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
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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 functionint 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 primeclass 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 primedef 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 functionprint("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 primeusing 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 primefunction 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 primefunction 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 of this solution is O(n)
Optimized School 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. (Source: wikipedia)
