Normal marching cubes finds 12 edges per cube, but you can do 3 edges per cube, save the edges inside an array, and then go through the cubes again, referencing the edges from the cubes adjacent rather than calculating them.
The process to reference adjacent cubes isn't clearly discussed on the Internet so anyone using marching cubes would be welcome to help find the details of the solution. do you know an implementation already?
here is a picture showing the 3 edges in yellow that you need for each cube, instead of 12.
EDIT- I just found this solution, although it's just a part of it:
Imagine 3 edges coming from the corner of the cube with lowerest coordinates. Then all other edges just belong to other cubes. If our cube has coordinates (x,y,z), the neiboring cubes have coordinates (x+1,y,z), (x,y+1,z), (x,y,z+1), (x+1,y+1,z), (x+1,y,z+1), (x,y+1,z+1). You can imagine the edge as a vector. Then the corner of the cube have edges (1,0,0), (0,1,0), (0,0,1). The cube with coordinates (x+1,y,z) have edges (0,1,0) and (0,0,1) that belong to our cube. The cube (x+1,y+1,z) has only one edge (0,0,1) that belongs to our cube. So if you store 4 elements for the cube you can access them like that:
edge1 = cube[x][y][z][0];
edge2 = cube[x][y][z][1];
edge3 = cube[x][y][z][2];
edge4 = cube[x+1][y][z][1];
edge5 = cube[x+1][y][z][2];
edge6 = cube[x][y+1][z][0];
edge7 = cube[x][y+1][z][2];
edge8 = cube[x][y][z+1][0];
edge9 = cube[x][y][z+1][1];
edge10 = cube[x+1][y+1][z][2];
edge11 = cube[x+1][y][z+1][1];
edge12 = cube[x][y+1][z+1][0];
Now which points edge7 connect? The answer is (x,y+1,z) and (x,y+1,z)+(0,0,1)=(x,y+1,z+1).
Now which cubes edge7 connect? It is more harder. We see that coordinate z is changes along the edge this means that neibour cube has the same z coordinate. Now all others coordinates change. Where we have +1, the cube has large coordinate. Where we have +0, the cube has smaller coordinates. So the edge connects cubes (x,y,z) and (x-1,y+1,z). Other 2 cubes that has the same edge are (x,y+1,z) and (x-1,y,z).
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EDIT2- So I am doing this, and it isn't so simple. I have a loop which simultaneously calculate 8 points, 12 edges, the interpolation of edges, the bit values and a vertex the values for the edges, all in one loop.
so I am doing a new loop previous to it to calculate as much as possible and place it in arrays to used in the complicated loop.
I can recycle the interpolated values of the intersection points along edges, in an array, although I will have to recalculate all the points again in the complicated loop, because the values of the points I used to decide bit numbers that reference values in the vertex table. That confuses me! I thought that once I have the edge intersection values, I could use those directly to get the triangle tables, without having to calculate the points all over again!
in fact no. anyway, here is another bit of information with someone that already did it, if only it was readable! http://www.new-npac.org/projects/sv2all/sv2/vtk/patented/vtkImageMarchingCubes.cxx scroll to this line: Cubes are responsible for edges on their min faces.