Simple square root calculator

Question

Very simple square root calculator. I am sure it's terrible and can be made much better.

I feel it's way too many lines.

import random
## ask user for a number to find sq root for
num1 = input("Please enter number for which we'll find sq root:\n")
## generate a random number and test it
randomNum1 = random.randint(1,9)
print("Random number is " + str(randomNum1))
ranMul1 = 0
while True:
 if float(ranMul1) == float(num1):
 print("square root of " + str(num1) + " is " + str(randomNum1))
 break
 else:
 randomNum1 = round((float(randomNum1) + (float(num1)/float(randomNum1))) / 2, 4)
 ranMul1 = round(randomNum1 * randomNum1,4)
 print("Trying number " + str(randomNum1))
 continue

• waltinator: Search for "Newton's method for square roots".

Answer 1

Welcome to the community :). A few pointers:

1. The code does not follow PEP-8 style guide. You should follow the snake_case naming convention for variables.
2. Split code into separate functions.
3. Instead of the code running as is, the execution condition should be placed inside an if __name__ block.
4. If you're on python 3.6+, make use of f-strings over concatenating separate variables and casting them to strings over and over.
5. Cast the input from user as float as soon as it is read.
6. I don't think you need to round the random variable calculated to 4 digits. For print purpose, you can use the precision specifier.
7. Again, if using python 3.6+, you can also provide type hinting.

---

The above mentioned points, and a few more changes, your code would be:

from math import isclose
from random import randint
def calculate_sq_root(num: float) -> float:
 "generate a random number and manipulate it until we calculate square root of input number"
 random_num = randint(1, 9)
 print(f"Random number is {random_num}.")
 while not isclose(num, random_num * random_num):
 random_num = (random_num ** 2 + num) / (random_num * 2)
 print(f"Trying number {random_num:.4f}")
 return random_num
def get_input() -> float:
 "ask user for a number to find sq root for"
 return float(input("Please enter number for which we'll find sq root:\n"))
def main():
 user_num = get_input()
 sq_root = calculate_sq_root(user_num)
 print(f"Square root of {user_num} is {sq_root}.")
if __name__ == "__main__":
 main()

• Reinderien: Your docstring for calculate_sq_root should use triple-quotes, not quotes.
• Reinderien: Also, the cast to float is redundant here: float(num / random_num) - Python will do type promotion.
• hjpotter92: @Reinderien yes. those points are mentioned in pep 8. I edited the code in tio.run page itself, so not much formatting could be done.

Answer 2

The great input from @hjpotter92 gets you 99% of the way there. The only additional comment I have is about this:

while num != random_num * random_num:

Exact comparison of floating-point numbers is problematic. You're going to want to pick some (very small) number, usually called epsilon, below which the error is insignificant and call that "close enough".

Read https://docs.python.org/3/library/math.html#math.isclose for more details.