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Subject: Re: (Supposed) Mate in 25 => Kasparov v. World (1999)

Author: Hans Havermann

Date: 21:19:03 02/07/01

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On February 07, 2001 at 22:43:33, I wrote:

>I took a slightly different tack. I let William Bryant's Screamer (b55), with
>5-piece TBs, run for 24 hours and took the position after 1. Kg5 Qd5+ 2. Qf5
>Qg2+ 3. Qg4 Qd5+ 4. Kf4 Qd6+ 5. Kf3 Qc6+ 6. Kf2 Qc2+ 7. Kg1 (a seemingly
>reasonable length of its output after that time) as input for new analysis. The
>remainder of the game would then run (I think) 7. ...d3 8. Qf4+ d2 9. g8=Q Qd1+
>10. Kg2 Qe2+ 11. Kh3 Qd3+ 12. Kh4 Kb2 13. Qb4+ Kc1 14. Qgc4+ Qc2 15. Qxc2+ Kxc2
>16. Qc4+ Kb2 17. Qd3 Kc1 18. Qc3+ Kd1 19. Kg4 Ke2 20. Qc2 Ke3 21. Kf5 Kd4 22.
>Qxd2+ Kc4 23. Ke6 Kb5 24. Kd6 Kc4 25. Kc6 Kb3 26. Kb5 Ka3 27. Kc4 Ka4 28. Qb4#

I realized (too late) that distance to mate within the TBs is a significant
unknown in such conjectures. After playing with the problem a bit, I propose the
following alternative:

1. Kg5 Qd5+ 2. Qf5 Qg2+ 3. Qg4 Qd5+ 4. Kf4 Qd6+ 5. Kf3 Qc6+ 6. Kf2 Qc2+ 7. Kg1
d3 8. g8=Q d2 9. Q8g5 Qc3 10. Q5f4 Qc5+ 11. Kh2 Qc3 12. Qc4 Qxc4 13. Qxc4+ Kb2
14. Qd3 Ka2 15. Qxd2+ Kb3 16. Qa5 Kc4 17. Kg3 Kd4 18. Kf4 Kc4 19. Ke4 Kb3 20.
Kd3 Kb2 21. Qb6+ Ka3 22. Kc3 Ka4 23. Qb4#



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