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 / / 2 ): # 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:
- 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.
- 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" ) # This code is contributed # |
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