Python program to check whether a number is Prime or not
Given a positive integer N. The task is to write a Python program to check if the number is prime or not.Definition: 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, ….}.
Examples :
Input: n =
11
Output: true
Input: n =
15
Output: false
Input: n = 1
Output: false
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.
Below is
the Python program to check if a number is prime:
# Python
program to check if
# given
number is prime or not
num = 11
# If
given number is greater than 1
if num
> 1:
# Iterate
from 2 to n / 2
for i in
range(2, num):
# 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
Optimized
Method
We can do following optimizations:
We can do following optimizations:
1. Instead of checking till n, we
can check till √n because a larger factor of n must be a multiple of smaller
factor that has been already checked.
2. The algorithm can be improved
further by observing that all primes are of the form 6k ± 1, with the exception
of 2 and 3. This is because all integers can be expressed as (6k + i) for some
integer k and for i = ?1, 0, 1, 2, 3, or 4; 2 divides (6k + 0), (6k + 2), (6k +
4); and 3 divides (6k + 3). So a more efficient method is to test if n is
divisible by 2 or 3, then to check through all the numbers of form 6k ± 1.
# A
optimized school method based
# Python3
program to check
# if a
number is prime
def
isPrime(n) :
# Corner cases
if (n <= 1) :
return False
if (n <= 3) :
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2)
== 0) :
return False
i = i + 6
return True
# Driver
Program
if
(isPrime(11)) :
print(" true")
else :
print(" false")
if(isPrime(15))
:
print(" true")
else :
print(" false")
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