astar.py

# Credit for this: Nicholas Swift
# as found at https://medium.com/@nicholas.w.swift/easy-a-star-pathfinding-7e6689c7f7b2
from warnings import warn
import heapq

class Node:
    """
    A node class for A* Pathfinding
    """

    def __init__(self, parent=None, position=None):
        self.parent = parent
        self.position = position

        self.g = 0
        self.h = 0
        self.f = 0

    def __eq__(self, other):
        return self.position == other.position
    
    def __repr__(self):
      return f"{self.position} - g: {self.g} h: {self.h} f: {self.f}"

    # defining less than for purposes of heap queue
    def __lt__(self, other):
      return self.f < other.f
    
    # defining greater than for purposes of heap queue
    def __gt__(self, other):
      return self.f > other.f

def return_path(current_node):
    path = []
    current = current_node
    while current is not None:
        path.append(current.position)
        current = current.parent
    return path[::-1]  # Return reversed path


def astar(maze, start, end, allow_diagonal_movement = False):
    """
    Returns a list of tuples as a path from the given start to the given end in the given maze
    :param maze:
    :param start:
    :param end:
    :return:
    """

    # Create start and end node
    start_node = Node(None, start)
    start_node.g = start_node.h = start_node.f = 0
    end_node = Node(None, end)
    end_node.g = end_node.h = end_node.f = 0

    # Initialize both open and closed list
    open_list = []
    closed_list = []

    # Heapify the open_list and Add the start node
    heapq.heapify(open_list) 
    heapq.heappush(open_list, start_node)

    # Adding a stop condition
    outer_iterations = 0
    max_iterations = (len(maze[0]) * len(maze) // 2)

    # what squares do we search
    adjacent_squares = ((0, -1), (0, 1), (-1, 0), (1, 0),)
    if allow_diagonal_movement:
        adjacent_squares = ((0, -1), (0, 1), (-1, 0), (1, 0), (-1, -1), (-1, 1), (1, -1), (1, 1),)

    # Loop until you find the end
    while len(open_list) > 0:
        outer_iterations += 1

        if outer_iterations > max_iterations:
          # if we hit this point return the path such as it is
          # it will not contain the destination
          warn("giving up on pathfinding too many iterations")
          return return_path(current_node)       
        
        # Get the current node
        current_node = heapq.heappop(open_list)
        closed_list.append(current_node)

        # Found the goal
        if current_node == end_node:
            return return_path(current_node)

        # Generate children
        children = []
        
        for new_position in adjacent_squares: # Adjacent squares

            # Get node position
            node_position = (current_node.position[0] + new_position[0], current_node.position[1] + new_position[1])

            # Make sure within range
            if node_position[0] > (len(maze) - 1) or node_position[0] < 0 or node_position[1] > (len(maze[len(maze)-1]) -1) or node_position[1] < 0:
                continue

            # Make sure walkable terrain
            if maze[node_position[0]][node_position[1]] != 0:
                continue

            # Create new node
            new_node = Node(current_node, node_position)

            # Append
            children.append(new_node)

        # Loop through children
        for child in children:
            # Child is on the closed list
            if len([closed_child for closed_child in closed_list if closed_child == child]) > 0:
                continue

            # Create the f, g, and h values
            child.g = current_node.g + 1
            child.h = ((child.position[0] - end_node.position[0]) ** 2) + ((child.position[1] - end_node.position[1]) ** 2)
            child.f = child.g + child.h

            # Child is already in the open list
            if len([open_node for open_node in open_list if child.position == open_node.position and child.g > open_node.g]) > 0:
                continue

            # Add the child to the open list
            heapq.heappush(open_list, child)

    warn("Couldn't get a path to destination")
    return None

def example(print_maze = True):

    maze = [[0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,] * 2,
            [0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,] * 2,
            [0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,] * 2,
            [0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,] * 2,
            [0,0,0,1,1,0,0,1,1,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,] * 2,
            [0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,] * 2,
            [0,0,0,1,0,1,1,1,1,0,1,1,0,0,1,1,1,0,0,0,1,1,1,1,1,1,1,0,0,0,] * 2,
            [0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,1,1,0,1,0,0,0,0,0,0,1,1,1,0,] * 2,
            [0,0,0,1,0,1,1,0,1,1,0,1,1,1,0,0,0,0,0,1,0,0,1,1,1,1,1,0,0,0,] * 2,
            [0,0,0,1,0,1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,0,0,0,0,1,0,1,0,1,1,] * 2,
            [0,0,0,1,0,1,0,1,1,0,1,1,1,1,0,0,1,1,1,1,1,1,1,0,1,0,1,0,0,0,] * 2,
            [0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,1,1,0,] * 2,
            [0,0,0,1,0,1,1,1,1,0,1,0,0,1,1,1,0,1,1,1,1,0,1,1,1,0,1,0,0,0,] * 2,
            [0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,1,1,] * 2,
            [0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,] * 2,
            [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,] * 2,]
    
    start = (0, 0)
    end = (len(maze)-1, len(maze[0])-1)

    path = astar(maze, start, end)

    if print_maze:
      for step in path:
        maze[step[0]][step[1]] = 2
      
      for row in maze:
        line = []
        for col in row:
          if col == 1:
            line.append("\u2588")
          elif col == 0:
            line.append(" ")
          elif col == 2:
            line.append(".")
        print("".join(line))

    print(path)

@frankhale @stabtazer @ryancollingwood I simplify your code as below:

           \# Child is already in the open list
            if child in open_list: 
                idx = open_list.index(child) 
                if child.g < open_list[idx].g:
                    # update the node in the open list
                    open_list[idx].g = child.g
                    open_list[idx].f = child.f
                    open_list[idx].h = child.h
            else:
                \# Add the child to the open list
                heapq.heappush(open_list, child)

Through in function, we can fastly check whether child has been in open list. Also, I think we only need to update the g,h,f of the node which has the same position as child rather than operating the heap. About the calculation of Euclidean distance, I agree to have a sqrt on h, otherwise, h will play the most role in wayfinding and the result will be a straight line (on weight map).

@Yichen-Lei that's a great solution, however you forgot to update the parent, this leads to suboptimal paths. You should just update the entire open_list[idx]

           # Child is already in the open list
            if child in open_list: 
                idx = open_list.index(child) 
                if child.g < open_list[idx].g:
                    # update the node in the open list
                    open_list[idx] = child
            else:
                # Add the child to the open list
                heapq.heappush(open_list, child)

After messing around with this code for a uni project, I perfected the code and created a version that works for both 2d and 3d

from warnings import warn
import heapq
from itertools import product

class Node:
    def __init__(self, parent=None, position=None):
        self.parent = parent
        self.position = position

        self.g = 0
        self.h = 0
        self.f = 0

    def __eq__(self, other):
        return self.position == other.position
    
    def __repr__(self):
      return f"{self.position} - g: {self.g} h: {self.h} f: {self.f}"

    # defining less than for purposes of heap queue
    def __lt__(self, other):
      return self.f < other.f
    
    # defining greater than for purposes of heap queue
    def __gt__(self, other):
      return self.f > other.f

    def return_path(self):
        path = []
        current = self
        while current is not None:
            path.append(current.position)
            current = current.parent
        return path[::-1]
    
    def euclidean_distance(self, other):
        return round(sum([(self.position[i] - other.position[i])**2 for i in range(len(self.position))]) ** 0.5, 3)

    def is_outside(self, maze):
        def check_dimension(dim_maze, position, dim=0):
            if dim >= len(position):
                return False
            
            if position[dim] < 0 or position[dim] >= len(dim_maze):
                return True
            
            return check_dimension(dim_maze[position[dim]], position, dim + 1)
        
        return check_dimension(maze, self.position)

    def is_wall(self, maze):
        def navigate_dimension(dim_maze, position, dim=0):
            if dim == len(position) - 1:
                return dim_maze[position[dim]] == 0
            
            return navigate_dimension(dim_maze[position[dim]], position, dim + 1)
        
        return navigate_dimension(maze, self.position)

def astar(maze, start, end):
    dimension = len(maze.shape)
    if len(start) != dimension or len(end) != dimension:
        raise ValueError("Start and end must have the same number of dimensions as the maze")

    # Create start and end node
    start_node = Node(None, start)
    end_node = Node(None, end)

    closed_list = []
    open_list = []
    heapq.heapify(open_list) #create priority queue for open list
    heapq.heappush(open_list, start_node)

    outer_iterations = 0
    max_iterations = (maze.size // 2)

    # Define the adjacent squares (including diagonals)
    adjacent_squares = [c for c in product((-1, 0, 1), repeat=dimension) if any(c)]

    # Loop until you find the end
    while len(open_list) > 0:
        outer_iterations += 1

        if outer_iterations > max_iterations:# if we cannot find by searching half the maze, we give up
            warn("giving up on pathfinding. too many iterations")
            return current_node.return_path()
        
        # Get the current node
        current_node = heapq.heappop(open_list)
        closed_list.append(current_node)

        if current_node == end_node and open_list[0].f >= current_node.f:
            #! we are done
            return current_node.return_path()

        for new_position in adjacent_squares:
            # Get node position
            node_position = tuple([current_node.position[i] + new_position[i] for i in range(dimension)])
            new_node = Node(current_node, node_position)

            if new_node.is_outside(maze):
                continue

            if new_node.is_wall(maze):
                continue

            if new_node in closed_list:
                continue

            child = new_node #the new node is a valid child of the current_node
            #calculate the heuristic
            cost = sum([(child.position[i] - current_node.position[i])**2 for i in range(dimension)]) ** 0.5
            child.g = current_node.g + cost
            child.h = child.euclidean_distance(end_node)
            child.f = child.g + child.h

            #add or update the child to the open list
            if child in open_list: 
                i = open_list.index(child) 
                if child.g < open_list[i].g:
                    # update the node in the open list
                    open_list[i] = child
            else:
                heapq.heappush(open_list, child)

    warn("Couldn't get a path to destination")
    return None

Heapifying an empty list is a no-op; if len(X)>0 can be simplified to if X.

You definitely should replace the closed_list with a set. You'd need to add a __hash__ method to Node:

def ___hash__(self):
    return hash(self.position)

The "add or update the child to the open list" part scans the open list twice. You might want to implement a NodeArray data structure that augments the array by remembering where its members are.

Finally, a major bug: replacing a heap member with a lower-weight one is likely to violate the heap invariants. After you replace the item at position n, you need to compare it with the one at (n-1)//2 and swap them if it's smaller (repeat until n is zero). heapq._siftdown does this for you. (In the generic case you'd also need to check that the items at 2n and 2n+1 are larger (repeat likewise), but that can't happen here.)