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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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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See P0003736 a companion problem.
milan: [-bPa7b6wPg3bSa1+bPg3bBa5-b4 bPc6-d6] h#2 1.2...bRh2? illegal move? M.Frelih (2023-03-30)
Mario Richter: (a) Because of bRh2 w0-0 impossible.
Solution: 1.Rh2xh3 Rxh3 2.0-0 Rh8#

(b) bRh2 can return to a8 only via e8, so b0-0 impossible.
Solution: 1.Ba4 0-0 2.Rf8 Re1#

But cooks:
1.Rh2xh3 Bb4 2.Rxg3 Rxh8#
1.Nd2 Bf6 2.Rf8 Nd6# (2009-01-30, hidden)
A.Buchanan: Suggested corrected version: 4k2r/p2p1Np1/1pp2p2/1b3p2/8/p4PPP/B4P1r/nn2K2R. Can add to the stipulation: "Which missing units were not captured by pawns?" (2022-03-15, hidden)
A.Buchanan: Suggested correction: 4k2r/p2p1Np1/1pp2p2/1b3p2/8/p4PPP/B4P1r/nn2K2R (2022-03-15, 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: In b) White may not castle, because last move was done by Ke1 or Th1 (2023-12-02)
A.Buchanan: Cute. Do wPe2 & bPe3 serve any purpose, however? (2023-12-03)
Henrik Juel: C+ Popeye 4.61 (2023-12-02, hidden) more ...
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See P0000899 a companion problem.
A.Buchanan: Something odd going on here. There are numerous h#2 in both a&b. PRA not normally found in a problem with single solution for each twin. White has 8 pawns so trivially wRh7 must be original. (2018-10-13)
A.Buchanan: There are three similar problems by Brogi which are all twinned RS rather than PRA, but don't have diagram error. P0003743, P0003746 & P0003747. (2018-10-13)
VL: Nothing to do with Retro Strategy (nor with PRA). One definite castling is legal in every case. (2021-02-09)
Ladislav Packa: Cooked a) and b):
1...b8S and 2...R:h8# (2021-02-10)
A.Buchanan: I think the composer simply forgot that wPb7 can promote. Most of the cooks come from S promotions, but it's also possible to have QR. I've fixed it, rejigging the captures to make them still add up. I've posted in Discord, as I did with the partner composition, and will create entries for them here. (2022-03-15)
milan: +sLb8 sBa7=sT M.Frelih (2023-12-02)
A.Buchanan: Hi Milan - I don't think your suggestion quite works for b). In a) there are 0+2 spare captures, so Black can certainly cross-capture. But in b) there is 1+1 so neither side can cross-capture, so there is no solution. Please compare with P1399806, in which there are 1+2 & 2+1 spare captures, so both twins are sound. (2023-12-03)
milan: Hi Andrew my correction works only with 2.1... solutions, black or white knights on a5. are not important. (2023-12-03)
A.Buchanan: Hi Milan not really clear what you are doing, but if as well as the piece changes you proposed, you also change the stipulation to 2.1... then there is still only one solution. Even if you remove Sa5 entirely as well, there is no White cross-capture possible. (2023-12-04)
milan: Hi Andrew my correction works only with 2.1... solutions, black or white knights on a5 are not important. (2023-12-03, hidden) more ...
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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)
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milan: i've tested by Linux 4.87 128 MB C+ M.Frelih (2022-12-11)
A.Buchanan: what's your suggested position, Milan? (2022-12-11)
milan: my purpose was to show difference beetwen gustav program and linux.i've solved with linux and found intended solution in 30 sec. Gustav found cooks? M.Frelih (2022-12-12)
A.Buchanan: When you say Linux, do you mean Popeye? When I solve with helpmate analyser, it finds a lot of cooks e.g. those shown here already (2022-12-13)
Ulrich Voigt: Just for the record, Popeye on Linux does find those cooks (11 solutions in total). (2022-12-14)
milan: i've tested in YACPDB site C+ M.Frelih (2022-12-17)
milan: my purpose was to show difference beetwen gustav program and linux.i've solved with linux and found intended solution in 30 sec. Gustav found cooks? M.Frelih (2022-12-12, hidden) more ...
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Yuri Bilokin: correction bBe6-f7 7K/5bRB/8/5p2/6p1/4n1kr/q4pp1/b4nr1 (3+12) (2023-05-22)
A.Buchanan: This exact correction was made by the authors in 2009, see P1094987. (2023-05-22)
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Yuri Bilokin: More economical -bPh7, then a1=b1 8/6p1/6K1/6N1/4pkq1/5p1p/3P3n/8 (3+7)
1... ... 2.Qg2 Kh5 3.e3 dxe3+ 4.Kg3 Se4# (MM)
1...d4 2.Qh4 Sh7 3.Sg4 Sf6 4.Qg3 Sd5# (MM) (2022-05-18)
Hiroaki Maeshima: Thank you, I added the version (P1404069). (2022-09-06)
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see P0003365
Henrik Juel: Intention:
White captured axSb and [Lc8]
If last move was b2-b4: 1.axb3ep bxc6+ 2.b5 cxb6ep#
If last move was d2-d4: 1.exd3ep Lb8+ 2.d5 cxd6ep# (2023-08-08)
Henrik Juel: Cook: last move could also be Kf6-e5
correction, e.g., move Sh3 to h5 (2023-08-08)
A.Buchanan: Good catch Henrik - I should have looked at this one more carefully! Luckily it admits an easy fix, but there is still the non-standard material. (2023-08-08)
Henrik Juel: I should have thought of a better birthday greeting, Andrew, but there you are (2023-08-08)
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