Sampling line segments within a box

Question

I'm at a bit of a loss for where to start with a sampling problem I'm having so any and all direction would be helpful. I essentially want to sample line segments of identical length within a bounding square. Imagine dropping a needle down a square cylinder and seeing how it lands (over and over).

If the sides of the square are smaller than the length of the segments, you could imagine the distribution is X-shaped. I want to understand how this changes as the square grows. How does convolution apply here if at all?

Edit: random/uniform in the sense that any valid segment position is equally likely to occur. Valid meaning both endpoints are within the boundary.

Could be related: https://math.stackexchange.com/questions/4796686/probability-of-line-segments-intersecting-on-a-plane-a-generalization-to-buffo?rq=1

Answer 1

While the comments are strictly correct that the answer could depend on details of the mechanism, it's easier to answer for some mechanisms than others.

I'm going to use the following model

• midpoints of segments are sampled uniformly at random from $[0,1]\times [0,1]$
• angles are sampled uniformly at random from $[0,\pi)$
• the segment is kept if it lies entirely in $[0,1]\times[0,1]$ and rejected otherwise

This samples uniformly from the subset of $[0,1]\times[0,1]\times[0,\pi)$ that gives segments entirely within the permitted square.

Here are segments of lengths 1.3, 1.2, 1.0, and 0.5

And here are the histograms of angles scaled to multiples of $\pi/4=90^\circ$

As you can see, you do start out with crosses for lengths longer than the side length -- the maximum possible length is $\sqrt{2}\approx 1.4$, but at 1.3 we're getting only about 1 in 1000 samples accepted. As the length decreases relative to the square dimensions, the peaks of the angle broaden and end up uniformish.

In the small-needle limit, the angle distribution will be uniform (because sampling acceptance is 100% at distance from the edge greater than half the needle length), and the endpoint distribution will be uniform over the square except near the edges.

I don't think there's any convolution going on with this model. You are sampling uniformly from a relatively complicated but well-defined subset of $[0,1]\times[0,1]\times[0,\pi)$ and the subset converges to that whole set as the needle length decreases.

I think this is actually the same distribution you'd get if you sampled one endpoint, then sampled the other endpoint from the circle whose radius was the length of the needle, then rejected if it wasn't in the square, but I'm not sure. It is not the same model as sampling one endpoint and then sampling the other endpoint from the points at the correct distance within the square (but I think that would be a bad model)

In any case, I think random centre/random angle is a reasonable model for dropping a needle.

Here's the sampling code

needle<-function(l,N){
    h<-l/2
x<-runif(N)
y<-runif(N)
angle<-runif(N,0,pi)
lx<-x-h*cos(angle)
ly<-y-h*sin(angle)
ux<-x+h*cos(angle)
uy<-y+h*sin(angle)
keep<- pmin(lx,ly,ux,ux)>0 & pmax(lx,ly,ux,uy)<1
data.frame(x,y,angle,lx,ly,ux,uy)[keep,]
}