Author: Tim Foden
Date: 07:34:59 10/04/01
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On October 04, 2001 at 09:59:49, Daniel Clausen wrote: >Hi > >On October 04, 2001 at 09:33:03, Colin Frayn wrote: > >>On October 04, 2001 at 07:03:14, Tim Foden wrote: >> >> >>>[D]5rk1/1pp2R1p/p1pb4/6q1/3P1p2/2P4r/PP1BQ1P1/5RKN w - - 2 0 >>> >>>This can be evaluated in 2 ways... >>> >>>(1) BxP (+1000) BxB (-3500) R1xB (+3500) QxR (-5500) RxQ (+10000) RxR (-5500) >>>.........+1000.......-2500........+1000.......-4500........+5500...........0 >>> >>>>>> value = 0 >>> >>>(2) BxP (+1000) BxB (-3500) R7xB (+3500) RxR (-5500) RxR (+5500) QxR (-5500) >>>.........+1000.......-2500........+1000.......-4500.......+1000.......-4500 >>> >>>>>> value = -2500 >> >>Damn that's clever. I'd never thought of that. I wonder if it's worth actually >>testing for these things or whether the extra CPU time would be better spent >>elsewhere? That's one of the things I want to know. :) I hadn't thought about this type of situation until I compared the output of 2 SEE's together while searching. >>Cheers, >>Col > >I'm wondering what the current SEE implementation does? I mean whether you >thought about this case or not, your implementation has to do something with it. >:) I suspect it simply chooses the rook move it considers first? Yes. Both of the SEE routines in GLC do this. It gets flagged because GLCs main routine chooses R1, and its secondary one chooses R7. :) >While in this example one could find out the better rook-move by looking ahead >one ply, it's easy to imagine there are situations where you'd have to look >ahead more than 1 ply. I guess. Cheers, Tim.
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