Circular primes are prime numbers which upon cyclic rotation
also give prime numbers. For example: 197 is prime number and its cyclic rotations 719 and
917 are also prime . Therefore 197 is a circular prime. It is to be noted that
cyclic rotations and permutations are different concepts. Cyclic rotations are
a subset of permutations of a given number.
There are 13 circular primes below 100:2 ,3 ,5 ,7 ,11 ,13
,17 ,31 ,37 ,71 ,73 ,79 ,97.
Here is an interesting program to find the circular prime
numbers between a given range of numbers input by the user. First of all, we’ll
check whether the number is prime or not. If the number is prime then its
cyclic rotations are further checked.
The circular primes with at least 2 digits can consist of
combinations of only 1,3,7 or 9 as the numbers consisting of digits 2,4,6,8,0
or 5 will be either divisible by 2 or 5 and hence are not prime. Therefore the
code can be shortened by checking that if any digit of the number is a multiple
of 2 or 5 then the number can be skipped. Since 2 and 5 are prime ,if they are
included in the range, they can be separately checked.
Source code # 1 (in C++)
/*****************************************************************************/
#include<iostream.h>
#include<conio.h>
#include<math.h>
void main()
{
clrscr();
int
c=0,r,flag,d;
unsigned
long num,num1,num2,div,i,j,n1,n2,nn;
cout<<"\n
Enter the starting number of the range:";
cin>>n1;
cout<<"\n
Enter the ending number of the range:";
cin>>n2;
for(i=n1;i<=n2;i++)
{
if(i==1)
goto
a;
if(i==2
|| i==5)
goto
b;
num2=i;
while(num2) //if number consists of a digit that is
multiple of 2 or 5
//then no need to check it.
{
d=num2%10;
if(d==5
|| d%2==0)
goto
a;
num2=num2/10;
}
num=i;
for(j=2;j<=sqrt(num);j++)
//to check whether number is prime or not.
{
if(num%j==0)
goto
a;
}
flag=0;
while(num>9) //to count the number of digits in the
number.
{
num=num/10;
flag++;
}
div=pow(10,flag);
num1=i;
while(flag)
//here cyclic rotations of the number are formed and
//checked whether they are prime or not.
{
r=num1%10;
nn=div*r+(num1/10);
for(j=2;j<=sqrt(nn);j++)
{
if(nn%j==0)
goto
a;
}
num1=nn;
flag=flag-1;
}
b:
c=c+1;
// to count the number of circular primes.
cout<<"\t"<<i;
a:
}
cout<<"\n
the number of circular primes between "
<<n1<<"
and "<<n2<<" are:"<<c;
getch();
}
/*****************************************************************************/
/*******************************************************************************/
#include<stdio.h>
#include<math.h>
#include<conio.h>
int isprime(long);
int main()
{
long int upper_limit,lower_limit,i,j,copy_i,new_num;
int ccounter=0,counter=0;
/*counter is used for counting the
total number of digits of any number
ccounter is used for counting
that how many numbers
that are circular permutations of
original number are primes
*/
clrscr();
printf("ENTER THE LOWER LIMIT : ");
scanf("%ld",&upper_limit);
printf("ENTER THE UPPERLIMIT : ");
scanf("%ld",&lower_limit);
for(i=upper_limit;i<=lower_limit;i++)
{
counter=0;ccounter=0;
copy_i=i;
while(copy_i)
{
counter++;
copy_i/=10;
}
copy_i=i;
if(isprime(i))
{
for(j=0;j<counter;j++)
{
new_num=copy_i%10;
copy_i/=10;
new_num*=(long)pow(10,(counter-1));
new_num+=copy_i;
if(isprime(new_num))
ccounter++;
copy_i=new_num;
}
if(ccounter==counter)
{
printf("%ld\t",i);
}
}
}
printf("\n\nDONE!!!");
getch();
return 1;
}
int isprime(long num)
{
long j;
for(j=2;j<=(num/2);j++)
{
if(num%j==0)
{
return 0;
}
}
return 1;
}
/******************************************************************************/
Output :-
