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Re: Noob dumb question (extending emacs)


From: Michael Heerdegen
Subject: Re: Noob dumb question (extending emacs)
Date: Sun, 31 Oct 2021 21:42:59 +0100
User-agent: Gnus/5.13 (Gnus v5.13) Emacs/29.0.50 (gnu/linux)

Emanuel Berg via Users list for the GNU Emacs text editor
<help-gnu-emacs@gnu.org> writes:

> > Or using the formula mentioned,
> >
> >   ln(N)     ln(10)
> >   ------ = -------- < 1.
> >   ln(b)     ln(60)
>
> I don't think I saw that, what does it say?

It's really only just counting.  The "10" above is the same as the 2^48
below - the total number of different passwords.

> I've seen the first one, this
>
>   48 bits, alphabet length n, password length l
>   2^48     = n^l            <=>
>   48*ln(2) = l*ln(n)        <=>
>   l        = 48*ln(2)/ln(n)
>
> which in Elisp is
>
> (defun epwgen-space-size-2 (bits abc-len)
>   (/ (* bits (log 2)) (log abc-len) ))
>
> ;; (epwgen-space-size-2 48 60) ; 8.13
>
> Are you saying, that it is saying, "with 48 bits passwords
> longer than 8.13 chars don't make it more secure"?

For brute force attacks and when you know how the passwords are
generated, then yes.  If you don't publish your algorithm then passwords
are good enough (unless for some reason one can guess or reconstruct the
algorithm).

Michael.



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