Author: Uri Blass
Date: 10:55:41 08/31/02
Go up one level in this thread
On August 31, 2002 at 11:11:50, Ralf Elvsén wrote: >On August 31, 2002 at 05:34:11, Uri Blass wrote: > >>On August 31, 2002 at 04:51:49, Ralf Elvsén wrote: >> >>>On August 30, 2002 at 23:00:30, Andreas Herrmann wrote: >>> >>>>On August 30, 2002 at 21:03:25, Uri Blass wrote: >>>> >>>>>>> >>>>>>>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)) >>> >>>Not if the numbers bf[i] are ratios of the type bf[i] = T[i]/T[i-1] (e.g.) >>>Then everything will cancel out except for the first and last T[i] >> >>I think that this is exactly the idea about branching factor. >>The question is if I need 1 second to get depth 1 how many seconds I need to get >>depth n. >> >>It is also possible to use the formula (T(n)/T(1))^(1/n-1) >> >>Uri > > >Well, it all depends on what you want. I personally wouldn't like this >measure to depend heavily on T(1) which I would expect to vary much. >And if you have a series for T which is > >T1 = 1 T2 = 2 T3 = 4 T4 = 16 T5 = 64 > >and another > >T1 = 1 T2 = 4 T3 = 16 T4 = 32 T5 = 64 > >then you have very different branching factors for low/high plies >(relatively speaking) but the proposed formula gives the same >overall value. So you are throwing away information and (in my >opinion) relies heavily on a suspect value: T(1). > >Just my opinion... ( bf[2] + bf[3] + bf[4] ... + bf[n] ) / (n - 1) is also the same in both examples so I see no advantage for using that formula. You cannot have all information in one number but if n is relatively big the first iterations do not change much. Uri
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