Emirp numbers - GeeksforGeeks

09 Mar, 2022
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 
Given a number n, the task is to print all Emirps 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++
Java
Python3
C#
PHP
Javascript
//C++ Program to print Emirp numbers less than n#include <bits/stdc++.h>using namespace std; // Function to find reverse of any numberint 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 emirp 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 programint main(){    int n = 40;    printEmirp(n);    return 0;}
// Java program to print Emirp// numbers less than nimport 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 emirp 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.
# Python3 Program to print Emirp numbers# less than n # Function to find reverse# of any numberdef 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 emirp 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 Coden = 100;printEmirp(n); # This code is contributed by mits
// C# program to print Emirp// numbers less than nusing 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 emirp 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.
<?php//PHP Program to print Emirp numbers// less than n // Function to find reverse// of any numberfunction 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 emirp 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?>
<script> // Javascript program to print Emirp// numbers less than n // Function to find reverse of any number    function reverse(x)    {        var rev = 0;        while (x > 0)        {            rev = (rev * 10) + x % 10;            x = parseInt(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.       var  prime=Array.from({length: n+1},                  (_, i) => true);                      for (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 (i = p * 2; i <= n; i += p)                    prime[i] = false;            }        }             // Traverse all prime numbers        for (p = 2; p <= n; p++)        {            if (prime[p])            {                // Find reverse a number                var rev = reverse(p);                     // A number is emirp if                // it is not a palindrome                // number and its reverse                // is also prime.                if (p != rev && rev <= n &&                    prime[rev])                {                    document.write(p + " " + rev + " ");                             // Mark reverse prime as false                    // so that it's                    // not printed again                    prime[rev] = false;                }            }        }    }         // Driver code    var n = 100;    printEmirp(n);  // This code contributed by Princi Singh </script>
Output : 
13 31 17 71 37 73 79 97
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