Die Schwalbe

252 problem(s) found in 4727 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20200919 AND NOT CPLUS 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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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 - P0000029
Walter Wittstock
5463 Die Schwalbe 98 04/1986
P0000029
(3+2) cooked
-1w -1s, dann h=1
R: 1. b2xBc3 b4xc3ep, dann bxa6 b3=
play all play one stop play next play all
Cook: R: 1. b2xDc3 Da3xSc3, dann 1. bxa6 bxa3=
Adrian Storisteanu: Possible fix: +bPc5. (2015-06-10)
Miguel Ambrona: The fix proposed by Adrian works.
Verified with Deadpos v.2.2. (2024-01-12)
comment
Keywords: En passant in the retro play, Help retractor, Kindergarten Problem, Minimal, Miniature, Ideal stalemate, Superseded by (P1360139)
Genre: Retro, Fairies
Computer test: cooked by Deadpos v2.3 14-Jan-2024
FEN: 8/1pK5/P7/k7/8/2P5/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-18 more...
3 - 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...
4 - 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...
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)
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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)
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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 - 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)
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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...
9 - 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...
10 - P0000615
Andrey Lobusov
4088 Die Schwalbe 77 10/1982
14.Lob
P0000615
(15+6) cooked
h#2* (AP)
1. ... Kg5 2. 0-0?? Se7#
1. ... Txh7 2. Kf8 Txh8#

1. cxb3ep Txh7 2. Kf8?? Txh8#
1. cxb3ep Kg5 2. 0-0! Se7#
play all play one stop play next play all
Wh captured 10 by pawns (including hxPg-g8) but not on the last move. wPa3 came from a2: wPa5 captured bPa.
Bl took only b3xa2.
In set play, Bl has no last move except for K or R.
For the main play, the intention is that if Bl retains castling rights, last moves were R: 1. b2-b4 b3xa2 e.g. 2. Bg1-d4 b4-b3 3. d4xc5 etc. However there are retro cooks e.g. R: 1. Bg1-d4 b3xa2 2. d4xc5 etc, with wPa5 the original wPb. wPb4 & wPb5 come from further East.
Cook: Retro logic for main play is unsound.
A.Buchanan: Not trivial to fix as shifting wPa5 to b6 leads to promotion cooks. Adding black material must be carefully done to avoid disrupting uniqueness of e.p. as tempo. (2022-03-11)
A.Buchanan: OK just to say I have found a form of this idea that works. This one was the very devil to fix - the final form includes different mates and retro logic, so becomes an "after" rather than a "correction", and I will find somewhere else to publish first. (2022-03-19)
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Keywords: Castling (sk), a posteriori (AP) (Type Petrovic), En passant as key, Tempo Move, Superseded by (P1409841)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/5p1p/5K1R/PPP2N2/1PpB4/P1P5/p1RNB1P1/Q7
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-20 more...
11 - 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...
12 - 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)
more ...
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...
13 - 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)
more ...
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...
14 - 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...
15 - P0000772
Klaus Wenda
1419 Die Schwalbe 30 12/1974
4. ehrende Erwähnung
P0000772
(4+7)
Weiß nimmt 1 Zug zurück, dann #2
b) sBh7 nach b6
a) R: 1. Kd5-e4, dann 1. Kd6
b) R: 1. dxc6ep, dann 1. d6
play all play one stop play next play all
In a) ist die s0-0-0 nicht mehr möglich, weil sich das Schach durch den sLg8 nur durch Kf7(x)e8 erklären läßt, in b) muß mindestens einer der sTT via e8 auf seinen Diagrammplatz gelangt sein.
In a) Entschlagdual R: Kd5xBe4, das wurde in der Lösungsbesprechung noch kritisiert und als NL vermerkt. Hat das der PR gelassener gesehen ('geduldeter Entschlagdual'), oder gibt's noch eine korrigierte Version dieses Problems?
Anton Baumann: Korrektur in 'Die Schwalbe' 12/1975 S.422: +sLh2, +sBe5;
nun geht in a) nur noch zurück: Kd5 x Be4.
Ausgezeichnet wurde gem. Preisbericht in 'Die Schwalbe' 06/1977 S.82 die korrigierte Version 1419v. (2022-12-08)
comment
Keywords: Castling (sg), Help retractor, En passant in the retro play, Cant Castler (sl)
Genre: Retro
FEN: r3k1bR/pr1p3p/2P5/8/2B1K3/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-10 more...
16 - 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...
17 - 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...
18 - 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...
19 - 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...
20 - 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)
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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...
21 - 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...
22 - 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.
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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...
23 - 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...
24 - P0001710
Hans Heinrich Schmitz
6901v Die Schwalbe 6 30/03/1944
P0001710
(3+11)
Schwarz und Weiß nehmen 1 Zug zurück, dann h#1
R: 1. Lb2-f6 exd3ep, dann Kf6 d5#
play all play one stop play next play all
Adrian Storisteanu: There is a set play (in the retro part): - 1... Kd7xQe6 & 1.Kd7-d8 Qe6xc8#. (2021-04-21)
comment
Keywords: En passant, Help retractor
Genre: Retro
FEN: 2n5/1p2n3/qp1NkBr1/6b1/1p6/3p4/5p2/3K4
Reprints: 323 FIDE Album 1914-1944/III 1975
Die Schwalbe 162 12/1996
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2014-03-13 more...
25 - 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...
26 - 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...
27 - 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)
more ...
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...
28 - P0002053
Francis Charles Collins
Land and Water 18/01/1879
P0002053
(16+3)
Weiß nimmt einen Zug zurück, dann #1
R: 1. Tg8xBg5, dann 1. hxg6ep#
play all play one stop play next play all
Adrian Storisteanu: As Mario remarked, see P0004708 for a more modern treatment. (2021-04-22)
more ...
comment
Keywords: En passant, Help retractor, En passant as key, Last Moves?, En passant as mating move
Genre: Retro
FEN: 4N3/P4p1p/PB5k/PB3KRP/1N4P1/8/PPP4Q/R7
Reprints: Huddersfield College Magazine , p. 161, 03/1879
607 Hartford Weekly Times 05/06/1879
101 A Selection of 107 Chess Problems [Collins] 1881
31 Retrograde Analysis 1915
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2020-11-08 more...
29 - 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...
30 - 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...
31 - 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
32 - 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
33 - 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...
34 - 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...
35 - 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...
36 - 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...
37 - 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)
more ...
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...
38 - 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...
39 - 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...
40 - 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...
41 - 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...
42 - 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...
43 - 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...
44 - 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...
45 - 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...
46 - 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...
47 - 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...
48 - 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...
49 - 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...
50 - 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...
51 - 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...
52 - 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...
53 - 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...
54 - 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...
55 - 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...
56 - 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#
more ...
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...
57 - P0003973
André Simonet
RA166 diagrammes 74 11-12/1985
Lob
P0003973
(6+8)
Die Position wurde nach 22 konsekutiven Schachgeboten erreicht. Finde die Anfangsposition heraus, dann #1! Kein Dual oder Vertauschen von Zügen ist erlaubt.
R: 1. ... b4xc3ep+ 2. c2-c4 Lf5xDd3+ 3. Df1xDd3 De4xDd3 5. Df3xDd3 De2xTd3 6. Te3-d3+ Td3-b3+ 7. Tc4xDc6+ Dh6xDc6+ 8. Dg6-c6+ De8xTd7+ 9. Tc7-d7+ Sd7xSb6+ 10. Sa4-b6 Ta7xDb7+ 11. Da8xDb7+ Dc8-b7+ 12. Tb7xSc7+ Sa6xSc7+ 13. Se6-c7+
play all play one stop play next play all
James Malcom: It seems likely that this could be cooked. (2021-01-23)
Hans-Jürgen Manthey: wenn man dem Player bis zum ende folgt und dies die Anfangsposition sein soll, ist die 2. Bedingung 'dann #1' nicht möglich ! Also ist es nicht gelöst und es muß eine andere Zugfolge nötig sein ! (2021-01-23)
Hans-Jürgen Manthey: mögliche Zugfolgen:
1. h4 b5 2. g4 Lb7 3. Sh3 La6 4. Lg2 Lc8 5. h5 c5 6. g5 c4 7. g6 c3 8. dxc3 d5 9. c4 e5 10. c5 f5 11. c6 f4 12. c7 f3 13. Kd2 d4 14. Kd3 b4 15. Kc4 fxe2 16. f4 e1D 17. f5 e4 18. f6 e3 19. a4 Lc5 20. a5 Lb6 21. axb6 Sh6 22. b7 Sg4 23. h6 e2 24. Ld5 Lf5 25. c8T a5 26. Kb5 Sa6 27. b8D Ta7 28. Da8 Se5 29. Sf4 a4 30. Sc3 Sd7 31. La2 a3 32. hxg7 Dg1 33. Tc7 Dc8 34. Tb7 Kd8 35. f7 De3 36. g8D+ De8 37. g7 h5 38. Dg1 d3 39. Dg6 h4 40. Sa4 Ke7 41. Se6 Kd6 42. Lg5 h3 43. Tag1 d2 44. Tg4 h2 45. Tc4 e1D 46. Lb1 Th3 47. Dh8 Td3 48. g8D Kd5 49. Dhh3 De4 50. Dgh8 a2 51. D8h5 a1D 52. La2 Dae1 53. Dg2 D1h4 54. Dhf3 Dh6 55. Te1 h1D 56. Te3 De1 57. Dgf1 De2 = Anfangsposition, aber kein Matt in 1 Zug
(57. ... Df2 58. Lf4 De2 = Anfangsposition mit Matt in 1 Zug: 59. Td4#)
es folgen die 22 Schachgebote bis zur Diagramstellung mit Lg5:
58. Sc7+ Sxc7+ 59. Tbxc7+ Db7+ 60. Dxb7+ Txb7+ 61. Sb6+ Sxb6+ 62. Td7+ Dxd7+ 63. Dc6+ Dhxc6+ 64. Txc6+ Tb3+ 65. Td3+ D2xd3+ 66. D3xd3+ Dxd3+ 67. Dxd3+ Lxd3+ 68. c4+ bxc3ep+ (ist auch möglich mit Lf4) (2021-01-24)
comment
Keywords: Consecutive checks (22), Help retractor, En passant in the retro play, Non-standard material, En passant
Genre: Retro
FEN: 8/1r1q1P2/1nR5/1K1k2B1/8/1rpb4/BP1p4/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-07-30 more...
58 - P0004036
Sergey A. Komarov
Andrey Frolkin

diagrammes 85bis 04-06/1988
3. Lob
P0004036
(5+15)
shc#7
1. a1=S 2. fxg3ep 3. Kxh5 4. Kxg6 5. Kf7 6. Ke8 7. 0-0-0 a8=D#
play all play one stop play next play all
8. TT
Henrik Juel: Solution: 1.a1S 2.fxg3ep 3.Kxh5xg6-f7-e8 7.0-0-0 a8Q#. Nice problem with open and hidden ep capture (exf3ep). (2003-04-28)
James Malcom: How is the key here justified? (2021-01-18)
James Malcom: I am still pondering. (2021-09-14)
Henrik Juel: James, I have forgotten all about this weird stipulation during the time since my last comment, and I don't know what I meant so say...
The definition of shc in the PDB is not complete; here is the Schwalbe definition:
Konsequenter Serienzüger: Ein Serienzüger, bei dem nach jedem Zug die Retroanalyse der Stellung neu durchgeführt wird (also ohne Kenntnis früherer Züge und Analysen). Beispielsweise wird eine Rochade wieder möglich, wenn König und Rochadeturm auf ihre entsprechenden Felder ziehen (weil in der neuen Analyse "vergessen" ist, dass beide bereits gezogen haben).
In english, something like
After each move the analysis of the position is done afresh, without knowledge of previous moves;
for instance, castling is possible when king and rook reach their original squares (e8 and a8), because it is 'forgotten' that they have already moved (2021-09-14)
Henrik Juel: You ask for a justification of the key, James; there is no need for justification, Black can do whatever he likes
In the position after 1.a1=S last move must have been g2-g4, so the ep capture 2.fxg3ep is legitimized
In the position after 2.fxg3ep last move must have been Ke3xPf3, with the double check being explained by exf3ep++; this is the 'hidden ep capture' I mentioned in my youthful comment (2021-09-14)
James Malcom: Thanks Henri! While the definition is incomplete, the reset each time is deducible from looking over shc problems. The justification here, however, was a level above me somehow. (2021-09-14)
comment
Keywords: Seriesmover, Consequent, En passant, Promotion (sD x2), Valladao Task, Castling, Promotion in the mating move
Genre: Retro, Fairies
FEN: r7/P1pp4/pp3pP1/4rpbP/5pPk/3n1K1b/p6q/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-09-14 more...
59 - P0004199
Werner Frangen
Problem 41-44, p. 52, 03/1957
P0004199
(14+10) cooked
#5
1. axb6ep+ Kb5 2. bxc7+ Tb6 3. 0-0-0 ...
play all play one stop play next play all
K. Fabel: "Von Interesse ist, dass diese strittige Idee [Beweis des e.p.-Schlagrechts durch spätere Ausführung der Rochade] auch im direkten Mattproblem dargestellt werden kann, vergl. das Diagramm [diese Aufgabe P0004199]. (Vielleicht ist diese Aufgabe noch nicht korrekt doch es wird nicht schwer sein, sie zu verbessern). Falls Schwarz zuletzt Bb6-b5, Tb6-c6 oder Lb6-c7 gezogen hat, ergibt sich, dass Weiss im letzten oder vorletzten Zug den K oder T bewegt haben muss. Falls jedoch K und T noch nicht gezogen haben, muss b7-b5 der letzte Zug gewesen sein. Weiss spielt daher 1. ab6e.p.+ Kb5 2. bc7+ Tb6. Jetzt könnte Weiss mit Dxb6 sofort mattsetzen, aber er muss ja den e.p.-Schlag noch legalisieren. Daher 3. 0-0-0! und Matt erst im 5. Zuge."
Cook: 1. Sxc7+!
1. Lxc7!
Innerhalb des 2. Teils einer Artikelserie "Die Konventionen im Problemschach" von Karl Fabel.
A.Buchanan: The retro logic is fine, but forward play is savagely cooked. I wonder about changing wSc8 to sS. The retro stuff still works OK, indeed sSc8 is a fourth thematic unit to retract in the try. Ignoring proof of ep legality, there is a unique #4 beginning with ep, and no #5. Promising: but when can we interpolate w000? Any solution must begin 1. axb6ep+ Kb5. There are now numerous #4, but none include w000, so I think we must stick with the still unique line beginning 2. Dxc8 thr 3. 0-0-0. There are 3 black defenses to refute this. 2. ... Txb6,axb3,d5. So maybe #6 needed for White to prevail? (2022-03-28)
A.Buchanan: But if we shift to #6 then 1. Sxc7!,Qxc8+! mate without ep. Can anyone retrieve the original diagram for this one, please? (2022-03-28)
Mario Richter: The position here is identical to the original diagram, but it should be regarded more as a schematic example than a "real problem". Fabel in his quote above: "... perhaps the problem is still not correct, but it should be easy to improve it ..." (2022-03-28)
Henrik Juel: unfortunately K. Fabel forgot to give the easy correction... (2022-03-28)
A.Buchanan: Thanks for this. Valery Liskovetz, an AP expert, was kind enough to send a pdf of the relevant Problem issue, so I can confirm. It's easy enough to remove the two cooks (e.g. sTc8) but the difficulty is in validation of the intended solution. Is there any "exact" option in Popeye that forces the solution to include a waiting move? The actual solution might be expected to be included in that set. (2022-03-29)
Henrik Juel: Yes, Andrew
'stip exact-#5' would disregard #4 as a solution, I believe (2022-03-29)
A.Buchanan: Even with the "exact", it's still very complicated to investigate. I think this kind of problem requires an engine feature to disallow any mate or pat if the specified castling has not been executed. Gazing at the 20,000 lines of output, I don't think sTc8 works as 1. axb6ep+ Kb5 2. Qb7 bxa3! threatens axb2+ after castling. I wonder if the board might be flipped left-right except for wK, but I haven't explored this. (2022-05-31)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Non-standard material (TLL)
Genre: Retro, n#
FEN: NQNB4/B1brpp2/k1rp4/Ppp5/Rp6/BP6/RPP5/R3K3
Reprints: (8) Problem 101-102 09/1966
(52) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-29 more...
60 - P0004296
Luis Alberto Garaza
1809 Problem 73-78 06/1961
P0004296
(16+13) cooked
h#1? h#2?
1. fxe3ep? Dxa8# No justification for the ep, so just retro try.
play all play one stop play next play all
White pcs: dxexf5,hxBg. Black none. Assume that Black can still castle. White can't have just played h4xg5 as sBh2 would be blocked. White might apparently just have moved D,Se5,Td4,Tc3,g4-g5,a5-a6,e3-e4 or e2-e4 but what might Black have played before? Only the last allows a move sDf1-e1. So set up for AP Type Petrovic is OK.
Cook: 446 candidate h#2
However 27 have no ep, while 59 have both ep & castling.
So a total of 86 viable solutions, 85 of which must be cooks.
s.a. Version P0004341
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004341)
Genre: h#, Retro
Computer test: Popeye v4.87 indicates cook
FEN: r3k3/p1ppp3/Pp6/4NPP1/2PRPp2/2RK2PN/1PBn1PQp/2Brq1b1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
61 - P0004341
Luis Alberto Garaza
(9) Problem 101-102 09/1966
P0004341
(15+12) cooked
h#1 oder h#2
1. fxe3ep Dxa8#? because no AP justification for ep

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

Autor: "In Nr.9, I intended to do something similiar to No. 8 [P0004199], but with a different key: the solution 1. PxP "e.p.", Qxa8 checkmate will not go, for if blackside does not castle, the "en-passant" capture cannot be justified. Though, well considering, there is the demolition 1. PxP "e.p." -any 2. 0-0-0! Qb7 or a8 checkmate.
A.Buchanan: Thanks Mario for retrieving the author's intent with this. So he knew it was cooked! Assume s000 rights remain. R: 1. b5xa6? as sBb promoted on b1. R: 1. h4xg5? as sBh retro-blocked. R: 1. Ke3-d3? impossible check from sBf4. R. 1. K~-d3 d3-d2? illegal check. I like these! So by elimination, R: 1. e2-e4 e3xTd2 2. T~d2 L~e1. So must have sLe1. (2022-03-22)
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004342)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k1N1/p1pp4/P5P1/4PPp1/2PNPprB/3K1QRp/P2p1PB1/4b2n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
62 - P0004342
Luis Alberto Garaza
(10) Problem 101-102 09/1966
P0004342
(15+11) cooked
h#2
1. bxc3ep Dg8#? (thematic try)
1. bxc3ep Dd5 2. 0-0-0 Db7#
play all play one stop play next play all
Cook: 1. bxc3ep Ld6 2. 0-0-0 Dc4#
Author: "From the former [d.h. P0004341], the No. 10 was born, something similar to No. 8 [P0004199] but with Black-castling for it is a help-mate. The mating move 1. ... Qg8? will not go."
more ...
comment
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P1399966)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k3/p2p4/8/NP6/BpPPN1PP/B2K1Ppp/QP1p1pP1/Rrb5
Reprints: (55) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
63 - 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...
64 - P0004481
Matjaz Zigman
Länderkampf Mazedonien-Slowenien 1969
7. Platz
P0004481
(13+10) cooked
ser-h#6** AP
* 1. ... Sb5#
* 1. ... dxc3#
1. fxe3ep 2. e2 3. exf1=S 4. Se3? 5. Sd5 Sb5# (must castle for AP)
1. fxe3ep 2. e2 3. exf1=S 4. Sxd2 5. Sxb1 6. Sxa3 0-0-0!#
play all play one stop play next play all
Luckily, illegal diagram can be fixed by removal of bPa4.
Then if White castling rights remain, ep is on. One ser-h#5 & 18 ser-h#6 tries exist, in which White does not bother to castle.
We don't know if this was a transcription error or a counting mistake by the composer.
Cook: Too many pawn captures required. For retro problem, illegal position implies unsound
Mike Neumeier: The solution appears to be 1.fxe4 e.p. 2.e2 3.exf1=L 4.Ld3 5.Le4 6.Ld5 dxc3#. Was that the intention? (2013-02-18)
Arno Tüngler: There would even be a solution in 5 moves by 3.exf1=S 5.Sd5 Sb5#
However, in order to justify the e.p.-key there must be an AP prove that e2-e4 (and not a move by the wK or wRa1) was played as White's last move. Thus the only solution giving this in 6 moves is 1.fxe3 e.p. 2.e2 3.exf1=S 4.Sxd2 5.Sxb1 6.Sxa3 0-0-0!# (2013-02-18)
Mike Neumeier: It struck me as odd there was no solution listed. (2013-02-18)
Mike Neumeier: And, if we let black be the idle side, there is 1.Txg1 2.Lxh1 3.Kd1 4.Kc1 hxg1T#. What does the asterisk(*) mean? (2013-02-18)
Henrik Juel: The asterisk means that there is also a white mate if he had the move, in this case two mates, so maybe the stipulation should be ser-h#6**:
1.dxc3,Sb5# (2013-02-18)
Ladislav Packa: It is clear that e.p. is possible only when the wK and wRa1 done neither move.
What convention is used here for the right to castling? (2013-02-18)
Henrik Juel: White may castle, unless you can show that he has lost the right to castle. In this problem there is a major difficulty, I think: the position seems illegal.
White pawns captured b2xc3xd4xe5xf6, g2xf3, and axb, promoting on b8; Black captured bxc, dxc, and g3xh2. We cannot explain the destiny of [Ph7].
Reversing wPh3 and bPh2 seems to handle the illegality, but then the problem can be solved in 5 moves (2013-02-18)
Mike Neumeier: Thanks, Henrik. Maybe the stipulation was a typo. Considering all comments together, perhaps the stip should have been ser-h#5**, with the one solution being the 5-mover Arno gave. I do not think it can be proven, except by a stipulation of ser-h#5 that en passant is possible. Which leads to the question of convention. Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible? Whether there is castling or not here seems immaterial. It is just another 6-move solution. There are 19 6-movers (Popeye) with one queenside castling among them. But only the one 5-mover. (2013-02-18)
Henrik Juel: By convention, an en passant capture as first move is not permitted, unless the pawn double step can be shown by some kind of retro analysis.
(Conversely, a castling is permitted, unless it can be shown by retro analysis that the right to castle has been lost, i.e., that king and/or rook has moved) (2013-02-18)
A.Buchanan: Suggested repair: Remove bPa4. Add AP to stipulation.
As Henrik points out, the diagram as it stands is illegal. We can't swap wPh3 & bPh2, because that allows 2 h#1 cooks.
Suppose we remove bPa4 instead. Then we still at least 5 White pawn captures: bxcxdxexf6 & gxf3. But wPh3 did still move from h2, so there are 3 Black pawn captures bxc, dxc & gxh2. So [bhP] died without capturing or promoting. [waP] must have promoted - so either [baP] was captured to clear the way, or [waP] captured to promote, and later [baP] promoted. Either way, all the numbers add up, and the position is legal. wPe4 cannot have just come from d3. So AP is triggered. Forward logic works just as before without bPa4. There is 1 5-move try, and 18 6-move tries, but AP eliminates them all because of need for castling to retrospectively justify the ep. (2013-02-23)
A.Buchanan: Mike asked: "Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible?"
Answer: No. The stipulation does imply a default player to move, but otherwise cannot be used as a premise to determine game state (castling, en passant). (2022-03-21)
A.Buchanan: WinChloe & Yacpdb have the same cooked diagram as PDB, so I think this cook was in the design. I would like to post a sound version with bPa4 removed, and I think it would just be “Zigman correction” with no one else’s name mentioned in the credits. Is everyone ok with this? (2022-05-30)
Mario Richter: 'Bilten 1970' (printed 1971) contains on pp. 62-81 a complete report on the "Problem Duel Slovenia - Macedonia 1969", the problem here can be found on page 79 and is printed there exactly as given here.
Andrew's handling of the suggested correction (remove black pawn a4) is o.k. for me. (2022-05-31)
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comment
Keywords: Castling (wg), Seriesmover, a posteriori (AP) (Type Petrovic), Illegal position, En passant as key, Promotion (s), Valladao Task, Superseded by (P1401546)
Genre: Retro, Fairies
Computer test: Popeye v4.87 but simple retro-logic shows illegality
FEN: 8/8/5P2/2p1p3/p1pkPp2/N1p2P1P/2PP1PBp/RN2KRbr
Reprints: 104 Bilten 1970 1971
(72) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-01 more...
65 - P0004647
Branko Pavlovic
International Solving Match 1947
P0004647
(7+3)
#2
1. Kg6,Sf8 g3 2. f7#
1. fxg6ep? because e.g. R: 1. Kf7-g8 gxSh8=S+ can be history
play all play one stop play next play all
Paulo Roque: NL : 1. Sf8! g3 2. f7# (2009-08-22)
Henrik Juel: The organizers of solving matches in the old days liked tricky problems. Here there are two tricks: The 'obvious' e.p. key does not work (because last move could be Kf7-g8), and there are two solutions. (2009-08-22)
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comment
Keywords: En passant as key
Genre: Retro, 2#
FEN: 6kN/3N3R/5P1B/5PpK/6p1/8/8/8
Reprints: (10) Problem 198-201 07/1980
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-01 more...
66 - P0004657
Mato Prikril
(27) Problem 206-210 07/1981
1. ehrende Erwähnung
P0004657
(5+7) cooked
h#2*
*) 1. ... Ta4 2. Kxg4 Txf4#
1. fxg3ep Ta4#? (no proof of ep right)
1. fxg3ep 0-0-0! 2. g2 Td4#
play all play one stop play next play all
A short retro try h#1 and 16 h#2 tries. B2 is unique tempo move
Cook: 1. Kxg4 Ta5 2. h4 Tg5#
1. Kxg4 Ta4 2. Kh4 Txf4#
61. TT (Pavlovic Memorial), Gruppe A
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comment
Keywords: Castling (wg), En passant as key, a posteriori (AP) (Type Petrovic), Tempo Move
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 8/6p1/6pB/7p/5pPk/5p1p/5P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-06 more...
67 - P0004707
Filip S. Bondarenko
1059 Die Schwalbe 1962
P0004707
(5+4) cooked
Weiß nimmt 1 Zug zurück, dann #1
R: 1. Ta8-f8, dann 1. fxg6ep#
play all play one stop play next play all
Aus der Lösungsbesprechung ('Schwalbe 14-15, Februar-März 1963): "Absicht - zurück Ta8-f8 und Matt durch den e.p.-Schlag, doch läßt sich der letzte Zug g7-g5 nicht beweisen, weil Schwarz Kh8(g8)xLh7(Sh7) gezogen haben kann. Diese Möglichkeit wird immer wieder übersehen ..."
Adrian Storisteanu: On second thought, the 'cooked' flag was probably justified. In the position after - 1.Ra8-f8, black's last move was not necessarily 0...g7-g5: 0...Kh8/g8xSh7, 0...Kh8xBh7. (2021-04-22)
Gerald Ettl: Verbesserungsvorschlag:
5R2/pp5k/5P1P/5PpK/8/PP6/1NPPP3/NB6/
Weiss:Bf6h6f5a3b3c2d2e2 Sb2a1 Lb1 Tf8 Kh5 Schwarz:Ba7b7g5 Kh7 Neutrale Figuren: (2021-04-23)
Gerald Ettl: von meinem Verbesserungsvorschlag muss noch der sBb7 beseitigt werden. (2021-04-23)
Gerald Ettl: von meinem Verbesserungsvorschlag kann der sBb7 aber auch auf dem Brett stehen. Mit der Verfuehrung R: 1.Tb8-f8 Kh8h7 2.Ta8xSb8+ (2021-04-23)
Adrian Storisteanu: Gerald, in your fix suggestion wBb1 should be moved to a2, otherwise cook - 1.f4-f5 & 1.c2-c3/c4#. (2021-04-23)
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comment
Keywords: Help retractor, En passant
Genre: Retro
FEN: 5R2/pp5k/5P1P/5PpK/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-04-22 more...
68 - P0004708
Filip S. Bondarenko
Die Schwalbe 1960
P0004708
(8+1)
Weiß nimmt 1 Zug zurück, dann #1
R: 1. Tg8xBg5, dann 1. hxg6ep#
play all play one stop play next play all
offensichtlich inspiriert von P0002053
Adrian Storisteanu: A try (failing due to retrostalemate): - 1.g2-g4?? & 1.Rg5-g4#. (2021-04-22)
A.Buchanan: If -wLh3 +wBh3 there would be a single clean cook R: 1. gxSh3, dann 1. g3# (2021-04-23)
comment
Keywords: Help retractor, En passant as key, En passant as mating move, En passant
Genre: Retro
FEN: 8/8/5P1R/5KRP/6Pk/7B/5P2/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-04-22 more...
69 - P0004711
Karl Fabel
Berliner Morgenpost 1943
P0004711
(8+7)
Weiß nimmt 1 Zug zurück, dann #1
R: 1. bxc6ep+, dann 1. dxc6ep#
play all play one stop play next play all
Adrian Storisteanu: A try may be worth mentioning: - 1.c5-c6+?? & 1.d5-d6# (fails due to retrostalemate). (2021-04-19)
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comment
Keywords: Help retractor, En passant, En passant in the retro play, En passant in the forward play
Genre: Retro
FEN: 2brrb2/1R1pkPp1/2P5/3P1K2/1B1P4/8/B7/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-04-19 more...
70 - P0004778
Hans Klüver
30 Schachkongress Teplitz-Schönau im Oktober 1922 , p. 484, 1923
P0004778
(13+14)
Weiß nimmt 1 Zug zurück, dann #2
R: 1. Lb6-d8, dann 1. bxc6ep
play all play one stop play next play all
Henrik Juel: White captured g6xDg7 and hxLg
Black captured b5xa4, fxe, and once more with an officer
Following the retraction, it is clear that last move was c7-c5
Possible retroplay R: 1.Lb6-b8 c7-c5 2.Kd4-d5 e6-e5+ 3.Sa2-b4 Kc6-b7 4.b4-b5+ etc. (2022-04-18)
comment
Keywords: Help retractor, En passant
Genre: Retro
FEN: nrbBB1n1/rk1pp2P/p4Ppp/PPpKpN2/pNP1P1P1/3P4/8/8
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2022-04-18 more...
71 - P0005275
Otto Kerekes
Tivadar Kardos

10 L'Echiquier de France 11/1956
P0005275
(11+15) cooked
h#2
b) sBg2 statt wBg2
a) 1. cxd4 Lxc4 2. Dd5 Ld3#

b)
play all play one stop play next play all
Paulo Roque: Illegale Stellung. Diagrammfehler? (2009-11-17)
Alfred Pfeiffer: außerdem in b) 9 schwarze Bauern. (2010-01-13)
A.Buchanan: Twin a) is also illegal, as too many Bl captures (2021-11-24)
more ...
comment
Keywords: En passant as key, Castling, Illegal position, Superseded by (P1400824)
Genre: h#, Retro
FEN: 8/p1p2p1b/1B3pp1/q1p1b3/r1pPk3/1Bp3P1/PPP2PPr/3n1RK1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-23 more...
72 - P0005379
Anatolii Vasylenko
6493 feenschach 107 01-09/1993
Lob
P0005379
(5+2)
a) h#2
b) -1(w+s), dann h#2
a) 1. Kf7 Lg6+ 2. Kg8 f7#
b) R: 1. exf6ep+ f7-f5, dann 1. Kg5 Sf6 [A] 2. Lh4 Lc1# [B]
Nicht: R: 1. g5xf6ep+ f7-f5, dann 1. Kh5 Lc1 [B] 2. Lh4 Sf6# [A], weil das Schach durch den wLb1 nach Rücknahme von Bf7-f5 illegal wäre.
play all play one stop play next play all
Autor: "Originelle Zwillinge mit 3 Modellmatts."
FM: "Hübsch, wenngleich die Zwillingsbildung wohl zwangsläufig den ep-Effekt erfordert."
Aber welchen? Das war hier die Frage! Die Verführung sah offenbar kein Löser?!
MR: "Beeindruckend ökonomischer ep-Witz mit hübschen Mustermatts." 2,5/I
Henrik Juel: part a) is C+ Popeye 4.61 (2021-04-14)
comment
Keywords: En passant in the retro play, Help retractor, Stipulation change, Miniature, Model mate (3)
Genre: Retro, h#
FEN: 8/3N4/5Pk1/8/8/8/KB6/1B2b3
Reprints: feenschach 142, p. 166, 07/2001
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-14 more...
73 - P0005849
E. A. von Vegesack
Danziger Neueste Nachrichten 1941
P0005849
(5+3)
Welches waren 4 Einzelzüge?
R: 1. fxe6ep+ e7-e5 2. Kd4xSc5 Se4-c5
play all play one stop play next play all
In 'Probleme für Tiger' nachgedruckt mit der Forderung: "Welches waren die letzten zwei Züge von Schwarz und Weiß?"
Axel Steinbrink: Hinweis von Bernd Schwarzkopf: Der letzte Zug ist nicht eindeutig: auf c5 hätte auch ein beliebiger schwarzer Stein geschlagen worden sein können. (2020-12-20)
Henrik Juel: I think that any white man (not black) could have been captured on c5
But the last 3 single moves are determined (2020-12-20)
Hans-Jürgen Manthey: die Lösung ist eindeutig!
Der wK muß d4xc5 geschlagen haben, und da der wK auf d4 im schach stand und der sT weder von a5 noch von a3 gekommen sein konnte, muß der zug sSe4-c5+ vor dem wK schlag gewesen sein. Also:
1... Se4-c5+ 2. Kd4xc5+ e7-e5 3. f5xe6ep+ (2020-12-21)
A.Buchanan: Hi Hans-Jürgen, Sie haben Recht mit den ersten "dreieinhalb" Zügen. Henriks Idee ist, dass Ne4-c5 ein Schlag eines unbekannten Steins sein könnte. Normalerweise müssen wir das genaue Diagramm vor dem Zug kennen, um einen Zug klar zu charakterisieren. (2020-12-21)
Hans-Jürgen Manthey: A.Buchanan ist es denn wichtig ob ein w Stein auf c5 geschlagen worden ist, oder nicht ? Auf jedenfall muß ein sS von e4 nach c5. (2020-12-21)
A.Buchanan: Hans-Jürgen, heute halten wir es für wichtig, weil Wir haben Rekorde für z.B. "Letzten 17 Einzelzüge?" brauche also eine einheitliche Definition von "Zug". Manchmal ist die Forderung eines älteren Problems auf diese Weise ungenau, aber nicht kaputt. (2020-12-22)
Hans-Jürgen Manthey: Ich hätte da eine Idee:
das Diagramm wird vervollständigt mit wBa2,b3,c2,d2,e2,g3 wLh1 wSb1,g1 wTc1 wDd1. Nun ist es nicht mehr möglich eine weiße Figur zu Schlagen! :-) (2020-12-23)
Mario Richter: Shouldn't the stipulation be: "Welches waren die letzten 4 Einzelzüge?" or "Letzte 4 Einzelzüge?
And, imho, the problem is cooked. Jans-Jürgen's suggestion would make the problem correct, but at the cost of much too much material. (2020-12-24)
A.Buchanan: The stipulation field in PDB has always been used to communicate the solvable intent of the composition. Stipulations have been rewritten by Gerd or others where necessary. What we lack is an Originalforderung field. But we can fake that in the Comment field. It hardly breaks the problem to change stip to 3 from 4 (2020-12-25)
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comment
Keywords: Last Moves?, En passant in the retro play
Genre: Retro
FEN: 8/8/4Pk2/2K4p/7r/7P/8/B4R2
Reprints: 255 Probleme für Tiger 2 04/1986
2318 Landeszeitung für die Lüneburger Heide 16/06/2012
Input: Gerd Wilts, 1995-06-21
Last update: A.Buchanan, 2020-12-25 more...
74 - P0006092
Bedrich Formanek
Breyer Gedenkturnier 1973
3. Preis
P0006092
(4+5)
h#1
1. "cxd6"ep b5 2. axb6"ep"#
play all play one stop play next play all
Henrik Juel: Quite cute: the two 'half' white moves add up to one white move
First White removes a black pawn, and after b7-b5 he does not (2021-04-30)
comment
Keywords: Joke, En passant, Complete an unfinished move, Start a move but do not finish it
Genre: h#, Retro
FEN: 8/kpK5/p2P4/P2p4/8/8/8/1r5B
Reprints: (28) diagrammes 97 04-06/1991
Input: Gerd Wilts, 1995-07-17
Last update: A.Buchanan, 2022-02-10 more...
75 - P0006274
Tivadar Kardos
6460 FEENSCHACH 09-10/1963
P0006274
(5+6)
-1s, dann h=2
R: 1. ... cxb3ep, dann 1. Sb1 0-0 2. Sa3 Txa1=
play all play one stop play next play all
more ...
comment
Keywords: En passant in the retro play, Help retractor, Castling, En passant
Genre: Retro, Fairies
FEN: 8/8/8/1p6/k7/1pP2p2/2Pn1P2/n3K2R
Input: Gerd Wilts, 1995-08-20
Last update: A.Buchanan, 2024-01-18 more...
76 - P0006423
Andrey Frolkin
9128 Die Schwalbe 157, p. 283, 02/1996
Leonid Borodatow gewidmet
P0006423
(9+4)
#3 (AP)
White pushes the move (Keym AP)
1. ... f5! (f6?,fxe6?,fxg6? 2.0-0! ~ 3.Tf3#) 2. gxf6ep! exf6 3. 0-0! f5 4. Tf3#
(2. 0-0=?)
Valladao Task via the try: 1. exf7? e5! 1. ... e6? 2. f8=D e5 3. Da3#
Another try shows fully differentiated black Albino:
1. Kf1? fxg6!
1. ... f5? 2. Lxf5! h3 3. Txh3#
1. ... fxe6? 2. Lxe6! h3 3. Txh3#
1. ... f6? 2. Lf5! h3,fxg5 3. Txh3#,Th3#
And retro tries:
1. 0-0?? f5,~ 2. Tf3# (short solution)
1. f5? f6,~ 2. 0-0?? fxg5,~ 3. Tf3#
play all play one stop play next play all
Wenn Schwarz zuletzt gezogen hat, dann kann Weiß nicht mehr rochieren: R: 1. Kh2-g3 Tf1(g1)-h1+. Da aber Weiß in einem direkten Matt beginnt, ist die Rochade nicht mehr zulässig. Der Zusatz "AP" in der Forderung ist überflüssig: es könnte höchstens der Anzug a posteriori dem Schwarzen übertragen werden. Das geht aber nicht, da Weiß in einem direkten Matt beginnt.
Guus Rol: This is apparently AP after Keym. The move goes to black:
0. ... f5! 1.gxf6ep exf6 2.0-0(justifies the handover) f5 3.Tf3 (2007-02-13)
A.Buchanan: There are two kinds of directmate Type Keym. In one Black unsuccessfully pulls the move, in the other (as here) White successfully pushes it to Black. (2022-04-16)
A.Buchanan: I disagree with the German comment in the solution text. I think including "AP" is advisable in the stipulation. Keym AP riffs off Codex Article 15, but it's not the default. (2022-04-17)
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comment
Keywords: a posteriori (AP) (Type Keym), Castling, En passant, Promotion (D), Valladao Task
Genre: Retro, 3#
FEN: 8/4pp2/4P1N1/6PP/5P1p/6kB/6P1/4K2R
Input: Gerd Wilts, 1996-06-12
Last update: A.Buchanan, 2022-04-18 more...
77 - 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...
78 - P0008246
Alexander Kislyak
4 Shakhmaty v SSSR 04/1974
P0008246
(12+12)
#2
1. bxc6ep
play all play one stop play next play all
Lösetip beim Originalabdruck: beim Abzählen der geschlagenen Figuren beachte man die Farbe der Felder, auf denen sie fielen.

AL
Le4 ist ein UW-L, der nur so entstanden sein kann: h2-h7xg8=L.
Schwarz zog h7xg6x5, um ihn vorbeizulassen.
Somit ergibt sich folgende Schlagbilanz:
Weiß: axb, dxc, fxg (als Schlagobjekt für den sBh7) [Anmerkung: statt fxg auch fxe mit UW auf e8 möglich, ändert aber nichts]
Schwarz: bxc, h7xg6xh5 sowie den wLf1, der nicht von den sBB geschlagen wurde

Unter den geschlagen weißen Figuren befindet sich der schwarzfeldrige Lc1.
Die Schläge h7xg6xh5 fanden auf weißen Feldern statt, also wurde der Lc1 mittels b4xc3 geschlagen. ("Effekt Zvetnosti")
Als letzte schwarze Züge kommen also nicht infrage: b3xc2, e7-e5 (sLf8 wird als Schlagobjekt gebraucht), g6xh5 (dann käme der wUW-L nicht von g8 nach e4).
Einziger legaler letzter schwarzer Zug war also c7-c5, und es löst: 1. bxc6ep
Des Autors (geb.27.12.1938)"`Erstling"', also mit 36 Jahren.

abgedruckt in der Rubrik: "Redkije Shanry"
Henrik Juel: Solution: 1.bxc6 ep (2.cxd7#). White captured axb, dxc, fxe, h7xg8=B and promoted on e8; Black captured orig. Bf1, bxc, h7xg6xh5 and promoted on a1. So last move must be c7-c5. (2003-05-27)
Mario Richter: @all PDB activists: Would the introduction of a new keyword: "Farbbalance / color balance"
or something similiar make any sense? (describing the fact, that a certain capture must have happened on a square of a determined color) (2022-04-26)
A.Buchanan: The pieces have colour: black & white. But the squares and the bishops have *shade*. Informally of course one can say anything, but if we are glossarizing then maybe be a bit more formal. And what is “balance” here? I think of balance in terms of pawn captures required vs available. Typically the issue you are talking about is where almost all pawn captures are in one shade square, allowing us say something interesting about a bishop capture. Maybe: “shade logic”? (2022-04-26)
Henrik Juel: I use Andrew's terminologi: men are white or black, and squares are light or dark
But my non-retro friends (both of them) use white/black for squares also
So 'Shade logic' may be too esoteric
What about 'Square color argument'?
It is used far more frequently in retro analysis than, say, 'Parity argument', so a keyword is needed
Thanks for the suggestion, Mario (2022-04-26)
A.Buchanan: I am convinced by Henrik, but I will continue to use “shade” in my own writing. (2022-04-27)
comment
Keywords: En passant as key, Obvious promotion (wLe4)
Genre: Retro
FEN: brkn2R1/Rn1p1pp1/N7/1PpKp2p/QPP1B3/2P5/2p1P1P1/8
Reprints: Caissas Schloßbewohner 3 1987
Input: Gerd Wilts, 1996-09-16
Last update: Mario Richter, 2022-04-26 more...
79 - P0008675
Valeriu Petrovici
(VII) Quartz 4 1997
P0008675
(10+8) cooked
Schwarz nimmt 3, Weiß 2 Züge zurück, dann h#1
R: 1. ... h4xg3ep 2. g2-g4 e2xTd1=L 3. Sc6-d4 0-0, dann 1. Tf7 Td8#
play all play one stop play next play all
Cook: R: 1. ... h4xg3ep+ 2. g2-g4 e2xTd1=L+ 3. e3-e4 Td8xTd7, dann 1. c1=T Dh7#
R: 1. ... h4xg3ep+ 2. g2-g4 e2xTd1=L+ 3. Sf5xLd4 Td8xTd7, dann 1. Lh8 Sxh6#
R: 1. ... h4xg3ep+ 2. g2-g4 d2-d1=L+ 3. e3-e4 Kf7xSg8, dann 1. c1=T Dg6#
Korrektur s. P1402012
Henrik Juel: -1... h4xPg4ep -2.g2 e2xTd1=L -3.Sc6 0-0, 0... Tf7 1.Td8#. (2004-09-09)
Mario Richter: Lots of Cooks, e.g.
R: 1. ... h4xg3ep+ 2. g2-g4 d2-d1=L+ 3. e3-e4 Kf7xSg8, dann 1. c1=T Dg6# (2022-06-17)
James Malcom: Seems the diagram is wrong then. Perhaps something like +sBd2 will fix it? (2022-06-17)
James Malcom: Wait, nevermind on the fix idea. (2022-06-17)
Mario Richter: Hi James, there is already a correction by the author himself, s. P1402012. (2022-06-18)
James Malcom: Thank you, Mario. It only took 25 years to find its way into PDB! The online Quartz archive dates back only to 2007. (2022-06-18)
Mario Richter: James, you may want to take a look at http://www.stere.ro/, where I think a complete run of all 'quartz' issues, starting with the very first number 1 of 1996 can be found ... (2022-06-18)
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comment
Keywords: Help retractor, Valladao Task, Promotion in the retro play (l), Castling in the retro play (sk), En passant in the retro play, Superseded by (P1402012), Castling (sk), Promotion (l), En passant
Genre: Retro
FEN: 5rk1/3r4/7b/7K/2PNPP2/2P3p1/PPp2P1q/1Q1b4
Input: Gerd Wilts, 1997-06-11
Last update: A.Buchanan, 2022-07-10 more...
80 - P0008793
Andrej N. Kornilow
Shakhmaty v SSSR 1978
P0008793
(15+8)
#1.5
a) 1. "Ta1-d1+?" ... Td4 2. Txd4# (Completing the illegal castling!)
b) 1. "xc5ep+?" ... Tb5 2. Txb5# (Completing the illegal e.p.!)
c) 1. "xe8=S+?" ... e6 2. Sf6# (Completing the illegal promotion!)

0) 1. Sxe7+! Lxe7 2. f8=B# (Truncating the promotion move!)
play all play one stop play next play all
n in directmate stipulation #n means that White has n moves to do the job. With n=1.5, therefore, one of the White moves is fractional, so we know we are in the realm of jokes, ho ho!

If White's first move is the fractional one, there are three retro tries which attempt to complete: castling, ep & promotion. However all are illegal:
castling: 8 white pawns, so wTa5 came from h1, dislodging wK.
ep: retracting sBc5 to c7 means wLb8 is promoted, but 8 white pawns.
promotion: white made 7 pawn captures, while sBgh were waylaid.

So we consider that White's second move was the fractional one, by omitting the replacement of wBf8 by an officer.
Henrik Juel: Completing the key isn't legal, 1.'Rd1'/'-bPc5'/'e8S'+, but omitting the promotion in the mating move is, 1.Sxe7+ Bxe7 2.'f8'#. Excellent joke. (2003-09-19)
James Malcom: A very witty joke Valladao! (2020-09-24)
A.Buchanan: Really like this joke (2020-09-25)
Henrik Juel: To see the illegality of completing 1.fxe8=S+?, note the captures:
Black captured fxDe, so [Pg7,h7] were captured on their files, while the other six missing black men were captured by white pawns
The illegality of completing 1.0-0-0,bxc6ep? is rather obvious (2020-09-26)
James Malcom: And for those you don't find it obvious: If the White king hasn't moved, then where did wRa5, and wBb8 if bPb5 has just done a double-step, come from? Neither can be promoted pieces, as White still has all eight pawns. Trying to finish castling and en passant therefore both produce illegal positions and thus cannot be the solution. (2020-09-26)
Henrik Juel: Continuing beating the dead horse...
How does the white player actually perform an entire move?
1.0-0-0+: Ke1-c1 and Ta1-d1
1.bxc6ep+: Pb5-c6 and remove sPc5
1.fxe8=S+: remove sYe8 and Pf7-e8 and replace wPe8 with wSe8 (three fractional actions)
2.f8=Y#: Pf7-f8 and replace wPf8 with wYf8
So a marginally better stipulation might be: 'White to move mates in less than 2 moves' (2020-09-26)
A.Buchanan: Hurray I've got the animation working! I agree with Henrik's stipulation. There is a dummy pawn on f8, but not by the Dummy Pawn rule. Instead it's the joke that does it. If Dummy Rule applied, the move would be full length! :) (2022-02-09)
more ...
comment
Keywords: En passant, Start a move but do not finish it, Castling, Promotion (S,B), Valladao Task (half!), Joke (End move, Start move), Dummy Pawn (not! Start move), Complete an unfinished move, waylaid (sBgh)
Genre: Retro, 2#
FEN: 1B1N1bB1/p2ppP2/2P5/R1pk1N2/1r2p3/1P2P1P1/4PP1P/R1K5
Reprints: (III) Quartz 4 1997
Input: Gerd Wilts, 1997-06-21
Last update: A.Buchanan, 2023-04-08 more...
81 - P0008794
Alexandr A. Klibanski
2 Shakhmaty v SSSR , p. 31, 07/1977
P0008794
(14+4)
#1
1. Tb1#
zuletzt geschah
R: 1. ... Kc2-b2 2. e5xf6ep+ f7-f5 3. e4-e5+
oder
R: 1. ... Kc1-b2 2. 0-0+
oder
R: 1. ... a2-a1=L
play all play one stop play next play all
weiße Bauernschläge: axb3,b2xc3xd4xe5xf6,c2xd3xe4xf5xg6xh7-h8=L,e2xfxg = 12 Schläge, alle fehlenden schwarzen Steine sind erklärt, insbesondere hat der wBh2 nie die h-Linie verlassen.
Der sBh4 stammt also entweder von f7 (2 Schläge) oder von h7 (ebenfalls 2 Schläge, um den wBh2 vorbeizulassen). Damit sind auch alle fehlenden weißen Steine erklärt. Da zu diesen auch der schwarzfeldrige wLc1 gehört, kann zuzletzt nicht h5-h4? geschehen sein ('Shakhmaty v SSSR' nennt das "Farbeffekt").
paul: La question est si les Blancs sont a jouer et le reponse est affirmative. Les seuls retro-jeux plausibles sont:
1.Rc1-b2 0-0;
1.Rc2-b2 e6xf6 e.p. 2.f7-f5 Tg6-a6+;
1.a2-a1F.
Donc le task Valladao. Le coup matant est 1.Tb1#
Est interessant a voir pourquoi ne va pas le retro-jeu h5-h4?.
L?analyse demontre que le seul Pion blanc transforme en en h8 a ete Pc2 et la sequence des coups a ete c2xd3xe4xf5xg6xh7-h8F. Le P blanc h6 vient de h2 sans captures, donc le P noir h4 a effectue deux prises, don?t une d?un F blanc de camps noirs. Donc h5-h4 est illegal. (Quartz 4/1997) (2010-03-22)
Henrik Juel: For the benefit of readers, whose French is a bit rusty:
Paul points out that the retroplay given on top is incorrect. Ph4 is [Pf7] who captured the dark-squared [Lc1] on its way to h4; so its latest move was g5xh4, not h5-h4. (2010-03-26)
Paulo Roque: Henrik:
Not R: 1.h5-h4. See also that [Pa7] was promoted in the square a1, at some point. But I think also possible Ph4 is [Ph7], by: h7xg6-g5xh4 or h7-h6xg5xh4 (2010-03-27)
Henrik Juel: You are right, Paulo; my mistake, not Paul's. I should have written: Ph4 is [Pf7] or [Ph7], who captured the dark-squared [Lc1] on its way to h4; so its latest move was g5xh4, not h5-h4. (2010-03-27)
James Malcom: Question: What prevents the Black king from going to c1 after castling is retracted? (2022-06-17)
comment
Keywords: En passant in the retro play, Castling, Promotion in the retro play, Valladao Task
Genre: Retro
FEN: 6NB/Q5pB/R4P1P/6P1/7p/1P4P1/1k1P1P2/b4RK1
Reprints: (V) Quartz 4 1997
Input: Gerd Wilts, 1997-06-21
Last update: Mario Richter, 2022-06-17 more...
82 - P0008971
Thierry Le Gleuher
645 Europe Echecs 460 10/1997
P0008971
(8+14) cooked
shc#16
1. Sc8 2. Sxb6 3. Sxa4 4. Sb2 5. Sd3 6. Se1 7. Sf3 8. exd3ep 9. Kd4 10. Ke4 11. Sd4 12. gxf3ep 13. Kf4 14. Kg4 15. f4 16. Sf5 exf3#
play all play one stop play next play all
NL. Verbesserung 1998 erschienen: P1000933
Cook: 1. Lf7 2. Txe5 3. Kd5 4. Ke6 5. Kf6 6. Sg7 7. Se6 8. Tg6 9. Dh7 10. Dg7 dxe5#

außerdem Dual in der AL
1. Sc8 2. Sxb6 3. Sxa4 4. Sc5
James Malcom: Where is the cook? (2021-09-14)
comment
Keywords: Seriesmover, Consequent, En passant (x2)
Genre: Retro, Fairies
FEN: 8/4n3/1P2b3/3rPprn/P1kPpPpq/2p1p1p1/4P1Pb/7K
Input: Gerd Wilts, 1998-06-26
Last update: Mario Richter, 2021-09-16 more...
83 - 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...
84 - P0505412
Martin Gohn
2456v Revista Romana de Sah 11-12/1947
3. Preis
P0505412
(3+13) cooked
h#5
1. Kd7 g4 2. Lh1 Lg2 3. Ke6 Lh3 4. Ld5 g5+ 5. f5 gxf6ep#

NL:
1. Td5 Lxb7 2. Sf6 g4 3. Sg8 Kxf8 4. Kd7 g5 5. Ke6 Lc8#
play all play one stop play next play all
Korrekturvorschlag HJS Co+: wKg7, wLa8, wBg2, sKc7, sDd6, sTf8h8, sLb7c1, sSa5, sBb2b5c4e2e5f7h6h7
Yuri Bilokin: Correction: bQb1-a4, bPe3-a6, -bRd3 B4n1r/1bk2pKn/p2p2p1/4p3/qp6/8/6P1/8 (3+12) (2021-03-03)
A.Buchanan: @Yuri: no improvement appears in WinChloe. The "Korrekturvorschlag HJS Co+" B4r1r/1bk2pKp/3q3p/np2p3/2p5/8/1p2p1P1/2b5 is sound, but at 3+15 much more expensive than your economical 3+12. To my surprise, I found I can improve your version still further as B4n2/1bk2pKn/3q4/4p3/8/8/6P1/8 3+7! bQd6 is a very effective cook-stopper. (2021-03-04)
Viktoras Paliulionis: Nice improvement. Compare with P1235831. (2021-03-05)
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comment
Keywords: Bahnung, En passant as mating move, Model mate, Superseded by (P1388090)
Genre: h#
FEN: B4n1r/1bk2pKn/3p2p1/4p3/1p6/3rp3/6P1/1q6
Input: hpr, 1996-12-27
Last update: A.Buchanan, 2021-04-03 more...
85 - P0540829
Filippo Minieri
H551 The Problemist 07-08/1977
P0540829
(5+9) cooked
h#2
1.2.1.1
1) 1. exd3ep b8=D 2. Kd4 Df4#
2) 1. exd3ep a8=S 2. d4 Sxb6#
play all play one stop play next play all
Cook: e.p. not justified as R: 0.e3xd4 possible
VL: Unjustified ep-capture because of –e3xd4. For correction it suffices e.g. to add bSe3.
This is possibly a unique 2-move helpmate with ep-key and 2 solutions of the type 1.2.1.1 (at least so currently in the PDB).
Publi?ation details. H551 in the Helpmate section of the July issue, with no preview. Here is the text of the solution published in the Nov.-77 issue: “…White may have just played P(e3)xd4. Very few solvers have spotted this. Full marks for claim of no solution with correct reasoning but, in fairness, one mark for composer's intention. White cannot have just played K(a5)a6 in reply to Pxb6+ as too many captures by Black are required.” (2023-12-01)
A.Buchanan: Yes Valery you are right this must be cooked. However, I'm not a fan of +bSe3 because it spoils one of the model mates. Separately, I mildly prefer bPf7 to bPe6. (2023-12-03)
VL: I do prefer bPf7, as well. I don't see any way to save at least one of the properties Kindergarten Problem and Model mate (2) after fixing (what for these keywords have been added just now?). Besides, Andrew, please, refine the specification ((full) number and publication month 07-08). (2023-12-03)
A.Buchanan: Ok have added the publication details. The keywords are useful to indicate the aim, even though the problem is cooked. I guess there is a way to keep the model mates but it’s very clunky and probably +bSe3 is better (2023-12-03)
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comment
Keywords: En passant as key, Kindergarten Problem, Model mate (2)
Genre: h#, Retro
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/PP6/Kpp1p3/2pp4/1pkPp3/2p5/P7/8
Input: Michal Dragoun, 1998-04-09
Last update: A.Buchanan, 2023-12-03 more...
86 - P0574383
Tivadar Kardos
The Fairy Chess Review 1956
P0574383
(9+14) cooked
h#3
1. cxb3ep+ Sb4+ 2. Kb1 Sxe2 3. b2 0-0#
play all play one stop play next play all
Cook: NL 1. Db2,Sb3 Sxe2 2. Sb3,Db2 0-0 3. Lxb4 Sxb4#
A.Buchanan: Two aspects of cookery here. First, NL 1. Db2,Sb3 Sxe2 2. Sb3,Db2 0-0 3. Lxb4 Sxb4# Second, White can retract c3xb4 (and earlier captured onto c-file). Black could have captured axbxcxdxe, exf, fxe, with wPg waylaid (2021-11-23)
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comment
Keywords: En passant as key, Superseded by (P1396163)
Genre: h#, Retro
Computer test: Popeye v4.87 says cooked
FEN: 8/8/1p6/b1P2P2/1PpqPpp1/r2ppb1p/k1N1p2P/n3K1NR
Input: Felber, Volker, 1999-12-20
Last update: A.Buchanan, 2021-11-25 more...
87 - P0577550
Johann J. Chaschtschanski
14447 Schach , p. 79, 03/2000
P0577550
(3+9)
h#7
b) sTc2 nach h6
a) 1. Txg2 Kxg2 2. Ld4 Kh1 3. Le5 fxe5 4. Kf4 e6 5. Kg5 e7 6. Kh4 e8=D 7. Kh3 Dxh5#
b) 1. gxf3ep gxf3 2. Ke5 Kg2 3. Le4 fxe4 4. Kf6 exf5 5. Kg7 f6 6. Kh8 f7 7. Th7 f8=D#
play all play one stop play next play all
Anton Baumann: C+ Gustav 4.1d (2020-11-24)
comment
Keywords: Switchback (K in a), Entblockung, Promotion (D), Promotion in the mating move (D in b), Twin, En passant as key, Selfblock (in b), Corner mate (in b), Model mate (in b)
Genre: h#, Retro
FEN: 8/8/8/5p1p/4kPp1/3bb1p1/2r3Pp/7K
Input: hpr, 2000-10-27
Last update: Gunter Jordan, 2020-11-25 more...
88 - P1000967
Thierry Le Gleuher
R008 Problem Paradise 14 09/1999
P1000967
(9+13)
shc#8
1. gxf3ep? Weiß hat keinen letzten Zug!

1. Le6 2. Tdf5 3. Lxa2 4. Lxb1 5. gxf3ep 6. Kf4 7. Kg4 8. Df4 h3#
play all play one stop play next play all
Henrik Juel: After 5.gxf3ep last move could be Ka2-a1 (2021-09-15)
comment
Keywords: En passant, Seriesmover, Consequent
Genre: Retro, Fairies
FEN: 8/8/p7/P2r1brp/pp2kPpb/p3q1p1/P1P1P1PP/KN6
Input: Gerd Wilts, 2000-10-15
Last update: James Malcom, 2021-09-14 more...
89 - P1012021
Jean-Francois Baudoin
Jean-Claude Gandy

diagrammes 1979
P1012021
(12+8)
#1 (wieviele?)
Si le roque B est jouable et si 0. ... g7g5, alors 1.O-O# et 1.fxg6 ep# en plus 1.Rf2#. Pour gagner c1, le RN n'a pu passer par a3, b3, c3, d3, ces cases ayant toujours été occupées ou défendues par des PB. Son passage par une autre case implique le jeu du RB, ce qui casse le roque. PNg4 vient de d7 (3 prise). PNa7 capturé sur sa colonne par une figure B tandis que les PB ont pris les 8 autres pièces N manquantes, y compris le FNf8 qui est sorti via g7. Donc 0. ... g7g5 nepeut pas être envisagé comme dernier coup possible. Si 0. g6g5 mais alors Dh6+ serait précédé par g7g6 qui serait illégal (FNf8 pris à dom.) Si trait N pas de mat après 1. ... g4g3! Trait aux B : 0.g6g5 f4f5+ -1.f5g4 etc. Donc seul 1.Rf2#.
play all play one stop play next play all
No. 1181 HN
paul: Duplicate of P0003893 (2011-08-05)
A.Buchanan: I guess this diagram is wrong, and the other (+wPa7,-bPb6) is the better (2021-06-30)
comment
Keywords: Cant Castler, En passant
Genre: Retro
FEN: 8/4pP1p/1p3p1Q/5Pp1/6p1/PPBP4/BPp1P3/2k1K2R
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2021-07-03 more...
90 - P1012052
Henri Nouguier
diagrammes 2004
P1012052
(9+15)
shc#6
1. dxc3ep 2. Kxb5 3. Kc6 4. Kd7 5. Ke8 6. 0-0 Txg7#
play all play one stop play next play all
Un des PNd a pris 1 pièce blanche. pour venir de e7. Donc les PB b, c et d ne peuvent se croiser. Le PBg6 ne peut venir de f5 car il faudrait 2 prises noires! Seule reste le dernier coup blanc 0... c2-c4 Ceci implique que le FNb1 est issue de promtion ! C'est le PNb7, il aurait pris 2 pièces blanches. Les PNg5 et h5 ont pris 2 pièces blanches.
No. 11661 HN
A.Buchanan: “Obvious promotion” is intended to refer to something visible in the initial diagram position, not what is reached during series play (2022-09-14)
A.Buchanan: Similarly “non-standard material (2022-09-14)
more ...
comment
Keywords: Seriesmover, Consequent, En passant as key, Promotion in the retro play (sLb1), Castling, Valladao Task
Genre: Retro, Fairies
FEN: 7r/5pnR/5PnR/1Pp2ppq/1kPp4/p2P4/P2pP3/Kbb4r
Input: Henri Nouguier, 2004-01-11
Last update: Mario Richter, 2022-09-14 more...
91 - P1012059
Ronald Turnbull
15 diagrammes 10/1990
P1012059
(4+12) cooked
h#3 (AP)
PRA
1. bxc3ep 0-0! (for AP reasons, not 1...Kf2?/Th2? etc) 2. Ta1 Txa1 3. Kc4 Txa4#
1. fxe3ep 0-0! (for AP reasons, not 1...Tf1?/Txh3? etc) b2 Txf3 3. Tc3 Tf4#
play all play one stop play next play all
Si les B. peuvent roquer, leur dernier coup est Pc2-c4 ou Pe2-e4. D'où les 2 solutions alternatives: 1.hc3:e.p!. O-O 2.Tal Tal: 3.Rc4 Ta4:# l.fe3:e.p. O-O 2.b2 Tf3: 3.Tc3 Tf4#
Cook: 1. Kxe4 Tg1/Txh3 2. Ke3,Sd4 Tg5/Th5 3. Sd4,Ke3 Txe5# (4 variants)
No. 11669 HN
paul: Cooked: 1.K×e4 Tg1 2.Ke3 Tg5 3.Sd4 T×e5# (2011-11-25)
A.Buchanan: This nice simple puzzle can be repaired by replacing pawn c5 with rook (C+). As often with cooks in PDB, one wonders if it's just a typo. I have an old email address of Ronald's, and I may ask him. (2012-02-20)
A.Buchanan: I met Ronald for dinner last year (it seems like a vanished age!) in Norwich, UK. He was there for a mountaineer's conference. The climbers selected Norwich as their venue for humorous reasons as it's in the flattest area of Britain. We discussed a bunch of compositions, including this one. He agreed it was cooked and agreed with my suggested fix, which I will post in a separate entry. Very nice chap - he doesn't compose much these days because he can get a bigger audience (and more money) from writing hill-walking books. My hill-walking friends say he's a good writer. (2020-12-08)
comment
Keywords: a posteriori (AP) (Type Petrovic - cee), En passant as key (2), Castling (wk), Superseded by (P1382808), Partial Retro Analysis (PRA)
Genre: Retro, h#
Computer test: cooked by Popeye v4.85
FEN: 8/8/8/1np1n3/ppPkPp2/rp1p1p1p/8/4K2R
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2022-02-18 more...
92 - P1012074
Henri Nouguier
diagrammes 1987
P1012074
(4+4)
shc#10
1. Td2 2. Tb2 3. bxc3ep 4. Th2 5. Th8 6. Kxb5 7. Kc6 8. Kd7 9. Ke8 10. 0-0 Lxb3#
play all play one stop play next play all
Following the ep, b4-b5 is the legitimate last move.
No. 1304 HN
A.Buchanan: Made changes, and checked that all moves comply with the condition. Following the e.p., b4-b5 is legitimate last move. (2021-01-19)
more ...
comment
Keywords: Seriesmover, Consequent, En passant, Castling, Grid Chess
Genre: Retro, Fairies
FEN: 8/8/8/1P6/kpP5/Kp1r4/B7/8
Input: Henri Nouguier, 2004-01-11
Last update: James Malcom, 2021-09-14 more...
93 - P1012345
Henri Nouguier
F979R The Problemist 11/1987
P1012345
(4+4)
shc#9
Gitterschach
1. Kd2 2. Ke3! 3. Kd4 4. Ke5 5. Kd6 6. Kd7 7. Ke8 8. 0-0 9. bxc3ep Lxb3#
1. Kd2 2. Kd3? - W hat keinen letzten Zug
play all play one stop play next play all
No. 1857 HN

Original ohne wBb2, mit wKb1 und sLa3 veröffentlicht, dann aber NL 1. bxc3ep, weil danach W als letzten Zug Kc2-b1 hatte.
In der korrigierten Stellung (03/1988, S.400) hat W nach bxc3ep nie einen letzten Zug, so daß dieser ep-Schlag erst am Ende der Lösung gespielt werden darf.

Außerdem bietet der Autor noch einen Zwilling an: Ks a5/a4, zusätzlich Monchromes Schach, shc#6
('The Problemist' (03/1988, S.400): "Composer's Mono-chromatic setting (1.bc ep.) saves another unit, but has less interest."
paul: Or directly 1.bxc3ep etc. (2011-10-24)
James Malcom: Is this cooked, as paul's comment seems to indicate? (2021-09-14)
Henrik Juel: No, I don't think so, James
After 1.bxc3ep White has no previous move (2021-09-15)
Mario Richter: Paul's comment was obviously referring to the origial position, not to the one now presented in the diagram.
(The germsn comment above reads: Originally published without white pawn b2, with wKb1 and black Bishop a3. But then cooked by 1. bxc3ep, because W has Kc2-b1 as last move [preceeded by c4xb3+]) (2021-09-16)
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comment
Keywords: En passant, Grid Chess, Seriesmover, Consequent, Castling
Genre: Retro, Fairies
FEN: 7r/8/8/8/1pP5/Kp6/BP6/4k3
Input: Henri Nouguier, 2004-01-11
Last update: Mario Richter, 2021-09-15 more...
94 - P1012536
Thomas R. Dawson
British Chess Magazine 06/1925
P1012536
(10+5) cooked
#2
Grashüpfer d5,d6
1. b8=S? droht 2. Sc6#
1. ... Gxa6!

1. bxc6ep! Gxd4,Gxb6,~ 2. Lb5#

R: 1. c7-c5 2. Gd7-d5+ Gb8-d6
play all play one stop play next play all
'The Problemist', p. 241, 09/1993: "The solution to this #2 with Grasshoppers is 1.bxc6
en passant. lf Black's last move had been 1...c6, then White's previous move would have been the illegal G(d7)-d5+.
Cook: 1. Sc4+ Kxa4 2. Sab6#
No. 2251 HN
A.Buchanan: Diagram error? Solution typo? (2023-01-06)
Mario Richter: Diagram error corrected (2023-01-07)
Ulrich Voigt: NL: 1. Sc4+ Kxa4 2. Sab6# (2023-01-08)
Mario Richter: Vielleicht sind Dawson und alle, die diese Aufgabe unkritisch nachgedruckt haben, einer typischen Halluzination erlegen: im Geiste nimmt man den Zug c7-c5 zurück, der erst den ep-Key ermöglicht, und meint dann, daß die NL: 1. Sc4+ Kxa4 2. Sab6 an 2. ... c7xb6 scheitert ... (2023-01-09)
A.Buchanan: Easily repaired e.g. by replacing wSa8 with wLa7 (2023-01-09)
comment
Keywords: En passant as key
Pieces: du = Grasshopper (G)
Genre: Retro, Fairies
FEN: N2K4/1P6/PN1*2q4/kPp*2Q4/Pp1R4/p2B4/8/8
Reprints: The Problemist , p. 241, 09/1993
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2023-01-09 more...
95 - P1013085
Wolfgang Dittmann
R232 Probleemblad 02/2004
1. Lob
P1013085
(5+5)
#1 vor 7
VRZ, Typ Proca ohne VV
Anticirce
R: 1. f2xDg3[+wBg2] Lg4-h3+ 2. Ke1-f1 d3-d2+ 3. e2xSf3[+wBf2] Db8-g3+ 4. d2xTe3[+wBe2] Lh8-a1+ 5. c5xBd5ep[+wBd2] d7-d5 6. Ka5xLb4[+wKe1] Df8-b8+ 7. c4-c5, dann 1. c4xSb5[+wBb2]#
play all play one stop play next play all
paul: Seems to be dualistic: 1. f2xLg3[+wBg2] Lg4-h3+ 2. Ke1-f1 d3-d2+ 3. e2xSf3[+wBf2] Lb8-g3+ 4. d2xTe3[+wBe2] Lh8-a1+ 5. c5xBd5ep[+wBd2] d7-d5 6. Ka5xLb6[+wKe1] La7-b6+ 7. c4-c5, dann 1.c4xSb5[+wBb2]# (2021-11-07)
more ...
comment
Keywords: Allentschlag, Circe (Anti-), En passant, Defensive Retractor, Type Proca
Genre: Retro, Fairies
FEN: 8/5P2/8/1n3P2/8/k6b/3p2P1/b1N2K2
Reprints: 218 Der Blick zurück 2006
Input: Gerd Wilts, 2004-02-20
Last update: A.Buchanan, 2018-12-04 more...
96 - P1013610
Hans Gruber
541 Problemkiste (12) 11/1983
P1013610
(11+7)
Letzter Zug?
Typ A
Längstzüger
R: 1. b5xa6ep!
play all play one stop play next play all
Mario Richter: Can somebody please provide an explanation? It's clear that Black didn't move last, but why not e.g. R: 1. Ke6xTf7 or even R: 1. b5xTa6 Ta3xDa6 2. Dc8-a6+ (2022-01-10)
comment
Keywords: En passant, Maximummer, Last Move?, Type A (BxB ep.)
Genre: Fairies, Retro
FEN: n7/2p2K1p/Pk6/7p/7p/8/PP1PPnPP/2B2BB1
Input: Hans Gruber, 2004-05-01
Last update: Erich Bartel, 2014-09-14 more...
97 - P1013826
Hans Gruber
Theodor Steudel

idee & form 07/1987
P1013826
(4+4)
-1w, dann +1
R: 1. Kc5xBb5, dann 1. axb6ep+ (letzter schwarzer Zug war nur b7-b5)
play all play one stop play next play all
Adrian Storisteanu: See also the (less subtle) P1013784. (2021-04-25)
comment
Keywords: Help retractor, En passant as key
Genre: Retro
FEN: rb6/k1p5/P1P5/PK6/8/8/8/8
Input: Hans Gruber, 2004-05-01
Last update: A.Buchanan, 2024-01-18 more...
98 - P1017717
Vincent Lanius Eaton
The Cincinnati Enquirer 1930
P1017717
(7+5)
#2
1. d4! droht 2. dxc5#
1. ... c1=D,L+, 2. Sf4#
1. ... c1=S 2. Sc3#
1. ... cxd3ep 2. Txd3#
1. ... cxd4 2. Txd4#
play all play one stop play next play all
Henrik Juel: duals after 1... c1=T (2024-01-26)
comment
Keywords: Brian Stephenson Collection (2548), En passant (s), Provoked check, Promotion (sD,sL), Blocknutzung, discovered checkmate, double-check mate, Cross check
Genre: 2#
FEN: 3R4/8/8/2p3K1/2p5/5B2/R1pPNB2/1b1k4
Reprints: 90 The American Two-move Chess Problem , p. 55, 1965
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2024-01-26 more...
99 - P1020144
Josef Kling
23 The Chess Euclid 1849
P1020144
(6+6)
#2
1. axb6ep+ Kb1 2. 0-0#
1. ... Da7 2. Kf2#
play all play one stop play next play all
in 'Le Pion' fälschlich wBa4 ("1 - R 1 R ... 5 - P 4 TD ...")

Nach heutigen Konventionen ist der Ep-Schlag unzulässig, man beachte aber:
Kling im 'Chess Euclid' zur Intention dieser Aufgabe: "This problem I have compossed expressly to shew the error the modern composers of problems have committed, in introducing the system of castling in the ends of games. In this problem I consider I have as good a reason to say that Black has just moved his Q. Kt. P. [b5] two squares as that White has not moved his K. R. or K."
James Malcom: I found this hidden ancient, another that rode the rails of the unjustified en passant key. (2020-12-28)
SP: Unsound in its own terms, as, whatever Black plays (Kb1/Qa7),
castling is never forced due to 2.Kf2#
I use the keyword "non-analytic en passant key" for such problems. (2023-08-21)
A.Buchanan: The composer's stated intention seems to be that (1) ep is ok (2) castling is not ok. In these terms, the problem is sound (2023-08-22)
comment
Keywords: Brian Stephenson Collection (4985), En passant as key, Castling (wk), En passant, Castling as mating move, Golden Age
Genre: 2#
FEN: R7/7q/7r/Pp6/5p1r/7P/4Q3/k3K2R
Reprints: 458 Dubuque Chess Journal 35 01/1873
P10 Le Pion 01/05/1873
Input: Brian Stephenson, 2004-08-12
Last update: A.Buchanan, 2023-08-22 more...
100 - P1023931
Sandor Hertmann
591 The Chess Amateur 1928
Ehrende Erwähnung
C.A. Valve-Turnier 1928
Bauern-Valve mit En Passant
P1023931
(10+9)
#2
1. d4! droht 2. Tc5#
1. ... c4xd3ep 2. Dxg2#
play all play one stop play next play all
SCHRECKE: 1. ... Sd3 2.D:g2,D:a4# (Mattdual) (2022-07-23)
comment
Keywords: Brian Stephenson Collection (8796), Valve (sB), En passant, Line closing
Genre: 2#
FEN: 5N1n/2B4p/BPk3qR/6R1/p1p3N1/2r5/1nQP2b1/1K6
Reprints: 116 Valves and Bi-Valves , p. 129, 1930
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2022-07-22 more...
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