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)
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Anton Baumann: Mattdual in b): 3.Tb8 axb8=D,T# (2022-12-16)
A.Buchanan: So Borodatow got it all working! In a) Black might have captured hxgxf and axPb. So castling rights might still be maintained with the ep. In b) on the other hand, it must be axb, bxa, exf and wPgxh6, so there was the cage. Promotion to TD is tolerated in the final move, although it may not be puristic, without the convention, too many mates would be excluded (2022-12-16)
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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)
Paulo Roque: obs: 28...S8d7 (2008-11-22, hidden) more ...
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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)
Paulo Roque: obs: 24.S1c3 (2008-12-16, hidden) more ...
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"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)
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Anton Baumann: Neufassung vergl. P0006288 (2023-01-06)
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Anton Baumann: Sollte eine Verbesserung von P0000777 sein.
Autorabsicht: "letzter Zug f7-f5 sonst retropatt! Jetzt ist aber die Rochade illegal, da der sTh8 nur über e8 aus seiner Ecke kommt. Daher Verführung 1. ... exf6 e.p. 2.0-0-0? Lxf4 3.Td7 a8=D#. Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 sg7#"
Aber: Der s h-Bauer kann auch auf h5 geschlagen worden sein; so konnte der sT über h6 aus seiner Ecke entkommen. Die vermeintliche Verführung ist also NL! (2022-12-14)
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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)
A.Buchanan: Isn't Guus' idea just AP-Prioritat? And even if it is "more logical" or "more interesting", it doesn't follow that previous interpretations are "incorrect". Once a "polar concept" has been defined, prior art remains, and is valid in its own context. "Freedom of interpretation" remains, although newer compositions may adopt the newer paradigm. If a new idea does succeed in displacing the old, then problems (old and new) which depend on the earlier interpretation should be preserved with the context in which they existed. I don't know whether the
"Timeline Gallery" will ever physically exist, but I hope that it will. This is basically the "Golden Age" argument, that says if we don't value & preserve the old, we cannot consistently move on to the new (2023-07-30, hidden)
A.Buchanan: Isn't Guus' idea just AP-Prioritat? And even if it were "more logical" or "more interesting", it doesn't follow that previous interpretations are "incorrect". Even if a "polar concept" has been defined, prior art remains, and is valid in the context in which it was formed. "Freedom of interpretation" remains, although newer compositions may preferentially follow the newer paradigm, in part because fresh design space is always desirable.
If a new idea does succeed in displacing the old, then problems (old and new) which depend on the earlier interpretation should be preserved in a "Timeline Gallery", each with the context in which it was created.
It is also a fallacy to say that a stipulation should "explain" the philosophy. Such problems have been defined as AP, and that's fine. I think AP-Prioritat (sorry for absence of umlaut) *does* deserve to be mentioned as part of the stip, but whether the problem is Keym or Petrovic or both or something else - or PRA or RS - all that is part of the solving fun: keywords can record the taxonomic details & they don't need to be part of the stip. (2023-07-30, hidden) more ...
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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)
A.Buchanan: How should we show this kind of correction in PDB? A separate diagram? Or if the same, what is the original date - a correction is not a reprint? Happy to follow whatever clear protocol can be defined (2022-12-07, hidden) more ...
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In a) Entschlagdual R: Kd5xBe4, das wurde in der Lösungsbesprechung noch kritisiert und als NL vermerkt. Hat das der PR gelassener gesehen ('geduldeter Entschlagdual'), oder gibt's noch eine korrigierte Version dieses Problems?
Anton Baumann: Korrektur in 'Die Schwalbe' 12/1975 S.422: +sLh2, +sBe5;
nun geht in a) nur noch zurück: Kd5 x Be4.
Ausgezeichnet wurde gem. Preisbericht in 'Die Schwalbe' 06/1977 S.82 die korrigierte Version 1419v. (2022-12-08)
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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)
A.Buchanan: The new form is now quite nice, and should be posted here in PDB. Alas I can't take much of the credit, all I spotted was that bPa doesn't have to be on the board: the number of black PCs is the same. (2022-12-15, hidden) more ...
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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)
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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)
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See P0003736 a companion problem.
milan: [-bPa7b6wPg3bSa1+bPg3bBa5-b4 bPc6-d6] h#2 1.2...bRh2? illegal move? M.Frelih (2023-03-30)
Mario Richter: (a) Because of bRh2 w0-0 impossible.
Solution: 1.Rh2xh3 Rxh3 2.0-0 Rh8#

(b) bRh2 can return to a8 only via e8, so b0-0 impossible.
Solution: 1.Ba4 0-0 2.Rf8 Re1#

But cooks:
1.Rh2xh3 Bb4 2.Rxg3 Rxh8#
1.Nd2 Bf6 2.Rf8 Nd6# (2009-01-30, hidden)
A.Buchanan: Suggested corrected version: 4k2r/p2p1Np1/1pp2p2/1b3p2/8/p4PPP/B4P1r/nn2K2R. Can add to the stipulation: "Which missing units were not captured by pawns?" (2022-03-15, hidden)
A.Buchanan: Suggested correction: 4k2r/p2p1Np1/1pp2p2/1b3p2/8/p4PPP/B4P1r/nn2K2R (2022-03-15, hidden) more ...
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Moldenhauer: Ergänzung Stelvio 1.2 C+: Keine Lösung: BP 8.5, 9.0.
10. dxc6ep++ Doppelschach. (2023-05-24, hidden) more ...
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Henrik Juel: the five missing black men were captured by white pawns (exfxgxh, fxgxj), so with Black to move last move was with Ta1 or Ke1, and White may not castle
C+ Popeye 4.61 (2022-11-26)
A.Buchanan: Pleasant White tempo play in both phases (2022-11-27)
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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)
A.Buchanan: Could we shift wPb2 to c2, and eliminate bR or bB? (2022-11-25, hidden) more ...
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A.Buchanan: White pawn caps: axb,dxe,gxf,hxg definite.
Black: fxg and two to resolve c-file. But that may be c&d cross-capture, so in set play last move might have been c6xd5. So I think this problem is cooked. What am I missing? (2022-03-21)
Mario Richter: If Black's last move was c6xd5, how do you get white Bishop a3 out of his cage? (In that case, black pawn b7 never left the b-file). (2022-03-21)
A.Buchanan: I agree Mario thanks (2022-03-21)
Hans-Jürgen Manthey: beide Rochaden und im Satz ep. ist möglich:
R.: 1. ... d7-d5 2. Sa7-b5 b5xTc4 3. Tg4-c4 Dd8-h4 4. c4-c5 Th8-h3 5. c2-c4 c4xDb3 6. Dd3-b3 h3-h2 7. Dd1-d3 h4-h3 8. Sc8-a7 h5-h4 9. Th4-g4 c5-c4 10. Th1-h4 h7-h5 11. h2xSg3 Sf5-g3 12. b3-b4 Sh6-f5 13. Lb4-a3 c7-c5 14. Ld2-b4 g3-g2 15. Lc1-d2 a3-a2 16. a2xSb3 Sc5-b3 17. Lb3-a4 Sa6-c5 18. Lc4-b3 Sb8-a6 19. d2xLe3 Lc5-e3 20. Ld3-c4 Lf8-c5 21. Lf5-d3 a4-a3 22. Lh3-f5 a5-a4 23. Lf1-h3 g4-g3 24. g2xLf3 Lb7-f3 25. Sb6-c8 Lc8-b7 26. Sa4-b6 a7-a5 27. Sc3-a4 b7-b5 28. Sb1-c3 f5xSg4 29. Se5-g4 Sg8-h6 30. Sf3-e5 e7-e6 31. Sg5-f3 f6-f5 32. Sh3-g5 f7-f6 33. Sg1-h3 (2023-02-23)
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hans: 1. Th2xf2 De1xf2#!
1. Th2xh3 0-0#? (Castling illegal)

R: -1. Kf1xSe1 Sg2-e1 -2. Ke1-f1 Sf4-g2+ -3. Tg1-h1 Sd5-f4 -4. Tf1-g1 Se3-d5 -5. Tg1-f1 Sf1-e3 -6. Tg1-h1 Tg2-h2 -7. Th2-h1-g1 Tg1-g2 -8. Th1-h2 g2-g1=T -9 h2-h3 h3xSg2 (2010-06-26)
HENRI: Hans solution has a tipo. Their is no D (=Queen in german) mating. One should read the solution in german : 1. Th2xf2 Ke1xf2#! 1.Th2xh3 0-0#?
or in english : 1. Rh2xf2 Ke1xf2#! 1.Rh2xh3 0-0#? (Castling illegal) (2023-10-25)
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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)
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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)
Moldenhauer: Computerprüfung: C+ Stelvio 1.2 2 Sekunden.
Keine Lösung: BP 16.0, BP 16.5. (2023-05-20, hidden) more ...
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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)
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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)
Henrik Juel: -1.Kg1:S Sf2:0 -2.0-0, 1.Rh8#; -1.. Bb8:S -2.Re7, 1.Sc6#; -1.. cSb6,dSb6,Se3:Q -2.Qc5, 1.Qf8#. Duplicated by P0006277 (2003-10-09, hidden) more ...
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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)
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Moldenhauer: Computerprüfung: C+ Stelvio 2.0 02:29:09 Stunden. (hh:mm:ss) (2023-12-23)
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"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)
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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)
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Hans-Jürgen Manthey: Nachdruck SExpress 1.Mai 1947 S.41 Nr.33 (2022-10-15)
E.Gon: In this line! Where is this mate? After 2.Dh8 Txh8! (2009-03-16, hidden)
E.Gon: Oh, sorry! It was my fault! 1...0-0-0? :) (2009-03-16, hidden)
Henrik Juel: C+ Popeye 4.61 (2022-04-26, hidden) more ...
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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)
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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)
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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:09:39 Minuten. (hh:mm:ss)
Keine Lösung: BP 22.0, BP 22.5. (2023-03-26, hidden) more ...
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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)
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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)
A.Buchanan: Can sBf capture to e-file to be captured directly by wBf rather than capture to g2 to promote? If so, then sBf could capture L on a dark square, meaning that last move could have been c6xb5 (2023-07-11, hidden) more ...
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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)
A.Buchanan: Surely 2. Rd1 works equally well? I don't see a retro reason why White has lost castling rights. I don't think this is AP. What's going on? (2021-11-26, hidden) more ...
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SCHRECKE: Nicht 1. Kb3 0-0-0?? 2. Ka2 Td3 3. Ka1 Txa3#, da Weiß zuletzt mit K oder T gezogen hat! (2023-06-28, hidden) more ...
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Mario Richter: Luboš Kekely (Slovakia) correctly points out that 1. ... Sf5! 2. 0-0-0 Sd6# is only a try and not a solution (last black move must have been by King or Rook, so the queenside castling is illegal).
I have changed the solution accordingly. (2023-03-10)
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SCHRECKE: NL: 1. Dg5,Kh4 gxf3 2. Kh4,Dg5 Sf1# (2023-09-13)
Ladislav Packa: Retro content is not needed, the solution is preserved even without it.
Pg4 Pg2 Sh2 Ke1 Rh1 (5)- Ph5 Kg3 (2) h#2* C+

1...0-0 2.h5-h4 Rf1-f3 #
1.Kg3-h4 Ke1-f2 2.h5*g4 Sh2-f3 # (2023-09-14)
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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)
Henrik Juel: HC+ Popeye 4.61 (2022-11-27, hidden) more ...
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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)
A.Buchanan: It's a bad habit for folk to post corrections without indicating where they were published. 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, the diagram is incorrect, one has no clue where to look in past issues of Die Schwalbe! It turns out that this 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 is bP missing from c7, but I hope you will agree that the real failure is the lack of indication of the true source, which rendered the detection of the innocent typo unnecessarily hard. You're welcome (2024-03-06, hidden)
A.Buchanan: I think the retro-logic works ok, but there is a single cook family. I think it can be fixed by swapping sDb6 & wBc6, and adding sBa4. So NBr5/N1p1p3/1Pqp2p1/2bb4/pp6/n1pP1rP1/1PP1P2k/R3KQ1n. Please confirm that this works (2024-03-06, hidden) more ...
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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)
Henrik Juel: C+ Popeye 4.61 (2023-12-02, hidden) more ...
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Henrik Juel: HC+ Popeye 4.61 (2022-11-28, hidden)
A.Buchanan: This is not a retro. There is no “can’t castle”. This is simple application of castling convention to a forward problem. This is not generally taken to make the problem “retro”. Castling convention is certainly used in many retros, but not here (2022-11-30, hidden)
A.Buchanan: Nice - could be done without twinning, just remove bPh7 for 2 solutions? (2022-11-30, hidden) more ...
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See P0000899 a companion problem.
A.Buchanan: Something odd going on here. There are numerous h#2 in both a&b. PRA not normally found in a problem with single solution for each twin. White has 8 pawns so trivially wRh7 must be original. (2018-10-13)
A.Buchanan: There are three similar problems by Brogi which are all twinned RS rather than PRA, but don't have diagram error. P0003743, P0003746 & P0003747. (2018-10-13)
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
Ladislav Packa: Cooked a) and b):
1...b8S and 2...R:h8# (2021-02-10)
A.Buchanan: I think the composer simply forgot that wPb7 can promote. Most of the cooks come from S promotions, but it's also possible to have QR. I've fixed it, rejigging the captures to make them still add up. I've posted in Discord, as I did with the partner composition, and will create entries for them here. (2022-03-15)
milan: +sLb8 sBa7=sT M.Frelih (2023-12-02)
A.Buchanan: Hi Milan - I don't think your suggestion quite works for b). In a) there are 0+2 spare captures, so Black can certainly cross-capture. But in b) there is 1+1 so neither side can cross-capture, so there is no solution. Please compare with P1399806, in which there are 1+2 & 2+1 spare captures, so both twins are sound. (2023-12-03)
milan: Hi Andrew my correction works only with 2.1... solutions, black or white knights on a5. are not important. (2023-12-03)
A.Buchanan: Hi Milan not really clear what you are doing, but if as well as the piece changes you proposed, you also change the stipulation to 2.1... then there is still only one solution. Even if you remove Sa5 entirely as well, there is no White cross-capture possible. (2023-12-04)
milan: Hi Andrew my correction works only with 2.1... solutions, black or white knights on a5 are not important. (2023-12-03, hidden) more ...
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Henrik Juel: The key R: 1.0-0 'threatens' with white retrostalemate, even though White seems to have many pawn retractions available
All missing men were captured by pawns (and White promoted on h8)
R: 1... Kd7-c8? 2.d5xLc6+ Tb8-d8 3.b4-b5 Ke8-d7 4.e4xd5 Ld7-c6 5.f3xe4 Lc8-d7 and now White is retrostalemate
not 6.g2xf3 because of [Lf1]
not 6.d2-d3 because of [Lc1]
not 6.a3xLb4 because of [Ta1]
and not 6.b3-b4 because [Lf8] was captured on a dark square
R: 1... 0-0-0 handles Td8, but it also fixes the black king, so Dd4 must retract to d8 before d7xTf6 can be retracted, but there is still just time enough:
R: 2.b4-b5 Dd8-d4 3.d5xLc6 Ld7-c6 4.e4xd5 Lc8-d7 5.f3xe4 d7xTe6
The rest is easy: Retract Te6 to a1, a3xLb4, Lb4 to f8, g7xLf6, Lf6 to c1, d2-d3, and now the road towards h7 is free for Pc2 (2023-04-08)
A.Buchanan: Great! So which Typ is this? (2023-04-08)
Henrik Juel: Any type, there are no uncaptures in the solution, so anything goes (2023-04-09)
A.Buchanan: Ok I see - the sequence of retro moves is not VRZ play, but history of the game. After the key, there is no choice for either player until wPe4xd5. But isn’t there some Typ where black can checkmate too? R: 1. 0-0 0-0-0 then c1=D#! (2023-04-09)
Henrik Juel: A black checkmate is a possibility in the tries of defensive retractors, regardless of type
When Black has completed a retraction, he has the right to mate White with a forward move, if this is possible
I should have added the testing of mating in my general story about Høeg retractors
1. White chooses a man and moves it back
2. Black chooses which man (if any) to supplement on the abandoned square
Now the white retraction is completed, and White may mate with a forward move, if this is possible
If so, a solution has been found
If not
3. Black chooses a man and moves it back
4. White chooses which man (if any) to supplement on the abandoned square
Now the black retraction is completed, and Black may mate with a forward move, if this is possible
If so, a try has been found
If not, go to step 1. (2023-04-09)
A.Buchanan: Thanks - so does that mean that the solution given here is just a try? (2023-04-09)
Henrik Juel: No, Andrew, in the solution given here White mates, so it is a solution
A try requires Black to mate (2023-04-09)
A.Buchanan: Sorry I am apparently being slow: isn't R: 1. 0-0 0-0-0, dann c1=D#! a mate for Black, so White never gets to retract further? (2023-04-09)
Henrik Juel: You are not slow, Andrew, but I never really saw the black mate in your first 04-09 comment, sorry
R: 1.0-0 0-0-0 2.b4-b5, then 1.Th8# is the intended solution (and not an intended try), but it fails because following R: 1.0-0 0-0-0, Black mates with 1.c1=D#, so you have cooked the problem
It is easily repaired by adding the condition 'Ohne Vorwärtsverteidigung' (without forward defense), but maybe the author implied this condition (or maybe he never saw your black mate) (2023-04-09)
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Henrik Juel: 1.0-0-0 (2.Txd7#). -1... Lh2 -2.Lg1, Sh5 uncaptures wPa6 and unpromotes on h1, then retract h2xg3, Lf2 via f4 to c1, d2xe3, Lg1 to f8, g7xf6 etc. (2004-01-12)
Henrik Juel: C+ Popeye 4.61 (2023-07-14)
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Henrik Juel: C+ Popeye 4.61 after analysis (2023-07-20, hidden) more ...
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A.Buchanan: A key feature of adversarial A Posteriori is that any castling must be forced in a finite number of moves (but not necessarily limited by the number of moves in the stipulation goal). If the other side can prevaricate indefinitely, then that is sufficient to defeat the A Posteriori "steal" (2022-02-16)
A.Buchanan: Why this would be "PRA"? Maybe the idea is that we don't know who is first to move, yet whoever it is, White wins. But that only applies to "pull" scenarios such as this, where Black snatches the move because otherwise the game is lost. In other situations where White to avoid loss must "push" the move, then there is no way this can be described as PRA. The fundamental push/pull thing has a unity, and I don't think it's helpful to use "PRA" which only describes half of this, and was really designed for a different context. Strategically, these push/pull adversarial battles are amongst the most interesting AP problems. (2023-07-22)
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Moldenhauer: Computerprüfung: C+ Stelvio NUPG cooked 1 Sekunde.
Keine Lösung: BP 23.5, BP 24.0.
Wenn Schwarz seinen 25. b3 spielt ist kein Matt in 2 möglich.
Wenn angenommen wird das Weiß aus dieser Stellung nochmals
ziehen darf dann Matt in 2 Zügen.
Notation: 1.Sc3 a5 2.Se4 Sa6 3.b4 a4 4.b5 a3 5.d4 b6 6.d5 Lb7 7.d6 exd6
8.Lf4 Dh4 9.bxa6 Dh3 10.c4 Sf6 11.Sxf6+ gxf6 12.c5 Lf3 13.exf3 Lh6
14.Lb5 c6 15.Kf1 cxb5 16.Se2 dxc5 17.Ld6 Ld2 18.Tg1 La5 19.gxh3 h6
20.Tg5 hxg5 21.Kg2 Th4 22.Db1 Ta4 23.Db4 cxb4 24.Sf4 gxf4 25.Tg1 (2023-04-06)
A.Buchanan: Should the stip be something like:
Position after White’s 25th move.
Can Black play to avoid being mated in 2?
But both b3 and Rc8 seem to let Black escape (2023-04-06)
A.Buchanan: The existence of mate in 2 is not contingent upon castling, but the soundness is. Was the stipulation really worded like this? (2012-04-07, hidden) more ...
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Moldenhauer: Ergänzung: Stelvio 1.2. Keine Lösung BP 15.0, BP 15.5. (2023-05-20)
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Moldenhauer: Computerprüfung: C+ Stelvio 1.2 03:15:11 Stunden. (hh:mm:ss)
C+ Euclide 1.11 01:00:47 Stunde.
Stelvio und Euclide finden nur 1 Lösung! Warum cooked?
Keine Lösung: BP12.5, BP 13.0.
Notation: 1.a4 Sf6 2.a5 Tg8 3.a6 bxa6 4.Txa6 Lb7 5.Txf6 a5 6.Txf7 Sa6
7.Txf8+ Kxf8 8.h4 De8 9.h5 Dxh5 10.f3 Dxf3 11.Sh3 Da3 12.e3 Te8
13.Lxa6 Lc8 14.0–0+#
Epaulettenmatt durch Rochade ist OK. (2023-05-28)
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Moldenhauer: Computerprüfung: C+ Stelvio 2.1 ca. 30 Minuten. 3S/12P 64 GB Hash.
Der Cook ist keine Cook weil die Endstellung nicht mit dem Diagramm
zusammenstimmt. Letzter Zug Lf3?
Nur die weiße kleine Rochade fand wirklich statt.
Keine Lösung: BP 19.5. (2024-03-24)
A.Buchanan: Please forgive me for being stubborn, but I don't think just checking BP20.0 & BP19.5 is sufficient for C+. The test must include BP19.0 & BP18.5, and these should be very quick to test compared to the first two. They can be done first, which would allow for some short cooks to be identified more quickly. Maybe Stelvio could offer an option for this "fourfold" test, to save on typing, which is the main inconvenience? (2024-03-25)
Moldenhauer: Hi Andrew, diese Frage hatten wir schon einmal. Damals antwortete Reto darauf.
In der Dokumentation zu Stelvio 2.1, Seite 6 unten und Seite 7 oben, steht die Erklärung. (2024-03-25)
A.Buchanan: Thanks Moldenhauer. Yes, Reto responded today to say: "Actually 20 and 19.5 are sufficient, Stelvio always checks for any solutions of same parity, so 20 includes 19, 18, ... and 19.5 includes 18.5, 17.5... I can see your point with adding an option for automatic double checking, I've added it to my list."
That's great, so in my opinion the PDB C+ report needs to indicate that Stelvio has also tested for the shorter solutions of the same parity, as this is tool-dependent behaviour which future readers might not know. Ideally this statement would appear at the end of the Stelvio output, as a kind of "certificate" that could be copied to PDB.
A problem with this whole issue is that there are different choices being made by different people. It complicates things unnecessarily. I think C+ for a PG should mean that the three shorter lengths have also been explored. This should be doable without getting into the fraught question of whether a shorter solution cooks a PG, although in my opinion it does, whatever the parity: a PG is just a diagram. Conceptually to me, PG in x.y is just SPG, with a courtesy hint to the solver of what length they are looking for. (2024-03-26)
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VL: Published in "Probl. Pribuzh." (or "Problemy Pribusch'ja"?),
1990, No.1, #12. An obscure Russian language chess problem
magazine issued in Nikolaev, the Ukraine (the South Bug
riverside region). (2006-01-27)
A.Buchanan: Lovely problem, but this is no more PRA than it is duplex. Rather, the push/pull of AP Type Keym gives a forfeit which must be discharged. (2023-07-22)
Andrey Zhuravlev: The correct name of this magazine is "Problemist Pribuzhya" (2006-04-10, hidden) more ...
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Vladimir Rey: C+ Natch 3.2 - Solution found: 1. Positions analysed: 51578. Resolution time: 5 h 33 mn 32.84 s. (2020-05-08)
Vladimir Rey: Check please: Keywords Tempo Loss. (2022-09-05)
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paul: Stelvio 2.0 found the solution in 2 minutes. (2024-04-02, hidden) more ...
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paul: Checked with Euclide for the first 19 moves (~45 min) (2014-07-04)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:02:47 Minuten. (hh:mm:ss)
Keine Lösung: BP 23.0, BP 23.5. (2023-04-01, hidden) more ...
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Henrik Juel: 0... fxe3ep 1.d8S Kxd5 2.0-0-0#. Not -1.f2? and Lg1 is caught. (2004-09-16)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
Henrik Juel: In principle I agree, Vaclav
But PDB does not (yet) allow HC+, so I find it acceptable to use the C+ label, when you also tell the whole story after 'Computer test:' below (2023-08-03)
A.Buchanan: Hopefully Gerd will have more time at some point, and can expand the functionality in this and other areas. In the meantime, engine solving of conventional retros including AP, is in its infancy. Retractor 2 has some effectiveness, but is still basic. And there is nothing that yet grasps the intricacies of castling/ep etc. However AP problems do often contain considerable forward chess, and the C+ tag is very useful to filter out those that have already been solved forwardly, without pretending that these are in any sense fully solved (2023-08-04)
Ladislav Packa: I know the definition of AP, but I don't understand the logic behind it. The move 1.fxe3 e.p. proves that White CAN castling. But the solution (2. ...Rd1#?) claims that castling is MANDATORY. From my point of view, AP is correct when only castling is necessary for the solution and the Rook move would be a dual. (2023-08-04)
A.Buchanan: @Ladislav: I am not sure how to help you. Maybe you can read this page from Retro Corner: https://www.janko.at/Retros/Glossary/APosteriori.htm (2023-08-05)
Ladislav Packa: Andrew, what should the article help me with? I quote the final sentence:
Some people still oppose this rule and argue that it should certainly not be the default convention. (2023-08-05)
Henrik Juel: You could view it this way, Ladislav
h#2 means that it is Black to move, so White made last move
What was last move? A little analysis shows just three possibilities: f2-f4, move by Ta1, or move by Ke1
So normally we cannot assume that last move was f2-f4
But if White can castle, then the last move was f2-f4
So if we could start with 0... 0-0-0, then 1.fxe3ep would be legitimate
AP says that you are allowed to reverse the sequence of events; first do the ep capture, then later legitimize it by castling
Was this helpful? (2023-08-05)
Ladislav Packa: Henrik, you don't have to explain that to me. I've done a few AP issues myself, like P1348357. But that doesn't mean I agree with AP's logic. I already wrote it - the term "you can castle" is applied as "you must castle". But these are only problems where, in addition to castling, the Rook move can also be used, I consider that a dual.
From that point of view, the P1000662 issue is perfectly fine for me. (2023-08-05)
Joost de Heer: AP: By castling, you prove a posteriori that the ep-capture was not just a try but the actual solution. Without castling, the ep-solution just is that: a try.
So: Try 1. fe3 ep e8=S 2. Kd5 Rd1 - but ep capture not allowed, as there is no proof that f2-f4 must've been the last move.
Solution 1. fe3 ep e8=S 2. Kd5 OOO - Now the ep capture was justified because white castled, thereby proving that the last move before the diagram position indeed was f2-f4. (2023-08-06)
Joost de Heer: See e.g. P1052919 : The try is an ep capture which is unjustified. (2023-08-06)
A.Buchanan: OK Ladislav: I think I get your point. If one solution with castling justifies the e.p., then based on that certainty, why shouldn't an alternative solution with no castling *then* be allowed as well? There are problems in which one twin shows 0-0-0 and the other shows 0-0. Each is based on the other in a similar way, so the idea of dependency is not new. Why are we not allowed to add other "parasitic" solutions as well? Why can we only have the "paying" solutions? We can't say that we are restricted to one solution: that's not the way chess problems operate! And this is just in the help world - in the adversarial world it might get even more complicated. Is this your issue, Ladislav? (2023-08-07)
Ladislav Packa: I don't want to unnecessarily prolong this discussion. However, I will add one more note: in this position, white castling is also possible without e.p. in Black's 1st move. If B1 were an indifferent move, then white can 1...0-0-0! The Codex of Chess Composition writes about it in Article 16 (1):
Castling convention. Casting is permitted unless it can be proven that it is not permissible.
In our case, 0-0-0 is possible because White's last move exists - e2-e4! It does not matter if it is this move or some a2-a3, both moves are equivalent. EP does not prove the possibility of casting, it would be legal even without it. (2023-08-07)
A.Buchanan: Ladislav was what I wrote your issue pls? Y/N :-) (2023-08-07)
Ladislav Packa: I have no problem, I'm just expressing my own opinion about the AP convention. (2023-08-07)
A.Buchanan: OK cos I think the point I raised is a real one that should be addressed by theory some day. Clearly from the nice problem that you composed Ladislav you understand the mechanics very well. From a justification perspective it's all a bit iffy, but that's why it's controversial. Under RS it's really the only way one can end up actually eping, and it's proved compositionally fertile. So that's enough to justify (2023-08-08)
Ladislav Packa: No need to apologize. This is a normal discussion with different views on the issue. Maybe it will come to some conclusion.
I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p. (2023-08-08)
Joost de Heer: "I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p."
You misinterpret AP. The e.p. capture does not authorize castling, castling provides a justification later on (hence the 'a posteriori') for the legality of ep.
Usually, for ep justification you need to examine all game trees that lead to the diagram, and only if all game trees end with the double-step, then ep is allowed.
With AP, you examine all the game trees including the actual play. If all those game trees have as last move before the diagram position the double step, then ep is possible.
In this case, if white doesn't castle, then there are game trees which don't have as last move the double step, and therefore AP logic dictates that the ep capture was illegal. However, all game trees which lead to the diagram and which have castling in the actual play have as last move before the diagram position the double step, hence AP dictates that the ep capture is legal. (2023-08-09)
Ladislav Packa: Joost: A simple question - is white allowed to castle after any 1st move by black (except e.p.)? (2023-08-09)
Joost de Heer: Of course he is. AP only is used to combine the ep justification with castling, not the castling right per se. (2023-08-09)
A.Buchanan: I agree with both. Also "cooked" is closer to "unsound". And we choose not to have "C+ cooked", which might legitimately mean "C-" (2023-08-04, hidden) more ...
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Anton Baumann: mögliche Korrektur: wBb4 nach b3, +sBd5, -wBg2
1.Sf7 Db4 2.0-0-0 Db7# (C+) (2024-01-03)
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Rochade
F135
Verst-4
Yuri Bilokin: correction bRh7-f7(-bQf7), bBa1-b6, -bPa7, +bSf4, +bSh6 r3k3/5r1B/1b2b2n/4Q3/5n2/8/8/3K4 (3+7) (2022-08-28)
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Eric Huber: Anticipated by 'P0534152' Erkki Heinonen, 3. Makuc-Moder-Gedenkturnier 1971-1973, 4. ehrende Erwähnung. (2008-10-01)
Viktoras Paliulionis: This problem shows paradoxical idea - two mutually exclusive tempo moves (B1 & B2). P0534152 has only one tempo move. (2023-10-09)
Eric Huber: Anticipated by http://www.softdecc.com/pdb/search.pdb?expression=PROBID='P0534152' Erkki Heinonen, 3. Makuc-Moder-Gedenkturnier 1971-1973, 4. ehrende Erwähnung. (2008-10-01, hidden) more ...
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in Vratnica von Datiaschwili/Gatseciladse steingetreu nachempfunden
Yuri Bilokin: you can see *) 1...Kc7 2.Rf8 Re7# (2022-09-11)
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Yuri Bilokin: more economical -wPf6, -wPg4, -bPa4, bBg6-f5, bPf7-d7 8/4p3/2R1p1p1/2P1Pb2/1Nppk3/2p3p1/rr2n1P1/R3K3 (7+12) h#3 2.1…
1.Sg1 0-0-0 2.Sf3 Sc2 3.Sxe5 Rxd4#
1.Sf4 Rd6 2.Sd5 Sc6 3.Se3 Rxd4#

AntiZielElement (B2, line obstruction)
Castling (white, long). Check prevention (W-W)
Exchange of functions (wRa1/wRc6, Mate / Passive)
Long-trip (bS, 3) × 2.Play on the same square (W3, 2)
Mates on the same square × 2 (2023-08-23)
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Auszeichnung wurde in Problemista 04/1964 annulliert.
Yuri Bilokin: correction wBc8-e8, bKb3-d4, bBf8-e3, bPc7-f7, +bQa6, -bPd3 4B3/5p2/q7/8/3k4/4b3/8/R3K3 (3+4) h#3

1.Kc3 Bb5 2.Bc5 0-0-0 3.Bb4 Rd3#

Castling (white, long) (2023-06-10)
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SCHRECKE: NL in b):
z.B.: 1. Ke7 Tb1 2. Ke6 d3 3. Kd5 dxe4+ 4. Kc4 b3# (2022-11-14)
HBae: Mögliche Korrektur sBd6 nach d3, ohne wBe2 (9+10 (2022-11-14)
HBae: Mit der (berichtigten) oben genannten Korrektur ist das Problem C+ gemäß Popeye v4.37. Die lange weiße Rochade ist nur in a) erlaubt. (2022-11-14)
HBae: Mögliche Korrektur sBd6 nach d3, ohne wBd2 (9+10) (2022-11-14, hidden) more ...
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Yuri Bilokin: you can see 1...Bh4 2.Se7 Be1 3.0-0-0 Ba5 4.Re8 Kb6 5.Kd8 Kb7#
1...Bh4 2.Se7 Be1 3.0-0-0 Kb6 4.Re8 Ba5 5.Kd8 Kb7#
1...Bh4 2.Se7 Kb6 3.0-0-0 Be1 4.Re8 Ba5 5.Kd8 Kb7#
1...Ba5 2.Se7 Be1 3.0-0-0 Ba5 4.Re8 Kb6 5.Kd8 Kb7# (2024-04-03)
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Yuri Bilokin: possibly -wPa2, -wPb6, wPh2-h3, bR7-a3, +bQf7, +bPb5, +bPb7, +bPg3 4n3/1p3q2/4r1P1/1p2k3/b7/r5pP/6P1/R3K2R (6+9) h+3 2.1…

1.Rb3 0-0-0 2.Kf6 Rhf1+ 3.Ke7 Rxf7#
1.Qc7 0-0 2.Kd6 Rad1+ 3.Ke7 Rf7#

AntiZielElement (B1, guard). Castling (white, long). Castling (white, short)
Check prevention (W-W) × 2. Delayed Umnov (bQ-wR)
Hideaway (bQ). Many-ways (bK, 2)
Epaulette mate × 2 (2023-08-23)
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klären autor?
Yuri Bilokin: author Yuri Bilikin

Castling (black, long) × 3 (2024-03-18)
Yuri Bilokin: slip of the pen, author Yuri Bilokin (2024-03-21)
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Dual in b)
Yuri Bilokin: possibly bRe8(-bBe8), +bNc7, +bPg6 4r1rn/2n1pk2/6p1/7N/6P1/6pp/R3K2R/8 (5+9) h#2 b) alles eine Reihe tiefer
a) 1.g2 Ra6 2.g1=R Rf2#
b) 1.Ke5 0-0-0 2.Ke4 Rhe1# (MM) (2022-12-16)
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klären: welches ist die NL?
Bernd Schwarzkopf: Ein Heft „Die Schwalbe 10-12/1968“ gibt es nicht.
Das Problem ist in „Die Schwalbe 10-11/1968“ oder „12/1968“ nicht zu finden.
Vielleicht war P0544925 gemeint. (2023-04-26)
Henrik Juel: The cook is easily removed by adding '2 solutions' to the stipulation
1.Kb3 0-0-0 etc. are tries, because White just moved Ta1 or Ke1, so he may not castle (2023-04-26)
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SCHRECKE: Braucht Weiß einen letzten Zug? Dann kann den nur der wTurm ausgeführt haben! (2023-01-25)
Henrik Juel: Good point, Michael
If this were a retro problem, there would be no solutions
But here you can also view it as a sequence of black moves, ending with DxPf5, say (2023-01-25)
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Henrik Juel: The paradox is explained by the need for Black to play De1xLc1
Natch 3.1 made no progress after an hour (2018-12-14)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 14:25:47 Stunden.
Keine Lösung: BP 14.0, BP 14.5. (2023-02-20)
TBr: In solution read 11.Taf1 (2008-09-20, hidden) more ...
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Henrik Juel: Natch 3.1 produced nothing in more than 3 hours (2018-12-12)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 01:02:15 Stunden.
Keine Lösung: BP 15.0, BP 15.5. Notation OK. (2023-02-18)
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Henrik Juel: Natch 3.1 did nothing in almost an hour (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 84:43:01 Stunden. (hh:mm:ss)
Keine Lösung: BP 22.0, BP 22.5. (2023-04-01, hidden) more ...
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Henrik Juel: typos in solution, should read: 5.Kg2 Kd6 6.Kf3 Sh4+ 7.Ke3
Fake castling, the hard way in 12 moves
First 21.5 moves found OK by Natch 3.1 (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 09:09:13 Stunden. (hh:mm:ss)
Keine Lösung: BP 21.0, BP 21.5. (2023-03-12, hidden) more ...
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Kommentare:
Als einziger letzter schwarzer Zug, der die weiße Rochade
erhält, kommt nur Bh7-h5 infrage, was dem Weißen den e.p.-Schlag
ermöglicht, um nicht die Rochade oder das Mattnetz zerstören zu
müssen (H.P.Suwe)
Gianni Donati: This intends to show the Valladao theme in the minimum number of moves. (2004-03-19)
VL: The waiting ep capture. Cf. also P1000260
(by T.Petrovic, 2000) somewhat enriched
thematically with the illegal try 2... Rd1#??. (2004-06-03)
A.Buchanan: The “illegal try” 2. Rd1+ is not actually mate because c2 is unprotected. This is kind of “logic dual” spoils the A.P. motivation for the castling (2020-05-20)
Henrik Juel: White captured h6xg7 and once more, e.g. axPb-b8=Y
Black captured cxdxe, dxexf, fxg, and once more, e.g. a2xb1=Y
Possible retroplay 1... h7-h5 2.h6xSg7 Df8-f4 3.h5-h6 Sf4-h3 4.Sh3-g1 etc., preserving the castling right
Any other black last move would force White to retract Ta1, as d2xc3, d2xe3, and h6xh7 would be illegal retractions
I agree with Andrew that the double motivation of 2... 000# is a weakness:
a. legitimizing h7-h5
b. mating (2020-05-21)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
A.Buchanan: See discussion at P0009121 (2023-08-04)
A.Buchanan: Maybe can replace wPf2 with wR, wRf1 with wB & shift bPg3 to g2? (2020-05-20, hidden) more ...
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Henrik Juel: Natch 3.1 did nothing in half an hour (2018-12-14)
paul: Correction of P1000629 (2022-10-12)
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AB: First proof of "can castle", without recourse to conventions. (2009-11-22)
A.Buchanan: deadpos doesn't understand about retro logic yet, so does not eliminate R: 1. Ke4xLSBe3? (2022-11-13)
AB: typo in diagram: Be6 is f5. (2001-12-29, hidden) more ...
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Henrik Juel: Natch 3.3 just needed less than a second to test this proof game (2023-06-12)
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VL: One solution consists of two partial ones, and there is another, castling independent, solution. A rare scenario; cf. P1091926. Presently, due to Art.16 (3) of the Codex, the mark "RV" may be deleted from the stipulation. (2017-11-25)
A.Buchanan: Methodologically, I think that one applies PRA first. So there are two parts, and each has two solutions. One solution is shared across both parts (2023-08-22)
A.Buchanan: I think "RV" used to mean *all* retro variants considered as separate histories, but since Werner Keym's revision, its meaning has been changed to be synonymous with PRA. The interpretation of this problem is now considerably less smooth than it was before. I would say that normally we must split into PRA parts before considering the number of solutions. So for me there are 2 parts, and each has 2 solutions, one of which is shared across the two parts, but that is a different way of fiddling around from VL's idea.
At what point was "2.1..." added to the stipulation? Of course, the problem works as PRA somehow, but I don't think this is a justification for changing the meaning of this term. And without an "originalforderung" field, we have no way to track edits to the stipulation.
I think this is a case for Golden Age: to preserve this old problem in its original form. (2023-08-21, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Stelvio 1.0 in 3 Sekunden.
Keine Lösung: BP 24.0, BP 24.5. (2023-02-14, hidden) more ...
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Henrik Juel: Natch 3.1 did little in an hour (2018-12-18, hidden)
Joost de Heer: C+ Stelvio 1.2 in about 30 minutes. (2023-05-10, hidden) more ...
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