立方体表面上的点之间的距离

Question

Let (x1, y1, z1) and (x2, y2, z2) be two points in Euclidian 3-space on the surface of an axis-aligned side-length-2 cube centered at the origin.

How do I efficiently compute the distance (or squared distance) between the points over the surface of the cube ?

Internally, I represent points as (offset1, offset2, faceNumber) but an (x,y,z) format (as referenced above) is readily available.

I prefer C or Python code but I'll happily accept pseudocode or anything, really.

EDIT:

Some facts:

1. Shortest paths are always monotone in x, y, and z.
2. If the points are on the same face then it's trivially just Euclidian distance.
3. If the points are not on the same face, the shortest path could involve either 2 or 3 faces.

Answer 1

EDIT: What I would do, is turn the 3d cube into a 2d plane. The caveat is that, if the point is on the opposite side of the cube, you need to place the final surface on all ends of the cross.

If a cube had sides like this that you could fold around so that 4 touched side 1.

5
1 2 3 4 
6

You would have a 2d plane that ultimately looked like this

         3
    4/5  5   5/2
3   4    1   2   3
    4/6  6   2/6
         3

So, I modified this. Now each of the corner panels represents the connections that can take place between both panels. When you initially lay out this array, each point on panels 2, 4, 5, and 6, will map to three points. The solution is then the shortest line to any of the given points, that represent point 2, in the event you need to map it to multiple points.

Mapping points from the 3d cube, to their initial 1 - 6 pains on the 2d graph is really quite simple. The only difficulty left is figuring out how to map points from the 2 plane, onto the "2/6" plane and so forth. This is just a matter of thinking through each situation. Ex: 2 -> 2/6 is different from 5 -> 5/2. My intuition is that it's either going to be 90 degree or -90 degree rotation, before shifting the width of the cube in the appropriate direction.

For example, to properly handle the situation you laid out we would have a value at the bottom left corner of plane one, and the bottom right corner of plane 2. After the following: '

points in plane 2/6 = rot90(points in plane 2) - width of the cube.  

We will have a point in the bottom left corner of plane 2/6. This will then appropriately be the shortest path, and appropriately this path crosses the face of plane 6.