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Re: Noob dumb question (extending emacs)

From: Michael Heerdegen
Subject: Re: Noob dumb question (extending emacs)
Date: Wed, 03 Nov 2021 22:24:55 +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
<> writes:

> What does that formula communicate?

For given entropy N, a given number, it returns for what word length
there are approximately N different (pass)words.

> The password length is a function of the number of bits of
> entropy and the alphabet size? What password is that?
> A password of just that length? ... ???

The functions counts words, all possible (pass)words, no password in

> And say that the alphabet is A = {a, b, c} then |A| = 3, the
> entropy bits is decided by /dev/urandom, so that doesn't
> change either, how can the formula hold for passwords of
> different lengths from that alphabet, e.g. "a" (length 1),
> "aa" (2), "abc" (3), and so on?

The formula returns a length.  So the question doesn't make much sense
to me.  Unless you ask why I don't say "maximum password length" and
don't include smaller passwords in the count.  That would not
significantly change the result though.

If these were not the answers you expect I think we are somehow
miscommunicating or I just don't express myself very well.

All I did was counting how many different words of a given length exist
for a fixed alphabet.  Or did you not understand why that matters here?


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