Td4 Kc2 4. Kc4 Se3#
Keywords: Interchange (kts,kls), Minimal
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/4p3/1pk5/1prn4/8/2N5/1Kb5
Input: hpr, 1996-08-28
Last update: Achim Schöneberg, 2021-12-09 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/4p3/1pk5/1prn4/8/2N5/1Kb5
Input: hpr, 1996-08-28
Last update: Achim Schöneberg, 2021-12-09 more...
1. Kb5 Kf5 2. Kb6 Kg6 3. Kc7 Kf7 4. Kd6 Lh2#
Cook: NL
1. Ta6 Sg8 2. Td6 Kf4 3. Kd5 Kf5 4. c4 Se7#
Cook: NL
1. Ta6 Sg8 2. Td6 Kf4 3. Kd5 Kf5 4. c4 Se7#
(KSp h)
Adrian Storisteanu: See P0502965 (failed correction attempt / self-anticipation). (2015-07-25)
milan: milan frelih:[Ke5 to h4 +bPh6] 1.Kb5 Kh5 2.Kb6 Kg6 3.Kc7 Kf7 4.Kd6 Bh2# (2015-07-26)
Yuri Bilokin: correction b1=a1, then wKd5-h4, -bPf7 8/8/1r2N3/1p6/1k5K/8/8/5B2 (3+3) h#4
1.Ka5 Kg5 2.Ka6 Kf6 3.Kb7 Ke7 4.Kc6 Bg2# (MM) (2023-05-24)
comment
Adrian Storisteanu: See P0502965 (failed correction attempt / self-anticipation). (2015-07-25)
milan: milan frelih:[Ke5 to h4 +bPh6] 1.Kb5 Kh5 2.Kb6 Kg6 3.Kc7 Kf7 4.Kd6 Bh2# (2015-07-26)
Yuri Bilokin: correction b1=a1, then wKd5-h4, -bPf7 8/8/1r2N3/1p6/1k5K/8/8/5B2 (3+3) h#4
1.Ka5 Kg5 2.Ka6 Kf6 3.Kb7 Ke7 4.Kc6 Bg2# (MM) (2023-05-24)
comment
Keywords: Figurenspiel
Genre: h#
FEN: 8/6p1/2r2N2/2p1K3/2k5/8/8/6B1
Input: hpr, 1996-08-30
Last update: Alfred Pfeiffer, 2015-07-25 more...
Genre: h#
FEN: 8/6p1/2r2N2/2p1K3/2k5/8/8/6B1
Input: hpr, 1996-08-30
Last update: Alfred Pfeiffer, 2015-07-25 more...
58 - P0503022
Claude Goumondy
Turnier des Tschechischen Sportbundes 1981-1982
1. Lob
(6+6) cooked
h#2
b) wTe1 nach a4
c) sDe5 tauschen mit sSe4
d) sSg6 nach h4
Claude Goumondy
Turnier des Tschechischen Sportbundes 1981-1982
1. Lob
(6+6) cooked
h#2
b) wTe1 nach a4
c) sDe5 tauschen mit sSe4
d) sSg6 nach h4
a) 1. Ke6 Lxe4 2. Df6 Ld5#
b) 1. Kc5 Lxe5 2. Sd6 Ld4#
c) 1. Kxc6 Txe4 2. Sd7 Te6#
d) 1. Kd6 Txe5 2. Sc5 Te6#
NL:
d) 1. Ke6 Tf5 2. Sd6 Txe5#
b) 1. Kc5 Lxe5 2. Sd6 Ld4#
c) 1. Kxc6 Txe4 2. Sd7 Te6#
d) 1. Kd6 Txe5 2. Sc5 Te6#
NL:
d) 1. Ke6 Tf5 2. Sd6 Txe5#
Keywords: Figurenspiel
Genre: h#
FEN: 8/8/2P3nK/3kq2R/4n1r1/5B1r/7B/4R3
Input: hpr, 1996-09-03
Last update: hpr, 1999-06-06 more...
Genre: h#
FEN: 8/8/2P3nK/3kq2R/4n1r1/5B1r/7B/4R3
Input: hpr, 1996-09-03
Last update: hpr, 1999-06-06 more...
59 - P0503025
Norman Alasdair Macleod
R. Tony Lewis
diagrammes 03-04/1982
2. Preis
6. Thematurnier
(3+5) C+
h#2
4.1...
Norman Alasdair Macleod
R. Tony Lewis
diagrammes 03-04/1982
2. Preis
6. Thematurnier
(3+5) C+
h#2
4.1...
1) 1. Kc7 e8=D 2. Db6 Dfe7#
2) 1. Kb7 e8=L 2. Ka8 Lc6#
3) 1. Kd7 e8=T 2. Dc6 De7#
4) 1. Dc4 e8=S 2. Kd5 Dd6#
2) 1. Kb7 e8=L 2. Ka8 Lc6#
3) 1. Kd7 e8=T 2. Dc6 De7#
4) 1. Dc4 e8=S 2. Kd5 Dd6#
(KSp)
t/Allumwandlung, u/61103, t/sK-Korrespondenz
A.Buchanan: A pity that 1 & 3 both end with 2...Dfe7# (2023-03-14)
more ...
comment
t/Allumwandlung, u/61103, t/sK-Korrespondenz
A.Buchanan: A pity that 1 & 3 both end with 2...Dfe7# (2023-03-14)
more ...
comment
Keywords: Allumwandlung, Figurenspiel
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 5Q2/n3P3/2k1q3/7b/4p3/K7/8/8
Reprints: feenschach 61 08/1982
13 The Modern Helpmate in two 1989
Input: Erich Bartel, 1996-09-03
Last update: A.Buchanan, 2023-03-14 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 5Q2/n3P3/2k1q3/7b/4p3/K7/8/8
Reprints: feenschach 61 08/1982
13 The Modern Helpmate in two 1989
Input: Erich Bartel, 1996-09-03
Last update: A.Buchanan, 2023-03-14 more...
1. Db6 Td7 2. Ld6 Ta7 3. Kd5 Ta5+ 4. Kc6 d5#
teilweiser Rundlauf T
Yuri Bilokin: Long-trip (wR, 3)
Umnov mate (bK-wP)
Ideal mate
Epaulette mate (2023-05-08)
comment
Yuri Bilokin: Long-trip (wR, 3)
Umnov mate (bK-wP)
Ideal mate
Epaulette mate (2023-05-08)
comment
Keywords: Pure Round Trip
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 2K5/8/3R4/4b3/3Pk3/1q6/8/8
Input: hpr, 1996-09-05
Last update: hpr, 1999-06-06 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 2K5/8/3R4/4b3/3Pk3/1q6/8/8
Input: hpr, 1996-09-05
Last update: hpr, 1999-06-06 more...
1. Th1 Sa1 2. Th4 Sh1 3. Tf4 Sc2#
Cook: NL
1. Tf1 Kh8 2. Tf2 Sa1 3. Df4 Sc2#
1. Df4 Sxa1 2. e1=L h6 3. Lf2 Sc2#
Cook: NL
1. Tf1 Kh8 2. Tf2 Sa1 3. Df4 Sc2#
1. Df4 Sxa1 2. e1=L h6 3. Lf2 Sc2#
teilweiser Rundlauf t
Adrian Storisteanu: Possible fix: bPe4 (instead of the bQ), +wPf5, +bPf6 (6+9). (2015-08-15)
Yuri Bilokin: correction bRd3(-bNd3), bBe4(-bQe4), bPf3(-bBf3) 6K1/8/8/7P/3Pb3/pN1rkpN1/3pp3/r7 (5+8) (2022-12-19)
A.Buchanan: This is very clean Yuri well done. I would shift bPa3 to a5. wK must be exactly on g8, else the problem is cooked (2022-12-20)
comment
Adrian Storisteanu: Possible fix: bPe4 (instead of the bQ), +wPf5, +bPf6 (6+9). (2015-08-15)
Yuri Bilokin: correction bRd3(-bNd3), bBe4(-bQe4), bPf3(-bBf3) 6K1/8/8/7P/3Pb3/pN1rkpN1/3pp3/r7 (5+8) (2022-12-19)
A.Buchanan: This is very clean Yuri well done. I would shift bPa3 to a5. wK must be exactly on g8, else the problem is cooked (2022-12-20)
comment
Keywords: Pure Round Trip
Genre: h#
FEN: 6K1/8/8/7P/3Pq3/pN1nkbN1/3pp3/r7
Input: hpr, 1996-09-05
Last update: Alfred Pfeiffer, 2015-08-15 more...
Genre: h#
FEN: 6K1/8/8/7P/3Pq3/pN1nkbN1/3pp3/r7
Input: hpr, 1996-09-05
Last update: Alfred Pfeiffer, 2015-08-15 more...
62 - P0503402
Laszlo Barna
3. Makuc-Moder-Gedenkturnier 03/1970
2. Preis
(4+9) cooked
h#2
b) wSc8 nach c6
c) wSc8 nach e6
d) wSc8 nach e4
Laszlo Barna
3. Makuc-Moder-Gedenkturnier 03/1970
2. Preis
(4+9) cooked
h#2
b) wSc8 nach c6
c) wSc8 nach e6
d) wSc8 nach e4
a) 1. Tf6 g8=D 2. Kc6 De8#
b) 1. Tc3 g8=L 2. Kc8 Le6#
c) 1. Te5 g8=S 2. Ke8 Sf6#
d) 1. Tc3 g8=T 2. Ke6 Te8#
NL:
1. Dh1 Sf6+ 2. Kd8 g8=D#
uvm
b) 1. Tc3 g8=L 2. Kc8 Le6#
c) 1. Te5 g8=S 2. Ke8 Sf6#
d) 1. Tc3 g8=T 2. Ke6 Te8#
NL:
1. Dh1 Sf6+ 2. Kd8 g8=D#
uvm
KSp Königsstern
t/Allumwandlung, u/61104
Daniel Novomesky: Add bPh5. C+ (2008-03-03)
Yuri Bilokin: correction +bPh5 2N4K/3k2PR/6pq/2rp1rpp/8/8/b7/b7 (4+10)
Allumwandlung (white)
JT Onkoud 50 theme (double)
Promotion (QRBS, 4)
Star (bK)
Mates on the same square × 2 (2023-06-26)
comment
t/Allumwandlung, u/61104
Daniel Novomesky: Add bPh5. C+ (2008-03-03)
Yuri Bilokin: correction +bPh5 2N4K/3k2PR/6pq/2rp1rpp/8/8/b7/b7 (4+10)
Allumwandlung (white)
JT Onkoud 50 theme (double)
Promotion (QRBS, 4)
Star (bK)
Mates on the same square × 2 (2023-06-26)
comment
Keywords: Allumwandlung, Figurenspiel
Genre: h#
FEN: 2N4K/3k2PR/6pq/2rp1rp1/8/8/b7/b7
Reprints: feenschach , p. 58, 04/1975
Input: Erich Bartel, 1996-09-17
Last update: hpr, 1999-06-12 more...
Genre: h#
FEN: 2N4K/3k2PR/6pq/2rp1rp1/8/8/b7/b7
Reprints: feenschach , p. 58, 04/1975
Input: Erich Bartel, 1996-09-17
Last update: hpr, 1999-06-12 more...
1. Kc3 De6 2. Kxd4 Dxd6+ 3. Ke4 De5#
NL
1. Kc3 Df5 2. Tf8 Kxd6 3. Kxd4 Dd3#
NL
1. Kc3 Df5 2. Tf8 Kxd6 3. Kxd4 Dd3#
Korrektur Hilmar Ebert: +sSe8 (erhält Quasi-Symmetrie)
Yuri Bilokin: Not all pieces in quasi-symmetry are necessary: the white pawn on h2 is superfluous. It is clear that more than 100 years ago there was such a direction in art, although some authors indulge in technical white pawns in our time. (2022-12-19)
A.Buchanan: The technical term for such a position is "symmetric constellation". "Quasi-symmetry" is where the position is nearly symmetric but not quite (e.g. game array except wPa2 is missing). In a symmetric constellation any redundant piece is a definite defect. A quasi-symmetric position is not an ornament, more of a consequence of the content. For example, a number of last move type DD problems are quasi-symmetric (2022-12-20)
Yuri Bilokin: Thanks for the interpretation (2022-12-20)
comment
Yuri Bilokin: Not all pieces in quasi-symmetry are necessary: the white pawn on h2 is superfluous. It is clear that more than 100 years ago there was such a direction in art, although some authors indulge in technical white pawns in our time. (2022-12-19)
A.Buchanan: The technical term for such a position is "symmetric constellation". "Quasi-symmetry" is where the position is nearly symmetric but not quite (e.g. game array except wPa2 is missing). In a symmetric constellation any redundant piece is a definite defect. A quasi-symmetric position is not an ornament, more of a consequence of the content. For example, a number of last move type DD problems are quasi-symmetric (2022-12-20)
Yuri Bilokin: Thanks for the interpretation (2022-12-20)
comment
Keywords: Pure Round Trip (D)
Genre: h#
FEN: 8/4K3/3p1r2/4Q3/3P1P2/8/1k2P2P/8
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 2001-12-21 more...
Genre: h#
FEN: 8/4K3/3p1r2/4Q3/3P1P2/8/1k2P2P/8
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 2001-12-21 more...
1. Tc2 Th1 2. Kb2 Th8 3. Kc1 Txa8 4. Sb2 Ta1#
Erich Bartel: Nachdruck:
1) V35) Problemkiste (125) X 1999.--- (2007-07-18)
Yuri Bilokin: Areal cycle (wR, with captures, 4)
Corner-to-corner (wR) × 4
Four corners (wR (2023-05-08)
comment
1) V35) Problemkiste (125) X 1999.--- (2007-07-18)
Yuri Bilokin: Areal cycle (wR, with captures, 4)
Corner-to-corner (wR) × 4
Four corners (wR (2023-05-08)
comment
Keywords: Pure Round Trip (D)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: r7/6K1/8/5q2/n3b3/1k6/r2p4/R7
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-13 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: r7/6K1/8/5q2/n3b3/1k6/r2p4/R7
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-13 more...
65 - P0503616
Johann Christoffel van Gool
Jacobus Theodorus Sanderse
876b The Problemist 01/1983
(4+5) C+
h#4
2.1...
Johann Christoffel van Gool
Jacobus Theodorus Sanderse
876b The Problemist 01/1983
(4+5) C+
h#4
2.1...
1) 1. Sc3 Th1 2. Kc2 Th3 3. Kd1 Txc3 4. Ke1 Tc1#
2) 1. Kxb3 Ta1 2. Kc2 Txa2+ 3. Kd1 Tc2 4. Ke1 Tc1#
2) 1. Kxb3 Ta1 2. Kc2 Txa2+ 3. Kd1 Tc2 4. Ke1 Tc1#
Keywords: Pure Round Trip (T)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/8/8/1p6/1P6/nk1npPK1/2R5
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-13 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/8/8/1p6/1P6/nk1npPK1/2R5
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-13 more...
66 - P0503644
Peter Kniest
3650v Deutsche Schachzeitung 11/1976
(3+11) C+
h#4
b) wBb3 nach c2, sBb4 nach c3
Peter Kniest
3650v Deutsche Schachzeitung 11/1976
(3+11) C+
h#4
b) wBb3 nach c2, sBb4 nach c3
a) 1. Dg4+ Lg2 2. De6 Lf1 3. Kd5 La6 4. c5 Lb7#
b) 1. De4 Ld5 2. c6 La2 3. Sb3 cxb3 4. Kd5 b4#
b) 1. De4 Ld5 2. c6 La2 3. Sb3 cxb3 4. Kd5 b4#
Keywords: Hilfsmatt-Inder (LB), Pure Round Trip (L)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/1Bp5/3p1p2/2k1r3/1p1p3q/1P5r/3n4/3n2K1
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-13 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/1Bp5/3p1p2/2k1r3/1p1p3q/1P5r/3n4/3n2K1
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-13 more...
1. Sf3 Sxf4 2. Kd2 Se6 3. Ke3 Sxc7 4. Kf4 Sd5#
NL
1. Dc2 Kh4 2. f3 Kg3 3. Kd1 Kf2 4. Dc1 Se3#
NL
1. Dc2 Kh4 2. f3 Kg3 3. Kd1 Kf2 4. Dc1 Se3#
Yuri Bilokin: correction bSe4 1B6/2p5/8/3N1p1K/4np2/8/3n4/2k5 (3+6) h#4
1.Sf3 Sxf4 2.Kd2 Se6 3.Ke3 Sxc7 4.Kf4 Sd5# (IM)
Annihilation × 2
Areal cycle (wS, with captures, 4)
Crusader theme. Kniest theme
Long-trip (wS, 4)
Ideal mate
Battery mate. Double-check mate (2023-05-23)
comment
1.Sf3 Sxf4 2.Kd2 Se6 3.Ke3 Sxc7 4.Kf4 Sd5# (IM)
Annihilation × 2
Areal cycle (wS, with captures, 4)
Crusader theme. Kniest theme
Long-trip (wS, 4)
Ideal mate
Battery mate. Double-check mate (2023-05-23)
comment
Keywords: Pure Round Trip (S)
Genre: h#
FEN: 1B6/2p5/8/3N1p1K/4qp2/8/3n4/2k5
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-13 more...
Genre: h#
FEN: 1B6/2p5/8/3N1p1K/4qp2/8/3n4/2k5
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-13 more...
1. Kc5 Ta2 2. Kb5 Kd5 3. Kb4 Kc6 4. Kc4 Ta4#
Keywords: Pure Round Trip (k)
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/8/8/2k1K3/2p5/1RP5/8
Reprints: Problemkiste 99 06/1995
1936 Ideal-Mate Encyclopedia Vol.1 1999
Problemkiste 121 02/1999
r628 Mirador 06/2002
153) Hobelspäne (2) 28/05/2017
Input: hpr, 1996-09-23
Last update: Alfred Pfeiffer, 2019-05-15 more...
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/8/8/2k1K3/2p5/1RP5/8
Reprints: Problemkiste 99 06/1995
1936 Ideal-Mate Encyclopedia Vol.1 1999
Problemkiste 121 02/1999
r628 Mirador 06/2002
153) Hobelspäne (2) 28/05/2017
Input: hpr, 1996-09-23
Last update: Alfred Pfeiffer, 2019-05-15 more...
1. Kf3 Td4 2. Kxe3 Txd3+ 3. Kxf4 Te3 4. Kg4 Te4#
Yuri Bilokin: Areal cycle (bK, with captures, 4)
Areal cycle (wR, with captures, 4) (2023-05-08)
comment
Areal cycle (wR, with captures, 4) (2023-05-08)
comment
Keywords: Pure Round Trip
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/7p/1p3n1p/bp2RPkP/pR1pP3/q2p4/rrnK4
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/7p/1p3n1p/bp2RPkP/pR1pP3/q2p4/rrnK4
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
1. Kd2 Lxb6 2. Kxc3 Lc7 3. Kxd4 Lxd6 4. Ke3 Lc5#
Yuri Bilokin: Areal cycle (bK, with captures, 4)
Areal cycle (wB, with captures, 4) (2023-05-08)
comment
Areal cycle (wB, with captures, 4) (2023-05-08)
comment
Keywords: Pure Round Trip
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 2b5/1p1R4/1p1p4/1PB4p/1PpP1p1P/2P1kPpK/2P3P1/3B4
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 2b5/1p1R4/1p1p4/1PB4p/1PpP1p1P/2P1kPpK/2P3P1/3B4
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
1) 1. Ke4 Txc4+ 2. Kxe5 Txc5+ 3. Kd4 Tb5 4. e5 Tb4#
2) 1. Kxe5 Tb5 2. Kd4 Txc5 3. Lb5+ Txb5 4. e5 Tb4#
2) 1. Kxe5 Tb5 2. Kd4 Txc5 3. Lb5+ Txb5 4. e5 Tb4#
Keywords: Pure Round Trip
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/p3p3/p1K1p3/P1p1P3/1Rbk4/2npp3/8/8
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/p3p3/p1K1p3/P1p1P3/1Rbk4/2npp3/8/8
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
1) 1. ... Txb8 2. Kc5 Txe8 3. Lc3 Te5+ 4. Kb4 Tb5#
2) 1. ... Te5 2. Dd6 Txe8 3. Kc5 Tb8 4. Ld4 Tb5#
NL:
1. ... Tb6 2. Kc3 Sb5+ 3. Kb4 Sd6+ 4. Ka5 Sc4#
2) 1. ... Te5 2. Dd6 Txe8 3. Kc5 Tb8 4. Ld4 Tb5#
NL:
1. ... Tb6 2. Kc3 Sb5+ 3. Kb4 Sd6+ 4. Ka5 Sc4#
Yuri Bilokin: correction +bNd2 1q2b2b/N7/8/1R6/3k4/pP6/3n4/7K (4+6)
AntiZielElement (W1, line obstruction)
Areal cycle (wR, with captures, 4) × 2
Exchange of moves (W1/W3)
JT Onkoud 50 theme
Long-trip (wR, 4) × 2. Many-ways (wR, 2) × 4
Zalokotsky theme (wR/wR, 3)
Model mate × 2.Mates on the same square × 2 (2023-05-08)
comment
AntiZielElement (W1, line obstruction)
Areal cycle (wR, with captures, 4) × 2
Exchange of moves (W1/W3)
JT Onkoud 50 theme
Long-trip (wR, 4) × 2. Many-ways (wR, 2) × 4
Zalokotsky theme (wR/wR, 3)
Model mate × 2.Mates on the same square × 2 (2023-05-08)
comment
Keywords: Pure Round Trip
Genre: h#
FEN: 1q2b2b/N7/8/1R6/3k4/pP6/8/7K
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
Genre: h#
FEN: 1q2b2b/N7/8/1R6/3k4/pP6/8/7K
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
1) 1. ... Lxb7 2. Le5 Lc8 3. Kd5 Lxd7 4. Ke4 Lc6#
2) 1. ... Th3 2. Ld2 Txh4 3. Kd3 Tf4 4. Ke3 Tf3#
NL:
1. ... Tf1 2. Kd3 Tf2 3. Sf3 Lb5+ 4. Ke3 Txe2#
2) 1. ... Th3 2. Ld2 Txh4 3. Kd3 Tf4 4. Ke3 Tf3#
NL:
1. ... Tf1 2. Kd3 Tf2 3. Sf3 Lb5+ 4. Ke3 Txe2#
Yuri Bilokin: correction +bQg4, +bPa6, -bNh4, -bPh2 8/1b1p4/p1Bn4/8/2kp1bq1/5R2/4p3/7K (3+9) h#4 0.2.1…
1...Rg3 2.Bd2 Rxg4 3.Kd3 Rf4 4.Ke3 Rf3# (MM)
1...Bxb7 2.Be5 Bc8 3.Kd5 Bxd7 4.Ke4 Bc6# (MM)
Areal cycle (wB, with captures, 4)
Areal cycle (wR, with captures, 4)
Exchange of functions (wRf3/wBc6, Mate / Passive guard)
Long-trip (wB, 4). Long-trip (wR, 4)
Model mate × 2 (2023-05-08)
comment
1...Rg3 2.Bd2 Rxg4 3.Kd3 Rf4 4.Ke3 Rf3# (MM)
1...Bxb7 2.Be5 Bc8 3.Kd5 Bxd7 4.Ke4 Bc6# (MM)
Areal cycle (wB, with captures, 4)
Areal cycle (wR, with captures, 4)
Exchange of functions (wRf3/wBc6, Mate / Passive guard)
Long-trip (wB, 4). Long-trip (wR, 4)
Model mate × 2 (2023-05-08)
comment
Keywords: Pure Round Trip
Genre: h#
FEN: 8/1b1p4/2Bn4/8/2kp1b1n/5R2/4p2p/7K
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
Genre: h#
FEN: 8/1b1p4/2Bn4/8/2kp1b1n/5R2/4p2p/7K
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-20 more...
1) 1. Lc6 De8 2. Kb7 Dxc6+ 3. Ka6 Da8#
2) 1. Ld5 Dd8+ 2. Kb7 Dxd5+ 3. Ka6 Da8#
3) 1. Le4 Da4 2. Kb7 Dxe4+ 3. Ka6 Da8#
4) 1. Lf3 Df8 2. Kb7 Dxf3+ 3. Ka6 Da8#
5) 1. Lg2 Dg8 2. Kb7 Dxg2+ 3. Ka6 Da8#
6) 1. e5 Dh8 2. Kb7 Dxh1+ 3. Ka6 Da8#
2) 1. Ld5 Dd8+ 2. Kb7 Dxd5+ 3. Ka6 Da8#
3) 1. Le4 Da4 2. Kb7 Dxe4+ 3. Ka6 Da8#
4) 1. Lf3 Df8 2. Kb7 Dxf3+ 3. Ka6 Da8#
5) 1. Lg2 Dg8 2. Kb7 Dxg2+ 3. Ka6 Da8#
6) 1. e5 Dh8 2. Kb7 Dxh1+ 3. Ka6 Da8#
Keywords: Pure Round Trip (D,D,D,D,D)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: Q7/2k5/1p2p3/1p6/8/K7/8/7b
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 2004-09-30 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: Q7/2k5/1p2p3/1p6/8/K7/8/7b
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 2004-09-30 more...
75 - P0503977
Leon Loewenton
H241 FIDE Olympia Turnier 14 Leipzig 1960
5. -9. Lob
(6+5) cooked
h#2
5.1...
Leon Loewenton
H241 FIDE Olympia Turnier 14 Leipzig 1960
5. -9. Lob
(6+5) cooked
h#2
5.1...
1) 1. f2 e8=S 2. Ld5 Txf2#
2) 1. Kd6 Tc2 2. Kc7 Le5#
3) 1. Kf6 e8=D 2. Kg7 Th2#
4) 1. Kf4 Th2 2. Kg3 Le5#
5) 1. Kd4 Le6 2. Kc3 Tb3#
NL:
1. Kxf5 e8=D 2. Kf6 Tb5#
2) 1. Kd6 Tc2 2. Kc7 Le5#
3) 1. Kf6 e8=D 2. Kg7 Th2#
4) 1. Kf4 Th2 2. Kg3 Le5#
5) 1. Kd4 Le6 2. Kc3 Tb3#
NL:
1. Kxf5 e8=D 2. Kf6 Tb5#
KSp Königsstern
Kees: Possible fix: +wBb4 -sLh1 (No 5.1... but 4.1... with only KSp Königsstern) (2023-06-05)
A.Buchanan: I agree with +wBb4 for soundness but let's keep sLh1. What's not to like about 5 well differentiated solutions? WinChloe has the same cooked diagram as we do here. (2023-06-06)
Yuri Bilokin: The fifth solution is not necessary, the author has kept it, so after adding the white pawn to the b4 square and without removing the black bishop, we fully preserve the author's intention.
8/4P3/1pP5/4kB2/1P6/4pp2/1R6/B2K3b (7+5) h#2 5.1...
Active sacrifice (black, delayed). BK moves only × 4. Delayed Umnov (bK-wB) × 2
Extended star (bK)
Promotion (QS, 2)
Battery mate × 3. Double-check mate. Mirror mate
Mate from initial bK square × 2 (2023-06-10)
more ...
comment
Kees: Possible fix: +wBb4 -sLh1 (No 5.1... but 4.1... with only KSp Königsstern) (2023-06-05)
A.Buchanan: I agree with +wBb4 for soundness but let's keep sLh1. What's not to like about 5 well differentiated solutions? WinChloe has the same cooked diagram as we do here. (2023-06-06)
Yuri Bilokin: The fifth solution is not necessary, the author has kept it, so after adding the white pawn to the b4 square and without removing the black bishop, we fully preserve the author's intention.
8/4P3/1pP5/4kB2/1P6/4pp2/1R6/B2K3b (7+5) h#2 5.1...
Active sacrifice (black, delayed). BK moves only × 4. Delayed Umnov (bK-wB) × 2
Extended star (bK)
Promotion (QS, 2)
Battery mate × 3. Double-check mate. Mirror mate
Mate from initial bK square × 2 (2023-06-10)
more ...
comment
Keywords: Figurenspiel
Genre: h#
FEN: 8/4P3/1pP5/4kB2/8/4pp2/1R6/B2K3b
Reprints: H241 Schach 19, p. 301, 10/1960
Input: hpr, 1996-10-03
Last update: A.Buchanan, 2023-06-06 more...
Genre: h#
FEN: 8/4P3/1pP5/4kB2/8/4pp2/1R6/B2K3b
Reprints: H241 Schach 19, p. 301, 10/1960
Input: hpr, 1996-10-03
Last update: A.Buchanan, 2023-06-06 more...
1) 1. Lf3 Le4+ 2. Kh1 Tb1#
2) 1. Tf3 Tb3 2. Kh3 Lf5#
Die Rollen der fesselnden und gefesselten Figur werden umgedreht. Grimshaw auf f3 und Funktionswechsel von T und L
2) 1. Tf3 Tb3 2. Kh3 Lf5#
Die Rollen der fesselnden und gefesselten Figur werden umgedreht. Grimshaw auf f3 und Funktionswechsel von T und L
Sally: Markant ist der Grimshaw auf "f3" (überraschend auf einem "Eck-Feld" des w. Königs), wobei die dennoch notwendige Fesselung prägend ist. [Eine der ersten Aufgaben des Erfolgsduos].
41 Mehr als ein Hobby ( F. Pachl 2011). (2016-12-07)
Achim Schöneberg: Knapper Vorgänger siehe P1413421 (2023-11-08)
comment
41 Mehr als ein Hobby ( F. Pachl 2011). (2016-12-07)
Achim Schöneberg: Knapper Vorgänger siehe P1413421 (2023-11-08)
comment
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/8/7p/KR3r1p/8/R1B3kP/3b4
Reprints: 41 Mehr als 1 Hobby [Pachl] 2011
Input: Markus Manhart, 1996-10-07
Last update: Alfred Pfeiffer, 2016-12-07 more...
1. Tg3 Tf3 2. Te3 hxg3 3. Ld3 Tf4#
1. Tg5 Lg8 2. De5 Df7#
NL:
1. Kg7 Dd2 2. Kh8 Dh6#
NL:
1. Kg7 Dd2 2. Kh8 Dh6#
79 - P0506014
Vladimir Korolkov
Lev Loshinsky
1806 Revista de Sah 11/1964
3. Preis
(8+7) C+
h#2
4.1...
Vladimir Korolkov
Lev Loshinsky
1806 Revista de Sah 11/1964
3. Preis
(8+7) C+
h#2
4.1...
1) 1. b5 Td7 2. bxc4 Tb7#
2) 1. b6 Txe4 2. b5 c5#
3) 1. bxa6 Sc5 2. a5 Sa6#
4) 1. bxc6 Sa5 2. c5 Sc6#
2) 1. b6 Txe4 2. b5 c5#
3) 1. bxa6 Sc5 2. a5 Sa6#
4) 1. bxc6 Sa5 2. c5 Sc6#
BSp
spiegelverkehrt von R.Senkus in Chess in Australia veröffentlicht
Yuri Bilokin: you can see 1...Rxe4 2.b5 c5# (2024-04-02)
comment
spiegelverkehrt von R.Senkus in Chess in Australia veröffentlicht
Yuri Bilokin: you can see 1...Rxe4 2.b5 c5# (2024-04-02)
comment
Keywords: Figurenspiel
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 3K4/1p6/P1P5/4p3/pkPRp3/pNp1P3/2P5/8
Reprints: 183 Chess in Australia 1992
Input: hpr, 1996-12-31
Last update: hpr, 2011-03-06 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 3K4/1p6/P1P5/4p3/pkPRp3/pNp1P3/2P5/8
Reprints: 183 Chess in Australia 1992
Input: hpr, 1996-12-31
Last update: hpr, 2011-03-06 more...
1) 1. e5 d7 2. e4 d8=D 3. d2+ Dxd2#
2) 1. exf6 d7 2. f5 d8=D 3. f4 Dxd3#
3) 1. e6 f7 2. e5 f8=D 3. e4 Df2#
4) 1. exd6 f7 2. d5 f8=D 3. d4 Df3#
2) 1. exf6 d7 2. f5 d8=D 3. f4 Dxd3#
3) 1. e6 f7 2. e5 f8=D 3. e4 Df2#
4) 1. exd6 f7 2. d5 f8=D 3. d4 Df3#
Keywords: Figurenspiel, Kindergarten Problem
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/4p3/3P1P2/8/8/3pk3/4P3/4K3
Input: hpr, 1996-12-31
Last update: Erich Bartel, 2012-05-24 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/4p3/3P1P2/8/8/3pk3/4P3/4K3
Input: hpr, 1996-12-31
Last update: Erich Bartel, 2012-05-24 more...
1) 1. e6 d7 2. e5 d8=D 3. e4 Dd2#
2) 1. exf6 d7 2. f5 d8=D 3. f4 Dd3#
3) 1. e5 f7 2. e4 f8=D 3. f2+ Dxf2#
4) 1. exd6 f7 2. d5 f8=D 3. d4 Dxf3#
2) 1. exf6 d7 2. f5 d8=D 3. f4 Dd3#
3) 1. e5 f7 2. e4 f8=D 3. f2+ Dxf2#
4) 1. exd6 f7 2. d5 f8=D 3. d4 Dxf3#
Keywords: Figurenspiel
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/4p3/3P1P2/8/8/4kp2/4P3/4K3
Input: hpr, 1996-12-31
Last update: hpr, 1999-08-29 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/4p3/3P1P2/8/8/4kp2/4P3/4K3
Input: hpr, 1996-12-31
Last update: hpr, 1999-08-29 more...
BSp
klären: Stellung falsch, wK im Schach!
Balsamo: The position is lacking a wLe1! (2001-12-21)
SCHRECKE: Mit wLe1 gibt es die Lösungen:
a) 1. Lf8 c4 2. Ld6 Se6#
b) 1. Da6 cxd3 2. Dd6 Sa4#
c) 1. Tg6 c3 2. Td6 Lf2# (2023-09-20)
comment
klären: Stellung falsch, wK im Schach!
Balsamo: The position is lacking a wLe1! (2001-12-21)
SCHRECKE: Mit wLe1 gibt es die Lösungen:
a) 1. Lf8 c4 2. Ld6 Se6#
b) 1. Da6 cxd3 2. Dd6 Sa4#
c) 1. Tg6 c3 2. Td6 Lf2# (2023-09-20)
comment
Keywords: Figurenspiel
Genre: h#
FEN: 4B3/8/7b/P1k5/5N2/q2p4/1NP5/2K3r1
Input: hpr, 1996-12-31
Last update: hpr, 1996-12-31 more...
Genre: h#
FEN: 4B3/8/7b/P1k5/5N2/q2p4/1NP5/2K3r1
Input: hpr, 1996-12-31
Last update: hpr, 1996-12-31 more...
83 - P0506144
Felix Alexander Sonnenfeld
Ricardo de Mattos Vieira
1895 diagrammes 85 04-06/1988
(4+5) C+
h#2
3.1...
Felix Alexander Sonnenfeld
Ricardo de Mattos Vieira
1895 diagrammes 85 04-06/1988
(4+5) C+
h#2
3.1...
1) 1. Sb5 Sd5+ 2. Kf5 Se7#
2) 1. Sc6 Sc4+ 2. Kf5 Sd6#
3) 1. Sc8 Sc2+ 2. Kf5 Sd4#
2) 1. Sc6 Sc4+ 2. Kf5 Sd6#
3) 1. Sc8 Sc2+ 2. Kf5 Sd4#
Yuri Bilokin: you can see 1...Sc2+ 2.Kf5 Sd4#
1...Sc4+ 2.Kf5 Sd6#
1...Sd5+ 2.Kf5 Se7# (2023-06-26)
comment
1...Sc4+ 2.Kf5 Sd6#
1...Sd5+ 2.Kf5 Se7# (2023-06-26)
comment
Keywords: Dual avoidance, Tempo Move
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/n7/4p1p1/4P3/4kb1K/4N3/8/4R3
Input: Markus Manhart, 1997-01-06
Last update: hpr, 1999-08-29 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/n7/4p1p1/4P3/4kb1K/4N3/8/4R3
Input: Markus Manhart, 1997-01-06
Last update: hpr, 1999-08-29 more...
1) 1. Da5 Tb5 2. Sd7 Tb6 3. Df5 Ld5#
2) 1. Da4 Lb5 2. Ld6 Lc4 3. Dd7 Te5#
NL:
1. Sa6 Txb5 2. Sc7 Tb6 3. Kd6 La4#
2) 1. Da4 Lb5 2. Ld6 Lc4 3. Dd7 Te5#
NL:
1. Sa6 Txb5 2. Sc7 Tb6 3. Kd6 La4#
Yuri Bilokin: Yuri Bilokin & Wenelin Alaikow correction a4=a1(-bPd), then +bPa4, +bPe5, +bPe6, bPg3-h2 8/8/4p3/4p3/pp2b3/2B1kn2/1q1R3p/1n1K4 (3+10) h#3 2.1...
1.Qa1 Bb2 2.Bd3 Bc1 3.Qd4 Re2#
1.Qa2 Rb2 2.Sd4 Rb3 3.Qf2 Bd2# (MM)
Ambush (wB). Ambush (wR). Anticipatory unpin
AntiZielElement (W1, line obstruction) × 2
Exchange of functions (wRd2/wBc3, Mating battery firing / Rear piece of mating battery + Line opening)
Play on the same square (W1, 2). Reciprocal batteries. Wigwag (bQ) × 2
Model mate × 1. Battery mate × 2. Double-check mate × 2 (2023-05-20)
comment
1.Qa1 Bb2 2.Bd3 Bc1 3.Qd4 Re2#
1.Qa2 Rb2 2.Sd4 Rb3 3.Qf2 Bd2# (MM)
Ambush (wB). Ambush (wR). Anticipatory unpin
AntiZielElement (W1, line obstruction) × 2
Exchange of functions (wRd2/wBc3, Mating battery firing / Rear piece of mating battery + Line opening)
Play on the same square (W1, 2). Reciprocal batteries. Wigwag (bQ) × 2
Model mate × 1. Battery mate × 2. Double-check mate × 2 (2023-05-20)
comment
Genre: h#
FEN: 8/1p2b3/2B1knp1/1q1R4/1n1K4/3p4/8/8
Input: Markus Manhart, 1997-01-06
Last update: hpr, 1999-08-29 more...
a) 1. Tb8 Th1 2. Tc8 Dh2#
b) 1. Tc8 Lb1 2. Tb8 Dc2#
NL:
a) 1. Kd6 Txf6+ 2. Ke7 Dxg7#
b) 1. Tc8 Lb1 2. Tb8 Dc2#
NL:
a) 1. Kd6 Txf6+ 2. Ke7 Dxg7#
Kees: Possible fix:Dd8-d7 Ld7-d8 +sSh8 -sLc1 -wBb2 h#1,5 ( a) 1. ...Th1 2.Tc8 Dh2# b) 1. ... Lb1 2. Tb8 2. Dc2#) (2023-06-05)
Gerald Ettl: Kees, das ist eine schlechte Verbesserung. Denn das Tempoverlustmanoever des sT ist hier eine wichtige Absicht des Autors, die beibehalten werden soll. (2023-06-05)
Ladislav Packa: White Rf7 Pg7 Qh7 Bg6 Pg5 Rh5 Pg4 Pb3 Pg2 Kg1 (10)
Black Ra8 Bd8 Bg8 Pb7 Kc7 Qd7 Pb6 Pc6 Pb4 (9)
b) bPc6-d6
C+ by Popeye v4.63 The author's idea is completely preserved. (2023-06-05)
A.Buchanan: I like Ladislav's fix a lot. I humbly suggest r2b2b1/1pkq1RPQ/1p1p2B1/6PR/3p4/8/4K3/6N1 saving two units. Some might say wSg1 is too passive for a White officer, but the way two units (bP & wK) block three rows is sufficiently amusing to offset this, imho. (2023-06-06)
A.Buchanan: The authors fixed it very well with P0523513 so any later fix is pointless. (2023-06-13)
more ...
comment
Gerald Ettl: Kees, das ist eine schlechte Verbesserung. Denn das Tempoverlustmanoever des sT ist hier eine wichtige Absicht des Autors, die beibehalten werden soll. (2023-06-05)
Ladislav Packa: White Rf7 Pg7 Qh7 Bg6 Pg5 Rh5 Pg4 Pb3 Pg2 Kg1 (10)
Black Ra8 Bd8 Bg8 Pb7 Kc7 Qd7 Pb6 Pc6 Pb4 (9)
b) bPc6-d6
C+ by Popeye v4.63 The author's idea is completely preserved. (2023-06-05)
A.Buchanan: I like Ladislav's fix a lot. I humbly suggest r2b2b1/1pkq1RPQ/1p1p2B1/6PR/3p4/8/4K3/6N1 saving two units. Some might say wSg1 is too passive for a White officer, but the way two units (bP & wK) block three rows is sufficiently amusing to offset this, imho. (2023-06-06)
A.Buchanan: The authors fixed it very well with P0523513 so any later fix is pointless. (2023-06-13)
more ...
comment
Keywords: Bahnung, Tempo Move (t t)
Genre: h#
FEN: r2q4/1pkb1RpQ/1p3nB1/6KR/1N4P1/5P2/1P1p4/2b5
Input: hpr, 1997-01-22
Last update: A.Buchanan, 2023-06-13 more...
Genre: h#
FEN: r2q4/1pkb1RpQ/1p3nB1/6KR/1N4P1/5P2/1P1p4/2b5
Input: hpr, 1997-01-22
Last update: A.Buchanan, 2023-06-13 more...
86 - P0506889
Claude Goumondy
3779 Die Schwalbe 73 02/1982
(5+7) C+
h#2
b) wSe8 nach g7
c) wSe8 nach e3
d) wLg5 nach g6
Claude Goumondy
3779 Die Schwalbe 73 02/1982
(5+7) C+
h#2
b) wSe8 nach g7
c) wSe8 nach e3
d) wLg5 nach g6
a) 1. Td4 Td5+ 2. Ke4 Sf6#
b) 1. Td5 Td6 2. Ld4 Te6#
c) 1. Td6 Td7 2. Ke6 Te7#
d) 1. Td3 Td4 2. Sd5 Te4#
b) 1. Td5 Td6 2. Ld4 Te6#
c) 1. Td6 Td7 2. Ke6 Te7#
d) 1. Td3 Td4 2. Sd5 Te4#
klären Jahr?
Anton Baumann: Korrektur in 'Die Schwalbe' 06/1983 S.87: +sDa5 zur Verhinderung der NL in c) (2022-12-21)
comment
Anton Baumann: Korrektur in 'Die Schwalbe' 06/1983 S.87: +sDa5 zur Verhinderung der NL in c) (2022-12-21)
comment
87 - P0506988
Vaclav Gebelt
Problem 07/1981
2. Preis
59. Thematurnier
(10+9) cooked
h#2
b) wLe1 nach h2
Vaclav Gebelt
Problem 07/1981
2. Preis
59. Thematurnier
(10+9) cooked
h#2
b) wLe1 nach h2
a) 1. Dxd5 Sd2 2. De4 Sd5#
b) 1. Txg4 Sg5+ 2. Te4 Sg4#
NL:
a) 1. Kd4 Sxd2 2. Sc3 Lf2#
b) 1. Txg4 Sg5+ 2. Te4 Sg4#
NL:
a) 1. Kd4 Sxd2 2. Sc3 Lf2#
Vitaly Medintsev: Correction
5K1Q/p6p/4RN1p/2nP1Pr1/4N1P1/b3k3/1pPqp3/3rB3 (10+11)
b) SLe1 nach h2
white Be1 Kf8 Qh8 Pd5g4c2f5 Sf6e4 Re6 (10)
black Ba3 Ke3 Qd2 Pa7h7h6e2b2 Sc5 Rg5d1 (11)
b) BBe1 move to h2
Version by Vitaly Medintsev, October 1st 2014 (2014-10-01)
SCHRECKE: Medintsev: P1288792 (2022-10-30)
more ...
comment
5K1Q/p6p/4RN1p/2nP1Pr1/4N1P1/b3k3/1pPqp3/3rB3 (10+11)
b) SLe1 nach h2
white Be1 Kf8 Qh8 Pd5g4c2f5 Sf6e4 Re6 (10)
black Ba3 Ke3 Qd2 Pa7h7h6e2b2 Sc5 Rg5d1 (11)
b) BBe1 move to h2
Version by Vitaly Medintsev, October 1st 2014 (2014-10-01)
SCHRECKE: Medintsev: P1288792 (2022-10-30)
more ...
comment
Keywords: Bahnung, Fesselungsspiel (212 202 000)
Genre: h#
FEN: 5K1Q/7p/3nRN2/3P1Pr1/4N1P1/b3k3/1pPqn3/3rB3
Input: hpr, 1997-01-22
Last update: Alfred Pfeiffer, 2014-10-01 more...
Genre: h#
FEN: 5K1Q/7p/3nRN2/3P1Pr1/4N1P1/b3k3/1pPqn3/3rB3
Input: hpr, 1997-01-22
Last update: Alfred Pfeiffer, 2014-10-01 more...
*) 1. ... f8=S 2. f1=L Le3#
1) 1. f1=T f8=D 2. Tf6 Dxg7#
1) 1. f1=T f8=D 2. Tf6 Dxg7#
t/Allumwandlung, u/61112, u/SDlt
Vorläufer: A.C.White,1930, s.Nr.1056
s. A. C. White 364v L'Echiquier 07-08/1930
Anton Baumann: Die Aufgabe stammt gem. 'Die Schwalbe' 05/1963 von André Simonet, Brüssel
Vorgänger vergl. P0542142 (2022-11-16)
comment
Vorläufer: A.C.White,1930, s.Nr.1056
s. A. C. White 364v L'Echiquier 07-08/1930
Anton Baumann: Die Aufgabe stammt gem. 'Die Schwalbe' 05/1963 von André Simonet, Brüssel
Vorgänger vergl. P0542142 (2022-11-16)
comment
Keywords: Allumwandlung, Promotion
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 4N3/5Pp1/6Pk/7p/4P2P/pp6/pqpp1p2/bbrr2BK
Input: Erich Bartel, 1997-05-23
Last update: hpr, 1999-10-16 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 4N3/5Pp1/6Pk/7p/4P2P/pp6/pqpp1p2/bbrr2BK
Input: Erich Bartel, 1997-05-23
Last update: hpr, 1999-10-16 more...
1) 1. cxb1=S g8=D 2. Sc3 Db3#
2) 1. cxb1=L g8=T 2. Ka2 Ta8#
2) 1. cxb1=L g8=T 2. Ka2 Ta8#
t/Allumwandlung, u/61113, u/DTsl
Yuri Bilokin: you can see 1...g8=Q 2.cxb1=B Qb3#
1...g8=Q 2.cxb1=S Qb3#
1...Bxc2 2.Ka2 Ra4# (2024-02-29)
comment
Yuri Bilokin: you can see 1...g8=Q 2.cxb1=B Qb3#
1...g8=Q 2.cxb1=S Qb3#
1...Bxc2 2.Ka2 Ra4# (2024-02-29)
comment
Keywords: Allumwandlung, Sacrifice of white pieces, Promotion
Genre: h#
Computer test: (Popeye C-Version 3. 47 (1024 KB))
FEN: 8/6P1/8/8/1R6/k2p4/2p5/1BK5
Input: Erich Bartel, 1997-05-23
Last update: hpr, 1999-10-16 more...
Genre: h#
Computer test: (Popeye C-Version 3. 47 (1024 KB))
FEN: 8/6P1/8/8/1R6/k2p4/2p5/1BK5
Input: Erich Bartel, 1997-05-23
Last update: hpr, 1999-10-16 more...
1. Da6+ Ta5 2. Tb6 Tb5 3. Th6 Txg5#
NL:
1. De6 Txf7 2. Dg4 Tb5 3. g6 Txh7#
NL:
1. De6 Txf7 2. Dg4 Tb5 3. g6 Txh7#
Yuri Bilokin: correction wKa1-f1, wRb1-f2, bRb7-b6, +bBh1, +bNg2, -bPf7 8/1r4pp/7q/5Rbk/7p/6p1/5Rn1/5K1b (3+10) h#3
1.Qa6+ Rb5 2.Rb6 Rff5 3.Rh6 Rxg5# (MM)
Blocking piece replacement (bQ-bR)
Helledie theme
Hideaway (bQ, checking)
Loyd's clearance (black, bQ-bR, pure, 7, 6)
Model mate × 1 (2023-05-02)
comment
1.Qa6+ Rb5 2.Rb6 Rff5 3.Rh6 Rxg5# (MM)
Blocking piece replacement (bQ-bR)
Helledie theme
Hideaway (bQ, checking)
Loyd's clearance (black, bQ-bR, pure, 7, 6)
Model mate × 1 (2023-05-02)
comment
Keywords: Check Protection
Genre: h#
FEN: 1r6/5ppp/7q/5Rbk/7p/6p1/8/KR6
Input: Markus Manhart, 1997-06-27
Last update: hpr, 1999-10-17 more...
Genre: h#
FEN: 1r6/5ppp/7q/5Rbk/7p/6p1/8/KR6
Input: Markus Manhart, 1997-06-27
Last update: hpr, 1999-10-17 more...
1. La7 Kg3 2. Kb6 Kf2 3. Kc5 Ke2 4. Kd4 Lh4 5. Lc5 Lf6#
NL:
1. Ld4 Kg4 2. Kb6 Kf4 3. Le5+ Ke3 4. Kc5 Kd2 5. Kd4 Lf2#
NL:
1. Ld4 Kg4 2. Kb6 Kf4 3. Le5+ Ke3 4. Kc5 Kd2 5. Kd4 Lf2#
Keywords: Check Protection
Genre: h#
FEN: 8/8/k7/3p4/4b2K/1P2b3/8/4B3
Input: Markus Manhart, 1997-06-26
Last update: hpr, 1999-10-18 more...
Genre: h#
FEN: 8/8/k7/3p4/4b2K/1P2b3/8/4B3
Input: Markus Manhart, 1997-06-26
Last update: hpr, 1999-10-18 more...
1. Kf8+ Df6+ 2. Sf7+ Kg6 3. Lc6 Df7#
NL:
1. Tc6 Kg4 2. Ke6 Kf4 3. Kd5 De5#
NL:
1. Tc6 Kg4 2. Ke6 Kf4 3. Kd5 De5#
Yuri Bilokin: correction bRb6-g8, +bQg3, +bNg4
3bb1rn/4k1p1/r5p1/6K1/2p3n1/5pq1/8/Q7 (2+12) h#3
1.Kf8+ Qf6+ 2.Sf7+ Kxg6 3.Bc6 Qxf7#
Active sacrifice (black, delayed)
AntiZielElement (B1, check)
Consecutive checks (3). Crosscheck (2)
Self-pin/self-unpin (white). Self-pin/unpin (white) (2024-02-09)
comment
3bb1rn/4k1p1/r5p1/6K1/2p3n1/5pq1/8/Q7 (2+12) h#3
1.Kf8+ Qf6+ 2.Sf7+ Kxg6 3.Bc6 Qxf7#
Active sacrifice (black, delayed)
AntiZielElement (B1, check)
Consecutive checks (3). Crosscheck (2)
Self-pin/self-unpin (white). Self-pin/unpin (white) (2024-02-09)
comment
Keywords: Check Protection
Genre: h#
FEN: 3bb2n/4k1p1/rr4p1/6K1/2p5/5p2/8/Q7
Input: Markus Manhart, 1997-06-28
Last update: hpr, 1999-10-19 more...
Genre: h#
FEN: 3bb2n/4k1p1/rr4p1/6K1/2p5/5p2/8/Q7
Input: Markus Manhart, 1997-06-28
Last update: hpr, 1999-10-19 more...
1. Sd7+ Kd5 2. La7 Kc4 3. Sc5 Kb4 4. Sb7 Sc7#
NL:
1. Sc4+ Kc7 2. Td8 Kxd8 3. Dc8+ Kxc8 4. La7 Sc7#
NL:
1. Sc4+ Kc7 2. Td8 Kxd8 3. Dc8+ Kxc8 4. La7 Sc7#
Yuri Bilokin: correction rotate 90, theh -bNg5, bpe7-e4, bPb6-a5, +bPa3, +bPe3 8/8/3p4/pp6/4p3/p1Kbp3/r1nN1p2/k1qrb3 (2+14) h#4
1.Sb4+ Kd4 2.Bb1 Kxe3 3.Sd3 Ke2 4.Sb2 Sb3# (MM) (2022-12-15)
comment
1.Sb4+ Kd4 2.Bb1 Kxe3 3.Sd3 Ke2 4.Sb2 Sb3# (MM) (2022-12-15)
comment
Keywords: Check Protection
Genre: h#
FEN: kr6/4pp2/qnKp4/rNb5/b5p1/1p6/4n3/8
Input: Markus Manhart, 1997-06-27
Last update: hpr, 1999-10-19 more...
Genre: h#
FEN: kr6/4pp2/qnKp4/rNb5/b5p1/1p6/4n3/8
Input: Markus Manhart, 1997-06-27
Last update: hpr, 1999-10-19 more...
1. Ka3 Sf6 2. Da8+ Se4 3. Dxa4 Sc5 4. Lb2 Txa4#
Autor klären
Felber, Volker: Ich gehe davon aus, dass hier die Korrektur zu P1406481 vorliegt. Dort ist im Original von SCHACH "Eduard Walzinsch" angegeben. In der PDB gibt es beide Schreibweisen für denselben Autor. (2022-12-07)
comment
Felber, Volker: Ich gehe davon aus, dass hier die Korrektur zu P1406481 vorliegt. Dort ist im Original von SCHACH "Eduard Walzinsch" angegeben. In der PDB gibt es beide Schreibweisen für denselben Autor. (2022-12-07)
comment
Keywords: Check Protection
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 6Nq/6b1/6p1/1n6/P6R/8/1k6/5b1K
Input: Markus Manhart, 1997-06-26
Last update: hpr, 1999-10-19 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 6Nq/6b1/6p1/1n6/P6R/8/1k6/5b1K
Input: Markus Manhart, 1997-06-26
Last update: hpr, 1999-10-19 more...
1. La1 Ke7 2. Sb2 Kf6 3. Sd3+ Kg5 4. Le5 Sd2#
NL:
1. Lc3 Kc7 2. Lb4 Lh2 3. Kd4 Sa3 4. Kc5 Lg1#
1. Kd3 Kc7 2. Kc4 Lh2 3. Ld4 Sa3+ 4. Kc5 Ld6#
NL:
1. Lc3 Kc7 2. Lb4 Lh2 3. Kd4 Sa3 4. Kc5 Lg1#
1. Kd3 Kc7 2. Kc4 Lh2 3. Ld4 Sa3+ 4. Kc5 Ld6#
Yuri Bilokin: correction wKe8-f8, +bSb7, +bPg4, +bPh4 5K1b/1n6/8/3b4/n3k1pp/8/8/1N4B1 (3+7) h#4
1.Ba1 Ke7 2.Sb2 Kf6 3.Sd3+ Kg5 4.Be5 Sd2# (MM)
AntiZielElement (B2, line obstruction)
Corner-to-corner (bB)
Hesitation (bB)
Indian (black)
Model mate × 1 (2023-05-24)
comment
1.Ba1 Ke7 2.Sb2 Kf6 3.Sd3+ Kg5 4.Be5 Sd2# (MM)
AntiZielElement (B2, line obstruction)
Corner-to-corner (bB)
Hesitation (bB)
Indian (black)
Model mate × 1 (2023-05-24)
comment
Keywords: Check Protection
Genre: h#
FEN: 3K3b/8/8/3b4/n3k3/8/8/1N4B1
Input: Markus Manhart, 1997-06-26
Last update: hpr, 1999-10-19 more...
Genre: h#
FEN: 3K3b/8/8/3b4/n3k3/8/8/1N4B1
Input: Markus Manhart, 1997-06-26
Last update: hpr, 1999-10-19 more...
96 - P0510656
Nikolay H. Dimitrov
Shakhmatna Misl 11/1966
1. ehrende Erwähnung
Bulgarian Chess Federation Ty
(3+8) cooked
h#3
Nikolay H. Dimitrov
Shakhmatna Misl 11/1966
1. ehrende Erwähnung
Bulgarian Chess Federation Ty
(3+8) cooked
h#3
1. Ka3 Txd3 2. Dg2+ T8d5 3. Db2 Ta5#
Cook: NL
1. Ka1 Tb8 2. Sd2 Txd3 3. Sd5 Ta3#
1. Sd4 Tb8 2. Dg4 Txb4+ 3. Ka1 Ta5#
Cook: NL
1. Ka1 Tb8 2. Sd2 Txd3 3. Sd5 Ta3#
1. Sd4 Tb8 2. Dg4 Txb4+ 3. Ka1 Ta5#
Adrian Storisteanu: Possible fix (adjusted raised position): Kb8 Rd6 Rd8 / Kb3 Qg8 Rc2 Rc5 Sb4 Sb5 pp.d4 f7 (3+8) -- 1.Ka4 Rxd4 2.Qg3+ R8d6 3.Qb3 Ra6#. (2015-09-19)
Yuri Bilokin: correction wRd5-d7, bRc4-b6, bRc2-f6(-bPf6) K2R2q1/3R4/1r3r2/8/1n6/1n1p4/1k6/8 (3+7) h#3
1.Ka3 Rxd3 2.Qg2+ R8d5 3.Qb2 Ra5# (MM)
AntiZielElement (B2, check)
Bristol (white, wR-wR, impure, 4, 3)
Chernous theme
Self-pin/unpin (white)
Model mate × 1
Pin-mate (2023-05-02)
comment
Yuri Bilokin: correction wRd5-d7, bRc4-b6, bRc2-f6(-bPf6) K2R2q1/3R4/1r3r2/8/1n6/1n1p4/1k6/8 (3+7) h#3
1.Ka3 Rxd3 2.Qg2+ R8d5 3.Qb2 Ra5# (MM)
AntiZielElement (B2, check)
Bristol (white, wR-wR, impure, 4, 3)
Chernous theme
Self-pin/unpin (white)
Model mate × 1
Pin-mate (2023-05-02)
comment
Keywords: Check Protection
Genre: h#
FEN: K2R2q1/8/5p2/3R4/1nr5/1n1p4/1k6/2r5
Input: Markus Manhart, 1997-06-26
Last update: Marcin Banaszek, 2020-06-19 more...
Genre: h#
FEN: K2R2q1/8/5p2/3R4/1nr5/1n1p4/1k6/2r5
Input: Markus Manhart, 1997-06-26
Last update: Marcin Banaszek, 2020-06-19 more...
1. Lc5 d4 2. Df7 d5 3. Le7 d6 4. Se8 d7 5. Ke6 dxe8=S
NL:
1. Kg5 La6 2. Kg4 Ld3 3. Kh3 Kf3 4. Kh2 Le4 5. Kh1 Kg3#
1. Kg5 Lb7 2. Kg4 La8 3. Kh3 Lc6 4. Kh2 Kf3 5. Kh1 Kg3#
NL:
1. Kg5 La6 2. Kg4 Ld3 3. Kh3 Kf3 4. Kh2 Le4 5. Kh1 Kg3#
1. Kg5 Lb7 2. Kg4 La8 3. Kh3 Lc6 4. Kh2 Kf3 5. Kh1 Kg3#
Verstellung BL
Yuri Bilokin: possibly bQa7-f7, bBg1-e1, bNg7-d5, +bPa5, +bPg3 2B5/5q2/5k2/p2n4/4K3/6p1/3P4/4b3 (3+6) h#5
1.Sc7 d4 2.Bb4 d5 3.Be7 d6 4.Se8 d7 5.Ke6 dxe8=S# (2023-01-11)
comment
Yuri Bilokin: possibly bQa7-f7, bBg1-e1, bNg7-d5, +bPa5, +bPg3 2B5/5q2/5k2/p2n4/4K3/6p1/3P4/4b3 (3+6) h#5
1.Sc7 d4 2.Bb4 d5 3.Be7 d6 4.Se8 d7 5.Ke6 dxe8=S# (2023-01-11)
comment
Keywords: Line closure, Excelsior white (auto key)
Genre: h#
FEN: 2B5/q5n1/5k2/8/4K3/8/3P4/6b1
Input: Andreas Mokosch, 1997-07-26
Last update: hpr, 1999-04-06 more...
Genre: h#
FEN: 2B5/q5n1/5k2/8/4K3/8/3P4/6b1
Input: Andreas Mokosch, 1997-07-26
Last update: hpr, 1999-04-06 more...
1. Sc6 Ke3 2. Se5 Tc6+ 3. Kd5 Le4#
NL:
1. Kf6 Ke3 2. Kg6 Kf4 3. Kh5 Th2#
NL:
1. Kf6 Ke3 2. Kg6 Kf4 3. Kh5 Th2#
Genre: h#
FEN: 8/n7/4k3/8/8/3K4/2R5/1B6
Input: Markus Manhart, 1997-09-30
Last update: hpr, 1999-10-21 more...
1. Kf4 Td2 2. Ke3 Kg3 3. Le4 Lg5#
NL:
1. Kf4 Lb6 2. Kf3 Kh3 3. Le4 Tf2#
NL:
1. Kf4 Lb6 2. Kf3 Kh3 3. Le4 Tf2#
Yuri Bilokin: correction wKh2-e2, wRe2-b2, wBd8-h1, bKf5-c5, bBa8-d2 8/8/8/2k5/8/8/1R1bK3/7B (3+2) h#3
1.Kc4 Ra2 2.Kb3 Kd3 3.Bb4 Bd5# (2023-06-09)
comment
1.Kc4 Ra2 2.Kb3 Kd3 3.Bb4 Bd5# (2023-06-09)
comment
Genre: h#
FEN: b2B4/8/8/5k2/8/8/4R2K/8
Input: Markus Manhart, 1997-09-30
Last update: hpr, 1999-10-21 more...
1. Sc8 Kd5 2. Te7 Se6+ 3. Kd7 Sf6#
NL:
1. Kc8 Kb6 2. Kb8 Sa6+ 3. Ka8 Sc7#
1. Ta4 Sg7 2. Ta8 Kd6 3. Tc8 Se6#
NL:
1. Kc8 Kb6 2. Kb8 Sa6+ 3. Ka8 Sc7#
1. Ta4 Sg7 2. Ta8 Kd6 3. Tc8 Se6#
Yuri Bilokin: possibly a2=a1, then wNc6-f3, wNe7-g3 8/3k4/4n3/8/2K5/4rNN1/8/8 (3+3) h#3
1.Sc7 Kd4 2.Re6 Se5+ 3.Kd6 Sf5# (IM)
Blocking piece replacement (bS-bR)
Ideal mate (2022-12-31)
comment
1.Sc7 Kd4 2.Re6 Se5+ 3.Kd6 Sf5# (IM)
Blocking piece replacement (bS-bR)
Ideal mate (2022-12-31)
comment
Genre: h#
FEN: 3kN3/2N1n3/8/2K5/4r3/8/8/8
Input: Markus Manhart, 1997-09-30
Last update: hpr, 1999-10-21 more...
Show statistic for complete result. Show search result faster by using ids.
https://pdb.dieschwalbe.de/search.jsp?expression=COMMENTDATE%3E%3D20220810+AND+G%3D%27h%23%27+AND+NOT+K%3D%27vorweggenommen%27+AND+G%3D%27h%23%27
The problems of this query have been registered by the following contributors:
Gerd Wilts (46)hpr (27)
Erich Bartel (4)
Hans-Jürgen Schäfer (9)
Markus Manhart (13)
Andreas Mokosch (1)
Preisbericht: 'Die Schwalbe' 06/2011 S.124 (2023-01-02)
Henrik Juel: How is the SE corner released, without ruining the castling? (2023-01-02)
Mario Richter: Good question, Henrik! I first thought that releasing the SE corner without ruining White's castling right is impossible, but the trick is to uncapture a black Queen in the SE corner at the right moment.
Perhaps Theodore Hwa can use ths problem as a test case for his latest improvement to Retractor 2 ... (2023-01-02)
Henrik Juel: Thanks, Mario
In view of the prize I suspected that the problem was correct, but I did not find the uncapture trick (2023-01-02)
Henrik Juel: C+ Popeye 4.61, because with Black to move White may not castle (2023-01-02)
comment