--- "Amara D. Angelica" <[EMAIL PROTECTED]> wrote: >
> > 1) The quote to totally wrong the the "^" should be a "," ?
>
> It's 10 to the 8432 power, according to the paper. This is the
> theoretical memory capacity, not its actual size, but no estimates are
> given for real-world typical size of memory, so the comparison with
> machine capacity seems unrealistic.
Really, they should have realised that 10^8432 bits makes no sense
at all.
> The number is derived as follows:
>
> "Assuming there are n neurons in the brain, and on average
> there are m connections between a given neuron and the rest of them, the
> magnitude of the brain memory capacity can be expressed by the following
> mathematical model, the human memory capacity model, as given below:
>
> n!/[m!(n-m)!]
>
> where n is the total number of neurons and m the number of average
> partial connections between neurons.
>
> However, this is "extremely hard to calculate and is almost intractable
> using a modern computer, because of the exponential complicity or the
> recursive computational costs for such large n and m," so they did some
> math tricks to estimate it.
Well the standard way to estimate (sorry about reverting to LaTeX here
for those of you who aren't mathematicians... ) this combination is
to use the log from of Sterling's equation:
\log_2 \binom{n}{m} = (n-m) \log_2 \frac{n}{n-m} + m \log_2 \frac{n}{m}
here n = 10^9 (neurons in brain)
and m = 10^4 (10,000 connections per neuron)
These are my numbers, if they use slightly different ones it won't
matter too much.
This gives approx,
10^9 \log_2 10^5 + 10^4 \log_2 10^5
which is about 10^10 bits which is about what I'd expect very roughly.
They should have talked to a mathematician first. Estimating basic
combinatorics like n!/[m!(n-m)!] is not hard.
Shane
________________________________________________________________________
Want to chat instantly with your online friends? Get the FREE Yahoo!
Messenger http://mail.messenger.yahoo.co.uk
-------
To unsubscribe, change your address, or temporarily deactivate your subscription,
please go to http://v2.listbox.com/member/[EMAIL PROTECTED]
