Die Schwalbe

16 problem(s) found in 7520 milliseconds (displaying 16 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT A='Petrovic, Tomislav' AND K='50-Züge-Regel'] [download as LaTeX]

1 - 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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comment
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...
2 - P0001967
Nenad Petrovic
628 Sahovski vjesnik 1950
Dr. Fabel und Dr. Ceriani gewidmet
2. Preis
P0001967
(15+15)
Längste Beweispartie?
(AL: 1021,0)
1. S S 50. S b6 100. S h6 250. S h3 300. a3 S 450. a6 S 500. b3 S 550. g3 S 600. 0-0 S 649. S hxSg 699. h3 S 899. h7 S 949. axSb Ke8 950 Tf1 Kd8 951. Tg1 Ke8 952. Tf1 Dd8 1020. Kd1 Kd8 1021. De1 Ke8=
play all play one stop play next play all
Henrik Juel: Note that in problems castling acts like capture and pawn move with respect to the 50 moves rule. After 949.a6xSb7 there are 4 moves left by [Pa7]; but each camp can shift only KDT, on d1-g1 and b8-e8, respectively, so the triple repetition rule now limits the length of the game. (2004-09-09)
A.Buchanan: In some problems it's certainly the case that the 50-move rule operates incorrectly in this way. Such problems are fine, but obviously wouldn't want to impose this as a standard. Different composers can make different assumptions here (2023-06-20)
comment
Keywords: 50 move rule, Non-Unique Proof Game, Longest Proof Game, Castling
Genre: Retro
FEN: Nrq1kb2/1PpppppP/1p6/8/8/pP4P1/BRPPPPp1/BrnKQ1Rb
Reprints: (I) Problem 5-6 12/1951
Problem 7-9 03/1952
1439 FIDE Album 1945-1955 1964
(130) Problem 91-94 04/1964
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2012-04-08 more...
3 - 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...
4 - P0008399
Thomas Volet
Rex Multiplex 1983
1. Preis
P0008399
(14+12)
=
R: 1. Th1-g1 Lb8-a7 2. Th5-h1 La7-b8 3. Td5-h5 Lb8-a7 4. Td2-d5 La7-b8 5. Tc2-d2 Lb8-a7 6. Tc1-c2 Kc2-b3 7. Ta1-c1 Kd2-c2 8. Lb4-a3 Ke1-d2 9. Ta3-a1 Kd2-e1 10. Tb3-a3 Ke1-d2 11. La3-b4 Kd2-e1 12. Tb5-b3 Ke1-d2 13. Ta5-b5 Kd2-e1 14. Ta7-a5 Ke1-d2 15. Tb7-a7 La7-b8 16. Tb8-b7 Kd2-e1 17. Kg8-f8 Ke1-d2 18. Tf8-b8 Te8-e7 19. Kh8-g8 Tb8-e8 20. Tc8-f8 Tb7-b8 21. Kg8-h8 Lb8-a7 22. Kf8-g8 Ta7-b7 23. Ke7-f8 Ta5-a7 24. Td8-c8 La7-b8 25. Tb8-d8 Tb5-a5 26. Tb7-b8 Lb8-a7 27. Ta7-b7 Tb3-b5 28. Lb4-a3 Ta3-b3 29. Ta5-a7 Ta1-a3 30. Tb5-a5 Tc1-a1 31. La5-b4 Tc2-c1 32. Tb3-b5 Td2-c2 33. Ta3-b3 Td6-d2 34. Ta1-a3 Td5-d6 35. Tc1-a1 Tf5-d5 36. Tc2-c1 Tf4-f5 37. Td2-c2 Te4-f4 38. Td6-d2 Td4-e4 39. Td5-d6 Td2-d4 40. Td4-d5 Tc2-d2 41. Td2-d4 Tc1-c2 42. Tc2-d2 Ta1-c1 43. Tc1-c2 Ta3-a1 44. Ta1-c1 Tb3-a3 45. Ta3-a1 Tb5-b3 46. Lb4-a5 Ta5-b5 47. Tb3-a3 Ta7-a5 48. La3-b4 Tb7-a7 49. Tb5-b3 La7-b8 50. Ta5-b5 Tb8-b7 51. Ta6-a5 Tg8-b8 52. Ta5-a6 Lb8-a7 53. Ta7-a5 Th8-g8 54. Tb7-a7 La7-b8 55. Tb8-b7 Tg8-h8 56. Tf8-b8 Lb8-a7 57. Kd8-e7 La7-b8 58. Kc8-d8 Th8-g8 59. Kb7-c8 Tg8-h8 60. Tb8-f8 Te8-g8 61. Kc8-b7 Te7-e8 62. Tb7-b8 Lb8-a7 63. Ta7-b7 Kd2-e1 64. Ta5-a7 Ke1-d2 65. Tb5-a5 Kd2-e1 66. Tb3-b5 Ke1-d2 67. Lb4-a3 Kd2-e1 68. Ta3-b3 Kc2-d2 69. Ta1-a3 Kb3-c2 70. Tc1-a1 Ka2-b3 71. Tc2-c1 Kb1-a2 72. Td2-c2 Ka2-b1 73. Td5-d2 Kb1-a2 74. Th5-d5 Ka2-b1 75. Th1-h5 Ka1-a2 76. h3xLg4
play all play one stop play next play all
Henrik Juel: Following R: 76.h3xLg4, the resolution could continue with
Retract wKc8 to b5, sLg4 to c8, b7-b6, sSa8 to g8, wSa4 to g1, wKb5 to e1, sLb8 to a5, b6xTc5, wTc5 to d4, sKa1 to b5, a2xTb3xDc4+, etc. (2020-08-13)
Henrik Juel: A little analysis may be in order
Pawns captured all missing men
Kb3 is now confined to the SW corner, but if we retract b3xc4 he is locked in for good, so [Lc8] was captured on g4
Before we retract h3xLg4, sLg4 to c8, and b7-b6, wKf8 must be liberated
The plan for this is: Retract wKg8-f8, sTg1 to f8, sTe7 out, wTf8 out past the sT, wKg8 to e7, sT to h8, wT to e8, and further as per my old comment (2022-01-22)
James Malcom: Corrected SPG: 1. Nh3 e6 2. Nc3 Qg5 3. Rg1 Qe3 4. dxe3 Bb4 5. Qd4 Nc6 6. Bd2 Na5 7. Kd1 Nb3 8.
axb3 Nf6 9. Ra6 Nd5 10. Rb6 axb6 11. Kc1 Ra4 12. Kb1 Ba5 13. Nd1 Rc4 14. Ka1 Ke7
15. Nf4 Kd6 16. Bb4+ Kc6 17. h3 Kb5 18. bxc4+ Ka4 19. c3 Kb3 20. Qc5 Kc2 21. Ka2
bxc5 22. Ka3 Bb6 23. Ka4 Ba7 24. Kb5 Nb6 25. Nd5 Na8 26. Nb6 Re8 27. Na4 b6 28.
Rh1 Bb7 29. Rh2 Bf3 30. Ka6 Bh5 31. Kb7 Re7 32. Kc8 Bg4 33. hxg4 Kb3 34. Rh5 Kc2
35. Rd5 Kb3 36. Rd2 Ka2 37. Rc2 Kb3 38. Rc1 Ka2 39. Ra1+ Kb3 40. Ra3+ Kc2 41.
Rb3 Bb8 42. Ba3 Ba7 43. Rb5 Bb8 44. Ra5 Kb3 45. Ra7 Kc2 46. Rb7 Ba7 47. Rb8 Re8+
48. Kb7 Rg8 49. Rf8 Kd2 50. Kc8 Ke1 51. Kd8 Rh8 52. Ke7 Rg8 53. Rb8 Rc8 54. Rb7
Bb8 55. Ra7 Rd8 56. Ra5 Ba7 57. Rb5 Rb8 58. Rb3 Rb7 59. Bb4 Bb8 60. Ra3 Ra7 61.
Ra1 Ra5 62. Rc1 Rb5 63. Ba5 Rb3 64. Rc2 Ra3 65. Rd2 Ra1 66. Rd6 Rc1 67. Kf8 Rc2
68. Kg8 Rd2 69. Rd5 Rd4 70. Rd6 Re4 71. Rd2 Rd4 72. Rc2 Rd2 73. Rc1 Rc2 74. Ra1
Rc1 75. Ra3 Ra1 76. Rb3 Ra3 77. Rb5 Rb3 78. Bb4 Kd2 79. Ra5 Kc2 80. Ba3 Rb5 81.
Ra7 Ra5 82. Rb7 Ba7 83. Rb8 Kb3 84. Rd8 Bb8 85. Re8 Ra7 86. Kh8 Rb7 87. Rf8 Ba7
88. Re8 Rb8 89. Rf8 Re8 90. Kg8 Re7 91. Rb8 Kc2 92. Rb7 Bb8 93. Ra7 Kb3 94. Ra5
Kc2 95. Rb5 Kb1 96. Rb3 Kc1 97. Bb4 Kb1 98. Ra3 Kc2 99. Ra1 Kb3 100. Rc1 Ba7
101. Rc2 Ka2 102. Rd2 Kb1 103. Rd5 Kc2 104. Rh5 Kc1 105. Rh1 Kb1 106. Rg1 Kc2
107. Ba3 Kb3 108. Kf8 (2023-03-11)
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comment
Keywords: 50 move rule, Move Length Record
Genre: Retro
FEN: n4K2/b1pprppp/1p2p3/2p5/N1P3P1/BkP1P3/1P2PPP1/3N1BR1
Reprints: Redkiye zhanry-plus 1996
Input: Gerd Wilts, 1996-09-17
Last update: James Malcom, 2023-03-11 more...
5 - P0008465
Thomas Volet
9375 Die Schwalbe 161 10/1996
Nikita Plaksin gewidmet
P0008465
(14+9) cooked
Die Partie ist beendet. Weshalb?
Thomas Volet: Problem P0008465 is faulty. (2000-11-27)
James Malcom: What is the author's solution and the cook? (2022-01-21)
Mario Richter: I do not the "official cook", but the following should work:
R: 1. Th1-g1 Kf5-e4 2. Th4-h1 Kg6-f5 3. Td4-h4 Kh6-g6 4. Td1-d4 Kg6-h6 5. Ta1-d1 Kh6-g6 6. Ta3-a1 Kg6-h6 7. Tb3-a3 Kh6-g6 8. Tb5-b3 Kg6-h6 9. Ta5-b5 Kh6-g6 10. Lb8-a7 Kg6-h6 11. Ta7-a5 Kh6-g6 12. Tb7-a7 Kg6-h6 13. La7-b8 Kh6-g6 14. Kg8-f8 Kg6-h6 15. Tb8-b7 Kh6-g6 16. Tf8-b8 Te8-e7 17. Lb8-a7 Tc8-e8 18. La7-b8 Tb8-c8 19. Te8-f8 Tb7-b8 20. Lb8-a7 Ta7-b7 21. Kf8-g8 Ta5-a7 22. Ke7-f8 Tb5-a5 23. Kd8-e7 Tb4-b5 24. Kc8-d8 Th4-b4 25. Kb7-c8 Th5-h4 26. Ka6-b7 Tg5-h5 27. Kb5-a6 Tf5-g5 28. Kc4-b5 Te5-f5 29. Kb3-c4 Te4-e5 30. Ka2-b3 Tb4-e4 31. Kb1-a2 Tb5-b4 32. Te7-e8 Ta5-b5 33. Kc1-b1 Ta7-a5 34. Kd1-c1 Tb7-a7 35. La7-b8 Tb8-b7 36. Ke1-d1 Th8-b8 37. Te8-e7 Kg5-h6 38. Tb8-e8 Tg8-h8 39. Tb7-b8 Th8-g8 40. Lb8-a7 Tg8-h8 41. Ta7-b7 Th8-g8 42. Ta5-a7 Tg8-h8 43. Tb5-a5 Th8-g8 44. Tb4-b5 Tg8-h8 45. Th4-b4 Tf8-g8 46. Th1-h4 Th8-f8 47. h2xLg3 (2022-01-23)
Thomas Volet: Mario, I do appreciate your tactfully lengthy unwind, but this effort at economy is hugely flawed. (2022-01-23)
comment
Keywords: 50 move rule
Genre: Retro
FEN: N4K2/B1pprPpp/1p2p3/2p5/N3k3/2P3P1/1PP1PPP1/5BR1
Input: Gerd Wilts, 1996-10-26
Last update: A.Buchanan, 2017-03-22 more...
6 - P0008784
Niels Høeg
The Chess Amateur 07/1926
P0008784
(1+1)
Längste BP ohne Schach. Welches war der letzte Zug?
R: 5899. Kg2xTh1
play all play one stop play next play all
The game ends after 50 consecutive moves without captures or pawn moves (loss of castling right is not included here), or when there is not enough material to mate (say K-K or K-KS). There are 30 captures and 96 pawn moves (including 8 pawn captures) available, so the longest game seems to last (30+96-8)x50=5900 moves. This cannot be achieved because of the move-loss when the draw-preventing move shifts between white and black. Niels Høeg believed that 2 moves were lost and stated the solution as 5898.- Kb7xTa8. Karl Fabel later showed that only 1.5 moves need be lost.

Since 1926, there have been some relevant innovations to rules and conventions
(1) 50 move rule applies only to retro compositions, and will trigger automatically (no issue with writing down the move). (Codex 1953?)
(2) Removal of rules about draw by insufficient material (Laws 1997)
(3) Dead position rule introduced (Laws 1997)
(4) 75 move rule introduced (Laws 2014)
(5) Dead position rule applies only to retro compositions (Codex 2015)
(6) Articles 9.2 & 9.3 apply to chess problems - this includes 50 move rule and excludes 75 move rule (Codex 2019)

This is certainly a composition rather than a question about over the board chess. And it is certainly a retro composition. So the 50 move rule will dominate the 75 rule. The standard interpretation of interaction between 50 move rule and Dead Position in compositions is that Dead Position assessment *is* aware of looming automatic draw by 50 moves. (Note there is a similar assessment for interaction between Dead Position and Draw by Repetition.)

So we can argue that the game cannot last to 5898.5 moves, because the final move leads to a mandatory draw: either the king captures the last officer, or the king avoids the capture and the game ends in draw under the 50 move rule. So the position is dead at 5898.0. But even 5898.0 is too long for the diagram position with the kings so far apart. In the alternate reality if the last capture of a rook does not take place, there must be sufficient moves left for the kings to come together so that the rook can deliver checkmate. This will take at least 6.0 moves.

There is also still an ambiguity in the rules as to whether checkmate overrides draw by 50 moves. This is explicitly mentioned in the 75 move rule, but not in 50 move rule. I assume that checkmate *does* take priority.

Or does a valid problem only exist in the context of the rules and conventions that pertained at the time of its composition? The Codex does not opine on this general point.

Compare P1331022

Duplicate Diagram: P1101148, P1189676, P1191185, P1304589

A.Buchanan: There's a question whether DP rule has visibility of 3Rep & 50M state. The current consensus among most of the tiny group who might care is that for retros, it does have visibility, but for purely forward problems, it does not. This align with the idea that by default 50M & DP rules apply only to retros (2023-09-06)
A.Buchanan: The old intended interpretation is protected under the Golden Age principle. Suppose we do apply modern rules, codex & clarifications. If Black just played Kb7xRa8, then the alternative leading to mate takes 7.0 moves, so this must have been Black's 5892nd move at the latest. If White just moved, then it was White's 5892nd at latest. So to maximize the length of the game, Black moved last. (2023-09-06) edit (2023-09-06)
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Keywords: Longest Proof Game, Last Move?, only Kings, Non-Unique Proof Game, Dead Position, 50 move rule, Constrained problem, Type A, Miniature, Golden Age (pre-dead), Aristocrat
Genre: Mathematics, Retro
FEN: k7/8/8/8/8/8/8/7K
Reprints: Schackproblemet 1928
Schach und Zahl 1966
Input: Gerd Wilts, 1997-06-20
Last update: A.Buchanan, 2024-01-18 more...
7 - P1000710
Thomas Volet
R023 Probleemblad 05-06/1998
P1000710
(14+11) cooked
Remis
R: 1. Lb1-a2 Kb2-a3 2. Tb4-b6 Ka3-b2 3. Tb6-a6 Kb2-a3 4. Ta4-b4 Kc1-b2 5. Tb4-b6 Kd1-c1 6. Ta2-a4 Kc1-d1 7. Ta4-b4 Kd1-c1 8. Tb2-a2 Ke1-d1 9. La2-b1 Ke2-e1 10. Tb1-b2 Ta6-a7 11. Th1-b1 Tb6-a6 12. Lb1-a2 Ta6-b6 13. Ta2-a4 Ta7-a6 14. Tb2-a2 Ta6-a7 15. La2-b1 Tb6-a6 16. Tb1-b2 Tb4-b6 17. Tc1-b1 Ta4-b4 18. Lb1-a2 Ta2-a4 19. Td1-c1 Tb2-a2 20. La2-b1 Tb1-b2 21. Tf1-d1 Te1-b1 22. Tg1-f1 Kd1-e2 23. Tf1-g1 Kc1-d1 24. Tg1-h1 Td1-e1 25. Te1-f1 Kb2-c1 26. Te6-e1 Te1-d1 27. Th1-g1 Te4-e1 28. Te1-h1 Kc1-b2 29. Te3-e1+ Kd1-c1 30. Kg8-h8 Ke1-d1 31. Td3-e3+ Kf1-e1 32. Kf8-g8 Kg1-f1 33. Ke8-f8 Te1-e4 34. Te3-d3 Tf1-e1 35. Te1-e3 Kh1-g1 36. Tb1-e1 Tc1-f1 37. Tb2-b1 Td1-c1 38. Lb1-a2 Kg1-h1 39. Ta2-b2 Tc1-d1 40. Ta4-a2 Td1-c1 41. Tb4-a4 Tc1-d1 42. Tb6-b4 Td1-c1 43. Ta6-b6 Tc1-d1 44. Ta8-a6 Tf1-c1 45. Td8-a8 Tc8-c7 46. Te5-e6 Ta8-c8 47. Tc8-d8 Tb8-a8 48. Kd8-e8 Ta8-b8 49. Kc7-d8 Tb8-a8 50. Kb6-c7 Ta8-b8 51. Kb5-b6 c7-c6+
play all play one stop play next play all
Cook: (Henrik Juel):
R: 1. Lb1-a2 Kb2-a3 2. Tb4-b6 Ka3-b2 3. Tb6-a6 Kb2-a3 4. Ta4-b4 Kc1-b2 5. Ta2-a4 Kd1-c1 6. Tb2-a2 Ke1-d1 7. La2-b1 Ke2-e1 8. Tb1-b2 Ta6-a7 9. Th1-b1 Ta7-a6 10. Lb1-a2 Ta6-a7 11. Tb4-b6 Ta7-a6 12. Ta4-b4 Ta6-a7 13. Ta2-a4 Ta7-a6 14. Tb2-a2 Ta6-a7 15. La2-b1 Tb6-a6 16. Tb1-b2 Tb4-b6 17. Td1-b1 Ta4-b4 18. Lb1-a2 Ta2-a4 19. Tg1-d1 Tb2-a2 20. La2-b1 Tb1-b2 21. Kg8-h8 Te1-b1 22. Kh8-g8 Kd1-e2 23. Kg8-h8 Kc1-d1 24. Kf8-g8 Td1-e1 25. Te1-g1 Kb2-c1 26. Te5-e1 Tb1-d1 27. Te1-h1 Kc1-b2 28. Te3-e1+ Kd1-c1 29. Td5-e5 Ke1-d1 30. Tc3-e3+ Kf1-e1 31. Ke8-f8 Kg1-f1 32. Te3-c3 Tf1-b1 33. Te1-e3 Kh1-g1 34. Tb1-e1 Kg1-h1 35. Tb2-b1 Kh1-g1 36. Lb1-a2 Kg1-h1 37. Ta2-b2 Kh1-g1 38. Ta4-a2 Kg1-h1 39. Tb4-a4 Kh1-g1 40. Tb6-b4 Kg1-h1 41. Ta6-b6 Kh1-g1 42. Ta8-a6 Kg1-h1 43. Td8-a8 Tc8-c7 44. Te5-d5 Ta8-c8 45. Tc8-d8 Kh1-g1 46. Kd8-e8 Kg1-h1 47. Kc7-d8 Kh1-g1 48. Kb6-c7 Kg1-h1 49. Kb5-b6 c7-c6+ 50. La2-b1
zunächst mit der 3. ehrenden Erwähnung ausgezeichnet, wegen der NL aber in 'Probleemblad' 09-10/2003 von PR Piet le Grand wieder aberkannt.
TV: Cooked by H. Juel, who found a 49.5 move retraction. (2001-09-21)
James Malcom: What is the author's solution and the cook? (2022-01-21)
Henrik Juel: Thanks for providing the retroplays, Mario (2022-03-22)
comment
Keywords: 50 move rule
Genre: Retro
FEN: 7K/rprppppp/RRp5/N1p5/2P5/kP3P2/B1PP1PPP/N7
Input: Gerd Wilts, 2000-08-13
Last update: Mario Richter, 2022-03-22 more...
8 - P1001097
Michel Caillaud
R121 Probleemblad 01/2001
Special Prize 2001
P1001097
(12+13) C+
h#2
1. cxb6 d6 2. bxc5 Txb8#
1. La7 Lxa7 2. 0-0-0? Tb8#
play all play one stop play next play all
Castling is impossible because, if black can still castle, the diagram position is draw due to the 50-move rule!
White captures: axb, bxc, exf. Black captures: hxg, gxf, exd, dxc. Promotions: a1=T, b1=T, h8=X. The position can only be unblocked by unpromoting White piece on h8. This can only be wRb5. wK must go to c6 to unpin wR. The journey is via g3, h2, e1, b2, & a3.The 50 white moves are: 34 by wK (screens on c8 [with careful step on a2 to let bB pass] then g8), 11 by wB6 (to e3 to exchange place with bB then screen on a7), 3 by wR (promotion on h8), 2 by wBa4.
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comment
Keywords: Non-standard material (tt), 50 move rule, Cant Castler
Genre: Retro, h#
Computer test: Popeye WINDOWS-32Bit-Version 3.65 (2048 KB) Popeye 4.61 (also)
FEN: rb2k3/2p2p1K/1B6/1RPP1PP1/B1Prbrp1/1Prp1p2/2p2P2/N7
Reprints: Probleemblad , p. 220-227, 12/2010
Input: Gerd Wilts, 2001-02-11
Last update: A.Buchanan, 2020-10-10 more...
9 - P1009492
Thomas Volet
R023c Probleemblad 07-08/2003
Henrik Juel gewidmet
P1009492
(14+12) cooked
Remis?

play all play one stop play next play all
Cook: R: 1. Lb1-a2 Db5-b6 2. Ta2-a3 Ka3-b4 3. Rb2-a2 Db4-b5 4. La2-b1 Dc3-b4 5. Tb1-b2 Kb2-a3 6. Th1-b1 Dd3-c3 7. Tb6-a6 Df1-d3 8. Tb4-b6 Kc1-b2 9. Ta4-b4 Kd1-c1 10. Lb1-a2 Ke2-d1 -11.Ra2-a4 Ra6-a7 -12.Rb2-a2 Rb6-a6 -13.Ba2-b1 Rb4-b6 -14.Rb1-b2 Dg1-f1 -15.Rf1-b1 Ra4-b4 -16.Bb1-a2 Ra2-a4 17.Kg8-h8 Rb2-a2 -18.Ba2-b1 Rb1-b2 -19.Kf8-g8 Re1-b1 -20.Ke8-f8 Kd1-e2 -21.Bb1-a2 Kc1-d1 -22.Ba2-b1 Kb2-c1 -23.Bb1-a2 Re2-e1 -24.Rd1-f1 Re1-e2 -25.Rc1-d1 Df1-g1 -26.Rd1-c1 De2-f1 -27.Rc1-d1 Rg1-e1 28.Rf1-c1 De1-e2 -29.Kf8-e8 Kc1-b2 -30.Ba2-b1 Kd1-c1 -31.Ke8-f8 Ke2-d1 -32.Kf8-e8 Db1-e1 -33.Ke8-f8 Db2-b1 -34.Rb1-f1 Df6-b2 -35.Rb2-b1 Dg5-f6 -36.Bb1-a2 Dh5-g5 -37.Ra2-b2 Dg5-h5 -38.Ra4-a2 Dh5-g5 -39.Rb4-a4 Dg5-h5 -40.Rb6-b4 Df5-g5 -41.Ra6-b6 Dg5-f5 -42.Ra8-a6 Dh5-g5 -43.Rd8-a8 Rc8-c7 -44.Ba2-b1 Ra8-c8 -45.Rb8-d8 Ra6-a8 -46.Kd8-e8 Rb6-a6 -47.Kc7-d8 Rb4-b6 -48.Kb6-c7 Ra4-b4 -49.Kb5-b6 c7-c6!
James Malcom: What is the author's solution? (2022-01-21)
comment
Keywords: 50 move rule
Genre: Retro
FEN: 7K/rprppppp/Rqp5/N1p5/1kP5/RP2P3/B1PP1PPP/N7
Input: Gerd Wilts, 2003-08-15
Last update: James Malcom, 2022-01-21 more...
10 - P1331867
Thomas Volet
Thierry Le Gleuher

16620 Die Schwalbe 277, p. 388, 02/2016
Henrik Juel gewidmet
P1331867
(14+14)
Gab es nach dem letzten Schlagfall einen Bauernzug?
Auflösung ohne Zuschlagen der 50-Züge-Regel (Mario Richter, PDB 15.03.2022):
R: 1. ... Da3-c1 2. Tb2-b1 Dc5-a3 3. Lb1-a2 Dd4-c5 4. Ta2-b2 De5-d4 5. Ta3-a2 Dg5-e5 6. La2-b1 De5-g5 7. Tb1-e1 Kf1-g2 8. Tb2-b1 Le4-h1 9. Kg8-f8 Kg2-f1 10. Tb1-b2 Db2-e5 11. Th1-b1 Dc1-b2 12. Kh8-g8 Dg1-c1 13. Lb8-a7 Kf1-g2 14. Lb1-a2 Ke1-f1 15. Ta2-a3 Df1-g1 16. Tg1-h1 Ta7-b7 17. Tg6-g1 Ta6-a7 18. Tb2-a2 Dh1-f1 19. La2-b1 Kf1-e1 20. Th6-g6 Kg2-f1 21. Tb1-b2 Dc1-h1 22. Th5-h6 Db2-c1 23. Th1-b1 Dc1-b2 24. Th6-h5 Dg1-c1 25. La7-b8 Kf1-g2 26. Th5-h6 Ke1-f1 27. Th6-h5 Kd1-e1 28. Th5-h6 Kc1-d1 29. Th6-h5 Kb2-c1 30. Th5-h6 Ka3-b2 31. Th6-h5 Kb4-a3 32. Th5-h6 Kb5-b4 33. Th6-h5 Kc6-b5 34. Th5-h6 Kb7-c6 35. Th6-h5 Kc8-b7 36. Th5-h6 Kd8-c8 37. Th6-h5 Ke8-d8 38. Th5-h6 Kf8-e8 39. Th6-h5 Te8-e7 40. Th5-h6 Tb8-e8 41. Th6-h5 Tb7-b8 42. Th4-h6 Ke7-f8 43. Kg8-h8 Kd6-e7 44. Kf8-g8 Kc5-d6 45. Ke7-f8 Tb8-b7 46. Th6-h4 Th8-b8 47. Lb8-a7 Ta7-a6 48. Tg6-h6 Tb7-a7 49. La7-b8 Tb8-b7 50. Th6-g6 Tg8-b8 51. Kd8xLe7
und weiter z.B.
51. ... Lf8-e7 52. Kc8-d8 Kd5-c5 53. Kb7-c8 Dg5-g1 54. Ka6-b7 Le7-f8 55. Kb5-a6 Tb8-g8 56. Kb4-b5 Kd6-d5 57. Kc3-b4 Lf8-e7 58. Tg6-h6 Dd8-g5 59. Tg1-g6 Ke7-d6 60. Tb1-g1 Lb7-e4 61. Kb2-c3 Lc8-b7 62. Kc1-b2 b7-b6 63. Ld4-a7 Ke8-e7 64. Kd1-c1 b5xDa4 65. Da3-a4 a6xSb5 66. Dc1-a3 a7-a6 67. Ke1-d1 Sb6-a8 68. Dd1-c1 Sd5-b6 69. Lb2-d4 Se7-d5 70. Lc1-b2 Sg8-e7 71. b2-b3 Ta8-b8 72. Ld5-a2 Sh6-g8 73. Lg2-d5 Sg8-h6 74. Lf1-g2 Sh6-g8 75. g2xSh3 Sf4-h3 76. Sb3-a1 Sd5-f4 77. Ta1-b1 Sb4-d5 78. Sd4-b3 Sa6-b4 79. Sc3-b5 Sb8-a6 80. Sf3-d4 Sg8-h6 81. Sg1-f3 Sh6-g8 82. Sb1-c3 Sg8-h6 83. a4-a5 e7-e6 84. a2-a4

AL R: 1. Tb2-b1 Dd1-c1 2. Lb1-a2 Dc1-d1 3. Ta2-b2 Dd1-c1 4. Ta3-a2 Dc1-d1 5. La2-b1 Db2-c1 6. Tb1-e1 Dc1-b2 7. Kg8-f8 Kg1-g2 8. Kf8-g8 Ld5-h1 9. Kg8-f8 Kg2-g1 10. Kf8-g8 Db2-c1 11. Th1-b1 Dc1-b2 12. Kg8-f8 Dg1-c1 13. Kf8-g8 Kf1-g2 14. Lb1-a2 Ke1-f1 15. Ta2-a3 Df1-g1 16. Lb8-a7 Ta7-b7 17. Tg1-h1 Ta6-a7 18. Tg6-g1 Dg1-f1 19. Tb2-a2 Kf1-e1 20. Th6-g6 Kg2-f1 21. La2-b1 Dc1-g1 22. Tb1-b2 Db2-c1 23. Th1-b1 Dc1-b2 24. Kg8-f8 Dg1-c1 25. Kf8-g8 Kf1-g2 26. Kg8-f8 Ke1-f1 27. Kf8-g8 Kd1-e1 28. Kg8-f8 Kc1-d1 29. Kf8-g8 Kb2-c1 30. Kg8-f8 Ka3-b2 31. Kf8-g8 Kb4-a3 32. Kg8-f8 Kb5-b4 33. Kf8-g8 Kc6-b5 34. Kg8-f8 Kb7-c6 35. Kf8-g8 Kc8-b7 36. Kg8-f8 Kd8-c8 37. Kh8-g8 Ke8-d8 38. Lb1-a2 Kf8-e8 39. La2-b1 Te8-e7 40. La7-b8 Tb8-e8 41. Lb1-a2 Tb7-b8 42. La2-b1 Ke7-f8 43. Kg8-h8 Kd6-e7 44. Kf8-g8 Kc5-d6 45. Ke7-f8 Tb8-b7 46. Lb1-a2 Th8-b8 47. Lb8-a7 Ta7-a6 48. La2-b1 Tb7-a7 49. La7-b8 Tb8-b7 50. Lb1-a2 Tg8-b8 51. Kd8xLe7
und die Diagrammposition ist wegen der 50-Züge-Regel remis.
play all play one stop play next play all
Die Antwort auf die Frage aus der Forderung lautet: Nein! mach dem letzten Schlagzug erfolgte kein Bauernzug mehr.

Löser Martin Hintz (https://thbrand.de/downloads/Hintz_Schwalbe_277_Loesungen.pdf)
"Ich möchte hervorheben, dass die 50-Züge-Regel für die Beantwortung der eigentlichen Aufgabenstellung keine Rolle spielt. Die Lösung wäre die gleiche, wenn es die 50-Züge-Regel nicht gäbe. Anders als es in vergleichbaren Aufgaben bereits dargestellt wurde, wird der Schlag des sL nicht aus dem Grund benötigt, um ein Remis infolge der 50-Züge-Regel zu vermeiden, sondern vielmehr aus dem Grund, dass sonst auf f8 ein Schachschutz fehlt. Die 50-Züge-Regel bildet damit "lediglich" eine zusätzliche Schwierigkeit, der man sich bei der Konstruktion einer konkreten Beweispartie gegenübersieht."
Korrektur zu 16238 aus Heft 271 (Februar 2015) [Pxxxxxxx]

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

1. Na3 Na6 2. Nb5 Nc5 3. Nd4 Ne6 4. Nb3 Ng5 5. Rb1 Nh3 6. Na1 Nf6 7. gxh3 Nd5 8. Bg2 Nb6 9. Bd5 Rb8 10. a4 Na8 11. Ba2 a6 12. b3 Rg8 13. Ba3 Rh8 14. Bc5 Rg8 15. Ba7 b6 16. Nf3 Bb7 17. Nd4 Bf3 18. Nb5 axb5 19. Qc1 Qc8 20. Qa3 Qb7 21. a5 Qc6 22. Qa4 Kd8 23. Kd1 Kc8 24. Kc1 Kb7 25. Kb2 Re8 26. Rbg1 Kc8 27. Ka3 Kd8 28. Kb4 Qh6 29. Rg6 bxa4 30. Kb5 Rh8 31. Ka6 Bg4 32. Kb7 Bf3+ 33. Rc6 Qg5 34. Kb8 Qg1 35. Rh6 e6 36. Rg6 Bd6 37. Rh6 Reg8 38. Rg6 Bf8 39. Rh6 Ke7 40. Kc8 Kd6 41. Kd8 Kc5 42. Rg6 Be7+ 43. Kxe7 Rb8 44. Rh6 Rb7 45. Bb8 Ra7 46. Rg6 Ra6 47. Ba7 Rb8 48. Rh6 Rb7 49. Kf8 Kd6 50. Kg8 Ke7 51. Kh8 Kf8 52. Rg6 Rb8 53. Rh6 Re8 54. Rg6 Re7 55. Rh6 Ke8 56. Rg6 Kd8 57. Rh6 Kc8 58. Rg6 Kb7 59. Rh6 Kc6 60. Rg6 Kb5 61. Rh6 Kb4 62. Rg6 Ka3 63. Rh6 Kb2 64. Rg6 Kc1 65. Rh6 Kd1 66. Rg6 Ke1 67. Rh6 Kf1 68. Bb8 Kg2 69. Kg8 Qb1 70. Kf8 Qb2 71. Rb1 Qf6 72. Rb2 Kf1 73. Bb1 Ke1 74. Ra2 Kd1 75. Ra3 Kc1 76. Rg6 Kb2 77. Rg1 Qg6 78. Rh1 Qg1 79. Ba7 Kc1 80. Bb8 Kd1 81. Ba7 Ke1 82. Bb8 Kf1 83. Ba7 Kg2 84. Bb8 Qc1 85. Ba2 Qb2 86. Rb1 Qc1 87. Rb2 Kf1 88. Rb1 Bh1 89. Rb2 Kg2 90. Rb1 Qb2 91. Re1 Qc1 92. Bb1 Qd1 93. Ra2 {50 moves rule} Qc1 94. Rb2 Qd1 95. Ba2 Ra7 96. Rb1 Rb7 97. Ba7 Qc1 (2022-03-14)
Thomas Volet: Apologies to Thierry. I should have addressed this earlier. My remark quoted above was (if recollection serves) made along with the submission of the composition for publication. Regardless, my belief at at the time was incorrect as I was then unaware of Nikita Plaksin's mirror compositions P0004364 and P0004395 (published in 1969). Both show the retroplay sequence ending (just short of the draw that the prevailing convention dictated, which saves the stipulated Mate in 3) in the uncapture by a King of a Bishop on the Bishop's home square, from which the Bishop never moved (exit Pawns occupying their home squares in the diagrams). The above 2016 joint composition with Thierry shows the 50 move retroplay sequence of non-Pawn, non-capturing moves being immediately followed by the uncapture by a King of a non-Pawn unit that has moved from its home square. That feature, which presents some additional constraints on realization, may also have been presented before, but obviously I should leave that to better historians. (2022-07-27)
comment
Keywords: 50 move rule
Genre: Retro
FEN: n4K2/Brpprppp/1p2p3/P7/p7/1P5P/B1PPPPkP/NRq1R2b
Input: A.Buchanan, 2017-02-25
Last update: Mario Richter, 2022-03-15 more...
11 - P1368554
Andrew Buchanan
OzProblems.com 16/10/2019
P1368554
(4+3) C+
Gewinn
1. Sd3! Kf8 2. Se3 Sg6+ 3. Kd4 Td8+ 4. Kc3 Tc8+ 5. Sc4 Kg7 6. Sf2 Sh8 7. Tf1 Sf7 8. Se4 Td8 9. Sf6 Sg5 10. Se3 Ta8 11. Kb4 Tb8+ 12. Kc4 Tc8+ 13. Kb5 Tb8+ 14. Kc6 Ta8 15. Sfd5 Ta3 16. Kb5 Se4 17. Te1 Sg5 18. Kb4 Ta2 19. Tf1 Kg6 20. Tf6+ Kh5 21. Kb3 Ta7 22. Sf4+ Kh4 23. Se2 Tb7+ 24. Kc4 Sf7 25. Tf1 Kg5 26. Sd4 Se5+ 27. Kd5 Sd3 28. Se6+ Kg6 29. Sf8+ Kg5 30. Ke4 Sc5+ 31. Kd4 Sd7 32. Se6+ Kg6 33. Sd5 Ta7 34. Se7+ Kh5 35. Tg1 Ta4+ 36. Ke3 Tg4 37. Ta1 Sb6 38. Kf2 Sd7 39. Ta8 Tc4 40. Tg8 Tc2+ 41. Ke3 Tc3+ 42. Kf4 Tc4+ 43. Kf5 Kh4 44. Sf4 Sf6 45. Tg2 (thr 46.Sg6#) Tc5+ 46. Kxf6 Tc6+ 47. Kf5 Tc5+ 48. Sed5 (thr 49.Tg4#) Txd5+ 49. Sxd5 (switch to Syzygy DTM) Kh3 50. Sf4+ Kh4 51. Tg4#
Compare 44. ... Tc5+ which gives a longer DTM at the cost of a slightly sub-optimal DTZ. 45. Sed5 (thr 46.Tg4#) Txd5+ 46. Sxd5 (switch to Syzygy DTM) Kh3 47. Tc8 Kg3 48. Ke6 Sb6 49. Sxb6 Kf2 50. Sc4 Ke1 51. Td8 Ke2 52. Td2+ Ke1 53. Kd5 Kf1 54. Ke4 Kg1 55. Kf3 Kh1 56. Kg3 Kg1 57. Td1#
play all play one stop play next play all
According to the Syzygy tablebase, *any* move wins for White except the silly 1. Sa2??
However, the tablebase does not know about castling!
The quickest winning move 1. Sd3!? would be defeated by 1. ... 0-0-0!? which attacks Sd3 and threatens the skewer Re8+. After this, White can only draw - or even loses after 2. Sc1??
But by a simple retro argument, Black cannot castle.
Since this retro point is essential to the solution, we have a retro problem, and hence the 50 move rule, which by Codex default is disengaged, is switched on.
When working with the 50 move rule, the key concept is not the number of moves to mate, but the number of moves to the first capture, pawn move or checkmate, known as DTZ.
All the alternatives to 1. Sd3 take more than 50 moves to force the first capture. The closest are 1. Sc3 & Se3 which according to Syzygy require exactly 50.5 moves, so seem *just* too slow. (For positions close to the cut-off, one has to be careful about *rounding*, see https://syzygy-tables.info/metrics. The number of half-moves for DTZ can be n or n+1. In fact here, after 1. Sc3/Sf3, the zeroing move is at its earliest White's 52nd move, not 51st.)
By the retro property therefore *only* 1. Sd3!, which requires 45 moves to force the first capture, can win. So this removes all the cooks for the first White move.

Lomonosov DTM #59: 1. Nd3!! Kf8!! 2. Ne3! Ng6+! 3. Kd5 Rd8+ 4. Kc4 Rc8+ 5. Kb5 Rb8+ 6. Ka6 Kf7 7. Nc5 Ne5 8. Nd5 Re8 9. Kb7 Nc4 10. Rf1+ Kg7 11. Kc6 Re5 12. Nd3 Re6+ 13. Kc5 Ne5 14. N3f4 Nd7+ 15. Kb5 Rd6 16. Nc7 Kf7 17. Nfd5+ Kg6 18. Ne7+ Kg5 19. Ncd5 Kg4 20. Rf4+ Kg5 21. Rf2 Kg4 22. Ne3+ Kg5 23. N7d5 Kg6 24. Rf1 Nf6 25. Nf4+ Kf7 26. Nc4 Rd8 27. Ne5+ Kg8 28. Rg1+ Kh7 29. Kc6 Rc8+ 30. Kd6 Re8 31. Re1 Ra8 32. Rh1+ Kg8 33. Ke7 Ra4 34. Rg1+ Kh7 35. Ne6 Nd5+ 36. Kd6 Nf4 37. Nc5 Rd4+ 38. Ke7 Rd2 39. Rg4 Rf2 40. Ncd7 Nd5+ 41. Ke6 Nf4+ 42. Kd6 Rd2+ 43. Ke7 Nd5+ 44. Kf8 Kh6 45. Rh4+ Kg5 46. Nf3+ Kf5 47. Nxd2 (zeroing 50M) Nc7 48. Re4 Ne6+ 49. Ke7 Nf4 50. Nf6 Ng6+ 51. Kf7 Ne5+ 52. Kg7 Kg5 53. Nf1 Kf5 54. Ne3+ Ke6 55. Nc4 Ke7 56. Rxe5+ Kd8 57. Rc5 Ke7 58. Rd5 Ke6 59. Re5# 1-0.
http://www.ozproblems.com/walkabout/walkabout2019#WA1610
https://www.chess.com/blog/Rocky64/a-heraldic-endgame-tablebase-composition

The school shield which inspired the problem:
http://www.calvert-trust.org.uk/images/colour.jpg
Henrik Juel: With White to move, last move was made with Ta8 or Ke8, so Black may not castle
An endgame connaisseur may supply the solution... (2019-10-25)
James Malcom: It is now supplied, Henrik. :) I worked through the Sygyzy tablebase, move by move, to get the computer recommended line. (2020-10-09)
A.Buchanan: Thanks James for this hard work to record a solution. Alas there is not just a single recommended line - White has a choice as early as W3. I wonder too if there is a DTM line which is faster than #57, and still clears the DTZ hurdle although maybe not as quickly as B45 as here. There is no DTM engines for 7 pieces yet.
What is certain is that the key works, and is unique. The exactly details of the best mating line remain to be revealed by improving technology. (2020-10-09)
A.Buchanan: For example, 2. Nc5 has DTZ 90, compared to the "superior" 2. Ne3 DTZ 88. However, if you keep picking the top move all the way through, then the final DTZ is in fact checkmate! I haven't all the Black choices though, but there aren't that many of them (2020-10-09)
A.Buchanan: The Lomonosov tablebase is DTM, and is available for free on Android. Today I found a PC app called Blue Stacks which emulates Android. According to Lomonosov, the position is #59, compared to Syzygy's DTZ offering #57 via zero at B45. When I run through the L. indicative solution (very quick to download and grab all of it), it is close to S.'s choices and all the mainline positions are safe from 50M. The zero is at W47, and the final mate is attractive in the middle of the board.
The real questions are:
(1) Is the #59 sound from 50M perspective? Or can Black force White to delay a few moves in order to avoid 50M trouble somewhere?
(2) How dualized is the solution? How does 50M play into pruning that?
What is clear is that both tablebases agree the solution begins 1. Nd3!! Kf8!! 2. Ne3! Ng6! (2020-10-10)
A.Buchanan: James' Syzygy solution had seemed a little odd, because it seemed to go against Syzygy's assertion that DTZ=n means that zeroing happens at n or n+1 ply (it's not exact because of Syzygy's rounding of the units bit of the integers, which apparently saves a vast amount of data, and is allegedly "safe"). But James's solution has zeroing at n-1. If Black was indeed forced to zero prematurely in this way, this would show a bug in Syzygy.

In no sense is this a criticism of James: for whose talent, hard work & existence I am incredibly grateful. But it turns out there is a tiny issue in the solution he posted, just before DTZ=0.

Quarantine lockdown in Hong Kong has given me a chance to look at the position after W44, https://syzygy-tables.info/?fen=6R1/3nN3/8/5K2/2r2N1k/8/8/8_b_-_-_0_1, Black has two choices Tc5+ & Sf6. James chose the first for simplicity, as sensibly he always did, but 45. Sed5 threatens 46. Tg4#. The only way to avoid this 45. ... Txd5+ prematurely zeroing. But Syzygy makes no promises about DTM - it just happens that the random path that James found has mate as the zeroing move! This is not typical, so a better illustrative solution has 44. ... Sf6. With best play by both sides, I think this results in mate at W51.

How to compare this with the Lomonosov solution with mate at W59? How is it possible that Syzygy, with the extra constraint of avoiding 50M draw, has found a *shorter* mate? If you think about it, there is no issue here. Syzygy is only focusing on DTZ, and there are many points in the long solution at which there are choices of equal DTZ. What Lomonosov tells us is that Black can somehow force a Syzygy branch which gives #59 or longer. For 6 or 7 pieces, we just don't have the visibility in Syzygy to find that DTM path. That is the explanation. However for 5 pieces, Syzygy also gives us the DTM value, and as long as one has escaped 50M range, it makes case to switch to that as soon as 5 pieces are reached. (2022-08-05)
A.Buchanan: As problemists, what we would really like to know is the minimum number of moves to mate, while not falling foul of 50M. Even with both DTZ & DTM values known, this is not easy to find. Here is a toy example to show one difficulty: https://syzygy-tables.info/?fen=7k/8/7K/4P2R/8/8/8/8_b_-_-_0_1 Black's forced move leads to DTZ=1 DTM=3, but White must choose which he wants to achieve, as they are mutually exclusive. If the 50M clock is low, 1.e6 gives DTM=16. But otherwise 1.Rf5 gives DTM=2. Now White will know which he wants, but Black is usually free to decide whether to defend the DTZ or the DTM or potentially to delay the commitment. The tactical richness of a position is not captured by simply the DTZ & DTM values (2022-08-05)
A.Buchanan: OK it turns out that what I'd been arguing for already exists to some extent: it's called DTM50 - see http://galen.metapath.org/egtb50/ (2022-08-05)
comment
Keywords: Cant Castler, Aristocrat, 50 move rule, Miniature
Genre: Retro, Studies
Computer test: www.syzygy.com
FEN: r3k2n/8/8/4K3/8/8/8/2NNR3
Reprints: www.chess.com 17/10/2019
Input: A.Buchanan, 2019-10-25
Last update: A.Buchanan, 2022-08-05 more...
12 - P1371439
Thierry Le Gleuher
R546 The Problemist 27-6, p. 229, 11/2019
P1371439
(13+12)
Last 102 single moves?
Duplex
BTM: R: 1. Lc2-b1 Kg1-f1 2. Lb1-c2 Tf1-f2 3. Lc2-b1 Tf2-f3 4. Lb1-c2 Tf3-g3 5. Lc2-b1 Tg3-h3 6. Th3-h2 Th2-h1 7. Lb1-c2 Kh1-g1 8. Lc2-b1 Tg1-f1 9. Lb1-c2 Tf1-f2 10. Lc2-b1 Tf2-f3 11. Lb1-c2 Tf3-g3 12. Tg3-h3 Th3-h2 13. Lc2-b1 Kh2-h1 14. Lb1-c2 Th1-g1 15. Lc2-b1 Kg1-h2 16. Lb1-c2 Th2-h1 17. Lc2-b1 Kh1-g1 18. Lb1-c2 Tg1-f1 19. Lc2-b1 Tf1-f2 20. Lb1-c2 Tf2-f3 21. Tf3-g3 Tg3-h3 22. Lc2-b1 Th3-h2 23. Lb1-c2 Kh2-h1 24. Lc2-b1 Th1-g1 25. Lb1-c2 Kg1-h2 26. Lc2-b1 Th2-h1 27. Lb1-c2 Kh1-g1 28. Lc2-b1 Tg1-f1 29. Lb1-c2 Tf1-f2 30. Tf2-f3 Tf3-g3 31. Lc2-b1 Tg3-h3 32. Lb1-c2 Th3-h2 33. Lc2-b1 Kh2-h1 34. Lb1-c2 Th1-g1 35. Lc2-b1 Tg1-f1 36. Tf1-f2 Lf2-e1 37. Te1-f1 Tf1-g1 38. Lb1-c2 Tg1-h1 39. Lc2-b1 Kh1-h2 40. Lb1-c2 Th2-h3 41. Lc2-b1 Th3-g3 42. Lb1-c2 Lg3-f2 43. Lc2-b1 Tf2-f1 44. Lb1-c2 Tf1-g1 45. Lc2-b1 Kg1-h1 46. Lb1-c2 Th1-h2 47. Lc2-b1 Kh2-g1 48. Lb1-c2 Tg1-h1 49. Lc2-b1 Kh1-h2 50. Lb1-c2 Th2-h3 51. h3-h4 Lh4-g3 52. Lc2-b1
52. ... Tg3-f3 53. Lb1-c2 Tf3-f2 54. Lc2-b1 Tf2-f1 55. Lb1-c2 Tf1-g1 56. Lc2-b1 Kg1-h1 57. Lb1-c2 Th1-h2 58. h2-h3 Th3-g3 59. Lc2-b1 Lg3-h4 60. Lb1-c2 Th8-h3
play all play one stop play next play all
Editor's comment: Do not be put off by the length of R546: there is a repeated sequence of moves.
James Malcom: Solution? Also, is this an Othoreconstrution? (2023-05-30)
Henrik Juel: Here is the solution from the May 2020 issue (from www.theproblemist.org , which is a treasure of info, althoug I find it hard to navigate)
R546 (Le Gleuher): Retroanalysis: The wPs captured the 4 missing black units.
Thus, wPb4 comes from b2, and (Pa2) promoted to the extra wR on a8 with (Pa7)
letting it pass by switching to the b file (first capture of the 3 missing white units).
(Pb7) must have promoted to a rook on a1 or c1 (second capture). (Ph7) must
have moved onto the g file (third and final capture) in order to be captured. Thus
(Pc7) and (Pe7) promoted to rooks on the c and e files, respectively. It is not
possible to immediately take back a4xb3 or c2xd3. The retro-move e4xf5 is
forbidden since then the bP which promoted on e1 cannot retract to e5. Black can
only move in the southeast cage, while White’s bishop is free to oscillate between
c2 and b1.
Black to move: back 1.Bc2-b1 Kg1-f1 2.Bb1-c2 Rf1-f2 3.Bc2-b1 Rf2-f3
4.Bb1-c2 Rf3-g3 5.Bc2-b1 Rg3-h3 6.Rh3-h2 Rh2-h1 7.Bb1-c2 Kh1-g1 8.Bc2-b1
Rg1-f1 9.Bb1-c2 Rf1-f2 10.Bc2-b1 Rf2-f3 11.Bb1-c2 Rf3-g3 12.Rg3-h3 Rh3-h2
13.Bc2-b1 Kh2-h1 14.Bb1-c2 Rh1-g1 15.Bc2-b1 Kg1-h2 16.Bb1-c2 Rh2-h1
17.Bc2-b1 Kh1-g1 18.Bb1-c2 Rg1-f1 19.Bc2-b1 Rf1-f2 20.Bb1-c2 Rf2-f3
21.Rf3-g3 Rg3-h3 22.Bc2-b1 Rh3-h2 23.Bb1-c2 Kh2-h1 24.Bc2-b1 Rh1-g1
25.Bb1-c2 Kg1-h2 26.Bc2-b1 Rh2-h1 27.Bb1-c2 Kh1-g1 28.Bc2-b1 Rg1-f1
29.Bb1-c2 Rf1-f2 30.Rf2-f3 Rf3-g3 31.Bc2-b1 Rg3-h3 32.Bb1-c2 Rh3-h2
33.Bc2-b1 Kh2-h1 34.Bb1-c2 Rh1-g1 35.Bc2-b1 Rg1-f1 36.Rf1-f2 Bf2-e1
37.Re1-f1 Rf1-g1 38.Bb1-c2 Rg1-h1 39.Bc2-b1 Kh1-h2 40.Bb1-c2 Rh2-h3
41.Bc2-b1 Rh3-g3 42.Bb1-c2 Bg3-f2 43.Bc2-b1 Rf2-f1 44.Bb1-c2 Rf1-g1
45.Bc2-b1 Kg1-h1 46.Bb1-c2 Rh1-h2 47.Bc2-b1 Kh2-g1 48.Bb1-c2 Rg1-h1
49.Bc2-b1 Kh1-h2 50.Bb1-c2 Rh2-h3 51.h3-h4 Bh4-g3 52.Bc2-b1
then 52…Rg3-f3 (or 52...Bg3-h4 53.Bb1-c2 Bh4-g3…) 53.Bb1-c2 Rf3-f2 54.Bc2-b1 Rf2-f1 55.Bb1-c2
Rf1-g1 56.Bc2-b1 Kg1-h1 57.Bb1-c2 Rh1-h2 58.h2-h3 Rh3-g3 59.Bc2-b1 Bg3-h4 60.Bb1-c2 Rh8-h3 etc…
The diagram position is a draw and the last 103 single moves are determined.
White to move: back 1…Kg1-f1 …
51.h3-h4 Bh4-g3 52.Bc2-b1 then the same… Here, there are 99 single moves between the diagram position and the retro-move h3-h4. Of course, trying to insert another manoeuvre in the solution would lead to an overshoot of the 101 single moves, which would be incompatible with the 50 moves rule. Here only 102 last single moves are determined.
So, we have a D25/102, a new record for last single moves for ‘D’ type (Duplex) with 25 pieces (It is also
the overall record for Duplex). With black to move, the same diagram is also a record with 25 pieces (B25/103) (2023-05-30)
comment
Keywords: Last Moves?, Non-standard material, 50 move rule
Genre: Retro
FEN: 8/6p1/8/5PP1/1P3pRP/1p1PPrrr/n2pRrPR/1BNKbk1r
Input: A.Buchanan, 2020-01-05
Last update: Mario Richter, 2023-05-30 more...
13 - P1384294
James Malcom
MatPlus.net Forum 27/12/2020
nach Bosko Miloseski
P1384294
(11+12)
#51
1. ... Td7 2. Lb1! Kd6 3. De5+ Kc6 4. De6+ Td6 5. De8+ Td7 6. Ka2 Kd6 7. De5+ Kc6 8. De6+ Td6 9. De8+ Td7 10. Ka1 Kd6 11. De5+ Kc6 12. De6+ Td6 13. De8+ Td7 14. La2 Kd6 15. De5+ Kc6 16. De6+ Td6 17. De8+ Td7 18. Kb1 Kd6 19. De5+ Kc6 20. De6+ Td6 21. De8+ Td7 22. Kc1 Kd6 23. De5+ Kc6 24. De6+ Td6 25. De8+ Td7 26. Kd1 Kd6 27. De5+ Kc6 28. De6+ Td6 29. De8+ Td7 30. Ke1 Kd6 31. De5+ Kc6 32. De6+ Td6 33. De8+ Td7 34. Kf2 Kd6 35. De5+ Kc6 36. De6+ Td6 37. De8+ Td7 38. Kg3 Kd6 39. De5+ Kc6 40. De6+ Td6 41. De8+ Td7 42. Kh4 Kd6 43. De5+ Kc6 44. De6+ Td6 45. De8+ Td7 46. Kg5 Kd6 47. De5+ Kc6 48. De6+ Td6 49. De8+ Td7 50. Kxf5! Kd6 51. De6#

50. Kf6? Kd6! Drawn by the 50 move rule!
play all play one stop play next play all
White's pre-diagram move couldn't have been a capture, so the 50-move counter has already been set to 1. Thus, White must play 50. Kxf5 in order to reset the counter, or else Black will draw.

See P1372258

http://matplus.net/start.php?px=1609121267&app=forum&act=posts&fid=gen&tid=2590
Henrik Juel: 3 and 5 captures by white and black pawns are obvious, leaving [Ph2,Ph7], so the remaining white capture was hxg (with promotions on g8 and h1) or officerxPh (with promotion on h8) (2022-11-05)
comment
Keywords: 50 move rule
Genre: Retro, n#
FEN: n3Q3/1pp5/brkr4/Bp1p1p2/1P1PpP2/KPp1P3/B1P1P3/8
Input: James Malcom, 2020-12-28
Last update: A.Buchanan, 2022-11-05 more...
14 - P1386410
Andrew Buchanan
chess.stackexchange.com 11/02/2021
P1386410
(16+16)
Find a diagram which can recur a maximum number of times, such that it also has a unique KBP (shortest proof game) to its first occurrence.
a) with 3-fold repetition & 50-move rules
b) with 5-fold repetition & 75-move rules
a) e.g. 1. a4 e5 2. a5 e4 3. a6 Sxa6 4. f4
b) e.g. 1. e3 a5 2. La6 e6 3. Se2
play all play one stop play next play all
a) 22 occurrences possible, with 44.5 moves, so no issue with 50-move rule even if it were mandatory.
b) 37 occurrences possible, with 74.0 moves + 0.5 for 3. Se2 in proof game.
The theoretical maximum of 42 is impossible, because it would take 84.5 moves, breaching the mandatory 75-move rule.
Other shortest unique SPGs are possible, but need:
(1) both sides have at least 2 of K,R & R able to move.
(2) both sides must have officer to triangulate.
(3) (for player on the move) triangulating officer must be B or Q.
(4) (for 3+50 case only) e.p. set up.
A.Buchanan: Without German KBP in the stip the animation won't work (2021-02-12)
comment
Keywords: Construction task (non-unique), Synthetic problem, 50 move rule, 75 move rule, Draw by repetition, En passant, Castling, Homebase
Genre: Retro
FEN: rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR
Input: A.Buchanan, 2021-02-12
Last update: A.Buchanan, 2024-03-19 more...
15 - P1398775
Hobacle
lichess.org 29/01/2022
P1398775
(10+10)
=
1. Sd7+? Kxc7 2. Sxb8 Sg3!

1. Tc6+? Ka5!

1. Tb7+! Dxb7 2. Sxb7 Kxb7 3. Kg6! Kc7 4. Lb2! Lxb2 5. Kxh5 Kd7 6. Kh6! Ke7 7. Kg6 Lc3 8. Kg7 Le1 9. Kg6 Lg3 10. Kg7 Lh4 11. Kg8 Ke8 12. Kg7 Lg5 13. Kg8 Lh6 14. Kh7 Lf8 15. Kg8 Ke7 16. Kh8 Kf7 17. Kh7 Le7 18. Kh6 Ld8 19. Kh7 La5 20. Kh6 Lb6 21. Kh7 Ld8 22. Kh8 Le7 23. Kh7 Lf8 24. Kh8 Lg7+ 25. Kh7 Kf8 26. Kg6 Kg8 27. Kh5 Kh7 28. Kh4 Lh6 29. Kg4 Lg5 30. Kh5 Kg7 31. Kg4 Kh6 32. Kh3 Kh5 33. Kg2 Lh4 34. Kf1! Lg3 35. Kg1 Kh4 36. Kg2 Le1 37. Kf1! Ld2 38. Kg2 Lc3 39. Kh2 Le1 40. Kg2 Lg3 41. Kf1 Kh3 42. Kg1 Lh4 43. Kh1 Le1 44. Kg1 Kg3 45. Kf1 Lf2 46. Ke2 Kg2 47. Kd2 Kf1 48. Kc2 Ke2 49. Kc1 Le3+ 50. Kc2 Ke1 51. Kb1 Kd1 52. Kb2 Kd2 53. Kb1 Kc3 54. Ka2 Kc2 55. Ka1
play all play one stop play next play all
Source Of Publication: https://lichess.org/study/su7muCoV
AB: An amazing young composer...
James Malcom: Thanks for adding this in Andrew. I have updated it to include the solution. (2022-01-31)
comment
Keywords: 50 move rule
Genre: Studies
FEN: 1q6/2R4K/1k1p1p2/2NPpP1n/1p1pBp2/1PbP1P2/8/B7
Input: A.Buchanan, 2022-01-31
Last update: James Malcom, 2022-01-31 more...
16 - P1401711
Andrew Buchanan
OP008 The Hopper Magazine I01 24/12/2021
"Ak ja, retten, retten; hvad hjælper det, at du har retten, når du ikke har nogen magt?"
P1401711
(3+3) C+
h#2.5 (2 solutions)
Position after Black's 5,696th move
1. ... Se6 2. Th2 Ta8 3. Th7 Txe8#
1. ... Txa2 2. Lg6 Tg2 3. Lh7 Sf7#
not 1. ... 0-0-0 2. Lg6 Tg1 3. Lh7 Sf7#?? (game just ended by 50M)
play all play one stop play next play all
A full solution is really too big for PDB, but it is available at https://www.thehoppermagazine.com/AA084
Henrik Juel: solutions
1...Sg5-e6 2.Ra2-h2 Ra1-a8 3.Rh2-h7 Ra8*e8 #
1...Ra1*a2 2.Be8-g6 Ra2-g2 3.Bg6-h7 Sg5-f7 #
not 1...0-0-0? 2.Be8-g6 Rd1-g1 3.Bg6-h7 Sg5-f7 #
because White has lost his right to castle, as Andrew will explain, I hope... (2023-07-29)
A.Buchanan: White has not necessarily lost the right to castle, but if he can castle then the 50M rule triggers before the mate can be executed. It is pretty complicated, sorry. (2023-07-29)
A.Buchanan: We don't do mottoes much these days. Mrs Baird was a big fan. On page 301 of the July 1916 Chess Amateur P.H. Williams wrote:
"I think it was Mrs Baird who did more to search Shakespeare for accidental (or deliberate) reference to chess, since all her retractors had Shakespearean mottoes, and her knowledge of the Avonian bard was obviously extensive. Her example was followed by other composers of retractors, who considered such positions would be incomplete without some reference to the poet."
But there are other great playwrights, and Asteroid 5696 was named after one such, whose bitter quotation accurately states this problem's theme. (2023-07-31)
Henrik Juel: The citation is by danish philosopher Søren Kierkegaard (1813-1855) and means something like
Oh yes, to be right, to be right; it does not help much that you are right, when you have no power. (2023-07-31)
Henrik Juel: Asteroid 5696 is named Ibsen after the norwegian playwright Henrik Ibsen (1828-1906), who must have borrowed the citation from Søren Kierkegaard (2023-07-31)
A.Buchanan: Thanks for this Henrik - I resolved the question here: https://philosophy.stackexchange.com/questions/101229/did-ibsen-originate-this-statement
Beyond the castling try, the other innovation here is in the solution beginning 1...Se6. Castling convention and 50M I think are trying to combine by Retro Strategy (being optimistic about 50M rights like we are with castling, rather than pessimistic like we are with ep). (2023-08-04)
more ...
comment
Keywords: Aristocrat, Miniature, 50 move rule, Castling, Exchange of roles (T/S, Guard/Mate), Chumakov theme (l/t, simplified), Retro Strategy (RS), Model mate (2), Constrained problem
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 and analysis
FEN: 4b2k/8/8/6N1/8/8/r7/R3K3
Reprints: AA084 The Hopper Magazine I04 13/07/2023
Input: A.Buchanan, 2022-06-09
Last update: A.Buchanan, 2023-08-27 more...
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The problems of this query have been registered by the following contributors:

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