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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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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Joost de Heer: The keyword is wrong (should be TxD, not BxS=T) (2023-05-20, hidden)
Henrik Juel: I believe that the condition 'Black to move' can be deleted (2023-05-21, hidden)
Adrian Storisteanu: This is not the magazine's informal tourney - p.137 of the indicated source says REZULTAT 16. TEMATSKOG TURNIRA "PROBLEMA". (2023-05-21, hidden) more ...
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paul: Correction of P0001580 (2010-09-25)
Silvio Baier: Mit 30....dc+ ist das C+ (Euclide 0.98, Retractor für den letzten Zug) (2011-07-08)
James Malcom: So this is fully C+? (2021-01-27)
Henrik Juel: No, another problem in 30.0 moves ending with 30. Tc2 dxc4+ is tested OK (2021-01-27)
Silvio Baier: Mit 30....dc+ ist das C+ (Euclide 0.98) (2011-07-08, hidden)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 65:30:51 Stunden. (hh:mm:ss)
Keine Lösung: BP 30.0, BP 30.5. (2023-05-08, hidden) more ...
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Moldenhauer: 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: Note that in problems castling acts like capture and pawn move with respect to the 50 moves rule. After 949.a6xSb7 there are 4 moves left by [Pa7]; but each camp can shift only KDT, on d1-g1 and b8-e8, respectively, so the triple repetition rule now limits the length of the game. (2004-09-09)
A.Buchanan: In some problems it's certainly the case that the 50-move rule operates incorrectly in this way. Such problems are fine, but obviously wouldn't want to impose this as a standard. Different composers can make different assumptions here (2023-06-20)
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Jeliss: "Obstruction of passage square f3 to Bishop of same colour."

"Version 'Pittsburgh Leader' 08.06.1913"
Yoav Ben-Zvi: Appears as the first problem (D445) in the booklet on Dawson's RA problems by G.P. Jellis. The obstruction that occurs in the Try -1.Pf3-f4?, by WP of WB, is described as "obstruction of passage square". It is not considered by Dawson and his disciples to be a Retro opposition. Dawson's conception of RO was quite broad, it included cases where the interference was not by occupation of the target square, so the only valid reason that I can see to exclude this case is that the 2 pieces involved are both of the same color. Fabel's definition explicitly excludes "Monochrome RO". I conclude that it would be preferrable to interpret RO as a bi-chromatic interference. The keyword Retro opposition should be removed. (2018-04-07)
A.Buchanan: To my mind, RO involves some kind of parity-tempo issue between the sides, not just some kind of race-tempo. If it was just about "bi-chromatic interference", one might say that bPe4 blocks wBh7 from an immediate retreat, so it has to be wPf4 that retreats first, legitimizing the ep key. So I agree this is not RO. (2024-01-06)
Henrik Juel: -1.f2 f5xTe4 -2.Tg4 f6 -3.Tg1 f7 -4.Le4 h5 -5.Lg2 h6 -6.Lf1 h7 -7.g2xDTSh3, unpromote DTS on c1, etc. (2004-03-18, hidden)
Henrik Juel: Nice analysis, Andrew, but your last comment seems incorrect.
Without Sb4, 1.b4 Txb4+ is not mate because of 2.Txb4#! (2013-07-26, hidden)
A.Buchanan: Thanks Henrik, I have fixed the solution. Forward chess is even harder than retro chess. And because Rb7 is retro-pinned, I overlooked that it's not forward-pinned! (2013-07-27, hidden) more ...
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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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Moldenhauer: Computerprüfung: C+ Stelvio 1.0 1 Sekunde.
Keine Lösung: BP 21.5, BP 22.0. (2023-02-17)
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 13 Sekunden.
Keine Lösung BP 19.0, BP 19.5 Cooked.
Beispiel BP 19.5:
1.d3 Sc6 2.Le3 Tb8 3.Ld4 Sf6 4.e3 Sh5 5.Dg4 b5 6.Ke2 La6 7.De6 b4
8.g4 Lb5 9.Lg2 b3 10.axb3 f6 11.Ta6 La4 12.bxa4 Sa5 13.Td6 Tb3
14.Lb7 Ta3 15.Kf3 Ta1 16.Sa3 Td1 17.Se2 Td2 18.Tb1 a6 19.Sc1 Te2 20.Lc8 (2023-05-03, hidden) more ...
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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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Adrian Storisteanu: Possible twin:
b)wSd2->d6
- 1.Sf7xRd6 Rd8xQd6 & 1.Rd8-f8 Qd6-g6# (2024-03-19)
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Henrik Juel: Cook: -1.a7xFb8=L Kb7xTa8, 1.Ka6 axb8DT#. (2004-04-01)
Mario Richter: vgl. mit P0006207 (2009-12-24)
Adrian Storisteanu: Possible (expensive, 50% more units) fix: Ka8 Bd8 / Kc8 (2+1) -- R: 1.e7xRd8=B Rd7-d8 & 1.Rc7 e8Q#. On the plus side, there is a wK in the picture now. (2016-02-10)
Adrian Storisteanu: See P1389595 (not clear which was first). (2024-03-19)
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Adrian Storisteanu: Is the star "*" in the stipulation supposed to indicate a kind of "set play" in the retro part -- in effect a "-1s dann h=1"? Because there is one:
- 1.Ke2xQf3 & 1.Ke2-e1 Qf3-d3=. (2015-08-02)
A.Buchanan: That's a clever interpretation Adrian, I am sure you are right, and it's sound according to Deadpos (2024-01-13)
Adrian Storisteanu: A more precise stipulation would be something like: (-w -s)*, dann h=1 (2024-01-14)
A.Buchanan: Good idea. The convention is that * promises at least one set play instance. (2024-01-15)
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Branko Koludrovic: P.S.
The black pawn a4(on the diagramm)came from c7 after capturing the white bishop on b6, then moving to b5 and capturing a white officer on a4. (2010-09-28)
A.Buchanan: I think sBf must have captured on g2 & then promoted on f1 to S, otherwise it’s impossible to unlock the southeast cage (2023-07-11)
A.Buchanan: So that means it’s sound. However I don’t see that it needs to have been sL shielding on f1 (2023-07-11)
A.Buchanan: Can sBf capture to e-file to be captured directly by wBf rather than capture to g2 to promote? If so, then sBf could capture L on a dark square, meaning that last move could have been c6xb5 (2023-07-11, hidden) more ...
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"Autor Ing. Rudolf Buljan, Zagreb"
AB: Have therefore marked him in the author table as Croatian. Yugoslavia has been removed from the PDB country list, I see.
A.Buchanan: Any ideas how to distinguish usage of “whose move?” and “no legal last move for white” (or black) keywords (2020-10-01)
Henrik Juel: This problem is clearly the latter
I would use 'Whose move?' in retro problems where a deeper analysis is required to determine the move, often with stipulations like '#1 (who?)' (2020-10-01)
A.Buchanan: I think "Whose move?" is where there is no forward stip, or where we have to determine if the forward stip should be interpreted as orthodox or as half-duplex. "no legal last move for..." is where the mater/stalemater/etc is the usual one, but we add or remove a single move at the beginning of the forward stip. I guess a third possibility is "last move?" where in a Type A position we also have to figure out who moved last.
With respect, I don't think the level of analysis can be relevant. (2023-12-06)
Henrik Juel: Analysis
White captured fxg and promoted [Pd2] on d8
Black captured axbxa2-a1=Y, b7xc6, d5xe4, and gxf
The retroplay was R: 1... f7-f5 2.f5xYg6
C+ Popeye 4.61 (2020-09-30, 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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SCHRECKE: NL: 1. Dg5,Kh4 gxf3 2. Kh4,Dg5 Sf1# (2023-09-13)
Ladislav Packa: Retro content is not needed, the solution is preserved even without it.
Pg4 Pg2 Sh2 Ke1 Rh1 (5)- Ph5 Kg3 (2) h#2* C+

1...0-0 2.h5-h4 Rf1-f3 #
1.Kg3-h4 Ke1-f2 2.h5*g4 Sh2-f3 # (2023-09-14)
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Henrik Juel: Diagram error? -1... g5xSf4, 0... Le5 1.Sd5#. But the symmetrical solution -1... e7xSd6 etc. seems legal too. (2004-03-22)
A.Buchanan: Yes bBf8 would have died at home but bPc could capture twice to promote on c1 or e1 (2023-03-31)
A.Buchanan: Odd. The diagram is thematically symmetric, so not just missing a unit. WinChloe has the same diagram. I wondering if we are missing something from the stipulation. Does anyone have access to an archive for this? It's also odd that it was reprinted: surely there would be a comment there? I suppose it can be rendered sound by placing wTTSS somewhere symmetric. Then Black runs out of pawn captures if sLd4 is promoted. In any case, wPh7xg8=S is affordable. (2023-04-01)
Mario Richter: The reprint in 'Problem' says: "Ne e7xSd6 radi nepravilne postave" ("Not e7xSd6 because of illegal Position.") The diagram in 'Problem' is the same as here in the PDB.
How about adding black pawns on b7 and g2? (2023-04-01)
A.Buchanan: Hi Mario. I considered adding sBb5e2. That gives 10 needed captures by Black and 6 by White, including sLf8. So the try fails for *two* reasons. sBb7g2 fails even more badly. I thought that also adding wTTSS although more uneconomical in pieces defeats the try for only one reason. Also I liked the trick h7xg8=S. And it must make it harder to find the solution. Does Rawbats do simple retractors? Can you try wTa0h1 wSa7g1 as a risky initial bid, and then back off until there are no cooks? (2023-04-01)
Mario Richter: Hi Andrew, I was well aware that with adding sBb5e2 or sBb7g2 the try fails for two reasons (for [subjective] optical reasons I preferred sBb7g2 to sBb5e2]). The reason I suggested the addition of 2 black pawns was the following consideration: I cannot believe that such an experienced Retro Composer like C.E. Kemp could make such a big mistake. It's much more likely that the printer made an error by omitting two pieces.
On the other hand, if we want to correct the problem including a unique reason for the try to fail, your suggestion of adding wTa8h1+wSa7g1 (FEN=R7/N4pp1/1P1pnkp1/5n2/3b1p2/1PP5/1PP2P2/K5NR) works - rawbats gives exactly R: 1. ... Bg5xSf4, dann 1. Ld4-e5 Sf4-d5# (and says about R: Be7xSd6 - "Illegal: Mehr schwarze Bauern-Schlaege notwendig als weisse Steine fehlen") (2023-04-01)
A.Buchanan: Hi Mario, thanks for this, I agree with everything you say. It's curious that the reprint in Problem did not pick up the error. Any overlooked black units are more likely to have been lurking on dark squares, maybe sBb4d2. This forces the try to only fail for 1 reason, albeit rather banal. Does this work. Or the +wTTSS mentioned forces the solver to do a little more counting, and there is a lot more solving space to search for the h#1. Rawbats' conversational skills seem to have improved: have you integrated it with ChatGPT maybe? (2023-04-02)
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Henrik Juel: C+ Popeye 4.61 after analysis (2023-07-20, hidden) more ...
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Juha Saukkola:: Solution not unique! (2000-09-29)
Moldenhauer: Computerprüfung: C+ NUPG Stelvio 1.11 00:02:12 Minuten. (hh:mm:ss)
Keine Lösung: BP 29.5, BP 30.0. (2023-04-24)
Moldenhauer: Beispiel Stelvio 1.11: 1.Sc3 Sc6 2.b3 Sf6 3.La3 a5 4.g4 a4 5.Lg2 Ta5 6.h4 Tf5
7.Lc5 axb3 8.Ld5 b6 9.a4 bxc5 10.gxf5 g5 11.hxg5 e5 12.Th4 h5 13.Tc4 h4
14.a5 h3 15.a6 h2 16.a7 h1D 17.Ta6 De4 18.Sf3 Le7 19.Sd4 exd4 20.a8D Se5
21.Te6 dxc3 22.Dc6 dxc6 23.d4 fxe6 24.dxe5 exd5 25.gxf6 Le6
26.fxe6 dxc4 27.fxe7 Ddd3 28.cxd3 Th3 29.Dc2 Te3 30.dxe4 bxc2 31.fxe3 (2023-04-24)
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paul: Correction of P1001275 (2010-05-08)
A.Buchanan: Does it have to end in a check which is in some sense a defect if unnecessary. Can it be 24. … Db3+ 25. Kd1 b5 26. Ke1 b4 (2023-06-13)
Joost de Heer: 24... Qb3+ 25. Kd1 is illegal. (2023-06-13)
A.Buchanan: Well clearly 24... Qxb2+ then same idea (2023-06-13)
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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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paul: Verified by Jacobi with this input:

stip dia 6
forsyth rnbqkbnr/pppp1ppp/8/8/4K3/5Q2/PPPP1PPP/RN3BNR
cond cavaliermaj (2023-06-16)
Mu-Tsun Tsai: Wait, 4.Ke2 walks into check by nightrider b8, isn't it? (2012-07-15, hidden)
Henrik Juel: You are right, Mu-Tsun, but this seems to work:
1.Nxe7 Nxe2 2.Nb1 Nxc1 3.Df3 Nca5 4.Ke2 Ngc6+ 5.Kd3 Nb8 6.Ke4 Ng8+
where the black nightriders switch places (2012-07-15, hidden)
Mu-Tsun Tsai: Yeah, I came up with the same thing. Computer check is still running to see if that's the only one. (2012-07-15, hidden)
Mu-Tsun Tsai: Hum... Probably have to give up computer check. The estimated run time is simply ridiculous. I'll just believe that this corrected solution is unique. (2012-07-15, 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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D. Meinking: Comm., USPB 1995-1997 (2002-07-06)
Moldenhauer: Computerprüfung: C+ Natch 3.3 in 45 Std. 46 Min.
Bei 16.5, 17.0 keine Lösung. BP 18.0 64 Lösungen.
C+ Euclide 1.11 in 44 Std. 07 Min. (2023-01-10)
A.Buchanan: This is a cool problem with tempo loss for both players (bK9 & wPf2). Noam likes to end proof games with checkmate, and in this case I guess it serves to eliminate shorter solutions which would otherwise save half a move. Checking 16.5 is sufficient for soundness. Thanks for taking the time to test. (2023-01-11)
A.Buchanan: Bit baffled how there can be 64 solutions in 18.0 moves since White just moved. 18.5 maybe? (2023-01-11)
ras@bsiag.com: Quickly checked because 45h seemed a lot for only 17.5 moves and only a few captures: Stelvio 0.98 is done in 210 seconds. (2023-01-11)
ras@bsiag.com: Well, Stelvio treats shorter solutions of same parity as cooks, so any shorter solution in 16.5, 15.5 etc would have been found by checking for 17.5. But anyhow, quickly checking 16.5 is done in a few seconds, finding no solution (of course). (2023-01-12)
A.Buchanan: Thanks for this. In order to be sure that no shorter solution exists, I’d also check 16.0. I agree it’s unlikely that there would be a solution, but it’s quick to check, for completeness (2023-01-11, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 5 Sekunden.
Keine Lösung: BP 26.0, BP 26.5. (2023-03-30)
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Moldenhauer: Ergänzung: Stelvio 1.2. Keine Lösung BP 15.0, BP 15.5. (2023-05-20)
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VL: Published in "Probl. Pribuzh." (or "Problemy Pribusch'ja"?),
1990, No.1, #12. An obscure Russian language chess problem
magazine issued in Nikolaev, the Ukraine (the South Bug
riverside region). (2006-01-27)
A.Buchanan: Lovely problem, but this is no more PRA than it is duplex. Rather, the push/pull of AP Type Keym gives a forfeit which must be discharged. (2023-07-22)
Andrey Zhuravlev: The correct name of this magazine is "Problemist Pribuzhya" (2006-04-10, hidden) more ...
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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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Gerald Ettl: Weiss jemand wie die Autorloesung geplant war? (2023-09-28)
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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: Natch 3.1 made no progress after half an hour (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 27:15:46 Stunden.
Keine Lösung: BP 21.5, BP 22.0. (2023-02-18)
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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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Unterverwandlung!Springer, Unterverwandlung!Läufer, Unterverwandlung!Turm, Umwandlungswechsel!3-fach
Adrian Storisteanu: See also slight variation P0577817. (2015-09-12)
Gunter Jordan: Dies ist kein Umwandlungswechsel, dazu müssten die UW auf dem selben Feld erfolgen. (2023-01-11)
A.Buchanan: Why would we talk of helpmate promotions (particularly black ones) as underpromotions. Is a knight ever an under-promotion? And we still with all this under-promotion effort don’t know what the pieces are: it might be rook or bishop. Just label them as promotion:lts and be done (2023-01-12)
SBD: Wechsel? (2012-12-25, hidden) more ...
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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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teilweiser Rundlauf T
Yuri Bilokin: Long-trip (wR, 3)
Umnov mate (bK-wP)
Ideal mate
Epaulette mate (2023-05-08)
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Erich Bartel: Nachdruck:
1) V35) Problemkiste (125) X 1999.--- (2007-07-18)
Yuri Bilokin: Areal cycle (wR, with captures, 4)
Corner-to-corner (wR) × 4
Four corners (wR (2023-05-08)
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Yuri Bilokin: you can see 1...Ra1 2.Kc2 Rxa2+ 3.Kd1 Rc2 4.Ke1 Rc1# (2023-05-08)
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Yuri Bilokin: correction bSe4 1B6/2p5/8/3N1p1K/4np2/8/3n4/2k5 (3+6) h#4

1.Sf3 Sxf4 2.Kd2 Se6 3.Ke3 Sxc7 4.Kf4 Sd5# (IM)

Annihilation × 2
Areal cycle (wS, with captures, 4)
Crusader theme. Kniest theme
Long-trip (wS, 4)
Ideal mate
Battery mate. Double-check mate (2023-05-23)
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Yuri Bilokin: you can see 1...Ra2 2.Kb5 Kd5 3.Kb4 Kc6 4.Kc4 Ra4# (2023-01-16)
Erich Bartel: Nachdrucke:
1) Problemkiste (99) VI 1995.---
2) Problemkiste (121) II 1999.---
3) r628 Mirador VI 2002.-- (2007-02-26, hidden) more ...
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Yuri Bilokin: Areal cycle (bK, with captures, 4)
Areal cycle (wB, with captures, 4) (2023-05-08)
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Vitaly Medintsev: Correction
5K1Q/p6p/4RN1p/2nP1Pr1/4N1P1/b3k3/1pPqp3/3rB3 (10+11)
b) SLe1 nach h2
white Be1 Kf8 Qh8 Pd5g4c2f5 Sf6e4 Re6 (10)
black Ba3 Ke3 Qd2 Pa7h7h6e2b2 Sc5 Rg5d1 (11)
b) BBe1 move to h2
Version by Vitaly Medintsev, October 1st 2014 (2014-10-01)
SCHRECKE: Medintsev: P1288792 (2022-10-30)
Vitaly Medintsev: Correction: 5K1Q/p6p/4RN1p/2nP1Pr1/4N1P1/b3k3/1pPqp3/3rB3 (10+11)
b) SLe1 nach h2
white: Be1 Kf8 Qh8 Pd5g4c2f5 Sf6e4 Re6 (10)
black: Ba3 Ke3 Qd2 Pa7h7h6e2b2 Sc5 Rg5d1 (11)
b) BBe1 move to h2
Version by Vitaly Medintsev, October 1st 2014 (2014-10-01, hidden) more ...
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Yuri Bilokin: correction bRb6-g8, +bQg3, +bNg4
3bb1rn/4k1p1/r5p1/6K1/2p3n1/5pq1/8/Q7 (2+12) h#3

1.Kf8+ Qf6+ 2.Sf7+ Kxg6 3.Bc6 Qxf7#

Active sacrifice (black, delayed)
AntiZielElement (B1, check)
Consecutive checks (3). Crosscheck (2)
Self-pin/self-unpin (white). Self-pin/unpin (white) (2024-02-09)
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Yuri Bilokin: possibly +bPa4 8/8/K7/8/p1n5/8/2b5/N2kN3 (3+4) h#3 b) wNa1-a5
a) 1.Bd3 Sac2 2.Be2 Sd3 3.Sd2 Se3# (MM)
b) 1.Sd2 Sc4 2.Kc1 Sa3 3.Bd1 Sd3# (MM) (2022-12-31)
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Yuri Bilokin: you can see b) 1...Sb6 2.Ke3 Kh4 3.Kf4 Sd5# (2022-12-28)
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Yuri Bilokin: correction bRd7-f7, bBg4-a4, -bN7 8/5r2/4p3/4N3/bKPkB2B/4N3/4p3/2nr4 (6+7) (2022-12-15)
A.Buchanan: The whole point of this ornamental problem is the symmetry. Any correction must preserve that. How about getting rid of the 2 rooks and the g-h-file bishops, then shift wBe4 to a4? 8/2n5/4p3/4N3/BKPk4/4N3/4p3/2n5? (2022-12-16)
Yuri Bilokin: I agree that the position should be symmetrical and economical, without unnecessary figures. The value of the positions reflected here is close to zero. (2022-12-16)
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Verst-4
Adrian Storisteanu: Possible fix: wSh4->b4, +bPh4 (2+10) -- 1.Bf5 Sc6 2.Sd3 Sd8 3.Bg7 Sf7#. (2015-08-26)
Yuri Bilokin: possibly bBb1-g6 6nk/7q/6bp/3Kn2r/3b3N/8/8/3r4 (2+9) (2022-12-19)
A.Buchanan: As well as bBb1-g6, one can shift bPh6-h7, i.e. removing bQ which serves no purpose (2022-12-20)
Yuri Bilokin: After correction, you can improve the efficiency (bPh6-h7), you can improve the sparsity (wNh4-b2). (2022-12-21)
A.Buchanan: I also wondered about shifting wSh4, but I thought it's good to have a knight start near to its destination square, because there are more convoluted paths for dual elimination, rather than pulling the line more taut. b4, b8 & e1 also work - it becomes a matter of taste. On balance I like Sh4 because it suggests 1. ... Sf5? unpinning the bS (2022-12-21)
A.Buchanan: Seems likely (2022-12-20, hidden) more ...
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Klären Autor
Verst-4

Original ohne sBg6 abgedruckt, Korrektur(versuch) 05/1967, S. 140
Als Autor ist angegeben: D.W.A. Brotherton
Yuri Bilokin: correction +bPg3 8/8/6pb/3r4/2k3B1/6p1/6N1/4K3 (3+5) (2023-05-24)
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Verst-4
klären Stellung
SCHRECKE: Unlösbar - wLb1 gehört vielleicht nach c2!? (2023-03-10)
HBae: habe korrigiert, auch das Datum war falsch (2023-03-10)
SCHRECKE: C+, popeye 4.87 (2023-03-10)
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klären Monat
Alfred Pfeiffer: No.1796 ist aus 04/1935 (vgl. P0003272) (2011-01-11)
Yuri Bilokin: possibly bBe4(-bNe4), +bPd3 8/3r2N1/6kN/8/4b3/3p3r/2R5/1K6 (4+5) (2022-12-16)
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Auszeichnung wurde in Problemista 04/1964 annulliert.
Yuri Bilokin: correction wBc8-e8, bKb3-d4, bBf8-e3, bPc7-f7, +bQa6, -bPd3 4B3/5p2/q7/8/3k4/4b3/8/R3K3 (3+4) h#3

1.Kc3 Bb5 2.Bc5 0-0-0 3.Bb4 Rd3#

Castling (white, long) (2023-06-10)
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Yuri Bilokin: anticipated (P0548424) (2024-03-29)
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VL: Unjustified ep-capture because of –e3xd4. For correction it suffices e.g. to add bSe3.
This is possibly a unique 2-move helpmate with ep-key and 2 solutions of the type 1.2.1.1 (at least so currently in the PDB).
Publi?ation details. H551 in the Helpmate section of the July issue, with no preview. Here is the text of the solution published in the Nov.-77 issue: “…White may have just played P(e3)xd4. Very few solvers have spotted this. Full marks for claim of no solution with correct reasoning but, in fairness, one mark for composer's intention. White cannot have just played K(a5)a6 in reply to Pxb6+ as too many captures by Black are required.” (2023-12-01)
A.Buchanan: Yes Valery you are right this must be cooked. However, I'm not a fan of +bSe3 because it spoils one of the model mates. Separately, I mildly prefer bPf7 to bPe6. (2023-12-03)
VL: I do prefer bPf7, as well. I don't see any way to save at least one of the properties Kindergarten Problem and Model mate (2) after fixing (what for these keywords have been added just now?). Besides, Andrew, please, refine the specification ((full) number and publication month 07-08). (2023-12-03)
A.Buchanan: Ok have added the publication details. The keywords are useful to indicate the aim, even though the problem is cooked. I guess there is a way to keep the model mates but it’s very clunky and probably +bSe3 is better (2023-12-03)
A.Buchanan: Yes Valery this is cooked. I'm not convinced by +bSe3 because it spoils one of the model mates. (2023-12-03, hidden) more ...
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A. Luxemburg-Gedenkturnier
Entfeßlung!indirekt, Mustermatts
Felber, Volker: Korrektur zu P1405949 (2022-11-16)
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Verst-4
Vorgänger: W. Hagemann
368. FEENSCHACH 1950
SCHRECKE: Hagemann: P0546092 (2023-05-23)
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Yuri Bilokin: you can see 1...Bb4 2.Ka4 Bxc3 3.Bb3 Ra5# (2024-04-17)
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Yuri Bilokin: possibly bNe7-f7, +bPf6, +bPg5 4Kn1B/5n2/4kpp1/6p1/6P1/8/8/8 (3+6) h#5
1.f5 Bf6 2.Kxf6 gxf5 3.Kg7 fxg6 4.Kh8 Kxf7 5.Sh7 g7# (MM) (2023-01-11)
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Yuri Bilokin: correction +wPb6, bPc7-b7 8/1p1pp3/1P1k4/R6K/4r3/3Np3/4P3/8 (5+6) h#2* b) wSd3-d8
a) 1...Sb4 2.Re6 Rd5#
1.Rc4 Sf4 2.Rc6 Rd5#
b) 1...Rc5 2.Re6 Sxb7#
1.Rc4 Re5 2.Rc6 Sf7# (2023-06-26)
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SCHRECKE: Das ist leicht zu reparieren, z.B. wKe1 (2023-09-18)
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Yuri Bilokin: possibly bNb3(-wPb3), bPc3(-bBc3), +bPc4, +bPf4 1B2R3/2p5/8/8/2pk1p2/1np1r3/2r3bK/8 (3+9) h#2 b) alles eine Reihe tiefer
a) 1.c5 Bxf4 2.Re4 Rd8# (MM)
b) 1.Sd1 Rxe2 2.Bd4 Ba6# (2022-12-16)
milan: a) -wPb3+bPa7wBf1
b)all pieces to the left
1.Ra6 B×b7 2.Bc5 Ba6#
1.Kc5 Rd5+ 2.Ka4 Ra5#
1.b5 Be4 2.Rd4 Rc8#
i'ts look like battle of the bishops
M.Frelih (2022-12-18)
Yuri Bilokin: Very bad. In a problem that has nothing to do with the author's intention, there is still an extra white bishop in the solution 1.Kc5 Rd5+ 2.Ka4 Ra5# (2022-12-19)
milan: +wBa6bPa5-wPb3 1.c5 Bf4 2.Re4 Rd8# 1.Rb6 B×c7 2.Bd5 B×b6#
your solution b) not C+... bRb5? M.Frelih (2022-12-19)
Yuri Bilokin: Sorry for your misunderstanding. Everything is fine in my edition. Twin FEN Code a) You see Twin FEN Code b) Showing 8/1B2R3/2p5/8/8/2pk1p2/1np1r3/2r3bK, solution
1.Sd1 Rxe2 2.Bd4 Ba6# (2022-12-20)
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klären AL ?
Yuri Bilokin: correction +bQh4(h3) 8/7k/1p3p1p/8/2n4q/p1pp2n1/p1p4P/1bK5 (2+13) (2023-05-08)
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Jakob Leck: Compare to P0559688 (2023-01-03)
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Yuri Bilokin: possibly +bPc4 8/2B1nK2/1r6/5kp1/2pn4/5B2/6b1/8 (3+7) (2022-12-16)
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Anton Baumann: 1. ... Lc4 2.Sb5 Ge6 3.Sa7 La6#
1.Sc6 Ld7 2.Da7 Ge6 3.Sb8 Lc8#
C+ Alybadix (2023-08-22)
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Adrian Storisteanu: Possible fix: wKg3->g2, bRe1->e3, +bPd6 (2+10) [2...Sxd6]. (2015-11-04)
Viktoras Paliulionis: Other possible fix: bPc5-d4, bBc1-h6, +bPf6. (2021-02-02)
Kenneth Solja: Best correction is: +bd6 and +bSg7(-bpg7) (2+10) No need to move white king. And initial position bL should be in h6 as it was in the original problem.
Solution: 1.Ld2 Sxe4 2.La5 Sxd6 3.Te6 Sxf5 4.Tb6 Sxg7 5.Dh7 Se6 6.Da7 Sxc5# (C+) (2024-04-04)
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Yuri Bilokin: correction bNb5-a1, bBh4-d8 3b4/3p4/2br4/4K3/8/k7/4P3/n7 (2+6)
1.Bb6 e4 2.Rd5+ Kf6 3.Kb4 e5 4.Kc5 e6 5.Kd6 e7 6.Bc5 e8=S# (MM)
the idea is not new P0564440 (2022-08-17)
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Adrian Storisteanu: Possible fix: +bSg8 (2+12). (2015-08-07)
Jakob Leck: Die Dawson-Aufgabe könnte P0555591 sein - zeitgleich erschienen. (2022-12-28)
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SCHRECKE: Nebenlösung
1. Na2 Te5 2. Nb4 Tg5 3. Nh7 Nf7# (2023-01-12)
HBae: Correction +wPawn d5 (5+2) C+ Popeye v4.37 (2023-01-12)
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Harry Fougiaxis: Cooked in 15 : 1.Txc1 Sxc1 2.Kb6 Sa2 3.Kc7 Sc1 4.Kd7 Sa2 5.Ke6 Sc1 6.Kxf6 Sa2 7.Kg5 Sc1 8.f5 Sa2 9.f4 Sc1 10.f3 exf3 11.e2 Kd2 12.e1=D+ Kxe1 13.Kf4 c4 14.dxc4 Se2+ 15.Ke3 dxc4# (2007-06-09, hidden)
Mirko Degenkolbe: Die Aufgabe muss falsch eingegeben worden sein. Wenn man den schwarzen Bauer von b7 nach d7 versetzt und einen schwarzen Läufer b7 ergänzt, ist sie korrekt. (2009-02-25, hidden)
Frank Richter: The problem was printed in "problem 152-156" (April 1973) with bBb7 and bPd7 (12+9). (2009-02-26, hidden)
SCHRECKE: C+, popeye 4.87 (2022-11-30, hidden) more ...
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Anton Baumann: C+ Alybadix (2023-09-13)
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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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SCHRECKE: C+, popeye 4.89 (2024-03-13, hidden) more ...
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SCHRECKE: C+, popeye 4.87 (2022-10-02)
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Joshua Green: I've been testing some modifications to Popeye with the goal of improving its performance against seriesmovers. Unfortunately, my improved -- assuming there are no bugs -- Popeye cooked this problem, offering:
1. Txd3 2. Th3 3. Kd5 4. Kc4 5. Le5 6. Lf5 7. Lxb1 8. La2 9. b1L 10. Da1 11. Kd3 12. Tc2 13. Lb2 14. La3 15. Tb2 16. Kc2 17. Th2 18. Te2+ Kxe2= (2022-09-13)
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paul: a) 1.Rh2 2.Bh1 3.Rfg2 4.Bf2 5.Rg1 6.Bg2 7.Rgh1 8.Kg1 9.Bf1 10.Rg2 11.Kh2 12.Bg1 13.Rf2 14.Qg2 15.Kh3 16.Rh2 17.Qh1 18.g2 Sf5=
b) 1.Bh3 2.Qg2 3.Rh2 4.Qh1 5.Bg2 6.Rh3 7.Bh2 8.Kg1 9.Bf1 10.Qg2 11.Kh1 12.Bg1 13.Kh2 14.Qh1 15.g2 16.Kg3 17.Rh2 18.Kh3 Sf5=
The judge was H.P.Rehm. (2013-05-06)
Joshua Green: This is C+ according to a special version of Popeye that I've modified to improve performance on ser-h= and ser-h# problems (hopefully without introducing any bugs). (2022-09-15)
A.Buchanan: Hi Joshua: great stuff, thanks! I guess you’ve been working with a test set of previously validated compositions? Do you plan to merge your branch with the main Popeye version? (2022-09-16)
Ian Shanahan: In FIDE Album 1992-1994 (2014-01-20, hidden) more ...
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Anton Baumann: C+ Alybadix (2023-02-23)
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Urversion NL mit wKe4!
Kevin Begley: Correction of 'P0577029' (was this actually published)? (2011-05-18)
paul: Cook: 1.Q×e7(Sd8) 2.K×d8(Sc7) 3.Qf8 4.S×c7(Sa8) 5.Se8 6.Q×g7(Qf8) 7.Qh8 8.Q×f6(Qh8) 9.Qe7 Qf6= (2024-02-28)
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Alfred Pfeiffer: Was ist "Iceberg"? Ist das eine Art Augsburger Schach? (2010-09-29)
Alfred Pfeiffer: "Iceberg" ist Peter Fayers' Version des Augsburger Schachs, bei folgenden drei Unterschiede gelten (Quelle: "http://www.bcvs.ukf.net/augsb.htm"):
E1: Pieces may combine, but not kings, and NOT PAWNS either (not even when moving to promote).
E2: The game-array queen is not separable, and Qs are distinct from (R+B)s.
E3: A pawn may promote to any combination seen in the diagram, or generated during play. (2010-10-11)
A.Buchanan: Link is now at https://www.mayhematics.com/s/augsb.htm. Piece images to make up the diagrams have been lost, but you can get a clue what the diagrams were by clicking in the html. (2019-02-04)
paul: c) 1.Lb2 (2022-08-17, hidden) more ...
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Gerald Ettl: Und nicht 14.h1D 15.Dg1 wegen Dxg7[Dd8] nette Rundlinien-Einmauerung(GE) (2009-02-10)
Ladislav Packa: Anticirce Cheylan (2023-02-28)
Erich Bartel: C+ PY V4.45.--- (2008-11-07, hidden) more ...
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Anton Baumann: a) & b) C+ Alybadix (2023-07-12)
Manfred Rittirsch: b) 1.- nDb8 2.nKf6+ nKe7 3.nBg5 nLh4 4.nBf5 nBgxf6 e.p.# (2003-11-22, hidden)
Ion Murarasu: correct in a) 1...nSeg4 2.nBf5..etc (2003-11-23, hidden) more ...
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Henrik Juel: The paradox is explained by the need for Black to play De1xLc1
Natch 3.1 made no progress after an hour (2018-12-14)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 14:25:47 Stunden.
Keine Lösung: BP 14.0, BP 14.5. (2023-02-20)
TBr: In solution read 11.Taf1 (2008-09-20, hidden) more ...
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Kommentare:
Als einziger letzter schwarzer Zug, der die weiße Rochade
erhält, kommt nur Bh7-h5 infrage, was dem Weißen den e.p.-Schlag
ermöglicht, um nicht die Rochade oder das Mattnetz zerstören zu
müssen (H.P.Suwe)
Gianni Donati: This intends to show the Valladao theme in the minimum number of moves. (2004-03-19)
VL: The waiting ep capture. Cf. also P1000260
(by T.Petrovic, 2000) somewhat enriched
thematically with the illegal try 2... Rd1#??. (2004-06-03)
A.Buchanan: The “illegal try” 2. Rd1+ is not actually mate because c2 is unprotected. This is kind of “logic dual” spoils the A.P. motivation for the castling (2020-05-20)
Henrik Juel: White captured h6xg7 and once more, e.g. axPb-b8=Y
Black captured cxdxe, dxexf, fxg, and once more, e.g. a2xb1=Y
Possible retroplay 1... h7-h5 2.h6xSg7 Df8-f4 3.h5-h6 Sf4-h3 4.Sh3-g1 etc., preserving the castling right
Any other black last move would force White to retract Ta1, as d2xc3, d2xe3, and h6xh7 would be illegal retractions
I agree with Andrew that the double motivation of 2... 000# is a weakness:
a. legitimizing h7-h5
b. mating (2020-05-21)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
A.Buchanan: See discussion at P0009121 (2023-08-04)
A.Buchanan: Maybe can replace wPf2 with wR, wRf1 with wB & shift bPg3 to g2? (2020-05-20, hidden) more ...
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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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Henrik Juel: Natch 3.1 produced nothing in some 8 hours (2018-12-16)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 3 Sekunden.
Keine Lösung: BP 23.0, BP 23.5. (2023-04-02)
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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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Moldenhauer: Computerprüfung: C+ Stelvio 1.0 in 3 Sekunden.
Keine Lösung: BP 24.0, BP 24.5. (2023-02-14, hidden) more ...
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Gerald Ettl: Wie lautet der Cook? (2023-09-28)
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Henrik Juel: C+ Popeye 4.61 (without the retraction, which is unique) (2022-11-08)
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Yuri Bilokin: you can see 1...Rg5 2.Sg3 Bf1 3.Sh1 Rg2# (MM)
1...Bd5 2.Sg3 Rg5 3.Sh1 Rg2# (MM)
1...Rg5 2.Sg3 Bd5 3.Sh1 Rg2# (MM) (2022-08-06)
Jean Marc Loustau: ****
Compare with P1100532. (2022-08-14)
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Henrik Juel: Natch 3.1 did almost nothing in half an hour (2018-12-18)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 39 Std. 40 Min.
Obwohl nur 2 Strategien im Histogramm standen! (2023-02-10)
Moldenhauer: Keine Lösung BP 24.0, BP 24.5, Zeit ca. 20:15 Stunden. (2023-02-11)
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