James Malcom: Where is the cook? (2021-09-14)
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Henrik Juel (PDB 2021-09-15): This is one of the few shc#'s without ep capture and/or castling
Henrik Juel: I don't agree entirely with you comment, James; this seems clearer:
1.Kf7? does not work, because now White has no last move ([Th1] was captured by a pawn)
1.Kf8 2.Kg7 3.Kg6? fails for the same reason
After 5.c1=S 6.Kf7 last move was h3-h4 or h2-h4
This is one of the few shc#'s without ep capture and/or castling (2021-09-15)
James Malcom: Henrik, what comment of mine? I merely edit swapped some text. (2021-09-15)
Henrik Juel: OK, so it was probably Gerd who wrote it in 1998; the early updates are no longer available, when you click on 'more...'
The german text roughly translates to:
Pg6xh7 happened before Kf7-e8, so first one of the black pawns must be eliminated by promotion.
I still like my comment better... (2021-09-15)
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Henrik Juel: After 5.gxf3ep last move could be Ka2-a1 (2021-09-15)
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Kees: .
-1.... Se1xc2# -2. Lg1-f2 Dd1-c1 -3. Lf2-g1 Tg1-h1 -4. Tc1-b1 Lh1-g2 -5. Tb1-c1 Tg1-g2 -6. Lg1-f2 Tf2-g2 -7. Tc1-b1 Pg2-e1 -8. Tb1-c1 Te1-f1 -9. Tc1-b1 Tf1-f2 -10. Tb1-c1 Dc1-d1 -11. Lf2-g1 Td1-e1 -12. Lg1-f2 Se1-g2 -13. Lf2-g1 Tg1-f1 -14. Lf1-h3 Tg2-g1 -15. Lg1-f2 Kh3-h4 -16. Lf2-g1 h4xSg3
This could be a solution, but black makes 4 captures exd, fxe, gxf, hxg, so white h-pawn must promote, so the solution failed. (2022-03-07)
Mario Richter: I do not know the intended solution, but the following plan works:
1. get bSe1 out of the Cage
2. get both black rooks out of the Cage
3. now there's enough space to get wRb1 out of the Cage - get it to h8 and unpromote it
4. bring back wPh7 to h3
5. retract bPh4xYg3 (e.g. bPh4xSg3), then everything resolves.
So in fact, Black can mate by 1. ... Se1xc2#! (2022-03-07)
Henrik Juel: But how do you get Se1 out of the cage, Mario? (2022-03-07)
Mario Richter: 1. get bSe1 out of the Cage
This requires a subplan:
1a. get bK to h3 (or lower)
1b. get bSe1 to g2
1c. step out by R: bSh4-g2 (2022-03-08)
A.Buchanan: Last pawn move / capture was 69 moves ago. Maybe Mario’s solution is not quite optimal but it would only be a couple of moves off. I suppose we can conclude that there was no draw claimed at any of the (69-50)x2 = 38 opportunities. On the other hand, if there was a way to retract to the last promotion in 50 moves, we would assume that this had happened because the 50 move rule would prevent anything longer? Is this how one would interpret these two cases? (2022-03-08)
Henrik Juel: Thanks Mario
A slightly different pla (2022-03-08)
Henrik Juel: ...
Andrew, I believe that Mario's solution is close to optimal, so it looks hopeless to quicken it to 50 moves
We have to assume that the 50 move is not in effect in this problem (2022-03-08)
Henrik Juel: ... should be '50 move rule', of course
My slightly different plan was to get wLh3 away early, but this would certainly not shorten the proof game significantly, if at all (2022-03-08)
A.Buchanan: Yes I was wondering how the conventions works in these two different situations. We basically want it to intervene only when convenient :) (2022-03-08)
Henrik Juel: I am afraid you are right, Andrew (2022-03-08)
A.Buchanan: Maybe the 50 move rule triggers unless we can prove that it didn't? Then we get the best of both worlds. And that's kind of the same template successfully used for castling & ep. (2022-03-09)
James Malcom: I think that it ought to not trigger unless you can prove it has. (2022-03-09)
A.Buchanan: In this problem we know that vast numbers of non-zeroing moves have passed. It’s a retro, so the 50 move convention would have terminated the game. The convention is bogus, but we can fix it a bit. Since under all histories the game would have terminated prior to the diagram position, the only answer is that the draw was never claimed. This might also “prove” that the move sequence was as short as possible, to reduce the number of missed claims.
It’s not clear from what James said whether he would agree with this, as he stated an opinion on the contrapositive case. And how could one "prove that the 50 move triggered" (i.e. that the game ended in a draw), without stating it as a fact in the stipulation? Even putting draw as the goal is arguably not enough. (2022-03-09)
Mario Richter: Would be interesting to know what Thierry really intended. Perhaps his idea was to fool the solver? (Letting him find a resolution similiar to mine, and therefore stating: "Yea,#1 is possible by 1. ... Sxc2#!", only to learn that the official solution says: "No, no #1 because of 50-moves-rule!")

btw., the unpromotion h7-h8=T can be done a little bit faster:
R: 1. Lg1-f2 Dd1-c1 2. Lf2-g1 Tg1-h1 3. Tc1-b1 Lh1-g2 4. Tb1-c1 Sg2-e1 5. Le1-f2 Dc1-d1 6. Lf2-e1 Td1-f1 7. Le1-f2 Tf1-g1 8. Lf2-e1 Se1-g2 9. Lg1-f2 Lg2-h1 10. Lf2-g1 Th1-f1 11. Lg1-f2 Lf1-g2 12. Lf2-g1 Tg1-h1 13. Lg2-h3 Kh3-h4 14. Lh1-g2 Sg2-e1 15. Le1-f2 Sh4-g2 16. Lf2-e1 Tg2-g1 17. Le1-f2 Lg1-h2 18. Lf2-e1 Kh2-h3 19. Le1-f2 Tf2-g2 20. Lg2-h1 Sf5-h4 21. Lh1-g2 Lh3-f1 22. Lg2-h1 Kh1-h2 23. Lf1-g2 Th2-f2 24. Lf2-e1 Lg2-h3 25. Le1-f2 Th8-h2 26. Lf2-e1 Kh2-h1 27. Le1-f2 Lh1-g2 28. Lh3-f1 Tb8-h8 29. Lf2-e1 Tf1-d1 30. Le1-f2 Tf2-f1 31. Lf1-h3 Dd1-c1 32. Tc1-b1 Tg2-f2 33. Kb1-a1 Lf2-g1 34. Ka1-b1 Kg1-h2 35. Kb1-a1 Th2-g2 36. Ka1-b1 Th8-h2 37. Lh3-f1 Tg8-h8 38. Kb1-a1 Kh2-g1 39. Ka1-b1 Lg1-f2 40. Lf2-e1 Lg2-h1 41. Kb1-a1 Lf1-g2 42. Lg2-h3 De1-d1 43. Lh1-g2 Lh3-f1 44. Td1-c1 Df1-e1 45. Le1-f2 Df2-f1 46. Tc1-d1 Lf1-h3 47. Td1-c1 Dg2-f2 48. Lf2-e1 Dh3-g2 49. Te1-d1 Lg2-f1 50. Tf1-e1 Dh6-h3 51. Le1-f2 Lh3-g2 52. Lg2-h1 Dc6-h6 53. Tf2-f1 Kh1-h2 54. Lf1-g2 Tf8-g8 55. Tg2-f2 Lf2-g1 56. Ka1-b1 Kg1-h1 57. Th2-g2 Lg2-h3 58. Th8-h2 Db7-c6 59. h7-h8=T

Anyone outthere who can squeeze out another 10 moves?! (2022-03-09)
Thomas Volet: I do not know what Thierry intended in this composition. However, it is the case that there are retro composers who are interested in long sequences of non-P non-capture moves without any interest in draw rules that might apply. Ceriani is an example. It is my belief that composers should be allowed to work with such sequences and perhaps even to share results with other composers and solvers, without constraints associated with draw rules. (2022-03-10)
A.Buchanan: I am aiming to reach a similar effect to Thomas of freeing composers and solvers. However rather than a confusing situation where contradictory interpretations of vague and poorly-worded conventions are randomly applied or ignored, I would like to express the underlying logic mathematically so we can all have our cake and eat it. Sometimes we want no 50M convention as we are looking for maximum length. Other times we do want 50M for dual elimination. We can have both. (2022-03-10)
A.Buchanan: Of course those like Thomas who as an escape from the rigours of work prefer to wander round in a swamp of vagueness are free to do so, there is no sense of “prohibition”/“allowing”. (2022-03-10)
Thomas Volet: Hopefully my colleagues can receive communication from the "swamp" to which I have been so publicly relegated. As to escaping from "the rigours of work", in the matter of sequences of non-P non-capture moves (the only subject of my comment above), I would have thought it clear that a substantial amount of work went into the composition of each of P0008399, P1202286, and P1331867. (2022-03-11)
A.Buchanan: Of course you're not "relegated", Thomas. These are great problems, but that's no excuse for discouraging others to form a clearer view of the underlying rules and conventions. You will always have your sarcasm to console you :-) (2022-03-11)
A.Buchanan: When I talk about "rigours of work", I mean your profession, not composing effort. You have explained to me before that the reason why you can't help yourself from discouraging others from sorting out the rules properly is that chess is an escape for you from an over-regulated real world. I can sympathize with that: you will always have that freedom. It's just such an unnecessary swamp you inhabit. As I have explained to you, I want to build a better foundation for newbies, programmers and fairy enthusiasts. This will enable more people to enjoy your compositions. Honestly, I don't know why you complain... :-) (2022-03-11)
A.Buchanan: In this problem we know that vast numbers of non-zeroing moves have passed, yet it’s a retro, so the convention would have terminated the game. Therefore the convention is bogus, but we can fix it a bit by saying that since all histories terminated earlier, the draw was never claimed. This might also “prove” that the move sequence was as short as possible, to reduce the number of missed claims.
It’s not clear from what James said whether he agrees with thus, as he comments on the contrapositive case. And how can you prove the game is actually ended, without stating it as a fact in the stipulation? Even putting draw as the goal is arguably not enough. (2022-03-09, hidden) more ...
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Henrik Juel: 1.d5-d4-d3-d2-d1B-b3xa2-c4-f1 10.Bg2-h1 12.fxg3ep+?? fxg3# seems to fail, because after 11.Bh1 last move could be g2xh3. So what is the solution? (2003-04-28)
Thierry LE GLEUHER: After 11.Bh1, the last move can't be g2xh3 because of the promoted Bishop h1 imcompatible with the bP structure! (2005-09-25)
Henrik Juel: Why couldn't Black have a promoted bishop at this point in time (where wPa2 and bPd7 are no longer present)? 12 captures by black pawns seem to match the 12 missing white men. (2005-09-26)
Joost de Heer: Besides, after g2-g4, the black bishop must've been promoted too. (2005-09-30)
James Malcom: Henrik, gxh3 is indeed impossible. The last move sequence must go something like 1. Kfg8 Se2+ 2. Kh2 Sg1. There cannot be a knight on h3, cutting off g8, or a queen/rook, cutting off h2. Thus, gxh3 has already been played and g2-g4 is the last move. (2021-09-14)
Mario Richter: Position after 11. Lh1 is 8/5p2/8/5ppp/5pPk/5p1P/5P1K/6nb.
It can be easily shown, that in this position g2xLh3! is indeed a legal retraction.
So the ep-capture is not justified and the problem is cooked!
(Perhaps Thierry forgot, that black pawn d7 "disappeared"? But Joost's argument raises [correctly] the question, why a promoted Bishop on h1 is ok for g2-g4, but not for g2xLh3.) (2021-09-15)
Henrik Juel: So the good guys (Mario, Joost, and me) win over the bad ones (Thierry and James) by 3-2... (2021-09-15)
Henrik Juel: or, less insultingly:
By a majority of 3 to 2 this problem has been deemed cooked (2021-09-15)
James Malcom: Make it 4 to 1: 1. Nc3 Nf6 2. Nd5 Rg8 3. Nxf6+ exf6 4. d4 Ke7 5. Bg5 fxg5 6. Nf3 Kf6 7. Ne5 Bd6 8. Nc6 Bg3 9. hxg3 bxc6 10. Rh4 gxh4 11. b4 Ba6 12. a4 Bd3 13. b5 Bf5 14. b6 h3 15. bxa7 h2 16. a5 Bh3 17. g4 Re8 18. a6 Re5 19. dxe5+ Kg5 20. Qd6 Qe7 21. c4 Qe6 22. Ra3 Qd5 23. cxd5 cxd5 24. Rf3 Nc6 25. e4 Re8 26. a8=Q Nd4 27. a7 cxd6 28. Qxe8 dxe4 29. Qe6 dxe6 30. a8=Q exf3 31. Bd3 Kh4 32. Bf5 exf5 33. Qa5 g5 34. Qd5 h5 35. Qe4 dxe5 36. Qf4 exf4 37. Kf1 h1=B 38. Kg1 Ne2+ 39. Kh2 Ng1 40. gxh3 (2021-09-16)
A.Buchanan: Is it also possible that in the critical position, the last moves were R: 1.gxSh3 exXf345 2.X~? I think I can fix the cook R: 1.gxLh3 nicely. (2021-09-17)
A.Buchanan: Can it be fixed? What’s the point of wPa2 elimination anyway? Move it to d5. Shift bPf3d7 to f6d6. Add bPb7. Then the cook becomes thematic try as we have 5.b1=L 7.Lxd6 8. Lh1 13.d1=L 14.Ldf3 15.e.p. fxg3# The exact capture count in the critical position prevents bPa from having promoted to L on b1. Does this work? (2021-09-16, hidden)
A.Buchanan: 8/1p1p2pp/8/3P2pp/P4pPk/1p5P/Pn3P1K/6n1 shc#13 fixes both gxLh3 & gxSh3 using an additional sS, which I've tried to make as active as possible. Please break it if you can :) (2021-09-17, hidden) more ...
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Henrik Juel: Commendation
(awarded by Hans Gruber, PB no.1 2005 p.4) (2019-03-07)
Joost de Heer: @Hans: First retraction move isn't a pawn move. (2024-02-01)
hans: R: 1. --- 0-0-0+ 2. Kg7-h8 f7-f6 3. Kf6-g7 e4-e3 4. Kf5-f6 e5-e4 5. Ke4-f5 e6-e5 6. Kd3-e4 e7-e6 7. Te6-e2 e2-e1=L 8. Db2-a1 e3-e2 9. Ke2-d3 e4-e3 10. Dc1-b2 e5-e4 11. Dd1-c1 d6xLe5 12. Lb2-e5 c6-c5 13. Lc1-b2 c7-c6 14. b2-b4 b3xTa2 15. Ta1-a2 b4-b3 16. La2-b1 b5-b4 17. Lb3-a2 b6-b5 18. La4-b3 b7-b6 19. Lb5xa4 a5-a4 20. Ke1-e2 a6-a5 21. Lf1-b5 a7-a6 22. e2xSf3 etc. (all black pawn-moves) (2011-03-16, hidden) more ...
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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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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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Record of the highest number of units to add so that a position becomes legal (Type A, number of units not specified, no promoted pieces).
See http://abrobecker.free.fr/chess/addunits.pdf (2011-12-22)
Alain Brobecker: See P1227635 for a version with promoted pieces. (2011-12-22)
Henrik Juel: Add five black pawns on a3,b3,c3,d3,e3 to provide the 18 tempo retractions needed.
-1.Sh3 a4 -2.Dh1 a5 -3.Tg1 a6 -4.Lb7 a7 -5.Lc8 b4 -6.Kb1 b5 -7.Kc1 b6 -8.Kd1 b7 -9.Ke1 c4 -10.Kf1 c5 -11.Kg2 c6 -12.Ta1 c7 -13.Dd1 d4 -14.Kf1 d5 -15.Ke1 d6 -16.Sg1 e4 -17.Lh3 e5 -18.Lf1 e6 -19.g2xDf3 Kg4 -20.Sf5 etc. (2011-12-22)
James Malcom: What formatting is required to add the five black pawns to the board with the solution? I know I have seen it done before somewhere. (2022-01-30)
A.Buchanan: You define another pieces list labelled e.g. b, then include that in the solution before you start moving pieces. Check out some other add pieces problems to see how it’s done. (2022-01-31)
James Malcom: Thanks! I have it figured out now. (2022-01-31)
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A.Buchanan: How about wLa1 instead? (2020-10-06)
Henrik Juel: Yes, that would improve the economy slightly (2020-10-06)
Ladislav Packa: wBa1 is illegal... (2020-10-07)
A.Buchanan: Hi Ladislav why is wLa1 illegal, please? That was essentially my question before (2020-10-07)
Mario Richter: wLa1 instead of wDa1 is of course legal too: one of the dark-squared wBishops unpromotes on h8, one wQueen on g8. Black pawns g7+h7 let the wPawns g2+h2 pass by cross-capturing the 2 original wRooks. (2020-10-07)
Henrik Juel: Add white knights on b2 and d2 and a white bishop on c4 (2015-07-08, hidden)
Henrik Juel: The man added on c4 should be a bishop
R: 1.Dd1-c2+ Kc2-c3 2.Dh5-d1+ Kc2-c3 etc. (2020-10-05, hidden) more ...
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Korrektur zu 16238 aus Heft 271 (Februar 2015) [Pxxxxxxx]

Autor Tom Volet: "I believe this to be the first published example of a 50 move draw sequence in which the move that immediately precedes the sequence must have been (not could have been), applying only the rules of forward chess without regard to problem conventions or stipulations as to forward play, a capture of a unit by a unit."
Kees: . from PAS

1. Na3 Na6 2. Nb5 Nc5 3. Nd4 Ne6 4. Nb3 Ng5 5. Rb1 Nh3 6. Na1 Nf6 7. gxh3 Nd5 8. Bg2 Nb6 9. Bd5 Rb8 10. a4 Na8 11. Ba2 a6 12. b3 Rg8 13. Ba3 Rh8 14. Bc5 Rg8 15. Ba7 b6 16. Nf3 Bb7 17. Nd4 Bf3 18. Nb5 axb5 19. Qc1 Qc8 20. Qa3 Qb7 21. a5 Qc6 22. Qa4 Kd8 23. Kd1 Kc8 24. Kc1 Kb7 25. Kb2 Re8 26. Rbg1 Kc8 27. Ka3 Kd8 28. Kb4 Qh6 29. Rg6 bxa4 30. Kb5 Rh8 31. Ka6 Bg4 32. Kb7 Bf3+ 33. Rc6 Qg5 34. Kb8 Qg1 35. Rh6 e6 36. Rg6 Bd6 37. Rh6 Reg8 38. Rg6 Bf8 39. Rh6 Ke7 40. Kc8 Kd6 41. Kd8 Kc5 42. Rg6 Be7+ 43. Kxe7 Rb8 44. Rh6 Rb7 45. Bb8 Ra7 46. Rg6 Ra6 47. Ba7 Rb8 48. Rh6 Rb7 49. Kf8 Kd6 50. Kg8 Ke7 51. Kh8 Kf8 52. Rg6 Rb8 53. Rh6 Re8 54. Rg6 Re7 55. Rh6 Ke8 56. Rg6 Kd8 57. Rh6 Kc8 58. Rg6 Kb7 59. Rh6 Kc6 60. Rg6 Kb5 61. Rh6 Kb4 62. Rg6 Ka3 63. Rh6 Kb2 64. Rg6 Kc1 65. Rh6 Kd1 66. Rg6 Ke1 67. Rh6 Kf1 68. Bb8 Kg2 69. Kg8 Qb1 70. Kf8 Qb2 71. Rb1 Qf6 72. Rb2 Kf1 73. Bb1 Ke1 74. Ra2 Kd1 75. Ra3 Kc1 76. Rg6 Kb2 77. Rg1 Qg6 78. Rh1 Qg1 79. Ba7 Kc1 80. Bb8 Kd1 81. Ba7 Ke1 82. Bb8 Kf1 83. Ba7 Kg2 84. Bb8 Qc1 85. Ba2 Qb2 86. Rb1 Qc1 87. Rb2 Kf1 88. Rb1 Bh1 89. Rb2 Kg2 90. Rb1 Qb2 91. Re1 Qc1 92. Bb1 Qd1 93. Ra2 {50 moves rule} Qc1 94. Rb2 Qd1 95. Ba2 Ra7 96. Rb1 Rb7 97. Ba7 Qc1 (2022-03-14)
Thomas Volet: Apologies to Thierry. I should have addressed this earlier. My remark quoted above was (if recollection serves) made along with the submission of the composition for publication. Regardless, my belief at at the time was incorrect as I was then unaware of Nikita Plaksin's mirror compositions P0004364 and P0004395 (published in 1969). Both show the retroplay sequence ending (just short of the draw that the prevailing convention dictated, which saves the stipulated Mate in 3) in the uncapture by a King of a Bishop on the Bishop's home square, from which the Bishop never moved (exit Pawns occupying their home squares in the diagrams). The above 2016 joint composition with Thierry shows the 50 move retroplay sequence of non-Pawn, non-capturing moves being immediately followed by the uncapture by a King of a non-Pawn unit that has moved from its home square. That feature, which presents some additional constraints on realization, may also have been presented before, but obviously I should leave that to better historians. (2022-07-27)
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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)
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