Rosalie Fay: White has lost only the bishops. So the pawn on c5 is not [Pa7] (because that entails 2 White units captured on black squares). White has played axbxcxdxe7, dxe, fxe, hxg, gxfxe. Black has 7 units, so white pawns have captured all missing Black units, but none on the a or h files.

Black has 2 pawns on the c-file, so one has captured. Thus [bPa7] and [bPh7] pawns have collectively captured no more than once. So at least one of them must have promoted, in order to either get to a file where White made a capture, or replace a captured unit; it didn't capture en route to promotion, so it displaced a white rook and thus spoilt one White castling right.

White would mate by 1 Rd1 & 2 Rd6 or 1 Rf1 & 2 Rf6, except that Black threatens Qxe2+. So either 1 0-0 or 1 0-0-0, though it's impossible to say which is legal. (2022-11-24)
Henrik Juel: one solution, but in two parts
if Ta1 has moved, 1.0-0 thr. 2.Dc8,Tf6#
if Th1 has moved, 1.0-0-0 thr. 2.Dg8,Td6# (2022-11-25)
Hans-Jürgen Manthey: nach der möglichen Zugfolge: 1. Sb1-c3 c7-c6 2. Sc3-d5 Dd8-b6 3. Sg1-f3 Db6-b3 4. a2xDb3 a7-a5 5. Sd5-b4 a5-a4 6. Sb4-a2 e7-e5 7. Sf3-h4 Lf8-c5 8. Sh4-g6 f7-f5 9. Sg6-f4 g7-g5 10. Sf4-g6 Lc5-e3 11. d2xLe3 f5-f4 12. g2-g3 Ta8-a5 13. g3xf4 g5-g4 14. f4xe5 g4-g3 15. h2xg3 h7-h5 16. Lf1-g2 h5-h4 17. Lc1-d2 h4-h3 18. Ld2-b4 Th8-h4 19. c2-c3 Th4-c4 20. Sa2-c1 a4-a3 21. Lb4-c5 a3-a2 22. Dd1-d4 Sb8-a6 23. Dd4-h4 Sa6-c7 24. Dh4-d8+ Ke8-f7 25. b3xTc4 Sc7-d5 26. c4xSd5 Sg8-e7 27. d5-d6 Ta5-b5 28. d6xSe7 d7-d6 --- folgt nun
29. Lg2-e4 Lc8-e6 30. Le4-b1 a2xLb1D 31. Th1-g1 Db1-d3 32. Sc1-b3 Le6-c4 33. Th1-g1 h3-h2 34. Dd8-e8+ Kf7-e6 35. Tg1-h1 Tb5-b6 36. Th1-g1 Le6-c4 37. Tg1-h1 h3-h2 38. Th1-g1 h2-h1D 39. Sc1-b3 Dh1-e4 40. f2-f3 Lc4-a6 41. f3xDe4 d6xLc5 42. Dd8-e8+ Kf7-e6 43. Tg1-h1 Dd3-b5 matt in 2:
1. OOO droht 2. De8-g8/Td1-d6# - 1. ... Db5-d3 2. Sb3xc5# oder:
29. Sc1-b3 h3xLg2 30. Sb3-d2 g2-g1D+ 31. Sd2-f1 Dg1-g2 32. Sf1-d2 Dg2-e4 33. f2-f3 Tb5-b6 34. f3xDe4 d6xLc5 35. Sd2-b3 Lc8-e6 36. Ta1-b1 Le6-c4 37. Ta1-b1 Kf7-e6 38. Tb1-a1 Lc4-a6 39. Ta1-b1 a2-a1D 40. Dd8-e8 Da1-a4 41. Tb1-a1 Da4-b5 matt in 2: 1. OO bel. 2. De8-c8/Tf1-f6# (2023-02-22)
A.Buchanan: Could we shift wPb2 to c2, and eliminate bR or bB? (2022-11-25, 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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VL: One solution consists of two partial ones, and there is another, castling independent, solution. A rare scenario; cf. P1091926. Presently, due to Art.16 (3) of the Codex, the mark "RV" may be deleted from the stipulation. (2017-11-25)
A.Buchanan: Methodologically, I think that one applies PRA first. So there are two parts, and each has two solutions. One solution is shared across both parts (2023-08-22)
A.Buchanan: I think "RV" used to mean *all* retro variants considered as separate histories, but since Werner Keym's revision, its meaning has been changed to be synonymous with PRA. The interpretation of this problem is now considerably less smooth than it was before. I would say that normally we must split into PRA parts before considering the number of solutions. So for me there are 2 parts, and each has 2 solutions, one of which is shared across the two parts, but that is a different way of fiddling around from VL's idea.
At what point was "2.1..." added to the stipulation? Of course, the problem works as PRA somehow, but I don't think this is a justification for changing the meaning of this term. And without an "originalforderung" field, we have no way to track edits to the stipulation.
I think this is a case for Golden Age: to preserve this old problem in its original form. (2023-08-21, hidden) more ...
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Henrik Juel: 1.0-0 thr. 2.Dxc6+,Td1 (2023-07-18)
A.Buchanan: I don't think it's as straightforward as this. Under PRA, there are 2 parts: a) wK & sD rights ok b) wD & sK ok. We tackle each separately. a) 1. 0-0? 0-0-0! 1. Td1! b) 1. 0-0-0? 0-0! 1. Tf1! The forward play is messy, and there are three promoted units, but the amazing paradox is that White wins by not castling. This is not a case of mutual exclusion. This is why it got first prize, I think. (2023-07-18)
Mu-Tsun Tsai: Yeah, on second thought I indeed got it the wrong way; it is in fact possible that Kq castling are both valid (I have the proof game in fact), and I take your word that the other side is also true. (2023-07-19)
A.Buchanan: For complete retro proof, need both retro logic & non-unique proof games. For the latter, maybe can show the legality of rn2k1nr/ppp1p1pp/4P2B/4p3/2P1Q3/2P5/PP2P1PP/R1b1KB1R. From here, can branch to the two ways of getting the two rook promotions: (a1+h8 vs h1+h8) (2023-07-19)
Henrik Juel: If Andrew's analysis holds, the keys are correct by Popeye 4.61
But there are duals in each part (2023-07-20)
Mu-Tsun Tsai: In this position, it is possible that Black can still castle queen side, and it is also possible that White can still castle king side, but it is not possible that both castling are still valid. Therefore 1. 0-0!! "steals" the castling rights of Black by chess problem conventions. (2023-07-18, hidden) more ...
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A.Buchanan: The retro content here is fun, but I don't see why it's suggested that this is AP. (a) is a cant castler, and (b) is a standard PRA matrix. Removing AP as a keyword for this one. (2023-07-21)
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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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See P0003365 (cooked) & P1411659 (fixably cooked but non-standard thematic material).
This new version has an obvious promoted unit (non-thematic), but Joaquim Crusats (a noted constructor) could not find any way to do better
Ladislav Packa: Mr. Bebesi's first name is Gyula (not Gyulia). It is the Hungarian version of the name Julius. (2023-12-05)
A.Buchanan: Thanks - edited (2023-12-06)
Henrik Juel: if last move was b2-b4: 1.axb3ep bxc6+ 2.b5 cxb6ep#
if last move was d2-d4: 1.exd3ep Dxg8+ 2.d5 cxd6ep# (2023-12-04, hidden)
Henrik Juel: HC+ Popeye 4.61 (2023-12-04, hidden) more ...
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