[* created with the free text-to-speech program "Balabolka", using the Microsoft Hazel voice, Pitch=-10, Rate=2]they were aiming at only three layers [of interdiction], which would require 43 interdictors laid out in a specific pattern to create a total barrier of three light years in any direction from the edge of interdicted space to Karis. If he wanted four light years, that number skyrocketed to 253

In the nerdy part of my brain, the following two questions arose:

- Are the 43 and 253 values accurate?

- What configuration should be used to minimize the number of interdictors required to cover a given volume?

I thought the answers to these questions would be fairly easy to find with a quick google search, but boy was I wrong.

You see, while there's plenty of information available about the best way to pack spheres into a given volume (known as the "sphere packing problem"), even the best packings utilizing uniform spheres leave over a quarter of the volume uncovered.

If we don't want to leave any gaps within our interdiction field, then we're really after a solution to the "3 dimensional ball covering problem". And there's a lot less information available on the web about covering problems than there is for packing problems. Finding an optimal covering is a non-trivial task, even in two dimensions (eg https://www2.stetson.edu/~efriedma/circovcir/), and the three dimensional case is even more difficult.

For anyone interested, here's what I discovered after looking into this topic over the course of several hours:

Firstly, I was fortunate to find a webpage which provides the configuration needed to generate a two light year radius interdiction field with the fewest possible interdictors: https://mathoverflow.net/questions/9800 ... the-radius. (You'll need to scale everything on that page up by a factor of 2, since the interdictors have a 1 LY radius, rather than the 1/2 unit used on the page).

Basically, the most optimal configuration requires 1 interdictor at the center, surrounded by a second shell of 20 interdictors placed sqrt(3) LYs from the origin, for a total of 21 interdictors.

Unfortunately, I couldn't find a description of how to optimally add on more shells to expand the field. I highly doubt 43 interdictors would be sufficient to generate a 3-shell 3 LY radius field, though, as a minimum of 21 interdictors is already required to generate a 2-shell 2 LY radius field. It might be possible to get close to 3 LY with a 2-shell set-up though, for example by using an inner shell at the vertexes of an icosahedron, and an outer shell a sufficient distance away from the inner shell to cover the rest of the volume. That's entirely my own speculation though, and I'm not really up to the task of finding out for sure.

Secondly, while my maths & google-fu skills are far from sufficient to find the minimum number of interdictors required to cover larger spherical volumes, I did find a less-than optimal but more accessible solution that can be scaled up to provide gap-free interdictor coverage for any rectangular volume.

Simply put, spheres don't tessellate, but cubes do. The largest cube that can be inscribed inside a sphere of radius 1 LY has edges of length 2/sqrt(3) LY, so we can pretend that each interdictor will cover a cube of side length 1.1547 LY, then ignore the bits of the sphere that project out beyond the surface of the cube.

Thus all we need to do is stack our cubes together to cover the volume we want to interdict. A 4 x 4 x 4 stack of interdictors will cover a cubic region of space 4 x 1.1547 LY = 4.6 LY across, which is sufficient to enclose a 2.3 LY radius sphere. A 5 x 5 x 5 stack will cover a 5.7 LY cube (enclosing a 2.85 LY radius sphere), a 6 x 6 x 6 stack covers a 6.9 LY cube (enclosing a 3.45 LY radius sphere), and a 7 x 7 x 7 stack covers an 8.08 LY cube (enclosing a 4.04 LY radius sphere).

Thus, even with our sub-optimal cube-stacking arrangement, 343 interdictors will easily cover a 4 LY radius sphere (actually, we can eliminate 7 interdictors at each of the 8 corners of the cube while still enclosing the 4 LY sphere, thus requiring only 343 - 56 = 287 interdictors). It is therefore plausible that the 253 interdictors mentioned in the story would be more than sufficient to cover a 4 LY radius sphere if we used a more optimal spherical shell stacking configuration.

* * *

Anyway, fun though it was to look into this topic, I think I'll leave the rest of the maths to Myli and Cybi. Or anyone else who feels their maths skills are up to the task.

Best of luck if you decide to take on the challenge!