Anton Baumann: Auszeichnung Informalturnier 1986: 2.Preis
Preisbericht: 'Die Schwalbe' 06/2011 S.124 (2023-01-02)
Henrik Juel: How is the SE corner released, without ruining the castling? (2023-01-02)
Mario Richter: Good question, Henrik! I first thought that releasing the SE corner without ruining White's castling right is impossible, but the trick is to uncapture a black Queen in the SE corner at the right moment.

Perhaps Theodore Hwa can use ths problem as a test case for his latest improvement to Retractor 2 ... (2023-01-02)
Henrik Juel: Thanks, Mario
In view of the prize I suspected that the problem was correct, but I did not find the uncapture trick (2023-01-02)
Henrik Juel: C+ Popeye 4.61, because with Black to move White may not castle (2023-01-02)
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

"das zeigt die Paradoxie der AP-Bedingung!" (MS). Gleich 1. bxc3 ep.? und dann 2. Kd4 geht natürlich nicht, weil zuvor außer K,T-Zug und c2-c4 auch S-Züge und c3-c4 möglich waren. "Interessanter und genau abgestimmter Lösungsverlauf" (GW) "Sehr schöne und ökonomische Verbindung zwischen konsequentem sh# und AP!" (TB) Mit 'Typ Tauber/Caillaud' ist gemeint, daß die einzelnen Stellungen doch als retroanlytischer Gesamtkomplex betrachtet werden dürfen/sollen (sonst wäre keine AP-Legalisierung möglich), obwohl sie ja aufgrund der retroanalytischen Neubewertung nach jedem Rückzug voneinander unabhängig sind. 6L./5+5P.
A.Buchanan: "By 'type Tauber/Caillaud' it is meant that the individual positions may/should be regarded as a retroanalytic overall complex (otherwise no AP legalization would be possible), although they are independent of each other due to the retroanalytic reassessment after each retraction." (2023-06-29)
comment

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

vergl. P0004915 (Hans Gruber, Schach 1979)
Brassaud: La solution proposée 1/Tg6# est possible
Mais il y a aussi le rétro jeu -1) Fa2-b1, Rg5g6 -2) Ta4-a5+, Rf4-f5 etc … et avec le trait aux noirs : 1) Fxe6 # est possible (2017-08-30)
A.Buchanan: @Brassaud: yes I agree. There is no reason why White should not have moved last. So both players can mate, but part (b) implies that the intended solution in (a) is 1 player. If the published stipulation for (a) was maybe just "#1", which by default is white to move, then there is a unique solution.
For (b) I am wondering about +sBg6, which would also stop the en passant trick, both by blocking sK from retreating there and also by locking sL in an impossible cage with sBf7. (2017-08-31)
Henrik Juel: Adding a black pawn on g6 of course prevents a black last move by Kf6, but it allows f7xg6 as last move; Lg8 is not locked, because Ph7 is white (2017-08-31)
A.Buchanan: Yes (2017-08-31)
Anton Baumann: vergl. P0004915 (Hans Gruber, Schach 1979) (2023-01-03)
Brassaud: Pour la position
a) il y a la possibilité 1) Tg6 # après 1)… Rg6xe6 ep , mais il y a également le rétro très simple :
1) Fa2-b1, Rg5g6 2) Ta4-a5+, Rf4-f5 etc … et 1) Fxe6 # est possible
b) 1) Tg6 # d’accord. (2015-09-14, hidden) more ...
comment

James Malcom: (Shortest) proof of legality: 1. d4 Nh6 2. Bxh6 gxh6 3. g4 Rg8 4. g5 Rg6 5. Nh3 Rf6 6. Nf4 Nc6 7. g6 Nb4 8. g7 Rg6 9. Nxg6 fxg6 10. Bh3 Nd5 11. Be6 dxe6 12. f4 Kd7 13. f5 Kc6 14. f6 Bd7 15. Rf1 Be8 16. a4 Bf7 17. a5 Bg8 18. f7 Qd6 19. Rf6 exf6 20. e4 Be7 21. e5 Bd8 22. exd6 Ne7 23. dxe7 Rc8 24. d5+ Kb5 25. d6 Kc6 26. d7 Kb5 27. Qd6 cxd6 28. a6 Kc6 29. c4 Kb6 30. c5+ Kb5 31. c6 Kc5 32. c7 Kb5 33. Na3+ Kc6 34. Nc4 Kb5 35. Nb6
axb6 36. Ra5+ Kb4 37. Rc5 bxc5 38. a7 Kb5 39. b4 Kc4 40. b5 Kd4 41. b6 Ke4 42. Kd2 Ke5 43. Kc3 Kd5 44. a8=B Kc6 45. Kc4 Rb8 46. c8=R+ Bc7 47. d8=N+ Kd7 48. e8=B+ Ke7 49. f8=Q#

Took me a bit to figure out the trick for maneuvering the Black bishops. (2022-08-28)
comment

Guus Rol: This is an incorrect interpretation of the AP-convention. Rules outrank goals in the definition of all GAMES. Therefore the legitimacy of a move cannot be restricted by the desire to achieve the goal (in this case: Remis). The proper way to view AP is that executing e.p. invalidates the legitimacy of all lines of future play that do not contain 0-0-0! In that sense black and white are forced to cooperate. In whatever freedom remains they can compete for the prize promised in the stipulation. By the way, this understanding of AP is not only more logical, it is also much more interesting as a playing field for AP-composition. (2005-09-21)
mri: citeWeiß kann aber so spielen, daß Schwarz nicht rochieren kann, z.B. 1. ... Lxc7 2. bxc7
/cite
How does white prevent black from castling after 1. cxb6 e.p. Ba7xb6+?
E.g. 2.Kxb6 a1=R 3.a7 Rxa7, or 2.Ka4 Bd8xc7 3.a7 Bb8. (2005-09-22)
VL: 1.c5xb6ep(??) a7xb6+ 2.Kxb6 a1R 3.Kb7 R1xa6 4.Rc8! (R6a7#?? illegal).
This study is correct under the generally accepted understanding of AP
a la' N.Petrovic'. Antiform (looking possibly somewhat strange):
Black's unsuccessful try. (2005-10-03)
Guus Rol: Generally accepted, true. Generally acceptable, false. Freedom of interpretation ceases for a concept once its polar concept (a priori validation) has been defined. "Goal induced AP" however, might be a passable stipulation for this type of problem. To keep this place from turning into a discussion forum I will discontinue comments on this issue. (2005-10-03)
paul: Author intention is: If black still can castle, his last move must have been b7-b5. However, to prove this, he has to castle (A Posteriori condition). So: 1.cxb6 e.p. axb6+ 2.Kxb6 a1R! (2...a1Q? 3.Kb7 Qxa6 mate obviously doesn`t prove b7-b5 as the last move) 3.Kb7 R1xa6 4.Rc8! and white has prevented black from castling. So black can`t prove the last move is b7-b5 and therefor is already stalemated in the initial position! (2011-07-12)
A.Buchanan: Isn't Guus' idea just AP-Prioritat? And even if it were "more logical" or "more interesting", it doesn't follow that other forms is AP are "incorrect" (2023-07-31)
A.Buchanan: Isn't Guus' idea just AP-Prioritat? And even if it is "more logical" or "more interesting", it doesn't follow that previous interpretations are "incorrect". Once a "polar concept" has been defined, prior art remains, and is valid in its own context. "Freedom of interpretation" remains, although newer compositions may adopt the newer paradigm. If a new idea does succeed in displacing the old, then problems (old and new) which depend on the earlier interpretation should be preserved with the context in which they existed. I don't know whether the
"Timeline Gallery" will ever physically exist, but I hope that it will. This is basically the "Golden Age" argument, that says if we don't value & preserve the old, we cannot consistently move on to the new (2023-07-30, hidden)
A.Buchanan: Isn't Guus' idea just AP-Prioritat? And even if it were "more logical" or "more interesting", it doesn't follow that previous interpretations are "incorrect". Even if a "polar concept" has been defined, prior art remains, and is valid in the context in which it was formed. "Freedom of interpretation" remains, although newer compositions may preferentially follow the newer paradigm, in part because fresh design space is always desirable.
If a new idea does succeed in displacing the old, then problems (old and new) which depend on the earlier interpretation should be preserved in a "Timeline Gallery", each with the context in which it was created.
It is also a fallacy to say that a stipulation should "explain" the philosophy. Such problems have been defined as AP, and that's fine. I think AP-Prioritat (sorry for absence of umlaut) *does* deserve to be mentioned as part of the stip, but whether the problem is Keym or Petrovic or both or something else - or PRA or RS - all that is part of the solving fun: keywords can record the taxonomic details & they don't need to be part of the stip. (2023-07-30, hidden) more ...
comment

In a) Entschlagdual R: Kd5xBe4, das wurde in der Lösungsbesprechung noch kritisiert und als NL vermerkt. Hat das der PR gelassener gesehen ('geduldeter Entschlagdual'), oder gibt's noch eine korrigierte Version dieses Problems?
Anton Baumann: Korrektur in 'Die Schwalbe' 12/1975 S.422: +sLh2, +sBe5;
nun geht in a) nur noch zurück: Kd5 x Be4.
Ausgezeichnet wurde gem. Preisbericht in 'Die Schwalbe' 06/1977 S.82 die korrigierte Version 1419v. (2022-12-08)
comment

Henrik Juel: The black men have made an even number of moves, so the white men (ending with Sf6+) have made an odd number of moves; hence [Ta1] has made an odd number of moves and was captured on b1; the fastest way of doing this is to let [Sb8] do all the black moves, incl. 5... SxTb1 and 10... Sb3-a1 (2023-04-20)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 KBP=10.5 in 1 Sekunde.
Keine Lösung: BP 9.5, BP 10.0.
Beispiel: 1.Sa3 Sa6 2.Tb1 Sb4 3.Ta1 Sd5 4.Tb1 Sc3 5.Sb5 Sxb1 6.Sa3 Sc3
7.Sb1 Sa4 8.Sf3 Sc5 9.Se5 Sb3 10.Sg4 Sa1 11.Sf6+ (2023-04-19, hidden) more ...
comment

Mario Richter: In der Lösungsbesprechung wurde die Forderung präzisiert: Wieviele (und welche) Steine zogen in der KBP mindestens?
Die richtige Antwort ist: 6 (sSb8, wSb1, wDd1, wKe1, wSg1, wTh1).
Die kürzeste BP braucht 12 Züge von Schwarz (Sb8xLc1-f3xTg1, dann zurück nach b8), wK und wSS können nur gerade Anzahlen von Zügen machen, wegen Th1-g1 muß also wD oder wTa1 ein Tempo verlieren. In Rahmen des 12-Züge-Limits schafft das nur die wD. (2010-05-24)
Moldenhauer: Computerprüfung: C+ NUPG cooked Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 11.0, BP 11.5. (2023-04-06)
comment

Anton Baumann: Autorabsicht: Die weiss-schwarzen Rochaden schliessen sich gegenseitig aus.
a) 1.O-O? Tf8! daher: 1.Tf1! O-O 2.Sxe7#
b) 1.Tf1? O-O! daher: 1.O-O! Tf8 2.Sxg7#
Aber in der Urfassung (= nebenstehendes Diagramm) geht in a) und b) die NL:
1.Tg1 O-O 2.Txg7,Sf5xh6#
Korrektur in 'Schwalbe' 04/1976 S.464: sLb7 nach g6, sBc5 nach b7
Ausgezeichnet wurde die korrigierte Fassung 1557v (vergl. 'Die Schwalbe' 06/1977 S.82) (2022-12-09)
comment

Henrik Juel: It is illegal for Black to supplement anything on b6, because [Ta8] was captured in its corner and the other missing black men were captured by white pawns (2016-03-28)
Henrik Juel: ... as wLb3 is a pawn promoted on e8 or g8
Nice type Høeg defensive retractor
Here are some other explanatory comments
In retraction 1 White chooses to move his king back to h3; Black could choose to supplement a black man on g3 (or nothing), but supplementing a pawn is the only way to maintain legality (Kh3 stands in double check from Lc8 and Dh8); again moving Pg3 back to h3 and White supplementing a pawn on g4 is forced (this e.p. case is the only one where the supplementing does no happen on the abandoned square)
In retraction 2 the white retraction is forced, and then moving Kd6 back to d7 to uncheck is illegal because of the double check from Sb6 and Pc6, so Black must uncheck by moving Kd6 back to e6 and White choose to supplement a pawn on the abandoned square
In retraction 3 White chooses to move Pd6 back to e5, forcing another e.p. situation (2023-04-08)
Henrik Juel: The Proca type is easy to define: White and Black alternate retractions, until White can mate with a forward move
The Høeg type is usually defined the same way, except that the other side decides which man (if any) was captured; but this can be detailed as follows:
1. White chooses a man and 'moves it back'
2. Black chooses which man (if any) to 'supplement' on the abandoned square
(only now is the white retraction complete)
3. Black chooses a man and 'moves it back'
4. White chooses which man (if any) to 'supplement' on the abandoned square
(only now is the black retraction complete)
etc. etc. until, following a white retraction, White can mate with a forward move
In tries, Black can ruin the white plan by mating White with a forward move after a black retraction
It goes without saying that the resulting retractions must be legal
'supplement' is my (poor) translation of the danish term 'supplere'; maybe 'add' would be better
'the abandoned square' needs a special interpretation in the e.p. case, which happens twice in this problem
These details may be the cause why new type Høeg defensive retractors are rarely seen, as type Proca is more natural and straightforward (2023-04-08)
A.Buchanan: Thanks Henrik. Yesterday, I went through all the defensive retractors to clear up keywords & genres. There were a very few where the stip did not specify the VRZ Type, and others where Anticirce did not specify Calvet vs Cheylan. The answers are probably obvious to you, and if you want to comment on those, then I will update the stips & keywords.
A more general question: Typ Friedlich appears to be the German for Type Pacific: can we standardize on one? (2023-04-08)
Henrik Juel: Thanks Andrew for enabling me to post my type Høeg spiel once again
Anticirce without specification usually means that both Calvet and Cheylan work
Friedlich is indeed german for Pacific, and as the PDB is a german product, I guess we must live with the present conditions (2023-04-08)
comment

Kees: +wKb5 1. Txc8 (2. Lxe7#)
0-0 is illegal for K or T must have made a move.
-1. a7-a5? Not possible with position of wL and bS (2022-11-23)
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

Datum der Originalpublikation nicht 100% sicher, laut ACM aus der "Jamaica Gleaner Christmas column".

Originalforderung: How has the position been arrived at and who is the winner, and in how many moves?

From the Jamaica Gleaner: "White mates in two moves. The last move made was by Black playing his Bishop and announcing mate. As it can be demonstrated that the Bishop is not a promoted Pawn and that Black's King's Knight's Pawn was captured on its original square by White's Queen's Knight's Pawn the Black Bishop must have been played from Bishop's square (f8) to Q3 (d6). This being an illegal move, White enforces the penalty of compelling Black to retract it and move his King whereupon White plays 1 PXB(Q) ch (1.gxf8=Q+) and mates next move by 2 Q-B4 (Qf4#). The following is a brief but pointed analysis, demonstrating the false move: White's Pawns have made six captures all on black squares. The Q Kt P (Pb2) made five of these and consequently captured the Kt P on the square upon which it now stands (g7). They could not have captured the Q B which is also lost. The White Bishop is the QRP (Pa2) promoted, the original KB having been captured on its own square as the unmoved Pawns show. To allow this promotion Black's QRP (Pa7) made two captures, the QKtP (Pb7) one, and the QBP (Pc7) two. The KRP (Ph7) has also made a capture, which accounts for the seven pieces White has lost. The Black Bishop is not a promoted Pawn, as if the Black KBP (Pf7) had played to the 7 th square (f2) and then captured a White piece on K or Kt square (e1 or g1) the captures by White Pawns cannot be accounted for without including the Black QB or KRP neither of which is available. As it can be demonstrated, then that the Black Bishop is not a promoted one, and that the KKtP was captured by the White Pawn which now stands on that square, in order to reach Q3 the Black Bishop must have an impossible move.

Der Kolumnist des ACM merkt aber zurecht an:
ACM: The above is a very fine piece of analytical work; but there is a slight flaw in connection with the minor condition, 'mate in two'. In a position of this kind we believe only that which can be proved; thus we do not think that White has any right to enact a penalty, as neither the analysis nor the conditions show that the Black Bishop came from Bishop's square on his last move; indeed, that Bishop may have played outside the Pawns on the very first move of the game which, being played, brought about the position.
HBae: White plays 1 PXB(Q) ch (1.gxf8=Q+). Muß der sK nicht auf f5 stehen? (2019-10-22)
Henrik Juel: Last move (supposedly) was Lf8-d6#, which is obviously impossible and hence illegal
The penalty for this is that Black must replace Ld6 on f8 AND instead make an arbitrary move with his king
So the forward play is
0... Kf5,Kf6,Kh6 1.gxf8=D+ Kg5,Ke5 2.Df4# (2019-10-22)
A.Buchanan: One long-standing approach to resolving illegal diagram jokes is to suppose that only the last move was illegal, with all prior play legal. The illegal move is then retracted, and play continues. Of course, the “illegal move” might in principle be from *any* legal position (even the game array!). So for sanity, we say the illegal move is a simple but somehow illegal shift of a single piece.

So here, candidates for the last move include Pe5-a4+, Sf4-e1+, Ke5-g5+ & B?-d6+. For all of these, White has 6 visible pawn captures, all on dark squares, so Black light-squared bishop is excluded. wPa must have promoted to light-squared bishop, so if the three Black pawns on a-file remain, there is only one unaccounted capture. Thus bPh could not promote, and must be bPg6 now. Thus bPg7 was captured at home, and bBf8 was thus locked in.

So Bf8-d6 is certainly a possible illegal move, but so are e.g. Be8-d6 (as the light-squared bishop is otherwise unexplained) and Pe5-a~. This is an example of an "implausible" joke according to Dawson & Hundsdorfer, because there is more than one retraction to the current position, and one just has to arbitrarily pick the one that makes the forward logic work. (2023-04-02)
A.Buchanan: Another issue is that according to the 1883 laws, White cannot force Black to move their king. The 1883 rules stated:
- If a player touches a piece or Pawn of his own he must move it.
- If he touches one of his adversary's he must take can be taken.
- If he touches plurality of pieces or Pawns of the same colour, in either of these instances his adversary may elect which such piece or Pawn he will call upon him to play or to take, as the case may be.
- If the rules governing the moves of pieces do not admit of the adversary exacting penalty as above, the player must move his King, but may not Castle. If the King cannot be moved without exposure to check, no penalty can then be exacted
So according to this, Black must play Bf8xPg7 as the penalty move.
Was there another revision to the rules between 1883 & 1891? (2023-04-02)
A.Buchanan: I don't think this is cooked, I think it's just a very bad joke. Most time when moves are retracted it's because of self-check or illegal castling, occasional illegal en passant. But here we are meant to conclude that the Bl bishop has hopped over Bl pawn. There are problems by Hieronymus Fischer e.g. P1309484 where we are meant to conclude that a unit has escaped its game array cage somehow, but the piece is simply replaced: there is no move to be retracted and supplanted. But as long as such a problem is true to its own (joke) internal logic, it can't be described as cooked. If it fails (e.g. P1246715) it can indeed be cooked and repaired. (2019-10-22, hidden) more ...
comment

Joost de Heer: The keyword is wrong (should be TxD, not BxS=T) (2023-05-20, hidden)
Henrik Juel: I believe that the condition 'Black to move' can be deleted (2023-05-21, hidden)
Adrian Storisteanu: This is not the magazine's informal tourney - p.137 of the indicated source says REZULTAT 16. TEMATSKOG TURNIRA "PROBLEMA". (2023-05-21, hidden) more ...
comment

Rosalie Fay: White has lost only the bishops. So the pawn on c5 is not [Pa7] (because that entails 2 White units captured on black squares). White has played axbxcxdxe7, dxe, fxe, hxg, gxfxe. Black has 7 units, so white pawns have captured all missing Black units, but none on the a or h files.

Black has 2 pawns on the c-file, so one has captured. Thus [bPa7] and [bPh7] pawns have collectively captured no more than once. So at least one of them must have promoted, in order to either get to a file where White made a capture, or replace a captured unit; it didn't capture en route to promotion, so it displaced a white rook and thus spoilt one White castling right.

White would mate by 1 Rd1 & 2 Rd6 or 1 Rf1 & 2 Rf6, except that Black threatens Qxe2+. So either 1 0-0 or 1 0-0-0, though it's impossible to say which is legal. (2022-11-24)
Henrik Juel: one solution, but in two parts
if Ta1 has moved, 1.0-0 thr. 2.Dc8,Tf6#
if Th1 has moved, 1.0-0-0 thr. 2.Dg8,Td6# (2022-11-25)
Hans-Jürgen Manthey: nach der möglichen Zugfolge: 1. Sb1-c3 c7-c6 2. Sc3-d5 Dd8-b6 3. Sg1-f3 Db6-b3 4. a2xDb3 a7-a5 5. Sd5-b4 a5-a4 6. Sb4-a2 e7-e5 7. Sf3-h4 Lf8-c5 8. Sh4-g6 f7-f5 9. Sg6-f4 g7-g5 10. Sf4-g6 Lc5-e3 11. d2xLe3 f5-f4 12. g2-g3 Ta8-a5 13. g3xf4 g5-g4 14. f4xe5 g4-g3 15. h2xg3 h7-h5 16. Lf1-g2 h5-h4 17. Lc1-d2 h4-h3 18. Ld2-b4 Th8-h4 19. c2-c3 Th4-c4 20. Sa2-c1 a4-a3 21. Lb4-c5 a3-a2 22. Dd1-d4 Sb8-a6 23. Dd4-h4 Sa6-c7 24. Dh4-d8+ Ke8-f7 25. b3xTc4 Sc7-d5 26. c4xSd5 Sg8-e7 27. d5-d6 Ta5-b5 28. d6xSe7 d7-d6 --- folgt nun
29. Lg2-e4 Lc8-e6 30. Le4-b1 a2xLb1D 31. Th1-g1 Db1-d3 32. Sc1-b3 Le6-c4 33. Th1-g1 h3-h2 34. Dd8-e8+ Kf7-e6 35. Tg1-h1 Tb5-b6 36. Th1-g1 Le6-c4 37. Tg1-h1 h3-h2 38. Th1-g1 h2-h1D 39. Sc1-b3 Dh1-e4 40. f2-f3 Lc4-a6 41. f3xDe4 d6xLc5 42. Dd8-e8+ Kf7-e6 43. Tg1-h1 Dd3-b5 matt in 2:
1. OOO droht 2. De8-g8/Td1-d6# - 1. ... Db5-d3 2. Sb3xc5# oder:
29. Sc1-b3 h3xLg2 30. Sb3-d2 g2-g1D+ 31. Sd2-f1 Dg1-g2 32. Sf1-d2 Dg2-e4 33. f2-f3 Tb5-b6 34. f3xDe4 d6xLc5 35. Sd2-b3 Lc8-e6 36. Ta1-b1 Le6-c4 37. Ta1-b1 Kf7-e6 38. Tb1-a1 Lc4-a6 39. Ta1-b1 a2-a1D 40. Dd8-e8 Da1-a4 41. Tb1-a1 Da4-b5 matt in 2: 1. OO bel. 2. De8-c8/Tf1-f6# (2023-02-22)
A.Buchanan: Could we shift wPb2 to c2, and eliminate bR or bB? (2022-11-25, hidden) more ...
comment

hans: 1. Th2xf2 De1xf2#!
1. Th2xh3 0-0#? (Castling illegal)

R: -1. Kf1xSe1 Sg2-e1 -2. Ke1-f1 Sf4-g2+ -3. Tg1-h1 Sd5-f4 -4. Tf1-g1 Se3-d5 -5. Tg1-f1 Sf1-e3 -6. Tg1-h1 Tg2-h2 -7. Th2-h1-g1 Tg1-g2 -8. Th1-h2 g2-g1=T -9 h2-h3 h3xSg2 (2010-06-26)
HENRI: Hans solution has a tipo. Their is no D (=Queen in german) mating. One should read the solution in german : 1. Th2xf2 Ke1xf2#! 1.Th2xh3 0-0#?
or in english : 1. Rh2xf2 Ke1xf2#! 1.Rh2xh3 0-0#? (Castling illegal) (2023-10-25)
comment

Henrik Juel: This problem demonstrates an advantage of type Høeg over type Proca:
another way of generating variations
2.0-0 cannot be an uncapture, of course
2.Te7-f7 and 2.Dc5-c8 etc. could be uncaptures, but no matter what Black supplements, he is mated (2023-04-08)
A.Buchanan: Thanks Henrik: is an alternative for White R: 1. Kg1-h1 Lb8xDa7+ 2. Dc5-a7/Da3-a7, dann 1. Df8# (2023-04-08)
Henrik Juel: No Andrew, when White moves his uncaptured queen back to c5 or a3, Black supplements a queen or bishop on a7, preventing the mate on f8 (2023-04-08)
A.Buchanan: Thanks! (2023-04-08)
Henrik Juel: -1.Kg1:S Sf2:0 -2.0-0, 1.Rh8#; -1.. Bb8:S -2.Re7, 1.Sc6#; -1.. cSb6,dSb6,Se3:Q -2.Qc5, 1.Qf8#. Duplicated by P0006277 (2003-10-09, hidden) more ...
comment

Henrik Juel: Last move is determined only if it is known to be white. (2003-11-20)
Henrik Juel: The coloring is White: Kh3, Pe4,f4,g2,g3,g4,h2 (7+9). (2003-11-18, hidden)
A.Buchanan: Yes the stipulation should always have stated "Schwarz am Zug". I've fixed it. Does that make it "Type E" or "Type F", I forget which is which (2023-06-03, hidden) more ...
comment

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

"Promenades to Power"

Es läßt sich beweisen, daß die UWs in D,T,L,S zwingend erfolgt sind (a2-b8=L,h2-b8=S und entweder b2-c8=D/c2-c8=T oder b2-c8=T/c2-c8=D).
Erich Bartel: weitere Nachdrucke:
3) 160 Die Allumwandlung im Problemschach VIII 1966a---
4) Schach ohne Grenzen 1969.-- (2007-01-09)
Sally: Vier erzwungene Umwandlungen. (2012-02-21)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 da NUPG sonst cooked in 00:03:51 Minuten. (hh:mm:ss)
Keine Lösung: BP 33.0, BP 33.5.
Beispiel: 1.Sf3 Sf6 2.a4 Sd5 3.a5 Sf4 4.Ta4 c5 5.Td4 Db6 6.b4 Sc6 7.b5 Tb8 8.bxc6 cxd4 9.c7 e5 10.d3 La3 11.Le3 Lc1 12.c4 Db2 13.c5 b5 14.c6 Tb6 15.axb6 Ke7 16.b7 Kf6 17.b8L Lb7 18.c8D b4 19.c7 d5 20.Dh3 b3 21.c8T h5 22.g4 h4 23.Dg3 hxg3 24.h4 Th6 25.Lh3 dxe3 26.0–0 Sg2 27.Sh2 Se1 28.h5 Kg5 29.Td8 Ta6 30.Ld6 Kh4 31.h6 Ta1 32.h7 a5 33.h8S a4 34.Sg6+ fxg6
Umwandlungen: 17.b8L, 18.c8D, 21.c8T, 33.h8S. (2023-04-19)
comment

Henrik Juel: Note that in problems castling acts like capture and pawn move with respect to the 50 moves rule. After 949.a6xSb7 there are 4 moves left by [Pa7]; but each camp can shift only KDT, on d1-g1 and b8-e8, respectively, so the triple repetition rule now limits the length of the game. (2004-09-09)
A.Buchanan: In some problems it's certainly the case that the 50-move rule operates incorrectly in this way. Such problems are fine, but obviously wouldn't want to impose this as a standard. Different composers can make different assumptions here (2023-06-20)
comment

Jeliss: "Obstruction of passage square f3 to Bishop of same colour."

"Version 'Pittsburgh Leader' 08.06.1913"
Yoav Ben-Zvi: Appears as the first problem (D445) in the booklet on Dawson's RA problems by G.P. Jellis. The obstruction that occurs in the Try -1.Pf3-f4?, by WP of WB, is described as "obstruction of passage square". It is not considered by Dawson and his disciples to be a Retro opposition. Dawson's conception of RO was quite broad, it included cases where the interference was not by occupation of the target square, so the only valid reason that I can see to exclude this case is that the 2 pieces involved are both of the same color. Fabel's definition explicitly excludes "Monochrome RO". I conclude that it would be preferrable to interpret RO as a bi-chromatic interference. The keyword Retro opposition should be removed. (2018-04-07)
A.Buchanan: To my mind, RO involves some kind of parity-tempo issue between the sides, not just some kind of race-tempo. If it was just about "bi-chromatic interference", one might say that bPe4 blocks wBh7 from an immediate retreat, so it has to be wPf4 that retreats first, legitimizing the ep key. So I agree this is not RO. (2024-01-06)
Henrik Juel: -1.f2 f5xTe4 -2.Tg4 f6 -3.Tg1 f7 -4.Le4 h5 -5.Lg2 h6 -6.Lf1 h7 -7.g2xDTSh3, unpromote DTS on c1, etc. (2004-03-18, hidden)
Henrik Juel: Nice analysis, Andrew, but your last comment seems incorrect.
Without Sb4, 1.b4 Txb4+ is not mate because of 2.Txb4#! (2013-07-26, hidden)
A.Buchanan: Thanks Henrik, I have fixed the solution. Forward chess is even harder than retro chess. And because Rb7 is retro-pinned, I overlooked that it's not forward-pinned! (2013-07-27, hidden) more ...
comment

Henrik Juel: Accprding to Retrograde Analysis, 1915, the intended solution does not involve adding pieces as such, rather:
Black retracts the illegal move Pd2xSe1=S+ (exposing Kb2 to a selfcheck from Dh2)
and instead pays the penalty of a king move, Kxc3 (only possibility)
Then White mates by 1.Dh8# (2023-04-13)
Mario Richter: @Andrew: Why did you remove the "illegal position" keyword?
Instead of removing this keyword, I suggest to remove the "Add pieces" KW ... (2023-04-14)
A.Buchanan: Hi Mario, yes good question. I removed "Illegal position" from some genuine "Add pieces" compositions, because it suggests some error in the composition. But on reflection, I think I will put the "Illegal position" back for these, and instead edit the description for the keyword.

Now this problem was in fact incorrectly marked as Add pieces. It's one of Dawson & Hundsdorfer's canonical examples of an "implausible" joke. I.e there is no reason why the intended retraction is the right one, except that it works. These too should be marked as illegal position. Sorry for all confusion (2023-04-14)
comment

Version in der 'Aarsskrift' 1935 innerhalb eines Artikels von K. Hannemann "Et Tema Fra den Retrograde Analyse". Im Original wDg4 statt h3 und wBh3 statt h4.
Henrik Juel: 1.exf6ep+. Not -1... a6?, requiring 3 black captures on light squares (incl. orig. Lf1); but missing orig. Lc1 was dark-squared. (2004-03-08)
Henrik Juel: C+ Popeye 4.61 and analysis (2022-06-11)
Hans-Jürgen Manthey: mögliche Zugfolge:
1. f2-f4 d7-d5 2. c2-c3 d5-d4 3. c3xd4 c7-c5 4. h2-h4 c5-c4 5. b2-b4 Lc8-d7 6. a2-a4 Ld7-b5 7. a4xLb5 Sb8-a6 8. Th1-h3 Sa6-c5 9. d4xSc5 Sg8-f6 10. d2-d4 Sf6-e4 11. c5-c6 Se4-g3 12. Lc1-d2 Se4xLf1 13. Sb1-c3 Sf1-g3 14. Sc3-a4 Sg3-e4 15. c6-c7 Ke8-d7 18. Sa4-c5+ Kd7-d6 19. Sc5-a6 Se4-c5 20. b4xSc5+ Kd6-e6 21. Ld2-a5 b7-b6 22. Sa6-b8 b6xLa5 23. Th3-e3+ Ke6-f6 24. Te3-e5 Dd8-c8 25. Dd1-d3 Dc8-a6 26. Dd3-h3 Da6-b6 27. Ta1-a3 Db6-c6 28. Ta3-g3 Dc6-d6 29. Te5-d5 Dd6-e6 30. c7-c8L De6-d6 31. Lc8-f5 Dd6-c6 32. Sg1-f3 Dc6-b6 33. Sf3-e5 Db6-a6 34. Se5-g6 Da6-b6 35. Sg6xTh8 Db6-c6 36. Tg3-g6+ h7xg6 37. Ke1-f2 Dc6-d6 38. Kf2-g3 Dd6-e6 39. Kg3-g4 De6-d6 40. Lf5-c2 Dd6-c7 41. b5-b6 Dc7-e5 42. f4xDe5+ Kf6-e6 43. Kg4-g5+ f7-f5 und nun :
1. e5xf6ep+ Ke6xTd5 2. Dh3-d7# (2023-02-24)
comment

Adrian Storisteanu: Possible twin:
b)wSd2->d6
- 1.Sf7xRd6 Rd8xQd6 & 1.Rd8-f8 Qd6-g6# (2024-03-19)
comment

Paulo Roque: obs: 23...Kd7 24.T8h3 f6 (2008-12-29)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 184:20:33 Stunden. (hh:mm:ss)
1.d4 Sa6 2.Ld2 Sb4 3.e3 Sf6 4.Ld3 c5 5.Ke2 c4 6.h4 c3 7.h5 cxd2 8.c4 Sc2 9.c5 Sg4
10.c6 e5 11.c7 Lc5 12.dxc5 d6 13.c6 Le6 14.c8=S Dg5 15.Se7 Sh2 16.Sf5 Sf1 17.c7 a5
18.c8=S a4 19.Sce7 a3 20.Sg6 axb2 21.a4 hxg6 22.a5 Th6 23.a6 gxf5 24.a7 Tg6
25.h6 Kd7 26.h7 f6 27.h8=T Sa3 28.T8h3 Th8 29.a8=T Thh6 30.Ta4 Lg8 31.Te4 fxe4
32.f4 Dg4+ 33.Tf3 exf3+
Keine Lösung: BP 32.0 wegen der Retraktion. Da NUPG C+. (2023-09-28)
comment

Branko Koludrovic: P.S.
The black pawn a4(on the diagramm)came from c7 after capturing the white bishop on b6, then moving to b5 and capturing a white officer on a4. (2010-09-28)
A.Buchanan: I think sBf must have captured on g2 & then promoted on f1 to S, otherwise it’s impossible to unlock the southeast cage (2023-07-11)
A.Buchanan: So that means it’s sound. However I don’t see that it needs to have been sL shielding on f1 (2023-07-11)
A.Buchanan: Can sBf capture to e-file to be captured directly by wBf rather than capture to g2 to promote? If so, then sBf could capture L on a dark square, meaning that last move could have been c6xb5 (2023-07-11, hidden) more ...
comment

Korrektur 1964, Seite 155: sBh3->f3
Henrik Juel: C+ Popeye 4.61 (2022-11-24)
comment

In 'Mattbilder eines Lebens' abgedruckt mit sTh7 statt h8 und der Quellenangabe: Europe Echecs, 1964
AB: (1) Where is wK?
(2) Why is 1.fxe3ep legal in b) but not a)? (2002-01-31)
Henrik Juel: wK is probably on g2. In part a) last move could have been Kh3-g2, I think (2002-02-01)
A.Buchanan: Very convincing, Henrik. I've repaired the diagram accordingly. (2023-05-28)
comment

SCHRECKE: NL: 1. Dg5,Kh4 gxf3 2. Kh4,Dg5 Sf1# (2023-09-13)
Ladislav Packa: Retro content is not needed, the solution is preserved even without it.
Pg4 Pg2 Sh2 Ke1 Rh1 (5)- Ph5 Kg3 (2) h#2* C+

1...0-0 2.h5-h4 Rf1-f3 #
1.Kg3-h4 Ke1-f2 2.h5*g4 Sh2-f3 # (2023-09-14)
comment

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 ...
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

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)
S N Ravi Shankar: The problem appears to be sound. 1.Ke2xLf1 2.Kd3xLe2 3.Kc4xLd3 4.Kb5xLc4 5.Kb6-b5! followed by 6.Sa6xSb8 and now 1.Sc7#. (2024-02-18)
A.Buchanan: WinChloe also gives the same composer. A curious situation, but at least the problem attributed to you appears to be sound! (2024-02-19)
Adrian Storisteanu: Above the "diagrammes" diagram there is: R. SHANKAR, Inde. In the volume index, under RETROS: this is the only problem of a composer named (just) "SHANKAR" (one original), while "RAVI SHANKAR" is a different author (with two originals). (2024-02-20)
A.Buchanan: I suggest that S.N.Ravishankar check the reaponse to the query (A='Shankar' or a='Ravishankar') AND NOT A='Shankar Ram, Narajan' as there are many other similar retros which are attributed to Ravishankar. If he can state here which ones are not his, I will happily modify the data. (2024-02-21)
A.Buchanan: I suggest that S.N.Ravi Shankar check the answers to (A='shankar' or a='Ravishankar') AND NOT A='Shankar Ram, Narajan' as there are many other retros which are attributed to ravishankar (2024-02-20, hidden) more ...
comment

Henrik Juel: The key R: 1.0-0 'threatens' with white retrostalemate, even though White seems to have many pawn retractions available
All missing men were captured by pawns (and White promoted on h8)
R: 1... Kd7-c8? 2.d5xLc6+ Tb8-d8 3.b4-b5 Ke8-d7 4.e4xd5 Ld7-c6 5.f3xe4 Lc8-d7 and now White is retrostalemate
not 6.g2xf3 because of [Lf1]
not 6.d2-d3 because of [Lc1]
not 6.a3xLb4 because of [Ta1]
and not 6.b3-b4 because [Lf8] was captured on a dark square
R: 1... 0-0-0 handles Td8, but it also fixes the black king, so Dd4 must retract to d8 before d7xTf6 can be retracted, but there is still just time enough:
R: 2.b4-b5 Dd8-d4 3.d5xLc6 Ld7-c6 4.e4xd5 Lc8-d7 5.f3xe4 d7xTe6
The rest is easy: Retract Te6 to a1, a3xLb4, Lb4 to f8, g7xLf6, Lf6 to c1, d2-d3, and now the road towards h7 is free for Pc2 (2023-04-08)
A.Buchanan: Great! So which Typ is this? (2023-04-08)
Henrik Juel: Any type, there are no uncaptures in the solution, so anything goes (2023-04-09)
A.Buchanan: Ok I see - the sequence of retro moves is not VRZ play, but history of the game. After the key, there is no choice for either player until wPe4xd5. But isn’t there some Typ where black can checkmate too? R: 1. 0-0 0-0-0 then c1=D#! (2023-04-09)
Henrik Juel: A black checkmate is a possibility in the tries of defensive retractors, regardless of type
When Black has completed a retraction, he has the right to mate White with a forward move, if this is possible
I should have added the testing of mating in my general story about Høeg retractors
1. White chooses a man and moves it back
2. Black chooses which man (if any) to supplement on the abandoned square
Now the white retraction is completed, and White may mate with a forward move, if this is possible
If so, a solution has been found
If not
3. Black chooses a man and moves it back
4. White chooses which man (if any) to supplement on the abandoned square
Now the black retraction is completed, and Black may mate with a forward move, if this is possible
If so, a try has been found
If not, go to step 1. (2023-04-09)
A.Buchanan: Thanks - so does that mean that the solution given here is just a try? (2023-04-09)
Henrik Juel: No, Andrew, in the solution given here White mates, so it is a solution
A try requires Black to mate (2023-04-09)
A.Buchanan: Sorry I am apparently being slow: isn't R: 1. 0-0 0-0-0, dann c1=D#! a mate for Black, so White never gets to retract further? (2023-04-09)
Henrik Juel: You are not slow, Andrew, but I never really saw the black mate in your first 04-09 comment, sorry
R: 1.0-0 0-0-0 2.b4-b5, then 1.Th8# is the intended solution (and not an intended try), but it fails because following R: 1.0-0 0-0-0, Black mates with 1.c1=D#, so you have cooked the problem
It is easily repaired by adding the condition 'Ohne Vorwärtsverteidigung' (without forward defense), but maybe the author implied this condition (or maybe he never saw your black mate) (2023-04-09)
comment

Henrik Juel: 1.0-0-0 (2.Txd7#). -1... Lh2 -2.Lg1, Sh5 uncaptures wPa6 and unpromotes on h1, then retract h2xg3, Lf2 via f4 to c1, d2xe3, Lg1 to f8, g7xf6 etc. (2004-01-12)
Henrik Juel: C+ Popeye 4.61 (2023-07-14)
comment

A.Buchanan: A key feature of adversarial A Posteriori is that any castling must be forced in a finite number of moves (but not necessarily limited by the number of moves in the stipulation goal). If the other side can prevaricate indefinitely, then that is sufficient to defeat the A Posteriori "steal" (2022-02-16)
A.Buchanan: Why this would be "PRA"? Maybe the idea is that we don't know who is first to move, yet whoever it is, White wins. But that only applies to "pull" scenarios such as this, where Black snatches the move because otherwise the game is lost. In other situations where White to avoid loss must "push" the move, then there is no way this can be described as PRA. The fundamental push/pull thing has a unity, and I don't think it's helpful to use "PRA" which only describes half of this, and was really designed for a different context. Strategically, these push/pull adversarial battles are amongst the most interesting AP problems. (2023-07-22)
comment

Henrik Juel: -1... Tc1xSd1 -2.Sc3 Tc2xLc1 etc.
The other black captures were f7xLg6 and g7xSh6.
White captured c2xPb3xPa4, g2xLf3, and h2xSg3xSf4xPe5xPd6xPc7.
Unusually easy for a Ceriani retro.
The record in this genre may be P1000077 ; does anyone know of a retro with less than 19 men in the diagram position and the stipulation 'What were the captures?' ? (2012-07-13)
James Malcom: What is the full solution? (2023-05-30)
Henrik Juel: The 12 captures are given in the first three lines of my 2012 comment
The further retroplay is easy and of little interest, I believe
James, you might want to change the initial retroplay to
R: 1. ... Tc1xSd1+ 2. Sc3-d1 Tc2xLc1+ etc.
(with uncheck plusses added) (2023-05-30)
comment

In "32 personaggi e 1 autore" (p.533) fälschlich ohne wBb7 abgedruckt (aber mit korrekter Steinkontrolle 10+11). Die Korrektur ist nachzulesen in "La genesi delle posizioni" (p.217).
Henrik Juel: The task of changing who has the move can be accomplished in 15 black moves:
1.La7 Kb2 .. 6.Lb8 Kb7 7.La7 Kc7 8.Lb8+ Kc8 9.La7 Kb7 .. 15.La7 Ka1 16.Lb8,
or, if it is Black to move:
1... Kb2 2.La7 Ka3 .. 6.La7 Kb7 7.Lb8 Kc8 8.La7 Kc7 9.Lb8+ Kb7 .. 15.Lb8 Ka1 (2012-07-26)
Mario Richter: Since the wL can lose a tempo on b6-d8, the OR can be done even faster:
WTM: 1.La7 Sb7 2.Lb6 Kb2 3.Ld8 Ka1 4.Lc7 Sd8 5.Lb8
BTM: 1. ... Sb7 2.La7 Sd8 3.Lb6 Sb7 4.Ld8 Kb2 5.Lc7 Sd8 6.Lb8 Ka1
My guess: diagram error. (2012-07-26)
TBr: Indeed, Mario, a diagram error just in "32 personaggi e 1 autore" (p.533), where the diagram is given as here -- but with piece count 10+11!
Correction in "La genesi delle posizioni" (p.217): Add white Pawn b7. (2012-07-26)
Henrik Juel: With a white pawn added on b7, the play by black king is a1-b2-a3-a4-b5 followed by a round trip, the orientation of which depends on who has the move, like in P0005205 (2012-07-26)
A.Buchanan: For an article I am writing, does anyone have any pithy quotes by Ceriani about ortho-reconstruction, preferably in Italian, please? (2023-08-10)
comment

Moldenhauer: Computerprüfung: Stelvio 1.11 C+ KBP 7.0, cooked in 1 Sekunde.
Keine Lösung: vor BP 7.0.
Beispiel: 1.Sf3 Sf6 2.Se5 h5 3.g4 Sxg4 4.Lh3 S+h2 5.Txh2 Th6
6.Le6 Tg6 7.Tg2 Txg2 #1. (2023-03-22)
comment

: ... Und jedem Anfang wohnt ein Zauber inne, der uns beschützt und der uns hilft zu leben. ... (2003-02-22)
Joost de Heer: This is C+ by Stelvio: 1 exact solution (starting with Sf3) and 1 dualistic cook (starting with 1. d4) (2023-08-02)
comment

in '64 Schach-Scherze' nachgedruckt mit der Forderung "Matt in gar keinem Zug"

vgl. P1309484
Erich Bartel: weiterer Nachdruck (oder Erstquelle?!):
3) 28) 64 Schachscherze 1916 (in dieser Quelle ist kein
Hinweis ob Urdruck oder Nachdruck) (2007-10-30)
Alain Brobecker: Same position and stipulation as P1265678, except the latter one is dated 1910. (2022-11-15)
comment

Konstruktionspreisausschreiben 'Die Schwalbe' 06/1973 Heft 21 S. 54,
Thema I (Dr. W. Dittmann):
Konstruiere, ausgehend von der Partieanfangsstellung, durch Versetzung von höchstens vier beliebigen Steinen eine partiemögliche Stellung mit Weiß am Zuge, deren kürzeste Beweispartie möglichst lang ist. Als Steinversetzung gilt die Postierung eines Steins auf einem anderen Feld oder auch die Entfernung eines Steins vom Brett. Gewertet wird nach der (möglichst hohen) Anzahl der Züge, die erforderlich sind, um die eingesandte Stellung von der Grundstellung aus zu erspielen.
Moldenhauer: Computerprüfung: C+ NUPG cooked Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 28.5, BP 29.0.
Notation: 1.Sa3 Sa6 2.Sb5 Tb8 3.Sd4 Ta8 4.Se6 Tb8 5.Sxf8 Ta8 6.Se6 Tb8 7.Sc5 Kf8 8.Sa4 Ke8 9.Sb6 Kf8 10.Sxc8 Ke8 11.Sb6 Kf8 12.Sa4 Ke8 13.Sc3 Kf8 14.b4 Ke8 15.La3 Kf8 16.Db1 Ke8 17.Kd1 Kf8 18.Kc1 Ke8 19.Kb2 Kf8 20.Dd1 Ke8 21.Sb1 Kf8 22.Kc3 Ke8 23.Kd3 Kf8 24.Ke3 Ke8 25.Kf3 Kf8 26.Kg3 Ke8 27.Kh3 Tc8 28.Lc1 Ta8 29.b5 Sb8 30.b6 (2023-04-09)
A.Buchanan: In order to (H)C+ a non-unique proof game, one should show there is no short solution, but also that all solutions of regular length exhibit the intended theme, whatever that was. As long as any strategy can be checked for compliance, Stelvio fine because one can set the parameter for Stelvio to make sure that all strategies are considered. But it’s not possible currently to view all transpositions of a strategy, if that matters (2023-04-09)
Moldenhauer: Man müsste natürlich dieses Konstruktionsausschreiben gelesen haben was hier die Vorgabe war.
Die Stellung an sich oder der Königsmarsch oder dass die Damen und Läufer die Ursprungsposition
wieder einnehmen, usw. Stelvio wird das erreichen der Stellung analysieren so wie Euclide und Natch.
Die Bewertung auf C+ überlasse ich auch den Experten. Da fehlt mir Wissen und jahrelange Erfahrung.
Deshalb nur Kommentare. Gibt es dieses Ausschreiben von 06/1973? (2023-04-27)
Mario Richter: Thema ergänzt - hilft das bei der Bewertung? (2023-04-28)
comment

Konstruktionsausschreiben 06/1973
Moldenhauer: Computerprüfung: C+ NUPG Stelvio 1.11 00:07:09 Minuten. (hh:mm:ss)
Keine Lösung: BP 28.5, BP 29.0.
Beispiel: 1.Sa3 Sa6 2.Sb5 Tb8 3.Sd4 Ta8 4.Sc6 Tb8 5.Sxb8 Sb4 6.Sc6 Sa6
7.Se5 Sb4 8.Sg6 Sa6 9.Sxh8 Sb4 10.Sg6 Sa6 11.Se5 Sb4 12.Sc4 Sa6 13.b4 Sb8
14.La3 Sa6 15.Db1 Sb8 16.Kd1 Sa6 17.Kc1 Sb8 18.Kb2 Sa6 19.Kc3 Sb8
20.Kd3 Sa6 21.Ke3 Sb8 22.Kf3 Sa6 23.Kg3 Sb8 24.Kh3 Sa6 25.Dd1 Sb8
26.Lc1 Sa6 27.Sa3 Sb8 28.Sb1 Sa6 29.b5 Sb8 30.b6 (2023-04-24)
comment

Konstruktionsausschreiben 06/1973
Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 4 Sekunden.
Keine Lösung: BP 27.5, BP 28.0, BP 29.0. BP 28.5 cooked.
Beispiel BP 29.5: 1.Sa3 Sa6 2.Sb5 Tb8 3.Sd4 Ta8 4.Sc6 Tb8 5.Sxd8 Ta8 6.Sc6 Tb8 7.Sa5 Kd8 8.Sc4 Ke8
9.b4 Kd8 10.La3 Ke8 11.Db1 Kd8 12.Kd1 Ke8 13.Kc1 Kd8 14.Kb2 Ke8 15.Kc3 Kd8 16.Kd3 Ke8 17.Ke3 Kd8
18.Kf3 Ke8 19.Kg3 Kd8 20.Dd1 Ke8 21.Lc1 Kd8 22.Sa3 Ke8 23.Sb1 Kd8 24.b5 Ke8 25.b6 Kd8 26.h4 Ke8
27.Kh2 Ta8 28.h5 Sb8 29.h6
Beispiel BP 28.5: 1.Sa3 Sa6 2.Sb5 Tb8 3.Sd4 Ta8 4.Sc6 Tb8 5.Sxd8 Ta8 6.Sc6 Tb8 7.Sa5 Kd8 8.Sc4 Ke8
9.b4 Kd8 10.La3 Ke8 11.Db1 Kd8 12.Kd1 Ke8 13.Kc1 Kd8 14.Kb2 Ke8 15.Kc3 Kd8 16.Kd3 Ke8 17.Ke3 Kd8
18.Kf3 Ke8 19.Kg3 Kd8 20.Dd1 Ke8 21.Lc1 Kd8 22.Sa3 Ke8 23.Sb1 Kd8 24.b5 Ke8 25.b6 Kd8 26.h4 Ke8
27.h5 Ta8 28.Kh2 Sb8 29.h6
Nach Konstruktionsauschreiben glaube ich wäre BP 28.5 richtig. (2023-05-12)
James Malcom: Fixed. (2023-05-13)
Henrik Juel: What was asked for in this construction tourney? (2023-05-13)
Moldenhauer: Bitte siehe P0005590, da gings schon mal darum. (2023-05-13)
comment

Konstruktionsausschreiben 06/1973
Moldenhauer: Computerprüfung: C+ NUPG BP 28.5 cooked in 1 Sekunde.
Keine Lösung: BP 26.5 bis BP 28.0.
Beispiel: 1.Sc3 Sa6 2.Sd5 Tb8 3.b4 Ta8 4.La3 Tb8 5.Db1 Ta8 6.Kd1 Tb8 7.Kc1 Ta8
8.Kb2 Tb8 9.Kc3 Ta8 10.Kd3 Tb8 11.Ke3 Ta8 12.Kf3 Tb8 13.Kg3 Ta8
14.Kh3 Tb8 15.b5 Ta8 16.b6 Sb8 17.Db5 Sc6 18.Da6 bxa6 19.b7 Sh6
20.b8D Tg8 21.Db1 Lb7 22.Dd1 Dc8 23.Sf6+ Kd8 24.Sxg8 Sb8 25.Sf6 Sg8
26.Sd5 Ke8 27.Lc1 Dd8 28.Sc3 Lc8 29.Sb1 (2023-04-02)
comment

Frank Müller: Originalforderung: In wieviel Zügen läßt sich vorstehende Position aus normaler Anfangsstellung entwickeln?
Eine Zugzahl war nicht angegeben! (2012-08-11)
Moldenhauer: Computerprüfung: Cooked Stelvio 2.0 in 1 Sek.
Keine Lösung: BP 17.0, BP 17.5.
C+ wegen der Forderung.

1.Sa3 Sa6 2.Sb5 Sc5 3.Sd4 Se4 4.Se6 Sg3 5.Sxf8 Sxh1 6.Sg6 Sg3 7.Sxh8 Sxf1
8.Sg6 Se3 9.Se5 Sf5 10.Sc4 Sd4 11.Sb6 Sb3 12.Sxa8 Sxa1 13.Sb6 Sb3
14.Sxc8 Sxc1 15.Sb6 Sb3 16.Sa4 Sd4 17.Sc3 Sc6 18.Sb1 Sb8 (2024-01-09)
A.Buchanan: If the composer makes no claim that the solution is unique, it’s unfair to call it “cooked” when multiple solutions are unearthed. “Cooked” means “Kaput”, “broken”. Please stop. (2024-01-09)
comment

Juha Saukkola:: Solution not unique! (2000-09-29)
Moldenhauer: Computerprüfung: C+ NUPG Stelvio 1.11 00:02:12 Minuten. (hh:mm:ss)
Keine Lösung: BP 29.5, BP 30.0. (2023-04-24)
Moldenhauer: Beispiel Stelvio 1.11: 1.Sc3 Sc6 2.b3 Sf6 3.La3 a5 4.g4 a4 5.Lg2 Ta5 6.h4 Tf5
7.Lc5 axb3 8.Ld5 b6 9.a4 bxc5 10.gxf5 g5 11.hxg5 e5 12.Th4 h5 13.Tc4 h4
14.a5 h3 15.a6 h2 16.a7 h1D 17.Ta6 De4 18.Sf3 Le7 19.Sd4 exd4 20.a8D Se5
21.Te6 dxc3 22.Dc6 dxc6 23.d4 fxe6 24.dxe5 exd5 25.gxf6 Le6
26.fxe6 dxc4 27.fxe7 Ddd3 28.cxd3 Th3 29.Dc2 Te3 30.dxe4 bxc2 31.fxe3 (2023-04-24)
comment

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)
comment

Bernd Schwarzkopf: zurück: 1.Kg4xBg3 Kd3-c2 2.Kg5-g4 Ke4-d3 3.Kh5xBg5 Kf5-e4; vor: 1.g3-g4# (2024-03-29)
comment

Henrik Juel: minor duals 0... hxg1=L 1.Sxf1+,Lh5 and 0.Kxg4 1.g8=DT+ (2022-08-27)
hans: black to move:
0. ... h1=D/T 1. Lh5! (2. Sg5#)
0. ... h1=L 1. g8=L! Kxg4 2. Le6#
0. ... h1=S 1. Lh5! (2. Sg5#) Sxg3 2. Sf4#
0. ... hxg1=D/T/L/S 1. Sfxg1+ Kxg4 2. g8=D/T#
0. ... Kxg4 1. g8=D/T+ Kh3/Kf5 2. Sf4/Sd2# (2015-08-19, hidden) more ...
comment

paul: Verified by Jacobi with this input:

stip dia 6
forsyth rnbqkbnr/pppp1ppp/8/8/4K3/5Q2/PPPP1PPP/RN3BNR
cond cavaliermaj (2023-06-16)
Mu-Tsun Tsai: Wait, 4.Ke2 walks into check by nightrider b8, isn't it? (2012-07-15, hidden)
Henrik Juel: You are right, Mu-Tsun, but this seems to work:
1.Nxe7 Nxe2 2.Nb1 Nxc1 3.Df3 Nca5 4.Ke2 Ngc6+ 5.Kd3 Nb8 6.Ke4 Ng8+
where the black nightriders switch places (2012-07-15, hidden)
Mu-Tsun Tsai: Yeah, I came up with the same thing. Computer check is still running to see if that's the only one. (2012-07-15, hidden)
Mu-Tsun Tsai: Hum... Probably have to give up computer check. The estimated run time is simply ridiculous. I'll just believe that this corrected solution is unique. (2012-07-15, hidden) more ...
comment

paul: Add wPa2b2b3e2, bPa7c7e7g7. (2022-09-18)
comment

SCHRECKE: In a) geschah zuletzt 0. ... Td8-a8, also ist die Rochade nicht mehr möglich.
In b) kam zuletzt die sD von h1, schwarze Rochade also noch spielbar.
Lösung:
a) 1. gxh5 Td8 2. Df7+ Kxf7#
b) 1. Dg6+ Dxg6 2. Sf5 0-0-0# (2023-11-07)
Joost de Heer: Slight error in previous comment: Last move was Td8xT/Da8, not Td8-a8 (as Td8-d1 would be longer) (2023-11-07)
comment

Olaf Jenkner: Wieso wird Weiß pattgesetzt? (2010-03-02)
Mario Richter: Weil der Komponist das so wollte :-)
In fs33 ist die Forderung wie folgt wiedergegeben: "Wei� nimmt einen Zug zurück und hilft, da� Schwarz sofort pattsetzen kann." (2010-03-03)
Olaf Jenkner: Ach so, h=1 bedeutet eigentlich, daß Schwarz beginnt. (2010-03-03)
Henrik Juel: Yes, it is a bit confusing; but for retros it is an old tradition that you should not interpret one move stipulations too rigorously. (2010-03-05)
Mario Richter: But at least retro play and forward play harmonize: the side that retracted last starts the forward play. (2010-03-05)
Adrian Storisteanu: Can easily make it conform to today's accepted standards (i.e., drop the half-duplex indication):

Ka1 pa2 / Kc1 pb3 (2+2)
-1b & h=1

- 1.a4xQb3 & 1.a4-a3 Qb3-d3= (2024-01-13)
Olaf Jenkner: Das wären dann viele Mütter. (2024-01-14)
A.Buchanan: @Olaf :-) (2024-01-15)
Adrian Storisteanu: :-) Just in reverse Polish notation...
(Didn't know the many fathers thing started that long ago. I totally missed both the obvious setting and the recorded keyword.) (2024-01-15)
A.Buchanan: The Originalforderung is quite clear. There is no need to represent this stipulation as a joke here. Popeye uses the unambiguous term "half-duplex" for this situation, which occurs in other help-retractors. (2024-01-12, hidden) more ...
comment

Henrik Juel: solution should be Kg1xSh2, I think (2023-05-14, hidden) more ...
comment

A.Buchanan: What is MsP, please? (2024-01-12)
Henrik Juel: Mate instead of stalemate (Mat statt Pat)
In the diagram position Black is stalemated; White retracts and mates (2024-01-12)
comment

Erich Bartel: Auszeichnung: 1.ehrende Erwähnung.--
Nachdruck: 1) Jaarboek 1970.-- (2006-11-26)
Henrik Juel: Something is wrong: De5 does not mate. And the intention certainly was not -1.S=a7 L=h2, 0... Kxc5 1.h8DL Kb6 2.DLd4#. (2006-11-27)
Erich Bartel: Ich habe die Aufgabe aus dem Jaarboek 1970 übernommen, wo
sie wie angegeben im Diagramm und Lösung so steht. Es
scheint also bereits hier ein Fehler vorzuliegen. Zur Klärung
ist es wohl notwendig in der Ur-Quelle Probleemblad -die ich
nicht besitze- nachzusehen. (2006-11-27)
Siegfried Hornecker: No matter how the position is changed, there can be no line where the king, coming from b6, is not allowed to return to b6 when the queen checks on e5. Also, there would be no reason to not play again 2.h2-h1~ as a waiting move.

The retroanalytics say that the last move must(!) have been 1.a7-a8S+ h2-h1L so I think the correct stipulation is:
a) Last move (w+s)
b) h#2

I have no source for this but if you think about it, the whole thing makes sense! The last move was 1.a8=S+ after 1.h1=L and from the diagram position the h#2 runs 2. Kxc5 h8=D 3. a1=T De5#. That's what I think. (2006-12-12)
Jan Hein Verduin: I got the Jaarboek too, and mr. Hornecker is right. The stipulation there is "h#2; what where the last w+b moves?" The way the solution then is given (the retro moves are given first) probably caused the confusion. (2007-01-26)
Adrian Storisteanu: Then, ceci n'est pas un ... retractor!? (2024-03-17)
Henrik Juel: Right, Adrian
The Help retractor keyword should be deleted (2024-03-17)
comment

Henrik Juel: 1.Sxb2#. -1... a3xDb2 -2.Sa2 b4xLa3 -3.c4 c5xSb4 -4.Da1 d6xTc5 -5.Lc1 Ka3 -6.Tb2 Ka4 -7.Lb1 Ka3 etc. All-uncapture. (2004-01-02)
James Malcom: Shortest possible proof game: 1. c4 c6 2. a4 Qc7 3. h4 Qh2 4. Nh3 Kd8 5. Ra2 Kc7 6. e4 Kb6 7. Nc3 Kc5 8. a5 Qg3 9. b4+ Kd4 10. b5 Kc5 11. Qa4 Qd3 12. g4 Qc2 13. Rb2 Qb3 14. Qa1 Kb4 15. Na2+ Ka3 16. Ke2 Qd1+ 17. Ke3 Qf3+ 18. Kd4 Na6 19. Bd3 Nc5 20. Bb1 Nd3 21. Nf4 Ne5 22. Kc5 Qh3 23. Nd3 a6 24. f4 Nf3 25. Kb6 Ng5 26. hxg5 d6 27. e5 Bd7 28. e6 Re8 29. exd7 e6 30. f5 Be7 31. f6 Rc8 32. fxe7 Nf6 33. e8=N Ng8 34. d8=B Qh4 35. Bf6 Qh5 36. Bc3 Qh6 37. Nf6 Qg6 38. Nd5 Nf6 39. gxf6 Rce8 40. N5b4 Re7 41. fxe7 Rf8 42. e8=R Rg8 43. Rd8 Rf8 44. Rh5 Re8 45. Rc5 Re7 46. g5 Qf6 47. gxf6 Ka4 48. fxe7 Ka3 49. e8=Q Ka4 50. Rd7 Ka3 51. Qe7 Ka4 52. Bc2+ Ka3 53. Rb3+ Ka4 54. Ba3 dxc5 55. Qb2 cxb4 56. c5 bxa3 57. Nab4 axb2 (2023-02-05)
comment

Henrik Juel: Following R: 76.h3xLg4, the resolution could continue with
Retract wKc8 to b5, sLg4 to c8, b7-b6, sSa8 to g8, wSa4 to g1, wKb5 to e1, sLb8 to a5, b6xTc5, wTc5 to d4, sKa1 to b5, a2xTb3xDc4+, etc. (2020-08-13)
Henrik Juel: A little analysis may be in order
Pawns captured all missing men
Kb3 is now confined to the SW corner, but if we retract b3xc4 he is locked in for good, so [Lc8] was captured on g4
Before we retract h3xLg4, sLg4 to c8, and b7-b6, wKf8 must be liberated
The plan for this is: Retract wKg8-f8, sTg1 to f8, sTe7 out, wTf8 out past the sT, wKg8 to e7, sT to h8, wT to e8, and further as per my old comment (2022-01-22)
James Malcom: Corrected SPG: 1. Nh3 e6 2. Nc3 Qg5 3. Rg1 Qe3 4. dxe3 Bb4 5. Qd4 Nc6 6. Bd2 Na5 7. Kd1 Nb3 8.
axb3 Nf6 9. Ra6 Nd5 10. Rb6 axb6 11. Kc1 Ra4 12. Kb1 Ba5 13. Nd1 Rc4 14. Ka1 Ke7
15. Nf4 Kd6 16. Bb4+ Kc6 17. h3 Kb5 18. bxc4+ Ka4 19. c3 Kb3 20. Qc5 Kc2 21. Ka2
bxc5 22. Ka3 Bb6 23. Ka4 Ba7 24. Kb5 Nb6 25. Nd5 Na8 26. Nb6 Re8 27. Na4 b6 28.
Rh1 Bb7 29. Rh2 Bf3 30. Ka6 Bh5 31. Kb7 Re7 32. Kc8 Bg4 33. hxg4 Kb3 34. Rh5 Kc2
35. Rd5 Kb3 36. Rd2 Ka2 37. Rc2 Kb3 38. Rc1 Ka2 39. Ra1+ Kb3 40. Ra3+ Kc2 41.
Rb3 Bb8 42. Ba3 Ba7 43. Rb5 Bb8 44. Ra5 Kb3 45. Ra7 Kc2 46. Rb7 Ba7 47. Rb8 Re8+
48. Kb7 Rg8 49. Rf8 Kd2 50. Kc8 Ke1 51. Kd8 Rh8 52. Ke7 Rg8 53. Rb8 Rc8 54. Rb7
Bb8 55. Ra7 Rd8 56. Ra5 Ba7 57. Rb5 Rb8 58. Rb3 Rb7 59. Bb4 Bb8 60. Ra3 Ra7 61.
Ra1 Ra5 62. Rc1 Rb5 63. Ba5 Rb3 64. Rc2 Ra3 65. Rd2 Ra1 66. Rd6 Rc1 67. Kf8 Rc2
68. Kg8 Rd2 69. Rd5 Rd4 70. Rd6 Re4 71. Rd2 Rd4 72. Rc2 Rd2 73. Rc1 Rc2 74. Ra1
Rc1 75. Ra3 Ra1 76. Rb3 Ra3 77. Rb5 Rb3 78. Bb4 Kd2 79. Ra5 Kc2 80. Ba3 Rb5 81.
Ra7 Ra5 82. Rb7 Ba7 83. Rb8 Kb3 84. Rd8 Bb8 85. Re8 Ra7 86. Kh8 Rb7 87. Rf8 Ba7
88. Re8 Rb8 89. Rf8 Re8 90. Kg8 Re7 91. Rb8 Kc2 92. Rb7 Bb8 93. Ra7 Kb3 94. Ra5
Kc2 95. Rb5 Kb1 96. Rb3 Kc1 97. Bb4 Kb1 98. Ra3 Kc2 99. Ra1 Kb3 100. Rc1 Ba7
101. Rc2 Ka2 102. Rd2 Kb1 103. Rd5 Kc2 104. Rh5 Kc1 105. Rh1 Kb1 106. Rg1 Kc2
107. Ba3 Kb3 108. Kf8 (2023-03-11)
AV: On part de la position Rb5, Th1, Fa3 et f1, Ca4 et d1, Pb2, c3, c4, e2, e3, f2, g2 et h2 (14) / Rb3, Th8, Fa7 et c8, Ca8, Pb7, c5, c7, d7, e6, f7, g7 et h7 (13). Et l'on joue :

1...b6 2.h3 Fb7 3.Fb4 Ff3 4.Fa5! Fh5 5.Ra6 Te8 6.Rb7 Te7 7.Rc8 Fg4 8.hxg4 Début du décompte. 8...Rc2 9.Th5 Rb3 10.Td5 Rc2 11.Td2+ Rb3 12.Tc2 Ra2 13.Tc1 Rb3 14.Ta1 Rc2 15.Ta3 Rc1 16.Tb3 Rc2 17.Tb5 Rc1 18.Fb4 Rc2 19.Ta5 Fb8 20.Ta7 Rb3 21.Tb7 Fa7 22.Tb8 Te8+ 23.Rb7 Tg8 24.Tf8 [24.Te8] 24...Rc2 25.Rc8 Rc1 26.Rd8 Rc2 27.Re7 Rc1 28.Tb8 Rd2! [28...Rb1?! 29.Tb7 Fb8 30.Ta7 Rc2 31.Ta5 Fa7 32.Tb5 Tb8 33.Fa5 Tb7 34.Tb3 Fb8 35.Ta3 perd un coup] 29.Tb7 Fb8 30.Ta7 Re1! 31.Ta5 Fa7 32.Tb5 Tb8 33.Fa5 Tb7 34.Tb3 Fb8 35.Ta3 Ta7 36.Fb4 Ta5 37.Ta1 Tb5 38.Fa5 Tb3 39.Tc1 Ta3 40.Tc2 Ta1 41.Td2 Tc1 42.Td4 Tc2 43.Te4 Td2 44.Tf4 Td5 45.Td4 Td6 46.Td2 Td5 47.Tc2 Td2 48.Tc1 Tc2 49.Ta1 Tc1 50.Ta3 Ta1 51.Tb3 Ta3 52.Tb5 Tb3 53.Fb4 Rd2 54.Fa3 Rc2 55.Ta5 Tb5 56.Ta7 Ta5 57.Tb7 Fa7 58.Tb8 Rb3 59.Td8 Fb8 60.Re8 Ta7 61.Rf8 Tb7 62.Rg8 Fa7 63.Tf8 Tb8 64.Rh8 Te8 65.Rg8 Te7 66.Tb8 Rc2 67.Tb7 Fb8 68.Ta7 Rc1 69.Ta5 Fa7 70.Tb5 Rb1 71.Tb3 Rc2 72.Fb4 Rc1 73.Ta3 Rc2 74.Ta1 Rb3 75.Tc1 Ra2 76.Tc2 Rb3 77.Td2 Ra2 78.Td5 Rb3 79.Th5 Ra2 80.Th1 Rb3 81.Tg1 Ra2 82.Fa3 Rb3 83.Rf8 . Donc 75 coups sans prise, roque ni mouvement de pion. (2005-06-25, hidden)
Henrik Juel: James, please note
a. AV's comment uses french notation, RDTFC for KQRBS (german KDTLS)
b. His play is forward from the given position (wKb5, wRh1, ...), not backwards from diagram position (2020-08-13, hidden)
James Malcom: I'm aware that it is French, Henrik, and thanks for explaining that it is foward play. I found a full solution in the Retro Mailing list, which I have modified to fit PDB requirements-https://pairlist1.pair.net/pipermail/retros/2004-December/001006.html (2020-08-13, hidden)
Henrik Juel: Very good, James
It is so nice to be able to use the animation in retros with a looong retroplay (2020-08-13, hidden)
James Malcom: Done and done, Henrik. (2020-08-14, hidden)
Henrik Juel: Fine, James, but you made a small but crucial error at the end
R: 108... e7-e6?? is illegal, because the tempo is wrong and there is no way to change it:
109.Sa3-b1 Sc4-e3 110.Sb1-a3 Sa5-c4 111.Sa3-b1 Sc6-a5 112.Sb1-a3 Sb8-c6
The initial array has been reached, but White should have made the last retraction
R: 108... Sc4-e3 or any other black retraction works (2020-08-15, hidden)
James Malcom: SPG: 1. Nc3 e6 2. Nf3 Qg5 3. h3 Qe3 4. dxe3 Nc6 5. Qd4 Na5 6. Qb6 Nb3 7. axb3 axb6 8. Bd2 Ra4 9. O-O-O Ke7 10. Ng5 Kd6 11. Be1+ Kc5 12. Rd6 Rc4 13. Nd1 Kb5 14. Rc6 Bb4 15. bxc4+ Ka4 16. Ne4 Ba5 17. Bb4 Nf6 18. c3 Kb3 19. Nc5+ Ka2 20. Kc2 Ka1 21. Kb3 Nd5 22. Ka4 Re8 23. Kb5 Ka2 24. Na4 Ka1 25. Rc5 bxc5 26. Rh2 Nb6 27. Rh1 Na8 28. Rh2 Bb6 29. Rh1 Ba7 30. Rh2 b6 31. Rh1 Bb7 32. Rh2 Bf3 33. Rh1 Bg4 34. Ka6 Re7 35. Kb7 Ka2 36. Kc8 Bb8 37. Kd8 Ka1 (2023-03-09, hidden)
James Malcom: SPG: 1. Nh3 e6 2. Nc3 Qg5 3. Rg1 Qe3 4. dxe3 Bb4 5. Qd4 Nc6 6. Bd2 Na5 7. Kd1 Nb3 8. axb3 Nf6 9. Ra6 Nd5 10. Rb6 axb6 11. Kc1 Ra4 12. Kb1 Ba5 13. Nd1 Rc4 14. Ka1 Ke7 15. Nf4 Kd6 16. Bb4+ Kc6 17. h3 Kb5 18. bxc4+ Ka4 19. c3 Kb3 20. Qc5 Kc2 21. Ka2 bxc5 22. Ka3 Bb6 23. Ka4 Ba7 24. Kb5 Nb6 25. Nd5 Na8 26. Nb6 Re8 27. Na4 b6 28. Rh1 Bb7 29. Rh2 Bf3 30. Ka6 Bh5 31. Kb7 Re7 32. Kc8 Bg4 33. hxg4 Kb3 34. Rh3 Kc2 35. Rh5 Kb3 36. Rd5 Kc2 37. Rd2+ Kb3 38. Rc2 Ka2 39. Rc1 Kb3 40. Ra1 Kc2 41. Ra3 Kb1 42. Rb3 Kc2 43. Ba3 Kb1 44. Rb5 Kc2 45. Ra5 Bb8 46. Ra7 Kd2 47. Rb7 Ba7 48. Rb8 Re8+ 49. Kb7 Rh8 50. Rf8 Rg8 51. Kc8 Rh8 52. Kd8 Rg8 53. Ke7 Rh8 54. Rb8 Rc8 55. Rb7 Bb8 56. Ra7 Ke1 57. Ra5 Ba7 58. Rb5 Rb8 59. Rb3 Rb7 60. Bb4 Bb8 61. Ra3 Ra7 62. Ra1 Ra5 63. Rc1 Rb5 64. Ba5 Rb3 65. Rc2 Ra3 66. Rd2 Ra1 67. Rd5 Rc1 68. Kf8 Rc2 69. Kg8 Rd2 70. Rd6 Rd4 71. Rd5 Re4 72. Rd2 Rd4 73. Rc2 Rd2 74. Rc1 Rc2 75. Ra1 Rc1 76. Ra3 Ra1 77. Rb3 Ra3 78. Rb5 Rb3 79.Bb4 Kd2 80. Ba3 Kc1 81. Ra5 Rb5 82. Ra7 Ra5 83. Rb7 Ba7 84. Rb8 Ra6 85. Rf8 Bb8 86. Kh8 Ra7 87. Re8 Rb7 88. Rd8 Ba7 89. Rc8 Rb8 90. Rf8 Re8 91. Kg8 Re7 92. Rb8 Kc2 93. Rb7 Bb8 94. Ra7 Kb3 95. Kf8 Kc2 96. Ra5 Ba7 97. Rb5 Kb1 98. Rb3 Kc2 99. Bb4 Kb1 100. Ra3 Kc2 101. Ra1 Kb3 102. Rc1 Ka2 103. Rc2 Ka1 104. Rd2 Ka2 105. Rd5 Kb1 106. Rh5 Ka2 107. Rh1 Kb3 108. Rg1 (2023-03-10, hidden) more ...
comment

"O, W, thuis..."
Henrik Juel: 1.Kxc2 Sb3 2.. Rc1#. -1.. Ra2 -2.a3:B Bf8 -3.Bb4 Kb3 -4.Be7 a7:S -5.Sa4 Qd5
-6.Sc3 Qd4 -7.Sb1 Ra1 -8.Bf6 Kb4 -9.a2 Kc5 -10.Bg5 Kd6 -11.Bf6 Kd7 -12.Be7 Kc8 -13.Bf6 Kb8 -14.Bd8 e7 -15.B=d7 Kc8 -16.e6:R Rd8 -17.f5:B 0-0-0 -18.f4 Bc8 -19.f3 Qd8 -20.f2 d7:B -21.Bf3 Sc6 etc. (2003-08-26)
Yoav Ben-Zvi: An attempt at a coherent argument that reveals the intricate sequence of events in the resolution while refuting substantial alternatives:
The promotion of wBc3, on d8 or f8, requires 2 captures by the promoting wP. These, together with the captures by wPd3 and wPb4, account for all 4 missing Black pieces. Black captured [wBc1] at home. The other 7 missing White pieces were captured by bPs. [wBf1] cannot be released (by wPe2xd3) until [bPh7] plays bPd3xc2 completing the capture of 5 White pieces. Therefore [wBf1] was captured by bPd7xBc6 (the remaining capture by Black on a light colored square), releasing [bBc8]. This means that [wBf1] is released (by wPe2xd3) before [bBc8] is released (by capturing the previously released wB on c6) so the bB could not have been the piece captured on d3. It also could not be captured on b4 (dark colored square) so it was captured by White's promoting pawn. This means the promotion could not have been played by capture from e7 as that would imply that the 2 captures by the promoting pawn were both on dark colored squares in contradiction with the requirement to capture [bBc8]. Therefore the promotion move was wPd7-d8=B with bK standing outside the squares c8,d7,d8,e8.
Looking backward: To avoid White retrostalemate the first 2 retractions must be -1...bRa2-a1+ -2.wPa3xb4.
After this, unlocking the South-West cage and extracting bRa2 requires retraction of wPe2xd3 which requires previous uncapture of wB by bPd7xBc6, so it can be retracted home to f1. bP standing on d7 implies that a bB must previously be uncaptured and retracted home to c8, preceded by retracting BPe7-e6 to clear the path of uncaptured bB to c8. Retreat of bP to e7 is preceded by uncapture and return home of [bBf8]. With both bBs locked at home and bP on e7 the Black Central cage is locked by retracting bPd7xBc6 so bK and bQ must also be retracted home first. It follows that bKa4 needs to exit the West cage before it has been unlocked (to get back home to e8 before the Black Central cage is locked). The bK cannot exit via b5 as this implies retraction of wPc3-c4+ but, with c3 occuppied by wP, bRa2 does not have a valid exit path (not via c3 as it is obstructed by wP or via f1 as it will be occupied by wB). wP standing on c4 prevents retracting bN away from a5 so bK cannot exit via a5. The only valid exit of the bK is via b4 implying retracton of wPa2-a3 which requires bRa2 to retract to a1 (vacating a2 for the wP) which requires provision of a shield, standing on b1, to protect wKc1 from bRa1.
After the unpromotion of wBc3, White has only pawn retractions left to make until a White officer is uncaptured. This is not enough to allow Black to complete the retro-maneuvers required in preparation for retracting bPd7xBc6 (getting bBs,bK,bQ home). It follows that, at some point in the middle of its preparation maneuver, Black will need to uncapture a White officer which provides the tempo retractions needed to complete the maneuver. This is only possible by bPa7xb6. With bP standing on a7, [bRa8] cannot be retracted home via the a file so it must retract to its original corner via row 8 before the bB is locked at c8. This implies that [bRa8] is uncaptured by a wP prior to unlocking of the cage.
Uncaptures by White prior to the unlocking of the South-West cage (one available on b4 and two by pawn unpromoted from wBc3) include uncaptures of both bBs and a bR since, as noted in the above analysis, all 3 must be uncaptured and retracted home before retracting bPd7xBc6 (all of which precedes the unlocking of the South-East cage). This means that the bN is uncaptured later (on d3) so it cannot provide the shield on b1. The shield also could not be provided by [bBc8] as this bishop must be returned home when it is needed to provide the shield. The only remaining possibility is that the shield on b1 is provided by a wN that is uncaptured by bPa7xNb6. The wN on b1 is immobile until the wK can be retracted away from c1 so here too, once the unpromotion occurs Black must perform the preparations for retraction of bPd7xBc6 (which now includes return of the uncaptured bR to a8 or b8) under time pressure, completing the required maneuvers before White runs out of tempo retractions. Having determined the structural elements it is not too difficult to construct a resolution that succeeds in preparing for retraction of bPd7xBc6 before White runs out of tempo retractions. The resolution maintains the following: delay the unpromotion (known from analysis above to be wPd7-d8=B) while retracting bK to b8; bQ to d4 and bB (uncaptured on b4) to f8 and continuing by retracting -1.wBd8-? Pe7-e6 -2.Pd7-d8=B Kc8-b8 -3.Pe6xRd7+ Rd8-d7 -4.Pf5xBe6 0-0-0 -5.Pf4-f5 Bc8-e6 -6.Pf3-f4 Qd8-d4 -7.Pf2-f3 Pd7xBf6 -8.Bc6-? Nc6-a5. Following this the wB is retracted to f1 allowing the unlocking of the cage by wPe2xNd3 and bRa1 is retracted back home to h8 via c1,c3. One of the White pieces uncaptured by bPc2 unpromotes on h8. The alternative of uncapturing the bR on b4 (instead of bB) fails to complete the preparations on time.
To draw attention to the rich, unconditional, content of the resolution the stipulation could be "Squares that must have been occupied by bRa1". This problem deserves a full SPG. The best I can manage is a game of 53.0 moves. (2018-12-23)
Zvi Mendlowitz: Proof game in 44.0 moves:
1. Nf3 Nc6 2. Nh4 Nb8 3. Ng6 hxg6 4. Na3 Rh3 5. c4 Rg3 6. h4 Nf6 7. h5 Nd5 8. h6 Nb4 9. Rh5 N8c6 10. Rf5 Rc3 11. h7 Rxc1 12. h8=Q Rc3 13. Rc1 Rb3 14. Rc3 gxf5 15. Rd3 Rc3 16. Qh4 Rc1 17. Qe4 Ra1 18. Nb1 fxe4 19. Qc2 exd3 20. Kd1 dxc2+ 21. Kc1 Nd3+ 22. exd3 Nb8 23. Be2 Nc6 24. Bf3 Na5 25. Bc6 dxc6 26. f3 Qd4 27. f4 Be6 28. f5 O-O-O 29. fxe6 Rd7 30. exd7+ Kb8 31. d8=B e6 32. Bf6 Kc8 33. Be7 Kd7 34. Bf6 Kd6 35. Bg5 Kc5 36. Bf6 Kb4 37. a3+ Kb3 38. Be7 Ra2 39. Nc3 Qd5 40. Na4 Qd4 41. Nb6 axb6 42. Bb4 Ka4 43. Bc3 Bb4 44. axb4 Ra1+ (2022-09-19)
comment

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)
comment

VL: Published in "Probl. Pribuzh." (or "Problemy Pribusch'ja"?),
1990, No.1, #12. An obscure Russian language chess problem
magazine issued in Nikolaev, the Ukraine (the South Bug
riverside region). (2006-01-27)
A.Buchanan: Lovely problem, but this is no more PRA than it is duplex. Rather, the push/pull of AP Type Keym gives a forfeit which must be discharged. (2023-07-22)
Andrey Zhuravlev: The correct name of this magazine is "Problemist Pribuzhya" (2006-04-10, hidden) more ...
comment

(2 sD)
Hans-Jürgen Manthey: Es sind 3 Damenzüge ! Ein Läufer kommt von b1, darum kann Bb2-b4-b5-b6-b7-b8 nicht geschehen sein. Es gibt die Schlagfälle bxsSc, hxsSg und axsL bei 13 Steinen. Darum verlässt die sD die D-spalte, damit der dBauer zum wS wird, der sich dann auf f6 opfert !
Vorwärts, damit Kurznotation besser lesbar:
1. g3 h5 2. b3 h4 3. b4 Sf6 4. a3 Sg4 5. a4 Sa6 6. a5 Sc5 7. bxSc5 b5 8. h3 c6 9. hxSg4 b4 10. g5 b3 11. g6 La6 12. g4 Lc4 13. g5 Le6 14. Sh3 Dc7(1.) 15. Lb2 Kd8 16. Le5 Kc8 17. Lh2 Kb7 18. Lg1 Ka6 19. Th2 Kb5 20. Tg2 Kc4 21. Sc3 b2 22. Sb5 b1=L 23. Sd4 Kd5 24. d3 La2 25. e3 Ke5 26. Sf4 L2d5 27. c4 h3 28. Ta2 h2 29. Te2 Th3 30. Sh5 Tf3 31. Kd2 h1=D 32. Te1 Dh4(2.) 33. Th2 Tf4 34. Lg2 Lf3 35. Se2 Kf5 36. Lh1 Kg4 37. Sg3 Le4 38. Tg2 Kf3 39. Sf5 L4d5 40. Kc3 Le4 41. Sd6 Lg4 42. Kd4 Lh5 43. Ke5 Lf5 44. d4 Lh3 45. d5 Tg4 46. Kf5 Tb8 47. Se4 Ta8 48. Sf6 gxSf6 49. g7 d6 50. g8=T Lg7 51. g6 Lh6 52. Tb8 Lf8 53. Tb2 Lh6 54. Td2 Lf4 55. Td4 Le5 56. Te4 La1 57. Te5 dxe5 58. d6 Ld4 59. d7 La1 60. d8=S Ld4 61. Se6 Td8 62. g7 Td6 63. Sf4 Te6 64. g8=T La1 65. Td8 Ld4 66. Td5 cxTd5 67. c6 Dd7(3.) 68. c7 La1 69. c8=S Ld4 70. Sd6 La1 71. Se4 Ld4 72. S4g3 La1 73. Sf1 Ld4 74. c5 La1 75. c6 Ld4 76. c7 La1 77. c8=S Ld4 78. Sd6 La1 79. Se4 Ld4 80. S4g3 Lb6 81. axLb6 exSf4 82. b7 a6 83. b8=T fxSg3 84. Sh2+ gxSh2 85. Tb4 a5 86. Te4 dxTe4 87. Tf1 (2021-08-02)
A.Buchanan: Two queen moves is enough, as the animation shows. I agree white captures are hxg,axb,bxa/c. bBh promoted on h1, then moved to h4. bBh2 captured in from the left. Black pawn captures something like cxBdxexfxgxh,gxf,dxe. White abcB all promoted and at least one wB on g8. No need for a third D move as far as I can see. (2021-08-03)
Mario Richter: Hi Andrew - the main question is: How do you get the black king home?
In the solution given, after R: 7. a5xLb6 there is a black pawn on e5.
The position can only be unlocked by retracting wKe5-f5. Where exactly is the mentioned black pawn e5 at the moment wKe5-f5 is retracted? (2021-08-03)
A.Buchanan: Hi Mario, I was just responding to H-J's text assertion that 3 D moves are needed. A PG with 3 D moves could not prove that 2 are insufficient, and I find long unanimated PGs too tedious to read through sorry. I now think that the animation must be wrong, but I haven't convinced myself that no solution exists. I must work, but will have a go later. (2021-08-03)
A.Buchanan: OK - we can make sL on queenside and shift to kingside, and I can see that knights (or bishops) can't reach g8 to retract there, so it must be rooks along the 8th rank. So we can't retract the sDd7, and can't shift sTe6. But few Black tempi then. But this is not my kind of retro, and I will give up here. What's the answer: 2 or 3? (2021-08-03)
Mario Richter: My guess is, that the intended solution was 2 Queen moves, but that it it is impossible to get the black king home: R: sLf8, g7xf6 locks e8 from the right side, R: sDd8-d7, sBd7-d6 locks e8 from the left side (2021-08-03)
A.Buchanan: WinChloe has two earlier versions, and the composer was very experienced (200+ retros just in PDB) but I have now looked enough to think this is cooked. Even if we shift sBa forward 2 squares to give us a couple more moves, then all goes well shifting sL & wT to kingside, but we reach a point where sBc7xwBd6, with still sBd5, so the two pawns are the wrong side of one another. What does Rawbats think? (2021-08-04)
Mario Richter: Perhaps it suffices to say what I think?!
With bPa2 instead of a5 it is easy to reach the position with 2 queen moves only:
1. retract R: 1. Te1-f1 g3xSh2 2. Sf1-h2+
2. retract d6xSe5(or d6xTe5)xSf4xBauer_g3 while W makes tempo moves e.g. with wLg1
3. unpromote the S or Te5 on b8 and retract the wPawn to b6, while bPawn a2 retracts to a6
4. uncapture a5xLb6 - now there is no time pressure
5. unpromote wS on c8 and retract the wPawn to c6
6. retract c7xBauer_d6(!)
[this idea was inspired by the question, how the situation mentioned in your comment "we reach a point where sBc7xwBd6, with still sBd5", can be avoided]
7. retract wPawn_d6 to e.g. d3 (but not d2)
8. retract d5xTe4
9. bring wT to g8, sL to f8
10. retract g7-g8=T Dd8-d7 g6-g7 g7xSf6 (or Tf6)
11. unpromote wS or wT on a8
12. the rest is easy to see

Since the author wanted to show a certain combination of C-F-pieces, the above mentioned choices in the uncapture of white pieces still make the modified version of the problem cooked, even if with bPa2 2 Queen-moves suffice.

Zolotarev used the matrix of this problem a lot, search the PDB with
A='Zolotarev' AND G='Retro' AND POSITION='wKf5 sKf3'

And yes, the Composer may be "very experienced" as you put it, but he was also very productive regarding cooks, check e.g. https://www.dieschwalbe.de/ergaenzungena2.htm ...

And one more mysterium: the Retro Corner gives 3 Queen-moves as the solution, s. https://www.janko.at/Retros/Schwalbe/Solutions.htm#9576 (2021-08-05)
Hans-Jürgen Manthey: Es sind 3 DAME-züge nötig !! (und nicht 2 wie A.Buchanan behauptet !)
Zuerst den wK nach f5 dann bK f3 nur möglich wenn der wK kurz auf g5 parkt. jedoch Dh4, Lh3, Tg4 verhindern diesen Plan. Es muß also zuerst bKf3 geschehen. Damit der wK nun nach f5 gelangen kann, darf e5 weder gedeckt (d6) noch besetzt sein (e5). Dank den Hinterstellungen Lh1-Tg2 & Lh3-Tg4 ist nach bxSf6 kein K-Zug mehr möglich ! Da ein g-Springer nicht die Ecke verlassen kann, muß es ein T sein. Auf c(oder a & c) entstandene T (oder S) können auf e5, f4 zwar geschlagen werden, aber es ist keine figur übrig um sich auf f6 zu opfern ! der bK muß aber über d7 oder d8 sein gefängnis verlassen, darum benötigt die ursprungsD 2 Züge(Dc7 oder Dc8 sowie Dd7) + 1 Zug die UmwandlunsDame(Dh4) = 3 Züge !!
Beispiel:
R.: 1. ... Te1-f1 2. d5xTe4 Tb4-e4 2. g3xSh2 Sf1-h2+ 3. f4xg3 Tb8-b4 4. a6-a5 b7-b8T 5. a7-a6 b6-b7 6. e5xTf4 Tb4-f4+ 7. d6xTe5 a5xLb6 8. Lc7-b6 Te5-e4 9. Ld8-c7 Tc4-e4 10. Lb6-d8 Tc8-c4 11. Ld4-b6 c7-c8T 12. Le5-d4 Tb8-b4 13. Lf4-e5 Tg8-b8 14. Dc8-d7 (1.) g7-g8T 15. Lh6-f4 g6-g7 16. Lf8-h6 g5-g6 17. g7xSf6 Se8-f6 18. d7-d6 c6-c7 19. Te4-e6 Ke5-f5 20. Tb4-e4+ c5-c6 21. Td4-g4 Sd6-e8 22. Kg4-f3 Th2-g2 23. Tf4-d4 Sd2-f1 24. c6xd5 d4-d5 25. Tf6-f4 Lg2-h1 26. Th6-f6 Lf1-g2 27. Lg2-h3 Sc4-d6 28. Ld5-g2 Sf3-d2 29. Kh3-g4 Tg2-h2+ 30. Kg4-h3 Sc3-e2 31. Lg6-h5 Te2-e1 32. Tb8-b4 Sb6-c4 33. Ta8-b8 Ta2-e2 34. Dh1-h4 (2.) Ta1-a2 35. h2-h1D e2-e3 36. h3-h2 Lh2-g1 37. Th8-h6 Tg1-g2 38. Lc4-d5 Th1-g1 39. h4-h3 g2-g3 40. La6-c4 Sg1-f3 41. h5-h4 Lf4-h2 42. Lb1-g6 Ke4-e5 43. b2-b1L+ Sb1-c3 44. b3-b2 Ke3-e4 45. Kf5-g4 Kd2-e3 46. b4-b3 Ke1-d2 47. b5-b4 Lc1-f4 48. Ke6-f5 g4-g5 49. Kd6-e6 c4-c5+ 50. Kc7-d6 d2-d4 51. Kd8-c7 c2-c4 52. Ke8-d8 a4-a5 53. Dd8-c8 (3.) a2-a4 54. h7-h5 h3xSg4 55. Sf6-g4 Sc8-b6 56. Sg8-f6 c7-c8S 57. Lc8-a6 h2-h3 58. b7-b5 b6xSc7 59. Sa6-c7 b5-b6 60. Sb8-a6 b4-b5 61. c7-c6 b2-b4 (2023-03-09)
comment

Moldenhauer: Computerprüfung: C+ da NUPG Stelvio 1.11 08:07:40 Stunden. (hh:mm:ss)
Keine Lösung: BP 21.5, BP 22.0.
Beispiel: 1.Sa3 b6 2.Sf3 Lb7 3.b4 c5 4.Lb2 c4 5.Lxg7 c3 6.Lxf8 e5 7.e3 Dh4
8.Lc4 Se7 9.0–0 Dxf2+ 10.Kh1 Tg8 11.Le6 cxd2 12.c3 Tg3 13.Da4 h6
14.Dxa7 dxe6 15.Dxb8+ Kd7 16.Tg1 Kc6 17.Sc2 Kb5 18.h4 Ta3
19.Scd4+ Ka4 20.b5 Th3+ 21.gxh3 Tb3 22.Tg6 fxg6 23.axb3# (2023-03-28)
comment

A.Buchanan: There's a question whether DP rule has visibility of 3Rep & 50M state. The current consensus among most of the tiny group who might care is that for retros, it does have visibility, but for purely forward problems, it does not. This align with the idea that by default 50M & DP rules apply only to retros (2023-09-06)
A.Buchanan: The old intended interpretation is protected under the Golden Age principle. Suppose we do apply modern rules, codex & clarifications. If Black just played Kb7xRa8, then the alternative leading to mate takes 7.0 moves, so this must have been Black's 5892nd move at the latest. If White just moved, then it was White's 5892nd at latest. So to maximize the length of the game, Black moved last. (2023-09-06) edit (2023-09-06)
A.Buchanan: The old intended interpretation is protected under the Golden Age principle. Suppose we do apply modern rules, codex & clarifications: (1) mate trumps 50M (2) 50M applies as this is retro (3) 75M rule exists but does not apply as this is problem not game (4) DP rule exists (5) DP rule applies as this is retro (6) DP has visibility of 50M state as this is retro. If Black just played Kb7xRa8, then the alternative leading to mate takes 7.0 moves, so this must have been Black's 5892nd move at the latest. If White just played Kg2xRh1, then again need another 7.0 moves for the half-duplex mate, which is at the latest Black's 5898th move, so this would have been White's 5892nd move at the latest. So to maximize the length of the game, Black moved last. (2023-09-06, hidden) more ...
comment

Nach Goldsteen: P0002345
Henrik Juel: White captured Sb5xQc7 and Be8xRf7. Black captured a7xPb6, c6xRd5, d7xRc6, e7xBd6, h7xPg6, Qc7xSc8, and Rf8xSf7 five times. (2003-10-14)
James Malcom: What is the full solution? (2023-05-30)
Henrik Juel: The retroplay is something like
R: 1. ... Sh7-f8+ 2. Le8xTf7 Tf8xSf7 3. Se5-f7 Tf7-f8+ 4. Sc4-e5 Tf8xSf7 5. Sh6-f7 Tf7-f8+ 6. Sa3-c4 Tf8xSf7 7. Sg5-f7 Tf7-f8+ 8. Sb5-a3 Tf8xSf7 9. Se5-f7 Tf7-f8+ 10. Sc7-b5 Tf8xSf7 11. Sb5xDc7 Dc8-c7 12. Dd8-d7 Dc7xSc8 13. Ld7-e8 Te8-f8 14. Sf5-h6 Tf8-e8 15. Sh6-f7 Tf7-f8+ 16. Df8-d8 Dd8-c7 17. Le8-d7 Kc7-b8 18. Sa3-b5+ Lb8-a7 19. Sa7-c8 Kc8-c7 20. Sb5-a7+ Lc7-b8 etc.
Earlier in the game Black captured a7xPb6, c6xTd5, d7xTc6, e7xLd6, h7xPg6, and White promoted on a8, c8, d8, h8.
The full retroplay back to the initial array can be constructed from this information; it has been done in a solution comment to Goldsteen's wonderful P0002345, which has no Pf2 in the diagram position (2023-05-30)
comment

Korrektur in "Die Schwalbe" 181, 02/2000, S.377
Henrik Juel: Natch 3.1 did nothing in more than 3 hours (2018-12-13)
Vladimir Rey: I guess, FEN-misprint. Correct e3 = e4, probably. If p. e3: dual 8. Da6-d3 (2022-11-06)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.5 wenn Bauer auf e3 steht.
Lösezeit: 70:23:07 Stunden. (hh:mm:ss) Keine Lösung: BP 20.5 ca. 22 Std.
Herr Vladimir Rey hat dies schon festgestellt.
8.Da6-d3 Ke8-d8 9.Dd3-h7 Kd8-c8
8.Da6-g6 Ke8-d8 9.Dg6-h7 Kd8-c8
Computerprüfung: C+ Stelvio 1.5 wenn Bauer auf e4 steht.
Lösezeit: 87:04:40 Stunden. Keine Lösung: BP 20.5 ca. 31Std.
Notation ist bereits für Bauer auf e4 richtig.
Richtige FEN: 4k2r/1pppppp1/B6p/Q6r/1B2P3/2KP4/PPPN1PPP/4R1NR
Die Strategien für e3 2944615/ e4 3379466 gerechnet mit 20/21 Playern. (2023-08-28)
comment

paul: Intention: 1.Sc3 f5 2.Se4 f4 3.Sg3 fg3 4.hg3 g5 5.Rh4 Bg7 6.Ra4 Bxb2 7.d4 h5 8.Bxg5 Sh6 9.e3 0-0! (9...Rf8? 10.Bc4 d6 11.Kf1 Bh3 12.Qxh5+ Kd7 13.Sh3??)10.Qxh5 Rf3 11.Bf4 Sc6 12.Qc5 Kf7 13.Sh3 Qh8 14.Bc4+ Ke8 15.Bg8 d5 16.Kf1 Be6 17.Kg1 Kd7 18.Kh1 Re8 19.Rg1 Kc8.
Cooked: 1.d3 g5 2.Bf4 Bg7 3.e3 h5 4.Qxh5 g4 5.Qc5 Bxb2 6.d4 Rh3 7.Bd3 f5 8.Bxf5 Rf3 9.Sh3 g3 10.0-0! d5 11.Sc3 Bxc3 12.Rb1 Be6 13.Rb4 Sh6 14.Ra4 Kd7 15.hxg3 Qh8 16.Bh7 Sc6 17.Bg8 Re8 18.Kh1 Bb2 19.Kg1 Kc8. (2010-07-02)
Henrik Juel: cooked, e.g.
1.Sb1-c3 Bg7-g5 2.Sc3-e4 Bg5-g4 3.Se4-g5 Bg4-g3
4.Sg1-f3 Bh7-h5 5.Bh2xg3 Bh5-h4 6.Th1xh4 Sb8-c6
7.Sg5-h3 Lf8-g7 8.Th4-a4 Lg7xb2 9.Bd2-d4 Sg8-h6
10.Dd1-d3 Bf7-f5 11.Dd3xf5 Th8-f8 12.Df5-c5 Tf8xf3
13.Lc1-f4 Bd7-d6 14.Be2-e3 Lc8-f5 15.Lf1-c4 Ke8-d7
16.Ke1-f1 Dd8-h8 17.Kf1-g1 Ta8-e8 18.Kg1-h1 Kd7-c8
19.Lc4-g8 Lf5-e6 20.Ta1-g1 Bd6-d5 (2018-12-13)
Henrik Juel: there even are shorter games, e.g.
1.Sb1-a3 Bg7-g6 2.Ta1-b1 Lf8-g7 3.Sg1-h3 Lg7xb2
4.Bd2-d4 Lb2xa3 5.Tb1-b4 La3-b2 6.Tb4-a4 Bf7-f5
7.Dd1-d3 Sg8-h6 8.Dd3xf5 Th8-f8 9.Df5-c5 Tf8-f3
10.Lc1-f4 Bd7-d5 11.Be2-e3 Lc8-e6 12.Lf1-d3 Ke8-d7
13.Ke1-f1 Dd8-h8 14.Kf1-g1 Bg6-g5 15.Ld3xh7 Bg5-g4
16.Lh7-g8 Bg4-g3 17.Bh2xg3 Sb8-c6 18.Kg1-h2 Ta8-e8
19.Th1-g1 Kd7-c8 20.Kh2-h1 (2018-12-13)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.31 53 Sekunden.
Keine Lösung: BP 17.5.
Beispiel Stelvio: 1.Sa3 g5 2.Tb1 Lg7 3.Sh3 Lxb2 4.d4 Lxa3 5.Dd3 Sc6
6.Tb4 Sh6 7.Ta4 f5 8.Dxf5 Tf8 9.Dc5 Tf3 10.Lf4 Lb2 11.e3 d5 12.Ld3 Le6
13.0–0 Kd7 14.Kh1 Dh8 15.Tg1 Te8 16.Lxh7 g4 17.Lg8 g3 18.hxg3 Kc8
Schwarz hat nicht rochiert und KBP = 18.0. (2023-06-20)
comment

Henrik Juel: Part b) is C+ by Natch 3.1 and no solution was found in 8.5
For part a) the program ran for an hour without finding the solution in 13.0 (2018-12-14)
Moldenhauer: Computerprüfung: C+ a und b Stelvio 1.2 a) 00:04:57 Minuten.(hh:mm:ss)
b) 1 Sekunde. Keine Lösung a) BP 12.0, b) BP 8.5.
KBP a) BP 13.0, b) BP 9.5. (2023-05-28)
comment

Moldenhauer: Computerprüfung: C+ Stelvio 1.5 in 147:43:15 Stunden.(hhh:mm:ss)
Keine Lösung: BP 27.0 in 49:30:01 Stunden. (2023-08-17)
comment

Henrik Juel: White has the move. -1... Sf8 -2.g5 Se6 -3.g4 Sc5 -4.Sd8 Sg7 -5.Se6 Ta2 ... -9.S=h7 Ta2 ... -13.h2! h3xSg2 -14.Sf4 Ta2 ... -17.Sb3 Ta1 -18.La2 Tg1 ... -22.La2 Te7 -23.Td3 Td7 -24.Tg3 d3 -25.Tg1 d4 -26.Ta1 d5 -27.Lb1 h4 -28.Sc1 h5 -29.Sd3 Lf8 -30.Sb4 e7xLf6 -31.Lg5 d6 -32.Lc1 Kd4 -33.d2xSc3 etc. Enjoyable. (2004-01-05)
Henrik Juel: In the official solution 16.Sd8-e6 is illegal, because wKc6 is checked (this seems fixable)
My 2004 solution also works, so the problem is almost cooked (2023-07-23)
comment

Henrik Juel: Natch 3.1 did nothing in half an hour (2018-12-14)
paul: Correction of P1000629 (2022-10-12)
comment

Moldenhauer: Computerprüfung: C+ Jacobi v0.7.6 beta1 PG demolition mode
Notation B: 8/8/8/8/8//8/r1N5/k1K5 (2022-08-11)
Wilfried: Position B is missing. It is: Kc1 Sc2; Ka1 Ta2
Then:
1... ... 2.Tg1-f1 + Kf2-e2 3.Tf1-d1 Sh1-f2 4.Td1-f1 Sf2-d1 + 5.Kb2-a1 Sd1-c3 6.Tf1-e1 + Ke2-d2 7.Te1-c1 Sc3-b5 8.Tc1-b1 Sb5-a3 9.Tb1-b2 Kd2-c1 10.Tb2-a2 Sa3-c2 #
1...Sh1-g3 2.Tg1-f1 + Kf2-e3 3.Tf1-f3 + Ke3-e2 4.Tf3-c3 Sg3-e4 5.Tc3-c2 Ke2-d1 6.Tc2-c4 Se4-d2 7.Tc4-c2 Sd2-c4 + 8.Kb2-a1 Sc4-a3 9.Tc2-d2 + Kd1-c1 10.Td2-a2 Sa3-c2 # (2008-05-15, hidden) more ...
comment

Henrik Juel: C+ Popeye 4.61 (without the retraction, which is unique) (2022-11-08)
comment

Henrik Juel: Natch 3.1 did little in an hour (2018-12-17)
Moldenhauer: Computerprüfung: C+ Euclide 1.11 5 Std. 32 Min. nur eine Lösung!
Keine Lösung bei BP 15.5 in 21 Min. und BP 16.0 in 45 Min.
Notation OK. (2022-12-21)
paul: 12. gxf8=T+ (2016-09-03, hidden) more ...
comment