906 problem(s) found in 4005 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT A='Stjopotschkin, Anatoli W.' AND S='The Problemist'] [download as LaTeX]
Keywords: Impostor (tt), Which are original men? (tt), Interchange (tt)
Genre: Retro
FEN: r1bbn2r/Kpp3p1/3pppp1/7B/5kP1/1P6/pPPPPP1P/NQB5
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-24 more...
Genre: Retro
FEN: r1bbn2r/Kpp3p1/3pppp1/7B/5kP1/1P6/pPPPPP1P/NQB5
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-24 more...
1. b3 h5 2. La3 h4 3. Dc1 h3 4. Kd1 hxg2 5. h4 e5 6. h5 e4 7. h6 e3 8. h7 exf2 9. e4 f5 10. Se2 g1=L 11. Lg2 f1=L 12. e5 Lb6 13. d4 Lfc5 14. dxc5 f4 15. cxb6 f3 16. Df4 f2 17. Sc1 La6 18. Lc5 f1=L 19. a3 Lfb5 20. c4 Se7 21. cxb5 Tg8 22. hxg8=L g5 23. Lc4 d5 24. bxa6 Lf5 25. e6 Sc8 26. e7 Kf7 27. e8=L+ Kg8 28. Leb5 c6 29. Ta2 cxb5 30. Tc2 Sc6 31. Sa2 dxc4+
paul: Correction of P0001580 (2010-09-25)
Silvio Baier: Mit 30....dc+ ist das C+ (Euclide 0.98, Retractor für den letzten Zug) (2011-07-08)
James Malcom: So this is fully C+? (2021-01-27)
Henrik Juel: No, another problem in 30.0 moves ending with 30. Tc2 dxc4+ is tested OK (2021-01-27)
more ...
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Silvio Baier: Mit 30....dc+ ist das C+ (Euclide 0.98, Retractor für den letzten Zug) (2011-07-08)
James Malcom: So this is fully C+? (2021-01-27)
Henrik Juel: No, another problem in 30.0 moves ending with 30. Tc2 dxc4+ is tested OK (2021-01-27)
more ...
comment
Keywords: Ceriani-Frolkin Theme (lllLL), Unique Proof Game, Promotion (lllLL)
Genre: Retro
Computer test: Computerprüfung: C+ Stelvio 1.11 65:30:51 Stunden. (hh:mm:ss) Keine Lösung: BP 30.0, BP 30.5.
FEN: r1nq2k1/pp6/PPn5/1pB2bp1/2p2Q2/PP6/N1R3B1/1N1K3R
Reprints: 580 Ukrainisches Album 1986-1990
26 Shortest Proof Games 11/1991
H25 FIDE Album 1986-1988 1995
(H25) Die Schwalbe 156, p. 237, 12/1995
feenschach 141 06/2001
Thema Danicum 105 2002
(C) Quartz 22 10-12/2002
H2 The Problemist 07/2010
(18) Die Schwalbe 250A 08/2011
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-08 more...
Genre: Retro
Computer test: Computerprüfung: C+ Stelvio 1.11 65:30:51 Stunden. (hh:mm:ss) Keine Lösung: BP 30.0, BP 30.5.
FEN: r1nq2k1/pp6/PPn5/1pB2bp1/2p2Q2/PP6/N1R3B1/1N1K3R
Reprints: 580 Ukrainisches Album 1986-1990
26 Shortest Proof Games 11/1991
H25 FIDE Album 1986-1988 1995
(H25) Die Schwalbe 156, p. 237, 12/1995
feenschach 141 06/2001
Thema Danicum 105 2002
(C) Quartz 22 10-12/2002
H2 The Problemist 07/2010
(18) Die Schwalbe 250A 08/2011
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-08 more...
1. h4 a5 2. h5 Ta6 3. Th4 Tb6 4. Sh3 Tb3 5. cxb3 h6 6. Dc2 Th7 7. Dxh7 Sc6 8. d3 Se5 9. Le3 Sg6 10. Sd2 Sh8 11. Tc1 g6 12. Tc5 Lg7 13. Te5 Lf6 14. Lc5 Lg5 15. e3 Lf4 16. Le2 Lg3 17. Lg4 Lh2 18. f3 Lf4 19. Kf2 Lg5 20. Kg3 Lf6 21. Sf2 Lg7 22. Lh3 Lf8 23. Tg4
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 1 Sekunde.
Keine Lösung: BP 21.5, BP 22.0. (2023-02-17)
comment
Keine Lösung: BP 21.5, BP 22.0. (2023-02-17)
comment
Keywords: Unique Proof Game, Tempo Loss (l11)
Genre: Retro
Computer test: Stelvio 1.0
FEN: 2bqkbnn/1ppppp1Q/6pp/p1B1R2P/6R1/1P1PPPKB/PP1N1NP1/8
Reprints: 62 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Silvio Baier, 2023-03-02 more...
Genre: Retro
Computer test: Stelvio 1.0
FEN: 2bqkbnn/1ppppp1Q/6pp/p1B1R2P/6R1/1P1PPPKB/PP1N1NP1/8
Reprints: 62 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Silvio Baier, 2023-03-02 more...
1. a4 Sf6 2. Ta3 Sh5 3. Td3 f6 4. Td6 Kf7 5. d3 Kg6 6. Le3 Kf5 7. Ld4 Kg5 8. e3 Kg6 9. Dg4+ Kf7 10. De6+ Ke8 11. g4 Sc6 12. Lg2 Sa5 13. Lxb7 Tb8 14. Lxc8 Tb3 15. Ke2 Ta3 16. Kf3 Ta1 17. Sa3 Td1 18. Se2 Td2 19. Tb1 a6 20. Sc1 Te2
Cook: 1. d3 b5 2. Le3 b4 3. Ld4 b3 4. axb3 Sf6 5. Ta6 Sh5 6. Td6 Sc6 7. e3 Sa5 8. Dg4 Tb8 9. De6 Tb4 10. g4 Ta4 11. Ke2 Ta1 12. Sa3 Td1 13. Lg2 La6 14. Lb7 Lc4 15. Lc8 Lb5 16. Kf3 La4 17. bxa4 Td2 18. Se2 f6 19. Tb1 a6 20. Sc1 Te2
1. d3 Sc6 2. Le3 Tb8 3. Ld4 Sf6 4. e3 Sh5 5. Dg4 b5 6. Ke2 La6 7. De6 b4 8. g4 Lb5 9. Lg2 b3 10. axb3 f6 11. Ta6 La4 12. bxa4 Sa5 13. Td6 Tb3 14. Lb7 Ta3 15. Kf3 Ta1 16. Sa3 Td1 17. Se2 Td2 18. Tb1 a6 19. Sc1 Te2 20. Lc8
Cook: 1. d3 b5 2. Le3 b4 3. Ld4 b3 4. axb3 Sf6 5. Ta6 Sh5 6. Td6 Sc6 7. e3 Sa5 8. Dg4 Tb8 9. De6 Tb4 10. g4 Ta4 11. Ke2 Ta1 12. Sa3 Td1 13. Lg2 La6 14. Lb7 Lc4 15. Lc8 Lb5 16. Kf3 La4 17. bxa4 Td2 18. Se2 f6 19. Tb1 a6 20. Sc1 Te2
1. d3 Sc6 2. Le3 Tb8 3. Ld4 Sf6 4. e3 Sh5 5. Dg4 b5 6. Ke2 La6 7. De6 b4 8. g4 Lb5 9. Lg2 b3 10. axb3 f6 11. Ta6 La4 12. bxa4 Sa5 13. Td6 Tb3 14. Lb7 Ta3 15. Kf3 Ta1 16. Sa3 Td1 17. Se2 Td2 18. Tb1 a6 19. Sc1 Te2 20. Lc8
Keywords: Unique Proof Game
Genre: Retro
Computer test: Moldenhauer: Cooked Stelvio 1.11:13 Keine Lösung BP 19.0, BP 19.5 Cooked
FEN: 2Bqkb1r/2ppp1pp/p2RQp2/n6n/P2B2P1/N2PPK2/1PP1rP1P/1RN5
Reprints: 168 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-03-06 more...
Genre: Retro
Computer test: Moldenhauer: Cooked Stelvio 1.11:13 Keine Lösung BP 19.0, BP 19.5 Cooked
FEN: 2Bqkb1r/2ppp1pp/p2RQp2/n6n/P2B2P1/N2PPK2/1PP1rP1P/1RN5
Reprints: 168 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-03-06 more...
1. dxe3ep 2. fxg1=L 3. 0-0 Tg4#
1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt
1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt
Henrik Juel: If Black may castle, his latest move must have been e3xf2, so last move was e2-e4. (Orig. wPg2 promoted on g8, so g7xh6 happened early). 1.dxe3 e.p. 2.fxg1=B 3.0-0 Tg4# (2003-03-21)
James Malcom: Is this the only know one-sided Valladao in a regular series-mover helpmate? Also, can wPh2 be shifted to h4, and the White king moved as well to h2, so that the solution must have 2. fxg1=S? (2020-12-08)
James Malcom: Well, the only known one that is in minimum form, i.e. 3 moves, I mean. (2020-12-08)
A.Buchanan: This is nice. I don't know of any other AP ser-h#3 with Valladao, but there could be. Just been cleansing PDB data today. Yes I think wKh3 and wBh2 can be transposed, and then promote to S instead of L. (2020-12-08)
A.Buchanan: The legality of intermediate positions in seriesmovers is of course irrelevant. However, each proof game that leads to the diagram position with BTM can be appended with a black move and then a "flip" that changes the player to move. We can repeat this process, and then the fact that Black retains castling rights implies that they had castling rights in the diagram position. I think this is the right formal way to handle AP with seriesmovers. (2022-03-21)
comment
James Malcom: Is this the only know one-sided Valladao in a regular series-mover helpmate? Also, can wPh2 be shifted to h4, and the White king moved as well to h2, so that the solution must have 2. fxg1=S? (2020-12-08)
James Malcom: Well, the only known one that is in minimum form, i.e. 3 moves, I mean. (2020-12-08)
A.Buchanan: This is nice. I don't know of any other AP ser-h#3 with Valladao, but there could be. Just been cleansing PDB data today. Yes I think wKh3 and wBh2 can be transposed, and then promote to S instead of L. (2020-12-08)
A.Buchanan: The legality of intermediate positions in seriesmovers is of course irrelevant. However, each proof game that leads to the diagram position with BTM can be appended with a black move and then a "flip" that changes the player to move. We can repeat this process, and then the fact that Black retains castling rights implies that they had castling rights in the diagram position. I think this is the right formal way to handle AP with seriesmovers. (2022-03-21)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sk), Seriesmover, Promotion, Valladao Task
Genre: Retro, Fairies
Computer test: C+ Popeye v4.85 + thinking
FEN: 4k2r/1pp2p1p/1N5p/3P4/2RpP2N/P2P1P1K/p1BP1p1P/B3b1R1
Reprints: Die Schwalbe 143 10/1993
Rex Multiplex 45-46 05/1994
(XII) Quartz 5 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
Genre: Retro, Fairies
Computer test: C+ Popeye v4.85 + thinking
FEN: 4k2r/1pp2p1p/1N5p/3P4/2RpP2N/P2P1P1K/p1BP1p1P/B3b1R1
Reprints: Die Schwalbe 143 10/1993
Rex Multiplex 45-46 05/1994
(XII) Quartz 5 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
1. hxg3ep Sc1 2. gxf2 Se3#
Welches ist die Originalquelle? Oder wurde es zweimal als Urdruck gebracht?
vgl. P0003291
Henrik Juel: White pawns captured all four missing black men
Last black move was not b7xPa6 ([Ta8]) nor e7-e6 ([Lf8]), so it was Lb7-c8, which is possible only if last white move was g2-g4
C+ Popeye 4.61 and above analysis (2021-04-23)
comment
vgl. P0003291
Henrik Juel: White pawns captured all four missing black men
Last black move was not b7xPa6 ([Ta8]) nor e7-e6 ([Lf8]), so it was Lb7-c8, which is possible only if last white move was g2-g4
C+ Popeye 4.61 and above analysis (2021-04-23)
comment
Keywords: En passant as key
Genre: h#, Retro
FEN: 2b5/p2p2pp/p3p2b/6Pr/5BPp/2P5/NRP1PP1P/3Nrk1K
Reprints: F84 The Problemist 05/1970
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-04-23 more...
Genre: h#, Retro
FEN: 2b5/p2p2pp/p3p2b/6Pr/5BPp/2P5/NRP1PP1P/3Nrk1K
Reprints: F84 The Problemist 05/1970
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-04-23 more...
7 - P0003417
John Frederick Keeble
2206 The Problemist Fairy Chess Supplement 16, p. 173, 02/1936
(4+5) C+
h#2.5 (AP)
John Frederick Keeble
2206 The Problemist Fairy Chess Supplement 16, p. 173, 02/1936
(4+5) C+
h#2.5 (AP)
1. ... exd6ep 2. 0-0-0 dxe7+ 3. Tf8 exf8=T,D#
H.Juel: If Black may castle, last move was not made by Ke8 or Ta8, but by Pd5, so it was d7-d5, and hence White may capture en passant. 1... exd6ep 2.0-0-0 dxe7 3.Tf8 exf8=D,T# where the castling (belatedly) justifies the e.p. capture.
Not 1... exd6ep 2.Kd8? dxe7 3.Kc8 e8=D,T
AB: Another 120 retro tries (=60x2 due to tolerated dual promotion) of general form 1. exd6ep 2. T~ dxc7 3. T~ c8=D,T.
No set play
H.Juel: If Black may castle, last move was not made by Ke8 or Ta8, but by Pd5, so it was d7-d5, and hence White may capture en passant. 1... exd6ep 2.0-0-0 dxe7 3.Tf8 exf8=D,T# where the castling (belatedly) justifies the e.p. capture.
Not 1... exd6ep 2.Kd8? dxe7 3.Kc8 e8=D,T
AB: Another 120 retro tries (=60x2 due to tolerated dual promotion) of general form 1. exd6ep 2. T~ dxc7 3. T~ c8=D,T.
No set play
Nachdruck in "Die Schwalbe" 33 mit Diagrammfehler (wBe6 fehlt).
VL: The first AP-problem! Thematic illegal try:
1. ... exd6ep 2.Kd8? dxe7+ 3.Kc8 e8=T(D)#?? (2007-02-11)
Mario Richter: 46 years between this Problem and P1366663 - perhaps this was not Keeble's first AP-Problem?! (2019-08-29)
Ladislav Packa: Of course I know the idea a posteriori. But let my heresy be forgiven - I do not consider it logical. In this particular example, I consider e.p. for proof that black can castling - not vice versa. (2019-08-29)
A.Buchanan: Hi Ladislav. If ep is ok, then Black just moved a pawn with double hop. That doesn't stop Black having e.g. moved bK right at the beginning of the game. So you can't prove that Black definitely can castle this way. What am I missing here? (2019-08-29)
Ladislav Packa: Your statement is perfectly fine. However, the AP condition is usually interpreted as castling is evidence of e.p. However, the second party does not have to do castling - the possibility has already been demonstrated by the existence of e.p. (2019-08-29)
A.Buchanan: Hi again Ladislav. Please try again, I am obviously being very stupid. I don't understand how 1. exd6ep is any more indicative that Black can castle than 1. K~. (2019-08-30)
Ladislav Packa: Hi, Andre!
e.p. is just a result of the d7-d5 move. Black can castle only if it was the last move of black (even without wPe5). Therefore, it does not have the logic of joining the e.p. with castling. For example, ask yourself if black is allowed to castle in the position without wPe5. (2019-08-30)
A.Buchanan: @Ladislav: ep is not "proof that black can castle", rather it is a *necessary* condition for Black to be able to castle. The logical implication is the other way round. We can castle because of the optimistic convention, and because sPd might just have moved. Without wPe, there is no solution, but whatever White plays, Black can castle (2020-12-11)
Hans-Jürgen Manthey: was heißt hier Thematisch Illegal ??
Mir völlig wurscht ob ep möglich oder nicht, nach dem weißen Zug ist natürlich Kd8 möglich !!
also ist auch 1. ... exd6ep 2. Kd8 dxe7+ 3. Kc8 e8DT# eine Lösung, sonst müßte auf die erzwungende Rochade in der Aufgabenstellung hingewiesen werden ! (2020-12-11)
A.Buchanan: Can one tolerate “tolerated dual promotion” in Valladao? Well Valladao himself did, so it must be ok (2022-03-26)
more ...
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VL: The first AP-problem! Thematic illegal try:
1. ... exd6ep 2.Kd8? dxe7+ 3.Kc8 e8=T(D)#?? (2007-02-11)
Mario Richter: 46 years between this Problem and P1366663 - perhaps this was not Keeble's first AP-Problem?! (2019-08-29)
Ladislav Packa: Of course I know the idea a posteriori. But let my heresy be forgiven - I do not consider it logical. In this particular example, I consider e.p. for proof that black can castling - not vice versa. (2019-08-29)
A.Buchanan: Hi Ladislav. If ep is ok, then Black just moved a pawn with double hop. That doesn't stop Black having e.g. moved bK right at the beginning of the game. So you can't prove that Black definitely can castle this way. What am I missing here? (2019-08-29)
Ladislav Packa: Your statement is perfectly fine. However, the AP condition is usually interpreted as castling is evidence of e.p. However, the second party does not have to do castling - the possibility has already been demonstrated by the existence of e.p. (2019-08-29)
A.Buchanan: Hi again Ladislav. Please try again, I am obviously being very stupid. I don't understand how 1. exd6ep is any more indicative that Black can castle than 1. K~. (2019-08-30)
Ladislav Packa: Hi, Andre!
e.p. is just a result of the d7-d5 move. Black can castle only if it was the last move of black (even without wPe5). Therefore, it does not have the logic of joining the e.p. with castling. For example, ask yourself if black is allowed to castle in the position without wPe5. (2019-08-30)
A.Buchanan: @Ladislav: ep is not "proof that black can castle", rather it is a *necessary* condition for Black to be able to castle. The logical implication is the other way round. We can castle because of the optimistic convention, and because sPd might just have moved. Without wPe, there is no solution, but whatever White plays, Black can castle (2020-12-11)
Hans-Jürgen Manthey: was heißt hier Thematisch Illegal ??
Mir völlig wurscht ob ep möglich oder nicht, nach dem weißen Zug ist natürlich Kd8 möglich !!
also ist auch 1. ... exd6ep 2. Kd8 dxe7+ 3. Kc8 e8DT# eine Lösung, sonst müßte auf die erzwungende Rochade in der Aufgabenstellung hingewiesen werden ! (2020-12-11)
A.Buchanan: Can one tolerate “tolerated dual promotion” in Valladao? Well Valladao himself did, so it must be ok (2022-03-26)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sg), Promotion (D/T), Tolerated dual promotion (D/T), Valladao Task, Retro Strategy (RS)
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: r3k3/2p1p3/2P1P3/2KpP3/8/8/8/8
Reprints: V. Die Schwalbe 33, p. 323, 06/1975
2001 CHM Themes 12/2000
3 Die Schwalbe 215, p. 240, 10/2005
A1 Problemkiste 169 02/2007
402 Eigenartige Schachprobleme , p. 130, 2010
14 Die Schwalbe 241, p. 374, 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-10 more...
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: r3k3/2p1p3/2P1P3/2KpP3/8/8/8/8
Reprints: V. Die Schwalbe 33, p. 323, 06/1975
2001 CHM Themes 12/2000
3 Die Schwalbe 215, p. 240, 10/2005
A1 Problemkiste 169 02/2007
402 Eigenartige Schachprobleme , p. 130, 2010
14 Die Schwalbe 241, p. 374, 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-10 more...
a) 1. Kf8 0-0-0 2. Te8 Tf1#
b) 1. Sd2 Le5 2. 0-0-0 Tb8#
In a&b have the same two forward solutions, one with White Castling the other with Black. So it must be retro concerns that eliminate one solution in each case. bRg8 is original in any case, as Black has 7 pawns, and bBa5 is promoted.
a) bhP must promote on g1 This allows for 4 captures of bPb3, but there is no chance for cross-capturing of b b&c pawns. White g and h pawn cannot reach Black cage to promote inside. So Black castling cannot be preserved. But bB once promoted can easily escape, so White capturing is unimpeded.
b) bhP can make 1 capture to promote, but now cannot escape without kicking wK, so White can't capture. Cross-capturing is still not possible, but wbP can promote on c8 without disrupting bK.
This is not PRA but twinned RS.
b) 1. Sd2 Le5 2. 0-0-0 Tb8#
In a&b have the same two forward solutions, one with White Castling the other with Black. So it must be retro concerns that eliminate one solution in each case. bRg8 is original in any case, as Black has 7 pawns, and bBa5 is promoted.
a) bhP must promote on g1 This allows for 4 captures of bPb3, but there is no chance for cross-capturing of b b&c pawns. White g and h pawn cannot reach Black cage to promote inside. So Black castling cannot be preserved. But bB once promoted can easily escape, so White capturing is unimpeded.
b) bhP can make 1 capture to promote, but now cannot escape without kicking wK, so White can't capture. Cross-capturing is still not possible, but wbP can promote on c8 without disrupting bK.
This is not PRA but twinned RS.
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
comment
comment
1. La7 Kxa7 2. c8=T Ka6 3. Ta8#
Max J. Meyer in 'Brighton and Hove Society', 1904: "It will be seen that the ideas of these two problems [dieses und P1143870] are the same. The addition of the White B. enables Mr. Mackenzie to get a much better key for this position, a point in which Steinweg's problem does not excel; but, of course, the use of an extra piece renders the miniature less remarkable from the point of view of smallness in size."
Paulo Roque: .
Solution:
1) 1. La7 KxLa7 2. c8=T! Ka6 3. Ta8#
2) Proof da legality of the answer:
Retro: 0... Ka7-a8 -1. b7-b8=L etc (2009-09-06)
Sally: Der letzte weiße Zug von Weiß kann nur Ba7xb8L gewesen sein, demgemäß enthält die Aufgabe 2 Bauernumwandlungen
in verschiedenen Figuren: in der lösung in einem turm und retrospektiv in einen Läufer.
406 Mattaufgaben mit 3 und 4 Steinen (Teil 1) Speckmann 1976 (2016-08-30)
Henrik Juel: There is even a third promotion in the threat 2.c8=D+
C+ by Popeye 4.61 (2016-08-30)
A.Buchanan: Here the threat play (1) is sound & (2) can never occur as Black has no neutral move. But if threat play is *unsound* yet can never occur, can it be safely ignored or will some directmate purists regard that as a defect? (2022-04-02)
Henrik Juel: I think not, Andrew
When I test problems, I ignore plays that do not occur (2022-04-02)
comment
Paulo Roque: .
Solution:
1) 1. La7 KxLa7 2. c8=T! Ka6 3. Ta8#
2) Proof da legality of the answer:
Retro: 0... Ka7-a8 -1. b7-b8=L etc (2009-09-06)
Sally: Der letzte weiße Zug von Weiß kann nur Ba7xb8L gewesen sein, demgemäß enthält die Aufgabe 2 Bauernumwandlungen
in verschiedenen Figuren: in der lösung in einem turm und retrospektiv in einen Läufer.
406 Mattaufgaben mit 3 und 4 Steinen (Teil 1) Speckmann 1976 (2016-08-30)
Henrik Juel: There is even a third promotion in the threat 2.c8=D+
C+ by Popeye 4.61 (2016-08-30)
A.Buchanan: Here the threat play (1) is sound & (2) can never occur as Black has no neutral move. But if threat play is *unsound* yet can never occur, can it be safely ignored or will some directmate purists regard that as a defect? (2022-04-02)
Henrik Juel: I think not, Andrew
When I test problems, I ignore plays that do not occur (2022-04-02)
comment
Keywords: under-promotion (T), Stalemate avoidance, Promotion in the retro play (L)
Genre: 3#, Retro
Computer test: Popeye 4.61
FEN: kB6/2P5/2K5/8/8/8/8/8
Reprints: 64 Chess Lyrics , p. 100, 1905
1359 Wiener Hausfrauen-Zeitung 5, p. 77, 29/01/1905
5 Vliegend Blaadje 22/07/1911
Szachy 01/1958
The Problemist (20) 03/1969
406 Mattaufgaben mit drei und vier Steinen 1976
diagrammes 64 01-02/1984
(1) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2020-02-26 more...
Genre: 3#, Retro
Computer test: Popeye 4.61
FEN: kB6/2P5/2K5/8/8/8/8/8
Reprints: 64 Chess Lyrics , p. 100, 1905
1359 Wiener Hausfrauen-Zeitung 5, p. 77, 29/01/1905
5 Vliegend Blaadje 22/07/1911
Szachy 01/1958
The Problemist (20) 03/1969
406 Mattaufgaben mit drei und vier Steinen 1976
diagrammes 64 01-02/1984
(1) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2020-02-26 more...
1. ... ... 2. 0-0?
1. ... Kxg2 2. e5#
1. ... Kxf4 2. Df6#
1. ... Kxg2 2. e5#
1. ... Kxf4 2. Df6#
Keywords: Castling (wk), No legal last move for Black, Castling in the forward play, Fabel-Opus (82)
Genre: Retro
Computer test: HC+ Popeye 4.61 with analysis
FEN: 8/8/2Q5/8/4PBN1/3P1k2/6R1/4K2R
Reprints: (IV) Problem 17-18 08/1952
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-08 more...
Genre: Retro
Computer test: HC+ Popeye 4.61 with analysis
FEN: 8/8/2Q5/8/4PBN1/3P1k2/6R1/4K2R
Reprints: (IV) Problem 17-18 08/1952
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-08 more...
1. h4 g5 2. hxg5 a5 3. b4 axb4 4. Th4 Ta3 5. Sf3 Te3 6. a4 Sf6 7. gxf6 e5 8. Tc4 h5 9. a5 h4 10. a6 h3 11. a7 h2 12. a8=L h1=L 13. Ta6 Th6 14. dxe3 Le7 15. Te6 fxe6 16. La3 b3 17. Lc5 Sc6 18. Sd4 exd4 19. Sc3 dxc3 20. g4 Se5 21. Lg2 b6 22. Lac6 dxc6 23. fxe7 Dd3 24. Ld5 exd5 25. cxd3 Lf5 26. gxf5 Te6 27. fxe6 Le4 28. dxe4 bxc5 29. f4 dxc4 30. Dc2 bxc2 31. fxe5
Juha Saukkola:: Solution not unique! (2000-09-29)
Moldenhauer: Computerprüfung: C+ NUPG Stelvio 1.11 00:02:12 Minuten. (hh:mm:ss)
Keine Lösung: BP 29.5, BP 30.0. (2023-04-24)
Moldenhauer: Beispiel Stelvio 1.11: 1.Sc3 Sc6 2.b3 Sf6 3.La3 a5 4.g4 a4 5.Lg2 Ta5 6.h4 Tf5
7.Lc5 axb3 8.Ld5 b6 9.a4 bxc5 10.gxf5 g5 11.hxg5 e5 12.Th4 h5 13.Tc4 h4
14.a5 h3 15.a6 h2 16.a7 h1D 17.Ta6 De4 18.Sf3 Le7 19.Sd4 exd4 20.a8D Se5
21.Te6 dxc3 22.Dc6 dxc6 23.d4 fxe6 24.dxe5 exd5 25.gxf6 Le6
26.fxe6 dxc4 27.fxe7 Ddd3 28.cxd3 Th3 29.Dc2 Te3 30.dxe4 bxc2 31.fxe3 (2023-04-24)
comment
Moldenhauer: Computerprüfung: C+ NUPG Stelvio 1.11 00:02:12 Minuten. (hh:mm:ss)
Keine Lösung: BP 29.5, BP 30.0. (2023-04-24)
Moldenhauer: Beispiel Stelvio 1.11: 1.Sc3 Sc6 2.b3 Sf6 3.La3 a5 4.g4 a4 5.Lg2 Ta5 6.h4 Tf5
7.Lc5 axb3 8.Ld5 b6 9.a4 bxc5 10.gxf5 g5 11.hxg5 e5 12.Th4 h5 13.Tc4 h4
14.a5 h3 15.a6 h2 16.a7 h1D 17.Ta6 De4 18.Sf3 Le7 19.Sd4 exd4 20.a8D Se5
21.Te6 dxc3 22.Dc6 dxc6 23.d4 fxe6 24.dxe5 exd5 25.gxf6 Le6
26.fxe6 dxc4 27.fxe7 Ddd3 28.cxd3 Th3 29.Dc2 Te3 30.dxe4 bxc2 31.fxe3 (2023-04-24)
comment
Keywords: Non-Unique Proof Game, Promotion
Genre: Retro
FEN: 4k3/2p1P3/2p1P3/2p1P3/2p1P3/2p1P3/2p1P3/4K3
Reprints: The Problemist 11/1993
Input: Gerd Wilts, 1995-06-26
Last update: A.Buchanan, 2012-04-10 more...
Genre: Retro
FEN: 4k3/2p1P3/2p1P3/2p1P3/2p1P3/2p1P3/2p1P3/4K3
Reprints: The Problemist 11/1993
Input: Gerd Wilts, 1995-06-26
Last update: A.Buchanan, 2012-04-10 more...
1. Sf3 a5 2. Sd4 a4 3. Sb3 axb3 4. Sc3 bxc2 5. Se4 cxd1=L 6. Sg5 Lb3 7. Sxh7 Lc4 8. Sg5 Th5 9. Sh7 Tha5 10. Kd1 d5 11. Kc2 Kd7 12. Kc3 Ke6 13. Kd4 Kf5 14. Ke3 Le6 15. Kf3 Sd7 16. Kg3 Tc8 17. Kh4 g5+ 18. Kh3 Lg7 19. Kg3 Le5+ 20. Kf3 Ld6 21. Ke3 Kg4 22. Kd4 c5+ 23. Kc3 Db6 24. Kc2 Db4 25. Kd1 b5 26. Ke1 Dxd2+
paul: Correction of P1001275 (2010-05-08)
A.Buchanan: Does it have to end in a check which is in some sense a defect if unnecessary. Can it be 24. … Db3+ 25. Kd1 b5 26. Ke1 b4 (2023-06-13)
Joost de Heer: 24... Qb3+ 25. Kd1 is illegal. (2023-06-13)
A.Buchanan: Well clearly 24... Qxb2+ then same idea (2023-06-13)
comment
A.Buchanan: Does it have to end in a check which is in some sense a defect if unnecessary. Can it be 24. … Db3+ 25. Kd1 b5 26. Ke1 b4 (2023-06-13)
Joost de Heer: 24... Qb3+ 25. Kd1 is illegal. (2023-06-13)
A.Buchanan: Well clearly 24... Qxb2+ then same idea (2023-06-13)
comment
Keywords: Unique Proof Game, Promenade (K), Non-standard material, Tempo Loss (K17)
Genre: Retro
Computer test: Stelvio 1.33, 1min
FEN: 2r3n1/3npp1N/3bb3/rppp2p1/2b3k1/8/PP1qPPPP/R1B1KB1R
Reprints: 3388 U.S. Problem Bulletin 102/103 07-10/1995
Input: Gerd Wilts, 1995-06-26
Last update: Reto, 2023-06-12 more...
Genre: Retro
Computer test: Stelvio 1.33, 1min
FEN: 2r3n1/3npp1N/3bb3/rppp2p1/2b3k1/8/PP1qPPPP/R1B1KB1R
Reprints: 3388 U.S. Problem Bulletin 102/103 07-10/1995
Input: Gerd Wilts, 1995-06-26
Last update: Reto, 2023-06-12 more...
a) 1. 0-0-0? illegal
1. Td1! droht 2. Dc4#
1. ... Se3 2. Dxe3#
1. ... Se5 2. Dxe5#
1. ... cxb5 2. Txb5#
1. ... Tf4 2. Dxe7#
1. ... Dxc2 2. Txc2#
1. ... Dd3 2. Sxd3#
1. ... De4+ 2. Sxe4#
b) 1. Td1? Te8! pinning
1. ... Te7? 2. Dxe7#
1. 0-0-0! then all variations as in a)
(a) sLa7 is promoted. If White castling rights remain, then bPf captured fxgxh2xg1=L, which with dxc6 makes 4 captures. White has lost Bd, Be, Bf/h & Lc. However, wBe can't have captured or promoted (h/fxg & Lf at home are the only Black casualties), so can't have contributed to Black captures. Therefore impossible, and wK did lose castling rights before. Now trivial for sBf to have promoted.
(b) sLa7 can be original, so castling is still ok. Even if sLa7 is promoted, wBe can have promoted, and again White can still castle.
1. Td1! droht 2. Dc4#
1. ... Se3 2. Dxe3#
1. ... Se5 2. Dxe5#
1. ... cxb5 2. Txb5#
1. ... Tf4 2. Dxe7#
1. ... Dxc2 2. Txc2#
1. ... Dd3 2. Sxd3#
1. ... De4+ 2. Sxe4#
b) 1. Td1? Te8! pinning
1. ... Te7? 2. Dxe7#
1. 0-0-0! then all variations as in a)
(a) sLa7 is promoted. If White castling rights remain, then bPf captured fxgxh2xg1=L, which with dxc6 makes 4 captures. White has lost Bd, Be, Bf/h & Lc. However, wBe can't have captured or promoted (h/fxg & Lf at home are the only Black casualties), so can't have contributed to Black captures. Therefore impossible, and wK did lose castling rights before. Now trivial for sBf to have promoted.
(b) sLa7 can be original, so castling is still ok. Even if sLa7 is promoted, wBe can have promoted, and again White can still castle.
14 - P0008879
Arpad Molnar
R259 The Problemist 01/1997
(10+9) cooked
Welches waren die letzten 16 Einzelzüge?
Arpad Molnar
R259 The Problemist 01/1997
(10+9) cooked
Welches waren die letzten 16 Einzelzüge?
Cook: R: 1. a4xSb3+ Lb8-c7 2. d6xSc5 Ld8-b6 3. b6xSa5 e7xTd8=L
Keywords: Last Moves? (16), Non-standard material
Genre: Retro
FEN: 8/p1Bp1ppp/RBR5/pBpP4/1kP5/Rp6/1PK5/8
Input: Gerd Wilts, 1998-06-26
Last update: Gerd Wilts, 2006-10-08 more...
Genre: Retro
FEN: 8/p1Bp1ppp/RBR5/pBpP4/1kP5/Rp6/1PK5/8
Input: Gerd Wilts, 1998-06-26
Last update: Gerd Wilts, 2006-10-08 more...
15 - P0501007
Thomas Brand
Jörg A. R. Kuhlmann
Bo Lindgren
Arno Tüngler
Kjell Widlert
3707v feenschach 63 12/1982
(1+8) cooked
h#8
Circe (nur Offiziere) Rex inklusive
Thomas Brand
Jörg A. R. Kuhlmann
Bo Lindgren
Arno Tüngler
Kjell Widlert
3707v feenschach 63 12/1982
(1+8) cooked
h#8
Circe (nur Offiziere) Rex inklusive
1. b1=S Kb2 2. a1=T Kxa1[+sTh8] 3. Te8 Kxb1[+sSg8] 4. c1=L Kxc1[+sLf8] 5. Ke1 Kc2 6. Se7 Kd3 7. f1=D+ Ke3 8. De2+ Kxe2[+sDd8]#
Korrektur in feenschach 1988
Cook: Michel Caillaud (2022-07-19): cooked using Jacobi 0.7.5 :
1. a1=L Kb3 2. b1=T Ka2 3. c1=D Kxa1[+sLf8] 4. Sb6 Kxb1[+sTa8] 5. Te8 Kxc1[+sDd8]+ 6. Ke1 Kc2 7. Sd5 Kd3 8. Se7+ Ke2#
Korrektur in feenschach 1988
Cook: Michel Caillaud (2022-07-19): cooked using Jacobi 0.7.5 :
1. a1=L Kb3 2. b1=T Ka2 3. c1=D Kxa1[+sLf8] 4. Sb6 Kxb1[+sTa8] 5. Te8 Kxc1[+sDd8]+ 6. Ke1 Kc2 7. Sd5 Kd3 8. Se7+ Ke2#
Michel Caillaud: cooked using Jacobi 0.7.5 :
1.a1=L Kb3 2.b1=T Ka2 3.c1=D Kxa1(Lf8) 4.Sb6 Kxb1(Ta8) 5.Te8 Kxc1(Dd8)+ 6.Ke1 Kc2 7.Sd5 Kd3 8.Se7+ Ke2# (2022-07-19)
more ...
comment
1.a1=L Kb3 2.b1=T Ka2 3.c1=D Kxa1(Lf8) 4.Sb6 Kxb1(Ta8) 5.Te8 Kxc1(Dd8)+ 6.Ke1 Kc2 7.Sd5 Kd3 8.Se7+ Ke2# (2022-07-19)
more ...
comment
Keywords: Circe (Rex inklusive), Allumwandlung
Genre: Fairies
FEN: n7/8/4pp2/8/8/2K5/pppk1p2/8
Reprints: The Problemist , p. 91, 09/1985
Rex Multiplex 12/1988
969 Minimalkunst im Schach 2006
My problems [Arno Tüngler] , p. 82, 06/2009
Input: Gerd Wilts, 1996-06-06
Last update: Mario Richter, 2022-07-19 more...
Genre: Fairies
FEN: n7/8/4pp2/8/8/2K5/pppk1p2/8
Reprints: The Problemist , p. 91, 09/1985
Rex Multiplex 12/1988
969 Minimalkunst im Schach 2006
My problems [Arno Tüngler] , p. 82, 06/2009
Input: Gerd Wilts, 1996-06-06
Last update: Mario Richter, 2022-07-19 more...
1. f3 Kb8 2. f2 Ka8 3. f1=L Kb8 4. Ld3 Ka8 5. Lb1 Kb8 6. La2 Ka8 7. Db1 Kb8 8. f5 Ka8 9. f4 Kb8 10. f3 Ka8 11. f2 Kb8 12. f1=L Ka8 13. Tf2 Kb8 14. Kf7 Ka8 15. Ke6 Kb8 16. Kd5 Ka8 17. Kc4 Kb8 18. Kc3 Ka8 19. Kb2 Kb8 20. Ka1 Ka8 21. Tb2 Kb8 22. T8f2 Ka8 23. Lf5 Kb8 24. Lc2 Ka8 25. d3 Kb8 26. Le3 Kc7 27. e5 Kxd6 28. Lc1 Ke6 29. Td2 Kf5 30. e4 Kg4 31. Le2+ Kxh3 32. e3 Kg2 33. Led1+ Kf1 34. e2+ Ke1=
Keywords: Rex solus (w), Superseded by (P0501023), Move Length Record
Genre: Fairies
FEN: K1b2rk1/4pr2/1b1n1p2/5q2/3p1p2/1p5n/8/8
Reprints: The Problemist 07/2008
Input: Gerd Wilts, 1996-06-06
Last update: James Malcom, 2022-04-27 more...
Genre: Fairies
FEN: K1b2rk1/4pr2/1b1n1p2/5q2/3p1p2/1p5n/8/8
Reprints: The Problemist 07/2008
Input: Gerd Wilts, 1996-06-06
Last update: James Malcom, 2022-04-27 more...
17 - P0501066
Henry Olof Axel Forsberg
599-603 Revista Romana de Sah 01/1935
1. Preis
Pauly Gedenkturnier 1935
(3+2) C+
h#2
b) sTa6 statt sDa6
c) sLa6 statt sDa6
d) sSa6 statt sDa6
e) sBa6 statt sDa6
Henry Olof Axel Forsberg
599-603 Revista Romana de Sah 01/1935
1. Preis
Pauly Gedenkturnier 1935
(3+2) C+
h#2
b) sTa6 statt sDa6
c) sLa6 statt sDa6
d) sSa6 statt sDa6
e) sBa6 statt sDa6
a) 1. Df6 Sc5 2. Db2 Ta4#
b) 1. Tb6 Tb1 2. Tb3 Ta1#
c) 1. Lc4 Se1 2. La2 Sc2#
d) 1. Sc5 Sc1 2. Sa4 Tb3#
e) 1. a5 Tb3+ 2. Ka4 Sc5#
b) 1. Tb6 Tb1 2. Tb3 Ta1#
c) 1. Lc4 Se1 2. La2 Sc2#
d) 1. Sc5 Sc1 2. Sa4 Tb3#
e) 1. a5 Tb3+ 2. Ka4 Sc5#
Das namensgebende [Stamm-]Problem für dieses Thema (=Forsberg-Zwilling).
Erich Bartel: weitere Nachdrucke:
13) 197 moving on 2008.---
14) Naef:Endspielstudien Hilfmattprobleme und Märchenschachaufgaben 1998,S.72.--- (2008-10-01)
Jacques Rotenberg: 5982, CHM 1995, G. Lestriguel
p.1315, Le Guide des Echecs 1993
5 p.8, The Modern Helpmate in Two, Z. Janevski & N. Stolev 1989
XIII p.151, Thèmes-64 10 (avr. 58)
XII bis p.38, Rex Multiplex 2 (avr. 82)
285, Svenska Miniatyrer I, A.Uddgren & A. Hildebrand 1996
p.76, Im Wunderland des Schachproblems, E. Ramin 1958
E-1 p.236, StrateGems 12 (oct. 00)
I33 p.1333, Phénix 14 (sept. 91)
3 p.219, Le Genre Aidé Supplément diagrammes 15 (juin 75)
161 p.241, Echecs, N. Ramini 1985
E p.14, Best Problems 21 (jan. 02)
259 p.238, Chess Problems : Introduction to an Art, M. Lipton, R.C.O. Matthews, J. Rice 1963
139, Skakopgaven 1942
135 p.31, Album FIDE 1914-44/III
9 p.403, feenschach 26 (déc. 74)
40 p.61, Europe Echecs 330 (juin 86)
p.185, feenschach 47 (juil. 79)
in WinChloe, you find also :
A p.5, Springaren 17 (mars 84)
p.50, feenschach 73 (jan. 85)
M p.171, The Problemist (sept. 81)
p.415, feenschach 97 (nov. 90)
133 p.104, Solving in Style, J. Nunn 1985
E-I p.43, Probleemblad 58/1 (jan. 00)
E-1 p.102, Probleemblad 58/3 (mai 00)
p.136, Springaren 70 (août 97)
4 p.4424, diagrammes 159 (oct. 06)
p.37, Problem-Forum 36 (déc. 08) (2012-03-10)
A.Buchanan: The position is unique (except for left-right reflection). In particular, it cannot be shifted up or down without breaking (b) and only the current location of wK makes (b) sound and (a)&(d) solvable. (2017-01-27)
Anton Baumann: in Revista de Sah 1936 soll Henry Forsberg eine weitere Möglichkeit gezeigt haben:
f) wSa6 statt sDa6: 1.Ka2 Sac5 2.Ka3 Ta4# (C+)
Ferner mit Märchenschachfiguren: vergl. P1349128 (2022-12-28)
more ...
comment
Erich Bartel: weitere Nachdrucke:
13) 197 moving on 2008.---
14) Naef:Endspielstudien Hilfmattprobleme und Märchenschachaufgaben 1998,S.72.--- (2008-10-01)
Jacques Rotenberg: 5982, CHM 1995, G. Lestriguel
p.1315, Le Guide des Echecs 1993
5 p.8, The Modern Helpmate in Two, Z. Janevski & N. Stolev 1989
XIII p.151, Thèmes-64 10 (avr. 58)
XII bis p.38, Rex Multiplex 2 (avr. 82)
285, Svenska Miniatyrer I, A.Uddgren & A. Hildebrand 1996
p.76, Im Wunderland des Schachproblems, E. Ramin 1958
E-1 p.236, StrateGems 12 (oct. 00)
I33 p.1333, Phénix 14 (sept. 91)
3 p.219, Le Genre Aidé Supplément diagrammes 15 (juin 75)
161 p.241, Echecs, N. Ramini 1985
E p.14, Best Problems 21 (jan. 02)
259 p.238, Chess Problems : Introduction to an Art, M. Lipton, R.C.O. Matthews, J. Rice 1963
139, Skakopgaven 1942
135 p.31, Album FIDE 1914-44/III
9 p.403, feenschach 26 (déc. 74)
40 p.61, Europe Echecs 330 (juin 86)
p.185, feenschach 47 (juil. 79)
in WinChloe, you find also :
A p.5, Springaren 17 (mars 84)
p.50, feenschach 73 (jan. 85)
M p.171, The Problemist (sept. 81)
p.415, feenschach 97 (nov. 90)
133 p.104, Solving in Style, J. Nunn 1985
E-I p.43, Probleemblad 58/1 (jan. 00)
E-1 p.102, Probleemblad 58/3 (mai 00)
p.136, Springaren 70 (août 97)
4 p.4424, diagrammes 159 (oct. 06)
p.37, Problem-Forum 36 (déc. 08) (2012-03-10)
A.Buchanan: The position is unique (except for left-right reflection). In particular, it cannot be shifted up or down without breaking (b) and only the current location of wK makes (b) sound and (a)&(d) solvable. (2017-01-27)
Anton Baumann: in Revista de Sah 1936 soll Henry Forsberg eine weitere Möglichkeit gezeigt haben:
f) wSa6 statt sDa6: 1.Ka2 Sac5 2.Ka3 Ta4# (C+)
Ferner mit Märchenschachfiguren: vergl. P1349128 (2022-12-28)
more ...
comment
Keywords: Forsberg Twins, Aristocrat, Model mate
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/q7/8/1R4K1/k2N4/8/8
Reprints: 6 The Chess Review , p. 25, 12/1945
141 Zwillinge und Mehrlinge 1952
Schach-Echo Vol.11, p. 151, 1953
5 The Problemist 01/1962
761 Szachy 06/1965
Problemschach [Sidler] , p. 140, 1968
Problem 127-132, p. 95, 09/1969
9) feenschach 26, p. 403, 12/1974
135 FIDE Album 1914-1944/III 1975
83 Schach ohne Partner [Grasemann] 1977
feenschach 47, p. 185, 07-09/1979
XII Rex Multiplex 2, p. 38, 04-06/1982
67 100 Classics of the Chessboard 1983
Belfort , p. 75, 1994
Tim Krabbé's Website (57) 2000
E-I Probleemblad 01-02/2000
harmonie 63 09/2000
König & Turm 4, p. 13, 03/2001
986 Minimalkunst im Schach 2006
Problemkiste 163 02/2006
633 Encyclopedia of Chess Problems 2012
Input: Gerd Wilts, 1996-06-06
Last update: Dieter Berlin, 2022-02-09 more...
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/q7/8/1R4K1/k2N4/8/8
Reprints: 6 The Chess Review , p. 25, 12/1945
141 Zwillinge und Mehrlinge 1952
Schach-Echo Vol.11, p. 151, 1953
5 The Problemist 01/1962
761 Szachy 06/1965
Problemschach [Sidler] , p. 140, 1968
Problem 127-132, p. 95, 09/1969
9) feenschach 26, p. 403, 12/1974
135 FIDE Album 1914-1944/III 1975
83 Schach ohne Partner [Grasemann] 1977
feenschach 47, p. 185, 07-09/1979
XII Rex Multiplex 2, p. 38, 04-06/1982
67 100 Classics of the Chessboard 1983
Belfort , p. 75, 1994
Tim Krabbé's Website (57) 2000
E-I Probleemblad 01-02/2000
harmonie 63 09/2000
König & Turm 4, p. 13, 03/2001
986 Minimalkunst im Schach 2006
Problemkiste 163 02/2006
633 Encyclopedia of Chess Problems 2012
Input: Gerd Wilts, 1996-06-06
Last update: Dieter Berlin, 2022-02-09 more...
1. Dxf4 0-0-0 2. Sf2 Kb1 3. Sd2+ Ka1 4. Td4 Te1#
Keywords: Interchange (KT (16)), Castling (wg), Tempo Move (K)
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/2r5/3b4/r1n1pP2/pp1pkp2/6pq/R3K1bn
Input: hpr, 1996-06-14
Last update: Alfred Pfeiffer, 2018-12-03 more...
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: 8/8/2r5/3b4/r1n1pP2/pp1pkp2/6pq/R3K1bn
Input: hpr, 1996-06-14
Last update: Alfred Pfeiffer, 2018-12-03 more...
1. Tc2 Th1 2. Kb2 Th8 3. Kc1 Txa8 4. Sb2 Ta1#
Erich Bartel: Nachdruck:
1) V35) Problemkiste (125) X 1999.--- (2007-07-18)
Yuri Bilokin: Areal cycle (wR, with captures, 4)
Corner-to-corner (wR) × 4
Four corners (wR (2023-05-08)
comment
1) V35) Problemkiste (125) X 1999.--- (2007-07-18)
Yuri Bilokin: Areal cycle (wR, with captures, 4)
Corner-to-corner (wR) × 4
Four corners (wR (2023-05-08)
comment
Keywords: Pure Round Trip (D)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: r7/6K1/8/5q2/n3b3/1k6/r2p4/R7
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-13 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: r7/6K1/8/5q2/n3b3/1k6/r2p4/R7
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-13 more...
20 - P0503616
Johann Christoffel van Gool
Jacobus Theodorus Sanderse
876b The Problemist 01/1983
(4+5) C+
h#4
2.1...
Johann Christoffel van Gool
Jacobus Theodorus Sanderse
876b The Problemist 01/1983
(4+5) C+
h#4
2.1...
1) 1. Sc3 Th1 2. Kc2 Th3 3. Kd1 Txc3 4. Ke1 Tc1#
2) 1. Kxb3 Ta1 2. Kc2 Txa2+ 3. Kd1 Tc2 4. Ke1 Tc1#
2) 1. Kxb3 Ta1 2. Kc2 Txa2+ 3. Kd1 Tc2 4. Ke1 Tc1#
Keywords: Pure Round Trip (T)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/8/8/1p6/1P6/nk1npPK1/2R5
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-13 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/8/8/1p6/1P6/nk1npPK1/2R5
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 1999-06-13 more...
1. Kb7 Lh3 2. Kc6 Lg4 3. Kd5 Lh3 4. Kc4 d3+ 5. Kxd3 Lg4 6. Kc3 Lh3 7. d3 Lg4 8. d2 Lh3 9. d1=L Lg4 10. Lb3 Lh3 11. Le6 fxe6 12. Kxb4 e7 13. Ka5 e8=D 14. Ka6 Da8#
Keywords: Pure Round Trip (k), Promotion
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/kp3p1p/1p3P1P/1P1p1pB1/5Pp1/3P2Pp/7K
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-13 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/8/kp3p1p/1p3P1P/1P1p1pB1/5Pp1/3P2Pp/7K
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-13 more...
22 - P0503912
Bernt Gösta Ahlgren
Christer Jonsson
Anders Lundström
616 The Problemist 09/1978
1. Preis
(3+9) C+
h#4
2.1...
Bernt Gösta Ahlgren
Christer Jonsson
Anders Lundström
616 The Problemist 09/1978
1. Preis
(3+9) C+
h#4
2.1...
1) 1. c2 Lxc2 2. Lc3 Lb3 3. Ke4 La2 4. Kd3 Lb1#
2) 1. f2 Lxf2 2. Lf3 Lg3+ 3. Ke3 Lh2 4. Df4 Lg1#
2) 1. f2 Lxf2 2. Lf3 Lg3+ 3. Ke3 Lh2 4. Df4 Lg1#
Keywords: Pure Round Trip (L4,L4)
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/K1p5/8/b7/2q2k2/2p2p2/3pp3/1B4Bb
Reprints: 608 FIDE Album 1977-1979
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: A.Buchanan, 2021-08-01 more...
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/K1p5/8/b7/2q2k2/2p2p2/3pp3/1B4Bb
Reprints: 608 FIDE Album 1977-1979
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: A.Buchanan, 2021-08-01 more...
23 - P0509780
Wolfgang Heidenfeld
352 The Problemist Fairy Chess Supplement 02/1932
T. R. Dawson gewidmet
(8+13) C+
Kürzestes h#? (wer?)
Wolfgang Heidenfeld
352 The Problemist Fairy Chess Supplement 02/1932
T. R. Dawson gewidmet
(8+13) C+
Kürzestes h#? (wer?)
1. ... Lxe1 2. Sd1 Lb4#
7 Black pawn captures: QRRBSSPP
2 White pawn captures: QSP
If White just moved, then d2xe3, and wBc1 retro blocked. This means all captures accounted for, but how did h pawns pass?
So, WTM.
There are numerous BWBW and BWB mates, but there is only one WBW mate.
7 Black pawn captures: QRRBSSPP
2 White pawn captures: QSP
If White just moved, then d2xe3, and wBc1 retro blocked. This means all captures accounted for, but how did h pawns pass?
So, WTM.
There are numerous BWBW and BWB mates, but there is only one WBW mate.
A.Buchanan: Odd. As h#2 there are many solutions, not just the nice slow-rolling wPb one. And these are all eliminated for both retropat reasons and because they are longer than the actual one. Why is the stipulation not just h#2? (2021-10-20)
comment
comment
Keywords: Check Protection, Whose move?
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic
FEN: 8/2p5/2r5/Ppkp4/p1b5/P1n1PPp1/pPp2BP1/2K1b1r1
Reprints: (V) Problem 29-30 05/1955
Input: Markus Manhart, 1997-06-27
Last update: A.Buchanan, 2021-10-21 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic
FEN: 8/2p5/2r5/Ppkp4/p1b5/P1n1PPp1/pPp2BP1/2K1b1r1
Reprints: (V) Problem 29-30 05/1955
Input: Markus Manhart, 1997-06-27
Last update: A.Buchanan, 2021-10-21 more...
*1) 1. ... Lg8 2. Dh7 Lxh7#
*2) 1. ... Th4 2. Th5 Txh5#
1) 1. Txh5 e8=S 2. Tg5 Sg7#
2) 1. Dxh7 g3 2. Dg6 g4#
*2) 1. ... Th4 2. Th5 Txh5#
1) 1. Txh5 e8=S 2. Tg5 Sg7#
2) 1. Dxh7 g3 2. Dg6 g4#
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/4Pp1B/3PppqP/5krR/3K1p2/7P/6P1/8
Input: hpr, 1997-10-09
Last update: hpr, 1999-10-24 more...
1. Tg5 Kd2 2. Lf8 Se1 3. Tb5 Sc2 4. Lc5 Le6#
NL:
1. Td1+ Kf2 2. Kd3 Le6 3. Ld2 Kg3 4. Ke2 Lc4#
NL:
1. Td1+ Kf2 2. Kd3 Le6 3. Ld2 Kg3 4. Ke2 Lc4#
Klären Autor
Verst-4
Original ohne sBg6 abgedruckt, Korrektur(versuch) 05/1967, S. 140
Als Autor ist angegeben: D.W.A. Brotherton
Yuri Bilokin: correction +bPg3 8/8/6pb/3r4/2k3B1/6p1/6N1/4K3 (3+5) (2023-05-24)
comment
Verst-4
Original ohne sBg6 abgedruckt, Korrektur(versuch) 05/1967, S. 140
Als Autor ist angegeben: D.W.A. Brotherton
Yuri Bilokin: correction +bPg3 8/8/6pb/3r4/2k3B1/6p1/6N1/4K3 (3+5) (2023-05-24)
comment
26 - P0531421
Thomas R. Dawson
1796 The Problemist Fairy Chess Supplement 02/1935
(4+4) cooked
h#2
b) sTd7 nach e7, sTh3 nach h4
Thomas R. Dawson
1796 The Problemist Fairy Chess Supplement 02/1935
(4+4) cooked
h#2
b) sTd7 nach e7, sTh3 nach h4
a) 1. Kg5 Tg2+ 2. Kh4 Tg4#
b) 1. Kxg7 Tg2+ 2. Kf8 Tg8#
NL:
a) 1. Kg5 Tg2+ 2. Kh4 Shf5#
b) 1. Kxg7 Tg2+ 2. Kf8 Tg8#
NL:
a) 1. Kg5 Tg2+ 2. Kh4 Shf5#
klären Monat
Alfred Pfeiffer: No.1796 ist aus 04/1935 (vgl. P0003272) (2011-01-11)
Yuri Bilokin: possibly bBe4(-bNe4), +bPd3 8/3r2N1/6kN/8/4b3/3p3r/2R5/1K6 (4+5) (2022-12-16)
comment
Alfred Pfeiffer: No.1796 ist aus 04/1935 (vgl. P0003272) (2011-01-11)
Yuri Bilokin: possibly bBe4(-bNe4), +bPd3 8/3r2N1/6kN/8/4b3/3p3r/2R5/1K6 (4+5) (2022-12-16)
comment
Keywords: Sacrifice of white pieces, Symmetrical position, Aristocrat, Aristocrat
Genre: h#
FEN: 8/3r2N1/6kN/8/4n3/7r/2R5/1K6
Reprints: 1796 The Problemist Fairy Chess Supplement 04/1935
Input: Ralf Krätschmer, 1998-03-30
Last update: Arnold Beine, 2022-12-17 more...
Genre: h#
FEN: 8/3r2N1/6kN/8/4n3/7r/2R5/1K6
Reprints: 1796 The Problemist Fairy Chess Supplement 04/1935
Input: Ralf Krätschmer, 1998-03-30
Last update: Arnold Beine, 2022-12-17 more...
a) 1. Tc4 Td4 2. Kb4 Lxd2#
b) 1. Tb4 Tc7 2. Tc4 Le5#
NL:
b) 1. Kb4 Tc7 2. Th5 Lxd2#
b) 1. Tb4 Tc7 2. Tc4 Le5#
NL:
b) 1. Kb4 Tc7 2. Th5 Lxd2#
Daniel Novomesky: In b) position is needed to take away even black Pawn a3. So, a) Diagram b) Kwg4-g3 and -Pba3. Problem is correct. C+ (2007-10-26)
Yuri Bilokin: Correction: -bBb8 8/3R4/8/1pr5/3r1BK1/p1kp4/3p4/3B4 (4+7) h#2 b) wPa3-e4
a) 1.Rdc4 Rd4 2.Kb4 Bxd2# (MM)
b) 1.Rb4 Rc7 2.Rcc4 Be5# (MM) (2021-05-20)
comment
Yuri Bilokin: Correction: -bBb8 8/3R4/8/1pr5/3r1BK1/p1kp4/3p4/3B4 (4+7) h#2 b) wPa3-e4
a) 1.Rdc4 Rd4 2.Kb4 Bxd2# (MM)
b) 1.Rb4 Rc7 2.Rcc4 Be5# (MM) (2021-05-20)
comment
Keywords: Fesselungsspiel (234 202 220)
Genre: h#
FEN: 1b6/3R4/8/1pr5/3r1BK1/p1kp4/3p4/3B4
Input: Markus Manhart, 1998-03-28
Last update: hpr, 1999-11-07 more...
Genre: h#
FEN: 1b6/3R4/8/1pr5/3r1BK1/p1kp4/3p4/3B4
Input: Markus Manhart, 1998-03-28
Last update: hpr, 1999-11-07 more...
1) 1. exd3ep b8=D 2. Kd4 Df4#
2) 1. exd3ep a8=S 2. d4 Sxb6#
Cook: e.p. not justified as R: 0.e3xd4 possible
2) 1. exd3ep a8=S 2. d4 Sxb6#
Cook: e.p. not justified as R: 0.e3xd4 possible
VL: Unjustified ep-capture because of –e3xd4. For correction it suffices e.g. to add bSe3.
This is possibly a unique 2-move helpmate with ep-key and 2 solutions of the type 1.2.1.1 (at least so currently in the PDB).
Publi?ation details. H551 in the Helpmate section of the July issue, with no preview. Here is the text of the solution published in the Nov.-77 issue: “…White may have just played P(e3)xd4. Very few solvers have spotted this. Full marks for claim of no solution with correct reasoning but, in fairness, one mark for composer's intention. White cannot have just played K(a5)a6 in reply to Pxb6+ as too many captures by Black are required.” (2023-12-01)
A.Buchanan: Yes Valery you are right this must be cooked. However, I'm not a fan of +bSe3 because it spoils one of the model mates. Separately, I mildly prefer bPf7 to bPe6. (2023-12-03)
VL: I do prefer bPf7, as well. I don't see any way to save at least one of the properties Kindergarten Problem and Model mate (2) after fixing (what for these keywords have been added just now?). Besides, Andrew, please, refine the specification ((full) number and publication month 07-08). (2023-12-03)
A.Buchanan: Ok have added the publication details. The keywords are useful to indicate the aim, even though the problem is cooked. I guess there is a way to keep the model mates but it’s very clunky and probably +bSe3 is better (2023-12-03)
more ...
comment
This is possibly a unique 2-move helpmate with ep-key and 2 solutions of the type 1.2.1.1 (at least so currently in the PDB).
Publi?ation details. H551 in the Helpmate section of the July issue, with no preview. Here is the text of the solution published in the Nov.-77 issue: “…White may have just played P(e3)xd4. Very few solvers have spotted this. Full marks for claim of no solution with correct reasoning but, in fairness, one mark for composer's intention. White cannot have just played K(a5)a6 in reply to Pxb6+ as too many captures by Black are required.” (2023-12-01)
A.Buchanan: Yes Valery you are right this must be cooked. However, I'm not a fan of +bSe3 because it spoils one of the model mates. Separately, I mildly prefer bPf7 to bPe6. (2023-12-03)
VL: I do prefer bPf7, as well. I don't see any way to save at least one of the properties Kindergarten Problem and Model mate (2) after fixing (what for these keywords have been added just now?). Besides, Andrew, please, refine the specification ((full) number and publication month 07-08). (2023-12-03)
A.Buchanan: Ok have added the publication details. The keywords are useful to indicate the aim, even though the problem is cooked. I guess there is a way to keep the model mates but it’s very clunky and probably +bSe3 is better (2023-12-03)
more ...
comment
Keywords: En passant as key, Kindergarten Problem, Model mate (2)
Genre: h#, Retro
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/PP6/Kpp1p3/2pp4/1pkPp3/2p5/P7/8
Input: Michal Dragoun, 1998-04-09
Last update: A.Buchanan, 2023-12-03 more...
Genre: h#, Retro
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/PP6/Kpp1p3/2pp4/1pkPp3/2p5/P7/8
Input: Michal Dragoun, 1998-04-09
Last update: A.Buchanan, 2023-12-03 more...
29 - P0546083
Frank Raven Adcock
1822 The Problemist Fairy Chess Supplement 06/1935
(4+10) cooked
h#3
Frank Raven Adcock
1822 The Problemist Fairy Chess Supplement 06/1935
(4+10) cooked
h#3
1. Txa7+ Kb6 2. Lf8 Ta8 3. Te7 Ld7#
Cook: NL 1. Lc5 Ta8 2. Lxa7 Ld7 3. Lb6 Kxb6#
Cook: NL 1. Lc5 Ta8 2. Lxa7 Ld7 3. Lb6 Kxb6#
Verst-4
Adrian Storisteanu: Possible fix: bRc2->g8, +bQc2 (4+11). (2015-08-22)
Yuri Bilokin: Correction: bSd3-e6, bPd4-b6, bPd5-c2(-bRc2) 4R3/B3r3/Kp2n3/8/B7/bp6/kpp5/1b6 (4+9) 1.Rxa7+ Kxb6 2.Bf8 Ra8 3.Re7 Bd7# (2021-05-23)
A.Buchanan: That's a very nice corrected version, Yuri (2021-05-24)
comment
Adrian Storisteanu: Possible fix: bRc2->g8, +bQc2 (4+11). (2015-08-22)
Yuri Bilokin: Correction: bSd3-e6, bPd4-b6, bPd5-c2(-bRc2) 4R3/B3r3/Kp2n3/8/B7/bp6/kpp5/1b6 (4+9) 1.Rxa7+ Kxb6 2.Bf8 Ra8 3.Re7 Bd7# (2021-05-23)
A.Buchanan: That's a very nice corrected version, Yuri (2021-05-24)
comment
Keywords: Sacrifice of white pieces
Genre: h#
FEN: 4R3/B3r3/K7/3p4/B2p4/bp1n4/kpr5/1b6
Input: HBae, 1998-01-19
Last update: Alfred Pfeiffer, 2015-08-22 more...
Genre: h#
FEN: 4R3/B3r3/K7/3p4/B2p4/bp1n4/kpr5/1b6
Input: HBae, 1998-01-19
Last update: Alfred Pfeiffer, 2015-08-22 more...
1. Da4 Te2 2. fxe2 f4 3. Kc7 f5 4. Kc8 f6 5. Thc7 f7 6. De8 fxe8=D#
Keywords: Sacrifice of white pieces, Excelsior white (auto key)
Genre: h#
Computer test: Anton Baumann: C+ Alybadix (2021-03-22)
FEN: 8/1r5r/pk6/8/8/b4p2/R1q2P1p/5b1K
Input: Hans-Jürgen Schäfer, 1998-07-01
Last update: Erich Bartel, 2021-03-23 more...
Genre: h#
Computer test: Anton Baumann: C+ Alybadix (2021-03-22)
FEN: 8/1r5r/pk6/8/8/b4p2/R1q2P1p/5b1K
Input: Hans-Jürgen Schäfer, 1998-07-01
Last update: Erich Bartel, 2021-03-23 more...
1. Kh7 Lb1 2. Kh6 La2 3. Kg5 Lb1 4. Kf4 La2 5. Ke3 Lb1 6. Kd2 La2 7. Kc1 Lb1 8. Kb2 La2 9. Kxa1 Lb1 10. Kb2 La2 11. Kxc2 Lb1+ 12. Kxb3 Lf5 13. gxf5 gxf5 14. Kxa4 f6 15. Kxb5 fxg7 16. Ka4 g8=D 17. b5 Da2#
Keywords: Sacrifice of white pieces
Genre: h#
Computer test: Gustav 4.1d
FEN: 7k/6p1/1p4p1/pP6/Pp4P1/1Pp3p1/B1P3p1/N5Kb
Input: Hans-Jürgen Schäfer, 1998-07-01
Last update: Arnold Beine, 2021-03-31 more...
Genre: h#
Computer test: Gustav 4.1d
FEN: 7k/6p1/1p4p1/pP6/Pp4P1/1Pp3p1/B1P3p1/N5Kb
Input: Hans-Jürgen Schäfer, 1998-07-01
Last update: Arnold Beine, 2021-03-31 more...
a) 1. Tf8 La5 2. Sf7 La4#
b) 1. Sfd7 Lb4 2. Dd8 Lg6#
b) 1. Sfd7 Lb4 2. Dd8 Lg6#
in 'The Problemist' abgedruckt mit der Zwillingsbildung b) sSc4 und den Lösungen a) 1. Tc2 Lh5 2. Sc3 Lh6 und b) 1. Se3 Lg6 2. De2 Lb4, die beide nicht zur Stellung passen.
Das 'Darnall Magazine' hat die falsche Zwillingsbildung von 'The Problemist' übernommen.
SCHRECKE: a)+b): C+, popeye 4.87 (2022-05-29)
comment
Das 'Darnall Magazine' hat die falsche Zwillingsbildung von 'The Problemist' übernommen.
SCHRECKE: a)+b): C+, popeye 4.87 (2022-05-29)
comment
Keywords: Fesselungsspiel (111 101 000), Aristocrat, Aristocrat
Genre: h#
FEN: 4k3/4q3/5r2/4n3/8/8/2B1R3/3KB3
Reprints: The Problemist , p. 284-285, 05/1987
L74 Darnall Magazine 19 14/07/1987
Input: Markus Manhart, 1998-06-10
Last update: Mario Richter, 2022-05-29 more...
Genre: h#
FEN: 4k3/4q3/5r2/4n3/8/8/2B1R3/3KB3
Reprints: The Problemist , p. 284-285, 05/1987
L74 Darnall Magazine 19 14/07/1987
Input: Markus Manhart, 1998-06-10
Last update: Mario Richter, 2022-05-29 more...
a) 1. Sec6 Lh5 2. Tg4 Lg6#
b) 1. Sdc6 Ld5 2. Le4 Le6#
NL in b):
1. Lh3 Le2 2. Lg4 Ld3#
b) 1. Sdc6 Ld5 2. Le4 Le6#
NL in b):
1. Lh3 Le2 2. Lg4 Ld3#
Genre: h#
FEN: 8/2B1nK2/1r6/5kp1/3n4/5B2/6r1/8
Input: Markus Manhart, 1998-10-20
Last update: hpr, 1999-03-10 more...
Anton Baumann: 1. ... Lc4 2.Sb5 Ge6 3.Sa7 La6#
1.Sc6 Ld7 2.Da7 Ge6 3.Sb8 Lc8#
C+ Alybadix (2023-08-22)
comment
1.Sc6 Ld7 2.Da7 Ge6 3.Sb8 Lc8#
C+ Alybadix (2023-08-22)
comment
Keywords: Axis Echo
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: *2Qq*2Q5/nk5K/*2Q2pB3/P7/8/6*2Q1/8/8
Input: Markus Manhart, 1999-01-26
Last update: hpr, 1999-12-04 more...
Pieces: = Grasshopper (G)
Genre: Fairies
FEN: *2Qq*2Q5/nk5K/*2Q2pB3/P7/8/6*2Q1/8/8
Input: Markus Manhart, 1999-01-26
Last update: hpr, 1999-12-04 more...
1. Ld2 Sxe4 2. La5 Sd6 3. Te6 Sxf5 4. Tb6 Sxg7 5. Dh7 Se6 6. Da7 Sxc5#
Cook: NL 1. La3 Kf4 2. Db3 Ke5 3. La4 Kd6 4. Ka5 Kc7 5. Kb4 Kb6 6. c4 Sd5#
Cook: NL 1. La3 Kf4 2. Db3 Ke5 3. La4 Kd6 4. Ka5 Kc7 5. Kb4 Kb6 6. c4 Sd5#
Adrian Storisteanu: Possible fix: wKg3->g2, bRe1->e3, +bPd6 (2+10) [2...Sxd6]. (2015-11-04)
Viktoras Paliulionis: Other possible fix: bPc5-d4, bBc1-h6, +bPf6. (2021-02-02)
Kenneth Solja: Best correction is: +bd6 and +bSg7(-bpg7) (2+10) No need to move white king. And initial position bL should be in h6 as it was in the original problem.
Solution: 1.Ld2 Sxe4 2.La5 Sxd6 3.Te6 Sxf5 4.Tb6 Sxg7 5.Dh7 Se6 6.Da7 Sxc5# (C+) (2024-04-04)
comment
Viktoras Paliulionis: Other possible fix: bPc5-d4, bBc1-h6, +bPf6. (2021-02-02)
Kenneth Solja: Best correction is: +bd6 and +bSg7(-bpg7) (2+10) No need to move white king. And initial position bL should be in h6 as it was in the original problem.
Solution: 1.Ld2 Sxe4 2.La5 Sxd6 3.Te6 Sxf5 4.Tb6 Sxg7 5.Dh7 Se6 6.Da7 Sxc5# (C+) (2024-04-04)
comment
Genre: h#
FEN: 8/6p1/k7/1bp2p2/4p3/2N3K1/8/1qb1r3
Input: Markus Manhart, 1999-02-17
Last update: Alfred Pfeiffer, 2015-11-04 more...
1. c6 Lh6 2. Da5+ Ld2 3. Dh5 Le3 4. Kd8 Lxa7 5. De8 Lb6#
Cook: NL
1. c6 Lh6 2. e6 Ke2 3. Dc7 Ke3 4. Ke7 Kd4 5. Kd6 Lf8#
Cook: NL
1. c6 Lh6 2. e6 Ke2 3. Dc7 Ke3 4. Ke7 Kd4 5. Kd6 Lf8#
Adrian Storisteanu: Possible fix: +bSg8 (2+12). (2015-08-07)
Jakob Leck: Die Dawson-Aufgabe könnte P0555591 sein - zeitgleich erschienen. (2022-12-28)
comment
Jakob Leck: Die Dawson-Aufgabe könnte P0555591 sein - zeitgleich erschienen. (2022-12-28)
comment
Genre: h#
FEN: 2bqkB2/ppppp3/5p2/8/8/6p1/6p1/4K3
Input: Markus Manhart, 1999-02-27
Last update: Alfred Pfeiffer, 2015-08-07 more...
*) 1. ... Te7 2. Ne2 Ne1 3. Ng6 Th7#
1) 1. Na2 Nf2 2. Kh5 Te6 3. Ng5 Th6#
1) 1. Na2 Nf2 2. Kh5 Te6 3. Ng5 Th6#
SCHRECKE: Nebenlösung
1. Na2 Te5 2. Nb4 Tg5 3. Nh7 Nf7# (2023-01-12)
HBae: Correction +wPawn d5 (5+2) C+ Popeye v4.37 (2023-01-12)
comment
1. Na2 Te5 2. Nb4 Tg5 3. Nh7 Nf7# (2023-01-12)
HBae: Correction +wPawn d5 (5+2) C+ Popeye v4.37 (2023-01-12)
comment
Keywords: Reihen-Echo
Pieces: = Nightrider (N)
Genre: Fairies
FEN: 8/8/1K5k/8/8/3*2N4/4R1P1/2*2n5
Input: Ralf Binnewirtz, 1998-10-22
Last update: hpr, 1999-12-12 more...
Pieces: = Nightrider (N)
Genre: Fairies
FEN: 8/8/1K5k/8/8/3*2N4/4R1P1/2*2n5
Input: Ralf Binnewirtz, 1998-10-22
Last update: hpr, 1999-12-12 more...
s) * 1. ... b8=S 2. f6 Sc6#
1. Td4 b8=D 2. Kd5 Db5#
w) * 1. ... ... Lc8 2. b8=T Ta5#
1. ... b8=L Tb5 2. La7 Ld5#
1. Td4 b8=D 2. Kd5 Db5#
w) * 1. ... ... Lc8 2. b8=T Ta5#
1. ... b8=L Tb5 2. La7 Ld5#
Allumwandlung!weiße, Kegel!motiv
t/Allumwandlung, t/Minimal-B,, u/61105, u/TL-SD
Allumwandlung!eines Bauern
paul: In 1944, Bo was 17 years old. (2022-07-04)
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t/Allumwandlung, t/Minimal-B,, u/61105, u/TL-SD
Allumwandlung!eines Bauern
paul: In 1944, Bo was 17 years old. (2022-07-04)
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comment
Keywords: Allumwandlung, Promotion
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: K7/1P3p2/3pb3/3rkp2/4pp2/8/8/8
Reprints: 1983 100 Classics of the Chessboard
4 varianter med 1 bonde 1972
143 FIDE Album 1914-1944/III 1975
The Problemist , p. 71, 10-12/1976
The Problemist , p. 171, 09/1981
(20) idee & form 47, p. 1293, 07/1995
Mat Plus 06-07/1995
080 Kegelschach [Ebert u. Wolfenter] 1997
305 Minimalkunst im Schach 2006
Input: Hans-Jürgen Schäfer, 1999-02-22
Last update: Alfred Pfeiffer, 2016-02-12 more...
Genre: h#
Computer test: Popeye C-Version 3.52 (2048 KB)
FEN: K7/1P3p2/3pb3/3rkp2/4pp2/8/8/8
Reprints: 1983 100 Classics of the Chessboard
4 varianter med 1 bonde 1972
143 FIDE Album 1914-1944/III 1975
The Problemist , p. 71, 10-12/1976
The Problemist , p. 171, 09/1981
(20) idee & form 47, p. 1293, 07/1995
Mat Plus 06-07/1995
080 Kegelschach [Ebert u. Wolfenter] 1997
305 Minimalkunst im Schach 2006
Input: Hans-Jürgen Schäfer, 1999-02-22
Last update: Alfred Pfeiffer, 2016-02-12 more...
s) 1. La3 2. Sb5 3. Tc3 4. Kb4 Sd3#
w) 1. ... Tg5 2. Se6 3. Lf4 4. Kf6 Sd5#
w) 1. ... Tg5 2. Se6 3. Lf4 4. Kf6 Sd5#
ANDROMEDA FAIRY CHESS MATCH GREAT BRITAIN v HUNGARY 1993-95
SECTION 1 (Serieshelpmates in 4-8 moves)
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SECTION 1 (Serieshelpmates in 4-8 moves)
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Keywords: Seriesmover
Genre: Fairies
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/6K1/6Pn/p7/k4NR1/2n2r2/1bPB4/8
Input: Torsten Linss, 1999-04-07
Last update: Arnold Beine, 2022-09-14 more...
Genre: Fairies
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/6K1/6Pn/p7/k4NR1/2n2r2/1bPB4/8
Input: Torsten Linss, 1999-04-07
Last update: Arnold Beine, 2022-09-14 more...
40 - P0572507
William Edmund Frank Fillery
Donald A. Smedley
875v The Problemist 01/1983
(3+5) cooked
h#4
3.1...
William Edmund Frank Fillery
Donald A. Smedley
875v The Problemist 01/1983
(3+5) cooked
h#4
3.1...
1) 1. Sd7 c5 2. Kb7 c6+ 3. Ka7 c7 4. Sb8 c8=S#
2) 1. Sh7 c5 2. Kb5 c6 3. Kb6 c7 4. Ka7 cxd8=D#
3) 1. Tc8 Ta7 2. Tc7 c5 3. Kd7 c6+ 4. Kc8 Ta8#
NL:
1. Ta7 Tg8 2. Kb7 c5 3. Ka8 c6 4. Sg6 Txd8#
1. Kd6 Td7+ 2. Ke5 Kg2 3. Lf6 Kf3 4. Se6 Td5#
2) 1. Sh7 c5 2. Kb5 c6 3. Kb6 c7 4. Ka7 cxd8=D#
3) 1. Tc8 Ta7 2. Tc7 c5 3. Kd7 c6+ 4. Kc8 Ta8#
NL:
1. Ta7 Tg8 2. Kb7 c5 3. Ka8 c6 4. Sg6 Txd8#
1. Kd6 Td7+ 2. Ke5 Kg2 3. Lf6 Kf3 4. Se6 Td5#
Yuri Bilokin: Correction: wRg7-f7, +bPg7, +bRg8, +bPh3 r2b1nr1/5Rp1/p1k5/8/2P5/7p/8/7K (3+8) (2022-04-20)
comment
comment
Genre: h#
FEN: r2b1n2/6R1/p1k5/8/2P5/8/8/7K
Input: Markus Manhart, 1999-03-28
Last update: Markus Manhart, 1999-03-29 more...
41 - P0576377
Bo Lindgren
67 The Problemist (10) 07/1967
G.W.Chandler gewidmet
5. ehrende Erwähnung
Informalturnier
(6+1) C+
#2
Zeroposition
a) wSf4 nach e8
b) wLf1 nach e2
c) wTh1 nach b5
d) wDb4 nach c2
Bo Lindgren
67 The Problemist (10) 07/1967
G.W.Chandler gewidmet
5. ehrende Erwähnung
Informalturnier
(6+1) C+
#2
Zeroposition
a) wSf4 nach e8
b) wLf1 nach e2
c) wTh1 nach b5
d) wDb4 nach c2
a) 1. h8=S+ Ke6 2. De4# 1. h8=S+ Kxe8 2. Lb5# 1. h8=S+ Kg8 2. Lc4#
b) 1. h8=L Ke8 2. Lh5# 1. h8=L Kg8 2. Lc4#
c) 1. h8=T Kg7 2. Df8# 1. h8=T Kf6 2. Df8#
d) 1. h8=D Ke7 2. Dc7#
Mit dieser Aufgabe hat BoL vermutlich erstmals den Begriff der ZeroPosition eingeführt. Hier zeigt die Verwendung der ZeroPosition auch thematische Geschlossenheit insofern, als bei Versetzung des wS auch UW in S erfolgt, bei wT auch UW in T etc. Es besteht also ein Zusammenhang versetzte Figur zu Umwandlung (eb).
Nachdrucke lt. Erich Bartel :
1) A) Die Schwalbe IV 1971
2) The Problemist V 1972,S.229
3) 16 Mat Plus (6-7) 1995
4) Sahovska Kompozicija I-XII 1995
5) Die Schwalbe (176) IV 1999,S.99
b) 1. h8=L Ke8 2. Lh5# 1. h8=L Kg8 2. Lc4#
c) 1. h8=T Kg7 2. Df8# 1. h8=T Kf6 2. Df8#
d) 1. h8=D Ke7 2. Dc7#
Mit dieser Aufgabe hat BoL vermutlich erstmals den Begriff der ZeroPosition eingeführt. Hier zeigt die Verwendung der ZeroPosition auch thematische Geschlossenheit insofern, als bei Versetzung des wS auch UW in S erfolgt, bei wT auch UW in T etc. Es besteht also ein Zusammenhang versetzte Figur zu Umwandlung (eb).
Nachdrucke lt. Erich Bartel :
1) A) Die Schwalbe IV 1971
2) The Problemist V 1972,S.229
3) 16 Mat Plus (6-7) 1995
4) Sahovska Kompozicija I-XII 1995
5) Die Schwalbe (176) IV 1999,S.99
Teppo Mänttä: reprint: nr. 14 "Svenska Miniatyrer i urval" (2nd edition), 1978. (2021-09-21)
Teppo Mänttä: double publication, see P1136428 (2021-09-21)
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Teppo Mänttä: double publication, see P1136428 (2021-09-21)
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Keywords: Rex solus (s), Allumwandlung
Genre: 2#
Computer test: Popeye Windows-32Bit v4.55
FEN: 2K5/5k1P/8/8/1Q3N2/8/8/5B1R
Input: Felber, Volker, 2000-01-02
Last update: Erich Bartel, 2014-11-15 more...
Genre: 2#
Computer test: Popeye Windows-32Bit v4.55
FEN: 2K5/5k1P/8/8/1Q3N2/8/8/5B1R
Input: Felber, Volker, 2000-01-02
Last update: Erich Bartel, 2014-11-15 more...
1. bxa2 Dxc1 2. Sdb3 Dd1 3. Sc1 Dxe2 4. Lf2+ Dxf2 5. g1=S Kxf1 6. Sge2 Dg2#
Genre: h#
FEN: 8/8/8/8/8/1pp1pPp1/Pprnp1pp/nrqQKbbk
Input: Felber, Volker, 2000-01-02
Last update: hpr, 2000-02-28 more...
43 - P0576518
Gabor Cseh
H2189 The Problemist 05/1998
1. Preis
(8+14) C+
h#5
0.1...
b) sDc6 tauschen mit sSd6
Gabor Cseh
H2189 The Problemist 05/1998
1. Preis
(8+14) C+
h#5
0.1...
b) sDc6 tauschen mit sSd6
a) 1. ... c8=D 2. Se8 fxe8=D 3. Dxe6 Dxc3 4. Dxe5 Dxb2+ 5. Kxb2 Dxe5#
b) 1. ... f8=D 2. Sd8 cxd8=D 3. Dxb6 Dxa3 4. Dxa5 Dxa2+ 5. Kxa2 Dxa5#
b) 1. ... f8=D 2. Sd8 cxd8=D 3. Dxb6 Dxa3 4. Dxa5 Dxa2+ 5. Kxa2 Dxa5#
Genre: h#
Computer test: Gustav 4.1c
FEN: 8/2P2Pp1/1PqnP3/P3P3/5P2/ppp2p2/bp1p1r2/krnK4
Reprints: E213 FIDE Album 1998-2000 2007
Input: Felber, Volker, 2000-01-02
Last update: Marcin Banaszek, 2020-02-02 more...
1) 1. ... Tf2 2. b5 Kg3 3. Kb6 Kf4 4. Kc7 Ke5 5. Kd8 Kd6 6. Tf8 Txf8#
2) 1. ... Tg2 2. Kb5 Kh3 3. Kc6 Kg4 4. Kd7 Kf5 5. Ke8 Ke6 6. Tg8 Txg8#
2) 1. ... Tg2 2. Kb5 Kh3 3. Kc6 Kg4 4. Kd7 Kf5 5. Ke8 Ke6 6. Tg8 Txg8#
Keywords: Minimal (T), Liniensperrung, Check Protection, Model mate, Echo
Genre: h#
FEN: 7r/7p/pp6/k7/q2p4/3p4/3R3K/4bb2
Input: Felber, Volker, 2000-01-02
Last update: Gunter Jordan, 2022-08-08 more...
Genre: h#
FEN: 7r/7p/pp6/k7/q2p4/3p4/3R3K/4bb2
Input: Felber, Volker, 2000-01-02
Last update: Gunter Jordan, 2022-08-08 more...
45 - P0576637
Mikael Grönroos
H2257 The Problemist 01/1999
E. Rautavaara gewidmet
1. Preis
(2+10)
h#8
Mikael Grönroos
H2257 The Problemist 01/1999
E. Rautavaara gewidmet
1. Preis
(2+10)
h#8
1. Dg2 Lg5 2. a2 Kg8 3. a3 Kf8 4. a4 Lxe7 5. a5 Ke8 6. a6 Kd8 7. Ka7 Kxc7 8. Da8 Lxc5#
Moldenhauer: Computerprüfung: C+ Gustav 4.1d in 10 Tagen 3 Std. 13 Min. (2022-07-03)
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Genre: h#
FEN: k6K/p1r1r3/p6B/p1b5/p7/p7/q7/8
Reprints: B03 Shift of groups of pawns in chess problems 28/04/2005
Input: Felber, Volker, 2000-01-02
Last update: Erich Bartel, 2018-01-24 more...
1. f1=L Ld5 2. Lc4 bxc4 3. d1=L c5 4. Lg4 c6 5. Ld7 cxd7 6. b1=L d8=L 7. Lh7 Lf6#
Genre: h#
FEN: 7k/2p5/2B5/8/1p1p4/1P6/1p1p1p2/7K
Input: Felber, Volker, 2000-01-02
Last update: James Malcom, 2021-02-18 more...
1) 1. Ta1+ Sd1 2. Ta2 Kg1 3. Txf2 Kh1 4. Ta2 Sb2 5. f2 Sc4#
2) 1. d5 Sd3 2. Te2 Sb4 3. Txe3 Sxd5 4. Te2 Se3 5. Ta2 Sc4#
2) 1. d5 Sd3 2. Te2 Sb4 3. Txe3 Sxd5 4. Te2 Se3 5. Ta2 Sc4#
Knickpendel
Yuri Bilokin: you can see 1...Sc4#
1...Sd3 2.Rd2 Se5 3.Ra2 Sc4#
1...Sd3 2.Rxf2 Se5 3.Ra2 Sc4#
1...Sd3 2.Re2 Se5 3.Ra2 Sc4#
1...Sd3 2.Rb2 Se5 3.Ra2 Sc4#
1...Sd3 2.Rc2 Se5 3.Ra2 Sc4# (2022-07-31)
comment
Yuri Bilokin: you can see 1...Sc4#
1...Sd3 2.Rd2 Se5 3.Ra2 Sc4#
1...Sd3 2.Rxf2 Se5 3.Ra2 Sc4#
1...Sd3 2.Re2 Se5 3.Ra2 Sc4#
1...Sd3 2.Rb2 Se5 3.Ra2 Sc4#
1...Sd3 2.Rc2 Se5 3.Ra2 Sc4# (2022-07-31)
comment
Genre: h#
Computer test: (Popeye C-Version 3.52 (2048 KB))
FEN: 8/1p6/1P1p4/5p2/p2PpP2/kpP1Pp1p/rN3P1P/7K
Input: Felber, Volker, 2000-01-02
Last update: hpr, 2003-01-10 more...
48 - P0576743
György Bakcsi
Laszlo Zoltan
PS849 The Problemist Supplement 38 01/1999
(8+8)
h#12
Ultraschachzwang
György Bakcsi
Laszlo Zoltan
PS849 The Problemist Supplement 38 01/1999
(8+8)
h#12
Ultraschachzwang
1. Lg5+ Kxg5 2. exf6 Kf4 3. fxe5 Ke3 4. exd4+ Kd2 5. dxc3+ Kc1 6. cxb2+ Kd2 7. c3+ Ke3 8. d4+ Kf4 9. e5+ Kg5 10. f6+ Kxg6 11. b1=D+ Kh6 12. Dg6+ Sxg6#
Keywords: Ultraschachzwang
Genre: Fairies
FEN: 5N1k/4pp2/4pPpK/3pP3/2pP3b/2P5/BP6/8
Input: Felber, Volker, 2000-01-02
Last update: Frank Müller, 2010-05-15 more...
Genre: Fairies
FEN: 5N1k/4pp2/4pPpK/3pP3/2pP3b/2P5/BP6/8
Input: Felber, Volker, 2000-01-02
Last update: Frank Müller, 2010-05-15 more...
49 - P0576961
Laszlo Zoltan
György Bakcsi
D The Problemist , p. 312, 03/1994
4. ehrende Erwähnung, Sherlock Holmes Turnier 1994
(2+3) C+
ser-h#3
b) -wDa1, -sLc2: ser-r#4
c) -wDa1, -sTh7: ser-r=5
d) -sLc2, -sTh7: ser-h=4
Laszlo Zoltan
György Bakcsi
D The Problemist , p. 312, 03/1994
4. ehrende Erwähnung, Sherlock Holmes Turnier 1994
(2+3) C+
ser-h#3
b) -wDa1, -sLc2: ser-r#4
c) -wDa1, -sTh7: ser-r=5
d) -sLc2, -sTh7: ser-h=4
a) 1. Th3 2. Tc3 3. Ld3 Da4#
b) 1. ... Kc6 2. Kb6 3. Ka5 4. Ka4 Ta7#
c) 1. ... Ke5 2. Kf4 3. Ke3 4. Kd2 5. Kc1 Kc3=
d) 1. Kb5 2. Kb6 3. Kb7 4. Kb8 Da6=
b) 1. ... Kc6 2. Kb6 3. Ka5 4. Ka4 Ta7#
c) 1. ... Ke5 2. Kf4 3. Ke3 4. Kd2 5. Kc1 Kc3=
d) 1. Kb5 2. Kb6 3. Kb7 4. Kb8 Da6=
Thema des TT ('The Problemist' 09/1992, S. 88): "... composers are required to combine 2 (or more) problems to make a distinct third."
(Im 'Problemist' haben die Teilaufgaben a),b),c) und d) die Labels: D,A,B und C.
PR Barry Barnes vermutet in seinem Kommentar zu diesem Problem, daß ein Computer bei der Erstellung geholfen haben könnte: "Essentially disparate parts (A),(B) and (C) come together as something quite new in (D). But note a minimal record: 5 pieces, cyclic absence of 2 of the pieces, 4 different series-demands, and only ideal mates! Was Holmes right in thinking that a later model of Prof Babbage's "engine" (Cambridge,1834) determined the solving conditions and lengths?"
Anton Baumann: a)-d): C+ Alybadix (2022-02-25)
comment
(Im 'Problemist' haben die Teilaufgaben a),b),c) und d) die Labels: D,A,B und C.
PR Barry Barnes vermutet in seinem Kommentar zu diesem Problem, daß ein Computer bei der Erstellung geholfen haben könnte: "Essentially disparate parts (A),(B) and (C) come together as something quite new in (D). But note a minimal record: 5 pieces, cyclic absence of 2 of the pieces, 4 different series-demands, and only ideal mates! Was Holmes right in thinking that a later model of Prof Babbage's "engine" (Cambridge,1834) determined the solving conditions and lengths?"
Anton Baumann: a)-d): C+ Alybadix (2022-02-25)
comment
Keywords: Seriesmover, Aristocrat, Minimal, Miniature, Aristocrat, Aristocrat, Aristocrat
Genre: Fairies
Computer test: Anton Baumann (2022-02-25): a)-d): C+ Alybadix
FEN: 8/7r/3K4/8/2k5/8/2b5/Q7
Input: Felber, Volker, 2000-01-03
Last update: Mario Richter, 2022-02-27 more...
Genre: Fairies
Computer test: Anton Baumann (2022-02-25): a)-d): C+ Alybadix
FEN: 8/7r/3K4/8/2k5/8/2b5/Q7
Input: Felber, Volker, 2000-01-03
Last update: Mario Richter, 2022-02-27 more...
50 - P0577047
Yves Cheylan
F1353 The Problemist 03/1993
2. ehrende Erwähnung
(4+8) C+
ser-h=8
Anticirce Cheylan
Yves Cheylan
F1353 The Problemist 03/1993
2. ehrende Erwähnung
(4+8) C+
ser-h=8
Anticirce Cheylan
1. Tcg1 2. Sc1 3. Ld1 4. Ta4 5. Ta1 6. Kd8 7. De4 8. De1 Ke8=
Gerald Ettl: Grundlinie-Einmauerung, sogar mit wK-Effekt. Mehr scheint wohl nicht zu gehen. (2009-02-10)
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Keywords: Circe (Anti-), Seriesmover
Genre: Fairies
Computer test: Alybadix
FEN: 2k2K2/2P2p2/1N3P2/8/b5r1/1n6/3p4/1qr5
Input: Felber, Volker, 2000-01-04
Last update: Arnold Beine, 2022-09-22 more...
Genre: Fairies
Computer test: Alybadix
FEN: 2k2K2/2P2p2/1N3P2/8/b5r1/1n6/3p4/1qr5
Input: Felber, Volker, 2000-01-04
Last update: Arnold Beine, 2022-09-22 more...
51 - P0577050
Franz Pachl
F1402 The Problemist 1993
4.-6. ehrende Erwähnung
(5+13) C+
ser-h#5
b) wKh6 nach b8
Franz Pachl
F1402 The Problemist 1993
4.-6. ehrende Erwähnung
(5+13) C+
ser-h#5
b) wKh6 nach b8
a) 1. Tb6 2. Se6 3. c5 4. c4 5. Sc7 Lxb6#
b) 1. Df4 2. Sd6 3. e4 4. e3 5. Sf5 Dxf4#
b) 1. Df4 2. Sd6 3. e4 4. e3 5. Sf5 Dxf4#
Anton Baumann: vermutlich Diagramm nicht korrekt:
a) unlösbar b) mehrere Lösungen (2020-12-08)
SCHRECKE: Der wBc2 muss offenbar auf b2 stehen!
Dann gibt es die Lösungen:
a) 1.Tb6 2.Se6 3.c5 4.c4 5.Sc7 L:b6#
b) 1.Df4 2.Sd6 3.e4 4.e3 5.Sf5 D:f4# (2020-12-08)
Mario Richter: Im Originaldiagramm wBb2 statt wBc2 - Stellung entsprechend korrigiert. (2020-12-09)
comment
a) unlösbar b) mehrere Lösungen (2020-12-08)
SCHRECKE: Der wBc2 muss offenbar auf b2 stehen!
Dann gibt es die Lösungen:
a) 1.Tb6 2.Se6 3.c5 4.c4 5.Sc7 L:b6#
b) 1.Df4 2.Sd6 3.e4 4.e3 5.Sf5 D:f4# (2020-12-08)
Mario Richter: Im Originaldiagramm wBb2 statt wBc2 - Stellung entsprechend korrigiert. (2020-12-09)
comment
Keywords: Seriesmover
Genre: Fairies
Computer test: Popeye Windows-64Bit v4.67
FEN: 3b4/B4r2/r1p4K/p1npp2p/3kn2Q/7p/1P2P2q/8
Input: Felber, Volker, 2000-01-04
Last update: Mario Richter, 2020-12-09 more...
Genre: Fairies
Computer test: Popeye Windows-64Bit v4.67
FEN: 3b4/B4r2/r1p4K/p1npp2p/3kn2Q/7p/1P2P2q/8
Input: Felber, Volker, 2000-01-04
Last update: Mario Richter, 2020-12-09 more...
1. Kb2 2. Kc3 3. Kd4 4. Kxe5 5. Kf6 6. e5 7. e4 8. e3 9. e2 10. e1=L 11. Lxh4 12. Lg5 13. Lxh6 14. Kg7 15. Kh8 16. Lg7 17. h5 18. h4 19. h3 20. h2 21. h1=L 22. Lxd5 23. Lg8 24. d5 25. d4 26. d3 27. d2 28. d1=T 29. Th1 30. Th7 Sg6#
Erich Bartel: im Problemist ist hierzu vermerkt:
Jonathan's Artikel has already apeared, in translation,in the
March 1995 issue of Rochade-Europa, and these originals
should compete in the informal tourney of that magazine.
--------
Die Frage ist nun welches Magazin als Erstqueelle gelten soll. (2011-11-28)
Anton Baumann: NL: 1.d6 2.dxe5 .. 6.e1=D .. 8.Dg8 .. 13.e1=L 14.Lxh4 15.Lg5 16.Lxh6 .. 23.Kh8 24.Lg7 .. 29.h1=T 30.Th7 Sg6# (2020-11-21)
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Jonathan's Artikel has already apeared, in translation,in the
March 1995 issue of Rochade-Europa, and these originals
should compete in the informal tourney of that magazine.
--------
Die Frage ist nun welches Magazin als Erstqueelle gelten soll. (2011-11-28)
Anton Baumann: NL: 1.d6 2.dxe5 .. 6.e1=D .. 8.Dg8 .. 13.e1=L 14.Lxh4 15.Lg5 16.Lxh6 .. 23.Kh8 24.Lg7 .. 29.h1=T 30.Th7 Sg6# (2020-11-21)
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Keywords: Seriesmover, corner to corner (sKa1h8), Excelsior (3-fach), under-promotion, konsekutive Umwandlungen 3 (llt)
Genre: Fairies
FEN: 8/3pp2p/7P/1K1PP3/5N1P/8/8/k7
Reprints: Rochade Europa 03/1995
Input: Felber, Volker, 2000-01-06
Last update: Olaf Jenkner, 2020-11-23 more...
Genre: Fairies
FEN: 8/3pp2p/7P/1K1PP3/5N1P/8/8/k7
Reprints: Rochade Europa 03/1995
Input: Felber, Volker, 2000-01-06
Last update: Olaf Jenkner, 2020-11-23 more...
1. Kb6 2. Ka5 3. Ka4 4. Kxa3 5. Kxa2 6. Kxa1 7. Kb2 8. Kc3 9. Kd4 19. Ke5 11. Kf6 12. Kg7 13. Kxh8 14. Kg7 15. Kf6 16. Ke5 17. Kxd6 18. Kxc7 19. Kb8 20. Kxa8 21. Kb7 22. Kc6 23. Kd5 24. Ke4 25. Kf3 26. Kg2 27. Kxh1 28. Lxf2+ Kxf2=
Keywords: Seriesmover
Genre: Fairies
Computer test: Gustav 4.1d
FEN: B6R/k1N5/3P4/8/8/P7/P4P1p/B3K1bR
Input: Felber, Volker, 2000-01-06
Last update: Arnold Beine, 2022-10-31 more...
Genre: Fairies
Computer test: Gustav 4.1d
FEN: B6R/k1N5/3P4/8/8/P7/P4P1p/B3K1bR
Input: Felber, Volker, 2000-01-06
Last update: Arnold Beine, 2022-10-31 more...
54 - P0577311
Satoshi Hashimoto
F1708v The Problemist 07/1997
5. ehrende Erwähnung
(8+3)
ser-h=11
Platzwechselcirce
Satoshi Hashimoto
F1708v The Problemist 07/1997
5. ehrende Erwähnung
(8+3)
ser-h=11
Platzwechselcirce
1. Dxd7[+wSd8] 2. Dxd8[+wSd7] 3. Dxe7[+wSd8] 4. Df8 5. Dxg7[+wSf8] 6. Dh8 7. Dxf6[+wSh8] 8. De7 9. Dxf8[+wSe7] 10. Dxf7[+wSf8] 11. De8 Shf7=
Urversion NL mit wKe4!
Kevin Begley: Correction of 'P0577029' (was this actually published)? (2011-05-18)
paul: Cook: 1.Q×e7(Sd8) 2.K×d8(Sc7) 3.Qf8 4.S×c7(Sa8) 5.Se8 6.Q×g7(Qf8) 7.Qh8 8.Q×f6(Qh8) 9.Qe7 Qf6= (2024-02-28)
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Kevin Begley: Correction of 'P0577029' (was this actually published)? (2011-05-18)
paul: Cook: 1.Q×e7(Sd8) 2.K×d8(Sc7) 3.Qf8 4.S×c7(Sa8) 5.Se8 6.Q×g7(Qf8) 7.Qh8 8.Q×f6(Qh8) 9.Qe7 Qf6= (2024-02-28)
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Keywords: Circe (Platzwechsel-), Seriesmover
Genre: Fairies
FEN: n2q4/2kPPPP1/1N2PP2/8/4K3/8/8/8
Input: Felber, Volker, 2000-01-06
Last update: Kevin Begley, 2011-09-12 more...
Genre: Fairies
FEN: n2q4/2kPPPP1/1N2PP2/8/4K3/8/8/8
Input: Felber, Volker, 2000-01-06
Last update: Kevin Begley, 2011-09-12 more...
55 - P0577314
Peter Fayers
F1715 The Problemist 07/1997
(5+5)
ser-h#3
Iceberg
Andernach
b) wSa1 statt sSa1
c) wTb1 statt sTb1
d) wLc1 statt sLc1
Peter Fayers
F1715 The Problemist 07/1997
(5+5)
ser-h#3
Iceberg
Andernach
b) wSa1 statt sSa1
c) wTb1 statt sTb1
d) wLc1 statt sLc1
a) 1. Ta1 2. (TS)c1 3. (TSL)xg5=w (TSB)e4#
b) 1. Tc1 2. (TL)c4 3. (TL)xe6=w Sb3#
c) 1. Lb2 2. La1 3. (LS)xe5=w Tb5#
d) 1. Ta1 2. (TS)b1 3. (TS)xb6=w Le3#
b) 1. Tc1 2. (TL)c4 3. (TL)xe6=w Sb3#
c) 1. Lb2 2. La1 3. (LS)xe5=w Tb5#
d) 1. Ta1 2. (TS)b1 3. (TS)xb6=w Le3#
Alfred Pfeiffer: Was ist "Iceberg"? Ist das eine Art Augsburger Schach? (2010-09-29)
Alfred Pfeiffer: "Iceberg" ist Peter Fayers' Version des Augsburger Schachs, bei folgenden drei Unterschiede gelten (Quelle: "http://www.bcvs.ukf.net/augsb.htm"):
E1: Pieces may combine, but not kings, and NOT PAWNS either (not even when moving to promote).
E2: The game-array queen is not separable, and Qs are distinct from (R+B)s.
E3: A pawn may promote to any combination seen in the diagram, or generated during play. (2010-10-11)
A.Buchanan: Link is now at https://www.mayhematics.com/s/augsb.htm. Piece images to make up the diagrams have been lost, but you can get a clue what the diagrams were by clicking in the html. (2019-02-04)
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Alfred Pfeiffer: "Iceberg" ist Peter Fayers' Version des Augsburger Schachs, bei folgenden drei Unterschiede gelten (Quelle: "http://www.bcvs.ukf.net/augsb.htm"):
E1: Pieces may combine, but not kings, and NOT PAWNS either (not even when moving to promote).
E2: The game-array queen is not separable, and Qs are distinct from (R+B)s.
E3: A pawn may promote to any combination seen in the diagram, or generated during play. (2010-10-11)
A.Buchanan: Link is now at https://www.mayhematics.com/s/augsb.htm. Piece images to make up the diagrams have been lost, but you can get a clue what the diagrams were by clicking in the html. (2019-02-04)
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Keywords: Seriesmover, Andernach Chess, Augsburg Chess (Iceberg)
Genre: Fairies
FEN: 8/8/1P2P3/K1k1P1P1/8/p7/8/nrb5
Input: Felber, Volker, 2000-01-06
Last update: Arnold Beine, 2022-09-19 more...
Genre: Fairies
FEN: 8/8/1P2P3/K1k1P1P1/8/p7/8/nrb5
Input: Felber, Volker, 2000-01-06
Last update: Arnold Beine, 2022-09-19 more...
1. Kh7 2. Kh6 3. Kh5 4. Kh4 5. Kh3 6. Kh2 7. Kxg1[+sKe8] 8. Ke7 9. Ke6 10. Ke5 11. Kxe4[+sKe8] 12. Ke7 13. Ke6 14. Ke5 15. Ke4 16. Kxe3[+sKe8] 17. Ke7 18. Ke6 19. Ke5 20. Ke4 21. Ke3 22. Kxd2[+sKe8] 23. Ke7 24. Ke6 25. Ke5 26. Ke4 27. Ke3 28. Kd2 29. Kc2 30. Kxb2[+sKe8] 31. Ke7 32. Ke6 33. Ke5 34. Ke4 35. Ke3 36. Kd2 37. Kc2 38. Kb2 39. Kxa3[+sKe8] 40. Ke7 41. Ke6 42. Ke5 43. Ke4 44. Ke3 45. Kd2 46. Kc2 47. Kb2 48. Ka3 49. Kxa4[+sKe8] 50. Ke7 51. Ke6 52. Ke5 53. Ke4 54. Ke3 55. Kd2 56. Kc2 57. Kb2 58. Ka3 59. Ka4 60. Kxa5[+sKe8] 61. Ke7 62. Ke6 63. Ke5 64. Ke4 65. Ke3 66. Kd2 67. Kc2 68. Kb2 69. Ka3 70. Kxb4[+sKe8] 71. Ke7 72. Ke6 73. Ke5 74. Ke4 75. Ke3 76. Kd2 77. Kc2 78. Kxb1[+sKe8] 79. Ke7 80. Ke6 81. Ke5 82. Ke4 83. Ke3 84. Kd2 85. Kc2 86. Kb2 87. Ka3 88. Ka4 Ta1#
Gerald Ettl: Man muss bevor man den wSb4 schlaegt, den wLa5 beseitigen, sonst kommt der sK nicht nach d2. Den wLb1 darf man erst nehmen, wenn der wSb4 weg ist, sonst kommt der sK nicht nach c2. (2009-02-10)
Anton Baumann: C+ Alybadix (2021-02-08)
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Anton Baumann: C+ Alybadix (2021-02-08)
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Keywords: Seriesmover, Circe (Anti-)
Genre: Fairies
Computer test: Anton Baumann: C+ Alybadix (2021-02-08) C+ Alybadix
FEN: 7k/1N1p4/3p4/B7/PNK1P3/P3P3/1P1PP3/1B3RR1
Input: Felber, Volker, 2000-01-06
Last update: Erich Bartel, 2021-02-09 more...
Genre: Fairies
Computer test: Anton Baumann: C+ Alybadix (2021-02-08) C+ Alybadix
FEN: 7k/1N1p4/3p4/B7/PNK1P3/P3P3/1P1PP3/1B3RR1
Input: Felber, Volker, 2000-01-06
Last update: Erich Bartel, 2021-02-09 more...
1. Kg8 2. Kxh8[+sKe8] 3. Kd8 4. Kc8 5. Kb8 6. Kxa8[+sKe8] 7. Kd8 8. Kc8 9. Kb8 10. Ka7 11. Kb6 12. Kc5 13. Kc4 14. Kd3 15. Kc2 16. Kb1 17. Kxa1[+sKe8] 18. Kd8 19. Kc8 20. Kb8 21. Ka7 22. Kb6 23. Kc5 24. Kd4 25. Ke3 26. Kf2 27. Kg1 28. Kxh1[+sKe8] Th8#
Gerald Ettl: Der sK muss alle Eckfiguren abraeumen. (GE) (2009-02-10)
K?vanç Çefle: Why 17. Th8 is not mate? After 17...Kd8(or f8) 18. Txd8(or f8)[+wTa1] seems legal to me... (2022-05-13)
SCHRECKE: 17.K:a1[+sKe8] 18.Th8= (Patt) (2022-05-13)
SCHRECKE: without 18. (2022-05-13)
Kivanç Çefle: So, 18. Th8 is not a checking move? That is strange... (2022-05-13)
paul: To Kivanç Çefle: 17... Th8 is not check because the square h1 is occupied. (2022-05-14)
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K?vanç Çefle: Why 17. Th8 is not mate? After 17...Kd8(or f8) 18. Txd8(or f8)[+wTa1] seems legal to me... (2022-05-13)
SCHRECKE: 17.K:a1[+sKe8] 18.Th8= (Patt) (2022-05-13)
SCHRECKE: without 18. (2022-05-13)
Kivanç Çefle: So, 18. Th8 is not a checking move? That is strange... (2022-05-13)
paul: To Kivanç Çefle: 17... Th8 is not check because the square h1 is occupied. (2022-05-14)
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Keywords: Seriesmover, Four corners (k), Circe (Anti-), Switchback (k), Model mate
Genre: Fairies
Computer test: Popeye 4.61
FEN: N4k1N/8/P2PP2R/7P/K7/8/8/B6B
Input: Felber, Volker, 2000-01-06
Last update: Arnold Beine, 2022-09-22 more...
Genre: Fairies
Computer test: Popeye 4.61
FEN: N4k1N/8/P2PP2R/7P/K7/8/8/B6B
Input: Felber, Volker, 2000-01-06
Last update: Arnold Beine, 2022-09-22 more...
1. f4 2. f3 3. f2 4. Kf7 5. Kg7 6. Kh8 7. f1=T 8. Tc1 9. Dd1 10. h5 11. h4 12. h3 13. h2 14. h1=T 15. Tg1 g7=
Gerald Ettl: Und nicht 14.h1D 15.Dg1 wegen Dxg7[Dd8] nette Rundlinien-Einmauerung(GE) (2009-02-10)
Ladislav Packa: Anticirce Cheylan (2023-02-28)
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Ladislav Packa: Anticirce Cheylan (2023-02-28)
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Keywords: Seriesmover, Miniature, Circe (Anticirce Cheylan)
Genre: Fairies
Computer test: PY V4.45
FEN: 8/8/4kNPp/5p1q/8/8/8/K7
Input: Felber, Volker, 2000-01-06
Last update: A.Buchanan, 2023-03-01 more...
Genre: Fairies
Computer test: PY V4.45
FEN: 8/8/4kNPp/5p1q/8/8/8/K7
Input: Felber, Volker, 2000-01-06
Last update: A.Buchanan, 2023-03-01 more...
59 - P0579877
Manfred Rittirsch
F1007v The Problemist 05/1988
Walter Wittstock gewidmet
1. Preis
(0+0+7)
h#4 0.1...
b) nSe3 nach f8
Manfred Rittirsch
F1007v The Problemist 05/1988
Walter Wittstock gewidmet
1. Preis
(0+0+7)
h#4 0.1...
b) nSe3 nach f8
a) 1. ... nSeg4 2. nBf5 nKg6+ 3. nKh7 nLd4 4. nBg5 nBfxg6ep#
b) 1. ... nDb8 2.nKf6+ nKe7 3.nBg5 nLh4 4.nBf5 nBgxf6ep#
b) 1. ... nDb8 2.nKf6+ nKe7 3.nBg5 nLh4 4.nBf5 nBgxf6ep#
Genre: Fairies
FEN: 8/5-P-P1/7-N/6-K1/8/4-N3/5-B2/1-Q6
Reprints: M13 mpk-Blätter 03/2011
Input: hpr, 2001-01-21
Last update: Frank Müller, 2011-03-12 more...
1. Kxa8 e5 2. Lg1 Kxg1 3. Kb8 Kh1 4. Kc8 Kg1 5. Kd8 Kh1 6. Ke8 Kg1 7. Kf8 Kh1 8. Kg7 Kg1 9. Kh6 Kh1 10. Kg5 Kg1 11. Kf4 Kh1 12. Ke4 Kg1 13. Kd5 Kh1 14. Kc5 Kg1 15. Kb5 Kh1 16. Ka4 Kg1 17. Kb3 Kh1 18. Kxc3 Kg1 19. Kd4 Kh1 20. c3 Kg1 21. c2 Kh1 22. c1=S Kg1 23. Sd3 exd3 24. Kd5 d4 25. Kxe6 Lc4#
Genre: h#
Computer test: C+ Gustav 4.1d
FEN: R7/k1p1p3/p1P1P1p1/P7/2p1P1P1/2P1p1p1/4P1Pb/5B1K
Input: hpr, 2001-08-12
Last update: A.Buchanan, 2021-04-07 more...
1. Sf3 a5 2. Sd4 a4 3. Sb3 axb3 4. Sc3 bxc2 5. Se4 cxd1=L 6. Sg5 Lb3 7. Sxh7 Lc4 8. Sg5 Th5 9. Sh7 Tha5 10. Kd1 d5 11. Kc2 Kd7 12. Kc3 Ke6 13. Kd4 Kf5 14. Ke3 Le6 15. Kf3 Sd7 16. Kg3 Tc8 17. Kh4 g5+ 18. Kh3 Lg7 19. Kg3 Le5+ 20. Kf3 Ld6 21. Ke3 Kg4 22. Kd4 c5+ 23. Kc3 Db6 24. Kc2 Db4 25. Kd1 b5 26. Ke1
17-zügige Wanderung des wK bis nach h3 zum Tempoverlust.
17-zügige Wanderung des wK bis nach h3 zum Tempoverlust.
paul: Un dernier coup noir s'impose. Voir P0005906. (2010-05-08)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 1 Sekunde.
Keine Lösung: BP 24.5, BP 25.0. (2023-02-17)
Reto: There are in fact solutions in exactly 25.0 moves and Stelvio finds those. So the above comment is incorrect. (2023-06-12)
Moldenhauer: Danke für die Korrektur. In meiner Worddatei im Bildschirmausschnitt steht cooked,
also habe ich die Eingabe hier falsch gemacht. Sorry! (2023-11-05)
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Moldenhauer: Computerprüfung: C+ Stelvio 1.0 1 Sekunde.
Keine Lösung: BP 24.5, BP 25.0. (2023-02-17)
Reto: There are in fact solutions in exactly 25.0 moves and Stelvio finds those. So the above comment is incorrect. (2023-06-12)
Moldenhauer: Danke für die Korrektur. In meiner Worddatei im Bildschirmausschnitt steht cooked,
also habe ich die Eingabe hier falsch gemacht. Sorry! (2023-11-05)
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Keywords: Unique Proof Game, Promenade (K), Tempo Loss (K17)
Genre: Retro
Computer test: Stelvio 1.0
FEN: 2r3n1/3npp1N/3bb3/rppp2p1/1qb3k1/8/PP1PPPPP/R1B1KB1R
Reprints: feenschach 137 08-09/2000
H25 FIDE Album Annexe 1992-1994 2003
Input: Gerd Wilts, 2001-08-29
Last update: Silvio Baier, 2023-03-02 more...
Genre: Retro
Computer test: Stelvio 1.0
FEN: 2r3n1/3npp1N/3bb3/rppp2p1/1qb3k1/8/PP1PPPPP/R1B1KB1R
Reprints: feenschach 137 08-09/2000
H25 FIDE Album Annexe 1992-1994 2003
Input: Gerd Wilts, 2001-08-29
Last update: Silvio Baier, 2023-03-02 more...
1) 1. gxh4 a8=L 2. h3 Lxg2 3. Dxg2 f8=D 4. Dg7 Dh8 5. Dxb2 Dxb2#
2) 1. gxf4 f8=D 2. f3 Db4 3. f2 Dxb3 4. Kxb3 a8=L 5. Ka2 Ld5#
2) 1. gxf4 f8=D 2. f3 Db4 3. f2 Dxb3 4. Kxb3 a8=L 5. Ka2 Ld5#
Anton Baumann: C+ Gustav 4.1d (2020-06-09)
Teppo Mänttä: reprint: Tehtäväniekka n:o 3, page 142, 14/09/2016. Published in Kari Valtonen's article about Hungarian themes and composers. (2021-04-28)
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Teppo Mänttä: reprint: Tehtäväniekka n:o 3, page 142, 14/09/2016. Published in Kari Valtonen's article about Hungarian themes and composers. (2021-04-28)
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Keywords: Capture key, Promotion (D), under-promotion (L), konsekutive Umwandlungen 2 (LD, DL), Square vacation, Opferbahnung (in 1), Model mate (in 2)
Genre: h#
FEN: 8/P4P2/4P3/1P4p1/5P1P/1p1p4/kPp3pp/rnK2nrq
Reprints: 5) idee & form 154, p. 5223, 04/2022
Input: hpr, 2002-03-02
Last update: Gunter Jordan, 2023-01-23 more...
Genre: h#
FEN: 8/P4P2/4P3/1P4p1/5P1P/1p1p4/kPp3pp/rnK2nrq
Reprints: 5) idee & form 154, p. 5223, 04/2022
Input: hpr, 2002-03-02
Last update: Gunter Jordan, 2023-01-23 more...
63 - P1004179
Arpad Molnar
R275 The Problemist 05/1998
4. ehrende Erwähnung
(11+10) cooked
Letzte 12 Einzelzüge?
Arpad Molnar
R275 The Problemist 05/1998
4. ehrende Erwähnung
(11+10) cooked
Letzte 12 Einzelzüge?
R: 1. a4xSb3+ Dd8-b6 2. b6xSa5 Tc8-c6 3. c7xSb6 Lc6-b5 4. b5xSa4 Sa8-b6 5. b6-b5 Lb7-c6 6. Kb5-b4 c3-c4+
Keywords: Ceriani-Frolkin Theme (SSSS), Last Moves? (12)
Genre: Retro
FEN: 1N6/p2p1ppp/RQRp4/pBNP4/1kP5/bp6/PPK5/8
Reprints: (1) Quartz 18 10-12/2001
Input: Gerd Wilts, 2002-09-05
Last update: Gerd Wilts, 2004-08-28 more...
Genre: Retro
FEN: 1N6/p2p1ppp/RQRp4/pBNP4/1kP5/bp6/PPK5/8
Reprints: (1) Quartz 18 10-12/2001
Input: Gerd Wilts, 2002-09-05
Last update: Gerd Wilts, 2004-08-28 more...
1) 1. Td6 Ld5+ 2. Kxd5 Sc7#
2) 1. Tf5 Le5 2. Kxe5 Te4#
2) 1. Tf5 Le5 2. Kxe5 Te4#
Harry Fougiaxis: Anticipated by P0583914 (2011-03-06)
Vitaly Medintsev: anticipated by P0579662 (2021-09-11)
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Vitaly Medintsev: anticipated by P0579662 (2021-09-11)
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Genre: h#
Computer test: (Popeye WINDOWS98-32Bit-Version 3.75 (2048 KB))
FEN: B3N2B/8/4k3/3r4/2R5/1K6/8/8
Input: hpr, 2002-05-20
Last update: hpr, 2012-11-01 more...
1) 1. Kg1 Ld5 2. Sf2 Th2 3. Sh1 Tg2#
2) 1. Kh1 Tg5 2. Sh2 Lf1 3. Sg1 Lg2#
2) 1. Kh1 Tg5 2. Sh2 Lf1 3. Sg1 Lg2#
Yuri Bilokin: you can see 1...Rg5 2.Sg3 Bf1 3.Sh1 Rg2# (MM)
1...Bd5 2.Sg3 Rg5 3.Sh1 Rg2# (MM)
1...Rg5 2.Sg3 Bd5 3.Sh1 Rg2# (MM) (2022-08-06)
Jean Marc Loustau: ****
Compare with P1100532. (2022-08-14)
comment
1...Bd5 2.Sg3 Rg5 3.Sh1 Rg2# (MM)
1...Rg5 2.Sg3 Bd5 3.Sh1 Rg2# (MM) (2022-08-06)
Jean Marc Loustau: ****
Compare with P1100532. (2022-08-14)
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Genre: h#
Computer test: (Popeye WINDOWS98-32Bit-Version 3.75 (2048 KB))
FEN: 8/1K6/8/7R/2B5/7n/7k/5n2
Input: hpr, 2002-05-20
Last update: hpr, 2012-11-01 more...
1) 1. Tc5 Tb6 2. Lb5 Txa6 3. Kc4 Ta3 4. Sd4 d3#
2) 1. Sd6 Tb7 2. f5 Tg7 3. Df6 Tg4+ 4. Ke5 d4#
2) 1. Sd6 Tb7 2. f5 Tg7 3. Df6 Tg4+ 4. Ke5 d4#
Yuri Bilokin: Version: bPa6-a5, -bPa7 1R1n1r2/7K/4pp2/pr1pqn2/1pbk4/8/3P4/8(3+12) h#4 2.1…
1.Rc5 Ra8 2.Bb5 Rxa5 3.Kc4 Ra3 4.Sd4 d3# (MM)
1.Sd6 Rb7 2.f5 Rg7 3.Qf6 Rg4+ 4.Ke5 d4# (MM) (2022-04-24)
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1.Rc5 Ra8 2.Bb5 Rxa5 3.Kc4 Ra3 4.Sd4 d3# (MM)
1.Sd6 Rb7 2.f5 Rg7 3.Qf6 Rg4+ 4.Ke5 d4# (MM) (2022-04-24)
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Genre: h#
Computer test: (Popeye WINDOWS98-32Bit-Version 3.75 (2048 KB))
FEN: 1R1n1r2/p6K/p3pp2/1r1pqn2/1pbk4/8/3P4/8
Input: hpr, 2002-05-20
Last update: hpr, 2012-11-01 more...
(I. Jarmonow). 1.Kb8 [2.a8D,L], 1.- Da4 2.c4+ T:c4/D:c4 3.Td4/a8D+ T:d4(L:d4) Dc6 4.Sc3(Df3) D:c6; 1.- Da1 2.Sc3+ L:c3/D:c3 3.Dd4+/a8D+ L:d4(T:d4) Dc6 4.c4(Tf5) D:c6 (1.- Dc5 2.a8D+ Kc4 3.Da4+ Kd5 4.D:e4; 1.- D:e3 2.a8D,L+ Kd4 3.D:d6). Nach Ablenkung der sD folgt deren Verbahnung sowie die Lenkung des L bzw. T über den Schnittpunkt d4, wonach der Nowotny erfolgen kann. Wie 9748 ein imponierendes Problem durch seine strategischen Dichte und Einheitlichkeit (MOe: 4,5/I-II). A good idea (LV).
Duplicate Diagram: P1058393
Felber, Volker: Duplikat unter P1058393 (2011-03-21)
Anton Baumann: Auszeichnung: 1. Preis
C+ Gustav 4.2a (2022-11-10)
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Anton Baumann: Auszeichnung: 1. Preis
C+ Gustav 4.2a (2022-11-10)
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Genre: n#
Computer test: Anton Baumann (2022-11-10): C+ Gustav 4.2a
FEN: K1R3bB/P4p2/1p1p1Q2/3kb1N1/3qrR2/4P3/2P1r3/1N6
Reprints: I The Problemist , p. 252, 11/2001
Input: Gerd Wilts, 2003-06-29
Last update: Mario Richter, 2022-11-10 more...
1. Dxc5! fxg1=D 2. f6 Df2 3. Tf5 Dxf5 4. Dxf5#
3. ... Df4,Df3 4. Txf4,Txf3#
1. ... fxg1=T 2. Sxg3+ hxg3 3. De7 e2+ 4. Dxe2#
1. ... fxg1=L 2. Dxa5 Lf2 3. Dxb4 Lg1,Le1 4. De1#
1. ... fxg1=S 2. Dxe3 Sxh3 3. Sxg3+ hxg3 4. De1#
2. ... Sxe2 3. Kd2! Sc3,S~ 4. De1#
1. Db8? fxg1=D!
1. Dc7? fxg1=D!
1. De6? fxg1=L!
1. f6? fxg1=L!
1. Dh8? fxg1=S!
Version aus Caissa's Schlossbewohner 4. Alle Varianten sind dualfrei. Original siehe P1185244
3. ... Df4,Df3 4. Txf4,Txf3#
1. ... fxg1=T 2. Sxg3+ hxg3 3. De7 e2+ 4. Dxe2#
1. ... fxg1=L 2. Dxa5 Lf2 3. Dxb4 Lg1,Le1 4. De1#
1. ... fxg1=S 2. Dxe3 Sxh3 3. Sxg3+ hxg3 4. De1#
2. ... Sxe2 3. Kd2! Sc3,S~ 4. De1#
1. Db8? fxg1=D!
1. Dc7? fxg1=D!
1. De6? fxg1=L!
1. f6? fxg1=L!
1. Dh8? fxg1=S!
Version aus Caissa's Schlossbewohner 4. Alle Varianten sind dualfrei. Original siehe P1185244
Felber, Volker: The Problemist gibt 'H. Bäcker' als Autor an. (2011-11-19)
Olaf Jenkner: Die beiden sind identisch. (2011-11-19)
Wilfried Neef: Horst Deichelbohrer has changed his name into Horst Bäcker some decades ago (2021-12-14)
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Olaf Jenkner: Die beiden sind identisch. (2011-11-19)
Wilfried Neef: Horst Deichelbohrer has changed his name into Horst Bäcker some decades ago (2021-12-14)
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Keywords: Allumwandlung, black
Genre: n#
Computer test: Popeye v4.55
FEN: 2Q5/8/8/p1pR1P2/PpP3Pp/1P1Np1pB/4Npp1/3K1kBb
Reprints: Caissas Schloßbewohner 4 1991
Problemkiste (74-75), p. 45, 05/1991
Q The Problemist 23-6, p. 225, 11/2011
Input: HBae, 2003-09-27
Last update: Alfred Pfeiffer, 2015-11-16 more...
Genre: n#
Computer test: Popeye v4.55
FEN: 2Q5/8/8/p1pR1P2/PpP3Pp/1P1Np1pB/4Npp1/3K1kBb
Reprints: Caissas Schloßbewohner 4 1991
Problemkiste (74-75), p. 45, 05/1991
Q The Problemist 23-6, p. 225, 11/2011
Input: HBae, 2003-09-27
Last update: Alfred Pfeiffer, 2015-11-16 more...
1.Sc3+ Ke8 2.Te4+ (2.Sb1? Kf8 3.Td8#) Kd7 (Kf8 3.Sb1 Kg8 4.Ke5 5.Tc4) 3.Sb1 Kd6 4.Td4+ Kc7 5.Ke5 (5.Tc4? Kd6!) 6.Tc4 7.Kd4 8.Kc3 9.Sd2 b1=D#. 1...Kc7 2.Sb1 Kb6 3.Te5 as above (3.Ke5? Kxb5 4.Kd5 Ka4 5.Tc4 Kb3 6.Kd4 Ka2 7.Kxc3 Kxb1)
Genre: r#
FEN: 3k4/8/5K2/1P1N4/1P1R4/8/1p6/b7
Input: Paul Valois, 2001-05-31
Last update: Paul Valois, 2001-05-31 more...
1. g4 b5 2. a4 h5 3. axb5 hxg4 4. b3 g6 5. La3 Lh6 6. Ld6 Le3 7. fxe3 cxd6 8. Ta4 Th5 9. Tf4 Tc5 10. b4 g5 11. bxc5 gxf4 12. h4 a5 13. h5 a4 14. h6 a3 15. h7 a2 16. h8=D a1=L 17. De5 Ld4 18. exd4 dxe5 19. Th3 Ta3 20. Te3 Td3 21. Sf3 Sc6 22. bxc6 gxf3 23. Lh3 fxe3 24. Le6 dxe6 25. exd3 Dd6 26. Sc3 Sf6 27. Se4 Sd5 28. c4 f5 29. De2 Ld7 30. cxd5 fxe4 31. cxd6 fxe2 32. cxd7+
No. 1462 HN
Moldenhauer: Computerprüfung: C+ NUPG Stelvio 1.11 00:05:52 Minuten. (hh:mm:ss)
Keine Lösung: BP 30.5, BP 31.0.
Beispiel: 1.Sc3 Sf6 2.a4 Sd5 3.b4 b5 4.Lb2 c5 5.axb5 h5 6.g4 hxg4 7.Ta6 Th3 8.Tf6 Td3 9.exd3 Sc6 10.bxc6 a5 11.h4 a4 12.h5 Ta5 13.h6 a3 14.h7 gxf6 15.h8D f5 16.Dd4 Lh6 17.Th3 Le3 18.fxe3 a2 19.e4 a1D 20.Te3 cxd4 21.Lh3 Tc5 22.bxc5 dxe3 23.exd5 d6 24.Se4 fxe4 25.Le5 Dd4 26.Df3 dxe5 27.Se2 Dd6 28.cxd6 Ld7 29.c3 gxf3 30.Le6 fxe6 31.cxd4 fxe2 32.cxd7+ (2023-04-24)
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Moldenhauer: Computerprüfung: C+ NUPG Stelvio 1.11 00:05:52 Minuten. (hh:mm:ss)
Keine Lösung: BP 30.5, BP 31.0.
Beispiel: 1.Sc3 Sf6 2.a4 Sd5 3.b4 b5 4.Lb2 c5 5.axb5 h5 6.g4 hxg4 7.Ta6 Th3 8.Tf6 Td3 9.exd3 Sc6 10.bxc6 a5 11.h4 a4 12.h5 Ta5 13.h6 a3 14.h7 gxf6 15.h8D f5 16.Dd4 Lh6 17.Th3 Le3 18.fxe3 a2 19.e4 a1D 20.Te3 cxd4 21.Lh3 Tc5 22.bxc5 dxe3 23.exd5 d6 24.Se4 fxe4 25.Le5 Dd4 26.Df3 dxe5 27.Se2 Dd6 28.cxd6 Ld7 29.c3 gxf3 30.Le6 fxe6 31.cxd4 fxe2 32.cxd7+ (2023-04-24)
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Keywords: Non-Unique Proof Game, Ornament, Kindergarten Problem
Genre: Retro
FEN: 4k3/3Pp3/3Pp3/3Pp3/3Pp3/3Pp3/3Pp3/4K3
Input: Henri Nouguier, 2004-01-11
Last update: James Malcom, 2021-02-25 more...
Genre: Retro
FEN: 4k3/3Pp3/3Pp3/3Pp3/3Pp3/3Pp3/3Pp3/4K3
Input: Henri Nouguier, 2004-01-11
Last update: James Malcom, 2021-02-25 more...
1. g4 g5 2. Lg2 e5 3. Lxb7 Lxb7 4. a4 Lxh1 5. e4 Lb4 6. a5 c5 7. a6 Da5 8. c4 Dxa1 9. Sc3 Dxc1 10. b3 Dxc3 11. Sh3 Dxh3 12. Df3 Lxf3=
Weiß ist patt.
Weiß ist patt.
No. 1473 HN
A.Buchanan: This is not intended to be a unique proof game, so is not cooked (2023-03-02)
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A.Buchanan: This is not intended to be a unique proof game, so is not cooked (2023-03-02)
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Keywords: Non-Unique Proof Game
Genre: Retro
Computer test: Computerprüfung: Stelvio 1.0 cooked in 1 Sekunde. Natch 3.3 64 Lösungen in 4 Sekunden. Da NUPG C+ Stelvio 1.0, Natch 3.3.
FEN: rn2k1nr/p2p1p1p/P7/2p1p1p1/1bP1P1P1/1P3b1q/3P1P1P/4K3
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2023-03-02 more...
Genre: Retro
Computer test: Computerprüfung: Stelvio 1.0 cooked in 1 Sekunde. Natch 3.3 64 Lösungen in 4 Sekunden. Da NUPG C+ Stelvio 1.0, Natch 3.3.
FEN: rn2k1nr/p2p1p1p/P7/2p1p1p1/1bP1P1P1/1P3b1q/3P1P1P/4K3
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2023-03-02 more...
1. Kd2 2. Ke3! 3. Kd4 4. Ke5 5. Kd6 6. Kd7 7. Ke8 8. 0-0 9. bxc3ep Lxb3#
1. Kd2 2. Kd3? - W hat keinen letzten Zug
1. Kd2 2. Kd3? - W hat keinen letzten Zug
No. 1857 HN
Original ohne wBb2, mit wKb1 und sLa3 veröffentlicht, dann aber NL 1. bxc3ep, weil danach W als letzten Zug Kc2-b1 hatte.
In der korrigierten Stellung (03/1988, S.400) hat W nach bxc3ep nie einen letzten Zug, so daß dieser ep-Schlag erst am Ende der Lösung gespielt werden darf.
Außerdem bietet der Autor noch einen Zwilling an: Ks a5/a4, zusätzlich Monchromes Schach, shc#6
('The Problemist' (03/1988, S.400): "Composer's Mono-chromatic setting (1.bc ep.) saves another unit, but has less interest."
paul: Or directly 1.bxc3ep etc. (2011-10-24)
James Malcom: Is this cooked, as paul's comment seems to indicate? (2021-09-14)
Henrik Juel: No, I don't think so, James
After 1.bxc3ep White has no previous move (2021-09-15)
Mario Richter: Paul's comment was obviously referring to the origial position, not to the one now presented in the diagram.
(The germsn comment above reads: Originally published without white pawn b2, with wKb1 and black Bishop a3. But then cooked by 1. bxc3ep, because W has Kc2-b1 as last move [preceeded by c4xb3+]) (2021-09-16)
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Original ohne wBb2, mit wKb1 und sLa3 veröffentlicht, dann aber NL 1. bxc3ep, weil danach W als letzten Zug Kc2-b1 hatte.
In der korrigierten Stellung (03/1988, S.400) hat W nach bxc3ep nie einen letzten Zug, so daß dieser ep-Schlag erst am Ende der Lösung gespielt werden darf.
Außerdem bietet der Autor noch einen Zwilling an: Ks a5/a4, zusätzlich Monchromes Schach, shc#6
('The Problemist' (03/1988, S.400): "Composer's Mono-chromatic setting (1.bc ep.) saves another unit, but has less interest."
paul: Or directly 1.bxc3ep etc. (2011-10-24)
James Malcom: Is this cooked, as paul's comment seems to indicate? (2021-09-14)
Henrik Juel: No, I don't think so, James
After 1.bxc3ep White has no previous move (2021-09-15)
Mario Richter: Paul's comment was obviously referring to the origial position, not to the one now presented in the diagram.
(The germsn comment above reads: Originally published without white pawn b2, with wKb1 and black Bishop a3. But then cooked by 1. bxc3ep, because W has Kc2-b1 as last move [preceeded by c4xb3+]) (2021-09-16)
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Keywords: En passant, Grid Chess, Seriesmover, Consequent, Castling
Genre: Retro, Fairies
FEN: 7r/8/8/8/1pP5/Kp6/BP6/4k3
Input: Henri Nouguier, 2004-01-11
Last update: Mario Richter, 2021-09-15 more...
Genre: Retro, Fairies
FEN: 7r/8/8/8/1pP5/Kp6/BP6/4k3
Input: Henri Nouguier, 2004-01-11
Last update: Mario Richter, 2021-09-15 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#
74 - P1012460
Andrey Frolkin
Sergiy O. Komarov
The Problemist 1/1990
2. ehrende Erwähnung
(11+7)
shc#7
Vertikaler Zylinder
Andrey Frolkin
Sergiy O. Komarov
The Problemist 1/1990
2. ehrende Erwähnung
(11+7)
shc#7
Vertikaler Zylinder
1. a5 2. a4 3. a3 4. a2 5. a1=T 6. Ta8 7. 0-0 (castling with a queenside rook!) Tg3#
The interest here is in the try 5. h1=T? 6. Th8, when, in a game to that position, the sK or sT must have moved to let out Ta8 to be captured (0-0?). The solution is a surprise, involving castling round the cylinder instead.
The interest here is in the try 5. h1=T? 6. Th8, when, in a game to that position, the sK or sT must have moved to let out Ta8 to be captured (0-0?). The solution is a surprise, involving castling round the cylinder instead.
No. 2142 HN
James Malcom: Due to the implementation of odds castling, the castling animation is off here. Is there any possible fix or circumvention? (2021-09-14)
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James Malcom: Due to the implementation of odds castling, the castling animation is off here. Is there any possible fix or circumvention? (2021-09-14)
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Keywords: Seriesmover, Consequent, Vertical Cylinder (Vertikaler), Homebase (s), Castling, Excelsior, Promotion (t)
Genre: Retro, Fairies
FEN: 4k3/pppp1p1p/1N3B2/4P3/4PK2/5R2/1PPPPP2/8
Input: Henri Nouguier, 2004-01-11
Last update: James Malcom, 2021-09-14 more...
Genre: Retro, Fairies
FEN: 4k3/pppp1p1p/1N3B2/4P3/4PK2/5R2/1PPPPP2/8
Input: Henri Nouguier, 2004-01-11
Last update: James Malcom, 2021-09-14 more...
(a) l. g2-glB+ by Black.
(b) 1. Kh3-h2 by White. This is the only way we can get back without having 2 consecutive moves from any man e.g. Kh3-h2, g2-g1B, h2xSg3, Se4-g3, Kh4-h3, h3xBg2, Bfl-g2 Sg3xRh1, g2xRf3 etc.
Fuddled Men is a recent Fairy form invented by John Beasley, in which all men (including Kings) are unable to make 2 moves in succession. This is a Fairy form with great potential both on its own, and combined with other Fairy rules.
(b) 1. Kh3-h2 by White. This is the only way we can get back without having 2 consecutive moves from any man e.g. Kh3-h2, g2-g1B, h2xSg3, Se4-g3, Kh4-h3, h3xBg2, Bfl-g2 Sg3xRh1, g2xRf3 etc.
Fuddled Men is a recent Fairy form invented by John Beasley, in which all men (including Kings) are unable to make 2 moves in succession. This is a Fairy form with great potential both on its own, and combined with other Fairy rules.
No. 2184 HN
Walter Lindenthal: Why not (b) as follows:
(b) 1. wKg2-h2 by White (provided bBg1 is 'fuddled' at that moment).
The wBf1 was captured possibly by a bS before (or the wK gave way for its capture), it need not have been ... Bfl-g2 h3xBg2 ...
So, the last moves: ... bPh3-h2 wKf1-g2 ... bPx?g1=B wKg2-h2
Yes, this will result in an 'automatic' Mate after black's next move, but it seems fully legal under Fuddled Men rules! (2021-01-17)
A.Buchanan: @Walter if "bPh3-h2 wKf1-g2 ... bPx?g1=B wKg2-h2" tHen what moves can be in the "..."? There are no spare black pieces to be uncaptured, as S&R are captured by wPg & wPh. (2021-01-17)
Walter Lindenthal: Oh yes, agreed. Thanks! (2021-01-18)
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Walter Lindenthal: Why not (b) as follows:
(b) 1. wKg2-h2 by White (provided bBg1 is 'fuddled' at that moment).
The wBf1 was captured possibly by a bS before (or the wK gave way for its capture), it need not have been ... Bfl-g2 h3xBg2 ...
So, the last moves: ... bPh3-h2 wKf1-g2 ... bPx?g1=B wKg2-h2
Yes, this will result in an 'automatic' Mate after black's next move, but it seems fully legal under Fuddled Men rules! (2021-01-17)
A.Buchanan: @Walter if "bPh3-h2 wKf1-g2 ... bPx?g1=B wKg2-h2" tHen what moves can be in the "..."? There are no spare black pieces to be uncaptured, as S&R are captured by wPg & wPh. (2021-01-17)
Walter Lindenthal: Oh yes, agreed. Thanks! (2021-01-18)
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Keywords: Fuddled Men, Last Move?
Genre: Retro, Fairies
FEN: 2bqkb2/ppppppp1/8/8/8/5PP1/4PP1K/6bn
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2016-09-23 more...
Genre: Retro, Fairies
FEN: 2bqkb2/ppppppp1/8/8/8/5PP1/4PP1K/6bn
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2016-09-23 more...
a) 1. De6? droht 2. Dc4#
1. ... Kxd3 2. 0-0-0?
1. Dd7! Kxb3 2. Da4#
b) 1. Ta3! Kb1 2. Kd2#
c) 1. De6! droht 2. Dc4#
1. ... Kxd3 2. 0-0-0#
1. Dd7? Kb3 2. Da4+? Kxa4!
1. ... Kxd3 2. 0-0-0?
1. Dd7! Kxb3 2. Da4#
b) 1. Ta3! Kb1 2. Kd2#
c) 1. De6! droht 2. Dc4#
1. ... Kxd3 2. 0-0-0#
1. Dd7? Kb3 2. Da4+? Kxa4!
No. 2220 HN
Henrik Juel: In parts a and b Ta1 has moved (because Kc2 just came from b1), so White may not castle
These parts are C+ Popeye 4.61
Part c is not clearly defined; if it is Bb3-a3, then it also is C+ Popeye 4.61 (2022-02-19)
A.Buchanan: I agree about part (c). I wonder though about (b) - it seems pointless. If there is also a part (d) which combines (b)&(c), then it does have a unique solution (with checking key) 1. De2+ Kb3 2. Sc5# (2022-02-20)
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Henrik Juel: In parts a and b Ta1 has moved (because Kc2 just came from b1), so White may not castle
These parts are C+ Popeye 4.61
Part c is not clearly defined; if it is Bb3-a3, then it also is C+ Popeye 4.61 (2022-02-19)
A.Buchanan: I agree about part (c). I wonder though about (b) - it seems pointless. If there is also a part (d) which combines (b)&(c), then it does have a unique solution (with checking key) 1. De2+ Kb3 2. Sc5# (2022-02-20)
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1. b8=S? droht 2. Sc6#
1. ... Gxa6!
1. bxc6ep! Gxd4,Gxb6,~ 2. Lb5#
R: 1. c7-c5 2. Gd7-d5+ Gb8-d6
'The Problemist', p. 241, 09/1993: "The solution to this #2 with Grasshoppers is 1.bxc6
en passant. lf Black's last move had been 1...c6, then White's previous move would have been the illegal G(d7)-d5+.
Cook: 1. Sc4+ Kxa4 2. Sab6#
1. ... Gxa6!
1. bxc6ep! Gxd4,Gxb6,~ 2. Lb5#
R: 1. c7-c5 2. Gd7-d5+ Gb8-d6
'The Problemist', p. 241, 09/1993: "The solution to this #2 with Grasshoppers is 1.bxc6
en passant. lf Black's last move had been 1...c6, then White's previous move would have been the illegal G(d7)-d5+.
Cook: 1. Sc4+ Kxa4 2. Sab6#
No. 2251 HN
A.Buchanan: Diagram error? Solution typo? (2023-01-06)
Mario Richter: Diagram error corrected (2023-01-07)
Ulrich Voigt: NL: 1. Sc4+ Kxa4 2. Sab6# (2023-01-08)
Mario Richter: Vielleicht sind Dawson und alle, die diese Aufgabe unkritisch nachgedruckt haben, einer typischen Halluzination erlegen: im Geiste nimmt man den Zug c7-c5 zurück, der erst den ep-Key ermöglicht, und meint dann, daß die NL: 1. Sc4+ Kxa4 2. Sab6 an 2. ... c7xb6 scheitert ... (2023-01-09)
A.Buchanan: Easily repaired e.g. by replacing wSa8 with wLa7 (2023-01-09)
comment
A.Buchanan: Diagram error? Solution typo? (2023-01-06)
Mario Richter: Diagram error corrected (2023-01-07)
Ulrich Voigt: NL: 1. Sc4+ Kxa4 2. Sab6# (2023-01-08)
Mario Richter: Vielleicht sind Dawson und alle, die diese Aufgabe unkritisch nachgedruckt haben, einer typischen Halluzination erlegen: im Geiste nimmt man den Zug c7-c5 zurück, der erst den ep-Key ermöglicht, und meint dann, daß die NL: 1. Sc4+ Kxa4 2. Sab6 an 2. ... c7xb6 scheitert ... (2023-01-09)
A.Buchanan: Easily repaired e.g. by replacing wSa8 with wLa7 (2023-01-09)
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Keywords: En passant as key
Pieces: = Grasshopper (G)
Genre: Retro, Fairies
FEN: N2K4/1P6/PN1*2q4/kPp*2Q4/Pp1R4/p2B4/8/8
Reprints: The Problemist , p. 241, 09/1993
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2023-01-09 more...
Pieces: = Grasshopper (G)
Genre: Retro, Fairies
FEN: N2K4/1P6/PN1*2q4/kPp*2Q4/Pp1R4/p2B4/8/8
Reprints: The Problemist , p. 241, 09/1993
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2023-01-09 more...
78 - P1012699
Dennis K. Hale
R57 The Problemist 05/1980
(12+10)
Ist die Stellung legal?
b) ferner sBg3 nach g7
c) ferner sBg4 nach h5
d) ferner +sBf4
e) ferner wBh3 nach h4
f) ferner h5 nach g6
g) ferner g6 nach g4
h) ferner a4 nach a3
i) ferner -sBf4
j) ferner b3 nach b2
k) ferner g7 nach g3
Dennis K. Hale
R57 The Problemist 05/1980
(12+10)
Ist die Stellung legal?
b) ferner sBg3 nach g7
c) ferner sBg4 nach h5
d) ferner +sBf4
e) ferner wBh3 nach h4
f) ferner h5 nach g6
g) ferner g6 nach g4
h) ferner a4 nach a3
i) ferner -sBf4
j) ferner b3 nach b2
k) ferner g7 nach g3
a) ja
b) nein
c) ja
d) nein
e) ja
f) nein
g) ja
h) nein
i) ja
j) nein
k) ja
Dans toutes les positions on doit reprendre PNcxb7 et on en déduit que TBb8 et 1TN est de promotion; PBf a fait 4 captures. a) Oui. PNc:Tb1 et les captures axb, bxa des PB ont laisser sortir la T. PNf:Tg1. b) Non. Captures B trop nombreuses. c) Oui. PNf:Te1 et PBa est en a4 d) Non.FB ne peut être pris par PNb ou c, donc PNc ne peut se promouvoir en a1. e) Oui. PNc:Ta1, PNb ou c xTBh1. f) Non. FBc1 ne peut être pris en c6 ou g6. g) Oui. FBc1 capturé en g5. h) Non. PBc promu ne pourrait plus sortir. i) Oui. PNf promu en T en g1. j) Non. TBa1, Fc1 non disponibles pour les prises de PN. k) Oui. PNh:Th1; les captures des PBh sont g3,h4.
b) nein
c) ja
d) nein
e) ja
f) nein
g) ja
h) nein
i) ja
j) nein
k) ja
Dans toutes les positions on doit reprendre PNcxb7 et on en déduit que TBb8 et 1TN est de promotion; PBf a fait 4 captures. a) Oui. PNc:Tb1 et les captures axb, bxa des PB ont laisser sortir la T. PNf:Tg1. b) Non. Captures B trop nombreuses. c) Oui. PNf:Te1 et PBa est en a4 d) Non.FB ne peut être pris par PNb ou c, donc PNc ne peut se promouvoir en a1. e) Oui. PNc:Ta1, PNb ou c xTBh1. f) Non. FBc1 ne peut être pris en c6 ou g6. g) Oui. FBc1 capturé en g5. h) Non. PBc promu ne pourrait plus sortir. i) Oui. PNf promu en T en g1. j) Non. TBa1, Fc1 non disponibles pour les prises de PN. k) Oui. PNh:Th1; les captures des PBh sont g3,h4.
No. 482 HN
Hans-Jürgen Manthey: d) ferner +sBf4
i) ferner +sBf4 ???
soll es i) - sBf4 heißen ? (2021-07-20)
Mario Richter: Ja! Diesen und ein paar Fehler bei der Zwillingsbildung korrigiert, Bitte nochmal prüfen! (2021-07-20)
Hans-Jürgen Manthey: b), d), f), h) und j) illegal
a) R: 1. b7xDc6 Dc5-c6 Tc6-c7 a2xSb3 Tb6-c6 c7xTb8=T Sd4-b3 Ke4-d5 Tb1-b6 Da3-c5 b2-b1=T Sh2-f3 b3-b2 Dc1-a3 Sf3-d4 Dd1-c1 c4xTb3 Tb1-b3 Sg1-f3 Ta1-b1 f2xTg1S Sb1-c3 f3-f2 Th1-g1 f4-f3 Sf3-h2 g5-g4 Sg1-f3 h4xLg3 Kf3-e4 f5-f4 Ld6-g3 f7-f5 La3-d6 c5-c4 d6xDc7 Dd8-c7 e5xSd6 Sf5-d6 f4xLe5 O-O Kf2-f3 Lg7-e5 Ke1-f2 Sh6-f5 f2-f4 h5-h4 b3xSa4 usw.
c) R: 1. b7xDc6 Dd6-c6 h7-h5 Se4-c3 Tc3-c7 b2-b3 Ta3-c3 c7xTb8 Ta1-a3 Sg1-f3 Te1-a1 Sg5-e4 f2xTe1=T Ta1-e1 O-O Se6-g5 f3-f2 Sf8-e6 f4-f3 Sg6xLf8 f5-f4 Dc5-d6 f7-f5 d6xDc7 Dd8-c7 e5xSd6 Se4-d6 Da3xc5 c7-c5 Da2-a3 Sc5-e4 Db1-a2 Sb3-c5 Dd1-b1 Sc1-b3 Ke4-d5 Sb3xLc1 Sf4-g6 Sa5-b3 Sd5-f4 Sc6-a5 Sc3-d5 Ta8-b8 Sb1-c3 Sb8-c6 f4xSe5 Sf3-e5 usw.
e) R: 1. b7xDc6 Db6-c6 f5-f4 Se4-c3 Tc3-c7 b2-b3 Ta3-c3 Sg1-f3 Ta1-a3 Df2-b6 Tc1-a1 De1-f2 Ta1xLc1 Dd1-e1 a2-a1=T Kd4-d5 b3xTa2 Ta1-a2 c4xTb3 Ke3-d4 c5-c4 c7xTb8=T Ta8-b8 d6xDc7 Dd8-c7 c7-c5 Sg5-e4 O-O Sh7-g5 f7-f5 Sf8-h7 h7-h5 Sg6xLf8 a7-a6 e5xSd6 Se4-d6 Kf2-e3 Sf6-e4+ Ke1-f2 Sg8-f6 f4xSe5 usw.
g) R: 1. b7xDc6 Db6-c6 f5-f4 Se4-c3 g5-g4 Df2-b6 h6xLg5 Lf6-g5 h7-h6 Lb2-f6 O-O Lc1-b2 Tc3-c7 b2-b3 Ta3-c3 Sg1-f3 Ta1-a3 Sg5-e4 a2-a1=T Se6-g5 b3xTa2 Ta1-a2 c4xTb3 Sf8-e6 c5-c4 Sg6xLf8 f7-f5 c7xTb8=T Ta8-b8 Th3-b3 a7-a6 d6xDc7 Dd8-c7 Th1-h3 c7-c5 e5xSd6 Se4-d6 h2-h4 Sc5-e4 a2-a4 Sa6-c5 Ke4-d5 Sb8-a6 f4xSe5 usw.
i) R: 1. b7xDc6 Db6-c6 g5-g4 Sb1-c3 Tc3-c7 Se5-f3 Th3-c3 c7xTb8=T Th1-h3 d6xDc7 Dd8-c7 Dc5-b6 Tg1-h1 Ke4-d5 f2xTg1=T Th1-g1 f3-f2 Df2xc5 c7-c5 De1-f2 f4-f3 Kf3-e4 f5-f4 Kf2-f3 O-O Sg6-e5 a7-a6 Sf8-g6 f7-f5 e5xSd6 h6xLg5 Lf6-g5 Ta8-b8 f4xSe5 Sc4-e5 Dd1-e1 Sb2-c4 Ke1-f2 Sc4xTb2 Ta2-b2 Sa5-c4 Lb2-f6 h7-h6 Sg6xLf8 usw.
k) R: 1. b7xDc6 Db6-c6 g5-g4 Sb1-c3 Tc3-c7 Se5-f3 g4-g3 Df2-b6 Th3-c3 De1-f2 Th1-h3 Kc4-d5 f5xTg4 Kd4-c4 h2-h1=T c7xTb8 h3-h2 g3xDh4 Dh5-h4 Kd3-d4 De8-h5 Dd1-e1 Dd8-e8 Ke3-d3 h4-h3 Kf2-e3 h5-h4 Th4-g4 O-O Sf3-e5 f7-f5 Th1-h4 h7-h5 h2xLg3 Le5-g3+ Sg1-f3 Lg7-e5 Ke1-f2 Lf8-g7 a2-a3 a7-a6 d6xc7 Ta8-b8 e5xSd6 Sf5-d6 f4xSe5 g7-g5 Sf3-g1 Sh6-f5 Sg5-f3 Sg8-h6 Sh3-g5 Sc4-e5 Sg1-h3 Sa5-c4 f2-f4 Sb3-a5 Sf3-g1 Sc1-b3 Sg5-f3 Sb3xLc1 Sh3-g5 Sa1-b3 Sb1-h3 Sb3xTa1 usw. (2021-07-21)
comment
Hans-Jürgen Manthey: d) ferner +sBf4
i) ferner +sBf4 ???
soll es i) - sBf4 heißen ? (2021-07-20)
Mario Richter: Ja! Diesen und ein paar Fehler bei der Zwillingsbildung korrigiert, Bitte nochmal prüfen! (2021-07-20)
Hans-Jürgen Manthey: b), d), f), h) und j) illegal
a) R: 1. b7xDc6 Dc5-c6 Tc6-c7 a2xSb3 Tb6-c6 c7xTb8=T Sd4-b3 Ke4-d5 Tb1-b6 Da3-c5 b2-b1=T Sh2-f3 b3-b2 Dc1-a3 Sf3-d4 Dd1-c1 c4xTb3 Tb1-b3 Sg1-f3 Ta1-b1 f2xTg1S Sb1-c3 f3-f2 Th1-g1 f4-f3 Sf3-h2 g5-g4 Sg1-f3 h4xLg3 Kf3-e4 f5-f4 Ld6-g3 f7-f5 La3-d6 c5-c4 d6xDc7 Dd8-c7 e5xSd6 Sf5-d6 f4xLe5 O-O Kf2-f3 Lg7-e5 Ke1-f2 Sh6-f5 f2-f4 h5-h4 b3xSa4 usw.
c) R: 1. b7xDc6 Dd6-c6 h7-h5 Se4-c3 Tc3-c7 b2-b3 Ta3-c3 c7xTb8 Ta1-a3 Sg1-f3 Te1-a1 Sg5-e4 f2xTe1=T Ta1-e1 O-O Se6-g5 f3-f2 Sf8-e6 f4-f3 Sg6xLf8 f5-f4 Dc5-d6 f7-f5 d6xDc7 Dd8-c7 e5xSd6 Se4-d6 Da3xc5 c7-c5 Da2-a3 Sc5-e4 Db1-a2 Sb3-c5 Dd1-b1 Sc1-b3 Ke4-d5 Sb3xLc1 Sf4-g6 Sa5-b3 Sd5-f4 Sc6-a5 Sc3-d5 Ta8-b8 Sb1-c3 Sb8-c6 f4xSe5 Sf3-e5 usw.
e) R: 1. b7xDc6 Db6-c6 f5-f4 Se4-c3 Tc3-c7 b2-b3 Ta3-c3 Sg1-f3 Ta1-a3 Df2-b6 Tc1-a1 De1-f2 Ta1xLc1 Dd1-e1 a2-a1=T Kd4-d5 b3xTa2 Ta1-a2 c4xTb3 Ke3-d4 c5-c4 c7xTb8=T Ta8-b8 d6xDc7 Dd8-c7 c7-c5 Sg5-e4 O-O Sh7-g5 f7-f5 Sf8-h7 h7-h5 Sg6xLf8 a7-a6 e5xSd6 Se4-d6 Kf2-e3 Sf6-e4+ Ke1-f2 Sg8-f6 f4xSe5 usw.
g) R: 1. b7xDc6 Db6-c6 f5-f4 Se4-c3 g5-g4 Df2-b6 h6xLg5 Lf6-g5 h7-h6 Lb2-f6 O-O Lc1-b2 Tc3-c7 b2-b3 Ta3-c3 Sg1-f3 Ta1-a3 Sg5-e4 a2-a1=T Se6-g5 b3xTa2 Ta1-a2 c4xTb3 Sf8-e6 c5-c4 Sg6xLf8 f7-f5 c7xTb8=T Ta8-b8 Th3-b3 a7-a6 d6xDc7 Dd8-c7 Th1-h3 c7-c5 e5xSd6 Se4-d6 h2-h4 Sc5-e4 a2-a4 Sa6-c5 Ke4-d5 Sb8-a6 f4xSe5 usw.
i) R: 1. b7xDc6 Db6-c6 g5-g4 Sb1-c3 Tc3-c7 Se5-f3 Th3-c3 c7xTb8=T Th1-h3 d6xDc7 Dd8-c7 Dc5-b6 Tg1-h1 Ke4-d5 f2xTg1=T Th1-g1 f3-f2 Df2xc5 c7-c5 De1-f2 f4-f3 Kf3-e4 f5-f4 Kf2-f3 O-O Sg6-e5 a7-a6 Sf8-g6 f7-f5 e5xSd6 h6xLg5 Lf6-g5 Ta8-b8 f4xSe5 Sc4-e5 Dd1-e1 Sb2-c4 Ke1-f2 Sc4xTb2 Ta2-b2 Sa5-c4 Lb2-f6 h7-h6 Sg6xLf8 usw.
k) R: 1. b7xDc6 Db6-c6 g5-g4 Sb1-c3 Tc3-c7 Se5-f3 g4-g3 Df2-b6 Th3-c3 De1-f2 Th1-h3 Kc4-d5 f5xTg4 Kd4-c4 h2-h1=T c7xTb8 h3-h2 g3xDh4 Dh5-h4 Kd3-d4 De8-h5 Dd1-e1 Dd8-e8 Ke3-d3 h4-h3 Kf2-e3 h5-h4 Th4-g4 O-O Sf3-e5 f7-f5 Th1-h4 h7-h5 h2xLg3 Le5-g3+ Sg1-f3 Lg7-e5 Ke1-f2 Lf8-g7 a2-a3 a7-a6 d6xc7 Ta8-b8 e5xSd6 Sf5-d6 f4xSe5 g7-g5 Sf3-g1 Sh6-f5 Sg5-f3 Sg8-h6 Sh3-g5 Sc4-e5 Sg1-h3 Sa5-c4 f2-f4 Sb3-a5 Sf3-g1 Sc1-b3 Sg5-f3 Sb3xLc1 Sh3-g5 Sa1-b3 Sb1-h3 Sb3xTa1 usw. (2021-07-21)
comment
Genre: Retro
FEN: 1Rb2rk1/2rpp3/p1p5/3K4/P5p1/1PN2NpP/2PPP1P1/5B2
Input: Henri Nouguier, 2004-01-11
Last update: Mario Richter, 2021-07-20 more...
1.Rh5 O-O 2.Th8+ Rf7 3.g6+ Re8 4.Fg4 Th8. Les noirs peuvent roquer car leurs dernier coup a pu être PxCf6 (le CB donnant échec au RN!).
Keywords: Maximummer
Genre: Retro, s#, Fairies
FEN: 4k2r/7R/5p2/5BP1/7K/8/8/8
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2023-06-03 more...
Genre: Retro, s#, Fairies
FEN: 4k2r/7R/5p2/5BP1/7K/8/8/8
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2023-06-03 more...
WTM: no solution
BTM:
1. ... Kxh1 2. Dd2 Kg1 3. 0-0-0#
1. ... Kxf3 2. 0-0+ Ke2/Ke3/Ke4 3. Tae1#
Across these two cases, both castling rights are demonstrated.
If we can assume that none of the White pieces in the first row has yet moved, then we must assign the move to Black. When this was published in The Problemist, opinions were divided. "White, to prove that the move is Black's, begin by claiming the right to castling on each side: and as they cannot in fact make both castles at once to establish this claim, it is sufficient to make one in each variation." "The solution consists of all the variations, and it is sufficient to have 0-0 in just one variation, since the small castling is enough to show that the move is Black's". "Since White claims the right to kingside castling, Black is forced to offer the opportunity" "No solution on 0. ... Rh1"
Si nous pouvons assumer qu'aucune des pièces blanches de la première rangée n'a encore joué, alors il faut attribuer le trait aux Noirs. D'où: 0. ... Rf3: 1,O-O+1 Re4 ou e3 2.Tae1# 0. ... Rhl: 1.Dd2I Rgl 2.O-O-O# Quand cela fut publié dans The Problemist, les avis furent partagés. "Les B., pour prouver que le trait est aux N., commencent par revendiquer le droit au roque de chaque côté: et comme ils ne peuvent de fait effectuer les deux roques à la fois pour établir cette revendication, il suffit d'en éxécuter un dans chaque variante". "La solution se compose de toutes les variantes, et il suffit d'avoir O-O dans juste une variante, puisque le petit roque suffit à montrer que le trait est aux Noirs''. "Comme les Blancs revendiquent le droit au petit roque, les Noirs sont forcés d'en offrir l'opportunité""Pas de solution sur 0. ... Rh1 :"
BTM:
1. ... Kxh1 2. Dd2 Kg1 3. 0-0-0#
1. ... Kxf3 2. 0-0+ Ke2/Ke3/Ke4 3. Tae1#
Across these two cases, both castling rights are demonstrated.
If we can assume that none of the White pieces in the first row has yet moved, then we must assign the move to Black. When this was published in The Problemist, opinions were divided. "White, to prove that the move is Black's, begin by claiming the right to castling on each side: and as they cannot in fact make both castles at once to establish this claim, it is sufficient to make one in each variation." "The solution consists of all the variations, and it is sufficient to have 0-0 in just one variation, since the small castling is enough to show that the move is Black's". "Since White claims the right to kingside castling, Black is forced to offer the opportunity" "No solution on 0. ... Rh1"
Si nous pouvons assumer qu'aucune des pièces blanches de la première rangée n'a encore joué, alors il faut attribuer le trait aux Noirs. D'où: 0. ... Rf3: 1,O-O+1 Re4 ou e3 2.Tae1# 0. ... Rhl: 1.Dd2I Rgl 2.O-O-O# Quand cela fut publié dans The Problemist, les avis furent partagés. "Les B., pour prouver que le trait est aux N., commencent par revendiquer le droit au roque de chaque côté: et comme ils ne peuvent de fait effectuer les deux roques à la fois pour établir cette revendication, il suffit d'en éxécuter un dans chaque variante". "La solution se compose de toutes les variantes, et il suffit d'avoir O-O dans juste une variante, puisque le petit roque suffit à montrer que le trait est aux Noirs''. "Comme les Blancs revendiquent le droit au petit roque, les Noirs sont forcés d'en offrir l'opportunité""Pas de solution sur 0. ... Rh1 :"
No. 8889 HN
A.Buchanan: My feeling is this doesn’t work, sadly. In an adversarial stipulation, Black can simply avoid ever playing the line that allows White to “prove” that Black had the move. In a helpmate on the other hand I suppose if PRA-AP is expressed properly, there should be no problem playing multiple parts. (2022-02-15)
comment
A.Buchanan: My feeling is this doesn’t work, sadly. In an adversarial stipulation, Black can simply avoid ever playing the line that allows White to “prove” that Black had the move. In a helpmate on the other hand I suppose if PRA-AP is expressed properly, there should be no problem playing multiple parts. (2022-02-15)
comment
Keywords: a posteriori (AP) (Type Keym), Castling, Rex solus
Genre: Retro, 2#
FEN: 8/8/3Q4/8/8/5PPP/6k1/R3K2R
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2022-02-15 more...
Genre: Retro, 2#
FEN: 8/8/3Q4/8/8/5PPP/6k1/R3K2R
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2022-02-15 more...
1. d4 Sc6 2. d5 Sd4 3. Lf4 Sxe2 4. Lg3 Sc1 5. La6 c6 6. c4 Da5+ 7. Sc3 Kd8 8. f4 Kc7 9. f5+ Kb6 10. f6 Kc5 11. fxg7 Sf6 12. g8=T Se4 13. Tg4 Sd2 14. Se4+ Kd4 15. Sc5+ Ke3 16. Td4 Sf1+ 17. Td2
First version had Sf1 on b1.
Moldenhauer: Computerprüfung: C+ Euclide 1.11 eindeutige Lösung
gekocht bei BP 17.0 und bei BP 16.0 Umwandlungsfehler (2022-08-21)
comment
Moldenhauer: Computerprüfung: C+ Euclide 1.11 eindeutige Lösung
gekocht bei BP 17.0 und bei BP 16.0 Umwandlungsfehler (2022-08-21)
comment
Keywords: Unique Proof Game, Non-standard material (T)
Genre: Retro
Computer test: Stelvio 1.2, 2sec
FEN: r1b2b1r/pp1ppp1p/B1p5/q1NP4/2P5/4k1B1/PP1R2PP/R1nQKnNR
Input: Gerd Wilts, 2004-02-04
Last update: Reto, 2023-05-19 more...
Genre: Retro
Computer test: Stelvio 1.2, 2sec
FEN: r1b2b1r/pp1ppp1p/B1p5/q1NP4/2P5/4k1B1/PP1R2PP/R1nQKnNR
Input: Gerd Wilts, 2004-02-04
Last update: Reto, 2023-05-19 more...
1. Db7 Ke8 2. Dg7 Kd8 3. a6 Kc8 4. a7 Kd8 5. Kf3 Kc8,Ke8 6. Ke4 Kd8 7. Kd5 Kc8,Ke8 8. Kc6 Kd8 9. Kb7 Ke8 10. Ka8 Kd8 11. Df7 Kc8 12. Dc7+ Kxc7=
Genre: Fairies
FEN: 3k4/8/8/PQ6/8/8/6K1/8
Reprints: 3188 Ideal-Mate Review 01-03/1989
Input: Hans Gruber, 2004-05-01
Last update: Alfred Pfeiffer, 2015-06-01 more...
1. Lh8! (2. Tg7#; 1. ... Tf7 2. Exf7#)
1. ... Te7 (verstellt Eb6) 2. Dxf6#
1. ... Ee7 (verstellt Cb4) 2. Eg8#
1. ... Ce7 (verstellt Te1) 2. Td6#
Tripel-Grimshaw mit zyklischer Verstellung.
1. ... Te7 (verstellt Eb6) 2. Dxf6#
1. ... Ee7 (verstellt Cb4) 2. Eg8#
1. ... Ce7 (verstellt Te1) 2. Td6#
Tripel-Grimshaw mit zyklischer Verstellung.
Genre: 2#
FEN: Kb6/1p1R2B1/1*1b3rkp/5p1p/*1B*3b5Q/5p2/*1B7/4r3
Input: Hans Gruber, 2004-05-01
Last update: Gerd Wilts, 2004-05-01 more...
* 1. ... bxa4 2. Dd4 a2#
* 1. ... b4 2. Dc6 a2#
1. Dd4 bxa4 2. Dc5 a2#
1. ... b4 2. Sb6 a2#
* 1. ... b4 2. Dc6 a2#
1. Dd4 bxa4 2. Dc5 a2#
1. ... b4 2. Sb6 a2#
Genre: s#
FEN: 8/8/8/1p6/N3Q3/pk6/1p6/1K6
Input: Selbstmatt-Miniaturen/Reflexmatt-Miniaturen, 2004-05-01
1. Db6 droht 2. Dd6+ Txd6#
1. ... T beliebig+ 2. Db2+ Lxb2#
1. ... T beliebig+ 2. Db2+ Lxb2#
Genre: s#
FEN: 7b/8/5r2/1Q6/8/k7/p1B5/K7
Input: Selbstmatt-Miniaturen/Reflexmatt-Miniaturen, 2004-05-01
a) 1. La7! f3 2. Td4 f2#
b) 1. Tc6! e3 2. Dh4 e2#
Zweimal Liniensperre.
b) 1. Tc6! e3 2. Dh4 e2#
Zweimal Liniensperre.
87 - P1014784
Valery Surkov
PS241 The Problemist 07/1994
(4+2)
s#2
b) wT d7 nach d6
c) ferner wB h7 nach g6
d) ferner ohne wD a7
Valery Surkov
PS241 The Problemist 07/1994
(4+2)
s#2
b) wT d7 nach d6
c) ferner wB h7 nach g6
d) ferner ohne wD a7
a) 1. Td8 Dxd8 2. Df7+ Kxf7#
b) 1. Tf6+ Df7 2. De3 Dxf6#
c) 1. Df7+ Dxf7 2. g7+ Dxg7#
d) 1. g7+ Kf7+ 2. g8=D+ Dxg8#
b) 1. Tf6+ Df7 2. De3 Dxf6#
c) 1. Df7+ Dxf7 2. g7+ Dxg7#
d) 1. g7+ Kf7+ 2. g8=D+ Dxg8#
Genre: s#
FEN: 4qk1K/Q2R3P/8/8/8/8/8/8
Input: Selbstmatt-Miniaturen/Reflexmatt-Miniaturen, 2004-05-01
Last update: Olaf Jenkner, 2012-08-25 more...
* 1. ... h5 2. Lg2 h4 3. 0-0 h3 4. Lh1 h2#
1. Lf1! h5 2. Lg2 h4 3. 0-0 h3 4. Lh1 h2#
1. Lf1! h5 2. Lg2 h4 3. 0-0 h3 4. Lh1 h2#
1. b8=L+ Kb6 2. Lb4 L beliebig 3. Db7+ Lxb7#
1. ... Kc8 2. La7 Kc7,L beliebig 3. Db7+ Lxb7#
1. ... Kc8 2. La7 Kc7,L beliebig 3. Db7+ Lxb7#
Genre: s#
FEN: K7/1Pk1B3/8/1P1Q4/8/8/8/7b
Input: Selbstmatt-Miniaturen/Reflexmatt-Miniaturen, 2004-05-01
1. Ka4+! Kc4,Ke4 2. Dc6+,Df5+ Kd4 3. Dc5+ Ke4 4. Dc4+ Dd4 5. g3 Dxc4#
1. ... Tg2 2.Lf3#
1. ... Lg2 2. Dxe3#
1. ... e2 2. Sd2#
1. Kh5! (Zugzwang)
1. ... Tg2 2. Df3#
1. ... Lg2 2. Dg4#
1. ... e2 2. Dxd3#
1. ... Txf1 2. Dg4#
1. ... Txg3 2. Sxg3#
1. ... Lf3+ 2. Dxf3#
1. ... d2 2. Lc2#
1. ... Kxf5 2. Df4#
BS Problemid: 2813
Authors: Penrose, Lionel Sharples
Sources: The Problemist 2001
Het Belgisch Schaakbord (L'Echiquier Belge) 1947
Mate in Two Moves 1931
Simple Two-move Themes 1924
Years: 1920
1. ... Lg2 2. Dxe3#
1. ... e2 2. Sd2#
1. Kh5! (Zugzwang)
1. ... Tg2 2. Df3#
1. ... Lg2 2. Dg4#
1. ... e2 2. Dxd3#
1. ... Txf1 2. Dg4#
1. ... Txg3 2. Sxg3#
1. ... Lf3+ 2. Dxf3#
1. ... d2 2. Lc2#
1. ... Kxf5 2. Df4#
BS Problemid: 2813
Authors: Penrose, Lionel Sharples
Sources: The Problemist 2001
Het Belgisch Schaakbord (L'Echiquier Belge) 1947
Mate in Two Moves 1931
Simple Two-move Themes 1924
Years: 1920
Keywords: Brian Stephenson Collection (2813)
Genre: 2#
Computer test: SCHRECKE: C+, popeye 4.85 (2020-11-17)
FEN: 8/2Np4/3P1p1B/3RpP2/4k1K1/3pp1Q1/5p2/3B1Nrb
Reprints: 32B Simple Two-move Themes 1924
Neues Wiener Tagblatt 16/07/1944
Het Belgisch Schaakbord 1947
The Problemist 2001
Input: Brian Stephenson, 2004-08-12
Last update: Erich Bartel, 2020-11-17 more...
Genre: 2#
Computer test: SCHRECKE: C+, popeye 4.85 (2020-11-17)
FEN: 8/2Np4/3P1p1B/3RpP2/4k1K1/3pp1Q1/5p2/3B1Nrb
Reprints: 32B Simple Two-move Themes 1924
Neues Wiener Tagblatt 16/07/1944
Het Belgisch Schaakbord 1947
The Problemist 2001
Input: Brian Stephenson, 2004-08-12
Last update: Erich Bartel, 2020-11-17 more...
1. Dxf4! droht 2. Se3#
Keywords: Brian Stephenson Collection (2849), Capture key, Unpinning (Schlüssel)
Genre: 2#
Computer test: Juel: Popeye 4.61
FEN: 2BNqr2/4r1n1/6p1/Rb1kn1Q1/2R2pp1/BP6/6N1/5K2
Reprints: Mate in two moves 1931
AJEC (Belgischer Fernschachbund) 05/1938
136 FIDE Album 1914-1944/I 1971
L The Problemist 25-3, p. 99, 05/2015
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2023-12-03 more...
Genre: 2#
Computer test: Juel: Popeye 4.61
FEN: 2BNqr2/4r1n1/6p1/Rb1kn1Q1/2R2pp1/BP6/6N1/5K2
Reprints: Mate in two moves 1931
AJEC (Belgischer Fernschachbund) 05/1938
136 FIDE Album 1914-1944/I 1971
L The Problemist 25-3, p. 99, 05/2015
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2023-12-03 more...
1. Dg4! droht 2. De6#
1. ... Kxe5 2. Dg5#
1. ... Kc4 2. Dxe4#
1. ... Sf5 2. Dg8#
BS Problemid: 3444
Authors: Bosetti, Fabio
Sources:
Years: 1993
1. ... Kxe5 2. Dg5#
1. ... Kc4 2. Dxe4#
1. ... Sf5 2. Dg8#
BS Problemid: 3444
Authors: Bosetti, Fabio
Sources:
Years: 1993
Abdruck im 'Problemist Supplement' mit der Autorangabe: B Fabio (Italy)
SCHRECKE: C+, popeye 4.87
1. Dg4! droht 2. De6#
1. ... Kxe5 2. Dg5#
1. ... Kc4 2. Dxe4#
1. ... Sf5 2. Dg8# (2023-04-01)
comment
SCHRECKE: C+, popeye 4.87
1. Dg4! droht 2. De6#
1. ... Kxe5 2. Dg5#
1. ... Kc4 2. Dxe4#
1. ... Sf5 2. Dg8# (2023-04-01)
comment
Keywords: Brian Stephenson Collection (3444)
Genre: 2#
Computer test: SCHRECKE (2023-04-01): C+, popeye 4.87
FEN: 4R2B/3Kn3/5b2/1pNkP3/4pQ2/1Nnb4/4B3/2R5
Input: Brian Stephenson, 2004-08-12
Last update: Mario Richter, 2023-04-01 more...
Genre: 2#
Computer test: SCHRECKE (2023-04-01): C+, popeye 4.87
FEN: 4R2B/3Kn3/5b2/1pNkP3/4pQ2/1Nnb4/4B3/2R5
Input: Brian Stephenson, 2004-08-12
Last update: Mario Richter, 2023-04-01 more...
BS Problemid: 3452
Authors: Kuligin, Nikolai V
Sources:
Years: 1993
SCHRECKE: C+, popeye 4.87
1. Te5! droht 2. Le7#
1. ... Sc6 2. Lc7#
1. ... Se6 2. Td5# (2022-12-31)
comment
1. Te5! droht 2. Le7#
1. ... Sc6 2. Lc7#
1. ... Se6 2. Td5# (2022-12-31)
comment
Keywords: Brian Stephenson Collection (3452)
Genre: 2#
FEN: 1nRBRn2/8/1N1k2p1/8/3P4/8/8/3K4
Input: Brian Stephenson, 2004-08-12
Genre: 2#
FEN: 1nRBRn2/8/1N1k2p1/8/3P4/8/8/3K4
Input: Brian Stephenson, 2004-08-12
BS Problemid: 3471
Authors: Yanuarta, Simadhinata
Sources:
Years: 1993
SCHRECKE: C+, popeye 4.87
1.d3! (Zugzwang)
1. ... S:d3,S~ 2.Dd7#
1. ... Tf1,T~ 2.Ld5#
1. ... Td5 2.L:d5#
1. ... Te5 2.S:d4#
1. ... Le5 2.Ld5#
1. ... Le7,Ld8 2.S:d4# (2021-07-03)
comment
1.d3! (Zugzwang)
1. ... S:d3,S~ 2.Dd7#
1. ... Tf1,T~ 2.Ld5#
1. ... Td5 2.L:d5#
1. ... Te5 2.S:d4#
1. ... Le5 2.Ld5#
1. ... Le7,Ld8 2.S:d4# (2021-07-03)
comment
Keywords: Brian Stephenson Collection (3471)
Genre: 2#
FEN: 8/2Q3p1/2B1kbP1/2n2rp1/2KpP1N1/8/3PN3/8
Input: Brian Stephenson, 2004-08-12
Last update: A.Buchanan, 2021-07-03 more...
Genre: 2#
FEN: 8/2Q3p1/2B1kbP1/2n2rp1/2KpP1N1/8/3PN3/8
Input: Brian Stephenson, 2004-08-12
Last update: A.Buchanan, 2021-07-03 more...
BS Problemid: 3484
Authors: Mayhew, John
Sources: BDS Website 2001
Years: 1994
Roland Ott: Vorgänger: Walter Gleave P1022462 (2021-12-28)
Henrik Juel: C+ Popeye 4.61 (2021-12-28)
Henrik Juel: 1.Ke7 thr. 2.Df5#
1... Kc5/Ke4/c5 2.Ke6/Kd6/Df3# (2021-12-28)
comment
Henrik Juel: C+ Popeye 4.61 (2021-12-28)
Henrik Juel: 1.Ke7 thr. 2.Df5#
1... Kc5/Ke4/c5 2.Ke6/Kd6/Df3# (2021-12-28)
comment
Keywords: Brian Stephenson Collection (3484)
Genre: 2#
FEN: 4RQ1B/1p3K2/1rp5/3k4/n7/8/8/5B2
Input: Brian Stephenson, 2004-08-12
Last update: Roland Ott, 2021-12-28 more...
Genre: 2#
FEN: 4RQ1B/1p3K2/1rp5/3k4/n7/8/8/5B2
Input: Brian Stephenson, 2004-08-12
Last update: Roland Ott, 2021-12-28 more...
1. b3! f1=D,T+ 2. Dxf1#
1. ... Sxd7 2. Dxg6#
1. ... Se7 2. Se5#
1. ... axb3 2. Dxb3#
1. ... bxa5 2. Sc5#
1. ... b4 2. Dc4#
1. ... g3 2. Df3#
1. ... g5 2. Df5#
1. ... Sxd7 2. Dxg6#
1. ... Se7 2. Se5#
1. ... axb3 2. Dxb3#
1. ... bxa5 2. Sc5#
1. ... b4 2. Dc4#
1. ... g3 2. Df3#
1. ... g5 2. Df5#
Keywords: Brian Stephenson Collection (3491), Zugzwang-Goal
Genre: 2#
Computer test: SCHRECKE: popeye 4.87
FEN: 5n2/3N1Q2/1pn3p1/Pp6/p2p2p1/P2k4/1P1P1p2/3K3B
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2022-12-02 more...
Genre: 2#
Computer test: SCHRECKE: popeye 4.87
FEN: 5n2/3N1Q2/1pn3p1/Pp6/p2p2p1/P2k4/1P1P1p2/3K3B
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2022-12-02 more...
BS Problemid: 3504
Authors: Nymian, HS
Sources:
Years: 1995
SCHRECKE: C+, popeye 4.87
1. Da8! (Zugzwang)
1. ... Kxh2 2. Dh8#
1. ... gxh2 2. g3# (2023-02-14)
comment
1. Da8! (Zugzwang)
1. ... Kxh2 2. Dh8#
1. ... gxh2 2. g3# (2023-02-14)
comment
Keywords: Brian Stephenson Collection (3504)
Genre: 2#
FEN: 8/8/8/8/6p1/3p2p1/3P2PP/Q4K1k
Input: Brian Stephenson, 2004-08-12
Genre: 2#
FEN: 8/8/8/8/6p1/3p2p1/3P2PP/Q4K1k
Input: Brian Stephenson, 2004-08-12
1. Se4? droht 2. Df2#
1. ... Db5 2. Sc5# 1. ... Db8 2. Sd6# 1. ... Dg3 2. Sexg3# 1. ... Dxe4 2. Txe4# 1. ... Dxh5!
1. Se6! droht 2. De2#
1. ... Db5 2. Sc5# 1. ... Db8 2. Sc7# 1. ... Dg3 2. Sef4# 1. ... Dxe6 2. Txe6#
1. ... Db5 2. Sc5# 1. ... Db8 2. Sd6# 1. ... Dg3 2. Sexg3# 1. ... Dxe4 2. Txe4# 1. ... Dxh5!
1. Se6! droht 2. De2#
1. ... Db5 2. Sc5# 1. ... Db8 2. Sc7# 1. ... Dg3 2. Sef4# 1. ... Dxe6 2. Txe6#
Keywords: Brian Stephenson Collection (3505), Mousetrap (doppelt), Changed mates
Genre: 2#
Computer test: SCHRECKE: popeye 4.87
FEN: 8/4R3/5p2/2N1q2N/p2P4/3Pkp1R/1Q6/1K3B1b
Reprints: Z25 Knobeln Sie auch gern? [Pachl] , p. 24, 2009
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2022-07-30 more...
Genre: 2#
Computer test: SCHRECKE: popeye 4.87
FEN: 8/4R3/5p2/2N1q2N/p2P4/3Pkp1R/1Q6/1K3B1b
Reprints: Z25 Knobeln Sie auch gern? [Pachl] , p. 24, 2009
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2022-07-30 more...
BS Problemid: 3582
Authors: Ouellet, Charles
Sources:
Years: 1999
SCHRECKE: C+, popeye 4.87
1. Ld7! (Zugzwang)
1. ... Sc2 2. Tcd1#
1. ... Se2 2. Ted1#
1. ... Sxb3 2. Sxb3#
1. ... Sxf3 2. Sxf3#
1. ... Sxb5 2. Lxb5#
1. ... Sf5 2. Lxf5#
1. ... Sc6 2. Lxc6#
1. ... Se6 2. Lxe6#
1. ... bxc5 2. Da5# (2023-07-20)
comment
1. Ld7! (Zugzwang)
1. ... Sc2 2. Tcd1#
1. ... Se2 2. Ted1#
1. ... Sxb3 2. Sxb3#
1. ... Sxf3 2. Sxf3#
1. ... Sxb5 2. Lxb5#
1. ... Sf5 2. Lxf5#
1. ... Sc6 2. Lxc6#
1. ... Se6 2. Lxe6#
1. ... bxc5 2. Da5# (2023-07-20)
comment
Keywords: Brian Stephenson Collection (3582)
Genre: 2#
FEN: 3QB2n/1p3p2/1p3Pp1/1PN1p1N1/3nKp2/1P3P2/1P1k4/2R1R3
Input: Brian Stephenson, 2004-08-12
Genre: 2#
FEN: 3QB2n/1p3p2/1p3Pp1/1PN1p1N1/3nKp2/1P3P2/1P1k4/2R1R3
Input: Brian Stephenson, 2004-08-12
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The problems of this query have been registered by the following contributors:
Gerd Wilts (21)hpr (9)
Hans-Jürgen Schäfer (6)
Markus Manhart (8)
HBae (3)
Ralf Krätschmer (1)
Michal Dragoun (1)
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Torsten Linss (1)
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Henri Nouguier (11)
Hans Gruber (2)
Selbstmatt-Miniaturen/Reflexmatt-Miniaturen (7)
Brian Stephenson (10)
Thomas Volet: The exception in this case to the otherwise observed constraint of not having a check in the diagram or stipulation as to who is on the move relates to a collegial colloquy at the time with with the composer J.G. Mauldon, who was active in this task. (2022-03-24)
A.Buchanan: Terrific composition! Often in a Type C composition, one might retract the check to give Type B. But the thematic Ta8 here could not exist without a check in diagram. There are different areas of design space are accessible depending upon level of self-imposed constraints. Some retro composers like Baibakov systematically explore Types B&C as well as Type A. I don't see this as any kind of defect.
I particularly like that the lock on h-file only allows one sT to squeeze itself out. (2022-03-24)
comment