All possible combinations of x, y, z values of two vectors?

Question

I'm using c# in Unity (though this question isn't specific to either) and need to get all possible combinations of the two vectors:

(x, y, z) and (-x, -y, -z)

so the result would be: (x, y, z) (-x, y, z) (-x, -y, z) (-x, -y, -z) (x, -y, z) (x, -y, -z) (x, y, -z) (-x, y, -z)

I know x, y and z. The origin is in the centre of the cube all I need is to get each corner of the cube's x, y, z position.

The points I need to get:

Answer 1

The absolutely simplest, fastest, and easiest to understand way to make the eight combinations is hard-coding the eight combinations of signs:

IList<int[]> MakeCombinations(int x, int y, int z) {
    return new[] {
        new[] { x,  y,  z}
    ,   new[] { x,  y, -z}
    ,   new[] { x, -y,  z}
    ,   new[] { x, -y, -z}
    ,   new[] {-x,  y,  z}
    ,   new[] {-x,  y, -z}
    ,   new[] {-x, -y,  z}
    ,   new[] {-x, -y, -z}
    };
}

If you prefer a solution with a little more programming, you can do it like this (I am using arrays for vectors; you could use your specific class):

int[] data = new[] {x, y, z};
IList<int[]> all = new List<int[]>();
for (int mask = 0 ; mask != 8 ; mask++) {
    int[] next = new[] {data[0], data[1], data[2]};
    if ((mask & 0x01) != 0) next[0] *= -1;
    if ((mask & 0x02) != 0) next[1] *= -1;
    if ((mask & 0x04) != 0) next[2] *= -1;
    all.add(next);
}

The combination of pluses and minuses in front of the vector component is modeled with a single three-bit number mask. When bit at position N is set, the N-th component is taken with the negative sign; otherwise, it is taken with a positive sign.