Poisson disc sampling in Python

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A set of points drawn from a uniform distribution on a two-dimensional domain typically display clustering:

import numpy as np
import matplotlib.pyplot as plt
width, height = 60, 45
N = width * height / 4
plt.scatter(np.random.uniform(0,width,N), np.random.uniform(0,height,N),
            c='g', alpha=0.6, lw=0)
plt.xlim(0,width)
plt.ylim(0,height)
plt.axis('off')
plt.show()

enter image description here

For computer graphics applications, it is often useful to obtain a sample of random points such that no one point is closer than some pre-determined distance from any other point, in order to avoid this clustering effect. This type of distribution, with no low-frequency components is sometimes called "blue noise".

A popular approach for obtaining non-clustered random sample of points is "poisson disc sampling"; an efficient ($O(n)$) algorithm to implement this approach was given by Bridson (ACM SIGGRAPH 2007 sketches, article 22)[pdf].

In a two-dimensional implementation of Bridson's algorithm, the sample $\boldsymbol{\mathrm{R}}^2$ domain is divided into square cells of side length $r/\sqrt{2}$ where $r$ is the minimum distance between samples, such that each cell can contain a maximum of one sample point. The sample points are stored in a list of $(x,y)$ coordinates, samples. The grid of cells are represented by a Python dictionary, cells, for which each key is the cell coordinates and the corresponding value is the index of the point in samples list (or None if the cell is empty).

We start by selecting an initial sample point (drawn at random uniformly from the domain), inserting it into samples and putting its index, 0, in the corresponding entry in the cells dictionary. We also initialize a separate list active with this index.

While the active list contains entries, we choose one at random, refpt, and generate up to $k$ (say, 30) points uniformly from the circular annulus around it of inner radius $r$ and outer radius $2r$. Any one of these points which is no closer than $r$ to any other in samples is "valid" and can be added to samples and active. We only need to check the surrounding cells in the local neighbourhood of refpt. If none of the $k$ points is valid, then refpt is removed from the active list: we will no longer search for points around this reference point.

Mike Bostock gives a nice animated demonstration of the Poisson disc sampling algorithm on his website.

Below is my Python code for Poisson disc sampling using Bridson's algorithm; a typical output is shown here:

enter image description here

Please see the next post for an object-oriented approach to this algorithm.

This code is also available on my github page.

import numpy as np
import matplotlib.pyplot as plt

# Choose up to k points around each reference point as candidates for a new
# sample point
k = 30

# Minimum distance between samples
r = 1.7

width, height = 60, 45

# Cell side length
a = r/np.sqrt(2)
# Number of cells in the x- and y-directions of the grid
nx, ny = int(width / a) + 1, int(height / a) + 1

# A list of coordinates in the grid of cells
coords_list = [(ix, iy) for ix in range(nx) for iy in range(ny)]
# Initilalize the dictionary of cells: each key is a cell's coordinates, the
# corresponding value is the index of that cell's point's coordinates in the
# samples list (or None if the cell is empty).
cells = {coords: None for coords in coords_list}

def get_cell_coords(pt):
    """Get the coordinates of the cell that pt = (x,y) falls in."""

    return int(pt[0] // a), int(pt[1] // a)

def get_neighbours(coords):
    """Return the indexes of points in cells neighbouring cell at coords.

    For the cell at coords = (x,y), return the indexes of points in the cells
    with neighbouring coordinates illustrated below: ie those cells that could 
    contain points closer than r.

                                     ooo
                                    ooooo
                                    ooXoo
                                    ooooo
                                     ooo

    """

    dxdy = [(-1,-2),(0,-2),(1,-2),(-2,-1),(-1,-1),(0,-1),(1,-1),(2,-1),
            (-2,0),(-1,0),(1,0),(2,0),(-2,1),(-1,1),(0,1),(1,1),(2,1),
            (-1,2),(0,2),(1,2),(0,0)]
    neighbours = []
    for dx, dy in dxdy:
        neighbour_coords = coords[0] + dx, coords[1] + dy
        if not (0 <= neighbour_coords[0] < nx and
                0 <= neighbour_coords[1] < ny):
            # We're off the grid: no neighbours here.
            continue
        neighbour_cell = cells[neighbour_coords]
        if neighbour_cell is not None:
            # This cell is occupied: store this index of the contained point.
            neighbours.append(neighbour_cell)
    return neighbours

def point_valid(pt):
    """Is pt a valid point to emit as a sample?

    It must be no closer than r from any other point: check the cells in its
    immediate neighbourhood.

    """

    cell_coords = get_cell_coords(pt)
    for idx in get_neighbours(cell_coords):
        nearby_pt = samples[idx]
        # Squared distance between or candidate point, pt, and this nearby_pt.
        distance2 = (nearby_pt[0]-pt[0])**2 + (nearby_pt[1]-pt[1])**2
        if distance2 < r**2:
            # The points are too close, so pt is not a candidate.
            return False
    # All points tested: if we're here, pt is valid
    return True

def get_point(k, refpt):
    """Try to find a candidate point relative to refpt to emit in the sample.

    We draw up to k points from the annulus of inner radius r, outer radius 2r
    around the reference point, refpt. If none of them are suitable (because
    they're too close to existing points in the sample), return False.
    Otherwise, return the pt.

    """
    i = 0
    while i < k:
        i += 1
        rho = np.sqrt(np.random.uniform(r**2, 4 * r**2))
        theta = np.random.uniform(0, 2*np.pi)
        pt = refpt[0] + rho*np.cos(theta), refpt[1] + rho*np.sin(theta)
        if not (0 <= pt[0] < width and 0 <= pt[1] < height):
            # This point falls outside the domain, so try again.
            continue
        if point_valid(pt):
            return pt

    # We failed to find a suitable point in the vicinity of refpt.
    return False

# Pick a random point to start with.
pt = (np.random.uniform(0, width), np.random.uniform(0, height))
samples = [pt]
# Our first sample is indexed at 0 in the samples list...
cells[get_cell_coords(pt)] = 0
# ... and it is active, in the sense that we're going to look for more points
# in its neighbourhood.
active = [0]

nsamples = 1
# As long as there are points in the active list, keep trying to find samples.
while active:
    # choose a random "reference" point from the active list.
    idx = np.random.choice(active)
    refpt = samples[idx]
    # Try to pick a new point relative to the reference point.
    pt = get_point(k, refpt)
    if pt:
        # Point pt is valid: add it to the samples list and mark it as active
        samples.append(pt)
        nsamples += 1
        active.append(len(samples)-1)
        cells[get_cell_coords(pt)] = len(samples) - 1
    else:
        # We had to give up looking for valid points near refpt, so remove it
        # from the list of "active" points.
        active.remove(idx)

plt.scatter(*zip(*samples), color='r', alpha=0.6, lw=0)
plt.xlim(0, width)
plt.ylim(0, height)
plt.axis('off')
plt.show()
Current rating: 4.4

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Brian Norman 6 years, 10 months ago

I know this is an old topic but it's just what I was looking for to generate a random field and it runs 10x faster than a pure python version I am currently using. You ought to put this on Github if not already (I couldn't find it)

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christian 6 years, 10 months ago

Thank you – I should probably get round to doing this. At the moment my github account is a bit of a graveyard.

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christian 6 years, 8 months ago

I've added the code from this article to my github page now: https://github.com/scipython/scipython_maths/tree/master/poisson_disc_sampled_noise

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Julian 2 years, 6 months ago

Thanks for this article, however we noticed two (minor) things one could possibly improve/do differently here:

1) To sample in the annulus uniformly you'd have to compute 'rho' as:

np.sqrt(np.random.uniform(r*r, 4*r*r))

2) Shortly after that line: when a point falls outside the domain you 'continue' and therefore don't increment 'i' in that iteration. This grants points near the boundary more attempts, but there is no reason to do that.

One could however think about constraining the sample space of boundary points to the part of the annuli that lies in our sample space and then sample with a fraction of k, for example:

When a point lies exactly in a corner, only a quarter of its annulus is inside our rectangle and k/4 sample attempts could suffice - not sure how much of a speedup that would give us, but it goes to show how in the current setting the number of sample attempts should be the same for all points.

I know this is an old topic, but one of the first results that pops up when looking for an implementation of Poisson Disc Sampling, so maybe someone sees and appreciates this.

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christian 2 years, 6 months ago

Hello Julian,
Many thanks for these improvements: the uniform sampling issue I should have caught already! I'll update the GitHub repository too.
Cheers,
Christian

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Rong 1 year, 8 months ago

I am a bit confused why we should use

rho = np.sqrt(np.random.uniform(r*r, 4*r*r))

Why we cannot directly use

rho = np.random.uniform(r, 2*r)

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christian 1 year, 8 months ago

Essentially, because for a fixed angular interval, dθ, the area defined by the interval dθ and dr is not independent of r: it increases the further from the origin you go. Think, e.g. of the area of an annulus (take "dθ" to be 2π) – even if its width is constant, its area increases as its distance from the centre increases. You need to take this into account to sample uniformly across the area of a circle or the "random" points will cluster towards the centre.

You can read more about this here: https://meyavuz.wordpress.com/2018/11/15/generate-uniform-random-points-within-a-circle/ and here: https://stackoverflow.com/questions/5837572/generate-a-random-point-within-a-circle-uniformly

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