Is this secure?

[Lie Ryan]

On 02/25/2010 06:16 AM, mk wrote:
> On 2010-02-24 20:01, Robert Kern wrote:
>> I will repeat my advice to just use random.SystemRandom.choice() instead
>> of trying to interpret the bytes from /dev/urandom directly.
> 
> Out of curiosity:
> 
> def gen_rand_string(length):
>     prng = random.SystemRandom()
>     chars = []
>     for i in range(length):
>         chars.append(prng.choice('abcdefghijklmnopqrstuvwxyz'))
>     return ''.join(chars)
> 
> if __name__ == "__main__":
>     chardict = {}
>     for i in range(10000):
> ##        w = gen_rand_word(10)
>         w = gen_rand_string(10)
>         count_chars(chardict, w)
>     counts = list(chardict.items())
>     counts.sort(key = operator.itemgetter(1), reverse = True)
>     for char, count in counts:
>         print char, count
> 
> 
> s 3966
> d 3912
> g 3909
> h 3905
> a 3901
> u 3900
> q 3891
> m 3888
> k 3884
> b 3878
> x 3875
> v 3867
> w 3864
> y 3851
> l 3825
> z 3821
> c 3819
> e 3819
> r 3816
> n 3808
> o 3797
> f 3795
> t 3784
> p 3765
> j 3730
> i 3704
> 
> Better, although still not perfect.
> 

I give you this:

I give you this:

import itertools
def gen():
    valid_chars = 'abcdefghijklmnopqrstuvwxyz'
    for char in itertools.repeat(valid_chars):
        yield char

gen = gen()
def gen_rand_string(length):
    chars = (next(gen) for i in range(length))
    return ''.join(chars)

since it gives me a perfect distribution of letters, it must be a very
secure random password generation scheme.