human
Pointer | Function to find the product of digits of a number N.
Function that returns true if n is prime else returns false.
Corner cases.
This is checked so that we can skip middle five numbers in below loop.
Function to return the smallest prime number greater than N.
Base case.
Loop continuously until isPrime returns true for a number greater than n.
Function to check Pointer - Prime numbers.
Driver Code.
Given number N.
Function call"
Solution in Python:

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Can you provide a code to pointer and the other steps required?
assistant
#Here is code to pointer and the other steps required.
 def digProduct ( n ) :
	product = 1
	while ( n != 0 ) :
		product = product * ( n % 10 )
		n = int ( n / 10 )
	return product
def isPrime ( n ) :
	if ( n <= 1 ) :
		return False
	if ( n <= 3 ) :
		return True
	if ( n % 2 == 0 or n % 3 == 0 ) :
		return False
	i = 5
	while ( i * i <= n ) :
		if ( n % i == 0 or n % ( i + 2 ) == 0 ) :
			return False
		i = i + 6
	return True
def nextPrime ( N ) :
	if ( N <= 1 ) :
		return 2 ;
	prime = N
	found = False
	while ( not found ) :
		prime = prime + 1
		if ( isPrime ( prime ) ) :
			found = True
	return prime
def isPointerPrime ( n ) :
	if ( isPrime ( n ) and ( n + digProduct ( n ) == nextPrime ( n ) ) ) :
		return True
	else :
		return False
if __name__ == " _ _ main _ _ " :
	N = 23
	if ( isPointerPrime ( N ) ) :
		print ( " Yes " )
	else :
		print ( " No " )