I want to learn how to implement a* path finding in Python3 - python-3.x

I found this code online and would really like for someone to explain it simply. I understand most of the initialisation of nodes, but don't understand as much once I get further down. If someone can explain the code near line by line, I would be grateful. The areas I find the most confusing are the calculation sections.
Thanks for any/all responses
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 astar(maze, start, end):
"""Returns a list of tuples as a path from the given start to the given end in the given maze"""
# Create start and end node
start_node = Node(None, start)
start_node.g = start_node.h = start_node.f = 0
end_node = Node(None, end)
end_node.g = end_node.h = end_node.f = 0
# Initialize both open and closed list
open_list = []
closed_list = []
# Add the start node
open_list.append(start_node)
# Loop until you find the end
while len(open_list) > 0:
# Get the current node
current_node = open_list[0]
current_index = 0
for index, item in enumerate(open_list):
if item.f < current_node.f:
current_node = item
current_index = index
# Pop current off open list, add to closed list
open_list.pop(current_index)
closed_list.append(current_node)
# Found the goal
if current_node == end_node:
path = []
current = current_node
while current is not None:
path.append(current.position)
current = current.parent
return path[::-1] # Return reversed path
# Generate children
children = []
for new_position in [(0, -1), (0, 1), (-1, 0), (1, 0), (-1, -1), (-1, 1), (1, -1), (1, 1)]: # Adjacent squares
# Get node position
node_position = (current_node.position[0] + new_position[0], current_node.position[1] + new_position[1])
# Make sure within range
if node_position[0] > (len(maze) - 1) or node_position[0] < 0 or node_position[1] > (len(maze[len(maze)-1]) -1) or node_position[1] < 0:
continue
# Make sure walkable terrain
if maze[node_position[0]][node_position[1]] != 0:
continue
# Create new node
new_node = Node(current_node, node_position)
# Append
children.append(new_node)
# Loop through children
for child in children:
# Child is on the closed list
for closed_child in closed_list:
if child == closed_child:
continue
# Create the f, g, and h values
child.g = current_node.g + 1
child.h = ((child.position[0] - end_node.position[0]) ** 2) + ((child.position[1] - end_node.position[1]) ** 2)
child.f = child.g + child.h
# Child is already in the open list
for open_node in open_list:
if child == open_node and child.g > open_node.g:
continue
# Add the child to the open list
open_list.append(child)
def main():
maze = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
start = (0, 0)
end = (7, 6)
path = astar(maze, start, end)
print(path)
if __name__ == '__main__':
main()

I suggest you first understand the A* algorithm itself, here is a good video i found, check here. After understanding this, try to code it yourself in whatever language you want and then if you can't understand something ask about that stuff on SO.

Related

Global variables and multiprocessing pools

In the python multiprocessing library, do processes spawned via Pool only have access to global variables bound at the time of Pool construction?
Why is this?
This appears to be the case, based on this experiment.
This code:
from multiprocessing import Pool
x = 0
class MyClass:
def get_x(self, i):
global x
return x
def foo():
global x
p = Pool(5)
for i in range(3):
x = i
c = MyClass()
print(list(p.imap(c.get_x, range(10*i, 10*i+10))))
foo()
produces the output
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
Whereas moving the Pool construction inside the loop, i.e.
from multiprocessing import Pool
x = 0
class MyClass:
def get_x(self, i):
global x
return x
def foo():
global x
for i in range(3):
p = Pool(5)
x = i
c = MyClass()
print(list(p.imap(c.get_x, range(10*i, 10*i+10))))
foo()
produces the output
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
(Note that my motivation for wanting to do this is that x is a large pandas DataFrame in my real application, that needs to be read, but not modified, by several processes.)
My two questions are:
Is it merely impossible to rebind x once the processes are created?
Or do the processes each make a copy of x, and modification is impossible?

Printing Paths in Dijkstra's shortest path Python3

I'm having issues with the code below where it's not displaying any paths for each vertex. Vertex and Distance from the source is displaying correctly, but for the path section of the output is blank. What am I missing? I would love to get some feedbacks or suggestions or even answer to this nightmare. I just can't seem to figure out what is causing the paths from displaying anything. I'm still fairly new to Python and I could really use some help!
Output of my current code
class Graph:
def minDistance(self, dist, queue):
# Initialize min value and min_index as -1
minimum = float("Inf")
min_index = -1
# from the dist array,pick one which
# has min value and is till in queue
for i in range(len(dist)):
if dist[i] < minimum and i in queue:
minimum = dist[i]
min_index = i
return min_index
def printPath(self, parent, j):
# Base Case : If j is source
if parent[j] == -1:
print()
j,
return
self.printPath(parent, parent[j])
print()
j,
def printSolution(self, dist, parent):
src = 0
print("Vertex \t\tDistance from Source\tPath")
for i in range(1, len(dist)):
print(("\n%d --> %d \t\t%d \t\t\t\t\t" % (src, i, dist[i])), end=' ')
self.printPath(parent, i)
def dijkstra(self, graph, src):
row = len(graph)
col = len(graph[0])
dist = [float("Inf")] * row
parent = [-1] * row
dist[src] = 0
queue = []
for i in range(row):
queue.append(i)
while queue:
u = self.minDistance(dist, queue)
queue.remove(u)
for i in range(col):
if graph[u][i] and i in queue:
if dist[u] + graph[u][i] < dist[i]:
dist[i] = dist[u] + graph[u][i]
parent[i] = u
self.printSolution(dist, parent)
g = Graph()
graph = [[0, 4, 0, 0, 0, 0, 0, 8, 0],
[4, 0, 8, 0, 0, 0, 0, 11, 0],
[0, 8, 0, 7, 0, 4, 0, 0, 2],
[0, 0, 7, 0, 9, 14, 0, 0, 0],
[0, 0, 0, 9, 0, 10, 0, 0, 0],
[0, 0, 4, 14, 10, 0, 2, 0, 0],
[0, 0, 0, 0, 0, 2, 0, 1, 6],
[8, 11, 0, 0, 0, 0, 1, 0, 7],
[0, 0, 2, 0, 0, 0, 6, 7, 0]
]
g.dijkstra(graph, 0)
The basic issue with your code is that the printPath method is only outputting blanks and secondly, the printSolution method is not including the results of the printPath method. To fix these problems:
1 rewrite the printPath function as below:
def printPath(self, parent, j, l):
# Returns a list from destination to source
# Base case when j = source
if parent[j] == -1:
return l
else:
l.append(j)
return self.printPath(parent, parent[j], l)
Then rewrite the printSolution method as follows:
Note: I used the f format structure to simplify my print statement
def printSolution(self, dist, parent):
src = 0
print("Vertex \t\tDistance from Source\tPath")
for i in range(1, len(dist)):
st = f"\n{src} --> {i}\t\t{dist[i]}\t\t\t\t\t{self.printPath(parent,i,[])[::-1]}"
#print(("\n%d --> %d \t\t%d \t\t\t\t\t" % (src, i, dist[i])), end=' ')
#self.printPath(parent, i)
print (st)

How to check the distance between a specific element an index? Python3

A = [1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0]
B = 9
I want to find out the distances between the index[9] (B) and each of it's closest's 1's. For example, If we look at list A, we see that index 9 is this:
A = [1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0]
^
I would like to figure out the distances between B(index 9) and it's nearest 1's. For example, the nearest 1's in this case would be this:
A = [1, 0, 0, 0, 1, 0, 0, 1, 0, B, 0, 0, 1, 0]
^ ^
So in this case the output would be:
>> [2, 3] ##Since the distance between 1, 0, B is 2 and the distance between B, 0, 0, 1 is 3.
So far I've come up with the following code:
def checkDistance(A, B):
for i in A:
if i == 1:
#Check if it is the closest to B
#Check distance
Sadly I'm a beginner in python and I am struggling to finish. Any help would be much appreciated :)
def distance(lst,index):
c=[i for i,j in enumerate(lst) if j==1]
for k,l in zip(c[:-1],c[1:]):
if k < index < l:
return [index-k, l-index]
a = [1, 0, 0, 0, 1, 0, 0, 1, 0, B, 0, 0, 1, 0]
b = 9
distance(a, b)
Out: [2, 3]
You could use the following function. In this case, to make the function more abstract, you needn't force the value for the comparison to be 1 in the function.
In the function below, you do a for loop starting at the position you specified (in Python indexes start at 0, not at 1) and finishing when the list finishes.
The if statement compares element with the value of the list at a given position i
def checkDistance(lst,index,element):
counter = 0
results = []
for i in range(index,len(lst)):
if lst[i] == element:
print("Detected element at distance: " + str(counter))
results.append(counter)
counter += 1
return results

Finding a sequence of numbers in a multi dimensional list

How do I find if a sequence of numbers exists in a two-dimensional list? i.e.
matrix: [[1, 1, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [0,
0, 0, 1, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 1, 0, 0, 0]]
if [1,1,1] in matrix:
print("It's in there")
else: print("It's not there")
I guess I could turn every int into a string but is there a slicker way?
Using an iterator over each cell of the matrix, I've managed to get a basic idea of what you wanted to achieve in Python script.
matrix = [[1, 1, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 1, 0, 0, 0]]
matchCount = 0
lastNumber = None
for cell in matrix:
for number in cell:
if number == 1 and lastNumber == 1 or lastNumber == None:
matchCount += 1
if matchCount >= 3:
print("MATCH in cell " + str(cell))
lastNumber = number
matchCount = 0
lastNumber = None
What happens is, it steps into the cell. It it's the first iteration then allow entry into our iterator. We don't know if it's a match list yet, so push it back in our little list.
Stepping over and over, if we get enough matches in a row, then wonderful! Print we found a match in our matrix's cell!

Tkinter, Draughts board

I'd like to create a Draughts' game board.
At first, Everything worked fine while I was drawing squares and pawns directly on the "board" canvas. As I have to manipulate those pawns afterward, I wanted it to be more explicite by creating each square and pawn as objects.
So I tried to create a Board which will contain black or white square which will contain circles (the pawns). Then everything, messed up. I don't know why. Even so it looks perfectly logical (for me).
I think it is related to the use of the pack method.
Here's the code :
from tkinter import *
class Parent(Tk):
def getRoot(self):
return(self.body)
def setTitle(self,title):
self.title(title)
def run(self):
self.mainloop()
class Drawing:
def __init__(self,root,width,height):
self.zone=Canvas(root, width=width, height=height)
def put(self,root,row,col):
self.zone.grid(root,row=row,column=col)
def getDrawing(self):
return(self.zone)
def rectangle(self,coordX1,coordY1,coordX2,coordY2,color):
self.zone.create_rectangle(coordX1,coordY1,coordX2,coordY2,fill=color, outline="black")
def circle(self,coordX1,coordY1,coordX2,coordY2,color):
self.zone.create_oval(coordX1,coordY1,coordX2,coordY2,fill=color,outline="black")
if __name__=="__main__":
root=Parent()
root.setTitle("Draughts")
board=Drawing(root,400,400)
size=40
logicBoard=[[0, -1, 0, -1, 0, -1, 0, -1, 0, -1],
[-1, 0, -1, 0, -1, 0, -1, 0, -1, 0],
[0, -1, 0, -1, 0, -1, 0, -1, 0, -1],
[-1, 0, -1, 0, -1, 0, -1, 0, -1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0]]
DIMENSION=10
for i in range(DIMENSION):
for j in range(DIMENSION):
coordX1 = (i * size)
coordY1 = (j * size)
coordX2 = coordX1 + size
coordY2 = coordY1 + size
if(not(i%2==j%2)):#if the square is black (on the board)
color="black"
else:
color="white"
case=Drawing(board.getDrawing(),40,40)
case.rectangle(coordX1,coordY1,coordX2,coordY2,color)
case.getDrawing().pack()
if(logicBoard[i][j]>0):
pawnColor="white"
elif(logicBoard[i][j]<0):
pawnColor="black"
if (not(i%2==j%2)):
pawn=Drawing(case.getDrawing(),40,40)
pawn.circle(0,0,30,30,pawnColor)
pawn.getDrawing().pack()
board.getDrawing().pack()
root.run()
Thank you !
EDIT:
This is what I get :
The problem is that you create a new Canvas and call pack() on it in each iteration, instead of using the one you create at the beginning. In the end, you are using the class Drawing to create new Drawings. I suggest you to use only one class to represent all the board, with the methods to draw the squares and the circles.
I've changed the colors of the ovals to have a better contrast:
from tkinter import Tk, Canvas
from itertools import product
class Board(Tk):
def __init__(self, width, height, cellsize):
Tk.__init__(self)
self.cellsize = cellsize
self.canvas = Canvas(self, width=width, height=height)
self.canvas.bind("<Button-1>", self.onclick)
self.canvas.pack()
def draw_rectangle(self, x1, y1, x2, y2, color):
self.canvas.create_rectangle(x1, y1, x2, y2, fill=color, outline="black")
def draw_circle(self, x1, y1, x2, y2, color):
self.canvas.create_oval(x1, y1, x2, y2, fill=color, outline="black")
def onclick(self, event):
i = int(event.x / self.cellsize)
j = int(event.y / self.cellsize)
print "You clicked on cell (%s, %s)" % (i, j)
if __name__=="__main__":
size = 40
board = Board(400, 400, size)
board.title("Draughts")
logicBoard = [[0, -1, 0, -1, 0, -1, 0, -1, 0, -1],
[-1, 0, -1, 0, -1, 0, -1, 0, -1, 0],
[0, -1, 0, -1, 0, -1, 0, -1, 0, -1],
[-1, 0, -1, 0, -1, 0, -1, 0, -1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0],
[0, 1, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0]]
for (i, j) in product(range(10), range(10)):
coordX1 = (i * size)
coordY1 = (j * size)
coordX2 = coordX1 + size
coordY2 = coordY1 + size
color = "white" if i%2 == j%2 else "black"
board.draw_rectangle(coordX1, coordY1, coordX2, coordY2, color)
cell = logicBoard[i][j]
if cell != 0:
pawnColor = "red" if cell > 0 else "blue"
board.draw_circle(coordX1, coordY1, coordX2, coordY2, pawnColor)
board.mainloop()
Edit:
If you want to keep track of the clicks, it is clearer and simpler to bind a handler function to the canvas and calculate the point, instead of creating lots of canvas with their own event handler.

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