Moldenhauer: Computerprüfung: C+ bei NUPG Stelvio 1.11 00:08:50 Minuten. (hh:mm:ss)
Beispiel Stelvio 1.11.
1.Sa3 Sf6 2.Sc4 d5 3.Sh3 Lg4 4.Sf4 Lf3 5.gxf3 Tg8 6.Lh3 dxc4 7.Lf5 Dd3 8.Le4 g5 9.cxd3 gxf4 10.Db3 Tg3 11.hxg3 a5 12.Th6 a4 13.gxf4 axb3 14.axb3 Ta3 15.bxc4 Tc3 16.bxc3 Sc6 17.La3 Sd4 18.cxd4 Sd5 19.Tc6 Lh6 20.Ld6 bxc6 21.Ta3 cxd6 22.cxd5 e5 23.fxe5 Le3 24.exd6 f5 25.fxe3 fxe4 26.fxe4
Keine Lösung: BP 24.5, BP 25.0. (2023-03-27)
Ryan McCracken: It seems there was no attempt made to force the move order for either side! For example, 1.Sa3 and 2.Sh3 are transposable, and this is only one of MANY examples. (2001-08-15, hidden) more ...
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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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Ryan McCracken: Cooked...Nf6/Nh6 is one of many move defects. (2001-09-13)
Moldenhauer: Computerprüfung: Cooked Stevio 1.2 1 Sekunde.
Keine Lösung: BP 16.0, BP 16.5.
Beispiel Notation:1.f4 Sc6 2.Sf3 Sf6 3.f5 Tg8 4.g4 Sxg4 5.f6 a5 6.fxe7 a4 7.exd8=S a3
8.Sxb7 axb2 9.Sa5 bxa1=S 10.Lb2 Lb7 11.Lxg7 Lc5 12.Sd4 Se3 13.c3 Sc4 14.Sc2 Lg1
15.Ld4 Tc8 16.Lf2 Sd4 17.Sb3 Lf3.
Die Idee mit den Buchstaben SL im Diagramm finde ich super! (2023-05-20)
A.Buchanan: Yes I like the letter idea, which can only work in German. Do you think this was intentionally a non-unique PG? Given the triumphs of the unique PG genre, such problems do not impress, but it wouldn't be "cooked". Or maybe we should classify all non-unique PGs here as cooked? What do you think? (2023-05-21)
Henrik Juel: Testing gives the result 'cooked'
This problem is correct in the sense that all shortest proof games last 17.0 moves
Still the current C+ is rather misleading
I suggest no label, neither C+ nor cooked (2023-05-21)
Moldenhauer: Ich nehme an, dass die Absicht es Sam Loyd zu widmen, das Ziel war.
Deshalb für diese Ausführung C+. Wenn der Autor selbst angibt das es nur
einen Lösungsweg gibt ist die Aufgabe cooked. NUPG haben andere
Schachproblemfreunde festgestellt und als Schlüsselwort angegeben. Oder irre ich? (2023-05-21)
Olaf Jenkner: Machine translation is very poor as far as problem chess is concerned. (2023-05-23)
A.Buchanan: Hi Moldenhauser. Non-unique PGs are uncooked if 1) the move total is correct, (2) the intended theme appears in every shortest solution. For a problem like this where the theme is the final diagram, the second is vacuous. In 1986 some unique PGs were being already published, and I t can be difficult to discern the intention of the composer but I think this one was intended to be non-unique. I don’t think any non-unique PG can be classed as C+. (2023-05-23)
A.Buchanan: Hallo Moldenhauser, Ich stimme mit Ihnen ein. Dieses alte Problem ist für sich genommen C+: (1) die Anzahl der Züge stimmt, (2) das Thema wird in jeder Partie erreicht. Nicht jedes nicht eindeutige Beweisspartie wird diese beiden Tests bestehen, obwohl der zweite Test in diesem Beispiel inhaltslos ist. Im Jahr 1986 wurden sicherlich einzigartige Beweisspartie entwickelt, und es ist nicht immer klar, ob die Absicht Einzigartigkeit war. Drei neuere Probleme, die Eindeutigkeit erreichen, sind k='einzigartiger Beweisspartie' und K='Buchstabenproblem'. (2023-05-22, hidden) more ...
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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: 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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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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Reto: This is now the new record for the longest computer tested SPG. Strategy seeking takes up all the time, playing the 800 strategies on the other hand is done within a few seconds. (2023-08-11)
Reto: With the upcoming v1.6, this is solved by Stelvio in around 20min. The improvement is due to the fact that the cage in the south-west is now detected (either the wRa1 is captured inside the cage or the wBc1 is captured at home) (2023-09-02)
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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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Joaquim Crusats: Can anyone explain the intention? (2013-01-06)
Henrik Juel: The eight white pawns captured the four missing black officers and promoted on b8, d8, f8, and h8; the eight black pawns captured the four missing white officers and promoted on a1, c1, e1, and g1.
If White had the move, 1.Ta5+? Kxb4 2.Sc6+ Kc4 3.Tc5 would mate in three.
But Black has no last move, so he has the move and mates in three by 1... T3d2+ 2.Lexd2 Txd2+ 3.Kb1 Txc1# (2013-01-06)
A.Buchanan: This is one of a number of problems which had been given both "No legal last move for..." & "Whose move?" keywords. These should correspond respectively to two situations: (1) Codex Article 15 where we add or remove a single move from the beginning of the solutions, but the mating party remains the same. (2) A kind of one-sided duplex, where who moved last implies who delivers the mate (or other final move).
So at most one of the two keyword forms can ever logically apply to a problem. Therefore I have removed the incorrect "No legal last move for Black" keyword.
We might replace "Whose move?" by "Halfduplex", which is related to the Popeye option. What do you think? (2023-11-28)
A.Buchanan: This is a nice problem. There are some final move duals for the actual solution, but all missing units are needed to justify promotions (2023-11-28)
Henrik Juel: Halfduplex is a bit too technical for my taste, although it describes the goings on well (2023-11-28)
A.Buchanan: Thanks Henrik - I'll leave the term "Whose move?" as it is then. (2023-11-28)
A.Buchanan: This is one of a number of problems which had been given both "Whose move?" and "No legal last move for..." keywords. There are basically two situations: (1) Codex Article 15 where we add or remove a single move from the beginning of the solutions, but the mating party remains the same. (2) A kind of one-sided duplex, where who moved last implies who delivers the mate (or other final move).
So at most one of the two keyword forms can ever logically apply to a problem. Therefore I have removed the incorrect "No legal last move" keyword. If you disagree with this interpretation, then please explain your view. (2023-11-28, 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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Compare P0000490 with equivalent material but 30 years later.
A.Buchanan: Solved in YouTube: https://www.youtube.com/watch?v=Do_rkLQnrpk (2022-02-17)
A.Buchanan: Solved in YouTube video: https://www.youtube.com/watch?v=Do_rkLQnrpk&list=PLYeOP7z8XTGTAF6LYk15wZgqD6-ZXgbM2&index=2 (2022-02-17, hidden)
A.Buchanan: Compare P0000490 with equivalent material but 30 years later. (2023-05-31, 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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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 63 Sekunden.
Keine Lösung: BP 29.5, BP 30.0. (2023-04-15, hidden) more ...
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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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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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Sally: Eine eindeutige BwP., die lange Zeit den Längenreord hielt und, wenn auch mit langer Verzögerung, wesentlich zum Boom dieser Gattung in den letzten 15 Jahren beigetragen hat. Im-
mer noch eine faszinierende Aufgabe. Nr. KF-13 Schwalbe Okt. 2005. (2011-12-05)
Moldenhauer: Ergänzung: Keine Lösung: BP 40.5, BP 41.0. (2023-05-20)
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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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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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Moldenhauer: Ergänzung: Stelvio 1.2. Keine Lösung BP 15.5, BP 16.0. (2023-05-20, hidden) more ...
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Silvio Baier: C+ (Natch 2.4) (2009-07-15, hidden)
Moldenhauer: Ergänzung Stelvio 1.2. Keine Lösung BP 23.0, BP 23.5. (2023-05-20, hidden) more ...
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Silvio Baier: C+ (Euclide 0.93) (2009-05-12, hidden)
Moldenhauer: Ergänzung Stelvio 1.2. Keine Lösung BP 25.5, BP 26.0. (2023-05-20, 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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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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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: C+ Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 25.5, BP 26.0. (2023-04-02)
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paul: Correction of P0001580 (2010-09-25)
Silvio Baier: Mit 30....dc+ ist das C+ (Euclide 0.98, Retractor für den letzten Zug) (2011-07-08)
James Malcom: So this is fully C+? (2021-01-27)
Henrik Juel: No, another problem in 30.0 moves ending with 30. Tc2 dxc4+ is tested OK (2021-01-27)
Silvio Baier: Mit 30....dc+ ist das C+ (Euclide 0.98) (2011-07-08, hidden)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 65:30:51 Stunden. (hh:mm:ss)
Keine Lösung: BP 30.0, BP 30.5. (2023-05-08, 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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paul: The last moves are: 14.Ta1 Ta4 15.Sh6 Sa6 16.b8=T Tf8 17.Tb4 Sh8 18.Th4 Tf7 19.Th1 Th4 20. Sg4 Sb4 21. Sf2 La6 22. Sh3 Ld3 23. Sg1 Kg5 24. Kf2 Lg6 25. Ke1
Phoenix-Pronkin theme: wNg1 & wRh1 are promoted, here combined with wK-switchback. (2008-03-24)
Moldenhauer: Ergänzung: Stelvio 1.2. Keine Lösung: BP 23.5, BP 24.0. (2023-05-20, hidden) more ...
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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: Computerprüfung: C+ Stelvio 1.2 18 Sekunden.
Keine Lösung: BP 23.5, BP 24.0. (2023-05-12)
Silvio Baier: C+ (Euclide 0.93) (2009-05-12, 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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Klären: Quelle = Schachmatt? - Felber, Volker: SCHACH ist korrekt, 6/1969, Seite 191 (2010-10-09)

Eliminierung von 4 Steinen zwischen sK und wT im h#2 (mit Hilfe von ep-Schlag)!
Felber, Volker: SCHACH ist korrekt, 6/1969, Seite 191 (2010-10-09, hidden)
Henrik Juel: HC+ Popeye 4.61 (2023-01-08, hidden) more ...
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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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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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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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