Die Schwalbe

54 problem(s) found in 3562 milliseconds (displaying 54 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND K='En passant' AND S='Die Schwalbe'] [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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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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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 - 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...
22 - 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...
23 - 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...
24 - 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...
25 - 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...
26 - 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
27 - 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...
28 - 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...
29 - 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...
30 - P0004346
Gerhard Paul Latzel
Die Schwalbe 1950
P0004346
(10+3)
#1,5
-sBg5 1. Lxb5,Lxb3 Ta8#
play all play one stop play next play all
-sBg5 (=Vollendung e.p.- Schlag). Nicht Ta1-d1 (=Vollendung w0-0-0), denn Schwarz hätte keinen letzten Zug
Henrik Juel: a slight flaw is that we cannot say whether the entire move was fxg6ep or hxg6ep (2022-07-05)
more ...
comment
Keywords: Castling (wg), Complete an unfinished move, En passant, Joke
Genre: Retro, h#
FEN: 7k/7P/2P1PPPP/1P4p1/b7/1P6/8/R1K5
Reprints: Problem 101-102 09/1966
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-07-10 more...
31 - P0006423
Andrey Frolkin
9128 Die Schwalbe 157, p. 283, 02/1996
Leonid Borodatow gewidmet
P0006423
(9+4)
#3 (AP)
White pushes the move (Keym AP)
1. ... f5! (f6?,fxe6?,fxg6? 2.0-0! ~ 3.Tf3#) 2. gxf6ep! exf6 3. 0-0! f5 4. Tf3#
(2. 0-0=?)
Valladao Task via the try: 1. exf7? e5! 1. ... e6? 2. f8=D e5 3. Da3#
Another try shows fully differentiated black Albino:
1. Kf1? fxg6!
1. ... f5? 2. Lxf5! h3 3. Txh3#
1. ... fxe6? 2. Lxe6! h3 3. Txh3#
1. ... f6? 2. Lf5! h3,fxg5 3. Txh3#,Th3#
And retro tries:
1. 0-0?? f5,~ 2. Tf3# (short solution)
1. f5? f6,~ 2. 0-0?? fxg5,~ 3. Tf3#
play all play one stop play next play all
Wenn Schwarz zuletzt gezogen hat, dann kann Weiß nicht mehr rochieren: R: 1. Kh2-g3 Tf1(g1)-h1+. Da aber Weiß in einem direkten Matt beginnt, ist die Rochade nicht mehr zulässig. Der Zusatz "AP" in der Forderung ist überflüssig: es könnte höchstens der Anzug a posteriori dem Schwarzen übertragen werden. Das geht aber nicht, da Weiß in einem direkten Matt beginnt.
Guus Rol: This is apparently AP after Keym. The move goes to black:
0. ... f5! 1.gxf6ep exf6 2.0-0(justifies the handover) f5 3.Tf3 (2007-02-13)
A.Buchanan: There are two kinds of directmate Type Keym. In one Black unsuccessfully pulls the move, in the other (as here) White successfully pushes it to Black. (2022-04-16)
A.Buchanan: I disagree with the German comment in the solution text. I think including "AP" is advisable in the stipulation. Keym AP riffs off Codex Article 15, but it's not the default. (2022-04-17)
more ...
comment
Keywords: a posteriori (AP) (Type Keym), Castling, En passant, Promotion (D), Valladao Task
Genre: Retro, 3#
FEN: 8/4pp2/4P1N1/6PP/5P1p/6kB/6P1/4K2R
Input: Gerd Wilts, 1996-06-12
Last update: A.Buchanan, 2022-04-18 more...
32 - P0006644
Chris Patzke
9260 Die Schwalbe 159, p. 391, 06/1996
2. Preis Abt. II
P0006644
(13+14)
BP in 16.0
Zeroposition
a) sD nach f3
b) +wBe3
a) 1. d3 a5 2. Lg5 a4 3. Lxe7 Kxe7 4. Kd2 Kd6 5. Kc3 Se7 6. Kb4 Sec6+ 7. Ka3 Le7 8. Sd2 Te8 9. Sf3 Lf8 10. Dd2 Txe2 11. Td1 Te5 12. Le2 Tea5 13. Se5 Df6 14. Lg4 Df3 15. Lxd7 Kd5+ 16. b4 axb3ep+
b) 1. b3 a5 2. La3 a4 3. Lxe7 axb3 4. Lb4 Df6 5. La5 Ke7 6. e3 Kd6 7. Lb5 Se7 8. Lxd7 Sec6 9. d3 Le7 10. Sd2 Te8 11. Sf3 Lf8 12. Dd2 Te5 13. 0-0-0 Tc5 14. Se5 Kd5 15. Kb2 Dd6 16. Ka3 Tcxa5+
play all play one stop play next play all
Henrik Juel: should the diagram have sD on d6? (2021-01-24)
Henrik Juel: With this change part a) (i.e. the current diagram position) is C+ Euclide 1.01
For part b) I stopped the test without results at pos. 62 after many hours (2021-01-25)
Hans-Jürgen Manthey: auch hier klappt der Player nur mit Änderungen:
bei a) 9. Sdf3
bei b) 11. Sdf3 (2021-01-25)
comment
Keywords: Unique Proof Game, En passant, Castling
Genre: Retro
FEN: rnb2b2/1ppB1ppp/2n5/r2kN3/8/Kp1P1q2/P1PQ1PPP/3R2NR
Input: Gerd Wilts, 1996-07-11
Last update: James Malcom, 2021-01-24 more...
33 - P0008985
Gianni Donati
9862 Die Schwalbe 168 12/1997
P0008985
(10+14)
ser-h#6 (AP)
1. gxf3ep 2. Dg4 3. 0-0-0 4. Te8 5. Kd8 6. Dxh5 0-0-0#
play all play one stop play next play all
Henrik Juel: 1.gxf3ep 2.Qg4 3.000 4.Re8 5.Kd8 6.Qxh5 000#. Both castlings are necessary to legitimize the en passant capture. (2003-09-29)
A.Buchanan: I don't think this is consequent series mover, under which every position would considered afresh from the perspective of legality i.e. possible histories. Instead, this is regular series mover where one player is simply saying "pass". It's hard to imagine how AP might operate in a consequent context. I will change the keyword suitably. (2021-11-08)
comment
Keywords: a posteriori (AP), Seriesmover, Castling, En passant as key
Genre: Retro, Fairies
FEN: r3k3/ppp1p3/2p5/6PP/5Ppq/1P4nB/PrP1P1p1/R3K1nb
Input: Gerd Wilts, 1998-06-26
Last update: A.Buchanan, 2021-11-08 more...
34 - P1007954
Peter Hoffmann
9345 Die Schwalbe 161 10/1996
P1007954
(10+7) C+
s#3*
*) 1. ... Lxe6 2. Sb3+ cxb3#

1. Td5! droht 2. Sd7+ Ka4 3. b3+ cxb3#
1. ... Lxe6 2. Sa4+ Lxd5 3. b4+! cxb3ep#
play all play one stop play next play all
Der Versuch mit der wD, 1.De5? scheitert an 1.- Sg4! - MS: Drei verschiedene weiße Züge erzwingen das Matt auf b3, witzig! - DrRJB: Gelungener Wechsel von Normal- zu e. ,p.-Schlag. - WW: Drohung und Lösung sind gut aufeinander abgestimmt, so daß ein Abspiel völlig ausreicht. - 2,8/II/4
more ...
comment
Keywords: En passant
Genre: s#
Computer test: SCHRECKE: popeye 4.89
FEN: 4B3/8/p2RQ3/k1N5/2p5/P1p2p1b/KP5n/RN2B3
Input: Gerd Wilts, 2003-06-29
Last update: Dieter Berlin, 2024-02-20 more...
35 - P1017730
Lev Loshinsky
Die Schwalbe 1930
P1017730
(7+10) C+
#2
1. d4! droht 2. Dc5#
1. ... cxd3ep 2. Sxe3#
1. ... exd3ep 2. Sxc3#
1. ... Sge6 2. Txf5#
1. ... Sfe6 2. Txd7#
play all play one stop play next play all
more ...
comment
Keywords: Brian Stephenson Collection (2561), Herpai, En passant (zweifach, s), Line closing
Genre: 2#
Computer test: SCHRECKE: popeye 4.87
FEN: 3N1n2/K2b1Rn1/1Q6/p2k1q2/2p1pP2/2r1r3/3P4/3N4
Reprints: 45 Chess Problems: Introduction to an Art , p. 58, 1963
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2023-02-08 more...
36 - P1021645
Siegfried Brehmer
359 Schach-Express 1948
2. Preis
P1021645
(8+8) C+
#3
1. Sc4? Tg6! 1. Td4? Tg5!

1. f4! droht 2. Tb6+ Kxd5 3. Sc7#
1. ... f5 2. Td4 Tb7 3. Tbc5#
1. ... f6 2. Sc4 Td7 3. Tdc5#

1. ... exf3ep 2. e4 ... Tb6#
play all play one stop play next play all
more ...
comment
Keywords: Brian Stephenson Collection (6499), Blockhamburger, En passant, Rook sacrifice (Drohung)
Genre: 3#
Computer test: Juel: Popeye 4.61
FEN: 7b/5pr1/N1k5/KR1R4/1P2p1b1/N1p1P3/5P1n/8
Reprints: 100 und ein Schachproblem von Siegfried Brehmer
Schach-Express 1948
Die Schwalbe 203, p. 397, 07-08/1949
Zadaniowiec 4 05-06/1956
Chess Wizardry [Rice] 1996
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2022-08-24 more...
37 - P1026547
Eeltje Visserman
Die Schwalbe 1964
1. Preis
P1026547
(9+10) C+
#3
1. a5! droht 2. De6+ Kc5 3. Dc4#
1. ... Sb6 2. Sc3+ dxc3 3. Se3#
1. ... Lg4 2. Se3+ dxe3 3. c4#
1. ... Te5 2. c4+ dxc3ep 3. Sxc3#
1. ... Dxa5 2. c4+ dxc3ep 3. Se3#
play all play one stop play next play all
more ...
comment
Keywords: Brian Stephenson Collection (11428), white move cycle (2. + 3. Züge), Deflection, Hinlenkung, En passant (s), Pawn mate, Knight sacrifice
Genre: 3#
Computer test: Juel: Popeye 4.61
FEN: n6b/q1pBQ3/2P3p1/1N1k2rb/P2p3r/3K1P2/2P5/5N2
Reprints: 1663 Beispiele zur Ideengeschichte des Schachproblems , p. 337, 1982
129b Michael Keller. Ein Meister der Schachkomposition [Chlubna] , p. 71, 1994
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2022-08-01 more...
38 - P1043951
Friedrich Köhnlein
Süddeutsche Schachblätter 1907
P1043951
(6+9)
#4
1. e4? dxe3ep!

1. Dd5+! Kh2 2. e4+ Kh1 3. Da5 Ta4 4. De1#
play all play one stop play next play all
Henrik Juel: dual 1.Dd5+ Kg1 2.Tc1+ Kh2 3.Df3,Th1+ (2022-01-24)
more ...
comment
Keywords: Brian Stephenson Collection (28947), En passant avoidance, Switchback (wD), Checking key
Genre: n#
FEN: 8/q4p2/p4b2/Q5pK/1r1p4/1P1P3p/2R1P3/7k
Reprints: Sammler 1916
976 FIDE Album 1914-1944/II 1972
B Die Schwalbe 61, p. 8, 02/1980
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2022-01-24 more...
39 - P1066685
Valery Liskovets
12365 Die Schwalbe 208 08/2004
P1066685
(14+5)
#1 (AP)
1. ... gxh3ep 2. 0-0-0#! (not 2.Sf3#?,Bh2#?)
to force
R: 1. h2-h4! h3xSg2 (not 1. Kd1-e1? Kf1-g1 2. Lf3-e2# )
play all play one stop play next play all
Diagram position is retropat so Article 15 give Black the move. Then White might have last move R: 1. h2-h4 or 1. Kd1-e1. To avoid stalemate in the diagram position, White must castle to demonstrate under AP the legality of 0. gxh3ep. So 1. 0-0-0#! not 1. Sf3#?,Bh2#?
VL: Solution. 1.0-0-0#/Bh2#?? - B. is on move.

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

W.Ps took 11: b*c*d*e*f*g, c*d*e*f, e*f*g and g*h.
11+5=16, hence b.Pa7 was also captured among them.
Thus it took once: 1+1(h*g)+14=16. Ph3 took on g2
when w.P stood on h2. No last W's move could be a
capture: e5 and f4 are occupied by w.pieces. Qf4
prevents from Ph3-h4 and Kh2-g1 before that. b.Ps
are blocked from above. Therefore, in his last move,
W. could retro-release B. only in two ways: Ph2-h4
with Ph3*Sg2 before that, or Kd1-e1 with Kf1-g1
and Bf3-e2+ before that. Thus, W. may castle, and
castling AP-legalizes ep (2004-12-09)
A.Buchanan: Article 15 (not AP Type Keym) drives this elegant problem, with usual AP Type Petrovic for ep/castling. (2022-02-16)
A.Buchanan: Have added genre as n#, even though n=1 here, because it's important that this is a #1 rather than a h#0.5, as they behave differently under Article 15. (2022-02-17)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, No legal last move for Black, Castling (wk)
Genre: Retro, n#
FEN: 8/6P1/5P2/4NpPP/5QpP/P5B1/3PBPp1/R3K1kb
Reprints: Die Schwalbe 228 12/2007
Input: Gerd Wilts, 2004-08-13
Last update: James Malcom, 2022-07-05 more...
40 - P1068032
Michael Grushko
Semion Shifrin

12772 Die Schwalbe 214 08/2005
P1068032
(16+16)
BP in 15.5
Circe Parrain
paul: Correction of P1067948 (2011-06-13)
James Malcom: Is this correct? (2021-01-24)
comment
Keywords: Circe (Parrain), Unique Proof Game, En passant
Genre: Retro, Fairies
FEN: 2kr1BNr/ppppp1pb/2nbBn2/q2Q4/8/5b1N/PP1PPPPP/RnB2RK1
Input: Gerd Wilts, 2005-09-04
Last update: James Malcom, 2021-01-24 more...
41 - P1068490
Kostas Prentos
Andrey Frolkin

13077 Die Schwalbe 219 06/2006
Werner Keym gewidmet
1. Preis
P1068490
(10+14) C+
BP in 26.0
1. h4 a5 2. h5 a4 3. h6 a3 4. hxg7 h5 5. g4 Sh6 6. g8=L Lg7 7. g5 Ld4 8. g6 f6 9. Ld5 Lc5 10. Lc6 0-0 11. g7 Kh7 12. g8=T bxc6 13. Tg5 La6 14. Te5 fxe5 15. f4 Tf6 16. f5 Td6 17. f6 Lc4 18. f7 Lxa2 19. f8=D Le6 20. Df3 a2 21. Dd5 axb1=S 22. Ta2 Sc3 23. dxc3 cxd5 24. Kd2 d4 25. Kd3 Lf5+ 26. e4 dxe3ep+
play all play one stop play next play all
The first known example of the Keym Task.
Henrik Juel: Stelvio 1.31 found the solution immediately and tested it unique in 9 minutes (for the first 25.0 moves) (2023-06-13)
more ...
comment
Keywords: Unique Proof Game, Ceriani-Frolkin Theme (DTLs), Castling, En passant, Valladao Task, Promotion (LTDs), Allumwandlung, Keym Task (The first known prob)
Genre: Retro
Computer test: Natch 3.1 (38 m, utilizing the final ep). Stelvio 1.33, 2min (retracting the e.p. move itself, so full C+)
FEN: rn1q4/2ppp2k/3r3n/2b1pb1p/8/2PKp3/RPP5/2BQ1BNR
Input: Gerd Wilts, 2006-08-05
Last update: Reto, 2023-06-13 more...
42 - P1095638
Andrej W. Borodulin
Sportklub Bogatyr Kriwoj Rog 1988
Spezielle ehrende Erwähnung (im Russ. Turnier)
P1095638
(12+6) C+
s#5
Zuletzt geschah zwingend e7-e5.
1. dxe6ep dxe6+ 2. Kf4 Ld7 3. c8=S+ Lxc8 4. Th8 Ld7 5. Td8 e5#
play all play one stop play next play all
more ...
comment
Keywords: Multiple publication, En passant as key, Stalemate dissolution
Genre: s#, Retro
Computer test: A. Baumann: Gustav 4.2a
FEN: 2b5/1pPp4/1P1k1P2/1B1PpKPR/1P2P1p1/4B1P1/8/8
Reprints: S1246 The Problemist 01/1989
6657 Die Schwalbe 08/1989
Input: Frank Müller, 2009-12-07
Last update: A.Buchanan, 2024-01-04 more...
43 - P1108454
Werner Keym
Schach-Echo 1967
P1108454
(15+5) cooked
#1
b) wDa5 nach e5, AP
a) BTM
1. ... bxa3ep! 2. Dc3#! (2.0-0#??)
1.0-0#?
b) BTM pushed by Wh
1. ... bxa3ep! 2. 0-0#! (2.Dc3#? AP needs the 00)
1.0-0#?
play all play one stop play next play all
(a) Wh made 11 pcs, right to left, which accounts for all missing Bl units. wBa never captured, so sBa must have captured a3xSb2, prior to Wh a2-a4. Since Wh is missing only 1 unit, bBh promoted on h1, disrupting Wh 00 rights. Moreover Bl has no last move and by Codex Article 15, it’s Bl to move. Last move by Wh must allow Bl a prior move. This can only be R: 1. a2-a4 a3xSb2. So ep is on and there is no alternative.
(b) Again Wh has made 11 pcs, but Black may now have made 2. If WTM, then Black's last move was R: 1. Kc2xSc1 Td1-d2+ 2. Kc3-c2 or R: 1. cxSb1=T or R: 1. cxSb2 (as wBc4 & wBc7 might be foreign to c-file and allow bBc a smooth retraction). In any case, sBa3xSb2 was played so sBh-h1= disrupting 00. So Wh can still not play 1. 0-0#! Now White pushes the move to Bl and also asserts that bxa3ep is on. Both claims will be "proved" by Wh 0-0. We know that with WTM, Wh can't castle, so that's half of it. But if BTM with Wh to castle, then bPaxb & bPhxg, and the only way we can retract is as in (a). The mating move is different, because we have that extra Bl pc to keep castling legal, but we must actually castle so that the whole AP logic can work!
However, I don't understand why Wh can't just play 1. Txb2#?
Cook: In (b) Wh can just play 1. Txb2#
b) Urdruck in Die Schwalbe 228, 12/2007
A.Buchanan: Any ideas about (b)? (2022-06-09)
Henrik Juel: Andrew, you can find Heft 228 on
https://www.dieschwalbe.de/archiv.htm
part b) is mentioned in an article by Valery Livkovets, p.299-304
diagram p.301, solution p.303 (2022-06-09)
A.Buchanan: Thanks Henrik - just the pointer I needed. Please check the detailed solution I'll post above. (2022-06-09)
A.Buchanan: OK the logic nearly works for me. But I don't understand why White can't just play 1. Txb2# in part b, because b2 is now covered by wD. What am I missing? (2022-06-09)
Henrik Juel: Indeed, 1.Txb2# looks like a cook (2022-06-09)
A.Buchanan: Werner agrees. He has fixed it with a better twinning condition (2022-06-12)
more ...
comment
Keywords: Cant Castler, En passant as key, No legal last move for Black, a posteriori (AP) (Type Petrovic), Castling, a posteriori (AP) (Type Keym), Superseded by (P1406456)
Genre: Retro
FEN: 8/2P5/3P4/QP2N1B1/PpP5/1P1P4/1p1RBP2/brk1K2R
Reprints: (10) Die Schwalbe 228, p. 301, 12/2007
Input: Gerd Wilts, 2010-06-23
Last update: A.Buchanan, 2022-12-05 more...
44 - P1231466
Otto Dehler
6594 Die Schwalbe 03/1942
P1231466
(8+10)
#3
1. e3? Se2 2. Kxa4 Sc3+
1. e4! droht 2. Tc6+ Lc5 3. Txc5#
1. ... fxe3ep 2. Kxa4
1. ... Le1+ 2. Kxa4
1. ... fxe4 2. Tc6+ Kd5 3. Sb4#
play all play one stop play next play all
Erschien erstmals im Lösungsturnier des Thüringer Schachbundes anlässlich des 48. Kongresses vom 26.12. - 29.12.1941 in Jena.
Henrik Juel: C+ Popeye 4.61, except for mate duals after 1.e4 Le1+ 2.Kxa4 (2022-01-10)
comment
Keywords: Double step of white Pawn, En passant (schwarz)
Genre: 3#
FEN: 8/5p2/N2R1B2/K4pp1/p1k2p1q/6Pr/1PP1Pb2/6n1
Input: Felber, Volker, 2012-02-01
Last update: Gunter Jordan, 2022-01-10 more...
45 - P1235808
Martin Walter
12171 Die Schwalbe 205 02/2004
P1235808
(5+7+1)
s#4
Längstzüger
Anticirce
1. a4+ bxa3ep[+nBa7] 2. a8=nT Txh8 3. Da6+ Kxa6[+sKe8] 4. Kxg1[+wKe1] 0-0#
play all play one stop play next play all
Steen Christensen: ... 4.Kxg1[Kh1] \oo\# should read ... 4.Kxg1[Ke1] 0-0# (2013-03-12)
SCHRECKE: C+, popeye 4.87 (2022-01-09)
more ...
comment
Keywords: Valladao Task, Maximummer, Circe (Anti), En passant, Castling, under-promotion
Genre: Fairies
FEN: 7B/8/4Q3/1kp5/1-P5p/4p3/P3P3/n4Knr
Reprints: F4 König & Turm (14), p. 57, 09/2004
Input: Erich Bartel, 2012-04-03
Last update: Alfred Pfeiffer, 2017-07-15 more...
46 - P1255845
Bernd Schwarzkopf
13750 Die Schwalbe 231 06/2008
P1255845
(2+3)
#6
Längstzüger
1. cxd6ep! g5 2. d7 Kg7 3. d8=D Kh6 4. Kg4 Kg7 5. Kf5 Kh6 6. Dh8#
play all play one stop play next play all
Anton Baumann: Retro: Unter der Annahme, dass die Längstzügerregel auch für den der Diagrammstellung vorangegangenen Zug gilt, kann der letzte Zug nur d7-d5 gewesen sein! Daher die Lösung:
1.cxd6 e.p.! g5 2.d7 Kg7 3.d8=D Kh6 4.Kg4 Kg7 5.Kf5 Kh6 6.Dh8# (2022-09-14)
comment
Keywords: Maximummer, Kindergarten Problem, Minimal, Miniature, En passant as key
Genre: Fairies
FEN: 7k/6p1/8/2Pp4/8/6K1/8/8
Input: Gerd Wilts, 2012-12-13
Last update: Mario Richter, 2022-09-14 more...
47 - P1272370
Wolfgang Hundsdorfer
Sammler 1906
P1272370
(8+7) C+
s#3
1. f4! gxf3ep 2. Ta5+ Txa5 3. Dd8+ Lxd8#
1. ... Lg5 2. Dh8+ Ld8+ 3. Ka6 Lxb5#
1. ... Lxf6 2. Tf8+ Ld8+ 3. Ka6 Lxb5#
play all play one stop play next play all
more ...
comment
Keywords: En passant
Genre: s#
Computer test: SCHRECKE: popeye 4.89
FEN: k7/3N1R2/1K3Q2/1R1r4/b5pb/1p4p1/1P3PB1/8
Reprints: 3 Die Schwalbe 276 12/2015
Input: Frank Müller, 2013-06-15
Last update: Dieter Berlin, 2024-01-16 more...
48 - P1287896
Jelisej K. Lebedkin
68 Die Schwalbe 29, p. 213, 10/1974
Preis, 154. TT., Abt. 3
Richter: H. P. Suwe
P1287896
(12+7) cooked
#3*
* 1. ... Sa3-c2+ 2. Db2xc2 droht 3. Dc2-g6#, 3. Dc2-f5#
2. ... f3-f2+ 3. Dc2xf2#
1. Th2-h4? droht 2. Th4-f4#
1. ... e7-e5!
1. 0-0-0! droht 2. Db2-b6 ... 3. Db6-g6,Db6-e6#, 2. Td1-f1 ... 3. Tf1xf3#
1. ... e7-e5 2. dxe6ep+ Kf7xe6 3. d7-d8=S#
1. ... g7-g5 2. hxg6ep+ Kf7xg6 3. h7-h8=S#
1. ... g7-g6 2. h5xg6+ Kf7xg6 3. h7-h8=S#
play all play one stop play next play all
Cook: NL:
1.Db2-b6 ! Drohung: 2.Db6-g6 # 2.Db6-e6 #
1...Sa3-c2 + 2.Ke1-d2 Drohung: 3.Db6-g6 # 3.Db6-e6 #
Von der NL mal abgesehen, dieser Wust von Nebenvarianten und "Verführungen" machen das Stück für meinen Geschmack schon fast zum Ärgernis. \eb 16.1.2001

'Die Schwalbe', Heft 33 (Juni 1975), S. 341: "Nr. 68/Lebedkin hat nach 1. ... e5 den Dual 2. Dxe5 und zu dem die NL 1. Dd6! [gemeint war wohl 1. Db6!].
Die Auszeichnungen der inkorrekten Aufgaben entfallen. Ein Nachrücken nachfolgender Preisprobleme an die Stelle entfallener Aufgaben findet nicht statt."
milan: m.j.frelih::sBb7 nach a7 1.0-0-0! (2014-09-14)
James Malcom: Is it possible that the bP on b7 was a misprint and it should have been on a7? Because it seems very bizarre that such an obvious cook could slip into an awarded problem. (2020-09-26)
comment
Keywords: Valladao Task, En passant, Castling
Genre: 3#
FEN: 2N2b2/1p1PpkpP/8/3P3P/8/n3BpN1/1Q4PR/R3K3
Input: Erich Bartel, 2014-09-10
Last update: Mario Richter, 2020-09-26 more...
49 - P1342118
Klaus Förster
16988 Die Schwalbe 284, p. 92, 04/2017
P1342118
(10+9) C+
#2vvv
1. Da2? A droht 2. cxd3# C, 2. c3# D, 2. c4# E
1. ... dxc2!

1. cxd3 C? droht 2. Da2# A, 2. Txe2# B
1. ... Lxd3!

1. c3? D droht 2. Da2# A
1. ... dxc3 2. Txe2# B
1. ... Le6 2. Se4#
1. ... b3!

1. c4! E Zugzwang
1. ... bxc3ep 2. Da2#
1. ... dxc3ep 2. Txe2#
1. ... La7,Lb6 2. Lxb4#
1. ... Le6,Lf5~ 2. Se4#
1. ... Txg2 2. Dxg2#
1. ... Tf2+ 2. Txf2#
play all play one stop play next play all
Nightrider: Original threat reduction from a double threat after 1.cxd3? via 1.c3? to zugzwang after 1.c4!, with re-appearance of the former threats after the ep captures. The basic matrix is closely related to P1008363, and this is why the problem was called anticipated by the judge. However, the content is very much different whence the problem should enjoy full right to exist. (2021-08-09)
comment
Keywords: Dreifachdrohung, threat reduction, threat-recurrence, En passant, Caprice
Genre: 2#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/8/3p2p1/B1bQ1bP1/Np1p1K2/3p2N1/2Pkr1R1/1B5R
Input: Felber, Volker, 2017-10-22
Last update: Nightrider, 2021-08-09 more...
50 - P1383322
Werner Keym
7 Die Schwalbe 215, p. 241, 10/2005
Hanspeter Suwe gewidmet
P1383322
(5+4) C+
#3
1. Th5! droht 2. Th8#
1. ... g5+ 2. f5xg6ep 0-0-0 3. a8=D#
play all play one stop play next play all
more ...
comment
Keywords: Valladao Task, En passant, Castling, Promotion
Genre: 3#
Computer test: popeye 4.85
FEN: r3k3/PR4p1/8/5PR1/7K/8/6p1/8
Input: Alfred Pfeiffer, 2020-12-24
Last update: Alfred Pfeiffer, 2020-12-24 more...
51 - P1391295
Andreas Thoma
17289 Die Schwalbe 288 12/2017
P1391295
(13+12) C+
BP in 13.5
1. b4 g5 2. b5 g4 3. b6 g3 4. bxa7 gxh2 5. axb8=S hxg1=S 6. Sxd7 Sxe2 7. Sc5 Sc3 8. Se4 Kd7 9. Dh5 Ke6 10. g4 Ld7 11. Lh3 Le8 12. 0-0 Dd7 13. g5+ f5 14. gxf6ep#
play all play one stop play next play all
more ...
comment
Keywords: Unique Proof Game, Valladao Task, Castling (wk), En passant, under-promotion (Ss), Phoenix (s), Prenix (S)
Genre: Retro
Computer test: C+ Euclide 1.01, no solution in 12.5
FEN: r3bbnr/1ppqp2p/4kP2/7Q/4N3/2n4B/P1PP1P2/RNB2RK1
Input: A.Buchanan, 2021-07-03
Last update: Alfred Pfeiffer, 2021-11-26 more...
52 - P1399606
Leonid I. Zagoruyko
Shakhmaty v SSSR 1985
3. Preis
P1399606
(10+12) C+
#4
1. d4+? cxd4 2. Sd3+ Kf5 3. e4+ dxe3ep! 1. e4? dxe4 2. Dg5+ Sf5 3. d4+ exd3ep!

1. h3! droht 2. Sd3+ Kf5 3. g4+
1. ... c4 2. d4 cxd3ep 3. Sxd3+
1. ... d4 2. e4 dxe3ep 3. Dg5+
play all play one stop play next play all
Henrik Juel: C+ Popeye 4.61, except for double threat 1.h3 thr. 2.Sd3+,g4 (2022-03-03)
comment
Keywords: Staggered preplans, En passant
Genre: n#
Computer test: Popeye 4.61
FEN: 1b4r1/4p1np/2pp3Q/R1ppk2p/5N2/n1P2PP1/3PP2P/6K1
Reprints: 12. Die Schwalbe 107, p. 324, 10/1987
Input: Dieter Berlin, 2022-03-03
Last update: Dieter Berlin, 2022-03-03 more...
53 - P1399755
Evgeny Bogdanov
Le Courrier des Echecs 1982-1983
1. Preis
P1399755
(8+13) C+
#3
* 1. ... c5 2. Txb6 1. ... e5 2. Df8+

1. Sf8! droht 2. Dxd7+
1. ... c5 2. dxc6ep
1. ... Sc5 2. Lxf4
1. ... e5 2. dxe6ep
1. ... Se5 2. Lb4+
play all play one stop play next play all
Henrik Juel: C+ Popeye 4.61 (2022-03-10)
comment
Keywords: white move cycle, half-battery, white, En passant
Genre: 3#
Computer test: Popeye 4.61
FEN: 3Q3b/1rppp2r/1p1k1pN1/1R1P1p2/5p2/3n1B2/3B4/1K1R1b2
Reprints: I Die Schwalbe 102, p. 184, 12/1986
Input: Dieter Berlin, 2022-03-10
Last update: Dieter Berlin, 2022-03-10 more...
54 - P1406482
Hermann Weissauer
(C) Die Schwalbe 29, p. 217, 10/1974
P1406482
(11+4)
#3
1. Sd3! droht 2. g4+ fxg3ep 3. Sxg3#
1. ... Txg2 2. Sd6+ Lxd6 3. Sf2#
1. ... Txd3 2. Sd6+ Lxd6 3. Lxd3#
1. ... Lc3 2. Lxc3 ... 3. Sd6#
play all play one stop play next play all
Pachl: Batteriewechselthema
Henrik Juel: double threat 2.g4+,La4 and several duals (2022-12-07)
comment
Keywords: En passant (s), Battery play
Genre: 3#
FEN: 8/8/5P1P/3PNk1P/1b2Np1K/5P2/1BBr2P1/8
Reprints: D42 Knobeln Sie auch gern? [Pachl] , p. 48, 2009
Input: Dieter Berlin, 2022-12-07
Last update: Dieter Berlin, 2022-12-07 more...
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The problems of this query have been registered by the following contributors:

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