The doubt of the famous #226 problem

[Hobacle]

According to:

They claim the following is a long mate with a unique solution but I suspect the solution is wrong and there is a shorter mate in less than 226 moves.

[fen]1N2Q3/p6n/p5n1/p1p2rpb/p4k1r/K1P4p/2PP4/6B1 w - - 6 1 [/fen]

They give 56.Kc1 in the solution but 56.c4 may be stronger.
[fen]8/p2N3n/p3Q1n1/p1p2rpb/6kr/2P4p/1KPP3B/8 w - - 23 56 [/fen]

after 56.c4 black has only 2 options not to lose by a fast checkmate:
56...a4 and 56...Kf3

After 56.c4 a4 57.c3 or 56.c4 Kf3 57.Qe3+ I could get by stockfish to see more than +6 by some backward analysis of not many moves so it seems that the solution is not unique and probably shorter than 226 moves even if stockfish did not see a mate score.

Then I suspect 19.c4 in this composition wins in less than 226-18=208 moves.
[pgn][Event ""]
[Site ""]
[Date "2022.02.06"]
[Round ""]
[White ""]
[Black ""]
[Result "*"]
[FEN "1N2Q3/p6n/p5n1/p1p2rpb/p4k1r/K1P4p/2PP4/6B1 w - - 0 1"]
[SetUp "1"]

1. Bh2+ Kf3 2. Qe3+ Kg2 3. Qg1+ Kf3 4. Qf1+ Kg4 5. Qe2+ Rf3 6. Qe6+ Rf5 7. Nd7
Kf3 8. Qe3+ Kg2 9. Qg1+ Kf3 10. Qf1+ Kg4 11. Qe2+ Rf3 12. Qe6+ Rf5 13. Kb2 Kf3
14. Qe3+ Kg2 15. Qg1+ Kf3 16. Qf1+ Kg4 17. Qe2+ Rf3 18. Qe6+ Rf5 19.c4 *
[/pgn]

Why?

The full pgn file of the #226:
[pgn][Event ""]
[Site ""]
[Date "2022.02.06"]
[Round ""]
[White ""]
[Black ""]
[Result "1-0"]
[FEN "1N2Q3/p6n/p5n1/p1p2rpb/p4k1r/K1P4p/2PP4/6B1 w - - 0 1"]
[SetUp "1"]

1. Bh2+ Kf3 2. Qe3+ Kg2 3. Qg1+ Kf3 4. Qf1+ Kg4 5. Qe2+ Rf3 6. Qe6+ Rf5 7. Nd7
Kf3 8. Qe3+ Kg2 9. Qg1+ Kf3 10. Qf1+ Kg4 11. Qe2+ Rf3 12. Qe6+ Rf5 13. Kb2 Kf3
14. Qe3+ Kg2 15. Qg1+ Kf3 16. Qf1+ Kg4 17. Qe2+ Rf3 18. Qe6+ Rf5 19. Kc1 Kf3
20. Qe3+ Kg2 21. Qg1+ Kf3 22. Qf1+ Kg4 23. Qe2+ Rf3 24. Qe6+ Rf5 25. Kd1 a3 26.
Kc1 Kf3 27. Qe3+ Kg2 28. Qg1+ Kf3 29. Qf1+ Kg4 30. Qe2+ Rf3 31. Qe6+ Rf5 32.
Kb1 Kf3 33. Qe3+ Kg2 34. Qg1+ Kf3 35. Qf1+ Kg4 36. Qe2+ Rf3 37. Qe6+ Rf5 38.
Ka2 Kf3 39. Qe3+ Kg2 40. Qg1+ Kf3 41. Qf1+ Kg4 42. Qe2+ Rf3 43. Qe6+ Rf5 44.
Kxa3 Kf3 45. Qe3+ Kg2 46. Qg1+ Kf3 47. Qf1+ Kg4 48. Qe2+ Rf3 49. Qe6+ Rf5 50.
Kb2 Kf3 51. Qe3+ Kg2 52. Qg1+ Kf3 53. Qf1+ Kg4 54. Qe2+ Rf3 55. Qe6+ Rf5 56.
Kc1 Kf3 57. Qe3+ Kg2 58. Qg1+ Kf3 59. Qf1+ Kg4 60. Qe2+ Rf3 61. Qe6+ Rf5 62.
Kd1 a4 63. Kc1 Kf3 64. Qe3+ Kg2 65. Qg1+ Kf3 66. Qf1+ Kg4 67. Qe2+ Rf3 68. Qe6+
Rf5 69. Kd1 a5 70. Kc1 Kf3 71. Qe3+ Kg2 72. Qg1+ Kf3 73. Qf1+ Kg4 74. Qe2+ Rf3
75. Qe6+ Rf5 76. Kd1 a6 77. Kc1 Kf3 78. Qe3+ Kg2 79. Qg1+ Kf3 80. Qf1+ Kg4 81.
Qe2+ Rf3 82. Qe6+ Rf5 83. Kd1 a3 84. Kc1 Kf3 85. Qe3+ Kg2 86. Qg1+ Kf3 87. Qf1+
Kg4 88. Qe2+ Rf3 89. Qe6+ Rf5 90. Kb1 Kf3 91. Qe3+ Kg2 92. Qg1+ Kf3 93. Qf1+
Kg4 94. Qe2+ Rf3 95. Qe6+ Rf5 96. Ka2 Kf3 97. Qe3+ Kg2 98. Qg1+ Kf3 99. Qf1+
Kg4 100. Qe2+ Rf3 101. Qe6+ Rf5 102. Kxa3 Kf3 103. Qe3+ Kg2 104. Qg1+ Kf3 105.
Qf1+ Kg4 106. Qe2+ Rf3 107. Qe6+ Rf5 108. Kb2 Kf3 109. Qe3+ Kg2 110. Qg1+ Kf3
111. Qf1+ Kg4 112. Qe2+ Rf3 113. Qe6+ Rf5 114. Kc1 Kf3 115. Qe3+ Kg2 116. Qg1+
Kf3 117. Qf1+ Kg4 118. Qe2+ Rf3 119. Qe6+ Rf5 120. Kd1 a4 121. Kc1 Kf3 122.
Qe3+ Kg2 123. Qg1+ Kf3 124. Qf1+ Kg4 125. Qe2+ Rf3 126. Qe6+ Rf5 127. Kd1 a5
128. Kc1 Kf3 129. Qe3+ Kg4 130. Qg1+ Kf3 131. Qf1+ Kg4 132. Qe2+ Rf3 133. Qe6+
Rf5 134. Kd1 a3 135. Kc1 Kf3 136. Qe3+ Kg2 137. Qg1+ Kf3 138. Qf1+ Kg4 139.
Qe2+ Rf3 140. Qe6+ Rf5 141. Kb1 Kf3 142. Qe3+ Kg2 143. Qg1+ Kf3 144. Qf1+ Kg4
145. Qe2+ Rf3 146. Qe6+ Rf5 147. Ka2 Kf3 148. Qe3+ Kg2 149. Qg1+ Kf3 150. Qf1+
Kg4 151. Qe2+ Rf3 152. Qe6+ Rf5 153. Kxa3 Kf3 154. Qe3+ Kg2 155. Qg1+ Kf3 156.
Qf1+ Kg4 157. Qe2+ Rf3 158. Qe6+ Rf5 159. Kb2 Kf3 160. Qe3+ Kg2 161. Qg1+ Kf3
162. Qf1+ Kg4 163. Qe2+ Rf3 164. Qe6+ Rf5 165. Kc1 Kf3 166. Qe3+ Kg2 167. Qg1+
Kf3 168. Qf1+ Kg4 169. Qe2+ Rf3 170. Qe6+ Rf5 171. Kd1 a4 172. Kc1 Kf3 173.
Qe3+ Kg2 174. Qg1+ Kf3 175. Qf1+ Kg4 176. Qe2+ Rf3 177. Qe6+ Rf5 178. Kd1 c4
179. Kc1 Kf3 180. Qe3+ Kg2 181. Qg1+ Kf3 182. Qf1+ Kg4 183. Qe2+ Rf3 184. Qe6+
Rf5 185. Kd1 a3 186. Kc1 Kf3 187. Qe3+ Kg2 188. Qg1+ Kf3 189. Qf1+ Kg4 190.
Qe2+ Rf3 191. Qe6+ Rf5 192. Kb1 Kf3 193. Qe3+ Kg2 194. Qg1+ Kf3 195. Qf1+ Kg4
196. Qe2+ Rf3 197. Qe6+ Rf5 198. Ka2 Kf3 199. Qe3+ Kg2 200. Qg1+ Kf3 201. Qf1+
Kg4 202. Qe2+ Rf3 203. Qe6+ Rf5 204. Kxa3 Kf3 205. Qe3+ Kg2 206. Qg1+ Kf3 207.
Qf1+ Kg4 208. Qe2+ Rf3 209. Qe6+ Rf5 210. Kb2 Kf3 211. Qe3+ Kg2 212. Qg1+ Kf3
213. Qf1+ Ke4 214. Qxc4+ Kf3 215. Qf1+ Kg4 216. Qe2+ Rf3 217. Qe6+ Rf5 218. Kc1
Kf3 219. Qe3+ Kg2 220. Qg1+ Kf3 221. Qf1+ Kg4 222. Qe2+ Rf3 223. Qe6+ Rf5 224.
Kd1 Nhf8 225. Nf6+ Kf3 226. Qe2# 1-0
[/pgn]

We see that in the range, when Black plays Kf3, White cannot capture the f5 rook because Black has Kg2 to get the h2 square, then the h pawn will inevitably promote or cause problems since Black's king is let loose.
But after c4, c3 and d4, everything changes. After Kf3 Qxf5+ Kg2, White has Qc2+! , then White has enough advantage to win. A possible line:
[pgn][Event ""]
[Site ""]
[Date "2022.02.06"]
[Round ""]
[White ""]
[Black ""]
[Result "*"]
[FEN "1N2Q3/p6n/p5n1/p1p2rpb/p4k1r/K1P4p/2PP4/6B1 w - - 0 1"]
[SetUp "1"]

1. Bh2+ Kf3 2. Qe3+ Kg2 3. Qg1+ Kf3 4. Qf1+ Kg4 5. Qe2+ Rf3 6. Qe6+ Rf5 7. Nd7
Kf3 8. Qe3+ Kg2 9. Qg1+ Kf3 10. Qf1+ Kg4 11. Qe2+ Rf3 12. Qe6+ Rf5 13. Kb2 Kf3
14. Qe3+ Kg2 15. Qg1+ Kf3 16. Qf1+ Kg4 17. Qe2+ Rf3 18. Qe6+ Rf5 19. c4 a3+ 20.
Ka2 Kf3 21. Qe3+ Kg2 22. Qg1+ Kf3 23. Qf1+ Kg4 24. Qe2+ Rf3 25. Qe6+ Rf5 26. c3
Kf3 27. Qe3+ Kg2 28. Qg1+ Kf3 29. Qf1+ Kg4 30. Qe2+ Rf3 31. Qe6+ Rf5 32. d4
cxd4 33. cxd4 Kf3 34. Qxf5+ *
[/pgn]



If 19.c4 really works, then it can easily cause a shorter mate, then this famous composition (the longest non-dual moremover over the world) will be cooked. Does 19.c4 really work? Can it lead a mate in shorter than 226-18=208 moves?

[peter]

If 19.c4 really works, then it can easily cause a shorter mate, then this famous composition (the longest non-dual moremover over the world) will be cooked. Does 19.c4 really work? Can it lead a mate in shorter than 226-18=208 moves?

I don't see why 19.c4 shouldn't work, but I wouldn't call the study cooked because of that. 208 is still rather nice, isn't it?

[Hobacle]

If 19.c4 really works, then it can easily cause a shorter mate, then this famous composition (the longest non-dual moremover over the world) will be cooked. Does 19.c4 really work? Can it lead a mate in shorter than 226-18=208 moves?

I don't see why 19.c4 shouldn't work, but I wouldn't call the study cooked because of that. 208 is still rather nice, isn't it?

So we should prove the win after 19.c4 is far shorter than 208 moves.

[peter]

If 19.c4 really works, then it can easily cause a shorter mate, then this famous composition (the longest non-dual moremover over the world) will be cooked. Does 19.c4 really work? Can it lead a mate in shorter than 226-18=208 moves?

I don't see why 19.c4 shouldn't work, but I wouldn't call the study cooked because of that. 208 is still rather nice, isn't it?

So we should prove the win after 19.c4 is far shorter than 208 moves.

Somethiong tells me, you have "proved" that on your own already since quite a while and just don't want to rely on it without engines' confirmation.
But that's something, you cannot really expect with such DsTM, at least I won't be the one trying on and on to come to that conclusion in the end again.

At the end of your line after 19.c4 I end up with about 70+, if that helps you in any way, here's one proposal, you can keep the #-score of in hash backward to about move nr. 45 rather easily, at 50th you have the 7some position reached regards

[pgn][Event "226 to 208 19.c4"]
[Site "?"]
[Date "2022.02.06"]
[Round "?"]
[White "Hobacle's , Shortcut"]
[Black "WorldRecord, MateLength"]
[Result "1-0"]
[SetUp "1"]
[FEN "1N2Q3/p6n/p5n1/p1p2rpb/p4k1r/K1P4p/2PP4/6B1 w - - 0 1"]
[PlyCount "147"]

1. Bh2+ Kf3 2. Qe3+ Kg2 3. Qg1+ Kf3 4. Qf1+ Kg4 5. Qe2+ Rf3 6. Qe6+ Rf5 7. Nd7
Kf3 8. Qe3+ Kg2 9. Qg1+ Kf3 10. Qf1+ Kg4 11. Qe2+ Rf3 12. Qe6+ Rf5 13. Kb2 Kf3
14. Qe3+ Kg2 15. Qg1+ Kf3 16. Qf1+ Kg4 17. Qe2+ Rf3 18. Qe6+ Rf5 19. c4 a3+ 20.
Ka2 Kf3 21. Qe3+ Kg2 22. Qg1+ Kf3 23. Qf1+ Kg4 24. Qe2+ Rf3 25. Qe6+ Rf5 26. c3
Kf3 27. Qe3+ Kg2 28. Qg1+ Kf3 29. Qf1+ Kg4 30. Qe2+ Rf3 31. Qe6+ Rf5 32. d4
cxd4 33. cxd4 Kf3 34. Qxf5+ Nf4 35. Qb1 Ne2 36. Qf1+ Ke3 37. Ne5 Kd2 38. d5 Re4
39. d6 Nc1+ 40. Kxa3 Re3+ 41. Ka4 Be8+ 42. d7 Rxe5 43. Qf2+ Ne2 44. Bxe5 Bxd7+
45. Kxa5 Be6 46. Bc3+ Kxc3 47. Qe3+ Kxc4 48. Qxe6+ Kd3 49. Qxh3+ Ke4 50. Qxh7+
Kf3 51. Qd3+ Kf2 52. Kxa6 g4 53. Qf5+ Ke3 54. Qxg4 Nd4 55. Kxa7 Nc6+ 56. Kb6
Ne5 57. Qh4 Nd3 58. Qc4 Ne5 59. Qc3+ Nd3 60. Kc6 Ke4 61. Kd6 Nf4 62. Qe5+ Kf3
63. Qd4 Nh3 64. Ke5 Nf2 65. Qf4+ Ke2 66. Kd4 Nd1 67. Qg3 Kd2 68. Qh2+ Kc1 69.
Qa2 Nb2 70. Kc3 Nd1+ 71. Kb3 Nb2 72. Qxb2+ Kd1 73. Qf2 Kc1 74. Qe1# 1-0
[/pgn]

[Hobacle]

So we should prove the win after 19.c4 is far shorter than 208 moves.

Somethiong tells me, you have "proved" that on your own already since quite a while and just don't want to rely on it without engines' confirmation.
But that's something, you cannot really expect with such DsTM, at least I won't be the one trying on and on to come to that conclusion in the end again.

At the end of your line after 19.c4 I end up with about 70+, if that helps you in any way, here's one proposal, you can keep the #-score of in hash backward to about move nr. 45 rather easily

[pgn][Event "226 to 208 19.c4"]
[Site "?"]
[Date "2022.02.06"]
[Round "?"]
[White "Hobacle's , Shortcut"]
[Black "WorldRecord, MateLength"]
[Result "1-0"]
[SetUp "1"]
[FEN "1N2Q3/p6n/p5n1/p1p2rpb/p4k1r/K1P4p/2PP4/6B1 w - - 0 1"]
[PlyCount "147"]

1. Bh2+ Kf3 2. Qe3+ Kg2 3. Qg1+ Kf3 4. Qf1+ Kg4 5. Qe2+ Rf3 6. Qe6+ Rf5 7. Nd7
Kf3 8. Qe3+ Kg2 9. Qg1+ Kf3 10. Qf1+ Kg4 11. Qe2+ Rf3 12. Qe6+ Rf5 13. Kb2 Kf3
14. Qe3+ Kg2 15. Qg1+ Kf3 16. Qf1+ Kg4 17. Qe2+ Rf3 18. Qe6+ Rf5 19. c4 a3+ 20.
Ka2 Kf3 21. Qe3+ Kg2 22. Qg1+ Kf3 23. Qf1+ Kg4 24. Qe2+ Rf3 25. Qe6+ Rf5 26. c3
Kf3 27. Qe3+ Kg2 28. Qg1+ Kf3 29. Qf1+ Kg4 30. Qe2+ Rf3 31. Qe6+ Rf5 32. d4
cxd4 33. cxd4 Kf3 34. Qxf5+ Nf4 35. Qb1 Ne2 36. Qf1+ Ke3 37. Ne5 Kd2 38. d5 Re4
39. d6 Nc1+ 40. Kxa3 Re3+ 41. Ka4 Be8+ 42. d7 Rxe5 43. Qf2+ Ne2 44. Bxe5 Bxd7+
45. Kxa5 Be6 46. Bc3+ Kxc3 47. Qe3+ Kxc4 48. Qxe6+ Kd3 49. Qxh3+ Ke4 50. Qxh7+
Kf3 51. Qd3+ Kf2 52. Kxa6 g4 53. Qf5+ Ke3 54. Qxg4 Nd4 55. Kxa7 Nc6+ 56. Kb6
Ne5 57. Qh4 Nd3 58. Qc4 Ne5 59. Qc3+ Nd3 60. Kc6 Ke4 61. Kd6 Nf4 62. Qe5+ Kf3
63. Qd4 Nh3 64. Ke5 Nf2 65. Qf4+ Ke2 66. Kd4 Nd1 67. Qg3 Kd2 68. Qh2+ Kc1 69.
Qa2 Nb2 70. Kc3 Nd1+ 71. Kb3 Nb2 72. Qxb2+ Kd1 73. Qf2 Kc1 74. Qe1# 1-0
[/pgn]

This time I am innocent... A composer told me that engines' scores cannot be the proof of the cook. We must find the shortest winning line and prove it. Humans cannot do it because the position is too complex. Only engines can.

[peter]

So we should prove the win after 19.c4 is far shorter than 208 moves.

Somethiong tells me, you have "proved" that on your own already since quite a while and just don't want to rely on it without engines' confirmation.
But that's something, you cannot really expect with such DsTM, at least I won't be the one trying on and on to come to that conclusion in the end again.

At the end of your line after 19.c4 I end up with about 70+, if that helps you in any way, here's one proposal, you can keep the #-score of in hash backward to about move nr. 45 rather easily

[pgn][Event "226 to 208 19.c4"]
[Site "?"]
[Date "2022.02.06"]
[Round "?"]
[White "Hobacle's , Shortcut"]
[Black "WorldRecord, MateLength"]
[Result "1-0"]
[SetUp "1"]
[FEN "1N2Q3/p6n/p5n1/p1p2rpb/p4k1r/K1P4p/2PP4/6B1 w - - 0 1"]
[PlyCount "147"]

1. Bh2+ Kf3 2. Qe3+ Kg2 3. Qg1+ Kf3 4. Qf1+ Kg4 5. Qe2+ Rf3 6. Qe6+ Rf5 7. Nd7
Kf3 8. Qe3+ Kg2 9. Qg1+ Kf3 10. Qf1+ Kg4 11. Qe2+ Rf3 12. Qe6+ Rf5 13. Kb2 Kf3
14. Qe3+ Kg2 15. Qg1+ Kf3 16. Qf1+ Kg4 17. Qe2+ Rf3 18. Qe6+ Rf5 19. c4 a3+ 20.
Ka2 Kf3 21. Qe3+ Kg2 22. Qg1+ Kf3 23. Qf1+ Kg4 24. Qe2+ Rf3 25. Qe6+ Rf5 26. c3
Kf3 27. Qe3+ Kg2 28. Qg1+ Kf3 29. Qf1+ Kg4 30. Qe2+ Rf3 31. Qe6+ Rf5 32. d4
cxd4 33. cxd4 Kf3 34. Qxf5+ Nf4 35. Qb1 Ne2 36. Qf1+ Ke3 37. Ne5 Kd2 38. d5 Re4
39. d6 Nc1+ 40. Kxa3 Re3+ 41. Ka4 Be8+ 42. d7 Rxe5 43. Qf2+ Ne2 44. Bxe5 Bxd7+
45. Kxa5 Be6 46. Bc3+ Kxc3 47. Qe3+ Kxc4 48. Qxe6+ Kd3 49. Qxh3+ Ke4 50. Qxh7+
Kf3 51. Qd3+ Kf2 52. Kxa6 g4 53. Qf5+ Ke3 54. Qxg4 Nd4 55. Kxa7 Nc6+ 56. Kb6
Ne5 57. Qh4 Nd3 58. Qc4 Ne5 59. Qc3+ Nd3 60. Kc6 Ke4 61. Kd6 Nf4 62. Qe5+ Kf3
63. Qd4 Nh3 64. Ke5 Nf2 65. Qf4+ Ke2 66. Kd4 Nd1 67. Qg3 Kd2 68. Qh2+ Kc1 69.
Qa2 Nb2 70. Kc3 Nd1+ 71. Kb3 Nb2 72. Qxb2+ Kd1 73. Qf2 Kc1 74. Qe1# 1-0
[/pgn]

This time I am innocent... A composer told me that engines' scores cannot be the proof of the cook. We must find the shortest winning line and prove it. Humans cannot do it because the position is too complex. Only engines can.

Maybe somebody with the according 7some postions stored on disk, edited my previous posting as for that after your answer, look up at move nr.50, I don't have Q against N+3P on disk, so further Backward with tbs loaded won't help from my side, sorry, I'm outa here again regards

[Hobacle]

Maybe somebody with the according 7some postions stored on disk, edited my previous posting as for that after your answer, look up at move nr.50, I don't have Q against N+3P on disk regards

Okay thanks, since this record is one of the most important records in chess composing.

[Ajedrecista]

Hello Peter:

If 19.c4 really works, then it can easily cause a shorter mate, then this famous composition (the longest non-dual moremover over the world) will be cooked. Does 19.c4 really work? Can it lead a mate in shorter than 226-18=208 moves?

I don't see why 19.c4 shouldn't work, but I wouldn't call the study cooked because of that. 208 is still rather nice, isn't it?

So we should prove the win after 19.c4 is far shorter than 208 moves.

Somethiong tells me, you have "proved" that on your own already since quite a while and just don't want to rely on it without engines' confirmation.
But that's something, you cannot really expect with such DsTM, at least I won't be the one trying on and on to come to that conclusion in the end again.

At the end of your line after 19.c4 I end up with about 70+, if that helps you in any way, here's one proposal, you can keep the #-score of in hash backward to about move nr. 45 rather easily, at 50th you have the 7some position reached regards

[pgn][Event "226 to 208 19.c4"]
[Site "?"]
[Date "2022.02.06"]
[Round "?"]
[White "Hobacle's , Shortcut"]
[Black "WorldRecord, MateLength"]
[Result "1-0"]
[SetUp "1"]
[FEN "1N2Q3/p6n/p5n1/p1p2rpb/p4k1r/K1P4p/2PP4/6B1 w - - 0 1"]
[PlyCount "147"]

1. Bh2+ Kf3 2. Qe3+ Kg2 3. Qg1+ Kf3 4. Qf1+ Kg4 5. Qe2+ Rf3 6. Qe6+ Rf5 7. Nd7
Kf3 8. Qe3+ Kg2 9. Qg1+ Kf3 10. Qf1+ Kg4 11. Qe2+ Rf3 12. Qe6+ Rf5 13. Kb2 Kf3
14. Qe3+ Kg2 15. Qg1+ Kf3 16. Qf1+ Kg4 17. Qe2+ Rf3 18. Qe6+ Rf5 19. c4 a3+ 20.
Ka2 Kf3 21. Qe3+ Kg2 22. Qg1+ Kf3 23. Qf1+ Kg4 24. Qe2+ Rf3 25. Qe6+ Rf5 26. c3
Kf3 27. Qe3+ Kg2 28. Qg1+ Kf3 29. Qf1+ Kg4 30. Qe2+ Rf3 31. Qe6+ Rf5 32. d4
cxd4 33. cxd4 Kf3 34. Qxf5+ Nf4 35. Qb1 Ne2 36. Qf1+ Ke3 37. Ne5 Kd2 38. d5 Re4
39. d6 Nc1+ 40. Kxa3 Re3+ 41. Ka4 Be8+ 42. d7 Rxe5 43. Qf2+ Ne2 44. Bxe5 Bxd7+
45. Kxa5 Be6 46. Bc3+ Kxc3 47. Qe3+ Kxc4 48. Qxe6+ Kd3 49. Qxh3+ Ke4 50. Qxh7+
Kf3 51. Qd3+ Kf2 52. Kxa6 g4 53. Qf5+ Ke3 54. Qxg4 Nd4 55. Kxa7 Nc6+ 56. Kb6
Ne5 57. Qh4 Nd3 58. Qc4 Ne5 59. Qc3+ Nd3 60. Kc6 Ke4 61. Kd6 Nf4 62. Qe5+ Kf3
63. Qd4 Nh3 64. Ke5 Nf2 65. Qf4+ Ke2 66. Kd4 Nd1 67. Qg3 Kd2 68. Qh2+ Kc1 69.
Qa2 Nb2 70. Kc3 Nd1+ 71. Kb3 Nb2 72. Qxb2+ Kd1 73. Qf2 Kc1 74. Qe1# 1-0
[/pgn]

I found three suboptimal moves with KQkn EGTB hosted at Shredder website:

------------------------

After 57.- ..., Nd3:

[d]8/8/1K6/8/7Q/3nk3/8/8 w - - 5 58

White plays 58.- Qc4 (#17 according to EGTB), while 58.- Qa4 is #16.

-----------------------

After 63.- Qd4:

[d]8/8/3K4/8/3Q1n2/5k2/8/8 b - - 16 63

Black plays 63.- ..., Nh3 (#10 according to EGTB), while 63.- ..., Ne2 is #11.

-----------------------

After 64.- ..., Nf2:

[d]8/8/8/4K3/3Q4/5k2/5n2/8 w - - 19 65

White plays 65.- Qf4 (#10 according to EGTB), while 65.- Qc3 is #9.

-----------------------

So, 58th and 64th white moves delay the checkmate one move each while 63th black move shortens the checkmate one move. Once your line reaches 6-man EGTB with 52.- Kxa6, the optimal line should end at 73th move, not at 74th.

I realised that this thread is a spin-off of this one and there were doubts of which position was the original. According to YACPDB (Yet Another Chess Problem Database):

https://yacpdb.org/#369143

This problem was published at 'Thema Danicum'. I found the exact issue were it was published, at page 36 of Nr. 26 April 1982 Nr. 2 Vol. 4, problem No. 1921:

Thema Danicum Nr. 26, Vol. 4-2 (1982)

Code: Select all

Walther Jørgensen
1921.              København (DK)
·———·———·———·———·———·———·———·———·
│   │░N░│   │░░░│ Q │░░░│   │░░░│
│———+———+———+———+———+———+———+———│
│░p░│   │░░░│   │░░░│   │░░░│ n │
│———+———+———+———+———+———+———+———│
│ p │░░░│   │░░░│   │░░░│ n │░░░│
│———+———+———+———+———+———+———+———│
│░p░│   │░p░│   │░░░│ r │░p░│ b │
│———+———+———+———+———+———+———+———│
│ p │░░░│   │░░░│   │░k░│   │░r░│
│———+———+———+———+———+———+———+———│
│░K░│   │░P░│   │░░░│   │░░░│ p │
│———+———+———+———+———+———+———+———│
│   │░░░│ P │░P░│   │░░░│   │░░░│
│———+———+———+———+———+———+———+———│
│░░░│   │░░░│   │░░░│   │░B░│   │
·———·———·———·———·———·———·———·———·
226≠                         7+13

The original intended solution can be found at page 88 of Nr. 28 October 1982 Nr. 4 Vol. 4, with text in Danish language:

Thema Danicum Nr. 28, Vol. 4-4 (1982)

1921. Walther Jørgensen:
1 Lh2+ Kf3 2 De3+ Kg2 3 Dg1+ Kf3
4 Df1+ Kg4 5 De2+ Tf3 6 De6+ Tf5
7 Sd7. Trækker sort nu med Sh7 er
der mat i få træk. Derfor 7 - Kf3.
Ved stadig gentagelse af 5-træks
tempomanøvren 8 De3+ Kg2 9 Dg1+ Kf3
10 Df1+ Kg4 11 De2+ Tf3 12 De6+ Tf5
spiller hvid kongen til d1, hvorved
sort tvinges til bondetræk. Sort
holder længst ud, når a-bønderne
efterhånden bortslås på a3. Trækket
c5-c4 (som må udføres, før den sid-
ste bonde betræder a3) forhaler med
yderligere to træk, idet hvid kan
tvinges til at slå bonden. Da De6xc4+
strander på Sf4! må det ske fra f1
når sort i tempomanøvren besvarer
Df1+med Ke4. (Sort må vente med Ke4,
indtil hvid konge har forladt dæk-
ningen af d2, ellers d3+!). Ved bed-
ste modspil kan sort først tvinges
til at trække med Sh7 efter 224 Kd1,
og så følger mat i to træk.
Denne "søslange" er en forbedring
af en 227≠ (THE PROBLEMIST 1981)
som har illegal stilling. Den må
være rekord for de opgaver, som
er bygget over WJ's eget geniale
skema fra DIE SCHWALBE 1976 (se
SKAKBLADET nr. 10109, sept. 1978,
og de udførlige artikler af C.J.
Morse: "The longest Problem" i
THE PROBLEMIST nov. 1979 og nov.
1981). Max. 4 p. til alle løsere,
som fandt tempomanøvren.

A very imperfect translation with the help of Google translator says:

1921. Walther Jørgensen:
1 Bh2+ Kf3 2 Qe3+ Kg2 3 Qg1+ Kf3 4 Qf1+ Kg4 5 Qe2+ Rf3 6 Qe6+ Rf5 7 Nd7. If black moves his kinght on Nh7 now, there is mate in a few moves. Therefore 7 - Kf3. By constantly repeating the 5-move tempo maneuver 8 Qe3+ Kg2 9 Qg1+ Kf3 10 Qf1+ Kg4 11 Qe2+ Rf3 12 Qe6+ Rf5 white plays the king to d1, forcing black to move pawns. Black lasts the longest when the a-pawns are eventually captured on a3. The move c5-c4 (which must be performed before the last pawn reaches a3) is delayed by two more moves, as White may be forced to capture the pawn. Since Qe6xc4+ is refuted with Nf4! it must happen from f1 when black in the tempo maneuver answers Df1+ with Ke4. (Black must wait with Ke4 until white king has left the cover of d2, otherwise d3+!). At best opposition, black can first be forced to move Nh7 after 224 Kd1, and then mate follows in two moves.
This 'sea snake' is an improvement of a #227 (THE PROBLEMIST 1981) which has an illegal position. It must be a record for the assignments built on WJ's own ingenious scheme from DIE SCHWALBE 1976 (see SKAKBLADET No. 10109, Sept. 1978, and the detailed articles by C.J. Morse: 'The longest Problem' in THE PROBLEMIST Nov. 1979 and Nov. 1981). Max. 4 p. to all solvers who found the tempo maneuver.

Regards from Spain.

Ajedrecista.

[peter]

I realised that this thread is a spin-off of this one and there were doubts of which position was the original. According to YACPDB (Yet Another Chess Problem Database):

https://yacpdb.org/#369143

This problem was published at 'Thema Danicum'. I found the exact issue were it was published, at page 36 of Nr. 26 April 1982 Nr. 2 Vol. 4, problem No. 1921:

Thema Danicum Nr. 26, Vol. 4-2 (1982)

Code: Select all

Walther Jørgensen
1921.              København (DK)
·———·———·———·———·———·———·———·———·
│   │░N░│   │░░░│ Q │░░░│   │░░░│
│———+———+———+———+———+———+———+———│
│░p░│   │░░░│   │░░░│   │░░░│ n │
│———+———+———+———+———+———+———+———│
│ p │░░░│   │░░░│   │░░░│ n │░░░│
│———+———+———+———+———+———+———+———│
│░p░│   │░p░│   │░░░│ r │░p░│ b │
│———+———+———+———+———+———+———+———│
│ p │░░░│   │░░░│   │░k░│   │░r░│
│———+———+———+———+———+———+———+———│
│░K░│   │░P░│   │░░░│   │░░░│ p │
│———+———+———+———+———+———+———+———│
│   │░░░│ P │░P░│   │░░░│   │░░░│
│———+———+———+———+———+———+———+———│
│░░░│   │░░░│   │░░░│   │░B░│   │
·———·———·———·———·———·———·———·———·
226≠                         7+13

The original intended solution can be found at page 88 of Nr. 28 October 1982 Nr. 4 Vol. 4, with text in Danish language:

Thema Danicum Nr. 28, Vol. 4-4 (1982)

1921. Walther Jørgensen:
1 Lh2+ Kf3 2 De3+ Kg2 3 Dg1+ Kf3
4 Df1+ Kg4 5 De2+ Tf3 6 De6+ Tf5
7 Sd7. Trækker sort nu med Sh7 er
der mat i få træk. Derfor 7 - Kf3.
Ved stadig gentagelse af 5-træks
tempomanøvren 8 De3+ Kg2 9 Dg1+ Kf3
10 Df1+ Kg4 11 De2+ Tf3 12 De6+ Tf5
spiller hvid kongen til d1, hvorved
sort tvinges til bondetræk. Sort
holder længst ud, når a-bønderne
efterhånden bortslås på a3. Trækket
c5-c4 (som må udføres, før den sid-
ste bonde betræder a3) forhaler med
yderligere to træk, idet hvid kan
tvinges til at slå bonden. Da De6xc4+
strander på Sf4! må det ske fra f1
når sort i tempomanøvren besvarer
Df1+med Ke4. (Sort må vente med Ke4,
indtil hvid konge har forladt dæk-
ningen af d2, ellers d3+!). Ved bed-
ste modspil kan sort først tvinges
til at trække med Sh7 efter 224 Kd1,
og så følger mat i to træk.
Denne "søslange" er en forbedring
af en 227≠ (THE PROBLEMIST 1981)
som har illegal stilling. Den må
være rekord for de opgaver, som
er bygget over WJ's eget geniale
skema fra DIE SCHWALBE 1976 (se
SKAKBLADET nr. 10109, sept. 1978,
og de udførlige artikler af C.J.
Morse: "The longest Problem" i
THE PROBLEMIST nov. 1979 og nov.
1981). Max. 4 p. til alle løsere,
som fandt tempomanøvren.

A very imperfect translation with the help of Google translator says:

1921. Walther Jørgensen:
1 Bh2+ Kf3 2 Qe3+ Kg2 3 Qg1+ Kf3 4 Qf1+ Kg4 5 Qe2+ Rf3 6 Qe6+ Rf5 7 Nd7. If black moves his kinght on Nh7 now, there is mate in a few moves. Therefore 7 - Kf3. By constantly repeating the 5-move tempo maneuver 8 Qe3+ Kg2 9 Qg1+ Kf3 10 Qf1+ Kg4 11 Qe2+ Rf3 12 Qe6+ Rf5 white plays the king to d1, forcing black to move pawns. Black lasts the longest when the a-pawns are eventually captured on a3. The move c5-c4 (which must be performed before the last pawn reaches a3) is delayed by two more moves, as White may be forced to capture the pawn. Since Qe6xc4+ is refuted with Nf4! it must happen from f1 when black in the tempo maneuver answers Df1+ with Ke4. (Black must wait with Ke4 until white king has left the cover of d2, otherwise d3+!). At best opposition, black can first be forced to move Nh7 after 224 Kd1, and then mate follows in two moves.
This 'sea snake' is an improvement of a #227 (THE PROBLEMIST 1981) which has an illegal position. It must be a record for the assignments built on WJ's own ingenious scheme from DIE SCHWALBE 1976 (see SKAKBLADET No. 10109, Sept. 1978, and the detailed articles by C.J. Morse: 'The longest Problem' in THE PROBLEMIST Nov. 1979 and Nov. 1981). Max. 4 p. to all solvers who found the tempo maneuver.

Thanks for your detailed input regards

[Uri Blass]

If 19.c4 really works, then it can easily cause a shorter mate, then this famous composition (the longest non-dual moremover over the world) will be cooked. Does 19.c4 really work? Can it lead a mate in shorter than 226-18=208 moves?

I don't see why 19.c4 shouldn't work, but I wouldn't call the study cooked because of that. 208 is still rather nice, isn't it?

So we should prove the win after 19.c4 is far shorter than 208 moves.

Somethiong tells me, you have "proved" that on your own already since quite a while and just don't want to rely on it without engines' confirmation.
But that's something, you cannot really expect with such DsTM, at least I won't be the one trying on and on to come to that conclusion in the end again.

At the end of your line after 19.c4 I end up with about 70+, if that helps you in any way, here's one proposal, you can keep the #-score of in hash backward to about move nr. 45 rather easily, at 50th you have the 7some position reached regards

[pgn][Event "226 to 208 19.c4"]
[Site "?"]
[Date "2022.02.06"]
[Round "?"]
[White "Hobacle's , Shortcut"]
[Black "WorldRecord, MateLength"]
[Result "1-0"]
[SetUp "1"]
[FEN "1N2Q3/p6n/p5n1/p1p2rpb/p4k1r/K1P4p/2PP4/6B1 w - - 0 1"]
[PlyCount "147"]

1. Bh2+ Kf3 2. Qe3+ Kg2 3. Qg1+ Kf3 4. Qf1+ Kg4 5. Qe2+ Rf3 6. Qe6+ Rf5 7. Nd7
Kf3 8. Qe3+ Kg2 9. Qg1+ Kf3 10. Qf1+ Kg4 11. Qe2+ Rf3 12. Qe6+ Rf5 13. Kb2 Kf3
14. Qe3+ Kg2 15. Qg1+ Kf3 16. Qf1+ Kg4 17. Qe2+ Rf3 18. Qe6+ Rf5 19. c4 a3+ 20.
Ka2 Kf3 21. Qe3+ Kg2 22. Qg1+ Kf3 23. Qf1+ Kg4 24. Qe2+ Rf3 25. Qe6+ Rf5 26. c3
Kf3 27. Qe3+ Kg2 28. Qg1+ Kf3 29. Qf1+ Kg4 30. Qe2+ Rf3 31. Qe6+ Rf5 32. d4
cxd4 33. cxd4 Kf3 34. Qxf5+ Nf4 35. Qb1 Ne2 36. Qf1+ Ke3 37. Ne5 Kd2 38. d5 Re4
39. d6 Nc1+ 40. Kxa3 Re3+ 41. Ka4 Be8+ 42. d7 Rxe5 43. Qf2+ Ne2 44. Bxe5 Bxd7+
45. Kxa5 Be6 46. Bc3+ Kxc3 47. Qe3+ Kxc4 48. Qxe6+ Kd3 49. Qxh3+ Ke4 50. Qxh7+
Kf3 51. Qd3+ Kf2 52. Kxa6 g4 53. Qf5+ Ke3 54. Qxg4 Nd4 55. Kxa7 Nc6+ 56. Kb6
Ne5 57. Qh4 Nd3 58. Qc4 Ne5 59. Qc3+ Nd3 60. Kc6 Ke4 61. Kd6 Nf4 62. Qe5+ Kf3
63. Qd4 Nh3 64. Ke5 Nf2 65. Qf4+ Ke2 66. Kd4 Nd1 67. Qg3 Kd2 68. Qh2+ Kc1 69.
Qa2 Nb2 70. Kc3 Nd1+ 71. Kb3 Nb2 72. Qxb2+ Kd1 73. Qf2 Kc1 74. Qe1# 1-0
[/pgn]

58.Qa4 mates one move faster than 58.Qc4 so this line is clearly not the best.