A.Buchanan: Not trivial to fix as shifting wPa5 to b6 leads to promotion cooks. Adding black material must be carefully done to avoid disrupting uniqueness of e.p. as tempo. (2022-03-11)
A.Buchanan: OK just to say I have found a form of this idea that works. This one was the very devil to fix - the final form includes different mates and retro logic, so becomes an "after" rather than a "correction", and I will find somewhere else to publish first. (2022-03-19)
VL: The author: Andrej Lobussow (Lobusov).

Solution:
1... Rxh7 2.Kf8 Rxh8#
(1... Kg5? 2.0-0?? Se7#;)
1.cxb3ep (waiting)! Kg5 2.0-0! Se7#
(1... Rxh7? 2.Kf8 Rxh8#??).

W. captured 10 by pawns (including hxPg-g8),
not on the last move (10+6=16).
Bl. took only b3xa2 with w.Pb4 on b2. (2004-06-03, hidden) more ...
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Mario Richter: Im Nachdruck in problem 144-147 steht nur auf e5 ein sB (also keiner auf e3). Allerdings gibt es in allen drei Varianten (sBe5+e3, nur sBe5, nur sBe3) jede Menge NL, die ohne Rochade und e.p.-Key auskommen - am wenigsten dann, wenn nur auf e3 ein sB steht. (2010-10-10)
VL: Cf. P0002475. (2012-08-26)
A.Buchanan: Possibly "Einheitlösung!" is a request to unify retro implications derived across separate parts of the PRA solution. But cooked chessically irrespective of these philosophical issues, which still wait to be resolved 60 years on. (2022-02-16)
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A.Buchanan: Easy enough to eliminate what I suppose is the intended solution (sL unblocking sD and blocking sT) by e.g. just removing sDf5. However I don't see a way to say goodbye to the cook instead. (2021-11-23)
James Malcom: I believe I may have a fix Andrew. If we substitute the Black queen with a Black rook, the same theme is kept even is if considered watered down.

h#3 8/3p4/3p1p2/3Kbr2/k1ppPp2/1n1r3P/P1p4P/N3nq1b (2021-11-24)
A.Buchanan: Hi James, hurray cool I thought I couldn't get through that way: I was confused in my mind about wPe2 blocking that line, but with the freshness of a new day, I can look at it again. The precise diagram that you propose doesn't quite work because R: 1. e3-e4 Sf3*e1+,Sg2*e1+,g2xf1=Q+,g2xh1=Q+, but this is all fixable with +bPd2,-bSe1. Now Popeye 4.87 says the forward solution is works uniquely, and Retractor 2.0 says R: 1. e2-e4 Rf3-d3+ 2. Ke4-d5 are unique. So for full C+ we only need some demo game that results in the position prior to Ke4-d5.
I don't see any "watering down" here: I think we have faithfully represented Tivadar's idea. Let's post it here as TK, corr JM&AB? (2021-11-24)
A.Buchanan: I have made an unedifying demo game with lichess, so our correction is fully C+. (2021-11-24)
James Malcom: "TK, corr JM&AB" sounds good to me. (2021-11-24)
A.Buchanan: Demo game: legal per Lichess.org, very unifying. 1. b4 a5 2. c3 axb4 3. Na3 bxc3 4. Rb1 c5 5. Rb6 Na6 6. Rc6 bxc6 7. Nc2 Nc7 8. Na1 Nb5 9. Ba3 Nd4 10. Qc2 Nb3 11. Bb4 Ra4 12. Ba5 Rf4 13. Bc7 Nf6 14. Bd6 exd6 15. Qf5 c2 16. Nf3 c4 17. Nd4 c5 18. Ne6 h6 19. d4 cxd4 20. Ng5 hxg5 21. f3 d3 22. Kf2 d2 23. Ke3 Ne4 24. Qd5 Nf2 25. Qf5 Nh3 26. gxh3 Qa5 27. Qd3 Rf6 28. f4 Qa3 29. Bg2 Qc1 30. Bd5 Qf1 31. Be6 fxe6 32. Rg1 Bb7 33. Rg4 Bh1 34. Qd5 exd5 35. Rh4 Rf5 36. Rh6 Be7 37. Rh5 Bf6 38. Rh6 Be5 39. Rf6 Rh4 40. Rg6 Rg4 41. Rf6 Rg2 42. Rg6 Rf6 43. Rh6 Rf8 44. Rg6 Rh8 45. Rf6 Rh4 46. Rg6 Rhg4 47. Rf6 d4+ 48. Ke4 R4g3 49. Rg6 Rf3 50. Rf6 Rg4 51. Rg6 Rh4 52. Rf6 Rh5 53. Rg6 gxf4 54. Rf6 Rf5 55. Rg6 Kd8 56. Rh6 Kc7 57. Rf6 Kb6 58. Rh6 Ka5 59. Rg6 Ka4 60. Rf6 gxf6 (2021-11-24, hidden) more ...
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A.Buchanan: Popeye v.487. WinChloe has the same diagram, and although it just claims 1 solution, it's C? Yet WinChloe engine gives the cooks right away (2021-10-27)
Mario Richter: Shouldn't this problem get the Label "No legal last move for White"?
(White pawn d2 has captured the missing black piece (Pdxe), so black pawn a7 had to promote to get to the capture square. This requires one capture (Pa3xb2), together with bxc,cxd,hxg,gxh,gxh this accounts for all missing white pieces.)

Btw. "for me, "C?" stands for "Not yet computer-tested or computer test didn't come to a conlusion" ... (2021-10-28)
A.Buchanan: Hi Mario, thank you.
(1) The stipulation is non-standard. I think this is why the WinChloe database doesn't include a solution, and just marks it as "C?" However if I run the WinChloe solving engine manually it (like Popeye) gives 121 h#3 solutions and 4 h#2.5 solutions. I think you, me and Christian would all agree on the meaning of "C?"
(2) The problem does not rely on Article 15 to eliminate the first single move, but implicitly requests the shortest. Currently the PDB usage of "No legal last move..." is to indicate that Codex Article 15 applies, so I did not add it. But I'm ok if you want me to add the keyword. In my correction, I shift the stipulation to h#3 so it's clear that Article 15 and not any desire for brevity which push the problem to h#2.5.
(3) However the PDB usage is confused. Ideallly think the two "No legal..." keywords should be replaced by "Retro-stalemate", while two new keywords "Article 15" & "Whose mate?" indicate the choice of implications of the retropat for forward play. "Board rotation" and other jokes are also implications of retropat (and of illegal diagram).
(4) But this is complicated to implement, requiring either database-level access or screen-scraping. The latter is quite practical for some future time when I have more leisure.
(5) https://www.stere.ro/biblioteca-digitala/reviste/revista-romana-de-sah/ is an excellent archive of Revista Romana de Sah. My sincere compliments and thanks to whoever prepared it. And the magazines themselves contain thorough annual indices. p225-6 of 6/1948, contains the solution to Kardos. I haven't translated it, nor have I found time to locate the prize announcement.
Comments welcome! (2021-10-29)
Mario Richter: I still believe, that the label "No legal last move for White" is justified here, because this is exactly what the author wanted to show.
(And this is too, what the reprint in 'Problem (Zagreb)' 37-40 09/1956 is all about - it appears there inside asn article

Btw., I wouldn't call the presentation of the solution in 'Revista Romana de Sah', p.225, 06/1948, a "reprint" - it lacks the presentation of the position ... (2021-10-30)
A.Buchanan: Added the keyword.
Yes I know that's the usage of "reprint", but then there should be a field for the location of the solution :-) It's idiotic that if I want to say where the solution is contained, I have to embed it in text like this: 2519 Revista Romana de Sah , p. 225-6, 06/1948 (2021-10-30)
A.Buchanan: A quick j'adoube to fix the cook is wPf2 to d2. (2021-10-28, hidden)
A.Buchanan: Hi Mario,
(1) The stipulation is non-standard. I think this is why the WinChloe database doesn't include a solution, and just marks it as "C?" However if I run the WinChloe solving engine manually it (like Popeye) gives 121 h#3 solutions and 4 h#2.5 solutions. I think you, me and Christian would all agree on the meaning of "C?"
(2) The problem does not rely on Article 15 to eliminate the first single move, but implicitly requests the shortest. Currently the PDB usage of "No legal last move..." is to indicate that Codex Article 15 applies, so I did not add it. But I'm not precious about it, if you want me to put it in. In ,y correction, I shift the stipulation to h#3 so it's clear that Article 15 as well as desire for brevity push the solution to h#2.5.
(3) However the PDB usage is confused. Ideallly think the two "No legal..." keywords should be replaced by "Retro-stalemate", while two new keywords "Article 15" & "Whose mate?" indicate the choice of implications of the retropat for forward play. "Board rotation" and other jokes are also implications of retropat (and of illegal diagram).
(4) But this is complicated to implement, requiring either database-level access or screen-scraping. The latter is quite practical for some future time when I have more leisure.
(5) https://www.stere.ro/biblioteca-digitala/reviste/revista-romana-de-sah/ is an excellent archive of Revista Romana de Sah. My sincere compliments and thanks to whoever prepared it. And the magazines themselves contain thorough annual indices. p225-6 of 6/1948, contains the solution to Kardos. I haven't translated it, nor have I found time to locate the prize announcement.
Comments welcome! (2021-10-29, hidden) more ...
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YM: The correct mechanism is in P1109109 (2010-07-13)
YM: Correction option: P1229434 (2021-06-28)
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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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Henrik Juel: The missing black man is dark-squared, so last move was not fxe4. In the solution Td6 should read Txd6. There seems to be a 'variation': 0... dxe3ep 1.0-0-0 Le4 2.Txd6 Sg4 3.Tf6#, so maybe the stipulation should have 1.1;2.1;1.1 added. (2004-09-23)
A.Buchanan: Well spotted Henrik. Your kindness is legendary, but in the harsh world of helpmates, this counts as a cooked I believe. In fact, it's the 2.Lf5/3.Sf3 version which is the cook. Yours is the actual solution! This is because 3.Sf3 would check if White hadn't castled, and we want the castling to be because of the AP condition only, not for 2 reasons, which one might term a "logical dual". The problem can be trivially fixed by sticking sSe5 on f6. I have no idea whether it's a typo or a cook: WinChloe has the same diagram, and the same confusion between cook & intended solution. The fact that AP does not appear in the stipulation might be a factor. WinChloe as far as I can see has no concept of AP as a keyword, and just picks it up based on a substring of the stipulation. In other compositions, T.Petrovic wasn't shy of flashing the AP acronym in the stipulation, but as usual the detective work peters out in the absence of primary records. (2020-12-08)
A.Buchanan: But maybe I'm being hasty. Perhaps the intention *is* to have two variations? Can anyone get at relevant issues of Problem, please? Or provide me with an online link to the relevant image files, and I'll check. What do you think Henrik? (2020-12-08, hidden) more ...
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A.Buchanan: This is a very heavy position to prevent the possibility of R: 1. c2xb3. I'm not sure why this was done. A much lighter position 8/8/8/5np1/1r1pPkr1/2Bp1p2/1p1P2P1/4K2R with 7 less units achieves the mates soundly. Am I missing something? (2022-02-16)
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AB: I think this is cooked. I don't see how AP castling can justify earlier ep here.
1. Kf3 Bf5 2. Kg2 Be4#
1. Kf3 Bxd7 2. Kg2 Bc6#
1. cxd3 Rf1 2. Qa4 Rxa4#
1. gxf3ep Ra5 2. Kf4 Rh4#
1. gxf3ep Rd1 2. Re7 Rh4#
Also, second solution given has typos 1. *c*xd3ep & 2. Q*a*4. (2002-03-21)
V.Liskovets: Indeed this problem is cooked, and I
failed to correct it preserving symmetry.
Here is a possible correction:
W: Ke1 Ra1 Rh1 Be6 Pd4 Pf4
B: Ke4 Qd7 Rf7 Pc3 Pc4 Pe3 Pg3 Pg4 Bh7 Rb2 Pb3

Another story, justifying its contents (e.p.).
In my opinion, all published treatments are
insatisfactory (the same concerns P0004295
(corrected) by Werner Kuntsche as well).
There is a way to make it sound under the
sophisticated HYBRID stipulation 'AP, pRA':
2 partial solutions legalized JOINTLY by
both castlings (details to be published). (2002-04-02)
VL: Sorry, Andrew, I see no issues with my version (excepting the mentioned special retro-convention/genre for justifying its soundness, of course!). h#2 (pRA&AP). The full solution consists of 2 partial AP-based ones: I 1.gxf3 e.p.(!?) 0-0-0! 2.Te7 Th4#; II 1.cxd3 e.p.(!?) 0-0! 2.Da4 Txa4#.
C+(popeye): h#2 & two h#1.5 after the keys (added Tb2 & Lh7 are cookstoppers).
The main features of the corresponding suitable AP-genre ("consolidated")are rather clear. However, not all subtle aspects have been analyzed thoroughly yet. I have got only a draft manuscript with few examples, still. (2022-02-15)
A.Buchanan: Yes Valery I think your correction is technically sound. Theological debate is something else, heh. Have posted it as separate entry. Please tell me if it was original here, or appeared somewhere else (2022-05-29)
Henrik Juel: I looked in Stella Polaris 1971 p.202-203, where the problem is given without AP under the diagram
It turns out that both the author and the editor (Jan Knöppel) consider the problem correct
After explaining the AP convention the latter writes [my translation]:
The author has elegantly applied the AP conventionn in combination with partial analysis. If we assume that last move was not f2-f4 or T-h1, the possibilities d2-d4, K-e1, and T-a1 remain, but the two latter disappear by 1... 0-0. The other assumption (last move was not d2-d4 or T-a1) is analogous.
So something like Valery's HYBRID stipulation was the intention, and the problem was considered correct in 1971; later developments do not change this fact (2022-05-29)
Henrik Juel: Of course, the obvious cooks 1.Kf3 etc. remain (2022-05-29)
A.Buchanan: I agree. Basic AP Petrovic does not allow multiple ep, nor does RS, so some alignment of AP with PRA is required. I would like to see this codified though. (2022-05-29)
VL: Reprints: (8) StrateGems SG19, p.156, 07/2002 (2017-12-01, hidden)
A.Buchanan: Valery's suggestion doesn't seem to work: 8/3q1r1b/4B3/8/2pPkPp1/1pp1p1p1/1r6/R3K2R. Have I got something wrong? (2022-02-14, hidden)
A.Buchanan: For Valery, words like "rich" and "sophisticated" are positive. For me, they are at best neutral. Instead for me, positive words are: e.g. "clear", "logical", "systematic". I can see that some statement is necessary for how AP scales in the context of both RS & PRA. The first thing I will check is whether these definitions work back in the simple case where we just have 1 castling & 1 e.p. And indeed when we just have 0 castling & 0 e.p., when PRA & RS should agree. This is a test of robustness for whether these definitions are seaworthy when we set sail in the stormy seas of fairy chess.
Actually, for PRA-AP I am missing even a basic 1 castling 1 e.p. definition.
No issue with "rich" and "sophisticated" once we've got the underlying definitions clear and published, but until that point, they are just unhelpful. (2022-02-15, hidden)
A.Buchanan: Hi Valery, thanks for checking - maybe I had got the stipulation wrong when I tested it. I think the following lighter correction also works: 8/3r1r2/4B3/8/2pPkPpn/2p1p1p1/8/R3K2R. From the original it's +bSh4 ~bRd7. Popeye v4.87 says:
1.c4*d3 ep. 0-0 2.Rd7-e7 Ra1-a4 #
1.g4*f3 ep. 0-0-0 2.Rf7-e7 Rh1*h4 #
With retro tries:
1.c4*d3 ep. Rh1-f1 2.Rd7-e7 Ra1-a4 #
1.g4*f3 ep. Ra1-a5 2.Ke4-f4 Rh1*h4 #
1.g4*f3 ep. Ra1-d1 2.Rf7-e7 Rh1*h4 #
See also P1399112 and P1382808.
If you are willing to share, I am very interested to see your draft manuscript. Maybe I can help it move forwards to publication after all this time? I have made my first AP Type Keym problem and will send it to you. (2022-02-16, hidden)
A.Buchanan: Hi Valery: I sent you an email couple of days ago - please check your spam folder if you haven’t received. It will shortly be the 20th anniversary of your 2002-04-02 post above where you talk about “details to be published”. It’s hard to get low urgency stuff done I know, but I’ve been patient all these decades. I want to help you! Please can you send me the 37 patterns that you have (another post of yours somewhere here). Happy to have a zoom call too if that works for you :-) (2022-02-17, hidden) more ...
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s.a. Version P0004341
A.Buchanan: Shifting a few of the White pieces (Ba6,Lc7,Td4,Se5 to Ba4,Lc6,Tc2,Sd4) gives this r3k3/p1ppp3/1pB5/5PP1/P1PNPp2/2RK2PN/1PRn1PQp/2Brq1b1. Does this work? (2022-02-28, hidden)
A.Buchanan: Shifting a few of the White pieces (Ba6,Lc2,Td4,Se5 to Ba4,Lc6,Tc2,Sd4) gives r3k3/p1ppp3/1pB5/5PP1/P1PNPp2/2RK2PN/1PRn1PQp/2Brq1b1. This works forwardly - any retro issues? (2022-02-28, hidden) more ...
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Version zu P0004296

Autor: "In Nr.9, I intended to do something similiar to No. 8 [P0004199], but with a different key: the solution 1. PxP "e.p.", Qxa8 checkmate will not go, for if blackside does not castle, the "en-passant" capture cannot be justified. Though, well considering, there is the demolition 1. PxP "e.p." -any 2. 0-0-0! Qb7 or a8 checkmate.
A.Buchanan: Thanks Mario for retrieving the author's intent with this. So he knew it was cooked! Assume s000 rights remain. R: 1. b5xa6? as sBb promoted on b1. R: 1. h4xg5? as sBh retro-blocked. R: 1. Ke3-d3? impossible check from sBf4. R. 1. K~-d3 d3-d2? illegal check. I like these! So by elimination, R: 1. e2-e4 e3xTd2 2. T~d2 L~e1. So must have sLe1. (2022-03-22)
A.Buchanan: Something is very wrong. I wonder if there is a diagram mistake: suppose we exchange wLh4 with sLe1. Suppose s000 rights remain. Then R: 1. e2-e4 e3xTd2 & White can't avoid retropat. So the only solution is 1. Kd8 Sc6+ 2. Kc8 Sge7#. The role of AP is to exclude all the ep solutions with no castling. Do you agree? What am I missing? (2022-03-22, hidden) more ...
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Author: "From the former [d.h. P0004341], the No. 10 was born, something similar to No. 8 [P0004199] but with Black-castling for it is a help-mate. The mating move 1. ... Qg8? will not go."
A.Buchanan: I think replacing wLc3 with wT works, but loses the h#1 try if that was intended (2022-03-22, hidden)
A.Buchanan: Well, now there is only 1 cook, compared to previous versions. [sBe] must promote, while [sBc] could not, so sBd2 originates to the left. The four pleasant wK retraction tries remain. Replacing wLc3 with wT works, but loses the intended h#1 try. (2022-03-22, hidden)
A.Buchanan: r3k3/3p4/P7/pP6/rpPPN1PP/B2K1P1B/QN1p1pPp/2bR4
I think this fixes it, with unique retro try h#1, and the sound h#2, yes? (2022-03-23, hidden)
A.Buchanan: r3k3/3p4/P7/pP5P/rpPPN1P1/B2K1P1B/QN1p1pPp/2bR4
I think this fixes it, with unique retro try h#1, and the sound h#2, yes? (2022-03-23, hidden) more ...
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Mike Neumeier: The solution appears to be 1.fxe4 e.p. 2.e2 3.exf1=L 4.Ld3 5.Le4 6.Ld5 dxc3#. Was that the intention? (2013-02-18)
Arno Tüngler: There would even be a solution in 5 moves by 3.exf1=S 5.Sd5 Sb5#
However, in order to justify the e.p.-key there must be an AP prove that e2-e4 (and not a move by the wK or wRa1) was played as White's last move. Thus the only solution giving this in 6 moves is 1.fxe3 e.p. 2.e2 3.exf1=S 4.Sxd2 5.Sxb1 6.Sxa3 0-0-0!# (2013-02-18)
Mike Neumeier: It struck me as odd there was no solution listed. (2013-02-18)
Mike Neumeier: And, if we let black be the idle side, there is 1.Txg1 2.Lxh1 3.Kd1 4.Kc1 hxg1T#. What does the asterisk(*) mean? (2013-02-18)
Henrik Juel: The asterisk means that there is also a white mate if he had the move, in this case two mates, so maybe the stipulation should be ser-h#6**:
1.dxc3,Sb5# (2013-02-18)
Ladislav Packa: It is clear that e.p. is possible only when the wK and wRa1 done neither move.
What convention is used here for the right to castling? (2013-02-18)
Henrik Juel: White may castle, unless you can show that he has lost the right to castle. In this problem there is a major difficulty, I think: the position seems illegal.
White pawns captured b2xc3xd4xe5xf6, g2xf3, and axb, promoting on b8; Black captured bxc, dxc, and g3xh2. We cannot explain the destiny of [Ph7].
Reversing wPh3 and bPh2 seems to handle the illegality, but then the problem can be solved in 5 moves (2013-02-18)
Mike Neumeier: Thanks, Henrik. Maybe the stipulation was a typo. Considering all comments together, perhaps the stip should have been ser-h#5**, with the one solution being the 5-mover Arno gave. I do not think it can be proven, except by a stipulation of ser-h#5 that en passant is possible. Which leads to the question of convention. Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible? Whether there is castling or not here seems immaterial. It is just another 6-move solution. There are 19 6-movers (Popeye) with one queenside castling among them. But only the one 5-mover. (2013-02-18)
Henrik Juel: By convention, an en passant capture as first move is not permitted, unless the pawn double step can be shown by some kind of retro analysis.
(Conversely, a castling is permitted, unless it can be shown by retro analysis that the right to castle has been lost, i.e., that king and/or rook has moved) (2013-02-18)
A.Buchanan: Suggested repair: Remove bPa4. Add AP to stipulation.
As Henrik points out, the diagram as it stands is illegal. We can't swap wPh3 & bPh2, because that allows 2 h#1 cooks.
Suppose we remove bPa4 instead. Then we still at least 5 White pawn captures: bxcxdxexf6 & gxf3. But wPh3 did still move from h2, so there are 3 Black pawn captures bxc, dxc & gxh2. So [bhP] died without capturing or promoting. [waP] must have promoted - so either [baP] was captured to clear the way, or [waP] captured to promote, and later [baP] promoted. Either way, all the numbers add up, and the position is legal. wPe4 cannot have just come from d3. So AP is triggered. Forward logic works just as before without bPa4. There is 1 5-move try, and 18 6-move tries, but AP eliminates them all because of need for castling to retrospectively justify the ep. (2013-02-23)
A.Buchanan: Mike asked: "Can the existence of a unique solution as implied only by the stipulation be used to determine that en passant is permissible?"
Answer: No. The stipulation does imply a default player to move, but otherwise cannot be used as a premise to determine game state (castling, en passant). (2022-03-21)
A.Buchanan: WinChloe & Yacpdb have the same cooked diagram as PDB, so I think this cook was in the design. I would like to post a sound version with bPa4 removed, and I think it would just be “Zigman correction” with no one else’s name mentioned in the credits. Is everyone ok with this? (2022-05-30)
Mario Richter: 'Bilten 1970' (printed 1971) contains on pp. 62-81 a complete report on the "Problem Duel Slovenia - Macedonia 1969", the problem here can be found on page 79 and is printed there exactly as given here.
Andrew's handling of the suggested correction (remove black pawn a4) is o.k. for me. (2022-05-31)
A.Buchanan: One question was not raised all those years ago. How does the "series mover" condition affect AP? How does flipping the player change our knowledge of the game history? If it's *consequent* series mover, then the flip erases all history: we use retro analysis to derive any information of the history of the game. But for regular series movers, the history is in some sense retained. Here, White's castling rights in the diagram position are the same as those after 6 single moves by Black. Is this correct? (2022-03-20, hidden) more ...
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Paulo Roque: Illegale Stellung. Diagrammfehler? (2009-11-17)
Alfred Pfeiffer: außerdem in b) 9 schwarze Bauern. (2010-01-13)
A.Buchanan: Twin a) is also illegal, as too many Bl captures (2021-11-24)
A.Buchanan: I couldn't find a single typo to fix, but 8/p1p2n2/1B3pp1/q1p1b2b/r1pPk3/1Bp3P1/PPP2PPr/3n1RK1 I think works. It changes sBf7 for sS, and shifts sLh7 to h5. Popeye v4.87 and Retractor 2.0 are happy with it. The last move in (a) is not unique, but that's ok in principle, because we just need to know that e.p. is not on. I suspect however that (a) is intended to have 0-0 as the definite last move. This problem is not to be found in WinChloe or YACPDB. Anyway, the new diagram is a good start. This also shows the practical utility of Retractor 2.0 (2021-11-24, hidden)
A.Buchanan: Replacing a sBf with sS sorts out the retro play. In (a) Retractor 2.0 suggests R: 1.Kh1-g1,0-0,d2-d4. The last is impossible, because wBe would promote on light square. But the e.p. convention is satisfied in any case. In (b) Retractor suggests only R: 1.d2-d4 and now wBh promotes to dark square bishop. With sSf7, the forward play in (a) is perfect, but (b) has 3 e.p. mates to choose from. (With sSf6, the situation seems less promising.) (2021-11-24, hidden)
A.Buchanan: I think 8/p1p2n1b/1B6/q1p1b3/r1pPkpp1/1Bp3P1/PPP2PPr/3n1RK works, forward and back, but it would be nice to have a definite last move in (a). (2021-11-24, hidden) more ...
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A.Buchanan: Two aspects of cookery here. First, NL 1. Db2,Sb3 Sxe2 2. Sb3,Db2 0-0 3. Lxb4 Sxb4# Second, White can retract c3xb4 (and earlier captured onto c-file). Black could have captured axbxcxdxe, exf, fxe, with wPg waylaid (2021-11-23)
A.Buchanan: Two aspects of cookery here. First, NL 1. Db2,Sb3 Sxe2 2. Sb3,Db2 0-0 3. Lxb4 Sxb4# Second, White can retract c3xb4 (and earlier captured onto c-file). Black could have captured axbxcxdxe, exf, fxe, with wPg waylaid. How about this to repair: 8/8/8/b1p1PP2/1PpqPpp1/r2ppb1p/k1N1p2P/n3K1NR (2021-11-23, hidden)
A.Buchanan: I think 8/8/1p6/b1pPp3/1PpqPp2/r2ppb1r/k1N1p2P/n3K1NR fixes this. (1) I have a (very non-unique, very inefficient) demo game in lichess.org (51.5 moves) which leads to the position prior to the last move, and castling is still legal (2) retractor 2.0 says that the only possible last move in the diagram is b2-b4 (3) Popeye says that the only solution is the same as Kardos' original intended solution. I think these three elements constitute complete C+ for such a problem. (2021-11-23, hidden) more ...
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No. 11669 HN
paul: Cooked: 1.K×e4 Tg1 2.Ke3 Tg5 3.Sd4 T×e5# (2011-11-25)
A.Buchanan: This nice simple puzzle can be repaired by replacing pawn c5 with rook (C+). As often with cooks in PDB, one wonders if it's just a typo. I have an old email address of Ronald's, and I may ask him. (2012-02-20)
A.Buchanan: I met Ronald for dinner last year (it seems like a vanished age!) in Norwich, UK. He was there for a mountaineer's conference. The climbers selected Norwich as their venue for humorous reasons as it's in the flattest area of Britain. We discussed a bunch of compositions, including this one. He agreed it was cooked and agreed with my suggested fix, which I will post in a separate entry. Very nice chap - he doesn't compose much these days because he can get a bigger audience (and more money) from writing hill-walking books. My hill-walking friends say he's a good writer. (2020-12-08)
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SCHRECKE: Konnte zuletzt nicht auch 0. ... Kg7-h6 geschehen sein? (2021-02-15)
A.Buchanan: I think the diagram must be wrong. Not only does the retro logic fail, but also there is a dual 2. Df3+,Dg4+. It's not in WinChloe. (2021-02-15)
Mario Richter: An additional white Pawn e7 would justify the epkey, but not prevent the mentioned dual. (2021-02-16)
VL: We might shift wPf2 to f3 against the dual. Or exchange Bf6 with Pf2 and add wP on f3 (instead of e7)... (2021-02-16)
A.Buchanan: Shifting wBf2 to f3 leads to short cook solutions 1.Dh2. However I think 8/5p1p/4NB1k/3PPKpP/4P3/8/5PP1/2Q5 fixes the retro issue, and it's C+ by Popeye, with still ideal mate! The composer Tomislav Petrovic is still alive but would be very old (90). Is he still active? (2021-02-17)
A.Buchanan: By the way, if you query for k='foo', that will always return uses of the keyword "foobar". It's just checking for the prefix. If you omit k='foo', then that will also block "foobar". Therefore if we tag with "en passant as key", then we should not tag with "en passant" too. Both keywords should exist, and over time we may refine uses of "en passant" to be more precise.
Because of the sensible way prefixes are handled in PDB, this is a different situation from tagging a problem as both "ceriani-frolkin" and "prentos", which we should do.
On this general subject, when counting instances within a single problem, we should always use Gerd's excellent mechanism of parameters "foo:1" etc, rather than having separate keywords "foo 1", "foo 2" etc. Otherwise the set of keywords becomes unmanageable. (2021-02-17)
A.Buchanan: wDc1 fixes the dual but the retro soundness is another matter. Current position has ideal mate, which it would be nice to retain (2021-02-17, hidden)
A.Buchanan: By the way, if you query for k='foo', that will always return uses of the keyword "foobar". It's just checking for the prefix. If you omit k='foo', then that will also block "foobar". Therefore if we are having "en passant as key" as a keyword, then we should have "en passant" as a keyword too. Because of the sensible way prefixes are handled in PDB, this is a different situation from having both "ceriani-frolkin" and "prentos" as keywords. (2021-02-17, hidden)
A.Buchanan: By the way, if you query for k='foo', that will always return uses of the keyword "foobar". It's just checking for the prefix. If you omit k='foo', then that will also block "foobar". Therefore if we are having "en passant as key" as a keyword, then we should not have "en passant" as a keyword too. Because of the sensible way prefixes are handled in PDB, this is a different situation from having both "ceriani-frolkin" and "prentos" as keywords. (2021-02-17, hidden)
A.Buchanan: By the way, if you query for k='foo', that will always return uses of the keyword "foobar". It's just checking for the prefix. If you omit k='foo', then that will also block "foobar". Therefore if we tag with "en passant as key", then we should not tag with "en passant" too. Both keywords should exist, and over time we may refine uses of "en passant" to be more precise. Because of the sensible way prefixes are handled in PDB, this is a different situation from having both "ceriani-frolkin" and "prentos" as keywords.
By the way, we should always use parameters "foo:1" etc when counting instances within a single problem, NEVER have separate keywords "foo 1", "foo 2" etc, or the set of keywords become unmanageable. (2021-02-17, hidden) more ...
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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)
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b) Urdruck in Die Schwalbe 228, 12/2007
A.Buchanan: Any ideas about (b)? (2022-06-09)
Henrik Juel: Andrew, you can find Heft 228 on
https://www.dieschwalbe.de/archiv.htm
part b) is mentioned in an article by Valery Livkovets, p.299-304
diagram p.301, solution p.303 (2022-06-09)
A.Buchanan: Thanks Henrik - just the pointer I needed. Please check the detailed solution I'll post above. (2022-06-09)
A.Buchanan: OK the logic nearly works for me. But I don't understand why White can't just play 1. Txb2# in part b, because b2 is now covered by wD. What am I missing? (2022-06-09)
Henrik Juel: Indeed, 1.Txb2# looks like a cook (2022-06-09)
A.Buchanan: Werner agrees. He has fixed it with a better twinning condition (2022-06-12)
A.Buchanan: I guess (b) is AP Type Keym but I don’t see it clearly (2022-06-09, hidden) more ...
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Henrik Juel: This un-dummy-pawn is like a pawn created on c8 by the Einstein condition
The retroplay looks wrong, what about 1... d8-d6!? 2.d7-d8=sB!? c6xYb5 (2020-04-30)
James Malcom: That makes much more sense Henrik. I have edited it accordingly. (2020-04-30)
A.Buchanan: You don't like Adrian's (: haha? What about :) then? In heraldic terms, it would be described as "regardant". See 270 on https://www.heraldica.org/cgi-bin/atlas.pl?12. I don't know the word for "non-regardant", which is the default direction for the head, e.g. most of the others on that page.
We could replace wBa6a7 with wSa6, but alas having the 3rd knight is non-thematic & distracting. The other thing I would've liked to do is empty b2 so the promotion is forced to be to D. Such a tiny problem should have precision promotion. Needs more thinking. (2020-05-01)
James Malcom: UPDATE: The position has been edited to have one less piece and a precise queen promotion. (2020-10-12)
A.Buchanan: There is dialogue here which is now orphaned by a change of diagram with new matrix. Suggest we revert to the older diagram and mark it superseded by a new PDB entry here (2020-10-12)
James Malcom: I agree-it is now done. (2020-10-12)
James Malcom: Oh, it’s (:! I remembered it wrong!

Also, here is one idea to make it so that queen promotion is the only one: k7/Pp1N4/1PpK4/Rp6/1P6/8/8/8
1. dxc7ep b6, and 2. c8=Q# to cover the newly vacated b7 square. Retro play is the same. (2020-05-01, hidden)
A.Buchanan: I don't think that works. Bl can retract a6xb5, and more comically: b8-b7 or c8xb7 (2020-05-01, hidden)
James Malcom: You’re right- it doesn’t work at all. (2020-05-01, hidden)
A.Buchanan: Let's just stick with the existing humble but functional version (2020-05-01, hidden)
James Malcom: Agreed. (2020-05-02, hidden) more ...
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01/1960, Seite 12: Der Autor ändert seine Aufgabe: wBf2->f5, sBh7->g6, sBh5->g3. So aber unlösbar weil ep nicht zulässig ist. (wBd2-e3-c4)
Felber, Volker: Der wBe4 kann von d2 über e3 nach d4 gekommen sein, somit war nicht zwingend d2-d4 der letzte weisse Zug. (2021-05-13)
A.Buchanan: sBs have captured 6 times e.g. fxgxh, exfxgxh, dxe so sBa, sBb & sBc never captured. wBs have captured visibly exfxg, but also e.g. axb=, bxa, cxd=. Thus all captures are accounted for, and Sf4 did not just capture. White's last move was not e3xd4, because too many pawn captures. So Wh *may8 e.p. The non-e.p. solution is probably best regarded as a cook. WinChloe marks this problem as cooked. I guess we can have a "C+ cooked" problem? (2021-05-13)
A.Buchanan: If my analysis is ok, then the problem can be fixed by transposing sBh5 & wBh3. (2021-05-14)
A.Buchanan: Volker usefully posted an update in the Comment field: "01/1960, Seite 12: Der Autor ändert seine Aufgabe: wBf2-f5, sBh7-g6, sBh5-g3. So aber unlösbar weil ep nicht zulässig ist." I agree with him that this new version 8/6p1/6pb/5P2/P1kPpNPR/n1p2BpP/p5Pp/2Kb4 has no solution: White might simply retract Th5-h4. More subtly, White might just have played Sf4 in response to f4xg3+! In the original, the ep *was* permitted (e3xd4 & Sxf4 impossible due to the 3 white captures on queenside). My earlier suggestion of swapping sBh5 & wBh3 remains, and has the advantage that the impossibility of g3-g4 now also depends on the capture count. (2021-05-15)
HBae: Lösung 1.exd3ep g5 2.Lb3 Se6# (2021-05-12, hidden)
A.Buchanan: This has the advantage that it's now *slightly* more obscure why White can't just retract g3-g4. (2021-05-15, hidden) more ...
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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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A.Buchanan: Found in WinChloe. It can be fixed by e.g. +sSa3. Is this a transcription error? What does feenschach reprint show? (2022-05-24, hidden) more ...
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Henrik Juel: a) C+ Popeye 4.61
1.fxg6 Txd5 2.Le5 0-0-0 3.g7 Th8 4.gxh8=DT#
b) no solution (2022-05-27)
Henrik Juel: I shall leave the analysis to you, Andrew... (2022-05-27)
A.Buchanan: OK, Henrik: please check my suggested solutions & rationale :) I find this a beautiful problem (2022-05-28)
A.Buchanan: In (b), is it possible that Black has just played RxQBh5? If this is feasible, it would apply even if Black 00 rights are still on. Cook? (2022-05-28)
Mario Richter: Looks indeed like a cook.

One can manually retract e.g. R: Th4xLh5 and use popeye to check if in the resulting position the reverse of this retraction is legal:
begin
remark Position after R: Th4xLh5
pieces white Bb8 Pd5 Pf5 Bh5 Ph3 Pd2 Ka1
black Ra8 Ke8 Pb7 Pd7 Ph7 Pa6 Pg5 Rh4
cond Maximummer
option HalfDuplex
stip h~0.5
end

This gives:
1...Rh4*h5
1...Ke8-e7
1...Ke8-d8
1...Ke8-f8

solution finished. Time = 0.013 s (2022-05-29)
A.Buchanan: Thanks Mario. I showed this problem on the international zoom call last night, and Siegfried spotted the cook immediately. Can g4 and h4/h6 be plugged suitably with pawns? Can't just shift bPh6 to h5, because it means that in a), bPg6 might have just captured checking Q/B, while not disrupting Black 000 rights. But I think can shift wPh3 to h4, together with adding Pg4 (Black or White - I prefer White aesthetically). Does this work? (2022-05-29)
Henrik Juel: Popeye 4.61 finds no solutions (in both twins) (2022-05-27, hidden)
A.Buchanan: Sorry my typo pls try again (2022-05-27, hidden) more ...
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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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