Die Schwalbe

508 problem(s) found in 7583 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT K='nur eine Steinart' AND K='En passant'] [download as LaTeX]

1 - P0000016
Nikita M. Plaksin
Alexander Kislyak

(F) Die Schwalbe 98 04/1986
P0000016
(11+9)
Welches war der erste Zug der beiden Könige?
R: 1. bxc3ep+ c2-c4 2. b5-b4+
play all play one stop play next play all
Erster K-Zug: Weiß: 0-0; Schwarz: 0-0
173. Thema-Turnier
Henrik Juel: Good motivation for the black castling
The further retroplay includes retracting Tf1-g1, Kd3 to g1, h2xTg3, Tg3 to f8, La6 to c8, d7xTSe6, and e6xPf7 (2021-04-19)
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comment
Keywords: En passant, First Move?, Last Moves? (3), Castling (wksk), Castling in the retro play, En passant in the retro play, Volet Pawn, Type C
Genre: Retro
FEN: 7k/1p2pPpp/b3p3/8/8/2pK2P1/pP1PPPP1/2B3RN
Reprints: 566 Ukrainisches Album 1986-1990
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-20 more...
2 - 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...
3 - P0000052
Andrey Frolkin
(V) Die Schwalbe 102 12/1986
P0000052
(16+11)
Vor mindestens 72 Einzelzügen mußte ep geschlagen werden!
R: 1. ... Sg2-h4+ 2. Sh4-g6+ h7-h6 3. Se2-g1 a3-a2 4. Tg1-f1 a4-a3 5. Sf1-h2 Sh2-g4 6. Lh6-g5 Sg4-h2 7. Lg7-h6 Sh2-g4 8. Lf8-g7! Sg4-h2 9. Ld6-f8 Sh2-g4 10. Lb8-d6 Sg4-h2 11. Sc3-e2 Sh2-g4 12. Sb5-c3 Sg4-h2 13. Sd6-b5 Sh2-g4 14. Sf7-d6 Sg4-h2 15. b7-b8=L Sh2-g4 16. b6-b7 Sg4-h2 17. b5-b6 Sh2-g4 18. b4-b5 Sg4-h2 19. b3-b4 Sh2-g4 20. Lg7-h8 Sg4-h2 21. Lf8-g7 Sh2-g4 22. Ld6-f8 Sg4-h2 23. Lb8-d6 Sh2-g4 24. b7-b8=L Sg4-h2 25. b6-b7 Sh2-g4 26. b5-b6 Sg4-h2 27. b4-b5 Sh2-g4 28. a3xBb4 Sg4-h2 29. Sg5-f7 Kh6-h5 30. Th2-h3 Kh5-h6 31. Sh3-g5 b5-b4 32. Sg6-h4 b6-b5 33. Se7-g6 b7-b6 34. Sg8-e7! a5-a4 35. g7-g8=S a6-a5 36. g6-g7 Kh6-h5 37. h5xg6ep+ g7-g5 38. Sg5-h3+ a7-a6 39. Th3-h2 Sh2-g4 40. Dh4-f4 Sf4-g2 41. Lg2-h1 Sg4-h2 42. Th1-h3 Sh2-g4 43. Lh3-g2 Sg4-h2 44. g2-g3 Sh2-g4 45. Dg4-h4 Tg3-f3 46. Dd1-g4
play all play one stop play next play all
5 weiße Schlagfälle durch Bauern: axb, c3xd4, exf, f5xe6 & hxg; zwei Umwandlungen auf b8 und eine auf g8. Eine schwarze Umwandlung auf c1. Die weißen Figuren, die zur Entwandlung zurückschreiten können, sind die schwarzfeldrigen Lh8 und g5 sowie der retrofreie Sg1. Aber vor jeder Entwandlung muß Weiß das Feld h2 räumen, um dem Sg4 das Pendeln zu ermöglichen und so ein schwarzes Retropatt zu verhindern.
Retro: 1. Sg2-h4+ Sh4-g6+ 2. h7-h6 Se2-g1 3. a3-a2 Tg1-f1 4. a4-a3 Sf1-h2 5. Sh2-g4 Lh6-g5 6. Sg4-h2 Lg7-h6 7. Sh2-g4 Lf8-g7! 8. Sg4-h2 Ld6-f8 9. ~ Lb8-c6 10. ~ Sc3-e2 11. ~ Sb5-c3 12. ~ Sd6-b5 13. ~ Sf7-d6 14. ~ b7-b8=L 15. ~ b6-b7 16. ~ b5-b6 17. ~ b4-b5 18. ~ b3-b4 19. ~ Lg7-h8 20. ~ Lf8-g7 21. ~ Ld6-f8 22. ~ Lb8-d6 23. ~ b7-b8=L 24. ~ b6-b7 25. ~ b5-b6 26. ~ b4-b5 27. Sh2-g4 a3xb4
Nach diesem Entschlag des schwarzen Bb führt die dritte Entwandlung auf g8 zu einem erzwungenen En-passant-Schlag. 28. Sg4-h2 Sg5-f7 29. Kh6-h5 Th2-h3 30. Kh5-h6 Sh3-g5 31. b5-b4 Sg6-h4 32. b6-b5 Se7-g6 33. b7-b6 Sg8-e7! 34. a5-a4 g7-g8=S 35. a6-a5 g6-g7 36. Kh6-h5 h5xg6ep+ 37. g7-g5 Sg5-h3+ 38. a7-a6 Th3-h2 39. Sh2-g4 Dh4-f4 40. Sf4-g2+ g2-g3 41. Tg3-f3
Diese Zugfolge mit möglichen Zugumstellungen kann nicht verkürzt werden. Mindestens 71 Einzelzüge sind nach dem En-passant-Schlag geschehen. Rekord für eines der Themen des 173. Thematurniers der "Schwalbe".
paul: This problem obtained first Prize in the informal tourney, see Die Schwalbe 249/2011 (judge M. Caillaud) (2020-01-06)
Henrik Juel: 41.g2-g3 is illegal, because now wLh1 cannot retract (2021-02-03)
Henrik Juel: 41.Lg2-h1 Sg4-h2 42.Th1-h3 Sh2-g4 43.Lh3-g2 Sg4-h2 44.g2-g3 Sh2-g4 45.Dg4-h4 Tg3-f3 46.Dd1-g4 seems to work (2021-02-03)
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Keywords: En passant, En passant in the retro play, Non-standard material, Promotion, Last Moves? (72)
Genre: Retro
FEN: 7B/8/4PPNp/4pKBk/3PrQnn/3pBrPR/p2P1p1N/4bRNB
Reprints: 568 Ukrainisches Album 1986-1990
H21 FIDE Album 1986-1988 1995
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-02-04 more...
4 - P0000058
Leonid M. Borodatow
5758v Die Schwalbe 103 02/1987
P0000058
(13+9) C+
h#3
b) sBa6 statt sLa6
a) 1. ... exf6ep 2. 0-0-0 Lxf4 3. Td7 a8=D#
b) 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
play all play one stop play next play all
Anton Baumann: Mattdual in b): 3.Tb8 axb8=D,T# (2022-12-16)
A.Buchanan: So Borodatow got it all working! In a) Black might have captured hxgxf and axPb. So castling rights might still be maintained with the ep. In b) on the other hand, it must be axb, bxa, exf and wPgxh6, so there was the cage. Promotion to TD is tolerated in the final move, although it may not be puristic, without the convention, too many mates would be excluded (2022-12-16)
comment
Keywords: En passant as key, Castling (sg), Cant Castler, Valladao Task
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & basic retro thinking
FEN: r3k3/P7/b3r1pP/4PpBP/3nnpKR/5PRB/5PP1/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
5 - 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...
6 - P0000254
Leonid M. Borodatow
7642 Die Schwalbe 133 02/1992
P0000254
(16+10)
Welches waren die letzten 7 Einzelzüge, wenn dabei keine Zugwiederholungen vorkamen?
R: 1. 0-0-0# Ke4-d4 2. e5xf6ep+ f7-f5 3. Tg6-b6+ Kf5-e4 4. c7-c8=L
play all play one stop play next play all
Die von einigen Lösern angeführte Abweichung 2. f5-f6+ Kd4-e4 3. Lh6-g7+ (und mehrdeutig weiter) ließe sich durch die Erweiterung '... keine Zugwiederholungen und keine Pendelzüge ...' (mühsam) kitten. Beim Autor hieß es bei dieser ich-weiß-nicht-wie-vielten Fassung nur 'letzte 9 (!) Einzelzüge ohne Wiederholung).
HHS meint ohnehin, daß es das ganze auch ohne die einengende Zusatzbedingung schon gibt.
Das von einem Löser angegebene 1. Ld3-h7# Th1-h8 2. Lh8-g7 Tg1-h1 3. Se1-g3 g2-g1=T 4. Th7-h8=L scheitert allerdings an der Schlagbilanz.
Anton Baumann: Neufassung vergl. P0006288 (2023-01-06)
comment
Keywords: En passant, Last Moves?, Non-standard material, Castling (wl), Promotion (L), Valladao Task (WWW)
Genre: Retro
FEN: qrB2brr/Bp2p1BB/pR3P2/1Q6/2Pk1P2/B1p2R2/2P3N1/2KR1N2
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2019-08-11 more...
7 - P0000324
Josef Haas
8259 Die Schwalbe 143 10/1993
P0000324
(7+5)
a) Wer setzt in 1 Zug matt?
b) Auf welchem Feld muß ein schwarzer Bauer eingefügt werden, damit die andere Partei als in a) mattsetzt?
b) (+sBc7) 1. ... Lg8xe6#
a) 1. Tg6#
1) R: 1. ... Kg6xBf6! 2. g5xf6ep++ f7-f5 3. La2-b1+
play all play one stop play next play all
"Vermutlich aus der Kleinkunstkiste des Autors hervorgekramt.
a) sollte einfach formuliert sein: 'Matt in 1 Zug' - denn wie es hier heißt, klingt es als ob nur einer mattsetzen kann. Das aber ist nicht der Fall, denn beide können's: 1. ... Lxe6# und 1. Tg6#. Üblicherweise hat Weiß das Prae und kann darauf pochen, den Schwarz hat einen altklassischen letzten Zug: 1. ... Kg6xBf6! (nebst 2. Bg5xBf6ep++ Bf7-f5 3. La2-(x)b1)" (HHS);
also ist Weiß am Zug und setzt matt mit 1. Tg6#.
b) Nach Einfügen eines sBc7 geht die o.g. Rückzugfolge nicht, weil der wK nicht auf die 8. Reihe gelangen kann. Also Schwarz am Zuge und 1. ... Lxe6#
"Allzubekanntes - kein Problem für Schwalbelöser" (HHS)
Wenn das alles so bekannt ist, erstaunt doch sehr, daß nur drei Löser die Autorintention nachvollziehen konnten. Alle anderen Löser (5) kamen zu genau entgegengesetzten Erkenntnissen (in a) setzt Schwarz matt, in b) Weiß), was wohl durch die nicht ganz konventionelle Formulierung suggeriert wurde. Ich find's ein interessantes Beispiel für Massenhypnose! (GL) 2/I/3L.
vergl. P0004915 (Hans Gruber, Schach 1979)
Brassaud: La solution proposée 1/Tg6# est possible
Mais il y a aussi le rétro jeu -1) Fa2-b1, Rg5g6 -2) Ta4-a5+, Rf4-f5 etc … et avec le trait aux noirs : 1) Fxe6 # est possible (2017-08-30)
A.Buchanan: @Brassaud: yes I agree. There is no reason why White should not have moved last. So both players can mate, but part (b) implies that the intended solution in (a) is 1 player. If the published stipulation for (a) was maybe just "#1", which by default is white to move, then there is a unique solution.
For (b) I am wondering about +sBg6, which would also stop the en passant trick, both by blocking sK from retreating there and also by locking sL in an impossible cage with sBf7. (2017-08-31)
Henrik Juel: Adding a black pawn on g6 of course prevents a black last move by Kf6, but it allows f7xg6 as last move; Lg8 is not locked, because Ph7 is white (2017-08-31)
A.Buchanan: Yes (2017-08-31)
Anton Baumann: vergl. P0004915 (Hans Gruber, Schach 1979) (2023-01-03)
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Keywords: Add pieces, No legal last move for Black, En passant in the retro play
Genre: Retro
FEN: 4K1br/1p4pP/4Pk2/R7/3P4/8/8/1B4R1
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-01-03 more...
8 - 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...
9 - 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...
10 - P0000604
Andrej N. Kornilow
3948 Die Schwalbe 75 06/1982
7. Lob
P0000604
(24+0)
Färbe die Steine!
Welches waren die letzten 11 Einzelzüge?
James Malcom: Solution? (2020-11-07)
Mario Richter: I do not know the official solution, but the following works:
wLf8 wBe7 wBg6 wBh6 wLh5 wBe4 wTf4 wKg4 wDh4 wSf3 wDg3 wLh3 wSf2 wBg2 wSh2
sBb7 sBc7 sTf7 sBh7 sBd6 sBe6 sKf6 sBe5 sBa3
R: 1. f5xg6ep g7-g5 2. Sg5-f3 a4-a3 3. Kf3-g4 a5-a4 4. Tg4-f4 a6-a5 5. f4-f5 Kf5-f6
6. e3-e4
wCaps: f5xg6ep d7xTe8=L c6xLd7 d6xDe7 b6xLa7 a5xSb6 b6xSa7
wProms: d7xTe8=L a7-a8=S a7-a8=D
sCaps: f6xTe5 (2020-11-10)
comment
Keywords: Colouring problem, En passant, Last Moves? (11)
Genre: Retro
FEN: 5B2/1PP1PR1P/3PPKPP/4P2B/4PRKQ/P4NQB/5NPN/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
11 - 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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comment
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...
12 - 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...
13 - 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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comment
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...
14 - 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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comment
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...
15 - 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...
16 - 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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comment
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...
17 - P0000775
Ferad Kakabadze
1422 Die Schwalbe 30 12/1974
P0000775
(11+5) C+
#3
Welches waren die letzten 2 Einzelzüge?
1. e8=D! f5 2. exf6ep e5 3. Dxe5#
R: 1. b2-b1=L 0-0
play all play one stop play next play all
Henrik Juel: The forward play is C+ Popeye 4.61
1.e8=D thr. 2.Dc8 thr. 3.Dc3#
1... f5 2.exf6ep e5 3.Dxe5#
(1... f6 does not prevent the threat, so 2.exf6 is not considered dualistic) (2020-09-26)
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comment
Keywords: Last Moves?, Valladao Task, En passant, Promotion, Castling in the retro play, Obvious promotion
Genre: Retro, 3#
FEN: 8/3pPp2/3Pp3/4P3/1N3P2/B7/P1P2P2/kb3RK1
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2020-09-27 more...
18 - 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)
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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...
19 - P0000819
Josef Haas
1893 Die Schwalbe 40 08/1976
1. Preis
P0000819
(9+6)
#1 vor 4 Zügen
VRZ, Typ Hoeg
R: 1. Kh3xBg3 hxg3ep+ 2. g2-g4 Ke6xBd6 3. exd6ep+ d7-d5 4. Sc4-b6, dann 1. Sd6#
play all play one stop play next play all
Henrik Juel: It is illegal for Black to supplement anything on b6, because [Ta8] was captured in its corner and the other missing black men were captured by white pawns (2016-03-28)
Henrik Juel: ... as wLb3 is a pawn promoted on e8 or g8
Nice type Høeg defensive retractor
Here are some other explanatory comments
In retraction 1 White chooses to move his king back to h3; Black could choose to supplement a black man on g3 (or nothing), but supplementing a pawn is the only way to maintain legality (Kh3 stands in double check from Lc8 and Dh8); again moving Pg3 back to h3 and White supplementing a pawn on g4 is forced (this e.p. case is the only one where the supplementing does no happen on the abandoned square)
In retraction 2 the white retraction is forced, and then moving Kd6 back to d7 to uncheck is illegal because of the double check from Sb6 and Pc6, so Black must uncheck by moving Kd6 back to e6 and White choose to supplement a pawn on the abandoned square
In retraction 3 White chooses to move Pd6 back to e5, forcing another e.p. situation (2023-04-08)
Henrik Juel: The Proca type is easy to define: White and Black alternate retractions, until White can mate with a forward move
The Høeg type is usually defined the same way, except that the other side decides which man (if any) was captured; but this can be detailed as follows:
1. White chooses a man and 'moves it back'
2. Black chooses which man (if any) to 'supplement' on the abandoned square
(only now is the white retraction complete)
3. Black chooses a man and 'moves it back'
4. White chooses which man (if any) to 'supplement' on the abandoned square
(only now is the black retraction complete)
etc. etc. until, following a white retraction, White can mate with a forward move
In tries, Black can ruin the white plan by mating White with a forward move after a black retraction
It goes without saying that the resulting retractions must be legal
'supplement' is my (poor) translation of the danish term 'supplere'; maybe 'add' would be better
'the abandoned square' needs a special interpretation in the e.p. case, which happens twice in this problem
These details may be the cause why new type Høeg defensive retractors are rarely seen, as type Proca is more natural and straightforward (2023-04-08)
A.Buchanan: Thanks Henrik. Yesterday, I went through all the defensive retractors to clear up keywords & genres. There were a very few where the stip did not specify the VRZ Type, and others where Anticirce did not specify Calvet vs Cheylan. The answers are probably obvious to you, and if you want to comment on those, then I will update the stips & keywords.
A more general question: Typ Friedlich appears to be the German for Type Pacific: can we standardize on one? (2023-04-08)
Henrik Juel: Thanks Andrew for enabling me to post my type Høeg spiel once again
Anticirce without specification usually means that both Calvet and Cheylan work
Friedlich is indeed german for Pacific, and as the PDB is a german product, I guess we must live with the present conditions (2023-04-08)
comment
Keywords: En passant, Promotion, Defensive Retractor, Type Høeg
Genre: Retro
FEN: 2b4q/1p2p3/pNPk4/8/8/1B2R1K1/1P2PP1P/8
Reprints: feenschach 42 04-07/1978
345 Europe Echecs 241 01/1979
(5) Die Schwalbe 163 02/1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-07 more...
20 - 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...
21 - P0001146
W. Wolf
Deutsches Wochenschach 23/04/1911
P0001146
(15+8)
#3
1. bxc6ep e4 2. Se3 Kxg5 3. Kxd6
play all play one stop play next play all
Henrik Juel: C+ Popeye 4.61 (the three threats never materialize)
It is obvious that last move was c7-c5 (2020-12-02)
comment
Keywords: En passant as key
Genre: Retro
FEN: 4B1R1/3NP1Pp/1Q1p1Prr/RPpKpNPk/6p1/6P1/P2B4/8
Reprints: 41 Volksgemeinschaft (Heidelberg) 19/01/1936
252 Comoedia 21/06/1936
22 Europe Echecs 18 02/1960
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2020-12-02 more...
22 - P0001226
Paul Vatarescu
99 Europe Echecs 53 05/1963
dedicated to L.Loewenton
P0001226
(16+9) C+
#1
1. dxc6ep#
R: 1. c7-c5
play all play one stop play next play all
hans: 1. dxc6ep.# R: -1. ... c7-c5 -2 Td2-d3 e7-e5 -3 Dh2xSg3+

1. dxe6ep.# R: -1. ... e7-e5 -2 Dh2xg3+ c7-c5 -3 Td2-d3 (2010-06-22)
Mario Richter: e.p. keys are not justified, Black's last move might even have been h2-h1=N. (2010-06-26)
Hans-Jürgen Manthey: +sPh2 und es geht nur noch die hans-Lösung

R: 1. c7-c5 Ka4-b5 2. Kc5-d6 Lg5-h4 3. Kd6-c5 Lc1-g5 4. Ke7-d6 c5xTb6 5. Th6-b6 Sc2-e1 6. Td1-f1 Ta3-d3 7. Th8-h6 Te1-e2 8. Se2-g1 Sb6-a8 9. Sf4-e2 Kb3-a4 10. Sg6-f4 Ta1-a3 11. f4-f3 Dc3-g3 12. Sg3-h1 Da5-c3 13. Sf5-g3 Da8-a5 14. Sh6-f5 a7-a8=D 15. Sg8-h6 a6-a7 16. Ke8-e7 a5-a5 17. Se7-g6 a4-a5 18. Sc6-e7 a2-a4 19. Td4-d1 Sa3-c2 20. Ta4-d4 Sc6-b6 21. Ta8-a4 Sa7xLc8 22. Sb8-c6 Sb5xa7 23. f5-f4 Th1-e1 24. f7-f5 Le2-g4 25. g4xDh3 h4-h5 26. g5-g4 Kc2-b3 27. g7-g5 Db3-h3 28. h3-h2 g3xDh4 29. Dd8-h4 h2xLg3 30. Lf4-g3 Kd1-c2 31. Lg5-f4 Ke1-d1 32. Lh4-g5 Lf1-e2 33. Lg5-h4 Dd1-b3 34. Lf6-g5 Sb1-a3 35. Le7-f6 Sd4-b5 36. Lf8-e7 Sf3-d4 37. h4-h3 e2-e4 38. h5-h4 d4-d5 39. h6-h5 d2-d4 40. h7-h6 c4-c5 41. e6-e5 c2-c4 42. e7-e6 Sg1-f3 (2021-07-06)
A.Buchanan: This one is not to be found in WinChloe or yacpdb (2021-07-07)
Ladislav Packa: Why not 1...f4-f3? (2021-07-07)
Mario Richter: Ladislav, good question!
The most likely answer is, that we have a diagram error, and that the correct position has a white Pf3 instead of a black one. With this modification, everything works fine.

Btw., adding a black pawn h2 still allows R: 1. ... f4-f3 as Black's last move, but even worse, it makes the position illegal (what happened to the missing white pawn c2 or a2?) (2021-07-09)
A.Buchanan: Yes it all clicks into place. bPgxh bPh waylaid bBc8 died at home. wPaxb, cxd, dxe, exf. Well done Ladislav and Mario. Let’s correct the diagram as this couldn’t have been a composition error. Mostly likely error in transcribing from EE (2021-07-09)
Ladislav Packa: bPh2 is not necessary, Sh1-h2 is enough. (2021-07-09)
Mario Richter: Perhaps Gerd can check the original source ... (2021-07-09)
VL: Originally wPf3. The problem is dedicated to L.Loewenton. (2021-07-10)
A.Buchanan: Thanks Valery! (2021-07-10)
Ladislav Packa: After wPb2-d2, it is possible to save bRf1 and one of the black Knights g1 or h1. (2021-07-11)
Hans-Jürgen Manthey: an Mario: von wegen ILLegal ! bei hinzufügen von sBh2:
der C-Bauer = 4. Ke7-d6 c5xTb6
der a-Bauer = 14. Sh6-f5 a7-a8=D
aber ...f4-f3 schließt natürlich eP aus.
mit wPf3 kann wegen Mangel an schlagfällen nicht Dh2x?g3. Darum
R: 1. dxc6ep# c7-c5 2. Lg5-h4 h4-h3 3. Sc2-e1 Ta1-f1 4. Td1-d3 a2-a1=T 5. Ta1-d1 Ke7-d6 6. Lc1-g5+ a3-a2 7. Te1-e2 a4-a3 8. Le6-g4 a5-a4 9. d4-d5 a7-a5 10. Lc4-e6 Se2-g1 11. a5xTb6 Tf6-b6+ 12. Sb6-a8 Tf8-f6 13. a4-a5 Ta8-f8 14. a2-a4 Sf4-e2 15. Sa3-c2 Sd5-f4 16. Df4-g3 Sf6-d5 17. Dd2-f4 Sg3-h1 18. Th1-e1 Sf5-g3 19. Lf1-c4 Ke8-e7 20. Kc4-b5 Sg8-f6 21. c3xLd4 Lc5-d4 22. Kd3-c4 Sd4-f5 23. c2-c3 Sc6-d4 24. Dd1-d2 Sb8-c6 25. e2xf3 Lf8-c5 26. Kd2-d3 f4-f3 27. d3xDe4 Dg6-e4 28. Ke1-d2 Dg5-g6 29. Sb1-a3 Dd8-g5 30. Sc8-b6 e7-e5 31. Sb6xc8 f5-f4 32. Sc4-b6 f7-f5 33. g4xTh5 Th8-h5 34. g3-g4 h5-h4 35. h2xg3 h7-h5 36. Se5-c4 g4-g3 37. Sf3-e5 g5-g4 38. Sg1-f3 g7-g5 39. d2-d3 (2021-07-12)
Hans-Jürgen Manthey: allerdings geht noch h2-h1=S... (2021-07-12)
Mario Richter: an Hans-Jürgen: bzgl. meiner Illegalitätsreklamation für die Stellung mit sBf3 und zusätzlichem sBh2 hast Du recht - da war ich zu sehr darauf fixiert, daß Schwarz auf a1 entwandeln muß, läßt man W auf a8 entwandeln, sieht man leicht, daß die Stellung legal ist ...

Btw. - worauf bezieht sich Deine letzte Anmerkung ("allerdings geht noch h2-h1=S")? (2021-07-13)
Hans-Jürgen Manthey: Mario da lag ich falsch... 16 weiße Steine, da geht nicht sBh2, sBh3, sorry (2021-07-14)
more ...
comment
Keywords: En passant
Genre: Retro
Computer test: HC+ counting captures + Retractor 2.0
FEN: N7/1p1p4/1P1k4/1KpPp2P/4P1BB/3R1PQp/1P2RPP1/4Nrnn
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-07-10 more...
23 - P0001228
Wolfgang Hundsdorfer
Deutsches Wochenschach 1909
1. Preis
En-passant-Turnier 1910
P0001228
(13+12)
#3
1. bxc6ep

R: 1. c7-c5 g5-g6 2. Tc6-c2 g4-g5 3. Tg6-c6 g3-g4 4. Tg8-g6 g2-g3 5. Ld4-a7 h4-h5 6. Lg7-d4 h3-h4 7. Lf8-g7 h2-h3 8. g7xLh6,g7xSh6
play all play one stop play next play all
Henrik Juel: minor dual in forward play
1.bxc6ep+ Db5 2.DLxb5+ Tc4 3.LDxc4#
1... b5,bxa6 2.Kxb4+ Tc3 3.Txc3# (2022-07-01)
Henrik Juel: Last move must be c7-c5, enabling Tc2 to reach g8 in just three retractions (2022-07-01)
comment
Keywords: En passant as key
Genre: Retro, 3#
FEN: 8/bp2p2p/B5Pp/RPp2q1P/Qp2P3/RK1kPP2/P1rppP2/8
Reprints: 139 Favorit-Schachaufgaben , p. 35, nach 1910
75 Retrograde Analysis 1915
101 Europe Echecs 53 05/1963
10 Die Schwalbe 276 12/2015
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-07-01 more...
24 - P0001270
Matti Arvo Myllyniemi
142 Europe Echecs 84 01/1966
P0001270
(15+15) C+
BP in 9.5
1. e3 d5 2. Lc4 d4 3. Se2 d3 4. 0-0 dxc2 5. d4 Kd7 6. d5 Kd6 7. Dd4 Sd7 8. Ld2 c1=L 9. Lb4+ c5 10. dxc6ep+
play all play one stop play next play all
more ...
comment
Keywords: Unique Proof Game, En passant, Non-standard material (l), Castling (wk), Promotion (l), Valladao Task (sww)
Genre: Retro
Computer test: Ergänzung Stelvio 1.2 C+: Keine Lösung: BP 8.5, 9.0. 10. dxc6ep++ Doppelschach.
FEN: r1bq1bnr/pp1npppp/2Pk4/8/1BBQ4/4P3/PP2NPPP/RNb2RK1
Reprints: 92 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-25 more...
25 - 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...
26 - P0001345
F. ben Galuth
219 Europe Echecs 126 07/1969
P0001345
(10+10) C+
h#2
1. cxd3ep? g3 2. Bd4 Lg2# (wLc1 retro-blocked)
1. c5! g3 2. cxd4 Lg2#
play all play one stop play next play all
James Malcom: Solution? (2020-12-29)
Hans-Jürgen Manthey: wohl beabsichtigt: 1. cxd3ep g3 2. Bd4 Lg2#
doch auch ein Dual: 1. c5 g3 2. cxd4 Lg2# (2020-12-29)
Mario Richter: rawbats says, that f2-f4 is White's only legal last move, and 1. c5 g3 2. cxd4 Lg2# the only solution, so 1. cxd3ep g3 2. d4 Lg2# might be a try, intended to fool the solvers ...
(notice that R: 1. d2-d4?? excludes wLc1 from the set of objects available for black pawn captures!) (2020-12-29)
James Malcom: Furthermore, the h pawn couldn't have moved last, as that rook is needed for the Black pawns. wPe6 takes all remaining captures of Black pieces, and bBf8 could never escape. So the d2 and h2 pawns did not move last, and gxf3ep does nothing, meaning that all en passants are not a solution. A very fine retro "joke." (2020-12-29)
James Malcom: This means that 1. c5 g3 2. cxd4 Bg2# is the only possible solution, and as such, this should be C+? (2020-12-29)
Hans-Jürgen Manthey: Stimmt Mario
habe übersehen das der Lc1 nicht gezogen haben kann bei 1. d2-d4; aber 6 schlagfälle von Schwarz nötig sind. (2020-12-30)
A.Buchanan: I think the composer here is exactly "trolling solvers' preoccupations for the lulz", to quote Hauke Reddmann in MatPlus today, which even my allegedly fine German skills are unable to translate. But Hauke is German, so maybe he could do it? (2020-12-30)
more ...
comment
Keywords: En passant, En passant as key (Tries)
Genre: h#, Retro
Computer test: Popeye v4.85 + thinking
FEN: 8/2p1p1p1/4P3/3p1p2/2pPkPpP/4p3/1PP1P1P1/Kb3B2
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2021-01-06 more...
27 - 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...
28 - 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...
29 - P0001551
Wolfgang Dittmann
2724 feenschach 46 04-06/1979
1. Preis
P0001551
(5+10)
#1 vor 7
VRZ, Typ Proca
R: 1. Kd2xLe1 e2-e1=L+ 2. Kc3-d2 e4xd3ep 3. d2-d4 e5-e4+ 4. Kd3xBc3 b4xc3ep 5. c2-c4 b5-b4+ 6. Kc4xTd3 c6xBb5 7. Kc5-c4, dann 1. b6#
play all play one stop play next play all
Mario Richter: For the retraction of ep-captures, the animation engine puts the uncaptured pawn on the wrong square. (2020-12-09)
A.Buchanan: Well spotted Mario. Suggest that you drop a quick email to Gerd? If that's not convenient for you, let me know and I will (2020-12-09)
Mario Richter: Gerd is informed. (2020-12-09)
comment
Keywords: En passant, Defensive Retractor, Type Proca
Genre: Retro
FEN: rn5b/kp1pp3/b7/8/8/3p1PP1/p4PP1/4K3
Reprints: feenschach 61 08/1982
422 Europe Echecs 295 07/1983
150 Der Blick zurück 2006
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2006-08-26 more...
30 - P0001561
Michel Caillaud
432 Europe Echecs 304 04/1984
Lob
P0001561
(8+13)
Kürzestes h#?
1. ... bxa6ep! 2. Txe2 axb7 3. Td2 b8=D 4. De2 Db1#
1. Dc6? bxc6 2. Ld7 cxd7 3. Ke1,Kxe2 d8=D 4. Kf1 Dd1# too slow
R: 1. a7-a5 e3xLf4 2. Ld6-f4 d2xTe3 3. Ta3-e3 b4-b5 4. Ta6-a3 b3-b4 5. Tc6xBa6 a5-a6 6. Tc8-c6 a4-a5 7. Tg8-c8 a3-a4 8. Lf8-d6 a2-a3 9. e7xDf6,e7xTf6
play all play one stop play next play all
The forward play gives shortest h# in 3.5 moves, *if* e.p. is on. Otherwise the shortest is a mildly dualized 4.0 moves. So our mission is clear: assuming that Black moved last, what was that move? The position looks very open - how can one ever know? There is a knot in the south-east corner, which will only be resolved by wQ/R visiting g1.

White pawns have captured dxexf & hxg, accounting for BSP. So wPa was waylaid while wPc promoted to B. Black pawns captured dxexfxgxh, exf as well as wBf1 captured at home and wPa. That's 7 units, but bPc must also have captured to allow wPc to promote (to B on c8). So all captured units are accounted for. bPc promoted (bPh7 came from h6, so is bPd). We can't undo any of the Black pawn captures now, but we can undo d2xe3xf4, to release 2 Black units.

We can unpromote bPc, but we can't undo its capture to release a White unit, until wBh3 has unpromoted. The only White unit we can get now is from e7xf6. So this means that the two Black units we can release must be B & R, so they can retreat to f8 & g8 respectively. The timing is very tight, and there is only one way to do it. The black rook must visit a6 to unwaylay wPa, which gives White 5 more tempi, just enough.

bPc promoted on b1 to R, so we would need to undo the cage to get that. Therefore all we can do is send wQg1, then Rf1-f2 f2-f3 etc. bPa must retract immediately tp a7, so that wPa when unwaylaid can make fully 4 unmoves. Black officers are arranged to give unique retro & forward play (although with minor dual for the try) with Sh8 not just retro dressing but ensuring unique retraction Tg8-c8 not Th8-c8.
A.Buchanan: Another surprising motivation for e.p. I love Sh8. Sorry for W3 dual in the try play else the stipulation could be h#4*. I can’t fix it but it’s hard to improve on MC. Very enjoyable (2021-10-23)
A.Buchanan: Non-standard material is where the diagram contains for a player more than 1 queen or bishop of a hue or more than 2 rooks or knights. Obtrusive material is standard material, but there is some cheap reason why the unit must be promoted, most commonly that a bishop's home square remains blockaded by pawns. These categories are disjoint: no piece is ever both. Many problems in PDB do not apply this consistently, but the distinction goes back a long way in chess problem history, and is discussed by Morse.
Honestly, I dislike the word "obtrusive" whose negativity (while perhaps valid in forward problems) is inappropriate for retros. One distinguished composer objected to this tag being assigned to one of his problems. Nevertheless the concept has some interest. Renaming is a perilous business, but I am looking for suggestions... :) (2021-10-24)
A.Buchanan: Another distinction that comes to mind between "non-standard material" and "obtrusive promotion" is that normally in the former, one can't immediately point to which of the non-standard pieces was promoted: it's just that there's too many; while for the latter, one can usually point to a specific unit immediately.
"Obvious" is a candidate replacement for "obtrusive", but this might commit a cardinal sin of trying to nail down a perfectly useful and inherently vague term. Both "obvious" & "obtrusive" begin with "ob" which is helpful. What do you think? (2021-10-24)
Henrik Juel: I am curious about the award; why did this problem with good forward play and excellent retro-play only obtain a Commendation?
Maybe the judge did not like that Lh3 obviously is obtrusive...
. (2021-10-24)
A.Buchanan: Yes, and the lovely P0001117 with a similar obtrusion only received 12th Lob! But there may have been other factors. I do observe that "non-standard material" is arguably a worse defect than "obtrusive material", but the term doesn't cause offence because it's objective and non-judgemental. "obtrusive" is more subjective and inherently pejorative. To have such terms in the glossary is to put curators in an invidious position. I think the concept has its place, but I would like to replace it with something less scornful. I'm up for "obvious". It's an easy and reversible change: let's do it and see if mobs of protesters form outside the gate :) (2021-10-24)
A.Buchanan: Have changed "obtrusive material" to the non-pejorative "obvious promotion". It may still be regarded as a defect. As a placeholder, I have also changed the unclear German "mit Umwandlungfigur(en)" to "augenscheinlich Umwandlungfigur". A native German speaker I'm sure will propose a better term. (2021-10-26)
A.Buchanan: “offensichtlich” it is thanks Mario (2021-10-26)
more ...
comment
Keywords: En passant, Last Moves?, Obvious promotion (L), En passant as key, Promotion (D)
Genre: h#, Retro
FEN: 7n/1p3pp1/4bp1p/pP6/4qPP1/5PrB/4PrPp/3k3K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-24 more...
31 - P0001617
Michel Caillaud
487 Europe Echecs 346 10/1987
P0001617
(9+7) cooked
#1 vor 12
VRZ, Typ Proca
paul: Cooked in 4: 1.Ke5-e6! b2×Ba1=B+ -2.Kf4×Pe5 e6-e5+ -3.Rc2×Sg2 Se1-g2+ -4.Re2×Pc2 & 1.Rxe1# (2023-06-14)
comment
Keywords: En passant, Defensive Retractor, Type Proca
Genre: Retro
FEN: 8/4p3/3PK1P1/8/1B5p/1P1qPP1p/P5Rp/b2k4
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2004-08-28 more...
32 - P0001622
Christian Poisson
491b Europe Echecs 355/356 07-08/1988
P0001622
(3+11)
shc#11
1. Lf1 2. Kh3 3. Kh4 4. Lh3 5. fxg3ep 6. Kh5 7. Kg6 8. Kf7 9. Ke8 10. 0-0-0 11. Td7 a8=D#
play all play one stop play next play all
The cluster of tries: Kh2-g1-f1-e1-d2-...-e8 0-0-0 Td7 is 12 moves: just too long.
In the diagram, e.p. is not permitted because last move might be g3-g4. So first shift sK to h4. Following the e.p., check that wK does now have prior move: Kf4-f3. Now sK scampers to e8 where through consequent forgetfulness it recovers castling rights.
more ...
comment
Keywords: En passant, Castling (sg), Seriesmover, Consequent, Valladao Task, Promotion in the mating move (D), Switchback (l), Promotion (D), Königswanderung
Genre: Retro, Fairies
FEN: r7/P1p5/pp6/5pp1/5pP1/5K1b/4rp1k/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
33 - P0001653
Philippe Leroy
522 Europe Echecs 373 01/1990
P0001653
(12+13)
BP in 60.0
Es fand ein ep-Schlag statt. Welcher?
1. a3 b5 2. a4 b4 3. a5 c5 4. a6 Lb7 5. axb7 a5 6. c4 bxc3ep 7. e4 f5 8. e5 Sf6 9. exf6 c2 10. De2 d5 11. De4 dxe4 12. g4 h5 13. g5 Th6 14. gxh6 g5 15. Se2 h4 16. Sg3 hxg3 17. f7+ Kd7 18. d4 c4 19. d5 Ta6 20. h7 g2 21. h8=L g1=L 22. Lc3 a4 23. La5 g4 24. h4 Lh2 25. h5 g3 26. h6 g2 27. h7 g1=L 28. h8=L a3 29. Lhc3 a2 30. f3 Ld4 31. Lcb4 Lh8 32. La3 c3 33. Le3 c1=L 34. Lg1 c2 35. Sc3 Tf6 36. Sa4 Sa6 37. b8=L Lch6 38. Lba7 Lb8 39. Th4 c1=L 40. d6 Kc6 41. d7 Db6 42. d8=L Lcg5 43. Lc7 e3 44. Lch2 e5 45. Sc5 Sc7 46. Tb4 e4 47. Kd1 e2+ 48. Kc2 e1=L 49. f4 Leh4 50. Kc3 e3 51. Kd4 e2 52. Ke5 e1=L 53. Td1 a1=L 54. Td7 Leg3 55. Sa4 Df2 56. Sb6 Lfg7 57. Lb5+ Kc5 58. Lc6 Sd5+ 59. Tc7 Sxf4 60. f8=L+ Td6
play all play one stop play next play all
James Malcom: A very interesting retro problem, with long-winded bishops promotions to cover its en passant tracks. (2021-01-24)
Hans-Jürgen Manthey: damit der Player klappt, muß 38. La7 in 38. Lba7 geändert werden. (2021-01-25)
James Malcom: Fixed. (2021-01-25)
comment
Keywords: Constrained problem, En passant, Non-standard material, Promotion, Non-Unique Proof Game
Genre: Retro
FEN: 1b3B1b/B1R3b1/1NBr3b/B1k1Kpb1/1R3n1b/B5b1/1P3q1B/b5B1
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-01-25 more...
34 - P0001764
Henri Nouguier
25 Phénix 1 05/1988
P0001764
(9+13) cooked
shc#6
1. dxc3ep 2. Kxb5 3. Kxc6 4. Kd7 5. Ke8 6. 0-0 Th8#
Aber es geht auch R: 1. Kb2-a1!?
play all play one stop play next play all
Henrik Juel: The intention may be 1.dxc3ep 2.Kxb5xc6-d7-e8 6.0-0 Rh8#, but -1.Kb2 seems possible. Should bBc1 be moved to a3? (2003-04-28)
GW: Yes, that's the solution, and the problem seems indeed to be cooked. I don't know if it has been corrected. (2003-04-28)
James Malcom: I believe that the intent was that the sLc1 must be the orginal one, and thus an impediment to wK foreplay, as the wBc6 "must" have captured the e7 Black pawn on its way from f2. The problem is that it can take the alternative path fxsLf3xDe3xTc5-c6, and there are enough White pieces to capture for the Black b pawn to make it d2 and for the d pawn to make it to c2. (2021-09-14)
James Malcom: It turns out this problem actually was corrected a long time later: P1012052 (2021-09-14)
comment
Keywords: En passant, Castling (sk), Seriesmover, Consequent, Non-standard material (sLb1), Valladao Task, Promotion in the retro play (sLb1), Obvious promotion ((sLb1)), Superseded by (P1012052)
Genre: Retro, Fairies
FEN: 7r/5pnR/2P3pR/pPpn3p/1kPp4/3P4/P2pP3/Kbb5
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-09-14 more...
35 - P0001859
William A. Langstaff
The Chess Amateur 1922
P0001859
(5+3) C+
#2
If Black can castle, e.p. is ok:
1. hxg6ep! 0-0 2. h7#
1. ... Tf8,Kf8 2. Td8#
Otherwise Black can't castle:
1. Ke6! ... 2. Td8#
play all play one stop play next play all
more ...
comment
Keywords: Cant Castler, Partial Retro Analysis (PRA), Castling (sk), En passant as key
Genre: Retro, 2#
Computer test: HC+ Popeye 4.61 & trivial retro-logic
FEN: 4k2r/8/5B1P/3R1KpP/8/8/8/8
Reprints: (D17) feenschach 27 04/1975
RA64 diagrammes 27 05-06/1977
(1) Die Schwalbe 86 04/1984
Razem 34 23/08/1987
(I) diagrammes 15 07-09/1994
(10) Die Schwalbe 241 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-30 more...
36 - P0001937
Werner Kuntsche
6025 Schach , p. 191, 06/1969
2. Preis
P0001937
(4+12) C+
h#2
1. exd3ep Lxg4 2. f3 Le6#
play all play one stop play next play all
Klären: Quelle = Schachmatt? - Felber, Volker: SCHACH ist korrekt, 6/1969, Seite 191 (2010-10-09)

Eliminierung von 4 Steinen zwischen sK und wT im h#2 (mit Hilfe von ep-Schlag)!
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comment
Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 8/8/8/1pn4b/2kPpppR/2q1rB1p/1p4r1/3K4
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-01-09 more...
37 - 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...
38 - P0001970
Thomas R. Dawson
26 Schachkongress Teplitz-Schönau im Oktober 1922 1923
P0001970
(13+13)
h#1
1. exf3ep Lc2#
R: 1. f2-f4 f5xTe4 2. Tg4-e4 & e.g. f6-f5 3. Tg1-g4 f7-f6 4. Le4-h7 h5-h4 5. Lg2-e4 h6-h5 6. Lf1-g2 h7-h6 7. g2xh3
play all play one stop play next play all
1.exf3e.p. Bc2# is the only possible solution, but this necessitates R: 1.f2-f4. Can we prove this?

(13+13) with 1+2 pawn captures. Bf8 captured at home, so to satisfy White appetite, the missing Black pawn (a or b) must have promoted via c2 on c1. Two more White units must be captured to allow this.

The kings cage can only be unlocked by retracting WPc2. But the clock is ticking as there are only 6 black moves which can be retracted.

The promoted piece was captured on e3 or h3. If either capture is undone, then a White bishop square is cut off, so WB must be replaced prior to this.

Now the order of the early moves is: WdP moves, WQB & WQR escape, BP promotes on c1 to X (capturing WR at some point), X captured by WP.

So the first White capture must be dxNe3 and the second White capture releases gxXh3. The second White capture releases WKB & WKR. WKR captured by original BfP.

The clock starts ticking with gxh3. Black has 6 pawn moves. WKB has 3 moves to reach h7. WR has 3 if it goes via d file, or 2 if it starts on g1 (in which case WfP or WQB must also move once). So certainly at least 6 White moves. Last move was therefore White (even if the stipulation didn't tell us), and it can only have been WfP coming from f3 or f4. If it had been coming from f3 it would have blocked WKB in its progress, so the last White move was indeed R: 1.f2-f4.

WKR did therefore move from g1-g4-e4, and R: 1. ... fxRe5 2. Rg4-e4. Prior to that, move order not unique, but counting still exact.

Note that WN loitering on b4, pretending to be part of the cage, is present on the board just to make up the numbers.
Jeliss: "Obstruction of passage square f3 to Bishop of same colour."

"Version 'Pittsburgh Leader' 08.06.1913"
Yoav Ben-Zvi: Appears as the first problem (D445) in the booklet on Dawson's RA problems by G.P. Jellis. The obstruction that occurs in the Try -1.Pf3-f4?, by WP of WB, is described as "obstruction of passage square". It is not considered by Dawson and his disciples to be a Retro opposition. Dawson's conception of RO was quite broad, it included cases where the interference was not by occupation of the target square, so the only valid reason that I can see to exclude this case is that the 2 pieces involved are both of the same color. Fabel's definition explicitly excludes "Monochrome RO". I conclude that it would be preferrable to interpret RO as a bi-chromatic interference. The keyword Retro opposition should be removed. (2018-04-07)
A.Buchanan: To my mind, RO involves some kind of parity-tempo issue between the sides, not just some kind of race-tempo. If it was just about "bi-chromatic interference", one might say that bPe4 blocks wBh7 from an immediate retreat, so it has to be wPf4 that retreats first, legitimizing the ep key. So I agree this is not RO. (2024-01-06)
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comment
Keywords: Last Moves?, En passant as key
Genre: h#, Retro
FEN: nqb5/1rrpp1pB/KRp5/1p4B1/kN2pP1p/2P1P2P/PP2P2P/8
Reprints: D445 Retro-Opposition & Other Retro-Analytical Chess Problems 1989
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-06 more...
39 - P0002100
Adolf Norlin
v Tidskrift för Schack , p. 34, 01/1907
P0002100
(13+10)
#2
1. exf6ep+!
play all play one stop play next play all
Version in der 'Aarsskrift' 1935 innerhalb eines Artikels von K. Hannemann "Et Tema Fra den Retrograde Analyse". Im Original wDg4 statt h3 und wBh3 statt h4.
Henrik Juel: 1.exf6ep+. Not -1... a6?, requiring 3 black captures on light squares (incl. orig. Lf1); but missing orig. Lc1 was dark-squared. (2004-03-08)
Henrik Juel: C+ Popeye 4.61 and analysis (2022-06-11)
Hans-Jürgen Manthey: mögliche Zugfolge:
1. f2-f4 d7-d5 2. c2-c3 d5-d4 3. c3xd4 c7-c5 4. h2-h4 c5-c4 5. b2-b4 Lc8-d7 6. a2-a4 Ld7-b5 7. a4xLb5 Sb8-a6 8. Th1-h3 Sa6-c5 9. d4xSc5 Sg8-f6 10. d2-d4 Sf6-e4 11. c5-c6 Se4-g3 12. Lc1-d2 Se4xLf1 13. Sb1-c3 Sf1-g3 14. Sc3-a4 Sg3-e4 15. c6-c7 Ke8-d7 18. Sa4-c5+ Kd7-d6 19. Sc5-a6 Se4-c5 20. b4xSc5+ Kd6-e6 21. Ld2-a5 b7-b6 22. Sa6-b8 b6xLa5 23. Th3-e3+ Ke6-f6 24. Te3-e5 Dd8-c8 25. Dd1-d3 Dc8-a6 26. Dd3-h3 Da6-b6 27. Ta1-a3 Db6-c6 28. Ta3-g3 Dc6-d6 29. Te5-d5 Dd6-e6 30. c7-c8L De6-d6 31. Lc8-f5 Dd6-c6 32. Sg1-f3 Dc6-b6 33. Sf3-e5 Db6-a6 34. Se5-g6 Da6-b6 35. Sg6xTh8 Db6-c6 36. Tg3-g6+ h7xg6 37. Ke1-f2 Dc6-d6 38. Kf2-g3 Dd6-e6 39. Kg3-g4 De6-d6 40. Lf5-c2 Dd6-c7 41. b5-b6 Dc7-e5 42. f4xDe5+ Kf6-e6 43. Kg4-g5+ f7-f5 und nun :
1. e5xf6ep+ Ke6xTd5 2. Dh3-d7# (2023-02-24)
comment
Keywords: En passant as key
Genre: Retro
FEN: rN3b1N/p3p1p1/1P2k1p1/p1PRPpK1/2pP3P/7Q/2B1P1P1/8
Reprints: 58 Retrograde Analysis 1915
Aarsskrift DSK , p. 14, 1935
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-06-11 more...
40 - 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...
41 - 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...
42 - P0002758
Andrej N. Kornilow
1078 Phénix 15-16 12/1991
P0002758
(25+0)
Färbe die Steine! Welches waren die letzten 9 Einzelzüge?
James Malcom: Solution? (2020-11-07)
Mario Richter: I do not know the official solution, but the following works:
wBb7 wBf7 wBb6 wBg6 wBc5 wTf5 wLf4 wBg4 wBe3 wKg3 wTh3 wDf2 wBh2 wSf1 wSh1
sLg8 sDh8 sBa7 sBc7 sBh7 sBe6 sBf6 sKh6 sBe5 sLg1
wCaps: h5xg6ep f3xTg4 g4xTh5 e6xSf7 d5xSe6 a4xBb5
sCaps: d7xLe6
R: 1. h5xg6ep g7-g5 2. Tg5-f5 d7xLe6 3. f3xTg4 Th4-g4 4. g4xTh5 Kg6-h6 5. Tf5-g5 (2020-11-10)
comment
Keywords: Colouring problem, Last Moves? (9), En passant
Genre: Retro
FEN: 6BQ/PPP2P1P/1P2PPPK/2P1PR2/5BP1/4P1KR/5Q1P/5NBN
Reprints: (34) Die Schwalbe 144 12/1993
Input: Gerd Wilts, 1995-06-03
43 - P0002853
Marcel Eugene Nordlohne
Umwandlungen in Märchenfiguren 2 1993
P0002853
(3+3) C+
s#4
Längstzüger
wDU,sDU=Grashüpfer
1. fxe6ep! Gf6 2. e7 Gd8 3. e8=G Gf8 4. Gg8 Gh8#
play all play one stop play next play all
Der letzte Zug kann unter der Längstzügerbedingung nur Be7-e5 gewesen sein. Also ist der En-passant-Schlüssel erlaubt. Erstaunlicherweise gelingt der En-passant-Nachweis in Miniaturfassung.
Henrik Juel: C+ Popeye 4.61 after analysis
Last move obviously was not made by Ga6
neither by Kf1 from e1, e2, f2, because K-f1 would not be a longest move
neither by Pe5 from d6, e6, f6, by a similar reason
So last move was e7-e5, and White may capture ep (2021-05-02)
comment
Keywords: En passant, Maximummer, Model mate
Pieces: du = Grasshopper (G)
Genre: Retro, Fairies
Computer test: HC+ Popeye 4.61 after analysis
FEN: 8/8/*2q7/4pP2/8/8/7*2Q/5k1K
Reprints: Probleemblad 05-06/196?
Input: Gerd Wilts, 1995-06-03
Last update: Marcin Banaszek, 2021-05-10 more...
44 - P0003100
John Derek Beasley
Cambridge 1985
P0003100
(3+8) C+
h#2
Zeroposition
a) -sBe4
b) -sSf5
a) 1. Sg3 Kxh4 2. Le4 hxg3#
b) 1. hxg3ep Kh4 2. Lf5 hxg3#
play all play one stop play next play all
Henrik Juel: in a) last move could be Kg2-h3
in b) last move must be g2-g4 (2024-01-14)
more ...
comment
Keywords: En passant as key, Zeroposition
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & trivial retro-logic
FEN: 8/7b/8/4pn2/4pkPp/4pp1K/7P/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-15 more...
45 - P0003122
Albert Zickermann
789 Kieler Neueste Nachrichten 18/03/1934
P0003122
(3+6)
h#2
1. fxg3ep Lxg7 2. Lg4 Lf6#
play all play one stop play next play all
A.Buchanan: Sg7 would be better than T, from both forward & retro perspectives I think? (2021-04-14)
comment
Keywords: En passant as key
Genre: h#, Retro
FEN: 7B/6rp/7K/8/5pPk/5b1r/8/8
Input: Gerd Wilts, 1995-06-03
46 - P0003132
Frank Bentler
10997 Schach-Echo , p. 228, 06/1985
P0003132
(9+13) C+
h#2
1. axb3ep Texf1 2. b2 Kd2#
play all play one stop play next play all
more ...
comment
Keywords: En passant as key, En passant
Genre: h#, Retro
FEN: n7/2ppp1p1/1p6/B7/pP6/5PP1/p4pPP/rk1KRrRn
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2021-02-11 more...
47 - P0003138
Branko Koludrovic
4208 Problem 12/1979
P0003138
(11+9)
h#3 (AP)
0.1...
1. ... axb6ep 2. 0-0-0 0-0-0 3. Td7 a8=D#
play all play one stop play next play all
This problem can be solved presuming that both castlings are executable. It implies that the last halfmoves have been.-1.Bb6xa7 sBb7-b5.
Bb5 could not come from c6 (sBc6xFb5), because all of the missing white pieces(5) have been captured on white squares except wLc1,which must have been captured by Bc7(Bc7xwLb6).Further,it has not played immediately before -1.Bb6-b5 because this would prevent white castling(forcing wT or wK to move)
Here is the proof that 4 remaining white pcs have been captured on white squares:
black d-pawn must have been captured on d4 and h,g f pawns had to promote after Bh3xwFg2,then wBh2moved to h4, enabling Bg4x Fh3 ..h1T, and finnaly sBf7
after fxg2... g1S. sLf1 prevented the check from Th1, and later has been captured by a white officer.
Conclusion:
Retro:-1.wBb6xFa7 Bb7-b5
Forward(AP) 1....axb6 e. p.2.0-0-0 0-0-0 3.Td7 a8Q# (Author)
Branko Koludrovic: P.S.
The black pawn a4(on the diagramm)came from c7 after capturing the white bishop on b6, then moving to b5 and capturing a white officer on a4. (2010-09-28)
A.Buchanan: I think sBf must have captured on g2 & then promoted on f1 to S, otherwise it’s impossible to unlock the southeast cage (2023-07-11)
A.Buchanan: So that means it’s sound. However I don’t see that it needs to have been sL shielding on f1 (2023-07-11)
more ...
comment
Keywords: a posteriori (AP), En passant as key, Castling (sgsgwg), Promotion (D), Valladao Task
Genre: h#, Retro
FEN: r3k3/P3p3/p7/Pp6/p6P/2P3PR/2PPP2b/R3K1nr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-11 more...
48 - P0003142
Marco Bonavoglia
3652 Sinfonie Scacchistiche 01-03/1982
P0003142
(6+5) C+
h#2
b) +sTd4
a) 1. Sxb4 Sxc3 2. Ka5 Sxc4#
b) 1. cxb3ep Sc4 2. Sb4 Sxc3#
play all play one stop play next play all
Henrik Juel: In a) last move could be Kb2-a1 (2022-12-05)
more ...
comment
Keywords: En passant as key, Cant ep, Zilahi
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 8/8/8/1P6/kPp5/N1p5/n1pP4/KN6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-06 more...
49 - P0003143
Zvi Roth
947 Shahmat 08/1972
P0003143
(6+7) C+
h#2*
*) 1. ... Kxb3 2. Txd5 Txd5#
1) 1. bxc3ep Kxb3 2. Txd5 Lxc3#
play all play one stop play next play all
A.Buchanan: Attractive. It reminded me of P0003347, which I have just discovered is sound as set play. (Just omit the first move!) Here part of the interest, although the intermediate moves are the same, the mating move is different, but in both cases has to cover d4 & e5. (2020-12-09)
more ...
comment
Keywords: En passant as key
Genre: h#, Retro
Computer test: C+ Popeye 4.61
FEN: 8/8/1P6/BRrP4/KpPkp3/1r1nq3/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-12-09 more...
50 - P0003151
Laszlo Talaber
Tivadar Kardos

3649 Arbejder-Skak 01/1954
1. Preis
P0003151
(9+16) C+
h#2.5*
* 1. ... ... 2. fxe3ep Sge4 3. Txe4 Sf3#
1. ... e5 2. dxe5 Sde4 3. fxe4 Sf5#
play all play one stop play next play all
Sally: Der letzte Zug war e2-e4! (2010-04-06)
A.Buchanan: Great harmony between the phases (2021-11-23)
comment
Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 3.52 & trivial retro-logic
FEN: 8/7p/3p3p/2nq1ppP/p1PkPp1r/P1p3N1/n2N2Pb/1b1r1B1K
Reprints: (XX) Problem 37-40 09/1956
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-23 more...
51 - P0003180
Jozsef Korponai
6875 Arbejder-Skak 02/1968
P0003180
(12+12)
h#2
1. hxg3ep Sc1 2. gxf2 Se3#
play all play one stop play next play all
Welches ist die Originalquelle? Oder wurde es zweimal als Urdruck gebracht?

vgl. P0003291
Henrik Juel: White pawns captured all four missing black men
Last black move was not b7xPa6 ([Ta8]) nor e7-e6 ([Lf8]), so it was Lb7-c8, which is possible only if last white move was g2-g4
C+ Popeye 4.61 and above analysis (2021-04-23)
comment
Keywords: En passant as key
Genre: h#, Retro
FEN: 2b5/p2p2pp/p3p2b/6Pr/5BPp/2P5/NRP1PP1P/3Nrk1K
Reprints: F84 The Problemist 05/1970
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-04-23 more...
52 - P0003187
Thomas R. Dawson
6577 British Chess Magazine 12/1944
P0003187
(12+14)
h#1.5
1. ... bxc6ep 2. b5 Txa3#
R: 1. ... c7-c5 2. f4xTg5 Tg6-g5 3. f3-f4 Ta6-g6 4. f2-f3 Sc6-a7 5. h4-h5 Ta8-a6 6. h3-h4 a6-a5 7. h2-h3 Ka5-a4 8. b3-b4 Ta4-a3
or R: 1. f4xTg5 Tg6-g5 2. f3-f4 Ta6-g6 3. f2-f3 Sc6-a7 4. h4-h5 Ta8-a6 5. h3-h4 a6-a5 6. h2-h3 Ka5-a4 7. b3-b4 Ta4-a3
but we are told that White has the move.
play all play one stop play next play all
Henrik Juel: Stipulation should probably be interpreted to mean h#1.5 . -1... c7 -2.f4xTg5 Tg6 -3.f3 Ta6 -4.f2 Sc6 -5.h4 Ta8 -6.h3 a6 -7.h2 Ka5 -8.b3 Ta4 etc. (2004-03-18)
A.Buchanan: Idea I suppose is that Black just moved, and moreover it was pawn double hop. I don't see clearly far enough back to confirm that. (2021-10-21)
Mario Richter: If Black moved last, the only legal retraction is R: 1. ... c7-c5! But the position can also be resolved if White moved last, starting with R: 1. f4xTg5 Tg6-g5 2. f3-f4 Ta6-g6 3. f2-f3 Sc6-a7 4. h4-h5 Ta8-a6 5. h3-h4 a6-a5 6. h2-h3 Ka5-a4 7. b3-b4 Ta4-a3.
Therefore the keyword "No legal last move for White" doesn't apply here; and to make the solution work, the Stipulation should be something like "h#2 0.1.1." or "h#1.5". (2021-10-21)
Henrik Juel: HC+ Popeye 4.61 (2021-10-22)
A.Buchanan: To my mind, the retro thinking is serious enough that we can’t claim HC+. If a tool like Replicator 2.0 was to validate the retro status I’d be ok. But I think it would be misleading to put HC+ to indicate the trivial h#1.5 is sound. We can certainly mark that Popeye has checked the h#, but not click the C+ flag. What do you think? (2021-10-22)
Henrik Juel: HC+ is not recognized (yet) by PDB, and I suppose it just meant that some sort of human analysis is needed in addition to a standard forward solver like Popeye
You are suggesting that we distinguish between easy and difficult human analysis and say HC+ for easy analysis only
I suggest saying HC+ for all problems where Popeye is not sufficient; you can write easy or difficult in the C+ comment
If and when a human analyst proves the analysis wrong, we delete the HC+ label, of course (2021-10-22)
A.Buchanan: What does Mario think? He invented the term, I seem to remember (2021-10-22)
comment
Keywords: En passant as key
Genre: h#, Retro
FEN: 2b5/np1p2pp/7q/pPp3PP/kP6/r1PPP3/RKpp2P1/BB2n3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-21 more...
53 - P0003188
Fritz Hoffmann
11614 Schach 04/1988
P0003188
(4+9)
h#2
b) Schlüsselfigur nach f3
a) 1. Sxg4 f3+ 2. Kh5 fxg4#
b) 1. fxg3ep fxg3 2. Kh5 g4#
play all play one stop play next play all
Henrik Juel: C+ Popeye 4.61 (2021-05-04)
comment
Keywords: En passant as key
Genre: h#, Retro
FEN: 8/4pp2/7p/5K2/5pPk/4r2r/3n1P1n/4B3
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2021-05-04 more...
54 - P0003189
Tivadar Kardos
2144 Diagramme und Figuren 19/09/1967
P0003189
(7+15) cooked
h#2
1. bxc3ep Lxe2 2. Sa3 0-0-0#
play all play one stop play next play all
Cook: 2. ... Td1#
Sally: Der letzte Zug war: Bc2 - c4!
Nr.138 200 Ausgewählte S .Probleme T. Kardos(W. Fentze1983) (2010-09-30)
A.Buchanan: All 9+1 captures are visible by pawns. Note R: 1. Kd1-e1 d3xe2 is illegal because it costs 2 more captures for Black. So e.p. is unconditionally legal, requiring no AP justification. I see no reason why White has lost castling rights. Indeed someone might make a demo game ending with c2-c4. So this one seems cooked. Neither WinChloe nor yacpdb contains this problem. (2022-01-08)
A.Buchanan: If this is AP, it will be rendered sound by removing wPa2. But is this the intention? Does anyone have access to the collection of selected Kardos problems mentioned? (2024-01-25)
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comment
Keywords: En passant as key, Castling (wg)
Genre: h#, Retro
FEN: 8/8/3pp3/2ppp3/1pPkrq2/4pb1P/P3p1rP/Rn2KBb1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-08 more...
55 - P0003194
Tivadar Kardos
1597 Schachmaty 05/1963
P0003194
(6+14) C+
h#2
1. cxd3ep? bxc3 2. Lxf4 Lxg2# Last move not d2-d4
1. gxf3ep! bxc3 2. Lf4 Lf5# Last move f2-f4
play all play one stop play next play all
White's last move was certainly one of the two double pawn hops. The last move couldn't be d2-d4 because Black's pawns had already captured 10 of White's pieces, and playing d2-d4 would have resulted in wBc1 being White's 17th piece. So, the only solution is 1.gxf3 e.p.
Viktoras Paliulionis: The last move couldn't be d2-d4 because Black's pawns had already captured 10 of White's pieces, and playing d2-d4 would have resulted in wBc1 being White's 17th piece. So, the only solution is 1.gxf3 e.p. (2023-12-30)
A.Buchanan: Yes that's right! (2023-12-31)
comment
Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 4.87 & trivial retro-logic
FEN: 2n1q3/5prp/5K1b/3b4/2pPkPp1/2p1p1pB/1P4pP/8
Reprints: 546 FIDE Album 1962-1964 1968
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-31 more...
56 - P0003195
Thomas R. Dawson
2130 Die Schwalbe 07/1932
P0003195
(14+14) C+
h#2
WTM
1. ... bxc6ep 2. Da7 c7#
retro tries:
1) 1. Lxg3 b6 2. Lxd6 Lxd6#
2) 1. Dc6 b6 2. Dc7 dxc7#
3) 1. Ta7 f8=D,f8=T 2. Ka8 Dxc8#,Txd8# ("untolerated" dual)
R: 1. ... c7-c5 2. b4-b5 Dc4-a6 (move order variable beyond this point) 3. b2-b3 Ta6-a8 4. d5-d6 Te6-a6 5. b3-b4 Te8-e6 6. a2xSb3 Th8-e8 7. d4-d5 Lb4-e1 8. d3-d4 Lf8-b4 9. Kg5-h5 e7xSf6+
play all play one stop play next play all
Missing: Wh: SS Bl: SS
Pawn captures: Wh: axb, hxg Bl: exf, fxg
bPf6 comes from e7, else collision with wPf7. So can't uncapture fxg2 until wP retracted. White has 6 pawn retractions before retro-pat, just giving Bl time to put Le1 & Ta8 (impostor) back home, and uncapture exf6. So Black must retract first, and it's WTM. sD must clear out of the way, and c4 is the only possible square, wPaxSb can't happen on b5, or it would block sD. And axSb4 would block sLe1. It was axSb3. So wPb5 must retreat to b4, and thus Black cannot play R: 1. Lb4-e1. So neatly R: 1. c7-c5 is forced.
Ladislav Packa: Is wRh1 needed? (2021-10-22)
Mario Richter: Yes, wRh1 is needed - without it the ep-key would not be justified (last moves could have been R: 1. ... Th1-g1 2. Lg1-h2 or even R: 1. ... Ld2xTe1). (2021-10-23)
A.Buchanan: + three mostly clean retro tries. (2021-10-24)
more ...
comment
Keywords: En passant as key, No legal last move for White, Impostor (t)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: rkb5/1p1p1Ppp/q2P1p2/pPp4K/6PR/1P4PQ/2P1P1pB/4bBrR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
57 - P0003200
William H. Reilly
British Chess Federation 05/1944
2. Preis
P0003200
(4+9) C+
h#4
1. bxc3ep Sxg2 2. cxd2 Sxh4 3. Ke3 Kg3 4. Te4 Sf5#
play all play one stop play next play all
A.Buchanan: So no twinning in the original? What's the point of sDb3? It can be replaced by sB without issue. This problem doesn't appear in WinChloe, so can't check (2021-02-12)
VL: The author could, e.g., prefer the position with black Queen as richer and more hidden for solution. (2021-02-12)
A.Buchanan: Hi Valery well dressing the board is a hypothesis. There is no try to compensate for loss of economy. Indeed here sD forms a distracting battery which “pins” sLd3. It’s all too easy to add this kind of “richness” (noise) to a problem but more is less in my humble opinion. Unless it was a thematic tourney or we are missing a twin, my top hypothesis is that the composer did very well without a computer to make a sound h#4, and it’s easy these days to see a minor improvement. (2021-02-13)
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comment
Keywords: En passant as key, En passant
Genre: h#, Retro
Computer test: C+ Gustav 4.1d
FEN: 8/8/8/8/rpPk3p/1q1b3K/3Pp1rn/4N3
Reprints: 49 PCFT 1944-1945
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-02-13 more...
58 - P0003206
Hans Joachim Schlüter
4443v Schach , p. 349, 11/1963
P0003206
(5+8)
h#2
b) wTf1 tauschen mit wLg1
a) 1. Kxb4 Lb6 2. a4 Tb1#
b) 1. cxb3ep gxf3 2. Sc1 Tg4#
play all play one stop play next play all
Korrektur 1964, Seite 155: sBh3->f3
Henrik Juel: C+ Popeye 4.61 (2022-11-24)
comment
Keywords: En passant as key
Genre: h#, Retro
FEN: 8/8/8/p7/kPp5/p1p2p2/4n1Pp/5RBK
Input: Gerd Wilts, 1995-06-03
Last update: Felber, Volker, 2022-11-24 more...
59 - P0003207
Klaus Gerber
4497 Schach , p. 30, 01/1964
P0003207
(4+9) C+
h#3
1. exf3ep c3 2. Df7 Ke4 3. Dh5 exf3#
play all play one stop play next play all
A.Buchanan: Can shift bQd5 to c4, while replacing bRg5 with bP to reach Meredith status. bBh4 can also be downgraded to bP. But probably even more economy is possible by shifting pieces one file to the right, keeping the ep, White tempo & model mate, which seem to be the main features (2022-11-30)
more ...
comment
Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 8/8/8/3q2r1/2p1pPkb/4K1pp/2PrP3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-30 more...
60 - P0003211
Tivadar Kardos
6476 Skakbladet 10/1957
P0003211
(7+14) cooked
h#3
1. dxe3ep+ Kxc4 2. Lc3 Sxc2 3. Da5 axb3#
play all play one stop play next play all
See the correction P1396159
Cook: 1. dxe3ep+ Kxc4 2. Td5 Sxc2 3. Ta5 axb3#
A.Buchanan: Easy enough to eliminate what I suppose is the intended solution (sL unblocking sD and blocking sT) by e.g. just removing sDf5. However I don't see a way to say goodbye to the cook instead. (2021-11-23)
James Malcom: I believe I may have a fix Andrew. If we substitute the Black queen with a Black rook, the same theme is kept even is if considered watered down.

h#3 8/3p4/3p1p2/3Kbr2/k1ppPp2/1n1r3P/P1p4P/N3nq1b (2021-11-24)
A.Buchanan: Hi James, hurray cool I thought I couldn't get through that way: I was confused in my mind about wPe2 blocking that line, but with the freshness of a new day, I can look at it again. The precise diagram that you propose doesn't quite work because R: 1. e3-e4 Sf3*e1+,Sg2*e1+,g2xf1=Q+,g2xh1=Q+, but this is all fixable with +bPd2,-bSe1. Now Popeye 4.87 says the forward solution is works uniquely, and Retractor 2.0 says R: 1. e2-e4 Rf3-d3+ 2. Ke4-d5 are unique. So for full C+ we only need some demo game that results in the position prior to Ke4-d5.
I don't see any "watering down" here: I think we have faithfully represented Tivadar's idea. Let's post it here as TK, corr JM&AB? (2021-11-24)
A.Buchanan: I have made an unedifying demo game with lichess, so our correction is fully C+. (2021-11-24)
James Malcom: "TK, corr JM&AB" sounds good to me. (2021-11-24)
more ...
comment
Keywords: En passant as key, Superseded by (P1396159)
Genre: h#, Retro
Computer test: Popeye C-Version 4.87 & trivial retro-logic
FEN: 8/3p4/3p1p2/3KbqP1/k1ppPpp1/1n1r3P/P1p4P/N6b
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-24 more...
61 - P0003221
Thomas R. Dawson
1511 The Chess Amateur 10/1929
P0003221
(9+16) C+
h#2
1. ... bxc6ep 2. Le7 cxd7#
1. Lc6? bxc6 2. Le7 cxd7# WTM
R: 1. c6-c5? Kd3-e3 2. Dc5-c4+ Ke3-d3 3. Dc4-c5+ retro-loop.
R: 1. c7-c5! Kd3-e3 2. Dc6-c4+ Ke3-d3! clear
play all play one stop play next play all
Easy to see no last move for White. Moreover, Black just double-hopped to allow sD/wK to avoid retro-loop.
Single clean retro try.
more ...
comment
Keywords: En passant as key, No legal last move for White
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & moderate retro-logic
FEN: 3nkr2/3p1r2/1p1npb2/PPpb4/Bpq2PP1/1p2Kpp1/2PPP3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
62 - P0003236
Thomas R. Dawson
10381 The Fairy Chess Review 12/1955
P0003236
(9+16) C+
h#2
1. ... bxc6ep 2. Dxa5 Db7#
1. Sc6? bxc6 2. Dxa5 Db7# but WTM
R: 1. ... c7-c5 2. g5-g6 Sc6-e7 3. De7-b4,D~ Sb4-c6+ unpromote wD on h8, etc.
play all play one stop play next play all
Clean retro try
Henrik Juel: Typo: 2.Dxa5 Db7#. wTa1 is missing, I think (10+15). -1... c7 -2.g5 Sc6 -3.De7 Sb4, unpromote wD on h8, etc. (2004-03-18)
A.Buchanan: I agree Henrik and have made changes. A number of these problems have been marked "Weisz zieht an", which is inappropriate as part of the point of the problem is to deduce this. I think it's 2. c2-c3! not 2. g5-g6? or the cross-captured pawns become retro-locked. (2021-10-21)
Henrik Juel: 2.g5-g6 also works, because the only cross-capture is axb,bxa
White captured [Pd7] with an officer (2021-10-22)
Henrik Juel: HC+ Popeye 4.61 (2021-10-22)
A.Buchanan: If g5-g6 also works, then can’t we have both? I.e. White moved last. I would be surprised if this was TRD’s intention, given the others of this general style (2021-10-22)
Henrik Juel: TRD died in 1951, so someone else may have messed up the problem
FRC is not generally available after 1953, so I cannot check the source (2021-10-22)
A.Buchanan: WinChloe has 29(!) posthumous problems by Dawson - the latest in 1958, FCR's final year. The composition in question is not one of them. (2021-10-23)
A.Buchanan: PROOF OF COOK + SUGGESTED FIX
I agree with much of what you say Henrik.
Capturing history. Basically 3 possibilities for the Black pawns.
Notation: ~,o,| denote "cross-capture", "original" & "waylaid"(=captured on home file by officer).
1) a~b co d~e f~g. This implies either (wPhxg & bPh=X) or (wPh=X & bPh|)
2) a~b co d~e hxgxf f|
3) a~b co hxgxfxexd d|
Suppose WTM. Then R: 1. c6-c5 c2-c3 2. c7-c6 g5-g6 3. Sc6-e7 De7-b4 4. Sb4-c6+ Kc5-c4 5. Lb2-d4+ etc. Now we're free to uncapture wDh8 (or g8 for scenario 1), and any of the three pawn scenarios can work. No e.p. rights means a cook.
If we fix it by e.g. shifting wPg6-g5 then I think that fixes this cook, and also it means that there is no solution if BTM.
Is there any possibility for premature capturing hxg? Not by Bl because scenario 1 says bPfxg, and the other two scenarios rely on wPh=X. If by Wh, then there's no waylaying on d or f, so we are back to scenario 1, but bPh is blocked from promotion, so that won't work.
But: what about R: 1. c6-c5 c2-c3 2. Sc3-b1 Th1-a1. Doesn't this completely cook it anyway?? How about k7/4n3/6P1/PPpPPP2/qQKbrrp1/ppRppp2/b1P5/1n6? h#2 (with Art 15 so it's really h#1.5)
Do you agree? (2021-10-23)
Mario Richter: In the diagram, both for WTM and BTM the retraction sequence c2-c3 Sc6-e7 De1-b4 Sb4-c6+ works and furthermore Andrew's R: 1. c6-c5 c2-c3 2. Sc3-b1 Th1-a1 shows, that Henrik's guess of the omitted wTa1 might be wrong. Instead, I think that Andrew's modification might be in fact the original Dawson diagram. (2021-10-23)
A.Buchanan: Yes that makes sense. I can well believe that there was a lot of constructions in flight when TRD passed away. It's not surprising that something appeared a few years later. Can we get access to FCR from that year to validate? (2021-10-23)
A.Buchanan: Brian Stephenson kindly checked the original magazine. He wrote:
"Eventually got round to looking out this TRD problem. It was published in the issue you quote as problem 10381. The diagram in PDB is wrong. There should be a bPc2. The solution was published in FCR April 1956 as: 1.Pb5xPc5ep Qxa5 2.Qb7#. Last moves must have been 1.Pc7-c5 Pg5-g6 2.Sc6-e7 Qe7-b4 Sb4-c6 ch etc. No flaws were noted, but I have looked later than that issue. Hope this helps." (2021-11-13)
A.Buchanan: The appearance of the 16th Black unit thanks to the offices of Brian Stephenson simplifies the retro logic substantially, and the solution seems sound. Amazing how much disruption a simple typo can cause, but at least this one I believe is laid to rest (2021-11-13)
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comment
Keywords: En passant as key, No legal last move for White
Genre: h#, Retro
Computer test: HC+ Popeye 4.6 + non-trivial retro
FEN: k7/4n3/6P1/PPpPPP2/qQKbrrp1/ppPppp2/b1p5/1n6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
63 - P0003240
Gabor Tar
Magyar Sakkelet 06/1972
P0003240
(4+10) C+
h#2
b) sLe3 nach f3
a) 1. f3 Lxa7 2. Lg5 Lxf2#
b) 1. fxg3ep Lxc7 2. Lh5 Ld8#
play all play one stop play next play all
Henrik Juel: a) C+ Popeye 4.61
b) the ep capture is unjustified, last move could be Ke2-f1 or f3xg4 (2023-08-06)
A.Buchanan: There was an error in the stipulation: "b) sLe3 nach h2 (-wSh2)" is completely incorrect. However "b) sLe3 nach f3" leads to a very nice problem which matches the solution already present. This problem is not in WinChloe, but I am sure this is the resolution of the mystery (2023-08-07)
Henrik Juel: You are probably right, Andrew; maybe there was a misprint in the source
Now part b) is C+ Popeye 4.61 after very simple analysis (2023-08-07)
comment
Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye v4.61 & simple retro-logic for b)
FEN: 1B6/npp5/8/8/5pPk/4b2p/3p1p1N/5Kn1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-08-07 more...
64 - P0003269
Tivadar Kardos
149 British Chess Magazine 10/1956
P0003269
(5+17) cooked
h#3
1. exf3ep e3 2. 0-0-0 Tb4 3. Tg4 Tb8#
I think that this cooked problem is an early A osteriori. I can see ways that it might be made sound, but I would like to see the published version. San anyone help please? Thanks.
play all play one stop play next play all
Cook: 17 Black pawns, retro cook and numerous forward cooks
klären wK im Schach, vielleicht wBb2?
Alfred Pfeiffer: 9 schwarze Bauern! (2012-02-07)
Ladislav Packa: Auch mit wBb2 NL, z.B. 1.exf3 e.p. Txg5 2.0-0-0 Tb5 3.Tg~ Tb8# (2012-02-07)
A.Buchanan: And adding to earlier comments, if we do swap sBb2 for wB, the retraction is still not unique with R: 1. f3-f4 Lh6xg5+ (2021-11-26)
A.Buchanan: I think that this cooked problem is an early A Posteriori. I can see ways that it might be made sound, but I would like to see the published version. Can anyone help please? Thanks. (2022-04-25)
Gerald Ettl: Verbesserungsvorschlag: -sBa2, -sBb2, -sSg7, +sBg7
r3k3/3b2p1/5p2/6b1/4pPRp/2pq2rp/2p1P1pB/2K3n1/ (2022-04-25)
Gerald Ettl: und +sSg8 (wegen exf4 Vermeidung) (2022-04-25)
Gerald Ettl: +sSa4 nicht g8 (2022-04-25)
A.Buchanan: Hi Gerald. Thanks for this. Your final proposed version is r3k3/3b2p1/5p2/6b1/n3pPRp/2pq2rp/2p1P1pB/2K3n1, yes? There are 16 candidate solutions, all using castling. But wPa cannot leave the a-file, and is required for capture balance, so Black cannot castle. Five of the candidates begin with e.p.: play might have just been R: 1. Kb1-c1 b3xPc2+, so e.p. is not permitted in any case. What am I missing? (2022-04-25)
Gerald Ettl: Hi Andrew,
Those were 2 mistakes I made. I improved it again.
r3k3/3b4/4pp2/6b1/4pPRp/n1pq2rp/PPp1P1pB/2K3n1/
(den wBa2 habe ich auf das Brett gestellt, da sonst wieder Lh6xXg5 geht.) (2022-04-26)
A.Buchanan: Hi Gerald: Deine Retro-Logik ist gut. Aber es gibt 18 Lösungsvorschläge für die Zukunft. Ich habe heute im Discord meinen eigenen Vorschlag zur AP-Korrektur von Kardos veröffentlicht. Ich werde ihn hier im PDB hinzufügen. (2022-04-26)
comment
Keywords: En passant as key, Castling
Genre: h#, Retro
FEN: r3k3/3b2n1/5p2/6b1/4pPRp/2pq2rp/ppp1P1pB/2K3n1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-26 more...
65 - P0003281
Fritz Hoffmann
Problem 08/1974
P0003281
(4+11) C+
h#2
3.1...
1) 1. Tb1 Lxf3 2. Tb3 Lc6#
2) 1. Lc6 Le4 2. Lb5 Lc2#
3) 1. cxb3ep Lxc3 2. Lc6 Lxc6#
play all play one stop play next play all
more ...
comment
Keywords: En passant as key
Genre: h#, Retro
Computer test: C+ Popeye 4.61
FEN: 8/8/1b6/B7/kPp3p1/p1p2b1p/6Bp/3r1n1K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-05 more...
66 - P0003287
Tomislav Petrovic
(VIII) Problem 37-40 09/1956
P0003287
(5+9) C+
h#2
1. fxg3ep h8=S 2. Sg4 Sxg6#
play all play one stop play next play all
A.Buchanan: The point I think is that the en passant is a tempo move. Otherwise wBf2 serves no purpose. How about sBh5 instead of sT? (2021-10-19)
more ...
comment
Keywords: En passant as key, Promotion (S), Tempo Move
Genre: h#, Retro
Computer test: HC+ Popeye v4.61 & trivial retro-logic
FEN: 8/7P/6pP/6pr/5pPk/4np1n/5P1p/7K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-19 more...
67 - P0003288
Vladimir Korolkov
211 Biuletyn 03/1960
P0003288
(9+10)
h#2
1. fxg3ep Lxe3 2. Ld5 cxd5#
play all play one stop play next play all
Henrik Juel: C+ Popeye 4.61 and very simple analysis (2022-08-09)
comment
Keywords: En passant as key
Genre: h#, Retro
FEN: 8/8/8/1PP1P2r/PRPBbpPk/ppKppp1r/8/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-08-09 more...
68 - P0003291
Jozsef Korponai
5412 Arbejder-Skak 02/1961
P0003291
(11+13) C+
h#2
1. hxg3ep Sd4 2. gxf2 Se3#
play all play one stop play next play all
vgl. P0003180
Henrik Juel: Last move was g2-g4 giving Black a preceeding move like Da8-b8 or Lb7-c8
C+ Popeye 4.61 (2021-04-24)
comment
Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 + simple retro thought
FEN: 1qb5/b1pp1ppp/pp6/6P1/5BPp/2P5/1PN1PP1P/3Nrk1K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-25 more...
69 - P0003297
Wilhelm Hagemann
Deutsche Nachrichten (Sao Paolo) 1960
P0003297
(5+9)
h#2
1. cxd3ep f4 2. Lf3 Kxc5#
play all play one stop play next play all
Henrik Juel: C+ Popeye 4.61 (2022-04-18)
Henrik Juel: Obviously, last move was d2-d4 (2022-04-18)
comment
Keywords: En passant as key
Genre: h#, Retro
FEN: 8/8/p7/P1r2p2/RKpPk3/p1p1p3/5P2/3b4
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2022-04-18 more...
70 - P0003312
Gerhard Paul Latzel
2680 Deutsche Schachzeitung 02/1970
P0003312
(7+14) C+
h#3
1. cxb3ep e5 2. Lh6 g5 3. Txh4 Txh4#
play all play one stop play next play all
Henrik Juel: The only possible last move is b2-b4, so the ep key is permissible
C+ Popeye 4.61 (2021-07-26)
A.Buchanan: The main idea is to unblock the cluttered 4th rank, and this is complemented thematically by *blocking* four other lines: g8-g4, h8-h4, h8-d4 & "blocking the blocker" h6-f4.
The last move was b2-b4, so by the ep rule, capture is definitely legal. The convention only serves to resolve ambiguous situations where multiple possible histories exist. In retro problems, such ambiguity is rare if the retro logic does its job. However the pessimistic nature of the e.p. convention motivates *why* the retro play must exist. (2021-07-26)
comment
Keywords: En passant as key
Genre: h#, Retro
Computer test: Last move was b2-b4, so by ep rule, capture ok as key. HC+ Popeye 4.61
FEN: 6rq/pn6/K7/8/kPprPbPP/p1pppppR/7P/8
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2021-07-26 more...
71 - P0003316
Eduard Schlatter
590v Schach-Echo 05/07/1955
P0003316
(4+15)
h#4
1. bxc3ep Kxa3 2. Ld5 Sg2 3. Tce1 Kb4 4. T1e4 bxc3#
play all play one stop play next play all
Henrik Juel: C+ Popeye 4.61
The only possible last move is c2-c4 (2021-08-04)
comment
Keywords: En passant as key
Genre: h#, Retro
FEN: 8/2p5/2p3pq/n3r3/KpPk1pp1/bp1p4/1P4b1/2r1N3
Input: Gerd Wilts, 1995-06-03
Last update: Dieter Berlin, 2021-08-04 more...
72 - P0003323
Peter Takacs
Imre Telkes

285 The Chess Amateur 12/1922
P0003323
(3+13) C+
h#2
1. gxf3ep Dxh2 2. Ld3 Dxh4#
play all play one stop play next play all
more ...
comment
Keywords: En passant as key, Selfblock, Line opening
Genre: h#, Retro
Computer test: C+ Popeye 4.61 and very simple analysis
FEN: 6n1/2r4q/3pK3/7p/4kPpp/4r1pQ/7n/b4b2
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-06 more...
73 - P0003325
Tivadar Kardos
4272 Arbejder-Skak 08/1956
P0003325
(6+13) C+
h#5
a) 1. gxf3ep g4 2. Kd5 gxf5 3. Kc6 fxe6 4. Kb7 exd7 5. Ka8 dxc8=D#
play all play one stop play next play all
A.Buchanan: Popeye v4.87 (via Olive v1.4 ) for PDB problem P0003325 h#5 delivers the correct solution with en passant set to f2f3f4. However no solution when the "intelligent" flag is also set. Reported to Popeye Github. (2021-11-23)
Henrik Juel: Popeye 4.61 with 'opt var int enp f3' works fine
Maybe you just have to change f2f3f4 to f3, Andrew (2021-11-23)
A.Buchanan: Thanks for the suggestion, Henrik, but even without the "intelligent", opti vari enpa f3 fails with v4.87. I think that the syntax has been rendered more complicated since v4.61 in order to support fancy fairy en passants. (2021-11-23)
comment
Keywords: En passant as key, Promotion (D), Excelsior, Pac-Man
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 3.52 & trivial retro-logic
FEN: 2n5/p2p4/4b3/p4p2/P3kPp1/Ppp5/Pp4P1/bK2n3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-23 more...
74 - P0003326
Tivadar Kardos
4252 Arbejder-Skak 07/1956
P0003326
(3+13) C+
h#5
1. dxc3ep Kxg7 2. f6 Kxf6 3. e5 Kxe5 4. Lc4 Kd4 5. Lb5 dxc3#
play all play one stop play next play all
A.Buchanan: Can be done with lighter material e.g. sLa5, sBa4, sSa3g7 (2021-11-23)
more ...
comment
Keywords: En passant as key
Genre: h#, Retro
Computer test: Popeye C-Version 4.87 & trivial retro-logic
FEN: 6bK/5pr1/2p1p3/p4p2/rkPp4/qp1p4/3P4/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-23 more...
75 - P0003328
Tivadar Kardos
4311 Arbejder-Skak 09/1956
P0003328
(7+16) C+
h#3
1. cxb3ep Sb4 2. Sc4 Sc2 3. Sa5 Sxc3#
play all play one stop play next play all
Henrik Juel: The only possible last move is b2-b4
C+ Popeye 4.61 (2021-04-24)
comment
Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & simple retro thought
FEN: 8/2p5/2P5/2p2p2/kPp2Pn1/n1pp1Kp1/N2P2pb/rNr1q2b
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-25 more...
76 - P0003335
Tomislav Petrovic
(XXI) Problem 37-40 09/1956
P0003335
(5+9) cooked
h#9
1. dxc3ep dxc3+ 2. Ka5 c4 3. d2 c5 4. d1=L c6 5. Lg4 hxg4 6. Kb6 g5 7. Ka7 g6 8. Kb8 g7 9. Kc8 g8=D/T#
play all play one stop play next play all
Cook: 1. b2 c5 2. b1=S c6 3. Sc3 dxc3+ 4. Ka5 cxd4 5. Kb6 d5 6. Ka7 d6 7. Kb8 dxc7+ 8. Ka7 c8=D 9. a5 Db7#
paul: See P0003212 as version. (2011-08-06)
Anton Baumann: C+?? Die Aufgabe ist eindeutig inkorrekt! (2021-01-27)
James Malcom: Fixed. (2021-01-27)
comment
Keywords: En passant as key, Promotion (D), konsekutive Umwandlungen 2 (L, D/T), En passant, Kindergarten Problem
Genre: h#, Retro
FEN: 8/2p5/p7/8/1kPp3p/1p1p2pP/3P2Pp/7K
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-01-27 more...
77 - P0003337
Rudolf Buljan
Problem 41-44 03/1957
2. Preis
15. Thematurnier
P0003337
(11+15) C+
h#2
1. ... exf6ep 2. Sf5 f7#
play all play one stop play next play all
White captured fxg and promoted [Pd2] on d8
Black captured axbxa2-a1=Y, b7xc6, d5xe4, and gxf
White cannot have moved last, and in fact Black's last move can only have been f7-f5 in order to give White a prior move. R: 1... d5-e4 would have retro-blocked wPd marching to its promotion.
So the retroplay was R: 1... f7-f5 2.f5xYg6
Numerous retro tries but none distinct.
"Autor Ing. Rudolf Buljan, Zagreb"
AB: Have therefore marked him in the author table as Croatian. Yugoslavia has been removed from the PDB country list, I see.
A.Buchanan: Any ideas how to distinguish usage of “whose move?” and “no legal last move for white” (or black) keywords (2020-10-01)
Henrik Juel: This problem is clearly the latter
I would use 'Whose move?' in retro problems where a deeper analysis is required to determine the move, often with stipulations like '#1 (who?)' (2020-10-01)
A.Buchanan: I think "Whose move?" is where there is no forward stip, or where we have to determine if the forward stip should be interpreted as orthodox or as half-duplex. "no legal last move for..." is where the mater/stalemater/etc is the usual one, but we add or remove a single move at the beginning of the forward stip. I guess a third possibility is "last move?" where in a Type A position we also have to figure out who moved last.
With respect, I don't think the level of analysis can be relevant. (2023-12-06)
more ...
comment
Keywords: En passant as key, No legal last move for White
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: 5nkb/1qp1p1n1/2p3PQ/4PpKb/4prRp/P4prB/1PP3PP/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
78 - P0003339
Tivadar Kardos
2519 Revista Romana de Sah 01/1948
2. Preis
P0003339
(10+15) cooked
h# in wieviel?
1. ... cxd6ep 2. Tc8 dxe7 3. Tf8 exf8=S#
play all play one stop play next play all
AL ('Revista Romana de Sah', 06/1948, S.225-226):
Bekanntlich beginnt im Hilfsmatt Schwarz, also hat in der Diagrammstellung Weiß den letzten Zug gemacht. Man findet aber, wenn man die Position analysiert, daß Weiß den letzten Zug nicht gemacht haben kann.
Der weiße Doppelbauer auf der d-Linie läßt sich nur durch Schlagen des einzig fehlenden schwarzen Steines, des sBa7, erklären. Dieser konnte aber erst nach Umwandlung geschlagen werden. Dazu mußte er zunächst bis a3 vorrücken, dann nach b2 schlagen, und dann umwandeln.
Also kann Weiß nicht zuletzt b2-b3 gezogen haben.
Die anderen schwarzen Bauern haben mindestens 5 Schläge gemacht, um die Position im Diagramm zu erreichen (man beachte, daß der sBh7 zweimal schlagen mußte, um den wBh2 durchzulassen)
also kann auch g2-g3 nicht der letzte Zug gewesen sein, denn sonst hätte der wLf1 nicht von einem sB geschlagen worden sein können.
Für alle anderen weißen Steine ist leicht zu sehen, daß sie den letzten Zug nicht gemacht haben können.
Es stellt sich heraus, dass der letzte Zug von Schwarz gemacht wurde, und eine einfache Analyse zeigt, daß dies nur möglich ist durch d7-d5 (sonst wäre es unmöglich, daß die weißen Steine ihre Position im Diagramm erreichen könnten).
In diesem Fall Weiß kann 1. ... c5:d6(e.p.) spielen, gefolgt von 2. Tc8 d:e7 3. Tf8 e:f8C matt!
Das kürzeste Hilfsmatt wird also in 2,5 Zügen erreicht!
Cook: 1. ... Txg5+ 2. Kxg5 f3 3. Lg6 gxf4#
1. ... Kxd5 2. Kf6 e4 3. Tg6,Dg6 e5#
A.Buchanan: Popeye v.487. WinChloe has the same diagram, and although it just claims 1 solution, it's C? Yet WinChloe engine gives the cooks right away (2021-10-27)
Mario Richter: Shouldn't this problem get the Label "No legal last move for White"?
(White pawn d2 has captured the missing black piece (Pdxe), so black pawn a7 had to promote to get to the capture square. This requires one capture (Pa3xb2), together with bxc,cxd,hxg,gxh,gxh this accounts for all missing white pieces.)

Btw. "for me, "C?" stands for "Not yet computer-tested or computer test didn't come to a conlusion" ... (2021-10-28)
A.Buchanan: Hi Mario, thank you.
(1) The stipulation is non-standard. I think this is why the WinChloe database doesn't include a solution, and just marks it as "C?" However if I run the WinChloe solving engine manually it (like Popeye) gives 121 h#3 solutions and 4 h#2.5 solutions. I think you, me and Christian would all agree on the meaning of "C?"
(2) The problem does not rely on Article 15 to eliminate the first single move, but implicitly requests the shortest. Currently the PDB usage of "No legal last move..." is to indicate that Codex Article 15 applies, so I did not add it. But I'm ok if you want me to add the keyword. In my correction, I shift the stipulation to h#3 so it's clear that Article 15 and not any desire for brevity which push the problem to h#2.5.
(3) However the PDB usage is confused. Ideallly think the two "No legal..." keywords should be replaced by "Retro-stalemate", while two new keywords "Article 15" & "Whose mate?" indicate the choice of implications of the retropat for forward play. "Board rotation" and other jokes are also implications of retropat (and of illegal diagram).
(4) But this is complicated to implement, requiring either database-level access or screen-scraping. The latter is quite practical for some future time when I have more leisure.
(5) https://www.stere.ro/biblioteca-digitala/reviste/revista-romana-de-sah/ is an excellent archive of Revista Romana de Sah. My sincere compliments and thanks to whoever prepared it. And the magazines themselves contain thorough annual indices. p225-6 of 6/1948, contains the solution to Kardos. I haven't translated it, nor have I found time to locate the prize announcement.
Comments welcome! (2021-10-29)
Mario Richter: I still believe, that the label "No legal last move for White" is justified here, because this is exactly what the author wanted to show.
(And this is too, what the reprint in 'Problem (Zagreb)' 37-40 09/1956 is all about - it appears there inside asn article

Btw., I wouldn't call the presentation of the solution in 'Revista Romana de Sah', p.225, 06/1948, a "reprint" - it lacks the presentation of the position ... (2021-10-30)
A.Buchanan: Added the keyword.
Yes I know that's the usage of "reprint", but then there should be a field for the location of the solution :-) It's idiotic that if I want to say where the solution is contained, I have to embed it in text like this: 2519 Revista Romana de Sah , p. 225-6, 06/1948 (2021-10-30)
more ...
comment
Keywords: En passant as key, Promotion (S), Superseded by (P1394978), No legal last move for White
Genre: h#, Retro
Computer test: Popeye v4.87 & WinChloe both say cooked
FEN: 7b/4p1nP/2r1P1kq/1nPpKbrp/2pp1pRp/1P4P1/P3PP2/8
Reprints: (III) Problem 37-40 09/1956
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-30 more...
79 - P0003343
Yaakov Mintz
344 Canadian Chess Chat 11-12/1980
P0003343
(4+15) cooked
h#4
1. hxg3ep Lxb1 2. Kf5 fxe4 3. Kg6 e5+ 4. f5 exf6ep#
play all play one stop play next play all
Cook: NL:
1. hxg3ep fxe4 2. f6 e5 3. Kf5 exf6 4. Kg6 Lxb1# uvm
YM: The correct mechanism is in P1109109 (2010-07-13)
YM: Correction option: P1229434 (2021-06-28)
comment
Keywords: En passant as key, Superseded by (P1109109)
Genre: h#, Retro
FEN: q2r4/4Kp2/3r3b/3p2pp/4nkPp/1n3P2/B4p1p/1b6
Reprints: 805 FIDE Album 1980-1982 1988
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-28 more...
80 - P0003347
Marko Klasinc
3. Makuc-Moder-Gedenkturnier 1971-1973
1. Lob
P0003347
(7+6) C+
h#2
1. axb3ep bxc6+ 2. b5 cxb6ep#
play all play one stop play next play all
Marko Klasinc: The most economical h#2 with two e.p. captures. (2002-01-28)
A.Buchanan: Very nice - and i think this is unbeatable at least in the case where ep is first and last move (2020-08-02)
A.Buchanan: Also works as set play! Just truncate Black's first move! :) (2020-12-09)
more ...
comment
Keywords: En passant as key, Economy record, En passant as mating move
Genre: h#, Retro
Computer test: C+ Popeye v4.61 + a little thinking
FEN: 8/1p6/B1p5/RPP5/pPkp4/K1p5/P7/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-12-09 more...
81 - 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...
82 - P0003365
Gyula Bebesi
41 Problemas 04-06/1962
P0003365
(8+14) cooked
h#2
1) 1. axb3ep bxc6+ 2. b5 cxb6ep#
2) 1. exd3ep g8=D,L+ 2. d5 cxd6ep#
play all play one stop play next play all
PRA: 1 solution with 2 parts
Henrik Juel: White captured [sLc8] on c8 and axb, so last move was either b2-b4 or d2-d4
C+ Popeye 4.61, except for the promotion dual (2020-08-03)
A.Buchanan: There's no keyword for an untolerated dual promotion! (2020-08-03)
Henrik Juel: Yes and no, Andrew
What about the cooked label? (2020-08-03)
A.Buchanan: Hi Henrik: this is obviously intentional. I tried to replace the pawn promotion with wDh8-h7+ but I ran out of pieces. I guess that Bebesi had the same difficulty. The word "cooked" seems too strong to me, but it's obviously a significant defect though (2020-08-04)
A.Buchanan: The bar is higher for helpmates than other problems, and this is unsound. However it’s not cooked but dualled. Yet all we have is “cooked” so I suppose it is in PDB-speak. I’ve added that and removed the 2.1… (2023-08-07)
Joost de Heer: Is there an 'unsound' keyword? (2023-08-07)
A.Buchanan: Hi Joost: no there isn't. It can be added, the challenge for any new common keyword would be populating it for more than a few values. (2023-08-07)
A.Buchanan: This time round, I've managed to find a fix. Bebesi's correction in WinChloe 66992 has non-standard wL. I have a feeling I've seen a similar one by Andrey Frolkin & someone else recently with a different matrix, and also non-standard wL. I've sent my effort to Joaquim Crusats at Problemas for his assessment (2023-08-08)
more ...
comment
Keywords: En passant as key (2), Partial Retro Analysis (PRA), En passant as mating move (2), Superseded by (P1411659, P1413906)
Genre: h#, Retro
FEN: 3qn3/1p1p2P1/B1r3p1/1PP1Kp2/pPkPp3/p1p4n/N7/2b5
Reprints: 233 Ungarische Schachproblem Anthologie , p. 188, 1983
1 Problemas 44, p. 1456, 10/2023
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-04 more...
83 - P0003366
Laszlo Lindner
3040v Europe Echecs 07-08/1982
P0003366
(8+16)
h#2
b) sBc7 nach f6
a) 1. Dc4 Lxb6+ 2. c5 dxc6ep#
b) 1. fxe3ep Lxb4 2. Sc4 bxc3#
play all play one stop play next play all
Lindner in 'Mattbilder eines Lebens':
In a) ist die Lösung der stellung b) nicht möglich, weil das e.p.-Schlagen durch Schwarz nicht legal ist. Der letzte Zug von Weiß muß nicht unbedingt e2-e4 geweseb sein. Es kommt als letzter zug auch Kh3-g2 in Betracht, mit den vorherigen Zügen h4:g3 e.p.+ und g2-g4.
In b) demgegenüber sind Kh3-g2 und vorher f4:g3 e.p. illegal, weil die Rücknahme von g2-g4 unmöglich ist: der sB würde 7 Schlagfälle benötigen, und es fehlen nur 6 weiße Steine. Der letzte weiße Zug muß also e2-e4 gewesen sein.
In 'Mattbilder eines Lebens' abgedruckt mit sTh7 statt h8 und der Quellenangabe: Europe Echecs, 1964
AB: (1) Where is wK?
(2) Why is 1.fxe3ep legal in b) but not a)? (2002-01-31)
Henrik Juel: wK is probably on g2. In part a) last move could have been Kh3-g2, I think (2002-02-01)
A.Buchanan: Very convincing, Henrik. I've repaired the diagram accordingly. (2023-05-28)
comment
Keywords: En passant as key, En passant in the retro play
Genre: h#, Retro
FEN: 7r/2pn4/1nqRb3/B2Pp3/pb1kPp2/2p2Pp1/1PP2pKp/7r
Reprints: 501 Mattbilder eines Lebens , p. 379, 1996
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-05-28 more...
84 - P0003369
Valerian Onitiu
6643 The Fairy Chess Review 02/1946
P0003369
(14+11) C+
h#1*
1. ... fxg6ep#
1. dxe3ep fxe3#
play all play one stop play next play all
Gerald Ettl: Der rumänische Autor zeigt, dass der letzte Zug jeweils ein Doppelschritt g7g5 oder e2e4 gewesen sein muss. Die wBc6,d7 schlugen ueber Kreuz und der wUW-L hat bxaB und a7xXb8 geschlagen. Die Forderung lautet wohl h#1 (2010-07-12)
Alfred Pfeiffer: Ja, die Forderung ist h#1*.
Online: "http://problem64.beda.cz/silo/fcr_4_1946.pdf" (2010-12-15)
more ...
comment
Keywords: En passant as key (2), Promotion (L), Non-standard material (L)
Genre: h#, Retro
Computer test: C+ Popeye 4.61 and analysis
FEN: 4b3/1p1P1p2/2PRBR1P/2prpPpK/3pPkr1/5pN1/P4P1B/6B1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-30 more...
85 - P0003384
Tivadar Kardos
4299 Stella Polaris 12/1971
P0003384
(9+16) C+
h#2
b) alles eine Reihe nach links
a)
1. gxf3ep Sxg7 2. d5 Lf5#
1. cxd3ep? Sxc7 2. f5 Ld5#
b) 1. bxc3ep Sxb7 2. e5 Lc5#
1. fxe3ep Sxf7 2. c5 Le5#
play all play one stop play next play all
Sally: a)Der letzte Zug war: Bf2 - f4!
b)Der lrtzte Zug war: Bc2 - c4!
Nr. 139 200 Ausgwwählte S. Pr. T. Kardos (W. Frentze 1983) (2010-09-30)
Henrik Juel: Black pawns captured all 7 missing white men, incl. [Pa1], which promoted on a8
Only possible last moves are d2-d4 and f2-f4
In each twin the closing off for an original white bishop determines the last move (2021-11-22)
Henrik Juel: HC+ Popeye 4.61 (2021-11-22)
comment
Keywords: En passant as key, Twin by board shift
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & Retractor 2.1.1 Human contribution is "positions prior to retractions look legal enough."
FEN: 2bKNb2/2rn1pqn/3pBp2/4B1P1/2pPkPpr/2p1p3/1P2p2P/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-27 more...
86 - P0003387
Tivadar Kardos
(4) Land og Folk 22/12/1973
P0003387
(6+15) C+
h#1
1. bxc3ep Lxc5#
1. fxe3ep Le5#
R: 1. c2-c4,e2-e4
play all play one stop play next play all
Originalforderung? "2 Lösungen"
A.Buchanan: The original stipulation here asks for 2 solutions. I don't think this would work as 2 solutions under Retro Strategy protocol which was the default in those days. Dubious e.p. captures are just not permitted. This is why AP was invented as a funny work-around.
However this problem works under the PRA protocol, which decomposes the history as "one solution, two parts". Note that this problem does not make use of the e.p. convention, as it is certain that the two e.p. cannot be simultaneously legal. (2021-12-22)
Henrik Juel: I can strengthen the human contribution
The positions before c2-c4 and e2-e4 are surely legal (2021-12-22)
more ...
comment
Keywords: En passant as key, Partial Retro Analysis (PRA)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 + Retractor 2.1.1 Human contribution is "position prior to retractions looks legal enough."
FEN: 3K4/1rp1n3/3Bbnq1/N1p2Qr1/1pPkPpp1/1p1p1p2/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-12-22 more...
87 - 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)
more ...
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...
88 - 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...
89 - 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...
90 - 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...
91 - 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...
92 - P0003431
Tomislav Petrovic
(XIV) Problem 37-40 09/1952
P0003431
(8+14) cooked
h#3
1. cxb3ep+ c3 2. Ld3 Tg4+ 3. Kh1 0-0-0#
play all play one stop play next play all
paul: White captures was axb and g2xh3, so the retro move c3xb4 is not possible. If b3-b4, the retro check is not justified. So last move was b2-b4 (preceded by Rc3-a3). (2011-08-06)
A.Buchanan: There's a lot to unpack here. Firstly, the intended solution has a definite transposition dual in the order of Black's 2nd & 3rd moves. Second, the pawn capture count is wrong. What happened to the original wBh, given sBh2 never left the h-file? Thirdly, and not a showstopper, White's only solution involves castling: so that's a logical dual. (2020-12-08)
more ...
comment
Keywords: En passant as key, Castling
Genre: h#, Retro
Computer test: Popeye v4.85 + retro thinking
FEN: 3r3n/p7/pp6/bP1q4/RPp5/r3p2P/p1P1P1kp/Rb2K3
Reprints: (61) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-02 more...
93 - 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...
94 - 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...
95 - 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...
96 - 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...
97 - P0003817
Jean Oudot
Nation Belge 1956
P0003817
(14+12) C+
#2
1. cxd6ep
play all play one stop play next play all
paul: Since the wB couldnt leave c1, the captured piece on b6 is the promoted wPh2. If last black move was f7-f5, then h-Pawn captured four pieces to promote on d8, among which bRh8, impossible. Thus, the h-Pawn could only have been promoted from f7 and therefore d7-d5 remains as the last move. (2011-07-02)
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comment
Keywords: En passant as key
Genre: Retro, 2#
Computer test: C+ Popeye 4.61
FEN: qrRNkb2/PpQ1p1pp/1pp1R3/2PpKpP1/6N1/7B/1P1PPP2/8
Reprints: RA15 diagrammes 15 06/1975
(51) Mat Plus 12 10-12/2009
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-27 more...
98 - P0003893
Jean-Francois Baudoin
Jean-Claude Gandy

RA112 diagrammes 38 03-04/1979
P0003893
(13+7)
Wieviele #1?
paul: The question is: How many #1? Must try: 1.0-0#, 1.f5xg6 e.p.# or 1.Kf2#. Obviously, 1.0-0#? is not possible be cause bK penetrated on first range only if wK was dislocated. 1.f5xg6 e.p.# is possible if last moves are g7-g5, f4-f5+, f5xg4. But in this case, wPs captured 9 pieces, bBf8 including and this is impossible. So, the unique mate is 1.Kf2# (2011-08-04)
paul: And the retro-play was Pg6-g5, f4-f5+, f5xS(R)g4, etc. (2011-08-04)
A.Buchanan: See P0003893 (2021-06-30)
comment
Keywords: Cant Castler, En passant
Genre: Retro
FEN: 8/P3pP1p/5p1Q/5Pp1/6p1/PPBP4/BPp1P3/2k1K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-30 more...
99 - P0003901
Michel Caillaud
diagrammes 41 09-10/1979
P0003901
(4+9) cooked
shc#6
1. Df2 2. Db6 3. Kd4 4. bxc3ep 5. Dg6 6. De4 Lb6
play all play one stop play next play all
Cook: 1. Kd4 2. Kc3 3. Db2 4. d2 5. Ld3 6. d4 Lxb4#
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comment
Keywords: Seriesmover, Consequent, En passant
Genre: Retro, Fairies
FEN: 8/8/p7/B2pp3/KpP1k3/Np1p4/q7/1b6
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
100 - 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)
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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...
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