Given (M, F) - With every step of a certain process, the tuple becomes either (M + F, F) or (M, M + F). So, starting with (1, 1), the following are possible:
The algorithm I've followed while drawing the tree is -
If not root,
left_ child = (parent0 + parent1, parent1)
and right_child = (parent0, parent0 + parent1)
where parent0 is referring to the first element of the parent tuple and parent1 second.
I am completely unsure on how I would create a tree that would follow the above algorithm for creation of the tree after n steps given any starting values to M and F.
When I decided to search online, I got something along the lines of :
class Tree(object):
def __init__(self):
self.left = None
self.right = None
self.data = None
root = Tree()
root.data = "root"
root.left = Tree()
root.left.data = "left"
root.right = Tree()
root.right.data = "right"
or
class Tree:
def __init__(self, cargo, left=None, right=None):
self.cargo = cargo
self.left = left
self.right = right
def __str__(self):
return str(self.cargo)
tree = Tree(1, Tree(2), Tree(3))
I couldn't think of a way to use the above piece of code to build the state space tree I wanted to. That is, how do I implement a tree that automatically creates subtrees with the calculated values when the given input is (M, F) and n which is the number of steps in the process (i.e. the number of levels in the tree).
All you need is a Tree that holds two values instead of just one, plus some properties that automatically generate subtrees:
class SpaceTree:
def __init__(self, M=1, F=1):
self.M= M
self.F= F
self._left= None
self._right= None
@property
def left(self):
if self._left is None:
self._left= SpaceTree(self.M+self.F, self.F)
return self._left
@property
def right(self):
if self._right is None:
self._right= SpaceTree(self.M, self.F+self.M)
return self._right
def __repr__(self):
return 'SpaceTree({}, {})'.format(self.M, self.F)
Now you can create a tree and traverse it as far as you want:
>>> SpaceTree()
SpaceTree(1, 1)
>>> SpaceTree().left
SpaceTree(2, 1)
>>> SpaceTree().left.right
SpaceTree(2, 3)