Adrian Storisteanu: Possible fix: +bPc5. (2015-06-10)
Miguel Ambrona: The fix proposed by Adrian works.
Verified with Deadpos v.2.2. (2024-01-12)
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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: NL: R: 1.Ke8-d8 g7-g6 V: 1.Th1..6 Le7#
Korrektur in 'Die Schwalbe' 06/1989 S.78: +sBh7 (2022-12-30)
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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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vergl. P0004915 (Hans Gruber, Schach 1979)
Brassaud: La solution proposée 1/Tg6# est possible
Mais il y a aussi le rétro jeu -1) Fa2-b1, Rg5g6 -2) Ta4-a5+, Rf4-f5 etc … et avec le trait aux noirs : 1) Fxe6 # est possible (2017-08-30)
A.Buchanan: @Brassaud: yes I agree. There is no reason why White should not have moved last. So both players can mate, but part (b) implies that the intended solution in (a) is 1 player. If the published stipulation for (a) was maybe just "#1", which by default is white to move, then there is a unique solution.
For (b) I am wondering about +sBg6, which would also stop the en passant trick, both by blocking sK from retreating there and also by locking sL in an impossible cage with sBf7. (2017-08-31)
Henrik Juel: Adding a black pawn on g6 of course prevents a black last move by Kf6, but it allows f7xg6 as last move; Lg8 is not locked, because Ph7 is white (2017-08-31)
A.Buchanan: Yes (2017-08-31)
Anton Baumann: vergl. P0004915 (Hans Gruber, Schach 1979) (2023-01-03)
Brassaud: Pour la position
a) il y a la possibilité 1) Tg6 # après 1)… Rg6xe6 ep , mais il y a également le rétro très simple :
1) Fa2-b1, Rg5g6 2) Ta4-a5+, Rf4-f5 etc … et 1) Fxe6 # est possible
b) 1) Tg6 # d’accord. (2015-09-14, hidden) more ...
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 29:20:32 Stunden. (hh:mm:ss)
Da NUPG C+
Notation: 1.b4 e5 2.d4 Le7 3.d5 f5 4.d6 h5 5.dxe7 d5 6.b5 Dd6 7.b6 Kd7 8.bxa7 b5
9.c4 b4 10.c5 d4 11.cxd6 c5 12.e8=T Sc6 13.Tf8 c4 14.e4 d3 15.Dg4 d2+ 16.Ke2 c3
17.Kd3 d1=L 18.Sd2 fxg4 19.Tf4 exf4 20.e5 Sd4 21.e6+ Kc6 22.e7 b3 23.e8=T b2
24.Te6 Lb3 25.Th6 Lf7 26.d7+ gxh6 27.d8=T Lce6 28.Td5 Te8 29.Kc4 b1=L 30.a8=T Lh7
31.Ta3 c2 32.Tg3 Se7 33.a4 Thf8 34.a5 Sg6 35.a6 Sh8 36.a7 fxg3 37.a8=T gxh2
38.T8a3 Lhg8 39.Th3 gxh3 40.Se4 Lxd5+
3 schwarze Läufer stehen zum Schluss auf den weißen Felder d5, f7, g8. (2023-09-21)
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Anton Baumann: 'Die Schwalbe' 08/1984 S.308: der Autor korrigiert: wBe5 statt sBe5 (2022-12-23)
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James Malcom: (Shortest) proof of legality: 1. d4 Nh6 2. Bxh6 gxh6 3. g4 Rg8 4. g5 Rg6 5. Nh3 Rf6 6. Nf4 Nc6 7. g6 Nb4 8. g7 Rg6 9. Nxg6 fxg6 10. Bh3 Nd5 11. Be6 dxe6 12. f4 Kd7 13. f5 Kc6 14. f6 Bd7 15. Rf1 Be8 16. a4 Bf7 17. a5 Bg8 18. f7 Qd6 19. Rf6 exf6 20. e4 Be7 21. e5 Bd8 22. exd6 Ne7 23. dxe7 Rc8 24. d5+ Kb5 25. d6 Kc6 26. d7 Kb5 27. Qd6 cxd6 28. a6 Kc6 29. c4 Kb6 30. c5+ Kb5 31. c6 Kc5 32. c7 Kb5 33. Na3+ Kc6 34. Nc4 Kb5 35. Nb6
axb6 36. Ra5+ Kb4 37. Rc5 bxc5 38. a7 Kb5 39. b4 Kc4 40. b5 Kd4 41. b6 Ke4 42. Kd2 Ke5 43. Kc3 Kd5 44. a8=B Kc6 45. Kc4 Rb8 46. c8=R+ Bc7 47. d8=N+ Kd7 48. e8=B+ Ke7 49. f8=Q#

Took me a bit to figure out the trick for maneuvering the Black bishops. (2022-08-28)
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Kees: possible fix: Lb1=Sb1 -De1 +Ld1
White begins: 1.Kxb7 Lxe2 2.Kc8 La6#
(1.Txd8+ Kxd8 2.Kf8 Th8# illegal for white has no last move) (2023-06-07)
A.Buchanan: Your fix is good, Kees. It removes the cook, and sLd1 denies R: 1. c2xb3 as well as sLb1 did. Note R: 1 Sc6-d8 0-0+? as black castling rights were lost to let wK enter the back rank. (2023-06-08)
A.Buchanan: From a forward angle, your solution is cool Kees, addressing the cook in one half of the duplex. And sLd1 denies R: 1. c2xb3 as well as sLb1 did.
Note R: 1 Sc6-d8 0-0+? as black castling rights were lost to let wK enter the back rank. (2023-06-07, hidden) more ...
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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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Moldenhauer: Computerprüfung: C+ NUPG cooked Stelvio 1.11 in 1 Sekunde.
Keine Lösung: BP 17.0. BP 17.5 cooked.
Beispiel BP 18.0: 1.d4 Sa6 2.Kd2 Sb8 3.Kc3 d6 4.e4 Le6 5.Se2 Sd7 6.e5 Db8 7.g4 Kd8
8.g5 f5 9.g6 Lf7 10.gxf7 Kc8 11.Kc4 g5 12.Kd5 Lg7 13.Ke6 d5 14.f8L g4 15.Kf7 g3
16.Ke8 g2 17.Lg5 gxf1L 18.e6 Lh3
Beispiel BP 17.5: 1.d4 Sa6 2.Kd2 Sc5 3.Ke3 d5 4.Kf4 Le6 5.Ke5 Kd7 6.e4 Db8
7.g4 Kc8 8.g5 f5 9.g6 Lf7 10.gxf7 Sd7+ 11.Ke6 g5 12.Se2 Lg7 13.e5 g4 14.Lg5 g3
15.f8L g2 16.Kf7 gxf1L 17.Ke8 Lh3 18.e6 (2023-05-07)
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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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Henrik Juel: The black men have made an even number of moves, so the white men (ending with Sf6+) have made an odd number of moves; hence [Ta1] has made an odd number of moves and was captured on b1; the fastest way of doing this is to let [Sb8] do all the black moves, incl. 5... SxTb1 and 10... Sb3-a1 (2023-04-20)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 KBP=10.5 in 1 Sekunde.
Keine Lösung: BP 9.5, BP 10.0.
Beispiel: 1.Sa3 Sa6 2.Tb1 Sb4 3.Ta1 Sd5 4.Tb1 Sc3 5.Sb5 Sxb1 6.Sa3 Sc3
7.Sb1 Sa4 8.Sf3 Sc5 9.Se5 Sb3 10.Sg4 Sa1 11.Sf6+ (2023-04-19, hidden) more ...
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Mario Richter: In der Lösungsbesprechung wurde die Forderung präzisiert: Wieviele (und welche) Steine zogen in der KBP mindestens?
Die richtige Antwort ist: 6 (sSb8, wSb1, wDd1, wKe1, wSg1, wTh1).
Die kürzeste BP braucht 12 Züge von Schwarz (Sb8xLc1-f3xTg1, dann zurück nach b8), wK und wSS können nur gerade Anzahlen von Zügen machen, wegen Th1-g1 muß also wD oder wTa1 ein Tempo verlieren. In Rahmen des 12-Züge-Limits schafft das nur die wD. (2010-05-24)
Moldenhauer: Computerprüfung: C+ NUPG cooked Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 11.0, BP 11.5. (2023-04-06)
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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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Henrik Juel: It is illegal for Black to supplement anything on b6, because [Ta8] was captured in its corner and the other missing black men were captured by white pawns (2016-03-28)
Henrik Juel: ... as wLb3 is a pawn promoted on e8 or g8
Nice type Høeg defensive retractor
Here are some other explanatory comments
In retraction 1 White chooses to move his king back to h3; Black could choose to supplement a black man on g3 (or nothing), but supplementing a pawn is the only way to maintain legality (Kh3 stands in double check from Lc8 and Dh8); again moving Pg3 back to h3 and White supplementing a pawn on g4 is forced (this e.p. case is the only one where the supplementing does no happen on the abandoned square)
In retraction 2 the white retraction is forced, and then moving Kd6 back to d7 to uncheck is illegal because of the double check from Sb6 and Pc6, so Black must uncheck by moving Kd6 back to e6 and White choose to supplement a pawn on the abandoned square
In retraction 3 White chooses to move Pd6 back to e5, forcing another e.p. situation (2023-04-08)
Henrik Juel: The Proca type is easy to define: White and Black alternate retractions, until White can mate with a forward move
The Høeg type is usually defined the same way, except that the other side decides which man (if any) was captured; but this can be detailed as follows:
1. White chooses a man and 'moves it back'
2. Black chooses which man (if any) to 'supplement' on the abandoned square
(only now is the white retraction complete)
3. Black chooses a man and 'moves it back'
4. White chooses which man (if any) to 'supplement' on the abandoned square
(only now is the black retraction complete)
etc. etc. until, following a white retraction, White can mate with a forward move
In tries, Black can ruin the white plan by mating White with a forward move after a black retraction
It goes without saying that the resulting retractions must be legal
'supplement' is my (poor) translation of the danish term 'supplere'; maybe 'add' would be better
'the abandoned square' needs a special interpretation in the e.p. case, which happens twice in this problem
These details may be the cause why new type Høeg defensive retractors are rarely seen, as type Proca is more natural and straightforward (2023-04-08)
A.Buchanan: Thanks Henrik. Yesterday, I went through all the defensive retractors to clear up keywords & genres. There were a very few where the stip did not specify the VRZ Type, and others where Anticirce did not specify Calvet vs Cheylan. The answers are probably obvious to you, and if you want to comment on those, then I will update the stips & keywords.
A more general question: Typ Friedlich appears to be the German for Type Pacific: can we standardize on one? (2023-04-08)
Henrik Juel: Thanks Andrew for enabling me to post my type Høeg spiel once again
Anticirce without specification usually means that both Calvet and Cheylan work
Friedlich is indeed german for Pacific, and as the PDB is a german product, I guess we must live with the present conditions (2023-04-08)
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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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Gerald Ettl: 1.h8D# (2023-04-03)
Gerald Ettl: die Stellung loest sich auf indem der sK nach b1 - g8 wandert und dann Sg5 weg zieht. Der wK kommt ueber g5 raus. (2023-04-03)
Michel Caillaud: The original stipulation is #1 (durch wen?).
As wPe2 cannot be captured on its file, the 4 white captures for 4 Pawns to promote to Knights are (a2)xb3, (e2)xf3 and (f2)xg3 2 times, and wPb2 was captured on its file by (Dd8)xb6.
As b3-b2 (before (a2xb3)) and c7-c6 (before (Dd8)xb6) cannot be immediately retracted, only bK and wSs can play the last moves.
As indicated by Gerald, bK has to go to g8 to unlock the position, freeing bSg5.
When Ka3-a2 is retracted, previous white move places the 6 white Knights on black squares; the resulting Retro-Opposition implies that black is to play in the diagram position.
1.b1S#! (1g8D#?) (2023-04-04)
Gerald Ettl: Danke Michel fuer Dein Erklärung.
Ich löse so auf, dass Schwarz am Rückzug ist:
R: 1.Kc1b1 Se4d6 2.Bb2b3 Sc5d3 3.Kb1a2 Sd3c1 4.Ka2b1 Sh1f2 5.Kb1a2 Sa4c5 6.Bb3b4 Sf5e3 7.Ka2a3 La1d4 8.Ka3a4 Sf2h1 9.Ka4a5 Sh1f2 10.Ka5b6 Sf2h1 11.Kb6c7 Sh1f2 12.Kc7d8 Sf2h1 13.Kd8e8 Sh1f2 14.Ke8f8 Sf2h1 15.Kf8g8 Sh1f2 16.Sg5f3 Kh6g5 17.Sb8a6 Sc5a4 18.Sa6c5 Sd6b5 19.Sc5b3 Sb5c7 20.Sb3a1 Sc7a6 21.Sa1b3 Sa6b8 22.Sb3a1 Sa4b6 23.Sa1b3 Sb6c8 24.Sb3a1 Sb8b7[+wBb7] 25.Sa1b3 Bb7a6[+sLb7] 26.Sb3a1 Sc8c7[+wBc7] 27.Lb7c8 Kg5h4 28.Sa1b3 Tg6g5 29.Sb3a1 Tg5f5 30.Sa1b3 Tf5f4 31.Tg7g5 Sf2h1 32.Tg5b5 Sh1f2 33.Tb5b8 Sf2h1 34.Bb4b5 Sh1f2 35.Bb5b7 Bc7b6[+sLc7] 36.Tb8a8 Tf4e4 37.Lc7f4 Sg3f5 38.Kg8f8 Sf2h3 39.Lf4h6 Sc1e2 40.Kf8e8 Se2g3 41.Lh6f8 Sf5h6 42.Sb3a1 Sh6g8 43.Sa1b3 Sg8g7[+wBg7] 44.Sb3a1 Bg7g6 45.Sa1b3 Bg6f5[+sTg6] 46.Tg6g8 Bf5f4 47.Tg8h8 Sg3f5 48.Sb3a1 Sf5h6 49.Sa1b3 Sh6g8 50.Sb3a1 Sg8g7[+wBg7] 51.Bg4g5 Bg7g6 52.Bf6g7[+wDf6] Lh5g4 53.Sa1b3 Bh7h6 54.Sb3a1 Bh6h5 55.Bg5h6[+wTg5]
warum geht das nicht? Den Zug b1S# habe ich vorher ueberhaupt nicht gesehen. (2023-04-04)
Gerald Ettl: Jetzt habe ich es gesehen: der wBa2 musste ja von a2 geschlagen haben. (2023-04-04)
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Datum der Originalpublikation nicht 100% sicher, laut ACM aus der "Jamaica Gleaner Christmas column".

Originalforderung: How has the position been arrived at and who is the winner, and in how many moves?

From the Jamaica Gleaner: "White mates in two moves. The last move made was by Black playing his Bishop and announcing mate. As it can be demonstrated that the Bishop is not a promoted Pawn and that Black's King's Knight's Pawn was captured on its original square by White's Queen's Knight's Pawn the Black Bishop must have been played from Bishop's square (f8) to Q3 (d6). This being an illegal move, White enforces the penalty of compelling Black to retract it and move his King whereupon White plays 1 PXB(Q) ch (1.gxf8=Q+) and mates next move by 2 Q-B4 (Qf4#). The following is a brief but pointed analysis, demonstrating the false move: White's Pawns have made six captures all on black squares. The Q Kt P (Pb2) made five of these and consequently captured the Kt P on the square upon which it now stands (g7). They could not have captured the Q B which is also lost. The White Bishop is the QRP (Pa2) promoted, the original KB having been captured on its own square as the unmoved Pawns show. To allow this promotion Black's QRP (Pa7) made two captures, the QKtP (Pb7) one, and the QBP (Pc7) two. The KRP (Ph7) has also made a capture, which accounts for the seven pieces White has lost. The Black Bishop is not a promoted Pawn, as if the Black KBP (Pf7) had played to the 7 th square (f2) and then captured a White piece on K or Kt square (e1 or g1) the captures by White Pawns cannot be accounted for without including the Black QB or KRP neither of which is available. As it can be demonstrated, then that the Black Bishop is not a promoted one, and that the KKtP was captured by the White Pawn which now stands on that square, in order to reach Q3 the Black Bishop must have an impossible move.

Der Kolumnist des ACM merkt aber zurecht an:
ACM: The above is a very fine piece of analytical work; but there is a slight flaw in connection with the minor condition, 'mate in two'. In a position of this kind we believe only that which can be proved; thus we do not think that White has any right to enact a penalty, as neither the analysis nor the conditions show that the Black Bishop came from Bishop's square on his last move; indeed, that Bishop may have played outside the Pawns on the very first move of the game which, being played, brought about the position.
HBae: White plays 1 PXB(Q) ch (1.gxf8=Q+). Muß der sK nicht auf f5 stehen? (2019-10-22)
Henrik Juel: Last move (supposedly) was Lf8-d6#, which is obviously impossible and hence illegal
The penalty for this is that Black must replace Ld6 on f8 AND instead make an arbitrary move with his king
So the forward play is
0... Kf5,Kf6,Kh6 1.gxf8=D+ Kg5,Ke5 2.Df4# (2019-10-22)
A.Buchanan: One long-standing approach to resolving illegal diagram jokes is to suppose that only the last move was illegal, with all prior play legal. The illegal move is then retracted, and play continues. Of course, the “illegal move” might in principle be from *any* legal position (even the game array!). So for sanity, we say the illegal move is a simple but somehow illegal shift of a single piece.

So here, candidates for the last move include Pe5-a4+, Sf4-e1+, Ke5-g5+ & B?-d6+. For all of these, White has 6 visible pawn captures, all on dark squares, so Black light-squared bishop is excluded. wPa must have promoted to light-squared bishop, so if the three Black pawns on a-file remain, there is only one unaccounted capture. Thus bPh could not promote, and must be bPg6 now. Thus bPg7 was captured at home, and bBf8 was thus locked in.

So Bf8-d6 is certainly a possible illegal move, but so are e.g. Be8-d6 (as the light-squared bishop is otherwise unexplained) and Pe5-a~. This is an example of an "implausible" joke according to Dawson & Hundsdorfer, because there is more than one retraction to the current position, and one just has to arbitrarily pick the one that makes the forward logic work. (2023-04-02)
A.Buchanan: Another issue is that according to the 1883 laws, White cannot force Black to move their king. The 1883 rules stated:
- If a player touches a piece or Pawn of his own he must move it.
- If he touches one of his adversary's he must take can be taken.
- If he touches plurality of pieces or Pawns of the same colour, in either of these instances his adversary may elect which such piece or Pawn he will call upon him to play or to take, as the case may be.
- If the rules governing the moves of pieces do not admit of the adversary exacting penalty as above, the player must move his King, but may not Castle. If the King cannot be moved without exposure to check, no penalty can then be exacted
So according to this, Black must play Bf8xPg7 as the penalty move.
Was there another revision to the rules between 1883 & 1891? (2023-04-02)
A.Buchanan: I don't think this is cooked, I think it's just a very bad joke. Most time when moves are retracted it's because of self-check or illegal castling, occasional illegal en passant. But here we are meant to conclude that the Bl bishop has hopped over Bl pawn. There are problems by Hieronymus Fischer e.g. P1309484 where we are meant to conclude that a unit has escaped its game array cage somehow, but the piece is simply replaced: there is no move to be retracted and supplanted. But as long as such a problem is true to its own (joke) internal logic, it can't be described as cooked. If it fails (e.g. P1246715) it can indeed be cooked and repaired. (2019-10-22, hidden) more ...
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Joost de Heer: The keyword is wrong (should be TxD, not BxS=T) (2023-05-20, hidden)
Henrik Juel: I believe that the condition 'Black to move' can be deleted (2023-05-21, hidden)
Adrian Storisteanu: This is not the magazine's informal tourney - p.137 of the indicated source says REZULTAT 16. TEMATSKOG TURNIRA "PROBLEMA". (2023-05-21, hidden) more ...
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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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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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Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 26.0, BP 26.5. (2023-05-08, hidden) more ...
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Paulo Roque: obs: Cook: 26...D6b6 (2008-12-24)
paul: See P0001634 as correction. (2010-09-25)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 in 31:54:04 Stunden. (hh:mm:ss).
1 Lösung mit 20 cooks. (2024-01-07)
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Henrik Juel: Dual: ... 4.d4 La6 5.d5 Tc8 6.d6 Tc7 7.dxc7 Db8 8.c8L d6 9.Lf5 ... (2004-02-12)
Moldenhauer: Ergänzung Stelvio 1.2. Keine Lösung BP 19.0, BP 19.5. (2023-05-20, hidden) more ...
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paul: Cooked in 4: 1.Ke5-e6! b2×Ba1=B+ -2.Kf4×Pe5 e6-e5+ -3.Rc2×Sg2 Se1-g2+ -4.Re2×Pc2 & 1.Rxe1# (2023-06-14)
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 01:00:21 Stunde. (hh:mm:ss)
1.Sf3 a5 2.b4 g6 3.h4 Lg7 4.Th3 Lc3 5.Sd4 Ta6 6.Te3 Tc6 7.g3 Tc4 8.Lh3 d6 9.Kf1 Dd7
10.Kg2 Dxh3+ 11.Kf3 Lg4+ 12.Ke4 Sd7 13.Kd5 b6 14.Sb3 Td4+ 15.Kc6 Se5+ 16.Kb7 Kd7
17.Dh1 Ke6 18.Kc8 Kf5 19.Da8 Lh5 20.g4+ Kf4 21.f3 Kg3 22.Kd8 Kf2 23.Ke8 Ke1
Keine Lösung: BP 22.5.
Wenn die Forderung KBP 25.0 richtig ist dann Cooked. (2023-09-28)
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Henrik Juel: Cooked acc. to Euclide 1.01, e.g.,
1. Bc2-c4 Ba7-a6 2. Be2-e4 Be7-e6 3. Bf2-f4 Bh7-h5 4. Dd1xh5 Ke8-e7 5. Dh5-b5 Th8-h4 6. Ke1-e2 Th4-g4 7. Bh2-h4 Ba6xb5 8. Bh4-h5 Bb5-b4 9. Bh5-h6 Ke7-d6 10. Bh6-h7 Kd6-c5 11. Th1-h6 Tg4-h4 12. Bh7-h8=D Bg7-g6 13. Dh8-c3 Bb4xc3 14. Ke2-f3 Bc3-c2 15. Bf4-f5 Bc2xb1=T 16. Bf5xg6 Kc5-b4 17. Bg6-g7 Be6-e5 18. Th6-c6 Bf7-f6 19. Lf1-e2 Sg8-h6 20. Bg7-g8=D (2017-11-22, hidden)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 2 Sekunden.
Keine Lösung: BP 18.5, BP 19.0. KBP = 19.5. (2023-04-15, hidden) more ...
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A.Buchanan: The unsoundness in this non-unique PG is that 6 promotions (apparently intended) are not forced (2023-05-11)
Moldenhauer: Computerprüfung: C+ Cooked Stelvio 1.11 48 Sekunden.
Keine Lösung: BP 36.0, BP 36.5. (2023-05-10, hidden) more ...
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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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Moldenhauer: Ergänzung: Stelvio 1.2. Keine Lösung: BP 14.0, BP 14.5.
BP 15.0 und BP 15.5 cooked.
Notation BP 15.0: 1.Sc3 Sc6 2.a4 b5 3.Ta3 La6 4.d3 Db8 5.Lg5 Db6 6.Lf6 0–0–0
7.e3 Kb7 8.Dg4 Dc5 9.Le2 Kb6 10.Lf3 Ka5 11.Sce2 Tb8 12.Tc3 Sd8 13.La8 De5
14.Tc6 Tb7 15.Tb6 c6
Notation BP 15.5: 1.Sc3 Sc6 2.a4 b5 3.Ta2 La6 4.Ta3 Db8 5.d3 Db6 6.Lg5 0–0–0
7.e3 Kb7 8.Dg4 Dc5 9.Le2 Kb6 10.Lf3 Ka5 11.Sce2 Tb8 12.Tc3 Sd8 13.La8 De5
14.Tc6 Tb7 15.Tb6 c6 16.Lf6.
Alles schlagfrei und ohne Umwandlung. (2023-05-20)
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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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Henrik Juel: Last move is determined only if it is known to be white. (2003-11-20)
Henrik Juel: The coloring is White: Kh3, Pe4,f4,g2,g3,g4,h2 (7+9). (2003-11-18, hidden)
A.Buchanan: Yes the stipulation should always have stated "Schwarz am Zug". I've fixed it. Does that make it "Type E" or "Type F", I forget which is which (2023-06-03, hidden) more ...
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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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Jeliss: "Obstruction of passage square f3 to Bishop of same colour."

"Version 'Pittsburgh Leader' 08.06.1913"
Yoav Ben-Zvi: Appears as the first problem (D445) in the booklet on Dawson's RA problems by G.P. Jellis. The obstruction that occurs in the Try -1.Pf3-f4?, by WP of WB, is described as "obstruction of passage square". It is not considered by Dawson and his disciples to be a Retro opposition. Dawson's conception of RO was quite broad, it included cases where the interference was not by occupation of the target square, so the only valid reason that I can see to exclude this case is that the 2 pieces involved are both of the same color. Fabel's definition explicitly excludes "Monochrome RO". I conclude that it would be preferrable to interpret RO as a bi-chromatic interference. The keyword Retro opposition should be removed. (2018-04-07)
A.Buchanan: To my mind, RO involves some kind of parity-tempo issue between the sides, not just some kind of race-tempo. If it was just about "bi-chromatic interference", one might say that bPe4 blocks wBh7 from an immediate retreat, so it has to be wPf4 that retreats first, legitimizing the ep key. So I agree this is not RO. (2024-01-06)
Henrik Juel: -1.f2 f5xTe4 -2.Tg4 f6 -3.Tg1 f7 -4.Le4 h5 -5.Lg2 h6 -6.Lf1 h7 -7.g2xDTSh3, unpromote DTS on c1, etc. (2004-03-18, hidden)
Henrik Juel: Nice analysis, Andrew, but your last comment seems incorrect.
Without Sb4, 1.b4 Txb4+ is not mate because of 2.Txb4#! (2013-07-26, hidden)
A.Buchanan: Thanks Henrik, I have fixed the solution. Forward chess is even harder than retro chess. And because Rb7 is retro-pinned, I overlooked that it's not forward-pinned! (2013-07-27, hidden) more ...
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Henrik Juel: Accprding to Retrograde Analysis, 1915, the intended solution does not involve adding pieces as such, rather:
Black retracts the illegal move Pd2xSe1=S+ (exposing Kb2 to a selfcheck from Dh2)
and instead pays the penalty of a king move, Kxc3 (only possibility)
Then White mates by 1.Dh8# (2023-04-13)
Mario Richter: @Andrew: Why did you remove the "illegal position" keyword?
Instead of removing this keyword, I suggest to remove the "Add pieces" KW ... (2023-04-14)
A.Buchanan: Hi Mario, yes good question. I removed "Illegal position" from some genuine "Add pieces" compositions, because it suggests some error in the composition. But on reflection, I think I will put the "Illegal position" back for these, and instead edit the description for the keyword.

Now this problem was in fact incorrectly marked as Add pieces. It's one of Dawson & Hundsdorfer's canonical examples of an "implausible" joke. I.e there is no reason why the intended retraction is the right one, except that it works. These too should be marked as illegal position. Sorry for all confusion (2023-04-14)
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Version in der 'Aarsskrift' 1935 innerhalb eines Artikels von K. Hannemann "Et Tema Fra den Retrograde Analyse". Im Original wDg4 statt h3 und wBh3 statt h4.
Henrik Juel: 1.exf6ep+. Not -1... a6?, requiring 3 black captures on light squares (incl. orig. Lf1); but missing orig. Lc1 was dark-squared. (2004-03-08)
Henrik Juel: C+ Popeye 4.61 and analysis (2022-06-11)
Hans-Jürgen Manthey: mögliche Zugfolge:
1. f2-f4 d7-d5 2. c2-c3 d5-d4 3. c3xd4 c7-c5 4. h2-h4 c5-c4 5. b2-b4 Lc8-d7 6. a2-a4 Ld7-b5 7. a4xLb5 Sb8-a6 8. Th1-h3 Sa6-c5 9. d4xSc5 Sg8-f6 10. d2-d4 Sf6-e4 11. c5-c6 Se4-g3 12. Lc1-d2 Se4xLf1 13. Sb1-c3 Sf1-g3 14. Sc3-a4 Sg3-e4 15. c6-c7 Ke8-d7 18. Sa4-c5+ Kd7-d6 19. Sc5-a6 Se4-c5 20. b4xSc5+ Kd6-e6 21. Ld2-a5 b7-b6 22. Sa6-b8 b6xLa5 23. Th3-e3+ Ke6-f6 24. Te3-e5 Dd8-c8 25. Dd1-d3 Dc8-a6 26. Dd3-h3 Da6-b6 27. Ta1-a3 Db6-c6 28. Ta3-g3 Dc6-d6 29. Te5-d5 Dd6-e6 30. c7-c8L De6-d6 31. Lc8-f5 Dd6-c6 32. Sg1-f3 Dc6-b6 33. Sf3-e5 Db6-a6 34. Se5-g6 Da6-b6 35. Sg6xTh8 Db6-c6 36. Tg3-g6+ h7xg6 37. Ke1-f2 Dc6-d6 38. Kf2-g3 Dd6-e6 39. Kg3-g4 De6-d6 40. Lf5-c2 Dd6-c7 41. b5-b6 Dc7-e5 42. f4xDe5+ Kf6-e6 43. Kg4-g5+ f7-f5 und nun :
1. e5xf6ep+ Ke6xTd5 2. Dh3-d7# (2023-02-24)
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Alfred Pfeiffer: Die Korrekturfassung 5674v mit gleicher Forderung (fs-99, S.38): 1ss5/d1b5/lbb1b1b1/1bl5/k2t2b1/2t5/BBB2BB1/TSLDKLST soll die Duale der Urfassung (4-fach mögliches Ziehen von g7-g6) nicht mehr enthalten. (2014-11-06)
Henrik Juel: The correction mentioned by Alfred is still cooked, acc. to Euclide 1.01 (2017-11-24)
Joost de Heer: For correction see P1067393. (2023-08-18)
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This is an tutorial example of a PG deliberately including a cook.
Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 13 Sekunden.
Keine Lösung: BP 14.0 bis 15.0 und 16.0. (2023-05-08, hidden) more ...
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 13 Sekunden.
Keine Lösung BP 19.0, BP 19.5 Cooked.
Beispiel BP 19.5:
1.d3 Sc6 2.Le3 Tb8 3.Ld4 Sf6 4.e3 Sh5 5.Dg4 b5 6.Ke2 La6 7.De6 b4
8.g4 Lb5 9.Lg2 b3 10.axb3 f6 11.Ta6 La4 12.bxa4 Sa5 13.Td6 Tb3
14.Lb7 Ta3 15.Kf3 Ta1 16.Sa3 Td1 17.Se2 Td2 18.Tb1 a6 19.Sc1 Te2 20.Lc8 (2023-05-03, hidden) more ...
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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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Adrian Storisteanu: Possible twin:
b)wSd2->d6
- 1.Sf7xRd6 Rd8xQd6 & 1.Rd8-f8 Qd6-g6# (2024-03-19)
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Henrik Juel: Cook: -1.a7xFb8=L Kb7xTa8, 1.Ka6 axb8DT#. (2004-04-01)
Mario Richter: vgl. mit P0006207 (2009-12-24)
Adrian Storisteanu: Possible (expensive, 50% more units) fix: Ka8 Bd8 / Kc8 (2+1) -- R: 1.e7xRd8=B Rd7-d8 & 1.Rc7 e8Q#. On the plus side, there is a wK in the picture now. (2016-02-10)
Adrian Storisteanu: See P1389595 (not clear which was first). (2024-03-19)
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Paulo Roque: obs: 23...Kd7 24.T8h3 f6 (2008-12-29)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 184:20:33 Stunden. (hh:mm:ss)
1.d4 Sa6 2.Ld2 Sb4 3.e3 Sf6 4.Ld3 c5 5.Ke2 c4 6.h4 c3 7.h5 cxd2 8.c4 Sc2 9.c5 Sg4
10.c6 e5 11.c7 Lc5 12.dxc5 d6 13.c6 Le6 14.c8=S Dg5 15.Se7 Sh2 16.Sf5 Sf1 17.c7 a5
18.c8=S a4 19.Sce7 a3 20.Sg6 axb2 21.a4 hxg6 22.a5 Th6 23.a6 gxf5 24.a7 Tg6
25.h6 Kd7 26.h7 f6 27.h8=T Sa3 28.T8h3 Th8 29.a8=T Thh6 30.Ta4 Lg8 31.Te4 fxe4
32.f4 Dg4+ 33.Tf3 exf3+
Keine Lösung: BP 32.0 wegen der Retraktion. Da NUPG C+. (2023-09-28)
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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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Korrektur 1964, Seite 155: sBh3->f3
Henrik Juel: C+ Popeye 4.61 (2022-11-24)
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Henrik Juel: White captured [sLc8] on c8 and axb, so last move was either b2-b4 or d2-d4
C+ Popeye 4.61, except for the promotion dual (2020-08-03)
A.Buchanan: There's no keyword for an untolerated dual promotion! (2020-08-03)
Henrik Juel: Yes and no, Andrew
What about the cooked label? (2020-08-03)
A.Buchanan: Hi Henrik: this is obviously intentional. I tried to replace the pawn promotion with wDh8-h7+ but I ran out of pieces. I guess that Bebesi had the same difficulty. The word "cooked" seems too strong to me, but it's obviously a significant defect though (2020-08-04)
A.Buchanan: The bar is higher for helpmates than other problems, and this is unsound. However it’s not cooked but dualled. Yet all we have is “cooked” so I suppose it is in PDB-speak. I’ve added that and removed the 2.1… (2023-08-07)
Joost de Heer: Is there an 'unsound' keyword? (2023-08-07)
A.Buchanan: Hi Joost: no there isn't. It can be added, the challenge for any new common keyword would be populating it for more than a few values. (2023-08-07)
A.Buchanan: This time round, I've managed to find a fix. Bebesi's correction in WinChloe 66992 has non-standard wL. I have a feeling I've seen a similar one by Andrey Frolkin & someone else recently with a different matrix, and also non-standard wL. I've sent my effort to Joaquim Crusats at Problemas for his assessment (2023-08-08)
Henrik Juel: better than 2.1...: one solution in each of two parts (2023-08-06, hidden)
A.Buchanan: This time round, I've managed to find a fix. As far as I can see, it hasn't been corrected hitherto. I'm offering the new one to Problemas again. I have a feeling there was one by Andrey Frolkin recently with a similar idea, but had non-standard material (wLL) (2023-08-07, hidden)
A.Buchanan: This time round, I've managed to find a fix. Bebesi's correction in WinChloe 66992 has non-standard wL, which is still not good: https://www.janko.at/Retros/d.php?ff=3r1b2/1p1p3B/B1r3p1/1PP1Kp2/pPkPp3/p1p4s/S7/2q5 I have a feeling there was one by Andrey Frolkin recently with a similar idea, and also non-standard wL. I've sent my effort to Joaquim Crusats at Problemas for his assessment (2023-08-07, hidden) more ...
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In 'Mattbilder eines Lebens' abgedruckt mit sTh7 statt h8 und der Quellenangabe: Europe Echecs, 1964
AB: (1) Where is wK?
(2) Why is 1.fxe3ep legal in b) but not a)? (2002-01-31)
Henrik Juel: wK is probably on g2. In part a) last move could have been Kh3-g2, I think (2002-02-01)
A.Buchanan: Very convincing, Henrik. I've repaired the diagram accordingly. (2023-05-28)
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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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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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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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A.Buchanan: In each of the 4 pairs of adjacent files, one cross-capture suffices. So removing 8 officers is enough. However, they need to be an even number from each side.
However it's possible to do better by removing pawns: wPadgh bPcf = 6 units. Now wPaxb bPcxb wPdxe bPfxe.
Is it possible with 5 or less? (2022-03-17)
Bob Baker: I find it interesting that the composer specified piece removals, since six captures suffice. Does that imply that 5 or less is possible by removing pieces? (2023-04-27)
A.Buchanan: I think the German term "Figuren" (like the English "pieces") can have two senses, either all 16 pieces or just the officers. Wikipidie.ge uses the phrase "im engeren Sinne" = "in the narrower sense" to distinguish. But maybe a native German speaker can be more authoritative here:

Auf dem Schachbrett befinden sich zu Beginn einer Partie insgesamt 32 Schachfiguren (auch als Steine bezeichnet), 16 weiße und 16 schwarze. Beide Spieler (bezeichnet als Weiß und Schwarz oder als Anziehender und Nachziehender) haben je folgende Schachfiguren zur Verfügung:
- acht Figuren im engeren Sinne:
= den König
= die Dame und zwei Türme (Schwerfiguren)
= zwei Springer und zwei Läufer (Leichtfiguren)
- acht Bauern (2023-04-27)
A.Buchanan: And the Originaldforderung would have been in French, surely (2023-04-27)
Bob Baker: A stipulation might have been "Reach the position with the greatest number of pieces still on the board." A position with six fewer men can be reached with no removals from the initial array. It would be a harder problem than if pieces can just be taken off the board. But maybe allowing removals could make 5 or less possible. (2023-04-28)
A.Buchanan: Not quite sure what you are saying. Generally, capturing pawns is more efficient than capturing officers to get pawns past one another, but here we have no doubled pawns. If we have a solution with 5 units including a missing officer, then might as well promote a pawn to replace it. So we can reduce our search to positions where 5 pawns have been removed (2023-04-28)
Bob Baker: I'm suggesting if six really is the answer, it would be a better (more challenging) problem to require a proof game with no removals except by capturing. (2023-04-28)
Bob Baker: Also, I think this may have been the composer's intent, but the stipulation was misunderstood as allowing pieces to be taken off the board without being captured. (2023-04-28)
Bob Baker: I just realized that the misunderstanding of the composer's intent was on my part, so all my previous comments should be disregarded. (2023-04-28)
Carot: I understand that all 16+16 figures are meant.
So for the e-row 1. e4 e5 2. e4f5 e5f4 3. f5e6 f4f3
This is illegal, so only the e-line is considered, one e-pawn too many to reach the given position legally.

I am just a beginner in Chess, but i think en passant is useable and therefore maybe less then 8 pieces have to removed, but i tried an hour and found no way for optimisation. :(..

I would be happy about an idea. Please share. I go study pawn endgames now.
I am happy to found this site and many old combinations of Max Lange. (2023-04-28)
Bob Baker: Play this game for a quick way to remove six pieces.
1. a4 a5 2. b4 b5 3. c4 c5 4. d4 d5 5. e4 e5 6. f4 f5 7. g4 g5 8. h4 h5 9. axb5
hxg4 10. bxc5 gxf4 11. exf5 dxc4 12. b6 a4 13. c6 a3 14. d5 c3 15. d6 e4 16. f6
e3 17. h5 f3 18. h6 g3 (2023-04-29)
A.Buchanan: In each of the 4 pairs of adjacent files, one cross-capture suffices. So a total of 8 captures is necessary. Hence removing 8 officers is enough. However, they need to be an even number from each side. (2022-03-17, hidden)
A.Buchanan: I think Figuren (like pieces) can have two senses, either all 16 pieces or just the officers. Wikipidie.ge says:
Auf dem Schachbrett befinden sich zu Beginn einer Partie insgesamt 32 Schachfiguren (auch als Steine bezeichnet),
16 weiße und 16 schwarze. Beide Spieler (bezeichnet als Weiß und Schwarz oder als Anziehender und Nachziehender)
haben je folgende Schachfiguren zur Verfügung:
- acht Figuren im engeren Sinne:
= den König
= die Dame und zwei Türme (Schwerfiguren)
= zwei Springer und zwei Läufer (Leichtfiguren)
- acht Bauern (2023-04-27, hidden) more ...
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Henrik Juel: 1... Txa2#, not 1.Dxb2? -1.Td1xSe1 Sf3 -2.Te1 h7 -3.Td1 Sh4 -4.Te1 Sf5 -5.Td1 Sh6 -6.Te1 Sg4xPh6 -7.Td1 Sf2 -8.Te1 Sd1 -9.h5 Kc1 -10.h4 Ld3 -11.h3 Kc2 -12.h2 Sf2 -13.Td1, retract sLd3 to c8, b7-b6, etc. Black promoted b2-b1=T and f2xTg1=L. The shortest proof game is rather long! (2003-12-19)
Moldenhauer: Histogramm Stelvio 1.2 128/0k bei BP 37.0.
Kein Histogramm bei BP 36.0, BP 36.5. Computerprüfung wird sehr lange dauern
da eine 5+1 auch nach 5 Stunden noch nicht fertig ist! Nur 49 Strategien sind
unter 5+1. Vielleicht kommt aber das "cooked" früher dazwischen. (2023-05-10)
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S N Ravi Shankar: This problem is not mine. (2019-01-02)
Mario Richter: My first guess for the solution was:
R: 1. Ke2xLf1 2. Kd3xLe2 3. Kc4xLd3 4. Kb5xLc4 5. Ka6xLb5 Lc6-b5+ 6. Sd7xLb8, dann 1. Sb6#, but that doesn't work because of 5. ... Ld7-b5+!
Does someone see (or know) the correct solution? (2019-01-03)
Henrik Juel: No, because with 1... Lh3-f1+, 5.Ka6-b5 thr. 6.Sd7xLb8 fails similarly on 5... Ld7-h3 (2019-01-03)
S N Ravi Shankar: Does adding a black knight on e6 cure? 1. Ke2xLf1 2. Kd3xLe2 3. Kc4xLd3 4. Kb5xLc4 5. Ka6-b5! followed by 6. Sd7xLb8 and now 1. Sb6#. (2023-02-20)
S N Ravi Shankar: The problem appears to be sound. 1.Ke2xLf1 2.Kd3xLe2 3.Kc4xLd3 4.Kb5xLc4 5.Kb6-b5! followed by 6.Sa6xSb8 and now 1.Sc7#. (2024-02-18)
A.Buchanan: WinChloe also gives the same composer. A curious situation, but at least the problem attributed to you appears to be sound! (2024-02-19)
Adrian Storisteanu: Above the "diagrammes" diagram there is: R. SHANKAR, Inde. In the volume index, under RETROS: this is the only problem of a composer named (just) "SHANKAR" (one original), while "RAVI SHANKAR" is a different author (with two originals). (2024-02-20)
A.Buchanan: I suggest that S.N.Ravishankar check the reaponse to the query (A='Shankar' or a='Ravishankar') AND NOT A='Shankar Ram, Narajan' as there are many other similar retros which are attributed to Ravishankar. If he can state here which ones are not his, I will happily modify the data. (2024-02-21)
A.Buchanan: I suggest that S.N.Ravi Shankar check the answers to (A='shankar' or a='Ravishankar') AND NOT A='Shankar Ram, Narajan' as there are many other retros which are attributed to ravishankar (2024-02-20, 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: Diagram error? -1... g5xSf4, 0... Le5 1.Sd5#. But the symmetrical solution -1... e7xSd6 etc. seems legal too. (2004-03-22)
A.Buchanan: Yes bBf8 would have died at home but bPc could capture twice to promote on c1 or e1 (2023-03-31)
A.Buchanan: Odd. The diagram is thematically symmetric, so not just missing a unit. WinChloe has the same diagram. I wondering if we are missing something from the stipulation. Does anyone have access to an archive for this? It's also odd that it was reprinted: surely there would be a comment there? I suppose it can be rendered sound by placing wTTSS somewhere symmetric. Then Black runs out of pawn captures if sLd4 is promoted. In any case, wPh7xg8=S is affordable. (2023-04-01)
Mario Richter: The reprint in 'Problem' says: "Ne e7xSd6 radi nepravilne postave" ("Not e7xSd6 because of illegal Position.") The diagram in 'Problem' is the same as here in the PDB.
How about adding black pawns on b7 and g2? (2023-04-01)
A.Buchanan: Hi Mario. I considered adding sBb5e2. That gives 10 needed captures by Black and 6 by White, including sLf8. So the try fails for *two* reasons. sBb7g2 fails even more badly. I thought that also adding wTTSS although more uneconomical in pieces defeats the try for only one reason. Also I liked the trick h7xg8=S. And it must make it harder to find the solution. Does Rawbats do simple retractors? Can you try wTa0h1 wSa7g1 as a risky initial bid, and then back off until there are no cooks? (2023-04-01)
Mario Richter: Hi Andrew, I was well aware that with adding sBb5e2 or sBb7g2 the try fails for two reasons (for [subjective] optical reasons I preferred sBb7g2 to sBb5e2]). The reason I suggested the addition of 2 black pawns was the following consideration: I cannot believe that such an experienced Retro Composer like C.E. Kemp could make such a big mistake. It's much more likely that the printer made an error by omitting two pieces.
On the other hand, if we want to correct the problem including a unique reason for the try to fail, your suggestion of adding wTa8h1+wSa7g1 (FEN=R7/N4pp1/1P1pnkp1/5n2/3b1p2/1PP5/1PP2P2/K5NR) works - rawbats gives exactly R: 1. ... Bg5xSf4, dann 1. Ld4-e5 Sf4-d5# (and says about R: Be7xSd6 - "Illegal: Mehr schwarze Bauern-Schlaege notwendig als weisse Steine fehlen") (2023-04-01)
A.Buchanan: Hi Mario, thanks for this, I agree with everything you say. It's curious that the reprint in Problem did not pick up the error. Any overlooked black units are more likely to have been lurking on dark squares, maybe sBb4d2. This forces the try to only fail for 1 reason, albeit rather banal. Does this work. Or the +wTTSS mentioned forces the solver to do a little more counting, and there is a lot more solving space to search for the h#1. Rawbats' conversational skills seem to have improved: have you integrated it with ChatGPT maybe? (2023-04-02)
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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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Korrekturversuch s. P0003009
Henrik Juel: If you want a great solving challenge, this is the retro for you.
If you need a hint:
[Dd8] never moved, and Black castled (as you may have guessed).
If you need another hint:
Last move was Ld8xLc7+.
I gave up, but Nikolai told me the solution:
-1... Ld8xLc7 -2.Lb6 Kb7 -3.Lc7 Kc8 -4.Lb6 Lc7 -5.Lc5 Lb8 -6.Lb6 Kb7 -7.Ld8 Kc7 -8.L=d7 Kd8 -9.e6xSd7 Sb6 -10.Sc5 Sa4 -11.Sd4 d7 -12.Sf6 Lh7 -13.Sg8 Lf4 -14.Sf6 Lh6 -15.Sg8 c7 -16.S=g7 Kc8 -17.f6xTg7 Tg8 -18.e5 Td8 -19.e4 0-0-0 -20.e3 Lf8 -21.f5 g7 -22.f4 Le4 -23.f3 Lb7 -24.f2 Lc8 -25.Kb5 b7xSa6 etc. (2012-07-22)
Mu-Tsun Tsai: Once I heard "great challenge" I started working. But I came to a complete different conclusion. Not only [Qd8] can move, but the first move of the black king need not be castling. Here's the proof game. After playing
1.c3 h5 2.b4 Rh6 3.Na3 Rg6 4.Nc2 Rg3 5.hxg3 Nf6 6.Rb1 Ng8 7.Na1 Nf6 8.e3 Ng8 9.Bd3 Nf6 10.Rb2 Ng8 11.Bb1 Nf6 12.Qb3 Ng8 13.Rc2 Nf6 14.Qb2 Ng8 15.Rh4 Nf6 16.Rd4 h4 17.Rd5 h3 18.Ra5 h2 19.Ra3 h1=Q 20.Rb3 Qh5 21.Nf3 Qa5 22.Ke2 Qa3 23.Kd3 Na6 24.Kc4 Nc5 25.Kb5 Na4 26.Ne5 Nh5 27.Nd3 Nf4 28.Nc5 Nd3 29.Na6 bxa6+ 30.Ka5 Bb7 31.e4 Qb8 32.e5 Bc8 33.e6 Qb7 34.g4 Qf3 35.g5 Qh3 36.g6 Qh7 37.gxh7 Nf4 38.h8=N Bb7 39.Ng6 Be4 40.Ne5 Bh7 41.Nc4 Bg8 42.Ne3 Ng6 43.Nc4 Nh8 44.Ne3 g6 45.Nc4 Bg7 46.Ne3 Be5 47.Nc4 c6 48.Ne3,
you could either play
48...O-O-O 49.Nc4 Bb8 50.Ne3 d6 51.Nc4 Rd7 52.exd7+ Kb7 53.d8=N+ Kc7 54.Ne6+ Kb7 55.Nb6 Nc5+ 56.Na4 Ne4 57.Nc5+ Kc7 58.Nd7 Kb7 59.Nb6 Ng3 60.Nd5 Bc7+ 61.Nb6 Bd8 62.fxg3 Kc7 63.Nd5+ Kb8+ 64.Nc7 Bxc7+,
which is castling version, or,
48...Kd8 49.Nc4 Kc7 50.Ne3 Rd8 51.Nc4 Kc8 52.Ne3 Bb8 53.Nc4 d6 54.Ne3 Rd7 55.exd7+ Kb7 56.d8=N+ Kc7 57.Ne6+ Kb7 58.Nc4 Bh7 59.Nb6 Nc5+ 60.Na4 Ne4 61.Nd4 Kc7 62.Ne6+ Kc8 63.Nd4 Kb7 64.Ne6 Bg8 65.Nc5+ Kc7 66.Nd7 Kb7 67.Nb6 Ng3 68.Nd5 Bc7+ 69.Nb6 Bd8 70.fxg3 Kc7 71.Nd5+ Kb8+ 72.Nc7 Bxc7+
which is none castling version. Both reach the diagram position.
Why am I feeling cooking Ceriani's problems too much lately? (2012-07-22)
Mu-Tsun Tsai: Also in your retraction, -11.Nd4 should be -11.Ne4 I believe? It seems like your retraction also works just fine, and looks like it should be the intended solution. (2012-07-22)
Henrik Juel: Yes, the intended solution should have -11.Se4, and it can be shortened a bit.
I am impressed by your cook, Mu-Tsun, with two white pawn captures on g3 and promotion on h8, but it is also a little sad that a seemingly fine problem has been rendered worthless.
Probably several more Ceriani problems will be cooked, because they were not scrutinized well enough by testers and solvers in the old days; now, when I finally have cracked a Ceriani nut, I have no energy left to search for errors (2012-07-23)
Mu-Tsun Tsai: I've been thinking about how this problem might be fixed, but unfortunately I cannot come up with anything other than adding extra assumptions in the stipulation, for example "g3 pawn came from h2". The structure of this one is good, and either method of releasing the position (mine or the intended one) is quite subtle, so I feel sad about have to cook this one as well. (2012-07-23)
Thomas Volet: In his 1961 book Ceriani discusses the cook with the WhP unpromoting on h8 and uncapturing to the g file and back to the h file, and gives P0003009 as the corrected diagram position. (2012-08-02)
Mu-Tsun Tsai: This is a really late comment, but I do think this will make a great problem by changing the stip to "You don't know the first move of the black queen nor the black king"! (2023-06-29)
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Henrik Juel: -1... Tc1xSd1 -2.Sc3 Tc2xLc1 etc.
The other black captures were f7xLg6 and g7xSh6.
White captured c2xPb3xPa4, g2xLf3, and h2xSg3xSf4xPe5xPd6xPc7.
Unusually easy for a Ceriani retro.
The record in this genre may be P1000077 ; does anyone know of a retro with less than 19 men in the diagram position and the stipulation 'What were the captures?' ? (2012-07-13)
James Malcom: What is the full solution? (2023-05-30)
Henrik Juel: The 12 captures are given in the first three lines of my 2012 comment
The further retroplay is easy and of little interest, I believe
James, you might want to change the initial retroplay to
R: 1. ... Tc1xSd1+ 2. Sc3-d1 Tc2xLc1+ etc.
(with uncheck plusses added) (2023-05-30)
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