How to fix this problem when generating a pseudorandom number in Python?

Question

import hashlib
import time
def random():
    hashtime = str(int(time.time() * (10 ** 6)))
    encoded = hashlib.new("sha3_512", hashtime.encode())
    decoded = int(encoded.hexdigest(), 16)
    
    dcsmall = decoded / (10 ** (len(str(decoded))))
    return (dcsmall)

I tried this code to simulate the function random.random() without the module random. I wanted to generate a random number between 0 and 1 but this code doesn’t output numbers between 0 and 0.1 because of the length of "decoded" and I have no idea how to fix it.
Please don’t make huge changes to my code but instead give me an idea how to solve the problem.
Thank you

Asked By: Linus 1804

||

Source

Answer 1

There is a discrepancy between the range of your decode value and the denominator of your division. One is a binary value and the other is decimal based. The potential range of the resulting fraction cannot be 0…1 because the maximum value of decode does not match the length of its string representation in number of digits.

What you need to do is take a chunk of the value where you can force or determine the minimum and maximum range of decode. I would suggest using a modulo of the decode value and use the same denominator to obtain a 0…1 fraction:

dcsmall = decoded%2**64 / 2**64

Answered By: Alain T.

Answer 2

This might do it:

import hashlib
import time

def random():
    hashtime = str(int(time.time() * (10 ** 6)))
    encoded = hashlib.new("sha3_512", hashtime.encode())
    decoded = int(encoded.hexdigest(), 16)
    
    dcsmall = decoded / (2 ** 512)
    return (dcsmall)

SHA3-512 hashes are 512 bits long – no need to worry about how long they are in decimal or hexadecimal notation.

Answered By: Lover of Structure

How to fix this problem when generating a pseudorandom number in Python?

Question:

Answers: