Develop Reference

foobar failing test case - a* with breakable wall

Working on the google foo-bar challenges and I am stuck on one failing test case (which is a hidden one - so I can't directly see what the problem is)
basically the test is an implementation of a maze solver with a single breakable wall.
I'm doing a modified a* search - with a Boolean flag for paths that contain a broken wall, so that when it reaches the second wall (along a path with a broken wall) it skips it for that path.
I've gone over this code and I can't seem to see the error (if there even is one - at this point I'm almost convinced the test case is somehow wrong)
the parameters:
given a grid of height x,y filled with zero or 1, representing spaces and walls respectively: find the shortest path if you can break exactly 1 wall (if needed)
an example:
[
[0,1,0,0,0,1,0,0,0,0,0]
[0,0,0,1,0,0,1,0,1,1,0]
[1,1,1,1,1,0,1,0,1,0,1]
[0,0,0,0,0,0,0,0,1,1,0]
]
22 is the shortest path breaking 1 wall.
I want a pointer in the right direction: I feel like whatever I'm missing is trivial.
below is the code.
from math import sqrt, ceil
cardinal_moves=[(0,1), (0,-1), (1,0), (-1,0)]
class Node:
def __init__(self, pos ,parent =None):
self.parent = parent
self.pos = pos
self.g = 0
self.h = 0
self.f = 0
self.wall_broken = False
def __eq__(self, other):
return ((self.pos == other.pos) and (self.wall_broken == other.wall_broken))
def update_heuristic(self,parent,end):
self.g = parent.g + 1
self.h = ceil(sqrt((end.pos[0] - self.pos[0])**2 + (end.pos[1] - self.pos[1])**2))
self.f = self.g + self.h
def not_inside_maze(pos,maze):
return pos[0] < 0 or pos[0] >= len(maze) or pos[1] < 0 or pos[1] >= len(maze[0])
def astar_with_1_breakable_wall(maze):
start_pos = (0,0)
end_pos = (len(maze)-1, len(maze[0])-1)
start = Node(start_pos)
end = Node(end_pos)
not_visted = []
visited = []
not_visted.append(start)
while not_visted:
current_node = not_visted[0]
for node in not_visted:
if current_node.f > node.f:
current_node = node
not_visted.remove(current_node)
visited.append(current_node)
if current_node.pos == end.pos:
temp = current_node
path = []
while temp:
path.append(temp.pos)
temp = temp.parent
return path[::-1]
children = []
for position in cardinal_moves:
new_pos = (current_node.pos[0] + position[0], current_node.pos[1] + position[1])
if not_inside_maze(new_pos,maze):
continue
if maze[new_pos[0]][new_pos[1]] == 1:
if current_node.wall_broken:
continue
else:
check = Node(new_pos,current_node)
check.wall_broken = True
already_visited = False
for node in visited:
if check == node:
already_visited = True
break
if not already_visited:
children.append(check)
continue
already_visited = False
check = Node(new_pos,current_node)
check.wall_broken = current_node.wall_broken
for node in visited:
if check == node:
already_visited = True
break
if already_visited:
continue
children.append(check)
for child in children:
child.update_heuristic(current_node,end)
for open_node in not_visted:
if open_node == child:
if open_node.g > child.g:
idx = not_visted.index(open_node)
not_visted[idx] = child
continue
else:
continue
not_visted.append(child)
def shortest_path(maze):
if (astar_with_1_breakable_wall(maze)):
return len(astar_with_1_breakable_wall(maze))
else:
return -1
but every check I make on my machine says this is correct:
#imports added
import sys
#then added this below the maze solver
test = [
[0,1,0,0,0,1,0,0,0,0,0],
[0,0,0,1,0,0,1,0,1,1,0],
[1,1,1,1,1,0,1,0,1,0,1],
[0,0,0,0,0,0,0,0,1,1,0],
]
test2=[
[0,1,0,0,0,0,1,0,0,0,0],
[0,1,0,1,1,0,1,0,1,0,0],
[0,0,0,1,0,0,1,0,1,0,0],
[1,1,1,1,0,1,1,0,1,0,1],
[0,0,0,0,0,0,0,0,1,0,1],
]
print("first test")
print(shortest_path(test))
print("second test")
print(shortest_path(test))
#both of these tests give the correct result
def generate_matrix(h,w,n):
sequence = "{0:b}".format(n).zfill(h*w)
maze =[]
if(sequence[0]=="1" or sequence[len(sequence)-1]=="1"):
return -1
i=0
for y in range(h):
maze.append([])
for x in range(w):
maze[y].append(int(sequence[i]))
i=i+1
return(maze)
for i in range(20):
for j in range(20):
if(i < 6 or j <6):
continue
shortest=j+i
for k in range(2**(i*j)):
if not k%2==0:
continue
if k> 2**(i*j-1):
continue
maze = generate_matrix(i,j,k)
if(not maze == -1):
path = astar_with_1_breakable_wall(maze)
res = shortest_path(maze)
if(res > shortest):
print("\n",res, "shortest was ", shortest)
for index,line in enumerate(maze):
for indx,cell in enumerate(line):
if ((index,indx) in path):
print("\033[94m"+str(cell),end="")
else:
print("\033[92m"+str(cell),end="")
print("\n")
for point in path:
print(point ," ", end="")
print("\n\n")
else:
sys.stdout.write("\r")
the above code makes every matrix possibility, and prints (with highlighting the path) the matrix if the path is longer than the base case.
every result is as I expect - returning the correct shortest path... I have not found the issue with why only the 3rd test case fails...

the answer was something specific to python 2.7.13 apparently - it took a bit of debugging to figure it out, but the not_visited.remove(current_node) was the thing that was failing in certain cases.

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.

How can I make my code shorter for a simple maze game?

Good Day. I'm a beginner in coding, and basically I made a simple maze game using python.
it's 3x3 so I made 9 dictionary variables that detailed the possible movement for each tile (North, south, east, west.)
my code consists of nested if-elif as well as a main while loop.
if level == 1: #level of difficulty.
i = 0
maplist1 = {'directions1':'south'}
maplist2 = {'directions1':'south','directions2':'east'}
maplist3 = {'directions1':'west'}
maplist4 = {'directions1':'north','directions2':'east','directions3':'south'}
....
....
current_position = 1 #local variable within the first if statement
Here's a snippet of the code block that would be repeated, with little to no variations (literally repeating) to it.
while i == 0: #to continuously loop the program
if current_position == 1:
directions = maplist1['directions1']
direction = input(f"What direction do you want to go: {directions} ").title() #choose the possible directions to go to.
if direction == "S" or direction == "South":
current_position = 4
print(f"You are in cell {current_position}.")
else:
print("Cannot go in that direction.")
if current_position == 2:
directions = maplist2['directions1']
directions2 = maplist2['directions2']
direction = input(f"What direction do you want to go: {directions} {directions2} ").title()
if direction == "S" or direction == "South":
current_position = 5
print(f"You are in cell {current_position}.")
elif direction == "E" or direction == "East":
current_position = 3
print(f"You are in cell {current_position}.")
else:
print("Cannot go in that direction.")
if current_position == 3:
directions = maplist3['directions1']
direction = input(f"What direction do you want to go: {directions} ").title()
if direction == "W" or direction == "West":
current_position = 2
print(f"You are in cell {current_position}.")
else:
print("Cannot go in that direction.")
if current_position == 4:
directions = maplist4['directions1']
directions2 = maplist4['directions2']
directions3 = maplist4['directions3']
direction = input(f"What direction do you want to go: {directions} {directions2} {directions3} ").title()
if direction == "N" or direction == "North":
current_position = 1
print(f"You are in cell {current_position}.")
elif direction == "S" or direction == "South":
current_position = 7
print(f"You are in cell {current_position}.")
elif direction == "E" or direction == "East":
current_position = 5
print(f"You are in cell {current_position}.")
else:
print("Cannot go in that direction.")
.......
......
.......
if current_position == 9:
print("Congratulations, you finished the game.")
i += 1 #to stop the loop
my question is how to make this simpler, and more compact?
Current location must be remembered every movement
I know how to use def but quite unsure on how to implement it in my maze game.
There are 3 levels to this game, basically 3x3, 4x4, and 5x5. that's the reason why I am asking for any ideas on how to shorten/compact the code.
I don't need you to give me your code, but some guidance would be great on how to proceed because i am feeling lost right now.

Try creating a maze data structure. One typical idea is to create a 2D array of cells, each with a north/west/south/east wall.
Let's create a class, just for better understanding. I want you to ignore everything except the can_move() method.
class Cell:
def __init__(self, walls):
self.walls = walls
def __str__(self):
"""Allows us to call print() on a Cell!"""
return '\n'.join(self.draw())
def _draw_wall(self, wall, s):
return s if wall in self.walls else ' ' * len(s)
def draw(self, inner=' '):
"""Draw top, mid, and bottom parts of the cell."""
n = self._draw_wall('N', '___')
w = self._draw_wall('W', '|')
e = self._draw_wall('E', '|')
s = self._draw_wall('S', '___')
top = ' ' + n + ' '
mid = w + inner + e
bot = w + s + e
return top, mid, bot
def can_move(self, direction):
return direction not in self.walls
Let's make some cells and see what they look like:
>>> Cell('NWSE')
___
| |
|___|
>>> Cell('NWS')
___
|
|___
Now, a maze is made out of a 2D list of cells. Here's a small maze:
maze = Maze(width=3, height=3, rows=[
[Cell('NWE'), Cell('NW'), Cell('NE')],
[Cell('WE'), Cell('WE'), Cell('WE')],
[Cell('WS'), Cell('SE'), Cell('WSE')]
])
Which looks like this:
>>> maze
___ ___ ___
| x || |
| || |
| || || |
| || || |
| || |
|___ ___||___|
But wait! We haven't defined what Maze is. Once again, ignore all the drawing related stuff and focus on move_direction().
class Maze:
def __init__(self, width, height, rows):
self.width = width
self.height = height
self.rows = rows
self.position = (0, 0)
def __str__(self):
return '\n'.join(self.draw_row(i) for i, _ in enumerate(self.rows))
def _inner(self, i, j):
return ' x ' if (i, j) == self.position else ' '
def draw_row(self, i):
triples = [
cell.draw(self._inner(i, j)) for j, cell in enumerate(self.rows[i])
]
return '\n'.join([
''.join(t for t, _, _ in triples),
''.join(m for _, m, _ in triples),
''.join(b for _, _, b in triples)])
def cell_at_position(self, i, j):
return self.rows[i][j]
def move_direction(self, direction):
curr_cell = self.cell_at_position(*self.position)
if not curr_cell.can_move(direction):
print("Can't go in that direction!")
return
deltas = {'N': (-1, 0), 'W': (0, -1), 'S': (1, 0), 'E': (0, 1)}
y, x = self.position
dy, dx = deltas[direction]
self.position = y + dy, x + dx
To use this, just create a simple small loop:
while True:
print(maze)
direction = input('What direction? ')
maze.move_direction(direction)
And watch the magic happen!
You can try it here.

Instead of thinking the maze in 1D i.e the current_position(1,2,3..,9), you should think and code it like a 2D matrix space((0,0),(0,1)).
And Instead of hard coding the base maze to 3, we should be thinking in terms of n x n maze, as now that you have implemented the logic for a 3 x 3, now you should be able to expand it to n x n.
And you handled the possible directions using your maps instead, we should write the logic for boundary conditions, like a method to check the directions possible on each step,
def possible_path(current_x, current_y, matrix_size=3):
# this returns like directions possible like current_x + 1 < matrix_size
# then the east direction is possible.
pass
Direction handling should be handled somewhat like this,
def position_handler(row, col, direction):
# can write the logic for mapping the direction to the movement in X and Y,
# if direction == 'NORTH'
# or can get the config from a map like,
# move = movement.get(direction)
# say NORTH config is =>movement= {'NORTH': {'x': 0, 'y':1}}
# and now move accordingly i.e
# row += move.get('x')
# col += move.get('y')
pass

Exploit rules instead of hardcoding anything.
If you have a 3x3 maze, your (x,y) indexes are somthing like
0,0 1,0 2,0
0,1 1,1 2,1
0,2 1,2 2,2
You can store this as list of lists:
field = [ ["one","two","three"],["four","five","six"],["seven","eight","nine"] ]
pos = (0,0)
p = field[pos[0]][pos[1]] # would be "one"
# rules for movement based on actual position:
move_right = pos[0] < (3-1)
move_left = pos[0] > 0
move_up = pos[1] > 0
move_down = pos[1] < (3-1)
You can use functions to code this:
# p is a coordinate (0,0) etc tuple of x and y
# as functions
def can_move_right(p, dim=3):
return p[0] < dim-1
def can_move_left(p,dim=3):
return p[0] > 0
def can_move_up(p, dim=3):
return p[1] > 0
def can_move_down(p,dim=3):
return p[1] < dim-1
def move_up(p):
if can_move_up(p):
return (p[0],p[1]-1)
# implicitly returns None if not possible
def move_down(p):
if can_move_down(p):
return (p[0],p[1]+1)
# etc
new_pos = move_up( (1,1) ) # => (1,0)
new_pos = move_down( (1,1) ) # => (1,2)
new_pos = move_up( (0,0) ) # None - need to handle it
if not new_pos:
print("Wrong move")
To adjust for n x n you just need to give other dim - the rules are the same.

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