Point in polygon

[Nicola_Piano]

Hello,
I am looking for an algorithm that allows me to understand if a point, of which I know the coordinates, is inside or outside a polygon of which I know the coordinates of the vertices.
On the internet I found this ...

The code below is from Wm. Randolph Franklin <w...@ecse.rpi.edu>
with some minor modifications for speed. It returns 1 for strictly
interior points, 0 for strictly exterior, and 0 or 1 for points on the boundary. The boundary behavior is complex but determined;
in particular, for a partition of a region into polygons, each point is "in" exactly one polygon. See the references below for more detail.

Code Select Expand


int pnpoly(int npol, float *xp, float *yp, float x, float y)
{
int i, j, c = 0;
for (i = 0, j = npol-1; i < npol; j = i++) {
if ((((yp[i]<=y) && (y<yp[j])) ||
((yp[j]<=y) && (y<yp[i]))) &&
(x < (xp[j] - xp[i]) * (y - yp[i]) / (yp[j] - yp[i]) + xp[i]))

c = !c;
}
return c;
}


How to translate it into O2?
Cheers

[Charles Pegge]

Hi Nicola,

Here is a partially readable translation:

Code Select Expand


function pnpoly(int npol, float *xp, float *yp, float x, float y) as int
=======================================================================
indexbase 0
int i, j, c = 0;
for i=1 to i<npol
  j=i-1
  'for (i = 0, j = npol-1; i < npol; j = i++) {
  if ((((yp[i]<=y) && (y<yp[j])) ||
    ((yp[j]<=y) && (y<yp[i]))) &&
    (x < (xp[j] - xp[i]) * (y - yp[i]) / (yp[j] - yp[i]) + xp[i]))
    c = not c;
  endif
next
return c;
end function

'TEST
float px={0,1,1,0}
float py={0,0,1,1}
float x,y
int n=4
float x=0.5, y=0.5
print pnpoly(n,px,py,x,y)
print pnpoly(n,px,py,x+0.6,y)
print pnpoly(n,px,py,x,y+0.6)


[Nicola_Piano]

Thanks Charles,
tomorrow I'll try it and tell you.

[Charles Pegge]

It splits into 3 useful functions which are much easier to understand:

Code Select Expand


'12:07 08/04/2022
'POINT INSIDE POLY

function inrange(float p,a,b) as int
====================================
if b<=p and p<a
  return -1
elseif a<=p and p<b
  return -1
endif
end function


function interpolate (float y,x1,x2,y1,y2) as float
===================================================
if y1=y2 'horizontal line
  y2=y1*+1.00001 'heuristic to avoid infinity issues
endif
return x1 + (x2 - x1) * (y - y1) / (y2 - y1)
end function


function pnpoly(int npol, float *xp,*yp, x, y) as int
=====================================================
indexbase 0
int i, j,c=0
for i=1 to i<npol
  j=i-1
  if inrange(y,yp[i],yp[j])
    if x < interpolate( y, xp[i], xp[j], yp[i], yp[j] )
      c= not c 'toggle
    endif
  endif
next
return c
end function

'TEST
uses console
float px={0,1,1,0}
float py={0,0,1,1}
float x,y
int n=4
def printi print "%1:   " cr : %1
def printd print "%1:   " %1 cr
printi ( float x=0.5, y=0.5)
printd ( pnpoly(n,px,py,x,y)     )
printd ( pnpoly(n,px,py,x+0.4,y) )
printd ( pnpoly(n,px,py,x,y+0.4) )
printd ( pnpoly(n,px,py,x+0.6,y) )
printd ( pnpoly(n,px,py,x,y+0.6) )
pause


[Nicola_Piano]

Hi Charles,
I tried the function, it seems to be fine, but doing a practical application with gps coordinates unfortunately is not good ... in my opinion the original algorithm was already wrong ...
See the attachment.
Hello

[Charles Pegge]

I recognize the algorithm as it is used to determine bounded areas for hatching or shading in CAD, but I think it is incomplete.

https://www.tutorialspoint.com/Check-if-a-given-point-lies-inside-a-Polygon

[Chris Chancellor]

Good this is interesting!
hope you guys can sort out this algo as it is fairly useful to have a routine like this ?

[Charles Pegge]

I have a solution using intersections in the inc/glo2/geoplanar.inc library. It is long but robust and tolerant of marginal cases. I'm still perfecting it.

The image below is a 4 sided shape with an inner shape. It is bombarded with 1000 points. Those which land 'inside' are highlighted yellow.

[Charles Pegge]

Using very similar techniques, we can do cross-hatching on the interior.

the shape is 'scanned' by each hatching line for intersection points. These are collected for each line scan, sorted into ascending order of x, then pairs of points are used to draw the line segments in the interior.

[Nicola_Piano]

Hi Charles,
it seems to me really fantastic what you managed to do. I saw the GeoPlanar.inc file, what is missing is an explanation of the various functions and input variables. Could you post an example?
Thanks.

[Charles Pegge]

Hi Nicola,

I need another day or two, since I am still working on geoplanar, and maybe a few more examples. I also wanted to demonstrate Delaunay triangles, and their complement Voronoi diagrams, but that may be too ambitious in a short space of time.

[Nicola_Piano]

Charles,
take your time ... the subject is quite peculiar and certainly needs special attention.

[Nicola_Piano]

Hi Charles,
I managed with the intersected function to find the points that are internal or external to the polygon.
I counted the total intersections with all sides and I verified that they were odd, if they were even the point is external ....
it seems to work ....
I try it a little.
The intersected algorithm is exceptional. Thanks.

[Charles Pegge]

Hi Nicola,

I'm still chasing a few anomalies like the missing hatch line:

[Nicola_Piano]

Hi Charles,
how is your research going? I am eager to see progress.