I'm stuck finding the path of the given node so I can find the right side but not the left side.
class Node:
def __init__(self, key):
self.left = None
self.right = None
self.val = key
def path(self, k):
if not self.val:
return []
if self.val == k:
return [self.val]
res = self.left.path(k)
if res:
return [self.val] + res
res = self.right.path(k)
if res:
return [self.val] + res
return []
For example, when I search for 5 in X="+ * + 5 7 4 1 6", the output is like ['+', '*', '+', '5'] but when I try to search for any number on the lift subtree, it gives me this error:
[Previous line repeated 1 more time]
AttributeError: 'NoneType' object has no attribute 'path'
Before accessing a method like path you should ensure that the object really is a Node instance, and not None.
Not the problem, but not self.val will also be a true expression when that val is 0, which seems like a valid key for a node... I suppose this test is there to detect a node with a None value. Usually such code is used when a tree is always created with at least one node instance, even when it is empty. This is not good practice. An empty tree should not have nodes at all, not even one with a None value.
So correct to:
class Node:
def __init__(self, key):
self.left = None
self.right = None
self.val = key
def path(self, k):
if self.val == k:
return [self.val]
if self.left:
res = self.left.path(k)
if res:
return [self.val] + res
if self.right:
res = self.right.path(k)
if res:
return [self.val] + res
return []
This is the python code which uses A* algorithm for finding solution for 8 puzzle problems, I got some error messages, how can I fix it?(The error message is under the code)
There are several object-oriented programming concepts for Problems class, Node class that are implemented to express the problem solution search that you need to understand in order to make the Python program complete. The priority queue is to make the nodes to be explored to be sorted according to their f-evaluation function score and return the min one as the first node to be searched next.
There is also a memorize function to memorize the heuristic value of state as a look-up table so that you don’t need to calculate the redundant computing of heuristic estimation value, so you can ignore it at this point if you don’t understand.
The components you need to implement is to make the abstract part of the program realizable for 8 -puzzle with the successor methods attached to a problem class which consists of initial state and goal state. Make sure the program can run correctly to generate the solution sequence that move the empty tile so that the 8-puzzle can move "Up", "Down", "Left", "Right", from initial state to goal state.
import math
infinity = math.inf
from itertools import chain
import numpy as np
import bisect
class memoize:
def __init__(self, f, memo={}):
self.f = f
self.memo = {}
def __call__(self, *args):
if not str(args) in self.memo:
self.memo[str(args)] = self.f(*args)
return self.memo[str(args)]
def coordinate(state):
index_state = {}
index = [[0,0], [0,1], [0,2], [1,0], [1,1], [1,2], [2,0], [2,1], [2,2]]
for i in range(len(state)):
index_state[state[i]] = index[i]
return index_state
def getInvCount(arr):
inv_count = 0
empty_value = -1
for i in range(0, 9):
for j in range(i + 1, 9):
if arr[j] != empty_value and arr[i] != empty_value and arr[i] > arr[j]:
inv_count += 1
return inv_count
def isSolvable(puzzle) :
inv_count = getInvCount([j for sub in puzzle for j in sub])
return (inv_count % 2 == 0)
def linear(state):
return sum([1 if state[i] != goal[i] else 0 for i in range(9)])
#memoize
def manhattan(state):
index_goal = coordinate(goal)
index_state = coordinate(state)
mhd = 0
for i in range(9):
for j in range(2):
mhd = abs(index_goal[i][j] - index_state[i][j]) + mhd
return mhd
#memoize
def sqrt_manhattan(state):
index_goal = coordinate(goal)
index_state = coordinate(state)
mhd = 0
for i in range(9):
for j in range(2):
mhd = (index_goal[i][j] - index_state[i][j])**2 + mhd
return math.sqrt(mhd)
#memoize
def max_heuristic(state):
score1 = manhattan(state)
score2 = linear(state)
return max(score1, score2)
class PriorityQueueElmt:
def __init__(self,val,e):
self.val = val
self.e = e
def __lt__(self,other):
return self.val < other.val
def value(self):
return self.val
def elem(self):
return self.e
class Queue:
def __init__(self):
pass
def extend(self, items):
for item in items: self.append(item)
class PriorityQueue(Queue):
def __init__(self, order=min, f=None):
self.A=[]
self.order=order
self.f=f
def append(self, item):
queueElmt = PriorityQueueElmt(self.f(item),item)
bisect.insort(self.A, queueElmt)
def __len__(self):
return len(self.A)
def pop(self):
if self.order == min:
return self.A.pop(0).elem()
else:
return self.A.pop().elem()
# Heuristics for 8 Puzzle Problem
class Problem:
def __init__(self, initial, goal=None):
self.initial = initial; self.goal = goal
def successor(self, state):
reachable = []
def get_key(val):
for key, value in index_state.items():
if val == value:
return key
return -1
def candidate(state, Position):
state = state.copy()
zero_index = state.index(0)
swap_index = state.index(get_key(Position))
state[zero_index], state[swap_index] = state[swap_index], state[zero_index]
return state
index_state = coordinate(state)
zero_position = index_state[0]
move_pair = {"left":[zero_position[0], zero_position[1] - 1],
"right":[zero_position[0], zero_position[1] + 1],
"up":[zero_position[0] - 1, zero_position[1]],
"down":[zero_position[0] + 1, zero_position[1]]
}
for action, position in move_pair.items():
#print(action, position)
if get_key(position) != -1:
reachable.append((action, candidate(state, position)))
#print(reachable)
return reachable
def goal_test(self, state):
return state == self.goal
def path_cost(self, c, state1, action, state2):
return c + 1
def value(self):
abstract
class Node:
def __init__(self, state, parent=None, action=None, path_cost=0, depth =0):
self.parent = parent
if parent:
self.depth = parent.depth + 1
else:
self.depth = 0
self.path_cost = path_cost
self.state = state
if action:
self.action = action
else: self.action = "init"
def __repr__(self):
return "Node state:\n " + str(np.array(self.state).reshape(3,3)) +"\n -> action: " + self.action + "\n -> depth: " + str(self.depth)
def path(self):
x, result = self, [self]
while x.parent:
result.append(x.parent)
x = x.parent
return result
def expand(self, problem):
for (act,n) in problem.successor(self.state):
if n not in [node.state for node in self.path()]:
yield Node(n, self, act,
problem.path_cost(self.path_cost, self.state, act, n))
def graph_search(problem, fringe):
closed = {}
fringe.append(Node(problem.initial,depth=0))
while fringe:
node = fringe.pop()
if problem.goal_test(node.state):
return node
if str(node.state) not in closed:
closed[str(node.state)] = True
fringe.extend(node.expand(problem))
return None
def best_first_graph_search(problem, f):
return graph_search(problem, PriorityQueue(min, f))
def astar_search(problem, h = None):
h = h or problem.h
def f(n):
return max(getattr(n, 'f', -infinity), n.path_cost + h(n.state))
return best_first_graph_search(problem, f)
def print_path(path, method):
print("*" * 30)
print("\nPath: (%s distance)" % method)
for i in range(len(path)-1, -1, -1):
print("-" * 15)
print(path[i])
goal = [1, 2, 3, 4, 5, 6, 7, 8, 0]
# Solving the puzzle
puzzle = [7, 2, 4, 5, 0, 6, 8, 3, 1]
if(isSolvable(np.array(puzzle).reshape(3,3))): # even true
# checks whether the initialized configuration is solvable or not
print("Solvable!")
problem = Problem(puzzle,goal)
path = astar_search(problem, manhattan).path()
print_path(path, "manhattan")
path = astar_search(problem, linear).path()
print_path(path, "linear")
path = astar_search(problem, sqrt_manhattan).path()
print_path(path, "sqrt_manhattan")
path = astar_search(problem, max_heuristic).path()
print_path(path, "max_heuristic")
else :
print("Not Solvable!") # non-even false
TypeError Traceback (most recent call last)
<ipython-input-124-2a60ddc8c009> in <module>
9 problem = Problem(puzzle,goal)
10
---> 11 path = astar_search(problem, manhattan).path()
12 print_path(path, "manhattan")
13
<ipython-input-123-caa97275712e> in astar_search(problem, h)
18 def f(n):
19 return max(getattr(n, 'f', -infinity), n.path_cost + h(n.state))
---> 20 return best_first_graph_search(problem, f)
21
22 def print_path(path, method):
<ipython-input-123-caa97275712e> in best_first_graph_search(problem, f)
12
13 def best_first_graph_search(problem, f):
---> 14 return graph_search(problem, PriorityQueue(min, f))
15
16 def astar_search(problem, h = None):
<ipython-input-123-caa97275712e> in graph_search(problem, fringe)
8 if str(node.state) not in closed:
9 closed[str(node.state)] = True
---> 10 fringe.extend(node.expand(problem))
11 return None
12
<ipython-input-121-e5a968bd54f0> in extend(self, items)
18
19 def extend(self, items):
---> 20 for item in items: self.append(item)
21
22 class PriorityQueue(Queue):
<ipython-input-122-db21613469b9> in expand(self, problem)
69
70 def expand(self, problem):
---> 71 for (act,n) in problem.successor(self.state):
72 if n not in [node.state for node in self.path()]:
73 yield Node(n, self, act,
TypeError: cannot unpack non-iterable int object
I got some error messages, how can I fix it?
There is one error message, The pieces of codes you get in the error message are the stack trace, which might help you to know how the execution got at the final point where the error occurred. In this case that is not so important. The essence of the error is this:
for (act,n) in problem.successor(self.state)
TypeError: cannot unpack non-iterable int object
So this means that the successor method returned an int instead of a list.
Looking at the code for successor, I notice that it intends to return a list called reachable, but there is a return statement right in the middle of the code, leaving the largest part of that code unexecuted (so-called "dead code"):
return state
This statement makes no sense where it is positioned. It seems to be an indentation problem: that return belongs inside the function just above it, like this:
def candidate(state, Position):
state = state.copy()
zero_index = state.index(0)
swap_index = state.index(get_key(Position))
state[zero_index], state[swap_index] = state[swap_index], state[zero_index]
return state # <-- indentation!
I'm trying to figure out a recursive solution to this problem. The main thing is to return the level in the binary tree where the node is.
def find_depth(tree, node):
if node == None:
return 0
else:
return max(find_depth(tree.left))
#recursive solution here
Using this class for the values:
class Tree:
def __init__(self, value, left=None, right=None):
self.value = value
self.left = left
self.right = right
Example: Calling find_depth(tree, 7) should return the level where 7 is in the tree. (level 2)
3
/ \
7 1 <------ return that 7 is at level 2
/ \
9 3
maybe this is what you are looking for
def find_depth(tree, node):
if node is None or tree is None:
return 0
if tree == node:
return 1
left = find_depth(tree.left, node)
if left != 0:
return 1 + left
right = find_depth(tree.right, node)
if right != 0:
return 1 + right
return 0
You need to provide information about depth in find_depth call. It might look like this (assuming 0 is a sentinel informing that node is not found):
def find_depth(tree, node, depth=1):
if node == None:
return 0
if tree.value == node:
return depth
left_depth = find_depth(tree.left, node, depth+1)
right_depth = find_depth(tree.right, node, depth+1)
return max(left_depth, right_depth)
Then you call it with two parameters: x = find_depth(tree, 7).
Recursion is a functional heritage and so using it with functional style yields the best results -
base case: if the input tree is empty, we cannot search, return None result
inductive, the input tree is not empty. if tree.data matches the search value, return the current depth, d
inductive, the input tree is not empty and tree.data does not match. return the recursive result of tree.left or the recursive result of find.right
def find (t = None, value = None, d = 1):
if not t:
return None # 1
elif t.value == value:
return d # 2
else:
return find(t.left, value, d + 1) or find(t.right, value, d + 1) # 3
class tree:
def __init__(self, value, left = None, right = None):
self.value = value
self.left = left
self.right = right
t = tree \
( 3
, tree(7, tree(9), tree(3))
, tree(1)
)
print(find(t, 7)) # 2
print(find(t, 99)) # None
You can implement the find method in your tree class too
def find (t = None, value = None, d = 1):
# ...
class tree
def __init__ #...
def find(self, value)
return find(self, value)
print(t.find(7)) # 2
print(t.find(99)) # None
I need to implement an insert method (insert(self, index, val)), that inserts val before index, and a pop method (pop(self)), that removes the last element from mylist, onto the MyList class. The behavior should be identical to the methods already available in python.
Note: For the insert method, similarly with the append method already done, the capacity of the array should be doubled if there is no room for an additional element. The pop method should return the element removed from the list, and put None
in its place in the array. If pop was called on an empty list, an IndexError
exception should be raised.
My code thus far:
import ctypes # provides low-level arrays
def make_array(n):
return (n * ctypes.py_object)()
class MyList:
def __init__(self):
self.data = make_array(1)
self.capacity = 1
self.n = 0
def __len__(self):
return self.n
def append(self, val):
if(self.n == self.capacity):
self.resize(2 * self.capacity)
self.data[self.n] = val
self.n += 1
def resize(self, new_size):
new_array = make_array(new_size)
for ind in range(self.n):
new_array[ind] = self.data[ind]
self.data = new_array
self.capacity = new_size
def extend(self, other):
for elem in other:
self.append(elem)
def __getitem__(self, ind):
if not(0 <= ind <= self.n - 1):
raise IndexError('MyList index is out of range')
return self.data[ind]
def __setitem__(self, ind, val):
if not(0 <= ind <= self.n - 1):
raise IndexError('MyList index is out of range')
self.data[ind] = val
mylst1 = MyList()
for i in range(5):
mylst1.append(i)
Situation: I'm trying to get a good handle on the doubly-linked structure. I've got a decent grip on the methods so far. I want to be able to create two objects for this class and check if every item in it is equal. I don't have any syntax errors, and the error I'm getting is kind of confusing. So here's what I have so far.
class LinkedList:
class Node:
def __init__(self, val, prior=None, next=None):
self.val = val
self.prior = prior
self.next = next
def __init__(self):
self.head = LinkedList.Node(None) # sentinel node (never to be removed)
self.head.prior = self.head.next = self.head # set up "circular" topology
self.length = 0
def append(self, value):
n = LinkedList.Node(value, prior=self.head.prior, next=self.head)
n.prior.next = n.next.prior = n
self.length += 1
def _normalize_idx(self, idx):
nidx = idx
if nidx < 0:
nidx += len(self)
if nidx < -1:
raise IndexError
return nidx
def __getitem__(self, idx):
"""Implements `x = self[idx]`"""
nidx = self._normalize_idx(idx)
currNode = self.head.next
for i in range(nidx):
currNode = currNode.next
if nidx >= len(self):
raise IndexError
return currNode.val
def __setitem__(self, idx, value):
"""Implements `self[idx] = x`"""
nidx = self._normalize_idx(idx)
currNode = self.head.next
if nidx >= len(self):
raise IndexError
for i in range(nidx):
currNode = currNode.next
currNode.val = value
def __iter__(self):
"""Supports iteration (via `iter(self)`)"""
cursor = self.head.next
while cursor is not self.head:
yield cursor.val
cursor = cursor.next
def __len__(self):
"""Implements `len(self)`"""
return self.length
def __eq__(self, other):
currNode = self.head.next
currNode2 = other.head.next
for currNode, currNode2 in zip(self, other):
if currNode.val != currNode2.val:
return False
return True
Test:
from unittest import TestCase
tc = TestCase()
lst = LinkedList()
lst2 = LinkedList()
tc.assertEqual(lst, lst2)
lst2.append(100)
tc.assertNotEqual(lst, lst2)
When I test this code I get I get an Assertion error saying " [] == [100] " I'm unsure why my code recognizes this as equal, when I want it to actually check specific values in the node.
zip only goes as far as the shortest list. You want itertools.zip_longest, and you don't want .val (your iterator returns the actual values already). Try this:
def __eq__(self, other):
for val1, val2 in zip_longest(self, other):
if val1 != val2:
return False
return True
or perhaps better?
def __eq__(self, other):
return all(val1 == val2 for val1, val2 in zip_longest(self, other))
EDIT
I like #BrenBarn's suggestion of checking length first. Here's a more efficient answer:
def __eq__(self, other):
return len(self) == len(other) and all(
val1 == val2 for val1, val2 in zip(self, other))
zip(self.other) only gives you as many elements as the shorter of the two lists. It discards the extra part of the longer list. So for [] == [100], zip doesn't give any elements, and your code returns True without checking anything.
You could just do a check at the beginning to see if the lists have different length. If they do, they can't be equal.