Plot random points a specified distance apart - numpy

Question

I'm trying to come up with a function that plots n points inside the unit circle, but I need them to be sufficiently spread out.
ie. something that looks like this:
Is it possible to write a function with two parameters, n (number of points) and min_d (minimum distance apart) such that the points are:
a) equidistant
b) no pairwise distance exceeds a given min_d
The problem with sampling from a uniform distribution is that it could happen that two points are almost on top of each other, which I do not want to happen. I need this kind of input for a network diagram representing node clusters.
EDIT: I have found an answer to a) here: Generator of evenly spaced points in a circle in python, but b) still eludes me.

Answer 1

At the time this answer was provided, the question asked for random numbers. This answer thus gives a solution drawing random numbers. It ignores any edits made to the question afterwards.
On may simply draw random points and for each one check if the condition of the minimum distance is fulfilled. If not, the point can be discarded. This can be done until a list is filled with enough points or some break condition is met.
import numpy as np
import matplotlib.pyplot as plt
class Points():
def __init__(self,n=10, r=1, center=(0,0), mindist=0.2, maxtrials=1000 ) :
self.success = False
self.n = n
self.r = r
self.center=np.array(center)
self.d = mindist
self.points = np.ones((self.n,2))*10*r+self.center
self.c = 0
self.trials = 0
self.maxtrials = maxtrials
self.tx = "rad: {}, center: {}, min. dist: {} ".format(self.r, center, self.d)
self.fill()
def dist(self, p, x):
if len(p.shape) >1:
return np.sqrt(np.sum((p-x)**2, axis=1))
else:
return np.sqrt(np.sum((p-x)**2))
def newpoint(self):
x = (np.random.rand(2)-0.5)*2
x = x*self.r-self.center
if self.dist(self.center, x) < self.r:
self.trials += 1
if np.all(self.dist(self.points, x) > self.d):
self.points[self.c,:] = x
self.c += 1
def fill(self):
while self.trials < self.maxtrials and self.c < self.n:
self.newpoint()
self.points = self.points[self.dist(self.points,self.center) < self.r,:]
if len(self.points) == self.n:
self.success = True
self.tx +="\n{} of {} found ({} trials)".format(len(self.points),self.n,self.trials)
def __repr__(self):
return self.tx
center =(0,0)
radius = 1
p = Points(n=40,r=radius, center=center)
fig, ax = plt.subplots()
x,y = p.points[:,0], p.points[:,1]
plt.scatter(x,y)
ax.add_patch(plt.Circle(center, radius, fill=False))
ax.set_title(p)
ax.relim()
ax.autoscale_view()
ax.set_aspect("equal")
plt.show()
If the number of points should be fixed, you may try to run find this number of points for decreasing distances until the desired number of points are found.
In the following case, we are looking for 60 points and start with a minimum distance of 0.6 which we decrease stepwise by 0.05 until there is a solution found. Note that this will not necessarily be the optimum solution, as there is only maxtrials of retries in each step. Increasing maxtrials will of course bring us closer to the optimum but requires more runtime.
center =(0,0)
radius = 1
mindist = 0.6
step = 0.05
success = False
while not success:
mindist -= step
p = Points(n=60,r=radius, center=center, mindist=mindist)
print p
if p.success:
break
fig, ax = plt.subplots()
x,y = p.points[:,0], p.points[:,1]
plt.scatter(x,y)
ax.add_patch(plt.Circle(center, radius, fill=False))
ax.set_title(p)
ax.relim()
ax.autoscale_view()
ax.set_aspect("equal")
plt.show()
Here the solution is found for a minimum distance of 0.15.