Given a cube with 8 vertex in 3D space. How could I determine the myPoint is inside the cube?

cube[0] = (x0, y0, z0);
cube[1] = (x1, y1, z1);
cube[2] = (x2, y2, z2);
cube[3] = (x3, y3, z3);
cube[4] = (x4, y4, z4);
cube[5] = (x5, y5, z5);
cube[6] = (x6, y6, z6);
cube[7] = (x7, y7, z7);

myPoint = (x, y, z); 

I found a solution for this on Stack overflow(below). This sample solution returns the indexes of all points outside the cube. I tried to remake it to make the function return the indexes of the points that are inside the cube but failed. I've been sitting on this for 2 days. Is anyone able to help?

import numpy as np

def inside_test(points , cube3d):
    """
    cube3d  =  numpy array of the shape (8,3) with coordinates in the clockwise order. first the bottom plane is considered then the top one.
    points = array of points with shape (N, 3).

    Returns the indices of the points array which are outside the cube3d
    """
    b1,b2,b3,b4,t1,t2,t3,t4 = cube3d

    dir1 = (t1-b1)
    size1 = np.linalg.norm(dir1)
    dir1 = dir1 / size1

    dir2 = (b2-b1)
    size2 = np.linalg.norm(dir2)
    dir2 = dir2 / size2

    dir3 = (b4-b1)
    size3 = np.linalg.norm(dir3)
    dir3 = dir3 / size3

    cube3d_center = (b1   t3)/2.0

    dir_vec = points - cube3d_center

    res1 = np.where( (np.absolute(np.dot(dir_vec, dir1)) * 2) > size1 )[0]
    res2 = np.where( (np.absolute(np.dot(dir_vec, dir2)) * 2) > size2 )[0]
    res3 = np.where( (np.absolute(np.dot(dir_vec, dir3)) * 2) > size3 )[0]

    return list( set().union(res1, res2, res3) )

CodePudding user response：

According to De Morgan's laws:

The complement of the union of two sets is the same as the intersection of their complements

The function returns the outside points, we want the inside – the complement set of points. The function returns the union of three sets. So, in order to get the complement, according to De Morgan's law we need to compute the intersection of the complements of these three sets.

So you need to reverse the unequality signs, and change union to intersection. This should work:

    ...
    res1 = np.where( (np.absolute(np.dot(dir_vec, dir1)) * 2) <= size1 )[0]
    res2 = np.where( (np.absolute(np.dot(dir_vec, dir2)) * 2) <= size2 )[0]
    res3 = np.where( (np.absolute(np.dot(dir_vec, dir3)) * 2) <= size3 )[0]

    return list( set().intersection(res1, res2, res3) )

CodePudding user response：

I don't mean to demean the question, but this can be seen as a learning opportunity with regards to understanding how to use the right function for the right job. Like other commenters have noted, the function that should be called in the last line is intersection, not union. To quickly describe the difference and its relevance:

Union: The union of two sets A and B is defined as the set of all elements that belong to A, B, or both.

Intersection: The intersection of two sets A and B is defined as the set of all elements that belong to both A and B at the same time.

Assuming the code to generate the final sets being compared is correct, the problem lies in the last line. Python's intersection function (link here for examples) needs to be called from a set onto another set. What this means, going back to sets A and B, is that an intersection call on these two sets would look like:

A.intersection(B)

There are several things that are important to note here. First, intersection can accept multiple sets to be compared to, so in the case of the original question, the line would look like:

res1.intersection(res2, res3)

The second thing to note is that numpy's where function returns as a tuple, and so the res1, res2, and res3 tuples need to be converted to a set. Thus, the last line would be replaced with the following:

res1 = set(res1)
res2 = set(res2)
res3 = set(res3)
return list(res1.intersection(res2, res3))