human
Can you provide a code to palindromic Primes and the other steps required given the following:
Palindromic Primes | A function that returns true only if num contains one digit.
comparison operation is faster than division operation . So using following instead of " return ▁ num ▁ / ▁ 10 ▁ = = ▁ 0.
".
A recursive function to find out whether num is palindrome or not . Initially , dupNum contains address of a copy of num ..
Base case ( needed for recursion termination ) : This statement / mainly compares the first digit with the last digit.
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.
The following statements are executed when we move up the recursion call tree.
At this point , if num % 10 contains ith digit from beginning , then ( dupNum ) % 10 contains ith digit from end.
The main function that uses recursive function isPalUtil ( ) to find out whether num is palindrome or not.
If num is negative , make it positive.
Create a separate copy of num , so that modifications made to address dupNum don 't change the input number.  dupNum = num.
dupNum = num.
Function to generate all primes and checking whether number is palindromic or not.
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 ..
If prime [ p ] is not changed , then it is a prime.
Update all multiples of p.
Print all palindromic prime numbers.
checking whether the given number is prime palindromic or not.
Driver Code"
How can the above be solved in Python?
assistant
#Here is code to palindromic Primes and the other steps required.
 def oneDigit ( num ) :
	return ( num >= 0 and num < 10 ) ;
def isPalUtil ( num , dupNum ) :
	if ( oneDigit ( num ) ) :
		return ( num == ( dupNum ) % 10 ) ;
	if ( not isPalUtil ( int ( num / 10 ) , dupNum ) ) :
		return False ;
	dupNum = int ( dupNum / 10 ) ;
	return ( num % 10 == ( dupNum ) % 10 ) ;
def isPal ( num ) :
	if ( num < 0 ) :
		num = - num ;
	return isPalUtil ( num , dupNum ) ;
def printPalPrimesLessThanN ( n ) :
	prime = [ True ] * ( n + 1 ) ;
	p = 2 ;
	while ( p * p <= n ) :
		if ( prime [ p ] ) :
			for i in range ( p * 2 , n + 1 , p ) :
				prime [ i ] = False ;
		p += 1 ;
	for p in range ( 2 , n + 1 ) :
		if ( prime [ p ] and isPal ( p ) ) :
			print ( p , end = " ▁ " ) ;
n = 100 ;
print ( " Palindromic ▁ primes ▁ smaller " , " than ▁ or ▁ equal ▁ to " , n , " are ▁ : " ) ;
printPalPrimesLessThanN ( n ) ;