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Re: Noob dumb question (extending emacs)

From: Michael Heerdegen
Subject: Re: Noob dumb question (extending emacs)
Date: Sun, 24 Oct 2021 10:16:29 +0200
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
<> 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

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,

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

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


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