## ISC Computer Science Practical 2016, Question 1

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.

### Vinay Singh

#### Latest posts by Vinay Singh (see all)

- Sorting non boundary elements - February 19, 2016
- Circular Prime – Solution of ISC Computer Practical 2016 - February 18, 2016
- Array Pattern - February 15, 2016

abyTry the number 70 for the circular prime program.

It won’t work.

Vinay SinghPost authorThank you for pointing out the mistake. I have modified the program to remove the anomaly.