VL: Solution:
I 1... c*d6 e.p. 2.0-0-0 Q*c7#
II 1... g*f6 e.p. 2.0-0-0 Rg8#

NOT:
2.Kf8? Q/Rg8#?? no AP-justification!

Castling implies e.p. 'a posteriori' and is necessary only to this end. (2002-04-03)
A.Buchanan: Solutions & tries (Popeye v.485) are:
1. ... cxd6ep 2. 0-0-0 Dxc7#
1. ... cxd6ep 2. Kf8 Dg8#
1. ... gxf6ep 2. d4 Tg8#
1. ... gxf6ep 2. 0-0-0 Tg8#
1. ... gxf6ep 2. Kd8 Tg8#
So after 1. ... gxf6ep:
2. Kf8 Rg8 is not even a mate (f7 flight)
and is 2. d4 a kind of dual thematic AP try? No-one's mentioned it. It would be easily fixable by shifting wBd2 to d4.
However, the main thing issue I don't understand PRA+AP logic (maybe I did once?). Under normal PRA algorithm, the "parts" for cxdep, gxfep & 0-0-0 respectively are YNY, NYY, NNN. Others are either impossible (3) or dominated by application of castling convention (2). OK so far.
There is no solution in the NNN part, so I suppose we must use AP in another part to eliminate it. But castling in another part where we are explicitly *assuming* that castling is ok seems to me to prove nothing. And if this inference did work somehow (which seems to me incredibly unlikely), why would we need to do it in *both* of these other parts? Surely one would be enough, but that means that the other part would be cooked.
Or maybe somehow AP "proves" that there are only 2 dimensions for PRA space, cxdep & gxfep, but that is so incredibly circular and horrible.
This kind of problem needs a proper step-by-step solution. Currently the solver can at best sort-of-vaguely-see-what-the-composer-might-be-alluding-to: i.e. it's just a joke.
Actually, I don't think that it can be PRA at all. I think it must be RS with two solutions supported by AP. But then there's still the question of why we need to castle in *both* solutions. I am sure I have seen an AP problem with two solutions where there is 0-0 in one solution and 0-0-0 in the other, which suggests that we *don't* need to castle twice in this problem. Which interpretation is correct, or do we have such a vague interpretation of these things (in order not to make independent spirits sad) that both are allowed to "work". (2020-12-30)
VL: Generally I agree with Andrew's analysis. The presented solution needs some corrections and clarification.

At first, the extra try 1.gxf6 e.p.? d4. I don't remember details but indeed, at present I'd prefer to exclude it by shifting wPd2 onto d4. One more thematic try enriches a problem but it is better to have an equal number of tries after both e.p.'s. In general, in AP-problems, thematic tries are highly desirable but not necessary formally for soundness. Dual avoidance in tries is here just due to the flight f7. Yes, 2.Kf8? Rg8#?? is indicated wrongly (unfortunately, H.103 with the published solution isn't available for me).

Now the most doubtful point: dual avoidance in the solution. Do we need castling in both partial solutions or at least one would suffice for justification? In my opinion, both approaches make sense in principle. In this problem implemented is the former, what means that every partial solution is permitted to be considered separately and, thus, should contain legalizing castling. This is similar to ordinary AP-problems, in which every line of the solution must contain castling. There is a distinct much subtler and rarer approach, which I call "Total AP", that considers all lines as a whole and thus requires to contain castling at least once (with subtle soundness issues). So, the current problem is treated (implicitly) not in the spirit of Total AP.

The problem implements a certain kind of triples of mutually retrodepending special moves (w/b castlings and e.p.). Namely, it is a "hybrid" AP-problem, where castling justifies a posteriori the typical pRA-choice between two e.p. captures. This is one of lucky triple kinds that admit (controversially) sound implementations based on hybrid logical conventions (logics for short). One more such a triple is represented in my earlier P0002474. Totally 37 distinct kinds of interacting triples can be identified, and only for few of them I know (or at least expect) reasonable implementations under special logics (whichever artificial and arguable they can be but consistent internally). Under an implementation I mean (loosely) a problem with an orthodox stipulation (#n, h#, etc.) supplemented by a suitable retro-convention like an ad hoc combination of pRA, RS or AP such that the position of the problem contains the corresponding triple of moves AND its full solution depends on all their combined (il)legalities. An extremely subtle matter. (2021-01-29)
A.Buchanan: Dear Valery, thanks for your response. We seem to differ fundamentally as to the minimum acceptable level of conceptual precision. My focus is always primarily on attracting new solvers (some of whom may then I hope become composers). Except for the occasional joke (alerted by "(-:") I want the mysteries to be in the chess, not in guessing which way certain terms are intended to be interpreted on a particular occasion. Where you write "subtle", I read "opaque". Subtlety becomes a virtue as soon as the concepts are accurately defined, written down, and made available to solvers, but until that point it is a major defect.

Terms must have meanings which are consistent from one problem to another (certainly within one composer's work!) to reward the solver for persisting through the initial learning curve. It's not OK (IMHO) for the composer to decide on the fly how terms in a problem will work, as when you write "both approaches make sense". The solver should be able to look at the tutorial, and see exactly how condition X works, and then can apply it to all problems which include condition X.

In fact, here neither approach makes sense, as I showed. If we have three PRA cases: YNY,NYY,NNN then even if in one or both of the first two solutions we demonstrate castling it does not allow us to eliminate the third case. The ability to castle was just an assumption in the first two cases, just as the inability to castle is an assumption in the last. There are no logical grounds to eliminate the third case. It has no solution, so the problem is broken. Thanks again for your time. (2021-01-31, hidden)
A.Buchanan: Dear Valery, thanks for your response. I wish I could be more enthusiastic. But at the moment I don’t see a perspective from which this problem can work. PRA simply does not give us a way to eliminate NNN. Given that NNN has no solution, Werner Keym's meta-meta-convention kicks in, and we abandon PRA and try to apply RS. The issue now is that castling won't tell us which ep was played.

Rather than reuse existing terms in vague ways, I suggest that you name and define a brand new logic, separate from PRA, RS & Touch Move, with precise semantics. I don’t know whether you would have AP built-in to this or an optional add-in. I think a lot of the "paradox" (i.e. confusion) disappears, and solvers having read a tutorial can apply your ideas repeatably and reliably to other problems, with no nuanced interpretations required. (2021-02-01, hidden) more ...
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"das zeigt die Paradoxie der AP-Bedingung!" (MS). Gleich 1. bxc3 ep.? und dann 2. Kd4 geht natürlich nicht, weil zuvor außer K,T-Zug und c2-c4 auch S-Züge und c3-c4 möglich waren. "Interessanter und genau abgestimmter Lösungsverlauf" (GW) "Sehr schöne und ökonomische Verbindung zwischen konsequentem sh# und AP!" (TB) Mit 'Typ Tauber/Caillaud' ist gemeint, daß die einzelnen Stellungen doch als retroanlytischer Gesamtkomplex betrachtet werden dürfen/sollen (sonst wäre keine AP-Legalisierung möglich), obwohl sie ja aufgrund der retroanalytischen Neubewertung nach jedem Rückzug voneinander unabhängig sind. 6L./5+5P.
A.Buchanan: "By 'type Tauber/Caillaud' it is meant that the individual positions may/should be regarded as a retroanalytic overall complex (otherwise no AP legalization would be possible), although they are independent of each other due to the retroanalytic reassessment after each retraction." (2023-06-29)
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Mario Richter: 1. exd3ep 0-0 2. cxb4 Tf4 ist nur Verfuehrung (scheitert an schwarzer Bauern-Schlagbilanz).
Loesung: 1. ... w0-0! (AP-Legalierung des weissen Anzugrechtes) 2.cxd4 Tc1# (2009-02-10)
A.Buchanan: Supersedes P0000553 (2022-01-08)
Mario Richter: Hi Andrew, I only quoted from the "official solution" ('Die Schwalbe' Heft 72, 12/1981, p.399).
Perhaps a better way to get anwers to your interesting questions is to ask the authors directly ... (2022-01-08)
A.Buchanan: AP Type Petrovic is a try, because wLc1 was captured at home, yet Black still made 6 pawn captures.
AP Type Keym is the actual solution, because if it was really BTM, then White would have lost castling rights.
So there are two kinds of AP here. (2022-03-21)
A.Buchanan: Hi Mario, can you give more details of AP-legalization of move rights, please:
(1) Can this happen any time a set play exists with AP?
(2) Should the stipulation show a second solution?
(3) What's to stop White doing the same thing again, and getting two (or more moves in a row)? How to count the number of moves in the solution?
(4) Isn't there a race condition here for any adversarial stipulation (#n, s#n). Does the actual player on move always have to wait for the opponent potentially to step in?
(5) Can a player by not moving when in checkmate assert by AP that it's not his move, and require the opponent to move under Lese Majeste?
(6) This isn't really "A Posteriori" - it's from itself. Nothing happens later. I would use the phrase "Ab Se". (2022-01-08, hidden) more ...
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Erich Bartel: vom Dual 2.-- Kf2/Kh2 abgesehen C+ PY V4.41.--- (2008-11-07)
A.Buchanan: This is cooked in Popeye because White can't castle as long as bK controls f1! This is quite reasonable behaviour in monochrome: compare to en passant! So no solution (2022-02-18)
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A.Buchanan: Not trivial to fix as shifting wPa5 to b6 leads to promotion cooks. Adding black material must be carefully done to avoid disrupting uniqueness of e.p. as tempo. (2022-03-11)
A.Buchanan: OK just to say I have found a form of this idea that works. This one was the very devil to fix - the final form includes different mates and retro logic, so becomes an "after" rather than a "correction", and I will find somewhere else to publish first. (2022-03-19)
VL: The author: Andrej Lobussow (Lobusov).

Solution:
1... Rxh7 2.Kf8 Rxh8#
(1... Kg5? 2.0-0?? Se7#;)
1.cxb3ep (waiting)! Kg5 2.0-0! Se7#
(1... Rxh7? 2.Kf8 Rxh8#??).

W. captured 10 by pawns (including hxPg-g8),
not on the last move (10+6=16).
Bl. took only b3xa2 with w.Pb4 on b2. (2004-06-03, hidden) more ...
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James Malcom: Full solution? (2021-02-05)
Henrik Juel: White pawns captured all 11 missing black men by axb, f2x..xa7, and g2x..xb7
Black captured hxg and once more, so Pc5 never captured
If last move was c6-c5, the preceding white move was with Ke1 or Th1
So if White can castle, last move was c7-c5
1.bxc6ep dxc6 2.0-0(AP) c5 3.Ta1,Sd7 and #4
2.Sd7 c5 3.0-0(AP) is also possible
I expect the full solution was omitted because it is very dualistic (2021-02-05)
Anton Baumann: Autorabsicht: 1.bxc6ep [2.0-0 3.Lb5#] bxa3 2.0-0 axb2 3.Tb3 nebst 4.Lb5,Tf4#
1. ... dxc6? 2.0-0 nebst 3.Tf4,Lb5#
Dual: 2.Tb3 dxc6 3.0-0 [4.Tf4#] c5 4.Lb5# 2. ... a2,axb2 3.0-0 [4.Lb5#] dxc6 4.Tf4# (2021-02-05)
A.Buchanan: C+ of helpmates with this kind of AP is relatively simple: one can just eliminate all lines without 0-0. But C+ of d# is harder.
(1) Henrik's retro logic is solid. There is no #4 unless 1.bxc6ep, so let's take that as a given: an AP debt is incurred.
(2) Black has 5 responses. Against 4 of them, 2.0-0 gives a genuine short mate with the AP debt paid. Such short mates don't impact soundness in a retro #4, to my mind.
(3) This leaves 1... bxa3. Following 2.0-0 axb2 (the only full length line) 3.Rb3 thr 4.Bb5,Rf4# but both mates are separately provided, so only minor duals.
(4) It remains to consider if White can delay castling after 1...axb3. Apart from 2.Tb3 nothing works, but 2.Tb3 thr 3.0-0 is unstoppable and is the major dual to my mind. I don't see any obvious fix. (2021-02-05)
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Henrik Juel: [I don't follow the silly convention of writing black moves first]. 1.0-0-0 0-0 2.dTg1+ Kh8 3.Txh6#. b) If Ke1,e8 and Th1 never moved, White captured f/hxPg and b6xLa7; if Ta1 also never moved, last white move was b6xLa7. 1.cxb6ep La6 2.0-0/0-0-0 0-0-0 3.a8Q#. The a) solution also works in b). (2003-12-18)
A.Buchanan: Candidates are 1. ... 0-0-0 2. 0-0 dTg1+ 3. Kh8 Txh6 & 1. ... cxb6ep 2. La6 ~ 3. 0-0-0 a8D#. There are 19 possible W2 waiting moves. Any ensemble of candidates that comprises a solution must include both White castlings. But why shouldn't the ensemble also include any number of the other 17 candidates? I don't understand how this could be prevented... (2022-03-04, hidden)
A.Buchanan: Where a single solution must justify itself, then the AP logic seems clear. But where evidence from different solutions must be Consolidated to justify themselves or other solutions, there are a number of unresolved issues. A general charitable approach: “this looks worthy enough, let’s say that it’s sound” is really counter-productive (2022-03-23, hidden) more ...
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hans: only black move is fxg3e.p.
1. fxg3+ Kh5 2. h8D/T#
Why the keyword 'castling'?
Castling is illegal, for there are 7 black captures, and because white's last move is g2-g4, needs the wK let pass Rh1 (2012-05-18)
Anton Baumann: g2-g4 war nicht zwingend der letzte Zug! Schwarz ist patt, also unlösbar!
Korrektur in 'die Schwalbe' 12/2074 S.264: "wTa1 nach h1. Nun gewinnt nach 1. - fxg3 Weiss mit 2.hxg3+ Kh5 3.f4 Kh6 4.h8=T+! Kg7 und der umgewandelte Turm holt die störenden sBB ab, sodass Weiss rochieren kann und beweist, dass Schwarz nicht patt war. Schwarz versucht die Rochade zu verhindern, ist aber machtlos." (2022-12-06)
A.Buchanan: The forward play is more complicated, as Black can play more aggressively than 3. … Kh6 (2022-12-07)
Anton Baumann: Wenn Schwarz nicht 3.-Kh6! zieht, holt sich Weiss auf h8 die Dame, mit einfacherer Fortsetzung.
3.-g4? 4.f5 Kg5 (4.-g5 5.f6 ~ 6.h8=D) 5.h8=D Kxf5 6.De8 +_
3.-Kg4? 4.h8=D Kf3 5.Dxh3 g4 6.Dh2 Ke4 7.Dg2+ +_ (2022-12-09)
A.Buchanan: How should we show this kind of correction in PDB? A separate diagram? Or if the same, what is the original date - a correction is not a reprint? Happy to follow whatever clear protocol can be defined (2022-12-07, hidden) more ...
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Henrik Juel: -1.c2 c5xLb4 -2.Lc3 c6 -3.Lb2 c7 -4.Lc1 a6! -5.b2 b2xDSa2 etc. White captured sDTSS by fxexd, gxfxe, allowing Black to capture f7xPe6xPd5 etc. (2003-12-18)
A.Buchanan: Why not e.g. wSb1? (2022-03-04)
Henrik Juel: That also seems to work (2022-03-04)
A.Buchanan: I really like the motivation for ep. I guess the motivation for Rb1 is to be inside the cage that forms around it, but it’s still a bit loose. Note there is no retro try. (2022-03-05)
A.Buchanan: I really like the motivation for ep. I like less the loose wKR in the Southwest corner. Note there is no retro try. (2022-03-05, hidden) more ...
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VL: A posteriori (AP)!
2.g8Q(R)+/Rg1+? Kf4 3.Qg5#??
Retro: d7-d5, Rc6xh6+ (unless Kg1-g2, Rh~-h1+).
One of the best AP-type problems. (2015-08-20)
James Malcom: The first Valladao AP? (2020-10-03)
A.Buchanan: Hi James, there are 6 earlier ones in PDB, not counting those which are not marked as Valladao (could quite a few, as basic AP is 2/3 of Valladao). Type k='a p' and not g='fairies' and k='valladao', and set sort order by date. The ur-problem is P0003417. (2020-10-03)
Henrik Juel: 1.cxd6ep+ Kxg3 2.00! Kg4/a1Q 3.g8QR/Qg5# (2003-11-20, hidden) more ...
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Henrik Juel: A possible retroplay is -1... g7 -2.f3xP(-Pg7) b5 -3.f2 c6xP(-Pb2) -4.b2 a6 -5.Rc3 a7 -6.Rc1 b6xB -7.Bf8 b7 -8.Bg7xB etc. (2003-04-22)
Michel Caillaud: The problem is not cooked as indicated, as under A Posteriori convention both castles must be played in the solution to justify the en passant key. (2022-02-04)
A.Buchanan: If wK never moved, then Black has not just moved another non-capturing pawn, as White has no prior move. The only possible exception is R: 1. g7-g5 f/h3xPg4[+Pg7]. (2022-02-04)
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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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Hans-Jürgen Manthey: 1. hxg3 O-O#(auch Tf1# !) mögliche zugfolge:
1. d2-d4 f7-f5 2. Lc1-d2 f5-f4 3. Ld2-c1 e7-e6 4. Lc1-d2 c7-c5 5. Ld2-c1 b7-b5
6. Lc1-d2 a7-a6 7. Ld2-c1 Lc8-b7 8. c2-c4 Ke8-f7 9. Lc1xf4 Sg8-f6 10. Lf4-h6 g7xh6
11. Dd1-d3 h6-h5 12. Dd3-d1 Lf8-h6 13. h2-h4 Lh6-g5 14. h4xLg5 h5-h4 15. Dd1-d3 h7-h5
16. Dd3-d1 b5xc4 17. a2-a4 Lb7-e4 18. d4-d5 e6xd5 19. Ta1-a3 Le4-c2 20. f2-f4 Th8-e8
21. Ta3-g3 Te8-e3 22. f4-f5 Sf6-e4 23. f5-f6 Kf7-e6 24. f6-f7 Se4-d2 25. Sg1-h3 Sd2xf1
26. Sh3-f2 Sf1-d2 27. Sf2-g4 Sd2-e4 28. Dd1-d3 Lc2-d1 29. Sg4-f6 c4-c3 30. Sf6-h7 c3-c2
31. b2-b4 c5-c4 32. b4-b5 Sb8-c6 33. b5-b6 Sc6-a5 34. b6-b7 c2-c1L 35. g5-g6 c4-c3
36. Sb1-a3 c3-c2 37. Sa3-b5 Ta8-c8 38. b7-b8L Tc8-c3 39. Lb8-d6 La5-c4 40. a4-a5 a6xSb5
41. a5-a6 Ke6-f5 42. a6-a7 Dd8-e8 43. a7-a8S De8-e5 44. Sa8-c7 Lc1-b2 45. Sc7-e6 Lb2-a1
46. Se6-f8 Kf5-f4 47. Tg3-g5 Se4-f6 48.Ld6-a3 Te3-h3 49. La3-b2 Sc4-a3 50. g2-g4 (2020-12-11)
Henrik Juel: The white castling is needed to prove (a posteriori) that last move was g2-g4, legitimizing the ep capture (2020-12-31)
Henrik Juel: 1.hxg3 0-0#, not 1... Tf1? (2020-12-08, hidden) more ...
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VL: AP after Keym. Solution:
I: 1.Sf6+.
II: Bl's try to be on move. 0... Bxb7+! 1.Ke3! O-O-O (legalizing!) 2.Sb6#. (2007-01-26)
A.Buchanan: Very nice problem. But why didn't VL say "Omigod! Sorry chaps: this problem doesn't work without PRA! You need to add it!" Answer: because PRA is irrelevant. So can't we get rid of it from most of the stipulations? (2022-02-15, hidden) more ...
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A.Buchanan: There was a diagram error here. It should be bPd7 not bPe7. The existing diagram is vastly cooked with e.g. 1. Ke8 Kd1 2. Te8 Lb6#. WinChloe has the correct diagram here. (2022-05-24)
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Mario Richter: How is the (AP) to be interpreted here? Is the intention as follows: Since in solution 1) White castles, Black is allowed to capture e.p. in solution 2)? (2011-05-28)
Henrik Juel: It seems to be an unusual situation, not covered by keywords like AP or PRA.
If last move was c2-c4, both solutions work; if not, there is no solution, because White may not castle.
(In the second solution, Kxc4 should be Kc4). (2011-05-28)
VL: This a generalized type of AP, which I call "consolidate AP": both solutions are considered as parts of one complete solution. However the order of both parts is significant. I know two similar problems: P0003437 and P0003186. (2011-06-01)
A.Buchanan: @Valery: I've started to classify all the AP problems in PDB. This and its kin are "Type Petrovic - cons" (standing for consolidated, but there's only a limited number of characters for a PDB parameter). I truly hope that the theory in the end will not have this as a special case, but rather is part of a more general pattern.
It's not just that PRA & RS each scales up within its own paradigm to handle seamlessly very complex collections of conditional move dependencies. They also scale *down* to handle situation when there is just one conditional move, or even zero conditional moves. And when n=0 or n=1, PRA & RS are in complete agreement. This is the kind of robust seaworthy behaviour we need if are going to set sail in the stormy seas of fairydom.
But small steps. The first thing is to classify all the problems in PDB into buckets, to find out what we have. I doubt there are 37 categories, so will probably need the results of your scholarship too. (2022-02-17)
A.Buchanan: Interesting position: does anyone have a document (even old or draft) about the theory of AP. Even Werner Keym's "Castling and E.p. Conventions" article is very skimpy on this subject.
The whole point of AP in a way is to say that the order of moves is unimportant. So for this topic more than any other topic, I would be surprised if it's necessary to say that the order of solutions matters!
The great thing about PRA or RS is that each is scalable in a systematic way to cover more and more conditional moves. So I would be very cautious about embracing an approach for AP which says that as soon as things get a little more complicated, we need a new type of theory. Valery, please, please send me the 39 patterns that you mention in another PDB comment, and I will try to make sense of them. (2022-02-17, hidden) more ...
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Henrik Juel: Analysis
If Black can castle both ways, last move must have been b7-b5
C+ Popeye 4.61 (2020-10-25)
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Henrik Juel: Black captured c7xd6x..x.h2 and once more with an officer
White captured a2xb3, b2xa3, f2xg3, and e.g. exf and once more
If Black may castle, last move was g7-g5 (not b7-b6, because of Lh3) (2022-04-26)
Henrik Juel: HC+ Popeye 4.61
The castling serves two purposes:
enabling the mate and legitimizing the ep capture
so the a posteriori legitimizing is impure (2022-04-26)
A.Buchanan: Yes Henrik. The absence of other candidate solutions (beginning e.p. but excluding 0-0) declines an opportunity to embed additional content, and is arguably an artistic defect. However this very early AP problem is sound, and purity of motive is just one school. IMHO, there is space for such problems, if other content compensates: e.g. 4k2r/p2ppp1p/p7/5PpK/8/1PBB2Pb/2PP2Pp/8 h#2* AP. (2022-04-27)
A.Buchanan: Yes Henrik. The absence of alternative candidate solutions (beginning e.p. but excluding 0-0) is misses the opportunity to embed additional content, and is arguably an artistic defect. However, the school which argues for purity of motive is just one school. The problem is sound, and there is space for such problems, particularly if fun retro logic compensates. Here, the AP solution is crying out to be the set play for a h#2.0, but that is not currently unique. (2022-04-27, hidden) more ...
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A.Buchanan: There is definitely something wrong here, with both (a) & (b). I think it's a simple diagram error: bPb7 has been misplaced on b6. Then the AP logic for (a) works great, and the castling "NL" for (b) is seen to be a thematic retro try. Can anyone confirm? (2022-03-21)
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A.Buchanan: This is a very heavy position to prevent the possibility of R: 1. c2xb3. I'm not sure why this was done. A much lighter position 8/8/8/5np1/1r1pPkr1/2Bp1p2/1p1P2P1/4K2R with 7 less units achieves the mates soundly. Am I missing something? (2022-02-16)
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A.Buchanan: Why is there sDe3? Isn't sL sufficient? (2022-05-27)
Ivan Skoba: a)
try: 1.dxc3 e.p. 2.Kc4 4.Rd5 5.Qc5 6.Ke4 - Rh4# ?? (white must legalizate e.p. capture by castling)
solution: 1.dxc3 e.p. 2.Rb2 5.Kc1 6.c2 O-O#
b)
try: 1.Rb2 4.Kc1 5.c2 O-O# ?? (illegal castling!)
solution: 1.Kc4 3.Rd5 4.Qc5 5.Ke4 - Rh4# (2000-11-03, hidden) more ...
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A.Buchanan: Ingenious play, but Black cannot execute the castling. Lines include a good try, checkmate by Black and pat by Black. (2022-04-16)
A.Buchanan: There are two kinds of directmate Type Keym. In one White successfully pushes the move to Black, in the other (as here) Black unsuccessfully pulls it. (2022-04-17)
Ladislav Packa: The logic of this problem is foreign to me, but it is incorrect: 2.Bxe6 Rg6+ 3.Kf5 0-0 (2022-04-18)
A.Buchanan: Hi Ladislav thanks for this. You're right. So 2. Lxe6? Tg6! 3. Kf5 0-0! is another try. The solution must be 2. Lg6! I'll post the solution above (2022-04-18)
A.Buchanan: White wins simply if on the move. But can Black steal the move? If Black moved last, it must be Kd8-e8 (or possibly capture). So by castling, Black will prove they are on the move. The only route is 0. … Tfxg8 1. Lxh7 Tg6+ 2. Lxh6 0-0 saving the game. What am I missing? (2022-04-16, hidden) more ...
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A.Buchanan: A key feature of adversarial A Posteriori is that any castling must be forced in a finite number of moves (but not necessarily limited by the number of moves in the stipulation goal). If the other side can prevaricate indefinitely, then that is sufficient to defeat the A Posteriori "steal" (2022-02-16)
A.Buchanan: Why this would be "PRA"? Maybe the idea is that we don't know who is first to move, yet whoever it is, White wins. But that only applies to "pull" scenarios such as this, where Black snatches the move because otherwise the game is lost. In other situations where White to avoid loss must "push" the move, then there is no way this can be described as PRA. The fundamental push/pull thing has a unity, and I don't think it's helpful to use "PRA" which only describes half of this, and was really designed for a different context. Strategically, these push/pull adversarial battles are amongst the most interesting AP problems. (2023-07-22)
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A.Buchanan: I don't get the regular part of this. White can mate in 3 in various ways: 1.Tf1,Se2+,Sf2. Black just played R: 1.Kh2g2 Td~h1+, so White can't castle but doesn't need to. What's going on? (2022-02-15)
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Guus Rol: This is apparently AP after Keym. The move goes to black:
0. ... f5! 1.gxf6ep exf6 2.0-0(justifies the handover) f5 3.Tf3 (2007-02-13)
A.Buchanan: There are two kinds of directmate Type Keym. In one Black unsuccessfully pulls the move, in the other (as here) White successfully pushes it to Black. (2022-04-16)
A.Buchanan: I disagree with the German comment in the solution text. I think including "AP" is advisable in the stipulation. Keym AP riffs off Codex Article 15, but it's not the default. (2022-04-17)
A.Buchanan: Can one shift wS to e6, thereby saving a unit and entering Meredith territory? I think so. (2022-04-16, hidden) more ...
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An earlier stipulation in PDB was "a) #3 b) Weiß setzt in 3 Zügen # (AP)"
hans: a) white pawns capture all black pieces, including h-pawn. If last move was fxLe4, the h-pawn promotes without capture, so 0-0 is illegal.
1. Da3!
1. … Dxb2 2. Ke2+ Kc2/Dc1 3. Sb4/Txc1#
1. … Dxa2 2. Ke2+ Kc2 3. Tc1#
1. … Kc2+ 2. Sxa1+ Kd3/Kb1 3. Da6/Ke2#

b) If last move was a white pawn capture, and black capture hxLg, there must be 0-0 in the solution to prove this.
0. … Dxb2 1. 0-0+ Dc1 2. Txc1+ Kb2 3. Le1#
0. … Dxa2 1. 0-0+ Kc2 2. Tc1+ Kxb2/Kd3 3. Le1/Db5#
0. … Kc2+ 1. Sxa1+ Kxb2/Kd3 2. Db3+/0-0 Ka1/Lxd7 3. 0-0/Sc1#
0. … Lxd7 1. 0-0+ Kc2 2. Lc1+ Kb1/Kd3 3. Sd2/Rd1#

0. … Kxb2+ 1. Lc1+ Kb1 2. 0-0 Dxc3/Dxa2 3. Sxc3/Lb2#
(1. Sxa1? Lxd7 2. Dc2+ Ka3 3. Lc1# but no castling, so illegal) (2015-09-08)
A.Buchanan: This is a good problem with varied and mostly accurate play in both a & b. How would one translate the stipulation of b into English, please? (2015-09-09)
Henrik Juel: What about the stipulation: #3
and the twinning: b) Black to move (AP)
(the German stipulation text is not clear, either) (2015-09-09)
VL: Stipulations like "Black to move (AP)" make no sense because using the AP-logic is a right rather than a duty: if Black to move were stipulated explicitly then nothing would need to be proven and therefore this twin would be cooked. In my opinion, the most appropriate stipulation
is the following paradoxical one: "#3. Two solutions (AP)". "AP" indicates that the AP-logic is permitted, and at least one solution (or maybe only a thematic try) does use it. An excellent problem, anyway. In Die Schwalbe, H.167, its detailed solution is published.

Here a kind of AP-logic, called sometimes "typ Keym" or "ad libitum", is employed: the justification of the improper side's turn to move (rather than of an e.p.-key) a posteriori. (2015-09-20)
A.Buchanan: In the V&V Encyclopedia, "Type Keym"/"ad libitum" is described in the context of PRA rather than AP. It is contrasted there with "Type Offner"/"a priori". I still feel it's all rather cloudy. How are these accurately defined, and exactly how does the distinction carry across to AP, please? (2022-02-15)
A.Buchanan: I've looked at V&V encyclopedia carefully, and in the absence of definitive information, I am going to make the assumption that for AP we distinguish between Types Petrovic & Keym, and this has nothing to do with the terms "Type Keym"="ad libitum"/"Type Offner"="a priori" in PRA. Werner Keym has two types, is all. If someone has an authoritative specification, then I would be grateful. This is sufficient to clear up the island which is A Posteriori to some extent. (2022-02-15)
A.Buchanan: I have adopted VL's suggestion for the stipulation. Apart from anything else, in set play, Black would *lose* a move, while in a retropat situation (like Codex Article 15, and here) Black *gains* a move. (2022-02-15)
A.Buchanan: Or see P1067403. Maybe if this problem is PRA, then it has "one solution, two parts". There is some beautiful chess underneath these AP stipulations, but voluntary failure to define terms clearly renders them increasingly unintelligible. What does "PRA-AP" really *mean*? (2022-02-15, hidden) more ...
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VL: Solution:
1.Rg2=wR Rb6=wR!! 2.Re8=bR! 0-0#! (2... Rf1#??)
2.Ra(c)8=wR??

Here only one w rook is volage since they both stand on light squares
of ODD ranks: any promotee had to visit a square of the opposite
colour at least once in order to arrive from the 8th rank! Similarly,
only one bl rook is volage (EVEN ranks). By the AP logic, we have a
right to justify the 1st W's move by subsequent castling, which shows
that it is Rh1 that is volage. Bl rooks play by the post factum
(subordination) logic: the first one turns out volage.

Cooked (feenschach, 126): 1.Ra2(Rc2,Rb3...)=wR Rh7+ 2.Kf8 Ra8#
Correction: Rh8- e8. (2002-04-05)
VL: Correction (H.130): Rh8 to d8 (rather than to the light square e8). (2021-02-12)
A.Buchanan: shifted sT to d8 as per request (2021-02-12)
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Henrik Juel: 1.gxf3ep 2.Qg4 3.000 4.Re8 5.Kd8 6.Qxh5 000#. Both castlings are necessary to legitimize the en passant capture. (2003-09-29)
A.Buchanan: I don't think this is consequent series mover, under which every position would considered afresh from the perspective of legality i.e. possible histories. Instead, this is regular series mover where one player is simply saying "pass". It's hard to imagine how AP might operate in a consequent context. I will change the keyword suitably. (2021-11-08)
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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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No. 11669 HN
paul: Cooked: 1.K×e4 Tg1 2.Ke3 Tg5 3.Sd4 T×e5# (2011-11-25)
A.Buchanan: This nice simple puzzle can be repaired by replacing pawn c5 with rook (C+). As often with cooks in PDB, one wonders if it's just a typo. I have an old email address of Ronald's, and I may ask him. (2012-02-20)
A.Buchanan: I met Ronald for dinner last year (it seems like a vanished age!) in Norwich, UK. He was there for a mountaineer's conference. The climbers selected Norwich as their venue for humorous reasons as it's in the flattest area of Britain. We discussed a bunch of compositions, including this one. He agreed it was cooked and agreed with my suggested fix, which I will post in a separate entry. Very nice chap - he doesn't compose much these days because he can get a bigger audience (and more money) from writing hill-walking books. My hill-walking friends say he's a good writer. (2020-12-08)
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VL: Solution. 1.0-0-0#/Bh2#?? - B. is on move.

0...g*h3 e.p. 1.0-0-0#! (1.Bh2#/Sf3#?? - illegal):
W. forces B. to capture e.p. and legalizes this
possibility a posteriori.

W.Ps took 11: b*c*d*e*f*g, c*d*e*f, e*f*g and g*h.
11+5=16, hence b.Pa7 was also captured among them.
Thus it took once: 1+1(h*g)+14=16. Ph3 took on g2
when w.P stood on h2. No last W's move could be a
capture: e5 and f4 are occupied by w.pieces. Qf4
prevents from Ph3-h4 and Kh2-g1 before that. b.Ps
are blocked from above. Therefore, in his last move,
W. could retro-release B. only in two ways: Ph2-h4
with Ph3*Sg2 before that, or Kd1-e1 with Kf1-g1
and Bf3-e2+ before that. Thus, W. may castle, and
castling AP-legalizes ep (2004-12-09)
A.Buchanan: Article 15 (not AP Type Keym) drives this elegant problem, with usual AP Type Petrovic for ep/castling. (2022-02-16)
A.Buchanan: Have added genre as n#, even though n=1 here, because it's important that this is a #1 rather than a h#0.5, as they behave differently under Article 15. (2022-02-17)
VL: Nachdruck: Die Schwalbe, Dec. 2007, h.228.
Forced e.p. as the key. Who is on move? (2009-01-14, hidden)
A.Buchanan: Article 15 (not AP Type Keym) drives this elegant problem, with usual AP Type Öffner for ep/castling. (2022-02-15, hidden)
A.Buchanan: As is usual, AP Type Öffner sits happily with Retro Strategy, which is all about shrinking the range of possible histories by playing moves, and was the default retro protocol when AP was conceived. It should be happy with PRA as well, but until someone has presented a coherent algorithm for PRA-AP, I will mark these as RS. (2022-02-15, hidden)
A.Buchanan: Article 15 (not AP Type Keym) drives this elegant problem, with usual AP Type Petrovic for ep/castling.
As is usual, AP Type Petrovic sits happily with Retro Strategy, which is all about shrinking the range of possible histories by playing moves, and was the default retro protocol when AP was conceived. It should be happy with PRA as well, but until someone has presented a coherent algorithm for PRA-AP, I will mark these as RS.
Stipulations do not distinguish between the two AP types, but the PDB keyword parameter should indicate. (2022-02-15, hidden) more ...
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A.Buchanan: Suppose White: dxexf,cxdxe,bxc=,a| Black: fxg. This means there is one capture by white unaccounted for. So can’t White have just played e.g. QxS? What am I missing? (2022-03-22)
Mario Richter: Regarding the "can’t White have just played e.g. QxS?-question", the following considerations may be helpful:
1. black Pawn f3 x Yg2 can only be retracted after wPf2 has returned home.
2. If at this moment the black K is still on g1, Y cannot be a white rook.
3. Assuming that W can still castle, wRg5 must be a promoted Piece, since wKe1+wPf2+wPg3+wPh2 form a cage from which the originalwRh1 could not escape ...

I think the same considerations can also be useful by answering the question about +sBc6 ... (2022-03-22)
A.Buchanan: Hi Mario, thanks I agree (2022-03-22)
A.Buchanan: How about sBc6 to fix it? (2022-03-22, hidden)
A.Buchanan: How about +sBc6 to fix it? (2022-03-22, hidden) more ...
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If we accept that this kind of AP plays with who has the move, then maybe we say that it has one solution two parts?
A.Buchanan: This is a really cool problem, despite being ruined by the failure to define AP-PRA. I think I despise poor definitions as much as Henrik hates duals. (2022-02-15, hidden) more ...
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VL: Solution: 1... Ke2# (1... 0-0#?? illegal).
1.cxd3 Sb3# (1... Ke2??, 0-0+?).
1.cxb3 ep! 0-0#! (1... Ke2#??, Sxb3#?? AP after N.Petrovic).
If castling is legal, then B. is on move and the last move
was b2-b4 (with b3x(B)c2 before that). Different mates. (2006-01-27)
A.Buchanan: If Wh 00 is ok, then bPh & hence bPa were waylaid. R: 1. Sa3-b1? Sb1-d2+ retropat, so R: 1. b2-b4 b3xBc2. (2022-05-24)
A.Buchanan: I am outwitted by this one. Black bxc, while 9 Wh pawn captures are clear: hxgxfxexdxc7 and e.g. exdxc5, dxcxb5. However can also have two more captures bxaxb, capturing the waylaid bPa. So Black might just have played bxc2, without retracting b2-b4. What am I missing? (2022-05-23, hidden)
A.Buchanan: Or even if we don't use bPa as pawn food, there is still e.g. R: 1. Sa3-b1 Sb1xDd2+ 2. Dd2+ etc (2022-05-24, hidden) more ...
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im Kongreßbuch St. Petersburg 1998 nur Nachdruck im Kapitel "Aus dem Schaffen der Kongreßteilnehmer"

Das Originaldiagramm in 'Zadachi i etyudy' ebenso wie der Nachdruck im Kongreßbuch St. Petersburg 1998 haben einen sTd8, aber der begleitende Lösungstext erwähnt im Zusammenhang mit der Frage, ob zuletzt axb3,cxb3 oder cxb6 möglich war, daß auf dem Brett folgende schwarze Steine stehen: eine Dame, zwei Türme, zwei Springer, der schwarzfeldrige Läufer und 8 Bauern.
Also offensichtlich Diagrammfehler, und sSd8 ist korrekt.
Ladislav Packa: Why NL? Castling is a posteriori proof for e.p.! (2018-08-28)
A.Buchanan: Why not a3xSb4 or c3xSb4 as last move? Where is the promotion? (2018-08-28)
VL: Yes, it definitely looks as AP (with thematic illegal castling avoiding tries). I suspect missing two b knights somewhere: "superfluous" similarly to the b rooks...
wBa5 proves to be a promotee. Bl to move of course. (2018-08-29)
A.Buchanan: I agree with VL. WinChloe adds sSh1 & sTd8 (the latter is illegal with 8sB+2sT on the board already). If we add sSh1d8, then the problem is sound forwardly and retroly. The missing light sL can't just have been captured. (2022-05-24)
Mario Richter: Adding black Sh1+Sd8 is correct: obviously the original diagram is misprinted, the accompanying solution text speaks of: one black Queen, two black Rooks, TWO black Knights, one black dark-squared Bishop and 8 black Pawns. (2022-05-24)
A.Buchanan: Issue with animation (2022-05-25)
paul: St Petersburg Congress was in 1998. (2018-08-28, hidden)
Henrik Juel: Is the stipulation h#2.5 or h#3.0? (2018-08-28, hidden) more ...
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VL: Keywords: Whose move? (Wer ist am Zug?); Forced en passant

(a) 1.Lf3?? gxf3 2.0-0-0#? (?? denotes illegal).
0... gxf3 e.p.(forced!) 1.Lxf3 Lf5 2.Txg2#! (2.0-0-0#??)
Bl has the move, e.p. capture is legal, and castling is illegal.

(b) 1.Lf3?? gxf3 2.0-0-0#?
0... gxf3 e.p.(forced) 1.Lxf3 Lf5 2.0-0-0#! (2.Txg2#?? - AP-illegal).
Bl still has the move (due to bPe7). Unlike (a), two different retro-moves are possible: i) f2-f4 (f3xSg2) or ii) b2xc3 (in which case castling is illegal due to the missing dark-squared w Bishop). AP after Petrovic in the reversed form "a la Abdurahmanovic": W forces Bl to capture e.p.

(c) 1.Lf3? gxf3 2.0-0-0#??
0... gxf3 e.p.(!) 1.Lxf3 Lf5 2.0-0-0#!! (2.Txg2#?? - AP-illegal).
W's turn to move is possible (in which case, however, castling is illegal). Executed castling justifies jointly Bl's turn to move (AP after Keym) and e.p.

The twins differ by the role of castling: it is illegal in (a), is legal and legalizes Bl's e.p.-key in (b) and legalizes both Bl's turn to move and e.p. in (c). Separately they have 1-mover predecessors: resp., P0005627, P1066685 and P1068112. (2009-06-22)
A.Buchanan: Very nice problem! Ke2# is a dual not provided separately, but I don't see a way to dispense with it (2023-07-22)
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Henrik Juel: 1...d5*c6ep[+bPc7] 2.Ra8*a4 Qa2*a4[+bRa8] 3.0-0-0 Qa4-a8#
1...Kb5-c6 2.Ra8*a4 Qa2*a4[+bRa8] 3.0-0-0 Qa4-a8# (2022-01-05)
A.Buchanan: I think that the solution with e.p. is 1. ... dxc6ep[+sBc7] 2. 0-0-0 c5 3. Tg8 Dxg8[+bTa8]# The one that you give Henrik doesn't work in Circe, because we don't know whether sTa8 had moved there prior to the diagram position (2022-01-07)
Henrik Juel: C+ Popeye 4.61 (2022-01-05, hidden) more ...
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Mario Richter: How is the ep-key justified here? (Perhaps some kind of AP?) (2010-06-15)
Gerd Wilts: The author's reasoning is 1. Sxc5? 0-0-0!, but if Black can castle, then White can capture en passant: 1. dxc6ep. But this if of course not correct according to the Codex. (2010-06-15)
A.Buchanan: The only problem with having this as PRA is that the part where ep & 000 are both off is cooked. That could be fixed by e.g. adding bPf6 & wPg7. But the composer is saying that this problem already works as AP, if we allow the try 1. Sxc5? (or equally 1. Se5?) to *prove* that 000 is ok, which implies that ep is ok too. (2022-05-24)
A.Buchanan: Why is this not regular PRA? "ep off & 000 on" are inconsistent. "ep on & 000 off" is against regular conventions. So have "ep/000 both on" or "ep/000 both off". In the former case have 1. dxc6ep! and in the latter 1. Sxc5! What am I missing? (2022-05-24, hidden)
A.Buchanan: Or is the composer saying: "Yes this *could* be just PRA, but I am looking for a different explanation."? (2022-05-24, hidden) more ...
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b) Urdruck in Die Schwalbe 228, 12/2007
A.Buchanan: Any ideas about (b)? (2022-06-09)
Henrik Juel: Andrew, you can find Heft 228 on
https://www.dieschwalbe.de/archiv.htm
part b) is mentioned in an article by Valery Livkovets, p.299-304
diagram p.301, solution p.303 (2022-06-09)
A.Buchanan: Thanks Henrik - just the pointer I needed. Please check the detailed solution I'll post above. (2022-06-09)
A.Buchanan: OK the logic nearly works for me. But I don't understand why White can't just play 1. Txb2# in part b, because b2 is now covered by wD. What am I missing? (2022-06-09)
Henrik Juel: Indeed, 1.Txb2# looks like a cook (2022-06-09)
A.Buchanan: Werner agrees. He has fixed it with a better twinning condition (2022-06-12)
A.Buchanan: I guess (b) is AP Type Keym but I don’t see it clearly (2022-06-09, hidden) more ...
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A.Buchanan: Twin b was marked as PRA, but I think it's AP. There is just one solution: ep justified by later castling. (2022-05-27)
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Anton Baumann: cook: 2 Lösungen in 9 Zügen: 1.a3 sowie 1.b3
1.a3 2.axb2 3.b1=S 4.b3 5.Kc3 6.Kb2 7.Ka3 8.Ka4 9.Sa3 axb3# (mit ZU) (2020-05-11)
A.Buchanan: How about R: 1. b3xc4? White has made 10 pawn captures which seems doable (2023-04-06, hidden) more ...
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Chessically corrects P1012059 following discussions. It's doubtful that AP logic works as simplistically described here, so this may still be unsound logically. Nevertheless, might as well fix the chess!
See P1399112, where the two parts each prove that one side of White castling rights remain. If this is correct, and only *one* part is required per castling, this implies that this Turnbull-Buchanan problem doesn’t work. For one part we have solution c_S and tries c_T1 & c_T2. In the other part, solution e_S and tries e_T1 & e_T2. Any of 5 combinations work:
c_S + e_S
c_S + e_T1
c_S + e_T2
c_T1 + e_S
c_T2 + e_S
because they all confirm that White retains castling rights.
Classify this problem as Golden Age because it’s on the wrong side of the evolving standard.
A.Buchanan: Let ??? Indicate respectively that w00, c ep & e ep are ok. Then YYN, YNY, NYN, NNY, NNN are the contenders.
Under PRA we reduce to YYN, YNY & NNN but the last has no solution so not the right paradigm.
Under SPRA with AP we reduce to YYN, YNY. Each considered as a separate part has kind of AP h#3, but I don’t understand how consolidation would take place over PRA anyway
Under RS with AP, we have 2 solutions assuming again that no consolidation. (2022-03-23)
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This is an important schema for AP. Previous cooked problems P0002476 & P0004295 (together with versions of each lacking sBe3) had multiple chess cooks that do not involve e.p. as key. This chessically sound composition allows us to focus on the fundamental theoretical issue of AP.

The double e.p. idea cannot work as an AP retro strategy problem, because one can never prove that one double pawn hop rather than the other occurred. PRA+AP seems potentially feasible, but I see no clear statement of how the normal PRA method should be extended to this case. All prescriptive suggestions are welcome.

In WinChloe, I found a correction from 2002 by Valery Liskovitz, just posted in PDB as P1401449. However it used three extra pawns, rather than 2, so the current version still has merit, but I will add Valery as co-corrector.

See also P1399178.
A.Buchanan: Marking this pattern as Type Petrovic - ccee (2022-02-17, hidden) more ...
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A.Buchanan: Found in WinChloe. It can be fixed by e.g. +sSa3. Is this a transcription error? What does feenschach reprint show? (2022-05-24, hidden) more ...
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Henrik Juel: a) 1.axb3ep 2.b2 3.Kb3 4.Kc2 5.Kc1 6.c2 0-0#, not 6... Kf2 because the ep capture must be legitimized
b) 1.b2 2.Kb3 3.Kc2 4.Kc1 5.c2 Kf2#, not 5... 0-0 because Ke1 or Th1 has moved (2022-05-26, hidden)
Henrik Juel: HC+ Popeye 4.61 and analysis (2022-05-26, hidden) more ...
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Henrik Juel: 1...h5*g6 ep. 2.0-0 Kh4-g5 3.Rf8*f6 Kg5*f6 =
The castling serves two purposes:
enabling the stalemate and legitimizing the ep capture (2022-05-26)
Henrik Juel: HC+ Popeye 4.61 (2022-05-26, hidden) more ...
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Henrik Juel: a) C+ Popeye 4.61
1.fxg6 Txd5 2.Le5 0-0-0 3.g7 Th8 4.gxh8=DT#
b) no solution (2022-05-27)
Henrik Juel: I shall leave the analysis to you, Andrew... (2022-05-27)
A.Buchanan: OK, Henrik: please check my suggested solutions & rationale :) I find this a beautiful problem (2022-05-28)
A.Buchanan: In (b), is it possible that Black has just played RxQBh5? If this is feasible, it would apply even if Black 00 rights are still on. Cook? (2022-05-28)
Mario Richter: Looks indeed like a cook.

One can manually retract e.g. R: Th4xLh5 and use popeye to check if in the resulting position the reverse of this retraction is legal:
begin
remark Position after R: Th4xLh5
pieces white Bb8 Pd5 Pf5 Bh5 Ph3 Pd2 Ka1
black Ra8 Ke8 Pb7 Pd7 Ph7 Pa6 Pg5 Rh4
cond Maximummer
option HalfDuplex
stip h~0.5
end

This gives:
1...Rh4*h5
1...Ke8-e7
1...Ke8-d8
1...Ke8-f8

solution finished. Time = 0.013 s (2022-05-29)
A.Buchanan: Thanks Mario. I showed this problem on the international zoom call last night, and Siegfried spotted the cook immediately. Can g4 and h4/h6 be plugged suitably with pawns? Can't just shift bPh6 to h5, because it means that in a), bPg6 might have just captured checking Q/B, while not disrupting Black 000 rights. But I think can shift wPh3 to h4, together with adding Pg4 (Black or White - I prefer White aesthetically). Does this work? (2022-05-29)
Henrik Juel: Popeye 4.61 finds no solutions (in both twins) (2022-05-27, hidden)
A.Buchanan: Sorry my typo pls try again (2022-05-27, hidden) more ...
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Henrik Juel: White pawns captured all seven missing black men (so Black promoted [Ph7] on h1)
The only way to preserve the black castling right is with a retroplay starting with R: 1.c2-c4 c3xDb2
So if Black may castle, he may also capture ep now
1.dxc3ep Lb5 does not work, because Black must castle to legitimize the ep capture
1.dxc3ep 2.0-0-0 3.Kc7 4.Ta8 5.Kd8 6.Ke8 Lb5#
Nice 'uncastling' (2022-05-27)
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Henrik Juel: 1.fxg3ep 2.0-0 3.Txf5 4.Kf8 5.Ke8 6.Tf8 7.Th8 Lh5#
Another 'uncastling' (2022-05-27)
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Henrik Juel: Not 1...c5*b6 ep. 2.Ra8-a7 b6*a7 3.Ke8-d8 a7*b8=Q
but 1...c5*b6 ep. 2.Sb8*a6 Bf8-d6 3.0-0-0 b6-b7#,
where the ep capture is legitimized by the castling (2022-05-27, hidden) more ...
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Corrects P1401496
Henrik Juel: analysis
Black captured [Pa2] with an officer and cxb, dxe, exf, and f3xg2
White captured g4xLh5 and h2xDg3, so if he may castle, Black may capture ep now (2022-05-30)
A.Buchanan: A detail I hadn't noticed: bPh never captured, so wPh5 must have arrived on h-file *above* the light square h3. If it was gxh4, then there was no light square to capture light bB, so it was gxh5. But the capture might have been h3xBg4xQh5. The problem is still sound: the last move can't have been h4-h5. (2022-05-30)
Henrik Juel: HC+ Popeye 4.61 and analysis (2022-05-30, hidden)
Henrik Juel: set play and solution
1...Ke1-d2 2.e4-e3 + f2*e3 #
1.b4*c3 ep. b2-b3 2.Kd4*d3 0-0-0 #
The castling is needed both for the mate and for the legitimization (2022-05-30, hidden)
A.Buchanan: A detail I hadn't noticed: bPh never captured, so wPh5 must have arrived on h-file *above* the light square h3. If it was gxh4, then there was no light square to capture light bB, so it was gxh5. But the capture might have been h3xBg4xQh5. The problem is still sound: the last move can't have been h4-h5.d (2022-05-30, hidden) more ...
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Economizes P1080354 by 2 units, adding a solution.
Henrik Juel: C+ Popeye 4.61 (2022-05-30, hidden)
Henrik Juel: solutions
1.Ra3-b3 Ba6*c4 2.Rb3*b4 Ra1*a2#
1.Ka4*b4 Ra1-b1+ 2.Kb4-a4 Ba6-b5#
1.c4*b3ep. 0-0-0 2.e5-e4 Rd1-d4# (not 1... Td1, of course) (2022-05-30, hidden) more ...
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corrects P1108454
Henrik Juel: here is my guess
a) 0... bxa3ep 1.Dc3#
b) 1.0-0# (2022-12-05)
A.Buchanan: Comments on the proposed solution welcome (2023-08-21)
A.Buchanan: Please can someone correct the solution by replacing … with three periods. Autocorrect from phone always replaces these with an ellipsis. (2023-02-08, hidden) more ...
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Corrects P0000615.
A.Buchanan: corrects P0000615 (2023-05-20, hidden) more ...
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