[rckoepke]

> The real problem is that in dimensions that high, the point set probably already is the hull and all this is a zero signal gain operation.

Well, if I have 10,000 samples of a 768-dimension volume, most of those points will probably be inside the volume, and not per se a vertex of the hull.

I’m very comfortable rolling my own solution, so thank you for pointing me to Jarvis’ algorithm!

[blamestross]

So, about that. Do the math on how many faces a 768-simplex has.

[rckoepke]

Revisiting this. Isn't it a bit of a red herring to enquire about the number of 2-faces that an n-simplex has? It still only has n+1 vertices. A 768-simplex may have 75.5 million faces but it will still only have 769 vertices which completely define the shape. So why would I expect a large number of the other >90% of the 10,000 samples I have to lie on the surface, rather than inside the interior volume?

To be more direct, what's the specific relevance of bringing up the number of 2-faces that an n-simplex has?

[rckoepke]

I believe the answer is:

(n+1)!/((k+1)!*(n-k)!) where n=768 and k=2

Or about 75.5 million triangular faces. Which explains a lot. Thanks for that.