Guus Rol: This is an incorrect interpretation of the AP-convention. Rules outrank goals in the definition of all GAMES. Therefore the legitimacy of a move cannot be restricted by the desire to achieve the goal (in this case: Remis). The proper way to view AP is that executing e.p. invalidates the legitimacy of all lines of future play that do not contain 0-0-0! In that sense black and white are forced to cooperate. In whatever freedom remains they can compete for the prize promised in the stipulation. By the way, this understanding of AP is not only more logical, it is also much more interesting as a playing field for AP-composition. (2005-09-21)
mri: citeWeiß kann aber so spielen, daß Schwarz nicht rochieren kann, z.B. 1. ... Lxc7 2. bxc7
/cite
How does white prevent black from castling after 1. cxb6 e.p. Ba7xb6+?
E.g. 2.Kxb6 a1=R 3.a7 Rxa7, or 2.Ka4 Bd8xc7 3.a7 Bb8. (2005-09-22)
VL: 1.c5xb6ep(??) a7xb6+ 2.Kxb6 a1R 3.Kb7 R1xa6 4.Rc8! (R6a7#?? illegal).
This study is correct under the generally accepted understanding of AP
a la' N.Petrovic'. Antiform (looking possibly somewhat strange):
Black's unsuccessful try. (2005-10-03)
Guus Rol: Generally accepted, true. Generally acceptable, false. Freedom of interpretation ceases for a concept once its polar concept (a priori validation) has been defined. "Goal induced AP" however, might be a passable stipulation for this type of problem. To keep this place from turning into a discussion forum I will discontinue comments on this issue. (2005-10-03)
paul: Author intention is: If black still can castle, his last move must have been b7-b5. However, to prove this, he has to castle (A Posteriori condition). So: 1.cxb6 e.p. axb6+ 2.Kxb6 a1R! (2...a1Q? 3.Kb7 Qxa6 mate obviously doesn`t prove b7-b5 as the last move) 3.Kb7 R1xa6 4.Rc8! and white has prevented black from castling. So black can`t prove the last move is b7-b5 and therefor is already stalemated in the initial position! (2011-07-12)
A.Buchanan: Isn't Guus' idea just AP-Prioritat? And even if it were "more logical" or "more interesting", it doesn't follow that other forms is AP are "incorrect" (2023-07-31)
A.Buchanan: Isn't Guus' idea just AP-Prioritat? And even if it is "more logical" or "more interesting", it doesn't follow that previous interpretations are "incorrect". Once a "polar concept" has been defined, prior art remains, and is valid in its own context. "Freedom of interpretation" remains, although newer compositions may adopt the newer paradigm. If a new idea does succeed in displacing the old, then problems (old and new) which depend on the earlier interpretation should be preserved with the context in which they existed. I don't know whether the
"Timeline Gallery" will ever physically exist, but I hope that it will. This is basically the "Golden Age" argument, that says if we don't value & preserve the old, we cannot consistently move on to the new (2023-07-30, hidden)
A.Buchanan: Isn't Guus' idea just AP-Prioritat? And even if it were "more logical" or "more interesting", it doesn't follow that previous interpretations are "incorrect". Even if a "polar concept" has been defined, prior art remains, and is valid in the context in which it was formed. "Freedom of interpretation" remains, although newer compositions may preferentially follow the newer paradigm, in part because fresh design space is always desirable.
If a new idea does succeed in displacing the old, then problems (old and new) which depend on the earlier interpretation should be preserved in a "Timeline Gallery", each with the context in which it was created.
It is also a fallacy to say that a stipulation should "explain" the philosophy. Such problems have been defined as AP, and that's fine. I think AP-Prioritat (sorry for absence of umlaut) *does* deserve to be mentioned as part of the stip, but whether the problem is Keym or Petrovic or both or something else - or PRA or RS - all that is part of the solving fun: keywords can record the taxonomic details & they don't need to be part of the stip. (2023-07-30, hidden) more ...
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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: 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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Nachdruck in "Die Schwalbe" 33 mit Diagrammfehler (wBe6 fehlt).
VL: The first AP-problem! Thematic illegal try:
1. ... exd6ep 2.Kd8? dxe7+ 3.Kc8 e8=T(D)#?? (2007-02-11)
Mario Richter: 46 years between this Problem and P1366663 - perhaps this was not Keeble's first AP-Problem?! (2019-08-29)
Ladislav Packa: Of course I know the idea a posteriori. But let my heresy be forgiven - I do not consider it logical. In this particular example, I consider e.p. for proof that black can castling - not vice versa. (2019-08-29)
A.Buchanan: Hi Ladislav. If ep is ok, then Black just moved a pawn with double hop. That doesn't stop Black having e.g. moved bK right at the beginning of the game. So you can't prove that Black definitely can castle this way. What am I missing here? (2019-08-29)
Ladislav Packa: Your statement is perfectly fine. However, the AP condition is usually interpreted as castling is evidence of e.p. However, the second party does not have to do castling - the possibility has already been demonstrated by the existence of e.p. (2019-08-29)
A.Buchanan: Hi again Ladislav. Please try again, I am obviously being very stupid. I don't understand how 1. exd6ep is any more indicative that Black can castle than 1. K~. (2019-08-30)
Ladislav Packa: Hi, Andre!
e.p. is just a result of the d7-d5 move. Black can castle only if it was the last move of black (even without wPe5). Therefore, it does not have the logic of joining the e.p. with castling. For example, ask yourself if black is allowed to castle in the position without wPe5. (2019-08-30)
A.Buchanan: @Ladislav: ep is not "proof that black can castle", rather it is a *necessary* condition for Black to be able to castle. The logical implication is the other way round. We can castle because of the optimistic convention, and because sPd might just have moved. Without wPe, there is no solution, but whatever White plays, Black can castle (2020-12-11)
Hans-Jürgen Manthey: was heißt hier Thematisch Illegal ??
Mir völlig wurscht ob ep möglich oder nicht, nach dem weißen Zug ist natürlich Kd8 möglich !!
also ist auch 1. ... exd6ep 2. Kd8 dxe7+ 3. Kc8 e8DT# eine Lösung, sonst müßte auf die erzwungende Rochade in der Aufgabenstellung hingewiesen werden ! (2020-12-11)
A.Buchanan: Can one tolerate “tolerated dual promotion” in Valladao? Well Valladao himself did, so it must be ok (2022-03-26)
Erich Bartel: Nachdrucke:
1) 3001 CHM Themes XII 2000.---
2) Die Schwalbe (215) X 2005.---
3) A1 Problemkiste (169) II 2007.-- (2007-02-23, hidden)
A.Buchanan: Hi Ladislav, I still don't understand you. Heresies must be explained extraordinarily carefully :-)
Following the conventions:
(1) Black retains castling rights because we cannot prove that bK or bR ever moved.
(2) If e.p. is now played, then we still require the same castling convention, as bK or bR might have moved earlier in the game.
(3) So the actual execution of the e.p. does not prove anything new about castling.
(4) Hence it is incorrect to say "e.p. [is] proof that black can [castle]".
Which step do you disagree with?
Yes by the convention, Black could castle in the position without wPe5. What does that prove? (2020-10-03, hidden)
A.Buchanan: OK, apart from the confusion of "A if B" with "A only if B" (which means "B if A") Ladislav said something that I overlooked: "However, the second party does not have to do castling - the possibility has already been demonstrated by the existence of e.p."
Castling & e.p. are handled very differently. Castling is considered to be "likely" by default, and so you do get to "prove it" just by playing it. However en passant is considered to be "unlikely" by default, and so does not get to be played. If you want to just do any possibly-legal e.p., try SPRA condition, but in AP the playing of ep is not sufficient to prove it. It's not a matter of "heresy", it's how the rules work. Define your own rules if you want. (2022-02-15, hidden)
A.Buchanan: Can one tolerate “tolerated dual promotion” in Valladao? I don’t think so (2022-03-25, hidden) more ...
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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: Popeye v4.87 finds 11 solutions to h004.5 following the mandatory ep. Only one of them can be followed by h#1. Any cook therefore would have to be of the form 6. ... 000# Searching now for h005.5, as can then easily check if there are any mates (2022-06-10)
Yuri Bilokin: H#1* *) 1. ... Sxb3#
1) 1. gxh3ep Sxb3# (2022-06-10)
A.Buchanan: Hi Yuri - you know that h#1 doesn't work, right? (2022-06-10)
Yuri Bilokin: Hi - don't know, please email polidox579@gmail.com (2022-06-11)
A.Buchanan: Hi there are over 110,000 ways to reach 6. 000, but *none* of them are also checkmate. Therefore I am happy to pronounce this problem sound. Happy, because (1) it deserves to be ok (2) I couldn't face trying to fix it. If White 000 rights remain, then the last move was certainly h2-h4. Orthodoxically, we cannot perform the e.p. but under the AP Type Petrovic we can ep, as long as we justify it with castling later in the game. Hurray! (2022-06-12)
A.Buchanan: 1. gxh3ep Sxb3#? fails because White never castles to justify the ep. It's like there is an additional check before granting that a position is checkmate or stalemate: have all AP debts been paid? If not, the move is illegal because the game would end with no chance to repay the AP debt later. Yes it's weird but that's AP. (2022-06-13)
Michel Caillaud: Now, the post en passant part of this kind of problem can be tested with Jacobi.
With the following set of data:
stip h#5.5 pieces
White Ke1 Qg1 Ra1f2 Be3h5 Sd2 Pa3b4c4d4f4g3g5
Black Kc1 Rb2 Bb1 Pa2b3c2d3g2h3
constraints Ke1!c1~ Ra1!d1~
Jacobi looks for helpmates in 5.5 moves including 0-0-0 in the play.
No need to scrutinize the 110000 ways found by Andrew; Jacobi makes all the work (but it takes some hours...) (2022-06-13)
A.Buchanan: Wh can't retract c3-c4 or c3xd4 because no prior Black retraction. Bl can't retract c3-c2 because prevents cross-capturing. Can't retract c3xb4 because the files need to be sorted out in pairs. Does the problem equally work if wPa3 is placed on a4? (2022-05-24, hidden)
A.Buchanan: 1. gxh3ep Sxb3#? fails because White never performs the c000 necessary to justify the ep (2022-06-12, 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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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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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: 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: 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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