On Wed, Jul 30, 2014 at 7:26 PM, Fabien COELHO 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 whe
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
On Wed, Jul 30, 2014 at 4:06 PM, Fabien COELHO wrote:
>
> double dx = 0.0, dy = 0.0;
>>
>> if (point->x < box->low.x)
>>dx = box->low.x - point->x;
>> if (point->x > box->high.x)
>>dx = point->x - box->high.x;
>> if (point->y < box->low.y)
>>dy = box->low.y - point->y;
>> if (point->
double dx = 0.0, dy = 0.0;
if (point->x < box->low.x)
dx = box->low.x - point->x;
if (point->x > box->high.x)
dx = point->x - box->high.x;
if (point->y < box->low.y)
dy = box->low.y - point->y;
if (point->y > box->high.y)
dy = point->y - box->high.y;
return HYPOT(dx, dy);
I feel my
Hackers,
while reading code written by my GSoC student for knn-spgist I faced many
questions about calculation distance from point to box.
1) dist_pb calls close_pb which calls on_pb, dist_ps_internal, close_ps and
so on. So, it's very complex way to calculate very simple value. I see this
way ha
