This may seem like a strange question, but I have been spending quite some time trying to find the answer to the following problem:
Given a Euclidean (L2) unit sphere and a point $p$, what is the minimum L1 distance of $p$ to that sphere?
I have had quite some success for circles, and the formulas tend to be elegant and simple. For instance, for points whose $x$ and $y$ coordinates are both outside of the range $[-1,1]$, the distance is simply $|x|+|y|-\sqrt2$. I am wondering if there are similarly elegant formulas in 3D. However, this is where my math skills are reaching their limits.
(To arrive at the 2D formulas, I minimized $|\cos\theta-x|+|\sin\theta-y|$ for $\theta$ to varying degrees of success. Depending on $x$ and $y$, the curve can become quite scary).
If you are wondering what this is for: I am writing a somewhat unconventional ray tracer that is using a discrete space. The L1 distance makes things a lot easier for many things, but as you may imagine circles and spheres take some extra tinkering.
It helps to look at the general case in two dimensions first, where the L1 circle $\Gamma_1$ is a diamond.
Suppose $|x|,|y|\ge\frac{\sqrt2}2$, i.e. $p=(x,y)$ is in the shaded area above. Then $\Gamma_1$ around $p$ can be enlarged until it is tangent to the L2 circle $\Gamma_2$ at $\left(\frac{\sqrt2}2,\frac{\sqrt2}2\right)$, which makes the minimal L1 distance $d=|x|+|y|-\sqrt2$. Otherwise, $\Gamma_1$ can only meet $\Gamma_2$ at a point, and simple geometry shows that $d=z_+-\sqrt{1-z_-^2}$ where $z_+=\max(|x|,|y|)$ and $z_-=\min(|x|,|y|)$.
For three dimensions, $\Gamma_1$ is an octahedron, but a similar idea applies. If $|x|,|y|,|z|\ge\frac1{\sqrt3}$, $\Gamma_1$ can be enlarged until one of its faces is tangent to $\Gamma_2$ at $\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right)$. Then $$d=|x|+|y|+|z|-\sqrt3$$ Now suppose $|z|\le\frac1{\sqrt3}$. $\Gamma_1$ can now only meet $\Gamma_2$ at an edge or a point. If we slice the entire set-up perpendicular to the $z$-axis at the given $z$, we get a scaled version of the two-dimensional case!
In conclusion, to find the minimal L1 distance $d$, flip and swap the coordinate axes so that $x\ge y\ge z\ge0$. Then
$$d=\begin{cases}
x+y+z-\sqrt3&\text{if }z\ge\frac1{\sqrt3}\\
x+y-\sqrt2\sqrt{1-z^2}&\text{if }z<\frac1{\sqrt3}\text{ and }y\ge\frac{\sqrt2}2\sqrt{1-z^2}\\
x-\sqrt{1-y^2-z^2}&\text{otherwise}
\end{cases}$$
This result naturally generalises to higher dimensions.