Date: Mon, 29 Apr 2013 09:38:14 -0500 From: Jeffrey Goldberg <jeffrey@...dmark.org> To: crypt-dev@...ts.openwall.com Subject: Re: Representing the crack resistance of a password. On 2013-04-29, at 7:26 AM, Solar Designer <solar@...nwall.com> wrote: > On Mon, Apr 29, 2013 at 01:05:30AM -0500, Jeffrey Goldberg wrote: >> I asked how we should characterize, or even name, this notion. I tossed out >> >> C(X, k) = 2 log_2 G(0.5, X, k) And now that I've had a few hours sleep, I see that even if that is the right concept, I got the math wrong. Should be C(X, k) = log_2 G(0.5, X, k) +1 (or log_2(2G(0.5, X, k)) Maybe I'll do even better if I get coffee. > Regarding your G() above, see: > > http://www.lysator.liu.se/~jc/mthesis/4_Entropy.html#SECTION00430000000000000000 > > for a formal definition of "Guessing entropy" and some discussion. Thank you! I would have gotten there eventually as I was starting to work through "Testing Metrics for Password Creation Policies by Attacking Large Sets of Revealed Passwords" (Weir et al.) http://goo.gl/YxRk Using Guessing Entropy, then my C (crack resistance) would be C(X) = log_2 (2G(X)). So here if X is a uniform distribution, C(X) == H(X). But I am looking for something more (and so maybe "entropy" isn't the right analogy). I want to be able to talk about the crack resistance of a particular password given a distribution. Suppose that a password policy is "at least 8 characters, at least one uppercase letter, at least one digit" but X is the actual distribution of what humans do when told to create a password under that policy. I would want C(X, 'Password1') < C(X, '2n8PGUoPeb') So instead of having G be the average number of guesses needed for a random x in X; I'd like to be able to talk about the strength of a particular password (with respect to a particular distribution of passwords). Cheers, -j
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