White non-gird lines on matplotlib plot - python

I have been working on a project for my math class in which I am creating a Julia set generator. I had finally succeeded in generating it, but when I show the plot and save the image generated there are white lines all over it. They do not line up with x ticks or y ticks. When I view the saved image it had even more white lines. I have been searching to find what these might be, but I have found nothing.
import matplotlib.pyplot as plt
#Set up window with a name and size.
plt.figure("Julia Set Generator by Eric Kapilik", figsize=(7.0,7.0))
#Set Range of axes
plt.xlim([-2,2])
plt.ylim([-2,2])
#Add labels to axes
plt.xlabel("Real Numbers")
plt.ylabel("Imaginary Numbers")
plt.grid(b=False, which = "major", axis = "both")
plt.grid(b=False, which = "minor", axis = "both")
name = input("Save file as... \n")
#Ask for maximum amount of iterations to base colour
#selection off of with fractions.
max = int(input("What is the maximum amount of iterations you wish to run? "))
#Generate an array of colour names to be used to plot points.
#Set seed of array.
colourArray = ["r"]
for colourNum in range (1, max):
#if the place in array is between 0% and 25% then set the colour to red.
#Same method used for other three colours.
if colourNum >= 0 and colourNum <= (max/4):
colourArray.append("r") #red
elif colourNum > (max/4) and colourNum <= (max/2):
colourArray.append("y") #yellow
elif colourNum > (max/2) and colourNum <= ((3*max)/4):
colourArray.append("g") #green
elif colourNum > ((3*max)/4) and colourNum <= max:
colourArray.append("c") #cyan
#Get constant value of which the julia set is based off of.
#The real number component is plotted on the horizontal axis
#of a complex number grid so we will use x.
xConstant = float(input("Enter real number constant component: "))
#The imaginary nuber compenent of a complex number is plotted on the vertical axis,
#so we will use y in our real number grid (for simplicity's sake).
yConstant = float(input("Enter imaginary number constant component: "))
#Title the graph based on the constatn complex number entered.
plt.title(str(xConstant) + " + " + str(yConstant) + "i")
#See the starting coordinates to be tested and conditions
xTest = float(-2)
yTest = float(2)
stop = False
i = 0
xPrevious = xTest
yPrevious = yTest
#Using an escape time algorith, determine the amout of iterations of the recursion
#are needed for the coordinate to be attarcted to infinity.
#Continue doing this while the y value of the coordinate being tested is less
#than or equal to -2.
while yTest >= -2:
#We are following the recursive function of
#f(Z-1) = Z^2 + C
#Where Z is current coordinate, and C is the constant value.
#Reminder: Both Z and C are actually complex numbers but in our case we
#are using them both as real number coordinates on a real number grid.
xCurrent = ((xPrevious**2) - (yPrevious**2)) + xConstant
yCurrent = (2 * xPrevious * yPrevious) + yConstant
#Points that surpass a circle of radius 2 with a centre point at the origin
#are considered to indefinitely escape to infinity.
#So when the radius of the recursive coordinate based off of the tested coordinate
#becomes greater or equal to two we know it will be attaracted to infinity.
radius = xCurrent**2 + yCurrent**2
#"Is the point an escapee?"
if radius >= 2:
#Since the point has been defined as one that esacpes to infintity
#it is considered an escapee, so set that to true.
escapee = True
#"Is the point a prisoner?"
if i == max:
#The point is considered a prisoner if max iterations is reached and
#the point is still within the circle of radius 2.
#The testeed point will be considered a prisoner based off of the amount
#of iterations we selected, it is possible that with more iterations
#that we would find this to be an escapee.
prisoner = True
#If we have not defined what kind of point this is yet, then go to the next
#iteration of the recursion. i is the number of iterations completed for
#the test point.
if escapee == False and prisoner == False:
i = i + 1
#Out with the old, in with the new. Set the current points to the previous
#points for the next iteration.
xPrevious = xCurrent
yPrevious= yCurrent
#If, however, we have defined the point, then colour it based off of
#the amount of iterations using the array of colours generated at the
#beginning to select the colour.
if escapee == True or prisoner == True:
#This sets the black points that are prisoners, this is the body
#of the julia set.
if i == max:
colourPoint = "k,"
else:
#Colour the point and concatenate a ",", which means to plot this point
#as a pixel.
colourPoint = colourArray[i] + ","
#Plot the point! (Most satisfying part)
plt.plot(xTest, yTest, colourPoint)
#Determine the percentage finished, to give user an idea of how the
#renderig is going. (Not nessecary, but appreciable)
percent = int(((yTest-2)/4) * (-100))
print(str(percent) + "%")
#After setting a colour and plotting the point, jump to the next test coordinate.
#Once the end of the line is reached, jump down one.
if xTest >= 2:
xTest = -2
yTest = yTest - 0.01
else:
xTest= xTest + 0.01
#Reset the starting conditions.
i = 0
escapee = False
prisoner = False
xPrevious = xTest
yPrevious = yTest
#Show the beauty.
print("100%")
print("Wait for matplotlib to finish things up...\nWill take a minute...")
plt.show()
plt.savefig(name)
Plot that was generated:
Saved image, with even more white lines:

Related

Add sprite to the group that it doesn't collide with any sprites in the group [duplicate]

A video by Sebastion Lague explained the Bridson's algorithm really well.
To oversimplify,
Create cell grid that has sides of radius/sqrt(2).
Place initial point and list as spawnpoint.
Place point into cell in grid.
For any spawnpoint, spawn a point between radius and 2*radius.
Look at the cells 2 units away from cell of new point.
If contains other points, compare distance.
If any point is closer to new point than the radius, new point is invalid.
If new point is valid, new point is listed as spawnpoint and placed into cell in grid.
If spawnpoint spawns too many invalid points, spawnpoint is removed and turns into point.
Repeat until no more spawnpoints exists.
Return points.
I basically written the same thing down in Python 3.7.2 and pygame 1.7~, but as said in the title, I'm stuck in recursive purgatory.
I used one Point() class for this algorithm, which might seem redundant given that pygame.Vector2() exists, but I needed some elements for a separate algorithm (Delaunay's with infinite vertices) that required this class to work.
For the sake of simplicity I'm going to cut away all the Delaunay-specific elements and show the bare-bones of this class that is needed for this algorithm:
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def DistanceToSquared(self,other):
return (self.x-other.x)**2 + (self.y-other.y)**2
The code that is related to the Bridson's algorithm is:
def PoissonDiskSampling(width, height, radius, startPos = None, spawnAttempts = 10):
if startPos == None:
startPos = [width//2,height//2]
cellSize = radius / math.sqrt(2)
cellNumberX = int(width // cellSize + 1) # Initialise a cells grid for optimisation
cellNumberY = int(height // cellSize + 1)
cellGrid = [[None for x in range(cellNumberX)] for y in range(cellNumberY)]
startingPoint = Point(startPos[0],startPos[1]) # Add an iniial point for spawning purposes
cellGrid[startingPoint.x//radius][startingPoint.y//radius] = startingPoint
points = [startingPoint] # Initialise 2 lists tracking all points and active points
spawnpoints = [startingPoint]
while len(spawnpoints) > 0:
spawnIndex = random.randint(0,len(spawnpoints)-1)
spawnpoint = spawnpoints[spawnIndex]
spawned = False
for i in range(spawnAttempts):
r = random.uniform(radius,2*radius)
radian = random.uniform(0,2*math.pi)
newPoint = Point(spawnpoint.x + r*math.cos(radian),
spawnpoint.y + r*math.sin(radian))
if 0 <= newPoint.x <= width and 0 <= newPoint.y <= height:
isValid = True
else:
continue
newPointIndex = [int(newPoint.x//cellSize), int(newPoint.y//cellSize)]
neighbours = FindNeighbours(cellNumberX,cellNumberY,newPointIndex,cellGrid)
for neighbour in neighbours:
if newPoint.DistanceToSquared(neighbour) < radius**2:
isValid = False
break
if isValid:
points.append(newPoint)
spawnpoints.append(newPoint)
spawned = True
break
else:
continue
if spawned == False:
spawnpoints.remove(spawnpoint)
return points
def FindNeighbours(cellNumberX, cellNumberY, index, cellGrid):
neighbours = []
for cellX in range(max(0,(index[0]-2)), min(cellNumberX,(index[1]+2))):
for cellY in range(max(0,(index[0]-2)), min(cellNumberY,(index[1]+2))):
if cellGrid[cellX][cellY] != None:
neighbours.append(cellGrid[cellX][cellY])
return neighbours
Please help.
The probably most important step is missing in your code:
If new point is valid, new point is listed as spawnpoint and placed into cell in grid.
I suggest to add the point to the cellGrid if it is valid:
if isValid:
cellGrid[newPointIndex[0]][newPointIndex[1]] = newPoint
points.append(newPoint)
spawnpoints.append(newPoint)
spawned = True
break
Further, you have to verify if the cell with the index newPointIndex is not already occupied before a point can be add:
newPointIndex = [int(newPoint.x/cellSize), int(newPoint.y/cellSize)]
if cellGrid[newPointIndex[0]][newPointIndex[1]] != None:
continue
neighbours = FindNeighbours(cellNumberX,cellNumberY,newPointIndex,cellGrid)
Finally there is an issue in the function FindNeighbours. range(start, stop) creates a range for x in start <= x < stop.
So the stop has to be index[0]+3 rather than index[0]+2.
Further the ranges which control the 2 nested for loops, run both from x-2 to y+2 rather than from x-2 to x+2 respectively from y-2 to y+2:
for cellX in range(max(0,(index[0]-2)), min(cellNumberX,(index[1]+2))):
for cellY in range(max(0,(index[0]-2)), min(cellNumberY,(index[1]+2)))
The fixed function has to be:
def FindNeighbours(cellNumberX, cellNumberY, index, cellGrid):
neighbours = []
for cellX in range(max(0, index[0]-2), min(cellNumberX, index[0]+3)):
for cellY in range(max(0, index[1]-2), min(cellNumberY, index[1]+3)):
if cellGrid[cellX][cellY] != None:
neighbours.append(cellGrid[cellX][cellY])
return neighbours
See the result, for a size of 300 x 300 and a radius of 15:
An even better result can be achieve, if always the 1st point of spawnpoints is used to rather than a random point:
# spawnIndex = random.randint(0,len(spawnpoints)-1)
spawnIndex = 0 # 0 rather than random
spawnpoint = spawnpoints[spawnIndex]

How do I program a spatially dependend random number distribution?

I wrote a routine that distributes circles randomly (uniformly) with an arbitrary diameter in my study area.
def no_nearby_dots(new_dot, dots_sim, min_distance):
for dot in dots_sim:
if np.sqrt((dot[0] - new_dot[0]) ** 2 + (dot[1] - new_dot[1]) ** 2) <= min_distance:
return False
return True
while realizations < simulations:
dots_sim = []
new_dot = True
dots_sim.append((np.random.uniform(xmin, xmax), np.random.uniform(ymin, ymax)))
failed_attempts = 0
while new_dot:
xp = np.random.uniform(xmin, xmax)
yp = np.random.uniform(ymin, ymax)
if no_nearby_dots((xp, yp), dots_sim, diameter):
dots_sim.append((xp, yp))
failed_attempts = 0
else:
failed_attempts += 1
if len(dots_sim) == n_sim:
new_dot = False
if failed_attempts > 2000:
new_dot = False
print('ERROR...exit loop')
break
x_sim = [dot[0] for dot in dots_sim]
y_sim = [dot[1] for dot in dots_sim]
I want to introduce a second circle around the initial ones where the possibility of distributing points reduces exponentially towards the inner border -> I want to prevent a "hard" border, the points are allowed to occur anywhere on the plane but not closer than diameter, additionally they can only occur to a certain degree between diameter and diameter2.
Are there any ideas how to do that?
Here is an idea.
Choose a random radius between diameter/2 and diameter2/2, then generate a random point in the circle formed by that radius. There are many ways to choose a radius that meets your requirements. For example, the following chooses a radius such that radii very close to diameter2/2 are much more likely to be chosen:
radius = (diameter1/2) + ((diameter2/2) - (diameter1/2)) * random.random()**(1/20)
Note that 1/20 is the 20th root of a uniform (0, 1) random number. Try changing 1/20 to a different value and see what happens.
There are other ways to choose a radius this way, and they can all be described by a probability density function (for more information, see the following answer: Generate a random point within a circle (uniformly), which shows how a linear density function leads to a uniform distribution of points in a circle).
I solved it, and this is what I did:
while realizations < simulations:
dots_sim = []
new_dot = True
dots_sim.append((np.random.uniform(x_min, x_max), np.random.uniform(y_min, y_max)))
failed_attempts = 0
while new_dot:
x = np.random.uniform(x_min, x_max)
y = np.random.uniform(y_min, y_max)
diameter_simulation = np.random.uniform(min_diameter, max_diameter)
if no_nearby_dots((x, y), dots_sim, diameter_simulation):
dots_sim.append((x, y))
failed_attempts = 0
else:
failed_attempts += 1
if len(dots_sim) == len(x_coordinate):
new_dot = False
if failed_attempts > 1000:
new_dot = False
print('ERROR... -> no more space to place QDs! -> exit loop!')
break
What I did was creating diameters for my circles also using uniformly distributed numbers in an arbitrary interval, which smoothes my cumulative distribution function. This is the solution I needed, but it might not fit the initial question very well (or the question was formulated inaccurately in the first place :p)

My implementation of Bridson's algorithm Poisson-Disk Sampling seems to be stuck in an infinite loop

A video by Sebastion Lague explained the Bridson's algorithm really well.
To oversimplify,
Create cell grid that has sides of radius/sqrt(2).
Place initial point and list as spawnpoint.
Place point into cell in grid.
For any spawnpoint, spawn a point between radius and 2*radius.
Look at the cells 2 units away from cell of new point.
If contains other points, compare distance.
If any point is closer to new point than the radius, new point is invalid.
If new point is valid, new point is listed as spawnpoint and placed into cell in grid.
If spawnpoint spawns too many invalid points, spawnpoint is removed and turns into point.
Repeat until no more spawnpoints exists.
Return points.
I basically written the same thing down in Python 3.7.2 and pygame 1.7~, but as said in the title, I'm stuck in recursive purgatory.
I used one Point() class for this algorithm, which might seem redundant given that pygame.Vector2() exists, but I needed some elements for a separate algorithm (Delaunay's with infinite vertices) that required this class to work.
For the sake of simplicity I'm going to cut away all the Delaunay-specific elements and show the bare-bones of this class that is needed for this algorithm:
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def DistanceToSquared(self,other):
return (self.x-other.x)**2 + (self.y-other.y)**2
The code that is related to the Bridson's algorithm is:
def PoissonDiskSampling(width, height, radius, startPos = None, spawnAttempts = 10):
if startPos == None:
startPos = [width//2,height//2]
cellSize = radius / math.sqrt(2)
cellNumberX = int(width // cellSize + 1) # Initialise a cells grid for optimisation
cellNumberY = int(height // cellSize + 1)
cellGrid = [[None for x in range(cellNumberX)] for y in range(cellNumberY)]
startingPoint = Point(startPos[0],startPos[1]) # Add an iniial point for spawning purposes
cellGrid[startingPoint.x//radius][startingPoint.y//radius] = startingPoint
points = [startingPoint] # Initialise 2 lists tracking all points and active points
spawnpoints = [startingPoint]
while len(spawnpoints) > 0:
spawnIndex = random.randint(0,len(spawnpoints)-1)
spawnpoint = spawnpoints[spawnIndex]
spawned = False
for i in range(spawnAttempts):
r = random.uniform(radius,2*radius)
radian = random.uniform(0,2*math.pi)
newPoint = Point(spawnpoint.x + r*math.cos(radian),
spawnpoint.y + r*math.sin(radian))
if 0 <= newPoint.x <= width and 0 <= newPoint.y <= height:
isValid = True
else:
continue
newPointIndex = [int(newPoint.x//cellSize), int(newPoint.y//cellSize)]
neighbours = FindNeighbours(cellNumberX,cellNumberY,newPointIndex,cellGrid)
for neighbour in neighbours:
if newPoint.DistanceToSquared(neighbour) < radius**2:
isValid = False
break
if isValid:
points.append(newPoint)
spawnpoints.append(newPoint)
spawned = True
break
else:
continue
if spawned == False:
spawnpoints.remove(spawnpoint)
return points
def FindNeighbours(cellNumberX, cellNumberY, index, cellGrid):
neighbours = []
for cellX in range(max(0,(index[0]-2)), min(cellNumberX,(index[1]+2))):
for cellY in range(max(0,(index[0]-2)), min(cellNumberY,(index[1]+2))):
if cellGrid[cellX][cellY] != None:
neighbours.append(cellGrid[cellX][cellY])
return neighbours
Please help.
The probably most important step is missing in your code:
If new point is valid, new point is listed as spawnpoint and placed into cell in grid.
I suggest to add the point to the cellGrid if it is valid:
if isValid:
cellGrid[newPointIndex[0]][newPointIndex[1]] = newPoint
points.append(newPoint)
spawnpoints.append(newPoint)
spawned = True
break
Further, you have to verify if the cell with the index newPointIndex is not already occupied before a point can be add:
newPointIndex = [int(newPoint.x/cellSize), int(newPoint.y/cellSize)]
if cellGrid[newPointIndex[0]][newPointIndex[1]] != None:
continue
neighbours = FindNeighbours(cellNumberX,cellNumberY,newPointIndex,cellGrid)
Finally there is an issue in the function FindNeighbours. range(start, stop) creates a range for x in start <= x < stop.
So the stop has to be index[0]+3 rather than index[0]+2.
Further the ranges which control the 2 nested for loops, run both from x-2 to y+2 rather than from x-2 to x+2 respectively from y-2 to y+2:
for cellX in range(max(0,(index[0]-2)), min(cellNumberX,(index[1]+2))):
for cellY in range(max(0,(index[0]-2)), min(cellNumberY,(index[1]+2)))
The fixed function has to be:
def FindNeighbours(cellNumberX, cellNumberY, index, cellGrid):
neighbours = []
for cellX in range(max(0, index[0]-2), min(cellNumberX, index[0]+3)):
for cellY in range(max(0, index[1]-2), min(cellNumberY, index[1]+3)):
if cellGrid[cellX][cellY] != None:
neighbours.append(cellGrid[cellX][cellY])
return neighbours
See the result, for a size of 300 x 300 and a radius of 15:
An even better result can be achieve, if always the 1st point of spawnpoints is used to rather than a random point:
# spawnIndex = random.randint(0,len(spawnpoints)-1)
spawnIndex = 0 # 0 rather than random
spawnpoint = spawnpoints[spawnIndex]

How to solve...ValueError: cannot convert float NaN to integer

I'm running quite a complex code so I won't bother with details as I've had it working before but now im getting this error.
Particle is a 3D tuple filled with 0 or 255, and I am using the scipy centre of mass function and then trying to turn the value into its closest integer (as I'm dealing with arrays). The error is found with on the last line... can anyone explain why this might be??
2nd line fills Particle
3rd line deletes any surrounding particles with a different label (This is in a for loop for all labels)
Particle = []
Particle = big_labelled_stack[x_start+20:x_stop+20,y_start+20:y_stop+20,z_start+20:z_stop+20]
Particle = np.where(Particle == i ,255,0)
CoM = scipy.ndimage.measurements.center_of_mass(Particle)
CoM = [ (int(round(x)) for x in CoM ]
Thanks in advance. If you need more code just ask but I dont think it will help you and its very messy.
################## MORE CODE
border = 30
[labelled_stack,no_of_label] = label(labelled,structure_array,output_type)
# RE-LABEL particles now no. of seeds has been reduced! LAST LABELLING
#Increase size of stack by increasing borders and equal them to 0; to allow us to cut out particles into cube shape which else might lye outside the border
h,w,l = labelled.shape
big_labelled_stack = np.zeros(shape=(h+60,w+60,l+60),dtype=np.uint32)
# Creates an empty border around labelled_stack full of zeros of size border
if (no_of_label > 0): #Small sample may return no particles.. so this stage not neccesary
info = np.zeros(shape=(no_of_label,19)) #Creates array to store coordinates of particles
for i in np.arange(1,no_of_label,1):
coordinates = find_objects(labelled_stack == i)[0] #Find coordinates of label i.
x_start = int(coordinates[0].start)
x_stop = int(coordinates[0].stop)
y_start = int(coordinates[1].start)
y_stop = int(coordinates[1].stop)
z_start = int(coordinates[2].start)
z_stop = int(coordinates[2].stop)
dx = (x_stop - x_start)
dy = (y_stop - y_start)
dz = (z_stop - z_start)
Particle = np.zeros(shape=(dy,dx,dz),dtype = np.uint16)
Particle = big_labelled_stack[x_start+30:x_start+dx+30,y_start+30:y_start+dy+30,z_start+30:z_start+dz+30]
Particle = np.where(Particle == i ,255,0)
big_labelled_stack[border:h+border,border:w+border,border:l+border] = labelled_stack
big_labelled_stack = np.where(big_labelled_stack == i , 255,0)
CoM_big_stack = scipy.ndimage.measurements.center_of_mass(big_labelled_stack)
C = np.asarray(CoM_big_stack) - border
if dx > dy:
b = dx
else: #Finds the largest of delta_x,y,z and saves as b, so that we create 'Cubic_Particle' of size 2bx2bx2b (cubic box)
b = dy
if dz > b:
b = dz
CoM = scipy.ndimage.measurements.center_of_mass(Particle)
CoM = [ (int(round(x))) for x in CoM ]
Cubic_Particle = np.zeros(shape=(2*b,2*b,2*b))
Cubic_Particle[(b-CoM[0]):(b+dx-CoM[0]),(b-CoM[1]):(b+dy-CoM[1]),(b-CoM[2]):(b+dz-CoM[2])] = Particle
volume = Cubic_Particle.size # Gives volume of the box in voxels
info[i-1,:] = [C[0],C[1],C[2],i,C[0]-b,C[1]-b,C[2]-b,C[0]+b,C[1]+b,C[2]+b,volume,0,0,0,0,0,0,0,0] # Fills an array with label.No., size of box, and co-ords
else:
print('No particles found, try increasing the sample size')
info = []
Ok, so I have a stack full of labelled particles, there are two things I am trying to do, first find the centre of masses of each particle with respect ot the labelled_stack which is what CoM_big_labelled_stack (and C) does. and stores the co-ords in a list (tuple) called info. I am also trying to create a cubic box around the particle, with its centre of mass as the centre (which is relating to the CoM variable), so first I use the find objects function in scipy to find a particle, i then use these coordinates to create a non-cubic box around the particle, and find its centre of mass.I then find the longest dimension of the box and call it b, creating a cubic box of size 2b and filling it with particle in the right position.
Sorry this code is a mess, I am very new to Python

Expanding blocks ( dilation ) in a binary grid

For an A* implementation (to generate a path for a 'car' robot), I need to adapt my model to take into account the car's 'width' and hence avoid obstacles.
One idea I got is to expand all obstacles by the car's width, that way all the cells that are too close to an obstacle will be also marked as obstacles.
I tried using two naive algorithms to do this, but it's still too slow (especially on big grids) because it goes through the same cells many times:
unreachable = set()
# I first add all the unreachables to a set to avoid 'propagation'
for line in self.grid:
for cell in line:
if not cell.reachable:
unreachable.add(cell)
for cell in unreachable:
# I set as unreachable all the cell's neighbours in a certain radius
for nCell in self.neighbours( cell, int(radius/division) ):
nCell.reachable = False
Here's the definition of neighbours:
def neighbours(self, cell, radius = 1, unreachables = False):
neighbours = set()
for i in xrange(-radius, radius + 1):
for j in xrange(-radius, radius + 1):
x = cell.x + j
y = cell.y + i
if 0 <= y < self.height and 0 <= x < self.width and (self.grid[y][x].reachable or unreachables )) :
neighbours.add(self.grid[y][x])
return neighbours
Is there any sequential algorithm (or O(n.log(n))) that could do the same thing ?
What you are looking for is what is known as Minkowski sum, and if your obstacles and car are convex, there is a linear algorithm to compute it.
I ended up using convolution product, with my 'map' (a matrix where '1' is an obstacle and '0' is a free cell) as the first operand and a matrix of the car's size and all filled with '1's as the second operand.
The convolution product of those two matrices gives a matrix where the cells that weren't in reach of any obstacle (that is: didn't have any obstacle in the neighberhood) have a value of '0', and those who had at least one obstacle in their neighberhood (that is a cell equal to '1') have a value != 0.
Here's the Python implementation (using scipy for the convolution product) :
# r: car's radius; 1 : Unreachable ; 0 : Reachable
car = scipy.array( [[1 for i in xrange(r)] for j in xrange(r)] )
# Converting my map to a binary matrix
grid = scipy.array( [[0 if self.grid[i][j].reachable else 1 for j in xrange(self.width)] for i in xrange(self.height)] )
result = scipy.signal.fftconvolve( grid, car, 'same' )
# Updating the map with the result
for i in xrange(self.height):
for j in xrange(self.width):
self.grid[i][j].reachable = int(result[i][j]) == 0

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