python how to generate random numbers based on given input without using random library?

Question

Imagine there is a rand8 function which returns a random number between [0 - 7]. Now I want to make a function namely as rand11 based on the input I get from rand8 . like this:

 Input from rand8     : 1, 3 , 5  , 7 , 0 , 7 , 2 , 1 , 6 , ...
Output given by rand11: 0, 5 , 10 , 7 , 6 , 0 , 2 , 9 , 8 , ...

So far I found this online:

def lcg(modulus, a, c, seed):
    while True:
        seed = (a * seed + c) % modulus
        yield seed

a = lcg(5, 0, 8, 1)
next(a)

But I am not familiar how to modify function to get numbers between 0 to 7 and return numbers between 0 to 10. Please note that I don't need to implement rand8 function. I only have to take care of rand11 function? And remember I am not allowed to use random library in python or any other random libraries like numpy.random()

Can anyone help me?

Answer 1

This will give a list result of the output you expect. As you can see if you ran this program it gives a uniform distribution of values between 0 and 10. The library random is only used to generate a larger list of rand8 seeds.

from random import randint
import collections

result = []

#seeds = [1, 3 , 5  , 7 , 0 , 7 , 2 , 1 , 6]
seeds = [randint(0,7) for i in range(1000000)]
gen = 1
for se in seeds:
    gen = (se  + gen) % 11
    result.append(gen)

counter =collections.Counter(result)
print(counter)

If you want the function to have different results each time it runs, you can add a multiplier. Final code without showing the uniform distribution:

result = []
c = int(input("Seed?"))
seeds = [1, 3 , 5  , 7 , 0 , 7 , 2 , 1 , 6]
gen = 1
for se in seeds:
    gen = (se * c + gen) % 11
    result.append(gen)

print(result)