How to check if a point (x,y) is inside a polygon in the Cartesian coordinate system?

Question

This question already has an answer here:
Point in Polygon aka hit test
C# Point in polygon

Given a random polygon formulated with N line equations in the Cartesian coordinate system, is there any standard formula that is used to check for membership of a point (x,y)?

The simple solution is to get all the line formulas and check if point X is below this line, above that line and to the right of the other line, etc. But this will probably be tedious.

I should note that the polygon can be of any shape with any number of sides and may concave or convex.

For convenience I have already added these utility functions:

float slope(CGPoint p1, CGPoint p2)
{
    return (p2.y - p1.y) / (p2.x - p1.x);
}

CGPoint pointOnLineWithY(CGPoint p, float m, float y)
{
    float x = (y - p.y)/m + p.x;
    return CGPointMake(x,y);
}

CGPoint pointOnLineWithX(CGPoint p, float m, float x)
{
    float y = m*(x - p.x) + p.y;
    return CGPointMake(x, y);
}

This was already discussed in length here, see http://stackoverflow.com/questions/217578/point-in-polygon-aka-hit-test — Stefan Haustein, May 20 '12 at 13:21

score 20 · Accepted Answer · answered May 20 '12 at 13:21

If you have the vertices, you can compute the sum of the angles made between the test point and each pair of points making up the polygon. If it is 2*pi, then it is an interior point. If it is 0, then it is an exterior point.

Some code:

    typedef struct {
   int h,v;
} Point;

int InsidePolygon(Point *polygon,int n,Point p)
{
   int i;
   double angle=0;
   Point p1,p2;

   for (i=0;i<n;i++) {
      p1.h = polygon[i].h - p.h;
      p1.v = polygon[i].v - p.v;
      p2.h = polygon[(i+1)%n].h - p.h;
      p2.v = polygon[(i+1)%n].v - p.v;
      angle += Angle2D(p1.h,p1.v,p2.h,p2.v);
   }

   if (ABS(angle) < PI)
      return(FALSE);
   else
      return(TRUE);
}

/*
   Return the angle between two vectors on a plane
   The angle is from vector 1 to vector 2, positive anticlockwise
   The result is between -pi -> pi
*/
double Angle2D(double x1, double y1, double x2, double y2)
{
   double dtheta,theta1,theta2;

   theta1 = atan2(y1,x1);
   theta2 = atan2(y2,x2);
   dtheta = theta2 - theta1;
   while (dtheta > PI)
      dtheta -= TWOPI;
   while (dtheta < -PI)
      dtheta += TWOPI;

   return(dtheta);
}

Source: http://paulbourke.net/geometry/insidepoly/

Other places you can take a look at: http://alienryderflex.com/polygon/

http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html

http://sidvind.com/wiki/Point-in-polygon:_Jordan_Curve_Theorem

Would this also work for concave polygons, like a "U"-shaped polygon? — Martijn, Sep 18 '14 at 10:38
@Martijn yes, according to this article: http://www.gamasutra.com/view/feature/131836/crashing_into_the_new_year_.php — , Feb 10 '15 at 04:43
@Martijn also note this is considered the "worst algorithm in the world" for this task, but only in terms of speed. i think it is best in terms of comprehension: http://erich.realtimerendering.com/ptinpoly/ — , Feb 10 '15 at 04:48
the links are old and expired. Does this apply to both concave and convex polygons ? — criticalth, May 29 '18 at 13:30

How to check if a point (x,y) is inside a polygon in the Cartesian coordinate system?

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