Generate random integer without an upper bound

Question

I want to generate a random seed in a predictable way.

I was hoping to do this

seed = 12345
prng_0 = random.Random(seed)
prng_1 = random.Random(prng_0.rand_int(0))

There, 0 is the lower bound, but it turns out I need to give it an upper bound as well. I don’t want to set a fixed upper bound.

If you are curious about my reason, I’m doing this because I need reproducibility when testing. Namely, this is a function receiving a seed and building its prng, prng_0, then calling multiple times another function that needs to receive a different seed every time.

def funct_a(seed=None):
    prng_1 = random.Random(seed)
    prng_2 = numpy.random.RandomState(prng_1.randint(0, 4294967296))
    print(prng_1.random())
    print(prng_2.random())

def funct_b(seed=None):
    prng_0 = random.Random(seed)
    for i in range(0, 5):
        seed = prng_0.randint(0)  # not working, needs upper bound
        funct_a(seed)

funct_b(12345)  # test call

EDIT: interestingly enough, numpy (which I’m also using) has a definite upper seed value, as testified by the doc and by this error

ValueError: Seed must be between 0 and 4294967295

Asked By: Agostino

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Answer 1

You can’t avoid an upper bound. How would the code work without one? This is how the code generates a random number between x and y:

0______________________________________________r__________________________________________1

r is a random decimal between 0 and 1. This is generated with a fixed algorithm.

Then, it takes r and multiplies it by the upper bound minus the lower bound. This pretty much means that 0 becomes x, and 1 becomes y. If rand is the random number, r : (1 - 0) :: rand : (y - x)

EDIT: There actually is a way to generate a random number without an upper bound, but it is not logarithmically and not uniformly distributed. Take a look at this python algorithm:

import random
def randint():
    i = 0
    while True:
        if random.random() < 0.5: # Or whatever other probability you want
            return i
        else:
            i += 1

Pretty much, what this is doing is starting from zero, and then every time it has a 0.5 probability of returning that number; otherwise it continues.

This means that there is a 0.5 probability of it being 0, 25% for 1, 12.5% for 2, 5.25% for 3, etc. This is logarithmic distribution “without an upper bound”.

Answered By: hyper-neutrino

Answer 2

When I don’t want an upper bound I’ll often use sys.maxint for the upper bound as an approximation

Answered By: Eric Renouf

Answer 3

Mathematically there does not exist a uniform distribution on an unbounded set of integers (this was correctly pointed out in the commends). +1 to that comment to Eric Renouf’s answer.

For posterity, the issue is that the probabilities of every possible outcome must collectively sum to a total value of 1 (that’s part of the definition of a probability distribution). If the chance of choosing any one integer is a positive value p>0, then when you sum up infinitely many of those (sum_np), you get a total of infinity, not 1. This is the outcome for any positive p>0. But if p=0, then you instead get a total of 0, not 1. You could say that "in theory" there is no such thing as random integer without an upper bound.

Answered By: Thomas Winckelman

Generate random integer without an upper bound

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