Author: Bo Persson
Date: 09:14:34 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 I believe the basic problem is that one processor is redoing work that has already been done by others, because it doesn't know exactly which nodes have been visited. Makes it hard to count them too... :-) Bo Persson bop@malmo.mail.telia.com
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