James Malcom: (Shortest) proof of legality: 1. d4 Nh6 2. Bxh6 gxh6 3. g4 Rg8 4. g5 Rg6 5. Nh3 Rf6 6. Nf4 Nc6 7. g6 Nb4 8. g7 Rg6 9. Nxg6 fxg6 10. Bh3 Nd5 11. Be6 dxe6 12. f4 Kd7 13. f5 Kc6 14. f6 Bd7 15. Rf1 Be8 16. a4 Bf7 17. a5 Bg8 18. f7 Qd6 19. Rf6 exf6 20. e4 Be7 21. e5 Bd8 22. exd6 Ne7 23. dxe7 Rc8 24. d5+ Kb5 25. d6 Kc6 26. d7 Kb5 27. Qd6 cxd6 28. a6 Kc6 29. c4 Kb6 30. c5+ Kb5 31. c6 Kc5 32. c7 Kb5 33. Na3+ Kc6 34. Nc4 Kb5 35. Nb6
axb6 36. Ra5+ Kb4 37. Rc5 bxc5 38. a7 Kb5 39. b4 Kc4 40. b5 Kd4 41. b6 Ke4 42. Kd2 Ke5 43. Kc3 Kd5 44. a8=B Kc6 45. Kc4 Rb8 46. c8=R+ Bc7 47. d8=N+ Kd7 48. e8=B+ Ke7 49. f8=Q#

Took me a bit to figure out the trick for maneuvering the Black bishops. (2022-08-28)
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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 63 Sekunden.
Keine Lösung: BP 29.5, BP 30.0. (2023-04-15, hidden) more ...
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Silvio Baier: C+ (Euclide 0.93) (2009-05-12, hidden)
Moldenhauer: Ergänzung Stelvio 1.2. Keine Lösung BP 25.5, BP 26.0. (2023-05-20, hidden) more ...
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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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Moldenhauer: Computerprüfung: C+ Stelvio 2.0 02:29:09 Stunden. (hh:mm:ss) (2023-12-23)
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"Promenades to Power"

Es läßt sich beweisen, daß die UWs in D,T,L,S zwingend erfolgt sind (a2-b8=L,h2-b8=S und entweder b2-c8=D/c2-c8=T oder b2-c8=T/c2-c8=D).
Erich Bartel: weitere Nachdrucke:
3) 160 Die Allumwandlung im Problemschach VIII 1966a---
4) Schach ohne Grenzen 1969.-- (2007-01-09)
Sally: Vier erzwungene Umwandlungen. (2012-02-21)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 da NUPG sonst cooked in 00:03:51 Minuten. (hh:mm:ss)
Keine Lösung: BP 33.0, BP 33.5.
Beispiel: 1.Sf3 Sf6 2.a4 Sd5 3.a5 Sf4 4.Ta4 c5 5.Td4 Db6 6.b4 Sc6 7.b5 Tb8 8.bxc6 cxd4 9.c7 e5 10.d3 La3 11.Le3 Lc1 12.c4 Db2 13.c5 b5 14.c6 Tb6 15.axb6 Ke7 16.b7 Kf6 17.b8L Lb7 18.c8D b4 19.c7 d5 20.Dh3 b3 21.c8T h5 22.g4 h4 23.Dg3 hxg3 24.h4 Th6 25.Lh3 dxe3 26.0–0 Sg2 27.Sh2 Se1 28.h5 Kg5 29.Td8 Ta6 30.Ld6 Kh4 31.h6 Ta1 32.h7 a5 33.h8S a4 34.Sg6+ fxg6
Umwandlungen: 17.b8L, 18.c8D, 21.c8T, 33.h8S. (2023-04-19)
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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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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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Erich Bartel: Auszeichnung: 1.ehrende Erwähnung.--
Nachdruck: 1) Jaarboek 1970.-- (2006-11-26)
Henrik Juel: Something is wrong: De5 does not mate. And the intention certainly was not -1.S=a7 L=h2, 0... Kxc5 1.h8DL Kb6 2.DLd4#. (2006-11-27)
Erich Bartel: Ich habe die Aufgabe aus dem Jaarboek 1970 übernommen, wo
sie wie angegeben im Diagramm und Lösung so steht. Es
scheint also bereits hier ein Fehler vorzuliegen. Zur Klärung
ist es wohl notwendig in der Ur-Quelle Probleemblad -die ich
nicht besitze- nachzusehen. (2006-11-27)
Siegfried Hornecker: No matter how the position is changed, there can be no line where the king, coming from b6, is not allowed to return to b6 when the queen checks on e5. Also, there would be no reason to not play again 2.h2-h1~ as a waiting move.

The retroanalytics say that the last move must(!) have been 1.a7-a8S+ h2-h1L so I think the correct stipulation is:
a) Last move (w+s)
b) h#2

I have no source for this but if you think about it, the whole thing makes sense! The last move was 1.a8=S+ after 1.h1=L and from the diagram position the h#2 runs 2. Kxc5 h8=D 3. a1=T De5#. That's what I think. (2006-12-12)
Jan Hein Verduin: I got the Jaarboek too, and mr. Hornecker is right. The stipulation there is "h#2; what where the last w+b moves?" The way the solution then is given (the retro moves are given first) probably caused the confusion. (2007-01-26)
Adrian Storisteanu: Then, ceci n'est pas un ... retractor!? (2024-03-17)
Henrik Juel: Right, Adrian
The Help retractor keyword should be deleted (2024-03-17)
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Henrik Juel: C+ Popeye 4.61 (without the retraction, which is unique) (2022-11-08)
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No. 1449 HN
AV: Je crois que c'est un Fou blanc en h8 et non un Cavalier, et les derniers coups sont gxFh8=F+ Rf6-e7 fxDe8=D+ (au lieu de fxCg8=C#). Avec l'antérieur cxTb8=T, cela donne une "Allumwandlung" avec une pièce noire prise de la même nature. (2005-06-09)
Henrik Juel: You must be right, AV, because with wSh8 the retroplay might be the bland -1.f7xDe8=D Kf6 etc. without unpromotion on h8. (2005-06-13)
Michel Caillaud: Indeed the diagram is wrong : there should be a white Bishop on h8
AUW spread over retro play (-1.fxDe8=D+ Kf6-e7 -2.gxLh8=L+ ...-n.cxTb8=T) and forward play 1.fxSg8=S#
Matching nature of promoted piece and piece captured by promotion. (2023-03-25)
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Henrik Juel: Stelvio 1.31 found the solution immediately and tested it unique in 9 minutes (for the first 25.0 moves) (2023-06-13)
Henrik Juel: C+ by Natch 3.1 (38 m, utilizing the final ep)
No solution in 24 or 25 (2018-12-23, hidden) more ...
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Mario Richter: Vermutlich fehlt ein Stein auf d7, andernfalls hat Schwarz sehr wohl einen letzten Zug (nämlich Bd7xc6). (2010-08-25)
Frank Richter: Warum ist die Allumwandlung hier unecht? Es gibt doch vier unterschiedliche Fortsetzungen. (2010-08-25, hidden)
Anton Baumann: Mit +wBd7 C+ Gustav 4.2a (2024-04-02, hidden) more ...
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Henrik Juel: Natch 3.1 did not find the solution in 4 hours
Stopped at pos. 409 (2019-01-01)
Henrik Juel: solution
1.e4 h5 2.Dg4 hxg4 3.b4 g3 4.b5 gxh2 5.b6 hxg1=S 6.bxa7 Txh1 7.axb8=T Ta5 8.a4 Tah5 9.a5 g5 10.a6 Lg7 11.a7 Le5 12.La6 bxa6 13.d4 Lb7 14.d5 Lc6 15.d6 La4 16.dxc7 Ld6 17.c8=L Db6 18.a8=D De3+ 19.fxe3 Kd8 20.Lb7+ Kc7 21.Tc8+ Kb6 22.Db8 Ka5 23.La8
Cyclical interchange by promoted Lc8-Da8-Tb8 (2019-01-01)
Moldenhauer: Computerprüfung: C+ Stelvio 1.4 53 Sekunden. 4736 Strategien zuvor
mit 8 Seekern auf Festplatte gespeichert 28:07:27 Stunden. (hh:mm:ss)
7747 Strategien mit Stelvio 1.11 berechnet in 169:58:46 Stunden.
Keine Lösung: BP 22.0 mit 16 Seekern 1 Player, 06:23:04.
Hier wurden 1150 Strategien berechnet.
Notation Stimmt! (2023-06-24)
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Henrik Juel: C++ Euclide 1.01
Euclide found the solution quickly and spent som 25 minutes to verify uniqueness
Here C++ means that Euclide found no solution in 12 or 11 moves
The testing times could have been reduced by utilizing that last move must be g2xLf1=T+ almost certainly (2022-10-30)
A.Buchanan: Thanks Henrik - unique PG in 13.0 implies there is no PG in 11.0. For comprehensive check of short solutions it suffices to check deltas 0.5, 1.0 & 1.5. Normally these are way easier to check than the intended length. And if there’s check in the final position as here, then half the tests are free! Is that C+++ in your notation? :) (2022-10-31)
Henrik Juel: No, Andrew, I think that C++ suffices (2022-10-31)
Henrik Juel: 1. a4 h5 2. a5 h4 3. a6 h3 4. axb7 hxg2 5. h4 a5 6. h5 a4 7. h6 a3 8. h7 a2 9. hxg8=S axb1=D 10. Txa8 Dxc1 11. Txb8 Da1 12. Txc8 Dxc8 13. bxc8=L gxf1=T+ (2022-10-30, hidden) more ...
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Michel Caillaud: 15...Dbxb2[+wBb6] and 16...Ta7 can be interverted (2022-12-28)
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Henrik Juel: Stelvio 1.0 found the solution in a couple of minutes
Testing was stopped after half an hour, and no cooks were found (2023-03-11)
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corrects P0003922
Henrik Juel: R: 1... h2-h1=S+ 2.a7-a8=L Kc8-d8, 1.h1=T a8=D# (2023-04-13, hidden)
A.Buchanan: Mario was kind enough to check this 2-Jul-2021. I hope it's ok if I quote his email:

The trick with the light-squared black Bishop is fine,
but makes things too complicated for rawbats.
It will happily retract e.g.
R: 1. ... h2-h1=S+ 2. Lb7-a8 Ke7xDd8, dann 1. Kd6 Df6#
without noticing that the retractions lead to a conflict
on the g-file: either wPh2 captured the original Bishop c8
on the g-file, then bPg7 was not available a a capture object
for the remaining white pawn captures, or wPh2 captured the
original bPg7 on the g-file, then the original light-squared
black bishop c8 was not available a capture object
for the remaining white pawn captures f2xe3xd4 (both on dark squares).
So brain-checking your retractor and your logical argumentation
I found no defects, and I think the problem is sound now. (2023-04-14, hidden) more ...
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Yuri Bilokin: more economical -wBg5, -wSh7, -wPa3, bQh8(-bRh8) 6kq/3P1NP1/8/8/8/1K6/p1P5/1b6 (5+4) H#1.5

1...d8=S 2.a1=R gxh8=Q# (MM)

Tempo promotion
Promotion (QSr, 2, 1)
Tempo move (bP, waiting, type 1)
Model mate × 1 (2023-08-04)
Henrik Juel: C+ Popeye 4.61 (2023-08-04)
A.Buchanan: heh Yuri the 4th promotion is in the retraction: R: 1. b2-b1=L. The other pieces serve to force retro-uniqueness. (2023-08-04)
Olaf Jenkner: Vergleiche mit P1190770 oder P1190771 (2023-08-04)
A.Buchanan: Very nice! These remind me of my own P1408845. (2023-08-05)
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See P0005542.
A.Buchanan: WinChloe incorrectly reports this as M.Rittirsch. But what is the true source & date of this correction? Obviously, it wasn't F.Lapya. (2024-01-19)
Adrian Storisteanu: I'd would have put a wB (instead of the wQ) on d8 -- a subtler non-standard material?!... (2024-01-19)
Adrian Storisteanu: (... Ooops, but then no 'AUW'.) (2024-01-19)
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