Die Schwalbe

1480 problem(s) found in 5018 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT S='Der Standard (Wien)' AND G='Retro'] [download as LaTeX]

1 - P0000006
Ulrich Ring
5407 Die Schwalbe 97 02/1986
P0000006
(10+3) C+
BP in 25.5
1. Sa3 d5 2. Sh3 Lg4 3. Sf4 Lf3 4. gxf3 g5 5. Lh3 gxf4 6. Lf5 Sf6 7. Le4 Tg8 8. Sc4 Tg3 9. hxg3 dxc4 10. gxf4 Dd3 11. cxd3 a5 12. Db3 a4 13. Th6 axb3 14. axb3 Ta3 15. bxc4 Tc3 16. bxc3 Sd5 17. cxd5 e5 18. fxe5 Sc6 19. La3 Sd4 20. Tc6 Lh6 21. Ld6 Le3 22. fxe3 f5 23. cxd4 fxe4 24. fxe4 bxc6 25. Ta3 cxd6 26. exd6
play all play one stop play next play all
Moldenhauer: Computerprüfung: C+ bei NUPG Stelvio 1.11 00:08:50 Minuten. (hh:mm:ss)
Beispiel Stelvio 1.11.
1.Sa3 Sf6 2.Sc4 d5 3.Sh3 Lg4 4.Sf4 Lf3 5.gxf3 Tg8 6.Lh3 dxc4 7.Lf5 Dd3 8.Le4 g5 9.cxd3 gxf4 10.Db3 Tg3 11.hxg3 a5 12.Th6 a4 13.gxf4 axb3 14.axb3 Ta3 15.bxc4 Tc3 16.bxc3 Sc6 17.La3 Sd4 18.cxd4 Sd5 19.Tc6 Lh6 20.Ld6 bxc6 21.Ta3 cxd6 22.cxd5 e5 23.fxe5 Le3 24.exd6 f5 25.fxe3 fxe4 26.fxe4
Keine Lösung: BP 24.5, BP 25.0. (2023-03-27)
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Keywords: Non-Unique Proof Game
Genre: Retro
Computer test: Stelvio 1.11
FEN: 4k3/7p/2pP4/3P4/3PP3/R2PP3/3PP3/4K3
Input: Gerd Wilts, 1995-06-03
Last update: Silvio Baier, 2023-08-15 more...
2 - P0000016
Nikita M. Plaksin
Alexander Kislyak

(F) Die Schwalbe 98 04/1986
P0000016
(11+9)
Welches war der erste Zug der beiden Könige?
R: 1. bxc3ep+ c2-c4 2. b5-b4+
play all play one stop play next play all
Erster K-Zug: Weiß: 0-0; Schwarz: 0-0
173. Thema-Turnier
Henrik Juel: Good motivation for the black castling
The further retroplay includes retracting Tf1-g1, Kd3 to g1, h2xTg3, Tg3 to f8, La6 to c8, d7xTSe6, and e6xPf7 (2021-04-19)
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Keywords: En passant, First Move?, Last Moves? (3), Castling (wksk), Castling in the retro play, En passant in the retro play, Volet Pawn, Type C
Genre: Retro
FEN: 7k/1p2pPpp/b3p3/8/8/2pK2P1/pP1PPPP1/2B3RN
Reprints: 566 Ukrainisches Album 1986-1990
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-20 more...
3 - P0000034
Joachim Brügge
5468 Die Schwalbe 98 04/1986
P0000034
(13+12) C+
BP in 17,0
Wem gewidmet?
1. f4 a5 2. f5 a4 3. f6 a3 4. fxe7 axb2 5. exd8=S bxa1=S 6. Sxb7 Sf6 7. g4 Sxg4 8. Lb2 Se3 9. Sf3 Sc4 10. Lxg7 Lc5 11. Sa5 Lg1 12. Ld4 Lb7 13. Lf2 Tg8 14. Sd4 Lf3 15. c3 Sc6 16. Sc2 Tc8 17. Sb3 Sd4
play all play one stop play next play all
6S make "S"; 4L make "L"
"SL" = Sam Loyd
Ryan McCracken: Cooked...Nf6/Nh6 is one of many move defects. (2001-09-13)
Moldenhauer: Computerprüfung: Cooked Stevio 1.2 1 Sekunde.
Keine Lösung: BP 16.0, BP 16.5.
Beispiel Notation:1.f4 Sc6 2.Sf3 Sf6 3.f5 Tg8 4.g4 Sxg4 5.f6 a5 6.fxe7 a4 7.exd8=S a3
8.Sxb7 axb2 9.Sa5 bxa1=S 10.Lb2 Lb7 11.Lxg7 Lc5 12.Sd4 Se3 13.c3 Sc4 14.Sc2 Lg1
15.Ld4 Tc8 16.Lf2 Sd4 17.Sb3 Lf3.
Die Idee mit den Buchstaben SL im Diagramm finde ich super! (2023-05-20)
A.Buchanan: Yes I like the letter idea, which can only work in German. Do you think this was intentionally a non-unique PG? Given the triumphs of the unique PG genre, such problems do not impress, but it wouldn't be "cooked". Or maybe we should classify all non-unique PGs here as cooked? What do you think? (2023-05-21)
Henrik Juel: Testing gives the result 'cooked'
This problem is correct in the sense that all shortest proof games last 17.0 moves
Still the current C+ is rather misleading
I suggest no label, neither C+ nor cooked (2023-05-21)
Moldenhauer: Ich nehme an, dass die Absicht es Sam Loyd zu widmen, das Ziel war.
Deshalb für diese Ausführung C+. Wenn der Autor selbst angibt das es nur
einen Lösungsweg gibt ist die Aufgabe cooked. NUPG haben andere
Schachproblemfreunde festgestellt und als Schlüsselwort angegeben. Oder irre ich? (2023-05-21)
Olaf Jenkner: Machine translation is very poor as far as problem chess is concerned. (2023-05-23)
A.Buchanan: Hi Moldenhauser. Non-unique PGs are uncooked if 1) the move total is correct, (2) the intended theme appears in every shortest solution. For a problem like this where the theme is the final diagram, the second is vacuous. In 1986 some unique PGs were being already published, and I t can be difficult to discern the intention of the composer but I think this one was intended to be non-unique. I don’t think any non-unique PG can be classed as C+. (2023-05-23)
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Keywords: Non-standard material, Promotion, Non-Unique Proof Game, Symbolproblem Endstellung (SL)
Genre: Retro
Computer test: Computerprüfung: Cooked Stevio 1.2 1 Sekunde. Keine Lösung: BP 16.0, BP 16.5.
FEN: 2r1k1r1/2pp1p1p/8/8/2nn4/1NP2b2/P1NPPB1P/nN1QKBbR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-23 more...
4 - P0000036
Valery Liskovets
Nikita M. Plaksin

5528 Die Schwalbe 99 06/1986
P0000036
(10+11)
h#1.5 (AP, RV)
1) 1. ... cxd6ep 2. 0-0-0 Dxc7#
Nicht 2. Kf8 Dg8? da der ep-Schlag AP bewiesen werden muss.
2) 1. ... gxf6ep 2. 0-0-0 Tg8#
Nicht 2. Kd8 Tg8#? da der ep-Schlag AP bewiesen werden muss.
play all play one stop play next play all
VL: Solution:
I 1... c*d6 e.p. 2.0-0-0 Q*c7#
II 1... g*f6 e.p. 2.0-0-0 Rg8#

NOT:
2.Kf8? Q/Rg8#?? no AP-justification!

Castling implies e.p. 'a posteriori' and is necessary only to this end. (2002-04-03)
A.Buchanan: Solutions & tries (Popeye v.485) are:
1. ... cxd6ep 2. 0-0-0 Dxc7#
1. ... cxd6ep 2. Kf8 Dg8#
1. ... gxf6ep 2. d4 Tg8#
1. ... gxf6ep 2. 0-0-0 Tg8#
1. ... gxf6ep 2. Kd8 Tg8#
So after 1. ... gxf6ep:
2. Kf8 Rg8 is not even a mate (f7 flight)
and is 2. d4 a kind of dual thematic AP try? No-one's mentioned it. It would be easily fixable by shifting wBd2 to d4.
However, the main thing issue I don't understand PRA+AP logic (maybe I did once?). Under normal PRA algorithm, the "parts" for cxdep, gxfep & 0-0-0 respectively are YNY, NYY, NNN. Others are either impossible (3) or dominated by application of castling convention (2). OK so far.
There is no solution in the NNN part, so I suppose we must use AP in another part to eliminate it. But castling in another part where we are explicitly *assuming* that castling is ok seems to me to prove nothing. And if this inference did work somehow (which seems to me incredibly unlikely), why would we need to do it in *both* of these other parts? Surely one would be enough, but that means that the other part would be cooked.
Or maybe somehow AP "proves" that there are only 2 dimensions for PRA space, cxdep & gxfep, but that is so incredibly circular and horrible.
This kind of problem needs a proper step-by-step solution. Currently the solver can at best sort-of-vaguely-see-what-the-composer-might-be-alluding-to: i.e. it's just a joke.
Actually, I don't think that it can be PRA at all. I think it must be RS with two solutions supported by AP. But then there's still the question of why we need to castle in *both* solutions. I am sure I have seen an AP problem with two solutions where there is 0-0 in one solution and 0-0-0 in the other, which suggests that we *don't* need to castle twice in this problem. Which interpretation is correct, or do we have such a vague interpretation of these things (in order not to make independent spirits sad) that both are allowed to "work". (2020-12-30)
VL: Generally I agree with Andrew's analysis. The presented solution needs some corrections and clarification.

At first, the extra try 1.gxf6 e.p.? d4. I don't remember details but indeed, at present I'd prefer to exclude it by shifting wPd2 onto d4. One more thematic try enriches a problem but it is better to have an equal number of tries after both e.p.'s. In general, in AP-problems, thematic tries are highly desirable but not necessary formally for soundness. Dual avoidance in tries is here just due to the flight f7. Yes, 2.Kf8? Rg8#?? is indicated wrongly (unfortunately, H.103 with the published solution isn't available for me).

Now the most doubtful point: dual avoidance in the solution. Do we need castling in both partial solutions or at least one would suffice for justification? In my opinion, both approaches make sense in principle. In this problem implemented is the former, what means that every partial solution is permitted to be considered separately and, thus, should contain legalizing castling. This is similar to ordinary AP-problems, in which every line of the solution must contain castling. There is a distinct much subtler and rarer approach, which I call "Total AP", that considers all lines as a whole and thus requires to contain castling at least once (with subtle soundness issues). So, the current problem is treated (implicitly) not in the spirit of Total AP.

The problem implements a certain kind of triples of mutually retrodepending special moves (w/b castlings and e.p.). Namely, it is a "hybrid" AP-problem, where castling justifies a posteriori the typical pRA-choice between two e.p. captures. This is one of lucky triple kinds that admit (controversially) sound implementations based on hybrid logical conventions (logics for short). One more such a triple is represented in my earlier P0002474. Totally 37 distinct kinds of interacting triples can be identified, and only for few of them I know (or at least expect) reasonable implementations under special logics (whichever artificial and arguable they can be but consistent internally). Under an implementation I mean (loosely) a problem with an orthodox stipulation (#n, h#, etc.) supplemented by a suitable retro-convention like an ad hoc combination of pRA, RS or AP such that the position of the problem contains the corresponding triple of moves AND its full solution depends on all their combined (il)legalities. An extremely subtle matter. (2021-01-29)
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Keywords: a posteriori (AP), En passant as key (2), Partial Retro Analysis (PRA), Castling (sg), Volet Pawn
Genre: h#, Retro
FEN: r3k2n/1pp4P/N6p/2PpKpP1/2Q1p1R1/4P1pB/3P1p2/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-23 more...
5 - P0000047
Nikita M. Plaksin
Faat Fatchullin

5646 Die Schwalbe 101 10/1986
2. Preis
P0000047
(11+10)
h#2*
*) 1. ... 0-0-0 2. Txf2 Dxg1#
1) 1. cxd2+ Kxd2 2. Txf2 Dxg1#
R: 1. ... Da8-h8 2. d3xTe4 Da2-a8 3. c2xSd3 Sc1-d3+ 4. b4xLc5 Db1-a2 5. b3-b4 Ba2xSb1=D 6. Sa3-b1 und weiter z.B. (Beispielauflösung mri)
6. ... Te8-e4 7. Sc4-a3 Sg4-h2 8. Se5-c4 Th2-g2 9. Sf3-e5 g2-g1=L 10. Sg1-f3 Sh6-g4 11. Sf3xDg1 Th8-e8 12. Sd4-f3 f3xSg2 13. Se3-g2 f4-f3 14. Sc6-d4 Tg2-h2 15. Sa7-c6 Kh2-h1 16. Sc8-a7 Dh1-g1 17. Sb6xLc8 Tg1-g2 18. Sf5-e3 Dc6-h1 19. Sh4-f5 Kh1-h2 20. h2-h3 a3-a2 21. Sa4-b6 Sa2-c1 22. Sb6-a4 Sb4-a2 23. Sa4-b6 Sd5-b4 24. Sb6-a4 Se3-d5 25. Dh3-f1 Sf1-e3 26. Dg4-h3 f5-f4 27. Df4-g4 a4-a3 28. Db4-f4 a5-a4 29. Da3-b4 Kg2-h1 30. Sf3-h4 Kh3-g2 31. Da2-a3 g2-g1=T 32. Db1-a2 Kg4-h3 33. Dd1-b1 h3xTg2 34. Tg1-g2 Kg5-g4 35. Sh4-f3 Kf6-g5 36. Th1-g1 Se3-f1 37. Sf3-h4 h4-h3 38. Sg1-f3 e6xLf5 39. Lh3-f5 Ke7-f6 40. Lf1-h3 Sc4-e3 41. g2-g3 Ke8-e7 42. Sa4-b6 Sg8-h6 43. a2xTb3 Tb6-b3 44. Sh3-g1 Ta6-b6 45. Sb6-a4 Ta8-a6 46. Sg1-h3 a7-a5 47. Sh3-g1 Sa5-c4 48. Sg1-h3 Sb3-a5 49. Sc4-b6 Sc1-b3 50. Sa3-c4 Sb3xLc1 51. Sb1-a3 Sa5-b3 52. Sh3-g1 h5-h4 53. Sg1-h3 c4-c3 54. Sh3-g1 Lf8-c5 55. Sg1-h3 Dc7-c6 56. Sh3-g1 Dd8-c7 57. Sg1-h3 c5-c4 58. Sh3-g1 h7-h5 59. Sg1-h3 e7-e6 60. Sh3-g1 Sc6-a5 61. Sg1-h3 Sb8-c6 62. Sh3-g1 c7-c5 63. Sg1-h3
play all play one stop play next play all
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)
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Keywords: Castling (wl)
Genre: h#, Retro
FEN: 7q/1p1p1pp1/8/2P5/4P3/2p3PP/1P1PPPrn/R3KQbk
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-01-02 more...
6 - P0000050
Andrey Frolkin
(T) Die Schwalbe 102 12/1986
P0000050
(13+13)
Vor mindestens 50 Einzelzügen mußte rochiert werden!
R: 1. ... La2-b1 2. Dh7-g8 Lb1-a2 3. Kg8-h8 La2-b1 4. Kf7-g8 Lb1-a2 5. Ke8-f7 La2-b1 6. Kd8-e8 Lb1-a2 7. Kc7-d8 La2-b1 8. Kb6-c7 Lb1-a2 9. Kc5-b6 La2-b1 10. Kd4-c5 Lb1-a2 11. Ke4-d4 La2-b1 12. Kf3-e4 Lb1-a2 13. Kg4-f3 La2-b1 14. Kh3-g4 Lb1-a2 15. Kh2-h3 La2-b1 16. Kg1-h2 Lb1-a2 17. h2xTg3 Th3-g3 18. Dg8-h7 Th8-h3 19. Sg3-h1 h7xTg6 20. Tg5-g6 La2-b1 21. Ta5-g5 Lb1-a2 22. Ta2-a5 f5-f4 23. Tb2-a2 La2-b1 24. Tb1-b2 f6-f5 25. Tf1-b1 Lb1-a2 26. 0-0
play all play one stop play next play all
James Malcom: Lastly, here is a PG that may or may not be the shortest: 1. Nf3 c5 2. Ne5 Qb6 3. Nc3 Qb3 4. axb3 c4 5. Nd5 c3 6. Ra6 Nf6 7. Rd6 Ng4 8. Re6 Ne3 9. Nf4 Nxf1 10. Nh5 Ne3 11. Ng3 Nc4 12. bxc4 dxe6 13. Nf5 Bd7 14. Ng3 Ba4 15. Nf5 Bb3 16. Ng3 Ba2 17. Nf5 Nc6 18. Ng3 Na5 19. Nf5 Nb3 20. Ng3 Na1 21. b3 a6 22. Ba3 Ra7 23. Bc5 Kd8 24. Qc1 Kc8 25. Qa3 Kc7 26. Qa4 Bb1 27. Qe8 f6 28. Qf7 Kc8 29. Qg8 Kb8 30. Bd6+ Ka8 31. Bb8 Ba2 32. Nd7 Bb1 33. O-O Ba2 34. Rb1 f5 35. Rb2 Bb1 36. Ra2 f4 37. Ra5 Ba2 38. Rg5 Bb1 39. Rg6 hxg6 40. Nh1 Rh3 41. Qh7 Rg3 42. hxg3 Ba2 43. Kh2 Bb1 44. Kh3 Ba2 45. Kg4 Bb1 46. Kf3 Ba2 47. Ke4 Bb1 48. Kd4 Ba2 49. Kc5 Bb1 50. Kb6 Ba2 51. Kc7 Bb1 52. Kd8 Ba2 53. Ke8 Bb1 54. Kf7 Ba2 55. Kg8 Bb1 56. Kh8 Ba2 57. Qg8 Bb1 (2020-11-08)
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Keywords: Castling in the retro play
Genre: Retro
FEN: kB3bQK/rp1Np1p1/p3p1p1/8/2P2p2/1Pp3P1/2PPPPP1/nb5N
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-02-02 more...
7 - P0000052
Andrey Frolkin
(V) Die Schwalbe 102 12/1986
P0000052
(16+11)
Vor mindestens 72 Einzelzügen mußte ep geschlagen werden!
R: 1. ... Sg2-h4+ 2. Sh4-g6+ h7-h6 3. Se2-g1 a3-a2 4. Tg1-f1 a4-a3 5. Sf1-h2 Sh2-g4 6. Lh6-g5 Sg4-h2 7. Lg7-h6 Sh2-g4 8. Lf8-g7! Sg4-h2 9. Ld6-f8 Sh2-g4 10. Lb8-d6 Sg4-h2 11. Sc3-e2 Sh2-g4 12. Sb5-c3 Sg4-h2 13. Sd6-b5 Sh2-g4 14. Sf7-d6 Sg4-h2 15. b7-b8=L Sh2-g4 16. b6-b7 Sg4-h2 17. b5-b6 Sh2-g4 18. b4-b5 Sg4-h2 19. b3-b4 Sh2-g4 20. Lg7-h8 Sg4-h2 21. Lf8-g7 Sh2-g4 22. Ld6-f8 Sg4-h2 23. Lb8-d6 Sh2-g4 24. b7-b8=L Sg4-h2 25. b6-b7 Sh2-g4 26. b5-b6 Sg4-h2 27. b4-b5 Sh2-g4 28. a3xBb4 Sg4-h2 29. Sg5-f7 Kh6-h5 30. Th2-h3 Kh5-h6 31. Sh3-g5 b5-b4 32. Sg6-h4 b6-b5 33. Se7-g6 b7-b6 34. Sg8-e7! a5-a4 35. g7-g8=S a6-a5 36. g6-g7 Kh6-h5 37. h5xg6ep+ g7-g5 38. Sg5-h3+ a7-a6 39. Th3-h2 Sh2-g4 40. Dh4-f4 Sf4-g2 41. Lg2-h1 Sg4-h2 42. Th1-h3 Sh2-g4 43. Lh3-g2 Sg4-h2 44. g2-g3 Sh2-g4 45. Dg4-h4 Tg3-f3 46. Dd1-g4
play all play one stop play next play all
5 weiße Schlagfälle durch Bauern: axb, c3xd4, exf, f5xe6 & hxg; zwei Umwandlungen auf b8 und eine auf g8. Eine schwarze Umwandlung auf c1. Die weißen Figuren, die zur Entwandlung zurückschreiten können, sind die schwarzfeldrigen Lh8 und g5 sowie der retrofreie Sg1. Aber vor jeder Entwandlung muß Weiß das Feld h2 räumen, um dem Sg4 das Pendeln zu ermöglichen und so ein schwarzes Retropatt zu verhindern.
Retro: 1. Sg2-h4+ Sh4-g6+ 2. h7-h6 Se2-g1 3. a3-a2 Tg1-f1 4. a4-a3 Sf1-h2 5. Sh2-g4 Lh6-g5 6. Sg4-h2 Lg7-h6 7. Sh2-g4 Lf8-g7! 8. Sg4-h2 Ld6-f8 9. ~ Lb8-c6 10. ~ Sc3-e2 11. ~ Sb5-c3 12. ~ Sd6-b5 13. ~ Sf7-d6 14. ~ b7-b8=L 15. ~ b6-b7 16. ~ b5-b6 17. ~ b4-b5 18. ~ b3-b4 19. ~ Lg7-h8 20. ~ Lf8-g7 21. ~ Ld6-f8 22. ~ Lb8-d6 23. ~ b7-b8=L 24. ~ b6-b7 25. ~ b5-b6 26. ~ b4-b5 27. Sh2-g4 a3xb4
Nach diesem Entschlag des schwarzen Bb führt die dritte Entwandlung auf g8 zu einem erzwungenen En-passant-Schlag. 28. Sg4-h2 Sg5-f7 29. Kh6-h5 Th2-h3 30. Kh5-h6 Sh3-g5 31. b5-b4 Sg6-h4 32. b6-b5 Se7-g6 33. b7-b6 Sg8-e7! 34. a5-a4 g7-g8=S 35. a6-a5 g6-g7 36. Kh6-h5 h5xg6ep+ 37. g7-g5 Sg5-h3+ 38. a7-a6 Th3-h2 39. Sh2-g4 Dh4-f4 40. Sf4-g2+ g2-g3 41. Tg3-f3
Diese Zugfolge mit möglichen Zugumstellungen kann nicht verkürzt werden. Mindestens 71 Einzelzüge sind nach dem En-passant-Schlag geschehen. Rekord für eines der Themen des 173. Thematurniers der "Schwalbe".
paul: This problem obtained first Prize in the informal tourney, see Die Schwalbe 249/2011 (judge M. Caillaud) (2020-01-06)
Henrik Juel: 41.g2-g3 is illegal, because now wLh1 cannot retract (2021-02-03)
Henrik Juel: 41.Lg2-h1 Sg4-h2 42.Th1-h3 Sh2-g4 43.Lh3-g2 Sg4-h2 44.g2-g3 Sh2-g4 45.Dg4-h4 Tg3-f3 46.Dd1-g4 seems to work (2021-02-03)
comment
Keywords: En passant, En passant in the retro play, Non-standard material, Promotion, Last Moves? (72)
Genre: Retro
FEN: 7B/8/4PPNp/4pKBk/3PrQnn/3pBrPR/p2P1p1N/4bRNB
Reprints: 568 Ukrainisches Album 1986-1990
H21 FIDE Album 1986-1988 1995
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-02-04 more...
8 - P0000058
Leonid M. Borodatow
5758v Die Schwalbe 103 02/1987
P0000058
(13+9) C+
h#3
b) sBa6 statt sLa6
a) 1. ... exf6ep 2. 0-0-0 Lxf4 3. Td7 a8=D#
b) 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
play all play one stop play next play all
Anton Baumann: Mattdual in b): 3.Tb8 axb8=D,T# (2022-12-16)
A.Buchanan: So Borodatow got it all working! In a) Black might have captured hxgxf and axPb. So castling rights might still be maintained with the ep. In b) on the other hand, it must be axb, bxa, exf and wPgxh6, so there was the cage. Promotion to TD is tolerated in the final move, although it may not be puristic, without the convention, too many mates would be excluded (2022-12-16)
comment
Keywords: En passant as key, Castling (sg), Cant Castler, Valladao Task
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & basic retro thinking
FEN: r3k3/P7/b3r1pP/4PpBP/3nnpKR/5PRB/5PP1/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
9 - P0000101
Leonid M. Borodatow
Evgeny V. Kharichev

6209v Die Schwalbe 110 04/1988
P0000101
(12+10) cooked
h#3
b) sBh5 nach h6
a) 1. 0-0-0 Sxc7 2. hxg3 Sd5 3. g2 Se7#

Beispielauflösung mri:
R: 1. Kf1-e1 Ta2-a1 2. Kg2-f1 Tb2-a2 3. La2-b1 Tb1-b2 4. Kf3-g2 Tf1-b1 5. Ke3-f3 g5xDh4 6. Lh2-g1 Tg1-f1 7. Dd4-h4 Tg2-g1 8. Lg1-h2 Th2-g2 9. Kd3-e3 Tg2-h2 10. Kc3-d3 Th2-g2 11. Kb2-c3 Tg2-h2 12. Kc1-b2 Th2-g2 13. Kd1-c1 Tg2-h2 14. Ke1-d1 Th2-g2 15. Da1-d4 Tg2-h2 16. Dd1-a1 d6xLc5 17. La3-c5 Th2-g2 18. Lc1-a3 Tg2-h2 19. b2-b3 Th2-g2 20. Ld5-a2 Tg2-h2 21. Sc5-a6 Th2-g2 22. Lg2-d5 h6-h5 23. Lf1-g2 Tg2-h2 24. Lh2-g1 Tg1-g2 25. Sa4-c5 g2-g1=T 26. Sb6-a4 f3xTg2 27. Tg1-g2 f4-f3 28. g2-g3 f5-f4 29. Le5-h2 h7-h6 30. Lg7-e5 f6-f5 31. Lf8-g7 f7-f6 32. e7xLf8=L Lg7-f8 33. e6-e7 Lf8-g7 34. d5xTe6 Tg6-e6 35. Sc8-b6 Tg8-g6 36. Sb6xLc8 g6-g5 37. c4xDd5 Dg5-d5 38. Sd5-b6 Dd8-g5 39. Sg3-h1 e7xTd6 40. Ta6-d6 Th8-g8 41. Ta1-a6 g7-g6 42. b3xSc4 Sa5-c4 43. a2xSb3 Sd4-b3 44. Sc3-d5 Sc6-a5 45. Sb1-c3 Sb8-c6 46. Th1-g1 Sf5-d4 47. Se4-g3 Sh6-f5 48. Sg5-e4 Sg8-h6 49. Sf3-g5 Sh6-g8 50. Sg1-f3 Sg8-h6 51. h2-h3


b) 1. Kf7 f4 2. Kg6 c3,c4+ 3. Kh5 g4#
play all play one stop play next play all
in b) fehlt Schwarz ein Retrotempo zur Aufrechterhaltung des Rochaderechts, es kann maximal eine Stellung wie z.B. r3kB2/pppp1p1p/3p2p1/8/8/7P/1PPPPPP1/1NBQKBRN erreicht werden, in der entweder sTh8 oder sLf8 nicht als Schlagobjekt für die notwendigen weißen Bauernschläge zur Verfügung stand.

Der Dual 2. ... c3+,c4+ in b) war von den Autoren in Kauf genommen worden., die (schon bei der Lösungsveröffentlichung bekannten) NLs aber natürlich nicht. Ein Schwalbe-Löser ("PS") kritisierte den "völlig unnötigen Vorwärtsballast" und schlug implizit vor, das Vorwärtsspiel komplett wegzulassen und stattdessen direkt zu fragen: "Ist die 0-0-0 erlaubt?"
Cook: a) 1. hxg3 Sxg3 2. c6 Sf5 3. 0-0-0 Sd6#
b) 1. Kf7 f4 2. Kg6 c4+ 3. Kh5 g4#
Neufassung 5949.
A.Buchanan: It would be great if someone can retrieve 6209v from Die Schwalbe, please, to confirm the diagram and intended solution. If castling right remains, then sBf must promote on g1=T. There is then some jiggling around because Tg1-h2 to unpin Lg1 and let it go b1. In (a) this is doable, but in (b), Black runs out of tempi however there is an alternative no-castling mate with sKh5. However that's just a rough sketch, and I don't want to spend more time until the actual published form is confirmed (2022-01-08)
Ladislav Packa: Is everything allright here? Because solution a) is not a solution - after 3 .... Se7 # ?? 4.Kc7 / b8! (2022-01-09)
Mario Richter: Indeed, the given solution only works with wLh2 instead of g1. I only have access to the solution, but not to original diagram. Perhaps someoe with access to it can give us the correct details? (2022-01-10)
A.Buchanan: Yes I didn't check the solution to a) matches one of the actual ones. With wLh2, it's still cooked, as is b) in any case. Possible to rearrange the free pieces to make b) sound, but I don't see how to fix a) at the same time. (2022-01-10)
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comment
Keywords: Castling (sg), Cant Castler (sg)
Genre: h#, Retro
FEN: r3k3/pppp4/N7/2p4p/7p/1P4PP/2PPPP2/rB2K1BN
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-01-10 more...
10 - P0000108
Andrey Frolkin
6272 Die Schwalbe 111 06/1988
1. Lob
P0000108
(30+0) C+
Färbe die Steine!
BP in 19.0
1. h4 d6 2. Th3 Sd7 3. Tb3 Sdf6 4. Tb6 axb6 5. f4 Ta3 6. Kf2 Th3 7. a4 Th1 8. Ta3 Lh3 9. Tg3 Dd7 10. Tg6 hxg6 11. a5 Th5 12. a6 Ta5 13. h5 Ta1 14. h6 Da4 15. h7 b5 16. h8=T Sh7 17. a7 f6 18. a8=T+ Kf7 19. Ta7 Dd4+
play all play one stop play next play all
Henrik Juel: C+ by Natch 3.1 (16 min.)
The colored position has black Ta1,h1 and Pb5,g6, and white Ta7,h8, otherwise 'natural' (2018-12-07)
François Labelle: Henrik, did you really test all 2^30 possible colorings with Natch, or did you use human deductions to reduce the possibilities (making this HC+), or did you only test the intended coloring (making this a partial C+)? (2018-12-09)
Mario Richter: I do not think that it is necessary to check all 2^30 colorings, since the color of the pawns on files c,d,e and f is completely determined. This and more restrictions on the set of potentially possible colorings may be derived from insights presented in an article "Aggregierte Schlagbilanz" by Frolkin & Kornilov, feenschach 130, p.411ff. Since the mechanisms presented there can at least partially be implemented in a little computer program, even a "full C+" label for this problem is not out of reach ... (2018-12-10)
Henrik Juel: I only tested the intended coloring, Francois,
so the C+ label is not justifiable
In the other coloring proof games I write something like
The colored position was C+ by Natch

Mario is right, of course, in that not all colorings need testing; but still the number is very large
This genre is somewhat messy; at first I thought that the solver could determine the coloring, but this is clearly not the case; also, the intention is not
'color the men such that a correct proof game in 19 results' (2018-12-11)
A.Buchanan: Henrik: so what exactly should the stipulation be for clarity? (2018-12-12)
Henrik Juel: Pretty much as it is now, Andrew
Color the men
PG19 (2021-02-20)
comment
Keywords: Unique Proof Game, Colouring problem, Promotion (TT), Phoenix
Genre: Retro
Computer test: Natch 3.1 (16 min.)
FEN: 5BNR/RPP1PKPN/3P1PP1/1P6/3Q1P2/7B/1PPPPKP1/RNBQ1BNR
Reprints: 158 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-02-08 more...
11 - P0000111
Andrey Frolkin
Leonid Lyubashevsky

6330 Die Schwalbe 112 08/1988
4. ehrende Erwähnung
P0000111
(28+0)
Färbe die Steine!
BP in 11.0
1. h4 f5 2. Th3 Kf7 3. Tb3 De8 4. Tb6 axb6 5. g3 Ta3 6. Lg2 Tc3 7. Lc6 dxc6 8. dxc3 Le6 9. Dd8 Lb3 10. axb3 Dd7 11. Ta8 Dd1+
play all play one stop play next play all
Henrik Juel: C+ by Natch 3.1
The colored position is 'natural' except for wTa8, wDd8, and sDd1 (2018-12-07)
paul: The problem is Jacobi+ with this code: stipulation PG 11 forsyth RS1Q1BSR/1PP1PKPP/1PP5/5P2/7P/1PP3P1/1PP1PP2/1SBQK1S1 ColorThePieces (2021-07-06)
comment
Keywords: Unique Proof Game, Colouring problem, Interchange
Genre: Retro
FEN: RN1Q1BNR/1PP1PKPP/1PP5/5P2/7P/1PP3P1/1PP1PP2/1NBQK1N1
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-02-08 more...
12 - P0000112
Dmitri W. Pronkin
Andrey Frolkin

6386v Die Schwalbe 113 10/1988
P0000112
(14+14) cooked
BP in 45.0
1. f4 g5 2. f5 g4 3. f6 g3 4. fxe7 gxh2 5. g4 d5 6. g5 d4 7. g6 d3 8. g7 dxc2 9. d4 f5 10. Lf4 c1=T 11. Lg3 Tc6 12. Dd3 Tg6 13. Da6 Sf6 14. g8=L c5 15. Lb3 c4 16. d5 Kf7 17. e8=T c3 18. Te4 c2 19. Ta4 c1=L 20. e4 Lf4 21. Sd2 h5 22. 0-0-0 h4 23. Te1 h3 24. Ld1 Th4 25. d6 L8h6 26. d7 Dh8 27. d8=T Le6 28. Tc8 S8d7 29. Tc2 Kg8 30. e5 Lf7 31. e6 Tf8 32. e7 Lb8 33. e8=S f4 34. Te7 f3 35. Se2 f2 36. Tg1 h1=S 37. b4 h2 38. Lh3 f1=S 39. b5 Sf2 40. b6 h1=L 41. bxa7 b6 42. a8=L Lb7 43. Th1 Lc8 44. Lag2 Se3 45. Lf1 S6e4
play all play one stop play next play all
Cook: 1. b4 c5 2. b5 c4 3. b6 c3 4. bxa7 d5 5. e4 d4 6. f4 d3 7. f5 dxc2 8. d4 g5 9. Lf4 c1=L 10. d5 g4 11. f6 g3 12. fxe7 gxh2 13. g4 c2 14. Lg3 Lf4 15. g5 c1=T 16. g6 Tc6 17. Dd3 f5 18. Sd2 h5 19. g7 Tg6 20. 0-0-0 Sf6 21. Te1 h4 22. d6 h3 23. g8=L Th4 24. Lb3 L8h6 25. Ld1 Kf7 26. d7 Dh8 27. d8=T Le6 28. Td4 Sbd7 29. e5 Tf8 30. e8=T Kg8 31. Tc8 Lf7 32. e6 b6 33. e7 Lb8 34. e8=S f4 35. Te7 f3 36. Se2 f2 37. Tg1 h1=S 38. Da6 h2 39. Lh3 f1=S 40. Ta4 Sf2 41. Tc2 h1=L
Michel Caillaud: cooked by Stelvio 0.93 :
1.b4 c5 2.b5 c4 3.b6 c3 4.bxa7 d5 5.e4 d4 6.f4 d3 7.f5 dxc2 8.d4 g5 9.Lf4 c1=L 10.d5 g4 11.f6 g3 12.fxe7 gxh2 13.g4 c2 14.Lg3 Lf4 15.g5 c1=T 16.g6 Tc6 17.Dd3 f5 18.Sd2 h5 19.g7 Tg6 20.0-0-0 Sf6 21.Te1 h4 22.d6 h3 23.g8=L Th4 24.Lb3 L8h6 25.Ld1 Kf7 26.d7 Dh8 27.d8=T Le6 28.Td4 Sbd7 29.e5 Tf8 30.e8=T Kg8 31.Tc8 Lf7 32.e6 b6 33.e7 Lb8 34.e8=S f4 35.Te7 f3 36.Se2 f2 37.Tg1 h1=S 38.Da6 h2 39.Lh3 f1=S 40.Ta4 Sf2 41.Tc2 h1=L... (2022-12-20)
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Keywords: Unique Proof Game, Move Length Record, Non-standard material, Castling, Promotion (tLTlTSsslL)
Genre: Retro
FEN: 1bb1Nrkq/3nRb2/Qp4rb/8/R3n2r/4n1BB/P1RNNn2/2KB1B1R
Reprints: 583 Ukrainisches Album 1986-1990
80 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Silvio Baier, 2023-03-02 more...
13 - P0000136
Dmitri W. Pronkin
Andrey Frolkin

6631v Die Schwalbe 117 06/1989
Preis
P0000136
(14+14)
BP in 57.5
1. a4 h5 2. a5 h4 3. a6 h3 4. axb7 hxg2 5. h4 d5 6. h5 d4 7. h6 d3 8. h7 dxc2 9. d4 a5 10. Lh6 c1=T 11. e4 Tc5 12. Se2 Th5 13. e5 c5 14. e6 Sc6 15. b8=T a4 16. Tb4 a3 17. Ta4 c4 18. b4 c3 19. b5 c2 20. b6 c1=T 21. b7 Tc4 22. b8=T Da5+ 23. Tbb4 Lb7 24. S1c3 0-0-0 25. exf7 e5 26. Tc1 Lc5 27. f8=T a2 28. Tf3 a1=T 29. Sa2 g1=T 30. Tfa3 Tg6 31. f4 Te6 32. f5 g5 33. f6 g4 34. f7 g3 35. f8=T g2 36. Tf5 g1=T 37. Lf8 Tg7 38. Sg3 e4 39. Ld3 e3 40. 0-0 e2 41. Tcc3 e1=T 42. Lc2 T1e3 43. d5 Tdd7 44. d6 Tdf7 45. d7+ Kb8 46. Dd6+ Ka8 47. Dc7 Sge7 48. d8=T+ Sc8 49. Tdd3 Thg8 50. h8=T Tae1 51. Th6 T1e2 52. T1f2 Tce4 53. Kf1 Ld4 54. Tfc5 Se5 55. Sf5 Sc4 56. Sd6 Sb2 57. Tbc4 Sb6 58. Db8+
play all play one stop play next play all
Der absolute KBP-Längenrekord.
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)
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Keywords: Unique Proof Game, Move Length Record, Non-standard material (TTTTTTtttttt), Castling, Aristocrat, Superseded by (P1397486)
Genre: Retro
FEN: kQ3Br1/1b3rr1/1n1Nr2R/q1R4r/R1Rbr3/R1RRr3/NnB1rR2/5K2
Reprints: 584 Ukrainisches Album 1986-1990
86 Shortest Proof Games 11/1991
(6) diagrammes 103 10-12/1992
H18 FIDE Album 1989-1991 1997
feenschach 137, p. 368, 08-09/2000
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-12-25 more...
14 - P0000172
Dmitri W. Pronkin
6991v Die Schwalbe 123 06/1990
P0000172
(13+13) cooked
BP in 24.5
1. Sc3 Sf6 2. Se4 Sd5 3. Sg5 Sc3 4. Sxh7 Sb1 5. Sxf8 Th5 6. h4 Tg5 7. hxg5 Kxf8 8. Th6 Kg8 9. Tb6 Kh7 10. g6+ Kh6 11. e4 Kg5 12. Lb5 Kf4 13. Ke2 Ke5 14. Kf3 Kd4 15. Kf4 Kc5 16. b4+ Kxb4+ 17. La3+ Kxa3 18. Df3+ Kb2 19. a4 Kc1 20. Se2+ Kd1 21. Sc1+ Ke1 22. Sd3+ Kf1 23. Sb2+ Kg1 24. Lf1 Kh1 25. Dd3
play all play one stop play next play all
Cook: 1. Sg1-f3 Sg8-f6 2. Sf3-g5 Sf6-d5 3. Sg5xh7 Sd5-c3 4. Sh7xf8 Th8-h5 5. Sf8-e6 Th5-g5 6. Se6-c5 Ke8-f8 7. b2-b4 Kf8-g8 8. e2-e4 Kg8-h7 9. Dd1-g4 Kh7-g6 10. h2-h4 Kg6-f6 11. h4xg5+ Kf6-e5 12. Dg4-g3+ Ke5-d4 13. Th1-h6 Sc3xb1 14. Ke1-e2 Kd4-c4 15. Ke2-e3+ Kc4xb4 16. Lc1-a3+ Kb4xa3 17. Ke3-f4+ Ka3-b2 18. Th6-b6+ Kb2-c1 19. Lf1-a6 Kc1-d1 20. a2-a4 Kd1-e1 21. Sc5-d3+ Ke1-e2 22. g5-g6 Ke2-f1 23. Sd3-b2+ Kf1-g1 24. La6-f1 Kg1-h1 25. Dg3-d3 (Stelvio v0.93, RA)
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Keywords: Unique Proof Game, Promenade (k)
Genre: Retro
FEN: rnbq4/ppppppp1/1R4P1/8/P3PK2/3Q4/1NPP1PP1/Rn3B1k
Reprints: 579 Ukrainisches Album 1986-1990
113 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2022-12-17 more...
15 - P0000195
Frank Christiaans
7167 Die Schwalbe 126 12/1990
P0000195
(13+10)
#3
b) wSe4 nach g8
a) 1. 0-0? droht 2. Tf8+ Kd7 3. Lxe6#
1. ... Lxc5! 2. Sxc5 0-0-0!
1. Thf1! 0-0-0 2. Lxe6+ Td7 3. Tf8#
b) 1. Thf1? Lxc5 2. Sxc5 0-0-0!
1. 0-0! Lxc5 2. Sxc5 ... (0-0-0?) 3. Tf8#
play all play one stop play next play all
a) zuletzt h6-h5 und beide Rochaden sind noch möglich. Schläge von Weiß: Bbxc, Bcxd, Bhxg, Bfxg-g8=L.
b) gegenseitiger Ausschluss der beiden Rochaden. 3 Fälle sind zu betrachten:
b1) zuletzt f7xDe6; Weiß muss 4x geschlagen haben: Bfxexdxc8=L, d.h. s0-0-0 ist unzulässig.
b2) zuletzt d7xDe6; zu den 4 Schlägen von a) wird auch der sBf gebraucht, der sich auf f1 umwandeln muss, also w0-0 unzulässig.
b3) zuletzt h6-h5: wBf zieht über f7 zum UW-Feld g8, g.h. s0-0-0 unzulässig.
Henrik Juel: C+ Popeye 4.61 after analysis (2020-10-30)
A.Buchanan: A very harmonious problem. Suppose that both sides retain castling rights. wLh3 is obtrusive, so wBf/h promoted via g7, and escaped via e6 so sBe6 came from d7 later. Wh has made at least 4 pcs, while Bl has made 2 (D & Lc) dxe6, gxh6. If last move was dxe6, then sD & sLc were captured by officers in cage. Thus sBf was captured by wB, and therefore promoted on f1, disrupting wK. Therefore the last move was h6-h5. In (a) there is no problem retaining both castling rights, but in (b) Sg8 must have reached that square via f6 *after* the promotion, disrupting bK. However in (b) the castling rights are incompatible. In (a) 1. 0-0? Lxc5! 1. Tf1!, while in (b) the reverse.
Neither twin is a PRA problem: in (a) there is no conflict between the castling, while in (b) RS applies: the pre-emptive Wh castling means we are in a reality where Bl cannot castle. (2020-10-31)
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Keywords: Castling key (wksg), Obvious promotion (L), Retro Strategy (RS)
Genre: Retro, 3#
FEN: r3k3/ppp1p2p/Nb2p3/P1PP2Pp/4N3/7B/3PPRP1/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2022-01-07 more...
16 - P0000250
Nikita M. Plaksin
Valery Liskovets

7577v Die Schwalbe 132 12/1991
P0000250
(5+14)
ser-h#5 (AP)
Typ Tauber/Caillaud
1. Kd4 2. bxc3ep (zuletzt nur K,T-Zug oder c2-c4) 3. e5! (nicht 4. Tb4? weil zuletzt K,T-Zug und 0-0-0 nicht mehr möglich) 4. Tb4 (zuletzt e7xLd8=S möglich) 5. Tc4 & 1. 0-0-0# nicht 1. Td1#? wegen fehlender AP-Legalisierung
play all play one stop play next play all
"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
Keywords: Castling (wl), a posteriori (AP), Promotion (S), Seriesmover, Consequent, En passant
Genre: Retro, Fairies
FEN: 1N6/1ppppppn/n7/1Pq1k3/1pP1r3/4p3/1r6/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-29 more...
17 - P0000254
Leonid M. Borodatow
7642 Die Schwalbe 133 02/1992
P0000254
(16+10)
Welches waren die letzten 7 Einzelzüge, wenn dabei keine Zugwiederholungen vorkamen?
R: 1. 0-0-0# Ke4-d4 2. e5xf6ep+ f7-f5 3. Tg6-b6+ Kf5-e4 4. c7-c8=L
play all play one stop play next play all
Die von einigen Lösern angeführte Abweichung 2. f5-f6+ Kd4-e4 3. Lh6-g7+ (und mehrdeutig weiter) ließe sich durch die Erweiterung '... keine Zugwiederholungen und keine Pendelzüge ...' (mühsam) kitten. Beim Autor hieß es bei dieser ich-weiß-nicht-wie-vielten Fassung nur 'letzte 9 (!) Einzelzüge ohne Wiederholung).
HHS meint ohnehin, daß es das ganze auch ohne die einengende Zusatzbedingung schon gibt.
Das von einem Löser angegebene 1. Ld3-h7# Th1-h8 2. Lh8-g7 Tg1-h1 3. Se1-g3 g2-g1=T 4. Th7-h8=L scheitert allerdings an der Schlagbilanz.
Anton Baumann: Neufassung vergl. P0006288 (2023-01-06)
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Keywords: En passant, Last Moves?, Non-standard material, Castling (wl), Promotion (L), Valladao Task (WWW)
Genre: Retro
FEN: qrB2brr/Bp2p1BB/pR3P2/1Q6/2Pk1P2/B1p2R2/2P3N1/2KR1N2
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2019-08-11 more...
18 - P0000324
Josef Haas
8259 Die Schwalbe 143 10/1993
P0000324
(7+5)
a) Wer setzt in 1 Zug matt?
b) Auf welchem Feld muß ein schwarzer Bauer eingefügt werden, damit die andere Partei als in a) mattsetzt?
b) (+sBc7) 1. ... Lg8xe6#
a) 1. Tg6#
1) R: 1. ... Kg6xBf6! 2. g5xf6ep++ f7-f5 3. La2-b1+
play all play one stop play next play all
"Vermutlich aus der Kleinkunstkiste des Autors hervorgekramt.
a) sollte einfach formuliert sein: 'Matt in 1 Zug' - denn wie es hier heißt, klingt es als ob nur einer mattsetzen kann. Das aber ist nicht der Fall, denn beide können's: 1. ... Lxe6# und 1. Tg6#. Üblicherweise hat Weiß das Prae und kann darauf pochen, den Schwarz hat einen altklassischen letzten Zug: 1. ... Kg6xBf6! (nebst 2. Bg5xBf6ep++ Bf7-f5 3. La2-(x)b1)" (HHS);
also ist Weiß am Zug und setzt matt mit 1. Tg6#.
b) Nach Einfügen eines sBc7 geht die o.g. Rückzugfolge nicht, weil der wK nicht auf die 8. Reihe gelangen kann. Also Schwarz am Zuge und 1. ... Lxe6#
"Allzubekanntes - kein Problem für Schwalbelöser" (HHS)
Wenn das alles so bekannt ist, erstaunt doch sehr, daß nur drei Löser die Autorintention nachvollziehen konnten. Alle anderen Löser (5) kamen zu genau entgegengesetzten Erkenntnissen (in a) setzt Schwarz matt, in b) Weiß), was wohl durch die nicht ganz konventionelle Formulierung suggeriert wurde. Ich find's ein interessantes Beispiel für Massenhypnose! (GL) 2/I/3L.
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)
more ...
comment
Keywords: Add pieces, No legal last move for Black, En passant in the retro play
Genre: Retro
FEN: 4K1br/1p4pP/4Pk2/R7/3P4/8/8/1B4R1
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-01-03 more...
19 - P0000404
Michel Caillaud
Die Schwalbe 80 04/1983
P0000404
(11+15) C+
BP in 45,0
1. Sf3 a5 2. Sd4 a4 3. Sb3 axb3 4. a3 Ta4 5. Ta2 bxa2 6. Tg1 axb1=S 7. Th1 Sc3 8. Tg1 Sd5 9. Th1 Sb6 10. Tg1 Sa8 11. Th1 b6 12. Tg1 La6 13. Th1 Dc8 14. Tg1 Db7 15. Th1 Df3 16. gxf3 h5 17. Lh3 h4 18. Le6 h3 19. La2 Lc4 20. Lb1 La2 21. b3 Thh4 22. Lb2 Thb4 23. Lf6 gxf6 24. Kf1 Lh6 25. Kg1 Kf8 26. Df1 Kg7 27. Dg2+ hxg2 28. h4 Kg6 29. Kh2 g1=S 30. Kg2 Lf4 31. h5+ Kg5 32. Th4 Lh2 33. Td4 Sh3 34. f4+ Kh4 35. Kf3 Sg5+ 36. Ke3 Kh3 37. h6 Kg2 38. h7 Kf1 39. h8=T Sh7 40. Kf3 Ke1 41. Kg2 Kd1 42. Kf1 Kc1 43. Ke1 Kb2 44. Kd1 Ka1 45. Kc1 Lg1
play all play one stop play next play all
Reto: This is now the new record for the longest computer tested SPG. Strategy seeking takes up all the time, playing the 800 strategies on the other hand is done within a few seconds. (2023-08-11)
Reto: With the upcoming v1.6, this is solved by Stelvio in around 20min. The improvement is due to the fact that the cage in the south-west is now detected (either the wRa1 is captured inside the cage or the wBc1 is captured at home) (2023-09-02)
comment
Keywords: Unique Proof Game, Move Length Record, Non-standard material (ss), Promotion (ssT)
Genre: Retro
Computer test: Stelvio 1.5 in about 2.5 days.
FEN: nn4nR/2pppp1n/1p3p2/8/rr1R1P2/PP6/b1PPPP2/kBK3b1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-08-12 more...
20 - P0000411
Andrey Frolkin
4429 Die Schwalbe 82 08/1983
2. Lob
P0000411
(9+13) cooked
BP in 40,0
1. d4 e5 2. d5 f5 3. d6 Le7 4. dxe7 d5 5. e4 d4 6. Dg4 fxg4 7. c4 d3 8. c5 Dd6 9. b4 Kd7 10. cxd6 c5 11. b5 c4 12. b6 Sc6 13. e8=T d2+ 14. Ke2 c3 15. Kd3 d1=L 16. Sd2 Sd4 17. Tf8 Kc6 18. Tf4 exf4 19. e5 h5 20. e6 Lb3 21. e7 Lf7 22. bxa7 b5 23. e8=T b4 24. Te6 b3 25. Th6 gxh6 26. d7 b2 27. d8=T Lce6 28. Td5 Te8 29. Kc4 b1=L 30. a8=T Lh7 31. Ta3 c2 32. Tg3 Se7 33. a4 Thf8 34. a5 Sg6 35. a6 Sh8 36. a7 fxg3 37. a8=T gxh2 38. T8a3 Lhg8 39. Th3 gxh3 40. Se4 Lxd5+
play all play one stop play next play all
Cook: 1. b4 e5 2. d4 Le7 3. d5 f5 4. d6 h5 5. dxe7 d5 6. b5 Dd6 7. b6 Kd7 8. bxa7 b5 9. c4 b4 10. c5 d4 11. cxd6 c5 12. e8=T Sc6 13. Tf8 c4 14. e4 d3 15. Dg4 d2+ 16. Ke2 c3 17. Kd3 d1=L 18. Sd2 fxg4 19. Tf4 exf4 20. e5 Sd4 21. e6+ Kc6 22. e7 b3 23. e8=T b2 24. Te6 Lb3 25. Th6 Lf7 26. d7+ gxh6 27. d8=T Lce6 28. Td5 Te8 29. Kc4 b1=L 30. a8=T Lh7 31. Ta3 c2 32. Tg3 Se7 33. a4 Thf8 34. a5 Sg6 35. a6 Sh8 36. a7 fxg3 37. a8=T gxh2
38. T8a3 Lhg8 39. Th3 gxh3 40. Se4 Lxd5+
Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 29:20:32 Stunden. (hh:mm:ss)
Da NUPG C+
Notation: 1.b4 e5 2.d4 Le7 3.d5 f5 4.d6 h5 5.dxe7 d5 6.b5 Dd6 7.b6 Kd7 8.bxa7 b5
9.c4 b4 10.c5 d4 11.cxd6 c5 12.e8=T Sc6 13.Tf8 c4 14.e4 d3 15.Dg4 d2+ 16.Ke2 c3
17.Kd3 d1=L 18.Sd2 fxg4 19.Tf4 exf4 20.e5 Sd4 21.e6+ Kc6 22.e7 b3 23.e8=T b2
24.Te6 Lb3 25.Th6 Lf7 26.d7+ gxh6 27.d8=T Lce6 28.Td5 Te8 29.Kc4 b1=L 30.a8=T Lh7
31.Ta3 c2 32.Tg3 Se7 33.a4 Thf8 34.a5 Sg6 35.a6 Sh8 36.a7 fxg3 37.a8=T gxh2
38.T8a3 Lhg8 39.Th3 gxh3 40.Se4 Lxd5+
3 schwarze Läufer stehen zum Schluss auf den weißen Felder d5, f7, g8. (2023-09-21)
comment
Keywords: Non-Unique Proof Game, Ceriani-Frolkin Theme (TTTTT), Non-standard material (ll), Promotion (TlTTlTT)
Genre: Retro
FEN: 4rrbn/5b2/2k4p/3b3p/2KnN3/7p/2p2PPp/R1B2BNR
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2023-09-22 more...
21 - P0000467
Mechislav Palevich
5151 Die Schwalbe 93 06/1985
1. ehrende Erwähnung
P0000467
(14+11)
Löse die Stellung auf!
R: 1. c2-c3+ a6-a5 2. La2-b1 Tb5-d5 3. Ld5-a2+ Tb1-b5 4. g5-g6 Th1-b1 5. g4-g5 h2-h1=T 6. g3-g4 h3-h2 7. h2xSg3 Se4-g3 8. a2-a3 Sc5-e4 9. Tb1-b7 Sb7-c5+ 10. Tf1-b1 h4-h3 11. Tf8-f1 h5-h4 12. f7-f8=T h6-h5 13. f6-f7 h7-h6 14. f5-f6 f6xLe5 15. Lg3-e5+ Ke5-d4 16. Lf2-g3+ Kf4-e5
play all play one stop play next play all
Henrik Juel: Solution: -1.c2 a6 -2.Ba2 Rb5 -3.Bd5 Rb1 -4.g5 Rh1 -5.g4 R=h2 -6.g3 h3 -7.h2:S Se4 -8.a2 Sc5 -9.Rb1 Sb7 -10.Rf1 h4 -11.Rf8 h5 -12.R=f7 h6 -13.f6 h7 -14.f5 f6:B -15.Bg3 Ke5 -16.Bf2 Kf4 etc. Last 14.0 moves determined. (2003-06-06)
Hans-Jürgen Manthey: R: 1. c2-c3# a6-a5 2. La2-b1 Tb5-d5 3. Ld5-a2+ Tb1-b5 4. g5-g6 Th1-b1 5. g4-g5 h2-h1=T 6. g3-g4 h3-h2 7. h2xSg3 Se4-g3 8. a2-a3 Sc5-e4 9. Tb1-b7 Sb7-c5+ 10. Tf1-b1 h4-h3 11. Tf8-f1 h5-h4 12. f7-f8=T h6-h5 13. f6-f7 h7-h6 14. f5-f6 f6xLe5 15. Lg3-e5+ Ke5-d4 16. Lf2-g3+ Kf4-e5 17. Lf3-d5 Kg5-f4 18. Td4-d6 Kh6-g5 19. Te4-d4 Td4-d7 20. Te1-e4 Tb4-d4 21. Ld4-f2 Tb1-b4 22. Kd7-c6 b2-b1=T 23. Sc6-b8 b3-b2 24. Lb2-d4 b4-b3 25. Le2-f3 De3-a7 26. Sd4-c6 Dg5-e3 27. Kc6-d7 a7-a6 28. Kb5-c6 Dh5-g5 29. Kh4-g5 Sd6-b7 30. d7-d8=L Lb7-a8 31. c6xTd7 Td8-d7 32. Kb3-a4 Th8-d8 33. Df8-c8 b5-b4 34. Da8xf8 De8-h5 35. La3-b2 Kg6-h6 36. f4-f5+ Kf7-g6 37. c5-c6 Se4-d6 38. b4xTc5 Td5-c5 39. Kb2-b3 Td8-d5 40. Sb3-a1 Tb8-d8 41. Kc1-b2 Dd8-e8 42. Th1-e1 Ke8-f7 43. Lf1-e2 d7xTe6 44. Sa5-b3 f7-f6 45. Sc4-a5 Sf6-e4 46. Te1-e6 Sg8-f6 47. Td1-e1 Lc8-b7 48. Df3-a8 Ta8-b8 49. O-O-O b6-b5 50. Dd1-f3 b7-b6 51. e2xSd3 Sf2-d3+ 52. Lc1-a3 Se4-f2 53. f2-f4 Sc5-e4 54. Sf3-d4 Sa6-c5 55. b2-b4 Sb4-a6 56. Sa3-c4 Sc6-b4 57. Sb1-a3 Sb8-c6 58. Sg1-f3 (2021-07-22)
comment
Keywords: Last Moves? (14), Retroablösung, Promotion in forward play
Genre: Retro
FEN: bNQB4/qRprp1p1/2KRp1P1/p2rp3/3k4/P1PP4/3P2P1/NB6
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2004-10-27 more...
22 - P0000498
Thomas Volet
3097 Die Schwalbe 62 04/1980
3. ehrende Erwähnung
P0000498
(10+13)
Wie weit muß der sK maximal nach links gelaufen sein?
Beispielauflösung mri:
R: 1. ... Ke1xSf1 2. Se3-f1 Kf1-e1 3. Sf5-e3 Ke1-f1 4. Sg7-f5 Kf1-e1 5. Sf5xTg7 Ke1-f1 6. Se3-f5 Tg8-g7 7. Sf1-e3 Tg7-g8 8. Tg1-g2 Tg8-g7 9. Lf3-h1 Tg7-g8 10. Tg2-g1 Tg8-g7 11. Se3-f1 Kf1-e1 12. Sc4-e3 Ke1-f1 13. Se5-c4 Kf1-e1 14. Sd3-e5 Tg7-g8 15. Kd1-c1 Tg8-g7 16. Sc1-d3 Tb1-b2 17. Le4-f3 Lb2-a1 18. Sd3-c1 Lc1-b2 19. Sb4-d3 Ta1-b1 20. Sd5-b4 Tb1-a1 21. Sc3-d5 Ta1-b1 22. Sb1-c3 Lb2-c1 23. Kc1-d1 Le5-b2 24. Kb2-c1 Ld6-e5 25. Sc3-b1 Ke1-f1 26. Sa4-c3 Kd1-e1 27. Sc5-a4 Tf8-g8 28. Sd3-c5 Tg8-f8 29. Se1-d3 Tc1-a1 30. Tg1-g2 Ta1-c1 31. Lg2-e4 Tc1-a1 32. Lf1-g2 Ta1-c1 33. Sf3-e1 Tc1-a1 34. Th1-g1 Ta1-c1 35. Sg1-f3 Tc1-a1 36. Lg2-f1 Ta1-c1 37. Ld5-g2 Ke1-d1 38. Lc4-d5 Kf1-e1 39. Ld3-c4 Kg2-f1 40. Sf3-g1 Kh3-g2 41. Sd4-f3 Kg4-h3 42. Sb5-d4 Kh5-g4 43. Sc3-b5 Kh6-h5 44. Sb1-c3 Kg7-h6 45. Kc1-b2 Kf8-g7 46. Kd1-c1 Ke8-f8 47. Ke1-d1 Le5-d6 48. Le4-d3 Lg7-e5 49. Lg2-e4 Lf8-g7 50. Lf1-g2 g7-g6 51. g2-g3 g3xBh2 52. Tg1-h1 f4xDg3 53. Dc3-g3 e5xTf4 54. Db2-c3 d6xLe5 55. Dc1-b2 Sg6-h8 56. Dd1-c1 Th8-g8 57. Lb2-e5 Se5-g6 58. Lc1-b2 Sc6-e5 59. b2xDa3 Da5-a3 60. Te4-f4 Dd8-a5 61. Td4-e4 a3-a2 62. Tc4-d4 a4-a3 63. Td4-c4 a5-a4 64. Tc4-d4 Ta4-a1 65. Td4-c4 Tb4-a4 66. Tc4-d4 Tb6-b4 67. Ta4-c4 Ta6-b6 68. Ta1-a4 Ta8-a6 69. a2xSb3 Sd4-b3 70. Th1-g1 c7xSd6 71. Sf5-d6+ Sb8-c6 72. Sh4-f5 Sf5-d4 73. Sf3-h4 Sh6-f5 74. Sg1-f3 Sg8-h6 75. Sh3-g1 a7-a5 76. Sg1-h3
play all play one stop play next play all
Henrik Juel: -1... Ke1:S, wS uncaptures a bR in the NE corner, screens on f1 to let wB out, then on c1 to let bBa1 out; retract this bB to f8 and g7-g6, use 2 S's to extract bK via h3 and c7 to e8, etc. (2003-11-27)
hans: .....Rb1+ Nc1 Ba1 ???? Solution don't work! (2010-05-23)
Henrik Juel: I think the solution I gave in 2003 works fine. Here it is spelled out in more detail.
Following -1... Ke1xSf1 -2.Se3 Kf1 -3.Sf5 Ke1 -4.Sg7 Kf1 -5.Sf5xRg7 Ke1, Black can make tempo retractions with Rg7 allowing -6.Se3, -7.Sf1, -8.Rg1, -9.Bf3, -10.Rg2, -11.Se3 Kf1 -12.Sc4, -13.Se5, -14.Sd3, -15.Kd1 Rg8 -16.Sc1 Rb1 -17.Be4 Bg7 -18.Sd3 Rb2 -19.Kc1 Bf8 -20.Sf4 g7 etc. (2010-05-24)
Mario Richter: Henrik, may I ask you to look at this problem again?
First, the intended solution:
The black King needs only to go to the d-file, not further to the left (mainly by avoiding an early retraction of g7-g6). This can indeed be done by using the two Bishops as shields:
1. Ke1xSf1
2. wSxTg7
3. wS shields on f1
4. now white Bishop get out, providing tempo moves for White
5. wSc1
6. black rook a1, black Bishop b2-c1
7. wS shields on b1
8. now black Bishop and wK can get out
9. sKd1,wSe1
10. wLf1, wTh1
11. wSg1
12. now white Bishop and sK can get out
13. sK back home via g7

So far, so good. But now my question: What happened to the white Pawn h2? (2022-03-21)
Mario Richter: I first thought that the provable retraction "black Pawn g3 x white Pawn h2" will introduce some difficulties in resolving the position. But as my sample resolution given above shows, this is not the case ... (2022-03-21)
Henrik Juel: I see your point, Mario (2022-03-22)
comment

Genre: Retro
FEN: 2b4n/1p1ppp1p/6p1/8/8/PP4P1/prPPPPRp/b1K2k1B
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2022-03-21 more...
23 - P0000541
Tivadar Kardos
3398 Die Schwalbe 67 02/1981
P0000541
(13+12) cooked
BP in 6,0
1. Sf3 e5 2. Sxe5 Se7 3. Sxd7 Sec6 4. Sxb8 Dxd2 5. Dxd2 Sxb8 6. Dd8+ Kxd8
play all play one stop play next play all
Cook: 1. Sh3 d5 2. Sf4 Sf6 3. Sxd5 Se4 4. Sxe7 Sxd2 5. Dxd2 Kxe7 6. Dxd8 Kxd8
(Can transpose B1&B2)
HHS ('Schwalbe' Heft 71, Oktober 1981, S. 349): "ein schwarzer Schuft - lässt seinen Zwillingsbruder von einem weissen Söldner killen, um dessen Platz einzunehmen: ein regelrechter Krimi."
Als Vergleichsaufgaben sind in der 'Schwalbe' angegeben: G. Schweig (P0002287), Mortimer (P1394750), W. Naef & H. Klüver (P0002832)
A.Buchanan: It's curious that if we shift sK to e8 (i.e. make the position fully homebase) then it is sound as BP in 6.0! The magic of homebase! :) I don't know if it was a typo - maybe someone can check the magazine archive? (2021-10-19)
A.Buchanan: WinChloe & YACPDB don't know this composition. (2021-10-20)
Mario Richter: No typo, the position with black Kd8 was intended, the cook was found by the 'Schwalbe'-solvers (Heft 71, Oktober 1981, p. 349) (2021-10-21)
more ...
comment
Keywords: Unique Proof Game, Homebase (W), Impostor (s), Superseded by (P1394733)
Genre: Retro
FEN: rnbk1b1r/ppp2ppp/8/8/8/8/PPP1PPPP/RNB1KB1R
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2021-10-21 more...
24 - P0000545
Andrej N. Kornilow
Andrey Frolkin

3460 Die Schwalbe 68 04/1981
P0000545
(9+12)
Welches waren die letzten 8 Einzelzüge?
R: 1. f7-f8=D# Kd7-e7 2. e7-e8=L+ Kc6-d7 3. d7-d8=S+ Ld8-c7 4. c7-c8=T+ Tc8-b8
play all play one stop play next play all
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
Keywords: Last Moves? (8), Allumwandlung
Genre: Retro
FEN: BrRNBQb1/1pb1k1Pp/1P1ppppp/2p5/2K5/8/7P/8
Input: Gerd Wilts, 1995-06-03
25 - P0000548
Alexander Kislyak
3463 Die Schwalbe 68 04/1981
Dr. L. Ceriani zum Gedenken
P0000548
(10+13) cooked
BP in 46,0
AL: 1. d4 a5 2. c4 a4 3. b4 a3 4. Lb2 axb2 5. a4 c5 6. a5 b5 7. a6 Lb7 8. a7 Da5 9. bxa5 Sc6 10. a6 0-0-0 11. a8=S Sa5 12. Sc7 b4 13. a7 b3 14. a8=S Sf6 15. Sb6 Kb8 16. Se6 dxe6 17. Sd7 Ka8 18. Sxf8 Td5 19. cxd5 c4 20. d6 c3 21. d7 c2 22. d5 c1=L 23. d6 Ld5 24. Ta4 Se8 25. Tf4 f6 26. Sc3 b1=L 27. d8=S b2 28. d7 Lg6 29. Sb7 b1=L 30. d8=S Lf7 31. Sg6 Lf5 32. Sd6 Lg4 33. Sf5 hxg6 34. Sb7 gxf5 35. Sc5 Th4 36. S5e4 fxe4 37. e3 Lf5 38. Ke2 Tg4 39. h4 Sb3 40. h5 Sa1 41. h6 Ld2 42. h7 Le1 43. h8=S Sc7 44. Sg6 Sb5 45. Sh4 Sa3 46. S4f3 exf3+
play all play one stop play next play all
Korrektur siehe 3463v
James Malcom: How in the world can a non-unique PG be cooked? I'm unmarking it, for now. (2021-01-25)
A.Buchanan: Maybe the intended theme was not forced? Haven’t looked at this problem though (2021-01-26)
more ...
comment
Keywords: Non-Unique Proof Game, Ceriani-Frolkin Theme, Non-standard material (ll), Castling, konsekutive Umwandlungen 8, Promotion, Superseded by (P0000582)
Genre: Retro
FEN: k7/4pbp1/4pp2/3b1b2/5Rr1/n1N1Pp2/4KPP1/n2QbBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-01-26 more...
26 - P0000574
Thomas Volet
The Problemist 11/1980
P0000574
(13+14)
Woher kamen die sTTa8 und h8?
Henrik Juel: The rooks have switched place. Following -1... Rb8:S -2.Sb6, bRb8 retracts further to g8, bBd8 retracts to f8 and bPe6 to e7, to allow wK to exit via e6; retract bRh8 to h6 and h7xRg6 etc. wPg4 cannot retract until orig. wRh1 is home. This is the only Volet retro with a K in check, composed in Tom's youth, before he saw the light! (2003-11-25)
Thomas Volet: The exception in this case to the otherwise observed constraint of not having a check in the diagram or stipulation as to who is on the move relates to a collegial colloquy at the time with with the composer J.G. Mauldon, who was active in this task. (2022-03-24)
A.Buchanan: Terrific composition! Often in a Type C composition, one might retract the check to give Type B. But the thematic Ta8 here could not exist without a check in diagram. There are different areas of design space are accessible depending upon level of self-imposed constraints. Some retro composers like Baibakov systematically explore Types B&C as well as Type A. I don't see this as any kind of defect.
I particularly like that the lock on h-file only allows one sT to squeeze itself out. (2022-03-24)
comment
Keywords: Impostor (tt), Which are original men? (tt), Interchange (tt)
Genre: Retro
FEN: r1bbn2r/Kpp3p1/3pppp1/7B/5kP1/1P6/pPPPPP1P/NQB5
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-24 more...
27 - P0000577
Tivadar Kardos
3742 Die Schwalbe 72 12/1981
P0000577
(8+12)
BP in 11,5
1. d4 e5 2. Lg5 Dxg5 3. Sf3 Dxg2 4. Sxe5 Dxh1 5. Sxd7 Dd5 6. Sxf8 Dxa2 7. Sg6 Dxa1 8. Sxh8 Da6 9. Sg6 Dxg6 10. Dd3 Dxd3 11. e3 Dxf1+ 12. Kd2
play all play one stop play next play all
paul: C+ pour les premiers dix coups (Popeye). (2011-09-17)
Sally: Die Reihenfolge der Züge ist festgelegt, aber eine bestimmte Thematik wird nicht gezeigt. Dieser Sachverhalt scheint sehr zur Schwierigkeut der Aufgabe beigetragen zu haben. Es gab nur 2 Löser! (2017-09-06)
A.Buchanan: Up to 10.5, the solution is unique (Jacobi, 2.5 hours on my little laptop). Maybe someone here can push it all the way? (2021-11-26)
Henrik Juel: Natch 3.1 could not even find pos. 1 in several hours (2021-11-26)
comment
Keywords: Unique Proof Game
Genre: Retro
FEN: rnb1k1n1/ppp2ppp/8/8/3P4/4P3/1PPK1P1P/1N3q2
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2005-12-27 more...
28 - P0000582
Alexander Kislyak
3463v Die Schwalbe 72 12/1981
Dr. L. Ceriani zum Gedenken
P0000582
(10+13) cooked
BP in 46,0
AL: 1. b4 a5 2. c4 a4 3. d4 a3 4. Lb2 axb2 5. a4 b5 6. a5 c5 7. a6 Da5 8. a7 Sc6 9. bxa5 Lb7 10. a6 0-0-0 11. a8=S b4 12. Sc7 Sa7 13. Se6 Sb5 14. Sxf8 Sh6 15. a7 f6 16. a8=S Sf7 17. Sc7 b3 18. Sce6 dxe6 19. Ta4 Td5 20. cxd5 c4 21. d6 c3 22. d5 c2 23. Sc3 Sg5 24. d7+ Kb8 25. d6 c1=L 26. d8=S b1=L 27. d7 b2 28. Sf7 Lf5 29. Sd6 Lh3 30. d8=S b1=L 31. S8f7 Lg6 32. Sh6 Lf7 33. Sg6 hxg6 34. Shf5 Th4 35. Tf4 gxf5 36. Sde4 fxe4 37. e3 Lf5 38. Ke2 Tg4 39. h4 Sa3 40. h5 Sb1 41. h6 Ka8 42. h7 Ld5 43. h8=S Ld2 44. Sg6 Le1 45. Sh4 Sh3 46. Shf3 exf3+
play all play one stop play next play all
Cook: NL: 1. b4 a5 2. c4 a4 3. d4 a3 4. Lb2 axb2 5. a4 b5 6. a5 c5 7. a6 Sc6 8. a7 Lb7 9. Ta6 Da5 10. bxa5 0-0-0 11. a8=S b4 12. Sc7 b3 13. Se6 Sb4 14. Sxf8 Sf6 15. Te6 dxe6 16. a6 Td5 17. cxd5 c4 18. a7 c3 19. a8=S c2 20. Sc3 c1=L 21. Sb6+ Kb8 22. Sc8 b1=L 23. d6 Lg6 24. d7 b2 25. d8=D b1=L 26. Da5 Lbf5 27. d5 Se4 28. d6 f6 29. d7 Lf7 30. Sg6 hxg6 31. d8=T Lh3 32. Td4 Th4 33. Df5 gxf5 34. Sd6 Sg5 35. Tf4 Ka8 36. S6e4 fxe4 37. e3 Lf5 38. Ke2 Tg4 39. h4 Sc2 40. h5 Sa3 41. h6 Sb1 42. h7 Ld5 43. h8=S Sh3 44. Sg6 Ld2 45. Sh4 Le1 46. Shf3 exf3+
James Malcom: Again, how is this cooked? (2021-01-25)
A.Buchanan: The AL has 5 Wh S promotions in 46.0 moves but the cook shows this is not necessary. (2021-01-26)
more ...
comment
Keywords: Non-Unique Proof Game, Ceriani-Frolkin Theme (SSSSS), Non-standard material (ll), Castling, Promotion, konsekutive Umwandlungen 8
Genre: Retro
FEN: k7/4pbp1/4pp2/3b1b2/5Rr1/2N1Pp1n/4KPP1/1n1QbBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-01-26 more...
29 - P0000583
Andrey Frolkin
Günter Lauinger

3468v Die Schwalbe 72 12/1981
P0000583
(10+12) C+
h#2 (AP)
Satzspiel:
*1. ... 0-0! 2. cxd4 Tc1#

Verführung:
1. exd3ep? 0-0 2. cxb4 Tf4# aber zuletzt R: 1. d2-d4?? ist illegal, da es den wLc1 als schwarzes Bauernschlagobjekt ausschließt

Lösung:
1. ... 0-0! (AP-Legalierung des weissen Anzugrechtes) 2. cxd4 Tc1#
play all play one stop play next play all
Alle Versuche, das Satzspiel 1. ... 0-0! 2. cxd4 Tc1# durch einen schwarzen Vorschaltzug aufrecht zu erhalten, scheitern.
Mario Richter: 1. exd3ep 0-0 2. cxb4 Tf4 ist nur Verfuehrung (scheitert an schwarzer Bauern-Schlagbilanz).
Loesung: 1. ... w0-0! (AP-Legalierung des weissen Anzugrechtes) 2.cxd4 Tc1# (2009-02-10)
A.Buchanan: Supersedes P0000553 (2022-01-08)
Mario Richter: Hi Andrew, I only quoted from the "official solution" ('Die Schwalbe' Heft 72, 12/1981, p.399).
Perhaps a better way to get anwers to your interesting questions is to ask the authors directly ... (2022-01-08)
A.Buchanan: AP Type Petrovic is a try, because wLc1 was captured at home, yet Black still made 6 pawn captures.
AP Type Keym is the actual solution, because if it was really BTM, then White would have lost castling rights.
So there are two kinds of AP here. (2022-03-21)
more ...
comment
Keywords: Castling (wk), a posteriori (AP) (Type Keym), En passant as key, a posteriori (AP) (Type Petrovic)
Genre: h#, Retro
Computer test: BC+ Popeye v4.87
FEN: 8/2p5/1pPp4/bRpP4/BPkPp3/qp2p2p/rP2P3/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
30 - P0000598
Nikita M. Plaksin
Andrej N. Kornilow

3876 Die Schwalbe 74 04/1982
P0000598
(4+3)
h=2 (AP)
Circe, Monochromes Schach
Intended solution:
1. dxc3ep[+wBc2] 0-0 2. axb1=S Kh2=
2. ... Kf2? 3. Kf3!
1. ... Tf1? 2. axb1=S Kf2? 3. Kf3!
play all play one stop play next play all
Erich Bartel: vom Dual 2.-- Kf2/Kh2 abgesehen C+ PY V4.41.--- (2008-11-07)
A.Buchanan: This is cooked in Popeye because White can't castle as long as bK controls f1! This is quite reasonable behaviour in monochrome: compare to en passant! So no solution (2022-02-18)
comment
Keywords: a posteriori (AP) (Type Petrovic), Circe, En passant as key, Monochromatic Chess, Castling (wk), Miniature, Golden Age (Monochrome castling)
Genre: Retro, Fairies
FEN: 8/8/8/8/2Pp4/8/p3k3/1N2K2R
Reprints: (IV) Quartz 4 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-18 more...
31 - P0000604
Andrej N. Kornilow
3948 Die Schwalbe 75 06/1982
7. Lob
P0000604
(24+0)
Färbe die Steine!
Welches waren die letzten 11 Einzelzüge?
James Malcom: Solution? (2020-11-07)
Mario Richter: I do not know the official solution, but the following works:
wLf8 wBe7 wBg6 wBh6 wLh5 wBe4 wTf4 wKg4 wDh4 wSf3 wDg3 wLh3 wSf2 wBg2 wSh2
sBb7 sBc7 sTf7 sBh7 sBd6 sBe6 sKf6 sBe5 sBa3
R: 1. f5xg6ep g7-g5 2. Sg5-f3 a4-a3 3. Kf3-g4 a5-a4 4. Tg4-f4 a6-a5 5. f4-f5 Kf5-f6
6. e3-e4
wCaps: f5xg6ep d7xTe8=L c6xLd7 d6xDe7 b6xLa7 a5xSb6 b6xSa7
wProms: d7xTe8=L a7-a8=S a7-a8=D
sCaps: f6xTe5 (2020-11-10)
comment
Keywords: Colouring problem, En passant, Last Moves? (11)
Genre: Retro
FEN: 5B2/1PP1PR1P/3PPKPP/4P2B/4PRKQ/P4NQB/5NPN/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
32 - P0000615
Andrey Lobusov
4088 Die Schwalbe 77 10/1982
14.Lob
P0000615
(15+6) cooked
h#2* (AP)
1. ... Kg5 2. 0-0?? Se7#
1. ... Txh7 2. Kf8 Txh8#

1. cxb3ep Txh7 2. Kf8?? Txh8#
1. cxb3ep Kg5 2. 0-0! Se7#
play all play one stop play next play all
Wh captured 10 by pawns (including hxPg-g8) but not on the last move. wPa3 came from a2: wPa5 captured bPa.
Bl took only b3xa2.
In set play, Bl has no last move except for K or R.
For the main play, the intention is that if Bl retains castling rights, last moves were R: 1. b2-b4 b3xa2 e.g. 2. Bg1-d4 b4-b3 3. d4xc5 etc. However there are retro cooks e.g. R: 1. Bg1-d4 b3xa2 2. d4xc5 etc, with wPa5 the original wPb. wPb4 & wPb5 come from further East.
Cook: Retro logic for main play is unsound.
A.Buchanan: Not trivial to fix as shifting wPa5 to b6 leads to promotion cooks. Adding black material must be carefully done to avoid disrupting uniqueness of e.p. as tempo. (2022-03-11)
A.Buchanan: OK just to say I have found a form of this idea that works. This one was the very devil to fix - the final form includes different mates and retro logic, so becomes an "after" rather than a "correction", and I will find somewhere else to publish first. (2022-03-19)
more ...
comment
Keywords: Castling (sk), a posteriori (AP) (Type Petrovic), En passant as key, Tempo Move, Superseded by (P1409841)
Genre: h#, Retro
Computer test: Popeye v4.87 & simple retro-logic
FEN: 4k2r/5p1p/5K1R/PPP2N2/1PpB4/P1P5/p1RNB1P1/Q7
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-20 more...
33 - P0000641
Werner Frangen
2201 Die Schwalbe 46 08/1977
P0000641
(14+5)
#4 (AP)
1. bxc6ep
play all play one stop play next play all
Nach 1. bxc6ep ist zwar ein Matt in 2 Zügen möglich, aber Weiß muss noch rochieren, um den ep-Schlag zu begründen.
James Malcom: Full solution? (2021-02-05)
Henrik Juel: White pawns captured all 11 missing black men by axb, f2x..xa7, and g2x..xb7
Black captured hxg and once more, so Pc5 never captured
If last move was c6-c5, the preceding white move was with Ke1 or Th1
So if White can castle, last move was c7-c5
1.bxc6ep dxc6 2.0-0(AP) c5 3.Ta1,Sd7 and #4
2.Sd7 c5 3.0-0(AP) is also possible
I expect the full solution was omitted because it is very dualistic (2021-02-05)
Anton Baumann: Autorabsicht: 1.bxc6ep [2.0-0 3.Lb5#] bxa3 2.0-0 axb2 3.Tb3 nebst 4.Lb5,Tf4#
1. ... dxc6? 2.0-0 nebst 3.Tf4,Lb5#
Dual: 2.Tb3 dxc6 3.0-0 [4.Tf4#] c5 4.Lb5# 2. ... a2,axb2 3.0-0 [4.Lb5#] dxc6 4.Tf4# (2021-02-05)
A.Buchanan: C+ of helpmates with this kind of AP is relatively simple: one can just eliminate all lines without 0-0. But C+ of d# is harder.
(1) Henrik's retro logic is solid. There is no #4 unless 1.bxc6ep, so let's take that as a given: an AP debt is incurred.
(2) Black has 5 responses. Against 4 of them, 2.0-0 gives a genuine short mate with the AP debt paid. Such short mates don't impact soundness in a retro #4, to my mind.
(3) This leaves 1... bxa3. Following 2.0-0 axb2 (the only full length line) 3.Rb3 thr 4.Bb5,Rf4# but both mates are separately provided, so only minor duals.
(4) It remains to consider if White can delay castling after 1...axb3. Apart from 2.Tb3 nothing works, but 2.Tb3 thr 3.0-0 is unstoppable and is the major dual to my mind. I don't see any obvious fix. (2021-02-05)
comment
Keywords: Castling (wk), En passant as key, a posteriori (AP)
Genre: Retro, n#
FEN: 1N6/PP1p4/BR6/pPp5/kp6/B2P4/1PP1P2P/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
34 - P0000649
André Hazebrouck
2261 Die Schwalbe 47 10/1977
4. Preis
P0000649
(8+14)
h#2.5
b) AP (2 Lösungen)
a) 1. ... 0-0-0 2. 0-0 Tdg1+ 3. Kh8 Txh6#
b) also 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#
is clearly the intention
play all play one stop play next play all
Can count 1+7 visible pawn captures, leaving 1+1 unexplained. Need to resolve wPfgh & bPg. Suppose bK never moved, then one of:
1) wPf waylaid, wPhxPg=, wPg=. Here w00 right can remain.
2) Or similarly, wPh waylaid, wPfxPg=, wPg=. Again w00 right can remain.
3) bPgxPh=, wPfxg= (or wPfxe then captured by bPf), wPg= w00 right lost.
1+1 captures explained in either case.
On the other hand if bK moved, then maybe wPf=, and only requires one more capture to resolve g&h files. The w00 right can remain. So there is a captured unit unaccounted for, and we can't validate the ep.
On the other hand if we are in case 3 above, then we can't know that wPbxa captured dark bishop, so White might retract a6-a7.
We also need to know that w00 rights remain. s000 rights (i.e. prior movement of sTa8) are not relevant.
If we know that w00, w000 & b000 remain, then we are in case 1 or 2 above with bPg captured unpromoted. Therefore wPb6xLa7 due to bishop shade. Therefore R: 1. b7-b5 b6xLa7 to unblock.

Only one candidate solution does not begin with ep: 1. ... 0-0-0 2. 0-0 Tdg1+ 3. Kh8 Txh6#. So this is the solution for a).

Let ????? denote validity of w00,w000,b00,b000,ep. Possibilities are: YYY?Y and all of YYN??, YNY??, NYY??, YNN??, NYN??, NNY??, NNN??. So there are 30 possibilities.
Under PRA, the solution parts would be YYYYY, YYNYN, YNYYN & NYYYN. For the first there are 20 solutions, while the second and third have 0 solutions. So this is not the right paradigm.
Under SPRA, there would be a single solution part YYYYY with 20 solutions. So this is not the right paradigm either.

Under RS with AP, the solution of a) still works. This comes from ?YY??. On the other hand, if the first move is ep, then we are in YYYYY. So all castlings are valid. But when we get to perform the mate, we need to know that based on castlings actually performed, the ep is valid. So the solution must include w00. So is 1. ... cxb6ep 2. La6 0-0 3. 0-0-0 a8=D#.

By combining the information of both of these solutions, we know we are in YYY?? so YYY?Y is the only possibility, and ep is legal. However, the combination of these two solutions would also validate any of the other 18 ep solutions that do not include 2. ... 0-0. They contribute no new evidence, but how to exclude them?

This is one of the issues with "AP Consolidation". Normally the number of solutions is not a constraint. But here perhaps we should insist that *only* two solutions are allowed?
Cook: 1. ... cxb6ep 2. La6 Tb1,~ 3. 0-0-0 a8=D#
18 different possibilities for W2 which are validated by the same logic that validates 2. ... 0-0.
Henrik Juel: [I don't follow the silly convention of writing black moves first]. 1.0-0-0 0-0 2.dTg1+ Kh8 3.Txh6#. b) If Ke1,e8 and Th1 never moved, White captured f/hxPg and b6xLa7; if Ta1 also never moved, last white move was b6xLa7. 1.cxb6ep La6 2.0-0/0-0-0 0-0-0 3.a8Q#. The a) solution also works in b). (2003-12-18)
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comment
Keywords: a posteriori (AP) (Type Petrovic), Castling (wkwlsk), En passant as key
Genre: h#, Retro
Computer test: Forward logic Popeye v4.87 & basic retro-logic works, but then AP Consolidated protocol seems to cook the problem.
FEN: r1b1k2r/P2p4/2p1n2p/ppPq4/1np5/p7/P2PP3/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
35 - P0000653
Eliahu Fasher
2311 Die Schwalbe 48 12/1977
P0000653
(12+14) cooked
h#2 (wer?)
Kees: possible fix: Lb1=Sb1 -De1 +Ld1
White begins: 1.Kxb7 Lxe2 2.Kc8 La6#
(1.Txd8+ Kxd8 2.Kf8 Th8# illegal for white has no last move) (2023-06-07)
A.Buchanan: Your fix is good, Kees. It removes the cook, and sLd1 denies R: 1. c2xb3 as well as sLb1 did. Note R: 1 Sc6-d8 0-0+? as black castling rights were lost to let wK enter the back rank. (2023-06-08)
more ...
comment

Genre: h#, Retro
FEN: 1bKN1rk1/1ppn1r1R/5p1P/4pP1p/3p1p2/1PP3P1/PP2P3/Rb2q3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-08 more...
36 - P0000656
Werner Frangen
2314 Die Schwalbe 48 12/1977
1. Preis
P0000656
(11+9)
RV.
h=2,5
(where here '=' means 'draw')
Text stipulation that PDB move engine can't understand: "RV. Weiß am Zug macht im 3. Zug Remis mit Hilfe von Schwarz."
Henrik Juel: Black pawns captured all missing white men by a7xb6xc5xd4xe3 and h2xg1=L; White captured [Lc8, Lf8] with officers and a2xb3, e2xd3, f2xe3xd4 (after Black played e3-e2), and h2xg3.
If last move was Ka2-a1, the play presumably is 1.Kc2 Ka2 2.Kc3 Ka1 3.Kc2;
if last move was Kb1-a1, the play presumably is 1.Kb3 Kb1 2.Kc3 Ka1 3.Kb3.
But where are the three-fold repetitions justifying remis? Last white move could be e3xd4. (2012-02-18)
Mario Richter: Genesis of the position:
bBa3 is bPh7 which promoted on g1 by Ph2xg1=B.
White pawn captures: Ph2xg3, Pf2xe3xd4, Pe2xd3, Pa2xb3
Black pawn captures: Pa7xb6xc5xd4xe3, Ph2xg1
Therefore Henrik's suggestion wPe3xd4 is illegal, since then the bB has no chance to reach g1. (2012-02-22)
Henrik Juel: Good point, Mario; this probably gives rise to three-fold repetition. (2012-02-22)
A.Buchanan: (1) R: 1. Ka2-a1 Kc2-c3 2. Ka1-a2 then e.g. c3-c4
Forwards: 1. ... Kc2 2. Ka2 Kc3 3. Ka1 Kc2=
(2) R: 1. Kb1-a1 Kb3-c3 2. Ka1-b1 then e.g. c3-c4
Forwards: 1. ... Kb3 2. Kb1 Kc3/Ka4 3. Ka1 Kb3= so dualized!
Is this right?
Trying h=3.0 for fun:
(1) R: 1. Kc2-c3 Ka2-a1 then e.g. 2. c3-c4
Forwards: 1. Ka2 Kc2 2. Ka1 Kc3 3. Ka2 Kc2=
(2) R: 1. Kb3-c3 Kb1-a1 then e.g. 2. c3-c4
Forwards: 1. Kb1 Kb3 2. Ka1 Kc3/Ka4 3. Kb1 Kb3= so again would be dualized! (2021-12-05)
more ...
comment
Keywords: Partial Retro Analysis (PRA), Draw by repetition
Genre: Retro
FEN: 8/1pppppp1/8/8/1PPP4/b1KP2P1/1P1Pp1P1/k1B2B2
Reprints: (C) feenschach 46 04-06/1979
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-08-21 more...
37 - P0000674
Leonid M. Borodatow
2475 Die Schwalbe 51 06/1978
P0000674
(12+11) cooked
h#3
1. ... Kgxf4 2. Tf6 e6 3. Tf8 Sg7#
play all play one stop play next play all
Cook: 1. ... exf6ep 2. 0-0-0 gxf4 3. Td7 a8=D
Failed try, as Bg4xBh5 allows bRh8 to have escaped the cage prior to ep.
Anton Baumann: Sollte eine Verbesserung von P0000777 sein.
Autorabsicht: "letzter Zug f7-f5 sonst retropatt! Jetzt ist aber die Rochade illegal, da der sTh8 nur über e8 aus seiner Ecke kommt. Daher Verführung 1. ... exf6 e.p. 2.0-0-0? Lxf4 3.Td7 a8=D#. Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 sg7#"
Aber: Der s h-Bauer kann auch auf h5 geschlagen worden sein; so konnte der sT über h6 aus seiner Ecke entkommen. Die vermeintliche Verführung ist also NL! (2022-12-14)
comment
Keywords: Castling (sg), Superseded by (P0000058)
Genre: h#, Retro
FEN: r3k3/P7/p1r3pP/4PpBN/3ppnKn/6P1/1PP2PPq/7N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
38 - P0000680
Tivadar Kardos
2527 Die Schwalbe 52 08/1978
P0000680
(4+2) C+
#1
b) Kb1 nach g1
a) 1. ... Kxa1 2. 0-0#
b) 1. ... Kxh1 2. 0-0-0#
play all play one stop play next play all
A.Buchanan: The term "retro" is jungle not garden - that means we should not expect an axiomatic definition. The current problem is a case in point. Case law has established that neither simple employment of the castling convention nor existence of check are sufficient to make a problem "retro". But all this problem has is the quirky use of Codex Article 15 to force BTM. So I think this problem has to be retro. The key point is that nothing hinges on the retro-ness. If the problem included 50M or DP, then one would expect a more solid foundation. As it is, all we need is the free direct mate in 1 that comes as part of the retro paradigm. (2021-11-26)
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comment
Keywords: Castling (wk), No legal last move for Black, Minimal, Miniature
Genre: Retro
Computer test: C+ Popeye v4.87
FEN: 8/8/8/8/8/4p3/4Q3/Rk2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-26 more...
39 - P0000684
Simo Ylikarjula
2531 Die Schwalbe 52 08/1978
P0000684
(15+15) cooked
BP in 18,0
NL: 1. d4 d6 2. Kd2 Le6 3. g4 f5 4. g5 Lf7 5. g6 Sd7 6. e4 Db8 7. gxf7+ Kd8 8. Kc3 g5 9. Kc4 g4 10. Kd5 Lg7 11. Ke6 Kc8 12. f8=L g3 13. Kf7 g2 14. Ke8 gxf1=L 15. e5 Lg2 16. Se2 d5 17. Lg5 Lh3 18. e6
Die Autorlösung wurde nie publiziert und ist unbekannt.
play all play one stop play next play all
Moldenhauer: Computerprüfung: C+ NUPG cooked Stelvio 1.11 in 1 Sekunde.
Keine Lösung: BP 17.0. BP 17.5 cooked.
Beispiel BP 18.0: 1.d4 Sa6 2.Kd2 Sb8 3.Kc3 d6 4.e4 Le6 5.Se2 Sd7 6.e5 Db8 7.g4 Kd8
8.g5 f5 9.g6 Lf7 10.gxf7 Kc8 11.Kc4 g5 12.Kd5 Lg7 13.Ke6 d5 14.f8L g4 15.Kf7 g3
16.Ke8 g2 17.Lg5 gxf1L 18.e6 Lh3
Beispiel BP 17.5: 1.d4 Sa6 2.Kd2 Sc5 3.Ke3 d5 4.Kf4 Le6 5.Ke5 Kd7 6.e4 Db8
7.g4 Kc8 8.g5 f5 9.g6 Lf7 10.gxf7 Sd7+ 11.Ke6 g5 12.Se2 Lg7 13.e5 g4 14.Lg5 g3
15.f8L g2 16.Kf7 gxf1L 17.Ke8 Lh3 18.e6 (2023-05-07)
comment
Keywords: Non-Unique Proof Game, Non-standard material
Genre: Retro
FEN: rqk1KBnr/pppnp1bp/4P3/3p1pB1/3P4/7b/PPP1NP1P/RN1Q3R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2014-05-20 more...
40 - P0000736
Tivadar Kardos
2935 Die Schwalbe 59 10/1979
P0000736
(10+14) C+
BP in 7,0
1. Sc3 Sf6 2. Se4 Sxe4 3. g3 Sxg3 4. Sf3 Sxh1 5. Se5 Sxf2 6. Sxd7 Sxd1 7. Sxb8 Txb8
play all play one stop play next play all
Sally: Eine eindeutige Zugfolge, was bei solchen Aufgaben nicht oft zutrifft. Nicht ganz leicht. Der Rekord steht bei 41,5 Zügen.
(BS). (2017-09-06)
Henrik Juel: 41.5 moves is a very old record
The current record length for proof games is 58.5 moves in a problem by Pronkin, Frolkin & Keym, problem C, p.7 in Die Schwalbe, heft 283, February 2017
The former record was 57.5 moves in a problem by Pronkin & Frolkin, Die Schwalbe, 1989 (V) (2017-09-06)
Olaf Jenkner: The current record length for proof games is 57.5 moves (P0000136) because P1338946 has been cooked. (2021-11-25)
comment
Keywords: Unique Proof Game, Superseded by (P1396199)
Genre: Retro
Computer test: (Natch 2.2Beta1 Copyright (C) 1997,98,99,2001,2002,2003 Pascal Wassong)
FEN: 1rbqkb1r/ppp1pppp/8/8/8/8/PPPPP2P/R1BnKB2
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-25 more...
41 - P0000758
Gerd Rinder
1033 Die Schwalbe 21 06/1973
1. Preis
P0000758
(7+11)
Remis (AP)
Weiß ist patt. 1. cxb6ep ist nur zulässig, wenn Schwarz diese a posteriori durch die Rochade rechtfertigt. Weiß kann aber so spielen, daß Schwarz nicht rochieren kann, z.B. 1. ... Lxc7 2. bxc7
play all play one stop play next play all
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)
more ...
comment
Keywords: En passant as key, Castling (sg), a posteriori (AP)
Genre: Retro, Studies
FEN: r2bk3/p1Rpp3/P1p2p2/KpP2P2/1P2p3/1P6/p7/8
Reprints: (4) Problem 161-164 11/1973
317 Europe Echecs 217 01/1977
(A) Die Schwalbe 80 04/1983
408 Eigenartige Schachprobleme 2010
M36 mpk-Blätter 12/2011
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-31 more...
42 - P0000759
Luis Alberto Garaza
1034 Die Schwalbe 21 06/1973
P0000759
(9+9) cooked
Schwarz am Zug, Weiß gewinnt
1. ... fxg3ep 2. hxg3+ Kh5 3. f4 Kh6 4. h8=T+! Kg7
play all play one stop play next play all
Cook: Intended a Posteriori combo of castling and ep rights retro-locks wTh in cage, contradicting pawn capture balance
hans: only black move is fxg3e.p.
1. fxg3+ Kh5 2. h8D/T#
Why the keyword 'castling'?
Castling is illegal, for there are 7 black captures, and because white's last move is g2-g4, needs the wK let pass Rh1 (2012-05-18)
Anton Baumann: g2-g4 war nicht zwingend der letzte Zug! Schwarz ist patt, also unlösbar!
Korrektur in 'die Schwalbe' 12/2074 S.264: "wTa1 nach h1. Nun gewinnt nach 1. - fxg3 Weiss mit 2.hxg3+ Kh5 3.f4 Kh6 4.h8=T+! Kg7 und der umgewandelte Turm holt die störenden sBB ab, sodass Weiss rochieren kann und beweist, dass Schwarz nicht patt war. Schwarz versucht die Rochade zu verhindern, ist aber machtlos." (2022-12-06)
A.Buchanan: The forward play is more complicated, as Black can play more aggressively than 3. … Kh6 (2022-12-07)
Anton Baumann: Wenn Schwarz nicht 3.-Kh6! zieht, holt sich Weiss auf h8 die Dame, mit einfacherer Fortsetzung.
3.-g4? 4.f5 Kg5 (4.-g5 5.f6 ~ 6.h8=D) 5.h8=D Kxf5 6.De8 +_
3.-Kg4? 4.h8=D Kf3 5.Dxh3 g4 6.Dh2 Ke4 7.Dg2+ +_ (2022-12-09)
more ...
comment
Keywords: Castling (wl), a posteriori (AP) (Type Petrovic), En passant as key
Genre: Retro, Studies
FEN: 8/7P/6p1/2p3p1/2p2pPk/2Pp1P1p/3PpP1P/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-07 more...
43 - P0000760
André Hazebrouck
1120 Die Schwalbe 23 10/1973
P0000760
(9+10) C+
h#2 (AP)
1. bxc3ep Sa6 2. 0-0-0 Tc4#
play all play one stop play next play all
Henrik Juel: -1.c2 c5xLb4 -2.Lc3 c6 -3.Lb2 c7 -4.Lc1 a6! -5.b2 b2xDSa2 etc. White captured sDTSS by fxexd, gxfxe, allowing Black to capture f7xPe6xPd5 etc. (2003-12-18)
A.Buchanan: Why not e.g. wSb1? (2022-03-04)
Henrik Juel: That also seems to work (2022-03-04)
A.Buchanan: I really like the motivation for ep. I guess the motivation for Rb1 is to be inside the cage that forms around it, but it’s still a bit loose. Note there is no retro try. (2022-03-05)
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comment
Keywords: En passant as key, Castling (sg), a posteriori (AP) (Type Petrovic), Volet Pawn
Genre: h#, Retro
Computer test: HC+ Popeye 4.87 + simple retro reflection
FEN: rN2k3/1p1pp1pp/8/p7/RpP5/PP6/p2PP3/KR6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-05 more...
44 - P0000763
Lothar Finzer
1159 Die Schwalbe 24 12/1973
P0000763
(12+12) C+
#3
Schwarz hat keinen letzten Zug, daher beginnt er (und setzt in 3 Zügen matt):
1. ... T3d2+ 2. Lexd2 Txd2+ 3. Lxd2 Ta1#,Db3#; 3. Lb2 Txb2#,Db3#; 3. Kb1 Txc1#
Verführung: 1. Ta5+? Kxb4 2. Sc6+ Kc4 3. Tc5#
play all play one stop play next play all
Die Umwandlung aller 16 Bauern in schwarzfeldrige Läufer bzw. in Türme erforderte alle 8 Schlagfälle, und zwar so, daß alle Umwandlungsfiguren auf schwarzen Feldern entstanden. Deshalb kann zuletzt auch keine schwarze UW erfolgt sein.
Joaquim Crusats: Can anyone explain the intention? (2013-01-06)
Henrik Juel: The eight white pawns captured the four missing black officers and promoted on b8, d8, f8, and h8; the eight black pawns captured the four missing white officers and promoted on a1, c1, e1, and g1.
If White had the move, 1.Ta5+? Kxb4 2.Sc6+ Kc4 3.Tc5 would mate in three.
But Black has no last move, so he has the move and mates in three by 1... T3d2+ 2.Lexd2 Txd2+ 3.Kb1 Txc1# (2013-01-06)
A.Buchanan: This is one of a number of problems which had been given both "No legal last move for..." & "Whose move?" keywords. These should correspond respectively to two situations: (1) Codex Article 15 where we add or remove a single move from the beginning of the solutions, but the mating party remains the same. (2) A kind of one-sided duplex, where who moved last implies who delivers the mate (or other final move).
So at most one of the two keyword forms can ever logically apply to a problem. Therefore I have removed the incorrect "No legal last move for Black" keyword.
We might replace "Whose move?" by "Halfduplex", which is related to the Popeye option. What do you think? (2023-11-28)
A.Buchanan: This is a nice problem. There are some final move duals for the actual solution, but all missing units are needed to justify promotions (2023-11-28)
Henrik Juel: Halfduplex is a bit too technical for my taste, although it describes the goings on well (2023-11-28)
A.Buchanan: Thanks Henrik - I'll leave the term "Whose move?" as it is then. (2023-11-28)
more ...
comment
Keywords: Non-standard material (LLLLLLLLtttttttt), Whose move?, Aristocrat
Genre: Retro, 3#
Computer test: HC+ Popeye v4.87 (subject to final move duals) with simple retro logic
FEN: 8/N7/8/2R1B1B1/kB1BrBrB/2qrrrrr/K7/2BrBrBr
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-11-28 more...
45 - P0000775
Ferad Kakabadze
1422 Die Schwalbe 30 12/1974
P0000775
(11+5) C+
#3
Welches waren die letzten 2 Einzelzüge?
1. e8=D! f5 2. exf6ep e5 3. Dxe5#
R: 1. b2-b1=L 0-0
play all play one stop play next play all
Henrik Juel: The forward play is C+ Popeye 4.61
1.e8=D thr. 2.Dc8 thr. 3.Dc3#
1... f5 2.exf6ep e5 3.Dxe5#
(1... f6 does not prevent the threat, so 2.exf6 is not considered dualistic) (2020-09-26)
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comment
Keywords: Last Moves?, Valladao Task, En passant, Promotion, Castling in the retro play, Obvious promotion
Genre: Retro, 3#
FEN: 8/3pPp2/3Pp3/4P3/1N3P2/B7/P1P2P2/kb3RK1
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2020-09-27 more...
46 - P0000777
Leonid M. Borodatow
1424 Die Schwalbe 30 12/1974
P0000777
(12+10) cooked
h#2.5
1. ... exf6ep 2. 0-0-0? Lxf4 3. Td7 a8=D# try
1. ... Kxf4 2. Tf6 e6 3. Tf8 Sg7# solution
play all play one stop play next play all
Cook: 1. ... Kxf4 2. Te7 e6 3. Tb8 axb8=D,T#
See P0000674
Henrik Juel: The stipulation should probably read h#2.5: 1.exf6ep 0-0-0 2.Bxf4 Rd7 3.a8Q# (2003-08-18)
Anton Baumann: Letzter Zug war f7-f5, sonst retropatt. Die weissen Bauern haben 6x geschlagen. Der s a-Bauer wurde so ebenfalls als Schlagopfer benötigt und musste 2x schlagen, um sich umwandeln zu können. Weiss hat also auf h6 den s Bauern geschlagen. Unabhängig davon ob der sTe6 ein Umwandlungsturm ist oder nicht, muss der sTh8 über e8 aus der Ecke gekommen sein. Also ist die Rochade illegal!
Autorabsicht: 1. ... exf6 e.p. 2.0-0-0 Lxf4 3.Td7 a8=D# ist die Verführung!
Lösung: 1. ... Kxf4 2.Tf6 e6 3.Tf8 Sg7#
leider geht auch: 1. ... Kxf4 2.Te7 e6 3.Tb8 axb8=D,T# (2022-12-13)
A.Buchanan: I agree Anton. I suggest as a fix -bBa6 +bDc2. I think retro, solution and try all work correctly now. Do you folks agree? (2022-12-13)
Anton Baumann: Der Vorschlag von Andrew ergibt eine korrekte Lösung, aber nicht im Sinne des Autors: ohne einen s Stein auf a6 ist f6-f5+ als letzter Zug möglich, d.h. kein Retropatt weil Weiss a6-a7 zurücknehmen kann! 1. ... exf6 e.p. ist deshalb nicht zulässig. LB wollte aber dass f7-f5 der letzte Zug war, damit aber die Rochade illegal ist! (2022-12-14)
Anton Baumann: Mögliche Korrektur: a6=sL, g2=sD (2022-12-15)
A.Buchanan: Yes I agree. I had overlooked the unmove of wPa7 (2022-12-15)
more ...
comment
Keywords: Castling (sg), Valladao Task, Superseded by (P0000058)
Genre: h#, Retro
FEN: r3k3/P7/p3r1pP/4PpBN/3nnpKP/6PB/1P2PPb1/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-12-16 more...
47 - P0000778
Gideon Husserl
1464 Die Schwalbe 31 02/1975
P0000778
(15+16)
Wieviele Züge hat die KBP?
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+ z.B.
play all play one stop play next play all
10,5
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)
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comment
Keywords: Non-Unique Proof Game
Genre: Retro
FEN: r1bqkbnr/pppppppp/5N2/8/8/8/PPPPPPPP/nNBQKB1R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-20 more...
48 - P0000790
Gideon Husserl
1555 Die Schwalbe 33 06/1975
P0000790
(14+16)
Wieviele und welche Steine zogen in der KBP?
1. Sa3 Sa6 2. Sh3 Sc5 3. Tg1 Sb3 4. Th1 Sxc1 5. Tg1 Sb3 6. Db1 Sd4 7. Dc1 Sf3+ 8. Kd1 Sxg1 9. Sb1 Sf3 10. Sg1 Se5 11. Ke1 Sc6 12. Dd1 Sb8
play all play one stop play next play all
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
Keywords: Non-Unique Proof Game, Tempo Loss, Homebase (2)
Genre: Retro
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RN1QKBN1
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-07 more...
49 - P0000792
Klaus Wenda
1557 Die Schwalbe 33 06/1975
2. Preis
P0000792
(13+12)
#2 Längstzüger
b) sTa7 nach d7
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
Keywords: Maximummer, Castling (wksk)
Genre: Retro, Fairies
FEN: 4k2r/rb2pNbp/1P5p/p1pppN2/8/8/PPPPP2P/2BQK2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-03 more...
50 - P0000793
Nikita M. Plaksin
Andrey Lobusov

1558 Die Schwalbe 33 06/1975
4. Preis
P0000793
(13+5) C+
#3 (AP)
1. cxd6ep+! Kxg3 2. 0-0! (proving ep ok)
2. ... Kg4 3. g8=D,T#
2. ... a1=D,L 3. Dg5#

R: 1. d7-d5 Tc6xh6 (unless R: Kg1-g2 Th~*h1)
play all play one stop play next play all
VL: A posteriori (AP)!
2.g8Q(R)+/Rg1+? Kf4 3.Qg5#??
Retro: d7-d5, Rc6xh6+ (unless Kg1-g2, Rh~-h1+).
One of the best AP-type problems. (2015-08-20)
James Malcom: The first Valladao AP? (2020-10-03)
A.Buchanan: Hi James, there are 6 earlier ones in PDB, not counting those which are not marked as Valladao (could quite a few, as basic AP is 2/3 of Valladao). Type k='a p' and not g='fairies' and k='valladao', and set sort order by date. The ur-problem is P0003417. (2020-10-03)
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comment
Keywords: Castling (wk), a posteriori (AP) (Type Petrovic), En passant as key, Promotion (ws), Valladao Task (w w w/s)
Genre: Retro, 3#
Computer test: Popeye v4.85 + minor retro/AP thought
FEN: B7/p5PP/p6R/Q1Pp4/8/PP1P2P1/p3P1k1/4K2R
Reprints: (B) Die Schwalbe 48 12/1976
feenschach 40 11-12/1977
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-11 more...
51 - P0000819
Josef Haas
1893 Die Schwalbe 40 08/1976
1. Preis
P0000819
(9+6)
#1 vor 4 Zügen
VRZ, Typ Hoeg
R: 1. Kh3xBg3 hxg3ep+ 2. g2-g4 Ke6xBd6 3. exd6ep+ d7-d5 4. Sc4-b6, dann 1. Sd6#
play all play one stop play next play all
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
Keywords: En passant, Promotion, Defensive Retractor, Type Høeg
Genre: Retro
FEN: 2b4q/1p2p3/pNPk4/8/8/1B2R1K1/1P2PP1P/8
Reprints: feenschach 42 04-07/1978
345 Europe Echecs 241 01/1979
(5) Die Schwalbe 163 02/1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-07 more...
52 - P0000822
Josef Haas
1938 Die Schwalbe 41 10/1976
P0000822
(12+11)
Ergänze den wK, dann #2
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
Keywords: Castling (sk), Add pieces
Genre: Retro
FEN: 2nBk2r/3pp3/1p1p2P1/p4NN1/PP4p1/7b/PP2P1Pp/2R2B2
Input: Gerd Wilts, 1995-06-03
53 - P0000846
Klaus Weisert
181 Die Schwalbe 04/1970
P0000846
(3+7) C+
h#1
wZug also
1. ... b3!#
1. d1=S?,d1=T? b3#
play all play one stop play next play all
Two indistinct retro tries.
Originalforderung: 'Hilfsmatt' in 1 Zuge

Die 'Schwalbe' (Heft 9, Juni 1971, S. 230) zum recht trivialen Inhalt des Problems: "Warum bringt der Redakteur so etwas? Erstens, um einen Verfasser, der sich zum ersten Mal an den Rand des Schachbretts [Name der Urdruck-Rubrik, in der solche Probleme publiziert wurden] begibt, zu ermuntern; und zweitens (indem er das Stück an den Anfang stellt), um Löser, die sich sonst nicht an den Rand des Schachbretts trauen, mit einigen leichten Punkten dorthin zu locken!"

Nicht 1. d1S(T) b3#, denn Schwarz ist nicht am Zuge; daher ganz simpel 1. b3#
A.Buchanan: I suspect I don't get the full idea here. (2021-10-20)
Henrik Juel: I do not either
The two possible promotions do not help (2021-10-20)
comment
Keywords: No legal last move for White
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & trivial retro-logic
FEN: 8/8/8/rr6/kp6/p7/KP1p4/Rn6
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-06 more...
54 - P0000899
Giuseppe Brogi
743 Die Schwalbe 06/1972
P0000899
(8+15) cooked
h#2
b) wSa1
a) 1. T2xh3 Txh3 2. 0-0 Th8#
b) 1. La4 0-0 2. Tf8 Te1#
play all play one stop play next play all
a) White cannot cross-capture, so because of sTh2, w0-0 impossible.
wBe captured sD, either to be captured on f-file or to promote without disrupting sKe8.
b) sTh2 can return to a8 only via e8 as Black cannot cross-capture, so s0-0 is impossible.
wBe was waylaid or promoted after disrupting sKe8.
Cook: 1. T2xh3 Lb4 2. Txg3 Txh8#
1. Sd2 Lf6 2. Tf8 Sd6#
(cooked by MR)
See P0003736 a companion problem.
milan: [-bPa7b6wPg3bSa1+bPg3bBa5-b4 bPc6-d6] h#2 1.2...bRh2? illegal move? M.Frelih (2023-03-30)
more ...
comment
Keywords: Cant Castler, Castling (wksk), Cross-capture (s,w), Superseded by (P1399805)
Genre: h#, Retro
Computer test: Popeye v4.87
FEN: B3k2r/pN1p1p2/1pp3p1/bb3p2/8/p1B3PP/5P1r/nn2K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-16 more...
55 - P0000914
Vladimir Archakov
852 Die Schwalbe 17 11/1972
P0000914
(14+3) cooked
#2*
Set Play
1. ... Kxh1 2. Kf2#
1. ... Kxh3 2. Sf4#
1. ... Kxf3 2. Lb7#
BTM
1. ... hxg5 2. Sc3
2. ... Kxh1 3. Kf2#
2. ... Kxh3 3. Lf1#
2. ... Kxf3 3. 0-0#
play all play one stop play next play all
Cook: 1. ... hxg5 2. Kd2 Kxh3 3. Df1# 2. ... Kxf3 3. Lb7#
A.Buchanan: We can +wBd2, but then need to e.g. transpose wLg7 & wBg5, otherwise the wBB are impossibly concentrated in corner. Then I think it's sound. (2021-10-08)
more ...
comment
Keywords: Castling (wk), No legal last move for Black, Superseded by (P1394496)
Genre: Retro
Computer test: C- Popeye v4.87
FEN: 8/6BR/B6p/6PP/3P2Np/5P1P/4N1kP/Q3K2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-12 more...
56 - P0000919
Marc Benoit
903 Die Schwalbe 18 12/1972
P0000919
(6+1)
#2
Kritisiert wurde sowohl von den Lösern als auch vom Retro-SB Frank Schützhold (die Aufgabe war ohne dessen Wissen der Retroabteilung vom Schriftleiter H.-D. Leiß zugeschlagen worden) die irreführende Forderung 'Matt in ZWEI Zügen'.
Kevin Begley: Best guess: #2.5?
wRa1-d1 (completing castling) [Kc2 [Ra1#]] (2009-01-22)
Henrik Juel: after the castling completion:
0... Kxa2 1.Kc2 Ka3 2.Ta1#
C+ Popeye 4.61 (2022-07-10)
comment
Keywords: No legal last move for Black, Complete an unfinished move (w0-0-0)
Genre: Retro
FEN: 8/8/8/8/P1R5/k1B5/P7/R1K5
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-07-10 more...
57 - P0000933
Luigi Ceriani
Sahovski vjesnik 12/1951
1. Preis
P0000933
(2+11) C+
Letzter Zug?
R: 1. Kg8xDh8
Solved in YouTubex httx//www. youtube. com/watch?v=Do_rkLQnrpk
play all play one stop play next play all
Compare P0000490 with equivalent material but 30 years later.
A.Buchanan: Solved in YouTube: https://www.youtube.com/watch?v=Do_rkLQnrpk (2022-02-17)
more ...
comment
Keywords: Type A, Last Move? (KxD), Economy record (Last Move? Type A), Type B (a fortiori), Economy record (Last Move? Type B), Minimal
Genre: Retro
Computer test: Solved in YouTube: https://www.youtube.com/watch?v=Do_rkLQnrpk
FEN: 3bkN1K/pppprp1p/4p1p1/8/8/8/8/8
Reprints: 1372 FEENSCHACH 08-09/1952
Problem 10-12 12/1952
1.2A Eigenartige Schachprobleme , p. 172, 2010
1.2B Eigenartige Schachprobleme , p. 172, 2010
YouTube 13/02/2022
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-01 more...
58 - P0000957
Theophilus Harding Willcocks
Die Schwalbe 1959
Willcocks 1959, version
P0000957
(4+8)
Letzter Zug?
R: 1. a7xTb8=L
play all play one stop play next play all
See P1409984.
Yoav Ben-Zvi: WBd8 can be replaced with a BN after moving BPg7 to f7 and BPh5 to h6. (2018-08-08)
A.Buchanan: Yes after all these years, I found this a few months ago, but when I told Thomas Brand, he said that Werner Keym had found it - one of a series of modifications through seeking to avoid non-standard material. (2018-08-12)
Henrik Juel: Werner's improvement can be found in his 'Eigenartige Schachprobleme' from 2010, p.196, dia 1.68 (2018-08-12)
A.Buchanan: I thought it was more recent (2018-08-12)
A.Buchanan: In fact Werner was not trying to avoid non-standard material, but to prefer "cheaper" knights over bishops. But this is not the canonical ordering, which regards knights and bishops as equivalent. So the older record will win out in classical terms. (2019-10-04)
A.Buchanan: And non-standard material is no defect, according to the 1977 grading criteria, so old Theophilus reigns supreme (2019-10-04)
Henrik Juel: The further retroplay is
-1... h6 -2.a6 and e.g. -2... Ka7 -3.a5 Ta8 -4.Tb8 Ka6 -5.Kc8
and wK out via g6 (2019-10-04)
A.Buchanan: I don't know if a bishop is more expensive than a knight for these economy records. If modifying the criteria, I personally would prefer to minimize non-standard material. But more generally, how to handle a later version of economy record, with similar force? In some cases, the change is trivial, just transforming one unit which might as well be a dummy. At the other extreme, the matrix is completely different. In "Eigenartige Schachprobleme", Werner Keym stoically prefers the earlier record. Even the original composer(s) of a record problem are forbidden from making any artistic improvement to a published economy record! Over the years, the only change to the grading criteria is that Type C problem now needs to contain a check. Any old check-free Type C would a fortiori have been a Type B economy record at least. Here in PDB, I've tagged some cases of "ex-aequo" (search k='economy:eq') but I fear that Werner would not approve of this. What should we do? (2022-03-31)
more ...
comment
Keywords: Type A, Last Move? (BxT=L), Promotion (L), Non-standard material (L)
Genre: Retro
FEN: kBRB4/1pKpp1p1/1pp5/7p/8/8/8/8
Reprints: 1.49A Eigenartige Schachprobleme , p. 190, 2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-02 more...
59 - P0001030
Jan R. Mortensen
Werner Keym

Die Schwalbe 1984
P0001030
(3+6)
Welches war der letzte Zug?
Schwarz am Zug
R: 1. f7xLg8=T
play all play one stop play next play all
siehe auch P0001031
Henrik Juel: Source is almost surely Caissa 1979 (CA stopped in 1930) (2020-12-17)
A.Buchanan: According to Werner Keym in ES, the original source is JM+WK Die Schwalbe 1984. That probably means that Werner modified P0001031 to tackle the different task here. I will adjust the source & authors accordingly. For these task problems, no-one tracks versioning, or afters. We just lump together all the composers who lent a hand. (2020-12-18)
Henrik Juel: OK, fine
Of course, this is also an example of First move by promoted Tg8, but not an economy record in this genre (2020-12-18)
A.Buchanan: Any last move which is promotion is implicitly a zero move for a the promoted unit. So I feel probably not worth adding the keyword of first move except for those which were intended as first move records. What do you think? (2020-12-19)
Henrik Juel: The same position with wSg8 was similar, but it was a double economy record, I think
The position with wTg8 is a last move record, but not a first move record, as removing Th7 and moving Ph6 to h7 is more economical
The first move problems with a zero move are not very impressive, so your approach is fine (2020-12-19)
comment
Keywords: Last Move? (BxL=T), Type B, Promotion (T), Economy record (Last Move? Type B)
Genre: Retro
FEN: k4bRK/4p1pR/5p1p/8/8/8/8/8
Reprints: 1.45B Eigenartige Schachprobleme , p. 188, 2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-30 more...
60 - P0001031
Jan R. Mortensen
Deutsche Schachrundschau Caissa 1979
P0001031
(3+6)
Welches war der letzte Zug?
Schwarz am Zug
(Erster Zug des wUWS?)
R: 1. f7xLg8=S
play all play one stop play next play all
Henrik Juel: Not R: 1.f7xSg8=S,
because the black knight can be extracted only by retracting f7-f6, trapping white king (2020-12-13)
A.Buchanan: Some brave person should sort out the keywords for record retros properly. Here's a good example. The last move is Type B, but the first move question is Type A. The original stipulation (which I think is just about the first move?) should stand. The first move question is a *deemed* stipulation (new terminology) and allows us to leverage an existing problem, without creating a new version or editing the prior stipulation. This is particularly true here, since the problems are of different type. (2020-12-14)
Hans-Jürgen Manthey: mögliche Retro-Zugfolge:
1.f7xLg8S Ka7-a8 2. e6xSf7 Sd6-f7+ 3. e5-e6 Le6-g8 4. Kg8-h8 Lc8-e6+ 5. Th8-h7 Ka8-a7 6. Kh7-g8 Kb7-a8 7. Kg6-h7 Kc7-b7 8. h7-h8T Kd8-c7 9. d4xSe5 Sd7-e5+ 10. Kf5-g6 Sf7-d6+ 11. g6xTh7 Th8-h7 12. g5-g6 h7-h6 13. Ke6-f5 Sh6-f7 14. g4-g5 Sg8-h6 15. e3xTd4 Ta4xSd4 16. f2xDe3 Dd2-e3+ 17. g2-g4 Ke8-d8 18. Kf5-e6 De2xd2 19. Ke4-f5 Dd1xe2+ 20. Kd3-e4 Df1xTd1 21. Sf3-d4 Dh1xLf1 22. Kc2-d3 Sb8-d7 23. Kc1-c2 Dh2xTh1 24. O-O-O Dc7xh2 25. Sg1-f3 Ta8xDa4 26. Dd1-a4+ Dc2-c7 27. Se6-c7+ Dc1xc2 28. Sc5-e6 Db2xLc1 29. Sd7-c5 Da2xb2 30. Se5xd7 Da5xa2 31. Sc6-e5 f7-f6 32. a7xc6 Dd8-a5 33. Sb5xa7 c7-c6 34. Sc3xb5 b7-b5 35. Sb1-c3 (2020-12-17)
Henrik Juel: 'The Chess Amateur 1979' must be wrong, as this magazine stopped in 1930 (2020-12-17)
Henrik Juel: .. and Eigenartige Schachprobleme just says Caissa 1979 (2020-12-17)
A.Buchanan: WinChloe does not have this problem at all! I am inclined to go with Eigenartige Schachprobleme. I suspect that Gerd was entering hundreds of compositions at scale, and a copy-paste error would be easy (2020-12-17)
A.Buchanan: Also P0001030 (2020-12-17)
A.Buchanan: Have removed "The Chess Amateur" as original source here. (2020-12-18)
comment
Keywords: Last Move? (BxL=S), Type B, Promotion (S), Economy record (Last move? Type B), First Move? (US0), Type A, Economy record (First move? Type A)
Genre: Retro
Computer test: RSP 1.2
FEN: k4bNK/4p1pR/5p1p/8/8/8/8/8
Reprints: The Chess Amateur 1979
1.55B Eigenartige Schachprobleme , p. 192, 2010
2.51 Eigenartige Schachprobleme , p. 202, 2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-30 more...
61 - P0001086
Peter Kahl
Jan R. Mortensen
Sveto Stambuk

Problem 5-6, p. 92, 12/1951
6. ehrende Erwähnung
P0001086
(4+1)
Welches war der letzte Zug?
R: 1. Dh3xSh2+
play all play one stop play next play all
im Original ist als Dritter Autor D. Suboticanec angegeben, habe entsprechend geändert.
Zu P. Kahl heißt es ergänzend: [aus] Osterholz-Scharmbeck
HBae: Ist "Klaus Peter Kahl" identisch zu "Peter Kahl" ? (2010-09-22)
Henrik Juel: Yes, probably
I know that 'Jan Robert Mortensen' is identical to 'Jan Mortensen'; he never used his middle name (2017-03-08)
A.Buchanan: It’s possible to put the middle name into the “addition” field in the author table, rather than have two names in the “first name” field. Then in overview list the middle name won’t appear, but it will still appear above each composition as here. Curiously the addition field is common across all four languages (German, English, French, Unicode) supported by the author table. There is no comment field to allow for random extra info to be stored, unlike the source table. So if we drop Robert, we lose it completely. (2022-03-31)
Henrik Juel: Thanks for the info, Andrew
I guess we will have to live with it
For Jan, there is really no risk of confusion
But for Knud Hannemann (son of the less illustrious composer Harald Hannemann) it is annoying to see him cast as Knud Harald Hannemann, just because it was a custom in his family that the sons were given the father's first name
They still use this custom in Russia, but with 'son of' added, e.g., Vladimir Vladimirovitch Putin (2022-03-31)
Henrik Juel: A chess problem example is Nikita Michaelovitch Plaksin
But the rendering of his name in the PBD also gives an obvious partial solution to my problem
Change the middle name of the two danish composers to the corresponding initial:
Jan R. Mortensen and Knud H. Hannemann (2022-03-31)
Henrik Juel: Thanks for the 'initialization', Andrew (2022-03-31)
A.Buchanan: OK fixed those, Henrik. HBae's original question concerned "Klaus Peter Kahl" vs "Peter Kahl". The former only has 2 problems here, and "both" collaborated with Jan Mortensen, so surely this is one person? (2022-03-31)
A.Buchanan: Kahls unified (2022-04-01)
Henrik Juel: I hear no protests from Osterholz-Scharmbeck, so that seems fine (2022-04-01)
comment
Keywords: Type C, Last Move? (DxS), Economy record (Last Move? Type C)
Genre: Retro
FEN: 8/8/8/8/7P/6P1/5K1Q/7k
Reprints: 1375 FEENSCHACH 08-09/1952
(C10) feenschach 160 07-08/2005
1.11C Eigenartige Schachprobleme , p. 176, 2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-01 more...
62 - P0001107
Michel Caillaud
3332 Die Schwalbe 66 12/1980
2. Preis
P0001107
(7+9) C+
h#4.5 (AP)
Circe
1. ... hxg6ep[+sBg7] 2. a1=T+ Txa1[+sTh8] 3. 0-0 Ta3 4. bxa3[+wTa1] 0-0-0 5. Td8 Txd8[+sTh8]#
play all play one stop play next play all
Wenn die beiden Könige noch nicht gezogen haben (was a.p. durch Ausführung beider Rochaden bewiesen wird!), kann der letzte Zug nur g7-g5 gewesen sein, Sonst ginge z.B. auch R: 1. Kd8-e8 c2xDb3
Henrik Juel: A possible retroplay is -1... g7 -2.f3xP(-Pg7) b5 -3.f2 c6xP(-Pb2) -4.b2 a6 -5.Rc3 a7 -6.Rc1 b6xB -7.Bf8 b7 -8.Bg7xB etc. (2003-04-22)
Michel Caillaud: The problem is not cooked as indicated, as under A Posteriori convention both castles must be played in the solution to justify the en passant key. (2022-02-04)
A.Buchanan: If wK never moved, then Black has not just moved another non-capturing pawn, as White has no prior move. The only possible exception is R: 1. g7-g5 f/h3xPg4[+Pg7]. (2022-02-04)
comment
Keywords: a posteriori (AP), Circe, En passant as key, Castling (wgsk), Promotion in forward play, Valladao Task
Genre: Retro, Fairies, h#
Computer test: HC+ all solutions begin with ep, but only one includes both white & black castling
FEN: 4k3/8/7p/p1p3pP/Pp4Pp/RP4p1/pP6/4K3
Reprints: feenschach 61, p. 534, 08/1982
RA152 diagrammes 72 07-08/1985
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-06-28 more...
63 - P0001114
Michel Caillaud
3676 Die Schwalbe 71 10/1981
3. ehrende Erwähnung
P0001114
(13+12)
#1
Mars-Circe
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
Keywords: Circe (Mars), Non-standard material, Promotion
Genre: Retro, Fairies
FEN: 1n6/p2ppprP/2p2pRK/2N2NnB/N3N1p1/6N1/1pPP4/B1k4N
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2019-02-03 more...
64 - P0001117
Michel Caillaud
3872 Die Schwalbe 74 04/1982
12. Lob
P0001117
(11+12)
h#2
b) sBh7 nach h6
a) 1. 0-0-0? Sg6 2. Lb8 Sxe7! illegal castling
1. Kf7! Sxh7 2. Ke6 Sg5#
b) 1. 0-0-0! Sg6 2. Lb8 Sxe7#
1. Kf7? Sh7 2. Ke6 Sg5+? 3. hxg5!
R: 1. Ke3-e4 Sd6-b7 2. h2xDg3 Dg2-g3+ 3. Kd2-e3 Db7-g2 4. Ke1-d2 Dc8-b7 5. e4-e5 Dd8-c8 6. f3xLe4 Lb7-e4 7. Kf1-e1 Lc8-b7 8. Ke1-f1 b7-b6 9. Kf1-e1 Lb6-a7 10. ... La5-b6 11. ... Ld2-a5 12. ... Lc1-d2 13. ... d2xTc1=L 14. ... e3xTd2 15. ... e4-e3 16. ... f5xLe4
play all play one stop play next play all
Missing: Wh: QRRBB Bl: QRBB
Captures: Wh: gxf3xe (inc B), hxg + [Bf8] Bl: axbxc (for Ba7) fxexdxc/e (g1 not possible)
Assume Black can castle: so neither bK nor bRa have moved. Before bPb7-b6 (which releases QB) *all* Bl captures have been made. wPd can freely advance, all Wh units released except for rooks & Bf1. gxRf3 is forced, and now all Wh units are free and can be captured. To avoid deadlock, wB was captured on e4 not e6. Sequence must be f7-f5 Rh8-f8-f6-...-f3gxRf3.
In this position wLe4 must be played back to f1. With bPh7, wBe4 must retract either by stopping on f7, (disrupting bK) or via f5 (implying retraction of f6-f5, in which case bRa8 is an imposter).
In the alternative route via h7, wB crosses over f7 harmlessly. While wB is on g5 & h4, f7 must be occupied by a static shielding knight, but there is no tempo issue. After all this excitement, b7-b6 if followed by simple and non-unique play to reach the diagram.

(Gerd's earlier solution: Weiße Schläge: h2xg3, gxfxe, sLf8. Schwarze Schläge: a7xb6xc5; fxexdxc1=L In dieser Stellung muß der wLe4 nach f1 zurückgespielt werden. Mit sBh7 kann der wLe4 nur entweder über f7 nach f1 zurück, so daß der sK bereits gezogen haben muß, oder der wLe4 kann über f5 zurück, wozu aber der sTa8 nach h8 zurückgezogen werden müsste, um f6-f5 zurücknehmen zu können.)
Henrik Juel: To make the retroplay plausible one should uncapture bQ early on g3 and retract it to d8. The wB could also get back to f1 via f5, but this would require retracting bRa8 to h8 before retracting bPf5, so castling is still illegal in part a). (2003-04-10)
Gerd Wilts: Hello Henrik, thank you for pointing out the inaccuracy of the solution, I will make the solution more precise soon. And thank you for adding all the other solutions! (2003-04-11)
A.Buchanan: Have posted a solution based on GW&HJ ideas. More often a j’adoube of a rook pawn signals a tempo idea, but not here. The surprising motivation harmonizes with accurate and varied forward play. (2021-10-24)
A.Buchanan: This problem was featured in yesterday's Monthly International Zoom Call, and was a great success, with audible gasps of appreciation, as the significance of h7 was realized :) (2021-10-24)
A.Buchanan: Oh dear, Alfred Pfeiffer has silently reverted the German keyword to "mit Umwandlungsfigur". I'm not going to enter an "edit war" with him, but I would appreciate if he can explain his position here. This is a problematic concept to keyword, but to me "mit Umwandlungfigur" is weak and inaccurate. What is the intended distinction with the existing keywords "Umwandlung" or "Umwandlungen"? It's hopeless. We have in PDB very many poor promotion keywords, and I would like to clean up progressively. I don't know if a native German speaker would care to engage with Alfred on this point. (2021-10-30)
A.Buchanan: To be clearer: to me the German definition seems pretty good. I think the term itself should give more of a clue what's happening :-) (2021-10-30)
more ...
comment
Keywords: Cant Castler, Castling (sg), Promotion (l), Obvious promotion (l), Corridor, Retro Shield
Genre: h#, Retro
Computer test: Forward: C+ Popeye V4.87 Retro: non-trivial reasoning
FEN: r3kN2/bnppp1pp/1p6/2p1P3/4K3/3P2P1/PPP1PP2/N6n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-30 more...
65 - P0001121
Michel Caillaud
4298v Die Schwalbe 80 04/1983
2. ehrende Erwähnung
P0001121
(12+14) C+
BP in 30,5
1. a4 c6 2. a5 Dc7 3. a6 De5 4. axb7 a5 5. b4 a4 6. b5 Ta5 7. b6 Sa6 8. b8=D a3 9. Dd6 a2 10. De6 dxe6 11. b7 Kd7 12. b8=T Kd6 13. Tb3 Kd5 14. Tf3 Ke4 15. Tf5 exf5 16. Sa3 Le6 17. Sb5 Lc4 18. La3 e6 19. Tc1 a1=S 20. d4 Sb3 21. d5 Sd2 22. d6 Sxf1 23. d7 Sg3 24. hxg3 Lb4+ 25. Kf1 Se7 26. Th6 Tb8 27. d8=L Sc8 28. Lh4 g5 29. g4 gxh4 30. g5 Le1 31. f3+
play all play one stop play next play all
more ...
comment
Keywords: Ceriani-Frolkin Theme (DTsL), Unique Proof Game, Allumwandlung, Promotion
Genre: Retro
Computer test: Moldenhauer: Computerprüfung: C+ Stelvio 1.11 63 Sekunden. Keine Lösung: BP 29.5, BP 30.0.
FEN: 1rn5/5p1p/n1p1p2R/rN2qpP1/2b1k2p/B4P2/2P1P1P1/2RQbKN1
Reprints: 12 Shortest Proof Games 11/1991
(A) Quartz 26 10-12/2004
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-15 more...
66 - P0001136
Edgar Fielder
The Fairy Chess Review 1941
P0001136
(13+10)
Darf Schwarz rochieren?
Schwarz hat bereits rochiert!
play all play one stop play next play all
See P1398939
Henrik Juel: Black captured b7xLc6 and axbxc; White captured [Lf8], e2xd3, and fxexdxc6xb7-b8=D.
Ke8 to b7, Th8 to b8, Kb7 to g8, Tb8 to f8, Da4 to b8 for unpromotion, -22.c6xDb7 Da8 -23.d5xLc6 Dd8 -24.d4 Lb7 -25.e3xTd4 Lc8 -26.f2xSe3 c6 -27.h5 b7xLc6, Lc6 to f1, f2xSe3 etc.
Black may not castle, because he already did.
It is not possible to avoid the early castling:
-11... Ka7 -12.Db4 Ka6 -13.Db8 Ka7 -14.D=b7 Kb8 -15.c6xDb7 Kc8 -16.d5xLc6 Kd8 -17.d4 Ke8 -18.e3xTd4 Da8 -19.f2xSe3 Dd8 -20.h5 Lb7 -21.h4 Lc8 and two white pawn retractions are missing (2012-07-23)
Yoav Ben-Zvi: The solution does not require a full analysis since if Black King and Rook never moved then Black is almost immediately in retro-stalemate. An alternative stipulation is "First move of Black King?". (2014-06-03)
A.Buchanan: Maybe there is intentional irony? (2014-06-03)
Henrik Juel: Nowadays composers are not afraid to use the real stipulation in this type of resolution retro, like 'Release the position'.
Formerly a formal stipulation was popular, like 'Mate in 1', with the real stipulation understood but not mentioned.
Here, the stipulation question may well be ironic, but you will get no solution points, if you just answer 'No'; you still need to explain how the position arose (2014-06-03)
more ...
comment
Keywords: Castling (sk), Castling Paradox (sk hidden)
Genre: Retro
FEN: 4k2r/2pppppp/7P/2p5/Q1P5/PPRP4/RBpP2P1/N1K5
Reprints: 71 32 personaggi e 1 autore 1955
12 Europe Echecs 12 08/1959
222 FIDE Album 1914-1944/III 1975
341 Eigenartige Schachprobleme , p. 110, 2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-23 more...
67 - P0001141
George Hume
Jamaica Gleaner 12/1891
Weihnachtsturnier 1891
1. Preis
P0001141
(9+9)
Auf welche Gedanken kommen Sie bei dieser Stellung?

Der Ld6 ist keine UWF und der sBg7 wurde auf seinem Ausgangsfeld geschlagen. Nach dem Autor muss der letzte Zug also Lf8-d6 gewesen sein, also illegale Stellung.
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)
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comment
Keywords: Illegal position, Joke, Retract illegal move (stuck at home), Touch Move, Volet Pawn, Obvious promotion (L)
Genre: Retro
FEN: 6B1/4p1P1/p2b2p1/p5kq/p7/4P1K1/2PPP1PP/3n4
Reprints: American Chess Monthly 1, p. 11, 03/1892
Jamaica Gleaner 30/04/1892
17 Europe Echecs 14 10/1959
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-02 more...
68 - P0001146
W. Wolf
Deutsches Wochenschach 23/04/1911
P0001146
(15+8)
#3
1. bxc6ep e4 2. Se3 Kxg5 3. Kxd6
play all play one stop play next play all
Henrik Juel: C+ Popeye 4.61 (the three threats never materialize)
It is obvious that last move was c7-c5 (2020-12-02)
comment
Keywords: En passant as key
Genre: Retro
FEN: 4B1R1/3NP1Pp/1Q1p1Prr/RPpKpNPk/6p1/6P1/P2B4/8
Reprints: 41 Volksgemeinschaft (Heidelberg) 19/01/1936
252 Comoedia 21/06/1936
22 Europe Echecs 18 02/1960
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2020-12-02 more...
69 - P0001157
Georges Glaeser
31v Europe Echecs 22 06/1960
P0001157
(13+11)
Löse die Stellung auf!
hans: R: -1 Kh7-g8 Tf8-f7
(captures w fxSexSdxDcxTb8=L and KxLf8 s hxg cxSb6 and Lc1 on homesquare (2010-06-21)
A.Buchanan: Couldn't it equally be R: 1. Th4-h8 h6xBg5 etc? Not sure what this problem is driving at, except for what would later be termed Donati theme. (2021-10-24)
Henrik Juel: Yes, it seems that there are several ways to skin this cat
The problem may be the first to show the Donati theme (2021-10-24)
Mario Richter: The stipulation did not ask for "last moves", so it seems sufficient to have the "main events" unique, and this is the case here:
White captured black Bishop f8 on its home square, White pawn f2 captured 4 times to promote to a Bishop (the original wBishop c1 was captured by a black officer), so among the capture objects was bRa8, so the promotion took place on b8. White pawn g2 never left its home file, so it was captured by black pawn h7, leaving only c7xSb6 to explain the b-double pawn.

Btw., in what sense is this Problem "C+"? (2021-10-24)
A.Buchanan: Hi Mario, this is not unsound, just loose. wDh2 prevents e.g. R: 1. Lg3-g1 Kb8-a8, so there seemed at least some interest in the last move. But 3 legal possibilities remain. Is there a typo in the diagram? With wKe8 instead of g8, I think the solution is R: 1. Th4-h8 h6xBg5 2. Lg3-b8 Kb8-a8 3. Tf4-a4+ which is a lot more satisfactory.
I've no idea why the problem has been marked as C+ (2021-10-25)
Mario Richter: I completely agree with Andrew, that e8 is a much better place for the wK (and only this way wQh2 makes any sense), so I'm supposing a typo.
Apropos, the third white retraction shoulkd read 3. Tf4-h4+ ... (2021-10-26)
A.Buchanan: Thanks Mario for correcting my spatio-algebraic dyslexia again :) We are still not using the fact that the White pieces are clustered tightly around wD. Any thoughts? Is there an (online) archive of this magazine? (2021-10-26)
more ...
comment
Keywords: Promotion (L), Donati Theme (L), Obvious promotion (L)
Genre: Retro
FEN: kBb3KR/pp1pprp1/1p3p2/6p1/8/7P/PPPPP1BQ/6NR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-26 more...
70 - P0001185
Zdravko Maslar
(291) Problem 45-48 11/1957
16. TEMATSKOG TURNIRA "PROBLEMA" 1.-2. Preis e.a.
P0001185
(4+9)
Welches war der letzte Zug?
R: 1. Tc8xDb8
play all play one stop play next play all
more ...
comment
Keywords: Last Move? (TxD), Type B (a fortiori), Type A
Genre: Retro
FEN: BR1rk3/1pKppRp1/1pp2p2/8/8/8/8/8
Reprints: 58 Europe Echecs 36 08/1961
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-22 more...
71 - P0001191
Leon Loewenton
64 Europe Echecs 38 10/1961
P0001191
(15+14) C+
h#2
b) wSd3 nach e1
a) 1. Kd8 Se5 2. Te8 Sxf7#
b) 1. 0-0 Sd5 2. Sh8 Sxe7#
play all play one stop play next play all
Originalquelle?

nicht sicher, ob Zwilling b) auch schon im Original.
Artikel von Petrovici zum "Thema Than" 03/2016 gibt als Originalquelle an: "Europe Echecs 10/1951"
hans: Counting problem
1. Kd8 Se5 2. Te8 Sxf7# !!
1. 0-0 Sd5 2. Sh8 Sxe7# ??
Black needs a tempo to get on move, and the only piece to do that is Th8, so 0-0 is illegal. (2010-06-20)
Mario Richter: The term usually used to describe this kind of problems is "Parity problem". The bRh8 might have been on h8 all the time, since the tempo move might also have been made by the black Queen, but in this case, the bK must have already moved, thereby losing the right to castle too.. (2010-06-21)
Ladislav Packa: You both are right. Black would be in this position made an uneven number moves.
This can be achieved by using Rook moves (eg Rg8-f8-h8), or by the Queen moves while she was still alive (eg Kf8-g8, then Qf8-e8-d8 and then the King back). But both possibilities preventing black castling. (2013-10-04)
A.Buchanan: Parity change could also be achieved without triangulation by e.g. SxDe8 after the black queen has moved once (2022-03-15)
comment
Keywords: Castling (sk), Parity Argument, Cant Castler, Than theme
Genre: h#, Retro
Computer test: rawbats
FEN: r1b1k2r/1ppppppp/p5n1/8/8/P1NN3P/1PPPPPPR/nRBK1B2
Reprints: 787 Themes-64 10-12/1961
(7) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2018-03-22 more...
72 - P0001198
Tivadar Kardos
1807 Problem 73-78 06/1961
P0001198
(8+15) C+
h#2
1. Dxc2 Tc1 2. 0-0-0? Txc2#
1. Sd7 0-0-0 2. 0-0 Tg1#
play all play one stop play next play all
Black qside castling is lost as wPa promotes but other rights can be maintained.
S: axb,cxb,dxcxb,fxexBd,gxfxe
W: a=X,g=X,fxe
A.Buchanan: wPh can be removed (2021-11-22)
comment
Keywords: Cant Castler, Castling (wgsk)
Genre: h#, Retro
Computer test: HC+ Popeye C-Version 3.54 & trivial retro-logic
FEN: r1q1kn1r/1p2p2p/bp2P3/bp1pB3/1p6/4p3/1PP1P2P/R3K3
Reprints: 71 Europe Echecs 41 01/1962
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-23 more...
73 - P0001226
Paul Vatarescu
99 Europe Echecs 53 05/1963
dedicated to L.Loewenton
P0001226
(16+9) C+
#1
1. dxc6ep#
R: 1. c7-c5
play all play one stop play next play all
hans: 1. dxc6ep.# R: -1. ... c7-c5 -2 Td2-d3 e7-e5 -3 Dh2xSg3+

1. dxe6ep.# R: -1. ... e7-e5 -2 Dh2xg3+ c7-c5 -3 Td2-d3 (2010-06-22)
Mario Richter: e.p. keys are not justified, Black's last move might even have been h2-h1=N. (2010-06-26)
Hans-Jürgen Manthey: +sPh2 und es geht nur noch die hans-Lösung

R: 1. c7-c5 Ka4-b5 2. Kc5-d6 Lg5-h4 3. Kd6-c5 Lc1-g5 4. Ke7-d6 c5xTb6 5. Th6-b6 Sc2-e1 6. Td1-f1 Ta3-d3 7. Th8-h6 Te1-e2 8. Se2-g1 Sb6-a8 9. Sf4-e2 Kb3-a4 10. Sg6-f4 Ta1-a3 11. f4-f3 Dc3-g3 12. Sg3-h1 Da5-c3 13. Sf5-g3 Da8-a5 14. Sh6-f5 a7-a8=D 15. Sg8-h6 a6-a7 16. Ke8-e7 a5-a5 17. Se7-g6 a4-a5 18. Sc6-e7 a2-a4 19. Td4-d1 Sa3-c2 20. Ta4-d4 Sc6-b6 21. Ta8-a4 Sa7xLc8 22. Sb8-c6 Sb5xa7 23. f5-f4 Th1-e1 24. f7-f5 Le2-g4 25. g4xDh3 h4-h5 26. g5-g4 Kc2-b3 27. g7-g5 Db3-h3 28. h3-h2 g3xDh4 29. Dd8-h4 h2xLg3 30. Lf4-g3 Kd1-c2 31. Lg5-f4 Ke1-d1 32. Lh4-g5 Lf1-e2 33. Lg5-h4 Dd1-b3 34. Lf6-g5 Sb1-a3 35. Le7-f6 Sd4-b5 36. Lf8-e7 Sf3-d4 37. h4-h3 e2-e4 38. h5-h4 d4-d5 39. h6-h5 d2-d4 40. h7-h6 c4-c5 41. e6-e5 c2-c4 42. e7-e6 Sg1-f3 (2021-07-06)
A.Buchanan: This one is not to be found in WinChloe or yacpdb (2021-07-07)
Ladislav Packa: Why not 1...f4-f3? (2021-07-07)
Mario Richter: Ladislav, good question!
The most likely answer is, that we have a diagram error, and that the correct position has a white Pf3 instead of a black one. With this modification, everything works fine.

Btw., adding a black pawn h2 still allows R: 1. ... f4-f3 as Black's last move, but even worse, it makes the position illegal (what happened to the missing white pawn c2 or a2?) (2021-07-09)
A.Buchanan: Yes it all clicks into place. bPgxh bPh waylaid bBc8 died at home. wPaxb, cxd, dxe, exf. Well done Ladislav and Mario. Let’s correct the diagram as this couldn’t have been a composition error. Mostly likely error in transcribing from EE (2021-07-09)
Ladislav Packa: bPh2 is not necessary, Sh1-h2 is enough. (2021-07-09)
Mario Richter: Perhaps Gerd can check the original source ... (2021-07-09)
VL: Originally wPf3. The problem is dedicated to L.Loewenton. (2021-07-10)
A.Buchanan: Thanks Valery! (2021-07-10)
Ladislav Packa: After wPb2-d2, it is possible to save bRf1 and one of the black Knights g1 or h1. (2021-07-11)
Hans-Jürgen Manthey: an Mario: von wegen ILLegal ! bei hinzufügen von sBh2:
der C-Bauer = 4. Ke7-d6 c5xTb6
der a-Bauer = 14. Sh6-f5 a7-a8=D
aber ...f4-f3 schließt natürlich eP aus.
mit wPf3 kann wegen Mangel an schlagfällen nicht Dh2x?g3. Darum
R: 1. dxc6ep# c7-c5 2. Lg5-h4 h4-h3 3. Sc2-e1 Ta1-f1 4. Td1-d3 a2-a1=T 5. Ta1-d1 Ke7-d6 6. Lc1-g5+ a3-a2 7. Te1-e2 a4-a3 8. Le6-g4 a5-a4 9. d4-d5 a7-a5 10. Lc4-e6 Se2-g1 11. a5xTb6 Tf6-b6+ 12. Sb6-a8 Tf8-f6 13. a4-a5 Ta8-f8 14. a2-a4 Sf4-e2 15. Sa3-c2 Sd5-f4 16. Df4-g3 Sf6-d5 17. Dd2-f4 Sg3-h1 18. Th1-e1 Sf5-g3 19. Lf1-c4 Ke8-e7 20. Kc4-b5 Sg8-f6 21. c3xLd4 Lc5-d4 22. Kd3-c4 Sd4-f5 23. c2-c3 Sc6-d4 24. Dd1-d2 Sb8-c6 25. e2xf3 Lf8-c5 26. Kd2-d3 f4-f3 27. d3xDe4 Dg6-e4 28. Ke1-d2 Dg5-g6 29. Sb1-a3 Dd8-g5 30. Sc8-b6 e7-e5 31. Sb6xc8 f5-f4 32. Sc4-b6 f7-f5 33. g4xTh5 Th8-h5 34. g3-g4 h5-h4 35. h2xg3 h7-h5 36. Se5-c4 g4-g3 37. Sf3-e5 g5-g4 38. Sg1-f3 g7-g5 39. d2-d3 (2021-07-12)
Hans-Jürgen Manthey: allerdings geht noch h2-h1=S... (2021-07-12)
Mario Richter: an Hans-Jürgen: bzgl. meiner Illegalitätsreklamation für die Stellung mit sBf3 und zusätzlichem sBh2 hast Du recht - da war ich zu sehr darauf fixiert, daß Schwarz auf a1 entwandeln muß, läßt man W auf a8 entwandeln, sieht man leicht, daß die Stellung legal ist ...

Btw. - worauf bezieht sich Deine letzte Anmerkung ("allerdings geht noch h2-h1=S")? (2021-07-13)
Hans-Jürgen Manthey: Mario da lag ich falsch... 16 weiße Steine, da geht nicht sBh2, sBh3, sorry (2021-07-14)
more ...
comment
Keywords: En passant
Genre: Retro
Computer test: HC+ counting captures + Retractor 2.0
FEN: N7/1p1p4/1P1k4/1KpPp2P/4P1BB/3R1PQp/1P2RPP1/4Nrnn
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-07-10 more...
74 - P0001228
Wolfgang Hundsdorfer
Deutsches Wochenschach 1909
1. Preis
En-passant-Turnier 1910
P0001228
(13+12)
#3
1. bxc6ep

R: 1. c7-c5 g5-g6 2. Tc6-c2 g4-g5 3. Tg6-c6 g3-g4 4. Tg8-g6 g2-g3 5. Ld4-a7 h4-h5 6. Lg7-d4 h3-h4 7. Lf8-g7 h2-h3 8. g7xLh6,g7xSh6
play all play one stop play next play all
Henrik Juel: minor dual in forward play
1.bxc6ep+ Db5 2.DLxb5+ Tc4 3.LDxc4#
1... b5,bxa6 2.Kxb4+ Tc3 3.Txc3# (2022-07-01)
Henrik Juel: Last move must be c7-c5, enabling Tc2 to reach g8 in just three retractions (2022-07-01)
comment
Keywords: En passant as key
Genre: Retro, 3#
FEN: 8/bp2p2p/B5Pp/RPp2q1P/Qp2P3/RK1kPP2/P1rppP2/8
Reprints: 139 Favorit-Schachaufgaben , p. 35, nach 1910
75 Retrograde Analysis 1915
101 Europe Echecs 53 05/1963
10 Die Schwalbe 276 12/2015
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-07-01 more...
75 - P0001263
Luigi Ceriani
135 Europe Echecs 78 06/1965
P0001263
(15+11)
h#2
1. bxa5 Sb6 2. axb4 Ta8#
play all play one stop play next play all
Henrik Juel: analysis
Black captured a3xb2 (not a7xb6, as the SW corner then cannot be released)
The missing white man is [Pe2], which must have promoted on b8
Now the white captures can be seen: axb, exdxcxb, and [Lf8] on f8
The only free white officer is Dd8, so the retroplay must include unpromotion of it on b8
It is impossible to shield Ke8 from check by Db8, so Ke8 has moved and Black may not castle (2022-04-27)
Henrik Juel: solution
1.bxa5 Sb6 2.axb4 Ta8#
not 1.0-0? Lxb6 2.Txa8 Txa8
HC+ Popeye 4.61 (2022-04-27)
Henrik Juel: Andrew, I should have been more precise in recommending your suggestion to introduce the new keyword Organ Pipes
I was talking about the standard meaning of Organ Pipes, which dates back to Sam Loyd, 1859: black LTTL in a problem where each L interferes with each T and vice versa, forming four Grimshaws
You have probably used a pattern search to find all occurrences of black or white LTTL, and you have neglected the interferences
This problem has LTTL in the wrong color, and there are no interferences (2022-04-27)
James Malcom: I agree with Henrik, although White organ pipes still exist, few and far. (2022-04-28)
James Malcom: Until Andrew bumbles back, I've updated the English definition to "A problem *utilizing* the pattern bishop, rook, rook, bishop in a straight line to create multiple Grimshaws." This specifies the purpose and formation of the Organ Pipes. (2022-04-28)
Henrik Juel: Thanks, James (2022-04-28)
A.Buchanan: Hi Guys. I wondered about other uses of organ pipes. In a first pass population of the 500-odd records with the current PDB interface, there’s not much time for thinking. I have multiple tabs open to eliminate the wait time associated with PDB refresh. One could go back and eliminate those which are not grimshaws. Alternatively (and this would be my preference) accept that this is a visual pattern which may occur in non-Grimshaw context. Then use the keyword Grimshaw(4x) to indicate when it’s really Grimshaw. I think it can also occur with white pattern, as a problem by Pal Benko shows. So I blitzed through the d# records, if someone wants to complete the rest that’s good. I think the existence of a few false transient positives is an acceptable price even if we take the narrow definition of Organ Pipes. (2022-04-28)
A.Buchanan: By the way, these days Deepl is good enough to give us decent translation to Feench & German and as a matter of policy whenever I make a change to a definition I try to align the other two. Other users who do not maintain the glossary are encouraged to propose definitions where there is a gap (E.g. Grimshaw). There are a lot of undefined terms, many very recent. (2022-04-28)
A.Buchanan: Hi Henrik feel free to respond to my response to your message, say thanks for the tags I have added, or add tags yourself to complete the work. The ones I’ve added were mostly the d# and it was a deliberate decision to blast through as a first pass and just add them for now anyway, not “neglected”. Now, I really don’t feel like continuing (2022-04-28)
Henrik Juel: Andrew, I believe that most PDB users appreciate your contributions to the site
I certainly do, so please continue your good work (2022-04-28)
comment
Keywords: Castling (sk), Cant Castler, Promotion
Genre: h#, Retro
FEN: N3k2r/2pppppp/1p6/B7/RP5P/RP3P2/BpPP2P1/NK1Q3r
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2022-04-28 more...
76 - P0001270
Matti Arvo Myllyniemi
142 Europe Echecs 84 01/1966
P0001270
(15+15) C+
BP in 9.5
1. e3 d5 2. Lc4 d4 3. Se2 d3 4. 0-0 dxc2 5. d4 Kd7 6. d5 Kd6 7. Dd4 Sd7 8. Ld2 c1=L 9. Lb4+ c5 10. dxc6ep+
play all play one stop play next play all
more ...
comment
Keywords: Unique Proof Game, En passant, Non-standard material (l), Castling (wk), Promotion (l), Valladao Task (sww)
Genre: Retro
Computer test: Ergänzung Stelvio 1.2 C+: Keine Lösung: BP 8.5, 9.0. 10. dxc6ep++ Doppelschach.
FEN: r1bq1bnr/pp1npppp/2Pk4/8/1BBQ4/4P3/PP2NPPP/RNb2RK1
Reprints: 92 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-25 more...
77 - P0001273
Luigi Ceriani
145 Europe Echecs 84 01/1965
P0001273
(6+7)
h#1 (wer? wie?)
Die F.P.I. wurde zweimal erreicht: das erstemal nach einer geraden, das zweitemal nach einer ungeraden Anzahl von Zügen.
Vertikaler Zylinder
Henrik Juel: The FPI (the initial array, but with Black to move) may be reached, e.g., by playing the white knights out, playing Ta1 to b1 and Th1 to g1, then correct the tempo by playing Tb1-a1-h1-b1, and finally moving rooks and knights back into the initial array.
The stipulation condition implies that all four castlings are illegal. Solution: 1.Txa7 Txh8#, not 1.0-0-0? Tc1 nor 0... 0-0-0? 1.Tc8. I do not see any tries involving king-side castling, although they would be nice to have also. (2010-12-08)
A.Buchanan: I am evidently being slow again: please explain. "The FPI was achieved twice: the first time after an even number of moves, the second time after an odd number of moves." To achieve FPI requires an odd number of single moves. Should the stip read "even number of *White* moves"? And then what does it mean to achieve FPI for the *second* time: does that mean game array with White to move again? I agree that the bottom line is to have eliminated all castling rights. (2020-12-25)
Henrik Juel: In P0001272 the FPI was reached once, implying that either White or Black has lost castling rights
I also do not understand this problem (2020-12-26)
Hans-Jürgen Manthey: da der lezte Zug b5x~a6 bzw b4x~a3 gewesen sein kann, sind alle Rochaden, sowie die Farbwahl zulässig.
Oder kann mir mal einer erklären was zum Teufel F.P.I. auf Deutsch heißt ?? (2020-12-26)
A.Buchanan: Siehe die Beschreibung des Schlüsselworts. Aber hier macht es nicht viel Sinn. Ich veröffentliche in ein paar Tagen einen Artikel in Problemas, der besser erklärt und einige neue Beispiele enthält. (2020-12-27)
Hans-Jürgen Manthey: Habe mich über F.P.I. schlau gemacht und FPI, Vertikale Zylinder sowie viele weitere Märchenschach-Bedingungen sind alsolut nicht mein Fall... (2020-12-28)
A.Buchanan: @H-JM: Ich bin kein großer Fan von Feenschach, aber es hat seinen Platz in der Welt der Komposition. Es vervielfacht den Gestaltungsraum für Schachkompositionen erheblich, und viele coole künstlerische Effekte können nur mit Feenschach erzielt werden. FPI ist bislang kein gutes Beispiel: Aber die zugrunde liegende Idee ist gut und reif für eine breitere Nutzung. (2020-12-28)
Henrik Juel: My goodness, Andrew, you are approaching Mario's command of all languages...
I believe, however, that Feenschach is an old-fashioned name for Märchenschach
It came about as a literal translation of Fairy Chess, but now lives on only in the magazine name feenschach
We did almost the same in Denmark: feskak in the 1930s, but now fantasiskak (2020-12-28)
A.Buchanan: Hi Henrik: thanks for your kind feedback. I try to draft in German, and then back translate in google translate to English to see what it thinks, and then go forwards again to correct dumb errors and give me better word choices. But there's still errors, which I am always keen to hear about, so thanks for reminding me about Märchenschach. I now understand that Märchen are "fairy tales" not "fairies" and that the term is used in English by folklorists. Encyclopedia Britannica characterizes them by: "Their usual theme is the triumph over difficulty, with or without supernatural aid, of the one least likely to succeed." I like the detail at the end, which suggests that the key move should be an unlikely one! (2020-12-29)
Henrik Juel: I had three years of german in middle school, but never used the language actively
I believe that most german problemists can read english as well as we can read german, so I do not make the extra effort to communicate in german; besides remembering to capitalize nouns, I would have to check masculinum, femininum, or neutrum (gender?), and check which prepositions entail accusative, dative, or both (kasus?) (2020-12-29)
Olaf Jenkner: (:-) (2020-12-29)
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comment
Keywords: Cant Castler (wbsb), Fake game array, Castling (wbsb), Constrained problem, Vertical Cylinder (Vertikaler)
Genre: Retro, Fairies
FEN: r3k2r/P3p3/P3P3/8/8/p3p3/p7/R3K2R
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2020-12-26 more...
78 - P0001285
Jean Oudot
370 Die Schwalbe 07/1960
R. Bédoni gewidmet
P0001285
(11+11) C+
h#2*
* 1. ... dxc5 2. Dxh3 0-0-0#
1. Db2 Le2+ 2. Kc2 Ld1#
play all play one stop play next play all
Henrik Juel: the five missing black men were captured by white pawns (exfxgxh, fxgxj), so with Black to move last move was with Ta1 or Ke1, and White may not castle
C+ Popeye 4.61 (2022-11-26)
A.Buchanan: Pleasant White tempo play in both phases (2022-11-27)
comment
Keywords: Cant Castler (wl), Castling (wl)
Genre: Retro, h#
Computer test: HC+ Popeye 4.61 with simple retro logic
FEN: 8/7p/7P/2pr2pP/2bP2Pb/2pk1BRP/6qN/R3KNrn
Reprints: 553 FIDE Album 1959-1961 1966
157 Europe Echecs 90 07/1966
(10) diagrammes 112 01-03/1995
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-27 more...
79 - P0001310
Dragan Petrovic
182 Europe Echecs 107 01/1968
4. Preis
P0001310
(13+14) C+
#2 (AP)
1. gxf6ep! droht 2. 0-0#! (2.Kf2#?)
R: 1. f7-f5 f6xDe7,f6xTe7 etc
play all play one stop play next play all
Wh caps: fxe7, hxg5
Bl caps: axbxc2, wBf1
If wK castling right remains, the only way to give White a prior move is to retract f7-f5, so ep would be on.
The key threatens (inter alia) 2. 0-0#/Kf2#, and both remain whatever Black plays, so the castling retro-justification for ep can never be disrupted.
R: 1. S~a3? a3-a4? blocks wT from retracting to home square if wK castling right remains.
Dragan Petrovic: Author is Dragan T. Petrovic (2007-12-02)
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comment
Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic)
Genre: Retro, 2#
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 4r3/1p2P1n1/4p2B/b1pp1pPb/P5pp/nPP5/Q1pPP1P1/N1k1K2R
Reprints: (50) Problem 144-147 12/1971
(38) Mat Plus 12 10-12/2009
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
80 - P0001341
Luigi Ceriani
213 Europe Echecs 122 03/1969
P0001341
(15+10)
Löse die Stellung auf!
AV: 1.b4 c6 2.Ch3 c5 3.b5 g6 4.Ca3 d6 5.Fb2 g5 6.f4 Ca6 7.Cf2 Cf6 8.Cc4 Cd7 9.f5 f6 10.e4 Tg8 11.Re2 Tg7 12.Rd3 Tg6 13.Ce5 Fg7 14.fxg6 f5 15.Cf7 Cf8 16.e5 f4 17.Ch8 Ff6 18.exf6 Fh3 19.Fe5 Tc8 20.Cg4 Rd7 21.Re4 Tc6 22.Rf5 d5 23.Fb8 Cc7 24.a4 Rd6 25.a5 Cd7 26.bxc6 b5 27.Ta4 b4 28.Fb5 b3 29.Te1 Dc8 30.Tee4 Db7 31.Df1 Cf8 32.Tab4 Db6 33.Fa4 Cd7 34.Da6 Db5 35.c4 Cb6 36.cxb5 La position critique. 36...Cc8 37.b6 f3 38.Tb5 f2 39.Teb4 c4 40.Tc5 h6 Tempo. 41.Db5 a6 42.Fa7 axb5 La fermeture de cage traditionnelle. 43.Fb8 Ca7 44.a6 Cc8 45.a7 h5 46.a8D Ca7 47.Db7 Cc8 48.Da6! Ca7 49.Dc8 h4 50.Dg8 Cc8 51.Fa7 Ce6 52.Dh7 Cf4 53.Dxh4 Ce6 54.Dg3+ Cf4 55.Df3 Ch5 56.Dc3 Cf4 57.Dc1 Ce6 58.Da3 Cg7+ 59.fxg7 f1C 60.Fb8+ . (2005-06-08)
Hans-Jürgen Manthey: es geht schneller (COOKed ?):
R: 1. La7-b8+ f2-f1=S 2. a5xTb6 Tb8-b6 4. Db2-a3 b7-b5 5. Lb5-a4 Ta8-b8 a4-a5 Kc7-d6 5. a2-a4 a4Xb3 6. La6-b5 Kd8-c7 7. Ke5-f5 f3-f2 8. Kd4-e5 Sd6-c8 9. Se5-g4 Lc8-h3 10. Dc2-b2 f4-f3 11. Sf3-e5 Sf7-d6 12. Lb5-a6 d7-d5 13. Te5-c5 Sh6-f7 14. Te1-e5 Sg8-h6 15. Th1-e1 a5-a4 16. e5xLf6 Lg7-f6 17. Sg1-f3 f5-f4 18. Ke3-d4 Lf8-g7 19. Ld4-a7 a7-a5 20. Lb2-d4 Ke8-d8 21. d5xSc6 c5-c4 22. Tf4-b4 Sb8-c6 23. Tf1-f4 f7-f5 24. f5xTg6 Th6-g6 25. Sg6-h8 Th8-h6 26. Sf4-g6 c6-c5 27. Sh3-f4 g6-g5 28. Sf2xh3 h4-h3 29. Ta1-f1 h5-h4 30. Lf1-b5 h7-h5 31. c4xDd5 Da5-d5 32. Sd1-f2 Dd8-a5 33. Sc3-d1 c7-c6 34. Sb1-c3 Lf8-g7 35. Lc1-b2 Lg7-f8 36. Dd1-c2 Lf8-g7 37. Ke2-e3 Lg7-f8 38. Ke1-e2 Lf8-g7 39. f4-f5 Lg7-f8 40. f2-f4 Lf8-g7 e4-e5 41. Lg7-f8 e2-e4 42. Lf8-g7 c2-c4 43. g7-g6 b2-b3 (2021-07-17)
comment
Keywords: Retro Opposition
Genre: Retro
FEN: 1Bn4N/4p1P1/1PPk2P1/1pRp1Kp1/BRp3N1/Qp5b/3P2PP/5n2
Input: Gerd Wilts, 1995-06-03
81 - P0001345
F. ben Galuth
219 Europe Echecs 126 07/1969
P0001345
(10+10) C+
h#2
1. cxd3ep? g3 2. Bd4 Lg2# (wLc1 retro-blocked)
1. c5! g3 2. cxd4 Lg2#
play all play one stop play next play all
James Malcom: Solution? (2020-12-29)
Hans-Jürgen Manthey: wohl beabsichtigt: 1. cxd3ep g3 2. Bd4 Lg2#
doch auch ein Dual: 1. c5 g3 2. cxd4 Lg2# (2020-12-29)
Mario Richter: rawbats says, that f2-f4 is White's only legal last move, and 1. c5 g3 2. cxd4 Lg2# the only solution, so 1. cxd3ep g3 2. d4 Lg2# might be a try, intended to fool the solvers ...
(notice that R: 1. d2-d4?? excludes wLc1 from the set of objects available for black pawn captures!) (2020-12-29)
James Malcom: Furthermore, the h pawn couldn't have moved last, as that rook is needed for the Black pawns. wPe6 takes all remaining captures of Black pieces, and bBf8 could never escape. So the d2 and h2 pawns did not move last, and gxf3ep does nothing, meaning that all en passants are not a solution. A very fine retro "joke." (2020-12-29)
James Malcom: This means that 1. c5 g3 2. cxd4 Bg2# is the only possible solution, and as such, this should be C+? (2020-12-29)
Hans-Jürgen Manthey: Stimmt Mario
habe übersehen das der Lc1 nicht gezogen haben kann bei 1. d2-d4; aber 6 schlagfälle von Schwarz nötig sind. (2020-12-30)
A.Buchanan: I think the composer here is exactly "trolling solvers' preoccupations for the lulz", to quote Hauke Reddmann in MatPlus today, which even my allegedly fine German skills are unable to translate. But Hauke is German, so maybe he could do it? (2020-12-30)
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comment
Keywords: En passant, En passant as key (Tries)
Genre: h#, Retro
Computer test: Popeye v4.85 + thinking
FEN: 8/2p1p1p1/4P3/3p1p2/2pPkPpP/4p3/1PP1P1P1/Kb3B2
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2021-01-06 more...
82 - P0001349
Jean Oudot
Echiquier de France 1957
P0001349
(14+7)
#2
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)
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comment
Keywords: Partial Retro Analysis (PRA), Castling (wb)
Genre: Retro, 2#
FEN: 4Q3/1p2P3/brp1k1N1/1qp1P3/4P3/1NP1P1P1/1P2P3/R3K2R
Reprints: 223 Europe Echecs 130 09/1969
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-11-26 more...
83 - P0001418
Janko Furman
292 Europe Echecs 194 01/1975
2. ehrende Erwähnung
P0001418
(14+14)
Löse die Stellung auf!
h#1 wer?
White has the move, so 1. ... Sxb6! Sa6# not 1. e5? Tcxd6#
R: 1. ... Lh1-f3 2. g3-g4 h2-h1=L 3. g2-g3 h3-h2 4. d4-d5 h4-h3 5. d3-d4 h5-h4 6. d2-d3 h6-h5 7. h5xDg6 Db1-g6 8. h4-h5 Da1-b1 9. h3-h4 a2-a1=D 10. h2-h3 a3-a2 11. c3-c4 a4-a3 12. c2-c3 a5-a4 13. b3-b4 a7-a5 14. a6xLb7 etc.
R: 1. ... Th8-f8 2. g3-g4 Sf8-h7 3. g2-g3 Th1-h8 4. d4-d5 Ta1-h1 5. d3-d4 a2-a1=T 6. d2-d3 a3-a2 7. c3-c4 a4-a3 8. c2-c3 a5-a4 9. b3-b4 a7-a5 10. a6xSb7 etc.
play all play one stop play next play all
bPa & bPh promoted with wPaxb7 & bPhxg6 to clear the way. There is an inner cage containing all the White officers, and White has only 6 current retractions. So Black must unpromote something. Both White captures were on light squares, so Lf4 is original.
1) Lf3 to h1. For speed, White must uncapture Ceriani-Frolkin D on g6, which arrives on a1 in 2 unmoves. When White uncaptures L on b7, it's Phoenix of the original.
2) Tf8 to a1. (It's too slow to unpromote on h1). When White must uncapture on b7, g2 is already blocked, so White must uncapture S, and unpromote that or one of the other Black S. However, the inner cage has been breached, so the clock has stopped now. Eventually, White will uncapture Phoenix T on g6.
The tempo is exact in both cases, so White, not Black must have the move.
A.Buchanan: Very nice - exactly matching lengths across the two lines and complementary non-trivial forward play (2021-10-21)
Henrik Juel: HC+ Popeye 4.61 (2021-10-22)
more ...
comment
Keywords: Allumwandlung (ld,ts), Whose move?, Phoenix (l,t), Ceriani-Frolkin Theme (d)
Genre: h#, Retro
FEN: NnQNqr2/1PKRPprn/1pRppkP1/2pP2p1/1PP2bP1/5b2/5P2/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-21 more...
84 - P0001420
Branko Koludrovic
294 Europe Echecs 197 04/1975
1. ehrende Erwähnung
P0001420
(13+12) C+
h#3*
1. ... cxd6ep 2. 0-0-0 0-0-0 3. Kd7 Sa7#
1. De7 c6 2. Th8 c7 3. Tf8 Sd6#
play all play one stop play next play all
White pawn caps: axb,dxe,gxf,hxg.
Black: fxg,bxc,cxb.
wPb4 came from b3 to release wBa3, so bPb3 captured to reach that square.
All pcs accounted for means bPd never captured.
In the set play, there are 13 retro tries in which one or both players do not castle. The intention is that both castling rights are needed in order to imply the pawn double hop.
A.Buchanan: White pawn caps: axb,dxe,gxf,hxg definite.
Black: fxg and two to resolve c-file. But that may be c&d cross-capture, so in set play last move might have been c6xd5. So I think this problem is cooked. What am I missing? (2022-03-21)
Mario Richter: If Black's last move was c6xd5, how do you get white Bishop a3 out of his cage? (In that case, black pawn b7 never left the b-file). (2022-03-21)
A.Buchanan: I agree Mario thanks (2022-03-21)
Hans-Jürgen Manthey: beide Rochaden und im Satz ep. ist möglich:
R.: 1. ... d7-d5 2. Sa7-b5 b5xTc4 3. Tg4-c4 Dd8-h4 4. c4-c5 Th8-h3 5. c2-c4 c4xDb3 6. Dd3-b3 h3-h2 7. Dd1-d3 h4-h3 8. Sc8-a7 h5-h4 9. Th4-g4 c5-c4 10. Th1-h4 h7-h5 11. h2xSg3 Sf5-g3 12. b3-b4 Sh6-f5 13. Lb4-a3 c7-c5 14. Ld2-b4 g3-g2 15. Lc1-d2 a3-a2 16. a2xSb3 Sc5-b3 17. Lb3-a4 Sa6-c5 18. Lc4-b3 Sb8-a6 19. d2xLe3 Lc5-e3 20. Ld3-c4 Lf8-c5 21. Lf5-d3 a4-a3 22. Lh3-f5 a5-a4 23. Lf1-h3 g4-g3 24. g2xLf3 Lb7-f3 25. Sb6-c8 Lc8-b7 26. Sa4-b6 a7-a5 27. Sc3-a4 b7-b5 28. Sb1-c3 f5xSg4 29. Se5-g4 Sg8-h6 30. Sf3-e5 e7-e6 31. Sg5-f3 f6-f5 32. Sh3-g5 f7-f6 33. Sg1-h3 (2023-02-23)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (wgsg)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & retro-logic.
FEN: r3k3/6p1/4p3/1NPp4/BPp4q/Bp2PPPr/pP2PPpp/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
85 - P0001427
Karl Fabel
Am Rande des Schachbretts 1947
P0001427
(13+14) C+
BP in 41,5
1. Sf3 a5 2. Sd4 a4 3. Sb3 axb3 4. a3 Ta4 5. Ta2 bxa2 6. Tg1 axb1=S 7. Th1 Sc3 8. Tg1 Sb5 9. Th1 Sa7 10. Tg1 b5 11. Th1 Lb7 12. Tg1 Ld5 13. Th1 La2 14. b3 h5 15. Lb2 h4 16. Lf6 h3 17. Lh4 Sf6 18. Da1 Sh5 19. Df6 Tf4 20. Da6 e6 21. Kd1 Dg5 22. Kc1 Dg3 23. hxg3 Lc5 24. gxf4 Le3 25. Lg3 Ke7 26. Lh2 Kf6 27. g3 Kf5 28. Lg2 Kg4 29. La8 Sbc6 30. Kb2 Tb8 31. Ta1 Lb1 32. Lg1 h2 33. Kc3 Kh3 34. Kd3 Kg2 35. Ke4 Kf1 36. Kf3 Ke1 37. Kg4 Kd1 38. Kg5 f6+ 39. Kg6 Kc1 40. Kf7 Sd8+ 41. Ke8 Kb2 42. Ta2+
play all play one stop play next play all
Sally: Eine eindeutige BwP., die lange Zeit den Längenreord hielt und, wenn auch mit langer Verzögerung, wesentlich zum Boom dieser Gattung in den letzten 15 Jahren beigetragen hat. Im-
mer noch eine faszinierende Aufgabe. Nr. KF-13 Schwalbe Okt. 2005. (2011-12-05)
Moldenhauer: Ergänzung: Keine Lösung: BP 40.5, BP 41.0. (2023-05-20)
comment
Keywords: Unique Proof Game, Move Length Record, Non-standard material, Promotion
Genre: Retro
Computer test: Stelvio 1.17, in 3h.
FEN: Br1nK3/n1pp2p1/Q3pp2/1p5n/5P2/PP2b1P1/RkPPPP1p/1b4B1
Reprints: 1366 FIDE Album 1945-1955 1964
301 Europe Echecs 200 07/1975
(VIII) feenschach 28 05-07/1975
feenschach 38 04-06/1977
78 Shortest Proof Games 11/1991
17a 64 Proof Games 2012
Input: Gerd Wilts, 1995-06-03
Last update: Reto, 2023-04-14 more...
86 - P0001450
Henri Nouguier
324 Europe Echecs 223 07/1977
P0001450
(13+3)
h#1
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
Keywords: Cant Castler, Castling (wk)
Genre: h#, Retro
FEN: 8/8/8/8/8/PPP3PP/B1rPPP1r/BNk1K2R
Input: Gerd Wilts, 1995-06-03
87 - P0001453
Luis Alberto Garaza
327v Europe Echecs 225/226 10/1977
P0001453
(11+14) C+
h#1
1. hxg3ep! 0-0#! (Tf1#?)
play all play one stop play next play all
Hans-Jürgen Manthey: 1. hxg3 O-O#(auch Tf1# !) mögliche zugfolge:
1. d2-d4 f7-f5 2. Lc1-d2 f5-f4 3. Ld2-c1 e7-e6 4. Lc1-d2 c7-c5 5. Ld2-c1 b7-b5
6. Lc1-d2 a7-a6 7. Ld2-c1 Lc8-b7 8. c2-c4 Ke8-f7 9. Lc1xf4 Sg8-f6 10. Lf4-h6 g7xh6
11. Dd1-d3 h6-h5 12. Dd3-d1 Lf8-h6 13. h2-h4 Lh6-g5 14. h4xLg5 h5-h4 15. Dd1-d3 h7-h5
16. Dd3-d1 b5xc4 17. a2-a4 Lb7-e4 18. d4-d5 e6xd5 19. Ta1-a3 Le4-c2 20. f2-f4 Th8-e8
21. Ta3-g3 Te8-e3 22. f4-f5 Sf6-e4 23. f5-f6 Kf7-e6 24. f6-f7 Se4-d2 25. Sg1-h3 Sd2xf1
26. Sh3-f2 Sf1-d2 27. Sf2-g4 Sd2-e4 28. Dd1-d3 Lc2-d1 29. Sg4-f6 c4-c3 30. Sf6-h7 c3-c2
31. b2-b4 c5-c4 32. b4-b5 Sb8-c6 33. b5-b6 Sc6-a5 34. b6-b7 c2-c1L 35. g5-g6 c4-c3
36. Sb1-a3 c3-c2 37. Sa3-b5 Ta8-c8 38. b7-b8L Tc8-c3 39. Lb8-d6 La5-c4 40. a4-a5 a6xSb5
41. a5-a6 Ke6-f5 42. a6-a7 Dd8-e8 43. a7-a8S De8-e5 44. Sa8-c7 Lc1-b2 45. Sc7-e6 Lb2-a1
46. Se6-f8 Kf5-f4 47. Tg3-g5 Se4-f6 48.Ld6-a3 Te3-h3 49. La3-b2 Sc4-a3 50. g2-g4 (2020-12-11)
Henrik Juel: The white castling is needed to prove (a posteriori) that last move was g2-g4, legitimizing the ep capture (2020-12-31)
more ...
comment
Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: 5N2/3p1P1N/5nP1/1p1pq1Rp/5kPp/n1rQ3r/1Bp1P3/b2bK2R
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-22 more...
88 - P0001468
Oskar E. Vinje
The Fairy Chess Review 1938
P0001468
(10+3)
Letzter Zug?
R: 1. 0-0-0
play all play one stop play next play all
Deemed stipulation: "Erster Zug des wTd1?"
Henrik Juel: White pawns captured all 13 missing black men
Retracting the castling is the only way to give Black a retraction, e.g. Kc2-b3 (2020-12-01)
comment
Keywords: Type A, Last Move? (0-0-0), Castling (wl), Economy record (Last Move? Type A), First Move? (T0), Economy record (First move)
Genre: Retro
FEN: 8/P1p5/PN6/1P6/P1N5/Pk6/pP6/2KR4
Reprints: 342 Europe Echecs 241 01/1979
1.57A Eigenartige Schachprobleme , p. 194, 2010
1 Die Schwalbe 360-1, p. 737, 12/2020
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-28 more...
89 - P0001550
Michel Caillaud
421 Europe Echecs 295 07/1983
A. Hazebrouck gewidmet
1. Preis
P0001550
(13+12) C+
BP in 35.5
1. h4 c5 2. h5 c4 3. Th4 c3 4. Tc4 b5 5. g4 b4 6. Lg2 b3 7. Lc6 bxa2 8. b4 a5 9. b5 a4 10. b6 a3 11. La4 Sc6 12. b7 d5 13. b8=D d4 14. Dd6 d3 15. Dg6 dxc2 16. d3 fxg6 17. Ld2 c1=L 18. Db3 c2 19. La5 Lh6 20. Sd2 c1=L 21. Sf1 Lcg5 22. f4 Kf7 23. 0-0-0 a1=L 24. fxg5 a2 25. gxh6 Kf6 26. Sh2 Kg5 27. Tf1 Lf6 28. Tff4 a1=L 29. Kb1 Lae5 30. d4 Lb7 31. dxe5 Dd2 32. exf6 Te8 33. f7 Sf6 34. Dd3 Sd7 35. Ld1 Sdb8 36. Sgf3+
play all play one stop play next play all
5 Frolkin-Ceriani-Umwandlungen: 4 schwarze Läufer und 1 weiße Dame! Eine der bahnbrechenden frühen KBPs.
Silvio Baier: Der wesentliche thematische Inhalt ist bereits nach 31,5 Zügen erreicht. Bis dahin ist es C+ (Euclide 0.98). (2010-08-04)
James Malcom: Is this fully C+ then? (2021-01-27)
Henrik Juel: No, only the first 31.5 moves are tested OK (2021-01-27)
James Malcom: No Henrik, as in is the entire problem testable. (2021-01-27)
Henrik Juel: I guess that testing the entire problem would take an unreasonably long time (2021-04-06)
A.Buchanan: It might be possible these days: the motivation for stopping at 31.5 is that the promotion theme had been demonstrated by then. But there’s still e.g. bSb8 as random impostor (2021-04-07)
Moldenhauer: Computerprüfung: C+ Stelvio 1.11 50:24:34 Stunden.(hh:mm:ss)
Keine Lösung: BP 34.5, BP 35.0. (2023-05-08)
Henrik Juel: Thanks for your patience, Moldenhauer; more than two days... (2023-05-08)
comment
Keywords: Ceriani-Frolkin Theme (llllD), Unique Proof Game, Castling, Promotion (llllD), Impostor (s)
Genre: Retro
Computer test: Computerprüfung: C+ Stelvio 1.11 50:24:34 Stunden.(hh:mm:ss) Keine Lösung: BP 34.5, BP 35.0.
FEN: 1n2rb1r/1b2pPpp/2n3pP/B5kP/2R2RP1/3Q1N2/3qP2N/1K1B4
Reprints: 25 Shortest Proof Games 11/1991
(1) diagrammes 103 10-12/1992
(A) Quartz 22 10-12/2002
(8-a) Die Schwalbe 250A 08/2011
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-04-07 more...
90 - P0001551
Wolfgang Dittmann
2724 feenschach 46 04-06/1979
1. Preis
P0001551
(5+10)
#1 vor 7
VRZ, Typ Proca
R: 1. Kd2xLe1 e2-e1=L+ 2. Kc3-d2 e4xd3ep 3. d2-d4 e5-e4+ 4. Kd3xBc3 b4xc3ep 5. c2-c4 b5-b4+ 6. Kc4xTd3 c6xBb5 7. Kc5-c4, dann 1. b6#
play all play one stop play next play all
Mario Richter: For the retraction of ep-captures, the animation engine puts the uncaptured pawn on the wrong square. (2020-12-09)
A.Buchanan: Well spotted Mario. Suggest that you drop a quick email to Gerd? If that's not convenient for you, let me know and I will (2020-12-09)
Mario Richter: Gerd is informed. (2020-12-09)
comment
Keywords: En passant, Defensive Retractor, Type Proca
Genre: Retro
FEN: rn5b/kp1pp3/b7/8/8/3p1PP1/p4PP1/4K3
Reprints: feenschach 61 08/1982
422 Europe Echecs 295 07/1983
150 Der Blick zurück 2006
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2006-08-26 more...
91 - P0001561
Michel Caillaud
432 Europe Echecs 304 04/1984
Lob
P0001561
(8+13)
Kürzestes h#?
1. ... bxa6ep! 2. Txe2 axb7 3. Td2 b8=D 4. De2 Db1#
1. Dc6? bxc6 2. Ld7 cxd7 3. Ke1,Kxe2 d8=D 4. Kf1 Dd1# too slow
R: 1. a7-a5 e3xLf4 2. Ld6-f4 d2xTe3 3. Ta3-e3 b4-b5 4. Ta6-a3 b3-b4 5. Tc6xBa6 a5-a6 6. Tc8-c6 a4-a5 7. Tg8-c8 a3-a4 8. Lf8-d6 a2-a3 9. e7xDf6,e7xTf6
play all play one stop play next play all
The forward play gives shortest h# in 3.5 moves, *if* e.p. is on. Otherwise the shortest is a mildly dualized 4.0 moves. So our mission is clear: assuming that Black moved last, what was that move? The position looks very open - how can one ever know? There is a knot in the south-east corner, which will only be resolved by wQ/R visiting g1.

White pawns have captured dxexf & hxg, accounting for BSP. So wPa was waylaid while wPc promoted to B. Black pawns captured dxexfxgxh, exf as well as wBf1 captured at home and wPa. That's 7 units, but bPc must also have captured to allow wPc to promote (to B on c8). So all captured units are accounted for. bPc promoted (bPh7 came from h6, so is bPd). We can't undo any of the Black pawn captures now, but we can undo d2xe3xf4, to release 2 Black units.

We can unpromote bPc, but we can't undo its capture to release a White unit, until wBh3 has unpromoted. The only White unit we can get now is from e7xf6. So this means that the two Black units we can release must be B & R, so they can retreat to f8 & g8 respectively. The timing is very tight, and there is only one way to do it. The black rook must visit a6 to unwaylay wPa, which gives White 5 more tempi, just enough.

bPc promoted on b1 to R, so we would need to undo the cage to get that. Therefore all we can do is send wQg1, then Rf1-f2 f2-f3 etc. bPa must retract immediately tp a7, so that wPa when unwaylaid can make fully 4 unmoves. Black officers are arranged to give unique retro & forward play (although with minor dual for the try) with Sh8 not just retro dressing but ensuring unique retraction Tg8-c8 not Th8-c8.
A.Buchanan: Another surprising motivation for e.p. I love Sh8. Sorry for W3 dual in the try play else the stipulation could be h#4*. I can’t fix it but it’s hard to improve on MC. Very enjoyable (2021-10-23)
A.Buchanan: Non-standard material is where the diagram contains for a player more than 1 queen or bishop of a hue or more than 2 rooks or knights. Obtrusive material is standard material, but there is some cheap reason why the unit must be promoted, most commonly that a bishop's home square remains blockaded by pawns. These categories are disjoint: no piece is ever both. Many problems in PDB do not apply this consistently, but the distinction goes back a long way in chess problem history, and is discussed by Morse.
Honestly, I dislike the word "obtrusive" whose negativity (while perhaps valid in forward problems) is inappropriate for retros. One distinguished composer objected to this tag being assigned to one of his problems. Nevertheless the concept has some interest. Renaming is a perilous business, but I am looking for suggestions... :) (2021-10-24)
A.Buchanan: Another distinction that comes to mind between "non-standard material" and "obtrusive promotion" is that normally in the former, one can't immediately point to which of the non-standard pieces was promoted: it's just that there's too many; while for the latter, one can usually point to a specific unit immediately.
"Obvious" is a candidate replacement for "obtrusive", but this might commit a cardinal sin of trying to nail down a perfectly useful and inherently vague term. Both "obvious" & "obtrusive" begin with "ob" which is helpful. What do you think? (2021-10-24)
Henrik Juel: I am curious about the award; why did this problem with good forward play and excellent retro-play only obtain a Commendation?
Maybe the judge did not like that Lh3 obviously is obtrusive...
. (2021-10-24)
A.Buchanan: Yes, and the lovely P0001117 with a similar obtrusion only received 12th Lob! But there may have been other factors. I do observe that "non-standard material" is arguably a worse defect than "obtrusive material", but the term doesn't cause offence because it's objective and non-judgemental. "obtrusive" is more subjective and inherently pejorative. To have such terms in the glossary is to put curators in an invidious position. I think the concept has its place, but I would like to replace it with something less scornful. I'm up for "obvious". It's an easy and reversible change: let's do it and see if mobs of protesters form outside the gate :) (2021-10-24)
A.Buchanan: Have changed "obtrusive material" to the non-pejorative "obvious promotion". It may still be regarded as a defect. As a placeholder, I have also changed the unclear German "mit Umwandlungfigur(en)" to "augenscheinlich Umwandlungfigur". A native German speaker I'm sure will propose a better term. (2021-10-26)
A.Buchanan: “offensichtlich” it is thanks Mario (2021-10-26)
more ...
comment
Keywords: En passant, Last Moves?, Obvious promotion (L), En passant as key, Promotion (D)
Genre: h#, Retro
FEN: 7n/1p3pp1/4bp1p/pP6/4qPP1/5PrB/4PrPp/3k3K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-10-24 more...
92 - P0001569
Andrey Frolkin
439v Europe Echecs 310 10/1984
Lob
P0001569
(12+14) cooked
BP in 27,0
1. f4 h5 2. f5 h4 3. f6 h3 4. fxe7 f5 5. d4 Kf7 6. e8=T Kg6 7. Te6+ Kh5 8. Ta6 bxa6 9. Dd2 Lb7 10. e4 Lc6 11. e5 La4 12. e6 Sc6 13. e7 Db8 14. e8=T Db3 15. Te5 Te8 16. d5 Te6 17. dxe6 Lb4 18. Ta5 Lc3 19. Lb5 f4 20. e7 f3 21. e8=T f2+ 22. Ke2 f1=S 23. Tb8 Sxh2 24. Tb6 cxb6 25. Sf3 bxa5 26. Se5 Sf3 27. Sg4 Sfd4+
play all play one stop play next play all
Cook: 1. f4 h5 2. f5 h4 3. f6 h3 4. fxe7 f5 5. e4 Kf7 6. e8=T f4 7. Te6 f3 8. Ta6 bxa6 9. e5 Kg6 10. e6 Lb7 11. e7 Lc6 12. e8=T La4 13. Te5 Sc6 14. Ta5 Db8 15. d4 Db3 16. Lb5 f2 17. Ke2 f1=S 18. Sf3 Te8 19. Se5 Kh5 20. d5 Te6 21. dxe6 Sxh2 22. e7 Sf3 23. e8=T Lb4 24. Tb8 Lc3 25. Tb6 cxb6 26. Sg4 bxa5 27. Dd2 Sd4
more ...
comment
Keywords: Ceriani-Frolkin Theme (TTT), Unique Proof Game, Non-standard material, Promotion (s)
Genre: Retro
Computer test: Computerprüfung: Cooked Stelvio 1.11 1 Sekunde. Keine Lösung: BP 26.0, BP 26.5.
FEN: 6nr/p2p2p1/p1n5/pB5k/b2n2N1/1qb4p/PPPQK1P1/RNB4R
Reprints: 21 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-08 more...
93 - P0001572
Andrey Frolkin
442 Europe Echecs 313 01/1985
P0001572
(15+13) C+
BP in 16,5
1. Sc3 Sf6 2. Sd5 Se4 3. Sxe7 Sg5 4. Sg6 Lb4 5. e3 d6 6. Se2 Kd7 7. Sc3 Kc6 8. Sb1 Sd7 9. Lb5+ Kxb5 10. Sxh8 c6 11. Sxf7 Da5 12. Se5 Sb6 13. Sf3 Ld7 14. Sg1 Te8 15. f3 Te4 16. Kf2 Le8 17. Kg3
play all play one stop play next play all
more ...
comment
Keywords: Unique Proof Game, Interchange (SS)
Genre: Retro
Computer test: Natch 2.3 Copyright (C) 1997,98,99,2001,2002,2003,2004 Pascal Wassong Ergänzung: Stelvio 1.2. Keine Lösung BP 15.5, BP 16.0.
FEN: 4b3/pp4pp/1npp4/qk4n1/1b2r3/4PPK1/PPPP2PP/RNBQ2NR
Reprints: 67 Shortest Proof Games 11/1991
(2) Quartz 21 07-09/2002
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-21 more...
94 - P0001573
Dmitri W. Pronkin
443 Europe Echecs 313 01/1985
P0001573
(16+14) C+
BP in 24,0
1. h4 g5 2. h5 Lg7 3. Th4 Lc3 4. bxc3 Sf6 5. La3 Sd5 6. Ld6 Sf4 7. Sa3 Sh3 8. gxh3 g4 9. Lg2 g3 10. Ld5 g2 11. Sf3 g1=L 12. Kf1 Lh2 13. Kg2 Le5 14. Dg1 Lg7 15. Kg3 b5 16. Kf4 b4 17. Dg6 b3 18. Kg5 b2 19. Lb3 b1=S 20. c4 Sc3 21. Tb1 Sd5 22. Tb2 Sf6 23. Sb1 Sg8 24. Df6 Lf8
play all play one stop play next play all
more ...
comment
Keywords: Unique Proof Game, Homebase (S), Pronkin Theme (ls)
Genre: Retro
Computer test: Natch 2.4 Ergänzung Stelvio 1.2. Keine Lösung BP 23.0, BP 23.5.
FEN: rnbqkbnr/p1pppp1p/3B1Q2/6KP/2P4R/1B3N1P/PRPPPP2/1N6
Reprints: 46 Shortest Proof Games 11/1991
(19a) Die Schwalbe 232 08/2008
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-21 more...
95 - P0001577
Michel Caillaud
447 Europe Echecs 316 04/1985
2. Preis
P0001577
(14+11) C+
BP in 26,5
1. f4 h5 2. f5 h4 3. f6 h3 4. fxe7 hxg2 5. h4 g5 6. h5 g4 7. Th4 g3 8. Sh3 g1=L 9. Lg2 Le3 10. Lc6 g2 11. dxe3 g1=T+ 12. Kf2 Tg3 13. e4 Tb3 14. axb3 f5 15. Ta6 f4 16. La4 f3 17. Kg3 f2 18. Th6 f1=D 19. Dd6 Df7 20. Lf4 Dc4 21. bxc4 a5 22. Lb3 a4 23. Da6 a3 24. Ld6 a2 25. Sf4 axb1=S 26. La2 Sc3 27. bxc3
play all play one stop play next play all
Erstdarstellung einer Frolkin-Ceriani-Allumwandlung in einer eindeutigen KBP.
more ...
comment
Keywords: Allumwandlung, Ceriani-Frolkin Theme (dtls), Unique Proof Game, Homebase (S)
Genre: Retro
Computer test: Euclide 0.93 Ergänzung Stelvio 1.2. Keine Lösung BP 25.5, BP 26.0.
FEN: rnbqkbnr/1pppP3/Q2B3R/7P/2P1PN1R/2P3K1/B1P1P3/8
Reprints: 13 Shortest Proof Games 11/1991
(19) diagrammes 15 07-09/1994
Thema Danicum 106 2002
(B) Quartz 26 10-12/2004
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-21 more...
96 - P0001580
Dmitri W. Pronkin
450 Europe Echecs 319/320 07-08/1985
P0001580
(12+11) cooked
BP in 31,5
1. b4 h5 2. Lb2 h4 3. Dc1 h3 4. Kd1 hxg2 5. h4 e5 6. h5 e4 7. h6 e3 8. h7 exf2 9. e4 f5 10. Se2 g1=L 11. Lg2 f1=L 12. e5 Lb6 13. d4 Lfc5 14. dxc5 f4 15. cxb6 f3 16. Df4 f2 17. Sc1 La6 18. Sd2 f1=L 19. Tb1 Lfb5 20. c4 Se7 21. cxb5 Tg8 22. hxg8=L g5 23. Lc4 d5 24. bxa6 Lf5 25. e6 Sc8 26. e7 Kf7 27. e8=L+ Kg8 28. Leb5 c6 29. Tf1 cxb5 30. Tf2 Sc6 31. Sf1 dxc4+ 32. Ld5+
play all play one stop play next play all
Cook: 1. b4 e5 2. Lb2 e4 3. Dc1 e3 4. Kd1 exf2 5. g4 Ld6 6. g5 Lxh2 7. g6 Kf8 8. gxf7 g5 9. Lg2 f1=L 10. e4 Lb5 11. Se2 Lg1 12. Th4 h5 13. e5 Kg7 14. c4 Th6 15. cxb5 Ta6 16. Tc4 Lc5 17. d4 h4 18. e6 h3 19. e7 h2 20. e8=D d5 21. bxa6 Se7 22. Db5 c6 23. Sd2 cxb5 24. Sf1 h1=D 25. f8=T Dh6 26. Tf2 Db6 27. Df4 Lf5 28. Tb1 Kg8 29. dxc5 Sc8 30. cxb6 Sc6 31. Sc1 dxc4+ 32. Ld5+
Paulo Roque: obs: Cook: 26...D6b6 (2008-12-24)
paul: See P0001634 as correction. (2010-09-25)
Moldenhauer: Computerprüfung: Cooked Stelvio 1.6 in 31:54:04 Stunden. (hh:mm:ss).
1 Lösung mit 20 cooks. (2024-01-07)
comment
Keywords: Ceriani-Frolkin Theme (lllLL), Unique Proof Game, Promotion (lllLL)
Genre: Retro
FEN: r1nq2k1/pp6/PPn5/1p1B1bp1/1Pp2Q2/8/PB3R2/1RNK1N2
Reprints: 26 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-06 more...
97 - P0001600
Pascal Wassong
470 Europe Echecs 334 10/1986
P0001600
(13+14) cooked
BP in 20,0
1. b4 c5 2. b5 Sc6 3. bxc6 b5 4. c7 La6 5. c8=L d6 6. Lf5 Tc8 7. d4 Tc6 8. d5 Db8 9. dxc6 e6 10. c7 Le7 11. c8=S Ld8 12. Se7 Lb6 13. Sg6 hxg6 14. h4 Ke7 15. h5 Kf6 16. h6 Se7 17. h7 Te8 18. h8=T gxf5 19. Th4 Sg6 20. Te4 fxe4
play all play one stop play next play all
Cook: 1. b4 c5 2. b5 Sc6 3. bxc6 b5 4. d4 La6 5. d5 Tc8 6. d6 Tc7 7. dxc7 Db8 8. c8=L d6 9. Lf5 e6 10. h4 Ke7 11. h5 Kf6 12. c7 Le7 13. c8=S Ld8 14. Se7 Lb6 15. Sg6 Se7 16. h6 hxg6 17. h7 Te8 18. h8=T gxf5 19. Th4 Sg6 20. Te4 fxe4
Henrik Juel: Dual: ... 4.d4 La6 5.d5 Tc8 6.d6 Tc7 7.dxc7 Db8 8.c8L d6 9.Lf5 ... (2004-02-12)
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comment
Keywords: Ceriani-Frolkin Theme (TLS), Unique Proof Game, Homebase (W), Promotion
Genre: Retro
Computer test: Ergänzung Stelvio 1.2. Keine Lösung BP 19.0, BP 19.5.
FEN: 1q2r3/p4pp1/bb1ppkn1/1pp5/4p3/8/P1P1PPP1/RNBQKBNR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-21 more...
98 - P0001601
Filip S. Bondarenko
471 Europe Echecs 334 10/1986
P0001601
(16+0)
Ergänze die 16 schwarzen Steine so, daß kein Stein angegriffen ist!
KBP
1. Sf3 b5 2. Sd4 e5 3. Sc6 g6 4. e3 Sf6 5. c4 Sd5 6. d4 Sb6 7. c5 a5 8. b4 a4 9. Ke2 Sa6 10. Kf3 e4+ 11. Kg4 Ld6 12. Sb8 Ke7 13. Sa3 Tf8 14. Kg5 f5 15. g4 h5 16. h4 Lh2 17. c6 d6 18. Lb2 f4 19. Kh6 Ke6 20. g5 Kf5 21. Db3 Kg4 22. Tc1 Kf3 23. Tc5 Lh3 24. Te5 Sc5 25. Lc4 Sd3 26. Kh7 Ke2 27. Te8 f3 28. Le6 Ta7 29. d5 Sa8 30. Sc4 Tf4 31. Kh8 Tg4 32. Lg7 Df6 33. Tc1 Da1 34. Tc3 Dh1 35. a3 Tg1 36. Sb6 Lg2 37. S6d7 Se1 38. Db2+ Kd1
play all play one stop play next play all
Originalforderung: Compléter le diagramme par les 16 pièces noires de façon à ce qu'aucune pièce des deux camps ne soit en prise.
Plus courte partie justificative aboutissant à cette position.
hans: Kd1, Dh1, Tg1, Ta7, Lg2, Lh2, Se1, Sa8, a4, b5, c7, d6, e4, f3, g6, h5. (2013-07-27)
Mario Richter: @Alain: with black Bishop on f1 instead of g2 white pawn g5 is attacked by black Rook g1 (2021-08-23)
Alain Brobecker: @Mario, you are right of course, thank you for correcting this! (2021-08-23)
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comment
Keywords: Add pieces, Non-Unique Proof Game, Construction task
Genre: Retro
FEN: 1N2R2K/3N2B1/2P1B3/3P2P1/1P5P/P1R1P3/1Q3P2/8
Input: Gerd Wilts, 1995-06-03
Last update: Alain Brobecker, 2021-08-23 more...
99 - P0001605
Dmitri W. Pronkin
Andrey Frolkin

475 Europe Echecs 337 01/1987
P0001605
(1+16) C+
BP in 17.0
1. d3 h6 2. Lg5 hxg5 3. h4 Txh4 4. a4 Txa4 5. Th6 Sxh6 6. g4 Sxg4 7. Lg2 Sxf2 8. Lc6 Sxc6 9. d4 Sxd4 10. c4 Sxe2 11. c5 Sxg1 12. c6 dxc6 13. Dd4 Dxd4 14. Ta3 Dxb2 15. Th3 Lxh3 16. Sc3 0-0-0 17. Sd5 Txd5= patt
play all play one stop play next play all
Henrik Juel: Black makes 15 captures.
The first 13.0 moves are correct by Euclide 1.0, but it takes some 4 minutes, so a complete test would probably take more than a day (2014-12-09)
paul: Jacobi checked the last 16 moves (in about 20 min). (2018-05-10)
Mario Richter: I would like to label this problem as a "Massacre PG", but the current definition of that term only knows of two-sided MPGs. Perhaps we could make the same differentiation as for "Homebase"?!
Btw., wasn't "popeye" best for such massacres?
rawbats confirms complete correctness of the problem after approx. 25 minutes. (2018-05-11)
Henrik Juel: Yes, Mario, Popeye is fairly good at this type
C+ by Popeye 4.61 in 57 minutes (2018-05-11)
A.Buchanan: @Mario: yes I agree it would be good to make massacre more specific. There are 69 massacres currently, almost all 2-sided, but maybe also some series-movers (unique or non-unique) which are currently not counted? (2018-05-12)
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comment
Keywords: Unique Proof Game, Castling, Homebase (w), Rex solus (w), Massacre PG
Genre: Retro
Computer test: C+ rawbats. C+ by Popeye 4.61 in 57 minutes. C+ Stelvio 1.2 2 Sekunden. Keine Lösung: BP 16.0, BP 16.5.
FEN: 2k2b2/ppp1ppp1/2p5/3r2p1/r7/7b/1q3n2/4K1n1
Reprints: 116 Shortest Proof Games 11/1991
20a 64 Proof Games 2012
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-21 more...
100 - P0001617
Michel Caillaud
487 Europe Echecs 346 10/1987
P0001617
(9+7) cooked
#1 vor 12
VRZ, Typ Proca
paul: Cooked in 4: 1.Ke5-e6! b2×Ba1=B+ -2.Kf4×Pe5 e6-e5+ -3.Rc2×Sg2 Se1-g2+ -4.Re2×Pc2 & 1.Rxe1# (2023-06-14)
comment
Keywords: En passant, Defensive Retractor, Type Proca
Genre: Retro
FEN: 8/4p3/3PK1P1/8/1B5p/1P1qPP1p/P5Rp/b2k4
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2004-08-28 more...
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