I'm trying to create a fast 2D point inside polygon algorithm, for use in hittesting (e.g. Polygon.contains(p:Point)
). Suggestions for effective techniques would be appreciated.

You forgot to tell us about your perceptions on the question of right or left handedness  which can also be interpreted as "inside" vs "outside"  RT– Richard TOct 20 '08 at 5:40

16Yes, I realize now the question leaves many details unspecified, but at this point I'm sorta interested in seeing the variety of responses.– Scott EverndenOct 20 '08 at 6:03

5A 90 sided polygon is called a enneacontagon and a 10,000 sided polygon is called a myriagon.– user263678Feb 1 '10 at 16:28

"Most elegant" is out of the target, since I have had trouble with finding a "work at all"algorithm. I must figure it out myself : stackoverflow.com/questions/14818567/…– davidkonradAug 17 '13 at 19:21
For graphics, I'd rather not prefer integers. Many systems use integers for UI painting (pixels are ints after all), but macOS, for example, uses float for everything. macOS only knows points and a point can translate to one pixel, but depending on monitor resolution, it might translate to something else. On retina screens half a point (0.5/0.5) is pixel. Still, I never noticed that macOS UIs are significantly slower than other UIs. After all, 3D APIs (OpenGL or Direct3D) also work with floats and modern graphics libraries very often take advantage of GPU acceleration.
Now you said speed is your main concern, okay, let's go for speed. Before you run any sophisticated algorithm, first do a simple test. Create an axis aligned bounding box around your polygon. This is very easy, fast and can already save you a lot of calculations. How does that work? Iterate over all points of the polygon and find the min/max values of X and Y.
E.g. you have the points (9/1), (4/3), (2/7), (8/2), (3/6)
. This means Xmin is 2, Xmax is 9, Ymin is 1 and Ymax is 7. A point outside of the rectangle with the two edges (2/1) and (9/7) cannot be within the polygon.
// p is your point, p.x is the x coord, p.y is the y coord
if (p.x < Xmin  p.x > Xmax  p.y < Ymin  p.y > Ymax) {
// Definitely not within the polygon!
}
This is the first test to run for any point. As you can see, this test is ultra fast but it's also very coarse. To handle points that are within the bounding rectangle, we need a more sophisticated algorithm. There are a couple of ways how this can be calculated. Which method works also depends on whether the polygon can have holes or will always be solid. Here are examples of solid ones (one convex, one concave):
And here's one with a hole:
The green one has a hole in the middle!
The easiest algorithm, that can handle all three cases above and is still pretty fast is named ray casting. The idea of the algorithm is pretty simple: Draw a virtual ray from anywhere outside the polygon to your point and count how often it hits a side of the polygon. If the number of hits is even, it's outside of the polygon, if it's odd, it's inside.
The winding number algorithm would be an alternative, it is more accurate for points being very close to a polygon line but it's also much slower. Ray casting may fail for points too close to a polygon side because of limited floating point precision and rounding issues, but in reality that is hardly a problem, as if a point lies that close to a side, it's often visually not even possible for a viewer to recognize if it is already inside or still outside.
You still have the bounding box of above, remember? Just pick a point outside the bounding box and use it as starting point for your ray. E.g. the point (Xmin  e/p.y)
is outside the polygon for sure.
But what is e
? Well, e
(actually epsilon) gives the bounding box some padding. As I said, ray tracing fails if we start too close to a polygon line. Since the bounding box might equal the polygon (if the polygon is an axis aligned rectangle, the bounding box is equal to the polygon itself!), we need some padding to make this safe, that's all. How big should you choose e
? Not too big. It depends on the coordinate system scale you use for drawing. If your pixel step width is 1.0, then just choose 1.0 (yet 0.1 would have worked as well)
Now that we have the ray with its start and end coordinates, the problem shifts from "is the point within the polygon" to "how often does the ray intersects a polygon side". Therefore we can't just work with the polygon points as before, now we need the actual sides. A side is always defined by two points.
side 1: (X1/Y1)(X2/Y2)
side 2: (X2/Y2)(X3/Y3)
side 3: (X3/Y3)(X4/Y4)
:
You need to test the ray against all sides. Consider the ray to be a vector and every side to be a vector. The ray has to hit each side exactly once or never at all. It can't hit the same side twice. Two lines in 2D space will always intersect exactly once, unless they are parallel, in which case they never intersect. However since vectors have a limited length, two vectors might not be parallel and still never intersect because they are too short to ever meet each other.
// Test the ray against all sides
int intersections = 0;
for (side = 0; side < numberOfSides; side++) {
// Test if current side intersects with ray.
// If yes, intersections++;
}
if ((intersections & 1) == 1) {
// Inside of polygon
} else {
// Outside of polygon
}
So far so well, but how do you test if two vectors intersect? Here's some C code (not tested), that should do the trick:
#define NO 0
#define YES 1
#define COLLINEAR 2
int areIntersecting(
float v1x1, float v1y1, float v1x2, float v1y2,
float v2x1, float v2y1, float v2x2, float v2y2
) {
float d1, d2;
float a1, a2, b1, b2, c1, c2;
// Convert vector 1 to a line (line 1) of infinite length.
// We want the line in linear equation standard form: A*x + B*y + C = 0
// See: http://en.wikipedia.org/wiki/Linear_equation
a1 = v1y2  v1y1;
b1 = v1x1  v1x2;
c1 = (v1x2 * v1y1)  (v1x1 * v1y2);
// Every point (x,y), that solves the equation above, is on the line,
// every point that does not solve it, is not. The equation will have a
// positive result if it is on one side of the line and a negative one
// if is on the other side of it. We insert (x1,y1) and (x2,y2) of vector
// 2 into the equation above.
d1 = (a1 * v2x1) + (b1 * v2y1) + c1;
d2 = (a1 * v2x2) + (b1 * v2y2) + c1;
// If d1 and d2 both have the same sign, they are both on the same side
// of our line 1 and in that case no intersection is possible. Careful,
// 0 is a special case, that's why we don't test ">=" and "<=",
// but "<" and ">".
if (d1 > 0 && d2 > 0) return NO;
if (d1 < 0 && d2 < 0) return NO;
// The fact that vector 2 intersected the infinite line 1 above doesn't
// mean it also intersects the vector 1. Vector 1 is only a subset of that
// infinite line 1, so it may have intersected that line before the vector
// started or after it ended. To know for sure, we have to repeat the
// the same test the other way round. We start by calculating the
// infinite line 2 in linear equation standard form.
a2 = v2y2  v2y1;
b2 = v2x1  v2x2;
c2 = (v2x2 * v2y1)  (v2x1 * v2y2);
// Calculate d1 and d2 again, this time using points of vector 1.
d1 = (a2 * v1x1) + (b2 * v1y1) + c2;
d2 = (a2 * v1x2) + (b2 * v1y2) + c2;
// Again, if both have the same sign (and neither one is 0),
// no intersection is possible.
if (d1 > 0 && d2 > 0) return NO;
if (d1 < 0 && d2 < 0) return NO;
// If we get here, only two possibilities are left. Either the two
// vectors intersect in exactly one point or they are collinear, which
// means they intersect in any number of points from zero to infinite.
if ((a1 * b2)  (a2 * b1) == 0.0f) return COLLINEAR;
// If they are not collinear, they must intersect in exactly one point.
return YES;
}
The input values are the two endpoints of vector 1 (v1x1/v1y1
and v1x2/v1y2
) and vector 2 (v2x1/v2y1
and v2x2/v2y2
). So you have 2 vectors, 4 points, 8 coordinates. YES
and NO
are clear. YES
increases intersections, NO
does nothing.
What about COLLINEAR? It means both vectors lie on the same infinite line, depending on position and length, they don't intersect at all or they intersect in an endless number of points. I'm not absolutely sure how to handle this case, I would not count it as intersection either way. Well, this case is rather rare in practice anyway because of floating point rounding errors; better code would probably not test for == 0.0f
but instead for something like < epsilon
, where epsilon is a rather small number.
If you need to test a larger number of points, you can certainly speed up the whole thing a bit by keeping the linear equation standard forms of the polygon sides in memory, so you don't have to recalculate these every time. This will save you two floating point multiplications and three floating point subtractions on every test in exchange for storing three floating point values per polygon side in memory. It's a typical memory vs computation time trade off.
Last but not least: If you may use 3D hardware to solve the problem, there is an interesting alternative. Just let the GPU do all the work for you. Create a painting surface that is off screen. Fill it completely with the color black. Now let OpenGL or Direct3D paint your polygon (or even all of your polygons if you just want to test if the point is within any of them, but you don't care for which one) and fill the polygon(s) with a different color, e.g. white. To check if a point is within the polygon, get the color of this point from the drawing surface. This is just a O(1) memory fetch.
Of course this method is only usable if your drawing surface doesn't have to be huge. If it cannot fit into the GPU memory, this method is slower than doing it on the CPU. If it would have to be huge and your GPU supports modern shaders, you can still use the GPU by implementing the ray casting shown above as a GPU shader, which absolutely is possible. For a larger number of polygons or a large number of points to test, this will pay off, consider some GPUs will be able to test 64 to 256 points in parallel. Note however that transferring data from CPU to GPU and back is always expensive, so for just testing a couple of points against a couple of simple polygons, where either the points or the polygons are dynamic and will change frequently, a GPU approach will rarely pay off.

29+1 Fantastic answer. On using the hardware to do it, I've read in other places that it can be slow because you have to get data back from the graphics card. But I do like the principle of taking load off the CPU a lot. Does anyone have any good references for how this might be done in OpenGL?– GavinDec 29 '09 at 4:32

6+1 because this is so simple! The main problem is if your polygon and test point line up on a grid (not uncommon) then you have to deal with "duplicitous" intersections, for example, straight through a polygon point! (yielding a parity of two instead of one). Gets into this weird area: stackoverflow.com/questions/2255842/… . Computer Graphics is full of these special cases: simple in theory, hairy in practice. Mar 26 '10 at 19:55

9@RMorrisey: Why do you think so? I fail to see how it would fail for a concave polygon. The ray may leave and reenter the polygon multiple times when the polygon is concave, but in the end, the hit counter will be odd if the point is within and even if it is outside, also for concave polygons.– MeckiMay 31 '11 at 14:47

9The 'Fast Winding Number Algorithm', described at softsurfer.com/Archive/algorithm_0103/algorithm_0103.htm works pretty fast...– S PJan 8 '12 at 14:35

17Your usage of / to separate x and y coordinates is confusing, it reads as x divided by y. It's much more clear to use x, y (i.e. x comma y) Overall, a useful answer.– AshSep 14 '13 at 8:56
I think the following piece of code is the best solution (taken from here):
int pnpoly(int nvert, float *vertx, float *verty, float testx, float testy)
{
int i, j, c = 0;
for (i = 0, j = nvert1; i < nvert; j = i++) {
if ( ((verty[i]>testy) != (verty[j]>testy)) &&
(testx < (vertx[j]vertx[i]) * (testyverty[i]) / (verty[j]verty[i]) + vertx[i]) )
c = !c;
}
return c;
}
Arguments
 nvert: Number of vertices in the polygon. Whether to repeat the first vertex at the end has been discussed in the article referred above.
 vertx, verty: Arrays containing the x and ycoordinates of the polygon's vertices.
 testx, testy: X and ycoordinate of the test point.
It's both short and efficient and works both for convex and concave polygons. As suggested before, you should check the bounding rectangle first and treat polygon holes separately.
The idea behind this is pretty simple. The author describes it as follows:
I run a semiinfinite ray horizontally (increasing x, fixed y) out from the test point, and count how many edges it crosses. At each crossing, the ray switches between inside and outside. This is called the Jordan curve theorem.
The variable c is switching from 0 to 1 and 1 to 0 each time the horizontal ray crosses any edge. So basically it's keeping track of whether the number of edges crossed are even or odd. 0 means even and 1 means odd.

5Question. What exactly are the variables that I pass it? What do they represent? Feb 2 '12 at 4:09

9@Mick It checks that
verty[i]
andverty[j]
are either side oftesty
, so they are never equal. Jun 14 '12 at 8:55 
4This code is not robust, and I would not recommend using it. Here is a link giving some detailed analysis: wwwma2.upc.es/geoc/SchirrapointPolygon.pdf– MikolaMay 6 '14 at 4:29

15This approach actually DOES have limitations (edgecases): Checking the Point (15,20) in the Polygon [(10,10),(10,20),(20,20),(20,10)] will return false instead of true. Same with (10,20) or (20,15). For all other cases, the algorithm works perfectly fine and the falsenegatives in edgecases are ok for my application. Oct 25 '14 at 13:34

14@Alexander, this is in fact by design: by handling left & bottom boundaries in the opposite sense to top & right boundaries, should two distinct polygons share an edge, any point along this edge will be located in one and only one polygon. ..a useful property.– wardwMay 3 '15 at 18:10
Here is a C# version of the answer given by nirg, which comes from this RPI professor. Note that use of the code from that RPI source requires attribution.
A bounding box check has been added at the top. However, as James Brown points out, the main code is almost as fast as the bounding box check itself, so the bounding box check can actually slow the overall operation, in the case that most of the points you are checking are inside the bounding box. So you could leave the bounding box check out, or an alternative would be to precompute the bounding boxes of your polygons if they don't change shape too often.
public bool IsPointInPolygon( Point p, Point[] polygon )
{
double minX = polygon[ 0 ].X;
double maxX = polygon[ 0 ].X;
double minY = polygon[ 0 ].Y;
double maxY = polygon[ 0 ].Y;
for ( int i = 1 ; i < polygon.Length ; i++ )
{
Point q = polygon[ i ];
minX = Math.Min( q.X, minX );
maxX = Math.Max( q.X, maxX );
minY = Math.Min( q.Y, minY );
maxY = Math.Max( q.Y, maxY );
}
if ( p.X < minX  p.X > maxX  p.Y < minY  p.Y > maxY )
{
return false;
}
// https://wrf.ecse.rpi.edu/Research/Short_Notes/pnpoly.html
bool inside = false;
for ( int i = 0, j = polygon.Length  1 ; i < polygon.Length ; j = i++ )
{
if ( ( polygon[ i ].Y > p.Y ) != ( polygon[ j ].Y > p.Y ) &&
p.X < ( polygon[ j ].X  polygon[ i ].X ) * ( p.Y  polygon[ i ].Y ) / ( polygon[ j ].Y  polygon[ i ].Y ) + polygon[ i ].X )
{
inside = !inside;
}
}
return inside;
}

5Works great, thanks, I converted to JavaScript. stackoverflow.com/questions/217578/… Jul 5 '13 at 14:12

2This is >1000x faster than using GraphicsPath.IsVisible!! The bounding box check makes the function about 70% slower. Oct 1 '15 at 3:06

Not only GraphicsPath.IsVisible() is way slower but also it doesn't work well with very small polygons with side in the 0.01f range Jan 21 '20 at 6:45

What's the point of the first
for
andif
? The lastfor
works just fine for all cases.– GDavoliJul 31 '20 at 6:47 
1@GDavoli It's an efficiency thing. If the point is not inside the bounding box of the polygon. it can't be in the polygon. This is based on the assumption that the loop to find the bounding box of the polygon is significantly faster than the second for loop. That might not be true on modern hardware. But in a real system it may make sense to cache the bounding box of each polygon, in which case the bounding box check definitely makes sense.– M KatzAug 1 '20 at 9:55
Here is a JavaScript variant of the answer by M. Katz based on Nirg's approach:
function pointIsInPoly(p, polygon) {
var isInside = false;
var minX = polygon[0].x, maxX = polygon[0].x;
var minY = polygon[0].y, maxY = polygon[0].y;
for (var n = 1; n < polygon.length; n++) {
var q = polygon[n];
minX = Math.min(q.x, minX);
maxX = Math.max(q.x, maxX);
minY = Math.min(q.y, minY);
maxY = Math.max(q.y, maxY);
}
if (p.x < minX  p.x > maxX  p.y < minY  p.y > maxY) {
return false;
}
var i = 0, j = polygon.length  1;
for (i, j; i < polygon.length; j = i++) {
if ( (polygon[i].y > p.y) != (polygon[j].y > p.y) &&
p.x < (polygon[j].x  polygon[i].x) * (p.y  polygon[i].y) / (polygon[j].y  polygon[i].y) + polygon[i].x ) {
isInside = !isInside;
}
}
return isInside;
}
Compute the oriented sum of angles between the point p and each of the polygon apices. If the total oriented angle is 360 degrees, the point is inside. If the total is 0, the point is outside.
I like this method better because it is more robust and less dependent on numerical precision.
Methods that compute evenness of number of intersections are limited because you can 'hit' an apex during the computation of the number of intersections.
EDIT: By The Way, this method works with concave and convex polygons.
EDIT: I recently found a whole Wikipedia article on the topic.

1Nope, this is not true. This works regardless of the convexity of the polygon. Oct 20 '08 at 13:08

2@DarenW: Only one acos per vertex; on the other hand, this algorithm should be the easiest to implement in SIMD because it has absolutely no branching. Jul 5 '09 at 22:38

2First use bounding box check to solve "point is far" cases. For trig, you could use pregenerated tables.– JOMJun 15 '10 at 19:17

2I realize this is old, I'm not sure if anyone will see it, but... David, what is the "oriented sum of angles"? Is this simply the angle between me and the point in question, 0..360? If so, don't you need to consider the number of points in the poly? Isn't it 360 only for fourpoint polys? Sep 16 '14 at 15:51

4This is the optimal solution, since it is O(n), and requires minimal computation. Works for all polygons. I researched this solution 30 years ago at my first IBM job. They signed off on it, and still use it today in their GIS technologies. Jul 21 '16 at 20:35
This question is so interesting. I have another workable idea different from other answers to this post. The idea is to use the sum of angles to decide whether the target is inside or outside. Better known as winding number.
Let x be the target point. Let array [0, 1, .... n] be the all the points of the area. Connect the target point with every border point with a line. If the target point is inside of this area. The sum of all angles will be 360 degrees. If not the angles will be less than 360.
Refer to this image to get a basic understanding of the idea:
My algorithm assumes the clockwise is the positive direction. Here is a potential input:
[[122.402015, 48.225216], [117.032049, 48.999931], [116.919132, 45.995175], [124.079107, 46.267259], [124.717175, 48.377557], [122.92315, 47.047963], [122.402015, 48.225216]]
The following is the python code that implements the idea:
def isInside(self, border, target):
degree = 0
for i in range(len(border)  1):
a = border[i]
b = border[i + 1]
# calculate distance of vector
A = getDistance(a[0], a[1], b[0], b[1]);
B = getDistance(target[0], target[1], a[0], a[1])
C = getDistance(target[0], target[1], b[0], b[1])
# calculate direction of vector
ta_x = a[0]  target[0]
ta_y = a[1]  target[1]
tb_x = b[0]  target[0]
tb_y = b[1]  target[1]
cross = tb_y * ta_x  tb_x * ta_y
clockwise = cross < 0
# calculate sum of angles
if(clockwise):
degree = degree + math.degrees(math.acos((B * B + C * C  A * A) / (2.0 * B * C)))
else:
degree = degree  math.degrees(math.acos((B * B + C * C  A * A) / (2.0 * B * C)))
if(abs(round(degree)  360) <= 3):
return True
return False
The Eric Haines article cited by bobobobo is really excellent. Particularly interesting are the tables comparing performance of the algorithms; the angle summation method is really bad compared to the others. Also interesting is that optimisations like using a lookup grid to further subdivide the polygon into "in" and "out" sectors can make the test incredibly fast even on polygons with > 1000 sides.
Anyway, it's early days but my vote goes to the "crossings" method, which is pretty much what Mecki describes I think. However I found it most succintly described and codified by David Bourke. I love that there is no real trigonometry required, and it works for convex and concave, and it performs reasonably well as the number of sides increases.
By the way, here's one of the performance tables from the Eric Haines' article for interest, testing on random polygons.
number of edges per polygon
3 4 10 100 1000
MacMartin 2.9 3.2 5.9 50.6 485
Crossings 3.1 3.4 6.8 60.0 624
Triangle Fan+edge sort 1.1 1.8 6.5 77.6 787
Triangle Fan 1.2 2.1 7.3 85.4 865
Barycentric 2.1 3.8 13.8 160.7 1665
Angle Summation 56.2 70.4 153.6 1403.8 14693
Grid (100x100) 1.5 1.5 1.6 2.1 9.8
Grid (20x20) 1.7 1.7 1.9 5.7 42.2
Bins (100) 1.8 1.9 2.7 15.1 117
Bins (20) 2.1 2.2 3.7 26.3 278
Really like the solution posted by Nirg and edited by bobobobo. I just made it javascript friendly and a little more legible for my use:
function insidePoly(poly, pointx, pointy) {
var i, j;
var inside = false;
for (i = 0, j = poly.length  1; i < poly.length; j = i++) {
if(((poly[i].y > pointy) != (poly[j].y > pointy)) && (pointx < (poly[j].xpoly[i].x) * (pointypoly[i].y) / (poly[j].ypoly[i].y) + poly[i].x) ) inside = !inside;
}
return inside;
}
Swift version of the answer by nirg:
extension CGPoint {
func isInsidePolygon(vertices: [CGPoint]) > Bool {
guard !vertices.isEmpty else { return false }
var j = vertices.last!, c = false
for i in vertices {
let a = (i.y > y) != (j.y > y)
let b = (x < (j.x  i.x) * (y  i.y) / (j.y  i.y) + i.x)
if a && b { c = !c }
j = i
}
return c
}
}

This has a potential divide by zero problem in calculating b. Need to only calculate "b" and the next line with "c" if the calculation for "a" shows that there is a possibility of intersecting. No possibility if both points are above, or both points below  which is described by the calculation for "a". Mar 23 '20 at 20:08
I did some work on this back when I was a researcher under Michael Stonebraker  you know, the professor who came up with Ingres, PostgreSQL, etc.
We realized that the fastest way was to first do a bounding box because it's SUPER fast. If it's outside the bounding box, it's outside. Otherwise, you do the harder work...
If you want a great algorithm, look to the open source project PostgreSQL source code for the geo work...
I want to point out, we never got any insight into right vs left handedness (also expressible as an "inside" vs "outside" problem...
UPDATE
BKB's link provided a good number of reasonable algorithms. I was working on Earth Science problems and therefore needed a solution that works in latitude/longitude, and it has the peculiar problem of handedness  is the area inside the smaller area or the bigger area? The answer is that the "direction" of the verticies matters  it's either lefthanded or right handed and in this way you can indicate either area as "inside" any given polygon. As such, my work used solution three enumerated on that page.
In addition, my work used separate functions for "on the line" tests.
...Since someone asked: we figured out that bounding box tests were best when the number of verticies went beyond some number  do a very quick test before doing the longer test if necessary... A bounding box is created by simply taking the largest x, smallest x, largest y and smallest y and putting them together to make four points of a box...
Another tip for those that follow: we did all our more sophisticated and "lightdimming" computing in a grid space all in positive points on a plane and then reprojected back into "real" longitude/latitude, thus avoiding possible errors of wrapping around when one crossed line 180 of longitude and when handling polar regions. Worked great!


8You can easily create a bounding box  it's just the four points which use the greatest and least x and greatest and least y. More soon. Oct 20 '08 at 6:09

"...avoiding possible errors of wrapping around when one crossed line 180 of longitude and when handling polar regions." can you perhaps describe this in more detail? I think I can figure out how to move everything 'up' to avoid my polygon crossing 0 longitude, but I'm not clear on how to handle polygons that contain either of the poles...– tiriteaMar 12 '19 at 1:39
David Segond's answer is pretty much the standard general answer, and Richard T's is the most common optimization, though therre are some others. Other strong optimizations are based on less general solutions. For example if you are going to check the same polygon with lots of points, triangulating the polygon can speed things up hugely as there are a number of very fast TIN searching algorithms. Another is if the polygon and points are on a limited plane at low resolution, say a screen display, you can paint the polygon onto a memory mapped display buffer in a given colour, and check the color of a given pixel to see if it lies in the polygons.
Like many optimizations, these are based on specific rather than general cases, and yield beneifits based on amortized time rather than single usage.
Working in this field, i found Joeseph O'Rourkes 'Computation Geometry in C' ISBN 0521440343 to be a great help.
The trivial solution would be to divide the polygon to triangles and hit test the triangles as explained here
If your polygon is CONVEX there might be a better approach though. Look at the polygon as a collection of infinite lines. Each line dividing space into two. for every point it's easy to say if its on the one side or the other side of the line. If a point is on the same side of all lines then it is inside the polygon.

very fast, and can be applied to more general shapes. back around 1990, we had "curvigons" where some sides were circular arcs. By analyzing those sides into circular wedges and a pair of triangles joining the wedge to the origin (polygon centroid), hit testing was fast and easy.– DarenWMar 11 '09 at 14:57

2


You missed an important solution for the case of convex polygons: by comparing the point against a diagonal, you can halve the number of vertices. And repeating this process, you reduce to a single triangle in Log(N) operations rather than N. Feb 12 '18 at 16:39
Java Version:
public class Geocode {
private float latitude;
private float longitude;
public Geocode() {
}
public Geocode(float latitude, float longitude) {
this.latitude = latitude;
this.longitude = longitude;
}
public float getLatitude() {
return latitude;
}
public void setLatitude(float latitude) {
this.latitude = latitude;
}
public float getLongitude() {
return longitude;
}
public void setLongitude(float longitude) {
this.longitude = longitude;
}
}
public class GeoPolygon {
private ArrayList<Geocode> points;
public GeoPolygon() {
this.points = new ArrayList<Geocode>();
}
public GeoPolygon(ArrayList<Geocode> points) {
this.points = points;
}
public GeoPolygon add(Geocode geo) {
points.add(geo);
return this;
}
public boolean inside(Geocode geo) {
int i, j;
boolean c = false;
for (i = 0, j = points.size()  1; i < points.size(); j = i++) {
if (((points.get(i).getLongitude() > geo.getLongitude()) != (points.get(j).getLongitude() > geo.getLongitude())) &&
(geo.getLatitude() < (points.get(j).getLatitude()  points.get(i).getLatitude()) * (geo.getLongitude()  points.get(i).getLongitude()) / (points.get(j).getLongitude()  points.get(i).getLongitude()) + points.get(i).getLatitude()))
c = !c;
}
return c;
}
}
I realize this is old, but here is a ray casting algorithm implemented in Cocoa, in case anyone is interested. Not sure it is the most efficient way to do things, but it may help someone out.
 (BOOL)shape:(NSBezierPath *)path containsPoint:(NSPoint)point
{
NSBezierPath *currentPath = [path bezierPathByFlatteningPath];
BOOL result;
float aggregateX = 0; //I use these to calculate the centroid of the shape
float aggregateY = 0;
NSPoint firstPoint[1];
[currentPath elementAtIndex:0 associatedPoints:firstPoint];
float olderX = firstPoint[0].x;
float olderY = firstPoint[0].y;
NSPoint interPoint;
int noOfIntersections = 0;
for (int n = 0; n < [currentPath elementCount]; n++) {
NSPoint points[1];
[currentPath elementAtIndex:n associatedPoints:points];
aggregateX += points[0].x;
aggregateY += points[0].y;
}
for (int n = 0; n < [currentPath elementCount]; n++) {
NSPoint points[1];
[currentPath elementAtIndex:n associatedPoints:points];
//line equations in Ax + By = C form
float _A_FOO = (aggregateY/[currentPath elementCount])  point.y;
float _B_FOO = point.x  (aggregateX/[currentPath elementCount]);
float _C_FOO = (_A_FOO * point.x) + (_B_FOO * point.y);
float _A_BAR = olderY  points[0].y;
float _B_BAR = points[0].x  olderX;
float _C_BAR = (_A_BAR * olderX) + (_B_BAR * olderY);
float det = (_A_FOO * _B_BAR)  (_A_BAR * _B_FOO);
if (det != 0) {
//intersection points with the edges
float xIntersectionPoint = ((_B_BAR * _C_FOO)  (_B_FOO * _C_BAR)) / det;
float yIntersectionPoint = ((_A_FOO * _C_BAR)  (_A_BAR * _C_FOO)) / det;
interPoint = NSMakePoint(xIntersectionPoint, yIntersectionPoint);
if (olderX <= points[0].x) {
//doesn't matter in which direction the ray goes, so I send it rightward.
if ((interPoint.x >= olderX && interPoint.x <= points[0].x) && (interPoint.x > point.x)) {
noOfIntersections++;
}
} else {
if ((interPoint.x >= points[0].x && interPoint.x <= olderX) && (interPoint.x > point.x)) {
noOfIntersections++;
}
}
}
olderX = points[0].x;
olderY = points[0].y;
}
if (noOfIntersections % 2 == 0) {
result = FALSE;
} else {
result = TRUE;
}
return result;
}

5Note that if you are really doing it in Cocoa, then you can use the [NSBezierPath containsPoint:] method.– ThomasWMar 7 '12 at 9:45
ObjC version of nirg's answer with sample method for testing points. Nirg's answer worked well for me.
 (BOOL)isPointInPolygon:(NSArray *)vertices point:(CGPoint)test {
NSUInteger nvert = [vertices count];
NSInteger i, j, c = 0;
CGPoint verti, vertj;
for (i = 0, j = nvert1; i < nvert; j = i++) {
verti = [(NSValue *)[vertices objectAtIndex:i] CGPointValue];
vertj = [(NSValue *)[vertices objectAtIndex:j] CGPointValue];
if (( (verti.y > test.y) != (vertj.y > test.y) ) &&
( test.x < ( vertj.x  verti.x ) * ( test.y  verti.y ) / ( vertj.y  verti.y ) + verti.x) )
c = !c;
}
return (c ? YES : NO);
}
 (void)testPoint {
NSArray *polygonVertices = [NSArray arrayWithObjects:
[NSValue valueWithCGPoint:CGPointMake(13.5, 41.5)],
[NSValue valueWithCGPoint:CGPointMake(42.5, 56.5)],
[NSValue valueWithCGPoint:CGPointMake(39.5, 69.5)],
[NSValue valueWithCGPoint:CGPointMake(42.5, 84.5)],
[NSValue valueWithCGPoint:CGPointMake(13.5, 100.0)],
[NSValue valueWithCGPoint:CGPointMake(6.0, 70.5)],
nil
];
CGPoint tappedPoint = CGPointMake(23.0, 70.0);
if ([self isPointInPolygon:polygonVertices point:tappedPoint]) {
NSLog(@"YES");
} else {
NSLog(@"NO");
}
}

2

@ZevEisenberg but that would be too easy! Thanks for the note. I'll dig up that project at some point to see why I used a custom solution. I likely didn't know about
CGPathContainsPoint()
– JonMay 15 '14 at 20:11
There is nothing more beutiful than an inductive definition of a problem. For the sake of completeness here you have a version in prolog which might also clarify the thoughs behind ray casting:
Based on the simulation of simplicity algorithm in http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html
Some helper predicates:
exor(A,B): \+A,B;A,\+B.
in_range(Coordinate,CA,CB) : exor((CA>Coordinate),(CB>Coordinate)).
inside(false).
inside(_,[_[]]).
inside(X:Y, [X1:Y1,X2:Y2R]) : in_range(Y,Y1,Y2), X > ( ((X2X1)*(YY1))/(Y2Y1) + X1),toggle_ray, inside(X:Y, [X2:Y2R]); inside(X:Y, [X2:Y2R]).
get_line(_,_,[]).
get_line([XA:YA,XB:YB],[X1:Y1,X2:Y2R]): [XA:YA,XB:YB]=[X1:Y1,X2:Y2]; get_line([XA:YA,XB:YB],[X2:Y2R]).
The equation of a line given 2 points A and B (Line(A,B)) is:
(YBYA)
Y  YA =  * (X  XA)
(XBYB)
It is important that the direction of rotation for the line is setted to clockwise for boundaries and anticlockwise for holes. We are going to check whether the point (X,Y), i.e the tested point is at the left halfplane of our line (it is a matter of taste, it could also be the right side, but also the direction of boundaries lines has to be changed in that case), this is to project the ray from the point to the right (or left) and acknowledge the intersection with the line. We have chosen to project the ray in the horizontal direction (again it is a matter of taste, it could also be done in vertical with similar restrictions), so we have:
(XBXA)
X <  * (Y  YA) + XA
(YBYA)
Now we need to know if the point is at the left (or right) side of the line segment only, not the entire plane, so we need to restrict the search only to this segment, but this is easy since to be inside the segment only one point in the line can be higher than Y in the vertical axis. As this is a stronger restriction it needs to be the first to check, so we take first only those lines meeting this requirement and then check its possition. By the Jordan Curve theorem any ray projected to a polygon must intersect at an even number of lines. So we are done, we will throw the ray to the right and then everytime it intersects a line, toggle its state. However in our implementation we are goint to check the lenght of the bag of solutions meeting the given restrictions and decide the innership upon it. for each line in the polygon this have to be done.
is_left_half_plane(_,[],[],_).
is_left_half_plane(X:Y,[XA:YA,XB:YB], [[X1:Y1,X2:Y2]R], Test) : [XA:YA, XB:YB] = [X1:Y1, X2:Y2], call(Test, X , (((XB  XA) * (Y  YA)) / (YB  YA) + XA));
is_left_half_plane(X:Y, [XA:YA, XB:YB], R, Test).
in_y_range_at_poly(Y,[XA:YA,XB:YB],Polygon) : get_line([XA:YA,XB:YB],Polygon), in_range(Y,YA,YB).
all_in_range(Coordinate,Polygon,Lines) : aggregate(bag(Line), in_y_range_at_poly(Coordinate,Line,Polygon), Lines).
traverses_ray(X:Y, Lines, Count) : aggregate(bag(Line), is_left_half_plane(X:Y, Line, Lines, <), IntersectingLines), length(IntersectingLines, Count).
% This is the entry point predicate
inside_poly(X:Y,Polygon,Answer) : all_in_range(Y,Polygon,Lines), traverses_ray(X:Y, Lines, Count), (1 is mod(Count,2)>Answer=inside;Answer=outside).
C# version of nirg's answer is here: I'll just share the code. It may save someone some time.
public static bool IsPointInPolygon(IList<Point> polygon, Point testPoint) {
bool result = false;
int j = polygon.Count()  1;
for (int i = 0; i < polygon.Count(); i++) {
if (polygon[i].Y < testPoint.Y && polygon[j].Y >= testPoint.Y  polygon[j].Y < testPoint.Y && polygon[i].Y >= testPoint.Y) {
if (polygon[i].X + (testPoint.Y  polygon[i].Y) / (polygon[j].Y  polygon[i].Y) * (polygon[j].X  polygon[i].X) < testPoint.X) {
result = !result;
}
}
j = i;
}
return result;
}

this works in most of the cases but it is wrong and doesnt work properly always ! use the solution from M Katz which is correct Jul 23 '13 at 13:58
I've made a Python implementation of nirg's c++ code:
Inputs
 bounding_points: nodes that make up the polygon.
bounding_box_positions: candidate points to filter. (In my implementation created from the bounding box.
(The inputs are lists of tuples in the format:
[(xcord, ycord), ...]
)
Returns
 All the points that are inside the polygon.
def polygon_ray_casting(self, bounding_points, bounding_box_positions):
# Arrays containing the x and ycoordinates of the polygon's vertices.
vertx = [point[0] for point in bounding_points]
verty = [point[1] for point in bounding_points]
# Number of vertices in the polygon
nvert = len(bounding_points)
# Points that are inside
points_inside = []
# For every candidate position within the bounding box
for idx, pos in enumerate(bounding_box_positions):
testx, testy = (pos[0], pos[1])
c = 0
for i in range(0, nvert):
j = i  1 if i != 0 else nvert  1
if( ((verty[i] > testy ) != (verty[j] > testy)) and
(testx < (vertx[j]  vertx[i]) * (testy  verty[i]) / (verty[j]  verty[i]) + vertx[i]) ):
c += 1
# If odd, that means that we are inside the polygon
if c % 2 == 1:
points_inside.append(pos)
return points_inside
Again, the idea is taken from here
Most of the answers in this question are not handling all corner cases well. Some subtle corner cases like below: This is a javascript version with all corner cases well handled.
/** Get relationship between a point and a polygon using raycasting algorithm
* @param {{x:number, y:number}} P: point to check
* @param {{x:number, y:number}[]} polygon: the polygon
* @returns 1: outside, 0: on edge, 1: inside
*/
function relationPP(P, polygon) {
const between = (p, a, b) => p >= a && p <= b  p <= a && p >= b
let inside = false
for (let i = polygon.length1, j = 0; j < polygon.length; i = j, j++) {
const A = polygon[i]
const B = polygon[j]
// corner cases
if (P.x == A.x && P.y == A.y  P.x == B.x && P.y == B.y) return 0
if (A.y == B.y && P.y == A.y && between(P.x, A.x, B.x)) return 0
if (between(P.y, A.y, B.y)) { // if P inside the vertical range
// filter out "ray pass vertex" problem by treating the line a little lower
if (P.y == A.y && B.y >= A.y  P.y == B.y && A.y >= B.y) continue
// calc cross product `PA X PB`, P lays on left side of AB if c > 0
const c = (A.x  P.x) * (B.y  P.y)  (B.x  P.x) * (A.y  P.y)
if (c == 0) return 0
if ((A.y < B.y) == (c > 0)) inside = !inside
}
}
return inside? 1 : 1
}
.Net port:
static void Main(string[] args)
{
Console.Write("Hola");
List<double> vertx = new List<double>();
List<double> verty = new List<double>();
int i, j, c = 0;
vertx.Add(1);
vertx.Add(2);
vertx.Add(1);
vertx.Add(4);
vertx.Add(4);
vertx.Add(1);
verty.Add(1);
verty.Add(2);
verty.Add(4);
verty.Add(4);
verty.Add(1);
verty.Add(1);
int nvert = 6; //Vértices del poligono
double testx = 2;
double testy = 5;
for (i = 0, j = nvert  1; i < nvert; j = i++)
{
if (((verty[i] > testy) != (verty[j] > testy)) &&
(testx < (vertx[j]  vertx[i]) * (testy  verty[i]) / (verty[j]  verty[i]) + vertx[i]))
c = 1;
}
}
VBA VERSION:
Note: Remember that if your polygon is an area within a map that Latitude/Longitude are Y/X values as opposed to X/Y (Latitude = Y, Longitude = X) due to from what I understand are historical implications from way back when Longitude was not a measurement.
CLASS MODULE: CPoint
Private pXValue As Double
Private pYValue As Double
'''''X Value Property'''''
Public Property Get X() As Double
X = pXValue
End Property
Public Property Let X(Value As Double)
pXValue = Value
End Property
'''''Y Value Property'''''
Public Property Get Y() As Double
Y = pYValue
End Property
Public Property Let Y(Value As Double)
pYValue = Value
End Property
MODULE:
Public Function isPointInPolygon(p As CPoint, polygon() As CPoint) As Boolean
Dim i As Integer
Dim j As Integer
Dim q As Object
Dim minX As Double
Dim maxX As Double
Dim minY As Double
Dim maxY As Double
minX = polygon(0).X
maxX = polygon(0).X
minY = polygon(0).Y
maxY = polygon(0).Y
For i = 1 To UBound(polygon)
Set q = polygon(i)
minX = vbMin(q.X, minX)
maxX = vbMax(q.X, maxX)
minY = vbMin(q.Y, minY)
maxY = vbMax(q.Y, maxY)
Next i
If p.X < minX Or p.X > maxX Or p.Y < minY Or p.Y > maxY Then
isPointInPolygon = False
Exit Function
End If
' SOURCE: http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html
isPointInPolygon = False
i = 0
j = UBound(polygon)
Do While i < UBound(polygon) + 1
If (polygon(i).Y > p.Y) Then
If (polygon(j).Y < p.Y) Then
If p.X < (polygon(j).X  polygon(i).X) * (p.Y  polygon(i).Y) / (polygon(j).Y  polygon(i).Y) + polygon(i).X Then
isPointInPolygon = True
Exit Function
End If
End If
ElseIf (polygon(i).Y < p.Y) Then
If (polygon(j).Y > p.Y) Then
If p.X < (polygon(j).X  polygon(i).X) * (p.Y  polygon(i).Y) / (polygon(j).Y  polygon(i).Y) + polygon(i).X Then
isPointInPolygon = True
Exit Function
End If
End If
End If
j = i
i = i + 1
Loop
End Function
Function vbMax(n1, n2) As Double
vbMax = IIf(n1 > n2, n1, n2)
End Function
Function vbMin(n1, n2) As Double
vbMin = IIf(n1 > n2, n2, n1)
End Function
Sub TestPointInPolygon()
Dim i As Integer
Dim InPolygon As Boolean
' MARKER Object
Dim p As CPoint
Set p = New CPoint
p.X = <ENTER X VALUE HERE>
p.Y = <ENTER Y VALUE HERE>
' POLYGON OBJECT
Dim polygon() As CPoint
ReDim polygon(<ENTER VALUE HERE>) 'Amount of vertices in polygon  1
For i = 0 To <ENTER VALUE HERE> 'Same value as above
Set polygon(i) = New CPoint
polygon(i).X = <ASSIGN X VALUE HERE> 'Source a list of values that can be looped through
polgyon(i).Y = <ASSIGN Y VALUE HERE> 'Source a list of values that can be looped through
Next i
InPolygon = isPointInPolygon(p, polygon)
MsgBox InPolygon
End Sub
Surprised nobody brought this up earlier, but for the pragmatists requiring a database: MongoDB has excellent support for Geo queries including this one.
What you are looking for is:
db.neighborhoods.findOne({ geometry: { $geoIntersects: { $geometry: { type: "Point", coordinates: [ "longitude", "latitude" ] } } } })
Neighborhoods
is the collection that stores one or more polygons in standard GeoJson format. If the query returns null it is not intersected otherwise it is.
Very well documented here: https://docs.mongodb.com/manual/tutorial/geospatialtutorial/
The performance for more than 6,000 points classified in a 330 irregular polygon grid was less than one minute with no optimization at all and including the time to update documents with their respective polygon.
Heres a point in polygon test in C that isn't using raycasting. And it can work for overlapping areas (self intersections), see the use_holes
argument.
/* math lib (defined below) */
static float dot_v2v2(const float a[2], const float b[2]);
static float angle_signed_v2v2(const float v1[2], const float v2[2]);
static void copy_v2_v2(float r[2], const float a[2]);
/* intersection function */
bool isect_point_poly_v2(const float pt[2], const float verts[][2], const unsigned int nr,
const bool use_holes)
{
/* we do the angle rule, define that all added angles should be about zero or (2 * PI) */
float angletot = 0.0;
float fp1[2], fp2[2];
unsigned int i;
const float *p1, *p2;
p1 = verts[nr  1];
/* first vector */
fp1[0] = p1[0]  pt[0];
fp1[1] = p1[1]  pt[1];
for (i = 0; i < nr; i++) {
p2 = verts[i];
/* second vector */
fp2[0] = p2[0]  pt[0];
fp2[1] = p2[1]  pt[1];
/* dot and angle and cross */
angletot += angle_signed_v2v2(fp1, fp2);
/* circulate */
copy_v2_v2(fp1, fp2);
p1 = p2;
}
angletot = fabsf(angletot);
if (use_holes) {
const float nested = floorf((angletot / (float)(M_PI * 2.0)) + 0.00001f);
angletot = nested * (float)(M_PI * 2.0);
return (angletot > 4.0f) != ((int)nested % 2);
}
else {
return (angletot > 4.0f);
}
}
/* math lib */
static float dot_v2v2(const float a[2], const float b[2])
{
return a[0] * b[0] + a[1] * b[1];
}
static float angle_signed_v2v2(const float v1[2], const float v2[2])
{
const float perp_dot = (v1[1] * v2[0])  (v1[0] * v2[1]);
return atan2f(perp_dot, dot_v2v2(v1, v2));
}
static void copy_v2_v2(float r[2], const float a[2])
{
r[0] = a[0];
r[1] = a[1];
}
Note: this is one of the less optimal methods since it includes a lot of calls to atan2f
, but it may be of interest to developers reading this thread (in my tests its ~23x slower then using the line intersection method).
This is a presumably slightly less optimized version of the C code from here which was sourced from this page.
My C++ version uses a std::vector<std::pair<double, double>>
and two doubles as an x and y. The logic should be exactly the same as the original C code, but I find mine easier to read. I can't speak for the performance.
bool point_in_poly(std::vector<std::pair<double, double>>& verts, double point_x, double point_y)
{
bool in_poly = false;
auto num_verts = verts.size();
for (int i = 0, j = num_verts  1; i < num_verts; j = i++) {
double x1 = verts[i].first;
double y1 = verts[i].second;
double x2 = verts[j].first;
double y2 = verts[j].second;
if (((y1 > point_y) != (y2 > point_y)) &&
(point_x < (x2  x1) * (point_y  y1) / (y2  y1) + x1))
in_poly = !in_poly;
}
return in_poly;
}
The original C code is
int pnpoly(int nvert, float *vertx, float *verty, float testx, float testy)
{
int i, j, c = 0;
for (i = 0, j = nvert1; i < nvert; j = i++) {
if ( ((verty[i]>testy) != (verty[j]>testy)) &&
(testx < (vertx[j]vertx[i]) * (testyverty[i]) / (verty[j]verty[i]) + vertx[i]) )
c = !c;
}
return c;
}
For Detecting hit on Polygon we need to test two things:
 If Point is inside polygon area. (can be accomplished by RayCasting Algorithm)
 If Point is on the polygon border(can be accomplished by same algorithm which is used for point detection on polyline(line)).
To deal with the following special cases in Ray casting algorithm:
 The ray overlaps one of the polygon's side.
 The point is inside of the polygon and the ray passes through a vertex of the polygon.
 The point is outside of the polygon and the ray just touches one of the polygon's angle.
Check Determining Whether A Point Is Inside A Complex Polygon. The article provides an easy way to resolve them so there will be no special treatment required for the above cases.
You can do this by checking if the area formed by connecting the desired point to the vertices of your polygon matches the area of the polygon itself.
Or you could check if the sum of the inner angles from your point to each pair of two consecutive polygon vertices to your check point sums to 360, but I have the feeling that the first option is quicker because it doesn't involve divisions nor calculations of inverse of trigonometric functions.
I don't know what happens if your polygon has a hole inside it but it seems to me that the main idea can be adapted to this situation
You can as well post the question in a math community. I bet they have one million ways of doing that
If you are looking for a javascript library there's a javascript google maps v3 extension for the Polygon class to detect whether or not a point resides within it.
var polygon = new google.maps.Polygon([], "#000000", 1, 1, "#336699", 0.3);
var isWithinPolygon = polygon.containsLatLng(40, 90);
The answer depends on if you have the simple or complex polygons. Simple polygons must not have any line segment intersections. So they can have the holes but lines can't cross each other. Complex regions can have the line intersections  so they can have the overlapping regions, or regions that touch each other just by a single point.
For simple polygons the best algorithm is Ray casting (Crossing number) algorithm. For complex polygons, this algorithm doesn't detect points that are inside the overlapping regions. So for complex polygons you have to use Winding number algorithm.
Here is an excellent article with C implementation of both algorithms. I tried them and they work well.