Die Schwalbe

89 problem(s) found in 3501 milliseconds (displaying 89 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT K='Springerrad' AND K='a posteriori (AP)'] [download as LaTeX]

1 - P0000036
Valery Liskovets
Nikita M. Plaksin

5528 Die Schwalbe 99 06/1986
P0000036
(10+11)
h#1.5 (AP, RV)
1) 1. ... cxd6ep 2. 0-0-0 Dxc7#
Nicht 2. Kf8 Dg8? da der ep-Schlag AP bewiesen werden muss.
2) 1. ... gxf6ep 2. 0-0-0 Tg8#
Nicht 2. Kd8 Tg8#? da der ep-Schlag AP bewiesen werden muss.
play all play one stop play next play all
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)
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Keywords: a posteriori (AP), En passant as key (2), Partial Retro Analysis (PRA), Castling (sg), Volet Pawn
Genre: h#, Retro
FEN: r3k2n/1pp4P/N6p/2PpKpP1/2Q1p1R1/4P1pB/3P1p2/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
2 - P0000250
Nikita M. Plaksin
Valery Liskovets

7577v Die Schwalbe 132 12/1991
P0000250
(5+14)
ser-h#5 (AP)
Typ Tauber/Caillaud
1. Kd4 2. bxc3ep (zuletzt nur K,T-Zug oder c2-c4) 3. e5! (nicht 4. Tb4? weil zuletzt K,T-Zug und 0-0-0 nicht mehr möglich) 4. Tb4 (zuletzt e7xLd8=S möglich) 5. Tc4 & 1. 0-0-0# nicht 1. Td1#? wegen fehlender AP-Legalisierung
play all play one stop play next play all
"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)
comment
Keywords: Castling (wl), a posteriori (AP), Promotion (S), Seriesmover, Consequent, En passant
Genre: Retro, Fairies
FEN: 1N6/1ppppppn/n7/1Pq1k3/1pP1r3/4p3/1r6/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-29 more...
3 - P0000583
Andrey Frolkin
Günter Lauinger

3468v Die Schwalbe 72 12/1981
P0000583
(10+12) C+
h#2 (AP)
Satzspiel:
*1. ... 0-0! 2. cxd4 Tc1#

Verführung:
1. exd3ep? 0-0 2. cxb4 Tf4# aber zuletzt R: 1. d2-d4?? ist illegal, da es den wLc1 als schwarzes Bauernschlagobjekt ausschließt

Lösung:
1. ... 0-0! (AP-Legalierung des weissen Anzugrechtes) 2. cxd4 Tc1#
play all play one stop play next play all
Alle Versuche, das Satzspiel 1. ... 0-0! 2. cxd4 Tc1# durch einen schwarzen Vorschaltzug aufrecht zu erhalten, scheitern.
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)
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Keywords: Castling (wk), a posteriori (AP) (Type Keym), En passant as key, a posteriori (AP) (Type Petrovic)
Genre: h#, Retro
Computer test: BC+ Popeye v4.87
FEN: 8/2p5/1pPp4/bRpP4/BPkPp3/qp2p2p/rP2P3/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
4 - P0000598
Nikita M. Plaksin
Andrej N. Kornilow

3876 Die Schwalbe 74 04/1982
P0000598
(4+3)
h=2 (AP)
Circe, Monochromes Schach
Intended solution:
1. dxc3ep[+wBc2] 0-0 2. axb1=S Kh2=
2. ... Kf2? 3. Kf3!
1. ... Tf1? 2. axb1=S Kf2? 3. Kf3!
play all play one stop play next play all
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)
comment
Keywords: a posteriori (AP) (Type Petrovic), Circe, En passant as key, Monochromatic Chess, Castling (wk), Miniature, Golden Age (Monochrome castling)
Genre: Retro, Fairies
FEN: 8/8/8/8/2Pp4/8/p3k3/1N2K2R
Reprints: (IV) Quartz 4 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-18 more...
5 - P0000615
Andrey Lobusov
4088 Die Schwalbe 77 10/1982
14.Lob
P0000615
(15+6) cooked
h#2* (AP)
1. ... Kg5 2. 0-0?? Se7#
1. ... Txh7 2. Kf8 Txh8#

1. cxb3ep Txh7 2. Kf8?? Txh8#
1. cxb3ep Kg5 2. 0-0! Se7#
play all play one stop play next play all
Wh captured 10 by pawns (including hxPg-g8) but not on the last move. wPa3 came from a2: wPa5 captured bPa.
Bl took only b3xa2.
In set play, Bl has no last move except for K or R.
For the main play, the intention is that if Bl retains castling rights, last moves were R: 1. b2-b4 b3xa2 e.g. 2. Bg1-d4 b4-b3 3. d4xc5 etc. However there are retro cooks e.g. R: 1. Bg1-d4 b3xa2 2. d4xc5 etc, with wPa5 the original wPb. wPb4 & wPb5 come from further East.
Cook: Retro logic for main play is unsound.
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)
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Keywords: Castling (sk), a posteriori (AP) (Type Petrovic), En passant as key, Tempo Move, Superseded by (P1409841)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/5p1p/5K1R/PPP2N2/1PpB4/P1P5/p1RNB1P1/Q7
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-20 more...
6 - P0000641
Werner Frangen
2201 Die Schwalbe 46 08/1977
P0000641
(14+5)
#4 (AP)
1. bxc6ep
play all play one stop play next play all
Nach 1. bxc6ep ist zwar ein Matt in 2 Zügen möglich, aber Weiß muss noch rochieren, um den ep-Schlag zu begründen.
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)
comment
Keywords: Castling (wk), En passant as key, a posteriori (AP)
Genre: Retro, n#
FEN: 1N6/PP1p4/BR6/pPp5/kp6/B2P4/1PP1P2P/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
7 - P0000649
André Hazebrouck
2261 Die Schwalbe 47 10/1977
4. Preis
P0000649
(8+14)
h#2.5
b) AP (2 Lösungen)
a) 1. ... 0-0-0 2. 0-0 Tdg1+ 3. Kh8 Txh6#
b) also 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#
is clearly the intention
play all play one stop play next play all
Can count 1+7 visible pawn captures, leaving 1+1 unexplained. Need to resolve wPfgh & bPg. Suppose bK never moved, then one of:
1) wPf waylaid, wPhxPg=, wPg=. Here w00 right can remain.
2) Or similarly, wPh waylaid, wPfxPg=, wPg=. Again w00 right can remain.
3) bPgxPh=, wPfxg= (or wPfxe then captured by bPf), wPg= w00 right lost.
1+1 captures explained in either case.
On the other hand if bK moved, then maybe wPf=, and only requires one more capture to resolve g&h files. The w00 right can remain. So there is a captured unit unaccounted for, and we can't validate the ep.
On the other hand if we are in case 3 above, then we can't know that wPbxa captured dark bishop, so White might retract a6-a7.
We also need to know that w00 rights remain. s000 rights (i.e. prior movement of sTa8) are not relevant.
If we know that w00, w000 & b000 remain, then we are in case 1 or 2 above with bPg captured unpromoted. Therefore wPb6xLa7 due to bishop shade. Therefore R: 1. b7-b5 b6xLa7 to unblock.

Only one candidate solution does not begin with ep: 1. ... 0-0-0 2. 0-0 Tdg1+ 3. Kh8 Txh6#. So this is the solution for a).

Let ????? denote validity of w00,w000,b00,b000,ep. Possibilities are: YYY?Y and all of YYN??, YNY??, NYY??, YNN??, NYN??, NNY??, NNN??. So there are 30 possibilities.
Under PRA, the solution parts would be YYYYY, YYNYN, YNYYN & NYYYN. For the first there are 20 solutions, while the second and third have 0 solutions. So this is not the right paradigm.
Under SPRA, there would be a single solution part YYYYY with 20 solutions. So this is not the right paradigm either.

Under RS with AP, the solution of a) still works. This comes from ?YY??. On the other hand, if the first move is ep, then we are in YYYYY. So all castlings are valid. But when we get to perform the mate, we need to know that based on castlings actually performed, the ep is valid. So the solution must include w00. So is 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#.

By combining the information of both of these solutions, we know we are in YYY?? so YYY?Y is the only possibility, and ep is legal. However, the combination of these two solutions would also validate any of the other 18 ep solutions that do not include 2. ... 0-0. They contribute no new evidence, but how to exclude them?

This is one of the issues with "AP Consolidation". Normally the number of solutions is not a constraint. But here perhaps we should insist that *only* two solutions are allowed?
Cook: 1. ... cxb6ep 2. La6 Tb1,~ 3. 0-0-0 a8=D#
18 different possibilities for W2 which are validated by the same logic that validates 2. ... 0-0.
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)
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Keywords: a posteriori (AP) (Type Petrovic), Castling (wkwlsk), En passant as key
Genre: h#, Retro
Computer test: Forward logic Popeye v4.87 & basic retro-logic works, but then AP Consolidated protocol seems to cook the problem.
FEN: r1b1k2r/P2p4/2p1n2p/ppPq4/1np5/p7/P2PP3/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
8 - P0000758
Gerd Rinder
1033 Die Schwalbe 21 06/1973
1. Preis
P0000758
(7+11)
Remis (AP)
Weiß ist patt. 1. cxb6ep ist nur zulässig, wenn Schwarz diese a posteriori durch die Rochade rechtfertigt. Weiß kann aber so spielen, daß Schwarz nicht rochieren kann, z.B. 1. ... Lxc7 2. bxc7
play all play one stop play next play all
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)
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Keywords: En passant as key, Castling (sg), a posteriori (AP)
Genre: Retro, Studies
FEN: r2bk3/p1Rpp3/P1p2p2/KpP2P2/1P2p3/1P6/p7/8
Reprints: (4) Problem 161-164 11/1973
317 Europe Echecs 217 01/1977
(A) Die Schwalbe 80 04/1983
408 Eigenartige Schachprobleme 2010
M36 mpk-Blätter 12/2011
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-31 more...
9 - P0000759
Luis Alberto Garaza
1034 Die Schwalbe 21 06/1973
P0000759
(9+9) cooked
Schwarz am Zug, Weiß gewinnt
1. ... fxg3ep 2. hxg3+ Kh5 3. f4 Kh6 4. h8=T+! Kg7
play all play one stop play next play all
Cook: Intended a Posteriori combo of castling and ep rights retro-locks wTh in cage, contradicting pawn capture balance
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)
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comment
Keywords: Castling (wl), a posteriori (AP) (Type Petrovic), En passant as key
Genre: Retro, Studies
FEN: 8/7P/6p1/2p3p1/2p2pPk/2Pp1P1p/3PpP1P/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-07 more...
10 - P0000760
André Hazebrouck
1120 Die Schwalbe 23 10/1973
P0000760
(9+10) C+
h#2 (AP)
1. bxc3ep Sa6 2. 0-0-0 Tc4#
play all play one stop play next play all
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)
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Keywords: En passant as key, Castling (sg), a posteriori (AP) (Type Petrovic), Volet Pawn
Genre: h#, Retro
Computer test: HC+ Popeye 4.87 + simple retro reflection
FEN: rN2k3/1p1pp1pp/8/p7/RpP5/PP6/p2PP3/KR6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-05 more...
11 - P0000793
Nikita M. Plaksin
Andrey Lobusov

1558 Die Schwalbe 33 06/1975
4. Preis
P0000793
(13+5) C+
#3 (AP)
1. cxd6ep+! Kxg3 2. 0-0! (proving ep ok)
2. ... Kg4 3. g8=D,T#
2. ... a1=D,L 3. Dg5#

R: 1. d7-d5 Tc6xh6 (unless R: Kg1-g2 Th~*h1)
play all play one stop play next play all
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)
more ...
comment
Keywords: Castling (wk), a posteriori (AP) (Type Petrovic), En passant as key, Promotion (ws), Valladao Task (w w w/s)
Genre: Retro, 3#
Computer test: Popeye v4.85 + minor retro/AP thought
FEN: B7/p5PP/p6R/Q1Pp4/8/PP1P2P1/p3P1k1/4K2R
Reprints: (B) Die Schwalbe 48 12/1976
feenschach 40 11-12/1977
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-11 more...
12 - P0001107
Michel Caillaud
3332 Die Schwalbe 66 12/1980
2. Preis
P0001107
(7+9) C+
h#4.5 (AP)
Circe
1. ... hxg6ep[+sBg7] 2. a1=T+ Txa1[+sTh8] 3. 0-0 Ta3 4. bxa3[+wTa1] 0-0-0 5. Td8 Txd8[+sTh8]#
play all play one stop play next play all
Wenn die beiden Könige noch nicht gezogen haben (was a.p. durch Ausführung beider Rochaden bewiesen wird!), kann der letzte Zug nur g7-g5 gewesen sein, Sonst ginge z.B. auch R: 1. Kd8-e8 c2xDb3
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)
comment
Keywords: a posteriori (AP), Circe, En passant as key, Castling (wgsk), Promotion in forward play, Valladao Task
Genre: Retro, Fairies, h#
Computer test: HC+ all solutions begin with ep, but only one includes both white & black castling
FEN: 4k3/8/7p/p1p3pP/Pp4Pp/RP4p1/pP6/4K3
Reprints: feenschach 61, p. 534, 08/1982
RA152 diagrammes 72 07-08/1985
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-28 more...
13 - P0001310
Dragan Petrovic
182 Europe Echecs 107 01/1968
4. Preis
P0001310
(13+14) C+
#2 (AP)
1. gxf6ep! droht 2. 0-0#! (2.Kf2#?)
R: 1. f7-f5 f6xDe7,f6xTe7 etc
play all play one stop play next play all
Wh caps: fxe7, hxg5
Bl caps: axbxc2, wBf1
If wK castling right remains, the only way to give White a prior move is to retract f7-f5, so ep would be on.
The key threatens (inter alia) 2. 0-0#/Kf2#, and both remain whatever Black plays, so the castling retro-justification for ep can never be disrupted.
R: 1. S~a3? a3-a4? blocks wT from retracting to home square if wK castling right remains.
Dragan Petrovic: Author is Dragan T. Petrovic (2007-12-02)
more ...
comment
Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic)
Genre: Retro, 2#
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 4r3/1p2P1n1/4p2B/b1pp1pPb/P5pp/nPP5/Q1pPP1P1/N1k1K2R
Reprints: (50) Problem 144-147 12/1971
(38) Mat Plus 12 10-12/2009
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
14 - 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...
15 - P0001453
Luis Alberto Garaza
327v Europe Echecs 225/226 10/1977
P0001453
(11+14) C+
h#1
1. hxg3ep! 0-0#! (Tf1#?)
play all play one stop play next play all
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)
more ...
comment
Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 5N2/3p1P1N/5nP1/1p1pq1Rp/5kPp/n1rQ3r/1Bp1P3/b2bK2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
16 - P0001941
Nenad Petrovic
2nd International Team Match 1967-1970
3. (2.?) Platz
P0001941
(10+12)
Gewinn
AP
Henrik Juel: If White may castle, last move was g7-g5 to avoid retrostalemate. 1.fxg6ep followed by 0-0 to legitimize the en passant capture. (2004-09-23)
A.Buchanan: In V&V Encyclopedia, which I admire more for its scope than for its precision, this problem is given to illustrate "Petrovic Theme". The definition given is: "PETROVIC THEME Also called 'retroproblem of Petrovic Type'. Mutual dependence of en passant capture and castling. By playing an en passant capture other retro elements of position are legalized (usually castling)."
The definition makes no reference to A Posteriori. I am trying to get my head around the text here, because the e.p. does not "legalize" castling, rather it mandates it. I also encountered Öffner for this castling/e.p. AP (but not in V&V), but that might be someone's confusion in that Types Öffner vs Keym exist in PRA. (2022-02-15)
comment
Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic), En passant
Genre: Retro, Studies
FEN: b7/p4P2/2kbPp1p/3ppPp1/n3pp2/8/P1PPP2P/4K2R
Reprints: (2) Problem 141-143 08/1971
(77) Problem 144-147 12/1971
(C) Die Schwalbe 16 10/1972
Encyclopedia of Chess Problems 2012
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-15 more...
17 - P0002353
Theodor Tauber
Michel Caillaud

The Problemist 1991
P0002353
(14+11) C+
ser-h#3 (AP)
1. dxe3ep 2. fxg1=L 3. 0-0 Tg4#
1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt
play all play one stop play next play all
Henrik Juel: If Black may castle, his latest move must have been e3xf2, so last move was e2-e4. (Orig. wPg2 promoted on g8, so g7xh6 happened early). 1.dxe3 e.p. 2.fxg1=B 3.0-0 Tg4# (2003-03-21)
James Malcom: Is this the only know one-sided Valladao in a regular series-mover helpmate? Also, can wPh2 be shifted to h4, and the White king moved as well to h2, so that the solution must have 2. fxg1=S? (2020-12-08)
James Malcom: Well, the only known one that is in minimum form, i.e. 3 moves, I mean. (2020-12-08)
A.Buchanan: This is nice. I don't know of any other AP ser-h#3 with Valladao, but there could be. Just been cleansing PDB data today. Yes I think wKh3 and wBh2 can be transposed, and then promote to S instead of L. (2020-12-08)
A.Buchanan: The legality of intermediate positions in seriesmovers is of course irrelevant. However, each proof game that leads to the diagram position with BTM can be appended with a black move and then a "flip" that changes the player to move. We can repeat this process, and then the fact that Black retains castling rights implies that they had castling rights in the diagram position. I think this is the right formal way to handle AP with seriesmovers. (2022-03-21)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sk), Seriesmover, Promotion, Valladao Task
Genre: Retro, Fairies
Computer test: C+ Popeye v4.85 + thinking
FEN: 4k2r/1pp2p1p/1N5p/3P4/2RpP2N/P2P1P1K/p1BP1p1P/B3b1R1
Reprints: Die Schwalbe 143 10/1993
Rex Multiplex 45-46 05/1994
(XII) Quartz 5 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
18 - P0002471
Gerd Rinder
(G) Die Schwalbe 48 12/1977
Lob
P0002471
(4+3)
#2 (AP)
BTM: 1. ... Lxb7+ 2. Ke3 0-0-0 3. Sb6#
WTM: 1. Sf6+! Kd8,Kf8 2. Dc7#,Ld6#
play all play one stop play next play all
If WTM, b000 rights are already lost. Under Keym AP, Black attempts to steal the move. White disruption of castling now counts as win for White, so the only chance is 1. … BxQb7+ 2. Ke3! thr 3. Sf6+ disrupting castling but 2. … 0-0-0 3. Sb6#. 2. Kd3,Kf5? Be4+ 3. ~ 0-0-0! as b7 is no longer occupied, or 2. Kd3,Kd4,Kd5? 0-0-0! pinning wS or 2. Ke5,Kf4? 0-0-0! as wL is blocked. As usual when flip of player to move in d#n, Black gets an extra move rather than White losing one (c.f. Codex Article 15). Sublime miniature!
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)
more ...
comment
Keywords: Castling (sg), a posteriori (AP) (Type Keym), Homebase (s), Aristocrat, Miniature
Genre: Retro, 2#
FEN: r1b1k3/1Q1N4/8/8/4K3/8/7B/8
Reprints: Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
19 - P0002476
Werner Kuntsche
1777v Problem 73-78 06/1961
P0002476
(5+7) cooked
h#3 AP
Einheitslösung!
1) 1. cxd3ep Ta5 2. Sf5 0-0 3. Sd4 Txe5#
2) 1. gxf3ep Th5 2. Sd5 0-0-0 3. Sf4 Txe5#
play all play one stop play next play all
Cook: 707 cooks which do not begin with e.p., e.g. 1.Sg3-e2 Ra1-a2 2.Ke4-d3 Rh1-h3 3.e5-e4 Ra2-d2 #
What is the obscure intent behind "Einheitslösung!"? (= "One size fits all solution"?)
Mario Richter: Im Nachdruck in problem 144-147 steht nur auf e5 ein sB (also keiner auf e3). Allerdings gibt es in allen drei Varianten (sBe5+e3, nur sBe5, nur sBe3) jede Menge NL, die ohne Rochade und e.p.-Key auskommen - am wenigsten dann, wenn nur auf e3 ein sB steht. (2010-10-10)
VL: Cf. P0002475. (2012-08-26)
A.Buchanan: Possibly "Einheitlösung!" is a request to unify retro implications derived across separate parts of the PRA solution. But cooked chessically irrespective of these philosophical issues, which still wait to be resolved 60 years on. (2022-02-16)
comment
Keywords: a posteriori (AP) (Type Petrovic - ccee), En passant as key (2), Castling (wb), Symmetrical position, Symmetrical solution, Superseded by (P1401449,P1399112)
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/8/8/4p3/2pPkPp1/2n1p1n1/8/R3K2R
Reprints: (66) Problem 144, p. 52, 12/1971
Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
20 - 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)
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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...
21 - P0003359
André Hazebrouck
3256 Themes-64 07-09/1977
P0003359
(14+10) C+
h#2
2.1...
1) 1. 0-0 Le6+ 2. Kh8 Sg6#
2) 1. bxc3ep 0-0 2. 0-0-0 Tc4#
play all play one stop play next play all
If all three castlings are legal, then the ep is on. So evidence must be accumulated from the two solutions. White 0-0 serves only in this retro role, eliminating 26 other candidate solutions.
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)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wksksg)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro logic
FEN: r3k2r/B2p2pp/8/NP6/RpP4N/pP5B/1p1PPPP1/n3K2R
Reprints: feenschach 54 04/1981
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-06 more...
22 - P0003411
Norman Alasdair Macleod
3970 Themes-64 04-06/1982
P0003411
(4+6) C+
h#2 (AP)
2.1...
1. Kc3 0-0-0 2. Txc4 Txd3#
1. bxc3ep e4 2. Kc4 Ta4#
play all play one stop play next play all
The idea is that the ep in one solution is validated by the castling in the other solution. Since no other solutions exist, there are no parasites which might "piggyback" off the proof given by the castling solution. This is not PRA: both solutions have the same history with both castling & hence ep legal.
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)
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comment
Keywords: a posteriori (AP) (Type Petrovic consol), En passant as key, Castling (wg)
Genre: h#, Retro
Computer test: HC+ Forward play proved by Popeye v4.87 AP logic requires some framework but I think we know we want this cool idea to end up being sound.
FEN: 8/8/8/1nr5/1pPk4/1p1p4/4P3/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-09-12 more...
23 - P0003417
John Frederick Keeble
2206 The Problemist Fairy Chess Supplement 16, p. 173, 02/1936
P0003417
(4+5) C+
h#2.5 (AP)
1. ... exd6ep 2. 0-0-0 dxe7+ 3. Tf8 exf8=T,D#
play all play one stop play next play all
H.Juel: If Black may castle, last move was not made by Ke8 or Ta8, but by Pd5, so it was d7-d5, and hence White may capture en passant. 1... exd6ep 2.0-0-0 dxe7 3.Tf8 exf8=D,T# where the castling (belatedly) justifies the e.p. capture.
Not 1... exd6ep 2.Kd8? dxe7 3.Kc8 e8=D,T
AB: Another 120 retro tries (=60x2 due to tolerated dual promotion) of general form 1. exd6ep 2. T~ dxc7 3. T~ c8=D,T.
No set play
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)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sg), Promotion (D/T), Tolerated dual promotion (D/T), Valladao Task, Retro Strategy (RS)
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: r3k3/2p1p3/2P1P3/2KpP3/8/8/8/8
Reprints: V. Die Schwalbe 33, p. 323, 06/1975
2001 CHM Themes 12/2000
3 Die Schwalbe 215, p. 240, 10/2005
A1 Problemkiste 169 02/2007
402 Eigenartige Schachprobleme , p. 130, 2010
14 Die Schwalbe 241, p. 374, 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-10 more...
24 - P0003423
Matti Arvo Myllyniemi
3975 Stella Polaris 01/1971
P0003423
(7+11)
h#3 (AP)
0.2.1...
1. ... g6 2. 0-0 gxf7 3. Kh8 Le5#
1. ... cxb6ep 2. 0-0-0 bxa7 3. Td7 a8=D#
play all play one stop play next play all
Henrik Juel: Analysis
If Black can castle both ways, last move must have been b7-b5
C+ Popeye 4.61 (2020-10-25)
comment
Keywords: a posteriori (AP), En passant as key, Castling (sksg), Promotion (D), Valladao Task
Genre: h#, Retro
FEN: r3k2r/p4p1p/8/KpP2PP1/4PBpP/5pp1/6p1/8
Reprints: Nordisches Turnier 1970-1971
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2020-10-25 more...
25 - P0003428
Imre Sztankovszky
130 Die Schwalbe 09-10/1953
P0003428
(10+11) C+
h#1.5 AP
1. ... fxg6ep 2. 0-0 gxh7#
play all play one stop play next play all
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)
more ...
comment
Keywords: En passant as key, Castling (sk), a posteriori (AP) (Type Petrovic), Volet Pawn
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 4k2r/3ppp1p/1p6/4BPpK/P7/pPPB2Pb/3P2Pp/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-27 more...
26 - P0003430
Tomislav Petrovic
(XIII) Problem 37-40 09/1956
P0003430
(4+15) cooked
h#3
1. dxe3ep Td1? 2. Le4 Txd6 3. Sg4 Tf6# usual AP try
1. dxe3ep 0-0-0! 2. Le4 Txd6 3. Sg4 Tf6#
play all play one stop play next play all
Cook: 1. dxe3ep 0-0-0 2. Lf5 Txd6 3. Sf3 Td4#
not
1. dxe3ep Td1? 2. Lf5 Txd6 3. Sf3+ ?? usual AP try and also check Bl 3rd move
Henrik Juel: The missing black man is dark-squared, so last move was not fxe4. In the solution Td6 should read Txd6. There seems to be a 'variation': 0... dxe3ep 1.0-0-0 Le4 2.Txd6 Sg4 3.Tf6#, so maybe the stipulation should have 1.1;2.1;1.1 added. (2004-09-23)
A.Buchanan: Well spotted Henrik. Your kindness is legendary, but in the harsh world of helpmates, this counts as a cooked I believe. In fact, it's the 2.Lf5/3.Sf3 version which is the cook. Yours is the actual solution! This is because 3.Sf3 would check if White hadn't castled, and we want the castling to be because of the AP condition only, not for 2 reasons, which one might term a "logical dual". The problem can be trivially fixed by sticking sSe5 on f6. I have no idea whether it's a typo or a cook: WinChloe has the same diagram, and the same confusion between cook & intended solution. The fact that AP does not appear in the stipulation might be a factor. WinChloe as far as I can see has no concept of AP as a keyword, and just picks it up based on a substring of the stipulation. In other compositions, T.Petrovic wasn't shy of flashing the AP acronym in the stipulation, but as usual the detective work peters out in the absence of primary records. (2020-12-08)
more ...
comment
Keywords: En passant as key, Castling (wg), a posteriori (AP) (Type Petrovic), Superseded by (P1382802)
Genre: h#, Retro
Computer test: Popeye v4.85 + retro thinking
FEN: 1n1r4/p1B4p/1p1p4/p3n1p1/3pPk2/3b4/pr5q/R3K3
Reprints: (60) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-16 more...
27 - P0003434
Jozsef Bajtay
2432 Problem 101-102 09/1966
P0003434
(10+11)
h#2
1. fxg3ep 0-0 2. Lg4 hxg3#
play all play one stop play next play all
A.Buchanan: Surely diagram typo. Change to sBh3, then everything works (2022-03-29)
Mario Richter: Yes, Pawn h3 is black (2022-03-29)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wk)
Genre: h#, Retro
FEN: 8/b3p3/4p3/6pp/2P2pPk/1pPP3p/2PP1P1P/r2bK2R
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-03-29 more...
28 - P0003442
Janko Furman
3. Makuc-Moder-Gedenkturnier 1971-1973
1. Preis
P0003442
(12+13) C+
h#2
b) wBd4 nach d5
a) 1. cxd3ep Sd5 2. 0-0 Se7#
NOT 1. ... Lf6? 2. Kf8 Tx8# because no AP justification
b) 1. Kd7 Lf6 2. Te8 Sxb6#
NOT 1. 0-0? Tf6 2. Kh8 Txf8# because rights lost
play all play one stop play next play all
Assume that bPb6 is really on b7:
Bl captures: dxe, exf, fxg, a|
Wh captures: bxc, g|, cxd=
So all pawn captures are accounted for.
a) If Bl 00 rights remain, then only way to give Black a prior move is by d2-d4.
b) No way to give Black a prior move, so Bl 00 rights must be lost.
Cook: NL
b) 1. 0-0 Tf6 2. Kh8 Txf8#
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)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sk)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/7p/1pR5/2P5/NNpP4/KB2PPPP/p3pppq/B4bnr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
29 - P0003444
Janko Furman
Miroslav Stosic

7273 Schach-Echo 11/1972
P0003444
(8+14) cooked
h#2
b) Gespiegelt (a1<->h1) & wKd1->e1
a) 1. dxe3ep 0-0 2. Te4 Txf3#
b) 1. exd3ep 0-0-0 2. dxe2 Ld5#
play all play one stop play next play all
Cook: a) 1. Kg3 Kf1 2. Df4 Th3#
1. Kg3 0-0 2. Dh4 Txf3#
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)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wk,wg), Superseded by (P1399967), Twinning by board reflection
Genre: h#, Retro
FEN: 8/1p1pp1p1/8/5nq1/1r1pPkr1/1PBp1p2/Pp1P2P1/2n1K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
30 - P0003451
Knud H. Hannemann
3823 Stella Polaris 03/1971
P0003451
(6+8) cooked
h#2 AP
1) 1. gxf3ep 0-0-0 2. Te7 Th4#
2) 1. cxd3ep 0-0 2. Da4 Txa4#
play all play one stop play next play all
AB: I think this is cooked. I don't see how AP castling can justify earlier ep here.
1. Kf3 Bf5 2. Kg2 Be4#
1. Kf3 Bxd7 2. Kg2 Bc6#
1. cxd3 Rf1 2. Qa4 Rxa4#
1. gxf3ep Ra5 2. Kf4 Rh4#
1. gxf3ep Rd1 2. Re7 Rh4#
Also, second solution given has typos 1. *c*xd3ep & 2. Q*a*4. (2002-03-21)
V.Liskovets: Indeed this problem is cooked, and I
failed to correct it preserving symmetry.
Here is a possible correction:
W: Ke1 Ra1 Rh1 Be6 Pd4 Pf4
B: Ke4 Qd7 Rf7 Pc3 Pc4 Pe3 Pg3 Pg4 Bh7 Rb2 Pb3

Another story, justifying its contents (e.p.).
In my opinion, all published treatments are
insatisfactory (the same concerns P0004295
(corrected) by Werner Kuntsche as well).
There is a way to make it sound under the
sophisticated HYBRID stipulation 'AP, pRA':
2 partial solutions legalized JOINTLY by
both castlings (details to be published). (2002-04-02)
VL: Sorry, Andrew, I see no issues with my version (excepting the mentioned special retro-convention/genre for justifying its soundness, of course!). h#2 (pRA&AP). The full solution consists of 2 partial AP-based ones: I 1.gxf3 e.p.(!?) 0-0-0! 2.Te7 Th4#; II 1.cxd3 e.p.(!?) 0-0! 2.Da4 Txa4#.
C+(popeye): h#2 & two h#1.5 after the keys (added Tb2 & Lh7 are cookstoppers).
The main features of the corresponding suitable AP-genre ("consolidated")are rather clear. However, not all subtle aspects have been analyzed thoroughly yet. I have got only a draft manuscript with few examples, still. (2022-02-15)
A.Buchanan: Yes Valery I think your correction is technically sound. Theological debate is something else, heh. Have posted it as separate entry. Please tell me if it was original here, or appeared somewhere else (2022-05-29)
Henrik Juel: I looked in Stella Polaris 1971 p.202-203, where the problem is given without AP under the diagram
It turns out that both the author and the editor (Jan Knöppel) consider the problem correct
After explaining the AP convention the latter writes [my translation]:
The author has elegantly applied the AP conventionn in combination with partial analysis. If we assume that last move was not f2-f4 or T-h1, the possibilities d2-d4, K-e1, and T-a1 remain, but the two latter disappear by 1... 0-0. The other assumption (last move was not d2-d4 or T-a1) is analogous.
So something like Valery's HYBRID stipulation was the intention, and the problem was considered correct in 1971; later developments do not change this fact (2022-05-29)
Henrik Juel: Of course, the obvious cooks 1.Kf3 etc. remain (2022-05-29)
A.Buchanan: I agree. Basic AP Petrovic does not allow multiple ep, nor does RS, so some alignment of AP with PRA is required. I would like to see this codified though. (2022-05-29)
more ...
comment
Keywords: Castling (wb), En passant as key (2), a posteriori (AP) (Type Petrovic), Superseded by (P1401508, P1399178), Quasi-symmetrical position, Partial Retro Analysis (PRA)
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/3q1r2/4B3/8/2pPkPp1/2p1p1p1/8/R3K2R
Reprints: (65) Problem 144-147 12/1971
(8) StrateGems SG19, p. 156, 07/2002
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-29 more...
31 - P0003925
Ivan Skoba
1009 diagrammes 47 09-10/1980
P0003925
(6+7) C+
a) ser-h#6 (AP)
b) nach dem Schlüssel von a): ser-h#5
a)
1. dxc3ep 2. Kc4? 3. Tb5 4. Td5 5. Dc5 6. Kd4 Th4#
2. Tb2! 3. Kb3 4. Kc2 5. Kc1 6. c2 0-0#
b)
1. Tb2? 2. Kb3 3. Kc2 4. Kc1 5. c2 0-0#
1. Kc4! 2. Tb5 3. Td5 4. Dc5 5. Kd4 Th4#
play all play one stop play next play all
There are basically two candidate solutions ending Th4# and 0-0#. In (a), the initial ep can only be justified by 0-0#. On the other hand in the diagram position in (b), White's last move must have been with wK or wR, so 0-0# is disrupted. Nice and paradoxical.
A.Buchanan: Why is there sDe3? Isn't sL sufficient? (2022-05-27)
more ...
comment
Keywords: Seriesmover, Castling (wk), Cant Castler, a posteriori (AP) (Type Petrovic), En passant as key
Genre: Retro, Fairies
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 8/8/8/8/bkPp4/pr1pq3/3PRP2/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-29 more...
32 - P0004199
Werner Frangen
Problem 41-44, p. 52, 03/1957
P0004199
(14+10) cooked
#5
1. axb6ep+ Kb5 2. bxc7+ Tb6 3. 0-0-0 ...
play all play one stop play next play all
K. Fabel: "Von Interesse ist, dass diese strittige Idee [Beweis des e.p.-Schlagrechts durch spätere Ausführung der Rochade] auch im direkten Mattproblem dargestellt werden kann, vergl. das Diagramm [diese Aufgabe P0004199]. (Vielleicht ist diese Aufgabe noch nicht korrekt doch es wird nicht schwer sein, sie zu verbessern). Falls Schwarz zuletzt Bb6-b5, Tb6-c6 oder Lb6-c7 gezogen hat, ergibt sich, dass Weiss im letzten oder vorletzten Zug den K oder T bewegt haben muss. Falls jedoch K und T noch nicht gezogen haben, muss b7-b5 der letzte Zug gewesen sein. Weiss spielt daher 1. ab6e.p.+ Kb5 2. bc7+ Tb6. Jetzt könnte Weiss mit Dxb6 sofort mattsetzen, aber er muss ja den e.p.-Schlag noch legalisieren. Daher 3. 0-0-0! und Matt erst im 5. Zuge."
Cook: 1. Sxc7+!
1. Lxc7!
Innerhalb des 2. Teils einer Artikelserie "Die Konventionen im Problemschach" von Karl Fabel.
A.Buchanan: The retro logic is fine, but forward play is savagely cooked. I wonder about changing wSc8 to sS. The retro stuff still works OK, indeed sSc8 is a fourth thematic unit to retract in the try. Ignoring proof of ep legality, there is a unique #4 beginning with ep, and no #5. Promising: but when can we interpolate w000? Any solution must begin 1. axb6ep+ Kb5. There are now numerous #4, but none include w000, so I think we must stick with the still unique line beginning 2. Dxc8 thr 3. 0-0-0. There are 3 black defenses to refute this. 2. ... Txb6,axb3,d5. So maybe #6 needed for White to prevail? (2022-03-28)
A.Buchanan: But if we shift to #6 then 1. Sxc7!,Qxc8+! mate without ep. Can anyone retrieve the original diagram for this one, please? (2022-03-28)
Mario Richter: The position here is identical to the original diagram, but it should be regarded more as a schematic example than a "real problem". Fabel in his quote above: "... perhaps the problem is still not correct, but it should be easy to improve it ..." (2022-03-28)
Henrik Juel: unfortunately K. Fabel forgot to give the easy correction... (2022-03-28)
A.Buchanan: Thanks for this. Valery Liskovetz, an AP expert, was kind enough to send a pdf of the relevant Problem issue, so I can confirm. It's easy enough to remove the two cooks (e.g. sTc8) but the difficulty is in validation of the intended solution. Is there any "exact" option in Popeye that forces the solution to include a waiting move? The actual solution might be expected to be included in that set. (2022-03-29)
Henrik Juel: Yes, Andrew
'stip exact-#5' would disregard #4 as a solution, I believe (2022-03-29)
A.Buchanan: Even with the "exact", it's still very complicated to investigate. I think this kind of problem requires an engine feature to disallow any mate or pat if the specified castling has not been executed. Gazing at the 20,000 lines of output, I don't think sTc8 works as 1. axb6ep+ Kb5 2. Qb7 bxa3! threatens axb2+ after castling. I wonder if the board might be flipped left-right except for wK, but I haven't explored this. (2022-05-31)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Non-standard material (TLL)
Genre: Retro, n#
FEN: NQNB4/B1brpp2/k1rp4/Ppp5/Rp6/BP6/RPP5/R3K3
Reprints: (8) Problem 101-102 09/1966
(52) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-29 more...
33 - P0004296
Luis Alberto Garaza
1809 Problem 73-78 06/1961
P0004296
(16+13) cooked
h#1? h#2?
1. fxe3ep? Dxa8# No justification for the ep, so just retro try.
play all play one stop play next play all
White pcs: dxexf5,hxBg. Black none. Assume that Black can still castle. White can't have just played h4xg5 as sBh2 would be blocked. White might apparently just have moved D,Se5,Td4,Tc3,g4-g5,a5-a6,e3-e4 or e2-e4 but what might Black have played before? Only the last allows a move sDf1-e1. So set up for AP Type Petrovic is OK.
Cook: 446 candidate h#2
However 27 have no ep, while 59 have both ep & castling.
So a total of 86 viable solutions, 85 of which must be cooks.
s.a. Version P0004341
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004341)
Genre: h#, Retro
Computer test: Popeye v4.87 indicates cook
FEN: r3k3/p1ppp3/Pp6/4NPP1/2PRPp2/2RK2PN/1PBn1PQp/2Brq1b1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
34 - P0004341
Luis Alberto Garaza
(9) Problem 101-102 09/1966
P0004341
(15+12) cooked
h#1 oder h#2
1. fxe3ep Dxa8#? because no AP justification for ep

1. 0-0-0 g7 2. Tf8 gxf8=D#
play all play one stop play next play all
62 apparent h#2, so seems cooked.
Cook: 415 candidate solutions for h#2. 401 begin with ep, of which 48 contain 0-0-0 to justify. There are also 14 solutions without ep, of which 13 begin with 0-0-0. The odd one is 1. Kd8 Sc6+ 2. Kc8 Sge7#
Version zu P0004296

Autor: "In Nr.9, I intended to do something similiar to No. 8 [P0004199], but with a different key: the solution 1. PxP "e.p.", Qxa8 checkmate will not go, for if blackside does not castle, the "en-passant" capture cannot be justified. Though, well considering, there is the demolition 1. PxP "e.p." -any 2. 0-0-0! Qb7 or a8 checkmate.
A.Buchanan: Thanks Mario for retrieving the author's intent with this. So he knew it was cooked! Assume s000 rights remain. R: 1. b5xa6? as sBb promoted on b1. R: 1. h4xg5? as sBh retro-blocked. R: 1. Ke3-d3? impossible check from sBf4. R. 1. K~-d3 d3-d2? illegal check. I like these! So by elimination, R: 1. e2-e4 e3xTd2 2. T~d2 L~e1. So must have sLe1. (2022-03-22)
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004342)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k1N1/p1pp4/P5P1/4PPp1/2PNPprB/3K1QRp/P2p1PB1/4b2n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
35 - P0004342
Luis Alberto Garaza
(10) Problem 101-102 09/1966
P0004342
(15+11) cooked
h#2
1. bxc3ep Dg8#? (thematic try)
1. bxc3ep Dd5 2. 0-0-0 Db7#
play all play one stop play next play all
Cook: 1. bxc3ep Ld6 2. 0-0-0 Dc4#
Author: "From the former [d.h. P0004341], the No. 10 was born, something similar to No. 8 [P0004199] but with Black-castling for it is a help-mate. The mating move 1. ... Qg8? will not go."
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P1399966)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k3/p2p4/8/NP6/BpPPN1PP/B2K1Ppp/QP1p1pP1/Rrb5
Reprints: (55) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
36 - P0004479
Matjaz Zigman
Länderkampf Mazedonien-Slowenien 1969
6. Platz
P0004479
(10+13) C+
ser-h#5* (AP)
* 1. ... Lxd2#
1. dxe3ep 2. Txf2 3. Txg2 4. Th2 5. Th3 0-0#
play all play one stop play next play all
more ...
comment
Keywords: Castling (wk), Seriesmover, En passant as key, a posteriori (AP) (Type Petrovic)
Genre: Retro, Fairies
Computer test: HC+ Popeye 4.61 after analysis
FEN: 8/8/8/4pPp1/1p1pPkpb/1P1P2p1/pPpr1PP1/rbB1K2R
Reprints: 103 Bilten 1970 1971
(70) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-20 more...
37 - P0004481
Matjaz Zigman
Länderkampf Mazedonien-Slowenien 1969
7. Platz
P0004481
(13+10) cooked
ser-h#6** AP
* 1. ... Sb5#
* 1. ... dxc3#
1. fxe3ep 2. e2 3. exf1=S 4. Se3? 5. Sd5 Sb5# (must castle for AP)
1. fxe3ep 2. e2 3. exf1=S 4. Sxd2 5. Sxb1 6. Sxa3 0-0-0!#
play all play one stop play next play all
Luckily, illegal diagram can be fixed by removal of bPa4.
Then if White castling rights remain, ep is on. One ser-h#5 & 18 ser-h#6 tries exist, in which White does not bother to castle.
We don't know if this was a transcription error or a counting mistake by the composer.
Cook: Too many pawn captures required. For retro problem, illegal position implies unsound
Mike Neumeier: The solution appears to be 1.fxe4 e.p. 2.e2 3.exf1=L 4.Ld3 5.Le4 6.Ld5 dxc3#. Was that the intention? (2013-02-18)
Arno Tüngler: There would even be a solution in 5 moves by 3.exf1=S 5.Sd5 Sb5#
However, in order to justify the e.p.-key there must be an AP prove that e2-e4 (and not a move by the wK or wRa1) was played as White's last move. Thus the only solution giving this in 6 moves is 1.fxe3 e.p. 2.e2 3.exf1=S 4.Sxd2 5.Sxb1 6.Sxa3 0-0-0!# (2013-02-18)
Mike Neumeier: It struck me as odd there was no solution listed. (2013-02-18)
Mike Neumeier: And, if we let black be the idle side, there is 1.Txg1 2.Lxh1 3.Kd1 4.Kc1 hxg1T#. What does the asterisk(*) mean? (2013-02-18)
Henrik Juel: The asterisk means that there is also a white mate if he had the move, in this case two mates, so maybe the stipulation should be ser-h#6**:
1.dxc3,Sb5# (2013-02-18)
Ladislav Packa: It is clear that e.p. is possible only when the wK and wRa1 done neither move.
What convention is used here for the right to castling? (2013-02-18)
Henrik Juel: White may castle, unless you can show that he has lost the right to castle. In this problem there is a major difficulty, I think: the position seems illegal.
White pawns captured b2xc3xd4xe5xf6, g2xf3, and axb, promoting on b8; Black captured bxc, dxc, and g3xh2. We cannot explain the destiny of [Ph7].
Reversing wPh3 and bPh2 seems to handle the illegality, but then the problem can be solved in 5 moves (2013-02-18)
Mike Neumeier: Thanks, Henrik. Maybe the stipulation was a typo. Considering all comments together, perhaps the stip should have been ser-h#5**, with the one solution being the 5-mover Arno gave. I do not think it can be proven, except by a stipulation of ser-h#5 that en passant is possible. Which leads to the question of convention. Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible? Whether there is castling or not here seems immaterial. It is just another 6-move solution. There are 19 6-movers (Popeye) with one queenside castling among them. But only the one 5-mover. (2013-02-18)
Henrik Juel: By convention, an en passant capture as first move is not permitted, unless the pawn double step can be shown by some kind of retro analysis.
(Conversely, a castling is permitted, unless it can be shown by retro analysis that the right to castle has been lost, i.e., that king and/or rook has moved) (2013-02-18)
A.Buchanan: Suggested repair: Remove bPa4. Add AP to stipulation.
As Henrik points out, the diagram as it stands is illegal. We can't swap wPh3 & bPh2, because that allows 2 h#1 cooks.
Suppose we remove bPa4 instead. Then we still at least 5 White pawn captures: bxcxdxexf6 & gxf3. But wPh3 did still move from h2, so there are 3 Black pawn captures bxc, dxc & gxh2. So [bhP] died without capturing or promoting. [waP] must have promoted - so either [baP] was captured to clear the way, or [waP] captured to promote, and later [baP] promoted. Either way, all the numbers add up, and the position is legal. wPe4 cannot have just come from d3. So AP is triggered. Forward logic works just as before without bPa4. There is 1 5-move try, and 18 6-move tries, but AP eliminates them all because of need for castling to retrospectively justify the ep. (2013-02-23)
A.Buchanan: Mike asked: "Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible?"
Answer: No. The stipulation does imply a default player to move, but otherwise cannot be used as a premise to determine game state (castling, en passant). (2022-03-21)
A.Buchanan: WinChloe & Yacpdb have the same cooked diagram as PDB, so I think this cook was in the design. I would like to post a sound version with bPa4 removed, and I think it would just be “Zigman correction” with no one else’s name mentioned in the credits. Is everyone ok with this? (2022-05-30)
Mario Richter: 'Bilten 1970' (printed 1971) contains on pp. 62-81 a complete report on the "Problem Duel Slovenia - Macedonia 1969", the problem here can be found on page 79 and is printed there exactly as given here.
Andrew's handling of the suggested correction (remove black pawn a4) is o.k. for me. (2022-05-31)
more ...
comment
Keywords: Castling (wg), Seriesmover, a posteriori (AP) (Type Petrovic), Illegal position, En passant as key, Promotion (s), Valladao Task, Superseded by (P1401546)
Genre: Retro, Fairies
Computer test: Popeye v4.87 but simple retro-logic shows illegality
FEN: 8/8/5P2/2p1p3/p1pkPp2/N1p2P1P/2PP1PBp/RN2KRbr
Reprints: 104 Bilten 1970 1971
(72) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-01 more...
38 - P0004657
Mato Prikril
(27) Problem 206-210 07/1981
1. ehrende Erwähnung
P0004657
(5+7) cooked
h#2*
*) 1. ... Ta4 2. Kxg4 Txf4#
1. fxg3ep Ta4#? (no proof of ep right)
1. fxg3ep 0-0-0! 2. g2 Td4#
play all play one stop play next play all
A short retro try h#1 and 16 h#2 tries. B2 is unique tempo move
Cook: 1. Kxg4 Ta5 2. h4 Tg5#
1. Kxg4 Ta4 2. Kh4 Txf4#
61. TT (Pavlovic Memorial), Gruppe A
more ...
comment
Keywords: Castling (wg), En passant as key, a posteriori (AP) (Type Petrovic), Tempo Move
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 8/6p1/6pB/7p/5pPk/5p1p/5P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-06 more...
39 - P0004917
Andrej N. Kornilow
(C) Die Schwalbe 80 04/1983
P0004917
(5+8)
#2 (AP)
1. ... Tfxg8 2. Lg6! h5 3. Lxf7+
2. ... fxg6=?,hxg6=?,Txg6#?
2. ... Tf8,Tg7 3. Lxf7+ Txf7#

2. Lxe6? Tg6+! 3. Kf5 0-0!
2. ... dxe6? 3. d7+!
2. ... fxe6=?

2. Lxh7? Tg6+! 3. Lxg6 0-0!

Therefore it's WTM
1. Dg2 h5,~ 2. Da8#
play all play one stop play next play all
Black cannot steal the move, as White can prevent the castling justification.
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)
more ...
comment
Keywords: Castling (sk), a posteriori (AP) (Type Keym)
Genre: Retro, 2#
FEN: 4krQr/3p1p1p/3PpK1p/4PB2/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
40 - P0004920
Valery Liskovets
(F) Die Schwalbe 80 04/1983
P0004920
(5+4)
#2 (AP, pRA)
BTM 1. ... Txh6? 2. Sd7! Txc6+ 3. Kb5! (Kb7?) Td6 4. Sf8 Tf6 5. Sh7! Th6 6. Sf8 no castling
5. a6? Tfxf8 6. a7 Tf5+ 7. Kc6,K~ 0-0!
1. ... 0-0? 2. Se7+! Kh8 3. S5g6#
WTM 1. Td6 droht 2. Td8#
play all play one stop play next play all
White to move has #2 since Black has lost castling rights. So Black pulls the move, but must castle at some point. If Black castles right away, then White has a different #2, so Black must be more subtle. 0... g6/g5/gxh6 leads to castling disruption, e.g. 1.Txg6/Te6+/Sg6. So Black only has 0... Txh6. This pins wSc6 and threatens 0-0, so 1.Sd7! (1. Sg6? Txg6 2. ~ 0-0) etc.
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)
comment
Keywords: a posteriori (AP) (Type Keym), Cant Castler, Castling
Genre: Retro, 2#
FEN: 4k2r/6pr/K1N4R/P3N3/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-22 more...
41 - P0005637
Osmo Ilmari Kaila
Hannu Harkola

(D) Die Schwalbe 48 12/1977
P0005637
(6+1) cooked
#3
AP
WTM: ???
BTM: 1. ... Kxf3 2. 0-0+! Ke3 3. Td1 Kf3 4. Td3#
not 2. Kd2? Kg2 3. Sf4+ Kf2,Kf3 4. Tf1#
play all play one stop play next play all
Cook: 1.Tf1,Se2+,Sf2 all lead to mate in 3, and none require castling
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)
comment
Keywords: a posteriori (AP) (Type Keym), Castling, Rex solus, Miniature
Genre: Retro, 3#
FEN: 8/8/6R1/8/8/5PNN/6k1/4K2R
Input: Gerd Wilts, 1995-06-06
Last update: A.Buchanan, 2023-07-30 more...
42 - P0006159
Marko Klasinc
8126 Schach-Echo 20, p. 319, 10/1974
P0006159
(15+9) C+
h#6* (AP)
*) 1. ... Sxb3#
1) 1. gxh3ep Ld1 2. h2 Lxc2 3. Kxc2 Df1 4. Kc3 Dxd3+ 5. Lxd3 0-0-0 6. Lxc4 Se4#
play all play one stop play next play all
Wh has made 6 visible pawn captures, Bl 1. If bPfxg, then bPh was waylaid, and bPe promoted, disrupting White's castling rights. If Wh 000 rights remain, therefore, bPhxg and bPe was waylaid instead. Since original g-pawns remain on g-file, they must be wPg3 & bPg4, and wPh must retract a double hop to allow bPh3xg2. So Black can ep, avoiding immediate pat. To castle then mate requires a lot of work.
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)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg)
Genre: h#, Retro
Computer test: C+ Jacobi v0.7.5
FEN: 8/8/8/6PB/1PPP1PpP/Pp1pB1P1/prpN1Rp1/Rbk1K1Q1
Reprints: 4113 Problem 05/1979
(19) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-07-19
Last update: A.Buchanan, 2022-07-10 more...
43 - P0006423
Andrey Frolkin
9128 Die Schwalbe 157, p. 283, 02/1996
Leonid Borodatow gewidmet
P0006423
(9+4)
#3 (AP)
White pushes the move (Keym AP)
1. ... f5! (f6?,fxe6?,fxg6? 2.0-0! ~ 3.Tf3#) 2. gxf6ep! exf6 3. 0-0! f5 4. Tf3#
(2. 0-0=?)
Valladao Task via the try: 1. exf7? e5! 1. ... e6? 2. f8=D e5 3. Da3#
Another try shows fully differentiated black Albino:
1. Kf1? fxg6!
1. ... f5? 2. Lxf5! h3 3. Txh3#
1. ... fxe6? 2. Lxe6! h3 3. Txh3#
1. ... f6? 2. Lf5! h3,fxg5 3. Txh3#,Th3#
And retro tries:
1. 0-0?? f5,~ 2. Tf3# (short solution)
1. f5? f6,~ 2. 0-0?? fxg5,~ 3. Tf3#
play all play one stop play next play all
Wenn Schwarz zuletzt gezogen hat, dann kann Weiß nicht mehr rochieren: R: 1. Kh2-g3 Tf1(g1)-h1+. Da aber Weiß in einem direkten Matt beginnt, ist die Rochade nicht mehr zulässig. Der Zusatz "AP" in der Forderung ist überflüssig: es könnte höchstens der Anzug a posteriori dem Schwarzen übertragen werden. Das geht aber nicht, da Weiß in einem direkten Matt beginnt.
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)
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comment
Keywords: a posteriori (AP) (Type Keym), Castling, En passant, Promotion (D), Valladao Task
Genre: Retro, 3#
FEN: 8/4pp2/4P1N1/6PP/5P1p/6kB/6P1/4K2R
Input: Gerd Wilts, 1996-06-12
Last update: A.Buchanan, 2022-04-18 more...
44 - P0008517
Valery Liskovets
12 Problemist Pribuzhya 1 1990
P0008517
(4+3)
#2 AP (pRA)
BTM 1. ... axb6+ 2. Ta5! 0-0-0 3. Ta8#
2. Ta7? 0-0-0! & no mate in 1
WTM 1. Lc5! (0-0-0??) droht 2. Tf8#
play all play one stop play next play all
VL: Published in "Probl. Pribuzh." (or "Problemy Pribusch'ja"?),
1990, No.1, #12. An obscure Russian language chess problem
magazine issued in Nikolaev, the Ukraine (the South Bug
riverside region). (2006-01-27)
A.Buchanan: Lovely problem, but this is no more PRA than it is duplex. Rather, the push/pull of AP Type Keym gives a forfeit which must be discharged. (2023-07-22)
more ...
comment
Keywords: a posteriori (AP) (Type Keym), Castling, Homebase (s), Miniature
Genre: Retro, 2#
FEN: r3k3/p6R/1B6/5R2/8/8/8/K7
Input: Gerd Wilts, 1996-12-14
Last update: A.Buchanan, 2022-02-15 more...
45 - P0008592
Leonid M. Borodatow
9504 Die Schwalbe 163 02/1997
P0008592
(15+7)
#3. Two solutions (AP)
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)
more ...
comment
Keywords: a posteriori (AP) (Type Keym), Castling
Genre: Retro, 3#
FEN: 4b3/p1pP1Pp1/3P4/4P3/Q2Pp3/1NP1P3/NP1B1R2/qk2K2R
Input: Gerd Wilts, 1997-03-19
Last update: A.Buchanan, 2022-02-16 more...
46 - P0008780
Valery Liskovets
7162 feenschach 123 01-06/1997
P0008780
(3+4)
h#2 AP(PF)
Pièces rétro-volages
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)
comment
Keywords: a posteriori (AP), Retro-volages, Post Factum (PF), Miniature, Castling, Castling as mating move
Genre: Retro, Fairies, h#
FEN: 3r4/4pk2/8/8/8/8/1r6/1R2K2R
Input: Gerd Wilts, 1997-06-16
Last update: James Malcom, 2021-02-12 more...
47 - P0008985
Gianni Donati
9862 Die Schwalbe 168 12/1997
P0008985
(10+14)
ser-h#6 (AP)
1. gxf3ep 2. Dg4 3. 0-0-0 4. Te8 5. Kd8 6. Dxh5 0-0-0#
play all play one stop play next play all
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)
comment
Keywords: a posteriori (AP), Seriesmover, Castling, En passant as key
Genre: Retro, Fairies
FEN: r3k3/ppp1p3/2p5/6PP/5Ppq/1P4nB/PrP1P1p1/R3K1nb
Input: Gerd Wilts, 1998-06-26
Last update: A.Buchanan, 2021-11-08 more...
48 - 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...
49 - P1000662
Gianni Donati
R074 Probleemblad 11/1999
P1000662
(10+14) C+
h#1.5 (AP)
1. ... gxh6ep 2. exf1=L 0-0-0#
play all play one stop play next play all
Kommentare:
Als einziger letzter schwarzer Zug, der die weiße Rochade
erhält, kommt nur Bh7-h5 infrage, was dem Weißen den e.p.-Schlag
ermöglicht, um nicht die Rochade oder das Mattnetz zerstören zu
müssen (H.P.Suwe)
Gianni Donati: This intends to show the Valladao theme in the minimum number of moves. (2004-03-19)
VL: The waiting ep capture. Cf. also P1000260
(by T.Petrovic, 2000) somewhat enriched
thematically with the illegal try 2... Rd1#??. (2004-06-03)
A.Buchanan: The “illegal try” 2. Rd1+ is not actually mate because c2 is unprotected. This is kind of “logic dual” spoils the A.P. motivation for the castling (2020-05-20)
Henrik Juel: White captured h6xg7 and once more, e.g. axPb-b8=Y
Black captured cxdxe, dxexf, fxg, and once more, e.g. a2xb1=Y
Possible retroplay 1... h7-h5 2.h6xSg7 Df8-f4 3.h5-h6 Sf4-h3 4.Sh3-g1 etc., preserving the castling right
Any other black last move would force White to retract Ta1, as d2xc3, d2xe3, and h6xh7 would be illegal retractions
I agree with Andrew that the double motivation of 2... 000# is a weakness:
a. legitimizing h7-h5
b. mating (2020-05-21)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
A.Buchanan: See discussion at P0009121 (2023-08-04)
more ...
comment
Keywords: a posteriori (AP), En passant, En passant as key, Castling, Valladao Task, Promotion (l)
Genre: Retro
Computer test: HC+ Popeye v4.87 + simple retro logic
FEN: 6n1/4p1P1/6p1/6Pp/2b1rqrb/2PkPppn/1P2pP2/R3KRN1
Reprints: König & Turm (4), p. 28, 03/2001
H Problemkiste (143) 10/2002
(V2) Problemkiste 156 12/2004
Input: Gerd Wilts, 2000-08-01
Last update: A.Buchanan, 2023-08-03 more...
50 - P1012059
Ronald Turnbull
15 diagrammes 10/1990
P1012059
(4+12) cooked
h#3 (AP)
PRA
1. bxc3ep 0-0! (for AP reasons, not 1...Kf2?/Th2? etc) 2. Ta1 Txa1 3. Kc4 Txa4#
1. fxe3ep 0-0! (for AP reasons, not 1...Tf1?/Txh3? etc) b2 Txf3 3. Tc3 Tf4#
play all play one stop play next play all
Si les B. peuvent roquer, leur dernier coup est Pc2-c4 ou Pe2-e4. D'où les 2 solutions alternatives: 1.hc3:e.p!. O-O 2.Tal Tal: 3.Rc4 Ta4:# l.fe3:e.p. O-O 2.b2 Tf3: 3.Tc3 Tf4#
Cook: 1. Kxe4 Tg1/Txh3 2. Ke3,Sd4 Tg5/Th5 3. Sd4,Ke3 Txe5# (4 variants)
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)
comment
Keywords: a posteriori (AP) (Type Petrovic - cee), En passant as key (2), Castling (wk), Superseded by (P1382808), Partial Retro Analysis (PRA)
Genre: Retro, h#
Computer test: cooked by Popeye v4.85
FEN: 8/8/8/1np1n3/ppPkPp2/rp1p1p1p/8/4K2R
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2022-02-18 more...
51 - P1012972
Yaakov Mintz
The Problemist 1985
P1012972
(7+1)
#2 (AP)
WTM: no solution
BTM:
1. ... Kxh1 2. Dd2 Kg1 3. 0-0-0#
1. ... Kxf3 2. 0-0+ Ke2/Ke3/Ke4 3. Tae1#
Across these two cases, both castling rights are demonstrated.
play all play one stop play next play all
If we can assume that none of the White pieces in the first row has yet moved, then we must assign the move to Black. When this was published in The Problemist, opinions were divided. "White, to prove that the move is Black's, begin by claiming the right to castling on each side: and as they cannot in fact make both castles at once to establish this claim, it is sufficient to make one in each variation." "The solution consists of all the variations, and it is sufficient to have 0-0 in just one variation, since the small castling is enough to show that the move is Black's". "Since White claims the right to kingside castling, Black is forced to offer the opportunity" "No solution on 0. ... Rh1"

Si nous pouvons assumer qu'aucune des pièces blanches de la première rangée n'a encore joué, alors il faut attribuer le trait aux Noirs. D'où: 0. ... Rf3: 1,O-O+1 Re4 ou e3 2.Tae1# 0. ... Rhl: 1.Dd2I Rgl 2.O-O-O# Quand cela fut publié dans The Problemist, les avis furent partagés. "Les B., pour prouver que le trait est aux N., commencent par revendiquer le droit au roque de chaque côté: et comme ils ne peuvent de fait effectuer les deux roques à la fois pour établir cette revendication, il suffit d'en éxécuter un dans chaque variante". "La solution se compose de toutes les variantes, et il suffit d'avoir O-O dans juste une variante, puisque le petit roque suffit à montrer que le trait est aux Noirs''. "Comme les Blancs revendiquent le droit au petit roque, les Noirs sont forcés d'en offrir l'opportunité""Pas de solution sur 0. ... Rh1 :"
No. 8889 HN
A.Buchanan: My feeling is this doesn’t work, sadly. In an adversarial stipulation, Black can simply avoid ever playing the line that allows White to “prove” that Black had the move. In a helpmate on the other hand I suppose if PRA-AP is expressed properly, there should be no problem playing multiple parts. (2022-02-15)
comment
Keywords: a posteriori (AP) (Type Keym), Castling, Rex solus
Genre: Retro, 2#
FEN: 8/8/3Q4/8/8/5PPP/6k1/R3K2R
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2022-02-15 more...
52 - P1066685
Valery Liskovets
12365 Die Schwalbe 208 08/2004
P1066685
(14+5)
#1 (AP)
1. ... gxh3ep 2. 0-0-0#! (not 2.Sf3#?,Bh2#?)
to force
R: 1. h2-h4! h3xSg2 (not 1. Kd1-e1? Kf1-g1 2. Lf3-e2# )
play all play one stop play next play all
Diagram position is retropat so Article 15 give Black the move. Then White might have last move R: 1. h2-h4 or 1. Kd1-e1. To avoid stalemate in the diagram position, White must castle to demonstrate under AP the legality of 0. gxh3ep. So 1. 0-0-0#! not 1. Sf3#?,Bh2#?
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)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, No legal last move for Black, Castling (wk)
Genre: Retro, n#
FEN: 8/6P1/5P2/4NpPP/5QpP/P5B1/3PBPp1/R3K1kb
Reprints: Die Schwalbe 228 12/2007
Input: Gerd Wilts, 2004-08-13
Last update: James Malcom, 2022-07-05 more...
53 - P1066741
Tomislav Petrovic
R033 Probleemblad 11-12/1998
Ing. R. Tomasevic gewidmet
P1066741
(14+10) C+
h#2 (AP)
1. gxf3ep 0-0-0 2. gxh1=T gxh4#
play all play one stop play next play all
White PCs: axb=, bxc=, dxexf, cxdxe
Black PCs: fxg2. wRh was captured by officer in cage, and since wK didn’t move wRg5 is promoted
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)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Valladao Task, Promotion (t)
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 & retro-logic
FEN: 1B6/p2p4/1p4Q1/5PRN/4PPpb/3Np1Pp/4P1pP/R3KbkB
Input: Gerd Wilts, 2004-09-16
Last update: A.Buchanan, 2023-09-11 more...
54 - P1067403
Alexander Kukush
8966 feenschach 152 07-09/2003
P1067403
(6+3)
#3 (AP, pRA)
BTM: 1. ... 0-0 2. Dh8+ Kxh8 3. Th3+ Kg8 4. Th8#
WTM: 1. Db8+! Kd7 (Ke7) 2. Dd6+ Ke8,Kc8 3. Te3#,Sa5#
play all play one stop play next play all
If BTM, then must castle immediately to prove that it's his right, but still #3
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?
more ...
comment
Keywords: a posteriori (AP) (Type Keym), Castling, Homebase (s)
Genre: Retro, 3#
FEN: 4k2r/5p2/8/6P1/2N5/2R5/KB5Q/8
Input: Gerd Wilts, 2005-01-09
Last update: A.Buchanan, 2022-02-15 more...
55 - P1068187
Valery Liskovets
9679 Thema Danicum 118 04/2005
P1068187
(15+5) C+
h#1* (AP)
2 Lösungen
1. ... Ke2#! (0-0#?? illegal)
1. cxd3 Sb3#! (Ke2??, 0-0+?)
1. cxb3ep 0-0#! (Ke2#?, Sxb3#?) AP
play all play one stop play next play all
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)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wk), Volet Pawn
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 + thinking
FEN: 8/2Pp4/3N4/1PP5/1PpB4/2PR4/P1pN1PP1/Qnk1K2R
Input: Gerd Wilts, 2005-12-21
Last update: A.Buchanan, 2022-06-06 more...
56 - P1072281
Werner Keym
Tomislav Petrovic

Hannoversche Allgemeine Zeitung 1999
P1072281
(4+3) C+
h#1.5 (AP)
1. ... cxb6ep 2. 0-0-0 b7#
play all play one stop play next play all
Very elegant representation of AP Type Petrovic. Keym himself established in the Codex that the stipulation should include "AP".
more ...
comment
Keywords: En passant as key, Miniature, a posteriori (AP) (Type Petrovic), Castling
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 together with elementary reflection
FEN: r3k3/8/N1P5/KpP5/8/8/8/8
Reprints: (A2) Problemkiste (169), p. 6, 02/2007
Input: hpr, 2007-03-04
Last update: A.Buchanan, 2023-09-11 more...
57 - P1079607
Igor Vereshchagin
29 Zadachi i etyudy 14, p. 52-53, 1997
3. Lob
Thematurnier "Uschol-Prischol-Vernulsa"
P1079607
(6+15)
h#3 (AP)
1. cxb3ep 0-0-0 (T~?,L~?) 2. Dxd2+ Kxd2 3. La1 Txa1#
play all play one stop play next play all
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)
more ...
comment
Keywords: Valladao Task, En passant as key, Castling (wg), Promotion (L), a posteriori (AP) (Type Petrovic)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 3n3b/1p5p/1P1p2p1/Bp6/kPp1p2r/3p3r/q2P4/R3K2n
Reprints: WCCC St. Petersburg 1998
Input: hpr, 2008-10-26
Last update: A.Buchanan, 2023-07-30 more...
58 - P1080354
Zoltan Laborczi
Gabor Tar

PS2104F The Problemist 09/2008
P1080354
(4+10) C+
h#2 (AP)
2.1...
1. Kxb4 Tb1+ 2. Ka4 Lb5#
1. cxb3ep 0-0-0! (Td1?) 2. Lh4 Txd4#
play all play one stop play next play all
See P1401524 & P1401526.
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg)
Genre: Retro, h#
Computer test: Popeye v4.87 & simple retro-logic
FEN: 8/1p2p3/B4b2/p7/kPpp4/p1p5/7r/R3K3
Input: Gerd Wilts, 2009-01-03
Last update: A.Buchanan, 2022-06-08 more...
59 - P1085404
Valery Liskovets
13336v Die Schwalbe 224 04/2007
Werner Keym zum 65. Geburtstag gewidmet
P1085404
(15+7) C+
#2
b) ohne wLb2, AP
c) ferner mit sBe7 nach d6, AP
a) 1. ... gxf3ep 2. Lxf3 Lf5 3. Txg2#! (3.0-0-0#??)
1. Lf3?? gxf3 2. 0-0-0#?

b) 1. ... gxf3ep 2. Lxf3 Lf5 3. 0-0-0#! (3.Txg2#??)
1. Lf3?? gxf3 2. 0-0-0#??

c) 1. ... gxf3ep 2. Lxf3 Lf5 3. 0-0-0#!! (3.Txg2#??)
1. Lf3? gxf3 2. 0-0-0#??
play all play one stop play next play all
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)
comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (wg), No legal last move for Black
Genre: Retro, 2#
Computer test: HC+ retro thinking & Popeye v4.87
FEN: 8/3PpP2/2Q1P3/3N2P1/5Pp1/2P1P1Pb/1B1R2pp/R2BK1kr
Input: Gerd Wilts, 2009-06-08
Last update: A.Buchanan, 2023-07-22 more...
60 - P1096298
Guus Rol
R362 Probleemblad 10-12/2009
P1096298
(6+3) C+
h#2.5
2 Lösungen
Circe (AP)
1. ... dxc6ep[+sBc7] 2. 0-0-0 c5 3. Tg8 Dxg8[+sTa8]#
1. ... Kc6 2. Txa4 Dxa4[+sTa8] 3. 0-0-0 Da8#
play all play one stop play next play all
1. ... dxc6ep[+sBc7] 2. Txa4? Dxa4[+sTa8] 3. 0-0-0 Da8# doesn't work because we don't know whether Ta8 in the diagram position had already moved.
There are another 11 retro tries where Black does not castle at all, so can be eliminated in the usual way.
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)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), Circe, Castling, En passant as key
Genre: Retro, Fairies, h#
Computer test: HC+ Popeye 4.61 + thinking about AP
FEN: r3k3/8/8/1KpP4/P1P5/8/QP6/8
Input: Gerd Wilts, 2009-12-19
Last update: A.Buchanan, 2022-03-21 more...
61 - P1107682
Matti Arvo Myllyniemi
Deutsche Schachzeitung 1966
P1107682
(7+8)
#2 AP
1. Sxc5,Se5? 0-0-0!
but by AP now
1. dxc6ep! exd6 2. Sg6#
play all play one stop play next play all
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)
more ...
comment
Keywords: En passant as key, Castling, a posteriori (AP) (Type Petrovic)
Genre: Retro
FEN: r3kNQ1/3Npp2/1P1Pp3/bKpP4/1p6/8/8/8
Reprints: (37) Mat Plus 12 10-12/2009
Input: Gerd Wilts, 2010-06-13
Last update: A.Buchanan, 2022-05-24 more...
62 - P1108454
Werner Keym
Schach-Echo 1967
P1108454
(15+5) cooked
#1
b) wDa5 nach e5, AP
a) BTM
1. ... bxa3ep! 2. Dc3#! (2.0-0#??)
1.0-0#?
b) BTM pushed by Wh
1. ... bxa3ep! 2. 0-0#! (2.Dc3#? AP needs the 00)
1.0-0#?
play all play one stop play next play all
(a) Wh made 11 pcs, right to left, which accounts for all missing Bl units. wBa never captured, so sBa must have captured a3xSb2, prior to Wh a2-a4. Since Wh is missing only 1 unit, bBh promoted on h1, disrupting Wh 00 rights. Moreover Bl has no last move and by Codex Article 15, it’s Bl to move. Last move by Wh must allow Bl a prior move. This can only be R: 1. a2-a4 a3xSb2. So ep is on and there is no alternative.
(b) Again Wh has made 11 pcs, but Black may now have made 2. If WTM, then Black's last move was R: 1. Kc2xSc1 Td1-d2+ 2. Kc3-c2 or R: 1. cxSb1=T or R: 1. cxSb2 (as wBc4 & wBc7 might be foreign to c-file and allow bBc a smooth retraction). In any case, sBa3xSb2 was played so sBh-h1= disrupting 00. So Wh can still not play 1. 0-0#! Now White pushes the move to Bl and also asserts that bxa3ep is on. Both claims will be "proved" by Wh 0-0. We know that with WTM, Wh can't castle, so that's half of it. But if BTM with Wh to castle, then bPaxb & bPhxg, and the only way we can retract is as in (a). The mating move is different, because we have that extra Bl pc to keep castling legal, but we must actually castle so that the whole AP logic can work!
However, I don't understand why Wh can't just play 1. Txb2#?
Cook: In (b) Wh can just play 1. Txb2#
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)
more ...
comment
Keywords: Cant Castler, En passant as key, No legal last move for Black, a posteriori (AP) (Type Petrovic), Castling, a posteriori (AP) (Type Keym), Superseded by (P1406456)
Genre: Retro
FEN: 8/2P5/3P4/QP2N1B1/PpP5/1P1P4/1p1RBP2/brk1K2R
Reprints: (10) Die Schwalbe 228, p. 301, 12/2007
Input: Gerd Wilts, 2010-06-23
Last update: A.Buchanan, 2022-12-05 more...
63 - P1186997
Erich Bartel
5552 Problemkiste (138) 12/2001
P1186997
(4+4) C+
h=3 (AP)
0.1...
1. ... hxg6ep 2. 0-0 g7 3. Kh7 gxf8=D=
play all play one stop play next play all
Wenn ep erlaubt sein soll, dann muß g7-g5 der letzte Zug gewesen sein. Dies wiederum bedingt, daß sK+sT noch nicht gezogen haben. Nur unter dieser Betrachtungsweise (AP) ist das Ding korrekt.\eb
Löserstimmen:
Diese AP-Idee ist ja nun hinlänglich bekannt, aber man muß auch darauf kommen, sie für den Valladao zu nutzen. Finde ich prima (MN).
ep, und dann kann Schwarz auch noch rochieren (HM).
Wenn g7-g5 letzter Zug war, darf Schwarz noch rochieren (KF)...
AP (= a posteriori) argumentiert andersrum: wenn Schwarz noch soll rochieren dürfen, muß g7-g5 der letzte schwarze Zug gewesen sein (\eb)
Unlösbar! Der Autor meint wohl, man könnte mit 1. ... hxg6ep 2. 0-0 loslegen; hier irrt Goethe. Die beiden Konventionen -ep-Schlag nur, wenn Zulässigkeit retroanalytisch beweisbar ist; Rochade stets, außer wenn Unzulässigkeit retroanalytisch beweisbar ist- sind zwar passend verknüpft. Was aber den Vorrang hat, zeigt die zeitlich Reihenfolge der Anwendung. Zuerst kommt der ep-Schlag, und diesbezüglich gilt: Zulässigkeit nicht beweisbar, also nicht erlaubt (LZ).
Lieber Herr Zagler, vielen Dank für Ihren Kommentar, aber ich denke nicht, daß dazu eine Stellungnahme erforderlich ist, denn dazu ist schon viel diskutiert und geschrieben worden, ichverweise nur auf G. Lauingers "a posteriori Zwischenbilanz" in feenschach (80) X-XI 1986, S.414 ff, wo die AP-Regelung allgemein wie folgt definiert wurde:
"Eine einleitende Handlung (z.B. ein ep-Schlag oder ein unkonventionell Beginn), deren Zulässigkeit nach gültigen Konventionen zunächst nicht nachgewiesen werden kann, wird im Verlauf der Lösung(en) nachträglich durch einen oder mehrere Sonderzüge legalisiert. Legalisieren bedeutet dabei, daß _nur_mit_der_Ausführung_ des Sonderzuges (der Sonderzüge) dessen (deren) Zulässigkeit nachgewiesen wird und gleichzeitig (i.a. aus retroanalytischen Gründen) damit auch nachträglich die Zulässigkeit der einleitenden Handlung."
Ob Sie nun mit dieser Regelung einverstanden sind hat für o.a. Aufgabe wenig Bedeutung, denn nach o.a. "AP"-Definition ist sie lösbar. Wenn Sie "AP" so nicht akzeptieren und einer anderen Retro-Philosophie zugetan sind, steht das auf einem anderen Blatt. Da müßten Sie sich dann mit den entsprechenden Experten auseinandersetzen. (\eb)
Da nur die Rochade den sK nach h7 zu manövrieren vermag, fällt manchem sicher gar nicht erst auf, daß allein sie (neben dem Autor, natürlich!) für die Legalisierung des e.p.-Schlages verantwortlich ist (MR).
Valladao-Task. Bei dieser Ausgangsstellung mit Ansage (HL).

Versions at P1382816 & P1276701
A.Buchanan: Removing the superfluous sBf7 adds various non-castling lines which the AP convention duly protects us from, and makes the composition a sound miniature (2020-10-03)
comment
Keywords: Valladao Task, Castling, En passant as key, Promotion, a posteriori (AP) (Type Petrovic)
Genre: Retro, Fairies
Computer test: %Popeye FreeBSD-4.3-RELEASE-32Bit-Version 3.73
FEN: 4k2r/5p2/5P1P/5KpP/8/8/8/8
Reprints: Springaren (116) 03/2010
K3966 Problemkiste (199/200) 03/2012
Input: Erich Bartel, 2011-02-09
Last update: A.Buchanan, 2022-03-21 more...
64 - P1235136
Manfred Nieroba
K3965 Problemkiste (199/200) 03/2012
P1235136
(2+5) C+
h#2.5 (AP)
Circe
1. ... exf6ep[+sBf7] 2. 0-0 fxe7 3. Kh8 exf8=D#
play all play one stop play next play all
In the usual way with Circe, can refute the idea that Black just captured, so if Black can castle, White can ep. So AP Type Petrovic triggers. No retro try, and no set play.
more ...
comment
Keywords: Valladao Task, Circe, En passant as key, Castling, a posteriori (AP) (Type Petrovic), Minimal, Miniature
Genre: Fairies, h#, Retro
Computer test: HC+ Popeye 4.61 and analysis
FEN: 4k2r/4p2p/8/4PpK1/8/8/8/8
Input: Erich Bartel, 2012-03-25
Last update: A.Buchanan, 2022-06-07 more...
65 - P1288128
Gianni Donati
C0029 StrateGems (20) 10-12/2002
P1288128
(5+9) C+
ser-h#6 (AP)
b) wLa4->a3
a) 1. Tf7-f2 2. Tf2xb2 3. Kd4-c3 4. Kc3-c2 5. Kc2-b1 6. Lb3-a2 0-0#
b) AP 1. bxc3ep 2. Ke5 3. Kf6 4. e5 5. Te7 6. Te6 0-0#
play all play one stop play next play all
In b), under AP Type Petrovic, the ep is enabled by the later castling.
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)
comment
Keywords: En passant as key, Castling (wk), Seriesmover, a posteriori (AP) (Type Petrovic)
Genre: Fairies, Retro
Computer test: %Popeye FreeBSD-4.3-RELEASE-32Bit-Version 3.75
FEN: 8/4prq1/6p1/6p1/BpPk4/1b1p4/1P6/4K2R
Input: Erich Bartel, 2014-09-15
Last update: A.Buchanan, 2022-05-28 more...
66 - P1288435
Manfred Nieroba
V5 Problemkiste (157) 02/2005
P1288435
(4+6) C+
h#1.5 (AP)
sDU=Grashüpfer
1. ... dxe6ep 2. 0-0-0 a8=D#
play all play one stop play next play all
Durch den ep-Schlag beweist Weiß, dass der letzte Zug von Schwarz
e7-e5 gewesen ist. Die sGG können mangels Bock nicht gezogen haben
und hätten sK oder sT gezogen, so wäre die {\ooo} nicht mehr
möglich (Autor).
A.Buchanan: sGh8 seems useless as it merely eliminates the thematic retro tries. What am I missing? I wonder if the composer really understood the nature of AP...? (2022-03-22)
more ...
comment
Keywords: Valladao Task, En passant as key, Castling, a posteriori (AP) (Type Petrovic), Promotion in the mating move (D)
Pieces: du = Grasshopper (G)
Genre: Fairies, h#, Retro
Computer test: %Popeye FreeBSD-4.3-RELEASE-32Bit-Version 3.81
FEN: r3k2*2q/P4*2q2/*2q2P4/3PpK2/8/8/8/8
Input: Erich Bartel, 2014-09-22
Last update: A.Buchanan, 2022-03-22 more...
67 - P1288499
Manfred Nieroba
6452 Problemkiste (159-160) 07/2005
P1288499
(3+4) C+
h#2 AP
0.1...
Strict-Circe
1. ... cxd6ep[+sBd7] 2. 0-0-0 a8=D#
play all play one stop play next play all
Der sSh8 kann wegen Schach zuletzt nicht gezogen haben. Er verhindert zudem NLen.
Mangels Schlagobjekt kann der sB nur von d7 gekommen sein (Autor).

t/nur Themazüge,
A.Buchanan: 1. ... Ke6 Tb8 2. axb8=D/T[-]#? won't work because under "strict" Circe, a capture can't take place at all if the rebirth square is occupied. (2022-05-23)
A.Buchanan: Pieces never leave the board in strict Circe. "Hotel California". The diagram position is hence illegal and it's interesting that one can nevertheless get some kind of retro logic going. It points out a generally unspoken difference between "local" & "global" illegality. (1) Can we reverse moves indefinitely, or do we get stuck in some kind of reverse pat? (Can't say "retropat", because that implies that it's only pat with one player to move, as Henrik explained to me a while back.) (2) If the answer is yes then can we get back to the game array? These are different questions, and problems like this one rely on local rather than global legality. (2022-05-24)
more ...
comment
Keywords: Valladao Task, En passant, Castling, Circe (Strict), Promotion (D), a posteriori (AP) (Type Petrovic)
Genre: Fairies, h#, Retro
Computer test: %Popeye FreeBSD-5.3-RELEASE-32Bit-Version 3.81
FEN: r3k2n/P7/8/2PpK3/8/8/8/8
Input: Erich Bartel, 2014-09-26
Last update: A.Buchanan, 2022-05-26 more...
68 - P1288715
Bernd Schwarzkopf
A12 Problemkiste (169) 02/2007
P1288715
(4+3) C+
=5
AP-Priorität
1. dxc6ep 0-0-0 2. b7+ Kb8 3. c7+ Kxb7 4. cxd8=D Ka7 5. Dc8=
3. ... Ka7 4. cxd8=S Kb8 5. Ka6,Kb6=
play all play one stop play next play all
A.Buchanan: Guessing so-called "AP-Priorität" is a variant of AP whereby Black is forced to legitimize (or at least forbidden to rule out the possibility of legitimizing) the earlier e.p. (2020-10-03)
comment
Keywords: En passant, a posteriori (AP) (Type Petrovic), Castling, En passant as key, Promotion, Valladao Task
Genre: Retro, Fairies
Computer test: Popeye v4.85 + minor retro/AP thinking
FEN: r3k3/8/1P1P4/1KpP4/8/8/8/8
Input: Erich Bartel, 2014-09-29
Last update: A.Buchanan, 2022-03-21 more...
69 - P1299824
Georgi Hadzi-Vaskov
Länderkampf Mazedonien-Slowenien 1969
4. Platz
P1299824
(9+6) cooked
ser-h#10
1. bxc3ep 2. cxb2 3. Kc3 4. Kc2 5. Kc1 6. b1=T 7. Tb2 8. Kb1 9. Ka1 10. Txa2 0-0#
play all play one stop play next play all
Cook: 1. a3 2. axb2 3. b1=S 4. b3 5. Kc3 6. Kb2 7. Ka3 8. Ka4 9. Sa3 axb3#
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)
more ...
comment
Keywords: Seriesmover, En passant as key, a posteriori (AP) (Type Petrovic), Castling (wk)
Genre: Fairies, Retro
FEN: 8/8/1P6/PRP5/ppPk4/3pp2p/PP6/4K2R
Reprints: 101 Bilten 1970 1971
Input: Frank Müller, 2015-03-22
Last update: A.Buchanan, 2023-04-06 more...
70 - P1359077
Joaquim Crusats
13 Internet 12/2018
2. ehrende Erwähnung
JT-Sergej-Volobujev-60
Abteilung Retro
P1359077
(11+14)
h#2.5 (AP)
1. ... fxg6ep 2. 0-0-0 gxf7 3. Te8 fxe8=D#
(+ hordes of tries that omit castling)

R: 1. ... g7-g5! 2. Lh1-g2 Lh3-f1 3. Se7-d5 Lf1-h3 4. Sg6-e7 Lh3-f1 5. Sh8-g6 Lf1-h3 6. h7-h8=S Lh3-f1 7. h6-h7 Lf1-h3 8. h5-h6 Lh3-f1 9. h4-h5 Lf1-h3 & e.g. 10.h2/h3-h4 h4xg3
play all play one stop play next play all
Link zum Preisbericht auf "http://www.selivanov.world/en/newss/detail.php?ID=1151"
Official solution text: This and the following tasks are connected by the search of retrogame, in which the position on the diagram allows the execution of moves that reach the goal (a posteriori).
1...f:g6 e.p. 2.0-0-0 g:f7 3./e8 f:/e8s# - the only solution with e.p. on the first move, leading to mate, requires preservation of the right to castling of Black. The problem is to prove that the position in the diagram is legal, provided that Black can castling and the last move was Black's g7-g5.
White's balance is closed: 11 + h:g3 + d:c2 + c:d + a:b + a-w! = 16
Black's balance closed: 14 + e:d3 + d:e4 = 16
Retro game: -1...g7-g5! -2.oh1-g2 +h3-f1 -3.me7-d5 +f1-h3 -4.mg6-e7 +h3-f1 -5.mh8-g6 +f1-h3 -6.h7-h8m +h3-f1 -7.h6-h7 +f1-h3 -8.h5-h6 +h3-f1 -9.h4-h5 +f1-h3 -10.h2/h3-h4 h4:Xg3 and the position is released. X is knight, bishop, or queen.
Retropositional play of the white knight and black bishop provides the possibility of ep. wPa2 is waylaid on the a-file. Example realization: 1. a4 d6 2. a5 Lh3 3. g4 Sc6 4. Ta3 Sxa5 5. Tg3 Sb3 6. Sf3 Sa1 7. Se5 a5 8. b3 a4 9. Sd3 a3 10. Sb2 axb2 11. Lg2 e6 12. Lf3 d5 13. Lg2 La3 14. b4 h5 15. Lf1 Th6 16. Lg2 Tf6 17. Lf1 Tf3 18. Lg2 Tb3 19. Lf1 e5 20. d3 e4 21. Le3 Sf6 22. f4 Sg8 23. Thg1 Se7 24. T1g2 Sg8 25. Tf2 Se7 26. Tff3 Sg8 27. Lg1 Se7 28. Lg2 Sg8 29. Lh1 Se7 30. f5 Sg8 31. Tf4 Sf6 32. Kf2 e3+ 33. Kf3 Se4 34. dxe4 c5 35. Lf2 c4 36. Lg1 c3 37. Sd2 cxd2 38. c3 d4 39. Dc2 d3 40. Lf2 dxc2 41. Lg1 Dd3 42. exd3 e2 43. Kf2 Lg2 44. Te3 Lh3 45. Kf3 Lf1 46. Lf2 h4 47. Le1 Lh3 48. Lg3 Lf1 49. h3 hxg3 50. h4 Lh3 51. h5 Lf1 52. h6 Lh3 53. h7 Lf1 54. h8=S Lh3 55. Sg6 Lf1 56. Se7 Lh3 57. Sd5 Lf1 58. Lg2 g5 59. fxg6 O-O-O 60. gxf7 Te8 61. fxe8=D#


Author's solution from Problemas reprint 2022: On previous occasions we have already dealt in these pages with the ep convention, which establishes that, for the first move in a direct problem to be a capture ep, it must be possible to demonstrate by retro analysis that the last move was the double pawn jump.
It may happen that this demonstration is subordinate to the fact that a certain castling right is preserved. In this case, for the ep capture to be legitimate, in the subsequent direct game castling must be performed, which will conclusively prove that the en passant capture was possible. The codex states that, in these situations, the stipulation should indicate that the capture en passant will be justified by the subsequent castling, which is indicated by the term "a posteriori" below the diagram. Problem 1, which we use as an example, illustrates this concept.
It is a helpmate, and we see that the first move is made by White. And, indeed, for checkmate to occur, the capture at the passage of the black pawn on g5 is necessary. There are several continuations that allow Black to checkmate, but we must not forget that Black must castle during the process and it is here that the solution becomes unique: 1...fxg6 e.p. 2.0-0-0 gxf7 3.Te8 fxe8=D# (in helpmate notation).
But problem 1 is mainly an exercise in retro analysis. We must show that, if Black maintains the castling right, his last move was ...g7-g5. Let us look at the balance of the pieces. There are 14 black pieces on the board and the two missing ones can be shown to have been captured with c~d (= "cxd & dxc") or d~e. Five white pieces are missing and were eaten as follows: the a2 pawn was eaten on its file (if Black keeps his castling rights) and the other four pieces with hxg3, axb, d~e or c~d respectively.
We must realize that to open the cage at the bottom we have to retract ...hxg3, but this is not going to be possible until the white pawn on the h-file has moved back to the origin. To do so, we must retract the following moves: -1...g7-g5! -2.Ah1-g2 Ah3-f1 -3.Ce7-d5 Bf1-h3 -4.CSg6-e7 Ah3-f1 -5.Ch8-g6 Bf1-h3 -6.h7-h8=C Ah3-f1 -7.h6-h7 Bf1-h3 -8.h5-h6 Ah3-f1 -9.h4-h5 Bf1-h3 -10.h2/h3-h4 h4xXg3 and the tangle unravels. There is a retro opposition between the white knight and the black bishop and, since the white knight must pass the g6-square to unpromote, if this square is occupied by the black pawn (in case the last move would have been -1...g6-g5), essential time is lost, and the reader can see that the cage cannot be opened. Thus it is demonstrated that, if Black retains his right to castle, his last move was -1...g7-g5 and, therefore, White can capture ep.
A.Buchanan: Sketch of solution: 7 units missing. Black captured axb & hxg. In addition there must be captures onto c,d&e files. These can't all be by White, so one was by Black which by file parity implies a 4th capture by a black pawn, but not a 5th. The other cde capture was by White, which implies another to cross-capture with. So WbcBde or WcdBde or WcdBef or WdeBbc or WdeBcd or WefBcd. But bc & cd can't cross-capture alone. So WcdBde or WdeBcd, I'm not sure which. wPh promoted, so can't retract h4xg3 any time soon. wPa was waylaid on a-file by some Black officer. The position must unlock by uncapturing wSh8. (2022-02-18)
Henrik Juel: White pawn captures: e2xd3 and d3xe4
Black pawn captures: a4xb3, cxd, d3xc2, and h4xg3; [Pa2] was captured by a black officer
The retroplay could be
R: 1... g7-g5 2.Lh1-g2 Lh3-f1, wSd5 unpromotes on h8 and the pawn retracts, while the black bishop pendulates on f1 and h3, 9.h4-h5 Lf1-h3 10.h2-h4 h4xLg3 11.Kf2-f3 Lg2-f1 12.Kg1-f2 Lh3-g2 13.Tf3-e3 e3-e2 14.e2xDd3 Dd8-d3 15.Tf1-f3 d2xDc2 16.Dd1-c2 Lg2-h3 17.c2-c3 Tc3-b3 18.Tf3-f4 b3-b2 19.Le5-g3 a4xSb3 etc. (2022-07-04)
James Malcom: I've taken the liberty to edit the text such that the PG in a recognizable format.

Here is the English notation of it: 1. a4 d6 2. a5 Bh3 3. g4 Nc6 4. Ra3 Nxa5 5. Rg3 Nb3 6. Nf3 Na1 7. Ne5 a5 8. b3 a4 9. Nd3 a3 10. Nb2 axb2 11. Bg2 e6 12. Bf3 d5 13. Bg2 Ba3 14. b4 h5 15. Bf1 Rh6 16. Bg2 Rf6 17. Bf1 Rf3 18. Bg2 Rb3 19. Bf1 e5 20. d3 e4 21. Be3 Nf6 22. f4 Ng8 23. Rhg1 Ne7 24. R1g2 Ng8 25. Rf2 Ne7 26. Rff3 Ng8 27. Bg1 Ne7 28. Bg2 Ng8 29. Bh1 Ne7 30. f5 Ng8 31. Rf4 Nf6 32. Kf2 e3+ 33. Kf3 Ne4 34. dxe4 c5 35. Bf2 c4 36. Bg1 c3 37. Nd2 cxd2 38. c3 d4 39. Qc2 d3 40. Bf2 dxc2 41. Bg1 Qd3 42. exd3 e2 43. Kf2 Bg2 44. Re3 Bh3 45. Kf3 Bf1 46. Bf2 h4 47. Be1 Bh3 48. Bg3 Bf1 49. h3 hxg3 50. h4 Bh3 51. h5 Bf1 52. h6 Bh3 53. h7 Bf1 54. h8=N Bh3 55. Ng6 Bf1 56. Ne7 Bh3 57. Nd5 Bf1 58. Bg2 g5 59. fxg6 O-O-O 60. gxf7 Re8 61. fxe8=Q#

Now, time to create a more streamlined PG... (2022-07-04)
James Malcom: 42.0 moves is optimal to reach the diagram; the proof speaks for itself, I believe: 1. a4 Nc6 2. a5 Nxa5 3. Ra3 h5 4. f4 Rh6 5. Nf3 Rc6 6. Ne5 Nf6 7. Rf3 Nb3 8. Nc4 Na1 9. Kf2 e5 10. Rg1 d5 11. b3 Bh3 12. g4 a5 13. Nb2 a4 14. f5 a3 15. Kg3 axb2 16. d3 Rc3 17. Bg2 Ba3 18. b4 Rb3 19. Bh1 c5 20. Bd2 c4 21. Be1 c3 22. Nd2 cxd2 23. c3 e4 24. Qc2 e3 25. Rf4 Ne4+ 26. dxe4 d4 27. Rgf1 d3 28. R1f2 dxc2 29. R2f3 Qd3 30. exd3 e2 31. Re3 h4+ 32. Kf3 Bf1 33. Bg3 hxg3 34. h4 Bh3 35. h5 Bf1 36. h6 Bh3 37. h7 Bf1 38. h8=N Bh3 39. Ng6 Bf1 40. Ne7 Bh3 41. Nd5 Bf1 42. Bg2 g5 (2022-07-04)
comment
Keywords: Valladao Task, Promotion in the mating move (D), a posteriori (AP) (Type Petrovic), En passant as key, Castling (sg)
Genre: h#, Retro
Computer test: Forward play Popeye v4.87 but retro play requires substantial thinking in order to establish retro-opposition. Believed to be sound.
FEN: r3k3/1p3p2/8/3N1Pp1/1P2PRP1/brPPRKp1/1pppp1B1/n4b2
Reprints: Problemas 07/2022
Input: Mario Richter, 2019-01-03
Last update: James Malcom, 2022-07-04 more...
71 - P1382808
Ronald Turnbull
Andrew Buchanan

PDB Website 20/01/2012
RT, correction AB
P1382808
(4+12)
h#3
AP
PRA
1. bxc3ep 0-0! (for AP reasons, not 1...Kf2?/Th2? etc) 2. Ta1 Txa1 3. Kc4 Txa4#
1. fxe3ep 0-0! (for AP reasons, not 1...Tf1?/Txh3? etc) b2 Txf3 3. Tc3 Tf4#
play all play one stop play next play all
For AP as for anything else, the default meta-convention is PRA before RS. Here we use AP to lock down wK & wTh1, so the last move was R: 1. c2-c4 or 1. e2-e4. There is a different h#3 in either case, beginning with ep and then 0-0.
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)
comment
Keywords: a posteriori (AP) (Type Petrovic - cee), En passant as key (2), Castling (wk), Retro Strategy (RS), Golden Age (AP unconsolidated)
Genre: h#, Retro
Computer test: Forward logic sound by Popeye v4.85 However, seems that AP logic is not sound
FEN: 8/8/8/1nr1n3/ppPkPp2/rp1p1p1p/8/4K2R
Input: A.Buchanan, 2020-12-08
Last update: A.Buchanan, 2023-04-06 more...
72 - P1399112
Werner Kuntsche
Valery Liskovets
Andrew Buchanan

PDB Website 14/02/2022
WK, correction AB
P1399112
(5+8) C+
h#3 AP

No solutions except for those beginning with e.p.
297 candidate solutions begin 1.gxf3ep, but only one includes castling. Queenside 2. ... 0-0-0
22 candidate solutions begin 1.fcxd3ep, but only one includes castling. Kingside 2. ... 0-0
In some way, these are trying to be combined so that from AP perspective, White's last move can only have been double hop with wBd or wBf.
We need a clear delineation of the precise PRA-AP algorithm to be followed.

The point: if this problem is "sound", then surely P1382808 is "unsound", because there only one of the two parts needs to prove that White castling rights remain, so there are 4 cooks.
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.
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key (2), Symmetrical position, Symmetrical solution, Castling (wb)
Genre: h#, Retro
Computer test: HC+ Forward play proved by Popeye v4.87 AP logic requires some framework but I think we know we want this cool idea to end up being sound.
FEN: 8/8/2p3p1/8/2pPkPp1/2n1p1n1/8/R3K2R
Input: A.Buchanan, 2022-02-14
Last update: A.Buchanan, 2022-05-24 more...
73 - P1400896
Tivadar Kardos
Andrew Buchanan

Discord Chess Problems & Studies Server 26/04/2022
TK, correction AB
P1400896
(7+15) C+
h#3 AP
1. gxf3ep+ e3 2. 0-0-0 Tb4 3. Tg4 Tb8#
play all play one stop play next play all
Black has visibly made 6 pawn captures right to left. Two captures are required to resolve the h-file (either two pcs by black, or wPgxh with wPh waylaid). If black preserves castling rights, then White pawns on a & d files require one more capture by either side to resolve. So all 10 captures accounted for. If wKc1 then bPdxc but this implies 2 more captures, only one of which can be taken from d file. Similarly wPfxexf too expensive. So If black can castle, then the last move was indeed pawn double hop.

The castling serves two purposes: justifying the ep capture and enabling the mate: there are no retro tries in this problem.
Author: Kardos' cooked P0003269 was from 1956 just after AP compositions began to appear, and I can't decide whether he intended to include AP but it seems to fit this corrected version.
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (sg), En passant as key
Genre: h#, Retro
Computer test: C+ Popeye 4.61 and analysis
FEN: r3k3/3q4/5p2/5pbP/n4PpR/1pp3rp/2p1P1pB/N1K3n1
Input: A.Buchanan, 2022-04-26
Last update: A.Buchanan, 2023-09-11 more...
74 - P1401450
Gerhard E. Schoen
511 Chessics 13 01/1982
P1401450
(4+5) cooked
h#3 A.P.
1. dxc3ep e3 (Td1?) 2. Db5 0-0-0 (Td1?) 3. Kc4 Td4#
play all play one stop play next play all
Cook: 1. Df3,Dh3 0-0-0 2. Ka3 Kc2 3. Tb4 Ta1#
1. Dxc4 Ta2,Kd1 2. Kc3 Kd1,Ta2 3. Db4 Tc2# etc
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (wg), En passant as key, Superseded by (P1000348)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro logic
FEN: 8/8/8/2n5/1kPp4/1r1q4/4P3/R3K3
Reprints: feenschach 137, p. 365, 08/2000
Input: A.Buchanan, 2022-05-24
Last update: A.Buchanan, 2022-05-24 more...
75 - P1401458
Miroslav Stosic
Delo-Tovaris 1970
P1401458
(4+5) C+
ser-h#6 (AP)
b) after the key: ser-h#5
a) 1. axb3ep 2. b2 3. Kb3 4. Kc2 5. Kc1 6. c2 0-0# not 6...Kf2 because the ep 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
play all play one stop play next play all
Paradox of different last move in solutions
more ...
comment
Keywords: Seriesmover, a posteriori (AP) (Type Petrovic), Castling (wK), En passant as key
Genre: Retro, Fairies
Computer test: HC+ Popeye 4.61 and analysis
FEN: 8/8/8/8/pPk5/p1pp4/8/1N2K2R
Input: A.Buchanan, 2022-05-26
Last update: A.Buchanan, 2022-05-27 more...
76 - P1401466
Gabor Tar
Best Problems 101 01/2022
P1401466
(4+8) C+
h#2 (AP)
Circe
2 solutions
1. fxg3ep[+wBg2] 0-0 (Tf1?) 2. Lxd5[+wBd2] Tf4#
1. Lfxg4[+wBg2] Kf2 2. f3 g3#
play all play one stop play next play all
One solution is orthodox, the other is AP. Under Type Petrovic, the ep is permitted by the later castling
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comment
Keywords: a posteriori (AP) (Type Petrovic), Circe, Castling (wk), En passant as key
Genre: h#, Retro, Fairies
Computer test: HC+ Popeye v4.87 & some thinking
FEN: 8/8/8/3P2np/3p1pPk/5b1r/8/2b1K2R
Input: A.Buchanan, 2022-05-26
Last update: A.Buchanan, 2022-05-28 more...
77 - P1401467
Rauf Aliovsadzade
Seven Chess Notes 2009
Special Mention of Honour
P1401467
(4+3) C+
h=2.5 (AP)
Monochromatic
1. ... hxg6ep 2. 0-0 Kg5 3. Txf6 Kxf6=
play all play one stop play next play all
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)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), Monochromatic Chess, Castling (sk), En passant as key, Miniature
Genre: Retro, Fairies
Computer test: HC+ Popeye 4.61
FEN: 4k2r/8/5P1R/6pP/7K/8/8/8
Input: A.Buchanan, 2022-05-26
Last update: A.Buchanan, 2022-05-27 more...
78 - 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...
79 - P1401480
Jean-Michel Trillon
v Rex Multiplex 2 04/1982
to C.Caminade
2nd Honourable Mention
P1401480
(6+8) cooked
#4 AP
Maximummer
b) g6->g5
a) 1. Ld6! Td8 2. fxg6 Txd5 3. g7 Th5 4. g8=D#
b) 1. fxg6ep Txd5 2. Le5 0-0-0 3. g7 Th8 4. gxh8=D/T#
play all play one stop play next play all
a) Black had a candidate last move b7-b5 of length 2. The only way this might not have been played is if bK was in check from e8 namely Kd8xDTe8,Kf7xDe8. So certainly Black castling rights are lost. With this known, there is still a unique solution.
b) Again, b7-b5 looms, and this time if Black castling rights are lost, there is no solution. The only possible way we can operate is if the last move was g7-g5. But we are only permitted to play this if we can show that the last move *was* not one of Kd8xDTe8,Kf7xDe8. I.e. under AP we need to castle after the fact. There is a unique solution after the ep, which does indeed include castling.
Cook: b) R: 1. Th4xDh5,Th4xLh5,Th6xDh5,Th6xLh5
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)
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comment
Keywords: a posteriori (AP) (Type Petrovic), Maximummer, Castling (sg), En passant as key, Promotion (D), Valladao Task, Superseded by (P1401566)
Genre: n#, Retro, Fairies
Computer test: Popeye v4.87 & not-so-simple retro-logic to identify the cook
FEN: rB2k3/1p1p3p/p5p1/3P1P1r/8/7P/3P4/K7
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2022-06-02 more...
80 - P1401481
Janko Furman
1502 Mat (Belgrade) 04/1976
P1401481
(15+9) C+
ser-h#6 (AP)
1. dxc3ep Lb5#? etc because under AP, only 0-0-0 justifies earlier ep
1. dxc3ep 2. 0-0-0! 3. Kc7 4. Ta8 5. Kd8 6. Ke8 Lb5#
play all play one stop play next play all
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 retro-play starting with R: 1.c2-c4 c3xDb2
So if Black may castle, he may also capture ep now.
Popeye quickly shows (1) there are no solutions without immediate ep, and (2) there is exactly one solution after 2. 0-0-0. However trickier to demonstrate that there is no solution in which the castling comes later (about 900K candidates searched).
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)
comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (sg), En passant as key, Seriesmover
Genre: Retro, Fairies
Computer test: HC+ Popeye v4.87 & simple retro-logic & MS Excel to search 900K candidate solutions
FEN: r3kb2/4p1p1/4P3/2P1PP2/1NPpKpN1/BP1B1R2/pp1RPP2/8
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2023-01-29 more...
81 - P1401482
Janko Furman
Ljubomir Ugren

British Chess Magazine 1971-72
2nd Prize
P1401482
(15+9)
ser-h#7 (AP)
1. fxg3ep 2. 0-0 3. Txf5 4. Kf8 5. Ke8 6. Tf8 7. Th8 Lh5#
play all play one stop play next play all
The sole Black pawn capture was gxh so White made 6 pawn captures in files a-f (even though bPd & bPe are missing). One missing Black unit was not captured by a pawn: what happened? Spot that wPb4 must have captured from c3, and wPa4 from b3. Then have two more captures axb & bxc. Meanwhile there are also two captures on e&f files. So bPd was waylaid!
White's last move must have been g2-g4 to permit bPg3xQh2 to be undone, so ep is on if Black retains 00 rights, demonstrated by castling in the forward play.
Henrik Juel: 1.fxg3ep 2.0-0 3.Txf5 4.Kf8 5.Ke8 6.Tf8 7.Th8 Lh5#
Another 'uncastling' (2022-05-27)
comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (sk), En passant as key, Castling Paradox (hidden), Seriesmover
Genre: Retro, Fairies
FEN: 4k2r/7p/8/4NPB1/PPP1KpP1/pprPP2P/1RpNBR1p/8
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2023-01-29 more...
82 - P1401494
Ivan Skoba
HC232 ChessProblems.ca 17 12/2019
P1401494
(6+5) cooked
ser-h#6 (AP)
b) after the key ser-h#5

play all play one stop play next play all
Cook: a) is cooked by e.g. 1. Th2 2. Txf2 3. Kg3 4. Kg2 5. Kg1 6. Th2 0-0-0#
This seems to be later version of P0003925. However it's cooked and I don't see any way to fix it except to revert to the earlier form, which was perfectly adequate.
A.Buchanan: Have asked Cornel Pacurar at ChessProblems.ca. (2022-06-01)
A.Buchanan: Cornel agrees that it's cooked. (2022-06-29)
comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (wg), En passant as key, Seriesmover
Genre: Retro, Fairies
Computer test: Popeye v4.87 & simple retro-logic to identify the cook
FEN: 8/8/8/8/5pPk/4bp1r/3PRP2/R3K3
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2023-01-29 more...
83 - 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...
84 - P1401497
Mario Velucchi
Scacco! 1998
P1401497
(5+5) C+
h#2.5 (AP)
1. ... cxb6ep
Not 2. Ta7? bxa7 3. Kd8 axb8=D#
But 2. Sxa6! Ld6 3. 0-0-0 b7# AP
play all play one stop play next play all
Under AP Type Petrovic, the ep is enabled by the later castling
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), Valladao Task, Castling (sg), En passant as key, Promotion (D)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: rn2kB2/3p4/P1P5/KpP5/8/8/8/8
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2022-05-28 more...
85 - P1401525
Joao Baptista Santiago
Andrew Buchanan

PDB Website 30/05/2022
JBS, correction AB
P1401525
(11+14) C+
h#2* (AP)
1. ... Kd2 2. e3+ fxe3#
1. bxc3ep b3 2. Kxd3 0-0-0#
play all play one stop play next play all
Black has no tempo moves to convert the set play to a main play solution. Rc5~,Se5~,e3,Sf1 all disrupt the set play.
The main play castling is needed both for the mate and for the legitimization.
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)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (wg), En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 and analysis
FEN: 8/8/pp2P3/brr1n2P/1pPkp1p1/B2P1pPp/1P3Ppn/R3K1N1
Input: A.Buchanan, 2022-05-30
Last update: A.Buchanan, 2022-05-30 more...
86 - P1401526
Zoltan Laborczi
Gabor Tar
Andrew Buchanan

PDB Website 30/05/2022
ZL & GT, version AB
P1401526
(4+8) C+
h#2 (AP)
3.1...
1. Tb3 Lxc4 2. Txb4 Txa2#
1. Kxb4 Tb1+ 2. Ka4 Lb5#
1. cxb3ep 0-0-0! (Td1?) 2. e4 Td4#
play all play one stop play next play all
Economizes P1080354 by 2 units, adding a solution.
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (wg), En passant as key
Genre: h#, Retro
Computer test: C+ Popeye 4.61
FEN: 8/1p6/B7/p3p3/kPp5/r1p5/p7/R3K3
Input: A.Buchanan, 2022-05-30
Last update: A.Buchanan, 2023-05-21 more...
87 - P1401566
Jean-Michel Trillon
Siegfried Hornecker
Andrew Buchanan

Discord Chess Problems & Studies Server 02/06/2022
J-MT, correction SH & AB
P1401566
(7+8) C+
#4 AP
Maximummer
b) g6->g5
a) 1. Ld6 Td8 2. fxg6 Txd5 3. g7 Th5 4. g8=D#
b) 1. fxg6ep Txd5 2. Le5 0-0-0 3. g7 Th8 4. gxh8=D,T#
play all play one stop play next play all
What did Black play last, and why not b7-b5?
a) Can only be Kxe8 to escape check. Thus castling rights are definitely lost.
b) Also the possibility of g7-g5, so 000 rights can remain. This breaks the earlier solution but allows another solution in which AP retro-justifies ep with 000, which is also necessary for the mate. The solution beginning 1.Ld6 does not work any more, because under RS convention, castling is legal and so must be played.
FYI: here are two non-unique games validated by Jacobi v0.7.5 to prove the legality of the diagram positions. 000 rights remain in b, while in a are retained until the penultimate move.
a) 1.e4 Sf6 2.a4 Sd5 3.exd5 Sc6 4.a5 Sb4 5.h4 Sd3+ 6.Ke2 Sxb2 7.g3 Sc4 8.Bb2 Sa3 9.Be5 Sxc2 10.Bxc7 Sb4 11.Sc3 Sa6 12.Se4 Sb8 13.Sg5 Sa6 14.Se6 Sb8 15.Sxd8 Sa6 16.Bb8 Sb4 17.Sxf7 Sa6 18.Sh6 Sc5 19.Ra4 Sa6 20.Rc4 Sc7 21.Rc6 Sa6 22.Rd6 Sb4 23.Sf3 Sa2 24.Sg5 Sb4 25.Se6 Sa2 26.Sxf8 Sb4 27.Sg6 Sa6 28.Sxe7 Sb4 29.Qc2 Sa6 30.Kd3 Sc7 31.Be2 Sa6 32.f4 Sc7 33.Rb1 Sa6 34.Rb5 Sc7 35.Kc3 Sa6 36.Kb2 Sc7 37.Ka1 Se6 38.Bg4 Sd8 39.Rf6 Sc6 40.Be6 Sd8 41.Sf7 Sc6 42.Sg5 Sd8 43.Qxc8 Rf8 44.Bg4 Rh8 45.Sf5 Rf8 46.Sd4 Rh8 47.Sde6 Rf8 48.Kb2 Rxf6 49.Ka1 Rh6 50.Bh5+ Rxh5 51.Sd4 g6 52.g4 a6 53.Qc1 Sc6 54.f5 Sxd4 55.Rb3 Sxb3+ 56.Ka2 Sxa5 57.Ka1 Sc6 58.Ka2 Se5 59.Ka1 Sf7 60.Ka2 Sxg5 61.Ka1 Sf7 62.Qc8+ Sd8 63.Qxd8+ Kf7 64. Qe8+ Kxe8
b) 51. … a6 52.Qc1 Se6 53.Qc2 Sxg5 54.f5 Sf3 55.g4 Sxd4 56.Rb3 Sxb3+ 57.Ka2 Sxa5 58.Qb1 Sb3 59.Qa1 Sxa1 60.Kxa1 g5
Corrects P1401480.
Henrik Juel: a) 1.fxg6 Txd5 2.Le5 0-0-0 3.g7 Th8 4.gxh8=DT#
b) 1.fxg6ep Txd5 2.Le5 0-0-0 3.g7 Th8 4.gxh8=DT# (2022-06-02)
Henrik Juel: a) is C+ Popeye 4.61, except for the promotion mate dual; no AP here
In b) Popeye with 'opt enp g6' gives the intended solution; here the castling is needed to rule out the possibility of Kf7xDe8 as last move
But how do we know that last move was not f6xg5 or h6xg5? (2022-06-02)
A.Buchanan: Hi Henrik thanks for examining this.
In a), the last move must have been Kxe8 to deal with checking piece. Otherwise black would have played e.g. b7-b5. If you forbid castling in Popeye you will see a completely different solution.
In b) hxg5 or fxg5 is too short to compete with b7-b5. (2022-06-02)
Henrik Juel: Yes, I missed b7-b5
Popeye with 'opt noc a8' gives this solution for a) 1.Ld6 Td8 2.fxg6 Txd5 3.g7 Th5 4.g8=D#
Very nice, with analysis in both parts (2022-06-02)
comment
Keywords: a posteriori (AP) (Type Petrovic), Maximummer, Castling (sg), En passant as key, Promotion (D,D/T), Valladao Task, Retro Strategy (RS)
Genre: h#, Retro, Fairies
Computer test: HC+ Popeye v4.87 for forward logic, Jacobi v0.7.5 to validate demo games & humans for non-trivial retro-logic
FEN: rB2k3/1p1p3p/p5p1/3P1P1r/6PP/8/3P4/K7
Input: A.Buchanan, 2022-06-02
Last update: A.Buchanan, 2022-06-03 more...
88 - P1406456
Werner Keym
OP016 The Hopper Magazine I02 25/06/2022
corrects Keym Schach-Echo 1967 / Die Schwalbe 2007
P1406456
(15+5)
#1 AP
b) - Lh8
a) 1. ... bxa3ep 2. Dc3#!
(2. 0-0#?)
1. 0-0#? (rights lost)

b) 1. ... bxa3ep 2. 0-0#!
(2. Dc3#?)
1. 0-0#? (Black to move if rights remain)
play all play one stop play next play all
a) White & Black have made 11+1 pawn captures, accounting for all missing material. bPa captured to b file (must be bPb2 now, to get behind wPb) so bPh promoted on h1, to supply balance. Therefore White cannot castle. Black has no last move, so by Article 15 we can conclude that Black has the move. The only way to give Black a move is R: 1. a2-a4 a3xNb2. Therefore ep is legal, and indeed that is Black's only move. After 0... bxa3ep, White mates with 1.Dc3#! not 1.0-0#? since castling is illegal. AP is not used in this twin, but is available as a condition in order to make twins as similar as possible, and show that the difference in behaviour comes just from wBh8, not from conditions.
b) Now there is a second white unit missing, which gives more retro possibilities. One case is that bPh might have captured to g-file, to promote (or to be captured if original bPg promoted), so White may retain castling rights, but in this case it must be Black to move. So 1.0-0#? doesn't work as a solution. Alternatively, if White has lost castling rights, Black might have just played c2xNb1=R, or c3xNb2 or a/c5xb4. So in this scenario we cannot prove unconditionally that it's really Black to play.
However, under adversarial AP, White makes a double claim: firstly that it is Black to move (AP Type Keym), and secondly that ep is legal (AP Type Petrovic). If White retains castling rights, then both of these items are correct, and the ep must be bxa3 not bxc3. So White mates with 1.0-0#! for AP justification, not 1.Dc3#?.
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)
more ...
comment
Keywords: a posteriori (AP) (Type Keym), a posteriori (AP) (type Petrovic), No legal last move for Black, En passant as key, Castling, Cant Castler (wk)
Genre: Retro
FEN: 7B/2P5/3P4/1P2Q3/PpP1N3/1P1P4/1p1RBP2/brk1K2R
Input: A.Buchanan, 2022-12-05
Last update: A.Buchanan, 2023-08-22 more...
89 - 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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