Henrik Juel: the five missing black men were captured by white pawns (exfxgxh, fxgxj), so with Black to move last move was with Ta1 or Ke1, and White may not castle
C+ Popeye 4.61 (2022-11-26)
A.Buchanan: Pleasant White tempo play in both phases (2022-11-27)
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Silvio Baier: Der wesentliche thematische Inhalt ist bereits nach 31,5 Zügen erreicht. Bis dahin ist es C+ (Euclide 0.98). (2010-08-04)
James Malcom: Is this fully C+ then? (2021-01-27)
Henrik Juel: No, only the first 31.5 moves are tested OK (2021-01-27)
James Malcom: No Henrik, as in is the entire problem testable. (2021-01-27)
Henrik Juel: I guess that testing the entire problem would take an unreasonably long time (2021-04-06)
A.Buchanan: It might be possible these days: the motivation for stopping at 31.5 is that the promotion theme had been demonstrated by then. But there’s still e.g. bSb8 as random impostor (2021-04-07)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 50:24:34 Stunden.(hh:mm:ss)
Keine Lösung: BP 34.5, BP 35.0. (2023-05-08)
Henrik Juel: Thanks for your patience, Moldenhauer; more than two days... (2023-05-08)
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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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Moldenhauer: Ergänzung: Stelvio 1.2. Keine Lösungen BP 12.5, BP 13.0.
Es ist wirklich eine Rarität dass es 2 komplette Lösungen gibt. (2023-05-20)
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Moldenhauer: Ergänzung: Stelvio 1.2. Keine Lösung: BP 14.0, BP 14.5.
BP 15.0 und BP 15.5 cooked.
Notation BP 15.0: 1.Sc3 Sc6 2.a4 b5 3.Ta3 La6 4.d3 Db8 5.Lg5 Db6 6.Lf6 0–0–0
7.e3 Kb7 8.Dg4 Dc5 9.Le2 Kb6 10.Lf3 Ka5 11.Sce2 Tb8 12.Tc3 Sd8 13.La8 De5
14.Tc6 Tb7 15.Tb6 c6
Notation BP 15.5: 1.Sc3 Sc6 2.a4 b5 3.Ta2 La6 4.Ta3 Db8 5.d3 Db6 6.Lg5 0–0–0
7.e3 Kb7 8.Dg4 Dc5 9.Le2 Kb6 10.Lf3 Ka5 11.Sce2 Tb8 12.Tc3 Sd8 13.La8 De5
14.Tc6 Tb7 15.Tb6 c6 16.Lf6.
Alles schlagfrei und ohne Umwandlung. (2023-05-20)
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Henrik Juel: Natch 3.1 tests the first 20.0 moves OK in 13 seconds, but the first 20.5 moves would take more than an hour, and the entire problem more than a couple of hours, because of the possibility of [Pd7] promoting
So we may never know for sure whether this proof game with 16 oscillations by Ta8 is correct (2018-12-08)
Mario Richter: Henrik, I think certainty about the correctness of this problem can be achieved, if we take into account that Black's last move must have been a capture. (2018-12-10)
A.Buchanan: Henrik: the times you propose for testing don't seem outrageous, even without Mario's sensible heuristic. Also, the last move must have been DxLd7 from one of 5 squares only. (2018-12-12)
Reto: Stelvio is now able to retract moves and solve the resulting problems individually. That helps a lot here, but still, the 5 reasonable last move retractions took over 13h in total to solve. (2023-06-11)
paul: 18.Tca3 & 19.T3a4 (2010-04-30, hidden) more ...
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paul: Checked for the first 19 moves, by Euclide. (2011-07-01)
Moldenhauer: Computerprüfung: C+ Stelvio 1. Std. 16 Min.
Keine Lösung BP 24.0 (2023-02-06)
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Henrik Juel: So Michel's proof game is not cooked; yet... (2019-09-10)
Reto: This is now the length record for a fully tested SPG. (2023-09-17)
paul: Cooked: 1.a3 g5 2.Sh3 a5 3.Sf4 a4 4.Sg6 hxg6 5.Ra2 Rh4 6.h3 Rb4 7.h4 Ra5 8.h5 Bg7 9.Rh4 Be5 10.Rd4 Bh2 11.g3 Kf8 12.Bg2 Kg7 13.Bc6 Kf6 14.Kf1 Qf8 15.Kg2 Qh6 16.Kf3 g4+ 17.Ke4 Qf4+ 18.Kd3 Kg5 19.gxf4+ Kh4 20.Qe1 Kh3 21.Qd1 g3 22.Ra1 g2 23.Qh1 g1=S 24.Qd5 Sf3 25.Qc5 Kg2 26.Qc4 Kf1 27.Qc5 Ke1 28.Qc4 Kd1 29.Qd5 Kxc1 30.Qb3 axb3 31.Ra2 bxa2 32.h6 axb1=S 33.h7 Sc3 34.h8=R Sa4 35.Rh5 Sb6 36.Rb5 Sa8 37.Rb6 Sg5 38.Ra6 Sh7 39.Ra7 b6 40.Rb7 Ra7 41.Bd5 Sa6 42.Rb8 Bb7 43.Ba2 Bd5 44.Bb1 Ba2 45.b3 Kb2 46.Rf8 Ka1 47.Kc3 Bg1.[N. Plaksin, Die Schwalbe 130/1991] (2019-09-09, hidden)
paul: Comparing the obtained positions, it was observed that the pawn in the g6 must be at f6. A mistake by the publishers, but mine! (2019-09-10, hidden) more ...
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Henrik Juel: with a black pawn on h3, we have all eight black pawns on the board, but none of them could come from a7 (2022-11-26)
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Henrik Juel: C+ Popeye 4.61 (2022-11-26, hidden) more ...
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paul: Since the wB couldnt leave c1, the captured piece on b6 is the promoted wPh2. If last black move was f7-f5, then h-Pawn captured four pieces to promote on d8, among which bRh8, impossible. Thus, the h-Pawn could only have been promoted from f7 and therefore d7-d5 remains as the last move. (2011-07-02)
Henrik Juel: C+ Popeye 4.61 (2022-11-26, hidden) more ...
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A.Buchanan: In each of the 4 pairs of adjacent files, one cross-capture suffices. So removing 8 officers is enough. However, they need to be an even number from each side.
However it's possible to do better by removing pawns: wPadgh bPcf = 6 units. Now wPaxb bPcxb wPdxe bPfxe.
Is it possible with 5 or less? (2022-03-17)
Bob Baker: I find it interesting that the composer specified piece removals, since six captures suffice. Does that imply that 5 or less is possible by removing pieces? (2023-04-27)
A.Buchanan: I think the German term "Figuren" (like the English "pieces") can have two senses, either all 16 pieces or just the officers. Wikipidie.ge uses the phrase "im engeren Sinne" = "in the narrower sense" to distinguish. But maybe a native German speaker can be more authoritative here:

Auf dem Schachbrett befinden sich zu Beginn einer Partie insgesamt 32 Schachfiguren (auch als Steine bezeichnet), 16 weiße und 16 schwarze. Beide Spieler (bezeichnet als Weiß und Schwarz oder als Anziehender und Nachziehender) haben je folgende Schachfiguren zur Verfügung:
- acht Figuren im engeren Sinne:
= den König
= die Dame und zwei Türme (Schwerfiguren)
= zwei Springer und zwei Läufer (Leichtfiguren)
- acht Bauern (2023-04-27)
A.Buchanan: And the Originaldforderung would have been in French, surely (2023-04-27)
Bob Baker: A stipulation might have been "Reach the position with the greatest number of pieces still on the board." A position with six fewer men can be reached with no removals from the initial array. It would be a harder problem than if pieces can just be taken off the board. But maybe allowing removals could make 5 or less possible. (2023-04-28)
A.Buchanan: Not quite sure what you are saying. Generally, capturing pawns is more efficient than capturing officers to get pawns past one another, but here we have no doubled pawns. If we have a solution with 5 units including a missing officer, then might as well promote a pawn to replace it. So we can reduce our search to positions where 5 pawns have been removed (2023-04-28)
Bob Baker: I'm suggesting if six really is the answer, it would be a better (more challenging) problem to require a proof game with no removals except by capturing. (2023-04-28)
Bob Baker: Also, I think this may have been the composer's intent, but the stipulation was misunderstood as allowing pieces to be taken off the board without being captured. (2023-04-28)
Bob Baker: I just realized that the misunderstanding of the composer's intent was on my part, so all my previous comments should be disregarded. (2023-04-28)
Carot: I understand that all 16+16 figures are meant.
So for the e-row 1. e4 e5 2. e4f5 e5f4 3. f5e6 f4f3
This is illegal, so only the e-line is considered, one e-pawn too many to reach the given position legally.

I am just a beginner in Chess, but i think en passant is useable and therefore maybe less then 8 pieces have to removed, but i tried an hour and found no way for optimisation. :(..

I would be happy about an idea. Please share. I go study pawn endgames now.
I am happy to found this site and many old combinations of Max Lange. (2023-04-28)
Bob Baker: Play this game for a quick way to remove six pieces.
1. a4 a5 2. b4 b5 3. c4 c5 4. d4 d5 5. e4 e5 6. f4 f5 7. g4 g5 8. h4 h5 9. axb5
hxg4 10. bxc5 gxf4 11. exf5 dxc4 12. b6 a4 13. c6 a3 14. d5 c3 15. d6 e4 16. f6
e3 17. h5 f3 18. h6 g3 (2023-04-29)
A.Buchanan: In each of the 4 pairs of adjacent files, one cross-capture suffices. So a total of 8 captures is necessary. Hence removing 8 officers is enough. However, they need to be an even number from each side. (2022-03-17, hidden)
A.Buchanan: I think Figuren (like pieces) can have two senses, either all 16 pieces or just the officers. Wikipidie.ge says:
Auf dem Schachbrett befinden sich zu Beginn einer Partie insgesamt 32 Schachfiguren (auch als Steine bezeichnet),
16 weiße und 16 schwarze. Beide Spieler (bezeichnet als Weiß und Schwarz oder als Anziehender und Nachziehender)
haben je folgende Schachfiguren zur Verfügung:
- acht Figuren im engeren Sinne:
= den König
= die Dame und zwei Türme (Schwerfiguren)
= zwei Springer und zwei Läufer (Leichtfiguren)
- acht Bauern (2023-04-27, hidden) more ...
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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)
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S N Ravi Shankar: This problem is not mine. (2019-01-02)
Mario Richter: My first guess for the solution was:
R: 1. Ke2xLf1 2. Kd3xLe2 3. Kc4xLd3 4. Kb5xLc4 5. Ka6xLb5 Lc6-b5+ 6. Sd7xLb8, dann 1. Sb6#, but that doesn't work because of 5. ... Ld7-b5+!
Does someone see (or know) the correct solution? (2019-01-03)
Henrik Juel: No, because with 1... Lh3-f1+, 5.Ka6-b5 thr. 6.Sd7xLb8 fails similarly on 5... Ld7-h3 (2019-01-03)
S N Ravi Shankar: Does adding a black knight on e6 cure? 1. Ke2xLf1 2. Kd3xLe2 3. Kc4xLd3 4. Kb5xLc4 5. Ka6-b5! followed by 6. Sd7xLb8 and now 1. Sb6#. (2023-02-20)
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Moldenhauer: Computerprüfung: C+ da NUPG Stelvio 1.11 47 Sekunden.
Keine Lösung: BP22.0, BP 22.5.
Auf b7 wird der wLf1 durch Sb8 der auf später auf c5 steht geschlagen. (2023-03-28)
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Moldenhauer: Ergänzung: C+ Stelvio 8 Sekunden.
Keine Lösung: BP 25.5. (2023-06-24)
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Frank Müller: Welche Bauern sind Berolinabauern? (2010-08-15)
Joost de Heer: No berolina pawns.
Solution: 1.é5 Rb8 2.é6 Ra8 3.é7 Rb8 4.é8=T Ra8 5.Th8 Rb8 6.Th1 Ra8 7.Té1 Rb8 8.Té2 Ra8 9.Té1(+é2) Rb8 10.Tb1+ Ra8 11.Tb8+ R×b8 12.é4 Ra8 13.é3 (2023-08-27)
A.Buchanan: Quite delightful. Can it be validated in Jacobi as an A-B PG? (2023-08-28)
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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 4 Sekunden.
Keine Lösung: BP 26.5, BP 27.0. (2023-03-30)
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No. 1111 HN
Correction of P0003822
A.Buchanan: Odd that the correction has fewer composers than the cooked original (2023-05-27)
HENRI: To be corrected :
The third author Jean-Claude Dumont must be add as mentioned in the award 1975-1976 page 539 diagrammes n° 31 jan - feb 1978. (2023-05-30)
paul: 1.Cg5? f7-f5! Correction of P0003822 (2011-08-05, hidden)
Henrik Juel: C+ Popeye 4.61 and (french) analysis (2023-05-26, hidden) more ...
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No. 11661 HN
A.Buchanan: “Obvious promotion” is intended to refer to something visible in the initial diagram position, not what is reached during series play (2022-09-14)
A.Buchanan: Similarly “non-standard material (2022-09-14)
paul: La solution commence avec 1.d4xc3 e.p.! La source est diagrammes 148/2004. (2011-07-03, hidden)
A.Buchanan: Similarly “non-standard material”, though this is not present in the problem at any point. (2022-09-14, hidden) more ...
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