What is the best way to check the strength of a password?
https://stackoverflow.com/questions/75057
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What is the best way of ensuring that a user supplied password is a strong password in a registration or change password form?
One idea I had (in python)
def validate_password(passwd):
conditions_met = 0
conditions_total = 3
if len(passwd) >= 6:
if passwd.lower() != passwd: conditions_met += 1
if len([x for x in passwd if x.isdigit()]) > 0: conditions_met += 1
if len([x for x in passwd if not x.isalnum()]) > 0: conditions_met += 1
result = False
print conditions_met
if conditions_met >= 2: result = True
return result
Solution
1: Eliminate often used passwords
Check the entered passwords against a list of often used passwords (see e.g. the top 100.000 passwords in the leaked LinkedIn password list: http://www.adeptus-mechanicus.com/codex/linkhap/combo_not.zip), make sure to include leetspeek substitutions:
A@, E3, B8, S5, etc.
Remove parts of the password that hit against this list from the entered phrase, before going to part 2 below.
2: Don't force any rules on the user
The golden rule of passwords is that longer is better.
Forget about forced use of caps, numbers, and symbols because (the vast majority of) users will:
- Make the first letter a capital;
- Put the number 1 at the end;
- Put a ! after that if a symbol is required.
Instead check password strength
For a decent starting point see: http://www.passwordmeter.com/
I suggest as a minimum the following rules:
Additions (better passwords)
-----------------------------
- Number of Characters Flat +(n*4)
- Uppercase Letters Cond/Incr +((len-n)*2)
- Lowercase Letters Cond/Incr +((len-n)*2)
- Numbers Cond +(n*4)
- Symbols Flat +(n*6)
- Middle Numbers or Symbols Flat +(n*2)
- Shannon Entropy Complex *EntropyScore
Deductions (worse passwords)
-----------------------------
- Letters Only Flat -n
- Numbers Only Flat -(n*16)
- Repeat Chars (Case Insensitive) Complex -
- Consecutive Uppercase Letters Flat -(n*2)
- Consecutive Lowercase Letters Flat -(n*2)
- Consecutive Numbers Flat -(n*2)
- Sequential Letters (3+) Flat -(n*3)
- Sequential Numbers (3+) Flat -(n*3)
- Sequential Symbols (3+) Flat -(n*3)
- Repeated words Complex -
- Only 1st char is uppercase Flat -n
- Last (non symbol) char is number Flat -n
- Only last char is symbol Flat -n
Just following passwordmeter is not enough, because sure enough its naive algorithm sees Password1! as good, whereas it is exceptionally weak. Make sure to disregard initial capital letters when scoring as well as trailing numbers and symbols (as per the last 3 rules).
Calculating Shannon entropy
See: Fastest way to compute entropy in Python
3: Don't allow any passwords that are too weak
Rather than forcing the user to bend to self-defeating rules, allow anything that will give a high enough score. How high depends on your use case.
And most importantly
When you accept the password and store it in a database, make sure to salt and hash it!.
OTHER TIPS
Depending on the language, I usually use regular expressions to check if it has:
- At least one uppercase and one lowercase letter
- At least one number
- At least one special character
- A length of at least six characters
You can require all of the above, or use a strength meter type of script. For my strength meter, if the password has the right length, it is evaluated as follows:
- One condition met: weak password
- Two conditions met: medium password
- All conditions met: strong password
You can adjust the above to meet your needs.
The object-oriented approach would be a set of rules. Assign a weight to each rule and iterate through them. In psuedo-code:
abstract class Rule {
float weight;
float calculateScore( string password );
}
Calculating the total score:
float getPasswordStrength( string password ) {
float totalWeight = 0.0f;
float totalScore = 0.0f;
foreach ( rule in rules ) {
totalWeight += weight;
totalScore += rule.calculateScore( password ) * rule.weight;
}
return (totalScore / totalWeight) / rules.count;
}
An example rule algorithm, based on number of character classes present:
float calculateScore( string password ) {
float score = 0.0f;
// NUMBER_CLASS is a constant char array { '0', '1', '2', ... }
if ( password.contains( NUMBER_CLASS ) )
score += 1.0f;
if ( password.contains( UPPERCASE_CLASS ) )
score += 1.0f;
if ( password.contains( LOWERCASE_CLASS ) )
score += 1.0f;
// Sub rule as private method
if ( containsPunctuation( password ) )
score += 1.0f;
return score / 4.0f;
}
The two simplest metrics to check for are:
- Length. I'd say 8 characters as a minimum.
- Number of different character classes the password contains. These are usually, lowercase letters, uppercase letters, numbers and punctuation and other symbols. A strong password will contain characters from at least three of these classes; if you force a number or other non-alphabetic character you significantly reduce the effectiveness of dictionary attacks.
Cracklib is great, and in newer packages there is a Python module available for it. However, on systems that don't yet have it, such as CentOS 5, I've written a ctypes wrapper for the system cryptlib. This would also work on a system that you can't install python-libcrypt. It does require python with ctypes available, so for CentOS 5 you have to install and use the python26 package.
It also has the advantage that it can take the username and check for passwords that contain it or are substantially similar, like the libcrypt "FascistGecos" function but without requiring the user to exist in /etc/passwd.
My ctypescracklib library is available on github
Some example uses:
>>> FascistCheck('jafo1234', 'jafo')
'it is based on your username'
>>> FascistCheck('myofaj123', 'jafo')
'it is based on your username'
>>> FascistCheck('jxayfoxo', 'jafo')
'it is too similar to your username'
>>> FascistCheck('cretse')
'it is based on a dictionary word'
after reading the other helpful answers, this is what i'm going with:
-1 same as username
+0 contains username
+1 more than 7 chars
+1 more than 11 chars
+1 contains digits
+1 mix of lower and uppercase
+1 contains punctuation
+1 non-printable char
pwscore.py:
import re
import string
max_score = 6
def score(username,passwd):
if passwd == username:
return -1
if username in passwd:
return 0
score = 0
if len(passwd) > 7:
score+=1
if len(passwd) > 11:
score+=1
if re.search('\d+',passwd):
score+=1
if re.search('[a-z]',passwd) and re.search('[A-Z]',passwd):
score+=1
if len([x for x in passwd if x in string.punctuation]) > 0:
score+=1
if len([x for x in passwd if x not in string.printable]) > 0:
score+=1
return score
example usage:
import pwscore
score = pwscore(username,passwd)
if score < 3:
return "weak password (score="
+ str(score) + "/"
+ str(pwscore.max_score)
+ "), try again."
probably not the most efficient, but seems reasonable. not sure FascistCheck => 'too similar to username' is worth it.
'abc123ABC!@£' = score 6/6 if not a superset of username
maybe that should score lower.
There's the open and free John the Ripper password cracker which is a great way to check an existing password database.
Well this is what I use:
var getStrength = function (passwd) {
intScore = 0;
intScore = (intScore + passwd.length);
if (passwd.match(/[a-z]/)) {
intScore = (intScore + 1);
}
if (passwd.match(/[A-Z]/)) {
intScore = (intScore + 5);
}
if (passwd.match(/\d+/)) {
intScore = (intScore + 5);
}
if (passwd.match(/(\d.*\d)/)) {
intScore = (intScore + 5);
}
if (passwd.match(/[!,@#$%^&*?_~]/)) {
intScore = (intScore + 5);
}
if (passwd.match(/([!,@#$%^&*?_~].*[!,@#$%^&*?_~])/)) {
intScore = (intScore + 5);
}
if (passwd.match(/[a-z]/) && passwd.match(/[A-Z]/)) {
intScore = (intScore + 2);
}
if (passwd.match(/\d/) && passwd.match(/\D/)) {
intScore = (intScore + 2);
}
if (passwd.match(/[a-z]/) && passwd.match(/[A-Z]/) && passwd.match(/\d/) && passwd.match(/[!,@#$%^&*?_~]/)) {
intScore = (intScore + 2);
}
return intScore;
}
In addition to the standard approach of mixing alpha,numeric and symbols, I noticed when I registered with MyOpenId last week, the password checker tells you if your password is based on a dictionary word, even if you add numbers or replace alphas with similar numbers (using zero instead of 'o', '1' instead of 'i', etc.).
I was quite impressed.
I wrote a small Javascript application. Take a look: Yet Another Password Meter. You can download the source and use/modify it under GPL. Have fun!
I don't know if anyone will find this useful, but I really liked the idea of a ruleset as suggested by phear so I went and wrote a rule Python 2.6 class (although it's probably compatible with 2.5):
import re
class SecurityException(Exception):
pass
class Rule:
"""Creates a rule to evaluate against a string.
Rules can be regex patterns or a boolean returning function.
Whether a rule is inclusive or exclusive is decided by the sign
of the weight. Positive weights are inclusive, negative weights are
exclusive.
Call score() to return either 0 or the weight if the rule
is fufilled.
Raises a SecurityException if a required rule is violated.
"""
def __init__(self,rule,weight=1,required=False,name=u"The Unnamed Rule"):
try:
getattr(rule,"__call__")
except AttributeError:
self.rule = re.compile(rule) # If a regex, compile
else:
self.rule = rule # Otherwise it's a function and it should be scored using it
if weight == 0:
return ValueError(u"Weights can not be 0")
self.weight = weight
self.required = required
self.name = name
def exclusive(self):
return self.weight < 0
def inclusive(self):
return self.weight >= 0
exclusive = property(exclusive)
inclusive = property(inclusive)
def _score_regex(self,password):
match = self.rule.search(password)
if match is None:
if self.exclusive: # didn't match an exclusive rule
return self.weight
elif self.inclusive and self.required: # didn't match on a required inclusive rule
raise SecurityException(u"Violation of Rule: %s by input \"%s\"" % (self.name.title(), password))
elif self.inclusive and not self.required:
return 0
else:
if self.inclusive:
return self.weight
elif self.exclusive and self.required:
raise SecurityException(u"Violation of Rule: %s by input \"%s\"" % (self.name,password))
elif self.exclusive and not self.required:
return 0
return 0
def score(self,password):
try:
getattr(self.rule,"__call__")
except AttributeError:
return self._score_regex(password)
else:
return self.rule(password) * self.weight
def __unicode__(self):
return u"%s (%i)" % (self.name.title(), self.weight)
def __str__(self):
return self.__unicode__()
I hope someone finds this useful!
Example Usage:
rules = [ Rule("^foobar",weight=20,required=True,name=u"The Fubared Rule"), ]
try:
score = 0
for rule in rules:
score += rule.score()
except SecurityException e:
print e
else:
print score
DISCLAIMER: Not unit tested
If you have the time, run a password cracker against it.
Password strength checkers, and if you have time+resources (its justified only if you are checking for more than a few passwords) use Rainbow Tables.
With a series of checks to ensure it meets minimum criteria:
- at least 8 characters long
- contains at least one non-alphanumeric symbol
- does not match or contain username/email/etc.
- etc
Here's a jQuery plugin that reports password strength (not tried it myself): http://phiras.wordpress.com/2007/04/08/password-strength-meter-a-jquery-plugin/
And the same thing ported to PHP: http://www.alixaxel.com/wordpress/2007/06/09/php-password-strength-algorithm/
What is the best way of ensuring that a user supplied password is a strong password in a registration or change password form?
Don't evaluate complexity and or strength, users will find a way to fool your system or get so frustrated that they will leave. That will only gets you situations like this. Just require certain length and that leaked passwords aren't used. Bonus points: make sure whatever you implement allows the use of password managers and/or 2FA.
