Is the following password-generator secure?

Question

i use the following password-generator (python) and would like to ask if the generated passwords (usually 32 characters and more) are random enough.

def pwgen(l=32):
    from random import Random
    rng = Random()

    a = "qwertzuiopasdfghjklyxcvbnm"
    n = "1234567890"
    r = "23456qwertasdfgzxcvbQWERTASDFGZXCVOB"
    l = "78901yuiophjklnmYUIPHJKLNM"

    all = a + n + r  + l

    pwl = 32
    try:
        pwl = int(l)
    except:
        pass    
    if pwl < 8  or pwl > 128:
        pwl = 32

    pw = []
    for j in range(pwl):
        pw.append(rng.choice(all))

    return("".join(pw))

the password produced look the following way:

cwVPBgYt5seoO97sg8fszNptAhnb60Jg
W3mkuLMH0HCiJ8L040pmMys8yvhncwI8
kro01d2Z2ft51WG8lmy9d5el50aZ7CUg
10QfutP1fvKdR8oVhdTt0xR0r5k7w7Md
OxRjsqs2rCjyVbalvhuw3nrjTu9096XF
6jjvkIvyeoei3YIz84XAgdsDRjIGjyRu
2bjnHvubpv2z2DOF9fjFDfLtdnf4mL3T
UhbhNvwdSKOj14hmbca7t2UiwRaJNwrC

Answer 1

a = "qwertzuiopasdfghjklyxcvbnm"
n = "1234567890"
r = "23456qwertasdfgzxcvbQWERTASDFGZXCVOB"
l = "78901yuiophjklnmYUIPHJKLNM"

all = a + n + r  + l

Why is this so complicated? By including some characters multiple times, you make those characters more likely to appear in passwords, which weakens the password. By making your password string so complicated, you make it hard to determine how secure the passwords are.

To get an even chance of any character, just list all of them once. Also, you can dramatically simplify this using the string module.

import string

all = string.ascii_letters + string.digits

Another problem is that the random module's documentation says it shouldn't be used for cryptography:

Python uses the Mersenne Twister as the core generator. It produces 53-bit precision floats and has a period of 2**19937-1. The underlying implementation in C is both fast and threadsafe. The Mersenne Twister is one of the most extensively tested random number generators in existence. However, being completely deterministic, it is not suitable for all purposes, and is completely unsuitable for cryptographic purposes.

[...]

Warning: The pseudo-random generators of this module should not be used for security purposes. Use os.urandom() or SystemRandom if you require a cryptographically secure pseudo-random number generator.**

So, use random.SystemRandom() instead:

import random
import string

# Consider adding string.punctuation
possible_characters = string.ascii_letters + string.digits

def generate_password(length=32):
    rng = random.SystemRandom()
    return "".join([rng.choice(possible_characters) for i in range(length)])

Since we know exactly how many possible characters there are (62), and we know that they should be randomly distributed, we can say with fairly high confidence that a password generated by this scheme will on average require (62^32) / 2 (about 10^57) attempts to guess.

That said, a 32 character password is probably secure enough, even if it's not optimally random.

Answer 2

Python's Random class is not a cryptographically secure random number generator, so it is probably too predictable for attackers. Furthermore, it is seeded by a timestamp (I guess in milliseconds), severely limiting the number of potential random streams and therefore the number of passwords. It would be very easy to brute-force the passwords for each possible millisecond at which the algorithm could have run, especially when the attacker can estimate around which time it happened.

Edit: after taking a closer look at the Python documentation, it seems that Python seeds the random generator with entropy from /dev/urandom (or a Windows equivalent thereof) rather than the system time, if available. So it is not nearly as insecure as I previously implied. However, you still definately should not rely on Python's default random number generator for password generation. I'd recommend using os.urandom or random.SystemRandom instead, as indicated by the previous poster.