When do points lie on a (hyper)hemisphere?

Each point on the surface of the unit hypersphere has a corresponding ray which starts at the origin and goes through the point. A set of points on the hypersphere will lie in a hemisphere iff the convex hull of the corresponding rays does not contain the whole space. This is because a (possibly unbounded) convex polyhedron in Euclidean space can be described either as a convex hull of points and rays, or as an intersection of half-spaces. So, if a set of rays lies in no half-space, the convex hull must be the intersection of no half-spaces, i.e., equal to all of space.

Another way of looking at the problem is that a half-space whose boundary plane goes through the origin can be expressed by a vector through the origin which is normal to the plane which bounds the half-space and points into the half-space. So, if we express the rays we are given as vectors v_1, ..., v_n, they will lie in a half-space iff there is a nonzero vector w such that <w, v_1> ≥ 0, ..., <w, v_n> ≥ 0. This is also a polyhedral computation problem as it is asking whether a polyhedron defined by a set of inequalities in w-space contains any points other than the origin.

There are a number of algorithms and programs which will go between the dual representations of a polyhedron. One usable system is cddlib (see below.)

For n=3, it seems that m points lie in the same hemisphere if and only if none of the m antipodes-points lie within any of the mC3 spherical triangles defined by the m points. Any algorithm to check this would need a sub-algorithm to decide if a point lies within a given spherical triangle. It seems this would be difficult but not impossible. The number of checks would be mC3 triangles * (m-3) antipodes points which is of order m^4.

It seems this might generalize up in dimension, but I can't visualize it.

Is your method more efficient that that?

Interesting question - how would an instance of the problem be given? Do we know the co-ordinates of points P1,...,Pm and an equation that represents the hypersphere?

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oh why don't you extend this, I'm still thinking about it

Edit: Josh, better pick a BA pretty soon, I don't think I'll be able to enter my answer in time.

Edit 2: If Josh misses the BA deadline, I'll vote for nudnik0's polyhedral answer as maybe the best way to go at present.