Geo Fencing - point inside/outside polygon

If i remember correctly, the algorithm is to draw a horizontal line through your test point. Count how many lines of of the polygon you intersect to reach your point.

If the answer is odd, you're inside. If the answer is even, you're outside.

Edit: Yeah, what he said (Wikipedia):

By far the best explanation and implementation can be found at Point In Polygon Winding Number Inclusion

There is even a C++ implementation at the end of the well explained article. This site also contains some great algorithms/solutions for other geometry based problems.

I have modified and used the C++ implementation and also created a C# implementation. You definitely want to use the Winding Number algorithm as it is more accurate than the edge crossing algorithm and it is very fast.

Relating to kobers answer I worked it out with more readable clean code and changed the longitudes that crosses the date border:

public bool IsPointInPolygon(List<PointPosition> polygon, double latitude, double longitude)
{
  bool isInIntersection = false;
  int actualPointIndex = 0;
  int pointIndexBeforeActual = polygon.Count - 1;

  var offset = calculateLonOffsetFromDateLine(polygon);
  longitude = longitude < 0.0 ? longitude + offset : longitude;

  foreach (var actualPointPosition in polygon)
  {
    var p1Lat = actualPointPosition.Latitude;
    var p1Lon = actualPointPosition.Longitude;

    var p0Lat = polygon[pointIndexBeforeActual].Latitude;
    var p0Lon = polygon[pointIndexBeforeActual].Longitude;

    if (p1Lon < 0.0) p1Lon += offset;
    if (p0Lon < 0.0) p0Lon += offset;

    // Jordan curve theorem - odd even rule algorithm
    if (isPointLatitudeBetweenPolyLine(p0Lat, p1Lat, latitude)
    && isPointRightFromPolyLine(p0Lat, p0Lon, p1Lat, p1Lon, latitude, longitude))
    {
      isInIntersection = !isInIntersection;
    }

    pointIndexBeforeActual = actualPointIndex;
    actualPointIndex++;
  }

  return isInIntersection;
}

private double calculateLonOffsetFromDateLine(List<PointPosition> polygon)
{
  double offset = 0.0;
  var maxLonPoly = polygon.Max(x => x.Longitude);
  var minLonPoly = polygon.Min(x => x.Longitude);
  if (Math.Abs(minLonPoly - maxLonPoly) > 180)
  {
    offset = 360.0;
  }

  return offset;
}

private bool isPointLatitudeBetweenPolyLine(double polyLinePoint1Lat, double polyLinePoint2Lat, double poiLat)
{
  return polyLinePoint2Lat <= poiLat && poiLat < polyLinePoint1Lat || polyLinePoint1Lat <= poiLat && poiLat < polyLinePoint2Lat;
}

private bool isPointRightFromPolyLine(double polyLinePoint1Lat, double polyLinePoint1Lon, double polyLinePoint2Lat, double polyLinePoint2Lon, double poiLat, double poiLon)
{
  // lon <(lon1-lon2)*(latp-lat2)/(lat1-lat2)+lon2
  return poiLon < (polyLinePoint1Lon - polyLinePoint2Lon) * (poiLat - polyLinePoint2Lat) / (polyLinePoint1Lat - polyLinePoint2Lat) + polyLinePoint2Lon;
}

Check if a point is inside a polygon or not -

Consider the polygon which has vertices a1,a2,a3,a4,a5. The following set of steps should help in ascertaining whether point P lies inside the polygon or outside.

Compute the vector area of the triangle formed by edge a1->a2 and the vectors connecting a2 to P and P to a1. Similarly, compute the vector area of the each of the possible triangles having one side as the side of the polygon and the other two connecting P to that side.

For a point to be inside a polygon, each of the triangles need to have positive area. Even if one of the triangles have a negative area then the point P stands out of the polygon.

In order to compute the area of a triangle given vectors representing its 3 edges, refer to http://www.jtaylor1142001.net/calcjat/Solutions/VCrossProduct/VCPATriangle.htm

The problem is easier if your polygon is convex. If so, you can do a simple test for each line to see if the point is on the inside or outside of that line (extending to infinity in both directions). Otherwise, for concave polygons, draw an imaginary ray from your point out to infinity (in any direction). Count how many times it crosses a boundary line. Odd means the point is inside, even means the point is outside.

This last algorithm is trickier than it looks. You will have to be very careful about what happens when your imaginary ray exactly hits one of the polygon's vertices.

If your imaginary ray goes in the -x direction, you can choose only to count lines that include at least one point whose y coordinate is strictly less than the y coordinate of your point. This is how you get most of the weird edge cases to work correctly.

If you have a simple polygon (none of the lines cross) and you don't have holes you can also triangulate the polygon, which you are probably going to do anyway in a GIS app to draw a TIN, then test for points in each triangle. If you have a small number of edges to the polygon but a large number of points this is fast.

For an interesting point in triangle see link text

Otherwise definately use the winding rule rather than edge crossing, edge crossing has real problems with points on edges, which if your data is generated form a GPS with limited precision is very likely.

Jan's answer is great.

Here is the same code using the GeoCoordinate class instead.

using System.Device.Location;

...

public static bool IsPointInPolygon(List<GeoCoordinate> poly, GeoCoordinate point)
{
    int i, j;
    bool c = false;
    for (i = 0, j = poly.Count - 1; i < poly.Count; j = i++)
    {
        if ((((poly[i].Latitude <= point.Latitude) && (point.Latitude < poly[j].Latitude))
                || ((poly[j].Latitude <= point.Latitude) && (point.Latitude < poly[i].Latitude)))
                && (point.Longitude < (poly[j].Longitude - poly[i].Longitude) * (point.Latitude - poly[i].Latitude)
                    / (poly[j].Latitude - poly[i].Latitude) + poly[i].Longitude))
            c = !c;
    }

    return c;
}

You can try this simple class https://github.com/xopbatgh/sb-polygon-pointer

It is easy to deal with it

1. You just insert polygon coordinates into array
2. Ask library is desired point with lat/lng inside the polygon

$polygonBox = [
    [55.761515, 37.600375],
    [55.759428, 37.651156],
    [55.737112, 37.649566],
    [55.737649, 37.597301],
];

$sbPolygonEngine = new sbPolygonEngine($polygonBox);

$isCrosses = $sbPolygonEngine->isCrossesWith(55.746768, 37.625605);

// $isCrosses is boolean

(answer was returned from deleted by myself because initially it was formatted wrong)