16

With forced move, it is meant that the player has only one legal move.

[fen "7k/q5Q1/p4PPK/6PP/8/5P2/8/8 b - - 0 1"]

1... Qxg7+ 2. fxg7+ Kg8 3. f4 a5 4. f5 a4 5. f6 a3 6. f7#

is an example with 10 half-moves.

Each half-move is the only legal one and the end is a checkmate.

  • 1
  • 3
    An inventive idea for a position, but I'm stumped as to where White's Q came from on the previous move, and how Qg7+ could have been a better choice than Qxa7 followed by mate on next move. – rolando2 Mar 14 '15 at 20:12
  • @rolando2 The white queen could have come from the squares c2-g2. Yes i could have mated, but that is not the point here. – Rewan Demontay May 7 at 13:05
  • Hey Peter, I have found a 15 ply sequencd made by someone else! Also, about that 10 ply example made in question, did you make that yourself or did you get it from somewhere? – Rewan Demontay Jun 11 at 1:22
  • @RewanDemontay The board is flipped (the white pawns are going down the board), so it would be c7-g7 – Acccumulation Jun 11 at 19:08
12

If you don't mind a bushel of promoted Queens, you can get 14 half-moves . . .

[Title "14 forced half-moves to checkmate"]
[Fen "3b3k/qqqqq3/rr4NK/7R/5N1Q/7Q/B6Q/7R b - - 0 0"]

1... Rxg6+ 2. Nxg6+ Rxg6+ 3. Kxg6+ Qh7+ 4. Rxh7+ Qxh7+ 5. Qxh7+ Qxh7+ 6. Qxh7+ Qxh7+ 7. Qxh7+ Qxh7+ 8. Rxh7#
2

Ah sorry, I missed your condition that each move be the only legal move. This answer is only the longest forced checkmate. The following is a forced checkmate in 262 moves (if both sides play perfectly).

[fen "6k1/5n2/8/8/8/5n2/1RK5/1N6 w - - 0 1"]

1. Kd3 N3g5 2. Kd4 Ne6+ 3. Kd5 Nc7+ 4. Kc6 Ne6 5. Kd7 Nfg5 6. Kd6 Kg7 7. Ke5 Nc5 8. Kf5 Nge4 9. Ra2 Nd6+ 10. Ke5 Nf7+ 11. Kd5 Nd3 12. Kd4 Nb4 13. Ra4 Nd8 14. Kc4 Nbc6 15. Kd5 Ne7+ 16. Kd6 Ng6 17. Ra8 Nf7+ 18. Ke6 Nf8+ 19. Kd5 Ng6 20. Ra6 Nf4+ 21. Kd4 Nd8 22. Nc3 Nde6+ 23. Ke3 Ng2+ 24. Kf2 Ngf4 25. Kf3 Kf7 26. Ra7+ Kg6 27. Ke4 Kg5 28. Ra5+ Kg4 29. Ke5  Nc5 30. Nd1 (30. Rxc5??... Nd3+ 31. Kf6 Nxc5 32. Ne2 Ne4+ =) Ng6+ 31. Kd5 Nb3 32. Ra2 Nf4+ 33. Ke5 Nc5 34. Ne3+ Kf3 35. Ra3 Ng6+ 36. Kd5 Nd3 37. Kd4 Nde5 38. Ra6 Nf4 39. Nf1 Nf7 40. Rf6 Ng5 41. Ke5 Ngh3 42. Rf7 Ke2 43. Ng3+ Ke3 44. Nf5+ Kf3 45. Nd6 Ke3 46. Nc4+ Kd3 47. Nb2+ Kc2 48. Na4 Nd3+ 49. Kd4 Ng5 50. Rd7 Ne6+ 51. Kd5 Ng5 52. Rg7 Kb3 53. Nb6 Nb4+ 54. Ke5 Nd3+ 55. Kf5 Nh3 56. Rc7 Nb4 57. Rc8 Nf2 58. Nd7 Nc2 59. Nc5+ Kb4 60. Ke5 Ng4+ 61. Kf4 Nge3 62. Nd3+ Kb3 63. Nc1+ Kb4 64. Ke4 Nc4 65. Rg8 Nb6 66. Rg2 Na3 67. Rb2+ Kc5 68. Nd3+ Kc6 69. Rg2 Na4 70. Rg5 Nc3+ 71. Ke5 Na4 72. Ke6 Nb5 73. Nb4+ Kb6 74. Nd5+ Ka5 75. Rh5 Nc5+ 76. Ke5 Nd3+ 77. Ke4 Nc5+ 78. Ke3 Na6 79. Rh1 Nbc7 80. Nf4 Kb4 81. Kd4 Nb5+ 82. Ke5 Nc5 83. Nd5+ Kb3 84. Rh3+ Kc4 85. Rh4+ Kd3 86. Nf4+ Kd2 87. Rh2+ Kc3 88. Kd5 Kb4 89. Rh4 Na4 90. Nd3+ Kb3 91. Rg4 Kc2 92. Nf4 Kd2 93. Rg7 Ke3 94. Ng2+ Kd2 95. Rh7 Nb2 96. Rd7 Ke2 97. Rb7 Nd1 98. Kc5 Nbc3 99. Kd4 Kd2 100. Rd7 Ke2 101. Nh4 Nb5+ 102. Kc4 Na3+ 103. Kb3 Nb1 104. Nf5 Nf2 105. Kc2 Na3+ 106. Kc3 Nb1+ 107. Kd4 Nd2 108. Re7+ Kf3 109. Rf7 Nh3 110. Nd6+ Ke2 111. Re7+ Kf3 112. Re3+ Kg4 113. Ra3 Nf1 114. Ra4 Ng5 115. Ke5+ Kf3 116. Nf5 Nd2 117. Ra3+ Kf2 118. Ra2 Ke2 119. Kd4 Nge4 120. Ne7 Kf3 121. Nd5 Nb3+ 122. Kc4 Nc1 123. Ra3+ Kf2 124. Ra8 Ne2 125. Rg8 Ng1 126. Kd4 Nd6 127. Ke5 Nc4+ 128. Ke4 Nd6+ 129. Kf4 Ne2+ 130. Ke5 Nb5 131. Rf8+ Ke1 132. Rd8 Nbc3 133. Nb4 Kf2 134. Nc2 Nc1 135. Rc8 Nb5 136. Rb8 Nc3 137. Rd8 Ke2 138. Kd4 Nd1 139. Na1 Kd2 140. Ke4+ Ke2 141. Rc8 Nf2+ 142. Kd4 Ncd3 143. Nb3 Nf4 144. Rg8 Kf3 145. Nd2+ Ke2 146. Nc4 Kf3 147. Ne5+ Ke2 148. Re8 N2h3 149. Nc4+ Kf3 150. Ra8 Ng5 151. Ra3+ Ke2 152. Ke5 Nd3+ 153. Kf5 Nh3 154. Ra2+ Kf3 155. Ra8 Nhf2 156. Re8 Nd1 157. Nd2+ Kf2 158. Ke4 Nc1 159. Nb1 Ke2 160. Kf4+ Kd3 161. Rd8+ Ke2 162. Rd2+ Ke1 163. Rh2 Ne2+ 164. Kf3 Nd4+ 165. Ke4 Nb3 166. Ra2 Nf2+ 167. Kf3 Nd4+ 168. Ke3 Nf5+ 169. Kf4 Nd4 170. Ra4 Nc2 171. Kf3 Nd3 172. Rh4 Kd1 173. Nc3+ Kd2 174. Ne4+ Kc1 175. Ke2 Ne5 176. Nc5 Na3 177. Re4 Nec4 178. Kd3 Nb2+ 179. Kc3 Nb1+ 180. Kd4 Na3 181. Rh4 Nb5+ 182. Ke5 Kc2 183. Rh2+ Kc3 184. Kd5 Nd1 185. Rh3+ Kb2 186. Nb7 Na3 187. Nd6 Nc3+ 188. Kc5 Kc2 189. Nf5 Nab1 190. Kd4 Nb5+ 191. Kc4 N5a3+ 192. Kc5 Kb2 193. Rh2+ Kb3 194. Nd4+ Kc3 195. Ne2+ Kb2 196. Ng3+ Kc3 197. Rh3 Kd3 198. Nf5+ Kc2 199. Rg3 Kb2 200. Ne3 Nd2 201. Rg2 Kc1 202. Kd4 Nf3+ 203. Kc3 Nb1+ 204. Kb4 Nbd2 205. Rg8 Ne4 206. Rd8 Nf2 207. Kc3 Ne4+ 208. Kd3 Nc5+ 209. Kc4 Ne4 210. Rd1+ Kb2 211. Rd5 Ne1 212. Rd8 Nf3 213. Rf8 Ne1 214. Rb8+ Kc1 215. Kd4 Nd2 216. Nd5 Kd1 217. Re8 Ng2 218. Kd3 Nb1 219. Re7 Ne1+ 220. Kc4 Ng2 221. Kb3 Nd2+ 222. Kb2 Nc4+ 223. Kc3 Nd2 224. Nf6 Nf4 225. Rd7 Ne2+ 226. Kd3 Nb1 227. Ke3+ Ke1 228. Re7 Kd1 229. Rh7 Nc1 230. Ne4 Na3 231. Nf2+ Kc2 232. Rc7+ Kb2 233. Nd1+ Kb1 234. Rc8 Nc2+ 235. Kd2 Nb3+ 236. Kd3 Ne1+ 237. Kc3 Kc1 238. Rd8 Nc5 239. Nf2 Na4+ 240. Kb3 Nb2 241. Rc8+ Kd2 242. Kxb2 Ke3 243. Ng4+ Kf4 244. Ne5 Ng2 245. Ng6+ Kg5 246. Rc3 Kg4 247. Rd3 Ne1 248. Rd4+ Kh5 249. Nh4 Nd3+ 250. Rxd3 [*]

http://kirill-kryukov.com/chess/longest-checkmates/longest-checkmates-sorted-by-length.shtml

  • 5
    This does not answer the question – Peter Feb 25 '17 at 17:48
0

Here's a game from 1912 between Edward Lasker and George Alan Thomas. All blacks moves are forced.

[fen "rn3rk1/pbppq1pp/1p2pb2/4N2Q/3PN3/3B4/PPP2PPP/R3K2R w KQ - 20 11"]

1... Qxh7+ Kxh7 Nxf6+ Kh6 Neg4+ Kg5 h4+ Kf4 g3+ Kf3 Be2+ Kg2 Rh2+ Kg1 Kd2# 1-0

In answer to your question, I'm sure there's an even better hypothetical board position, but this was a really game played by competent players all be it in a casual setting.

Sources:

https://youtu.be/QIiaeY76VY8?t=3m45s

http://www.chessgames.com/perl/chessgame?gid=1259009

  • 3
    That's a classic combination, but only Black's moves are legally forced (and even that starting only after Kh6, since Black could have allowed mate in one with Kh8 (Ng6#)). – Noam D. Elkies Dec 15 '17 at 15:44
-1

UPDATE #2: Now I have discovered a case of 16 ply for a sequence of only one legal moves that results in checkmate! It can be found here in a PDF of Feenschach, the 1980 edition.

Enjoy!

[Title "Bernd Schwarzkopf nach Karl Scherer (Urdruck), Mate In 8"]
[FEN "1KN4r/QRRRRRRr/kq6/p1q1R3/PP1q1b2/4q3/5q2/1r4q1 b KQkq - 0 1"]

1... Qxa7+ 2. Rxa7+ Qxa7+ 3. Rxa7+ Qxa7+ 4. Rxa7+ Qxa7+ 5. Rxa7+ Qxa7+ 6. Rxa7+ Qxa7+ 7. Rxa7+ Rxa7 8. b5+ Rxb5+ 9. axb5#

UPDATE: Sorry Professor Elkies, but it looks like someone beat your 14 with a 15 two years before you in the chess.com forums!

I found this the following, and amazing, sequence here in the chess.com forums. This was created by a person with the username of shoopi

Enjoy!

[Title "shoopi, 2013"]
[FEN "rk1KB3/b1q1QQQR/bnqQ1q2/1Rq1Q1q1/2q1rQ1q/6B1/6QQ/8 w - - 0 1"]

1. Qxc7+ Qxc7+ 2. Qxc7+ Qxc7+ 3. Qxc7+ Qxc7+ 4. Bxc7+ Kb7+ 5. Bb8+ Qxe7+ 6. Qxe7+ Qxe7+ 7. Qxe7+ Qxe7+ 8. Rxe7#

If you are wondering if this is legal, let's look at the position. Black has 6 promoted and white has 7. Using a trick that I leaned from Professor Elkies (This was when I read from Tim Krabbe'a diary entry of his evidence for the legality of his 19 ply no-brainer puzzle. Thanks for that Mr. Elkies!), here is a short proof game of legality that I made myself.

[FEN ""]

1. d4 c5 2. b4 a5 3. dxc5 axb4 4. Nc3 e6 5. Nd5 exd5 6. g4 h5 7. gxh5 Nf6 8. f3 Ne4 9. fxe4 

And now black and white are capable of promoting to all of the queens seen in the diagram.


Here is another 14-mover that ties with Noam D. Elkies’s composition of 14 half-moves. Of course, it is based off of the position given by Hauke Reddmann in his answer to this similar question here on CSE.

[Title "Rewan Demontay, 2019"]
[FEN "1Qb5/1pQp4/1P1Q4/3PQ2p/5Q1P/6QK/Rrrrrrrq/R4nk1 w - - 0 1"]

1. Qxh2+ Rxh2+ 2. Qxh2+ Rxh2+ 3. Qxh2+ Rxh2+ 4. Qxh2+ Rxh2+ 5. Qxh2+ Rxh2+ 6. Qxh2+ Rxh2+ 7. Rxh2 d6#

It may not look like a legal position, but I assure you that it is legal. A proof game made by SpiderUnicorn can be found here in the chess.com forums. (I basically turned it into a puzzle because I had NO idea myself whether or not legal position!)

I believe that this shows that 14 is the definitive maximum for this kind of problem.


If you want to know the record for one legal move regarding discovered checks, I can give you this.

I found this funny example on Tim Krabe’s website (Journal Entry #265.)

He gives this series of 7 mutual discovered checks. All of the moves, minus the first, are forced, which is what makes it unique.

[Title "V. Korolkov, 1940"]
[FEN "6B1/5Nb1/3p4/q2krP1R/Nn2p3/pPKnr3/Q1PB4/3R4 w - - 0 1"]

1. Nd8+ Re6+ 2. f6+ Ne5+ 3. Bxe3+ Nd3+ 4. b4#
  • 1
    Explained already: the "x" in "Rgxg3+" means that there was a White piece or pawn on g3 and the Black rook captured it from somewhere on the g-file. Because of the White unit on g3, White was not in check. Similarly for the retraction of Nxb3+: if there was a Black unit on b3 (other than another Q or R) then White was not in check and the position is retro-quiescent. – Noam D. Elkies Apr 7 at 3:35
  • Yes, that's an example of what I meant (though I'd probably make it a Bishop on g3 instead of yet another promoted Queen). – Noam D. Elkies Apr 7 at 4:42
  • In problem chess, the stipulation is known (not necessarily leading to mate - stalemate or dead would also count) as "Problem Without Words". HTH for research - without promoted material, the problem by Röpke mentioned in another thread may be still optimal: chess.stackexchange.com/questions/4963/… – Hauke Reddmann Apr 8 at 8:32
  • In your last update with 17 queens, it seems to me that Black moves are not forced, e.g. 2...Qe5xg7+ is legal instead of 2.Qe7xg7+. – Evargalo Apr 8 at 9:24
  • @HaukeReddmann Actually, the answer you linked is beaten by this question’s given example. 10 here, 9 over there. – Rewan Demontay May 7 at 19:57

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