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Subject: Re: Christmas Tactical Test - results for F8, S7, F7, S6.02 & Ruffian (LONG)

Author: Steve Maughan

Date: 01:11:44 12/24/02

Go up one level in this thread


Dieter,

>>6r1/1p3k2/pPp4R/K1P1p1p1/1P2Pp1p/5P1P/6P1/8 w - - bm Rxd6; id "BBC630"
>
>This should be "bm Rxc6"

Yes - I corrected this in the version I used then pasted in the wrong version -
apologies!


>>r5r1/n1q1p2k/3pPpp1/P1pP4/2P4N/R1B5/2Q3PP/7K w - - bm Bd2; id "BBC958"
>
>Yace finds Bd2 fast with score mate in 10, then switches to Be1 with score mate
>in 9. In the automatically generated table below, it is considered unsolved.

WOW - I hadn't realised that there was a shorter mate. In BBC they state that
the best move is Bxf6 but CM 8000 found Bd2 really quickly.  I've always
considered this a good test for threat detection / extension.

>Yace, P4 2.53, 300M hash, 3 min test time. Not surprisingly, it solved less
>positions than the programs you tested (especially when considering the
>hardware difference). From the positions it found, for quite a few, it was
>rather fast compared to the others. Some positions seem to be at the edge of
>finding. It sees compensation for a sac, but just a bit too little
>compensation.

I chose the positions so that they hopefully would be "at the edge of finding"
them - so that's good!

>                   5s        20s       60s       120s      180s
>          BBC55:   1 (Rxg7)  1 (Rxg7)  1 (Rxg7)  1 (Rxg7)  1 (Rxg7)   3.370
>         BBC116:   0 (Qd3)   0 (Qd3)   0 (Qd3)   0 (Qd3)   0 (Qd3)       nf
>         BBC142:   1 (Nxf7)  1 (Nxf7)  1 (Nxf7)  1 (Nxf7)  1 (Nxf7)   4.151
>         BBC158:   1 (Qh6+)  1 (Qh6+)  1 (Qh6+)  1 (Qh6+)  1 (Qh6+)   4.043
>         BBC161:   1 (Nf6+)  1 (Nf6+)  1 (Nf6+)  1 (Nf6+)  1 (Nf6+)   4.962
>         BBC167:   1 (Rxa6)  1 (Rxa6)  1 (Rxa6)  1 (Rxa6)  1 (Rxa6)   0.978
>         BBC175:   1 (Bh6)   1 (Bh6)   1 (Bh6)   1 (Bh6)   1 (Bh6)    3.705
>         BBC189:   1 (Nxh6)  1 (Nxh6)  1 (Nxh6)  1 (Nxh6)  1 (Nxh6)   2.655
>         BBC192:   0 (Ne5)   0 (Ne5)   0 (Ne5)   0 (Ne5)   0 (Ne5)       nf
>         BBC346:   1 (Ba4)   1 (Ba4)   1 (Ba4)   1 (Ba4)   1 (Ba4)    1.238
>         BBC354:   0 (N3d4)  0 (Qe2)   0 (Qe2)   0 (Qe2)   0 (Qe2)       nf
>         BBC403:   0 (O-O)   0 (O-O)   0 (O-O)   1 (Nxh7)  1 (Nxh7)  63.790
>         BBC405:   0 (Rac1)  1 (Rxd4)  1 (Rxd4)  1 (Rxd4)  1 (Rxd4)  17.046
>         BBC442:   0 (cxd5)  0 (cxd5)  1 (Nxg6)  1 (Nxg6)  1 (Nxg6)  25.040
>         BBC459:   0 (Nxf6)  0 (Nxf6)  0 (Nxf6)  0 (Nxf6)  0 (Nxf6)      nf
>         BBC539:   0 (Qh4)   0 (Qh4)   0 (Rb6)   1 (f4)    1 (f4)    92.582
>         BBC547:   0 (Bxf8)  0 (Bxf8)  1 (Nd5)   0 (Bc5)   0 (Ne2)       nf
>         BBC630:   1 (Rxc6)  1 (Rxc6)  1 (Rxc6)  1 (Rxc6)  1 (Rxc6)   3.498
>         BBC958:   1 (Bd2)   1 (Bd2)   1 (Bd2)   0 (Be1)   0 (Be1)       nf
>         BBC770:   0 (Bxe5)  0 (Bxe5)  0 (Bxe6)  0 (Bxe6)  0 (Bxe6)      nf
>
>        Summary:  10        11        13        13        13        227.058

Nice output.  I do prefer the exponential decay method - you can even express
the final score as a percentage.

Regards,

Steve



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