spreadtallent
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re:Longest Forced Mate - 2006/06/30 18:03
Binder provided the following position, but was missing the complete solution.
White: Rb1, Kb6, Bb8, Bf8 Pawns at c2, e2, b3, f3, f4, f5, e6, a7 Black: Kd5, Rc8, Re7, Ng1, Ba8, Bd8, Bf6 Pawns at: b2, e3, c5, h5, c6, h6, h7
1. Td1+ Bd4 2. c4+ Kd6 3. Rxg1 Be5 4. Rd1+ Bd4 5. Ka5 Bb7 6. Ka4 Ba8 7. Ka3 Bb7 8. Ka2 Ba8 9. Kb1 Bb7 10. Kc2 Ba8 11. Kd3 Bb7 12. Rb1 Ba8 13. Rf1 Bb7 14. Rd1 Ba8 15-20 Ka5 Ba8 21. Kb6 h4
The sequence from 5 through 21 is a 17-move maneuver to lose a tempo. It is repeated 15 more times as the h-file pawns inch toward promotion where they are captured in the two-move combination Rxh1 Be5 Rd1+ Bd4. Therefore we have 15x17 + 3x2 + 21 = 282, and play then continues:
282. . . . . . Bb7 283. Kxb7 b1Q 284. Rxb1 Be5 285. Rd1+ Bd4 286. Rxd4+ cxd4 287. Kb6 d3 288. a8Q Rxb8+ 289. Qxb8 any 290. Qxd8 mate.
Source: 96 Citaat-Problemen by Dr. M. Niemeijer, Wassenaar, 1960. Attribution is given as Christmas-card, December 25, 1929.
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