## Simulating a Poisson point process on a triangle

The title gives it away. But yes, after  two posts about simulating a Poisson point process  on a rectangle and disk, the next shape is a triangle. Useful, right?

Well, I actually had to do this once for a part of something larger. You can divide  polygons, regular or irregular, into triangles, which is often called triangulation, and there is much code that does triangulation.  Using the independence property of the Poisson process, you can then simulate a Poisson point process on each triangle, and you end up with a Poisson point process on a polygon.

But the first step is to do it on a triangle. Consider a general triangle with its three corners labelled $$\textbf{A}$$, $$\textbf{B}$$ and $$\textbf{C}$$. Again, simulating a Poisson point process comes down to the number of points and the locations of points.

### Method

##### Number of points

The number of points of a homogeneous Poisson point process  in any shape with with area $$A$$ is simply a Poisson random variable with mean  $$\lambda A$$, where $$A$$ is the area of the shape. For the triangle’s area, we just uses Herron’s formula, which says that a triangle with sides $$a$$, $$b$$, and $$c$$  has the area $$A=\sqrt{s(s-a)(s-b)(s-c)}$$, where $$s=(a+b+c)/2$$, which means you just need to use Pythagoras’ theorem for  the lengths $$a$$, $$b$$, and $$c$$. Interestingly, this standard formula can be prone to numerical error if the triangle is very thin or needle-shaped. A more secure and stable expression is

$$A=\frac{1}{4}\sqrt{ (a+(b+c)) (c-(a-b)) (c+(a-b)) (a+(b-c)) },$$

where the brackets do matter; see Section 2.5 in the book by Higham.

Of course in MATLAB you can just use the function polyarea or the function with the same name in R.

Now just generate or simulate Poisson random variables with mean (or parameter)  $$\lambda A$$. In MATLAB,  use the poissrnd function with the argument $$\lambda A$$. In R, it is done similarly with the standard  function rpois . In Python, we can use either the scipy.stats.poisson or numpy.random.poisson function from the SciPy or NumPy libraries.

##### Locations of points

We need to position all the points randomly and uniformly on a triangle.  As with the previous two simulation cases, to position a single point $$(x, y)$$, you first need to produce two random uniform variables on the unit interval $$(0,1)$$, say $$U$$ and $$V$$. I’ll denote the $$x$$ and $$y$$ coordinates of point by $$x_{\textbf{A}}$$ and $$y_{\textbf{A}}$$, and similarly for the other two points.  To get the random $$x$$ and $$y$$ values, you use these two formulas:

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

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

Done. A Poisson point process simulated on a triangle .

### Code

I now present some code in MATLAB, R and Python, which you can see are all very similar.  To avoid using a for-loop and employing instead MATLAB’s inbuilt vectorization, I use the dot notation for the product $$\sqrt{U}V$$. In R and Python (using SciPy), that’s done automatically.

MATLAB


%Simulation window parameters -- points A,B,C of a triangle
xA=0;xB=1;xC=1; %x values of three points
yA=0;yB=0;yC=1; %y values of three points

%Point process parameters
lambda=100; %intensity (ie mean density) of the Poisson process

%calculate sides of trinagle
a=sqrt((xA-xB)^2+(yA-yB)^2);
b=sqrt((xB-xC)^2+(yB-yC)^2);
c=sqrt((xC-xA)^2+(yC-yA)^2);
s=(a+b+c)/2; %calculate semi-perimeter

%Use Herron's forumula -- or use polyarea
areaTotal=sqrt(s*(s-a)*(s-b)*(s-c)); %area of triangle

%Simulate Poisson point process
numbPoints=poissrnd(areaTotal*lambda);%Poisson number of points
U=(rand(numbPoints,1));%uniform random variables
V=(rand(numbPoints,1));%uniform random variables

xx=sqrt(U)*xA+sqrt(U).*(1-V)*xB+sqrt(U).*V*xC;%x coordinates of points
yy=sqrt(U)*yA+sqrt(U).*(1-V)*yB+sqrt(U).*V*yC;%y coordinates of points

%Plotting
scatter(xx,yy);
xlabel('x');ylabel('y');


R

Note: it is a bit tricky to write “<-” in the R code (as it automatically changes to the html equivalent in the HTML editor I am using), so I have usually used “=” instead of the usual “<-”.


#Simulation window parameters -- points A,B,C of a triangle
xA=0;xB=1;xC=1; #x values of three points
yA=0;yB=0;yC=1; #y values of three points

#Point process parameters
lambda=100; #intensity (ie mean density) of the Poisson process

#calculate sides of trinagle
a=sqrt((xA-xB)^2+(yA-yB)^2);
b=sqrt((xB-xC)^2+(yB-yC)^2);
c=sqrt((xC-xA)^2+(yC-yA)^2);
s=(a+b+c)/2; #calculate semi-perimeter

#Use Herron's forumula to calculate area
areaTotal=sqrt(s*(s-a)*(s-b)*(s-c)); #area of triangle

#Simulate a Poisson point process
numbPoints=rpois(1,areaTotal*lambda);#Poisson number of points
U=runif(numbPoints);#uniform random variables
V=runif(numbPoints);#uniform random variables

xx=sqrt(U)*xA+sqrt(U)*(1-V)*xB+sqrt(U)*V*xC;#x coordinates of points
yy=sqrt(U)*yA+sqrt(U)*(1-V)*yB+sqrt(U)*V*yC;#y coordinates of points

#Plotting
plot(xx,yy,'p',xlab='x',ylab='y',col='blue');


Simulating a Poisson point process in R is even easier, with the amazing spatial statistics library spatstat. You just need to define the triangular window.

#Simulation window parameters -- points A,B,C of a triangle
xA=0;xB=1;xC=1; #x values of three points
yA=0;yB=0;yC=1; #y values of three points

#Point process parameters
lambda=100; #intensity (ie mean density) of the Poisson process

library("spatstat");
#create triangle window object
windowTriangle=owin(poly=list(x=c(xA,xB,xC), y=c(yA,yB,yC)));
#create Poisson "point pattern" object
ppPoisson=rpoispp(lambda,win=windowTriangle)
plot(ppPoisson); #Plot point pattern object
#retrieve x/y values from point pattern object
xx=ppPoisson$x; yy=ppPoisson$y;


Python

Note: “lambda” is a reserved word in Python (and other languages), so I have used “lambda0” instead.

#import libraries
import numpy as np
import scipy.stats
import matplotlib.pyplot as plt

#Simulation window parameters -- points A,B,C of a triangle
xA=0;xB=1;xC=1; #x values of three points
yA=0;yB=0;yC=1; #y values of three points

#Point process parameters
lambda0=100; #intensity (ie mean density) of the Poisson process

#calculate sides of trinagle
a=np.sqrt((xA-xB)**2+(yA-yB)**2);
b=np.sqrt((xB-xC)**2+(yB-yC)**2);
c=np.sqrt((xC-xA)**2+(yC-yA)**2);
s=(a+b+c)/2; #calculate semi-perimeter

#Use Herron's forumula to calculate area -- or use polyarea
areaTotal=np.sqrt(s*(s-a)*(s-b)*(s-c)); #area of triangle

#Simulate a Poisson point process
numbPoints = scipy.stats.poisson(lambda0*areaTotal).rvs();#Poisson number of points
U = scipy.stats.uniform.rvs(0,1,((numbPoints,1)));#uniform random variables
V= scipy.stats.uniform.rvs(0,1,((numbPoints,1)));#uniform random variables

xx=np.sqrt(U)*xA+np.sqrt(U)*(1-V)*xB+np.sqrt(U)*V*xC;#x coordinates of points
yy=np.sqrt(U)*yA+np.sqrt(U)*(1-V)*yB+np.sqrt(U)*V*yC;#y coordinates of points

#Plotting
plt.scatter(xx,yy, edgecolor='b', facecolor='none', alpha=0.5 );
plt.xlabel("x"); plt.ylabel("y");


### Results

MATLAB

R

Python

The topic of placing a single point uniformly on a general triangle is discussed in this StackExchange post.  For the formulas, it cites the paper “Shape distributions” by Osada, Funkhouser, Chazelle and Dobkin”, where no proof is given.

I originally looked at placing single points in cells of a Dirichlet or Voronoi tesselation — terms vary. There is a lot of literature on this topic, particularly when the seeds of the cells form a Poisson point process. The references in the articles on Wikipedia and MathWorld are good starting points.

### Correction

In a previous version of this blog, there was an error in the two Cartesian formula for randomly simulating a point in a triangle. This has been fixed, but the error never existed in the code.

## Simulating a Poisson point process on a disk

Sometimes one needs to simulate a Poisson point process on a disk. I know I often do. A disk or disc, depending on your spelling preference, is isotropic or rotationally-invariant, so a lot of my simulations of a Poisson point process happen in a circular simulation window when I am considering such a setting. For example, maybe you want to consider a single wireless receiver in a Poisson network of wireless transmitters, which only cares about the distance to a transmitter. Alternatively, maybe you want to randomly sprinkle a virtual cake. What to do? A Poisson point process on a disk is the answer.

I will simulate a Poisson point process with intensity $$\lambda>0$$ on a disk with radius $$r>0$$. The simulation steps are very similar to those in the previous post where I simulated a  homogeneous Poisson point process on a rectangle, and I suggest going back to that post if you are not familiar with the material. The main difference between simulation on a rectangle and a disk is the positioning of the points, but first we’ll look at the number of points.

## Steps

##### Number of points

The number of points of a Poisson point process falling within a circle of radius $$r>0$$ is a Poisson random variable with mean  $$\lambda A$$, where $$A=\pi r^2$$ is the area of the disk. As in the rectangular case, this is the most complicated part of the simulation procedure. But as long as your preferred programming language can produce (pseudo-)random numbers according to a Poisson distribution, you can simulate a homogeneous Poisson point process on a disk.

To do this in MATLAB,  use the poissrnd function with the argument $$\lambda A$$. In R, it is done similarly with the standard  function rpois . In Python, we can use either the scipy.stats.poisson or numpy.random.poisson function from the SciPy or NumPy libraries.

### Locations of points

We need to position all the points randomly and uniformly on a disk. In the case of the rectangle, we worked in Cartesian coordinates. It is then natural that we now work in polar coordinates.  I’ll denote the angular and radial coordinate respectively by $$\theta$$ and $$\rho$$. To generate the random angular (or $$\theta$$) values, we simply produce uniform random variables between zero and one, which is what all standard (pseudo-)random number generators do in programming languages. But we then multiply all these numbers by $$2\pi$$, meaning that all the numbers now fall between $$0$$ and $$2\pi$$.

To generate the random radial (or $$\rho$$) values, a reasonable guess would be to do the same as before and generate uniform random variables between zero and one, and then multiply them by the disk radius $$r$$. But that would be wrong.

Before multiplying uniform random variables by the radius, we must take the square root of all the random numbers. We then multiply them by the radius, generating random variables between $$0$$ and $$r$$. (We must take the square root because the area element of a sector or disk is proportional to the radius squared, and not the radius.) These random numbers do not have a uniform distribution, due to the square root, but in fact their distribution is an example of the triangular distribution, which is defined with three real-valued parameters $$a$$, $$b$$ and $$c$$, and for our case, set $$a=0$$ and $$b=c=r$$.

In summary, if $$U$$ and $$V$$ are two independent uniform random variables on $$(0,1)$$, then random point located uniformly on a disk of radius $$r$$ has the polar coordinates $$(r\sqrt(U), 2\pi V)$$.

From polar to Cartesian coordinates

That’s it. We have generated polar coordinates for points randomly and uniformly located in a disk. But to plot the points, often we have to convert coordinates back to Cartesian form. This is easily done in MATLAB with the pol2cart function. In languages without such a function, trigonometry comes to the rescue: $$x=\rho\cos(\theta)$$ and $$y=\rho\sin(\theta)$$.

Equal x and y axes

Sometimes the plotted points more resemble points on an ellipse than a disk due to the different scaling of the x and y axes. To fix this in MATLAB, run the command: axis square. In Python, set axis(‘equal’) in your plot; see this page for a demonstration.

## Code

I’ll now give some code in MATLAB, R and Python, which you can see are all very similar

##### MATLAB
%Simulation window parameters
xx0=0; yy0=0; %centre of disk

areaTotal=pi*r^2; %area of disk

%Point process parameters
lambda=100; %intensity (ie mean density) of the Poisson process

%Simulate Poisson point process
numbPoints=poissrnd(areaTotal*lambda);%Poisson number of points
theta=2*pi*(rand(numbPoints,1)); %angular coordinates

%Convert from polar to Cartesian coordinates
[xx,yy]=pol2cart(theta,rho); %x/y coordinates of Poisson points

%Shift centre of disk to (xx0,yy0)
xx=xx+xx0;
yy=yy+yy0;

%Plotting
scatter(xx,yy);
xlabel('x');ylabel('y');
axis square;

##### R

Note: it is a bit tricky to write “<-” in the R code (as it automatically changes to the html equivalent in the HTML editor I am using), so I have usually used “=” instead of the usual “<-”.

#Simulation window parameters
xx0=0; yy0=0; #centre of disk

areaTotal=pi*r^2; #area of disk

#Point process parameters
lambda=100; #intensity (ie mean density) of the Poisson process

#Simulate Poisson point process
numbPoints=rpois(1,areaTotal*lambda);#Poisson number of points
theta=2*pi*runif(numbPoints);#angular  of Poisson points

#Convert from polar to Cartesian coordinates
xx=rho*cos(theta);
yy=rho*sin(theta);

#Shift centre of disk to (xx0,yy0)
xx=xx+xx0;
yy=yy+yy0;

#Plotting
par(pty="s")
plot(xx,yy,'p',xlab='x',ylab='y',col='blue');


Of course, with the amazing spatial statistics library spatstat, simulating a Poisson point process in R is even easier.

library("spatstat"); #load spatial statistics library
#create Poisson "point pattern" object
plot(ppPoisson); #Plot point pattern object
#retrieve x/y values from point pattern object
xx=ppPoisson$x; yy=ppPoisson$y;


Actually, you can even do it all in two lines: one for loading the spatstat library and one for creating and plotting the point pattern object.

##### Python

Note: “lambda” is a reserved word in Python (and other languages), so I have used “lambda0” instead.

import numpy as np; #NumPy package for arrays, random number generation, etc
import matplotlib.pyplot as plt #for plotting

#Simulation window parameters
xx0=0; yy0=0; #centre of disk
areaTotal=np.pi*r**2; #area of disk

#Point process parameters
lambda0=100; #intensity (ie mean density) of the Poisson process

#Simulate Poisson point process
numbPoints = np.random.poisson(lambda0*areaTotal);#Poisson number of points
theta=2*np.pi*np.random.uniform(0,1,numbPoints); #angular coordinates

#Convert from polar to Cartesian coordinates
xx = rho * np.cos(theta);
yy = rho * np.sin(theta);

#Shift centre of disk to (xx0,yy0)
xx=xx+xx0; yy=yy+yy0;

#Plotting
plt.scatter(xx,yy, edgecolor='b', facecolor='none', alpha=0.5 );
plt.xlabel("x"); plt.ylabel("y");
plt.axis('equal');

##### Julia

After writing this post, I later wrote the code in Julia. The code is here and my thoughts about Julia are here.

## Results

##### Python

The third edition of the classic book Stochastic Geometry and its Applications by Chiu, Stoyan, Kendall and Mecke details on page 54 how to uniformly place points on a disk, which they call the radial way. The same simulation section appears in the previous edition by Stoyan, Kendall and Mecke (Chiu didn’t appear as an author until the current edition), though these books in general have little material on simulation methods. There is the book Spatial Point Patterns: Methodology and Applications with R written by spatial statistics experts  Baddeley, Rubak and Turner, which covers the spatial statistics (and point process simulation) R-package spatstat.

## Simulating a homogeneous Poisson point process on a rectangle

This is the first of a series of posts about simulating Poisson point processes. I’ll start with arguably the simplest Poisson point process on two-dimensional space, which is the homogeneous one defined on a rectangle. Let’s say that we we want to simulate a Poisson point process with intensity $$\lambda>0$$ on a (bounded) rectangular region, for example, the rectangle $$[0,w]\times[0,h]$$ with dimensions $$w>0$$ and $$h>0$$ and area $$A=wh$$. We assume for now that the bottom left corner of the rectangle is at the origin.

## Steps

##### Number of points

The number of points in the rectangle  $$[0,w]\times[0,h]$$ is a Poisson random variable with mean $$\lambda A$$. In other words, this random variable is distributed according to the Poisson distribution with parameter $$\lambda A$$, and not just $$\lambda$$, because the number of points depends on the size of the simulation region.

This is the most complicated part of the simulation procedure. As long as your preferred programming language can produce (pseudo-)random numbers according to a Poisson distribution, you can simulate a homogeneous Poisson point process. There’s a couple of different ways used to simulate Poisson random variables, but we will skip the details. In MATLAB, it is done by using the poissrnd function with the argument $$\lambda A$$. In R, it is done similarly with the standard  function rpois . In Python, we can use either the scipy.stats.poisson or numpy.random.poisson function from the SciPy or NumPy libraries.

##### Location of points

The points now need to be positioned randomly, which is done by using Cartesian coordinates. For a homogeneous Poisson point process, the $$x$$ and $$y$$ coordinates of each point are independent uniform points, which is also the case for the binomial point process, covered in a previous post. For the rectangle $$[0,w]\times[0,h]$$, the $$x$$ coordinates are uniformly sampled on the interval $$[0,w]$$, and similarly for the $$y$$ coordinates. If the bottom left corner of rectangle is located at the point $$(x_0,y_0)$$, then we just have to shift the random $$x$$ and $$y$$ coordinates by respectively adding $$x_0$$ and $$y_0$$.

Every scientific programming language has a random uniform number generator because it is the default random number generator. In MATLAB, R and SciPy, it is respectively rand, runif and scipy.stats.uniform.

## Code

Here is some code that I wrote for simulating a homogeneous Poisson point process on a rectangle. You will notice that in all the code samples the part that simulates the Poisson point process requires only three lines of code: one line for the number of points and two lines lines for the $$x$$ and $$y$$ coordinates  of the points.

##### MATLAB
%Simulation window parameters
xMin=0;xMax=1;
yMin=0;yMax=1;
xDelta=xMax-xMin;yDelta=yMax-yMin; %rectangle dimensions
areaTotal=xDelta*yDelta;

%Point process parameters
lambda=100; %intensity (ie mean density) of the Poisson process

%Simulate Poisson point process
numbPoints=poissrnd(areaTotal*lambda);%Poisson number of points
xx=xDelta*(rand(numbPoints,1))+xMin;%x coordinates of Poisson points
yy=yDelta*(rand(numbPoints,1))+yMin;%y coordinates of Poisson points

%Plotting
scatter(xx,yy);
xlabel('x');ylabel('y');

##### R

Note: it is a bit tricky to write “<-” in the R code (as it automatically changes to the html equivalent in the HTML editor I am using), so I have usually used “=” instead of the usual “<-”.

#Simulation window parameters
xMin=0;xMax=1;
yMin=0;yMax=1;
xDelta=xMax-xMin;yDelta=yMax-yMin; #rectangle dimensions
areaTotal=xDelta*yDelta;

#Point process parameters
lambda=100; #intensity (ie mean density) of the Poisson process

#Simulate Poisson point process
numbPoints=rpois(1,areaTotal*lambda);#Poisson number of points
xx=xDelta*runif(numbPoints)+xMin;#x coordinates of Poisson points
yy=yDelta*runif(numbPoints)+yMin;#y coordinates of Poisson points

#Plotting
plot(xx,yy,'p',xlab='x',ylab='y',col='blue');

##### Python

Note: “lambda” is a reserved word in Python (and other languages), so I have used “lambda0” instead.

import numpy as np
import scipy.stats
import matplotlib.pyplot as plt

#Simulation window parameters
xMin=0;xMax=1;
yMin=0;yMax=1;
xDelta=xMax-xMin;yDelta=yMax-yMin; #rectangle dimensions
areaTotal=xDelta*yDelta;

#Point process parameters
lambda0=100; #intensity (ie mean density) of the Poisson process

#Simulate Poisson point process
numbPoints = scipy.stats.poisson( lambda0*areaTotal ).rvs()#Poisson number of points
xx = xDelta*scipy.stats.uniform.rvs(0,1,((numbPoints,1)))+xMin#x coordinates of Poisson points
yy = yDelta*scipy.stats.uniform.rvs(0,1,((numbPoints,1)))+yMin#y coordinates of Poisson points
#Plotting
plt.scatter(xx,yy, edgecolor='b', facecolor='none', alpha=0.5 )
plt.xlabel("x"); plt.ylabel("y")

##### Julia

After writing this post, I later wrote the code in Julia. The code is here and my thoughts about Julia are here.

## Higher dimensions

If you want to simulate a Poisson point process in a three-dimensional box (typically called a cuboid or rectangular prism), you just need two modifications.

For a box $$[0,w]\times[0,h]\times[0,\ell]$$, the number of points now a Poisson random variable with mean $$\lambda V$$, where $$V= wh\ell$$ is the volume of the box. (For higher dimensions, you need to use $n$-dimensional volume.)

To position the points in the box, you just need an additional uniform variable for the extra coordinate. In other words, the $$x$$, $$y$$ and $$z$$ coordinates are uniformly and independently sampled on the respective intervals $$[0,w]$$, $$[0,h]$$, $$[0,\ell]$$. The more dimensions, the more uniform random variables.

And that’s it.