SV: Brute-forcing cached Windows login password hashes



Sorry, my bad. Anyway; doing a dictionary/hybrid attack will probably give
you access a lot faster. I've done quite a bit of password audits on Windows
systems over the last 9 years or so, and based on my experience you should
get 3-10% of all passwords in a domain within a few minutes of running a
simple dictionary logon attack.

Then again; why break the passwords, as pass-the-hash is fully possible in
most Windows environments?

Regards,
Per Thorsheim


-----Opprinnelig melding-----
Fra: listbounce@xxxxxxxxxxxxxxxxx [mailto:listbounce@xxxxxxxxxxxxxxxxx] På
vegne av Carl Livitt
Sendt: 26. juli 2007 16:39
Til: Ben Greenberg; pen-test@xxxxxxxxxxxxxxxxx
Emne: Re: Brute-forcing cached Windows login password hashes


The hash algorithm is a salted MD4. It's impossible (ok, to be pedantic it's
mathematically infeasible) to use rainbow tables because of the salting, so
that leaves you with dictionary and brute-force.

The latest version of John and the MS Cache Hash patches are all available
from http://openwall.com/john/. I believe v1.7.2 is the latest version.

Regards,
Carl


Ben Greenberg wrote:
Greetings all,

My question is regarding the encrypted password hashes that Windows
stores in the registry of the last 10 logins to a workstation.

I read the original white paper written by Arnaud Pilon and I've used
his cachedump tool to extract the password hashes from the registry.
What I'm wondering is what type of hash those passwords use. Is it
straight MD4? I know that each hash is salted with a machine-specific
unique string. What I am unclear on is what exactly the password hash
is and how it can be brute-forced. I know that there is a patch for
John the Ripper, but every mention I can find refers to a two year old
version of John. Does anyone know if the most recent version has this
patch in it already? Also, is anyone familiar with any rainbow tables
for cracking these passwords? Are rainbow tables possible for these hashes
because of the salting?

Thanks all.

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