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 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 beiginning, 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 ?> |
Output:
Palindromic primes smaller than or equal to 100 are : 2 3 5 7 11
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