有没有办法在给定顶点的情况下找到 3D 形状的所有边缘？

Question

I have a list of vertices of a simple 3D shape like a pyramid, a cube or a dodecahedron, is there an algorithm to find all the connections between the "outer vertices" that make up a face?

for example, once projected a pyramid to 2D I have a matrix of 8 coordinates x,y for each vertice: int[] coords = new int [8][2] and came up with this way of calculating them

for(int i = 0; i<4; i++){
    for(int a = 1; a<=4; a++){
        if(i+a!=3){
            if(a>i){
                edges.add(new Line( coords[i][0] * GROWTH + displacement ,
                                    coords[i][1] * GROWTH + displacement,
                                    coords[a][0] * GROWTH + displacement,
                                    coords[a][1] * GROWTH + displacement));
                                    }
                                }
                            }
                        }

This only works with pyramids, I wonder if there's a way of calculating all the edges of a given [n][2] set of coordinates representing a projected 3D shape.

Answer 1

There is no unique answer to your question, check Four ways to crate a mesh for a sphere

Here I translate the first approach, that is similar to geographical coordinates, based on latitude and longitude.

for(int i = 0; i<num_parallels; i++){
    for(int a = 1; a<=4; num_meridians++){
        double y1 = PI * i / num_parallels;
        double y2 = l1 + PI / num_parallels;
        double x1 = 2.0 * PI * i / num_meridians;
        double x2 = x1 + 2.0 * PI / num_meridians;
        add_sphere_line(x1, y1, x1, y2); // a parallel
        add_sphere_line(x1, y1, x2, y1); // a meridian
   }
}

The function to add a sphere line

void add_sphere_line(double x1, double y1, double x2, double y2){
  add_3d_line(
   // starting point in 3D coordinates
   Math.cos(x1) * Math.sin(y1), Math.cos(y1), Math.sin(x1) * Math.sin(y1),
   // end point in 3D coordinates
   Math.cos(x2) * Math.sin(y2), Math.cos(y2), Math.sin(x2) * Math.sin(y2)
  );
}

Since you mention you projected the pyramid to 2D I imagine you can workout the projection of these arbitrary lines to 2D as well.