Die Schwalbe

1 problem(s) found in 2328 milliseconds (displaying 1 problem(s)). [PROBID='P0003447'] [download as LaTeX]

1 - P0003447
Gerd Rinder
(O) Die Schwalbe 16, p. 393, 10/1972
(14+13) C+
1. Txg3 Sxg3+ 2. Kf3 0-0#
1. cxd3ep Lxg1 2. e2 Sd2#
1. gxf3ep Lxg1 2. Dd3 Txh4#
play all play one stop play next play all
W. Keym (26.03.2020): The remarks of VL (below) are history. According to the Codex for Chess Composition (2009) the cases in which several move rights (castlings and e.p. captures) are mutually dependent are solved. However, in the Codex it is not regulated how to find out the partial problems of a PRA composition. Here is a formal method that is suitable particularly for complex cases.

1) In P0003447 there are four special move rights; the opposite rights are marked with ’.
A = 0-0-0 is permitted         A’ = 0-0-0 is not permitted	 
B = 0-0 is permitted B’ = 0-0 is not permitted
C = c4xd3ep is not permitted C’ = c4xd3ep is permitted
D = g4xf3ep is not permitted D’ = g4xf3ep is permitted

2) The calculation results into 24 = 16 combinations of special move rights:

3) The combinations that are not legal are eliminated. These are the eight underlined ones.
4) The combinations that do not correspond with the castling or en-passant convention are eliminated. These are the five in italics.
5) The remaining combinations form the partial problems. These are the three in bold.

6) The first partial problem (AB’CD’) has the solution 1. g4xf3ep!, the second (AB’C’D) 1. c4xd3ep!, the third (A’BCD) 1. Txg3!.

This method works with all PRA problems.
The supplementary stipulation PRA is not necessary and should not be published any longer.
See the article ‘Castling and En-passant capture in the Codex 2009’ in the web.

VL: This problem was first published in the article entitled "Konstruktive Kritik am Kodex von Piran" (by W.Keym; H.16). The author is a strong opponent of the retro-synthese (including AP a la Keeble--Petrovic = "Typ Petrovic"). Here we meet a retro-situation resembling that of AP, which, however, is interpreted and resolved purely in the framework of pRA. Namely, the problem contains three retro-variants: (a) 0-0; (b) 0-0-0 & cxd3 ep; (c) 0-0-0 & gxf3 ep. Each of them results in a unique (partial) solution (see above), so that the problem is sound. The absence of 0-0-0 in the latter two solutions is a witty feature that emphasizes a fundamental distinction from AP (where only actual castling can legalize the ep-key). Judging from the article "PRA oder Kodex" by B.Pavlovic in the "Problem", 152-6, where this problem is reproduced, it was initially stipulated "h#2, PRA-Typ Rinder" (this is presently known under the name "Typ Keym"). It is reproduced also in the book "Eigenartige Schachprobleme" by W.Keym, 2010, Pr.385. (2012-08-26)
Sally: Hier schlugen die w. Bauern dreimal, unter anderem eine UW-Figur von "h1 oder "a1",(entweder ist 000 oder 00 zulässig).
Der letzte Zug war 1. K-e1, oder T-a1, oder "d2-d4". oder
"f2-f4". Falls 000 zulässig ist, dann ist 00 unzulässig und jeweils einer der E.-P.-Schläge erlaubt, Also löst entweder:
1. Bg4xf3 E.p.! - Lxg1, 2. Dd3 - Txh4#. oder :
1. Bc4xd3 E.p.! - Lxg1, 2. Be2 - Sd2#.
Beide Male wird das Zugrecht 000 anerkannt, aber nicht aufgeführt! Falls 00 zulässig ist, dann ist 000 unzulässig;
in diesem Fall war zuletzt der Zug "T-a1" möglich, daher kein E.p.-Schlag erlaubt. Also löst :
1. Txg3! - Sxg3+, 2. Kf3 - 00 #. (2018-06-15)
Henrik Juel: C+ by Popeye 4.61, after analysis (2018-06-15)
more ...
Keywords: Castling, En passant as key, Partial Retro Analysis (PRA)
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: 8/3p1b2/1Pp1B3/nPP2PP1/rnpPkPpb/1q2p1P1/7B/R3KNrR
Reprints: (II) Problem 152-156 04/1973
XI Caissas Schloßbewohner 1, p. 77, 1983
385 Eigenartige Schachprobleme 2010
13 Die Schwalbe 241, p. 374, 02/2010
366 Chess Problems Out of the Box 2018
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2021-09-23 more...
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