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I've been trying to create an algorithm for finding the order of points in a simple polygon.
The aim is when given points on a 2D plane (there is always possible to form a simple polygon btw) I output the order of points in a valid simple polygon. All points must be part of said polygon.
I've somewhat achieved this, but it fails for some test cases. The way I have done this is by finding the geometrical centre
int centerX = (lowX + highX) / 2;
int centerY = (lowY + highY) / 2;
Point center = new Point(centerX, centerY, -1);
and then sorting all points by their polar angle.
Collections.sort(points, (a, b) -> {
if(a == b || a.equals(b)) {
return 0;
}
double aTheta = Math.atan2((long)a.y - center.y, (long)a.x - center.x);
double bTheta = Math.atan2((long)b.y - center.y, (long)b.x - center.x);
if(aTheta < bTheta) {
return -1;
}
else if(aTheta > bTheta) {
return 1;
}
else {
double aDist = Math.sqrt((((long)center.x - a.x) * ((long)center.x - a.x)) +
(((long)center.y - a.y) * ((long)center.y - a.y)));
double bDist = Math.sqrt((((long)center.x - b.x) * ((long)center.x - b.x)) +
(((long)center.y - b.y) * ((long)center.y - b.y)));
if (aDist < bDist) {
return -1;
}
else {
return 1;
}
}
});
I'm struggling with finding out what makes this break for some of the test cases. Any help or pointers are greatly appreciated! Also wondering if there are any efficient, yet not overly complicated algorithms that can perform this.
UPDATE
I've found one of the failing test cases: When given the points (101, 101), (100, 100), (105, 100), (103, 100), (107, 100), (102, 100), (109, 100) Labeled 0 to 9 respectively
My program outputs 2 4 6 0 3 5 1 but it is not a valid simple polygon
It should be a permutation of 1 0 6 4 2 3 5
Here's a Java implementation of the nice answer already provided by Reblochon Masque.
Note that instead of using any trig functions to calculate angles we use a comparison of the relative orientation (or turn direction) from the min point to each of the two points being compared. Personally I find this more elegant than using angles, but others may disagree. However, as with any calculations based on double the orient2D method is susceptible to errors in rounding.
Also, when there's a tie based on orientation, because the min point and the two points are collinear, we break the tie by considering the relative ordering of the two points. This means that we'll visit points in order, with no "switchbacks", which I think is preferable.
static List<Point2D> simplePolygon(Collection<Point2D> points)
{
final Point2D min = minPoint2D(points);
List<Point2D> simple = new ArrayList<>(points);
Collections.sort(simple, (p1, p2) ->
{
int cmp = orient2D(min, p2, p1);
if(cmp == 0)
cmp = order2D(p1, p2);
return cmp;
});
return simple;
}
// return lowest, leftmost point
static Point2D minPoint2D(Collection<Point2D> points)
{
Point2D min = null;
for(Point2D p : points)
if(min == null || order2D(p, min) < 0) min = p;
return min;
}
// order points by increasing y, break ties by increasing x
static int order2D(Point2D p1, Point2D p2)
{
if(p1.getY() < p2.getY()) return -1;
else if(p1.getY() > p2.getY()) return 1;
else if(p1.getX() < p2.getX()) return -1;
else if(p1.getX() > p2.getX()) return 1;
else return 0;
}
// Does p involve a CCW(+1), CW(-1) or No(0) turn from the line p1-p2
static int orient2D(Point2D p1, Point2D p2, Point2D p)
{
double dx = p2.getX() - p1.getX();
double dy = p2.getY() - p1.getY();
double px = p.getX() - p1.getX();
double py = p.getY() - p1.getY();
double dot = py * dx - px * dy;
return dot < 0 ? -1 : dot > 0 ? 1 : 0;
}
Test:
int[] a = {101, 101, 100, 100, 105, 100, 103, 100, 107, 100, 102, 100, 109, 100};
List<Point2D> points = new ArrayList<>();
for(int i=0; i<a.length; i+=2)
points.add(new Point2D.Double(a[i], a[i+1]));
List<Point2D> simple = simplePolygon(points);
for(Point2D p : simple) System.out.println(p);
Output:
Point2D.Double[100.0, 100.0]
Point2D.Double[102.0, 100.0]
Point2D.Double[103.0, 100.0]
Point2D.Double[105.0, 100.0]
Point2D.Double[107.0, 100.0]
Point2D.Double[109.0, 100.0]
Point2D.Double[101.0, 101.0]
Which I believe is correct.
here is an easy to implement O(n logn) algorithm that is guaranteed to produce a simple polygon (no edge crossings)
1- find the point the most south, (and the most westwards if you have a tie with the y values).
2- Sort all points based on their angle between this most south point, and the horizontal line.
3- the ordered sequence is a simple polygon.
In some rare cases, some points may not form a vertex, but be included in an edge, if they are collinear at the same angle.
I am looking for some help with some game code i have inherited from a flight sim. The code below simulates bombs exploding on the ground, it works fine but i am trying to refine it.
At the moment it takes a random value for x and y as a start point and then adds another random value between -20 and 20 to this. It works ok, but doesn't simulate bombs dropping very well as the pattern does not lay along a straight line/
What i would like to achieve though is all x and y points after the first random values, to lay along a straight line, so that the effects called for all appear to lay in a line. It doesn't matter which way the line is orientated.
Thanks for any help
slipper
public static class BombUnit extends CandCGeneric
{
public boolean danger()
{
Point3d point3d = new Point3d();
pos.getAbs(point3d);
Vector3d vector3d = new Vector3d();
Random random = new Random();
Aircraft aircraft = War.GetNearestEnemyAircraft(this, 10000F, 9);
if(counter > 10)
{
counter = 0;
startpoint.set(point3d.x + (double)(random.nextInt(1000) - 500), point3d.y + (double)(random.nextInt(1000) - 500), point3d.z);
}
if(aircraft != null && (aircraft instanceof TypeBomber) && aircraft.getArmy() != myArmy)
{
World.MaxVisualDistance = 50000F;
counter++;
String s = "weapon.bomb_std";
startpoint.x += random.nextInt(40) - 20;
startpoint.y += random.nextInt(40) - 20;
Explosions.generate(this, startpoint, 7F, 0, 30F, !Mission.isNet());
startpoint.z = World.land().HQ(startpoint.x, startpoint.y);
MsgExplosion.send(this, s, startpoint, getOwner(), 0.0F, 7F, 0, 30F);
Engine.land();
int i = Landscape.getPixelMapT(Engine.land().WORLD2PIXX(startpoint.x), Engine.land().WORLD2PIXY(startpoint.y));
if(firecounter < 100 && i >= 16 && i < 20)
{
Eff3DActor.New(null, null, new Loc(startpoint.x, startpoint.y, startpoint.z + 5D, 0.0F, 90F, 0.0F), 1.0F, "Effects/Smokes/CityFire3.eff", 300F);
firecounter++;
}
super.setTimer(15);
}
return true;
}
private static Point3d startpoint = new Point3d();
private int counter;
private int firecounter;
public BombUnit()
{
counter = 11;
firecounter = 0;
Timer1 = Timer2 = 0.05F;
}
}
The code in the question is a mess, but ignoring this and trying to focus on the relevant parts: You can generate a random position for the first point, and a random direction, and then walk along this direction in several steps.
(This still raises the question of whether the direction is really not important. Wouldn't it matter if only the first bomb was dropped in the "valid" area, and the remaining ones outside of the screen?)
However, the relevant code could roughly look like this:
class Bombs
{
private final Random random = new Random(0);
int getScreenSizeX() { ... }
int getScreenSizeY() { ... }
// Method to drop a single bomb at the given position
void dropBombAt(double x, double y) { ... }
void dropBombs(int numberOfBombs, double distanceBetweenBombs)
{
// Create a random position in the screen
double currentX = random.nextDouble() * getScreenSizeX();
double currentY = random.nextDouble() * getScreenSizeY();
// Create a random step size
double directionX = random.nextDouble();
double directionY = random.nextDouble();
double invLength = 1.0 / Math.hypot(directionX, directionY);
double stepX = directionX * invLength * distanceBetweenBombs;
double stepY = directionY * invLength * distanceBetweenBombs;
// Drop the bombs
for (int i=0; i<numberOfBombs; i++)
{
dropBombAt(currentX, currentY);
currentX += stepX;
currentY += stepY;
}
}
}
I am assuming your startpoint is a StartPoint class with x,y,z coordinates as integers in it.
I hope I have understood your problem correctly. It looks like you either want to create a vertical explosion or a horizontal explosion. Since an explosion always occurs on ground, the z coordinate will be zero. Now you can vary one of x or y coordinate to give you a random explosion along a straight line. Whether you choose x or y could be fixed or could be randomized itself. A potential randomized solution below:
public boolean danger() {
// stuff
int orientation = Random.nextInt(2);
if(aircraft != null && (aircraft instanceof TypeBomber) && aircraft.getArmy() != myArmy)
{
// stuff
startPoint = randomizeStartPoint(orientation, startPoint);
// stuff
}
}
StartPoint randomizeStartPoint(int orientation, StartPoint startPoint) {
if(orientation == 0) {
startPoint.x += random.nextInt(40) - 20;
}
else {
startPoint.y += random.nextInt(40) - 20;
}
return startPoint;
}
In response to the image you uploaded, it seems that the orientation of the explosion need not necessarily be horizontal or vertical. So the code I posted above gives a limited solution to your problem.
Since you want any random straight line, your problem boils down to two sub parts:
1. Generate a random straight line equation.
2. Generate random point along this line.
Now, a straight line equation in coordinate geometry is y = mx + c where m is the slope and c is the constant where the line crosses the y-axis. The problem with c is that it gives rise to irrational coordinates. I am assuming you are looking for integer coordinates only, since this will ensure that your points are accurately plotted. (You could do with rational fractions, but then a fraction like 1/3 will still result in loss of accuracy). The best way to get rid of this irrational problem is to get rid of c. So now your straight line always looks like y = mx. So for step one, you have to generate a random m.
Then for step 2, you can either generate a random x or random y. It doesn't matter which one, since either one will result in random coordinates.
Here is a possible code for the solution:
int generateRandomSlope() {
return Random.nextInt(100); // arbitrarily chose 100.
}
int randomizeStartPoint(int m, StartPoint startPoint) { // takes the slope we generated earlier. without the slope, your points will never be on a straight line!
startPoint.x += random.nextInt(40) - 20;
startPoint.y += x * m; // because a line equation is y = mx
return startPoint;
}
public boolean danger() {
// stuff
int m = generateRandomSlope(); // you may want to generate this elsewhere so that it doesn't change each time danger() is called.
if(aircraft != null && (aircraft instanceof TypeBomber) && aircraft.getArmy() != myArmy)
{
// stuff
startPoint = randomizeStartPoint(m, startPoint);
// stuff
}
}
Again, this is not a complete or the best solution.
Rotating Asteroids ( Polygons )
I am trying to rotate asteroids(polygons) so that they look nice. I am doing this through multiple mathematical equations. To start I give the individual asteroid a rotation velocity:
rotVel = ((Math.random()-0.5)*Math.PI/16);
Then I create the polygon shape,
this.shape = new Polygon();
Followed by generating the points,
for (j = 0; j < s; j++) {
theta = 2 * Math.PI / s * j;
r = MIN_ROCK_SIZE + (int) (Math.random() * (MAX_ROCK_SIZE - MIN_ROCK_SIZE));
x = (int) -Math.round(r * Math.sin(theta)) + asteroidData[0];
y = (int) Math.round(r * Math.cos(theta)) + asteroidData[1];
shape.addPoint(x, y);
}
Finally, in a loop a method is being called in which it attempts to move the polygon and its points down as well as rotating them. (I'm just pasting the rotating part as the other one is working)
for (int i = 0; i < shape.npoints; i++) {
// Subtract asteroid's x and y position
double x = shape.xpoints[i] - asteroidData[0];
double y = shape.ypoints[i] - asteroidData[1];
double temp_x = ((x * Math.cos(rotVel)) - (y * Math.sin(rotVel)));
double temp_y = ((x * Math.sin(rotVel)) + (y * Math.cos(rotVel)));
shape.xpoints[i] = (int) Math.round(temp_x + asteroidData[0]);
shape.ypoints[i] = (int) Math.round(temp_y + asteroidData[1]);
}
now, the problem is that when it prints to the screen the asteroids appear to 'warp' or rather the x and y positions on some of the polygon points 'float' off course.
I've noticed that when I make 'rotVel' be a whole number the problem is solved however the asteroid will rotate at mach speeds. So I've concluded that the problem has to be in the rounding but no matter what I do I can't seem to find a way to get it to work as the Polygon object requires an array of ints.
Does anyone know how to fix this?
Currently your asteroids rotate around (0 , 0) as far as i can see. Correct would be to rotate them around the center of the shape, which would be (n , m), where n is the average of all x-coordinates of the shape, and m is the average of all y-coordinates of the shape.
Your problem is definitely caused by rounding to int! The first improvement is to make all shape coordinates to be of type double. This will solve most of your unwanted 'effects'.
But even with double you might experience nasty rounding errors in case you do a lot of very small updates of the coordinates. The solution is simple: Just avoid iterative updates of the asteroid points. Every time, you update the coordinates based on the previous coordinates, the rounding error will get worse.
Instead, add a field for the rotation angle to the shape and increment it instead of the points themselves. Not until drawing the shape, you compute the final positions by applying the rotation to the points. Note that this will never change the points themselves.
You can extend this concept to other transformations (e.g. translation) too. What you get is some kind of local coordinate system for every shape/object. The points of the shape are defined in the local coordinate system. By moving and rotating this system, you can reposition the entire object anywhere in space.
public class Shape {
// rotation and position of the local coordinate system
private double rot, x, y;
// points of the shape in local coordinate system
private double[] xp, yp;
private int npoints;
// points of the shape in world coordinates
private int[][] wxp, wyp;
private boolean valid;
public void setRotation(double r) { this.rot = r; valid = false; }
public void setPosition(double x, double y) { this.x = x; this.y = y; valid = false; }
public void addPoint(double x, double y) {
// TODO: add point to xp, yp
valid = false;
}
public void draw(...) {
if (!valid) {
computeWorldCoordinates(wxp, wyp);
valid = true;
}
// TODO: draw shape at world coordaintes wxp and wyp
}
protected void computeWorldCoordinates(int[] xcoord, int[] ycoord) {
for (int i = 0; i < npoints; i++) {
double temp_x = xp[i] * Math.cos(rot) - yp[i] * Math.sin(rot);
double temp_y = xp[i] * Math.sin(rot) + yp[i] * Math.cos(rot);
xcoord[i] = (int) Math.round(x + temp_x);
ycoord[i] = (int) Math.round(y + temp_y);
}
}
}
I have a problem that I can't seem to get a working algorithm for, I've been trying to days and get so close but yet so far.
I want to draw a triangle defined by 3 points (p0, p1, p2). This triangle can be any shape, size, and orientation. The triangle must also be filled inside.
Here's a few things I've tried and why they've failed:
1
Drawing lines along the triangle from side to side
Failed because the triangle would have holes and would not be flat due to the awkwardness of drawing lines across the angled surface with changing locations
2
Iterate for an area and test if the point falls past the plane parallel to the triangle and 3 other planes projected onto the XY, ZY, and XZ plane that cover the area of the triangle
Failed because for certain triangles (that have very close sides) there would be unpredictable results, e.g. voxels floating around not connected to anything
3
Iterate for an area along the sides of the triangle (line algorithm) and test to see if a point goes past a parallel plane
Failed because drawing a line from p0 to p1 is not the same as a line from p1 to p0 and any attempt to rearrange either doesn't help, or causes more problems. Asymmetry is the problem with this one.
This is all with the intent of making polygons and flat surfaces. 3 has given me the most success and makes accurate triangles, but when I try to connect these together everything falls apart and I get issues with things not connecting, asymmetry, etc. I believe 3 will work with some tweaking but I'm just worn out from trying to make this work for so long and need help.
There's a lot of small details in my algorithms that aren't really relevant so I left them out. For number 3 it might be a problem with my implementation and not the algorithm itself. If you want code I'll try and clean it up enough to be understandable, it will take me a few minutes though. But I'm looking for algorithms that are known to work. I can't seem to find any voxel shape making algorithms anywhere, I've been doing everything from scratch.
EDIT:
Here's the third attempt. It's a mess, but I tried to clean it up.
// Point3i is a class I made, however the Vector3fs you'll see are from lwjgl
public void drawTriangle (Point3i r0, Point3i r1, Point3i r2)
{
// Util is a class I made with some useful stuff inside
// Starting values for iteration
int sx = (int) Util.min(r0.x, r1.x, r2.x);
int sy = (int) Util.min(r0.y, r1.y, r2.y);
int sz = (int) Util.min(r0.z, r1.z, r2.z);
// Ending values for iteration
int ex = (int) Util.max(r0.x, r1.x, r2.x);
int ey = (int) Util.max(r0.y, r1.y, r2.y);
int ez = (int) Util.max(r0.z, r1.z, r2.z);
// Side lengths
float l0 = Util.distance(r0.x, r1.x, r0.y, r1.y, r0.z, r1.z);
float l1 = Util.distance(r2.x, r1.x, r2.y, r1.y, r2.z, r1.z);
float l2 = Util.distance(r0.x, r2.x, r0.y, r2.y, r0.z, r2.z);
// Calculate the normal vector
Vector3f nn = new Vector3f(r1.x - r0.x, r1.y - r0.y, r1.z - r0.z);
Vector3f n = new Vector3f(r2.x - r0.x, r2.y - r0.y, r2.z - r0.z);
Vector3f.cross(nn, n, n);
// Determines which direction we increment for
int iz = n.z >= 0 ? 1 : -1;
int iy = n.y >= 0 ? 1 : -1;
int ix = n.x >= 0 ? 1 : -1;
// Reorganize for the direction of iteration
if (iz < 0) {
int tmp = sz;
sz = ez;
ez = tmp;
}
if (iy < 0) {
int tmp = sy;
sy = ey;
ey = tmp;
}
if (ix < 0) {
int tmp = sx;
sx = ex;
ex = tmp;
}
// We're we want to iterate over the end vars so we change the value
// by their incrementors/decrementors
ex += ix;
ey += iy;
ez += iz;
// Maximum length
float lmax = Util.max(l0, l1, l2);
// This is a class I made which manually iterates over a line, I already
// know that this class is working
GeneratorLine3d g0, g1, g2;
// This is a vector for the longest side
Vector3f v = new Vector3f();
// make the generators
if (lmax == l0) {
v.x = r1.x - r0.x;
v.y = r1.y - r0.y;
v.z = r1.z - r0.z;
g0 = new GeneratorLine3d(r0, r1);
g1 = new GeneratorLine3d(r0, r2);
g2 = new GeneratorLine3d(r2, r1);
}
else if (lmax == l1) {
v.x = r1.x - r2.x;
v.y = r1.y - r2.y;
v.z = r1.z - r2.z;
g0 = new GeneratorLine3d(r2, r1);
g1 = new GeneratorLine3d(r2, r0);
g2 = new GeneratorLine3d(r0, r1);
}
else {
v.x = r2.x - r0.x;
v.y = r2.y - r0.y;
v.z = r2.z - r0.z;
g0 = new GeneratorLine3d(r0, r2);
g1 = new GeneratorLine3d(r0, r1);
g2 = new GeneratorLine3d(r1, r2);
}
// Absolute values for the normal
float anx = Math.abs(n.x);
float any = Math.abs(n.y);
float anz = Math.abs(n.z);
int i, o;
int si, so;
int ii, io;
int ei, eo;
boolean maxx, maxy, maxz,
midy, midz, midx,
minx, miny, minz;
maxx = maxy = maxz =
midy = midz = midx =
minx = miny = minz = false;
// Absolute values for the longest side vector
float rnx = Math.abs(v.x);
float rny = Math.abs(v.y);
float rnz = Math.abs(v.z);
int rmid = Util.max(rnx, rny, rnz);
if (rmid == rnz) midz = true;
else if (rmid == rny) midy = true;
midx = !midz && !midy;
// Determine the inner and outer loop directions
if (midz) {
if (any > anx)
{
maxy = true;
si = sy;
ii = iy;
ei = ey;
}
else {
maxx = true;
si = sx;
ii = ix;
ei = ex;
}
}
else {
if (anz > anx) {
maxz = true;
si = sz;
ii = iz;
ei = ez;
}
else {
maxx = true;
si = sx;
ii = ix;
ei = ex;
}
}
if (!midz && !maxz) {
minz = true;
so = sz;
eo = ez;
}
else if (!midy && !maxy) {
miny = true;
so = sy;
eo = ey;
}
else {
minx = true;
so = sx;
eo = ex;
}
// GeneratorLine3d is iterable
Point3i p1;
for (Point3i p0 : g0) {
// Make sure the two 'mid' coordinate correspond for the area inside the triangle
if (midz)
do p1 = g1.hasNext() ? g1.next() : g2.next();
while (p1.z != p0.z);
else if (midy)
do p1 = g1.hasNext() ? g1.next() : g2.next();
while (p1.y != p0.y);
else
do p1 = g1.hasNext() ? g1.next() : g2.next();
while (p1.x != p0.x);
eo = (minx ? p0.x : miny ? p0.y : p0.z);
so = (minx ? p1.x : miny ? p1.y : p1.z);
io = eo - so >= 0 ? 1 : -1;
for (o = so; o != eo; o += io) {
for (i = si; i != ei; i += ii) {
int x = maxx ? i : midx ? p0.x : o;
int y = maxy ? i : midy ? p0.y : o;
int z = maxz ? i : midz ? p0.z : o;
// isPassing tests to see if a point goes past a plane
// I know it's working, so no code
// voxels is a member that is an arraylist of Point3i
if (isPassing(x, y, z, r0, n.x, n.y, n.z)) {
voxels.add(new Point3i(x, y, z));
break;
}
}
}
}
}
You could use something like Besenham's line algorithm, but extended into three dimensions. The two main ideas we want to take from it are:
rotate the initial line so its slope isn't too steep.
for any given x value, find an integer value that is closest to the ideal y value.
Just as Bresenham's algorithm prevents gaps by performing an initial rotation, we'll avoid holes by performing two initial rotations.
Get the normal vector and point that represent the plane your triangle lies on. Hint: use the cross product of (line from p0 to p1) and (line from p0 to p2) for the vector, and use any of your corner points for the point.
You want the plane to be sufficiently not-steep, to avoid holes. You must satisfy these conditions:
-1 >= norm.x / norm.y >= 1
-1 >= norm.z / norm.y >= 1
Rotate your normal vector and initial points 90 degrees about the x axis and 90 degrees about the z axis until these conditions are satisfied. I'm not sure how to do this in the fewest number of rotations, but I'm fairly sure you can satisfy these conditions for any plane.
Create a function f(x,z) which represents the plane your rotated triangle now lies on. It should return the Y value of any pair of X and Z values.
Project your triangle onto the XZ plane (i.e., set all the y values to 0), and use your favorite 2d triangle drawing algorithm to get a collection of x-and-z coordinates.
For each pixel value from step 4, pass the x and z values into your function f(x,z) from step 3. Round the result to the nearest integer, and store the x, y, and z values as a voxel somewhere.
If you performed any rotations in step 2, perform the opposite of those rotations in reverse order on your voxel collection.
Start with a function that checks for triangle/voxel intersection. Now you can scan a volume and find the voxels that intersect the triangle - these are the ones you're interested in. This is a lousy algorithm but is also a regression test for anything else you try. This test is easy to implement using SAT (separating axis theorem) and considering the triangle a degenerate volume (1 face, 3 edges) and considering the voxels symmetry (only 3 face normals).
I use octtrees, so my preferred method is to test a triangle against a large voxel and figure out which of the 8 child octants it intersects. Then use recursion on the intersected children until the desired level of subdivision is attained. Hint: at most 6 of the children can be intersected by the triangle and often fewer than that. This is tricky but will produce the same results as the first method but much quicker.
Rasterization in 3d is probably fastest, but IMHO is even harder to guarantee no holes in all cases. Again, use the first method for comparison.
Is there a reason that they decided not to add the contains method (for Path) in Android?
I'm wanting to know what points I have in a Path and hoped it was easier than seen here:
How can I tell if a closed path contains a given point?
Would it be better for me to create an ArrayList and add the integers into the array? (I only check the points once in a control statement) Ie. if(myPath.contains(x,y)
So far my options are:
Using a Region
Using an ArrayList
Extending the Class
Your suggestion
I'm just looking for the most efficient way I should go about this
I came up against this same problem a little while ago, and after some searching, I found this to be the best solution.
Java has a Polygon class with a contains() method that would make things really simple. Unfortunately, the java.awt.Polygonclass is not supported in Android. However, I was able to find someone who wrote an equivalent class.
I don't think you can get the individual points that make up the path from the Android Path class, so you will have to store the data in a different way.
The class uses a Crossing Number algorithm to determine whether or not the point is inside of the given list of points.
/**
* Minimum Polygon class for Android.
*/
public class Polygon
{
// Polygon coodinates.
private int[] polyY, polyX;
// Number of sides in the polygon.
private int polySides;
/**
* Default constructor.
* #param px Polygon y coods.
* #param py Polygon x coods.
* #param ps Polygon sides count.
*/
public Polygon( int[] px, int[] py, int ps )
{
polyX = px;
polyY = py;
polySides = ps;
}
/**
* Checks if the Polygon contains a point.
* #see "http://alienryderflex.com/polygon/"
* #param x Point horizontal pos.
* #param y Point vertical pos.
* #return Point is in Poly flag.
*/
public boolean contains( int x, int y )
{
boolean oddTransitions = false;
for( int i = 0, j = polySides -1; i < polySides; j = i++ )
{
if( ( polyY[ i ] < y && polyY[ j ] >= y ) || ( polyY[ j ] < y && polyY[ i ] >= y ) )
{
if( polyX[ i ] + ( y - polyY[ i ] ) / ( polyY[ j ] - polyY[ i ] ) * ( polyX[ j ] - polyX[ i ] ) < x )
{
oddTransitions = !oddTransitions;
}
}
}
return oddTransitions;
}
}
I would just like to comment on #theisenp answer: The code has integer arrays and if you look on the algorithm description webpage it warns against using integers instead of floating point.
I copied your code above and it seemed to work fine except for some corner cases when I made lines that didnt connect to themselves very well.
By changing everything to floating point, I got rid of this bug.
Tried the other answer, but it gave an erroneous outcome for my case. Didn't bother to find the exact cause, but made my own direct translation from the algorithm on:
http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html
Now the code reads:
/**
* Minimum Polygon class for Android.
*/
public class Polygon
{
// Polygon coodinates.
private int[] polyY, polyX;
// Number of sides in the polygon.
private int polySides;
/**
* Default constructor.
* #param px Polygon y coods.
* #param py Polygon x coods.
* #param ps Polygon sides count.
*/
public Polygon( int[] px, int[] py, int ps )
{
polyX = px;
polyY = py;
polySides = ps;
}
/**
* Checks if the Polygon contains a point.
* #see "http://alienryderflex.com/polygon/"
* #param x Point horizontal pos.
* #param y Point vertical pos.
* #return Point is in Poly flag.
*/
public boolean contains( int x, int y )
{
boolean c = false;
int i, j = 0;
for (i = 0, j = polySides - 1; i < polySides; j = i++) {
if (((polyY[i] > y) != (polyY[j] > y))
&& (x < (polyX[j] - polyX[i]) * (y - polyY[i]) / (polyY[j] - polyY[i]) + polyX[i]))
c = !c;
}
return c;
}
}
For completeness, I want to make a couple notes here:
As of API 19, there is an intersection operation for Paths. You could create a very small square path around your test point, intersect it with the Path, and see if the result is empty or not.
You can convert Paths to Regions and do a contains() operation. However Regions work in integer coordinates, and I think they use transformed (pixel) coordinates, so you'll have to work with that. I also suspect that the conversion process is computationally intensive.
The edge-crossing algorithm that Hans posted is good and quick, but you have to be very careful for certain corner cases such as when the ray passes directly through a vertex, or intersects a horizontal edge, or when round-off error is a problem, which it always is.
The winding number method is pretty much fool proof, but involves a lot of trig and is computationally expensive.
This paper by Dan Sunday gives a hybrid algorithm that's as accurate as the winding number but as computationally simple as the ray-casting algorithm. It blew me away how elegant it was.
My code
This is some code I wrote recently in Java which handles a path made out of both line segments and arcs. (Also circles, but those are complete paths on their own, so it's sort of a degenerate case.)
package org.efalk.util;
/**
* Utility: determine if a point is inside a path.
*/
public class PathUtil {
static final double RAD = (Math.PI/180.);
static final double DEG = (180./Math.PI);
protected static final int LINE = 0;
protected static final int ARC = 1;
protected static final int CIRCLE = 2;
/**
* Used to cache the contents of a path for pick testing. For a
* line segment, x0,y0,x1,y1 are the endpoints of the line. For
* a circle (ellipse, actually), x0,y0,x1,y1 are the bounding box
* of the circle (this is how Android and X11 like to represent
* circles). For an arc, x0,y0,x1,y1 are the bounding box, a1 is
* the start angle (degrees CCW from the +X direction) and a1 is
* the sweep angle (degrees CCW).
*/
public static class PathElement {
public int type;
public float x0,y0,x1,y1; // Endpoints or bounding box
public float a0,a1; // Arcs and circles
}
/**
* Determine if the given point is inside the given path.
*/
public static boolean inside(float x, float y, PathElement[] path) {
// Based on algorithm by Dan Sunday, but allows for arc segments too.
// http://geomalgorithms.com/a03-_inclusion.html
int wn = 0;
// loop through all edges of the polygon
// An upward crossing requires y0 <= y and y1 > y
// A downward crossing requires y0 > y and y1 <= y
for (PathElement pe : path) {
switch (pe.type) {
case LINE:
if (pe.x0 < x && pe.x1 < x) // left
break;
if (pe.y0 <= y) { // start y <= P.y
if (pe.y1 > y) { // an upward crossing
if (isLeft(pe, x, y) > 0) // P left of edge
++wn; // have a valid up intersect
}
}
else { // start y > P.y
if (pe.y1 <= y) { // a downward crossing
if (isLeft(pe, x, y) < 0) // P right of edge
--wn; // have a valid down intersect
}
}
break;
case ARC:
wn += arcCrossing(pe, x, y);
break;
case CIRCLE:
// This should be the only element in the path, so test it
// and get out.
float rx = (pe.x1-pe.x0)/2;
float ry = (pe.y1-pe.y0)/2;
float xc = (pe.x1+pe.x0)/2;
float yc = (pe.y1+pe.y0)/2;
return (x-xc)*(x-xc)/rx*rx + (y-yc)*(y-yc)/ry*ry <= 1;
}
}
return wn != 0;
}
/**
* Return >0 if p is left of line p0-p1; <0 if to the right; 0 if
* on the line.
*/
private static float
isLeft(float x0, float y0, float x1, float y1, float x, float y)
{
return (x1 - x0) * (y - y0) - (x - x0) * (y1 - y0);
}
private static float isLeft(PathElement pe, float x, float y) {
return isLeft(pe.x0,pe.y0, pe.x1,pe.y1, x,y);
}
/**
* Determine if an arc segment crosses the test ray up or down, or not
* at all.
* #return winding number increment:
* +1 upward crossing
* 0 no crossing
* -1 downward crossing
*/
private static int arcCrossing(PathElement pe, float x, float y) {
// Look for trivial reject cases first.
if (pe.x1 < x || pe.y1 < y || pe.y0 > y) return 0;
// Find the intersection of the test ray with the arc. This consists
// of finding the intersection(s) of the line with the ellipse that
// contains the arc, then determining if the intersection(s)
// are within the limits of the arc.
// Since we're mostly concerned with whether or not there *is* an
// intersection, we have several opportunities to punt.
// An upward crossing requires y0 <= y and y1 > y
// A downward crossing requires y0 > y and y1 <= y
float rx = (pe.x1-pe.x0)/2;
float ry = (pe.y1-pe.y0)/2;
float xc = (pe.x1+pe.x0)/2;
float yc = (pe.y1+pe.y0)/2;
if (rx == 0 || ry == 0) return 0;
if (rx < 0) rx = -rx;
if (ry < 0) ry = -ry;
// We start by transforming everything so the ellipse is the unit
// circle; this simplifies the math.
x -= xc;
y -= yc;
if (x > rx || y > ry || y < -ry) return 0;
x /= rx;
y /= ry;
// Now find the points of intersection. This is simplified by the
// fact that our line is horizontal. Also, by the time we get here,
// we know there *is* an intersection.
// The equation for the circle is x²+y² = 1. We have y, so solve
// for x = ±sqrt(1 - y²)
double x0 = 1 - y*y;
if (x0 <= 0) return 0;
x0 = Math.sqrt(x0);
// We only care about intersections to the right of x, so
// that's another opportunity to punt. For a CCW arc, The right
// intersection is an upward crossing and the left intersection
// is a downward crossing. The reverse is true for a CW arc.
if (x > x0) return 0;
int wn = arcXing1(x0,y, pe.a0, pe.a1);
if (x < -x0) wn -= arcXing1(-x0,y, pe.a0, pe.a1);
return wn;
}
/**
* Return the winding number of the point x,y on the unit circle
* which passes through the arc segment defined by a0,a1.
*/
private static int arcXing1(double x, float y, float a0, float a1) {
double a = Math.atan2(y,x) * DEG;
if (a < 0) a += 360;
if (a1 > 0) { // CCW
if (a < a0) a += 360;
return a0 + a1 > a ? 1 : 0;
} else { // CW
if (a0 < a) a0 += 360;
return a0 + a1 <= a ? -1 : 0;
}
}
}
Edit: by request, adding some sample code that makes use of this.
import PathUtil;
import PathUtil.PathElement;
/**
* This class represents a single geographic area defined by a
* circle or a list of line segments and arcs.
*/
public class Area {
public float lat0, lon0, lat1, lon1; // bounds
Path path = null;
PathElement[] pathList;
/**
* Return true if this point is inside the area bounds. This is
* used to confirm touch events and may be computationally expensive.
*/
public boolean pointInBounds(float lat, float lon) {
if (lat < lat0 || lat > lat1 || lon < lon0 || lon > lon1)
return false;
return PathUtil.inside(lon, lat, pathList);
}
static void loadBounds() {
int n = number_of_elements_in_input;
path = new Path();
pathList = new PathElement[n];
for (Element element : elements_in_input) {
PathElement pe = new PathElement();
pathList[i] = pe;
pe.type = element.type;
switch (element.type) {
case LINE: // Line segment
pe.x0 = element.x0;
pe.y0 = element.y0;
pe.x1 = element.x1;
pe.y1 = element.y1;
// Add to path, not shown here
break;
case ARC: // Arc segment
pe.x0 = element.xmin; // Bounds of arc ellipse
pe.y0 = element.ymin;
pe.x1 = element.xmax;
pe.y1 = element.ymax;
pe.a0 = a0; pe.a1 = a1;
break;
case CIRCLE: // Circle; hopefully the only entry here
pe.x0 = element.xmin; // Bounds of ellipse
pe.y0 = element.ymin;
pe.x1 = element.xmax;
pe.y1 = element.ymax;
// Add to path, not shown here
break;
}
}
path.close();
}