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)
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Kostas Prentos: A correction was published in Phenix, 2009 (No.186/Pg.7979)
Solution:
1. Sf3 f5 2. Sd4 Kf7 3. Sb5 Kg6 4. Sd6 exd6 5. e4 Le7 6. e5 Lh4 7. e6 Dg5 8. e7 Dg3 9. e8=T Se7 10. Lc4 Tf8 11. 0-0 Tf6 12. f3 Kh5 13. Te1 Th6 14. Lf7+ Sg6 15. T1e6 dxe6 16. Txc8 Sd7 17. Th8 Txh8 18. Le8 (2022-12-08)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.2 00:33:56 Minuten. (hh:mm:ss)
Keine Lösung: BP 17.5, BP 18.0.
Beispiel: 1.Sf3 f5 2.d4 Kf7 3.e4 g5 4.Lc4+ Kg6 5.e5 Lg7 6.e6 Lxd4
7.0–0 Lxb2 8.Dd6 exd6 9.e7 Kh5 10.e8S Se7 11.Te1 Tf8 12.Sxg5 Tf6
13.f3 Th6 14.Lf7+ Sg6 15.Te7 Lf6 16.Se6 dxe6 17.Td7 Lh4 18.Lg5 Dxg5
19.Sg7# (2023-05-30)
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James Malcom: Solution? (2020-11-07)
Mario Richter: I do not know the official solution, but the following works:
wBb7 wBf7 wBb6 wBg6 wBc5 wTf5 wLf4 wBg4 wBe3 wKg3 wTh3 wDf2 wBh2 wSf1 wSh1
sLg8 sDh8 sBa7 sBc7 sBh7 sBe6 sBf6 sKh6 sBe5 sLg1
wCaps: h5xg6ep f3xTg4 g4xTh5 e6xSf7 d5xSe6 a4xBb5
sCaps: d7xLe6
R: 1. h5xg6ep g7-g5 2. Tg5-f5 d7xLe6 3. f3xTg4 Th4-g4 4. g4xTh5 Kg6-h6 5. Tf5-g5 (2020-11-10)
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Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:09:39 Minuten. (hh:mm:ss)
Keine Lösung: BP 22.0, BP 22.5. (2023-03-26, hidden) more ...
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Yoav Ben-Zvi: The cook found in P0002554 re-opens the question of the current length record for unique capture-free SPG. A search in PDB finds this problem, assuming it is sound, to be the record holder, until Unto Heinonen publishes a correction (expected very soon). Prior to publication of this problem the record was 38.5, in Michel Caillaud's P0002187. (2019-07-10)
Michel Caillaud: cooked (testing last moves with Jacobi 0.7.5 and constraints) :
16...Sd7 17.h4 Tc8 18.Th3 Tc6 19.Kf4 Te6 20.Tf3 Te3 21.Kg5 Db8 22.Tf6 Th3 23.Tc6 Sgf6 24.Kh6 Th1 25.Kg7 Dh2 26.g3 Se5 27.Lg2 Kd7 28.Tc8 Ke6 29.Kf8... (2021-02-13)
James Malcom: And thus, another giant falls. (2021-02-13)
A.Buchanan: Did Unto find a fix? (2021-02-14)
Michel Caillaud: Unto published a correction of P0002554 in StrateGems (42,5 moves, january 2020), not C+ of course... (2021-02-14)
James Malcom: May you input it into PDB please, Michel? (2021-02-14)
James Malcom: Thanks for the email, Michel. Unto's correction can now be seen in the database: P1386520 (2021-02-15)
Paulo Roque: Dual:
a) 18...Qb8 19.Rc6 Qh2
b) 35...Nf3 36.Kd7 Ke1 37.Kc7 Ba5+ 38.Kb8 Kd1 39.Re8 Re1 (2008-11-27, hidden)
Paulo Roque: No cook. No dual.
The final positions are different.
In the authorâ??s solution is the bishop at square b6.
Right solution:
1.a4 h5 2.Ra3 Rh6 3.Rc3 Rb6 4.Rc4 Rb3 5.Nc3 Ra3 6.b3 Ra1 7.Ba3 d5 8.Qb1 Bf5 9.Kd1 Bh7 10.Kc1 g6 11.Kb2 Bh6 12.Qa2 Rc1 13.Nb1 Be3 14.Kc3 Bb6 15.Kd3 c5 16.Ke3 Nc6 17.Kf4 Qc7+ 18.Kg5 000 19.h4 Rd6 20.Rh3 Rf6 21.Re3 Rf3 22.Re6 Nf6 23.Kh6 Rh3 24.Kg7 Rh1 25.Kf8 Qh2 26.g3 Ne5 27.Rc6+ Kd7 28.Rc8 Ke6 29.Bg2 Kf5 30.Be4+ Kg4 31.Nf3 Kh3 32.Bf5+ Kg2 33.Rg4 Bg8 34.Nd4 Nh7+ 35.Ke8 Nf3 36.Kd7 Kf1 37.Re8 Ke1 38.Kc8 Kd1 39.Kb8 Re1 40.Bc8 Ng1 41.f3 (2008-11-28, hidden)
Gerd Wilts: The previously given cook was indeed wrong because black Bishop must finally be on b6 instead of a5. (2008-11-28, hidden)
paul: 33. Tg4 (2014-07-04, hidden) more ...
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See P0003960
Henrik Juel: The forward part is C+ Popeye 4.61 (2022-06-01)
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paul: Cooked by: 1.h3 d5 2.Rh2 Bxh3(ph2) 3.Sxh3(Bc8) Bxh3 4.g4 a5 5.Bg2 Bxg4(pg2) 6.c4 a4 7.b4 e6 8.Bb2 Bxb4(pb2) 9.Sa3 Ra5 10.Qxa4(pa7) Qc8 11.QxKe8(Ke8) (suicide of wQ)b5 12.Sc2 dxc4(pc2) 13.a3 Bxe2(pe2) 14.Ra2 Bxa3(pa2) 15.Kd1 b4. (2010-06-19)
Arnold Beine: Here is something wrong. Either Paul's cook does not work, because 10.Qxa4(Pa7)+ is a check (10...Qc8??) or in the given condition "Rex inclusive" must be added. (2023-07-26)
paul: Right, Arnold. Anyway, the proof game is cooked: 1.a4 e6 2.Sa3 B×a3 3.b4 a5 4.Bb2 a×b4(b2) 5.Ra2 R×a4(a2) 6.e4 Ra5 etc, as in author's solution: 7.Ba6 b×a6(Bf1) 8.B×a6(a7) d5 9.B×c8(bBc8) Bb7 10.c4 Qc8 11.Qc2 d×c4(c2) 12.Se2 B×e4(e2) 13.h3 Bf5 14.Rh2 B×h3(h2) 15.Kd1 B×g2(g2). (2024-03-13)
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Henrik Juel: This is one of the proof games that is solved faster by hand than by computer
The two captures are obvious, and the interchanged rooks make a guess like Th1-g1-g3-h3-h5-g5-g8-h8-a8 reasonable
Natch 3.1 got nowhere in an hour (2018-12-10)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 40 Sekunden.
Keine Lösung: BP 21.5, BP 22.0. (2023-02-26)
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paul: Sol.: 1.g3 h5 2.Bh3 Rh6 3.Kf1 Re6 4.Kg2 f6 5.Kf3 Re3+ 6.Kf4 e5+ 7.Kf5 Bb4 8.Kg6 Qe7 9.Kh7 Qc5 10.Kxg8 d6 11.Kxg7 Bg4 12.Kg6 Bf3 13.exf3 Re1 14.Se2 Rg1 15.Kf5 Rg2 16.Re1 a5 17.Sg1 a4 18.Re3 Ra5 19.Rb3 axb3 20.Ke4 bxa2 21.Kd3 e4+ 22.Ke2 e3 23.Ke1 (Ra4 is in a5)
The King captures an officer on its place and return.
For comparison, see P1013070 and P1007320 (2008-04-25)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 09:10:17 Stunden.
Keine Lösung: BP 21.5, BP 22.0, Prüfzeit 02:01:51 Stunden. (2023-02-17)
Reto: A good example of Stelvio's progress: Stelvio 1.5 only needs 40min on my notebook. (2023-08-19)
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Henrik Juel: 0... fxe3ep 1.d8S Kxd5 2.0-0-0#. Not -1.f2? and Lg1 is caught. (2004-09-16)
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)
Henrik Juel: In principle I agree, Vaclav
But PDB does not (yet) allow HC+, so I find it acceptable to use the C+ label, when you also tell the whole story after 'Computer test:' below (2023-08-03)
A.Buchanan: Hopefully Gerd will have more time at some point, and can expand the functionality in this and other areas. In the meantime, engine solving of conventional retros including AP, is in its infancy. Retractor 2 has some effectiveness, but is still basic. And there is nothing that yet grasps the intricacies of castling/ep etc. However AP problems do often contain considerable forward chess, and the C+ tag is very useful to filter out those that have already been solved forwardly, without pretending that these are in any sense fully solved (2023-08-04)
Ladislav Packa: I know the definition of AP, but I don't understand the logic behind it. The move 1.fxe3 e.p. proves that White CAN castling. But the solution (2. ...Rd1#?) claims that castling is MANDATORY. From my point of view, AP is correct when only castling is necessary for the solution and the Rook move would be a dual. (2023-08-04)
A.Buchanan: @Ladislav: I am not sure how to help you. Maybe you can read this page from Retro Corner: https://www.janko.at/Retros/Glossary/APosteriori.htm (2023-08-05)
Ladislav Packa: Andrew, what should the article help me with? I quote the final sentence:
Some people still oppose this rule and argue that it should certainly not be the default convention. (2023-08-05)
Henrik Juel: You could view it this way, Ladislav
h#2 means that it is Black to move, so White made last move
What was last move? A little analysis shows just three possibilities: f2-f4, move by Ta1, or move by Ke1
So normally we cannot assume that last move was f2-f4
But if White can castle, then the last move was f2-f4
So if we could start with 0... 0-0-0, then 1.fxe3ep would be legitimate
AP says that you are allowed to reverse the sequence of events; first do the ep capture, then later legitimize it by castling
Was this helpful? (2023-08-05)
Ladislav Packa: Henrik, you don't have to explain that to me. I've done a few AP issues myself, like P1348357. But that doesn't mean I agree with AP's logic. I already wrote it - the term "you can castle" is applied as "you must castle". But these are only problems where, in addition to castling, the Rook move can also be used, I consider that a dual.
From that point of view, the P1000662 issue is perfectly fine for me. (2023-08-05)
Joost de Heer: AP: By castling, you prove a posteriori that the ep-capture was not just a try but the actual solution. Without castling, the ep-solution just is that: a try.
So: Try 1. fe3 ep e8=S 2. Kd5 Rd1 - but ep capture not allowed, as there is no proof that f2-f4 must've been the last move.
Solution 1. fe3 ep e8=S 2. Kd5 OOO - Now the ep capture was justified because white castled, thereby proving that the last move before the diagram position indeed was f2-f4. (2023-08-06)
Joost de Heer: See e.g. P1052919 : The try is an ep capture which is unjustified. (2023-08-06)
A.Buchanan: OK Ladislav: I think I get your point. If one solution with castling justifies the e.p., then based on that certainty, why shouldn't an alternative solution with no castling *then* be allowed as well? There are problems in which one twin shows 0-0-0 and the other shows 0-0. Each is based on the other in a similar way, so the idea of dependency is not new. Why are we not allowed to add other "parasitic" solutions as well? Why can we only have the "paying" solutions? We can't say that we are restricted to one solution: that's not the way chess problems operate! And this is just in the help world - in the adversarial world it might get even more complicated. Is this your issue, Ladislav? (2023-08-07)
Ladislav Packa: I don't want to unnecessarily prolong this discussion. However, I will add one more note: in this position, white castling is also possible without e.p. in Black's 1st move. If B1 were an indifferent move, then white can 1...0-0-0! The Codex of Chess Composition writes about it in Article 16 (1):
Castling convention. Casting is permitted unless it can be proven that it is not permissible.
In our case, 0-0-0 is possible because White's last move exists - e2-e4! It does not matter if it is this move or some a2-a3, both moves are equivalent. EP does not prove the possibility of casting, it would be legal even without it. (2023-08-07)
A.Buchanan: Ladislav was what I wrote your issue pls? Y/N :-) (2023-08-07)
Ladislav Packa: I have no problem, I'm just expressing my own opinion about the AP convention. (2023-08-07)
A.Buchanan: OK cos I think the point I raised is a real one that should be addressed by theory some day. Clearly from the nice problem that you composed Ladislav you understand the mechanics very well. From a justification perspective it's all a bit iffy, but that's why it's controversial. Under RS it's really the only way one can end up actually eping, and it's proved compositionally fertile. So that's enough to justify (2023-08-08)
Ladislav Packa: No need to apologize. This is a normal discussion with different views on the issue. Maybe it will come to some conclusion.
I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p. (2023-08-08)
Joost de Heer: "I just want to point out the fact that it is not the e.p. that authorizes castling, because according to the Codex castling is possible regardless of the e.p. It's the exact opposite: castling authorizes the possibility of e.p."
You misinterpret AP. The e.p. capture does not authorize castling, castling provides a justification later on (hence the 'a posteriori') for the legality of ep.
Usually, for ep justification you need to examine all game trees that lead to the diagram, and only if all game trees end with the double-step, then ep is allowed.
With AP, you examine all the game trees including the actual play. If all those game trees have as last move before the diagram position the double step, then ep is possible.
In this case, if white doesn't castle, then there are game trees which don't have as last move the double step, and therefore AP logic dictates that the ep capture was illegal. However, all game trees which lead to the diagram and which have castling in the actual play have as last move before the diagram position the double step, hence AP dictates that the ep capture is legal. (2023-08-09)
Ladislav Packa: Joost: A simple question - is white allowed to castle after any 1st move by black (except e.p.)? (2023-08-09)
Joost de Heer: Of course he is. AP only is used to combine the ep justification with castling, not the castling right per se. (2023-08-09)
A.Buchanan: I agree with both. Also "cooked" is closer to "unsound". And we choose not to have "C+ cooked", which might legitimately mean "C-" (2023-08-04, hidden) more ...
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Moldenhauer: Computerprüfung: Stelvio 1.4 Keine Lösung: BP 18.5. (2023-08-17)
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{ nDR} = Chamäleon
SCHRECKE: Aus der Lösung geht hervor, dass ein wBe6 fehlt, aber das scheint mir hoffnungslos kaputt zu sein. (2023-07-09)
Anton Baumann: mögliche Korrektur: wKh4 +wBe2 +sBc3 nCH-D nach b6 (2+4+1) s#9 Längstzüger
mit der Lösung: 1.e4 nCHg1(=nS)+ 2.nCHe2(=nL) nCHa6(=nT) 3.e5 nCHg6(=nD) 4.e6 nCHb1(=nS) 5.e7 nCHa3(=nL) 6.e8=S nCHf8(=nT) 7.Sg7 nCHa8(=nD) 8.nCHa1(=nS) nCHc2(=nL) 9.Kh5 nCHh7(=nT)#
5. ... nCHd2(=nL)? geht nicht wegen Selbstschach
C+ Alybadix (2023-07-12)
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Yuri Bilokin: anticipated P1097369 (2023-03-21)
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paul: With bPd3 to c4, the checkmate looks better. (2022-03-30)
A.Buchanan: I guess the smothered mate is thematic, at the cost of the more common model mate (2022-03-31)
Erich Bartel: Nachdrucke:
1) Suomen Tehtäväniekat 1.11.1994.---
2) Phenix (32) VIII 1995.---
3) 072 Kegelschach 1997.---
4) 846 Minimalkunst im Schach 2006.--- (2008-01-28, hidden)
SCHRECKE: C+, Gustav 4.1d, Brute Force (2020-05-08, hidden) more ...
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Hans-Jürgen Manthey: R: 1. c2xDd3 De3-d3 2. e4xTd5 Dg1-e3 3. d4xLc5 g2-g1=D 4. e3-e4 g3-g2 5. a4-a5 h4xLg3 6. Le5-g3 f2-f1=S 7. Ld6-e5 Tg5-d5 8. Le5-d6 Tg1-g5 9. Lf6-e5 g2-g1=T 10. Lg7-f6 g3-g2 11. g2xDf3 Df8-f3+ 12. Lh8-g7 g4-g3 13. Lg7-h8 Ld6-c5 14. Lf6-g7 Lg3-d6 15. Le5-f6 f3-f2 16. Lf6-e5 Le1-g3 17. f2xDe3 De7-e3 18. Lg7-f6 Db4-e7 19. Lh8-g7 Db1-b4 20. Lg7-h8 b2-b1=D 21. Lf6-g7 b3-b2 22. Le5-f6 Da3-f8 23. Lf6-e5 Da1-a3 24. Lg7-f6 Db1-a1 25. Lh8-g7 b2-b1=D 26. Lg7-h8 b4-b3 27. Lf8-g7 28. a5xDb4 De7-b4+ 29. b3-b2 De8-e7 30. h5-h4 La3-f8 31. h6-h5 Lc1-a3 32. h7-h6 b2xDc3 33. De3-c3 Dd8-e8 34. g5-g4 De8-d8 35. g7-g5 Dd8-e8 36. f4-f3 De8-d8 37. f5-f4 Dd8-e8 38. f7-f5 De8-d8 39. SF6-D7 Dd8-e8 40. Ld3-e2 De8-d8 41. e2-e2=L De4-e8 42. Dg5-e3 Dg6-e4 43. Kb4-c4 Tf5-b5+ 44. e3-e2 Tf4-f5 45. Lf5-d3 Lc4-a6 46. e4-e3 Ta6-a7 47. e5-e4 Te4-f4 48. La7-b8 Te1-e4 49. Tb8-b7 Th1-e1 50. Th8-b8 Lf1-c4 51. Lc8-f5 Kd5-c6 52. SG8-F6+ Ke4xd5 53. d7-d5+ e3xSd4 54. SC6-D4 Dh5-g6 55. Kc5-b4 Kf3-e4 56. b4-b3 Sg1-h3 57. Kd6-c5 Sc3-d1 58. Ke7-d6 Sb1-c3 59. Ke8-e7 Ke2-f3 60. Dd8-g5 Ke1-e2 61. SB8-C6 Dd1-h5 62. b5-b4 e2-e3 63. b7-b5 Td6-a6 64. Lc5-a7 Td3-d6 65. Lf8-c5 Ta3-d3 66. e7-e5 Ta1-a3 67. a7-a5 a2-a4 (2021-07-10)
Henrik Juel: These solution corrections should help the animation:
3... g2-g1=D, 6... f2-f1=S, 24... Db1-b2
Hans-Jürgen's numbering is a single move off from retraction 28.
(1.c2xDd3 is white, 28.a5xDb4 is black)
The stipulation seems to ask for the minimal number of black knight moves
(usually SS means knights, but here it may mean 'schwarze Springer') (2021-07-10)
Mario Richter: Henrik, if the question were about black Knight moves only, wSh3 would be able to make tempo moves, allowing e.g. to start with c2xTd3 instead of c2xDd3.
I do not have the official solution, but it obviously is: 4 Knight moves - wSb1-c3-d1,wSg1-h3 and bSb8-d7. bSg8 didn't move at all and was captured by a White officer. (2021-07-11)
Henrik Juel: Thanks for your new solution, Mario, and for your enlightenments
I was misled by the originally stated solution and by Hans-Jürgen's retro game
The german convention that D means Dame=queen and DD means Damen=queens etc. may not be generally understood, so for clarification I suggest that the stipulation is spelled out, something like
Welches war die minimale Anzahl von Springerzüge?
I totally agree with your comment on the strong condition implied by the stipulation (2021-07-11)
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paul: 15...Kc2 (2021-09-27)
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Jean Marc Loustau: Cycle of moves:
s) 1.A B 2.C D#
w) 1.B C 2.D A# (2022-05-03)
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Jean Marc Loustau: Cycle of moves:
s) 1.A B 2.C D#
w) 1.B C 2.D A# (2022-05-03)
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Jean Marc Loustau: Cycle of moves:
s) 1.A B 2.C D#
w) 1.B C 2.D A# (2022-05-03)
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Jean Marc Loustau: Complete reversal of the order of moves:
s)1.A B 2.C D#
w)1.D C 2.B A#
A kind of B/W D/O echo... (2022-05-03)
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Jean Marc Loustau: Double exchange of the White and Black moves:
1.A B 2.C D#
1.B A 2.D C# (2022-05-03)
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Described as a correction of his 2856, Themes-64, 1975 (P1969) - what is wrong with that problem ??!?)
Keyword=Roman Theme
Keyword=Duel - Bishops
Anton Baumann: C+ Alybadix (2020-05-30)
Anton Baumann: ursprüngliche Aufgabe vergl. P1010228 (2021-01-05)
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Henrik Juel: The retroplay does not show
Addition of wSh8 is missing (2024-01-14, hidden) more ...
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Moldenhauer: Computerprüfung: C+ Stevio 1.0 34 Sekunden.
Keine Lösung: BP 21.0, BP 21.5. (2023-03-12, hidden) more ...
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SH: cooked? 1. b4 d5 2. Ba3 Kd7 3. Qc1 Kc6 4. Qb2 Kb5 5. Qxg7 Ka4 6. Nc3+ Kxa3 7. Rc1 Kb2 8. Nb1+ Kxc1 9. Qa1 Bh6 10. f4 d4 11. Kf2 Kd1 12. Kg3 Ke1 13. Kh4 Kxf1 14. Nf3+ Kf2 15. Rd1 Bxf4 16. Ne1 Kg1 17. Kh5 Kh1 18. Nc3 Be3 19. Nd3+ Bg1 (2006-06-06)
Henrik Juel: Yes, the move order is not unique after 10.f4, so the problem is indeed cooked (2018-12-17)
Moldenhauer: Computerprüfung: C- oder cooked Stelvio 1.4 48 Sekunden.
Keine Lösung: BP 18.5. (2023-07-29, hidden) more ...
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Anton Baumann: C+ Gustav 4.1d (2020-12-03)
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Anton Baumann: C+ Alybadix (2020-11-11)
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bg: Marscirce !! (2005-08-22)
SCHRECKE: C+, popeye 4.87 (2023-09-25)
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1053
Anton Baumann: a) sowie b) C+ Gustav 4.2a (2022-03-10)
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Cf: P1347359
paul: Checked with stip ser-dia 31 forsyt rsbqkbsr/pppppppp/1RR5/BPPP1S2/1QB4K/5PP1/P3P2P/3S4 by Jacobi. (2021-09-27)
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SCHRECKE: C+, popeye 4.87
1. Se5! droht 2. Sf7 droht 3. Sxh6#
2. ... Lxd6 3. Sbxd6#
1. ... Txe5 2. Dd3+ Lxd3 3. Sd6#
2. ... Te4 3. Dxe4#
1. ... fxe5 2. Df8+ Sxf8 3. Sd6#
2. ... Sf6 3. Dxf6#
1. ... Lxe5 2. Dd5 droht 3. Sd6#
2. ... exd5 3. Lxd7#
1. ... Sxe5 2. Dxc5 droht 3. Sd6#
2. ... Td1 3. Le4# (2023-06-18, hidden) more ...
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Henrik Juel: damaged by double threat 1.g4 thr. 2.Lg5,Lf6
and the dual 1... d4 2.Le7 thr. 3.Sc4+ Kb5 4. Sa3+,Lb4 (2021-03-29)
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SCHRECKE: C+, Gustav 4.1d, Brute Force
außerdem: 4. ... L:g4 5.Sc4# (2021-01-23)
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SCHRECKE: C+, Gustav 4.1d, Brute Force
1.Sf4+? K:c5+!
1.Tf2! e4! 2.Tf5+ De5 3.Sf4+! K:c5 4.Sge6# (2021-01-20)
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Henrik Juel: Natch 3.1 did little in half an hour (2018-12-19)
Moldenhauer: Computerprüfung: C+ Stelvio 1.2 02:04:15 Stunden. (hh:mm:ss)
Keine Lösung: BP 23.5, BP 24.0. (2023-05-11)
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paul: C+ for the first 20.5 moves.(Euclide) (2017-07-24)
P.R.: Solution: 1.c4 Cf6 2.Db3 Ce4 3.Db6 axb6 4.c5 Ta3 5.c6 bxc6 6.d4 Fa6 7.Ff4 Cd2 8.e4 Fe2 9.Fd6 Td3 10.a4 cxd6 11.a5 bxa5 12.h4 Db6 13.Db3 14.h6 gxh6 15.Th5 Fg7 16.Tb5 cxb5 17.d5 Fd4 18.Ch3 f6 19.Cf4 Rf7 20.Ce6 Tc8 21.f4 Tc2 22.Cc5 dxc5
Les trois Pions noirs sont restes dans leur colonne mais apres avoir effectue deux prises chacun. (2006-05-31, hidden)
paul: 13. h5 (2017-07-24, hidden)
Kostas Prentos: Fully tested with Stelvio 1.0 (on two computers) in combined time of approximately 667 hours. (2023-03-20, hidden) more ...
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Henrik Juel: Natch 3.1 did not even find the solution in an hour (2018-12-15)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.2 296:36:23 Stunden. (hh:mm:ss)
Keine Lösung BP 20.5, BP 21.0. Histogramm 355767/10739k
Cooked bei 59917/922k. Die letzten 295850 Strategien wären sicher
schneller geprüft worden da die meisten großen Werte bereits waren.
Notations Beispiel: 1.b4 Sf6 2.c4 b6 3.Sc3 c5 4.bxc5 Lb7 5.cxb6 Lf3
6.gxf3 Sd5 7.Lh3 Se3 8.Lf5 Sf1 9.Lc2 g5 10.Sh3 g4 11.d3 g3 12.Dd2 g2
13.Dh6 g1D 14.Ld2 Dg3 15.Tc1 Se3 16.Lb1 Dgc7 17.bxc7 Sd1 18.0–0 Se3
19.Tfd1 Sd5 20.Kf1 Sf6 21.Le1 Sg8 22.Td2
oder wie vor bis 19. dann 20.Le1 Sf6 21.Kf1 Sg8 22.Td2 (2023-05-24)
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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 ...
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paul: Cook: 1.Sf3 [-Ra1h1, -Sb1, -Pa2a7h7, -Ra8h8, -Sb8g8] g5 2.e4 [-Pe4] d5 3.d4 g4 4.Bc4 [-Pg2] g3 5.B×d5 g×h2 6.Ke2 h1=R 7.Sg5 c5 8.d×c5 [-Pc5] Qa5 [-Qa5, -Bc8] 9.Se6 Bg7 10.Bf3 Rh8. (2024-03-13, hidden)
A.Buchanan: what's Berkeley Chess, please? (2024-03-14, hidden) more ...
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Thomas Volet: TV: solution given is slightly incorrect. Black's first retraction cannot be Se6-d4. It should be Sf5-d4, and the next two Black retromoves would then be Sd6-f5 and Sb7-d6. GW is so reliable I never check the solutions -- just happened to look at this one in connection with some correspondence. (2010-02-07)
Gerd Wilts: Thank you, Tom, I have now corrected the solution. (2010-02-07)
Mario Richter: Instead of ... 8.Lg8-e6 Dd4-d5 9. f7xLg8=L De3-d4 a.s.o. the unpromotion can also be done on e8: 8.Lf7-e6 Dd4-d5 9.Le8-f7+ De3-d4 10.f7xLe8=L a.s.o. Does that make the problem dualistic? (2010-02-08)
Thomas Volet: White also might have begun the game with either 1.Sg1-f3 or 1.Sg1-h3. The point was to show reciprocal retroscreens (here by the WBb7 and BQb4, each of which screens for and is screened by the other) involving both Kings. Beyond that .... (2010-02-12)
Henrik Juel: This problem was awarded 1st HM by judge Klaus Wenda in the informal fairy tourney (including retros) for 2005
Klaus wrote:
Nach meinem Geschmack die anspruchsvollste der klassischen Retroaufgaben. Die 14-zügige Auflösung der Stellung mit wechselseitigem Schachschutz durch retroentfesselte Figuren, garniert mit der Entwandlung eines wL auf g8, wobei der entwandelte Bauer af sein Grundfeld f2 zurück zieht (Retro-Exzelsior), präsentiert sich als spannender und geistreicher Ablauf.
I expect that Mario's observation (that the unpromotion could also happen on e8) would not have changed the award; if the wording had been 'auf e8 oder g8', we would know for sure (2021-06-18)
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Henrik Juel: d2xDc3xTd4 was too tricky for Natch 3.1, which did not find the solution in half an hour
Stopped at pos. 917 (2018-12-22)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 00:08:50 Minuten.
Keine Lösung: BP 19.0, BP 19.5. (2023-03-01)
paul: 1.f3 c6 2.Kf2 Qa5 3.Kg3 Qc3 4.dxc3 h5 5.Qd6 h4+ 6.Kg4 h3 7.Qc7 hxg2 8.Sh3 g1=Q 9.Bg2 Qb6 10.Qxc8+ Qd8 11.Qc7 Rh5 12.Qg3 Qb6 13.Rd1 Qg1 14.Rd6 Rd5 15.Be3 g5 16.Sd2 Qxa1 17.Bc5 Qg1 18.Ba3 Qb6 19.Qe1 Rd4+ 20.cxd4 Qd8. (2010-07-07, hidden) more ...
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Henrik Juel: Natch 3.1 did little in more than an hour
Stopped at position 1 (2018-12-22)
Moldenhauer: Computerprüfung: C+ Stevio 1.0 7:08:52 Stunden.
Keine Lösung: BP 24.5, BP 25.0. (2023-02-14)
paul: 1.e3 h5 2.Lb5 Sh6 3.Lc6 dxc6 4.a4 Dd3 5.Ta3 Dh7 6.Td3 Lg4 7.Td6 f5 8.Tf6 exf6 9.e4 Lb4 10.e5 0-0 11.e6 Td8 12.e7 Td3 13.e8=L Tc3 14.d4 b5 15.d5 bxa4 16.d6 a3 17.d7 a2 18.d8=T a1=D 19.Td7 Da6 20.Tf7 Dd3 21.Ld7 a5 22.Le6 Sa6 23.La2 Td8 24.Te7+ Td5 25.Te6 Tdc5 26.Te8# Écrans successifs entre Fou et Tour. (2011-06-12, hidden) more ...
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Henrik Juel: Natch 3.1 did not find the solution in half an hour
Stopped at pos. 4961 (2018-12-22)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 14:12:45 Stunden. (hh:mm:ss)
Keine Lösung: BP22.0, BP 22.5.
Beispiel: 1.Sa3 a5 2.Sb5 e5 3.a3 Lb4 4.axb4 h5 5.Ta3 h4 6.Tg3 hxg3 7.hxg3 Ta6
8.Th7 Th6 9.Sf3 Th1 10.Th2 Txf1+ 11.Kxf1 Th3 12.Kg1 e4 13.Th1 f5
14.Sh4 f4 15.gxf4 Ta3 16.Kh2 Ta1 17.Kg3 Txc1 18.Kg4 Ta1 19.Dg1 Ta3
20.Dh2 Tb3 21.Sa3 Sf6+ 22.Kg5 d6 23.Dh3 Tg3+ (2023-03-28)
paul: 1.a3 e5 2.Ta2 Lb4 3.ab3 h5 4.Ta3 h4 5.Tg3 hg3 6.hg3 f5 7.Th3 f4 8.gf4 a5 9.Ta3 Th1 10.Sf3 Txf1+ 11.Kxf1 Ta6 12.Kg1 Tg6 13.Kh2 d6 14.Dh1 Lg4 15.Kg3 Sf6 16.Dh7 Th6 17.Ta1 Th1 18.Sa3 Txc1 19.Kh4 Th1+ 20.Kg5 Th3 21.Th1 Tg3 22.Dh3 e4 23.Sh4 Lc8+
Carré Th1-h3-a3-a1-h1 sans capture. (2010-07-16, hidden) more ...
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Henrik Juel: Sb1 is actually [Sg1], and Sg1 is [Pa2] promoted on c8.
Testing with Euclide ran out of time. (2011-02-27)
paul: The first 19 moves checked by Euclide in half a day. (2020-12-02)
Moldenhauer: Computerprüfung: C+ Stelvio 1.0 01:04:22 Stunde. (hh:mm:ss)
Keine Lösung: BP 19.0, BP 19.5. (2023-03-02)
hans: 1. Nc3 Nc6 2. Nd5 Rb8 3. Nxe7 Kxe7 4. a4 Kf6 5. a5 Kg6 6. a6 Qh4 7. axb7 Nf6 8. bxc8=N Rb3 9. h3 Bb4 10. Nd6 Re8 11. Nf5 Ree3 12. Ng3 Ne5 13. f4 Rf3 14. e4 Rf2 15. N1e2 Rbf3 16. Nc3 c6 17. d3 Rd2 18. Nb1 Rff2 19. Ne2 Kh5 20. Ng1+ Rfe2# (2011-02-27, hidden) more ...
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Henrik Juel: Natch 3.1 did little in more than half an hour
Stopped at pos. 15 (2018-12-22)
Moldenhauer: Computerprüfung: Stelvio 1.0 20:35:18 Stunden. (hh:mm:ss)
Keine Lösung: BP 21.0, BP 21.5.
Notation: 1.e3 h5 2.Ld3 Th6 3.Lh7 e5 4. d3 Dh4 5.Ld2 g5 6.La5 Lb4+
7.Sc3 c5 8.Lc7 a5 9.g3 Taa6 10.f3 Tab6 11.Kf2 Sa6 12.Sd5 Ld2
13.Se7 d5 14.Sh3 Lxh3 15.c3 Lf5 16.h3 Lg6 17.Th2 f5
18.Da4+ Kf7 19.Dc6 Tb3 20.axb3 Db4 21.Ta4 Da3 22.bxa3 b5 (2023-03-10)
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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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hans: a) 1. Ke5 Td1 (0-0-0?) 2. f5 Lxg7# b) 1. 0-0-0! Dd8 2. Dxd1#
Some comment: The composer shows how important a tempo can be. Masterly accomplished! (HM). A very intensive study is needed to find the solutions: in fact not the solutions but the motivation. Has cost me days to solve this fine, extremely difficult and profound problem. In (a) 0-0-0 is not allowed., in (b) it is. In my opinion a 1st Prize Retros 2006 (GHB)

R: -1. … Dd8-b8 -2. a5-a6 Kf6-f5 -3. a4-a5 Ke7-f6 -4. a3-a4 Ke8-e7 -5. a2-a3 Lh2-g1 -6. h7-h8=L Ld6-h2 -7. h6-h7 Lf8-d6 -8. h5-h6 e7-e6 -9. h4-h5 Le6-c4 -10. h3-h4 Lc8-e6 -11. h2-h3 d7xDc6 -12. Dc5-c6 f5xSe4 -13. c6xb7 g6xTf5 -14. d5xTc6 etc.
bCap: axbxc2-c1, dxc6, hxg6xf5xe4, wCap: dxe3, gxf3xe4xd5xc6xb7, (2013-08-30)
hans: *2. Kc1-b1 (2013-08-30)
Mario Richter: I've found a cook for a), based on the idea to unpromote the black queen on c1 (shielded by the black bishop c4 on d1). The following PG shows, that the position can be reached with White still able to castle queenside:
1.Ng1-f3 a7-a6 2.Nb1-a3 a6-a5 3.Na3-c4 Ra8-a6 4.Rh1-g1 Ra6-f6 5.Nf3-h4 Rf6-f3 6.Nh4-g6 h7xg6 7.g2xf3 Ng8-f6 8.Bf1-h3 Nf6-e4 9.f3xe4 Nb8-c6 10.Bh3-f5 Nc6-b4 11.Rg1-g4 Nb4-d5 12.e4xd5 Rh8-h3 13.Rg4-e4 Rh3-c3 14.Nc4-e5 Rc3-c6 15.d5xc6 g6xf5 16.c6xb7 f5xe4 17.Ne5-c6 d7xc6 18.h2-h3 Qd8-d4 19.h3-h4 Qd4-e3 20.d2xe3 Bc8-e6 21.Bc1-d2 Be6-b3 22.Bd2-b4 a5xb4 23.Qd1-c1 e7-e6 24.Qc1-d1 Ke8-e7 25.Qd1-c1 Ke7-f6 26.Qc1-d1 Kf6-f5 27.Qd1-c1 Bf8-d6 28.Qc1-d1 Bd6-h2 29.c2-c3 Bh2-g1 30.Qd1-c1 Bb3-d1 31.Qc1-d2 b4-b3 32.Qd2-c2 b3xc2 33.h4-h5 c2-c1Q 34.h5-h6 Qc1-c2 35.h6-h7 Qc2-a4 36.h7-h8B Qa4-a8 37.a2-a3 Qa8-b8 38.a3-a4 Bd1-b3 39.a4-a5 Bb3-c4 40.a5-a6 (and now) Kf5-e5 43.0-0-0! (still possible!) f7-f5 44.Bh8xg7# (2013-08-31)
Henrik Juel: Nicely found, Mario.
So it was fortunate that the problem was not awarded in the tourney, despite the enthusiastic solver comments (2013-08-31)
Henrik Juel: According to judge Bernd Gräfrath, he did not see the cook and awarded the problem 1st-2nd prize (2013-09-01)
A.Buchanan: Was Thierry made aware of the cook? Did this ever get fixed? (2021-06-29)
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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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Hier liegt Keine Vorgängerschaft vor, denn es wird ein interessanter 6-zügiger Rundlauf des wS gezeigt!
paul: Anticipated by P1186964 (Schildberg) (2022-09-14)
Jacques Rotenberg: not at all anticipated !!
the solution is completely different (2022-12-04)
Jacques Rotenberg: the P1186964 itself is anticipated! (2022-12-04)
Jacques Rotenberg: about P1186964 see P1186601 (2022-12-04)
Iwan Soroka: @paul. P1186964 is kein Vorgänger zu P1090925. Was genau ist anticipated? (2022-12-04)
Iwan Soroka: C+, Gustav 3.2d Brute Force (2012-01-18, hidden) more ...
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Anton Baumann: C+ Gustav 4.2a (2023-07-05)
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Anton Baumann: scheitert ferner auch an 6.Lf6+ Se5!
Mögliche Korrektur: sSg4 nach g8, -wBe7, wSg7 nach e7 (10+8)
Mit diesen Korrekturen C+ Gustav 4.2a (2022-04-29)
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paul: The source is Phenix 09/1990 (2022-09-13)
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Erich Bartel: Nachdrucke:
1) Phenix (RM) V 1994,S.2451
2) 79 Das Matt des weißen Königs 1995
3) Problem Paradise II 1997,S.39
4) feenschach (146) III-I 2002 (2009-09-14)
paul: The source is Phenix 10/1990 (2022-09-13)
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SCHRECKE: Vorgänger: siehe P1264474 (2021-07-26)
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Anton Baumann: C+ Gustav 4.1d (2020-11-20)
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Anton Baumann: C+ Gustav 4.2a (2023-12-07, hidden) more ...
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Anton Baumann: C+ Gustav 4.2a (2023-12-15, hidden) more ...
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SCHRECKE: C+, Gustav 4.2a, Brute Force (2022-03-19)
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SCHRECKE: C+, Gustav 4.1d, Brute Force (2020-12-30)
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paul: Stockfish+ (2021-05-07)
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Anton Baumann: C+ Gustav 4.2a (2022-05-06)
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SCHRECKE: C+, Gustav 4.1d, Brute Force (2020-09-30, hidden) more ...
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Anton Baumann: C+ Gustav 4.1d (2020-12-17)
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Moldenhauer: Computerprüfung: C+ Gustav 4.1d in 51 Tagen 21 Std. 33 Min.
Keine Nebenlösungen! (2022-10-04)
Henrik Juel: An interesting comparison:
Popeye 4.61 with 'opt non 0 1 thr 0' found no flaws in 0.2 seconds (2022-10-04)
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SCHRECKE: a)+b): C+, Gustav 4.1d, Brute Force (2020-12-09)
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Henrik Juel: Natch 3.1 did little in half an hour
Stopped at pos. 1 (2018-12-27)
paul: 1.h4 g5 2.hxg5 c5 3.Rh6 c4 4.Rb6 axb6 5.Sc3 Ra3 6.Sb5 Re3 7.fxe3 c3 8.dxc3 h5 9.Qd5 Rh6 10.Bd2 Rf6 11.O-O-O Rxf1 12.Be1 Rf4 13.Qf3 f5 14.Rd6 Rd4 15.Rf6 exf6 16.Sa3 Qe7 17.Sb1 Qa3 18.Bd2 Qa8 19.a4 Ba3 20.Kd1 d6 21.Ke1 Kd7 22.Bc1. Hidden white castling. (2020-12-01)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 00:25:05 Minuten. (hh:mm:ss)
Keine Lösung: BP 20.5, BP 21.0. (2023-03-12)
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Yuri Bilokin: the idea is not new P1005163 (2022-08-06)
Jean Marc Loustau: ***
The matrix and the mates are indeed the same; but in this problem (P1100532) there is one more significant feature which is, in fact the main idea (which is not on the P1005163):
The 2 white thematic pieces can already play on the mate square in the diagram position: thus we have a kind of "peri-white Grimwhaw" (a 3 moves manoeuvre to reach the mate square which was already reachable).
Of course I did no know the P1005163 when I have composed this one... I would say "fortunately", because if I had knew it, probably I would not have worked on this matrix: my final result, in my opinion, is interesting enough with its new feature to deserve the publication. (2022-08-14)
Yuri Bilokin: you can see 1...Bd5 2.Sa7 Rc6 3.Ka8 Rc8#
1...Be4 2.Sa7 Rc6 3.Ka8 Rc8#
1...Bf3 2.Sa7 Rc6 3.Ka8 Rc8#
1...Bg2 2.Sa7 Rc6 3.Ka8 Rc8# (2024-03-18)
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YM: Allumwandlung (2021-06-28)
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Anton Baumann: a)-d): C+ Gustav 4.2a (2022-06-13)
YM: In each of the four variations, an additional sacrifice of the white piece in the middle of the combination. (2022-06-13)
YM: In c) 6. ... Kc4 and 10. ... Txc5 (2010-07-13, hidden) more ...
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Yuri Bilokin: Correction: bRc3, +bBa8, +bPa4, +bPc6, +bPe5 b7/8/2p5/4p3/p2p4/2p5/1pp1P1P1/1k4K1 (3+9) (2021-05-02)
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paul: Self-anticipated by P1271622 (2022-09-26)
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Anton Baumann: Informalturnier 1990: 2. ehrende Erwähnung (2023-02-17)
Erich Bartel: Nachdruck:
1) Phenix (37) I 1996 (2010-07-09, hidden) more ...
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Erich Bartel: Ein Wahnsinnsproblem!!! (2012-06-13)
Arnold Beine: Leider kaputt, denn am Ende kann man das Selbstmatt auch mit einem Batterieschach erzwingen. Die thematisch vergleichbare P1278403 hat aber gute Chancen, korrekt zu bleiben. (2022-10-22)
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LX8L
SCHRECKE: C+, Gustav 4.2b, Brute Force (2022-10-16)
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SfK m
Rosalie Fay: 1. Kg3 (droht 2. Kh3) e6 2. Ld4 (droht 3. Kh3) e5 3. Lf2 ~ 4. Kh3 ~ 5. Sg3#
C+ Popeye Windows-64Bit v4.69 (2023-12-20)
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A12-34+ mm Echo
SCHRECKE: C+, popeye 4.87 (2021-11-09)
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A4-TLL 4m
SCHRECKE: C+, popeye 4.89 (2023-11-09, hidden) more ...
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ASb5
vf: a1<>a8, Nikolai Chartschinschin, 07.2005, Kudesnik (94.95)
SCHRECKE: C+, Gustav 4.2f, Brute Force
Der "Vorgänger" (P1123589) hat einen Dual! (2024-03-08)
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m22 5m
Vendelbo: 1. e2d4
a) 1. -, c4c3 2. f5c5 c3b2 3. c5c2 b2a1 4. d4b3#
b) 1. -, c4c3 2. f5c5 c3d2 3. c5c2 d2e3 4. d4f5#
c) 1. -, c4b4 2. f5c2 b4a5 3. c2b3 a5a6 4. b3b5# (2024-02-15, hidden)
Henrik Juel: C+ Popeye 4.61 (2024-02-16, hidden)
Henrik Juel: 1.Sd4 zz
1... Kb4 2.Dc2 Ka5! 3.Db3 thr. 4.Db5#
1... Kc3 2.Dc5+ Kd3,Kd2,Kb2 3.Dc2+ (2024-02-16, hidden) more ...
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Henrik Juel: White has the move
-1... f4xLg3 -2.Lh4 f5 -3.Lf6 h3 -4.Lb2 h4 -5.Lc1 h5 -6.b2xLa3 Lc5 -7.e3 Ld4 -8.c5 Lf6 -9.c4 Le7 -10.c3 Lf8 -11.c2 e7xTd6 -12.Td4 Kd8 -13.d6xTc7, retract Td4 to h1, Ka8 via g6 to e1, Pd6 to h2, g3-g2, g2xLf3 etc. (2013-08-28)
James Malcom: Is the given retraction unique? (2020-11-08)
Henrik Juel: Unique up to 11.c2-c3, I think (2020-11-08)
Henrik Juel: One of my favorite retros with just kings and pawns
even though the retroplay is not completely unique (so my previous comment is imprecise)
R: 2... h3-h2 3.Lf6-h4 h4-h3 or 3... f5-f4 also work
but not R: 1.e3-e4? or 1.c5-c6?, and White runs out of pawn retractions
nor 2. e3-e4? or 2. c5-c6? f5-f4 3. Le5-g3, and same difficulty (2020-11-09)
Henrik Juel: In the forward play we have several temporal implications, e.g.,
e2-e3 before b7xLa6 before g2xLf3 before g3-g2 before h2xg3
d6xc7 before e7xd6 before b2xLa3 before f4xLg3
Thierry was inspired by P0008574 (2020-11-09)
James Malcom: I see the similarities as well. If Thierry's last 21 moves, stopping at 11. c2-c3 of course, are indeed unique, it seems to be the record for a kindergarten position? (2020-11-09)
A.Buchanan: I think the last 2 single moves are determined. Prior to that it could be f5-f4 or h3-h2, no? (2020-11-10)
Henrik Juel: You are right, Andrew
So, no record, James (2020-11-10)
James Malcom: Ah well then-the linked problem with 16 moves by Gerd oughtta be it then, from what I've searched so far. (2020-11-10)
A.Buchanan: Isn’t the retro record one of the Breyer versions by Thierry Le Gleuher? (2020-11-10)
James Malcom: As far as I know, the TOTAL retro record is, all from P1339006, by Theierry indeed: Type A, 100 moves, Type B, 101 with bP6-h6 & White to move, and, finally Type C, 102 with bP6-h6 and wSc7-a8. (2020-11-10)
James Malcom: For Type B, I meant "bPh5-h6." (2020-11-10)
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Anton Baumann: kürzer: 5.Kb1 bxa4 6.Ka1 axb3 7.Df1+ Kd2 8.Df2+ Kc1 9.Db2+ cxb2# (2023-03-07)
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Yuri Bilokin: version given by me SuperProblem (Website) 30/10/2017 wKa3-a2 8/7k/5BN1/8/8/8/K7/8 (2022-12-12)
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SCHRECKE: a)+b): C+, popeye 4.87 (2023-09-12)
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Henrik Juel: The double CF is even double Prentos, as the promoted men are captured by officers (2018-12-29)
Henrik Juel: Natch 3.1 found the solution at pos. 2, but did not conclude in an hour
Stopped at pos. 850 (2018-12-29, hidden)
Henrik Juel: I let Natch run another hour, stopping at pos. 2090, still without conclusion (2018-12-29, hidden)
Moldenhauer: Computerprüfung: C+ Stelvio 1.4 08:35:12 Stunden. (hh:mm:ss)
Keine Lösung: BP 9.0. (2023-07-11, hidden) more ...
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