Python Program to Check Prime Number

Given a positive integer N, The task is to write a Python program to check if the number is Prime or not in Python.

Examples: 

Input:  n = 11
Output: True


Input:  n = 1
Output: False

Explanation: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The first few prime numbers are {2, 3, 5, 7, 11, ….}. 

Prime Number Program in Python 

The idea to solve this problem is to iterate through all the numbers starting from 2 to (N/2) using a for loop and for every number check if it divides N. If we find any number that divides, we return false. If we did not find any number between 2 and N/2 which divides N then it means that N is prime and we will return True.

Python3

num = 11
# If given number is greater than 1
if num > 1:
    # Iterate from 2 to n / 2
    for i in range(2, int(num/2)+1):
        # If num is divisible by any number between
        # 2 and n / 2, it is not prime
        if (num % i) == 0:
            print(num, "is not a prime number")
            break
    else:
        print(num, "is a prime number")
else:
    print(num, "is not a prime number")

Output

11 is a prime number

Time complexity: O(n)
Auxiliary space: O(1)

Fastest Algorithm to Find Prime Numbers

Instead of checking till n, we can check till √n because a larger factor of n must be a multiple of a smaller factor that has been already checked. Now let’s see the code for the first optimization method ( i.e. checking till √n )

Python3

from math import sqrt
# n is the number to be check whether it is prime or not
n = 1

# this flag maintains status whether the n is prime or not
prime_flag = 0

if(n > 1):
    for i in range(2, int(sqrt(n)) + 1):
        if (n % i == 0):
            prime_flag = 1
            break
    if (prime_flag == 0):
        print("True")
    else:
        print("False")
else:
    print("False")

Output

False

Time complexity: O(sqrt(n))
Auxiliary space: O(1)

Check Prime Numbers Using recursion

We can also find the number prime or not using recursion. We can use the exact logic shown in method 2 but in a recursive way.

Python3

from math import sqrt

def Prime(number,itr):  #prime function to check given number prime or not
  if itr == 1:   #base condition
    return True
  if number % itr == 0:  #if given number divided by itr or not
    return False
  if Prime(number,itr-1) == False:   #Recursive function Call
    return False

  return True

num = 13

itr = int(sqrt(num)+1)

print(Prime(num,itr))

Output

True

Time complexity: O(sqrt(n))
Auxiliary space: O(sqrt(n))

Check the Prime Trial Division Method

Python3

def is_prime_trial_division(n):
    # Check if the number is less than or equal to 1, return False if it is
    if n <= 1:
        return False
    # Loop through all numbers from 2 to
    # the square root of n (rounded down to the nearest integer)
    for i in range(2, int(n**0.5)+1):
        # If n is divisible by any of these numbers, return False
        if n % i == 0:
            return False
    # If n is not divisible by any of these numbers, return True
    return True

# Test the function with n = 11
print(is_prime_trial_division(11))

Output

True

Time complexity: O(sqrt(n))
Auxiliary space: O(sqrt(n))

Using a while loop to check divisibility:

Approach:

Initialize a variable i to 2.
While i squared is less than or equal to n, check if n is divisible by i.
If n is divisible by i, return False.
Otherwise, increment i by 1.
If the loop finishes without finding a divisor, return True.

Python3

import math

def is_prime(n):
    if n < 2:
        return False
    i = 2
    while i*i <= n:
        if n % i == 0:
            return False
        i += 1
    return True

print(is_prime(11))  # True
print(is_prime(1))   # False

Output

True
False

Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)

METHOD:USING MATH

APPROACH:

The code implements a basic approach to check if a number is prime or not, by traversing all the numbers from 2 to sqrt(n)+1 and checking if n is divisible by any of those numbers.

ALGORITHM:

1.Check if the given number n is less than or equal to 1, if yes, return False.
2.Traverse all numbers in the range of 2 to sqrt(n)+1.
3.Check if n is divisible by any of the numbers from 2 to sqrt(n)+1, if yes, return False.
4.If n is not divisible by any of the numbers from 2 to sqrt(n)+1, return True.

Python3

import math

def is_prime(n):
    if n <= 1:
        return False
    for i in range(2, int(math.sqrt(n)) + 1):
        if n % i == 0:
            return False
    return True

n = 11
print(is_prime(n))

Output

True

Time complexity: O(sqrt(n))
The time complexity of the code is O(sqrt(n)) because we traverse all numbers in the range of 2 to sqrt(n)+1 to check if n is divisible by any of them.

Auxiliary Space: O(1)
The space complexity of the code is O(1) because we are only using a constant amount of memory to store the input number n and the loop variables.

Method: Using sympy.isprime() method

In the sympy module, we can test whether a given number n is prime or not using sympy.isprime() function. For n < 2⁶⁴ the answer is definitive; larger n values have a small probability of actually being pseudoprimes.

N.B.: Negative numbers (e.g. -13) are not considered prime number.

Syntax: 

sympy.isprime()

Python3

# Python program to check prime number
# using sympy.isprime() method

# importing sympy module
from sympy import *

# calling isprime function on different numbers
geek1 = isprime(30)
geek2 = isprime(13)
geek3 = isprime(2)

print(geek1) # check for 30 is prime or not
print(geek2) # check for 13 is prime or not
print(geek3) # check for 2 is prime or not

# This code is contributed by Susobhan Akhuli

Output

False
True
True

Time Complexity: O(sqrt(n)), where n is the input number.
Auxiliary Space: O(1)