Author: Dieter Buerssner
Date: 09:27:43 08/17/01
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On August 15, 2001 at 12:13:04, Uri Blass wrote: >The problems are >Mate in 10,mate in 11,mate in 12,mate in 13,mate in 17,mate in 18 I started with the mate in 18 position. I made the suggested moves by Yace rather fast, until a mate was found (which was after your mate in 10 position). >I told my opponent that it is a forced mate only after he blundered by 28.Qxe3 >and if Rxe3 then 29.Rxh3(it is a typical computer error at short time control >and the best move in order to lose more slowly was 28.Ke1) I went back, and also Yace only found now, that Ke1 leads to a longer mate. At the end, I get the following PV: 2k1r3/ppp5/5P2/8/3P1P1q/3P2Nr/PP1Q1K1n/6RR b - - 11.01 3:33 +M18 1...Sg4+ 2.Kf1 Se3+ 3.Ke1 Txh1 4.Df2 Sg4+ 5.Kd2 Sxf2 6.Sxh1 Dh2 7.Sxf2 Dxf2+ 8.Kc3 Dxg1 9.f7 Td8 10.d5 Dc5+ 11.Kd2 Dxd5 12.Ke3 Dxd3+ 13.Kf2 Tf8 14.f5 Txf7 (47.739.002) 223.8 To translate to English notation, substitute S with N, T with R and D with Q (Under the S5 GUI, I was using, not more is displayed of the PV). Of course, this is no proof, that this is the shortest mate possible. Regards, Dieter
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