# Median on 1-Sphere

Given a set of points $$X \subset \mathbb{R}$$, one can define the median $$m^*$$ using optimization.

Namely $$m^* = \textbf{argmin}_{m \in \mathbb{R}} \sum_{i \in X} |i - m|$$. Here $$|.|$$ is absolute value. As $$m^*$$ could not be uniquely define, there could be an interval of $$m$$ all minimize the function, then apply a $$min$$ would be my fix to it. The reason behind that is i also care about the minimum value of such objective function, all the $$m^*$$ would gives the same value.

What if a set of points $$Y \subset S^1$$ from 1-sphere ?

I can still define the median $$n^*$$ as $$n^* = \textbf{argmin}_{n \in S^1} \sum_{j \in Y} |j - n|$$. Here $$|.|$$ is the shortest distance along the circle between two points. Similarly if there exists an interval of $$n$$ all minimize the objective, pick the minimum as the desired value. (or any feasible $$n$$ if that is more efficient)

Is the $$n^*$$ going to be the conventional median? How can I efficiently find such median?

Your definition of the median isn't actually uniquely defined even in the $$X \subset \mathbb{R}$$ case if $$X$$ has a (finite) even number of points. Any number $$m$$ between the middle two values of $$X$$ will minimise your optimisation problem. Similarly, the median isn't necessarily uniquely defined in $$S^1$$ when you define it that way. In particular, if you have an even number of evenly spaced points, then every point in $$S^1$$ is a median by your definition.
In general, I think you can say that $$m\in S^1$$ is a local minimum of your objective if and only if there are the same number of points in the clockwise half-circle from $$m$$ as there are in the anticlockwise half-circle. This will be constant on intervals where there aren't any points in $$X$$ or any points opposite a point in $$X$$. So you just need to create a copy $$X'$$ of $$X$$ shifted halfway around the circle, then for each interval between points in $$X \cup X'$$, count the points of $$X$$ in the half-circles on each side of (any point in) that interval.