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@jdp jdp/LICENSE
Last active Apr 20, 2018

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What would you like to do?
A* pathfinding over any arbitrary graph structure, with example Cartesian grid implementation
class AStar(object):
def __init__(self, graph):
self.graph = graph
def heuristic(self, node, start, end):
raise NotImplementedError
def search(self, start, end):
openset = set()
closedset = set()
current = start
openset.add(current)
while openset:
current = min(openset, key=lambda o:o.g + o.h)
if current == end:
path = []
while current.parent:
path.append(current)
current = current.parent
path.append(current)
return path[::-1]
openset.remove(current)
closedset.add(current)
for node in self.graph[current]:
if node in closedset:
continue
if node in openset:
new_g = current.g + current.move_cost(node)
if node.g > new_g:
node.g = new_g
node.parent = current
else:
node.g = current.g + current.move_cost(node)
node.h = self.heuristic(node, start, end)
node.parent = current
openset.add(node)
return None
class AStarNode(object):
def __init__(self):
self.g = 0
self.h = 0
self.parent = None
def move_cost(self, other):
raise NotImplementedError
from astar import AStar, AStarNode
from math import sqrt
class AStarGrid(AStar):
def heuristic(self, node, start, end):
return sqrt((end.x - node.x)**2 + (end.y - node.y)**2)
class AStarGridNode(AStarNode):
def __init__(self, x, y):
self.x, self.y = x, y
super(AStarGridNode, self).__init__()
def move_cost(self, other):
diagonal = abs(self.x - other.x) == 1 and abs(self.y - other.y) == 1
return 14 if diagonal else 10
from astar_grid import AStarGrid, AStarGridNode
from itertools import product
def make_graph(width, height):
nodes = [[AStarGridNode(x, y) for y in range(height)] for x in range(width)]
graph = {}
for x, y in product(range(width), range(height)):
node = nodes[x][y]
graph[node] = []
for i, j in product([-1, 0, 1], [-1, 0, 1]):
if not (0 <= x + i < width):
continue
if not (0 <= y + j < height):
continue
graph[nodes[x][y]].append(nodes[x+i][y+j])
return graph, nodes
graph, nodes = make_graph(8, 8)
paths = AStarGrid(graph)
start, end = nodes[1][1], nodes[5][7]
path = paths.search(start, end)
if path is None:
print "No path found"
else:
print "Path found:", path
Copyright (C) 2012 Justin Poliey <justin.d.poliey@gmail.com>
Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions:
The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software.
THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
@zmyspc

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zmyspc commented Mar 3, 2013

Great code!

@vindolin

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vindolin commented Aug 13, 2013

mapinfo.height/width should probably be mapinfo['height/width'] ?

@benaxelrod

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benaxelrod commented May 29, 2014

i made the bugfix that vindolin suggested as well as improved pretty printing and added an obstacle to the gridworld example in my fork: https://gist.github.com/benaxelrod/57dd9202a93e8295c2c1

@KasunRama

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KasunRama commented May 31, 2014

It always returns "No path found" for me. and it doesn't go inside the for loop- "for node in self.graph[current]: " (checked)
What may be the reason..

@kwrl

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kwrl commented Sep 26, 2014

Damn, these are the prettiest lines of code that I've seen in a long time.

Great work!

@kamisori

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kamisori commented Jun 16, 2015

hm, does it affect the results at all that you say each node is adjacent to itself?

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