Re: [gentoo-user] [OT] Script to crack gpg passphrase

[Michael Orlitzky <michael@orlitzky.com>]

On 04/28/2011 10:04 AM, Mick wrote:
> This finds the passphrase and prints it out on the terminal.  However, its
> success depends on the dictionary file I use.  Also, it's not particularly
> fast ...
> 
> Any idea how I can create a dictionary file?  I've used apg but it's <aheam!>
> too random.  :-)
> 
> I have been given something like 6 passphrases that may have been used.  The
> problem is that at the time of creation the passphrase was typed in
> incorrectly (twice!)  So I would need to use some method of generating a
> dictionary with potential typos of these known passphrases (pretty much how
> the rephrase application works).  What is a good way to generate such a file
> by imputing a range of candidate characters?
> 
> Finally, is there a way or parallelising the run so that it speeds up?
> 

If you know the actual passphrase is "close" to some known passphrase,
read up on the concept of Levenshtein distance:

  http://en.wikipedia.org/wiki/Levenshtein_distance

If you can define a function that computes all passwords of Levenshtein
distance 1 from a given password, then you can run that function twice
to get the password of distance 2,3... Three or four is probably a good
place to stop.

Then you can take those six known passwords, and compute all words of
distance 1,2,3 from all of them. Stick those in a list, and try them.