Convex-Hull | convex hull by using Graham 's Scan Algorithm | 3D Printing library

This Answer was the key to solve this problem.

Coords:

[[(38, 11), (251, 364)], [(254, 62), (592, 266)], [(254, 312),

(592, 518)], [(46, 456), (247, 797)], [(346, 557), (526, 797)]]

tl;dr: It returns the index of a triangle containing the point. And it does not always pick the same index if there are multiple triangles containing it.

I think you misunderstood "It returns:Indices of simplices containing each point. Points outside the triangulation get the value -1."

My interpretation is that Delaunay(pts_outer) triangulates your rectangle with two triangles with the indices 0 and 1 respectively. Then hull.find_simplex(pts_inner) retruning [0, 0, 1, 1] means that your first two points are in triangle 0 and the 2nd two are in triangle one.

Finally it is a bit odd that find_simplex now tells you your point [5, 5] is in triangle 1. But that is not incorrect since the point [5, 5] is in both triangles.

Half a solution... Not an exact solution to the original question, but a different way of achieving the same outcome. Any triangular prism can be split into exactly three tetrahedra (see http://www.alecjacobson.com/weblog/?p=1888). This is a specific case of the fact that any polyhedron may be split into tetrahedra by connecting all faces to one vertex, if the faces does not already include it.

Knowing exactly how I would like the face triangles of my prism to be arranged, I can work out what three tetrahedra would reproduce the same configuration of triangles (with extra faces of course being added inside the original prism itself). I then form Delaunay triangulations around each of these three tetrahedra (ie, collections of 4 points) in turn and perform the original interior point tests: if it matches on any then I have a positive result for the whole prism. The key point is that by only giving four points to the Delaunay constructor at a time, I know exactly what triangulation it will return as there is only one way of forming such a tetrahedra (assuming no geometric degeneracy).

It's a bit longwinded, and involves 3x as many tests as I would like, but it's a start. If anyone in the future does know how I could do this better please do let me know.

Thanks to shapely polygons, on which trimesh is built you don't need to go over the vertices. You can use some of the inbuilt functions. After creating your 2D path you are almost there. The path has two attributes: polygons_full and polygons_closed. The first one is the outmost polygon without the inner ones and the second one are all polygons of your path.
You can simply do:

I found a file which was causing the CMAKE to include 10 sm_arch in compatibility list. Here's the link. I will re-compile after editing the file for just 1 sm_arch and compare the size of binaries generated. – Anuj Patil Jan 22 at 18:10

So findCUDA was the culprit here. Editing the files to required sm_arch does the trick!

We can use PyGEL 3d python library for this.

First, install it with pip install PyGEL3D

Second, the code:

Your config path contains an ASCII Backspace, the \b in \bin, which pdfkit appears to be stripping out and converting C:\Program Files\wkhtmltopdf\bin\wkhtmltopdf.exe to C:\Program Files\wkhtmltopdf\wkhtmltopdf.exe.

This can be resolved by using r, which makes it a raw literal

It"s the classical problem of linear programming. Given a set of linear inequalities (represented by planes in your case, say ax+by+cz+d>=0) and a linear function (say f(x,y,z)=x), find a point that minimises the function while satisfying all the inequalities. The AABB is the solution of 6 such problems, for functions x, -x, y, -y, z and -z.

There are several methods of solving this problem, the most known being the simplex algorithm, and many ready-made libraries (including some that utilise GPUs).

UPDATE: Based on the comments by @AndreasFabri, my solution is too convoluted. You can find a much shorter solution based on what is already implemented in CGAL https://gist.github.com/afabri/a32f0d1ac5af1a99491d62786a0d5559. I am copying the same code below: