On Wed, Jul 30, 2014 at 7:26 PM, Fabien COELHO <coe...@cri.ensmp.fr> wrote:

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>
> 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 :)
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With best regards,
Alexander Korotkov.