Solved – How to generate random points in the volume of a sphere with uniform nearest neighbour distances

clusteringgeometryrandomnessuniform distribution

With respect to post (1) and post (2), I generated a large number of uniformly distributed points inside the ball of radius $R$ using $\frac{R_s U^{1/3}}{\sqrt{X_1^2 + X_2^2 + X_3^2}} (X_1, X_2, X_3)$, where $U$ is uniformly distributed between 0 and 1, and $X_1, X_2, X_3$ are independent normal random variables with mean 0 and variance 1.
The following figure shows a uniform spherical distribution obtained by this method using 10000 independent draws in a sphere of radius 10.
enter image description here

By computing the nearest neighbour distance $d_i$ of every point, I observed that the diagnostic plot of nearest neighbour distances does not follow a uniform distribution. Does this non-uniform distribution mean that one can cluster the points? Does it mean points dont have spatial randomness? If so, then how can I generate random points with uniform nearest neighbour distances.

Temporary images for @Anony-Mousse consideration:
enter image description here

Best Answer

I wouldn't expect the distribution of the nearest neighbour distances to be uniform under spatial randomness.

According to Wikipedia (http://en.wikipedia.org/wiki/Complete_spatial_randomness), the distance of the first neighbour in your case has the following distribution:

$P_1(r) = 3\lambda r^2\exp(-\lambda r^3)$

where $\lambda$ is ta density dependent parameter. This is obviously non-uniform!

Concerning your clustering question: You can always cluster points, independently of their distribution.

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Solved – Estimate center and radius of a sphere from points on the surface

Here is some R code that shows one approach using least squares:

# set parameters

mu.x <- 8
mu.y <- 13
mu.z <- 20
mu.r <- 5
sigma <- 0.5

# create data
tmp <- matrix(rnorm(300), ncol=3)
tmp <- tmp/apply(tmp,1,function(x) sqrt(sum(x^2)))

r <- rnorm(100, mu.r, sigma)

tmp2 <- tmp*r

x <- tmp2[,1] + mu.x
y <- tmp2[,2] + mu.y
z <- tmp2[,3] + mu.z


# function to minimize
tmpfun <- function(pars) {
    x.center <- pars[1]
    y.center <- pars[2]
    z.center <- pars[3]
    rhat <- pars[4]

    r <- sqrt( (x-x.center)^2 + (y-y.center)^2 + (z-z.center)^2 )
    sum( (r-rhat)^2 )
}

# run optim
out <- optim( c(mean(x),mean(y),mean(z),diff(range(x))/2), tmpfun )
out


# now try a hemisphere (harder problem)

tmp <- matrix(rnorm(300), ncol=3)
tmp[,1] <- abs(tmp[,1])
tmp <- tmp/apply(tmp,1,function(x) sqrt(sum(x^2)))

r <- rnorm(100, mu.r, sigma)

tmp2 <- tmp*r

x <- tmp2[,1] + mu.x
y <- tmp2[,2] + mu.y
z <- tmp2[,3] + mu.z

out <- optim( c(mean(x),mean(y),mean(z),diff(range(y))/2), tmpfun )
out

If you don't use R then you should still be able to follow the logic and traslate it into another language.

Technically the radius parameter should be bounded by 0, but if the variability is small relative to the true radius then the unbounded method should work fine, or optim has options for doing the bounded optimization, (or you could just do the absolute value of the radius in the function to minimize).

Best Answer

Related Solutions

Solved – Estimate center and radius of a sphere from points on the surface

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