Palindromic Primes
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
- To find if a given number is prime or not- sieve-of-eratosthenes method
- To check whether the given number is palindromic number or not: Recursive function for checking palindrome
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 returns true only if num// contains one digitint 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 notint 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 notvoid 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 Programint 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 returns 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 returns true only if# num contains one digitdef 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 notdef 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 notdef 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 Coden = 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 returns 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 returns true only// if num contains one digitfunction 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 notfunction 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, $dupNum);}// Function to generate all primes and checking// whether number is palindromic or notfunction printPalPrimesLessThanN($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 print all palindromic primes// smaller than or equal to n. // A function that returns 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;} } } // 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)){ 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
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