clipka <ano### [at] anonymousorg> wrote:
> Am 27.01.2013 21:00, schrieb Trevor G Quayle:
>
> > Stop thinking of everything!
>
> Sorry, but I just can't help it ;-)
>
> > OK one more think at this:
> >
> > 1) check outer boundary sphere intersection (radius = center to vertex).  If no
> > intersection (distance > r1 + r2), cubes don't intersect,
> >
> > else
> >
> > 2) check inner boundary sphere intersection (radius = center to face).  If
> > intersection (distance < r1 + r2), cubes intersect or nest.
> >
> > else
> >
> > 3) determine closest vertex for each cube to the center of other (don't need to
> > check each vertex). if either vertex is inside other cube, intersected or nested
> >
> > else
> >
> > 4) for one cube, check each edge from closest vertex for intersection with each
> > face from each vertex of other cube, if intersect, cubes intersect. (don't need
> > to check every single one, only 3 edges from 1 and 3 faces from other)
> >
> > else
> >
> > 5) cubes don't intersect!
> >
> > Does that cover it all?  Or is there some other scenario that falls outside
> > these rules?
>
> You're forgetting that the vertex of cube A closest to the /center/ of
> cube B isn't necessarily the one closest to the /surface/ of the other.

Perhaps, but it will be on the other end of one of the edges that get tested (if
not the same vertex, the vertex closet the center should be on the other end of
an edge shared with the vertex closet to the surface) and should still be valid
for detecting the intersection.  I can't envision any scenarios that don't get
captured with this process.

-tgq

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