Placing a random point uniformly in a Voronoi cell

In the previous post, I discussed how Voronoi or Dirichlet tesselations are useful and how they can be calculated or estimated with many scientific programming languages by using standard libraries usually based on Qhull. The cells of such a tessellation divide the underlying space. Now I want to randomly place a single point in a uniform manner in each bounded Voronoi cell.

But why?

Well, this task arises occasionally, particularly when developing mathematical models of wireless networks, such as mobile or cellular phone networks. A former colleague of mine had to do this, which inspired me to write some MATLAB code a couple of years ago. And I’ve seen the question posed a couple of times on the web . So I thought: I can do that.

Overview

For this problem, I see two obvious methods.

Simple but crude

The simplest method is to cover each Voronoi cell with a rectangle or disk. Then randomly place a point uniformly on the rectangle or disk. If it doesn’t randomly land inside the rectangle or disk, then do it again until it does. Crude, slightly inefficient, but simple.

Elegant but slightly tricky

A more elegant way, which I will implement, is to partition (or divide) each Voronoi cell into triangles. Then randomly choose a triangle based on its area and, finally, uniformly position a point on that triangle.

Method

Partition cells into triangles

It is straightforward to divide a Voronoi cell into triangles. Each side of the cell corresponds to one side of a triangle (that is, two points). The third point of the triangle is the original point corresponding to the Voronoi cell.

Choose a triangle

Randomly choosing a triangle is also easy. For a given cell, number the triangles. Which random triangle is chosen is simply a discrete random variable whose probability mass function is formed from triangle areas normalized (or divided) by the total area of the Voronoi cell. In other words, the probability of randomly choosing triangle \(i\) with area \(A_i\) from \(m\) triangles is simply

\(P_i=\frac{A_i}{\sum_{j=1}^m A_j}.\)

To calculate the area of the triangles, I use the shoelace formula , which for a triangle with corners labelled \(\textbf{A}\), \(\textbf{B}\) and \(\textbf{C}\) gives

\(A= \frac{1}{2} |(x_{\textbf{B}}-x_{\textbf{A}})(y_{\textbf{C}}-y_{\textbf{A}})-(x_{\textbf{C}}-x_{\textbf{A}})(y_{\textbf{B}}-y_{\textbf{A}})|.\)

But you can also use Herron’s formula.

Then the random variable is sampled using the probabilities based on the triangle areas.

Place point on chosen triangle

Given a triangle, the final step is also easy, if you know how, which is often the case in mathematics. I have already covered this step in a previous post, but I’ll give some details here.

To position a single point uniformly in the triangle, generate two random uniform variables on the unit interval \((0,1)\), say \(U\) and \(V\). The random \(X\) and \(Y\) coordinates of the single point are given by the expressions:

\(X=\sqrt{U} x_{\textbf{A}}+\sqrt{U}(1-V x_{\textbf{B}})+\sqrt{U}V x_{\textbf{C}}\)

\(Y=\sqrt{U} y_{\textbf{A}}+\sqrt{U}(1-V y_{\textbf{B}})+\sqrt{U}V y_{\textbf{C}}\)

Results

The blue points are the points of the underlying point pattern that was used to generate the Voronoi tesselation. (These points have also been arbitrarily numbered.) The red points are the random points that have been uniformly placed in all the bounded Voronoi cells.

MATLAB

Python

Empirical validation

We can empirically validate that the points are being placed uniformly on the bounded Voronoi cells. For a given (that is, non-random) Voronoi cell, we can repeatedly place (or sample) a random point uniformly in the cell. Increasing the number of randomly placed points, the respective averages of the \(x\) and \(y\) coordinates of the points will converge to the centroids (or geometric centres) of the Voronoi cell, which can be calculated with simple formulas.

Code

The code for all my posts is located online here. For this post, the code in MATLAB and Python is here.

I have also written in MATLAB and Python the code as functions (in files funVoronoiUniform.m and funVoronoiUniform.py, respectively), so people can use it more easily. The function files are located here, where I have also included an implementation of the aforementioned empirical test involving centroids. You should be able to use those functions and for any two-dimensional point patterns.

Further reading

Much has been written on Voronoi or Dirichlet tessellations, particularly when the seeds of the cells form a Poisson point process. For references, I recommend my previous post, where I say that the references in the articles on Wikipedia and MathWorld are good starting points.

In this StackExchange post, people discuss how to place a single point uniformly on a general triangle. For the formulas, the thread cites the paper Shape distributions by Osada, Funkhouser, Chazelle and Dobkin, where no proof is given.

Simulating a Cox point process based on a Poisson line process

In the previous post, I described how to simulate a Poisson line process, which in turn was done by using insight from an earlier post on the Bertrand paradox.

Now, given a Poisson line process, for each line, if we generate an independent one-dimensional Poisson point point process on each line, then we obtain an example of a Cox point process. Cox point processes are also known as doubly stochastic Poisson point processes. On the topic of names, Guttorp and Thorarinsdottir argue that it should be called the Quenouille point process, as Maurice Quenouille introduced an example of it before Sir David Cox, but I opt for the more common name.

Cox point proceesses

A Cox point process is a generalization of a Poisson point process. It is created by first considering a non-negative random measure, sometimes called a driving measure. Then a Poisson point process, which is independent of the random driving measure, is generated by using the random measure as its intensity or mean measure.

The driving measure of a Cox point process can be, for example, a non-negative random variable or field multiplied by a Lebesgue measure. In our case, the random measure is the underlying Poisson line process coupled with the Lebesgue measure on the line (that is, length).

Cox processes form a very large and general family of point processes, which exhibit clustering. In previous posts, I have covered two special cases of Cox point processes: the Matérn and Thomas cluster point processes. These are, more specifically, examples of a Neyman-Scott point process, which is a special case of a shot noise Cox point process. These two point processes are fairly easy to simulate, but that’s not the case for Cox point processes in general. Some are considerably easier than others.

Motivation

I will focus on simulating the Cox point process formed from a Poisson line process with homogeneous Poisson point processes. I do this for two reasons. First, it’s easy to simulate, given we can simulate a Poisson line process. Second, it has been used and studied recently in the mathematics and engineering literature for investigating wireless communication networks in cities, where the streets correspond to Poisson lines; for example, see these two preprints:

  1. Continuum percolation for Cox point processes
  2. Poisson Cox Point Processes for Vehicular Networks

Incidentally, I don’t know what to call this particular Cox point process. A Cox line-point process? A Cox-Poisson line-point process? But it doesn’t matter for simulation purposes.

Method

We will simulate the Cox (-Poisson line-) point process on a disk. Why a disk? I suggest reading the previous posts on the Poisson line process the Bertrand paradox for why the disk is a natural simulation window for line processes.

Provided we can simulate a Poisson line process, the simulation method is quite straightforward, as I have essentially already described it.

Line process

First simulate a Poisson line process on a disk. We recall that for each line of the line process, we need to generate two independent random variables \(\Theta\) and \(P\) describing the position of the line. The first random variable \(\Theta\) gives the line orientation, and it is a uniform random variable on the interval \((0,2\pi)\).

The second random variables \(P\) gives the distance from the origin to the disk edge, and it is a uniform random variable on the interval \((0,r)\), where \(r\) is the radius of the disk. The distance from the point \((\Theta, P)\) to the disk edge (that is, the circle) along the chord is:

$$Q=\sqrt{r^2-P^2}.$$

The endpoints of the chord (that is, the points on the disk edge) are then:

Point 1: \(X_1=P \cos \Theta+ Q\sin \Theta\), \(Y_1= P \sin \Theta- Q\cos \Theta\),

Point 2: \(X_2=P \cos \Theta- Q\sin \Theta\), \(Y_2= P \sin \Theta+Q \cos \Theta\).

The length of the line segment is \(2 Q\). We can say this random line is described by the point \((\Theta,P)\).

One-dimensional Poisson point process

For each line (segment) in the line process, simulate a one-dimensional Poisson point process on it. Although I have never discussed how to simulate a one-dimensional (homogeneous) Poisson point process, it’s essentially one dimension less than simulating a homogeneous Poisson point process on a rectangle.

More specifically, given a line segment \((\Theta,P)=(q,\theta)\), you simulate a homogeneous Poisson point process with intensity \(\mu\) on a line segment with length \(2 q\), where \(q=\sqrt{r^2-p^2}\). (I am now using lowercase letters to stress that the line is no longer random.) To simulate the homogeneous Poisson point process, you generate a Poisson random variable with parameter \(2 \mu q\).

Now you need to place the points uniformly on the line segment. To do this, consider a single point on a single line. For this point, generate a single uniform variable \(U\) on the interval \((-1,1)\). The tricky part is now getting the Cartesian coordinates right. But the above expressions for the endpoints suggest that the single random point has the Cartesian coordinates:

\(x=p \cos \Theta+ U q\sin \theta\), \(y=p \sin \theta- U q\cos \theta\).

The two extreme cases of the uniform random variable \(U\) (that is, \(U=-1\) and \(U=1\)) correspond to the two endpoints of the line segment. We recall that \(Q\) is the distance from the midpoint of the line segment to the disk edge along the line segment, so it makes sense that we want to vary this distance uniformly in order to uniformly place a point on the line segment. This uniform placement step is done for all the points of the homogeneous Point process on that line segment.

You repeat this procedure for every line segment. And that’s it: a Cox point process built upon a Poisson line process.

Results

MATLAB

R

Python

Code

As always, the code from all my posts is online. For this post, I have written the code in MATLAB, R and Python.

Further reading

For the first step, the reading material is basically the same as that for the Poisson line process, which overlaps with that of the Bertrand paradox. For the one-dimensional Poisson point process, we can use the reading material on the homogeneous Poisson point process on a rectangle.

For general Cox point processes, I recommend starting with the following: Chapter 6 in the monograph Poisson processes by Kingman; Chapter 5 in Statistical Inference and Simulation for Spatial Point Processes by Møller and Waagepetersen; and Section 5.2 in Stochastic Geometry and its Applications by Chiu, Stoyan, Kendall and Mecke. For a much more mathematical treatment, see Chapter 13 in Lectures on the Poisson Process by Last and Penrose.

For this particularly Cox point process, see the two aforementioned preprints, located here and here.

Simulating a Poisson line process

Instead of points, we can consider other objects randomly scattered on some underlying mathematical space. If we take a Poisson point process, then we can use (infinitely long) straight lines instead of points, giving a Poisson line process. Researchers have studied and used this random object to model physical phenomena. In this post I’ll cover how to simulate a homogeneous Poisson line process in MATLAB, R and Python. The code which can be downloaded from here

Overview

For simulating a Poisson line process, the key question is how to randomly position the lines.  This is related to a classic problem in probability theory called the Bertrand paradox.  I discussed this illustration in probability in a previous post, where I also included code for simulating it. I highly recommend reading that post first.

The Bertrand paradox involves three methods for positioning random lines. We use Method 2 to achieve a uniform positioning of lines, meaning the number of lines and orientation of lines is statistically uniform. Then it also makes sense to use this method for simulating a homogeneous (or uniform) Poisson line process.  

We can interpret a line process as a point process. For a line process on the plane \(\textbf{R}^2\), it can be described by a point process on \((0,\infty)\times (0,2\pi)\), which is an an infinitely long cylinder. In other words, the Poisson line process can be described as a Poisson point process.

For simulating a Poisson line process, it turns out the disk is the most natural setting. (Again, this goes back to the Bertrand paradox.) More specifically, how the (infinitely long) lines intersect a disk of a fixed radius \(r>0\). The problem of simulating a Poisson line process reduces to randomly placing chords in a disk. For other simulation windows in the plane, we can always bound any non-circular region with a sufficiently large disk.

Steps

Number of lines

Of course, with all things Poisson, the number of lines will be  a Poisson random variable, but what will its parameter be? This homogeneous (or uniform) Poisson line process forms a one-dimensional homogeneous (or uniform) Poisson point process around the edge of the disk with a circumference \(2 \pi r \). Then the number of lines is simply a Poisson variable with parameter \(\lambda 2 \pi r \).

Locations of points

To position a single line uniformly in a disk, we need to generate two uniform random variables. One random variable gives the angle describing orientation of the line, so it’s a uniform random variable \(\Theta\) on the interval \((0,2\pi)\). 

The other random variable gives the distance from the origin to the disk edge, meaning it’s a uniform random variable \(P\) on the interval \((0,r)\), where \(r\) is the radius of the disk.  The random radius and its perpendicular chord create a right-angle triangle.  The distance from the point \((\Theta, P)\) to the disk edge (that is, the circle) along the chord is:

$$Q=\sqrt{r^2-P^2}.$$

The endpoints of the chord (that is, the points on the disk edge) are then:

Point 1: \(X_1=P \cos \Theta+ Q\sin \Theta\), \(Y_1= P \sin \Theta- Q\cos \Theta\),

Point 2: \(X_2=P \cos \Theta- Q\sin \Theta\), \(Y_2= P \sin \Theta+Q \cos \Theta\).

Code

I have implemented the simulation procedure in MATLAB, R and Python, which, as usual, are all very similar. I haven’t put my code here, because the software behind my website keeps mangling it.  As always, I have uploaded my code to a repository; for this post, it’s in this directory.

I have written the code in R, but I wouldn’t use it in general. That’s because if you’re using R, then, as I have said before, I strongly recommend using the powerful spatial statistics library spatstat. For a simulating Poisson line process, there’s the function rpoisline.  

The chief author of spatstat, Adrian Baddeley, has written various lectures and books on the related topics of point processes, spatial statistics, and geometric probability. In this post, he answered why the angular coordinates have to be chosen uniformly. 

Results

MATLAB

R

Python

Further reading

To read about the Poisson line process, it’s best to start with the Bertrand problem, which is covered in many works on geometric probability and related fields. A good and recent introduction is given by Calka in (Section 1.3) of the lectures titled Stochastic Geometry: Modern Research Frontiers, which were edited by Coupier and published by Springer.  Also see, for example, problem 1.2 in Geometrical Probability by Kendall and Moran or page 44 in Geometric Probability by Solomon.  

For the Poisson line process, I highly recommend Section 7.2 in the monograph Poisson processes by Kingman. Also see Example 8.2 in the standard textbook Stochastic Geometry and its Applications by Chiu, Stoyan, Kendall and Mecke. The Poisson line process book also appears in Exercise 5.2 in the book Stochastic Simulation – Algorithms and Analysis by  Asmussen and Glynn. 

For online resources, this set of lectures by Christian Lantuéjoul
covers the Poisson line process. Wilfrid Kendall covers the Poisson line process in this talk in relation to so-called Poisson cities. 

The Bertrand paradox

Mathematical paradoxes are results or observations in mathematics that are (seemingly) conflicting, unintuitive, incomprehensible, or just plain bizarre. They come in different flavours, such as those that play with notions of infinity, which means they often make little or no sense in a physical world. Other paradoxes, particularly those in probability, serve as a lesson that the problem needs to be posed in a precise manner. The Bertrand paradox is one of these.

Joseph Bertrand posed the original problem in his 1889 book Calcul des probabilités, which is available online (albeit in French). It’s a great illustrative problem involving simple probability and geometry, so it often appears in literature on the (closely related) mathematical fields of geometric probability and integral geometry.

Based on constructing a random chord in a circle, the paradox involves a single mathematical problem with three reasonable but different solutions. It’s less a paradox and more a cautionary tale. It’s really asking the question: What do you mean by random?

Consequently, over the years the Bertrand paradox has inspired debate, with papers arguing what the true solution is. I recently discovered it has even inspired some passionate remarks on the internet; read the comments at www.bertrands-paradox.com.

But I am less interested in the different interpretations or philosophies of the problem. Rather, I want to simulate the three solutions. This is not very difficult, provided some trigonometry and knowledge from a previous post, where I describe how to simulate a (homogeneous) Poisson point process on the disk.

I won’t try to give a thorough analysis of the solutions, as there are much better websites doing that. For example, this MIT website gives a colourful explanation with pizza and fire-breathing monsters. The Wikipedia article also gives a detailed though less creative explanation for the three solutions.

My final code in MATLAB, R and Python code is located here.

The Problem

Bertrand considered a circle with an equilateral triangle inscribed it. If a chord in the circle is randomly chosen, what is the probability that the chord is longer than a side of the equilateral triangle?

The Solution(s)

Bertrand argued that there are three natural but different methods to randomly choose a chord, giving three distinct answers. (Of course, there are other methods, but these are arguably not the natural ones that first come to mind.)

Method 1: Random endpoints

On the circumference of the circle two points are randomly (that is, uniformly and independently) chosen, which are then used as the two endpoints of the chord.

The probability of this random chord being longer than a side of the triangle is one third.

Method 2: Random radius

A radius of the circle is randomly chosen (so the angle is chosen uniformly), then a point is randomly (also uniformly) chosen along the radius, and then a chord is constructed at this point so it is perpendicular to the radius.

The probability of this random chord being longer than a side of the triangle is one half.

Method 3: Random midpoint

A point is randomly (so uniformly) chosen in the circle, which is used as the midpoint of the chord, and the chord is randomly (also uniformly) rotated.

The probability of this random chord being longer than a side of the triangle is one quarter.

Simulation

All three answers involve randomly and independently sampling two random variables, and then doing some simple trigonometry. The setting naturally inspires the use of polar coordinates. I assume the circle has a radius \(r\) and a centre at the point origin \(o\). I’ll number the end points one and two.

In all three solutions we need to generate uniform random variables on the interval \((0, 2\pi)\) to simulate random angles. I have already done this a couple of times in previous posts such as this one.

Method 1: Random endpoints

This is probably the most straightforward solution to simulate. We just need to simulate two uniform random variables \(\Theta_1\) and \(\Theta_2\) on the interval \((0, 2\pi)\) to describe the angles of the two points.

The end points of the chord (in Cartesian coordinates) are then simply:

Point 1: \(X_1=r \cos \Theta_1\), \(Y_1=r \sin \Theta_1\),

Point 2: \(X_2=r \cos \Theta_2\), \(Y_2=r \sin \Theta_2\).

Method 2: Random radius

This method also involves generating two uniform random variables. One random variable \(\Theta\) is for the angle, while the other \(P\) is the random radius, which means generating the random variable \(P\) on the interval \((0, r)\).

I won’t go into the trigonometry, but the random radius and its perpendicular chord create a right-angle triangle. The distance from the point \((\Theta, P)\) to the circle along the chord is:

$$Q=\sqrt{r^2-P^2}.$$

The endpoints of the chord are then:

Point 1: \(X_1=P \cos \Theta+ Q\sin \Theta\), \(Y_1= P \sin \Theta- Q\cos \Theta\),

Point 2: \(X_2=P \cos \Theta- Q\sin \Theta\), \(Y_2= P \sin \Theta+Q \cos \Theta\).

Take note of the signs in these expressions.

Method 3: Random midpoint

This method requires placing a point uniformly on a disk, which is also done when simulating a homogeneous Poisson point process on a disk, and requires two random variables \(\Theta’\) and \(P’\). Again, the angular random variable \(\Theta’\) is uniform.

The other random variable \(P’\) is not uniform. For \(P’\), we generate a random uniform variable on the unit interval \((0,1)\), and then we take the square root of it. We then multiply it by the radius, generating a random variable between \(0\) and \(r\). (We must take the square root because the area element of a sector is proportional to the radius squared, and not the radius.) The distribution of this random variable is an example of the triangular distribution.

The same trigonometry from Method 2 applies here, which gives the endpoints of the chord as:

Point 1: \(X_1=P’ \cos \Theta’+ Q’\sin \Theta’\), \(Y_1= P’ \sin \Theta’- Q’\cos \Theta’\),

Point 2: \(X_2=P’\cos \Theta’- Q’\sin \Theta’\), \(Y_2= P’\sin \Theta’+Q’ \cos \Theta’\),

where \(Q’:=\sqrt{r^2-{P’}^2}\). Again, take note of the signs in these expressions.

Results

To illustrate how the three solutions are different, I’ve plotted a hundred random line segments and their midpoints side by side. Similar plots are in the Wikipedia article.

Method 1: Random endpoints
Method 2: Random radius

Method 3: Random midpoint

Conclusion

For the chord midpoints, we know and can see that Method 3 gives uniform points, while Method 2 has a concentration of midpoints around the circle centre. Method 1 gives results that seem to somewhere between Method 2 and 3 in terms of clustering around the circle centre.

For the chords, we see that Method 3 results in fewer chords passing through the circle centre. Methods 1 and 2 seem to give a similar number of lines passing through this central region.

It’s perhaps hard to see, but it can be shown that Method 2 gives the most uniform results. By this, I mean that the number of lines and their orientations statistically do not vary in different regions of the circle.

We can now position random lines in uniform manner. All we need now is a Poisson number of lines to generate something known as a Poisson line process, which will be the focus of the next post.

Further reading

I’ve already mentioned that there are some good websites on the topic of the Bertrand paradox. For example:

www.bertrands-paradox.com

web.mit.edu/tee/www/bertrand

www.cut-the-knot.org/bertrand.shtml

mathworld.wolfram.com/BertrandsProblem.html

Various authors have mentioned or discussed the Bertrand paradox in books on the related subjects of geometric probability, integral geometry and stochastic geometry. A good and recent introduction is given by Calka in Section 1.3 of the published lectures Stochastic Geometry: Modern Research Frontiers.

Other classic books that cover the topic including, for example, see Problem 1.2 in Geometrical Probability by Kendall and Moran. (Despite Maurice G. Kendall writing a book on geometric probability, he was not related to stochastic geometry pioneer David G Kendall.) It’s also discussed on page 44 in Geometric Probability by Solomon. For a book that involves more advance knowledge of geometry and (abstract) algebra, see Chapter 3 in Integral Geometry and Geometric Probability by Santaló.

The Bertrand paradox is also in The Pleasures of Probability by Isaac. It’s covered in a non-mathematical way in the book Paradoxes from A to Z by Clark. Edwin Jaynes studied the problem and proposed a solution in a somewhat famous 1973 paper, titled The Well-Posed Problem.

The original problem can be read in French in Bertrand’s work, which is available online here or here (starting at the bottom of page 4).

Code

The MATLAB, R and Python code can be found here. In the code, I have labelled the methods A, B and C instead 1, 2 and 3.