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# 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) |
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.)
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