On Wed, Jul 30, 2014 at 7:26 PM, Fabien COELHO <coe...@cri.ensmp.fr> wrote:
> > ISTM that you miss the projection on the segment if dx=0 or dy=0. >>> >> >> I don't need to find projection itself, I need only distance. When dx = 0 >> then nearest point is on horizontal line of box, so distance to it is dy. >> Same when dy = 0. When both of them are 0 then point is in the box. >> > > Indeed. I thought that the box sides where not parallel to the axis, but > they are. So I do not see why it should be more complex. Maybe they is a > general algorithm which works for polygons, and they use it for boxes? Yeah, this answers question #1, but not #2 and #3 :) ------ With best regards, Alexander Korotkov.
