human
Emirp numbers | Function to find reverse of any number.
Sieve method used for generating emirp number ( use of sieve ofEratosthenes ).
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.
Traverse all prime numbers.
Find reverse a number.
A number is emrip if it is not a palindrome number and its reverse is also prime ..
Mark reverse prime as false so that it 's  not printed again.
Driver Code"
How can the above be solved in Python?
 ... Give me a code to emirp numbers and the other steps required?
assistant
#Here is code to emirp numbers and the other steps required.
 def reverse ( x ) :
	rev = 0 ;
	while ( x > 0 ) :
		rev = ( rev * 10 ) + x % 10 ;
		x = int ( x / 10 ) ;
	return rev ;
def printEmirp ( n ) :
	prime = [ 1 ] * ( n + 1 ) ;
	p = 2 ;
	while ( p * p <= n ) :
		if ( prime [ p ] == 1 ) :
			for i in range ( p * 2 , n + 1 , p ) :
				prime [ i ] = 0 ;
		p += 1 ;
	for p in range ( 2 , n + 1 ) :
		if ( prime [ p ] == 1 ) :
			rev = reverse ( p ) ;
			if ( p != rev and rev <= n and prime [ rev ] == 1 ) :
				print ( p , rev , end = " ▁ " ) ;
				prime [ rev ] = 0 ;
n = 100 ;
printEmirp ( n ) ;