Die Schwalbe

25 problem(s) found in 6900 milliseconds (displaying 25 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT S='Klassik pur! [Böhringer]' AND A='Liskowez, Waleri A.'] [download as LaTeX]

1 - 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)
more ...
comment
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...
2 - P0000250
Nikita M. Plaksin
Valery Liskovets

7577v Die Schwalbe 132 12/1991
P0000250
(5+14)
ser-h#5 (AP)
Typ Tauber/Caillaud
1. Kd4 2. bxc3ep (zuletzt nur K,T-Zug oder c2-c4) 3. e5! (nicht 4. Tb4? weil zuletzt K,T-Zug und 0-0-0 nicht mehr möglich) 4. Tb4 (zuletzt e7xLd8=S möglich) 5. Tc4 & 1. 0-0-0# nicht 1. Td1#? wegen fehlender AP-Legalisierung
play all play one stop play next play all
"das zeigt die Paradoxie der AP-Bedingung!" (MS). Gleich 1. bxc3 ep.? und dann 2. Kd4 geht natürlich nicht, weil zuvor außer K,T-Zug und c2-c4 auch S-Züge und c3-c4 möglich waren. "Interessanter und genau abgestimmter Lösungsverlauf" (GW) "Sehr schöne und ökonomische Verbindung zwischen konsequentem sh# und AP!" (TB) Mit 'Typ Tauber/Caillaud' ist gemeint, daß die einzelnen Stellungen doch als retroanlytischer Gesamtkomplex betrachtet werden dürfen/sollen (sonst wäre keine AP-Legalisierung möglich), obwohl sie ja aufgrund der retroanalytischen Neubewertung nach jedem Rückzug voneinander unabhängig sind. 6L./5+5P.
A.Buchanan: "By 'type Tauber/Caillaud' it is meant that the individual positions may/should be regarded as a retroanalytic overall complex (otherwise no AP legalization would be possible), although they are independent of each other due to the retroanalytic reassessment after each retraction." (2023-06-29)
comment
Keywords: Castling (wl), a posteriori (AP), Promotion (S), Seriesmover, Consequent, En passant
Genre: Retro, Fairies
FEN: 1N6/1ppppppn/n7/1Pq1k3/1pP1r3/4p3/1r6/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-29 more...
3 - P0004920
Valery Liskovets
(F) Die Schwalbe 80 04/1983
P0004920
(5+4)
#2 (AP, pRA)
BTM 1. ... Txh6? 2. Sd7! Txc6+ 3. Kb5! (Kb7?) Td6 4. Sf8 Tf6 5. Sh7! Th6 6. Sf8 no castling
5. a6? Tfxf8 6. a7 Tf5+ 7. Kc6,K~ 0-0!
1. ... 0-0? 2. Se7+! Kh8 3. S5g6#
WTM 1. Td6 droht 2. Td8#
play all play one stop play next play all
White to move has #2 since Black has lost castling rights. So Black pulls the move, but must castle at some point. If Black castles right away, then White has a different #2, so Black must be more subtle. 0... g6/g5/gxh6 leads to castling disruption, e.g. 1.Txg6/Te6+/Sg6. So Black only has 0... Txh6. This pins wSc6 and threatens 0-0, so 1.Sd7! (1. Sg6? Txg6 2. ~ 0-0) etc.
A.Buchanan: A key feature of adversarial A Posteriori is that any castling must be forced in a finite number of moves (but not necessarily limited by the number of moves in the stipulation goal). If the other side can prevaricate indefinitely, then that is sufficient to defeat the A Posteriori "steal" (2022-02-16)
A.Buchanan: Why this would be "PRA"? Maybe the idea is that we don't know who is first to move, yet whoever it is, White wins. But that only applies to "pull" scenarios such as this, where Black snatches the move because otherwise the game is lost. In other situations where White to avoid loss must "push" the move, then there is no way this can be described as PRA. The fundamental push/pull thing has a unity, and I don't think it's helpful to use "PRA" which only describes half of this, and was really designed for a different context. Strategically, these push/pull adversarial battles are amongst the most interesting AP problems. (2023-07-22)
comment
Keywords: a posteriori (AP) (Type Keym), Cant Castler, Castling
Genre: Retro, 2#
FEN: 4k2r/6pr/K1N4R/P3N3/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-22 more...
4 - P0008499
Valery Liskovets
Valery Voynov

Shakhmaty v SSSR 02/1974
after Arthur Ford Mackenzie
Speziallob
P0008499
(4+1)
#3
Last 4 single moves?
1. La7 Kxa7 2. c8=T! Kxa6 3. Ta8#
R: 1. Ka7-a8 b7-b8=L+ 2. Ka8-a7 b6-b7+
play all play one stop play next play all
The theme: under-promotions both in the solution and the retro-play.
See P0003960
Henrik Juel: The forward part is C+ Popeye 4.61 (2022-06-01)
comment
Keywords: Last Moves? (4), Type B, Move Length Record, Rex solus, Miniature
Genre: Retro, 3#
FEN: kB6/2P5/N1K5/8/8/8/8/8
Reprints: B5b Length Records in Last Single Moves? , p. 7, 11/2009
B05b Phénix 282, p. 11063, 02/2018
Input: Gerd Wilts, 1996-12-14
Last update: Rainer Staudte, 2022-06-01 more...
5 - P0008517
Valery Liskovets
12 Problemist Pribuzhya 1 1990
P0008517
(4+3)
#2 AP (pRA)
BTM 1. ... axb6+ 2. Ta5! 0-0-0 3. Ta8#
2. Ta7? 0-0-0! & no mate in 1
WTM 1. Lc5! (0-0-0??) droht 2. Tf8#
play all play one stop play next play all
VL: Published in "Probl. Pribuzh." (or "Problemy Pribusch'ja"?),
1990, No.1, #12. An obscure Russian language chess problem
magazine issued in Nikolaev, the Ukraine (the South Bug
riverside region). (2006-01-27)
A.Buchanan: Lovely problem, but this is no more PRA than it is duplex. Rather, the push/pull of AP Type Keym gives a forfeit which must be discharged. (2023-07-22)
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Keywords: a posteriori (AP) (Type Keym), Castling, Homebase (s), Miniature
Genre: Retro, 2#
FEN: r3k3/p6R/1B6/5R2/8/8/8/K7
Input: Gerd Wilts, 1996-12-14
Last update: A.Buchanan, 2022-02-15 more...
6 - P0008780
Valery Liskovets
7162 feenschach 123 01-06/1997
P0008780
(3+4)
h#2 AP(PF)
Pièces rétro-volages
VL: Solution:
1.Rg2=wR Rb6=wR!! 2.Re8=bR! 0-0#! (2... Rf1#??)
2.Ra(c)8=wR??

Here only one w rook is volage since they both stand on light squares
of ODD ranks: any promotee had to visit a square of the opposite
colour at least once in order to arrive from the 8th rank! Similarly,
only one bl rook is volage (EVEN ranks). By the AP logic, we have a
right to justify the 1st W's move by subsequent castling, which shows
that it is Rh1 that is volage. Bl rooks play by the post factum
(subordination) logic: the first one turns out volage.

Cooked (feenschach, 126): 1.Ra2(Rc2,Rb3...)=wR Rh7+ 2.Kf8 Ra8#
Correction: Rh8- e8. (2002-04-05)
VL: Correction (H.130): Rh8 to d8 (rather than to the light square e8). (2021-02-12)
A.Buchanan: shifted sT to d8 as per request (2021-02-12)
comment
Keywords: a posteriori (AP), Retro-volages, Post Factum (PF), Miniature, Castling, Castling as mating move
Genre: Retro, Fairies, h#
FEN: 3r4/4pk2/8/8/8/8/1r6/1R2K2R
Input: Gerd Wilts, 1997-06-16
Last update: James Malcom, 2021-02-12 more...
7 - P1017589
Valery Liskovets
120 Redkiye zhanry-plus 73 1997
P1017589
(7+5)
h#3 (pRA)
I) 1. Sg4 f4+ 2. Kxf4 0-0+ 3. Kg3 Tf3#
II) 1. Sc4 d4+ 2. Kxd4 0-0-0+ 3. Kc3 Td3#
play all play one stop play next play all
Symmetrical (both the position and partial solutions). The last move?
Henrik Juel: C+ Popeye 4.61 (2021-01-23)
comment
Keywords: Partial Retro Analysis (PRA), Symmetrical position, Castling
Genre: Retro
FEN: 8/8/8/4k3/1p5p/4n2p/P2PPP2/R3K2R
Input: Valery Liskovets, 2004-06-25
Last update: James Malcom, 2021-01-23 more...
8 - P1066685
Valery Liskovets
12365 Die Schwalbe 208 08/2004
P1066685
(14+5)
#1 (AP)
1. ... gxh3ep 2. 0-0-0#! (not 2.Sf3#?,Bh2#?)
to force
R: 1. h2-h4! h3xSg2 (not 1. Kd1-e1? Kf1-g1 2. Lf3-e2# )
play all play one stop play next play all
Diagram position is retropat so Article 15 give Black the move. Then White might have last move R: 1. h2-h4 or 1. Kd1-e1. To avoid stalemate in the diagram position, White must castle to demonstrate under AP the legality of 0. gxh3ep. So 1. 0-0-0#! not 1. Sf3#?,Bh2#?
VL: Solution. 1.0-0-0#/Bh2#?? - B. is on move.

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

W.Ps took 11: b*c*d*e*f*g, c*d*e*f, e*f*g and g*h.
11+5=16, hence b.Pa7 was also captured among them.
Thus it took once: 1+1(h*g)+14=16. Ph3 took on g2
when w.P stood on h2. No last W's move could be a
capture: e5 and f4 are occupied by w.pieces. Qf4
prevents from Ph3-h4 and Kh2-g1 before that. b.Ps
are blocked from above. Therefore, in his last move,
W. could retro-release B. only in two ways: Ph2-h4
with Ph3*Sg2 before that, or Kd1-e1 with Kf1-g1
and Bf3-e2+ before that. Thus, W. may castle, and
castling AP-legalizes ep (2004-12-09)
A.Buchanan: Article 15 (not AP Type Keym) drives this elegant problem, with usual AP Type Petrovic for ep/castling. (2022-02-16)
A.Buchanan: Have added genre as n#, even though n=1 here, because it's important that this is a #1 rather than a h#0.5, as they behave differently under Article 15. (2022-02-17)
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comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, No legal last move for Black, Castling (wk)
Genre: Retro, n#
FEN: 8/6P1/5P2/4NpPP/5QpP/P5B1/3PBPp1/R3K1kb
Reprints: Die Schwalbe 228 12/2007
Input: Gerd Wilts, 2004-08-13
Last update: James Malcom, 2022-07-05 more...
9 - P1068187
Valery Liskovets
9679 Thema Danicum 118 04/2005
P1068187
(15+5) C+
h#1* (AP)
2 Lösungen
1. ... Ke2#! (0-0#?? illegal)
1. cxd3 Sb3#! (Ke2??, 0-0+?)
1. cxb3ep 0-0#! (Ke2#?, Sxb3#?) AP
play all play one stop play next play all
VL: Solution: 1... Ke2# (1... 0-0#?? illegal).
1.cxd3 Sb3# (1... Ke2??, 0-0+?).
1.cxb3 ep! 0-0#! (1... Ke2#??, Sxb3#?? AP after N.Petrovic).
If castling is legal, then B. is on move and the last move
was b2-b4 (with b3x(B)c2 before that). Different mates. (2006-01-27)
A.Buchanan: If Wh 00 is ok, then bPh & hence bPa were waylaid. R: 1. Sa3-b1? Sb1-d2+ retropat, so R: 1. b2-b4 b3xBc2. (2022-05-24)
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comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wk), Volet Pawn
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 + thinking
FEN: 8/2Pp4/3N4/1PP5/1PpB4/2PR4/P1pN1PP1/Qnk1K2R
Input: Gerd Wilts, 2005-12-21
Last update: A.Buchanan, 2022-06-06 more...
10 - P1085404
Valery Liskovets
13336v Die Schwalbe 224 04/2007
Werner Keym zum 65. Geburtstag gewidmet
P1085404
(15+7) C+
#2
b) ohne wLb2, AP
c) ferner mit sBe7 nach d6, AP
a) 1. ... gxf3ep 2. Lxf3 Lf5 3. Txg2#! (3.0-0-0#??)
1. Lf3?? gxf3 2. 0-0-0#?

b) 1. ... gxf3ep 2. Lxf3 Lf5 3. 0-0-0#! (3.Txg2#??)
1. Lf3?? gxf3 2. 0-0-0#??

c) 1. ... gxf3ep 2. Lxf3 Lf5 3. 0-0-0#!! (3.Txg2#??)
1. Lf3? gxf3 2. 0-0-0#??
play all play one stop play next play all
VL: Keywords: Whose move? (Wer ist am Zug?); Forced en passant

(a) 1.Lf3?? gxf3 2.0-0-0#? (?? denotes illegal).
0... gxf3 e.p.(forced!) 1.Lxf3 Lf5 2.Txg2#! (2.0-0-0#??)
Bl has the move, e.p. capture is legal, and castling is illegal.

(b) 1.Lf3?? gxf3 2.0-0-0#?
0... gxf3 e.p.(forced) 1.Lxf3 Lf5 2.0-0-0#! (2.Txg2#?? - AP-illegal).
Bl still has the move (due to bPe7). Unlike (a), two different retro-moves are possible: i) f2-f4 (f3xSg2) or ii) b2xc3 (in which case castling is illegal due to the missing dark-squared w Bishop). AP after Petrovic in the reversed form "a la Abdurahmanovic": W forces Bl to capture e.p.

(c) 1.Lf3? gxf3 2.0-0-0#??
0... gxf3 e.p.(!) 1.Lxf3 Lf5 2.0-0-0#!! (2.Txg2#?? - AP-illegal).
W's turn to move is possible (in which case, however, castling is illegal). Executed castling justifies jointly Bl's turn to move (AP after Keym) and e.p.

The twins differ by the role of castling: it is illegal in (a), is legal and legalizes Bl's e.p.-key in (b) and legalizes both Bl's turn to move and e.p. in (c). Separately they have 1-mover predecessors: resp., P0005627, P1066685 and P1068112. (2009-06-22)
A.Buchanan: Very nice problem! Ke2# is a dual not provided separately, but I don't see a way to dispense with it (2023-07-22)
comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (wg), No legal last move for Black
Genre: Retro, 2#
Computer test: HC+ retro thinking & Popeye v4.87
FEN: 8/3PpP2/2Q1P3/3N2P1/5Pp1/2P1P1Pb/1B1R2pp/R2BK1kr
Input: Gerd Wilts, 2009-06-08
Last update: A.Buchanan, 2023-07-22 more...
11 - P1300392
Valery Liskovets
Valery Voynov

B5a Length Records in Last Single Moves? , p. 7, 11/2009
version by Dmitrij Baibakov
P1300392
(4+1)
Black played last.
Last 4 single moves?
R: 1. ... Ka7-a8 2. b7-b8=L+ Ka8-a7 3. b6-b7+
play all play one stop play next play all
Valery Liskovets writes: A misattribution appeared already in the original publication of P1300392 in 2009: it ought to be attributed more properly as "D.Baibikow after V.Liskovets and Ju.Wojnow."
Not merely it is stipulated distinctly and with a slightly changed position but my/our intention was very different: under-promotions both in the solution and the retro-play. In the PDB description, a direct reference to our initial P0008499 with almost the same position and solution but non-rigorous last move) should also be added.
more ...
comment
Keywords: Last Moves? (4), Type B, Move Length Record (ex-aequo), Miniature
Genre: n#, Retro
FEN: kB6/2K5/P1P5/8/8/8/8/8
Input: A.Buchanan, 2015-04-01
Last update: A.Buchanan, 2022-04-03 more...
12 - P1366493
Valery Liskovets
SuperProblem (Website) 02/11/2015
4. Preis
154. Thematurnier
P1366493
(9+7) C+
s#6
1. Lc2? a5,h3 2. Le1 d5 3. Sd2 d4 4. Ld1 d3 5. Lb3 h3,a5 6. Sxe7 Lxe7#
1. ... d5!

1. Le1? d5 2. Sd2 a5,h3 3. Lc2 d4 4. Ld1 d3 5. Lb3 h3,a5 6. Sxe7 Lxe7#
1. ... Kb1!

1. Sd2! d5 2. La5 d4 3. Lc7 d3 4. Lg3 h3 5. Le1 a5 6. Sxe7 Lxe7#
4. ... hxg3 5. hxg3 a5 6. Sxe7 Lxe7#
play all play one stop play next play all
VL: A long pericritical hideaway maneuver (wB). (2020-12-11)
comment

Genre: s#
Computer test: Gustav 4.0i
FEN: 5bN1/4p1p1/p2pP1P1/8/PBN4p/KB6/7P/k7
Input: Marcin Banaszek, 2019-08-20
Last update: Marcin Banaszek, 2019-08-20 more...
13 - P1381088
Valery Liskovets
18145 Die Schwalbe 302 04/2020
P1381088
(4+13) C+
h#6
0.1.1...
1. ... a4 2. Lb8 a5 3. Lf4 a6 4. Lxg5 axb7 5. Lh6 b8=D 6. g5 Dxb6#
play all play one stop play next play all
Hierzu gibt es eine Version vom gleichen Autor, siehe P1381089 sowie P1401805
VL: A long hideaway maneuver. (2020-10-21)
comment
Keywords: blocking of a piece (Selbst-), Excelsior white, Model mate, Promotion
Genre: h#
FEN: 2K5/bp4p1/1p4p1/1B4P1/6pp/6pr/P5pn/6kb
Input: Gunter Jordan, 2020-10-18
Last update: Felber, Volker, 2022-06-13 more...
14 - P1381117
Valery Liskovets
Problemist Ukrainy Website 10/2020
1.-2. Lob
Stanislaw Kirilitschenko MT
Sektion h#3 4-5 Steine
P1381117
(3+1) C+
h#3*
b) wKf1->c5
a) * 1. ... Se1-g2 2. Kc3-c2 Kf1-f2 3. Kc2-d1 Sg2-e3#
1. Kc3-b2 Se1-g2 2. Kb2-c2 Kf1-f2 3. Kc2-d1 Sg2-e3#
b) * 1. ... Kc5-b6 2. Kc3-b4 Se1-c2+ 3. Kb4-a4 Sb3-c5#
1. Kc3-b2 Kc5-b6 2. Kb2-a3 Se1-c2+ 3. Ka3-a4 Sb3-c5#
play all play one stop play next play all
PR Valery Kopyl
VL: Two waiting moves in a): B1 and W2. (2020-10-21)
comment
Keywords: Rex solus (s), Aristocrat
Genre: h#
Computer test: Popeye 4.83
FEN: 8/8/8/8/8/1Nk5/8/4NK2
Input: Marcin Banaszek, 2020-10-18
Last update: Alfred Pfeiffer, 2020-10-21 more...
15 - P1382214
Valery Liskovets
H4302 The Problemist 27-7, p. 275, 01/2020
P1382214
(10+9) C+
h#5
0.1.1...
1. ... e6 2. Lf8 exf7 3. Lc5 f8=D 4. Lb6 c5 5. Lb1 Dxf6#
play all play one stop play next play all
VL: A hideaway maneuver (bB).
Tempo-try: 1... exf6/c5? 2.Lf8 c5/exf6 3.Le7 fxe7 4.Lb1 e8=D 5.?? De5# (2020-12-04)
comment
Keywords: Promotion (D), Verstecken (l), Corner mate, Model mate
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/p1p2pb1/p1P2p1p/P2BPP1P/1PP1P3/8/b7/k4K2
Input: Felber, Volker, 2020-11-21
Last update: Gunter Jordan, 2020-11-21 more...
16 - P1382489
Valery Liskovets
S2801 The Problemist 27-9, p. 354, 05/2020
P1382489
(10+8) C+
s#5
1. Lb7! a2 2. Lf3 a3 3. Ld1 h4 4. La4 h5 5. La7 g1=D,T,L,S#
1. ... h4 2. Lf3 a2 3. Lh5 a3 4. Lg6 h5 5. La7 g1=D,T,L,S#
2. ... h5 3. Lxh5 a2 4. Lg6
play all play one stop play next play all
VL: A long branching hideaway maneuver (wB).
1.Lc8? a2? 2.d8=L+! Kxc8 3.La5 a3/h4 4.Sxh6 h4/a3 5.La7 g1=~# but 1.h4! (2020-12-04)
comment

Genre: s#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: K7/2kP1N2/B3P2p/1P3P1p/p7/p6p/6pP/N5Bb
Input: Felber, Volker, 2020-11-26
Last update: Felber, Volker, 2020-11-26 more...
17 - P1386352
Valery Liskovets
PS3556 The Problemist Supplement 167, p. 91, 07/2020
P1386352
(7+11) C+
h#5
0.1.1...
b) wBc2->h3
c) wBe2->h3
a) 1. ... cxb6 2. bxc6 b7 3. Sf1 b8=D 4. Kh2 Df4 5. Lg1 Dxh4#
b) 1. ... cxb6 2. Kf1 bxa7 3. g1=T a8=D 4. Tg2 Da4 5. Tf2 Dd1#
c) 1. ... Lxb5 2. Lb8 c6 3. Lf4 cxb7 4. Lc1 b8=D 5. Lb2 Dxb6#
play all play one stop play next play all
VL: A long hideaway maneuver of black Bishop in c) that contrasts with preceding bB's roles in the twins a) (self-block) and b) (annihilation). One could prefer the multi-solution version without wPs c2 and e2 stipulated h#5, 0.1.1...+0.1.2.1... (2021-03-15)
comment

Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 2K5/bp6/1pB5/1pP5/1P5n/p1P3p1/2P1P1pn/6kb
Input: Felber, Volker, 2021-02-09
Last update: Felber, Volker, 2021-02-09 more...
18 - P1386381
Geir Sune Tallaksen Østmoe
Valery Liskovets

feenschach 2014
P1386381
(6+5)
s###10
1. Ld5! a5 2. e3 a4 3. e4 a3 4. b4 a2 5. Le6 d5 6. e5 d4 7. b5 d3 8. b6 cxb6 9. Ld7 b5 10. e6 b4###
play all play one stop play next play all
Reprint: http://matplus.net/start.php?px=1612896936&app=forum&act=posts&tid=2298&fid=xshows&page=1

"Some years ago, Valery Liskovets suggested the stipulation «CU-mate», or «completely unavoidable mate», with the notation ###."
VL: Publication: feenschach 06/2014 H.207 #10961

“###” denotes a completely unavoidable mate (=CU-mate; self-CU-mate in this problem), that is a position after one side's move that is a mate or ends inevitably with a mate to the opponent's King in all variations of the subsequent play, no matter how many variations are possible and how long they are. The game stops immediately in such a deadly won/lost position. The subsequent (legal no-brainer) variations form the so-called “proof-play” (not to be confused with the famous proof game, of course). The proof-play lies formally outside of the intended solution, and all its elements should be verified in order to confirm the declared CU-mate.
This is one of the first self-CU-mate problems.
Here in the final position, white Bishop has two moves to intervene and prevent the mate. It fails (as well as promoted Queen): the proof-play is trivially 11.~ b3 12.~ b2#. The square d7 is the unique suitable place for wB to that end (locked by wPe6). This is a generalization of an ordinary hideaway. Later I called such a CU-motivated maneuver “an exile"; it is most fruitful for self-CU-mates.
Computer test. HC+ by Popeye 4.67. There is an appropriate sstipulation clause, namely,
‘sstip white 20ad[1d[1d[1d[1d[#]a]a]a]a]’. Quite decisively in this case. (2021-02-21)
comment

Genre: Fairies
FEN: 8/2p5/p1Pp4/8/8/8/1PpPP1B1/k1K5
Input: James Malcom, 2021-02-11
Last update: James Malcom, 2021-02-11 more...
19 - P1393982
Valery Liskovets
8813 Phénix 310-311 09-10/2020
nach Roland Lecomte
P1393982
(3+1) C+
h#3
b) wSa6-->c4
c)=b) wKa1-->f1
d)=c) wKf1-->c7
e)=d) wSc4-->d3
f)=e) wKc7-->h1
a) 1. Kc3-c4 Sa6-c5 2. Kc4-b4 Ka1-b1 3. Kb4-a3 Sd4-c2#
b) 1. Kc3-d3 Sc4-e3 2. Kd3-d2 Ka1-a2 3. Kd2-c1 Sd4-b3#+
c) 1. Kc3-d3 Sd4-b3 2. Kd3-c2 Kf1-f2 3. Kc2-d1 Sc4-e3#+
d) 1. Kc3-b4 Sd4-e6 2. Kb4-b5 Kc7-b8 3. Kb5-a6 Se6-c7#+
e) 1. Kc3-c4 Sd4-c2 2. Kc4-b3 Kc7-b6 3. Kb3-a4 Sd3-c5#+
f) 1. Kc3-d2 Sd4-f5 2. Kd2-e2 Kh1-h2 3. Ke2-f1 Sf5-g3#
play all play one stop play next play all
VL: Cf. P1204543.
Only wK and one wS change their places in the twins. (2021-11-03)
comment
Keywords: Aristocrat, Miniature
Genre: h#
Computer test: Popeye 4.83
FEN: 8/8/N7/8/3N4/2k5/8/K7
Input: Marcin Banaszek, 2021-09-23
Last update: Marcin Banaszek, 2021-09-23 more...
20 - P1395560
Valery Liskovets
E1133 SuperProblem (Website) 10/2021
P1395560
(2+3) C+
h#3
b) +sLb5
a) 1. b2-b1=T Kc2xc3 2. Ke2-e1 Kc3-d3 3. Ke1-d1 Ta1xb1#
b) 1. b2-b1=L+ Kc2-b3 2. Ke2-d1 Kb3xc3 3. Lb5-e2 Ta1xb1#
play all play one stop play next play all
VL: Schnoebelen Theme (t/l) (2022-01-23)
comment
Keywords: Minimal (T), Miniature
Genre: h#
Computer test: Popeye4.83
FEN: 8/8/8/8/8/2p5/1pK1k3/R7
Input: Marcin Banaszek, 2021-11-07
Last update: Gunter Jordan, 2021-11-07 more...
21 - P1395561
Valery Liskovets
E1134 SuperProblem (Website) 10/2021
P1395561
(2+4) C+
h#3
b) sBe3-->d3
a) 1. b2-b1=L Kc4-b3 2. Kd2-d1 Kb3-c3 3. e3-e2 Ta1xb1#
b) 1. b2-b1=S Kc4-b3 2. Kd2-c1 Kb3-a2 3. d3-d2 Ta1xb1#
play all play one stop play next play all
VL: Reprint: (9) Problemas 37, p.1233, 01/2022
Schnoebelen Theme (l/s) (2022-01-23)
comment
Keywords: Minimal, Miniature
Genre: h#
Computer test: Popeye 4.83
FEN: 8/8/8/8/2K5/4p3/1pnk4/R7
Input: Marcin Banaszek, 2021-11-07
Last update: Marcin Banaszek, 2021-11-07 more...
22 - P1395562
Valery Liskovets
E1135 SuperProblem (Website) 10/2021
P1395562
(4+5) C+
h#3
b) sSc4-->c2
a) 1. b2-b1=T d3xc4 2. Ke2-e1 Kc3-d3 3. Ke1-d1 Ta1xb1#
b) 1. b2-b1=S+ Kc3-b3 2. Ke2-d1 Kb3-a2 3. Kd1-c1 Ta1xb1#
play all play one stop play next play all
VL: Schnoebelen Theme (t/s) (2022-01-23)
comment

Genre: h#
Computer test: Popeye 4.83
FEN: 8/8/8/8/2nP4/p1KP4/1p1pk3/R7
Input: Marcin Banaszek, 2021-11-07
Last update: Marcin Banaszek, 2021-11-07 more...
23 - P1396048
Valery Liskovets
E912 SuperProblem (Website) 03/2020
P1396048
(6+9) C+
h#2
4.1...
1) 1. b4-b3 c2xb3 2. Tg4-g1 Tb2-h2#
2) 1. b4xc3 Tb2xb5 2. Te7-a7 Tb5-b8#
3) 1. Tg4-e4 Tb2-b1 2. Te4-e6 Tb1xh1#
4) 1. Te7-e4 Tb2-a2 2. Te4-f4 Ta2xa8#
play all play one stop play next play all
VL: Hideaway twice and hideaway maneuver twice (sTT).
1 HM. (2022-07-15)
comment

Genre: h#
Computer test: Popeye 4.83
FEN: n6k/4r1p1/5N2/1p1K4/1p1P2r1/2P1p3/1RP5/7n
Input: Marcin Banaszek, 2021-11-21
Last update: Marcin Banaszek, 2021-11-21 more...
24 - P1399112
Werner Kuntsche
Valery Liskovets
Andrew Buchanan

PDB Website 14/02/2022
WK, correction AB
P1399112
(5+8) C+
h#3 AP

No solutions except for those beginning with e.p.
297 candidate solutions begin 1.gxf3ep, but only one includes castling. Queenside 2. ... 0-0-0
22 candidate solutions begin 1.fcxd3ep, but only one includes castling. Kingside 2. ... 0-0
In some way, these are trying to be combined so that from AP perspective, White's last move can only have been double hop with wBd or wBf.
We need a clear delineation of the precise PRA-AP algorithm to be followed.

The point: if this problem is "sound", then surely P1382808 is "unsound", because there only one of the two parts needs to prove that White castling rights remain, so there are 4 cooks.
This is an important schema for AP. Previous cooked problems P0002476 & P0004295 (together with versions of each lacking sBe3) had multiple chess cooks that do not involve e.p. as key. This chessically sound composition allows us to focus on the fundamental theoretical issue of AP.

The double e.p. idea cannot work as an AP retro strategy problem, because one can never prove that one double pawn hop rather than the other occurred. PRA+AP seems potentially feasible, but I see no clear statement of how the normal PRA method should be extended to this case. All prescriptive suggestions are welcome.

In WinChloe, I found a correction from 2002 by Valery Liskovitz, just posted in PDB as P1401449. However it used three extra pawns, rather than 2, so the current version still has merit, but I will add Valery as co-corrector.

See also P1399178.
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key (2), Symmetrical position, Symmetrical solution, Castling (wb)
Genre: h#, Retro
Computer test: HC+ Forward play proved by Popeye v4.87 AP logic requires some framework but I think we know we want this cool idea to end up being sound.
FEN: 8/8/2p3p1/8/2pPkPp1/2n1p1n1/8/R3K2R
Input: A.Buchanan, 2022-02-14
Last update: A.Buchanan, 2022-05-24 more...
25 - P1401805
Valery Liskovets
18145V Die Schwalbe 310, p. 238, 08/2021
Version
P1401805
(3+11) C+
h#6
1. Ta4 b4 2. Ta5 bxa5 3. Lb8 a6 4. Le5 axb7 5. Lb2 b8=D 6. Lc1 Dxb6#
play all play one stop play next play all
Dies ist eine Version zu P1381088. Sie erschien unter "Bemerkungen und Berichtigungen"
VL: Long hideaway maneuver (sL). Excelsior. (2022-07-15)
comment

Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 2K5/bp6/1p6/1B6/1r6/p6p/1P1n2pr/6kb
Input: Felber, Volker, 2022-06-13
Last update: Felber, Volker, 2022-06-13 more...
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The problems of this query have been registered by the following contributors:

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Valery Liskovets (1)
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