I've got a set of data points in $\mathit n$-dimensional space. I choose a specific data point $\mathit p$ and ask the question: What is the smallest ball that will enclose $\mathit m$ data points centered on $\mathit p$?
To solve it, I calculate the distance between $\mathit p$ and all other points, then sort the results and choose the $\mathit m$th smallest distance, $\mathit d$. Then I can calculate the $\mathit n$-dimensional volume with $\mathit d$ as the radius of the ball.
So far so good, but for the specific data points I'm working with, I know that the data is bounded between $-1$ and $1$ in each dimension. So really I am interested in the "corrected volume" which excludes the volume of the region that cannot contain any points given the constraint on the data. For clarity, in the 2-dimensional case, finding the "corrected volume" amounts to measuring the area of a circle centered at $\mathit p$ with radius $\mathit d$ bounded by a square centered at the origin with unit length.
I'm interested in solving this problem using the $\mathit L_1$ norm and the $\mathit L_\infty$ norm. I've solved the case for the $\mathit L_\infty$ norm, but I am having trouble for the $\mathit L_1$ norm.
For the $\mathit L_\infty$ norm, geometrically, the "ball" is a rectangle. I check for when the ball goes beyond the boundary in each dimension by comparing $\mathit p$ +/- $\mathit d$ to +/-1 and adjust the lengths accordingly. Then I multiply these adjusted lengths together to arrive at the $\mathit n$-dimensional volume.
For the case of the $\mathit L_1$ norm, I know the general formula for the volume is $\frac{2^n}{n!} r^n$ (which I picked up from https://en.wikipedia.org/wiki/Volume_of_an_n-ball). I know that the geometry is diamond shaped, but I'm struggling to arrive at a simple algebraic formula that will give me the corrected volume.