CGAL users discussion list

["J.L.M." <>]

For some reason I didn't get that post in email, but I do see it in the 
forum interface. I don't believe you will need step three: E = D * A, 
and step four can be result = A - D. Note that the part of D that is 
subtracted from A is precisely that part which intersects it, so there 
is no need to obtain that part separately.

Also, if O is not symmetric about the origin, you do need to mirror it 
about each axis. Maybe this isn't crucial for your implementation, and 
if not I suggest against it because it is among the things I have to do 
which leads me into trouble with CGAL.

On 12/28/2010 09:31 PM, johnzjq wrote:
> Hi,
>
> Many thanks for your detailed explaination. And your suggestions are really
> helpful.
>
> I have implmented the code I attached in the post "Any idea about Minkowski
> difference?" and it works fine on many complex CAD models (which should be
> 2-manifold) , but a bit slow.
>
> Regards,
>
> John