Author: Robert Hyatt
Date: 14:32:26 04/13/00
Go up one level in this thread
On April 13, 2000 at 11:26:19, Pete Galati wrote: >On April 13, 2000 at 09:12:43, Robert Hyatt wrote: > >>On April 13, 2000 at 03:02:09, blass uri wrote: >> >>>On April 12, 2000 at 23:29:54, Robert Hyatt wrote: >>> >>>>On April 12, 2000 at 17:46:51, Dann Corbit wrote: >>>> >>>>>On April 12, 2000 at 17:32:44, Derrick Williams wrote: >>>>> >>>>>>On April 12, 2000 at 16:48:15, Dann Corbit wrote: >>>>>> >>>>>>>On April 12, 2000 at 16:36:09, Derrick Williams wrote: >>>>>>> >>>>>>>>I would like to simulate the expierence of playing against Deepblue. How long >>>>>>>>would I have to let fritz6 think per move on a pent 450 to simulate playing >>>>>>>>deepblue at 40/2 hrs? Should I let fritz6 think one hour per move or what? >>>>>>> >>>>>>>Does fritz6 have a 40 moves / 2000 hrs setting? >>>>>>>That should be about right, as far as NPS. >>>>>> >>>>>> >>>>>> You are exaggerating just a bit aren't you? >>>>> >>>>>No. >>>>>DB calculates 200M NPS, micros about 200K NPS. (roughly speaking -- might be >>>>>off by a factor of 2 or so for what fritz 6 can do on a PIII 450, which would >>>>>reduce it to 1000 hours instead of 2000). >>>>> >>>>>DB was one heck of a machine. >>>> >>>> >>>>Yes.... and it could peak at 1B nodes per second, with 200M being the typical >>>>lower bound... 480 chess processors at 2 to 2.4M nodes per second each... >>> >>>480 chess processors at 2 to 2.4M nodes can be the same as 200M with one >>>processor if you consider loss of speed from parallel search. >>> >>>Uri >> >> >>Parallel search doesn't lose speed. It just searches extra nodes. But the NPS >>value goes up fairly linearly. Try crafty on a quad xeon using 1cpu, 2cpus, >>etc. At 4 cpus the NPS is pretty much 4x. But roughly 25% of the search space >>is redundant... >> >>I haven't seen anyone adjust the NPS to reflect search efficiency, since it is >>impossible to determine exactly how many nodes are 'extra overhead'... > >Are you talking about in relation to an individual position? In otherwards, >that the amount of redundant search would change with each position, making it >basically imposible to have an etched in stone formula? > >Pete absolutely. And not even by position, but by "run". IE run the same position 10 times and you will get 10 different node counts...
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