[CentOS] PAM configuration?

Thomas Harold thomas-lists at nybeta.com
Tue Jan 5 11:20:52 UTC 2010

On 1/4/2010 12:42 PM, Roland Roland wrote:
> also is there a way I could enable the PAM module which uses crack
> library to check the strength of a users password?
> any help with this is truly appreciated...


The default is:

password    requisite     pam_cracklib.so try_first_pass retry=3
password    sufficient    pam_unix.so md5 shadow nullok try_first_pass 
password    required      pam_deny.so

A stronger version is:

# See: http://www.deer-run.com/~hal/sysadmin/pam_cracklib.html
password    requisite     pam_cracklib.so try_first_pass retry=3 
minlen=20 difok=5
password    sufficient    pam_unix.so md5 shadow nullok try_first_pass 
use_authtok remember=36
password    required      pam_deny.so

(Note that the entire line that starts with "password" should be all on 
one line.  It'll make sense when you view the system-auth-ac file. 
There are 3 password lines by default, and all you're going to do is add 
options to the first two lines.)

The key changes that I made to my setup are:

- Added "minlen=20", note that cracklib gives a bonus point if you use a 
number, symbol or different case character.  So the minimum length is as 
short as 16 characters (4 less then what you set minlen to).  But that 
minimum length is only achievable if you use both upper and lower case 
letters along with at least one symbol and at least one number.  Minimum 
length should never be below 10 or 12 (in my opinion).

- The 'try_first_pass' tells cracklib to do some checking of the 
supplied password against a built-in word list.  I don't remember the 
details behind how it works.

- Added "difok=5", which says that the new password (when the user 
changes it) has to be at least 5 letters different from the old 
password.  In general, you'll want this to be about 1/3 to 1/4 of the 
minlen value.

- Added remember=36, which tells pam_unix to remember the last 36 
passwords.  So if a user wants to change their password, it can't be any 
of the past 36 passwords.  Which is probably overkill to the Nth degree.


(Minor ramble about why I say minlen should be set to at least 10-12.)

My current estimate is that a $1500 PC can brute-force about 2-4 billion 
MD5 password hashes per second now.  (Using NVIDIA Cuda in a 4-way SLI 
setup.)  That's an offline attack where the attacker has a copy of your 
password hash.  A completely random 8-char password can be found in 
about half a day.  Add 2 more characters and it'll takes more like 2 

Things go a lot slower if they're simply doing a dictionary attack on 
the SSH port (where they don't have the MD5 hash).  Those attacks 
typically go after low-hanging fruit using common usernames + common 
passwords.  Plus you can throttle the SSH logins or do other things (at 
the risk of locking yourself out).  Keep in mind that modern SSH 
dictionary attacks ping your machine about once every few minutes from 
thousands of different IP addresses.  Locking out an IP address after 3 
incorrect attempts only works against attackers that aren't using a 
slow-attack botnet.

So in situations where you can throttle the attacker, 8 random chars is 
probably enough.  But if you want something a little easier to type, 
you'd best go for 10-12 chars and assume that the attacker has the hash.

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