human
A backtracking approach to generate n bit Gray Codes | we have 2 choices for each of the n bits either we can include i . e invert the bit or we can exclude the bit i . e we can leave the number as it is ..
base case when we run out bits to process we simply include it in gray code sequence ..
ignore the bit ..
invert the bit ..
returns the vector containing the gray code sequence of n bits ..
num is passed by reference to keep track of current code ..
Driver Code"
How can the above be solved in Python?

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Write me a code to a backtracking approach to generate n bit Gray Codes and the other steps required?
assistant
#Here is code to a backtracking approach to generate n bit Gray Codes and the other steps required.
 def grayCodeUtil ( res , n , num ) :
	if ( n == 0 ) :
		res . append ( num [ 0 ] )
		return
	grayCodeUtil ( res , n - 1 , num )
	num [ 0 ] = num [ 0 ] ^ ( 1 << ( n - 1 ) )
	grayCodeUtil ( res , n - 1 , num )
def grayCodes ( n ) :
	res = [ ]
	num = [ 0 ]
	grayCodeUtil ( res , n , num )
	return res
n = 3
code = grayCodes ( n )
for i in range ( len ( code ) ) :
	print ( code [ i ] )