Circular Prime – Solution of ISC Computer Practical 2016

ISC Computer Science Practical 2016, Question 1

circular prime

Circular prime

A Circular Prime is a prime number that remains prime under cyclic shifts of its digits. When the leftmost digit is removed and replaced at the end of the remaining string of digits, the generated number is still prime. The process is repeated until the original number is reached again.
A number is said to be prime if it has only two factors 1 and itself.

Example:
131
311
113
Hence, 131 is a circular prime.

Test your program with the sample data and some random data:

 

Example 1

INPUT:  N= 197
OUTPUT: 
        197
        971
        719

        197 IS A CIRCULAR PRIME

Example 2

INPUT:  N= 1193
OUTPUT: 
        1193
        1931
        9311
        3119

        1193 IS A CIRCULAR PRIME

Example 3

INPUT:  N= 29
OUTPUT: 
        29
        92

        29 IS NOT A CIRCULAR PRIME

An efficient Java implementation for circular prime problem is as follows:

import java.util.*;
class CircularPrime{
    public static boolean isPrime( int number ){
        int factorCount = 0;
        if( number < 2 ) return false;
        else if( number == 2 ) return true;
        else if( number % 2 == 0 ) return false;
        else{
            int limit = (int) Math.sqrt( number );
            for( int i = 3 ; i <= limit ; i+=2 ){
                if( number %i == 0 ){
                    return false;
                }
            }
        }
        return true;
    }

    public static int circulate( int n, int divisor ){
        //left most digit is n/divisor;
        //remainder after removing left most is n%divisor;
        if( n < 10 ) 
            return n; 
        else return ( n % divisor ) * 10 + n / divisor; 
    } 
    public static void main( String args[] ){ 
        int N, digit; 
        Scanner sc = new Scanner( System.in ); 
        System.out.print("INPUT: N= "); 
        N = sc.nextInt(); 
        int numOfDigits = 0, divisor=1, circular=N; 
        boolean allPrime = true, isValidDigit=true; 
        for( int temp = N; temp > 0; temp /= 10 ){
            numOfDigits++;
            divisor *=10;
            digit=temp%10;
            if( !(digit==1 || digit==3 || digit==7 || digit==9) ) {
                isValidDigit=false;
                break;
            }
        }
        if(!isValidDigit && numOfDigits>=2){
            System.out.println( N + " IS NOT A CIRCULAR PRIME" );
        }else{
            divisor /=10;
            System.out.println( "OUTPUT: " );
            do{
                System.out.println( "        "+circular );
                circular = circulate( circular, divisor );
                if( !isPrime( circular ) ) allPrime=false;
            }while( circular != N );
            System.out.print( "\n        " );
            if( allPrime ) System.out.println( N + " IS A CIRCULAR PRIME" );
            else System.out.println( N + " IS NOT A CIRCULAR PRIME" );
        }
    }
}

Note: A circular prime with at least two digits can only consist of combinations of the digits 1, 3, 7 or 9, because having 0, 2, 4, 6 or 8 as the last digit makes the number divisible by 2, and having 0 or 5 as the last digit makes it divisible by 5.

Please note that in the above solution the number N is an integer and all operations are performed which is computationally better (efficient) than converting between string and integer for circulating the number.

The function isPrime also uses a number of optimization for efficiently determining if the number is a prime number or not. For details of the isPrime() function refer to my earlier post on primality i.e. Prime time.

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Vinay Singh

Computer Teacher at Saint John's Academy
I am a GNU/Linux enthusiast and a computer science teacher by profession based at Allahabad, India. My hobbies include reading (mostly computer related stuff) and photography.
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