Die Schwalbe

18 problem(s) found in 3315 milliseconds (displaying 18 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND S='Problemkiste' AND G='Retro'] [download as LaTeX]

1 - P0003417
John Frederick Keeble
2206 The Problemist Fairy Chess Supplement 16, p. 173, 02/1936
P0003417
(4+5) C+
h#2.5 (AP)
1. ... exd6ep 2. 0-0-0 dxe7+ 3. Tf8 exf8=T,D#
play all play one stop play next play all
H.Juel: If Black may castle, last move was not made by Ke8 or Ta8, but by Pd5, so it was d7-d5, and hence White may capture en passant. 1... exd6ep 2.0-0-0 dxe7 3.Tf8 exf8=D,T# where the castling (belatedly) justifies the e.p. capture.
Not 1... exd6ep 2.Kd8? dxe7 3.Kc8 e8=D,T
AB: Another 120 retro tries (=60x2 due to tolerated dual promotion) of general form 1. exd6ep 2. T~ dxc7 3. T~ c8=D,T.
No set play
Nachdruck in "Die Schwalbe" 33 mit Diagrammfehler (wBe6 fehlt).
VL: The first AP-problem! Thematic illegal try:
1. ... exd6ep 2.Kd8? dxe7+ 3.Kc8 e8=T(D)#?? (2007-02-11)
Mario Richter: 46 years between this Problem and P1366663 - perhaps this was not Keeble's first AP-Problem?! (2019-08-29)
Ladislav Packa: Of course I know the idea a posteriori. But let my heresy be forgiven - I do not consider it logical. In this particular example, I consider e.p. for proof that black can castling - not vice versa. (2019-08-29)
A.Buchanan: Hi Ladislav. If ep is ok, then Black just moved a pawn with double hop. That doesn't stop Black having e.g. moved bK right at the beginning of the game. So you can't prove that Black definitely can castle this way. What am I missing here? (2019-08-29)
Ladislav Packa: Your statement is perfectly fine. However, the AP condition is usually interpreted as castling is evidence of e.p. However, the second party does not have to do castling - the possibility has already been demonstrated by the existence of e.p. (2019-08-29)
A.Buchanan: Hi again Ladislav. Please try again, I am obviously being very stupid. I don't understand how 1. exd6ep is any more indicative that Black can castle than 1. K~. (2019-08-30)
Ladislav Packa: Hi, Andre!
e.p. is just a result of the d7-d5 move. Black can castle only if it was the last move of black (even without wPe5). Therefore, it does not have the logic of joining the e.p. with castling. For example, ask yourself if black is allowed to castle in the position without wPe5. (2019-08-30)
A.Buchanan: @Ladislav: ep is not "proof that black can castle", rather it is a *necessary* condition for Black to be able to castle. The logical implication is the other way round. We can castle because of the optimistic convention, and because sPd might just have moved. Without wPe, there is no solution, but whatever White plays, Black can castle (2020-12-11)
Hans-Jürgen Manthey: was heißt hier Thematisch Illegal ??
Mir völlig wurscht ob ep möglich oder nicht, nach dem weißen Zug ist natürlich Kd8 möglich !!
also ist auch 1. ... exd6ep 2. Kd8 dxe7+ 3. Kc8 e8DT# eine Lösung, sonst müßte auf die erzwungende Rochade in der Aufgabenstellung hingewiesen werden ! (2020-12-11)
A.Buchanan: Can one tolerate “tolerated dual promotion” in Valladao? Well Valladao himself did, so it must be ok (2022-03-26)
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comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sg), Promotion (D/T), Tolerated dual promotion (D/T), Valladao Task, Retro Strategy (RS)
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: r3k3/2p1p3/2P1P3/2KpP3/8/8/8/8
Reprints: V. Die Schwalbe 33, p. 323, 06/1975
2001 CHM Themes 12/2000
3 Die Schwalbe 215, p. 240, 10/2005
A1 Problemkiste 169 02/2007
402 Eigenartige Schachprobleme , p. 130, 2010
14 Die Schwalbe 241, p. 374, 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-10 more...
2 - P0006467
Michel Caillaud
Jacques Rotenberg

3927 Problemkiste 102, p. 168, 12/1995
P0006467
(14+15)
BP in 6.0
N statt S in der PAS
wSU,sSU=Nachtreiter
a) 1. Nxe7 Nxe2 2. Nb1 Nxc1 3. Df3 Nca5 4. Ke2 Ngc6+ 5. Kd3 Nb8 6. Ke4 Ng8+
play all play one stop play next play all
paul: Verified by Jacobi with this input:

stip dia 6
forsyth rnbqkbnr/pppp1ppp/8/8/4K3/5Q2/PPPP1PPP/RN3BNR
cond cavaliermaj (2023-06-16)
more ...
comment
Keywords: Unique Proof Game, Cavalier majeur
Pieces: su = Nightrider (N)
Genre: Retro, Fairies
FEN: r*2nbqkb*2nr/pppp1ppp/8/8/4K3/5Q2/PPPP1PPP/R*2N3B*2NR
Input: Gerd Wilts, 1996-06-16
Last update: A.Buchanan, 2021-02-19 more...
3 - P1000662
Gianni Donati
R074 Probleemblad 11/1999
P1000662
(10+14) C+
h#1.5 (AP)
1. ... gxh6ep 2. exf1=L 0-0-0#
play all play one stop play next play all
Kommentare:
Als einziger letzter schwarzer Zug, der die weiße Rochade
erhält, kommt nur Bh7-h5 infrage, was dem Weißen den e.p.-Schlag
ermöglicht, um nicht die Rochade oder das Mattnetz zerstören zu
müssen (H.P.Suwe)
Gianni Donati: This intends to show the Valladao theme in the minimum number of moves. (2004-03-19)
VL: The waiting ep capture. Cf. also P1000260
(by T.Petrovic, 2000) somewhat enriched
thematically with the illegal try 2... Rd1#??. (2004-06-03)
A.Buchanan: The “illegal try” 2. Rd1+ is not actually mate because c2 is unprotected. This is kind of “logic dual” spoils the A.P. motivation for the castling (2020-05-20)
Henrik Juel: White captured h6xg7 and once more, e.g. axPb-b8=Y
Black captured cxdxe, dxexf, fxg, and once more, e.g. a2xb1=Y
Possible retroplay 1... h7-h5 2.h6xSg7 Df8-f4 3.h5-h6 Sf4-h3 4.Sh3-g1 etc., preserving the castling right
Any other black last move would force White to retract Ta1, as d2xc3, d2xe3, and h6xh7 would be illegal retractions
I agree with Andrew that the double motivation of 2... 000# is a weakness:
a. legitimizing h7-h5
b. mating (2020-05-21)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
A.Buchanan: See discussion at P0009121 (2023-08-04)
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Keywords: a posteriori (AP), En passant, En passant as key, Castling, Valladao Task, Promotion (l)
Genre: Retro
Computer test: HC+ Popeye v4.87 + simple retro logic
FEN: 6n1/4p1P1/6p1/6Pp/2b1rqrb/2PkPppn/1P2pP2/R3KRN1
Reprints: König & Turm (4), p. 28, 03/2001
H Problemkiste (143) 10/2002
(V2) Problemkiste 156 12/2004
Input: Gerd Wilts, 2000-08-01
Last update: A.Buchanan, 2023-08-03 more...
4 - P1004380
Bror Larsson
Tidskrift för Schack 1953
P1004380
(8+0)
Auf welchen Feldern darf der sK nicht stehen, da die Stellung dann illegal wäre?
Es sind die 12 Felder a2,a3,a4,a6,d2,d3,e2,f2,h2,h3,h5,h6.
Auf Feld f3 steht der schwarze König legal weil das Doppelschach
Se2-g1 möglich wäre.
play all play one stop play next play all

Duplicate Diagram: P1386681

t/One-Liner 8-feldrig a1h1
Adrian Storisteanu: And not to be missed, P0007560 (wK vs the black home-based officers)... (2021-05-07)
comment
Keywords: Add pieces, Oneliner (8-feldrig a1h1)
Genre: Retro
FEN: 8/8/8/8/8/8/8/RNBQKBNR
Reprints: feenschach (46), p. 81, 04-06/1979
O4 Problemkiste 06/1999
Input: Gerd Wilts, 2003-01-10
Last update: A.Buchanan, 2023-04-13 more...
5 - P1009495
Peter Rösler
B4 Problemkiste 148 08/2003
P1009495
(15+16) C+
BP in 5,0
Stafettenschach
1. e4 e5 2. Ke2 Dh4 3. De1 Dxe4+ 4. Kd1 Dh4 5. d4 Dd8
play all play one stop play next play all
Henrik Juel: Without the condition the solution also works, but there are many cooks (2020-11-01)
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comment
Keywords: Unique Proof Game, Relay Chess, Interchange (KD)
Genre: Retro, Fairies
Computer test: paul: Checked by Jacobi (with relay condition) (2020-10-31)
FEN: rnbqkbnr/pppp1ppp/8/4p3/3P4/8/PPP2PPP/RNBKQBNR
Input: Gerd Wilts, 2003-08-15
Last update: Erich Bartel, 2020-11-01 more...
6 - 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...
7 - P1013968
Hans Gruber
Frank Müller

3312c Problemkiste 10/1993
Erich Bartel zum (Schachbrett-1)ten Geburtstag gewidmet
P1013968
(16+16)
KBP zu einem Legal Cluster?
Kamikaze rex inklusiv
0 moves
play all play one stop play next play all
Null Züge! Denn schon die PAS ist ein Legal Cluster!
Die Autoren schreiben:
Correction:
Problemkiste, Issue 91, Februar 1994, Seite 128, Abteilung Bemerkungen & Berichtigungen
"Nr. 3212 [sic!] v.Gruber/Müller: wie HG mitteilt muß die Bedingung "KAMIKAZE REX INCLUSIV" heißen, nicht Kamikaze-Circe rex inclusiv. Letzteres wäre nicht nur ein Zuviel an Bedingungen, sondern auch noch UL, weil da auch 31-steinige Stellungen möglich sind."
Dieser Eintrag wird also mit der Korrektur aktualisiert!
Erich Bartel: Das war ein schöner Tag am 21.8.1993 bei Hans Gruber zusammen mit Frank Müller und ich bedanke mich recht herzlich für die Widmung, die mir ausgezeichnet schmeckt, fast so gut wie der super delikate Original-Zwetschgendatschi von Hansens Mutter!!!(\eb) (2007-03-04)
A.Buchanan: I don't get this: please can someone explain (2020-12-09)
Henrik Juel: Under the Kamikaze-Circe condition, where both captured and capturing man are reborn after a capture, the initial game array is legal, but removing one man gives an illegal position, so the IGA is a legal cluster
I have no info about the super delicate original Zwetschgendatschi from Hansen's mother (2020-12-09)
A.Buchanan: Hi Henrik, please clarify the rules for the middle part. Why is e.g. 1. a4 b5 axb5[sBb7] with the white pawn disappearing because it can't be reborn on occupied b2? Is it because the rebirth square for capturer is taken to be the start square not the end square of the move? That seems unlikely because I read in PDB that promotion counts before capturer's rebirth. The other possibility is that capture is illegal if the capturer cannot be reborn, but that seems inconsistent with the spirit of Kamikaze. Normally in Circe, a captured piece doesn't come back if the rebirth square is occupied. That surely would still apply here. Glossary is silent on all this. Thanks! (2020-12-09)
Henrik Juel: I think you are right, Andrew, so the IGA is not a legal cluster under Kamikaze-Circe, and the problem is cooked
In the more serious matter at hand, Zwetschgendatschi means plum cake... (2020-12-09)
A.Buchanan: I can't believe that serious fairy heavyweights such as were engaged in this problem would have made something cooked. And I still don't know what K-C actually is. I will ask HG. Maybe he has some comments on plum cake too? :-) (2020-12-09)
A.Buchanan: Henrik has also been copied on an email from Hans Gruber which states in part:
"The problem by Frank and me (PDB: P1013968) is cooked - the intended solution does not work, because the initial game array is NOT a legal cluster under Kamikaze-Circe rex inclusive. Of course this was unavoidable, as dedication problems always are unsound. We should have known that before but did not see in-how-far we were wrong. It is good to learn about that a bit more than a quarter of a century after. I dare to state, without consulting Frank, that there will be no correction to the problem. The basic idea "initial game array = legal cluster" just does not work." (2020-12-09)
Arnold Beine: Perhaps the problem can be fixed by changing the fairy condition into "Platzwechselcirce rex inklusiv". Then every position with 31 pieces is illegal and the IGA should be legal. (2020-12-09)
Henrik Juel: The problem can probably be saved just by changing the condition to Strict Kamikaze-Circe RI (2020-12-09)
A.Buchanan: What is the exact rule for Kamikaze-Circe please? And how does strict differ? (2020-12-09)
Henrik Juel: In Kamikaze-Circe, as part of a capture, the capturer is reborn if its rebirth square is empty (otherwise it vanishes), and the captured is reborn if its rebirth square is empty (otherwise it vanishes). The rebirth squares are the standard Circe ones (based on the capture square).
Strict Kamikaze-Circe additionally stipulates that a capture is possible only when both rebirth squares are empty; so the piece count never changes (2020-12-09)
A.Buchanan: Thanks Henrik: that doesn't sound very "strict" Kamikaze, sounds like "no-sharp-edges video-game" Kamikaze, where no-one actually can get hurt :) Regular Kamikaze RI is trivially a legal cluster, no? And more interestingly, ordinary Kamikaze is a legal cluster too, because if just one piece is removed, that must have been captured by the *other* king, who is still stuck in his cage. That would be my suggested fix (2020-12-10)
Henrik Juel: Yes, any condition where 31 men cannot occur in a game would do
I chose strict K-C, because that was the most simple change to K-C (2020-12-10)
A.Buchanan: Strict K-C RI the number of pieces for each side must always be the same. K RX you can certainly have 31 pieces on the board but not in game array position. It’s still very simplistic but it’s a bit more subtle (2020-12-10)
comment
Keywords: Unique Proof Game, Kamikaze (RI), Initial Game Array (2), Homebase (2), Constrained problem, Legal cluster
Genre: Fairies, Retro
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Input: Hans Gruber, 2004-05-01
Last update: A.Buchanan, 2020-12-22 more...
8 - P1072281
Werner Keym
Tomislav Petrovic

Hannoversche Allgemeine Zeitung 1999
P1072281
(4+3) C+
h#1.5 (AP)
1. ... cxb6ep 2. 0-0-0 b7#
play all play one stop play next play all
Very elegant representation of AP Type Petrovic. Keym himself established in the Codex that the stipulation should include "AP".
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comment
Keywords: En passant as key, Miniature, a posteriori (AP) (Type Petrovic), Castling
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 together with elementary reflection
FEN: r3k3/8/N1P5/KpP5/8/8/8/8
Reprints: (A2) Problemkiste (169), p. 6, 02/2007
Input: hpr, 2007-03-04
Last update: A.Buchanan, 2023-09-11 more...
9 - P1072354
Tomislav Petrovic
(H17) Problemkiste 165, p. 214, 06/2006
P1072354
(5+1) C+
h#2.5
1. ... c3+ 2. Kxa4 Lb5+ 3. Ka5 b4#
play all play one stop play next play all
Letzter Zug: Ka5-b4
A.Buchanan: A small group of similar problems forces some short Round Trip across past and forward play. The last move is not uniquely defined (it may be a capture) and is not mentioned in the stipulation. However the genre includes retro. This means that an aesthetic criterion (e.g. existence of Round Trip) can define whether a composition is subject to 50 move rule or dead position rule! (2021-10-31)
comment
Keywords: Pure Round Trip (k), Miniature, Rex solus
Genre: h#, Retro
Computer test: (Popeye WINDOWS-32Bit V4.05 (1708292 KB))
FEN: 8/8/2K5/8/PkB5/8/1PP5/8
Input: hpr, 2007-03-14
Last update: A.Buchanan, 2021-10-31 more...
10 - P1186997
Erich Bartel
5552 Problemkiste (138) 12/2001
P1186997
(4+4) C+
h=3 (AP)
0.1...
1. ... hxg6ep 2. 0-0 g7 3. Kh7 gxf8=D=
play all play one stop play next play all
Wenn ep erlaubt sein soll, dann muß g7-g5 der letzte Zug gewesen sein. Dies wiederum bedingt, daß sK+sT noch nicht gezogen haben. Nur unter dieser Betrachtungsweise (AP) ist das Ding korrekt.\eb
Löserstimmen:
Diese AP-Idee ist ja nun hinlänglich bekannt, aber man muß auch darauf kommen, sie für den Valladao zu nutzen. Finde ich prima (MN).
ep, und dann kann Schwarz auch noch rochieren (HM).
Wenn g7-g5 letzter Zug war, darf Schwarz noch rochieren (KF)...
AP (= a posteriori) argumentiert andersrum: wenn Schwarz noch soll rochieren dürfen, muß g7-g5 der letzte schwarze Zug gewesen sein (\eb)
Unlösbar! Der Autor meint wohl, man könnte mit 1. ... hxg6ep 2. 0-0 loslegen; hier irrt Goethe. Die beiden Konventionen -ep-Schlag nur, wenn Zulässigkeit retroanalytisch beweisbar ist; Rochade stets, außer wenn Unzulässigkeit retroanalytisch beweisbar ist- sind zwar passend verknüpft. Was aber den Vorrang hat, zeigt die zeitlich Reihenfolge der Anwendung. Zuerst kommt der ep-Schlag, und diesbezüglich gilt: Zulässigkeit nicht beweisbar, also nicht erlaubt (LZ).
Lieber Herr Zagler, vielen Dank für Ihren Kommentar, aber ich denke nicht, daß dazu eine Stellungnahme erforderlich ist, denn dazu ist schon viel diskutiert und geschrieben worden, ichverweise nur auf G. Lauingers "a posteriori Zwischenbilanz" in feenschach (80) X-XI 1986, S.414 ff, wo die AP-Regelung allgemein wie folgt definiert wurde:
"Eine einleitende Handlung (z.B. ein ep-Schlag oder ein unkonventionell Beginn), deren Zulässigkeit nach gültigen Konventionen zunächst nicht nachgewiesen werden kann, wird im Verlauf der Lösung(en) nachträglich durch einen oder mehrere Sonderzüge legalisiert. Legalisieren bedeutet dabei, daß _nur_mit_der_Ausführung_ des Sonderzuges (der Sonderzüge) dessen (deren) Zulässigkeit nachgewiesen wird und gleichzeitig (i.a. aus retroanalytischen Gründen) damit auch nachträglich die Zulässigkeit der einleitenden Handlung."
Ob Sie nun mit dieser Regelung einverstanden sind hat für o.a. Aufgabe wenig Bedeutung, denn nach o.a. "AP"-Definition ist sie lösbar. Wenn Sie "AP" so nicht akzeptieren und einer anderen Retro-Philosophie zugetan sind, steht das auf einem anderen Blatt. Da müßten Sie sich dann mit den entsprechenden Experten auseinandersetzen. (\eb)
Da nur die Rochade den sK nach h7 zu manövrieren vermag, fällt manchem sicher gar nicht erst auf, daß allein sie (neben dem Autor, natürlich!) für die Legalisierung des e.p.-Schlages verantwortlich ist (MR).
Valladao-Task. Bei dieser Ausgangsstellung mit Ansage (HL).

Versions at P1382816 & P1276701
A.Buchanan: Removing the superfluous sBf7 adds various non-castling lines which the AP convention duly protects us from, and makes the composition a sound miniature (2020-10-03)
comment
Keywords: Valladao Task, Castling, En passant as key, Promotion, a posteriori (AP) (Type Petrovic)
Genre: Retro, Fairies
Computer test: %Popeye FreeBSD-4.3-RELEASE-32Bit-Version 3.73
FEN: 4k2r/5p2/5P1P/5KpP/8/8/8/8
Reprints: Springaren (116) 03/2010
K3966 Problemkiste (199/200) 03/2012
Input: Erich Bartel, 2011-02-09
Last update: A.Buchanan, 2022-03-21 more...
11 - P1233676
Sergio Orce
5421 Problemkiste (136) 08/2001
P1233676
(12+10)
#2
(RV)
b) wBb3->b4
a) 1. Ke6! (0-0-0 illegal!) ... 2. h8=D,T#
b) 1. dxe6ep! 0-0-0 2. Dxb7#
oder 1. Ke6 ... 2. h8=D,T#
play all play one stop play next play all
a) 0-0-0 ist illegal. Die wBB haben 5-mal geschlagen, wBBf3/h3 also nicht.
-> sBf2h2 haben 3-mal geschlagen, entweder 3-mal auf schwarz oder nach Rücknahme von g2-g4
-> Lf1 ist nicht unter den Schlagopfern.
-> wBa2 mußte umwandeln. Falls 0-0-0 noch gehen soll, muß die UW auf h8 geschehen sein
-> letzter Schlagfall W×S ist sowieso klar
-> sBa7 muß als Schlagfall für wB dienen. Da e5 es selbst nicht kann, muß er umgewandelt haben, kann dazu aber nur wLf1 schlagen
-> Widerspruch. -> 0-0-0 ist illegal.
b) dito, aber es geht sBa7-a4xLb3-b1 usw.
-> 0-0-0 ist legal. Aber nur, wenn zuletzt e7-e5.
-> 1. dxe6ep, oder falls zuletzt sK/sT-Zug 1. Ke6.
-> "Retrovarianten" (RV) wohl besser als "A posteriori" (AP)
(Übersetzung, Anmerkung und Prüfung durch Hans Gruber).
A.Buchanan: The retro content here is fun, but I don't see why it's suggested that this is AP. (a) is a cant castler, and (b) is a standard PRA matrix. Removing AP as a keyword for this one. (2023-07-21)
comment
Keywords: Valladao Task, Castling, En passant as key, Partial Retro Analysis (PRA)
Genre: Retro, 2#
FEN: r3k3/1pp4P/3p2N1/2NPpK2/4QpP1/1P3PRP/4Pp1p/6n1
Input: Erich Bartel, 2012-03-04
Last update: A.Buchanan, 2023-07-21 more...
12 - P1235136
Manfred Nieroba
K3965 Problemkiste (199/200) 03/2012
P1235136
(2+5) C+
h#2.5 (AP)
Circe
1. ... exf6ep[+sBf7] 2. 0-0 fxe7 3. Kh8 exf8=D#
play all play one stop play next play all
In the usual way with Circe, can refute the idea that Black just captured, so if Black can castle, White can ep. So AP Type Petrovic triggers. No retro try, and no set play.
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comment
Keywords: Valladao Task, Circe, En passant as key, Castling, a posteriori (AP) (Type Petrovic), Minimal, Miniature
Genre: Fairies, h#, Retro
Computer test: HC+ Popeye 4.61 and analysis
FEN: 4k2r/4p2p/8/4PpK1/8/8/8/8
Input: Erich Bartel, 2012-03-25
Last update: A.Buchanan, 2022-06-07 more...
13 - P1278097
Michael Schlosser
V46) Problemkiste (207) 09/2013
P1278097
(5+3) cooked
#3
1. ... bxa6 La7 a5,Ka5 2. Tc5#
1. ... bxc6 La7 c5 2. Tb6+ Ka5 3. Sc6#
1. Ta5#?
1. Tc5#?
play all play one stop play next play all
Schwarz beginnt, da er keinen letzten Zug hat.
A.Buchanan: What's going on here? (2021-10-06)
Henrik Juel: Maybe Pb4 should be removed (2021-10-06)
A.Buchanan: WinChloe has the same diagram, and the solution given here (with incorrect move counting) assumes sBb4 is present. Without it, there is a cook: 1. ... bxc6 2. Kc3 c5 3. Ld8,Lc7 c4 4. Ta5# (2021-10-07)
comment
Keywords: Symmetrical position, Superseded by (P1394575), Symbol problem (Kreuz)
Genre: 3#, Retro
Computer test: %Popeye Windows-32Bit v4.61
FEN: 1N6/1p6/RBR5/1k6/1p6/1K6/8/8
Reprints: V46 Bulletin Michael Schlosser-60-Jubiläumsturnier , p. 38, 2016
Input: Erich Bartel, 2013-09-20
Last update: Gunter Jordan, 2023-07-21 more...
14 - P1288169
Kjell Widlert
3027 Problemkiste (82) 08/1992
P1288169
(4+9)
h#2
Gitterschach
1. cxb3ep! (Tempo) Txg6 2. Le2 0-0#
2. ... Tf1+? 3. Kf2!
play all play one stop play next play all
Der letzte Zug war b2-b4
En passant als Tempozug. Die Grundidee (Einfachschritt des wB ist im Gitterbrett automatisch ausgeschlossen) kann man schon in Fairy Chess Review finden. Hier ist sie mit Rochade garniert in offener Stellung. die jedoch keinen anderen letzten Zug erlaubt. (Autor).
Ganz toll gemacht! (HB).
Raffinierter Zugzwang (KF).
Das ist schon gemein, was der Autor mit den Lösern vorhat (HM).
Viel Gitterschach (Doppelschritt, Retropatt des Tg7 und Rochade trotz Le2) läßt den bekannten e.p.-Trick in neuem Licht erscheinen (MR).
Henrik Juel: Where is the promotion needed for Valladao? (2024-02-04)
comment
Keywords: Valladao Task, En passant, Castling, Grid Chess
Genre: Fairies, Retro
FEN: 8/4p1R1/4p1p1/8/1Pp4p/2p2k2/2p5/3bK2R
Input: Erich Bartel, 2014-09-16
Last update: Olaf Jenkner, 2015-03-19 more...
15 - P1288435
Manfred Nieroba
V5 Problemkiste (157) 02/2005
P1288435
(4+6) C+
h#1.5 (AP)
sDU=Grashüpfer
1. ... dxe6ep 2. 0-0-0 a8=D#
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Durch den ep-Schlag beweist Weiß, dass der letzte Zug von Schwarz
e7-e5 gewesen ist. Die sGG können mangels Bock nicht gezogen haben
und hätten sK oder sT gezogen, so wäre die {\ooo} nicht mehr
möglich (Autor).
A.Buchanan: sGh8 seems useless as it merely eliminates the thematic retro tries. What am I missing? I wonder if the composer really understood the nature of AP...? (2022-03-22)
more ...
comment
Keywords: Valladao Task, En passant as key, Castling, a posteriori (AP) (Type Petrovic), Promotion in the mating move (D)
Pieces: du = Grasshopper (G)
Genre: Fairies, h#, Retro
Computer test: %Popeye FreeBSD-4.3-RELEASE-32Bit-Version 3.81
FEN: r3k2*2q/P4*2q2/*2q2P4/3PpK2/8/8/8/8
Input: Erich Bartel, 2014-09-22
Last update: A.Buchanan, 2022-03-22 more...
16 - P1288499
Manfred Nieroba
6452 Problemkiste (159-160) 07/2005
P1288499
(3+4) C+
h#2 AP
0.1...
Strict-Circe
1. ... cxd6ep[+sBd7] 2. 0-0-0 a8=D#
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Der sSh8 kann wegen Schach zuletzt nicht gezogen haben. Er verhindert zudem NLen.
Mangels Schlagobjekt kann der sB nur von d7 gekommen sein (Autor).

t/nur Themazüge,
A.Buchanan: 1. ... Ke6 Tb8 2. axb8=D/T[-]#? won't work because under "strict" Circe, a capture can't take place at all if the rebirth square is occupied. (2022-05-23)
A.Buchanan: Pieces never leave the board in strict Circe. "Hotel California". The diagram position is hence illegal and it's interesting that one can nevertheless get some kind of retro logic going. It points out a generally unspoken difference between "local" & "global" illegality. (1) Can we reverse moves indefinitely, or do we get stuck in some kind of reverse pat? (Can't say "retropat", because that implies that it's only pat with one player to move, as Henrik explained to me a while back.) (2) If the answer is yes then can we get back to the game array? These are different questions, and problems like this one rely on local rather than global legality. (2022-05-24)
more ...
comment
Keywords: Valladao Task, En passant, Castling, Circe (Strict), Promotion (D), a posteriori (AP) (Type Petrovic)
Genre: Fairies, h#, Retro
Computer test: %Popeye FreeBSD-5.3-RELEASE-32Bit-Version 3.81
FEN: r3k2n/P7/8/2PpK3/8/8/8/8
Input: Erich Bartel, 2014-09-26
Last update: A.Buchanan, 2022-05-26 more...
17 - P1288715
Bernd Schwarzkopf
A12 Problemkiste (169) 02/2007
P1288715
(4+3) C+
=5
AP-Priorität
1. dxc6ep 0-0-0 2. b7+ Kb8 3. c7+ Kxb7 4. cxd8=D Ka7 5. Dc8=
3. ... Ka7 4. cxd8=S Kb8 5. Ka6,Kb6=
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A.Buchanan: Guessing so-called "AP-Priorität" is a variant of AP whereby Black is forced to legitimize (or at least forbidden to rule out the possibility of legitimizing) the earlier e.p. (2020-10-03)
comment
Keywords: En passant, a posteriori (AP) (Type Petrovic), Castling, En passant as key, Promotion, Valladao Task
Genre: Retro, Fairies
Computer test: Popeye v4.85 + minor retro/AP thinking
FEN: r3k3/8/1P1P4/1KpP4/8/8/8/8
Input: Erich Bartel, 2014-09-29
Last update: A.Buchanan, 2022-03-21 more...
18 - P1346371
Tomislav Petrovic
H3 Problemkiste (159-160) 07/2005
P1346371
(9+4)
h#0,5
3 Lösungen
1) 0,5 ... -sBb5# (Teil von axb6ep)
2) 0,5 ... -sBf5# (Teil von gxf6ep)
3) 0,5 ... Tad1# (2. Teil von 0-0-0)
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A.Buchanan: This is definitely not h#0.5 rather it's h#0.25 or #0.5.
This is retro joke not fairy, imho (2022-03-03)
comment
Keywords: Half-moves, Castling, En passant, Complete an unfinished move, Joke
Genre: Retro
FEN: 8/7Q/BP3P2/1p2Rp2/3p4/1P1k4/1P6/R1K5
Input: Erich Bartel, 2018-01-31
Last update: A.Buchanan, 2022-03-03 more...
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