human
Number of different cyclic paths of length N in a tetrahedron | Function to count the number of steps in a tetrahedron.
initially coming to B is B -> B.
cannot reach A , D or C.
iterate for all steps.
recurrence relation.
memoize previous values.
returns steps.
Driver code"
How can the above be solved in Python?
assistant
#Here is code to number of different cyclic paths of length N in a tetrahedron and the other steps required.
 def countPaths ( n ) :
	zB = 1
	zADC = 0
	for i in range ( 1 , n + 1 ) :
		nzB = zADC * 3
		nzADC = ( zADC * 2 + zB )
		zB = nzB
		zADC = nzADC
	return zB
n = 3
print ( countPaths ( n ) )