Author: Robert Hyatt
Date: 07:15:36 12/04/03
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On December 02, 2003 at 21:55:34, Russell Reagan wrote: >On December 02, 2003 at 21:00:03, Robert Hyatt wrote: > >>Makes sense, and points out that there is a _big_ cost for doing that still. >>The EBF is meaningless when N-1 and N are _both_ doing something that is "bad" >>in terms of tree search space. > >Why is it bad? If program A has an EBF of 2, and program B has an EBF of 3, >wouldn't you want program A even if program A was 16x slower than program B? > >Depth A Nodes B Nodes Who finishes first? >1 100 100 B >2 200 300 B >3 400 900 B >4 800 2700 B >5 1600 8100 B >6 3200 24300 B >7 6400 72900 B >8 12800 218700 A >... No. Because A is going to be slower for a _long_ while before it catches up. But if you do the N-best search as given, your EBF should not change whether N=1 (normal) or N=99. But the time to complete a single iteration will certainly change by roughly a factor of N. So I'll take N=1 myself. :)
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