Die Schwalbe

22 problem(s) found in 3267 milliseconds (displaying 22 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND K='a posteriori (AP)' AND K='Umwandlung'] [download as LaTeX]

1 - 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...
2 - P0000793
Nikita M. Plaksin
Andrey Lobusov

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

R: 1. d7-d5 Tc6xh6 (unless R: Kg1-g2 Th~*h1)
play all play one stop play next play all
VL: A posteriori (AP)!
2.g8Q(R)+/Rg1+? Kf4 3.Qg5#??
Retro: d7-d5, Rc6xh6+ (unless Kg1-g2, Rh~-h1+).
One of the best AP-type problems. (2015-08-20)
James Malcom: The first Valladao AP? (2020-10-03)
A.Buchanan: Hi James, there are 6 earlier ones in PDB, not counting those which are not marked as Valladao (could quite a few, as basic AP is 2/3 of Valladao). Type k='a p' and not g='fairies' and k='valladao', and set sort order by date. The ur-problem is P0003417. (2020-10-03)
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Keywords: Castling (wk), a posteriori (AP) (Type Petrovic), En passant as key, Promotion (ws), Valladao Task (w w w/s)
Genre: Retro, 3#
Computer test: Popeye v4.85 + minor retro/AP thought
FEN: B7/p5PP/p6R/Q1Pp4/8/PP1P2P1/p3P1k1/4K2R
Reprints: (B) Die Schwalbe 48 12/1976
feenschach 40 11-12/1977
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-11 more...
3 - P0001107
Michel Caillaud
3332 Die Schwalbe 66 12/1980
2. Preis
P0001107
(7+9) C+
h#4.5 (AP)
Circe
1. ... hxg6ep[+sBg7] 2. a1=T+ Txa1[+sTh8] 3. 0-0 Ta3 4. bxa3[+wTa1] 0-0-0 5. Td8 Txd8[+sTh8]#
play all play one stop play next play all
Wenn die beiden Könige noch nicht gezogen haben (was a.p. durch Ausführung beider Rochaden bewiesen wird!), kann der letzte Zug nur g7-g5 gewesen sein, Sonst ginge z.B. auch R: 1. Kd8-e8 c2xDb3
Henrik Juel: A possible retroplay is -1... g7 -2.f3xP(-Pg7) b5 -3.f2 c6xP(-Pb2) -4.b2 a6 -5.Rc3 a7 -6.Rc1 b6xB -7.Bf8 b7 -8.Bg7xB etc. (2003-04-22)
Michel Caillaud: The problem is not cooked as indicated, as under A Posteriori convention both castles must be played in the solution to justify the en passant key. (2022-02-04)
A.Buchanan: If wK never moved, then Black has not just moved another non-capturing pawn, as White has no prior move. The only possible exception is R: 1. g7-g5 f/h3xPg4[+Pg7]. (2022-02-04)
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Keywords: a posteriori (AP), Circe, En passant as key, Castling (wgsk), Promotion in forward play, Valladao Task
Genre: Retro, Fairies, h#
Computer test: HC+ all solutions begin with ep, but only one includes both white & black castling
FEN: 4k3/8/7p/p1p3pP/Pp4Pp/RP4p1/pP6/4K3
Reprints: feenschach 61, p. 534, 08/1982
RA152 diagrammes 72 07-08/1985
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-28 more...
4 - P0002353
Theodor Tauber
Michel Caillaud

The Problemist 1991
P0002353
(14+11) C+
ser-h#3 (AP)
1. dxe3ep 2. fxg1=L 3. 0-0 Tg4#
1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt
play all play one stop play next play all
Henrik Juel: If Black may castle, his latest move must have been e3xf2, so last move was e2-e4. (Orig. wPg2 promoted on g8, so g7xh6 happened early). 1.dxe3 e.p. 2.fxg1=B 3.0-0 Tg4# (2003-03-21)
James Malcom: Is this the only know one-sided Valladao in a regular series-mover helpmate? Also, can wPh2 be shifted to h4, and the White king moved as well to h2, so that the solution must have 2. fxg1=S? (2020-12-08)
James Malcom: Well, the only known one that is in minimum form, i.e. 3 moves, I mean. (2020-12-08)
A.Buchanan: This is nice. I don't know of any other AP ser-h#3 with Valladao, but there could be. Just been cleansing PDB data today. Yes I think wKh3 and wBh2 can be transposed, and then promote to S instead of L. (2020-12-08)
A.Buchanan: The legality of intermediate positions in seriesmovers is of course irrelevant. However, each proof game that leads to the diagram position with BTM can be appended with a black move and then a "flip" that changes the player to move. We can repeat this process, and then the fact that Black retains castling rights implies that they had castling rights in the diagram position. I think this is the right formal way to handle AP with seriesmovers. (2022-03-21)
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Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sk), Seriesmover, Promotion, Valladao Task
Genre: Retro, Fairies
Computer test: C+ Popeye v4.85 + thinking
FEN: 4k2r/1pp2p1p/1N5p/3P4/2RpP2N/P2P1P1K/p1BP1p1P/B3b1R1
Reprints: Die Schwalbe 143 10/1993
Rex Multiplex 45-46 05/1994
(XII) Quartz 5 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
5 - P0003138
Branko Koludrovic
4208 Problem 12/1979
P0003138
(11+9)
h#3 (AP)
0.1...
1. ... axb6ep 2. 0-0-0 0-0-0 3. Td7 a8=D#
play all play one stop play next play all
This problem can be solved presuming that both castlings are executable. It implies that the last halfmoves have been.-1.Bb6xa7 sBb7-b5.
Bb5 could not come from c6 (sBc6xFb5), because all of the missing white pieces(5) have been captured on white squares except wLc1,which must have been captured by Bc7(Bc7xwLb6).Further,it has not played immediately before -1.Bb6-b5 because this would prevent white castling(forcing wT or wK to move)
Here is the proof that 4 remaining white pcs have been captured on white squares:
black d-pawn must have been captured on d4 and h,g f pawns had to promote after Bh3xwFg2,then wBh2moved to h4, enabling Bg4x Fh3 ..h1T, and finnaly sBf7
after fxg2... g1S. sLf1 prevented the check from Th1, and later has been captured by a white officer.
Conclusion:
Retro:-1.wBb6xFa7 Bb7-b5
Forward(AP) 1....axb6 e. p.2.0-0-0 0-0-0 3.Td7 a8Q# (Author)
Branko Koludrovic: P.S.
The black pawn a4(on the diagramm)came from c7 after capturing the white bishop on b6, then moving to b5 and capturing a white officer on a4. (2010-09-28)
A.Buchanan: I think sBf must have captured on g2 & then promoted on f1 to S, otherwise it’s impossible to unlock the southeast cage (2023-07-11)
A.Buchanan: So that means it’s sound. However I don’t see that it needs to have been sL shielding on f1 (2023-07-11)
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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...
6 - 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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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...
7 - 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)
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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...
8 - 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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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...
9 - 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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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...
10 - P0009121
Tomislav Petrovic
2949 Phénix 69 12/1998
P0009121
(10+9) C+
h#2 (AP)
1. fxe3ep d8=S 2. Kxd5 0-0-0# (Td1#?)
play all play one stop play next play all
White has made 6 pawn captures with one missing black piece unaccounted for, that by parity can only have been captured by an officer. White's last move cannot have been a pawn capture. If R: 1.f2-f3, then sLg1 was promoted, implying 7 captures by black pawns - one too many. So if White can prove they retain castling rights, then the ep is on. Hence AP Petrovic is valid.
Henrik Juel: 0... fxe3ep 1.d8S Kxd5 2.0-0-0#. Not -1.f2? and Lg1 is caught. (2004-09-16)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
Henrik Juel: In principle I agree, Vaclav
But PDB does not (yet) allow HC+, so I find it acceptable to use the C+ label, when you also tell the whole story after 'Computer test:' below (2023-08-03)
A.Buchanan: Hopefully Gerd will have more time at some point, and can expand the functionality in this and other areas. In the meantime, engine solving of conventional retros including AP, is in its infancy. Retractor 2 has some effectiveness, but is still basic. And there is nothing that yet grasps the intricacies of castling/ep etc. However AP problems do often contain considerable forward chess, and the C+ tag is very useful to filter out those that have already been solved forwardly, without pretending that these are in any sense fully solved (2023-08-04)
Ladislav Packa: I know the definition of AP, but I don't understand the logic behind it. The move 1.fxe3 e.p. proves that White CAN castling. But the solution (2. ...Rd1#?) claims that castling is MANDATORY. From my point of view, AP is correct when only castling is necessary for the solution and the Rook move would be a dual. (2023-08-04)
A.Buchanan: @Ladislav: I am not sure how to help you. Maybe you can read this page from Retro Corner: https://www.janko.at/Retros/Glossary/APosteriori.htm (2023-08-05)
Ladislav Packa: Andrew, what should the article help me with? I quote the final sentence:
Some people still oppose this rule and argue that it should certainly not be the default convention. (2023-08-05)
Henrik Juel: You could view it this way, Ladislav
h#2 means that it is Black to move, so White made last move
What was last move? A little analysis shows just three possibilities: f2-f4, move by Ta1, or move by Ke1
So normally we cannot assume that last move was f2-f4
But if White can castle, then the last move was f2-f4
So if we could start with 0... 0-0-0, then 1.fxe3ep would be legitimate
AP says that you are allowed to reverse the sequence of events; first do the ep capture, then later legitimize it by castling
Was this helpful? (2023-08-05)
Ladislav Packa: Henrik, you don't have to explain that to me. I've done a few AP issues myself, like P1348357. But that doesn't mean I agree with AP's logic. I already wrote it - the term "you can castle" is applied as "you must castle". But these are only problems where, in addition to castling, the Rook move can also be used, I consider that a dual.
From that point of view, the P1000662 issue is perfectly fine for me. (2023-08-05)
Joost de Heer: AP: By castling, you prove a posteriori that the ep-capture was not just a try but the actual solution. Without castling, the ep-solution just is that: a try.
So: Try 1. fe3 ep e8=S 2. Kd5 Rd1 - but ep capture not allowed, as there is no proof that f2-f4 must've been the last move.
Solution 1. fe3 ep e8=S 2. Kd5 OOO - Now the ep capture was justified because white castled, thereby proving that the last move before the diagram position indeed was f2-f4. (2023-08-06)
Joost de Heer: See e.g. P1052919 : The try is an ep capture which is unjustified. (2023-08-06)
A.Buchanan: OK Ladislav: I think I get your point. If one solution with castling justifies the e.p., then based on that certainty, why shouldn't an alternative solution with no castling *then* be allowed as well? There are problems in which one twin shows 0-0-0 and the other shows 0-0. Each is based on the other in a similar way, so the idea of dependency is not new. Why are we not allowed to add other "parasitic" solutions as well? Why can we only have the "paying" solutions? We can't say that we are restricted to one solution: that's not the way chess problems operate! And this is just in the help world - in the adversarial world it might get even more complicated. Is this your issue, Ladislav? (2023-08-07)
Ladislav Packa: I don't want to unnecessarily prolong this discussion. However, I will add one more note: in this position, white castling is also possible without e.p. in Black's 1st move. If B1 were an indifferent move, then white can 1...0-0-0! The Codex of Chess Composition writes about it in Article 16 (1):
Castling convention. Casting is permitted unless it can be proven that it is not permissible.
In our case, 0-0-0 is possible because White's last move exists - e2-e4! It does not matter if it is this move or some a2-a3, both moves are equivalent. EP does not prove the possibility of casting, it would be legal even without it. (2023-08-07)
A.Buchanan: Ladislav was what I wrote your issue pls? Y/N :-) (2023-08-07)
Ladislav Packa: I have no problem, I'm just expressing my own opinion about the AP convention. (2023-08-07)
A.Buchanan: OK cos I think the point I raised is a real one that should be addressed by theory some day. Clearly from the nice problem that you composed Ladislav you understand the mechanics very well. From a justification perspective it's all a bit iffy, but that's why it's controversial. Under RS it's really the only way one can end up actually eping, and it's proved compositionally fertile. So that's enough to justify (2023-08-08)
Ladislav Packa: No need to apologize. This is a normal discussion with different views on the issue. Maybe it will come to some conclusion.
I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p. (2023-08-08)
Joost de Heer: "I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p."
You misinterpret AP. The e.p. capture does not authorize castling, castling provides a justification later on (hence the 'a posteriori') for the legality of ep.
Usually, for ep justification you need to examine all game trees that lead to the diagram, and only if all game trees end with the double-step, then ep is allowed.
With AP, you examine all the game trees including the actual play. If all those game trees have as last move before the diagram position the double step, then ep is possible.
In this case, if white doesn't castle, then there are game trees which don't have as last move the double step, and therefore AP logic dictates that the ep capture was illegal. However, all game trees which lead to the diagram and which have castling in the actual play have as last move before the diagram position the double step, hence AP dictates that the ep capture is legal. (2023-08-09)
Ladislav Packa: Joost: A simple question - is white allowed to castle after any 1st move by black (except e.p.)? (2023-08-09)
Joost de Heer: Of course he is. AP only is used to combine the ep justification with castling, not the castling right per se. (2023-08-09)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Promotion (S), Valladao Task
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 + simple retro-logic
FEN: 8/3P2p1/2PP4/1ppPp3/2pkPp2/5PP1/6Pp/R3K1b1
Input: Gerd Wilts, 1999-02-27
Last update: A.Buchanan, 2023-09-11 more...
11 - P1000662
Gianni Donati
R074 Probleemblad 11/1999
P1000662
(10+14) C+
h#1.5 (AP)
1. ... gxh6ep 2. exf1=L 0-0-0#
play all play one stop play next play all
Kommentare:
Als einziger letzter schwarzer Zug, der die weiße Rochade
erhält, kommt nur Bh7-h5 infrage, was dem Weißen den e.p.-Schlag
ermöglicht, um nicht die Rochade oder das Mattnetz zerstören zu
müssen (H.P.Suwe)
Gianni Donati: This intends to show the Valladao theme in the minimum number of moves. (2004-03-19)
VL: The waiting ep capture. Cf. also P1000260
(by T.Petrovic, 2000) somewhat enriched
thematically with the illegal try 2... Rd1#??. (2004-06-03)
A.Buchanan: The “illegal try” 2. Rd1+ is not actually mate because c2 is unprotected. This is kind of “logic dual” spoils the A.P. motivation for the castling (2020-05-20)
Henrik Juel: White captured h6xg7 and once more, e.g. axPb-b8=Y
Black captured cxdxe, dxexf, fxg, and once more, e.g. a2xb1=Y
Possible retroplay 1... h7-h5 2.h6xSg7 Df8-f4 3.h5-h6 Sf4-h3 4.Sh3-g1 etc., preserving the castling right
Any other black last move would force White to retract Ta1, as d2xc3, d2xe3, and h6xh7 would be illegal retractions
I agree with Andrew that the double motivation of 2... 000# is a weakness:
a. legitimizing h7-h5
b. mating (2020-05-21)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
A.Buchanan: See discussion at P0009121 (2023-08-04)
more ...
comment
Keywords: a posteriori (AP), En passant, En passant as key, Castling, Valladao Task, Promotion (l)
Genre: Retro
Computer test: HC+ Popeye v4.87 + simple retro logic
FEN: 6n1/4p1P1/6p1/6Pp/2b1rqrb/2PkPppn/1P2pP2/R3KRN1
Reprints: König & Turm (4), p. 28, 03/2001
H Problemkiste (143) 10/2002
(V2) Problemkiste 156 12/2004
Input: Gerd Wilts, 2000-08-01
Last update: A.Buchanan, 2023-08-03 more...
12 - P1066741
Tomislav Petrovic
R033 Probleemblad 11-12/1998
Ing. R. Tomasevic gewidmet
P1066741
(14+10) C+
h#2 (AP)
1. gxf3ep 0-0-0 2. gxh1=T gxh4#
play all play one stop play next play all
White PCs: axb=, bxc=, dxexf, cxdxe
Black PCs: fxg2. wRh was captured by officer in cage, and since wK didn’t move wRg5 is promoted
A.Buchanan: Suppose White: dxexf,cxdxe,bxc=,a| Black: fxg. This means there is one capture by white unaccounted for. So can’t White have just played e.g. QxS? What am I missing? (2022-03-22)
Mario Richter: Regarding the "can’t White have just played e.g. QxS?-question", the following considerations may be helpful:
1. black Pawn f3 x Yg2 can only be retracted after wPf2 has returned home.
2. If at this moment the black K is still on g1, Y cannot be a white rook.
3. Assuming that W can still castle, wRg5 must be a promoted Piece, since wKe1+wPf2+wPg3+wPh2 form a cage from which the originalwRh1 could not escape ...

I think the same considerations can also be useful by answering the question about +sBc6 ... (2022-03-22)
A.Buchanan: Hi Mario, thanks I agree (2022-03-22)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Valladao Task, Promotion (t)
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 & retro-logic
FEN: 1B6/p2p4/1p4Q1/5PRN/4PPpb/3Np1Pp/4P1pP/R3KbkB
Input: Gerd Wilts, 2004-09-16
Last update: A.Buchanan, 2023-09-11 more...
13 - P1079607
Igor Vereshchagin
29 Zadachi i etyudy 14, p. 52-53, 1997
3. Lob
Thematurnier "Uschol-Prischol-Vernulsa"
P1079607
(6+15)
h#3 (AP)
1. cxb3ep 0-0-0 (T~?,L~?) 2. Dxd2+ Kxd2 3. La1 Txa1#
play all play one stop play next play all
im Kongreßbuch St. Petersburg 1998 nur Nachdruck im Kapitel "Aus dem Schaffen der Kongreßteilnehmer"

Das Originaldiagramm in 'Zadachi i etyudy' ebenso wie der Nachdruck im Kongreßbuch St. Petersburg 1998 haben einen sTd8, aber der begleitende Lösungstext erwähnt im Zusammenhang mit der Frage, ob zuletzt axb3,cxb3 oder cxb6 möglich war, daß auf dem Brett folgende schwarze Steine stehen: eine Dame, zwei Türme, zwei Springer, der schwarzfeldrige Läufer und 8 Bauern.
Also offensichtlich Diagrammfehler, und sSd8 ist korrekt.
Ladislav Packa: Why NL? Castling is a posteriori proof for e.p.! (2018-08-28)
A.Buchanan: Why not a3xSb4 or c3xSb4 as last move? Where is the promotion? (2018-08-28)
VL: Yes, it definitely looks as AP (with thematic illegal castling avoiding tries). I suspect missing two b knights somewhere: "superfluous" similarly to the b rooks...
wBa5 proves to be a promotee. Bl to move of course. (2018-08-29)
A.Buchanan: I agree with VL. WinChloe adds sSh1 & sTd8 (the latter is illegal with 8sB+2sT on the board already). If we add sSh1d8, then the problem is sound forwardly and retroly. The missing light sL can't just have been captured. (2022-05-24)
Mario Richter: Adding black Sh1+Sd8 is correct: obviously the original diagram is misprinted, the accompanying solution text speaks of: one black Queen, two black Rooks, TWO black Knights, one black dark-squared Bishop and 8 black Pawns. (2022-05-24)
A.Buchanan: Issue with animation (2022-05-25)
more ...
comment
Keywords: Valladao Task, En passant as key, Castling (wg), Promotion (L), a posteriori (AP) (Type Petrovic)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 3n3b/1p5p/1P1p2p1/Bp6/kPp1p2r/3p3r/q2P4/R3K2n
Reprints: WCCC St. Petersburg 1998
Input: hpr, 2008-10-26
Last update: A.Buchanan, 2023-07-30 more...
14 - 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...
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#
play all play one stop play next play all
Durch den ep-Schlag beweist Weiß, dass der letzte Zug von Schwarz
e7-e5 gewesen ist. Die sGG können mangels Bock nicht gezogen haben
und hätten sK oder sT gezogen, so wäre die {\ooo} nicht mehr
möglich (Autor).
A.Buchanan: sGh8 seems useless as it merely eliminates the thematic retro tries. What am I missing? I wonder if the composer really understood the nature of AP...? (2022-03-22)
more ...
comment
Keywords: Valladao Task, En passant as key, Castling, a posteriori (AP) (Type Petrovic), Promotion in the mating move (D)
Pieces: du = Grasshopper (G)
Genre: Fairies, h#, Retro
Computer test: %Popeye FreeBSD-4.3-RELEASE-32Bit-Version 3.81
FEN: r3k2*2q/P4*2q2/*2q2P4/3PpK2/8/8/8/8
Input: Erich Bartel, 2014-09-22
Last update: A.Buchanan, 2022-03-22 more...
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#
play all play one stop play next play all
Der sSh8 kann wegen Schach zuletzt nicht gezogen haben. Er verhindert zudem NLen.
Mangels Schlagobjekt kann der sB nur von d7 gekommen sein (Autor).

t/nur Themazüge,
A.Buchanan: 1. ... Ke6 Tb8 2. axb8=D/T[-]#? won't work because under "strict" Circe, a capture can't take place at all if the rebirth square is occupied. (2022-05-23)
A.Buchanan: Pieces never leave the board in strict Circe. "Hotel California". The diagram position is hence illegal and it's interesting that one can nevertheless get some kind of retro logic going. It points out a generally unspoken difference between "local" & "global" illegality. (1) Can we reverse moves indefinitely, or do we get stuck in some kind of reverse pat? (Can't say "retropat", because that implies that it's only pat with one player to move, as Henrik explained to me a while back.) (2) If the answer is yes then can we get back to the game array? These are different questions, and problems like this one rely on local rather than global legality. (2022-05-24)
more ...
comment
Keywords: Valladao Task, En passant, Castling, Circe (Strict), Promotion (D), a posteriori (AP) (Type Petrovic)
Genre: Fairies, h#, Retro
Computer test: %Popeye FreeBSD-5.3-RELEASE-32Bit-Version 3.81
FEN: r3k2n/P7/8/2PpK3/8/8/8/8
Input: Erich Bartel, 2014-09-26
Last update: A.Buchanan, 2022-05-26 more...
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=
play all play one stop play next play all
A.Buchanan: Guessing so-called "AP-Priorität" is a variant of AP whereby Black is forced to legitimize (or at least forbidden to rule out the possibility of legitimizing) the earlier e.p. (2020-10-03)
comment
Keywords: En passant, a posteriori (AP) (Type Petrovic), Castling, En passant as key, Promotion, Valladao Task
Genre: Retro, Fairies
Computer test: Popeye v4.85 + minor retro/AP thinking
FEN: r3k3/8/1P1P4/1KpP4/8/8/8/8
Input: Erich Bartel, 2014-09-29
Last update: A.Buchanan, 2022-03-21 more...
18 - P1359077
Joaquim Crusats
13 Internet 12/2018
2. ehrende Erwähnung
JT-Sergej-Volobujev-60
Abteilung Retro
P1359077
(11+14)
h#2.5 (AP)
1. ... fxg6ep 2. 0-0-0 gxf7 3. Te8 fxe8=D#
(+ hordes of tries that omit castling)

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


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

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

Now, time to create a more streamlined PG... (2022-07-04)
James Malcom: 42.0 moves is optimal to reach the diagram; the proof speaks for itself, I believe: 1. a4 Nc6 2. a5 Nxa5 3. Ra3 h5 4. f4 Rh6 5. Nf3 Rc6 6. Ne5 Nf6 7. Rf3 Nb3 8. Nc4 Na1 9. Kf2 e5 10. Rg1 d5 11. b3 Bh3 12. g4 a5 13. Nb2 a4 14. f5 a3 15. Kg3 axb2 16. d3 Rc3 17. Bg2 Ba3 18. b4 Rb3 19. Bh1 c5 20. Bd2 c4 21. Be1 c3 22. Nd2 cxd2 23. c3 e4 24. Qc2 e3 25. Rf4 Ne4+ 26. dxe4 d4 27. Rgf1 d3 28. R1f2 dxc2 29. R2f3 Qd3 30. exd3 e2 31. Re3 h4+ 32. Kf3 Bf1 33. Bg3 hxg3 34. h4 Bh3 35. h5 Bf1 36. h6 Bh3 37. h7 Bf1 38. h8=N Bh3 39. Ng6 Bf1 40. Ne7 Bh3 41. Nd5 Bf1 42. Bg2 g5 (2022-07-04)
comment
Keywords: Valladao Task, Promotion in the mating move (D), a posteriori (AP) (Type Petrovic), En passant as key, Castling (sg)
Genre: h#, Retro
Computer test: Forward play Popeye v4.87 but retro play requires substantial thinking in order to establish retro-opposition. Believed to be sound.
FEN: r3k3/1p3p2/8/3N1Pp1/1P2PRP1/brPPRKp1/1pppp1B1/n4b2
Reprints: Problemas 07/2022
Input: Mario Richter, 2019-01-03
Last update: James Malcom, 2022-07-04 more...
19 - P1401480
Jean-Michel Trillon
v Rex Multiplex 2 04/1982
to C.Caminade
2nd Honourable Mention
P1401480
(6+8) cooked
#4 AP
Maximummer
b) g6->g5
a) 1. Ld6! Td8 2. fxg6 Txd5 3. g7 Th5 4. g8=D#
b) 1. fxg6ep Txd5 2. Le5 0-0-0 3. g7 Th8 4. gxh8=D/T#
play all play one stop play next play all
a) Black had a candidate last move b7-b5 of length 2. The only way this might not have been played is if bK was in check from e8 namely Kd8xDTe8,Kf7xDe8. So certainly Black castling rights are lost. With this known, there is still a unique solution.
b) Again, b7-b5 looms, and this time if Black castling rights are lost, there is no solution. The only possible way we can operate is if the last move was g7-g5. But we are only permitted to play this if we can show that the last move *was* not one of Kd8xDTe8,Kf7xDe8. I.e. under AP we need to castle after the fact. There is a unique solution after the ep, which does indeed include castling.
Cook: b) R: 1. Th4xDh5,Th4xLh5,Th6xDh5,Th6xLh5
Henrik Juel: a) C+ Popeye 4.61
1.fxg6 Txd5 2.Le5 0-0-0 3.g7 Th8 4.gxh8=DT#
b) no solution (2022-05-27)
Henrik Juel: I shall leave the analysis to you, Andrew... (2022-05-27)
A.Buchanan: OK, Henrik: please check my suggested solutions & rationale :) I find this a beautiful problem (2022-05-28)
A.Buchanan: In (b), is it possible that Black has just played RxQBh5? If this is feasible, it would apply even if Black 00 rights are still on. Cook? (2022-05-28)
Mario Richter: Looks indeed like a cook.

One can manually retract e.g. R: Th4xLh5 and use popeye to check if in the resulting position the reverse of this retraction is legal:
begin
remark Position after R: Th4xLh5
pieces white Bb8 Pd5 Pf5 Bh5 Ph3 Pd2 Ka1
black Ra8 Ke8 Pb7 Pd7 Ph7 Pa6 Pg5 Rh4
cond Maximummer
option HalfDuplex
stip h~0.5
end

This gives:
1...Rh4*h5
1...Ke8-e7
1...Ke8-d8
1...Ke8-f8

solution finished. Time = 0.013 s (2022-05-29)
A.Buchanan: Thanks Mario. I showed this problem on the international zoom call last night, and Siegfried spotted the cook immediately. Can g4 and h4/h6 be plugged suitably with pawns? Can't just shift bPh6 to h5, because it means that in a), bPg6 might have just captured checking Q/B, while not disrupting Black 000 rights. But I think can shift wPh3 to h4, together with adding Pg4 (Black or White - I prefer White aesthetically). Does this work? (2022-05-29)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), Maximummer, Castling (sg), En passant as key, Promotion (D), Valladao Task, Superseded by (P1401566)
Genre: n#, Retro, Fairies
Computer test: Popeye v4.87 & not-so-simple retro-logic to identify the cook
FEN: rB2k3/1p1p3p/p5p1/3P1P1r/8/7P/3P4/K7
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2022-06-02 more...
20 - P1401495
Igor Vereshchagin
Odessa 1997
Special commendation
P1401495
(10+14) C+
h#2 (AP)
1. a4xb3ep+ Sxa5+ 2. Kd5 0-0-0# (by AP not Td1#?)
R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7,Sd6xLc7
play all play one stop play next play all
With the typo wPd5 corrected to e5, the two forward candidate solutions are correct. How about the retro? Assume that White castling rights remain. bBg5 is promoted, and did so on c1 or g1. In either case, there were 6 pawn caps by Black, but in the former case, the blockaded wRh was not consumable. So the promotion must have been on g8, and White can't retract h2-h3. The missing White unit is light B, so was not just captured on e5 or b4. wRf6 is promoted, say wPd, which would have involved no captures.

This leaves b2-b4, preceded by Kb4-c4+ and before that Sd6-b7. Prior to b2-b4, why couldn't bQ have just played e.g. Qa8-a5+ instead? Because there would still be no prior move: wSb7 can't have just come from h5 or e6. Similarly Rc3-a3 has no precedent.

We don't know exactly what happened to sLc, but that doesn’t affect the soundness. It might have enabled wPc or wPd to capture. wPd did promote, but maybe it was wPc that after capturing promoted to T. Alternatively, Sd6xLc7 was possible.
Henrik Juel: Popeye 4.61 with 'opt enp b3' found no solution (2022-05-27)
Gerald Ettl: Schaut so aus, als ob die sDa5 nach e5 muss. 1.axb3 ep Sa5+ 2.Kxd5 0-0-0# (2022-05-27)
Gerald Ettl: aber auch 2.-Td1# ? (2022-05-27)
Henrik Juel: Andrew, can you throw any light on this problem? (2022-05-28)
Henrik Juel: Gerald, in AP problems like this one White must castle to legitimize the ep capture key (2022-05-28)
A.Buchanan: Hi Henrik, Gerald. I found this one in WinChloe with the current diagram. It gave the "solution" 1. axb3ep+ Sxa5+ 2. Kd5?? 0-0-0#. However when I tried running the WinChloe solver just now, it said there's no solution. Quite right: Rf5 controls d5. I don't think we can shift Qa5 to e5 though.
My first thought is that wPd5 should be on e5. Check my suggested solution in the solution text. However, as you will see, I think we need a second thought as well! Hope the finest retro minds can figure this out. Failing that, we could ask Igor. (2022-05-28)
Mario Richter: The diagram in the reprint in 'Uralsky Problemist' is exactly as given here. The solution given there without further explanation is: 1. ab+ (e.p.) Sxa5+ 2. Kd5 0-0-0#!
Notice that the second black move doesn't contain a capture sign!

Some typos in Andrew's analysis:
The black potential promotion squares should be c1 and g1.
White's last move couldn't have been b3-b4?? simply because that would give an illegal check. (2022-05-28)
A.Buchanan: Thanks Mario for the careful reading. I’ve fixed the errors. Pd5 may be a typo in the reprint then. It also creates a bogus retraction d4-d5, so I don’t think it can be right. Now for Igor. (2022-05-29)
A.Buchanan: Joaquim Crusats spotted R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7, which I'm sure is the intended retraction. So we diagnose a single typo that occurred in the Uralsky reprint diagram: wPe5 appeared on d5. I suggest correcting it here. (2022-05-30)
A.Buchanan: Author confirmed this was a typo (2022-11-23)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Promotion (Tl), Valladao Task
Genre: h#, Retro
Computer test: Once the diagram typo was fixed, C+ Popeye v4.87 & basic but tricky retro logic
FEN: 3n4/1Np1p1p1/4pn2/q1p1PRb1/pPk1pr2/r6P/P4PP1/R3K3
Reprints: 18 The Ural's Problemist 13, p. 5, 02/1998
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2022-05-30 more...
21 - P1401497
Mario Velucchi
Scacco! 1998
P1401497
(5+5) C+
h#2.5 (AP)
1. ... cxb6ep
Not 2. Ta7? bxa7 3. Kd8 axb8=D#
But 2. Sxa6! Ld6 3. 0-0-0 b7# AP
play all play one stop play next play all
Under AP Type Petrovic, the ep is enabled by the later castling
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), Valladao Task, Castling (sg), En passant as key, Promotion (D)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: rn2kB2/3p4/P1P5/KpP5/8/8/8/8
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2022-05-28 more...
22 - P1401566
Jean-Michel Trillon
Siegfried Hornecker
Andrew Buchanan

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