Anton Baumann: Mattdual in b): 3.Tb8 axb8=D,T# (2022-12-16)
A.Buchanan: So Borodatow got it all working! In a) Black might have captured hxgxf and axPb. So castling rights might still be maintained with the ep. In b) on the other hand, it must be axb, bxa, exf and wPgxh6, so there was the cage. Promotion to TD is tolerated in the final move, although it may not be puristic, without the convention, too many mates would be excluded (2022-12-16)
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Henrik Juel: the five missing black men were captured by white pawns (exfxgxh, fxgxj), so with Black to move last move was with Ta1 or Ke1, and White may not castle
C+ Popeye 4.61 (2022-11-26)
A.Buchanan: Pleasant White tempo play in both phases (2022-11-27)
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A.Buchanan: White pawn caps: axb,dxe,gxf,hxg definite.
Black: fxg and two to resolve c-file. But that may be c&d cross-capture, so in set play last move might have been c6xd5. So I think this problem is cooked. What am I missing? (2022-03-21)
Mario Richter: If Black's last move was c6xd5, how do you get white Bishop a3 out of his cage? (In that case, black pawn b7 never left the b-file). (2022-03-21)
A.Buchanan: I agree Mario thanks (2022-03-21)
Hans-Jürgen Manthey: beide Rochaden und im Satz ep. ist möglich:
R.: 1. ... d7-d5 2. Sa7-b5 b5xTc4 3. Tg4-c4 Dd8-h4 4. c4-c5 Th8-h3 5. c2-c4 c4xDb3 6. Dd3-b3 h3-h2 7. Dd1-d3 h4-h3 8. Sc8-a7 h5-h4 9. Th4-g4 c5-c4 10. Th1-h4 h7-h5 11. h2xSg3 Sf5-g3 12. b3-b4 Sh6-f5 13. Lb4-a3 c7-c5 14. Ld2-b4 g3-g2 15. Lc1-d2 a3-a2 16. a2xSb3 Sc5-b3 17. Lb3-a4 Sa6-c5 18. Lc4-b3 Sb8-a6 19. d2xLe3 Lc5-e3 20. Ld3-c4 Lf8-c5 21. Lf5-d3 a4-a3 22. Lh3-f5 a5-a4 23. Lf1-h3 g4-g3 24. g2xLf3 Lb7-f3 25. Sb6-c8 Lc8-b7 26. Sa4-b6 a7-a5 27. Sc3-a4 b7-b5 28. Sb1-c3 f5xSg4 29. Se5-g4 Sg8-h6 30. Sf3-e5 e7-e6 31. Sg5-f3 f6-f5 32. Sh3-g5 f7-f6 33. Sg1-h3 (2023-02-23)
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Klären: Quelle = Schachmatt? - Felber, Volker: SCHACH ist korrekt, 6/1969, Seite 191 (2010-10-09)

Eliminierung von 4 Steinen zwischen sK und wT im h#2 (mit Hilfe von ep-Schlag)!
Felber, Volker: SCHACH ist korrekt, 6/1969, Seite 191 (2010-10-09, hidden)
Henrik Juel: HC+ Popeye 4.61 (2023-01-08, hidden) more ...
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Henrik Juel: in a) last move could be Kg2-h3
in b) last move must be g2-g4 (2024-01-14)
Henrik Juel: C+ Popeye 4.61 (2024-01-14, hidden) more ...
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Henrik Juel: In a) last move could be Kb2-a1 (2022-12-05)
Henrik Juel: HC+ Popeye 4.61 (2022-12-05, hidden) more ...
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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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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: a) C+ Popeye 4.61
b) the ep capture is unjustified, last move could be Ke2-f1 or f3xg4 (2023-08-06)
A.Buchanan: There was an error in the stipulation: "b) sLe3 nach h2 (-wSh2)" is completely incorrect. However "b) sLe3 nach f3" leads to a very nice problem which matches the solution already present. This problem is not in WinChloe, but I am sure this is the resolution of the mystery (2023-08-07)
Henrik Juel: You are probably right, Andrew; maybe there was a misprint in the source
Now part b) is C+ Popeye 4.61 after very simple analysis (2023-08-07)
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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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SCHRECKE: Nicht 1. Kb3 0-0-0?? 2. Ka2 Td3 3. Ka1 Txa3#, da Weiß zuletzt mit K oder T gezogen hat! (2023-06-28, hidden) more ...
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Mario Richter: Luboš Kekely (Slovakia) correctly points out that 1. ... Sf5! 2. 0-0-0 Sd6# is only a try and not a solution (last black move must have been by King or Rook, so the queenside castling is illegal).
I have changed the solution accordingly. (2023-03-10)
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A.Buchanan: White must cover g2. If not by check, then by rook. However Rf1 is check, so both Black's moves are forced to shield bK (2022-11-27)
Henrik Juel: HC+ Popeye 4.61 (2022-11-27, hidden) more ...
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Henrik Juel: In b) White may not castle, because last move was done by Ke1 or Th1 (2023-12-02)
A.Buchanan: Cute. Do wPe2 & bPe3 serve any purpose, however? (2023-12-03)
Henrik Juel: C+ Popeye 4.61 (2023-12-02, 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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Retro
b) Weiß am Zug
Adrian Storisteanu: Two knight promotions differently motivated.
The technical challenge was in averting the short second solution in a) as set play, in order to keep the two phases nicely apart, and also, of course, preventing black tempi and potential white retro moves. (2000-11-20)
A.Buchanan: Could wBc3 simply be removed for b)? I guess author's preference is to just slide the pawn over, to keep the twins as similar as possible (2023-12-06, hidden) more ...
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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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Henrik Juel: 0... fxe3ep 1.d8S Kxd5 2.0-0-0#. Not -1.f2? and Lg1 is caught. (2004-09-16)
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)
Henrik Juel: In principle I agree, Vaclav
But PDB does not (yet) allow HC+, so I find it acceptable to use the C+ label, when you also tell the whole story after 'Computer test:' below (2023-08-03)
A.Buchanan: Hopefully Gerd will have more time at some point, and can expand the functionality in this and other areas. In the meantime, engine solving of conventional retros including AP, is in its infancy. Retractor 2 has some effectiveness, but is still basic. And there is nothing that yet grasps the intricacies of castling/ep etc. However AP problems do often contain considerable forward chess, and the C+ tag is very useful to filter out those that have already been solved forwardly, without pretending that these are in any sense fully solved (2023-08-04)
Ladislav Packa: I know the definition of AP, but I don't understand the logic behind it. The move 1.fxe3 e.p. proves that White CAN castling. But the solution (2. ...Rd1#?) claims that castling is MANDATORY. From my point of view, AP is correct when only castling is necessary for the solution and the Rook move would be a dual. (2023-08-04)
A.Buchanan: @Ladislav: I am not sure how to help you. Maybe you can read this page from Retro Corner: https://www.janko.at/Retros/Glossary/APosteriori.htm (2023-08-05)
Ladislav Packa: Andrew, what should the article help me with? I quote the final sentence:
Some people still oppose this rule and argue that it should certainly not be the default convention. (2023-08-05)
Henrik Juel: You could view it this way, Ladislav
h#2 means that it is Black to move, so White made last move
What was last move? A little analysis shows just three possibilities: f2-f4, move by Ta1, or move by Ke1
So normally we cannot assume that last move was f2-f4
But if White can castle, then the last move was f2-f4
So if we could start with 0... 0-0-0, then 1.fxe3ep would be legitimate
AP says that you are allowed to reverse the sequence of events; first do the ep capture, then later legitimize it by castling
Was this helpful? (2023-08-05)
Ladislav Packa: Henrik, you don't have to explain that to me. I've done a few AP issues myself, like P1348357. But that doesn't mean I agree with AP's logic. I already wrote it - the term "you can castle" is applied as "you must castle". But these are only problems where, in addition to castling, the Rook move can also be used, I consider that a dual.
From that point of view, the P1000662 issue is perfectly fine for me. (2023-08-05)
Joost de Heer: AP: By castling, you prove a posteriori that the ep-capture was not just a try but the actual solution. Without castling, the ep-solution just is that: a try.
So: Try 1. fe3 ep e8=S 2. Kd5 Rd1 - but ep capture not allowed, as there is no proof that f2-f4 must've been the last move.
Solution 1. fe3 ep e8=S 2. Kd5 OOO - Now the ep capture was justified because white castled, thereby proving that the last move before the diagram position indeed was f2-f4. (2023-08-06)
Joost de Heer: See e.g. P1052919 : The try is an ep capture which is unjustified. (2023-08-06)
A.Buchanan: OK Ladislav: I think I get your point. If one solution with castling justifies the e.p., then based on that certainty, why shouldn't an alternative solution with no castling *then* be allowed as well? There are problems in which one twin shows 0-0-0 and the other shows 0-0. Each is based on the other in a similar way, so the idea of dependency is not new. Why are we not allowed to add other "parasitic" solutions as well? Why can we only have the "paying" solutions? We can't say that we are restricted to one solution: that's not the way chess problems operate! And this is just in the help world - in the adversarial world it might get even more complicated. Is this your issue, Ladislav? (2023-08-07)
Ladislav Packa: I don't want to unnecessarily prolong this discussion. However, I will add one more note: in this position, white castling is also possible without e.p. in Black's 1st move. If B1 were an indifferent move, then white can 1...0-0-0! The Codex of Chess Composition writes about it in Article 16 (1):
Castling convention. Casting is permitted unless it can be proven that it is not permissible.
In our case, 0-0-0 is possible because White's last move exists - e2-e4! It does not matter if it is this move or some a2-a3, both moves are equivalent. EP does not prove the possibility of casting, it would be legal even without it. (2023-08-07)
A.Buchanan: Ladislav was what I wrote your issue pls? Y/N :-) (2023-08-07)
Ladislav Packa: I have no problem, I'm just expressing my own opinion about the AP convention. (2023-08-07)
A.Buchanan: OK cos I think the point I raised is a real one that should be addressed by theory some day. Clearly from the nice problem that you composed Ladislav you understand the mechanics very well. From a justification perspective it's all a bit iffy, but that's why it's controversial. Under RS it's really the only way one can end up actually eping, and it's proved compositionally fertile. So that's enough to justify (2023-08-08)
Ladislav Packa: No need to apologize. This is a normal discussion with different views on the issue. Maybe it will come to some conclusion.
I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p. (2023-08-08)
Joost de Heer: "I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p."
You misinterpret AP. The e.p. capture does not authorize castling, castling provides a justification later on (hence the 'a posteriori') for the legality of ep.
Usually, for ep justification you need to examine all game trees that lead to the diagram, and only if all game trees end with the double-step, then ep is allowed.
With AP, you examine all the game trees including the actual play. If all those game trees have as last move before the diagram position the double step, then ep is possible.
In this case, if white doesn't castle, then there are game trees which don't have as last move the double step, and therefore AP logic dictates that the ep capture was illegal. However, all game trees which lead to the diagram and which have castling in the actual play have as last move before the diagram position the double step, hence AP dictates that the ep capture is legal. (2023-08-09)
Ladislav Packa: Joost: A simple question - is white allowed to castle after any 1st move by black (except e.p.)? (2023-08-09)
Joost de Heer: Of course he is. AP only is used to combine the ep justification with castling, not the castling right per se. (2023-08-09)
A.Buchanan: I agree with both. Also "cooked" is closer to "unsound". And we choose not to have "C+ cooked", which might legitimately mean "C-" (2023-08-04, hidden) more ...
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zunächst für die 3. ehrende Erwähnung vorgesehen, wegen Disqualifikation von P0532932 hochgerutscht (vergl. 'Die Schwalbe' 08/1984 S.308)
Anton Baumann: Nachträglich zur 2. ehrenden Erwähnung aufgerückt (vergl. 'Die Schwalbe' 08/1984 S.308) (2022-12-23)
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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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Unterverwandlung!Springer!3-fach
Yuri Bilokin: more economical bNh6-f6(-bPf6), bPh5-d5, then a1=b1 6n1/6b1/6nk/4p3/8/4p1K1/4P1p1/8 (2+7) h#7
1.g1=S Kg4 2.Sf3 exf3 3.e2 f4 4.e1=S f5 5.Sf3 f6 6.Sg5 f7 7.Sh7 fxg8=S# (MM)

Slow Excelsior. Active sacrifice (black). Annihilation.
Excelsior (white). Long-trip (bS, promoted, 3) Phoenix. Promotion (Sss, 1, 2)
Model mate × 1 (2023-05-10)
Erich Bartel: C+ PY V3.52.--- (2008-04-28, hidden) more ...
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Grimshaw
Yuri Bilokin: more economical bPg2-f3, bRd8-g2, bBg8-h5(-bNh5), -bPb5 7K/2qr4/8/4Q2b/3b4/3kpp2/3pn1r1/8 (2+10) 2.1…

1.Bc3 Qf6 2.Rd4 Qb6 3.Qc4 Qb1# (MM)
1.Bf7 Kh7 2.Qc3 Kh6 3.Bc4 Qf5# (MM)

Amazon theme. Blocking piece replacement (bB-bR)
Check prevention (W-W). Hideaway maneuver (wK, 2)
Long-trip (wQ, 3). Pelle move (white)
Play on the same square (B3, 2)
Model mate × 2 (2024-02-09)
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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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Yuri Bilokin: you can see *) 1...Sf1 2.Rc3 Sxe3 3.Kb3+ Sc4 4.Sb4 Sd2# (2022-12-10)
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Yuri Bilokin: you can see
1...Rxb1 2.a1=Q Rxa1 3.b1=Q+ Rxb1 4.Ra1 Rxa1#
1...Rxb1 2.a1=Q Rxa1 3.b1=R Rxb1 4.Ra1 Rxa1#
1...Rxb1 2.a1=Q Rxa1 3.b1=B+ Rxb1 4.Ra1 Rxa1#
1...Rxb1 2.a1=Q Rxa1 3.b1=S Rxb1 4.Ra1 Rxa1# (2023-05-08)
Harry Fougiaxis: 1 Hon. Mention BCM 1986-87 (2005-08-31, 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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Achim Schöneberg: Vergleiche P1414844 (2024-02-07)
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klären Datum
Erich Bartel: die erste Zeile stimmt.Dort steht die Aufgabe auf Seite 459
als Urdruck. (2007-10-12)
Achim Schöneberg: Siehe P0502915 (2022-10-05)
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Achim Schöneberg: AS: Siehe P0502798 (2022-10-05)
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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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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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Jakob Leck: Version zu P0575203 (2022-10-08)
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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 (wR, with captures, 4) (2023-05-08)
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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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Yuri Bilokin: you can see 1...Rxc4+ 2.Kxe5 Rxc5+ 3.Kd4 Rb5 4.e5 Rb4# (MM) (2023-05-08)
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Yuri Bilokin: you can see 1...Qh8 2.Kb7 Qxh1+ 3.Ka6 Qa8# (2024-02-09)
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Sally: Markant ist der Grimshaw auf "f3" (überraschend auf einem "Eck-Feld" des w. Königs), wobei die dennoch notwendige Fesselung prägend ist. [Eine der ersten Aufgaben des Erfolgsduos].
41 Mehr als ein Hobby ( F. Pachl 2011). (2016-12-07)
Achim Schöneberg: Knapper Vorgänger siehe P1413421 (2023-11-08)
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udo.de: das richtige Datum ist 12. April 1933 (2024-01-17, hidden) more ...
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BSp

spiegelverkehrt von R.Senkus in Chess in Australia veröffentlicht
Yuri Bilokin: you can see 1...Rxe4 2.b5 c5# (2024-04-02)
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BSp
siehe P0506028
Yuri Bilokin: you can see 1...f7 2.e5 f8=Q 3.e4 Qf2# (2024-03-10)
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BSp
siehe P0506027
Yuri Bilokin: you can see 1...d7 2.e5 d8=Q 3.e4 Qd2# (2024-03-10)
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Yuri Bilokin: you can see 1...Sc2+ 2.Kf5 Sd4#
1...Sc4+ 2.Kf5 Sd6#
1...Sd5+ 2.Kf5 Se7# (2023-06-26)
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klären Jahr?
Anton Baumann: Korrektur in 'Die Schwalbe' 06/1983 S.87: +sDa5 zur Verhinderung der NL in c) (2022-12-21)
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t/Allumwandlung, u/61112, u/SDlt
Vorläufer: A.C.White,1930, s.Nr.1056
s. A. C. White 364v L'Echiquier 07-08/1930
Anton Baumann: Die Aufgabe stammt gem. 'Die Schwalbe' 05/1963 von André Simonet, Brüssel
Vorgänger vergl. P0542142 (2022-11-16)
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t/Allumwandlung, u/61113, u/DTsl
Yuri Bilokin: you can see 1...g8=Q 2.cxb1=B Qb3#
1...g8=Q 2.cxb1=S Qb3#
1...Bxc2 2.Ka2 Ra4# (2024-02-29)
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Autor klären
Felber, Volker: Ich gehe davon aus, dass hier die Korrektur zu P1406481 vorliegt. Dort ist im Original von SCHACH "Eduard Walzinsch" angegeben. In der PDB gibt es beide Schreibweisen für denselben Autor. (2022-12-07)
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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: you can see 1...Kc6 2.Bf4 Bc3# (2023-06-04)
A.Buchanan: Yuri if the composer wanted set play, they would have specified it. Instead there is a pleasant pure tempo first move in one solution which is much more interesting (2023-06-05, hidden) more ...
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Jean Marc Loustau: The setting is not legal, but it is without impotance here... Let us say it is "fairy" (2003-05-08)
A.Buchanan: Legality/illegality is orthogonal to fairy. Let us say here it is "not fairy". (2023-07-09)
Adrian Storisteanu: Add, say, "circe" under the diagram, and all is pretty cool. Even cooler, "chess960". (2023-07-10)
paul: cond einstein
stip h#2.5 forsyth 8/8/8/7k/8/5K1B/3R2P1/8

Two solutions (2023-07-10)
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Eric Huber: Anticipated by 'P0534152' Erkki Heinonen, 3. Makuc-Moder-Gedenkturnier 1971-1973, 4. ehrende Erwähnung. (2008-10-01)
Viktoras Paliulionis: This problem shows paradoxical idea - two mutually exclusive tempo moves (B1 & B2). P0534152 has only one tempo move. (2023-10-09)
Eric Huber: Anticipated by http://www.softdecc.com/pdb/search.pdb?expression=PROBID='P0534152' Erkki Heinonen, 3. Makuc-Moder-Gedenkturnier 1971-1973, 4. ehrende Erwähnung. (2008-10-01, hidden) more ...
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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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Frank Richter: Correction of P0506581 (2023-06-13)
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Yuri Bilokin: you can see *) 1...Bc1 2.cxb1=R Ra7# (2022-12-15)
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Yuri Bilokin: you can see 1...Sd1 2.c3 Sxe5#
1...Sxa2 2.Rc3 Sb4#
1...Sa4 2.Qc3 Sc5# (2022-09-11)
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in Vratnica von Datiaschwili/Gatseciladse steingetreu nachempfunden
Yuri Bilokin: you can see *) 1...Kc7 2.Rf8 Re7# (2022-09-11)
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Yuri Bilokin: anticipated P1097369 (2023-03-21)
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Yuri Bilokin: more economical -wPf6, -wPg4, -bPa4, bBg6-f5, bPf7-d7 8/4p3/2R1p1p1/2P1Pb2/1Nppk3/2p3p1/rr2n1P1/R3K3 (7+12) h#3 2.1…
1.Sg1 0-0-0 2.Sf3 Sc2 3.Sxe5 Rxd4#
1.Sf4 Rd6 2.Sd5 Sc6 3.Se3 Rxd4#

AntiZielElement (B2, line obstruction)
Castling (white, long). Check prevention (W-W)
Exchange of functions (wRa1/wRc6, Mate / Passive)
Long-trip (bS, 3) × 2.Play on the same square (W3, 2)
Mates on the same square × 2 (2023-08-23)
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Jean Haymann: Not simply "Fesselungsspiel" but Goethard. There are
some better presentations of that basic shema but not
presented as "after A.Uddgren". (2007-01-02)
Yuri Bilokin: more economical bQh8-g6(-bPg6) 1r6/K4pp1/R3Nkq1/4N3/7P/8/8/Brnb4 (6+8) (2023-01-03)
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Klären Quelle
Yuri Bilokin: you can see 1...Rxe7 2.Re4 Rd7# (2023-06-24)
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Yuri Bilokin: 1. ... Sf3#, white pawn extra (2023-06-24)
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Yuri Bilokin: 1. ... g4#, white rook extra (2023-06-24)
Ladislav Packa: There is one interesting possibility, wRg3 instead of wPg3. The set play is then replaced by the second solution. Zilahi theme, C+ by Popeye v4.63
1.Sh1xg3 Kh3xg3 2.g5-g4 Rf2-f5#
1.Sh1xf2 + Kh3-g2 2.Sf2-g4 Rg3-h3# (2023-06-24)
Yuri Bilokin: Ladislav, in the proposed edition can be seen 1...Rh2 2.Sxg3 Kxg3#. There is a better implementation of the Zilahi theme for this white material P0527728 (2023-06-26)
Ladislav Packa: Zilahi is just one part of the theme, besides it contains multiple tempo play.
1. ...Rh2 (thr. 2.Kg2#) 2.Sxg3 Kxg3#
1.g4+? Rxg4 (thr. 2.Rf5#) but 2.Sxg3, Sxf2+
1.Sxg3 Kxg3 (thr. 2.Rh2#) 2.g4 Rh5#
1.Sxf2+ Kg2 2.Sg4 Rh3# (2023-06-27)
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zunächst für die 2. ehrende Erwähnung vorgesehen, wegen Vorgänger (P0565488) Auszeichnung aberkannt (vergl. 'Die Schwalbe' 08/1984 S.308)
Anton Baumann: wegen Vorgänger (P0565488) Auszeichnung aberkannt (vergl. 'Die Schwalbe' 08/1984 S.308) (2022-12-23)
Yuri Bilokin: Model mats, other finals. The task has the main reward, it is published. (2022-12-25)
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Yuri Bilokin: you can see 1...Bh4 2.Se7 Be1 3.0-0-0 Ba5 4.Re8 Kb6 5.Kd8 Kb7#
1...Bh4 2.Se7 Be1 3.0-0-0 Kb6 4.Re8 Ba5 5.Kd8 Kb7#
1...Bh4 2.Se7 Kb6 3.0-0-0 Be1 4.Re8 Ba5 5.Kd8 Kb7#
1...Ba5 2.Se7 Be1 3.0-0-0 Ba5 4.Re8 Kb6 5.Kd8 Kb7# (2024-04-03)
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klären autor?
Yuri Bilokin: author Yuri Bilikin

Castling (black, long) × 3 (2024-03-18)
Yuri Bilokin: slip of the pen, author Yuri Bilokin (2024-03-21)
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Anton Baumann: Informalturnier 1991: 2. ehrende Erwähnung
PB in 'Die Schwalbe' Heft 171A (06/1998 S.489) (2023-01-21)
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Yuri Bilokin: anticipated (P0548424) (2024-03-29)
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Yuri Bilokin: you can see 1...Qg6+ 2.Kd5 Qc6#
1...Qe8 2.Kd5 Qc6# (2023-10-16)
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Yuri Bilokin: you can see 1...Bc6 2.Rb6 Qa8# (2023-10-16)
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Yuri Bilokin: you can see *) 1...Re1# (2022-09-11)
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Anton Baumann: C+ Alybadix (2023-09-05, hidden) more ...
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Yuri Bilokin: you can see 1...Qg6 2.Bd5 Bd6# (MM) (2023-10-16)
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Yuri Bilokin: anticipated P1097369 (2023-03-21)
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A. Luxemburg-Gedenkturnier
Entfeßlung!indirekt, Mustermatts
Felber, Volker: Korrektur zu P1405949 (2022-11-16)
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Yuri Bilokin: possibly Exchange bQa4-bBg8 6qb/1p3p2/R7/4N3/bk1K4/5r2/8/8 (3+7) h#2 2.1…
1.f6 Sc4 2.Kb5 Rb6# (MM)
1.Rf6 Rb6+ 2.Ka5 Sc4# (MM) (2022-12-17)
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F123 Verst-4
Felber, Volker: Wegen NL wurde sBb5 in der LB vorgeschlagen. (2023-02-08)
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Tempozug
Yuri Bilokin: you can see 1...Rc6 2.exd4 Rxe6+ 3.Kd5 Bc4#
1...Rc6 2.Kd5 Rxe6 3.exd4 Bc4#
1...Rc7 2.Kd5 Bb5 3.e4 Rd7#
1...Rc7 2.Kd5 Kb5 3.e4 Rd7# (2024-03-09)
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Yuri Bilokin: you can see in a) 1...Kc5 2.Kd3 Kb4 3.Kc2 Ka3 4.Kb1 Bf5# Gracefully and more meaningfully done later P1005864 (2022-12-12)
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SCHRECKE: Umwandlungsläufer sLh6! (2023-07-24, hidden) more ...
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Sally: Miniatur mit viermaliger Abwandlung von Turmblocks. Alle Lösungen enden zwar mattrein, sin aber alles andere als offensichtlich. (2018-10-25)
Yuri Bilokin: you can see 1...Bd1 2.Ke4 Rg2 3.Kf4 Rg4# (2023-06-07)
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Yuri Bilokin: you can see 1...Be7 2.Kh3 Rh1+ 3.Kg4 Rh4# (2024-03-09)
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Rainer Staudte: Autor: Tibor Flórián (Feldmann war sein Pseudonym) (2008-06-07)
SP: Most certainly not a pseudonym but his real name.
Like most Jews and others whose name was not sufficiently
Magyar, he was forcibly renamed by the Hungarian Nazis. (2023-03-24)
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Benkö in 'My Life': "Creating a double Queen sacrifice like this is quite an achievement, yet the judge in the Schach Echo helpmate contest didn't appreciate it and only awarded the problem an honorable mention. Later, this problem became one of the most popular examples of the bad judging that is so common in publications."
TLi: a) There are no sacrifices, only captures.
b) The wQ can be omitted without changing the solutions.

Bad judging, indeed! (2023-02-16)
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s. Uddgren, 11/1966
Anton Baumann: vergl. P0527589 (2022-12-27)
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Yuri Bilokin: you can see 1...Bh1 2.Kd6 Rg2 3.Kd5 Rg6#
1...Ba8 2.Kd6 Rb7 3.Kd5 Rb6# (2023-01-02)
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Yuri Bilokin: you can see 1...Be4 2.Sh2 Rf3 3.Kh1 Rf1# (2023-06-12)
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Jakob Leck: Compare to P0559688 (2023-01-03)
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Branko Koludrovic: This task has been improved, and the new setting has been printed in"Die Schwalbe"Feb 2002:
White: Ke8,Bc7d6e7f6g7(6)
Black:Ke6,Sd5e5 Bc2d7f4f5f7(8) H#3
1.c1D c8S 2.Dc7 g8S 3.Dd8+ exd8S#
Task Type dSSS (2010-04-05)
Anton Baumann: auch oben genannte Version ist C+ (Gustav 4.2a) (2023-02-18)
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Harry Fougiaxis: The composer is Vratislav HADAC (2006-08-11)
Felber, Volker: Schach schreibt "V. Michal-Hadac" (2023-08-04)
Vaclav Kotesovec: The names of the authors are correct: Vratislav Hadac and Dr. Václav Míchal. See also other problem by these authors http://problem64.beda.cz/silo/ceskoslovensky_sach_1968.pdf, p.6 (in PDF). (2023-08-04)
Felber, Volker: Autor ist geklärt! (2010-10-09, hidden) more ...
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SP: Published as an original 33 years later:
275 feenschach ix/1971 (p. 141)
A double publication if the source given
in the Niemann collection is right. (2023-02-08)
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Quelle klären
Yuri Bilokin: you can see 1...Rb2+ 2.Kh1 Bf4 3.Rg1 Rh2# (2023-06-12)
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Yuri Bilokin: you can see 1...Qd5 2.Sf3+ Qd1 3.Re1 Qxe1+ 4.f1=B Qd1 5.Se1 Qd4#
1...Qd5 2.Sf3+ Qd1 3.Re1 Qxe1+ 4.f1=B Qd1 5.Sd4 Qxd4#
1...Qd5 2.Sf3+ Qd1 3.Re1 Qxe1+ 4.f1=B Qd1 5.Sh4 Qd4#
1...Qd5 2.Sf3+ Qd1 3.Re1 Qxe1+ 4.f1=B Qd1 5.Se5 Qd4# (2024-02-09)
milan: wQa8-a7 C+ by PY 4.61 M.Frelih (2024-03-28)
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SCHRECKE: Die Inkorrektheit besteht darin, dass im Satz 2. Kf5 und 2. Kd3 geht. (2022-10-23)
HBae: Mögliche Korrektur wSd3 statt wLg3, C+ Popeye v4.37 (2022-10-23)
Dieter Berlin: Bitte NL angeben (2022-10-23, hidden) more ...
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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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James Malcom: I don’t believe that the great Fabel, with his keen eye, would have not seen such a simply cook as the taking of Black’s c5 pawn. I suspect that a Black bishop was on d4 but was forgotten placing the problem in the database, or it was a misprint. I am changing this problem to not cooked for now. However, someone should check the original source in the event I am wrong, which is always a possibility. (2020-03-15)
Mario Richter: Hi, James, there's indeed a black bishop on d4. Original stipluation: "h#48 (nicht exakt!)"
A scan for some issues of 'Schachmatt' is available in the archive at stere.ro, see
www.stere.ro/wp-content/uploads/2014/06/stere_sah_istoria_sahului-1947-1949-Schachmatt1.pdf (2020-03-26)
Joost de Heer: Stupid question, but won't moving Bh3 to g4 add another halfmove? (2023-01-30)
Henrik Juel: Yes, Joost, but this would make the solution even less exact
0... Lf3,Lh3,Lh5 1.g4 Lxg4 etc. (2023-01-30)
Joost de Heer: @Henrik: So? This is an attempt at a non-unique (exact is something else) h# length record, the level of non-uniqueness is irrelevant. And as James pointed out, adding Kg4-f4 will add yet another move.
Only possible cooks are when white starts with 1... Bh5 2. g4 Be8 (so not capturing Pg4) but I don't see how that could lead to a quicker mate. (2023-01-31)
James Malcom: I double-checked that myself, Joost, and found that going around for wPg2 actually takes far longer, which would be required any PxB action to take place, but that takes far longer than the normal solution. I know Olaf's Gustav would check, since it C+ the h#111. (2023-01-31)
James Malcom: Is this the record for the longest legal, but dualed, helpmate? It is as far as I know. Is it possible to correct this problem with a black bishop placed on d4 (this is legally possible as Black has a pawn to spare), which would uphold the original solution? (2020-01-27, hidden) more ...
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vergl. P0539482
Anton Baumann: vergl. P0539482 (2023-01-10)
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Yuri Bilokin: yoy can see 1...c8=Q 2.f1=S Qc6#
1...c8=Q 2.f1=S Qb7#
1...c8=Q 2.f1=S Qa8#
1...c8=B 2.f1=S Bb7# (2024-02-29)
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