Introduction to Primality Test 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: trueInput: n = 15
Output: falseInput: 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> |
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 whether 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> |
true false
Time Complexity: O(√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 |
C#
// C# program to check the given number // is prime or not using System; class GFG { static bool 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)) Console.WriteLine( " true" ); else Console.WriteLine( " false" ); if (isPrime(15)) Console.WriteLine( " true" ); else Console.WriteLine( " false" ); } } // This code is contributed by Aman Kumar |
Javascript
<script> // JavaScript program to check the given number // is prime or not // Function to check if the given number // is prime or not. function isPrime(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 (let i = 5; i * i <= n; i += 6) { if (n % i == 0 || n % (i + 2) == 0) return false ; } return true ; } // Driver Code 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 Pushpesh Raj. </script> |
true false
Time Complexity: O(√n)
Auxiliary Space: O(1)
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