Emirp numbers

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 the implementation of above idea.

C++

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// 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;
}

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Java

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// Java program to print Emirp 
// numbers less than n
import java.util.Arrays;
  
class GFG
{
    // Function to find reverse of any number
    static 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)
    static 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.
        boolean prime[]=new boolean[n + 1];
        Arrays.fill(prime,true);
      
        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])
                {
                    System.out.print(p + " " + rev + " ");
          
                    // Mark reverse prime as false so that it's
                    // not printed again
                    prime[rev] = false;
                }
            }
        }
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int n = 100;
        printEmirp(n);
    }
}
  
// This code is contributed by Anant Agarwal.

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Python3

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# Program to print Emirp numbers
# less than n
  
# Function to find reverse 
# of any number
def reverse(x):
  
    rev = 0;
    while (x > 0):
        rev = (rev * 10) + x % 10;
        x = int(x / 10);
  
    return rev;
  
# Sieve method used for generating 
# emirp number(use of sieve of 
# Eratosthenes)
def printEmirp(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.
    prime = [1] * (n + 1);
    p = 2;
    while (p * p <= n):
          
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p] == 1):
              
            # Update all multiples of p
            for i in range(p * 2, n + 1, p):
                prime[i] = 0;
        p += 1;
  
    # Traverse all prime numbers
    for p in range(2, n + 1):
        if (prime[p] == 1):
              
            # Find reverse a number
            rev = reverse(p);
  
            # A number is emrip if it is not 
            # a palindrome number and its 
            # reverse is also prime.
            if (p != rev and rev <= n and
                       prime[rev] == 1):
                print(p, rev, end = " ");
      
                # Mark reverse prime as 
                # false so that it's
                # not printed again
                prime[rev] = 0;
  
# Driver Code
n = 100;
printEmirp(n);
  
# This code is contributed by mits

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C#

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// C# program to print Emirp 
// numbers less than n
using System;
  
class GFG
{
    // Function to find 
    // reverse of any number
    static 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)
    static 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 = new bool[n + 1];
        for(int i = 0; i < n + 1; i++)
        prime[i] = true;
      
      
        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])
                {
                    Console.Write(p + " " + rev + " ");
          
                    // Mark reverse prime as false 
                    // so that it's not printed again
                    prime[rev] = false;
                }
            }
        }
    }
      
    // Driver code
    public static void Main ()
    {
        int n = 100;
        printEmirp(n);
    }
}
  
// This code is contributed by nitin mittal.

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PHP

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<?php
// Program to print Emirp numbers
// less than n
  
// Function to find reverse 
// of any number
function reverse($x)
{
    $rev = 0;
    while ($x > 0)
    {
         $rev = ($rev  * 10) + $x % 10;
        $x = (int)($x / 10);
    }
    return $rev;
}
  
// Sieve method used for generating 
// emirp number(use of sieve of 
// Eratosthenes)
function printEmirp($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.
    $prime = array_fill(0, ($n + 1), 1);
  
    for ($p = 2; $p * $p <= $n; $p++)
    {
        // If prime[p] is not changed,
        // then it is a prime
        if ($prime[$p] == 1)
        {
            // Update all multiples of p
            for ($i = $p * 2; $i <= $n; $i += $p)
                $prime[$i] = 0;
        }
    }
  
    // Traverse all prime numbers
    for ($p = 2; $p <= $n; $p++)
    {
        if ($prime[$p] == 1)
        {
            // Find reverse a number
            $rev = reverse($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] == 1)
            {
                echo $p . " " . $rev . " ";
      
                // Mark reverse prime as 
                // false so that it's
                // not printed again
                $prime[$rev] = 0;
            }
        }
    }
}
  
// Driver Code
$n = 100;
printEmirp($n);
  
// This code is contributed by mits
?>

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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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Improved By : nitin mittal, Mithun Kumar