Tree structure in minimax (python) - python

I am trying to make AI for board game with minimax algorithm, but I have problem with creating a tree structure. I know that you actually don't need to build tree for minimax but I have to. I want to build Tree separately from minimax and then use minimax on that tree. I don't know how to do it so that is what I need help for. Thanks in advance.
So I tried this:
from queue import Queue
class TreeNode(object):
__slots__ = 'parent', 'children', 'data', 'player'
def __init__(self, data, player):
self.parent = None
self.player = player
self.children = []
self.data = data
def is_root(self):
return self.parent is None
def is_leaf(self):
return len(self.children) == 0
def add_child(self, x):
x.parent = self
self.children.append(x)
def print_tree(self):
print(' '*self.get_level() + '|--', end = '')
print(self.data)
if self.children:
for each in self.children:
each.print_tree()
def get_level(self):
level = 0
p = self.parent
while p :
p = p.parent
level += 1
return level
class Tree(object):
def __init__(self):
self.root = None
And I tried with recursion to create Tree. Board is Hash map representing state of board(initially is empty and it should be root, root children should be all possible moves on empty board and so on)
def make_tree(self,board, player, depth):
list_possible_moves = self.possible_moves()
last_move = []
t = Tree()
t.root = TreeNode(board,None)
if depth > 0:
node = TreeNode(board,player)
board2 = deepcopy(self._board)
player = player * -1
for i in range(len(list_possible_moves)):
board2 = deepcopy(board2)
if len(last_move) != 0 :
board2.set_val(last_move[-1],"0")
if player == 1:
board2.set_val(i,"W")
else:
board2.set_val(i,"B")
child = TreeNode(board2,player)
if child.get_level == 1:
t.root.add_child()
else:
node.add_child(child)
if player == 1 :
self.make_tree(board2,1,depth-1)
if player == -1 :
self.make_tree(child,-1,depth-1)
else :
return None
But it is not working, and it takes 6-10 seconds to make tree.

Related

Trouble balancing a binary search tree using AVL

I'm unable to identify where I'm going wrong with my AVL implementation for balancing an existing binary search tree. I'm not getting any errors but my binary search tree does not come out to be properly balanced. After insertion, my binary search tree looks like (it would be prudent here to mention that my display_keys method gives us a visualization that is rotated by 90 degrees):
∅
The-Dreamers
∅
Saint-Laurent
∅
Pierrot-le-Fou
∅
Contempt
Cold-War
Before-Sunrise
∅
Basic-Instinct
∅
This is correct as it seems be to be following the rules for a BST.
But after calling the BalanceTree() method on my binary search tree, I seem to get:
∅
The-Dreamers
Saint-Laurent
Pierrot-le-Fou
Contempt
Cold-War
Before-Sunrise
Basic-Instinct
∅
which as you can see, is not Balanced. But wait, and here's the catch, if I call BalanceTree() again, the tree comes out to be perfectly balanced and isBSTBalanced returns True also. After the second call to BalanceTree(), our tree looks like:
∅
The-Dreamers
Saint-Laurent
Pierrot-le-Fou
Contempt
Cold-War
Before-Sunrise
Basic-Instinct
∅
I am adding the complete source code for my BST class for clarity and if somebody wants to execute the code, but I have added a comment (#Addition of new methods for AVL starts here) in the BST class to indicate where the methods for AVL start. You need only concern yourself with them. I would like for you help me pinpoint what exactly is going wrong in my code.
class BST:
class TreeNode:
def __init__(self, key, value, left=None, right=None, parent=None):
self.key = key
self.value = value
self.left = left
self.right = right
self.parent = parent
self.height = 1
def __init__(self):
self.root = None
self.size = 0
def __len__(self):
return self.size
def insert(self, key, value):
if self.root == None:
self.root = self.TreeNode(key, value)
else:
self._insert(key, value, self.root)
self.size += 1
def _insert(self, key, value, curr_node):
if key < curr_node.key:
if curr_node.left is not None:
self._insert(key, value, curr_node.left)
else:
curr_node.left = self.TreeNode(key, value, parent=curr_node)
elif key > curr_node.key:
if curr_node.right is not None:
self._insert(key, value, curr_node.right)
else:
curr_node.right = self.TreeNode(key, value, parent=curr_node)
def search(self, key):
if self.root:
found = self._search(key, self.root)
if found:
return found.value
else:
return None
else:
return None
def _search(self, key, curr_node):
if not curr_node:
return None
elif curr_node.key == key:
return curr_node
elif key < curr_node.key:
return self._search(key, curr_node.left)
else:
return self._search(key, curr_node.right)
def find_min(self):
curr = self.root
while curr.left is not None:
curr = curr.left
return curr
def find(self, node):
curr = node
while curr.left is not None:
curr = curr.left
return curr
def delete(self, key):
node_to_remove = self._search(key, self.root)
if node_to_remove.left is None and node_to_remove.right is None:
#Then we identify this as a leaf node
if node_to_remove is node_to_remove.parent.left:
#Setting the parent's reference to this to None
node_to_remove.parent.left = None
elif node_to_remove is node_to_remove.parent.right:
node_to_remove.parent.right = None
#2nd Case --> Two child
elif node_to_remove.left and node_to_remove.right:
minimum = self.find(node_to_remove.right)
self.delete(minimum.key) #We will still have a ref to this node afterwards
node_to_remove.key, node_to_remove.value = minimum.key, minimum.value
#3rd Case -> One child
else:
if node_to_remove.left:
node_to_remove.left.parent = node_to_remove.parent
node_to_remove.parent.left = node_to_remove.left
elif node_to_remove.right:
node_to_remove.right.parent = node_to_remove.parent
node_to_remove.parent.right = node_to_remove.right
def traversal(self, root):
res = []
if root:
res = self.traversal(root.left)
res.append(root)
res = res + self.traversal(root.right)
return res
def inorder_traversal(self, root):
if root:
self.inorder_traversal(root.left)
print(root.key)
self.inorder_traversal(root.right)
#Addition of new methods for AVL starts here
def display_keys(self, node, space='\t', level=0):
"""
Allows us to visualize the tree (albiet rotated by 90 degrees)
"""
# print(node.key if node else None, level)
# If the node is empty
if node is None:
print(space*level + '∅')
return
# If the node is a leaf
if node.left is None and node.right is None:
print(space*level + str(node.key))
return
# If the node has children
self.display_keys(node.right, space, level+1)
print(space*level + str(node.key))
self.display_keys(node.left,space, level+1)
def height(self):
return self._height(self.root)
def _height(self, curr_node):
if curr_node is None:
return -1 #since we are counting number of edges, we will return -1
else:
return 1 + max(self._height(curr_node.left), self._height(curr_node.right))
def isBSTBalanced(self):
return self._isBSTBalanced(self.root)
def _isBSTBalanced(self, curr_node):
if curr_node is None:
return True
hleft_subtree = self._height(curr_node.left)
hright_subtree = self._height(curr_node.right)
if hleft_subtree - hright_subtree in [-1,0,1]:
return self._isBSTBalanced(curr_node.left) and self._isBSTBalanced(curr_node.right)
else:
return False
def balance_factor(self):
if self.root is not None:
return self._balance_factor(self.root)
else:
return 0
def _balance_factor(self, curr_node):
if curr_node is None:
return
hleft_subtree = self._height(curr_node.left)
hright_subtree = self._height(curr_node.right)
b_factor = hleft_subtree - hright_subtree
return b_factor
def BalanceTree(self):
if self.isBSTBalanced() == False:
return self._rebalance(self.root)
def _rebalance(self, curr_node):
if curr_node is None:
return None
curr_node.left = self._rebalance(curr_node.left)
curr_node.right = self._rebalance(curr_node.right)
curr_node.height = 1 + max(self._height(curr_node.left), self._height(curr_node.right))
#print(curr_node.height)
if self._balance_factor(curr_node) > 1 and self._balance_factor(curr_node.left) >= 0:
#left heavy subtree
return self._rotate_right(curr_node)
if self._balance_factor(curr_node) < -1 and self._balance_factor(curr_node.right) <= 0:
#right heavy subtree
return self._rotate_left(curr_node)
if self._balance_factor(curr_node) < 0 and self._balance_factor(curr_node.right) > 0:
self._rotate_right(curr_node.right)
return self._rotate_left(curr_node)
if self._balance_factor(curr_node) > 0 and self._balance_factor(curr_node.left) < 0:
self._rotate_left(curr_node.left)
return self._rotate_right(curr_node)
return curr_node
def _rotate_left(self, oldRoot):
newRoot = oldRoot.right #the newRoot is the right child of the previous root
oldRoot.right = newRoot.left #replacing right child of the old root with the left child of the new
if newRoot.left is not None:
newRoot.left.parent = oldRoot
newRoot.parent = oldRoot.parent
if oldRoot == self.root:
self.root = newRoot
else:
if oldRoot.parent.left is oldRoot: #Checking isLeftChild
oldRoot.parent.left = newRoot
else:
oldRoot.parent.right = newRoot
newRoot.left = oldRoot
oldRoot.parent = newRoot
oldRoot.height = 1 + max(self._height(oldRoot.left), self._height(oldRoot.right))
newRoot.height = 1 + max(self._height(newRoot.left), self._height(newRoot.right))
return newRoot
def _rotate_right(self, oldRoot):
newRoot = oldRoot.left #the newRoot is the left child of the previous root
oldRoot.left = newRoot.right #replacing left child of the old root with the right child of the new
if newRoot.right is not None:
newRoot.right.parent = oldRoot
newRoot.parent = oldRoot.parent
if oldRoot == self.root:
self.root = newRoot
else:
if oldRoot.parent.right is oldRoot: #Checking isRightChild
oldRoot.parent.right = newRoot
else:
oldRoot.parent.left = newRoot
newRoot.right = oldRoot
oldRoot.parent = newRoot
oldRoot.height = 1 + max(self._height(oldRoot.left), self._height(oldRoot.right))
newRoot.height = 1 + max(self._height(newRoot.left), self._height(newRoot.right))
return newRoot
if __name__ == '__main__':
obj = BST()
obj.insert('Basic-Instinct', 0)
obj.insert('The-Dreamers', 1)
obj.insert('Saint-Laurent', 2)
obj.insert('Pierrot-le-Fou', 3)
obj.insert('Contempt', 4)
obj.insert('Before-Sunrise', 5)
obj.insert('Cold-War', 8)
obj.display_keys(obj.root) #displays a visual representation of our tree, albeit rotated by 90 degrees
print()
print("isBSTBalanced:", obj.isBSTBalanced())
obj.BalanceTree()
print("isBSTBalanced:", obj.isBSTBalanced()) #After executing BalanceTree(), isBSTBalanced still returns False
print()
obj.display_keys(obj.root)
Progress: Revamped _isBSTBalanced method so that it visits every node recursively and not just the root node. The final outcome, however, remains the same.
Progress: I was able to identify one of the major issues being that while I was calling _rotate_left and _rotate_right methods in the _rebalance method, I was not returning them. In addition to this, I was not recursively visiting the left and right subtrees of curr_node, which was initially set to the root of the tree, to be able to traverse the tree in a bottom up manner. I have resolved this too. I have added a display_keys method which allows us to visualize the tree, albeit rotated by 90 degrees. I'm updating the code and prompt in this post accordingly. The problem that still remains is that I have to call the BalanceTree() method more than once in some cases for isBSTBalanced to return True.

Trouble calling height() on a binary tree

I'm writing a function to check if a binary tree satisfies the Height-Balance Property. This is my code but I'm having trouble calling the height function for left and right from my given LinkedBinaryTree class. The main thing that's confusing me is that the nested function takes the root a parameter but height() doesn't. For reference, bin_tree is a LinkedBinaryTree() not a node. Thank you in advance for any help!
My Code
from LinkedBinaryTree import LinkedBinaryTree
def is_height_balanced(bin_tree):
if bin_tree.root is None:
return True
left = bin_tree.height()
right = bin_tree.height()
if abs(left - right) <= 1:
if is_height_balanced(bin_tree.root.left) is True and is_height_balanced(bin_tree.root.right) is True:
return True
return False
Portion of LinkedBinaryTree class
class LinkedBinaryTree:
class Node:
def __init__(self, data, left=None, right=None):
self.data = data
self.parent = None
self.left = left
if (self.left is not None):
self.left.parent = self
self.right = right
if (self.right is not None):
self.right.parent = self
def __init__(self, root=None):
self.root = root
self.size = self.count_nodes()
# assuming count_nodes() and is_empty() works as expected
def height(self):
def subtree_height(root):
if (root.left is None and root.right is None):
return 0
elif (root.left is None):
return 1 + subtree_height(root.right)
elif (root.right is None):
return 1 + subtree_height(root.left)
else:
left_height = subtree_height(root.left)
right_height = subtree_height(root.right)
return 1 + max(left_height, right_height)
if(self.is_empty()):
raise Exception("Tree is empty")
return subtree_height(self.root)

what is wrong with my invert tree function?

i am trying to invert a binary tree, but when i print, nothing changes.
my strategy is to the nextSibling a prevSibling. that means, i need to reverse the linked-list, but here i am just using nodes, not linked-lists. please help
class treeNode:
def __init__(self,data):
self.data = data
self.children = []
self.parent = None
self.nextSibling = None
self.level = 0
self.prevSibling = None
def add_children(self,*child):
for i in child:
i.parent = self
self.children.append(i)
if len(self.children) > 1:
self.children[-2].nextSibling = i
i.prevSibling = self.children[-2]
class Tree:
def __init__(self,root:treeNode):
self.root = root
self.depth = 0
def invert_tree(self):
self.root.children = self.root.children[::-1]
kid = self.root.children[0]
while kid != self.root:
if len(kid.children) == 0:
while kid.parent:
if kid.prevSibling:
kid = kid.prevSibling
break
else:
kid = kid.parent
else:
kid.children = kid.children[::-1]
kid = kid.children[0]
parent = treeNode(1)
child = treeNode(2)
grand = treeNode(3)
parent.add_children(child,treeNode(22))
child.add_children(grand,treeNode(33))
grand.add_children(treeNode(4))
tree = Tree(parent)
tree.print_tree()
tree.invert_tree()
this is the output i get:
1
|__ 22
thanks in advance :)

What does assigning self to a variable in Python Class mean?

While I was learning Tree data structure in python, I came across this code:
class TreeNode:
def __init__(self, data):
self.data = data
self.children = []
self.parent = None
def get_level(self):
level = 0
p = self.parent
while p:
level += 1
p = p.parent
return level
def print_tree(self):
spaces = ' ' * self.get_level() * 3
prefix = spaces + "|__" if self.parent else ""
print(prefix + self.data)
if self.children:
for child in self.children:
child.print_tree()
def add_child(self, child):
child.parent = self
self.children.append(child)
def build_product_tree():
root = TreeNode("Electronics")
laptop = TreeNode("Laptop")
laptop.add_child(TreeNode("Mac"))
laptop.add_child(TreeNode("Surface"))
laptop.add_child(TreeNode("Thinkpad"))
cellphone = TreeNode("Cell Phone")
cellphone.add_child(TreeNode("iPhone"))
cellphone.add_child(TreeNode("Google Pixel"))
cellphone.add_child(TreeNode("Vivo"))
tv = TreeNode("TV")
tv.add_child(TreeNode("Samsung"))
tv.add_child(TreeNode("LG"))
root.add_child(laptop)
root.add_child(cellphone)
root.add_child(tv)
root.print_tree()
if __name__ == '__main__':
build_product_tree()
It was a tutorial from Codebasics: https://www.youtube.com/watch?v=4r_XR9fUPhQ&list=PLeo1K3hjS3uu_n_a__MI_KktGTLYopZ12&index=9
Now my question is why in the 25th line there is child.parent = self what does it mean or why should I have to do that.
self references the current instance. This line of code will tell the child that it's parent is the current instance.

How to check if a Binary tree is full?

Student here. I have coded a method to add to a BinaryTree class that checks to see if the binary tree is full (the rest of the class code came from a Python textbook). I believe it is working as it should but am new to coding and I do not know how to check without the visual of the tree to see if all of the leaves are on the same level. Can anyone take a look at the is isFullBinaryTree(self) method is correct? It is near the end of the class. The test program seems to be working, but I am not positive.
My thought was that if I coded a count variable for each side of the tree and if the count is the same on both sides, then the leaves must all be on the same level which would make it a full tree.
Here is the full code:
class TreeNode:
def __init__(self, e):
self.element = e
self.left = None # Point to the left node, default None
self.right = None # Point to the right node, default None
class BinaryTree:
def __init__(self):
self.root = None
self.size = 0
# Insert element e into the binary search tree
# Return True if the element is inserted successfully
def insert(self, e):
if self.root == None:
self.root = self.createNewNode(e) # Create a new root/Create the node for e as the root
else:
# Locate the parent node
parent = None
current = self.root
while current != None:
if e < current.element:
parent = current # Keep the parent
current = current.left # Go left
elif e > current.element:
parent = current # Keep the parent
current = current.right # Go right
else:
return False # Duplicate node not inserted
# Create a new node for e and attach it to parent
if e < parent.element:
parent.left = self.createNewNode(e)
else:
parent.right = self.createNewNode(e)
self.size += 1 # Increase tree size
return True # Element inserted
# Create a new TreeNode for element e
def createNewNode(self, e):
return TreeNode(e)
# Returns true if the tree is a full binary tree
def isFullBinaryTree(self):
current = self.root # Start from the root
while current != None:
leftNode = current.left
rightNode = current.right
leftCount = 0
rightCount = 0
while leftNode != None:
current = leftNode
leftNode = current.left
leftCount += 1 # add 1 because we are moving from current one time
while rightNode != None:
current = rightNode
rightNode = current.right
rightCount += 1 # add 1 because we are moving from current one time
if leftCount == rightCount:
return True
else:
return False
return False
def main():
numbers = [2, 4, 3, 1, 8, 5, 6, 7, 0]
intTree1 = BinaryTree()
for e in numbers:
intTree1.insert(e)
print("\nIs intTree1 full? ", end = "")
print(intTree1.isFullBinaryTree())
numbers2 = [2, 4, 3, 1, 8, 5, 6, 7]
intTree2 = BinaryTree()
for e in numbers2:
intTree2.insert(e)
print("\nIs intTree2 full? ", end = "")
print(intTree2.isFullBinaryTree())
main()
If your tree looked like /\, then I think it would return True because you're only iterating the outside of the tree, not checking the fullness of the inner branches
Instead of looping at all, or even counting, I would suggest recursion
You'll need to implement the isFullBinaryTree methods on the TreeNode class for this to work, though.
class TreeNode:
def __init__(self, e):
self.element = e
self.left = None # Point to the left node, default None
self.right = None # Point to the right node, default None
def isFullBinaryTree(self):
# check if we are a leaf
if self.left is None and self.right is None:
return True
# recursively check the left fullness
full_left = self.left.isFullBinaryTree() if self.left else False
# recursively check the right fullness
full_right = self.right.isFullBinaryTree() if self.right else False
# return True if checked that both left and right are full
return full_left and full_right
Once you do this, then the BinaryTree class can simply be checking if the root exists and if the root TreeNode is considered full itself.
e.g.
def isFullBinaryTree(self):
return self.root.isFullBinaryTree() if self.root else False

Categories