**Efficient Algorithm To Test Primality**

A number which is greater than 1 is called prime if its only
positive divisors are 1 and the number itself. A number which is not prime is
called composite number. Suppose we have following numbers:

1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9

Then the prime numbers are : 2 , 3 , 5 , 7

and composite numbers are: 1
, 4 , 6 , 8 , 9

So to test whether a number is prime or not, the first employed
approach is to divide the given number(n) by all numbers from 2 up to n-1 . If
the number is divided perfectly then the number is not prime otherwise the
number is prime. Implementing this logic
needs n-2 comparison, So the complexity is O(n).

Another approach to test primality is to divide n by all numbers
from 2 upto n/2. Also this is not a good approach. It just reduces the number
of comparison by half compared to previous approach.

An efficient algorithm to test primality is as follows:

1) If n is less than 2
then n is invalid number.

2) If n is equal to 2
then it is a prime number.

3) If n is divisible by
2 then it is not prime. If n is not divisible by 2 then it won’t be divided by
any even number.

4) If n is prime then
there must exist a divisor of n which is less than or equal to square root(n).

So, if n is less than 2 then it is not prime. If n is equal to 2
then it is a prime number. If it is not
divisible by 2 then it will not be divided by any even number. So we need to
check only for odd numbers from 3 upto square root(n). Of course if n is prime
then a divisor of n must be less than or equal to square root(n). In this
approach we need to do (square root(n)/2)-1 comparison. So it is quite fast
compared to other two approaches.

Here goes implementation of this efficient primality test algorithm in various languages:

JAVA implementation of primality test:

C implementation of primality test:

C++ implementation of primality test:

Here goes implementation of this efficient primality test algorithm in various languages:

JAVA implementation of primality test:

//PrimeTest.java import java.util.Scanner; /** * * @author VIK VIKKU VIKASH VIKASHVVERMA */ public class PrimeTest { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); System.out.println("Enter a number:"); int num = scanner.nextInt(); PrimeTest test = new PrimeTest(); if (test.isPrime(num)) { System.out.println(num + " is a prime number!"); } else { System.out.println(num + " is not a prime number!"); } } private boolean isPrime(int num) { boolean flag = false; if (num < 2) { return !flag; } if (num == 2) { return true; } if (num % 2 == 0) { return flag; } int i = 3; int rootn = (int) Math.sqrt(num); for (; i <= rootn; i += 2) { if (num % i == 0) { break; } } if (i > rootn) { flag = true; } return flag; } }

C implementation of primality test:

#include<stdio.h> #include<math.h> int isPrime(int num) { int i=3,rootn=sqrt(num); if(num<2) { return 0; } if(num==2) { return 1; } if(num%2==0) { return 0; } for(;i<=rootn;i+=2) { if(num%i==0) return 0; } return 1; } int main() { int num; printf("Enter a number:"); scanf("%d",&num); if(isPrime(num)) { printf("%d is a prime number!",num); }else { printf("%d is not a prime number!",num); } }

C++ implementation of primality test:

#include<iostream.h> #include<math.h> int isPrime(int num) { int i=3,rootn=sqrt(num); if(num<2) { return 0; } if(num==2) { return 1; } if(num%2==0) { return 0; } for(;i<=rootn;i+=2) { if(num%i==0) return 0; } return 1; } int main() { int num; cout<<"Enter a number:"; cin>>num; if(isPrime(num)) { cout<<num<<" is a prime number!"; }else { cout<<num<<" is not a prime number!"; } }

nice..

ReplyDeleteeasy to understand.. :)