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 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.

// 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;
}

// 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.

# 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

// 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
// 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
?>

true
false

Time complexity of this solution is O(n)

Optimized School Method
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. (Source: wikipedia)

Below is the implementation of this solution.


C++

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

bool isPrime(int 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 || n%3 == 0) return false;

    for (int i=5; i*i<=n; i=i+6)
        if (n%i == 0 || n%(i+2) == 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 optimized school method based Java 
// program to check if a number is prime
import java.io.*;

class GFG {
    
    static boolean isPrime(int 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 || n % 3 == 0) return false;
    
        for (int i = 5; i * i <= n; i = 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 Nikita Tiwari.*/

Python3

# 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 
# by Nikita Tiwari.


Output:

true
false

Time complexity of this solution is O(√n)

Primality Test | Set 2 (Fermat Method)

References:
https://en.wikipedia.org/wiki/Prime_number
http://www.cse.iitk.ac.in/users/manindra/presentations/FLTBasedTests.pdf
https://en.wikipedia.org/wiki/Primality_test

This article is contributed by Ajay. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above