Compare P0000490 with equivalent material but 30 years later.
A.Buchanan: Solved in YouTube: https://www.youtube.com/watch?v=Do_rkLQnrpk (2022-02-17)
A.Buchanan: Solved in YouTube video: https://www.youtube.com/watch?v=Do_rkLQnrpk&list=PLYeOP7z8XTGTAF6LYk15wZgqD6-ZXgbM2&index=2 (2022-02-17, hidden)
A.Buchanan: Compare P0000490 with equivalent material but 30 years later. (2023-05-31, hidden) more ...
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im Original ist als Dritter Autor D. Suboticanec angegeben, habe entsprechend geändert.
Zu P. Kahl heißt es ergänzend: [aus] Osterholz-Scharmbeck
HBae: Ist "Klaus Peter Kahl" identisch zu "Peter Kahl" ? (2010-09-22)
Henrik Juel: Yes, probably
I know that 'Jan Robert Mortensen' is identical to 'Jan Mortensen'; he never used his middle name (2017-03-08)
A.Buchanan: It’s possible to put the middle name into the “addition” field in the author table, rather than have two names in the “first name” field. Then in overview list the middle name won’t appear, but it will still appear above each composition as here. Curiously the addition field is common across all four languages (German, English, French, Unicode) supported by the author table. There is no comment field to allow for random extra info to be stored, unlike the source table. So if we drop Robert, we lose it completely. (2022-03-31)
Henrik Juel: Thanks for the info, Andrew
I guess we will have to live with it
For Jan, there is really no risk of confusion
But for Knud Hannemann (son of the less illustrious composer Harald Hannemann) it is annoying to see him cast as Knud Harald Hannemann, just because it was a custom in his family that the sons were given the father's first name
They still use this custom in Russia, but with 'son of' added, e.g., Vladimir Vladimirovitch Putin (2022-03-31)
Henrik Juel: A chess problem example is Nikita Michaelovitch Plaksin
But the rendering of his name in the PBD also gives an obvious partial solution to my problem
Change the middle name of the two danish composers to the corresponding initial:
Jan R. Mortensen and Knud H. Hannemann (2022-03-31)
Henrik Juel: Thanks for the 'initialization', Andrew (2022-03-31)
A.Buchanan: OK fixed those, Henrik. HBae's original question concerned "Klaus Peter Kahl" vs "Peter Kahl". The former only has 2 problems here, and "both" collaborated with Jan Mortensen, so surely this is one person? (2022-03-31)
A.Buchanan: Kahls unified (2022-04-01)
Henrik Juel: I hear no protests from Osterholz-Scharmbeck, so that seems fine (2022-04-01)
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Henrik Juel: This problem demonstrates an advantage of type Høeg over type Proca:
another way of generating variations
2.0-0 cannot be an uncapture, of course
2.Te7-f7 and 2.Dc5-c8 etc. could be uncaptures, but no matter what Black supplements, he is mated (2023-04-08)
A.Buchanan: Thanks Henrik: is an alternative for White R: 1. Kg1-h1 Lb8xDa7+ 2. Dc5-a7/Da3-a7, dann 1. Df8# (2023-04-08)
Henrik Juel: No Andrew, when White moves his uncaptured queen back to c5 or a3, Black supplements a queen or bishop on a7, preventing the mate on f8 (2023-04-08)
A.Buchanan: Thanks! (2023-04-08)
Henrik Juel: -1.Kg1:S Sf2:0 -2.0-0, 1.Rh8#; -1.. Bb8:S -2.Re7, 1.Sc6#; -1.. cSb6,dSb6,Se3:Q -2.Qc5, 1.Qf8#. Duplicated by P0006277 (2003-10-09, hidden) more ...
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A.Buchanan: Nice unique try! (2021-10-16, hidden)
Stefan Felber: Umwandlungsläufer b1! (2021-10-16, hidden) more ...
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A.Buchanan: Setplay doesn't care whether the position is legal. (2023-12-06)
A.Buchanan: So is this a retro problem? I think not. (2023-12-06)
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A.Buchanan: This is an earliest position in which the last two moves can be determined. (2021-06-29)
A.Buchanan: This problem is a good example of one that has lost its intent, its meaning. (2012-04-08, hidden) more ...
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Gerald Ettl: 1.Kb6xSc6 Sd4xSc6
dann 1.Kc4 Sxe5# (2009-06-08)
Adrian Storisteanu: Set play: - 1... Ke4xSd3 & 1.d5-d4 Sd3-c5# (2021-04-21)
Adrian Storisteanu: See also P0006009!? (2021-04-21)
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Mario Richter: vgl. P1011972 (dort sLa3 statt sTa3) (2020-09-23)
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klären Autorlösung
Anton Baumann: Autorabsicht: 1.Lg2 Th1 2.Lg1 Th8 3.Lh2 Th1 4.Lg1 Th7#
Korrektur Version A. Bulawka: -sSg6, sKg7 nach f7, +sBe6, +sBf5, +sBg6 (vergl. yacpdb 320494) (2020-07-15)
Ladislav Packa: The modified version has an illegal position. (2020-07-15)
A.Buchanan: A good, sound, economical version is 320495 in YACPDB (2020-07-15)
A.Buchanan: Bulawka's correction 320494 is sound and legal (6 sB captures + 2 wL died at home). But Smirnov made a much lighter sound version 320495. (2021-05-16)
A.Buchanan: Bulawka's correction 320494 is correct and legal (6 sB captures + 2 wL died at home). But it's heavy and wooden. I'll post Smirnov's much lighter corrected version 320495. (2021-05-15, hidden) more ...
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{ NL}
klären AL
SCHRECKE: Geplant war wohl:
1. Kd6 Lh2+ 2. e5 dxe6 e.p. 3. Ke7 Sd5#
Nebenlösungen, z.B.:
1. Df6 Kb4 2. De6 fxe6 3. Kd6 Lh2#
1. Kd7 Lh2 2. Sc7 Sc4 3. De8 Sb6# (2023-09-20)
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{ NL}
klären AL
SCHRECKE: Satz:
1. ... c8=T+ 2. Kd7 Le6#
Es lösen:
1) 1. Ld8 cxd8=T+ 2. Ke7 Lh4#
2) 1. Kd7 cxb8=D 2. Kc6 Dc7#
3) 1. Ld6 c8=D+ 2. Ke7 Lh4#
4) 1. Kf8 Lf4 2. De8 Lh6# (2023-09-20)
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HBae: Lösung in feenschach 11/1971 S.192, Lösungsstellung siehe P0530804 (2021-05-24)
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Yuri Bilokin: Correction: sNd5-g8 6n1/3b4/6K1/4pR2/4k3/8/5P2/8 (3+4) h#4
1.Sh6 Rf7 2.Sg4 Rxd7 3.Kf4 Rd5 4.e4 Rf5# (2022-04-20)
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DSp
Adrian Storisteanu: Possible fix: wPd4->d2,+bPd3 (11+8). (2021-09-17)
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Yuri Bilokin: correction wKa7-b1, +bNd1 7k/5p1b/3p4/8/8/8/1Pn5/1K1n4 (2+6) h#7

1.f5 b4 2.f4 b5 3.f3 b6 4.f2 b7 5.f1=B b8=B 6.Bc4 Bxd6 7.Bcg8 Be5# (MM)

Excelsior (black)
Excelsior (white)
Promotion (Bb, 1, 1)
Model mate × 1 (2023-05-10)
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F111
Verst-4
Adrian Storisteanu: Possible fixes: wSc5 (instead of the wP) [1... Sb7]; or, +bRd1 (6+10). (2015-09-07)
Yuri Bilokin: Correction: wKa6, wRe3, wBe4, wPc6, wPd3, wPg6 – bKe6, bQb2, bBf1, bBh2, bSg3, bPd5, bPd6, bPf6 8/8/K1PpkpP1/3p4/4B3/3PR1n1/1q5b/5b2 (6+8) 1.Se2 d4 2.Sf4+ Bd3# (2021-05-10)
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Yuri Bilokin: you can see 1...e4 2.Kd1 e5 3.Ke1 e6 4.Kd1 e7 5.Ke1 e8=Q+ 6.Kd1 Qe2# (2023-01-14)
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Adrian Storisteanu: The problem is correct with wPa3->b2. Typo!? (2021-09-17)
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Adrian Storisteanu: Possible (quick) fix: bBd7->f1/d1,+bQd7 (5+10). (2021-09-17)
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klären Zwilling
Ladislav Packa: Lösung b) 1.kBa3-a2 kBc2-c4 2.kBa2-a1=kS! Le1-c3# (2023-02-25)
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Yuri Bilokin: correction wKg4-f4, -bBa6, +bRe7, then a1=c1 6N1/5Pr1/4k3/8/2P4K/8/8/8 (4+2) h#2*

1...f8=Q 2.Rd7 Qf6#
1.Rg5 f8=Q 2.Rf5 Qe7#

Echo mates (mirrored diagonally, 0, 0)
Promotion (QQ, 2)
Echo (2023-06-04)
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Yuri Bilokin: possibly wKg8-h8 7K/4p3/7k/6p1/8/5p2/6P1/8 (2+4) (2023-02-01)
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Yuri Bilokin: wBc5-b6, +bSg2, +bPb5, +bPf6 8/8/1B3p2/1p1P4/6K1/1n2b3/1p2k1n1/3bq3 (3+9)
1.b1=B d6 2.Bg6 d7 3.Be8 dxe8=B 4.Bd2 Bxb5# (MM) (2021-06-12)
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Erich Bartel: Nachdruck:
Die Schwalbe (2) XII 1969 (2013-02-22)
Yuri Bilokin: possibly wKb3-a2 7k/1p4pn/8/8/8/8/K4P2/8 (2+4) h#7
1.b5 f4 2.b4 f5 3.b3+ Ka3 4.b2 f6 5.b1=B f7 6.Ba2 f8=S 7.Bg8 Sg6# (MM) (2023-02-01)
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James Malcom: The engine seems to not like this edge case and refuses to animate it. Anybody any idea why? (2022-06-11, hidden)
James Malcom: Thanks for fixing, Olaf. (2022-06-12, hidden) more ...
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Anton Baumann: C+ Gustav 4.1d (2021-04-16, hidden) more ...
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No. 1126 HN
A.Buchanan: Consequential series movers were invented in 1979. (2015-11-11)
A.Buchanan: Although there is trivial retro consideration at the beginning, this is not a consequent problem. After Black's first four series moves, the position has no last move for White, which violates the consequent condition. This is just a regular seriesmover, with ep convention in the diagram, contrary to the initial post in 2004. (2020-12-10)
A.Buchanan: Not consequential series mover. The order of the first two moves is because Black thinks he must capture e.p. immediately. And then after e.p., White has no prior move. This is just a series mover. (2015-11-11, hidden)
A.Buchanan: Actually found one from 1976 now. (2015-11-11, hidden)
Anton Baumann: C+ Gustav 4.1d (2020-12-10, hidden) more ...
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0218
SCHRECKE: C+, popeye 4.87 (2022-09-12)
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0505 und 0862
SCHRECKE: C+, popeye 4.87 (2022-09-12)
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0457
SCHRECKE: C+, popeye 4.87 (2023-09-30)
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0493
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-01-20)
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0376
SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-01-20)
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0508
SCHRECKE: C+, Gustav 4.2a, Brute Force (2021-12-15)
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Anton Baumann: C+ Gustav 4.1d (2021-03-23, hidden) more ...
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Anton Baumann: C+ Gustav 4.2a (2022-03-12)
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Anton Baumann: C+ Alybadix (2021-12-28)
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T12T IM
vf: sBh4->h5, Sergej Radtschenko, 08.1990, Ceskoslovensky Sach, #7793
SCHRECKE: C+, popeye 4.87 (2023-03-16)
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SCHRECKE: Unlösbar (2023-07-14)
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Anton Baumann: C+ Alybadix (2023-03-29)
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SCHRECKE: C+, Gustav 4.1d, Brute Force (2021-01-12)
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shankar ram: Both here and in P1285517, Black's 6th move with mate is wrongly shown as 6.Db7. It should be 6.Dh7# (2021-05-28)
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Steen Christensen: Anticipated by / identical to Nixon (P0575588):
Turn the board 90 degrees anticlockwise. {wKb3,sKb5} becomes {wKf2,sKd2}. (2023-10-01)
Henrik Juel: 1... Kf3 2. Ke1 Ke2=
1. Ke1 Ke3 2. Kd1 Kd2=
C+ by Popeye 4.69 (2015-03-11, hidden) more ...
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Gunter Jordan: a) und letztlich auch c) sind vorweggenommen durch Nixon 1953 (P0575588) und auch durch Steudel 1958 (P1207519). (2015-03-09)
Steen Christensen: I do not see similarities between Nixon (P0575588) and a), except for the obvious ones: Identical material, stipulation and stale mate position. The problem is correctly not assigned the keyword “anticipated”, but “vgl. Nixon 1953” might be added as a comment. (2023-10-01)
Henrik Juel: a) and b) are C+ by Popeye 4.69
The first solution in part b) should have 2. Ka6, not Kc8
What does the stipulation in part c) mean? (2015-03-09, hidden)
Henrik Juel: Thanks for the clarification, Alfred
Part c) is also C+ by Popeye 4.69 using 'bed git hor' (2015-03-09, hidden) more ...
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prüfen?
SCHRECKE: Auf a7 muss offenbar ein wBauer stehen. (2024-03-24)
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SCHRECKE: C+, Gustav 4.1d, Brute Force (2021-08-10)
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SCHRECKE: Forderung: Grazer h=7 (2023-07-25)
Henrik Juel: C+ Popeye 4.61
(Popeye stipulation: reci-h=7) (2023-07-25)
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t/Sternblocks
t/4 differente S-Matts
Gefällt mir! -eb-
Anton Baumann: C+ Gustav 4.2a (2022-03-09)
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SCHRECKE: C+, Gustav 4.2e, Brute Force (2023-08-11)
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SCHRECKE: C+, Gustav 4.1d, Brute Force (2021-07-06)
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Anton Baumann: Dual: 5.Sf3+ Kf6 6.Kh4 d2 7.Kh5 nebst 8.Dg6+ hxg6# (2021-04-10)
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Henrik Juel: Part a is C+ by Popeye 4.61 (2016-03-04)
Steen Christensen: Cook part b: 1.Kf3 g8DT 2.Tf5~ DTg3# (2021-07-20)
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Anton Baumann: C+ Alybadix (2023-03-31)
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Klären ob Originalforderung h#2
Alfred Pfeiffer: Originalforderung war h#2. In der Lösungsbesprechung (fs-1, 01/1971, S.17) steht: "Korrekterweise sollte bei Aufgaben dieser Art ein Hinweis gegeben werden, daß sie Retroanalyse erfordern!".
Bei der Scheinlösung hat weiß keinen letzten Zug. (2017-11-23)
Henrik Juel: Giving a hint about retro analysis would spoil the fun (2017-11-23)
A.Buchanan: Quite right Henrik. Solvers should always be on their toes (2023-12-07)
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