Emirp numbers

3

Emirp is the word “prime” spelled backwards, and it refers to a prime number that becomes a new prime number when you reverse its digits. Emirps do not include palindromic primes (like 151 or 787) nor 1-digit primes like 7. 107, 113, 149, and 157 – reverse them and you’ve got a new prime number on your hands. Source: Wiki

emirp numbers

Given a number n, the task is to print all Emrips smaller than or equal to n.
Examples :

Input  : n = 40
Output : 13 31 

Input  : n = 100
Output : 13 31 17 71 37 73 79 97

Below are the steps :
1) Use Sieve of Eratosthenes to generate all primes smaller than or equal to n. We can also use sieve of sundaram.

2) Traverse all generated prime numbers. For every traversed prime number print this number and its reverse if following conditions are satisfied.
………….a) If reverse is also prime.
………….b) Reverse is not same as prime (palindromes are not allowed)
………….c) Reverse is smaller than or equal to n.

Below is C++ implementation of above idea.

// Program to print Emirp numbers less than n
#include <bits/stdc++.h>
using namespace std;

// Function to find reverse of any number
int reverse(int x)
{
    int rev = 0;
    while (x > 0)
    {
        rev = (rev*10) + x%10;
        x = x/10;
    }
    return rev;
}

// Sieve method used for generating emirp number
// (use of sieve of Eratosthenes)
void printEmirp(int n)
{
    // Create a boolean array "prime[0..n]" and initialize
    // all entries it as true. A value in prime[i] will
    // finally be false if i is Not a prime, else true.
    bool prime[n+1];
    memset(prime, true, sizeof(prime));

    for (int p=2; p*p<=n; p++)
    {
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true)
        {
            // Update all multiples of p
            for (int i=p*2; i<=n; i += p)
                prime[i] = false;
        }
    }

    // Traverse all prime numbers
    for (int p=2; p<=n; p++)
    {
        if (prime[p])
        {
            // Find reverse a number
            int rev = reverse(p);

            // A number is emrip if it is not a palindrome
            // number and its reverse is also prime.
            if (p != rev && rev <= n && prime[rev])
            {
               cout << p << " " << rev << " ";

               // Mark reverse prime as false so that it's
               // not printed again
               prime[rev] = false;
            }
        }
    }
}

// Driver program
int main()
{
    int n = 40;
    printEmirp(n);
    return 0;
}

Output:

13 31 17 71 37 73 79 97

This article is contributed by Shivam Pradhan ( anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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