Primality Test | Set 1 (Introduction and School Method)

Given a positive integer, check if the number is prime or not. A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself. Examples of the first few prime numbers are {2, 3, 5, …}
Examples :

Input:  n = 11
Output: true

Input:  n = 15
Output: false

Input:  n = 1
Output: false

School Method: A simple solution is to iterate through all numbers from 2 to n-1 and for every number check if it divides n. If we find any number that divides, we return false. 

Below is the implementation of this method. 

C++

// A school method based C++ program to check if a
// number is prime
#include <bits/stdc++.h>
using namespace std;

bool isPrime(int n)
{
    // Corner case
    if (n <= 1)
        return false;

    // Check from 2 to n-1
    for (int i = 2; i < n; i++)
        if (n % i == 0)
            return false;

    return true;
}

// Driver Program to test above function
int main()
{
    isPrime(11) ? cout << " true\n" : cout << " false\n";
    isPrime(15) ? cout << " true\n" : cout << " false\n";
    return 0;
}

Java

// A school method based JAVA program
// to check if a number is prime
class GFG {

    static boolean isPrime(int n)
    {
        // Corner case
        if (n <= 1)
            return false;

        // Check from 2 to n-1
        for (int i = 2; i < n; i++)
            if (n % i == 0)
                return false;

        return true;
    }

    // Driver Program
    public static void main(String args[])
    {
        if (isPrime(11))
            System.out.println(" true");
        else
            System.out.println(" false");
        if (isPrime(15))
            System.out.println(" true");
        else
            System.out.println(" false");
    }
}

// This code is contributed
// by Nikita Tiwari.

Python3

# A school method based Python3
# program to check if a number
# is prime


def isPrime(n):

    # Corner case
    if n <= 1:
        return False

    # Check from 2 to n-1
    for i in range(2, n):
        if n % i == 0:
            return False

    return True


# Driver Program to test above function
print("true") if isPrime(11) else print("false")
print("true") if isPrime(14) else print("false")

# This code is contributed by Smitha Dinesh Semwal

C#

// A optimized school method based C#
// program to check if a number is prime
using System;

namespace prime {
public class GFG {
    public static bool isprime(int n)
    {
        // Corner cases
        if (n <= 1)
            return false;

        for (int i = 2; i < n; i++)
            if (n % i == 0)
                return false;

        return true;
    }

    // Driver program
    public static void Main()
    {
        if (isprime(11))
            Console.WriteLine("true");
        else
            Console.WriteLine("false");

        if (isprime(15))
            Console.WriteLine("true");
        else
            Console.WriteLine("false");
    }
}
}

// This code is contributed by Sam007

PHP

<?php
// A school method based PHP
// program to check if a number
// is prime

function isPrime($n)
{
    // Corner case
    if ($n <= 1) return false;

    // Check from 2 to n-1
    for ($i = 2; $i < $n; $i++)
        if ($n % $i == 0)
            return false;

    return true;
}

// Driver Code
$tet = isPrime(11) ? " true\n" :
                     " false\n";
echo $tet;
$tet = isPrime(15) ? " true\n" :
                     " false\n";
echo $tet;

// This code is contributed by m_kit
?>

Javascript

<script>

// A school method based Javascript program to check if a
// number is prime
function isPrime(n)
{
    // Corner case
    if (n <= 1) return false;

    // Check from 2 to n-1
    for (let i = 2; i < n; i++)
        if (n % i == 0)
            return false;
    return true;
}

// Driver Program to test above function
    isPrime(11)? document.write(" true" + "<br>"): document.write(" false" + "<br>");
    isPrime(15)? document.write(" true" + "<br>"): document.write(" false" + "<br>");

// This code is contributed by Mayank Tyagi

</script>

Output

 true
 false

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

Optimized School Method: We can do the following optimizations: 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. The implementation of this method is as follows:

C++

// Optimised school method based C++ program to check if a
// number is prime
#include <bits/stdc++.h>
using namespace std;

bool isPrime(int n)
{
    // Corner case
    if (n <= 1)
        return false;

    // Check from 2 to square root of n
    for (int i = 2; i <= sqrt(n); i++)
        if (n % i == 0)
            return false;

    return true;
}

// Driver Program to test above function
int main()
{
    isPrime(11) ? cout << " true\n" : cout << " false\n";
    isPrime(15) ? cout << " true\n" : cout << " false\n";
    return 0;
}

// This code is contributed by Vikash Sangai

Java

// Optimised school method based JAVA program
// to check if a number is prime
class GFG {

    static boolean isPrime(int n)
    {
        // Corner case
        if (n <= 1)
            return false;

        // Check from 2 to square root of n
        for (int i = 2; i * i <= n; i++)
            if (n % i == 0)
                return false;

        return true;
    }

    // Driver Program
    public static void main(String args[])
    {
        if (isPrime(11))
            System.out.println(" true");
        else
            System.out.println(" false");
        if (isPrime(15))
            System.out.println(" true");
        else
            System.out.println(" false");
    }
}

// This code is contributed by Vikash Sangai

Python3

# Optimised school method based PYTHON program
# to check if a number is prime
# import the math module
import math

# function to check weather the number is prime or not


def isPrime(n):

    # Corner case
    if (n <= 1):
        return False

    # Check from 2 to square root of n
    for i in range(2, int(math.sqrt(n)) + 1):
        if (n % i == 0):
            return False
    return True


# Driver Program to test above function
print("true") if isPrime(11) else print("false")
print("true") if isPrime(15) else print("false")

# This code is contributed by bhoomikavemula

C#

// Optimised school method based C#
// program to check if a number is prime
using System;

namespace prime {
public class GFG {
    public static bool isprime(int n)
    {
        // Corner cases
        if (n <= 1)
            return false;

        for (int i = 2; i * i <= n; i++)
            if (n % i == 0)
                return false;

        return true;
    }

    // Driver program
    public static void Main()
    {
        if (isprime(11))
            Console.WriteLine("true");
        else
            Console.WriteLine("false");

        if (isprime(15))
            Console.WriteLine("true");
        else
            Console.WriteLine("false");
    }
}
}

// This code is contributed by Vikash Sangai

Javascript

<script>
        // JavaScript code for the above approach

  function isPrime(n)
    {
        // Corner case
        if (n <= 1) return false;

        // Check from 2 to square root of n
        for (let i = 2; i*i <= n; i++)
            if (n % i == 0)
                return false;

        return true;
    }

        // Driver Code

        if(isPrime(11))
            document.write(" true" + "<br/>");
        else
            document.write(" false" + "<br/>");
        if(isPrime(15))
            document.write(" true" + "<br/>");
        else
           document.write(" false" + "<br/>");

// This code is contributed by sanjoy_62.
    </script>

Output

 true
 false

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

Another approach: It is based on the fact that all primes greater than 3 are of the form 6k ± 1, where k is any integer greater than 0. This is because all integers can be expressed as (6k + i), where i = −1, 0, 1, 2, 3, or 4. And note that 2 divides (6k + 0), (6k + 2), and (6k + 4) and 3 divides (6k + 3). So, a more efficient method is to test whether n is divisible by 2 or 3, then to check through all numbers of the form 6k ± 1 <= √n. This is 3 times faster than testing all numbers up to √n. (Source: wikipedia).  

Below is the implementation of the above approach:

C++

// C++ program to check the given number
// is prime or not
#include <bits/stdc++.h>
using namespace std;

// Function to check if the given number
// is prime or not.
bool isPrime(int n)
{
    if (n == 2 || n == 3)
        return true;

    if (n <= 1 || n % 2 == 0 || n % 3 == 0)
        return false;

    // To check through all numbers of the form 6k ± 1
    for (int i = 5; i * i <= n; i += 6) {
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
    }

    return true;
}

// Driver Code

int main()
{
    isPrime(11) ? cout << " true\n" : cout << " false\n";
    isPrime(15) ? cout << " true\n" : cout << " false\n";
    return 0;
}

Java

// JAVA program to check the given number
// is prime or not
class GFG {

  static boolean isPrime(int n)
  {
    // since 2 and 3 are prime
    if (n == 2 || n == 3)
      return true;

    // if n<=1 or divided by 2 or 3 then it can not be prime
    if (n <= 1 || n % 2 == 0 || n % 3 == 0)
      return false;

    // To check through all numbers of the form 6k ± 1
    for (int i = 5; i * i <= n; i += 6)
    {
      if (n % i == 0 || n % (i + 2) == 0)
        return false;
    }

    return true;
  }

  // Driver Program
  public static void main(String args[])
  {
    if (isPrime(11))
      System.out.println(" true");
    else
      System.out.println(" false");
    if (isPrime(15))
      System.out.println(" true");
    else
      System.out.println(" false");
  }
}

// This code is contributed by Ujjwal Kumar Bhardwaj

Python3

# Python program to check the given number
# is prime or not

# Function to check if the given number
# is prime or not.
import math

def isPrime(n):
    if n == 2 or n == 3:
        return True
    elif n <= 1 or n % 2 == 0 or n % 3 == 0:
        return False

        # To check through all numbers of the form 6k ± 1
    # until i <= square root of n, with step value 6
    for i in range(5, int(math.sqrt(n))+1, 6):
        if n % i == 0 or n % (i+2) == 0:
            return False

    return True

# # Driver code
print(isPrime(11))
print(isPrime(20))

# # This code is contributed by Harsh Master

Output

 true
 false

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