Is this code capable of generating Cryptographically Secure Pseudo-Random Number Generator

Question

I have a piece of code which generates a string of length 50 to 100 with random characters

the code is :

CharLIST = list('abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890')
def NEW(id_len):
    _id = ''
    while len(_id) < id_len:
        _id += random.choice(CharLIST)
    return _id
NEW(random.randint(50, 100))

I expect it to generate a completely random string which is not unpredictable
I used a code analyser which said it is not Cryptographically Secure
but it generate random strings which I think are unpredictable based on seeing 200 results

Asked By: shashank A

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Answer 1

No, none of the utilities in random should be considered secure. Their output is completely deterministic.
This is emphasized in warnings at the top of the module’s documentation:

Python uses the Mersenne Twister as the core generator …. [which], being completely deterministic, 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. For security or cryptographic uses, see the secrets module.

As suggested, for cryptographically secure random number generation, you should use the secrets module:

The secrets module is used for generating cryptographically strong random numbers suitable for managing data such as passwords, account authentication, security tokens, and related secrets.

In particular, secrets should be used in preference to the default pseudo-random number generator in the random module, which is designed for modelling and simulation, not security or cryptography.

For your particular use case, the secrets module provides a secrets.choice function that works analogously to random.choice.

Answered By: Brian61354270

Answer 2

Like @Brian said, random isn’t secure with encrypting. Use the secrets module with the built in choice function instead of random. You can also use the string module to shorten your combinations:

import string
import secrets


CharLIST = string.digits + string.ascii_letters

def NEW():
  myPass = "".join(secrets.choice(CharLIST) for i in range(50,100))
  return myPass


print(NEW())

Answered By: Ray953

Is this code capable of generating Cryptographically Secure Pseudo-Random Number Generator

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