Die Schwalbe

79 problem(s) found in 3457 milliseconds (displaying 79 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT A='Roscher, Willy' AND K='Überholt'] [download as LaTeX]

1 - P0000136
Dmitri W. Pronkin
Andrey Frolkin

6631v Die Schwalbe 117 06/1989
Preis
P0000136
(14+14)
BP in 57.5
1. a4 h5 2. a5 h4 3. a6 h3 4. axb7 hxg2 5. h4 d5 6. h5 d4 7. h6 d3 8. h7 dxc2 9. d4 a5 10. Lh6 c1=T 11. e4 Tc5 12. Se2 Th5 13. e5 c5 14. e6 Sc6 15. b8=T a4 16. Tb4 a3 17. Ta4 c4 18. b4 c3 19. b5 c2 20. b6 c1=T 21. b7 Tc4 22. b8=T Da5+ 23. Tbb4 Lb7 24. S1c3 0-0-0 25. exf7 e5 26. Tc1 Lc5 27. f8=T a2 28. Tf3 a1=T 29. Sa2 g1=T 30. Tfa3 Tg6 31. f4 Te6 32. f5 g5 33. f6 g4 34. f7 g3 35. f8=T g2 36. Tf5 g1=T 37. Lf8 Tg7 38. Sg3 e4 39. Ld3 e3 40. 0-0 e2 41. Tcc3 e1=T 42. Lc2 T1e3 43. d5 Tdd7 44. d6 Tdf7 45. d7+ Kb8 46. Dd6+ Ka8 47. Dc7 Sge7 48. d8=T+ Sc8 49. Tdd3 Thg8 50. h8=T Tae1 51. Th6 T1e2 52. T1f2 Tce4 53. Kf1 Ld4 54. Tfc5 Se5 55. Sf5 Sc4 56. Sd6 Sb2 57. Tbc4 Sb6 58. Db8+
play all play one stop play next play all
Der absolute KBP-Längenrekord.
See P1338946 cooked.
paul: Compare with P0002278 & P0002279 (2010-04-30)
Mu-Tsun Tsai: This one is by far the toughest retro I've ever solved. Very little certain information can be determined by structural consideration alone, even with long and complicated argument. It took me five days to complete solving this. (2012-07-22)
A.Buchanan: @Mu-Tsun: that's an interesting data point - thanks for posting. (2017-09-07)
Henrik Juel: The current record is 58.5 moves in a proof game problem by the authors + Keym, Die Schwalbe 2017 (2017-09-07)
Henrik Juel: I just learned that the 58.5 move proof game has been cooked... (2017-09-07)
A.Buchanan: In retrospect, my earlier comment about "interesting data point" is a bit weak. It's actually great that for such an extreme problem, someone took substantial time to independently validate it. It's like doing science: people want to do their own new stuff, and are unwilling to take the time to validate what's already been claimed. This one has survived 30+ years, and maybe the use of constraints e.g. in Jacobi can eventually allow it to be HC+. (2021-05-29)
Olaf Jenkner: This problem is the current record, because P1338946 (58.5 moves) has been cooked. (2021-11-25)
Reto: This is C+ up to 51.0 moves with Stelvio 2.0. This ties the record for partial testing of an SPG. Took 1200 CPU hours of strategy seeking (finding 378 0+0 strategies) and another 13h of strategy playing these strategies. If this can ever be completely solved, then it needs to be the case that all strategies have 0+0 free moves, otherwise playing is utterly hopeless.
@Andrew: There is absolutely no way a brute-force based program like Jacobi ever stands a chance at solving something like this, no matter how many conditions you add. (2023-12-14)
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Keywords: Unique Proof Game, Move Length Record, Non-standard material (TTTTTTtttttt), Castling, Aristocrat, Superseded by (P1397486)
Genre: Retro
FEN: kQ3Br1/1b3rr1/1n1Nr2R/q1R4r/R1Rbr3/R1RRr3/NnB1rR2/5K2
Reprints: 584 Ukrainisches Album 1986-1990
86 Shortest Proof Games 11/1991
(6) diagrammes 103 10-12/1992
H18 FIDE Album 1989-1991 1997
feenschach 137, p. 368, 08-09/2000
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-12-25 more...
2 - P0000541
Tivadar Kardos
3398 Die Schwalbe 67 02/1981
P0000541
(13+12) cooked
BP in 6,0
1. Sf3 e5 2. Sxe5 Se7 3. Sxd7 Sec6 4. Sxb8 Dxd2 5. Dxd2 Sxb8 6. Dd8+ Kxd8
play all play one stop play next play all
Cook: 1. Sh3 d5 2. Sf4 Sf6 3. Sxd5 Se4 4. Sxe7 Sxd2 5. Dxd2 Kxe7 6. Dxd8 Kxd8
(Can transpose B1&B2)
HHS ('Schwalbe' Heft 71, Oktober 1981, S. 349): "ein schwarzer Schuft - lässt seinen Zwillingsbruder von einem weissen Söldner killen, um dessen Platz einzunehmen: ein regelrechter Krimi."
Als Vergleichsaufgaben sind in der 'Schwalbe' angegeben: G. Schweig (P0002287), Mortimer (P1394750), W. Naef & H. Klüver (P0002832)
A.Buchanan: It's curious that if we shift sK to e8 (i.e. make the position fully homebase) then it is sound as BP in 6.0! The magic of homebase! :) I don't know if it was a typo - maybe someone can check the magazine archive? (2021-10-19)
A.Buchanan: WinChloe & YACPDB don't know this composition. (2021-10-20)
Mario Richter: No typo, the position with black Kd8 was intended, the cook was found by the 'Schwalbe'-solvers (Heft 71, Oktober 1981, p. 349) (2021-10-21)
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Keywords: Unique Proof Game, Homebase (W), Impostor (s), Superseded by (P1394733)
Genre: Retro
FEN: rnbk1b1r/ppp2ppp/8/8/8/8/PPP1PPPP/RNB1KB1R
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-10-21 more...
3 - P0000548
Alexander Kislyak
3463 Die Schwalbe 68 04/1981
Dr. L. Ceriani zum Gedenken
P0000548
(10+13) cooked
BP in 46,0
AL: 1. d4 a5 2. c4 a4 3. b4 a3 4. Lb2 axb2 5. a4 c5 6. a5 b5 7. a6 Lb7 8. a7 Da5 9. bxa5 Sc6 10. a6 0-0-0 11. a8=S Sa5 12. Sc7 b4 13. a7 b3 14. a8=S Sf6 15. Sb6 Kb8 16. Se6 dxe6 17. Sd7 Ka8 18. Sxf8 Td5 19. cxd5 c4 20. d6 c3 21. d7 c2 22. d5 c1=L 23. d6 Ld5 24. Ta4 Se8 25. Tf4 f6 26. Sc3 b1=L 27. d8=S b2 28. d7 Lg6 29. Sb7 b1=L 30. d8=S Lf7 31. Sg6 Lf5 32. Sd6 Lg4 33. Sf5 hxg6 34. Sb7 gxf5 35. Sc5 Th4 36. S5e4 fxe4 37. e3 Lf5 38. Ke2 Tg4 39. h4 Sb3 40. h5 Sa1 41. h6 Ld2 42. h7 Le1 43. h8=S Sc7 44. Sg6 Sb5 45. Sh4 Sa3 46. S4f3 exf3+
play all play one stop play next play all
Korrektur siehe 3463v
James Malcom: How in the world can a non-unique PG be cooked? I'm unmarking it, for now. (2021-01-25)
A.Buchanan: Maybe the intended theme was not forced? Haven’t looked at this problem though (2021-01-26)
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Keywords: Non-Unique Proof Game, Ceriani-Frolkin Theme, Non-standard material (ll), Castling, konsekutive Umwandlungen 8, Promotion, Superseded by (P0000582)
Genre: Retro
FEN: k7/4pbp1/4pp2/3b1b2/5Rr1/n1N1Pp2/4KPP1/n2QbBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-01-26 more...
4 - P0000615
Andrey Lobusov
4088 Die Schwalbe 77 10/1982
14.Lob
P0000615
(15+6) cooked
h#2* (AP)
1. ... Kg5 2. 0-0?? Se7#
1. ... Txh7 2. Kf8 Txh8#

1. cxb3ep Txh7 2. Kf8?? Txh8#
1. cxb3ep Kg5 2. 0-0! Se7#
play all play one stop play next play all
Wh captured 10 by pawns (including hxPg-g8) but not on the last move. wPa3 came from a2: wPa5 captured bPa.
Bl took only b3xa2.
In set play, Bl has no last move except for K or R.
For the main play, the intention is that if Bl retains castling rights, last moves were R: 1. b2-b4 b3xa2 e.g. 2. Bg1-d4 b4-b3 3. d4xc5 etc. However there are retro cooks e.g. R: 1. Bg1-d4 b3xa2 2. d4xc5 etc, with wPa5 the original wPb. wPb4 & wPb5 come from further East.
Cook: Retro logic for main play is unsound.
A.Buchanan: Not trivial to fix as shifting wPa5 to b6 leads to promotion cooks. Adding black material must be carefully done to avoid disrupting uniqueness of e.p. as tempo. (2022-03-11)
A.Buchanan: OK just to say I have found a form of this idea that works. This one was the very devil to fix - the final form includes different mates and retro logic, so becomes an "after" rather than a "correction", and I will find somewhere else to publish first. (2022-03-19)
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Keywords: Castling (sk), a posteriori (AP) (Type Petrovic), En passant as key, Tempo Move, Superseded by (P1409841)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/5p1p/5K1R/PPP2N2/1PpB4/P1P5/p1RNB1P1/Q7
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-20 more...
5 - P0000674
Leonid M. Borodatow
2475 Die Schwalbe 51 06/1978
P0000674
(12+11) cooked
h#3
1. ... Kgxf4 2. Tf6 e6 3. Tf8 Sg7#
play all play one stop play next play all
Cook: 1. ... exf6ep 2. 0-0-0 gxf4 3. Td7 a8=D
Failed try, as Bg4xBh5 allows bRh8 to have escaped the cage prior to ep.
Anton Baumann: Sollte eine Verbesserung von P0000777 sein.
Autorabsicht: "letzter Zug f7-f5 sonst retropatt! Jetzt ist aber die Rochade illegal, da der sTh8 nur über e8 aus seiner Ecke kommt. Daher Verführung 1. ... exf6 e.p. 2.0-0-0? Lxf4 3.Td7 a8=D#. Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 sg7#"
Aber: Der s h-Bauer kann auch auf h5 geschlagen worden sein; so konnte der sT über h6 aus seiner Ecke entkommen. Die vermeintliche Verführung ist also NL! (2022-12-14)
comment
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...
6 - P0000736
Tivadar Kardos
2935 Die Schwalbe 59 10/1979
P0000736
(10+14) C+
BP in 7,0
1. Sc3 Sf6 2. Se4 Sxe4 3. g3 Sxg3 4. Sf3 Sxh1 5. Se5 Sxf2 6. Sxd7 Sxd1 7. Sxb8 Txb8
play all play one stop play next play all
Sally: Eine eindeutige Zugfolge, was bei solchen Aufgaben nicht oft zutrifft. Nicht ganz leicht. Der Rekord steht bei 41,5 Zügen.
(BS). (2017-09-06)
Henrik Juel: 41.5 moves is a very old record
The current record length for proof games is 58.5 moves in a problem by Pronkin, Frolkin & Keym, problem C, p.7 in Die Schwalbe, heft 283, February 2017
The former record was 57.5 moves in a problem by Pronkin & Frolkin, Die Schwalbe, 1989 (V) (2017-09-06)
Olaf Jenkner: The current record length for proof games is 57.5 moves (P0000136) because P1338946 has been cooked. (2021-11-25)
comment
Keywords: Unique Proof Game, Superseded by (P1396199)
Genre: Retro
Computer test: (Natch 2.2Beta1 Copyright (C) 1997,98,99,2001,2002,2003 Pascal Wassong)
FEN: 1rbqkb1r/ppp1pppp/8/8/8/8/PPPPP2P/R1BnKB2
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-25 more...
7 - P0000777
Leonid M. Borodatow
1424 Die Schwalbe 30 12/1974
P0000777
(12+10) cooked
h#2.5
1. ... exf6ep 2. 0-0-0? Lxf4 3. Td7 a8=D# try
1. ... Kxf4 2. Tf6 e6 3. Tf8 Sg7# solution
play all play one stop play next play all
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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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...
8 - P0000899
Giuseppe Brogi
743 Die Schwalbe 06/1972
P0000899
(8+15) cooked
h#2
b) wSa1
a) 1. T2xh3 Txh3 2. 0-0 Th8#
b) 1. La4 0-0 2. Tf8 Te1#
play all play one stop play next play all
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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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...
9 - P0000914
Vladimir Archakov
852 Die Schwalbe 17 11/1972
P0000914
(14+3) cooked
#2*
Set Play
1. ... Kxh1 2. Kf2#
1. ... Kxh3 2. Sf4#
1. ... Kxf3 2. Lb7#
BTM
1. ... hxg5 2. Sc3
2. ... Kxh1 3. Kf2#
2. ... Kxh3 3. Lf1#
2. ... Kxf3 3. 0-0#
play all play one stop play next play all
Cook: 1. ... hxg5 2. Kd2 Kxh3 3. Df1# 2. ... Kxf3 3. Lb7#
A.Buchanan: We can +wBd2, but then need to e.g. transpose wLg7 & wBg5, otherwise the wBB are impossibly concentrated in corner. Then I think it's sound. (2021-10-08)
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Keywords: Castling (wk), No legal last move for Black, Superseded by (P1394496)
Genre: Retro
Computer test: C- Popeye v4.87
FEN: 8/6BR/B6p/6PP/3P2Np/5P1P/4N1kP/Q3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-12 more...
10 - P0001764
Henri Nouguier
25 Phénix 1 05/1988
P0001764
(9+13) cooked
shc#6
1. dxc3ep 2. Kxb5 3. Kxc6 4. Kd7 5. Ke8 6. 0-0 Th8#
Aber es geht auch R: 1. Kb2-a1!?
play all play one stop play next play all
Henrik Juel: The intention may be 1.dxc3ep 2.Kxb5xc6-d7-e8 6.0-0 Rh8#, but -1.Kb2 seems possible. Should bBc1 be moved to a3? (2003-04-28)
GW: Yes, that's the solution, and the problem seems indeed to be cooked. I don't know if it has been corrected. (2003-04-28)
James Malcom: I believe that the intent was that the sLc1 must be the orginal one, and thus an impediment to wK foreplay, as the wBc6 "must" have captured the e7 Black pawn on its way from f2. The problem is that it can take the alternative path fxsLf3xDe3xTc5-c6, and there are enough White pieces to capture for the Black b pawn to make it d2 and for the d pawn to make it to c2. (2021-09-14)
James Malcom: It turns out this problem actually was corrected a long time later: P1012052 (2021-09-14)
comment
Keywords: En passant, Castling (sk), Seriesmover, Consequent, Non-standard material (sLb1), Valladao Task, Promotion in the retro play (sLb1), Obvious promotion ((sLb1)), Superseded by (P1012052)
Genre: Retro, Fairies
FEN: 7r/5pnR/2P3pR/pPpn3p/1kPp4/3P4/P2pP3/Kbb5
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-09-14 more...
11 - P0002476
Werner Kuntsche
1777v Problem 73-78 06/1961
P0002476
(5+7) cooked
h#3 AP
Einheitslösung!
1) 1. cxd3ep Ta5 2. Sf5 0-0 3. Sd4 Txe5#
2) 1. gxf3ep Th5 2. Sd5 0-0-0 3. Sf4 Txe5#
play all play one stop play next play all
Cook: 707 cooks which do not begin with e.p., e.g. 1.Sg3-e2 Ra1-a2 2.Ke4-d3 Rh1-h3 3.e5-e4 Ra2-d2 #
What is the obscure intent behind "Einheitslösung!"? (= "One size fits all solution"?)
Mario Richter: Im Nachdruck in problem 144-147 steht nur auf e5 ein sB (also keiner auf e3). Allerdings gibt es in allen drei Varianten (sBe5+e3, nur sBe5, nur sBe3) jede Menge NL, die ohne Rochade und e.p.-Key auskommen - am wenigsten dann, wenn nur auf e3 ein sB steht. (2010-10-10)
VL: Cf. P0002475. (2012-08-26)
A.Buchanan: Possibly "Einheitlösung!" is a request to unify retro implications derived across separate parts of the PRA solution. But cooked chessically irrespective of these philosophical issues, which still wait to be resolved 60 years on. (2022-02-16)
comment
Keywords: a posteriori (AP) (Type Petrovic - ccee), En passant as key (2), Castling (wb), Symmetrical position, Symmetrical solution, Superseded by (P1401449,P1399112)
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/8/8/4p3/2pPkPp1/2n1p1n1/8/R3K2R
Reprints: (66) Problem 144, p. 52, 12/1971
Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
12 - P0002554
Unto Heinonen
Probleemblad 1991
1. Preis
P0002554
(16+16) cooked
BP in 43.0
1. h4 a5 2. Th3 Ta6 3. Tf3 Tg6 4. Sh3 Tg3 5. Sf4 Th3 6. g3 Th2 7. Lh3 Sh6 8. Le6 Sf5 9. Lc4 d5 10. Kf1 Th1+ 11. Kg2 Tg1+ 12. Kh3 Kd7 13. Kg4 h5+ 14. Kg5 Th6 15. a4 Tb6 16. T1a3 Kc6 17. Tac3 Tb3 18. Sg6 Ta3 19. b3 Ta2 20. La3 Dd6 21. Lc5 Sd7 22. La7 Sb6 23. Sh8 Kc5 24. La6+ Kb4 25. Tc5 g6 26. Sc3 Lg7 27. Se4 Lc3 28. Da1 Sd4 29. Kh6 Ld7 30. Kg7 Lb5 31. Kf8 Ld3 32. Ke8 Dc6+ 33. Kd8 Sc4 34. Kc8 Ka3 35. Kb8 Lb4 36. Ka8 De8+ 37. Lb8 Sb6+ 38. Ka7 Sa8 39. c4 Sc2 40. Df6 Sa1 41. Db6 Kb2 42. Sf6 Ta3 43. Sd7 Ka2
play all play one stop play next play all
Cook: NL
1. h4 a5 2. Th3 Ta6 3. Tf3 Te6 4. g3 Te3 5. Lh3 Sh6 6. Le6 Sf5 7. Lc4 d5 8. Kf1 Dd6 9. Kg2 Da6 10. Kh3 Kd7 11. Kg4 h5+ 12. Kg5 Th6 13. a4 Tb6 14. Ta3 Kc6 15. Tc3 Tb3 16. Sh3 Ta3 17. b3 Ta2 18. La3 Sd4 19. Lc5 Lf5 20. La7 Db6 21. Sf4 Kc5 22. Sg6 Kb4 23. Sh8 g6 24. La6 Lg7 25. Tc5 Db5 26. c4 Sc2 27. Sc3 Ka3 28. Se4 Lc3 29. Da1 Lb4 30. Df6 Tc3 31. Kh6 Kb2 32. Kg7 Ta3 33. Kf8 Sd7+ 34. Ke8 Ka2 35. Kd8 Sa1 36. Kc8 Sf8+ 37. Kb8 Sd7+ 38. Ka8 Tc1 39. Lb8 Sb6+ 40. Ka7 Sa8 41. Db6 Tg1 42. Sf6 De8 43. Sd7 Ld3 (Sebastian Natschke)
Henrik Juel: Very good, Sebastian (although not for the author...) (2019-06-17)
paul: Probably the cook of the year, because the problem is well known. Which is now the length record for a capture free proof game? (2019-07-16)
paul: The length record for a capture-free PG is now 40.5 moves by Th. le Gleuher (Phenix50/1997) (2019-07-16)
Mario Richter: The problem Paul was referring to is P0006623. So far it has resisted all my and rawbats' attempts to cook it, and I think that chances are high that it is indeed correct ... (2019-07-17)
James Malcom: See the correction P1386520 (2021-02-15)
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Keywords: Unique Proof Game, Move Length Record, Capture-free, Superseded by (P1386520)
Genre: Retro
FEN: nB2q2N/KppNpp2/BQ4p1/p1Rp3p/PbP4P/rP1b1RP1/k2PPP2/n5r1
Reprints: H17 FIDE Album 1989-1991 1997
feenschach 137 08-09/2000
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-28 more...
13 - P0003211
Tivadar Kardos
6476 Skakbladet 10/1957
P0003211
(7+14) cooked
h#3
1. dxe3ep+ Kxc4 2. Lc3 Sxc2 3. Da5 axb3#
play all play one stop play next play all
See the correction P1396159
Cook: 1. dxe3ep+ Kxc4 2. Td5 Sxc2 3. Ta5 axb3#
A.Buchanan: Easy enough to eliminate what I suppose is the intended solution (sL unblocking sD and blocking sT) by e.g. just removing sDf5. However I don't see a way to say goodbye to the cook instead. (2021-11-23)
James Malcom: I believe I may have a fix Andrew. If we substitute the Black queen with a Black rook, the same theme is kept even is if considered watered down.

h#3 8/3p4/3p1p2/3Kbr2/k1ppPp2/1n1r3P/P1p4P/N3nq1b (2021-11-24)
A.Buchanan: Hi James, hurray cool I thought I couldn't get through that way: I was confused in my mind about wPe2 blocking that line, but with the freshness of a new day, I can look at it again. The precise diagram that you propose doesn't quite work because R: 1. e3-e4 Sf3*e1+,Sg2*e1+,g2xf1=Q+,g2xh1=Q+, but this is all fixable with +bPd2,-bSe1. Now Popeye 4.87 says the forward solution is works uniquely, and Retractor 2.0 says R: 1. e2-e4 Rf3-d3+ 2. Ke4-d5 are unique. So for full C+ we only need some demo game that results in the position prior to Ke4-d5.
I don't see any "watering down" here: I think we have faithfully represented Tivadar's idea. Let's post it here as TK, corr JM&AB? (2021-11-24)
A.Buchanan: I have made an unedifying demo game with lichess, so our correction is fully C+. (2021-11-24)
James Malcom: "TK, corr JM&AB" sounds good to me. (2021-11-24)
more ...
comment
Keywords: En passant as key, Superseded by (P1396159)
Genre: h#, Retro
Computer test: Popeye C-Version 4.87 & trivial retro-logic
FEN: 8/3p4/3p1p2/3KbqP1/k1ppPpp1/1n1r3P/P1p4P/N6b
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-24 more...
14 - P0003339
Tivadar Kardos
2519 Revista Romana de Sah 01/1948
2. Preis
P0003339
(10+15) cooked
h# in wieviel?
1. ... cxd6ep 2. Tc8 dxe7 3. Tf8 exf8=S#
play all play one stop play next play all
AL ('Revista Romana de Sah', 06/1948, S.225-226):
Bekanntlich beginnt im Hilfsmatt Schwarz, also hat in der Diagrammstellung Weiß den letzten Zug gemacht. Man findet aber, wenn man die Position analysiert, daß Weiß den letzten Zug nicht gemacht haben kann.
Der weiße Doppelbauer auf der d-Linie läßt sich nur durch Schlagen des einzig fehlenden schwarzen Steines, des sBa7, erklären. Dieser konnte aber erst nach Umwandlung geschlagen werden. Dazu mußte er zunächst bis a3 vorrücken, dann nach b2 schlagen, und dann umwandeln.
Also kann Weiß nicht zuletzt b2-b3 gezogen haben.
Die anderen schwarzen Bauern haben mindestens 5 Schläge gemacht, um die Position im Diagramm zu erreichen (man beachte, daß der sBh7 zweimal schlagen mußte, um den wBh2 durchzulassen)
also kann auch g2-g3 nicht der letzte Zug gewesen sein, denn sonst hätte der wLf1 nicht von einem sB geschlagen worden sein können.
Für alle anderen weißen Steine ist leicht zu sehen, daß sie den letzten Zug nicht gemacht haben können.
Es stellt sich heraus, dass der letzte Zug von Schwarz gemacht wurde, und eine einfache Analyse zeigt, daß dies nur möglich ist durch d7-d5 (sonst wäre es unmöglich, daß die weißen Steine ihre Position im Diagramm erreichen könnten).
In diesem Fall Weiß kann 1. ... c5:d6(e.p.) spielen, gefolgt von 2. Tc8 d:e7 3. Tf8 e:f8C matt!
Das kürzeste Hilfsmatt wird also in 2,5 Zügen erreicht!
Cook: 1. ... Txg5+ 2. Kxg5 f3 3. Lg6 gxf4#
1. ... Kxd5 2. Kf6 e4 3. Tg6,Dg6 e5#
A.Buchanan: Popeye v.487. WinChloe has the same diagram, and although it just claims 1 solution, it's C? Yet WinChloe engine gives the cooks right away (2021-10-27)
Mario Richter: Shouldn't this problem get the Label "No legal last move for White"?
(White pawn d2 has captured the missing black piece (Pdxe), so black pawn a7 had to promote to get to the capture square. This requires one capture (Pa3xb2), together with bxc,cxd,hxg,gxh,gxh this accounts for all missing white pieces.)

Btw. "for me, "C?" stands for "Not yet computer-tested or computer test didn't come to a conlusion" ... (2021-10-28)
A.Buchanan: Hi Mario, thank you.
(1) The stipulation is non-standard. I think this is why the WinChloe database doesn't include a solution, and just marks it as "C?" However if I run the WinChloe solving engine manually it (like Popeye) gives 121 h#3 solutions and 4 h#2.5 solutions. I think you, me and Christian would all agree on the meaning of "C?"
(2) The problem does not rely on Article 15 to eliminate the first single move, but implicitly requests the shortest. Currently the PDB usage of "No legal last move..." is to indicate that Codex Article 15 applies, so I did not add it. But I'm ok if you want me to add the keyword. In my correction, I shift the stipulation to h#3 so it's clear that Article 15 and not any desire for brevity which push the problem to h#2.5.
(3) However the PDB usage is confused. Ideallly think the two "No legal..." keywords should be replaced by "Retro-stalemate", while two new keywords "Article 15" & "Whose mate?" indicate the choice of implications of the retropat for forward play. "Board rotation" and other jokes are also implications of retropat (and of illegal diagram).
(4) But this is complicated to implement, requiring either database-level access or screen-scraping. The latter is quite practical for some future time when I have more leisure.
(5) https://www.stere.ro/biblioteca-digitala/reviste/revista-romana-de-sah/ is an excellent archive of Revista Romana de Sah. My sincere compliments and thanks to whoever prepared it. And the magazines themselves contain thorough annual indices. p225-6 of 6/1948, contains the solution to Kardos. I haven't translated it, nor have I found time to locate the prize announcement.
Comments welcome! (2021-10-29)
Mario Richter: I still believe, that the label "No legal last move for White" is justified here, because this is exactly what the author wanted to show.
(And this is too, what the reprint in 'Problem (Zagreb)' 37-40 09/1956 is all about - it appears there inside asn article

Btw., I wouldn't call the presentation of the solution in 'Revista Romana de Sah', p.225, 06/1948, a "reprint" - it lacks the presentation of the position ... (2021-10-30)
A.Buchanan: Added the keyword.
Yes I know that's the usage of "reprint", but then there should be a field for the location of the solution :-) It's idiotic that if I want to say where the solution is contained, I have to embed it in text like this: 2519 Revista Romana de Sah , p. 225-6, 06/1948 (2021-10-30)
more ...
comment
Keywords: En passant as key, Promotion (S), Superseded by (P1394978), No legal last move for White
Genre: h#, Retro
Computer test: Popeye v4.87 & WinChloe both say cooked
FEN: 7b/4p1nP/2r1P1kq/1nPpKbrp/2pp1pRp/1P4P1/P3PP2/8
Reprints: (III) Problem 37-40 09/1956
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-30 more...
15 - P0003343
Yaakov Mintz
344 Canadian Chess Chat 11-12/1980
P0003343
(4+15) cooked
h#4
1. hxg3ep Lxb1 2. Kf5 fxe4 3. Kg6 e5+ 4. f5 exf6ep#
play all play one stop play next play all
Cook: NL:
1. hxg3ep fxe4 2. f6 e5 3. Kf5 exf6 4. Kg6 Lxb1# uvm
YM: The correct mechanism is in P1109109 (2010-07-13)
YM: Correction option: P1229434 (2021-06-28)
comment
Keywords: En passant as key, Superseded by (P1109109)
Genre: h#, Retro
FEN: q2r4/4Kp2/3r3b/3p2pp/4nkPp/1n3P2/B4p1p/1b6
Reprints: 805 FIDE Album 1980-1982 1988
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-28 more...
16 - P0003365
Gyula Bebesi
41 Problemas 04-06/1962
P0003365
(8+14) cooked
h#2
1) 1. axb3ep bxc6+ 2. b5 cxb6ep#
2) 1. exd3ep g8=D,L+ 2. d5 cxd6ep#
play all play one stop play next play all
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)
more ...
comment
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...
17 - P0003430
Tomislav Petrovic
(XIII) Problem 37-40 09/1956
P0003430
(4+15) cooked
h#3
1. dxe3ep Td1? 2. Le4 Txd6 3. Sg4 Tf6# usual AP try
1. dxe3ep 0-0-0! 2. Le4 Txd6 3. Sg4 Tf6#
play all play one stop play next play all
Cook: 1. dxe3ep 0-0-0 2. Lf5 Txd6 3. Sf3 Td4#
not
1. dxe3ep Td1? 2. Lf5 Txd6 3. Sf3+ ?? usual AP try and also check Bl 3rd move
Henrik Juel: The missing black man is dark-squared, so last move was not fxe4. In the solution Td6 should read Txd6. There seems to be a 'variation': 0... dxe3ep 1.0-0-0 Le4 2.Txd6 Sg4 3.Tf6#, so maybe the stipulation should have 1.1;2.1;1.1 added. (2004-09-23)
A.Buchanan: Well spotted Henrik. Your kindness is legendary, but in the harsh world of helpmates, this counts as a cooked I believe. In fact, it's the 2.Lf5/3.Sf3 version which is the cook. Yours is the actual solution! This is because 3.Sf3 would check if White hadn't castled, and we want the castling to be because of the AP condition only, not for 2 reasons, which one might term a "logical dual". The problem can be trivially fixed by sticking sSe5 on f6. I have no idea whether it's a typo or a cook: WinChloe has the same diagram, and the same confusion between cook & intended solution. The fact that AP does not appear in the stipulation might be a factor. WinChloe as far as I can see has no concept of AP as a keyword, and just picks it up based on a substring of the stipulation. In other compositions, T.Petrovic wasn't shy of flashing the AP acronym in the stipulation, but as usual the detective work peters out in the absence of primary records. (2020-12-08)
more ...
comment
Keywords: En passant as key, Castling (wg), a posteriori (AP) (Type Petrovic), Superseded by (P1382802)
Genre: h#, Retro
Computer test: Popeye v4.85 + retro thinking
FEN: 1n1r4/p1B4p/1p1p4/p3n1p1/3pPk2/3b4/pr5q/R3K3
Reprints: (60) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-16 more...
18 - P0003444
Janko Furman
Miroslav Stosic

7273 Schach-Echo 11/1972
P0003444
(8+14) cooked
h#2
b) Gespiegelt (a1<->h1) & wKd1->e1
a) 1. dxe3ep 0-0 2. Te4 Txf3#
b) 1. exd3ep 0-0-0 2. dxe2 Ld5#
play all play one stop play next play all
Cook: a) 1. Kg3 Kf1 2. Df4 Th3#
1. Kg3 0-0 2. Dh4 Txf3#
A.Buchanan: This is a very heavy position to prevent the possibility of R: 1. c2xb3. I'm not sure why this was done. A much lighter position 8/8/8/5np1/1r1pPkr1/2Bp1p2/1p1P2P1/4K2R with 7 less units achieves the mates soundly. Am I missing something? (2022-02-16)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wk,wg), Superseded by (P1399967), Twinning by board reflection
Genre: h#, Retro
FEN: 8/1p1pp1p1/8/5nq1/1r1pPkr1/1PBp1p2/Pp1P2P1/2n1K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
19 - P0003451
Knud H. Hannemann
3823 Stella Polaris 03/1971
P0003451
(6+8) cooked
h#2 AP
1) 1. gxf3ep 0-0-0 2. Te7 Th4#
2) 1. cxd3ep 0-0 2. Da4 Txa4#
play all play one stop play next play all
AB: I think this is cooked. I don't see how AP castling can justify earlier ep here.
1. Kf3 Bf5 2. Kg2 Be4#
1. Kf3 Bxd7 2. Kg2 Bc6#
1. cxd3 Rf1 2. Qa4 Rxa4#
1. gxf3ep Ra5 2. Kf4 Rh4#
1. gxf3ep Rd1 2. Re7 Rh4#
Also, second solution given has typos 1. *c*xd3ep & 2. Q*a*4. (2002-03-21)
V.Liskovets: Indeed this problem is cooked, and I
failed to correct it preserving symmetry.
Here is a possible correction:
W: Ke1 Ra1 Rh1 Be6 Pd4 Pf4
B: Ke4 Qd7 Rf7 Pc3 Pc4 Pe3 Pg3 Pg4 Bh7 Rb2 Pb3

Another story, justifying its contents (e.p.).
In my opinion, all published treatments are
insatisfactory (the same concerns P0004295
(corrected) by Werner Kuntsche as well).
There is a way to make it sound under the
sophisticated HYBRID stipulation 'AP, pRA':
2 partial solutions legalized JOINTLY by
both castlings (details to be published). (2002-04-02)
VL: Sorry, Andrew, I see no issues with my version (excepting the mentioned special retro-convention/genre for justifying its soundness, of course!). h#2 (pRA&AP). The full solution consists of 2 partial AP-based ones: I 1.gxf3 e.p.(!?) 0-0-0! 2.Te7 Th4#; II 1.cxd3 e.p.(!?) 0-0! 2.Da4 Txa4#.
C+(popeye): h#2 & two h#1.5 after the keys (added Tb2 & Lh7 are cookstoppers).
The main features of the corresponding suitable AP-genre ("consolidated")are rather clear. However, not all subtle aspects have been analyzed thoroughly yet. I have got only a draft manuscript with few examples, still. (2022-02-15)
A.Buchanan: Yes Valery I think your correction is technically sound. Theological debate is something else, heh. Have posted it as separate entry. Please tell me if it was original here, or appeared somewhere else (2022-05-29)
Henrik Juel: I looked in Stella Polaris 1971 p.202-203, where the problem is given without AP under the diagram
It turns out that both the author and the editor (Jan Knöppel) consider the problem correct
After explaining the AP convention the latter writes [my translation]:
The author has elegantly applied the AP conventionn in combination with partial analysis. If we assume that last move was not f2-f4 or T-h1, the possibilities d2-d4, K-e1, and T-a1 remain, but the two latter disappear by 1... 0-0. The other assumption (last move was not d2-d4 or T-a1) is analogous.
So something like Valery's HYBRID stipulation was the intention, and the problem was considered correct in 1971; later developments do not change this fact (2022-05-29)
Henrik Juel: Of course, the obvious cooks 1.Kf3 etc. remain (2022-05-29)
A.Buchanan: I agree. Basic AP Petrovic does not allow multiple ep, nor does RS, so some alignment of AP with PRA is required. I would like to see this codified though. (2022-05-29)
more ...
comment
Keywords: Castling (wb), En passant as key (2), a posteriori (AP) (Type Petrovic), Superseded by (P1401508, P1399178), Quasi-symmetrical position, Partial Retro Analysis (PRA)
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/3q1r2/4B3/8/2pPkPp1/2p1p1p1/8/R3K2R
Reprints: (65) Problem 144-147 12/1971
(8) StrateGems SG19, p. 156, 07/2002
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-29 more...
20 - P0003659
Maurice Jago
8202v Die Schwalbe , p. 148, 10/1980
version of cooked 8202 Die Schwalbe 08/1977
P0003659
(12+14) cooked
h#2
b) sBb4->b5
a) 1. Sf2 Dxf2+ 2. Kh1 0-0-0#
b) 1. Sb1 Ta4 2. Sxg3 Th4#
play all play one stop play next play all
Cook: 1. Txg3 0-0-0 2. La2/Lb3/Lc4 Dxh1#
1. La2/Lb3/Lc4 0-0-0 2. Txg3 Dxh1#
See P0000642
Mario Richter: Hier stimmt was nicht: In der Schwalbe 08/1977 erschien als Nr. 2202 die folgende Aufgabe: 8/Pp2p1p1/Bq1p3p/4p3/1p6/n4pP1/1PPPPr1k/R3KQ1n/; h#2, b) sBb4 nach b5 und der AL a) 1.Tf2-g2 0-0-0 2.Tg2xg3 Df1xh1#; b) 1.Sa3-b1 Ta1-a4 2.Sh1xg3 Ta4-h4#, die aber in a) und b) durch 1.f3xe2 La6xb7 2.Tf2-g2 Df1xg2# nebenlösig war.
Ferner ist in nebenstehendem Diagramm sowohl in a) als auch in b) sehr wohl die w0-0-0 möglich
(Weiß hat als letzten Zug Sc7-a8, Schwarz entwandelt auf b1 und h1, Weiß auf f8); und selbst unter der Annahme, daß die w0-0-0 nur mit sBb4 möglich sei, wäre die Aufgabe mehrfach nebenlösig, z.B. durch 1.Ld5-c4 0-0-0 2.Bd6-d5 Df1xh1# (2010-05-29)
A.Buchanan: Here, the source was incorrectly given as "8202v Die Schwalbe 08/1977". No, that was where the original problem appeared, with a "v" added. So when, as here, we suspect the diagram has a typo, we would have to look through many back issues of Die Schwalbe! It turns out that this correction (itself cooked) was published (as text) in p.148 Die Schwalbe 65 10/1980. I have corrected the source. The reference to the earlier buggy problem can then cleanly be put in the "after" field. The actual typo was missing bPc7. (2024-03-06)
A.Buchanan: With the typo fixed, I think the retro-logic works ok, but there is a single cook family, fortunately repairable (2024-03-06)
more ...
comment
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...
21 - P0003666
Johannes Jacob Burbach
Alain C. White

392 Probleemblad 03/1947
P0003666
(5+8) C+
h#2
b) +sBh6
a) 1. Ta6! Lxc3 2. Ta2 0-0#
b) 1. Ta5! Kd1 2. Ta2 Kc2#
play all play one stop play next play all
Henrik Juel: In b) White may not castle, because last move was done by Ke1 or Th1 (2023-12-02)
A.Buchanan: Cute. Do wPe2 & bPe3 serve any purpose, however? (2023-12-03)
more ...
comment
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...
22 - P0003736
Giuseppe Brogi
1249 Sinfonie Scacchistiche 07-09/1972
P0003736
(12+12) cooked
h#2
b) wSa5
a) 1. f5 Le5 2. 0-0 Th8#
b) 1. Kd8 0-0-0 2. Te8 Lf8#
play all play one stop play next play all
Try to deduce the diagram error, as we do have the intended solutions!
a) wRh7 is original, but cannot have escaped from cage by white cross-capture as there are too many captures needed. On the other hand, Black only has to make 4 captures so the cross-capture is ok.
b) changing the colour of Sa5 shifts the capture balance: now White can cross-capture but not Black, so White but not Black can castle.
Thus the retro-logic works with the diagram as it is, but the forward logic does not. How can we repair the diagram?
See P0000899 a companion problem.
A.Buchanan: Something odd going on here. There are numerous h#2 in both a&b. PRA not normally found in a problem with single solution for each twin. White has 8 pawns so trivially wRh7 must be original. (2018-10-13)
A.Buchanan: There are three similar problems by Brogi which are all twinned RS rather than PRA, but don't have diagram error. P0003743, P0003746 & P0003747. (2018-10-13)
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
Ladislav Packa: Cooked a) and b):
1...b8S and 2...R:h8# (2021-02-10)
A.Buchanan: I think the composer simply forgot that wPb7 can promote. Most of the cooks come from S promotions, but it's also possible to have QR. I've fixed it, rejigging the captures to make them still add up. I've posted in Discord, as I did with the partner composition, and will create entries for them here. (2022-03-15)
milan: +sLb8 sBa7=sT M.Frelih (2023-12-02)
A.Buchanan: Hi Milan - I don't think your suggestion quite works for b). In a) there are 0+2 spare captures, so Black can certainly cross-capture. But in b) there is 1+1 so neither side can cross-capture, so there is no solution. Please compare with P1399806, in which there are 1+2 & 2+1 spare captures, so both twins are sound. (2023-12-03)
milan: Hi Andrew my correction works only with 2.1... solutions, black or white knights on a5. are not important. (2023-12-03)
A.Buchanan: Hi Milan not really clear what you are doing, but if as well as the piece changes you proposed, you also change the stipulation to 2.1... then there is still only one solution. Even if you remove Sa5 entirely as well, there is no White cross-capture possible. (2023-12-04)
more ...
comment
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...
23 - P0004296
Luis Alberto Garaza
1809 Problem 73-78 06/1961
P0004296
(16+13) cooked
h#1? h#2?
1. fxe3ep? Dxa8# No justification for the ep, so just retro try.
play all play one stop play next play all
White pcs: dxexf5,hxBg. Black none. Assume that Black can still castle. White can't have just played h4xg5 as sBh2 would be blocked. White might apparently just have moved D,Se5,Td4,Tc3,g4-g5,a5-a6,e3-e4 or e2-e4 but what might Black have played before? Only the last allows a move sDf1-e1. So set up for AP Type Petrovic is OK.
Cook: 446 candidate h#2
However 27 have no ep, while 59 have both ep & castling.
So a total of 86 viable solutions, 85 of which must be cooks.
s.a. Version P0004341
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004341)
Genre: h#, Retro
Computer test: Popeye v4.87 indicates cook
FEN: r3k3/p1ppp3/Pp6/4NPP1/2PRPp2/2RK2PN/1PBn1PQp/2Brq1b1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
24 - P0004341
Luis Alberto Garaza
(9) Problem 101-102 09/1966
P0004341
(15+12) cooked
h#1 oder h#2
1. fxe3ep Dxa8#? because no AP justification for ep

1. 0-0-0 g7 2. Tf8 gxf8=D#
play all play one stop play next play all
62 apparent h#2, so seems cooked.
Cook: 415 candidate solutions for h#2. 401 begin with ep, of which 48 contain 0-0-0 to justify. There are also 14 solutions without ep, of which 13 begin with 0-0-0. The odd one is 1. Kd8 Sc6+ 2. Kc8 Sge7#
Version zu P0004296

Autor: "In Nr.9, I intended to do something similiar to No. 8 [P0004199], but with a different key: the solution 1. PxP "e.p.", Qxa8 checkmate will not go, for if blackside does not castle, the "en-passant" capture cannot be justified. Though, well considering, there is the demolition 1. PxP "e.p." -any 2. 0-0-0! Qb7 or a8 checkmate.
A.Buchanan: Thanks Mario for retrieving the author's intent with this. So he knew it was cooked! Assume s000 rights remain. R: 1. b5xa6? as sBb promoted on b1. R: 1. h4xg5? as sBh retro-blocked. R: 1. Ke3-d3? impossible check from sBf4. R. 1. K~-d3 d3-d2? illegal check. I like these! So by elimination, R: 1. e2-e4 e3xTd2 2. T~d2 L~e1. So must have sLe1. (2022-03-22)
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004342)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k1N1/p1pp4/P5P1/4PPp1/2PNPprB/3K1QRp/P2p1PB1/4b2n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
25 - P0004342
Luis Alberto Garaza
(10) Problem 101-102 09/1966
P0004342
(15+11) cooked
h#2
1. bxc3ep Dg8#? (thematic try)
1. bxc3ep Dd5 2. 0-0-0 Db7#
play all play one stop play next play all
Cook: 1. bxc3ep Ld6 2. 0-0-0 Dc4#
Author: "From the former [d.h. P0004341], the No. 10 was born, something similar to No. 8 [P0004199] but with Black-castling for it is a help-mate. The mating move 1. ... Qg8? will not go."
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P1399966)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k3/p2p4/8/NP6/BpPPN1PP/B2K1Ppp/QP1p1pP1/Rrb5
Reprints: (55) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
26 - P0004481
Matjaz Zigman
Länderkampf Mazedonien-Slowenien 1969
7. Platz
P0004481
(13+10) cooked
ser-h#6** AP
* 1. ... Sb5#
* 1. ... dxc3#
1. fxe3ep 2. e2 3. exf1=S 4. Se3? 5. Sd5 Sb5# (must castle for AP)
1. fxe3ep 2. e2 3. exf1=S 4. Sxd2 5. Sxb1 6. Sxa3 0-0-0!#
play all play one stop play next play all
Luckily, illegal diagram can be fixed by removal of bPa4.
Then if White castling rights remain, ep is on. One ser-h#5 & 18 ser-h#6 tries exist, in which White does not bother to castle.
We don't know if this was a transcription error or a counting mistake by the composer.
Cook: Too many pawn captures required. For retro problem, illegal position implies unsound
Mike Neumeier: The solution appears to be 1.fxe4 e.p. 2.e2 3.exf1=L 4.Ld3 5.Le4 6.Ld5 dxc3#. Was that the intention? (2013-02-18)
Arno Tüngler: There would even be a solution in 5 moves by 3.exf1=S 5.Sd5 Sb5#
However, in order to justify the e.p.-key there must be an AP prove that e2-e4 (and not a move by the wK or wRa1) was played as White's last move. Thus the only solution giving this in 6 moves is 1.fxe3 e.p. 2.e2 3.exf1=S 4.Sxd2 5.Sxb1 6.Sxa3 0-0-0!# (2013-02-18)
Mike Neumeier: It struck me as odd there was no solution listed. (2013-02-18)
Mike Neumeier: And, if we let black be the idle side, there is 1.Txg1 2.Lxh1 3.Kd1 4.Kc1 hxg1T#. What does the asterisk(*) mean? (2013-02-18)
Henrik Juel: The asterisk means that there is also a white mate if he had the move, in this case two mates, so maybe the stipulation should be ser-h#6**:
1.dxc3,Sb5# (2013-02-18)
Ladislav Packa: It is clear that e.p. is possible only when the wK and wRa1 done neither move.
What convention is used here for the right to castling? (2013-02-18)
Henrik Juel: White may castle, unless you can show that he has lost the right to castle. In this problem there is a major difficulty, I think: the position seems illegal.
White pawns captured b2xc3xd4xe5xf6, g2xf3, and axb, promoting on b8; Black captured bxc, dxc, and g3xh2. We cannot explain the destiny of [Ph7].
Reversing wPh3 and bPh2 seems to handle the illegality, but then the problem can be solved in 5 moves (2013-02-18)
Mike Neumeier: Thanks, Henrik. Maybe the stipulation was a typo. Considering all comments together, perhaps the stip should have been ser-h#5**, with the one solution being the 5-mover Arno gave. I do not think it can be proven, except by a stipulation of ser-h#5 that en passant is possible. Which leads to the question of convention. Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible? Whether there is castling or not here seems immaterial. It is just another 6-move solution. There are 19 6-movers (Popeye) with one queenside castling among them. But only the one 5-mover. (2013-02-18)
Henrik Juel: By convention, an en passant capture as first move is not permitted, unless the pawn double step can be shown by some kind of retro analysis.
(Conversely, a castling is permitted, unless it can be shown by retro analysis that the right to castle has been lost, i.e., that king and/or rook has moved) (2013-02-18)
A.Buchanan: Suggested repair: Remove bPa4. Add AP to stipulation.
As Henrik points out, the diagram as it stands is illegal. We can't swap wPh3 & bPh2, because that allows 2 h#1 cooks.
Suppose we remove bPa4 instead. Then we still at least 5 White pawn captures: bxcxdxexf6 & gxf3. But wPh3 did still move from h2, so there are 3 Black pawn captures bxc, dxc & gxh2. So [bhP] died without capturing or promoting. [waP] must have promoted - so either [baP] was captured to clear the way, or [waP] captured to promote, and later [baP] promoted. Either way, all the numbers add up, and the position is legal. wPe4 cannot have just come from d3. So AP is triggered. Forward logic works just as before without bPa4. There is 1 5-move try, and 18 6-move tries, but AP eliminates them all because of need for castling to retrospectively justify the ep. (2013-02-23)
A.Buchanan: Mike asked: "Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible?"
Answer: No. The stipulation does imply a default player to move, but otherwise cannot be used as a premise to determine game state (castling, en passant). (2022-03-21)
A.Buchanan: WinChloe & Yacpdb have the same cooked diagram as PDB, so I think this cook was in the design. I would like to post a sound version with bPa4 removed, and I think it would just be “Zigman correction” with no one else’s name mentioned in the credits. Is everyone ok with this? (2022-05-30)
Mario Richter: 'Bilten 1970' (printed 1971) contains on pp. 62-81 a complete report on the "Problem Duel Slovenia - Macedonia 1969", the problem here can be found on page 79 and is printed there exactly as given here.
Andrew's handling of the suggested correction (remove black pawn a4) is o.k. for me. (2022-05-31)
more ...
comment
Keywords: Castling (wg), Seriesmover, a posteriori (AP) (Type Petrovic), Illegal position, En passant as key, Promotion (s), Valladao Task, Superseded by (P1401546)
Genre: Retro, Fairies
Computer test: Popeye v4.87 but simple retro-logic shows illegality
FEN: 8/8/5P2/2p1p3/p1pkPp2/N1p2P1P/2PP1PBp/RN2KRbr
Reprints: 104 Bilten 1970 1971
(72) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-01 more...
27 - P0005036
Luigi Ceriani
36 32 personaggi e 1 autore 1955
P0005036
(14+12) cooked
Welches war der erste Zug der sD und des sK?
R: 1. ... Ld8xLc7 2. Lb6-c7 Kb7-b8 3. Lc7-b6 Kc8-b7 4. Lb6-c7 Lc7-d8 5. Lc5-b6 Lb8-c7 6. Lb6-c5 Kb7-c8 7. Ld8-b6 Kc7-b7 8. d7-d8=L Kd8-c7 9. e6xSd7 Sb6-d7 10. Sc5-a4 Sa4-b6 11. Se4-c5 d7-d6 12. Sf6-e4 Lh7-g8 13. Sg8-f6 Lf4-b8 14. Sf6-g8 Lh6-f4 15. Sg8-f6 c7-c6 16. g7-g8=S Kc8-d8 17. f6xTg7 Tg8-g7 18. e5-e6 Td8-g8 19. e4-e5 0-0-0 20. e3-e4 Lf8-h6 21. f5-f6 g7-g6 22. f4-f5 Le4-h7 23. f3-f4 Lb7-e4 24. f2-f3 Lc8-b7 25. Kb5-a5 b7xSa6
play all play one stop play next play all
alternative Auflösung (Mu-Tsun Tsai, PDB 2012-07-22, leicht gekürzt)
R: 1. ... Ld8xSc7+ 2. Sd5-c7 Kc7-b8 3. Sb6-d5 Kb7-c7 4. f2xSg3 Lc7-d8 5. Sd5-b6 Lb8-c7 6. Sf4-d5 Se4-g3 7. Se6-f4 Sc5-e4 8. Sb6-a4 Sa4-c5 9. Sc4-b6 Kc8-b7 10. Sd8-e6 Kc7-c8 11. d7-d8=S Kd8-c7 12. e6xTd7 Tb7-d7 13. Se3-c4 d7-d6 14. Sf5-e3 Le5-b8 15. Sg3-f5 Lg7-e5 16. Sf5-g3 Lf8-g7 17. Sh4-f5 g7-g6 18. Sf3-h4 Sg6-h8 19. Sh4-f3 Se5-g6 20. Sg6-h4 Lh7-g8 21. Sh8-g6 Le4-h7 22. Sg6-h8 Ke8-d8 23. Sh8-g6 Sf3-e5 24. h7-h8=S Ld5-e4 25. g6xDh7 Dh2-h7 26. g5-g6 Db8-h2 27. g4-g5 c7-c6 28. g3-g4 Dd8-b8 29. e5-e6 Tb8-b7 30. e4-e5 Lb7-d5 31. e3-e4 Lc8-b7 32. Kb5-a5 b7xSa6 33. Kc4-b5 Sd4-f3 34. Kd3-c4 Sc6-d4 35. Sc5-a6 Ta8-b8 36. Se4-c5 Sb8-c6 37. Ke2-d3 Sc5-a4 38. Ke1-e2 Da6-a3 39. Sg5-e4 Dh6-a6 40. Ta3-b3 Se4-c5 41. Db3-b2 Sf6-e4 42. Tb2-c2 Sg8-f6 43. Dd1-b3 Sf6-g8 44. Sf3-g5 Sg8-f6 45. Sg1-f3 Dh1-h6 46. Ld3-b1 h2-h1=D 47. Ta4-a3 h3-h2 48. Tb1-b2 h4-h3 49. b2-b4 h5-h4 50. Th4-a4 Sf6-g8 51. Th1-h4 Sg8-f6 52. h2xTg3 Tg6-g3 53. Lf1-d3 Th6-g6 54. e2-e3 Th7-h6 55. Sc2-a1 Th6-h7 56. Sa3-c2 Th7-h6 57. Ta1-b1 Th8-h7 58. Sb1-a3 h7-h5 59. c2-c3
Korrekturversuch s. P0003009
Henrik Juel: If you want a great solving challenge, this is the retro for you.
If you need a hint:
[Dd8] never moved, and Black castled (as you may have guessed).
If you need another hint:
Last move was Ld8xLc7+.
I gave up, but Nikolai told me the solution:
-1... Ld8xLc7 -2.Lb6 Kb7 -3.Lc7 Kc8 -4.Lb6 Lc7 -5.Lc5 Lb8 -6.Lb6 Kb7 -7.Ld8 Kc7 -8.L=d7 Kd8 -9.e6xSd7 Sb6 -10.Sc5 Sa4 -11.Sd4 d7 -12.Sf6 Lh7 -13.Sg8 Lf4 -14.Sf6 Lh6 -15.Sg8 c7 -16.S=g7 Kc8 -17.f6xTg7 Tg8 -18.e5 Td8 -19.e4 0-0-0 -20.e3 Lf8 -21.f5 g7 -22.f4 Le4 -23.f3 Lb7 -24.f2 Lc8 -25.Kb5 b7xSa6 etc. (2012-07-22)
Mu-Tsun Tsai: Once I heard "great challenge" I started working. But I came to a complete different conclusion. Not only [Qd8] can move, but the first move of the black king need not be castling. Here's the proof game. After playing
1.c3 h5 2.b4 Rh6 3.Na3 Rg6 4.Nc2 Rg3 5.hxg3 Nf6 6.Rb1 Ng8 7.Na1 Nf6 8.e3 Ng8 9.Bd3 Nf6 10.Rb2 Ng8 11.Bb1 Nf6 12.Qb3 Ng8 13.Rc2 Nf6 14.Qb2 Ng8 15.Rh4 Nf6 16.Rd4 h4 17.Rd5 h3 18.Ra5 h2 19.Ra3 h1=Q 20.Rb3 Qh5 21.Nf3 Qa5 22.Ke2 Qa3 23.Kd3 Na6 24.Kc4 Nc5 25.Kb5 Na4 26.Ne5 Nh5 27.Nd3 Nf4 28.Nc5 Nd3 29.Na6 bxa6+ 30.Ka5 Bb7 31.e4 Qb8 32.e5 Bc8 33.e6 Qb7 34.g4 Qf3 35.g5 Qh3 36.g6 Qh7 37.gxh7 Nf4 38.h8=N Bb7 39.Ng6 Be4 40.Ne5 Bh7 41.Nc4 Bg8 42.Ne3 Ng6 43.Nc4 Nh8 44.Ne3 g6 45.Nc4 Bg7 46.Ne3 Be5 47.Nc4 c6 48.Ne3,
you could either play
48...O-O-O 49.Nc4 Bb8 50.Ne3 d6 51.Nc4 Rd7 52.exd7+ Kb7 53.d8=N+ Kc7 54.Ne6+ Kb7 55.Nb6 Nc5+ 56.Na4 Ne4 57.Nc5+ Kc7 58.Nd7 Kb7 59.Nb6 Ng3 60.Nd5 Bc7+ 61.Nb6 Bd8 62.fxg3 Kc7 63.Nd5+ Kb8+ 64.Nc7 Bxc7+,
which is castling version, or,
48...Kd8 49.Nc4 Kc7 50.Ne3 Rd8 51.Nc4 Kc8 52.Ne3 Bb8 53.Nc4 d6 54.Ne3 Rd7 55.exd7+ Kb7 56.d8=N+ Kc7 57.Ne6+ Kb7 58.Nc4 Bh7 59.Nb6 Nc5+ 60.Na4 Ne4 61.Nd4 Kc7 62.Ne6+ Kc8 63.Nd4 Kb7 64.Ne6 Bg8 65.Nc5+ Kc7 66.Nd7 Kb7 67.Nb6 Ng3 68.Nd5 Bc7+ 69.Nb6 Bd8 70.fxg3 Kc7 71.Nd5+ Kb8+ 72.Nc7 Bxc7+
which is none castling version. Both reach the diagram position.
Why am I feeling cooking Ceriani's problems too much lately? (2012-07-22)
Mu-Tsun Tsai: Also in your retraction, -11.Nd4 should be -11.Ne4 I believe? It seems like your retraction also works just fine, and looks like it should be the intended solution. (2012-07-22)
Henrik Juel: Yes, the intended solution should have -11.Se4, and it can be shortened a bit.
I am impressed by your cook, Mu-Tsun, with two white pawn captures on g3 and promotion on h8, but it is also a little sad that a seemingly fine problem has been rendered worthless.
Probably several more Ceriani problems will be cooked, because they were not scrutinized well enough by testers and solvers in the old days; now, when I finally have cracked a Ceriani nut, I have no energy left to search for errors (2012-07-23)
Mu-Tsun Tsai: I've been thinking about how this problem might be fixed, but unfortunately I cannot come up with anything other than adding extra assumptions in the stipulation, for example "g3 pawn came from h2". The structure of this one is good, and either method of releasing the position (mine or the intended one) is quite subtle, so I feel sad about have to cook this one as well. (2012-07-23)
Thomas Volet: In his 1961 book Ceriani discusses the cook with the WhP unpromoting on h8 and uncapturing to the g file and back to the h file, and gives P0003009 as the corrected diagram position. (2012-08-02)
Mu-Tsun Tsai: This is a really late comment, but I do think this will make a great problem by changing the stip to "You don't know the first move of the black queen nor the black king"! (2023-06-29)
comment
Keywords: First Move? (kd), Superseded by (P0003009)
Genre: Retro
FEN: 1k4bn/p1b1pp2/p1pp2p1/K7/NP6/qRP3P1/PQRP2P1/NBB5
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-07-02 more...
28 - P0005275
Otto Kerekes
Tivadar Kardos

10 L'Echiquier de France 11/1956
P0005275
(11+15) cooked
h#2
b) sBg2 statt wBg2
a) 1. cxd4 Lxc4 2. Dd5 Ld3#

b)
play all play one stop play next play all
Paulo Roque: Illegale Stellung. Diagrammfehler? (2009-11-17)
Alfred Pfeiffer: außerdem in b) 9 schwarze Bauern. (2010-01-13)
A.Buchanan: Twin a) is also illegal, as too many Bl captures (2021-11-24)
more ...
comment
Keywords: En passant as key, Castling, Illegal position, Superseded by (P1400824)
Genre: h#, Retro
FEN: 8/p1p2p1b/1B3pp1/q1p1b3/r1pPk3/1Bp3P1/PPP2PPr/3n1RK1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-23 more...
29 - P0005367
Ted Brandhorst
3500 British Chess Magazine 12/1973
P0005367
(0+0)
BP in which both sides get their pieces all on squares of the same colour?
(AL: 12,5)
1. a3 a5 2. c3 b6 3. d4 d6 4. e3 e5 5. f4 f6 6. g3 g5 7. h4 h6 8. Sd2 Ta7 9. Kf2 Lg7 10. De1 Kf8 11. Lh3 Lxh3 12. Txh3 Se7 13. Th2
play all play one stop play next play all
Hauke Reddmann: Beim Vorfuehren (im Club vor voellig problemunerfahrenen
Patznasen) kam einer auf
folgende Idee:

1.c3 e5 2.a3 Bxa3 3.Nxa3 a5 4.e3 b6 5.f4 d6 6.g3
Ne7 7.h4 f6 8.Bh3 Bxh3 9.Rxh3 Ra7 10.Rh2 Kf8 11.Kf2 h6 12.Qe1

2 Halbzüge gespart! (2002-11-19)
Ryan McCracken: It's possible to do this faster with a good capturing scheme:
1.e3 d6 2.Ba6 Bh3 3.Bxb7 Bxg2 4.Bxa8 Bxh1 5.Be4 Bf3 6.Bxh7 Bxd1 7.Bxg8 Bxc2 8.Nc3 Bb3 9.Bxf7+ Kxf7 10.axb3 Kf6 11.b4 reaching a monochrome position in a mere 10.5 moves. (2004-10-01)
A.Buchanan: I don't think this "cooked". It's not even unsound: just the author's effort to solve a synthetic challenge, and twice the solution has been superseded, that's all. (2020-11-03)
comment
Keywords: Synthetic problem, Superseded by (P1381496), Non-Unique Proof Game, Ornament
Genre: Retro
FEN: 8/8/8/8/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-11-04 more...
30 - P0005911
Unto Heinonen
2375 U.S. Problem Bulletin 01-02/1992
P0005911
(13+16) cooked
BP in 34,5
1. Sc3 h5 2. Se4 h4 3. Sg3 hxg3 4. h4 Th5 5. Th2 gxh2 6. Tb1 hxg1=S 7. Ta1 Sh3 8. Tb1 Sf4 9. Ta1 Sd5 10. Tb1 Sb4 11. Ta1 Td5 12. Tb1 g5 13. Ta1 Lg7 14. Tb1 Ld4 15. Ta1 Lb6 16. Tb1 c5 17. Ta1 Dc7 18. Tb1 De5 19. Ta1 Ld8 20. Tb1 b6 21. Ta1 La6 22. Tb1 Lc4 23. Ta1 a6 24. Tb1 Ta7 25. Ta1 Tc7 26. Tb1 Tc6 27. Ta1 Tf6 28. Tb1 d6 29. Ta1 Kd7 30. Tb1 Ke6 31. Ta1 Kf5 32. Tb1 Kg4 33. Ta1 Tf5 34. Tb1 Sf6 35. Ta1
play all play one stop play next play all
Korrektur siehe P0008426.
James Malcom: What is the cook? (2023-05-02)
more ...
comment
Keywords: Unique Proof Game, Non-standard material, Pendulum (T x30), Superseded by (P0008426)
Genre: Retro
FEN: 1n1b4/4pp2/pp1p1n2/2prqrp1/1nb3kP/8/PPPPPPP1/R1BQKB2
Reprints: (12) feenschach 117 11/1995
Input: Gerd Wilts, 1995-06-26
Last update: James Malcom, 2023-05-02 more...
31 - P0006972
Tivadar Kardos
1357 feenschach 24 08/1974
P0006972
(11+12) cooked
h#3
1. Lxa2 Txa2 2. 0-0-0 Txa7 3. Td7 Ta8#
1. Lxa2 0-0-0? 2. Dxg2 Tg1 3. 0-0 Txg2# but White can't castle
play all play one stop play next play all
Cook: 1. Lxa2 Lf6 2. Lb1 Txa7 3. Kf8 Txa8#
1. Lxa2 Lf6 2. Lxb3 Txa7 3. Kf8 Txa8#
1. Dxg2 Txb1 2. Dxf2+ Kxf2 3. 0-0 Tg1#
1. Dg3 Txb1 2. Dxb3 Txb3 3. 0-0 Tg3#
paul: White Rook h1 left the S-E camp, so the white castle is no more possible.
Intention: 1.L×a2 T×a2 2.0-0-0 T×a7 3.Td7 Ta8#
1.L×a2 0-0-0 2.D×g2 Tg1 3.0-0 T×g2#
Cooked by 1.L×a2 Lf6 2.Lb1 T×a7 3.Rf8 T×a8# or
1.D×g2 T×b1 2.D×f2+ K×f2 3.0-0 Tg1# (2011-09-15)
milan: wBe5-f4 M.Frelih (2021-12-01)
A.Buchanan: @Milan: that removes all 4 cooks, and alas the intended try also. (2021-12-01)
comment
Keywords: Cant Castler, Obvious promotion (l), Superseded by (P1396304)
Genre: Retro
FEN: r3k1qr/pp3p1p/2p5/3pB2p/8/1P6/P1PPPPPP/Rb2K3
Input: Gerd Wilts, 1996-08-11
Last update: A.Buchanan, 2021-12-01 more...
32 - P0501022
Zdravko Maslar
Problem 1958
Sonderpreis
P0501022
(1+13)
h=34
1. f3 Kb8 2. f2 Ka8 3. f1=L Kb8 4. Ld3 Ka8 5. Lb1 Kb8 6. La2 Ka8 7. Db1 Kb8 8. f5 Ka8 9. f4 Kb8 10. f3 Ka8 11. f2 Kb8 12. f1=L Ka8 13. Tf2 Kb8 14. Kf7 Ka8 15. Ke6 Kb8 16. Kd5 Ka8 17. Kc4 Kb8 18. Kc3 Ka8 19. Kb2 Kb8 20. Ka1 Ka8 21. Tb2 Kb8 22. T8f2 Ka8 23. Lf5 Kb8 24. Lc2 Ka8 25. d3 Kb8 26. Le3 Kc7 27. e5 Kxd6 28. Lc1 Ke6 29. Td2 Kf5 30. e4 Kg4 31. Le2+ Kxh3 32. e3 Kg2 33. Led1+ Kf1 34. e2+ Ke1=
play all play one stop play next play all
James Malcom: This is outdone by P0501023. (2022-04-27)
more ...
comment
Keywords: Rex solus (w), Superseded by (P0501023), Move Length Record
Genre: Fairies
FEN: K1b2rk1/4pr2/1b1n1p2/5q2/3p1p2/1p5n/8/8
Reprints: The Problemist 07/2008
Input: Gerd Wilts, 1996-06-06
Last update: James Malcom, 2022-04-27 more...
33 - P0502612
John Niemann
1259 FEENSCHACH 102 05/1952
3. Preis
P0502612
(8+6) cooked
h#4
Autorabsicht: 1. Lh2 Th1 2. Lg1 Th8 3. Lh2 Th1 4. Lg1 T1h7#
play all play one stop play next play all
Cook: Viele NL, zB: 1. Sh4 Txg1 2. Sf3 gxf3 3. Kh6 Txg3 4. Kh5 Th1#
klären Autorlösung
Anton Baumann: Autorabsicht: 1.Lg2 Th1 2.Lg1 Th8 3.Lh2 Th1 4.Lg1 Th7#
Korrektur Version A. Bulawka: -sSg6, sKg7 nach f7, +sBe6, +sBf5, +sBg6 (vergl. yacpdb 320494) (2020-07-15)
Ladislav Packa: The modified version has an illegal position. (2020-07-15)
A.Buchanan: A good, sound, economical version is 320495 in YACPDB (2020-07-15)
A.Buchanan: Bulawka's correction 320494 is sound and legal (6 sB captures + 2 wL died at home). But Smirnov made a much lighter sound version 320495. (2021-05-16)
more ...
comment
Keywords: Superseded by (P1389613)
Genre: h#
FEN: 8/6k1/5pn1/8/8/6p1/PPKPPpP1/RR4b1
Input: hpr, 1996-07-23
Last update: A.Buchanan, 2021-05-16 more...
34 - P0505412
Martin Gohn
2456v Revista Romana de Sah 11-12/1947
3. Preis
P0505412
(3+13) cooked
h#5
1. Kd7 g4 2. Lh1 Lg2 3. Ke6 Lh3 4. Ld5 g5+ 5. f5 gxf6ep#

NL:
1. Td5 Lxb7 2. Sf6 g4 3. Sg8 Kxf8 4. Kd7 g5 5. Ke6 Lc8#
play all play one stop play next play all
Korrekturvorschlag HJS Co+: wKg7, wLa8, wBg2, sKc7, sDd6, sTf8h8, sLb7c1, sSa5, sBb2b5c4e2e5f7h6h7
Yuri Bilokin: Correction: bQb1-a4, bPe3-a6, -bRd3 B4n1r/1bk2pKn/p2p2p1/4p3/qp6/8/6P1/8 (3+12) (2021-03-03)
A.Buchanan: @Yuri: no improvement appears in WinChloe. The "Korrekturvorschlag HJS Co+" B4r1r/1bk2pKp/3q3p/np2p3/2p5/8/1p2p1P1/2b5 is sound, but at 3+15 much more expensive than your economical 3+12. To my surprise, I found I can improve your version still further as B4n2/1bk2pKn/3q4/4p3/8/8/6P1/8 3+7! bQd6 is a very effective cook-stopper. (2021-03-04)
Viktoras Paliulionis: Nice improvement. Compare with P1235831. (2021-03-05)
more ...
comment
Keywords: Bahnung, En passant as mating move, Model mate, Superseded by (P1388090)
Genre: h#
FEN: B4n1r/1bk2pKn/3p2p1/4p3/1p6/3rp3/6P1/1q6
Input: hpr, 1996-12-27
Last update: A.Buchanan, 2021-04-03 more...
35 - P0510541
Dragomir Rusin
Problem 41-44 03/1957
3. Preis
15. Thematurnier
P0510541
(3+14) cooked
h#4
1. ... Txg7 2. Td3 Txg6 3. Sd4 Txg4 4. Sb3 Ta4#
play all play one stop play next play all
Cook: NL
1. ... Txg7 2. Ka1 Txf7 3. Sg3 Tf2 4. Dc8 Lxd4#
Yuri Bilokin: Correction: 4nrRB/2p2pbK/p3p1p1/5n2/3r2q1/5p2/k7/1b6 (3+14) (2019-01-19)
A.Buchanan: The diagram matches the published one (2020-10-01)
A.Buchanan: Yuri’s correction is sound and seems economical (2020-10-01)
more ...
comment
Keywords: Check Protection, No legal last move for White, Superseded by (P1389481)
Genre: h#, Retro
FEN: 5rRB/pp2ppbK/2pp2p1/5n2/3r2q1/8/k7/1b6
Input: Markus Manhart, 1997-06-26
Last update: A.Buchanan, 2021-05-12 more...
36 - P0510770
Christopher John Feather
342v diagrammes 11-12/1976
P0510770
(3+15) cooked
h#5
1. c1=L Lxd3 2. Ld2+ Lb1 3. Le1 Lxe4 4. Lg3+ Lb1 5. Se1 Lf5#

NL:
1. c1=L Lxd3 2. Td1 Kb1 3. Ld2+ Kc2 4. Le1 Lxe4 5. Lg3 Lf5#
play all play one stop play next play all
Yuri Bilokin: Correction: wPa2-b2, bPa3-b3, bBe4, -bPf6, bPg2 8/8/8/8/1n2bp1p/1p1prn1k/1Pp3pp/KB4rq (3+14) (2021-03-04)
A.Buchanan: According to WinChloe, 8/8/5p2/8/1n2pp1p/1p1prn1k/1Pp3bp/KB4rq was a 1998 repair by Chris Feather himself, but Yuri's fix is more economical: bravo! (2021-03-04)
Viktoras Paliulionis: Correction: P0577628. Capturing a pawn is better than capturing a bishop. In the Yuri's version bSb4 can be changed with bPd5. (2021-03-05)
A.Buchanan: Hi Viktoras, thanks for this. Taken in isolation, capturing a pawn is of course better than capturing a bishop, but is it worth an extra cook-blocker on the board? The problem to compare with CJF's fix is your version 8/8/8/3p4/4bp1p/1p1prn1k/1Pp3pp/KB4rq 3+14, with a knight less. Personally, I might actively prefer that Le4 has another role as well as being a blocker, but maybe this isn't the common view. If Le4 moved around, I might have a different opinion. Does CJF opine on this point in "Black to move?" (2021-03-05)
Viktoras Paliulionis: Capturing of black guarding piece is always looks rude (bB guards the mating square), in addition, we get double motivation for the move. CJF justifies even two completely unnecessary white pawns on the board to emphasize the idea. (see 167). For 143, he suggests a less economical position to make motivation cleaner. (2021-03-05)
A.Buchanan: Ok very clear thanks! (2021-03-05)
comment
Keywords: Check Protection, Superseded by (P0577628)
Genre: h#
FEN: 8/8/5p2/8/1n2pp1p/p2prn1k/P1p3bp/KB4rq
Input: Markus Manhart, 1997-06-27
Last update: A.Buchanan, 2021-03-05 more...
37 - P0515072
Thomas R. Dawson
4056 Stratford Express 18/09/1937
P0515072
(4+5) cooked
h#2
1. d4 Sh6 2. Tg5 Ta6#
play all play one stop play next play all
Cook: NL:
1. d4 Kg8 2. Tf5 Ta6#
Adrian Storisteanu: Possible fix (a bit rough, but the wK is still employed):
Kg7 Ra1 Se6 Sf5 / Ke5 Rb4 Rd2 Sh8 pc4 pd3 (4+6)
1.c3 Sg5 2.Rf4 Ra5#. (2019-10-12)
A.Buchanan: Pushing wS to f8 & h6 seems to fix the cook. Maybe that's where they were meant to be all along: hard to imagine old TRD making such an error. WinChloe hasn't got this one, so can't confirm. (2019-10-14)
Adrian Storisteanu: Nice workaround, Andrew! Dawson had his share of cooks... (2019-10-14)
Yuri Bilokin: possibly wNf7-f5, bRe2-f2, -bPe4 7K/8/5kN1/1r1p1N2/8/8/5r2/R7 (4+4) (2022-12-16)
comment
Keywords: Symmetrical position, Asymmetrical solution, Superseded by (P1389169)
Genre: h#
FEN: 7K/5N2/5kN1/1r1p4/4p3/8/4r3/R7
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: A.Buchanan, 2021-05-02 more...
38 - P0530949
Felix Seidemann
1740 Teplitz-Schönauer Anzeiger 30/03/1930
P0530949
(4+3) C+ cooked
h#2*
* 1. ... Sf3,Sg6 2. fxg5 g4#
1. fxg5 g3 2. gxh4 g4#
play all play one stop play next play all
Adrian Storisteanu: Possible fix: +wPg6 (5+3) [*1...Sf3]. (2019-10-11)
Adrian Storisteanu: (Or a diagram typo.) (2019-10-12)
A.Buchanan: Hard to guess the idea. WinChloe hasn't heard of this one. (2019-10-14)
SCHRECKE: simple correction: -bPf6, -wPg5, +bPg7, (3+3)!
1. ... Sg6? stalemate!, *) 1. ... Sf3 2.g5 g4#, I) 1.g5 g3 2.g:h4 g4# (2019-10-15)
A.Buchanan: Sorry: I never saw SCHRECKE's reply. This is very clean, and the witty idea now shines out. Although the difference between the two phases is one *Black* move, it's *White* who has the tempo play. (2021-04-30)
Adrian Storisteanu: Not quite. White in fact never plays a waiting move. In the set play Sh4 must move in order to protect h4, while 1...g3 in the solution is not tempo - a straight 2...g2-g4#? would not be mate: 3.hxg3e.p.!
In SCHRECKE's correction - Set play try: 1...Sg6? 2.what?? g4#; Set play: 1...Sf3! 2.g5 tempo g4#; Tries: 1.g6+? Sxg6 2.what?? g4#, 1.g5! Sg6/f3? 2.what?? g4#; Solution: 1.g5 g3! 2.gxh4 g4#. (2021-05-02)
A.Buchanan: There was a discussion in MatPlus a while back about tempo loss, which was the theme of a tourney. I entered with the same perspective as you. It's only tempo loss if the move could be replaced by a "what". But other folks said no that's not how tempo loss is usually seen: it's possible to have impure motivation, as we see here with the desire to avoid e.p. (2021-05-02)
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comment
Keywords: Sacrifice of white pieces, Tempo Loss (w), Superseded by (P1389113)
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/5p1p/5KPk/7N/8/6P1/8
Input: Ralf Krätschmer, 1998-03-30
Last update: A.Buchanan, 2021-04-30 more...
39 - P0552475
Aurel M. Karpati
Raketa 07/1941
P0552475
(8+7) cooked
h#5
1. bxa1=S Sg6 2. Sxb3 Sh4 3. Sd4 Lc4 4. Sdf3 Lf1 5. Sh2 Lg2#
play all play one stop play next play all
Cook: NL
1. Se2 Sg6 2. Sd4 Sh4 3. Sxc2 Kxc2+ 4. b1=T Sf3 5. Tg1 Txg1#
Adrian Storisteanu: Possible fixes: wSa1 (instead of the wR); or bBg1 (instead of the bS), +wPf3 (9+7). (2015-08-07)
Yuri Bilokin: Correction: wSa1, wSf8-g8, -bPb4, -bPf5 7N/8/3p4/3P4/5P2/1Pp5/BpP5/NK4nk (8+5) (2021-02-11)
Hans-Jürgen Manthey: die Fen von Yuri (wSf8-h8 nicht wSf8-g8): 1. b2xa1S Sh8-g6 2. Sa1xb3 Sg6-h4 3. Sb3-d4 La2-c4 4. Sd4-f3 Lc4-f1 5. Sf3-h2 Lf1-g2# (2021-02-11)
Viktoras Paliulionis: Another possible fix: wSa1 wSf8-b6 -bPb4 -bPd6 -wPf4 -bPf5 (7+4) (8/8/1N6/3P4/8/1Pp5/BpP5/NK4nk) (2021-02-12)
A.Buchanan: Viktoras’ version is much more economical with an open board, and I think no more pawns can be trimmed. I prefer bS starting on h2 rather than g1 because 3. Sc1 is also a try. (2021-02-14)
A.Buchanan: So shall we publish this new version here in PDB? Five people to repair one person’s composition :-) (2021-02-14)
more ...
comment
Keywords: Promotion (s), Superseded by (P1386489)
Genre: h#
FEN: 5N2/8/3p4/3P1p2/1p3P2/1Pp5/BpP5/RK4nk
Input: Hans-Jürgen Schäfer, 1998-07-01
Last update: A.Buchanan, 2021-02-15 more...
40 - P0574383
Tivadar Kardos
The Fairy Chess Review 1956
P0574383
(9+14) cooked
h#3
1. cxb3ep+ Sb4+ 2. Kb1 Sxe2 3. b2 0-0#
play all play one stop play next play all
Cook: NL 1. Db2,Sb3 Sxe2 2. Sb3,Db2 0-0 3. Lxb4 Sxb4#
A.Buchanan: Two aspects of cookery here. First, NL 1. Db2,Sb3 Sxe2 2. Sb3,Db2 0-0 3. Lxb4 Sxb4# Second, White can retract c3xb4 (and earlier captured onto c-file). Black could have captured axbxcxdxe, exf, fxe, with wPg waylaid (2021-11-23)
more ...
comment
Keywords: En passant as key, Superseded by (P1396163)
Genre: h#, Retro
Computer test: Popeye v4.87 says cooked
FEN: 8/8/1p6/b1P2P2/1PpqPpp1/r2ppb1p/k1N1p2P/n3K1NR
Input: Felber, Volker, 1999-12-20
Last update: A.Buchanan, 2021-11-25 more...
41 - P0582614
Branko Koludrovic
Mat (Belgrade) 1984
P0582614
(5+15) cooked
h#3
1. fxg1=S c8=S 2. Sh3 hxg8=S 3. b1=S Txe1#

NL:
1. Ke5 c8=D 2. Dd6 Dc1 3. f6 De3#
1. Se7 Kh3 2. Sf6 Lh2 3. Td8 cxd8=S#
play all play one stop play next play all
milan: i've tested by Linux 4.87 128 MB C+ M.Frelih (2022-12-11)
A.Buchanan: what's your suggested position, Milan? (2022-12-11)
milan: my purpose was to show difference beetwen gustav program and linux.i've solved with linux and found intended solution in 30 sec. Gustav found cooks? M.Frelih (2022-12-12)
A.Buchanan: When you say Linux, do you mean Popeye? When I solve with helpmate analyser, it finds a lot of cooks e.g. those shown here already (2022-12-13)
Ulrich Voigt: Just for the record, Popeye on Linux does find those cooks (11 solutions in total). (2022-12-14)
milan: i've tested in YACPDB site C+ M.Frelih (2022-12-17)
more ...
comment
Keywords: Superseded by (P0559785)
Genre: h#
FEN: 4n1nr/2Pb1p1P/2q1k3/3p1p2/8/8/pp3ppK/r3b1BR
Input: hpr, 2001-03-25
Last update: A.Buchanan, 2022-12-13 more...
42 - P1012059
Ronald Turnbull
15 diagrammes 10/1990
P1012059
(4+12) cooked
h#3 (AP)
PRA
1. bxc3ep 0-0! (for AP reasons, not 1...Kf2?/Th2? etc) 2. Ta1 Txa1 3. Kc4 Txa4#
1. fxe3ep 0-0! (for AP reasons, not 1...Tf1?/Txh3? etc) b2 Txf3 3. Tc3 Tf4#
play all play one stop play next play all
Si les B. peuvent roquer, leur dernier coup est Pc2-c4 ou Pe2-e4. D'où les 2 solutions alternatives: 1.hc3:e.p!. O-O 2.Tal Tal: 3.Rc4 Ta4:# l.fe3:e.p. O-O 2.b2 Tf3: 3.Tc3 Tf4#
Cook: 1. Kxe4 Tg1/Txh3 2. Ke3,Sd4 Tg5/Th5 3. Sd4,Ke3 Txe5# (4 variants)
No. 11669 HN
paul: Cooked: 1.K×e4 Tg1 2.Ke3 Tg5 3.Sd4 T×e5# (2011-11-25)
A.Buchanan: This nice simple puzzle can be repaired by replacing pawn c5 with rook (C+). As often with cooks in PDB, one wonders if it's just a typo. I have an old email address of Ronald's, and I may ask him. (2012-02-20)
A.Buchanan: I met Ronald for dinner last year (it seems like a vanished age!) in Norwich, UK. He was there for a mountaineer's conference. The climbers selected Norwich as their venue for humorous reasons as it's in the flattest area of Britain. We discussed a bunch of compositions, including this one. He agreed it was cooked and agreed with my suggested fix, which I will post in a separate entry. Very nice chap - he doesn't compose much these days because he can get a bigger audience (and more money) from writing hill-walking books. My hill-walking friends say he's a good writer. (2020-12-08)
comment
Keywords: a posteriori (AP) (Type Petrovic - cee), En passant as key (2), Castling (wk), Superseded by (P1382808), Partial Retro Analysis (PRA)
Genre: Retro, h#
Computer test: cooked by Popeye v4.85
FEN: 8/8/8/1np1n3/ppPkPp2/rp1p1p1p/8/4K2R
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2022-02-18 more...
43 - P1013759
Hans Gruber
1470 Problemkiste 04/1986
P1013759
(6+6) cooked
h=4
1. e1=D g7 2. Dh1 g8=S 3. Db7+ axb7 4. a1=L bxc8=T=
{ NL} 1. a1=L a7 2. e1=L a8=D 3. La5 Dxa5 4. Kc6 Kxc8=
Nachdrucke:
1) 5 Grüße aus Augsburg, 30. 9. 1998
play all play one stop play next play all
{ NL}
Anton Baumann: ergänzen: +wBb4 so C+ Gustav 4.1d (2020-11-26)
A.Buchanan: Not in WinChloe. Have emailed Hans (2020-11-26)
A.Buchanan: Managed to find fix with no extra material - Hans is ok if I post mildly changed material here, see P1382844. (2020-12-10)
comment
Keywords: Allumwandlung, Superseded by (P1382844)
Genre: Fairies
Computer test: C- Popeye v4.85
FEN: 1Kn5/3k4/P3p1P1/4P3/8/4P3/pp2p3/1N6
Input: Hans Gruber, 2004-05-01
Last update: A.Buchanan, 2020-12-10 more...
44 - P1066823
Tomislav Petrovic
The Ural's Problemist 04-06/2004
2. Thematurnier
4. ehrende Erwähnung
P1066823
(10+4) cooked
s#6
1. hxg6ep Kh5 2. Dg4+ Kh6 3. Dh5+ Kxh5 4. g4+ Kh6 5. f4 fxe6+ 6. dxe6 hxg6#
play all play one stop play next play all
Cook: Retro logic doesn’t work
(As well as major dual)
So unsound
SCHRECKE: Konnte zuletzt nicht auch 0. ... Kg7-h6 geschehen sein? (2021-02-15)
A.Buchanan: I think the diagram must be wrong. Not only does the retro logic fail, but also there is a dual 2. Df3+,Dg4+. It's not in WinChloe. (2021-02-15)
Mario Richter: An additional white Pawn e7 would justify the epkey, but not prevent the mentioned dual. (2021-02-16)
VL: We might shift wPf2 to f3 against the dual. Or exchange Bf6 with Pf2 and add wP on f3 (instead of e7)... (2021-02-16)
A.Buchanan: Shifting wBf2 to f3 leads to short cook solutions 1.Dh2. However I think 8/5p1p/4NB1k/3PPKpP/4P3/8/5PP1/2Q5 fixes the retro issue, and it's C+ by Popeye, with still ideal mate! The composer Tomislav Petrovic is still alive but would be very old (90). Is he still active? (2021-02-17)
A.Buchanan: By the way, if you query for k='foo', that will always return uses of the keyword "foobar". It's just checking for the prefix. If you omit k='foo', then that will also block "foobar". Therefore if we tag with "en passant as key", then we should not tag with "en passant" too. Both keywords should exist, and over time we may refine uses of "en passant" to be more precise.
Because of the sensible way prefixes are handled in PDB, this is a different situation from tagging a problem as both "ceriani-frolkin" and "prentos", which we should do.
On this general subject, when counting instances within a single problem, we should always use Gerd's excellent mechanism of parameters "foo:1" etc, rather than having separate keywords "foo 1", "foo 2" etc. Otherwise the set of keywords becomes unmanageable. (2021-02-17)
more ...
comment
Keywords: En passant as key, Ideal mate, Superseded by (P1386589)
Genre: Retro, s#
FEN: 8/5p1p/4PB1k/3PPKpP/4PQ2/8/5PP1/8
Input: Frank Müller, 2004-11-01
Last update: A.Buchanan, 2021-02-18 more...
45 - P1067894
Gerald Irsigler
12643 Die Schwalbe 212 04/2005
P1067894
(12+12) C+
26 konsekutive Schachgebote mit Umwandlungsfiguren
1. Dgh5+ Deh4+ 2. Dhxh4+ Ddxh4+ 3. Dcxh4+ Dhxh4+ 4. Dbxh4+ Dhxh4+ 5. Daxh4+ Dhxh4+ 6. Dg3+ Ddxg3+ 7. Dxg3+ Dhxg3+ 8. Dxg3+ Dxg3+ 9. Kf1+ Dgf2+ 10. Dexf2+ Ddxf2+ 11. Dfxf2+ Dxf2+ 12. Txf2+ Txf2+ 13. Txf2+ Txf2+
play all play one stop play next play all
James Malcom: I've looked at the source on page 42: https://www.dieschwalbe.de/hefte/schwalbe_212_April_2005.pdf
And the solution on page 53: https://www.dieschwalbe.de/hefte/schwalbe_215_Oktober_2005.pdf

The diagram and solution match exactly. Where is the mistake, or perhaps there was a misprint of some kind? (2021-02-06)
Mario Richter: White's first move is 1. Dh5+!, not 1. Dh4+! (2021-02-06)
comment
Keywords: Consecutive checks (26), Construction task, Non-standard material, Aristocrat, Superseded by (P1386254)
Genre: Fairies
Computer test: rawbats
FEN: 5R1q/5R1q/5Q1q/6Q1/QQQqq3/qQQqQ3/rrqqQK1k/8
Input: Gerd Wilts, 2005-05-21
Last update: Mario Richter, 2021-02-06 more...
46 - P1094839
Vladislav Nefyodov
Rashid Usmanov

Shakhmatnaya Poeziya 42 2008
5th Honourable Mention
P1094839
(3+12) cooked
h#3
b) sBg2 nach h2
a) 1. Txh7+ Kxh7 2. Kh2 Txg4 3. Sg3 Th4#
b) 1. Lxg7+ Kxg7 2. Kg2 Lxf5 3. g3 Le4#

NL:
a) 1. Txh7+ Kxh7 2. Kh2 Tg5 3. g3 Th5#
b) 1. Txh7+ Kxh7 2. Kh4 Kg6 3. Tg3 Th7#
play all play one stop play next play all
Yuri Bilokin: correction bBe6-f7 7K/5bRB/8/5p2/6p1/4n1kr/q4pp1/b4nr1 (3+12) (2023-05-22)
A.Buchanan: This exact correction was made by the authors in 2009, see P1094987. (2023-05-22)
comment
Keywords: Superseded by (P1094987)
Genre: h#
Computer test: Popeye Windows-32Bit v4.51 (100000 KB)
FEN: 7K/6RB/4b3/5p2/6p1/4n1kr/q4pp1/b4nr1
Input: hpr, 2009-11-21
Last update: A.Buchanan, 2023-05-22 more...
47 - P1106295
Thomas R. Dawson
776 Deutsche Schachblätter 15/12/1912
P1106295
(14+10)
#2
1. hxg6ep Lg5 2. Dxg5#
R: 1. g7-g5 Te3-e7+ 2. Kg5-h6+ Lf5-h7+
play all play one stop play next play all
A.Buchanan: I posted this in Facebook today - Lion Xray asked if wBh2 can be removed - I think it can. (2021-02-16)
Henrik Juel: Without Ph2 last move could also be f6xg5, it seems (2021-02-16)
Mario Richter: I too think the wPh2 can be removed, since then f6xg5 as Black's last move is still impossible (think of wPf2+h2) (2021-02-16)
A.Buchanan: With only one Wh unit unaccounted for, bPg & bPh never left their files. So wPh5 must have captured in. Either fxgxh or hxgxh. In the former case, the original wPh was "waylaid" captured unpromoted on its own file. In the latter case, wPf must have promoted on f8 without capturing. I.e. bPf captured to clear the way. However Black’s last move can’t have been f6xg5, because wPf would not have had a free path to march to f8. The count is correct in either case. (2021-02-17)
A.Buchanan: Another possibility is that wPf was waylaid, to allow bPf to promote on f1. Again, Black’s last move was not f6xg5. (2021-02-17)
A.Buchanan: "Superseded by" was replaced by "better version". I am not averse to improvements in terminology, but here I have reverted, for the following reasons: (1) A keyword should describe the compositions which are tagged by it. "Better version" does not describe a problem it tags, but indicates that *another* problem is the "better version". (2) A superseding problem may not be a version, but could be radically different matrix tackling a record or task. I have left the new German term "bessere Version" alone, and put this comment in the keyword text. (2021-02-18)
comment
Keywords: En passant as key, Superseded by (P1386614)
Genre: 2#, Retro
FEN: 3KN3/qrp1R2B/pp5k/Q5pP/PPP4b/1p4Pp/1PP4P/2B5
Input: Frank Müller, 2010-05-29
Last update: A.Buchanan, 2021-02-18 more...
48 - P1108454
Werner Keym
Schach-Echo 1967
P1108454
(15+5) cooked
#1
b) wDa5 nach e5, AP
a) BTM
1. ... bxa3ep! 2. Dc3#! (2.0-0#??)
1.0-0#?
b) BTM pushed by Wh
1. ... bxa3ep! 2. 0-0#! (2.Dc3#? AP needs the 00)
1.0-0#?
play all play one stop play next play all
(a) Wh made 11 pcs, right to left, which accounts for all missing Bl units. wBa never captured, so sBa must have captured a3xSb2, prior to Wh a2-a4. Since Wh is missing only 1 unit, bBh promoted on h1, disrupting Wh 00 rights. Moreover Bl has no last move and by Codex Article 15, it’s Bl to move. Last move by Wh must allow Bl a prior move. This can only be R: 1. a2-a4 a3xSb2. So ep is on and there is no alternative.
(b) Again Wh has made 11 pcs, but Black may now have made 2. If WTM, then Black's last move was R: 1. Kc2xSc1 Td1-d2+ 2. Kc3-c2 or R: 1. cxSb1=T or R: 1. cxSb2 (as wBc4 & wBc7 might be foreign to c-file and allow bBc a smooth retraction). In any case, sBa3xSb2 was played so sBh-h1= disrupting 00. So Wh can still not play 1. 0-0#! Now White pushes the move to Bl and also asserts that bxa3ep is on. Both claims will be "proved" by Wh 0-0. We know that with WTM, Wh can't castle, so that's half of it. But if BTM with Wh to castle, then bPaxb & bPhxg, and the only way we can retract is as in (a). The mating move is different, because we have that extra Bl pc to keep castling legal, but we must actually castle so that the whole AP logic can work!
However, I don't understand why Wh can't just play 1. Txb2#?
Cook: In (b) Wh can just play 1. Txb2#
b) Urdruck in Die Schwalbe 228, 12/2007
A.Buchanan: Any ideas about (b)? (2022-06-09)
Henrik Juel: Andrew, you can find Heft 228 on
https://www.dieschwalbe.de/archiv.htm
part b) is mentioned in an article by Valery Livkovets, p.299-304
diagram p.301, solution p.303 (2022-06-09)
A.Buchanan: Thanks Henrik - just the pointer I needed. Please check the detailed solution I'll post above. (2022-06-09)
A.Buchanan: OK the logic nearly works for me. But I don't understand why White can't just play 1. Txb2# in part b, because b2 is now covered by wD. What am I missing? (2022-06-09)
Henrik Juel: Indeed, 1.Txb2# looks like a cook (2022-06-09)
A.Buchanan: Werner agrees. He has fixed it with a better twinning condition (2022-06-12)
more ...
comment
Keywords: Cant Castler, En passant as key, No legal last move for Black, a posteriori (AP) (Type Petrovic), Castling, a posteriori (AP) (Type Keym), Superseded by (P1406456)
Genre: Retro
FEN: 8/2P5/3P4/QP2N1B1/PpP5/1P1P4/1p1RBP2/brk1K2R
Reprints: (10) Die Schwalbe 228, p. 301, 12/2007
Input: Gerd Wilts, 2010-06-23
Last update: A.Buchanan, 2022-12-05 more...
49 - P1109508
Valery Kirillov
Mykhailo Myshko

1234 Pat a Mat 54 2006
Informalturnier 2006/07
Lob
P1109508
(15+3) cooked
s#6
Schwarz hat keinen letzten Zug, also
1. ... h5! 2. Tc3 h4 3. Tf7+ Kh6 4. Le3+ Kh5 5. Tf5+ Kg4 6. h3+ Kxg3 7. Lg1+ Lxc3#
mit Schwarz am Zug:
1. Te4? h5 2. Le3 h4 3. Seg6 h3 4. Tg4 hxg2 5. Lh6+ Kxh6 6. Sf4+ Lxf6#
play all play one stop play next play all
Cook: 1) NL: 1. ... h5 2. Df3,Df4,Df5 h4 3. Dg4+ Kxf6 4. Tf3+ Ke5 5. Tf7 Lg7,Lf6,h3,hxg3 6. Ld6+ Kxd6+ 7. Dd4+ Lxd4# (bad wBb6 blocks Kc7).
Anton Baumann: ohne den wBb6 ist die Aufgabe C+ (Gustav 4.1d) (2021-05-12)
A.Buchanan: WinChloe has the identical diagram with wBb6, but doesn't pick up the cook (2021-05-12)
A.Buchanan: Note need to have something like "1)" at the beginning of the cook, else it assumes that the play is a variation from the retro try which began 1. Te4? (2021-10-09)
Mario Richter: The award ('Pat a Mat', issue 61 (April 2008), pp. 190-193) too has the "bad wBb6"; the distinction given is "1. Lob" ["1. cestné uznanie"] (2021-10-09)
A.Buchanan: Pat A Mat site has Tourney records only back to 2010. I've sent emails to request that the authors be contacted. I have given Anton's name as the corrector. (2021-10-10)
A.Buchanan: Resolved, see P1394513 (2021-10-12)
more ...
comment
Keywords: No legal last move for Black, Superseded by (P1394513)
Genre: s#, Retro
FEN: 5N1b/4N1kP/1P2PR1p/2B5/8/4R1P1/P5PP/KB3Q2
Input: Frank Müller, 2010-07-03
Last update: A.Buchanan, 2021-10-11 more...
50 - P1113604
Erich Bartel
Jugendschach (46) 06/01/1985
P1113604
(6+3) cooked
Illegal position.
White to move and h#1.5*
*) Null-Satz!

1. ... Lb6 2. e5 Lc5#
play all play one stop play next play all
Cook: 1. ... Ka7,Ka8+ 2. Kxc7 b8=D#
A.Buchanan: Can probably be fixed by shunting everything to the left by 1 file. Alternatively, if one of the cooks can be suppressed, the other could represent a second solution. (2018-04-29)
Erich Bartel: keine cooks,
siehie dazu in Jugendschach (46) 6.01.1985, S.66-67,72,73
Theodor Steudel:Das Thema Null bei der Mattforderung,
Elmar Bartel & Erich Bartel:Nachbetrachtungen zum Thema Null,
Hans Gruber:Noch ein Wort zum Thema Null (2018-04-29)
A.Buchanan: OK danke, aber können Sie bitte hier schreiben, warum Weiss kann nicht 1... Ka7/8+ spielen. (2018-04-30)
A.Buchanan: Not reasonable to ask people to refer to 3 articles from a minor journal 33 years ago. Erich can you please give a simple explanation (English or German) for why 1... Ka7/8+ aren't cooks. Note that Thema Null is not a fairy condition but a theme: so what's going on? (2018-11-23)
A.Buchanan: By the way: I REALLY LIKE Thema Null. It is a wonderful theme. But I need clarity how it is intended to operate here. This was the only one of the Lese Majeste problems solved by Popeye v4.81 which was cooked. (2018-11-23)
A.Buchanan: Failing an explanation, I've marked this as cooked, following both Popeye & Jacobi (2021-05-11)
more ...
comment
Keywords: Thema Null, Lese Majeste, Illegal position, Fata Morgana, Superseded by (P1399587)
Genre: h#
Computer test: Popeye v4.87 & Jacobi v0.6.1
FEN: 1K6/1PBbp3/3k4/3P4/1P2P3/8/8/8
Input: Erich Bartel, 2010-08-26
Last update: A.Buchanan, 2022-03-02 more...
51 - P1180287
Karl Hermann Schwarz
R427 feenschach 21 04/1974
P1180287
(14+13)
42 konsekutive Schachgebote mit Umwandlungsfiguren
1. Sed3+ Kf1+ 2. Lg2+ Sxg2+ 3. Tf4+ Sxf4+ 4. Lg2+ Sxg2+ 5. Tf4+ Sxf4+ 6. Lg2+ Sxg2+ 7. Tf4+ Sxf4+ 8. Lg2+ Sxg2+ 9. Tf4+ Sxf4+ 10. Lg2+ Sxg2+ 11. Sf4+ Te2+ 12. Scxe2+ Dc1+ 13. Sxc1+ Te2+ 14. Scxe2+ Dc1+ 15. Sxc1+ Sc4+ 16. Dxc4+ Td3+ 17. Scxd3+ Lc1+ 18. Txc1+ Lxc1+ 19. Dxc1+ Dd1+ 20. Dxd1+ Se1+ 21. De2+ Txe2+
play all play one stop play next play all
Dr. Zagler schreibt hierzu in fs-21: "Nur einem extremen Zufall ist es zuzuschreiben, daß diese Rekordstellung von Dr. Schwarz nicht verschollen blieb. Erhalten habe ich sie von dem Münchener Schachfreund und Partiespieler S. Rathgeber, dem sie vor über 20 Jahren gezeigt wurde, als er noch ein Schüler war und in Stuttgart lebte. Er hat sie sich damals notiert und seither aufbewahrt. Durch ein beiläufiges Gespräch ist diese Angelegenheit nun doch noch ans Tageslicht gekommen."
Autor: Dr. Schwarz
HBae: Autor unklar, welcher Schwarz ? (2010-11-15)
Henrik Juel: 'Dr. Schwarz' may well be a pseudonym.
The position is legal, but just barely so. (2010-11-15)
Olli Heimo: The current record is 51, look at http://www.xs4all.nl/~timkr/chess2/diary.htm
item 378.

378. 25 March 2008: New record by Alexey Khanyan - 51 consecutive checks (2010-11-16)
HBae: Thank You Olli (2010-11-17)
Mario Richter: The author might have been Dr. Karl Hermann Schwarz.

The position has some interesting features:
- With Black to move, there are 43 consecutive checks: 1. ... Rh3-d3+ 2.Nc1xd3+ Kf2-f1+ 3.Be4-g2+ Nf4xg2+ 4.Rd4-f4+ Ng2xf4+ 5.Bd5-g2+ Nf4xg2+ 6.Rc4-f4+ Ng2xf4+ 7.Bc6-g2+ Nf4xg2+ 8.Rb4-f4+ Ng2xf4+ 9.Bb7-g2+ Nf4xg2+ 10.Ra4-f4+ Ng2xf4+ 11.Ba8-g2+ Rh2xg2+ 12.Ne1xg2+ Qc7-c1+ 13.Nd3xc1+ Re7-e2+ 14.Nc1xe2+ Qc8-c1+ 15.Ne2xc1+ Re8-e2+ 16.Nc1xe2+ Bb2-c1+ 17.Ne2xc1+ Nd6-c4+ 18.Qa6xc4+ Qh5-e2+ 19.Nc1xe2+ Ba3-c1+ 20.Ne2xc1+ Nf4-d3+ 21.Qc4-f4+ Bh6xf4+ 22.Ng2-e3+ Bf4xe3+
- slightly modified (if the bQ is put on g1 instead of h5), then with WTM 44 consecutive checks are possible. (2010-11-17)
James Malcom: The current record is 54, achieved in 2013: P1386254 (2021-02-06)
comment
Keywords: Consecutive checks (42), Construction task, Non-standard material, Aristocrat, Superseded by (P1386254)
Genre: Fairies
FEN: B1q1r3/1Bq1r3/Q1Bn3b/3B3q/RRRRBn2/b6r/1b1K1k1r/R1N1N3
Input: HBae, 2010-11-15
Last update: James Malcom, 2021-02-06 more...
52 - P1240697
Dmitri Baibikow
C22 Length Records in Last Single Moves? 11/2009
P1240697
(12+10) cooked
Last 46 single moves?
R: 1. Tc3xDc2 h5xSg4 2. Se5-g4 h6-h5 3. Sf3-e5 h7-h6 4. Se1-f3 Dd1-c2 5. Sc2-e1 Df1-d1 6. g5-g6 f2-f1=D 7. g4-g5 f3-f2 8. g3-g4 g4xSf3 9. Se5-f3 g5-g4 10. Sd7-e5 g6-g5 11. Sc5-d7 g7-g6 12. Sa4-c5 Sd1-b2 13. Sb2-a4 Sf2-d1 14. f5-f6 Sh3-f2 15. f4-f5 Sg1-h3 16. f3-f4 g2-g1=S 17. f2-f3 f3xDg2 18. Dh3-g2 f4-f3 19. Dc8-h3 f5-f4 20. Da8-c8 f6-f5 21. a7-a8=D f7-f6 22. a6-a7 a7xDb6 23. Kc5-b5 La5-b4
play all play one stop play next play all
Cook: alternative Auflösung (mri, PDB 03.07.2023)
R: 1. Tc3xDc2 g5-g4 2. f5-f6 f6xSg5 3. Sf3-g5 f7-f6 4. Se1-f3 Dd1-c2 5. Sc2-e1 Dg1-d1 6. g5-g6 g2-g1=D 7. g4-g5 h3xDg2 8. Dg3-g2 h4-h3 9. Db8-g3 h5-h4 10. Da8-b8 h6-h5 11. a7-a8=D h7-h6 12. a6-a7 a7xSb6 13. Kc5-b5 La5-b4 14. Kd5-c5 Da4-b3 15. Ke5-d5 Sd6-c4 16. Kf4-e5 Se8-d6 17. Kg3-f4 Lb5-d3 18. Le7-a3 Sd3-b2 19. Ta3-c3 Kb2-c1 20. Td1-d2 Sc1-d3 21. Th1-d1 Sd2-b1 22. Kg2-g3 Sf1-d2 23. Se1-c2 Sf6-e8 24. g3-g4 f2-f1=S 25. Sf3-e1 c2-c1=S 26. d2xTe3 Tc3-e3 27. Sg1-f3 Tc8-c3 28. Kf1-g2 Ld7-b5 29. Le6-a2 Th8-c8 30. Ta1-a3 Sg8-f6 31. f4-f5 f3-f2 32. c3xBd4 Kb3-b2 33. Lh3-e6 g4xDf3 34. La3-e7 g5-g4 35. Lc1-a3 g7-g5 36. b2xTc3 Tc8-c3 37. Df2-f3 Ta8-c8 38. De1-f2 Lc8-d7 39. d3xBe4 c3-c2 40. f3-f4 Kb4-b3 41. c2xSd3 Dd7-a4 42. Dd1-e1 Kc5-b4 43. Sc4-b6 Kd6-c5 44. Sa3-c4 Ke7-d6 45. Sb1-a3 Ke8-e7 46. Ke1-f1 Sc5-d3 47. Lf1-h3 Lb4-a5 48. a5-a6 Sa6-c5 49. a4-a5 Sb8-a6 50. g2-g3 Dd8-d7 51. f2-f3 c4-c3 52. Sf3-g1 Lf8-b4 53. Sg1-f3 e5-e4 54. Sf3-g1 d5-d4 55. Sg1-f3 c5-c4 56. Sf3-g1 e7-e5 57. Sg1-f3 d7-d5 58. a3-a4 c7-c5 59. a2-a3
Gerald Ettl: Die Stellung ist falsch. wTc2 auf c3 und +sDc2 (2023-07-02)
Gerald Ettl: Bzw. vielleicht ist die Stellung richtig, aber dann stimmt die Lösungsangabe und Forderung nicht. (2023-07-02)
Mario Richter: Hallo Gerald, Type C heißt: ein K muß im Schach stehen, also ist das Diagramm wohl richtig. Die Originalforderung war "Last 46 single moves?" (nicht ... 45 ...) - da hat wohl jemand den einleitenden schachbietenden Zug vergessen. (2023-07-03)
more ...
comment
Keywords: Last Moves? (45), Non-standard material (s), Type C, Superseded by (P1353149)
Genre: Retro
FEN: 8/1p6/1p3PP1/1K6/1bnPP1p1/Bq1bP3/BnRRP2P/1nk5
Input: Gerd Wilts, 2012-06-04
Last update: A.Buchanan, 2023-07-05 more...
53 - P1278097
Michael Schlosser
V46) Problemkiste (207) 09/2013
P1278097
(5+3) cooked
#3
1. ... bxa6 La7 a5,Ka5 2. Tc5#
1. ... bxc6 La7 c5 2. Tb6+ Ka5 3. Sc6#
1. Ta5#?
1. Tc5#?
play all play one stop play next play all
Schwarz beginnt, da er keinen letzten Zug hat.
A.Buchanan: What's going on here? (2021-10-06)
Henrik Juel: Maybe Pb4 should be removed (2021-10-06)
A.Buchanan: WinChloe has the same diagram, and the solution given here (with incorrect move counting) assumes sBb4 is present. Without it, there is a cook: 1. ... bxc6 2. Kc3 c5 3. Ld8,Lc7 c4 4. Ta5# (2021-10-07)
comment
Keywords: Symmetrical position, Superseded by (P1394575), Symbol problem (Kreuz)
Genre: 3#, Retro
Computer test: %Popeye Windows-32Bit v4.61
FEN: 1N6/1p6/RBR5/1k6/1p6/1K6/8/8
Reprints: V46 Bulletin Michael Schlosser-60-Jubiläumsturnier , p. 38, 2016
Input: Erich Bartel, 2013-09-20
Last update: Gunter Jordan, 2023-07-21 more...
54 - P1280468
Michael Schlosser
c 90 Jahre Schachkomposition in Chemnitz 31/05/2008
150 Jahre Vereinsschach in Chemnitz
P1280468
(6+3) cooked
#3
3 Lösungen
1) 1. Tg1 b6 2. Lf6 gxf6 3. g7#
2) 1. Tf1 b6 2. Tf6 gxf6 3. Lxf6#
3) 1. Lf7 b6 2. Lxg7+ Kxg7 3. h8=D#
play all play one stop play next play all
Cook: Black has no last move. And there's no solution with Black to move.
Stefan Felber: Schwarz hatte keinen letzten Zug, muß also anziehen. Kein Matt in 3 Zügen möglich! (2013-11-16)
A.Buchanan: This problem has no intention of being retro. Fortunately, it is easily fixable: shift wK to c4 & sBb7 to c6. (a-file will work as well, but surely c-file is more entertaining!) (2021-10-10)
more ...
comment
Keywords: Superseded by (P1394606), No legal last move for Black (unintended), no solution
Genre: 3#
Computer test: C+ Popeye v4.87
FEN: 6Bk/1p4pP/6P1/1K6/8/8/8/B6R
Input: Rainer Staudte, 2013-11-16
Last update: A.Buchanan, 2023-12-04 more...
55 - P1323157
Lord Dunsany
3813 The Times Literary Supplement , p. 1028, 27/12/1928
P1323157
(10+9) cooked
#3
"This problem is the complement of the one by the same author, published last week [i.e., #3809]."
See P1323156
Henrik Juel: I fail to see the intention
1.Txa7? thr. 2.Ta8#, but 1... 0-0!
Th7 may be a promoted pawn, or it may be [Ta8] with black cross-capture b7xc6 and c7xb6, so even if White may castle there seems to be no #2 (2016-08-02)
A.Buchanan: I agree Henrik, but I didn't want to say anything to allow people to come at this afresh. My theory is that this is really a #3, *and* sBc6 should be on c7. Then 1. 0-0-0! 0-0? for retro reasons, and 2. Txa7 3. Ta8# can't be prevented. I have sent an email to Mr Gullette, asking him about this, inter alia. (2016-08-02)
A.Buchanan: Pushing the pawn back to c7 doesn't work. Maybe no-one had realized about cross-capturing in 1928? :) The simplest way I can see to fix it, is to push sBa7b6 to a6b7, push wTa6 to b6, remove wL, and add wSa7. (2016-08-02)
A.Buchanan: The stipulation was #2 obviously wrong, and so I have fixed it to #3. However the problem remains cooked and it can't simply be put down to a typo. I will post my correction as discussed. (2021-05-11)
comment
Keywords: Castling, Superseded by (P1389458)
Genre: Retro, 3#
FEN: 4k2r/pB1p1p1r/RppP1P2/7p/8/7P/1PP3P1/R3K3
Reprints: www.alangullette.com/lit/dunsany/ 1/6/1997
Input: A.Buchanan, 2016-08-02
Last update: A.Buchanan, 2021-05-11 more...
56 - P1370911
James Malcom
Andrew Buchanan

(4) MatPlus.net Forum Retro/Math 21/12/2019
P1370911
(12+14)
PG in 12.5
(-:
1. e4 d5 2. e5 Dd6 3. exd6 Kd7 4. dxe7 Kc6 5. Lb5+ Kxb5 6. c4+ Kxc4 7. e8=T Kd3 8. Df3+ Kc2 9. d3 Lh3 10. Ld2 Lxg2 11. Lb4 Lxh1 12. Sc3 Lg2 13. 0-0-0-0-0-0#
play all play one stop play next play all
http://matplus.net/start.php?px=1609818957&app=forum&act=posts&fid=xshowr&tid=2402&pid=20379#n20379
Objective: Staugaard (Pam-Krabbé) castling as mating move.
more ...
comment
Keywords: Unique Proof Game, Staugaard castling, Joke, Prenix, Promotion (T), Castling as mating move, Superseded by (P1385595)
Genre: Retro, Fairies
Computer test: C+ Natch 3.1 has no solutions in 12.5 moves & 1 solution for first 12.0 moves. (H.Juel)
FEN: rn3bnr/ppp2ppp/8/3p4/1B6/2NPKQ2/PPk1RPbP/R5N1
Input: A.Buchanan, 2019-12-23
Last update: James Malcom, 2022-02-26 more...
57 - P1375286
Andrew Buchanan
James Malcom

PDB Website 30/4/2020
P1375286
(6+3)
#2
(-:
1. dxc7ep! b4 2. c8=D,T#

R: 1. ... c8-c6?! and earlier c8=sB?!
play all play one stop play next play all
Supersession of P1380851

Superseded by P1380851

The 2018 FIDE laws state: "3.7.2 on its first move the pawn may move as in 3.7.1 or alternatively it may advance two squares along the same file, provided that both squares are unoccupied."

Hence Black's newborn 8th rank pawn, which has never been moved before, reserves the right to commence a double-step on its first move, and likewise White receives the right to capture it en passant as they do with a 7th rank pawn.
Henrik Juel: This un-dummy-pawn is like a pawn created on c8 by the Einstein condition
The retroplay looks wrong, what about 1... d8-d6!? 2.d7-d8=sB!? c6xYb5 (2020-04-30)
James Malcom: That makes much more sense Henrik. I have edited it accordingly. (2020-04-30)
A.Buchanan: You don't like Adrian's (: haha? What about :) then? In heraldic terms, it would be described as "regardant". See 270 on https://www.heraldica.org/cgi-bin/atlas.pl?12. I don't know the word for "non-regardant", which is the default direction for the head, e.g. most of the others on that page.
We could replace wBa6a7 with wSa6, but alas having the 3rd knight is non-thematic & distracting. The other thing I would've liked to do is empty b2 so the promotion is forced to be to D. Such a tiny problem should have precision promotion. Needs more thinking. (2020-05-01)
James Malcom: UPDATE: The position has been edited to have one less piece and a precise queen promotion. (2020-10-12)
A.Buchanan: There is dialogue here which is now orphaned by a change of diagram with new matrix. Suggest we revert to the older diagram and mark it superseded by a new PDB entry here (2020-10-12)
James Malcom: I agree-it is now done. (2020-10-12)
more ...
comment
Keywords: En passant as key, Joke, Joke promotion (b), Tolerated dual promotion (D/T), Promotion (D/T), Superseded by (P1380851)
Genre: 2#, Retro
FEN: k7/PN1N4/PKpP4/1p6/8/8/8/8
Input: James Malcom, 2020-04-30
Last update: A.Buchanan, 2022-09-18 more...
58 - P1381070
James Malcom
SuperProblem (Website) 09/04/2020
P1381070
(10+5) C+
s#8
1. Ta1+! Kg2 2. Tb2+ Kf3 3. Ta3+ Ke4 4. Tb4+ Kd5 5. Ta5+ Kc6 6. Tc4+ Kb7 7. Tc7+ Kxa8 8. Ke7+ Dxf8#
play all play one stop play next play all
7 square route of the Black king

http://superproblem.ru/archive/probl/Sn/Malcom-s8-2020.gif
Olaf Jenkner: wBg3, +wPh2 you get a model mate (2020-10-18)
A.Buchanan: Can one begin with wTa3b3 so there is a genuine try 1. Tb1? (2020-10-19)
James Malcom: It turns out that this record was actually done, doubled, by P1371538. So a 14 square route of the Black king appears to be the record, for now anyway. (2021-01-01)
more ...
comment
Keywords: corner to corner, Checking key, Superseded by (P1371538)
Genre: s#
Computer test: Gustav 4.1d, Brute Force
FEN: N3KQqr/P2P4/3BP3/5P2/6p1/1R5p/R7/7k
Input: James Malcom, 2020-10-17
Last update: James Malcom, 2021-01-01 more...
59 - P1381369
Andrew Buchanan
MatPlus.net Forum 28/10/2020
P1381369
(14+14) C+
KBP in 4.0
1. e3 a6 2. Lxa6 bxa6 3. De2 Lb7 4. Dxa6 Lxa6
play all play one stop play next play all
shortest PG with 4 units captured on 1 square
Henrik Juel: 1.Be2-e3 Ba7-a6 2.Lf1xa6 Bb7xa6 3.Dd1-e2 Lc8-b7 4.De2xa6 Lb7xa6
Very easy and C+ by Natch 3.1
Nice tempo move 3... Lb7 (2020-10-29)
A.Buchanan: Thanks - tempo is to force unique order on Wh moves (2020-10-30)
comment
Keywords: Unique Proof Game, Tempo Move, Task (Construction) (4), Superseded by (P1381879), x pieces capture on one square (4)
Genre: Retro
Computer test: C+ by Natch 3.1
FEN: rn1qkbnr/2pppppp/b7/8/8/4P3/PPPP1PPP/RNB1K1NR
Input: A.Buchanan, 2020-10-29
Last update: A.Buchanan, 2021-04-18 more...
60 - P1381496
Ryan McCracken
PDB Website 01/10/2004
P1381496
(0+0)
BP in which both sides get their pieces all on squares of the same colour?
1. e3 d6 2. La6 Lh3 3. Lxb7 Lxg2 4. Lxa8 Lxh1 5. Le4 Lf3 6. Lxh7 Lxd1 7. Lxg8 Lxc2 8. Sc3 Lb3 9. Lxf7+ Kxf7 10. axb3 Kf6 11. b4
play all play one stop play next play all
A.Buchanan: Improves on P0005367 (2020-11-03)
comment
Keywords: Synthetic problem, Non-Unique Proof Game, Ornament, Superseded by (P1381525)
Genre: Retro
FEN: 8/8/8/8/8/8/8/8
Input: A.Buchanan, 2020-11-03
Last update: A.Buchanan, 2020-11-04 more...
61 - P1381529
James Malcom
SuperProblem (Website) 05/11/2020
P1381529
(11+14)
#22
1. 0-0! droht 2. Tf8#
1. ... 0-0-0 2. Txb6+ Kc7 3. Tb7+ Kc8 4. Txb5+ Kc7 5. Tb7+ Kc8 6. Txb4+ Kc7 7. Tb7+ Kc8 8. Txb3+ Kc7 9. Tb7+ Kc8 10. Txb2+ Kc7 11. Tb7+ Kc8 12. Txb1+ Kc7 13. Tb7+ Kc8 14. Txd1 Tg8+ 15. Sg5 Kd8 16. Tdb1 Ta1 17. Txa1 exd5 18. Tb8+ Kc7 19. Txg8 dxe4 20. Tb1 d5 21. Tc8+ Kd6 22. Tb6#
play all play one stop play next play all
The longest problem with both sides castling

http://superproblem.ru/archive/probl/N/Malcom-22x-2020.gif
more ...
comment
Keywords: Zwickmühle, Move Length Record, Superseded by (P1396373), Castling
Genre: n#
FEN: r3k3/1R1pp2N/Bq1pp3/1n1P4/Pp1PP2P/rp2P3/1b6/1b1nK2R
Input: James Malcom, 2020-11-04
Last update: James Malcom, 2021-11-28 more...
62 - P1381888
Ion Murarasu
Probleemblad 1993
2. Preis #2
151. Thematurnier
P1381888
(14+10) cooked
#2
Bosma
1. Tc3! zz
1. ... Txf4 2. Tc5#
1. ... Tg4 2. Sc7#
1. ... Txh3 2. Sc7#
1. ... Lxf3 2. Sf6#
1. ... Lg4 2. Sc7#
1. ... dxc3 2. Td1#
1. ... d3 2. Txd3#
1. ... Txb3 2. Le6#
1. ... Dxb4 2. Te5#
1. ... Dxb3 2. Le6#
1. ... Dxb5 2. Se7#
1. ... axb4 2. Te5#
play all play one stop play next play all
Cook: 1. Td1! droht 2. Sf6,Sc7,Tc5#
1. ... d3 2. Txd3#

1. Td2! droht 2. Le6,Se7,Te5#
1. ... d3 2. Txd3#
Preisrichter: "An impressive problem and a multitude of variants. The therefore also somewhat heavy position and the one dual 1 ... Kc5 or Ke5 is forgiven. A problem to play."

Correction at P1381912
A.Buchanan: This is a delightful problem, but cooked in the original. However it can be smoothly corrected by shifting wTc2 to c7 (and I don't see any other ways to do it). Alas the composer is no longer with us. (2020-11-14)
comment
Keywords: Bosma, Superseded by (P1381912)
Genre: 2#, Fairies
Computer test: C- WinChloe 3.51
FEN: 2NQN3/8/1P1p2p1/pP1k2Pb/qP1p1P1r/rP3P1B/p1R5/K3R3
Input: A.Buchanan, 2020-11-14
Last update: A.Buchanan, 2020-11-18 more...
63 - P1385596
Nikola Predrag
MatPlus.net Forum 08/01/2021
P1385596
(14+14)
PG in 7.5
Staugaard castling
b) 1. e4 Sf6 2. Df3 Sh5 3. Df6 exf6 4. e5 Le7 5. exf6 0-0 6. fxe7 Sg3 7. e8=T Sxh1 8. 0-0-0-0-0-0
play all play one stop play next play all
Objective: fastest Staugaard (Pam-Krabbé) + kingside castling.
more ...
comment
Keywords: Unique Proof Game, Staugaard castling, Castling (2), Prenix (T), Promotion (T), Superseded by (P1389375)
Genre: Retro, Fairies
FEN: rnbq1rk1/pppp1ppp/8/8/8/4K3/PPPPRPPP/RNB2BNn
Input: A.Buchanan, 2021-01-21
Last update: Rainer Staudte, 2021-05-10 more...
64 - P1386064
James Malcom
SuperProblem (Website) 15/01/2020
P1386064
(8+11) C+
#12
1. hxg8=D! a1=D+ 2. Kh2 Dg1+ 3. Kxg1 b1=D+ 4. Kh2 Dg1+ 5. Kxg1 c1=D+ 6. Kh2 Dg1+ 7. Kxg1 d1=D+ 8. Kh2 Dg1+ 9. Kxg1 e1=D+ 10. Kh2 Dg1+ 11. Kxg1 Sf4 12. f8=D#
play all play one stop play next play all
My first queening record

Record of 2+5 queen promotions

http://superproblem.ru/archive/probl/N/Malcom-12x-2020.gif
more ...
comment
Keywords: Promotion (Dd), Promotion key, Promotion in the mating move, Superseded by (P1378033), Capture key, Pendulum (K), Promotions to queens, konsekutive Umwandlungen 7 (DdddddD)
Genre: n#
Computer test: a. Baumann: Gustav 4.2a
FEN: 2n3r1/3BkP1P/4Pp2/8/8/6BP/pppppPn1/6Kb
Input: James Malcom, 2021-02-01
Last update: Dieter Berlin, 2023-12-08 more...
65 - P1389486
Werner Kuntsche
3241 Schach 15, p. 235, 08/1959
P1389486
(10+11) cooked
h#2
1. exd3ep g5 2. Lb3 Se6#
play all play one stop play next play all
Cook: 1. Kxd4 Txh5 2. Sc4 Td5#
01/1960, Seite 12: Der Autor ändert seine Aufgabe: wBf2->f5, sBh7->g6, sBh5->g3. So aber unlösbar weil ep nicht zulässig ist. (wBd2-e3-c4)
Felber, Volker: Der wBe4 kann von d2 über e3 nach d4 gekommen sein, somit war nicht zwingend d2-d4 der letzte weisse Zug. (2021-05-13)
A.Buchanan: sBs have captured 6 times e.g. fxgxh, exfxgxh, dxe so sBa, sBb & sBc never captured. wBs have captured visibly exfxg, but also e.g. axb=, bxa, cxd=. Thus all captures are accounted for, and Sf4 did not just capture. White's last move was not e3xd4, because too many pawn captures. So Wh *may8 e.p. The non-e.p. solution is probably best regarded as a cook. WinChloe marks this problem as cooked. I guess we can have a "C+ cooked" problem? (2021-05-13)
A.Buchanan: If my analysis is ok, then the problem can be fixed by transposing sBh5 & wBh3. (2021-05-14)
A.Buchanan: Volker usefully posted an update in the Comment field: "01/1960, Seite 12: Der Autor ändert seine Aufgabe: wBf2-f5, sBh7-g6, sBh5-g3. So aber unlösbar weil ep nicht zulässig ist." I agree with him that this new version 8/6p1/6pb/5P2/P1kPpNPR/n1p2BpP/p5Pp/2Kb4 has no solution: White might simply retract Th5-h4. More subtly, White might just have played Sf4 in response to f4xg3+! In the original, the ep *was* permitted (e3xd4 & Sxf4 impossible due to the 3 white captures on queenside). My earlier suggestion of swapping sBh5 & wBh3 remains, and has the advantage that the impossibility of g3-g4 now also depends on the capture count. (2021-05-15)
more ...
comment
Keywords: Capture key, En passant as key, Superseded by (P1389660)
Genre: h#, Retro
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/6pp/7b/7p/P1kPpNPR/n1p2B1P/p4PPp/2Kb4
Input: Felber, Volker, 2021-05-12
Last update: Arnold Beine, 2021-05-17 more...
66 - P1390750
Thomas R. Dawson

P1390750
(14+13) cooked
#3
1. Txg7 Sf4+ 2. Kh7 ... 3. Thxg8#
1. ... 0-0-0 illegal
play all play one stop play next play all
Cook: Unlösbar wegen 1. Txg7 0-0-0! (legal!)

BP (Ing. O. Koštál):
1. f3 e6 2. Sa3 Df6 3. Sc4 Dc3 4. bxc3 f5 5. Se5 d6 6. Sg6 Sh6 7. La3 Le7 8. Lc5 Tf8 9. Le3 f4 10. Tb1 Tf5 11. Kf2 Tg5 12. Tb5 Tg4 13. Db1 Ld7 14. fxg4 f3 15. Kg3 Lc6 16. Kf4 Lh4 17. Db4 Le1 18. Tf5 f2 19. Sh3 Lf3 20. gxf3 Sf7 21. Tg1 Se5 22. Tg3 Sec6 23. Sg1 Sd8 24. Th3 Sf7 25. Sf8 Se5 26. Th6 Sec6 27. Lh3 Sd8 28. Tg6 h5 29. Th6 Sf7 30. Tff6 Se5 31. Th8 Sec6 32. Tfh6 Sd8 33. Tg8 Sf7 34. Thh8 Se5 35. Kg5 Sec6 36. Kg6 Sd8 37. Kh7 Sf7 38. Lh6 Sd7 39. g5 Sb6 40. Lf5 Sd5 41. Dg4 hxg4 42. Lg6 Sf4 43. Lh5 Sh3 44. Kg6 Sd8 45. Th7 Sc6 46. Thh8 Sb4 47. Th7 c6 48. Thh8 Sd5 49. Th7 Sf6 50. Tgh8 Sg8 51. Sd7 a6 52. Sc5 dxc5
In der 'Prager Presse' mit der Information nachgedruckt: "Die Schachrubrik des 'Cas' macht folgende Aufgaben [diese und P0001348] zum Gegenstande eines Lösungsturniers"; Forderung "Matt in 3 Zügen. (Rückläufige Analyse)."

Korrektur s. P1390813
Henrik Juel: C+ Popeye 4.61 and analysis (except for mate dual after 2... Kf8)
1.Txg7 Sf4+ 2.Kh7+ any 3.Thxg8# (2021-06-12)
Henrik Juel: White captured b2xc3, g2xf3, and fxg; Black captured dxc and hxg
If Ke8 remains on e8 there is no way to release the eastern cage, so Ke8 has moved, and Black may not castle (2021-06-12)
Mario Richter: Sorry, Henrik, but you fell into the same trap than Dawson: by uncapturing a wS on c5 White can provide a retro-shield on f8, and then everyting easily resolves (making it possible for the wK to temporarily visit h7) ... (2021-06-12)
VL: Mario, the (parity changing) tempomove a7-a6 looks unavoidable for such a castling preserving genesis; accordingly, the shift of bPa6 to a7 would repair the problem. (2021-06-14)
Mario Richter: Valery, I think your correction would work too, but Dawson found another way to repair the Problem. (2021-06-14)
A.Buchanan: Mario: did really Dawson not architect the retro-shield? It's so nice, I'm sure he must have! sS can only reach f7 via d8, so c7-c6 must be retracted too. This throws the parity, which was otherwise good, so Black must spend his other parity move a7-a6 to get back in the game. Then he should have no waiting retractions while Lg5-h6. I think that Dawson didn't spot until too late that Sf4-h3 is always available once wK is off g6. That's what breaks the problem. Certainly shifting sBa6-a7 will render the #3 sound by eliminating the castling defence, however it costs retro content, because the intention (I believe) is that it's parity that costs Black his essential waiting retraction. Note that in the official repair, P1390813, the Black qside pawn configuration is unchanged: but Sh3 is no longer to be seen. (2021-06-15)
Mario Richter: Yes, of course - the retro-shield on f8 is part of the intended solution. The problematic piece is black Sh3, which has tempo moves after R: Kh7-g6 ... (2021-06-15)
comment
Keywords: Cant Castler (sg), Superseded by (P1390813), Retro Opposition
Genre: 3#, Retro
FEN: r3k1nR/1p4pR/p1p1p1KB/2p3PB/6p1/2P2P1n/P1PPPp1P/4b1N1
Reprints: 232 Cas 01/01/1922
159 Prager Presse 15/01/1922
Input: Mario Richter, 2021-06-12
Last update: A.Buchanan, 2021-06-16 more...
67 - P1391856
James Malcom
548 OzProblems.com 22/5/2021
P1391856
(4+9) C+
#8
Two Solutions
2.1...
1. f8=D! g5 2. Tg8 Kh1 3. Dh6+ Kg1 4. Dxg5 Kh1 5. Dh4+ Kg1 6. Dxg3 Kh1 7. Th8+ Kg1 8. Dh2#

1. f8=S! g5 2. Sh7 Kh1 3. Sxg5+ Kg1 4. Sh3+ Kh1 5. Sxf2+ Kg1 6. Sh3+ Kh1 7. Sg5+ Kg1 8. Sxf3#
play all play one stop play next play all
SCHRECKE: C+, Gustav 4.1d, Brute Force (2021-07-17)
Henrik Juel: C+ Popeye 4.61
1.f8=D g5 2.Tg8 Kh1 3.Dh6+ Kg1 4.Dxg5 Kh1 5.Dh4+ Kg1 6.Dxg3 Kh1 7.Th8+ Kg1 8.Dh2#
1.f8=S g5 2.Sh7 Kh1 3.Sxg5+ Kg1 4.Sh3+ Kh1 5.Sxf2+ Kg1 6.Sh3+ Kh1 7.Sg5+ Kg1 8.Sxf3#
Two very different solutions (2021-07-17)
James Malcom: See P1401781 for an improvement. (2022-06-12)
comment
Keywords: Promotion, Promotion key (D,S), Stalemate avoidance, Switchback (T), Superseded by (P1401781)
Computer test: Gustav 4.1d, Brute Force Popeye 4.61
FEN: 7R/5P2/6p1/8/6P1/4ppp1/4prp1/4Kbk1
Input: James Malcom, 2021-07-17
Last update: A.Buchanan, 2022-06-13 more...
68 - P1394920
Andrey Frolkin
B3 WCCC Rhodes 10/2021
P1394920
(15+10)
Last 4 single moves?
R: 1. Lf4xSg3# Sh5-g3 2. Le3xLf4+ Sg3-h5
play all play one stop play next play all
then Lf4 to a1; a3-a2-a1=L; a2×Db3; Sg3 to b1; b3-b2-b1=S; b2×Lc3; c3×Td2
Judge Michel Caillaud: "The theme [of one piece capturing two promoted ones] in classical retro form, with the capturing piece not being a Pawn. The nice point is that -2.Lé3×Pf4+? doesn't work. The promoted Knight cannot reach a1 so it has to unpromote on b1 to unlock the cage and another promoted piece has to be produced to unpromote on a1."

Author Andrey Frolkin: "This one is superseded by P1400748 (Q+B) which is the official 1st HM. The predecessor (B+S) should be stored “for posterity” in case someone wants to compose more problems in which an officer captures 2 promoted pieces."
Henrik Juel: Black never captured, so the only way to explain the check from Lg3 seems to be
R: 1.Lf4xSg3+ Sh5-g3 2.Lg5xLf4+ Sg3-h5
and later the two uncaptured officers are unpromoted on a1 and b1 (2021-10-26)
A.Buchanan: new version in final report (2021-10-28)
A.Buchanan: Henrik may wish to update his analysis. My question: can't we just remove sBd7? (2021-10-29)
Henrik Juel: Now the solution is R: 1.Lf4xSg3+ Sh5-g3 2.Le3xLf4+ Sg3-h5
I see no immediate purpose for Pd7, but since both Andrej and Michel have accepted it, there probably is some reason for it (2021-10-29)
A.Buchanan: It’s not necessary, according to Andriy (2021-10-29)
Henrik Juel: OK
Another matter: I believe that Andrej prefers his first name to be spelled Andrey in english
This spelling is used in the WCCC-11 announcement and in the FIDE Album 2010-12 (2021-10-30)
A.Buchanan: Thanks Henrik. In the German field of the PDB author table, he is Andrij, and I've now changed the English name to match the French as Andrey. As recently as a year ago, I have seen him as Andriy in English articles which he has authored in Problemas (2021-10-30)
Henrik Juel: Rendering cyrillic names in a western language is extremely messy
Each western language has a 'standard' transliteration from russian cyrillic, but ukrainian and bulgarian cyrillic are slightly different from russian cyrillic
My bulgarian chess friend Nikolai/Nikolaj/Nikolay accepts any of those spellings, but he prefers Nikolay
Next time you write to Andrej/Andrij/Andrey/Andriy, you should ask him what he prefers, Andy...
I shall probably continue to use Nikolaj and Andrej, which seem most natural for me (2021-10-30)
A.Buchanan: I have a feeling that I did ask him before and he said that his favoured version is Andrey. On the receiving side, my only comment is that “iy” is a very rare bigram in English and almost never occurs. At least we all agree on the first 4 letters! (2021-10-30)
A.Buchanan: He should pick whichever spelling gives him the best anagram :) (2021-10-30)
more ...
comment
Keywords: Last Moves? (4), Champagne-TT (2021), Type C, Ceriani-Frolkin Theme (sl), Prentos Theme (sl), Superseded by (P1400748)
Genre: Retro
FEN: 8/3p2p1/5NB1/4QRN1/4p2K/1PPP1PBP/2PpPPbk/4nrnr
Input: A.Buchanan, 2021-10-26
Last update: A.Buchanan, 2022-04-21 more...
69 - P1395488
Thomas R. Dawson
Tidskrift för Schack 1923
P1395488
(15+9) cooked
#2
1. axb6ep+! Kxe3 2. De2#
play all play one stop play next play all
Cooked by the possibility of e2-e1=S!

https://timkr.home.xs4all.nl/admag/promo.htm Tim Krabbe writes in "PROMOTIEMOTIVATIE": Een van de mooiste uitspraken in de Nederlandse schaakliteratuur werd gedaan door Johan Barendregt (1924-1982), in een interview met Max Pam: 'Mijn leven is bepaald door de zet e2-e1P.'

Hij doelde daarmee op een ontdekking die hij had gedaan in een stelling die in 1937 aan de lezers van het blad De Schaakwereld ter oplossing was voorgelegd. (Zie diagram.)
Het mat op zich was niet moeilijk, dat kon alleen 1.axb6+ Kxe3 2.De2 mat zijn, maar het ging om het bewijs dat alleen b7-b5 Zwarts laatste zet geweest kon zijn, en niet b6-b5 of Kc4-d3. Toen een paar weken later de oplossing werd gepubliceerd, bleek dat de 13-jarige Barendregt alle oplossers de baas was geweest, omdat hij had aangetoond dat het probleem van de grootheid Dawson incorrect was. Met een bewijspartij van 48 zetten liet hij zien dat ook e2-e1P Zwarts laatste zet geweest kon zijn, wat 1.axb6 illegaal maakte. En passant repareerde hij het probleem ook, door de pion van f5 naar e7 te verplaatsen - dàn moet Zwart zojuist b7-b5 gespeeld hebben.
Dat bewijs laat ik hier voor wat het is - het gaat me om de verrukking die Barendregt moet hebben gevoeld toen hem, temidden van de partijen van Euwe, Aljechin en Keres, lof werd toegezwaaid, maar die hij vooral moet hebben gevoeld bij de ontdekking van e2-e1P zèlf. Dat geluksgevoel bond hem voor de rest van zijn leven aan het schaken - hij was jarenlang een van de sterkste Nederlanders, werd Internationaal Meester toen dat nog iets betekende, en won partijen tegen Botwinnik en Portisch.

In English: One of the most beautiful statements in Dutch chess literature was made by Johan Barendregt (1924-1982), in an interview with Max Pam: 'My life is determined by the move e2-e1P.'

He was referring to a discovery he had made in a proposition that had been submitted to the readers of the magazine De Schaakwereld in 1937 for a solution. ( See diagram. )
The mate itself was not difficult, only 1.axb6+ Kxe3 2.De2 ??mate but it was to prove that only b7-b5 could have been Black's last move, and not b6-b5 or Kc4-d3. When the solution was published a few weeks later, it turned out that 13-year-old Barendregt had beaten all the solvers because he had shown that the problem of the great Dawson was incorrect. With a proof game of 48 moves, he showed that e2-e1P could also have been Black's last move, making 1.axb6 illegal. In the meantime, he also solved the problem by moving the pawn from f5 to e7 - then Black must have just played b7-b5.
I will leave that proof for what it is here - I am concerned with the delight that Barendregt must have felt when he was praised among the parties of Euwe, Aljechin and Keres, but which he must have felt above all when he discovered e2-e1P itself. That happiness tied him to playing chess for the rest of his life - for years he was one of the strongest Dutchmen, became an International Master when it still meant something, and won games against Botvinnik and Portisch.
Cook: Possible proof game by James Malcom in 42.0 moves: 1. b4 a5 2. bxa5 g5 3. Lb2 g4 4. Sc3 g3 5. hxg3 Sf6 6. Th6 Se4 7. Sd5 Sc3 8. Tc1 Sb1 9. Ta6 e5 10. Sf6+ Ke7 11. e4 d5 12. c3 d4 13. Tc2 d3 14. Ta7 dxc2 15. Sh3 Sa6 16. d4 c6 17. d5 Sc7 18. La6 Kd6 19. Dd3 Kc5 20. d6 Lh6 21. Ke2 Ld2 22. dxc7 h5 23. Sg5 Dd5 24. exd5 e4 25. a3 e3 26. Kf3 h4 27. De2 Lf5 28. Kf4 Tae8 29. Ta8 h3 30. Tc8 h2 31. d6 h1=T 32. Sd5 Tc1 33. f3 Le4 34. fxe4 b5 35. Dd1 Th4+ 36. Kf5 Tf4+ 37. gxf4 Kc4 38. Sh7 Kd3 39. Kg5 f5 40. Shf6 e2 41. Se3 Te5 42. fxe5 e1=S
means e.p. convention won't fire
Henrik Juel: James, when I played your game I did not reach the diagram position
Is the diagram or your game wrong? (2021-11-06)
James Malcom: Gahhhhh, I got myself again Henrik. I garbled some pawns, which should be fixed now. (2021-11-06)
Henrik Juel: Thanks
The intended retroplay was
R: 1... b7-b5! 2.Sc4-d3+ Ke3-d3+ 3.Sb6-c4+ d3xTc2 (2021-11-07)
Henrik Juel: The cooking last move e2-e1=S is rather obvious
I would have thought that it was found by the TfS solvers and corrected by TRD
Can anyone check in TfS 1923-24? (2021-11-07)
Henrik Juel: Never mind...
The problem appeared in October 1923, and there is no mention of e2-e1=S in the solution in the March 1924 issue (2021-11-07)
comment
Keywords: En passant as key, Superseded by (P1395505)
Computer test: Popeye v4.87 & simple retro-logic & demonstrative proof game
FEN: 2R5/2P5/B1pP1N2/Pp2PpK1/4P3/P1PkN3/1Bpb2P1/1nrQn3
Reprints: De Schaakwereld 1937
Tim Krabbé's Website 2001
MatPlus.net Forum 2021
Input: James Malcom, 2021-11-06
Last update: A.Buchanan, 2021-11-07 more...
70 - P1398779
Andrew Buchanan
AA006 The Hopper Magazine I01 24/12/2021
P1398779
(12+5) cooked
h=3*

play all play one stop play next play all
Cook: Set: 1. ... Th1 2. Ka1 Dg1/Tg1 3. Ka2 Df1=/Th1=
Main: 1. Ka1 Th1 2. Ka2 Dg1/Tg1 3. Ka1 Df1=/Th1=
A.Buchanan: Oops! Vlaicu Crisan informed me of a bug in the forward play. Not so easy to fix, but think I've managed. Sent back to Vlaicu, and will republish in The Hopper. (2022-12-03)
comment
Keywords: Draw by repetition, Superseded by (P1408741)
Genre: Retro, Fairies
FEN: 8/p7/8/8/8/P4P1P/kPPPPRPr/2KnrQR1
Input: A.Buchanan, 2022-01-31
Last update: Miguel Ambrona, 2023-07-26 more...
71 - P1400191
James Malcom
Andrew Buchanan

PDB Website 02/04/2022
P1400191
(8+7) C+
r#4
1. bxa8=B! f1=B 2. e8=B d1=B 3. Le1 Sc2 4. Lb1 Te3#
play all play one stop play next play all
More economical version of P1375039
James Malcom: I had a thought looking at this again. Does n7/1P2P3/Q3P3/1r6/N1k1KP2/Pr6/B2p1p2/n7 work to save another piece? (2022-04-02)
A.Buchanan: James, yes your version does save another unit for your lovely idea. Beyond this, a version n7/1P2P3/Q7/Br6/2kPKP2/1r6/B2p1p2/n7 saves another unit, and forces a unique order on the Black promotions too. To avoid distracting unachievable threats, set NoThreat as well as the usual NoShortVariations. Your composition has driven me finally to report a long-standing Olive bug that cruelly punishes we Dummy Pawn enthusiasts: https://github.com/dturevski/olive-gui/issues/54. (2022-04-03)
James Malcom: Thanks for the verification and shaving off a unit, Andrew. (2022-04-03)
more ...
comment
Keywords: Dummy Pawn, Golden Age (Dummy Pawn), Superseded by (P1411297)
Genre: Fairies, r#
Computer test: Henrik Juel: C+ Popeye 4.87 (2022-04-02)
FEN: n7/1P2P3/Q7/Br6/2kPKP2/1r6/B2p1p2/n7
Input: James Malcom, 2022-04-02
Last update: A.Buchanan, 2023-12-14 more...
72 - P1401254
Hiroaki Maeshima
PS3693 The Problemist Supplement 173, p. 163, 07/2021
P1401254
(3+8) C+
h#4*
0.1.1...
* 1. ... ... 2. Df2 Kg5 3. d3 cxd3 4. Kf3 Sd4#
1. ... c4 2. Dg4 Sg7 3. Sf4 Se6 4. Df3 Sc5#
play all play one stop play next play all
Yuri Bilokin: More economical -bPh7, then a1=b1 8/6p1/6K1/6N1/4pkq1/5p1p/3P3n/8 (3+7)
1... ... 2.Qg2 Kh5 3.e3 dxe3+ 4.Kg3 Se4# (MM)
1...d4 2.Qh4 Sh7 3.Sg4 Sf6 4.Qg3 Sd5# (MM) (2022-05-18)
Hiroaki Maeshima: Thank you, I added the version (P1404069). (2022-09-06)
comment
Keywords: Superseded by (P1404069)
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/5p1p/5K2/5N2/3pkq2/4p1p1/2P3n1/8
Input: Felber, Volker, 2022-05-11
Last update: A.Buchanan, 2022-09-06 more...
73 - P1401450
Gerhard E. Schoen
511 Chessics 13 01/1982
P1401450
(4+5) cooked
h#3 A.P.
1. dxc3ep e3 (Td1?) 2. Db5 0-0-0 (Td1?) 3. Kc4 Td4#
play all play one stop play next play all
Cook: 1. Df3,Dh3 0-0-0 2. Ka3 Kc2 3. Tb4 Ta1#
1. Dxc4 Ta2,Kd1 2. Kc3 Kd1,Ta2 3. Db4 Tc2# etc
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (wg), En passant as key, Superseded by (P1000348)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro logic
FEN: 8/8/8/2n5/1kPp4/1r1q4/4P3/R3K3
Reprints: feenschach 137, p. 365, 08/2000
Input: A.Buchanan, 2022-05-24
Last update: A.Buchanan, 2022-05-24 more...
74 - P1401480
Jean-Michel Trillon
v Rex Multiplex 2 04/1982
to C.Caminade
2nd Honourable Mention
P1401480
(6+8) cooked
#4 AP
Maximummer
b) g6->g5
a) 1. Ld6! Td8 2. fxg6 Txd5 3. g7 Th5 4. g8=D#
b) 1. fxg6ep Txd5 2. Le5 0-0-0 3. g7 Th8 4. gxh8=D/T#
play all play one stop play next play all
a) Black had a candidate last move b7-b5 of length 2. The only way this might not have been played is if bK was in check from e8 namely Kd8xDTe8,Kf7xDe8. So certainly Black castling rights are lost. With this known, there is still a unique solution.
b) Again, b7-b5 looms, and this time if Black castling rights are lost, there is no solution. The only possible way we can operate is if the last move was g7-g5. But we are only permitted to play this if we can show that the last move *was* not one of Kd8xDTe8,Kf7xDe8. I.e. under AP we need to castle after the fact. There is a unique solution after the ep, which does indeed include castling.
Cook: b) R: 1. Th4xDh5,Th4xLh5,Th6xDh5,Th6xLh5
Henrik Juel: a) C+ Popeye 4.61
1.fxg6 Txd5 2.Le5 0-0-0 3.g7 Th8 4.gxh8=DT#
b) no solution (2022-05-27)
Henrik Juel: I shall leave the analysis to you, Andrew... (2022-05-27)
A.Buchanan: OK, Henrik: please check my suggested solutions & rationale :) I find this a beautiful problem (2022-05-28)
A.Buchanan: In (b), is it possible that Black has just played RxQBh5? If this is feasible, it would apply even if Black 00 rights are still on. Cook? (2022-05-28)
Mario Richter: Looks indeed like a cook.

One can manually retract e.g. R: Th4xLh5 and use popeye to check if in the resulting position the reverse of this retraction is legal:
begin
remark Position after R: Th4xLh5
pieces white Bb8 Pd5 Pf5 Bh5 Ph3 Pd2 Ka1
black Ra8 Ke8 Pb7 Pd7 Ph7 Pa6 Pg5 Rh4
cond Maximummer
option HalfDuplex
stip h~0.5
end

This gives:
1...Rh4*h5
1...Ke8-e7
1...Ke8-d8
1...Ke8-f8

solution finished. Time = 0.013 s (2022-05-29)
A.Buchanan: Thanks Mario. I showed this problem on the international zoom call last night, and Siegfried spotted the cook immediately. Can g4 and h4/h6 be plugged suitably with pawns? Can't just shift bPh6 to h5, because it means that in a), bPg6 might have just captured checking Q/B, while not disrupting Black 000 rights. But I think can shift wPh3 to h4, together with adding Pg4 (Black or White - I prefer White aesthetically). Does this work? (2022-05-29)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), Maximummer, Castling (sg), En passant as key, Promotion (D), Valladao Task, Superseded by (P1401566)
Genre: n#, Retro, Fairies
Computer test: Popeye v4.87 & not-so-simple retro-logic to identify the cook
FEN: rB2k3/1p1p3p/p5p1/3P1P1r/8/7P/3P4/K7
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2022-06-02 more...
75 - P1404281
Mikhail Mikhaylov
3 Shakhmatna Misl , p. 160, 10/1956
P1404281
(8+6) cooked
#2
1. d8=sS! droht 2. g8=S#
1. ... Sb7 2. Sd7#
1. ... Sc6 2. Te6#
1. ... Se6 2. Se4#
1. ... Sf7 2. Sg4#
play all play one stop play next play all
Cook: (found by Henrik Juel, 2022-09-12)
1. d8=sL! droht 2. g8=S#
1. ... Lc7 2. Sd7#
Henrik Juel: Popeye 4.61 finds no solution (2022-09-11)
Henrik Juel: Joke problem
1.d8=sL?? thr. 2.g8=S#
1... Lc7 2.Sd7# (2022-09-12)
Mario Richter: Does Henrik's answer show that the problem is cooked? (2022-09-12)
Henrik Juel: Yes, Mario (2022-09-12)
A.Buchanan: Could it be that bRb6 is meant to be on a6? (2022-09-13)
A.Buchanan: The point being that with bRa6, 1. d8=sL? La5! (not Lc7/Lb6 2. Sd7#/Te6#) So bishop promotion becomes a thematic try. (2022-09-13)
Mario Richter: I have access to the original issue of 'Shakhmatna Misl', so I can confirm that the black rooks are on a7 and b6. I checked some of the following issues to see, if there was a mentioning a of the cook and/or a correction, but without success.
Changing the cook 1. d8=sL! into a try would be nice, but I think the purpose of black Tb6 is to prevent the defense 1. ... Db3!, so unless I'm missing something, Andrew's suggestion doesn't work. (2022-09-13)
A.Buchanan: Thanks Mario - I didn't check enough. If further one swaps bQb8 with bRa7, and pushes wK to h2 further out of harms way, then I think the =S solution & =L try are sound: 1r6/q2Pp1P1/r4k1N/2N3R1/5p1P/8/4R2K/8 (2022-09-13)
James Malcom: Nice find, Mario; sweet fix, Andrew. (2022-09-14)
A.Buchanan: Thanks James. r7/q2Pp1P1/r4k1N/2N3R1/7P/8/8/4RK2 further saves a pawn and has better positions for the remote units. I suggest attribution is MM correction HJ,MR,AB. (2022-09-15)
A.Buchanan: Alternatively r7/q2Pp1P1/r4k1N/2N3R1/5p1K/8/8/4R3 economizes a wP rather than a bP. I don't know which version my esteemed co-creators would prefer to record this momentous achievement for posterity. I don't see a way to reach Meredith (2022-09-17)
comment
Keywords: Joke promotion, Joke, Superseded by (P1404422)
Genre: 2#
FEN: 1q6/r2Pp1P1/1r3k1N/2N3R1/5p1P/8/4RK2/8
Input: Mario Richter, 2022-09-11
Last update: A.Buchanan, 2022-09-18 more...
76 - P1405053
Norman L. Guinasso
1662 Los Angeles Times 18/11/1945
P1405053
(5+2) cooked
#3
1. S4b6? axb6 2. c4 b5 3. cxb5#
play all play one stop play next play all
Cook: Schwarz hat keinen letzten Zug, und mit Schwarz am Zug gibt es kein 3#
SCHRECKE: Schwarz hat keinen letzten Zug! (2022-10-10)
Henrik Juel: So it is Black to move, and he is stalemated (2022-10-10)
Rainer Staudte: Eine Korrektur wäre einfach, z. B. durch Verschiebung um eine Reihe nach unten. Aber wäre das Autorabsicht? (2022-10-10)
A.Buchanan: According to the Codex, mere illegality does not cook a problem or even make it unsound. But in the grunting world of PDB data fields, we only have "cooked" to express all degrees of structural displeasure. If I was fixing the problem, I would stick wLa8 on h1 instead. For then we have the tiny satisfaction of knowing that Black's last move was R: 1. Kb7xTa6.
This author is unknown to WinChloe, and only has one other problem in YAC. (I have added it here.) Amazingly, WinChloe has not heard of the LA Times either! (2022-10-11)
A.Buchanan: It's a rare name: https://www.findagrave.com/memorial/62652231/norman-l-guinasso 1920-1978 might fit as a California memorial. There's then a Norman L. Guinasso Jr, whose age might fit as a son, who retired as a Professor of Oceanography in 2015 (2022-10-11)
A.Buchanan: Or replace wLa8 with wBc6. (sBb7 also works, but is still illegal!) (2022-10-11)
more ...
comment
Keywords: Minimal, Miniature, No legal last move for Black (unintended), Superseded by (P1413729)
Genre: 3#
FEN: B1N5/p7/k7/8/1KN5/2P5/8/8
Input: Rainer Staudte, 2022-10-10
Last update: A.Buchanan, 2023-11-28 more...
77 - P1410861
James Malcom
Siegfried Hornecker
Joaquim Crusats
Andrew Buchanan

12 MatPlus.net Forum 25/06/2023
P1410861
(16+4)
What is curious about this position?

Fewest Black pieces requiring all 16 White pieces in a legal checkmate position

https://matplus.net/start.php?px=1689032781&app=forum&act=posts&fid=gen&tid=3043
A.Buchanan: Mu-Tsun Tsai suggests that for construction problems such as this, the stipulation might be "What is curious about this position?" (2023-07-13)
James Malcom: That feels apt. (2023-07-14)
more ...
comment
Keywords: Superseded by (P1410903)
Genre: Retro, Fairies
FEN: B1Qr2RK/3P2N1/1P4RB/1rP5/4kP2/PPP5/q2P4/2N5
Input: James Malcom, 2023-07-11
Last update: A.Buchanan, 2023-07-27 more...
78 - P1411257
James Malcom
Andrew Buchanan

OP029 The Hopper Magazine I03 30/12/2022
P1411257
(8+7) C+
S#2 (-:
1. Tg8! (Zugzwang)
1. ... Txg8 2. hxg8
1. ... Tf8 2. gxf8
1. ... Td8 2. cxd8
play all play one stop play next play all
SCHRECKE: Have a look at this: R4q1k/pPP3R1/rp3PKn/p5PQ/P7/8/8/8
s#2, 1. f7! - 6(!) dummy promotions (2023-07-28)
SCHRECKE: In my work a black rook on f8 is better! (2023-07-28)
A.Buchanan: Yes this is a great improvement. I can't see how one might get a 7th (2023-07-28)
more ...
comment
Keywords: Dummy Pawn, Golden Age (Dummy Pawn), Superseded by (P1411367)
Genre: s#
Computer test: C+, popeye 4.87
FEN: 2b1r2R/1pPpk1PP/1P3pB1/4pK2/4N3/8/8/8
Input: Miguel Ambrona, 2023-07-26
Last update: A.Buchanan, 2023-07-29 more...
79 - P1411659
Gyula Bebesi
v Problemas 1962
P1411659
(8+14) cooked
h#2
1. exd3ep Lg8+ 2. d5 cxd6ep#
1. axb3ep bxc6+ 2. b5 cxb6ep#
play all play one stop play next play all
PRA means one problem two parts
Cook: R: 1. Kf6-e5 Ld6-f8+
see P0003365
Henrik Juel: Intention:
White captured axSb and [Lc8]
If last move was b2-b4: 1.axb3ep bxc6+ 2.b5 cxb6ep#
If last move was d2-d4: 1.exd3ep Lb8+ 2.d5 cxd6ep# (2023-08-08)
Henrik Juel: Cook: last move could also be Kf6-e5
correction, e.g., move Sh3 to h5 (2023-08-08)
A.Buchanan: Good catch Henrik - I should have looked at this one more carefully! Luckily it admits an easy fix, but there is still the non-standard material. (2023-08-08)
Henrik Juel: I should have thought of a better birthday greeting, Andrew, but there you are (2023-08-08)
comment
Keywords: Partial Retro Analysis (PRA), En passant as key (2), En passant as mating move (2), Non-standard material (L), Superseded by (P1413906)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 3r1b2/1p1p3B/B1r3p1/1PP1Kp2/pPkPp3/p1p4n/N7/2q5
Reprints: Schach-Echo , p. 324, 08/1984
23 125 ausgewählte Schachprobleme , p. 10, 1985
2 Problemas 44, p. 1456, 10/2023
Input: A.Buchanan, 2023-08-08
Last update: A.Buchanan, 2023-12-04 more...
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