Die Schwalbe

3403 problem(s) found in 4133 milliseconds (displaying 100 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT A='Dittrich, Stefan' AND S='Die Schwalbe'] [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 - 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...
27 - 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)
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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...
28 - 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...
29 - 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...
30 - 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...
31 - 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)
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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...
32 - 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...
33 - 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...
34 - 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)
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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...
35 - 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)
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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...
36 - 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...
37 - 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)
more ...
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...
38 - 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...
39 - 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...
40 - 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)
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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...
41 - 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)
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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...
42 - 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...
43 - 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)
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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...
44 - 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...
45 - 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)
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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...
46 - 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...
47 - 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...
48 - 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...
49 - 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...
50 - 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...
51 - 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
52 - 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...
53 - 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)
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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...
54 - 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...
55 - 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...
56 - 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...
57 - 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...
58 - 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...
59 - 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...
60 - 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...
61 - 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...
62 - 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...
63 - 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...
64 - 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...
65 - 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...
66 - 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...
67 - P0001634
Dmitri W. Pronkin
503v Europe Echecs 358 04/1989
5. Preis
P0001634
(12+11) C+
BP in 31,0
1. b3 h5 2. La3 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. Lc5 f1=L 19. a3 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. Ta2 cxb5 30. Tc2 Sc6 31. Sa2 dxc4+
play all play one stop play next play all
paul: Correction of P0001580 (2010-09-25)
Silvio Baier: Mit 30....dc+ ist das C+ (Euclide 0.98, Retractor für den letzten Zug) (2011-07-08)
James Malcom: So this is fully C+? (2021-01-27)
Henrik Juel: No, another problem in 30.0 moves ending with 30. Tc2 dxc4+ is tested OK (2021-01-27)
more ...
comment
Keywords: Ceriani-Frolkin Theme (lllLL), Unique Proof Game, Promotion (lllLL)
Genre: Retro
Computer test: Computerprüfung: C+ Stelvio 1.11 65:30:51 Stunden. (hh:mm:ss) Keine Lösung: BP 30.0, BP 30.5.
FEN: r1nq2k1/pp6/PPn5/1pB2bp1/2p2Q2/PP6/N1R3B1/1N1K3R
Reprints: 580 Ukrainisches Album 1986-1990
26 Shortest Proof Games 11/1991
H25 FIDE Album 1986-1988 1995
(H25) Die Schwalbe 156, p. 237, 12/1995
feenschach 141 06/2001
Thema Danicum 105 2002
(C) Quartz 22 10-12/2002
H2 The Problemist 07/2010
(18) Die Schwalbe 250A 08/2011
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-05-08 more...
68 - P0001859
William A. Langstaff
The Chess Amateur 1922
P0001859
(5+3) C+
#2
If Black can castle, e.p. is ok:
1. hxg6ep! 0-0 2. h7#
1. ... Tf8,Kf8 2. Td8#
Otherwise Black can't castle:
1. Ke6! ... 2. Td8#
play all play one stop play next play all
more ...
comment
Keywords: Cant Castler, Partial Retro Analysis (PRA), Castling (sk), En passant as key
Genre: Retro, 2#
Computer test: HC+ Popeye 4.61 & trivial retro-logic
FEN: 4k2r/8/5B1P/3R1KpP/8/8/8/8
Reprints: (D17) feenschach 27 04/1975
RA64 diagrammes 27 05-06/1977
(1) Die Schwalbe 86 04/1984
Razem 34 23/08/1987
(I) diagrammes 15 07-09/1994
(10) Die Schwalbe 241 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-11-30 more...
69 - P0001941
Nenad Petrovic
2nd International Team Match 1967-1970
3. (2.?) Platz
P0001941
(10+12)
Gewinn
AP
Henrik Juel: If White may castle, last move was g7-g5 to avoid retrostalemate. 1.fxg6ep followed by 0-0 to legitimize the en passant capture. (2004-09-23)
A.Buchanan: In V&V Encyclopedia, which I admire more for its scope than for its precision, this problem is given to illustrate "Petrovic Theme". The definition given is: "PETROVIC THEME Also called 'retroproblem of Petrovic Type'. Mutual dependence of en passant capture and castling. By playing an en passant capture other retro elements of position are legalized (usually castling)."
The definition makes no reference to A Posteriori. I am trying to get my head around the text here, because the e.p. does not "legalize" castling, rather it mandates it. I also encountered Öffner for this castling/e.p. AP (but not in V&V), but that might be someone's confusion in that Types Öffner vs Keym exist in PRA. (2022-02-15)
comment
Keywords: En passant as key, Castling (wk), a posteriori (AP) (Type Petrovic), En passant
Genre: Retro, Studies
FEN: b7/p4P2/2kbPp1p/3ppPp1/n3pp2/8/P1PPP2P/4K2R
Reprints: (2) Problem 141-143 08/1971
(77) Problem 144-147 12/1971
(C) Die Schwalbe 16 10/1972
Encyclopedia of Chess Problems 2012
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-15 more...
70 - P0002056
Samuel Loyd
Musical World 1859
P0002056
(2+4) C+
#2
1. Da1! droht (unparierbar) 2. Dh8#
1. ... 0-0-0? ist illegal, da sK oder sTa8 schon gezogen haben muss
play all play one stop play next play all
Hans-Jürgen Manthey: Nachdruck SExpress 1.Mai 1947 S.41 Nr.33 (2022-10-15)
more ...
comment
Keywords: Cant Castler, Castling (sg), Minimal, Miniature, Homebase
Genre: Retro, 2#
Computer test: C+ Popeye 4.61
FEN: r3k3/p1p5/Q3K3/8/8/8/8/8
Reprints: 304 Chess Strategy (Loyd) 1878
73 150 Schachkuriositäten 1910
63 Sam Loyd and his Chess Problems 1913
43 64 Schach-Scherze 1915
32 Retrograde Analysis 1915
168 Allgemeine Zeitung Chemnitz 27/11/1927
Arbeiter-Zeitung (Wien) 27/11/1932
8 Comoedia 09/07/1933
(II) Problem 37-40 09/1956
(D10) feenschach 27 04/1975
84 100 Classics of the Chessboard 1983
(7a) Die Schwalbe 145 08/1995
Thema Danicum 95 1999
52 Opfer-Opfer-Matt Gaudium 21 10/2000
Outrageous Chess Problems 2005
Input: Gerd Wilts, 1995-06-03
Last update: Mario Richter, 2023-05-26 more...
71 - P0002277
Karlheinz Bachmann
v Die Schwalbe 104, p. 217, 04/1987
1. ehrende Erwähnung
P0002277
(11+15) C+
BP in 48,0
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 Sa6 22. Lb2 Sc5 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. h6 Lg3 33. h7 Sh6 34. Th4 Tb8 35. Tb4 Tb7 36. f4+ Kh4 37. h8=T Sf3 38. Tg8 Se5 39. Tg5 Sg6 40. Kf3 Kh3 41. Ke3 Kg2 42. Te5 Kf1 43. Kf3 Ke1 44. Kg2 Kd1 45. Kf1 Kc1 46. Ke1 Kb2 47. Kd1 Ka1 48. Kc1 Sh8
play all play one stop play next play all
paul: Correction of P0000067. (2018-03-03)
Reto: The new length record for a fully tested SPG. (2023-09-18)
Olaf Jenkner: How long does it take with one core and how many cores did you use? (2023-09-18)
Reto: I used 15 seekers and 15 players, but not all of them are active all the time. My guess is that it takes about 7-10h single-threaded. (2023-09-19)
comment
Keywords: Unique Proof Game, Move Length Record, Non-standard material, Promotion (Tss)
Genre: Retro
Computer test: Stelvio 1.66 in 23min (heavily parallelized)
FEN: n6n/1rpppp2/1p3p1n/2n1R3/rR3P2/PP4b1/b1PPPP2/kBK5
Reprints: 82 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: Reto, 2023-09-18 more...
72 - P0002337
Michel Caillaud
4019 Die Schwalbe 76 08/1982
Dr. Karl Fabel zum Gedenken
1. Preis
P0002337
(11+15) C+
BP in 47,0
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 Ke1 40. Th5 Sh7 41. Ta5 Kd1 42. Ta7 Kc1 43. Tb7 Ta7 44. Kd3 Sa6 45. Tb8 Kb2 46. Tf8 Ka1 47. Kc3 Lg1
play all play one stop play next play all
Henrik Juel: So Michel's proof game is not cooked; yet... (2019-09-10)
Reto: This is now the length record for a fully tested SPG. (2023-09-17)
Moldenhauer: Keine Lösung: BP 46.5. (2024-01-05)
more ...
comment
Keywords: Unique Proof Game, Move Length Record, Non-standard material (ss), Promotion (ssT)
Genre: Retro
Computer test: Stelvio 1.64 in around 20min (highly parallelized).
FEN: n4Rn1/r1pppp1n/np3p2/8/1r1R1P2/PPK5/b1PPPP2/kB4b1
Reprints: diagrammes 69 01-02/1985
989 FIDE Album 1980-1982 1988
81 Shortest Proof Games 11/1991
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-09-17 more...
73 - P0002353
Theodor Tauber
Michel Caillaud

The Problemist 1991
P0002353
(14+11) C+
ser-h#3 (AP)
1. dxe3ep 2. fxg1=L 3. 0-0 Tg4#
1. dxe3ep 2. Ke7? 3. Kd6 Sf5# doesn't pay AP debt
play all play one stop play next play all
Henrik Juel: If Black may castle, his latest move must have been e3xf2, so last move was e2-e4. (Orig. wPg2 promoted on g8, so g7xh6 happened early). 1.dxe3 e.p. 2.fxg1=B 3.0-0 Tg4# (2003-03-21)
James Malcom: Is this the only know one-sided Valladao in a regular series-mover helpmate? Also, can wPh2 be shifted to h4, and the White king moved as well to h2, so that the solution must have 2. fxg1=S? (2020-12-08)
James Malcom: Well, the only known one that is in minimum form, i.e. 3 moves, I mean. (2020-12-08)
A.Buchanan: This is nice. I don't know of any other AP ser-h#3 with Valladao, but there could be. Just been cleansing PDB data today. Yes I think wKh3 and wBh2 can be transposed, and then promote to S instead of L. (2020-12-08)
A.Buchanan: The legality of intermediate positions in seriesmovers is of course irrelevant. However, each proof game that leads to the diagram position with BTM can be appended with a black move and then a "flip" that changes the player to move. We can repeat this process, and then the fact that Black retains castling rights implies that they had castling rights in the diagram position. I think this is the right formal way to handle AP with seriesmovers. (2022-03-21)
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sk), Seriesmover, Promotion, Valladao Task
Genre: Retro, Fairies
Computer test: C+ Popeye v4.85 + thinking
FEN: 4k2r/1pp2p1p/1N5p/3P4/2RpP2N/P2P1P1K/p1BP1p1P/B3b1R1
Reprints: Die Schwalbe 143 10/1993
Rex Multiplex 45-46 05/1994
(XII) Quartz 5 1997
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-03-21 more...
74 - P0002372
Alexander Kislyak
Die Schwalbe 136 08/1992
P0002372
(15+10) cooked
BP in 10,5
1. Sf3 e6 2. Se5 Df6 3. Sxd7 Sxd7 4. a4 Sb6 5. a5 Ld7 6. axb6 0-0-0 7. Txa7 La4 8. Txa4 Td4 9. Txd4 Kb8 10. bxc7+ Ka8 11. Ta4#
play all play one stop play next play all
Cook: 1. Sf3 e6 2. Se5 Df6 3. Sxd7 Kd8 4. Sxb8 Ld7 5. a4 Lb5 6. axb5 Kc8 7. Txa7 Kxb8 8. Ta3 c6 9. bxc6 Ta4 10. c7 Ka8 11. Txa4
Moldenhauer: Computerprüfung: Cooked Stellung Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 9.5, BP 10.0.
Eine Notation von Stelvio:
1.Sf3 e6 2.Se5 Df6 3.Sxd7 Kd8 4.Sxb8 Ld7 5.Sc6+ Kc8 6.Sxa7+ Kb8
7.a4 Lb5 8.axb5 Txa7 9.b6 Ka8 10.bxc7 Ta4 11.Txa4#
Schlüsselwort Rochade? (2023-05-02)
comment
Keywords: Unique Proof Game, Castling
Genre: Retro
FEN: k4bnr/1pP2ppp/4pq2/8/R7/8/1PPPPPPP/1NBQKB1R
Input: Gerd Wilts, 1995-06-03
Last update: Gerd Wilts, 2006-10-15 more...
75 - P0002381
Jan Rusinek
New Statesman 1971
1. Preis
P0002381
(5+4)
Remis
1. g8=D? Lxg8 2. a7 Ld5 3. a8=D Lxa8 4. b7 Se4

1. a7! La6+ 2. b7 Se4 3. g8=S+ Ke8 4. Sf6+ Sexf6 5. a8=L Se5 6. Kb8 Sc6+ 7. Kc8 Lf1 8. b8=T La6+ 9. Tb7 Se4 patt
9. ... Sa5 10. Kb8 Sc6+ 11. Kc8

9. Lb7? Se4!
8. b8=D? La6+ 9. Db7 Se4 10. Dxa6 Sd6#
8. b8=S? Se7+ 9. Kb7 Lg2+ 10. Sc6 Lxc6+ 11. Ka7 Ld7
5. a8=D? Sd5 6. Dxa6 Se7#
play all play one stop play next play all
more ...
comment
Keywords: Stalemate defense
Genre: Studies
FEN: 2K5/2Pnk1P1/PP6/8/2b5/2n5/8/8
Reprints: 1307 Problemista 140-144 05-09/1973
772 FIDE Album 1971-1973 1978
Die Schwalbe 127 02/1991
Input: Gerd Wilts, 1995-06-03
Last update: Marcin Banaszek, 2022-03-31 more...
76 - P0002471
Gerd Rinder
(G) Die Schwalbe 48 12/1977
Lob
P0002471
(4+3)
#2 (AP)
BTM: 1. ... Lxb7+ 2. Ke3 0-0-0 3. Sb6#
WTM: 1. Sf6+! Kd8,Kf8 2. Dc7#,Ld6#
play all play one stop play next play all
If WTM, b000 rights are already lost. Under Keym AP, Black attempts to steal the move. White disruption of castling now counts as win for White, so the only chance is 1. … BxQb7+ 2. Ke3! thr 3. Sf6+ disrupting castling but 2. … 0-0-0 3. Sb6#. 2. Kd3,Kf5? Be4+ 3. ~ 0-0-0! as b7 is no longer occupied, or 2. Kd3,Kd4,Kd5? 0-0-0! pinning wS or 2. Ke5,Kf4? 0-0-0! as wL is blocked. As usual when flip of player to move in d#n, Black gets an extra move rather than White losing one (c.f. Codex Article 15). Sublime miniature!
VL: AP after Keym. Solution:
I: 1.Sf6+.
II: Bl's try to be on move. 0... Bxb7+! 1.Ke3! O-O-O (legalizing!) 2.Sb6#. (2007-01-26)
more ...
comment
Keywords: Castling (sg), a posteriori (AP) (Type Keym), Homebase (s), Aristocrat, Miniature
Genre: Retro, 2#
FEN: r1b1k3/1Q1N4/8/8/4K3/8/7B/8
Reprints: Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
77 - P0002476
Werner Kuntsche
1777v Problem 73-78 06/1961
P0002476
(5+7) cooked
h#3 AP
Einheitslösung!
1) 1. cxd3ep Ta5 2. Sf5 0-0 3. Sd4 Txe5#
2) 1. gxf3ep Th5 2. Sd5 0-0-0 3. Sf4 Txe5#
play all play one stop play next play all
Cook: 707 cooks which do not begin with e.p., e.g. 1.Sg3-e2 Ra1-a2 2.Ke4-d3 Rh1-h3 3.e5-e4 Ra2-d2 #
What is the obscure intent behind "Einheitslösung!"? (= "One size fits all solution"?)
Mario Richter: Im Nachdruck in problem 144-147 steht nur auf e5 ein sB (also keiner auf e3). Allerdings gibt es in allen drei Varianten (sBe5+e3, nur sBe5, nur sBe3) jede Menge NL, die ohne Rochade und e.p.-Key auskommen - am wenigsten dann, wenn nur auf e3 ein sB steht. (2010-10-10)
VL: Cf. P0002475. (2012-08-26)
A.Buchanan: Possibly "Einheitlösung!" is a request to unify retro implications derived across separate parts of the PRA solution. But cooked chessically irrespective of these philosophical issues, which still wait to be resolved 60 years on. (2022-02-16)
comment
Keywords: a posteriori (AP) (Type Petrovic - ccee), En passant as key (2), Castling (wb), Symmetrical position, Symmetrical solution, Superseded by (P1401449,P1399112)
Genre: h#, Retro
Computer test: C- Forward play cooked by Popeye v4.87. However the basic idea is good, if a suitable theoretical framework can be found.
FEN: 8/8/8/4p3/2pPkPp1/2n1p1n1/8/R3K2R
Reprints: (66) Problem 144, p. 52, 12/1971
Die Schwalbe 99 06/1986
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-05-24 more...
78 - P0002758
Andrej N. Kornilow
1078 Phénix 15-16 12/1991
P0002758
(25+0)
Färbe die Steine! Welches waren die letzten 9 Einzelzüge?
James Malcom: Solution? (2020-11-07)
Mario Richter: I do not know the official solution, but the following works:
wBb7 wBf7 wBb6 wBg6 wBc5 wTf5 wLf4 wBg4 wBe3 wKg3 wTh3 wDf2 wBh2 wSf1 wSh1
sLg8 sDh8 sBa7 sBc7 sBh7 sBe6 sBf6 sKh6 sBe5 sLg1
wCaps: h5xg6ep f3xTg4 g4xTh5 e6xSf7 d5xSe6 a4xBb5
sCaps: d7xLe6
R: 1. h5xg6ep g7-g5 2. Tg5-f5 d7xLe6 3. f3xTg4 Th4-g4 4. g4xTh5 Kg6-h6 5. Tf5-g5 (2020-11-10)
comment
Keywords: Colouring problem, Last Moves? (9), En passant
Genre: Retro
FEN: 6BQ/PPP2P1P/1P2PPPK/2P1PR2/5BP1/4P1KR/5Q1P/5NBN
Reprints: (34) Die Schwalbe 144 12/1993
Input: Gerd Wilts, 1995-06-03
79 - P0003195
Thomas R. Dawson
2130 Die Schwalbe 07/1932
P0003195
(14+14) C+
h#2
WTM
1. ... bxc6ep 2. Da7 c7#
retro tries:
1) 1. Lxg3 b6 2. Lxd6 Lxd6#
2) 1. Dc6 b6 2. Dc7 dxc7#
3) 1. Ta7 f8=D,f8=T 2. Ka8 Dxc8#,Txd8# ("untolerated" dual)
R: 1. ... c7-c5 2. b4-b5 Dc4-a6 (move order variable beyond this point) 3. b2-b3 Ta6-a8 4. d5-d6 Te6-a6 5. b3-b4 Te8-e6 6. a2xSb3 Th8-e8 7. d4-d5 Lb4-e1 8. d3-d4 Lf8-b4 9. Kg5-h5 e7xSf6+
play all play one stop play next play all
Missing: Wh: SS Bl: SS
Pawn captures: Wh: axb, hxg Bl: exf, fxg
bPf6 comes from e7, else collision with wPf7. So can't uncapture fxg2 until wP retracted. White has 6 pawn retractions before retro-pat, just giving Bl time to put Le1 & Ta8 (impostor) back home, and uncapture exf6. So Black must retract first, and it's WTM. sD must clear out of the way, and c4 is the only possible square, wPaxSb can't happen on b5, or it would block sD. And axSb4 would block sLe1. It was axSb3. So wPb5 must retreat to b4, and thus Black cannot play R: 1. Lb4-e1. So neatly R: 1. c7-c5 is forced.
Ladislav Packa: Is wRh1 needed? (2021-10-22)
Mario Richter: Yes, wRh1 is needed - without it the ep-key would not be justified (last moves could have been R: 1. ... Th1-g1 2. Lg1-h2 or even R: 1. ... Ld2xTe1). (2021-10-23)
A.Buchanan: + three mostly clean retro tries. (2021-10-24)
more ...
comment
Keywords: En passant as key, No legal last move for White, Impostor (t)
Genre: h#, Retro
Computer test: HC+ Popeye v4.87 & simple retro-logic
FEN: rkb5/1p1p1Ppp/q2P1p2/pPp4K/6PR/1P4PQ/2P1P1pB/4bBrR
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-12-08 more...
80 - P0003417
John Frederick Keeble
2206 The Problemist Fairy Chess Supplement 16, p. 173, 02/1936
P0003417
(4+5) C+
h#2.5 (AP)
1. ... exd6ep 2. 0-0-0 dxe7+ 3. Tf8 exf8=T,D#
play all play one stop play next play all
H.Juel: If Black may castle, last move was not made by Ke8 or Ta8, but by Pd5, so it was d7-d5, and hence White may capture en passant. 1... exd6ep 2.0-0-0 dxe7 3.Tf8 exf8=D,T# where the castling (belatedly) justifies the e.p. capture.
Not 1... exd6ep 2.Kd8? dxe7 3.Kc8 e8=D,T
AB: Another 120 retro tries (=60x2 due to tolerated dual promotion) of general form 1. exd6ep 2. T~ dxc7 3. T~ c8=D,T.
No set play
Nachdruck in "Die Schwalbe" 33 mit Diagrammfehler (wBe6 fehlt).
VL: The first AP-problem! Thematic illegal try:
1. ... exd6ep 2.Kd8? dxe7+ 3.Kc8 e8=T(D)#?? (2007-02-11)
Mario Richter: 46 years between this Problem and P1366663 - perhaps this was not Keeble's first AP-Problem?! (2019-08-29)
Ladislav Packa: Of course I know the idea a posteriori. But let my heresy be forgiven - I do not consider it logical. In this particular example, I consider e.p. for proof that black can castling - not vice versa. (2019-08-29)
A.Buchanan: Hi Ladislav. If ep is ok, then Black just moved a pawn with double hop. That doesn't stop Black having e.g. moved bK right at the beginning of the game. So you can't prove that Black definitely can castle this way. What am I missing here? (2019-08-29)
Ladislav Packa: Your statement is perfectly fine. However, the AP condition is usually interpreted as castling is evidence of e.p. However, the second party does not have to do castling - the possibility has already been demonstrated by the existence of e.p. (2019-08-29)
A.Buchanan: Hi again Ladislav. Please try again, I am obviously being very stupid. I don't understand how 1. exd6ep is any more indicative that Black can castle than 1. K~. (2019-08-30)
Ladislav Packa: Hi, Andre!
e.p. is just a result of the d7-d5 move. Black can castle only if it was the last move of black (even without wPe5). Therefore, it does not have the logic of joining the e.p. with castling. For example, ask yourself if black is allowed to castle in the position without wPe5. (2019-08-30)
A.Buchanan: @Ladislav: ep is not "proof that black can castle", rather it is a *necessary* condition for Black to be able to castle. The logical implication is the other way round. We can castle because of the optimistic convention, and because sPd might just have moved. Without wPe, there is no solution, but whatever White plays, Black can castle (2020-12-11)
Hans-Jürgen Manthey: was heißt hier Thematisch Illegal ??
Mir völlig wurscht ob ep möglich oder nicht, nach dem weißen Zug ist natürlich Kd8 möglich !!
also ist auch 1. ... exd6ep 2. Kd8 dxe7+ 3. Kc8 e8DT# eine Lösung, sonst müßte auf die erzwungende Rochade in der Aufgabenstellung hingewiesen werden ! (2020-12-11)
A.Buchanan: Can one tolerate “tolerated dual promotion” in Valladao? Well Valladao himself did, so it must be ok (2022-03-26)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sg), Promotion (D/T), Tolerated dual promotion (D/T), Valladao Task, Retro Strategy (RS)
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: r3k3/2p1p3/2P1P3/2KpP3/8/8/8/8
Reprints: V. Die Schwalbe 33, p. 323, 06/1975
2001 CHM Themes 12/2000
3 Die Schwalbe 215, p. 240, 10/2005
A1 Problemkiste 169 02/2007
402 Eigenartige Schachprobleme , p. 130, 2010
14 Die Schwalbe 241, p. 374, 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-10 more...
81 - P0003428
Imre Sztankovszky
130 Die Schwalbe 09-10/1953
P0003428
(10+11) C+
h#1.5 AP
1. ... fxg6ep 2. 0-0 gxh7#
play all play one stop play next play all
Henrik Juel: Black captured c7xd6x..x.h2 and once more with an officer
White captured a2xb3, b2xa3, f2xg3, and e.g. exf and once more
If Black may castle, last move was g7-g5 (not b7-b6, because of Lh3) (2022-04-26)
Henrik Juel: HC+ Popeye 4.61
The castling serves two purposes:
enabling the mate and legitimizing the ep capture
so the a posteriori legitimizing is impure (2022-04-26)
A.Buchanan: Yes Henrik. The absence of other candidate solutions (beginning e.p. but excluding 0-0) declines an opportunity to embed additional content, and is arguably an artistic defect. However this very early AP problem is sound, and purity of motive is just one school. IMHO, there is space for such problems, if other content compensates: e.g. 4k2r/p2ppp1p/p7/5PpK/8/1PBB2Pb/2PP2Pp/8 h#2* AP. (2022-04-27)
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comment
Keywords: En passant as key, Castling (sk), a posteriori (AP) (Type Petrovic), Volet Pawn
Genre: h#, Retro
Computer test: HC+ Popeye 4.61
FEN: 4k2r/3ppp1p/1p6/4BPpK/P7/pPPB2Pb/3P2Pp/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-04-27 more...
82 - P0003531
Tivadar Kardos
3563 Schach 24, p. 379, 12/1960
P0003531
(10+12)
h#2
1. Kd8 Txh8 2. Kc8 Txg8#
play all play one stop play next play all
Henrik Juel: The diagram pawns captured all missing men, so the missing pawns [Pa7,Ph2] promoted on a1,h8, and Ta1,Th8 have moved, ruling out the apparent solutions 1.Ke7 0-0-0 2.Kxe6 The1#, 1.Dg1+ Ke2 2.0-0 Taxg1#
C+ by Popeye 4.61 and analysis (2021-08-31)
comment
Keywords: Cant Castler, Castling (wbsg)
Genre: h#, Retro
FEN: r3k1qr/1bp2p2/2p1Pp2/5p2/1P3p2/1Pp2P2/1PP2P2/R3K2R
Reprints: 664 Die Schwalbe 05/1961
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-10 more...
83 - P0003659
Maurice Jago
8202v Die Schwalbe , p. 148, 10/1980
version of cooked 8202 Die Schwalbe 08/1977
P0003659
(12+14) cooked
h#2
b) sBb4->b5
a) 1. Sf2 Dxf2+ 2. Kh1 0-0-0#
b) 1. Sb1 Ta4 2. Sxg3 Th4#
play all play one stop play next play all
Cook: 1. Txg3 0-0-0 2. La2/Lb3/Lc4 Dxh1#
1. La2/Lb3/Lc4 0-0-0 2. Txg3 Dxh1#
See P0000642
Mario Richter: Hier stimmt was nicht: In der Schwalbe 08/1977 erschien als Nr. 2202 die folgende Aufgabe: 8/Pp2p1p1/Bq1p3p/4p3/1p6/n4pP1/1PPPPr1k/R3KQ1n/; h#2, b) sBb4 nach b5 und der AL a) 1.Tf2-g2 0-0-0 2.Tg2xg3 Df1xh1#; b) 1.Sa3-b1 Ta1-a4 2.Sh1xg3 Ta4-h4#, die aber in a) und b) durch 1.f3xe2 La6xb7 2.Tf2-g2 Df1xg2# nebenlösig war.
Ferner ist in nebenstehendem Diagramm sowohl in a) als auch in b) sehr wohl die w0-0-0 möglich
(Weiß hat als letzten Zug Sc7-a8, Schwarz entwandelt auf b1 und h1, Weiß auf f8); und selbst unter der Annahme, daß die w0-0-0 nur mit sBb4 möglich sei, wäre die Aufgabe mehrfach nebenlösig, z.B. durch 1.Ld5-c4 0-0-0 2.Bd6-d5 Df1xh1# (2010-05-29)
A.Buchanan: Here, the source was incorrectly given as "8202v Die Schwalbe 08/1977". No, that was where the original problem appeared, with a "v" added. So when, as here, we suspect the diagram has a typo, we would have to look through many back issues of Die Schwalbe! It turns out that this correction (itself cooked) was published (as text) in p.148 Die Schwalbe 65 10/1980. I have corrected the source. The reference to the earlier buggy problem can then cleanly be put in the "after" field. The actual typo was missing bPc7. (2024-03-06)
A.Buchanan: With the typo fixed, I think the retro-logic works ok, but there is a single cook family, fortunately repairable (2024-03-06)
more ...
comment
Keywords: Cant Castler, Castling (wg), Superseded by (P1415606)
Genre: h#, Retro
Computer test: both twins cooked by Popeye v4.87
FEN: NBr5/N1p3p1/1qPpp3/2bb4/1p6/n1pP1rP1/1PP1P2k/R3KQ1n
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-03-06 more...
84 - P0004137
Hans Stempel
Die Schwalbe 1952
2. Preis
105. TT
P0004137
(12+15)
Längste Beweispartie?
a) Da li su tocno ordedena crna uzimanja?
R: 1. Tb8-b5 h4-h5 2. b7xLa6 Lb5-a6 3. Sd2-f1 Lc4-b5 4. Kf1-f2 f2-f3 5. Se4-d2 Ld5-c4 6. g6-g5 Sf3-e1 7. g7-g6 Sd4-f3 8. f3xTe2 Kd1-c1 9. f4-f3 Dc1-b1 10. Sb3-a1 Dd2-c1 11. Sc5-b3 Sf3-g1 12. Sa6-c5 Se1-f3 13. c4-c3 Kc1-d1 14. Ta8-b8 Dc3-d2 15. Sb8-a6 Td2-e2 16. c5-c4 Sb5-d4 17. Ke2-f1 Td1-d2+ 18. e5xLf4 Sa3-b5 19. f3xTg2 Tg1-g2 20. e6-e5 Lg5-f4 21. f4-f3 Sg2-e1 22. Kf3-e2 Dd2-c3 23. Kg4-f3 Se1-g2 24. Df3-h1 Th1-g1 25. De2-f3 Sf3-e1 26. De1-e2 De2-d2 27. Kf5-g4 Sg1-f3 28. Sf6-e4 Lg2-d5 29. Da5-e1 Dh5-e2 30. Dd8-a5 0-0-0 31. c7-c5 Dd1xh5 32. Sg8-f6 Lf1-g2 33. Kg6-f5 g2-g3 34. Kf7-g6 e3-e2 35. f5-f4 Lc1-g5 36. Ld6-h2 Lh6-c1 37. Tg3-h3 Lg5-h6 38. Ke8-f7 Lf4-g5 39. Lf8-d6 Le3-f4 40. Tg6-g3 Ld2-e3 41. Th6-g6 Lc1-d2 42. Th8-h6 Sb1-a3 43. e7-e6 d2-d3 44. f7-f5 h3-h4 45. h7-h5 h2-h3
play all play one stop play next play all
1646 Züge
Hans-Jürgen Manthey: R: 1. Tb8-b5 h4-h5 2. b7XLa6 Lb5-a6 3. Sd2-f1 Lc4-b5 4. Kf1-f2 f2-f3 5. Se4-d2 Ld5-c4 6. g6-g5 Sf3-e1 7. g7-g6 Sd4-f3 8. f3xTe2 Kd1-c1 9. f4-f3 Dc1-b1 10. Sb3-a1 Dd2-c1 11. Sc5-b3 Sf3-g1 12. Sa6-c5 Se1-f3 13. c4-c3 Kc1-d1 14. Ta8-b8 Dc3-d2 15. Sb8-a6 Td2-e2 16. c5-c4 Sb5-d4 17. Ke2-f1 Td1-d2+ 18. e5xLf4 Sa3-b5 19. f3xTg2 Tg1-g2 20. e6-e5 Lg5-f4 21. f4-f3 Sg2-e1 22. Kf3-e2 Dd2-c3 23. Kg4-f3 Se1-g2 24. Df3-h1 Th1-g1 25. De2-f3 Sf3-e1 26. De1-e2 De2-d2 27. Kf5-g4 Sg1-f3 28. Sf6-e4 Lg2-d5 29. Da5-e1 Dh5-d2 30. Dd8-a5 O-O-O 31. c7-c5 Dd1xh5 32. Sg8-f6 Lf1-g2 33. Kg6-f5 g2-g3 34. Kf7-g6 e3-e2 35. f5-f4 Lc1-g5 36. Ld6-h2 Lh6-c1 37. Tg3-h3 Lg5-h6 38. Ke8-f7 Lf4-g5 39. Lf8-d6 Le3-f4 40. Tg6-g3 Ld2-e3 41. Th6-g6 Lc1-d2 42. Th8-h6 Sb1-a3 43. e7-e6 d2-d3 44. f7-f5 h3-h4 45. h7-h5 h2-h3 (2021-07-16)
Henrik Juel: The stipulation 'Longest proof game' makes little sense here
The serbo-kroatian question in a) probably means something like
What is the exact order of the captures by Black? (2021-07-16)
Mario Richter: This problem was awarded the second prize in the 105. TT of 'Die Schwalbe', which had the theme "Longest Proof Games" (s. also P0004136, P0004139). I do not have the original german stipulation, but I think the serbo-kroatian question means something like "What black captures are exactly determined?"
(The order of the captures is not unique - one can first retract f3xTe2 or f3xTg2 ...) (2021-07-16)
comment
Keywords: Longest Proof Game, Non-Unique Proof Game
Genre: Retro
FEN: 2b5/p2p4/p7/1r4pP/8/2pPPPPr/PPP1pkpb/nQK1NnNq
Reprints: Problem 15-16 05/1952
Input: Gerd Wilts, 1995-06-03
Last update: James Malcom, 2021-07-17 more...
85 - P0004208
Bruno Oswald Sommer
9222 Die Schwalbe 243-244 11-12/1953
P0004208
(9+11)
#1 vor 2
VRZ
R: 1. 0-0! 0-0-0 (erzwungen wegen Retropattvermeidung) 2. b4-b5!, dann 1. Th8#

Eine mögliche Auflösung:
R: 1. 0-0 0-0-0 2. b4-b5! (nur so wird der sD nicht der schnellste Weg nach d8 versperrt) Dd8-d4 3. d5xLc6 Ld7-c6 4. e4xSd5 Lc8-d7 5. f3xSe4 d7xTe6 6. Ta6-e6 Se3-d5 7. Ta1-a6 Sc4-e3 8. a3xLb4 Ld6-b4 9. a2-a3 Lf4-d6 10. Tb1-a1 Lh6-f4 11. Tc1-b1 Lf8-h6 12. Ta1-c1 g7xLf6 13. Lg5-f6 Sf6-e4 14. Lc1-g5 Sa5-c4 15. d2-d3 d3xSc2 16. Sa3-c2 e4xDd3 17. Dc2-d3 f5xTe4 18. Th4-e4 Sg8-f6 19. Th8-h4 Sc6-a5 20. h7-h8=T Sb8-c6 21. h6-h7 g6xLf5 22. Lh3-f5 Sf6-g8 23. Lf1-h3 Sg8-f6 24. g2xTf3 Tf4-f3 25. h5-h6 Tf5-f4 26. h4-h5 Th5-f5 27. Sb1-a3 Th8-h5 28. Dd1-c2 h7xSg6 29. Sf4-g6 Sf6-g8 30. c2-c3 Sg8-f6 31. Sh3-f4 Sf6-g8 32. Sg1-h3 Sg8-f6 33. h2-h4
play all play one stop play next play all
Henrik Juel: The key R: 1.0-0 'threatens' with white retrostalemate, even though White seems to have many pawn retractions available
All missing men were captured by pawns (and White promoted on h8)
R: 1... Kd7-c8? 2.d5xLc6+ Tb8-d8 3.b4-b5 Ke8-d7 4.e4xd5 Ld7-c6 5.f3xe4 Lc8-d7 and now White is retrostalemate
not 6.g2xf3 because of [Lf1]
not 6.d2-d3 because of [Lc1]
not 6.a3xLb4 because of [Ta1]
and not 6.b3-b4 because [Lf8] was captured on a dark square
R: 1... 0-0-0 handles Td8, but it also fixes the black king, so Dd4 must retract to d8 before d7xTf6 can be retracted, but there is still just time enough:
R: 2.b4-b5 Dd8-d4 3.d5xLc6 Ld7-c6 4.e4xd5 Lc8-d7 5.f3xe4 d7xTe6
The rest is easy: Retract Te6 to a1, a3xLb4, Lb4 to f8, g7xLf6, Lf6 to c1, d2-d3, and now the road towards h7 is free for Pc2 (2023-04-08)
A.Buchanan: Great! So which Typ is this? (2023-04-08)
Henrik Juel: Any type, there are no uncaptures in the solution, so anything goes (2023-04-09)
A.Buchanan: Ok I see - the sequence of retro moves is not VRZ play, but history of the game. After the key, there is no choice for either player until wPe4xd5. But isn’t there some Typ where black can checkmate too? R: 1. 0-0 0-0-0 then c1=D#! (2023-04-09)
Henrik Juel: A black checkmate is a possibility in the tries of defensive retractors, regardless of type
When Black has completed a retraction, he has the right to mate White with a forward move, if this is possible
I should have added the testing of mating in my general story about Høeg retractors
1. White chooses a man and moves it back
2. Black chooses which man (if any) to supplement on the abandoned square
Now the white retraction is completed, and White may mate with a forward move, if this is possible
If so, a solution has been found
If not
3. Black chooses a man and moves it back
4. White chooses which man (if any) to supplement on the abandoned square
Now the black retraction is completed, and Black may mate with a forward move, if this is possible
If so, a try has been found
If not, go to step 1. (2023-04-09)
A.Buchanan: Thanks - so does that mean that the solution given here is just a try? (2023-04-09)
Henrik Juel: No, Andrew, in the solution given here White mates, so it is a solution
A try requires Black to mate (2023-04-09)
A.Buchanan: Sorry I am apparently being slow: isn't R: 1. 0-0 0-0-0, dann c1=D#! a mate for Black, so White never gets to retract further? (2023-04-09)
Henrik Juel: You are not slow, Andrew, but I never really saw the black mate in your first 04-09 comment, sorry
R: 1.0-0 0-0-0 2.b4-b5, then 1.Th8# is the intended solution (and not an intended try), but it fails because following R: 1.0-0 0-0-0, Black mates with 1.c1=D#, so you have cooked the problem
It is easily repaired by adding the condition 'Ohne Vorwärtsverteidigung' (without forward defense), but maybe the author implied this condition (or maybe he never saw your black mate) (2023-04-09)
comment
Keywords: Castling (wk,sg), Defensive Retractor
Genre: Retro
FEN: 2kr4/ppp1pp2/2P1pp2/1P6/3q4/2PP4/1Pp1PP2/5RK1
Reprints: (5) Die Schwalbe 67 02/1981
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-04-07 more...
86 - P0004346
Gerhard Paul Latzel
Die Schwalbe 1950
P0004346
(10+3)
#1,5
-sBg5 1. Lxb5,Lxb3 Ta8#
play all play one stop play next play all
-sBg5 (=Vollendung e.p.- Schlag). Nicht Ta1-d1 (=Vollendung w0-0-0), denn Schwarz hätte keinen letzten Zug
Henrik Juel: a slight flaw is that we cannot say whether the entire move was fxg6ep or hxg6ep (2022-07-05)
more ...
comment
Keywords: Castling (wg), Complete an unfinished move, En passant, Joke
Genre: Retro, h#
FEN: 7k/7P/2P1PPPP/1P4p1/b7/1P6/8/R1K5
Reprints: Problem 101-102 09/1966
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-07-10 more...
87 - P0004454
Immo Fuß
Die Schwalbe , p. 463, 03/1939
Allen Teilnehmern herzlich gewidmet
Internationaler Lösungswettkampf 1938
P0004454
(13+11)
#3
1. Tf1? 0-0! Thematic retro try
1. Dxd7+,Sf6+,exd7+,exf7+? Kf8!
1. Dxa8+? Sd8!
1. Sh7? Txh7!

1. 0-0! (droht 2. Txf7,Dxa8+) 0-0? illegal
1. ... Lxb2,Lxb4,b5,c6 2. Dxa8+ Sd8 3. Sxc7#
1. ... Kf8 2. Txf7+
1. ... Tg8 2. Dxd7+,Dxa8+
play all play one stop play next play all
Suppose both players can castle, and derive a contradiction.
White captures: a3xb4,fxe,g2xh3,Bc8 & QR in cage a8--d8.
Black captures: a7xb6,fxe/g,Bc1. bPg promoted. bPf promoted or was captured by wPfxe.
So all captures accounted for. Pieces captured by pawns were wRB & bQXX
b6 is dark, so light wBf1 was not captured there. By elimination, a7xRb6, which released bQ.
Was an original officer captured on b4 to release wR?
bQ not yet released
bRh never moved, bRa captured in cage
bB wrong shade, bBc8 captured in cage
bS couldn't escape g1, and two others on board.
So it must have been a promoted officer captured on b4 earlier.
What was captured on h3, to open the line for promotion on g1? Must be original as wPa & bPa have not yet captured. For the same reasons as axb4, we can eliminate all 4 possible officer types.
Contradiction! So at least one player cannot castle.

We apply the Retro Strategy (RS) convention, and White is permitted to castle, while the range of possible histories shrinks to exclude all cases where Black might have castled. This common special case of RS is also known as Mutual Exclusion.
Kees: Only one castling is legal With black castling there's no #3
1. 0-0!
1. ... Kf8 2.Txf7+ Kg8 3 Dxa8#
1. ... c6 2. Dxa8+ Pd8 3. Pc7#

(axNb3, and for f1=N or g1=N wK must move. so bD must pass bK)
Somebody can better explain than me. (2022-02-14)
A.Buchanan: Hi Kees thanks for the solution which grabs the essence - I have used more words, please point out any slips I might have made! :) (2022-02-15)
comment
Keywords: Retro Strategy (RS), Castling, mutual exclusive (wksk)
Genre: Retro, 3#
Computer test: Popeye v4.87 for forward play Non-trivial thinking for retro logic
FEN: n3k2r/1ppppn2/1p2P3/3N2Np/QP6/b6P/1PPPP2P/4K2R
Reprints: (14) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-15 more...
88 - P0004870
Hans Apholte
Mannheimer Morgen 1961
P0004870
(5+3)
#0
R: 1. 0-0?
play all play one stop play next play all
See P1388763
James Malcom: Black has no last move, so they must have castled illegally out of checkmate. (2021-04-18)
A.Buchanan: “dann” is used in retractor animation when shifting from retro to forward moves (2021-04-19)
comment
Keywords: Joke, Retract illegal move, Castling
Genre: Retro
FEN: 3R1rk1/5N1n/4KB1P/8/8/8/8/8
Reprints: (XV) Die Schwalbe 8 04/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-04-19 more...
89 - P0004917
Andrej N. Kornilow
(C) Die Schwalbe 80 04/1983
P0004917
(5+8)
#2 (AP)
1. ... Tfxg8 2. Lg6! h5 3. Lxf7+
2. ... fxg6=?,hxg6=?,Txg6#?
2. ... Tf8,Tg7 3. Lxf7+ Txf7#

2. Lxe6? Tg6+! 3. Kf5 0-0!
2. ... dxe6? 3. d7+!
2. ... fxe6=?

2. Lxh7? Tg6+! 3. Lxg6 0-0!

Therefore it's WTM
1. Dg2 h5,~ 2. Da8#
play all play one stop play next play all
Black cannot steal the move, as White can prevent the castling justification.
A.Buchanan: Ingenious play, but Black cannot execute the castling. Lines include a good try, checkmate by Black and pat by Black. (2022-04-16)
A.Buchanan: There are two kinds of directmate Type Keym. In one White successfully pushes the move to Black, in the other (as here) Black unsuccessfully pulls it. (2022-04-17)
Ladislav Packa: The logic of this problem is foreign to me, but it is incorrect: 2.Bxe6 Rg6+ 3.Kf5 0-0 (2022-04-18)
A.Buchanan: Hi Ladislav thanks for this. You're right. So 2. Lxe6? Tg6! 3. Kf5 0-0! is another try. The solution must be 2. Lg6! I'll post the solution above (2022-04-18)
more ...
comment
Keywords: Castling (sk), a posteriori (AP) (Type Keym)
Genre: Retro, 2#
FEN: 4krQr/3p1p1p/3PpK1p/4PB2/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-21 more...
90 - P0004920
Valery Liskovets
(F) Die Schwalbe 80 04/1983
P0004920
(5+4)
#2 (AP, pRA)
BTM 1. ... Txh6? 2. Sd7! Txc6+ 3. Kb5! (Kb7?) Td6 4. Sf8 Tf6 5. Sh7! Th6 6. Sf8 no castling
5. a6? Tfxf8 6. a7 Tf5+ 7. Kc6,K~ 0-0!
1. ... 0-0? 2. Se7+! Kh8 3. S5g6#
WTM 1. Td6 droht 2. Td8#
play all play one stop play next play all
White to move has #2 since Black has lost castling rights. So Black pulls the move, but must castle at some point. If Black castles right away, then White has a different #2, so Black must be more subtle. 0... g6/g5/gxh6 leads to castling disruption, e.g. 1.Txg6/Te6+/Sg6. So Black only has 0... Txh6. This pins wSc6 and threatens 0-0, so 1.Sd7! (1. Sg6? Txg6 2. ~ 0-0) etc.
A.Buchanan: A key feature of adversarial A Posteriori is that any castling must be forced in a finite number of moves (but not necessarily limited by the number of moves in the stipulation goal). If the other side can prevaricate indefinitely, then that is sufficient to defeat the A Posteriori "steal" (2022-02-16)
A.Buchanan: Why this would be "PRA"? Maybe the idea is that we don't know who is first to move, yet whoever it is, White wins. But that only applies to "pull" scenarios such as this, where Black snatches the move because otherwise the game is lost. In other situations where White to avoid loss must "push" the move, then there is no way this can be described as PRA. The fundamental push/pull thing has a unity, and I don't think it's helpful to use "PRA" which only describes half of this, and was really designed for a different context. Strategically, these push/pull adversarial battles are amongst the most interesting AP problems. (2023-07-22)
comment
Keywords: a posteriori (AP) (Type Keym), Cant Castler, Castling
Genre: Retro, 2#
FEN: 4k2r/6pr/K1N4R/P3N3/8/8/8/8
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2023-07-22 more...
91 - P0005310
Karlheinz Bachmann
8761 Die Schwalbe 151 02/1995
P0005310
(16+16) cooked
BP in 34,0
1. Sa3 Sa6 2. Sc4 Sc5 3. Se5 Se4 4. c4 c5 5. Da4 Da5 6. b3 b6 7. Lb2 Lb7 8. Tc1 Tc8 9. Tc3 Tc6 10. Th3 Th6 11. e3 e6 12. Le2 Le7 13. Lh5 Lh4 14. Sgf3 Sgf6 15. 0-0 0-0 16. Tc1 Tc8 17. Tc3 Tc6 18. Td3 Td6 19. Td4 Td5 20. d3 d6 21. Dd7 Dd2 22. De7 De2 23. Sd7 Sd2 24. Tg4 Tg5 25. Le5 Le4 26. Lf4 Lf5 27. Sfe5 Sfe4 28. f3 f6 29. Le8 Le1 30. Lg3 Lg6 31. Lf2 Lf7 32. Thg3 Thg6 33. h3 h6 34. Kh2 Kh7
play all play one stop play next play all
Cook: 1. Sa3 Sa6 2. Sc4 Sc5 3. Se5 Se4 4. c4 c5 5. Da4 Da5 6. b3 b6 7. Lb2 Lb7 8. Tc1 Tc8 9. Tc3 Tc6 10. Th3 Th6 11. e3 e6 12. Le2 Le7 13. Lh5 Lh4 14. Sgf3 Sgf6 15. 0-0 0-0 16. Tc1 Tc8 17. Tc3 Tc6 18. Td3 Td6 19. Td4 Td5 20. d3 d6 21. Dd7 Dd2 22. De7 De2 23. Sd7 Sd2 24. Tg4 Tg5 25. Le5 Le4 26. Lf4 Lf5 27. Sfe5 Sfe4 28. f3 Le1 29. Lg3 f6 30. Le8 Lg6 31. Lf2 Lf7 32. Thg3 Thg6 33. h3 h6 34. Kh2 Kh7
more ...
comment
Keywords: Unique Proof Game, Castling, Symmetrical position, Capture-free
Genre: Retro
FEN: 4B3/p2NQbpk/1p1ppprp/2p1N1r1/2P1n1R1/1P1PPPRP/P2nqBPK/4b3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-05-13 more...
92 - P0005409
Erich Bartel
1482 Die Schwalbe 12/1963
P0005409
(10+8) C+
#2 (wer?)
1. ... Txe8 zz Ld7,Le6 2. Txb8#
1. Sd6? zz Te8,T~ 2. Sxf7# geht nicht, denn Weiß ist nicht am Zug
play all play one stop play next play all
Henrik Juel: C+ Popeye 4.61 and noting that it is Black to move, because he has no last move (2021-01-10)
A.Buchanan: This is really fun. Note the stipulation “wer?” indicates that the default convention (Black gets an extra zeroth move, but it’s still white to mate) doesn’t apply here. (2021-01-11)
more ...
comment
Keywords: Whose move?
Genre: Retro, 2#
Computer test: HC+ Popeye 4.61 and noting that it is Black to move, because he has no last move
FEN: KRB1Nrrk/RP2Ppbp/P4p1p/5P1P/8/8/8/8
Reprints: 1.Wolfschach-Club 1962 (Notizbuch) , p. 14, 1963
13 Brisbane Xmas Card , p. 14, 1967
7 Krumme Hunde 01/12/1970
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-01-11 more...
93 - P0005581
Leonid M. Borodatow
5647v Die Schwalbe 105 06/1987
Günter Lauinger gewidmet
P0005581
(11+13) cooked
Wie oft stand der wK mindestens auf c7?
R: 1. ... De5-b8+ 2. Kc7xSc8 Sd6-c8+ 3. Kc8-c7 Sc4-d6+ 4. Kc7xSc8 Sd6xLc4+
play all play one stop play next play all
Cook: R: 1. ... Dd6:Sb8+ 2. Kc7-c8 ...
Nikolai Beluhov: Diagram is now correct (missing wRe8 and bBf8 restored). (2011-05-05)
Nikolai Beluhov: This problem seems to be cooked by 1. ... Qd6:Nb8+ 2. Kc7-c8 Q~d6+ ...
Fortunately, this flaw is very easy to fix: just relocate bQb8 to c7, as in
L. Borodatov (correction)
k1KRRb2/P1qpp1p1/rpp5/p7/5p2/3PP3/P1Pp1PP1/4Bb2
(http://www.janko.at/Retros/d.php?ff=k1KRRb2/P1qpp1p1/rpp5/p7/5p2/3PP3/P1Pp1PP1/4Bb2)
(11 + 13) How often did the wK visit c7 at least? (2011-05-06)
Henrik Juel: The wK visited c7 at least four times, twice as shown in the indicated retroplay, and twice to let wK pass the two white rooks en route to f7, with Le1 screening on b8. (2011-05-06)
Anton Baumann: Informalturnier 1986 (PB in 'Die Schwalbe' 06/2011 S.124)
Auszeichnung: 2.Lob; ausgezeichnete Fassung: mit sDc7 statt b8 (2023-01-04)
comment

Genre: Retro
FEN: kqKRRb2/P2pp1p1/rpp5/p7/5p2/3PP3/P1Pp1PP1/4Bb2
Input: Gerd Wilts, 1995-06-04
Last update: Nikolai Beluhov, 2011-05-06 more...
94 - P0005585
Hieronymus Fischer
56 Fern vom Alltag 1925
P0005585
(5+12)
#0
Einer der sTT muß nach a8 zurückgestellt werden.
play all play one stop play next play all
Lösung aus '64 Schach-Scherze': "Da noch sämtliche schwarze Bauern vorhanden sind, kann keiner der beiden schwarzen Türme durch Umwandlung entstanden sein.
Nun kann aber der schwarze Turm von a8 unmöglich 'herausgekommen' sein. Mithin ist einer der beiden schwarzen Türme nach a8 (oder a7 oder b8) zurückzustellen.
Nach dieser Korrektur ist Schwarz von selbst matt."
in '64 Schach-Scherze' nachgedruckt mit der Forderung "Matt in gar keinem Zug"

vgl. P1309484
Erich Bartel: weiterer Nachdruck (oder Erstquelle?!):
3) 28) 64 Schachscherze 1916 (in dieser Quelle ist kein
Hinweis ob Urdruck oder Nachdruck) (2007-10-30)
Alain Brobecker: Same position and stipulation as P1265678, except the latter one is dated 1910. (2022-11-15)
comment
Keywords: Joke, Illegal position (can't leave home)
Genre: Retro
FEN: 2b5/1pppR1p1/p4p2/3p3p/4r3/4kr1R/2P1N3/4K3
Reprints: 28 64 Schach-Scherze 1915
Das Geheimnis des schwarzen Königs 1960
(III) Die Schwalbe 22 08/1973
Input: Gerd Wilts, 1995-06-05
Last update: A.Buchanan, 2019-10-22 more...
95 - P0005590
William Cross
Hans Hofmann
Josef Kutscher
John Niemann
Hansjörg Schiegl
Bernd Ellinghoven

(1) Die Schwalbe 25 02/1974
P0005590
(16+14)
BP in 29,5
1. Sc3 Sh6 2. Se4 Sg8 3. Sg5 Sh6 4. Se6 Sg8 5. Sxf8 Sh6 6. Se6 Sg8 7. Sc5 Sh6 8. Sa4 Sg8 9. Sb6 Sh6 10. Sxc8 Sg8 11. Sb6 Sh6 12. Sa4 Sg8 13. b4 Sh6 14. b5 Sg8 15. b6 Sh6 16. La3 Sg8 17. Db1 Sh6 18. Kd1 Sg8 19. Kc1 Sh6 20. Kb2 Sg8 21. Kc3 Sh6 22. Kd3 Sg8 23. Ke3 Sh6 24. Kf3 Sg8 25. Kg3 Sh6 26. Kh3 Tf8 27. Dd1 Tg8 28. Lc1 Th8 29. Sc3 Sg8 30. Sb1
play all play one stop play next play all
Konstruktionspreisausschreiben 'Die Schwalbe' 06/1973 Heft 21 S. 54,
Thema I (Dr. W. Dittmann):
Konstruiere, ausgehend von der Partieanfangsstellung, durch Versetzung von höchstens vier beliebigen Steinen eine partiemögliche Stellung mit Weiß am Zuge, deren kürzeste Beweispartie möglichst lang ist. Als Steinversetzung gilt die Postierung eines Steins auf einem anderen Feld oder auch die Entfernung eines Steins vom Brett. Gewertet wird nach der (möglichst hohen) Anzahl der Züge, die erforderlich sind, um die eingesandte Stellung von der Grundstellung aus zu erspielen.
Moldenhauer: Computerprüfung: C+ NUPG cooked Stelvio 1.11 1 Sekunde.
Keine Lösung: BP 28.5, BP 29.0.
Notation: 1.Sa3 Sa6 2.Sb5 Tb8 3.Sd4 Ta8 4.Se6 Tb8 5.Sxf8 Ta8 6.Se6 Tb8 7.Sc5 Kf8 8.Sa4 Ke8 9.Sb6 Kf8 10.Sxc8 Ke8 11.Sb6 Kf8 12.Sa4 Ke8 13.Sc3 Kf8 14.b4 Ke8 15.La3 Kf8 16.Db1 Ke8 17.Kd1 Kf8 18.Kc1 Ke8 19.Kb2 Kf8 20.Dd1 Ke8 21.Sb1 Kf8 22.Kc3 Ke8 23.Kd3 Kf8 24.Ke3 Ke8 25.Kf3 Kf8 26.Kg3 Ke8 27.Kh3 Tc8 28.Lc1 Ta8 29.b5 Sb8 30.b6 (2023-04-09)
A.Buchanan: In order to (H)C+ a non-unique proof game, one should show there is no short solution, but also that all solutions of regular length exhibit the intended theme, whatever that was. As long as any strategy can be checked for compliance, Stelvio fine because one can set the parameter for Stelvio to make sure that all strategies are considered. But it’s not possible currently to view all transpositions of a strategy, if that matters (2023-04-09)
Moldenhauer: Man müsste natürlich dieses Konstruktionsausschreiben gelesen haben was hier die Vorgabe war.
Die Stellung an sich oder der Königsmarsch oder dass die Damen und Läufer die Ursprungsposition
wieder einnehmen, usw. Stelvio wird das erreichen der Stellung analysieren so wie Euclide und Natch.
Die Bewertung auf C+ überlasse ich auch den Experten. Da fehlt mir Wissen und jahrelange Erfahrung.
Deshalb nur Kommentare. Gibt es dieses Ausschreiben von 06/1973? (2023-04-27)
Mario Richter: Thema ergänzt - hilft das bei der Bewertung? (2023-04-28)
comment
Keywords: Non-Unique Proof Game, Construction task, Homebase
Genre: Retro
FEN: rn1qk1nr/pppppppp/1P6/8/8/7K/P1PPPPPP/RNBQ1BNR
Input: Gerd Wilts, 1995-06-05
Last update: Mario Richter, 2023-04-28 more...
96 - P0005591
Alfred Gschwend
Jorge Joaquin Lois
Julio Alberto Pancaldo
Pedro Gomez Masia
Emiliano F. Ruth

Die Schwalbe 25 02/1974
P0005591
(16+14)
BP in 29,5
1. Sa3 Sa6 2. Sc4 Tb8 3. Se5 Sh6 4. Sg6 Sg8 5. Sxh8 Sh6 6. Sg6 Sg8 7. Se5 Sh6 8. Sc6 Sg8 9. Sxb8 Sh6 10. Sc6 Sg8 11. Sa5 Sh6 12. Sc4 Sg8 13. b4 Sh6 14. b5 Sg8 15. b6 Sh6 16. La3 Sg8 17. Db1 Sh6 18. Kd1 Sg8 19. Kc1 Sh6 20. Kb2 Sg8 21. Kc3 Sh6 22. Kd3 Sg8 23. Ke3 Sh6 24. Kf3 Sg8 25. Kg3 Sh6 26. Kh3 Sg8 27. Dd1 Sh6 28. Lc1 Sg8 29. Sa3 Sb8 30. Sb1
play all play one stop play next play all
Konstruktionsausschreiben 06/1973
Moldenhauer: Computerprüfung: C+ NUPG Stelvio 1.11 00:07:09 Minuten. (hh:mm:ss)
Keine Lösung: BP 28.5, BP 29.0.
Beispiel: 1.Sa3 Sa6 2.Sb5 Tb8 3.Sd4 Ta8 4.Sc6 Tb8 5.Sxb8 Sb4 6.Sc6 Sa6
7.Se5 Sb4 8.Sg6 Sa6 9.Sxh8 Sb4 10.Sg6 Sa6 11.Se5 Sb4 12.Sc4 Sa6 13.b4 Sb8
14.La3 Sa6 15.Db1 Sb8 16.Kd1 Sa6 17.Kc1 Sb8 18.Kb2 Sa6 19.Kc3 Sb8
20.Kd3 Sa6 21.Ke3 Sb8 22.Kf3 Sa6 23.Kg3 Sb8 24.Kh3 Sa6 25.Dd1 Sb8
26.Lc1 Sa6 27.Sa3 Sb8 28.Sb1 Sa6 29.b5 Sb8 30.b6 (2023-04-24)
comment
Keywords: Construction task, Non-Unique Proof Game
Genre: Retro
FEN: 1nbqkbn1/pppppppp/1P6/8/8/7K/P1PPPPPP/RNBQ1BNR
Input: Gerd Wilts, 1995-06-05
Last update: A.Buchanan, 2012-04-08 more...
97 - P0005593
Hector Guillermo Zucal
Horacio Tomas Amil Meylan

(B) Die Schwalbe 25 02/1974
P0005593
(16+15)
BP in 28.5
1. Sa3 Sa6 2. Sc4 Sb8 3. Se5 Sa6 4. Sc6 Sb8 5. Sxd8 Sa6 6. Sc6 Sb8 7. Se5 Sa6 8. Sc4 Sb8 9. b4 Sa6 10. b5 Sb8 11. b6 Sa6 12. h4 Sb8 13. h5 Sa6 14. h6 Sb8 15. La3 Sa6 16. Db1 Sb8 17. Kd1 Sa6 18. Kc1 Sb8 19. Kb2 Sa6 20. Kc3 Sb8 21. Kd3 Sa6 22. Ke3 Sb8 23. Kf3 Sa6 24. Kg3 Sb8 25. Kh2 Sa6 26. Dd1 Sb8 27. Lc1 Sa6 28. Sa3 Sb8 29. Sb1
play all play one stop play next play all
Konstruktionsausschreiben 06/1973
Moldenhauer: Computerprüfung: Cooked Stelvio 1.11 4 Sekunden.
Keine Lösung: BP 27.5, BP 28.0, BP 29.0. BP 28.5 cooked.
Beispiel BP 29.5: 1.Sa3 Sa6 2.Sb5 Tb8 3.Sd4 Ta8 4.Sc6 Tb8 5.Sxd8 Ta8 6.Sc6 Tb8 7.Sa5 Kd8 8.Sc4 Ke8
9.b4 Kd8 10.La3 Ke8 11.Db1 Kd8 12.Kd1 Ke8 13.Kc1 Kd8 14.Kb2 Ke8 15.Kc3 Kd8 16.Kd3 Ke8 17.Ke3 Kd8
18.Kf3 Ke8 19.Kg3 Kd8 20.Dd1 Ke8 21.Lc1 Kd8 22.Sa3 Ke8 23.Sb1 Kd8 24.b5 Ke8 25.b6 Kd8 26.h4 Ke8
27.Kh2 Ta8 28.h5 Sb8 29.h6
Beispiel BP 28.5: 1.Sa3 Sa6 2.Sb5 Tb8 3.Sd4 Ta8 4.Sc6 Tb8 5.Sxd8 Ta8 6.Sc6 Tb8 7.Sa5 Kd8 8.Sc4 Ke8
9.b4 Kd8 10.La3 Ke8 11.Db1 Kd8 12.Kd1 Ke8 13.Kc1 Kd8 14.Kb2 Ke8 15.Kc3 Kd8 16.Kd3 Ke8 17.Ke3 Kd8
18.Kf3 Ke8 19.Kg3 Kd8 20.Dd1 Ke8 21.Lc1 Kd8 22.Sa3 Ke8 23.Sb1 Kd8 24.b5 Ke8 25.b6 Kd8 26.h4 Ke8
27.h5 Ta8 28.Kh2 Sb8 29.h6
Nach Konstruktionsauschreiben glaube ich wäre BP 28.5 richtig. (2023-05-12)
James Malcom: Fixed. (2023-05-13)
Henrik Juel: What was asked for in this construction tourney? (2023-05-13)
Moldenhauer: Bitte siehe P0005590, da gings schon mal darum. (2023-05-13)
comment
Keywords: Non-Unique Proof Game, Construction task, Homebase
Genre: Retro
FEN: rnb1kbnr/pppppppp/1P5P/8/8/8/P1PPPPPK/RNBQ1BNR
Input: Gerd Wilts, 1995-06-05
Last update: James Malcom, 2023-05-13 more...
98 - P0005597
Frank Schützhold
Jorge Abel Barros
Raul Ocampo
Jorge Joaquin Lois
Julio Alberto Pancaldo

(2) Die Schwalbe 25 02/1974
P0005597
(15+15)
BP in 28,5
(27,5?)
1. Sc3 Sh6 2. Sd5 Tg8 3. b4 Sc6 4. b5 Sb8 5. b6 Sc6 6. La3 Sb8 7. Db1 Sc6 8. Db5 Sb8 9. Da6 bxa6 10. Kd1 Lb7 11. Kc1 Dc8 12. Sf6+ Kd8 13. Sxg8 Sc6 14. Sf6 Sg8 15. Sd5 Ke8 16. Kb2 Dd8 17. Kc3 Lc8 18. b7 Sh6 19. b8=D Sg8 20. Db1 Sh6 21. Dd1 Sg8 22. Lc1 Sh6 23. Kd3 Sg8 24. Ke3 Sh6 25. Kf3 Sg8 26. Kg3 Sh6 27. Kh3 Sg8 28. Sc3 Sb8 29. Sb1
play all play one stop play next play all
Konstruktionsausschreiben 06/1973
Moldenhauer: Computerprüfung: C+ NUPG BP 28.5 cooked in 1 Sekunde.
Keine Lösung: BP 26.5 bis BP 28.0.
Beispiel: 1.Sc3 Sa6 2.Sd5 Tb8 3.b4 Ta8 4.La3 Tb8 5.Db1 Ta8 6.Kd1 Tb8 7.Kc1 Ta8
8.Kb2 Tb8 9.Kc3 Ta8 10.Kd3 Tb8 11.Ke3 Ta8 12.Kf3 Tb8 13.Kg3 Ta8
14.Kh3 Tb8 15.b5 Ta8 16.b6 Sb8 17.Db5 Sc6 18.Da6 bxa6 19.b7 Sh6
20.b8D Tg8 21.Db1 Lb7 22.Dd1 Dc8 23.Sf6+ Kd8 24.Sxg8 Sb8 25.Sf6 Sg8
26.Sd5 Ke8 27.Lc1 Dd8 28.Sc3 Lc8 29.Sb1 (2023-04-02)
comment
Keywords: Non-Unique Proof Game, Construction task, Promotion
Genre: Retro
FEN: rnbqkbn1/p1pppppp/p7/8/8/7K/P1PPPPPP/RNBQ1BNR
Input: Gerd Wilts, 1995-06-05
Last update: A.Buchanan, 2012-04-08 more...
99 - P0005637
Osmo Ilmari Kaila
Hannu Harkola

(D) Die Schwalbe 48 12/1977
P0005637
(6+1) cooked
#3
AP
WTM: ???
BTM: 1. ... Kxf3 2. 0-0+! Ke3 3. Td1 Kf3 4. Td3#
not 2. Kd2? Kg2 3. Sf4+ Kf2,Kf3 4. Tf1#
play all play one stop play next play all
Cook: 1.Tf1,Se2+,Sf2 all lead to mate in 3, and none require castling
A.Buchanan: I don't get the regular part of this. White can mate in 3 in various ways: 1.Tf1,Se2+,Sf2. Black just played R: 1.Kh2g2 Td~h1+, so White can't castle but doesn't need to. What's going on? (2022-02-15)
comment
Keywords: a posteriori (AP) (Type Keym), Castling, Rex solus, Miniature
Genre: Retro, 3#
FEN: 8/8/6R1/8/8/5PNN/6k1/4K2R
Input: Gerd Wilts, 1995-06-06
Last update: A.Buchanan, 2023-07-30 more...
100 - P0005649
Andrey Frolkin
(10) Die Schwalbe 59 10/1979
P0005649
(12+10)
Ist die Rochade zulässig?
R: 0-0! Kb6xc7 1. Ld8xc7+ Lf4-d6 2. Sb1-a3 De5-c7 3. Lg5-d8+ Da1-e5 3. Sd2-b1 Dc3-a1+ 4. Sf1-d2 Dg3-c3+ 5. Lf6-g5 Ld6-f4 6. Lc3-f6 Lb4-d6 7. c7xDb8=T Dg1-g3 8. Sg3-f1 g2-g1=D+ 9. Se4-g3 g3-g2 10. d6xDc7 g4-g3 11. Sd2-e4 g5-g4 12. Sf1-d2 g7-g5 13. d5-d6 Dg3-c7 14. Ld2-c3 Dg1-g3 15. Se3-f1 g2-g1=D+ 16. Lc3-d2 g3-g2 17. Sf1-e3 g4-g3 18. Sg3-f1 g5-g4 19. Se4-g3 g6-g5 20. Sd2-e4 h7xSg6 21. c7xDb8=T Lb4-a3 22. d6xDc7 Dg8-d8 23. Sf1-d2 Dg1-g8 24. Sg3-f1 g2-g1=D+ 25. Se4-g3 g3-g2 26. Sd2-e4 g4-g3 27. d5-d6 Dg3-c7 28. Sf1-d2 Dg1-g3 29. Se3-f1 g2-g1=D+ 30. Sf5-e3 g5-g4 31. Sd4-f5 g3-g2 32. Sb3-d4 Kc5-b6 33. Sd2-b3+ g4-g3 34. Sf1-d2 Dg3-b8 35. Tb8-a8 g7-g5 36. c7xDb8=T g5-g4 37. b6xDc7 Dg1-g3 38. Se3-f1 g2-g1=D+ 39. Sc4-e3 g6-g5 40. Sa3-c4 g3-g2 41. Sb1-a3 g4-g3 42. Sd2-b1 Dg3-c7 43. Sf1-d2 Dg1-g3 44. Ld2-c3 g2-g1=D 45. Lc3-d2 f3xLg2 46. Se3-f1 e4xTf3 47. Sc4-e3 Kd6-c5 48. Sa3-c4+ h7xSg6 49. Sb1-a3 Ke7-d6 50. Sf4-g6+ Ke8-e7 51. Lf1-g2 f5xDg4 52. g2xTh3 Th6-h3 53. Sd2-b1 e5-e4 54. Lb2-c3 Tf6-h6 55. Tc3-f3 Tf8-f6 56. Sh3-f4 Th8-f8 57. Sg1-h3 f7-f5 58. Sc4-d2 Sc5-a4 59. Sa3-c4 Se4-c5 60. Td3-c3 Lf8-b4+ 61. Td1-d3 Sf6-e4 62. Ta1-d1 Sg8-f6 63. Sb1-a3 e7-e5 64. Lc1-b2 Dc7-b8 65. b5-b6 Dd8-c7 66. b4-b5 c7-c6 67. b2-b4 Sc6-a5 68. Dd4-g4 Sb8-c6 69. Dd1-d4 Ta8-a7 70. d4-d5 a7-a6 71. d2-d4
play all play one stop play next play all
hans: 4 Q-promotions on g1 and captured on b8 and c7. Sf1 for shield, so 0-0 is legal. (2010-04-30)
Hans-Jürgen Manthey: R: 0. O-O ! Kb6xc7 1. Ld8xc7+ Lf4-d6 2. Sb1-a3 De5-c7 3. Lg5-d8+ Da1-e5 3. Sd2-b1 Dc3-a1+ 4. Sf1-d2 Dg3-c3+ 5. Lf6-g5 Ld6-f4 6. Lc3-f6 Lb4-d6 7. c7xDb8=T Dg1-g3 8. Sg3-f1 g2-g1=D+ 9. Se4-g3 g3-g2 10. d6xDc7 g4-g3 11. Sd2-e4 g5-g4 12. Sf1-d2 g7-g5 13. d5-d6 Dg3-c7 14. Ld2-c3 Dg1-g3 15. Se3-f1 g2-g1=D+ 16. Lc3-d2 g3-g2 17. Sf1-e3 g4-g3 18. Sg3-f1 g5-g4 19. Se4-g3 g6-g5 20. Sd2-e4 h7xSg6 21. c7xDb8=T Lb4-a3 22. d6xDc7 Dg8-d8 23. Sf1-d2 Dg1-g8 24. Sg3-f1 g2-g1=D+ 25. Se4-g3 g3-g2 26. Sd2-e4 g4-g3 27. d5-d6 Dg3-c7 28. Sf1-d2 Dg1-g3 29. Se3-f1 g2-g1=D+ 30. Sf5-e3 g5-g4 31. Sd4-f5 g3-g2 32. Sb3-d4 Kc5-b6 33. Sd2-b3+ g4-g3 34. Sf1-d2 Dg3-b8 35. Tb8-a8 g7-g5 36. c7xDb8=T g5-g4 37. b6xDc7 Dg1-g3 38. Se3-f1 g2-g1=D+ 39. Sc4-e3 g6-g5 40. Sa3-c4 g3-g2 41. Sb1-a3 g4-g3 42. Sd2-b1 Dg3-c7 43. Sf1-d2 Dg1-g3 44. Ld2-c3 g2-g1=D 45. Lc3-d2 f3xLg2 46. Se3-f1 e4xTf3 47. Sc4-e3 Kd6-c5 48. Sa3-c4+ h7xSg6 49. Sb1-a3 Ke7-d6 50. Sf4-g6+ Ke8-e7 51. Lf1-g2 f5xDg4 52. g2xTh3 Th6-h3 53. Sd2-b1 e5-e4 54. Lb2-c3 Tf6-h6 55. Tc3-f3 Tf8-f6 56. Sh3-f4 Th8-f8 57. Sg1-h3 f7-f5 58. Sc4-d2 Sc5-a4 59. Sa3-c4 Se4-c5 60. Td3-c3 Lf8-b4+ 61. Td1-d3 Sf6-e4 62. Ta1-d1 Sg8-f6 63. Sb1-a3 e7-e5 64. Lc1-b2 Dc7-b8 65. b5-b6 Dd8-c7 66. b4-b5 c7-c6 67. b2-b4 Sc6-a5 68. Dd4-g4 Sb8-c6 69. Dd1-d4 Ta8-a7 70. d4-d5 a7-a6 71. d2-d4 (2021-07-17)
comment
Keywords: Ceriani-Frolkin Theme, Castling, Non-standard material
Genre: Retro
FEN: RRb5/rpBp4/pkpb4/n7/n7/N6P/P1P1PP1P/4K2R
Reprints: (17) diagrammes 15 07-09/1994
Input: Gerd Wilts, 1995-06-06
Last update: James Malcom, 2021-07-17 more...
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Gerd Wilts (100)