Author: blass uri
Date: 17:25:54 07/29/98
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On July 29, 1998 at 09:42:08, Steven J. Edwards wrote: >On July 28, 1998 at 23:07:40, Steven J. Edwards wrote: > >>Here are the numbers for trajectory enumerations (distinct legal pathways) for >>the first few ply depth values: >> >>ply: 0 count: 1 >>ply: 1 count: 20 >>ply: 2 count: 400 >>ply: 3 count: 8902 >>ply: 4 count: 197281 >>ply: 5 count: 4865609 >> >>For ply=6, the number is a little under 120 million and for ply=7 it's about 3.1 >>billion. I haven't calculated ply=8 yet, but is should be just under 100 >>billion. > >My other machine freed up, so here are the next two values: > >ply: 6 count: 119060324 >ply: 7 count: 3195901860 How much time did you use to compute ply 7? Did you use hash tables to save time? can you compute the number of different positions at every ply? Uri > >Who wants to try for ply=8? > >-- Steven (sje@mv.mv.com)
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