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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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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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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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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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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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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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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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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"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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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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Renny Bosch: This problem together with P0002238 and P0002243 form a lovely triad in each of which the rightmost three white pawns promote to a common target (Q's, R's, or B's) which subsequently get captured. Studying their similarities and differences, and what it took to make each one work, gives one a good appreciation of the art of SPG composition.

To see them all at once, enter
PROBID='P0002238' OR PROBID='P0002241' OR PROBID='P0002243'
into the search window on the home page.  (2008-03-23)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00;56:06 Minuten. (hh:mm:ss)
Keine Lösung: BP 33.0, BP 33.5. (2023-04-19)
Reto: With Stelvio 1.33, solving time drops to 20sec (2023-06-10)
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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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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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Korrektur 1964, Seite 155: sBh3->f3
Henrik Juel: C+ Popeye 4.61 (2022-11-24)
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A.Buchanan: Can shift bQd5 to c4, while replacing bRg5 with bP to reach Meredith status. bBh4 can also be downgraded to bP. But probably even more economy is possible by shifting pieces one file to the right, keeping the ep, White tempo & model mate, which seem to be the main features (2022-11-30)
Henrik Juel: HC+ Popeye 4.61 (2022-11-28, hidden) more ...
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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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Henrik Juel: HC+ Popeye 4.61 (2022-11-28, hidden)
A.Buchanan: This is not a retro. There is no “can’t castle”. This is simple application of castling convention to a forward problem. This is not generally taken to make the problem “retro”. Castling convention is certainly used in many retros, but not here (2022-11-30, hidden)
A.Buchanan: Nice - could be done without twinning, just remove bPh7 for 2 solutions? (2022-11-30, hidden) more ...
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: ... Und jedem Anfang wohnt ein Zauber inne, der uns beschützt und der uns hilft zu leben. ... (2003-02-22)
Joost de Heer: This is C+ by Stelvio: 1 exact solution (starting with Sf3) and 1 dualistic cook (starting with 1. d4) (2023-08-02)
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in '64 Schach-Scherze' nachgedruckt mit der Forderung "Matt in gar keinem Zug"

vgl. P1309484
Erich Bartel: weiterer Nachdruck (oder Erstquelle?!):
3) 28) 64 Schachscherze 1916 (in dieser Quelle ist kein
Hinweis ob Urdruck oder Nachdruck) (2007-10-30)
Alain Brobecker: Same position and stipulation as P1265678, except the latter one is dated 1910. (2022-11-15)
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Korrektur siehe P0008426.
James Malcom: What is the cook? (2023-05-02)
paul: Missing wQd1. (2010-05-04, hidden) more ...
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Henrik Juel: minor duals 0... hxg1=L 1.Sxf1+,Lh5 and 0.Kxg4 1.g8=DT+ (2022-08-27)
hans: black to move:
0. ... h1=D/T 1. Lh5! (2. Sg5#)
0. ... h1=L 1. g8=L! Kxg4 2. Le6#
0. ... h1=S 1. Lh5! (2. Sg5#) Sxg3 2. Sf4#
0. ... hxg1=D/T/L/S 1. Sfxg1+ Kxg4 2. g8=D/T#
0. ... Kxg4 1. g8=D/T+ Kh3/Kf5 2. Sf4/Sd2# (2015-08-19, hidden) more ...
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paul: Add wPa2b2b3e2, bPa7c7e7g7. (2022-09-18)
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Henrik Juel: solution should be Kg1xSh2, I think (2023-05-14, hidden) more ...
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Kostas Prentos: Here is a computer tested correction: PG 12,5

1. a4 e6 2. Ta3 Ld6 3. Tb3 Lg3 4. Tb6 axb6 5. hxg3 Txa4 6. Th6 Ta1 7. Tg6 hxg6 8. Sa3 Th1 9. Sc4 Dh4 10. Sa5 bxa5 11. e3 b6 12. Dh5 gxh5 13. gxh4

C+ Euclide v1.11 (42 h 2 min 37 sec} (2022-12-08)
Michel Caillaud: cooked by Stelvio 0.92 :
1.a3 e6 2.h4 Lxa3 3.Sxa3 h5 4.Sb5 Th6 5.Ta6 Tf6 6.Tb6 Tf3 7.Sd4 Th3 8.Sc6 Txh1 9.Sxd8 axb6 10.Sc6 Ta1 11.Sa5 bxa5 12.g3 b6 13.Lg2 Lb7 14.Kf1 Lxg2+ 15.Ke1 Lh3 (2022-12-03, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Stelvio 1.2 03:15:11 Stunden. (hh:mm:ss)
C+ Euclide 1.11 01:00:47 Stunde.
Stelvio und Euclide finden nur 1 Lösung! Warum cooked?
Keine Lösung: BP12.5, BP 13.0.
Notation: 1.a4 Sf6 2.a5 Tg8 3.a6 bxa6 4.Txa6 Lb7 5.Txf6 a5 6.Txf7 Sa6
7.Txf8+ Kxf8 8.h4 De8 9.h5 Dxh5 10.f3 Dxf3 11.Sh3 Da3 12.e3 Te8
13.Lxa6 Lc8 14.0–0+#
Epaulettenmatt durch Rochade ist OK. (2023-05-28)
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Henrik Juel: 1.Sxb2#. -1... a3xDb2 -2.Sa2 b4xLa3 -3.c4 c5xSb4 -4.Da1 d6xTc5 -5.Lc1 Ka3 -6.Tb2 Ka4 -7.Lb1 Ka3 etc. All-uncapture. (2004-01-02)
James Malcom: Shortest possible proof game: 1. c4 c6 2. a4 Qc7 3. h4 Qh2 4. Nh3 Kd8 5. Ra2 Kc7 6. e4 Kb6 7. Nc3 Kc5 8. a5 Qg3 9. b4+ Kd4 10. b5 Kc5 11. Qa4 Qd3 12. g4 Qc2 13. Rb2 Qb3 14. Qa1 Kb4 15. Na2+ Ka3 16. Ke2 Qd1+ 17. Ke3 Qf3+ 18. Kd4 Na6 19. Bd3 Nc5 20. Bb1 Nd3 21. Nf4 Ne5 22. Kc5 Qh3 23. Nd3 a6 24. f4 Nf3 25. Kb6 Ng5 26. hxg5 d6 27. e5 Bd7 28. e6 Re8 29. exd7 e6 30. f5 Be7 31. f6 Rc8 32. fxe7 Nf6 33. e8=N Ng8 34. d8=B Qh4 35. Bf6 Qh5 36. Bc3 Qh6 37. Nf6 Qg6 38. Nd5 Nf6 39. gxf6 Rce8 40. N5b4 Re7 41. fxe7 Rf8 42. e8=R Rg8 43. Rd8 Rf8 44. Rh5 Re8 45. Rc5 Re7 46. g5 Qf6 47. gxf6 Ka4 48. fxe7 Ka3 49. e8=Q Ka4 50. Rd7 Ka3 51. Qe7 Ka4 52. Bc2+ Ka3 53. Rb3+ Ka4 54. Ba3 dxc5 55. Qb2 cxb4 56. c5 bxa3 57. Nab4 axb2 (2023-02-05)
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Moldenhauer: Computerprüfung: C+ da NUPG Stelvio 1.11 08:07:40 Stunden. (hh:mm:ss)
Keine Lösung: BP 21.5, BP 22.0.
Beispiel: 1.Sa3 b6 2.Sf3 Lb7 3.b4 c5 4.Lb2 c4 5.Lxg7 c3 6.Lxf8 e5 7.e3 Dh4
8.Lc4 Se7 9.0–0 Dxf2+ 10.Kh1 Tg8 11.Le6 cxd2 12.c3 Tg3 13.Da4 h6
14.Dxa7 dxe6 15.Dxb8+ Kd7 16.Tg1 Kc6 17.Sc2 Kb5 18.h4 Ta3
19.Scd4+ Ka4 20.b5 Th3+ 21.gxh3 Tb3 22.Tg6 fxg6 23.axb3# (2023-03-28)
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A.Buchanan: There's a question whether DP rule has visibility of 3Rep & 50M state. The current consensus among most of the tiny group who might care is that for retros, it does have visibility, but for purely forward problems, it does not. This align with the idea that by default 50M & DP rules apply only to retros (2023-09-06)
A.Buchanan: The old intended interpretation is protected under the Golden Age principle. Suppose we do apply modern rules, codex & clarifications. If Black just played Kb7xRa8, then the alternative leading to mate takes 7.0 moves, so this must have been Black's 5892nd move at the latest. If White just moved, then it was White's 5892nd at latest. So to maximize the length of the game, Black moved last. (2023-09-06) edit (2023-09-06)
A.Buchanan: The old intended interpretation is protected under the Golden Age principle. Suppose we do apply modern rules, codex & clarifications: (1) mate trumps 50M (2) 50M applies as this is retro (3) 75M rule exists but does not apply as this is problem not game (4) DP rule exists (5) DP rule applies as this is retro (6) DP has visibility of 50M state as this is retro. If Black just played Kb7xRa8, then the alternative leading to mate takes 7.0 moves, so this must have been Black's 5892nd move at the latest. If White just played Kg2xRh1, then again need another 7.0 moves for the half-duplex mate, which is at the latest Black's 5898th move, so this would have been White's 5892nd move at the latest. So to maximize the length of the game, Black moved last. (2023-09-06, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 in 1 Sekunde.
Keine Lösung: BP 22.0, BP 22.5. (2023-03-27)
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Henrik Juel: Natch 3.1 made no progress after half an hour (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:06:29 Minuten. (hh:mm:ss)
Keine Lösung: BP 20.5, BP 21.0. (2023-03-16)
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Henrik Juel: dual 2... Lc6 3.Kxc6,De7 (2023-04-10)
Sally: S.150 Ideengeschichte (Breuer).
S. 174 Problemschach Jahrbuch 2001 (W.E. Kuhn) (2010-08-09, hidden) more ...
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Henrik Juel: duals 1... Tb8 2.Txg6 Kb1 3.Tf1+,Tf2,Dd2,Dd3+ (2022-09-12)
Sally: Nach 1. Tg6! (droht 2. Txf6) hebt sich aus dem Wust der Ne-
benspielen - dem Preis, für die Verwirklichung der Idee
in einer so gewagten Aufstellung zu zahlen war - als Haupt-
spiel strahlend heraus. Kritische Lenkung des sL. mit seiner nchfolgenden Verstellung durch den wT. Schwarz -
weißer Grimshaw mit den Wirkungsfeldern e5/d4 und c3 für die wD. Nr. 1164 Ideengeschichte ( Breuer 1982).
Nr. 146 Das log. S. -Pr. (Speckmann 1980).
S. 501 Schwalbe Aug 2000. (2012-04-19, hidden) more ...
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Das namensgebende [Stamm-]Problem für dieses Thema (=Forsberg-Zwilling).
Erich Bartel: weitere Nachdrucke:
13) 197 moving on 2008.---
14) Naef:Endspielstudien Hilfmattprobleme und Märchenschachaufgaben 1998,S.72.--- (2008-10-01)
Jacques Rotenberg: 5982, CHM 1995, G. Lestriguel
p.1315, Le Guide des Echecs 1993
5 p.8, The Modern Helpmate in Two, Z. Janevski & N. Stolev 1989
XIII p.151, Thèmes-64 10 (avr. 58)
XII bis p.38, Rex Multiplex 2 (avr. 82)
285, Svenska Miniatyrer I, A.Uddgren & A. Hildebrand 1996
p.76, Im Wunderland des Schachproblems, E. Ramin 1958
E-1 p.236, StrateGems 12 (oct. 00)
I33 p.1333, Phénix 14 (sept. 91)
3 p.219, Le Genre Aidé Supplément diagrammes 15 (juin 75)
161 p.241, Echecs, N. Ramini 1985
E p.14, Best Problems 21 (jan. 02)
259 p.238, Chess Problems : Introduction to an Art, M. Lipton, R.C.O. Matthews, J. Rice 1963
139, Skakopgaven 1942
135 p.31, Album FIDE 1914-44/III
9 p.403, feenschach 26 (déc. 74)
40 p.61, Europe Echecs 330 (juin 86)
p.185, feenschach 47 (juil. 79)
in WinChloe, you find also :
A p.5, Springaren 17 (mars 84)
p.50, feenschach 73 (jan. 85)
M p.171, The Problemist (sept. 81)
p.415, feenschach 97 (nov. 90)
133 p.104, Solving in Style, J. Nunn 1985
E-I p.43, Probleemblad 58/1 (jan. 00)
E-1 p.102, Probleemblad 58/3 (mai 00)
p.136, Springaren 70 (août 97)
4 p.4424, diagrammes 159 (oct. 06)
p.37, Problem-Forum 36 (déc. 08) (2012-03-10)
A.Buchanan: The position is unique (except for left-right reflection). In particular, it cannot be shifted up or down without breaking (b) and only the current location of wK makes (b) sound and (a)&(d) solvable. (2017-01-27)
Anton Baumann: in Revista de Sah 1936 soll Henry Forsberg eine weitere Möglichkeit gezeigt haben:
f) wSa6 statt sDa6: 1.Ka2 Sac5 2.Ka3 Ta4# (C+)
Ferner mit Märchenschachfiguren: vergl. P1349128 (2022-12-28)
Erich Bartel: Das namensgebende [Stamm]problem für dieses Thema (=Forsberg-Thema).--
Nachdrucke:
1) 141 Zwillinge und Mehrlinge 1952.---
2) Schach-Echo 11.Jahrgang,S.151, 1953.---
3) Problem (127-132) IX 1969,S.95.---
4) 9 feenschach (26) XII 1974.---
5) 83 Schach ohne Partner 1977.---
6) 67 100 classics of the chessboard.---
7) Belfort 1994,S.75.---
8) E-I Probleemblad I-II 2000.---
9) Harmonie (63) IX 2000.---
10) König und Turm (4) III 2001,S.13.---
11) Problemkiste (163) II 2006.---
12) 986 Minimalkunst im Schach 2006.--- (2007-08-20, hidden)
Erich Bartel: weitere Nachdrucke:
15) 5 The Problemist I 1962.--
16) feenschach (47) VII-IX 1979,S.185.---
17) XII Rex Multiplex (2) IV-VI 1982,S.38.--- (2009-01-28, hidden)
Alfred Pfeiffer: noch ein Nachdruck:
18) 6 "The Chess Review" 12/1945, S.25 (2012-03-08, hidden) more ...
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There was an error in position, corrected.
Henrik Juel: C+ Popeye 4.61 also (2023-04-14)
Gerald Ettl: b.)wLf6 c.)wSf6 (2009-03-06, hidden)
Mario Richter: An interesting coincidence: s. P0501247 (2009-10-08, hidden) more ...
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Yuri Bilokin: correction bQf1-c1, then a1=b1 8/8/4k3/8/3B3p/2bK4/7N/2q5 (3+4) h#4

1.Qg5 Bg1 2.Bf6 Sf3 3.Qd5+ Sd4+ 4.Ke5 Bh2# (MM) (2023-05-24)
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SCHRECKE: Das ist kein "Model mate", da wLh1 am Mattbild unbeteiligt ist! (2023-05-12, hidden) more ...
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{ NL}
klären AL
SCHRECKE: Geplant war wohl:
1. Kd6 Lh2+ 2. e5 dxe6 e.p. 3. Ke7 Sd5#
Nebenlösungen, z.B.:
1. Df6 Kb4 2. De6 fxe6 3. Kd6 Lh2#
1. Kd7 Lh2 2. Sc7 Sc4 3. De8 Sb6# (2023-09-20)
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{ NL}
klären AL
SCHRECKE: Satz:
1. ... c8=T+ 2. Kd7 Le6#
Es lösen:
1) 1. Ld8 cxd8=T+ 2. Ke7 Lh4#
2) 1. Kd7 cxb8=D 2. Kc6 Dc7#
3) 1. Ld6 c8=D+ 2. Ke7 Lh4#
4) 1. Kf8 Lf4 2. De8 Lh6# (2023-09-20)
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(KSp)
t/Allumwandlung, u/61103, t/sK-Korrespondenz
A.Buchanan: A pity that 1 & 3 both end with 2...Dfe7# (2023-03-14)
SCHRECKE: 1) exakt: 1. Kc7 e8=D 2. Db6 Dfe7# (2023-03-13, hidden) more ...
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KSp Königsstern
t/Allumwandlung, u/61104
Daniel Novomesky: Add bPh5. C+ (2008-03-03)
Yuri Bilokin: correction +bPh5 2N4K/3k2PR/6pq/2rp1rpp/8/8/b7/b7 (4+10)

Allumwandlung (white)
JT Onkoud 50 theme (double)
Promotion (QRBS, 4)
Star (bK)
Mates on the same square × 2 (2023-06-26)
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Jakob Leck: Version zu P0575203 (2022-10-08)
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Yuri Bilokin: correction +bQg4, +bPa6, -bNh4, -bPh2 8/1b1p4/p1Bn4/8/2kp1bq1/5R2/4p3/7K (3+9) h#4 0.2.1…
1...Rg3 2.Bd2 Rxg4 3.Kd3 Rf4 4.Ke3 Rf3# (MM)
1...Bxb7 2.Be5 Bc8 3.Kd5 Bxd7 4.Ke4 Bc6# (MM)

Areal cycle (wB, with captures, 4)
Areal cycle (wR, with captures, 4)
Exchange of functions (wRf3/wBc6, Mate / Passive guard)
Long-trip (wB, 4). Long-trip (wR, 4)
Model mate × 2 (2023-05-08)
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KSp Königsstern
Kees: Possible fix: +wBb4 -sLh1 (No 5.1... but 4.1... with only KSp Königsstern) (2023-06-05)
A.Buchanan: I agree with +wBb4 for soundness but let's keep sLh1. What's not to like about 5 well differentiated solutions? WinChloe has the same cooked diagram as we do here. (2023-06-06)
Yuri Bilokin: The fifth solution is not necessary, the author has kept it, so after adding the white pawn to the b4 square and without removing the black bishop, we fully preserve the author's intention.
8/4P3/1pP5/4kB2/1P6/4pp2/1R6/B2K3b (7+5) h#2 5.1...

Active sacrifice (black, delayed). BK moves only × 4. Delayed Umnov (bK-wB) × 2
Extended star (bK)
Promotion (QS, 2)
Battery mate × 3. Double-check mate. Mirror mate
Mate from initial bK square × 2 (2023-06-10)
A.Buchanan: What’s not to like about 5 solutions? :-) (2023-06-05, hidden) more ...
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Adrian Storisteanu: Possible fix (adjusted raised position): Kb8 Rd6 Rd8 / Kb3 Qg8 Rc2 Rc5 Sb4 Sb5 pp.d4 f7 (3+8) -- 1.Ka4 Rxd4 2.Qg3+ R8d6 3.Qb3 Ra6#. (2015-09-19)
Yuri Bilokin: correction wRd5-d7, bRc4-b6, bRc2-f6(-bPf6) K2R2q1/3R4/1r3r2/8/1n6/1n1p4/1k6/8 (3+7) h#3
1.Ka3 Rxd3 2.Qg2+ R8d5 3.Qb2 Ra5# (MM)

AntiZielElement (B2, check)
Bristol (white, wR-wR, impure, 4, 3)
Chernous theme
Self-pin/unpin (white)
Model mate × 1
Pin-mate (2023-05-02)
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Yuri Bilokin: correction wKa7-b1, +bNd1 7k/5p1b/3p4/8/8/8/1Pn5/1K1n4 (2+6) h#7

1.f5 b4 2.f4 b5 3.f3 b6 4.f2 b7 5.f1=B b8=B 6.Bc4 Bxd6 7.Bcg8 Be5# (MM)

Excelsior (black)
Excelsior (white)
Promotion (Bb, 1, 1)
Model mate × 1 (2023-05-10)
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Adrian Storisteanu: Possible fix: wKh1->a1, wSa7->b6. (2015-07-25)
Felber, Volker: Der Korrekturversuch aus der LB Seite 318 (wKa1-h1) reicht nicht aus. (2023-03-31)
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Verstellung TcL
Yuri Bilokin: possibly wPa2-a3 8/7p/6pr/1P1BP2b/2p4K/P6R/7P/Bk6 (8+6) (2022-12-19)
A.Buchanan: Yes looks like a typo (2022-12-20)
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F111
Verst-4
Felber, Volker: Die 'Verbesserung' reicht nicht. Original war wSe4 statt d5. (2023-02-21)
Rolf Wiehagen: C+ mit wLd2--a5, allerdings ohne die ws Linienöffnung bei 2. - La5-b4. (2023-02-21)
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Doppelveröffentlichung
Adrian Storisteanu: Possible fix: -bPe5, +wPf4 (7+6). (2015-11-27)
Yuri Bilokin: possibly wNc1-a7 8/N7/8/4p3/p2rk1P1/P1b3P1/3pNp2/3K4 (6+7) (2022-12-19)
HBae: Bei dem Korrekturvorschlag von Adrian wäre auch -wPa3 +bPa3 möglich, C+ Popeye v4.37 (2022-12-20)
Yuri Bilokin: Yes, it's better
There is no need to strain yourself by dragging the white knight to the a7 square :) (2022-12-20)
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Yuri Bilokin: you can see 1...e4 2.Kd1 e5 3.Ke1 e6 4.Kd1 e7 5.Ke1 e8=Q+ 6.Kd1 Qe2# (2023-01-14)
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Yuri Bilokin: you can see *) 1...Bc1 2.cxb1=R Ra7# (2022-12-15)
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Beim Original fehlte der Hinweis zu b) -sBf6 der auf Seite 347 ergänzt wurde.
Felber, Volker: Bei der Erfassung der Aufgabe wurde der sBe2 nicht berücksichtigt. Aufgabe ist nach dessen Ergänzung korrekt und die Fehlerhinweise habe ich gelöscht. (2023-04-15)
Henrik Juel: C+ Popeye 4.61 also (2023-04-15)
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Anton Baumann: Informalturnier 1991: 2. ehrende Erwähnung
PB in 'Die Schwalbe' Heft 171A (06/1998 S.489) (2023-01-21)
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klären : Lösung d?
SCHRECKE: a)-d): C+, popeye 4.87
a) 1. Th6+ Sxg5 2. Th5 Sxf3# (korrigiert)
b) 1. Lf4 Sxd4[+sTh8] 2. Th5 Sxf3[+sBf7]#
c) 1. Dh8 Sf4 2. Dh5 Sxg2#
d) 1. Lf4 Kxf6 2. Kh5 Kg5# (2023-03-06)
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klären Zwilling
Ladislav Packa: Lösung b) 1.kBa3-a2 kBc2-c4 2.kBa2-a1=kS! Le1-c3# (2023-02-25)
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Verst-4
Yuri Bilokin: possibly a1=c1, then wRe1-e3, bRd3-d1, bPd2-b3, +bBb6, +bPg7, +bPh7 8/6pp/1b1kbB2/5pK1/8/1p2R3/8/3r4 (3+8) h#3

1.Bc4 Re8 2.Rd5 Be7+ 3.Ke5 Bc5#

Cheney-Loyd theme (black)
Hideaway (bB, critical)
Indian (white)
Battery mate (2023-06-14)
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F123 Verst-4
Felber, Volker: Wegen NL wurde sBb5 in der LB vorgeschlagen. (2023-02-08)
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Yuri Bilokin: you can see in a) 1...Kc5 2.Kd3 Kb4 3.Kc2 Ka3 4.Kb1 Bf5# Gracefully and more meaningfully done later P1005864 (2022-12-12)
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SCHRECKE: Umwandlungsläufer sLh6! (2023-07-24, hidden) more ...
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Sally: Miniatur mit viermaliger Abwandlung von Turmblocks. Alle Lösungen enden zwar mattrein, sin aber alles andere als offensichtlich. (2018-10-25)
Yuri Bilokin: you can see 1...Bd1 2.Ke4 Rg2 3.Kf4 Rg4# (2023-06-07)
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Benkö in 'My Life': "Creating a double Queen sacrifice like this is quite an achievement, yet the judge in the Schach Echo helpmate contest didn't appreciate it and only awarded the problem an honorable mention. Later, this problem became one of the most popular examples of the bad judging that is so common in publications."
TLi: a) There are no sacrifices, only captures.
b) The wQ can be omitted without changing the solutions.

Bad judging, indeed! (2023-02-16)
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Yuri Bilokin: you can see 1...Be4 2.Sh2 Rf3 3.Kh1 Rf1# (2023-06-12)
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Yuri Bilokin: correction wKg4-f4, -bBa6, +bRe7, then a1=c1 6N1/5Pr1/4k3/8/2P4K/8/8/8 (4+2) h#2*

1...f8=Q 2.Rd7 Qf6#
1.Rg5 f8=Q 2.Rf5 Qe7#

Echo mates (mirrored diagonally, 0, 0)
Promotion (QQ, 2)
Echo (2023-06-04)
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Yuri Bilokin: possibly wKg8-h8 7K/4p3/7k/6p1/8/5p2/6P1/8 (2+4) (2023-02-01)
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Erich Bartel: Nachdruck:
Die Schwalbe (2) XII 1969 (2013-02-22)
Yuri Bilokin: possibly wKb3-a2 7k/1p4pn/8/8/8/8/K4P2/8 (2+4) h#7
1.b5 f4 2.b4 f5 3.b3+ Ka3 4.b2 f6 5.b1=B f7 6.Ba2 f8=S 7.Bg8 Sg6# (MM) (2023-02-01)
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Yuri Bilokin: correction wKd1-h1, bBe1-a7, bBf1-a2, bPg2-h2, +bNb8, +bNe7, +bPa6, + bPd7, +bPf6 1nR5/b2pn3/pp3p2/3k4/R1r3pq/8/b3r2p/7K (3+14) (2023-05-02)
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Harry Fougiaxis: The composer is Vratislav HADAC (2006-08-11)
Felber, Volker: Schach schreibt "V. Michal-Hadac" (2023-08-04)
Vaclav Kotesovec: The names of the authors are correct: Vratislav Hadac and Dr. Václav Míchal. See also other problem by these authors http://problem64.beda.cz/silo/ceskoslovensky_sach_1968.pdf, p.6 (in PDF). (2023-08-04)
Felber, Volker: Autor ist geklärt! (2010-10-09, hidden) more ...
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SCHRECKE: C+, popeye 4.87
1. ... Sb6 2. Gca8 Sd7+ 3. Ka7 Tc7#
1. Ka7 Sc7 2. Gaa8 Sb5+ 3. Kb8 Tb6# (2022-11-30)
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Standfeld des sK klären
SCHRECKE: C+, popeye 4.87
1) 1. Td2 Le5 2. Tf3 Lc2+ 3. Tdd3 Sc3#
2) 1. Td3 Lc2 2. Tb5 Le5 3. Tbd5 Sd2# (2023-09-20)
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James Malcom: I don’t believe that the great Fabel, with his keen eye, would have not seen such a simply cook as the taking of Black’s c5 pawn. I suspect that a Black bishop was on d4 but was forgotten placing the problem in the database, or it was a misprint. I am changing this problem to not cooked for now. However, someone should check the original source in the event I am wrong, which is always a possibility. (2020-03-15)
Mario Richter: Hi, James, there's indeed a black bishop on d4. Original stipluation: "h#48 (nicht exakt!)"
A scan for some issues of 'Schachmatt' is available in the archive at stere.ro, see
www.stere.ro/wp-content/uploads/2014/06/stere_sah_istoria_sahului-1947-1949-Schachmatt1.pdf (2020-03-26)
Joost de Heer: Stupid question, but won't moving Bh3 to g4 add another halfmove? (2023-01-30)
Henrik Juel: Yes, Joost, but this would make the solution even less exact
0... Lf3,Lh3,Lh5 1.g4 Lxg4 etc. (2023-01-30)
Joost de Heer: @Henrik: So? This is an attempt at a non-unique (exact is something else) h# length record, the level of non-uniqueness is irrelevant. And as James pointed out, adding Kg4-f4 will add yet another move.
Only possible cooks are when white starts with 1... Bh5 2. g4 Be8 (so not capturing Pg4) but I don't see how that could lead to a quicker mate. (2023-01-31)
James Malcom: I double-checked that myself, Joost, and found that going around for wPg2 actually takes far longer, which would be required any PxB action to take place, but that takes far longer than the normal solution. I know Olaf's Gustav would check, since it C+ the h#111. (2023-01-31)
James Malcom: Is this the record for the longest legal, but dualed, helpmate? It is as far as I know. Is it possible to correct this problem with a black bishop placed on d4 (this is legally possible as Black has a pawn to spare), which would uphold the original solution? (2020-01-27, hidden) more ...
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Jakob Leck: Nachgedruckt in Bakcsi (Hrsg.): Ungarische Schachprobleme mit wenigen Steinen, dort lauten die Quelle und Auszeichnung für das Problem (Nr. 537 im Buch): Imre Tóth, Thematurnier des Budapester Schachverbands, 1958, 3. Preis (2023-05-29, hidden) more ...
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mecislovas: das Berichtung: Tf1-e1,Tg1-f1,Dh1-g1(!)- ?+ (2012-09-25)
Yuri Bilokin: correctiom wRf1-g1, bRg1-h1(-bQh1), bNa2, bNe2, -bBh6 2R5/8/3p4/8/8/p2k4/n3nppp/K5Rr (3+9) h#3

1.Kd2 Rb1 2.Rd1 Rcc1 3.Ke1 Rxd1# (MM)

Active sacrifice (black, delayed)
Bristol (bicolor, wR-bR, impure, 5, 4)
Brunner-Turton doubling
Pelle move (white)
Model mate × 1 (2023-05-02)
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Yuri Bilokin: correction bOb6-c7 8/2q5/3k1b2/3p4/2pn4/pp3N2/p1p1n3/Kb2r3 (2+13) h#6
1.Se6+ Sd4 2.Ke5 Sxc2 3.S2f4 Sxa3 4.Bf5+ Sb1 5.Re4 Sd2 6.Qd6 Sf3# (MM)

Annihilation
AntiZielElement (B1, check). AntiZielElement (B4, check)
Areal cycle (wS, with captures, 6)
Chernous theme. Crusader theme
Long-trip (wS, 6)
Self-pin/unpin (white) × 2
Model mate × 1 (2023-05-05)
A.Buchanan: sTd7 works as well as sDc7, I think? (2023-05-06)
A.Buchanan: It would be great to integrate the Helpmate Analyzer keywords, definitions and discovery some time. (2023-05-06)
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Jakob Leck: Siehe Version P0503644 (2022-10-08)
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A.Buchanan: If one wanted a unique solution, -wLc4 +wBd6 seems the only way (2023-04-24)
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Henrik Juel: The set play is C+ Popeye 4.61 (2022-09-15, hidden)
paul: The stipulation tested by Jacobi (in about 10 hours) with this code:
stip ser-~1 forsyth br6/bk1K4/p3p3/1P6/3p4/8/8/5q1r
stip ser-h#7
The output: 1.b6 Rc8 2...Kb8 3...Bg2 4...Qf3 5...Qa8 6...Bb7 7...Rc7+ 8.bxc7# (2022-09-17, hidden) more ...
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Henrik Juel: C+ Popeye 4.61 (2023-02-02)
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SCHRECKE: C+, popeye 4.87 (2023-02-02)
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Erich Bartel: das ursprüngliche Diagramm (s.4735 feenschach (76) 8.11.1985
war vielfach NL. Dieses Diagramm, wohl als Korrektur gedacht,
ist dafür unlösbar. (2006-12-01)
Anton Baumann: die Autorabsicht scheitert an 4.Le2!
Mögliche Korrektur: -sLa6 +sDa7 (C+ Alybadix);
oder wenn 2 sD unbeliebt: +wLf1 +sBf6 (C+ Alybadix) (2023-09-06)
Gerald Ettl: Bei mir ist als Korrektur sLa6 nach a7 angegeben. (2023-09-06)
Anton Baumann: sLa7 ist C+; es ist anzunehmen, dass dies die Absicht der Autoren war! (2023-09-06)
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Anton Baumann: C+ Alybadix (2023-06-24)
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.31 00:06:21 Minuten. (hh:mm:ss)
Keine Lösung: BP 36.0.
Beispiel Stelvio: 1.c3 a5 2.d3 a4 3.Le3 a3 4.Sd2 Ta4 5.Tb1 Tc4 6.dxc4 Sa6 7.g4 b5
8.g5 b4 9.g6 b3 10.gxh7 bxa2 11.b4 a1=S 12.b5 a2 13.b6 Sb3 14.h4 g5 15.h5 g4
16.h6 g3 17.Th5 g2 18.Sh3 g1=D 19.b7 Dg5 20.b8=D Db5 21.Tf5 c5 22.Df4 a1=D 23.Dg4 Sa5
24.f4 Sb7 25.Lg1 Daa5 26.e3 e6 27.Le2 Ld6 28.Sf1 Lb8 29.Dd2 Se7 30.Kd1 Tg8
31.h8=S d6 32.Sg6 Ld7 33.Sh4 Sc8 34.Sg2 De7 35.h7 f6 36.h8=D exf5 37.Dhh4 (2023-06-20)
Henrik Juel: cooked, e.g.
1.Bd2-d3 Ba7-a5 2.Lc1-e3 Ba5-a4 3.Sb1-d2 Ba4-a3
4.Bg2-g4 Ta8-a4 5.Bg4-g5 Ta4-c4 6.Bd3xc4 Bb7-b5
7.Bg5-g6 Bb5-b4 8.Bg6xh7 Bb4-b3 9.Bh2-h4 Bb3xa2
10.Bb2-b4 Be7-e6 11.Bb4-b5 Bg7-g5 12.Bb5-b6 Bg5-g4
13.Bb6-b7 Bg4-g3 14.Bh4-h5 Bg3-g2 15.Bh5-h6 Sb8-a6
16.Th1-h5 Bf7-f6 17.Ta1-b1 Ba2-a1=S 18.Sg1-h3 Bg2-g1=D
19.Bb7-b8=D Dg1-g5 20.Bc2-c3 Dg5-b5 21.Th5-f5 Bc7-c5
22.Db8-f4 Lf8-d6 23.Df4-g4 Ld6-b8 24.Bf2-f4 Sg8-e7
25.Le3-g1 Ba3-a2 26.Be2-e3 Sa1-b3 27.Lf1-e2 Ba2-a1=D
28.Sd2-f1 Sb3-a5 29.Dd1-d2 Sa5-b7 30.Ke1-d1 Th8-g8
31.Bh7-h8=S Da1-a5 32.Sh8-g6 Be6xf5 33.Sg6-h4 Bd7-d6
34.Sh4-g2 Lc8-d7 35.Bh6-h7 Se7-c8 36.Bh7-h8=D Dd8-e7
37.Dh8-h4 (2018-12-10, hidden) more ...
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Henrik Juel: This fine proof game by Gianni is faster to solve by hand than by computer
Last move was not 19.Lb5xPd7+, because Sa1 did an uneven number of moves; 18.Lb5xPd7+ fails, because White needs 19 moves, since Td1-d5 must happen early. So [Pd7] has promoted, and Black must play 1... d6 and 2... Dd7 or Ld7; only the latter works
Partially tested by Natch 3.1 in 18.0 moves with wLb5 and sDTLSPd7; only L works (2018-12-14)
Henrik Juel: Typos: solution should end 18.Td3 L6d7 19.Lxd7+ (2018-12-14, hidden)
Moldenhauer: Computerprüfung: C+ Euclide 1.11 nach 237 Std. 28 Min.
Keine Lösung bei BP 17,5 , geprüft in 1 Min. (2023-01-22, hidden)
Moldenhauer: Computerprüfung: C+ auch Stevio 1.11 00:27:56 Minuten. (hh:mm:ss)
Keine Lösung: BP 18.0. 4994/419k Histogramm. (2023-03-16, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Euclide 1.11 in 16 Std. 23 Min.
Keine Lösung in BP 16,5, Zeit 50 Min.
Notationsfehler: 6. Tg2 muss 6. Lg2 heissen. (2023-01-23, hidden)
Bob Baker: C+ Stelvio 1.4 in 17 seconds. (2023-08-29, hidden)
Bob Baker: In listed proof game: Change 6. Tg2 to 6. Lg2. Change 13. Sc3 to 13. Sbc3. (2023-08-29, hidden) more ...
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Henrik Juel: Partial testing with Natch 3.1
18.0 moves with sDTLSPe7 and wTd5; only sSe7 works (2018-12-14)
Mario Richter: @Henrik: Did you mean "... and wTe4"? (2018-12-15)
Henrik Juel: Yes, I did, Mario (2018-12-16)
paul: Checked by Stelvio 1.2 in about 5 hours. (2023-05-29, hidden)
Moldenhauer: Computerprüfung: C+ Stelvio 1.2 04:02:35 Stunden. (hh:mm:ss)
Keine Lösung BP 17.5, BP 18.0. (2023-05-30, hidden) more ...
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Henrik Juel: Natch 3.1 did nothing for an hour (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 © Reto Aschwanden
33 Minuten, 1 Lösung, Notation OK.
Stelvio Histogramme/Strategien 40, vor der Lösung wegen der Lösezeit
mit Stelvio berechnet. (2023-02-08)
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