that which you call “a convex hull”, I apparently know how (in logic) to create in “an infinite number of dimensions” (i.e. SPACE (itself) becomes “relative” (unless there are _________________ (censored) in which situation space is computed (= maximum efficiency (in theory) (noise considerations can be factored in via a more sophisticated technique I reckon) of storing data (as “real-time “objects”)(the whole thing becomes a _______________ (the _____________________)(curiously it is a “perpetual motion machine” in that everything (can) rotate around everything else (“perpetual motion”) within internally (or sometimes externally in other versions e.g. teleportation dynamic scenario ..) generated “limits” (“machine” like..) – the whole thing becomes like a solid object, … an algorithm partial ‘interferometer’ i.e. you don’t need an algorithm to store the data, the data itself ” becomes ” an ____________(quantum dynamical; inter-space (a between space betwen space: a hyperspace bypass!!!!!!!! )

Well, this problem is so important in Windows update for the screen shots, fast games, and 3D cg scenes!

To elaborate on the point-in-convex-hull test. This can be done in O(log n):

Given a point (x,y) look whether it is in the lower or upper half-space, using the same test as applied in partition_points().
Then lookup the index ‘a’ of the first coordinate with x-value larger than x in lower_partition_points if (x,y) is in the lower half-space.
Lookup the index ‘a’ of the first coordinate with x-value larger than x
in upper_partition_points if (x,y) is in the upper half-space.
This lookup can be done using a binary search (http://en.wikipedia.org/wiki/Binary_search_algorithm) as the arrays are sorted by x-coordinate, So in O(log n) time.
Then get the vector from the coordinate at index ‘a-1′ to ‘a’. If (x,y) is in the lower half-space and on or left of the vector then it is inside the convex hull.
Similarly if (x,y) is is in the upper half-space and on or right of the vector then it is inside the convex hull.
You can use the direction() function from the c++ code for the vector side test.
If index ‘a’ or ‘a-1′ does not exist then (x,y) is outside the convex hull or on the hull at right-most hull point. To check against the latter case, test if x equals ‘right’ (from the c++ class) and if true,
set ‘a’ = last index of lower_partition_points (or upper_partition_points depending on whether (x,y) is in the lower or upper half-space) and test if (x,y) is on the vector from ‘a-1′ to ‘a’ using direction()

Thanks, I updated the link!

- Mark

Antoon

I don’t know if there is a sublinear solution to this problem, but people love their convex hulls, so it should be easy to find.

My solution: be the point, turn in a circle 360 degrees. If all you see is hull all the way around, you’re inside.