Check whether a number is circular prime or not
We are given a number n. Our task is to check whether the number is circular prime or not.
Circular Prime : A prime number is said to be a circular prime if after any cyclic permutations of the digits, it remains a prime.
Examples:
Input : n = 113 Output : Yes All cyclic permutations of 113 (311 and 131) are prime. Input : 1193 Output : Yes
The idea is simple, we generate all permutations of given number and check each permutation for prime. To generate permutations, we one by one move last digit to first position.
C++
// Program to check if a number is circular // prime or not. #include <iostream> #include <cmath> using namespace std; // Function to check if a number is prime or not. bool isPrime( int n) { // Corner cases if (n <= 1) return false ; if (n <= 3) return true ; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false ; for ( int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false ; return true ; } // Function to check if the number is circular // prime or not. bool checkCircular( int N) { // Count digits. int count = 0, temp = N; while (temp) { count++; temp /= 10; } int num = N; while (isPrime(num)) { // Following three lines generate the next // circular permutation of a number. We move // last digit to first position. int rem = num % 10; int div = num / 10; num = ( pow (10, count - 1)) * rem + div ; // If all the permutations are checked and // we obtain original number exit from loop. if (num == N) return true ; } return false ; } // Driver Program int main() { int N = 1193; if (checkCircular(N)) cout << "Yes" << endl; else cout << "No" << endl; return 0; } |
Java
// Java Program to check if a number // is circular prime or not. import java.lang.*; class GFG { // Function to check if a number is prime or not. static boolean isPrime( int n) { // Corner cases if (n <= 1 ) return false ; if (n <= 3 ) return true ; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0 ) return false ; for ( int i = 5 ; i * i <= n; i = i + 6 ) if (n % i == 0 || n % (i + 2 ) == 0 ) return false ; return true ; } // Function to check if the number is circular // prime or not. static boolean checkCircular( int N) { // Count digits. int count = 0 , temp = N; while (temp> 0 ) { count++; temp /= 10 ; } int num = N; while (isPrime(num)) { // Following three lines generate the next // circular permutation of a number. We // move last digit to first position. int rem = num % 10 ; int div = num / 10 ; num = ( int )((Math.pow( 10 , count - 1 )) * rem) + div; // If all the permutations are checked and // we obtain original number exit from loop. if (num == N) return true ; } return false ; } // Driver Program public static void main (String[] args) { int N = 1193 ; if (checkCircular(N)) System.out.println( "Yes" ); else System.out.println( "No" ); } } /* This code is contributed by Kriti Shukla */ |
Python
# Python Program to check if a number # is circular prime or not. import math # Function to check if a number is prime # or not. def isPrime(n) : # Corner cases if (n < = 1 ) : return False if (n < = 3 ) : return True # This is checked so that we can skip # middle five numbers in below loop if (n % 2 = = 0 or n % 3 = = 0 ) : return False i = 5 while i * i < = n : if (n % i = = 0 or n % (i + 2 ) = = 0 ) : return False i = i + 6 return True # Function to check if the number is # circular prime or not. def checkCircular(N) : #Count digits. count = 0 temp = N while (temp > 0 ) : count = count + 1 temp = temp / 10 num = N; while (isPrime(num)) : # Following three lines generate the # next circular permutation of a # number. We move last digit to # first position. rem = num % 10 div = num / 10 num = ( int )((math. pow ( 10 , count - 1 )) * rem) + div # If all the permutations are checked # and we obtain original number exit # from loop. if (num = = N) : return True return False # Driver Program N = 1193 ; if (checkCircular(N)) : print "Yes" else : print "No" # This code is contributed by Nikita Tiwari |
C#
// C# Program to check if a number // is circular prime or not. using System; class GFG { // Function to check if a // number is prime or not. static bool isPrime( int n) { // Corner cases if (n <= 1) return false ; if (n <= 3) return true ; // This is checked so that we // can skip middle five numbers // in below loop if (n % 2 == 0 || n % 3 == 0) return false ; for ( int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false ; return true ; } // Function to check if the number // is circular prime or not. static bool checkCircular( int N) { // Count digits. int count = 0, temp = N; while (temp > 0) { count++; temp /= 10; } int num = N; while (isPrime(num)) { // Following three lines generate // the next circular permutation // of a number. We move last digit // to first position. int rem = num % 10; int div = num / 10; num = ( int )((Math.Pow(10, count - 1)) * rem) + div; // If all the permutations are // checked and we obtain original // number exit from loop. if (num == N) return true ; } return false ; } // Driver code public static void Main () { int N = 1193; if (checkCircular(N)) Console.Write( "Yes" ); else Console.Write( "No" ); } } // This code is contributed by Nitin Mittal. |
PHP
<?php // Program to check if // a number is circular // prime or not. // Function to check if a // number is prime or not. function isPrime( $n ) { // Corner cases if ( $n <= 1) return false; if ( $n <= 3) return true; // This is checked so that // we can skip middle five // numbers in below loop if ( $n % 2 == 0 || $n % 3 == 0) return false; for ( $i = 5; $i * $i <= $n ; $i = $i + 6) if ( $n % $i == 0 || $n % ( $i + 2) == 0) return -1; return true; } // Function to check if // the number is circular // prime or not. function checkCircular( $N ) { // Count digits. $count = 0; $temp = $N ; while ( $temp ) { $count ++; $temp =(int) $temp / 10; } $num = $N ; while (isPrime( $num )) { // Following three lines generate // the next circular permutation // of a number. We move last digit // to first position. $rem = $num % 10; $div = (int) $num / 10; $num = (pow(10, $count - 1)) * $rem + $div ; // If all the permutations are // checked and we obtain original // number exit from loop. if ( $num == $N ) return true; } return -1; } // Driver Code $N = 1193; if (checkCircular( $N )) echo "Yes" , "\n" ; else echo "No" , "\n" ; // This code is contributed // by akt_mit ?> |
Javascript
<script> // Javascript Program to check if a number // is circular prime or not. // Function to check if a // number is prime or not. function isPrime(n) { // Corner cases if (n <= 1) return false ; if (n <= 3) return true ; // This is checked so that we // can skip middle five numbers // in below loop if (n % 2 == 0 || n % 3 == 0) return false ; for (let i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false ; return true ; } // Function to check if the number // is circular prime or not. function checkCircular(N) { // Count digits. let count = 0, temp = N; while (temp > 0) { count++; temp = parseInt(temp / 10, 10); } let num = N; while (isPrime(num)) { // Following three lines generate // the next circular permutation // of a number. We move last digit // to first position. let rem = num % 10; let div = parseInt(num / 10, 10); num = ((Math.pow(10, count - 1)) * rem) + div; // If all the permutations are // checked and we obtain original // number exit from loop. if (num == N) return true ; } return false ; } let N = 1193; if (checkCircular(N)) document.write( "Yes" ); else document.write( "No" ); </script> |
Output:
Yes
Time Complexity: O(N*N1/2)
Auxiliary Space: O(1)
Optimizations:
Clearly numbers containing 0, 2, 4, 5, 6, or 8 can never be circular primes as numbers ending with these will always be divisible by 2 or 5. Hence one of their permutation will not be a prime. While counting digits in first step, we can also check if current digit is one of these.
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