Author: Omid David
Date: 11:02:14 08/31/02
Go up one level in this thread
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. > >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)).
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