Adrian Storisteanu: Possible fix: +bPc5. (2015-06-10)
Miguel Ambrona: The fix proposed by Adrian works.
Verified with Deadpos v.2.2. (2024-01-12)
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Anton Baumann: NL: R: 1.Ke8-d8 g7-g6 V: 1.Th1..6 Le7#
Korrektur in 'Die Schwalbe' 06/1989 S.78: +sBh7 (2022-12-30)
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Michel Caillaud: cooked by Stelvio 0.93 :
1.b4 c5 2.b5 c4 3.b6 c3 4.bxa7 d5 5.e4 d4 6.f4 d3 7.f5 dxc2 8.d4 g5 9.Lf4 c1=L 10.d5 g4 11.f6 g3 12.fxe7 gxh2 13.g4 c2 14.Lg3 Lf4 15.g5 c1=T 16.g6 Tc6 17.Dd3 f5 18.Sd2 h5 19.g7 Tg6 20.0-0-0 Sf6 21.Te1 h4 22.d6 h3 23.g8=L Th4 24.Lb3 L8h6 25.Ld1 Kf7 26.d7 Dh8 27.d8=T Le6 28.Td4 Sbd7 29.e5 Tf8 30.e8=T Kg8 31.Tc8 Lf7 32.e6 b6 33.e7 Lb8 34.e8=S f4 35.Te7 f3 36.Se2 f2 37.Tg1 h1=S 38.Da6 h2 39.Lh3 f1=S 40.Ta4 Sf2 41.Tc2 h1=L... (2022-12-20)
Paulo Roque: obs: 28...S8d7 (2008-11-22, hidden) more ...
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 29:20:32 Stunden. (hh:mm:ss)
Da NUPG C+
Notation: 1.b4 e5 2.d4 Le7 3.d5 f5 4.d6 h5 5.dxe7 d5 6.b5 Dd6 7.b6 Kd7 8.bxa7 b5
9.c4 b4 10.c5 d4 11.cxd6 c5 12.e8=T Sc6 13.Tf8 c4 14.e4 d3 15.Dg4 d2+ 16.Ke2 c3
17.Kd3 d1=L 18.Sd2 fxg4 19.Tf4 exf4 20.e5 Sd4 21.e6+ Kc6 22.e7 b3 23.e8=T b2
24.Te6 Lb3 25.Th6 Lf7 26.d7+ gxh6 27.d8=T Lce6 28.Td5 Te8 29.Kc4 b1=L 30.a8=T Lh7
31.Ta3 c2 32.Tg3 Se7 33.a4 Thf8 34.a5 Sg6 35.a6 Sh8 36.a7 fxg3 37.a8=T gxh2
38.T8a3 Lhg8 39.Th3 gxh3 40.Se4 Lxd5+
3 schwarze Läufer stehen zum Schluss auf den weißen Felder d5, f7, g8. (2023-09-21)
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Anton Baumann: 'Die Schwalbe' 08/1984 S.308: der Autor korrigiert: wBe5 statt sBe5 (2022-12-23)
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Kees: possible fix: Lb1=Sb1 -De1 +Ld1
White begins: 1.Kxb7 Lxe2 2.Kc8 La6#
(1.Txd8+ Kxd8 2.Kf8 Th8# illegal for white has no last move) (2023-06-07)
A.Buchanan: Your fix is good, Kees. It removes the cook, and sLd1 denies R: 1. c2xb3 as well as sLb1 did. Note R: 1 Sc6-d8 0-0+? as black castling rights were lost to let wK enter the back rank. (2023-06-08)
A.Buchanan: From a forward angle, your solution is cool Kees, addressing the cook in one half of the duplex. And sLd1 denies R: 1. c2xb3 as well as sLb1 did.
Note R: 1 Sc6-d8 0-0+? as black castling rights were lost to let wK enter the back rank. (2023-06-07, hidden) more ...
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Anton Baumann: Sollte eine Verbesserung von P0000777 sein.
Autorabsicht: "letzter Zug f7-f5 sonst retropatt! Jetzt ist aber die Rochade illegal, da der sTh8 nur über e8 aus seiner Ecke kommt. Daher Verführung 1. ... exf6 e.p. 2.0-0-0? Lxf4 3.Td7 a8=D#. Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 sg7#"
Aber: Der s h-Bauer kann auch auf h5 geschlagen worden sein; so konnte der sT über h6 aus seiner Ecke entkommen. Die vermeintliche Verführung ist also NL! (2022-12-14)
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Moldenhauer: Computerprüfung: C+ NUPG cooked Stelvio 1.11 in 1 Sekunde.
Keine Lösung: BP 17.0. BP 17.5 cooked.
Beispiel BP 18.0: 1.d4 Sa6 2.Kd2 Sb8 3.Kc3 d6 4.e4 Le6 5.Se2 Sd7 6.e5 Db8 7.g4 Kd8
8.g5 f5 9.g6 Lf7 10.gxf7 Kc8 11.Kc4 g5 12.Kd5 Lg7 13.Ke6 d5 14.f8L g4 15.Kf7 g3
16.Ke8 g2 17.Lg5 gxf1L 18.e6 Lh3
Beispiel BP 17.5: 1.d4 Sa6 2.Kd2 Sc5 3.Ke3 d5 4.Kf4 Le6 5.Ke5 Kd7 6.e4 Db8
7.g4 Kc8 8.g5 f5 9.g6 Lf7 10.gxf7 Sd7+ 11.Ke6 g5 12.Se2 Lg7 13.e5 g4 14.Lg5 g3
15.f8L g2 16.Kf7 gxf1L 17.Ke8 Lh3 18.e6 (2023-05-07)
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hans: only black move is fxg3e.p.
1. fxg3+ Kh5 2. h8D/T#
Why the keyword 'castling'?
Castling is illegal, for there are 7 black captures, and because white's last move is g2-g4, needs the wK let pass Rh1 (2012-05-18)
Anton Baumann: g2-g4 war nicht zwingend der letzte Zug! Schwarz ist patt, also unlösbar!
Korrektur in 'die Schwalbe' 12/2074 S.264: "wTa1 nach h1. Nun gewinnt nach 1. - fxg3 Weiss mit 2.hxg3+ Kh5 3.f4 Kh6 4.h8=T+! Kg7 und der umgewandelte Turm holt die störenden sBB ab, sodass Weiss rochieren kann und beweist, dass Schwarz nicht patt war. Schwarz versucht die Rochade zu verhindern, ist aber machtlos." (2022-12-06)
A.Buchanan: The forward play is more complicated, as Black can play more aggressively than 3. … Kh6 (2022-12-07)
Anton Baumann: Wenn Schwarz nicht 3.-Kh6! zieht, holt sich Weiss auf h8 die Dame, mit einfacherer Fortsetzung.
3.-g4? 4.f5 Kg5 (4.-g5 5.f6 ~ 6.h8=D) 5.h8=D Kxf5 6.De8 +_
3.-Kg4? 4.h8=D Kf3 5.Dxh3 g4 6.Dh2 Ke4 7.Dg2+ +_ (2022-12-09)
A.Buchanan: How should we show this kind of correction in PDB? A separate diagram? Or if the same, what is the original date - a correction is not a reprint? Happy to follow whatever clear protocol can be defined (2022-12-07, hidden) more ...
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See P0000674
Henrik Juel: The stipulation should probably read h#2.5: 1.exf6ep 0-0-0 2.Bxf4 Rd7 3.a8Q# (2003-08-18)
Anton Baumann: Letzter Zug war f7-f5, sonst retropatt. Die weissen Bauern haben 6x geschlagen. Der s a-Bauer wurde so ebenfalls als Schlagopfer benötigt und musste 2x schlagen, um sich umwandeln zu können. Weiss hat also auf h6 den s Bauern geschlagen. Unabhängig davon ob der sTe6 ein Umwandlungsturm ist oder nicht, muss der sTh8 über e8 aus der Ecke gekommen sein. Also ist die Rochade illegal!
Autorabsicht: 1. ... exf6 e.p. 2.0-0-0 Lxf4 3.Td7 a8=D# ist die Verführung!
Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 Sg7#
leider geht auch: 1. ... Kxf4 2.Te7 e6 3.Tb8 axb8=D,T# (2022-12-13)
A.Buchanan: I agree Anton. I suggest as a fix -bBa6 +bDc2. I think retro, solution and try all work correctly now. Do you folks agree? (2022-12-13)
Anton Baumann: Der Vorschlag von Andrew ergibt eine korrekte Lösung, aber nicht im Sinne des Autors: ohne einen s Stein auf a6 ist f6-f5+ als letzter Zug möglich, d.h. kein Retropatt weil Weiss a6-a7 zurücknehmen kann! 1. ... exf6 e.p. ist deshalb nicht zulässig. LB wollte aber dass f7-f5 der letzte Zug war, damit aber die Rochade illegal ist! (2022-12-14)
Anton Baumann: Mögliche Korrektur: a6=sL, g2=sD (2022-12-15)
A.Buchanan: Yes I agree. I had overlooked the unmove of wPa7 (2022-12-15)
A.Buchanan: The new form is now quite nice, and should be posted here in PDB. Alas I can't take much of the credit, all I spotted was that bPa doesn't have to be on the board: the number of black PCs is the same. (2022-12-15, hidden) more ...
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 26.0, BP 26.5. (2023-05-08, hidden) more ...
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Paulo Roque: obs: Cook: 26...D6b6 (2008-12-24)
paul: See P0001634 as correction. (2010-09-25)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 in 31:54:04 Stunden. (hh:mm:ss).
1 Lösung mit 20 cooks. (2024-01-07)
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Henrik Juel: Dual: ... 4.d4 La6 5.d5 Tc8 6.d6 Tc7 7.dxc7 Db8 8.c8L d6 9.Lf5 ... (2004-02-12)
Moldenhauer: Ergänzung Stelvio 1.2. Keine Lösung BP 19.0, BP 19.5. (2023-05-20, hidden) more ...
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paul: Cooked in 4: 1.Ke5-e6! b2×Ba1=B+ -2.Kf4×Pe5 e6-e5+ -3.Rc2×Sg2 Se1-g2+ -4.Re2×Pc2 & 1.Rxe1# (2023-06-14)
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 01:00:21 Stunde. (hh:mm:ss)
1.Sf3 a5 2.b4 g6 3.h4 Lg7 4.Th3 Lc3 5.Sd4 Ta6 6.Te3 Tc6 7.g3 Tc4 8.Lh3 d6 9.Kf1 Dd7
10.Kg2 Dxh3+ 11.Kf3 Lg4+ 12.Ke4 Sd7 13.Kd5 b6 14.Sb3 Td4+ 15.Kc6 Se5+ 16.Kb7 Kd7
17.Dh1 Ke6 18.Kc8 Kf5 19.Da8 Lh5 20.g4+ Kf4 21.f3 Kg3 22.Kd8 Kf2 23.Ke8 Ke1
Keine Lösung: BP 22.5.
Wenn die Forderung KBP 25.0 richtig ist dann Cooked. (2023-09-28)
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Henrik Juel: Cooked acc. to Euclide 1.01, e.g.,
1. Bc2-c4 Ba7-a6 2. Be2-e4 Be7-e6 3. Bf2-f4 Bh7-h5 4. Dd1xh5 Ke8-e7 5. Dh5-b5 Th8-h4 6. Ke1-e2 Th4-g4 7. Bh2-h4 Ba6xb5 8. Bh4-h5 Bb5-b4 9. Bh5-h6 Ke7-d6 10. Bh6-h7 Kd6-c5 11. Th1-h6 Tg4-h4 12. Bh7-h8=D Bg7-g6 13. Dh8-c3 Bb4xc3 14. Ke2-f3 Bc3-c2 15. Bf4-f5 Bc2xb1=T 16. Bf5xg6 Kc5-b4 17. Bg6-g7 Be6-e5 18. Th6-c6 Bf7-f6 19. Lf1-e2 Sg8-h6 20. Bg7-g8=D (2017-11-22, hidden)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 2 Sekunden.
Keine Lösung: BP 18.5, BP 19.0. KBP = 19.5. (2023-04-15, hidden) more ...
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A.Buchanan: The unsoundness in this non-unique PG is that 6 promotions (apparently intended) are not forced (2023-05-11)
Moldenhauer: Computerprüfung: C+ Cooked Stelvio 1.11 48 Sekunden.
Keine Lösung: BP 36.0, BP 36.5. (2023-05-10, hidden) more ...
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Moldenhauer: Ergänzung: Stelvio 1.2. Keine Lösung: BP 14.0, BP 14.5.
BP 15.0 und BP 15.5 cooked.
Notation BP 15.0: 1.Sc3 Sc6 2.a4 b5 3.Ta3 La6 4.d3 Db8 5.Lg5 Db6 6.Lf6 0–0–0
7.e3 Kb7 8.Dg4 Dc5 9.Le2 Kb6 10.Lf3 Ka5 11.Sce2 Tb8 12.Tc3 Sd8 13.La8 De5
14.Tc6 Tb7 15.Tb6 c6
Notation BP 15.5: 1.Sc3 Sc6 2.a4 b5 3.Ta2 La6 4.Ta3 Db8 5.d3 Db6 6.Lg5 0–0–0
7.e3 Kb7 8.Dg4 Dc5 9.Le2 Kb6 10.Lf3 Ka5 11.Sce2 Tb8 12.Tc3 Sd8 13.La8 De5
14.Tc6 Tb7 15.Tb6 c6 16.Lf6.
Alles schlagfrei und ohne Umwandlung. (2023-05-20)
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Kostas Prentos: A correction was published in Phenix, 2009 (No.186/Pg.7979)
Solution:
1. Sf3 f5 2. Sd4 Kf7 3. Sb5 Kg6 4. Sd6 exd6 5. e4 Le7 6. e5 Lh4 7. e6 Dg5 8. e7 Dg3 9. e8=T Se7 10. Lc4 Tf8 11. 0-0 Tf6 12. f3 Kh5 13. Te1 Th6 14. Lf7+ Sg6 15. T1e6 dxe6 16. Txc8 Sd7 17. Th8 Txh8 18. Le8 (2022-12-08)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.2 00:33:56 Minuten. (hh:mm:ss)
Keine Lösung: BP 17.5, BP 18.0.
Beispiel: 1.Sf3 f5 2.d4 Kf7 3.e4 g5 4.Lc4+ Kg6 5.e5 Lg7 6.e6 Lxd4
7.0–0 Lxb2 8.Dd6 exd6 9.e7 Kh5 10.e8S Se7 11.Te1 Tf8 12.Sxg5 Tf6
13.f3 Th6 14.Lf7+ Sg6 15.Te7 Lf6 16.Se6 dxe6 17.Td7 Lh4 18.Lg5 Dxg5
19.Sg7# (2023-05-30)
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Alfred Pfeiffer: Die Korrekturfassung 5674v mit gleicher Forderung (fs-99, S.38): 1ss5/d1b5/lbb1b1b1/1bl5/k2t2b1/2t5/BBB2BB1/TSLDKLST soll die Duale der Urfassung (4-fach mögliches Ziehen von g7-g6) nicht mehr enthalten. (2014-11-06)
Henrik Juel: The correction mentioned by Alfred is still cooked, acc. to Euclide 1.01 (2017-11-24)
Joost de Heer: For correction see P1067393. (2023-08-18)
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This is an tutorial example of a PG deliberately including a cook.
Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 13 Sekunden.
Keine Lösung: BP 14.0 bis 15.0 und 16.0. (2023-05-08, hidden) more ...
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 13 Sekunden.
Keine Lösung BP 19.0, BP 19.5 Cooked.
Beispiel BP 19.5:
1.d3 Sc6 2.Le3 Tb8 3.Ld4 Sf6 4.e3 Sh5 5.Dg4 b5 6.Ke2 La6 7.De6 b4
8.g4 Lb5 9.Lg2 b3 10.axb3 f6 11.Ta6 La4 12.bxa4 Sa5 13.Td6 Tb3
14.Lb7 Ta3 15.Kf3 Ta1 16.Sa3 Td1 17.Se2 Td2 18.Tb1 a6 19.Sc1 Te2 20.Lc8 (2023-05-03, hidden) more ...
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Moldenhauer: Computerprüfung: Cooked Stellung Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 9.5, BP 10.0.
Eine Notation von Stelvio:
1.Sf3 e6 2.Se5 Df6 3.Sxd7 Kd8 4.Sxb8 Ld7 5.Sc6+ Kc8 6.Sxa7+ Kb8
7.a4 Lb5 8.axb5 Txa7 9.b6 Ka8 10.bxc7 Ta4 11.Txa4#
Schlüsselwort Rochade? (2023-05-02)
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Henrik Juel: Cook: -1.a7xFb8=L Kb7xTa8, 1.Ka6 axb8DT#. (2004-04-01)
Mario Richter: vgl. mit P0006207 (2009-12-24)
Adrian Storisteanu: Possible (expensive, 50% more units) fix: Ka8 Bd8 / Kc8 (2+1) -- R: 1.e7xRd8=B Rd7-d8 & 1.Rc7 e8Q#. On the plus side, there is a wK in the picture now. (2016-02-10)
Adrian Storisteanu: See P1389595 (not clear which was first). (2024-03-19)
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s.a. 32Pe1A

Korrekturversuch zu P0005036
hans: Good motivation to capture Qd8 on the spot, to make long castling possible, which is needed to retrack the captured bRg7 just on time. Also a minor-promotion. I like this one, and I think the stipulation asks for which move black queen makes.
R: -1. …Ld8xLc7+ -2. Lb6c7+ Kc8b8 -3. Lc7b6 Kb7c8 -4. Lb6c7 Lc7d8 -5. Lg1b6 Lb8c7+ -6.Lb6g1 Kc8b7 -7. Ld8b6 Kb7c8 -8. d7d8=L Lh7g8 -9. e6xSd7 Sc5d7 -10. Sc3a4 Sa4c5+ -11.Sd1c3 d7d6 -12.Sc3d1 Le5b8 -13.Se4c3 Lg7e5 -14.Sf6e4 c7c6 -15.Sg8f6 Lh6g7 -16.g7g8=S Kc8b7 -17.f6xTg7 Tg8g7 -18.e5e6 Td8g8 -19.e4e5 0-0-0!! -20.e3e4 Lf8h6 -21.f6f5 g7g6 -21.f4f5 Le4h7 -22.f3f4 Lb7e4 -23.f2f3 Lc8b7 -24. Kb5a5 b7xSa6 and cage can be undone while black plays only with Ta3 and Sa4.
captures white axSb, SxDd8, exSd7, fxTg7, black hxSg, bxSa6, LxLc7 (2012-07-26)
Henrik Juel: Very similar to P0005036 and with identical stipulation question:
What was the first move by black queen and by black king.
Ceriani's abbreviations for move, queen, king, black, (and white) are t., D, R, n, (and b); in his ortho reconstruction problems the color abbreviations are capitalized, e.g. N=11 meaning 11 black moves (2012-07-26)
Thomas Volet: This composition appears on p.197 of Ceriani's 1961 book as his correction of P0005036 (which appeared in his earlier book). On that page, he discusses the clever cook in P0005036 ("ma questa bella posizione e demolita") with the WhP unpromoting at h8, uncapturing to the g file, and uncapturing back to the h file. (2012-08-02)
Mario Richter: I'm sorry to say this, but Ceriani's correction attempt still leaves room for cooks ... (2023-06-30)
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Sally: Der letzte Zug war: Bc2 - c4!
Nr.138 200 Ausgewählte S .Probleme T. Kardos(W. Fentze1983) (2010-09-30)
A.Buchanan: All 9+1 captures are visible by pawns. Note R: 1. Kd1-e1 d3xe2 is illegal because it costs 2 more captures for Black. So e.p. is unconditionally legal, requiring no AP justification. I see no reason why White has lost castling rights. Indeed someone might make a demo game ending with c2-c4. So this one seems cooked. Neither WinChloe nor yacpdb contains this problem. (2022-01-08)
A.Buchanan: If this is AP, it will be rendered sound by removing wPa2. But is this the intention? Does anyone have access to the collection of selected Kardos problems mentioned? (2024-01-25)
A.Buchanan: Surely 2. Rd1 works equally well? I don't see a retro reason why White has lost castling rights. I don't think this is AP. What's going on? (2021-11-26, hidden) more ...
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Henrik Juel: White captured [sLc8] on c8 and axb, so last move was either b2-b4 or d2-d4
C+ Popeye 4.61, except for the promotion dual (2020-08-03)
A.Buchanan: There's no keyword for an untolerated dual promotion! (2020-08-03)
Henrik Juel: Yes and no, Andrew
What about the cooked label? (2020-08-03)
A.Buchanan: Hi Henrik: this is obviously intentional. I tried to replace the pawn promotion with wDh8-h7+ but I ran out of pieces. I guess that Bebesi had the same difficulty. The word "cooked" seems too strong to me, but it's obviously a significant defect though (2020-08-04)
A.Buchanan: The bar is higher for helpmates than other problems, and this is unsound. However it’s not cooked but dualled. Yet all we have is “cooked” so I suppose it is in PDB-speak. I’ve added that and removed the 2.1… (2023-08-07)
Joost de Heer: Is there an 'unsound' keyword? (2023-08-07)
A.Buchanan: Hi Joost: no there isn't. It can be added, the challenge for any new common keyword would be populating it for more than a few values. (2023-08-07)
A.Buchanan: This time round, I've managed to find a fix. Bebesi's correction in WinChloe 66992 has non-standard wL. I have a feeling I've seen a similar one by Andrey Frolkin & someone else recently with a different matrix, and also non-standard wL. I've sent my effort to Joaquim Crusats at Problemas for his assessment (2023-08-08)
Henrik Juel: better than 2.1...: one solution in each of two parts (2023-08-06, hidden)
A.Buchanan: This time round, I've managed to find a fix. As far as I can see, it hasn't been corrected hitherto. I'm offering the new one to Problemas again. I have a feeling there was one by Andrey Frolkin recently with a similar idea, but had non-standard material (wLL) (2023-08-07, hidden)
A.Buchanan: This time round, I've managed to find a fix. Bebesi's correction in WinChloe 66992 has non-standard wL, which is still not good: https://www.janko.at/Retros/d.php?ff=3r1b2/1p1p3B/B1r3p1/1PP1Kp2/pPkPp3/p1p4s/S7/2q5 I have a feeling there was one by Andrey Frolkin recently with a similar idea, and also non-standard wL. I've sent my effort to Joaquim Crusats at Problemas for his assessment (2023-08-07, hidden) more ...
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See P0000642
Mario Richter: Hier stimmt was nicht: In der Schwalbe 08/1977 erschien als Nr. 2202 die folgende Aufgabe: 8/Pp2p1p1/Bq1p3p/4p3/1p6/n4pP1/1PPPPr1k/R3KQ1n/; h#2, b) sBb4 nach b5 und der AL a) 1.Tf2-g2 0-0-0 2.Tg2xg3 Df1xh1#; b) 1.Sa3-b1 Ta1-a4 2.Sh1xg3 Ta4-h4#, die aber in a) und b) durch 1.f3xe2 La6xb7 2.Tf2-g2 Df1xg2# nebenlösig war.
Ferner ist in nebenstehendem Diagramm sowohl in a) als auch in b) sehr wohl die w0-0-0 möglich
(Weiß hat als letzten Zug Sc7-a8, Schwarz entwandelt auf b1 und h1, Weiß auf f8); und selbst unter der Annahme, daß die w0-0-0 nur mit sBb4 möglich sei, wäre die Aufgabe mehrfach nebenlösig, z.B. durch 1.Ld5-c4 0-0-0 2.Bd6-d5 Df1xh1# (2010-05-29)
A.Buchanan: Here, the source was incorrectly given as "8202v Die Schwalbe 08/1977". No, that was where the original problem appeared, with a "v" added. So when, as here, we suspect the diagram has a typo, we would have to look through many back issues of Die Schwalbe! It turns out that this correction (itself cooked) was published (as text) in p.148 Die Schwalbe 65 10/1980. I have corrected the source. The reference to the earlier buggy problem can then cleanly be put in the "after" field. The actual typo was missing bPc7. (2024-03-06)
A.Buchanan: With the typo fixed, I think the retro-logic works ok, but there is a single cook family, fortunately repairable (2024-03-06)
A.Buchanan: It's a bad habit for folk to post corrections without indicating where they were published. Here, the source was incorrectly given as "8202v Die Schwalbe 08/1977". No, that was where the original problem appeared, with a "v" added. So when, as here, the diagram is incorrect, one has no clue where to look in past issues of Die Schwalbe! It turns out that this was published as text in p.148 Die Schwalbe 65 10/1980. I have corrected the source. The reference to the earlier buggy problem can then cleanly be put in the "after" field. The actual typo is bP missing from c7, but I hope you will agree that the real failure is the lack of indication of the true source, which rendered the detection of the innocent typo unnecessarily hard. You're welcome (2024-03-06, hidden)
A.Buchanan: I think the retro-logic works ok, but there is a single cook family. I think it can be fixed by swapping sDb6 & wBc6, and adding sBa4. So NBr5/N1p1p3/1Pqp2p1/2bb4/pp6/n1pP1rP1/1PP1P2k/R3KQ1n. Please confirm that this works (2024-03-06, hidden) more ...
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Henrik Juel: Diagram error? -1... g5xSf4, 0... Le5 1.Sd5#. But the symmetrical solution -1... e7xSd6 etc. seems legal too. (2004-03-22)
A.Buchanan: Yes bBf8 would have died at home but bPc could capture twice to promote on c1 or e1 (2023-03-31)
A.Buchanan: Odd. The diagram is thematically symmetric, so not just missing a unit. WinChloe has the same diagram. I wondering if we are missing something from the stipulation. Does anyone have access to an archive for this? It's also odd that it was reprinted: surely there would be a comment there? I suppose it can be rendered sound by placing wTTSS somewhere symmetric. Then Black runs out of pawn captures if sLd4 is promoted. In any case, wPh7xg8=S is affordable. (2023-04-01)
Mario Richter: The reprint in 'Problem' says: "Ne e7xSd6 radi nepravilne postave" ("Not e7xSd6 because of illegal Position.") The diagram in 'Problem' is the same as here in the PDB.
How about adding black pawns on b7 and g2? (2023-04-01)
A.Buchanan: Hi Mario. I considered adding sBb5e2. That gives 10 needed captures by Black and 6 by White, including sLf8. So the try fails for *two* reasons. sBb7g2 fails even more badly. I thought that also adding wTTSS although more uneconomical in pieces defeats the try for only one reason. Also I liked the trick h7xg8=S. And it must make it harder to find the solution. Does Rawbats do simple retractors? Can you try wTa0h1 wSa7g1 as a risky initial bid, and then back off until there are no cooks? (2023-04-01)
Mario Richter: Hi Andrew, I was well aware that with adding sBb5e2 or sBb7g2 the try fails for two reasons (for [subjective] optical reasons I preferred sBb7g2 to sBb5e2]). The reason I suggested the addition of 2 black pawns was the following consideration: I cannot believe that such an experienced Retro Composer like C.E. Kemp could make such a big mistake. It's much more likely that the printer made an error by omitting two pieces.
On the other hand, if we want to correct the problem including a unique reason for the try to fail, your suggestion of adding wTa8h1+wSa7g1 (FEN=R7/N4pp1/1P1pnkp1/5n2/3b1p2/1PP5/1PP2P2/K5NR) works - rawbats gives exactly R: 1. ... Bg5xSf4, dann 1. Ld4-e5 Sf4-d5# (and says about R: Be7xSd6 - "Illegal: Mehr schwarze Bauern-Schlaege notwendig als weisse Steine fehlen") (2023-04-01)
A.Buchanan: Hi Mario, thanks for this, I agree with everything you say. It's curious that the reprint in Problem did not pick up the error. Any overlooked black units are more likely to have been lurking on dark squares, maybe sBb4d2. This forces the try to only fail for 1 reason, albeit rather banal. Does this work. Or the +wTTSS mentioned forces the solver to do a little more counting, and there is a lot more solving space to search for the h#1. Rawbats' conversational skills seem to have improved: have you integrated it with ChatGPT maybe? (2023-04-02)
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Korrekturversuch s. P0003009
Henrik Juel: If you want a great solving challenge, this is the retro for you.
If you need a hint:
[Dd8] never moved, and Black castled (as you may have guessed).
If you need another hint:
Last move was Ld8xLc7+.
I gave up, but Nikolai told me the solution:
-1... Ld8xLc7 -2.Lb6 Kb7 -3.Lc7 Kc8 -4.Lb6 Lc7 -5.Lc5 Lb8 -6.Lb6 Kb7 -7.Ld8 Kc7 -8.L=d7 Kd8 -9.e6xSd7 Sb6 -10.Sc5 Sa4 -11.Sd4 d7 -12.Sf6 Lh7 -13.Sg8 Lf4 -14.Sf6 Lh6 -15.Sg8 c7 -16.S=g7 Kc8 -17.f6xTg7 Tg8 -18.e5 Td8 -19.e4 0-0-0 -20.e3 Lf8 -21.f5 g7 -22.f4 Le4 -23.f3 Lb7 -24.f2 Lc8 -25.Kb5 b7xSa6 etc. (2012-07-22)
Mu-Tsun Tsai: Once I heard "great challenge" I started working. But I came to a complete different conclusion. Not only [Qd8] can move, but the first move of the black king need not be castling. Here's the proof game. After playing
1.c3 h5 2.b4 Rh6 3.Na3 Rg6 4.Nc2 Rg3 5.hxg3 Nf6 6.Rb1 Ng8 7.Na1 Nf6 8.e3 Ng8 9.Bd3 Nf6 10.Rb2 Ng8 11.Bb1 Nf6 12.Qb3 Ng8 13.Rc2 Nf6 14.Qb2 Ng8 15.Rh4 Nf6 16.Rd4 h4 17.Rd5 h3 18.Ra5 h2 19.Ra3 h1=Q 20.Rb3 Qh5 21.Nf3 Qa5 22.Ke2 Qa3 23.Kd3 Na6 24.Kc4 Nc5 25.Kb5 Na4 26.Ne5 Nh5 27.Nd3 Nf4 28.Nc5 Nd3 29.Na6 bxa6+ 30.Ka5 Bb7 31.e4 Qb8 32.e5 Bc8 33.e6 Qb7 34.g4 Qf3 35.g5 Qh3 36.g6 Qh7 37.gxh7 Nf4 38.h8=N Bb7 39.Ng6 Be4 40.Ne5 Bh7 41.Nc4 Bg8 42.Ne3 Ng6 43.Nc4 Nh8 44.Ne3 g6 45.Nc4 Bg7 46.Ne3 Be5 47.Nc4 c6 48.Ne3,
you could either play
48...O-O-O 49.Nc4 Bb8 50.Ne3 d6 51.Nc4 Rd7 52.exd7+ Kb7 53.d8=N+ Kc7 54.Ne6+ Kb7 55.Nb6 Nc5+ 56.Na4 Ne4 57.Nc5+ Kc7 58.Nd7 Kb7 59.Nb6 Ng3 60.Nd5 Bc7+ 61.Nb6 Bd8 62.fxg3 Kc7 63.Nd5+ Kb8+ 64.Nc7 Bxc7+,
which is castling version, or,
48...Kd8 49.Nc4 Kc7 50.Ne3 Rd8 51.Nc4 Kc8 52.Ne3 Bb8 53.Nc4 d6 54.Ne3 Rd7 55.exd7+ Kb7 56.d8=N+ Kc7 57.Ne6+ Kb7 58.Nc4 Bh7 59.Nb6 Nc5+ 60.Na4 Ne4 61.Nd4 Kc7 62.Ne6+ Kc8 63.Nd4 Kb7 64.Ne6 Bg8 65.Nc5+ Kc7 66.Nd7 Kb7 67.Nb6 Ng3 68.Nd5 Bc7+ 69.Nb6 Bd8 70.fxg3 Kc7 71.Nd5+ Kb8+ 72.Nc7 Bxc7+
which is none castling version. Both reach the diagram position.
Why am I feeling cooking Ceriani's problems too much lately? (2012-07-22)
Mu-Tsun Tsai: Also in your retraction, -11.Nd4 should be -11.Ne4 I believe? It seems like your retraction also works just fine, and looks like it should be the intended solution. (2012-07-22)
Henrik Juel: Yes, the intended solution should have -11.Se4, and it can be shortened a bit.
I am impressed by your cook, Mu-Tsun, with two white pawn captures on g3 and promotion on h8, but it is also a little sad that a seemingly fine problem has been rendered worthless.
Probably several more Ceriani problems will be cooked, because they were not scrutinized well enough by testers and solvers in the old days; now, when I finally have cracked a Ceriani nut, I have no energy left to search for errors (2012-07-23)
Mu-Tsun Tsai: I've been thinking about how this problem might be fixed, but unfortunately I cannot come up with anything other than adding extra assumptions in the stipulation, for example "g3 pawn came from h2". The structure of this one is good, and either method of releasing the position (mine or the intended one) is quite subtle, so I feel sad about have to cook this one as well. (2012-07-23)
Thomas Volet: In his 1961 book Ceriani discusses the cook with the WhP unpromoting on h8 and uncapturing to the g file and back to the h file, and gives P0003009 as the corrected diagram position. (2012-08-02)
Mu-Tsun Tsai: This is a really late comment, but I do think this will make a great problem by changing the stip to "You don't know the first move of the black queen nor the black king"! (2023-06-29)
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"These promoted piece games are anticipated in general idea by CMF" (TRD)
Moldenhauer: Computerprüfung: C+ KBP 5.5 cooked in 1 Sekunde.
Keine Lösung: vor BP 5.5.
Beispiellösung:1.g4 d5 2.g5 h6 3.gxh6 Dd6 4.hxg7 Ddxh2
5.gxh8=T Dxh1 6.Txh1 (2023-03-22)
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Mario Richter: Ist die Stellung korrekt? Es geht auch (mit vielen Variationen): R: 1.Df7xDf8 Bg2xDh1, dann 1.g2xh1=T e7xf8=L/S= (2009-12-08)
A.Buchanan: I don't see any obvious typo or fix. This is not in WinChloe (2024-01-14)
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A.Buchanan: In both candidate solutions, counting of wPs after retraction on means it must be an original unit uncaptured on h1 so can only be wS.
The solution for b) does not work for a) because, if one retracts two promoting moves, then wPs have captured 9/10 missing Black units, while Black has 1/2 accounted for by wBc1. However wPg & bPg passed one another, which requires two more captures by a single player.
In b) the exciting retraction Qa8xh8 is blocked, however now the free pawn captures to explain g-file are shifted from 1+1 to 2+0, so bPg can pass wPg. But the vacation of d2 allows a new cook. This is fixable by shifting wPb instead of wPd. We don't need to shift it to an adjacent file: the solution is still ok if the balance is 2+1.
So "b) wPb2 to b7" fixes the cook, and it's validated by deadpos. (2024-01-15)
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Nikolai Beluhov: Diagram is now correct (missing wRe8 and bBf8 restored). (2011-05-05)
Nikolai Beluhov: This problem seems to be cooked by 1. ... Qd6:Nb8+ 2. Kc7-c8 Q~d6+ ...
Fortunately, this flaw is very easy to fix: just relocate bQb8 to c7, as in
L. Borodatov (correction)
k1KRRb2/P1qpp1p1/rpp5/p7/5p2/3PP3/P1Pp1PP1/4Bb2
(http://www.janko.at/Retros/d.php?ff=k1KRRb2/P1qpp1p1/rpp5/p7/5p2/3PP3/P1Pp1PP1/4Bb2)
(11 + 13) How often did the wK visit c7 at least? (2011-05-06)
Henrik Juel: The wK visited c7 at least four times, twice as shown in the indicated retroplay, and twice to let wK pass the two white rooks en route to f7, with Le1 screening on b8. (2011-05-06)
Anton Baumann: Informalturnier 1986 (PB in 'Die Schwalbe' 06/2011 S.124)
Auszeichnung: 2.Lob; ausgezeichnete Fassung: mit sDc7 statt b8 (2023-01-04)
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Korrektur siehe P0008426.
James Malcom: What is the cook? (2023-05-02)
paul: Missing wQd1. (2010-05-04, hidden) more ...
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siehe 9200
mri: Mehrere NL, z.B. R: Kc6xLd6 Lc5xDd6 V: 1.Ka5-a6 Dd6xc5 oder R: Kc5xTd6 Tb6xDd6 V: 1.Tb6-c6 Dd6xc6 (2007-10-25)
Mario Richter: s. P0005054 (2009-12-31)
A.Buchanan: Reasonable to suppose that P0005054 is the correction (2024-01-15)
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Adrian Storisteanu: Two wPs (e.g. +wPe4,+wPh4) can restore the glory to this nice, thoroughly mixed AUW... (2015-06-10)
A.Buchanan: Or instead +wPa5 or +wPc5. In either case, free to "downgrade" wBg1 to wS. This is more economical than adding two pawns, but separates the two islands conceptually, as b6 no longer needs to be controlled by wBg1. (2024-01-15)
mri: cooks: R: Da4-a8 Sf2xLh1 V: 1.Kxb6 Lxg2

R: Da4-a8 Sf2xTh1 V: 1.Kxb6 Th7 (2004-11-17, hidden) more ...
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Adrian Storisteanu: Possible fix: +wPh5 (4+4). (2015-08-05)
A.Buchanan: A cool fix which keeps miniature status is -bSd8 + bRd8 (2024-01-15)
A.Buchanan: Alas my suggestion is anticipated by P1202424, ah well. (2024-01-16)
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paul: Correction of P0005875. (2011-06-02)
paul: Dans une version ultérieure, Mr.Leroy a remplacé 30.-a6 avec 30.-a5. Dans un e-mail, il m'a expliqué: La version de ce problème avec 30...a6 a été démolie par Thierry le Gleuher, d'une manière presque plus compliquée que la solution
de l'auteur. En résumé de sa démolition, le PN en a6 peut n'avoir joué qu'un seul coup, mais par b7xa6!
La version avec le PN en a5 est donc une correction de ce grave défaut.
Donc, Thierry, quel était cette démolition explicitement? (2011-06-09)
paul: Démoli(T. le Gleuher) : 1.c4 g5 2.Da4 g4 3.D×a7 g3 4.f4 g×h2 5.Df2 Ta3 6.Rd1 Tf3 7.a4 e5 8.a5 e4 9.a6 e3 10.Ta5 e×f2 11.Ch3 Cf6 12.Tg1 h×g1=C 13.Th5 Ce4 14.c5 Cg5 15.c6 Fc5 16.e4 0-0 17.Fc4 f1=F 18.F×f7+ Rg7 19.e5 Fb5 20.Fc4 Tf6 21.Ff1 Ce2 22.Cg1 Dg8 23.e6 Th6 24.Th1 Th2 25.e×d7 h5 26.d8=D Fe6 27.Dd3 Fa2 28.Dc2 Db3 29.Re1 Rh6 30.Dd1 b×a6 31.Ca3 Cd7 32.Cc2 Ff2?
La correction avec 30.-a5 a été publié dans Europe-Échecs 452/1997. (2011-06-12)
Joost de Heer: The position with a5 is also cooked:
1. h4 Nf6 2. h5 Ne4 3. h6 Ng5 4. hxg7 h5 5. a4 h4 6. a5 h3 7. a6 h2 8. axb7 hxg1=N 9. Ra6 Rh2 10. Rg6 a5 11. e4 fxg6 12. Qh5 Kf7 13. e5 gxh5 14. e6+ Kg6 15. exd7 e5 16. Na3 e4 17. c4 e3 18. c5 e2 19. c6 Bc5 20. f4 Qg8 21. d8=N Be6 22. Nf7 Nd7 23. b8=Q Ba2 24. Qb3 exf1=B 25. Ne5+ Kh6 26. Qd1 Qb3 27. g8=B Bb5 28. Bc4 Re8 29. Bf1 Ne2 30. Nf3 Re3 31. Ng1 Rf3 32. Nc2 Bf2# (2023-07-26)
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Kostas Prentos: Here is a computer tested correction: PG 12,5

1. a4 e6 2. Ta3 Ld6 3. Tb3 Lg3 4. Tb6 axb6 5. hxg3 Txa4 6. Th6 Ta1 7. Tg6 hxg6 8. Sa3 Th1 9. Sc4 Dh4 10. Sa5 bxa5 11. e3 b6 12. Dh5 gxh5 13. gxh4

C+ Euclide v1.11 (42 h 2 min 37 sec} (2022-12-08)
Michel Caillaud: cooked by Stelvio 0.92 :
1.a3 e6 2.h4 Lxa3 3.Sxa3 h5 4.Sb5 Th6 5.Ta6 Tf6 6.Tb6 Tf3 7.Sd4 Th3 8.Sc6 Txh1 9.Sxd8 axb6 10.Sc6 Ta1 11.Sa5 bxa5 12.g3 b6 13.Lg2 Lb7 14.Kf1 Lxg2+ 15.Ke1 Lh3 (2022-12-03, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Stelvio 1.2 03:15:11 Stunden. (hh:mm:ss)
C+ Euclide 1.11 01:00:47 Stunde.
Stelvio und Euclide finden nur 1 Lösung! Warum cooked?
Keine Lösung: BP12.5, BP 13.0.
Notation: 1.a4 Sf6 2.a5 Tg8 3.a6 bxa6 4.Txa6 Lb7 5.Txf6 a5 6.Txf7 Sa6
7.Txf8+ Kxf8 8.h4 De8 9.h5 Dxh5 10.f3 Dxf3 11.Sh3 Da3 12.e3 Te8
13.Lxa6 Lc8 14.0–0+#
Epaulettenmatt durch Rochade ist OK. (2023-05-28)
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Moldenhauer: Die hier genannte Lösung Cook: hat am Ende die schwarze Dame auf h4
und den schwarzen Springer auf e7. Die FEN sind anders. Bitte löschen.
Autorenlösung? FEN?
Computerprüfung: FEN:1Q4nb/2p2rrp/2p1kpbp/5p1q/p1n5/8/2P1PP1P/RNBQKBNR
Cooked Stelvio 1.4 nach 3 Sekunden. Keine Lösung: BP 24.5.
Stelvio Beispiel: 1.a4 Sc6 2.d4 Sa5 3.d5 Sc4 4.d6 a5 5.g4 b5 6.axb5 a4
7.b6 f5 8.b7 Kf7 9.b8=S Ke6 10.Sc6 De8 11.g5 dxc6 12.d7 Dh5 13.d8=D Ld7
14.g6 Le8 15.D8d2 Sf6 16.Dh6 gxh6 17.g7 Lg6 18.g8=T Lg7 19.Tf8 Tg8
20.b4 Lh8 21.b5 Tg7 22.b6 Sg8 23.Tf6+ exf6 24.b7 Tf8 25.b8D Tff7 (2023-07-29)
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paul: Cooked by: 1.h3 d5 2.Rh2 Bxh3(ph2) 3.Sxh3(Bc8) Bxh3 4.g4 a5 5.Bg2 Bxg4(pg2) 6.c4 a4 7.b4 e6 8.Bb2 Bxb4(pb2) 9.Sa3 Ra5 10.Qxa4(pa7) Qc8 11.QxKe8(Ke8) (suicide of wQ)b5 12.Sc2 dxc4(pc2) 13.a3 Bxe2(pe2) 14.Ra2 Bxa3(pa2) 15.Kd1 b4. (2010-06-19)
Arnold Beine: Here is something wrong. Either Paul's cook does not work, because 10.Qxa4(Pa7)+ is a check (10...Qc8??) or in the given condition "Rex inclusive" must be added. (2023-07-26)
paul: Right, Arnold. Anyway, the proof game is cooked: 1.a4 e6 2.Sa3 B×a3 3.b4 a5 4.Bb2 a×b4(b2) 5.Ra2 R×a4(a2) 6.e4 Ra5 etc, as in author's solution: 7.Ba6 b×a6(Bf1) 8.B×a6(a7) d5 9.B×c8(bBc8) Bb7 10.c4 Qc8 11.Qc2 d×c4(c2) 12.Se2 B×e4(e2) 13.h3 Bf5 14.Rh2 B×h3(h2) 15.Kd1 B×g2(g2). (2024-03-13)
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Michel Caillaud: cooked by Stelvio 0.93 :
1.a4 Sa6 2.Ta3 Sb8 3.Tg3 c6 4.Tg6 hxg6 5.Sh3 Th4 6.Sf4 Tg4 7.h4 Tg3 8.Th3 Ta3 9.h5 Ta1 10.Ta3 Da5 11.c3 Db4 12.Db3 Dc5 13.Db6 axb6 14.Kd1 Ta5 15.Ta2 Tb5 16.h6 Tb3 17.h7 Ta3 18.b3 Dxf2 19.Tc2 T3a2 20.h8=T Tb2 21.Th3… (2022-12-19)
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Gerald Ettl: Weiss jemand wie die Autorloesung geplant war? (2023-09-28)
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VL: Unjustified ep-capture because of –e3xd4. For correction it suffices e.g. to add bSe3.
This is possibly a unique 2-move helpmate with ep-key and 2 solutions of the type 1.2.1.1 (at least so currently in the PDB).
Publi?ation details. H551 in the Helpmate section of the July issue, with no preview. Here is the text of the solution published in the Nov.-77 issue: “…White may have just played P(e3)xd4. Very few solvers have spotted this. Full marks for claim of no solution with correct reasoning but, in fairness, one mark for composer's intention. White cannot have just played K(a5)a6 in reply to Pxb6+ as too many captures by Black are required.” (2023-12-01)
A.Buchanan: Yes Valery you are right this must be cooked. However, I'm not a fan of +bSe3 because it spoils one of the model mates. Separately, I mildly prefer bPf7 to bPe6. (2023-12-03)
VL: I do prefer bPf7, as well. I don't see any way to save at least one of the properties Kindergarten Problem and Model mate (2) after fixing (what for these keywords have been added just now?). Besides, Andrew, please, refine the specification ((full) number and publication month 07-08). (2023-12-03)
A.Buchanan: Ok have added the publication details. The keywords are useful to indicate the aim, even though the problem is cooked. I guess there is a way to keep the model mates but it’s very clunky and probably +bSe3 is better (2023-12-03)
A.Buchanan: Yes Valery this is cooked. I'm not convinced by +bSe3 because it spoils one of the model mates. (2023-12-03, hidden) more ...
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klären: welches ist die NL?
Bernd Schwarzkopf: Ein Heft „Die Schwalbe 10-12/1968“ gibt es nicht.
Das Problem ist in „Die Schwalbe 10-11/1968“ oder „12/1968“ nicht zu finden.
Vielleicht war P0544925 gemeint. (2023-04-26)
Henrik Juel: The cook is easily removed by adding '2 solutions' to the stipulation
1.Kb3 0-0-0 etc. are tries, because White just moved Ta1 or Ke1, so he may not castle (2023-04-26)
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.31 108 Sekunden.
Keine Lösung: BP 25.0. KBP = 25.5! (2023-06-20)
Henrik Juel: cooked, e.g.
1.Sg1-f3 Sg8-f6 2.Be2-e4 Sf6xe4 3.Lf1-c4 Bh7-h5
4.Ke1-e2 Se4-g3 5.Ke2-d3 Th8-h6 6.Lc4-b3 Th6-a6
7.Th1-e1 Sb8-c6 8.Te1-e6 Ta6-a3 9.Te6-h6 Bg7xh6
10.Sf3-e5 Lf8-g7 11.Se5xd7 Lg7-d4 12.Sd7-b6 Lc8-f5
13.Kd3-c4 Dd8-d7 14.Bb2xa3 Ta8-d8 15.Ba3-a4 Dd7-c8
16.Lc1-a3 Td8-d6 17.Dd1-f1 Td6-e6 18.La3-d6 Bf7-f6
19.Sb1-a3 Ke8-f7 20.Ta1-e1 Kf7-g6 21.Te1-e3 Kg6-g5
22.Te3-d3 Ld4-e3 23.Bc2-c3 Sc6-d4 24.Kc4-b4 Lf5-g6
25.Sa3-c4 Sd4-f5 26.Ba2-a3 (2018-12-10, hidden) more ...
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.31 00:06:21 Minuten. (hh:mm:ss)
Keine Lösung: BP 36.0.
Beispiel Stelvio: 1.c3 a5 2.d3 a4 3.Le3 a3 4.Sd2 Ta4 5.Tb1 Tc4 6.dxc4 Sa6 7.g4 b5
8.g5 b4 9.g6 b3 10.gxh7 bxa2 11.b4 a1=S 12.b5 a2 13.b6 Sb3 14.h4 g5 15.h5 g4
16.h6 g3 17.Th5 g2 18.Sh3 g1=D 19.b7 Dg5 20.b8=D Db5 21.Tf5 c5 22.Df4 a1=D 23.Dg4 Sa5
24.f4 Sb7 25.Lg1 Daa5 26.e3 e6 27.Le2 Ld6 28.Sf1 Lb8 29.Dd2 Se7 30.Kd1 Tg8
31.h8=S d6 32.Sg6 Ld7 33.Sh4 Sc8 34.Sg2 De7 35.h7 f6 36.h8=D exf5 37.Dhh4 (2023-06-20)
Henrik Juel: cooked, e.g.
1.Bd2-d3 Ba7-a5 2.Lc1-e3 Ba5-a4 3.Sb1-d2 Ba4-a3
4.Bg2-g4 Ta8-a4 5.Bg4-g5 Ta4-c4 6.Bd3xc4 Bb7-b5
7.Bg5-g6 Bb5-b4 8.Bg6xh7 Bb4-b3 9.Bh2-h4 Bb3xa2
10.Bb2-b4 Be7-e6 11.Bb4-b5 Bg7-g5 12.Bb5-b6 Bg5-g4
13.Bb6-b7 Bg4-g3 14.Bh4-h5 Bg3-g2 15.Bh5-h6 Sb8-a6
16.Th1-h5 Bf7-f6 17.Ta1-b1 Ba2-a1=S 18.Sg1-h3 Bg2-g1=D
19.Bb7-b8=D Dg1-g5 20.Bc2-c3 Dg5-b5 21.Th5-f5 Bc7-c5
22.Db8-f4 Lf8-d6 23.Df4-g4 Ld6-b8 24.Bf2-f4 Sg8-e7
25.Le3-g1 Ba3-a2 26.Be2-e3 Sa1-b3 27.Lf1-e2 Ba2-a1=D
28.Sd2-f1 Sb3-a5 29.Dd1-d2 Sa5-b7 30.Ke1-d1 Th8-g8
31.Bh7-h8=S Da1-a5 32.Sh8-g6 Be6xf5 33.Sg6-h4 Bd7-d6
34.Sh4-g2 Lc8-d7 35.Bh6-h7 Se7-c8 36.Bh7-h8=D Dd8-e7
37.Dh8-h4 (2018-12-10, hidden) more ...
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A.Buchanan: the authors fixed this in 1998 C+ with 1sbq1ksr/2p1ppbp/p2R2p1/1Q2B3/PpP1BP1K/3P1SP1/1P2P2P/R4S1r
16. Lg2 Te3 17. Le4 Tf3 18. Sd2 Tf2 19. Sgf3 Tg2 20. Ta1 Tg1 21. Sf1 Th1 (2023-02-04)
Henrik Juel: Cooked, e.g.,
1.Bc2-c4 Bb7-b5 2.Dd1-a4 Bb5-b4 3.Da4-b5 Bg7-g6
4.Ba2-a4 Lf8-g7 5.Ta1-a3 Ke8-f8 6.Ta3-d3 Bd7-d6
7.Td3xd6 Sb8-a6 8.Bd2-d3 Ta8-b8 9.Lc1-f4 Tb8-b6
10.Lf4-e5 Tb6-c6 11.Bf2-f4 Tc6-c5 12.Ke1-f2 Tc5-d5
13.Kf2-g3 Td5-d4 14.Kg3-h4 Td4-e4 15.Bg2-g3 Te4-e3
16.Lf1-h3 Te3-f3 17.Lh3-g4 Tf3-f1 18.Sg1-h3 Tf1-e1
19.Th1-f1 Sa6-b8 20.Tf1-f3 Te1-h1 21.Tf3-e3 Ba7-a6
22.Te3-e4 (2018-12-12, hidden) more ...
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Moldenhauer: Computerprüfung: Cooked Stelvio 1.31 8 Sekunden.
Keine Lösung: BP 19.0.
Beispiel: 1.a4 h5 2.a5 h4 3.a6 h3 4.axb7 hxg2 5.h4 g5 6.Th3 g4 7.Td3 g3
8.Sf3 g1=S 9.Txd7 Sh3 10.Lxh3 g2 11.d3 g1=S 12.Kf1 a5 13.Se1 Sf3 14.Le3 Sd2+
15.Dxd2 a4 16.Dc3 a3 17.Sd2 a2 18.Tc1 a1=S 19.f3 Sb3 20.Sxb3 (2023-06-19)
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paul: Cook: 1.b4 d5 2. h4 c6 3. Nf3 h6 4. Ng5 Qc7 5. Bb2 Nd7 6. b5 hg5 7. Be5 cb5 8. Rg1 g4 9. Bh2 g3 10. Na3 gh2 11. Rb1 hg1=B 12. Nc4 Bh2 13.Rb3 Bd6 14. Rh3 Ba3 15. Rh1 Bc1 16. Nb6 Qh2 17. Rg1 e5 18. Rh1 Bfa3 19. Rg1 Ne7 20. Rh1 0-0 21. Rg1 Kh8 22. Rh1 Rg8 23. Rg1 Nf8 24. Rh1 Bd7 25. Rg1 Rc8 26. Rh1 Rc3 27. Nc4 Nc8 28. Rg1 bxc4 29.Rh1. (2010-09-05)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.4 1 Sekunde.
KBP = 27.0, keine Lösung BP 26.5.
Beispiel: 1.Sa3 d5 2.Sc4 Dd6 3.Se3 Sd7 4.Sg4 h5 5.h4 hxg4 6.Th3 gxh3 7.Tb1 h2 8.b4 hxg1=L
9.Tb2 Lh2 10.Tb3 Le5 11.La3 Lb2 12.Th3 Lc1 13.Th1 Dh2 14.Lb2 e5 15.b5 La3 16.Tg1 Se7
17.Th1 0–0 18.Tg1 Kh8 19.Th1 Tg8 20.Tg1 Sf8 21.Th1 Ld7 22.La1 Tc8 23.b6 cxb6 24.Ld4 Tc3
25.Tg1 Sc8 26.Lc5 bxc5 27.Th1 c4
Die Notation ganz oben ist falsch! 1.Sc3 h5 2.Sd4 ist nicht möglich. (2023-06-23)
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Moldenhauer: Computerprüfung: C- oder Cooked Stelvio 1.4 in 48 Sekunden.
Keine Lösung: BP 28.5.
1.a4 Sa6 2.a5 Sc5 3.a6 Sh6 4.axb7 Sf5 5.b8=S La6 6.Sc6 Lc4 7.Ta6 Db8 8.Tb6 a5
9.g4 a4 10.Sa5 Kd8 11.g5 a3 12.g6 c6 13.gxf7 g5 14.Sc3 Lg7 15.Sd5 Lc3
16.Sf3 Lb4 17.c3 g4 18.d4 g3 19.Lh6 g2 20.Db1 Dg3 21.Se5 Dg8 22.fxg8=S g1=S
23.Sgf6 Sh3 24.Sh5 Sg5 25.Lh3 a2 26.0–0 Tc8 27.Td1 axb1=S 28.Td3 Sa3 29.Te3 Sb5 (2023-08-01)
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Gerald Ettl: Wie lautet der Cook? (2023-09-28)
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Henrik Juel: Natch 3.1 did nothing in an hour (2018-12-18)
A.Buchanan: Suggests flipping white & black with 2.Ld3 (2023-03-01)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.0 01:03:04 Stunde.
Keine Lösung: BP 13.0, BP 13.5.
Cooked Euclide 1.11 22:03:22 Stunden (2023-02-28, hidden)
A.Buchanan: Suggests flipping white & black with 2.Ld3 but maybe this would go against the theme (2023-03-01, hidden) more ...
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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)
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SH: cooked? 1. b4 d5 2. Ba3 Kd7 3. Qc1 Kc6 4. Qb2 Kb5 5. Qxg7 Ka4 6. Nc3+ Kxa3 7. Rc1 Kb2 8. Nb1+ Kxc1 9. Qa1 Bh6 10. f4 d4 11. Kf2 Kd1 12. Kg3 Ke1 13. Kh4 Kxf1 14. Nf3+ Kf2 15. Rd1 Bxf4 16. Ne1 Kg1 17. Kh5 Kh1 18. Nc3 Be3 19. Nd3+ Bg1 (2006-06-06)
Henrik Juel: Yes, the move order is not unique after 10.f4, so the problem is indeed cooked (2018-12-17)
Moldenhauer: Computerprüfung: C- oder cooked Stelvio 1.4 48 Sekunden.
Keine Lösung: BP 18.5. (2023-07-29, hidden) more ...
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paul: First 22.5 moves tested with Euclide 1.11 (~11h) (2017-12-29)
Henrik Juel: Now things are OK, I think
Earlier the solution given was for a BP in 19, starting 1.b4 c5 (2017-12-29)
Olaf Jenkner: Yes, I overlooked the date of the first comment. I would not have asked. (2017-12-29)
P. Rãican: 1.b4 c5 2.b5 Dc7 3.b6 Dg3 4.hg3 h6 5.Txh6 axb6 6.Tc6 Txa2 7.Sa3 Txc2 8.Lb2 Tc4 9.Sc2 Tch4 10.e4 g6 11.Lc4 Lh6 12.Se2 Le3 13.de3 e6 14.Dd3 Se7 15.0-0-0 0-0 16.Txc8 Sbc6 17.Ta8 Th8 18.Ta1 Ta8 19.Th1 Sb8. Platzwechsel der weisen Turme und platzwechsel der schwarzen Turme. Terrific! (it got the second prize only..) (2007-06-08, hidden)
P. Raican: The previous solution is for 2Prize of 1996 Springaren tournement! P1017682 has the following solution: 1.b4 Sh6 2.Lb2 Tg8 3.Lf6 gf6 4.Sc3 Tg3 5.Tb1 Th3 6.g4 Sa6 7.Lg2 Tb8 8.Lc6 bc6 9.g5 Tb5 10.g6 Tah5 11.b5 Sg4 12.b6 Lh6 13.Tb5 Sb4 14.Tg5 La6 15.b7 Lb5 16.b8=L a6 17.La7 Da8 18.Le3 c5 19.g7 Dg2 20.g8=L Sc6 21.Tg7 Lg5 22.Sf3 h6 23.Lh7 Sb8 24.Le4 f5 25.Lb7 Sf6 26.Sd5 Sg8. Zwei Springerrundlaufe, zwei Lauferphonixe. PR: H.P.Rehm (2007-06-08, hidden)
Olaf Jenkner: The solution fits the diagram. What's wrong?
The link to 2Prize of 1996 Springaren tournement shows the problem above. (2017-12-29, hidden)
Reto: Cooked by Stelvio 1.11 (2023-03-31, hidden) more ...
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Henrik Juel: Natch 3.1 did not even find the solution in an hour (2018-12-15)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.2 296:36:23 Stunden. (hh:mm:ss)
Keine Lösung BP 20.5, BP 21.0. Histogramm 355767/10739k
Cooked bei 59917/922k. Die letzten 295850 Strategien wären sicher
schneller geprüft worden da die meisten großen Werte bereits waren.
Notations Beispiel: 1.b4 Sf6 2.c4 b6 3.Sc3 c5 4.bxc5 Lb7 5.cxb6 Lf3
6.gxf3 Sd5 7.Lh3 Se3 8.Lf5 Sf1 9.Lc2 g5 10.Sh3 g4 11.d3 g3 12.Dd2 g2
13.Dh6 g1D 14.Ld2 Dg3 15.Tc1 Se3 16.Lb1 Dgc7 17.bxc7 Sd1 18.0–0 Se3
19.Tfd1 Sd5 20.Kf1 Sf6 21.Le1 Sg8 22.Td2
oder wie vor bis 19. dann 20.Le1 Sf6 21.Kf1 Sg8 22.Td2 (2023-05-24)
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paul: Cook: 1.Sf3 [-Ra1h1, -Sb1, -Pa2a7h7, -Ra8h8, -Sb8g8] g5 2.e4 [-Pe4] d5 3.d4 g4 4.Bc4 [-Pg2] g3 5.B×d5 g×h2 6.Ke2 h1=R 7.Sg5 c5 8.d×c5 [-Pc5] Qa5 [-Qa5, -Bc8] 9.Se6 Bg7 10.Bf3 Rh8. (2024-03-13, hidden)
A.Buchanan: what's Berkeley Chess, please? (2024-03-14, hidden) more ...
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