How to find the seed of the random number generator?

Question

I'm using the following lines of code to populate the entries of l_new from l.

import random

l = [11.1, 22.2, 33.3, 11.1, 33.3, 33.3, 22.2, 55.5]

l_new = random.choices(l, k=30)
print(l_new)

I could generate a particular l_new always by fixing the seed using

random.seed(1)
l_new = random.choices(l, k=30)

Now, I would like to know if it is possible to find the seed for a given l_new when the seed is not fixed prior to the generation of l_new.

EDIT:

From what has been suggested below can I do

import sys
import random
l = [11.1, 22.2, 33.3, 11.1, 33.3, 33.3, 22.2, 55.5]

for i in range(10):
    seed = random.randrange(sys.maxsize)
    print("Seed:", seed)
    random.seed(seed)
    l_new = random.choices(l, k=30)
    print(l_new)

and save `seed` to retrieve the required `l_new` later on?

user_7505395 · Answer

It is not ~~possible~~ feasible to get "the seed" from a certain sequence of random numbers. See below how big the internal state of a mersenne twister ist.

Theoretically you could have two different seeds that produce the first 10 numbers identically to only then deviate - the more numbers you draw the less likely it is but the possibility is non-0 so no inverse possible.

If you do not set the seed it will use a seed created from the datetime:

random.seed(a=None, version=2)

Initialize the random number generator.

If a is omitted or None, the current system time is used [...]

seed() is just a way to set the internal start state of the mersenne twister that creates the random numbers.

However: you can get this state and store it and reset it:

import random

random.seed(42)

numbers = [random.randint(2,22) for _ in range(15)] 
print(numbers) # now you have another interal state 

d = random.getstate()
numbers = [random.randint(2,22) for _ in range(15)] 
print(numbers) # after drawing more numbers you now have another interal state 

# go back to the last state
random.setstate(d)
numbers = [random.randint(2,22) for _ in range(15)] 
print(numbers) # you now draw the same exact numbers

Output:

# first numbers, using system time of 1st use as seed
[22, 5, 2, 10, 9, 9, 6, 5, 19, 4, 20, 15, 3, 2, 4]
# we remebered state and create another 15
[8, 9, 18, 21, 2, 19, 8, 22, 19, 15, 9, 16, 20, 10, 2]
# we reset state to last and create the same 15 again
[8, 9, 18, 21, 2, 19, 8, 22, 19, 15, 9, 16, 20, 10, 2]

Which kind of random method you use will influence the internal state changes.

Just in case you wondered how the state of random.seed(42) looks for me:

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     624),
 None)

user_3767239 · Answer

This is not possible due to the fact that the random output of the RNG gets mapped via floor and that relationship is not an injection:

return [population[floor(random() * n)] for i in _repeat(None, k)]

So even if you knew which elements got selected from the original list l, you cannot tell what output random produced and hence you cannot deduce the internal state of the RNG.

How to find the seed of the random number generator?

2 Answers