86 problem(s) found in 4162 milliseconds (displaying 86 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT A='Porten, Fritz' AND A='Kardos, Tivadar'] [download as LaTeX]
1. Sf3 e5 2. Sxe5 Se7 3. Sxd7 Sec6 4. Sxb8 Dxd2 5. Dxd2 Sxb8 6. Dd8+ Kxd8
Cook: 1. Sh3 d5 2. Sf4 Sf6 3. Sxd5 Se4 4. Sxe7 Sxd2 5. Dxd2 Kxe7 6. Dxd8 Kxd8
(Can transpose B1&B2)
Cook: 1. Sh3 d5 2. Sf4 Sf6 3. Sxd5 Se4 4. Sxe7 Sxd2 5. Dxd2 Kxe7 6. Dxd8 Kxd8
(Can transpose B1&B2)
Keywords: Unique Proof Game, Homebase (W), Impostor (s), Superseded by (P1394733)
Genre: Retro
FEN: rnbk1b1r/ppp2ppp/8/8/8/8/PPP1PPPP/RNB1KB1R
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-10-21 more...
Genre: Retro
FEN: rnbk1b1r/ppp2ppp/8/8/8/8/PPP1PPPP/RNB1KB1R
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-10-21 more...
1. d4 e5 2. Lg5 Dxg5 3. Sf3 Dxg2 4. Sxe5 Dxh1 5. Sxd7 Dd5 6. Sxf8 Dxa2 7. Sg6 Dxa1 8. Sxh8 Da6 9. Sg6 Dxg6 10. Dd3 Dxd3 11. e3 Dxf1+ 12. Kd2
paul: C+ pour les premiers dix coups (Popeye). (2011-09-17)
Sally: Die Reihenfolge der Züge ist festgelegt, aber eine bestimmte Thematik wird nicht gezeigt. Dieser Sachverhalt scheint sehr zur Schwierigkeut der Aufgabe beigetragen zu haben. Es gab nur 2 Löser! (2017-09-06)
A.Buchanan: Up to 10.5, the solution is unique (Jacobi, 2.5 hours on my little laptop). Maybe someone here can push it all the way? (2021-11-26)
Henrik Juel: Natch 3.1 could not even find pos. 1 in several hours (2021-11-26)
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Sally: Die Reihenfolge der Züge ist festgelegt, aber eine bestimmte Thematik wird nicht gezeigt. Dieser Sachverhalt scheint sehr zur Schwierigkeut der Aufgabe beigetragen zu haben. Es gab nur 2 Löser! (2017-09-06)
A.Buchanan: Up to 10.5, the solution is unique (Jacobi, 2.5 hours on my little laptop). Maybe someone here can push it all the way? (2021-11-26)
Henrik Juel: Natch 3.1 could not even find pos. 1 in several hours (2021-11-26)
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Keywords: Unique Proof Game
Genre: Retro
FEN: rnb1k1n1/ppp2ppp/8/8/3P4/4P3/1PPK1P1P/1N3q2
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2005-12-27 more...
Genre: Retro
FEN: rnb1k1n1/ppp2ppp/8/8/3P4/4P3/1PPK1P1P/1N3q2
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2005-12-27 more...
a) 1. ... Kxa1 2. 0-0#
b) 1. ... Kxh1 2. 0-0-0#
b) 1. ... Kxh1 2. 0-0-0#
A.Buchanan: The term "retro" is jungle not garden - that means we should not expect an axiomatic definition. The current problem is a case in point. Case law has established that neither simple employment of the castling convention nor existence of check are sufficient to make a problem "retro". But all this problem has is the quirky use of Codex Article 15 to force BTM. So I think this problem has to be retro. The key point is that nothing hinges on the retro-ness. If the problem included 50M or DP, then one would expect a more solid foundation. As it is, all we need is the free direct mate in 1 that comes as part of the retro paradigm. (2021-11-26)
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Keywords: Castling (wk), No legal last move for Black, Minimal, Miniature
Genre: Retro
Computer test: C+ Popeye v4.87
FEN: 8/8/8/8/8/4p3/4Q3/Rk2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-26 more...
Genre: Retro
Computer test: C+ Popeye v4.87
FEN: 8/8/8/8/8/4p3/4Q3/Rk2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-26 more...
1. Sc3 Sf6 2. Se4 Sxe4 3. g3 Sxg3 4. Sf3 Sxh1 5. Se5 Sxf2 6. Sxd7 Sxd1 7. Sxb8 Txb8
Sally: Eine eindeutige Zugfolge, was bei solchen Aufgaben nicht oft zutrifft. Nicht ganz leicht. Der Rekord steht bei 41,5 Zügen.
(BS). (2017-09-06)
Henrik Juel: 41.5 moves is a very old record
The current record length for proof games is 58.5 moves in a problem by Pronkin, Frolkin & Keym, problem C, p.7 in Die Schwalbe, heft 283, February 2017
The former record was 57.5 moves in a problem by Pronkin & Frolkin, Die Schwalbe, 1989 (V) (2017-09-06)
Olaf Jenkner: The current record length for proof games is 57.5 moves (P0000136) because P1338946 has been cooked. (2021-11-25)
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(BS). (2017-09-06)
Henrik Juel: 41.5 moves is a very old record
The current record length for proof games is 58.5 moves in a problem by Pronkin, Frolkin & Keym, problem C, p.7 in Die Schwalbe, heft 283, February 2017
The former record was 57.5 moves in a problem by Pronkin & Frolkin, Die Schwalbe, 1989 (V) (2017-09-06)
Olaf Jenkner: The current record length for proof games is 57.5 moves (P0000136) because P1338946 has been cooked. (2021-11-25)
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Keywords: Unique Proof Game, Superseded by (P1396199)
Genre: Retro
Computer test: (Natch 2.2Beta1 Copyright (C) 1997,98,99,2001,2002,2003 Pascal Wassong)
FEN: 1rbqkb1r/ppp1pppp/8/8/8/8/PPPPP2P/R1BnKB2
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-25 more...
Genre: Retro
Computer test: (Natch 2.2Beta1 Copyright (C) 1997,98,99,2001,2002,2003 Pascal Wassong)
FEN: 1rbqkb1r/ppp1pppp/8/8/8/8/PPPPP2P/R1BnKB2
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-25 more...
1. Dxc2 Tc1 2. 0-0-0? Txc2#
1. Sd7 0-0-0 2. 0-0 Tg1#
Black qside castling is lost as wPa promotes but other rights can be maintained.
S: axb,cxb,dxcxb,fxexBd,gxfxe
W: a=X,g=X,fxe
1. Sd7 0-0-0 2. 0-0 Tg1#
Black qside castling is lost as wPa promotes but other rights can be maintained.
S: axb,cxb,dxcxb,fxexBd,gxfxe
W: a=X,g=X,fxe
* 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. 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...
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. 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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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...
14 - 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...
15 - 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)
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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)
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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...
a) 1. Df7! h4 2. 0-0 h5 3. Lh8 hxg6 4. Sg7 gxh7#
b) 1. De6! h4 2. 0-0 h5 3. Kh8 hxg6 4. Tg8 Txh7#
1. 0-0-0?? 0-0 2. Kb8 Txc1 3. Ka8 Txb1 4. Tb8 Ta1# (0-0-0 ist illegal)
1. 0-0-0? ist nicht gestattet, da wegen der schwarzen Bauernstellung der weiße Bauer a2 auf a8 umgewandelt wurde!
b) 1. De6! h4 2. 0-0 h5 3. Kh8 hxg6 4. Tg8 Txh7#
1. 0-0-0?? 0-0 2. Kb8 Txc1 3. Ka8 Txb1 4. Tb8 Ta1# (0-0-0 ist illegal)
1. 0-0-0? ist nicht gestattet, da wegen der schwarzen Bauernstellung der weiße Bauer a2 auf a8 umgewandelt wurde!
Marcin Banaszek: Die Urdruck-Stellung ist mit sSe6 (nicht sSd5) und über die Zwilling-Stellung ist dort keine Erwähnung. (2022-06-09)
A.Buchanan: Yes I don't know where the twin concept was introduced to this record (the visible "Last Update" history does not mention it). The twin adds very little: the retro try is identical, and the solution shares most of the White moves, ending in a duller mate. Did the 1983 reprint include it? (2022-06-10)
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A.Buchanan: Yes I don't know where the twin concept was introduced to this record (the visible "Last Update" history does not mention it). The twin adds very little: the retro try is identical, and the solution shares most of the White moves, ending in a duller mate. Did the 1983 reprint include it? (2022-06-10)
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1. Kxd8 Ta7 2. Te8 Td7#
1. 0-0 0-0-0? 2. Th8 Txg1# try
1. 0-0 0-0-0? 2. Th8 Txg1# try
White castling not possible as sBa= on a1. (Also sBgxh= or sBhxg=)
A.Buchanan: Retractor 2.1.1 does not notice that promotion happened on a1 (2021-11-25)
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A.Buchanan: Retractor 2.1.1 does not notice that promotion happened on a1 (2021-11-25)
comment
1. Kd8 Txh8 2. Kc8 Txg8#
Henrik Juel: The diagram pawns captured all missing men, so the missing pawns [Pa7,Ph2] promoted on a1,h8, and Ta1,Th8 have moved, ruling out the apparent solutions 1.Ke7 0-0-0 2.Kxe6 The1#, 1.Dg1+ Ke2 2.0-0 Taxg1#
C+ by Popeye 4.61 and analysis (2021-08-31)
comment
C+ by Popeye 4.61 and analysis (2021-08-31)
comment
not: 1. Sd7 0-0-0 2. 0-0 Tdg1#
but: 1. Lxe6 Th7 2. 0-0-0 Ta8#
and: 1. Ke7 0-0 2. Kxe6 Tae1#
but: 1. Lxe6 Th7 2. 0-0-0 Ta8#
and: 1. Ke7 0-0 2. Kxe6 Tae1#
A.Buchanan: Hard to see this as anything but a version of P0563989. Popeye reports again three apparent mates, with all 4 castlings. However pawn capture analysis indicates that Black qside castling is illegal, while the two kingside castlings are mutually exclusive. (2022-01-08)
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Keywords: Cant Castler (sg), Castling, Partial Retro Analysis (PRA)
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: r1b1kn1r/1n6/1pPpP3/1p1p4/1bp5/1qpp4/1PP5/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-08 more...
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: r1b1kn1r/1n6/1pPpP3/1p1p4/1bp5/1qpp4/1PP5/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-08 more...
1. Th8 0-0-0 2. Txf8 The1#
1. Sa6 0-0 2. Txd8 Tae1#
1. Th7 Sxg6 2. Td7 Th8#
1. Sa6 0-0 2. Txd8 Tae1#
1. Th7 Sxg6 2. Td7 Th8#
wie lautet die Retroanalyse? Warum sollten die Rochaden nicht möglich sein?
Henrik Juel: White pawns captured all 8 missing black men, incl. [Pa7,e7,h7]
For both white castlings to be possible, 4 black pawn captures are needed (axb, dxc, e2xd1/f1, and hxg)
but only 3 white men are missing, so the the white castlings are mutually exclusive (2021-11-22)
Henrik Juel: HC+ Popeye 4.61, if '2 solutions' is added to the stipulation
The 2 solutions are 1.Th7... and the one with the permissible castling (2021-11-22)
comment
Henrik Juel: White pawns captured all 8 missing black men, incl. [Pa7,e7,h7]
For both white castlings to be possible, 4 black pawn captures are needed (axb, dxc, e2xd1/f1, and hxg)
but only 3 white men are missing, so the the white castlings are mutually exclusive (2021-11-22)
Henrik Juel: HC+ Popeye 4.61, if '2 solutions' is added to the stipulation
The 2 solutions are 1.Th7... and the one with the permissible castling (2021-11-22)
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22 - 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. Lxa2 Txa2 2. 0-0-0 Txa7 3. Td7 Ta8#
1. Lxa2 0-0-0? 2. Dxg2 Tg1 3. 0-0 Txg2# but White can't castle
Cook: 1. Lxa2 Lf6 2. Lb1 Txa7 3. Kf8 Txa8#
1. Lxa2 Lf6 2. Lxb3 Txa7 3. Kf8 Txa8#
1. Dxg2 Txb1 2. Dxf2+ Kxf2 3. 0-0 Tg1#
1. Dg3 Txb1 2. Dxb3 Txb3 3. 0-0 Tg3#
1. Lxa2 0-0-0? 2. Dxg2 Tg1 3. 0-0 Txg2# but White can't castle
Cook: 1. Lxa2 Lf6 2. Lb1 Txa7 3. Kf8 Txa8#
1. Lxa2 Lf6 2. Lxb3 Txa7 3. Kf8 Txa8#
1. Dxg2 Txb1 2. Dxf2+ Kxf2 3. 0-0 Tg1#
1. Dg3 Txb1 2. Dxb3 Txb3 3. 0-0 Tg3#
paul: White Rook h1 left the S-E camp, so the white castle is no more possible.
Intention: 1.L×a2 T×a2 2.0-0-0 T×a7 3.Td7 Ta8#
1.L×a2 0-0-0 2.D×g2 Tg1 3.0-0 T×g2#
Cooked by 1.L×a2 Lf6 2.Lb1 T×a7 3.Rf8 T×a8# or
1.D×g2 T×b1 2.D×f2+ K×f2 3.0-0 Tg1# (2011-09-15)
milan: wBe5-f4 M.Frelih (2021-12-01)
A.Buchanan: @Milan: that removes all 4 cooks, and alas the intended try also. (2021-12-01)
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Intention: 1.L×a2 T×a2 2.0-0-0 T×a7 3.Td7 Ta8#
1.L×a2 0-0-0 2.D×g2 Tg1 3.0-0 T×g2#
Cooked by 1.L×a2 Lf6 2.Lb1 T×a7 3.Rf8 T×a8# or
1.D×g2 T×b1 2.D×f2+ K×f2 3.0-0 Tg1# (2011-09-15)
milan: wBe5-f4 M.Frelih (2021-12-01)
A.Buchanan: @Milan: that removes all 4 cooks, and alas the intended try also. (2021-12-01)
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Keywords: Cant Castler, Obvious promotion (l), Superseded by (P1396304)
Genre: Retro
FEN: r3k1qr/pp3p1p/2p5/3pB2p/8/1P6/P1PPPPPP/Rb2K3
Input: Gerd Wilts, 1996-08-11
Last update: A.Buchanan, 2021-12-01 more...
Genre: Retro
FEN: r3k1qr/pp3p1p/2p5/3pB2p/8/1P6/P1PPPPPP/Rb2K3
Input: Gerd Wilts, 1996-08-11
Last update: A.Buchanan, 2021-12-01 more...
1. ... Sxb6 Sxc7#
1. ... axb6 Dxa8#
1. ... bxa6 Tb8#
1. ... bxc6 Dc8#
1. ... cxb6 Sxd6#
1. ... dxc5 Td8#
1. ... d5 f7#
1. ... Sxe7 Txe7#
1. ... Sxf6 Sxf6#
1. ... Dxg7 Sxg7#
1. ... Dxh7 gxh7#
1. ... Dxg6 Lxg6#
1. ... Dxh5 Txg8#
BTM but for every move there’s a unique mate
1. ... axb6 Dxa8#
1. ... bxa6 Tb8#
1. ... bxc6 Dc8#
1. ... cxb6 Sxd6#
1. ... dxc5 Td8#
1. ... d5 f7#
1. ... Sxe7 Txe7#
1. ... Sxf6 Sxf6#
1. ... Dxg7 Sxg7#
1. ... Dxh7 gxh7#
1. ... Dxg6 Lxg6#
1. ... Dxh5 Txg8#
BTM but for every move there’s a unique mate
Keywords: Non-standard material (TT), No legal last move for Black
Genre: Retro
Computer test: C+ Popeye v4.87 edit
FEN: n3k1n1/ppp1P1RR/QRPpNPPq/1PB1pKpB/4N1P1/3R4/8/8
Input: Gerd Wilts, 1998-06-26
Last update: A.Buchanan, 2021-11-26 more...
Genre: Retro
Computer test: C+ Popeye v4.87 edit
FEN: n3k1n1/ppp1P1RR/QRPpNPPq/1PB1pKpB/4N1P1/3R4/8/8
Input: Gerd Wilts, 1998-06-26
Last update: A.Buchanan, 2021-11-26 more...
1. f1=D+ Kg3 2. Dc1 h4 3. Dg5+ hxg5 4. b1=D g6 5. Db4 g7 6. Df8 gxf8=D#
Yuri Bilokin: more economical -bPe4 3k4/2pp4/8/1p3p1p/8/7K/1p3p1P/8 (2+8) (2022-07-29)
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Keywords: konsekutive Umwandlungen 3 (ddD), Excelsior white, Kindergarten Problem
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 3k4/2pp4/8/1p3p1p/4p3/7K/1p3p1P/8
Reprints: 125 200 Ausgewählte Schachaufgaben 1983
723 Ungarische Schachprobleme mit wenigen Steinen [Bakcsi] 1985
406 Minimalkunst im Schach 2006
Input: Gerd Wilts, 1996-06-06
Last update: Rainer Staudte, 2019-12-25 more...
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 3k4/2pp4/8/1p3p1p/4p3/7K/1p3p1P/8
Reprints: 125 200 Ausgewählte Schachaufgaben 1983
723 Ungarische Schachprobleme mit wenigen Steinen [Bakcsi] 1985
406 Minimalkunst im Schach 2006
Input: Gerd Wilts, 1996-06-06
Last update: Rainer Staudte, 2019-12-25 more...
1. La1 Ke7 2. Sb2 Kf6 3. Sd3+ Kg5 4. Le5 Sd2#
NL:
1. Lc3 Kc7 2. Lb4 Lh2 3. Kd4 Sa3 4. Kc5 Lg1#
1. Kd3 Kc7 2. Kc4 Lh2 3. Ld4 Sa3+ 4. Kc5 Ld6#
NL:
1. Lc3 Kc7 2. Lb4 Lh2 3. Kd4 Sa3 4. Kc5 Lg1#
1. Kd3 Kc7 2. Kc4 Lh2 3. Ld4 Sa3+ 4. Kc5 Ld6#
Yuri Bilokin: correction wKe8-f8, +bSb7, +bPg4, +bPh4 5K1b/1n6/8/3b4/n3k1pp/8/8/1N4B1 (3+7) h#4
1.Ba1 Ke7 2.Sb2 Kf6 3.Sd3+ Kg5 4.Be5 Sd2# (MM)
AntiZielElement (B2, line obstruction)
Corner-to-corner (bB)
Hesitation (bB)
Indian (black)
Model mate × 1 (2023-05-24)
comment
1.Ba1 Ke7 2.Sb2 Kf6 3.Sd3+ Kg5 4.Be5 Sd2# (MM)
AntiZielElement (B2, line obstruction)
Corner-to-corner (bB)
Hesitation (bB)
Indian (black)
Model mate × 1 (2023-05-24)
comment
Keywords: Check Protection
Genre: h#
FEN: 3K3b/8/8/3b4/n3k3/8/8/1N4B1
Input: Markus Manhart, 1997-06-26
Last update: hpr, 1999-10-19 more...
Genre: h#
FEN: 3K3b/8/8/3b4/n3k3/8/8/1N4B1
Input: Markus Manhart, 1997-06-26
Last update: hpr, 1999-10-19 more...
1. Kb8 Ka4 2. Kc8 Ka3 3. Kd8 Ka4 4. Ke8 Ka3 5. Kf8 Ka4 6. Kg7 Ka3 7. Kf6 Ka4 8. Kg5 Ka3 9. Kf4 Ka4 10. Kxe4 Ka3 11. Kd3 Ka4 12. e4 Ka3 13. e3 Ka4 14. e2 Ka3 15. e1=D Ka4 16. Dxe6 Ka3 17. Dd7 cxd7 18. Kc2 d8=D 19. Kb1 Dd1#
Keywords: Promotion (Dd), Entblockung, Ceriani-Frolkin Theme (d), Model mate, Königswanderung, Switchback (K), Kindergarten Problem, Pronkin Theme (D)
Genre: h#
Computer test: Gustav 4.1d
FEN: 8/k1p1p3/1pP1P2p/1P2p2P/1Pp1P3/KpP5/1P6/8
Input: Hans-Jürgen Schäfer, 1999-01-01
Last update: A.Buchanan, 2021-04-06 more...
Genre: h#
Computer test: Gustav 4.1d
FEN: 8/k1p1p3/1pP1P2p/1P2p2P/1Pp1P3/KpP5/1P6/8
Input: Hans-Jürgen Schäfer, 1999-01-01
Last update: A.Buchanan, 2021-04-06 more...
1. e1=D Kg2 2. De3 h4 3. Dg5+ hxg5 4. d1=D g6 5. Dd6 g7 6. Df8 gxf8=D#
La grande releve 06/06/1953
Yuri Bilokin: more economical -bPb4, bPe4-f4 3k4/2pp4/8/5p1p/5p2/7K/3pp2P/8 (3+8) h#6
1.e1=Q Kg2 2.Qe7 h4 3.Qg5+ hxg5 4.d1=Q g6 5.Qd6 g7 6.Qf8 gxf8=Q# (MM) (2022-07-30)
comment
Yuri Bilokin: more economical -bPb4, bPe4-f4 3k4/2pp4/8/5p1p/5p2/7K/3pp2P/8 (3+8) h#6
1.e1=Q Kg2 2.Qe7 h4 3.Qg5+ hxg5 4.d1=Q g6 5.Qd6 g7 6.Qf8 gxf8=Q# (MM) (2022-07-30)
comment
not: 1. Dd7 Txa6 2. 0-0-0? Ta8#
but: 1. Ke7 0-0 2. Kxe6 Tae1#
and: 1. Sd7 0-0-0 2. 0-0 Tdg1#
but: 1. Ke7 0-0 2. Kxe6 Tae1#
and: 1. Sd7 0-0-0 2. 0-0 Tdg1#
someone wrote here a long time ago "Autorlösung klären. Steht im Original ein sB auf e6?"
Henrik Juel: solution 1 is a try, right? (2022-01-08)
A.Buchanan: I can't find this problem in WinChloe or yacpdb. Compare with P0003561 which has a similar matrix including sBe6. The three apparent mates clearly form a set. So I am willing to trust the board. Pawn captures w: dxc b: axb gxfxexd fxexdxc exd account for all missing units except one White unit. What of wPa, wPh & bPh?
bPh must have promoted: either on h1 after wPh is waylaid or by capturing hxg allowing wPh to promote on h8. So the last White capture is needed here, and one player has lost kingside castling rights. Therefore wPa must have promoted on a8, and Black has lost queenside castling rights.
So 1. Dd7 Txa6 2. 0-0-0 Ta8# is a try. The other two ways to mate form two parts of a single solution under PRA (2022-01-08)
A.Buchanan: Yes solution 1 is a try: Black can't 0-0-0 (2022-01-08)
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comment
Henrik Juel: solution 1 is a try, right? (2022-01-08)
A.Buchanan: I can't find this problem in WinChloe or yacpdb. Compare with P0003561 which has a similar matrix including sBe6. The three apparent mates clearly form a set. So I am willing to trust the board. Pawn captures w: dxc b: axb gxfxexd fxexdxc exd account for all missing units except one White unit. What of wPa, wPh & bPh?
bPh must have promoted: either on h1 after wPh is waylaid or by capturing hxg allowing wPh to promote on h8. So the last White capture is needed here, and one player has lost kingside castling rights. Therefore wPa must have promoted on a8, and Black has lost queenside castling rights.
So 1. Dd7 Txa6 2. 0-0-0 Ta8# is a try. The other two ways to mate form two parts of a single solution under PRA (2022-01-08)
A.Buchanan: Yes solution 1 is a try: Black can't 0-0-0 (2022-01-08)
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Keywords: Castling, Partial Retro Analysis (PRA)
Genre: h#, Retro
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: r1q1kn1r/1pp5/bpPpP3/1n1p4/1b6/2pp4/1PP5/R3K2R
Input: Markus Manhart, 1999-02-08
Last update: A.Buchanan, 2022-01-08 more...
Genre: h#, Retro
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: r1q1kn1r/1pp5/bpPpP3/1n1p4/1b6/2pp4/1PP5/R3K2R
Input: Markus Manhart, 1999-02-08
Last update: A.Buchanan, 2022-01-08 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...
*) 1. ... Sf7=
1) 1. Kxh8 Ka4 2. Kg8 Ka3 3. Kh8 Kb2 4. Kg8 Kc1 5. Kh8 Kd1 6. Kg8 Ke1 7. Kh8 Kf2 8. Kg8 Kg3 9. Kh8 Kf4 10. Kg8 Kxe4 11. Kh8 Kd5 12. Kg8 Kxc4 13. Kh8 Kxd3 14. Kg8 Kc4 15. Kh8 d4 16. Kg8 d5 17. Kh8 d6 18. Kg8 d7 19. Kh8 d8=S 20. Kg8 Sf7=
1) 1. Kxh8 Ka4 2. Kg8 Ka3 3. Kh8 Kb2 4. Kg8 Kc1 5. Kh8 Kd1 6. Kg8 Ke1 7. Kh8 Kf2 8. Kg8 Kg3 9. Kh8 Kf4 10. Kg8 Kxe4 11. Kh8 Kd5 12. Kg8 Kxc4 13. Kh8 Kxd3 14. Kg8 Kc4 15. Kh8 d4 16. Kg8 d5 17. Kh8 d6 18. Kg8 d7 19. Kh8 d8=S 20. Kg8 Sf7=
Genre: Fairies
FEN: 5bkN/1p2p1p1/1P2P1P1/KPp1P3/2p1p3/2PpP3/3P4/8
Input: Felber, Volker, 1999-12-20
Last update: hpr, 2000-01-30 more...
32 - P0574405
Tivadar Kardos
Pavlos N. Moutecidis
Feladvanykedvelök Lapja 1970
5. ehrende Erwähnung
(2+5) cooked
ser-h=20
Tivadar Kardos
Pavlos N. Moutecidis
Feladvanykedvelök Lapja 1970
5. ehrende Erwähnung
(2+5) cooked
ser-h=20
Erich Bartel (2008-01-25):
1. Kb2 2. Ka3 3. Kb4 4. Kc4 5. Kd3 6. Ke3 7. Kf2 8. Kf1 9. h1=D 10. Dh8 11. Kf2 12. Ke3 13. Kd3 14. Kc4 15. Kb4 16. Ka3 17. Kb2 18. Kb1 19. Da1 20. b2 Sd4=
Cook: Anton Baumann (2020-11-20):
1. a1=L 2. Ld4 3. Lg1 4. h1=T 5. Th8 6. Le3 7. Lc1 8. Kb2 9. Ta8 10. Ta1 11. Tb1 12. Ka1 13. b2 14. d4 15. d3 16. d2 Sc3=
1. Kb2 2. Ka3 3. Kb4 4. Kc4 5. Kd3 6. Ke3 7. Kf2 8. Kf1 9. h1=D 10. Dh8 11. Kf2 12. Ke3 13. Kd3 14. Kc4 15. Kb4 16. Ka3 17. Kb2 18. Kb1 19. Da1 20. b2 Sd4=
Cook: Anton Baumann (2020-11-20):
1. a1=L 2. Ld4 3. Lg1 4. h1=T 5. Th8 6. Le3 7. Lc1 8. Kb2 9. Ta8 10. Ta1 11. Tb1 12. Ka1 13. b2 14. d4 15. d3 16. d2 Sc3=
Anton Baumann: kürzer, wenn eine Umwandlungsfigur rechts abschirmt:
1.a1=L 3.Lg1 4.h1=T 6.Ta8 8.Lc1 9.Kb2 11.Tb1 12.Ka1 13.b2 16.d2 Sc3= (2020-11-20)
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comment
1.a1=L 3.Lg1 4.h1=T 6.Ta8 8.Lc1 9.Kb2 11.Tb1 12.Ka1 13.b2 16.d2 Sc3= (2020-11-20)
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Keywords: Seriesmover
Genre: Fairies
FEN: 8/8/8/3p4/8/1p6/p3N2p/1k1K4
Input: Felber, Volker, 1999-12-20
Last update: Mario Richter, 2020-11-20 more...
Genre: Fairies
FEN: 8/8/8/3p4/8/1p6/p3N2p/1k1K4
Input: Felber, Volker, 1999-12-20
Last update: Mario Richter, 2020-11-20 more...
17. Kxf8 35. Kxd8 50. Kxe6 51. Kd7 56. e1=T 58. Tg8 fxg8=D=
Kr4Bzzz: NL: 51. Kf5 56. e1=T 57. Te8 fxe8=D= (2010-04-15)
Mario Richter: Wenn die Forderung lautete "ser-h=57", dann gäbe es genau eine Lösung. (2010-04-28)
Anton Baumann: als ser-h=57 C+ (Gustav 4.2a) (2023-10-30)
SCHRECKE: Ohne den wBh3 geht nur die AL! (2023-10-30)
comment
Mario Richter: Wenn die Forderung lautete "ser-h=57", dann gäbe es genau eine Lösung. (2010-04-28)
Anton Baumann: als ser-h=57 C+ (Gustav 4.2a) (2023-10-30)
SCHRECKE: Ohne den wBh3 geht nur die AL! (2023-10-30)
comment
Keywords: Seriesmover
Genre: Fairies
FEN: 2kN1R2/2p1pP2/2p1P3/2P5/2PP3B/2K3PP/6P1/8
Input: Felber, Volker, 2000-01-03
Last update: hpr, 2000-03-04 more...
Genre: Fairies
FEN: 2kN1R2/2p1pP2/2p1P3/2P5/2PP3B/2K3PP/6P1/8
Input: Felber, Volker, 2000-01-03
Last update: hpr, 2000-03-04 more...
34 - P0576860
Tivadar Kardos
Magyar Sakkelet 1975
(2+3)
Weiß und Schwarz nehmen je einen Zug zurück, dann h#1
b) wLa1 nach h1
Tivadar Kardos
Magyar Sakkelet 1975
(2+3)
Weiß und Schwarz nehmen je einen Zug zurück, dann h#1
b) wLa1 nach h1
a) Le5xTa1 Ta8xDa1 1. 0-0-0 Da6#
b) Le4xTh1 Th8xDh1 1. 0-0 Dh7#
b) Le4xTh1 Th8xDh1 1. 0-0 Dh7#
Genre: Fairies, h#
FEN: 4k3/3n4/8/3p4/8/8/4K3/B7
Input: Felber, Volker, 2000-01-03
Last update: hpr, 2000-03-04 more...
1. h1=D 2. Dh3 3. De6 4. h3 5. h2 6. h1=D 7. Dh4 8. Dd8 9. h4 10. h3 11. h2 12. h1=D 13. Dhh4 14. Dhe7 Lc6#
Keywords: Seriesmover (konsekutiv ddd), Minimal
Genre: Fairies
Computer test: (Popeye WINDOWS-32Bit-Version 3.65 (2048 KB))
FEN: 8/1K1k3p/3p4/7p/4B2p/8/7p/8
Input: Felber, Volker, 2000-01-06
Last update: James Malcom, 2022-06-21 more...
Genre: Fairies
Computer test: (Popeye WINDOWS-32Bit-Version 3.65 (2048 KB))
FEN: 8/1K1k3p/3p4/7p/4B2p/8/7p/8
Input: Felber, Volker, 2000-01-06
Last update: James Malcom, 2022-06-21 more...
1. Dc7 droht 2. De7#
1. ... 0-0 2. Dg7#
aber 1. ... Th7!
Also 1. Dg7!
1. ... 0-0 2. Dg7#
aber 1. ... Th7!
Also 1. Dg7!
1. ... Kxa3 2. Da1#
1. ... Kxa5 2. Da7#
1. ... Kxa5 2. Da7#
No. 1701 HN
A.Buchanan: I'm inclined to say that "Retro" genre can include a free directmate in 1, since it's so common, with no other genre mention necessary. (2021-11-26)
comment
A.Buchanan: I'm inclined to say that "Retro" genre can include a free directmate in 1, since it's so common, with no other genre mention necessary. (2021-11-26)
comment
Keywords: No legal last move for Black, Miniature
Genre: Retro
Computer test: C+ Popeye v4.87 + Retractor v2.1.1
FEN: 8/8/8/P7/k1KQ4/P7/8/8
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2021-11-26 more...
Genre: Retro
Computer test: C+ Popeye v4.87 + Retractor v2.1.1
FEN: 8/8/8/P7/k1KQ4/P7/8/8
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2021-11-26 more...
1. Te2+ Kxf3 2. Ta3#
1. ... Kg3,Kh3 2. Dg4#
1. ... Kg1,Kh1 2. 0-0-0#
1. ... Kg3,Kh3 2. Dg4#
1. ... Kg1,Kh1 2. 0-0-0#
1. 0-0 Td1 2. Kh7 Td4 3. Tfg8 Txh4#
1. 0-0 0-0-0? 2. Kh7 Td4,Th1 3. Tfg8 Txh4#
Les N ont pris 4 P.B. Les B une P.N. en g3 promue en a1 (2 P.B.) ou d1 (3 P.B.).Le roque B est cassé sauf si PNe3 vient de a7 (4 p.B.) et PNh7 est aller en h1. Donc les B ont pas joués les derniers. Le O-O-O est cassé. S'il n'est pas casséles N ont joués les derniers.
1. 0-0 0-0-0? 2. Kh7 Td4,Th1 3. Tfg8 Txh4#
Les N ont pris 4 P.B. Les B une P.N. en g3 promue en a1 (2 P.B.) ou d1 (3 P.B.).Le roque B est cassé sauf si PNe3 vient de a7 (4 p.B.) et PNh7 est aller en h1. Donc les B ont pas joués les derniers. Le O-O-O est cassé. S'il n'est pas casséles N ont joués les derniers.
No. 438 HN
Henrik Juel: White captured h2xg3, so [Pa7] must have promoted
Black captured [Lf1], hxgxf, and 3 more white men, incl. [Pd2]
The promotion was either axb-b3xa2-a1 or axbxcxPd-d1, so White may not castle
1.0-0 Ra1-d1 2.Kg8-h7 Rd1-d4 3.Rf8-g8 Rd4*h4#
HC+ Popeye 4.61 (2021-11-26)
A.Buchanan: If sBe3 is [sBa], then [sBh] either captured to g3, where it was captured in turn, or it promoted on h1. In either case, White castling rights may endure, so the intended retro try is a cook. There is also a dual. (2021-11-28)
A.Buchanan: Nouguier (in French) noted the possibility of the Western origin of sBe3, which I think the composer overlooked. But (I think) Nouguier did not see that White castling rights remain even with Black to move. He seemed to be concerned about possibility of WTM, but this is not relevant for this problem. (2021-11-28)
A.Buchanan: This can be quickly patched by e.g. shifting bPe3 to e4, and bSc4 to b4, but this is the kind of cook which invites the problem’s expansion, to include both possible origins of bPe3 as thematic retro lines to be explored (2021-11-29)
comment
Henrik Juel: White captured h2xg3, so [Pa7] must have promoted
Black captured [Lf1], hxgxf, and 3 more white men, incl. [Pd2]
The promotion was either axb-b3xa2-a1 or axbxcxPd-d1, so White may not castle
1.0-0 Ra1-d1 2.Kg8-h7 Rd1-d4 3.Rf8-g8 Rd4*h4#
HC+ Popeye 4.61 (2021-11-26)
A.Buchanan: If sBe3 is [sBa], then [sBh] either captured to g3, where it was captured in turn, or it promoted on h1. In either case, White castling rights may endure, so the intended retro try is a cook. There is also a dual. (2021-11-28)
A.Buchanan: Nouguier (in French) noted the possibility of the Western origin of sBe3, which I think the composer overlooked. But (I think) Nouguier did not see that White castling rights remain even with Black to move. He seemed to be concerned about possibility of WTM, but this is not relevant for this problem. (2021-11-28)
A.Buchanan: This can be quickly patched by e.g. shifting bPe3 to e4, and bSc4 to b4, but this is the kind of cook which invites the problem’s expansion, to include both possible origins of bPe3 as thematic retro lines to be explored (2021-11-29)
comment
Keywords: Cant Castler
Genre: Retro, h#
FEN: 4k2r/1p4r1/1bpp2p1/1b2pp2/n1n4q/P3pPP1/1PP1P1P1/R3K3
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2021-11-28 more...
Genre: Retro, h#
FEN: 4k2r/1p4r1/1bpp2p1/1b2pp2/n1n4q/P3pPP1/1PP1P1P1/R3K3
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2021-11-28 more...
a) 1.O-O! b) 1.O-O-O!
Genre: 2#
Computer test: C+ Popeye 4.61
FEN: 8/4k3/p6p/8/8/8/3Q4/R3K2R
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2021-11-26 more...
1. 0-0 dr 2. Tf8# 1. ... 0-0-0 2. Tc1#
1. Txh6 dr 2. Th8# 1. ... 0-0-0 2. Tc6#
1. Txh6 dr 2. Th8# 1. ... 0-0-0 2. Tc6#
1. Dh1 ... 2. Da8#
1. ... Le4,Lf5 2. Dxh8#
1. ... 0-0 2. Dxh7#
1. ... Le4,Lf5 2. Dxh8#
1. ... 0-0 2. Dxh7#
Keywords: Aristocrat, Miniature, Four corners
Genre: 2#
Computer test: C+, popeye 4.87
FEN: 4k2r/R6b/8/4K3/8/8/8/Q7
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2021-11-26 more...
Genre: 2#
Computer test: C+, popeye 4.87
FEN: 4k2r/R6b/8/4K3/8/8/8/Q7
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2021-11-26 more...
1. axb3ep g8=T 2. bxc2 Txg6 3. c1=S e4+ 4. Sd3 Txh6=
Le Pb6 venant de f2 et g7 de h2 (5 prises au total, soit le maximum possible). Le dernier coup blanc ne peut être une prise.
Le Pb6 venant de f2 et g7 de h2 (5 prises au total, soit le maximum possible). Le dernier coup blanc ne peut être une prise.
BS Problemid: 2723
Authors: Kardos, Tividar
Sources:
Years: 1950
SCHRECKE: C+, popeye 4.87
1. Se4! (Zugzwang)
1. ... Sxg6,S~ 2. Dxd5#
1. ... Lh6,Lg5 2. Sg5#
1. ... Lg1,Lf2 2. Sf2#
1. ... Lxd2 2. Sxd2#
1. ... Lf4 2. gxf4#
1. ... dxe4 2. Sc4#
1. ... Kxe4 2. Df4# (2023-02-09)
comment
1. Se4! (Zugzwang)
1. ... Sxg6,S~ 2. Dxd5#
1. ... Lh6,Lg5 2. Sg5#
1. ... Lg1,Lf2 2. Sf2#
1. ... Lxd2 2. Sxd2#
1. ... Lf4 2. gxf4#
1. ... dxe4 2. Sc4#
1. ... Kxe4 2. Df4# (2023-02-09)
comment
Keywords: Brian Stephenson Collection (2723)
Genre: 2#
FEN: 8/4nQp1/6P1/N1Kpk3/3p2p1/2Npb1P1/3P4/4R3
Input: Brian Stephenson, 2004-08-12
Last update: Frank Müller, 2012-03-15 more...
Genre: 2#
FEN: 8/4nQp1/6P1/N1Kpk3/3p2p1/2Npb1P1/3P4/4R3
Input: Brian Stephenson, 2004-08-12
Last update: Frank Müller, 2012-03-15 more...
1. Ld5-a2!
1. ... f7-f6 2. Kd7-c8 Da8xb8+ 3. Kc8xb8 Ta7-a8#
1. ... f7xe6 2. La2xe6 Da8xb8 3. Ld6-c7+ Db8xc7#
1. ... Da8xb8 2. De6-b3+ a4xb3 3. Ld6-c7+ Db8xc7#
1. ... f7-f6 2. Kd7-c8 Da8xb8+ 3. Kc8xb8 Ta7-a8#
1. ... f7xe6 2. La2xe6 Da8xb8 3. Ld6-c7+ Db8xc7#
1. ... Da8xb8 2. De6-b3+ a4xb3 3. Ld6-c7+ Db8xc7#
Keywords: Brian Stephenson Collection (4318)
Genre: s#
FEN: qN1NR3/rp1K1p2/rk1BQ3/p2B1R2/p7/P7/8/8
Input: Brian Stephenson, 2004-08-12
Last update: Marcin Banaszek, 2021-08-13 more...
Genre: s#
FEN: qN1NR3/rp1K1p2/rk1BQ3/p2B1R2/p7/P7/8/8
Input: Brian Stephenson, 2004-08-12
Last update: Marcin Banaszek, 2021-08-13 more...
* 1. ... e3 2. Ld5#
* 1. ... Sh4,Sg6~ 2. Sf4#
1. g5! (Zugzwang)
1. ... e3 2. Tf6#
1. ... Sh4,Sg6~ 2. Te5#
1. ... c4 2. Sd4#
1. ... Kxf5 2. Dd7#
1. ... Sb5,Sc8,Sc6 2. Dc8#
* 1. ... Sh4,Sg6~ 2. Sf4#
1. g5! (Zugzwang)
1. ... e3 2. Tf6#
1. ... Sh4,Sg6~ 2. Te5#
1. ... c4 2. Sd4#
1. ... Kxf5 2. Dd7#
1. ... Sb5,Sc8,Sc6 2. Dc8#
Keywords: Brian Stephenson Collection (20148)
Genre: 2#
FEN: 4Q3/n3bN2/4k1n1/2p2R2/K3p1P1/8/4N2p/7B
Input: Brian Stephenson, 2004-08-12
Last update: Mario Richter, 2021-07-24 more...
Genre: 2#
FEN: 4Q3/n3bN2/4k1n1/2p2R2/K3p1P1/8/4N2p/7B
Input: Brian Stephenson, 2004-08-12
Last update: Mario Richter, 2021-07-24 more...
1. Tf2! h5 2. 0-0 h4 3. Kh1 h3 4. Tg2 hxg2#
1. Lb7+! Ka7 2. Kb3 Lh7 3. Kc2 Lg8 4. Kd1 Lh7 5. Ke1 Lg8 6. Kf1 Lh7 7. Kg1 Lg8 8. Kh2 Lh7 9. Kg3 Lg8 10. Kxf3 Lh7 11. Kf4 Lg8 12. f3 Lh7 13. Sxc7 Lg8 14. Sxg8 Lxc7#
Keywords: Königswanderung
Genre: s#
Computer test: Gustav 4.1d
FEN: kb2N1b1/2p2p2/BpP2PpN/1P1p2Pp/K2p3P/3P1p2/3P1P2/8
Input: HBae, 2009-02-06
Last update: Marcin Banaszek, 2021-03-23 more...
Genre: s#
Computer test: Gustav 4.1d
FEN: kb2N1b1/2p2p2/BpP2PpN/1P1p2Pp/K2p3P/3P1p2/3P1P2/8
Input: HBae, 2009-02-06
Last update: Marcin Banaszek, 2021-03-23 more...
1. Th3 axb7 2. Sh1 b8=T 3. Dh2 Txb3 4. Tg2 Tf2 5. exf2 Txd3=
Genre: Fairies
Computer test: C+ Gustav 4.1d
FEN: 8/1p2p3/P3N3/8/3p2pp/1p1ppnk1/R1rq1n2/5Kbr
Reprints: Magyar Sakkelet 1981
Input: HBae, 2009-02-06
Last update: Gerd Wilts, 2020-12-28 more...
1. Sb8 cxb8=S 2. Sc6 Sxc6 3. Td8+ exd8=S 4. Te8+ fxe8=S 5. Df8+ gxf8=S=
Cook: Dual 4. ... fxe8=D 5. Df7 Dxf7=
NL: 1.Dd3 e8D+ 2.Kb7 D×e3 3.Sc6 D×d2 4.Sa7 D×d3 5.Ka8 D×a6+
mit vielen Umstellungen.
Cook: Dual 4. ... fxe8=D 5. Df7 Dxf7=
NL: 1.Dd3 e8D+ 2.Kb7 D×e3 3.Sc6 D×d2 4.Sa7 D×d3 5.Ka8 D×a6+
mit vielen Umstellungen.
1. Sg8 hxg8=S 2. Th8 gxh8=S 3. Lb8 axb8=S 4. Ta8 bxa8=S 5. Db6 Sxb6 6. Sc4 Sxc4=
Korrektur Schwalbe 10/1989 S. 130
Cook: 1. Sd2-c4 b7-b8=D 2. Th3-h6 h7-h8=D 3. Db4-b3 Db8xb3 4. Ta5-b5 Dh8xh6 5. Tb5-b8 a7xb8=D 6. Lh2-c7 Db8xc7=
Korrektur Schwalbe 10/1989 S. 130
Cook: 1. Sd2-c4 b7-b8=D 2. Th3-h6 h7-h8=D 3. Db4-b3 Db8xb3 4. Ta5-b5 Dh8xh6 5. Tb5-b8 a7xb8=D 6. Lh2-c7 Db8xc7=
Anton Baumann: Arnold Beine fand den Dual: 2.Db3 b8=D 3.Sc4 Dxb3 4.Th8 gxh8=D 5.Ta7 Dxh2 6.Tc7 Dxc7 = mit ZU (vergl. BuB in 'Die Schwalbe' 12/2010 S.743
Mögliche Korrektur: sBa3 nach a4, wBa2 nach a3 (2023-04-07)
Vaclav Kotesovec: Correction above (sBa3 nach a4, wBa2 nach a3) is C+, Popeye 4.87 65:22:47 h:m:s (2023-06-26)
comment
Mögliche Korrektur: sBa3 nach a4, wBa2 nach a3 (2023-04-07)
Vaclav Kotesovec: Correction above (sBa3 nach a4, wBa2 nach a3) is C+, Popeye 4.87 65:22:47 h:m:s (2023-06-26)
comment
Keywords: konsekutive Umwandlungen 4 (SSSS)
Genre: Fairies
FEN: 8/PP4PP/4kn2/r1p5/1q2P3/p4p1r/P2n1P1b/7K
Input: HBae, 2010-07-07
Last update: Vaclav Kotesovec, 2023-06-23 more...
Genre: Fairies
FEN: 8/PP4PP/4kn2/r1p5/1q2P3/p4p1r/P2n1P1b/7K
Input: HBae, 2010-07-07
Last update: Vaclav Kotesovec, 2023-06-23 more...
Mario Richter: Hier scheint etwas zu fehlen ... (2023-05-13)
SCHRECKE: siehe P1160618 (2023-05-13)
comment
SCHRECKE: siehe P1160618 (2023-05-13)
comment
Keywords: Miniature Collection (0003568)
Genre: 2#
FEN: 8/8/K6k/R7/8/8/8/8
Input: Felber, Volker, 2010-09-11
Last update: Felber, Volker, 2010-09-11 more...
Genre: 2#
FEN: 8/8/K6k/R7/8/8/8/8
Input: Felber, Volker, 2010-09-11
Last update: Felber, Volker, 2010-09-11 more...
1. ... Kxe8 2. Dg8#
1. ... Kxc8 2. Da8#
1. ... Kxc8 2. Da8#
Keywords: Miniature Collection (0321180), Aristocrat, Miniature, No legal last move for Black
Genre: n#, Retro
Computer test: Juel: C+ Popeye 4.61
FEN: 2NkN3/8/3K4/3Q4/8/8/8/8
Input: Zuncke/Bruder, 2010-09-11
Last update: Dieter Berlin, 2023-08-15 more...
Genre: n#, Retro
Computer test: Juel: C+ Popeye 4.61
FEN: 2NkN3/8/3K4/3Q4/8/8/8/8
Input: Zuncke/Bruder, 2010-09-11
Last update: Dieter Berlin, 2023-08-15 more...
* 1. ... 0-0-0 2. Dc3#
1. Dh7! droht 2. Dxg8#
1. ... 0-0-0 2. Dc2#
1. ... Sh6,Sf6 2. De7#
1. ... Kf8 2. Df7#
1. ... Kd8 2. Dd7#
1. Dh7! droht 2. Dxg8#
1. ... 0-0-0 2. Dc2#
1. ... Sh6,Sf6 2. De7#
1. ... Kf8 2. Df7#
1. ... Kd8 2. Dd7#
* 1. ... Lxc6#
1. Ld5 Lc6 2. Th8+ Kb7 3. a8=S Lxd5#
1. Ld5 Lc6 2. Th8+ Kb7 3. a8=S Lxd5#
Keywords: under-promotion (wS)
Genre: s#
FEN: k7/Pb5R/2B5/8/8/8/5p1P/Q4N1K
Reprints: 493 Ungarische Schachprobleme mit wenigen Steinen [Bakcsi] , p. 143, 1985
Input: Frank Müller, 2011-10-11
Last update: Dieter Berlin, 2023-08-16 more...
Genre: s#
FEN: k7/Pb5R/2B5/8/8/8/5p1P/Q4N1K
Reprints: 493 Ungarische Schachprobleme mit wenigen Steinen [Bakcsi] , p. 143, 1985
Input: Frank Müller, 2011-10-11
Last update: Dieter Berlin, 2023-08-16 more...
1. bxc8=S+! Kb7 2. cxd8=S+ Kc7 3. dxe8=S+ Kd7 4. gxf8=S+ Kxe8 5. Sf5+ Dxe2#
Cook: Dual
Cook: Dual
Keywords: Promotions to Knights, konsekutive Umwandlungen 4 (SSSS), Multiple publication
Genre: s#
FEN: rnrnbbQ1/kPPPNNP1/p6p/6p1/5pK1/7P/4R3/3q4
Reprints: 1734 Schach 2 01/1954
Input: Frank Müller, 2012-03-04
Last update: Marcin Banaszek, 2021-09-24 more...
Genre: s#
FEN: rnrnbbQ1/kPPPNNP1/p6p/6p1/5pK1/7P/4R3/3q4
Reprints: 1734 Schach 2 01/1954
Input: Frank Müller, 2012-03-04
Last update: Marcin Banaszek, 2021-09-24 more...
* 1. ... Se8,Sh5#
1. Dd1 Lxd5 2. Dxd5 Se8,Sh5#
1. Dd1 Ld3,Lc2,Lb1 2. Dxd3,Dxc2,Dxb1 Se8,Sh5#
1. Dd1 Lf3 2. gxf3 Se8,Sh5#
1. Dd1 Lxg2 2. Lxg2 Se8,Sh5#
1. Dd1 Lxd5 2. Dxd5 Se8,Sh5#
1. Dd1 Ld3,Lc2,Lb1 2. Dxd3,Dxc2,Dxb1 Se8,Sh5#
1. Dd1 Lf3 2. gxf3 Se8,Sh5#
1. Dd1 Lxg2 2. Lxg2 Se8,Sh5#
* 1. ... De6=
1. dxe4 d5 2. e3 d6 3. e2 d7 4. e1=T d8=S 5. Te6 Sxe6=
1. dxe4 d5 2. e3 d6 3. e2 d7 4. e1=T d8=S 5. Te6 Sxe6=
Beispielaufgabe zum 112. Thematurnier:
Hilfspatt mit einem einzügigen Satzspiel. Zügezahl unbegrenzt, jedoch mindestens dreizügig.
Anton Baumann: C+ Gustav 4.1d (2021-02-11)
comment
Hilfspatt mit einem einzügigen Satzspiel. Zügezahl unbegrenzt, jedoch mindestens dreizügig.
Anton Baumann: C+ Gustav 4.1d (2021-02-11)
comment
* 1. ... Sd7=
1. Kxb8 Kg1 2. Kc8 Kh1 3. Kd8 Kg1 4. Ke8 Kh1 5. Kf8 Kg1 6. Kg7 Kh1 7. Kf6 Kg1 8. Ke5 Kh1 9. Kd4 Kg1 10. Kc3 Kh1 11. Kd2 Kg1 12. Ke1 Kh1 13. Kxf1 c3 14. Kxe2 Kg1 15. Kd3 Kf1 16. Ke4 Ke2 17. Kf5 Kxe3 18. Kg6 Ke4 19. Kh5 Kf5 20. Kh4 Kg6=
1. Kxb8 Kg1 2. Kc8 Kh1 3. Kd8 Kg1 4. Ke8 Kh1 5. Kf8 Kg1 6. Kg7 Kh1 7. Kf6 Kg1 8. Ke5 Kh1 9. Kd4 Kg1 10. Kc3 Kh1 11. Kd2 Kg1 12. Ke1 Kh1 13. Kxf1 c3 14. Kxe2 Kg1 15. Kd3 Kf1 16. Ke4 Ke2 17. Kf5 Kxe3 18. Kg6 Ke4 19. Kh5 Kf5 20. Kh4 Kg6=
Genre: Fairies
FEN: kN6/p1p1p3/P1P1P3/2p3p1/2P3p1/4p1p1/2P1P1P1/5B1K
Input: Felber, Volker, 2012-11-28
Last update: Felber, Volker, 2012-11-28 more...
1. Sg8 hxg8=S 2. Th6 Sxh6 3. Ld8 cxd8=S 4. Se6 Sxe6 5. Tc8 bxc8=S 6. Db8 axb8=S=
1. Dh1 Kg4! 2. Dh2 Kg5! 3. Dh3 Kg6! 4. Dh4 Kf5 5. Dh5 6. Dh6 7. Dg6+ 8. Df5 9. Df6 10. De5 11. De6 12. Dd5 13. Dd6 14. Dd8 Ke6! 15. De7 Kd5 16. Df6 Ke5 17. De6 Kb5 18. Dd6 Ka5 19. Dc6 Kb4 20. Db6+ Ka4 21. Db8 Ka5 22. Db7 Ka4 23. Db6 Ka3 24. Db3 cxb3#
Lösung falsch, aber lt Alybadix lösbar in 25 Zügen. Aber Duale!
Lösung falsch, aber lt Alybadix lösbar in 25 Zügen. Aber Duale!
Anton Baumann: Die s Königszüge dürfen nicht weggelassen werden; die w Züge sind abhängig vom Weg des sK!
Mögliche Lösung: .. 5.Dh5 Kf6 6.Dh6+ Kf7 (6. .. Ke7? 7.Dg6 - s#24) 7.Dg5! Ke6 8.Dg6+ Ke7 9.Df5 Kd6 10.Df6+ Kd7 11.De5 Kd8 12.De6 Kc7 13.Dd5 Kc8 14.Dd6 Kb7 15.Dd8 Kc6 16.De7 Kd5 17.Df6 Kc5 18.De6 Kb5 19.Dd6 Ka5 20.Dc6 Kb4 21.Db6+ Ka4 22.Db8 Ka5 23.Db7 Ka4 24.Db6 Ka3 25.Db3 cxb3# (2020-12-05)
comment
Mögliche Lösung: .. 5.Dh5 Kf6 6.Dh6+ Kf7 (6. .. Ke7? 7.Dg6 - s#24) 7.Dg5! Ke6 8.Dg6+ Ke7 9.Df5 Kd6 10.Df6+ Kd7 11.De5 Kd8 12.De6 Kc7 13.Dd5 Kc8 14.Dd6 Kb7 15.Dd8 Kc6 16.De7 Kd5 17.Df6 Kc5 18.De6 Kb5 19.Dd6 Ka5 20.Dc6 Kb4 21.Db6+ Ka4 22.Db8 Ka5 23.Db7 Ka4 24.Db6 Ka3 25.Db3 cxb3# (2020-12-05)
comment
Keywords: Alleinunterhalter, Königslenkung
Genre: s#
FEN: 8/8/8/8/2ppp3/2prp1k1/p1KpP3/B2N1Q2
Input: Frank Müller, 2013-03-08
Last update: Frank Müller, 2013-03-08 more...
Genre: s#
FEN: 8/8/8/8/2ppp3/2prp1k1/p1KpP3/B2N1Q2
Input: Frank Müller, 2013-03-08
Last update: Frank Müller, 2013-03-08 more...
* 1. ... fxe2#
1. Te4 Kxf2 2. Da1 Kf1 3. Da6+ Kf2 4. Tcc4 Kf1 5. Tc1+ Kf2 6. Tec4 Kf1 7. T4c2+ Kf2 8. De2+ fxe2#
Vorgänger Limbach 1958. Der erkannte sogar, dass Bb2 überflüssig ist.
1. Te4 Kxf2 2. Da1 Kf1 3. Da6+ Kf2 4. Tcc4 Kf1 5. Tc1+ Kf2 6. Tec4 Kf1 7. T4c2+ Kf2 8. De2+ fxe2#
Vorgänger Limbach 1958. Der erkannte sogar, dass Bb2 überflüssig ist.
Keywords: zyklischer Platzwechsel, anticipated, Obvious promotion (Lh2)
Genre: s#
Computer test: Gustav 4.1c
FEN: 8/8/8/8/1p6/1P3p2/1PRPRP1B/2QK1k1B
Input: Frank Müller, 2013-03-08
Last update: Marcin Banaszek, 2021-04-01 more...
Genre: s#
Computer test: Gustav 4.1c
FEN: 8/8/8/8/1p6/1P3p2/1PRPRP1B/2QK1k1B
Input: Frank Müller, 2013-03-08
Last update: Marcin Banaszek, 2021-04-01 more...
1. b8=T+ Kc6 2. c8=T+ Kd6 3. d8=T+ Ke6 4. e8=T+ Kxf6 5. f8=D+ Kxg6 6. g8=T+ Lg7#
fehlender sBa3 war Druckfehler.
fehlender sBa3 war Druckfehler.
Keywords: Promotions to rooks, konsekutive Umwandlungen 6 (TTTTDT)
Genre: s#
FEN: 8/RPPPPPP1/1k3NN1/Rpp2ppp/8/p7/rb1K2n1/2n1r3
Input: Frank Müller, 2013-03-13
Last update: Marcin Banaszek, 2020-11-12 more...
Genre: s#
FEN: 8/RPPPPPP1/1k3NN1/Rpp2ppp/8/p7/rb1K2n1/2n1r3
Input: Frank Müller, 2013-03-13
Last update: Marcin Banaszek, 2020-11-12 more...
1. c8=T+ Kd6 2. d8=T+ Ke6 3. e8=T+ Kf6 4. f8=T+ Kxg6 5. g8=T+ Kh6 6. h8=T+ Lh7#
Keywords: Promotions to rooks, konsekutive Umwandlungen 6
Genre: s#
FEN: 8/R1PPPPPP/1Nk3N1/Ppp3p1/1p4Pn/8/r1bK1b2/2n1r3
Input: Frank Müller, 2013-03-13
Last update: Marcin Banaszek, 2020-11-12 more...
Genre: s#
FEN: 8/R1PPPPPP/1Nk3N1/Ppp3p1/1p4Pn/8/r1bK1b2/2n1r3
Input: Frank Müller, 2013-03-13
Last update: Marcin Banaszek, 2020-11-12 more...
65 - P1268134
Tivadar Kardos
12984v Schweizerische Schachzeitung 09/1987
Informalturnier 1987/88
1. Lob
(11+5) cooked
s#12
Tivadar Kardos
12984v Schweizerische Schachzeitung 09/1987
Informalturnier 1987/88
1. Lob
(11+5) cooked
s#12
Korrektur in 05/1988, Seite 187
1. De7+ 2. Dd6+ 3. Dd4+ 4. Tb1+ 5. Dd5+ Sb3 6. Dg2+ Sd2 7. Ta1+ Kb3 8. Dd5+ Sc4 9. Df3+ Se3 10. Ta3+ Kc4 11. Df7+ Sd5 12. Df4+ Sxf4#
Cook: Gustav findet 4 Nebenlösungen in 11 Zügen
1. Dc6+
1. Df7+
1. Dc8+
1. Lf5+
1. De7+ 2. Dd6+ 3. Dd4+ 4. Tb1+ 5. Dd5+ Sb3 6. Dg2+ Sd2 7. Ta1+ Kb3 8. Dd5+ Sc4 9. Df3+ Se3 10. Ta3+ Kc4 11. Df7+ Sd5 12. Df4+ Sxf4#
Cook: Gustav findet 4 Nebenlösungen in 11 Zügen
1. Dc6+
1. Df7+
1. Dc8+
1. Lf5+
Anton Baumann: Mit Stellung gem. Diagramm (wBh2 sBh4) unlösbar!
Original: ohne wBh2 ; Korrektur: "im s#129984 versetzt der Verfasser den sBh4 nach h2 und fügt einen wBh4 hinzu". Die oben angegebenen NL beziehen sich auf diese Korrektur-Stellung!
Verbesserungsvorschlag ABm: wDb7 nach a7, -wBh2, -wBh6, sKe5, +sBg2, +sBg7, h4=sB (9+7)
Lösung gemäss Autorabsicht: 1.De7+ Kd5 2.Dd6+ Kc4 3.Dd4+ ... (2023-09-12)
comment
Original: ohne wBh2 ; Korrektur: "im s#129984 versetzt der Verfasser den sBh4 nach h2 und fügt einen wBh4 hinzu". Die oben angegebenen NL beziehen sich auf diese Korrektur-Stellung!
Verbesserungsvorschlag ABm: wDb7 nach a7, -wBh2, -wBh6, sKe5, +sBg2, +sBg7, h4=sB (9+7)
Lösung gemäss Autorabsicht: 1.De7+ Kd5 2.Dd6+ Kc4 3.Dd4+ ... (2023-09-12)
comment
Genre: s#
FEN: 8/1Q5B/4k2P/1pB3PK/N5Pp/8/7P/n4RRr
Input: Frank Müller, 2013-04-21
Last update: Frank Müller, 2018-02-04 more...
1. a1=L f4 2. Le5 fxe5 3. 0-0-0 e6 4. Kb8 e7 5. Ka8 e8=S 6. Tb8 Sc7#
Cook: vielfach NL, bereits in 5 Zügen:
1. 0-0-0 f2-f4 2. Td8-d7 f4-f5 3. Kc8-d8 f5-f6 4. La7-b8 f6-f7 5. Lb8-c7 f7-f8=D#
1. Ke8-f7 f2-f4 2. Ta8-h8 f4-f5 3. Kf7-g8 f5-f6 4. Th8-h7 f6-f7+ 5. Kg8-h8 f7-f8=D# u.a.
Cook: vielfach NL, bereits in 5 Zügen:
1. 0-0-0 f2-f4 2. Td8-d7 f4-f5 3. Kc8-d8 f5-f6 4. La7-b8 f6-f7 5. Lb8-c7 f7-f8=D#
1. Ke8-f7 f2-f4 2. Ta8-h8 f4-f5 3. Kf7-g8 f5-f6 4. Th8-h7 f6-f7+ 5. Kg8-h8 f7-f8=D# u.a.
1. b3 axb3 2. a4 bxa4 3. Sb5 axb5 4. Ld5 bxa6 5. 0-0 axb7 6. Ta8 bxa8=L 7. Kh8 Lxd5=
Genre: Fairies
FEN: 4k1br/np6/p2p2K1/p7/1p6/8/P7/8
Input: Frank Müller, 2015-04-30
Last update: Frank Müller, 2015-05-01 more...
* 1. ... Txh8#
1. Txh7+ Kxh7 2. Lf6 Kh6 3. Ke7 Kh7 4. Kd6 Kh6 5. Kc5 Kh7 6. Kxd4 Kh6 7. Ke4 Kh7 8. Te3 Kh6 9. f8=S g3#
1. Txh7+ Kxh7 2. Lf6 Kh6 3. Ke7 Kh7 4. Kd6 Kh6 5. Kc5 Kh7 6. Kxd4 Kh6 7. Ke4 Kh7 8. Te3 Kh6 9. f8=S g3#
Keywords: Schlag Satzmattfigur, Capture key, under-promotion (S), Selfblock, Königswanderung, Fata Morgana (Typ Weber)
Genre: s#
FEN: 5K1R/4BP1r/4p1pk/4P1pb/3p2pr/3P2Rp/7P/8
Reprints: 3322 Revista Romana de Sah 05-07/1949
Input: Frank Müller, 2015-05-15
Last update: Gunter Jordan, 2023-11-16 more...
Genre: s#
FEN: 5K1R/4BP1r/4p1pk/4P1pb/3p2pr/3P2Rp/7P/8
Reprints: 3322 Revista Romana de Sah 05-07/1949
Input: Frank Müller, 2015-05-15
Last update: Gunter Jordan, 2023-11-16 more...
a) 1. Kd4 bxc5 2. Ke5 d4#
b) 1. axb3ep Txe1 2. Sd3 Txe4#
b) 1. axb3ep Txe1 2. Sd3 Txe4#
in der mir vorliegenden Sekundärquelle noch zusätzlicher wBa2 und Steinkontrolle 8+16, aber dann jede Menge NL mit 1. ... Txe1 2. Sd3 Txe4#;
s.a. P0003396 (mit wBa5 aber Teil b) illegal)
TO DO: Wer kann in der Originalquelle nachschauen?
Henrik Juel: C+ by Popeye 4.61 (and analysis)
In part a the ep capture cannot be legitimized, as last move could be g7-g8=S (2018-03-22)
more ...
comment
s.a. P0003396 (mit wBa5 aber Teil b) illegal)
TO DO: Wer kann in der Originalquelle nachschauen?
Henrik Juel: C+ by Popeye 4.61 (and analysis)
In part a the ep capture cannot be legitimized, as last move could be g7-g8=S (2018-03-22)
more ...
comment
Keywords: Than theme, En passant as key
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 4.61 & trivial retro-logic
FEN: 6N1/4p3/4pP1p/1prn1rqp/pPk1bp2/7K/3P2pP/2nRb3
Input: Mario Richter, 2018-03-22
Last update: A.Buchanan, 2021-11-24 more...
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 4.61 & trivial retro-logic
FEN: 6N1/4p3/4pP1p/1prn1rqp/pPk1bp2/7K/3P2pP/2nRb3
Input: Mario Richter, 2018-03-22
Last update: A.Buchanan, 2021-11-24 more...
1. Db5+! Kd4 2. Sf5#
1. ... Kd6 2. Se8#
1. ... Kd6 2. Se8#
Keywords: Mirror mate, Miniature
Genre: 2#
Computer test: popeye 4.87
FEN: 8/5KN1/8/3k4/1Q6/5P2/1P6/8
Input: Rainer Staudte, 2021-09-15
Last update: Alfred Pfeiffer, 2021-09-16 more...
Genre: 2#
Computer test: popeye 4.87
FEN: 8/5KN1/8/3k4/1Q6/5P2/1P6/8
Input: Rainer Staudte, 2021-09-15
Last update: Alfred Pfeiffer, 2021-09-16 more...
1. c8=S! Ka6 2. Dxb6#
1. ... Kc6 2. Dxb6#
1. ... Kxc8 2. a8=D#
1. ... Ka8 2. Dg2#
1. ... Kc6 2. Dxb6#
1. ... Kxc8 2. a8=D#
1. ... Ka8 2. Dg2#
Keywords: X flights, under-promotion key (S), konsekutive Umwandlungen 2 (SD)
Genre: 2#
FEN: 4N3/PkP5/1b2K3/8/8/8/1Q6/8
Input: Rainer Staudte, 2021-09-15
Last update: Rainer Staudte, 2021-09-15 more...
Genre: 2#
FEN: 4N3/PkP5/1b2K3/8/8/8/1Q6/8
Input: Rainer Staudte, 2021-09-15
Last update: Rainer Staudte, 2021-09-15 more...
1. Dh2?
1. ... b5!
1. Kg6? droht 2. Dg7#
1. ... Kf8!
1. Dg2+! Kh7 2. Dg7#
1. ... Kf8 2. Da8#
1. ... Kh8 2. Dg7#
1. ... b5!
1. Kg6? droht 2. Dg7#
1. ... Kf8!
1. Dg2+! Kh7 2. Dg7#
1. ... Kf8 2. Da8#
1. ... Kh8 2. Dg7#
Genre: 2#
FEN: 6k1/8/1p3K2/8/8/8/1Q6/8
Input: Rainer Staudte, 2021-09-15
Last update: Rainer Staudte, 2021-09-15 more...
73 - P1396159
Tivadar Kardos
James Malcom
Andrew Buchanan
PDB Website 24/11/2021
TK, correction JM & AB
(6+15) C+
h#3
Tivadar Kardos
James Malcom
Andrew Buchanan
PDB Website 24/11/2021
TK, correction JM & AB
(6+15) C+
h#3
1. dxe3ep+ Kxc4 2. Lc3 Sxc2 3. Ta5 axb3#
R: 1. e2-e4 Tf3-d3+ 2. Ke4-d5
See P0003211
Example Proof Game: 1. b3 a5 2. c4 a4 3. Na3 b5 4. d4 c5 5. Bg5 h6 6. f4 hxg5 7. Nf3 gxf4 8. Qd3 bxc4 9. Nc2 axb3 10. Rc1 Bb7 11. Na1 Nf6 12. Rc3 Nh5 13. Qe3 fxe3 14. Nd2 exd2+ 15. Kf2 Nf4 16. Rg1 Nh3+ 17. gxh3 Bh1 18. Rcg3 Ra6 19. Rg6 Rf6+ 20. Ke3 Rh5 21. Bg2 Nc6 22. Be4 Na5 23. Bc2 bxc2 24. R1g4 Rf3+ 25. Ke4 Nb3 26. Rd6 exd6 27. Rf4 Qh4 28. Rg4 Qe1 29. Rf4 Kd8 30. Rg4 Kc7 31. Rf4 Kb6 32. Rg4 Ka5 33. Rf4 Ka4 34. Rg4 Rhf5 35. Rh4 g5 36. Rg4 Bg7 37. Rh4 Be5 38. Rf4 cxd4 39. Rg4 f6 40. Rh4 Qf1 41. Rf4 gxf4 42. Kd5 Rd3+ 43. e4
R: 1. e2-e4 Tf3-d3+ 2. Ke4-d5
See P0003211
Example Proof Game: 1. b3 a5 2. c4 a4 3. Na3 b5 4. d4 c5 5. Bg5 h6 6. f4 hxg5 7. Nf3 gxf4 8. Qd3 bxc4 9. Nc2 axb3 10. Rc1 Bb7 11. Na1 Nf6 12. Rc3 Nh5 13. Qe3 fxe3 14. Nd2 exd2+ 15. Kf2 Nf4 16. Rg1 Nh3+ 17. gxh3 Bh1 18. Rcg3 Ra6 19. Rg6 Rf6+ 20. Ke3 Rh5 21. Bg2 Nc6 22. Be4 Na5 23. Bc2 bxc2 24. R1g4 Rf3+ 25. Ke4 Nb3 26. Rd6 exd6 27. Rf4 Qh4 28. Rg4 Qe1 29. Rf4 Kd8 30. Rg4 Kc7 31. Rf4 Kb6 32. Rg4 Ka5 33. Rf4 Ka4 34. Rg4 Rhf5 35. Rh4 g5 36. Rg4 Bg7 37. Rh4 Be5 38. Rf4 cxd4 39. Rg4 f6 40. Rh4 Qf1 41. Rf4 gxf4 42. Kd5 Rd3+ 43. e4
Henrik Juel: Black pawns captured all 10 missing white men, and White captured g2xSh3
Last move was not e3-e4 (because of the check from Lh1), but e2-e4
HC+ Popeye 4.61 (2021-11-24)
A.Buchanan: I’d like to claim that Popeye + Retractor + example proof game means this problem is C+. If not, what more would be required. (2021-11-24)
Henrik Juel: I guess that suffices (2021-11-24)
comment
Last move was not e3-e4 (because of the check from Lh1), but e2-e4
HC+ Popeye 4.61 (2021-11-24)
A.Buchanan: I’d like to claim that Popeye + Retractor + example proof game means this problem is C+. If not, what more would be required. (2021-11-24)
Henrik Juel: I guess that suffices (2021-11-24)
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Keywords: En passant as key, Checking key
Genre: h#, Retro
Computer test: C+ Popeye 4.61 for forward play Retractor 2.0 for retraction Lichess to check demo game
FEN: 8/3p4/3p1p2/3Kbr2/k1ppPp2/1n1r3P/P1pp3P/N4q1b
Input: James Malcom, 2021-11-24
Last update: A.Buchanan, 2023-07-13 more...
Genre: h#, Retro
Computer test: C+ Popeye 4.61 for forward play Retractor 2.0 for retraction Lichess to check demo game
FEN: 8/3p4/3p1p2/3Kbr2/k1ppPp2/1n1r3P/P1pp3P/N4q1b
Input: James Malcom, 2021-11-24
Last update: A.Buchanan, 2023-07-13 more...
1. cxb3ep+ Sb4+ 2. Kb1 Sxe2 3. b2 0-0#
R: 1. b2-b4 Tc3-a3+ 2. Sa3-c2
Black captured 8 white men with pawns. Critically sBd & sBe never captured, therefore White pawns captured onto d- & e-files (to explain the pawn inversions there). Therefore wBb4 did not come from c3.
R: 1. b2-b4 Tc3-a3+ 2. Sa3-c2
Black captured 8 white men with pawns. Critically sBd & sBe never captured, therefore White pawns captured onto d- & e-files (to explain the pawn inversions there). Therefore wBb4 did not come from c3.
See P0574383.
With the ink still wet on this correction, I would like to remove sD. Sound by Popeye & Retractor, but needs an adapted proof game for C+.
Proof Game: 1. c4 a6 2. g3 Nc6 3. Bg2 Na5 4. Bc6 h5 5. Bb5 axb5 6. g4 c5 7. d4 cxd4 8. e4 f5 9. Be3 fxe4 10. f4 g5 11. Qf3 Nb3 12. Na3 hxg4 13. Nc2 gxf3 14. Rd1 Na1 15. Rd2 Ra3 16. Rg2 Rc3 17. a4 d3 18. Bf2 e3 19. Bg3 e5 20. Re2 e4 21. a5 d5 22. a6 d4 23. a7 Qd5 24. a8=R Qe5 25. Rf2 Kd7 26. Re2 Nf6 27. Rd2 Nd5 28. cxd5 Kd6 29. Rf2 Kc5 30. Re2 b6 31. Rd2 Kc4 32. fxe5 Kb3 33. Re2 fxe2 34. Rb8 Bg4 35. Rb7 Bf3 36. Rc7 Ka2 37. Bf4 gxf4 38. Rc6 Bb4 39. Rc5 Rh3 40. Na3 Ba5 41. Rc4 bxc4 42. Nc2 Ra3+ 43. b4
A.Buchanan: 6n1/8/1q6/b7/1Ppppp2/r2ppp1r/k1N1p2P/n3K1NR has one less piece again, but I prefer the current version as retro logic is more interesting. (2021-11-26)
James Malcom: Enjoy an optimal PG Andrew. The total sum of the queen & knight journeys is 8 no matter which knight ends up on a1. (2021-11-27)
A.Buchanan: :-) thanks James (2021-11-27)
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With the ink still wet on this correction, I would like to remove sD. Sound by Popeye & Retractor, but needs an adapted proof game for C+.
Proof Game: 1. c4 a6 2. g3 Nc6 3. Bg2 Na5 4. Bc6 h5 5. Bb5 axb5 6. g4 c5 7. d4 cxd4 8. e4 f5 9. Be3 fxe4 10. f4 g5 11. Qf3 Nb3 12. Na3 hxg4 13. Nc2 gxf3 14. Rd1 Na1 15. Rd2 Ra3 16. Rg2 Rc3 17. a4 d3 18. Bf2 e3 19. Bg3 e5 20. Re2 e4 21. a5 d5 22. a6 d4 23. a7 Qd5 24. a8=R Qe5 25. Rf2 Kd7 26. Re2 Nf6 27. Rd2 Nd5 28. cxd5 Kd6 29. Rf2 Kc5 30. Re2 b6 31. Rd2 Kc4 32. fxe5 Kb3 33. Re2 fxe2 34. Rb8 Bg4 35. Rb7 Bf3 36. Rc7 Ka2 37. Bf4 gxf4 38. Rc6 Bb4 39. Rc5 Rh3 40. Na3 Ba5 41. Rc4 bxc4 42. Nc2 Ra3+ 43. b4
A.Buchanan: 6n1/8/1q6/b7/1Ppppp2/r2ppp1r/k1N1p2P/n3K1NR has one less piece again, but I prefer the current version as retro logic is more interesting. (2021-11-26)
James Malcom: Enjoy an optimal PG Andrew. The total sum of the queen & knight journeys is 8 no matter which knight ends up on a1. (2021-11-27)
A.Buchanan: :-) thanks James (2021-11-27)
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75 - P1396199
Tivadar Kardos
Andrew Buchanan
PDB Website 25/11/2021
TK, version AB
(10+13) C+
PG in 8.5
Tivadar Kardos
Andrew Buchanan
PDB Website 25/11/2021
TK, version AB
(10+13) C+
PG in 8.5
1. Sc3 Sf6 2. Se4 Sxe4 3. g3 Sxg3 4. Sf3 Sxh1 5. Se5 Sxf2 6. Sxd7 Sxd1 7. Sxb8 Txb8 8. Kxd1 Ta8 9. Ke1
P0000736 plus the 1.5 obvious moves. Why did he stop? :-)
Henrik Juel: He probably stopped because
1. he was afraid of cooks
2. the Homebase theme was unknown in 1979
Is the extension to 8.5 C+? (2021-11-25)
Olaf Jenkner: This version is in my opinion 100 times better than P0000736, a homebase with 9 missing men in only 8.5 moves.
Is there a record list of proof games "homebase with x missung men" in a minimum number of moves? (2021-11-25)
A.Buchanan: Thanks Olaf for your kind words. I don't think there is such a list. PDB has today 19 homebase unique PGs which are no longer than this with no fewer captures. Most are 9 captures but P1000453 by Le Gleuher alone manages to get a 10th capture. He also manages the shortest (6.5 moves P1000944), with 7.5 moves by Mintz P1329742 as runner-up, and 4 others in 8.0. To be fair, the aesthetic for the Future Proof Game gang has generally been to *minimize* the number of captures. Except for Mintz and one other, all involve promotions. (2021-11-26)
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Henrik Juel: He probably stopped because
1. he was afraid of cooks
2. the Homebase theme was unknown in 1979
Is the extension to 8.5 C+? (2021-11-25)
Olaf Jenkner: This version is in my opinion 100 times better than P0000736, a homebase with 9 missing men in only 8.5 moves.
Is there a record list of proof games "homebase with x missung men" in a minimum number of moves? (2021-11-25)
A.Buchanan: Thanks Olaf for your kind words. I don't think there is such a list. PDB has today 19 homebase unique PGs which are no longer than this with no fewer captures. Most are 9 captures but P1000453 by Le Gleuher alone manages to get a 10th capture. He also manages the shortest (6.5 moves P1000944), with 7.5 moves by Mintz P1329742 as runner-up, and 4 others in 8.0. To be fair, the aesthetic for the Future Proof Game gang has generally been to *minimize* the number of captures. Except for Mintz and one other, all involve promotions. (2021-11-26)
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Keywords: Unique Proof Game, Homebase (2), Switchback (Kt)
Genre: Retro
Computer test: C+ Natch 3.1 in 53 minutes
FEN: r1bqkb1r/ppp1pppp/8/8/8/8/PPPPP2P/R1B1KB2
Input: A.Buchanan, 2021-11-25
Last update: A.Buchanan, 2021-11-26 more...
Genre: Retro
Computer test: C+ Natch 3.1 in 53 minutes
FEN: r1bqkb1r/ppp1pppp/8/8/8/8/PPPPP2P/R1B1KB2
Input: A.Buchanan, 2021-11-25
Last update: A.Buchanan, 2021-11-26 more...
76 - P1396304
Tivadar Kardos
Andrew Buchanan
PDB Website 27/11/2021
Happy Birthday to James Malcom!
A gift from Tivadar Kardos (helped a little by Andrew Buchanan)
I don't think this one needs a proof game :)
TK, version AB
(11+13) C+
h#3
2 solutions
Tivadar Kardos
Andrew Buchanan
PDB Website 27/11/2021
Happy Birthday to James Malcom!
A gift from Tivadar Kardos (helped a little by Andrew Buchanan)
I don't think this one needs a proof game :)
TK, version AB
(11+13) C+
h#3
2 solutions
1. Lxa2 Txa2 2. 0-0-0 Txa7 3. Td7 Ta8#
1. Dg3 Txb1 2. Dxb3 Txb3 3. 0-0 Tg3#
1. Lxa2 0-0-0? 2. Dxg2 Tg1 3. 0-0 Txg2# (White can't castle)
1. Dg3 Txb1 2. Dxb3 Txb3 3. 0-0 Tg3#
1. Lxa2 0-0-0? 2. Dxg2 Tg1 3. 0-0 Txg2# (White can't castle)
The cooked P0006972 can be fixed many ways. I wanted to find a single touch that would keep the intended solution, the intended try, and exactly one of the cooks which seemed worth saving, while eliminating all the other cooks.
Henrik Juel: Black captured [Lf1] and the remaining 4 missing white men with exdxcxTb2-b1=L and gxh
1.Bb1*a2 Ra1*a2 2.0-0-0 Ra2*a7 3.Rd8-d7 Ra7-a8 #
1.Qg8-g3 Ra1*b1 2.Qg3*b3 Rb1*b3 3.0-0 Rb3-g3 #
Not 1.Bb1*a2 0-0-0? 2.Qg8*g2 Rd1-g1 3.0-0 Rg1*g2 # (White may not castle, as he has moved his king)
Neat birthday greeting (2021-11-27)
James Malcom: Thanks for the warm gift Andrew. Huzzah! (2021-11-28)
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Henrik Juel: Black captured [Lf1] and the remaining 4 missing white men with exdxcxTb2-b1=L and gxh
1.Bb1*a2 Ra1*a2 2.0-0-0 Ra2*a7 3.Rd8-d7 Ra7-a8 #
1.Qg8-g3 Ra1*b1 2.Qg3*b3 Rb1*b3 3.0-0 Rb3-g3 #
Not 1.Bb1*a2 0-0-0? 2.Qg8*g2 Rd1-g1 3.0-0 Rg1*g2 # (White may not castle, as he has moved his king)
Neat birthday greeting (2021-11-27)
James Malcom: Thanks for the warm gift Andrew. Huzzah! (2021-11-28)
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Keywords: Castling (sgsk), Cant Castler (wg), Obvious promotion (l)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 + trivial retro-logic
FEN: r3k1qr/pp3p1p/1bp5/3pB2p/8/1P6/P1PPPPPP/Rb2K3
Input: A.Buchanan, 2021-11-27
Last update: A.Buchanan, 2021-11-28 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 + trivial retro-logic
FEN: r3k1qr/pp3p1p/1bp5/3pB2p/8/1P6/P1PPPPPP/Rb2K3
Input: A.Buchanan, 2021-11-27
Last update: A.Buchanan, 2021-11-28 more...
77 - P1396401
Laszlo Lindner
Jozsef Bajtay
Tivadar Kardos
Andrew Buchanan
Discord Chess Problems & Studies Server 29/11/2021
LL,JB,TK, correction AB
(6+15) C+
h#3
Laszlo Lindner
Jozsef Bajtay
Tivadar Kardos
Andrew Buchanan
Discord Chess Problems & Studies Server 29/11/2021
LL,JB,TK, correction AB
(6+15) C+
h#3
1. axb3ep a3 2. Sa2 Kxg4 3. Sc3 d3#
AB: The en passant is pure tempo. *All* other moves interfere with the solution. After first clearing a2 with 1. ... a3, White's second move is also pure tempo. sDf2 is a "value weasel" - the problem would still be sound with sB, but it's entertaining that sD cannot move to any of 9 apparently available squares.
AB: The en passant is pure tempo. *All* other moves interfere with the solution. After first clearing a2 with 1. ... a3, White's second move is also pure tempo. sDf2 is a "value weasel" - the problem would still be sound with sB, but it's entertaining that sD cannot move to any of 9 apparently available squares.
Corrects P0574612.
A.Buchanan: All I have done is fix a bug. Apart from applause at the motive for B1 & W2, also note the way all pieces are necessary for for unique forward play: there is zero "retro-dressing" (2021-11-29)
Henrik Juel: White has captured once only, so last move was not a3xb4 or c3xb4, but b2-b4, legitimizing the ep key (2021-11-29)
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A.Buchanan: All I have done is fix a bug. Apart from applause at the motive for B1 & W2, also note the way all pieces are necessary for for unique forward play: there is zero "retro-dressing" (2021-11-29)
Henrik Juel: White has captured once only, so last move was not a3xb4 or c3xb4, but b2-b4, legitimizing the ep key (2021-11-29)
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Keywords: En passant as key, Tempo Move (sw)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & Retractor 2.1.1 Human contribution is "position prior to retractions looks legal enough."
FEN: 8/7p/5p1P/1ppb2pK/pPkn2r1/8/P1PPpq2/2n1br2
Input: A.Buchanan, 2021-11-29
Last update: A.Buchanan, 2021-11-29 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: 8/7p/5p1P/1ppb2pK/pPkn2r1/8/P1PPpq2/2n1br2
Input: A.Buchanan, 2021-11-29
Last update: A.Buchanan, 2021-11-29 more...
1. Txc4!
Der Rekord für Mattänderungen im Zugwechsel: Vier Satzmatts werden ersetzt
durch 8(!) neue und noch 3 zugefügte Matts (H. Albrecht).
Der Rekord für Mattänderungen im Zugwechsel: Vier Satzmatts werden ersetzt
durch 8(!) neue und noch 3 zugefügte Matts (H. Albrecht).
1. d4 Td2 2. Lc3 Lc1 3. d5+ e5+ 4. dxe6ep+ Td4#
Keywords: En passant
Genre: Fairies
Computer test: C+, Gustav 4.2a, Brute Force
FEN: 4R3/4p3/5k2/8/5K2/b4PR1/1r1P4/B7
Input: A.Buchanan, 2021-12-31
Last update: James Malcom, 2022-01-02 more...
Genre: Fairies
Computer test: C+, Gustav 4.2a, Brute Force
FEN: 4R3/4p3/5k2/8/5K2/b4PR1/1r1P4/B7
Input: A.Buchanan, 2021-12-31
Last update: James Malcom, 2022-01-02 more...
1. bxc3ep+ d3 2. 0-0 Tf4 3. Kh8 Txf8#
Not R: 1. c3-c4? La6xb5? because Black's 10 captures are already accounted for: 6 pawn captures on files a-d, e.g.: bax, cxbxa, dxcxbxa + 4 waylaid pawns on files e-h or maybe 2 more pawn captures if bPe & bPf cross-capture.
Not R: 1. c3-c4? La6xb5? because Black's 10 captures are already accounted for: 6 pawn captures on files a-d, e.g.: bax, cxbxa, dxcxbxa + 4 waylaid pawns on files e-h or maybe 2 more pawn captures if bPe & bPf cross-capture.
81 - P1400083
Tivadar Kardos
3283 Mat-Pat 50 12/1996
3. ehrende Erwähnung
Martin 1994-1995
(10+10)
s#13
Tivadar Kardos
3283 Mat-Pat 50 12/1996
3. ehrende Erwähnung
Martin 1994-1995
(10+10)
s#13
1. h7-h8=D+! Kh6-g6 2. g7-g8=D+ Kg6-f5 3. Dh8xf6+ Kf5xf6 4. f7-f8=D+ Kf6-e5 5. Dg8xe6+ Ke5xe6 6. e7-e8=D+ Ke6-d5 7. Df8xd6+ Kd5xd6 8. d7-d8=D+ Kd6-c5 9. De8xc6+ Kc5xc6 10. c7-c8=D+ Kc6-b5 11. Dd8xb6+ Kb5xb6 12. b7-b8=D+ Kb6-a5 13. Db8-b4+ De1xb4#
Judge Vladislav Bunka 12/1996
Judge Vladislav Bunka 12/1996
Henrik Juel: Popeye 4.61 with 'opt non 0 1 thr 0' found no flaws (2022-03-28)
SCHRECKE: Kurzlösung in 12 Zügen: 5.Dg7+,Dh8+ Kd5 6.Df3+ Kc5 7.Dd4+ Kb5 8.D:b6+ K:b6 9.b8D+ Ka5! 10.Db4+ Ka6 11.c8D+ Ka7 12.Da5+ D:a5# (2022-03-29)
Henrik Juel: Thanks for pointing out this cook, Michael
I have not seen something like that before, so maybe I should write '... found no flaws in 13 moves' or similarly in the future
Of course Popeye can 'handle' this with 'stip s#12 opt non 1 1 thr 0', but using a dedicated solver is easier (2022-03-29)
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SCHRECKE: Kurzlösung in 12 Zügen: 5.Dg7+,Dh8+ Kd5 6.Df3+ Kc5 7.Dd4+ Kb5 8.D:b6+ K:b6 9.b8D+ Ka5! 10.Db4+ Ka6 11.c8D+ Ka7 12.Da5+ D:a5# (2022-03-29)
Henrik Juel: Thanks for pointing out this cook, Michael
I have not seen something like that before, so maybe I should write '... found no flaws in 13 moves' or similarly in the future
Of course Popeye can 'handle' this with 'stip s#12 opt non 1 1 thr 0', but using a dedicated solver is easier (2022-03-29)
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Genre: s#
FEN: 8/1PPPPPPP/1ppppp1k/6p1/2p5/K3P3/2r2N2/4q3
Input: Marcin Banaszek, 2022-03-28
Last update: Marcin Banaszek, 2022-03-28 more...
82 - P1400824
Otto Kerekes
Tivadar Kardos
Andrew Buchanan
France-Echecs 23/04/2022
OK & TK, correction AB
(11+15)
h#2
b) wBg2 -> sB
Otto Kerekes
Tivadar Kardos
Andrew Buchanan
France-Echecs 23/04/2022
OK & TK, correction AB
(11+15)
h#2
b) wBg2 -> sB
a) 1. cxd4 Lxc4 2. Dd5 Ld3#
not 1. cxd3ep? Txd1 2. dxc2 Lxc2# because last move was 0-0.
b) 1. cxd3ep cxd3+ 2. Kf3 Lxd1# because last move was d2-d4.
not 1. cxd3ep? Txd1 2. dxc2 Lxc2# because last move was 0-0.
b) 1. cxd3ep cxd3+ 2. Kf3 Lxd1# because last move was d2-d4.
Corrects P0005275
Henrik Juel: C+ Popeye 4.61 and analysis (2022-04-23)
Henrik Juel: solutions
a) 1.cxd4 Lxc4 2.Dd5 Lxc2#
not 1.cxd3ep? Txd1 2.dxc2 Lxc2, because last move could be 0-0
b) 1.cxd3ep cxd3+ 2.Kf3 Lxd1# (2022-04-23)
Mario Richter: In a), why not 1. cxd3ep?
Andrew: "because last move was 0-0."
Henrik: "because last move could be 0-0"
Mario: "because d2-d4 definitively was not White's last move" (2022-04-24)
Henrik Juel: You are right, Mario
But my argument suffices to rule out the ep capture, which is allowed only if you can show that last move was d2-d4 (2022-04-24)
A.Buchanan: Thanks Mario. I think that our opinions are consistent. I wanted to emphasise the castling because of Valladao. In b), h8=L is accurate, even if we never know the result of e8=X in either twin. (2022-04-25)
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Henrik Juel: C+ Popeye 4.61 and analysis (2022-04-23)
Henrik Juel: solutions
a) 1.cxd4 Lxc4 2.Dd5 Lxc2#
not 1.cxd3ep? Txd1 2.dxc2 Lxc2, because last move could be 0-0
b) 1.cxd3ep cxd3+ 2.Kf3 Lxd1# (2022-04-23)
Mario Richter: In a), why not 1. cxd3ep?
Andrew: "because last move was 0-0."
Henrik: "because last move could be 0-0"
Mario: "because d2-d4 definitively was not White's last move" (2022-04-24)
Henrik Juel: You are right, Mario
But my argument suffices to rule out the ep capture, which is allowed only if you can show that last move was d2-d4 (2022-04-24)
A.Buchanan: Thanks Mario. I think that our opinions are consistent. I wanted to emphasise the castling because of Valladao. In b), h8=L is accurate, even if we never know the result of e8=X in either twin. (2022-04-25)
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Keywords: Valladao Task, En passant as key, Castling in the retro play, Promotion (L), Castling (wk)
Genre: h#, Retro
Computer test: HC+ Popeye v4.61 & Retractor 2 which determines that only candidate last move are a) d2-d4 & 0-0 b) d2–d4. Manual logic determines in a) that last move was not in fact d2-d4 because wPe would promote on wrong colour square. Also C+ with rawbats
FEN: 8/p1p5/1B6/q1p1b3/r1pPkpp1/1Bp3Pb/PPP2PPr/1n1n1RK1
Input: A.Buchanan, 2022-04-23
Last update: A.Buchanan, 2023-08-15 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.61 & Retractor 2 which determines that only candidate last move are a) d2-d4 & 0-0 b) d2–d4. Manual logic determines in a) that last move was not in fact d2-d4 because wPe would promote on wrong colour square. Also C+ with rawbats
FEN: 8/p1p5/1B6/q1p1b3/r1pPkpp1/1Bp3Pb/PPP2PPr/1n1n1RK1
Input: A.Buchanan, 2022-04-23
Last update: A.Buchanan, 2023-08-15 more...
83 - P1400896
Tivadar Kardos
Andrew Buchanan
Discord Chess Problems & Studies Server 26/04/2022
TK, correction AB
(7+15) C+
h#3 AP
Tivadar Kardos
Andrew Buchanan
Discord Chess Problems & Studies Server 26/04/2022
TK, correction AB
(7+15) C+
h#3 AP
1. gxf3ep+ e3 2. 0-0-0 Tb4 3. Tg4 Tb8#
Black has visibly made 6 pawn captures right to left. Two captures are required to resolve the h-file (either two pcs by black, or wPgxh with wPh waylaid). If black preserves castling rights, then White pawns on a & d files require one more capture by either side to resolve. So all 10 captures accounted for. If wKc1 then bPdxc but this implies 2 more captures, only one of which can be taken from d file. Similarly wPfxexf too expensive. So If black can castle, then the last move was indeed pawn double hop.
The castling serves two purposes: justifying the ep capture and enabling the mate: there are no retro tries in this problem.
Black has visibly made 6 pawn captures right to left. Two captures are required to resolve the h-file (either two pcs by black, or wPgxh with wPh waylaid). If black preserves castling rights, then White pawns on a & d files require one more capture by either side to resolve. So all 10 captures accounted for. If wKc1 then bPdxc but this implies 2 more captures, only one of which can be taken from d file. Similarly wPfxexf too expensive. So If black can castle, then the last move was indeed pawn double hop.
The castling serves two purposes: justifying the ep capture and enabling the mate: there are no retro tries in this problem.
Author: Kardos' cooked P0003269 was from 1956 just after AP compositions began to appear, and I can't decide whether he intended to include AP but it seems to fit this corrected version.
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Keywords: a posteriori (AP) (Type Petrovic), Castling (sg), En passant as key
Genre: h#, Retro
Computer test: C+ Popeye 4.61 and analysis
FEN: r3k3/3q4/5p2/5pbP/n4PpR/1pp3rp/2p1P1pB/N1K3n1
Input: A.Buchanan, 2022-04-26
Last update: A.Buchanan, 2023-09-11 more...
Genre: h#, Retro
Computer test: C+ Popeye 4.61 and analysis
FEN: r3k3/3q4/5p2/5pbP/n4PpR/1pp3rp/2p1P1pB/N1K3n1
Input: A.Buchanan, 2022-04-26
Last update: A.Buchanan, 2023-09-11 more...
1. De1-d1 Sh1-f2 2. Dd1-a4 Sf2-d3+ 3. Kb4-a5 b2-b4#
Cook: NL 1. De1-h4 Ta1-g1 2. Kb4-a5 Tg1-g8 3. Dh4-b4 Tg8-a8#
1. De1-g3 Ta1-c1 2. Kb4-a5 b2-b3 3. b5-b4 Tc1-c5#
1. De1-c3 Ta1-g1 2. Kb4-a5 Tg1-g6 3. Dc3-b4 Tg6-a6#
und viele andere
Cook: NL 1. De1-h4 Ta1-g1 2. Kb4-a5 Tg1-g8 3. Dh4-b4 Tg8-a8#
1. De1-g3 Ta1-c1 2. Kb4-a5 b2-b3 3. b5-b4 Tc1-c5#
1. De1-c3 Ta1-g1 2. Kb4-a5 Tg1-g6 3. Dc3-b4 Tg6-a6#
und viele andere
Yuri Bilokin: correction wRa1-a2, bQe1-h5, + bPd8, +bPd7, +bPg3 3b4/1K1p4/8/1p5q/1k6/6p1/RP6/7N (4+6) (2022-06-25)
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1. g1=D gxf7 2. Lh7 fxg8=D 3. Dg6 Dxh8 4. Db1 Dxf6 5. Da1 Dxc6 6. Lb1 Dxb5 7. c2 Df1=
* 1. ... Ta2 2. Kb4 Ta3 3. Kxa3 Lb7 4. Tf1#
1. Kd4! Ta2 2. Ke5 Ta3 3. Kf6 Ta2 4. Ke7 Ta3 5. Ke8 Ta2 6. Kd8 Ta3 7. Kc7 Ta2
8. Kb6 Ta3 9. Kc5 Ta2 10. Kb4 Ta3 11. Kxa3 Lb7 12. Tf1#
1. Kd4! Ta2 2. Ke5 Ta3 3. Kf6 Ta2 4. Ke7 Ta3 5. Ke8 Ta2 6. Kd8 Ta3 7. Kc7 Ta2
8. Kb6 Ta3 9. Kc5 Ta2 10. Kb4 Ta3 11. Kxa3 Lb7 12. Tf1#
Henrik Juel: White king corrects the tempo by passing e8, the only square where he is safe from Lg2
He does a long march, but not a round trip (2023-08-15)
SCHRECKE: 10. ... a3 11. Kc3,Kc5 (Dual) (2023-08-15)
msterkowiec: Maybe it will be interesting (for me it is) that this position can be easily transformed into #34. This is because pawns wh.a5 and bl.a6 are only to avoid later/minor dual (there would be also 9.Ka5 without these pawns).
If remove these pawns (and accept minor later duals) and add black pawn a7 (8/3p3p/3P4/8/p4N2/rpK1p1pp/1p2Bkbr/bR5n) we would have the following #34: 1. Kd4 Ra2 2. Ke5 Ra3 3. Kf6 Ra2 4. Ke7 Ra3 5. Ke8 Ra2 6. Kd8 Ra3 7. Kc7 Ra2 8. Kb6 Ra3 9. Ka5 (also 9.Kc5) 9... Ra2 10. Kb4 h6 11. Kc5 Ra3 12. Kd4 Ra2 13. Ke5 Ra3 14. Kf6 Ra2 15. Kg7 h5 16. Kf6 Ra3 17. Ke5 Ra2 18. Kd4 Ra3 19. Kc3 (also 19.Kc5 )19... Ra2 20. Kb4 h4 21. Ka5 Ra3 22. Kb6 Ra2 23. Kc7 Ra3 24. Kd8 Ra2 25. Ke8 Ra3 26. Ke7 Ra2 27. Kd8 (also 27.Kf6) 27...Ra3 28. Kc7 Ra2 29. Kb6 Ra3 30. Ka5 Ra2 31. Kb4 a3 32. Kc3 (also 32.Ka5 and 32.Kc5) 32... Bd5 33. Rf1x (2023-08-15)
msterkowiec: Erratum: Add black p.h7 (not a7) (2023-08-15)
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He does a long march, but not a round trip (2023-08-15)
SCHRECKE: 10. ... a3 11. Kc3,Kc5 (Dual) (2023-08-15)
msterkowiec: Maybe it will be interesting (for me it is) that this position can be easily transformed into #34. This is because pawns wh.a5 and bl.a6 are only to avoid later/minor dual (there would be also 9.Ka5 without these pawns).
If remove these pawns (and accept minor later duals) and add black pawn a7 (8/3p3p/3P4/8/p4N2/rpK1p1pp/1p2Bkbr/bR5n) we would have the following #34: 1. Kd4 Ra2 2. Ke5 Ra3 3. Kf6 Ra2 4. Ke7 Ra3 5. Ke8 Ra2 6. Kd8 Ra3 7. Kc7 Ra2 8. Kb6 Ra3 9. Ka5 (also 9.Kc5) 9... Ra2 10. Kb4 h6 11. Kc5 Ra3 12. Kd4 Ra2 13. Ke5 Ra3 14. Kf6 Ra2 15. Kg7 h5 16. Kf6 Ra3 17. Ke5 Ra2 18. Kd4 Ra3 19. Kc3 (also 19.Kc5 )19... Ra2 20. Kb4 h4 21. Ka5 Ra3 22. Kb6 Ra2 23. Kc7 Ra3 24. Kd8 Ra2 25. Ke8 Ra3 26. Ke7 Ra2 27. Kd8 (also 27.Kf6) 27...Ra3 28. Kc7 Ra2 29. Kb6 Ra3 30. Ka5 Ra2 31. Kb4 a3 32. Kc3 (also 32.Ka5 and 32.Kc5) 32... Bd5 33. Rf1x (2023-08-15)
msterkowiec: Erratum: Add black p.h7 (not a7) (2023-08-15)
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Keywords: Tempo Move
Genre: n#
FEN: 8/3p4/p2P4/P7/p4N2/rpK1p1pp/1p2Bkbr/bR5n
Reprints: 194 Ungarische Schachproblem Anthologie , p. 177, 1983
Input: Dieter Berlin, 2023-08-15
Last update: Dieter Berlin, 2023-08-15 more...
Genre: n#
FEN: 8/3p4/p2P4/P7/p4N2/rpK1p1pp/1p2Bkbr/bR5n
Reprints: 194 Ungarische Schachproblem Anthologie , p. 177, 1983
Input: Dieter Berlin, 2023-08-15
Last update: Dieter Berlin, 2023-08-15 more...
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The problems of this query have been registered by the following contributors:
Gerd Wilts (25)Markus Manhart (2)
Hans-Jürgen Schäfer (2)
Felber, Volker (11)
Henri Nouguier (8)
Brian Stephenson (3)
HBae (5)
Zuncke/Bruder (1)
Frank Müller (10)
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Mario Richter (1)
Rainer Staudte (3)
James Malcom (1)
A.Buchanan (8)
Dieter Berlin (2)
Marcin Banaszek (2)
Als Vergleichsaufgaben sind in der 'Schwalbe' angegeben: G. Schweig (P0002287), Mortimer (P1394750), W. Naef & H. Klüver (P0002832)
A.Buchanan: It's curious that if we shift sK to e8 (i.e. make the position fully homebase) then it is sound as BP in 6.0! The magic of homebase! :) I don't know if it was a typo - maybe someone can check the magazine archive? (2021-10-19)
A.Buchanan: WinChloe & YACPDB don't know this composition. (2021-10-20)
Mario Richter: No typo, the position with black Kd8 was intended, the cook was found by the 'Schwalbe'-solvers (Heft 71, Oktober 1981, p. 349) (2021-10-21)
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