195 problem(s) found in 2972 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT K='Selbstblock' AND K='En passant als Schlüssel'] [download as LaTeX]
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.
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.
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...
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...
a) 1. ... exf6ep 2. 0-0-0 Lxf4 3. Td7 a8=D#
b) 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
b) 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
Anton Baumann: Mattdual in b): 3.Tb8 axb8=D,T# (2022-12-16)
A.Buchanan: So Borodatow got it all working! In a) Black might have captured hxgxf and axPb. So castling rights might still be maintained with the ep. In b) on the other hand, it must be axb, bxa, exf and wPgxh6, so there was the cage. Promotion to TD is tolerated in the final move, although it may not be puristic, without the convention, too many mates would be excluded (2022-12-16)
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A.Buchanan: So Borodatow got it all working! In a) Black might have captured hxgxf and axPb. So castling rights might still be maintained with the ep. In b) on the other hand, it must be axb, bxa, exf and wPgxh6, so there was the cage. Promotion to TD is tolerated in the final move, although it may not be puristic, without the convention, too many mates would be excluded (2022-12-16)
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Keywords: En passant as key, Castling (sg), Cant Castler, Valladao Task
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & basic retro thinking
FEN: r3k3/P7/b3r1pP/4PpBP/3nnpKR/5PRB/5PP1/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & basic retro thinking
FEN: r3k3/P7/b3r1pP/4PpBP/3nnpKR/5PRB/5PP1/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
Satzspiel:
*1. ... 0-0! 2. cxd4 Tc1#
Verführung:
1. exd3ep? 0-0 2. cxb4 Tf4# aber zuletzt R: 1. d2-d4?? ist illegal, da es den wLc1 als schwarzes Bauernschlagobjekt ausschließt
Lösung:
1. ... 0-0! (AP-Legalierung des weissen Anzugrechtes) 2. cxd4 Tc1#
Alle Versuche, das Satzspiel 1. ... 0-0! 2. cxd4 Tc1# durch einen schwarzen Vorschaltzug aufrecht zu erhalten, scheitern.
*1. ... 0-0! 2. cxd4 Tc1#
Verführung:
1. exd3ep? 0-0 2. cxb4 Tf4# aber zuletzt R: 1. d2-d4?? ist illegal, da es den wLc1 als schwarzes Bauernschlagobjekt ausschließt
Lösung:
1. ... 0-0! (AP-Legalierung des weissen Anzugrechtes) 2. cxd4 Tc1#
Alle Versuche, das Satzspiel 1. ... 0-0! 2. cxd4 Tc1# durch einen schwarzen Vorschaltzug aufrecht zu erhalten, scheitern.
Mario Richter: 1. exd3ep 0-0 2. cxb4 Tf4 ist nur Verfuehrung (scheitert an schwarzer Bauern-Schlagbilanz).
Loesung: 1. ... w0-0! (AP-Legalierung des weissen Anzugrechtes) 2.cxd4 Tc1# (2009-02-10)
A.Buchanan: Supersedes P0000553 (2022-01-08)
Mario Richter: Hi Andrew, I only quoted from the "official solution" ('Die Schwalbe' Heft 72, 12/1981, p.399).
Perhaps a better way to get anwers to your interesting questions is to ask the authors directly ... (2022-01-08)
A.Buchanan: AP Type Petrovic is a try, because wLc1 was captured at home, yet Black still made 6 pawn captures.
AP Type Keym is the actual solution, because if it was really BTM, then White would have lost castling rights.
So there are two kinds of AP here. (2022-03-21)
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Loesung: 1. ... w0-0! (AP-Legalierung des weissen Anzugrechtes) 2.cxd4 Tc1# (2009-02-10)
A.Buchanan: Supersedes P0000553 (2022-01-08)
Mario Richter: Hi Andrew, I only quoted from the "official solution" ('Die Schwalbe' Heft 72, 12/1981, p.399).
Perhaps a better way to get anwers to your interesting questions is to ask the authors directly ... (2022-01-08)
A.Buchanan: AP Type Petrovic is a try, because wLc1 was captured at home, yet Black still made 6 pawn captures.
AP Type Keym is the actual solution, because if it was really BTM, then White would have lost castling rights.
So there are two kinds of AP here. (2022-03-21)
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Keywords: Castling (wk), a posteriori (AP) (Type Keym), En passant as key, a posteriori (AP) (Type Petrovic)
Genre: h#, Retro
Computer test: BC+ Popeye v4.87
FEN: 8/2p5/1pPp4/bRpP4/BPkPp3/qp2p2p/rP2P3/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
Genre: h#, Retro
Computer test: BC+ Popeye v4.87
FEN: 8/2p5/1pPp4/bRpP4/BPkPp3/qp2p2p/rP2P3/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
4 - P0000598
Nikita M. Plaksin
Andrej N. Kornilow
3876 Die Schwalbe 74 04/1982
(4+3)
h=2 (AP)
Circe, Monochromes Schach
Nikita M. Plaksin
Andrej N. Kornilow
3876 Die Schwalbe 74 04/1982
(4+3)
h=2 (AP)
Circe, Monochromes Schach
Intended solution:
1. dxc3ep[+wBc2] 0-0 2. axb1=S Kh2=
2. ... Kf2? 3. Kf3!
1. ... Tf1? 2. axb1=S Kf2? 3. Kf3!
1. dxc3ep[+wBc2] 0-0 2. axb1=S Kh2=
2. ... Kf2? 3. Kf3!
1. ... Tf1? 2. axb1=S Kf2? 3. Kf3!
Erich Bartel: vom Dual 2.-- Kf2/Kh2 abgesehen C+ PY V4.41.--- (2008-11-07)
A.Buchanan: This is cooked in Popeye because White can't castle as long as bK controls f1! This is quite reasonable behaviour in monochrome: compare to en passant! So no solution (2022-02-18)
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A.Buchanan: This is cooked in Popeye because White can't castle as long as bK controls f1! This is quite reasonable behaviour in monochrome: compare to en passant! So no solution (2022-02-18)
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Keywords: a posteriori (AP) (Type Petrovic), Circe, En passant as key, Monochromatic Chess, Castling (wk), Miniature, Golden Age (Monochrome castling)
Genre: Retro, Fairies
FEN: 8/8/8/8/2Pp4/8/p3k3/1N2K2R
Reprints: (IV) Quartz 4 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-18 more...
Genre: Retro, Fairies
FEN: 8/8/8/8/2Pp4/8/p3k3/1N2K2R
Reprints: (IV) Quartz 4 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-18 more...
1. ... Kg5 2. 0-0?? Se7#
1. ... Txh7 2. Kf8 Txh8#
1. cxb3ep Txh7 2. Kf8?? Txh8#
1. cxb3ep Kg5 2. 0-0! Se7#
Wh captured 10 by pawns (including hxPg-g8) but not on the last move. wPa3 came from a2: wPa5 captured bPa.
Bl took only b3xa2.
In set play, Bl has no last move except for K or R.
For the main play, the intention is that if Bl retains castling rights, last moves were R: 1. b2-b4 b3xa2 e.g. 2. Bg1-d4 b4-b3 3. d4xc5 etc. However there are retro cooks e.g. R: 1. Bg1-d4 b3xa2 2. d4xc5 etc, with wPa5 the original wPb. wPb4 & wPb5 come from further East.
Cook: Retro logic for main play is unsound.
1. ... Txh7 2. Kf8 Txh8#
1. cxb3ep Txh7 2. Kf8?? Txh8#
1. cxb3ep Kg5 2. 0-0! Se7#
Wh captured 10 by pawns (including hxPg-g8) but not on the last move. wPa3 came from a2: wPa5 captured bPa.
Bl took only b3xa2.
In set play, Bl has no last move except for K or R.
For the main play, the intention is that if Bl retains castling rights, last moves were R: 1. b2-b4 b3xa2 e.g. 2. Bg1-d4 b4-b3 3. d4xc5 etc. However there are retro cooks e.g. R: 1. Bg1-d4 b3xa2 2. d4xc5 etc, with wPa5 the original wPb. wPb4 & wPb5 come from further East.
Cook: Retro logic for main play is unsound.
A.Buchanan: Not trivial to fix as shifting wPa5 to b6 leads to promotion cooks. Adding black material must be carefully done to avoid disrupting uniqueness of e.p. as tempo. (2022-03-11)
A.Buchanan: OK just to say I have found a form of this idea that works. This one was the very devil to fix - the final form includes different mates and retro logic, so becomes an "after" rather than a "correction", and I will find somewhere else to publish first. (2022-03-19)
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A.Buchanan: OK just to say I have found a form of this idea that works. This one was the very devil to fix - the final form includes different mates and retro logic, so becomes an "after" rather than a "correction", and I will find somewhere else to publish first. (2022-03-19)
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Keywords: Castling (sk), a posteriori (AP) (Type Petrovic), En passant as key, Tempo Move, Superseded by (P1409841)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/5p1p/5K1R/PPP2N2/1PpB4/P1P5/p1RNB1P1/Q7
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-20 more...
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/5p1p/5K1R/PPP2N2/1PpB4/P1P5/p1RNB1P1/Q7
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-20 more...
1. bxc6ep
Nach 1. bxc6ep ist zwar ein Matt in 2 Zügen möglich, aber Weiß muss noch rochieren, um den ep-Schlag zu begründen.
Nach 1. bxc6ep ist zwar ein Matt in 2 Zügen möglich, aber Weiß muss noch rochieren, um den ep-Schlag zu begründen.
James Malcom: Full solution? (2021-02-05)
Henrik Juel: White pawns captured all 11 missing black men by axb, f2x..xa7, and g2x..xb7
Black captured hxg and once more, so Pc5 never captured
If last move was c6-c5, the preceding white move was with Ke1 or Th1
So if White can castle, last move was c7-c5
1.bxc6ep dxc6 2.0-0(AP) c5 3.Ta1,Sd7 and #4
2.Sd7 c5 3.0-0(AP) is also possible
I expect the full solution was omitted because it is very dualistic (2021-02-05)
Anton Baumann: Autorabsicht: 1.bxc6ep [2.0-0 3.Lb5#] bxa3 2.0-0 axb2 3.Tb3 nebst 4.Lb5,Tf4#
1. ... dxc6? 2.0-0 nebst 3.Tf4,Lb5#
Dual: 2.Tb3 dxc6 3.0-0 [4.Tf4#] c5 4.Lb5# 2. ... a2,axb2 3.0-0 [4.Lb5#] dxc6 4.Tf4# (2021-02-05)
A.Buchanan: C+ of helpmates with this kind of AP is relatively simple: one can just eliminate all lines without 0-0. But C+ of d# is harder.
(1) Henrik's retro logic is solid. There is no #4 unless 1.bxc6ep, so let's take that as a given: an AP debt is incurred.
(2) Black has 5 responses. Against 4 of them, 2.0-0 gives a genuine short mate with the AP debt paid. Such short mates don't impact soundness in a retro #4, to my mind.
(3) This leaves 1... bxa3. Following 2.0-0 axb2 (the only full length line) 3.Rb3 thr 4.Bb5,Rf4# but both mates are separately provided, so only minor duals.
(4) It remains to consider if White can delay castling after 1...axb3. Apart from 2.Tb3 nothing works, but 2.Tb3 thr 3.0-0 is unstoppable and is the major dual to my mind. I don't see any obvious fix. (2021-02-05)
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Henrik Juel: White pawns captured all 11 missing black men by axb, f2x..xa7, and g2x..xb7
Black captured hxg and once more, so Pc5 never captured
If last move was c6-c5, the preceding white move was with Ke1 or Th1
So if White can castle, last move was c7-c5
1.bxc6ep dxc6 2.0-0(AP) c5 3.Ta1,Sd7 and #4
2.Sd7 c5 3.0-0(AP) is also possible
I expect the full solution was omitted because it is very dualistic (2021-02-05)
Anton Baumann: Autorabsicht: 1.bxc6ep [2.0-0 3.Lb5#] bxa3 2.0-0 axb2 3.Tb3 nebst 4.Lb5,Tf4#
1. ... dxc6? 2.0-0 nebst 3.Tf4,Lb5#
Dual: 2.Tb3 dxc6 3.0-0 [4.Tf4#] c5 4.Lb5# 2. ... a2,axb2 3.0-0 [4.Lb5#] dxc6 4.Tf4# (2021-02-05)
A.Buchanan: C+ of helpmates with this kind of AP is relatively simple: one can just eliminate all lines without 0-0. But C+ of d# is harder.
(1) Henrik's retro logic is solid. There is no #4 unless 1.bxc6ep, so let's take that as a given: an AP debt is incurred.
(2) Black has 5 responses. Against 4 of them, 2.0-0 gives a genuine short mate with the AP debt paid. Such short mates don't impact soundness in a retro #4, to my mind.
(3) This leaves 1... bxa3. Following 2.0-0 axb2 (the only full length line) 3.Rb3 thr 4.Bb5,Rf4# but both mates are separately provided, so only minor duals.
(4) It remains to consider if White can delay castling after 1...axb3. Apart from 2.Tb3 nothing works, but 2.Tb3 thr 3.0-0 is unstoppable and is the major dual to my mind. I don't see any obvious fix. (2021-02-05)
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Keywords: Castling (wk), En passant as key, a posteriori (AP)
Genre: Retro, n#
FEN: 1N6/PP1p4/BR6/pPp5/kp6/B2P4/1PP1P2P/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
Genre: Retro, n#
FEN: 1N6/PP1p4/BR6/pPp5/kp6/B2P4/1PP1P2P/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
a) 1. ... 0-0-0 2. 0-0 Tdg1+ 3. Kh8 Txh6#
b) also 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#
is clearly the intention
Can count 1+7 visible pawn captures, leaving 1+1 unexplained. Need to resolve wPfgh & bPg. Suppose bK never moved, then one of:
1) wPf waylaid, wPhxPg=, wPg=. Here w00 right can remain.
2) Or similarly, wPh waylaid, wPfxPg=, wPg=. Again w00 right can remain.
3) bPgxPh=, wPfxg= (or wPfxe then captured by bPf), wPg= w00 right lost.
1+1 captures explained in either case.
On the other hand if bK moved, then maybe wPf=, and only requires one more capture to resolve g&h files. The w00 right can remain. So there is a captured unit unaccounted for, and we can't validate the ep.
On the other hand if we are in case 3 above, then we can't know that wPbxa captured dark bishop, so White might retract a6-a7.
We also need to know that w00 rights remain. s000 rights (i.e. prior movement of sTa8) are not relevant.
If we know that w00, w000 & b000 remain, then we are in case 1 or 2 above with bPg captured unpromoted. Therefore wPb6xLa7 due to bishop shade. Therefore R: 1. b7-b5 b6xLa7 to unblock.
Only one candidate solution does not begin with ep: 1. ... 0-0-0 2. 0-0 Tdg1+ 3. Kh8 Txh6#. So this is the solution for a).
Let ????? denote validity of w00,w000,b00,b000,ep. Possibilities are: YYY?Y and all of YYN??, YNY??, NYY??, YNN??, NYN??, NNY??, NNN??. So there are 30 possibilities.
Under PRA, the solution parts would be YYYYY, YYNYN, YNYYN & NYYYN. For the first there are 20 solutions, while the second and third have 0 solutions. So this is not the right paradigm.
Under SPRA, there would be a single solution part YYYYY with 20 solutions. So this is not the right paradigm either.
Under RS with AP, the solution of a) still works. This comes from ?YY??. On the other hand, if the first move is ep, then we are in YYYYY. So all castlings are valid. But when we get to perform the mate, we need to know that based on castlings actually performed, the ep is valid. So the solution must include w00. So is 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#.
By combining the information of both of these solutions, we know we are in YYY?? so YYY?Y is the only possibility, and ep is legal. However, the combination of these two solutions would also validate any of the other 18 ep solutions that do not include 2. ... 0-0. They contribute no new evidence, but how to exclude them?
This is one of the issues with "AP Consolidation". Normally the number of solutions is not a constraint. But here perhaps we should insist that *only* two solutions are allowed?
Cook: 1. ... cxb6ep 2. La6 Tb1,~ 3. 0-0-0 a8=D#
18 different possibilities for W2 which are validated by the same logic that validates 2. ... 0-0.
b) also 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#
is clearly the intention
Can count 1+7 visible pawn captures, leaving 1+1 unexplained. Need to resolve wPfgh & bPg. Suppose bK never moved, then one of:
1) wPf waylaid, wPhxPg=, wPg=. Here w00 right can remain.
2) Or similarly, wPh waylaid, wPfxPg=, wPg=. Again w00 right can remain.
3) bPgxPh=, wPfxg= (or wPfxe then captured by bPf), wPg= w00 right lost.
1+1 captures explained in either case.
On the other hand if bK moved, then maybe wPf=, and only requires one more capture to resolve g&h files. The w00 right can remain. So there is a captured unit unaccounted for, and we can't validate the ep.
On the other hand if we are in case 3 above, then we can't know that wPbxa captured dark bishop, so White might retract a6-a7.
We also need to know that w00 rights remain. s000 rights (i.e. prior movement of sTa8) are not relevant.
If we know that w00, w000 & b000 remain, then we are in case 1 or 2 above with bPg captured unpromoted. Therefore wPb6xLa7 due to bishop shade. Therefore R: 1. b7-b5 b6xLa7 to unblock.
Only one candidate solution does not begin with ep: 1. ... 0-0-0 2. 0-0 Tdg1+ 3. Kh8 Txh6#. So this is the solution for a).
Let ????? denote validity of w00,w000,b00,b000,ep. Possibilities are: YYY?Y and all of YYN??, YNY??, NYY??, YNN??, NYN??, NNY??, NNN??. So there are 30 possibilities.
Under PRA, the solution parts would be YYYYY, YYNYN, YNYYN & NYYYN. For the first there are 20 solutions, while the second and third have 0 solutions. So this is not the right paradigm.
Under SPRA, there would be a single solution part YYYYY with 20 solutions. So this is not the right paradigm either.
Under RS with AP, the solution of a) still works. This comes from ?YY??. On the other hand, if the first move is ep, then we are in YYYYY. So all castlings are valid. But when we get to perform the mate, we need to know that based on castlings actually performed, the ep is valid. So the solution must include w00. So is 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#.
By combining the information of both of these solutions, we know we are in YYY?? so YYY?Y is the only possibility, and ep is legal. However, the combination of these two solutions would also validate any of the other 18 ep solutions that do not include 2. ... 0-0. They contribute no new evidence, but how to exclude them?
This is one of the issues with "AP Consolidation". Normally the number of solutions is not a constraint. But here perhaps we should insist that *only* two solutions are allowed?
Cook: 1. ... cxb6ep 2. La6 Tb1,~ 3. 0-0-0 a8=D#
18 different possibilities for W2 which are validated by the same logic that validates 2. ... 0-0.
Henrik Juel: [I don't follow the silly convention of writing black moves first]. 1.0-0-0 0-0 2.dTg1+ Kh8 3.Txh6#. b) If Ke1,e8 and Th1 never moved, White captured f/hxPg and b6xLa7; if Ta1 also never moved, last white move was b6xLa7. 1.cxb6ep La6 2.0-0/0-0-0 0-0-0 3.a8Q#. The a) solution also works in b). (2003-12-18)
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Keywords: a posteriori (AP) (Type Petrovic), Castling (wkwlsk), En passant as key
Genre: h#, Retro
Computer test: Forward logic Popeye v4.87 & basic retro-logic works, but then AP Consolidated protocol seems to cook the problem.
FEN: r1b1k2r/P2p4/2p1n2p/ppPq4/1np5/p7/P2PP3/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
Genre: h#, Retro
Computer test: Forward logic Popeye v4.87 & basic retro-logic works, but then AP Consolidated protocol seems to cook the problem.
FEN: r1b1k2r/P2p4/2p1n2p/ppPq4/1np5/p7/P2PP3/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
Weiß ist patt. 1. cxb6ep ist nur zulässig, wenn Schwarz diese a posteriori durch die Rochade rechtfertigt. Weiß kann aber so spielen, daß Schwarz nicht rochieren kann, z.B. 1. ... Lxc7 2. bxc7
Guus Rol: This is an incorrect interpretation of the AP-convention. Rules outrank goals in the definition of all GAMES. Therefore the legitimacy of a move cannot be restricted by the desire to achieve the goal (in this case: Remis). The proper way to view AP is that executing e.p. invalidates the legitimacy of all lines of future play that do not contain 0-0-0! In that sense black and white are forced to cooperate. In whatever freedom remains they can compete for the prize promised in the stipulation. By the way, this understanding of AP is not only more logical, it is also much more interesting as a playing field for AP-composition. (2005-09-21)
mri: citeWeiß kann aber so spielen, daß Schwarz nicht rochieren kann, z.B. 1. ... Lxc7 2. bxc7
/cite
How does white prevent black from castling after 1. cxb6 e.p. Ba7xb6+?
E.g. 2.Kxb6 a1=R 3.a7 Rxa7, or 2.Ka4 Bd8xc7 3.a7 Bb8. (2005-09-22)
VL: 1.c5xb6ep(??) a7xb6+ 2.Kxb6 a1R 3.Kb7 R1xa6 4.Rc8! (R6a7#?? illegal).
This study is correct under the generally accepted understanding of AP
a la' N.Petrovic'. Antiform (looking possibly somewhat strange):
Black's unsuccessful try. (2005-10-03)
Guus Rol: Generally accepted, true. Generally acceptable, false. Freedom of interpretation ceases for a concept once its polar concept (a priori validation) has been defined. "Goal induced AP" however, might be a passable stipulation for this type of problem. To keep this place from turning into a discussion forum I will discontinue comments on this issue. (2005-10-03)
paul: Author intention is: If black still can castle, his last move must have been b7-b5. However, to prove this, he has to castle (A Posteriori condition). So: 1.cxb6 e.p. axb6+ 2.Kxb6 a1R! (2...a1Q? 3.Kb7 Qxa6 mate obviously doesn`t prove b7-b5 as the last move) 3.Kb7 R1xa6 4.Rc8! and white has prevented black from castling. So black can`t prove the last move is b7-b5 and therefor is already stalemated in the initial position! (2011-07-12)
A.Buchanan: Isn't Guus' idea just AP-Prioritat? And even if it were "more logical" or "more interesting", it doesn't follow that other forms is AP are "incorrect" (2023-07-31)
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mri: citeWeiß kann aber so spielen, daß Schwarz nicht rochieren kann, z.B. 1. ... Lxc7 2. bxc7
/cite
How does white prevent black from castling after 1. cxb6 e.p. Ba7xb6+?
E.g. 2.Kxb6 a1=R 3.a7 Rxa7, or 2.Ka4 Bd8xc7 3.a7 Bb8. (2005-09-22)
VL: 1.c5xb6ep(??) a7xb6+ 2.Kxb6 a1R 3.Kb7 R1xa6 4.Rc8! (R6a7#?? illegal).
This study is correct under the generally accepted understanding of AP
a la' N.Petrovic'. Antiform (looking possibly somewhat strange):
Black's unsuccessful try. (2005-10-03)
Guus Rol: Generally accepted, true. Generally acceptable, false. Freedom of interpretation ceases for a concept once its polar concept (a priori validation) has been defined. "Goal induced AP" however, might be a passable stipulation for this type of problem. To keep this place from turning into a discussion forum I will discontinue comments on this issue. (2005-10-03)
paul: Author intention is: If black still can castle, his last move must have been b7-b5. However, to prove this, he has to castle (A Posteriori condition). So: 1.cxb6 e.p. axb6+ 2.Kxb6 a1R! (2...a1Q? 3.Kb7 Qxa6 mate obviously doesn`t prove b7-b5 as the last move) 3.Kb7 R1xa6 4.Rc8! and white has prevented black from castling. So black can`t prove the last move is b7-b5 and therefor is already stalemated in the initial position! (2011-07-12)
A.Buchanan: Isn't Guus' idea just AP-Prioritat? And even if it were "more logical" or "more interesting", it doesn't follow that other forms is AP are "incorrect" (2023-07-31)
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Keywords: En passant as key, Castling (sg), a posteriori (AP)
Genre: Retro, Studies
FEN: r2bk3/p1Rpp3/P1p2p2/KpP2P2/1P2p3/1P6/p7/8
Reprints: (4) Problem 161-164 11/1973
317 Europe Echecs 217 01/1977
(A) Die Schwalbe 80 04/1983
408 Eigenartige Schachprobleme 2010
M36 mpk-Blätter 12/2011
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-31 more...
Genre: Retro, Studies
FEN: r2bk3/p1Rpp3/P1p2p2/KpP2P2/1P2p3/1P6/p7/8
Reprints: (4) Problem 161-164 11/1973
317 Europe Echecs 217 01/1977
(A) Die Schwalbe 80 04/1983
408 Eigenartige Schachprobleme 2010
M36 mpk-Blätter 12/2011
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-31 more...
9 - P0000759
Luis Alberto Garaza
1034 Die Schwalbe 21 06/1973
(9+9) cooked
Schwarz am Zug, Weiß gewinnt
Luis Alberto Garaza
1034 Die Schwalbe 21 06/1973
(9+9) cooked
Schwarz am Zug, Weiß gewinnt
1. ... fxg3ep 2. hxg3+ Kh5 3. f4 Kh6 4. h8=T+! Kg7
Cook: Intended a Posteriori combo of castling and ep rights retro-locks wTh in cage, contradicting pawn capture balance
Cook: Intended a Posteriori combo of castling and ep rights retro-locks wTh in cage, contradicting pawn capture balance
hans: only black move is fxg3e.p.
1. fxg3+ Kh5 2. h8D/T#
Why the keyword 'castling'?
Castling is illegal, for there are 7 black captures, and because white's last move is g2-g4, needs the wK let pass Rh1 (2012-05-18)
Anton Baumann: g2-g4 war nicht zwingend der letzte Zug! Schwarz ist patt, also unlösbar!
Korrektur in 'die Schwalbe' 12/2074 S.264: "wTa1 nach h1. Nun gewinnt nach 1. - fxg3 Weiss mit 2.hxg3+ Kh5 3.f4 Kh6 4.h8=T+! Kg7 und der umgewandelte Turm holt die störenden sBB ab, sodass Weiss rochieren kann und beweist, dass Schwarz nicht patt war. Schwarz versucht die Rochade zu verhindern, ist aber machtlos." (2022-12-06)
A.Buchanan: The forward play is more complicated, as Black can play more aggressively than 3. … Kh6 (2022-12-07)
Anton Baumann: Wenn Schwarz nicht 3.-Kh6! zieht, holt sich Weiss auf h8 die Dame, mit einfacherer Fortsetzung.
3.-g4? 4.f5 Kg5 (4.-g5 5.f6 ~ 6.h8=D) 5.h8=D Kxf5 6.De8 +_
3.-Kg4? 4.h8=D Kf3 5.Dxh3 g4 6.Dh2 Ke4 7.Dg2+ +_ (2022-12-09)
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1. fxg3+ Kh5 2. h8D/T#
Why the keyword 'castling'?
Castling is illegal, for there are 7 black captures, and because white's last move is g2-g4, needs the wK let pass Rh1 (2012-05-18)
Anton Baumann: g2-g4 war nicht zwingend der letzte Zug! Schwarz ist patt, also unlösbar!
Korrektur in 'die Schwalbe' 12/2074 S.264: "wTa1 nach h1. Nun gewinnt nach 1. - fxg3 Weiss mit 2.hxg3+ Kh5 3.f4 Kh6 4.h8=T+! Kg7 und der umgewandelte Turm holt die störenden sBB ab, sodass Weiss rochieren kann und beweist, dass Schwarz nicht patt war. Schwarz versucht die Rochade zu verhindern, ist aber machtlos." (2022-12-06)
A.Buchanan: The forward play is more complicated, as Black can play more aggressively than 3. … Kh6 (2022-12-07)
Anton Baumann: Wenn Schwarz nicht 3.-Kh6! zieht, holt sich Weiss auf h8 die Dame, mit einfacherer Fortsetzung.
3.-g4? 4.f5 Kg5 (4.-g5 5.f6 ~ 6.h8=D) 5.h8=D Kxf5 6.De8 +_
3.-Kg4? 4.h8=D Kf3 5.Dxh3 g4 6.Dh2 Ke4 7.Dg2+ +_ (2022-12-09)
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Keywords: Castling (wl), a posteriori (AP) (Type Petrovic), En passant as key
Genre: Retro, Studies
FEN: 8/7P/6p1/2p3p1/2p2pPk/2Pp1P1p/3PpP1P/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-07 more...
Genre: Retro, Studies
FEN: 8/7P/6p1/2p3p1/2p2pPk/2Pp1P1p/3PpP1P/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-07 more...
1. bxc3ep Sa6 2. 0-0-0 Tc4#
Henrik Juel: -1.c2 c5xLb4 -2.Lc3 c6 -3.Lb2 c7 -4.Lc1 a6! -5.b2 b2xDSa2 etc. White captured sDTSS by fxexd, gxfxe, allowing Black to capture f7xPe6xPd5 etc. (2003-12-18)
A.Buchanan: Why not e.g. wSb1? (2022-03-04)
Henrik Juel: That also seems to work (2022-03-04)
A.Buchanan: I really like the motivation for ep. I guess the motivation for Rb1 is to be inside the cage that forms around it, but it’s still a bit loose. Note there is no retro try. (2022-03-05)
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A.Buchanan: Why not e.g. wSb1? (2022-03-04)
Henrik Juel: That also seems to work (2022-03-04)
A.Buchanan: I really like the motivation for ep. I guess the motivation for Rb1 is to be inside the cage that forms around it, but it’s still a bit loose. Note there is no retro try. (2022-03-05)
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Keywords: En passant as key, Castling (sg), a posteriori (AP) (Type Petrovic), Volet Pawn
Genre: h#, Retro
Computer test: HC+ Popeye 4.87 + simple retro reflection
FEN: rN2k3/1p1pp1pp/8/p7/RpP5/PP6/p2PP3/KR6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-05 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.87 + simple retro reflection
FEN: rN2k3/1p1pp1pp/8/p7/RpP5/PP6/p2PP3/KR6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-05 more...
11 - P0000793
Nikita M. Plaksin
Andrey Lobusov
1558 Die Schwalbe 33 06/1975
4. Preis
(13+5) C+
#3 (AP)
Nikita M. Plaksin
Andrey Lobusov
1558 Die Schwalbe 33 06/1975
4. Preis
(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)
2. ... Kg4 3. g8=D,T#
2. ... a1=D,L 3. Dg5#
R: 1. d7-d5 Tc6xh6 (unless R: Kg1-g2 Th~*h1)
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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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...
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...
1. ... hxg6ep[+sBg7] 2. a1=T+ Txa1[+sTh8] 3. 0-0 Ta3 4. bxa3[+wTa1] 0-0-0 5. Td8 Txd8[+sTh8]#
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
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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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...
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...
1. bxc6ep e4 2. Se3 Kxg5 3. Kxd6
Henrik Juel: C+ Popeye 4.61 (the three threats never materialize)
It is obvious that last move was c7-c5 (2020-12-02)
comment
It is obvious that last move was c7-c5 (2020-12-02)
comment
Keywords: En passant as key
Genre: Retro
FEN: 4B1R1/3NP1Pp/1Q1p1Prr/RPpKpNPk/6p1/6P1/P2B4/8
Reprints: 41 Volksgemeinschaft (Heidelberg) 19/01/1936
252 Comoedia 21/06/1936
22 Europe Echecs 18 02/1960
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2020-12-02 more...
Genre: Retro
FEN: 4B1R1/3NP1Pp/1Q1p1Prr/RPpKpNPk/6p1/6P1/P2B4/8
Reprints: 41 Volksgemeinschaft (Heidelberg) 19/01/1936
252 Comoedia 21/06/1936
22 Europe Echecs 18 02/1960
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2020-12-02 more...
14 - P0001228
Wolfgang Hundsdorfer
Deutsches Wochenschach 1909
1. Preis
En-passant-Turnier 1910
(13+12)
#3
Wolfgang Hundsdorfer
Deutsches Wochenschach 1909
1. Preis
En-passant-Turnier 1910
(13+12)
#3
1. bxc6ep
R: 1. c7-c5 g5-g6 2. Tc6-c2 g4-g5 3. Tg6-c6 g3-g4 4. Tg8-g6 g2-g3 5. Ld4-a7 h4-h5 6. Lg7-d4 h3-h4 7. Lf8-g7 h2-h3 8. g7xLh6,g7xSh6
R: 1. c7-c5 g5-g6 2. Tc6-c2 g4-g5 3. Tg6-c6 g3-g4 4. Tg8-g6 g2-g3 5. Ld4-a7 h4-h5 6. Lg7-d4 h3-h4 7. Lf8-g7 h2-h3 8. g7xLh6,g7xSh6
Henrik Juel: minor dual in forward play
1.bxc6ep+ Db5 2.DLxb5+ Tc4 3.LDxc4#
1... b5,bxa6 2.Kxb4+ Tc3 3.Txc3# (2022-07-01)
Henrik Juel: Last move must be c7-c5, enabling Tc2 to reach g8 in just three retractions (2022-07-01)
comment
1.bxc6ep+ Db5 2.DLxb5+ Tc4 3.LDxc4#
1... b5,bxa6 2.Kxb4+ Tc3 3.Txc3# (2022-07-01)
Henrik Juel: Last move must be c7-c5, enabling Tc2 to reach g8 in just three retractions (2022-07-01)
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Keywords: En passant as key
Genre: Retro, 3#
FEN: 8/bp2p2p/B5Pp/RPp2q1P/Qp2P3/RK1kPP2/P1rppP2/8
Reprints: 139 Favorit-Schachaufgaben , p. 35, nach 1910
75 Retrograde Analysis 1915
101 Europe Echecs 53 05/1963
10 Die Schwalbe 276 12/2015
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-07-01 more...
Genre: Retro, 3#
FEN: 8/bp2p2p/B5Pp/RPp2q1P/Qp2P3/RK1kPP2/P1rppP2/8
Reprints: 139 Favorit-Schachaufgaben , p. 35, nach 1910
75 Retrograde Analysis 1915
101 Europe Echecs 53 05/1963
10 Die Schwalbe 276 12/2015
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-07-01 more...
1. gxf6ep! droht 2. 0-0#! (2.Kf2#?)
R: 1. f7-f5 f6xDe7,f6xTe7 etc
Wh caps: fxe7, hxg5
Bl caps: axbxc2, wBf1
If wK castling right remains, the only way to give White a prior move is to retract f7-f5, so ep would be on.
The key threatens (inter alia) 2. 0-0#/Kf2#, and both remain whatever Black plays, so the castling retro-justification for ep can never be disrupted.
R: 1. S~a3? a3-a4? blocks wT from retracting to home square if wK castling right remains.
R: 1. f7-f5 f6xDe7,f6xTe7 etc
Wh caps: fxe7, hxg5
Bl caps: axbxc2, wBf1
If wK castling right remains, the only way to give White a prior move is to retract f7-f5, so ep would be on.
The key threatens (inter alia) 2. 0-0#/Kf2#, and both remain whatever Black plays, so the castling retro-justification for ep can never be disrupted.
R: 1. S~a3? a3-a4? blocks wT from retracting to home square if wK castling right remains.
Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic)
Genre: Retro, 2#
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 4r3/1p2P1n1/4p2B/b1pp1pPb/P5pp/nPP5/Q1pPP1P1/N1k1K2R
Reprints: (50) Problem 144-147 12/1971
(38) Mat Plus 12 10-12/2009
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
Genre: Retro, 2#
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 4r3/1p2P1n1/4p2B/b1pp1pPb/P5pp/nPP5/Q1pPP1P1/N1k1K2R
Reprints: (50) Problem 144-147 12/1971
(38) Mat Plus 12 10-12/2009
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
1. cxd3ep? g3 2. Bd4 Lg2# (wLc1 retro-blocked)
1. c5! g3 2. cxd4 Lg2#
1. c5! g3 2. cxd4 Lg2#
James Malcom: Solution? (2020-12-29)
Hans-Jürgen Manthey: wohl beabsichtigt: 1. cxd3ep g3 2. Bd4 Lg2#
doch auch ein Dual: 1. c5 g3 2. cxd4 Lg2# (2020-12-29)
Mario Richter: rawbats says, that f2-f4 is White's only legal last move, and 1. c5 g3 2. cxd4 Lg2# the only solution, so 1. cxd3ep g3 2. d4 Lg2# might be a try, intended to fool the solvers ...
(notice that R: 1. d2-d4?? excludes wLc1 from the set of objects available for black pawn captures!) (2020-12-29)
James Malcom: Furthermore, the h pawn couldn't have moved last, as that rook is needed for the Black pawns. wPe6 takes all remaining captures of Black pieces, and bBf8 could never escape. So the d2 and h2 pawns did not move last, and gxf3ep does nothing, meaning that all en passants are not a solution. A very fine retro "joke." (2020-12-29)
James Malcom: This means that 1. c5 g3 2. cxd4 Bg2# is the only possible solution, and as such, this should be C+? (2020-12-29)
Hans-Jürgen Manthey: Stimmt Mario
habe übersehen das der Lc1 nicht gezogen haben kann bei 1. d2-d4; aber 6 schlagfälle von Schwarz nötig sind. (2020-12-30)
A.Buchanan: I think the composer here is exactly "trolling solvers' preoccupations for the lulz", to quote Hauke Reddmann in MatPlus today, which even my allegedly fine German skills are unable to translate. But Hauke is German, so maybe he could do it? (2020-12-30)
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Hans-Jürgen Manthey: wohl beabsichtigt: 1. cxd3ep g3 2. Bd4 Lg2#
doch auch ein Dual: 1. c5 g3 2. cxd4 Lg2# (2020-12-29)
Mario Richter: rawbats says, that f2-f4 is White's only legal last move, and 1. c5 g3 2. cxd4 Lg2# the only solution, so 1. cxd3ep g3 2. d4 Lg2# might be a try, intended to fool the solvers ...
(notice that R: 1. d2-d4?? excludes wLc1 from the set of objects available for black pawn captures!) (2020-12-29)
James Malcom: Furthermore, the h pawn couldn't have moved last, as that rook is needed for the Black pawns. wPe6 takes all remaining captures of Black pieces, and bBf8 could never escape. So the d2 and h2 pawns did not move last, and gxf3ep does nothing, meaning that all en passants are not a solution. A very fine retro "joke." (2020-12-29)
James Malcom: This means that 1. c5 g3 2. cxd4 Bg2# is the only possible solution, and as such, this should be C+? (2020-12-29)
Hans-Jürgen Manthey: Stimmt Mario
habe übersehen das der Lc1 nicht gezogen haben kann bei 1. d2-d4; aber 6 schlagfälle von Schwarz nötig sind. (2020-12-30)
A.Buchanan: I think the composer here is exactly "trolling solvers' preoccupations for the lulz", to quote Hauke Reddmann in MatPlus today, which even my allegedly fine German skills are unable to translate. But Hauke is German, so maybe he could do it? (2020-12-30)
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Keywords: En passant, En passant as key (Tries)
Genre: h#, Retro
Computer test: Popeye v4.85 + thinking
FEN: 8/2p1p1p1/4P3/3p1p2/2pPkPpP/4p3/1PP1P1P1/Kb3B2
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2021-01-06 more...
Genre: h#, Retro
Computer test: Popeye v4.85 + thinking
FEN: 8/2p1p1p1/4P3/3p1p2/2pPkPpP/4p3/1PP1P1P1/Kb3B2
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2021-01-06 more...
1. ... cxd6ep 2. 0-0-0 0-0-0 3. Kd7 Sa7#
1. De7 c6 2. Th8 c7 3. Tf8 Sd6#
White pawn caps: axb,dxe,gxf,hxg.
Black: fxg,bxc,cxb.
wPb4 came from b3 to release wBa3, so bPb3 captured to reach that square.
All pcs accounted for means bPd never captured.
In the set play, there are 13 retro tries in which one or both players do not castle. The intention is that both castling rights are needed in order to imply the pawn double hop.
1. De7 c6 2. Th8 c7 3. Tf8 Sd6#
White pawn caps: axb,dxe,gxf,hxg.
Black: fxg,bxc,cxb.
wPb4 came from b3 to release wBa3, so bPb3 captured to reach that square.
All pcs accounted for means bPd never captured.
In the set play, there are 13 retro tries in which one or both players do not castle. The intention is that both castling rights are needed in order to imply the pawn double hop.
A.Buchanan: White pawn caps: axb,dxe,gxf,hxg definite.
Black: fxg and two to resolve c-file. But that may be c&d cross-capture, so in set play last move might have been c6xd5. So I think this problem is cooked. What am I missing? (2022-03-21)
Mario Richter: If Black's last move was c6xd5, how do you get white Bishop a3 out of his cage? (In that case, black pawn b7 never left the b-file). (2022-03-21)
A.Buchanan: I agree Mario thanks (2022-03-21)
Hans-Jürgen Manthey: beide Rochaden und im Satz ep. ist möglich:
R.: 1. ... d7-d5 2. Sa7-b5 b5xTc4 3. Tg4-c4 Dd8-h4 4. c4-c5 Th8-h3 5. c2-c4 c4xDb3 6. Dd3-b3 h3-h2 7. Dd1-d3 h4-h3 8. Sc8-a7 h5-h4 9. Th4-g4 c5-c4 10. Th1-h4 h7-h5 11. h2xSg3 Sf5-g3 12. b3-b4 Sh6-f5 13. Lb4-a3 c7-c5 14. Ld2-b4 g3-g2 15. Lc1-d2 a3-a2 16. a2xSb3 Sc5-b3 17. Lb3-a4 Sa6-c5 18. Lc4-b3 Sb8-a6 19. d2xLe3 Lc5-e3 20. Ld3-c4 Lf8-c5 21. Lf5-d3 a4-a3 22. Lh3-f5 a5-a4 23. Lf1-h3 g4-g3 24. g2xLf3 Lb7-f3 25. Sb6-c8 Lc8-b7 26. Sa4-b6 a7-a5 27. Sc3-a4 b7-b5 28. Sb1-c3 f5xSg4 29. Se5-g4 Sg8-h6 30. Sf3-e5 e7-e6 31. Sg5-f3 f6-f5 32. Sh3-g5 f7-f6 33. Sg1-h3 (2023-02-23)
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Black: fxg and two to resolve c-file. But that may be c&d cross-capture, so in set play last move might have been c6xd5. So I think this problem is cooked. What am I missing? (2022-03-21)
Mario Richter: If Black's last move was c6xd5, how do you get white Bishop a3 out of his cage? (In that case, black pawn b7 never left the b-file). (2022-03-21)
A.Buchanan: I agree Mario thanks (2022-03-21)
Hans-Jürgen Manthey: beide Rochaden und im Satz ep. ist möglich:
R.: 1. ... d7-d5 2. Sa7-b5 b5xTc4 3. Tg4-c4 Dd8-h4 4. c4-c5 Th8-h3 5. c2-c4 c4xDb3 6. Dd3-b3 h3-h2 7. Dd1-d3 h4-h3 8. Sc8-a7 h5-h4 9. Th4-g4 c5-c4 10. Th1-h4 h7-h5 11. h2xSg3 Sf5-g3 12. b3-b4 Sh6-f5 13. Lb4-a3 c7-c5 14. Ld2-b4 g3-g2 15. Lc1-d2 a3-a2 16. a2xSb3 Sc5-b3 17. Lb3-a4 Sa6-c5 18. Lc4-b3 Sb8-a6 19. d2xLe3 Lc5-e3 20. Ld3-c4 Lf8-c5 21. Lf5-d3 a4-a3 22. Lh3-f5 a5-a4 23. Lf1-h3 g4-g3 24. g2xLf3 Lb7-f3 25. Sb6-c8 Lc8-b7 26. Sa4-b6 a7-a5 27. Sc3-a4 b7-b5 28. Sb1-c3 f5xSg4 29. Se5-g4 Sg8-h6 30. Sf3-e5 e7-e6 31. Sg5-f3 f6-f5 32. Sh3-g5 f7-f6 33. Sg1-h3 (2023-02-23)
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Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wgsg)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro-logic.
FEN: r3k3/6p1/4p3/1NPp4/BPp4q/Bp2PPPr/pP2PPpp/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro-logic.
FEN: r3k3/6p1/4p3/1NPp4/BPp4q/Bp2PPPr/pP2PPpp/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
1. hxg3ep! 0-0#! (Tf1#?)
Hans-Jürgen Manthey: 1. hxg3 O-O#(auch Tf1# !) mögliche zugfolge:
1. d2-d4 f7-f5 2. Lc1-d2 f5-f4 3. Ld2-c1 e7-e6 4. Lc1-d2 c7-c5 5. Ld2-c1 b7-b5
6. Lc1-d2 a7-a6 7. Ld2-c1 Lc8-b7 8. c2-c4 Ke8-f7 9. Lc1xf4 Sg8-f6 10. Lf4-h6 g7xh6
11. Dd1-d3 h6-h5 12. Dd3-d1 Lf8-h6 13. h2-h4 Lh6-g5 14. h4xLg5 h5-h4 15. Dd1-d3 h7-h5
16. Dd3-d1 b5xc4 17. a2-a4 Lb7-e4 18. d4-d5 e6xd5 19. Ta1-a3 Le4-c2 20. f2-f4 Th8-e8
21. Ta3-g3 Te8-e3 22. f4-f5 Sf6-e4 23. f5-f6 Kf7-e6 24. f6-f7 Se4-d2 25. Sg1-h3 Sd2xf1
26. Sh3-f2 Sf1-d2 27. Sf2-g4 Sd2-e4 28. Dd1-d3 Lc2-d1 29. Sg4-f6 c4-c3 30. Sf6-h7 c3-c2
31. b2-b4 c5-c4 32. b4-b5 Sb8-c6 33. b5-b6 Sc6-a5 34. b6-b7 c2-c1L 35. g5-g6 c4-c3
36. Sb1-a3 c3-c2 37. Sa3-b5 Ta8-c8 38. b7-b8L Tc8-c3 39. Lb8-d6 La5-c4 40. a4-a5 a6xSb5
41. a5-a6 Ke6-f5 42. a6-a7 Dd8-e8 43. a7-a8S De8-e5 44. Sa8-c7 Lc1-b2 45. Sc7-e6 Lb2-a1
46. Se6-f8 Kf5-f4 47. Tg3-g5 Se4-f6 48.Ld6-a3 Te3-h3 49. La3-b2 Sc4-a3 50. g2-g4 (2020-12-11)
Henrik Juel: The white castling is needed to prove (a posteriori) that last move was g2-g4, legitimizing the ep capture (2020-12-31)
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1. d2-d4 f7-f5 2. Lc1-d2 f5-f4 3. Ld2-c1 e7-e6 4. Lc1-d2 c7-c5 5. Ld2-c1 b7-b5
6. Lc1-d2 a7-a6 7. Ld2-c1 Lc8-b7 8. c2-c4 Ke8-f7 9. Lc1xf4 Sg8-f6 10. Lf4-h6 g7xh6
11. Dd1-d3 h6-h5 12. Dd3-d1 Lf8-h6 13. h2-h4 Lh6-g5 14. h4xLg5 h5-h4 15. Dd1-d3 h7-h5
16. Dd3-d1 b5xc4 17. a2-a4 Lb7-e4 18. d4-d5 e6xd5 19. Ta1-a3 Le4-c2 20. f2-f4 Th8-e8
21. Ta3-g3 Te8-e3 22. f4-f5 Sf6-e4 23. f5-f6 Kf7-e6 24. f6-f7 Se4-d2 25. Sg1-h3 Sd2xf1
26. Sh3-f2 Sf1-d2 27. Sf2-g4 Sd2-e4 28. Dd1-d3 Lc2-d1 29. Sg4-f6 c4-c3 30. Sf6-h7 c3-c2
31. b2-b4 c5-c4 32. b4-b5 Sb8-c6 33. b5-b6 Sc6-a5 34. b6-b7 c2-c1L 35. g5-g6 c4-c3
36. Sb1-a3 c3-c2 37. Sa3-b5 Ta8-c8 38. b7-b8L Tc8-c3 39. Lb8-d6 La5-c4 40. a4-a5 a6xSb5
41. a5-a6 Ke6-f5 42. a6-a7 Dd8-e8 43. a7-a8S De8-e5 44. Sa8-c7 Lc1-b2 45. Sc7-e6 Lb2-a1
46. Se6-f8 Kf5-f4 47. Tg3-g5 Se4-f6 48.Ld6-a3 Te3-h3 49. La3-b2 Sc4-a3 50. g2-g4 (2020-12-11)
Henrik Juel: The white castling is needed to prove (a posteriori) that last move was g2-g4, legitimizing the ep capture (2020-12-31)
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Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 5N2/3p1P1N/5nP1/1p1pq1Rp/5kPp/n1rQ3r/1Bp1P3/b2bK2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 5N2/3p1P1N/5nP1/1p1pq1Rp/5kPp/n1rQ3r/1Bp1P3/b2bK2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
1. ... bxa6ep! 2. Txe2 axb7 3. Td2 b8=D 4. De2 Db1#
1. Dc6? bxc6 2. Ld7 cxd7 3. Ke1,Kxe2 d8=D 4. Kf1 Dd1# too slow
R: 1. a7-a5 e3xLf4 2. Ld6-f4 d2xTe3 3. Ta3-e3 b4-b5 4. Ta6-a3 b3-b4 5. Tc6xBa6 a5-a6 6. Tc8-c6 a4-a5 7. Tg8-c8 a3-a4 8. Lf8-d6 a2-a3 9. e7xDf6,e7xTf6
The forward play gives shortest h# in 3.5 moves, *if* e.p. is on. Otherwise the shortest is a mildly dualized 4.0 moves. So our mission is clear: assuming that Black moved last, what was that move? The position looks very open - how can one ever know? There is a knot in the south-east corner, which will only be resolved by wQ/R visiting g1.
White pawns have captured dxexf & hxg, accounting for BSP. So wPa was waylaid while wPc promoted to B. Black pawns captured dxexfxgxh, exf as well as wBf1 captured at home and wPa. That's 7 units, but bPc must also have captured to allow wPc to promote (to B on c8). So all captured units are accounted for. bPc promoted (bPh7 came from h6, so is bPd). We can't undo any of the Black pawn captures now, but we can undo d2xe3xf4, to release 2 Black units.
We can unpromote bPc, but we can't undo its capture to release a White unit, until wBh3 has unpromoted. The only White unit we can get now is from e7xf6. So this means that the two Black units we can release must be B & R, so they can retreat to f8 & g8 respectively. The timing is very tight, and there is only one way to do it. The black rook must visit a6 to unwaylay wPa, which gives White 5 more tempi, just enough.
bPc promoted on b1 to R, so we would need to undo the cage to get that. Therefore all we can do is send wQg1, then Rf1-f2 f2-f3 etc. bPa must retract immediately tp a7, so that wPa when unwaylaid can make fully 4 unmoves. Black officers are arranged to give unique retro & forward play (although with minor dual for the try) with Sh8 not just retro dressing but ensuring unique retraction Tg8-c8 not Th8-c8.
1. Dc6? bxc6 2. Ld7 cxd7 3. Ke1,Kxe2 d8=D 4. Kf1 Dd1# too slow
R: 1. a7-a5 e3xLf4 2. Ld6-f4 d2xTe3 3. Ta3-e3 b4-b5 4. Ta6-a3 b3-b4 5. Tc6xBa6 a5-a6 6. Tc8-c6 a4-a5 7. Tg8-c8 a3-a4 8. Lf8-d6 a2-a3 9. e7xDf6,e7xTf6
The forward play gives shortest h# in 3.5 moves, *if* e.p. is on. Otherwise the shortest is a mildly dualized 4.0 moves. So our mission is clear: assuming that Black moved last, what was that move? The position looks very open - how can one ever know? There is a knot in the south-east corner, which will only be resolved by wQ/R visiting g1.
White pawns have captured dxexf & hxg, accounting for BSP. So wPa was waylaid while wPc promoted to B. Black pawns captured dxexfxgxh, exf as well as wBf1 captured at home and wPa. That's 7 units, but bPc must also have captured to allow wPc to promote (to B on c8). So all captured units are accounted for. bPc promoted (bPh7 came from h6, so is bPd). We can't undo any of the Black pawn captures now, but we can undo d2xe3xf4, to release 2 Black units.
We can unpromote bPc, but we can't undo its capture to release a White unit, until wBh3 has unpromoted. The only White unit we can get now is from e7xf6. So this means that the two Black units we can release must be B & R, so they can retreat to f8 & g8 respectively. The timing is very tight, and there is only one way to do it. The black rook must visit a6 to unwaylay wPa, which gives White 5 more tempi, just enough.
bPc promoted on b1 to R, so we would need to undo the cage to get that. Therefore all we can do is send wQg1, then Rf1-f2 f2-f3 etc. bPa must retract immediately tp a7, so that wPa when unwaylaid can make fully 4 unmoves. Black officers are arranged to give unique retro & forward play (although with minor dual for the try) with Sh8 not just retro dressing but ensuring unique retraction Tg8-c8 not Th8-c8.
A.Buchanan: Another surprising motivation for e.p. I love Sh8. Sorry for W3 dual in the try play else the stipulation could be h#4*. I can’t fix it but it’s hard to improve on MC. Very enjoyable (2021-10-23)
A.Buchanan: Non-standard material is where the diagram contains for a player more than 1 queen or bishop of a hue or more than 2 rooks or knights. Obtrusive material is standard material, but there is some cheap reason why the unit must be promoted, most commonly that a bishop's home square remains blockaded by pawns. These categories are disjoint: no piece is ever both. Many problems in PDB do not apply this consistently, but the distinction goes back a long way in chess problem history, and is discussed by Morse.
Honestly, I dislike the word "obtrusive" whose negativity (while perhaps valid in forward problems) is inappropriate for retros. One distinguished composer objected to this tag being assigned to one of his problems. Nevertheless the concept has some interest. Renaming is a perilous business, but I am looking for suggestions... :) (2021-10-24)
A.Buchanan: Another distinction that comes to mind between "non-standard material" and "obtrusive promotion" is that normally in the former, one can't immediately point to which of the non-standard pieces was promoted: it's just that there's too many; while for the latter, one can usually point to a specific unit immediately.
"Obvious" is a candidate replacement for "obtrusive", but this might commit a cardinal sin of trying to nail down a perfectly useful and inherently vague term. Both "obvious" & "obtrusive" begin with "ob" which is helpful. What do you think? (2021-10-24)
Henrik Juel: I am curious about the award; why did this problem with good forward play and excellent retro-play only obtain a Commendation?
Maybe the judge did not like that Lh3 obviously is obtrusive...
. (2021-10-24)
A.Buchanan: Yes, and the lovely P0001117 with a similar obtrusion only received 12th Lob! But there may have been other factors. I do observe that "non-standard material" is arguably a worse defect than "obtrusive material", but the term doesn't cause offence because it's objective and non-judgemental. "obtrusive" is more subjective and inherently pejorative. To have such terms in the glossary is to put curators in an invidious position. I think the concept has its place, but I would like to replace it with something less scornful. I'm up for "obvious". It's an easy and reversible change: let's do it and see if mobs of protesters form outside the gate :) (2021-10-24)
A.Buchanan: Have changed "obtrusive material" to the non-pejorative "obvious promotion". It may still be regarded as a defect. As a placeholder, I have also changed the unclear German "mit Umwandlungfigur(en)" to "augenscheinlich Umwandlungfigur". A native German speaker I'm sure will propose a better term. (2021-10-26)
A.Buchanan: “offensichtlich” it is thanks Mario (2021-10-26)
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A.Buchanan: Non-standard material is where the diagram contains for a player more than 1 queen or bishop of a hue or more than 2 rooks or knights. Obtrusive material is standard material, but there is some cheap reason why the unit must be promoted, most commonly that a bishop's home square remains blockaded by pawns. These categories are disjoint: no piece is ever both. Many problems in PDB do not apply this consistently, but the distinction goes back a long way in chess problem history, and is discussed by Morse.
Honestly, I dislike the word "obtrusive" whose negativity (while perhaps valid in forward problems) is inappropriate for retros. One distinguished composer objected to this tag being assigned to one of his problems. Nevertheless the concept has some interest. Renaming is a perilous business, but I am looking for suggestions... :) (2021-10-24)
A.Buchanan: Another distinction that comes to mind between "non-standard material" and "obtrusive promotion" is that normally in the former, one can't immediately point to which of the non-standard pieces was promoted: it's just that there's too many; while for the latter, one can usually point to a specific unit immediately.
"Obvious" is a candidate replacement for "obtrusive", but this might commit a cardinal sin of trying to nail down a perfectly useful and inherently vague term. Both "obvious" & "obtrusive" begin with "ob" which is helpful. What do you think? (2021-10-24)
Henrik Juel: I am curious about the award; why did this problem with good forward play and excellent retro-play only obtain a Commendation?
Maybe the judge did not like that Lh3 obviously is obtrusive...
. (2021-10-24)
A.Buchanan: Yes, and the lovely P0001117 with a similar obtrusion only received 12th Lob! But there may have been other factors. I do observe that "non-standard material" is arguably a worse defect than "obtrusive material", but the term doesn't cause offence because it's objective and non-judgemental. "obtrusive" is more subjective and inherently pejorative. To have such terms in the glossary is to put curators in an invidious position. I think the concept has its place, but I would like to replace it with something less scornful. I'm up for "obvious". It's an easy and reversible change: let's do it and see if mobs of protesters form outside the gate :) (2021-10-24)
A.Buchanan: Have changed "obtrusive material" to the non-pejorative "obvious promotion". It may still be regarded as a defect. As a placeholder, I have also changed the unclear German "mit Umwandlungfigur(en)" to "augenscheinlich Umwandlungfigur". A native German speaker I'm sure will propose a better term. (2021-10-26)
A.Buchanan: “offensichtlich” it is thanks Mario (2021-10-26)
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Keywords: En passant, Last Moves?, Obvious promotion (L), En passant as key, Promotion (D)
Genre: h#, Retro
FEN: 7n/1p3pp1/4bp1p/pP6/4qPP1/5PrB/4PrPp/3k3K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-24 more...
Genre: h#, Retro
FEN: 7n/1p3pp1/4bp1p/pP6/4qPP1/5PrB/4PrPp/3k3K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-24 more...
If Black can castle, e.p. is ok:
1. hxg6ep! 0-0 2. h7#
1. ... Tf8,Kf8 2. Td8#
Otherwise Black can't castle:
1. Ke6! ... 2. Td8#
1. hxg6ep! 0-0 2. h7#
1. ... Tf8,Kf8 2. Td8#
Otherwise Black can't castle:
1. Ke6! ... 2. Td8#
Keywords: Cant Castler, Partial Retro Analysis (PRA), Castling (sk), En passant as key
Genre: Retro, 2#
Computer test: HC+ Popeye 4.61 & trivial retro-logic
FEN: 4k2r/8/5B1P/3R1KpP/8/8/8/8
Reprints: (D17) feenschach 27 04/1975
RA64 diagrammes 27 05-06/1977
(1) Die Schwalbe 86 04/1984
Razem 34 23/08/1987
(I) diagrammes 15 07-09/1994
(10) Die Schwalbe 241 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-30 more...
Genre: Retro, 2#
Computer test: HC+ Popeye 4.61 & trivial retro-logic
FEN: 4k2r/8/5B1P/3R1KpP/8/8/8/8
Reprints: (D17) feenschach 27 04/1975
RA64 diagrammes 27 05-06/1977
(1) Die Schwalbe 86 04/1984
Razem 34 23/08/1987
(I) diagrammes 15 07-09/1994
(10) Die Schwalbe 241 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-30 more...
1. exd3ep Lxg4 2. f3 Le6#
Klären: Quelle = Schachmatt? - Felber, Volker: SCHACH ist korrekt, 6/1969, Seite 191 (2010-10-09)
Eliminierung von 4 Steinen zwischen sK und wT im h#2 (mit Hilfe von ep-Schlag)!
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Eliminierung von 4 Steinen zwischen sK und wT im h#2 (mit Hilfe von ep-Schlag)!
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Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 8/8/8/1pn4b/2kPpppR/2q1rB1p/1p4r1/3K4
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-01-09 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 8/8/8/1pn4b/2kPpppR/2q1rB1p/1p4r1/3K4
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-01-09 more...
Henrik Juel: If White may castle, last move was g7-g5 to avoid retrostalemate. 1.fxg6ep followed by 0-0 to legitimize the en passant capture. (2004-09-23)
A.Buchanan: In V&V Encyclopedia, which I admire more for its scope than for its precision, this problem is given to illustrate "Petrovic Theme". The definition given is: "PETROVIC THEME Also called 'retroproblem of Petrovic Type'. Mutual dependence of en passant capture and castling. By playing an en passant capture other retro elements of position are legalized (usually castling)."
The definition makes no reference to A Posteriori. I am trying to get my head around the text here, because the e.p. does not "legalize" castling, rather it mandates it. I also encountered Öffner for this castling/e.p. AP (but not in V&V), but that might be someone's confusion in that Types Öffner vs Keym exist in PRA. (2022-02-15)
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A.Buchanan: In V&V Encyclopedia, which I admire more for its scope than for its precision, this problem is given to illustrate "Petrovic Theme". The definition given is: "PETROVIC THEME Also called 'retroproblem of Petrovic Type'. Mutual dependence of en passant capture and castling. By playing an en passant capture other retro elements of position are legalized (usually castling)."
The definition makes no reference to A Posteriori. I am trying to get my head around the text here, because the e.p. does not "legalize" castling, rather it mandates it. I also encountered Öffner for this castling/e.p. AP (but not in V&V), but that might be someone's confusion in that Types Öffner vs Keym exist in PRA. (2022-02-15)
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Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic), En passant
Genre: Retro, Studies
FEN: b7/p4P2/2kbPp1p/3ppPp1/n3pp2/8/P1PPP2P/4K2R
Reprints: (2) Problem 141-143 08/1971
(77) Problem 144-147 12/1971
(C) Die Schwalbe 16 10/1972
Encyclopedia of Chess Problems 2012
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-15 more...
Genre: Retro, Studies
FEN: b7/p4P2/2kbPp1p/3ppPp1/n3pp2/8/P1PPP2P/4K2R
Reprints: (2) Problem 141-143 08/1971
(77) Problem 144-147 12/1971
(C) Die Schwalbe 16 10/1972
Encyclopedia of Chess Problems 2012
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-15 more...
1. exf3ep Lc2#
R: 1. f2-f4 f5xTe4 2. Tg4-e4 & e.g. f6-f5 3. Tg1-g4 f7-f6 4. Le4-h7 h5-h4 5. Lg2-e4 h6-h5 6. Lf1-g2 h7-h6 7. g2xh3
1.exf3e.p. Bc2# is the only possible solution, but this necessitates R: 1.f2-f4. Can we prove this?
(13+13) with 1+2 pawn captures. Bf8 captured at home, so to satisfy White appetite, the missing Black pawn (a or b) must have promoted via c2 on c1. Two more White units must be captured to allow this.
The kings cage can only be unlocked by retracting WPc2. But the clock is ticking as there are only 6 black moves which can be retracted.
The promoted piece was captured on e3 or h3. If either capture is undone, then a White bishop square is cut off, so WB must be replaced prior to this.
Now the order of the early moves is: WdP moves, WQB & WQR escape, BP promotes on c1 to X (capturing WR at some point), X captured by WP.
So the first White capture must be dxNe3 and the second White capture releases gxXh3. The second White capture releases WKB & WKR. WKR captured by original BfP.
The clock starts ticking with gxh3. Black has 6 pawn moves. WKB has 3 moves to reach h7. WR has 3 if it goes via d file, or 2 if it starts on g1 (in which case WfP or WQB must also move once). So certainly at least 6 White moves. Last move was therefore White (even if the stipulation didn't tell us), and it can only have been WfP coming from f3 or f4. If it had been coming from f3 it would have blocked WKB in its progress, so the last White move was indeed R: 1.f2-f4.
WKR did therefore move from g1-g4-e4, and R: 1. ... fxRe5 2. Rg4-e4. Prior to that, move order not unique, but counting still exact.
Note that WN loitering on b4, pretending to be part of the cage, is present on the board just to make up the numbers.
R: 1. f2-f4 f5xTe4 2. Tg4-e4 & e.g. f6-f5 3. Tg1-g4 f7-f6 4. Le4-h7 h5-h4 5. Lg2-e4 h6-h5 6. Lf1-g2 h7-h6 7. g2xh3
1.exf3e.p. Bc2# is the only possible solution, but this necessitates R: 1.f2-f4. Can we prove this?
(13+13) with 1+2 pawn captures. Bf8 captured at home, so to satisfy White appetite, the missing Black pawn (a or b) must have promoted via c2 on c1. Two more White units must be captured to allow this.
The kings cage can only be unlocked by retracting WPc2. But the clock is ticking as there are only 6 black moves which can be retracted.
The promoted piece was captured on e3 or h3. If either capture is undone, then a White bishop square is cut off, so WB must be replaced prior to this.
Now the order of the early moves is: WdP moves, WQB & WQR escape, BP promotes on c1 to X (capturing WR at some point), X captured by WP.
So the first White capture must be dxNe3 and the second White capture releases gxXh3. The second White capture releases WKB & WKR. WKR captured by original BfP.
The clock starts ticking with gxh3. Black has 6 pawn moves. WKB has 3 moves to reach h7. WR has 3 if it goes via d file, or 2 if it starts on g1 (in which case WfP or WQB must also move once). So certainly at least 6 White moves. Last move was therefore White (even if the stipulation didn't tell us), and it can only have been WfP coming from f3 or f4. If it had been coming from f3 it would have blocked WKB in its progress, so the last White move was indeed R: 1.f2-f4.
WKR did therefore move from g1-g4-e4, and R: 1. ... fxRe5 2. Rg4-e4. Prior to that, move order not unique, but counting still exact.
Note that WN loitering on b4, pretending to be part of the cage, is present on the board just to make up the numbers.
Jeliss: "Obstruction of passage square f3 to Bishop of same colour."
"Version 'Pittsburgh Leader' 08.06.1913"
Yoav Ben-Zvi: Appears as the first problem (D445) in the booklet on Dawson's RA problems by G.P. Jellis. The obstruction that occurs in the Try -1.Pf3-f4?, by WP of WB, is described as "obstruction of passage square". It is not considered by Dawson and his disciples to be a Retro opposition. Dawson's conception of RO was quite broad, it included cases where the interference was not by occupation of the target square, so the only valid reason that I can see to exclude this case is that the 2 pieces involved are both of the same color. Fabel's definition explicitly excludes "Monochrome RO". I conclude that it would be preferrable to interpret RO as a bi-chromatic interference. The keyword Retro opposition should be removed. (2018-04-07)
A.Buchanan: To my mind, RO involves some kind of parity-tempo issue between the sides, not just some kind of race-tempo. If it was just about "bi-chromatic interference", one might say that bPe4 blocks wBh7 from an immediate retreat, so it has to be wPf4 that retreats first, legitimizing the ep key. So I agree this is not RO. (2024-01-06)
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"Version 'Pittsburgh Leader' 08.06.1913"
Yoav Ben-Zvi: Appears as the first problem (D445) in the booklet on Dawson's RA problems by G.P. Jellis. The obstruction that occurs in the Try -1.Pf3-f4?, by WP of WB, is described as "obstruction of passage square". It is not considered by Dawson and his disciples to be a Retro opposition. Dawson's conception of RO was quite broad, it included cases where the interference was not by occupation of the target square, so the only valid reason that I can see to exclude this case is that the 2 pieces involved are both of the same color. Fabel's definition explicitly excludes "Monochrome RO". I conclude that it would be preferrable to interpret RO as a bi-chromatic interference. The keyword Retro opposition should be removed. (2018-04-07)
A.Buchanan: To my mind, RO involves some kind of parity-tempo issue between the sides, not just some kind of race-tempo. If it was just about "bi-chromatic interference", one might say that bPe4 blocks wBh7 from an immediate retreat, so it has to be wPf4 that retreats first, legitimizing the ep key. So I agree this is not RO. (2024-01-06)
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Keywords: Last Moves?, En passant as key
Genre: h#, Retro
FEN: nqb5/1rrpp1pB/KRp5/1p4B1/kN2pP1p/2P1P2P/PP2P2P/8
Reprints: D445 Retro-Opposition & Other Retro-Analytical Chess Problems 1989
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-06 more...
Genre: h#, Retro
FEN: nqb5/1rrpp1pB/KRp5/1p4B1/kN2pP1p/2P1P2P/PP2P2P/8
Reprints: D445 Retro-Opposition & Other Retro-Analytical Chess Problems 1989
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-06 more...
1. exf6ep+!
Version in der 'Aarsskrift' 1935 innerhalb eines Artikels von K. Hannemann "Et Tema Fra den Retrograde Analyse". Im Original wDg4 statt h3 und wBh3 statt h4.
Henrik Juel: 1.exf6ep+. Not -1... a6?, requiring 3 black captures on light squares (incl. orig. Lf1); but missing orig. Lc1 was dark-squared. (2004-03-08)
Henrik Juel: C+ Popeye 4.61 and analysis (2022-06-11)
Hans-Jürgen Manthey: mögliche Zugfolge:
1. f2-f4 d7-d5 2. c2-c3 d5-d4 3. c3xd4 c7-c5 4. h2-h4 c5-c4 5. b2-b4 Lc8-d7 6. a2-a4 Ld7-b5 7. a4xLb5 Sb8-a6 8. Th1-h3 Sa6-c5 9. d4xSc5 Sg8-f6 10. d2-d4 Sf6-e4 11. c5-c6 Se4-g3 12. Lc1-d2 Se4xLf1 13. Sb1-c3 Sf1-g3 14. Sc3-a4 Sg3-e4 15. c6-c7 Ke8-d7 18. Sa4-c5+ Kd7-d6 19. Sc5-a6 Se4-c5 20. b4xSc5+ Kd6-e6 21. Ld2-a5 b7-b6 22. Sa6-b8 b6xLa5 23. Th3-e3+ Ke6-f6 24. Te3-e5 Dd8-c8 25. Dd1-d3 Dc8-a6 26. Dd3-h3 Da6-b6 27. Ta1-a3 Db6-c6 28. Ta3-g3 Dc6-d6 29. Te5-d5 Dd6-e6 30. c7-c8L De6-d6 31. Lc8-f5 Dd6-c6 32. Sg1-f3 Dc6-b6 33. Sf3-e5 Db6-a6 34. Se5-g6 Da6-b6 35. Sg6xTh8 Db6-c6 36. Tg3-g6+ h7xg6 37. Ke1-f2 Dc6-d6 38. Kf2-g3 Dd6-e6 39. Kg3-g4 De6-d6 40. Lf5-c2 Dd6-c7 41. b5-b6 Dc7-e5 42. f4xDe5+ Kf6-e6 43. Kg4-g5+ f7-f5 und nun :
1. e5xf6ep+ Ke6xTd5 2. Dh3-d7# (2023-02-24)
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Henrik Juel: 1.exf6ep+. Not -1... a6?, requiring 3 black captures on light squares (incl. orig. Lf1); but missing orig. Lc1 was dark-squared. (2004-03-08)
Henrik Juel: C+ Popeye 4.61 and analysis (2022-06-11)
Hans-Jürgen Manthey: mögliche Zugfolge:
1. f2-f4 d7-d5 2. c2-c3 d5-d4 3. c3xd4 c7-c5 4. h2-h4 c5-c4 5. b2-b4 Lc8-d7 6. a2-a4 Ld7-b5 7. a4xLb5 Sb8-a6 8. Th1-h3 Sa6-c5 9. d4xSc5 Sg8-f6 10. d2-d4 Sf6-e4 11. c5-c6 Se4-g3 12. Lc1-d2 Se4xLf1 13. Sb1-c3 Sf1-g3 14. Sc3-a4 Sg3-e4 15. c6-c7 Ke8-d7 18. Sa4-c5+ Kd7-d6 19. Sc5-a6 Se4-c5 20. b4xSc5+ Kd6-e6 21. Ld2-a5 b7-b6 22. Sa6-b8 b6xLa5 23. Th3-e3+ Ke6-f6 24. Te3-e5 Dd8-c8 25. Dd1-d3 Dc8-a6 26. Dd3-h3 Da6-b6 27. Ta1-a3 Db6-c6 28. Ta3-g3 Dc6-d6 29. Te5-d5 Dd6-e6 30. c7-c8L De6-d6 31. Lc8-f5 Dd6-c6 32. Sg1-f3 Dc6-b6 33. Sf3-e5 Db6-a6 34. Se5-g6 Da6-b6 35. Sg6xTh8 Db6-c6 36. Tg3-g6+ h7xg6 37. Ke1-f2 Dc6-d6 38. Kf2-g3 Dd6-e6 39. Kg3-g4 De6-d6 40. Lf5-c2 Dd6-c7 41. b5-b6 Dc7-e5 42. f4xDe5+ Kf6-e6 43. Kg4-g5+ f7-f5 und nun :
1. e5xf6ep+ Ke6xTd5 2. Dh3-d7# (2023-02-24)
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Keywords: En passant as key
Genre: Retro
FEN: rN3b1N/p3p1p1/1P2k1p1/p1PRPpK1/2pP3P/7Q/2B1P1P1/8
Reprints: 58 Retrograde Analysis 1915
Aarsskrift DSK , p. 14, 1935
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-06-11 more...
Genre: Retro
FEN: rN3b1N/p3p1p1/1P2k1p1/p1PRPpK1/2pP3P/7Q/2B1P1P1/8
Reprints: 58 Retrograde Analysis 1915
Aarsskrift DSK , p. 14, 1935
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-06-11 more...
1. dxe3ep 2. fxg1=L 3. 0-0 Tg4#
1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt
1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt
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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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...
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...
1) 1. cxd3ep Ta5 2. Sf5 0-0 3. Sd4 Txe5#
2) 1. gxf3ep Th5 2. Sd5 0-0-0 3. Sf4 Txe5#
Cook: 707 cooks which do not begin with e.p., e.g. 1.Sg3-e2 Ra1-a2 2.Ke4-d3 Rh1-h3 3.e5-e4 Ra2-d2 #
What is the obscure intent behind "Einheitslösung!"? (= "One size fits all solution"?)
2) 1. gxf3ep Th5 2. Sd5 0-0-0 3. Sf4 Txe5#
Cook: 707 cooks which do not begin with e.p., e.g. 1.Sg3-e2 Ra1-a2 2.Ke4-d3 Rh1-h3 3.e5-e4 Ra2-d2 #
What is the obscure intent behind "Einheitslösung!"? (= "One size fits all solution"?)
Mario Richter: Im Nachdruck in problem 144-147 steht nur auf e5 ein sB (also keiner auf e3). Allerdings gibt es in allen drei Varianten (sBe5+e3, nur sBe5, nur sBe3) jede Menge NL, die ohne Rochade und e.p.-Key auskommen - am wenigsten dann, wenn nur auf e3 ein sB steht. (2010-10-10)
VL: Cf. P0002475. (2012-08-26)
A.Buchanan: Possibly "Einheitlösung!" is a request to unify retro implications derived across separate parts of the PRA solution. But cooked chessically irrespective of these philosophical issues, which still wait to be resolved 60 years on. (2022-02-16)
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VL: Cf. P0002475. (2012-08-26)
A.Buchanan: Possibly "Einheitlösung!" is a request to unify retro implications derived across separate parts of the PRA solution. But cooked chessically irrespective of these philosophical issues, which still wait to be resolved 60 years on. (2022-02-16)
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Keywords: a posteriori (AP) (Type Petrovic - ccee), En passant as key (2), Castling (wb), Symmetrical position, Symmetrical solution, Superseded by (P1401449,P1399112)
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/8/8/4p3/2pPkPp1/2n1p1n1/8/R3K2R
Reprints: (66) Problem 144, p. 52, 12/1971
Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/8/8/4p3/2pPkPp1/2n1p1n1/8/R3K2R
Reprints: (66) Problem 144, p. 52, 12/1971
Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
a) 1. Sg3 Kxh4 2. Le4 hxg3#
b) 1. hxg3ep Kh4 2. Lf5 hxg3#
b) 1. hxg3ep Kh4 2. Lf5 hxg3#
Henrik Juel: in a) last move could be Kg2-h3
in b) last move must be g2-g4 (2024-01-14)
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in b) last move must be g2-g4 (2024-01-14)
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Keywords: En passant as key, Zeroposition
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & trivial retro-logic
FEN: 8/7b/8/4pn2/4pkPp/4pp1K/7P/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-15 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & trivial retro-logic
FEN: 8/7b/8/4pn2/4pkPp/4pp1K/7P/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-15 more...
1. fxg3ep Lxg7 2. Lg4 Lf6#
A.Buchanan: Sg7 would be better than T, from both forward & retro perspectives I think? (2021-04-14)
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Keywords: En passant as key
Genre: h#, Retro
FEN: 7B/6rp/7K/8/5pPk/5b1r/8/8
Input: Gerd Wilts, 1995-06-03
Genre: h#, Retro
FEN: 7B/6rp/7K/8/5pPk/5b1r/8/8
Input: Gerd Wilts, 1995-06-03
1. axb3ep Texf1 2. b2 Kd2#
Keywords: En passant as key, En passant
Genre: h#, Retro
FEN: n7/2ppp1p1/1p6/B7/pP6/5PP1/p4pPP/rk1KRrRn
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2021-02-11 more...
Genre: h#, Retro
FEN: n7/2ppp1p1/1p6/B7/pP6/5PP1/p4pPP/rk1KRrRn
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2021-02-11 more...
1. ... axb6ep 2. 0-0-0 0-0-0 3. Td7 a8=D#
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)
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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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...
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...
a) 1. Sxb4 Sxc3 2. Ka5 Sxc4#
b) 1. cxb3ep Sc4 2. Sb4 Sxc3#
b) 1. cxb3ep Sc4 2. Sb4 Sxc3#
*) 1. ... Kxb3 2. Txd5 Txd5#
1) 1. bxc3ep Kxb3 2. Txd5 Lxc3#
1) 1. bxc3ep Kxb3 2. Txd5 Lxc3#
A.Buchanan: Attractive. It reminded me of P0003347, which I have just discovered is sound as set play. (Just omit the first move!) Here part of the interest, although the intermediate moves are the same, the mating move is different, but in both cases has to cover d4 & e5. (2020-12-09)
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Keywords: En passant as key
Genre: h#, Retro
Computer test: C+ Popeye 4.61
FEN: 8/8/1P6/BRrP4/KpPkp3/1r1nq3/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-12-09 more...
Genre: h#, Retro
Computer test: C+ Popeye 4.61
FEN: 8/8/1P6/BRrP4/KpPkp3/1r1nq3/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-12-09 more...
* 1. ... ... 2. fxe3ep Sge4 3. Txe4 Sf3#
1. ... e5 2. dxe5 Sde4 3. fxe4 Sf5#
1. ... e5 2. dxe5 Sde4 3. fxe4 Sf5#
Sally: Der letzte Zug war e2-e4! (2010-04-06)
A.Buchanan: Great harmony between the phases (2021-11-23)
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A.Buchanan: Great harmony between the phases (2021-11-23)
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Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 3.52 & trivial retro-logic
FEN: 8/7p/3p3p/2nq1ppP/p1PkPp1r/P1p3N1/n2N2Pb/1b1r1B1K
Reprints: (XX) Problem 37-40 09/1956
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-23 more...
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 3.52 & trivial retro-logic
FEN: 8/7p/3p3p/2nq1ppP/p1PkPp1r/P1p3N1/n2N2Pb/1b1r1B1K
Reprints: (XX) Problem 37-40 09/1956
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-23 more...
1. hxg3ep Sc1 2. gxf2 Se3#
Welches ist die Originalquelle? Oder wurde es zweimal als Urdruck gebracht?
vgl. P0003291
Henrik Juel: White pawns captured all four missing black men
Last black move was not b7xPa6 ([Ta8]) nor e7-e6 ([Lf8]), so it was Lb7-c8, which is possible only if last white move was g2-g4
C+ Popeye 4.61 and above analysis (2021-04-23)
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vgl. P0003291
Henrik Juel: White pawns captured all four missing black men
Last black move was not b7xPa6 ([Ta8]) nor e7-e6 ([Lf8]), so it was Lb7-c8, which is possible only if last white move was g2-g4
C+ Popeye 4.61 and above analysis (2021-04-23)
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Keywords: En passant as key
Genre: h#, Retro
FEN: 2b5/p2p2pp/p3p2b/6Pr/5BPp/2P5/NRP1PP1P/3Nrk1K
Reprints: F84 The Problemist 05/1970
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-04-23 more...
Genre: h#, Retro
FEN: 2b5/p2p2pp/p3p2b/6Pr/5BPp/2P5/NRP1PP1P/3Nrk1K
Reprints: F84 The Problemist 05/1970
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-04-23 more...
1. ... bxc6ep 2. b5 Txa3#
R: 1. ... c7-c5 2. f4xTg5 Tg6-g5 3. f3-f4 Ta6-g6 4. f2-f3 Sc6-a7 5. h4-h5 Ta8-a6 6. h3-h4 a6-a5 7. h2-h3 Ka5-a4 8. b3-b4 Ta4-a3
or R: 1. f4xTg5 Tg6-g5 2. f3-f4 Ta6-g6 3. f2-f3 Sc6-a7 4. h4-h5 Ta8-a6 5. h3-h4 a6-a5 6. h2-h3 Ka5-a4 7. b3-b4 Ta4-a3
but we are told that White has the move.
R: 1. ... c7-c5 2. f4xTg5 Tg6-g5 3. f3-f4 Ta6-g6 4. f2-f3 Sc6-a7 5. h4-h5 Ta8-a6 6. h3-h4 a6-a5 7. h2-h3 Ka5-a4 8. b3-b4 Ta4-a3
or R: 1. f4xTg5 Tg6-g5 2. f3-f4 Ta6-g6 3. f2-f3 Sc6-a7 4. h4-h5 Ta8-a6 5. h3-h4 a6-a5 6. h2-h3 Ka5-a4 7. b3-b4 Ta4-a3
but we are told that White has the move.
Henrik Juel: Stipulation should probably be interpreted to mean h#1.5 . -1... c7 -2.f4xTg5 Tg6 -3.f3 Ta6 -4.f2 Sc6 -5.h4 Ta8 -6.h3 a6 -7.h2 Ka5 -8.b3 Ta4 etc. (2004-03-18)
A.Buchanan: Idea I suppose is that Black just moved, and moreover it was pawn double hop. I don't see clearly far enough back to confirm that. (2021-10-21)
Mario Richter: If Black moved last, the only legal retraction is R: 1. ... c7-c5! But the position can also be resolved if White moved last, starting with R: 1. f4xTg5 Tg6-g5 2. f3-f4 Ta6-g6 3. f2-f3 Sc6-a7 4. h4-h5 Ta8-a6 5. h3-h4 a6-a5 6. h2-h3 Ka5-a4 7. b3-b4 Ta4-a3.
Therefore the keyword "No legal last move for White" doesn't apply here; and to make the solution work, the Stipulation should be something like "h#2 0.1.1." or "h#1.5". (2021-10-21)
Henrik Juel: HC+ Popeye 4.61 (2021-10-22)
A.Buchanan: To my mind, the retro thinking is serious enough that we can’t claim HC+. If a tool like Replicator 2.0 was to validate the retro status I’d be ok. But I think it would be misleading to put HC+ to indicate the trivial h#1.5 is sound. We can certainly mark that Popeye has checked the h#, but not click the C+ flag. What do you think? (2021-10-22)
Henrik Juel: HC+ is not recognized (yet) by PDB, and I suppose it just meant that some sort of human analysis is needed in addition to a standard forward solver like Popeye
You are suggesting that we distinguish between easy and difficult human analysis and say HC+ for easy analysis only
I suggest saying HC+ for all problems where Popeye is not sufficient; you can write easy or difficult in the C+ comment
If and when a human analyst proves the analysis wrong, we delete the HC+ label, of course (2021-10-22)
A.Buchanan: What does Mario think? He invented the term, I seem to remember (2021-10-22)
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A.Buchanan: Idea I suppose is that Black just moved, and moreover it was pawn double hop. I don't see clearly far enough back to confirm that. (2021-10-21)
Mario Richter: If Black moved last, the only legal retraction is R: 1. ... c7-c5! But the position can also be resolved if White moved last, starting with R: 1. f4xTg5 Tg6-g5 2. f3-f4 Ta6-g6 3. f2-f3 Sc6-a7 4. h4-h5 Ta8-a6 5. h3-h4 a6-a5 6. h2-h3 Ka5-a4 7. b3-b4 Ta4-a3.
Therefore the keyword "No legal last move for White" doesn't apply here; and to make the solution work, the Stipulation should be something like "h#2 0.1.1." or "h#1.5". (2021-10-21)
Henrik Juel: HC+ Popeye 4.61 (2021-10-22)
A.Buchanan: To my mind, the retro thinking is serious enough that we can’t claim HC+. If a tool like Replicator 2.0 was to validate the retro status I’d be ok. But I think it would be misleading to put HC+ to indicate the trivial h#1.5 is sound. We can certainly mark that Popeye has checked the h#, but not click the C+ flag. What do you think? (2021-10-22)
Henrik Juel: HC+ is not recognized (yet) by PDB, and I suppose it just meant that some sort of human analysis is needed in addition to a standard forward solver like Popeye
You are suggesting that we distinguish between easy and difficult human analysis and say HC+ for easy analysis only
I suggest saying HC+ for all problems where Popeye is not sufficient; you can write easy or difficult in the C+ comment
If and when a human analyst proves the analysis wrong, we delete the HC+ label, of course (2021-10-22)
A.Buchanan: What does Mario think? He invented the term, I seem to remember (2021-10-22)
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Keywords: En passant as key
Genre: h#, Retro
FEN: 2b5/np1p2pp/7q/pPp3PP/kP6/r1PPP3/RKpp2P1/BB2n3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-21 more...
Genre: h#, Retro
FEN: 2b5/np1p2pp/7q/pPp3PP/kP6/r1PPP3/RKpp2P1/BB2n3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-21 more...
a) 1. Sxg4 f3+ 2. Kh5 fxg4#
b) 1. fxg3ep fxg3 2. Kh5 g4#
b) 1. fxg3ep fxg3 2. Kh5 g4#
Keywords: En passant as key
Genre: h#, Retro
FEN: 8/4pp2/7p/5K2/5pPk/4r2r/3n1P1n/4B3
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2021-05-04 more...
Genre: h#, Retro
FEN: 8/4pp2/7p/5K2/5pPk/4r2r/3n1P1n/4B3
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2021-05-04 more...
1. bxc3ep Lxe2 2. Sa3 0-0-0#
Cook: 2. ... Td1#
Cook: 2. ... Td1#
Sally: Der letzte Zug war: Bc2 - c4!
Nr.138 200 Ausgewählte S .Probleme T. Kardos(W. Fentze1983) (2010-09-30)
A.Buchanan: All 9+1 captures are visible by pawns. Note R: 1. Kd1-e1 d3xe2 is illegal because it costs 2 more captures for Black. So e.p. is unconditionally legal, requiring no AP justification. I see no reason why White has lost castling rights. Indeed someone might make a demo game ending with c2-c4. So this one seems cooked. Neither WinChloe nor yacpdb contains this problem. (2022-01-08)
A.Buchanan: If this is AP, it will be rendered sound by removing wPa2. But is this the intention? Does anyone have access to the collection of selected Kardos problems mentioned? (2024-01-25)
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Nr.138 200 Ausgewählte S .Probleme T. Kardos(W. Fentze1983) (2010-09-30)
A.Buchanan: All 9+1 captures are visible by pawns. Note R: 1. Kd1-e1 d3xe2 is illegal because it costs 2 more captures for Black. So e.p. is unconditionally legal, requiring no AP justification. I see no reason why White has lost castling rights. Indeed someone might make a demo game ending with c2-c4. So this one seems cooked. Neither WinChloe nor yacpdb contains this problem. (2022-01-08)
A.Buchanan: If this is AP, it will be rendered sound by removing wPa2. But is this the intention? Does anyone have access to the collection of selected Kardos problems mentioned? (2024-01-25)
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1. cxd3ep? bxc3 2. Lxf4 Lxg2# Last move not d2-d4
1. gxf3ep! bxc3 2. Lf4 Lf5# Last move f2-f4
White's last move was certainly one of the two double pawn hops. The last move couldn't be d2-d4 because Black's pawns had already captured 10 of White's pieces, and playing d2-d4 would have resulted in wBc1 being White's 17th piece. So, the only solution is 1.gxf3 e.p.
1. gxf3ep! bxc3 2. Lf4 Lf5# Last move f2-f4
White's last move was certainly one of the two double pawn hops. The last move couldn't be d2-d4 because Black's pawns had already captured 10 of White's pieces, and playing d2-d4 would have resulted in wBc1 being White's 17th piece. So, the only solution is 1.gxf3 e.p.
Viktoras Paliulionis: The last move couldn't be d2-d4 because Black's pawns had already captured 10 of White's pieces, and playing d2-d4 would have resulted in wBc1 being White's 17th piece. So, the only solution is 1.gxf3 e.p. (2023-12-30)
A.Buchanan: Yes that's right! (2023-12-31)
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A.Buchanan: Yes that's right! (2023-12-31)
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Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 4.87 & trivial retro-logic
FEN: 2n1q3/5prp/5K1b/3b4/2pPkPp1/2p1p1pB/1P4pP/8
Reprints: 546 FIDE Album 1962-1964 1968
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-31 more...
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 4.87 & trivial retro-logic
FEN: 2n1q3/5prp/5K1b/3b4/2pPkPp1/2p1p1pB/1P4pP/8
Reprints: 546 FIDE Album 1962-1964 1968
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-31 more...
WTM
1. ... bxc6ep 2. Da7 c7#
retro tries:
1) 1. Lxg3 b6 2. Lxd6 Lxd6#
2) 1. Dc6 b6 2. Dc7 dxc7#
3) 1. Ta7 f8=D,f8=T 2. Ka8 Dxc8#,Txd8# ("untolerated" dual)
R: 1. ... c7-c5 2. b4-b5 Dc4-a6 (move order variable beyond this point) 3. b2-b3 Ta6-a8 4. d5-d6 Te6-a6 5. b3-b4 Te8-e6 6. a2xSb3 Th8-e8 7. d4-d5 Lb4-e1 8. d3-d4 Lf8-b4 9. Kg5-h5 e7xSf6+
Missing: Wh: SS Bl: SS
Pawn captures: Wh: axb, hxg Bl: exf, fxg
bPf6 comes from e7, else collision with wPf7. So can't uncapture fxg2 until wP retracted. White has 6 pawn retractions before retro-pat, just giving Bl time to put Le1 & Ta8 (impostor) back home, and uncapture exf6. So Black must retract first, and it's WTM. sD must clear out of the way, and c4 is the only possible square, wPaxSb can't happen on b5, or it would block sD. And axSb4 would block sLe1. It was axSb3. So wPb5 must retreat to b4, and thus Black cannot play R: 1. Lb4-e1. So neatly R: 1. c7-c5 is forced.
1. ... bxc6ep 2. Da7 c7#
retro tries:
1) 1. Lxg3 b6 2. Lxd6 Lxd6#
2) 1. Dc6 b6 2. Dc7 dxc7#
3) 1. Ta7 f8=D,f8=T 2. Ka8 Dxc8#,Txd8# ("untolerated" dual)
R: 1. ... c7-c5 2. b4-b5 Dc4-a6 (move order variable beyond this point) 3. b2-b3 Ta6-a8 4. d5-d6 Te6-a6 5. b3-b4 Te8-e6 6. a2xSb3 Th8-e8 7. d4-d5 Lb4-e1 8. d3-d4 Lf8-b4 9. Kg5-h5 e7xSf6+
Missing: Wh: SS Bl: SS
Pawn captures: Wh: axb, hxg Bl: exf, fxg
bPf6 comes from e7, else collision with wPf7. So can't uncapture fxg2 until wP retracted. White has 6 pawn retractions before retro-pat, just giving Bl time to put Le1 & Ta8 (impostor) back home, and uncapture exf6. So Black must retract first, and it's WTM. sD must clear out of the way, and c4 is the only possible square, wPaxSb can't happen on b5, or it would block sD. And axSb4 would block sLe1. It was axSb3. So wPb5 must retreat to b4, and thus Black cannot play R: 1. Lb4-e1. So neatly R: 1. c7-c5 is forced.
Ladislav Packa: Is wRh1 needed? (2021-10-22)
Mario Richter: Yes, wRh1 is needed - without it the ep-key would not be justified (last moves could have been R: 1. ... Th1-g1 2. Lg1-h2 or even R: 1. ... Ld2xTe1). (2021-10-23)
A.Buchanan: + three mostly clean retro tries. (2021-10-24)
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Mario Richter: Yes, wRh1 is needed - without it the ep-key would not be justified (last moves could have been R: 1. ... Th1-g1 2. Lg1-h2 or even R: 1. ... Ld2xTe1). (2021-10-23)
A.Buchanan: + three mostly clean retro tries. (2021-10-24)
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Keywords: En passant as key, No legal last move for White, Impostor (t)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: rkb5/1p1p1Ppp/q2P1p2/pPp4K/6PR/1P4PQ/2P1P1pB/4bBrR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: rkb5/1p1p1Ppp/q2P1p2/pPp4K/6PR/1P4PQ/2P1P1pB/4bBrR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
1. bxc3ep Sxg2 2. cxd2 Sxh4 3. Ke3 Kg3 4. Te4 Sf5#
A.Buchanan: So no twinning in the original? What's the point of sDb3? It can be replaced by sB without issue. This problem doesn't appear in WinChloe, so can't check (2021-02-12)
VL: The author could, e.g., prefer the position with black Queen as richer and more hidden for solution. (2021-02-12)
A.Buchanan: Hi Valery well dressing the board is a hypothesis. There is no try to compensate for loss of economy. Indeed here sD forms a distracting battery which “pins” sLd3. It’s all too easy to add this kind of “richness” (noise) to a problem but more is less in my humble opinion. Unless it was a thematic tourney or we are missing a twin, my top hypothesis is that the composer did very well without a computer to make a sound h#4, and it’s easy these days to see a minor improvement. (2021-02-13)
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VL: The author could, e.g., prefer the position with black Queen as richer and more hidden for solution. (2021-02-12)
A.Buchanan: Hi Valery well dressing the board is a hypothesis. There is no try to compensate for loss of economy. Indeed here sD forms a distracting battery which “pins” sLd3. It’s all too easy to add this kind of “richness” (noise) to a problem but more is less in my humble opinion. Unless it was a thematic tourney or we are missing a twin, my top hypothesis is that the composer did very well without a computer to make a sound h#4, and it’s easy these days to see a minor improvement. (2021-02-13)
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Keywords: En passant as key, En passant
Genre: h#, Retro
Computer test: C+ Gustav 4.1d
FEN: 8/8/8/8/rpPk3p/1q1b3K/3Pp1rn/4N3
Reprints: 49 PCFT 1944-1945
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-02-13 more...
Genre: h#, Retro
Computer test: C+ Gustav 4.1d
FEN: 8/8/8/8/rpPk3p/1q1b3K/3Pp1rn/4N3
Reprints: 49 PCFT 1944-1945
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-02-13 more...
41 - P0003206
Hans Joachim Schlüter
4443v Schach , p. 349, 11/1963
(5+8)
h#2
b) wTf1 tauschen mit wLg1
Hans Joachim Schlüter
4443v Schach , p. 349, 11/1963
(5+8)
h#2
b) wTf1 tauschen mit wLg1
a) 1. Kxb4 Lb6 2. a4 Tb1#
b) 1. cxb3ep gxf3 2. Sc1 Tg4#
b) 1. cxb3ep gxf3 2. Sc1 Tg4#
Keywords: En passant as key
Genre: h#, Retro
FEN: 8/8/8/p7/kPp5/p1p2p2/4n1Pp/5RBK
Input: Gerd Wilts, 1995-06-03
Last update: Felber, Volker, 2022-11-24 more...
Genre: h#, Retro
FEN: 8/8/8/p7/kPp5/p1p2p2/4n1Pp/5RBK
Input: Gerd Wilts, 1995-06-03
Last update: Felber, Volker, 2022-11-24 more...
1. exf3ep c3 2. Df7 Ke4 3. Dh5 exf3#
A.Buchanan: Can shift bQd5 to c4, while replacing bRg5 with bP to reach Meredith status. bBh4 can also be downgraded to bP. But probably even more economy is possible by shifting pieces one file to the right, keeping the ep, White tempo & model mate, which seem to be the main features (2022-11-30)
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Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 8/8/8/3q2r1/2p1pPkb/4K1pp/2PrP3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-30 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 8/8/8/3q2r1/2p1pPkb/4K1pp/2PrP3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-30 more...
1. dxe3ep+ Kxc4 2. Lc3 Sxc2 3. Da5 axb3#
See the correction P1396159
Cook: 1. dxe3ep+ Kxc4 2. Td5 Sxc2 3. Ta5 axb3#
See the correction P1396159
Cook: 1. dxe3ep+ Kxc4 2. Td5 Sxc2 3. Ta5 axb3#
A.Buchanan: Easy enough to eliminate what I suppose is the intended solution (sL unblocking sD and blocking sT) by e.g. just removing sDf5. However I don't see a way to say goodbye to the cook instead. (2021-11-23)
James Malcom: I believe I may have a fix Andrew. If we substitute the Black queen with a Black rook, the same theme is kept even is if considered watered down.
h#3 8/3p4/3p1p2/3Kbr2/k1ppPp2/1n1r3P/P1p4P/N3nq1b (2021-11-24)
A.Buchanan: Hi James, hurray cool I thought I couldn't get through that way: I was confused in my mind about wPe2 blocking that line, but with the freshness of a new day, I can look at it again. The precise diagram that you propose doesn't quite work because R: 1. e3-e4 Sf3*e1+,Sg2*e1+,g2xf1=Q+,g2xh1=Q+, but this is all fixable with +bPd2,-bSe1. Now Popeye 4.87 says the forward solution is works uniquely, and Retractor 2.0 says R: 1. e2-e4 Rf3-d3+ 2. Ke4-d5 are unique. So for full C+ we only need some demo game that results in the position prior to Ke4-d5.
I don't see any "watering down" here: I think we have faithfully represented Tivadar's idea. Let's post it here as TK, corr JM&AB? (2021-11-24)
A.Buchanan: I have made an unedifying demo game with lichess, so our correction is fully C+. (2021-11-24)
James Malcom: "TK, corr JM&AB" sounds good to me. (2021-11-24)
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James Malcom: I believe I may have a fix Andrew. If we substitute the Black queen with a Black rook, the same theme is kept even is if considered watered down.
h#3 8/3p4/3p1p2/3Kbr2/k1ppPp2/1n1r3P/P1p4P/N3nq1b (2021-11-24)
A.Buchanan: Hi James, hurray cool I thought I couldn't get through that way: I was confused in my mind about wPe2 blocking that line, but with the freshness of a new day, I can look at it again. The precise diagram that you propose doesn't quite work because R: 1. e3-e4 Sf3*e1+,Sg2*e1+,g2xf1=Q+,g2xh1=Q+, but this is all fixable with +bPd2,-bSe1. Now Popeye 4.87 says the forward solution is works uniquely, and Retractor 2.0 says R: 1. e2-e4 Rf3-d3+ 2. Ke4-d5 are unique. So for full C+ we only need some demo game that results in the position prior to Ke4-d5.
I don't see any "watering down" here: I think we have faithfully represented Tivadar's idea. Let's post it here as TK, corr JM&AB? (2021-11-24)
A.Buchanan: I have made an unedifying demo game with lichess, so our correction is fully C+. (2021-11-24)
James Malcom: "TK, corr JM&AB" sounds good to me. (2021-11-24)
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Keywords: En passant as key, Superseded by (P1396159)
Genre: h#, Retro
Computer test: Popeye C-Version 4.87 & trivial retro-logic
FEN: 8/3p4/3p1p2/3KbqP1/k1ppPpp1/1n1r3P/P1p4P/N6b
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-24 more...
Genre: h#, Retro
Computer test: Popeye C-Version 4.87 & trivial retro-logic
FEN: 8/3p4/3p1p2/3KbqP1/k1ppPpp1/1n1r3P/P1p4P/N6b
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-24 more...
1. ... bxc6ep 2. Le7 cxd7#
1. Lc6? bxc6 2. Le7 cxd7# WTM
R: 1. c6-c5? Kd3-e3 2. Dc5-c4+ Ke3-d3 3. Dc4-c5+ retro-loop.
R: 1. c7-c5! Kd3-e3 2. Dc6-c4+ Ke3-d3! clear
Easy to see no last move for White. Moreover, Black just double-hopped to allow sD/wK to avoid retro-loop.
Single clean retro try.
1. Lc6? bxc6 2. Le7 cxd7# WTM
R: 1. c6-c5? Kd3-e3 2. Dc5-c4+ Ke3-d3 3. Dc4-c5+ retro-loop.
R: 1. c7-c5! Kd3-e3 2. Dc6-c4+ Ke3-d3! clear
Easy to see no last move for White. Moreover, Black just double-hopped to allow sD/wK to avoid retro-loop.
Single clean retro try.
Keywords: En passant as key, No legal last move for White
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & moderate retro-logic
FEN: 3nkr2/3p1r2/1p1npb2/PPpb4/Bpq2PP1/1p2Kpp1/2PPP3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & moderate retro-logic
FEN: 3nkr2/3p1r2/1p1npb2/PPpb4/Bpq2PP1/1p2Kpp1/2PPP3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
1. ... bxc6ep 2. Dxa5 Db7#
1. Sc6? bxc6 2. Dxa5 Db7# but WTM
R: 1. ... c7-c5 2. g5-g6 Sc6-e7 3. De7-b4,D~ Sb4-c6+ unpromote wD on h8, etc.
Clean retro try
1. Sc6? bxc6 2. Dxa5 Db7# but WTM
R: 1. ... c7-c5 2. g5-g6 Sc6-e7 3. De7-b4,D~ Sb4-c6+ unpromote wD on h8, etc.
Clean retro try
Henrik Juel: Typo: 2.Dxa5 Db7#. wTa1 is missing, I think (10+15). -1... c7 -2.g5 Sc6 -3.De7 Sb4, unpromote wD on h8, etc. (2004-03-18)
A.Buchanan: I agree Henrik and have made changes. A number of these problems have been marked "Weisz zieht an", which is inappropriate as part of the point of the problem is to deduce this. I think it's 2. c2-c3! not 2. g5-g6? or the cross-captured pawns become retro-locked. (2021-10-21)
Henrik Juel: 2.g5-g6 also works, because the only cross-capture is axb,bxa
White captured [Pd7] with an officer (2021-10-22)
Henrik Juel: HC+ Popeye 4.61 (2021-10-22)
A.Buchanan: If g5-g6 also works, then can’t we have both? I.e. White moved last. I would be surprised if this was TRD’s intention, given the others of this general style (2021-10-22)
Henrik Juel: TRD died in 1951, so someone else may have messed up the problem
FRC is not generally available after 1953, so I cannot check the source (2021-10-22)
A.Buchanan: WinChloe has 29(!) posthumous problems by Dawson - the latest in 1958, FCR's final year. The composition in question is not one of them. (2021-10-23)
A.Buchanan: PROOF OF COOK + SUGGESTED FIX
I agree with much of what you say Henrik.
Capturing history. Basically 3 possibilities for the Black pawns.
Notation: ~,o,| denote "cross-capture", "original" & "waylaid"(=captured on home file by officer).
1) a~b co d~e f~g. This implies either (wPhxg & bPh=X) or (wPh=X & bPh|)
2) a~b co d~e hxgxf f|
3) a~b co hxgxfxexd d|
Suppose WTM. Then R: 1. c6-c5 c2-c3 2. c7-c6 g5-g6 3. Sc6-e7 De7-b4 4. Sb4-c6+ Kc5-c4 5. Lb2-d4+ etc. Now we're free to uncapture wDh8 (or g8 for scenario 1), and any of the three pawn scenarios can work. No e.p. rights means a cook.
If we fix it by e.g. shifting wPg6-g5 then I think that fixes this cook, and also it means that there is no solution if BTM.
Is there any possibility for premature capturing hxg? Not by Bl because scenario 1 says bPfxg, and the other two scenarios rely on wPh=X. If by Wh, then there's no waylaying on d or f, so we are back to scenario 1, but bPh is blocked from promotion, so that won't work.
But: what about R: 1. c6-c5 c2-c3 2. Sc3-b1 Th1-a1. Doesn't this completely cook it anyway?? How about k7/4n3/6P1/PPpPPP2/qQKbrrp1/ppRppp2/b1P5/1n6? h#2 (with Art 15 so it's really h#1.5)
Do you agree? (2021-10-23)
Mario Richter: In the diagram, both for WTM and BTM the retraction sequence c2-c3 Sc6-e7 De1-b4 Sb4-c6+ works and furthermore Andrew's R: 1. c6-c5 c2-c3 2. Sc3-b1 Th1-a1 shows, that Henrik's guess of the omitted wTa1 might be wrong. Instead, I think that Andrew's modification might be in fact the original Dawson diagram. (2021-10-23)
A.Buchanan: Yes that makes sense. I can well believe that there was a lot of constructions in flight when TRD passed away. It's not surprising that something appeared a few years later. Can we get access to FCR from that year to validate? (2021-10-23)
A.Buchanan: Brian Stephenson kindly checked the original magazine. He wrote:
"Eventually got round to looking out this TRD problem. It was published in the issue you quote as problem 10381. The diagram in PDB is wrong. There should be a bPc2. The solution was published in FCR April 1956 as: 1.Pb5xPc5ep Qxa5 2.Qb7#. Last moves must have been 1.Pc7-c5 Pg5-g6 2.Sc6-e7 Qe7-b4 Sb4-c6 ch etc. No flaws were noted, but I have looked later than that issue. Hope this helps." (2021-11-13)
A.Buchanan: The appearance of the 16th Black unit thanks to the offices of Brian Stephenson simplifies the retro logic substantially, and the solution seems sound. Amazing how much disruption a simple typo can cause, but at least this one I believe is laid to rest (2021-11-13)
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A.Buchanan: I agree Henrik and have made changes. A number of these problems have been marked "Weisz zieht an", which is inappropriate as part of the point of the problem is to deduce this. I think it's 2. c2-c3! not 2. g5-g6? or the cross-captured pawns become retro-locked. (2021-10-21)
Henrik Juel: 2.g5-g6 also works, because the only cross-capture is axb,bxa
White captured [Pd7] with an officer (2021-10-22)
Henrik Juel: HC+ Popeye 4.61 (2021-10-22)
A.Buchanan: If g5-g6 also works, then can’t we have both? I.e. White moved last. I would be surprised if this was TRD’s intention, given the others of this general style (2021-10-22)
Henrik Juel: TRD died in 1951, so someone else may have messed up the problem
FRC is not generally available after 1953, so I cannot check the source (2021-10-22)
A.Buchanan: WinChloe has 29(!) posthumous problems by Dawson - the latest in 1958, FCR's final year. The composition in question is not one of them. (2021-10-23)
A.Buchanan: PROOF OF COOK + SUGGESTED FIX
I agree with much of what you say Henrik.
Capturing history. Basically 3 possibilities for the Black pawns.
Notation: ~,o,| denote "cross-capture", "original" & "waylaid"(=captured on home file by officer).
1) a~b co d~e f~g. This implies either (wPhxg & bPh=X) or (wPh=X & bPh|)
2) a~b co d~e hxgxf f|
3) a~b co hxgxfxexd d|
Suppose WTM. Then R: 1. c6-c5 c2-c3 2. c7-c6 g5-g6 3. Sc6-e7 De7-b4 4. Sb4-c6+ Kc5-c4 5. Lb2-d4+ etc. Now we're free to uncapture wDh8 (or g8 for scenario 1), and any of the three pawn scenarios can work. No e.p. rights means a cook.
If we fix it by e.g. shifting wPg6-g5 then I think that fixes this cook, and also it means that there is no solution if BTM.
Is there any possibility for premature capturing hxg? Not by Bl because scenario 1 says bPfxg, and the other two scenarios rely on wPh=X. If by Wh, then there's no waylaying on d or f, so we are back to scenario 1, but bPh is blocked from promotion, so that won't work.
But: what about R: 1. c6-c5 c2-c3 2. Sc3-b1 Th1-a1. Doesn't this completely cook it anyway?? How about k7/4n3/6P1/PPpPPP2/qQKbrrp1/ppRppp2/b1P5/1n6? h#2 (with Art 15 so it's really h#1.5)
Do you agree? (2021-10-23)
Mario Richter: In the diagram, both for WTM and BTM the retraction sequence c2-c3 Sc6-e7 De1-b4 Sb4-c6+ works and furthermore Andrew's R: 1. c6-c5 c2-c3 2. Sc3-b1 Th1-a1 shows, that Henrik's guess of the omitted wTa1 might be wrong. Instead, I think that Andrew's modification might be in fact the original Dawson diagram. (2021-10-23)
A.Buchanan: Yes that makes sense. I can well believe that there was a lot of constructions in flight when TRD passed away. It's not surprising that something appeared a few years later. Can we get access to FCR from that year to validate? (2021-10-23)
A.Buchanan: Brian Stephenson kindly checked the original magazine. He wrote:
"Eventually got round to looking out this TRD problem. It was published in the issue you quote as problem 10381. The diagram in PDB is wrong. There should be a bPc2. The solution was published in FCR April 1956 as: 1.Pb5xPc5ep Qxa5 2.Qb7#. Last moves must have been 1.Pc7-c5 Pg5-g6 2.Sc6-e7 Qe7-b4 Sb4-c6 ch etc. No flaws were noted, but I have looked later than that issue. Hope this helps." (2021-11-13)
A.Buchanan: The appearance of the 16th Black unit thanks to the offices of Brian Stephenson simplifies the retro logic substantially, and the solution seems sound. Amazing how much disruption a simple typo can cause, but at least this one I believe is laid to rest (2021-11-13)
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Keywords: En passant as key, No legal last move for White
Genre: h#, Retro
Computer test: HC+ Popeye 4.6 + non-trivial retro
FEN: k7/4n3/6P1/PPpPPP2/qQKbrrp1/ppPppp2/b1p5/1n6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.6 + non-trivial retro
FEN: k7/4n3/6P1/PPpPPP2/qQKbrrp1/ppPppp2/b1p5/1n6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
a) 1. f3 Lxa7 2. Lg5 Lxf2#
b) 1. fxg3ep Lxc7 2. Lh5 Ld8#
b) 1. fxg3ep Lxc7 2. Lh5 Ld8#
Henrik Juel: a) C+ Popeye 4.61
b) the ep capture is unjustified, last move could be Ke2-f1 or f3xg4 (2023-08-06)
A.Buchanan: There was an error in the stipulation: "b) sLe3 nach h2 (-wSh2)" is completely incorrect. However "b) sLe3 nach f3" leads to a very nice problem which matches the solution already present. This problem is not in WinChloe, but I am sure this is the resolution of the mystery (2023-08-07)
Henrik Juel: You are probably right, Andrew; maybe there was a misprint in the source
Now part b) is C+ Popeye 4.61 after very simple analysis (2023-08-07)
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b) the ep capture is unjustified, last move could be Ke2-f1 or f3xg4 (2023-08-06)
A.Buchanan: There was an error in the stipulation: "b) sLe3 nach h2 (-wSh2)" is completely incorrect. However "b) sLe3 nach f3" leads to a very nice problem which matches the solution already present. This problem is not in WinChloe, but I am sure this is the resolution of the mystery (2023-08-07)
Henrik Juel: You are probably right, Andrew; maybe there was a misprint in the source
Now part b) is C+ Popeye 4.61 after very simple analysis (2023-08-07)
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Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye v4.61 & simple retro-logic for b)
FEN: 1B6/npp5/8/8/5pPk/4b2p/3p1p1N/5Kn1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-08-07 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.61 & simple retro-logic for b)
FEN: 1B6/npp5/8/8/5pPk/4b2p/3p1p1N/5Kn1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-08-07 more...
1. exf3ep e3 2. 0-0-0 Tb4 3. Tg4 Tb8#
I think that this cooked problem is an early A osteriori. I can see ways that it might be made sound, but I would like to see the published version. San anyone help please? Thanks.
Cook: 17 Black pawns, retro cook and numerous forward cooks
I think that this cooked problem is an early A osteriori. I can see ways that it might be made sound, but I would like to see the published version. San anyone help please? Thanks.
Cook: 17 Black pawns, retro cook and numerous forward cooks
klären wK im Schach, vielleicht wBb2?
Alfred Pfeiffer: 9 schwarze Bauern! (2012-02-07)
Ladislav Packa: Auch mit wBb2 NL, z.B. 1.exf3 e.p. Txg5 2.0-0-0 Tb5 3.Tg~ Tb8# (2012-02-07)
A.Buchanan: And adding to earlier comments, if we do swap sBb2 for wB, the retraction is still not unique with R: 1. f3-f4 Lh6xg5+ (2021-11-26)
A.Buchanan: I think that this cooked problem is an early A Posteriori. I can see ways that it might be made sound, but I would like to see the published version. Can anyone help please? Thanks. (2022-04-25)
Gerald Ettl: Verbesserungsvorschlag: -sBa2, -sBb2, -sSg7, +sBg7
r3k3/3b2p1/5p2/6b1/4pPRp/2pq2rp/2p1P1pB/2K3n1/ (2022-04-25)
Gerald Ettl: und +sSg8 (wegen exf4 Vermeidung) (2022-04-25)
Gerald Ettl: +sSa4 nicht g8 (2022-04-25)
A.Buchanan: Hi Gerald. Thanks for this. Your final proposed version is r3k3/3b2p1/5p2/6b1/n3pPRp/2pq2rp/2p1P1pB/2K3n1, yes? There are 16 candidate solutions, all using castling. But wPa cannot leave the a-file, and is required for capture balance, so Black cannot castle. Five of the candidates begin with e.p.: play might have just been R: 1. Kb1-c1 b3xPc2+, so e.p. is not permitted in any case. What am I missing? (2022-04-25)
Gerald Ettl: Hi Andrew,
Those were 2 mistakes I made. I improved it again.
r3k3/3b4/4pp2/6b1/4pPRp/n1pq2rp/PPp1P1pB/2K3n1/
(den wBa2 habe ich auf das Brett gestellt, da sonst wieder Lh6xXg5 geht.) (2022-04-26)
A.Buchanan: Hi Gerald: Deine Retro-Logik ist gut. Aber es gibt 18 Lösungsvorschläge für die Zukunft. Ich habe heute im Discord meinen eigenen Vorschlag zur AP-Korrektur von Kardos veröffentlicht. Ich werde ihn hier im PDB hinzufügen. (2022-04-26)
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Alfred Pfeiffer: 9 schwarze Bauern! (2012-02-07)
Ladislav Packa: Auch mit wBb2 NL, z.B. 1.exf3 e.p. Txg5 2.0-0-0 Tb5 3.Tg~ Tb8# (2012-02-07)
A.Buchanan: And adding to earlier comments, if we do swap sBb2 for wB, the retraction is still not unique with R: 1. f3-f4 Lh6xg5+ (2021-11-26)
A.Buchanan: I think that this cooked problem is an early A Posteriori. I can see ways that it might be made sound, but I would like to see the published version. Can anyone help please? Thanks. (2022-04-25)
Gerald Ettl: Verbesserungsvorschlag: -sBa2, -sBb2, -sSg7, +sBg7
r3k3/3b2p1/5p2/6b1/4pPRp/2pq2rp/2p1P1pB/2K3n1/ (2022-04-25)
Gerald Ettl: und +sSg8 (wegen exf4 Vermeidung) (2022-04-25)
Gerald Ettl: +sSa4 nicht g8 (2022-04-25)
A.Buchanan: Hi Gerald. Thanks for this. Your final proposed version is r3k3/3b2p1/5p2/6b1/n3pPRp/2pq2rp/2p1P1pB/2K3n1, yes? There are 16 candidate solutions, all using castling. But wPa cannot leave the a-file, and is required for capture balance, so Black cannot castle. Five of the candidates begin with e.p.: play might have just been R: 1. Kb1-c1 b3xPc2+, so e.p. is not permitted in any case. What am I missing? (2022-04-25)
Gerald Ettl: Hi Andrew,
Those were 2 mistakes I made. I improved it again.
r3k3/3b4/4pp2/6b1/4pPRp/n1pq2rp/PPp1P1pB/2K3n1/
(den wBa2 habe ich auf das Brett gestellt, da sonst wieder Lh6xXg5 geht.) (2022-04-26)
A.Buchanan: Hi Gerald: Deine Retro-Logik ist gut. Aber es gibt 18 Lösungsvorschläge für die Zukunft. Ich habe heute im Discord meinen eigenen Vorschlag zur AP-Korrektur von Kardos veröffentlicht. Ich werde ihn hier im PDB hinzufügen. (2022-04-26)
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1) 1. Tb1 Lxf3 2. Tb3 Lc6#
2) 1. Lc6 Le4 2. Lb5 Lc2#
3) 1. cxb3ep Lxc3 2. Lc6 Lxc6#
2) 1. Lc6 Le4 2. Lb5 Lc2#
3) 1. cxb3ep Lxc3 2. Lc6 Lxc6#
Keywords: En passant as key
Genre: h#, Retro
Computer test: C+ Popeye 4.61
FEN: 8/8/1b6/B7/kPp3p1/p1p2b1p/6Bp/3r1n1K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-05 more...
Genre: h#, Retro
Computer test: C+ Popeye 4.61
FEN: 8/8/1b6/B7/kPp3p1/p1p2b1p/6Bp/3r1n1K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-05 more...
1. fxg3ep h8=S 2. Sg4 Sxg6#
A.Buchanan: The point I think is that the en passant is a tempo move. Otherwise wBf2 serves no purpose. How about sBh5 instead of sT? (2021-10-19)
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Keywords: En passant as key, Promotion (S), Tempo Move
Genre: h#, Retro
Computer test: HC+ Popeye v4.61 & trivial retro-logic
FEN: 8/7P/6pP/6pr/5pPk/4np1n/5P1p/7K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-19 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.61 & trivial retro-logic
FEN: 8/7P/6pP/6pr/5pPk/4np1n/5P1p/7K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-19 more...
1. fxg3ep Lxe3 2. Ld5 cxd5#
Keywords: En passant as key
Genre: h#, Retro
FEN: 8/8/8/1PP1P2r/PRPBbpPk/ppKppp1r/8/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-08-09 more...
Genre: h#, Retro
FEN: 8/8/8/1PP1P2r/PRPBbpPk/ppKppp1r/8/8
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-08-09 more...
1. hxg3ep Sd4 2. gxf2 Se3#
vgl. P0003180
Henrik Juel: Last move was g2-g4 giving Black a preceeding move like Da8-b8 or Lb7-c8
C+ Popeye 4.61 (2021-04-24)
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Henrik Juel: Last move was g2-g4 giving Black a preceeding move like Da8-b8 or Lb7-c8
C+ Popeye 4.61 (2021-04-24)
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Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 + simple retro thought
FEN: 1qb5/b1pp1ppp/pp6/6P1/5BPp/2P5/1PN1PP1P/3Nrk1K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-25 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 + simple retro thought
FEN: 1qb5/b1pp1ppp/pp6/6P1/5BPp/2P5/1PN1PP1P/3Nrk1K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-25 more...
1. cxd3ep f4 2. Lf3 Kxc5#
Henrik Juel: C+ Popeye 4.61 (2022-04-18)
Henrik Juel: Obviously, last move was d2-d4 (2022-04-18)
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Henrik Juel: Obviously, last move was d2-d4 (2022-04-18)
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Keywords: En passant as key
Genre: h#, Retro
FEN: 8/8/p7/P1r2p2/RKpPk3/p1p1p3/5P2/3b4
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2022-04-18 more...
Genre: h#, Retro
FEN: 8/8/p7/P1r2p2/RKpPk3/p1p1p3/5P2/3b4
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2022-04-18 more...
1. cxb3ep e5 2. Lh6 g5 3. Txh4 Txh4#
Henrik Juel: The only possible last move is b2-b4, so the ep key is permissible
C+ Popeye 4.61 (2021-07-26)
A.Buchanan: The main idea is to unblock the cluttered 4th rank, and this is complemented thematically by *blocking* four other lines: g8-g4, h8-h4, h8-d4 & "blocking the blocker" h6-f4.
The last move was b2-b4, so by the ep rule, capture is definitely legal. The convention only serves to resolve ambiguous situations where multiple possible histories exist. In retro problems, such ambiguity is rare if the retro logic does its job. However the pessimistic nature of the e.p. convention motivates *why* the retro play must exist. (2021-07-26)
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C+ Popeye 4.61 (2021-07-26)
A.Buchanan: The main idea is to unblock the cluttered 4th rank, and this is complemented thematically by *blocking* four other lines: g8-g4, h8-h4, h8-d4 & "blocking the blocker" h6-f4.
The last move was b2-b4, so by the ep rule, capture is definitely legal. The convention only serves to resolve ambiguous situations where multiple possible histories exist. In retro problems, such ambiguity is rare if the retro logic does its job. However the pessimistic nature of the e.p. convention motivates *why* the retro play must exist. (2021-07-26)
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Keywords: En passant as key
Genre: h#, Retro
Computer test: Last move was b2-b4, so by ep rule, capture ok as key. HC+ Popeye 4.61
FEN: 6rq/pn6/K7/8/kPprPbPP/p1pppppR/7P/8
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2021-07-26 more...
Genre: h#, Retro
Computer test: Last move was b2-b4, so by ep rule, capture ok as key. HC+ Popeye 4.61
FEN: 6rq/pn6/K7/8/kPprPbPP/p1pppppR/7P/8
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2021-07-26 more...
1. bxc3ep Kxa3 2. Ld5 Sg2 3. Tce1 Kb4 4. T1e4 bxc3#
Keywords: En passant as key
Genre: h#, Retro
FEN: 8/2p5/2p3pq/n3r3/KpPk1pp1/bp1p4/1P4b1/2r1N3
Input: Gerd Wilts, 1995-06-03
Last update: Dieter Berlin, 2021-08-04 more...
Genre: h#, Retro
FEN: 8/2p5/2p3pq/n3r3/KpPk1pp1/bp1p4/1P4b1/2r1N3
Input: Gerd Wilts, 1995-06-03
Last update: Dieter Berlin, 2021-08-04 more...
a) 1. gxf3ep g4 2. Kd5 gxf5 3. Kc6 fxe6 4. Kb7 exd7 5. Ka8 dxc8=D#
A.Buchanan: Popeye v4.87 (via Olive v1.4 ) for PDB problem P0003325 h#5 delivers the correct solution with en passant set to f2f3f4. However no solution when the "intelligent" flag is also set. Reported to Popeye Github. (2021-11-23)
Henrik Juel: Popeye 4.61 with 'opt var int enp f3' works fine
Maybe you just have to change f2f3f4 to f3, Andrew (2021-11-23)
A.Buchanan: Thanks for the suggestion, Henrik, but even without the "intelligent", opti vari enpa f3 fails with v4.87. I think that the syntax has been rendered more complicated since v4.61 in order to support fancy fairy en passants. (2021-11-23)
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Henrik Juel: Popeye 4.61 with 'opt var int enp f3' works fine
Maybe you just have to change f2f3f4 to f3, Andrew (2021-11-23)
A.Buchanan: Thanks for the suggestion, Henrik, but even without the "intelligent", opti vari enpa f3 fails with v4.87. I think that the syntax has been rendered more complicated since v4.61 in order to support fancy fairy en passants. (2021-11-23)
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1. dxc3ep Kxg7 2. f6 Kxf6 3. e5 Kxe5 4. Lc4 Kd4 5. Lb5 dxc3#
Keywords: En passant as key
Genre: h#, Retro
Computer test: Popeye C-Version 4.87 & trivial retro-logic
FEN: 6bK/5pr1/2p1p3/p4p2/rkPp4/qp1p4/3P4/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-23 more...
Genre: h#, Retro
Computer test: Popeye C-Version 4.87 & trivial retro-logic
FEN: 6bK/5pr1/2p1p3/p4p2/rkPp4/qp1p4/3P4/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-23 more...
1. cxb3ep Sb4 2. Sc4 Sc2 3. Sa5 Sxc3#
Keywords: En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & simple retro thought
FEN: 8/2p5/2P5/2p2p2/kPp2Pn1/n1pp1Kp1/N2P2pb/rNr1q2b
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-25 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 & simple retro thought
FEN: 8/2p5/2P5/2p2p2/kPp2Pn1/n1pp1Kp1/N2P2pb/rNr1q2b
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-25 more...
1. dxc3ep dxc3+ 2. Ka5 c4 3. d2 c5 4. d1=L c6 5. Lg4 hxg4 6. Kb6 g5 7. Ka7 g6 8. Kb8 g7 9. Kc8 g8=D/T#
Cook: 1. b2 c5 2. b1=S c6 3. Sc3 dxc3+ 4. Ka5 cxd4 5. Kb6 d5 6. Ka7 d6 7. Kb8 dxc7+ 8. Ka7 c8=D 9. a5 Db7#
Cook: 1. b2 c5 2. b1=S c6 3. Sc3 dxc3+ 4. Ka5 cxd4 5. Kb6 d5 6. Ka7 d6 7. Kb8 dxc7+ 8. Ka7 c8=D 9. a5 Db7#
paul: See P0003212 as version. (2011-08-06)
Anton Baumann: C+?? Die Aufgabe ist eindeutig inkorrekt! (2021-01-27)
James Malcom: Fixed. (2021-01-27)
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Anton Baumann: C+?? Die Aufgabe ist eindeutig inkorrekt! (2021-01-27)
James Malcom: Fixed. (2021-01-27)
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Keywords: En passant as key, Promotion (D), konsekutive Umwandlungen 2 (L, D/T), En passant, Kindergarten Problem
Genre: h#, Retro
FEN: 8/2p5/p7/8/1kPp3p/1p1p2pP/3P2Pp/7K
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-01-27 more...
Genre: h#, Retro
FEN: 8/2p5/p7/8/1kPp3p/1p1p2pP/3P2Pp/7K
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-01-27 more...
1. ... exf6ep 2. Sf5 f7#
White captured fxg and promoted [Pd2] on d8
Black captured axbxa2-a1=Y, b7xc6, d5xe4, and gxf
White cannot have moved last, and in fact Black's last move can only have been f7-f5 in order to give White a prior move. R: 1... d5-e4 would have retro-blocked wPd marching to its promotion.
So the retroplay was R: 1... f7-f5 2.f5xYg6
Numerous retro tries but none distinct.
White captured fxg and promoted [Pd2] on d8
Black captured axbxa2-a1=Y, b7xc6, d5xe4, and gxf
White cannot have moved last, and in fact Black's last move can only have been f7-f5 in order to give White a prior move. R: 1... d5-e4 would have retro-blocked wPd marching to its promotion.
So the retroplay was R: 1... f7-f5 2.f5xYg6
Numerous retro tries but none distinct.
"Autor Ing. Rudolf Buljan, Zagreb"
AB: Have therefore marked him in the author table as Croatian. Yugoslavia has been removed from the PDB country list, I see.
A.Buchanan: Any ideas how to distinguish usage of “whose move?” and “no legal last move for white” (or black) keywords (2020-10-01)
Henrik Juel: This problem is clearly the latter
I would use 'Whose move?' in retro problems where a deeper analysis is required to determine the move, often with stipulations like '#1 (who?)' (2020-10-01)
A.Buchanan: I think "Whose move?" is where there is no forward stip, or where we have to determine if the forward stip should be interpreted as orthodox or as half-duplex. "no legal last move for..." is where the mater/stalemater/etc is the usual one, but we add or remove a single move at the beginning of the forward stip. I guess a third possibility is "last move?" where in a Type A position we also have to figure out who moved last.
With respect, I don't think the level of analysis can be relevant. (2023-12-06)
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AB: Have therefore marked him in the author table as Croatian. Yugoslavia has been removed from the PDB country list, I see.
A.Buchanan: Any ideas how to distinguish usage of “whose move?” and “no legal last move for white” (or black) keywords (2020-10-01)
Henrik Juel: This problem is clearly the latter
I would use 'Whose move?' in retro problems where a deeper analysis is required to determine the move, often with stipulations like '#1 (who?)' (2020-10-01)
A.Buchanan: I think "Whose move?" is where there is no forward stip, or where we have to determine if the forward stip should be interpreted as orthodox or as half-duplex. "no legal last move for..." is where the mater/stalemater/etc is the usual one, but we add or remove a single move at the beginning of the forward stip. I guess a third possibility is "last move?" where in a Type A position we also have to figure out who moved last.
With respect, I don't think the level of analysis can be relevant. (2023-12-06)
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Keywords: En passant as key, No legal last move for White
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: 5nkb/1qp1p1n1/2p3PQ/4PpKb/4prRp/P4prB/1PP3PP/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: 5nkb/1qp1p1n1/2p3PQ/4PpKb/4prRp/P4prB/1PP3PP/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
60 - P0003339
Tivadar Kardos
2519 Revista Romana de Sah 01/1948
2. Preis
(10+15) cooked
h# in wieviel?
Tivadar Kardos
2519 Revista Romana de Sah 01/1948
2. Preis
(10+15) cooked
h# in wieviel?
1. ... cxd6ep 2. Tc8 dxe7 3. Tf8 exf8=S#
AL ('Revista Romana de Sah', 06/1948, S.225-226):
Bekanntlich beginnt im Hilfsmatt Schwarz, also hat in der Diagrammstellung Weiß den letzten Zug gemacht. Man findet aber, wenn man die Position analysiert, daß Weiß den letzten Zug nicht gemacht haben kann.
Der weiße Doppelbauer auf der d-Linie läßt sich nur durch Schlagen des einzig fehlenden schwarzen Steines, des sBa7, erklären. Dieser konnte aber erst nach Umwandlung geschlagen werden. Dazu mußte er zunächst bis a3 vorrücken, dann nach b2 schlagen, und dann umwandeln.
Also kann Weiß nicht zuletzt b2-b3 gezogen haben.
Die anderen schwarzen Bauern haben mindestens 5 Schläge gemacht, um die Position im Diagramm zu erreichen (man beachte, daß der sBh7 zweimal schlagen mußte, um den wBh2 durchzulassen)
also kann auch g2-g3 nicht der letzte Zug gewesen sein, denn sonst hätte der wLf1 nicht von einem sB geschlagen worden sein können.
Für alle anderen weißen Steine ist leicht zu sehen, daß sie den letzten Zug nicht gemacht haben können.
Es stellt sich heraus, dass der letzte Zug von Schwarz gemacht wurde, und eine einfache Analyse zeigt, daß dies nur möglich ist durch d7-d5 (sonst wäre es unmöglich, daß die weißen Steine ihre Position im Diagramm erreichen könnten).
In diesem Fall Weiß kann 1. ... c5:d6(e.p.) spielen, gefolgt von 2. Tc8 d:e7 3. Tf8 e:f8C matt!
Das kürzeste Hilfsmatt wird also in 2,5 Zügen erreicht!
Cook: 1. ... Txg5+ 2. Kxg5 f3 3. Lg6 gxf4#
1. ... Kxd5 2. Kf6 e4 3. Tg6,Dg6 e5#
AL ('Revista Romana de Sah', 06/1948, S.225-226):
Bekanntlich beginnt im Hilfsmatt Schwarz, also hat in der Diagrammstellung Weiß den letzten Zug gemacht. Man findet aber, wenn man die Position analysiert, daß Weiß den letzten Zug nicht gemacht haben kann.
Der weiße Doppelbauer auf der d-Linie läßt sich nur durch Schlagen des einzig fehlenden schwarzen Steines, des sBa7, erklären. Dieser konnte aber erst nach Umwandlung geschlagen werden. Dazu mußte er zunächst bis a3 vorrücken, dann nach b2 schlagen, und dann umwandeln.
Also kann Weiß nicht zuletzt b2-b3 gezogen haben.
Die anderen schwarzen Bauern haben mindestens 5 Schläge gemacht, um die Position im Diagramm zu erreichen (man beachte, daß der sBh7 zweimal schlagen mußte, um den wBh2 durchzulassen)
also kann auch g2-g3 nicht der letzte Zug gewesen sein, denn sonst hätte der wLf1 nicht von einem sB geschlagen worden sein können.
Für alle anderen weißen Steine ist leicht zu sehen, daß sie den letzten Zug nicht gemacht haben können.
Es stellt sich heraus, dass der letzte Zug von Schwarz gemacht wurde, und eine einfache Analyse zeigt, daß dies nur möglich ist durch d7-d5 (sonst wäre es unmöglich, daß die weißen Steine ihre Position im Diagramm erreichen könnten).
In diesem Fall Weiß kann 1. ... c5:d6(e.p.) spielen, gefolgt von 2. Tc8 d:e7 3. Tf8 e:f8C matt!
Das kürzeste Hilfsmatt wird also in 2,5 Zügen erreicht!
Cook: 1. ... Txg5+ 2. Kxg5 f3 3. Lg6 gxf4#
1. ... Kxd5 2. Kf6 e4 3. Tg6,Dg6 e5#
A.Buchanan: Popeye v.487. WinChloe has the same diagram, and although it just claims 1 solution, it's C? Yet WinChloe engine gives the cooks right away (2021-10-27)
Mario Richter: Shouldn't this problem get the Label "No legal last move for White"?
(White pawn d2 has captured the missing black piece (Pdxe), so black pawn a7 had to promote to get to the capture square. This requires one capture (Pa3xb2), together with bxc,cxd,hxg,gxh,gxh this accounts for all missing white pieces.)
Btw. "for me, "C?" stands for "Not yet computer-tested or computer test didn't come to a conlusion" ... (2021-10-28)
A.Buchanan: Hi Mario, thank you.
(1) The stipulation is non-standard. I think this is why the WinChloe database doesn't include a solution, and just marks it as "C?" However if I run the WinChloe solving engine manually it (like Popeye) gives 121 h#3 solutions and 4 h#2.5 solutions. I think you, me and Christian would all agree on the meaning of "C?"
(2) The problem does not rely on Article 15 to eliminate the first single move, but implicitly requests the shortest. Currently the PDB usage of "No legal last move..." is to indicate that Codex Article 15 applies, so I did not add it. But I'm ok if you want me to add the keyword. In my correction, I shift the stipulation to h#3 so it's clear that Article 15 and not any desire for brevity which push the problem to h#2.5.
(3) However the PDB usage is confused. Ideallly think the two "No legal..." keywords should be replaced by "Retro-stalemate", while two new keywords "Article 15" & "Whose mate?" indicate the choice of implications of the retropat for forward play. "Board rotation" and other jokes are also implications of retropat (and of illegal diagram).
(4) But this is complicated to implement, requiring either database-level access or screen-scraping. The latter is quite practical for some future time when I have more leisure.
(5) https://www.stere.ro/biblioteca-digitala/reviste/revista-romana-de-sah/ is an excellent archive of Revista Romana de Sah. My sincere compliments and thanks to whoever prepared it. And the magazines themselves contain thorough annual indices. p225-6 of 6/1948, contains the solution to Kardos. I haven't translated it, nor have I found time to locate the prize announcement.
Comments welcome! (2021-10-29)
Mario Richter: I still believe, that the label "No legal last move for White" is justified here, because this is exactly what the author wanted to show.
(And this is too, what the reprint in 'Problem (Zagreb)' 37-40 09/1956 is all about - it appears there inside asn article
Btw., I wouldn't call the presentation of the solution in 'Revista Romana de Sah', p.225, 06/1948, a "reprint" - it lacks the presentation of the position ... (2021-10-30)
A.Buchanan: Added the keyword.
Yes I know that's the usage of "reprint", but then there should be a field for the location of the solution :-) It's idiotic that if I want to say where the solution is contained, I have to embed it in text like this: 2519 Revista Romana de Sah , p. 225-6, 06/1948 (2021-10-30)
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Mario Richter: Shouldn't this problem get the Label "No legal last move for White"?
(White pawn d2 has captured the missing black piece (Pdxe), so black pawn a7 had to promote to get to the capture square. This requires one capture (Pa3xb2), together with bxc,cxd,hxg,gxh,gxh this accounts for all missing white pieces.)
Btw. "for me, "C?" stands for "Not yet computer-tested or computer test didn't come to a conlusion" ... (2021-10-28)
A.Buchanan: Hi Mario, thank you.
(1) The stipulation is non-standard. I think this is why the WinChloe database doesn't include a solution, and just marks it as "C?" However if I run the WinChloe solving engine manually it (like Popeye) gives 121 h#3 solutions and 4 h#2.5 solutions. I think you, me and Christian would all agree on the meaning of "C?"
(2) The problem does not rely on Article 15 to eliminate the first single move, but implicitly requests the shortest. Currently the PDB usage of "No legal last move..." is to indicate that Codex Article 15 applies, so I did not add it. But I'm ok if you want me to add the keyword. In my correction, I shift the stipulation to h#3 so it's clear that Article 15 and not any desire for brevity which push the problem to h#2.5.
(3) However the PDB usage is confused. Ideallly think the two "No legal..." keywords should be replaced by "Retro-stalemate", while two new keywords "Article 15" & "Whose mate?" indicate the choice of implications of the retropat for forward play. "Board rotation" and other jokes are also implications of retropat (and of illegal diagram).
(4) But this is complicated to implement, requiring either database-level access or screen-scraping. The latter is quite practical for some future time when I have more leisure.
(5) https://www.stere.ro/biblioteca-digitala/reviste/revista-romana-de-sah/ is an excellent archive of Revista Romana de Sah. My sincere compliments and thanks to whoever prepared it. And the magazines themselves contain thorough annual indices. p225-6 of 6/1948, contains the solution to Kardos. I haven't translated it, nor have I found time to locate the prize announcement.
Comments welcome! (2021-10-29)
Mario Richter: I still believe, that the label "No legal last move for White" is justified here, because this is exactly what the author wanted to show.
(And this is too, what the reprint in 'Problem (Zagreb)' 37-40 09/1956 is all about - it appears there inside asn article
Btw., I wouldn't call the presentation of the solution in 'Revista Romana de Sah', p.225, 06/1948, a "reprint" - it lacks the presentation of the position ... (2021-10-30)
A.Buchanan: Added the keyword.
Yes I know that's the usage of "reprint", but then there should be a field for the location of the solution :-) It's idiotic that if I want to say where the solution is contained, I have to embed it in text like this: 2519 Revista Romana de Sah , p. 225-6, 06/1948 (2021-10-30)
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Keywords: En passant as key, Promotion (S), Superseded by (P1394978), No legal last move for White
Genre: h#, Retro
Computer test: Popeye v4.87 & WinChloe both say cooked
FEN: 7b/4p1nP/2r1P1kq/1nPpKbrp/2pp1pRp/1P4P1/P3PP2/8
Reprints: (III) Problem 37-40 09/1956
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-30 more...
Genre: h#, Retro
Computer test: Popeye v4.87 & WinChloe both say cooked
FEN: 7b/4p1nP/2r1P1kq/1nPpKbrp/2pp1pRp/1P4P1/P3PP2/8
Reprints: (III) Problem 37-40 09/1956
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-30 more...
1. hxg3ep Lxb1 2. Kf5 fxe4 3. Kg6 e5+ 4. f5 exf6ep#
Cook: NL:
1. hxg3ep fxe4 2. f6 e5 3. Kf5 exf6 4. Kg6 Lxb1# uvm
Cook: NL:
1. hxg3ep fxe4 2. f6 e5 3. Kf5 exf6 4. Kg6 Lxb1# uvm
YM: The correct mechanism is in P1109109 (2010-07-13)
YM: Correction option: P1229434 (2021-06-28)
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YM: Correction option: P1229434 (2021-06-28)
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Keywords: En passant as key, Superseded by (P1109109)
Genre: h#, Retro
FEN: q2r4/4Kp2/3r3b/3p2pp/4nkPp/1n3P2/B4p1p/1b6
Reprints: 805 FIDE Album 1980-1982 1988
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-28 more...
Genre: h#, Retro
FEN: q2r4/4Kp2/3r3b/3p2pp/4nkPp/1n3P2/B4p1p/1b6
Reprints: 805 FIDE Album 1980-1982 1988
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-28 more...
1. axb3ep bxc6+ 2. b5 cxb6ep#
Marko Klasinc: The most economical h#2 with two e.p. captures. (2002-01-28)
A.Buchanan: Very nice - and i think this is unbeatable at least in the case where ep is first and last move (2020-08-02)
A.Buchanan: Also works as set play! Just truncate Black's first move! :) (2020-12-09)
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A.Buchanan: Very nice - and i think this is unbeatable at least in the case where ep is first and last move (2020-08-02)
A.Buchanan: Also works as set play! Just truncate Black's first move! :) (2020-12-09)
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Keywords: En passant as key, Economy record, En passant as mating move
Genre: h#, Retro
Computer test: C+ Popeye v4.61 + a little thinking
FEN: 8/1p6/B1p5/RPP5/pPkp4/K1p5/P7/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-12-09 more...
Genre: h#, Retro
Computer test: C+ Popeye v4.61 + a little thinking
FEN: 8/1p6/B1p5/RPP5/pPkp4/K1p5/P7/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-12-09 more...
1) 1. 0-0 Le6+ 2. Kh8 Sg6#
2) 1. bxc3ep 0-0 2. 0-0-0 Tc4#
If all three castlings are legal, then the ep is on. So evidence must be accumulated from the two solutions. White 0-0 serves only in this retro role, eliminating 26 other candidate solutions.
2) 1. bxc3ep 0-0 2. 0-0-0 Tc4#
If all three castlings are legal, then the ep is on. So evidence must be accumulated from the two solutions. White 0-0 serves only in this retro role, eliminating 26 other candidate solutions.
A.Buchanan: There was a diagram error here. It should be bPd7 not bPe7. The existing diagram is vastly cooked with e.g. 1. Ke8 Kd1 2. Te8 Lb6#. WinChloe has the correct diagram here. (2022-05-24)
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Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wksksg)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro logic
FEN: r3k2r/B2p2pp/8/NP6/RpP4N/pP5B/1p1PPPP1/n3K2R
Reprints: feenschach 54 04/1981
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-06 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro logic
FEN: r3k2r/B2p2pp/8/NP6/RpP4N/pP5B/1p1PPPP1/n3K2R
Reprints: feenschach 54 04/1981
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-06 more...
1) 1. axb3ep bxc6+ 2. b5 cxb6ep#
2) 1. exd3ep g8=D,L+ 2. d5 cxd6ep#
PRA: 1 solution with 2 parts
2) 1. exd3ep g8=D,L+ 2. d5 cxd6ep#
PRA: 1 solution with 2 parts
Henrik Juel: White captured [sLc8] on c8 and axb, so last move was either b2-b4 or d2-d4
C+ Popeye 4.61, except for the promotion dual (2020-08-03)
A.Buchanan: There's no keyword for an untolerated dual promotion! (2020-08-03)
Henrik Juel: Yes and no, Andrew
What about the cooked label? (2020-08-03)
A.Buchanan: Hi Henrik: this is obviously intentional. I tried to replace the pawn promotion with wDh8-h7+ but I ran out of pieces. I guess that Bebesi had the same difficulty. The word "cooked" seems too strong to me, but it's obviously a significant defect though (2020-08-04)
A.Buchanan: The bar is higher for helpmates than other problems, and this is unsound. However it’s not cooked but dualled. Yet all we have is “cooked” so I suppose it is in PDB-speak. I’ve added that and removed the 2.1… (2023-08-07)
Joost de Heer: Is there an 'unsound' keyword? (2023-08-07)
A.Buchanan: Hi Joost: no there isn't. It can be added, the challenge for any new common keyword would be populating it for more than a few values. (2023-08-07)
A.Buchanan: This time round, I've managed to find a fix. Bebesi's correction in WinChloe 66992 has non-standard wL. I have a feeling I've seen a similar one by Andrey Frolkin & someone else recently with a different matrix, and also non-standard wL. I've sent my effort to Joaquim Crusats at Problemas for his assessment (2023-08-08)
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C+ Popeye 4.61, except for the promotion dual (2020-08-03)
A.Buchanan: There's no keyword for an untolerated dual promotion! (2020-08-03)
Henrik Juel: Yes and no, Andrew
What about the cooked label? (2020-08-03)
A.Buchanan: Hi Henrik: this is obviously intentional. I tried to replace the pawn promotion with wDh8-h7+ but I ran out of pieces. I guess that Bebesi had the same difficulty. The word "cooked" seems too strong to me, but it's obviously a significant defect though (2020-08-04)
A.Buchanan: The bar is higher for helpmates than other problems, and this is unsound. However it’s not cooked but dualled. Yet all we have is “cooked” so I suppose it is in PDB-speak. I’ve added that and removed the 2.1… (2023-08-07)
Joost de Heer: Is there an 'unsound' keyword? (2023-08-07)
A.Buchanan: Hi Joost: no there isn't. It can be added, the challenge for any new common keyword would be populating it for more than a few values. (2023-08-07)
A.Buchanan: This time round, I've managed to find a fix. Bebesi's correction in WinChloe 66992 has non-standard wL. I have a feeling I've seen a similar one by Andrey Frolkin & someone else recently with a different matrix, and also non-standard wL. I've sent my effort to Joaquim Crusats at Problemas for his assessment (2023-08-08)
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Keywords: En passant as key (2), Partial Retro Analysis (PRA), En passant as mating move (2), Superseded by (P1411659, P1413906)
Genre: h#, Retro
FEN: 3qn3/1p1p2P1/B1r3p1/1PP1Kp2/pPkPp3/p1p4n/N7/2b5
Reprints: 233 Ungarische Schachproblem Anthologie , p. 188, 1983
1 Problemas 44, p. 1456, 10/2023
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-04 more...
Genre: h#, Retro
FEN: 3qn3/1p1p2P1/B1r3p1/1PP1Kp2/pPkPp3/p1p4n/N7/2b5
Reprints: 233 Ungarische Schachproblem Anthologie , p. 188, 1983
1 Problemas 44, p. 1456, 10/2023
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-04 more...
a) 1. Dc4 Lxb6+ 2. c5 dxc6ep#
b) 1. fxe3ep Lxb4 2. Sc4 bxc3#
Lindner in 'Mattbilder eines Lebens':
In a) ist die Lösung der stellung b) nicht möglich, weil das e.p.-Schlagen durch Schwarz nicht legal ist. Der letzte Zug von Weiß muß nicht unbedingt e2-e4 geweseb sein. Es kommt als letzter zug auch Kh3-g2 in Betracht, mit den vorherigen Zügen h4:g3 e.p.+ und g2-g4.
In b) demgegenüber sind Kh3-g2 und vorher f4:g3 e.p. illegal, weil die Rücknahme von g2-g4 unmöglich ist: der sB würde 7 Schlagfälle benötigen, und es fehlen nur 6 weiße Steine. Der letzte weiße Zug muß also e2-e4 gewesen sein.
b) 1. fxe3ep Lxb4 2. Sc4 bxc3#
Lindner in 'Mattbilder eines Lebens':
In a) ist die Lösung der stellung b) nicht möglich, weil das e.p.-Schlagen durch Schwarz nicht legal ist. Der letzte Zug von Weiß muß nicht unbedingt e2-e4 geweseb sein. Es kommt als letzter zug auch Kh3-g2 in Betracht, mit den vorherigen Zügen h4:g3 e.p.+ und g2-g4.
In b) demgegenüber sind Kh3-g2 und vorher f4:g3 e.p. illegal, weil die Rücknahme von g2-g4 unmöglich ist: der sB würde 7 Schlagfälle benötigen, und es fehlen nur 6 weiße Steine. Der letzte weiße Zug muß also e2-e4 gewesen sein.
In 'Mattbilder eines Lebens' abgedruckt mit sTh7 statt h8 und der Quellenangabe: Europe Echecs, 1964
AB: (1) Where is wK?
(2) Why is 1.fxe3ep legal in b) but not a)? (2002-01-31)
Henrik Juel: wK is probably on g2. In part a) last move could have been Kh3-g2, I think (2002-02-01)
A.Buchanan: Very convincing, Henrik. I've repaired the diagram accordingly. (2023-05-28)
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AB: (1) Where is wK?
(2) Why is 1.fxe3ep legal in b) but not a)? (2002-01-31)
Henrik Juel: wK is probably on g2. In part a) last move could have been Kh3-g2, I think (2002-02-01)
A.Buchanan: Very convincing, Henrik. I've repaired the diagram accordingly. (2023-05-28)
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Keywords: En passant as key, En passant in the retro play
Genre: h#, Retro
FEN: 7r/2pn4/1nqRb3/B2Pp3/pb1kPp2/2p2Pp1/1PP2pKp/7r
Reprints: 501 Mattbilder eines Lebens , p. 379, 1996
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-05-28 more...
Genre: h#, Retro
FEN: 7r/2pn4/1nqRb3/B2Pp3/pb1kPp2/2p2Pp1/1PP2pKp/7r
Reprints: 501 Mattbilder eines Lebens , p. 379, 1996
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-05-28 more...
1. ... fxg6ep#
1. dxe3ep fxe3#
1. dxe3ep fxe3#
Gerald Ettl: Der rumänische Autor zeigt, dass der letzte Zug jeweils ein Doppelschritt g7g5 oder e2e4 gewesen sein muss. Die wBc6,d7 schlugen ueber Kreuz und der wUW-L hat bxaB und a7xXb8 geschlagen. Die Forderung lautet wohl h#1 (2010-07-12)
Alfred Pfeiffer: Ja, die Forderung ist h#1*.
Online: "http://problem64.beda.cz/silo/fcr_4_1946.pdf" (2010-12-15)
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Alfred Pfeiffer: Ja, die Forderung ist h#1*.
Online: "http://problem64.beda.cz/silo/fcr_4_1946.pdf" (2010-12-15)
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Keywords: En passant as key (2), Promotion (L), Non-standard material (L)
Genre: h#, Retro
Computer test: C+ Popeye 4.61 and analysis
FEN: 4b3/1p1P1p2/2PRBR1P/2prpPpK/3pPkr1/5pN1/P4P1B/6B1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-30 more...
Genre: h#, Retro
Computer test: C+ Popeye 4.61 and analysis
FEN: 4b3/1p1P1p2/2PRBR1P/2prpPpK/3pPkr1/5pN1/P4P1B/6B1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-30 more...
67 - P0003384
Tivadar Kardos
4299 Stella Polaris 12/1971
(9+16) C+
h#2
b) alles eine Reihe nach links
Tivadar Kardos
4299 Stella Polaris 12/1971
(9+16) C+
h#2
b) alles eine Reihe nach links
a)
1. gxf3ep Sxg7 2. d5 Lf5#
1. cxd3ep? Sxc7 2. f5 Ld5#
b) 1. bxc3ep Sxb7 2. e5 Lc5#
1. fxe3ep Sxf7 2. c5 Le5#
1. gxf3ep Sxg7 2. d5 Lf5#
1. cxd3ep? Sxc7 2. f5 Ld5#
b) 1. bxc3ep Sxb7 2. e5 Lc5#
1. fxe3ep Sxf7 2. c5 Le5#
Sally: a)Der letzte Zug war: Bf2 - f4!
b)Der lrtzte Zug war: Bc2 - c4!
Nr. 139 200 Ausgwwählte S. Pr. T. Kardos (W. Frentze 1983) (2010-09-30)
Henrik Juel: Black pawns captured all 7 missing white men, incl. [Pa1], which promoted on a8
Only possible last moves are d2-d4 and f2-f4
In each twin the closing off for an original white bishop determines the last move (2021-11-22)
Henrik Juel: HC+ Popeye 4.61 (2021-11-22)
comment
b)Der lrtzte Zug war: Bc2 - c4!
Nr. 139 200 Ausgwwählte S. Pr. T. Kardos (W. Frentze 1983) (2010-09-30)
Henrik Juel: Black pawns captured all 7 missing white men, incl. [Pa1], which promoted on a8
Only possible last moves are d2-d4 and f2-f4
In each twin the closing off for an original white bishop determines the last move (2021-11-22)
Henrik Juel: HC+ Popeye 4.61 (2021-11-22)
comment
Keywords: En passant as key, Twin by board shift
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & Retractor 2.1.1 Human contribution is "positions prior to retractions look legal enough."
FEN: 2bKNb2/2rn1pqn/3pBp2/4B1P1/2pPkPpr/2p1p3/1P2p2P/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-27 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & Retractor 2.1.1 Human contribution is "positions prior to retractions look legal enough."
FEN: 2bKNb2/2rn1pqn/3pBp2/4B1P1/2pPkPpr/2p1p3/1P2p2P/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-27 more...
1. bxc3ep Lxc5#
1. fxe3ep Le5#
R: 1. c2-c4,e2-e4
1. fxe3ep Le5#
R: 1. c2-c4,e2-e4
Originalforderung? "2 Lösungen"
A.Buchanan: The original stipulation here asks for 2 solutions. I don't think this would work as 2 solutions under Retro Strategy protocol which was the default in those days. Dubious e.p. captures are just not permitted. This is why AP was invented as a funny work-around.
However this problem works under the PRA protocol, which decomposes the history as "one solution, two parts". Note that this problem does not make use of the e.p. convention, as it is certain that the two e.p. cannot be simultaneously legal. (2021-12-22)
Henrik Juel: I can strengthen the human contribution
The positions before c2-c4 and e2-e4 are surely legal (2021-12-22)
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A.Buchanan: The original stipulation here asks for 2 solutions. I don't think this would work as 2 solutions under Retro Strategy protocol which was the default in those days. Dubious e.p. captures are just not permitted. This is why AP was invented as a funny work-around.
However this problem works under the PRA protocol, which decomposes the history as "one solution, two parts". Note that this problem does not make use of the e.p. convention, as it is certain that the two e.p. cannot be simultaneously legal. (2021-12-22)
Henrik Juel: I can strengthen the human contribution
The positions before c2-c4 and e2-e4 are surely legal (2021-12-22)
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Keywords: En passant as key, Partial Retro Analysis (PRA)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 + Retractor 2.1.1 Human contribution is "position prior to retractions looks legal enough."
FEN: 3K4/1rp1n3/3Bbnq1/N1p2Qr1/1pPkPpp1/1p1p1p2/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-12-22 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 + Retractor 2.1.1 Human contribution is "position prior to retractions looks legal enough."
FEN: 3K4/1rp1n3/3Bbnq1/N1p2Qr1/1pPkPpp1/1p1p1p2/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-12-22 more...
1. Kc3 0-0-0 2. Txc4 Txd3#
1. bxc3ep e4 2. Kc4 Ta4#
The idea is that the ep in one solution is validated by the castling in the other solution. Since no other solutions exist, there are no parasites which might "piggyback" off the proof given by the castling solution. This is not PRA: both solutions have the same history with both castling & hence ep legal.
1. bxc3ep e4 2. Kc4 Ta4#
The idea is that the ep in one solution is validated by the castling in the other solution. Since no other solutions exist, there are no parasites which might "piggyback" off the proof given by the castling solution. This is not PRA: both solutions have the same history with both castling & hence ep legal.
Mario Richter: How is the (AP) to be interpreted here? Is the intention as follows: Since in solution 1) White castles, Black is allowed to capture e.p. in solution 2)? (2011-05-28)
Henrik Juel: It seems to be an unusual situation, not covered by keywords like AP or PRA.
If last move was c2-c4, both solutions work; if not, there is no solution, because White may not castle.
(In the second solution, Kxc4 should be Kc4). (2011-05-28)
VL: This a generalized type of AP, which I call "consolidate AP": both solutions are considered as parts of one complete solution. However the order of both parts is significant. I know two similar problems: P0003437 and P0003186. (2011-06-01)
A.Buchanan: @Valery: I've started to classify all the AP problems in PDB. This and its kin are "Type Petrovic - cons" (standing for consolidated, but there's only a limited number of characters for a PDB parameter). I truly hope that the theory in the end will not have this as a special case, but rather is part of a more general pattern.
It's not just that PRA & RS each scales up within its own paradigm to handle seamlessly very complex collections of conditional move dependencies. They also scale *down* to handle situation when there is just one conditional move, or even zero conditional moves. And when n=0 or n=1, PRA & RS are in complete agreement. This is the kind of robust seaworthy behaviour we need if are going to set sail in the stormy seas of fairydom.
But small steps. The first thing is to classify all the problems in PDB into buckets, to find out what we have. I doubt there are 37 categories, so will probably need the results of your scholarship too. (2022-02-17)
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Henrik Juel: It seems to be an unusual situation, not covered by keywords like AP or PRA.
If last move was c2-c4, both solutions work; if not, there is no solution, because White may not castle.
(In the second solution, Kxc4 should be Kc4). (2011-05-28)
VL: This a generalized type of AP, which I call "consolidate AP": both solutions are considered as parts of one complete solution. However the order of both parts is significant. I know two similar problems: P0003437 and P0003186. (2011-06-01)
A.Buchanan: @Valery: I've started to classify all the AP problems in PDB. This and its kin are "Type Petrovic - cons" (standing for consolidated, but there's only a limited number of characters for a PDB parameter). I truly hope that the theory in the end will not have this as a special case, but rather is part of a more general pattern.
It's not just that PRA & RS each scales up within its own paradigm to handle seamlessly very complex collections of conditional move dependencies. They also scale *down* to handle situation when there is just one conditional move, or even zero conditional moves. And when n=0 or n=1, PRA & RS are in complete agreement. This is the kind of robust seaworthy behaviour we need if are going to set sail in the stormy seas of fairydom.
But small steps. The first thing is to classify all the problems in PDB into buckets, to find out what we have. I doubt there are 37 categories, so will probably need the results of your scholarship too. (2022-02-17)
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Keywords: a posteriori (AP) (Type Petrovic consol), En passant as key, Castling (wg)
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: 8/8/8/1nr5/1pPk4/1p1p4/4P3/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-09-12 more...
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: 8/8/8/1nr5/1pPk4/1p1p4/4P3/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-09-12 more...
70 - P0003417
John Frederick Keeble
2206 The Problemist Fairy Chess Supplement 16, p. 173, 02/1936
(4+5) C+
h#2.5 (AP)
John Frederick Keeble
2206 The Problemist Fairy Chess Supplement 16, p. 173, 02/1936
(4+5) C+
h#2.5 (AP)
1. ... exd6ep 2. 0-0-0 dxe7+ 3. Tf8 exf8=T,D#
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
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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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...
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...
1. ... g6 2. 0-0 gxf7 3. Kh8 Le5#
1. ... cxb6ep 2. 0-0-0 bxa7 3. Td7 a8=D#
1. ... cxb6ep 2. 0-0-0 bxa7 3. Td7 a8=D#
Henrik Juel: Analysis
If Black can castle both ways, last move must have been b7-b5
C+ Popeye 4.61 (2020-10-25)
comment
If Black can castle both ways, last move must have been b7-b5
C+ Popeye 4.61 (2020-10-25)
comment
Keywords: a posteriori (AP), En passant as key, Castling (sksg), Promotion (D), Valladao Task
Genre: h#, Retro
FEN: r3k2r/p4p1p/8/KpP2PP1/4PBpP/5pp1/6p1/8
Reprints: Nordisches Turnier 1970-1971
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2020-10-25 more...
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...
1. ... fxg6ep 2. 0-0 gxh7#
Henrik Juel: Black captured c7xd6x..x.h2 and once more with an officer
White captured a2xb3, b2xa3, f2xg3, and e.g. exf and once more
If Black may castle, last move was g7-g5 (not b7-b6, because of Lh3) (2022-04-26)
Henrik Juel: HC+ Popeye 4.61
The castling serves two purposes:
enabling the mate and legitimizing the ep capture
so the a posteriori legitimizing is impure (2022-04-26)
A.Buchanan: Yes Henrik. The absence of other candidate solutions (beginning e.p. but excluding 0-0) declines an opportunity to embed additional content, and is arguably an artistic defect. However this very early AP problem is sound, and purity of motive is just one school. IMHO, there is space for such problems, if other content compensates: e.g. 4k2r/p2ppp1p/p7/5PpK/8/1PBB2Pb/2PP2Pp/8 h#2* AP. (2022-04-27)
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White captured a2xb3, b2xa3, f2xg3, and e.g. exf and once more
If Black may castle, last move was g7-g5 (not b7-b6, because of Lh3) (2022-04-26)
Henrik Juel: HC+ Popeye 4.61
The castling serves two purposes:
enabling the mate and legitimizing the ep capture
so the a posteriori legitimizing is impure (2022-04-26)
A.Buchanan: Yes Henrik. The absence of other candidate solutions (beginning e.p. but excluding 0-0) declines an opportunity to embed additional content, and is arguably an artistic defect. However this very early AP problem is sound, and purity of motive is just one school. IMHO, there is space for such problems, if other content compensates: e.g. 4k2r/p2ppp1p/p7/5PpK/8/1PBB2Pb/2PP2Pp/8 h#2* AP. (2022-04-27)
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Keywords: En passant as key, Castling (sk), a posteriori (AP) (Type Petrovic), Volet Pawn
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 4k2r/3ppp1p/1p6/4BPpK/P7/pPPB2Pb/3P2Pp/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-27 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 4k2r/3ppp1p/1p6/4BPpK/P7/pPPB2Pb/3P2Pp/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-27 more...
1. dxe3ep Td1? 2. Le4 Txd6 3. Sg4 Tf6# usual AP try
1. dxe3ep 0-0-0! 2. Le4 Txd6 3. Sg4 Tf6#
Cook: 1. dxe3ep 0-0-0 2. Lf5 Txd6 3. Sf3 Td4#
not
1. dxe3ep Td1? 2. Lf5 Txd6 3. Sf3+ ?? usual AP try and also check Bl 3rd move
1. dxe3ep 0-0-0! 2. Le4 Txd6 3. Sg4 Tf6#
Cook: 1. dxe3ep 0-0-0 2. Lf5 Txd6 3. Sf3 Td4#
not
1. dxe3ep Td1? 2. Lf5 Txd6 3. Sf3+ ?? usual AP try and also check Bl 3rd move
Henrik Juel: The missing black man is dark-squared, so last move was not fxe4. In the solution Td6 should read Txd6. There seems to be a 'variation': 0... dxe3ep 1.0-0-0 Le4 2.Txd6 Sg4 3.Tf6#, so maybe the stipulation should have 1.1;2.1;1.1 added. (2004-09-23)
A.Buchanan: Well spotted Henrik. Your kindness is legendary, but in the harsh world of helpmates, this counts as a cooked I believe. In fact, it's the 2.Lf5/3.Sf3 version which is the cook. Yours is the actual solution! This is because 3.Sf3 would check if White hadn't castled, and we want the castling to be because of the AP condition only, not for 2 reasons, which one might term a "logical dual". The problem can be trivially fixed by sticking sSe5 on f6. I have no idea whether it's a typo or a cook: WinChloe has the same diagram, and the same confusion between cook & intended solution. The fact that AP does not appear in the stipulation might be a factor. WinChloe as far as I can see has no concept of AP as a keyword, and just picks it up based on a substring of the stipulation. In other compositions, T.Petrovic wasn't shy of flashing the AP acronym in the stipulation, but as usual the detective work peters out in the absence of primary records. (2020-12-08)
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A.Buchanan: Well spotted Henrik. Your kindness is legendary, but in the harsh world of helpmates, this counts as a cooked I believe. In fact, it's the 2.Lf5/3.Sf3 version which is the cook. Yours is the actual solution! This is because 3.Sf3 would check if White hadn't castled, and we want the castling to be because of the AP condition only, not for 2 reasons, which one might term a "logical dual". The problem can be trivially fixed by sticking sSe5 on f6. I have no idea whether it's a typo or a cook: WinChloe has the same diagram, and the same confusion between cook & intended solution. The fact that AP does not appear in the stipulation might be a factor. WinChloe as far as I can see has no concept of AP as a keyword, and just picks it up based on a substring of the stipulation. In other compositions, T.Petrovic wasn't shy of flashing the AP acronym in the stipulation, but as usual the detective work peters out in the absence of primary records. (2020-12-08)
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Keywords: En passant as key, Castling (wg), a posteriori (AP) (Type Petrovic), Superseded by (P1382802)
Genre: h#, Retro
Computer test: Popeye v4.85 + retro thinking
FEN: 1n1r4/p1B4p/1p1p4/p3n1p1/3pPk2/3b4/pr5q/R3K3
Reprints: (60) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-16 more...
Genre: h#, Retro
Computer test: Popeye v4.85 + retro thinking
FEN: 1n1r4/p1B4p/1p1p4/p3n1p1/3pPk2/3b4/pr5q/R3K3
Reprints: (60) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-16 more...
1. cxb3ep+ c3 2. Ld3 Tg4+ 3. Kh1 0-0-0#
paul: White captures was axb and g2xh3, so the retro move c3xb4 is not possible. If b3-b4, the retro check is not justified. So last move was b2-b4 (preceded by Rc3-a3). (2011-08-06)
A.Buchanan: There's a lot to unpack here. Firstly, the intended solution has a definite transposition dual in the order of Black's 2nd & 3rd moves. Second, the pawn capture count is wrong. What happened to the original wBh, given sBh2 never left the h-file? Thirdly, and not a showstopper, White's only solution involves castling: so that's a logical dual. (2020-12-08)
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A.Buchanan: There's a lot to unpack here. Firstly, the intended solution has a definite transposition dual in the order of Black's 2nd & 3rd moves. Second, the pawn capture count is wrong. What happened to the original wBh, given sBh2 never left the h-file? Thirdly, and not a showstopper, White's only solution involves castling: so that's a logical dual. (2020-12-08)
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1. fxg3ep 0-0 2. Lg4 hxg3#
A.Buchanan: Surely diagram typo. Change to sBh3, then everything works (2022-03-29)
Mario Richter: Yes, Pawn h3 is black (2022-03-29)
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Mario Richter: Yes, Pawn h3 is black (2022-03-29)
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Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wk)
Genre: h#, Retro
FEN: 8/b3p3/4p3/6pp/2P2pPk/1pPP3p/2PP1P1P/r2bK2R
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-03-29 more...
Genre: h#, Retro
FEN: 8/b3p3/4p3/6pp/2P2pPk/1pPP3p/2PP1P1P/r2bK2R
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-03-29 more...
76 - P0003442
Janko Furman
3. Makuc-Moder-Gedenkturnier 1971-1973
1. Preis
(12+13) C+
h#2
b) wBd4 nach d5
Janko Furman
3. Makuc-Moder-Gedenkturnier 1971-1973
1. Preis
(12+13) C+
h#2
b) wBd4 nach d5
a) 1. cxd3ep Sd5 2. 0-0 Se7#
NOT 1. ... Lf6? 2. Kf8 Tx8# because no AP justification
b) 1. Kd7 Lf6 2. Te8 Sxb6#
NOT 1. 0-0? Tf6 2. Kh8 Txf8# because rights lost
Assume that bPb6 is really on b7:
Bl captures: dxe, exf, fxg, a|
Wh captures: bxc, g|, cxd=
So all pawn captures are accounted for.
a) If Bl 00 rights remain, then only way to give Black a prior move is by d2-d4.
b) No way to give Black a prior move, so Bl 00 rights must be lost.
Cook: NL
b) 1. 0-0 Tf6 2. Kh8 Txf8#
NOT 1. ... Lf6? 2. Kf8 Tx8# because no AP justification
b) 1. Kd7 Lf6 2. Te8 Sxb6#
NOT 1. 0-0? Tf6 2. Kh8 Txf8# because rights lost
Assume that bPb6 is really on b7:
Bl captures: dxe, exf, fxg, a|
Wh captures: bxc, g|, cxd=
So all pawn captures are accounted for.
a) If Bl 00 rights remain, then only way to give Black a prior move is by d2-d4.
b) No way to give Black a prior move, so Bl 00 rights must be lost.
Cook: NL
b) 1. 0-0 Tf6 2. Kh8 Txf8#
A.Buchanan: There is definitely something wrong here, with both (a) & (b). I think it's a simple diagram error: bPb7 has been misplaced on b6. Then the AP logic for (a) works great, and the castling "NL" for (b) is seen to be a thematic retro try. Can anyone confirm? (2022-03-21)
comment
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sk)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/7p/1pR5/2P5/NNpP4/KB2PPPP/p3pppq/B4bnr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/7p/1pR5/2P5/NNpP4/KB2PPPP/p3pppq/B4bnr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
77 - P0003444
Janko Furman
Miroslav Stosic
7273 Schach-Echo 11/1972
(8+14) cooked
h#2
b) Gespiegelt (a1<->h1) & wKd1->e1
Janko Furman
Miroslav Stosic
7273 Schach-Echo 11/1972
(8+14) cooked
h#2
b) Gespiegelt (a1<->h1) & wKd1->e1
a) 1. dxe3ep 0-0 2. Te4 Txf3#
b) 1. exd3ep 0-0-0 2. dxe2 Ld5#
Cook: a) 1. Kg3 Kf1 2. Df4 Th3#
1. Kg3 0-0 2. Dh4 Txf3#
b) 1. exd3ep 0-0-0 2. dxe2 Ld5#
Cook: a) 1. Kg3 Kf1 2. Df4 Th3#
1. Kg3 0-0 2. Dh4 Txf3#
A.Buchanan: This is a very heavy position to prevent the possibility of R: 1. c2xb3. I'm not sure why this was done. A much lighter position 8/8/8/5np1/1r1pPkr1/2Bp1p2/1p1P2P1/4K2R with 7 less units achieves the mates soundly. Am I missing something? (2022-02-16)
comment
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wk,wg), Superseded by (P1399967), Twinning by board reflection
Genre: h#, Retro
FEN: 8/1p1pp1p1/8/5nq1/1r1pPkr1/1PBp1p2/Pp1P2P1/2n1K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
Genre: h#, Retro
FEN: 8/1p1pp1p1/8/5nq1/1r1pPkr1/1PBp1p2/Pp1P2P1/2n1K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
1) 1. gxf3ep 0-0-0 2. Te7 Th4#
2) 1. cxd3ep 0-0 2. Da4 Txa4#
2) 1. cxd3ep 0-0 2. Da4 Txa4#
AB: I think this is cooked. I don't see how AP castling can justify earlier ep here.
1. Kf3 Bf5 2. Kg2 Be4#
1. Kf3 Bxd7 2. Kg2 Bc6#
1. cxd3 Rf1 2. Qa4 Rxa4#
1. gxf3ep Ra5 2. Kf4 Rh4#
1. gxf3ep Rd1 2. Re7 Rh4#
Also, second solution given has typos 1. *c*xd3ep & 2. Q*a*4. (2002-03-21)
V.Liskovets: Indeed this problem is cooked, and I
failed to correct it preserving symmetry.
Here is a possible correction:
W: Ke1 Ra1 Rh1 Be6 Pd4 Pf4
B: Ke4 Qd7 Rf7 Pc3 Pc4 Pe3 Pg3 Pg4 Bh7 Rb2 Pb3
Another story, justifying its contents (e.p.).
In my opinion, all published treatments are
insatisfactory (the same concerns P0004295
(corrected) by Werner Kuntsche as well).
There is a way to make it sound under the
sophisticated HYBRID stipulation 'AP, pRA':
2 partial solutions legalized JOINTLY by
both castlings (details to be published). (2002-04-02)
VL: Sorry, Andrew, I see no issues with my version (excepting the mentioned special retro-convention/genre for justifying its soundness, of course!). h#2 (pRA&AP). The full solution consists of 2 partial AP-based ones: I 1.gxf3 e.p.(!?) 0-0-0! 2.Te7 Th4#; II 1.cxd3 e.p.(!?) 0-0! 2.Da4 Txa4#.
C+(popeye): h#2 & two h#1.5 after the keys (added Tb2 & Lh7 are cookstoppers).
The main features of the corresponding suitable AP-genre ("consolidated")are rather clear. However, not all subtle aspects have been analyzed thoroughly yet. I have got only a draft manuscript with few examples, still. (2022-02-15)
A.Buchanan: Yes Valery I think your correction is technically sound. Theological debate is something else, heh. Have posted it as separate entry. Please tell me if it was original here, or appeared somewhere else (2022-05-29)
Henrik Juel: I looked in Stella Polaris 1971 p.202-203, where the problem is given without AP under the diagram
It turns out that both the author and the editor (Jan Knöppel) consider the problem correct
After explaining the AP convention the latter writes [my translation]:
The author has elegantly applied the AP conventionn in combination with partial analysis. If we assume that last move was not f2-f4 or T-h1, the possibilities d2-d4, K-e1, and T-a1 remain, but the two latter disappear by 1... 0-0. The other assumption (last move was not d2-d4 or T-a1) is analogous.
So something like Valery's HYBRID stipulation was the intention, and the problem was considered correct in 1971; later developments do not change this fact (2022-05-29)
Henrik Juel: Of course, the obvious cooks 1.Kf3 etc. remain (2022-05-29)
A.Buchanan: I agree. Basic AP Petrovic does not allow multiple ep, nor does RS, so some alignment of AP with PRA is required. I would like to see this codified though. (2022-05-29)
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1. Kf3 Bf5 2. Kg2 Be4#
1. Kf3 Bxd7 2. Kg2 Bc6#
1. cxd3 Rf1 2. Qa4 Rxa4#
1. gxf3ep Ra5 2. Kf4 Rh4#
1. gxf3ep Rd1 2. Re7 Rh4#
Also, second solution given has typos 1. *c*xd3ep & 2. Q*a*4. (2002-03-21)
V.Liskovets: Indeed this problem is cooked, and I
failed to correct it preserving symmetry.
Here is a possible correction:
W: Ke1 Ra1 Rh1 Be6 Pd4 Pf4
B: Ke4 Qd7 Rf7 Pc3 Pc4 Pe3 Pg3 Pg4 Bh7 Rb2 Pb3
Another story, justifying its contents (e.p.).
In my opinion, all published treatments are
insatisfactory (the same concerns P0004295
(corrected) by Werner Kuntsche as well).
There is a way to make it sound under the
sophisticated HYBRID stipulation 'AP, pRA':
2 partial solutions legalized JOINTLY by
both castlings (details to be published). (2002-04-02)
VL: Sorry, Andrew, I see no issues with my version (excepting the mentioned special retro-convention/genre for justifying its soundness, of course!). h#2 (pRA&AP). The full solution consists of 2 partial AP-based ones: I 1.gxf3 e.p.(!?) 0-0-0! 2.Te7 Th4#; II 1.cxd3 e.p.(!?) 0-0! 2.Da4 Txa4#.
C+(popeye): h#2 & two h#1.5 after the keys (added Tb2 & Lh7 are cookstoppers).
The main features of the corresponding suitable AP-genre ("consolidated")are rather clear. However, not all subtle aspects have been analyzed thoroughly yet. I have got only a draft manuscript with few examples, still. (2022-02-15)
A.Buchanan: Yes Valery I think your correction is technically sound. Theological debate is something else, heh. Have posted it as separate entry. Please tell me if it was original here, or appeared somewhere else (2022-05-29)
Henrik Juel: I looked in Stella Polaris 1971 p.202-203, where the problem is given without AP under the diagram
It turns out that both the author and the editor (Jan Knöppel) consider the problem correct
After explaining the AP convention the latter writes [my translation]:
The author has elegantly applied the AP conventionn in combination with partial analysis. If we assume that last move was not f2-f4 or T-h1, the possibilities d2-d4, K-e1, and T-a1 remain, but the two latter disappear by 1... 0-0. The other assumption (last move was not d2-d4 or T-a1) is analogous.
So something like Valery's HYBRID stipulation was the intention, and the problem was considered correct in 1971; later developments do not change this fact (2022-05-29)
Henrik Juel: Of course, the obvious cooks 1.Kf3 etc. remain (2022-05-29)
A.Buchanan: I agree. Basic AP Petrovic does not allow multiple ep, nor does RS, so some alignment of AP with PRA is required. I would like to see this codified though. (2022-05-29)
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Keywords: Castling (wb), En passant as key (2), a posteriori (AP) (Type Petrovic), Superseded by (P1401508, P1399178), Quasi-symmetrical position, Partial Retro Analysis (PRA)
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/3q1r2/4B3/8/2pPkPp1/2p1p1p1/8/R3K2R
Reprints: (65) Problem 144-147 12/1971
(8) StrateGems SG19, p. 156, 07/2002
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-29 more...
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/3q1r2/4B3/8/2pPkPp1/2p1p1p1/8/R3K2R
Reprints: (65) Problem 144-147 12/1971
(8) StrateGems SG19, p. 156, 07/2002
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-29 more...
1. cxd6ep
paul: Since the wB couldnt leave c1, the captured piece on b6 is the promoted wPh2. If last black move was f7-f5, then h-Pawn captured four pieces to promote on d8, among which bRh8, impossible. Thus, the h-Pawn could only have been promoted from f7 and therefore d7-d5 remains as the last move. (2011-07-02)
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more ...
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Keywords: En passant as key
Genre: Retro, 2#
Computer test: C+ Popeye 4.61
FEN: qrRNkb2/PpQ1p1pp/1pp1R3/2PpKpP1/6N1/7B/1P1PPP2/8
Reprints: RA15 diagrammes 15 06/1975
(51) Mat Plus 12 10-12/2009
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-27 more...
Genre: Retro, 2#
Computer test: C+ Popeye 4.61
FEN: qrRNkb2/PpQ1p1pp/1pp1R3/2PpKpP1/6N1/7B/1P1PPP2/8
Reprints: RA15 diagrammes 15 06/1975
(51) Mat Plus 12 10-12/2009
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-27 more...
80 - P0003925
Ivan Skoba
1009 diagrammes 47 09-10/1980
(6+7) C+
a) ser-h#6 (AP)
b) nach dem Schlüssel von a): ser-h#5
Ivan Skoba
1009 diagrammes 47 09-10/1980
(6+7) C+
a) ser-h#6 (AP)
b) nach dem Schlüssel von a): ser-h#5
a)
1. dxc3ep 2. Kc4? 3. Tb5 4. Td5 5. Dc5 6. Kd4 Th4#
2. Tb2! 3. Kb3 4. Kc2 5. Kc1 6. c2 0-0#
b)
1. Tb2? 2. Kb3 3. Kc2 4. Kc1 5. c2 0-0#
1. Kc4! 2. Tb5 3. Td5 4. Dc5 5. Kd4 Th4#
There are basically two candidate solutions ending Th4# and 0-0#. In (a), the initial ep can only be justified by 0-0#. On the other hand in the diagram position in (b), White's last move must have been with wK or wR, so 0-0# is disrupted. Nice and paradoxical.
1. dxc3ep 2. Kc4? 3. Tb5 4. Td5 5. Dc5 6. Kd4 Th4#
2. Tb2! 3. Kb3 4. Kc2 5. Kc1 6. c2 0-0#
b)
1. Tb2? 2. Kb3 3. Kc2 4. Kc1 5. c2 0-0#
1. Kc4! 2. Tb5 3. Td5 4. Dc5 5. Kd4 Th4#
There are basically two candidate solutions ending Th4# and 0-0#. In (a), the initial ep can only be justified by 0-0#. On the other hand in the diagram position in (b), White's last move must have been with wK or wR, so 0-0# is disrupted. Nice and paradoxical.
Keywords: Seriesmover, Castling (wk), Cant Castler, a posteriori (AP) (Type Petrovic), En passant as key
Genre: Retro, Fairies
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 8/8/8/8/bkPp4/pr1pq3/3PRP2/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-29 more...
Genre: Retro, Fairies
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 8/8/8/8/bkPp4/pr1pq3/3PRP2/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-29 more...
1. ... dxc6ep 2. dxc3 Lxb6#
1. dxe3 Sd4 2. e2 Sc6#
1. dxe3 Sd4 2. e2 Sc6#
Keywords: En passant as key, Interchange (Sb)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic
FEN: B5Rr/k4pPp/1p3P1p/1KpPP2N/2Pp4/2PPB3/2P1N3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-23 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic
FEN: B5Rr/k4pPp/1p3P1p/1KpPP2N/2Pp4/2PPB3/2P1N3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-23 more...
1. axb6ep+ Kb5 2. bxc7+ Tb6 3. 0-0-0 ...
K. Fabel: "Von Interesse ist, dass diese strittige Idee [Beweis des e.p.-Schlagrechts durch spätere Ausführung der Rochade] auch im direkten Mattproblem dargestellt werden kann, vergl. das Diagramm [diese Aufgabe P0004199]. (Vielleicht ist diese Aufgabe noch nicht korrekt doch es wird nicht schwer sein, sie zu verbessern). Falls Schwarz zuletzt Bb6-b5, Tb6-c6 oder Lb6-c7 gezogen hat, ergibt sich, dass Weiss im letzten oder vorletzten Zug den K oder T bewegt haben muss. Falls jedoch K und T noch nicht gezogen haben, muss b7-b5 der letzte Zug gewesen sein. Weiss spielt daher 1. ab6e.p.+ Kb5 2. bc7+ Tb6. Jetzt könnte Weiss mit Dxb6 sofort mattsetzen, aber er muss ja den e.p.-Schlag noch legalisieren. Daher 3. 0-0-0! und Matt erst im 5. Zuge."
Cook: 1. Sxc7+!
1. Lxc7!
K. Fabel: "Von Interesse ist, dass diese strittige Idee [Beweis des e.p.-Schlagrechts durch spätere Ausführung der Rochade] auch im direkten Mattproblem dargestellt werden kann, vergl. das Diagramm [diese Aufgabe P0004199]. (Vielleicht ist diese Aufgabe noch nicht korrekt doch es wird nicht schwer sein, sie zu verbessern). Falls Schwarz zuletzt Bb6-b5, Tb6-c6 oder Lb6-c7 gezogen hat, ergibt sich, dass Weiss im letzten oder vorletzten Zug den K oder T bewegt haben muss. Falls jedoch K und T noch nicht gezogen haben, muss b7-b5 der letzte Zug gewesen sein. Weiss spielt daher 1. ab6e.p.+ Kb5 2. bc7+ Tb6. Jetzt könnte Weiss mit Dxb6 sofort mattsetzen, aber er muss ja den e.p.-Schlag noch legalisieren. Daher 3. 0-0-0! und Matt erst im 5. Zuge."
Cook: 1. Sxc7+!
1. Lxc7!
Innerhalb des 2. Teils einer Artikelserie "Die Konventionen im Problemschach" von Karl Fabel.
A.Buchanan: The retro logic is fine, but forward play is savagely cooked. I wonder about changing wSc8 to sS. The retro stuff still works OK, indeed sSc8 is a fourth thematic unit to retract in the try. Ignoring proof of ep legality, there is a unique #4 beginning with ep, and no #5. Promising: but when can we interpolate w000? Any solution must begin 1. axb6ep+ Kb5. There are now numerous #4, but none include w000, so I think we must stick with the still unique line beginning 2. Dxc8 thr 3. 0-0-0. There are 3 black defenses to refute this. 2. ... Txb6,axb3,d5. So maybe #6 needed for White to prevail? (2022-03-28)
A.Buchanan: But if we shift to #6 then 1. Sxc7!,Qxc8+! mate without ep. Can anyone retrieve the original diagram for this one, please? (2022-03-28)
Mario Richter: The position here is identical to the original diagram, but it should be regarded more as a schematic example than a "real problem". Fabel in his quote above: "... perhaps the problem is still not correct, but it should be easy to improve it ..." (2022-03-28)
Henrik Juel: unfortunately K. Fabel forgot to give the easy correction... (2022-03-28)
A.Buchanan: Thanks for this. Valery Liskovetz, an AP expert, was kind enough to send a pdf of the relevant Problem issue, so I can confirm. It's easy enough to remove the two cooks (e.g. sTc8) but the difficulty is in validation of the intended solution. Is there any "exact" option in Popeye that forces the solution to include a waiting move? The actual solution might be expected to be included in that set. (2022-03-29)
Henrik Juel: Yes, Andrew
'stip exact-#5' would disregard #4 as a solution, I believe (2022-03-29)
A.Buchanan: Even with the "exact", it's still very complicated to investigate. I think this kind of problem requires an engine feature to disallow any mate or pat if the specified castling has not been executed. Gazing at the 20,000 lines of output, I don't think sTc8 works as 1. axb6ep+ Kb5 2. Qb7 bxa3! threatens axb2+ after castling. I wonder if the board might be flipped left-right except for wK, but I haven't explored this. (2022-05-31)
comment
A.Buchanan: The retro logic is fine, but forward play is savagely cooked. I wonder about changing wSc8 to sS. The retro stuff still works OK, indeed sSc8 is a fourth thematic unit to retract in the try. Ignoring proof of ep legality, there is a unique #4 beginning with ep, and no #5. Promising: but when can we interpolate w000? Any solution must begin 1. axb6ep+ Kb5. There are now numerous #4, but none include w000, so I think we must stick with the still unique line beginning 2. Dxc8 thr 3. 0-0-0. There are 3 black defenses to refute this. 2. ... Txb6,axb3,d5. So maybe #6 needed for White to prevail? (2022-03-28)
A.Buchanan: But if we shift to #6 then 1. Sxc7!,Qxc8+! mate without ep. Can anyone retrieve the original diagram for this one, please? (2022-03-28)
Mario Richter: The position here is identical to the original diagram, but it should be regarded more as a schematic example than a "real problem". Fabel in his quote above: "... perhaps the problem is still not correct, but it should be easy to improve it ..." (2022-03-28)
Henrik Juel: unfortunately K. Fabel forgot to give the easy correction... (2022-03-28)
A.Buchanan: Thanks for this. Valery Liskovetz, an AP expert, was kind enough to send a pdf of the relevant Problem issue, so I can confirm. It's easy enough to remove the two cooks (e.g. sTc8) but the difficulty is in validation of the intended solution. Is there any "exact" option in Popeye that forces the solution to include a waiting move? The actual solution might be expected to be included in that set. (2022-03-29)
Henrik Juel: Yes, Andrew
'stip exact-#5' would disregard #4 as a solution, I believe (2022-03-29)
A.Buchanan: Even with the "exact", it's still very complicated to investigate. I think this kind of problem requires an engine feature to disallow any mate or pat if the specified castling has not been executed. Gazing at the 20,000 lines of output, I don't think sTc8 works as 1. axb6ep+ Kb5 2. Qb7 bxa3! threatens axb2+ after castling. I wonder if the board might be flipped left-right except for wK, but I haven't explored this. (2022-05-31)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Non-standard material (TLL)
Genre: Retro, n#
FEN: NQNB4/B1brpp2/k1rp4/Ppp5/Rp6/BP6/RPP5/R3K3
Reprints: (8) Problem 101-102 09/1966
(52) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-29 more...
Genre: Retro, n#
FEN: NQNB4/B1brpp2/k1rp4/Ppp5/Rp6/BP6/RPP5/R3K3
Reprints: (8) Problem 101-102 09/1966
(52) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-29 more...
1. exf6ep d5,g5 2. Db7 g5,d5 3. Dg7#
Henrik Juel: The longest among the possible last moves is f7-f5, so 1.exf6ep d5 2.f7 g5 3.f8Q!?, but 3... Kh7!, so something is wrong. (2003-11-11)
HBae: Habe korrigiert. Es fehlte der wKa1 und die wDb2. (2019-10-11)
Henrik Juel: C+ by Popeye 4.61 (assuming that the condition was in effect also in last move)
Pd7 and Pg4 can be removed without affecting correctness or symmetry (2019-10-11)
Bernd Schwarzkopf: Pd7 and Pg4 are necessary. Without them last move could have been: 1.Pg6xXf5. (2021-02-06)
Mario Richter: Interestingly, the Editors of 'TfS' too thought that Pd7 and Pg4 can be omitted (s. 'TfS' 06/1938, p.124: "Går det inte lika bra utan bönderna på d7 och g4? (BL). Jo, det går visst lika bra. Men förf. har väl haft någon mening med dem också (Red.)."
I think, pawn d7 serves to prevent retractions like R: 1.Pg6xXf5, but pawn d4 is only there to complete the symmetry (i.e. Pd7 is necessary, Pg4 is not). (2021-02-06)
A.Buchanan: Doesn’t bPg4 stop 2.Dg2 as a dual? (2021-02-07)
comment
HBae: Habe korrigiert. Es fehlte der wKa1 und die wDb2. (2019-10-11)
Henrik Juel: C+ by Popeye 4.61 (assuming that the condition was in effect also in last move)
Pd7 and Pg4 can be removed without affecting correctness or symmetry (2019-10-11)
Bernd Schwarzkopf: Pd7 and Pg4 are necessary. Without them last move could have been: 1.Pg6xXf5. (2021-02-06)
Mario Richter: Interestingly, the Editors of 'TfS' too thought that Pd7 and Pg4 can be omitted (s. 'TfS' 06/1938, p.124: "Går det inte lika bra utan bönderna på d7 och g4? (BL). Jo, det går visst lika bra. Men förf. har väl haft någon mening med dem också (Red.)."
I think, pawn d7 serves to prevent retractions like R: 1.Pg6xXf5, but pawn d4 is only there to complete the symmetry (i.e. Pd7 is necessary, Pg4 is not). (2021-02-06)
A.Buchanan: Doesn’t bPg4 stop 2.Dg2 as a dual? (2021-02-07)
comment
Keywords: Maximummer, En passant as key, Asymmetrical solution, Symmetrical position
Genre: Retro, 3#, Fairies
FEN: 7k/3p2p1/4p3/4Pp2/6p1/8/1Q6/K7
Reprints: (II) Problem 65-68 01/1960
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
Genre: Retro, 3#, Fairies
FEN: 7k/3p2p1/4p3/4Pp2/6p1/8/1Q6/K7
Reprints: (II) Problem 65-68 01/1960
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
1. fxe3ep? Dxa8# No justification for the ep, so just retro try.
White pcs: dxexf5,hxBg. Black none. Assume that Black can still castle. White can't have just played h4xg5 as sBh2 would be blocked. White might apparently just have moved D,Se5,Td4,Tc3,g4-g5,a5-a6,e3-e4 or e2-e4 but what might Black have played before? Only the last allows a move sDf1-e1. So set up for AP Type Petrovic is OK.
Cook: 446 candidate h#2
However 27 have no ep, while 59 have both ep & castling.
So a total of 86 viable solutions, 85 of which must be cooks.
White pcs: dxexf5,hxBg. Black none. Assume that Black can still castle. White can't have just played h4xg5 as sBh2 would be blocked. White might apparently just have moved D,Se5,Td4,Tc3,g4-g5,a5-a6,e3-e4 or e2-e4 but what might Black have played before? Only the last allows a move sDf1-e1. So set up for AP Type Petrovic is OK.
Cook: 446 candidate h#2
However 27 have no ep, while 59 have both ep & castling.
So a total of 86 viable solutions, 85 of which must be cooks.
Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004341)
Genre: h#, Retro
Computer test: Popeye v4.87 indicates cook
FEN: r3k3/p1ppp3/Pp6/4NPP1/2PRPp2/2RK2PN/1PBn1PQp/2Brq1b1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
Genre: h#, Retro
Computer test: Popeye v4.87 indicates cook
FEN: r3k3/p1ppp3/Pp6/4NPP1/2PRPp2/2RK2PN/1PBn1PQp/2Brq1b1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
1. fxe3ep Dxa8#? because no AP justification for ep
1. 0-0-0 g7 2. Tf8 gxf8=D#
62 apparent h#2, so seems cooked.
Cook: 415 candidate solutions for h#2. 401 begin with ep, of which 48 contain 0-0-0 to justify. There are also 14 solutions without ep, of which 13 begin with 0-0-0. The odd one is 1. Kd8 Sc6+ 2. Kc8 Sge7#
1. 0-0-0 g7 2. Tf8 gxf8=D#
62 apparent h#2, so seems cooked.
Cook: 415 candidate solutions for h#2. 401 begin with ep, of which 48 contain 0-0-0 to justify. There are also 14 solutions without ep, of which 13 begin with 0-0-0. The odd one is 1. Kd8 Sc6+ 2. Kc8 Sge7#
Version zu P0004296
Autor: "In Nr.9, I intended to do something similiar to No. 8 [P0004199], but with a different key: the solution 1. PxP "e.p.", Qxa8 checkmate will not go, for if blackside does not castle, the "en-passant" capture cannot be justified. Though, well considering, there is the demolition 1. PxP "e.p." -any 2. 0-0-0! Qb7 or a8 checkmate.
A.Buchanan: Thanks Mario for retrieving the author's intent with this. So he knew it was cooked! Assume s000 rights remain. R: 1. b5xa6? as sBb promoted on b1. R: 1. h4xg5? as sBh retro-blocked. R: 1. Ke3-d3? impossible check from sBf4. R. 1. K~-d3 d3-d2? illegal check. I like these! So by elimination, R: 1. e2-e4 e3xTd2 2. T~d2 L~e1. So must have sLe1. (2022-03-22)
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Autor: "In Nr.9, I intended to do something similiar to No. 8 [P0004199], but with a different key: the solution 1. PxP "e.p.", Qxa8 checkmate will not go, for if blackside does not castle, the "en-passant" capture cannot be justified. Though, well considering, there is the demolition 1. PxP "e.p." -any 2. 0-0-0! Qb7 or a8 checkmate.
A.Buchanan: Thanks Mario for retrieving the author's intent with this. So he knew it was cooked! Assume s000 rights remain. R: 1. b5xa6? as sBb promoted on b1. R: 1. h4xg5? as sBh retro-blocked. R: 1. Ke3-d3? impossible check from sBf4. R. 1. K~-d3 d3-d2? illegal check. I like these! So by elimination, R: 1. e2-e4 e3xTd2 2. T~d2 L~e1. So must have sLe1. (2022-03-22)
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Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P0004342)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k1N1/p1pp4/P5P1/4PPp1/2PNPprB/3K1QRp/P2p1PB1/4b2n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k1N1/p1pp4/P5P1/4PPp1/2PNPprB/3K1QRp/P2p1PB1/4b2n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
1. bxc3ep Dg8#? (thematic try)
1. bxc3ep Dd5 2. 0-0-0 Db7#
Cook: 1. bxc3ep Ld6 2. 0-0-0 Dc4#
1. bxc3ep Dd5 2. 0-0-0 Db7#
Cook: 1. bxc3ep Ld6 2. 0-0-0 Dc4#
Author: "From the former [d.h. P0004341], the No. 10 was born, something similar to No. 8 [P0004199] but with Black-castling for it is a help-mate. The mating move 1. ... Qg8? will not go."
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Keywords: Castling (sg), En passant as key, a posteriori (AP) (Type Petrovic), Superseded by (P1399966)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k3/p2p4/8/NP6/BpPPN1PP/B2K1Ppp/QP1p1pP1/Rrb5
Reprints: (55) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: r3k3/p2p4/8/NP6/BpPPN1PP/B2K1Ppp/QP1p1pP1/Rrb5
Reprints: (55) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
* 1. ... Lxd2#
1. dxe3ep 2. Txf2 3. Txg2 4. Th2 5. Th3 0-0#
1. dxe3ep 2. Txf2 3. Txg2 4. Th2 5. Th3 0-0#
Keywords: Castling (wk), Seriesmover, En passant as key, a posteriori (AP) (Type Petrovic)
Genre: Retro, Fairies
Computer test: HC+ Popeye 4.61 after analysis
FEN: 8/8/8/4pPp1/1p1pPkpb/1P1P2p1/pPpr1PP1/rbB1K2R
Reprints: 103 Bilten 1970 1971
(70) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-20 more...
Genre: Retro, Fairies
Computer test: HC+ Popeye 4.61 after analysis
FEN: 8/8/8/4pPp1/1p1pPkpb/1P1P2p1/pPpr1PP1/rbB1K2R
Reprints: 103 Bilten 1970 1971
(70) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-20 more...
88 - P0004481
Matjaz Zigman
Länderkampf Mazedonien-Slowenien 1969
7. Platz
(13+10) cooked
ser-h#6** AP
Matjaz Zigman
Länderkampf Mazedonien-Slowenien 1969
7. Platz
(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!#
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
* 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!#
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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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...
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...
1. Kg6,Sf8 g3 2. f7#
1. fxg6ep? because e.g. R: 1. Kf7-g8 gxSh8=S+ can be history
1. fxg6ep? because e.g. R: 1. Kf7-g8 gxSh8=S+ can be history
Paulo Roque: NL : 1. Sf8! g3 2. f7# (2009-08-22)
Henrik Juel: The organizers of solving matches in the old days liked tricky problems. Here there are two tricks: The 'obvious' e.p. key does not work (because last move could be Kf7-g8), and there are two solutions. (2009-08-22)
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Henrik Juel: The organizers of solving matches in the old days liked tricky problems. Here there are two tricks: The 'obvious' e.p. key does not work (because last move could be Kf7-g8), and there are two solutions. (2009-08-22)
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Keywords: En passant as key
Genre: Retro, 2#
FEN: 6kN/3N3R/5P1B/5PpK/6p1/8/8/8
Reprints: (10) Problem 198-201 07/1980
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-01 more...
Genre: Retro, 2#
FEN: 6kN/3N3R/5P1B/5PpK/6p1/8/8/8
Reprints: (10) Problem 198-201 07/1980
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-01 more...
*) 1. ... Ta4 2. Kxg4 Txf4#
1. fxg3ep Ta4#? (no proof of ep right)
1. fxg3ep 0-0-0! 2. g2 Td4#
A short retro try h#1 and 16 h#2 tries. B2 is unique tempo move
Cook: 1. Kxg4 Ta5 2. h4 Tg5#
1. Kxg4 Ta4 2. Kh4 Txf4#
1. fxg3ep Ta4#? (no proof of ep right)
1. fxg3ep 0-0-0! 2. g2 Td4#
A short retro try h#1 and 16 h#2 tries. B2 is unique tempo move
Cook: 1. Kxg4 Ta5 2. h4 Tg5#
1. Kxg4 Ta4 2. Kh4 Txf4#
Keywords: Castling (wg), En passant as key, a posteriori (AP) (Type Petrovic), Tempo Move
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 8/6p1/6pB/7p/5pPk/5p1p/5P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-06 more...
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 8/6p1/6pB/7p/5pPk/5p1p/5P2/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-06 more...
91 - P0005275
Otto Kerekes
Tivadar Kardos
10 L'Echiquier de France 11/1956
(11+15) cooked
h#2
b) sBg2 statt wBg2
Otto Kerekes
Tivadar Kardos
10 L'Echiquier de France 11/1956
(11+15) cooked
h#2
b) sBg2 statt wBg2
a) 1. cxd4 Lxc4 2. Dd5 Ld3#
b)
b)
Paulo Roque: Illegale Stellung. Diagrammfehler? (2009-11-17)
Alfred Pfeiffer: außerdem in b) 9 schwarze Bauern. (2010-01-13)
A.Buchanan: Twin a) is also illegal, as too many Bl captures (2021-11-24)
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Alfred Pfeiffer: außerdem in b) 9 schwarze Bauern. (2010-01-13)
A.Buchanan: Twin a) is also illegal, as too many Bl captures (2021-11-24)
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Keywords: En passant as key, Castling, Illegal position, Superseded by (P1400824)
Genre: h#, Retro
FEN: 8/p1p2p1b/1B3pp1/q1p1b3/r1pPk3/1Bp3P1/PPP2PPr/3n1RK1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-23 more...
Genre: h#, Retro
FEN: 8/p1p2p1b/1B3pp1/q1p1b3/r1pPk3/1Bp3P1/PPP2PPr/3n1RK1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-23 more...
*) 1. ... Sxb3#
1) 1. gxh3ep Ld1 2. h2 Lxc2 3. Kxc2 Df1 4. Kc3 Dxd3+ 5. Lxd3 0-0-0 6. Lxc4 Se4#
Wh has made 6 visible pawn captures, Bl 1. If bPfxg, then bPh was waylaid, and bPe promoted, disrupting White's castling rights. If Wh 000 rights remain, therefore, bPhxg and bPe was waylaid instead. Since original g-pawns remain on g-file, they must be wPg3 & bPg4, and wPh must retract a double hop to allow bPh3xg2. So Black can ep, avoiding immediate pat. To castle then mate requires a lot of work.
1) 1. gxh3ep Ld1 2. h2 Lxc2 3. Kxc2 Df1 4. Kc3 Dxd3+ 5. Lxd3 0-0-0 6. Lxc4 Se4#
Wh has made 6 visible pawn captures, Bl 1. If bPfxg, then bPh was waylaid, and bPe promoted, disrupting White's castling rights. If Wh 000 rights remain, therefore, bPhxg and bPe was waylaid instead. Since original g-pawns remain on g-file, they must be wPg3 & bPg4, and wPh must retract a double hop to allow bPh3xg2. So Black can ep, avoiding immediate pat. To castle then mate requires a lot of work.
A.Buchanan: Popeye v4.87 finds 11 solutions to h004.5 following the mandatory ep. Only one of them can be followed by h#1. Any cook therefore would have to be of the form 6. ... 000# Searching now for h005.5, as can then easily check if there are any mates (2022-06-10)
Yuri Bilokin: H#1* *) 1. ... Sxb3#
1) 1. gxh3ep Sxb3# (2022-06-10)
A.Buchanan: Hi Yuri - you know that h#1 doesn't work, right? (2022-06-10)
Yuri Bilokin: Hi - don't know, please email polidox579@gmail.com (2022-06-11)
A.Buchanan: Hi there are over 110,000 ways to reach 6. 000, but *none* of them are also checkmate. Therefore I am happy to pronounce this problem sound. Happy, because (1) it deserves to be ok (2) I couldn't face trying to fix it. If White 000 rights remain, then the last move was certainly h2-h4. Orthodoxically, we cannot perform the e.p. but under the AP Type Petrovic we can ep, as long as we justify it with castling later in the game. Hurray! (2022-06-12)
A.Buchanan: 1. gxh3ep Sxb3#? fails because White never castles to justify the ep. It's like there is an additional check before granting that a position is checkmate or stalemate: have all AP debts been paid? If not, the move is illegal because the game would end with no chance to repay the AP debt later. Yes it's weird but that's AP. (2022-06-13)
Michel Caillaud: Now, the post en passant part of this kind of problem can be tested with Jacobi.
With the following set of data:
stip h#5.5 pieces
White Ke1 Qg1 Ra1f2 Be3h5 Sd2 Pa3b4c4d4f4g3g5
Black Kc1 Rb2 Bb1 Pa2b3c2d3g2h3
constraints Ke1!c1~ Ra1!d1~
Jacobi looks for helpmates in 5.5 moves including 0-0-0 in the play.
No need to scrutinize the 110000 ways found by Andrew; Jacobi makes all the work (but it takes some hours...) (2022-06-13)
more ...
comment
Yuri Bilokin: H#1* *) 1. ... Sxb3#
1) 1. gxh3ep Sxb3# (2022-06-10)
A.Buchanan: Hi Yuri - you know that h#1 doesn't work, right? (2022-06-10)
Yuri Bilokin: Hi - don't know, please email polidox579@gmail.com (2022-06-11)
A.Buchanan: Hi there are over 110,000 ways to reach 6. 000, but *none* of them are also checkmate. Therefore I am happy to pronounce this problem sound. Happy, because (1) it deserves to be ok (2) I couldn't face trying to fix it. If White 000 rights remain, then the last move was certainly h2-h4. Orthodoxically, we cannot perform the e.p. but under the AP Type Petrovic we can ep, as long as we justify it with castling later in the game. Hurray! (2022-06-12)
A.Buchanan: 1. gxh3ep Sxb3#? fails because White never castles to justify the ep. It's like there is an additional check before granting that a position is checkmate or stalemate: have all AP debts been paid? If not, the move is illegal because the game would end with no chance to repay the AP debt later. Yes it's weird but that's AP. (2022-06-13)
Michel Caillaud: Now, the post en passant part of this kind of problem can be tested with Jacobi.
With the following set of data:
stip h#5.5 pieces
White Ke1 Qg1 Ra1f2 Be3h5 Sd2 Pa3b4c4d4f4g3g5
Black Kc1 Rb2 Bb1 Pa2b3c2d3g2h3
constraints Ke1!c1~ Ra1!d1~
Jacobi looks for helpmates in 5.5 moves including 0-0-0 in the play.
No need to scrutinize the 110000 ways found by Andrew; Jacobi makes all the work (but it takes some hours...) (2022-06-13)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg)
Genre: h#, Retro
Computer test: C+ Jacobi v0.7.5
FEN: 8/8/8/6PB/1PPP1PpP/Pp1pB1P1/prpN1Rp1/Rbk1K1Q1
Reprints: 4113 Problem 05/1979
(19) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-07-19
Last update: A.Buchanan, 2022-07-10 more...
Genre: h#, Retro
Computer test: C+ Jacobi v0.7.5
FEN: 8/8/8/6PB/1PPP1PpP/Pp1pB1P1/prpN1Rp1/Rbk1K1Q1
Reprints: 4113 Problem 05/1979
(19) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-07-19
Last update: A.Buchanan, 2022-07-10 more...
1. fxg6ep+ Kg5 2. Lh4#
R: 1. g7-g5
R: 1. g7-g5
Keywords: En passant as key
Pieces: = Grasshopper (G)
Genre: Retro, 2#, Fairies
Computer test: HC+ Popeye 4.61 after very simple analysis
FEN: 8/5*2Q1K/5P1P/5Ppk/6p1/6B1/6N1/8
Reprints: 115 Caissa's Wild Roses 1935
Input: Gerd Wilts, 1996-07-16
Last update: A.Buchanan, 2021-05-03 more...
Pieces: = Grasshopper (G)
Genre: Retro, 2#, Fairies
Computer test: HC+ Popeye 4.61 after very simple analysis
FEN: 8/5*2Q1K/5P1P/5Ppk/6p1/6B1/6N1/8
Reprints: 115 Caissa's Wild Roses 1935
Input: Gerd Wilts, 1996-07-16
Last update: A.Buchanan, 2021-05-03 more...
1. ... hxg6ep#, 1. gxh4 2. h3 3. h2 4. h1=L 5. Lc6 6. Le8 7. Lxh5 g5#
Keywords: En passant as key, Seriesmover, Promotion in forward play
Genre: Retro
FEN: 8/7p/4NP1k/5KpP/6PR/8/8/8
Input: Gerd Wilts, 1996-07-24
Last update: James Malcom, 2021-01-29 more...
Genre: Retro
FEN: 8/7p/4NP1k/5KpP/6PR/8/8/8
Input: Gerd Wilts, 1996-07-24
Last update: James Malcom, 2021-01-29 more...
1. bxc6ep
Lösetip beim Originalabdruck: beim Abzählen der geschlagenen Figuren beachte man die Farbe der Felder, auf denen sie fielen.
AL
Le4 ist ein UW-L, der nur so entstanden sein kann: h2-h7xg8=L.
Schwarz zog h7xg6x5, um ihn vorbeizulassen.
Somit ergibt sich folgende Schlagbilanz:
Weiß: axb, dxc, fxg (als Schlagobjekt für den sBh7) [Anmerkung: statt fxg auch fxe mit UW auf e8 möglich, ändert aber nichts]
Schwarz: bxc, h7xg6xh5 sowie den wLf1, der nicht von den sBB geschlagen wurde
Unter den geschlagen weißen Figuren befindet sich der schwarzfeldrige Lc1.
Die Schläge h7xg6xh5 fanden auf weißen Feldern statt, also wurde der Lc1 mittels b4xc3 geschlagen. ("Effekt Zvetnosti")
Als letzte schwarze Züge kommen also nicht infrage: b3xc2, e7-e5 (sLf8 wird als Schlagobjekt gebraucht), g6xh5 (dann käme der wUW-L nicht von g8 nach e4).
Einziger legaler letzter schwarzer Zug war also c7-c5, und es löst: 1. bxc6ep
Lösetip beim Originalabdruck: beim Abzählen der geschlagenen Figuren beachte man die Farbe der Felder, auf denen sie fielen.
AL
Le4 ist ein UW-L, der nur so entstanden sein kann: h2-h7xg8=L.
Schwarz zog h7xg6x5, um ihn vorbeizulassen.
Somit ergibt sich folgende Schlagbilanz:
Weiß: axb, dxc, fxg (als Schlagobjekt für den sBh7) [Anmerkung: statt fxg auch fxe mit UW auf e8 möglich, ändert aber nichts]
Schwarz: bxc, h7xg6xh5 sowie den wLf1, der nicht von den sBB geschlagen wurde
Unter den geschlagen weißen Figuren befindet sich der schwarzfeldrige Lc1.
Die Schläge h7xg6xh5 fanden auf weißen Feldern statt, also wurde der Lc1 mittels b4xc3 geschlagen. ("Effekt Zvetnosti")
Als letzte schwarze Züge kommen also nicht infrage: b3xc2, e7-e5 (sLf8 wird als Schlagobjekt gebraucht), g6xh5 (dann käme der wUW-L nicht von g8 nach e4).
Einziger legaler letzter schwarzer Zug war also c7-c5, und es löst: 1. bxc6ep
Des Autors (geb.27.12.1938)"`Erstling"', also mit 36 Jahren.
abgedruckt in der Rubrik: "Redkije Shanry"
Henrik Juel: Solution: 1.bxc6 ep (2.cxd7#). White captured axb, dxc, fxe, h7xg8=B and promoted on e8; Black captured orig. Bf1, bxc, h7xg6xh5 and promoted on a1. So last move must be c7-c5. (2003-05-27)
Mario Richter: @all PDB activists: Would the introduction of a new keyword: "Farbbalance / color balance"
or something similiar make any sense? (describing the fact, that a certain capture must have happened on a square of a determined color) (2022-04-26)
A.Buchanan: The pieces have colour: black & white. But the squares and the bishops have *shade*. Informally of course one can say anything, but if we are glossarizing then maybe be a bit more formal. And what is “balance” here? I think of balance in terms of pawn captures required vs available. Typically the issue you are talking about is where almost all pawn captures are in one shade square, allowing us say something interesting about a bishop capture. Maybe: “shade logic”? (2022-04-26)
Henrik Juel: I use Andrew's terminologi: men are white or black, and squares are light or dark
But my non-retro friends (both of them) use white/black for squares also
So 'Shade logic' may be too esoteric
What about 'Square color argument'?
It is used far more frequently in retro analysis than, say, 'Parity argument', so a keyword is needed
Thanks for the suggestion, Mario (2022-04-26)
A.Buchanan: I am convinced by Henrik, but I will continue to use “shade” in my own writing. (2022-04-27)
comment
abgedruckt in der Rubrik: "Redkije Shanry"
Henrik Juel: Solution: 1.bxc6 ep (2.cxd7#). White captured axb, dxc, fxe, h7xg8=B and promoted on e8; Black captured orig. Bf1, bxc, h7xg6xh5 and promoted on a1. So last move must be c7-c5. (2003-05-27)
Mario Richter: @all PDB activists: Would the introduction of a new keyword: "Farbbalance / color balance"
or something similiar make any sense? (describing the fact, that a certain capture must have happened on a square of a determined color) (2022-04-26)
A.Buchanan: The pieces have colour: black & white. But the squares and the bishops have *shade*. Informally of course one can say anything, but if we are glossarizing then maybe be a bit more formal. And what is “balance” here? I think of balance in terms of pawn captures required vs available. Typically the issue you are talking about is where almost all pawn captures are in one shade square, allowing us say something interesting about a bishop capture. Maybe: “shade logic”? (2022-04-26)
Henrik Juel: I use Andrew's terminologi: men are white or black, and squares are light or dark
But my non-retro friends (both of them) use white/black for squares also
So 'Shade logic' may be too esoteric
What about 'Square color argument'?
It is used far more frequently in retro analysis than, say, 'Parity argument', so a keyword is needed
Thanks for the suggestion, Mario (2022-04-26)
A.Buchanan: I am convinced by Henrik, but I will continue to use “shade” in my own writing. (2022-04-27)
comment
Keywords: En passant as key, Obvious promotion (wLe4)
Genre: Retro
FEN: brkn2R1/Rn1p1pp1/N7/1PpKp2p/QPP1B3/2P5/2p1P1P1/8
Reprints: Caissas Schloßbewohner 3 1987
Input: Gerd Wilts, 1996-09-16
Last update: Mario Richter, 2022-04-26 more...
Genre: Retro
FEN: brkn2R1/Rn1p1pp1/N7/1PpKp2p/QPP1B3/2P5/2p1P1P1/8
Reprints: Caissas Schloßbewohner 3 1987
Input: Gerd Wilts, 1996-09-16
Last update: Mario Richter, 2022-04-26 more...
1. gxf3ep 2. Dg4 3. 0-0-0 4. Te8 5. Kd8 6. Dxh5 0-0-0#
Henrik Juel: 1.gxf3ep 2.Qg4 3.000 4.Re8 5.Kd8 6.Qxh5 000#. Both castlings are necessary to legitimize the en passant capture. (2003-09-29)
A.Buchanan: I don't think this is consequent series mover, under which every position would considered afresh from the perspective of legality i.e. possible histories. Instead, this is regular series mover where one player is simply saying "pass". It's hard to imagine how AP might operate in a consequent context. I will change the keyword suitably. (2021-11-08)
comment
A.Buchanan: I don't think this is consequent series mover, under which every position would considered afresh from the perspective of legality i.e. possible histories. Instead, this is regular series mover where one player is simply saying "pass". It's hard to imagine how AP might operate in a consequent context. I will change the keyword suitably. (2021-11-08)
comment
Keywords: a posteriori (AP), Seriesmover, Castling, En passant as key
Genre: Retro, Fairies
FEN: r3k3/ppp1p3/2p5/6PP/5Ppq/1P4nB/PrP1P1p1/R3K1nb
Input: Gerd Wilts, 1998-06-26
Last update: A.Buchanan, 2021-11-08 more...
Genre: Retro, Fairies
FEN: r3k3/ppp1p3/2p5/6PP/5Ppq/1P4nB/PrP1P1p1/R3K1nb
Input: Gerd Wilts, 1998-06-26
Last update: A.Buchanan, 2021-11-08 more...
1. fxe3ep d8=S 2. Kxd5 0-0-0# (Td1#?)
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.
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
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)
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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...
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...
*) 1. ... dxc6ep 2. dxc3 Lxb6#
1) 1. dxe3 Dxg7 2. Kxb7 Dxd7#
2) 1. Lxb7 Sc6+ 2. Ka8 Sc7#
pawns history:
Bl: axb,exd,f=
Wh: hxgxh,gxh,fxgxh,cxd,a=,e=
All accounted for.
Bl can't uncapture axb or exd else block promotion.
If WTM, Bl just played c7-c5 so ep is on
1) 1. dxe3 Dxg7 2. Kxb7 Dxd7#
2) 1. Lxb7 Sc6+ 2. Ka8 Sc7#
pawns history:
Bl: axb,exd,f=
Wh: hxgxh,gxh,fxgxh,cxd,a=,e=
All accounted for.
Bl can't uncapture axb or exd else block promotion.
If WTM, Bl just played c7-c5 so ep is on
A.Buchanan: WinChloe doesn't think this problem is cooked at all. It says there are two full solutions, with "Echange de place de pièces noires entre les positions finales" i.e. Kb7 La8 vs Ka8 Lb7. WinChloe does not cover the (single) set play. Must the number of set plays normally match the number of full solutions? I will mark this problem HC+ uncooked. (2021-10-31)
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Keywords: En passant as key, Sacrifice of white pieces
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 + simple retro logic.
FEN: bN6/kB1p2p1/Np1P4/1KpP3P/1prp3P/1PQRB2P/2R4p/8
Input: Hans-Jürgen Schäfer, 1997-10-11
Last update: A.Buchanan, 2021-10-31 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 + simple retro logic.
FEN: bN6/kB1p2p1/Np1P4/1KpP3P/1prp3P/1PQRB2P/2R4p/8
Input: Hans-Jürgen Schäfer, 1997-10-11
Last update: A.Buchanan, 2021-10-31 more...
1) 1. exd3ep b8=D 2. Kd4 Df4#
2) 1. exd3ep a8=S 2. d4 Sxb6#
Cook: e.p. not justified as R: 0.e3xd4 possible
2) 1. exd3ep a8=S 2. d4 Sxb6#
Cook: e.p. not justified as R: 0.e3xd4 possible
VL: Unjustified ep-capture because of –e3xd4. For correction it suffices e.g. to add bSe3.
This is possibly a unique 2-move helpmate with ep-key and 2 solutions of the type 1.2.1.1 (at least so currently in the PDB).
Publi?ation details. H551 in the Helpmate section of the July issue, with no preview. Here is the text of the solution published in the Nov.-77 issue: “…White may have just played P(e3)xd4. Very few solvers have spotted this. Full marks for claim of no solution with correct reasoning but, in fairness, one mark for composer's intention. White cannot have just played K(a5)a6 in reply to Pxb6+ as too many captures by Black are required.” (2023-12-01)
A.Buchanan: Yes Valery you are right this must be cooked. However, I'm not a fan of +bSe3 because it spoils one of the model mates. Separately, I mildly prefer bPf7 to bPe6. (2023-12-03)
VL: I do prefer bPf7, as well. I don't see any way to save at least one of the properties Kindergarten Problem and Model mate (2) after fixing (what for these keywords have been added just now?). Besides, Andrew, please, refine the specification ((full) number and publication month 07-08). (2023-12-03)
A.Buchanan: Ok have added the publication details. The keywords are useful to indicate the aim, even though the problem is cooked. I guess there is a way to keep the model mates but it’s very clunky and probably +bSe3 is better (2023-12-03)
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This is possibly a unique 2-move helpmate with ep-key and 2 solutions of the type 1.2.1.1 (at least so currently in the PDB).
Publi?ation details. H551 in the Helpmate section of the July issue, with no preview. Here is the text of the solution published in the Nov.-77 issue: “…White may have just played P(e3)xd4. Very few solvers have spotted this. Full marks for claim of no solution with correct reasoning but, in fairness, one mark for composer's intention. White cannot have just played K(a5)a6 in reply to Pxb6+ as too many captures by Black are required.” (2023-12-01)
A.Buchanan: Yes Valery you are right this must be cooked. However, I'm not a fan of +bSe3 because it spoils one of the model mates. Separately, I mildly prefer bPf7 to bPe6. (2023-12-03)
VL: I do prefer bPf7, as well. I don't see any way to save at least one of the properties Kindergarten Problem and Model mate (2) after fixing (what for these keywords have been added just now?). Besides, Andrew, please, refine the specification ((full) number and publication month 07-08). (2023-12-03)
A.Buchanan: Ok have added the publication details. The keywords are useful to indicate the aim, even though the problem is cooked. I guess there is a way to keep the model mates but it’s very clunky and probably +bSe3 is better (2023-12-03)
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Keywords: En passant as key, Kindergarten Problem, Model mate (2)
Genre: h#, Retro
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/PP6/Kpp1p3/2pp4/1pkPp3/2p5/P7/8
Input: Michal Dragoun, 1998-04-09
Last update: A.Buchanan, 2023-12-03 more...
Genre: h#, Retro
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/PP6/Kpp1p3/2pp4/1pkPp3/2p5/P7/8
Input: Michal Dragoun, 1998-04-09
Last update: A.Buchanan, 2023-12-03 more...
1. cxb3ep+ Sb4+ 2. Kb1 Sxe2 3. b2 0-0#
Cook: NL 1. Db2,Sb3 Sxe2 2. Sb3,Db2 0-0 3. Lxb4 Sxb4#
Cook: NL 1. Db2,Sb3 Sxe2 2. Sb3,Db2 0-0 3. Lxb4 Sxb4#
A.Buchanan: Two aspects of cookery here. First, NL 1. Db2,Sb3 Sxe2 2. Sb3,Db2 0-0 3. Lxb4 Sxb4# Second, White can retract c3xb4 (and earlier captured onto c-file). Black could have captured axbxcxdxe, exf, fxe, with wPg waylaid (2021-11-23)
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Keywords: En passant as key, Superseded by (P1396163)
Genre: h#, Retro
Computer test: Popeye v4.87 says cooked
FEN: 8/8/1p6/b1P2P2/1PpqPpp1/r2ppb1p/k1N1p2P/n3K1NR
Input: Felber, Volker, 1999-12-20
Last update: A.Buchanan, 2021-11-25 more...
Genre: h#, Retro
Computer test: Popeye v4.87 says cooked
FEN: 8/8/1p6/b1P2P2/1PpqPpp1/r2ppb1p/k1N1p2P/n3K1NR
Input: Felber, Volker, 1999-12-20
Last update: A.Buchanan, 2021-11-25 more...
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https://pdb.dieschwalbe.de/search.jsp?expression=COMMENTDATE%3E%3D20200919+AND+NOT+K%3D%27Hilfsr%C3%BCckz%C3%BCger%27+AND+NOT+K%3D%27Selbstblock%27+AND+K%3D%27En+passant+als+Schl%C3%BCssel%27
The problems of this query have been registered by the following contributors:
Gerd Wilts (97)Hans-Jürgen Schäfer (1)
Michal Dragoun (1)
Felber, Volker (1)
I 1... c*d6 e.p. 2.0-0-0 Q*c7#
II 1... g*f6 e.p. 2.0-0-0 Rg8#
NOT:
2.Kf8? Q/Rg8#?? no AP-justification!
Castling implies e.p. 'a posteriori' and is necessary only to this end. (2002-04-03)
A.Buchanan: Solutions & tries (Popeye v.485) are:
1. ... cxd6ep 2. 0-0-0 Dxc7#
1. ... cxd6ep 2. Kf8 Dg8#
1. ... gxf6ep 2. d4 Tg8#
1. ... gxf6ep 2. 0-0-0 Tg8#
1. ... gxf6ep 2. Kd8 Tg8#
So after 1. ... gxf6ep:
2. Kf8 Rg8 is not even a mate (f7 flight)
and is 2. d4 a kind of dual thematic AP try? No-one's mentioned it. It would be easily fixable by shifting wBd2 to d4.
However, the main thing issue I don't understand PRA+AP logic (maybe I did once?). Under normal PRA algorithm, the "parts" for cxdep, gxfep & 0-0-0 respectively are YNY, NYY, NNN. Others are either impossible (3) or dominated by application of castling convention (2). OK so far.
There is no solution in the NNN part, so I suppose we must use AP in another part to eliminate it. But castling in another part where we are explicitly *assuming* that castling is ok seems to me to prove nothing. And if this inference did work somehow (which seems to me incredibly unlikely), why would we need to do it in *both* of these other parts? Surely one would be enough, but that means that the other part would be cooked.
Or maybe somehow AP "proves" that there are only 2 dimensions for PRA space, cxdep & gxfep, but that is so incredibly circular and horrible.
This kind of problem needs a proper step-by-step solution. Currently the solver can at best sort-of-vaguely-see-what-the-composer-might-be-alluding-to: i.e. it's just a joke.
Actually, I don't think that it can be PRA at all. I think it must be RS with two solutions supported by AP. But then there's still the question of why we need to castle in *both* solutions. I am sure I have seen an AP problem with two solutions where there is 0-0 in one solution and 0-0-0 in the other, which suggests that we *don't* need to castle twice in this problem. Which interpretation is correct, or do we have such a vague interpretation of these things (in order not to make independent spirits sad) that both are allowed to "work". (2020-12-30)
VL: Generally I agree with Andrew's analysis. The presented solution needs some corrections and clarification.
At first, the extra try 1.gxf6 e.p.? d4. I don't remember details but indeed, at present I'd prefer to exclude it by shifting wPd2 onto d4. One more thematic try enriches a problem but it is better to have an equal number of tries after both e.p.'s. In general, in AP-problems, thematic tries are highly desirable but not necessary formally for soundness. Dual avoidance in tries is here just due to the flight f7. Yes, 2.Kf8? Rg8#?? is indicated wrongly (unfortunately, H.103 with the published solution isn't available for me).
Now the most doubtful point: dual avoidance in the solution. Do we need castling in both partial solutions or at least one would suffice for justification? In my opinion, both approaches make sense in principle. In this problem implemented is the former, what means that every partial solution is permitted to be considered separately and, thus, should contain legalizing castling. This is similar to ordinary AP-problems, in which every line of the solution must contain castling. There is a distinct much subtler and rarer approach, which I call "Total AP", that considers all lines as a whole and thus requires to contain castling at least once (with subtle soundness issues). So, the current problem is treated (implicitly) not in the spirit of Total AP.
The problem implements a certain kind of triples of mutually retrodepending special moves (w/b castlings and e.p.). Namely, it is a "hybrid" AP-problem, where castling justifies a posteriori the typical pRA-choice between two e.p. captures. This is one of lucky triple kinds that admit (controversially) sound implementations based on hybrid logical conventions (logics for short). One more such a triple is represented in my earlier P0002474. Totally 37 distinct kinds of interacting triples can be identified, and only for few of them I know (or at least expect) reasonable implementations under special logics (whichever artificial and arguable they can be but consistent internally). Under an implementation I mean (loosely) a problem with an orthodox stipulation (#n, h#, etc.) supplemented by a suitable retro-convention like an ad hoc combination of pRA, RS or AP such that the position of the problem contains the corresponding triple of moves AND its full solution depends on all their combined (il)legalities. An extremely subtle matter. (2021-01-29)
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