72 problem(s) found in 2642 milliseconds (displaying 72 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT A='Seliwanow, Andrej W.' AND A='Dawson, Thomas R.'] [download as LaTeX]
1. a4 c5 2. a5 Db6 3. Ta4 Sc6 4. Td4 Tb8 5. b4 cxd4 6. b5 e5 7. d3 La3 8. Le3 Lc1 9. bxc6 Db2 10. c7 b5 11. a6 Tb7 12. axb7 h5 13. b8=L Lb7 14. c8=D+ Ke7 15. Sf3 d5 16. Dh3 Th6 17. g4 Ta6 18. Dg3 h4 19. c4 hxg3 20. h4 Ta1 21. h5 Kf6 22. Sh2 dxe3 23. h6 Kg5 24. h7 Kh4 25. c5 a5 26. c6 a4 27. c7 b4 28. c8=T b3 29. h8=S Se7 30. Sg6 fxg6 31. Lh3 Sc6 32. 0-0 Sb4 33. Ld6 Sc2 34. Td8 Se1
Keywords: Allumwandlung, Castling (wk), Non-Unique Proof Game
Genre: Retro
FEN: 3R4/1b4p1/3B2p1/3pp3/p5Pk/1p1Pp1pB/1q2PP1N/rNbQnRK1
Reprints: 53 Caissa's Wild Roses 1935
Chess unlimited 1969
150 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Gunter Jordan, 2020-07-01 more...
Genre: Retro
FEN: 3R4/1b4p1/3B2p1/3pp3/p5Pk/1p1Pp1pB/1q2PP1N/rNbQnRK1
Reprints: 53 Caissa's Wild Roses 1935
Chess unlimited 1969
150 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Gunter Jordan, 2020-07-01 more...
1. exf3ep Lc2#
R: 1. f2-f4 f5xTe4 2. Tg4-e4 & e.g. f6-f5 3. Tg1-g4 f7-f6 4. Le4-h7 h5-h4 5. Lg2-e4 h6-h5 6. Lf1-g2 h7-h6 7. g2xh3
1.exf3e.p. Bc2# is the only possible solution, but this necessitates R: 1.f2-f4. Can we prove this?
(13+13) with 1+2 pawn captures. Bf8 captured at home, so to satisfy White appetite, the missing Black pawn (a or b) must have promoted via c2 on c1. Two more White units must be captured to allow this.
The kings cage can only be unlocked by retracting WPc2. But the clock is ticking as there are only 6 black moves which can be retracted.
The promoted piece was captured on e3 or h3. If either capture is undone, then a White bishop square is cut off, so WB must be replaced prior to this.
Now the order of the early moves is: WdP moves, WQB & WQR escape, BP promotes on c1 to X (capturing WR at some point), X captured by WP.
So the first White capture must be dxNe3 and the second White capture releases gxXh3. The second White capture releases WKB & WKR. WKR captured by original BfP.
The clock starts ticking with gxh3. Black has 6 pawn moves. WKB has 3 moves to reach h7. WR has 3 if it goes via d file, or 2 if it starts on g1 (in which case WfP or WQB must also move once). So certainly at least 6 White moves. Last move was therefore White (even if the stipulation didn't tell us), and it can only have been WfP coming from f3 or f4. If it had been coming from f3 it would have blocked WKB in its progress, so the last White move was indeed R: 1.f2-f4.
WKR did therefore move from g1-g4-e4, and R: 1. ... fxRe5 2. Rg4-e4. Prior to that, move order not unique, but counting still exact.
Note that WN loitering on b4, pretending to be part of the cage, is present on the board just to make up the numbers.
R: 1. f2-f4 f5xTe4 2. Tg4-e4 & e.g. f6-f5 3. Tg1-g4 f7-f6 4. Le4-h7 h5-h4 5. Lg2-e4 h6-h5 6. Lf1-g2 h7-h6 7. g2xh3
1.exf3e.p. Bc2# is the only possible solution, but this necessitates R: 1.f2-f4. Can we prove this?
(13+13) with 1+2 pawn captures. Bf8 captured at home, so to satisfy White appetite, the missing Black pawn (a or b) must have promoted via c2 on c1. Two more White units must be captured to allow this.
The kings cage can only be unlocked by retracting WPc2. But the clock is ticking as there are only 6 black moves which can be retracted.
The promoted piece was captured on e3 or h3. If either capture is undone, then a White bishop square is cut off, so WB must be replaced prior to this.
Now the order of the early moves is: WdP moves, WQB & WQR escape, BP promotes on c1 to X (capturing WR at some point), X captured by WP.
So the first White capture must be dxNe3 and the second White capture releases gxXh3. The second White capture releases WKB & WKR. WKR captured by original BfP.
The clock starts ticking with gxh3. Black has 6 pawn moves. WKB has 3 moves to reach h7. WR has 3 if it goes via d file, or 2 if it starts on g1 (in which case WfP or WQB must also move once). So certainly at least 6 White moves. Last move was therefore White (even if the stipulation didn't tell us), and it can only have been WfP coming from f3 or f4. If it had been coming from f3 it would have blocked WKB in its progress, so the last White move was indeed R: 1.f2-f4.
WKR did therefore move from g1-g4-e4, and R: 1. ... fxRe5 2. Rg4-e4. Prior to that, move order not unique, but counting still exact.
Note that WN loitering on b4, pretending to be part of the cage, is present on the board just to make up the numbers.
Jeliss: "Obstruction of passage square f3 to Bishop of same colour."
"Version 'Pittsburgh Leader' 08.06.1913"
Yoav Ben-Zvi: Appears as the first problem (D445) in the booklet on Dawson's RA problems by G.P. Jellis. The obstruction that occurs in the Try -1.Pf3-f4?, by WP of WB, is described as "obstruction of passage square". It is not considered by Dawson and his disciples to be a Retro opposition. Dawson's conception of RO was quite broad, it included cases where the interference was not by occupation of the target square, so the only valid reason that I can see to exclude this case is that the 2 pieces involved are both of the same color. Fabel's definition explicitly excludes "Monochrome RO". I conclude that it would be preferrable to interpret RO as a bi-chromatic interference. The keyword Retro opposition should be removed. (2018-04-07)
A.Buchanan: To my mind, RO involves some kind of parity-tempo issue between the sides, not just some kind of race-tempo. If it was just about "bi-chromatic interference", one might say that bPe4 blocks wBh7 from an immediate retreat, so it has to be wPf4 that retreats first, legitimizing the ep key. So I agree this is not RO. (2024-01-06)
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"Version 'Pittsburgh Leader' 08.06.1913"
Yoav Ben-Zvi: Appears as the first problem (D445) in the booklet on Dawson's RA problems by G.P. Jellis. The obstruction that occurs in the Try -1.Pf3-f4?, by WP of WB, is described as "obstruction of passage square". It is not considered by Dawson and his disciples to be a Retro opposition. Dawson's conception of RO was quite broad, it included cases where the interference was not by occupation of the target square, so the only valid reason that I can see to exclude this case is that the 2 pieces involved are both of the same color. Fabel's definition explicitly excludes "Monochrome RO". I conclude that it would be preferrable to interpret RO as a bi-chromatic interference. The keyword Retro opposition should be removed. (2018-04-07)
A.Buchanan: To my mind, RO involves some kind of parity-tempo issue between the sides, not just some kind of race-tempo. If it was just about "bi-chromatic interference", one might say that bPe4 blocks wBh7 from an immediate retreat, so it has to be wPf4 that retreats first, legitimizing the ep key. So I agree this is not RO. (2024-01-06)
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Keywords: Last Moves?, En passant as key
Genre: h#, Retro
FEN: nqb5/1rrpp1pB/KRp5/1p4B1/kN2pP1p/2P1P2P/PP2P2P/8
Reprints: D445 Retro-Opposition & Other Retro-Analytical Chess Problems 1989
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-06 more...
Genre: h#, Retro
FEN: nqb5/1rrpp1pB/KRp5/1p4B1/kN2pP1p/2P1P2P/PP2P2P/8
Reprints: D445 Retro-Opposition & Other Retro-Analytical Chess Problems 1989
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-06 more...
1. Lf7! droht 2. Txc3 Kxh7 3. Th3#
1. ... c2 2. Txc2 Kxh7 3. Th2#
1. ... Kxh7 2. Tc5 Kh6,~ 3. Th5#
weisser Anzug ist legal, zuletzt z.B.:
R: 1. Kg7-h8 Be7-e8=S+
1. ... c2 2. Txc2 Kxh7 3. Th2#
1. ... Kxh7 2. Tc5 Kh6,~ 3. Th5#
weisser Anzug ist legal, zuletzt z.B.:
R: 1. Kg7-h8 Be7-e8=S+
Gerd Wilts (2001-10-17): The retro play given in the solution serves only to justify that it may indeed be White's turn. The forward play indeed starts with 1. Bf7. The idea of TRD was to compose a problem which is legal but in which one of the knights must be a promoted pawn. TRD called this an "obtrusive Knight".
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Keywords: Promotion (S)
Genre: Retro, 3#
Computer test: C+ Popeye 4.61
FEN: 4N1Bk/2R4P/8/8/1P1P4/1Kp5/8/8
Reprints: 17B Retrograde Analysis 1915
476 Hamburger Problem-Nachrichten 09-10/1949
E1 Problem 5-6 12/1951
RA69 diagrammes 28 07-08/1977
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-11-01 more...
Genre: Retro, 3#
Computer test: C+ Popeye 4.61
FEN: 4N1Bk/2R4P/8/8/1P1P4/1Kp5/8/8
Reprints: 17B Retrograde Analysis 1915
476 Hamburger Problem-Nachrichten 09-10/1949
E1 Problem 5-6 12/1951
RA69 diagrammes 28 07-08/1977
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2020-11-01 more...
1. 0-0-0?? illegal
1. Td1! ... 2. Td3#
1. Td1! ... 2. Td3#
1. e4 h5 2. Lb5 h4 3. Lc6 bxc6 4. Dh5 Lb7 5. Dd5 cxd5 6. Sf3 dxe4 7. 0-0 exf3 8. Te1 fxg2 9. Te6 dxe6 10. b4 Kd7 11. Lb2 Kc6 12. Lf6 Kb5 13. c4+ Ka4 14. Sc3+ Ka3 15. Te1 gxf6 and then 16. Tb1! ... 17. Tb3#
TRD: "The point here is that every move of the game is strictly forced in order and kind, the longest such game existant at present. I expect this task to go much further."
TRD: "The point here is that every move of the game is strictly forced in order and kind, the longest such game existant at present. I expect this task to go much further."
Original stipulation was: "White here mated in two and noticed on examining his score that the mate was administered on his 17th move. What was the game?"
A.Buchanan: This is the earliest unique PG recorded in PDB, but Dawson implies earlier, shorter ones exist. Can any of these be traced? (2021-01-22)
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A.Buchanan: This is the earliest unique PG recorded in PDB, but Dawson implies earlier, shorter ones exist. Can any of these be traced? (2021-01-22)
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Keywords: Unique Proof Game, Move Length Record, Volet Pawn
Genre: Retro
Computer test: Euclide 0.99
FEN: rn1q1bnr/pbp1pp2/4pp2/8/1PP4p/k1N5/P2P1PpP/4R1K1
Reprints: (7) The Chess Amateur 08/1914
136 Ultimate Themes 1938
76 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-01-22 more...
Genre: Retro
Computer test: Euclide 0.99
FEN: rn1q1bnr/pbp1pp2/4pp2/8/1PP4p/k1N5/P2P1PpP/4R1K1
Reprints: (7) The Chess Amateur 08/1914
136 Ultimate Themes 1938
76 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-01-22 more...
1. ... bxc6ep 2. b5 Txa3#
R: 1. ... c7-c5 2. f4xTg5 Tg6-g5 3. f3-f4 Ta6-g6 4. f2-f3 Sc6-a7 5. h4-h5 Ta8-a6 6. h3-h4 a6-a5 7. h2-h3 Ka5-a4 8. b3-b4 Ta4-a3
or R: 1. f4xTg5 Tg6-g5 2. f3-f4 Ta6-g6 3. f2-f3 Sc6-a7 4. h4-h5 Ta8-a6 5. h3-h4 a6-a5 6. h2-h3 Ka5-a4 7. b3-b4 Ta4-a3
but we are told that White has the move.
R: 1. ... c7-c5 2. f4xTg5 Tg6-g5 3. f3-f4 Ta6-g6 4. f2-f3 Sc6-a7 5. h4-h5 Ta8-a6 6. h3-h4 a6-a5 7. h2-h3 Ka5-a4 8. b3-b4 Ta4-a3
or R: 1. f4xTg5 Tg6-g5 2. f3-f4 Ta6-g6 3. f2-f3 Sc6-a7 4. h4-h5 Ta8-a6 5. h3-h4 a6-a5 6. h2-h3 Ka5-a4 7. b3-b4 Ta4-a3
but we are told that White has the move.
Henrik Juel: Stipulation should probably be interpreted to mean h#1.5 . -1... c7 -2.f4xTg5 Tg6 -3.f3 Ta6 -4.f2 Sc6 -5.h4 Ta8 -6.h3 a6 -7.h2 Ka5 -8.b3 Ta4 etc. (2004-03-18)
A.Buchanan: Idea I suppose is that Black just moved, and moreover it was pawn double hop. I don't see clearly far enough back to confirm that. (2021-10-21)
Mario Richter: If Black moved last, the only legal retraction is R: 1. ... c7-c5! But the position can also be resolved if White moved last, starting with R: 1. f4xTg5 Tg6-g5 2. f3-f4 Ta6-g6 3. f2-f3 Sc6-a7 4. h4-h5 Ta8-a6 5. h3-h4 a6-a5 6. h2-h3 Ka5-a4 7. b3-b4 Ta4-a3.
Therefore the keyword "No legal last move for White" doesn't apply here; and to make the solution work, the Stipulation should be something like "h#2 0.1.1." or "h#1.5". (2021-10-21)
Henrik Juel: HC+ Popeye 4.61 (2021-10-22)
A.Buchanan: To my mind, the retro thinking is serious enough that we can’t claim HC+. If a tool like Replicator 2.0 was to validate the retro status I’d be ok. But I think it would be misleading to put HC+ to indicate the trivial h#1.5 is sound. We can certainly mark that Popeye has checked the h#, but not click the C+ flag. What do you think? (2021-10-22)
Henrik Juel: HC+ is not recognized (yet) by PDB, and I suppose it just meant that some sort of human analysis is needed in addition to a standard forward solver like Popeye
You are suggesting that we distinguish between easy and difficult human analysis and say HC+ for easy analysis only
I suggest saying HC+ for all problems where Popeye is not sufficient; you can write easy or difficult in the C+ comment
If and when a human analyst proves the analysis wrong, we delete the HC+ label, of course (2021-10-22)
A.Buchanan: What does Mario think? He invented the term, I seem to remember (2021-10-22)
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A.Buchanan: Idea I suppose is that Black just moved, and moreover it was pawn double hop. I don't see clearly far enough back to confirm that. (2021-10-21)
Mario Richter: If Black moved last, the only legal retraction is R: 1. ... c7-c5! But the position can also be resolved if White moved last, starting with R: 1. f4xTg5 Tg6-g5 2. f3-f4 Ta6-g6 3. f2-f3 Sc6-a7 4. h4-h5 Ta8-a6 5. h3-h4 a6-a5 6. h2-h3 Ka5-a4 7. b3-b4 Ta4-a3.
Therefore the keyword "No legal last move for White" doesn't apply here; and to make the solution work, the Stipulation should be something like "h#2 0.1.1." or "h#1.5". (2021-10-21)
Henrik Juel: HC+ Popeye 4.61 (2021-10-22)
A.Buchanan: To my mind, the retro thinking is serious enough that we can’t claim HC+. If a tool like Replicator 2.0 was to validate the retro status I’d be ok. But I think it would be misleading to put HC+ to indicate the trivial h#1.5 is sound. We can certainly mark that Popeye has checked the h#, but not click the C+ flag. What do you think? (2021-10-22)
Henrik Juel: HC+ is not recognized (yet) by PDB, and I suppose it just meant that some sort of human analysis is needed in addition to a standard forward solver like Popeye
You are suggesting that we distinguish between easy and difficult human analysis and say HC+ for easy analysis only
I suggest saying HC+ for all problems where Popeye is not sufficient; you can write easy or difficult in the C+ comment
If and when a human analyst proves the analysis wrong, we delete the HC+ label, of course (2021-10-22)
A.Buchanan: What does Mario think? He invented the term, I seem to remember (2021-10-22)
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Keywords: En passant as key
Genre: h#, Retro
FEN: 2b5/np1p2pp/7q/pPp3PP/kP6/r1PPP3/RKpp2P1/BB2n3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-21 more...
Genre: h#, Retro
FEN: 2b5/np1p2pp/7q/pPp3PP/kP6/r1PPP3/RKpp2P1/BB2n3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-21 more...
WTM
1. ... bxc6ep 2. Da7 c7#
retro tries:
1) 1. Lxg3 b6 2. Lxd6 Lxd6#
2) 1. Dc6 b6 2. Dc7 dxc7#
3) 1. Ta7 f8=D,f8=T 2. Ka8 Dxc8#,Txd8# ("untolerated" dual)
R: 1. ... c7-c5 2. b4-b5 Dc4-a6 (move order variable beyond this point) 3. b2-b3 Ta6-a8 4. d5-d6 Te6-a6 5. b3-b4 Te8-e6 6. a2xSb3 Th8-e8 7. d4-d5 Lb4-e1 8. d3-d4 Lf8-b4 9. Kg5-h5 e7xSf6+
Missing: Wh: SS Bl: SS
Pawn captures: Wh: axb, hxg Bl: exf, fxg
bPf6 comes from e7, else collision with wPf7. So can't uncapture fxg2 until wP retracted. White has 6 pawn retractions before retro-pat, just giving Bl time to put Le1 & Ta8 (impostor) back home, and uncapture exf6. So Black must retract first, and it's WTM. sD must clear out of the way, and c4 is the only possible square, wPaxSb can't happen on b5, or it would block sD. And axSb4 would block sLe1. It was axSb3. So wPb5 must retreat to b4, and thus Black cannot play R: 1. Lb4-e1. So neatly R: 1. c7-c5 is forced.
1. ... bxc6ep 2. Da7 c7#
retro tries:
1) 1. Lxg3 b6 2. Lxd6 Lxd6#
2) 1. Dc6 b6 2. Dc7 dxc7#
3) 1. Ta7 f8=D,f8=T 2. Ka8 Dxc8#,Txd8# ("untolerated" dual)
R: 1. ... c7-c5 2. b4-b5 Dc4-a6 (move order variable beyond this point) 3. b2-b3 Ta6-a8 4. d5-d6 Te6-a6 5. b3-b4 Te8-e6 6. a2xSb3 Th8-e8 7. d4-d5 Lb4-e1 8. d3-d4 Lf8-b4 9. Kg5-h5 e7xSf6+
Missing: Wh: SS Bl: SS
Pawn captures: Wh: axb, hxg Bl: exf, fxg
bPf6 comes from e7, else collision with wPf7. So can't uncapture fxg2 until wP retracted. White has 6 pawn retractions before retro-pat, just giving Bl time to put Le1 & Ta8 (impostor) back home, and uncapture exf6. So Black must retract first, and it's WTM. sD must clear out of the way, and c4 is the only possible square, wPaxSb can't happen on b5, or it would block sD. And axSb4 would block sLe1. It was axSb3. So wPb5 must retreat to b4, and thus Black cannot play R: 1. Lb4-e1. So neatly R: 1. c7-c5 is forced.
Ladislav Packa: Is wRh1 needed? (2021-10-22)
Mario Richter: Yes, wRh1 is needed - without it the ep-key would not be justified (last moves could have been R: 1. ... Th1-g1 2. Lg1-h2 or even R: 1. ... Ld2xTe1). (2021-10-23)
A.Buchanan: + three mostly clean retro tries. (2021-10-24)
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Mario Richter: Yes, wRh1 is needed - without it the ep-key would not be justified (last moves could have been R: 1. ... Th1-g1 2. Lg1-h2 or even R: 1. ... Ld2xTe1). (2021-10-23)
A.Buchanan: + three mostly clean retro tries. (2021-10-24)
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Keywords: En passant as key, No legal last move for White, Impostor (t)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: rkb5/1p1p1Ppp/q2P1p2/pPp4K/6PR/1P4PQ/2P1P1pB/4bBrR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: rkb5/1p1p1Ppp/q2P1p2/pPp4K/6PR/1P4PQ/2P1P1pB/4bBrR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
1. ... bxc6ep 2. Le7 cxd7#
1. Lc6? bxc6 2. Le7 cxd7# WTM
R: 1. c6-c5? Kd3-e3 2. Dc5-c4+ Ke3-d3 3. Dc4-c5+ retro-loop.
R: 1. c7-c5! Kd3-e3 2. Dc6-c4+ Ke3-d3! clear
Easy to see no last move for White. Moreover, Black just double-hopped to allow sD/wK to avoid retro-loop.
Single clean retro try.
1. Lc6? bxc6 2. Le7 cxd7# WTM
R: 1. c6-c5? Kd3-e3 2. Dc5-c4+ Ke3-d3 3. Dc4-c5+ retro-loop.
R: 1. c7-c5! Kd3-e3 2. Dc6-c4+ Ke3-d3! clear
Easy to see no last move for White. Moreover, Black just double-hopped to allow sD/wK to avoid retro-loop.
Single clean retro try.
Keywords: En passant as key, No legal last move for White
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & moderate retro-logic
FEN: 3nkr2/3p1r2/1p1npb2/PPpb4/Bpq2PP1/1p2Kpp1/2PPP3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & moderate retro-logic
FEN: 3nkr2/3p1r2/1p1npb2/PPpb4/Bpq2PP1/1p2Kpp1/2PPP3/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
1. ... bxc6ep 2. Dxa5 Db7#
1. Sc6? bxc6 2. Dxa5 Db7# but WTM
R: 1. ... c7-c5 2. g5-g6 Sc6-e7 3. De7-b4,D~ Sb4-c6+ unpromote wD on h8, etc.
Clean retro try
1. Sc6? bxc6 2. Dxa5 Db7# but WTM
R: 1. ... c7-c5 2. g5-g6 Sc6-e7 3. De7-b4,D~ Sb4-c6+ unpromote wD on h8, etc.
Clean retro try
Henrik Juel: Typo: 2.Dxa5 Db7#. wTa1 is missing, I think (10+15). -1... c7 -2.g5 Sc6 -3.De7 Sb4, unpromote wD on h8, etc. (2004-03-18)
A.Buchanan: I agree Henrik and have made changes. A number of these problems have been marked "Weisz zieht an", which is inappropriate as part of the point of the problem is to deduce this. I think it's 2. c2-c3! not 2. g5-g6? or the cross-captured pawns become retro-locked. (2021-10-21)
Henrik Juel: 2.g5-g6 also works, because the only cross-capture is axb,bxa
White captured [Pd7] with an officer (2021-10-22)
Henrik Juel: HC+ Popeye 4.61 (2021-10-22)
A.Buchanan: If g5-g6 also works, then can’t we have both? I.e. White moved last. I would be surprised if this was TRD’s intention, given the others of this general style (2021-10-22)
Henrik Juel: TRD died in 1951, so someone else may have messed up the problem
FRC is not generally available after 1953, so I cannot check the source (2021-10-22)
A.Buchanan: WinChloe has 29(!) posthumous problems by Dawson - the latest in 1958, FCR's final year. The composition in question is not one of them. (2021-10-23)
A.Buchanan: PROOF OF COOK + SUGGESTED FIX
I agree with much of what you say Henrik.
Capturing history. Basically 3 possibilities for the Black pawns.
Notation: ~,o,| denote "cross-capture", "original" & "waylaid"(=captured on home file by officer).
1) a~b co d~e f~g. This implies either (wPhxg & bPh=X) or (wPh=X & bPh|)
2) a~b co d~e hxgxf f|
3) a~b co hxgxfxexd d|
Suppose WTM. Then R: 1. c6-c5 c2-c3 2. c7-c6 g5-g6 3. Sc6-e7 De7-b4 4. Sb4-c6+ Kc5-c4 5. Lb2-d4+ etc. Now we're free to uncapture wDh8 (or g8 for scenario 1), and any of the three pawn scenarios can work. No e.p. rights means a cook.
If we fix it by e.g. shifting wPg6-g5 then I think that fixes this cook, and also it means that there is no solution if BTM.
Is there any possibility for premature capturing hxg? Not by Bl because scenario 1 says bPfxg, and the other two scenarios rely on wPh=X. If by Wh, then there's no waylaying on d or f, so we are back to scenario 1, but bPh is blocked from promotion, so that won't work.
But: what about R: 1. c6-c5 c2-c3 2. Sc3-b1 Th1-a1. Doesn't this completely cook it anyway?? How about k7/4n3/6P1/PPpPPP2/qQKbrrp1/ppRppp2/b1P5/1n6? h#2 (with Art 15 so it's really h#1.5)
Do you agree? (2021-10-23)
Mario Richter: In the diagram, both for WTM and BTM the retraction sequence c2-c3 Sc6-e7 De1-b4 Sb4-c6+ works and furthermore Andrew's R: 1. c6-c5 c2-c3 2. Sc3-b1 Th1-a1 shows, that Henrik's guess of the omitted wTa1 might be wrong. Instead, I think that Andrew's modification might be in fact the original Dawson diagram. (2021-10-23)
A.Buchanan: Yes that makes sense. I can well believe that there was a lot of constructions in flight when TRD passed away. It's not surprising that something appeared a few years later. Can we get access to FCR from that year to validate? (2021-10-23)
A.Buchanan: Brian Stephenson kindly checked the original magazine. He wrote:
"Eventually got round to looking out this TRD problem. It was published in the issue you quote as problem 10381. The diagram in PDB is wrong. There should be a bPc2. The solution was published in FCR April 1956 as: 1.Pb5xPc5ep Qxa5 2.Qb7#. Last moves must have been 1.Pc7-c5 Pg5-g6 2.Sc6-e7 Qe7-b4 Sb4-c6 ch etc. No flaws were noted, but I have looked later than that issue. Hope this helps." (2021-11-13)
A.Buchanan: The appearance of the 16th Black unit thanks to the offices of Brian Stephenson simplifies the retro logic substantially, and the solution seems sound. Amazing how much disruption a simple typo can cause, but at least this one I believe is laid to rest (2021-11-13)
more ...
comment
A.Buchanan: I agree Henrik and have made changes. A number of these problems have been marked "Weisz zieht an", which is inappropriate as part of the point of the problem is to deduce this. I think it's 2. c2-c3! not 2. g5-g6? or the cross-captured pawns become retro-locked. (2021-10-21)
Henrik Juel: 2.g5-g6 also works, because the only cross-capture is axb,bxa
White captured [Pd7] with an officer (2021-10-22)
Henrik Juel: HC+ Popeye 4.61 (2021-10-22)
A.Buchanan: If g5-g6 also works, then can’t we have both? I.e. White moved last. I would be surprised if this was TRD’s intention, given the others of this general style (2021-10-22)
Henrik Juel: TRD died in 1951, so someone else may have messed up the problem
FRC is not generally available after 1953, so I cannot check the source (2021-10-22)
A.Buchanan: WinChloe has 29(!) posthumous problems by Dawson - the latest in 1958, FCR's final year. The composition in question is not one of them. (2021-10-23)
A.Buchanan: PROOF OF COOK + SUGGESTED FIX
I agree with much of what you say Henrik.
Capturing history. Basically 3 possibilities for the Black pawns.
Notation: ~,o,| denote "cross-capture", "original" & "waylaid"(=captured on home file by officer).
1) a~b co d~e f~g. This implies either (wPhxg & bPh=X) or (wPh=X & bPh|)
2) a~b co d~e hxgxf f|
3) a~b co hxgxfxexd d|
Suppose WTM. Then R: 1. c6-c5 c2-c3 2. c7-c6 g5-g6 3. Sc6-e7 De7-b4 4. Sb4-c6+ Kc5-c4 5. Lb2-d4+ etc. Now we're free to uncapture wDh8 (or g8 for scenario 1), and any of the three pawn scenarios can work. No e.p. rights means a cook.
If we fix it by e.g. shifting wPg6-g5 then I think that fixes this cook, and also it means that there is no solution if BTM.
Is there any possibility for premature capturing hxg? Not by Bl because scenario 1 says bPfxg, and the other two scenarios rely on wPh=X. If by Wh, then there's no waylaying on d or f, so we are back to scenario 1, but bPh is blocked from promotion, so that won't work.
But: what about R: 1. c6-c5 c2-c3 2. Sc3-b1 Th1-a1. Doesn't this completely cook it anyway?? How about k7/4n3/6P1/PPpPPP2/qQKbrrp1/ppRppp2/b1P5/1n6? h#2 (with Art 15 so it's really h#1.5)
Do you agree? (2021-10-23)
Mario Richter: In the diagram, both for WTM and BTM the retraction sequence c2-c3 Sc6-e7 De1-b4 Sb4-c6+ works and furthermore Andrew's R: 1. c6-c5 c2-c3 2. Sc3-b1 Th1-a1 shows, that Henrik's guess of the omitted wTa1 might be wrong. Instead, I think that Andrew's modification might be in fact the original Dawson diagram. (2021-10-23)
A.Buchanan: Yes that makes sense. I can well believe that there was a lot of constructions in flight when TRD passed away. It's not surprising that something appeared a few years later. Can we get access to FCR from that year to validate? (2021-10-23)
A.Buchanan: Brian Stephenson kindly checked the original magazine. He wrote:
"Eventually got round to looking out this TRD problem. It was published in the issue you quote as problem 10381. The diagram in PDB is wrong. There should be a bPc2. The solution was published in FCR April 1956 as: 1.Pb5xPc5ep Qxa5 2.Qb7#. Last moves must have been 1.Pc7-c5 Pg5-g6 2.Sc6-e7 Qe7-b4 Sb4-c6 ch etc. No flaws were noted, but I have looked later than that issue. Hope this helps." (2021-11-13)
A.Buchanan: The appearance of the 16th Black unit thanks to the offices of Brian Stephenson simplifies the retro logic substantially, and the solution seems sound. Amazing how much disruption a simple typo can cause, but at least this one I believe is laid to rest (2021-11-13)
more ...
comment
Keywords: En passant as key, No legal last move for White
Genre: h#, Retro
Computer test: HC+ Popeye 4.6 + non-trivial retro
FEN: k7/4n3/6P1/PPpPPP2/qQKbrrp1/ppPppp2/b1p5/1n6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
Genre: h#, Retro
Computer test: HC+ Popeye 4.6 + non-trivial retro
FEN: k7/4n3/6P1/PPpPPP2/qQKbrrp1/ppPppp2/b1p5/1n6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
10 - P0005358
Thomas R. Dawson
(4) The Chess Amateur 08/1914
(9+12) C+
Black, having just made hist 25th move, is it possible for White to mate in 2?
(BP in 24,5)
Thomas R. Dawson
(4) The Chess Amateur 08/1914
(9+12) C+
Black, having just made hist 25th move, is it possible for White to mate in 2?
(BP in 24,5)
1. Sc3 a5 2. Se4 Sa6 3. b4 a4 4. b5 a3 5. d4 b6 6. d5 Lb7 7. d6 exd6 8. Lf4 Dh4 9. bxa6 Dh3 10. c4 Sf6 11. Sxf6+ gxf6 12. c5 Lf3 13. exf3 Lh6 14. Lb5 c6 15. Kf1 cxb5 16. Se2 dxc5 17. Ld6 Ld2 18. Tg1 La5 19. gxh3 h6 20. Tg5 hxg5 21. Kg2 Th4 22. Db1 Ta4 23. Db4 cxb4 24. Sf4 gxf4 25. Tg1
Black can have wasted only one move and so cannot have moved K or KR, which would require a second waste move returning to diagram. Hence Black is legally entitled to castle. Therefore White mates in 2 by: 1. Kh1! 0-0-0 2. Tc1, else 2. Tg8
Black can have wasted only one move and so cannot have moved K or KR, which would require a second waste move returning to diagram. Hence Black is legally entitled to castle. Therefore White mates in 2 by: 1. Kh1! 0-0-0 2. Tc1, else 2. Tg8
Moldenhauer: Computerprüfung: C+ Stelvio NUPG cooked 1 Sekunde.
Keine Lösung: BP 23.5, BP 24.0.
Wenn Schwarz seinen 25. b3 spielt ist kein Matt in 2 möglich.
Wenn angenommen wird das Weiß aus dieser Stellung nochmals
ziehen darf dann Matt in 2 Zügen.
Notation: 1.Sc3 a5 2.Se4 Sa6 3.b4 a4 4.b5 a3 5.d4 b6 6.d5 Lb7 7.d6 exd6
8.Lf4 Dh4 9.bxa6 Dh3 10.c4 Sf6 11.Sxf6+ gxf6 12.c5 Lf3 13.exf3 Lh6
14.Lb5 c6 15.Kf1 cxb5 16.Se2 dxc5 17.Ld6 Ld2 18.Tg1 La5 19.gxh3 h6
20.Tg5 hxg5 21.Kg2 Th4 22.Db1 Ta4 23.Db4 cxb4 24.Sf4 gxf4 25.Tg1 (2023-04-06)
A.Buchanan: Should the stip be something like:
Position after White’s 25th move.
Can Black play to avoid being mated in 2?
But both b3 and Rc8 seem to let Black escape (2023-04-06)
more ...
comment
Keine Lösung: BP 23.5, BP 24.0.
Wenn Schwarz seinen 25. b3 spielt ist kein Matt in 2 möglich.
Wenn angenommen wird das Weiß aus dieser Stellung nochmals
ziehen darf dann Matt in 2 Zügen.
Notation: 1.Sc3 a5 2.Se4 Sa6 3.b4 a4 4.b5 a3 5.d4 b6 6.d5 Lb7 7.d6 exd6
8.Lf4 Dh4 9.bxa6 Dh3 10.c4 Sf6 11.Sxf6+ gxf6 12.c5 Lf3 13.exf3 Lh6
14.Lb5 c6 15.Kf1 cxb5 16.Se2 dxc5 17.Ld6 Ld2 18.Tg1 La5 19.gxh3 h6
20.Tg5 hxg5 21.Kg2 Th4 22.Db1 Ta4 23.Db4 cxb4 24.Sf4 gxf4 25.Tg1 (2023-04-06)
A.Buchanan: Should the stip be something like:
Position after White’s 25th move.
Can Black play to avoid being mated in 2?
But both b3 and Rc8 seem to let Black escape (2023-04-06)
more ...
comment
Keywords: Castling, Non-Unique Proof Game
Genre: Retro
Computer test: Computerprüfung: C+ Stelvio NUPG cooked 1 Sekunde. Keine Lösung: BP 23.5, BP 24.0.
FEN: r3k3/3p1p2/Pp1B1p2/bp6/rp3p2/p4P1P/P4PKP/6R1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-06 more...
Genre: Retro
Computer test: Computerprüfung: C+ Stelvio NUPG cooked 1 Sekunde. Keine Lösung: BP 23.5, BP 24.0.
FEN: r3k3/3p1p2/Pp1B1p2/bp6/rp3p2/p4P1P/P4PKP/6R1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-06 more...
1. fxg6ep+ Kg5 2. Lh4#
R: 1. g7-g5
R: 1. g7-g5
Keywords: En passant as key
Pieces: = Grasshopper (G)
Genre: Retro, 2#, Fairies
Computer test: HC+ Popeye 4.61 after very simple analysis
FEN: 8/5*2Q1K/5P1P/5Ppk/6p1/6B1/6N1/8
Reprints: 115 Caissa's Wild Roses 1935
Input: Gerd Wilts, 1996-07-16
Last update: A.Buchanan, 2021-05-03 more...
Pieces: = Grasshopper (G)
Genre: Retro, 2#, Fairies
Computer test: HC+ Popeye 4.61 after very simple analysis
FEN: 8/5*2Q1K/5P1P/5Ppk/6p1/6B1/6N1/8
Reprints: 115 Caissa's Wild Roses 1935
Input: Gerd Wilts, 1996-07-16
Last update: A.Buchanan, 2021-05-03 more...
1. Th1 Sa1 2. Th4 Sh1 3. Tf4 Sc2#
Cook: NL
1. Tf1 Kh8 2. Tf2 Sa1 3. Df4 Sc2#
1. Df4 Sxa1 2. e1=L h6 3. Lf2 Sc2#
Cook: NL
1. Tf1 Kh8 2. Tf2 Sa1 3. Df4 Sc2#
1. Df4 Sxa1 2. e1=L h6 3. Lf2 Sc2#
teilweiser Rundlauf t
Adrian Storisteanu: Possible fix: bPe4 (instead of the bQ), +wPf5, +bPf6 (6+9). (2015-08-15)
Yuri Bilokin: correction bRd3(-bNd3), bBe4(-bQe4), bPf3(-bBf3) 6K1/8/8/7P/3Pb3/pN1rkpN1/3pp3/r7 (5+8) (2022-12-19)
A.Buchanan: This is very clean Yuri well done. I would shift bPa3 to a5. wK must be exactly on g8, else the problem is cooked (2022-12-20)
comment
Adrian Storisteanu: Possible fix: bPe4 (instead of the bQ), +wPf5, +bPf6 (6+9). (2015-08-15)
Yuri Bilokin: correction bRd3(-bNd3), bBe4(-bQe4), bPf3(-bBf3) 6K1/8/8/7P/3Pb3/pN1rkpN1/3pp3/r7 (5+8) (2022-12-19)
A.Buchanan: This is very clean Yuri well done. I would shift bPa3 to a5. wK must be exactly on g8, else the problem is cooked (2022-12-20)
comment
Keywords: Pure Round Trip
Genre: h#
FEN: 6K1/8/8/7P/3Pq3/pN1nkbN1/3pp3/r7
Input: hpr, 1996-09-05
Last update: Alfred Pfeiffer, 2015-08-15 more...
Genre: h#
FEN: 6K1/8/8/7P/3Pq3/pN1nkbN1/3pp3/r7
Input: hpr, 1996-09-05
Last update: Alfred Pfeiffer, 2015-08-15 more...
1. Kc3 De6 2. Kxd4 Dxd6+ 3. Ke4 De5#
NL
1. Kc3 Df5 2. Tf8 Kxd6 3. Kxd4 Dd3#
NL
1. Kc3 Df5 2. Tf8 Kxd6 3. Kxd4 Dd3#
Korrektur Hilmar Ebert: +sSe8 (erhält Quasi-Symmetrie)
Yuri Bilokin: Not all pieces in quasi-symmetry are necessary: the white pawn on h2 is superfluous. It is clear that more than 100 years ago there was such a direction in art, although some authors indulge in technical white pawns in our time. (2022-12-19)
A.Buchanan: The technical term for such a position is "symmetric constellation". "Quasi-symmetry" is where the position is nearly symmetric but not quite (e.g. game array except wPa2 is missing). In a symmetric constellation any redundant piece is a definite defect. A quasi-symmetric position is not an ornament, more of a consequence of the content. For example, a number of last move type DD problems are quasi-symmetric (2022-12-20)
Yuri Bilokin: Thanks for the interpretation (2022-12-20)
comment
Yuri Bilokin: Not all pieces in quasi-symmetry are necessary: the white pawn on h2 is superfluous. It is clear that more than 100 years ago there was such a direction in art, although some authors indulge in technical white pawns in our time. (2022-12-19)
A.Buchanan: The technical term for such a position is "symmetric constellation". "Quasi-symmetry" is where the position is nearly symmetric but not quite (e.g. game array except wPa2 is missing). In a symmetric constellation any redundant piece is a definite defect. A quasi-symmetric position is not an ornament, more of a consequence of the content. For example, a number of last move type DD problems are quasi-symmetric (2022-12-20)
Yuri Bilokin: Thanks for the interpretation (2022-12-20)
comment
Keywords: Pure Round Trip (D)
Genre: h#
FEN: 8/4K3/3p1r2/4Q3/3P1P2/8/1k2P2P/8
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 2001-12-21 more...
Genre: h#
FEN: 8/4K3/3p1r2/4Q3/3P1P2/8/1k2P2P/8
Input: Hans-Jürgen Schäfer, 1996-09-23
Last update: hpr, 2001-12-21 more...
1. Dxe5+ Lf5 2. Dxd4 Lc2 3. De4 Lb3#
NL:
1. d1=S Le7 2. Sxb2 Sb3 3. f1=S Sc1#
NL:
1. d1=S Le7 2. Sxb2 Sb3 3. f1=S Sc1#
Korrekturvorschlag HJS Co+:wKg5, wLc2f6, wSc5, wBb2d4e5g4, sKa2, sDe4, sTa1h1, sLb1d8, sBb4d5e6f2f7
milan: +wPd6 M.Frelih (2021-12-01)
A.Buchanan: 3b4/5p2/4pB2/2NpP1K1/1p1Pq1P1/8/kPB2p2/rb5r as specified in Korrekturvorschlag HJS is C+ with 2 fewer units (2021-12-02)
comment
milan: +wPd6 M.Frelih (2021-12-01)
A.Buchanan: 3b4/5p2/4pB2/2NpP1K1/1p1Pq1P1/8/kPB2p2/rb5r as specified in Korrekturvorschlag HJS is C+ with 2 fewer units (2021-12-02)
comment
Keywords: Pure Round Trip (d)
Genre: h#
FEN: 3b2n1/5p2/4pB2/1pNpP1K1/3Pq1P1/8/kPBp1p1r/rb6
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-19 more...
Genre: h#
FEN: 3b2n1/5p2/4pB2/1pNpP1K1/3Pq1P1/8/kPBp1p1r/rb6
Input: hpr, 1996-09-23
Last update: hpr, 1999-06-19 more...
1. Kxe3 c5 2. Kd2 Sc4#
NL:
1. Td5 Lf5 2. Txe5 Sc2#
1. Lf3 S3g4 2. Lxe4 Lf2#
NL:
1. Td5 Lf5 2. Txe5 Sc2#
1. Lf3 S3g4 2. Lxe4 Lf2#
Yuri Bilokin: correction bRd7-f7, bBg4-a4, -bN7 8/5r2/4p3/4N3/bKPkB2B/4N3/4p3/2nr4 (6+7) (2022-12-15)
A.Buchanan: The whole point of this ornamental problem is the symmetry. Any correction must preserve that. How about getting rid of the 2 rooks and the g-h-file bishops, then shift wBe4 to a4? 8/2n5/4p3/4N3/BKPk4/4N3/4p3/2n5? (2022-12-16)
Yuri Bilokin: I agree that the position should be symmetrical and economical, without unnecessary figures. The value of the positions reflected here is close to zero. (2022-12-16)
comment
A.Buchanan: The whole point of this ornamental problem is the symmetry. Any correction must preserve that. How about getting rid of the 2 rooks and the g-h-file bishops, then shift wBe4 to a4? 8/2n5/4p3/4N3/BKPk4/4N3/4p3/2n5? (2022-12-16)
Yuri Bilokin: I agree that the position should be symmetrical and economical, without unnecessary figures. The value of the positions reflected here is close to zero. (2022-12-16)
comment
Keywords: Sacrifice of white pieces, Symmetrical position
Genre: h#
FEN: 8/2nr4/4p3/4N3/1KPkB1bB/4N3/4p3/2nr4
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: hpr, 1999-10-24 more...
Genre: h#
FEN: 8/2nr4/4p3/4N3/1KPkB1bB/4N3/4p3/2nr4
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: hpr, 1999-10-24 more...
1. Sxe6 Sf3 2. Sg7 Tf8#
NL:
1. Sf7 exf7 2. Kg7 f8=D#
NL:
1. Sf7 exf7 2. Kg7 f8=D#
Yuri Bilokin: correction wKa1-b2, then a2=a1 8/7R/4n3/4Pk2/4NPn1/8/8/1K6 (5+3) h#2
1.Sxe5 Sf2 2.Sg6 Rf7# (MM) (2022-12-15)
comment
1.Sxe5 Sf2 2.Sg6 Rf7# (MM) (2022-12-15)
comment
Keywords: Symmetrical position
Genre: h#
FEN: 7R/4n3/4Pk2/4NPn1/8/8/8/K7
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: hpr, 1999-10-24 more...
Genre: h#
FEN: 7R/4n3/4Pk2/4NPn1/8/8/8/K7
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: hpr, 1999-10-24 more...
1. d4 Sh6 2. Tg5 Ta6#
Cook: NL:
1. d4 Kg8 2. Tf5 Ta6#
Cook: NL:
1. d4 Kg8 2. Tf5 Ta6#
Adrian Storisteanu: Possible fix (a bit rough, but the wK is still employed):
Kg7 Ra1 Se6 Sf5 / Ke5 Rb4 Rd2 Sh8 pc4 pd3 (4+6)
1.c3 Sg5 2.Rf4 Ra5#. (2019-10-12)
A.Buchanan: Pushing wS to f8 & h6 seems to fix the cook. Maybe that's where they were meant to be all along: hard to imagine old TRD making such an error. WinChloe hasn't got this one, so can't confirm. (2019-10-14)
Adrian Storisteanu: Nice workaround, Andrew! Dawson had his share of cooks... (2019-10-14)
Yuri Bilokin: possibly wNf7-f5, bRe2-f2, -bPe4 7K/8/5kN1/1r1p1N2/8/8/5r2/R7 (4+4) (2022-12-16)
comment
Kg7 Ra1 Se6 Sf5 / Ke5 Rb4 Rd2 Sh8 pc4 pd3 (4+6)
1.c3 Sg5 2.Rf4 Ra5#. (2019-10-12)
A.Buchanan: Pushing wS to f8 & h6 seems to fix the cook. Maybe that's where they were meant to be all along: hard to imagine old TRD making such an error. WinChloe hasn't got this one, so can't confirm. (2019-10-14)
Adrian Storisteanu: Nice workaround, Andrew! Dawson had his share of cooks... (2019-10-14)
Yuri Bilokin: possibly wNf7-f5, bRe2-f2, -bPe4 7K/8/5kN1/1r1p1N2/8/8/5r2/R7 (4+4) (2022-12-16)
comment
Keywords: Symmetrical position, Asymmetrical solution, Superseded by (P1389169)
Genre: h#
FEN: 7K/5N2/5kN1/1r1p4/4p3/8/4r3/R7
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: A.Buchanan, 2021-05-02 more...
Genre: h#
FEN: 7K/5N2/5kN1/1r1p4/4p3/8/4r3/R7
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: A.Buchanan, 2021-05-02 more...
*) 1. ... Ld4 2. c3 Txc3 3. Ka1 Tc1#
1) 1. c3 a4 2. c2 Ta3 3. c1=L Le4#
NL:
*) 1. ... Td3 2. Kc2 Td1 3. c3 Lb3#
1. Kc1 Td3 2. Kc2 Td1 3. c3 Lb3#
1) 1. c3 a4 2. c2 Ta3 3. c1=L Le4#
NL:
*) 1. ... Td3 2. Kc2 Td1 3. c3 Lb3#
1. Kc1 Td3 2. Kc2 Td1 3. c3 Lb3#
Verstellung TcL
Yuri Bilokin: possibly wPa2-a3 8/7p/6pr/1P1BP2b/2p4K/P6R/7P/Bk6 (8+6) (2022-12-19)
A.Buchanan: Yes looks like a typo (2022-12-20)
comment
Yuri Bilokin: possibly wPa2-a3 8/7p/6pr/1P1BP2b/2p4K/P6R/7P/Bk6 (8+6) (2022-12-19)
A.Buchanan: Yes looks like a typo (2022-12-20)
comment
Keywords: Line closure
Genre: h#
FEN: 8/7p/6pr/1P1BP2b/2p4K/7R/P6P/Bk6
Input: Andreas Mokosch, 1997-07-23
Last update: hpr, 1999-10-26 more...
Genre: h#
FEN: 8/7p/6pr/1P1BP2b/2p4K/7R/P6P/Bk6
Input: Andreas Mokosch, 1997-07-23
Last update: hpr, 1999-10-26 more...
1. Lf5 Sf3 2. Sd3 Sg5 3. Lg7 Sf7#
Cook: NL 1. Sd3+ Sf5 2. Th1 Sxh6 3. Lg7 Sf7#
Cook: NL 1. Sd3+ Sf5 2. Th1 Sxh6 3. Lg7 Sf7#
Verst-4
Adrian Storisteanu: Possible fix: wSh4->b4, +bPh4 (2+10) -- 1.Bf5 Sc6 2.Sd3 Sd8 3.Bg7 Sf7#. (2015-08-26)
Yuri Bilokin: possibly bBb1-g6 6nk/7q/6bp/3Kn2r/3b3N/8/8/3r4 (2+9) (2022-12-19)
A.Buchanan: As well as bBb1-g6, one can shift bPh6-h7, i.e. removing bQ which serves no purpose (2022-12-20)
Yuri Bilokin: After correction, you can improve the efficiency (bPh6-h7), you can improve the sparsity (wNh4-b2). (2022-12-21)
A.Buchanan: I also wondered about shifting wSh4, but I thought it's good to have a knight start near to its destination square, because there are more convoluted paths for dual elimination, rather than pulling the line more taut. b4, b8 & e1 also work - it becomes a matter of taste. On balance I like Sh4 because it suggests 1. ... Sf5? unpinning the bS (2022-12-21)
more ...
comment
Adrian Storisteanu: Possible fix: wSh4->b4, +bPh4 (2+10) -- 1.Bf5 Sc6 2.Sd3 Sd8 3.Bg7 Sf7#. (2015-08-26)
Yuri Bilokin: possibly bBb1-g6 6nk/7q/6bp/3Kn2r/3b3N/8/8/3r4 (2+9) (2022-12-19)
A.Buchanan: As well as bBb1-g6, one can shift bPh6-h7, i.e. removing bQ which serves no purpose (2022-12-20)
Yuri Bilokin: After correction, you can improve the efficiency (bPh6-h7), you can improve the sparsity (wNh4-b2). (2022-12-21)
A.Buchanan: I also wondered about shifting wSh4, but I thought it's good to have a knight start near to its destination square, because there are more convoluted paths for dual elimination, rather than pulling the line more taut. b4, b8 & e1 also work - it becomes a matter of taste. On balance I like Sh4 because it suggests 1. ... Sf5? unpinning the bS (2022-12-21)
more ...
comment
Genre: h#
FEN: 6nk/7q/7p/3Kn2r/3b3N/8/8/1b1r4
Input: HBae, 1997-10-18
Last update: A.Buchanan, 2022-12-20 more...
1. 0-0-0+ Ld3 2. Ld7 La6#
NL:
1. Dg8/Db7/Dc7 Dxe6+ 2. Te7 Dg8#
1. Df8 Dd4/Dc7/Dd6/Dd5 2. Lf7 Dd7#
NL:
1. Dg8/Db7/Dc7 Dxe6+ 2. Te7 Dg8#
1. Df8 Dd4/Dc7/Dd6/Dd5 2. Lf7 Dd7#
Rochade
F135
Verst-4
Yuri Bilokin: correction bRh7-f7(-bQf7), bBa1-b6, -bPa7, +bSf4, +bSh6 r3k3/5r1B/1b2b2n/4Q3/5n2/8/8/3K4 (3+7) (2022-08-28)
comment
F135
Verst-4
Yuri Bilokin: correction bRh7-f7(-bQf7), bBa1-b6, -bPa7, +bSf4, +bSh6 r3k3/5r1B/1b2b2n/4Q3/5n2/8/8/3K4 (3+7) (2022-08-28)
comment
1. Tb3 a3 2. Kxa3 Dxa1#
NL:
1. De1 Dxd5 2. d2 Dxc4#
NL:
1. De1 Dxd5 2. d2 Dxc4#
klären Quelle ?
Bath+Welts Chronicle+Herald ?
Yuri Bilokin: correction wKg4-f4, -wPa6, bBd3(-bPd3), bNa7-c6, -bNa1, -bPd5 8/8/2n5/r7/kqpQ1K2/r2b4/P7/8 (3+7) (2022-12-15)
A.Buchanan: Any acceptable correction must surely retain the thematic symmetry? (2022-12-16)
A.Buchanan: Bath & Wells (not Welts) are two small neighbouring cities in the West of England which are kind of paired together. E.g. there is a Bishop of B&W (which it occurs for me might stand for Black & White, suggesting a thematic problem with a neutral bishop). (2022-12-16)
milan: +wPb3b5 1.T×b3 a3 2. K×a3 D×a1# simply correction M.Frelih (2022-12-18)
comment
Bath+Welts Chronicle+Herald ?
Yuri Bilokin: correction wKg4-f4, -wPa6, bBd3(-bPd3), bNa7-c6, -bNa1, -bPd5 8/8/2n5/r7/kqpQ1K2/r2b4/P7/8 (3+7) (2022-12-15)
A.Buchanan: Any acceptable correction must surely retain the thematic symmetry? (2022-12-16)
A.Buchanan: Bath & Wells (not Welts) are two small neighbouring cities in the West of England which are kind of paired together. E.g. there is a Bishop of B&W (which it occurs for me might stand for Black & White, suggesting a thematic problem with a neutral bishop). (2022-12-16)
milan: +wPb3b5 1.T×b3 a3 2. K×a3 D×a1# simply correction M.Frelih (2022-12-18)
comment
Keywords: Symmetrical position
Genre: h#
FEN: 8/n7/P7/r2p4/kqpQ2K1/r2p4/P7/n7
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: hpr, 1999-10-28 more...
Genre: h#
FEN: 8/n7/P7/r2p4/kqpQ2K1/r2p4/P7/n7
Input: Hans-Jürgen Schäfer, 1997-09-01
Last update: hpr, 1999-10-28 more...
1. f1=L Sa2 2. Lxe2+ Kxe2 3. d1=T Sxc3#
Cook: NL 1. Td3 Sf4 2. Ld4 Sb3 3. Te3 Sxd2#
Cook: NL 1. Td3 Sf4 2. Ld4 Sb3 3. Te3 Sxd2#
Doppelveröffentlichung
Adrian Storisteanu: Possible fix: -bPe5, +wPf4 (7+6). (2015-11-27)
Yuri Bilokin: possibly wNc1-a7 8/N7/8/4p3/p2rk1P1/P1b3P1/3pNp2/3K4 (6+7) (2022-12-19)
HBae: Bei dem Korrekturvorschlag von Adrian wäre auch -wPa3 +bPa3 möglich, C+ Popeye v4.37 (2022-12-20)
Yuri Bilokin: Yes, it's better
There is no need to strain yourself by dragging the white knight to the a7 square :) (2022-12-20)
comment
Adrian Storisteanu: Possible fix: -bPe5, +wPf4 (7+6). (2015-11-27)
Yuri Bilokin: possibly wNc1-a7 8/N7/8/4p3/p2rk1P1/P1b3P1/3pNp2/3K4 (6+7) (2022-12-19)
HBae: Bei dem Korrekturvorschlag von Adrian wäre auch -wPa3 +bPa3 möglich, C+ Popeye v4.37 (2022-12-20)
Yuri Bilokin: Yes, it's better
There is no need to strain yourself by dragging the white knight to the a7 square :) (2022-12-20)
comment
Keywords: Sacrifice of white pieces, Tempo Move (b(t))
Genre: h#
FEN: 8/8/8/4p3/p2rk1P1/P1b3P1/3pNp2/2NK4
Reprints: 3605 Falkirk Herald 09/05/1934
Input: Markus Manhart, 1998-02-10
Last update: Alfred Pfeiffer, 2015-11-30 more...
Genre: h#
FEN: 8/8/8/4p3/p2rk1P1/P1b3P1/3pNp2/2NK4
Reprints: 3605 Falkirk Herald 09/05/1934
Input: Markus Manhart, 1998-02-10
Last update: Alfred Pfeiffer, 2015-11-30 more...
23 - 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...
24 - P0531454
Thomas R. Dawson
65 British Chess Federation 07/1945
5. ehrende Erwähnug
(2+11) cooked
h#4*
Thomas R. Dawson
65 British Chess Federation 07/1945
5. ehrende Erwähnug
(2+11) cooked
h#4*
*) 1. ... fxe5 2. Sd6 exd6 3. Sc7 d7 4. Se8 d8=S#
1) 1. g5 fxg5 2. Lh6 gxh6 3. Sg7 h7 4. Te6 h8=S#
NL:
1. Sc7 fxe5 2. Sd6 exd6 3. f4 d7 4. Se8 d8=S#
1) 1. g5 fxg5 2. Lh6 gxh6 3. Sg7 h7 4. Te6 h8=S#
NL:
1. Sc7 fxe5 2. Sd6 exd6 3. f4 d7 4. Se8 d8=S#
Yuri Bilokin: Correction: -bPf5, wKa2-g2, bRe5-e1 4nrb1/4pkb1/4npp1/8/5P2/8/6K1/4r3 (2+10) h#4 2.1…
1.Re5 fxe5 2.Sd6 exd6 3.Sc7 d7 4.Se8 d8=S# (MM)
1.g5 fxg5 2.Bh6 gxh6 3.S6g7 h7 4.Re6 h8=S# (MM) (2021-11-26)
comment
1.Re5 fxe5 2.Sd6 exd6 3.Sc7 d7 4.Se8 d8=S# (MM)
1.g5 fxg5 2.Bh6 gxh6 3.S6g7 h7 4.Re6 h8=S# (MM) (2021-11-26)
comment
Genre: h#
FEN: 4nrb1/4pkb1/4npp1/4rp2/5P2/8/K7/8
Input: Ralf Krätschmer, 1998-03-30
Last update: hpr, 1999-11-07 more...
25 - P0534722
Thomas R. Dawson
2828v Stratford Express 15/12/1934
(5+7)
h#2
b) alles eine Reihe tiefer
Thomas R. Dawson
2828v Stratford Express 15/12/1934
(5+7)
h#2
b) alles eine Reihe tiefer
a) 1. g2 Ta6 2. g1=T Tf2#
b) 1. Ke5 0-0-0 2. Ke4/Lf6 Te1#
NL in a)
1. Kf8 Sf4 /Sg7 2. Sf7 Se6#
1. Kf8 Tf2+ 2. Lf7 Ta8#
b) 1. Ke5 0-0-0 2. Ke4/Lf6 Te1#
NL in a)
1. Kf8 Sf4 /Sg7 2. Sf7 Se6#
1. Kf8 Tf2+ 2. Lf7 Ta8#
Dual in b)
Yuri Bilokin: possibly bRe8(-bBe8), +bNc7, +bPg6 4r1rn/2n1pk2/6p1/7N/6P1/6pp/R3K2R/8 (5+9) h#2 b) alles eine Reihe tiefer
a) 1.g2 Ra6 2.g1=R Rf2#
b) 1.Ke5 0-0-0 2.Ke4 Rhe1# (MM) (2022-12-16)
comment
Yuri Bilokin: possibly bRe8(-bBe8), +bNc7, +bPg6 4r1rn/2n1pk2/6p1/7N/6P1/6pp/R3K2R/8 (5+9) h#2 b) alles eine Reihe tiefer
a) 1.g2 Ra6 2.g1=R Rf2#
b) 1.Ke5 0-0-0 2.Ke4 Rhe1# (MM) (2022-12-16)
comment
*) 1. ... Kd7 2. Sd4 b4#
1) 1. Sb4 Td2 2. Lc6 d4#
NL:
1. Se3 fxe3 2. Lc6 b4#
1) 1. Sb4 Td2 2. Lc6 d4#
NL:
1. Se3 fxe3 2. Lc6 b4#
Yuri Bilokin: correction a1=a2, then wRh6-g6, wPf3-f2, wPh5-g5, bPf7-d7, +bPf3 8/1p1pK1p1/1pkn1nR1/4b1P1/1P1P1p2/1R3p2/5P2/8 (7+10) h#2*
1...Kd8 2.Sd5 b5#
1.Sb5 Rd3 2.Bc7 d5# (2022-12-16)
comment
1...Kd8 2.Sd5 b5#
1.Sb5 Rd3 2.Bc7 d5# (2022-12-16)
comment
Keywords: Fesselungsspiel (212 202 000)
Genre: h#
FEN: 8/8/1p2Kpp1/1pkn1n1R/4b2P/1P1P1p2/1R3P2/8
Input: Markus Manhart, 1998-04-17
Last update: hpr, 1999-11-09 more...
Genre: h#
FEN: 8/8/1p2Kpp1/1pkn1n1R/4b2P/1P1P1p2/1R3P2/8
Input: Markus Manhart, 1998-04-17
Last update: hpr, 1999-11-09 more...
a) 1. La7 Ta1 2. Db8 Sb6#
b) 1. Da1 Tg8 2. Da7 Sc7#
NL in b):
1. Dd4 Ta1+ 2. Da7 Sb6#
1. Dd4 Tg8 2. Da7 Sc7#
b) 1. Da1 Tg8 2. Da7 Sc7#
NL in b):
1. Dd4 Ta1+ 2. Da7 Sb6#
1. Dd4 Tg8 2. Da7 Sc7#
Yuri Bilokin: redaction Thomas R. Dawson & Yuri Bilokin wRg1-h1, bQg8-h8, +bPc5 kb5q/1n1p4/3p4/2pN4/8/1K6/8/7R (3+7) h#2 2.1…
1.Ba7 Ra1 2.Qb8 Sb6# (MM)
1.Qa1 Rh8 2.Qa7 Sc7# (MM)
Blocking piece replacement (bB-bQ)
Corner-to-corner (bQ). Corner-to-corner (wR) × 2
Echo mates (mirrored diagonally, 0, 0). JT Onkoud 50 theme
Place exchange in the final positions (black, qb). Transferred pin (bB)
Model mate × 2. Echo. Pin-mate × 2. Smothered mate × 2 (2022-12-16)
comment
1.Ba7 Ra1 2.Qb8 Sb6# (MM)
1.Qa1 Rh8 2.Qa7 Sc7# (MM)
Blocking piece replacement (bB-bQ)
Corner-to-corner (bQ). Corner-to-corner (wR) × 2
Echo mates (mirrored diagonally, 0, 0). JT Onkoud 50 theme
Place exchange in the final positions (black, qb). Transferred pin (bB)
Model mate × 2. Echo. Pin-mate × 2. Smothered mate × 2 (2022-12-16)
comment
Keywords: Fesselungsspiel (222 202 000)
Genre: h#
FEN: kb4q1/1n1p4/3p4/3N4/8/1K6/8/6R1
Input: Markus Manhart, 1998-05-09
Last update: hpr, 1999-11-11 more...
Genre: h#
FEN: kb4q1/1n1p4/3p4/3N4/8/1K6/8/6R1
Input: Markus Manhart, 1998-05-09
Last update: hpr, 1999-11-11 more...
28 - P0547948
Thomas R. Dawson
831 Queenstown Daily Reporter 05/09/1939
(4+3) cooked
h#2
b) wKe7 nach g1
Thomas R. Dawson
831 Queenstown Daily Reporter 05/09/1939
(4+3) cooked
h#2
b) wKe7 nach g1
a) 1. Td4 Le6 2. Ke5 Sd3#
b) 1. Td3 Ld1 2. Ke3 Sd5#
NL:
a) 1. Td1 Dg3+ 2. Kf5 Le6#
1. Td1 Df3+ 2. Ke5 Sc6#
b) 1. Td3 Ld1 2. Ke3 Sd5#
NL:
a) 1. Td1 Dg3+ 2. Kf5 Le6#
1. Td1 Df3+ 2. Ke5 Sc6#
Yuri Bilokin: posibly b1=a1, then wKd7-f1, +wPe5, +bPc6, +bPc7, +bPe6
8/2p5/2p1p3/4P3/N2nk3/BQ6/2r5/5K2 (5+6) h#2 2.1…
1.Rc3 Bc1 2.Kd3 Sc5# (MM)
1.Rc4 Bd6 2.Kd5 Sc3# (MM)
Analogy (complete)
Anticipatory self-pin × 2
Brochettes theme
Transferred pin (bR)
Model mate × 2
Pin-mate × 2 (2022-12-16)
comment
8/2p5/2p1p3/4P3/N2nk3/BQ6/2r5/5K2 (5+6) h#2 2.1…
1.Rc3 Bc1 2.Kd3 Sc5# (MM)
1.Rc4 Bd6 2.Kd5 Sc3# (MM)
Analogy (complete)
Anticipatory self-pin × 2
Brochettes theme
Transferred pin (bR)
Model mate × 2
Pin-mate × 2 (2022-12-16)
comment
Keywords: Fesselungsspiel (222 202 000)
Genre: h#
FEN: 8/4K3/8/8/1N2nk2/1BQ5/3r4/8
Input: Markus Manhart, 1998-05-11
Last update: hpr, 1999-02-19 more...
Genre: h#
FEN: 8/4K3/8/8/1N2nk2/1BQ5/3r4/8
Input: Markus Manhart, 1998-05-11
Last update: hpr, 1999-02-19 more...
29 - P0552366
Thomas R. Dawson
Charles Masson Fox
3277v The Fairy Chess Review 08/1938
(5+5) cooked
h#2
b) wSf7 nach e5
Thomas R. Dawson
Charles Masson Fox
3277v The Fairy Chess Review 08/1938
(5+5) cooked
h#2
b) wSf7 nach e5
a) 1. Kxf7 Tf2+ 2. Ke8 Sg7#
b) 1. Kxe5 Kd2+ 2. Kd6 Kxe1#
Cook: NL
a) 1. Kg6 Tg2+ 2. Kh7 Tg7#
b) 1. Kxe5 Kd2+ 2. Kd6 Kxe1#
Cook: NL
a) 1. Kg6 Tg2+ 2. Kh7 Tg7#
Verstellung KT
Adrian Storisteanu: Possible fix: +bPh7 (5+6). (2015-08-23)
Yuri Bilokin: possibly b1=a1 bNh8 7n/3bN3/1p1Nk3/8/8/3K4/3R4/2Rn4 (5+5) h#2 b) wNe7-d5
a) 1.Kxe7 Re2+ 2.Kd8 Sb7# (MM)
b) 1.Kxd5 Kc2+ 2.Kc6 Kxd1# (MM) (2022-12-16)
comment
Adrian Storisteanu: Possible fix: +bPh7 (5+6). (2015-08-23)
Yuri Bilokin: possibly b1=a1 bNh8 7n/3bN3/1p1Nk3/8/8/3K4/3R4/2Rn4 (5+5) h#2 b) wNe7-d5
a) 1.Kxe7 Re2+ 2.Kd8 Sb7# (MM)
b) 1.Kxd5 Kc2+ 2.Kc6 Kxd1# (MM) (2022-12-16)
comment
Keywords: Sacrifice of white pieces, Line closure
Genre: h#
FEN: n7/4bN2/2p1Nk2/8/8/4K3/4R3/3Rn3
Input: Hans-Jürgen Schäfer, 1998-01-20
Last update: Alfred Pfeiffer, 2015-08-24 more...
Genre: h#
FEN: n7/4bN2/2p1Nk2/8/8/4K3/4R3/3Rn3
Input: Hans-Jürgen Schäfer, 1998-01-20
Last update: Alfred Pfeiffer, 2015-08-24 more...
1. Kxa8 Lb1 2. Kb8 La2 3. Kc8 Lb1 4. Kd7 La2 5. Ke6 Lb1 6. Kf5 La2 7. Kg4 Lb1 8. Kh3 La2 9. Kg2 Lb1 10. Kf1 La2 11. Ke1 Lb1 12. Kd1 La2 13. Kc1 Lb1 14. Kb2 La2 15. Kxa1 Lb1 16. Kb2 La2 17. Kxc2 Lb1+ 18. Kxb3 g4 19. Kc4 La2#
Keywords: Sacrifice of white pieces
Genre: h#
FEN: Rk6/1p6/1P6/Bp1pp1p1/1P2p3/1Pp1KpP1/B1P2n1n/N5b1
Input: Hans-Jürgen Schäfer, 1998-07-01
Last update: hpr, 1999-03-30 more...
Genre: h#
FEN: Rk6/1p6/1P6/Bp1pp1p1/1P2p3/1Pp1KpP1/B1P2n1n/N5b1
Input: Hans-Jürgen Schäfer, 1998-07-01
Last update: hpr, 1999-03-30 more...
31 - P0556423
Thomas R. Dawson
Israel Problem Association 1948
3. Lob
Seidemann Memorial
(4+6) cooked
h#2
b) alles eine Reihe tiefer
Thomas R. Dawson
Israel Problem Association 1948
3. Lob
Seidemann Memorial
(4+6) cooked
h#2
b) alles eine Reihe tiefer
a) 1. c5 Lf4 2. Te4 Td8#
b) 1. Ld1 Txe2 2. Ld4 La6#
NL:
b) 1. Td2 Te1 2. Ld4 La6#
b) 1. Ld1 Txe2 2. Ld4 La6#
NL:
b) 1. Td2 Te1 2. Ld4 La6#
Yuri Bilokin: possibly bNb3(-wPb3), bPc3(-bBc3), +bPc4, +bPf4 1B2R3/2p5/8/8/2pk1p2/1np1r3/2r3bK/8 (3+9) h#2 b) alles eine Reihe tiefer
a) 1.c5 Bxf4 2.Re4 Rd8# (MM)
b) 1.Sd1 Rxe2 2.Bd4 Ba6# (2022-12-16)
milan: a) -wPb3+bPa7wBf1
b)all pieces to the left
1.Ra6 B×b7 2.Bc5 Ba6#
1.Kc5 Rd5+ 2.Ka4 Ra5#
1.b5 Be4 2.Rd4 Rc8#
i'ts look like battle of the bishops
M.Frelih (2022-12-18)
Yuri Bilokin: Very bad. In a problem that has nothing to do with the author's intention, there is still an extra white bishop in the solution 1.Kc5 Rd5+ 2.Ka4 Ra5# (2022-12-19)
milan: +wBa6bPa5-wPb3 1.c5 Bf4 2.Re4 Rd8# 1.Rb6 B×c7 2.Bd5 B×b6#
your solution b) not C+... bRb5? M.Frelih (2022-12-19)
Yuri Bilokin: Sorry for your misunderstanding. Everything is fine in my edition. Twin FEN Code a) You see Twin FEN Code b) Showing 8/1B2R3/2p5/8/8/2pk1p2/1np1r3/2r3bK, solution
1.Sd1 Rxe2 2.Bd4 Ba6# (2022-12-20)
comment
a) 1.c5 Bxf4 2.Re4 Rd8# (MM)
b) 1.Sd1 Rxe2 2.Bd4 Ba6# (2022-12-16)
milan: a) -wPb3+bPa7wBf1
b)all pieces to the left
1.Ra6 B×b7 2.Bc5 Ba6#
1.Kc5 Rd5+ 2.Ka4 Ra5#
1.b5 Be4 2.Rd4 Rc8#
i'ts look like battle of the bishops
M.Frelih (2022-12-18)
Yuri Bilokin: Very bad. In a problem that has nothing to do with the author's intention, there is still an extra white bishop in the solution 1.Kc5 Rd5+ 2.Ka4 Ra5# (2022-12-19)
milan: +wBa6bPa5-wPb3 1.c5 Bf4 2.Re4 Rd8# 1.Rb6 B×c7 2.Bd5 B×b6#
your solution b) not C+... bRb5? M.Frelih (2022-12-19)
Yuri Bilokin: Sorry for your misunderstanding. Everything is fine in my edition. Twin FEN Code a) You see Twin FEN Code b) Showing 8/1B2R3/2p5/8/8/2pk1p2/1np1r3/2r3bK, solution
1.Sd1 Rxe2 2.Bd4 Ba6# (2022-12-20)
comment
Genre: h#
FEN: 1B2R3/2p5/8/8/3k4/1Pb1r3/1r4bK/8
Input: Franz Pachl, 1998-08-31
Last update: hpr, 1999-04-04 more...
32 - P0559805
Thomas R. Dawson
35 Schachkongress Teplitz-Schönau im Oktober 1922 , p. 505, 1923
(3+9) cooked
h#3
Thomas R. Dawson
35 Schachkongress Teplitz-Schönau im Oktober 1922 , p. 505, 1923
(3+9) cooked
h#3
1. Tc6 Tf4 2. Tce6 Tcc4 3. T2e5 Tfd4#
NL:
1. Kd4 Txc4+ 2. Kd3 Tb4 3. Lg3 Ta3#
uvm
NL:
1. Kd4 Txc4+ 2. Kd3 Tb4 3. Lg3 Ta3#
uvm
Yuri Bilokin: correction wKd1-h1, bBe1-a7, bBf1-a2, bPg2-h2, +bNb8, +bNe7, +bPa6, + bPd7, +bPf6 1nR5/b2pn3/pp3p2/3k4/R1r3pq/8/b3r2p/7K (3+14) (2023-05-02)
comment
comment
Genre: h#
FEN: 2R5/8/1p6/3k4/R1r3pq/8/4r1p1/3Kbb2
Input: hpr, 1999-02-17
Last update: Rainer Staudte, 2022-04-18 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...
*) 1. ... Tc3 2. Dxc2 Dxh1#
1) 1. Dxe2 Kb3 2. Kd1 Tf1#
NL:
*) 1. ... Dxf5 2. Df1 Txf1#
1. Dg1 Dxf5 2. Df1 Txf1#
1) 1. Dxe2 Kb3 2. Kd1 Tf1#
NL:
*) 1. ... Dxf5 2. Df1 Txf1#
1. Dg1 Dxf5 2. Df1 Txf1#
Felber, Volker: Original mit wT auf h3 (auch NL).
Heft 84, Dezember 1934, Seite 193, II. (2011-02-10)
Yuri Bilokin: possibly wRf3-h3, +wPb3 8/3p1p2/3pbp2/3r1p1Q/2K2P2/1P5R/1bPPP3/1Nkq3n (9+11) h#2*
1...Rc3 2.Qxc2 Qxh1# (MM)
1.Qxc2+ Kb4 2.Kd1 Rxh1# (MM)
Bi-valve (wR-wQ-bQ)
Check prevention (W-W)
Hideaway (bQ, self-pinning)
Play on the same square (W2, 2)
Model mate × 2. Pin-mate. Mates on the same square × 2 (2022-12-17)
comment
Heft 84, Dezember 1934, Seite 193, II. (2011-02-10)
Yuri Bilokin: possibly wRf3-h3, +wPb3 8/3p1p2/3pbp2/3r1p1Q/2K2P2/1P5R/1bPPP3/1Nkq3n (9+11) h#2*
1...Rc3 2.Qxc2 Qxh1# (MM)
1.Qxc2+ Kb4 2.Kd1 Rxh1# (MM)
Bi-valve (wR-wQ-bQ)
Check prevention (W-W)
Hideaway (bQ, self-pinning)
Play on the same square (W2, 2)
Model mate × 2. Pin-mate. Mates on the same square × 2 (2022-12-17)
comment
Genre: h#
FEN: 8/3p1p2/3pbp2/3r1p1Q/2K2P2/5R2/1bPPP3/1Nkq3n
Input: Felber, Volker, 1999-12-23
Last update: hpr, 2000-01-21 more...
1. ... Ld2 2. Sa4 Lxf4 3. Ka5 Ld6 4. Lb5 Lb4#
NL:
1. ... Le1 2. Te6 Tg2 3. Sa4 Td2 4. Ka5 Txd5#
NL:
1. ... Le1 2. Te6 Tg2 3. Sa4 Td2 4. Ka5 Txd5#
Korrektur Hilmar Ebert: -sBd5, sDc8 nach f5
Jakob Leck: vgl. P1071030 (2019-12-27)
Yuri Bilokin: possibly bQc8-e6, -bPd5, -bPf3 8/1p1b4/pp2qb2/1kn5/1B3pR1/7K/4r3/1n6 (3+11) (2022-12-15)
comment
Jakob Leck: vgl. P1071030 (2019-12-27)
Yuri Bilokin: possibly bQc8-e6, -bPd5, -bPf3 8/1p1b4/pp2qb2/1kn5/1B3pR1/7K/4r3/1n6 (3+11) (2022-12-15)
comment
Genre: h#
FEN: 2q5/1p1b4/pp3b2/1knp4/1B3pR1/5p1K/4r3/1n6
Reprints: 2244 Bolton Football Field 01/06/1912
Input: hpr, 2001-12-21
Last update: hpr, 2001-12-21 more...
*) 1. ... Sf5#
1) 1. Lf6 Kxh2 2. Le7 Sc4#
1) 1. Ke7 Dg8 2. Kf6 Sd5#
1) 1. Lf6 Kxh2 2. Le7 Sc4#
1) 1. Ke7 Dg8 2. Kf6 Sd5#
Korrektur John NIemann: wKh1 nach b1, sBh2 nach b2, sBh3 nach b3
Yuri Bilokin: Der Autor könnte eine weitere Lösung sparen, indem er den Inhalt des Problems erweitert: Allowing tempo (B-W)
Apparent mate
BK moves only
Echo mates (mirrored vertically, 2, 0)
Tempo move (wK, waiting)
Zalokotsky theme (bB/bK, reduced, 2)
Model mate × 3
Echo (2022-12-17)
comment
Yuri Bilokin: Der Autor könnte eine weitere Lösung sparen, indem er den Inhalt des Problems erweitert: Allowing tempo (B-W)
Apparent mate
BK moves only
Echo mates (mirrored vertically, 2, 0)
Tempo move (wK, waiting)
Zalokotsky theme (bB/bK, reduced, 2)
Model mate × 3
Echo (2022-12-17)
comment
Genre: h#
Computer test: (Popeye WINDOWS-32Bit-Version 3.70 (2048 KB))
FEN: 2Q5/8/3k4/4b3/4P3/4N2p/7p/7K
Input: hpr, 2001-12-25
Last update: hpr, 2001-12-25 more...
1. Lg6! droht 2. Dd8#
1. ... 0-0 2. Lh7#
1. ... Txg5 2. Txh8#
1. ... 0-0 2. Lh7#
1. ... Txg5 2. Txh8#
No. 278 HN
A.Buchanan: Without sTa5, Black castling rights would be gone. But I don’t think that observation is sufficient to make the problem retro, even if sTa5 had no role in the forward play. Have changed the genre from
Retro to #2. (2022-04-26)
more ...
comment
A.Buchanan: Without sTa5, Black castling rights would be gone. But I don’t think that observation is sufficient to make the problem retro, even if sTa5 had no role in the forward play. Have changed the genre from
Retro to #2. (2022-04-26)
more ...
comment
Keywords: Flight-giving key, Castling, Switchback (wL)
Genre: 2#
Computer test: popeye 4.87 C+, popeye 4.87
FEN: 4k2r/1p2pNpB/7R/1K1Q2Pr/8/8/8/8
Reprints: Am Rande des Schachbretts , p. 14, 1947
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2022-04-26 more...
Genre: 2#
Computer test: popeye 4.87 C+, popeye 4.87
FEN: 4k2r/1p2pNpB/7R/1K1Q2Pr/8/8/8/8
Reprints: Am Rande des Schachbretts , p. 14, 1947
Input: Henri Nouguier, 2004-01-11
Last update: A.Buchanan, 2022-04-26 more...
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)
comment
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...
39 - P1012897
Thomas R. Dawson
Die Schwalbe 1935
(16+15)
Stellung nach dem 10. Zug von Schwarz. Mit Weiss am zug, welches war der letzte Zug?
Thomas R. Dawson
Die Schwalbe 1935
(16+15)
Stellung nach dem 10. Zug von Schwarz. Mit Weiss am zug, welches war der letzte Zug?
Beispielauflösung:
R: 1. ... f7-f5 2. Ke4-e5 Sg6-h8 3. Lf1-a6 Sh4-g6 4. Kd3-e4 Sf5-h4 5. Ke2-d3 Sh6-f5 6. h4xTg5 Tg6-g5 7. e5-e6 Sg8-h6 8. e4-e5 Th6-g6 9. Ke1-e2 Th8-h6 10. e2-e4 h7-h5 11. h2-h4
zuletzt kann nur R: 1. ... f7-f5 geschehen sein, nicht R: 1. ... h6-h5,h7-h5??, denn dann hätte der sTh8, der auf g3,g4 oder g5 vom wBh2 geschlagen wurde, nicht auf das Schlagfeld gelangen können.
R: 1. ... f7-f5 2. Ke4-e5 Sg6-h8 3. Lf1-a6 Sh4-g6 4. Kd3-e4 Sf5-h4 5. Ke2-d3 Sh6-f5 6. h4xTg5 Tg6-g5 7. e5-e6 Sg8-h6 8. e4-e5 Th6-g6 9. Ke1-e2 Th8-h6 10. e2-e4 h7-h5 11. h2-h4
zuletzt kann nur R: 1. ... f7-f5 geschehen sein, nicht R: 1. ... h6-h5,h7-h5??, denn dann hätte der sTh8, der auf g3,g4 oder g5 vom wBh2 geschlagen wurde, nicht auf das Schlagfeld gelangen können.
No. 791 HN
Kees: 1. h4 h5 2. e4 Rh6 3. e5 Rg6 4. e6 Rg5 5. hxg5 Nh6 6. Ke2 Ng4 7. Kf3 Ne5+
8. Ke4 Ng6 9. Ba6 Nh8 10. Ke5 f5 * (2022-02-12)
comment
Kees: 1. h4 h5 2. e4 Rh6 3. e5 Rg6 4. e6 Rg5 5. hxg5 Nh6 6. Ke2 Ng4 7. Kf3 Ne5+
8. Ke4 Ng6 9. Ba6 Nh8 10. Ke5 f5 * (2022-02-12)
comment
Keywords: Last Move?, Non-Unique Proof Game, Type B
Genre: Retro
FEN: rnbqkb1n/ppppp1p1/B3P3/4KpPp/8/8/PPPP1PP1/RNBQ2NR
Input: Henri Nouguier, 2004-01-11
Last update: Mario Richter, 2022-02-13 more...
Genre: Retro
FEN: rnbqkb1n/ppppp1p1/B3P3/4KpPp/8/8/PPPP1PP1/RNBQ2NR
Input: Henri Nouguier, 2004-01-11
Last update: Mario Richter, 2022-02-13 more...
BS Problemid: 6433
Authors: Dawson, Thomas Rayner
Sources:
Years: 1947
Hans-Jürgen Manthey: in Schach-Express seite 78 5/1947 steht:
Manchester Guardian 1947 (2020-05-30)
Henrik Juel: C+ by Popeye 4.61
1.Sg1 thr. 2.Sg3#
1... Dd6/Dc3/Dc7 2.Lxe6/Sxd4/Lh7+
Holzhausen tripled (2020-05-30)
Hans-Jürgen Manthey: Nachdruck SExpress 1.Jun 1947 S.78 Nr.58 (2022-10-20)
comment
Manchester Guardian 1947 (2020-05-30)
Henrik Juel: C+ by Popeye 4.61
1.Sg1 thr. 2.Sg3#
1... Dd6/Dc3/Dc7 2.Lxe6/Sxd4/Lh7+
Holzhausen tripled (2020-05-30)
Hans-Jürgen Manthey: Nachdruck SExpress 1.Jun 1947 S.78 Nr.58 (2022-10-20)
comment
Keywords: Brian Stephenson Collection (6433)
Genre: 3#
FEN: 6B1/r7/r3p2n/2qp1kBK/3p4/1N6/5N2/b3R3
Input: Brian Stephenson, 2004-08-12
Genre: 3#
FEN: 6B1/r7/r3p2n/2qp1kBK/3p4/1N6/5N2/b3R3
Input: Brian Stephenson, 2004-08-12
1. Sc7! droht 2. Ld5+
1. ... Sge3 2. Sb5
1. ... Sf4 2. Sa6
1. ... Sxh4 2. Se6
1. ... Sge3 2. Sb5
1. ... Sf4 2. Sa6
1. ... Sxh4 2. Se6
Keywords: Brian Stephenson Collection (16743)
Genre: 3#
Computer test: Juel: Popeye 4.61
FEN: K3N3/1N1p1B2/2k5/B7/1pP4P/5b2/2n3n1/4b2q
Reprints: 28 Valves and Bi-Valves , p. 40, 1930
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2024-03-17 more...
Genre: 3#
Computer test: Juel: Popeye 4.61
FEN: K3N3/1N1p1B2/2k5/B7/1pP4P/5b2/2n3n1/4b2q
Reprints: 28 Valves and Bi-Valves , p. 40, 1930
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2024-03-17 more...
1. Sf6! droht 2. Sxd5 droht 3. d7#/Se7#/Sb6#
2. ... Lh4 3. d7#/Sb6#
2. ... Lh3 3. Se7#/Sb6#
2. ... d3 3. d7#/Se7#
1. ... Td3,Te3 2. Sd7 droht 3. Sb6#
1. ... Tg3 2. Se8 droht 3. d7#
1. ... Th3 2. Sg8 droht 3. Se7#
2. ... Lh4 3. d7#/Sb6#
2. ... Lh3 3. Se7#/Sb6#
2. ... d3 3. d7#/Se7#
1. ... Td3,Te3 2. Sd7 droht 3. Sb6#
1. ... Tg3 2. Se8 droht 3. d7#
1. ... Th3 2. Sg8 droht 3. Se7#
A.Buchanan: I guess the theme is representation of all combinations of the three mates. But I don't see the purpose of bS & bD. Why are they needed, please? (2024-03-17)
more ...
comment
more ...
comment
Keywords: Brian Stephenson Collection (16865)
Genre: 3#
Computer test: Juel: Popeye 4.61
FEN: 2k2K2/2P4N/2PP2Pp/3pp3/3p1p2/2p2r2/5bb1/2n3q1
Reprints: 19 Valves and Bi-Valves , p. 31, 1930
Input: Brian Stephenson, 2004-08-12
Last update: A.Buchanan, 2024-03-17 more...
Genre: 3#
Computer test: Juel: Popeye 4.61
FEN: 2k2K2/2P4N/2PP2Pp/3pp3/3p1p2/2p2r2/5bb1/2n3q1
Reprints: 19 Valves and Bi-Valves , p. 31, 1930
Input: Brian Stephenson, 2004-08-12
Last update: A.Buchanan, 2024-03-17 more...
1. Td1? Dxh4! 1. Kd1? c3!
1. 0-0-0! Dxh4 2. Le5#
1. ... c3 2. dxc3#
1. ... cxd3 2. Ta4#
1. 0-0-0! Dxh4 2. Le5#
1. ... c3 2. dxc3#
1. ... cxd3 2. Ta4#
Keywords: Brian Stephenson Collection (21742), Castling key, Knight sacrifice, Pawn mate, Line opening, Zugzwang-Goal
Genre: 2#
Computer test: SCHRECKE: popeye 4.87
FEN: 8/p5p1/P2QPqP1/3P1P2/2pk3N/R2P2B1/3P2Nn/R3K1b1
Reprints: 73 Versunkene Schätze , p. 37, 1998
A The Problemist 25-3, p. 125, 05/2015
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2022-10-11 more...
Genre: 2#
Computer test: SCHRECKE: popeye 4.87
FEN: 8/p5p1/P2QPqP1/3P1P2/2pk3N/R2P2B1/3P2Nn/R3K1b1
Reprints: 73 Versunkene Schätze , p. 37, 1998
A The Problemist 25-3, p. 125, 05/2015
Input: Brian Stephenson, 2004-08-12
Last update: Dieter Berlin, 2022-10-11 more...
1. Te1! droht 2. Dc3#
1. ... e3 2. Dc6#
1. ... exf3 2. Te6#
1. ... gxf3 2. Tg6#
1. ... e3 2. Dc6#
1. ... exf3 2. Te6#
1. ... gxf3 2. Tg6#
Keywords: Brian Stephenson Collection (23707), Symmetrical position, Asymmetrical solution
Genre: 2#
FEN: 5K2/5B2/5k2/5p2/4pPp1/4RQR1/5P2/5r2
Reprints: 50 Bote von der Ybbs 24/10/1930
Input: Brian Stephenson, 2004-08-12
Last update: Mario Richter, 2021-05-16 more...
Genre: 2#
FEN: 5K2/5B2/5k2/5p2/4pPp1/4RQR1/5P2/5r2
Reprints: 50 Bote von der Ybbs 24/10/1930
Input: Brian Stephenson, 2004-08-12
Last update: Mario Richter, 2021-05-16 more...
BS Problemid: 40893
Authors: Dawson, Thomas Rayner
Sources: Systematic Terminology 1984
Years: 1937
SCHRECKE: C+, popeye 4.87
1. Sxd7! droht 2. Sf6#
1. ... Taxd5 2. S3c5#
1. ... Thxd5 2. De6#
1. ... Dxd5 2. Tc4#
1. ... Tf5 2. Df3#
1. ... Lxd3 2. exd3#
1. ... Sxg1,Sf2 2. Dxf4# (2023-08-21)
comment
1. Sxd7! droht 2. Sf6#
1. ... Taxd5 2. S3c5#
1. ... Thxd5 2. De6#
1. ... Dxd5 2. Tc4#
1. ... Tf5 2. Df3#
1. ... Lxd3 2. exd3#
1. ... Sxg1,Sf2 2. Dxf4# (2023-08-21)
comment
Keywords: Brian Stephenson Collection (40893)
Genre: 2#
FEN: 3K4/1Bpp3p/p6p/r2PN2r/4kpQp/2RN3n/q1b1P3/6B1
Input: Brian Stephenson, 2004-08-12
Last update: Rainer Staudte, 2023-08-19 more...
Genre: 2#
FEN: 3K4/1Bpp3p/p6p/r2PN2r/4kpQp/2RN3n/q1b1P3/6B1
Input: Brian Stephenson, 2004-08-12
Last update: Rainer Staudte, 2023-08-19 more...
1. ... De5 2. Kxd3 Dd5+ 3. Kxe3 De4#
NL:
1. ... Dh7 2. Td1 Dh1 3. Kb1 Dxd1#
1. ... Dd5 2. Sxc6 Dxb5 3. Kc3 Dc4#
NL:
1. ... Dh7 2. Td1 Dh1 3. Kb1 Dxd1#
1. ... Dd5 2. Sxc6 Dxb5 3. Kc3 Dc4#
Yuri Bilokin: correction wKa3-g1, wPc6-d5, bNd4-h2, bPb5-c4, -bBa1, -bPa4, -bNa5, -bPc7, - bPf2 4r3/4q3/4p3/3P4/2p1QP2/3PPP2/2kr3n/6K1 (7+7) h#3 0.1…
1...Qe5 2.Kxd3 Qf5+ 3.Kxe3 Qe4# (MM)
Areal cycle (wQ, captureless, 3) (2022-12-20)
comment
1...Qe5 2.Kxd3 Qf5+ 3.Kxe3 Qe4# (MM)
Areal cycle (wQ, captureless, 3) (2022-12-20)
comment
Genre: h#
Computer test: (Popeye WINDOWS-32Bit-Version 3.75 (2048 KB))
FEN: 4r3/2p1q3/2P1p3/np6/pn2QP2/K2PPP2/2kr1p2/b7
Input: hpr, 2004-09-28
Last update: hpr, 2004-09-28 more...
1. ... Sf5 2. Ke4 Sxd4 3. Tf4 Sf5 4. d5 Sg3#
NL:
1. ... Sf1 2. d3 Ka3 3. Kc4 Ka4 4. c5 Sxe3#
1. ... Sh5 2. Ke4 Kb3 3. d3 Kc4 4. Tf4 Sg3#
NL:
1. ... Sf1 2. d3 Ka3 3. Kc4 Ka4 4. c5 Sxe3#
1. ... Sh5 2. Ke4 Kb3 3. d3 Kc4 4. Tf4 Sg3#
Yuri Bilokin: correction wKa2-g1, wBb2-a1, bPc7-f5, bRd6(-bPd6), bRf6-f7 8/5r2/3r4/3k1p2/3p4/4p1N1/4P3/B6K (4+6)
1...Sxf5 2.Ke4 Sxd4 3.Rf4 Sf5 4.Rd5 Sg3# (MM)
Annihilation × 2
Crusader theme
Klasinc theme (wS-bR)
Long-trip (wS, 4)
Switchback (wS, with captures, 2)
Model mate × 1 (2022-11-21)
comment
1...Sxf5 2.Ke4 Sxd4 3.Rf4 Sf5 4.Rd5 Sg3# (MM)
Annihilation × 2
Crusader theme
Klasinc theme (wS-bR)
Long-trip (wS, 4)
Switchback (wS, with captures, 2)
Model mate × 1 (2022-11-21)
comment
Genre: h#
Computer test: (Popeye WINDOWS-32Bit-Version 3.75 (2048 KB))
FEN: 8/2p5/3p1r2/3k4/3p4/4p1N1/KB2P3/8
Input: hpr, 2004-09-28
Last update: hpr, 2004-09-28 more...
1. dxe6ep dxe6 2. Lg4#
Henrik Juel: Last move was not e6-e5 (exd5 is longer) nor b3xc2 or d3xc2 (Txc2), nor f5xe4/f3xg2 (Txe4/Txg2), nor Kg3-h3/Kh2-h3, nor c3-c2 or g3-g2
So last move was e7-e5, legitimizing the ep capture
C+ Popeye 4.61 (2021-02-16)
comment
So last move was e7-e5, legitimizing the ep capture
C+ Popeye 4.61 (2021-02-16)
comment
Keywords: En passant as key, Maximummer
Genre: Retro, Fairies
FEN: 8/3p4/3B4/3Pp1KB/4p3/7k/2p1r1p1/8
Input: Frank Müller, 2010-05-06
Last update: A.Buchanan, 2023-06-03 more...
Genre: Retro, Fairies
FEN: 8/3p4/3B4/3Pp1KB/4p3/7k/2p1r1p1/8
Input: Frank Müller, 2010-05-06
Last update: A.Buchanan, 2023-06-03 more...
1. hxg6ep Lg5 2. Dxg5#
R: 1. g7-g5 Te3-e7+ 2. Kg5-h6+ Lf5-h7+
R: 1. g7-g5 Te3-e7+ 2. Kg5-h6+ Lf5-h7+
A.Buchanan: I posted this in Facebook today - Lion Xray asked if wBh2 can be removed - I think it can. (2021-02-16)
Henrik Juel: Without Ph2 last move could also be f6xg5, it seems (2021-02-16)
Mario Richter: I too think the wPh2 can be removed, since then f6xg5 as Black's last move is still impossible (think of wPf2+h2) (2021-02-16)
A.Buchanan: With only one Wh unit unaccounted for, bPg & bPh never left their files. So wPh5 must have captured in. Either fxgxh or hxgxh. In the former case, the original wPh was "waylaid" captured unpromoted on its own file. In the latter case, wPf must have promoted on f8 without capturing. I.e. bPf captured to clear the way. However Black’s last move can’t have been f6xg5, because wPf would not have had a free path to march to f8. The count is correct in either case. (2021-02-17)
A.Buchanan: Another possibility is that wPf was waylaid, to allow bPf to promote on f1. Again, Black’s last move was not f6xg5. (2021-02-17)
A.Buchanan: "Superseded by" was replaced by "better version". I am not averse to improvements in terminology, but here I have reverted, for the following reasons: (1) A keyword should describe the compositions which are tagged by it. "Better version" does not describe a problem it tags, but indicates that *another* problem is the "better version". (2) A superseding problem may not be a version, but could be radically different matrix tackling a record or task. I have left the new German term "bessere Version" alone, and put this comment in the keyword text. (2021-02-18)
comment
Henrik Juel: Without Ph2 last move could also be f6xg5, it seems (2021-02-16)
Mario Richter: I too think the wPh2 can be removed, since then f6xg5 as Black's last move is still impossible (think of wPf2+h2) (2021-02-16)
A.Buchanan: With only one Wh unit unaccounted for, bPg & bPh never left their files. So wPh5 must have captured in. Either fxgxh or hxgxh. In the former case, the original wPh was "waylaid" captured unpromoted on its own file. In the latter case, wPf must have promoted on f8 without capturing. I.e. bPf captured to clear the way. However Black’s last move can’t have been f6xg5, because wPf would not have had a free path to march to f8. The count is correct in either case. (2021-02-17)
A.Buchanan: Another possibility is that wPf was waylaid, to allow bPf to promote on f1. Again, Black’s last move was not f6xg5. (2021-02-17)
A.Buchanan: "Superseded by" was replaced by "better version". I am not averse to improvements in terminology, but here I have reverted, for the following reasons: (1) A keyword should describe the compositions which are tagged by it. "Better version" does not describe a problem it tags, but indicates that *another* problem is the "better version". (2) A superseding problem may not be a version, but could be radically different matrix tackling a record or task. I have left the new German term "bessere Version" alone, and put this comment in the keyword text. (2021-02-18)
comment
Keywords: En passant as key, Superseded by (P1386614)
Genre: 2#, Retro
FEN: 3KN3/qrp1R2B/pp5k/Q5pP/PPP4b/1p4Pp/1PP4P/2B5
Input: Frank Müller, 2010-05-29
Last update: A.Buchanan, 2021-02-18 more...
Genre: 2#, Retro
FEN: 3KN3/qrp1R2B/pp5k/Q5pP/PPP4b/1p4Pp/1PP4P/2B5
Input: Frank Müller, 2010-05-29
Last update: A.Buchanan, 2021-02-18 more...
* 1. ... d5 2. Sc4+ dxc4 3. Sb2 c3 4. Lb1 cxb2#
1. Se4! d5 2. Sc3 d4 3. Lb1 dxc3 4. Sb2 cxb2#
1. Se4! d5 2. Sc3 d4 3. Lb1 dxc3 4. Sb2 cxb2#
Keywords: Opferfeldwechsel
Genre: s#
FEN: 8/8/3p4/B7/p7/k2N4/B1PN4/K7
Input: Frank Müller, 2010-06-05
Last update: Marcin Banaszek, 2021-11-29 more...
Genre: s#
FEN: 8/8/3p4/B7/p7/k2N4/B1PN4/K7
Input: Frank Müller, 2010-06-05
Last update: Marcin Banaszek, 2021-11-29 more...
1. a5 2. a4 3. a3 4. a2 5. a1=T 6. Ta7 Sb6#
set play:Sc7#
set play:Sc7#
Yuri Bilokin: Maybe the black pawn on the field d6 is superfluous, with the last move White took the black piece on the field d5. (2022-12-21)
comment
comment
Keywords: Seriesmover
Genre: Fairies
Computer test: Popeye WINDOWS-32Bit V4.37 (374384 KB)
FEN: k1K5/p7/3p4/3N4/8/8/8/8
Input: Frank Müller, 2010-09-08
Last update: Erich Bartel, 2010-09-08 more...
Genre: Fairies
Computer test: Popeye WINDOWS-32Bit V4.37 (374384 KB)
FEN: k1K5/p7/3p4/3N4/8/8/8/8
Input: Frank Müller, 2010-09-08
Last update: Erich Bartel, 2010-09-08 more...
* 1. ... Kxe8 2. b8=D#
1. b8=S! Kxe8 2. Dg8#
1. b8=S! Kxe8 2. Dg8#
Keywords: Miniature Collection (0005846), Minimal, Miniature, Exchange of Promotions, under-promotion key, Promotion in the mating move (D im Satz)
Genre: 2#
Computer test: Juel: Popeye 4.61
FEN: K3Nk2/PP2p3/4Q3/8/8/8/8/8
Reprints: 3 Arbeiter-Schachzeitung (Chemnitz) 8, p. 233, 08/1928
23 Schachminiaturen Zweizüger , p. 16, 1965
Input: Felber, Volker, 2010-09-11
Last update: Dieter Berlin, 2023-01-12 more...
Genre: 2#
Computer test: Juel: Popeye 4.61
FEN: K3Nk2/PP2p3/4Q3/8/8/8/8/8
Reprints: 3 Arbeiter-Schachzeitung (Chemnitz) 8, p. 233, 08/1928
23 Schachminiaturen Zweizüger , p. 16, 1965
Input: Felber, Volker, 2010-09-11
Last update: Dieter Berlin, 2023-01-12 more...
53 - P1133982
Thomas R. Dawson
25 Deutsches Wochenschach 29/03/1914
(3+1)
#2
b) Verschiebung a1=c3: #3
Thomas R. Dawson
25 Deutsches Wochenschach 29/03/1914
(3+1)
#2
b) Verschiebung a1=c3: #3
a) 1. Tb2 Kf1 2. Td1#
b) 1. Td8 Kg4,K~ 2. Tg8+ Kh5,K~ 3. Th7#
b) 1. Td8 Kg4,K~ 2. Tg8+ Kh5,K~ 3. Th7#
Keywords: Miniature Collection (0012554)
Genre: 2#, 3#
FEN: 8/8/8/1RKR4/8/8/8/4k3
Input: Felber, Volker, 2010-09-11
Last update: A.Buchanan, 2021-10-07 more...
Genre: 2#, 3#
FEN: 8/8/8/1RKR4/8/8/8/4k3
Input: Felber, Volker, 2010-09-11
Last update: A.Buchanan, 2021-10-07 more...
1. Dc1! droht 2. Dh1#
1. ... g6 2. T6e7#
1. ... g5 2. Dc7#
1. ... g6 2. T6e7#
1. ... g5 2. Dc7#
Keywords: Miniature Collection (0308098), Minimal, Miniature, Switchback
Genre: 2#
FEN: 4R3/2Q3pk/4R3/4P3/4K3/8/8/8
Reprints: 18 Veckans Krönika 19/10/1912
Input: Zuncke/Bruder, 2010-09-12
Last update: Mario Richter, 2022-04-07 more...
Genre: 2#
FEN: 4R3/2Q3pk/4R3/4P3/4K3/8/8/8
Reprints: 18 Veckans Krönika 19/10/1912
Input: Zuncke/Bruder, 2010-09-12
Last update: Mario Richter, 2022-04-07 more...
1. Th7! (Zugzwang)
1. ... Kxa4 2. Tb7 Ka3 3. Ta7#
1. ... h5 2. Txh5 Kxa4 3. Ta5#
1. ... Kxa4 2. Tb7 Ka3 3. Ta7#
1. ... h5 2. Txh5 Kxa4 3. Ta5#
56 - P1180459
Thomas R. Dawson
903 American Chess Bulletin 12/1913
Holiday Problem
(9+12)
s#25
Schlagzickzack
Thomas R. Dawson
903 American Chess Bulletin 12/1913
Holiday Problem
(9+12)
s#25
Schlagzickzack
paul: Every black move is unique, so the problem can be partially checked by Jacobi using:
forsyth K1k1s3/5p2/2S5/b7/1ppppppp/7r/PPPPPPP1/8
stipulation u#25
condition Blackcap Zigzag
Sol: 1.Ka7 2.Ka6 3.Kb5 4.Kc5 5.Kd5 6.Ke5 7.Kf5 8.Kg5 9.Kh6 10.Kh7 11.Kg8 12.Kf8 13.Ke7 14.Sd8 Bxd8+ 15.Kf8 16.Kg8 17.Kh7 18.Kh6 19.a4 bxa3 e.p. 20.b4 cxb3 e.p. 21.c4 dxc3 e.p. 22.d4 exd3 e.p. 23.e4 fxe3 e.p. 24.f4 gxf3 e.p. 25.g4 hxg3 e.p.# (2022-08-24)
Henrik Juel: Ka8 needs to go to e7 to avoid KxSd8
Later, White must use double pawn steps to avoid TxPa3 etc. (2022-08-24)
comment
forsyth K1k1s3/5p2/2S5/b7/1ppppppp/7r/PPPPPPP1/8
stipulation u#25
condition Blackcap Zigzag
Sol: 1.Ka7 2.Ka6 3.Kb5 4.Kc5 5.Kd5 6.Ke5 7.Kf5 8.Kg5 9.Kh6 10.Kh7 11.Kg8 12.Kf8 13.Ke7 14.Sd8 Bxd8+ 15.Kf8 16.Kg8 17.Kh7 18.Kh6 19.a4 bxa3 e.p. 20.b4 cxb3 e.p. 21.c4 dxc3 e.p. 22.d4 exd3 e.p. 23.e4 fxe3 e.p. 24.f4 gxf3 e.p. 25.g4 hxg3 e.p.# (2022-08-24)
Henrik Juel: Ka8 needs to go to e7 to avoid KxSd8
Later, White must use double pawn steps to avoid TxPa3 etc. (2022-08-24)
comment
Keywords: Blackcap zigzag
Genre: Fairies
FEN: K1k1n3/5p2/2N5/b7/1ppppppp/7r/PPPPPPP1/8
Input: Frank Müller, 2010-11-16
Last update: Frank Müller, 2010-11-17 more...
Genre: Fairies
FEN: K1k1n3/5p2/2N5/b7/1ppppppp/7r/PPPPPPP1/8
Input: Frank Müller, 2010-11-16
Last update: Frank Müller, 2010-11-17 more...
57 - P1181239
Thomas R. Dawson
The Chess Amateur 12/1923
Asymmetrie-Turnier 1924
1. Preis
(15+10) C+
s#5
Thomas R. Dawson
The Chess Amateur 12/1923
Asymmetrie-Turnier 1924
1. Preis
(15+10) C+
s#5
1. gxf6ep+ Kxe8 2. Dg6+ hxg6 3. cxd8=S ... 4. f7+ Ke7 5. Sxg6+ Sxg6#
Schwarz hat zuletzt f7-f5 gezogen, da d7-d5 an der retrograden Einsperrung des Lc8 scheitert.
Schwarz hat zuletzt f7-f5 gezogen, da d7-d5 an der retrograden Einsperrung des Lc8 scheitert.
Henrik Juel: C+ Popeye 4.61 after analysis
Analysis
White is missing one man, so last move was not a pawn capture
White pawns captured all six missing black men, so last move was not d7-d5
Obviously last move was not made by Ke7 or Sd8f8, so last move was f7-f5 legitimizing the ep key (2020-11-02)
comment
Analysis
White is missing one man, so last move was not a pawn capture
White pawns captured all six missing black men, so last move was not d7-d5
Obviously last move was not made by Ke7 or Sd8f8, so last move was f7-f5 legitimizing the ep key (2020-11-02)
comment
Keywords: Asymmetrical solution, En passant as key, Symmetrical position, Promotion (S)
Genre: Retro, s#
Computer test: C+ Popeye 4.61 after analysis
FEN: 1NRnBnRN/1pP1k1Pp/4p3/2PpKpP1/2pPQPp1/8/2P3P1/8
Reprints: 87 Asymmetry 1927
199 Mat Plus Review 01-06/2010
H2 ASymmetrie 2013
Input: Frank Müller, 2010-11-25
Last update: A.Buchanan, 2020-11-03 more...
Genre: Retro, s#
Computer test: C+ Popeye 4.61 after analysis
FEN: 1NRnBnRN/1pP1k1Pp/4p3/2PpKpP1/2pPQPp1/8/2P3P1/8
Reprints: 87 Asymmetry 1927
199 Mat Plus Review 01-06/2010
H2 ASymmetrie 2013
Input: Frank Müller, 2010-11-25
Last update: A.Buchanan, 2020-11-03 more...
58 - P1193258
Thomas R. Dawson
1539 Pittsburgh Gazette Times 22/12/1914
(5+6) C+
a) #3
b) Alles 1 nach rechts. Then what result?
Thomas R. Dawson
1539 Pittsburgh Gazette Times 22/12/1914
(5+6) C+
a) #3
b) Alles 1 nach rechts. Then what result?
a) 1. e4 Th6 2. e5+ Kg6 3. Lf5#
Keywords: No legal last move for Black (b))
Genre: 3#, Retro
Computer test: a) is C+ Popeye 4.61 b) is trivial retro logic
FEN: 8/1p1B1pp1/1P1K1kr1/6p1/6P1/4P3/8/8
Input: Frank Müller, 2011-04-23
Last update: A.Buchanan, 2021-11-02 more...
Genre: 3#, Retro
Computer test: a) is C+ Popeye 4.61 b) is trivial retro logic
FEN: 8/1p1B1pp1/1P1K1kr1/6p1/6P1/4P3/8/8
Input: Frank Müller, 2011-04-23
Last update: A.Buchanan, 2021-11-02 more...
1. Tf8 Txa1 2. Lg8 Le8+ 3. Df7 Ta1xa6#
Cook: 1. Lxg4 Lf3,Tf1 2. De6,Kf5 Txf1,Lf3 3. Kf5,De6 Le4#
1. Tf4 Lf3,Tf1 2. Txg4 Tf1,Lf3 3. Kf5 Le4#
1. Te4 Txf1 2. Txg4 Lf3 3. Kf5 Le4#
1. Da4 Lf3,Txf1 2. Dxg4 Txf1,Lf3 3. Kf5 Le4#
Cook: 1. Lxg4 Lf3,Tf1 2. De6,Kf5 Txf1,Lf3 3. Kf5,De6 Le4#
1. Tf4 Lf3,Tf1 2. Txg4 Tf1,Lf3 3. Kf5 Le4#
1. Te4 Txf1 2. Txg4 Lf3 3. Kf5 Le4#
1. Da4 Lf3,Txf1 2. Dxg4 Txf1,Lf3 3. Kf5 Le4#
8 NL
Yuri Bilokin: Correction: wBc6-b5, wKd3-c2, bbe6-d5, bpc5d2(-bnd2)
8/7p/p5kP/1B1bp1p1/6P1/8/qpKpr1p1/b4rR1 (5+13) (2019-03-02)
milan: wBc6-a4 coorrection must be close to original as is possible M.Frelih (2019-03-03)
Yuri Bilokin: more economical wBc6-b5, wKd3-c2, wRg1-h1, bPc5-d2(-bNd2), bRe2-e3, -bPg2 8/7p/p3b1kP/1B2p1p1/6P1/4r3/qpKp4/b4r1R (5+12) (2022-12-20)
comment
Yuri Bilokin: Correction: wBc6-b5, wKd3-c2, bbe6-d5, bpc5d2(-bnd2)
8/7p/p5kP/1B1bp1p1/6P1/8/qpKpr1p1/b4rR1 (5+13) (2019-03-02)
milan: wBc6-a4 coorrection must be close to original as is possible M.Frelih (2019-03-03)
Yuri Bilokin: more economical wBc6-b5, wKd3-c2, wRg1-h1, bPc5-d2(-bNd2), bRe2-e3, -bPg2 8/7p/p3b1kP/1B2p1p1/6P1/4r3/qpKp4/b4r1R (5+12) (2022-12-20)
comment
Genre: h#
FEN: 8/7p/p1B1b1kP/2p1p1p1/6P1/3K4/qp1nr1p1/b4rR1
Input: Felber, Volker, 2011-11-08
Last update: A.Buchanan, 2022-12-20 more...
1. Sf3 Lc3 2. Sh4 (droht 3.Sg2 nebst 4.Sf4 oder Se1) Le5 3. Sf5 (droht 4.Se7 nebst 5.Sc6 oder Sd5) Lf6 4. Sd6 Ld4 5. Sxb5 Le5 6. Sa7 Lf6 7. Sc8 Ld8 8. Sd6 Lg5 9. Sc4 Lf4 10. Sxb6 Le3 11. Sc4 Ld4 12. Sa5 nebst 13. Sxb5#
6. ... Le5 7. Se7 Lc3 8. Sd5 Ld2 9. Sxb6 nebst 10. Sc4 und 11. Sa5
WSZ nennt 1930 als Erscheinungsjahr.
6. ... Le5 7. Se7 Lc3 8. Sd5 Ld2 9. Sxb6 nebst 10. Sc4 und 11. Sa5
WSZ nennt 1930 als Erscheinungsjahr.
Anton Baumann: Dual: 2.Sg5!(droht 3.Sf3) Ld4 3.Se4 (droht 4.Sd2) Le3 4.Sd6 Ld4 5.Sxb5 ... (2023-11-08)
comment
comment
* 1. ... Kb7 2. Tc8 Kxc8 3. Ta7 a2#
* 1. ... Kb7 2. Tc8 Kxa6 2. Tb8 a2#
1. Sd4 Kxd8 2. Ta7 Kc8,Ke8 3. Sc6,Se6 a2#
1. Sd4 Kb7 2. Sb5 Kxa6 3. Tb8 a2#
* 1. ... Kb7 2. Tc8 Kxa6 2. Tb8 a2#
1. Sd4 Kxd8 2. Ta7 Kc8,Ke8 3. Sc6,Se6 a2#
1. Sd4 Kb7 2. Sb5 Kxa6 3. Tb8 a2#
Genre: s#
FEN: 3R4/2k5/R1N5/p7/p7/pp6/1p6/bK6
Input: Frank Müller, 2011-12-28
Last update: Frank Müller, 2012-09-01 more...
SCHRECKE: Unlösbar
Die Stellung ist offensichtlich verdruckt.
Mit sBa2 statt c4 geht:
1.Ld3! e5 2.Lf5 e4 3.b5 a:b5#
1. ... e6 2.Le4 e5 3.b5 a:b5# (2021-05-11)
comment
Die Stellung ist offensichtlich verdruckt.
Mit sBa2 statt c4 geht:
1.Ld3! e5 2.Lf5 e4 3.b5 a:b5#
1. ... e6 2.Le4 e5 3.b5 a:b5# (2021-05-11)
comment
Genre: s#
FEN: 8/4p3/p6p/P6R/KPp5/P1B1N3/1r1N2P1/k4B2
Input: Frank Müller, 2011-12-31
Last update: Frank Müller, 2011-12-31 more...
63 - P1345893
Thomas R. Dawson
Geoff Foster
7 The Problemist Supplement 151, p. 608, 2017
T.R.Dawson, v G.Foster
(2+6) C+
h#4.5
Thomas R. Dawson
Geoff Foster
7 The Problemist Supplement 151, p. 608, 2017
T.R.Dawson, v G.Foster
(2+6) C+
h#4.5
1. ... Lf4 2. Ke7 Lxc7 3. De8 Lf4 4. Kd8+ Le3 5. Le7 Lb6#
correct version of P0555591
Viktoras Paliulionis: I changed to "after TR Dawson" because that's the usual heading for such cases in other sources and databases. All databases should use unified headings for problems. "Correction", "version", and "after ..." are different things. (2021-02-23)
Viktoras Paliulionis: The person who corrected the problem or submitted a version of it is not considered a co-author; the name of the first author with a note (version ..., or correction ...) should remain above the diagram. If the problem is sufficiently reworked (as this problem), only the new author is indicated with a note "after ...". (2021-02-23)
A.Buchanan: Hi Viktoras,
Thanks for your interest in this. Oh dear, there are several things to unpack here.
(0) I do follow the Codex as exactly as possible, however...
(1) The list of names above a PDB entry is NOT necessarily a list of equal co-authors. There is no separate PDB field to record the persons who correct or improve a problem. If the names only appear in other (pure text) fields, (a) you can't validate them against the author table and (b) you can't search for them properly. So the names at the top indicate only that these individuals were associated with the problem, and in all the problems which I've ever touched the "after" field then clarifies who did what, if that's necessary, e.g. "TRD, version GF".
(2) It was genuinely an "after TRD", then it's right that TRD wouldn't appear as a name at the top.
(3) However in this case, Geoff made the new version, and was also the magazine editor. If Geoff chose to show TRD the respect of publishing his new idea as a version, then we in PDB have absolutely no right to tamper with that. Similarly, there may be cases where we disagree with what the original author and the improver have agreed together, or what an improver has unilaterally decided. But in PDB we should just report what was actually published. If you feel strongly that a classification is inappropriate, take it up with the composers first rather than going behind their backs, as you apparently have here.
(4) In many cases there is a welcome blurring over time. If "PQ, version RS" exists, then what happens if TU comes along and makes a new version? I've never seen a situation of "PQ, version RS, version TU". Pickiness of fractional ownership just doesn't matter for these things. Particularly for retros, one often finds a bunch of names all together. It's essential that each names be there, but implying a tug-of-war for fractional ownership would just be too tedious, and people often don't bother. (2021-02-23)
Viktoras Paliulionis: Andrew, of course, the database must provide accurate information. My notes:
1. In Winchloe (Christian Poisson is always very accurate in presenting information), above the diagram of this problem it says: Jeffrey Foster, after Thomas R. Dawson, The Problemist 2017. Above diagram of problem Pxxxx is written: Thomas R. Dawson, correction A. Buchanan, The Problemist 1944 (v). (Note: correction: The Problemist Supplement 2017).
2. I think Christian Poisson is right, because in the magazine there is only one author listed above each diagram.
3. How can other people know that not only the author can be specified in the "author" field?
4. Associated individuals should be noted in some other fields, because problems can be reprinted (copy-pasted) in other sources with incorrect information.
5. Instead of "PQ, RS version, TU version", it should be "PQ, TU version".
6. "The correction should be published as a joint composition only if A agrees" (codex).
7. "Bunch of names all together" often occur when they all contribute to the creation together rather than after a single author’s publication. (2021-02-23)
Viktoras Paliulionis: Mentioned problem number is P1345892. (2021-02-23)
A.Buchanan: Sorry that this is long.
PDB is not WinChloe, which has a different data model, and unilateral decisions are made by one person.
PDB is looser, a shared artifact, and is under-documented. Fields and keywords are used in varying ways. In everything I have done I have aimed for an individual’s contributions to problems to be accessible fairly, never over-stating their role.
In PDB this can only be via the “a” field and then detailed annotation elsewhere in a text field. Sometimes in the past the “comment” field has been used by people foe this but I always use the “after” field. Otherwise it is impossible to track people’s involvement and this seems to me hugely unfair, far worse than some trivial confusion over roles for what are generally not earth-shaking problems.
Ideally there would be roles of “corrector”, “versioner” etc in PDB but that’s not the case, so we have to model things with what we have.
In any area of PDB data modelling (e.g. keywords), I try identifying stakeholders and their needs and interests, and how best to serve them. Here they are old and new composers, readers, searchers, etc.
It is of course not the case that PQ v TU would always be the answer. It depends. Your suggested approach is Procrustean - we need something softer and kinder here. :-) The rules in the Codex as specified are simply not scalable to multiple applications. This is independent of the fact that PDB does not represent these rules directly.
We are not talking about album-level problems usually, we are talking about the humble, organic sharing of constructions which to me is the lifeblood of chess problem composition particularly for the newcomers who I try to introduce to PDB from social media. Let’s not over-complicate! Thanks :-) (2021-02-23)
A.Buchanan: Sorry for length, but want a single place to log my current position.
I spend some time correcting cooked old problems which are otherwise of little interest. I enjoy the connection with history, and am scrupulous never to overstate my contribution (although personally I feel that a manual correction may sometimes be more worthy than one simply using without acknowledgement a computer program to trawl for optimal arrangements of pieces). I have an open request with YACPDB to recognize roles and dependencies in their data model. I have also requested through MatPlus that WFCC regularize the unsatisfactory Codex clause handling this area, however due to well-known disruptions in the world, this has not yet happened. It is particularly an issue where the original author is deceased. But other issues are (1) non-scalability of versions (2) unilateral "aftering" (3) lack of documentation of the implications of versioning (e.g. FIDE Album) (4) consistent annotation of versions (sources, dates) etc (5) knock-on effect that databases then don't have a clear model to implement.
A little shocked to read here that Christian Poisson apparently changed a published assignment from "TRD, v GF" to "GF, after TRD". This is not what the corrector (also the editor in this case) chose to publish. While I applaud Christian's terrific work over the years, I think this example shows why it's essential to have multiple databases, so that one perspective does not predominate, and that history is not rewritten. (2022-05-15)
more ...
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Viktoras Paliulionis: I changed to "after TR Dawson" because that's the usual heading for such cases in other sources and databases. All databases should use unified headings for problems. "Correction", "version", and "after ..." are different things. (2021-02-23)
Viktoras Paliulionis: The person who corrected the problem or submitted a version of it is not considered a co-author; the name of the first author with a note (version ..., or correction ...) should remain above the diagram. If the problem is sufficiently reworked (as this problem), only the new author is indicated with a note "after ...". (2021-02-23)
A.Buchanan: Hi Viktoras,
Thanks for your interest in this. Oh dear, there are several things to unpack here.
(0) I do follow the Codex as exactly as possible, however...
(1) The list of names above a PDB entry is NOT necessarily a list of equal co-authors. There is no separate PDB field to record the persons who correct or improve a problem. If the names only appear in other (pure text) fields, (a) you can't validate them against the author table and (b) you can't search for them properly. So the names at the top indicate only that these individuals were associated with the problem, and in all the problems which I've ever touched the "after" field then clarifies who did what, if that's necessary, e.g. "TRD, version GF".
(2) It was genuinely an "after TRD", then it's right that TRD wouldn't appear as a name at the top.
(3) However in this case, Geoff made the new version, and was also the magazine editor. If Geoff chose to show TRD the respect of publishing his new idea as a version, then we in PDB have absolutely no right to tamper with that. Similarly, there may be cases where we disagree with what the original author and the improver have agreed together, or what an improver has unilaterally decided. But in PDB we should just report what was actually published. If you feel strongly that a classification is inappropriate, take it up with the composers first rather than going behind their backs, as you apparently have here.
(4) In many cases there is a welcome blurring over time. If "PQ, version RS" exists, then what happens if TU comes along and makes a new version? I've never seen a situation of "PQ, version RS, version TU". Pickiness of fractional ownership just doesn't matter for these things. Particularly for retros, one often finds a bunch of names all together. It's essential that each names be there, but implying a tug-of-war for fractional ownership would just be too tedious, and people often don't bother. (2021-02-23)
Viktoras Paliulionis: Andrew, of course, the database must provide accurate information. My notes:
1. In Winchloe (Christian Poisson is always very accurate in presenting information), above the diagram of this problem it says: Jeffrey Foster, after Thomas R. Dawson, The Problemist 2017. Above diagram of problem Pxxxx is written: Thomas R. Dawson, correction A. Buchanan, The Problemist 1944 (v). (Note: correction: The Problemist Supplement 2017).
2. I think Christian Poisson is right, because in the magazine there is only one author listed above each diagram.
3. How can other people know that not only the author can be specified in the "author" field?
4. Associated individuals should be noted in some other fields, because problems can be reprinted (copy-pasted) in other sources with incorrect information.
5. Instead of "PQ, RS version, TU version", it should be "PQ, TU version".
6. "The correction should be published as a joint composition only if A agrees" (codex).
7. "Bunch of names all together" often occur when they all contribute to the creation together rather than after a single author’s publication. (2021-02-23)
Viktoras Paliulionis: Mentioned problem number is P1345892. (2021-02-23)
A.Buchanan: Sorry that this is long.
PDB is not WinChloe, which has a different data model, and unilateral decisions are made by one person.
PDB is looser, a shared artifact, and is under-documented. Fields and keywords are used in varying ways. In everything I have done I have aimed for an individual’s contributions to problems to be accessible fairly, never over-stating their role.
In PDB this can only be via the “a” field and then detailed annotation elsewhere in a text field. Sometimes in the past the “comment” field has been used by people foe this but I always use the “after” field. Otherwise it is impossible to track people’s involvement and this seems to me hugely unfair, far worse than some trivial confusion over roles for what are generally not earth-shaking problems.
Ideally there would be roles of “corrector”, “versioner” etc in PDB but that’s not the case, so we have to model things with what we have.
In any area of PDB data modelling (e.g. keywords), I try identifying stakeholders and their needs and interests, and how best to serve them. Here they are old and new composers, readers, searchers, etc.
It is of course not the case that PQ v TU would always be the answer. It depends. Your suggested approach is Procrustean - we need something softer and kinder here. :-) The rules in the Codex as specified are simply not scalable to multiple applications. This is independent of the fact that PDB does not represent these rules directly.
We are not talking about album-level problems usually, we are talking about the humble, organic sharing of constructions which to me is the lifeblood of chess problem composition particularly for the newcomers who I try to introduce to PDB from social media. Let’s not over-complicate! Thanks :-) (2021-02-23)
A.Buchanan: Sorry for length, but want a single place to log my current position.
I spend some time correcting cooked old problems which are otherwise of little interest. I enjoy the connection with history, and am scrupulous never to overstate my contribution (although personally I feel that a manual correction may sometimes be more worthy than one simply using without acknowledgement a computer program to trawl for optimal arrangements of pieces). I have an open request with YACPDB to recognize roles and dependencies in their data model. I have also requested through MatPlus that WFCC regularize the unsatisfactory Codex clause handling this area, however due to well-known disruptions in the world, this has not yet happened. It is particularly an issue where the original author is deceased. But other issues are (1) non-scalability of versions (2) unilateral "aftering" (3) lack of documentation of the implications of versioning (e.g. FIDE Album) (4) consistent annotation of versions (sources, dates) etc (5) knock-on effect that databases then don't have a clear model to implement.
A little shocked to read here that Christian Poisson apparently changed a published assignment from "TRD, v GF" to "GF, after TRD". This is not what the corrector (also the editor in this case) chose to publish. While I applaud Christian's terrific work over the years, I think this example shows why it's essential to have multiple databases, so that one perspective does not predominate, and that history is not rewritten. (2022-05-15)
more ...
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Keywords: Homebase (2), Minimal, Interchange (kd), Alleinunterhalter
Genre: h#
Computer test: Popeye 4.61
FEN: 2bqkb2/2pp4/8/8/8/8/8/2B1K3
Input: A.Buchanan, 2018-01-21
Last update: A.Buchanan, 2021-02-23 more...
Genre: h#
Computer test: Popeye 4.61
FEN: 2bqkb2/2pp4/8/8/8/8/8/2B1K3
Input: A.Buchanan, 2018-01-21
Last update: A.Buchanan, 2021-02-23 more...
1. La5! Lxa5+ 2. Kd3 Lc7 3. Ke4 +-
1. ... Ke8 2. Lxc7 Kf7 3. Kd3,Lf4 Kg6 4. Lf4,Kd3 e5 5. Lc1,Ld2 Kf7 6. Ke4 Kf6 7. Lg5+
'Bote von der Ybbs': "Der Aufgabe liegt der Gedanke zu Grunde, den schwarzen Läufer von seiner Wirkungslinie abzulenken, damit der kleine Bauer flugs zur Dame eilen kann. Eine leichte, aber lehrreiche Aufgabe, die man sich merken soll."
1. ... Ke8 2. Lxc7 Kf7 3. Kd3,Lf4 Kg6 4. Lf4,Kd3 e5 5. Lc1,Ld2 Kf7 6. Ke4 Kf6 7. Lg5+
'Bote von der Ybbs': "Der Aufgabe liegt der Gedanke zu Grunde, den schwarzen Läufer von seiner Wirkungslinie abzulenken, damit der kleine Bauer flugs zur Dame eilen kann. Eine leichte, aber lehrreiche Aufgabe, die man sich merken soll."
65 - P1386614
Thomas R. Dawson
Lion Xray
Andrew Buchanan
PDB Website 19/02/2021
TRD, version LX & AB
(13+9) C+
#2
Thomas R. Dawson
Lion Xray
Andrew Buchanan
PDB Website 19/02/2021
TRD, version LX & AB
(13+9) C+
#2
1. hxg6ep+ Lg5 2. Dxg5#
R: 1. g7-g5 Te3-e7+ 2. Kg5-h6+ Lf5-h7+ 3. Kf6-g5 Sd6-e8+
Black has captured 3 white units. dxcxb accounts for 2. With only one left over, we know that bPh3 never captured. So to explain wPh4, White must have played fxgxh or hxgxh. Combined with dxcxb & exdxcxb, that accounts for all of White's captures.
(a) If White played fxgxh, then the original [wPh] was "waylaid" by an officer, without promoting or capturing.
(b) If White played hxgxh, then the original [wPf] did not capture. We know that [bPf] or [bPg] promoted, using at most one capture.
(b.1) If Black did capture fxg or fxe, then [wPf] did promote to make the balance, and Black's last move can't have been fxg, or the f-file would be blocked to wP.
(b.2) If Black didn't capture fxg or fxe, then [wPf] was waylaid to clear the way for [bPf] to promote.
We don't know which of these three possibilities happened, but in every case:
(1) All captures are accounted for (so e.g. Kg5xSh6 was not possible).
(2) Black did not just play f6xg5.
If Black just played g6-g5, retract prior checking moves and discover retropat as g6 is blocked preventing discovered check by wB now on h7. By elimination, last move was g7-g5 and ep key is ok.
R: 1. g7-g5 Te3-e7+ 2. Kg5-h6+ Lf5-h7+ 3. Kf6-g5 Sd6-e8+
Black has captured 3 white units. dxcxb accounts for 2. With only one left over, we know that bPh3 never captured. So to explain wPh4, White must have played fxgxh or hxgxh. Combined with dxcxb & exdxcxb, that accounts for all of White's captures.
(a) If White played fxgxh, then the original [wPh] was "waylaid" by an officer, without promoting or capturing.
(b) If White played hxgxh, then the original [wPf] did not capture. We know that [bPf] or [bPg] promoted, using at most one capture.
(b.1) If Black did capture fxg or fxe, then [wPf] did promote to make the balance, and Black's last move can't have been fxg, or the f-file would be blocked to wP.
(b.2) If Black didn't capture fxg or fxe, then [wPf] was waylaid to clear the way for [bPf] to promote.
We don't know which of these three possibilities happened, but in every case:
(1) All captures are accounted for (so e.g. Kg5xSh6 was not possible).
(2) Black did not just play f6xg5.
If Black just played g6-g5, retract prior checking moves and discover retropat as g6 is blocked preventing discovered check by wB now on h7. By elimination, last move was g7-g5 and ep key is ok.
Removes 2 pieces from P1106295, and enriches the retro logic
Henrik Juel: Nice clean-up and careful retro argumentation (2021-02-18)
A.Buchanan: Thanks Henrik - in fact I've just seen that one can uniquely retract a couple of single moves further, see the animation (2021-02-19)
comment
Henrik Juel: Nice clean-up and careful retro argumentation (2021-02-18)
A.Buchanan: Thanks Henrik - in fact I've just seen that one can uniquely retract a couple of single moves further, see the animation (2021-02-19)
comment
Keywords: En passant as key, Last Moves? (6)
Genre: 2#, Retro
Computer test: HC+ retro musing + Popeye v4.85
FEN: 3KN3/rPp1R2B/pp5k/Q5pP/PPP4b/1p4Pp/1P6/2B5
Input: A.Buchanan, 2021-02-18
Last update: A.Buchanan, 2021-04-16 more...
Genre: 2#, Retro
Computer test: HC+ retro musing + Popeye v4.85
FEN: 3KN3/rPp1R2B/pp5k/Q5pP/PPP4b/1p4Pp/1P6/2B5
Input: A.Buchanan, 2021-02-18
Last update: A.Buchanan, 2021-04-16 more...
1. Le5! droht 2. Kc4,Kc3 Ka3 3. Ta5#
2. Lc3 ... 3. Ta5#
1. ... Ka3 2. Ta5+ Kb4 3. Lc3#
1. ... Kb4 2. Lc3+ Ka3,Ka4 3. Ta5#
2. Lc3 ... 3. Ta5#
1. ... Ka3 2. Ta5+ Kb4 3. Lc3#
1. ... Kb4 2. Lc3+ Ka3,Ka4 3. Ta5#
67 - P1389169
Thomas R. Dawson
Andrew Buchanan
PDB Website 14/10/2019
TRD, correction AB
(4+5) C+
h#2
Thomas R. Dawson
Andrew Buchanan
PDB Website 14/10/2019
TRD, correction AB
(4+5) C+
h#2
1. d4 Sg6 2. Tg5 Ta6#
See P0515072
Henrik Juel: C+ Popeye 4.61
1.d4 Sg6 2.Tg5 Ta6#
pawn asymmetry (2021-05-02)
A.Buchanan: Yes Henrik exactly - every symmetry has to be broken somehow to make the solution unique, and here the point is that the symmetry is just visual, broken by the pawn movement rules (2021-05-02)
comment
Henrik Juel: C+ Popeye 4.61
1.d4 Sg6 2.Tg5 Ta6#
pawn asymmetry (2021-05-02)
A.Buchanan: Yes Henrik exactly - every symmetry has to be broken somehow to make the solution unique, and here the point is that the symmetry is just visual, broken by the pawn movement rules (2021-05-02)
comment
Keywords: Symmetrical position, Asymmetrical solution, Model mate
Genre: h#
Computer test: C+ Popeye 4.61
FEN: 5N1K/8/5k1N/1r1p4/4p3/8/4r3/R7
Input: A.Buchanan, 2021-05-02
Last update: A.Buchanan, 2021-05-08 more...
Genre: h#
Computer test: C+ Popeye 4.61
FEN: 5N1K/8/5k1N/1r1p4/4p3/8/4r3/R7
Input: A.Buchanan, 2021-05-02
Last update: A.Buchanan, 2021-05-08 more...
1. Txg7 Sf4+ 2. Kh7 ... 3. Thxg8#
1. ... 0-0-0 illegal
Cook: Unlösbar wegen 1. Txg7 0-0-0! (legal!)
BP (Ing. O. Koštál):
1. f3 e6 2. Sa3 Df6 3. Sc4 Dc3 4. bxc3 f5 5. Se5 d6 6. Sg6 Sh6 7. La3 Le7 8. Lc5 Tf8 9. Le3 f4 10. Tb1 Tf5 11. Kf2 Tg5 12. Tb5 Tg4 13. Db1 Ld7 14. fxg4 f3 15. Kg3 Lc6 16. Kf4 Lh4 17. Db4 Le1 18. Tf5 f2 19. Sh3 Lf3 20. gxf3 Sf7 21. Tg1 Se5 22. Tg3 Sec6 23. Sg1 Sd8 24. Th3 Sf7 25. Sf8 Se5 26. Th6 Sec6 27. Lh3 Sd8 28. Tg6 h5 29. Th6 Sf7 30. Tff6 Se5 31. Th8 Sec6 32. Tfh6 Sd8 33. Tg8 Sf7 34. Thh8 Se5 35. Kg5 Sec6 36. Kg6 Sd8 37. Kh7 Sf7 38. Lh6 Sd7 39. g5 Sb6 40. Lf5 Sd5 41. Dg4 hxg4 42. Lg6 Sf4 43. Lh5 Sh3 44. Kg6 Sd8 45. Th7 Sc6 46. Thh8 Sb4 47. Th7 c6 48. Thh8 Sd5 49. Th7 Sf6 50. Tgh8 Sg8 51. Sd7 a6 52. Sc5 dxc5
1. ... 0-0-0 illegal
Cook: Unlösbar wegen 1. Txg7 0-0-0! (legal!)
BP (Ing. O. Koštál):
1. f3 e6 2. Sa3 Df6 3. Sc4 Dc3 4. bxc3 f5 5. Se5 d6 6. Sg6 Sh6 7. La3 Le7 8. Lc5 Tf8 9. Le3 f4 10. Tb1 Tf5 11. Kf2 Tg5 12. Tb5 Tg4 13. Db1 Ld7 14. fxg4 f3 15. Kg3 Lc6 16. Kf4 Lh4 17. Db4 Le1 18. Tf5 f2 19. Sh3 Lf3 20. gxf3 Sf7 21. Tg1 Se5 22. Tg3 Sec6 23. Sg1 Sd8 24. Th3 Sf7 25. Sf8 Se5 26. Th6 Sec6 27. Lh3 Sd8 28. Tg6 h5 29. Th6 Sf7 30. Tff6 Se5 31. Th8 Sec6 32. Tfh6 Sd8 33. Tg8 Sf7 34. Thh8 Se5 35. Kg5 Sec6 36. Kg6 Sd8 37. Kh7 Sf7 38. Lh6 Sd7 39. g5 Sb6 40. Lf5 Sd5 41. Dg4 hxg4 42. Lg6 Sf4 43. Lh5 Sh3 44. Kg6 Sd8 45. Th7 Sc6 46. Thh8 Sb4 47. Th7 c6 48. Thh8 Sd5 49. Th7 Sf6 50. Tgh8 Sg8 51. Sd7 a6 52. Sc5 dxc5
In der 'Prager Presse' mit der Information nachgedruckt: "Die Schachrubrik des 'Cas' macht folgende Aufgaben [diese und P0001348] zum Gegenstande eines Lösungsturniers"; Forderung "Matt in 3 Zügen. (Rückläufige Analyse)."
Korrektur s. P1390813
Henrik Juel: C+ Popeye 4.61 and analysis (except for mate dual after 2... Kf8)
1.Txg7 Sf4+ 2.Kh7+ any 3.Thxg8# (2021-06-12)
Henrik Juel: White captured b2xc3, g2xf3, and fxg; Black captured dxc and hxg
If Ke8 remains on e8 there is no way to release the eastern cage, so Ke8 has moved, and Black may not castle (2021-06-12)
Mario Richter: Sorry, Henrik, but you fell into the same trap than Dawson: by uncapturing a wS on c5 White can provide a retro-shield on f8, and then everyting easily resolves (making it possible for the wK to temporarily visit h7) ... (2021-06-12)
VL: Mario, the (parity changing) tempomove a7-a6 looks unavoidable for such a castling preserving genesis; accordingly, the shift of bPa6 to a7 would repair the problem. (2021-06-14)
Mario Richter: Valery, I think your correction would work too, but Dawson found another way to repair the Problem. (2021-06-14)
A.Buchanan: Mario: did really Dawson not architect the retro-shield? It's so nice, I'm sure he must have! sS can only reach f7 via d8, so c7-c6 must be retracted too. This throws the parity, which was otherwise good, so Black must spend his other parity move a7-a6 to get back in the game. Then he should have no waiting retractions while Lg5-h6. I think that Dawson didn't spot until too late that Sf4-h3 is always available once wK is off g6. That's what breaks the problem. Certainly shifting sBa6-a7 will render the #3 sound by eliminating the castling defence, however it costs retro content, because the intention (I believe) is that it's parity that costs Black his essential waiting retraction. Note that in the official repair, P1390813, the Black qside pawn configuration is unchanged: but Sh3 is no longer to be seen. (2021-06-15)
Mario Richter: Yes, of course - the retro-shield on f8 is part of the intended solution. The problematic piece is black Sh3, which has tempo moves after R: Kh7-g6 ... (2021-06-15)
comment
Korrektur s. P1390813
Henrik Juel: C+ Popeye 4.61 and analysis (except for mate dual after 2... Kf8)
1.Txg7 Sf4+ 2.Kh7+ any 3.Thxg8# (2021-06-12)
Henrik Juel: White captured b2xc3, g2xf3, and fxg; Black captured dxc and hxg
If Ke8 remains on e8 there is no way to release the eastern cage, so Ke8 has moved, and Black may not castle (2021-06-12)
Mario Richter: Sorry, Henrik, but you fell into the same trap than Dawson: by uncapturing a wS on c5 White can provide a retro-shield on f8, and then everyting easily resolves (making it possible for the wK to temporarily visit h7) ... (2021-06-12)
VL: Mario, the (parity changing) tempomove a7-a6 looks unavoidable for such a castling preserving genesis; accordingly, the shift of bPa6 to a7 would repair the problem. (2021-06-14)
Mario Richter: Valery, I think your correction would work too, but Dawson found another way to repair the Problem. (2021-06-14)
A.Buchanan: Mario: did really Dawson not architect the retro-shield? It's so nice, I'm sure he must have! sS can only reach f7 via d8, so c7-c6 must be retracted too. This throws the parity, which was otherwise good, so Black must spend his other parity move a7-a6 to get back in the game. Then he should have no waiting retractions while Lg5-h6. I think that Dawson didn't spot until too late that Sf4-h3 is always available once wK is off g6. That's what breaks the problem. Certainly shifting sBa6-a7 will render the #3 sound by eliminating the castling defence, however it costs retro content, because the intention (I believe) is that it's parity that costs Black his essential waiting retraction. Note that in the official repair, P1390813, the Black qside pawn configuration is unchanged: but Sh3 is no longer to be seen. (2021-06-15)
Mario Richter: Yes, of course - the retro-shield on f8 is part of the intended solution. The problematic piece is black Sh3, which has tempo moves after R: Kh7-g6 ... (2021-06-15)
comment
Keywords: Cant Castler (sg), Superseded by (P1390813), Retro Opposition
Genre: 3#, Retro
FEN: r3k1nR/1p4pR/p1p1p1KB/2p3PB/6p1/2P2P1n/P1PPPp1P/4b1N1
Reprints: 232 Cas 01/01/1922
159 Prager Presse 15/01/1922
Input: Mario Richter, 2021-06-12
Last update: A.Buchanan, 2021-06-16 more...
Genre: 3#, Retro
FEN: r3k1nR/1p4pR/p1p1p1KB/2p3PB/6p1/2P2P1n/P1PPPp1P/4b1N1
Reprints: 232 Cas 01/01/1922
159 Prager Presse 15/01/1922
Input: Mario Richter, 2021-06-12
Last update: A.Buchanan, 2021-06-16 more...
1. Txg7 ... 2. Thxg8#
1. ... 0-0-0?? illegal
R: 1. d6xSc5 Sd7-c5 2. Sf6-g8 Sf8-d7+ 3. Sd5-f6 Tg8-h8 4. Sb4-d5 Th8-h7 5. c7-c6 Th7-h8 6. Sc6-b4 Th8-h7 7. Sd8-c6 Th7-h8 8. a7-a6 Th8-h7 9. Sf7-d8 Kh7-g6 10. ??? Lg5-h6 11. Sd5-f7 L~-g5+
If Black preserves castling rights, there is insoluble retro-opposition between sSg8 (unpinned by wS uncaptured at c5) and wTTK, none of whom can triangulate in this position. Apparently "the stars are right" for the sSf7 to allow wKg6 to shift, but the sS route must pass through c6, and retracting c7-c6 means "the stars are wrong". Black must spend his last spare tempo a7-a6 to rectify the situation, and that means there is no possible move while waiting for wLh5-g6 to be retracted.
Cook: There is an easily fixed error in this version: wTa1 can't retract home at all if Black castling rights are retained, rendering irrelevant the subtle timing considerations.
1. ... 0-0-0?? illegal
R: 1. d6xSc5 Sd7-c5 2. Sf6-g8 Sf8-d7+ 3. Sd5-f6 Tg8-h8 4. Sb4-d5 Th8-h7 5. c7-c6 Th7-h8 6. Sc6-b4 Th8-h7 7. Sd8-c6 Th7-h8 8. a7-a6 Th8-h7 9. Sf7-d8 Kh7-g6 10. ??? Lg5-h6 11. Sd5-f7 L~-g5+
If Black preserves castling rights, there is insoluble retro-opposition between sSg8 (unpinned by wS uncaptured at c5) and wTTK, none of whom can triangulate in this position. Apparently "the stars are right" for the sSf7 to allow wKg6 to shift, but the sS route must pass through c6, and retracting c7-c6 means "the stars are wrong". Black must spend his last spare tempo a7-a6 to rectify the situation, and that means there is no possible move while waiting for wLh5-g6 to be retracted.
Cook: There is an easily fixed error in this version: wTa1 can't retract home at all if Black castling rights are retained, rendering irrelevant the subtle timing considerations.
Korrektur zu P1390750
A.Buchanan: Under opposition idea, if Black had an extra tempo (e.g. sBa6 on a5 instead) it would be ok for Black to castle. But that still wouldn't work. We need to unlock the cage with e2xf3, following the retraction wLf1. However wRa1 is locked out! Am I right? Can this be fixed e.g. by starting wBa2 on a4 instead? (2021-06-15)
Mario Richter: So the correction still is not perfect, because the need to bring wRa1 back home already makes black 0-0-0 illegal. But with your suggested modification everything seems to be ok. (2021-06-15)
A.Buchanan: Thank you, Mario. I considered today some new possibilities for Black queenside, but the fundamental challenge always seems to be: prevent Black retracting c7-c6 *before* the retro-opposition phase is entered. So the version I propose is more modest: r3k1nR/1p4pR/p1p1p1KB/2p3PB/6Pb/5Ppb/2P2pP1/7N. It does at least save three pawns, but does it work? (2021-06-16)
Mario Richter: I think r3k1nR/1p4pR/p1p1p1KB/2p3PB/6Pb/5Ppb/2P2pP1/7N doesn't work, because the saved pawns can be used to provide an extra tempo for White: before the black Knight shields on f7, he can uncapture a White pawn eg. on b6 ... (2021-06-16)
A.Buchanan: Oops haha and I was doing so well on this one - thanks. And there is no restriction on sS moving around, because the wT can vibrate indefinitely. Well in that case the earlier suggestion of wBa2 on a4 instead is the simple correction. (2021-06-16)
more ...
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A.Buchanan: Under opposition idea, if Black had an extra tempo (e.g. sBa6 on a5 instead) it would be ok for Black to castle. But that still wouldn't work. We need to unlock the cage with e2xf3, following the retraction wLf1. However wRa1 is locked out! Am I right? Can this be fixed e.g. by starting wBa2 on a4 instead? (2021-06-15)
Mario Richter: So the correction still is not perfect, because the need to bring wRa1 back home already makes black 0-0-0 illegal. But with your suggested modification everything seems to be ok. (2021-06-15)
A.Buchanan: Thank you, Mario. I considered today some new possibilities for Black queenside, but the fundamental challenge always seems to be: prevent Black retracting c7-c6 *before* the retro-opposition phase is entered. So the version I propose is more modest: r3k1nR/1p4pR/p1p1p1KB/2p3PB/6Pb/5Ppb/2P2pP1/7N. It does at least save three pawns, but does it work? (2021-06-16)
Mario Richter: I think r3k1nR/1p4pR/p1p1p1KB/2p3PB/6Pb/5Ppb/2P2pP1/7N doesn't work, because the saved pawns can be used to provide an extra tempo for White: before the black Knight shields on f7, he can uncapture a White pawn eg. on b6 ... (2021-06-16)
A.Buchanan: Oops haha and I was doing so well on this one - thanks. And there is no restriction on sS moving around, because the wT can vibrate indefinitely. Well in that case the earlier suggestion of wBa2 on a4 instead is the simple correction. (2021-06-16)
more ...
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Keywords: Cant Castler (sl), Retro Opposition
Genre: 2#, Retro
FEN: r3k1nR/1p4pR/p1p1p1KB/2p3PB/6Pb/1P3Ppb/P1PP1pP1/N7
Reprints: D1668 Retro-Opposition & Other Retro-Analytical Chess Problems , p. 13, 1989
Input: Mario Richter, 2021-06-14
Last update: A.Buchanan, 2021-06-16 more...
Genre: 2#, Retro
FEN: r3k1nR/1p4pR/p1p1p1KB/2p3PB/6Pb/1P3Ppb/P1PP1pP1/N7
Reprints: D1668 Retro-Opposition & Other Retro-Analytical Chess Problems , p. 13, 1989
Input: Mario Richter, 2021-06-14
Last update: A.Buchanan, 2021-06-16 more...
1. axb6ep+! Kxe3 2. De2#
Cooked by the possibility of e2-e1=S!
https://timkr.home.xs4all.nl/admag/promo.htm Tim Krabbe writes in "PROMOTIEMOTIVATIE": Een van de mooiste uitspraken in de Nederlandse schaakliteratuur werd gedaan door Johan Barendregt (1924-1982), in een interview met Max Pam: 'Mijn leven is bepaald door de zet e2-e1P.'
Hij doelde daarmee op een ontdekking die hij had gedaan in een stelling die in 1937 aan de lezers van het blad De Schaakwereld ter oplossing was voorgelegd. (Zie diagram.)
Het mat op zich was niet moeilijk, dat kon alleen 1.axb6+ Kxe3 2.De2 mat zijn, maar het ging om het bewijs dat alleen b7-b5 Zwarts laatste zet geweest kon zijn, en niet b6-b5 of Kc4-d3. Toen een paar weken later de oplossing werd gepubliceerd, bleek dat de 13-jarige Barendregt alle oplossers de baas was geweest, omdat hij had aangetoond dat het probleem van de grootheid Dawson incorrect was. Met een bewijspartij van 48 zetten liet hij zien dat ook e2-e1P Zwarts laatste zet geweest kon zijn, wat 1.axb6 illegaal maakte. En passant repareerde hij het probleem ook, door de pion van f5 naar e7 te verplaatsen - dàn moet Zwart zojuist b7-b5 gespeeld hebben.
Dat bewijs laat ik hier voor wat het is - het gaat me om de verrukking die Barendregt moet hebben gevoeld toen hem, temidden van de partijen van Euwe, Aljechin en Keres, lof werd toegezwaaid, maar die hij vooral moet hebben gevoeld bij de ontdekking van e2-e1P zèlf. Dat geluksgevoel bond hem voor de rest van zijn leven aan het schaken - hij was jarenlang een van de sterkste Nederlanders, werd Internationaal Meester toen dat nog iets betekende, en won partijen tegen Botwinnik en Portisch.
In English: One of the most beautiful statements in Dutch chess literature was made by Johan Barendregt (1924-1982), in an interview with Max Pam: 'My life is determined by the move e2-e1P.'
He was referring to a discovery he had made in a proposition that had been submitted to the readers of the magazine De Schaakwereld in 1937 for a solution. ( See diagram. )
The mate itself was not difficult, only 1.axb6+ Kxe3 2.De2 ??mate but it was to prove that only b7-b5 could have been Black's last move, and not b6-b5 or Kc4-d3. When the solution was published a few weeks later, it turned out that 13-year-old Barendregt had beaten all the solvers because he had shown that the problem of the great Dawson was incorrect. With a proof game of 48 moves, he showed that e2-e1P could also have been Black's last move, making 1.axb6 illegal. In the meantime, he also solved the problem by moving the pawn from f5 to e7 - then Black must have just played b7-b5.
I will leave that proof for what it is here - I am concerned with the delight that Barendregt must have felt when he was praised among the parties of Euwe, Aljechin and Keres, but which he must have felt above all when he discovered e2-e1P itself. That happiness tied him to playing chess for the rest of his life - for years he was one of the strongest Dutchmen, became an International Master when it still meant something, and won games against Botvinnik and Portisch.
Cook: Possible proof game by James Malcom in 42.0 moves: 1. b4 a5 2. bxa5 g5 3. Lb2 g4 4. Sc3 g3 5. hxg3 Sf6 6. Th6 Se4 7. Sd5 Sc3 8. Tc1 Sb1 9. Ta6 e5 10. Sf6+ Ke7 11. e4 d5 12. c3 d4 13. Tc2 d3 14. Ta7 dxc2 15. Sh3 Sa6 16. d4 c6 17. d5 Sc7 18. La6 Kd6 19. Dd3 Kc5 20. d6 Lh6 21. Ke2 Ld2 22. dxc7 h5 23. Sg5 Dd5 24. exd5 e4 25. a3 e3 26. Kf3 h4 27. De2 Lf5 28. Kf4 Tae8 29. Ta8 h3 30. Tc8 h2 31. d6 h1=T 32. Sd5 Tc1 33. f3 Le4 34. fxe4 b5 35. Dd1 Th4+ 36. Kf5 Tf4+ 37. gxf4 Kc4 38. Sh7 Kd3 39. Kg5 f5 40. Shf6 e2 41. Se3 Te5 42. fxe5 e1=S
means e.p. convention won't fire
Cooked by the possibility of e2-e1=S!
https://timkr.home.xs4all.nl/admag/promo.htm Tim Krabbe writes in "PROMOTIEMOTIVATIE": Een van de mooiste uitspraken in de Nederlandse schaakliteratuur werd gedaan door Johan Barendregt (1924-1982), in een interview met Max Pam: 'Mijn leven is bepaald door de zet e2-e1P.'
Hij doelde daarmee op een ontdekking die hij had gedaan in een stelling die in 1937 aan de lezers van het blad De Schaakwereld ter oplossing was voorgelegd. (Zie diagram.)
Het mat op zich was niet moeilijk, dat kon alleen 1.axb6+ Kxe3 2.De2 mat zijn, maar het ging om het bewijs dat alleen b7-b5 Zwarts laatste zet geweest kon zijn, en niet b6-b5 of Kc4-d3. Toen een paar weken later de oplossing werd gepubliceerd, bleek dat de 13-jarige Barendregt alle oplossers de baas was geweest, omdat hij had aangetoond dat het probleem van de grootheid Dawson incorrect was. Met een bewijspartij van 48 zetten liet hij zien dat ook e2-e1P Zwarts laatste zet geweest kon zijn, wat 1.axb6 illegaal maakte. En passant repareerde hij het probleem ook, door de pion van f5 naar e7 te verplaatsen - dàn moet Zwart zojuist b7-b5 gespeeld hebben.
Dat bewijs laat ik hier voor wat het is - het gaat me om de verrukking die Barendregt moet hebben gevoeld toen hem, temidden van de partijen van Euwe, Aljechin en Keres, lof werd toegezwaaid, maar die hij vooral moet hebben gevoeld bij de ontdekking van e2-e1P zèlf. Dat geluksgevoel bond hem voor de rest van zijn leven aan het schaken - hij was jarenlang een van de sterkste Nederlanders, werd Internationaal Meester toen dat nog iets betekende, en won partijen tegen Botwinnik en Portisch.
In English: One of the most beautiful statements in Dutch chess literature was made by Johan Barendregt (1924-1982), in an interview with Max Pam: 'My life is determined by the move e2-e1P.'
He was referring to a discovery he had made in a proposition that had been submitted to the readers of the magazine De Schaakwereld in 1937 for a solution. ( See diagram. )
The mate itself was not difficult, only 1.axb6+ Kxe3 2.De2 ??mate but it was to prove that only b7-b5 could have been Black's last move, and not b6-b5 or Kc4-d3. When the solution was published a few weeks later, it turned out that 13-year-old Barendregt had beaten all the solvers because he had shown that the problem of the great Dawson was incorrect. With a proof game of 48 moves, he showed that e2-e1P could also have been Black's last move, making 1.axb6 illegal. In the meantime, he also solved the problem by moving the pawn from f5 to e7 - then Black must have just played b7-b5.
I will leave that proof for what it is here - I am concerned with the delight that Barendregt must have felt when he was praised among the parties of Euwe, Aljechin and Keres, but which he must have felt above all when he discovered e2-e1P itself. That happiness tied him to playing chess for the rest of his life - for years he was one of the strongest Dutchmen, became an International Master when it still meant something, and won games against Botvinnik and Portisch.
Cook: Possible proof game by James Malcom in 42.0 moves: 1. b4 a5 2. bxa5 g5 3. Lb2 g4 4. Sc3 g3 5. hxg3 Sf6 6. Th6 Se4 7. Sd5 Sc3 8. Tc1 Sb1 9. Ta6 e5 10. Sf6+ Ke7 11. e4 d5 12. c3 d4 13. Tc2 d3 14. Ta7 dxc2 15. Sh3 Sa6 16. d4 c6 17. d5 Sc7 18. La6 Kd6 19. Dd3 Kc5 20. d6 Lh6 21. Ke2 Ld2 22. dxc7 h5 23. Sg5 Dd5 24. exd5 e4 25. a3 e3 26. Kf3 h4 27. De2 Lf5 28. Kf4 Tae8 29. Ta8 h3 30. Tc8 h2 31. d6 h1=T 32. Sd5 Tc1 33. f3 Le4 34. fxe4 b5 35. Dd1 Th4+ 36. Kf5 Tf4+ 37. gxf4 Kc4 38. Sh7 Kd3 39. Kg5 f5 40. Shf6 e2 41. Se3 Te5 42. fxe5 e1=S
means e.p. convention won't fire
Henrik Juel: James, when I played your game I did not reach the diagram position
Is the diagram or your game wrong? (2021-11-06)
James Malcom: Gahhhhh, I got myself again Henrik. I garbled some pawns, which should be fixed now. (2021-11-06)
Henrik Juel: Thanks
The intended retroplay was
R: 1... b7-b5! 2.Sc4-d3+ Ke3-d3+ 3.Sb6-c4+ d3xTc2 (2021-11-07)
Henrik Juel: The cooking last move e2-e1=S is rather obvious
I would have thought that it was found by the TfS solvers and corrected by TRD
Can anyone check in TfS 1923-24? (2021-11-07)
Henrik Juel: Never mind...
The problem appeared in October 1923, and there is no mention of e2-e1=S in the solution in the March 1924 issue (2021-11-07)
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Is the diagram or your game wrong? (2021-11-06)
James Malcom: Gahhhhh, I got myself again Henrik. I garbled some pawns, which should be fixed now. (2021-11-06)
Henrik Juel: Thanks
The intended retroplay was
R: 1... b7-b5! 2.Sc4-d3+ Ke3-d3+ 3.Sb6-c4+ d3xTc2 (2021-11-07)
Henrik Juel: The cooking last move e2-e1=S is rather obvious
I would have thought that it was found by the TfS solvers and corrected by TRD
Can anyone check in TfS 1923-24? (2021-11-07)
Henrik Juel: Never mind...
The problem appeared in October 1923, and there is no mention of e2-e1=S in the solution in the March 1924 issue (2021-11-07)
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Keywords: En passant as key, Superseded by (P1395505)
Computer test: Popeye v4.87 & simple retro-logic & demonstrative proof game
FEN: 2R5/2P5/B1pP1N2/Pp2PpK1/4P3/P1PkN3/1Bpb2P1/1nrQn3
Reprints: De Schaakwereld 1937
Tim Krabbé's Website 2001
MatPlus.net Forum 2021
Input: James Malcom, 2021-11-06
Last update: A.Buchanan, 2021-11-07 more...
Computer test: Popeye v4.87 & simple retro-logic & demonstrative proof game
FEN: 2R5/2P5/B1pP1N2/Pp2PpK1/4P3/P1PkN3/1Bpb2P1/1nrQn3
Reprints: De Schaakwereld 1937
Tim Krabbé's Website 2001
MatPlus.net Forum 2021
Input: James Malcom, 2021-11-06
Last update: A.Buchanan, 2021-11-07 more...
1. axb6ep+ Kxe3 2. De2#
R: 1. … b7-b5 2. Sc4-e3+ Ke3-d3+ 3. Sb6-c4+ d3xTc2
R: 1. … b7-b5 2. Sc4-e3+ Ke3-d3+ 3. Sb6-c4+ d3xTc2
Correction of P1395488
Henrik Juel: Black captured d3xTc2 and promoted on h1
White captured bxPa, d6xc7, exd, fxe, and h2xPg3xf4xe5
R: 1... b7-b5 2.Sc4-e3+ Ke3-d3+ 3.Sb6-c4+ d3xTc2
Yet another way to motivate the double step b7-b5
Solution 1.axb6ep+ Kxe3 2.De2# (2021-11-07)
comment
Henrik Juel: Black captured d3xTc2 and promoted on h1
White captured bxPa, d6xc7, exd, fxe, and h2xPg3xf4xe5
R: 1... b7-b5 2.Sc4-e3+ Ke3-d3+ 3.Sb6-c4+ d3xTc2
Yet another way to motivate the double step b7-b5
Solution 1.axb6ep+ Kxe3 2.De2# (2021-11-07)
comment
Keywords: En passant as key
Genre: 2#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 2R5/2P1p3/B1pP1N2/Pp2P1K1/4P3/P1PkN3/1Bpb2P1/1nrQn3
Reprints: Tim Krabbé's Website 2001
MatPlus.net Forum 2021
Input: A.Buchanan, 2021-11-07
Last update: A.Buchanan, 2021-11-07 more...
Genre: 2#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 2R5/2P1p3/B1pP1N2/Pp2P1K1/4P3/P1PkN3/1Bpb2P1/1nrQn3
Reprints: Tim Krabbé's Website 2001
MatPlus.net Forum 2021
Input: A.Buchanan, 2021-11-07
Last update: A.Buchanan, 2021-11-07 more...
1. Ta1! a3 2. Sb1 a2 3. Lc1
Keywords: Zugzwang-Goal
Genre: n#
Computer test: Popeye 4.61
FEN: 7k/6pP/6P1/8/p1PPP1p1/1p2BpP1/1P1N1P2/1R2K3
Reprints: Am Rande des Schachbretts , p. 18, 1947
Input: Dieter Berlin, 2022-04-23
Last update: Dieter Berlin, 2022-04-23 more...
Genre: n#
Computer test: Popeye 4.61
FEN: 7k/6pP/6P1/8/p1PPP1p1/1p2BpP1/1P1N1P2/1R2K3
Reprints: Am Rande des Schachbretts , p. 18, 1947
Input: Dieter Berlin, 2022-04-23
Last update: Dieter Berlin, 2022-04-23 more...
Show statistic for complete result. Show search result faster by using ids.
https://pdb.dieschwalbe.de/search.jsp?expression=COMMENTDATE%3E%3D20200919+AND+NOT+K%3D%27Hilfsr%C3%BCckz%C3%BCger%27+AND+NOT+A%3D%27Seliwanow%2C+Andrej+W.%27+AND+A%3D%27Dawson%2C+Thomas+R.%27
The problems of this query have been registered by the following contributors:
Gerd Wilts (11)hpr (7)
Hans-Jürgen Schäfer (7)
Andreas Mokosch (2)
HBae (2)
Markus Manhart (5)
Ralf Krätschmer (2)
Franz Pachl (1)
Felber, Volker (5)
Henri Nouguier (3)
Brian Stephenson (6)
Frank Müller (9)
Zuncke/Bruder (2)
A.Buchanan (4)
Mario Richter (4)
James Malcom (1)
Dieter Berlin (1)
Es läßt sich beweisen, daß die UWs in D,T,L,S zwingend erfolgt sind (a2-b8=L,h2-b8=S und entweder b2-c8=D/c2-c8=T oder b2-c8=T/c2-c8=D).
Erich Bartel: weitere Nachdrucke:
3) 160 Die Allumwandlung im Problemschach VIII 1966a---
4) Schach ohne Grenzen 1969.-- (2007-01-09)
Sally: Vier erzwungene Umwandlungen. (2012-02-21)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 da NUPG sonst cooked in 00:03:51 Minuten. (hh:mm:ss)
Keine Lösung: BP 33.0, BP 33.5.
Beispiel: 1.Sf3 Sf6 2.a4 Sd5 3.a5 Sf4 4.Ta4 c5 5.Td4 Db6 6.b4 Sc6 7.b5 Tb8 8.bxc6 cxd4 9.c7 e5 10.d3 La3 11.Le3 Lc1 12.c4 Db2 13.c5 b5 14.c6 Tb6 15.axb6 Ke7 16.b7 Kf6 17.b8L Lb7 18.c8D b4 19.c7 d5 20.Dh3 b3 21.c8T h5 22.g4 h4 23.Dg3 hxg3 24.h4 Th6 25.Lh3 dxe3 26.0–0 Sg2 27.Sh2 Se1 28.h5 Kg5 29.Td8 Ta6 30.Ld6 Kh4 31.h6 Ta1 32.h7 a5 33.h8S a4 34.Sg6+ fxg6
Umwandlungen: 17.b8L, 18.c8D, 21.c8T, 33.h8S. (2023-04-19)
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