Introduction to Primality Test and School Method
Last Updated : 13 Feb, 2025
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.
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; } // 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"); } } # 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") // 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"); } } } <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>"); </script> <?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; ?> 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 // 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 # 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 // 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 <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> 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.
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 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 # 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# 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 <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/>"); </script> Time Complexity: O(√n)
Auxiliary Space: O(1)
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