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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Henrik Juel: 1.0-0-0 (2.Txd7#). -1... Lh2 -2.Lg1, Sh5 uncaptures wPa6 and unpromotes on h1, then retract h2xg3, Lf2 via f4 to c1, d2xe3, Lg1 to f8, g7xf6 etc. (2004-01-12)
Henrik Juel: C+ Popeye 4.61 (2023-07-14)
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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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"O, W, thuis..."
Henrik Juel: 1.Kxc2 Sb3 2.. Rc1#. -1.. Ra2 -2.a3:B Bf8 -3.Bb4 Kb3 -4.Be7 a7:S -5.Sa4 Qd5
-6.Sc3 Qd4 -7.Sb1 Ra1 -8.Bf6 Kb4 -9.a2 Kc5 -10.Bg5 Kd6 -11.Bf6 Kd7 -12.Be7 Kc8 -13.Bf6 Kb8 -14.Bd8 e7 -15.B=d7 Kc8 -16.e6:R Rd8 -17.f5:B 0-0-0 -18.f4 Bc8 -19.f3 Qd8 -20.f2 d7:B -21.Bf3 Sc6 etc. (2003-08-26)
Yoav Ben-Zvi: An attempt at a coherent argument that reveals the intricate sequence of events in the resolution while refuting substantial alternatives:
The promotion of wBc3, on d8 or f8, requires 2 captures by the promoting wP. These, together with the captures by wPd3 and wPb4, account for all 4 missing Black pieces. Black captured [wBc1] at home. The other 7 missing White pieces were captured by bPs. [wBf1] cannot be released (by wPe2xd3) until [bPh7] plays bPd3xc2 completing the capture of 5 White pieces. Therefore [wBf1] was captured by bPd7xBc6 (the remaining capture by Black on a light colored square), releasing [bBc8]. This means that [wBf1] is released (by wPe2xd3) before [bBc8] is released (by capturing the previously released wB on c6) so the bB could not have been the piece captured on d3. It also could not be captured on b4 (dark colored square) so it was captured by White's promoting pawn. This means the promotion could not have been played by capture from e7 as that would imply that the 2 captures by the promoting pawn were both on dark colored squares in contradiction with the requirement to capture [bBc8]. Therefore the promotion move was wPd7-d8=B with bK standing outside the squares c8,d7,d8,e8.
Looking backward: To avoid White retrostalemate the first 2 retractions must be -1...bRa2-a1+ -2.wPa3xb4.
After this, unlocking the South-West cage and extracting bRa2 requires retraction of wPe2xd3 which requires previous uncapture of wB by bPd7xBc6, so it can be retracted home to f1. bP standing on d7 implies that a bB must previously be uncaptured and retracted home to c8, preceded by retracting BPe7-e6 to clear the path of uncaptured bB to c8. Retreat of bP to e7 is preceded by uncapture and return home of [bBf8]. With both bBs locked at home and bP on e7 the Black Central cage is locked by retracting bPd7xBc6 so bK and bQ must also be retracted home first. It follows that bKa4 needs to exit the West cage before it has been unlocked (to get back home to e8 before the Black Central cage is locked). The bK cannot exit via b5 as this implies retraction of wPc3-c4+ but, with c3 occuppied by wP, bRa2 does not have a valid exit path (not via c3 as it is obstructed by wP or via f1 as it will be occupied by wB). wP standing on c4 prevents retracting bN away from a5 so bK cannot exit via a5. The only valid exit of the bK is via b4 implying retracton of wPa2-a3 which requires bRa2 to retract to a1 (vacating a2 for the wP) which requires provision of a shield, standing on b1, to protect wKc1 from bRa1.
After the unpromotion of wBc3, White has only pawn retractions left to make until a White officer is uncaptured. This is not enough to allow Black to complete the retro-maneuvers required in preparation for retracting bPd7xBc6 (getting bBs,bK,bQ home). It follows that, at some point in the middle of its preparation maneuver, Black will need to uncapture a White officer which provides the tempo retractions needed to complete the maneuver. This is only possible by bPa7xb6. With bP standing on a7, [bRa8] cannot be retracted home via the a file so it must retract to its original corner via row 8 before the bB is locked at c8. This implies that [bRa8] is uncaptured by a wP prior to unlocking of the cage.
Uncaptures by White prior to the unlocking of the South-West cage (one available on b4 and two by pawn unpromoted from wBc3) include uncaptures of both bBs and a bR since, as noted in the above analysis, all 3 must be uncaptured and retracted home before retracting bPd7xBc6 (all of which precedes the unlocking of the South-East cage). This means that the bN is uncaptured later (on d3) so it cannot provide the shield on b1. The shield also could not be provided by [bBc8] as this bishop must be returned home when it is needed to provide the shield. The only remaining possibility is that the shield on b1 is provided by a wN that is uncaptured by bPa7xNb6. The wN on b1 is immobile until the wK can be retracted away from c1 so here too, once the unpromotion occurs Black must perform the preparations for retraction of bPd7xBc6 (which now includes return of the uncaptured bR to a8 or b8) under time pressure, completing the required maneuvers before White runs out of tempo retractions. Having determined the structural elements it is not too difficult to construct a resolution that succeeds in preparing for retraction of bPd7xBc6 before White runs out of tempo retractions. The resolution maintains the following: delay the unpromotion (known from analysis above to be wPd7-d8=B) while retracting bK to b8; bQ to d4 and bB (uncaptured on b4) to f8 and continuing by retracting -1.wBd8-? Pe7-e6 -2.Pd7-d8=B Kc8-b8 -3.Pe6xRd7+ Rd8-d7 -4.Pf5xBe6 0-0-0 -5.Pf4-f5 Bc8-e6 -6.Pf3-f4 Qd8-d4 -7.Pf2-f3 Pd7xBf6 -8.Bc6-? Nc6-a5. Following this the wB is retracted to f1 allowing the unlocking of the cage by wPe2xNd3 and bRa1 is retracted back home to h8 via c1,c3. One of the White pieces uncaptured by bPc2 unpromotes on h8. The alternative of uncapturing the bR on b4 (instead of bB) fails to complete the preparations on time.
To draw attention to the rich, unconditional, content of the resolution the stipulation could be "Squares that must have been occupied by bRa1". This problem deserves a full SPG. The best I can manage is a game of 53.0 moves. (2018-12-23)
Zvi Mendlowitz: Proof game in 44.0 moves:
1. Nf3 Nc6 2. Nh4 Nb8 3. Ng6 hxg6 4. Na3 Rh3 5. c4 Rg3 6. h4 Nf6 7. h5 Nd5 8. h6 Nb4 9. Rh5 N8c6 10. Rf5 Rc3 11. h7 Rxc1 12. h8=Q Rc3 13. Rc1 Rb3 14. Rc3 gxf5 15. Rd3 Rc3 16. Qh4 Rc1 17. Qe4 Ra1 18. Nb1 fxe4 19. Qc2 exd3 20. Kd1 dxc2+ 21. Kc1 Nd3+ 22. exd3 Nb8 23. Be2 Nc6 24. Bf3 Na5 25. Bc6 dxc6 26. f3 Qd4 27. f4 Be6 28. f5 O-O-O 29. fxe6 Rd7 30. exd7+ Kb8 31. d8=B e6 32. Bf6 Kc8 33. Be7 Kd7 34. Bf6 Kd6 35. Bg5 Kc5 36. Bf6 Kb4 37. a3+ Kb3 38. Be7 Ra2 39. Nc3 Qd5 40. Na4 Qd4 41. Nb6 axb6 42. Bb4 Ka4 43. Bc3 Bb4 44. axb4 Ra1+ (2022-09-19)
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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: Checked the first 16.5 moves with Euclide. (2014-12-04)
paul: Thr first 17.5 moves checked by Euclide (~29 hours) (2014-12-06)
Joost de Heer: Solution:
1. Nf3 b5 2. Nd4 Bb7 3. Nxb5 Bxg2 4. Nxa7 c6 5. Nc8 Rxa2 6. Nc3 Rxb2 7. Ra8
Qa5 8. Na2 Qxd2+ 9. Kxd2 Na6 10. Kc3 Nc5 11. Qd4 Rxc2+ 12. Kb4 Rxe2 13. Ba3
Rxf2 14. Ba6 h5 15. Ra1 Bf1 16. Qxg7 Rxh2 17. Qxg8 Rh4+ 18. Qg4 Bh6 19. Ka5
Bc1 20. Qa4 Rh6 21. Na7# (2001-11-05, hidden)
Moldenhauer: Computerprüfung: C+ Natch 3.3 in 302 Std. 48 Min.
Lösung 1 (Stellung 3391) Notation OK. (2022-11-24, hidden) more ...
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Henrik Juel: Natch 3.1 made no progress after half an hour (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:03:45 Minuten. (hh:mm:ss)
Keine Lösung: BP 23.0, BP 23.5. (2023-04-01, hidden) more ...
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paul: Here, we have also an interchange between Bc1 and Bf8. (2018-03-21)
Henrik Juel: The months should be 11-12, not 09 (2018-03-21)
Mario Richter: @Henrik: Is my guess correct that "9" might be the issue number? (i.e. issue 9 of 1998 was for the months Nov.+Dec.?) (2018-03-21)
Henrik Juel: No, Mario; I think the error might have occurred, because Andrej also had a proof game in the preceding issue
Back then PB had six issues per year, numbered 1-6 for each year (unlike Die Schwalbe), so the issue number contains little information
This problem appeared in issue 6, Nov-Dec 1998 (2018-03-21)
Henrik Juel: Natch 3.1 did not even find the solution in an hour (2018-12-10)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:49:22 Minuten. (hh:mm:ss)
Keine Lösung: BP 25.0, BP 25.5. (2023-04-06, hidden) more ...
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Henrik Juel: Natch 3.1 did little in an hour and a half (2018-12-10, hidden)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 00:08:49 Minuten. (hh:mm:ss)
Keine Lösung: BP 21.0, BP 21.5. (2023-03-12, hidden) more ...
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Das namensgebende [Stamm-]Problem für dieses Thema (=Forsberg-Zwilling).
Erich Bartel: weitere Nachdrucke:
13) 197 moving on 2008.---
14) Naef:Endspielstudien Hilfmattprobleme und Märchenschachaufgaben 1998,S.72.--- (2008-10-01)
Jacques Rotenberg: 5982, CHM 1995, G. Lestriguel
p.1315, Le Guide des Echecs 1993
5 p.8, The Modern Helpmate in Two, Z. Janevski & N. Stolev 1989
XIII p.151, Thèmes-64 10 (avr. 58)
XII bis p.38, Rex Multiplex 2 (avr. 82)
285, Svenska Miniatyrer I, A.Uddgren & A. Hildebrand 1996
p.76, Im Wunderland des Schachproblems, E. Ramin 1958
E-1 p.236, StrateGems 12 (oct. 00)
I33 p.1333, Phénix 14 (sept. 91)
3 p.219, Le Genre Aidé Supplément diagrammes 15 (juin 75)
161 p.241, Echecs, N. Ramini 1985
E p.14, Best Problems 21 (jan. 02)
259 p.238, Chess Problems : Introduction to an Art, M. Lipton, R.C.O. Matthews, J. Rice 1963
139, Skakopgaven 1942
135 p.31, Album FIDE 1914-44/III
9 p.403, feenschach 26 (déc. 74)
40 p.61, Europe Echecs 330 (juin 86)
p.185, feenschach 47 (juil. 79)
in WinChloe, you find also :
A p.5, Springaren 17 (mars 84)
p.50, feenschach 73 (jan. 85)
M p.171, The Problemist (sept. 81)
p.415, feenschach 97 (nov. 90)
133 p.104, Solving in Style, J. Nunn 1985
E-I p.43, Probleemblad 58/1 (jan. 00)
E-1 p.102, Probleemblad 58/3 (mai 00)
p.136, Springaren 70 (août 97)
4 p.4424, diagrammes 159 (oct. 06)
p.37, Problem-Forum 36 (déc. 08) (2012-03-10)
A.Buchanan: The position is unique (except for left-right reflection). In particular, it cannot be shifted up or down without breaking (b) and only the current location of wK makes (b) sound and (a)&(d) solvable. (2017-01-27)
Anton Baumann: in Revista de Sah 1936 soll Henry Forsberg eine weitere Möglichkeit gezeigt haben:
f) wSa6 statt sDa6: 1.Ka2 Sac5 2.Ka3 Ta4# (C+)
Ferner mit Märchenschachfiguren: vergl. P1349128 (2022-12-28)
Erich Bartel: Das namensgebende [Stamm]problem für dieses Thema (=Forsberg-Thema).--
Nachdrucke:
1) 141 Zwillinge und Mehrlinge 1952.---
2) Schach-Echo 11.Jahrgang,S.151, 1953.---
3) Problem (127-132) IX 1969,S.95.---
4) 9 feenschach (26) XII 1974.---
5) 83 Schach ohne Partner 1977.---
6) 67 100 classics of the chessboard.---
7) Belfort 1994,S.75.---
8) E-I Probleemblad I-II 2000.---
9) Harmonie (63) IX 2000.---
10) König und Turm (4) III 2001,S.13.---
11) Problemkiste (163) II 2006.---
12) 986 Minimalkunst im Schach 2006.--- (2007-08-20, hidden)
Erich Bartel: weitere Nachdrucke:
15) 5 The Problemist I 1962.--
16) feenschach (47) VII-IX 1979,S.185.---
17) XII Rex Multiplex (2) IV-VI 1982,S.38.--- (2009-01-28, hidden)
Alfred Pfeiffer: noch ein Nachdruck:
18) 6 "The Chess Review" 12/1945, S.25 (2012-03-08, hidden) more ...
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Yuri Bilokin: you can see *) 1...Sf1 2.Rc3 Sxe3 3.Kb3+ Sc4 4.Sb4 Sd2# (2022-12-10)
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Sally: Markant ist der Grimshaw auf "f3" (überraschend auf einem "Eck-Feld" des w. Königs), wobei die dennoch notwendige Fesselung prägend ist. [Eine der ersten Aufgaben des Erfolgsduos].
41 Mehr als ein Hobby ( F. Pachl 2011). (2016-12-07)
Achim Schöneberg: Knapper Vorgänger siehe P1413421 (2023-11-08)
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Yuri Bilokin: you can see 1...Sd1 2.c3 Sxe5#
1...Sxa2 2.Rc3 Sb4#
1...Sa4 2.Qc3 Sc5# (2022-09-11)
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Yuri Bilokin: more economical -wPf6, -wPg4, -bPa4, bBg6-f5, bPf7-d7 8/4p3/2R1p1p1/2P1Pb2/1Nppk3/2p3p1/rr2n1P1/R3K3 (7+12) h#3 2.1…
1.Sg1 0-0-0 2.Sf3 Sc2 3.Sxe5 Rxd4#
1.Sf4 Rd6 2.Sd5 Sc6 3.Se3 Rxd4#

AntiZielElement (B2, line obstruction)
Castling (white, long). Check prevention (W-W)
Exchange of functions (wRa1/wRc6, Mate / Passive)
Long-trip (bS, 3) × 2.Play on the same square (W3, 2)
Mates on the same square × 2 (2023-08-23)
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klären Satz UL verdruckt ?
VL: Represented by the author (in an article with M.Klasinc) in The Problemist, July 2017, p.128 (C) in the sound form with 5 additional pieces: 2B5/1p1R4/bp1P4/1p2Bpp1/1P4kb/4ppN1/4p1P1/4K3. Set: 1...Txb7 2.Lxb7 (tempo) Lxf5# and "hidden zugzwang": 1.f4 ?? 2.fxg3 Th7# and 1.f4 Lxf4 2.?? Tf7#. (2023-04-07)
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SCHRECKE: a)-d): C+, popeye 4.87
a) 1. Tf4+ Sxg5 2. Te4 Sxf3# (korrigiert)
b) 1. Dxe5[+wLc1] Lxe3 2. Tf4 Lf2#
c) 1. Sd6 Lc3 2. Dd5 Le1#
d) 1. Tf8+ Lf6 2. Lg3 Lxg5# (2023-03-06)
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Thema: Verstellung
Yuri Bilokin: correction bRh8-c8, +bBe8 q1r1b3/1p2KppR/Bn1np1k1/4b1p1/5pP1/2Rr4/8/8 (5+14) h#2 b) wKe7-a1

a) 1.Bd4 Kxd6 2.Bf6+ Bxd3#
b) 1.Rd4 Rc7 2.f6 Rcxg7#

Ambush (wR). Bi-valve (bP-wR-bB). Check prevention (W-W). Epaulette interferences
Exchange of functions (wRc3/wBa6, Mate / Passive)
Grimshaw (black). Hesitation (bB). Pickabish
Play on the same square (B1, 2). Play on the same square (B2, 2)
Wigwag (bB) (2023-06-20)
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Anton Baumann: a) 1.Txh6[wTa1] Tc1 2.Lg3 Lxe3#
b) 1.gxh6[wTa1] Th1 2.Tg3 Txc2#
C+ Alybadix (2023-06-23)
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Henrik Juel: Part b) is C+ by Natch 3.1 and no solution was found in 8.5
For part a) the program ran for an hour without finding the solution in 13.0 (2018-12-14)
Moldenhauer: Computerprüfung: C+ a und b Stelvio 1.2 a) 00:04:57 Minuten.(hh:mm:ss)
b) 1 Sekunde. Keine Lösung a) BP 12.0, b) BP 8.5.
KBP a) BP 13.0, b) BP 9.5. (2023-05-28)
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Moldenhauer: Computerprüfung: Keine Lösung BP 17.5. (2023-08-17)
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ras@bsiag.com: Stelvio 1.0 ist gut in solchen Dingen, C+ in 15sec. (2023-02-10)
Moldenhauer: Computerprüfung Ergänzung: Keine Lösung: BP 29.0. (2023-07-29)
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Henrik Juel: Natch 3.1 found nothing in an hour (2018-12-14)
Paulo Roque: obs: 11.Thh8
obs: 18.Taa8 (2008-12-01, hidden)
Reto: This is C+ in about 56h using Stelvio 1.11 and lots of RAM. (2023-03-13, 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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Henrik Juel: Natch 3.1 did nothing in almost an hour (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 84:43:01 Stunden. (hh:mm:ss)
Keine Lösung: BP 22.0, BP 22.5. (2023-04-01, hidden) more ...
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Henrik Juel: typos in solution, should read: 5.Kg2 Kd6 6.Kf3 Sh4+ 7.Ke3
Fake castling, the hard way in 12 moves
First 21.5 moves found OK by Natch 3.1 (2018-12-13)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 09:09:13 Stunden. (hh:mm:ss)
Keine Lösung: BP 21.0, BP 21.5. (2023-03-12, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Stelvio 1.5 in 147:43:15 Stunden.(hhh:mm:ss)
Keine Lösung: BP 27.0 in 49:30:01 Stunden. (2023-08-17)
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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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Kommentare:
Als einziger letzter schwarzer Zug, der die weiße Rochade
erhält, kommt nur Bh7-h5 infrage, was dem Weißen den e.p.-Schlag
ermöglicht, um nicht die Rochade oder das Mattnetz zerstören zu
müssen (H.P.Suwe)
Gianni Donati: This intends to show the Valladao theme in the minimum number of moves. (2004-03-19)
VL: The waiting ep capture. Cf. also P1000260
(by T.Petrovic, 2000) somewhat enriched
thematically with the illegal try 2... Rd1#??. (2004-06-03)
A.Buchanan: The “illegal try” 2. Rd1+ is not actually mate because c2 is unprotected. This is kind of “logic dual” spoils the A.P. motivation for the castling (2020-05-20)
Henrik Juel: White captured h6xg7 and once more, e.g. axPb-b8=Y
Black captured cxdxe, dxexf, fxg, and once more, e.g. a2xb1=Y
Possible retroplay 1... h7-h5 2.h6xSg7 Df8-f4 3.h5-h6 Sf4-h3 4.Sh3-g1 etc., preserving the castling right
Any other black last move would force White to retract Ta1, as d2xc3, d2xe3, and h6xh7 would be illegal retractions
I agree with Andrew that the double motivation of 2... 000# is a weakness:
a. legitimizing h7-h5
b. mating (2020-05-21)
Vaclav Kotesovec: Similar problems should not be labeled as "C+". Such a designation is only acceptable if the entire analysis was performed by a computer program. (2023-08-03)
A.Buchanan: See discussion at P0009121 (2023-08-04)
A.Buchanan: Maybe can replace wPf2 with wR, wRf1 with wB & shift bPg3 to g2? (2020-05-20, hidden) more ...
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paul: Stelvio 2.0 solves this PG in 3 minutes. (2024-04-02, hidden) more ...
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Henrik Juel: Natch 3.1 did nothing in half an hour (2018-12-14)
paul: Correction of P1000629 (2022-10-12)
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Henrik Juel: C+ by Euclide 1.01 also, but in just 71 seconds (2016-06-16)
Reto: C+ also by Stelvio 2.0 in under 10sec (2024-01-07)
Per Olin: The 7th move by Black is 7. - Thh6. The automatic move sequence is disturbed as is chosen 7. - Tfh6. (2016-06-16, hidden) more ...
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Mario Richter: brauchen wir wirklich zwei verschiedene Keywords - hier "Tempoverlustmanöver", bei z.B. P1007491 hingegen "Loss of tempo"? (2018-10-21)
Henrik Juel: Good point, Mario
Other things in the never-ending Keyword discussion include:
Tempoverlustmanöver is currently called the same in english and french, and has no definitions
Loss of tempo (Tempoverlust in german) is defined as
'With White to play, a sequence of moves to reach the same position with Black to play. (Help game, abbr .: TV)'
which makes little sense in proof games
In this problem [Pe2] goes e2-e3-e4 ('wasting' a move)
In P1007491 [Pc2] similarly goes c2-c3-c4-... and [Lc8] goes (among other things) c8-b7-a6-c8 ('triangulation') (2018-10-21)
Henrik Juel: There are many solutions in 6.0 moves, like
1.Sb1-c3 Sg8-f6 2.Pe2-e4 Sf6xe4 3.Sc3-a4 Se4-c3 4.Ra1-b1 Sc3xb1 5.Sa4-c3 Sb1-a3 6.Sc3-b1 Sa3-c4
Correction: Change the stipulation to
BP in exactly 6.5 (2024-03-10)
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Moldenhauer: Computerprüfung: C+ Stelvio 1.0 in 3 Sekunden.
Keine Lösung: BP 24.0, BP 24.5. (2023-02-14, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Stelvio 1.0 1 Sekunde.
Keine Lösung: BP 24.0, BP 24.5. (2023-02-14)
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Henrik Juel: Commendation
(awarded by Hans Gruber, PB no.1 2005 p.4) (2019-03-07)
Joost de Heer: @Hans: First retraction move isn't a pawn move. (2024-02-01)
hans: R: 1. --- 0-0-0+ 2. Kg7-h8 f7-f6 3. Kf6-g7 e4-e3 4. Kf5-f6 e5-e4 5. Ke4-f5 e6-e5 6. Kd3-e4 e7-e6 7. Te6-e2 e2-e1=L 8. Db2-a1 e3-e2 9. Ke2-d3 e4-e3 10. Dc1-b2 e5-e4 11. Dd1-c1 d6xLe5 12. Lb2-e5 c6-c5 13. Lc1-b2 c7-c6 14. b2-b4 b3xTa2 15. Ta1-a2 b4-b3 16. La2-b1 b5-b4 17. Lb3-a2 b6-b5 18. La4-b3 b7-b6 19. Lb5xa4 a5-a4 20. Ke1-e2 a6-a5 21. Lf1-b5 a7-a6 22. e2xSf3 etc. (all black pawn-moves) (2011-03-16, hidden) more ...
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Henrik Juel: tiny typos: 9... Lg7+ and 19.Sge2 (2018-12-18, hidden)
Henrik Juel: Natch 3.1 found nothing in half an hour (2018-12-18, hidden)
Henrik Juel: I forgot to kill Natch and discovered later that the solution was found (2018-12-18, hidden)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 04:24:38 Stunden. (hh:mm:ss)
Keine Lösung: BP 19.0, BP 19.5.
Notationsfehler: richtig 19.Sge2 (2023-03-24, hidden) more ...
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Henrik Juel: Natch 3.1 did nothing in more than an hour (2018-12-19)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:08:53 Minuten. (hh:mm:ss)
Keine Lösung: BP 20.5, BP 21.0. (2023-03-22)
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Reto: Solving in parallel with Stelvio v1.25 at brings this down to 24h (with room for improvement still) (2023-05-28)
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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:10:35 Minuten. (hh:mm:ss)
Keine Lösung: BP 20.5, BP 21.0. (2023-03-16)
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Henrik Juel: dual 1.Tb3 d2 2.De3,Te3+ (2024-03-03)
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SCHRECKE: C+, Gustav 4.2b, Brute Force
1. Lg8! (Zugzwang)
1. ... gxh6 2. Sf3 exf3 3. Tc4 Kxa2 4. Ta4#
1. ... g6 2. Tg5 gxh5 3. Td5 Kxa2 4. Ta5#
1. ... gxf6 2. Tf4 f5 3. Te6 Kxa2 4. Ta6#
1. ... g5 2. Th7 gxh4 3. Tf7 Kxa2 4. Ta7# (2022-11-24)
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SCHRECKE: C+, popeye 4.87
Satz:
1. ... Se5 2. Txe5#
1. ... Dxe3 2. Sd6#
Lösung:
1. Sf4! droht 2. Lf3#
1. ... Se5,Sd2 2. Sd6#
1. ... Dxe3 2. Te5#
1. ... Dxf6+ 2. Sxf6# (2023-06-23)
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SCHRECKE: C+, popeye 4.87
1. Sf6! droht 2. Lc4#
1. ... Ke6 2. Ld7#
1. ... Ld5 2. Sd7#
1. ... Le6 2. d4# (2023-06-25)
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SCHRECKE: C+, popeye 4.87
1. Lf8! droht 2. De7#
1. ... Tf2 2. Txd3#
1. ... Tf3 2. Txf3#
1. ... Sc3 2. Lxc5#
1. ... Tg5 2. Tf3#
1. ... f5 2. De5# (2023-04-28)
Sally: 1.Dd6?- Tf2!(und 3. Dxd3#)ist durchdas Fluchtfeld nach f4
unterbunden) 1. Td6? -Sc3!(2. Lxc5 geht nicht)1.Ld6? -Bf5!
(De5 geht nicht)
1. Lf8! vemeidet diese Behinderungen! (2010-06-25, hidden) more ...
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SCHRECKE: 1. De4? droht 2. c5(A)#
1. ... exd4(x) 2. Le7(B)#
aber 1. ... Sb7!
1. Df5! droht 2. Le7(B)#
1. ... exd4(x) 2. c5(A)#
1. ... Th1 2. Df8#
1. ... Sxc6 2. De6#
1. ... Txf5+ 2. Sxf5#
1. ... Sxc4,Sd5 2. Dd7#
1. ... Sc8 2. Dd7,Txd8# (Mattdual)
Le Grand-Thema (2023-06-27)
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Henrik Juel: semi-dual 1... dxe2 2.Sb4,Dc5 (2024-01-07)
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SCHRECKE: C+, popeye 4.87
1. Sxf6! droht 2. Sxb5#
1. ... Txf6+ 2. Lxf6#
1. ... fxe4 2. Td5#
1. ... Txe4 2. Dd5#
1. ... c4 2. Lb6# (2023-08-02)
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Ion Murarasu: Source: Probleemblad 3/2000, 11284, p.111
Sol.
1.Qd4!
1…h4 2.Qg4‡
1..f5 2.Rh6#
1..Kf5 2.Qe4#
1..Kg7 2.Qg1#
Try
1.Kf3?
1…f5 2.Qg5#
1…Kf5 2.Qd3#
1…Kg7 2.Qg8#
but 1…h4! (2007-03-07)
SCHRECKE: C+, popeye 4.87 (2023-02-14)
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Reduzierung der Mattduale, je nachdem welche s UW gewählt wird.
Hier wird also der Dual zum Thema erhoben. \eb
milan: [-wSf4 +bPd4e3] M.Frelih (2024-03-28)
Erich Bartel: ND 1) 4 varianter med 1 bonde,1972
2) Probleemblad 1992,S.128
3) 185 Die Allumwandlung im Problemschach VIII 1966 (2012-08-24, hidden) more ...
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Michael Lipton: Set Ka4 2.Kc3. Kc5 2.Kb3.
1.S~+? Kc4! 1.Sxd6+!? 2.Kc5!
1.Sa3+! Ka4 2.Kb2. Kc5 2.Kd3. (Kxa3 2.Ta1.)
White correction leads to changed WK mates from two batteries.
FIDE Album, 1956-58. (2011-08-03)
SCHRECKE: C+, popeye 4.87 (2022-08-17)
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SCHRECKE: C+, popeye 4.87
1. Sg6! droht 2. Se5#
1. ... Lxd5+ 2. Dxd5#
1. ... Sdc6 2. Te5#
1. ... Sac6 2. Tc5#
1. ... Sf3 2. Td4# (2022-12-20)
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SCHRECKE: C+, popeye 4.89
1. d6! (Zugzwang)
1. ... cxd6 2. Tc3 d5 3. Sc2#
1. ... c5 2. La5 c4 3. Lb6#
1. ... c6 2. La3 c5 3. Lb2# (2023-12-23, hidden) more ...
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SCHRECKE: C+, popeye 4.87
1. Kb2! (Zugzwang)
1. ... Kxd3 2. g3 Kxc4 3. Lb5#
1. ... Kd1 2. g4 hxg4 3. Td2#
1. ... Kf1 2. g4 hxg4 3. Tf2# (2023-06-25)
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Henrik Juel: C+ Popeye 4.61 (2023-03-01, hidden) more ...
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Vendelbo: 1. b2d1 droht f5h4 f2h4 c4e2 f3e2 h7e4#
1. -, c6c5 2. h7f7 (droht f5d4 und f7f4#) f2g3 3. f5g3 und 4. f7f4#
1. -, a7a4 2. h7d7 (droht f5d4 und d7g4#) f2g3 3. f5d4#
1. -, a3c5 2. f8e8 (droht f5h4 und h7e4#) f2e1 3. f5h4 e1h4 4. h7e4# (2024-02-11)
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Henrik Juel: C+ Popeye 4.61 (except for promotion mate duals) (2023-10-05)
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Henrik Juel: dual in last variation (2023-05-06)
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Harry Fougiaxis: Duplicate of P1011573 (2008-08-08)
Henrik Juel: C+ Popeye 4.61 (2022-09-07)
Henrik Juel: 1.e8=T thr. 2.Da3 Tb1#
1... Tb1+ 2.Ka3 Dc1#
1... e1=S/L 2.Kc1/Ka1 (2022-09-07)
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Sally: Problemblad Juli/Aug. 1959, Meindert Niemeijer gew. 2.Pr.
1. Lb1! (2. Sc7 -Bf5, 3. Txf5#) - Ke6, 2. Sxf6 - Ke5,
3. Sg4+ - Ke6, 4. Lf5#. Dieses feine Mustermatt wiederholt sich echoartig an anderer Stelle des Brettes im Abspiel :
1. - Kd4, 2. Sb6+ - Ke3,(2..- Kc3, 3. Tf2 -; Td3#; bzw. 2..- Ke5, 3. Sc4+; 4. Lf5#) 3. Sc4+ - Ke2, 4. Ld3#.
1. - Bf5, 2. Sc7 - Kf6/f4, 3. Txf5+/Te1+; 4. T..#.
Ein prächtiges Mattbilderproblem mit fein begründeten Schlüsselzug: 1. Lc2/Lxh5? - Kd4, 2. Sb6+ - Kc3!
Nr. 225 Rätselvolle S. Aufg. ( G. Murkisch 1980). (2012-02-18, hidden)
Henrik Juel: C+ Popeye 4.61, except for a mate trial (2023-11-22, hidden) more ...
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SCHRECKE: C+, popeye 4.89 (2023-10-17, hidden) more ...
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SCHRECKE: C+, popeye 4.89 (2023-10-17, hidden) more ...
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Henrik Juel: C+ Popeye 4.61 (2024-03-21)
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Henrik Juel: several duals, e.g., 1.Se8 Dg3 2.Le6,Lxg3+ (2022-09-07)
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Henrik Juel: dual 1.Df1+ Sxf1 2.Lc2 Sf4 3.Txf4,Tg6+ (2022-08-17)
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Henrik Juel: C+ Popeye 4.61 (2023-08-04)
Henrik Juel: 1.Se3 thr. 2.Dxf4#
1... Tf6 2.Ld5#
1... Te6 2.Df5# (2023-08-04)
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SCHRECKE: C+, popeye 4.87
1. Lh3! (Zugzwang)
1. ... Tg3 2. Tf3#
1. ... Tg4 2. Tf4#
1. ... Tg5 2. Tf5#
1. ... Tg6 2. Tf6#
1. ... Tg7 2. Tf7#
1. ... Tg8+ 2. Tf8#
1. ... Th2 2. Txh2#
1. ... Txf2 2. Lxf2# (2023-02-08, hidden) more ...
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SCHRECKE: C+, popeye 4.87
1. Tf7? Lh7!
1. Te7! droht 2. Dd1#
1. ... Ta1 2. Lf6#
1. ... Lh7 2. Se6#
1. ... Se4 2. Dxe4# (2022-09-06)
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Henrik Juel: C+ Popeye 4.61, except for some mate duals (2022-09-20, hidden)
Henrik Juel: 1.0-0-0 thr. 2.Td8+ Lxd8 3.Txd8#
1... Dd4,Dxf6,Dd6,Dd5 2.exf7+
1... 0-0 2.Se7+ Kh8 3.Txh1# (2022-09-20, hidden) more ...
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SCHRECKE: C+, popeye 4.89
1. Td5! (Zugzwang)
1. ... Ke2 2. Ld1+ Kf2,K~ 3. Tf5#
1. ... Kf2 2. Ld1 Kf1 3. Tf5#
1. ... Kxe4 2. Lc6 Kf3 3. Td2# (2023-12-23, hidden) more ...
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Thomas Brieden: 1) Th5! [droht 2) Te4#]
1) --- Le3/Te3 2) Sf3#/d4# (2014-03-10)
SCHRECKE: C+, popeye 4.87 (2023-01-03)
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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:05:59 Minuten. (hh:mm:ss)
keine Lösung: BP 28.5, BP 29.0. (2023-04-15, hidden) more ...
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SCHRECKE: C+, Gustav 4.2b, Brute Force (2022-12-14)
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Henrik Juel: Natch 3.1 found little in half an hour
Stopped at position 4 (2018-12-21)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 00:24:45 Minuten (hh:mm:ss).
Keine Lösung: BP 19.0, BP 19.5. (2023-03-01)
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Henrik Juel: Natch 3.1 found the solution, but did not conclude in an hour (2018-12-20)
paul: Stelvio verified in ~ 25 minutes. (2023-02-01)
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Henrik Juel: Natch 3.1 did little in half an hour
Stopped at pos. 1 (2018-12-22)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:05:37 Minuten. (hh:mm:ss)
Keine Lösung: BP 23.0, BP 23.5. (2023-04-02)
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paul: Five times delayed ascension of wPawns. See P0000309, the previous record. (2010-05-10)
Vladimir Rey: Probably, cook: Natch 3.3 test, first 28 moves. (2023-01-24)
A.Buchanan: According to Reto, Stelvio indicates this is sound (2023-01-25)
Vladimir Rey: OK! Wrong cook (white king moves: Kd1,Ke1 and castling is impossible, 30. 0-0-0+!) (2023-01-25)
paul: 1.Pe2-e3 Ph7-h5 2.Qd1-g4 Ph5xg4 3.Ng1-f3 Pg4xf3 4.Bf1-e2 Pf3xe2 5.Rh1-f1 Pe2xf1=B 6.Nb1-c3 Bf1-c4 7. Nc3-a4 Bc4-e6 8.Na4-b6 Pa7xb6 9. Pb2-b3 Ra8-a5 10.Bc1-a3 Ra5-g5 11.Ba3-c5 Pb6xc5 12.Pa2-a3 Pb7-b5 13.Pa3-a4 Bc8-b7 14.Pa4-a5 Bb7-e4 15.Pa5-a6 Pc7-c6 16.Pa6-a7 Nb8-a6 17.Pf2-f3 Qd8-b8 18.Pf3-f4 Ke8-d8 19.Pf4-f5 Kd8-c7 20.Pf5-f6 Be4-f5 21.Pe3-e4 Kc7-d6 22.Pe4-e5 Kd6xe5 23.Ph2-h3 Ke5-f4 24.Ph3-h4 Kf4-g3
25.Ph4-h5 Kg3-h2 26.Ph5-h6 Kh2-g1 27.Ph6xg7 Rh8-h1 28.c3 Qh2 29.c4 Qh8 30.0-0-0 (2010-05-08, hidden)
paul: 20.- Le4-f5 (2017-04-08, hidden) more ...
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Henrik Juel: Cyclical shift by three white knights.
Sf8 is [Sb1], Sb1 is [Sg1], Sg1 is a promotee from f8 (2014-08-20)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 2 Sekunden.
Keine Lösung: BP 19.5, BP 20.0. (2023-03-02)
hans: 1. Nc3 f5 2. Nd5 Nf6 3. Nxe7 Nd5 4. Ng6 Qe7 5. a3 Qxa3 6. e4 d6 7. Ne2 Kd7 8. Nc3 Kc6 9. Nb1 Nd7 10. Bb5+ Kxb5 11. e5 Ka4 12. e6 b5 13. e7 N7b6 14. exf8=N Bd7 15. Nxh7 Raf8 16. Ng5 Rh4 17. Nh3 Rb4 18. Qg4 Be8 19. Qd4 f4 20. Ng1 Rf5 21. Nf8 (2014-08-20, hidden) more ...
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Reto: The 21.5 move paths leading to this position are known to me and they are not cooks in my eyes, as move parity is part of the position to find. Anyone who likes can think "exact spg" or similar if that helps them. (2023-03-13)
A.Buchanan: A number of different opinions exist. To me:
(1) a shorter PG, unique or not, does not cook a PG in x.y.
(2) This is true whether the delta is z.0 or z.5 moves.
(3) A composer may choose to include "exactly" in the stipulation, but if they choose not to then the composer's stipulation "Originalforderung" may not be fiddled with by later generations.
(4) However, there may be a "database stipulation" added by curators for uniformity within a catalogue, and if the database stipulation does not match the Originalforderung, then the Originalforderung must appear in an Originalforderung field, not just as a comment. [PDB does not support this feature.]
(5) Accidental shorter solutions do constitute a defect, in my view.
Again to say: this is just my personal opinion :) (2023-03-14)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 00:30:02 Minuten. (hh:mm:ss)
Keine Lösung: BP 21.0, BP 21.5 cooked 00:09:02 Minuten. (2023-03-09, hidden) more ...
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paul: 1. e3 a5 2. Qh5 a4 3. Qa5 h5 4. Ke2 Rh6 5. Kd3 Re6 6. Kc4 Re4+ 7. Kb5 e6 8. h4 Ke7 9. Rh3 Kf6 10. Rg3 Kf5 11. Rg6 Qf6 12. Rh6 c6+ 13. Kb6+ Kg4 14. Kc7 Qf3 15. Kd8 Qd1 16. f3+ Kg3 17. Ke8 Kf2 18. Qd8 b6 19. g3 Ba6 20. Bh3 Bf1 21. d3 Ba3 22. Bd2 Ne7 23. Bb4 Ng6 24. Bf8 Ke1 25. b4 Bc1 26. Rh8. Belfort 1994 theme. (2013-08-12)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0, 87 Sekunden.
Keine Lösung: BP 25.0 und BP 24.5. (2023-02-10)
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paul: -1.Bg7-h8 Bg1-h2+ -2.Kf2×Pg2(Ke1) Bh2-g1+ -3.Ke1-f2 Bg1-h2+ -4.Ke2×Pd2(Ke1) f4-f3+ -5.Kf2-e2 Bh2-g1+ -6.Kg3-f2 Bg1-h2+ -7.Bh8-g7 f5-f4+ -8.Kf2-g3 Bh2-g1+ -9.Ke1-f2 d3-d2+ -10.h7-h8=B Bg1-h2+ -11.Kh4×Bh5(Ke1) Bh2-g1+ (h2-h1=R? illegal) -12.Kg3-h4 Bg1-h2+ -13.Kf2-g3 Bh2-g1+ -14.Ke1-f2 Bg1-h2+ -15.Kg8×Qh8(Ke1) Qf6-h8+ -16.Bf4-e5 & 1.h8=S+ Q×h8(Qd8)# (2023-06-12)
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Henrik Juel: Natch 3.1 did not find the solution in 4 hours
Although the last move is obvious, testing the first 28.0 moves does help much, because after a couple of hours thousands of solutions appear with [Ta1,Ke1] having moved
The problems shows five hesitating white pawns (a2-a3-a4 instead of a2-a4, etc.)
The author claimed that this was a record in 2006 (2018-12-25)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:01:10 Minute. (hh:mm:ss)
Keine Lösung: BP 27.5, BP 28.0. (2023-04-08)
Ryan McCracken: Solution:
1. e3 h5 2. Qg4 hxg4 3. Nf3 gxf3 4. Be2 fxe2 5. Rf1 exf1=B 6. Nc3 Bc4 7. Na4 Be6 8. Nb6 axb6 9. b3 Ra5 10. Ba3 Rg5 11. Bc5 bxc5 12. a3 b5 13. a4 Bb7 14. a5 Be4 15. a6 c6 16. a7 Na6 17. f3 Qb8 18. f4 Kd8 19. f5 Kc7 20. f6 B4f5 21. e4 Kd6 22. e5+ Kxe5 23. h3 Kf4 24. h4 Kg3 25. h5 Kh2 26. h6 Kh1 27. h7 Nh6 28. b4 Rg8 29. O-O-O+ (2011-08-23, hidden) more ...
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SCHRECKE: C+, popeye 4.89 (2024-03-05, hidden) more ...
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Silvio Baier: C+ (Natch 2.4) (2009-11-09)
Joost de Heer: In the records PDF at the Die Schwalbe site (https://dieschwalbe.de/download/artikel/Pronkin.pdf) there's a version with 25. Sh3 replaced by 25. Rh6, but this is cooked:
1. Nc3 e5 2. Rb1 Qf6 3. Nd5 Ba3 4. bxa3 Qf3 5. Rb6 h5 6. Rh6 b5 7. c4 b4 8.
c5 b3 9. c6 b2 10. exf3 b1=Q 11. Bb2 Qg6 12. Qc1 Qg3 13. Bd3 e4 14. hxg3 e3
15. Rh4 e2 16. Rf4 h4 17. Bb1 h3 18. Ne3 h2 19. d3 h1=Q 20. Kd2 e1=B+ 21.
Ke2 Bb4 22. g4 Bf8 23. Qc5 Qh4 24. Bc3 Qd8 25. Rff6 d5 (2023-07-22)
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