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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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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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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A.Buchanan: Setplay doesn't care whether the position is legal. (2023-12-06)
A.Buchanan: So is this a retro problem? I think not. (2023-12-06)
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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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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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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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paul: Stelvio 2.0 found the solution in 2 minutes. (2024-04-02, hidden) more ...
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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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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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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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{ 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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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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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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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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SCHRECKE: Umwandlungsläufer sLh6! (2023-07-24, hidden) more ...
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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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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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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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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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SCHRECKE: C+, popeye 4.87 (2023-01-17)
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