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6 problem(s) found in 7732 milliseconds (displaying 6 problem(s)). [COMMENTDATE>=20220810 AND G='h#' AND NOT A='Hage, Poul' AND K='a posteriori (AP)'] [download as LaTeX]

1 - P0001420
Branko Koludrovic
294 Europe Echecs 197 04/1975
1. ehrende Erwähnung
P0001420
(13+12) C+
h#3*
1. ... cxd6ep 2. 0-0-0 0-0-0 3. Kd7 Sa7#
1. De7 c6 2. Th8 c7 3. Tf8 Sd6#
play all play one stop play next play all
White pawn caps: axb,dxe,gxf,hxg.
Black: fxg,bxc,cxb.
wPb4 came from b3 to release wBa3, so bPb3 captured to reach that square.
All pcs accounted for means bPd never captured.
In the set play, there are 13 retro tries in which one or both players do not castle. The intention is that both castling rights are needed in order to imply the pawn double hop.
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)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wgsg)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro-logic.
FEN: r3k3/6p1/4p3/1NPp4/BPp4q/Bp2PPPr/pP2PPpp/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
2 - P0003138
Branko Koludrovic
4208 Problem 12/1979
P0003138
(11+9)
h#3 (AP)
0.1...
1. ... axb6ep 2. 0-0-0 0-0-0 3. Td7 a8=D#
play all play one stop play next play all
This problem can be solved presuming that both castlings are executable. It implies that the last halfmoves have been.-1.Bb6xa7 sBb7-b5.
Bb5 could not come from c6 (sBc6xFb5), because all of the missing white pieces(5) have been captured on white squares except wLc1,which must have been captured by Bc7(Bc7xwLb6).Further,it has not played immediately before -1.Bb6-b5 because this would prevent white castling(forcing wT or wK to move)
Here is the proof that 4 remaining white pcs have been captured on white squares:
black d-pawn must have been captured on d4 and h,g f pawns had to promote after Bh3xwFg2,then wBh2moved to h4, enabling Bg4x Fh3 ..h1T, and finnaly sBf7
after fxg2... g1S. sLf1 prevented the check from Th1, and later has been captured by a white officer.
Conclusion:
Retro:-1.wBb6xFa7 Bb7-b5
Forward(AP) 1....axb6 e. p.2.0-0-0 0-0-0 3.Td7 a8Q# (Author)
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)
more ...
comment
Keywords: a posteriori (AP), En passant as key, Castling (sgsgwg), Promotion (D), Valladao Task
Genre: h#, Retro
FEN: r3k3/P3p3/p7/Pp6/p6P/2P3PR/2PPP2b/R3K1nr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-11 more...
3 - P0009121
Tomislav Petrovic
2949 Phénix 69 12/1998
P0009121
(10+9) C+
h#2 (AP)
1. fxe3ep d8=S 2. Kxd5 0-0-0# (Td1#?)
play all play one stop play next play all
White has made 6 pawn captures with one missing black piece unaccounted for, that by parity can only have been captured by an officer. White's last move cannot have been a pawn capture. If R: 1.f2-f3, then sLg1 was promoted, implying 7 captures by black pawns - one too many. So if White can prove they retain castling rights, then the ep is on. Hence AP Petrovic is valid.
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)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Promotion (S), Valladao Task
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 + simple retro-logic
FEN: 8/3P2p1/2PP4/1ppPp3/2pkPp2/5PP1/6Pp/R3K1b1
Input: Gerd Wilts, 1999-02-27
Last update: A.Buchanan, 2023-09-11 more...
4 - P1401468
Branko Pavlovic
Problem 1977
P1401468
(4+7)
h#2 (AP)
Maximum exact
1. Txf3 0-0 2. Tf8 Txf8#

Thematischer Fehlversuch:
1. 0-0-0? Txh3 2. Td1 Th8+ 3. Td8!

Illegal:
1. Txf3 Txh3 2. Tf8 Te3#
play all play one stop play next play all
Autor: Rochaden schließen sich gegenseitig aus. Die schwarze wird durch das Retrospiel f2-f3 Tf3xf7 ausgeschlossen, die weiße durch KxLe1. ... 'a posteriori'-Legalisierung durch Ausführung der Rochade.
Henrik Juel: I do not know what Maximum exact means; neither do Popeye 4.61 and Märchenschachlexikon
Popeye 4.61 with 'con max' or 'con max ultra' both produced no solutions (2022-05-26)
Mario Richter: KW "en Passant as key" deleted, since I do not see any epkey here.

Classification after Branko Pavlovic:
1. Classical Maximummer:
Black has to make the geometrical longest move with the following restrictions:
a) Black attacks the white king according to the orthodox rules
b) If the absolutely longest moves are not executable, Black makes the relatively longest moves, i.e. one of the executable ones with maximum length.

2. Ultra-Maximummer:
As the classical Maximummer, but without a)

3. Exact-Maximummer:
Only the longest moves are "active". (2022-06-11)
A.Buchanan: Thanks Mario. I would prefer if this kind of distinction between checking, control & move was defined abstractly, independent of any particular fairy condition, to which it can then easily be applied. Also, prefixes "ultra & exact" are misleading (2023-06-29)
comment
Keywords: a posteriori (AP) (Type Petrovic), Maximummer (exact), Castling (wk)
Genre: h#, Retro, Fairies
FEN: r3k3/p4r2/8/8/8/p4P1p/4p1P1/4K2R
Input: A.Buchanan, 2022-05-26
Last update: Mario Richter, 2022-06-11 more...
5 - P1401495
Igor Vereshchagin
Odessa 1997
Special commendation
P1401495
(10+14) C+
h#2 (AP)
1. a4xb3ep+ Sxa5+ 2. Kd5 0-0-0# (by AP not Td1#?)
R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7,Sd6xLc7
play all play one stop play next play all
With the typo wPd5 corrected to e5, the two forward candidate solutions are correct. How about the retro? Assume that White castling rights remain. bBg5 is promoted, and did so on c1 or g1. In either case, there were 6 pawn caps by Black, but in the former case, the blockaded wRh was not consumable. So the promotion must have been on g8, and White can't retract h2-h3. The missing White unit is light B, so was not just captured on e5 or b4. wRf6 is promoted, say wPd, which would have involved no captures.

This leaves b2-b4, preceded by Kb4-c4+ and before that Sd6-b7. Prior to b2-b4, why couldn't bQ have just played e.g. Qa8-a5+ instead? Because there would still be no prior move: wSb7 can't have just come from h5 or e6. Similarly Rc3-a3 has no precedent.

We don't know exactly what happened to sLc, but that doesn’t affect the soundness. It might have enabled wPc or wPd to capture. wPd did promote, but maybe it was wPc that after capturing promoted to T. Alternatively, Sd6xLc7 was possible.
Henrik Juel: Popeye 4.61 with 'opt enp b3' found no solution (2022-05-27)
Gerald Ettl: Schaut so aus, als ob die sDa5 nach e5 muss. 1.axb3 ep Sa5+ 2.Kxd5 0-0-0# (2022-05-27)
Gerald Ettl: aber auch 2.-Td1# ? (2022-05-27)
Henrik Juel: Andrew, can you throw any light on this problem? (2022-05-28)
Henrik Juel: Gerald, in AP problems like this one White must castle to legitimize the ep capture key (2022-05-28)
A.Buchanan: Hi Henrik, Gerald. I found this one in WinChloe with the current diagram. It gave the "solution" 1. axb3ep+ Sxa5+ 2. Kd5?? 0-0-0#. However when I tried running the WinChloe solver just now, it said there's no solution. Quite right: Rf5 controls d5. I don't think we can shift Qa5 to e5 though.
My first thought is that wPd5 should be on e5. Check my suggested solution in the solution text. However, as you will see, I think we need a second thought as well! Hope the finest retro minds can figure this out. Failing that, we could ask Igor. (2022-05-28)
Mario Richter: The diagram in the reprint in 'Uralsky Problemist' is exactly as given here. The solution given there without further explanation is: 1. ab+ (e.p.) Sxa5+ 2. Kd5 0-0-0#!
Notice that the second black move doesn't contain a capture sign!

Some typos in Andrew's analysis:
The black potential promotion squares should be c1 and g1.
White's last move couldn't have been b3-b4?? simply because that would give an illegal check. (2022-05-28)
A.Buchanan: Thanks Mario for the careful reading. I’ve fixed the errors. Pd5 may be a typo in the reprint then. It also creates a bogus retraction d4-d5, so I don’t think it can be right. Now for Igor. (2022-05-29)
A.Buchanan: Joaquim Crusats spotted R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7, which I'm sure is the intended retraction. So we diagnose a single typo that occurred in the Uralsky reprint diagram: wPe5 appeared on d5. I suggest correcting it here. (2022-05-30)
A.Buchanan: Author confirmed this was a typo (2022-11-23)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Promotion (Tl), Valladao Task
Genre: h#, Retro
Computer test: Once the diagram typo was fixed, C+ Popeye v4.87 & basic but tricky retro logic
FEN: 3n4/1Np1p1p1/4pn2/q1p1PRb1/pPk1pr2/r6P/P4PP1/R3K3
Reprints: 18 The Ural's Problemist 13, p. 5, 02/1998
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2022-05-30 more...
6 - P1409841
Andrew Buchanan
1 Phénix 331, p. 12922, 06/2022
after A.Lubusov
P1409841
(15+6) C+
h#2* (AP)
1. ... e6 2. 0-0? Lxh7# (castling rights lost)
1. ... Txh7 2. Tf8 Te7#

1. cxb3ep Txh7 2. Tf8? Te7# (ep needs AP justification)
1. cxb3ep e6 2. 0-0! Lxh7#
play all play one stop play next play all
White pawns have captured 9 times right-to-left, accounting for all but one missing Black unit. So by parity, there were no other pawn captures by White. So wBb never left that file, and Black cannot have just played Bb3xa2 behind it. sBa was waylaid on a-file by an officer.

So if Black moved last, it must have been Ke8 or Th8, and Black has lost castling rights.

On the other hand, if White moved last, then Black needs a tempo move. The only possible one is ep capture. (1. Tg8+? is check.) But the ep capture is only legal if Black retains castling rights, so Black's second move must indeed be 0-0 for A Posteriori justification.

In this case, last moves were R: 1. b2-b4 b3xDa2 & e.g. 2. Le3-d4,~ b4-b3 3. d4xSc5,~ Sa6-c5,~ with many ways for White to release Black.
Corrects P0000615.
more ...
comment
Keywords: a posteriori (AP), RIFACE Retro Solving Tourney (2022), En passant as key, Castling (sk), Tempo Move, waylaid (sBa)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro thinking
FEN: 4k2r/1N1p3p/3P4/1PPPP3/1PpBBP2/P1N3K1/p7/R6R
Input: A.Buchanan, 2023-05-20
Last update: A.Buchanan, 2023-09-11 more...
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