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Date: Sun, 28 Jan 2007 11:58:59 +0100
From: "Frank Dittrich" <frank_dittrich@...mail.com>
To: john-users@...ts.openwall.com
Subject: Re: how to find a password of 16 digits
Solar Designer wrote:
--
frank_dittrich@...mail.com
>From: Solar Designer <solar@...nwall.com>
>Reply-To: john-users@...ts.openwall.com
>To: john-users@...ts.openwall.com
>Subject: Re: [john-users] how to find a password of 16 digits
>Date: Sun, 28 Jan 2007 10:11:11 +0300
>MIME-Version: 1.0
>And generate a new .chr file:
>
> cd ../run
> ./john --make-charset=digits16.chr
>
[...]
>Now you can define a new "incremental" mode:
>
>[Incremental:Digits16]
>File = $JOHN/digits16.chr
>MinLen = 16
>MaxLen = 16
>CharCount = 10
>
>Note that this new digits16.chr will also work for lengths 0 to 15, if
>you ever need that.
Why do you need a new chr file?
Wouldn't the original digits.chr file work as well (or a poorly,
for that matter)?
Not that I think trying all password sequentially is any better.
But I doubt you gathered any useful statistical information for
16 digit "passwords".
I didn't recompile JtR for incremental mode with length 16, but your
example looks like John tries 2222222222222222, then all passwords
containing just the digits 2 and 4 (and at least one digit 4), then
all passwords containing at least one digit 0 and only digits 2, 4,
and 0, and so on.
You should get similar results by just recompiling JtR, and
using the original digits.chr file
[Incremental:Digits16]
File = $JOHN/digits.chr
MinLen = 16
MaxLen = 16
CharCount = 10
The only difference is the somewhat different character frequency
and frequency of character combinations in the john.pot file generated
according to your suggestion, compared to the john.pot file that has
been used to generate the original digits.chr file.
You can reproduce this behaviour without recompiling john.
Just assume you want to crack case insensitive passwords of length 8.
There's no suitable .chr file, so you could use:
[Incremental:lanman8]
File = $JOHN/lanman.chr
MinLen = 8
MaxLen = 8
CharCount = 69
Or, you could generate a dummy john.pot file
(BTW: make sure you don't accidently overwrite an existing
john.pot file you might still need)
./john -i=lanman --stdout | sed 's/^/:/; 1000000q' > john.pot
generate a new .chr file
./john --make-charset=lanman8.chr
and define a section
[Incremental:myy_lanman8]
File = $JOHN/lanman8.chr
MinLen = 8
MaxLen = 8
CharCount = 69
Now, both
./john -i=lanman8 --stdout|head
and
./john -i=my_lanman8 --stdout|head
produce the same output:
./john -i=lanman8 --stdout|head
There's, however a difference.
./john -i=lanman8 --stdout | grep -n -C 3 "[^AE]"|head
254-AAAAAAEA
255-AAAAAAEE
256-AAAAAAAE
257:RAAAAAAA
258:RAAAAAAE
259:RAAAAAAR
260:RAAAAAEA
261:RAAAAAEE
262:RAAAAAER
263:RAAAAARA
vs.
./john -i=my_lanman8 --stdout | grep -n -C 3 "[^AE]"|head
254-AAAAAAEA
255-AAAAAAEE
256-AAAAAAAE
257:SAAAAAAA
258:SAAAAAAE
259:SAAAAAAS
260:SAAAAAEA
261:SAAAAAEE
262:SAAAAAES
263:SAAAAASA
And:
./john -i=lanman8 --stdout | grep -n -C 3 "[^AER]"|head
6559-EEEEEAER
6560-EEEEEEAR
6561-EEEEEEER
6562:OAAAAAAA
6563:OAAAAAAE
6564:OAAAAAAR
6565:OAAAAAAO
6566:OAAAAAEA
6567:OAAAAAEE
6568:OAAAAAER
vs.
./john -i=my_lanman8 --stdout | grep -n -C 3 "[^AES]"|head
6559-EEEEEAES
6560-EEEEEEAS
6561-EEEEEEES
6562:OAAAAAAA
6563:OAAAAAAE
6564:OAAAAAAS
6565:OAAAAAAO
6566:OAAAAAEA
6567:OAAAAAEE
6568:OAAAAAES
In my_lanman8.chr, letter R has a higher probability than
letter O (and, among others, letter S), while in lanman.chr,
letter S has a higher probability than letter O (and, among others,
letter R).
But, in both cases, prior to generating any password which contains
a new letter, john first generates all passwords containing only
characters which have been used in passwords generated earlier.
Passwords up to number 1**8 contasin just 1 character (obviously),
all passwords up to number 2**8 contain just 2 different characters,
all passwords up to 3**8 (6561) contain just 3 different characters,
all passwords up to 4**8 (65536) contain just 4 different characters.
For the 16 digits example, that means, the first password which
doesn't contain just the 2 most probable digits (2 amd 4, in our
case), is password number 2**16 + 1 = 65537, the first password
containing any other character than 2, 4, and 0, will be password
number 3**16 + 1 = 43046722.
IMO this proves that such statistical information from shorter
passwords is of limited use for longer passwords.
I just want to mention this to emphasize that, without additional
knowledge about the password, you shouldn't hope to find the
correct password easily, no matter what method you use,
because the key space of 16 digit passwords is larger than the
key space of DES passwords (up to 8 characters, with CharCount 95).
Regards, Frank
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