Author: Ralf Elvsén
Date: 12:40:40 08/31/02
Go up one level in this thread
On August 31, 2002 at 14:02:14, Omid David wrote: >On August 31, 2002 at 13:55:41, Uri Blass wrote: > >>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. I never advocated that. >> >>You cannot have all information in one number but if n is relatively big the >>first iterations do not change much. >> >>Uri > >Once I thought of the following: > >Just take all nodes of the last iteration (iteration n), and calculate its >"n-1"th root: > >b-factor = (nodes in iteration n) ^ (1 / (n - 1)). For practical purpuses I could imagine this: make a little plot of log (Time[n]) vs n. Fit a straight line and let that determine the average BF. Then, by eye, one can see if something strange is going on at various depths. Ralf
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