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A* path finding algo gives path but it's not the shortest path

I am trying to implement a* pathfinding in my program, but it is returning the following output.
Here is the output image
In the image blue blocks are visited blocks, the one with circles are yet to visit blocks and the yellow blocks are path retuned to us.
I could not figure out the problem in my code here is the following code the I used.
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 search(maze, cost, start, end): # Hear maze is a 2D list with 1 means a wall and 0 is a clear block
start_node = Node(None, tuple(start))
start_node.g = start_node.h = start_node.f = 0
end_node = Node(None, tuple(end))
end_node.g = end_node.h = end_node.f = 0
yet_to_visit_list = []
visited_list = []
yet_to_visit_list.append(start_node)
outer_iterations = 0
max_iterations = (len(maze) // 2) ** 10
move = [[-1, 0], [0, -1], [1, 0], [0, 1]]
no_rows, no_columns = np.shape(maze)
while len(yet_to_visit_list) > 0:
outer_iterations += 1
current_node = yet_to_visit_list[0]
current_index = 0
for index, item in enumerate(yet_to_visit_list):
if item.f < current_node.f:
current_node = item
current_index = index
if outer_iterations > max_iterations:
print("Cann't find the path... too many iterations")
return return_path(current_node, maze)
yet_to_visit_list.pop(current_index)
visited_list.append(current_node)
maze[current_node.position[0]][current_node.position[1]] = 2 # 1 wall 2 visited 3 yet to visit
if current_node == end_node:
return return_path(current_node, maze)
children = []
for new_position in move:
node_position = (current_node.position[0]+ new_position[0], current_node.position[1]+ new_position[1])
if (node_position[0] > (no_rows - 1) or node_position[0] < 0 or node_position[1]>(no_columns-1) or node_position[1] < 0):
continue
if maze[node_position[0]][node_position[1]] != 0:
continue
new_node = Node(current_node, node_position)
children.append(new_node)
for child in children:
if len([visited_child for visited_child in visited_list if visited_child == child]) > 0:
continue
child.g = current_node.g + cost
child.h = (((child.position[0] - end_node.position[0]) ** 2 ) +
((child.position[0] - end_node.position[0])) ** 2)
child.f = child.g + child.h
if len([i for i in yet_to_visit_list if child == i and child.g>i.g]) > 0:
continue
yet_to_visit_list.append(child)
And the following function returns the path
def return_path(current_node, maze):
path = []
no_rows, no_columns = np.shape(maze)
result = maze
current = current_node
while current is not None:
path.append(current.position)
current = current.parent
path = path[::-1]
start_value = 0
for i in range(len(path)):
result[path[i][0]][path[i][1]] = start_value
start_value += 1
return result

In your for loop over move, you exclude neighbors with maze[.][.] != 0: i.e., neighbors that are either "wall" or "visited". The original A* algorithm requires you to reconsider visited nodes to see if the cost can be further reduced.
Also, I get the sense that you should be removing items from the visited_list at some point in your program but I don't see this happening.
It would be easier to tell exactly what's wrong once you make the program input available, since it would then become possible to step through your code with a debugger.
Another problem is that you are using the squared distance for your heuristic cost, which likely overestimates the cost. With up-down-left-right neighbors, use the Manhattan distance function (see the warning about using squared Euclidean distance in the link). I don't think this is at the heart of what's wrong with your program, though.

A* path finding algo sometimes doesn't find path even when there is one (Python)

I made a custom path A* Pathfinding Algorithm in Python, but it sometimes doesn't even find path to the end node even when there clearly is a path. Here is my implementation.
# this is my Node class. I am representing the whole maze as a matrix and every cell
# of that matrix is a Node Object
class Node():
def __init__(self, i, j):
self.i = i
self.j = j
self.isWall = False
self.isOpen = None
self.f = 0
self.g = 0
self.h = 0
self.neighbors = []
self.previous = None
def __repr__(self):
return f'< i = {self.i}, j = {self.j}, previous = {self.previous} >'
def add_neighbors(self, grid, diagonal):
i = self.i
j = self.j
if i > 0:
self.neighbors.append(grid[i - 1][j])
if i < len(grid) - 1:
self.neighbors.append(grid[i + 1][j])
if j > 0:
self.neighbors.append(grid[i][j - 1])
if j < len(grid) - 1:
self.neighbors.append(grid[i][j + 1])
if diagonal:
# for diagonal neighbors
# down and right
if i < len(grid) - 1 and j < len(grid) - 1:
self.neighbors.append(grid[i + 1][j + 1])
# up and right
if i > 0 and j < len(grid) - 1:
self.neighbors.append(grid[i - 1][j + 1])
#down and left
if i < len(grid) - 1 and j > 0:
self.neighbors.append(grid[i + 1][j - 1])
#up and left
if i > 0 and j > 0:
self.neighbors.append(grid[i - 1][j - 1])
iterate through and set up the nodes
def make_grid(length):
main_grid = []
for i in range(length):
lst = []
for j in range(length):
node = Node(i, j)
# 30 % chance that the current node will be set as a wall
if random.randrange(1, 101) > 70 and i != 0 and j != 0: node.isWall = True
lst.append(node)
main_grid.append(lst)
for i in range(length):
for j in range(length):
main_grid[i][j].add_neighbors(main_grid, diagonal = True)
return main_grid
# Below is how the above function 'make_grid' is called
# making the grid
grid_len = 25
main_grid = make_grid(grid_len)
path = [] # to reconstruct the optimal path
Below is the HScore function I'm using
# HScore function
def getHScore(node, endNode):
return sqrt(abs(node.i - endNode.i)**2 + abs(node.j - endNode.j)**2)
Below is the actual Algorithm implementation
# A* PATHFINDING ALGORITHM
def aStar(start_node, end_node):
# node.f = node.g + node.h
# node.g = distance of current node from the starting node
# node.h = distance of current node from the end node
start_node.g = 0
start_node.h = getHScore(start_node, end_node)
start_node.f = start_node.g + start_node.h
open_set = [start_node]
closed_set = []
if start_node.isWall:
print("The start node is a wall")
return
while True:
if len(open_set) < 1:
print('No Solutions Found')
break
current_node = open_set[0]
for node in open_set:
if node.f < current_node.f:
current_node = node
current_node.isOpen = True
# print(f'current_node = {current_node.i, current_node.j}', end = " ")
if current_node == end_node:
temp = end_node
path.append(temp)
while temp.previous is not None:
path.append(temp.previous)
temp = temp.previous
print("DONE")
colorFinalPath(main_grid)
break
# current_node.isPath = True
current_node.isOpen = False
open_set.remove(current_node)
closed_set.append(current_node)
for neighbor in current_node.neighbors:
# assuming 1 as the distance btw two neighbouring points that aren't diagonally
# neighbors
# need to add 1.14 if neighbor is diagonal. add propery to node class to check if neighbor is diagonal
if neighbor in closed_set:
continue
tempG = current_node.g + getHScore(current_node, neighbor)
if neighbor not in open_set and not neighbor.isWall:
neighbor.g = tempG
open_set.append(neighbor)
neighbor.isOpen = True
neighbor.previous = current_node
if tempG >= neighbor.g:
continue # there is no better path
# neighbor was found in the open set, so we check if we can get to it in
# a better way as tempG is now less than neighbor.g
neighbor.previous = current_node
neighbor.g = tempG
neighbor.h = getHScore(neighbor, end_node)
neighbor.f = neighbor.g + neighbor.h
show_steps(main_grid, start_node, end_node)
Some Screen shots
In the third picture there is clearly a path between the starting Node (top left) and the end Node (bottom right) but it doesn't find any solutions.
Please tell me what's wrong with my implementation. Any help is appreciated

I see some issues in this piece of code:
tempG = current_node.g + getHScore(current_node, neighbor)
if neighbor not in open_set and not neighbor.isWall:
neighbor.g = tempG
open_set.append(neighbor)
neighbor.isOpen = True
neighbor.previous = current_node
if tempG >= neighbor.g:
continue # there is no better path
When the neighbor is a wall, you should skip it immediately. So add at the top:
if neighbor.isWall:
continue
This also means you can remove the wall check from the if you already had
The condition to check there is no better path will also be true when you just had set the g component for the first time, i.e. when the middle part was executed. This should not happen. So change that if to an elif:
if neighbor not in open_set:
# ... etc ...
elif tempG >= neighbor.g:
continue # there is no better path
Your make_grid code could mark the end node as a wall. You do not reject that situation, and then your code will continue and skip that as a neighbor to put in the open set. It is not clear from your image whether this happened, as you colored the ending node in blue.
Then, less of a problem, but you will sometimes call getHScore multiple times for the same node. Obviously, that function will return the same value for each of those calls. So you could improve on that. For example, by moving that line in an if condition:
if neighbor.h == 0:
neighbor.h = getHScore(neighbor, end_node)
I don't know whether it was intended but a diagonal step gets a cost of 2 (1²+1²), and it has no advantage over a 2-step walk to that same square. This is a tiny detail as you would visit those nodes first via the diagonal step and then ignore paths with the same cost.

Avoiding dead ends in snake game with moving food in python

I'm trying to do a snake game, where 2 snakes compete between each other. One snake simply follows the food, and avoids obstacles, the other, is the one, for which i'm writing the code, and is supposed to find the best way to get to the food. The food position, every bit of the map and the position of the other snake is known, and the position of the food changes, with every movement of the snakes.
If the map allows it, if there is no obstacle, the snake can traverse through the walls, to go to the other side of the map, like the map is a donut. The snake doesn't move diagonally, only vertically and horizontally, and it can't move backwards.
I'm using jump point search to find a way to the food, and it's working fine, although at 50fps some times the game slows down a bit.
The major problem i'm having, is finding a way to avoid dead ends. If the food gets in a dead end, i want to wait that it leaves the dead end, but what happens is that my snake, goes there, and then dies. Because i'm not avoiding dead ends, when my snake get's big enough, sometimes it crashes in its own body.
This is the code of the agent of my snake.
class AgentStudent(Snake, SearchDomain):
def __init__(self, body=[(0, 0)], direction=(1, 0), name="punkJD"):
super().__init__(body, direction, name=name)
self.count = 0;
#given the current state, and the next state, it returns a direction ( (1,0), (-1,0), (0,1), (0,-1) )
def dir(self, state, n_state):
if state[0] == 0 and n_state[0] == (self.mapsize[0] - 1):
return left
elif state[0] == (self.mapsize[0] - 1) and n_state[0] == 0:
return right
elif state[1] == 0 and n_state[1] == (self.mapsize[1] - 1):
return up
elif state[1] == (self.mapsize[1] - 1) and n_state == 0:
return down
return n_state[0] - state[0], n_state[1] - state[1]
#doesn't matter for the question
def update(self, points=None, mapsize=None, count=None, agent_time=None):
self.mapsize = mapsize
return None
#given current position and food position, it will create a class that will do the search. Seach code bellow
def search_food(self, pos, foodpos):
prob = SearchProblem(self, pos, foodpos, self.olddir)
my_tree = SearchTree(prob, self.mapsize, self.maze)
#doesn't matter, before i was using A*, but then i changed my whole search class
my_tree.strategy = 'A*'
return my_tree.search()
#given the current position and the direction the snake is faced it returns a list of all the possible directions the snake can take. If the current direction is still possible it will be put first in the list to be the first to be considered
def actions(self, pos, dir):
dirTemp = dir
invaliddir = [x for (x, y) in self.complement if y == dir]
validdir = [dir for dir in directions if not (dir in invaliddir)]
validdir = [dir for dir in validdir if
not (self.result(pos, dir) in self.maze.obstacles or self.result(pos, dir) in self.maze.playerpos)]
dirList = [dirTemp] if dirTemp in validdir else []
if dirList != []:
for a in range(len(validdir)):
if validdir[a] != dirTemp:
dirList.append(validdir[a])
return dirList
return validdir
#given the current position and the current direction, it returns the new position
def result(self, a, b):
n_pos = a[0] + b[0], a[1] + b[1]
if n_pos[0] == -1:
n_pos = (self.mapsize[0] - 1), a[1] + b[1]
if n_pos[1] == -1:
n_pos = a[0] + b[0], (self.mapsize[1] - 1)
if n_pos[0] == (self.mapsize[0]):
n_pos = 0, a[1] + b[1]
if n_pos[1] == (self.mapsize[1]):
n_pos = a[0] + b[0], 0
return n_pos
#given the current position and food position it returns the manhattan distance heuristic
def heuristic(self, position, foodpos):
distancex = min(abs(position[0] - foodpos[0]), self.mapsize[0] - abs(position[0] - foodpos[0]))
distancey = min(abs(position[1] - foodpos[1]), self.mapsize[1] - abs(position[1] - foodpos[1]))
return distancex + distancey
#this function is called by the main module of the game, to update the position of the snake
def updateDirection(self, maze):
# this is the brain of the snake player
self.olddir = self.direction
position = self.body[0]
self.maze = maze
# new direction can't be up if current direction is down...and so on
self.complement = [(up, down), (down, up), (right, left), (left, right)]
self.direction = self.search_food(position, self.maze.foodpos)
Bellow is the code to do the search.
I reused a file i had with some classes to do a tree search, and changed it to use jump point search. And for every jump point i find i expand a node in the tree.
class SearchDomain:
def __init__(self):
abstract
def actions(self, state):
abstract
def result(self, state, action):
abstract
def cost(self, state, action):
abstract
def heuristic(self, state, goal_state):
abstract
class SearchProblem:
def __init__(self, domain, initial, goal,dir):
self.domain = domain
self.initial = initial
self.goal = goal
self.dir = dir
def goal_test(self, state):
return state == self.goal
# class that defines the nodes in the tree. It has some attributes that are not used due to my old aproach.
class SearchNode:
def __init__(self,state,parent,heuristic,dir,cost=0,depth=0):
self.state = state
self.parent = parent
self.heuristic = heuristic
self.depth = depth
self.dir = dir
self.cost = cost
if parent!=None:
self.cost = cost + parent.cost
def __str__(self):
return "no(" + str(self.state) + "," + str(self.parent) + "," + str(self.heuristic) + ")"
def __repr__(self):
return str(self)
class SearchTree:
def __init__(self,problem, mapsize, maze, strategy='breadth'):
#attributes used to represent the map in a matrix
#represents obstacle
self.OBS = -1
#represents all the positions occupied by both snakes
self.PPOS = -2
#represents food position
self.FOODPOS = -3
#represents not explored
self.UNIN = -4
self.problem = problem
h = self.problem.domain.heuristic(self.problem.initial,self.problem.goal)
self.root = SearchNode(problem.initial, None,h,self.problem.dir)
self.open_nodes = [self.root]
self.strategy = strategy
self.blacklist = []
self.pqueue = FastPriorityQueue()
self.mapa = maze
#here i initialize the matrix to represent the map
self.field = []
for a in range(mapsize[0]):
self.field.append([])
for b in range(mapsize[1]):
self.field[a].append(self.UNIN)
for a,b in maze.obstacles:
self.field[a][b] = self.OBS
for a,b in maze.playerpos:
self.field[a][b] = self.PPOS
self.field[maze.foodpos[0]][maze.foodpos[1]] = self.FOODPOS
self.field[self.root.state[0]][self.root.state[1]] = self.UNIN
#function to add a jump point to the priority queue
def queue_jumppoint(self,node):
if node is not None:
self.pqueue.add_task(node, self.problem.domain.heuristic(node.state,self.problem.goal)+node.cost)
# given a node it returns the path until the root of the tree
def get_path(self,node):
if node.parent == None:
return [node]
path = self.get_path(node.parent)
path += [node]
return(path)
#Not used in this approach
def remove(self,node):
if node.parent != None:
a = self.problem.domain.actions(node.parent.state, node.dir)
self.blacklist+=node.state
if a == []:
self.remove(node.parent)
node = None
#Function that searches for the food
def search(self):
tempNode = self.root
self.queue_jumppoint(self.root)
count = 0
while not self.pqueue.empty():
node = self.pqueue.pop_task()
actions = self.problem.domain.actions(node.state,node.dir)
if count == 1:
tempNode = node
count+=1
#for every possible direction i call the explore function that finds a jump point in a given direction
for a in range(len(actions)):
print (a)
print (actions[a])
jumpPoint = self.explore(node,actions[a])
if jumpPoint != None:
newnode = SearchNode((jumpPoint[0],jumpPoint[1]),node,self.problem.domain.heuristic(node.state,self.problem.goal),actions[a],jumpPoint[2])
if newnode.state == self.problem.goal:
return self.get_path(newnode)[1].dir
self.queue_jumppoint(newnode)
dirTemp = tempNode.dir
return dirTemp
#Explores the given direction, starting in the position of the given node, to find a jump point
def explore(self,node,dir):
pos = node.state
cost = 0
while (self.problem.domain.result(node.state,dir)) != node.state:
pos = self.problem.domain.result(pos, dir)
cost += 1
#Marking a position as explored
if self.field[pos[0]][pos[1]] == self.UNIN or self.field[pos[0]][pos[1]] == self.PPOS:
self.field[pos[0]][pos[1]] = 20
elif pos[0] == self.problem.goal[0] and pos[1] == self.problem.goal[1]: # destination found
return pos[0],pos[1],cost
else:
return None
#if the snake is going up or down
if dir[0] == 0:
#if there is no obstacle/(or body of any snake) at the right but in the previous position there was, then this is a jump point
if (self.field [self.problem.domain.result(pos,(1,0))[0]] [pos[1]] != self.OBS and self.field [self.problem.domain.result(pos,(1,0))[0]] [self.problem.domain.result(pos,(1,-dir[1]))[1]] == self.OBS) or \
(self.field [self.problem.domain.result(pos,(1,0))[0]] [pos[1]] != self.PPOS and self.field [self.problem.domain.result(pos,(1,0))[0]] [self.problem.domain.result(pos,(1,-dir[1]))[1]] == self.PPOS):
return pos[0], pos[1],cost
#if there is no obstacle/(or body of any snake) at the left but in the previous position there was, then this is a jump point
if (self.field [self.problem.domain.result(pos,(-1,0))[0]] [pos[1]] != self.OBS and self.field [self.problem.domain.result(pos,(-1,0))[0]] [self.problem.domain.result(pos,(1,-dir[1]))[1]] == self.OBS) or \
(self.field [self.problem.domain.result(pos,(-1,0))[0]] [pos[1]] != self.PPOS and self.field [self.problem.domain.result(pos,(-1,0))[0]] [self.problem.domain.result(pos,(1,-dir[1]))[1]] == self.PPOS):
return pos[0], pos[1],cost
#if the snake is going right or left
elif dir[1] == 0:
#if there is no obstacle/(or body of any snake) at the upper part but in the previous position there was, then this is a jump point
if (self.field [pos[0]][self.problem.domain.result(pos,(1,1))[1]] != self.OBS and self.field [self.problem.domain.result(pos,(-dir[0],dir[1]))[0]] [self.problem.domain.result(pos,(1,1))[1]] == self.OBS) or \
(self.field [pos[0]][self.problem.domain.result(pos,(1,1))[1]] != self.PPOS and self.field [self.problem.domain.result(pos,(-dir[0],dir[1]))[0]] [self.problem.domain.result(pos,(1,1))[1]] == self.PPOS):
return pos[0], pos[1],cost
#if there is no obstacle/(or body of any snake) at the down part but in the previous position there was, then this is a jump point
if (self.field [pos[0]] [self.problem.domain.result(pos,(-1,-1))[1]] != self.OBS and self.field [self.problem.domain.result(pos,(-dir[0],dir[1]))[0]] [self.problem.domain.result(pos,(-1,-1))[1]] == self.OBS) or \
(self.field [pos[0]] [self.problem.domain.result(pos,(-1,-1))[1]] != self.PPOS and self.field [self.problem.domain.result(pos,(-dir[0],dir[1]))[0]] [self.problem.domain.result(pos,(-1,-1))[1]] == self.PPOS):
return pos[0], pos[1],cost
#if the food is aligned in some way with the snake head, then this is a jump point
if (pos[0] == self.mapa.foodpos[0] and node.state[0] != self.mapa.foodpos[0]) or \
(pos[1] == self.mapa.foodpos[1] and node.state[1] != self.mapa.foodpos[1]):
return pos[0], pos[1],cost
#if the food is in front of the head of the snake, right next to it, then this is a jump point
if self.field[self.problem.domain.result(pos,(dir[0],dir[1]))[0]][self.problem.domain.result(pos,(1,dir[1]))[1]] == self.FOODPOS:
return pos[0], pos[1],cost
##if an obstacle is in front of the head of the snake, right next to it, then this is a jump point
if self.field[self.problem.domain.result(pos,(dir[0],dir[1]))[0]][ self.problem.domain.result(pos,(1,dir[1]))[1]] == self.OBS:
return pos[0], pos[1],cost
return None
class FastPriorityQueue:
def __init__(self):
self.pq = [] # list of entries arranged in a heap
self.counter = 0 # unique sequence count
def add_task(self, task, priority=0):
self.counter+=1
entry = [priority, self.counter, task]
heapq.heappush(self.pq, entry)
def pop_task(self):
while self.pq:
priority, count, task = heapq.heappop(self.pq)
return task
raise KeyError('pop from an empty priority queue')
def empty(self):
return len(self.pq) == 0
This is my code. I would appreciate any help to be able to avoid dead ends.
I searched for similar problems but couldn't find any that helped me.

StackOverflow is not a coding service, so I'm not going to write your code for you, but I can most definitely tell you what steps you would need to take to solve your issue.
In your comments, you said it would be nice if you could check for dead ends before the game starts. A dead end can be classified as any point that has three or more orthogonally adjacent walls. I'm assuming you want every point leading up to a dead end that is inescapable. Here is how you would check:
Check every point starting from one corner and moving to the other, in either rows or columns, it doesn't matter. Once you reach a point that has three or more orthogonally adjacent walls, mark that point as a dead end, and go to 2.
Find the direction of the empty space next to this point (if any), and check every point in that direction. For each of those points: if it has two or more adjacent walls, mark it as a dead end. If it has only one wall, go to 3. If it has no walls, stop checking in this direction and continue with number 1.
In every direction that does not have a wall, repeat number 2.
Follow these steps until step 1 has checked every tile on the grid.
If you need a programming example, just ask for one in the comments. I didn't have time to make one, but I can make one later if needed. Also, if you need extra clarification, just ask!

Solving a maze using recursion in python

So, I have an assignment which asks me to solve a maze using recursion. I will post the assignment guidelines so you can see what I am talking about. The professor didn't explain recursion that much, he gave us examples of recursion, which I will post, but I was hoping someone might be able to give me a more in depth explanation of the recursion, and how I would apply this to solving a maze. I'm not asking for anyone to write the code, I'm just hoping some explanations would put me on the right path. Thank you to anyone who answers.
Here are the examples I have:
def foo():
print("Before")
bar()
print("After")
def bar():
print("During")
def factorial(n):
"""n!"""
product = 1
for i in range(n,0,-1):
product *= i
return product
def recFac(n):
"""n! = n * (n-1)!"""
if(n == 1):
return 1
return n * recFac(n-1)
def hello():
"""Stack overflow!"""
hello()
def fib(n):
"""f(n) = f(n-1) + f(n-2)
f(0) = 0
f(1) = 1"""
if n == 0 or n == 1: #base case
return n
return fib(n-1) + fib(n-2) #recursive case
def mult(a,b):
"""a*b = a + a + a + a ..."""
#base case
if (b == 1):
return a
#recursive case
prod = mult(a,b-1)
prod *= a
return prod
def exp(a,b):
"""a ** b = a* a * a * a * a *.... 'b times'"""
#base case
if (b==0):
return 1
if (b == 1):
return a
#recursive case
return exp(a,b-1)*a
def pallindrome(word):
"""Returns True if word is a pallindrome, False otherwise"""
#base case
if word == "" or len(word)==1:
return True
#recursive case
if word[0] == word[len(word)-1]:
word = word[1:len(word)-1]
return pallindrome(word)
else:
return False
Here are the guidelines:
You are going to create a maze crawler capable of solving any maze you give it with the power of recursion!
Question 1 - Loading the maze
Before you can solve a maze you will have to load it. For this assignment you will use a simple text format for the maze. You may use this sample maze or create your own.
Your objective for this question is to load any given maze file, and read it into a 2-dimensional list.
E.g.: loadMaze("somemaze.maze") should load the somemaze.maze file and create a list like the following...
[['#','#','#','#','#','#','#','#','#'],
['#','S','#',' ',' ',' ','#','E','#'],
['#',' ','#',' ','#',' ',' ',' ','#'],
['#',' ',' ',' ','#',' ','#',' ','#'],
['#', #','#','#','#','#','#','#','#']]
Note that the lists have been stripped of all '\r' and '\n' characters. In order to make the next question simpler you may make this list a global variable.
Next write a function that prints out the maze in a much nicer format:
E.g.,
####################################
#S# ## ######## # # # # #
# # # # # # #
# # ##### ## ###### # ####### # #
### # ## ## # # # #### #
# # # ####### # ### #E#
####################################
Test your code with different mazes before proceeding.
Question 2 - Preparing to solve the maze
Before you can solve the maze you need to find the starting point! Add a function to your code called findStart() that will search the maze (character-by-character) and return the x and y coordinate of the 'S' character. You may assume that at most one such character exists in the maze. If no 'S' is found in the maze return -1 as both the x and y coordinates.
Test your code with the 'S' in multiple locations (including no location) before proceeding.
Question 3 - Solving the maze!
Finally, you are ready to solve the maze recursively! Your solution should only require a single method: solve(y,x)
A single instance of the solve method should solve a single location in your maze. The parameters y and x are the current coordinates to be solved. There are a few things your solve method should accomplish. It should check if it is currently solving the location of the 'E'. In that case your solve method has completed successfully. Otherwise it should try to recursively solve the space to the right. Note, your method should only try to solve spaces, not walls ('#'). If that recursion doesn't lead to the end, then try down, then left, and up. If all that fails, your code should backtrack a step, and try another direction.
Lastly, while solving the maze, your code should leave indicators of its progress. If it is searching to the right, the current location should have a '>' in place of the empty space. If searching down put a 'v'. If searching left '<', and if searching up '^'. If your code has to backtrack remove the direction arrow, and set the location back to a ' '.
Once your maze is solved print out the maze again. You should a see step-by-step guide to walking the maze.
E.g.,
main("somemaze.maze")
#########
#S# #E#
# # # #
# # # #
#########
S is at (1,1)
#########
#S#>>v#E#
#v#^#>>^#
#>>^# # #
#########
Test your code with different different start and end locations, and optionally over a variety of mazes.
Here is the code I have so far:
But the code is not actually printing the track in the maze, and I'm not sure why.
def loadMaze():
readIt = open('Maze.txt', 'r')
readLines = readIt.readlines()
global mazeList
mazeList = [list(i.strip()) for i in readLines]
def showMaze():
for i in mazeList:
mazeprint = ''
for j in i:
mazeprint = mazeprint + j
print(mazeprint)
print('\n')
def solve(x,y, mazeList):
mazeList[x][y] = "o"
#Base case
if y > len(mazeList) or x > len(mazeList[y]):
return False
if mazeList[y][x] == "E":
return True
if mazeList[y][x] != " ":
return False
#marking
if solve(x+1,y) == True: #right
mazeList[x][y]= '>'
elif solve(x,y+1) == True: #down
mazeList[x][y]= 'v'
elif solve(x-1,y) == True: #left
mazeList[x][y]= '<'
elif solve(x,y-1) == True: #up
mazeList[x][y]= '^'
else:
mazeList[x][y]= ' '
return (mazeList[x][y]!= ' ')

(Dating myself, I actually did this problem in COBOL, in high-school.)
You can think of solving the maze as taking steps.
When you take a step, the same rules apply every time. Because the same rules apply every time, you can use the exact same set of instructions for each step. When you take a step, you just call the same routine again, changing the parameters to indicate the new step. That's recursion. You break the problem down by taking it one step at a time.
Note: Some recursion solutions break the problem in half, solving each half independent of the other, that works when the two solutions are actually independent. It doesn't work here because each step (solution) depends on the previous steps.
If you hit a dead end, you back out of the dead end, until you find a step where there are still viable squares to check.
Helpful Hint: You don't mark the correct path on the way to the exit, because you don't know that the step you're taking right now is part of the path to the exit. You mark the path on the way back, when you know that each step is indeed part of the path. You can do this because each step remembers which square it was in before it took the next step.
Instead, you put a mark in each square you've tried that only says: I've been here, no need to check this one again. Clean those up before you print the solution.

Here is my solution of CodeEval's The Labirynth challenge:
import sys
sys.setrecursionlimit(5000)
class Maze(object):
FLOOR = ' '
WALLS = '*'
PATH = '+'
def __init__(self):
self.cols = 0
self.rows = 0
self.maze = []
def walk_forward(self, current_k, r, c):
self.maze[r][c] = current_k
next_k = current_k + 1
# up
if r > 1:
up = self.maze[r - 1][c]
if up != self.WALLS:
if up == self.FLOOR or int(up) > current_k:
self.walk_forward(next_k, r - 1, c)
# down
if r < self.rows - 1:
down = self.maze[r + 1][c]
if down != self.WALLS:
if down == self.FLOOR or int(down) > current_k:
self.walk_forward(next_k, r + 1, c)
# left
if c > 1:
left = self.maze[r][c - 1]
if left != self.WALLS:
if left == self.FLOOR or int(left) > current_k:
self.walk_forward(next_k, r, c - 1)
# right
if c < self.cols - 1:
right = self.maze[r][c + 1]
if right != self.WALLS:
if right == self.FLOOR or int(right) > current_k:
self.walk_forward(next_k, r, c + 1)
def walk_backward(self, r, c):
current_k = self.maze[r][c]
if not isinstance(current_k, int):
return False
self.maze[r][c] = self.PATH
up = self.maze[r - 1][c] if r > 0 else None
down = self.maze[r + 1][c] if r < self.rows - 1 else None
left = self.maze[r][c - 1] if c > 1 else None
right = self.maze[r][c + 1] if c < self.cols else None
passed = False
if up and isinstance(up, int) and up == current_k - 1:
self.walk_backward(r - 1, c)
passed = True
if down and isinstance(down, int) and down == current_k - 1:
self.walk_backward(r + 1, c)
passed = True
if left and isinstance(left, int) and left == current_k - 1:
self.walk_backward(r, c - 1)
passed = True
if right and isinstance(right, int) and right == current_k - 1:
self.walk_backward(r, c + 1)
def cleanup(self, cleanup_path=False):
for r in range(0, self.rows):
for c in range(0, self.cols):
if isinstance(self.maze[r][c], int):
self.maze[r][c] = self.FLOOR
if cleanup_path and self.maze[r][c] == self.PATH:
self.maze[r][c] = self.FLOOR
def solve(self, start='up', show_path=True):
# finding start and finish points
upper = lower = None
for c in range(0, self.cols):
if self.maze[0][c] == self.FLOOR:
upper = (0, c)
break
for c in range(0, self.cols):
if self.maze[self.rows - 1][c] == self.FLOOR:
lower = (self.rows - 1, c)
break
if start == 'up':
start = upper
finish = lower
else:
start = lower
finish = upper
self.cleanup(cleanup_path=True)
self.walk_forward(1, start[0], start[1])
length = self.maze[finish[0]][finish[1]]
if not isinstance(length, int):
length = 0
if show_path:
self.walk_backward(finish[0], finish[1])
self.cleanup(cleanup_path=False)
else:
self.cleanup(cleanup_path=True)
return length
def save_to_file(self, filename):
with open(filename, 'w') as f:
f.writelines(str(self))
def load_from_file(self, filename):
self.maze = []
with open(filename, 'r') as f:
lines = f.readlines()
for line in lines:
row = []
for c in line.strip():
row.append(c)
self.maze.append(row)
self.rows = len(self.maze)
self.cols = len(self.maze[0]) if self.rows > 0 else 0
def get_maze(self):
return copy.copy(self.maze)
def __str__(self):
as_string = u''
for row in self.maze:
as_string += u''.join([str(s)[-1] for s in row]) + "\n"
return as_string
maze = Maze()
maze.load_from_file(sys.argv[1])
maze.solve(show_path=True)
print str(maze)

import os
class Maze_Crawler:
def __init__(self):
self.maze = []
def load_maze(self, path):
rows = []
with open(path, 'r') as f:
rows = f.readlines()
for i in range(len(rows)):
self.maze.append([])
for j in range(len(rows[i])-1):
self.maze[i].append(rows[i][j])
return self.maze
def get_start_coor(self):
for i in range(len(self.maze)):
for j in range(len(self.maze[i])):
if self.maze[i][j] == 'S':
return i, j
return -1, -1
def solve_maze(self, coor):
x, y = coor
if self.maze[x][y] == '#' or self.maze[x][y] == 'X':
return False
if self.maze[x][y] == 'E':
return True
if self.maze[x][y] != 'S':
self.maze[x][y] = 'X'
if self.solve_maze((x+1, y)):
if self.maze[x][y] != 'S':
self.maze[x][y] = 'v'
elif self.solve_maze((x-1, y)):
if self.maze[x][y] != 'S':
self.maze[x][y] = '^'
elif self.solve_maze((x, y+1)):
if self.maze[x][y] != 'S':
self.maze[x][y] = '>'
elif self.solve_maze((x, y-1)):
if self.maze[x][y] != 'S':
self.maze[x][y] = '<'
else:
return False
return True
def show_solution(self):
for i in range(len(self.maze)):
r = ''
for j in range(len(self.maze[i])):
if self.maze[i][j] == 'X':
r += ' '
else:
r += self.maze[i][j]
print(r)

Maze solving with python shows my answer. However, if you want to do the code yourself the steps are.
1. Start at the entrance.
2. Call the function solve(x,y) with the entrance co-ordinates
3. in solve, return false if the input point has already been handled or is a wall.
4. Mark the current point as handled (tag = 'o')
5. go to the right and call solve on that point. If it returns true, set tag to '>'
6 elif do the same for left and '<'
7 elif do the same for up and '^'
8 elif do the same for down and 'v'
9 else this is a false path, set tag = ' '
10 set the current maze point to tag
11 return (tag != ' ')
Alternatively leave step 9 out and make step 11
return(tag != 'o')
Then search through the maze and replace every 'o' with ' '
You can display the maze both ways so that it will show how you tried to solve it as well as the final answer. This has been used as a Solaris screensaver with the potential paths showing in one color and the actual path in a different color so that you can see it trying and then succeeding.

Recursion is actually a simple idea: to solve a problem, you shrink the problem by one step, then solve the reduced problem. This continues until you reach a "base problem" that you know how to solve completely. You return the base solution, then add to the solution returned at each step until you have the full solution.
So to solve n!, we remember n and solve for (n-1)!. The base case is 1!, for which we return 1; then at each return step we multiply by the remembered number (2 * 1! is 2, 3 * 2! is 6, 4 * 3! is 24, 5 * 4! is 120) until we multiply by n and have the full solution. This is actually a pretty pale and anemic sort of recursion; there is only one possible decision at each step. Known as "tail recursion", this is very easy to turn inside-out and convert to an iterative solution (start at 1 and multiply by each number up to n).
A more interesting sort of recursion is where you split the problem in half, solve each half, then combine the two half-solutions; for example quicksort sorts a list by picking one item, dividing the list into "everything smaller than item" and "everything bigger than item", quicksorting each half, then returning quicksorted(smaller) + item + quicksorted(larger). The base case is "when my list is only one item, it is sorted".
For the maze, we are going to split the problem four ways - all solutions possible if I go right, left, up, and down from my current location - with the special feature that only one of the recursive searches will actually find a solution. The base case is "I am standing on E", and a failure is "I am in a wall" or "I am on a space I have already visited".
Edit: for interest's sake, here is an OO solution (compatible with both Python 2.x and 3.x):
from collections import namedtuple
Dir = namedtuple("Dir", ["char", "dy", "dx"])
class Maze:
START = "S"
END = "E"
WALL = "#"
PATH = " "
OPEN = {PATH, END} # map locations you can move to (not WALL or already explored)
RIGHT = Dir(">", 0, 1)
DOWN = Dir("v", 1, 0)
LEFT = Dir("<", 0, -1)
UP = Dir("^", -1, 0)
DIRS = [RIGHT, DOWN, LEFT, UP]
#classmethod
def load_maze(cls, fname):
with open(fname) as inf:
lines = (line.rstrip("\r\n") for line in inf)
maze = [list(line) for line in lines]
return cls(maze)
def __init__(self, maze):
self.maze = maze
def __str__(self):
return "\n".join(''.join(line) for line in self.maze)
def find_start(self):
for y,line in enumerate(self.maze):
try:
x = line.index("S")
return y, x
except ValueError:
pass
# not found!
raise ValueError("Start location not found")
def solve(self, y, x):
if self.maze[y][x] == Maze.END:
# base case - endpoint has been found
return True
else:
# search recursively in each direction from here
for dir in Maze.DIRS:
ny, nx = y + dir.dy, x + dir.dx
if self.maze[ny][nx] in Maze.OPEN: # can I go this way?
if self.maze[y][x] != Maze.START: # don't overwrite Maze.START
self.maze[y][x] = dir.char # mark direction chosen
if self.solve(ny, nx): # recurse...
return True # solution found!
# no solution found from this location
if self.maze[y][x] != Maze.START: # don't overwrite Maze.START
self.maze[y][x] = Maze.PATH # clear failed search from map
return False
def main():
maze = Maze.load_maze("somemaze.txt")
print("Maze loaded:")
print(maze)
try:
sy, sx = maze.find_start()
print("solving...")
if maze.solve(sy, sx):
print(maze)
else:
print(" no solution found")
except ValueError:
print("No start point found.")
if __name__=="__main__":
main()
and when run produces:
Maze loaded:
####################################
#S# ## ######## # # # # #
# # # # # # #
# # ##### ## ###### # ####### # #
### # ## ## # # # #### #
# # # ####### # ### #E#
####################################
solving...
####################################
#S# ## ######## # #>>>>>v# >>v# #
#v#>>v# >>>v #^# >>>>^#>>v#
#>>^#v#####^##v######^# ####### #v#
### #v##>>>^##>>>>>v#^# # ####v#
# #>>>^# #######>>^# ### #E#
####################################
Note that the assignment as given has a few unPythonic elements:
it asks for camelCase function names rather than underscore_separated
it suggests using a global variable rather than passing data explicitly
it asks for find_start to return flag values on failure rather than raising an exception

Python A* implementation

I am currently working on my Python game, in ika, which uses python 2.5
I decided to use A* pathfinding for the AI. However, I find it too slow for my needs (3-4 enemies can lag the game, but I would like to supply up to 4-5 without problems). I know, that such complex search like A* is not mean to be scripted in python, but I am pretty sure, that my pathfinder is also implemented in the wrong way.
My question is: How can I speed up this algorithm?
I wrote my own binary heap, and there are some try: except: lines inside some functions. Those lines can create large overhead? Are there better methods maintaining the open list?
I supplied the algorithm with graphics interface, for testing purposes (when the pathfinder finishes searching, it will write the number of iterations and seconds it takes to find the path, inside the ika.txt file. Also, Pressing A will do a complete search, and S does that step by step.)
Graphical version:
http://data.hu/get/6084681/A_star.rar
Also, here is a pastebin version:
http://pastebin.com/9N8ybX5F
Here is the main code I use for pathfinding:
import ika
import time
class Node:
def __init__(self,x,y,parent=None,g=0,h=0):
self.x = x
self.y = y
self.parent = parent
self.g = g
self.h = h
def cost(self):
return self.g + self.h
def equal(self,node):
if self.x == node.x and self.y == node.y:
return True
else:
return False
class Emerald_Pathfinder:
def __init__(self):
pass
def setup(self,start,goal):
self.start = start
self.goal = goal
self.openlist = [None,start] # Implemented as binary heap
self.closedlist = {} # Implemented as hash
self.onopenlist = {} # Hash, for searching the openlist
self.found = False
self.current = None
self.iterations = 0
def lowest_cost(self):
pass
def add_nodes(self,current):
nodes = []
x = current.x
y = current.y
self.add_node(x+1,y,current,10,nodes)
self.add_node(x-1,y,current,10,nodes)
self.add_node(x,y+1,current,10,nodes)
self.add_node(x,y-1,current,10,nodes)
# Dont cut across corners
up = map.is_obstacle((x,y-1),x,y-1)
down = map.is_obstacle((x,y+1),x,y+1)
left = map.is_obstacle((x-1,y),x-1,y)
right = map.is_obstacle((x+1,y),x+1,y)
if right == False and down == False:
self.add_node(x+1,y+1,current,14,nodes)
if left == False and up == False:
self.add_node(x-1,y-1,current,14,nodes)
if right == False and up == False:
self.add_node(x+1,y-1,current,14,nodes)
if left == False and down == False:
self.add_node(x-1,y+1,current,14,nodes)
return nodes
def heuristic(self,x1,y1,x2,y2):
return (abs(x1-x2)+abs(y1-y2))*10
def add_node(self,x,y,parent,cost,list):
# If not obstructed
if map.is_obstacle((x,y),x,y) == False:
g = parent.g + cost
h = self.heuristic(x,y,self.goal.x,self.goal.y)
node = Node(x,y,parent,g,h)
list.append(node)
def ignore(self,node,current):
# If its on the closed list, or open list, ignore
try:
if self.closedlist[(node.x,node.y)] == True:
return True
except:
pass
# If the node is on the openlist, do the following
try:
# If its on the open list
if self.onopenlist[(node.x,node.y)] != None:
# Get the id number of the item on the real open list
index = self.openlist.index(self.onopenlist[(node.x,node.y)])
# If one of the coordinates equal, its not diagonal.
if node.x == current.x or node.y == current.y:
cost = 10
else:
cost = 14
# Check, is this items G cost is higher, than the current G + cost
if self.openlist[index].g > (current.g + cost):
# If so, then, make the list items parent, the current node.
self.openlist[index].g = current.g + cost
self.openlist[index].parent = current
# Now resort the binary heap, in the right order.
self.resort_binary_heap(index)
# And ignore the node
return True
except:
pass
return False
def resort_binary_heap(self,index):
m = index
while m > 1:
if self.openlist[m/2].cost() > self.openlist[m].cost():
temp = self.openlist[m/2]
self.openlist[m/2] = self.openlist[m]
self.openlist[m] = temp
m = m / 2
else:
break
def heap_add(self,node):
self.openlist.append(node)
# Add item to the onopenlist.
self.onopenlist[(node.x,node.y)] = node
m = len(self.openlist)-1
while m > 1:
if self.openlist[m/2].cost() > self.openlist[m].cost():
temp = self.openlist[m/2]
self.openlist[m/2] = self.openlist[m]
self.openlist[m] = temp
m = m / 2
else:
break
def heap_remove(self):
if len(self.openlist) == 1:
return
first = self.openlist[1]
# Remove the first item from the onopenlist
self.onopenlist[(self.openlist[1].x,self.openlist[1].y)] = None
last = self.openlist.pop(len(self.openlist)-1)
if len(self.openlist) == 1:
return last
else:
self.openlist[1] = last
v = 1
while True:
u = v
# If there is two children
if (2*u)+1 < len(self.openlist):
if self.openlist[2*u].cost() <= self.openlist[u].cost():
v = 2*u
if self.openlist[(2*u)+1].cost() <= self.openlist[v].cost():
v = (2*u)+1
# If there is only one children
elif 2*u < len(self.openlist):
if self.openlist[2*u].cost() <= self.openlist[u].cost():
v = 2*u
# If at least one child is smaller, than parent, swap them
if u != v:
temp = self.openlist[u]
self.openlist[u] = self.openlist[v]
self.openlist[v] = temp
else:
break
return first
def iterate(self):
# If the open list is empty, exit the game
if len(self.openlist) == 1:
ika.Exit("no path found")
# Expand iteration by one
self.iterations += 1
# Make the current node the lowest cost
self.current = self.heap_remove()
# Add it to the closed list
self.closedlist[(self.current.x,self.current.y)] = True
# Are we there yet?
if self.current.equal(self.goal) == True:
# Target reached
self.goal = self.current
self.found = True
print self.iterations
else:
# Add the adjacent nodes, and check them
nodes_around = self.add_nodes(self.current)
for na in nodes_around:
if self.ignore(na,self.current) == False:
self.heap_add(na)
def iterateloop(self):
time1 = time.clock()
while 1:
# If the open list is empty, exit the game
if len(self.openlist) == 1:
ika.Exit("no path found")
# Expand iteration by one
self.iterations += 1
# Make the current node the lowest cost
self.current = self.heap_remove()
# Add it to the closed list
self.closedlist[(self.current.x,self.current.y)] = True
# Are we there yet?
if self.current.equal(self.goal) == True:
# Target reached
self.goal = self.current
self.found = True
print "Number of iterations"
print self.iterations
break
else:
# Add the adjacent nodes, and check them
nodes_around = self.add_nodes(self.current)
for na in nodes_around:
if self.ignore(na,self.current) == False:
self.heap_add(na)
time2 = time.clock()
time3 = time2-time1
print "Seconds to find path:"
print time3
class Map:
def __init__(self):
self.map_size_x = 20
self.map_size_y = 15
self.obstructed = {} # Library, containing x,y couples
self.start = [2*40,3*40]
self.unit = [16*40,8*40]
def is_obstacle(self,couple,x,y):
if (x >= self.map_size_x or x < 0) or (y >= self.map_size_y or y < 0):
return True
try:
if self.obstructed[(couple)] != None:
return True
except:
return False
def render_screen():
# Draw the Character
ika.Video.DrawRect(map.start[0],map.start[1],map.start[0]+40,map.start[1]+40,ika.RGB(40,200,10),1)
# Draw walls
for x in range(0,map.map_size_x):
for y in range(0,map.map_size_y):
if map.is_obstacle((x,y),x,y) == True:
ika.Video.DrawRect(x*40,y*40,(x*40)+40,(y*40)+40,ika.RGB(168,44,0),1)
# Draw openlist items
for node in path.openlist:
if node == None:
continue
x = node.x
y = node.y
ika.Video.DrawRect(x*40,y*40,(x*40)+40,(y*40)+40,ika.RGB(100,100,100,50),1)
# Draw closedlist items
for x in range(0,map.map_size_x):
for y in range(0,map.map_size_y):
try:
if path.closedlist[(x,y)] == True:
ika.Video.DrawRect(x*40,y*40,(x*40)+20,(y*40)+20,ika.RGB(0,0,255))
except:
pass
# Draw the current square
try:
ika.Video.DrawRect(path.current.x*40,path.current.y*40,(path.current.x*40)+40,(path.current.y*40)+40,ika.RGB(128,128,128), 1)
except:
pass
ika.Video.DrawRect(mouse_x.Position(),mouse_y.Position(),mouse_x.Position()+8,mouse_y.Position()+8,ika.RGB(128,128,128), 1)
# Draw the path, if reached
if path.found == True:
node = path.goal
while node.parent:
ika.Video.DrawRect(node.x*40,node.y*40,(node.x*40)+40,(node.y*40)+40,ika.RGB(40,200,200),1)
node = node.parent
# Draw the Target
ika.Video.DrawRect(map.unit[0],map.unit[1],map.unit[0]+40,map.unit[1]+40,ika.RGB(128,40,200),1)
def mainloop():
while 1:
render_screen()
if mouse_middle.Pressed():
# Iterate pathfinder
if path.found == False:
path.iterateloop()
elif mouse_right.Pressed():
# Iterate pathfinder by one
if path.found == False:
path.iterate()
elif ika.Input.keyboard["A"].Pressed():
# Iterate pathfinder
if path.found == False:
path.iterateloop()
elif ika.Input.keyboard["S"].Pressed():
# Iterate pathfinder by one
if path.found == False:
path.iterate()
elif mouse_left.Position():
# Add a square to the map, to be obstructed
if path.iterations == 0:
x = mouse_x.Position()
y = mouse_y.Position()
map.obstructed[(int(x/40),int(y/40))] = True
# Mouse preview
x = mouse_x.Position()
y = mouse_y.Position()
mx = int(x/40)*40
my = int(y/40)*40
ika.Video.DrawRect(mx,my,mx+40,my+40,ika.RGB(150,150,150,70),1)
ika.Video.ShowPage()
ika.Input.Update()
map = Map()
path = Emerald_Pathfinder()
path.setup(Node(map.start[0]/40,map.start[1]/40),Node(map.unit[0]/40,map.unit[1]/40))
mouse_middle = ika.Input.mouse.middle
mouse_right = ika.Input.mouse.right
mouse_left = ika.Input.mouse.left
mouse_x = ika.Input.mouse.x
mouse_y = ika.Input.mouse.y
# Initialize loop
mainloop()
I appreciate any help!
(sorry for any spelling mistakes, English is not my native language)

I think a proper implementation in python will be fast enough for your purposes. But the boost library has an astar implementation and python bindings. https://github.com/erwinvaneijk/bgl-python