# Continuous dispersal on a discrete lattice

Dispersal is a key process in many domains, and particularly in ecology. Individuals move in space, and this movement can be modelled as a random process following some kernel. The dispersal kernel is simply a probability distribution describing the distance travelled in a given time frame. Since space is continuous, it is natural to use a continuous kernel. However, some modelling frameworks are formulated on a lattice, or discrete array of patches.

So how can we implement a continuous kernel in discrete space?

As with many modelling situations, we could approach this in a number of ways.  Here is the implementation that I can up with, and I welcome your suggestions, dear reader, for alternatives or improvements to this approach.

1. Draw a random variate d from the dispersal kernel.
2. Draw a uniform random number θ between 0 and 2π, which we will use to choose a direction.
3. Calculate the relative location (in continuous space) of the dispersed individuals using some trig:
x  = cos(θ) d
y = sin(θ) d
4. Determine the new location on the lattice for each individual by adding the relative x and y positions to the current location. Round these locations to the nearest integer and take the modulo of this integer and the lattice size. This creates a torus out of the lattice such that the outer edges loop back on each other. If you remember the old Donkey Kong games, you can think of this like how when you leave out the right side of the screen you enter from the left.

I implemented this approach in R as a function that takes in a population in a lattice, and returns a lattice with the dispersed population. The user can also specify which dispersal kernel to use. Here is the result using a negative-exponential kernel on a 50×50 lattice. Created by iterating over the dispersal function:

```## General function to take in a lattice and disperse
## according to a user provided dispersal kernel
## Author: Corey Chivers
lat_disp<-function(pop,kernel,...)
{
lattice_size<-dim(pop)
new_pop<-array(0,dim=lattice_size)
for(i in 1:lattice_size)
{
for(j in 1:lattice_size)
{
N<-pop[i,j]
dist<-kernel(N,...)
theta<-runif(N,0,2*pi)
x<-cos(theta)*dist
y<-sin(theta)*dist

for(k in 1:N)
{
x_ind<-(round(i+x[k])-1) %% lattice_size + 1
y_ind<-(round(j+y[k])-1) %% lattice_size + 1
new_pop[x_ind,y_ind]<-new_pop[x_ind,y_ind]+1
}
}
}
return(new_pop)
}
```

For comparison, I also ran the same population using a Gaussian kernel. I defined the parameters of both kernels to have a mean dispersal distance of 1.

Here is the result using a Gaussian kernel: The resulting population after 35 time steps has a smaller range than when using the exponential kernel, highlighting the importance of the shape of the dispersal kernel for spreading populations (remember that in both cases the average dispersal distance is the same).

Code for generating the plots:

```############## Run and plot #######################

## Custom colour ramp
colours<-c('grey','blue','black')
cus_col<-colorRampPalette(colors=colours, space = c("rgb", "Lab"),interpolate = c("linear", "spline"))

## Initialize population array
Time=35
pop<-array(0,dim=c(Time,50,50))
pop[1,25,25]<-10000

### Normal Kernel ###
par(mfrow=c(1,1))
for(i in 2:Time)
{
image(pop[i-1,,],col=cus_col(100),xaxt='n',yaxt='n')
pop[i,,]<-lat_disp(pop[i-1,,],kernel=rnorm,mean=0,sd=1)
}

## Plot
png('normal_kern.png', width = 800, height = 800)
par(mfrow=c(2,2),pty='s',omi=c(0.1,0.1,0.5,0.1),mar=c(2,0,2,0))
times<-c(5,15,25,35)
for(i in times)
image(pop[i-1,,],
col=cus_col(100),
xaxt='n',
yaxt='n',
useRaster=TRUE,
main=paste("Time =",i))

mtext("Gaussian Kernel",outer=TRUE,cex=1.5)
dev.off()

### Exponential Kernel ###
par(mfrow=c(1,1))
for(i in 2:Time)
{
image(pop[i-1,,],col=cus_col(100),xaxt='n',yaxt='n')
pop[i,,]<-lat_disp(pop[i-1,,],kernel=rexp,rate=1)
}

## Plot
png('exp_kern.png',  width = 800, height = 800)
par(mfrow=c(2,2),pty='s',omi=c(0.1,0.1,0.5,0.1),mar=c(2,0,2,0))
times<-c(5,15,25,35)
for(i in times)
image(pop[i-1,,],
col=cus_col(100),
xaxt='n',
yaxt='n',
useRaster=TRUE,
main=paste("Time =",i))

mtext("Exponential Kernel",outer=TRUE,cex=1.5)
dev.off()

############################################################
```

## 12 thoughts on “Continuous dispersal on a discrete lattice”

1. stevencarlislewalker says:

Fun stuff. Instead of donkey kong boundaries, it would seem more biologically reasonable to me to truncate the kernel at the boundary. This would be easy for kernels centered near the center of the lattice with fairly small spread. In these cases jumps outside of the lattice would be rare and rejection sampling ideas would allow you to ignore those. But if the chances of jumping off the edge increased you might need to do something more sophisticated in order to be reasonably efficient (e.g. rejection sampling with a mixture of uniforms as the proposal???).

• Corey Chivers says:

Ya, good call. You could also just increase the lattice size to ensure that dispersal would rarely exeed the boundaries. The reason I like the donkey kong world (it is also kind of like a globe) is that after enough time, the population will become well mixed (a la a cosmopolitan species).

• stevencarlislewalker says:

Ya actually…now that I think about it…truncating isn’t very biologically reasonable either! What time am I living in again? We know that the world isn’t flat right? 8)

2. Alex says:

Nice, but won’t there be an issue with indexing if those modulo operations return 0?

• Corey Chivers says:

Good catch. I’ve edited the code to account for this. There was also a shift error that I have fixed in the same lines:

x_ind<-(round(i+x[k])-1) %% lattice_size + 1
y_ind<-(round(j+y[k])-1) %% lattice_size + 1

Thanks!

3. C. Liam Brown says:

Which part of the code results in the torus-shaped world? I’m not clever enough to find it.

Question the second: Why does the Gaussian kernel result in less spread with the same mean? Does the exponential one have a larger standard deviation?

• Corey Chivers says:

It is when I take the modulus of the x,y locations:

x_ind<-(round(i+x[k])-1) %% lattice_size + 1
y_ind<-(round(j+y[k])-1) %% lattice_size + 1

%% is the modulus operator.

Yes, the difference is totally due to the sd of the two kernels. The average jump length of each kernal is the same, yet the variance in jump lengths is larger for the exponential kernel. This leads to more long distance jumps, and the effect compounds over time.

• C. Liam Brown says:

Cool, thanks!

4. seeddispersal says:

This is great–thank you. I was wondering a further application, however. I’ve only performed statistical modeling, and never any spatial simulations in R. It appears that the kernel is drawn from kernel=rexp. If a user had a custom-fit probability density function (for example, a power exponential: (1/(2*pi*(a^2)))*exp(-((x^b)/(a^b))), with already derived parameter estimates of a and b), would that be straight-forward to implement?

• Corey Chivers says:

As long as you can convert your probability density function (pdf) into an inverse cumulative distribution function (icdf) (ie the inverse of the integral of the pdf), than you’d be good to go. Don’t worry, this isn’t usually that hard. You can then simulate random outcomes from any arbitrary icdf by first generating uniform outcomes with runif(), and then evaluating the icdf at the resulting values(s). Try googling Inverse transform sampling for more info about how to do this.

Good luck!

• seeddispersal says:

I’ll see what I can do with that! Thanks for the direction and quick response!