Re: Noob dumb question (extending emacs)

[Michael Heerdegen]

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

> Cool, how did you compute that?

Not hard to do (unless I'm missing something): For which password length
l exist approximately 2^48 different passwords using an alphabet of n
characters?

2^48 = n^l

48 ln 2 = l ln n

l = 48 ln 2 / (ln n)

If you use lowercase and uppercase letters and, say, 8 other symbols,
n=26*2+8=60

    48 ln(2)
l = -------- ~ 8.13.
     ln(60)

That would mean that already for a length of 9 only a small fraction of
passwords are computable.


Michael.