resolution lower bound

resolution_lower.py

import numpy as np

# generate a random number between 0 and 1
# calculate its distance (min) to either 0 and 1 (absolute value of difference)
def do_once():
    x = np.random.rand()
    return min(abs(x - 0), abs(x - 1))

# do it many times and get the average value
def do_many_times(n):
    return np.mean([do_once() for _ in range(n)])


# generalize do_once to take in a parameter k, if k is 1, the distance is measured to 0/1 and 1/1
# if k is 2, the distance is 0/2, 1/2, 2/2, and if k is 3, 0/3, 1/3, 2/3, 3/3, so on so forth
def do_once(k):
    x = np.random.rand()
    return min(abs(x - i/k) for i in range(k+1))

# do it many times and get the average value
def do_many_times(n, k):
    return np.mean([do_once(k) for _ in range(n)])

print (f'denominator = 1', "resolution lower bound = ",  do_many_times(1000000, 1))
print (f'denominator = 2', "resolution lower bound = ",  do_many_times(1000000, 2))
print (f'denominator = 3', "resolution lower bound = ",  do_many_times(1000000, 3))
print (f'denominator = 4', "resolution lower bound = ",  do_many_times(1000000, 4))
print (f'denominator = 5', "resolution lower bound = ",  do_many_times(1000000, 5))
print (f'denominator = 6', "resolution lower bound = ",  do_many_times(1000000, 6))
print (f'denominator = 7', "resolution lower bound = ",  do_many_times(1000000, 7))
print (f'denominator = 8', "resolution lower bound = ",  do_many_times(1000000, 8))