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Create a list of N random (x,y,z) points that are at least a distance r away from each other

I'm trying to create a list of N random (x,y,z) points using python, in a way that each point is at least a distance r away from any other point.
I'm super new to programming and so far I'm only able to generate x, y, and z separately (then put together) by using
import random
def RandX(start, end, num):
res = []
for j in range(num):
res.append(random.randint(start, end))
return res
num = N
start = 0
end = 100
print(RandX(start, end, num))
but I have no idea how to control or check the positions of the points(x, y, z) so that the points are a distance away from each other.

To check the distance between two points (x,y,z) and (a,b,c) (stored as tuples), you can try:
def distance(p1,p2):
d=0
for i in range(3):
d+=(p1[i]-p2[i])**2
return d**(1/2)
Once you generate xyz randomly, you can set the following:
p1=x,y,z
p2=a,b,c
If your numbers are not too large, while this is inefficient, you can generate random numbers until they satisfy the distance condition.

Here is my solution: all we need is a distance function and a loop to generate random points and check for minimum distance criteria within our already-generated list:
def dist(new_point, points, r_threshold):
for point in points:
dist = np.sqrt(np.sum(np.square(new_point-point)))
if dist < r_threshold:
return False
return True
def RandX(N, r_threshold):
points = []
scope = np.arange(0,10,0.1)
while len(points) < N:
new_point = np.random.choice(scope, 3)
if dist(new_point, points, r_threshold):
points.append(new_point)
return points
For example:
RandX(5, 4)
[array([3.5, 2.6, 7.6]),
array([9.9, 0.1, 7.2]),
array([4. , 2.8, 0.3]),
array([0.2, 7.4, 5.1]),
array([7.4, 6.3, 5.2])]

Something like this. (It can be optimised but should serve you as very first version)
from collections import namedtuple
import random
import math
Point = namedtuple('Point', ' x y z')
MIN = 0
MAX = 1000
def fill_points_list(points, number_of_required_points, min_distance):
def _get_distance(p1, p2):
return math.sqrt(sum([(a - b) ** 2 for a, b in zip(p1, p2)]))
while len(points) < number_of_required_points:
temp = Point(random.randint(MIN, MAX), random.randint(MIN, MAX), random.randint(MIN, MAX))
count = 0
for p in points:
if _get_distance(temp, p) > min_distance:
count += 1
else:
break
if len(points) == count:
points.append(temp)
number_of_required_points = 9
min_distance = 51
points = []
fill_points_list(points, number_of_required_points, min_distance)
print(points)
output
[Point(x=771, y=590, z=226), Point(x=385, y=835, z=900), Point(x=551, y=294, z=800), Point(x=824, y=306, z=333), Point(x=892, y=548, z=879), Point(x=520, y=660, z=384), Point(x=409, y=193, z=331), Point(x=411, y=706, z=300), Point(x=272, y=116, z=719)]

You could try to generate randomly a number of points, and then filter them based on the distance criteria. The numpy and sklearn packages can be helpful make the process more efficient. You could imagine something like this:
import numpy as np
from sklearn.metrics.pairwise import euclidean_distances
r = 2
# Generate 100 points (3-tuples) between 0 and 10
points = np.random.randint(0,100,[1000,3])
# Pairwise distances between points
distances = euclidean_distances(points)
# "Remove" distance to itself by setting to a distance of r+1 (to discard it later)
distances += np.identity(len(distances)) * (r+1)
# Retrieve the distance to the closest point
min_dist = np.min(distances,axis=1)
# Filter your set of points
filtered_points = points[min_dist>r]
This should run pretty fast.

Strange result sorting points by distance in python

I got a list of points by extracting the edge of a image, like that:But it is not well ordered, so if I connect it as a line, it will be:
Thus I want to sort this list if points. Like, start with point_0, find which one has the shortest distance with it, say, point_3, then find which one's closest to point_3 then continue...
To sort the points, I wrote this:
import matplotlib.pyplot as plt
import numpy as np
import math
def dist(now, seek):
return math.sqrt((now[0] - seek[0])**2 + (now[1] - seek[1])**2)
def sortNearest(x, y):
if len(x) != len(y):
raise Exception('Error! Array length do not match!')
return False
xNew = []; yNew = []
nearest = 0 #record which point is nearest
now = [x[0], y[0]] #start point index
seekValue = 0
while len(x) > 0:
distance = (max(x) - min(x)) + (max(y) - min(y))
for seek in range(len(x)): # other
temp = dist(now, [x[seek], y[seek]])
if temp < distance and temp != 0.0:
distance = temp
seekValue = x[seek]
xNew.append(now[0]);
yNew.append(now[1]);
if len(x) > 0:
x.remove(now[0])
y.remove(now[1])
if len(x) > 0:
nearest = x.index(seekValue)
now = [x[nearest], y[nearest]]
x = list(xNew); y = list(yNew)
return xNew, yNew
x, y = getBorder('large.png', maxRes = 125)
x, y = sortNearest(x, y)
But that doesn't work well, I came up with this:
Which is obviously incorrect, if I zoom in, see:
If my code runs what I want, point_644 should connect 620 or 675, any but 645... What's wrong with it?

Well, point 644 cannot connect to point 620, because 620 is already part of your path.
As for why it connects to 645 instead of the closer 675: in your loop, you aren't actually remembering the index of the closest point, you're only remembering its x coordinate. After the loop, you then locate an arbitrary point with the same x coordinate - it could be anywhere on a vertical line going through the desired point.

I don't know how I would do this in python 3.x, so please forgive changes that I have not made from python 2.7. You'll also want to figure out what point you'd like to start with:
def find_distance(point1, point2):
distance = sqrt(square(point1[0]-point2[0]) + square(point1[1] - point2[1]))
return distance
x, y = getBorder('large.png', maxRes = 125)
points_in_border = [(i,j) for i, j in zip(x,y)]
current_point = points_in_border.pop([0])
points_in_order = [current_point]
while len(points_in_border) > 0:
min_distance = 10000
for point in points_in_border:
if find_distance(current_point, point) < min_distance:
closest_point = point
min_distance = find_distance(current_point, point)
points_in_border.remove(closest_point)
current_point = closest_point
points_in_order.append(closest_point)

I think what you want to do can be optimized with numpy and scipy:
import numpy as np
import scipy.spatial.distance as distance
import matplotlib.pyplot as plt
points = np.random.random((6,2))
dists =distance.pdist(points)
m=np.argsort(distance.squareform(dists))[:,1:]
order = [0,m[0,0]]
next_point = order[-1]
while len(order)<len(points):
row = m[next_point]
i = 0
while row[i] in order:
i += 1
order.append(row[i])
next_point = order[-1]
order.append(0)
ordered=points[order]
plt.plot(ordered[:,0], ordered[:,1], 'o-')
The idea underlying this code is the following. First you calculate all the distances. Then you use argsort to get the indices that would order each row. You can remove the first column, as each point is closest to itself. We know that. Then you look which is the next closest point and you add it to the list order if the point is not there yet. You then go to the row corresponding to this point, and look for the next point. And so on.
If what you are only interested in is just sorting the enclosing set of points, you can use ConvexHull to find them:
ch = ConvexHull(points)
plt.plot(points[ch.vertices,0], points[ch.vertices,1], 'o-')

Buffon's needle simulation in python

import numpy as np
import matplotlib.pylab as plt
class Buffon_needle_problem:
def __init__(self,x,y,n,m):
self.x = x #width of the needle
self.y = y #witdh of the space
self.r = []#coordinated of the centre of the needle
self.z = []#measure of the alingment of the needle
self.n = n#no of throws
self.m = m#no of simulations
self.pi_approx = []
def samples(self):
# throwing the needles
for i in range(self.n):
self.r.append(np.random.uniform(0,self.y))
self.z.append(np.random.uniform(0,self.x/2.0))
return [self.r,self.z]
def simulation(self):
self.samples()
# m simulation
for j in range(self.m):
# n throw
hits = 0 #setting the succes to 0
for i in range(self.n):
#condition for a hit
if self.r[i]+self.z[i]>=self.y or self.r[i]-self.z[i] <= 0.0:
hits += 1
else:
continue
hits = 2*(self.x/self.y)*float(self.n/hits)
self.pi_approx.append(hits)
return self.pi_approx
y = Buffon_needle_problem(1,2,40000,5)
print (y.simulation())
For those who unfamiliar with Buffon's problem, here is the http://mathworld.wolfram.com/BuffonsNeedleProblem.html
or
implementing the same idea (and output)
http://pythonfiddle.com/historically-accurate-buffons-needle/
My expected output should be the value of pi but my code give me around 4. Can anyone point out the logical error?

The sampling of the needle's alignment should be a uniform cosine. See the following link for the method: http://pdg.lbl.gov/2012/reviews/rpp2012-rev-monte-carlo-techniques.pdf
Also, there were a few logical problems with the program. Here is a working version.
#!/bin/python
import numpy as np
def sample_cosine():
rr=2.
while rr > 1.:
u1=np.random.uniform(0,1.)
u2=np.random.uniform(0,1.)
v1=2*u1-1.
rr=v1*v1+u2*u2
cc=(v1*v1-u2*u2)/rr
return cc
class Buffon_needle_problem:
def __init__(self,x,y,n,m):
self.x = float(x) #width of the needle
self.y = float(y) #witdh of the space
self.r = [] #coordinated of the centre of the needle
self.z = [] #measure of the alignment of the needle
self.n = n #no of throws
self.m = m #no of simulations
self.p = self.x/self.y
self.pi_approx = []
def samples(self):
# throwing the needles
for i in range(self.n):
self.r.append(np.random.uniform(0,self.y))
C=sample_cosine()
self.z.append(C*self.x/2.)
return [self.r,self.z]
def simulation(self):
# m simulation
for j in range(self.m):
self.r=[]
self.z=[]
self.samples()
# n throw
hits = 0 #setting the success to 0
for i in range(self.n):
#condition for a hit
if self.r[i]+self.z[i]>=self.y or self.r[i]-self.z[i]<0.:
hits += 1
else:
continue
est =self.p*float(self.n)/float(hits)
self.pi_approx.append(est)
return self.pi_approx
y = Buffon_needle_problem(1,2,80000,5)
print (y.simulation())

Buffon's needle work accurately only when the distance between the two lines is double the length of needle. Make sure to cross check it.
I have seen many baffon's online simulation which are doing this mistake. They just take the distance between two adjacent lines to be equal to the needle's length. That's their main logical errors.

I would say that the problem is that you are defining the alignment of the needle by a simple linear function, when in fact the effective length of the needle from its centre is defined by a sinusoidal function.
You want to calculate the effective length of the needle (at 90° to the lines) by using a function that will calculate it from its angle.
Something like:
self.z.append(np.cos(np.random.uniform(-np.pi/2, np.pi/2))*self.x)
This will give the cosine of a random angle between -90° and +90°, times the length of the needle.
For reference, cos(+/-90) = 0 and cos(0) = 1, so at 90°, the needle with have effectively zero length, and at 0°, its full length.
I have neither mathplotlib or numpy installed on this machine, so I can't see if this fixes it, but it's definitely necessary.

Looks like you were committing a simple rounding error. The code below works, though the results are not very close to pi...
import numpy as np
import matplotlib.pylab as plt
class Buffon_needle_problem:
def __init__(self,x,y,n,m):
self.x = x #width of the needle
self.y = y #witdh of the space
self.r = []#coordinated of the centre of the needle
self.z = []#measure of the alingment of the needle
self.n = n#no of throws
self.m = m#no of simulations
self.pi_approx = []
def samples(self):
# throwing the needles
for i in range(self.n):
self.r.append(np.random.uniform(0,self.y))
self.z.append(np.random.uniform(0,self.x/2.0))
return [self.r,self.z]
def simulation(self):
#self.samples()
# m simulations
for j in range(self.m):
self.r=[]
self.z=[]
for i in range(self.n):
self.r.append(np.random.uniform(0,self.y))
self.z.append(np.random.uniform(0,self.x/2.0))
# n throws
hits = 0 # setting the succes to 0
for i in range(self.n):
# condition for a hit
if self.r[i]+self.z[i]>=self.y or self.r[i]-self.z[i] <= 0.0:
hits += 1
else:
continue
hits = 2.0*(float(self.x)/self.y)*float(self.n)/float(hits)
self.pi_approx.append(hits)
return self.pi_approx
y = Buffon_needle_problem(1,2,40000,5)
print (y.simulation())
Also note that you were using the same sample for all simulations!

I used Python turtle to approximate the value of Pi:
from turtle import *
from random import *
setworldcoordinates(-100, -200, 200, 200)
ht(); speed(0); color('blue')
drops = 20 # increase number of drops for better approximation
hits = 0 # hits counter
# draw parallel lines with distance 20 between adjacent lines
for i in range(0, 120, 20):
pu(); setpos(0, i); pd()
fd(100) # length of line
# throw needles
color('red')
for j in range(drops):
pu()
goto(randrange(10, 90), randrange(0,100))
y1 = ycor() # keep ycor of start point
seth(360*random())
pd(); fd(20) # draw needle of length 20
y2 = ycor() # keep ycor of end point
if y1//20 != y2//20: # decisive test: if it is a hit then ...
hits += 1 # increase the hits counter by 1
print(2 * drops / hits)
Output samples
With 50 drops 3.225806451612903
with 200 drops 3.3057851239669422
with 1000 drops 3.1645569620253164

NOT answer to original question, if you just want the pi estimate, here's some simple code from I did in a computational revision exercise yesterday at Uni Sydney (Aust), against my early inclinations, to reduce complexity, we only modelled for a random point between zero and distance between lines and a random angle from zero to 90 degrees.
import random
from numpy import pi, sin
def buffon(L, D, N):
'''
BUFFON takes L (needle length),
D = distance between lines and N = number of drops,
returns probability of hitting line
generate random number 'd' between 0 and D
generate theta between 0 and pi/2
hit when L*sin(theta)) - d is great than D
'''
hit = 0;
for loop in range(N) :
theta = pi*random.random()/2
if L * sin(theta) > D - D*random.random(): # d = random*D
hit += 1
return hit/N
#% Prob_hit = 2*L/(D*pi) hence: Pi_est = 2*L / (P_hit*D);
L = 1
D = 4
N = int(1e8)
Pi_est = 2*L / (buffon(L,D,N)*D)
It was in MatLab, I wanted to try it in Python, see if I could use any comprehension lists, any ideas to speed this up WELCOME.

It should be noted that the Monte Carlo method is not the best for this kind of calculation (calculating the number pi). One way or another, it is necessary to throw quite a lot of needles (or points, in the case of a quarter circle) in order to get a more accurate pi. The main disadvantage of the Monte Carlo method is its unpredictability.
https://github.com/Battle-Of-Two-K/Buffon-s-Noodle-Problem
https://github.com/Battle-Of-Two-K/Calculating-Pi-by-Monte-Carlo-Method
enter image description here

Many particles in box - physics simulation

I'm currently trying to simulate many particles in a box bouncing around.
I've taken into account #kalhartt's suggestions and this is the improved code to initialize the particles inside the box:
import numpy as np
import scipy.spatial.distance as d
import matplotlib.pyplot as plt
# 2D container parameters
# Actual container is 50x50 but chose 49x49 to account for particle radius.
limit_x = 20
limit_y = 20
#Number and radius of particles
number_of_particles = 350
radius = 1
def force_init(n):
# equivalent to np.array(list(range(number_of_particles)))
count = np.linspace(0, number_of_particles-1, number_of_particles)
x = (count + 2) % (limit_x-1) + radius
y = (count + 2) / (limit_x-1) + radius
return np.column_stack((x, y))
position = force_init(number_of_particles)
velocity = np.random.randn(number_of_particles, 2)
The initialized positions look like this:
Once I have the particles initialized I'd like to update them at each time-step. The code for updating follows the previous code immediately and is as follows:
# Updating
while np.amax(abs(velocity)) > 0.01:
# Assume that velocity slowly dying out
position += velocity
velocity *= 0.995
#Get pair-wise distance matrix
pair_dist = d.cdist(position, position)
pair_d = pair_dist<=4
#If pdist [i,j] is <=4 then the particles are too close and so treat as collision
for i in range(len(pair_d)):
for j in range(i):
# Only looking at upper triangular matrix (not inc. diagonal)
if pair_d[i,j] ==True:
# If two particles are too close then swap velocities
# It's a bad hack but it'll work for now.
vel_1 = velocity[j][:]
velocity[j] = velocity[i][:]*0.9
velocity[i] = vel_1*0.9
# Masks for particles beyond the boundary
xmax = position[:, 0] > limit_x
xmin = position[:, 0] < 0
ymax = position[:, 1] > limit_y
ymin = position[:, 1] < 0
# flip velocity and assume that it looses 10% of energy
velocity[xmax | xmin, 0] *= -0.9
velocity[ymax | ymin, 1] *= -0.9
# Force maximum positions of being +/- 2*radius from edge
position[xmax, 0] = limit_x-2*radius
position[xmin, 0] = 2*radius
position[ymax, 0] = limit_y-2*radius
position[ymin, 0] = 2*radius
After updating it and letting it run to completion I get this result:
This is infinitely better than before but there are still patches that are too close together - such as:
Too close together. I think the updating works... and thanks to #kalhartt my code is wayyyy better and faster (and I learnt some things about numpy... props #kalhartt) but I still don't know where it's screwing up. I've tried changing the order of the actual updates with the pair-wise distance going last or the position +=velocity going last but to no avail. I added the *0.9 to make the entire thing die down faster and I tried it with 4 to make sure that 2*radius (=2) wasn't too tight a criteria... but nothing seems to work.
Any and all help would be appreciated.

There are just two typos standing in your way. First for i in range(len(positions)/2): only iterates over half of your particles. This is why half the particles stay in the x bounds (if you watch for large iterations its more clear). Second, the second y condition should be a minimum (I assume) position[i][1] < 0. The following block works to bound the particles for me (I didn't test with the collision code so there could be problems there).
for i in range(len(position)):
if position[i][0] > limit_x or position[i][0] < 0:
velocity[i][0] = -velocity[i][0]
if position[i][1] > limit_y or position[i][1] < 0:
velocity[i][1] = -velocity[i][1]
As an aside, try to leverage numpy to eliminate loops when possible. It is faster, more efficient, and in my opinion more readable. For example force_init would look like this:
def force_init(n):
# equivalent to np.array(list(range(number_of_particles)))
count = np.linspace(0, number_of_particles-1, number_of_particles)
x = (count * 2) % limit_x + radius
y = (count * 2) / limit_x + radius
return np.column_stack((x, y))
And your boundary conditions would look like this:
while np.amax(abs(velocity)) > 0.01:
position += velocity
velocity *= 0.995
# Masks for particles beyond the boundary
xmax = position[:, 0] > limit_x
xmin = position[:, 0] < 0
ymax = position[:, 1] > limit_y
ymin = position[:, 1] < 0
# flip velocity
velocity[xmax | xmin, 0] *= -1
velocity[ymax | ymin, 1] *= -1
Final note, it is probably a good idea to hard clip position to the bounding box with something like position[xmax, 0] = limit_x; position[xmin, 0] = 0. There may be cases where velocity is small and a particle outside the box will be reflected but not make it inside in the next iteration. So it will just sit outside the box being reflected forever.
EDIT: Collision
The collision detection is a much harder problem, but lets see what we can do. Lets take a look at your current implementation.
pair_dist = d.cdist(position, position)
pair_d = pair_dist<=4
for i in range(len(pair_d)):
for j in range(i):
# Only looking at upper triangular matrix (not inc. diagonal)
if pair_d[i,j] ==True:
# If two particles are too close then swap velocities
# It's a bad hack but it'll work for now.
vel_1 = velocity[j][:]
velocity[j] = velocity[i][:]*0.9
velocity[i] = vel_1*0.9
Overall a very good approach, cdist will efficiently calculate the distance
between sets of points and you find which points collide with pair_d = pair_dist<=4.
The nested for loops are the first problem. We need to iterate over True values of pair_d where j > i. First your code actually iterate over the lower triangular region by using for j in range(i) so that j < i, not particularly important in this instance as long since i,j pairs are not repeated. However Numpy has two builtins we can use instead, np.triu lets us set all values below a diagonal to 0 and np.nonzero will give us the indices of non-zero elements in a matrix. So this:
pair_dist = d.cdist(position, position)
pair_d = pair_dist<=4
for i in range(len(pair_d)):
for j in range(i+1, len(pair_d)):
if pair_d[i, j]:
...
is equivalent to
pair_dist = d.cdist(position, position)
pair_d = np.triu(pair_dist<=4, k=1) # k=1 to exclude the diagonal
for i, j in zip(*np.nonzero(pair_d)):
...
The second problem (as you noted) is that the velocities are just switched and scaled instead of reflected. What we really want to do is negate and scale the component of each particles velocity along the axis that connects them. Note that to do this we will need the vector connecting them position[j] - position[i] and the length of the vector connecting them (which we already calculated). So unfortunately part of the cdist calculation gets repeated. Lets quit using cdist and do it ourselves instead. The goal here is to make two arrays diff and norm where diff[i][j] is a vector pointing from particle i to j (so diff is a 3D array) and norm[i][j] is the distance between particles i and j. We can do this with numpy like so:
nop = number_of_particles
# Give pos a 3rd index so we can use np.repeat below
# equivalent to `pos3d = np.array([ position ])
pos3d = position.reshape(1, nop, 2)
# 3D arras with a repeated index so we can form combinations
# diff_i[i][j] = position[i] (for all j)
# diff_j[i][j] = position[j] (for all i)
diff_i = np.repeat(pos3d, nop, axis=1).reshape(nop, nop, 2)
diff_j = np.repeat(pos3d, nop, axis=0)
# diff[i][j] = vector pointing from position[i] to position[j]
diff = diff_j - diff_i
# norm[i][j] = sqrt( diff[i][j]**2 )
norm = np.linalg.norm(diff, axis=2)
# check for collisions and take the region above the diagonal
collided = np.triu(norm < radius, k=1)
for i, j in zip(*np.nonzero(collided)):
# unit vector from i to j
unit = diff[i][j] / norm[i][j]
# flip velocity
velocity[i] -= 1.9 * np.dot(unit, velocity[i]) * unit
velocity[j] -= 1.9 * np.dot(unit, velocity[j]) * unit
# push particle j to be radius units from i
# This isn't particularly effective when 3+ points are close together
position[j] += (radius - norm[i][j]) * unit
...
Since this post is long enough already, here is a gist of the code with my modifications.

Checking if a point is inside a polygon

I have a class describing a Point (has 2 coordinates x and y) and a class describing a Polygon which has a list of Points which correspond to corners (self.corners)
I need to check if a Point is in a Polygon
Here is the function that is supposed to check if the Point is in the Polygon. I am using the Ray Casting Method
def in_me(self, point):
result = False
n = len(self.corners)
p1x = int(self.corners[0].x)
p1y = int(self.corners[0].y)
for i in range(n+1):
p2x = int(self.corners[i % n].x)
p2y = int(self.corners[i % n].y)
if point.y > min(p1y,p2y):
if point.x <= max(p1x,p2x):
if p1y != p2y:
xinters = (point.y-p1y)*(p2x-p1x)/(p2y-p1y)+p1x
print xinters
if p1x == p2x or point.x <= xinters:
result = not result
p1x,p1y = p2x,p2y
return result
I run a test with following shape and point:
PG1 = (0,0), (0,2), (2,2), (2,0)
point = (1,1)
The script happily returns False even though the point it within the line. I am unable to find the mistake

I would suggest using the Path class from matplotlib
import matplotlib.path as mplPath
import numpy as np
poly = [190, 50, 500, 310]
bbPath = mplPath.Path(np.array([[poly[0], poly[1]],
[poly[1], poly[2]],
[poly[2], poly[3]],
[poly[3], poly[0]]]))
bbPath.contains_point((200, 100))
(There is also a contains_points function if you want to test for multiple points)

I'd like to suggest some other changes there:
def contains(self, point):
if not self.corners:
return False
def lines():
p0 = self.corners[-1]
for p1 in self.corners:
yield p0, p1
p0 = p1
for p1, p2 in lines():
... # perform actual checks here
Notes:
A polygon with 5 corners also has 5 bounding lines, not 6, your loop is one off.
Using a separate generator expression makes clear that you are checking each line in turn.
Checking for an empty number of lines was added. However, how to treat zero-length lines and polygons with a single corner is still open.
I'd also consider making the lines() function a normal member instead of a nested utility.
Instead of the many nested if structures, you could also check for the inverse and then continue or use and.

I was trying to solve the same problem for my project and I got this code from someone in my network.
#!/usr/bin/env python
#
# routine for performing the "point in polygon" inclusion test
# Copyright 2001, softSurfer (www.softsurfer.com)
# This code may be freely used and modified for any purpose
# providing that this copyright notice is included with it.
# SoftSurfer makes no warranty for this code, and cannot be held
# liable for any real or imagined damage resulting from its use.
# Users of this code must verify correctness for their application.
# translated to Python by Maciej Kalisiak <mac#dgp.toronto.edu>
# a Point is represented as a tuple: (x,y)
#===================================================================
# is_left(): tests if a point is Left|On|Right of an infinite line.
# Input: three points P0, P1, and P2
# Return: >0 for P2 left of the line through P0 and P1
# =0 for P2 on the line
# <0 for P2 right of the line
# See: the January 2001 Algorithm "Area of 2D and 3D Triangles and Polygons"
def is_left(P0, P1, P2):
return (P1[0] - P0[0]) * (P2[1] - P0[1]) - (P2[0] - P0[0]) * (P1[1] - P0[1])
#===================================================================
# cn_PnPoly(): crossing number test for a point in a polygon
# Input: P = a point,
# V[] = vertex points of a polygon
# Return: 0 = outside, 1 = inside
# This code is patterned after [Franklin, 2000]
def cn_PnPoly(P, V):
cn = 0 # the crossing number counter
# repeat the first vertex at end
V = tuple(V[:])+(V[0],)
# loop through all edges of the polygon
for i in range(len(V)-1): # edge from V[i] to V[i+1]
if ((V[i][1] <= P[1] and V[i+1][1] > P[1]) # an upward crossing
or (V[i][1] > P[1] and V[i+1][1] <= P[1])): # a downward crossing
# compute the actual edge-ray intersect x-coordinate
vt = (P[1] - V[i][1]) / float(V[i+1][1] - V[i][1])
if P[0] < V[i][0] + vt * (V[i+1][0] - V[i][0]): # P[0] < intersect
cn += 1 # a valid crossing of y=P[1] right of P[0]
return cn % 2 # 0 if even (out), and 1 if odd (in)
#===================================================================
# wn_PnPoly(): winding number test for a point in a polygon
# Input: P = a point,
# V[] = vertex points of a polygon
# Return: wn = the winding number (=0 only if P is outside V[])
def wn_PnPoly(P, V):
wn = 0 # the winding number counter
# repeat the first vertex at end
V = tuple(V[:]) + (V[0],)
# loop through all edges of the polygon
for i in range(len(V)-1): # edge from V[i] to V[i+1]
if V[i][1] <= P[1]: # start y <= P[1]
if V[i+1][1] > P[1]: # an upward crossing
if is_left(V[i], V[i+1], P) > 0: # P left of edge
wn += 1 # have a valid up intersect
else: # start y > P[1] (no test needed)
if V[i+1][1] <= P[1]: # a downward crossing
if is_left(V[i], V[i+1], P) < 0: # P right of edge
wn -= 1 # have a valid down intersect
return wn

Steps:
Iterate over all the segments in the polygon
Check whether they intersect with a ray going in the increasing-x direction
Using the intersect function from This SO Question
def ccw(A,B,C):
return (C.y-A.y) * (B.x-A.x) > (B.y-A.y) * (C.x-A.x)
# Return true if line segments AB and CD intersect
def intersect(A,B,C,D):
return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)
def point_in_polygon(pt, poly, inf):
result = False
for i in range(len(poly.corners)-1):
if intersect((poly.corners[i].x, poly.corners[i].y), ( poly.corners[i+1].x, poly.corners[i+1].y), (pt.x, pt.y), (inf, pt.y)):
result = not result
if intersect((poly.corners[-1].x, poly.corners[-1].y), (poly.corners[0].x, poly.corners[0].y), (pt.x, pt.y), (inf, pt.y)):
result = not result
return result
Please note that the inf parameter should be the maximum point in the x axis in your figure.