Die Schwalbe

667 problem(s) found in 4158 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT A='Brjuchanow, Iwan O.' AND K='Rochade'] [download as LaTeX]

1 - P0000016
Nikita M. Plaksin
Alexander Kislyak

(F) Die Schwalbe 98 04/1986
P0000016
(11+9)
Welches war der erste Zug der beiden Könige?
R: 1. bxc3ep+ c2-c4 2. b5-b4+
play all play one stop play next play all
Erster K-Zug: Weiß: 0-0; Schwarz: 0-0
173. Thema-Turnier
Henrik Juel: Good motivation for the black castling
The further retroplay includes retracting Tf1-g1, Kd3 to g1, h2xTg3, Tg3 to f8, La6 to c8, d7xTSe6, and e6xPf7 (2021-04-19)
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comment
Keywords: En passant, First Move?, Last Moves? (3), Castling (wksk), Castling in the retro play, En passant in the retro play, Volet Pawn, Type C
Genre: Retro
FEN: 7k/1p2pPpp/b3p3/8/8/2pK2P1/pP1PPPP1/2B3RN
Reprints: 566 Ukrainisches Album 1986-1990
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-20 more...
2 - P0000036
Valery Liskovets
Nikita M. Plaksin

5528 Die Schwalbe 99 06/1986
P0000036
(10+11)
h#1.5 (AP, RV)
1) 1. ... cxd6ep 2. 0-0-0 Dxc7#
Nicht 2. Kf8 Dg8? da der ep-Schlag AP bewiesen werden muss.
2) 1. ... gxf6ep 2. 0-0-0 Tg8#
Nicht 2. Kd8 Tg8#? da der ep-Schlag AP bewiesen werden muss.
play all play one stop play next play all
VL: Solution:
I 1... c*d6 e.p. 2.0-0-0 Q*c7#
II 1... g*f6 e.p. 2.0-0-0 Rg8#

NOT:
2.Kf8? Q/Rg8#?? no AP-justification!

Castling implies e.p. 'a posteriori' and is necessary only to this end. (2002-04-03)
A.Buchanan: Solutions & tries (Popeye v.485) are:
1. ... cxd6ep 2. 0-0-0 Dxc7#
1. ... cxd6ep 2. Kf8 Dg8#
1. ... gxf6ep 2. d4 Tg8#
1. ... gxf6ep 2. 0-0-0 Tg8#
1. ... gxf6ep 2. Kd8 Tg8#
So after 1. ... gxf6ep:
2. Kf8 Rg8 is not even a mate (f7 flight)
and is 2. d4 a kind of dual thematic AP try? No-one's mentioned it. It would be easily fixable by shifting wBd2 to d4.
However, the main thing issue I don't understand PRA+AP logic (maybe I did once?). Under normal PRA algorithm, the "parts" for cxdep, gxfep & 0-0-0 respectively are YNY, NYY, NNN. Others are either impossible (3) or dominated by application of castling convention (2). OK so far.
There is no solution in the NNN part, so I suppose we must use AP in another part to eliminate it. But castling in another part where we are explicitly *assuming* that castling is ok seems to me to prove nothing. And if this inference did work somehow (which seems to me incredibly unlikely), why would we need to do it in *both* of these other parts? Surely one would be enough, but that means that the other part would be cooked.
Or maybe somehow AP "proves" that there are only 2 dimensions for PRA space, cxdep & gxfep, but that is so incredibly circular and horrible.
This kind of problem needs a proper step-by-step solution. Currently the solver can at best sort-of-vaguely-see-what-the-composer-might-be-alluding-to: i.e. it's just a joke.
Actually, I don't think that it can be PRA at all. I think it must be RS with two solutions supported by AP. But then there's still the question of why we need to castle in *both* solutions. I am sure I have seen an AP problem with two solutions where there is 0-0 in one solution and 0-0-0 in the other, which suggests that we *don't* need to castle twice in this problem. Which interpretation is correct, or do we have such a vague interpretation of these things (in order not to make independent spirits sad) that both are allowed to "work". (2020-12-30)
VL: Generally I agree with Andrew's analysis. The presented solution needs some corrections and clarification.

At first, the extra try 1.gxf6 e.p.? d4. I don't remember details but indeed, at present I'd prefer to exclude it by shifting wPd2 onto d4. One more thematic try enriches a problem but it is better to have an equal number of tries after both e.p.'s. In general, in AP-problems, thematic tries are highly desirable but not necessary formally for soundness. Dual avoidance in tries is here just due to the flight f7. Yes, 2.Kf8? Rg8#?? is indicated wrongly (unfortunately, H.103 with the published solution isn't available for me).

Now the most doubtful point: dual avoidance in the solution. Do we need castling in both partial solutions or at least one would suffice for justification? In my opinion, both approaches make sense in principle. In this problem implemented is the former, what means that every partial solution is permitted to be considered separately and, thus, should contain legalizing castling. This is similar to ordinary AP-problems, in which every line of the solution must contain castling. There is a distinct much subtler and rarer approach, which I call "Total AP", that considers all lines as a whole and thus requires to contain castling at least once (with subtle soundness issues). So, the current problem is treated (implicitly) not in the spirit of Total AP.

The problem implements a certain kind of triples of mutually retrodepending special moves (w/b castlings and e.p.). Namely, it is a "hybrid" AP-problem, where castling justifies a posteriori the typical pRA-choice between two e.p. captures. This is one of lucky triple kinds that admit (controversially) sound implementations based on hybrid logical conventions (logics for short). One more such a triple is represented in my earlier P0002474. Totally 37 distinct kinds of interacting triples can be identified, and only for few of them I know (or at least expect) reasonable implementations under special logics (whichever artificial and arguable they can be but consistent internally). Under an implementation I mean (loosely) a problem with an orthodox stipulation (#n, h#, etc.) supplemented by a suitable retro-convention like an ad hoc combination of pRA, RS or AP such that the position of the problem contains the corresponding triple of moves AND its full solution depends on all their combined (il)legalities. An extremely subtle matter. (2021-01-29)
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Keywords: a posteriori (AP), En passant as key (2), Partial Retro Analysis (PRA), Castling (sg), Volet Pawn
Genre: h#, Retro
FEN: r3k2n/1pp4P/N6p/2PpKpP1/2Q1p1R1/4P1pB/3P1p2/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
3 - P0000047
Nikita M. Plaksin
Faat Fatchullin

5646 Die Schwalbe 101 10/1986
2. Preis
P0000047
(11+10)
h#2*
*) 1. ... 0-0-0 2. Txf2 Dxg1#
1) 1. cxd2+ Kxd2 2. Txf2 Dxg1#
R: 1. ... Da8-h8 2. d3xTe4 Da2-a8 3. c2xSd3 Sc1-d3+ 4. b4xLc5 Db1-a2 5. b3-b4 Ba2xSb1=D 6. Sa3-b1 und weiter z.B. (Beispielauflösung mri)
6. ... Te8-e4 7. Sc4-a3 Sg4-h2 8. Se5-c4 Th2-g2 9. Sf3-e5 g2-g1=L 10. Sg1-f3 Sh6-g4 11. Sf3xDg1 Th8-e8 12. Sd4-f3 f3xSg2 13. Se3-g2 f4-f3 14. Sc6-d4 Tg2-h2 15. Sa7-c6 Kh2-h1 16. Sc8-a7 Dh1-g1 17. Sb6xLc8 Tg1-g2 18. Sf5-e3 Dc6-h1 19. Sh4-f5 Kh1-h2 20. h2-h3 a3-a2 21. Sa4-b6 Sa2-c1 22. Sb6-a4 Sb4-a2 23. Sa4-b6 Sd5-b4 24. Sb6-a4 Se3-d5 25. Dh3-f1 Sf1-e3 26. Dg4-h3 f5-f4 27. Df4-g4 a4-a3 28. Db4-f4 a5-a4 29. Da3-b4 Kg2-h1 30. Sf3-h4 Kh3-g2 31. Da2-a3 g2-g1=T 32. Db1-a2 Kg4-h3 33. Dd1-b1 h3xTg2 34. Tg1-g2 Kg5-g4 35. Sh4-f3 Kf6-g5 36. Th1-g1 Se3-f1 37. Sf3-h4 h4-h3 38. Sg1-f3 e6xLf5 39. Lh3-f5 Ke7-f6 40. Lf1-h3 Sc4-e3 41. g2-g3 Ke8-e7 42. Sa4-b6 Sg8-h6 43. a2xTb3 Tb6-b3 44. Sh3-g1 Ta6-b6 45. Sb6-a4 Ta8-a6 46. Sg1-h3 a7-a5 47. Sh3-g1 Sa5-c4 48. Sg1-h3 Sb3-a5 49. Sc4-b6 Sc1-b3 50. Sa3-c4 Sb3xLc1 51. Sb1-a3 Sa5-b3 52. Sh3-g1 h5-h4 53. Sg1-h3 c4-c3 54. Sh3-g1 Lf8-c5 55. Sg1-h3 Dc7-c6 56. Sh3-g1 Dd8-c7 57. Sg1-h3 c5-c4 58. Sh3-g1 h7-h5 59. Sg1-h3 e7-e6 60. Sh3-g1 Sc6-a5 61. Sg1-h3 Sb8-c6 62. Sh3-g1 c7-c5 63. Sg1-h3
play all play one stop play next play all
Anton Baumann: Auszeichnung Informalturnier 1986: 2.Preis
Preisbericht: 'Die Schwalbe' 06/2011 S.124 (2023-01-02)
Henrik Juel: How is the SE corner released, without ruining the castling? (2023-01-02)
Mario Richter: Good question, Henrik! I first thought that releasing the SE corner without ruining White's castling right is impossible, but the trick is to uncapture a black Queen in the SE corner at the right moment.

Perhaps Theodore Hwa can use ths problem as a test case for his latest improvement to Retractor 2 ... (2023-01-02)
Henrik Juel: Thanks, Mario
In view of the prize I suspected that the problem was correct, but I did not find the uncapture trick (2023-01-02)
Henrik Juel: C+ Popeye 4.61, because with Black to move White may not castle (2023-01-02)
comment
Keywords: Castling (wl)
Genre: h#, Retro
FEN: 7q/1p1p1pp1/8/2P5/4P3/2p3PP/1P1PPPrn/R3KQbk
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-01-02 more...
4 - P0000050
Andrey Frolkin
(T) Die Schwalbe 102 12/1986
P0000050
(13+13)
Vor mindestens 50 Einzelzügen mußte rochiert werden!
R: 1. ... La2-b1 2. Dh7-g8 Lb1-a2 3. Kg8-h8 La2-b1 4. Kf7-g8 Lb1-a2 5. Ke8-f7 La2-b1 6. Kd8-e8 Lb1-a2 7. Kc7-d8 La2-b1 8. Kb6-c7 Lb1-a2 9. Kc5-b6 La2-b1 10. Kd4-c5 Lb1-a2 11. Ke4-d4 La2-b1 12. Kf3-e4 Lb1-a2 13. Kg4-f3 La2-b1 14. Kh3-g4 Lb1-a2 15. Kh2-h3 La2-b1 16. Kg1-h2 Lb1-a2 17. h2xTg3 Th3-g3 18. Dg8-h7 Th8-h3 19. Sg3-h1 h7xTg6 20. Tg5-g6 La2-b1 21. Ta5-g5 Lb1-a2 22. Ta2-a5 f5-f4 23. Tb2-a2 La2-b1 24. Tb1-b2 f6-f5 25. Tf1-b1 Lb1-a2 26. 0-0
play all play one stop play next play all
James Malcom: Lastly, here is a PG that may or may not be the shortest: 1. Nf3 c5 2. Ne5 Qb6 3. Nc3 Qb3 4. axb3 c4 5. Nd5 c3 6. Ra6 Nf6 7. Rd6 Ng4 8. Re6 Ne3 9. Nf4 Nxf1 10. Nh5 Ne3 11. Ng3 Nc4 12. bxc4 dxe6 13. Nf5 Bd7 14. Ng3 Ba4 15. Nf5 Bb3 16. Ng3 Ba2 17. Nf5 Nc6 18. Ng3 Na5 19. Nf5 Nb3 20. Ng3 Na1 21. b3 a6 22. Ba3 Ra7 23. Bc5 Kd8 24. Qc1 Kc8 25. Qa3 Kc7 26. Qa4 Bb1 27. Qe8 f6 28. Qf7 Kc8 29. Qg8 Kb8 30. Bd6+ Ka8 31. Bb8 Ba2 32. Nd7 Bb1 33. O-O Ba2 34. Rb1 f5 35. Rb2 Bb1 36. Ra2 f4 37. Ra5 Ba2 38. Rg5 Bb1 39. Rg6 hxg6 40. Nh1 Rh3 41. Qh7 Rg3 42. hxg3 Ba2 43. Kh2 Bb1 44. Kh3 Ba2 45. Kg4 Bb1 46. Kf3 Ba2 47. Ke4 Bb1 48. Kd4 Ba2 49. Kc5 Bb1 50. Kb6 Ba2 51. Kc7 Bb1 52. Kd8 Ba2 53. Ke8 Bb1 54. Kf7 Ba2 55. Kg8 Bb1 56. Kh8 Ba2 57. Qg8 Bb1 (2020-11-08)
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Keywords: Castling in the retro play
Genre: Retro
FEN: kB3bQK/rp1Np1p1/p3p1p1/8/2P2p2/1Pp3P1/2PPPPP1/nb5N
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-02-02 more...
5 - P0000058
Leonid M. Borodatow
5758v Die Schwalbe 103 02/1987
P0000058
(13+9) C+
h#3
b) sBa6 statt sLa6
a) 1. ... exf6ep 2. 0-0-0 Lxf4 3. Td7 a8=D#
b) 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
play all play one stop play next play all
Anton Baumann: Mattdual in b): 3.Tb8 axb8=D,T# (2022-12-16)
A.Buchanan: So Borodatow got it all working! In a) Black might have captured hxgxf and axPb. So castling rights might still be maintained with the ep. In b) on the other hand, it must be axb, bxa, exf and wPgxh6, so there was the cage. Promotion to TD is tolerated in the final move, although it may not be puristic, without the convention, too many mates would be excluded (2022-12-16)
comment
Keywords: En passant as key, Castling (sg), Cant Castler, Valladao Task
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & basic retro thinking
FEN: r3k3/P7/b3r1pP/4PpBP/3nnpKR/5PRB/5PP1/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
6 - P0000101
Leonid M. Borodatow
Evgeny V. Kharichev

6209v Die Schwalbe 110 04/1988
P0000101
(12+10) cooked
h#3
b) sBh5 nach h6
a) 1. 0-0-0 Sxc7 2. hxg3 Sd5 3. g2 Se7#

Beispielauflösung mri:
R: 1. Kf1-e1 Ta2-a1 2. Kg2-f1 Tb2-a2 3. La2-b1 Tb1-b2 4. Kf3-g2 Tf1-b1 5. Ke3-f3 g5xDh4 6. Lh2-g1 Tg1-f1 7. Dd4-h4 Tg2-g1 8. Lg1-h2 Th2-g2 9. Kd3-e3 Tg2-h2 10. Kc3-d3 Th2-g2 11. Kb2-c3 Tg2-h2 12. Kc1-b2 Th2-g2 13. Kd1-c1 Tg2-h2 14. Ke1-d1 Th2-g2 15. Da1-d4 Tg2-h2 16. Dd1-a1 d6xLc5 17. La3-c5 Th2-g2 18. Lc1-a3 Tg2-h2 19. b2-b3 Th2-g2 20. Ld5-a2 Tg2-h2 21. Sc5-a6 Th2-g2 22. Lg2-d5 h6-h5 23. Lf1-g2 Tg2-h2 24. Lh2-g1 Tg1-g2 25. Sa4-c5 g2-g1=T 26. Sb6-a4 f3xTg2 27. Tg1-g2 f4-f3 28. g2-g3 f5-f4 29. Le5-h2 h7-h6 30. Lg7-e5 f6-f5 31. Lf8-g7 f7-f6 32. e7xLf8=L Lg7-f8 33. e6-e7 Lf8-g7 34. d5xTe6 Tg6-e6 35. Sc8-b6 Tg8-g6 36. Sb6xLc8 g6-g5 37. c4xDd5 Dg5-d5 38. Sd5-b6 Dd8-g5 39. Sg3-h1 e7xTd6 40. Ta6-d6 Th8-g8 41. Ta1-a6 g7-g6 42. b3xSc4 Sa5-c4 43. a2xSb3 Sd4-b3 44. Sc3-d5 Sc6-a5 45. Sb1-c3 Sb8-c6 46. Th1-g1 Sf5-d4 47. Se4-g3 Sh6-f5 48. Sg5-e4 Sg8-h6 49. Sf3-g5 Sh6-g8 50. Sg1-f3 Sg8-h6 51. h2-h3


b) 1. Kf7 f4 2. Kg6 c3,c4+ 3. Kh5 g4#
play all play one stop play next play all
in b) fehlt Schwarz ein Retrotempo zur Aufrechterhaltung des Rochaderechts, es kann maximal eine Stellung wie z.B. r3kB2/pppp1p1p/3p2p1/8/8/7P/1PPPPPP1/1NBQKBRN erreicht werden, in der entweder sTh8 oder sLf8 nicht als Schlagobjekt für die notwendigen weißen Bauernschläge zur Verfügung stand.

Der Dual 2. ... c3+,c4+ in b) war von den Autoren in Kauf genommen worden., die (schon bei der Lösungsveröffentlichung bekannten) NLs aber natürlich nicht. Ein Schwalbe-Löser ("PS") kritisierte den "völlig unnötigen Vorwärtsballast" und schlug implizit vor, das Vorwärtsspiel komplett wegzulassen und stattdessen direkt zu fragen: "Ist die 0-0-0 erlaubt?"
Cook: a) 1. hxg3 Sxg3 2. c6 Sf5 3. 0-0-0 Sd6#
b) 1. Kf7 f4 2. Kg6 c4+ 3. Kh5 g4#
Neufassung 5949.
A.Buchanan: It would be great if someone can retrieve 6209v from Die Schwalbe, please, to confirm the diagram and intended solution. If castling right remains, then sBf must promote on g1=T. There is then some jiggling around because Tg1-h2 to unpin Lg1 and let it go b1. In (a) this is doable, but in (b), Black runs out of tempi however there is an alternative no-castling mate with sKh5. However that's just a rough sketch, and I don't want to spend more time until the actual published form is confirmed (2022-01-08)
Ladislav Packa: Is everything allright here? Because solution a) is not a solution - after 3 .... Se7 # ?? 4.Kc7 / b8! (2022-01-09)
Mario Richter: Indeed, the given solution only works with wLh2 instead of g1. I only have access to the solution, but not to original diagram. Perhaps someoe with access to it can give us the correct details? (2022-01-10)
A.Buchanan: Yes I didn't check the solution to a) matches one of the actual ones. With wLh2, it's still cooked, as is b) in any case. Possible to rearrange the free pieces to make b) sound, but I don't see how to fix a) at the same time. (2022-01-10)
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Keywords: Castling (sg), Cant Castler (sg)
Genre: h#, Retro
FEN: r3k3/pppp4/N7/2p4p/7p/1P4PP/2PPPP2/rB2K1BN
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-01-10 more...
7 - P0000112
Dmitri W. Pronkin
Andrey Frolkin

6386v Die Schwalbe 113 10/1988
P0000112
(14+14) cooked
BP in 45.0
1. f4 g5 2. f5 g4 3. f6 g3 4. fxe7 gxh2 5. g4 d5 6. g5 d4 7. g6 d3 8. g7 dxc2 9. d4 f5 10. Lf4 c1=T 11. Lg3 Tc6 12. Dd3 Tg6 13. Da6 Sf6 14. g8=L c5 15. Lb3 c4 16. d5 Kf7 17. e8=T c3 18. Te4 c2 19. Ta4 c1=L 20. e4 Lf4 21. Sd2 h5 22. 0-0-0 h4 23. Te1 h3 24. Ld1 Th4 25. d6 L8h6 26. d7 Dh8 27. d8=T Le6 28. Tc8 S8d7 29. Tc2 Kg8 30. e5 Lf7 31. e6 Tf8 32. e7 Lb8 33. e8=S f4 34. Te7 f3 35. Se2 f2 36. Tg1 h1=S 37. b4 h2 38. Lh3 f1=S 39. b5 Sf2 40. b6 h1=L 41. bxa7 b6 42. a8=L Lb7 43. Th1 Lc8 44. Lag2 Se3 45. Lf1 S6e4
play all play one stop play next play all
Cook: 1. b4 c5 2. b5 c4 3. b6 c3 4. bxa7 d5 5. e4 d4 6. f4 d3 7. f5 dxc2 8. d4 g5 9. Lf4 c1=L 10. d5 g4 11. f6 g3 12. fxe7 gxh2 13. g4 c2 14. Lg3 Lf4 15. g5 c1=T 16. g6 Tc6 17. Dd3 f5 18. Sd2 h5 19. g7 Tg6 20. 0-0-0 Sf6 21. Te1 h4 22. d6 h3 23. g8=L Th4 24. Lb3 L8h6 25. Ld1 Kf7 26. d7 Dh8 27. d8=T Le6 28. Td4 Sbd7 29. e5 Tf8 30. e8=T Kg8 31. Tc8 Lf7 32. e6 b6 33. e7 Lb8 34. e8=S f4 35. Te7 f3 36. Se2 f2 37. Tg1 h1=S 38. Da6 h2 39. Lh3 f1=S 40. Ta4 Sf2 41. Tc2 h1=L
Michel Caillaud: cooked by Stelvio 0.93 :
1.b4 c5 2.b5 c4 3.b6 c3 4.bxa7 d5 5.e4 d4 6.f4 d3 7.f5 dxc2 8.d4 g5 9.Lf4 c1=L 10.d5 g4 11.f6 g3 12.fxe7 gxh2 13.g4 c2 14.Lg3 Lf4 15.g5 c1=T 16.g6 Tc6 17.Dd3 f5 18.Sd2 h5 19.g7 Tg6 20.0-0-0 Sf6 21.Te1 h4 22.d6 h3 23.g8=L Th4 24.Lb3 L8h6 25.Ld1 Kf7 26.d7 Dh8 27.d8=T Le6 28.Td4 Sbd7 29.e5 Tf8 30.e8=T Kg8 31.Tc8 Lf7 32.e6 b6 33.e7 Lb8 34.e8=S f4 35.Te7 f3 36.Se2 f2 37.Tg1 h1=S 38.Da6 h2 39.Lh3 f1=S 40.Ta4 Sf2 41.Tc2 h1=L... (2022-12-20)
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Keywords: Unique Proof Game, Move Length Record, Non-standard material, Castling, Promotion (tLTlTSsslL)
Genre: Retro
FEN: 1bb1Nrkq/3nRb2/Qp4rb/8/R3n2r/4n1BB/P1RNNn2/2KB1B1R
Reprints: 583 Ukrainisches Album 1986-1990
80 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Silvio Baier, 2023-03-02 more...
8 - 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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comment
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...
9 - P0000195
Frank Christiaans
7167 Die Schwalbe 126 12/1990
P0000195
(13+10)
#3
b) wSe4 nach g8
a) 1. 0-0? droht 2. Tf8+ Kd7 3. Lxe6#
1. ... Lxc5! 2. Sxc5 0-0-0!
1. Thf1! 0-0-0 2. Lxe6+ Td7 3. Tf8#
b) 1. Thf1? Lxc5 2. Sxc5 0-0-0!
1. 0-0! Lxc5 2. Sxc5 ... (0-0-0?) 3. Tf8#
play all play one stop play next play all
a) zuletzt h6-h5 und beide Rochaden sind noch möglich. Schläge von Weiß: Bbxc, Bcxd, Bhxg, Bfxg-g8=L.
b) gegenseitiger Ausschluss der beiden Rochaden. 3 Fälle sind zu betrachten:
b1) zuletzt f7xDe6; Weiß muss 4x geschlagen haben: Bfxexdxc8=L, d.h. s0-0-0 ist unzulässig.
b2) zuletzt d7xDe6; zu den 4 Schlägen von a) wird auch der sBf gebraucht, der sich auf f1 umwandeln muss, also w0-0 unzulässig.
b3) zuletzt h6-h5: wBf zieht über f7 zum UW-Feld g8, g.h. s0-0-0 unzulässig.
Henrik Juel: C+ Popeye 4.61 after analysis (2020-10-30)
A.Buchanan: A very harmonious problem. Suppose that both sides retain castling rights. wLh3 is obtrusive, so wBf/h promoted via g7, and escaped via e6 so sBe6 came from d7 later. Wh has made at least 4 pcs, while Bl has made 2 (D & Lc) dxe6, gxh6. If last move was dxe6, then sD & sLc were captured by officers in cage. Thus sBf was captured by wB, and therefore promoted on f1, disrupting wK. Therefore the last move was h6-h5. In (a) there is no problem retaining both castling rights, but in (b) Sg8 must have reached that square via f6 *after* the promotion, disrupting bK. However in (b) the castling rights are incompatible. In (a) 1. 0-0? Lxc5! 1. Tf1!, while in (b) the reverse.
Neither twin is a PRA problem: in (a) there is no conflict between the castling, while in (b) RS applies: the pre-emptive Wh castling means we are in a reality where Bl cannot castle. (2020-10-31)
comment
Keywords: Castling key (wksg), Obvious promotion (L), Retro Strategy (RS)
Genre: Retro, 3#
FEN: r3k3/ppp1p2p/Nb2p3/P1PP2Pp/4N3/7B/3PPRP1/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2022-01-07 more...
10 - P0000250
Nikita M. Plaksin
Valery Liskovets

7577v Die Schwalbe 132 12/1991
P0000250
(5+14)
ser-h#5 (AP)
Typ Tauber/Caillaud
1. Kd4 2. bxc3ep (zuletzt nur K,T-Zug oder c2-c4) 3. e5! (nicht 4. Tb4? weil zuletzt K,T-Zug und 0-0-0 nicht mehr möglich) 4. Tb4 (zuletzt e7xLd8=S möglich) 5. Tc4 & 1. 0-0-0# nicht 1. Td1#? wegen fehlender AP-Legalisierung
play all play one stop play next play all
"das zeigt die Paradoxie der AP-Bedingung!" (MS). Gleich 1. bxc3 ep.? und dann 2. Kd4 geht natürlich nicht, weil zuvor außer K,T-Zug und c2-c4 auch S-Züge und c3-c4 möglich waren. "Interessanter und genau abgestimmter Lösungsverlauf" (GW) "Sehr schöne und ökonomische Verbindung zwischen konsequentem sh# und AP!" (TB) Mit 'Typ Tauber/Caillaud' ist gemeint, daß die einzelnen Stellungen doch als retroanlytischer Gesamtkomplex betrachtet werden dürfen/sollen (sonst wäre keine AP-Legalisierung möglich), obwohl sie ja aufgrund der retroanalytischen Neubewertung nach jedem Rückzug voneinander unabhängig sind. 6L./5+5P.
A.Buchanan: "By 'type Tauber/Caillaud' it is meant that the individual positions may/should be regarded as a retroanalytic overall complex (otherwise no AP legalization would be possible), although they are independent of each other due to the retroanalytic reassessment after each retraction." (2023-06-29)
comment
Keywords: Castling (wl), a posteriori (AP), Promotion (S), Seriesmover, Consequent, En passant
Genre: Retro, Fairies
FEN: 1N6/1ppppppn/n7/1Pq1k3/1pP1r3/4p3/1r6/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-29 more...
11 - P0000254
Leonid M. Borodatow
7642 Die Schwalbe 133 02/1992
P0000254
(16+10)
Welches waren die letzten 7 Einzelzüge, wenn dabei keine Zugwiederholungen vorkamen?
R: 1. 0-0-0# Ke4-d4 2. e5xf6ep+ f7-f5 3. Tg6-b6+ Kf5-e4 4. c7-c8=L
play all play one stop play next play all
Die von einigen Lösern angeführte Abweichung 2. f5-f6+ Kd4-e4 3. Lh6-g7+ (und mehrdeutig weiter) ließe sich durch die Erweiterung '... keine Zugwiederholungen und keine Pendelzüge ...' (mühsam) kitten. Beim Autor hieß es bei dieser ich-weiß-nicht-wie-vielten Fassung nur 'letzte 9 (!) Einzelzüge ohne Wiederholung).
HHS meint ohnehin, daß es das ganze auch ohne die einengende Zusatzbedingung schon gibt.
Das von einem Löser angegebene 1. Ld3-h7# Th1-h8 2. Lh8-g7 Tg1-h1 3. Se1-g3 g2-g1=T 4. Th7-h8=L scheitert allerdings an der Schlagbilanz.
Anton Baumann: Neufassung vergl. P0006288 (2023-01-06)
comment
Keywords: En passant, Last Moves?, Non-standard material, Castling (wl), Promotion (L), Valladao Task (WWW)
Genre: Retro
FEN: qrB2brr/Bp2p1BB/pR3P2/1Q6/2Pk1P2/B1p2R2/2P3N1/2KR1N2
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2019-08-11 more...
12 - 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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comment
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...
13 - P0000582
Alexander Kislyak
3463v Die Schwalbe 72 12/1981
Dr. L. Ceriani zum Gedenken
P0000582
(10+13) cooked
BP in 46,0
AL: 1. b4 a5 2. c4 a4 3. d4 a3 4. Lb2 axb2 5. a4 b5 6. a5 c5 7. a6 Da5 8. a7 Sc6 9. bxa5 Lb7 10. a6 0-0-0 11. a8=S b4 12. Sc7 Sa7 13. Se6 Sb5 14. Sxf8 Sh6 15. a7 f6 16. a8=S Sf7 17. Sc7 b3 18. Sce6 dxe6 19. Ta4 Td5 20. cxd5 c4 21. d6 c3 22. d5 c2 23. Sc3 Sg5 24. d7+ Kb8 25. d6 c1=L 26. d8=S b1=L 27. d7 b2 28. Sf7 Lf5 29. Sd6 Lh3 30. d8=S b1=L 31. S8f7 Lg6 32. Sh6 Lf7 33. Sg6 hxg6 34. Shf5 Th4 35. Tf4 gxf5 36. Sde4 fxe4 37. e3 Lf5 38. Ke2 Tg4 39. h4 Sa3 40. h5 Sb1 41. h6 Ka8 42. h7 Ld5 43. h8=S Ld2 44. Sg6 Le1 45. Sh4 Sh3 46. Shf3 exf3+
play all play one stop play next play all
Cook: NL: 1. b4 a5 2. c4 a4 3. d4 a3 4. Lb2 axb2 5. a4 b5 6. a5 c5 7. a6 Sc6 8. a7 Lb7 9. Ta6 Da5 10. bxa5 0-0-0 11. a8=S b4 12. Sc7 b3 13. Se6 Sb4 14. Sxf8 Sf6 15. Te6 dxe6 16. a6 Td5 17. cxd5 c4 18. a7 c3 19. a8=S c2 20. Sc3 c1=L 21. Sb6+ Kb8 22. Sc8 b1=L 23. d6 Lg6 24. d7 b2 25. d8=D b1=L 26. Da5 Lbf5 27. d5 Se4 28. d6 f6 29. d7 Lf7 30. Sg6 hxg6 31. d8=T Lh3 32. Td4 Th4 33. Df5 gxf5 34. Sd6 Sg5 35. Tf4 Ka8 36. S6e4 fxe4 37. e3 Lf5 38. Ke2 Tg4 39. h4 Sc2 40. h5 Sa3 41. h6 Sb1 42. h7 Ld5 43. h8=S Sh3 44. Sg6 Ld2 45. Sh4 Le1 46. Shf3 exf3+
James Malcom: Again, how is this cooked? (2021-01-25)
A.Buchanan: The AL has 5 Wh S promotions in 46.0 moves but the cook shows this is not necessary. (2021-01-26)
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comment
Keywords: Non-Unique Proof Game, Ceriani-Frolkin Theme (SSSSS), Non-standard material (ll), Castling, Promotion, konsekutive Umwandlungen 8
Genre: Retro
FEN: k7/4pbp1/4pp2/3b1b2/5Rr1/2N1Pp1n/4KPP1/1n1QbBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-01-26 more...
14 - P0000583
Andrey Frolkin
Günter Lauinger

3468v Die Schwalbe 72 12/1981
P0000583
(10+12) C+
h#2 (AP)
Satzspiel:
*1. ... 0-0! 2. cxd4 Tc1#

Verführung:
1. exd3ep? 0-0 2. cxb4 Tf4# aber zuletzt R: 1. d2-d4?? ist illegal, da es den wLc1 als schwarzes Bauernschlagobjekt ausschließt

Lösung:
1. ... 0-0! (AP-Legalierung des weissen Anzugrechtes) 2. cxd4 Tc1#
play all play one stop play next play all
Alle Versuche, das Satzspiel 1. ... 0-0! 2. cxd4 Tc1# durch einen schwarzen Vorschaltzug aufrecht zu erhalten, scheitern.
Mario Richter: 1. exd3ep 0-0 2. cxb4 Tf4 ist nur Verfuehrung (scheitert an schwarzer Bauern-Schlagbilanz).
Loesung: 1. ... w0-0! (AP-Legalierung des weissen Anzugrechtes) 2.cxd4 Tc1# (2009-02-10)
A.Buchanan: Supersedes P0000553 (2022-01-08)
Mario Richter: Hi Andrew, I only quoted from the "official solution" ('Die Schwalbe' Heft 72, 12/1981, p.399).
Perhaps a better way to get anwers to your interesting questions is to ask the authors directly ... (2022-01-08)
A.Buchanan: AP Type Petrovic is a try, because wLc1 was captured at home, yet Black still made 6 pawn captures.
AP Type Keym is the actual solution, because if it was really BTM, then White would have lost castling rights.
So there are two kinds of AP here. (2022-03-21)
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comment
Keywords: Castling (wk), a posteriori (AP) (Type Keym), En passant as key, a posteriori (AP) (Type Petrovic)
Genre: h#, Retro
Computer test: BC+ Popeye v4.87
FEN: 8/2p5/1pPp4/bRpP4/BPkPp3/qp2p2p/rP2P3/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
15 - P0000598
Nikita M. Plaksin
Andrej N. Kornilow

3876 Die Schwalbe 74 04/1982
P0000598
(4+3)
h=2 (AP)
Circe, Monochromes Schach
Intended solution:
1. dxc3ep[+wBc2] 0-0 2. axb1=S Kh2=
2. ... Kf2? 3. Kf3!
1. ... Tf1? 2. axb1=S Kf2? 3. Kf3!
play all play one stop play next play all
Erich Bartel: vom Dual 2.-- Kf2/Kh2 abgesehen C+ PY V4.41.--- (2008-11-07)
A.Buchanan: This is cooked in Popeye because White can't castle as long as bK controls f1! This is quite reasonable behaviour in monochrome: compare to en passant! So no solution (2022-02-18)
comment
Keywords: a posteriori (AP) (Type Petrovic), Circe, En passant as key, Monochromatic Chess, Castling (wk), Miniature, Golden Age (Monochrome castling)
Genre: Retro, Fairies
FEN: 8/8/8/8/2Pp4/8/p3k3/1N2K2R
Reprints: (IV) Quartz 4 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-18 more...
16 - 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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comment
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...
17 - P0000641
Werner Frangen
2201 Die Schwalbe 46 08/1977
P0000641
(14+5)
#4 (AP)
1. bxc6ep
play all play one stop play next play all
Nach 1. bxc6ep ist zwar ein Matt in 2 Zügen möglich, aber Weiß muss noch rochieren, um den ep-Schlag zu begründen.
James Malcom: Full solution? (2021-02-05)
Henrik Juel: White pawns captured all 11 missing black men by axb, f2x..xa7, and g2x..xb7
Black captured hxg and once more, so Pc5 never captured
If last move was c6-c5, the preceding white move was with Ke1 or Th1
So if White can castle, last move was c7-c5
1.bxc6ep dxc6 2.0-0(AP) c5 3.Ta1,Sd7 and #4
2.Sd7 c5 3.0-0(AP) is also possible
I expect the full solution was omitted because it is very dualistic (2021-02-05)
Anton Baumann: Autorabsicht: 1.bxc6ep [2.0-0 3.Lb5#] bxa3 2.0-0 axb2 3.Tb3 nebst 4.Lb5,Tf4#
1. ... dxc6? 2.0-0 nebst 3.Tf4,Lb5#
Dual: 2.Tb3 dxc6 3.0-0 [4.Tf4#] c5 4.Lb5# 2. ... a2,axb2 3.0-0 [4.Lb5#] dxc6 4.Tf4# (2021-02-05)
A.Buchanan: C+ of helpmates with this kind of AP is relatively simple: one can just eliminate all lines without 0-0. But C+ of d# is harder.
(1) Henrik's retro logic is solid. There is no #4 unless 1.bxc6ep, so let's take that as a given: an AP debt is incurred.
(2) Black has 5 responses. Against 4 of them, 2.0-0 gives a genuine short mate with the AP debt paid. Such short mates don't impact soundness in a retro #4, to my mind.
(3) This leaves 1... bxa3. Following 2.0-0 axb2 (the only full length line) 3.Rb3 thr 4.Bb5,Rf4# but both mates are separately provided, so only minor duals.
(4) It remains to consider if White can delay castling after 1...axb3. Apart from 2.Tb3 nothing works, but 2.Tb3 thr 3.0-0 is unstoppable and is the major dual to my mind. I don't see any obvious fix. (2021-02-05)
comment
Keywords: Castling (wk), En passant as key, a posteriori (AP)
Genre: Retro, n#
FEN: 1N6/PP1p4/BR6/pPp5/kp6/B2P4/1PP1P2P/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
18 - P0000649
André Hazebrouck
2261 Die Schwalbe 47 10/1977
4. Preis
P0000649
(8+14)
h#2.5
b) AP (2 Lösungen)
a) 1. ... 0-0-0 2. 0-0 Tdg1+ 3. Kh8 Txh6#
b) also 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#
is clearly the intention
play all play one stop play next play all
Can count 1+7 visible pawn captures, leaving 1+1 unexplained. Need to resolve wPfgh & bPg. Suppose bK never moved, then one of:
1) wPf waylaid, wPhxPg=, wPg=. Here w00 right can remain.
2) Or similarly, wPh waylaid, wPfxPg=, wPg=. Again w00 right can remain.
3) bPgxPh=, wPfxg= (or wPfxe then captured by bPf), wPg= w00 right lost.
1+1 captures explained in either case.
On the other hand if bK moved, then maybe wPf=, and only requires one more capture to resolve g&h files. The w00 right can remain. So there is a captured unit unaccounted for, and we can't validate the ep.
On the other hand if we are in case 3 above, then we can't know that wPbxa captured dark bishop, so White might retract a6-a7.
We also need to know that w00 rights remain. s000 rights (i.e. prior movement of sTa8) are not relevant.
If we know that w00, w000 & b000 remain, then we are in case 1 or 2 above with bPg captured unpromoted. Therefore wPb6xLa7 due to bishop shade. Therefore R: 1. b7-b5 b6xLa7 to unblock.

Only one candidate solution does not begin with ep: 1. ... 0-0-0 2. 0-0 Tdg1+ 3. Kh8 Txh6#. So this is the solution for a).

Let ????? denote validity of w00,w000,b00,b000,ep. Possibilities are: YYY?Y and all of YYN??, YNY??, NYY??, YNN??, NYN??, NNY??, NNN??. So there are 30 possibilities.
Under PRA, the solution parts would be YYYYY, YYNYN, YNYYN & NYYYN. For the first there are 20 solutions, while the second and third have 0 solutions. So this is not the right paradigm.
Under SPRA, there would be a single solution part YYYYY with 20 solutions. So this is not the right paradigm either.

Under RS with AP, the solution of a) still works. This comes from ?YY??. On the other hand, if the first move is ep, then we are in YYYYY. So all castlings are valid. But when we get to perform the mate, we need to know that based on castlings actually performed, the ep is valid. So the solution must include w00. So is 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#.

By combining the information of both of these solutions, we know we are in YYY?? so YYY?Y is the only possibility, and ep is legal. However, the combination of these two solutions would also validate any of the other 18 ep solutions that do not include 2. ... 0-0. They contribute no new evidence, but how to exclude them?

This is one of the issues with "AP Consolidation". Normally the number of solutions is not a constraint. But here perhaps we should insist that *only* two solutions are allowed?
Cook: 1. ... cxb6ep 2. La6 Tb1,~ 3. 0-0-0 a8=D#
18 different possibilities for W2 which are validated by the same logic that validates 2. ... 0-0.
Henrik Juel: [I don't follow the silly convention of writing black moves first]. 1.0-0-0 0-0 2.dTg1+ Kh8 3.Txh6#. b) If Ke1,e8 and Th1 never moved, White captured f/hxPg and b6xLa7; if Ta1 also never moved, last white move was b6xLa7. 1.cxb6ep La6 2.0-0/0-0-0 0-0-0 3.a8Q#. The a) solution also works in b). (2003-12-18)
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comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (wkwlsk), En passant as key
Genre: h#, Retro
Computer test: Forward logic Popeye v4.87 & basic retro-logic works, but then AP Consolidated protocol seems to cook the problem.
FEN: r1b1k2r/P2p4/2p1n2p/ppPq4/1np5/p7/P2PP3/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
19 - 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...
20 - P0000680
Tivadar Kardos
2527 Die Schwalbe 52 08/1978
P0000680
(4+2) C+
#1
b) Kb1 nach g1
a) 1. ... Kxa1 2. 0-0#
b) 1. ... Kxh1 2. 0-0-0#
play all play one stop play next play all
A.Buchanan: The term "retro" is jungle not garden - that means we should not expect an axiomatic definition. The current problem is a case in point. Case law has established that neither simple employment of the castling convention nor existence of check are sufficient to make a problem "retro". But all this problem has is the quirky use of Codex Article 15 to force BTM. So I think this problem has to be retro. The key point is that nothing hinges on the retro-ness. If the problem included 50M or DP, then one would expect a more solid foundation. As it is, all we need is the free direct mate in 1 that comes as part of the retro paradigm. (2021-11-26)
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comment
Keywords: Castling (wk), No legal last move for Black, Minimal, Miniature
Genre: Retro
Computer test: C+ Popeye v4.87
FEN: 8/8/8/8/8/4p3/4Q3/Rk2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-26 more...
21 - P0000758
Gerd Rinder
1033 Die Schwalbe 21 06/1973
1. Preis
P0000758
(7+11)
Remis (AP)
Weiß ist patt. 1. cxb6ep ist nur zulässig, wenn Schwarz diese a posteriori durch die Rochade rechtfertigt. Weiß kann aber so spielen, daß Schwarz nicht rochieren kann, z.B. 1. ... Lxc7 2. bxc7
play all play one stop play next play all
Guus Rol: This is an incorrect interpretation of the AP-convention. Rules outrank goals in the definition of all GAMES. Therefore the legitimacy of a move cannot be restricted by the desire to achieve the goal (in this case: Remis). The proper way to view AP is that executing e.p. invalidates the legitimacy of all lines of future play that do not contain 0-0-0! In that sense black and white are forced to cooperate. In whatever freedom remains they can compete for the prize promised in the stipulation. By the way, this understanding of AP is not only more logical, it is also much more interesting as a playing field for AP-composition. (2005-09-21)
mri: citeWeiß kann aber so spielen, daß Schwarz nicht rochieren kann, z.B. 1. ... Lxc7 2. bxc7
/cite
How does white prevent black from castling after 1. cxb6 e.p. Ba7xb6+?
E.g. 2.Kxb6 a1=R 3.a7 Rxa7, or 2.Ka4 Bd8xc7 3.a7 Bb8. (2005-09-22)
VL: 1.c5xb6ep(??) a7xb6+ 2.Kxb6 a1R 3.Kb7 R1xa6 4.Rc8! (R6a7#?? illegal).
This study is correct under the generally accepted understanding of AP
a la' N.Petrovic'. Antiform (looking possibly somewhat strange):
Black's unsuccessful try. (2005-10-03)
Guus Rol: Generally accepted, true. Generally acceptable, false. Freedom of interpretation ceases for a concept once its polar concept (a priori validation) has been defined. "Goal induced AP" however, might be a passable stipulation for this type of problem. To keep this place from turning into a discussion forum I will discontinue comments on this issue. (2005-10-03)
paul: Author intention is: If black still can castle, his last move must have been b7-b5. However, to prove this, he has to castle (A Posteriori condition). So: 1.cxb6 e.p. axb6+ 2.Kxb6 a1R! (2...a1Q? 3.Kb7 Qxa6 mate obviously doesn`t prove b7-b5 as the last move) 3.Kb7 R1xa6 4.Rc8! and white has prevented black from castling. So black can`t prove the last move is b7-b5 and therefor is already stalemated in the initial position! (2011-07-12)
A.Buchanan: Isn't Guus' idea just AP-Prioritat? And even if it were "more logical" or "more interesting", it doesn't follow that other forms is AP are "incorrect" (2023-07-31)
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comment
Keywords: En passant as key, Castling (sg), a posteriori (AP)
Genre: Retro, Studies
FEN: r2bk3/p1Rpp3/P1p2p2/KpP2P2/1P2p3/1P6/p7/8
Reprints: (4) Problem 161-164 11/1973
317 Europe Echecs 217 01/1977
(A) Die Schwalbe 80 04/1983
408 Eigenartige Schachprobleme 2010
M36 mpk-Blätter 12/2011
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-31 more...
22 - P0000759
Luis Alberto Garaza
1034 Die Schwalbe 21 06/1973
P0000759
(9+9) cooked
Schwarz am Zug, Weiß gewinnt
1. ... fxg3ep 2. hxg3+ Kh5 3. f4 Kh6 4. h8=T+! Kg7
play all play one stop play next play all
Cook: Intended a Posteriori combo of castling and ep rights retro-locks wTh in cage, contradicting pawn capture balance
hans: only black move is fxg3e.p.
1. fxg3+ Kh5 2. h8D/T#
Why the keyword 'castling'?
Castling is illegal, for there are 7 black captures, and because white's last move is g2-g4, needs the wK let pass Rh1 (2012-05-18)
Anton Baumann: g2-g4 war nicht zwingend der letzte Zug! Schwarz ist patt, also unlösbar!
Korrektur in 'die Schwalbe' 12/2074 S.264: "wTa1 nach h1. Nun gewinnt nach 1. - fxg3 Weiss mit 2.hxg3+ Kh5 3.f4 Kh6 4.h8=T+! Kg7 und der umgewandelte Turm holt die störenden sBB ab, sodass Weiss rochieren kann und beweist, dass Schwarz nicht patt war. Schwarz versucht die Rochade zu verhindern, ist aber machtlos." (2022-12-06)
A.Buchanan: The forward play is more complicated, as Black can play more aggressively than 3. … Kh6 (2022-12-07)
Anton Baumann: Wenn Schwarz nicht 3.-Kh6! zieht, holt sich Weiss auf h8 die Dame, mit einfacherer Fortsetzung.
3.-g4? 4.f5 Kg5 (4.-g5 5.f6 ~ 6.h8=D) 5.h8=D Kxf5 6.De8 +_
3.-Kg4? 4.h8=D Kf3 5.Dxh3 g4 6.Dh2 Ke4 7.Dg2+ +_ (2022-12-09)
more ...
comment
Keywords: Castling (wl), a posteriori (AP) (Type Petrovic), En passant as key
Genre: Retro, Studies
FEN: 8/7P/6p1/2p3p1/2p2pPk/2Pp1P1p/3PpP1P/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-07 more...
23 - P0000760
André Hazebrouck
1120 Die Schwalbe 23 10/1973
P0000760
(9+10) C+
h#2 (AP)
1. bxc3ep Sa6 2. 0-0-0 Tc4#
play all play one stop play next play all
Henrik Juel: -1.c2 c5xLb4 -2.Lc3 c6 -3.Lb2 c7 -4.Lc1 a6! -5.b2 b2xDSa2 etc. White captured sDTSS by fxexd, gxfxe, allowing Black to capture f7xPe6xPd5 etc. (2003-12-18)
A.Buchanan: Why not e.g. wSb1? (2022-03-04)
Henrik Juel: That also seems to work (2022-03-04)
A.Buchanan: I really like the motivation for ep. I guess the motivation for Rb1 is to be inside the cage that forms around it, but it’s still a bit loose. Note there is no retro try. (2022-03-05)
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comment
Keywords: En passant as key, Castling (sg), a posteriori (AP) (Type Petrovic), Volet Pawn
Genre: h#, Retro
Computer test: HC+ Popeye 4.87 + simple retro reflection
FEN: rN2k3/1p1pp1pp/8/p7/RpP5/PP6/p2PP3/KR6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-05 more...
24 - P0000775
Ferad Kakabadze
1422 Die Schwalbe 30 12/1974
P0000775
(11+5) C+
#3
Welches waren die letzten 2 Einzelzüge?
1. e8=D! f5 2. exf6ep e5 3. Dxe5#
R: 1. b2-b1=L 0-0
play all play one stop play next play all
Henrik Juel: The forward play is C+ Popeye 4.61
1.e8=D thr. 2.Dc8 thr. 3.Dc3#
1... f5 2.exf6ep e5 3.Dxe5#
(1... f6 does not prevent the threat, so 2.exf6 is not considered dualistic) (2020-09-26)
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comment
Keywords: Last Moves?, Valladao Task, En passant, Promotion, Castling in the retro play, Obvious promotion
Genre: Retro, 3#
FEN: 8/3pPp2/3Pp3/4P3/1N3P2/B7/P1P2P2/kb3RK1
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2020-09-27 more...
25 - 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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comment
Keywords: Castling (sg), Valladao Task, Superseded by (P0000058)
Genre: h#, Retro
FEN: r3k3/P7/p3r1pP/4PpBN/3nnpKP/6PB/1P2PPb1/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
26 - P0000792
Klaus Wenda
1557 Die Schwalbe 33 06/1975
2. Preis
P0000792
(13+12)
#2 Längstzüger
b) sTa7 nach d7
Anton Baumann: Autorabsicht: Die weiss-schwarzen Rochaden schliessen sich gegenseitig aus.
a) 1.O-O? Tf8! daher: 1.Tf1! O-O 2.Sxe7#
b) 1.Tf1? O-O! daher: 1.O-O! Tf8 2.Sxg7#
Aber in der Urfassung (= nebenstehendes Diagramm) geht in a) und b) die NL:
1.Tg1 O-O 2.Txg7,Sf5xh6#
Korrektur in 'Schwalbe' 04/1976 S.464: sLb7 nach g6, sBc5 nach b7
Ausgezeichnet wurde die korrigierte Fassung 1557v (vergl. 'Die Schwalbe' 06/1977 S.82) (2022-12-09)
comment
Keywords: Maximummer, Castling (wksk)
Genre: Retro, Fairies
FEN: 4k2r/rb2pNbp/1P5p/p1pppN2/8/8/PPPPP2P/2BQK2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
27 - P0000793
Nikita M. Plaksin
Andrey Lobusov

1558 Die Schwalbe 33 06/1975
4. Preis
P0000793
(13+5) C+
#3 (AP)
1. cxd6ep+! Kxg3 2. 0-0! (proving ep ok)
2. ... Kg4 3. g8=D,T#
2. ... a1=D,L 3. Dg5#

R: 1. d7-d5 Tc6xh6 (unless R: Kg1-g2 Th~*h1)
play all play one stop play next play all
VL: A posteriori (AP)!
2.g8Q(R)+/Rg1+? Kf4 3.Qg5#??
Retro: d7-d5, Rc6xh6+ (unless Kg1-g2, Rh~-h1+).
One of the best AP-type problems. (2015-08-20)
James Malcom: The first Valladao AP? (2020-10-03)
A.Buchanan: Hi James, there are 6 earlier ones in PDB, not counting those which are not marked as Valladao (could quite a few, as basic AP is 2/3 of Valladao). Type k='a p' and not g='fairies' and k='valladao', and set sort order by date. The ur-problem is P0003417. (2020-10-03)
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comment
Keywords: Castling (wk), a posteriori (AP) (Type Petrovic), En passant as key, Promotion (ws), Valladao Task (w w w/s)
Genre: Retro, 3#
Computer test: Popeye v4.85 + minor retro/AP thought
FEN: B7/p5PP/p6R/Q1Pp4/8/PP1P2P1/p3P1k1/4K2R
Reprints: (B) Die Schwalbe 48 12/1976
feenschach 40 11-12/1977
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-11 more...
28 - P0000822
Josef Haas
1938 Die Schwalbe 41 10/1976
P0000822
(12+11)
Ergänze den wK, dann #2
Kees: +wKb5 1. Txc8 (2. Lxe7#)
0-0 is illegal for K or T must have made a move.
-1. a7-a5? Not possible with position of wL and bS (2022-11-23)
comment
Keywords: Castling (sk), Add pieces
Genre: Retro
FEN: 2nBk2r/3pp3/1p1p2P1/p4NN1/PP4p1/7b/PP2P1Pp/2R2B2
Input: Gerd Wilts, 1995-06-03
29 - 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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comment
Keywords: Cant Castler, Castling (wksk), Cross-capture (s,w), Superseded by (P1399805)
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: B3k2r/pN1p1p2/1pp3p1/bb3p2/8/p1B3PP/5P1r/nn2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-16 more...
30 - 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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comment
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...
31 - P0001107
Michel Caillaud
3332 Die Schwalbe 66 12/1980
2. Preis
P0001107
(7+9) C+
h#4.5 (AP)
Circe
1. ... hxg6ep[+sBg7] 2. a1=T+ Txa1[+sTh8] 3. 0-0 Ta3 4. bxa3[+wTa1] 0-0-0 5. Td8 Txd8[+sTh8]#
play all play one stop play next play all
Wenn die beiden Könige noch nicht gezogen haben (was a.p. durch Ausführung beider Rochaden bewiesen wird!), kann der letzte Zug nur g7-g5 gewesen sein, Sonst ginge z.B. auch R: 1. Kd8-e8 c2xDb3
Henrik Juel: A possible retroplay is -1... g7 -2.f3xP(-Pg7) b5 -3.f2 c6xP(-Pb2) -4.b2 a6 -5.Rc3 a7 -6.Rc1 b6xB -7.Bf8 b7 -8.Bg7xB etc. (2003-04-22)
Michel Caillaud: The problem is not cooked as indicated, as under A Posteriori convention both castles must be played in the solution to justify the en passant key. (2022-02-04)
A.Buchanan: If wK never moved, then Black has not just moved another non-capturing pawn, as White has no prior move. The only possible exception is R: 1. g7-g5 f/h3xPg4[+Pg7]. (2022-02-04)
comment
Keywords: a posteriori (AP), Circe, En passant as key, Castling (wgsk), Promotion in forward play, Valladao Task
Genre: Retro, Fairies, h#
Computer test: HC+ all solutions begin with ep, but only one includes both white & black castling
FEN: 4k3/8/7p/p1p3pP/Pp4Pp/RP4p1/pP6/4K3
Reprints: feenschach 61, p. 534, 08/1982
RA152 diagrammes 72 07-08/1985
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-28 more...
32 - P0001117
Michel Caillaud
3872 Die Schwalbe 74 04/1982
12. Lob
P0001117
(11+12)
h#2
b) sBh7 nach h6
a) 1. 0-0-0? Sg6 2. Lb8 Sxe7! illegal castling
1. Kf7! Sxh7 2. Ke6 Sg5#
b) 1. 0-0-0! Sg6 2. Lb8 Sxe7#
1. Kf7? Sh7 2. Ke6 Sg5+? 3. hxg5!
R: 1. Ke3-e4 Sd6-b7 2. h2xDg3 Dg2-g3+ 3. Kd2-e3 Db7-g2 4. Ke1-d2 Dc8-b7 5. e4-e5 Dd8-c8 6. f3xLe4 Lb7-e4 7. Kf1-e1 Lc8-b7 8. Ke1-f1 b7-b6 9. Kf1-e1 Lb6-a7 10. ... La5-b6 11. ... Ld2-a5 12. ... Lc1-d2 13. ... d2xTc1=L 14. ... e3xTd2 15. ... e4-e3 16. ... f5xLe4
play all play one stop play next play all
Missing: Wh: QRRBB Bl: QRBB
Captures: Wh: gxf3xe (inc B), hxg + [Bf8] Bl: axbxc (for Ba7) fxexdxc/e (g1 not possible)
Assume Black can castle: so neither bK nor bRa have moved. Before bPb7-b6 (which releases QB) *all* Bl captures have been made. wPd can freely advance, all Wh units released except for rooks & Bf1. gxRf3 is forced, and now all Wh units are free and can be captured. To avoid deadlock, wB was captured on e4 not e6. Sequence must be f7-f5 Rh8-f8-f6-...-f3gxRf3.
In this position wLe4 must be played back to f1. With bPh7, wBe4 must retract either by stopping on f7, (disrupting bK) or via f5 (implying retraction of f6-f5, in which case bRa8 is an imposter).
In the alternative route via h7, wB crosses over f7 harmlessly. While wB is on g5 & h4, f7 must be occupied by a static shielding knight, but there is no tempo issue. After all this excitement, b7-b6 if followed by simple and non-unique play to reach the diagram.

(Gerd's earlier solution: Weiße Schläge: h2xg3, gxfxe, sLf8. Schwarze Schläge: a7xb6xc5; fxexdxc1=L In dieser Stellung muß der wLe4 nach f1 zurückgespielt werden. Mit sBh7 kann der wLe4 nur entweder über f7 nach f1 zurück, so daß der sK bereits gezogen haben muß, oder der wLe4 kann über f5 zurück, wozu aber der sTa8 nach h8 zurückgezogen werden müsste, um f6-f5 zurücknehmen zu können.)
Henrik Juel: To make the retroplay plausible one should uncapture bQ early on g3 and retract it to d8. The wB could also get back to f1 via f5, but this would require retracting bRa8 to h8 before retracting bPf5, so castling is still illegal in part a). (2003-04-10)
Gerd Wilts: Hello Henrik, thank you for pointing out the inaccuracy of the solution, I will make the solution more precise soon. And thank you for adding all the other solutions! (2003-04-11)
A.Buchanan: Have posted a solution based on GW&HJ ideas. More often a j’adoube of a rook pawn signals a tempo idea, but not here. The surprising motivation harmonizes with accurate and varied forward play. (2021-10-24)
A.Buchanan: This problem was featured in yesterday's Monthly International Zoom Call, and was a great success, with audible gasps of appreciation, as the significance of h7 was realized :) (2021-10-24)
A.Buchanan: Oh dear, Alfred Pfeiffer has silently reverted the German keyword to "mit Umwandlungsfigur". I'm not going to enter an "edit war" with him, but I would appreciate if he can explain his position here. This is a problematic concept to keyword, but to me "mit Umwandlungfigur" is weak and inaccurate. What is the intended distinction with the existing keywords "Umwandlung" or "Umwandlungen"? It's hopeless. We have in PDB very many poor promotion keywords, and I would like to clean up progressively. I don't know if a native German speaker would care to engage with Alfred on this point. (2021-10-30)
A.Buchanan: To be clearer: to me the German definition seems pretty good. I think the term itself should give more of a clue what's happening :-) (2021-10-30)
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comment
Keywords: Cant Castler, Castling (sg), Promotion (l), Obvious promotion (l), Corridor, Retro Shield
Genre: h#, Retro
Computer test: Forward: C+ Popeye V4.87 Retro: non-trivial reasoning
FEN: r3kN2/bnppp1pp/1p6/2p1P3/4K3/3P2P1/PPP1PP2/N6n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-30 more...
33 - P0001136
Edgar Fielder
The Fairy Chess Review 1941
P0001136
(13+10)
Darf Schwarz rochieren?
Schwarz hat bereits rochiert!
play all play one stop play next play all
See P1398939
Henrik Juel: Black captured b7xLc6 and axbxc; White captured [Lf8], e2xd3, and fxexdxc6xb7-b8=D.
Ke8 to b7, Th8 to b8, Kb7 to g8, Tb8 to f8, Da4 to b8 for unpromotion, -22.c6xDb7 Da8 -23.d5xLc6 Dd8 -24.d4 Lb7 -25.e3xTd4 Lc8 -26.f2xSe3 c6 -27.h5 b7xLc6, Lc6 to f1, f2xSe3 etc.
Black may not castle, because he already did.
It is not possible to avoid the early castling:
-11... Ka7 -12.Db4 Ka6 -13.Db8 Ka7 -14.D=b7 Kb8 -15.c6xDb7 Kc8 -16.d5xLc6 Kd8 -17.d4 Ke8 -18.e3xTd4 Da8 -19.f2xSe3 Dd8 -20.h5 Lb7 -21.h4 Lc8 and two white pawn retractions are missing (2012-07-23)
Yoav Ben-Zvi: The solution does not require a full analysis since if Black King and Rook never moved then Black is almost immediately in retro-stalemate. An alternative stipulation is "First move of Black King?". (2014-06-03)
A.Buchanan: Maybe there is intentional irony? (2014-06-03)
Henrik Juel: Nowadays composers are not afraid to use the real stipulation in this type of resolution retro, like 'Release the position'.
Formerly a formal stipulation was popular, like 'Mate in 1', with the real stipulation understood but not mentioned.
Here, the stipulation question may well be ironic, but you will get no solution points, if you just answer 'No'; you still need to explain how the position arose (2014-06-03)
more ...
comment
Keywords: Castling (sk), Castling Paradox (sk hidden)
Genre: Retro
FEN: 4k2r/2pppppp/7P/2p5/Q1P5/PPRP4/RBpP2P1/N1K5
Reprints: 71 32 personaggi e 1 autore 1955
12 Europe Echecs 12 08/1959
222 FIDE Album 1914-1944/III 1975
341 Eigenartige Schachprobleme , p. 110, 2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-23 more...
34 - P0001191
Leon Loewenton
64 Europe Echecs 38 10/1961
P0001191
(15+14) C+
h#2
b) wSd3 nach e1
a) 1. Kd8 Se5 2. Te8 Sxf7#
b) 1. 0-0 Sd5 2. Sh8 Sxe7#
play all play one stop play next play all
Originalquelle?

nicht sicher, ob Zwilling b) auch schon im Original.
Artikel von Petrovici zum "Thema Than" 03/2016 gibt als Originalquelle an: "Europe Echecs 10/1951"
hans: Counting problem
1. Kd8 Se5 2. Te8 Sxf7# !!
1. 0-0 Sd5 2. Sh8 Sxe7# ??
Black needs a tempo to get on move, and the only piece to do that is Th8, so 0-0 is illegal. (2010-06-20)
Mario Richter: The term usually used to describe this kind of problems is "Parity problem". The bRh8 might have been on h8 all the time, since the tempo move might also have been made by the black Queen, but in this case, the bK must have already moved, thereby losing the right to castle too.. (2010-06-21)
Ladislav Packa: You both are right. Black would be in this position made an uneven number moves.
This can be achieved by using Rook moves (eg Rg8-f8-h8), or by the Queen moves while she was still alive (eg Kf8-g8, then Qf8-e8-d8 and then the King back). But both possibilities preventing black castling. (2013-10-04)
A.Buchanan: Parity change could also be achieved without triangulation by e.g. SxDe8 after the black queen has moved once (2022-03-15)
comment
Keywords: Castling (sk), Parity Argument, Cant Castler, Than theme
Genre: h#, Retro
Computer test: rawbats
FEN: r1b1k2r/1ppppppp/p5n1/8/8/P1NN3P/1PPPPPPR/nRBK1B2
Reprints: 787 Themes-64 10-12/1961
(7) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2018-03-22 more...
35 - P0001198
Tivadar Kardos
1807 Problem 73-78 06/1961
P0001198
(8+15) C+
h#2
1. Dxc2 Tc1 2. 0-0-0? Txc2#
1. Sd7 0-0-0 2. 0-0 Tg1#
play all play one stop play next play all
Black qside castling is lost as wPa promotes but other rights can be maintained.
S: axb,cxb,dxcxb,fxexBd,gxfxe
W: a=X,g=X,fxe
A.Buchanan: wPh can be removed (2021-11-22)
comment
Keywords: Cant Castler, Castling (wgsk)
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 3.54 & trivial retro-logic
FEN: r1q1kn1r/1p2p2p/bp2P3/bp1pB3/1p6/4p3/1PP1P2P/R3K3
Reprints: 71 Europe Echecs 41 01/1962
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-23 more...
36 - P0001263
Luigi Ceriani
135 Europe Echecs 78 06/1965
P0001263
(15+11)
h#2
1. bxa5 Sb6 2. axb4 Ta8#
play all play one stop play next play all
Henrik Juel: analysis
Black captured a3xb2 (not a7xb6, as the SW corner then cannot be released)
The missing white man is [Pe2], which must have promoted on b8
Now the white captures can be seen: axb, exdxcxb, and [Lf8] on f8
The only free white officer is Dd8, so the retroplay must include unpromotion of it on b8
It is impossible to shield Ke8 from check by Db8, so Ke8 has moved and Black may not castle (2022-04-27)
Henrik Juel: solution
1.bxa5 Sb6 2.axb4 Ta8#
not 1.0-0? Lxb6 2.Txa8 Txa8
HC+ Popeye 4.61 (2022-04-27)
Henrik Juel: Andrew, I should have been more precise in recommending your suggestion to introduce the new keyword Organ Pipes
I was talking about the standard meaning of Organ Pipes, which dates back to Sam Loyd, 1859: black LTTL in a problem where each L interferes with each T and vice versa, forming four Grimshaws
You have probably used a pattern search to find all occurrences of black or white LTTL, and you have neglected the interferences
This problem has LTTL in the wrong color, and there are no interferences (2022-04-27)
James Malcom: I agree with Henrik, although White organ pipes still exist, few and far. (2022-04-28)
James Malcom: Until Andrew bumbles back, I've updated the English definition to "A problem *utilizing* the pattern bishop, rook, rook, bishop in a straight line to create multiple Grimshaws." This specifies the purpose and formation of the Organ Pipes. (2022-04-28)
Henrik Juel: Thanks, James (2022-04-28)
A.Buchanan: Hi Guys. I wondered about other uses of organ pipes. In a first pass population of the 500-odd records with the current PDB interface, there’s not much time for thinking. I have multiple tabs open to eliminate the wait time associated with PDB refresh. One could go back and eliminate those which are not grimshaws. Alternatively (and this would be my preference) accept that this is a visual pattern which may occur in non-Grimshaw context. Then use the keyword Grimshaw(4x) to indicate when it’s really Grimshaw. I think it can also occur with white pattern, as a problem by Pal Benko shows. So I blitzed through the d# records, if someone wants to complete the rest that’s good. I think the existence of a few false transient positives is an acceptable price even if we take the narrow definition of Organ Pipes. (2022-04-28)
A.Buchanan: By the way, these days Deepl is good enough to give us decent translation to Feench & German and as a matter of policy whenever I make a change to a definition I try to align the other two. Other users who do not maintain the glossary are encouraged to propose definitions where there is a gap (E.g. Grimshaw). There are a lot of undefined terms, many very recent. (2022-04-28)
A.Buchanan: Hi Henrik feel free to respond to my response to your message, say thanks for the tags I have added, or add tags yourself to complete the work. The ones I’ve added were mostly the d# and it was a deliberate decision to blast through as a first pass and just add them for now anyway, not “neglected”. Now, I really don’t feel like continuing (2022-04-28)
Henrik Juel: Andrew, I believe that most PDB users appreciate your contributions to the site
I certainly do, so please continue your good work (2022-04-28)
comment
Keywords: Castling (sk), Cant Castler, Promotion
Genre: h#, Retro
FEN: N3k2r/2pppppp/1p6/B7/RP5P/RP3P2/BpPP2P1/NK1Q3r
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-04-28 more...
37 - P0001270
Matti Arvo Myllyniemi
142 Europe Echecs 84 01/1966
P0001270
(15+15) C+
BP in 9.5
1. e3 d5 2. Lc4 d4 3. Se2 d3 4. 0-0 dxc2 5. d4 Kd7 6. d5 Kd6 7. Dd4 Sd7 8. Ld2 c1=L 9. Lb4+ c5 10. dxc6ep+
play all play one stop play next play all
more ...
comment
Keywords: Unique Proof Game, En passant, Non-standard material (l), Castling (wk), Promotion (l), Valladao Task (sww)
Genre: Retro
Computer test: Ergänzung Stelvio 1.2 C+: Keine Lösung: BP 8.5, 9.0. 10. dxc6ep++ Doppelschach.
FEN: r1bq1bnr/pp1npppp/2Pk4/8/1BBQ4/4P3/PP2NPPP/RNb2RK1
Reprints: 92 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-25 more...
38 - P0001273
Luigi Ceriani
145 Europe Echecs 84 01/1965
P0001273
(6+7)
h#1 (wer? wie?)
Die F.P.I. wurde zweimal erreicht: das erstemal nach einer geraden, das zweitemal nach einer ungeraden Anzahl von Zügen.
Vertikaler Zylinder
Henrik Juel: The FPI (the initial array, but with Black to move) may be reached, e.g., by playing the white knights out, playing Ta1 to b1 and Th1 to g1, then correct the tempo by playing Tb1-a1-h1-b1, and finally moving rooks and knights back into the initial array.
The stipulation condition implies that all four castlings are illegal. Solution: 1.Txa7 Txh8#, not 1.0-0-0? Tc1 nor 0... 0-0-0? 1.Tc8. I do not see any tries involving king-side castling, although they would be nice to have also. (2010-12-08)
A.Buchanan: I am evidently being slow again: please explain. "The FPI was achieved twice: the first time after an even number of moves, the second time after an odd number of moves." To achieve FPI requires an odd number of single moves. Should the stip read "even number of *White* moves"? And then what does it mean to achieve FPI for the *second* time: does that mean game array with White to move again? I agree that the bottom line is to have eliminated all castling rights. (2020-12-25)
Henrik Juel: In P0001272 the FPI was reached once, implying that either White or Black has lost castling rights
I also do not understand this problem (2020-12-26)
Hans-Jürgen Manthey: da der lezte Zug b5x~a6 bzw b4x~a3 gewesen sein kann, sind alle Rochaden, sowie die Farbwahl zulässig.
Oder kann mir mal einer erklären was zum Teufel F.P.I. auf Deutsch heißt ?? (2020-12-26)
A.Buchanan: Siehe die Beschreibung des Schlüsselworts. Aber hier macht es nicht viel Sinn. Ich veröffentliche in ein paar Tagen einen Artikel in Problemas, der besser erklärt und einige neue Beispiele enthält. (2020-12-27)
Hans-Jürgen Manthey: Habe mich über F.P.I. schlau gemacht und FPI, Vertikale Zylinder sowie viele weitere Märchenschach-Bedingungen sind alsolut nicht mein Fall... (2020-12-28)
A.Buchanan: @H-JM: Ich bin kein großer Fan von Feenschach, aber es hat seinen Platz in der Welt der Komposition. Es vervielfacht den Gestaltungsraum für Schachkompositionen erheblich, und viele coole künstlerische Effekte können nur mit Feenschach erzielt werden. FPI ist bislang kein gutes Beispiel: Aber die zugrunde liegende Idee ist gut und reif für eine breitere Nutzung. (2020-12-28)
Henrik Juel: My goodness, Andrew, you are approaching Mario's command of all languages...
I believe, however, that Feenschach is an old-fashioned name for Märchenschach
It came about as a literal translation of Fairy Chess, but now lives on only in the magazine name feenschach
We did almost the same in Denmark: feskak in the 1930s, but now fantasiskak (2020-12-28)
A.Buchanan: Hi Henrik: thanks for your kind feedback. I try to draft in German, and then back translate in google translate to English to see what it thinks, and then go forwards again to correct dumb errors and give me better word choices. But there's still errors, which I am always keen to hear about, so thanks for reminding me about Märchenschach. I now understand that Märchen are "fairy tales" not "fairies" and that the term is used in English by folklorists. Encyclopedia Britannica characterizes them by: "Their usual theme is the triumph over difficulty, with or without supernatural aid, of the one least likely to succeed." I like the detail at the end, which suggests that the key move should be an unlikely one! (2020-12-29)
Henrik Juel: I had three years of german in middle school, but never used the language actively
I believe that most german problemists can read english as well as we can read german, so I do not make the extra effort to communicate in german; besides remembering to capitalize nouns, I would have to check masculinum, femininum, or neutrum (gender?), and check which prepositions entail accusative, dative, or both (kasus?) (2020-12-29)
Olaf Jenkner: (:-) (2020-12-29)
more ...
comment
Keywords: Cant Castler (wbsb), Fake game array, Castling (wbsb), Constrained problem, Vertical Cylinder (Vertikaler)
Genre: Retro, Fairies
FEN: r3k2r/P3p3/P3P3/8/8/p3p3/p7/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2020-12-26 more...
39 - P0001285
Jean Oudot
370 Die Schwalbe 07/1960
R. Bédoni gewidmet
P0001285
(11+11) C+
h#2*
* 1. ... dxc5 2. Dxh3 0-0-0#
1. Db2 Le2+ 2. Kc2 Ld1#
play all play one stop play next play all
Henrik Juel: the five missing black men were captured by white pawns (exfxgxh, fxgxj), so with Black to move last move was with Ta1 or Ke1, and White may not castle
C+ Popeye 4.61 (2022-11-26)
A.Buchanan: Pleasant White tempo play in both phases (2022-11-27)
comment
Keywords: Cant Castler (wl), Castling (wl)
Genre: Retro, h#
Computer test: HC+ Popeye 4.61 with simple retro logic
FEN: 8/7p/7P/2pr2pP/2bP2Pb/2pk1BRP/6qN/R3KNrn
Reprints: 553 FIDE Album 1959-1961 1966
157 Europe Echecs 90 07/1966
(10) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-27 more...
40 - P0001310
Dragan Petrovic
182 Europe Echecs 107 01/1968
4. Preis
P0001310
(13+14) C+
#2 (AP)
1. gxf6ep! droht 2. 0-0#! (2.Kf2#?)
R: 1. f7-f5 f6xDe7,f6xTe7 etc
play all play one stop play next play all
Wh caps: fxe7, hxg5
Bl caps: axbxc2, wBf1
If wK castling right remains, the only way to give White a prior move is to retract f7-f5, so ep would be on.
The key threatens (inter alia) 2. 0-0#/Kf2#, and both remain whatever Black plays, so the castling retro-justification for ep can never be disrupted.
R: 1. S~a3? a3-a4? blocks wT from retracting to home square if wK castling right remains.
Dragan Petrovic: Author is Dragan T. Petrovic (2007-12-02)
more ...
comment
Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic)
Genre: Retro, 2#
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 4r3/1p2P1n1/4p2B/b1pp1pPb/P5pp/nPP5/Q1pPP1P1/N1k1K2R
Reprints: (50) Problem 144-147 12/1971
(38) Mat Plus 12 10-12/2009
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
41 - P0001349
Jean Oudot
Echiquier de France 1957
P0001349
(14+7)
#2
Rosalie Fay: White has lost only the bishops. So the pawn on c5 is not [Pa7] (because that entails 2 White units captured on black squares). White has played axbxcxdxe7, dxe, fxe, hxg, gxfxe. Black has 7 units, so white pawns have captured all missing Black units, but none on the a or h files.

Black has 2 pawns on the c-file, so one has captured. Thus [bPa7] and [bPh7] pawns have collectively captured no more than once. So at least one of them must have promoted, in order to either get to a file where White made a capture, or replace a captured unit; it didn't capture en route to promotion, so it displaced a white rook and thus spoilt one White castling right.

White would mate by 1 Rd1 & 2 Rd6 or 1 Rf1 & 2 Rf6, except that Black threatens Qxe2+. So either 1 0-0 or 1 0-0-0, though it's impossible to say which is legal. (2022-11-24)
Henrik Juel: one solution, but in two parts
if Ta1 has moved, 1.0-0 thr. 2.Dc8,Tf6#
if Th1 has moved, 1.0-0-0 thr. 2.Dg8,Td6# (2022-11-25)
Hans-Jürgen Manthey: nach der möglichen Zugfolge: 1. Sb1-c3 c7-c6 2. Sc3-d5 Dd8-b6 3. Sg1-f3 Db6-b3 4. a2xDb3 a7-a5 5. Sd5-b4 a5-a4 6. Sb4-a2 e7-e5 7. Sf3-h4 Lf8-c5 8. Sh4-g6 f7-f5 9. Sg6-f4 g7-g5 10. Sf4-g6 Lc5-e3 11. d2xLe3 f5-f4 12. g2-g3 Ta8-a5 13. g3xf4 g5-g4 14. f4xe5 g4-g3 15. h2xg3 h7-h5 16. Lf1-g2 h5-h4 17. Lc1-d2 h4-h3 18. Ld2-b4 Th8-h4 19. c2-c3 Th4-c4 20. Sa2-c1 a4-a3 21. Lb4-c5 a3-a2 22. Dd1-d4 Sb8-a6 23. Dd4-h4 Sa6-c7 24. Dh4-d8+ Ke8-f7 25. b3xTc4 Sc7-d5 26. c4xSd5 Sg8-e7 27. d5-d6 Ta5-b5 28. d6xSe7 d7-d6 --- folgt nun
29. Lg2-e4 Lc8-e6 30. Le4-b1 a2xLb1D 31. Th1-g1 Db1-d3 32. Sc1-b3 Le6-c4 33. Th1-g1 h3-h2 34. Dd8-e8+ Kf7-e6 35. Tg1-h1 Tb5-b6 36. Th1-g1 Le6-c4 37. Tg1-h1 h3-h2 38. Th1-g1 h2-h1D 39. Sc1-b3 Dh1-e4 40. f2-f3 Lc4-a6 41. f3xDe4 d6xLc5 42. Dd8-e8+ Kf7-e6 43. Tg1-h1 Dd3-b5 matt in 2:
1. OOO droht 2. De8-g8/Td1-d6# - 1. ... Db5-d3 2. Sb3xc5# oder:
29. Sc1-b3 h3xLg2 30. Sb3-d2 g2-g1D+ 31. Sd2-f1 Dg1-g2 32. Sf1-d2 Dg2-e4 33. f2-f3 Tb5-b6 34. f3xDe4 d6xLc5 35. Sd2-b3 Lc8-e6 36. Ta1-b1 Le6-c4 37. Ta1-b1 Kf7-e6 38. Tb1-a1 Lc4-a6 39. Ta1-b1 a2-a1D 40. Dd8-e8 Da1-a4 41. Tb1-a1 Da4-b5 matt in 2: 1. OO bel. 2. De8-c8/Tf1-f6# (2023-02-22)
more ...
comment
Keywords: Partial Retro Analysis (PRA), Castling (wb)
Genre: Retro, 2#
FEN: 4Q3/1p2P3/brp1k1N1/1qp1P3/4P3/1NP1P1P1/1P2P3/R3K2R
Reprints: 223 Europe Echecs 130 09/1969
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-26 more...
42 - P0001420
Branko Koludrovic
294 Europe Echecs 197 04/1975
1. ehrende Erwähnung
P0001420
(13+12) C+
h#3*
1. ... cxd6ep 2. 0-0-0 0-0-0 3. Kd7 Sa7#
1. De7 c6 2. Th8 c7 3. Tf8 Sd6#
play all play one stop play next play all
White pawn caps: axb,dxe,gxf,hxg.
Black: fxg,bxc,cxb.
wPb4 came from b3 to release wBa3, so bPb3 captured to reach that square.
All pcs accounted for means bPd never captured.
In the set play, there are 13 retro tries in which one or both players do not castle. The intention is that both castling rights are needed in order to imply the pawn double hop.
A.Buchanan: White pawn caps: axb,dxe,gxf,hxg definite.
Black: fxg and two to resolve c-file. But that may be c&d cross-capture, so in set play last move might have been c6xd5. So I think this problem is cooked. What am I missing? (2022-03-21)
Mario Richter: If Black's last move was c6xd5, how do you get white Bishop a3 out of his cage? (In that case, black pawn b7 never left the b-file). (2022-03-21)
A.Buchanan: I agree Mario thanks (2022-03-21)
Hans-Jürgen Manthey: beide Rochaden und im Satz ep. ist möglich:
R.: 1. ... d7-d5 2. Sa7-b5 b5xTc4 3. Tg4-c4 Dd8-h4 4. c4-c5 Th8-h3 5. c2-c4 c4xDb3 6. Dd3-b3 h3-h2 7. Dd1-d3 h4-h3 8. Sc8-a7 h5-h4 9. Th4-g4 c5-c4 10. Th1-h4 h7-h5 11. h2xSg3 Sf5-g3 12. b3-b4 Sh6-f5 13. Lb4-a3 c7-c5 14. Ld2-b4 g3-g2 15. Lc1-d2 a3-a2 16. a2xSb3 Sc5-b3 17. Lb3-a4 Sa6-c5 18. Lc4-b3 Sb8-a6 19. d2xLe3 Lc5-e3 20. Ld3-c4 Lf8-c5 21. Lf5-d3 a4-a3 22. Lh3-f5 a5-a4 23. Lf1-h3 g4-g3 24. g2xLf3 Lb7-f3 25. Sb6-c8 Lc8-b7 26. Sa4-b6 a7-a5 27. Sc3-a4 b7-b5 28. Sb1-c3 f5xSg4 29. Se5-g4 Sg8-h6 30. Sf3-e5 e7-e6 31. Sg5-f3 f6-f5 32. Sh3-g5 f7-f6 33. Sg1-h3 (2023-02-23)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wgsg)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro-logic.
FEN: r3k3/6p1/4p3/1NPp4/BPp4q/Bp2PPPr/pP2PPpp/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
43 - P0001450
Henri Nouguier
324 Europe Echecs 223 07/1977
P0001450
(13+3)
h#1
hans: 1. Th2xf2 De1xf2#!
1. Th2xh3 0-0#? (Castling illegal)

R: -1. Kf1xSe1 Sg2-e1 -2. Ke1-f1 Sf4-g2+ -3. Tg1-h1 Sd5-f4 -4. Tf1-g1 Se3-d5 -5. Tg1-f1 Sf1-e3 -6. Tg1-h1 Tg2-h2 -7. Th2-h1-g1 Tg1-g2 -8. Th1-h2 g2-g1=T -9 h2-h3 h3xSg2 (2010-06-26)
HENRI: Hans solution has a tipo. Their is no D (=Queen in german) mating. One should read the solution in german : 1. Th2xf2 Ke1xf2#! 1.Th2xh3 0-0#?
or in english : 1. Rh2xf2 Ke1xf2#! 1.Rh2xh3 0-0#? (Castling illegal) (2023-10-25)
comment
Keywords: Cant Castler, Castling (wk)
Genre: h#, Retro
FEN: 8/8/8/8/8/PPP3PP/B1rPPP1r/BNk1K2R
Input: Gerd Wilts, 1995-06-03
44 - P0001453
Luis Alberto Garaza
327v Europe Echecs 225/226 10/1977
P0001453
(11+14) C+
h#1
1. hxg3ep! 0-0#! (Tf1#?)
play all play one stop play next play all
Hans-Jürgen Manthey: 1. hxg3 O-O#(auch Tf1# !) mögliche zugfolge:
1. d2-d4 f7-f5 2. Lc1-d2 f5-f4 3. Ld2-c1 e7-e6 4. Lc1-d2 c7-c5 5. Ld2-c1 b7-b5
6. Lc1-d2 a7-a6 7. Ld2-c1 Lc8-b7 8. c2-c4 Ke8-f7 9. Lc1xf4 Sg8-f6 10. Lf4-h6 g7xh6
11. Dd1-d3 h6-h5 12. Dd3-d1 Lf8-h6 13. h2-h4 Lh6-g5 14. h4xLg5 h5-h4 15. Dd1-d3 h7-h5
16. Dd3-d1 b5xc4 17. a2-a4 Lb7-e4 18. d4-d5 e6xd5 19. Ta1-a3 Le4-c2 20. f2-f4 Th8-e8
21. Ta3-g3 Te8-e3 22. f4-f5 Sf6-e4 23. f5-f6 Kf7-e6 24. f6-f7 Se4-d2 25. Sg1-h3 Sd2xf1
26. Sh3-f2 Sf1-d2 27. Sf2-g4 Sd2-e4 28. Dd1-d3 Lc2-d1 29. Sg4-f6 c4-c3 30. Sf6-h7 c3-c2
31. b2-b4 c5-c4 32. b4-b5 Sb8-c6 33. b5-b6 Sc6-a5 34. b6-b7 c2-c1L 35. g5-g6 c4-c3
36. Sb1-a3 c3-c2 37. Sa3-b5 Ta8-c8 38. b7-b8L Tc8-c3 39. Lb8-d6 La5-c4 40. a4-a5 a6xSb5
41. a5-a6 Ke6-f5 42. a6-a7 Dd8-e8 43. a7-a8S De8-e5 44. Sa8-c7 Lc1-b2 45. Sc7-e6 Lb2-a1
46. Se6-f8 Kf5-f4 47. Tg3-g5 Se4-f6 48.Ld6-a3 Te3-h3 49. La3-b2 Sc4-a3 50. g2-g4 (2020-12-11)
Henrik Juel: The white castling is needed to prove (a posteriori) that last move was g2-g4, legitimizing the ep capture (2020-12-31)
more ...
comment
Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 5N2/3p1P1N/5nP1/1p1pq1Rp/5kPp/n1rQ3r/1Bp1P3/b2bK2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
45 - P0001468
Oskar E. Vinje
The Fairy Chess Review 1938
P0001468
(10+3)
Letzter Zug?
R: 1. 0-0-0
play all play one stop play next play all
Deemed stipulation: "Erster Zug des wTd1?"
Henrik Juel: White pawns captured all 13 missing black men
Retracting the castling is the only way to give Black a retraction, e.g. Kc2-b3 (2020-12-01)
comment
Keywords: Type A, Last Move? (0-0-0), Castling (wl), Economy record (Last Move? Type A), First Move? (T0), Economy record (First move)
Genre: Retro
FEN: 8/P1p5/PN6/1P6/P1N5/Pk6/pP6/2KR4
Reprints: 342 Europe Echecs 241 01/1979
1.57A Eigenartige Schachprobleme , p. 194, 2010
1 Die Schwalbe 360-1, p. 737, 12/2020
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-28 more...
46 - P0001550
Michel Caillaud
421 Europe Echecs 295 07/1983
A. Hazebrouck gewidmet
1. Preis
P0001550
(13+12) C+
BP in 35.5
1. h4 c5 2. h5 c4 3. Th4 c3 4. Tc4 b5 5. g4 b4 6. Lg2 b3 7. Lc6 bxa2 8. b4 a5 9. b5 a4 10. b6 a3 11. La4 Sc6 12. b7 d5 13. b8=D d4 14. Dd6 d3 15. Dg6 dxc2 16. d3 fxg6 17. Ld2 c1=L 18. Db3 c2 19. La5 Lh6 20. Sd2 c1=L 21. Sf1 Lcg5 22. f4 Kf7 23. 0-0-0 a1=L 24. fxg5 a2 25. gxh6 Kf6 26. Sh2 Kg5 27. Tf1 Lf6 28. Tff4 a1=L 29. Kb1 Lae5 30. d4 Lb7 31. dxe5 Dd2 32. exf6 Te8 33. f7 Sf6 34. Dd3 Sd7 35. Ld1 Sdb8 36. Sgf3+
play all play one stop play next play all
5 Frolkin-Ceriani-Umwandlungen: 4 schwarze Läufer und 1 weiße Dame! Eine der bahnbrechenden frühen KBPs.
Silvio Baier: Der wesentliche thematische Inhalt ist bereits nach 31,5 Zügen erreicht. Bis dahin ist es C+ (Euclide 0.98). (2010-08-04)
James Malcom: Is this fully C+ then? (2021-01-27)
Henrik Juel: No, only the first 31.5 moves are tested OK (2021-01-27)
James Malcom: No Henrik, as in is the entire problem testable. (2021-01-27)
Henrik Juel: I guess that testing the entire problem would take an unreasonably long time (2021-04-06)
A.Buchanan: It might be possible these days: the motivation for stopping at 31.5 is that the promotion theme had been demonstrated by then. But there’s still e.g. bSb8 as random impostor (2021-04-07)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 50:24:34 Stunden.(hh:mm:ss)
Keine Lösung: BP 34.5, BP 35.0. (2023-05-08)
Henrik Juel: Thanks for your patience, Moldenhauer; more than two days... (2023-05-08)
comment
Keywords: Ceriani-Frolkin Theme (llllD), Unique Proof Game, Castling, Promotion (llllD), Impostor (s)
Genre: Retro
Computer test: Computerprüfung: C+ Stelvio 1.11 50:24:34 Stunden.(hh:mm:ss) Keine Lösung: BP 34.5, BP 35.0.
FEN: 1n2rb1r/1b2pPpp/2n3pP/B5kP/2R2RP1/3Q1N2/3qP2N/1K1B4
Reprints: 25 Shortest Proof Games 11/1991
(1) diagrammes 103 10-12/1992
(A) Quartz 22 10-12/2002
(8-a) Die Schwalbe 250A 08/2011
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-04-07 more...
47 - P0001605
Dmitri W. Pronkin
Andrey Frolkin

475 Europe Echecs 337 01/1987
P0001605
(1+16) C+
BP in 17.0
1. d3 h6 2. Lg5 hxg5 3. h4 Txh4 4. a4 Txa4 5. Th6 Sxh6 6. g4 Sxg4 7. Lg2 Sxf2 8. Lc6 Sxc6 9. d4 Sxd4 10. c4 Sxe2 11. c5 Sxg1 12. c6 dxc6 13. Dd4 Dxd4 14. Ta3 Dxb2 15. Th3 Lxh3 16. Sc3 0-0-0 17. Sd5 Txd5= patt
play all play one stop play next play all
Henrik Juel: Black makes 15 captures.
The first 13.0 moves are correct by Euclide 1.0, but it takes some 4 minutes, so a complete test would probably take more than a day (2014-12-09)
paul: Jacobi checked the last 16 moves (in about 20 min). (2018-05-10)
Mario Richter: I would like to label this problem as a "Massacre PG", but the current definition of that term only knows of two-sided MPGs. Perhaps we could make the same differentiation as for "Homebase"?!
Btw., wasn't "popeye" best for such massacres?
rawbats confirms complete correctness of the problem after approx. 25 minutes. (2018-05-11)
Henrik Juel: Yes, Mario, Popeye is fairly good at this type
C+ by Popeye 4.61 in 57 minutes (2018-05-11)
A.Buchanan: @Mario: yes I agree it would be good to make massacre more specific. There are 69 massacres currently, almost all 2-sided, but maybe also some series-movers (unique or non-unique) which are currently not counted? (2018-05-12)
more ...
comment
Keywords: Unique Proof Game, Castling, Homebase (w), Rex solus (w), Massacre PG
Genre: Retro
Computer test: C+ rawbats. C+ by Popeye 4.61 in 57 minutes. C+ Stelvio 1.2 2 Sekunden. Keine Lösung: BP 16.0, BP 16.5.
FEN: 2k2b2/ppp1ppp1/2p5/3r2p1/r7/7b/1q3n2/4K1n1
Reprints: 116 Shortest Proof Games 11/1991
20a 64 Proof Games 2012
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-21 more...
48 - P0001622
Christian Poisson
491b Europe Echecs 355/356 07-08/1988
P0001622
(3+11)
shc#11
1. Lf1 2. Kh3 3. Kh4 4. Lh3 5. fxg3ep 6. Kh5 7. Kg6 8. Kf7 9. Ke8 10. 0-0-0 11. Td7 a8=D#
play all play one stop play next play all
The cluster of tries: Kh2-g1-f1-e1-d2-...-e8 0-0-0 Td7 is 12 moves: just too long.
In the diagram, e.p. is not permitted because last move might be g3-g4. So first shift sK to h4. Following the e.p., check that wK does now have prior move: Kf4-f3. Now sK scampers to e8 where through consequent forgetfulness it recovers castling rights.
more ...
comment
Keywords: En passant, Castling (sg), Seriesmover, Consequent, Valladao Task, Promotion in the mating move (D), Switchback (l), Promotion (D), Königswanderung
Genre: Retro, Fairies
FEN: r7/P1p5/pp6/5pp1/5pP1/5K1b/4rp1k/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
49 - P0001637
Andrej N. Kornilow
Andrey Frolkin

506v Europe Echecs 364 04/1989
P0001637
(16+10) C+
BP in 13,5
2 solutions
1. Sc3 c5 2. Sd5 c4 3. Sxe7 c3 4. Sxg8 Dh4 5. Sh6 g5 6. Sxf7 Lg7 7. Se5 0-0 8. Sxd7 Lh8 9. Sxb8 Ld7 10. Sc6 Tc8 11. Sxa7 La4 12. Sb5 Tc6 13. Sa3 b5 14. Sb1
und
1. Sf3 c5 2. Se5 c4 3. Sxd7 c3 4. Sxb8 Ld7 5. Sc6 Tc8 6. Sxa7 La4 7. Sc6 b5 8. Sxe7 Tc6 9. Sxg8 Dh4 10. Sh6 g5 11. Sxf7 Lg7 12. Se5 0-0 13. Sf3 Lh8 14. Sg1
play all play one stop play next play all
Moldenhauer: Ergänzung: Stelvio 1.2. Keine Lösungen BP 12.5, BP 13.0.
Es ist wirklich eine Rarität dass es 2 komplette Lösungen gibt. (2023-05-20)
comment
Keywords: Unique Proof Game, Homebase (W), Initial Game Array (W), Castling
Genre: Retro
Computer test: Natch 2.2 Copyright (C) 1997,98,99,2001,2002,2003 Pascal Wassong Ergänzung: Stelvio 1.2. Keine Lösungen BP 12.5, BP 13.0.
FEN: 5rkb/7p/2r5/1p4p1/b6q/2p5/PPPPPPPP/RNBQKBNR
Reprints: 569 Ukrainisches Album 1986-1990
140 Shortest Proof Games 11/1991
(10) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-21 more...
50 - P0001740
Vladimir Korolkov
6545 FEENSCHACH 11-12/1963
P0001740
(3+6)
#1 vor 2
VRZ, Typ Hoeg
R: 1. Kg1xSh1 Sf2-h1+ 2. 0-0, dann 1. Th8#
R: 1. Kg1-h1 Lb8xSa7+ 2. Te7-f7, dann 1. Sc6#
R: 1. Kg1-h1 Sb6xDc8,Sb6xDd5,Se3xDd5+ 2. Dc5-c8,Dc5-d5,Dc5-d5+, dann 1. Df8#
play all play one stop play next play all
Henrik Juel: This problem demonstrates an advantage of type Høeg over type Proca:
another way of generating variations
2.0-0 cannot be an uncapture, of course
2.Te7-f7 and 2.Dc5-c8 etc. could be uncaptures, but no matter what Black supplements, he is mated (2023-04-08)
A.Buchanan: Thanks Henrik: is an alternative for White R: 1. Kg1-h1 Lb8xDa7+ 2. Dc5-a7/Da3-a7, dann 1. Df8# (2023-04-08)
Henrik Juel: No Andrew, when White moves his uncaptured queen back to c5 or a3, Black supplements a queen or bishop on a7, preventing the mate on f8 (2023-04-08)
A.Buchanan: Thanks! (2023-04-08)
more ...
comment
Keywords: Defensive Retractor, Type Høeg, Castling (wk)
Genre: Retro
FEN: 2nk4/b4Rp1/8/3n1q2/8/8/8/5R1K
Reprints: 729 FIDE Album 1962-1964 1968
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-07 more...
51 - 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...
52 - P0001765
Kostas Prentos
26 Phénix 1 05/1988
P0001765
(11+15) cooked
BP in 18,5
1. Sf3 f5 2. Sd4 Kf7 3. Sb5 Kg6 4. Sd6 exd6 5. e4 Le7 6. e5 Lh4 7. e6 Df6 8. e7 Dxb2 9. e8=S Se7 10. Lc4 Tf8 11. 0-0 Tf6 12. f3 Kh5 13. Te1 Th6 14. Lf7+ Sg6 15. Te7 Dxc1 16. De1 Dxd2 17. De6 dxe6 18. Td7 Dg5 19. Sxg7#
play all play one stop play next play all
Cook: 1. d4 g5 2. Sf3 Lg7 3. e4 Lxd4 4. e5 Lxb2 5. Dd6 exd6 6. e6 f5 7. e7 Kf7 8. e8=S Se7 9. Sxg5+ Kg6 10. Lc4 Kh5 11. 0-0 Tg8 12. Te1 Tg6 13. f3 Th6 14. Lf7+ Sg6 15. Te7 Lf6 16. Se6 dxe6 17. Td7 Lh4 18. Lg5 Dxg5 19. Sg7#
Kostas Prentos: A correction was published in Phenix, 2009 (No.186/Pg.7979)
Solution:
1. Sf3 f5 2. Sd4 Kf7 3. Sb5 Kg6 4. Sd6 exd6 5. e4 Le7 6. e5 Lh4 7. e6 Dg5 8. e7 Dg3 9. e8=T Se7 10. Lc4 Tf8 11. 0-0 Tf6 12. f3 Kh5 13. Te1 Th6 14. Lf7+ Sg6 15. T1e6 dxe6 16. Txc8 Sd7 17. Th8 Txh8 18. Le8 (2022-12-08)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.2 00:33:56 Minuten. (hh:mm:ss)
Keine Lösung: BP 17.5, BP 18.0.
Beispiel: 1.Sf3 f5 2.d4 Kf7 3.e4 g5 4.Lc4+ Kg6 5.e5 Lg7 6.e6 Lxd4
7.0–0 Lxb2 8.Dd6 exd6 9.e7 Kh5 10.e8S Se7 11.Te1 Tf8 12.Sxg5 Tf6
13.f3 Th6 14.Lf7+ Sg6 15.Te7 Lf6 16.Se6 dxe6 17.Td7 Lh4 18.Lg5 Dxg5
19.Sg7# (2023-05-30)
comment
Keywords: Unique Proof Game, Castling, Promotion
Genre: Retro
FEN: rnb5/pppR1BNp/3pp1nr/5pqk/7b/5P2/P1P3PP/RN4K1
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2006-10-15 more...
53 - P0001859
William A. Langstaff
The Chess Amateur 1922
P0001859
(5+3) C+
#2
If Black can castle, e.p. is ok:
1. hxg6ep! 0-0 2. h7#
1. ... Tf8,Kf8 2. Td8#
Otherwise Black can't castle:
1. Ke6! ... 2. Td8#
play all play one stop play next play all
more ...
comment
Keywords: Cant Castler, Partial Retro Analysis (PRA), Castling (sk), En passant as key
Genre: Retro, 2#
Computer test: HC+ Popeye 4.61 & trivial retro-logic
FEN: 4k2r/8/5B1P/3R1KpP/8/8/8/8
Reprints: (D17) feenschach 27 04/1975
RA64 diagrammes 27 05-06/1977
(1) Die Schwalbe 86 04/1984
Razem 34 23/08/1987
(I) diagrammes 15 07-09/1994
(10) Die Schwalbe 241 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-30 more...
54 - P0001870
Dmitri W. Pronkin
Rex Multiplex 1985
3. ehrende Erwähnung
P0001870
(14+10)
BP in 29,5
1. Sf3 e5 2. Sd4 Dg5 3. Sc6 De3 4. fxe3 h5 5. Kf2 h4 6. De1 h3 7. Kg3 Th4 8. Df2 Ta4 9. Df4 hxg2 10. h4 dxc6 11. h5 Sd7 12. h6 Sb6 13. h7 Ld7 14. h8=T 0-0-0 15. T8h6 Le8 16. Tf6 Tdd4 17. exd4 Lc5 18. dxc5 g1=L 19. cxb6 Lc5 20. bxa7 La3 21. bxa3 g5 22. Lb2 g4 23. Kh4 g3 24. Ld4 g2 25. Sc3 g1=S 26. Tb1 Sh3 27. Tb6 Sf2 28. Ta6 Sd1 29. Lf2 b6 30. a8=D+
play all play one stop play next play all
Moldenhauer: Computerprüfung: C+ Stelvio 2.0 02:29:09 Stunden. (hh:mm:ss) (2023-12-23)
comment
Keywords: Ceriani-Frolkin Theme (Tls), Unique Proof Game, Non-standard material, Castling, Allumwandlung (DTls)
Genre: Retro
FEN: Q1k1b1n1/2p2p2/Rpp2R2/4p3/r4Q1K/P1N5/P1PPPB2/3n1B1R
Reprints: 136 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Kevin Begley, 2011-05-18 more...
55 - P0001885
Thomas R. Dawson
Hampstead and Highgate Express 1912
P0001885
(12+14)
BP in 34,0
1. a4 c5 2. a5 Db6 3. Ta4 Sc6 4. Td4 Tb8 5. b4 cxd4 6. b5 e5 7. d3 La3 8. Le3 Lc1 9. bxc6 Db2 10. c7 b5 11. a6 Tb7 12. axb7 h5 13. b8=L Lb7 14. c8=D+ Ke7 15. Sf3 d5 16. Dh3 Th6 17. g4 Ta6 18. Dg3 h4 19. c4 hxg3 20. h4 Ta1 21. h5 Kf6 22. Sh2 dxe3 23. h6 Kg5 24. h7 Kh4 25. c5 a5 26. c6 a4 27. c7 b4 28. c8=T b3 29. h8=S Se7 30. Sg6 fxg6 31. Lh3 Sc6 32. 0-0 Sb4 33. Ld6 Sc2 34. Td8 Se1
play all play one stop play next play all
"Promenades to Power"

Es läßt sich beweisen, daß die UWs in D,T,L,S zwingend erfolgt sind (a2-b8=L,h2-b8=S und entweder b2-c8=D/c2-c8=T oder b2-c8=T/c2-c8=D).
Erich Bartel: weitere Nachdrucke:
3) 160 Die Allumwandlung im Problemschach VIII 1966a---
4) Schach ohne Grenzen 1969.-- (2007-01-09)
Sally: Vier erzwungene Umwandlungen. (2012-02-21)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 da NUPG sonst cooked in 00:03:51 Minuten. (hh:mm:ss)
Keine Lösung: BP 33.0, BP 33.5.
Beispiel: 1.Sf3 Sf6 2.a4 Sd5 3.a5 Sf4 4.Ta4 c5 5.Td4 Db6 6.b4 Sc6 7.b5 Tb8 8.bxc6 cxd4 9.c7 e5 10.d3 La3 11.Le3 Lc1 12.c4 Db2 13.c5 b5 14.c6 Tb6 15.axb6 Ke7 16.b7 Kf6 17.b8L Lb7 18.c8D b4 19.c7 d5 20.Dh3 b3 21.c8T h5 22.g4 h4 23.Dg3 hxg3 24.h4 Th6 25.Lh3 dxe3 26.0–0 Sg2 27.Sh2 Se1 28.h5 Kg5 29.Td8 Ta6 30.Ld6 Kh4 31.h6 Ta1 32.h7 a5 33.h8S a4 34.Sg6+ fxg6
Umwandlungen: 17.b8L, 18.c8D, 21.c8T, 33.h8S. (2023-04-19)
comment
Keywords: Allumwandlung, Castling (wk), Non-Unique Proof Game
Genre: Retro
FEN: 3R4/1b4p1/3B2p1/3pp3/p5Pk/1p1Pp1pB/1q2PP1N/rNbQnRK1
Reprints: 53 Caissa's Wild Roses 1935
Chess unlimited 1969
150 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Gunter Jordan, 2020-07-01 more...
56 - P0001938
Hendrik Hermanus Kamstra
Tijdschrift van den NSB 05/1929
P0001938
(4+3) C+
#2
b) Bh7 nach h6
a) 1. Ta1! 0-0? illegal 1. ... hxg6,~ 2. Ta8#
1. Tff7? hxg6!
b) 1. Tff7! h5,~ Tb8#
1. Ta1? 0-0!
play all play one stop play next play all
In a), Schwarz muss zuletzt mit K oder T gezogen haben, deshalb s0-0 illegal.
Wikipedianer: Quelle: Tijdschrift van den NSB 05/1929, S. 162, Aufgaben Nr. 5220 (a) und 5221 (b). Teil eines Artikels "Rochade in Retros" (2021-05-02)
more ...
comment
Keywords: Cant Castler, Castling (sk), Miniature, Homebase
Genre: Retro, 2#
FEN: 4k2r/1R5p/6P1/8/8/8/8/5RK1
Reprints: Razem 34 23/08/1987
Outrageous Chess Problems 2005
Input: Gerd Wilts, 1995-06-03
Last update: Marcin Banaszek, 2021-11-29 more...
57 - P0001940
Nikita M. Plaksin
Shakhmaty v SSSR 1980
Spezialpreis
P0001940
(13+16)
Remis
hans: 1. b4 a5 2. Bb2 Na6 3. Bd4 Nc5 4. Nc3 Na4 5. Bc5 Nb6 6. b5 Nh6 7. a4 Nf5 8.
Na2 Nh4 9. Nb4 axb4 10. a5 Rb8 11. a6 Na8 12. a7 Nb6 13. a8=Q Rg8 14. Qa2
Ra8 15. Qe6 Ra3 16. Qg6 hxg6 17. Qb1 Na8 18. Ba7 b6 19. Qb3 Bb7 20. Qd3 Bf3
21. Qf5 gxf5 22. Nh3 Bh5 23. Nf4 Bg6 24. e3 Rh8 25. Bc4 Rh6 26. f3 Bh7 27.
Ng6 Bg8 28. Nh8 Rd6 29. h3 g6 30. Rf1 Bh6 31. Rf2 Bf4 32. Rf1 Bh2 33. f4
Rdd3 34. Rf3 Rac3 35. Rg3 Kf8 36. Rg5 Kg7 37. Rh5 Kf6 38. Rh7 Qc8 39. Rg7
Bh7 40. Rg8 Qa6 41. Rb8 Qa3 42. Rb7 Qb2 43. Bb8 Rb3 44. Rba7 Qd4 45. R7a2
Ra3 46. Rb2 Ra7 47. Rba2 Rb7 48. Ba7 Rb8 49. Rb2 Rg8 50. Rba2 Rg7 51. Bb8
Bg8 52. Rb2 Rh7 53. Rba2 Rh5 54. Rb2 Rg5 55. Rba2 Rg3 56. Rb2 Rf3 57. Rba2
Rf2 58. Rb2 Ra3 59. Rba2 Ra7 60. Rb2 Rb7 61. Ba7 Rb8 62. Rba2 Bh7 63. Ra6
Rg8 64. Bb8 Rg7 65. Ra7 Bg8 66. Rb7 Rh7 67. Ba7 Rh5 68. Rb8 Bh7 69. Rg8 Rg5
70. Rg7 Bg8 71. Rh7 Rg3 72. Rh6 Rgf3 73. Bb3 Kg7 74. Ba4 Kf8 75. Rh7 Ke8
76. Rg7 Bh7 77. Bb8 Kd8 78. Bb3 Kc8 79. Rg8+ Kb7 80. Ba4 Qg7 81. Re8 Bg8
82. Rf8 Bg3 83. O-O-O {50 moves rule} 1/2-1/2 (2012-11-14)
Olaf Jenkner: Warum nicht z.B. 83. Te8 remis?
Was bedeutet das Schlüsselwort unused? (2012-11-14)
Henrik Juel: The castling shows that White never moved his king before; otherwise the position could be reached faster, and the 50 moves rule could not be applied.
The keyword unused seems non-sensical here and should be deleted (2012-11-14)
A.Buchanan: This problem is very interesting. It's orthodox 50M, so that castling does not reset the counter, and the timing works out nicely. But it makes me wonder... suppose we have a problem where it is B32 which was the last reset. Then W83 0-0-0 would prove that the position had already been at 50.0. If 50M convention (which needs to be rewritten because it's a mess) operated like 3Rep, then the game end would have been mandatory after B82. So castling W83 would be illegal. Is this how one would want 50M convention to work? Or should there be a carve-out to say: if you can definitely prove that no-one claimed, then no-one claimed. And should there be a similar carve-out for 3Rep rule? See https://www.thehoppermagazine.com/AA010 (2022-01-20)
A.Buchanan: Personally, I think no carve-outs. The 50M convention should be rewritten as: "A position is considered as a draw if it can be proved that the last 50.0 moves in the proof game combined with the solution did not contain a capture or pawn move. Unless expressly stipulated, this applies only to retro-problems." (2022-01-20)
Thomas Volet: What if the composer intends the retroplay to go beyond 50 non-P and non-capturing moves? (2022-01-20)
A.Buchanan: Hi Thomas - thanks for your question. I think it’s best if we take this offline. I will email you, if that’s ok (2022-01-21)
James Malcom: This is the Volet problem in question: P0008399 In hindsight, it is quite humorous that the 75 move rule was later introduced in the 21st century. (2022-01-21)
Thomas Volet: The question was not directed specifically to P0008399, which is just one of several compositions with the property at issue. (2022-01-22)
A.Buchanan: I've emailed Thomas. My earlier comment should have include checkmate as a third mechanism to zero the count. (2022-01-22)
James Malcom: You can always carry it on over to MatPlus. (2022-01-24)
more ...
comment
Keywords: 50 move rule, Castling (wl)
Genre: Retro
FEN: nB3RbN/1kppppq1/1p4p1/1P3p2/Bp3P1n/4PrbP/2PP1rP1/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-25 more...
58 - P0001941
Nenad Petrovic
2nd International Team Match 1967-1970
3. (2.?) Platz
P0001941
(10+12)
Gewinn
AP
Henrik Juel: If White may castle, last move was g7-g5 to avoid retrostalemate. 1.fxg6ep followed by 0-0 to legitimize the en passant capture. (2004-09-23)
A.Buchanan: In V&V Encyclopedia, which I admire more for its scope than for its precision, this problem is given to illustrate "Petrovic Theme". The definition given is: "PETROVIC THEME Also called 'retroproblem of Petrovic Type'. Mutual dependence of en passant capture and castling. By playing an en passant capture other retro elements of position are legalized (usually castling)."
The definition makes no reference to A Posteriori. I am trying to get my head around the text here, because the e.p. does not "legalize" castling, rather it mandates it. I also encountered Öffner for this castling/e.p. AP (but not in V&V), but that might be someone's confusion in that Types Öffner vs Keym exist in PRA. (2022-02-15)
comment
Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic), En passant
Genre: Retro, Studies
FEN: b7/p4P2/2kbPp1p/3ppPp1/n3pp2/8/P1PPP2P/4K2R
Reprints: (2) Problem 141-143 08/1971
(77) Problem 144-147 12/1971
(C) Die Schwalbe 16 10/1972
Encyclopedia of Chess Problems 2012
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-15 more...
59 - P0001967
Nenad Petrovic
628 Sahovski vjesnik 1950
Dr. Fabel und Dr. Ceriani gewidmet
2. Preis
P0001967
(15+15)
Längste Beweispartie?
(AL: 1021,0)
1. S S 50. S b6 100. S h6 250. S h3 300. a3 S 450. a6 S 500. b3 S 550. g3 S 600. 0-0 S 649. S hxSg 699. h3 S 899. h7 S 949. axSb Ke8 950 Tf1 Kd8 951. Tg1 Ke8 952. Tf1 Dd8 1020. Kd1 Kd8 1021. De1 Ke8=
play all play one stop play next play all
Henrik Juel: Note that in problems castling acts like capture and pawn move with respect to the 50 moves rule. After 949.a6xSb7 there are 4 moves left by [Pa7]; but each camp can shift only KDT, on d1-g1 and b8-e8, respectively, so the triple repetition rule now limits the length of the game. (2004-09-09)
A.Buchanan: In some problems it's certainly the case that the 50-move rule operates incorrectly in this way. Such problems are fine, but obviously wouldn't want to impose this as a standard. Different composers can make different assumptions here (2023-06-20)
comment
Keywords: 50 move rule, Non-Unique Proof Game, Longest Proof Game, Castling
Genre: Retro
FEN: Nrq1kb2/1PpppppP/1p6/8/8/pP4P1/BRPPPPp1/BrnKQ1Rb
Reprints: (I) Problem 5-6 12/1951
Problem 7-9 03/1952
1439 FIDE Album 1945-1955 1964
(130) Problem 91-94 04/1964
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2012-04-08 more...
60 - P0002056
Samuel Loyd
Musical World 1859
P0002056
(2+4) C+
#2
1. Da1! droht (unparierbar) 2. Dh8#
1. ... 0-0-0? ist illegal, da sK oder sTa8 schon gezogen haben muss
play all play one stop play next play all
Hans-Jürgen Manthey: Nachdruck SExpress 1.Mai 1947 S.41 Nr.33 (2022-10-15)
more ...
comment
Keywords: Cant Castler, Castling (sg), Minimal, Miniature, Homebase
Genre: Retro, 2#
Computer test: C+ Popeye 4.61
FEN: r3k3/p1p5/Q3K3/8/8/8/8/8
Reprints: 304 Chess Strategy (Loyd) 1878
73 150 Schachkuriositäten 1910
63 Sam Loyd and his Chess Problems 1913
43 64 Schach-Scherze 1915
32 Retrograde Analysis 1915
168 Allgemeine Zeitung Chemnitz 27/11/1927
Arbeiter-Zeitung (Wien) 27/11/1932
8 Comoedia 09/07/1933
(II) Problem 37-40 09/1956
(D10) feenschach 27 04/1975
84 100 Classics of the Chessboard 1983
(7a) Die Schwalbe 145 08/1995
Thema Danicum 95 1999
52 Opfer-Opfer-Matt Gaudium 21 10/2000
Outrageous Chess Problems 2005
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-05-26 more...
61 - P0002075
Thomas R. Dawson
Pittsburgh Gazette Times 21/12/1913
P0002075
(16+10)
#2
1. 0-0-0?? illegal
1. Td1! ... 2. Td3#
play all play one stop play next play all
Henrik Juel: 1.Td1, not 1.0-0-0? -1... c7 -2.Kd2 Tg3 etc. (2004-03-09)
more ...
comment
Keywords: Cant Castler (wl), Castling (wl)
Genre: Retro
FEN: 8/1p1ppppp/2p1B3/P6R/pP4PP/2N4k/PP2PPrN/R3KQB1
Reprints: 43 Retrograde Analysis 1915
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-08 more...
62 - P0002296
Andrey Frolkin
Dmitri W. Pronkin

2179 diagrammes 91 10-12/1989
Lob
P0002296
(13+15) C+
BP in 25.0
1. a4 h5 2. Ta3 h4 3. Tc3 h3 4. Tc6 bxc6 5. g3 La6 6. Lg2 Lc4 7. Le4 Sa6 8. f3 Db8 9. Kf2 Db4 10. Ke3 0-0-0 11. Kf4 Dxd2+ 12. Kf5 Dxc2 13. Dd5 Kb7 14. Dxf7 Kb6 15. a5+ Kc5 16. b4+ Kd4 17. Lb2+ Ke3 18. La1 Kf2 19. e3 d5 20. Se2 Td6 21. Td1 Tdh6 22. Td4 g6+ 23. Ke6 Ke1 24. Kd7 e6 25. Ke8 Sf6+
play all play one stop play next play all
paul: Checked for the first 19 moves, by Euclide. (2011-07-01)
Moldenhauer: Computerprüfung: C+ Stelvio 1. Std. 16 Min.
Keine Lösung BP 24.0 (2023-02-06)
comment
Keywords: Unique Proof Game, Promenade (K), Castling, Belfort (Kk)
Genre: Retro
Computer test: Stelvio 1.0
FEN: 4Kb1r/p1p2Q2/n1p1pnpr/P2p4/1PbRB3/4PPPp/2q1N2P/BN2k3
Reprints: 114 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Silvio Baier, 2023-03-02 more...
63 - P0002316
Alexander Kislyak
3587 feenschach 60 05/1982
A. G. Kuznecow gewidmet
1. Preis
P0002316
(15+13)
BP in 44.0
1. a4 Sa6 2. a5 Sc5 3. a6 b6 4. Ta4 Sb7 5. axb7 c5 6. b8=L g6 7. Lg3 Dc7 8. Lh4 Dg3 9. hxg3 Lh6 10. Tg4 f5 11. Sf3 f4 12. Sd4 f3 13. Sc3 Lf4 14. gxf4 a5 15. Lg3 a4 16. Lh2 a3 17. g3 a2 18. Lg2 a1=D 19. 0-0 Da2 20. Kh1 Sf6 21. Lg1 Sd5 22. Kh2 Se3 23. Kh3 Sxd1 24. Kh4 Se3 25. Kg5 Tf8 26. Kh6 Tf5 27. Kg7 c4 28. Kh8 Kf7 29. Sd1 Kf6 30. Kg8 c3+ 31. Kf8 De6 32. Ke8 Kg7 33. Kd8 Kh6 34. Kc7 Ta7+ 35. Kb8 Tb7+ 36. Ka8 Tb8+ 37. Ka7 Tb7+ 38. Ka6 Tf7 39. Kb5 Kh5 40. Ka4 h6 41. Ka3 Sd5 42. Ka2 Sc7+ 43. Ka1 Ta7+ 44. Kb1 Da2
play all play one stop play next play all
Paulo Roque: Is not possible: 38... Kh5 39.Kb5 (because black rook at square f5).
Probably the continuation of the author:
38...Tf7 39.Kb5 Kh5 40.Ka4 h6 41.Ka3 Sd5 42.Ka2 Sc7+ 43.Ka1 Ta7+ 44.Kb1 Da2# (2008-12-11)
James Malcom: Fixed. (2021-02-15)
comment
Keywords: Promotion (L), Promenade (K), Castling (wk), Non-standard material, Non-Unique Proof Game
Genre: Retro
FEN: 2b5/r1nppr2/1p4pp/7k/3N1PR1/2p2pP1/qPPPPPB1/1KBN1RB1
Reprints: 152 Shortest Proof Games 11/1991
(7) feenschach 103 01-09/1992
(8) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-02-15 more...
64 - P0002353
Theodor Tauber
Michel Caillaud

The Problemist 1991
P0002353
(14+11) C+
ser-h#3 (AP)
1. dxe3ep 2. fxg1=L 3. 0-0 Tg4#
1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt
play all play one stop play next play all
Henrik Juel: If Black may castle, his latest move must have been e3xf2, so last move was e2-e4. (Orig. wPg2 promoted on g8, so g7xh6 happened early). 1.dxe3 e.p. 2.fxg1=B 3.0-0 Tg4# (2003-03-21)
James Malcom: Is this the only know one-sided Valladao in a regular series-mover helpmate? Also, can wPh2 be shifted to h4, and the White king moved as well to h2, so that the solution must have 2. fxg1=S? (2020-12-08)
James Malcom: Well, the only known one that is in minimum form, i.e. 3 moves, I mean. (2020-12-08)
A.Buchanan: This is nice. I don't know of any other AP ser-h#3 with Valladao, but there could be. Just been cleansing PDB data today. Yes I think wKh3 and wBh2 can be transposed, and then promote to S instead of L. (2020-12-08)
A.Buchanan: The legality of intermediate positions in seriesmovers is of course irrelevant. However, each proof game that leads to the diagram position with BTM can be appended with a black move and then a "flip" that changes the player to move. We can repeat this process, and then the fact that Black retains castling rights implies that they had castling rights in the diagram position. I think this is the right formal way to handle AP with seriesmovers. (2022-03-21)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sk), Seriesmover, Promotion, Valladao Task
Genre: Retro, Fairies
Computer test: C+ Popeye v4.85 + thinking
FEN: 4k2r/1pp2p1p/1N5p/3P4/2RpP2N/P2P1P1K/p1BP1p1P/B3b1R1
Reprints: Die Schwalbe 143 10/1993
Rex Multiplex 45-46 05/1994
(XII) Quartz 5 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
65 - P0002372
Alexander Kislyak
Die Schwalbe 136 08/1992
P0002372
(15+10) cooked
BP in 10,5
1. Sf3 e6 2. Se5 Df6 3. Sxd7 Sxd7 4. a4 Sb6 5. a5 Ld7 6. axb6 0-0-0 7. Txa7 La4 8. Txa4 Td4 9. Txd4 Kb8 10. bxc7+ Ka8 11. Ta4#
play all play one stop play next play all
Cook: 1. Sf3 e6 2. Se5 Df6 3. Sxd7 Kd8 4. Sxb8 Ld7 5. a4 Lb5 6. axb5 Kc8 7. Txa7 Kxb8 8. Ta3 c6 9. bxc6 Ta4 10. c7 Ka8 11. Txa4
Moldenhauer: Computerprüfung: Cooked Stellung Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 9.5, BP 10.0.
Eine Notation von Stelvio:
1.Sf3 e6 2.Se5 Df6 3.Sxd7 Kd8 4.Sxb8 Ld7 5.Sc6+ Kc8 6.Sxa7+ Kb8
7.a4 Lb5 8.axb5 Txa7 9.b6 Ka8 10.bxc7 Ta4 11.Txa4#
Schlüsselwort Rochade? (2023-05-02)
comment
Keywords: Unique Proof Game, Castling
Genre: Retro
FEN: k4bnr/1pP2ppp/4pq2/8/R7/8/1PPPPPPP/1NBQKB1R
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2006-10-15 more...
66 - P0002471
Gerd Rinder
(G) Die Schwalbe 48 12/1977
Lob
P0002471
(4+3)
#2 (AP)
BTM: 1. ... Lxb7+ 2. Ke3 0-0-0 3. Sb6#
WTM: 1. Sf6+! Kd8,Kf8 2. Dc7#,Ld6#
play all play one stop play next play all
If WTM, b000 rights are already lost. Under Keym AP, Black attempts to steal the move. White disruption of castling now counts as win for White, so the only chance is 1. … BxQb7+ 2. Ke3! thr 3. Sf6+ disrupting castling but 2. … 0-0-0 3. Sb6#. 2. Kd3,Kf5? Be4+ 3. ~ 0-0-0! as b7 is no longer occupied, or 2. Kd3,Kd4,Kd5? 0-0-0! pinning wS or 2. Ke5,Kf4? 0-0-0! as wL is blocked. As usual when flip of player to move in d#n, Black gets an extra move rather than White losing one (c.f. Codex Article 15). Sublime miniature!
VL: AP after Keym. Solution:
I: 1.Sf6+.
II: Bl's try to be on move. 0... Bxb7+! 1.Ke3! O-O-O (legalizing!) 2.Sb6#. (2007-01-26)
more ...
comment
Keywords: Castling (sg), a posteriori (AP) (Type Keym), Homebase (s), Aristocrat, Miniature
Genre: Retro, 2#
FEN: r1b1k3/1Q1N4/8/8/4K3/8/7B/8
Reprints: Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
67 - 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...
68 - P0002665
Bernd Schwarzkopf
120v feenschach 2 03/1971
P0002665
(1+1)
a) Diagramm
b) wKh8, sLa7
Ergänze zu einem IC!
a) sK, 2 sTT, 3 sBB
b) sK, 2 sTT, 4 sBB
a) sKg8, sTg7,f8, sBf7,g6,h6
b) sKc8 sTb7,d8, sBa6,b6,c7,d7
play all play one stop play next play all
Henrik Juel: In part b) the set of added men should be changed to sK, 2 sTT, 4 sBB
Both parts rely on the last move being a black castling with check (2021-10-31)
comment
Keywords: Illegal cluster, Castling (sksg), Aristocrat, Miniature, Aristocrat, Miniature
Genre: Retro
FEN: K7/7b/8/8/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-11-01 more...
69 - P0002762
Alexander Kislyak
1180 Phénix 17 06/1992
P0002762
(16+16) C+
BP in 23,0
1. c4 c5 2. Db3 Db6 3. Dh3 Dh6 4. b3 b6 5. La3 La6 6. Lb4 Lb5 7. Sa3 Sa6 8. 0-0-0 0-0-0 9. Kb1 Kb8 10. Tc1 Tc8 11. Tc3 Tc6 12. Tg3 Tg6 13. Tg4 Tg5 14. g3 g6 15. Lg2 Lg7 16. Ld5 Ld4 17. Sf3 Sf6 18. Tc1 Tc8 19. Tc3 Tc6 20. Te3 Te6 21. Te4 Te5 22. Tef4 Tef5 23. e4 e5
play all play one stop play next play all
more ...
comment
Keywords: Unique Proof Game, Castling (wgsg), Symmetrical position, Capture-free
Genre: Retro
Computer test: Mondenhauer C+ Stelvio 1.11 00:09:39 Minuten. (hh:mm:ss) Keine Lösung: BP 22.0, BP 22.5.
FEN: 1k6/p2p1p1p/np3npq/1bpBprr1/1BPbPRR1/NP3NPQ/P2P1P1P/1K6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-03-27 more...
70 - P0003009
Luigi Ceriani
117 La Genesi delle Posizioni 1961
P0003009
(13+12) cooked
Welches war der erste Zug der sD und des sK?
AL in der Version von "hans" (PDB 2012-07-26):
R: 1. ... Ld8xLc7 2. Lb6-c7 Kc8-b8 3. Lc7-b6 Kb7-c8 4. Lb6-c7 Lc7-d8 5. Lg1-b6 Lb8-c7 6. Lb6-g1 Kc8-b7 7. Ld8-b6 Kb7-c8 8. d7-d8=L Lh7-g8 9. e6xSd7 Sc5-d7 10. Sc3-a4 Sa4-c5 11. Sd1-c3 d7-d6 12. Sc3-d1 Le5-b8 13. Se4-c3 Lg7-e5 14. Sf6-e4 c7-c6 15. Sg8-f6 Lh6-g7 16. g7-g8=S Kc8-b7 17. f6xTg7 Tg8-g7 18. e5-e6 Td8-g8 19. e4-e5 0-0-0 20. e3-e4 Lf8-h6 21. f5-f6 g7-g6 22. f4-f5 Le4-h7 23. f3-f4 Lb7-e4 24. f2-f3 Lc8-b7 25. Kb5-a5 b7xSa6 26. Kc4-b5 Sc5-a4 27. Kd3-c4 Se4-c5 28. Ke2-d3 Ta5-a3 29. Ta4-a2 Sf6-e4 30. La2-b1 Sg8-f6 31. Tb1-b2 Tf5-a5 32. Lb2-c1 Tf6-f5 33. Th1-b1 Tg6-f6 34. Ke1-e2 Th6-g6 35. Dd1-a1 Th7-h6 36. Lc1-b2 Th8-h7 37. b2-b3 Th7-h8 38. Lc4-a2 Th6-h7 39. Lf1-c4 Th7-h6 40. Ta1-a4 Th8-h7 41. Sc5-a6 h6xSg5 42. Se6-g5 h7-h6 43. a3xSb4 Sa6-b4 44. Sd8-e6 Sb8-a6 45. Se6xDd8 Sh6-g8 46. Sf4-e6 Sg8-h6 47. Sh3-f4 Sh6-g8 48. Sa4-c5 Sg8-h6 49. Sc3-a4 Sh6-g8 50. a2-a3 Sg8-h6 51. Sb1-c3 Sh6-g8 52. e2-e3 Sg8-h6 53. Sg1-h3
play all play one stop play next play all
Cook: (Mario Richter, PDB 2023-06-30)
R: 1. ... Ld8xLc7 2. Lb6-c7 Kc8-b8 3. Lc7-b6 Lh7-g8 4. Lb6-c7 Lc7-d8 5. Ld4-b6 Lb8-c7 6. Lb6-d4 Kb7-c8 7. Ld8-b6 Kc7-b7 8. d7-d8=L Kd8-c7 9. e6xSd7 Sb6-d7 10. Sc5-a4 Sa4-b6 11. Se4-c5 d7-d6 12. Sf6-e4 Ke8-d8 13. Sg8-f6 Le5-b8 14. Sh6-g8 Lg7-e5 15. Sg8-h6 Lf8-g7 16. g7-g8=S c7-c6 17. f6xDg7 Dh6-g7 18. f5-f6 g7-g6 19. e5-e6 Dc6-h6 20. e4-e5 Da8-c6 21. e3-e4 Dd8-a8 22. f4-f5 Le4-h7 23. f3-f4 Lb7-e4 24. f2-f3 Lc8-b7 25. Kb5-a5 b7xSa6 26. Sb8-a6 Sc5-a4 27. Sc6xTb8 Se4-c5 28. Se5-c6 Sf6-e4 29. Kc4-b5 Ta6-a3 30. Kd3-c4 Sg8-f6 31. Ta4-a2 Tb6-a6 32. La2-b1 Tc6-b6 33. Tb1-b2 Te6-c6 34. Lb2-c1 Tf6-e6 35. Th1-b1 Tg6-f6 36. Dd1-a1 Th6-g6 37. Ke2-d3 Th5-h6 38. Ke1-e2 Th6-h5 39. Lc1-b2 Th7-h6 40. b2-b3 Th8-h7 41. Lc4-a2 Th7-h8 42. Lf1-c4 Th8-h7 43. Sc4-e5 Th7-h8 44. Ta1-a4 Th6-h7 45. Sa3-c4 Ta8-b8 46. Sb1-a3 Th8-h6 47. a3xSb4 h6xSg5 48. Sf3-g5 h7-h6 49. Sg1-f3 Sa6-b4 50. e2-e3 Sb8-a6 51. a2-a3
s.a. 32Pe1A

Korrekturversuch zu P0005036
hans: Good motivation to capture Qd8 on the spot, to make long castling possible, which is needed to retrack the captured bRg7 just on time. Also a minor-promotion. I like this one, and I think the stipulation asks for which move black queen makes.
R: -1. …Ld8xLc7+ -2. Lb6c7+ Kc8b8 -3. Lc7b6 Kb7c8 -4. Lb6c7 Lc7d8 -5. Lg1b6 Lb8c7+ -6.Lb6g1 Kc8b7 -7. Ld8b6 Kb7c8 -8. d7d8=L Lh7g8 -9. e6xSd7 Sc5d7 -10. Sc3a4 Sa4c5+ -11.Sd1c3 d7d6 -12.Sc3d1 Le5b8 -13.Se4c3 Lg7e5 -14.Sf6e4 c7c6 -15.Sg8f6 Lh6g7 -16.g7g8=S Kc8b7 -17.f6xTg7 Tg8g7 -18.e5e6 Td8g8 -19.e4e5 0-0-0!! -20.e3e4 Lf8h6 -21.f6f5 g7g6 -21.f4f5 Le4h7 -22.f3f4 Lb7e4 -23.f2f3 Lc8b7 -24. Kb5a5 b7xSa6 and cage can be undone while black plays only with Ta3 and Sa4.
captures white axSb, SxDd8, exSd7, fxTg7, black hxSg, bxSa6, LxLc7 (2012-07-26)
Henrik Juel: Very similar to P0005036 and with identical stipulation question:
What was the first move by black queen and by black king.
Ceriani's abbreviations for move, queen, king, black, (and white) are t., D, R, n, (and b); in his ortho reconstruction problems the color abbreviations are capitalized, e.g. N=11 meaning 11 black moves (2012-07-26)
Thomas Volet: This composition appears on p.197 of Ceriani's 1961 book as his correction of P0005036 (which appeared in his earlier book). On that page, he discusses the clever cook in P0005036 ("ma questa bella posizione e demolita") with the WhP unpromoting at h8, uncapturing to the g file, and uncapturing back to the h file. (2012-08-02)
Mario Richter: I'm sorry to say this, but Ceriani's correction attempt still leaves room for cooks ... (2023-06-30)
comment
Keywords: First Move? (kd), Castling in the retro play (sg)
Genre: Retro
FEN: 1k4b1/p1b1pp2/p1pp2p1/K5p1/NP6/rP6/RRPP2PP/QBB5
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-07-02 more...
71 - P0003138
Branko Koludrovic
4208 Problem 12/1979
P0003138
(11+9)
h#3 (AP)
0.1...
1. ... axb6ep 2. 0-0-0 0-0-0 3. Td7 a8=D#
play all play one stop play next play all
This problem can be solved presuming that both castlings are executable. It implies that the last halfmoves have been.-1.Bb6xa7 sBb7-b5.
Bb5 could not come from c6 (sBc6xFb5), because all of the missing white pieces(5) have been captured on white squares except wLc1,which must have been captured by Bc7(Bc7xwLb6).Further,it has not played immediately before -1.Bb6-b5 because this would prevent white castling(forcing wT or wK to move)
Here is the proof that 4 remaining white pcs have been captured on white squares:
black d-pawn must have been captured on d4 and h,g f pawns had to promote after Bh3xwFg2,then wBh2moved to h4, enabling Bg4x Fh3 ..h1T, and finnaly sBf7
after fxg2... g1S. sLf1 prevented the check from Th1, and later has been captured by a white officer.
Conclusion:
Retro:-1.wBb6xFa7 Bb7-b5
Forward(AP) 1....axb6 e. p.2.0-0-0 0-0-0 3.Td7 a8Q# (Author)
Branko Koludrovic: P.S.
The black pawn a4(on the diagramm)came from c7 after capturing the white bishop on b6, then moving to b5 and capturing a white officer on a4. (2010-09-28)
A.Buchanan: I think sBf must have captured on g2 & then promoted on f1 to S, otherwise it’s impossible to unlock the southeast cage (2023-07-11)
A.Buchanan: So that means it’s sound. However I don’t see that it needs to have been sL shielding on f1 (2023-07-11)
more ...
comment
Keywords: a posteriori (AP), En passant as key, Castling (sgsgwg), Promotion (D), Valladao Task
Genre: h#, Retro
FEN: r3k3/P3p3/p7/Pp6/p6P/2P3PR/2PPP2b/R3K1nr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-11 more...
72 - P0003189
Tivadar Kardos
2144 Diagramme und Figuren 19/09/1967
P0003189
(7+15) cooked
h#2
1. bxc3ep Lxe2 2. Sa3 0-0-0#
play all play one stop play next play all
Cook: 2. ... Td1#
Sally: Der letzte Zug war: Bc2 - c4!
Nr.138 200 Ausgewählte S .Probleme T. Kardos(W. Fentze1983) (2010-09-30)
A.Buchanan: All 9+1 captures are visible by pawns. Note R: 1. Kd1-e1 d3xe2 is illegal because it costs 2 more captures for Black. So e.p. is unconditionally legal, requiring no AP justification. I see no reason why White has lost castling rights. Indeed someone might make a demo game ending with c2-c4. So this one seems cooked. Neither WinChloe nor yacpdb contains this problem. (2022-01-08)
A.Buchanan: If this is AP, it will be rendered sound by removing wPa2. But is this the intention? Does anyone have access to the collection of selected Kardos problems mentioned? (2024-01-25)
more ...
comment
Keywords: En passant as key, Castling (wg)
Genre: h#, Retro
FEN: 8/8/3pp3/2ppp3/1pPkrq2/4pb1P/P3p1rP/Rn2KBb1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-08 more...
73 - P0003269
Tivadar Kardos
149 British Chess Magazine 10/1956
P0003269
(5+17) cooked
h#3
1. exf3ep e3 2. 0-0-0 Tb4 3. Tg4 Tb8#
I think that this cooked problem is an early A osteriori. I can see ways that it might be made sound, but I would like to see the published version. San anyone help please? Thanks.
play all play one stop play next play all
Cook: 17 Black pawns, retro cook and numerous forward cooks
klären wK im Schach, vielleicht wBb2?
Alfred Pfeiffer: 9 schwarze Bauern! (2012-02-07)
Ladislav Packa: Auch mit wBb2 NL, z.B. 1.exf3 e.p. Txg5 2.0-0-0 Tb5 3.Tg~ Tb8# (2012-02-07)
A.Buchanan: And adding to earlier comments, if we do swap sBb2 for wB, the retraction is still not unique with R: 1. f3-f4 Lh6xg5+ (2021-11-26)
A.Buchanan: I think that this cooked problem is an early A Posteriori. I can see ways that it might be made sound, but I would like to see the published version. Can anyone help please? Thanks. (2022-04-25)
Gerald Ettl: Verbesserungsvorschlag: -sBa2, -sBb2, -sSg7, +sBg7
r3k3/3b2p1/5p2/6b1/4pPRp/2pq2rp/2p1P1pB/2K3n1/ (2022-04-25)
Gerald Ettl: und +sSg8 (wegen exf4 Vermeidung) (2022-04-25)
Gerald Ettl: +sSa4 nicht g8 (2022-04-25)
A.Buchanan: Hi Gerald. Thanks for this. Your final proposed version is r3k3/3b2p1/5p2/6b1/n3pPRp/2pq2rp/2p1P1pB/2K3n1, yes? There are 16 candidate solutions, all using castling. But wPa cannot leave the a-file, and is required for capture balance, so Black cannot castle. Five of the candidates begin with e.p.: play might have just been R: 1. Kb1-c1 b3xPc2+, so e.p. is not permitted in any case. What am I missing? (2022-04-25)
Gerald Ettl: Hi Andrew,
Those were 2 mistakes I made. I improved it again.
r3k3/3b4/4pp2/6b1/4pPRp/n1pq2rp/PPp1P1pB/2K3n1/
(den wBa2 habe ich auf das Brett gestellt, da sonst wieder Lh6xXg5 geht.) (2022-04-26)
A.Buchanan: Hi Gerald: Deine Retro-Logik ist gut. Aber es gibt 18 Lösungsvorschläge für die Zukunft. Ich habe heute im Discord meinen eigenen Vorschlag zur AP-Korrektur von Kardos veröffentlicht. Ich werde ihn hier im PDB hinzufügen. (2022-04-26)
comment
Keywords: En passant as key, Castling
Genre: h#, Retro
FEN: r3k3/3b2n1/5p2/6b1/4pPRp/2pq2rp/ppp1P1pB/2K3n1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-26 more...
74 - P0003359
André Hazebrouck
3256 Themes-64 07-09/1977
P0003359
(14+10) C+
h#2
2.1...
1) 1. 0-0 Le6+ 2. Kh8 Sg6#
2) 1. bxc3ep 0-0 2. 0-0-0 Tc4#
play all play one stop play next play all
If all three castlings are legal, then the ep is on. So evidence must be accumulated from the two solutions. White 0-0 serves only in this retro role, eliminating 26 other candidate solutions.
A.Buchanan: There was a diagram error here. It should be bPd7 not bPe7. The existing diagram is vastly cooked with e.g. 1. Ke8 Kd1 2. Te8 Lb6#. WinChloe has the correct diagram here. (2022-05-24)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wksksg)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro logic
FEN: r3k2r/B2p2pp/8/NP6/RpP4N/pP5B/1p1PPPP1/n3K2R
Reprints: feenschach 54 04/1981
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-06 more...
75 - P0003411
Norman Alasdair Macleod
3970 Themes-64 04-06/1982
P0003411
(4+6) C+
h#2 (AP)
2.1...
1. Kc3 0-0-0 2. Txc4 Txd3#
1. bxc3ep e4 2. Kc4 Ta4#
play all play one stop play next play all
The idea is that the ep in one solution is validated by the castling in the other solution. Since no other solutions exist, there are no parasites which might "piggyback" off the proof given by the castling solution. This is not PRA: both solutions have the same history with both castling & hence ep legal.
Mario Richter: How is the (AP) to be interpreted here? Is the intention as follows: Since in solution 1) White castles, Black is allowed to capture e.p. in solution 2)? (2011-05-28)
Henrik Juel: It seems to be an unusual situation, not covered by keywords like AP or PRA.
If last move was c2-c4, both solutions work; if not, there is no solution, because White may not castle.
(In the second solution, Kxc4 should be Kc4). (2011-05-28)
VL: This a generalized type of AP, which I call "consolidate AP": both solutions are considered as parts of one complete solution. However the order of both parts is significant. I know two similar problems: P0003437 and P0003186. (2011-06-01)
A.Buchanan: @Valery: I've started to classify all the AP problems in PDB. This and its kin are "Type Petrovic - cons" (standing for consolidated, but there's only a limited number of characters for a PDB parameter). I truly hope that the theory in the end will not have this as a special case, but rather is part of a more general pattern.
It's not just that PRA & RS each scales up within its own paradigm to handle seamlessly very complex collections of conditional move dependencies. They also scale *down* to handle situation when there is just one conditional move, or even zero conditional moves. And when n=0 or n=1, PRA & RS are in complete agreement. This is the kind of robust seaworthy behaviour we need if are going to set sail in the stormy seas of fairydom.
But small steps. The first thing is to classify all the problems in PDB into buckets, to find out what we have. I doubt there are 37 categories, so will probably need the results of your scholarship too. (2022-02-17)
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comment
Keywords: a posteriori (AP) (Type Petrovic consol), En passant as key, Castling (wg)
Genre: h#, Retro
Computer test: HC+ Forward play proved by Popeye v4.87 AP logic requires some framework but I think we know we want this cool idea to end up being sound.
FEN: 8/8/8/1nr5/1pPk4/1p1p4/4P3/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-09-12 more...
76 - P0003417
John Frederick Keeble
2206 The Problemist Fairy Chess Supplement 16, p. 173, 02/1936
P0003417
(4+5) C+
h#2.5 (AP)
1. ... exd6ep 2. 0-0-0 dxe7+ 3. Tf8 exf8=T,D#
play all play one stop play next play all
H.Juel: If Black may castle, last move was not made by Ke8 or Ta8, but by Pd5, so it was d7-d5, and hence White may capture en passant. 1... exd6ep 2.0-0-0 dxe7 3.Tf8 exf8=D,T# where the castling (belatedly) justifies the e.p. capture.
Not 1... exd6ep 2.Kd8? dxe7 3.Kc8 e8=D,T
AB: Another 120 retro tries (=60x2 due to tolerated dual promotion) of general form 1. exd6ep 2. T~ dxc7 3. T~ c8=D,T.
No set play
Nachdruck in "Die Schwalbe" 33 mit Diagrammfehler (wBe6 fehlt).
VL: The first AP-problem! Thematic illegal try:
1. ... exd6ep 2.Kd8? dxe7+ 3.Kc8 e8=T(D)#?? (2007-02-11)
Mario Richter: 46 years between this Problem and P1366663 - perhaps this was not Keeble's first AP-Problem?! (2019-08-29)
Ladislav Packa: Of course I know the idea a posteriori. But let my heresy be forgiven - I do not consider it logical. In this particular example, I consider e.p. for proof that black can castling - not vice versa. (2019-08-29)
A.Buchanan: Hi Ladislav. If ep is ok, then Black just moved a pawn with double hop. That doesn't stop Black having e.g. moved bK right at the beginning of the game. So you can't prove that Black definitely can castle this way. What am I missing here? (2019-08-29)
Ladislav Packa: Your statement is perfectly fine. However, the AP condition is usually interpreted as castling is evidence of e.p. However, the second party does not have to do castling - the possibility has already been demonstrated by the existence of e.p. (2019-08-29)
A.Buchanan: Hi again Ladislav. Please try again, I am obviously being very stupid. I don't understand how 1. exd6ep is any more indicative that Black can castle than 1. K~. (2019-08-30)
Ladislav Packa: Hi, Andre!
e.p. is just a result of the d7-d5 move. Black can castle only if it was the last move of black (even without wPe5). Therefore, it does not have the logic of joining the e.p. with castling. For example, ask yourself if black is allowed to castle in the position without wPe5. (2019-08-30)
A.Buchanan: @Ladislav: ep is not "proof that black can castle", rather it is a *necessary* condition for Black to be able to castle. The logical implication is the other way round. We can castle because of the optimistic convention, and because sPd might just have moved. Without wPe, there is no solution, but whatever White plays, Black can castle (2020-12-11)
Hans-Jürgen Manthey: was heißt hier Thematisch Illegal ??
Mir völlig wurscht ob ep möglich oder nicht, nach dem weißen Zug ist natürlich Kd8 möglich !!
also ist auch 1. ... exd6ep 2. Kd8 dxe7+ 3. Kc8 e8DT# eine Lösung, sonst müßte auf die erzwungende Rochade in der Aufgabenstellung hingewiesen werden ! (2020-12-11)
A.Buchanan: Can one tolerate “tolerated dual promotion” in Valladao? Well Valladao himself did, so it must be ok (2022-03-26)
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comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sg), Promotion (D/T), Tolerated dual promotion (D/T), Valladao Task, Retro Strategy (RS)
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: r3k3/2p1p3/2P1P3/2KpP3/8/8/8/8
Reprints: V. Die Schwalbe 33, p. 323, 06/1975
2001 CHM Themes 12/2000
3 Die Schwalbe 215, p. 240, 10/2005
A1 Problemkiste 169 02/2007
402 Eigenartige Schachprobleme , p. 130, 2010
14 Die Schwalbe 241, p. 374, 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-10 more...
77 - P0003423
Matti Arvo Myllyniemi
3975 Stella Polaris 01/1971
P0003423
(7+11)
h#3 (AP)
0.2.1...
1. ... g6 2. 0-0 gxf7 3. Kh8 Le5#
1. ... cxb6ep 2. 0-0-0 bxa7 3. Td7 a8=D#
play all play one stop play next play all
Henrik Juel: Analysis
If Black can castle both ways, last move must have been b7-b5
C+ Popeye 4.61 (2020-10-25)
comment
Keywords: a posteriori (AP), En passant as key, Castling (sksg), Promotion (D), Valladao Task
Genre: h#, Retro
FEN: r3k2r/p4p1p/8/KpP2PP1/4PBpP/5pp1/6p1/8
Reprints: Nordisches Turnier 1970-1971
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2020-10-25 more...
78 - P0003428
Imre Sztankovszky
130 Die Schwalbe 09-10/1953
P0003428
(10+11) C+
h#1.5 AP
1. ... fxg6ep 2. 0-0 gxh7#
play all play one stop play next play all
Henrik Juel: Black captured c7xd6x..x.h2 and once more with an officer
White captured a2xb3, b2xa3, f2xg3, and e.g. exf and once more
If Black may castle, last move was g7-g5 (not b7-b6, because of Lh3) (2022-04-26)
Henrik Juel: HC+ Popeye 4.61
The castling serves two purposes:
enabling the mate and legitimizing the ep capture
so the a posteriori legitimizing is impure (2022-04-26)
A.Buchanan: Yes Henrik. The absence of other candidate solutions (beginning e.p. but excluding 0-0) declines an opportunity to embed additional content, and is arguably an artistic defect. However this very early AP problem is sound, and purity of motive is just one school. IMHO, there is space for such problems, if other content compensates: e.g. 4k2r/p2ppp1p/p7/5PpK/8/1PBB2Pb/2PP2Pp/8 h#2* AP. (2022-04-27)
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comment
Keywords: En passant as key, Castling (sk), a posteriori (AP) (Type Petrovic), Volet Pawn
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 4k2r/3ppp1p/1p6/4BPpK/P7/pPPB2Pb/3P2Pp/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-27 more...
79 - 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)
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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...
80 - P0003431
Tomislav Petrovic
(XIV) Problem 37-40 09/1952
P0003431
(8+14) cooked
h#3
1. cxb3ep+ c3 2. Ld3 Tg4+ 3. Kh1 0-0-0#
play all play one stop play next play all
paul: White captures was axb and g2xh3, so the retro move c3xb4 is not possible. If b3-b4, the retro check is not justified. So last move was b2-b4 (preceded by Rc3-a3). (2011-08-06)
A.Buchanan: There's a lot to unpack here. Firstly, the intended solution has a definite transposition dual in the order of Black's 2nd & 3rd moves. Second, the pawn capture count is wrong. What happened to the original wBh, given sBh2 never left the h-file? Thirdly, and not a showstopper, White's only solution involves castling: so that's a logical dual. (2020-12-08)
more ...
comment
Keywords: En passant as key, Castling
Genre: h#, Retro
Computer test: Popeye v4.85 + retro thinking
FEN: 3r3n/p7/pp6/bP1q4/RPp5/r3p2P/p1P1P1kp/Rb2K3
Reprints: (61) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-02 more...
81 - P0003434
Jozsef Bajtay
2432 Problem 101-102 09/1966
P0003434
(10+11)
h#2
1. fxg3ep 0-0 2. Lg4 hxg3#
play all play one stop play next play all
A.Buchanan: Surely diagram typo. Change to sBh3, then everything works (2022-03-29)
Mario Richter: Yes, Pawn h3 is black (2022-03-29)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wk)
Genre: h#, Retro
FEN: 8/b3p3/4p3/6pp/2P2pPk/1pPP3p/2PP1P1P/r2bK2R
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-03-29 more...
82 - P0003442
Janko Furman
3. Makuc-Moder-Gedenkturnier 1971-1973
1. Preis
P0003442
(12+13) C+
h#2
b) wBd4 nach d5
a) 1. cxd3ep Sd5 2. 0-0 Se7#
NOT 1. ... Lf6? 2. Kf8 Tx8# because no AP justification
b) 1. Kd7 Lf6 2. Te8 Sxb6#
NOT 1. 0-0? Tf6 2. Kh8 Txf8# because rights lost
play all play one stop play next play all
Assume that bPb6 is really on b7:
Bl captures: dxe, exf, fxg, a|
Wh captures: bxc, g|, cxd=
So all pawn captures are accounted for.
a) If Bl 00 rights remain, then only way to give Black a prior move is by d2-d4.
b) No way to give Black a prior move, so Bl 00 rights must be lost.
Cook: NL
b) 1. 0-0 Tf6 2. Kh8 Txf8#
A.Buchanan: There is definitely something wrong here, with both (a) & (b). I think it's a simple diagram error: bPb7 has been misplaced on b6. Then the AP logic for (a) works great, and the castling "NL" for (b) is seen to be a thematic retro try. Can anyone confirm? (2022-03-21)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sk)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/7p/1pR5/2P5/NNpP4/KB2PPPP/p3pppq/B4bnr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
83 - 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...
84 - 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...
85 - P0003460
Johannes Jacob Burbach
1089 Probleemblad 06-08/1950
P0003460
(5+4) C+
h#3
1. 0-0-0 Ke2 2. Kb8 Td1 3. Ka8 Txd8#
play all play one stop play next play all
1. 0-0-0 0-0? usw. ist nicht erlaubt, da Weiß im letzten Zug einen schwarzen Stein geschlagen haben muss, was nur mit K oder T ging.
A.Buchanan: Tried some experiments to see if other formats of position entry are possible, but having reverted those, the solution animation no longer works. The usual PDB problem of hidden & undocumented variables between stipulation, diagrams & solution animation, sigh. (2021-04-01)
A.Buchanan: I found a "fix" - I added a bogus empty diagram b! How random! (2021-04-02)
A.Buchanan: For h#, I run Popeye without opti noca, so can see all the retro tries explicitly. For d#, I would apply opti noca if relevant, else lines of play might be blocked (2021-04-04)
more ...
comment
Keywords: Cant Castler, Castling (wksg), Homebase (s)
Genre: h#, Retro
Computer test: Gustav 4.1d C+ Popeye 4.61 also (with 'opt noc h1')
FEN: r3k3/pp6/8/8/8/6P1/5P1P/4K2R
Reprints: (27) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-04 more...
86 - P0003463
Tivadar Kardos
1501 Problemista 68 03/1967
P0003463
(9+16) C+
h#4
b) sSe6->d5
a) 1. Df7! h4 2. 0-0 h5 3. Lh8 hxg6 4. Sg7 gxh7#
b) 1. De6! h4 2. 0-0 h5 3. Kh8 hxg6 4. Tg8 Txh7#
play all play one stop play next play all
1. 0-0-0?? 0-0 2. Kb8 Txc1 3. Ka8 Txb1 4. Tb8 Ta1# (0-0-0 ist illegal)
1. 0-0-0? ist nicht gestattet, da wegen der schwarzen Bauernstellung der weiße Bauer a2 auf a8 umgewandelt wurde!
Marcin Banaszek: Die Urdruck-Stellung ist mit sSe6 (nicht sSd5) und über die Zwilling-Stellung ist dort keine Erwähnung. (2022-06-09)
A.Buchanan: Yes I don't know where the twin concept was introduced to this record (the visible "Last Update" history does not mention it). The twin adds very little: the retro try is identical, and the solution shares most of the White moves, ending in a duller mate. Did the 1983 reprint include it? (2022-06-10)
more ...
comment
Keywords: Cant Castler, Castling (wksg)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic
FEN: r3k1qr/1p5p/4n1p1/4bp2/6p1/1pPpPpP1/1P1P1P1P/1bn1K2R
Reprints: 144 200 Ausgewählte Schachaufgaben 1983
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-10 more...
87 - P0003501
Jozsef Korponai
1224 Sahovski glasnik 01/1970
P0003501
(6+3)
h#2*
2 solutions
set play
1. ... a3,a4? 2. cxb2 0-0?# castling illegal
1. ... Kf1! 2. cxb2 Kg2#
1. ... Kd1! 2. cxb2 Kc2#
actual play
1. c3xb2? 0-0#? castling illegal
1. Ka1! Kd1 2. c2+ Kxc2#
1. Kxb2! Kd1 2. Ka1 Kc2#
play all play one stop play next play all
A.Buchanan: Not sure what is intended here: should one include set play? It's a bit unclear. (2021-10-22)
A.Buchanan: I have marked stip from "h#2" to "h#2* 2 solutions" (2021-10-22)
comment
Keywords: Cant Castler, Castling (wk)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic
FEN: 8/8/8/8/1N6/2p1p3/PP2P3/1k2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-22 more...
88 - P0003505
Herbert Hultberg
7153 The Fairy Chess Review 02/1947
P0003505
(2+3) C+
h#3
1. Kb3 Kd2 2. Ka4 Kc3 3. a2 Txa2#
Nicht 1. Kb3 0-0-0?? 2. Ka2 Td3 3. Ka1 Txa3#, da Weiß zuletzt mit K oder T gezogen hat
play all play one stop play next play all
more ...
comment
Keywords: Cant Castler, Castling (wg), Minimal, Miniature, Homebase
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 + trivial retro logic
FEN: 8/8/8/1p6/8/p7/2k5/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-29 more...
89 - P0003506
Leon Loewenton
4495 Arbejder-Skak 06/1957
nach T.R. Dawson
P0003506
(7+5) C+
h#2*
* 1. ... Tg1 2. e4 g3#
1. Kg3 0-0? 2. h4 Tf3#
1. Kg3 Tg1! 2. Kh4 g3#
play all play one stop play next play all
A.Buchanan: I suppose set play is intended, because it's the only function of sBe5. But I think it's fine without. (2021-10-22)
Mario Richter: The set play is not mentioned in the "official solution" ('Arbejder-Skak' 08/1957 p.231) (2021-10-23)
A.Buchanan: Thanks. Odd! (2021-10-23)
comment
Keywords: Cant Castler, Castling (wk)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic I suppose set play is intended, because it's the only function of sBe5. But I think it's fine without.
FEN: 8/8/8/4p1pp/6Pk/4p2P/4P1PR/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-24 more...
90 - P0003529
Tivadar Kardos
2290 Magyar Sakkelet 05/1957
P0003529
(10+11) C+
h#2
1. Kxd8 Ta7 2. Te8 Td7#
1. 0-0 0-0-0? 2. Th8 Txg1# try
play all play one stop play next play all
White castling not possible as sBa= on a1. (Also sBgxh= or sBhxg=)
A.Buchanan: Retractor 2.1.1 does not notice that promotion happened on a1 (2021-11-25)
comment
Keywords: Cant Castler, Castling (wgsk), Sacrifice of white pieces
Genre: h#, Retro
Computer test: HC+ Popeye 4.87 & trivial retro-logic
FEN: b1nNk2r/8/3p4/2Pp1B2/3p4/1PpPp3/1PPP4/R3K1nr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-25 more...
91 - P0003531
Tivadar Kardos
3563 Schach 24, p. 379, 12/1960
P0003531
(10+12)
h#2
1. Kd8 Txh8 2. Kc8 Txg8#
play all play one stop play next play all
Henrik Juel: The diagram pawns captured all missing men, so the missing pawns [Pa7,Ph2] promoted on a1,h8, and Ta1,Th8 have moved, ruling out the apparent solutions 1.Ke7 0-0-0 2.Kxe6 The1#, 1.Dg1+ Ke2 2.0-0 Taxg1#
C+ by Popeye 4.61 and analysis (2021-08-31)
comment
Keywords: Cant Castler, Castling (wbsg)
Genre: h#, Retro
FEN: r3k1qr/1bp2p2/2p1Pp2/5p2/1P3p2/1Pp2P2/1PP2P2/R3K2R
Reprints: 664 Die Schwalbe 05/1961
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-10 more...
92 - P0003539
Hilding Fröberg
332 Land og Folk 01/03/1970
P0003539
(4+2) C+
h#1.5
1. ... Sf5? 2. 0-0-0?? Sd6# - aber die s0-0-0 ist illegal, den zuletzt muß Schwarz mit K oder T gezogen haben.

1. ... Sg8! 2. Td8 Sc7#
play all play one stop play next play all
Mario Richter: Luboš Kekely (Slovakia) correctly points out that 1. ... Sf5! 2. 0-0-0 Sd6# is only a try and not a solution (last black move must have been by King or Rook, so the queenside castling is illegal).
I have changed the solution accordingly. (2023-03-10)
comment
Keywords: Cant Castler, Castling (sg), Aristocrat, Minimal, Miniature, Homebase
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic | C+ rawbats
FEN: r3k3/6K1/N6N/8/8/7B/8/8
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-03-10 more...
93 - P0003561
Tivadar Kardos
5396 Arbejder-Skak 01/1961
P0003561
(7+15) C+
h#2
not: 1. Sd7 0-0-0 2. 0-0 Tdg1#
but: 1. Lxe6 Th7 2. 0-0-0 Ta8#
and: 1. Ke7 0-0 2. Kxe6 Tae1#
play all play one stop play next play all
A.Buchanan: Hard to see this as anything but a version of P0563989. Popeye reports again three apparent mates, with all 4 castlings. However pawn capture analysis indicates that Black qside castling is illegal, while the two kingside castlings are mutually exclusive. (2022-01-08)
comment
Keywords: Cant Castler (sg), Castling, Partial Retro Analysis (PRA)
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: r1b1kn1r/1n6/1pPpP3/1p1p4/1bp5/1qpp4/1PP5/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-08 more...
94 - P0003574
Juan Ruiz Luque
Jean-Claude Gandy

RA37 diagrammes 21 05-06/1976
P0003574
(8+8)
h#2
Duplex
s) 1. Kf8 g6 2. Se7 Th8#
w) 1. ... Kg1 Td8 2. Lh2 Td1#
play all play one stop play next play all
Rochaderecht
Henrik Juel: C+ by Popeye 4.61 and analysis
If Black could castle, both parts would be cooked:
1.0-0-0?? Txg7 2.Sb8 Tc7#
1... Kg1 0-0-0?? 2.Lh2 Td1#
But he cannot, because explaining wTh7 by h7-h8=T would require nine white pawn captures, and only eight black men are missing; hxg8=T is even worse, requiring 10 captures (Th7 being original or promoted further towards the west obviously would require Ke8 to have moved) (2021-07-04)
comment
Keywords: Cant Castler, Castling (sg)
Genre: h#, Retro
FEN: r3k3/6pR/2n4p/5PP1/pp5b/5PBP/6PK/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-07-03 more...
95 - P0003607
Ljubomir Ugren
anonymous

881 Sahovski glasnik 07-08/1964
P0003607
(6+8)
h#3*
*) 1. ... 0-0 2. Tf8 Tc1 3. Tf7 Tc8#
1) 1. 0-0 Tf1 2. Txf2 Txf2 3. Kh8 Tf8#
play all play one stop play next play all
Henrik Juel: Ljubomir had an anonymous co-author?? (2021-10-31)
A.Buchanan: I've no idea. There were three users called "anonymous" of different kinds, so I merged them into one, that's all. (2021-10-31)
A.Buchanan: I think anticipated in 1948 by Peter Kniest P0003606 (2021-11-01)
comment
Keywords: Cant Castler, Castling (wksk), Homebase (w), anticipated (P0003606)
Genre: h#, Retro
FEN: 4k2r/3pp1pp/pp6/8/8/8/3PPP1P/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-01 more...
96 - P0003622
Janez Moder
Problem 07/1962
P0003622
(4+3)
h#2*
*) 1. ... 0-0 2. Dh4 Txf3#
1) 1. Kh4 Kf2 2. fxg2 Sf3#
play all play one stop play next play all
SCHRECKE: NL: 1. Dg5,Kh4 gxf3 2. Kh4,Dg5 Sf1# (2023-09-13)
Ladislav Packa: Retro content is not needed, the solution is preserved even without it.
Pg4 Pg2 Sh2 Ke1 Rh1 (5)- Ph5 Kg3 (2) h#2* C+

1...0-0 2.h5-h4 Rf1-f3 #
1.Kg3-h4 Ke1-f2 2.h5*g4 Sh2-f3 # (2023-09-14)
comment
Keywords: Cant Castler, Castling (wk)
Genre: h#, Retro
FEN: 8/8/8/8/6q1/5pk1/6PN/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: hpr, 1997-05-07 more...
97 - P0003630
Jean Oudot
652 Themes-64 10-12/1962
P0003630
(9+6) C+
h#2*
*) 1. ... 0-0#
1) 1. Dc5 Tf1+ 2. Df2 Txf2#
play all play one stop play next play all
A.Buchanan: White must cover g2. If not by check, then by rook. However Rf1 is check, so both Black's moves are forced to shield bK (2022-11-27)
more ...
comment
Keywords: Cant Castler, Castling (wk)
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: b7/7P/6p1/6qP/6Pb/3p1kNP/3P3P/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-27 more...
98 - P0003657
Vladislav Bunka
4204 Problem 12/1979
P0003657
(4+13)
h#3
b) -wBh3
a) 1. Txh3 0-0 2. Sd3 Txf4 3. Sd6 Ta4#
b) 1. Sd1 Txh4 2. Se3 Th8 3. Sd6 Ta8#
play all play one stop play next play all
Adrian Storisteanu: Fix for P0534003. (2015-10-17)
Yuri Bilokin: version: -wPh3 8/3pq3/5r2/5p2/4nppp/5r2/kpP2n2/1b2K2R (3+13) h#3 2 Sol
1.Rh3 0-0 2.Sd3 Rxf4 3.Sd6 Ra4# (MM)
1.Sd1 Rxh4 2.Se3 Rh8 3.Sd6 Ra8# (MM) (2021-06-24)
Ladislav Packa: Yuri: Without wPh3, the castling is impossible (theme Cant Castler). But after wPh3-d5, two solutions are OK. (2021-06-24)
Yuri Bilokin: Ladislav, sorry, was wrong. Thank you for your vigilance. A slightly different edition, too, with two solutions: 8/2p1q3/4r3/5pPp/4npp1/5r2/kpP2n2/1b2K2R (4+13) h#3 2 Sol. (2021-06-25)
comment
Keywords: Cant Castler, Castling (wk)
Genre: h#, Retro
FEN: 8/3pq3/5r2/5p2/4nppp/5r1P/kpP2n2/1b2K2R
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 1999-04-24 more...
99 - 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...
100 - 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...
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