Question

We will now simulate numpy.random(), which produces floating-point values in the range [0..1).

To do this, it is simply necessary to convert integers in the range [0..?)[0..m) returned by my_random_int() to floating point numbers in the range [0..1).[0..1).

Hint: Easily done by division!

# my_random returns a random floating-point number within [0, 1)

def my_random():
# your code here
return 0   # just to get it to compile

# Test it!
my_seed(0)
for x in range(0,10):
print(my_random())

CODE GIVEN:

a = 914334
m = 2**22 - 3

#a = 3 # You will use these in part (b)
#m = 7

def hash(x):
return (a * x) % m

# Test it!
X = [231,45,123,87,133,123]
for x in X:
print(hash(x))

next_value = 1 # just to create the variable

# seed next_value with the hash of (n+1)
def my_seed(n):
global next_value
next_value = hash(n+1) # so values do not start with seed and are not 0   

# my_random_int() returns a random number generated by the hash function, in the range [0..(m-1)].

def my_random_int():
global next_value
next_value = hash(next_value)
return next_value
  
# Test it

my_seed(0)

for x in range(10):
print(my_random_int())

Answer 1

python code

import numpy as np

# my_random returns a random floating-point number within (0, 1)

def my_random():
# return random float numbers
return np.random.uniform(0, 1)   

# Test it!
#my_seed(0)
for x in range(0,10):
print(my_random())#print numbers

code &output:

//i have given the needed code alone, please do comments if you need any clarification or modification.