Does the Ray Casting Algorithm works in Poincare's Disk to detect if point is inside Polygone?

Question

As the Ray Casting Algorithm looks to me like a geometric construction on geodesics, and geodesics are redefined in Poincare's Disk, I feel this method would also work in hyperbolic geometry.

Is this true ? It seems visually correct to me. Is there any proof ?

Another method I can think of would be to check if the point is at the right to every side (clock-wise) or at the left to every side (anticlock-wise).

Is this true too ? Again it seems visually correct to me. Is there any proof ?

Answer 1

Yes, it is certainly true. The ray-casting algorithm should work for any uniquely geodesic metric space which is diffeomorphic to $\mathbb R^2$.

Not sure exactly how rigorous you want it, but I find this proof adequate:

Outline of proof:

First suppose that the point $P$ does not lie on the extension of any of the geodesic edges. Then when a ray from $P$ intersects the boundary of the shape, it is either entering or exiting the polygon.

(one side of the ray is inside, while the other is outside. This is because the only way for this not to be the case is if the ray was locally following the edge, in which case it would be exactly the same geodesic because a geodesic is determined entirely from its local trajectory).

Ray always leaves the polygon eventually:

Let $\{C_i\}$ be a compact polygon. Let $P$ be a non-boundary point of this polygon. Any ray drawn from a point $P$ is not compact (as it extends infinitely far) but is closed, and so eventually will be outside the polygon (intersection of a closed set with a compact set is compact I believe. So intersection of the polygon with the ray is compact, hence there exists a point in the ray which is outside the polygon).

Tying it up:

We know that one of the intersections indicates an exit from the polygon. It follows easily that if the number of intersections is odd then the ray's first and last intersections were exits. It must have started on the inside. If the number of intersections is even then the ray's first and last intersections were an entry and an exit respectively. It must have started on the outside.