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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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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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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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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Adrian Storisteanu: See twin P1216951. Another possible twin (see Thomas Brand's Retroblog 11/2019):
wKc7->.h5
- 1.Kg4xRh5 Re3xQb3 & 1.Re5 Qc4# (2024-03-19)
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Paulo Roque: obs: 23...Kd7 24.T8h3 f6 (2008-12-29)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 184:20:33 Stunden. (hh:mm:ss)
1.d4 Sa6 2.Ld2 Sb4 3.e3 Sf6 4.Ld3 c5 5.Ke2 c4 6.h4 c3 7.h5 cxd2 8.c4 Sc2 9.c5 Sg4
10.c6 e5 11.c7 Lc5 12.dxc5 d6 13.c6 Le6 14.c8=S Dg5 15.Se7 Sh2 16.Sf5 Sf1 17.c7 a5
18.c8=S a4 19.Sce7 a3 20.Sg6 axb2 21.a4 hxg6 22.a5 Th6 23.a6 gxf5 24.a7 Tg6
25.h6 Kd7 26.h7 f6 27.h8=T Sa3 28.T8h3 Th8 29.a8=T Thh6 30.Ta4 Lg8 31.Te4 fxe4
32.f4 Dg4+ 33.Tf3 exf3+
Keine Lösung: BP 32.0 wegen der Retraktion. Da NUPG C+. (2023-09-28)
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Viktoras Paliulionis: The last move couldn't be d2-d4 because Black's pawns had already captured 10 of White's pieces, and playing d2-d4 would have resulted in wBc1 being White's 17th piece. So, the only solution is 1.gxf3 e.p. (2023-12-30)
A.Buchanan: Yes that's right! (2023-12-31)
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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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A harbinger of Dead Reckoning-style reasoning, before that rule ever existed. As sound today in 2018, as when it was composed under "insufficient material for mate" rule.
Henrik Juel: Nice to see that I am not the only freak who mixes english text with german notation (to avoid the ambiguous B)
Were you inspired by this problem in your work on Dead Reckoning, Andrew? (2020-05-04)
A.Buchanan: Ha no: when I stumbled across DP rule I knew even less about chess problems than I know now, if you can imagine, and had never encountered this one which Werner Keym unearthed and published it in "Out Of the Box". Ecclesiastes 1:9 says "There is nothing new under the sun." Mario made this point about origin of Schnoebelen promotions too. Gollon's intepretation of Chaturanga (the earliest known forebear of chess) offers fascinating possibilities for Schnoebelen promotion, which I am sure the Indian retrograde analysts of 7th Century CE were adept at exploiting, in tablets alas now lost. See https://www.chessvariants.com/historic.dir/chaturanga.html.

The "insufficient material" composition does perhaps explain Plaksin's openness to the concept of DP, at a time when perhaps some noses elsewhere were being turned up, as he later composed with Kornilov *27* other DP problems, all to be found in PDB. (2020-05-04)
Ladislav Packa: Interesting, but on YouTube I saw a game (one of the actors was Magnus Carlsson), which ended in a position with only two kings and one of the players still made a move. (2020-05-05)
A.Buchanan: http://wismuth.com/chess/illegal-moves.html contains many other games where GMs were unable to tear themselves away from playing the dead game. I think it's kind of sweet that the GMs just can't stop playing. Have you got a link to the Carlsen game please, Ladislav? (2020-05-06)
Ladislav Packa: I'm sorry, Andrew, but I don't remember. I've seen hundreds of games on youtube. But I encountered similar cases in my own practice, mostly a joke, and the game immediately ended with a smile and a handshake. (2020-05-06)
A.Buchanan: A completely different stipulation: -w, h=1 half-duplex (2024-01-11)
A.Buchanan: http://wismuth.com/chess/illegal-moves.html contains many other games where GMs were unable to tear themselves away from playing the dead game. Actually, I got this wrong for many years, but now I realize now that these moves aren't "illegal". They are impossible, because the game is already over. Legality/llegality is defined in the FIDE rules - basically just about the moves of the pieces and staying out of check. You list the legal moves, and if there are none the game is over, otherwise if there's no later checkmate possible, the position is dead. So Francois should write about "impossible" moves. But this doesn't affect the logic, just the terminology. I think it's kind of sweet that the GMs just can't stop playing, and I'm happy no arbiter stepped in, because maybe if one did then they would think about changing the rule. (2020-05-05, hidden) more ...
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A.Buchanan: In both candidate solutions, counting of wPs after retraction on means it must be an original unit uncaptured on h1 so can only be wS.
The solution for b) does not work for a) because, if one retracts two promoting moves, then wPs have captured 9/10 missing Black units, while Black has 1/2 accounted for by wBc1. However wPg & bPg passed one another, which requires two more captures by a single player.
In b) the exciting retraction Qa8xh8 is blocked, however now the free pawn captures to explain g-file are shifted from 1+1 to 2+0, so bPg can pass wPg. But the vacation of d2 allows a new cook. This is fixable by shifting wPb instead of wPd. We don't need to shift it to an adjacent file: the solution is still ok if the balance is 2+1.
So "b) wPb2 to b7" fixes the cook, and it's validated by deadpos. (2024-01-15)
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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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Frank Müller: Originalforderung: In wieviel Zügen läßt sich vorstehende Position aus normaler Anfangsstellung entwickeln?
Eine Zugzahl war nicht angegeben! (2012-08-11)
Moldenhauer: Computerprüfung: Cooked Stelvio 2.0 in 1 Sek.
Keine Lösung: BP 17.0, BP 17.5.
C+ wegen der Forderung.

1.Sa3 Sa6 2.Sb5 Sc5 3.Sd4 Se4 4.Se6 Sg3 5.Sxf8 Sxh1 6.Sg6 Sg3 7.Sxh8 Sxf1
8.Sg6 Se3 9.Se5 Sf5 10.Sc4 Sd4 11.Sb6 Sb3 12.Sxa8 Sxa1 13.Sb6 Sb3
14.Sxc8 Sxc1 15.Sb6 Sb3 16.Sa4 Sd4 17.Sc3 Sc6 18.Sb1 Sb8 (2024-01-09)
A.Buchanan: If the composer makes no claim that the solution is unique, it’s unfair to call it “cooked” when multiple solutions are unearthed. “Cooked” means “Kaput”, “broken”. Please stop. (2024-01-09)
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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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Bernd Schwarzkopf: zurück: 1.Kg4xBg3 Kd3-c2 2.Kg5-g4 Ke4-d3 3.Kh5xBg5 Kf5-e4; vor: 1.g3-g4# (2024-03-29)
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siehe 9200
mri: Mehrere NL, z.B. R: Kc6xLd6 Lc5xDd6 V: 1.Ka5-a6 Dd6xc5 oder R: Kc5xTd6 Tb6xDd6 V: 1.Tb6-c6 Dd6xc6 (2007-10-25)
Mario Richter: s. P0005054 (2009-12-31)
A.Buchanan: Reasonable to suppose that P0005054 is the correction (2024-01-15)
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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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Adrian Storisteanu: Two wPs (e.g. +wPe4,+wPh4) can restore the glory to this nice, thoroughly mixed AUW... (2015-06-10)
A.Buchanan: Or instead +wPa5 or +wPc5. In either case, free to "downgrade" wBg1 to wS. This is more economical than adding two pawns, but separates the two islands conceptually, as b6 no longer needs to be controlled by wBg1. (2024-01-15)
mri: cooks: R: Da4-a8 Sf2xLh1 V: 1.Kxb6 Lxg2

R: Da4-a8 Sf2xTh1 V: 1.Kxb6 Th7 (2004-11-17, hidden) more ...
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Adrian Storisteanu: Possible fix: +wPh5 (4+4). (2015-08-05)
A.Buchanan: A cool fix which keeps miniature status is -bSd8 + bRd8 (2024-01-15)
A.Buchanan: Alas my suggestion is anticipated by P1202424, ah well. (2024-01-16)
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paul: Add wPa2b2b3e2, bPa7c7e7g7. (2022-09-18)
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SCHRECKE: In a) geschah zuletzt 0. ... Td8-a8, also ist die Rochade nicht mehr möglich.
In b) kam zuletzt die sD von h1, schwarze Rochade also noch spielbar.
Lösung:
a) 1. gxh5 Td8 2. Df7+ Kxf7#
b) 1. Dg6+ Dxg6 2. Sf5 0-0-0# (2023-11-07)
Joost de Heer: Slight error in previous comment: Last move was Td8xT/Da8, not Td8-a8 (as Td8-d1 would be longer) (2023-11-07)
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Olaf Jenkner: Wieso wird Weiß pattgesetzt? (2010-03-02)
Mario Richter: Weil der Komponist das so wollte :-)
In fs33 ist die Forderung wie folgt wiedergegeben: "Wei� nimmt einen Zug zurück und hilft, da� Schwarz sofort pattsetzen kann." (2010-03-03)
Olaf Jenkner: Ach so, h=1 bedeutet eigentlich, daß Schwarz beginnt. (2010-03-03)
Henrik Juel: Yes, it is a bit confusing; but for retros it is an old tradition that you should not interpret one move stipulations too rigorously. (2010-03-05)
Mario Richter: But at least retro play and forward play harmonize: the side that retracted last starts the forward play. (2010-03-05)
Adrian Storisteanu: Can easily make it conform to today's accepted standards (i.e., drop the half-duplex indication):

Ka1 pa2 / Kc1 pb3 (2+2)
-1b & h=1

- 1.a4xQb3 & 1.a4-a3 Qb3-d3= (2024-01-13)
Olaf Jenkner: Das wären dann viele Mütter. (2024-01-14)
A.Buchanan: @Olaf :-) (2024-01-15)
Adrian Storisteanu: :-) Just in reverse Polish notation...
(Didn't know the many fathers thing started that long ago. I totally missed both the obvious setting and the recorded keyword.) (2024-01-15)
A.Buchanan: The Originalforderung is quite clear. There is no need to represent this stipulation as a joke here. Popeye uses the unambiguous term "half-duplex" for this situation, which occurs in other help-retractors. (2024-01-12, hidden) more ...
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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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Moldenhauer: Computerprüfung: C+ Stelvio 2.1 ca. 30 Minuten. 3S/12P 64 GB Hash.
Der Cook ist keine Cook weil die Endstellung nicht mit dem Diagramm
zusammenstimmt. Letzter Zug Lf3?
Nur die weiße kleine Rochade fand wirklich statt.
Keine Lösung: BP 19.5. (2024-03-24)
A.Buchanan: Please forgive me for being stubborn, but I don't think just checking BP20.0 & BP19.5 is sufficient for C+. The test must include BP19.0 & BP18.5, and these should be very quick to test compared to the first two. They can be done first, which would allow for some short cooks to be identified more quickly. Maybe Stelvio could offer an option for this "fourfold" test, to save on typing, which is the main inconvenience? (2024-03-25)
Moldenhauer: Hi Andrew, diese Frage hatten wir schon einmal. Damals antwortete Reto darauf.
In der Dokumentation zu Stelvio 2.1, Seite 6 unten und Seite 7 oben, steht die Erklärung. (2024-03-25)
A.Buchanan: Thanks Moldenhauer. Yes, Reto responded today to say: "Actually 20 and 19.5 are sufficient, Stelvio always checks for any solutions of same parity, so 20 includes 19, 18, ... and 19.5 includes 18.5, 17.5... I can see your point with adding an option for automatic double checking, I've added it to my list."
That's great, so in my opinion the PDB C+ report needs to indicate that Stelvio has also tested for the shorter solutions of the same parity, as this is tool-dependent behaviour which future readers might not know. Ideally this statement would appear at the end of the Stelvio output, as a kind of "certificate" that could be copied to PDB.
A problem with this whole issue is that there are different choices being made by different people. It complicates things unnecessarily. I think C+ for a PG should mean that the three shorter lengths have also been explored. This should be doable without getting into the fraught question of whether a shorter solution cooks a PG, although in my opinion it does, whatever the parity: a PG is just a diagram. Conceptually to me, PG in x.y is just SPG, with a courtesy hint to the solver of what length they are looking for. (2024-03-26)
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A.Buchanan: What is MsP, please? (2024-01-12)
Henrik Juel: Mate instead of stalemate (Mat statt Pat)
In the diagram position Black is stalemated; White retracts and mates (2024-01-12)
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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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Retro!-patt, Aufspaltung Doppelschritt!, Excelsior, Umwandlung
Erich Bartel: Nachdruck:
1) 365 Minimalkunst im Schach 2006.--- (2008-03-08, hidden)
A.Buchanan: Currently, the stip is "h#5 0.2.1..." Really not a fan of the "0." usage. Normally, I would prefer "h#4.5 2.1..." but here it should be "h#5 2.1...". (2023-12-07, hidden) more ...
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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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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: 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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A.Buchanan: the authors fixed this in 1998 C+ with 1sbq1ksr/2p1ppbp/p2R2p1/1Q2B3/PpP1BP1K/3P1SP1/1P2P2P/R4S1r
16. Lg2 Te3 17. Le4 Tf3 18. Sd2 Tf2 19. Sgf3 Tg2 20. Ta1 Tg1 21. Sf1 Th1 (2023-02-04)
Henrik Juel: Cooked, e.g.,
1.Bc2-c4 Bb7-b5 2.Dd1-a4 Bb5-b4 3.Da4-b5 Bg7-g6
4.Ba2-a4 Lf8-g7 5.Ta1-a3 Ke8-f8 6.Ta3-d3 Bd7-d6
7.Td3xd6 Sb8-a6 8.Bd2-d3 Ta8-b8 9.Lc1-f4 Tb8-b6
10.Lf4-e5 Tb6-c6 11.Bf2-f4 Tc6-c5 12.Ke1-f2 Tc5-d5
13.Kf2-g3 Td5-d4 14.Kg3-h4 Td4-e4 15.Bg2-g3 Te4-e3
16.Lf1-h3 Te3-f3 17.Lh3-g4 Tf3-f1 18.Sg1-h3 Tf1-e1
19.Th1-f1 Sa6-b8 20.Tf1-f3 Te1-h1 21.Tf3-e3 Ba7-a6
22.Te3-e4 (2018-12-12, hidden) more ...
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Henrik Juel: Natch 3.1 did nothing in almost an hour (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 84:43:01 Stunden. (hh:mm:ss)
Keine Lösung: BP 22.0, BP 22.5. (2023-04-01, hidden) more ...
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paul: Stelvio 2.0 solves this PG in 3 minutes. (2024-04-02, hidden) more ...
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Henrik Juel: White has the move. -1... Sf8 -2.g5 Se6 -3.g4 Sc5 -4.Sd8 Sg7 -5.Se6 Ta2 ... -9.S=h7 Ta2 ... -13.h2! h3xSg2 -14.Sf4 Ta2 ... -17.Sb3 Ta1 -18.La2 Tg1 ... -22.La2 Te7 -23.Td3 Td7 -24.Tg3 d3 -25.Tg1 d4 -26.Ta1 d5 -27.Lb1 h4 -28.Sc1 h5 -29.Sd3 Lf8 -30.Sb4 e7xLf6 -31.Lg5 d6 -32.Lc1 Kd4 -33.d2xSc3 etc. Enjoyable. (2004-01-05)
Henrik Juel: In the official solution 16.Sd8-e6 is illegal, because wKc6 is checked (this seems fixable)
My 2004 solution also works, so the problem is almost cooked (2023-07-23)
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Henrik Juel: Natch 3.1 did nothing in half an hour (2018-12-14)
paul: Correction of P1000629 (2022-10-12)
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Mario Richter: brauchen wir wirklich zwei verschiedene Keywords - hier "Tempoverlustmanöver", bei z.B. P1007491 hingegen "Loss of tempo"? (2018-10-21)
Henrik Juel: Good point, Mario
Other things in the never-ending Keyword discussion include:
Tempoverlustmanöver is currently called the same in english and french, and has no definitions
Loss of tempo (Tempoverlust in german) is defined as
'With White to play, a sequence of moves to reach the same position with Black to play. (Help game, abbr .: TV)'
which makes little sense in proof games
In this problem [Pe2] goes e2-e3-e4 ('wasting' a move)
In P1007491 [Pc2] similarly goes c2-c3-c4-... and [Lc8] goes (among other things) c8-b7-a6-c8 ('triangulation') (2018-10-21)
Henrik Juel: There are many solutions in 6.0 moves, like
1.Sb1-c3 Sg8-f6 2.Pe2-e4 Sf6xe4 3.Sc3-a4 Se4-c3 4.Ra1-b1 Sc3xb1 5.Sa4-c3 Sb1-a3 6.Sc3-b1 Sa3-c4
Correction: Change the stipulation to
BP in exactly 6.5 (2024-03-10)
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paul: Un dernier coup noir s'impose. Voir P0005906. (2010-05-08)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 1 Sekunde.
Keine Lösung: BP 24.5, BP 25.0. (2023-02-17)
Reto: There are in fact solutions in exactly 25.0 moves and Stelvio finds those. So the above comment is incorrect. (2023-06-12)
Moldenhauer: Danke für die Korrektur. In meiner Worddatei im Bildschirmausschnitt steht cooked,
also habe ich die Eingabe hier falsch gemacht. Sorry! (2023-11-05)
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Moldenhauer: Computerprüfung: C+ Jacobi v0.7.6 beta1 PG demolition mode
Notation B: 8/8/8/8/8//8/r1N5/k1K5 (2022-08-11)
Wilfried: Position B is missing. It is: Kc1 Sc2; Ka1 Ta2
Then:
1... ... 2.Tg1-f1 + Kf2-e2 3.Tf1-d1 Sh1-f2 4.Td1-f1 Sf2-d1 + 5.Kb2-a1 Sd1-c3 6.Tf1-e1 + Ke2-d2 7.Te1-c1 Sc3-b5 8.Tc1-b1 Sb5-a3 9.Tb1-b2 Kd2-c1 10.Tb2-a2 Sa3-c2 #
1...Sh1-g3 2.Tg1-f1 + Kf2-e3 3.Tf1-f3 + Ke3-e2 4.Tf3-c3 Sg3-e4 5.Tc3-c2 Ke2-d1 6.Tc2-c4 Se4-d2 7.Tc4-c2 Sd2-c4 + 8.Kb2-a1 Sc4-a3 9.Tc2-d2 + Kd1-c1 10.Td2-a2 Sa3-c2 # (2008-05-15, hidden) more ...
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Henrik Juel: Natch 3.1 found the solution quickly, but did not finish in more than an hour (2018-12-16)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 nach nur 1 Sekunde. (2023-02-09)
Henrik Juel: and no solution in 17.5 moves, so C++ Stelvio 1.0 (2023-02-09)
Henrik Juel: Fine motivation for the round trip: making room for white Sd7-b8-a6 (2023-02-09)
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Henrik Juel: Natch 3.1 did not find the solution in half an hour (2018-12-16)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 01:37:02 Stunde.
Keine Lösung: BP 19.0, BP 19.5. (2023-02-28, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Stelvio 1.0 33:18 Minuten.
Keine Lösung BP 20.0, BP 19.5.
Bitte Notation richtigstellen: 19.Sc4 Tg5 (2023-02-13)
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Moldenhauer: Computerprüfung: C+ Euclide 1.11 in 9 Std. 5 Min. 37 Sek.
Notation OK. Nur eine Lösung! Keine Lösung in BP22.0 und BP 22.5. (2022-12-23)
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Henrik Juel: Natch 3.1 did little in an hour (2018-12-16)
Vladimir Rey: Fix: P0115 StrateGems; 10. a5 Dxa5+ (2023-01-21)
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Moldenhauer: Computerprüfung: C+ Stelvio 1.0 5 Sekunden.
Keine Lösung: BP 21.5, BP 22.0. (2023-02-18)
paul: Correction of P1003987 (2010-05-15, hidden) more ...
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paul: wPd4 is superfluous. (2023-12-07)
paul: C+ (pacemaker) (2010-10-02, hidden) more ...
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Kostas Prentos: This version is cooked: 1. h4 g5 2. hxg5 b6 3. Th5 La6 4. b3 Ld3 5. exd3 Lg7 6. Df3 Lxa1 7. Dxf7+ Kxf7 8. Le2 Ke6 9. Ld1 Kd5 10. Lb2 Kc5 11. Ld4 Kb4 12. Sc3 Ka3 13. g6 Kb2 14. a3 hxg6 15. Sa2+ Kxa2 16. Lb2 Th7 17. Lc1 gxh5 18. c3

A correction was published in StrateGems, 2018 (No.82/Pg.92, P0118v): PG 17,5:
1. h4 b6 2. h5 Lb7 3. h6 Lf3 4. hxg7 Sc6 5. gxf8=S Tb8 6. Se6 fxe6 7. Th5 Kf7 8. Td5 exd5 9. exf3 Ke6 10. Lc4 dxc4 11. c3 Kd5 12. Db3 cxb3 13. a3 Kc4 14. Ta2 bxa2 15. Ke2 Kb3 16. Kd3 a1=L 17. Se2 Ka2 18. b3 (C+) (2022-12-08)
Joost de Heer: Solution:
1. h4 b6 2. h5 Ba6 3. h6 Bd3 4. hg7 h5 5. gf8=S Rh7 6. Se6 fe6 7. Sc3 Kf7 8. Sd5 ed5 9. Rh4 Ke6 10. Rc4 dc4 11. c3 Kd5 12. Qb3 cb3 13. a3 Kc4 14. Ra2 ba2 15. ed3 Kb3 16. Be2 a1=B 17. Bd1 Ka2 18. b3.

Black king follows a pawn. (2003-08-12, hidden)
Henrik Juel: Natch 3.1 spent more than 2 hours on the full PG
The first 17.0 moves were tested correct (0.5 sec.) (2018-12-17, hidden) more ...
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Henrik Juel: Natch 3.1 found the solution but did not finish testing in more than an hour (2018-12-18)
Henrik Juel: Stelvio 2.0 is slightly faster than Stelvio 1.14: 55 sec. (2024-02-04)
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paul: Tested by Euclide 1.11 (2020-02-26)
Henrik Juel: wPa2 promotes on b8 to S, which goes to g1 to screen and then back to b8, where it is captured
Besides CF and Pronkin, also Donati theme (away from and back to b8) (2020-02-26)
paul: Stelvio 2.0 solve it in few seconds. (2024-04-08)
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TT: proof game with capture of Anti-Pronkin piece
paul: Stelvio 2.0 found the solution in few seconds. (2024-04-08)
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Henrik Juel: C+ Stelvio 2.0, 19 seconds
No shorter solutions (2024-01-21)
Henrik Juel: Euclide 1.01 did not even find the solution in 10 minutes (2024-01-21)
Henrik Juel: ditto in some 6 hours, so Stelvio is the clear winner here (2024-01-21)
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Henrik Juel: The retroplay does not show
Addition of wSh8 is missing (2024-01-14, hidden) more ...
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