Author: Andreas Herrmann
Date: 20:00:30 08/30/02
Go up one level in this thread
On August 30, 2002 at 21:03:25, Uri Blass wrote: >On August 30, 2002 at 17:24:34, Andreas Herrmann wrote: > >>On August 30, 2002 at 14:33:08, Omid David wrote: >> >>>On August 30, 2002 at 14:27:53, Andreas Herrmann wrote: >>> >>>>On August 30, 2002 at 09:56:21, Omid David wrote: >>>> >>>>>On August 29, 2002 at 23:03:43, Dann Corbit wrote: >>>>> >>>>>>On August 29, 2002 at 22:50:53, Brian Richardson wrote: >>>>>>[snip] >>>>>>>Is move sorting turned off in Yace, GLC and Gnu for the depth = 6 searchs? >>>>>> >>>>>>Pretty irrelevant since all of them show a branching factor between 2 and 3 for >>>>>>the opening position. >>>>>> >>>>>>The program described by the OP had a branching factor of 6-8, IIRC. >>>>>> >>>>>>IOW, something is clearly amiss. >>>>>> >>>>>>I wonder how often the OP finds the requested position in the hash table. >>>>>>Usually, hash table alone would be enough to prevent a branch factor that >>>>>>terrible. >>>>> >>>>>How do you calculate branching factor here? >>>> >>>>A branching factor of 3 means that each node has in the average 3 child nodes >>>>(This is the description i have found on an internet page). So the formula must >>>>be: >>>> >>>>bf = ( Nodes [ply n] - nodes [ply n-1] ) / nodes [ply n-1] >>>> >>>>Excample: >>>>Whole nodes until ply 5 = 4000 and whole nodes until ply 6 = 20000. >>>>Then you got a branching factor of >>>>bf [ply 6] = ( 20000 - 4000 ) / 4000 = 4.0 >>>> >>>>have a nice day >>>>Andreas >>> >>>I know this :-) >>> >>>But there is the odd/even issue, so the b-factor can change drastically while >>>moving from an odd ply to an even ply, and vice versa. >> >>I think the best is to calculate an average branching factor from all plys. >> >>bf[avg] = ( bf[2] + bf[3] + bf[4] ... + bf[n] ) / (n - 1) >> >>Andreas > >It is better to use >( bf[2] * bf[3] * bf[4] ... * bf[n] )^(1/(n-1)) > >Uri Hi Uri, have you found an official description of the calculation of the branching factor in search trees? Because i don't know how to calculate the branching factor correct. My formulas i have created only based on the description "A branching factor of 3 means that each node has in the average 3 child nodes" that i have found on an internet page. Can you tell me where your formula is from (perhaps an internet link), because i want to implement a calculation in Holmes. I got every time smaller values with your formula: Excample: bf[2] = 2; bf[3] = 4; bf[4] = 3; bf[5] = 7 bf[avg a] = 4.0 bf[avg b] = 3.6 I got only the same average bf if there is in all plys the same bf because then is bf[avg] = bf[n]^(n-1)^(1/(n-1)) = bf[n] = maximum Andreas
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