Distance from point to box
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 has some mathematical beauty, but is it acceptable in practice?
2) computeDistance in knn-gist uses it's own quite simplified
implementation of this calculation. But why has it own implementation
instead on simplifying dist_pb?
3) computeDistance implementation looks still very complex for me. Right
way (simple and fast) to calculate point to box distance seems for me to be
so:
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 myself quite tangled.
Could anybody clarify it for me? Did I miss something? Thanks.
------
With best regards,
Alexander Korotkov.
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 myself quite tangled.
Could anybody clarify it for me? Did I miss something? Thanks.
ISTM that you miss the projection on the segment if dx=0 or dy=0.
--
Fabien.
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On Wed, Jul 30, 2014 at 4:06 PM, Fabien COELHO <coelho@cri.ensmp.fr> 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->y > box->high.y)
dy = point->y - box->high.y;
return HYPOT(dx, dy);I feel myself quite tangled.
Could anybody clarify it for me? Did I miss something? Thanks.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.
------
With best regards,
Alexander Korotkov.
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?
--
Fabien.
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On Wed, Jul 30, 2014 at 7:26 PM, Fabien COELHO <coelho@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.