Die Schwalbe

14 problem(s) found in 3971 milliseconds (displaying 14 problem(s)). [COMMENTDATE>=20200919 AND NOT S='Sammlung leichterer Schachaufgaben III' AND A='Plaksin, Nikita M.'] [download as LaTeX]

1 - 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...
2 - 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...
3 - 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...
4 - 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)
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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...
5 - 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)
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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...
6 - 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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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...
7 - P0001940
Nikita M. Plaksin
Shakhmaty v SSSR 1980
Spezialpreis
P0001940
(13+16)
Remis
hans: 1. b4 a5 2. Bb2 Na6 3. Bd4 Nc5 4. Nc3 Na4 5. Bc5 Nb6 6. b5 Nh6 7. a4 Nf5 8.
Na2 Nh4 9. Nb4 axb4 10. a5 Rb8 11. a6 Na8 12. a7 Nb6 13. a8=Q Rg8 14. Qa2
Ra8 15. Qe6 Ra3 16. Qg6 hxg6 17. Qb1 Na8 18. Ba7 b6 19. Qb3 Bb7 20. Qd3 Bf3
21. Qf5 gxf5 22. Nh3 Bh5 23. Nf4 Bg6 24. e3 Rh8 25. Bc4 Rh6 26. f3 Bh7 27.
Ng6 Bg8 28. Nh8 Rd6 29. h3 g6 30. Rf1 Bh6 31. Rf2 Bf4 32. Rf1 Bh2 33. f4
Rdd3 34. Rf3 Rac3 35. Rg3 Kf8 36. Rg5 Kg7 37. Rh5 Kf6 38. Rh7 Qc8 39. Rg7
Bh7 40. Rg8 Qa6 41. Rb8 Qa3 42. Rb7 Qb2 43. Bb8 Rb3 44. Rba7 Qd4 45. R7a2
Ra3 46. Rb2 Ra7 47. Rba2 Rb7 48. Ba7 Rb8 49. Rb2 Rg8 50. Rba2 Rg7 51. Bb8
Bg8 52. Rb2 Rh7 53. Rba2 Rh5 54. Rb2 Rg5 55. Rba2 Rg3 56. Rb2 Rf3 57. Rba2
Rf2 58. Rb2 Ra3 59. Rba2 Ra7 60. Rb2 Rb7 61. Ba7 Rb8 62. Rba2 Bh7 63. Ra6
Rg8 64. Bb8 Rg7 65. Ra7 Bg8 66. Rb7 Rh7 67. Ba7 Rh5 68. Rb8 Bh7 69. Rg8 Rg5
70. Rg7 Bg8 71. Rh7 Rg3 72. Rh6 Rgf3 73. Bb3 Kg7 74. Ba4 Kf8 75. Rh7 Ke8
76. Rg7 Bh7 77. Bb8 Kd8 78. Bb3 Kc8 79. Rg8+ Kb7 80. Ba4 Qg7 81. Re8 Bg8
82. Rf8 Bg3 83. O-O-O {50 moves rule} 1/2-1/2 (2012-11-14)
Olaf Jenkner: Warum nicht z.B. 83. Te8 remis?
Was bedeutet das Schlüsselwort unused? (2012-11-14)
Henrik Juel: The castling shows that White never moved his king before; otherwise the position could be reached faster, and the 50 moves rule could not be applied.
The keyword unused seems non-sensical here and should be deleted (2012-11-14)
A.Buchanan: This problem is very interesting. It's orthodox 50M, so that castling does not reset the counter, and the timing works out nicely. But it makes me wonder... suppose we have a problem where it is B32 which was the last reset. Then W83 0-0-0 would prove that the position had already been at 50.0. If 50M convention (which needs to be rewritten because it's a mess) operated like 3Rep, then the game end would have been mandatory after B82. So castling W83 would be illegal. Is this how one would want 50M convention to work? Or should there be a carve-out to say: if you can definitely prove that no-one claimed, then no-one claimed. And should there be a similar carve-out for 3Rep rule? See https://www.thehoppermagazine.com/AA010 (2022-01-20)
A.Buchanan: Personally, I think no carve-outs. The 50M convention should be rewritten as: "A position is considered as a draw if it can be proved that the last 50.0 moves in the proof game combined with the solution did not contain a capture or pawn move. Unless expressly stipulated, this applies only to retro-problems." (2022-01-20)
Thomas Volet: What if the composer intends the retroplay to go beyond 50 non-P and non-capturing moves? (2022-01-20)
A.Buchanan: Hi Thomas - thanks for your question. I think it’s best if we take this offline. I will email you, if that’s ok (2022-01-21)
James Malcom: This is the Volet problem in question: P0008399 In hindsight, it is quite humorous that the 75 move rule was later introduced in the 21st century. (2022-01-21)
Thomas Volet: The question was not directed specifically to P0008399, which is just one of several compositions with the property at issue. (2022-01-22)
A.Buchanan: I've emailed Thomas. My earlier comment should have include checkmate as a third mechanism to zero the count. (2022-01-22)
James Malcom: You can always carry it on over to MatPlus. (2022-01-24)
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Keywords: 50 move rule, Castling (wl)
Genre: Retro
FEN: nB3RbN/1kppppq1/1p4p1/1P3p2/Bp3P1n/4PrbP/2PP1rP1/R3K3
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-01-25 more...
8 - P0004587
Nikita M. Plaksin
(4) Problem 188-193 05/1979
P0004587
(12+16)
#1 (How many solutions?)
1. ... Sxd2#
1. ... Sg3 Matt oder Remis?

Beispielauflösung mit Schwarz am Zug (mri):
R: 1. Ta7-a6 Tb1-c1 2. Ta6-a7 Dc1-d1 3. Ta7-a6 Td1-e1 4. Ta6-a7 Ke1-f2 5. Kg1-h1 Sg3-f1 6. Kh1-g1 Se4-g3 7. Kg1-h1 Sg5-e4 8. Kh1-g1 Sh3-g5 9. Ta7-a6 Sg1-h3 10. Ta6-a7 Kf2-e1 11. Ta7-a6 Tf1-d1 12. Ta6-a7 Ke1-f2 13. Ta7-a6 Tf3-f1 14. Tb7-a7 Kf1-e1 15. Ta7-b7 De1-c1 16. Ta6-a7 Dh4-e1 17. Ta7-a6 Th3-f3 18. Ta6-a7 Kf2-f1 19. Ta7-a6 Tf1-b1 20. Ta6-a7 Ke1-f2 21. Ta7-a6 Tf3-f1 22. Ta6-a7 Tg3-f3 23. Ta7-a6 Kf2-e1 24. Tb1-b2 Sf3-g1 25. Tg1-b1 Df6-h4 26. Ta6-a7 Db2-f6 27. Ta7-a6 Dc1-b2 28. Ta6-a7 Df1-c1 29. Ta7-a6 Ke1-f2 30. Ta6-a7 Kd1-e1 31. Ta7-a6 Kc1-d1 32. Ta6-a7 Kb2-c1 33. Ta7-a6 Ka3-b2 34. Ta6-a7 Kb4-a3 35. Ta7-a6 Kc5-b4 36. Ta6-a7 Kc6-c5 37. Ta7-a6 Sh4-f3 38. Ta6-a7 Df3-f1 39. Tb1-g1 Kb7-c6 40. Tc1-b1 Kc8-b7 41. Tb1-c1 Kd8-c8 42. Tc1-b1 Ke8-d8 43. Tb1-c1 Db7-f3 44. Ta7-a6 Dc8-b7 45. Ta6-a7 Dd8-c8 46. Ta7-a6 Lh7-g8 47. Ta6-a7 Le4-h7 48. Ta7-a6 Lb7-e4 49. Ta6-a7 Lc8-b7 50. Ta7-a6 b7-b6
play all play one stop play next play all
Mario Richter: Das Diagramm ist offensichtlich verdruckt (s. sBBa5a6b7c7). Wie lautet die korrekte Stellung? (2010-01-23)
TBr: Habe die Stellung korrigiert laut Problem 188-189. (2020-03-13)
Henrik Juel: Thanks, Thomas
Black captured [Lf1] on f1 and hxgxfxe
Retracting e4-e3 is illegal as [Ph7] captured [Lc1] on a dark square
Outline of retroplay:
Retract -1.Ta7-a6, sTDT to the left, sKe1-f2, wKg1-h1, sSg3-f1, wKh1-g1, sSg3 to g1, extract sTDT via f1 and f3, sSh3-g1, wTb2 to g1, sD to f1, sKf2 to e8, sDf1 to d8, sLg8 to c8, and finally probably -50... b7-b6 to avoid draw by the 50 move rule (2020-03-13)
Mario Richter: Henrik, I don't think that "-50... b7-b6 to avoid draw by the 50 move rule" is the point of this problem. I think the author's intention was to raise the question what has higher priority: 50-moves-rule or checkmate.
So 1. ... Sxd2 mates, the question is, if the non-capturing non-pawn move 1. ... Sd2 too is checkmate.
Furthermore, I do not like the stipulation: White has no mating move, so the question "#1 (wer?)" doesn't make any sense to me. Imho a better stipulation would be "#1, How many solutions?" (2020-03-26)
Henrik Juel: Thanks for providing a retroplay, Mario
You are probably right about the author intention; I think that Plaksin would argue that 1... Sg3 stops the game immediately, with draw by the 50 move rule
The stipulation may have been translated incorrectly from serbo-kroatian (2020-03-27)
A.Buchanan: I’ve modified the stipulation. What was Plaksin’s published solution? The FIDE rules unfortunately allow for arbiter judgement, something that really only should be needed for tournament guidelines. Today if you work through the cases, it seems to be up to the mating player to decide if he wants to claim a draw before he plays the move. The Laws expect rationality. If the rule would cut in 1 single move later, the mated player definitely does not get a chance to use this rule to escape the obligation to declare there is no flight. The Codex simply says if it’s retro, apply the rule - but scoresheets, clocks, arbiters etc don’t exist for a composition so this is meaningless (2022-07-23)
A.Buchanan: It would be possible to redraft the Codex along the lines of the draw by repetition rule, to clarify that pawn move, capture and checkmate all zero the move count.
However there are two other points of view. One is the historical idea that castling (but not other king/rook moves?) also zeroes the count. Even under the current form of the Laws & Codex, the few problems based on this are currently unsound. They deserve to be better protected by redrafting the Codex and then calling the problems out as based on a clear historical interpretation at the time. As with Dummy Pawn, contemporary composers can still “go back in time” to make compositions based on the old interpretation. (2022-07-23)
A.Buchanan: The second objection is more serious, and is an issue in common with draw by repetition. This is whether a player must claim if possible. If you say “no” then there are longer retro sequences possible. But there are also problems which rely on the answer “yes” for soundness. This is a complicated area, and I don’t have an answer currently, although as always I want to protect the integrity and accessibility of prior art, while also giving a clear set of rules for newcomers to work with (2022-07-23)
A.Buchanan: Another “philosophical” issue is how we regard rules at all. I see them as being mathematical, and want to clear them up so engine developers, computer scientists and mathematicians can fully engage with retro chess.
But I must acknowledge that Tom Volet (who I respect very much) comes from a background of law, where interpretation of words is vital. He sees the vagueness as part of the artistry while I see it as a boring barrier to enjoying the artistry that is elsewhere. (2022-07-23)
A.Buchanan: I would like to explore a clause for both 3Rep and 50M, that claiming is automatic unless you can show that it didn’t happen (e.g. if a move was played, and there’s no shorter history which avoids the issue). I think this can potentially solve everything but it involves a precision that may be anathema to Tom.
Other fine points are whether stalemate and dead position like checkmate also trump 50M. And how about 50M vs 3Rep? And a separate more difficult point as to whether DP role has visibility over 50M & 3Rep.
And these posts summarise my complete understanding of the issues with the 50M rule. (2022-07-23)
comment
Keywords: 50 move rule
Genre: Retro
FEN: n4bb1/2ppppp1/Rp6/p7/5P2/1P2p3/PRPPPkPP/N1rqrn1K
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-07-23 more...
9 - P0005387
Nikita M. Plaksin
6557 feenschach 109 11/1993
P0005387
(2+2) C+
-1w, dann =1
R: 1. Ka3-a4?, dann 1. Lxa2=
R: 1. Ka3xBa4!, dann 1. Lxa2=
play all play one stop play next play all
Mit folgender witziger Begründung des Autors: "If on the board are only KL-KL (and LL on squares of the same colour), by the rules of chess an immediate draw occurs! Therefore, the retro move Ka3-a4 is illegal." Ein Fehllöser: "Reichlich trivial." Ein Ratloser: "... oder wie darf ich das sehen?" TK hatte den Durchblick: "Die doppelte Rücknahmemöglichkeit kann man aber gar nicht übersehen (und schon gar nicht der Autor). Einzige Möglichkeit: nur Ka3xBa4, weil andernfalls die Stellung schon vorher patt wäre. Aber, aber: das ist remis, kein Patt!!" FR: "Meine 'Ehrenlösung' für diese Gruppe (hoffentlich stimmt sie auch)." WM: "Nette Pointe. Das erzähl' ich meinen Kindern, daß ich einen PLAKSIN gelöst habe." 1,2/I
A harbinger of Dead Reckoning-style reasoning, before that rule ever existed. As sound today in 2018, as when it was composed under "insufficient material for mate" rule.
Henrik Juel: Nice to see that I am not the only freak who mixes english text with german notation (to avoid the ambiguous B)
Were you inspired by this problem in your work on Dead Reckoning, Andrew? (2020-05-04)
A.Buchanan: Ha no: when I stumbled across DP rule I knew even less about chess problems than I know now, if you can imagine, and had never encountered this one which Werner Keym unearthed and published it in "Out Of the Box". Ecclesiastes 1:9 says "There is nothing new under the sun." Mario made this point about origin of Schnoebelen promotions too. Gollon's intepretation of Chaturanga (the earliest known forebear of chess) offers fascinating possibilities for Schnoebelen promotion, which I am sure the Indian retrograde analysts of 7th Century CE were adept at exploiting, in tablets alas now lost. See https://www.chessvariants.com/historic.dir/chaturanga.html.

The "insufficient material" composition does perhaps explain Plaksin's openness to the concept of DP, at a time when perhaps some noses elsewhere were being turned up, as he later composed with Kornilov *27* other DP problems, all to be found in PDB. (2020-05-04)
Ladislav Packa: Interesting, but on YouTube I saw a game (one of the actors was Magnus Carlsson), which ended in a position with only two kings and one of the players still made a move. (2020-05-05)
A.Buchanan: http://wismuth.com/chess/illegal-moves.html contains many other games where GMs were unable to tear themselves away from playing the dead game. I think it's kind of sweet that the GMs just can't stop playing. Have you got a link to the Carlsen game please, Ladislav? (2020-05-06)
Ladislav Packa: I'm sorry, Andrew, but I don't remember. I've seen hundreds of games on youtube. But I encountered similar cases in my own practice, mostly a joke, and the game immediately ended with a smile and a handshake. (2020-05-06)
A.Buchanan: A completely different stipulation: -w, h=1 half-duplex (2024-01-11)
more ...
comment
Keywords: Help retractor, Dead Position, Aristocrat, Miniature, Minimal
Genre: Retro, Fairies
Computer test: C+ deadpos v2.1 8-Jan-2024
FEN: 8/8/8/8/K7/8/b7/kB6
Reprints: 440 Out Of the Box , p. 141, 2018
www.thbrand.de 14/01/2018
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2024-01-18 more...
10 - P0006067
Nikita M. Plaksin
Andrej N. Kornilow

(2) diagrammes 15 07-09/1994
P0006067
(2+7)
Ist die Stellung legal?
Monochromes Schach
R: 1. ... 0-0 2. Kf8-e7 Lg8-h7 3. Kg7-f8 Lh7-g8 4. Le7-d8 Lg8-h7 5. La3-e7 Lh7-g8 6. Kh6-g7 e7xTf6 7. Kg5-h6
play all play one stop play next play all
Kees: R: -1. … 0-0 -2. Kf8-e7 Lg8-h7 -3. Kg7-f8 Lh7-g8 -4. Le7-d8 Lg8-h7 -5.Le7-a3 Lh7-g8 -6. Kh6-g7 e7xTf6 Kg5-h6 (2022-02-16)
A.Buchanan: The final position in Kees' solution has wTf6 which must be promoted. This could have been wBb, which captured e.g. bBa5, bBb6 e.p., bLa7, bSb8.
wTf7 is also promoted, and might be sBh having capturing wBg5, wBh6 e.p., wLg2, wTh1. sBfxLg6 completes the picture. I don't see any difficult captures e.g. of S remaining, so the position looks to be legal. (2022-02-16)
comment
Keywords: Monochromatic Chess, Minimal
Genre: Retro, Fairies
FEN: 3B1rk1/2p1Kr1b/5pp1/8/8/8/8/8
Input: Gerd Wilts, 1995-07-16
Last update: A.Buchanan, 2022-02-16 more...
11 - P0006406
Nikita M. Plaksin
4.B Caissas Schloßbewohner 1, p. 73, 1983
P0006406
(4+2)
Welches waren die letzten 3 Einzelzüge?
Alle Bauern sind Berolinabauern
Teilaufgabe B von P1393956
Henrik Juel: R: 1.BEb7-a8+ Ka8-a7 2.BEc7-b8+, not 1... Ka8xYa7? (2021-09-23)
comment
Keywords: Last Moves?, Berolina chess
Pieces: bu = Berolina Pawn (BE)
Genre: Retro, Fairies
FEN: RRK5/k7/*2p*2P6/8/8/8/8/8
Reprints: (26) Die Schwalbe 156, p. 227, 12/1995
Input: Gerd Wilts, 1996-06-12
Last update: A.Buchanan, 2022-09-13 more...
12 - P1012581
Nikita M. Plaksin
Themes-64 1980
1. Lob
P1012581
(14+10)
#1?
1. ... g6#!

1. Db2#??, aber Schwarz hat keinen legalen letzten Zug

Les 6 pièces noires manquantes ont été prises par pions B : bxcxdxexfxTg8:T pourla promotion de la TBh8 et gxh. Les 2 pièces blanches manquantes sont le FB sur cases noires et une TB. Les N ne peuvent avoir joué le dernier coup : 1.Rb1xFa1?Fb2a1+ 2.Ra1b1 Fc1b2+ rétro-échec perpétuel 1.g2xTh1? échec illégal au RN. Le trait est donc aux N et mat par 1.g6# et non 1.Db2#?
play all play one stop play next play all
wBb2 hat auf seinem Weg nach g8 fünf fehlende schwarzen Steine geschlagen, den letzten fehlenden schwarzen Stein der wBg2

zuletzt nicht:
R: 1. ... Kb1xLa1?? 2. Lb2-a1+ Ka1-b1 3. Lc1-b2+ Kb1-a1 4. Lb2-c1+ forciertes ewiges Retroschach
R: 1. ... g2xTh1=S?? illegales Schach durch wTh1

Schwarz hat keinen letzten Zug, darf also durch 1. ... g6# mattsetzen
No. 319 HN
Hans-Jürgen Manthey: 1. ... g6#
R: 1. Tb8-b7 La6-c8 2. Dh3-a3 Lb7-a6 3. Tg8-h8 Kb2-a1 4. Se3-g4 Ka3-b2 5. Sf3-g5 Lc8-b7 6. Tb1-b8 Ka4-a3 7. Ta1-b1 g2xTh1=S 8. Sg1-f3 g3-g2 9. Df1-h3 f4xLg3 10. f7xTg8=T Th8-g8 11. Kg4-h5 Ka5-a4 12. Kf3-g4 Kb6-a5 13. e6xSf7 Sh6-f7 14. Dd1-f1 Sg8-h6 15. Kg2-f3 La6-c8 16. Kf1-g2 f5-f4 17. Lg2-a8 f7-f5 18. d5xDe6 Kb7-b6 19. Ld6-g3 Kc8-b7 20. Ke1-f1 Kd8-c8 21. La3-d6 Ke8-d8 22. Lf1-g2 Db6-e6 23. c4xDd5 Db1-b6 24. h3-h4 Da8-d5 25. Sd5-e3 b2-b1=D 26. Sc3-d5 b3-b2 27. Sb1-c3 b4-b3 28. b3xTc4 Dd8-a8 29. g2xSh3 Tc6-c4 30. Lc1-a3 Tb6-c6 31. Lb2-c1 Tb8-b6 32. Lc3-b2 Ta8-b8 33. Ld4-c3 Lc8-a6 34. Le5-d4 b5-b4 35. Ld6-e5 b7-b5 36. Lc5-d6 Sf4-h3 37. Lb4-c5 Sd5-f4 38. La3-b4 Sb4-d5 39. Lb2-a3 Sc6-b4 40. Lc1-b2 Sb8-c6 41. b2-b3 (2021-06-28)
comment

Genre: Retro
FEN: B1b2b1R/pRppp1pp/8/6NK/6NP/Q7/P1PPPP1P/k6n
Input: Henri Nouguier, 2004-01-11
Last update: Mario Richter, 2021-06-28 more...
13 - P1135833
Nikita M. Plaksin
Alexej P. Schitow

Tscherkaska Prawda 1990
P1135833
(4+1)
#8
1. La7 Kxa7 2. Sc5 Ka8 3. b3 Ka7 4. b4 Ka8 5. b5 Ka7 6. b6+ Ka8 7. b7+ Ka7 8. b8=D#
play all play one stop play next play all
BUD1 EXD BRL TEB LO8 NJL
SCHRECKE: C+, Gustav 4.2e, Brute Force (2023-10-27)
comment
Keywords: Miniature Collection (0341331), Excelsior (D)
Genre: n#
FEN: kB6/2K5/N7/8/8/8/1P6/8
Input: Zuncke/Bruder, 2010-09-11
Last update: Rainer Staudte, 2023-10-27 more...
14 - P1293419
Nikita M. Plaksin
3735 Die Schwalbe (72) 12/1981
P1293419
(9+12) C+
h=4
Circe
1. exd1=L Txd1[+sLc8] 2. fxe1=S Txe1[+sSb8] 3. gxf1=T[+wTh1] Txf1[+sTa8] 4. Dg8+ hxg8=D=
play all play one stop play next play all
Anton Baumann: C+ Alybadix (2021-02-04)
comment
Keywords: Allumwandlung, Circe, konsekutive Umwandlungen 4 (lstd), Promotion
Genre: Fairies
FEN: 3K4/rbrp3P/p1pP3k/P1P4n/8/8/4ppp1/2RBNRq1
Input: Erich Bartel, 2014-12-28
Last update: James Malcom, 2021-02-04 more...
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