paul: This problem obtained first Prize in the informal tourney, see Die Schwalbe 249/2011 (judge M. Caillaud) (2020-01-06)
Henrik Juel: 41.g2-g3 is illegal, because now wLh1 cannot retract (2021-02-03)
Henrik Juel: 41.Lg2-h1 Sg4-h2 42.Th1-h3 Sh2-g4 43.Lh3-g2 Sg4-h2 44.g2-g3 Sh2-g4 45.Dg4-h4 Tg3-f3 46.Dd1-g4 seems to work (2021-02-03)
comment

Michel Caillaud: cooked by Stelvio 0.93 :
1.b4 c5 2.b5 c4 3.b6 c3 4.bxa7 d5 5.e4 d4 6.f4 d3 7.f5 dxc2 8.d4 g5 9.Lf4 c1=L 10.d5 g4 11.f6 g3 12.fxe7 gxh2 13.g4 c2 14.Lg3 Lf4 15.g5 c1=T 16.g6 Tc6 17.Dd3 f5 18.Sd2 h5 19.g7 Tg6 20.0-0-0 Sf6 21.Te1 h4 22.d6 h3 23.g8=L Th4 24.Lb3 L8h6 25.Ld1 Kf7 26.d7 Dh8 27.d8=T Le6 28.Td4 Sbd7 29.e5 Tf8 30.e8=T Kg8 31.Tc8 Lf7 32.e6 b6 33.e7 Lb8 34.e8=S f4 35.Te7 f3 36.Se2 f2 37.Tg1 h1=S 38.Da6 h2 39.Lh3 f1=S 40.Ta4 Sf2 41.Tc2 h1=L... (2022-12-20)
Paulo Roque: obs: 28...S8d7 (2008-11-22, hidden) more ...
comment

See P1338946 cooked.
paul: Compare with P0002278 & P0002279 (2010-04-30)
Mu-Tsun Tsai: This one is by far the toughest retro I've ever solved. Very little certain information can be determined by structural consideration alone, even with long and complicated argument. It took me five days to complete solving this. (2012-07-22)
A.Buchanan: @Mu-Tsun: that's an interesting data point - thanks for posting. (2017-09-07)
Henrik Juel: The current record is 58.5 moves in a proof game problem by the authors + Keym, Die Schwalbe 2017 (2017-09-07)
Henrik Juel: I just learned that the 58.5 move proof game has been cooked... (2017-09-07)
A.Buchanan: In retrospect, my earlier comment about "interesting data point" is a bit weak. It's actually great that for such an extreme problem, someone took substantial time to independently validate it. It's like doing science: people want to do their own new stuff, and are unwilling to take the time to validate what's already been claimed. This one has survived 30+ years, and maybe the use of constraints e.g. in Jacobi can eventually allow it to be HC+. (2021-05-29)
Olaf Jenkner: This problem is the current record, because P1338946 (58.5 moves) has been cooked. (2021-11-25)
Reto: This is C+ up to 51.0 moves with Stelvio 2.0. This ties the record for partial testing of an SPG. Took 1200 CPU hours of strategy seeking (finding 378 0+0 strategies) and another 13h of strategy playing these strategies. If this can ever be completely solved, then it needs to be the case that all strategies have 0+0 free moves, otherwise playing is utterly hopeless.
@Andrew: There is absolutely no way a brute-force based program like Jacobi ever stands a chance at solving something like this, no matter how many conditions you add. (2023-12-14)
Paulo Roque: obs: 24.S1c3 (2008-12-16, hidden) more ...
comment

Anton Baumann: Neufassung vergl. P0006288 (2023-01-06)
comment

Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 29:20:32 Stunden. (hh:mm:ss)
Da NUPG C+
Notation: 1.b4 e5 2.d4 Le7 3.d5 f5 4.d6 h5 5.dxe7 d5 6.b5 Dd6 7.b6 Kd7 8.bxa7 b5
9.c4 b4 10.c5 d4 11.cxd6 c5 12.e8=T Sc6 13.Tf8 c4 14.e4 d3 15.Dg4 d2+ 16.Ke2 c3
17.Kd3 d1=L 18.Sd2 fxg4 19.Tf4 exf4 20.e5 Sd4 21.e6+ Kc6 22.e7 b3 23.e8=T b2
24.Te6 Lb3 25.Th6 Lf7 26.d7+ gxh6 27.d8=T Lce6 28.Td5 Te8 29.Kc4 b1=L 30.a8=T Lh7
31.Ta3 c2 32.Tg3 Se7 33.a4 Thf8 34.a5 Sg6 35.a6 Sh8 36.a7 fxg3 37.a8=T gxh2
38.T8a3 Lhg8 39.Th3 gxh3 40.Se4 Lxd5+
3 schwarze Läufer stehen zum Schluss auf den weißen Felder d5, f7, g8. (2023-09-21)
comment

Korrektur siehe 3463v
James Malcom: How in the world can a non-unique PG be cooked? I'm unmarking it, for now. (2021-01-25)
A.Buchanan: Maybe the intended theme was not forced? Haven’t looked at this problem though (2021-01-26)
Paulo Roque: obs: 36.S5e4
obs: 46.S1f3 (2008-12-08, hidden)
Paulo Roque: obs: (No 46.S1f3) 46.S4f3 (2008-12-08, hidden) more ...
comment

James Malcom: Again, how is this cooked? (2021-01-25)
A.Buchanan: The AL has 5 Wh S promotions in 46.0 moves but the cook shows this is not necessary. (2021-01-26)
Paulo Roque: obs: NL: 36.S6e4

A very good cook! (2008-12-09, hidden) more ...
comment

Moldenhauer: Computerprüfung: C+ NUPG cooked Stelvio 1.11 in 1 Sekunde.
Keine Lösung: BP 17.0. BP 17.5 cooked.
Beispiel BP 18.0: 1.d4 Sa6 2.Kd2 Sb8 3.Kc3 d6 4.e4 Le6 5.Se2 Sd7 6.e5 Db8 7.g4 Kd8
8.g5 f5 9.g6 Lf7 10.gxf7 Kc8 11.Kc4 g5 12.Kd5 Lg7 13.Ke6 d5 14.f8L g4 15.Kf7 g3
16.Ke8 g2 17.Lg5 gxf1L 18.e6 Lh3
Beispiel BP 17.5: 1.d4 Sa6 2.Kd2 Sc5 3.Ke3 d5 4.Kf4 Le6 5.Ke5 Kd7 6.e4 Db8
7.g4 Kc8 8.g5 f5 9.g6 Lf7 10.gxf7 Sd7+ 11.Ke6 g5 12.Se2 Lg7 13.e5 g4 14.Lg5 g3
15.f8L g2 16.Kf7 gxf1L 17.Ke8 Lh3 18.e6 (2023-05-07)
comment

See P1409984.
Yoav Ben-Zvi: WBd8 can be replaced with a BN after moving BPg7 to f7 and BPh5 to h6. (2018-08-08)
A.Buchanan: Yes after all these years, I found this a few months ago, but when I told Thomas Brand, he said that Werner Keym had found it - one of a series of modifications through seeking to avoid non-standard material. (2018-08-12)
Henrik Juel: Werner's improvement can be found in his 'Eigenartige Schachprobleme' from 2010, p.196, dia 1.68 (2018-08-12)
A.Buchanan: I thought it was more recent (2018-08-12)
A.Buchanan: In fact Werner was not trying to avoid non-standard material, but to prefer "cheaper" knights over bishops. But this is not the canonical ordering, which regards knights and bishops as equivalent. So the older record will win out in classical terms. (2019-10-04)
A.Buchanan: And non-standard material is no defect, according to the 1977 grading criteria, so old Theophilus reigns supreme (2019-10-04)
Henrik Juel: The further retroplay is
-1... h6 -2.a6 and e.g. -2... Ka7 -3.a5 Ta8 -4.Tb8 Ka6 -5.Kc8
and wK out via g6 (2019-10-04)
A.Buchanan: I don't know if a bishop is more expensive than a knight for these economy records. If modifying the criteria, I personally would prefer to minimize non-standard material. But more generally, how to handle a later version of economy record, with similar force? In some cases, the change is trivial, just transforming one unit which might as well be a dummy. At the other extreme, the matrix is completely different. In "Eigenartige Schachprobleme", Werner Keym stoically prefers the earlier record. Even the original composer(s) of a record problem are forbidden from making any artistic improvement to a published economy record! Over the years, the only change to the grading criteria is that Type C problem now needs to contain a check. Any old check-free Type C would a fortiori have been a Type B economy record at least. Here in PDB, I've tagged some cases of "ex-aequo" (search k='economy:eq') but I fear that Werner would not approve of this. What should we do? (2022-03-31)
Zvi Mendlowitz: See P1012544. Very strange. (2004-12-11, hidden) more ...
comment

Gerald Ettl: 1.h8D# (2023-04-03)
Gerald Ettl: die Stellung loest sich auf indem der sK nach b1 - g8 wandert und dann Sg5 weg zieht. Der wK kommt ueber g5 raus. (2023-04-03)
Michel Caillaud: The original stipulation is #1 (durch wen?).
As wPe2 cannot be captured on its file, the 4 white captures for 4 Pawns to promote to Knights are (a2)xb3, (e2)xf3 and (f2)xg3 2 times, and wPb2 was captured on its file by (Dd8)xb6.
As b3-b2 (before (a2xb3)) and c7-c6 (before (Dd8)xb6) cannot be immediately retracted, only bK and wSs can play the last moves.
As indicated by Gerald, bK has to go to g8 to unlock the position, freeing bSg5.
When Ka3-a2 is retracted, previous white move places the 6 white Knights on black squares; the resulting Retro-Opposition implies that black is to play in the diagram position.
1.b1S#! (1g8D#?) (2023-04-04)
Gerald Ettl: Danke Michel fuer Dein Erklärung.
Ich löse so auf, dass Schwarz am Rückzug ist:
R: 1.Kc1b1 Se4d6 2.Bb2b3 Sc5d3 3.Kb1a2 Sd3c1 4.Ka2b1 Sh1f2 5.Kb1a2 Sa4c5 6.Bb3b4 Sf5e3 7.Ka2a3 La1d4 8.Ka3a4 Sf2h1 9.Ka4a5 Sh1f2 10.Ka5b6 Sf2h1 11.Kb6c7 Sh1f2 12.Kc7d8 Sf2h1 13.Kd8e8 Sh1f2 14.Ke8f8 Sf2h1 15.Kf8g8 Sh1f2 16.Sg5f3 Kh6g5 17.Sb8a6 Sc5a4 18.Sa6c5 Sd6b5 19.Sc5b3 Sb5c7 20.Sb3a1 Sc7a6 21.Sa1b3 Sa6b8 22.Sb3a1 Sa4b6 23.Sa1b3 Sb6c8 24.Sb3a1 Sb8b7[+wBb7] 25.Sa1b3 Bb7a6[+sLb7] 26.Sb3a1 Sc8c7[+wBc7] 27.Lb7c8 Kg5h4 28.Sa1b3 Tg6g5 29.Sb3a1 Tg5f5 30.Sa1b3 Tf5f4 31.Tg7g5 Sf2h1 32.Tg5b5 Sh1f2 33.Tb5b8 Sf2h1 34.Bb4b5 Sh1f2 35.Bb5b7 Bc7b6[+sLc7] 36.Tb8a8 Tf4e4 37.Lc7f4 Sg3f5 38.Kg8f8 Sf2h3 39.Lf4h6 Sc1e2 40.Kf8e8 Se2g3 41.Lh6f8 Sf5h6 42.Sb3a1 Sh6g8 43.Sa1b3 Sg8g7[+wBg7] 44.Sb3a1 Bg7g6 45.Sa1b3 Bg6f5[+sTg6] 46.Tg6g8 Bf5f4 47.Tg8h8 Sg3f5 48.Sb3a1 Sf5h6 49.Sa1b3 Sh6g8 50.Sb3a1 Sg8g7[+wBg7] 51.Bg4g5 Bg7g6 52.Bf6g7[+wDf6] Lh5g4 53.Sa1b3 Bh7h6 54.Sb3a1 Bh6h5 55.Bg5h6[+wTg5]
warum geht das nicht? Den Zug b1S# habe ich vorher ueberhaupt nicht gesehen. (2023-04-04)
Gerald Ettl: Jetzt habe ich es gesehen: der wBa2 musste ja von a2 geschlagen haben. (2023-04-04)
comment

Adrian Storisteanu: I may be wrong, but it looks like this problem works without all those pawns:
  Kb4 Rd6 Bb8 Bc8 Sa8 Sd7 / Ka7 (6+1)
(Tries such as - 1.c7xS/Bb8=B+?? & 1.c7xb8=Q# fail due to retrostalemate.) (2021-04-23)
comment

Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 26.0, BP 26.5. (2023-05-08, hidden) more ...
comment

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 ...
comment

James Malcom: A very interesting retro problem, with long-winded bishops promotions to cover its en passant tracks. (2021-01-24)
Hans-Jürgen Manthey: damit der Player klappt, muß 38. La7 in 38. Lba7 geändert werden. (2021-01-25)
James Malcom: Fixed. (2021-01-25)
comment

Henrik Juel: The intention may be 1.dxc3ep 2.Kxb5xc6-d7-e8 6.0-0 Rh8#, but -1.Kb2 seems possible. Should bBc1 be moved to a3? (2003-04-28)
GW: Yes, that's the solution, and the problem seems indeed to be cooked. I don't know if it has been corrected. (2003-04-28)
James Malcom: I believe that the intent was that the sLc1 must be the orginal one, and thus an impediment to wK foreplay, as the wBc6 "must" have captured the e7 Black pawn on its way from f2. The problem is that it can take the alternative path fxsLf3xDe3xTc5-c6, and there are enough White pieces to capture for the Black b pawn to make it d2 and for the d pawn to make it to c2. (2021-09-14)
James Malcom: It turns out this problem actually was corrected a long time later: P1012052 (2021-09-14)
comment

James Malcom: What is the author's solution and the cook? (2021-07-30)
A.Buchanan: One cook is R: 1. b4xBc3ep+ c2-c4+ 2. Le4xDd3+ Dg3-d3+, dann 1. Dd6#.
Maybe the intention was 2. Sc4xD/Tb6+ etc, but the branching is hard to analyze even with Retractor 2.0. (2021-07-31)
A.Buchanan: a cook is R: 1. b4xBc3ep+ c2-c4+ 2. Le4xDd3+ Dg3-d3+, dann 1. Dd6‡ (2021-07-31, hidden) more ...
comment

Moldenhauer: Computerprüfung: C+ Stelvio 2.0 02:29:09 Stunden. (hh:mm:ss) (2023-12-23)
comment

Paulo Roque: Is not possible: 38... Kh5 39.Kb5 (because black rook at square f5).
Probably the continuation of the author:
38...Tf7 39.Kb5 Kh5 40.Ka4 h6 41.Ka3 Sd5 42.Ka2 Sc7+ 43.Ka1 Ta7+ 44.Kb1 Da2# (2008-12-11)
James Malcom: Fixed. (2021-02-15)
comment

Hans-Jürgen Manthey: Zurück: 0. ... h2-h1=S+ a7-a8=L 1. ... Kc8-d8
vor 1. a7-a8=D#

R: 1. Lc6-d7 Ke3-f2 2. Lg2-c6 b6xTa7 3. Lf1-g2 c5xSb6 4. f2-f1=L d4xLc5 5. f3-f2 Kf2-e3 6. Lf8-c5 Td2-d1 7. Kd8-c8 Tf1-d1 8. Sc8-b6 d2-d3 9. d3xLc2 e3xDd4 10. Ke8-d8 Th1-f1 11. f4-f3 Kf1-f2 12. d4-d3 Le4-c2 13. Db6-d4 Lg2-e4 14. Lf5-b1 Sb1-a3 15. Db8-b6 a3-a4 16. a4xDb3 Dd1-b3 17. Ta8-a7 c2-c3 18. Tb3-b2 Ke1-f1 19. Tc3-b3 b2-b4 20. Sb3-a1 f2xe3 21. Sa5-b3 Lf1-g2 22. b3xTa2 Ta1-a2 23. Se7-c8 g2-g3 24. g3xSh2 Sf3-h2 25. Dd8-b8 Sg5-f3 26. Tc6-c3 Sf7-g5 27. Th6-c6 Sh8-f7 28. Sc6-a4 Sf7-h8 29. Th8-h6 Sg5-f7 30. Sg8-e7 Sh7-g5 31. Lc8-f5 Sg5xh7 32. a5-a4 Sf3-g5 33. Sb8-c6 Sg5-f3 34. e4-e3 h2-h3 35. g4-g3 a2-a3 36. b4-b3 Se6-g5 37. d5-d4 Sd4-e6 38. e5-e4 Sf3-d4 39. f5-f4 Sg5-f3 40. a7-a5 Se6-g5 41. b5-b4 Sd4-e6 42. g5-g4 Sf3-d4 43. g7-g5 Sg5-f3 44. f7-f5 Se6-g5 45. d7-d5 Sd4-e6 46. e7-e5 Sf3-d4 47. b7-b5 Sg1-f3 (zB) (2021-06-27)
A.Buchanan: Without getting into the distant retro history, I am guessing that the intended solution is R: 1. h2-h1=S+ a7-d8=L 2. Kc8-d8, then 1. h1=T a8=D#, so it's an allumwandlung. (2021-06-28)
A.Buchanan: If we retract the last three single moves in this way, the position is legal. White Volet pawn captures f2xPe3xdxcxbxa7, while bPh was waylaid. Black captures were bxa,axb,dxc2,gxh2. bPf promoted to B. Can we release this position? There are no tempo issues, the White officers float free, then e.g. d2-d3 unlocks everything.
Is the solution unique? If wBa8 is original, or promoted further to the right, this saves captures of black officers, which allows wPab to cross-capture instead of bPab. This means bQ or bB might uncapture wQ or wR, however I don't see any h#1 forward from that. An engine supporting simple retractions, without regard to global legality, could allow an HC+ solution as exceptions can be checked by hand. I hand the baton to Mario Richter at this point, for further validation :) (2021-06-28)
Mario Richter: rawbats finds many solutions, the most trivial one supports your observation that claiming that wLa8 is an original Bishop, gives White new opportunities:
R: 1. ... Bh2-h1=S 2. Lc6-a8 Le8xBd7, dann 1. h1=T dxe8=D# (2021-06-28)
A.Buchanan: Thanks, Mario. WinChloe has Jean-Claude Gandy as co-author, who I will add. However it is listed in WinChloe as "analytic retractor" rather than "help-retractor". What is the distinction, and does it affect soundness? (2021-06-30)
Henrik Juel: I think there is no real distinction, Andrew
This problem is a help retractor, because White and Black help each other in the retroplay (as opposed to a defensive retractor, where they fight), not because of the forward stipulation
For some people this may be confusing, and the term 'analytic retractor' might 'help' those few, although any problem with retroplay necessarily involves som retro analysis
Analytic retractor can also be used as opposed to the Mrs. Baird type retractor, which involves little analysis (2021-06-30)
A.Buchanan: Thanks for the helpful clarification, Henrik (2021-06-30)
A.Buchanan: A minor question is whether one might replace bBb1 by e.g. bR, in order to avoid the non-standard material, bPf might have promoted instead to S,R or Q, to balance the pawn capture books (2021-06-28, hidden) more ...
comment

Henrik Juel: 1... Txa2#, not 1.Dxb2? -1.Td1xSe1 Sf3 -2.Te1 h7 -3.Td1 Sh4 -4.Te1 Sf5 -5.Td1 Sh6 -6.Te1 Sg4xPh6 -7.Td1 Sf2 -8.Te1 Sd1 -9.h5 Kc1 -10.h4 Ld3 -11.h3 Kc2 -12.h2 Sf2 -13.Td1, retract sLd3 to c8, b7-b6, etc. Black promoted b2-b1=T and f2xTg1=L. The shortest proof game is rather long! (2003-12-19)
Moldenhauer: Histogramm Stelvio 1.2 128/0k bei BP 37.0.
Kein Histogramm bei BP 36.0, BP 36.5. Computerprüfung wird sehr lange dauern
da eine 5+1 auch nach 5 Stunden noch nicht fertig ist! Nur 49 Strategien sind
unter 5+1. Vielleicht kommt aber das "cooked" früher dazwischen. (2023-05-10)
comment

James Malcom: It seems likely that this could be cooked. (2021-01-23)
Hans-Jürgen Manthey: wenn man dem Player bis zum ende folgt und dies die Anfangsposition sein soll, ist die 2. Bedingung 'dann #1' nicht möglich ! Also ist es nicht gelöst und es muß eine andere Zugfolge nötig sein ! (2021-01-23)
Hans-Jürgen Manthey: mögliche Zugfolgen:
1. h4 b5 2. g4 Lb7 3. Sh3 La6 4. Lg2 Lc8 5. h5 c5 6. g5 c4 7. g6 c3 8. dxc3 d5 9. c4 e5 10. c5 f5 11. c6 f4 12. c7 f3 13. Kd2 d4 14. Kd3 b4 15. Kc4 fxe2 16. f4 e1D 17. f5 e4 18. f6 e3 19. a4 Lc5 20. a5 Lb6 21. axb6 Sh6 22. b7 Sg4 23. h6 e2 24. Ld5 Lf5 25. c8T a5 26. Kb5 Sa6 27. b8D Ta7 28. Da8 Se5 29. Sf4 a4 30. Sc3 Sd7 31. La2 a3 32. hxg7 Dg1 33. Tc7 Dc8 34. Tb7 Kd8 35. f7 De3 36. g8D+ De8 37. g7 h5 38. Dg1 d3 39. Dg6 h4 40. Sa4 Ke7 41. Se6 Kd6 42. Lg5 h3 43. Tag1 d2 44. Tg4 h2 45. Tc4 e1D 46. Lb1 Th3 47. Dh8 Td3 48. g8D Kd5 49. Dhh3 De4 50. Dgh8 a2 51. D8h5 a1D 52. La2 Dae1 53. Dg2 D1h4 54. Dhf3 Dh6 55. Te1 h1D 56. Te3 De1 57. Dgf1 De2 = Anfangsposition, aber kein Matt in 1 Zug
(57. ... Df2 58. Lf4 De2 = Anfangsposition mit Matt in 1 Zug: 59. Td4#)
es folgen die 22 Schachgebote bis zur Diagramstellung mit Lg5:
58. Sc7+ Sxc7+ 59. Tbxc7+ Db7+ 60. Dxb7+ Txb7+ 61. Sb6+ Sxb6+ 62. Td7+ Dxd7+ 63. Dc6+ Dhxc6+ 64. Txc6+ Tb3+ 65. Td3+ D2xd3+ 66. D3xd3+ Dxd3+ 67. Dxd3+ Lxd3+ 68. c4+ bxc3ep+ (ist auch möglich mit Lf4) (2021-01-24)
comment

Innerhalb des 2. Teils einer Artikelserie "Die Konventionen im Problemschach" von Karl Fabel.
A.Buchanan: The retro logic is fine, but forward play is savagely cooked. I wonder about changing wSc8 to sS. The retro stuff still works OK, indeed sSc8 is a fourth thematic unit to retract in the try. Ignoring proof of ep legality, there is a unique #4 beginning with ep, and no #5. Promising: but when can we interpolate w000? Any solution must begin 1. axb6ep+ Kb5. There are now numerous #4, but none include w000, so I think we must stick with the still unique line beginning 2. Dxc8 thr 3. 0-0-0. There are 3 black defenses to refute this. 2. ... Txb6,axb3,d5. So maybe #6 needed for White to prevail? (2022-03-28)
A.Buchanan: But if we shift to #6 then 1. Sxc7!,Qxc8+! mate without ep. Can anyone retrieve the original diagram for this one, please? (2022-03-28)
Mario Richter: The position here is identical to the original diagram, but it should be regarded more as a schematic example than a "real problem". Fabel in his quote above: "... perhaps the problem is still not correct, but it should be easy to improve it ..." (2022-03-28)
Henrik Juel: unfortunately K. Fabel forgot to give the easy correction... (2022-03-28)
A.Buchanan: Thanks for this. Valery Liskovetz, an AP expert, was kind enough to send a pdf of the relevant Problem issue, so I can confirm. It's easy enough to remove the two cooks (e.g. sTc8) but the difficulty is in validation of the intended solution. Is there any "exact" option in Popeye that forces the solution to include a waiting move? The actual solution might be expected to be included in that set. (2022-03-29)
Henrik Juel: Yes, Andrew
'stip exact-#5' would disregard #4 as a solution, I believe (2022-03-29)
A.Buchanan: Even with the "exact", it's still very complicated to investigate. I think this kind of problem requires an engine feature to disallow any mate or pat if the specified castling has not been executed. Gazing at the 20,000 lines of output, I don't think sTc8 works as 1. axb6ep+ Kb5 2. Qb7 bxa3! threatens axb2+ after castling. I wonder if the board might be flipped left-right except for wK, but I haven't explored this. (2022-05-31)
comment

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)
comment

hans: 4 Q-promotions on g1 and captured on b8 and c7. Sf1 for shield, so 0-0 is legal. (2010-04-30)
Hans-Jürgen Manthey: R: 0. O-O ! Kb6xc7 1. Ld8xc7+ Lf4-d6 2. Sb1-a3 De5-c7 3. Lg5-d8+ Da1-e5 3. Sd2-b1 Dc3-a1+ 4. Sf1-d2 Dg3-c3+ 5. Lf6-g5 Ld6-f4 6. Lc3-f6 Lb4-d6 7. c7xDb8=T Dg1-g3 8. Sg3-f1 g2-g1=D+ 9. Se4-g3 g3-g2 10. d6xDc7 g4-g3 11. Sd2-e4 g5-g4 12. Sf1-d2 g7-g5 13. d5-d6 Dg3-c7 14. Ld2-c3 Dg1-g3 15. Se3-f1 g2-g1=D+ 16. Lc3-d2 g3-g2 17. Sf1-e3 g4-g3 18. Sg3-f1 g5-g4 19. Se4-g3 g6-g5 20. Sd2-e4 h7xSg6 21. c7xDb8=T Lb4-a3 22. d6xDc7 Dg8-d8 23. Sf1-d2 Dg1-g8 24. Sg3-f1 g2-g1=D+ 25. Se4-g3 g3-g2 26. Sd2-e4 g4-g3 27. d5-d6 Dg3-c7 28. Sf1-d2 Dg1-g3 29. Se3-f1 g2-g1=D+ 30. Sf5-e3 g5-g4 31. Sd4-f5 g3-g2 32. Sb3-d4 Kc5-b6 33. Sd2-b3+ g4-g3 34. Sf1-d2 Dg3-b8 35. Tb8-a8 g7-g5 36. c7xDb8=T g5-g4 37. b6xDc7 Dg1-g3 38. Se3-f1 g2-g1=D+ 39. Sc4-e3 g6-g5 40. Sa3-c4 g3-g2 41. Sb1-a3 g4-g3 42. Sd2-b1 Dg3-c7 43. Sf1-d2 Dg1-g3 44. Ld2-c3 g2-g1=D 45. Lc3-d2 f3xLg2 46. Se3-f1 e4xTf3 47. Sc4-e3 Kd6-c5 48. Sa3-c4+ h7xSg6 49. Sb1-a3 Ke7-d6 50. Sf4-g6+ Ke8-e7 51. Lf1-g2 f5xDg4 52. g2xTh3 Th6-h3 53. Sd2-b1 e5-e4 54. Lb2-c3 Tf6-h6 55. Tc3-f3 Tf8-f6 56. Sh3-f4 Th8-f8 57. Sg1-h3 f7-f5 58. Sc4-d2 Sc5-a4 59. Sa3-c4 Se4-c5 60. Td3-c3 Lf8-b4+ 61. Td1-d3 Sf6-e4 62. Ta1-d1 Sg8-f6 63. Sb1-a3 e7-e5 64. Lc1-b2 Dc7-b8 65. b5-b6 Dd8-c7 66. b4-b5 c7-c6 67. b2-b4 Sc6-a5 68. Dd4-g4 Sb8-c6 69. Dd1-d4 Ta8-a7 70. d4-d5 a7-a6 71. d2-d4 (2021-07-17)
comment

Korrektur siehe P0008426.
James Malcom: What is the cook? (2023-05-02)
paul: Missing wQd1. (2010-05-04, hidden) more ...
comment

Henrik Juel: The problems in this issue were dedicated to T.R. Dawson to celebrate his 60th birthday on 28/11/1949
The solution (FCR 06/1950 p.109) offered no real explanation
Dawson just wrote: Cooks claimed to this masterpiece involve illegal retractions (2022-04-28)
Erich Bartel: weiterer Nachdruck:
4) 7 Rund um das Schachbrett 1955.--- (2008-11-10, hidden) more ...
comment

Mario Richter: vgl. P1011972 (dort sLa3 statt sTa3) (2020-09-23)
comment

Henrik Juel: C+ Popeye 4.61
It is obvious that if Ld8 is not a promoted pawn then Black has the move (2021-12-26)
Henrik Juel: Typo: In 2) the first variation should start 1... Kxd8. (2003-11-07, hidden) more ...
comment

Henrik Juel: C+ Popeye 4.61 (2022-01-02)
comment

Henrik Juel: C+ Popeye 4.61 (2022-01-02)
comment

Henrik Juel: Next retraction is not unique: 13... g6xTLh7 (2020-11-26)
comment

Gerald Ettl: Weiss jemand wie die Autorloesung geplant war? (2023-09-28)
comment

Moldenhauer: Computerprüfung: Cooked Stelvio 1.31 00:06:21 Minuten. (hh:mm:ss)
Keine Lösung: BP 36.0.
Beispiel Stelvio: 1.c3 a5 2.d3 a4 3.Le3 a3 4.Sd2 Ta4 5.Tb1 Tc4 6.dxc4 Sa6 7.g4 b5
8.g5 b4 9.g6 b3 10.gxh7 bxa2 11.b4 a1=S 12.b5 a2 13.b6 Sb3 14.h4 g5 15.h5 g4
16.h6 g3 17.Th5 g2 18.Sh3 g1=D 19.b7 Dg5 20.b8=D Db5 21.Tf5 c5 22.Df4 a1=D 23.Dg4 Sa5
24.f4 Sb7 25.Lg1 Daa5 26.e3 e6 27.Le2 Ld6 28.Sf1 Lb8 29.Dd2 Se7 30.Kd1 Tg8
31.h8=S d6 32.Sg6 Ld7 33.Sh4 Sc8 34.Sg2 De7 35.h7 f6 36.h8=D exf5 37.Dhh4 (2023-06-20)
Henrik Juel: cooked, e.g.
1.Bd2-d3 Ba7-a5 2.Lc1-e3 Ba5-a4 3.Sb1-d2 Ba4-a3
4.Bg2-g4 Ta8-a4 5.Bg4-g5 Ta4-c4 6.Bd3xc4 Bb7-b5
7.Bg5-g6 Bb5-b4 8.Bg6xh7 Bb4-b3 9.Bh2-h4 Bb3xa2
10.Bb2-b4 Be7-e6 11.Bb4-b5 Bg7-g5 12.Bb5-b6 Bg5-g4
13.Bb6-b7 Bg4-g3 14.Bh4-h5 Bg3-g2 15.Bh5-h6 Sb8-a6
16.Th1-h5 Bf7-f6 17.Ta1-b1 Ba2-a1=S 18.Sg1-h3 Bg2-g1=D
19.Bb7-b8=D Dg1-g5 20.Bc2-c3 Dg5-b5 21.Th5-f5 Bc7-c5
22.Db8-f4 Lf8-d6 23.Df4-g4 Ld6-b8 24.Bf2-f4 Sg8-e7
25.Le3-g1 Ba3-a2 26.Be2-e3 Sa1-b3 27.Lf1-e2 Ba2-a1=D
28.Sd2-f1 Sb3-a5 29.Dd1-d2 Sa5-b7 30.Ke1-d1 Th8-g8
31.Bh7-h8=S Da1-a5 32.Sh8-g6 Be6xf5 33.Sg6-h4 Bd7-d6
34.Sh4-g2 Lc8-d7 35.Bh6-h7 Se7-c8 36.Bh7-h8=D Dd8-e7
37.Dh8-h4 (2018-12-10, hidden) more ...
comment

Anton Baumann: Informalturnier 2000: 1. Preis (PR: Jörg Kuhlmann) (2023-09-26)
comment

Silvio Baier: Kann noch nicht komplett getestet werden, aber die Version 12. h7 a4 13. hxg8=L a3 14. Lh7 a2 15. Lg8 La3 16. Lh7 Ke7 17. Lg8 Kd6 18. Lh7 Kc5 19. Lg8 Kb4 20. Lh7 c5 21. Lg8 Qa5 22. Lh7 b6 23. Lg8 Lb7 24. Lh7 Lf3 25. Lg8 Sc6 26. Lh7 Re8 27. Lg8 Te4 28. Lh7 Se5 29. Lg8 ist C+ (Stelvio 1.5) (2023-08-18)
comment

No. 928 HN
Mario Richter: My assumption: a record construction for the ratio g/f with f=number of uniquely determined last moves, g=length of the shortest proof game (s. Fairy Chess Review No.9 28.Nov.1949, p.69 for an explanation), here we have 42/19 (2012-07-16)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 1 Sekunde.
BP 21.0 cooked da NUPG C+.
Keine Lösung: BP 20.0, BP 20.5, BP 21.5, BP 22.0 cooked.
Beispiel BP 21.0: 1. a4 Sa6 2. a5 Sb8 3.Ta4 Sf6 4. Tf4 Sd5 5. h3 Sb6
6. Th2 a6 7. axb6 Ta7 8. bxa7 h5 9. axb8=L Th6 10. La7 Tc6 11. Le3 Tc3
12. dxc3 a5 13. Dd3 a4 14. Led2 a3 15. e3 a2 16. Le2 a1=L 17. Lg4 hxg4
18. Ke2 g3 19. Kf3 gxh2 20. Ke4 h1=L 21. Ke5 f6+ (2023-03-24)
A.Buchanan: I think it's perhaps a little disrespectful to describe the problem is "cooked" when there is no intention for the PG to be solution to be unique. It's certainly good to know that there is no shorter solution. (2023-03-25)
comment

Rainer Staudte: In feenschach sollte nach 2012 eine Korrektur dieser Aufgabe erschienen sein. Wer kann diese finden? (2022-08-10)
Henrik Juel: Popeye 4.61 with 'opt non 0 1 thr 0' found no flaws (2023-06-12)
Vaclav Kotesovec: Gustav 4.2e (Lösezeit: 25 : 12 : 21 : 12,11 d, Hashtables: 350000 MB). Brute force, but with an error message "NICHT ALLE WEISSEN ZUEGE WURDEN GEPRUEFT" (Gustav has only room for 126 white moves), so not 100% C+, but C+ with high probability. (2024-03-23)
Rainer Staudte: Keywords: Promoted pieces on the board, Non-standard material (2005-01-01, hidden) more ...
comment

1211
SCHRECKE: C+, popeye 4.87 (2023-09-25)
comment

James Malcom: I wonder if the newly placed wBb2 and sBd7 could be considered as imposters. (2021-06-27)
A.Buchanan: I think they are. This is not so hard a problem to solve. (2021-06-28)
paul: 11. - Lg7 (2018-10-17, hidden) more ...
comment

Anton Baumann: kürzer nach 1. - g1=S 2.Dc3 Sf3,Sxh3 3.Db4+ Ke5 4.Dg7+ Sxg7#
U-DL nach 1. - g1=L 2.Sc4+ Kd5 3.a8=D,L .. (2023-12-29)
comment

Anton Baumann: ausser bei 1. - Le3 differenzieren die angegebenen Gegenzüge die beiden Drohungen. Zu ergänzen ist die Fortsetzung nach 1. - La7, denn nur wegen dieser Parade ist der sLh3 erforderlich!
1. ... La7! 2.Tb2+ Ka1 3.Txg2+ Kxb1 4.Tb2+ Ka1 5.Df1+ Lxf1 6.Tb5+ Ka2 7.Sc1+ Ka3 8.Ta5+ Txa5# (2023-11-28)
SBD: "1. Lc3 Lf2 2. Dg1+ wird s#7" ist inkorrekt oben,

1. Lc3 Lf2/d4/b6 2. Da7+! Lxa7 3. Tb2+ Ka1 4. Txg2+ Kb1 5. Tb2+ Ka1 6. Tb5+ Ka2 7. Sc1+ Ka3 8. Ta5 Txa5#

Aber 1. .. Le3 ist dualistisch, dann geht 2. Tb2 (oder auch 2. Da7+) Ka1 3. Txg2+ Kb1 4. Tb2+ Ka1 5. Tb5+ Ka2
6. Da7+ Lxa7 7. Sc1+ Ka3 8. Ta5+ Txa5#

1. .. Lxg4 ist auch dualistisch.

Eine Variante mit 8. .. Te1# findet Gustav nicht. 3. Txb7+ Kxa2 ist unmoeglich und auch 7. Dg1+.

1. Lc3 Lh2 wird aber s#7 da Lh2 b8 deckt. (2011-12-02, hidden) more ...
comment

G4
Henrik Juel: C+ Popeye 4.61 (except for late double threats) (2020-12-21)
comment

A.Buchanan: A tiny economy possible e.g. Dg2 to L, Td2 to D. I don't think one can dispense with some black unit for the last move (2020-12-13, hidden) more ...
comment

Autor: Dr. Schwarz
HBae: Autor unklar, welcher Schwarz ? (2010-11-15)
Henrik Juel: 'Dr. Schwarz' may well be a pseudonym.
The position is legal, but just barely so. (2010-11-15)
Olli Heimo: The current record is 51, look at http://www.xs4all.nl/~timkr/chess2/diary.htm
item 378.

378. 25 March 2008: New record by Alexey Khanyan - 51 consecutive checks (2010-11-16)
HBae: Thank You Olli (2010-11-17)
Mario Richter: The author might have been Dr. Karl Hermann Schwarz.

The position has some interesting features:
- With Black to move, there are 43 consecutive checks: 1. ... Rh3-d3+ 2.Nc1xd3+ Kf2-f1+ 3.Be4-g2+ Nf4xg2+ 4.Rd4-f4+ Ng2xf4+ 5.Bd5-g2+ Nf4xg2+ 6.Rc4-f4+ Ng2xf4+ 7.Bc6-g2+ Nf4xg2+ 8.Rb4-f4+ Ng2xf4+ 9.Bb7-g2+ Nf4xg2+ 10.Ra4-f4+ Ng2xf4+ 11.Ba8-g2+ Rh2xg2+ 12.Ne1xg2+ Qc7-c1+ 13.Nd3xc1+ Re7-e2+ 14.Nc1xe2+ Qc8-c1+ 15.Ne2xc1+ Re8-e2+ 16.Nc1xe2+ Bb2-c1+ 17.Ne2xc1+ Nd6-c4+ 18.Qa6xc4+ Qh5-e2+ 19.Nc1xe2+ Ba3-c1+ 20.Ne2xc1+ Nf4-d3+ 21.Qc4-f4+ Bh6xf4+ 22.Ng2-e3+ Bf4xe3+
- slightly modified (if the bQ is put on g1 instead of h5), then with WTM 44 consecutive checks are possible. (2010-11-17)
James Malcom: The current record is 54, achieved in 2013: P1386254 (2021-02-06)
comment

Henrik Juel: The position is indeed legal, but Noam's explanation of seven white and seven black promotions might be clarified
By 'pawn capture' he means PxP, and by 'piece capture' he means that an officer is captured by a pawn
e6xLf7 and g6xLh7 are not both possible here, as the two missing black officers have different square colors, but the latter capture could be g5xLh6 after Black has played Ph7-h5 (2018-11-02)
James Malcom: See the 3 ply longer P1181668. (2021-02-06)
Yuri Bilokin: 2 weitere Lösungen: 1.Tf5-e5
1.Df4-e5 (2018-11-02, hidden) more ...
comment

Henrik Juel: sKf5 is in check from wTf3, so it is Black to move
With 11 white moves to reach mate, the stipulation should read #11, and the conventional numbering is 0... Sf4+ .. 11.Dxe5# (2020-04-19)
Henrik Juel: The missing men are wLLS and sPPP
8+5 promotions have happened, after the needed 3+2 pawn captures, e.g.
wPaxPb, wPcxPd, wPexPf, sPgxL, sPhxL
(The position would still be legal with one white light officer added to the diagram) (2020-04-19)
Henrik Juel: It may be that the 'conventional' numbering would ruin the PDB animation... (2020-04-19)
James Malcom: See the earlier 19 ply P1181666. (2021-02-06)
VL: My version of this problem, stipulated ###1, with wRf3 shifted to e3 was published in my article “Completely unavoidable mate” (The Problemist, v.24, No.6, November 2013; Pr.No.7), where ### denotes a CU-mate introduced in the article. (2021-02-09)
HBae: Autor nicht in Namenstabelle (2010-12-04, hidden) more ...
comment

Nikolai Beluhov: Throughout the game, three different positions had to be repeated two times each (with the other side on the move) in order for the position to be formed.
The theme of loss-of-tempo-motivated position repetition was deeply explored by Luigi Ceriani (1894-1969) and is present in many of his works, including numerous ortho-reconstructions.
I only know of two twofold implementations: P1183992 by Ceriani himself (possibly a version by N. Plaksin or A. Kuznetsov of P0002948) and P0005418 by S. Volobuev (with a "#1" stipulation which determines the side to move in the diagram position). D. Baibikov's problem presents the first ever threefold implementation. (2011-01-02)
A.Buchanan: “fake game array” (FPI) is another special case of Ceriani’s player inversion (2021-02-24)
A.Buchanan: This is a wonderful problem. The repeated diagrams are where wDb8 & sLb7. There are sequences e.g. n. Db8~ Lb7-c8 (*) n+1. Dd8-b8 La8-b7 n+2. Dc8-d8 Lb7-a8 n+3. Db8-c8 (*) La8-b7. Need to ensure that over this sL shifts from c8 to a8, otherwise there are unnecessary additional repeated diagrams. Have fixed the solution given above. (2021-02-24)
A.Buchanan: “Position” is diagram + the move + castling rights + e.p. status. What is repeated is a diagram. Given that the stipulation was in Russian originally, I dare to change the stipulation here... (2021-02-25)
Henrik Juel: 27. Db2-e2 should make the animation work (2021-02-23, hidden)
James Malcom: Thanks Henrik, fixed. (2021-02-24, hidden)
A.Buchanan: And the c8/a8 business needs to be the other way for the third repetition when b7xDc6, else sL is trapped on a8 (2021-02-24, hidden) more ...
comment

Hans-Jürgen Manthey: R: 1. Lh5xTg4+ Sg7-f5 2. g3xf4 f5-f4 3. Sc1xLe2 Td4-g4 4. Sg4xf2+ Se4-f2+ 5. f2xDe3+ Kf4-f3 6. g2-g3+ Lf3-e2+ 7. Sh2-c1 Te2-d2 8. Lb2-c3 Lc3-e1 9. Sf6-g4 Te1-e2 10. Le2-d1 Sd2-e4 11. La3-b2 c5-c4 12. e5xd6 c7-c5 13. Lb2-a3 Tb1-e1 14. La3-b2 Tb8-b1 15. Ta1-f1 Sb1-d2 16. Sh7-f6 b2-b1=S 17. Dd1-g1 b3-b2 18. Lf1-e2 f7-f5 19. Lg4-h5 Tb4-d4 20. Ld7-g4 Th8-b8 21. Lc6-d7 Da7-e3 22. La8-c6 Db8-a7 23. a7-a8=L Dd8-b8 24. a6-a7 Tb8-b4 25. a5-a6 b4-b3 26. e4-e5 b5-b4 27. a4-a5 b7-b5 28. Ta5-g5 Se6-g7 29. Ta8-a5 Sd4-e6 30. a7-a8=T Sc6-d4 31. e2-e4 h5-h4 32. Kh4-h3 Lg4-f3 33. Sg5-h7 Lc8-g4 34. Sf3-g5 Kf5-f4 35. Lc1-a3 Lg7-c3 36. Sg1-f3 Ke6-f5 37. Kg5-h4 Kd7-e6 38. Kf4-g5 Ke8-d7 39. Ke3-f4 Sg8-h6 40. Kd2-e3 h7-h5 41. Ke1-d2 Lf8-g7 42. d2-d3 g7-g6 43. Sc3-a2 d7-d6 44. a2-a4 Ta8-b8 45. a6-a7 Sb8-c6 46. a5-a6 Ta6-a8 47. b4xa5 Ta8-a6 48. b2-b4 a7-a5 49. Sb1-c3 (2021-07-19)
Henrik Juel: The official solution should be
R: 1. Lh5xTg4+ Sg7-f5 2. g3xBf4 f5-f4 3. Sc1xLe2 Td4-g4 4. Sg4xSf2+ Se4-f2+ 5. f2xDe3+ Kf4-f3 6. g2-g3+ (2021-07-19, hidden) more ...
comment

Gerald Ettl: Die Stellung ist falsch. wTc2 auf c3 und +sDc2 (2023-07-02)
Gerald Ettl: Bzw. vielleicht ist die Stellung richtig, aber dann stimmt die Lösungsangabe und Forderung nicht. (2023-07-02)
Mario Richter: Hallo Gerald, Type C heißt: ein K muß im Schach stehen, also ist das Diagramm wohl richtig. Die Originalforderung war "Last 46 single moves?" (nicht ... 45 ...) - da hat wohl jemand den einleitenden schachbietenden Zug vergessen. (2023-07-03)
A.Buchanan: Is C20-60 available in PDB? I try to populate "superseded by" parameter with the PDB Probid (2023-07-04, hidden)
A.Buchanan: Is C22-60 available in PDB? I try to populate "superseded by" parameter with the PDB Probid (2023-07-04, hidden)
Mario Richter: s. P1353149 (2023-07-04, hidden) more ...
comment

James Malcom: How would one even begin to approach solving this, along with its compatriot P1299316? (2020-11-20)
comment

Henrik Juel: Solution by author
Retract: -1…Qb7-c8+ -2.a7-a8=B Bc4-a6 -3.a6-a7 Qb1-b7 -4.a5-a6 b2-b1=Q 5.a4-a5 b3-b2 -6.a3-a4 a4xSb3! These 11 single moves were unique ones. -7.Sd4-b3 Bc3-a1 -8.Sb5-d4 Rd1-d3 -9.Sd6-b5 Ra1-d1 -10.Se8-d6 a2-a1=R -11.Sg7-e8 B~ -12.Se8xRg7 Rh7-g7 -13.Qh8-g8 B -14.Kg8-f8 B~ -15.Sd6-e8 B~ … -20.b7-b8=S … -24.b2-b4 b3xRa2! (b3xQa2?) -25.Ra1-a2 … -30.Rf8-a8(b8) Re8-e7 -31.a2-a3! Ra8-e8
The “mobile part” of the upper cage has been released, but the “immovable part” – on the squares e5, e6, f5,
f6, f7, g4, g5, g6, h5 – is yet to be released.
-32.Re8-f8 Ra5-a8 -33.e7-e8=R! ~ -34.d6xSe7 (d6xQe7 is also possible) Sc6-e7+ -35.~ e7-e6. (2015-12-04)
James Malcom: Shortest proof game: 1. Nc3 a5 2. Nd5 a4 3. Nf6+ gxf6 4. e4 Bh6 5. Be2 b5 6. h4 Nc6 7. h5 Ne5 8. d4 Ng6 9. hxg6 Bf4 10. Bh5 Be5 11. Bg5 h6 12. Nf3 hxg5 13. Kd2 Ba6 14. Re1 Bh2 15. e5 Kf8 16. e6 Kg7 17. Re5 g4 18. Rg5 f5 19. Ne5 f6 20. Qf3 Ra7 21. Qa8 fxe5 22. d5 b4 23. Re1 b3 24. Re4 Kh6 25. Rb4 Rh7 26. Rb8 Qf8 27. Kc3 Qg7 28. Rf8 Bc4 29. Rf6 Qf7 30. exf7 e6 31. d6 Ne7 32. Kb4 Bd5 33. Kb5 Rb7+ 34. Ka6 Rb4 35. Qh8 Rc4 36. dxe7 Rc6+ 37. Kb7 Rb6+ 38. Kc8 Bc4 39. Kd8 Bf4 40. Ke8 Bd2 41. Kf8 Bc3 42. Kg8 Rb8+ 43. e8=R Rd8 44. Rf8 Re8 45. a3 Re7 46. Rb8 Bb4 47. Rb5 Ba5 48. Rd5 Bc3 49. Rd1 Bd5 50. Ra1 Bc4 51. Ra2 bxa2 52. b4 Bd4 53. b5 a1=R 54. b6 Rd1 55. b7 Rd3 56. b8=N Ba1 57. Nc6 Ba6 58. Na7 Bb5 59. Nc8 Ba6 60. Nd6 Bb5 61. Ne8 Ba6 62. Kf8 Bc4 63. Qg8 Rg7 64. Nxg7 Bb5 65. Ne8 Bc4 66. Nd6 Bb5 67. Ne4 Ba6 68. Nd2 Bc4 69. Nb3 axb3 70. a4 b2 71. a5 b1=Q 72. a6 Qb7 73. a7 Ba6 74. a8=B Qc8+ (2023-02-11)
Bob Baker: Proof game in 71.0 posted in a chess.com forum by EvinSung (screen name): 1. d4 h6 2. Bg5 hxg5 3. h4 f6 4. e4 Kf7 5. e5 Kg6 6. e6 b5 7. h5+ Kh6 8. Qf3 Nc6 9. Kd2 Ne5 10. Kc3 Rb8 11. Qa8 Bb7 12. Kb4 Bf3 13. Nc3 Bg4 14. Re1 Ng6 15. hxg6+ Bh3 16. Be2 Rb6 17. Bh5 Qe8 18. Nf3 Qf7 19. exf7 Bf5 20. Re6 Bd3 21. Rhe1 Rd6 22. R1e5 g4 23. Rg5 f5 24. Nd5 Be2 25. Nf6 gxf6 26. Ne5 fxe5 27. Rf6 e6 28. Ka5 Rd5 29. Ka6 Rd6+ 30. Kb7 Rc6 31. d5 Bc5 32. d6 Ne7 33. dxe7 Rh7 34. Qh8 Bc4 35. Kc8 Rd6 36. Kd8 Rc6 37. Ke8 Rd6 38. Kf8 Rb6 39. Kg8 Rb8+ 40. e8=R a5 41. Rf8 Re8 42. a3 Re7 43. Rb8 a4 44. Rb6 b4 45. Rd6 b3 46. Rd1 Bd4 47. Ra1 Bc3 48. Ra2 bxa2 49. b4 Bb2 50. b5 a1=R 51. b6 Rb1 52. b7 Rc1 53. b8=N Bc3 54. Nc6 Ba1 55. Nd4 Rd1 56. Nb5 Rd2 57. Nd6 Rd3 58. Ne8 Rd4 59. Kf8 Rd5 60. Qg8 Rg7 61. Nxg7 Rd6 62. Ne8 Rd5 63. Nd6 Rd4 64. Nb5 Rd2 65. Nd4 Rd3 66. Nb3 axb3 67. a4 b2 68. a5 b1=Q 69. a6 Qb7 70. a7 Ba6 71. a8=B Qc8+ (2024-02-02)
comment

Henrik Juel: 1.Sf3 a5 2.Sd4 a4 3.Sb3 axb3 4.a4 Ta5 5.Ta2 bxa2 6.h4 a1=S 7.h5 Sb3 8.h6 Sd4 9.hxg7 Sf5 10.gxh8=L Sh4 11.Le5 f5 12.Lh2 e5 13.g3 La3 14.Lg2 Se7 15.Lc6 dxc6 16.O-O Sd7 17.Kh1 Sf8 18.Lg1 Dd5+ 19.Kh2 Kd7 20.Kh3 Dh1+ 21.Lh2 e4 22.Tg1 Te5 23.Df1 e5 24.Dg2 Kc6 25.Df3 Kb6 26.Tg2 Dd1 27.Tg1 Ka5 28.Te1 b6 29.Lg1 La6 30.Kh2 Ld3 31.Kh1 c4 32.Lh2 c5 33.Kg1 Dxe1#
Judge J. Lois:
This problem presents the theme Lois between the wK1-WBh2(2) with a switchback of both pieces.
In addition, change of place between WKe1-wRh1 in move 31. (2017-08-30, hidden)
Henrik Juel: In the award this proof game is not labelled C+ (2017-08-30, hidden)
Michel Caillaud: reported as cooked in Winchloé :
1.a4 b5 2.Ta3 b4 3.h4 bxa3 4.h5 a2 5.h6 a1=S 6.hxg7 Sb3 7.gxh8=L Sd4 8.Le5 Sf5 9.Lh2 e5 10.g3 La3 11.Lg2 La6 12.Lc6 dxc6 13.Kf1 c5 14.Kg2 Kd7 15.Sh3 Kc6 16.Sf4 Kb6 17.Sd5+ Ka5 18.Sb6 axb6 19.Kh3 Dd5 20.Tg1 Dh1 21.Df1 e4 22.Dg2 Sd7 23.Df3 Te8 24.Tg2 Dd1 25.Tg1 Te5 26.Te1 Sf8 27.Lg1 Sge7 28.Lh2 Sh4 29.Lg1 f5… (2020-10-14, hidden) more ...
comment

Joaquim Crusats: ?1.Sg6-h8 Bh7-g8 (1) ?2.Se7/h4-g6 Bg8-h7+ ?3.Sf5-e7/h4 Bh7-g8 (2) ?4.Sd6-f5 Bg8-h7+ ?5.Se4-d6 Bh7-g8 (3) ?6.Sc5-e4 Bg8-h7+ ?7.Sd3-c5 Bh7-g8 (4) ?8.Sb4-d3 Bg8-h7+ ?9.Sc2-b4 Bh7-g8 (5) ?10.Sa3-c2 Bg8-h7+ ?11.Sc2xPa3 Bh7-g8 (6) ?12.Sb4-c2 Bg8-h7+ ?13.Sd3-b4 Bh7-g8 (7) ?14.Sc5-d3 Bg8-h7+ ?15.Se4-c5 Bh7-g8 (8) ?16.Sd6-f5 Bg8-h7+ ?17.Sf5-d7 Bh7-g8 (9) ?18.Se7/h4-f5 Bg8-h7+ ?19.Sg6-e7/h4 Bh7-g8 (10) ?20.Sh8-g7 Bg8-h7+ ?21.h7?h8=S a4-a3 ?22.h6?h7 a5-a4 ?23.h5?h6 h6xSg5 ?24.Se4-g5 Bh7-g8 (11); then e.g.: ?25.Sd2-e4 Bg8-h7+ ?26.Sb3-d2 a7/a6-a5 ?27.Sc1-b3 Kd2-d1 (12) ?28.h4-h5 Ke3-d2 ?29.Kc2-b1 etc.

Full both ways stair walk of the wS (unavoidable).
12 screenings.

White balance: 14 + gxf + hxg = 16 bPb7?b1=R
Black balance: 12 + bxc + cxd + dxe + a-bP = 16 wPh2?h8 (2023-10-03)
comment

paul: Checked by Jacobi. (2021-12-08)
comment

Editor's comment: Do not be put off by the length of R546: there is a repeated sequence of moves.
James Malcom: Solution? Also, is this an Othoreconstrution? (2023-05-30)
Henrik Juel: Here is the solution from the May 2020 issue (from www.theproblemist.org , which is a treasure of info, althoug I find it hard to navigate)
R546 (Le Gleuher): Retroanalysis: The wPs captured the 4 missing black units.
Thus, wPb4 comes from b2, and (Pa2) promoted to the extra wR on a8 with (Pa7)
letting it pass by switching to the b file (first capture of the 3 missing white units).
(Pb7) must have promoted to a rook on a1 or c1 (second capture). (Ph7) must
have moved onto the g file (third and final capture) in order to be captured. Thus
(Pc7) and (Pe7) promoted to rooks on the c and e files, respectively. It is not
possible to immediately take back a4xb3 or c2xd3. The retro-move e4xf5 is
forbidden since then the bP which promoted on e1 cannot retract to e5. Black can
only move in the southeast cage, while White’s bishop is free to oscillate between
c2 and b1.
Black to move: back 1.Bc2-b1 Kg1-f1 2.Bb1-c2 Rf1-f2 3.Bc2-b1 Rf2-f3
4.Bb1-c2 Rf3-g3 5.Bc2-b1 Rg3-h3 6.Rh3-h2 Rh2-h1 7.Bb1-c2 Kh1-g1 8.Bc2-b1
Rg1-f1 9.Bb1-c2 Rf1-f2 10.Bc2-b1 Rf2-f3 11.Bb1-c2 Rf3-g3 12.Rg3-h3 Rh3-h2
13.Bc2-b1 Kh2-h1 14.Bb1-c2 Rh1-g1 15.Bc2-b1 Kg1-h2 16.Bb1-c2 Rh2-h1
17.Bc2-b1 Kh1-g1 18.Bb1-c2 Rg1-f1 19.Bc2-b1 Rf1-f2 20.Bb1-c2 Rf2-f3
21.Rf3-g3 Rg3-h3 22.Bc2-b1 Rh3-h2 23.Bb1-c2 Kh2-h1 24.Bc2-b1 Rh1-g1
25.Bb1-c2 Kg1-h2 26.Bc2-b1 Rh2-h1 27.Bb1-c2 Kh1-g1 28.Bc2-b1 Rg1-f1
29.Bb1-c2 Rf1-f2 30.Rf2-f3 Rf3-g3 31.Bc2-b1 Rg3-h3 32.Bb1-c2 Rh3-h2
33.Bc2-b1 Kh2-h1 34.Bb1-c2 Rh1-g1 35.Bc2-b1 Rg1-f1 36.Rf1-f2 Bf2-e1
37.Re1-f1 Rf1-g1 38.Bb1-c2 Rg1-h1 39.Bc2-b1 Kh1-h2 40.Bb1-c2 Rh2-h3
41.Bc2-b1 Rh3-g3 42.Bb1-c2 Bg3-f2 43.Bc2-b1 Rf2-f1 44.Bb1-c2 Rf1-g1
45.Bc2-b1 Kg1-h1 46.Bb1-c2 Rh1-h2 47.Bc2-b1 Kh2-g1 48.Bb1-c2 Rg1-h1
49.Bc2-b1 Kh1-h2 50.Bb1-c2 Rh2-h3 51.h3-h4 Bh4-g3 52.Bc2-b1
then 52…Rg3-f3 (or 52...Bg3-h4 53.Bb1-c2 Bh4-g3…) 53.Bb1-c2 Rf3-f2 54.Bc2-b1 Rf2-f1 55.Bb1-c2
Rf1-g1 56.Bc2-b1 Kg1-h1 57.Bb1-c2 Rh1-h2 58.h2-h3 Rh3-g3 59.Bc2-b1 Bg3-h4 60.Bb1-c2 Rh8-h3 etc…
The diagram position is a draw and the last 103 single moves are determined.
White to move: back 1…Kg1-f1 …
51.h3-h4 Bh4-g3 52.Bc2-b1 then the same… Here, there are 99 single moves between the diagram position and the retro-move h3-h4. Of course, trying to insert another manoeuvre in the solution would lead to an overshoot of the 101 single moves, which would be incompatible with the 50 moves rule. Here only 102 last single moves are determined.
So, we have a D25/102, a new record for last single moves for ‘D’ type (Duplex) with 25 pieces (It is also
the overall record for Duplex). With black to move, the same diagram is also a record with 25 pieces (B25/103) (2023-05-30)
comment

Henrik Juel: Correct key, but damaged by multiple threats and duals, e.g.
1.Kd3 thr. 2.e8=DTLS,Ke4 (2020-09-27)
James Malcom: Typo fixed. And Henrik, the threats are not the point here. The point here is how the different play of the Black knight changes White's promotions. (2020-09-27)
A.Buchanan: It's a beautiful idea, still looking for its rendering though as the many duals and the obstreperous wL are quite distracting (2020-09-28)
James Malcom: There aren't any duals in full-length variations, though. The extra bishop is indeed annoying, but unfixable so far. (2020-09-28)
A.Buchanan: There are still some trivial duals in the mating move, but yes. This problem is trying to ignore threats: you can see that the computer solution. The solution is much more coherent with "nosh noth" switched on. So when are the cases, Henrik, when you switch those options on? It would be event better if one was able to switch threats on for the 4th move. This underscores that there should be a Popeye option for any white move "noth" is self-selecting, depending upon the number of variations. (2020-09-29)
Henrik Juel: Andrew, I use the Popeye option 'nosh' when the output without them is overwhelmingly large
The 'not' option is tricky, and usually expands the output; in some cases it shows apparent duals that are not real ones, when you apply the rule that dualistic play should be ignored, when it appears after black moves that do not prevent the threat
In the case of multiple threats things get messy, like in this problem, as I recall (2020-09-29)
Henrik Juel: tiny typo
1... Sf5 2.Lxf5, not Lxe5 (2020-09-27, hidden) more ...
comment

Henrik Juel: Ruined by duals after 1... Kd4,Kc2,f6 (2020-10-13)
comment

See P1234039, P1375777 and P1380852.
The joke here is that pawns may promote to *any* unit (friend or enemy; officer, king or pawns).
Henrik Juel: R: 1... 0-0-0+ 2.a7-a8=sT
not 2.e7-e8=sK, because then sTa8 would have moved, and castling not allowed (2020-10-14)
A.Buchanan: Exactly, Henrik. Also: sB on 7th rank could retract to 8th rank, so we need sT instead (2020-10-14)
Henrik Juel: A non-joke solution is
R: 1... 0-0-0+ 2.h5-h6 (2020-10-14, hidden)
A.Buchanan: Sorry Henrik typo have fixed (2020-10-14, hidden) more ...
comment

Henrik Juel: Popeye 4.61 with 'opt thr 0' found no flaws in part a (2020-11-12)
Henrik Juel: ... and with 'opt thr 0 non 0 1' no flaws in part b (2020-11-12)
comment

James Malcom: How does this work exactly? What forbids the rook from just making some random tempi move in a game that leads to this position? (2020-12-12)
Mario Richter: James, think of time pressure. In the diagram, Black has only two retractions (a7xSb6xSa5). The only way to unlock the position is the retraction of e2-e3.
Little hint: castling is considered as a king m,opve ... (2020-12-12)
Henrik Juel: By now there are so many clues revealed for this fine retro that I dare to give a retroplay despite the Byzantine dating, in this case indicating an ongoing solving competition
R: 1.Tf1xPf6 b6xSa5 2.0-0 a7xSb6 3.Lf1-e2 Kd4,Kf4 4.e2-e3+ etc.
Incidentally, TxP is not the first move, but the last... (2020-12-12)
Henrik Juel: I missed the from square in the retraction by the black king, sorry
3... Kd4-e5,Kf4-e5 (2020-12-12)
A.Buchanan: Ha, there is no solving competition for these either. But it's to avoid Gerd's solving competition protection algorithm picking up false positives (2020-12-12)
A.Buchanan: TxP happens to be the last move, but it's also the first effective move of the rook, as it is considered to be a *passenger* in the castling, and not to have moved yet. Otherwise can't get first move of a rook fixed if it capturing (except possibly by DP logic). See the explanatory article: https://www.dieschwalbe.de/download/artikel/Passenger_Rooks.pdf (2020-12-12)
A.Buchanan: Your solution is completely correct, Henrik! (2020-12-12)
comment

Henrik Juel: Oh, I was confused by the LLLLl, which should be TLLLLl, I believe
R: 1.Lg2-f3 b4-b3 2.Tf1xSf6! b5-b4 3.0-0 b6-b5 4.Lf1-g2 Kf4-g5,Kh4-g5 5.g2-g3+ (2020-12-12)
A.Buchanan: Thanks for the catch. I copied the diagrams straight from the article FEN, so should be clean of transcription errors, but then I put in the non-standard material field by hand & eye, so could be other errors as here. (2020-12-12)
A.Buchanan: Your move sequence is again correct! :D (2020-12-12)
Henrik Juel: Is there a diagram error?
White has three rooks (2020-12-12, hidden) more ...
comment

Henrik Juel: Good motivation for TxL
R: 1.Tf1xLf8! e4-e3 2.0-0 e5-e4 3.Lf1-g2 e6-e5 4.g2xPh3 h4-h3 and e.g.
5.d4-d5 h5-h4 6.Le4-c6 h6-h5 7.c7-c8=L! h7-h6 8.c6-c7 c7xLb6
f7xTg8=L happened early (2020-12-12)
comment

Hans-Jürgen Manthey: 1. Lg2-a8 c3-c2 2. Tf1xTf6 c4-c3 3. O-O ??? (Rochade nicht möglich wegen sTf6 !)

mögliche Lösung:
1. Tf1xTf6 c3-c2 2. Lf3-a8 c4-c3 3. O-O c5-c4 4. Lg2-f3 c6-c5 5. Lf1-g2 Kf4-g5 6. g2-g3+ Sg5-h7 7. Ld3-g6 Se4-g5 8. Th7-g7 Dg6-f7 9. f7-f8L Tf5-f6 10. Le5-h8 Lf8-h6 11. Lh4-e7 Dg5-g6 12. La7-e5 Tf6-f5 13. Lb8-a7 Th6-f6 14. Lf6-d8 Sc5-e4 15. Lb2-f6 Dd8-g5 16. Lf2-h4 Kg5-f4 17. f6-f7 Sa6-c5 18. Te7-h7 Th8-h6 19. f5-f6 Kh6-g5 20. La3-b2 Kg7-h6 21. Lf7-e8+ Kf6-g7 22. Lc1-a3 Kg7-f6 23. b7-b8L h7-h5 24. La7-f2 Kf6-g7 25. Lb8-a7 Kg7-f6 26. b6-b7 Kf6-g7 27. b7-b8L Kg7-f6 28. Te8-e7 Df6-d8 29. Ta8-e8 g5-g4 30. Le8-f7 f7xSe6 31. Sf4-e6+ Sb8-a6 32. Ta1-a8 Dd8-f6 33. e7-e8L Kf6-g7 34. e6-e7 Ke7-f6 35. b5-b6 g6-g5 36. b6-b7 g7-g6 37. c4xLb5 La6-b5 38. c2-c4 Lc8-a6 39. La6-d3 Ke8-e7 40. Lb7-a6 Ke7-e8 41. La8-b7 Ke8-e7 42. a7-a8L Ke7-e8 43. d5xTe6 Te3-e6 44. a6-a7 Ke8-e7 45. a5-a6 Ke7-e8 46. a4-a5 Ke8-e7 47. a2-a4 Ta3-e3 48. Sh3-f4 Ta8-a3 49. Sg1-h3 b7xSc6 50. Sa7-c6 Sc6-b8 51. Sb5xa7 e7xDd6 52. Dh6-d6 Sb8-c6 53. Dd2-h6 Sc6-b8 54. Dd1-d2 Sb8-c6 55. f4-f5 Sc6-b8 56. f2-f4 Sb8-c6 57. d4-d5 Sc6-b8 58. d2-d4 Sb8-c6 59. Sc3-b5 Sc6-b8 60. Sb1-c3 Sb8-c6 61. b5-b6 Sc6-b8 62. b4-b5 Sb8-c6 63. b2-b4 (2020-12-14)
Henrik Juel: e.g. R: 1.Lg2-a8 c3-c2 2.Tf1xTf6 c4-c3 3.0-0 c5-c4 4.Lf1-g2 Kf4-g5 5.g2-g3+ (2020-12-12, hidden)
A.Buchanan: wLf3 must shield uncastling from sTf6. (2020-12-13, hidden)
Henrik Juel: Yes, of course (2020-12-13, hidden) more ...
comment