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Palindromic Primes

  • Difficulty Level : Medium
  • Last Updated : 22 Jun, 2021

A palindromic prime (sometimes called a palprime) is a prime number that is also a palindromic number. 
Given a number n, print all palindromic primes smaller than or equal to n. For example, If n is 10, the output should be “2, 3, 5, 7′. And if n is 20, the output should be “2, 3, 5, 7, 11′.
Idea is to generate all prime numbers smaller than or equal to given number n and checking every prime number whether it is palindromic or not.
Methods used

Below is the implementation of above algorithm: 
 

C++




// C++ Program to print all palindromic primes
// smaller than or equal to n.
#include<bits/stdc++.h>
using namespace std;
 
// A function that reurns true only if num
// contains one digit
int oneDigit(int num)
{
    // comparison operation is faster than
    // division operation. So using following
    // instead of "return num / 10 == 0;"
    return (num >= 0 && num < 10);
}
 
// A recursive function to find out whether
// num is palindrome or not. Initially, dupNum
// contains address of a copy of num.
bool isPalUtil(int num, int* dupNum)
{
    // Base case (needed for recursion termination):
    // This statement/ mainly compares the first
    // digit with the last digit
    if (oneDigit(num))
        return (num == (*dupNum) % 10);
 
    // This is the key line in this method. Note
    // that all recursive/ calls have a separate
    // copy of num, but they all share same copy
    // of *dupNum. We divide num while moving up
    // the recursion tree
    if (!isPalUtil(num/10, dupNum))
        return false;
 
    // The following statements are executed when
    // we move up the recursion call tree
    *dupNum /= 10;
 
    // At this point, if num%10 contains i'th
    // digit from beginning, then (*dupNum)%10
    // contains i'th digit from end
    return (num % 10 == (*dupNum) % 10);
}
 
// The main function that uses recursive function
// isPalUtil() to find out whether num is palindrome
// or not
int isPal(int num)
{
    // If num is negative, make it positive
    if (num < 0)
       num = -num;
 
    // Create a separate copy of num, so that
    // modifications made to address dupNum don't
    // change the input number.
    int *dupNum = new int(num); // *dupNum = num
 
    return isPalUtil(num, dupNum);
}
 
// Function to generate all primes and checking
// whether number is palindromic or not
void printPalPrimesLessThanN(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;
        }
    }
 
    // Print all palindromic prime numbers
    for (int p=2; p<=n; p++)
 
       // checking whether the given number is
       // prime palindromic or not
       if (prime[p] && isPal(p))
          cout << p << " ";
}
 
// Driver Program
int main()
{
    int n = 100;
    printf("Palindromic primes smaller than or "
           "equal to %d are :\n", n);
    printPalPrimesLessThanN(n);
}

Java




// Java Program to print all palindromic primes
// smaller than or equal to n.
import java.util.*;
 
class GFG {
     
    // A function that reurns true only if num
    // contains one digit
    static boolean oneDigit(int num)
    {
        // comparison operation is faster than
        // division operation. So using following
        // instead of "return num / 10 == 0;"
        return (num >= 0 && num < 10);
    }
      
    // A recursive function to find out whether
    // num is palindrome or not. Initially, dupNum
    // contains address of a copy of num.
    static boolean isPalUtil(int num, int dupNum)
    {
        // Base case (needed for recursion termination):
        // This statement/ mainly compares the first
        // digit with the last digit
        if (oneDigit(num))
            return (num == (dupNum) % 10);
      
        // This is the key line in this method. Note
        // that all recursive/ calls have a separate
        // copy of num, but they all share same copy
        // of dupNum. We divide num while moving up
        // the recursion tree
        if (!isPalUtil(num/10, dupNum))
            return false;
      
        // The following statements are executed when
        // we move up the recursion call tree
        dupNum /= 10;
      
        // At this point, if num%10 contains ith
        // digit from beginning, then (dupNum)%10
        // contains ith digit from end
        return (num % 10 == (dupNum) % 10);
    }
      
    // The main function that uses recursive function
    // isPalUtil() to find out whether num is palindrome
    // or not
    static boolean isPal(int num)
    {
        // If num is negative, make it positive
        if (num < 0)
           num = -num;
      
        // Create a separate copy of num, so that
        // modifications made to address dupNum don't
        // change the input number.
        int dupNum = num; // dupNum = num
      
        return isPalUtil(num, dupNum);
    }
      
    // Function to generate all primes and checking
    // whether number is palindromic or not
    static void printPalPrimesLessThanN(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])
            {
                // Update all multiples of p
                for (int i = p*2; i <= n; i += p){
                    prime[i] = false;}
            }
        }
      
        // Print all palindromic prime numbers
        for (int p = 2; p <= n; p++){
      
           // checking whether the given number is
           // prime palindromic or not
           if (prime[p] && isPal(p)){
              System.out.print(p + " ");
              }
           }
    }
     
    // Driver function
    public static void main(String[] args)
    {
         int n = 100;
            System.out.printf("Palindromic primes smaller than or "
                   +"equal to %d are :\n", n);
            printPalPrimesLessThanN(n);
        }
    }
         
// This code is contributed by Arnav Kr. Mandal.

Python3




# Python3 Program to print all palindromic
# primes smaller than or equal to n.
     
# A function that reurns true only if
# num contains one digit
def oneDigit(num):
     
    # comparison operation is faster than
    # division operation. So using following
    # instead of "return num / 10 == 0;"
    return (num >= 0 and num < 10);
     
# A recursive function to find out whether
# num is palindrome or not. Initially, dupNum
# contains address of a copy of num.
def isPalUtil(num, dupNum):
     
    # Base case (needed for recursion termination):
    # This statement/ mainly compares the first
    # digit with the last digit
    if (oneDigit(num)):
        return (num == (dupNum) % 10);
     
    # This is the key line in this method. Note
    # that all recursive/ calls have a separate
    # copy of num, but they all share same copy
    # of dupNum. We divide num while moving up
    # the recursion tree
    if (not isPalUtil(int(num / 10), dupNum)):
        return False;
     
    # The following statements are executed
    # when we move up the recursion call tree
    dupNum =int(dupNum/10);
     
    # At this point, if num%10 contains ith
    # digit from beginning, then (dupNum)%10
    # contains ith digit from end
    return (num % 10 == (dupNum) % 10);
     
# The main function that uses recursive
# function isPalUtil() to find out whether
# num is palindrome or not
def isPal(num):
     
    # If num is negative, make it positive
    if (num < 0):
        num = -num;
     
    # Create a separate copy of num, so that
    # modifications made to address dupNum
    # don't change the input number.
    dupNum = num; # dupNum = num
     
    return isPalUtil(num, dupNum);
     
# Function to generate all primes and checking
# whether number is palindromic or not
def printPalPrimesLessThanN(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 = [True] * (n + 1);
    p = 2;
    while (p * p <= n):
         
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p]):
             
            # Update all multiples of p
            for i in range(p * 2, n + 1, p):
                prime[i] = False;
        p += 1;
         
    # Print all palindromic prime numbers
    for p in range(2, n + 1):
         
        # checking whether the given number
        # is prime palindromic or not
        if (prime[p] and isPal(p)):
            print(p, end = " ");
     
# Driver Code
n = 100;
print("Palindromic primes smaller",
      "than or equal to", n, "are :");
printPalPrimesLessThanN(n);
 
# This code is contributed by chandan_jnu

C#




// C# Program to print all palindromic
// primes smaller than or equal to n.
using System;
 
class GFG {
     
    // A function that reurns true only
    // if num contains one digit
    static bool oneDigit(int num)
    {
        // comparison operation is faster than
        // division operation. So using following
        // instead of "return num / 10 == 0;"
        return (num >= 0 && num < 10);
    }
     
    // A recursive function to find out whether
    // num is palindrome or not. Initially, dupNum
    // contains address of a copy of num.
    static bool isPalUtil(int num, int dupNum)
    {
        // Base case (needed for recursion termination):
        // This statement/ mainly compares the first
        // digit with the last digit
        if (oneDigit(num))
            return (num == (dupNum) % 10);
     
        // This is the key line in this method. Note
        // that all recursive/ calls have a separate
        // copy of num, but they all share same copy
        // of dupNum. We divide num while moving up
        // the recursion tree
        if (!isPalUtil(num/10, dupNum))
            return false;
     
        // The following statements are executed when
        // we move up the recursion call tree
        dupNum /= 10;
     
        // At this point, if num%10 contains ith
        // digit from beginning, then (dupNum)%10
        // contains ith digit from end
        return (num % 10 == (dupNum) % 10);
    }
     
    // The main function that uses recursive
    // function isPalUtil() to find out
    // whether num is palindrome or not
    static bool isPal(int num)
    {
        // If num is negative, make it positive
        if (num < 0)
        num = -num;
     
        // Create a separate copy of num, so that
        // modifications made to address dupNum don't
        // change the input number.
        int dupNum = num; // dupNum = num
     
        return isPalUtil(num, dupNum);
    }
     
    // Function to generate all primes and checking
    // whether number is palindromic or not
    static void printPalPrimesLessThanN(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])
            {
                // Update all multiples of p
                for (int i = p*2; i <= n; i += p){
                    prime[i] = false;}
            }
        }
     
        // Print all palindromic prime numbers
        for (int p = 2; p <= n; p++){
     
        // checking whether the given number is
        // prime palindromic or not
        if (prime[p] && isPal(p)){
            Console.Write(p + " ");
            }
        }
    }
     
    // Driver function
    public static void Main()
    {
        int n = 100;
        Console.Write("Palindromic primes smaller than or " +
                      "equal to are :\n", n);
        printPalPrimesLessThanN(n);
    }
}
         
// This code is contributed by nitin mittal.

PHP




<?php
// PHP Program to print all palindromic
// primes smaller than or equal to n.
     
// A function that reurns true only
// if num contains one digit
function oneDigit($num)
{
    // comparison operation is faster than
    // division operation. So using following
    // instead of "return num / 10 == 0;"
    return ($num >= 0 && $num < 10);
}
 
// A recursive function to find out whether
// num is palindrome or not. Initially,
// dupNum contains address of a copy of num.
function isPalUtil($num, $dupNum)
{
    // Base case (needed for recursion termination):
    // This statement/ mainly compares the first
    // digit with the last digit
    if (oneDigit($num))
        return ($num == ($dupNum) % 10);
 
    // This is the key line in this method. Note
    // that all recursive/ calls have a separate
    // copy of num, but they all share same copy
    // of dupNum. We divide num while moving up
    // the recursion tree
    if (!isPalUtil((int)($num/10), $dupNum))
        return false;
 
    // The following statements are executed 
    // when we move up the recursion call tree
    $dupNum = (int)($dupNum / 10);
 
    // At this point, if num%10 contains ith
    // digit from beginning, then (dupNum)%10
    // contains ith digit from end
    return ($num % 10 == ($dupNum) % 10);
}
 
// The main function that uses recursive
// function isPalUtil() to find out whether
// num is palindrome or not
function isPal($num)
{
    // If num is negative, make it positive
    if ($num < 0)
    $num = -$num;
 
    // Create a separate copy of num, so that
    // modifications made to address dupNum
    // don't change the input number.
    $dupNum = $num; // dupNum = num
 
    return isPalUtil($num, $dupNum);
}
 
// Function to generate all primes and checking
// whether number is palindromic or not
function printPalPrimesLessThanN($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, true);
     
    for ($p = 2; $p * $p <= $n; $p++)
    {
        // If prime[p] is not changed, then
        // it is a prime
        if ($prime[$p])
        {
            // Update all multiples of p
            for ($i = $p * 2; $i <= $n; $i += $p)
            {
                $prime[$i] = false;
            }
        }
    }
 
    // Print all palindromic prime numbers
    for ($p = 2; $p <= $n; $p++)
    {
 
    // checking whether the given number 
    // is prime palindromic or not
    if ($prime[$p] && isPal($p))
    {
        print($p . " ");
    }
    }
}
 
// Driver Code
$n = 100;
print("Palindromic primes smaller " .
      "than or equal to ".$n." are :\n");
printPalPrimesLessThanN($n);
 
// This code is contributed by mits
?>

Javascript




<script>
 
// javascript Program to prvar all palindromic primes
// smaller than or equal to n.
    
    // A function that reurns true only if num
    // contains one digit
    function oneDigit(num)
    {
        // comparison operation is faster than
        // division operation. So using following
        // instead of "return num / 10 == 0;"
        return (num >= 0 && num < 10);
    }
      
    // A recursive function to find out whether
    // num is palindrome or not. Initially, dupNum
    // contains address of a copy of num.
    function isPalUtil(num , dupNum)
    {
        // Base case (needed for recursion termination):
        // This statement/ mainly compares the first
        // digit with the last digit
        if (oneDigit(num))
            return (num == (dupNum) % 10);
      
        // This is the key line in this method. Note
        // that all recursive/ calls have a separate
        // copy of num, but they all share same copy
        // of dupNum. We divide num while moving up
        // the recursion tree
        if (!isPalUtil(parseInt(num/10), dupNum))
            return false;
      
        // The following statements are executed when
        // we move up the recursion call tree
        dupNum = parseInt(dupNum/10);
      
        // At this point, if num%10 contains ith
        // digit from beginning, then (dupNum)%10
        // contains ith digit from end
        return (num % 10 == (dupNum) % 10);
    }
      
    // The main function that uses recursive function
    // isPalUtil() to find out whether num is palindrome
    // or not
    function isPal(num)
    {
        // If num is negative, make it positive
        if (num < 0)
           num = -num;
      
        // Create a separate copy of num, so that
        // modifications made to address dupNum don't
        // change the input number.
        var dupNum = num; // dupNum = num
      
        return isPalUtil(num, dupNum);
    }
      
    // Function to generate all primes and checking
    // whether number is palindromic or not
    function printPalPrimesLessThanN(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])
            {
                // Update all multiples of p
                for (i = p*2; i <= n; i += p){
                    prime[i] = false;}
            }
        }
      
        // Prvar all palindromic prime numbers
        for (p = 2; p <= n; p++){
      
           // checking whether the given number is
           // prime palindromic or not
           if (prime[p] && isPal(p)){
              document.write(p + " ");
              }
           }
    }
     
    // Driver function
    var n = 100;
    document.write('Palindromic primes smaller than or equal to '+n+' are :<br>');
    printPalPrimesLessThanN(n);
 
// This code is contributed by Princi Singh
</script>

Output: 
 

Palindromic primes smaller than or equal to 100 are :
2 3 5 7 11 

This article is contributed by Rahul Agrawal .If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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