63 problem(s) found in 3655 milliseconds (displaying 63 problem(s)). [COMMENTDATE>=20220810 AND G='h#' AND NOT A='Pitkänen, Jorma' AND K='Umwandlung'] [download as LaTeX]
1. ... axb6ep 2. 0-0-0 0-0-0 3. Td7 a8=D#
This problem can be solved presuming that both castlings are executable. It implies that the last halfmoves have been.-1.Bb6xa7 sBb7-b5.
Bb5 could not come from c6 (sBc6xFb5), because all of the missing white pieces(5) have been captured on white squares except wLc1,which must have been captured by Bc7(Bc7xwLb6).Further,it has not played immediately before -1.Bb6-b5 because this would prevent white castling(forcing wT or wK to move)
Here is the proof that 4 remaining white pcs have been captured on white squares:
black d-pawn must have been captured on d4 and h,g f pawns had to promote after Bh3xwFg2,then wBh2moved to h4, enabling Bg4x Fh3 ..h1T, and finnaly sBf7
after fxg2... g1S. sLf1 prevented the check from Th1, and later has been captured by a white officer.
Conclusion:
Retro:-1.wBb6xFa7 Bb7-b5
Forward(AP) 1....axb6 e. p.2.0-0-0 0-0-0 3.Td7 a8Q# (Author)
This problem can be solved presuming that both castlings are executable. It implies that the last halfmoves have been.-1.Bb6xa7 sBb7-b5.
Bb5 could not come from c6 (sBc6xFb5), because all of the missing white pieces(5) have been captured on white squares except wLc1,which must have been captured by Bc7(Bc7xwLb6).Further,it has not played immediately before -1.Bb6-b5 because this would prevent white castling(forcing wT or wK to move)
Here is the proof that 4 remaining white pcs have been captured on white squares:
black d-pawn must have been captured on d4 and h,g f pawns had to promote after Bh3xwFg2,then wBh2moved to h4, enabling Bg4x Fh3 ..h1T, and finnaly sBf7
after fxg2... g1S. sLf1 prevented the check from Th1, and later has been captured by a white officer.
Conclusion:
Retro:-1.wBb6xFa7 Bb7-b5
Forward(AP) 1....axb6 e. p.2.0-0-0 0-0-0 3.Td7 a8Q# (Author)
Keywords: a posteriori (AP), En passant as key, Castling (sgsgwg), Promotion (D), Valladao Task
Genre: h#, Retro
FEN: r3k3/P3p3/p7/Pp6/p6P/2P3PR/2PPP2b/R3K1nr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-11 more...
Genre: h#, Retro
FEN: r3k3/P3p3/p7/Pp6/p6P/2P3PR/2PPP2b/R3K1nr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-11 more...
a) 1. fxg1=S Dxe3 2. Sd2 Df2#
b) 1. ... Dxe3 2. c1=S Dc3#
a) R: 1. d2xc3
b) WTM, so solution in a) can't work here.
b) 1. ... Dxe3 2. c1=S Dc3#
a) R: 1. d2xc3
b) WTM, so solution in a) can't work here.
Retro
b) Weiß am Zug
Adrian Storisteanu: Two knight promotions differently motivated.
The technical challenge was in averting the short second solution in a) as set play, in order to keep the two phases nicely apart, and also, of course, preventing black tempi and potential white retro moves. (2000-11-20)
more ...
comment
b) Weiß am Zug
Adrian Storisteanu: Two knight promotions differently motivated.
The technical challenge was in averting the short second solution in a) as set play, in order to keep the two phases nicely apart, and also, of course, preventing black tempi and potential white retro moves. (2000-11-20)
more ...
comment
Keywords: No legal last move for White, Promotion (s,s)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro logic
FEN: 8/8/8/8/8/2P1p1P1/ppp1ppRP/bqQbknRK
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro logic
FEN: 8/8/8/8/8/2P1p1P1/ppp1ppRP/bqQbknRK
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
1. fxe3ep d8=S 2. Kxd5 0-0-0# (Td1#?)
White has made 6 pawn captures with one missing black piece unaccounted for, that by parity can only have been captured by an officer. White's last move cannot have been a pawn capture. If R: 1.f2-f3, then sLg1 was promoted, implying 7 captures by black pawns - one too many. So if White can prove they retain castling rights, then the ep is on. Hence AP Petrovic is valid.
White has made 6 pawn captures with one missing black piece unaccounted for, that by parity can only have been captured by an officer. White's last move cannot have been a pawn capture. If R: 1.f2-f3, then sLg1 was promoted, implying 7 captures by black pawns - one too many. So if White can prove they retain castling rights, then the ep is on. Hence AP Petrovic is valid.
Henrik Juel: 0... fxe3ep 1.d8S Kxd5 2.0-0-0#. Not -1.f2? and Lg1 is caught. (2004-09-16)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
Henrik Juel: In principle I agree, Vaclav
But PDB does not (yet) allow HC+, so I find it acceptable to use the C+ label, when you also tell the whole story after 'Computer test:' below (2023-08-03)
A.Buchanan: Hopefully Gerd will have more time at some point, and can expand the functionality in this and other areas. In the meantime, engine solving of conventional retros including AP, is in its infancy. Retractor 2 has some effectiveness, but is still basic. And there is nothing that yet grasps the intricacies of castling/ep etc. However AP problems do often contain considerable forward chess, and the C+ tag is very useful to filter out those that have already been solved forwardly, without pretending that these are in any sense fully solved (2023-08-04)
Ladislav Packa: I know the definition of AP, but I don't understand the logic behind it. The move 1.fxe3 e.p. proves that White CAN castling. But the solution (2. ...Rd1#?) claims that castling is MANDATORY. From my point of view, AP is correct when only castling is necessary for the solution and the Rook move would be a dual. (2023-08-04)
A.Buchanan: @Ladislav: I am not sure how to help you. Maybe you can read this page from Retro Corner: https://www.janko.at/Retros/Glossary/APosteriori.htm (2023-08-05)
Ladislav Packa: Andrew, what should the article help me with? I quote the final sentence:
Some people still oppose this rule and argue that it should certainly not be the default convention. (2023-08-05)
Henrik Juel: You could view it this way, Ladislav
h#2 means that it is Black to move, so White made last move
What was last move? A little analysis shows just three possibilities: f2-f4, move by Ta1, or move by Ke1
So normally we cannot assume that last move was f2-f4
But if White can castle, then the last move was f2-f4
So if we could start with 0... 0-0-0, then 1.fxe3ep would be legitimate
AP says that you are allowed to reverse the sequence of events; first do the ep capture, then later legitimize it by castling
Was this helpful? (2023-08-05)
Ladislav Packa: Henrik, you don't have to explain that to me. I've done a few AP issues myself, like P1348357. But that doesn't mean I agree with AP's logic. I already wrote it - the term "you can castle" is applied as "you must castle". But these are only problems where, in addition to castling, the Rook move can also be used, I consider that a dual.
From that point of view, the P1000662 issue is perfectly fine for me. (2023-08-05)
Joost de Heer: AP: By castling, you prove a posteriori that the ep-capture was not just a try but the actual solution. Without castling, the ep-solution just is that: a try.
So: Try 1. fe3 ep e8=S 2. Kd5 Rd1 - but ep capture not allowed, as there is no proof that f2-f4 must've been the last move.
Solution 1. fe3 ep e8=S 2. Kd5 OOO - Now the ep capture was justified because white castled, thereby proving that the last move before the diagram position indeed was f2-f4. (2023-08-06)
Joost de Heer: See e.g. P1052919 : The try is an ep capture which is unjustified. (2023-08-06)
A.Buchanan: OK Ladislav: I think I get your point. If one solution with castling justifies the e.p., then based on that certainty, why shouldn't an alternative solution with no castling *then* be allowed as well? There are problems in which one twin shows 0-0-0 and the other shows 0-0. Each is based on the other in a similar way, so the idea of dependency is not new. Why are we not allowed to add other "parasitic" solutions as well? Why can we only have the "paying" solutions? We can't say that we are restricted to one solution: that's not the way chess problems operate! And this is just in the help world - in the adversarial world it might get even more complicated. Is this your issue, Ladislav? (2023-08-07)
Ladislav Packa: I don't want to unnecessarily prolong this discussion. However, I will add one more note: in this position, white castling is also possible without e.p. in Black's 1st move. If B1 were an indifferent move, then white can 1...0-0-0! The Codex of Chess Composition writes about it in Article 16 (1):
Castling convention. Casting is permitted unless it can be proven that it is not permissible.
In our case, 0-0-0 is possible because White's last move exists - e2-e4! It does not matter if it is this move or some a2-a3, both moves are equivalent. EP does not prove the possibility of casting, it would be legal even without it. (2023-08-07)
A.Buchanan: Ladislav was what I wrote your issue pls? Y/N :-) (2023-08-07)
Ladislav Packa: I have no problem, I'm just expressing my own opinion about the AP convention. (2023-08-07)
A.Buchanan: OK cos I think the point I raised is a real one that should be addressed by theory some day. Clearly from the nice problem that you composed Ladislav you understand the mechanics very well. From a justification perspective it's all a bit iffy, but that's why it's controversial. Under RS it's really the only way one can end up actually eping, and it's proved compositionally fertile. So that's enough to justify (2023-08-08)
Ladislav Packa: No need to apologize. This is a normal discussion with different views on the issue. Maybe it will come to some conclusion.
I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p. (2023-08-08)
Joost de Heer: "I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p."
You misinterpret AP. The e.p. capture does not authorize castling, castling provides a justification later on (hence the 'a posteriori') for the legality of ep.
Usually, for ep justification you need to examine all game trees that lead to the diagram, and only if all game trees end with the double-step, then ep is allowed.
With AP, you examine all the game trees including the actual play. If all those game trees have as last move before the diagram position the double step, then ep is possible.
In this case, if white doesn't castle, then there are game trees which don't have as last move the double step, and therefore AP logic dictates that the ep capture was illegal. However, all game trees which lead to the diagram and which have castling in the actual play have as last move before the diagram position the double step, hence AP dictates that the ep capture is legal. (2023-08-09)
Ladislav Packa: Joost: A simple question - is white allowed to castle after any 1st move by black (except e.p.)? (2023-08-09)
Joost de Heer: Of course he is. AP only is used to combine the ep justification with castling, not the castling right per se. (2023-08-09)
more ...
comment
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
Henrik Juel: In principle I agree, Vaclav
But PDB does not (yet) allow HC+, so I find it acceptable to use the C+ label, when you also tell the whole story after 'Computer test:' below (2023-08-03)
A.Buchanan: Hopefully Gerd will have more time at some point, and can expand the functionality in this and other areas. In the meantime, engine solving of conventional retros including AP, is in its infancy. Retractor 2 has some effectiveness, but is still basic. And there is nothing that yet grasps the intricacies of castling/ep etc. However AP problems do often contain considerable forward chess, and the C+ tag is very useful to filter out those that have already been solved forwardly, without pretending that these are in any sense fully solved (2023-08-04)
Ladislav Packa: I know the definition of AP, but I don't understand the logic behind it. The move 1.fxe3 e.p. proves that White CAN castling. But the solution (2. ...Rd1#?) claims that castling is MANDATORY. From my point of view, AP is correct when only castling is necessary for the solution and the Rook move would be a dual. (2023-08-04)
A.Buchanan: @Ladislav: I am not sure how to help you. Maybe you can read this page from Retro Corner: https://www.janko.at/Retros/Glossary/APosteriori.htm (2023-08-05)
Ladislav Packa: Andrew, what should the article help me with? I quote the final sentence:
Some people still oppose this rule and argue that it should certainly not be the default convention. (2023-08-05)
Henrik Juel: You could view it this way, Ladislav
h#2 means that it is Black to move, so White made last move
What was last move? A little analysis shows just three possibilities: f2-f4, move by Ta1, or move by Ke1
So normally we cannot assume that last move was f2-f4
But if White can castle, then the last move was f2-f4
So if we could start with 0... 0-0-0, then 1.fxe3ep would be legitimate
AP says that you are allowed to reverse the sequence of events; first do the ep capture, then later legitimize it by castling
Was this helpful? (2023-08-05)
Ladislav Packa: Henrik, you don't have to explain that to me. I've done a few AP issues myself, like P1348357. But that doesn't mean I agree with AP's logic. I already wrote it - the term "you can castle" is applied as "you must castle". But these are only problems where, in addition to castling, the Rook move can also be used, I consider that a dual.
From that point of view, the P1000662 issue is perfectly fine for me. (2023-08-05)
Joost de Heer: AP: By castling, you prove a posteriori that the ep-capture was not just a try but the actual solution. Without castling, the ep-solution just is that: a try.
So: Try 1. fe3 ep e8=S 2. Kd5 Rd1 - but ep capture not allowed, as there is no proof that f2-f4 must've been the last move.
Solution 1. fe3 ep e8=S 2. Kd5 OOO - Now the ep capture was justified because white castled, thereby proving that the last move before the diagram position indeed was f2-f4. (2023-08-06)
Joost de Heer: See e.g. P1052919 : The try is an ep capture which is unjustified. (2023-08-06)
A.Buchanan: OK Ladislav: I think I get your point. If one solution with castling justifies the e.p., then based on that certainty, why shouldn't an alternative solution with no castling *then* be allowed as well? There are problems in which one twin shows 0-0-0 and the other shows 0-0. Each is based on the other in a similar way, so the idea of dependency is not new. Why are we not allowed to add other "parasitic" solutions as well? Why can we only have the "paying" solutions? We can't say that we are restricted to one solution: that's not the way chess problems operate! And this is just in the help world - in the adversarial world it might get even more complicated. Is this your issue, Ladislav? (2023-08-07)
Ladislav Packa: I don't want to unnecessarily prolong this discussion. However, I will add one more note: in this position, white castling is also possible without e.p. in Black's 1st move. If B1 were an indifferent move, then white can 1...0-0-0! The Codex of Chess Composition writes about it in Article 16 (1):
Castling convention. Casting is permitted unless it can be proven that it is not permissible.
In our case, 0-0-0 is possible because White's last move exists - e2-e4! It does not matter if it is this move or some a2-a3, both moves are equivalent. EP does not prove the possibility of casting, it would be legal even without it. (2023-08-07)
A.Buchanan: Ladislav was what I wrote your issue pls? Y/N :-) (2023-08-07)
Ladislav Packa: I have no problem, I'm just expressing my own opinion about the AP convention. (2023-08-07)
A.Buchanan: OK cos I think the point I raised is a real one that should be addressed by theory some day. Clearly from the nice problem that you composed Ladislav you understand the mechanics very well. From a justification perspective it's all a bit iffy, but that's why it's controversial. Under RS it's really the only way one can end up actually eping, and it's proved compositionally fertile. So that's enough to justify (2023-08-08)
Ladislav Packa: No need to apologize. This is a normal discussion with different views on the issue. Maybe it will come to some conclusion.
I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p. (2023-08-08)
Joost de Heer: "I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p."
You misinterpret AP. The e.p. capture does not authorize castling, castling provides a justification later on (hence the 'a posteriori') for the legality of ep.
Usually, for ep justification you need to examine all game trees that lead to the diagram, and only if all game trees end with the double-step, then ep is allowed.
With AP, you examine all the game trees including the actual play. If all those game trees have as last move before the diagram position the double step, then ep is possible.
In this case, if white doesn't castle, then there are game trees which don't have as last move the double step, and therefore AP logic dictates that the ep capture was illegal. However, all game trees which lead to the diagram and which have castling in the actual play have as last move before the diagram position the double step, hence AP dictates that the ep capture is legal. (2023-08-09)
Ladislav Packa: Joost: A simple question - is white allowed to castle after any 1st move by black (except e.p.)? (2023-08-09)
Joost de Heer: Of course he is. AP only is used to combine the ep justification with castling, not the castling right per se. (2023-08-09)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Promotion (S), Valladao Task
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 + simple retro-logic
FEN: 8/3P2p1/2PP4/1ppPp3/2pkPp2/5PP1/6Pp/R3K1b1
Input: Gerd Wilts, 1999-02-27
Last update: A.Buchanan, 2023-09-11 more...
Genre: Retro, h#
Computer test: HC+ Popeye v4.87 + simple retro-logic
FEN: 8/3P2p1/2PP4/1ppPp3/2pkPp2/5PP1/6Pp/R3K1b1
Input: Gerd Wilts, 1999-02-27
Last update: A.Buchanan, 2023-09-11 more...
1) 1. ... g4 2. e6 g5 3. Ke7 g6 4. Td8 gxf7 5. Td7 f8=D#
2) 1. ... g3 2. Kd7 g4 3. Td8 g5 4. Ke8 g6 5. Sd7 gxf7#
Numerous retro tries but none distinct
2) 1. ... g3 2. Kd7 g4 3. Td8 g5 4. Ke8 g6 5. Sd7 gxf7#
Numerous retro tries but none distinct
Keywords: Promotion in the mating move, Excelsior white (auto key), No legal last move for White, Minimal
Genre: h#, Retro
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 4k2r/4ppK1/5n1p/8/8/8/6P1/8
Reprints: 365 Minimalkunst im Schach 2006
Input: Gerd Wilts, 1996-06-06
Last update: A.Buchanan, 2023-12-08 more...
Genre: h#, Retro
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 4k2r/4ppK1/5n1p/8/8/8/6P1/8
Reprints: 365 Minimalkunst im Schach 2006
Input: Gerd Wilts, 1996-06-06
Last update: A.Buchanan, 2023-12-08 more...
1. Tg8 h8=D 2. Kg6 Df6+ 3. Kh7 Dxf8 4. Kh8 Dh6#
*) 1. ... f8=S 2. f1=L Le3#
1) 1. f1=T f8=D 2. Tf6 Dxg7#
1) 1. f1=T f8=D 2. Tf6 Dxg7#
t/Allumwandlung, u/61112, u/SDlt
Vorläufer: A.C.White,1930, s.Nr.1056
s. A. C. White 364v L'Echiquier 07-08/1930
Anton Baumann: Die Aufgabe stammt gem. 'Die Schwalbe' 05/1963 von André Simonet, Brüssel
Vorgänger vergl. P0542142 (2022-11-16)
comment
Vorläufer: A.C.White,1930, s.Nr.1056
s. A. C. White 364v L'Echiquier 07-08/1930
Anton Baumann: Die Aufgabe stammt gem. 'Die Schwalbe' 05/1963 von André Simonet, Brüssel
Vorgänger vergl. P0542142 (2022-11-16)
comment
Keywords: Allumwandlung, Promotion
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 4N3/5Pp1/6Pk/7p/4P2P/pp6/pqpp1p2/bbrr2BK
Input: Erich Bartel, 1997-05-23
Last update: hpr, 1999-10-16 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 4N3/5Pp1/6Pk/7p/4P2P/pp6/pqpp1p2/bbrr2BK
Input: Erich Bartel, 1997-05-23
Last update: hpr, 1999-10-16 more...
1) 1. cxb1=S g8=D 2. Sc3 Db3#
2) 1. cxb1=L g8=T 2. Ka2 Ta8#
2) 1. cxb1=L g8=T 2. Ka2 Ta8#
t/Allumwandlung, u/61113, u/DTsl
Yuri Bilokin: you can see 1...g8=Q 2.cxb1=B Qb3#
1...g8=Q 2.cxb1=S Qb3#
1...Bxc2 2.Ka2 Ra4# (2024-02-29)
comment
Yuri Bilokin: you can see 1...g8=Q 2.cxb1=B Qb3#
1...g8=Q 2.cxb1=S Qb3#
1...Bxc2 2.Ka2 Ra4# (2024-02-29)
comment
Keywords: Allumwandlung, Sacrifice of white pieces, Promotion
Genre: h#
Computer test: (Popeye C-Version 3. 47 (1024 KB))
FEN: 8/6P1/8/8/1R6/k2p4/2p5/1BK5
Input: Erich Bartel, 1997-05-23
Last update: hpr, 1999-10-16 more...
Genre: h#
Computer test: (Popeye C-Version 3. 47 (1024 KB))
FEN: 8/6P1/8/8/1R6/k2p4/2p5/1BK5
Input: Erich Bartel, 1997-05-23
Last update: hpr, 1999-10-16 more...
1. f5 b4 2. f4 b5 3. f3 b6 4. f2 b7 5. f1=L b8=L 6. Lc4 Lxd6 7. Lcg8 Le5#
NL:
1. Le4 b4 2. Ld5 b5 3. Kg8 b6 4. Kf8 b7 5. Ke8 b8=D+ 6. Kd7 Ka6 7. Kc6 Dc8#
NL:
1. Le4 b4 2. Ld5 b5 3. Kg8 b6 4. Kf8 b7 5. Ke8 b8=D+ 6. Kd7 Ka6 7. Kc6 Dc8#
Yuri Bilokin: correction wKa7-b1, +bNd1 7k/5p1b/3p4/8/8/8/1Pn5/1K1n4 (2+6) h#7
1.f5 b4 2.f4 b5 3.f3 b6 4.f2 b7 5.f1=B b8=B 6.Bc4 Bxd6 7.Bcg8 Be5# (MM)
Excelsior (black)
Excelsior (white)
Promotion (Bb, 1, 1)
Model mate × 1 (2023-05-10)
comment
1.f5 b4 2.f4 b5 3.f3 b6 4.f2 b7 5.f1=B b8=B 6.Bc4 Bxd6 7.Bcg8 Be5# (MM)
Excelsior (black)
Excelsior (white)
Promotion (Bb, 1, 1)
Model mate × 1 (2023-05-10)
comment
Keywords: Symmetrical position, Promotion, Excelsior black (auto key), Excelsior white (auto key), Excelsior (sw)
Genre: h#
FEN: 7k/K4p1b/3p4/8/8/8/1Pn5/8
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: Rainer Staudte, 2014-06-21 more...
Genre: h#
FEN: 7k/K4p1b/3p4/8/8/8/1Pn5/8
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: Rainer Staudte, 2014-06-21 more...
1. d1=T+ Le1 2. Td4 Lc3#
Verst-4
klären Stellung
SCHRECKE: Unlösbar - wLb1 gehört vielleicht nach c2!? (2023-03-10)
HBae: habe korrigiert, auch das Datum war falsch (2023-03-10)
SCHRECKE: C+, popeye 4.87 (2023-03-10)
comment
klären Stellung
SCHRECKE: Unlösbar - wLb1 gehört vielleicht nach c2!? (2023-03-10)
HBae: habe korrigiert, auch das Datum war falsch (2023-03-10)
SCHRECKE: C+, popeye 4.87 (2023-03-10)
comment
*) 1. ... d8=D 2. Lb7 Dd6#
1) 1. Lc4 d8=D 2. Ld5 Dc7#
NL:
1. Kc5 d8=D 2. Lc4 Dd6#
1) 1. Lc4 d8=D 2. Ld5 Dc7#
NL:
1. Kc5 d8=D 2. Lc4 Dd6#
Yuri Bilokin: correction wKg4-f4, -bBa6, +bRe7, then a1=c1 6N1/5Pr1/4k3/8/2P4K/8/8/8 (4+2) h#2*
1...f8=Q 2.Rd7 Qf6#
1.Rg5 f8=Q 2.Rf5 Qe7#
Echo mates (mirrored diagonally, 0, 0)
Promotion (QQ, 2)
Echo (2023-06-04)
comment
1...f8=Q 2.Rd7 Qf6#
1.Rg5 f8=Q 2.Rf5 Qe7#
Echo mates (mirrored diagonally, 0, 0)
Promotion (QQ, 2)
Echo (2023-06-04)
comment
11 - P0559261
Jacobus Theodorus Sanderse
861 Boletim da UBP 10-11/1967
1. ehrende Erwähnung
(2+5) cooked
h#7
Jacobus Theodorus Sanderse
861 Boletim da UBP 10-11/1967
1. ehrende Erwähnung
(2+5) cooked
h#7
1. Sc4 d3 2. e5 dxc4 3. e4 c5 4. e3 c6 5. e2 c7 6. e1=T c8=D 7. Tg1 Dh3#
NL:
1. Kg2 d4 2. Kf3 d5 3. Sc6+ Kd7 4. Dd4 dxc6 5. Ke4 c7 6. Kd5 c8=D 7. e5 Dc6#
NL:
1. Kg2 d4 2. Kf3 d5 3. Sc6+ Kd7 4. Dd4 dxc6 5. Ke4 c7 6. Kd5 c8=D 7. e5 Dc6#
Yuri Bilokin: correction bQa1-b7, -bPd6 3K4/1q2p3/8/4n3/8/8/3P4/7k (2+4) h#3
1.Sc4 d3 2.e5 dxc4 3.e4 c5 4.e3 c6 5.e2 c7 6.e1=R c8=Q 7.Rg1 Qh3# (MM)
Slow Excelsior
Active sacrifice (black, delayed)
Excelsior (black)
Excelsior (white)
Promotion (Qr, 1, 1)
Model mate × 1 (2023-05-10)
comment
1.Sc4 d3 2.e5 dxc4 3.e4 c5 4.e3 c6 5.e2 c7 6.e1=R c8=Q 7.Rg1 Qh3# (MM)
Slow Excelsior
Active sacrifice (black, delayed)
Excelsior (black)
Excelsior (white)
Promotion (Qr, 1, 1)
Model mate × 1 (2023-05-10)
comment
1. a1=T hxg3 2. Ta2 g4 3. Tb2 g5 4. Tb5 gxf6 5. Tg5 f7 6. Tg8 f8=D 7. Th8 Df7#
NL:
1. a1=T hxg3 2. Ta2 g4 3. Tb2 g5 4. Tb6 gxf6 5. Kh8 f7 6. Tg6 f8=D+ 7. Tg8 Dxh6#
NL:
1. a1=T hxg3 2. Ta2 g4 3. Tb2 g5 4. Tb6 gxf6 5. Kh8 f7 6. Tg6 f8=D+ 7. Tg8 Dxh6#
klären AL ?
Yuri Bilokin: correction +bQh4(h3) 8/7k/1p3p1p/8/2n4q/p1pp2n1/p1p4P/1bK5 (2+13) (2023-05-08)
comment
Yuri Bilokin: correction +bQh4(h3) 8/7k/1p3p1p/8/2n4q/p1pp2n1/p1p4P/1bK5 (2+13) (2023-05-08)
comment
1. e5 gxf3 2. e4 f4 3. e3 f5 4. e2 f6 5. e1=T f7 6. Th1 f8=T 7. Th5 Tf6#
NL:
1. f2 Kf7 2. f1=S g4 3. Sg3 Ke6 4. Sh5 gxh5 5. Kg7 h6+ 6. Kf8 h7 7. Ke8 h8=T#
NL:
1. f2 Kf7 2. f1=S g4 3. Sg3 Ke6 4. Sh5 gxh5 5. Kg7 h6+ 6. Kf8 h7 7. Ke8 h8=T#
1. b5 f4 2. b4 Ka4 3. b3 f5 4. b2 f6 5. b1=L f7 6. La2 f8=S 7. Lg8 Sg6#
Cook: 1. Kg8 Kc2 2. Kf8 f4 3. Ke7 f5 4. Kd6 f6 5. Kc5 fxg7 6. Kb4 g8=D 7. Ka3 Db3#
Cook: 1. Kg8 Kc2 2. Kf8 f4 3. Ke7 f5 4. Kd6 f6 5. Kc5 fxg7 6. Kb4 g8=D 7. Ka3 Db3#
Erich Bartel: Nachdruck:
Die Schwalbe (2) XII 1969 (2013-02-22)
Yuri Bilokin: possibly wKb3-a2 7k/1p4pn/8/8/8/8/K4P2/8 (2+4) h#7
1.b5 f4 2.b4 f5 3.b3+ Ka3 4.b2 f6 5.b1=B f7 6.Ba2 f8=S 7.Bg8 Sg6# (MM) (2023-02-01)
comment
Die Schwalbe (2) XII 1969 (2013-02-22)
Yuri Bilokin: possibly wKb3-a2 7k/1p4pn/8/8/8/8/K4P2/8 (2+4) h#7
1.b5 f4 2.b4 f5 3.b3+ Ka3 4.b2 f6 5.b1=B f7 6.Ba2 f8=S 7.Bg8 Sg6# (MM) (2023-02-01)
comment
1. d1=L Kb1 2. Lb3 axb3 3. f1=L b4 4. La6 b5 5. h1=L bxa6 6. Lb7 axb7 7. Kh4 b8=D 8. g5 Dh2#
1. a1=D a8=S 2. Da7 e8=S 3. Db8 cxb8=S#
Branko Koludrovic: This task has been improved, and the new setting has been printed in"Die Schwalbe"Feb 2002:
White: Ke8,Bc7d6e7f6g7(6)
Black:Ke6,Sd5e5 Bc2d7f4f5f7(8) H#3
1.c1D c8S 2.Dc7 g8S 3.Dd8+ exd8S#
Task Type dSSS (2010-04-05)
Anton Baumann: auch oben genannte Version ist C+ (Gustav 4.2a) (2023-02-18)
comment
White: Ke8,Bc7d6e7f6g7(6)
Black:Ke6,Sd5e5 Bc2d7f4f5f7(8) H#3
1.c1D c8S 2.Dc7 g8S 3.Dd8+ exd8S#
Task Type dSSS (2010-04-05)
Anton Baumann: auch oben genannte Version ist C+ (Gustav 4.2a) (2023-02-18)
comment
1. g4 Lxg4 2. Lg5 Lf3 3. Lh6 ... 4. Lf8 ... 5. Ld6 ... 6. Lc7 ... 7. La5 ... 8. Lb4 ... 9. La3 ... 10. Lc1 ... 11. Ld2 ... 12. Le1 ... 13. Lf2 ... 14. Lxg1 ... 15. Kg5 ... 16. Kh6 ... 17. Kg7 ... 18. Kf8 ... 19. Ke7 ... 20. Kd6 ... 21. Kc7 ... 22. Kb6 ... 23. Ka5 ... 24. Kb4 ... 25. Ka3 ... 26. Kb2 ... 27. Kc1 ... 28. Kd2 ... 29. Ke1 ... 30. Kxf1 ... 31. Ke1 Kxg1 32. Kd1 Kf1 33. Kc1 Ke1 34. Kb2 Kd1 35. Ka3 Kc1 36. Ka4 Kb1 37. Ka5 Ka2 38. Kb6 Ka3 39. Kc6 Ka4 40. Kd6 Kb5 41. Ke7 Kc6 42. Kf8 Kd6 43. Kg7 Ke6 44. Kh8 Kxf6 45. Kg8 Kg6 46. Kh8 f6 47. Kg8 f7+ 48. Kh8 f8=D,T#
James Malcom: I don’t believe that the great Fabel, with his keen eye, would have not seen such a simply cook as the taking of Black’s c5 pawn. I suspect that a Black bishop was on d4 but was forgotten placing the problem in the database, or it was a misprint. I am changing this problem to not cooked for now. However, someone should check the original source in the event I am wrong, which is always a possibility. (2020-03-15)
Mario Richter: Hi, James, there's indeed a black bishop on d4. Original stipluation: "h#48 (nicht exakt!)"
A scan for some issues of 'Schachmatt' is available in the archive at stere.ro, see
www.stere.ro/wp-content/uploads/2014/06/stere_sah_istoria_sahului-1947-1949-Schachmatt1.pdf (2020-03-26)
Joost de Heer: Stupid question, but won't moving Bh3 to g4 add another halfmove? (2023-01-30)
Henrik Juel: Yes, Joost, but this would make the solution even less exact
0... Lf3,Lh3,Lh5 1.g4 Lxg4 etc. (2023-01-30)
Joost de Heer: @Henrik: So? This is an attempt at a non-unique (exact is something else) h# length record, the level of non-uniqueness is irrelevant. And as James pointed out, adding Kg4-f4 will add yet another move.
Only possible cooks are when white starts with 1... Bh5 2. g4 Be8 (so not capturing Pg4) but I don't see how that could lead to a quicker mate. (2023-01-31)
James Malcom: I double-checked that myself, Joost, and found that going around for wPg2 actually takes far longer, which would be required any PxB action to take place, but that takes far longer than the normal solution. I know Olaf's Gustav would check, since it C+ the h#111. (2023-01-31)
more ...
comment
Mario Richter: Hi, James, there's indeed a black bishop on d4. Original stipluation: "h#48 (nicht exakt!)"
A scan for some issues of 'Schachmatt' is available in the archive at stere.ro, see
www.stere.ro/wp-content/uploads/2014/06/stere_sah_istoria_sahului-1947-1949-Schachmatt1.pdf (2020-03-26)
Joost de Heer: Stupid question, but won't moving Bh3 to g4 add another halfmove? (2023-01-30)
Henrik Juel: Yes, Joost, but this would make the solution even less exact
0... Lf3,Lh3,Lh5 1.g4 Lxg4 etc. (2023-01-30)
Joost de Heer: @Henrik: So? This is an attempt at a non-unique (exact is something else) h# length record, the level of non-uniqueness is irrelevant. And as James pointed out, adding Kg4-f4 will add yet another move.
Only possible cooks are when white starts with 1... Bh5 2. g4 Be8 (so not capturing Pg4) but I don't see how that could lead to a quicker mate. (2023-01-31)
James Malcom: I double-checked that myself, Joost, and found that going around for wPg2 actually takes far longer, which would be required any PxB action to take place, but that takes far longer than the normal solution. I know Olaf's Gustav would check, since it C+ the h#111. (2023-01-31)
more ...
comment
Keywords: Non-standard material, Promotion in the mating move
Genre: h#
Computer test: Gustav 4.2 b: Es gibt kein kürzeres Matt
FEN: 8/8/5p2/2p1pPp1/2PbPk1b/2pBp1pB/2P1P1P1/5BRK
Input: hpr, 1999-03-31
Last update: Olaf Jenkner, 2023-01-31 more...
Genre: h#
Computer test: Gustav 4.2 b: Es gibt kein kürzeres Matt
FEN: 8/8/5p2/2p1pPp1/2PbPk1b/2pBp1pB/2P1P1P1/5BRK
Input: hpr, 1999-03-31
Last update: Olaf Jenkner, 2023-01-31 more...
1. ... a8=D+ 2. Kc7 Da7+ 3. Tb7 Da8 4. Ta7 Dxa7#
Yuri Bilokin: you can see
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.d5 Qb8#
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.g5 Qb8#
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.Sg4 Qb8#
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.Sf5 Qb8# (2023-12-28)
comment
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.d5 Qb8#
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.g5 Qb8#
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.Sg4 Qb8#
1... ... 2.Rb1 a8=Q+ 3.Kb6 Qxc8 4.Sf5 Qb8# (2023-12-28)
comment
R: 1. f7xDe8=D+, dann 1. fxg8=S#
R: 1. f7xDe8=D+ Kf6-e7 g7xLh8=L+ ... Tb8-a8 ... c7xTb8=T
R: 1. f7xDe8=D+ Kf6-e7 g7xLh8=L+ ... Tb8-a8 ... c7xTb8=T
No. 1449 HN
AV: Je crois que c'est un Fou blanc en h8 et non un Cavalier, et les derniers coups sont gxFh8=F+ Rf6-e7 fxDe8=D+ (au lieu de fxCg8=C#). Avec l'antérieur cxTb8=T, cela donne une "Allumwandlung" avec une pièce noire prise de la même nature. (2005-06-09)
Henrik Juel: You must be right, AV, because with wSh8 the retroplay might be the bland -1.f7xDe8=D Kf6 etc. without unpromotion on h8. (2005-06-13)
Michel Caillaud: Indeed the diagram is wrong : there should be a white Bishop on h8
AUW spread over retro play (-1.fxDe8=D+ Kf6-e7 -2.gxLh8=L+ ...-n.cxTb8=T) and forward play 1.fxSg8=S#
Matching nature of promoted piece and piece captured by promotion. (2023-03-25)
comment
AV: Je crois que c'est un Fou blanc en h8 et non un Cavalier, et les derniers coups sont gxFh8=F+ Rf6-e7 fxDe8=D+ (au lieu de fxCg8=C#). Avec l'antérieur cxTb8=T, cela donne une "Allumwandlung" avec une pièce noire prise de la même nature. (2005-06-09)
Henrik Juel: You must be right, AV, because with wSh8 the retroplay might be the bland -1.f7xDe8=D Kf6 etc. without unpromotion on h8. (2005-06-13)
Michel Caillaud: Indeed the diagram is wrong : there should be a white Bishop on h8
AUW spread over retro play (-1.fxDe8=D+ Kf6-e7 -2.gxLh8=L+ ...-n.cxTb8=T) and forward play 1.fxSg8=S#
Matching nature of promoted piece and piece captured by promotion. (2023-03-25)
comment
Keywords: Help retractor, Promotion in the retro play, Promotion in forward play, Allumwandlung
Genre: Retro, h#
FEN: R1brQnnB/pp1pk2p/3pp1p1/6N1/6Kp/8/8/8
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2024-01-21 more...
Genre: Retro, h#
FEN: R1brQnnB/pp1pk2p/3pp1p1/6N1/6Kp/8/8/8
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2024-01-21 more...
20 - P1074078
Fadil Abdurahmanovic
Michael Rittmann
13217v Die Schwalbe , p. 163, 07/2005
3. ehrende Erwähnung
(2+11) C+
h#6
0.2.1...
Fadil Abdurahmanovic
Michael Rittmann
13217v Die Schwalbe , p. 163, 07/2005
3. ehrende Erwähnung
(2+11) C+
h#6
0.2.1...
1) 1. ... e4 2. Lb8 e5 3. Ld6+ exd6 4. c1=L d7 5. Lb2 d8=D 6. Lg7 Dd3#
2) 1. ... e3 2. Ta5 e4 3. Tf5 exf5 4. c1=T f6 5. Tg1 f7 6. Tg7 f8=S#
2) 1. ... e3 2. Ta5 e4 3. Tf5 exf5 4. c1=T f6 5. Tg1 f7 6. Tg7 f8=S#
Yuri Bilokin: you can see *) 1.Ra5 e4 2.Rf5 exf5 3.c1=R f6 4.Rg1 f7 5.Rg7 f8=S# (2022-09-30)
comment
comment
Keywords: Allumwandlung, Promotion, under-promotion, under-promotion in mating move (in 2), Excelsior white, Model mate, Selfblock, changed blocks, Minimal (B), Pawn step variations
Genre: h#
Computer test: (Popeye WINDOWS-32Bit V4.37 (1580608 KB))
FEN: 6nq/b6k/1p2p2p/8/1K6/rp6/b1p1P3/8
Input: hpr, 2007-09-09
Last update: Gunter Jordan, 2020-03-04 more...
Genre: h#
Computer test: (Popeye WINDOWS-32Bit V4.37 (1580608 KB))
FEN: 6nq/b6k/1p2p2p/8/1K6/rp6/b1p1P3/8
Input: hpr, 2007-09-09
Last update: Gunter Jordan, 2020-03-04 more...
a) 1. c1=T Lg6 2. Tc4 Td3#
b) 1. c1=L Td3 2. Lf4 Lg6#
b) 1. c1=L Td3 2. Lf4 Lg6#
Keywords: Reciprocal moves by White, Exchange of Promotions
Genre: h#
FEN: 8/8/1K6/4p2B/3k4/5R2/2p5/8
Input: Gunter Jordan, 2010-06-02
Last update: Frank Müller, 2010-06-02 more...
Genre: h#
FEN: 8/8/1K6/4p2B/3k4/5R2/2p5/8
Input: Gunter Jordan, 2010-06-02
Last update: Frank Müller, 2010-06-02 more...
22 - P1196784
Gunter Jordan
209 Pat a Mat 71 06/2010
(2+2)
h#3
2 Lösungen
Vogtländerschach
Weißer Längstzüger
Gunter Jordan
209 Pat a Mat 71 06/2010
(2+2)
h#3
2 Lösungen
Vogtländerschach
Weißer Längstzüger
1. c1=L Sa3 2. Lxa3 Kb6 3. Kb2 Kc5#
1. Kd3 Sd2 2. Kxd2 Kb4 3. c1=L Ka3#
1. Kd3 Sd2 2. Kxd2 Kb4 3. c1=L Ka3#
Keywords: Echo, Maximummer (Weiss), Promotion key, under-promotion, Vogtländer Chess
Genre: h#, Fairies
FEN: 8/8/8/K7/2N5/2k5/2p5/8
Input: Gunter Jordan, 2011-06-03
Last update: A.Buchanan, 2024-01-18 more...
Genre: h#, Fairies
FEN: 8/8/8/K7/2N5/2k5/2p5/8
Input: Gunter Jordan, 2011-06-03
Last update: A.Buchanan, 2024-01-18 more...
1) 1. b1=T Kg2 2. Td1 Sc2 3. Kd3 Kf3 4. Td2 Se1#
2) 1. b1=S Sc2 2. Kd3 Kg2 3. Dd2+ Kf3 4. Sc3 Sb4#
2) 1. b1=S Sc2 2. Kd3 Kg2 3. Dd2+ Kf3 4. Sc3 Sb4#
A.Buchanan: I think we can safely downgrade the Black force: D to L and L to B (2022-12-10)
more ...
comment
more ...
comment
Keywords: Minimal, Miniature, Ideal mate (2), Promotion (t,s)
Genre: h#
Computer test: Popeye Windows-32Bit v4.51 (100000 KB)
FEN: 8/8/8/8/2pb4/2q5/1p1k4/4N2K
Input: hpr, 2011-12-16
Last update: A.Buchanan, 2022-12-10 more...
Genre: h#
Computer test: Popeye Windows-32Bit v4.51 (100000 KB)
FEN: 8/8/8/8/2pb4/2q5/1p1k4/4N2K
Input: hpr, 2011-12-16
Last update: A.Buchanan, 2022-12-10 more...
1. Lg3 hxg3 2. e1=L g4 3. Da1 gxf5 4. Da7 f6 5. La5 fxe7 6. Ld8 exd8=D#
Keywords: Minimal, under-promotion (sL), Promotion in the mating move (wD), Excelsior white
Genre: h#
Computer test: A. Baumann: Gustav 4.2a
FEN: 1k6/1p2p1q1/8/2p2p2/7p/8/4p1pP/4b1K1
Input: Felber, Volker, 2012-04-15
Last update: Dieter Berlin, 2024-01-25 more...
Genre: h#
Computer test: A. Baumann: Gustav 4.2a
FEN: 1k6/1p2p1q1/8/2p2p2/7p/8/4p1pP/4b1K1
Input: Felber, Volker, 2012-04-15
Last update: Dieter Berlin, 2024-01-25 more...
1. Lf2 a3 2. Le1 Kg1 3. Ld2 Kh1 4. Lc1 Kg1 5. Lxa3 Kh1 6. Lc5 Kg1 7. a3 Kh1 8. a2 Kg1 9. a1=S Kh1 10. Sb3 cxb3 11. c2 b4 12. c1=T bxc5 13. Tb1 c6 14. Tb7 c7 15. Ta7 c8=D#
Schachecke in SE ab 01/1992 eingestellt. Lösung bis 12/1991 nicht veröffentlicht.
Schachecke in SE ab 01/1992 eingestellt. Lösung bis 12/1991 nicht veröffentlicht.
Henrik Juel: solution
1.Lf2 a3 2.Le1 Kg1 3.Ld2 Kh1 4.Lc1 Kg1 5.Lxa3 Kh1 6.Lc5 Kg1 7.a3 Kh1 8.a2 Kg1 9.a1=S Kh1 10.Sb3 cxb3 11.c2 b4 12.c1=T bxc5 13.Tb1 c6 14.Tb7 c7 15.Ta7 c8=D#
Three promotions (2016-01-13)
SCHRECKE: Ab 6. Zug dualistisch, z.B.: 6. Lb4 Kg1 7. a3 Kh1 8. a2 Kg1 9. a1=S Kh1 10. Sb3 cxb3 11. Ld6 b4 12. c2 b5 13. c1=S b6 14. Sxe2 Lxe2 15. Lb8 Lf3# (2023-07-24)
comment
1.Lf2 a3 2.Le1 Kg1 3.Ld2 Kh1 4.Lc1 Kg1 5.Lxa3 Kh1 6.Lc5 Kg1 7.a3 Kh1 8.a2 Kg1 9.a1=S Kh1 10.Sb3 cxb3 11.c2 b4 12.c1=T bxc5 13.Tb1 c6 14.Tb7 c7 15.Ta7 c8=D#
Three promotions (2016-01-13)
SCHRECKE: Ab 6. Zug dualistisch, z.B.: 6. Lb4 Kg1 7. a3 Kh1 8. a2 Kg1 9. a1=S Kh1 10. Sb3 cxb3 11. Ld6 b4 12. c2 b5 13. c1=S b6 14. Sxe2 Lxe2 15. Lb8 Lf3# (2023-07-24)
comment
Keywords: Promotion in the mating move, konsekutive Umwandlungen 3 (stD), Switchback (K), Entblockung, Selfblock, Corner mate, Excelsior white, Antiphönix
Genre: h#
FEN: k7/8/p7/8/p2p4/2pPp1p1/P1P1P1P1/5BbK
Input: Felber, Volker, 2016-01-13
Last update: Gunter Jordan, 2023-07-23 more...
Genre: h#
FEN: k7/8/p7/8/p2p4/2pPp1p1/P1P1P1P1/5BbK
Input: Felber, Volker, 2016-01-13
Last update: Gunter Jordan, 2023-07-23 more...
26 - P1338989
Manfred Seidel
Frank Fiedler
2848 Problemkiste (78) 12/1991
(3+3)
h#2
b) Tauschen wBe6<->wSb7 und sSd6<->sBb6
Manfred Seidel
Frank Fiedler
2848 Problemkiste (78) 12/1991
(3+3)
h#2
b) Tauschen wBe6<->wSb7 und sSd6<->sBb6
a) 1. Sf7 e7 2. Sd8 exd8=D#
b) 1. Kd7 Kf7 2. Sc8 b8=S#
b) 1. Kd7 Kf7 2. Sc8 b8=S#
Keywords: Exchange of Promotions
Genre: h#
FEN: 4k1K1/1N6/1p1nP3/8/8/8/8/8
Input: Erich Bartel, 2017-09-07
Last update: Alfred Pfeiffer, 2017-09-08 more...
Genre: h#
FEN: 4k1K1/1N6/1p1nP3/8/8/8/8/8
Input: Erich Bartel, 2017-09-07
Last update: Alfred Pfeiffer, 2017-09-08 more...
27 - P1355253
Mario Parrinello
17328 Die Schwalbe 289, p. 379, 02/2018
Informalturnier 2018
5. ehrende Erwähnung
(9+8) C+
h#2
3.1...
Mario Parrinello
17328 Die Schwalbe 289, p. 379, 02/2018
Informalturnier 2018
5. ehrende Erwähnung
(9+8) C+
h#2
3.1...
1) 1. Sxb2 bxa8=D 2. Sxc4 Da1#
2) 1. Sxe5 bxc8=D 2. Sxc4 Dh8#
3) 1. Dxd5 b8=D 2. Dxc4 Dxd6#
2) 1. Sxe5 bxc8=D 2. Sxc4 Dh8#
3) 1. Dxd5 b8=D 2. Dxc4 Dxd6#
1) 1. Lb3 g8=D 2. Tf4 Dg3#
2) 1. Df8+ gxf8=D 2. Kh4 Dh6#
2) 1. Df8+ gxf8=D 2. Kh4 Dh6#
2 Mustermatts
statt des wLh2 ginge auch wBh2
uschol-prischol
SCHRECKE: C+, popeye 4.87 (2022-10-15)
comment
statt des wLh2 ginge auch wBh2
uschol-prischol
SCHRECKE: C+, popeye 4.87 (2022-10-15)
comment
Keywords: Model mate, Promotion
Genre: h#
FEN: 1b1K2b1/6P1/8/8/6r1/q6k/7B/8
Input: Mario Richter, 2019-01-08
Last update: Alfred Pfeiffer, 2019-01-08 more...
Genre: h#
FEN: 1b1K2b1/6P1/8/8/6r1/q6k/7B/8
Input: Mario Richter, 2019-01-08
Last update: Alfred Pfeiffer, 2019-01-08 more...
1. Lg7 Ke8 2. Lf8 Kd8 3. Lg7 Kd7 4. Lf8 Kc6 5. Lg7 Kb5 6. Lf8 Kb4 7. Lg7 Kc3 8. Lf8 Kd2 9. Lg7 Ke1 10. Lf8 Kf1 11. Lg7 Kg1 12. Lf8 Kh2 13. Lg7 Kh3 14. Lf8 Kg4 15. Lg7 Kxf3 16. Lf8 Ke2 17. Lg7 Kd2 18. Lf8 Kc3 19. Lg7 Kc4 20. Lf8 Kb5 21. Lg7 Kc6 22. Lf8 Kd7 23. Lg7 Ke8 24. Lf8 Kxf8 25. f3 Ke8 26. Kg7 Kd8 27. Kf8 Kc8 28. Ke8 Kb7 29. Kd8 Kb6 30. Kc8 Kc6 31. Kb8 Kd7 32. Ka8 Ke8 33. Kb8 Kf8 34. Kc8 Kxg8 35. Kd8 Kf8 36. Kc8 Kxe7 37. Kb8 Kd7 38. Ka8 Kc6 39. Kb8 e7 40. Kc8 e8=D#,e8=T#
Henrik Juel: The late part of the solution is dualistic (2020-08-14)
Joost de Heer: The first dual is already at move 6 (Kb4 or Kc4), so it's not just the late part that's dualistic. (2023-01-30)
comment
Joost de Heer: The first dual is already at move 6 (Kb4 or Kc4), so it's not just the late part that's dualistic. (2023-01-30)
comment
Keywords: Promotion in the mating move, under-promotion in mating move (T)
Genre: h#
FEN: 5bnk/4pK1p/3pPp1p/3P1P1P/3P1p2/3P1p2/5P2/8
Input: Rainer Staudte, 2020-08-14
Last update: James Malcom, 2023-01-31 more...
Genre: h#
FEN: 5bnk/4pK1p/3pPp1p/3P1P1P/3P1p2/3P1p2/5P2/8
Input: Rainer Staudte, 2020-08-14
Last update: James Malcom, 2023-01-31 more...
1. f3-f2 h2-h3 2. f2-f1=S h3-h4 3. Sf1-g3 h4-h5 4. Sg3-f5 h5-h6 5. Sf5-g7 h6xg7 6. Kb2-b1 g7-g8=D 7. a2-a1=T Dg8-b3#
Viktoras Paliulionis: Illegal position because White has no last move. (2023-02-02)
Anton Baumann: mögliche Korrektur: wBf3 nach f2, mit Forderung h#6: 1.f1=S h4 ... (2023-02-13)
comment
Anton Baumann: mögliche Korrektur: wBf3 nach f2, mit Forderung h#6: 1.f1=S h4 ... (2023-02-13)
comment
Keywords: under-promotion (st), Promotion (D), Selfblock, Antiphönix, Excelsior white, Kindergarten Problem, Minimal (B)
Genre: h#
Computer test: Gustav 4.1c
FEN: 8/7p/8/8/8/2pppp2/pk1p3P/3K4
Input: Marcin Banaszek, 2020-12-28
Last update: Gunter Jordan, 2020-12-28 more...
Genre: h#
Computer test: Gustav 4.1c
FEN: 8/7p/8/8/8/2pppp2/pk1p3P/3K4
Input: Marcin Banaszek, 2020-12-28
Last update: Gunter Jordan, 2020-12-28 more...
31 - P1387867
Reinhardt Fiebig
19333 Schach , p. 78, 12/2020
Lob
PR Torsten Linß
(2+12)
h#6
b) sKa3->a5
Reinhardt Fiebig
19333 Schach , p. 78, 12/2020
Lob
PR Torsten Linß
(2+12)
h#6
b) sKa3->a5
a) 1. e1=D g4 2. Dh4 g5 3. Dh6 gxh6 4. c1=D h7 5. Dc3 h8=D 6. Db4 Da1#
b) 1. e1=T g4 2. Tf1 g5 3. Tf6 gxf6 4. c1=T f7 5. Ta1 f8=D 6. Ta4 Dc5#
b) 1. e1=T g4 2. Tf1 g5 3. Tf6 gxf6 4. c1=T f7 5. Ta1 f8=D 6. Ta4 Dc5#
Keywords: Minimal, Exchange of Promotions, Model mate
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/8/p5p1/8/2p1p3/kp2p1P1/2pnp1K1/3b3n
Input: Felber, Volker, 2021-03-26
Last update: Gunter Jordan, 2024-01-30 more...
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/8/p5p1/8/2p1p3/kp2p1P1/2pnp1K1/3b3n
Input: Felber, Volker, 2021-03-26
Last update: Gunter Jordan, 2024-01-30 more...
32 - P1396733
Mykhailo Gershynskyi
8555 Seven Chess Notes 170 03/2020
3. ehrende Erwähnung
Abt. h#>3
PR Alexander Semenenko
(2+4) C+
h#7
0.1.1...
Mykhailo Gershynskyi
8555 Seven Chess Notes 170 03/2020
3. ehrende Erwähnung
Abt. h#>3
PR Alexander Semenenko
(2+4) C+
h#7
0.1.1...
1. ... Kf2 2. Lc2 d4 3. Kc3 d5 4. Kd3 d6 5. Sc3 d7 6. Sd5 d8=D 7. c3 Dxd5#
Yuri Bilokin: you can see 1.Bf1 d4 2.Sd2 d5 3.Be2 d6 4.Bd1 d7 5.Bb3 d8=Q 6.Kc3 Qxd2#
1.Bf1 d4 2.Sd2 d5 3.Be2 d6 4.Bd1 d7 5.Kc3 d8=Q 6.Bb3 Qxd2#
1.Bf1 d4 2.Sd2 d5 3.Be2 d6 4.Kc3 d7 5.Bd1 d8=Q 6.Bb3 Qxd2#
1.Bf1 d4 2.Sd2 d5 3.Bd3 d6 4.Bc2 d7 5.Bb3 d8=Q 6.Kc3 Qxd2# (2023-02-01)
comment
1.Bf1 d4 2.Sd2 d5 3.Be2 d6 4.Bd1 d7 5.Kc3 d8=Q 6.Bb3 Qxd2#
1.Bf1 d4 2.Sd2 d5 3.Be2 d6 4.Kc3 d7 5.Bd1 d8=Q 6.Bb3 Qxd2#
1.Bf1 d4 2.Sd2 d5 3.Bd3 d6 4.Bc2 d7 5.Bb3 d8=Q 6.Kc3 Qxd2# (2023-02-01)
comment
Keywords: Entblockung, Excelsior white, Remote selfblock, Ideal mate, Liniensperrung, Miniature, Minimal (B), Check Protection, Selfblock, Promotion (D), Homebase
Genre: h#
FEN: 8/8/8/8/1kp5/3b4/3P4/1n2K3
Reprints: Seven Chess Notes 184, p. 10, 05/2021
Input: Gunter Jordan, 2021-12-08
Last update: Marcin Banaszek, 2022-01-04 more...
Genre: h#
FEN: 8/8/8/8/1kp5/3b4/3P4/1n2K3
Reprints: Seven Chess Notes 184, p. 10, 05/2021
Input: Gunter Jordan, 2021-12-08
Last update: Marcin Banaszek, 2022-01-04 more...
33 - P1396990
Anatoly Kirichenko
Aleksandr Pankratiev
1147 Pat a Mat 116 06/2021
(2+5) C+
h#6*
0.1.1...
Anatoly Kirichenko
Aleksandr Pankratiev
1147 Pat a Mat 116 06/2021
(2+5) C+
h#6*
0.1.1...
* 1. ... ... 2. Kd4 c4 3. Sd5+ cxd5 4. Kc4 d6 5. Kd5 d7 6. Kd6 d8=D#
1. ... c3 2. Ta6 cxb4 3. Kc6 bxc5 4. Kb7 c6+ 5. Ka8 c7 6. Ta7 c8=D#
1. ... c3 2. Ta6 cxb4 3. Kc6 bxc5 4. Kb7 c6+ 5. Ka8 c7 6. Ta7 c8=D#
Keywords: Corner mate (in Los.), Excelsior white, Ideal mate (im *), Minimal (B), Model mate (in Lös.), Round Trip (k-3 im *), Selfblock (in Lös.), Promotion (D), Promotion in the mating move
Genre: h#
FEN: 8/8/2r1p3/2pk4/1n3K2/8/2P5/8
Input: Gunter Jordan, 2021-12-13
Last update: Gunter Jordan, 2021-12-13 more...
Genre: h#
FEN: 8/8/2r1p3/2pk4/1n3K2/8/2P5/8
Input: Gunter Jordan, 2021-12-13
Last update: Gunter Jordan, 2021-12-13 more...
34 - P1397020
Anatoly Kirichenko
Aleksandr Pankratiev
4674 Best Problems 100 10/2021
(2+4) C+
h#6
0.2.1...
Anatoly Kirichenko
Aleksandr Pankratiev
4674 Best Problems 100 10/2021
(2+4) C+
h#6
0.2.1...
1) 1. ... c3 2. Sd4+ cxd4 3. Kc4 d5 4. Kd4 d6 5. Kd5 d7 6. Kd6 d8=D#
2) 1. ... cxb3 2. Ta6 b4 3. Kc6 bxc5 4. Kb7 c6+ 5. Ka8 c7 6. Ta7 c8=D#
2) 1. ... cxb3 2. Ta6 b4 3. Kc6 bxc5 4. Kb7 c6+ 5. Ka8 c7 6. Ta7 c8=D#
Keywords: Corner mate (in 2), Excelsior white, Ideal mate (in 1), Klasinc theme (k-B in 1), Model mate (in 2), Round Trip (k-3 in 1), Capture key (in 2), Selfblock, Promotion (D), Promotion in the mating move
Genre: h#
FEN: 8/8/2r5/2pk1K2/8/1n6/2P5/8
Input: Gunter Jordan, 2021-12-14
Last update: Gunter Jordan, 2021-12-14 more...
Genre: h#
FEN: 8/8/2r5/2pk1K2/8/1n6/2P5/8
Input: Gunter Jordan, 2021-12-14
Last update: Gunter Jordan, 2021-12-14 more...
35 - P1398028
Rolf Wiehagen
SuperProblem (Website) 20/12/2021
after Ed Hoes and Hans Uitenbroek
5. Platz TT-263
(2+6) C+
h#7
0.1.1...
Rolf Wiehagen
SuperProblem (Website) 20/12/2021
after Ed Hoes and Hans Uitenbroek
5. Platz TT-263
(2+6) C+
h#7
0.1.1...
1. ... Kg3-g4 2. Sd4-f3 g2xf3 3. e5-e4 f3xe4 4. Lb3-d5 e4xd5 5. Dh6-c6 d5xc6 6. Ka7-a8 c6-c7 7. Ta2-a7 c7-c8=D#
Vergleiche mit P1373236
Yuri Bilokin: you can see 1.Sf3 gxf3 2.e4 fxe4 3.Bd5 exd5 4.Qc6 dxc6 5.Ka8 c7 6.Ra7 c8=Q#
1.Sf3 gxf3 2.Be6 f4 3.Bc8 fxe5 4.Qd6 exd6 5.Ka8 d7 6.Ra7 dxc8=Q#
1.Sf3 gxf3 2.Qd6 f4 3.Be6 fxe5 4.Bc8 exd6 5.Ka8 d7 6.Ra7 dxc8=Q#
1.Sf3 gxf3 2.Qd6 f4 3.Be6 fxe5 4.Ka8 exd6 5.Bc8 d7 6.Ra7 dxc8=Q# (2023-08-15)
comment
Yuri Bilokin: you can see 1.Sf3 gxf3 2.e4 fxe4 3.Bd5 exd5 4.Qc6 dxc6 5.Ka8 c7 6.Ra7 c8=Q#
1.Sf3 gxf3 2.Be6 f4 3.Bc8 fxe5 4.Qd6 exd6 5.Ka8 d7 6.Ra7 dxc8=Q#
1.Sf3 gxf3 2.Qd6 f4 3.Be6 fxe5 4.Bc8 exd6 5.Ka8 d7 6.Ra7 dxc8=Q#
1.Sf3 gxf3 2.Qd6 f4 3.Be6 fxe5 4.Ka8 exd6 5.Bc8 d7 6.Ra7 dxc8=Q# (2023-08-15)
comment
Keywords: Minimal (B), Promotion in the mating move, Excelsior white, Promotion (D), Selfblock, Corner mate, Model mate
Genre: h#
Computer test: Gustav 4.1c
FEN: 8/k7/7q/4p3/3n4/1b4K1/r5P1/8
Input: Marcin Banaszek, 2022-01-07
Last update: Gunter Jordan, 2022-05-07 more...
Genre: h#
Computer test: Gustav 4.1c
FEN: 8/k7/7q/4p3/3n4/1b4K1/r5P1/8
Input: Marcin Banaszek, 2022-01-07
Last update: Gunter Jordan, 2022-05-07 more...
36 - P1398030
Zoltan Laborczi
SuperProblem (Website) 20/12/2021
7. Platz TT-263
(2+6) C+
h#7
0.1.1...
Zoltan Laborczi
SuperProblem (Website) 20/12/2021
7. Platz TT-263
(2+6) C+
h#7
0.1.1...
1. ... Kg7-h8 2. Sb4-a6 b3-b4 3. De3-e2 b4-b5 4. Tf3-c3 b5xa6 5. e4-e3 a6-a7 6. Lh7-c2 a7-a8=D 7. Kd4-d3 Da8-d5#
Yuri Bilokin: you can see 1.Sa6 b4 2.Qe2 b5 3.Rc3 bxa6 4.e3 a7 5.Bc2 a8=Q 6.Kd3 Qd5#
1.Sc2 b4 2.Ke5 b5 3.Kf4 b6 4.Qg1+ Kf6 5.Se3 b7 6.Qg4 b8=Q#
1.Sc2 b4 2.Ke5 b5 3.Kf4 b6 4.Qg1+ Kf6 5.Qg4 b7 6.Se3 b8=Q#
1.Sc2 b4 2.Ke5 b5 3.Kf4 Kf6 4.Qg1 b6 5.Se3 b7 6.Qg4 b8=Q# (2023-08-15)
comment
1.Sc2 b4 2.Ke5 b5 3.Kf4 b6 4.Qg1+ Kf6 5.Se3 b7 6.Qg4 b8=Q#
1.Sc2 b4 2.Ke5 b5 3.Kf4 b6 4.Qg1+ Kf6 5.Qg4 b7 6.Se3 b8=Q#
1.Sc2 b4 2.Ke5 b5 3.Kf4 Kf6 4.Qg1 b6 5.Se3 b7 6.Qg4 b8=Q# (2023-08-15)
comment
Keywords: Minimal (B), Schlüssel = Eckzug, Promotion (D), Model mate, Epaulettenmatt, Entblockung, Tempo Loss
Genre: h#
Computer test: Gustav 4.1c
FEN: 8/6Kb/8/8/1n1kp3/1P2qr2/8/8
Input: Marcin Banaszek, 2022-01-07
Last update: Gunter Jordan, 2022-01-07 more...
Genre: h#
Computer test: Gustav 4.1c
FEN: 8/6Kb/8/8/1n1kp3/1P2qr2/8/8
Input: Marcin Banaszek, 2022-01-07
Last update: Gunter Jordan, 2022-01-07 more...
1. a4xb3ep+ Sxa5+ 2. Kd5 0-0-0# (by AP not Td1#?)
R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7,Sd6xLc7
With the typo wPd5 corrected to e5, the two forward candidate solutions are correct. How about the retro? Assume that White castling rights remain. bBg5 is promoted, and did so on c1 or g1. In either case, there were 6 pawn caps by Black, but in the former case, the blockaded wRh was not consumable. So the promotion must have been on g8, and White can't retract h2-h3. The missing White unit is light B, so was not just captured on e5 or b4. wRf6 is promoted, say wPd, which would have involved no captures.
This leaves b2-b4, preceded by Kb4-c4+ and before that Sd6-b7. Prior to b2-b4, why couldn't bQ have just played e.g. Qa8-a5+ instead? Because there would still be no prior move: wSb7 can't have just come from h5 or e6. Similarly Rc3-a3 has no precedent.
We don't know exactly what happened to sLc, but that doesn’t affect the soundness. It might have enabled wPc or wPd to capture. wPd did promote, but maybe it was wPc that after capturing promoted to T. Alternatively, Sd6xLc7 was possible.
R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7,Sd6xLc7
With the typo wPd5 corrected to e5, the two forward candidate solutions are correct. How about the retro? Assume that White castling rights remain. bBg5 is promoted, and did so on c1 or g1. In either case, there were 6 pawn caps by Black, but in the former case, the blockaded wRh was not consumable. So the promotion must have been on g8, and White can't retract h2-h3. The missing White unit is light B, so was not just captured on e5 or b4. wRf6 is promoted, say wPd, which would have involved no captures.
This leaves b2-b4, preceded by Kb4-c4+ and before that Sd6-b7. Prior to b2-b4, why couldn't bQ have just played e.g. Qa8-a5+ instead? Because there would still be no prior move: wSb7 can't have just come from h5 or e6. Similarly Rc3-a3 has no precedent.
We don't know exactly what happened to sLc, but that doesn’t affect the soundness. It might have enabled wPc or wPd to capture. wPd did promote, but maybe it was wPc that after capturing promoted to T. Alternatively, Sd6xLc7 was possible.
Henrik Juel: Popeye 4.61 with 'opt enp b3' found no solution (2022-05-27)
Gerald Ettl: Schaut so aus, als ob die sDa5 nach e5 muss. 1.axb3 ep Sa5+ 2.Kxd5 0-0-0# (2022-05-27)
Gerald Ettl: aber auch 2.-Td1# ? (2022-05-27)
Henrik Juel: Andrew, can you throw any light on this problem? (2022-05-28)
Henrik Juel: Gerald, in AP problems like this one White must castle to legitimize the ep capture key (2022-05-28)
A.Buchanan: Hi Henrik, Gerald. I found this one in WinChloe with the current diagram. It gave the "solution" 1. axb3ep+ Sxa5+ 2. Kd5?? 0-0-0#. However when I tried running the WinChloe solver just now, it said there's no solution. Quite right: Rf5 controls d5. I don't think we can shift Qa5 to e5 though.
My first thought is that wPd5 should be on e5. Check my suggested solution in the solution text. However, as you will see, I think we need a second thought as well! Hope the finest retro minds can figure this out. Failing that, we could ask Igor. (2022-05-28)
Mario Richter: The diagram in the reprint in 'Uralsky Problemist' is exactly as given here. The solution given there without further explanation is: 1. ab+ (e.p.) Sxa5+ 2. Kd5 0-0-0#!
Notice that the second black move doesn't contain a capture sign!
Some typos in Andrew's analysis:
The black potential promotion squares should be c1 and g1.
White's last move couldn't have been b3-b4?? simply because that would give an illegal check. (2022-05-28)
A.Buchanan: Thanks Mario for the careful reading. I’ve fixed the errors. Pd5 may be a typo in the reprint then. It also creates a bogus retraction d4-d5, so I don’t think it can be right. Now for Igor. (2022-05-29)
A.Buchanan: Joaquim Crusats spotted R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7, which I'm sure is the intended retraction. So we diagnose a single typo that occurred in the Uralsky reprint diagram: wPe5 appeared on d5. I suggest correcting it here. (2022-05-30)
A.Buchanan: Author confirmed this was a typo (2022-11-23)
more ...
comment
Gerald Ettl: Schaut so aus, als ob die sDa5 nach e5 muss. 1.axb3 ep Sa5+ 2.Kxd5 0-0-0# (2022-05-27)
Gerald Ettl: aber auch 2.-Td1# ? (2022-05-27)
Henrik Juel: Andrew, can you throw any light on this problem? (2022-05-28)
Henrik Juel: Gerald, in AP problems like this one White must castle to legitimize the ep capture key (2022-05-28)
A.Buchanan: Hi Henrik, Gerald. I found this one in WinChloe with the current diagram. It gave the "solution" 1. axb3ep+ Sxa5+ 2. Kd5?? 0-0-0#. However when I tried running the WinChloe solver just now, it said there's no solution. Quite right: Rf5 controls d5. I don't think we can shift Qa5 to e5 though.
My first thought is that wPd5 should be on e5. Check my suggested solution in the solution text. However, as you will see, I think we need a second thought as well! Hope the finest retro minds can figure this out. Failing that, we could ask Igor. (2022-05-28)
Mario Richter: The diagram in the reprint in 'Uralsky Problemist' is exactly as given here. The solution given there without further explanation is: 1. ab+ (e.p.) Sxa5+ 2. Kd5 0-0-0#!
Notice that the second black move doesn't contain a capture sign!
Some typos in Andrew's analysis:
The black potential promotion squares should be c1 and g1.
White's last move couldn't have been b3-b4?? simply because that would give an illegal check. (2022-05-28)
A.Buchanan: Thanks Mario for the careful reading. I’ve fixed the errors. Pd5 may be a typo in the reprint then. It also creates a bogus retraction d4-d5, so I don’t think it can be right. Now for Igor. (2022-05-29)
A.Buchanan: Joaquim Crusats spotted R: 1. b2-b4 Kb4-c4+ 2. Sd6-b7, which I'm sure is the intended retraction. So we diagnose a single typo that occurred in the Uralsky reprint diagram: wPe5 appeared on d5. I suggest correcting it here. (2022-05-30)
A.Buchanan: Author confirmed this was a typo (2022-11-23)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wg), Promotion (Tl), Valladao Task
Genre: h#, Retro
Computer test: Once the diagram typo was fixed, C+ Popeye v4.87 & basic but tricky retro logic
FEN: 3n4/1Np1p1p1/4pn2/q1p1PRb1/pPk1pr2/r6P/P4PP1/R3K3
Reprints: 18 The Ural's Problemist 13, p. 5, 02/1998
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2022-05-30 more...
Genre: h#, Retro
Computer test: Once the diagram typo was fixed, C+ Popeye v4.87 & basic but tricky retro logic
FEN: 3n4/1Np1p1p1/4pn2/q1p1PRb1/pPk1pr2/r6P/P4PP1/R3K3
Reprints: 18 The Ural's Problemist 13, p. 5, 02/1998
Input: A.Buchanan, 2022-05-27
Last update: A.Buchanan, 2022-05-30 more...
38 - P1402506
Reinhardt Fiebig
Mirko Degenkolbe
10042 Seven Chess Notes 198, p. 19, 07/2022
(2+4) C+
h#6
Reinhardt Fiebig
Mirko Degenkolbe
10042 Seven Chess Notes 198, p. 19, 07/2022
(2+4) C+
h#6
1. e1=L a3 2. Lb4 axb4 3. h1=L b5 4. Lc6 bxc6 5. Kb8 c7+ 6. Ka8 c8=D#
Lösung:
1.e1L! [e1D?] a3! [a4?]
2.Lb4 axb4
3.h1L! [h1D?] b5
4.Lc6 bxc6
5.Kb8! c7
6.Ka8 [Rückkehr] c8D# - Mustermatt
Lösung:
1.e1L! [e1D?] a3! [a4?]
2.Lb4 axb4
3.h1L! [h1D?] b5
4.Lc6 bxc6
5.Kb8! c7
6.Ka8 [Rückkehr] c8D# - Mustermatt
(MD1790)
Eile-mit-Weile.
Weiße Auswahl im ersten Zug.
Schwarze Auswahl im ersten und dritten Zug.
Schwarzer Wartezug mit Rückkehr.
Weißer Excelsior.
Zweifache schwarze Läufer-Unterverwandlung.
Mustermatt.
Miniatur.
Sechssteiner.
Weißes Minimal.
Kindergarten-Problem
Klaus Funk: identischer Vorgänger: Ulf Hammerström 1994, P1095198 (2023-01-24)
comment
Eile-mit-Weile.
Weiße Auswahl im ersten Zug.
Schwarze Auswahl im ersten und dritten Zug.
Schwarzer Wartezug mit Rückkehr.
Weißer Excelsior.
Zweifache schwarze Läufer-Unterverwandlung.
Mustermatt.
Miniatur.
Sechssteiner.
Weißes Minimal.
Kindergarten-Problem
Klaus Funk: identischer Vorgänger: Ulf Hammerström 1994, P1095198 (2023-01-24)
comment
Keywords: Echo, Excelsior white, under-promotion (ll), Model mate, Minimal (B), Miniature, under-promotion key, Promotions to bishops, Promotion (D), konsekutive Umwandlungen 3 (llD), Switchback (k)
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: k7/p7/8/8/8/8/P3p2p/2K5
Input: Felber, Volker, 2022-07-07
Last update: Rainer Staudte, 2022-11-07 more...
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: k7/p7/8/8/8/8/P3p2p/2K5
Input: Felber, Volker, 2022-07-07
Last update: Rainer Staudte, 2022-11-07 more...
1. ... Le5 2. Tc7 Ld5 3. Tb7 axb7 4. Kxe5 b8=D 5. Kxd5 Dxb5#
Keywords: Entblockung, Drag-in sacrifice (2-fach), Model mate, Promotion
Genre: h#
FEN: 7B/p7/P2pp3/1p3k2/2rpp3/8/B3r3/1K6
Input: Gunter Jordan, 2022-09-25
Last update: Gunter Jordan, 2022-09-26 more...
Genre: h#
FEN: 7B/p7/P2pp3/1p3k2/2rpp3/8/B3r3/1K6
Input: Gunter Jordan, 2022-09-25
Last update: Gunter Jordan, 2022-09-26 more...
Versuch: 1. ... a3? 2. Le2 a4 3. Lb5 axb5 4. Lh2 b6 5. Lc7 bxc7 6. ??
1. ... a4 2. Le2 Kh3 3. Lb5 axb5 4. Lh2 b6 5. Lc7 bxc7 6. g5 c8=D#
LB aus SCHACH: 19.714 (Fiebig/Degenkolbe) Verf.: 1… a3? 2.Le2 a4 3.Lb5 a:b5 4.Lh2 b6 5.Lc7 b:c7 6.??? c8D matt scheitert einzig daran, dass Schwarz keinen Wartezug hat. Deshalb muss Weiß umdenken. 1… a4! 2.Le2! (Ld7?) Kh3! (Kg3?, Kg5?) 3.Lb5 a:b5 4.Lh2! (Lb6?) b6 5.Lc7 b:c7 8.g5! (Tempo) c8D matt (Mustermatt) (4 P.). „Der Clou ist der Tempozug g5, weshalb der wK ein Fluchtfeld finden muss, was wiederum den Schlüsselzug 2.Le2 begründet. Super“ (JG). „Perfekt“ (HR). „Witzig! Der wK muss zeitig genau so ziehen, dass Schwarz am Ende einen unschädlichen Tempozug machen kann“ (GJ).
LB der Autoren:
Verführung:
1. … a3?
2.Le2 a4
3.Lb5 axb5
4.Lh2 b6
5.Lc7 bxc7
6.??? c8D# scheitert einzig daran, dass Schwarz keinen Wartezug hat.
Deshalb muss Weiß umdenken.
Lösung:
1. ... a4! [a3?]
2.Le2! [Ld7?] Kh3! [Kg3?, Kg5?]
3.Lb5 axb5
4.Lh2! [Lb6?] b6
5.Lc7 bxc7
6.g5! [Tempo] c8D# - Mustermatt
Vermeidung des Eile-mit-Weile-Schlüsselzuges 1. …a3 mit logischer Begründung.
Weiße Auswahl im ersten und zweiten Zug.
Schwarze Auswahl im zweiten und vierten Zug.
Schwarzer Tempozug.
Weißer Excelsior.
Mustermatt.
Miniatur.
Weißes Minimal.
1. ... a4 2. Le2 Kh3 3. Lb5 axb5 4. Lh2 b6 5. Lc7 bxc7 6. g5 c8=D#
LB aus SCHACH: 19.714 (Fiebig/Degenkolbe) Verf.: 1… a3? 2.Le2 a4 3.Lb5 a:b5 4.Lh2 b6 5.Lc7 b:c7 6.??? c8D matt scheitert einzig daran, dass Schwarz keinen Wartezug hat. Deshalb muss Weiß umdenken. 1… a4! 2.Le2! (Ld7?) Kh3! (Kg3?, Kg5?) 3.Lb5 a:b5 4.Lh2! (Lb6?) b6 5.Lc7 b:c7 8.g5! (Tempo) c8D matt (Mustermatt) (4 P.). „Der Clou ist der Tempozug g5, weshalb der wK ein Fluchtfeld finden muss, was wiederum den Schlüsselzug 2.Le2 begründet. Super“ (JG). „Perfekt“ (HR). „Witzig! Der wK muss zeitig genau so ziehen, dass Schwarz am Ende einen unschädlichen Tempozug machen kann“ (GJ).
LB der Autoren:
Verführung:
1. … a3?
2.Le2 a4
3.Lb5 axb5
4.Lh2 b6
5.Lc7 bxc7
6.??? c8D# scheitert einzig daran, dass Schwarz keinen Wartezug hat.
Deshalb muss Weiß umdenken.
Lösung:
1. ... a4! [a3?]
2.Le2! [Ld7?] Kh3! [Kg3?, Kg5?]
3.Lb5 axb5
4.Lh2! [Lb6?] b6
5.Lc7 bxc7
6.g5! [Tempo] c8D# - Mustermatt
Vermeidung des Eile-mit-Weile-Schlüsselzuges 1. …a3 mit logischer Begründung.
Weiße Auswahl im ersten und zweiten Zug.
Schwarze Auswahl im zweiten und vierten Zug.
Schwarzer Tempozug.
Weißer Excelsior.
Mustermatt.
Miniatur.
Weißes Minimal.
Keywords: Minimal (B), Miniature, Tempo Loss, Excelsior white, Corner mate, Promotion in the mating move, Promotion (D), Model mate
Genre: h#
Computer test: SCHRECKE: C+, Gustav 4.2b, Brute Force (2023-01-28)
FEN: k7/p7/6p1/8/6bK/8/P7/6b1
Input: Felber, Volker, 2022-10-14
Last update: Felber, Volker, 2023-01-28 more...
Genre: h#
Computer test: SCHRECKE: C+, Gustav 4.2b, Brute Force (2023-01-28)
FEN: k7/p7/6p1/8/6bK/8/P7/6b1
Input: Felber, Volker, 2022-10-14
Last update: Felber, Volker, 2023-01-28 more...
41 - P1407382
Christer Jonsson
H924 Problem-Forum 06/2018
1. Lob
PR Hans Gruber
(5+11) C+
h#3
0.3.1...
Christer Jonsson
H924 Problem-Forum 06/2018
1. Lob
PR Hans Gruber
(5+11) C+
h#3
0.3.1...
1) 1. ... f7 2. Sxe3 f8=D 3. Sd5 Df1#
2) 1. ... fxg7 2. bxa6 gxh8=D 3. axb5 Dd4#
3) 1. ... fxe7 2. Kxb5 e8=S 3. Kxa6 Sc7#
Der Preisbericht erschien in Gaudium 262, 07/2022.
2) 1. ... fxg7 2. bxa6 gxh8=D 3. axb5 Dd4#
3) 1. ... fxe7 2. Kxb5 e8=S 3. Kxa6 Sc7#
Der Preisbericht erschien in Gaudium 262, 07/2022.
Yuri Bilokin: more economical bBa7(-bRa7) 7n/bp2r1p1/Np3P2/pN5K/2k1p3/1p2P3/8/5n2 (5+11) (2023-01-24)
comment
comment
Keywords: Epaulettenmatt (in 2), Model mate, Capture key (in 2+3), Selfblock (in 1+2), Promotion (D in 1+2), under-promotion (S in 3)
Genre: h#
FEN: 7n/rp2r1p1/Np3P2/pN5K/2k1p3/1p2P3/8/5n2
Reprints: Gaudium 262, p. 2, 15/07/2022
Input: Gunter Jordan, 2023-01-21
Last update: Gunter Jordan, 2023-01-21 more...
Genre: h#
FEN: 7n/rp2r1p1/Np3P2/pN5K/2k1p3/1p2P3/8/5n2
Reprints: Gaudium 262, p. 2, 15/07/2022
Input: Gunter Jordan, 2023-01-21
Last update: Gunter Jordan, 2023-01-21 more...
1) 1. Kb2 Kd7 2. e5 fxe5 3. Kc3 e6 4. Kd4 e7 5. Ke5 e8=D 6. Kf6 De7#
2) 1. g5 fxg5 2. f4 g6 3. f3 g7 4. f2 g8=D 5. f1=T Dg3 6. Tb1 Da3#
2) 1. g5 fxg5 2. f4 g6 3. f3 g7 4. f2 g8=D 5. f1=T Dg3 6. Tb1 Da3#
Keywords: Kindergarten Problem, Miniature, Minimal (B), Promotion (D), under-promotion (t in 2), konsekutive Umwandlungen 2 (Dt in 2), Selfblock (in 2), Corner mate (in 2), Model mate (in 2), Entblockung
Genre: h#
FEN: 8/4p3/4K1p1/5p2/5P2/8/8/k7
Input: Gunter Jordan, 2023-01-28
Last update: Gerd Wilts, 2023-01-29 more...
Genre: h#
FEN: 8/4p3/4K1p1/5p2/5P2/8/8/k7
Input: Gunter Jordan, 2023-01-28
Last update: Gerd Wilts, 2023-01-29 more...
1. c1=S Kb1 2. Sb3 axb3 3. Se4 b4 4. d2 b5 5. d1=S b6 6. Se3 b7 7. S3f5 b8=S 8. Sg5 Sd7#
Turnier-Thema: h#8, one solution, with wS(s) + bS(s).
Turnier-Thema: h#8, one solution, with wS(s) + bS(s).
Keywords: Entblockung, Excelsior white, konsekutive Umwandlungen 3 (ssS), Liniensperrung, Model mate, Check Protection, Selfblock, Promotions to Knights, under-promotion (Sss), under-promotion key, Minimal (B)
Genre: h#
FEN: 8/4nqb1/4pkb1/8/8/p2p4/P1p2n2/K7
Reprints: 4 The Problemist 283 01/2022
Input: Gunter Jordan, 2023-02-09
Last update: Gunter Jordan, 2023-02-09 more...
Genre: h#
FEN: 8/4nqb1/4pkb1/8/8/p2p4/P1p2n2/K7
Reprints: 4 The Problemist 283 01/2022
Input: Gunter Jordan, 2023-02-09
Last update: Gunter Jordan, 2023-02-09 more...
1. f2 Ka3 2. f1=S Ka4 3. Sg3 hxg3 4. Lf5 g4 5. h2 gxf5 6. h1=D fxe6 7. Dh2 e7 8. Dd6 exd8=S#
Turnier-Thema: h#8, one solution, with wS(s) + bS(s).
Turnier-Thema: h#8, one solution, with wS(s) + bS(s).
Keywords: Antiphönix, Entblockung, Excelsior white, konsekutive Umwandlungen 3 (sdS), Model mate, Phoenix, Rückkehr nach Umwandlung (b/d), Selfblock, Promotion (d), Promotions to Knights, under-promotion (Ss), under-promotion in mating move, Minimal (B)
Genre: h#
FEN: 3b4/2nn4/1rk1p3/2pp4/6bp/5p1p/K6P/8
Reprints: 9 The Problemist , p. 306, 01/2022
Input: Gunter Jordan, 2023-02-09
Last update: Gunter Jordan, 2023-02-09 more...
Genre: h#
FEN: 3b4/2nn4/1rk1p3/2pp4/6bp/5p1p/K6P/8
Reprints: 9 The Problemist , p. 306, 01/2022
Input: Gunter Jordan, 2023-02-09
Last update: Gunter Jordan, 2023-02-09 more...
a) 1. Kc5 f8=D+ 2. Kd4 Da8 3. Dc3 Dxg2 4. Kd3+ Kf1 5. Ld4 De2#
b) 1. Sg8 fxg8=D 2. Db4 Dxg3 3. Kc5 Kf2 4. Kc4+ Ke2 5. Lc5 Dd3#
b) 1. Sg8 fxg8=D 2. Db4 Dxg3 3. Kc5 Kf2 4. Kc4+ Ke2 5. Lc5 Dd3#
Keywords: Minimal (B), Promotion (D), Discovered check, Epaulettenmatt, Echo, Model mate, Selfblock, Remote selfblock (in b)
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/b4P2/1kq2n2/8/8/6p1/6p1/6Kb
Input: Felber, Volker, 2023-03-09
Last update: Felber, Volker, 2023-03-10 more...
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/b4P2/1kq2n2/8/8/6p1/6p1/6Kb
Input: Felber, Volker, 2023-03-09
Last update: Felber, Volker, 2023-03-10 more...
1) 1. ... b8=D+ 2. Kg4+ Df4+ 3. Kh5 Dd2 4. Le3 fxe3 5. Kg4 Dd1 6. Kg3 Dg1#
2) 1. ... b8=T 2. b2+ Txb2 3. d2+ Kxc2 4. d1=S Kd3 5. Se3 Te2 6. Sf5 Te4#
2) 1. ... b8=T 2. b2+ Txb2 3. d2+ Kxc2 4. d1=S Kd3 5. Se3 Te2 6. Sf5 Te4#
Yuri Bilokin: you can see 1.h2 b8=Q+ 2.Kg4+ Qf4+ 3.Kh3 Qe3 4.Bf6 Qxf3#
1.h2 b8=Q+ 2.Kg4+ Qf4+ 3.Kh3 Qe3 4.Be7 Qxf3#
1.h2 b8=Q+ 2.Kg4+ Qf4+ 3.Kh3 Qe3 4.Bd8 Qxf3#
1.Bf6 b8=Q+ 2.Kg4 Qxb3 3.h2 Qd5 4.Kh3 Qxf3# (2024-02-29)
comment
1.h2 b8=Q+ 2.Kg4+ Qf4+ 3.Kh3 Qe3 4.Be7 Qxf3#
1.h2 b8=Q+ 2.Kg4+ Qf4+ 3.Kh3 Qe3 4.Bd8 Qxf3#
1.Bf6 b8=Q+ 2.Kg4 Qxb3 3.h2 Qd5 4.Kh3 Qxf3# (2024-02-29)
comment
Keywords: Checking key (in 1), Promotion key (in 1), under-promotion key (in 2), under-promotion (ts in 2), konsekutive Umwandlungen 2 (ts in 2), bi-coloured Bristol (in 1), Cross check (in 1), Discovered check (in 1), selfpinning (in 1), Unpinning (in 1), Switchback (k in 1), Model mate (in 1), Exchange of Promotions
Genre: h#
FEN: 5n1n/1P6/8/6b1/5k1p/pppp1p1p/2p2P2/2K5
Input: Gunter Jordan, 2023-04-10
Last update: Felber, Volker, 2023-10-15 more...
Genre: h#
FEN: 5n1n/1P6/8/6b1/5k1p/pppp1p1p/2p2P2/2K5
Input: Gunter Jordan, 2023-04-10
Last update: Felber, Volker, 2023-10-15 more...
1. Lg3 Le3 2. g1=D Kg5 3. Df1 Kf5 4. Da6 Kxe4 5. h1=L+ Kd4 6. Lc6 Kc5 7. La4 Ld2#
Keywords: Minimal (L), Promotion (d), under-promotion (l), konsekutive Umwandlungen 2 (dl), Selfblock, Liniensperrung, Check Protection (Voraus-)
Genre: h#
FEN: 7n/8/7K/k7/4pp1b/8/6pp/6B1
Input: Gunter Jordan, 2023-04-16
Last update: Gunter Jordan, 2023-07-25 more...
Genre: h#
FEN: 7n/8/7K/k7/4pp1b/8/6pp/6B1
Input: Gunter Jordan, 2023-04-16
Last update: Gunter Jordan, 2023-07-25 more...
48 - P1409605
Reinhardt Fiebig
Eberhard Schulze
Problem-Forum 15/03/2014
(3+8) C+
h#5
0.1.1...
b) sSg7->h4
Reinhardt Fiebig
Eberhard Schulze
Problem-Forum 15/03/2014
(3+8) C+
h#5
0.1.1...
b) sSg7->h4
a) 1. ... a8=S 2. Tf2 Sb6 3. Tf7 exf7+ 4. Kd8 f8=S 5. Se8 Se6#
b) 1. ... a8=L 2. Sg6 Le4 3. Sf8 Lh7 4. Ta8 Lg8 5. Td8 Lf7#
b) 1. ... a8=L 2. Sg6 Le4 3. Sf8 Lh7 4. Ta8 Lg8 5. Td8 Lf7#
Keywords: under-promotion (SS,L), under-promotion key, Promotions to Knights (in a), konsekutive Umwandlungen 2 (SS in a), Twin, Model mate, Selfblock
Genre: h#
Computer test: Gustav 4.2
FEN: 4k3/P3p1n1/4P3/8/3p4/6pp/r5p1/6K1
Input: Eberhard Schulze, 2023-05-14
Last update: Gunter Jordan, 2023-06-28 more...
Genre: h#
Computer test: Gustav 4.2
FEN: 4k3/P3p1n1/4P3/8/3p4/6pp/r5p1/6K1
Input: Eberhard Schulze, 2023-05-14
Last update: Gunter Jordan, 2023-06-28 more...
49 - P1410865
Mirko Degenkolbe
10582 Seven Chess Notes 210, p. 23, 07/2023
spezielles Lob
Abt. h#5,5-n
PR Alexej Iwunin
(5+1) C+
h#6
0.2.1...
b) wKf1<->sKc1
Mirko Degenkolbe
10582 Seven Chess Notes 210, p. 23, 07/2023
spezielles Lob
Abt. h#5,5-n
PR Alexej Iwunin
(5+1) C+
h#6
0.2.1...
b) wKf1<->sKc1
a1) 1. ... d4 2. Kd2 d5 3. Ke3 d6 4. Kf4 d7 5. Kg3 d8=D 6. Kh2 Dh4#
a2) 1. ... g4 2. Kxd2 g5 3. Ke3 g6 4. Kf4 g7 5. Kg3 g8=D+ 6. Kh2 Dg2#
b1) 1. ... b4 2. Kxe2 b5 3. Kd3 b6 4. Kc4 b7 5. Kb3 b8=D+ 6. Ka2 Db2#
b2) 1. ... e4 2. Ke2 e5 3. Kd3 e6 4. Kc4 e7 5. Kb3 e8=D 6. Ka2 Da4#
Komplett gleichbleibende schwarze Züge innerhalb der Lösung a) und innerhalb der Lösung b) mit völlig anderen Mattbildern.
Vier verschiedene Excelsiors. Jeder Bauer, der sich auf dem Brett befindet, führt einen Excelsior aus. Zweimal Rückkehr im Mattzug. Zweimal Chamäleonechomatts.
Rex solus. Sechssteiner
a2) 1. ... g4 2. Kxd2 g5 3. Ke3 g6 4. Kf4 g7 5. Kg3 g8=D+ 6. Kh2 Dg2#
b1) 1. ... b4 2. Kxe2 b5 3. Kd3 b6 4. Kc4 b7 5. Kb3 b8=D+ 6. Ka2 Db2#
b2) 1. ... e4 2. Ke2 e5 3. Kd3 e6 4. Kc4 e7 5. Kb3 e8=D 6. Ka2 Da4#
Komplett gleichbleibende schwarze Züge innerhalb der Lösung a) und innerhalb der Lösung b) mit völlig anderen Mattbildern.
Vier verschiedene Excelsiors. Jeder Bauer, der sich auf dem Brett befindet, führt einen Excelsior aus. Zweimal Rückkehr im Mattzug. Zweimal Chamäleonechomatts.
Rex solus. Sechssteiner
Keywords: Miniature, Kindergarten Problem, Rex solus (s), Excelsior white, Promotion (D), Chameleon Echo, Rückkehr nach Umwandlung (in a2+b1), Round Trip (B/D in a2+b1), Twin, Symmetrical solution
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/8/8/8/8/8/1P1PP1P1/2k2K2
Input: Felber, Volker, 2023-07-11
Last update: Gunter Jordan, 2024-02-04 more...
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/8/8/8/8/8/1P1PP1P1/2k2K2
Input: Felber, Volker, 2023-07-11
Last update: Gunter Jordan, 2024-02-04 more...
1. Le1 c5 2. Lf2 c4 3. Lf2xg1 Kxg1 4. Kg7 Kh1 5. Kf6 Kg1 6. Ke5 Kh1 7. Kd4 Kg1 8. Kxc4 Kh1 9. Kxc5 Kg1 10. Kd6 Kh1 11. c5 Kg1 12. c4 Kh1 13. c3 Kg1 14. c2 Kh1 15. c1=S Kg1 16. Sxe2 Kh1 17. Sf4 Lc4 18. Sh5 gxh5 19. Kd7 h6 20. Ke8 h7 21. Ke8-f8 h8=D#
PAS-Läufer-Hilfsmatt ab dem 4. Zug.
Die letzten 14 Züge sind identisch mit Kardos, P0502947.
Außerdem Selbstplagiat, V. M. veröffentlichte diese Aufgabe bereits 2008 in Schachmatnaja Komposizija (ganz leicht modifiziert, siehe P1080117).
Die Angabe "harmonie 2009" ist unsicher, denn sie wurde dem Hopper Magazine 03 (Zweitquelle) entnommen.
PAS-Läufer-Hilfsmatt ab dem 4. Zug.
Die letzten 14 Züge sind identisch mit Kardos, P0502947.
Außerdem Selbstplagiat, V. M. veröffentlichte diese Aufgabe bereits 2008 in Schachmatnaja Komposizija (ganz leicht modifiziert, siehe P1080117).
Die Angabe "harmonie 2009" ist unsicher, denn sie wurde dem Hopper Magazine 03 (Zweitquelle) entnommen.
Keywords: Promotion (D), under-promotion (s), Entblockung, konsekutive Umwandlungen 2 (sD), Switchback (K), Antiphönix
Genre: h#
Computer test: Gustav 4.0 i
FEN: 7k/4p3/2p5/6p1/2P3P1/2P1p1p1/3bP1P1/5BRK
Reprints: OP024 The Hopper Magazine I03 12/2022
Input: Miguel Ambrona, 2023-07-26
Last update: A.Buchanan, 2023-07-27 more...
Genre: h#
Computer test: Gustav 4.0 i
FEN: 7k/4p3/2p5/6p1/2P3P1/2P1p1p1/3bP1P1/5BRK
Reprints: OP024 The Hopper Magazine I03 12/2022
Input: Miguel Ambrona, 2023-07-26
Last update: A.Buchanan, 2023-07-27 more...
1) 1. ... d8=D+ 2. Ke6+ Df6+ 3. Kd7 Dd4 4. De5 fxe5 5. Ke6 Dxd6#
2) 1. ... d8=S 2. Dg2 Sc6 3. Te8 Sxb4 4. Te6 Ka2 5. Dg6 Sd5#
3) 1. ... d8=T 2. Tg5 Tg8 3. Kg6+ Txg7+ 4. Kh6 Tg8 5. Lh5 fxg5#
2) 1. ... d8=S 2. Dg2 Sc6 3. Te8 Sxb4 4. Te6 Ka2 5. Dg6 Sd5#
3) 1. ... d8=T 2. Tg5 Tg8 3. Kg6+ Txg7+ 4. Kh6 Tg8 5. Lh5 fxg5#
Yuri Bilokin: you can see 1.c1=B d8=Q+ 2.Kg6+ Kb1 3.Rg5 Qxg5#
1.c1=B d8=Q+ 2.Kg6+ Kb1 3.Rh6 Qg5#
1.c1=B Kb1 2.Rh1 d8=Q+ 3.Kg6 Qg5#
1.c1=B Kb1 2.Rh2 d8=Q+ 3.Kg6 Qg5# (2023-08-04)
comment
1.c1=B d8=Q+ 2.Kg6+ Kb1 3.Rh6 Qg5#
1.c1=B Kb1 2.Rh1 d8=Q+ 3.Kg6 Qg5#
1.c1=B Kb1 2.Rh2 d8=Q+ 3.Kg6 Qg5# (2023-08-04)
comment
Keywords: Antiphönix (in 1), Pawn mate (in 3), Model mate (in 2+3), Promotion (D in 1), under-promotion (-,S,T), under-promotion key (in 2+3)
Genre: h#
Computer test: Popeye Windows-64Bit v4.85 (4096 KB)
FEN: r7/3P1bbp/3p1k2/3q1p1r/1n3P2/p7/2p5/K7
Input: Gunter Jordan, 2023-07-29
Last update: Gunter Jordan, 2023-07-29 more...
Genre: h#
Computer test: Popeye Windows-64Bit v4.85 (4096 KB)
FEN: r7/3P1bbp/3p1k2/3q1p1r/1n3P2/p7/2p5/K7
Input: Gunter Jordan, 2023-07-29
Last update: Gunter Jordan, 2023-07-29 more...
1) 1. Kxc2 d4 2. Lc1 d5 3. Te6 dxe6 4. Lb2 e7 5. Kc1 e8=D 6. c2 De1#
2) 1. Lc5+ Kxc3 2. Te5 Kb3 3. Kd2 Kb2 4. Ke3 Kc1 5. Kd4 Kd2 6. d5 c3#
2) 1. Lc5+ Kxc3 2. Te5 Kb3 3. Kd2 Kb2 4. Ke3 Kc1 5. Kd4 Kd2 6. d5 c3#
Keywords: Capture key (in 1), Entblockung, Promotion (D), Selfblock, Square vacation, Switchback (k), Klasinc theme (k-l), Model mate (in 1), Checking key (in 2), Ideal mate (in 2), Remote selfblock (in 2), Tschumakow Theme, Pawn mate (in 2)
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/8/3p4/8/1K6/2pPb3/2P5/2k1r3
Input: Felber, Volker, 2023-11-29
Last update: Gunter Jordan, 2024-03-26 more...
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/8/3p4/8/1K6/2pPb3/2P5/2k1r3
Input: Felber, Volker, 2023-11-29
Last update: Gunter Jordan, 2024-03-26 more...
1. a1=T e4 2. Ta2 e5 3. Ta1 exf6 4. Ta2 f7 5. Ta1 f8=D 6. Ta2 Dxf5#
Yuri Bilokin: you can see 1...e4 2.fxe4 Bc1 3.Ka1 Kc2 4.e3 Bb2#
1...e4 2.fxe4 Bc1 3.Ka1 Kc2 4.f5 Bb2#
1...Bc1 2.Ka1 e4 3.fxe4 Kc2 4.e3 Bb2#
1...Bc1 2.Ka1 e4 3.fxe4 Kc2 4.f5 Bb2# (2023-12-20)
comment
1...e4 2.fxe4 Bc1 3.Ka1 Kc2 4.f5 Bb2#
1...Bc1 2.Ka1 e4 3.fxe4 Kc2 4.e3 Bb2#
1...Bc1 2.Ka1 e4 3.fxe4 Kc2 4.f5 Bb2# (2023-12-20)
comment
Keywords: Pendulum (b/t), under-promotion (t), Promotion (D), konsekutive Umwandlungen 2 (tD), under-promotion key, Rückkehr nach Umwandlung
Genre: h#
Computer test: C+ Popeye 4.89
FEN: 8/8/5p2/p1p2p2/P1P2P2/PKP1P3/pB2P3/1k6
Input: Achim Schöneberg, 2023-12-12
Last update: Gunter Jordan, 2023-12-13 more...
Genre: h#
Computer test: C+ Popeye 4.89
FEN: 8/8/5p2/p1p2p2/P1P2P2/PKP1P3/pB2P3/1k6
Input: Achim Schöneberg, 2023-12-12
Last update: Gunter Jordan, 2023-12-13 more...
54 - P1414321
Rolf Wiehagen
1858 Springaren 167 0972023
nach D. Elekes & G. Schweig
(3+5) C+
h#4
0.1.1...
Rolf Wiehagen
1858 Springaren 167 0972023
nach D. Elekes & G. Schweig
(3+5) C+
h#4
0.1.1...
1. ... f8=S (f8=~?) 2. De6 fxe6 3. Kh8 e7 4. Lxf8 exf8=D#
Veröffentlicht in der Rubrik "Samasaker och hugskott".
Vergleiche P0552543. (Autor)
Veröffentlicht in der Rubrik "Samasaker och hugskott".
Vergleiche P0552543. (Autor)
Yuri Bilokin: you can see 1.Qb1+ Kxb1 2.Bc1 Ka1 3.Kh8 f8=Q#
1.Qb1+ Kxb1 2.Bc1 Kxc1 3.Kh8 f8=Q#
1.Qb1+ Kxb1 2.Bc1 Ka2 3.Kh8 f8=Q#
1.Qb1+ Kxb1 2.Bc1 Kc2 3.Kh8 f8=Q# (2024-01-13)
comment
1.Qb1+ Kxb1 2.Bc1 Kxc1 3.Kh8 f8=Q#
1.Qb1+ Kxb1 2.Bc1 Ka2 3.Kh8 f8=Q#
1.Qb1+ Kxb1 2.Bc1 Kc2 3.Kh8 f8=Q# (2024-01-13)
comment
Keywords: Corner mate, Model mate, Phoenix (DS), Schnoebelen Theme (S), Promotion (D), Promotion in the mating move, under-promotion (S), under-promotion key
Genre: h#
FEN: 8/5Pkp/5p1b/5P2/8/1q6/8/K7
Input: Gunter Jordan, 2024-01-02
Last update: Gunter Jordan, 2024-01-02 more...
Genre: h#
FEN: 8/5Pkp/5p1b/5P2/8/1q6/8/K7
Input: Gunter Jordan, 2024-01-02
Last update: Gunter Jordan, 2024-01-02 more...
1. Kg8 Kg2 2. Kh8 Kh3 3. g4+ Kxg4 4. Kg8 Kh3! (K~?) 5. Kh8 g4 6. Kg8 g5 7. hxg5 h6 8. gxh6 g7 9. Kh7 gxf8=D 10. Kg6 Df7#
PAS-Läufer-Hilfsmatt
PAS-Läufer-Hilfsmatt
Yuri Bilokin: you can see 1...Kg2 2.Kg8 Kh3 3.g4+ Kxg4 4.Kh8 Kh3 5.Kg8 g4 6.Kh8 g5 7.hxg5 h6 8.gxh6 g7+ 9.Kh7 gxf8=Q 10.Kg6 Qf7# (2024-01-03)
comment
comment
Keywords: changed blocks, Entblockung, Epaulettenmatt, Remote selfblock, Model mate, Switchback (Kk), Promotion (D), x pieces move to one square
Genre: h#
Computer test: Popeye Windows-64Bit v4.85 (4096 KB)
FEN: 5b1k/4p1p1/4P1Pp/6pP/4p3/4P1P1/8/5K2
Input: Gunter Jordan, 2024-01-03
Last update: Gunter Jordan, 2024-01-03 more...
Genre: h#
Computer test: Popeye Windows-64Bit v4.85 (4096 KB)
FEN: 5b1k/4p1p1/4P1Pp/6pP/4p3/4P1P1/8/5K2
Input: Gunter Jordan, 2024-01-03
Last update: Gunter Jordan, 2024-01-03 more...
56 - P1414428
Mirko Degenkolbe
3154 Problemskak 65, p. 14, 01/2024
nach Zlatko Mihajloski
(2+7)
h#6
2.1...
Mirko Degenkolbe
3154 Problemskak 65, p. 14, 01/2024
nach Zlatko Mihajloski
(2+7)
h#6
2.1...
1) 1. Sf2 g4 2. Se4 g5 3. Sf6 gxf6 4. Ta6 f7 5. Ka5 f8=D 6. Ta4 Dc5#
2) 1. Se5 g4 2. Sf7 g5 3. Sh6 gxh6 4. Ta2 h7 5. Ka3 h8=D 6. Ta4 Dc3#
Versuche:
1. Sf6+??
1. Sh6+??
2) 1. Se5 g4 2. Sf7 g5 3. Sh6 gxh6 4. Ta2 h7 5. Ka3 h8=D 6. Ta4 Dc3#
Versuche:
1. Sf6+??
1. Sh6+??
(MDxxxx)
Mihajloski = P1412069
Herstellung von Echo-Mustermatts mittels entgegengesetzten Umnow-Bahnungen.
Wenn man die Verführungen mit einbezieht, sehen wir zwei virtuelle Rundläufe des schwarzen Springers. Außerdem zweimal weißer Excelsior.
Henrik Juel: C+ Popeye 4.61 also (2024-01-11)
comment
Mihajloski = P1412069
Herstellung von Echo-Mustermatts mittels entgegengesetzten Umnow-Bahnungen.
Wenn man die Verführungen mit einbezieht, sehen wir zwei virtuelle Rundläufe des schwarzen Springers. Außerdem zweimal weißer Excelsior.
Henrik Juel: C+ Popeye 4.61 also (2024-01-11)
comment
Keywords: Minimal (B), Promotion (D), Excelsior white, Epaulettenmatt, Echo, Selfblock, Model mate, Umnov, Square vacation, Entblockung
Genre: h#
Computer test: Gustav 4.1d
FEN: 6K1/3p4/6p1/r7/k5n1/r3p3/6P1/8
Input: Felber, Volker, 2024-01-11
Last update: Gunter Jordan, 2024-01-11 more...
Genre: h#
Computer test: Gustav 4.1d
FEN: 6K1/3p4/6p1/r7/k5n1/r3p3/6P1/8
Input: Felber, Volker, 2024-01-11
Last update: Gunter Jordan, 2024-01-11 more...
a) 1. ... c3 2. Lg1 c4 3. f2 c5 4. f1=T c6 5. Tf8 c7 6. Tb8 c8=D 7. Sd4 Da6#
b) 1. ... c8=D+ 2. Db8 Dd8 3. Lg3 Kg1 4. Lc7 Kf2 5. Kb7 Ke3 6. Kc6 Ke4 7. Db6 Dd5#
b) 1. ... c8=D+ 2. Db8 Dd8 3. Lg3 Kg1 4. Lc7 Kf2 5. Kb7 Ke3 6. Kc6 Ke4 7. Db6 Dd5#
Yuri Bilokin: you can see a) 1.Bg1 c4 2.f2 c5 3.f1=R c6 4.Rf8 c7 5.Rb8 c8=Q 6.Sd4 Qa6#
1.Bg1 c3 2.f2 c4 3.f1=R c5 4.Rf7 c6 5.Ra7 c7 6.Sf4 c8=Q#
1.Bg1 c3 2.f2 c4 3.f1=R c5 4.Rf7 c6 5.Ra7 c7 6.Qe1 c8=Q#
1.Bg1 c3 2.f2 c4 3.f1=R c5 4.Rf7 c6 5.Ra7 c7 6.Qf2 c8=Q#
b) 1.Qg5 Kxh2 2.Qc5 Kh3 3.Kb7 Kg4 4.Kc6 c8=Q+ 5.Kd5 Kf5 6.Bd4 Qxe6#
1.Qg5 Kxh2 2.Kb7 Kh3 3.Qc5 Kg4 4.Kc6 c8=Q+ 5.Kd5 Kf5 6.Bd4 Qxe6#
1.Qg7 Kxh2 2.Kb7 Kh3 3.Kc6 c8=Q+ 4.Qc7 Kg4 5.Bc5 Kf5 6.Kd6 Qxe6#
1.Qg7 Kxh2 2.Kb7 Kh3 3.Kc6 c8=Q+ 4.Qc7 Kg4 5.Kd6 Kf5 6.Bc5 Qxe6# (2024-01-30)
comment
1.Bg1 c3 2.f2 c4 3.f1=R c5 4.Rf7 c6 5.Ra7 c7 6.Sf4 c8=Q#
1.Bg1 c3 2.f2 c4 3.f1=R c5 4.Rf7 c6 5.Ra7 c7 6.Qe1 c8=Q#
1.Bg1 c3 2.f2 c4 3.f1=R c5 4.Rf7 c6 5.Ra7 c7 6.Qf2 c8=Q#
b) 1.Qg5 Kxh2 2.Qc5 Kh3 3.Kb7 Kg4 4.Kc6 c8=Q+ 5.Kd5 Kf5 6.Bd4 Qxe6#
1.Qg5 Kxh2 2.Kb7 Kh3 3.Qc5 Kg4 4.Kc6 c8=Q+ 5.Kd5 Kf5 6.Bd4 Qxe6#
1.Qg7 Kxh2 2.Kb7 Kh3 3.Kc6 c8=Q+ 4.Qc7 Kg4 5.Bc5 Kf5 6.Kd6 Qxe6#
1.Qg7 Kxh2 2.Kb7 Kh3 3.Kc6 c8=Q+ 4.Qc7 Kg4 5.Kd6 Kf5 6.Bc5 Qxe6# (2024-01-30)
comment
Keywords: Corner mate (in a), Epaulettenmatt (in b), Excelsior white (in a), Remote selfblock (in b), konsekutive Umwandlungen 2 (tD in a), Line closing (in a), Model mate, Checking key (in b), Selfblock, Promotion (D), Promotion key (in b), under-promotion (t in a), Minimal
Genre: h#
Computer test: Popeye Windows-64Bit v4.85 (4096 KB)
FEN: k7/8/4n3/8/8/5pq1/2P2b1n/3b3K
Input: Gunter Jordan, 2024-01-11
Last update: Gunter Jordan, 2024-01-11 more...
Genre: h#
Computer test: Popeye Windows-64Bit v4.85 (4096 KB)
FEN: k7/8/4n3/8/8/5pq1/2P2b1n/3b3K
Input: Gunter Jordan, 2024-01-11
Last update: Gunter Jordan, 2024-01-11 more...
Versuch: 1. ... d6? 2. Kf4 Ke2 3. Ld2 ?? 4. Te1+ Kxf2 5. Te7 dxe7 6. Kg4 e8=D 7. Kh3 Dh5#
1. ... Kc1 2. Kf3 Kb1 3. Ld2+ Ka2 4. Te1 d6 5. Te7 dxe7 6. Lf4 e8=D 7. Lg3 De4#
1. ... Kc1 2. Kf3 Kb1 3. Ld2+ Ka2 4. Te1 d6 5. Te7 dxe7 6. Lf4 e8=D 7. Lg3 De4#
Yuri Bilokin: you can see 1.Kf4 d6 2.Bd2+ Ke2 3.Re1+ Kxf2 4.Re7 dxe7 5.Kg4 e8=Q 6.Kh3 Qh5#
1.Kf4 Ke2 2.Bd2 d6 3.Re1+ Kxf2 4.Re7 dxe7 5.Kg4 e8=Q 6.Kh3 Qh5# (2024-01-30)
comment
1.Kf4 Ke2 2.Bd2 d6 3.Re1+ Kxf2 4.Re7 dxe7 5.Kg4 e8=Q 6.Kh3 Qh5# (2024-01-30)
comment
1. Sd7 Ke2 2. Se5 fxe5 3. Kg5 e6 4. Kf4 e7 5. g5 e8=D 6. g4 De3#
Gunter Jordan: Ohne die beiden Bauern g2,g3 und mit wKd1 entsteht eine Miniatur mit Idealmatt bei gleicher Lösung. (2024-01-15)
Ladislav Packa: Gunter: The king should stand on c1, otherwise the dual is 1...Kd2, Ke2. (2024-01-15)
Gunter Jordan: ... o.k. Ladislav, c1 ist richtig. War ein Schreibfehler. Danke für den Hinweis. (2024-01-18)
comment
Ladislav Packa: Gunter: The king should stand on c1, otherwise the dual is 1...Kd2, Ke2. (2024-01-15)
Gunter Jordan: ... o.k. Ladislav, c1 ist richtig. War ein Schreibfehler. Danke für den Hinweis. (2024-01-18)
comment
Keywords: Attack number, Entblockung, Selfblock, Promotion (D)
Genre: h#
FEN: 5n2/6p1/6k1/5p2/5P2/6p1/6P1/5K2
Input: Gunter Jordan, 2024-01-14
Last update: Gunter Jordan, 2024-01-15 more...
Genre: h#
FEN: 5n2/6p1/6k1/5p2/5P2/6p1/6P1/5K2
Input: Gunter Jordan, 2024-01-14
Last update: Gunter Jordan, 2024-01-15 more...
1) 1. Sd2 Txh2 2. Lf1 Tg2 3. Dh2 Txf2+ 4. Kg3 Txe2 5. Sf3+ Kxf1 6. Dg2+ Txg2#
2) 1. Txh1 a4 2. Lh2 a5 3. Sg1 a6 4. Dh3 a7 5. Kg3 a8=D 6. f3 Db8#
2) 1. Txh1 a4 2. Lh2 a5 3. Sg1 a6 4. Dh3 a7 5. Kg3 a8=D 6. f3 Db8#
Yuri Bilokin: you can see 1...Rxg1 2.Bh1 Rxf1 3.Kg2 Rxf2+ 4.Kg1 Kxe2 5.Rg2 Ke1 6.Qh2 Rf1#
1...Rxh2 2.Sd2 Rxg2 3.Qh2 Rxf2+ 4.Kg3 Rxe2 5.Sf3+ Kf1 6.Qg2+ Rxg2#
1...Rxh2 2.Sd2 Rxg2 3.Qh2 Rxf2+ 4.Kg3 Kxe2 5.Sf3 Kf1 6.Qg2+ Rxg2# (2024-01-30)
comment
1...Rxh2 2.Sd2 Rxg2 3.Qh2 Rxf2+ 4.Kg3 Rxe2 5.Sf3+ Kf1 6.Qg2+ Rxg2#
1...Rxh2 2.Sd2 Rxg2 3.Qh2 Rxf2+ 4.Kg3 Kxe2 5.Sf3 Kf1 6.Qg2+ Rxg2# (2024-01-30)
comment
Keywords: Excelsior white (in 2), Castling position, Selfblock, Umnov, Promotion (D in 2)
Genre: h#
FEN: 8/8/8/6p1/4pppp/P3pkqn/4prbr/4KnbR
Input: Gunter Jordan, 2024-01-20
Last update: Gunter Jordan, 2024-01-20 more...
Genre: h#
FEN: 8/8/8/6p1/4pppp/P3pkqn/4prbr/4KnbR
Input: Gunter Jordan, 2024-01-20
Last update: Gunter Jordan, 2024-01-20 more...
a) 1. ... c4 2. Dh2+ Kxh2 3. g3+ Kxh3 4. g2 c5 5. g1=T c6 6. Ta1 c7 7. Ta7 c8=D#
b) 1. ... c8=S 2. La7 Sd6 3. Sb6 Sxe8 4. Lg2+ Kg1 5. Lb7+ Kf1 6. Sd7 Ke2 7. Sb8 Sc7#
b) 1. ... c8=S 2. La7 Sd6 3. Sb6 Sxe8 4. Lg2+ Kg1 5. Lb7+ Kf1 6. Sd7 Ke2 7. Sb8 Sc7#
Yuri Bilokin: you can see b) 1.Ba7 c8=S 2.Sb6 Sd6 3.Bg2+ Kg1 4.Bb7+ Kf1 5.Sd7 Sxe8 6.Sb8 Sc7#
1.Sb4 c8=S 2.Sa6 Sd6 3.Sb8 Sc4 4.Sd6 Sa3 5.Sb7 Sb5 6.Ba7 Sc7#
1.Sb4 c8=S 2.Sa6 Sd6 3.Sb8 Sc4 4.Sd6 Sa3 5.Ba7 Sb5 6.Sb7 Sc7#
1.Sb4 c8=S 2.Sa6 Sd6 3.Sb8 Sc4 4.Sd6 Se3 5.Sb7 Sd5 6.Ba7 Sc7# (2024-02-18)
comment
1.Sb4 c8=S 2.Sa6 Sd6 3.Sb8 Sc4 4.Sd6 Sa3 5.Sb7 Sb5 6.Ba7 Sc7#
1.Sb4 c8=S 2.Sa6 Sd6 3.Sb8 Sc4 4.Sd6 Sa3 5.Ba7 Sb5 6.Sb7 Sc7#
1.Sb4 c8=S 2.Sa6 Sd6 3.Sb8 Sc4 4.Sd6 Se3 5.Sb7 Sd5 6.Ba7 Sc7# (2024-02-18)
comment
Keywords: Corner mate, Smothered mate (in b), Excelsior white (in a), konsekutive Umwandlungen 2 (tD), Minimal (B), Model mate, Phoenix (in a), Rückkehr nach Umwandlung (in b), Round Trip (B/S in b), Selfblock, Promotion (D in a), Promotion in the mating move (in a), Exchange of Promotions, under-promotion (t,S), under-promotion key (in b)
Genre: h#
FEN: k3n3/8/8/3n4/5pp1/6qb/2P2b2/7K
Input: Gunter Jordan, 2024-02-13
Last update: Gunter Jordan, 2024-02-13 more...
Genre: h#
FEN: k3n3/8/8/3n4/5pp1/6qb/2P2b2/7K
Input: Gunter Jordan, 2024-02-13
Last update: Gunter Jordan, 2024-02-13 more...
62 - P1415229
Claus Grupen
Andreas Thoma
2015/60 Schach in Schleswig-Holstein 11/2015
2. ehrende Erwähnung
(2+5) C+
h#6
b) sLc2
Claus Grupen
Andreas Thoma
2015/60 Schach in Schleswig-Holstein 11/2015
2. ehrende Erwähnung
(2+5) C+
h#6
b) sLc2
a) 1. c1=L h4 2. Tg2 h5 3. Tg6 h6 4. Lg5 h7 5. Kg7 Kb2 6. Kh6 h8=D#
b) 1. Lg6 h4 2. Tf2 h5 3. Tf7 Kb2 4. a1=T hxg6 5. Th1 gxf7 6. Th7 f8=D#
b) 1. Lg6 h4 2. Tf2 h5 3. Tf7 Kb2 4. a1=T hxg6 5. Th1 gxf7 6. Th7 f8=D#
Keywords: corner to corner (t in b), Corner mate (in b), Entblockung, Excelsior white, Remote selfblock (2-fach in a), Forsberg Twins, konsekutive Umwandlungen 2 (lD,tD), Line opening, Minimal (B), Model mate, Selfblock (in b), Tempo Loss (in a), Tschumakow Theme, Promotion (D), Promotion in the mating move, under-promotion (l,t), under-promotion key (in a), Miniature
Genre: h#
Computer test: Gustav 4.1 a
FEN: 7k/8/8/8/8/1p6/prp4P/K7
Input: Gunter Jordan, 2024-02-21
Last update: Gunter Jordan, 2024-02-21 more...
Genre: h#
Computer test: Gustav 4.1 a
FEN: 7k/8/8/8/8/1p6/prp4P/K7
Input: Gunter Jordan, 2024-02-21
Last update: Gunter Jordan, 2024-02-21 more...
1. Kf1-e2 c4-c5 2. Ke2-d3 c5-c6 3. Kd3-c4 c6-c7 4. Kc4-b5 c7-c8=D 5. Kb5-a5 Dc8-a6#
Keywords: Minimal (B), Kindergarten Problem, Miniature, Promotion (D)
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/K7/8/8/1pP5/p7/8/5k2
Input: Felber, Volker, 2024-04-02
Last update: Gunter Jordan, 2024-04-02 more...
Genre: h#
Computer test: Popeye WINDOWS98-32Bit-Version 3.77 (8192KB)
FEN: 8/K7/8/8/1pP5/p7/8/5k2
Input: Felber, Volker, 2024-04-02
Last update: Gunter Jordan, 2024-04-02 more...
Show statistic for complete result. Show search result faster by using ids.
https://pdb.dieschwalbe.de/search.jsp?expression=COMMENTDATE%3E%3D20220810+AND+G%3D%27h%23%27+AND+NOT+A%3D%27Pitk%C3%A4nen%2C+Jorma%27+AND+K%3D%27Umwandlung%27
The problems of this query have been registered by the following contributors:
Gerd Wilts (4)hpr (5)
Erich Bartel (2)
Hans-Jürgen Schäfer (8)
HBae (1)
Henri Nouguier (1)
Gunter Jordan (21)
Felber, Volker (11)
Erich Bartel (1)
Mario Richter (1)
Rainer Staudte (1)
Marcin Banaszek (3)
A.Buchanan (1)
Eberhard Schulze (1)
Miguel Ambrona (1)
Achim Schöneberg (1)
The black pawn a4(on the diagramm)came from c7 after capturing the white bishop on b6, then moving to b5 and capturing a white officer on a4. (2010-09-28)
A.Buchanan: I think sBf must have captured on g2 & then promoted on f1 to S, otherwise it’s impossible to unlock the southeast cage (2023-07-11)
A.Buchanan: So that means it’s sound. However I don’t see that it needs to have been sL shielding on f1 (2023-07-11)
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