# How to choose your next holidays destination – Uniform distribution on a sphere

**ProbaPerception**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

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If you want to choose randomly your next holidays destination, you are likely to process in a way which is certainly biased. Especially if you choose randomly the latitude and the longitude. A bit like they do in this lovely advertising (For those of you who do not speak French, this is about a couple who have won the national gamble prize and have to decide their next travel. The husband randomly picks Australia and the wife is complaining : “Not again!”). So let me help you to choose your next travel!

If we were able to generate uniformly distributed variables on [0 ; 1], we could easily generate variables on other spaces such as [0 ; 1] x [0 ; 1], which is a square of side 1. It can be simply done by using two independant variables X1 and X2, uniformly distributed on [0 ; 1], and by considering X = (X1 , X2). Then X is a random variable evenly distributed on the square [0 ; 1] x [0 ; 1]

However, this generation may be a bit more complicated if we work with more complicated shapes, such as a sphere for example. Indeed, for a sphere of a given radius, say R = 1, it is possible to project the variables of a square on the sphere.

The first method I thought of, is wrong. I was tempted to generate two variables Y1 and Y2 where Y1 follows a uniform on [0 ; Pi] and Y2 follows a uniform on [-Pi ; Pi]. In other words :

– Y1 = Pi * X1

– Y2 = 2Pi * (X2 – 0.5).

Then I wanted to consider (Y1, Y2) as the spherical coordinated of the sphere.

However, this method does NOT generate a uniform on the sphere. Indeed, it has a tendency to over generate north and south poles. The reason is simple, this method generates, in average the same amount of variable in each latitude. North and South poles are of smaller area than Equator. Therefore, the closer we are from the Equator latitude the less variables there are.

On the following graph, we observe a high density of point in the “middle” of the sphere, this is the North Pole. It shows that this method does not offer a uniform distribution. By the way, the graph is computed with the library rgl, which provides a display device for 3 dimensions objects. And then a little code provided in the rgl help allows to move automatically the device and take a snapshot for each view. You can eventually generate a GIF on gifmake (like for the map in Spatial segregation in cities – An explanation by a neural network model).

A wrong method to generate evenly distributed variables on the sphere. The North Pole and the South Pole are over represented. |

Many methods have been proposed to generate evenly distributed random variables on a sphere. We propose one of them here. We consider the couple z = (u,v) defined as :

– u = 2 * Pi * X1

– v = arc-cos(2 * (X2 – 0.5))

In this case, a theorem shows that z = (u , v) generates evenly distributed variables. It can be observed on the next graph. There is no irregularity in the distribution of the random variables.

A correct simulation of a uniform distribution on a sphere. There is no over represented area. |

Other methods exist. I like this one since it is really simple and uses a uniform distribution at the beginning. The idea of this post is to show that generating a uniform distribution can be adapted to many shapes and cases. However, to do so, a previous analytical study has to be done to find the correct transformation.

So this method would help you to avoid being too many times in the chilly places such as North Pole and South Pole since it does not overrepresent the extreme latitudes.

**The program (R) : **

# import package to plot in 3D

install.packages(“rgl”, dependencies = TRUE)

library(rgl)

################################################################

# Uniform distribution in a square

################################################################

size = 10000

x1 = runif(size)

x2 = runif(size)

# the option pch = ‘.’ change the symbol for the graph into dot.

# cex = 2 doubles the size of the dots

plot(x1,x2, col = ‘blue’, pch = ‘.’, cex = 2)

################################################################

# Wrong solution for the sphere

################################################################

y1 = pi * x1

y2 = 2* pi * (x2-0.5)

y = matrix(0, nrow = 2, ncol = size)

y[1,] = y1

y[2,] = y2

plot(y1, y2)

# This function transform the spherical coordinates into cartesian coordinates

sphereToCartesian = function(matrice){

x= matrix(0,nrow = 3, ncol = length(matrice[1,]))

x[1,] = sin(matrice[2,]) * cos(matrice[1,])

x[2,] = sin(matrice[2,]) * sin(matrice[1,])

x[3,] = cos(matrice[2,])

return(x)

}

a = sphereToCartesian(y)

plot3d(x = a[1,], y = a[2,], z = a[3,])

#you should enlarge the device window, before running this, if you want to have a meaningful graph

rgl.bringtotop()

rgl.viewpoint(0,20)

for (i in 1:45) {

rgl.viewpoint(i,20)

filename <- paste(“pic”,i,”.png”,sep=””)

rgl.snapshot(filename, fmt=”png”)

}

################################################################

# Correct solution for the sphere

################################################################

uniformSphere = function(length){

x1 = runif(length, 0,1)

x2 = runif(length, 0,1)

u = 2*pi*x1

v = acos(2*x2- 1)

z =matrix(0, ncol = length, nrow = 2)

z[1,] = u

z[2,] = v

return(z)

}

z = uniformSphere(size)

b = sphereToCartesian(z)

plot3d(x = b[1,], y = b[2,], z = b[3,])

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