We Keep Coding

How do I fix this heart?

Seeing as Valentine's Day is fast approaching, I decided to create a heart. So I found this heart from mathematica.se:
I played around in Mathematica (solved for z, switching some variables around) to get this equation for the z-value of the heart, given the x and y values (click for full-size):
I faithfully ported this equation to Java, dealing with a couple out-of-bounds cases:
import static java.lang.Math.cbrt;
import static java.lang.Math.pow;
import static java.lang.Math.sqrt;
...
public static double heart(double xi, double yi) {
double x = xi;
double y = -yi;
double temp = 5739562800L * pow(y, 3) + 109051693200L * pow(x, 2) * pow(y, 3)
- 5739562800L * pow(y, 5);
double temp1 = -244019119519584000L * pow(y, 9) + pow(temp, 2);
//
if (temp1 < 0) {
return -1; // this is one possible out of bounds location
// this spot is the location of the problem
}
//
double temp2 = sqrt(temp1);
double temp3 = cbrt(temp + temp2);
if (temp3 != 0) {
double part1 = (36 * cbrt(2) * pow(y, 3)) / temp3;
double part2 = 1 / (10935 * cbrt(2)) * temp3;
double looseparts = 4.0 / 9 - 4.0 / 9 * pow(x, 2) - 4.0 / 9 * pow(y, 2);
double sqrt_body = looseparts + part1 + part2;
if (sqrt_body >= 0) {
return sqrt(sqrt_body);
} else {
return -1; // this works; returns -1 if we are outside the heart
}
} else {
// through trial and error, I discovered that this should
// be an ellipse (or that it is close enough)
return Math.sqrt(Math.pow(2.0 / 3, 2) * (1 - Math.pow(x, 2)));
}
}
The only problem is that when temp1 < 0, I cannot simply return -1, like I do:
if (temp1 < 0) {
return -1; // this is one possible out of bounds location
// this spot is the location of the problem
}
That's not the behavior of the heart at that point. As it is, when I try to make my image:
import java.awt.image.BufferedImage;
import java.io.File;
import java.io.IOException;
import javax.imageio.ImageIO;
import static java.lang.Math.cbrt;
import static java.lang.Math.pow;
import static java.lang.Math.sqrt;
public class Heart {
public static double scale(int x, int range, double l, double r) {
double width = r - l;
return (double) x / (range - 1) * width + l;
}
public static void main(String[] args) throws IOException {
BufferedImage img = new BufferedImage(1000, 1000, BufferedImage.TYPE_INT_RGB);
// this is actually larger than the max heart value
final double max_heart = 0.679;
double max = 0.0;
for (int x = 0; x < img.getWidth(); x++) {
for (int y = 0; y < img.getHeight(); y++) {
double xv = scale(x, img.getWidth(), -1.2, 1.2);
double yv = scale(y, img.getHeight(), -1.3, 1);
double heart = heart(xv, yv); //this isn't an accident
// yes I don't check for the return of -1, but still
// the -1 values return a nice shade of pink: 0xFFADAD
// None of the other values should be negative, as I did
// step through from -1000 to 1000 in python, and there
// were no negatives that were not -1
int r = 0xFF;
int gb = (int) (0xFF * (max_heart - heart));
int rgb = (r << 16) | (gb << 8) | gb;
img.setRGB(x, y, rgb);
}
}
ImageIO.write(img, "png", new File("location"));
}
// heart function clipped; it belongs here
}
I get this:
Look at that dip at the top! I tried changing that problematic -1 to a .5, resulting in this:
Now the heart has horns. But it becomes clear where that problematic if's condition is met.
How can I fix this problem? I don't want a hole in my heart at the top, and I don't want a horned heart. If I could clip the horns to the shape of a heart, and color the rest appropriately, that would be perfectly fine. Ideally, the two sides of the heart would come together as a point (hearts have a little point at the join), but if they curve together like shown in the horns, that would be fine too. How can I achieve this?

The problem is simple. If we look at that horseshoe region, we get imaginary numbers. For part of it, it should belong to our heart. In that region, if we were to evaluate our function (by math, not by programming), the imaginary parts of the function cancel. So it should look like this (generated in Mathematica):
Basically, the function for that part is almost identical; we just have to do arithmetic with complex numbers instead of real numbers. Here's a function that does exactly that:
private static double topOfHeart(double x, double y, double temp, double temp1) {
//complex arithmetic; each double[] is a single number
double[] temp3 = cbrt_complex(temp, sqrt(-temp1));
double[] part1 = polar_reciprocal(temp3);
part1[0] *= 36 * cbrt(2) * pow(y, 3);
double[] part2 = temp3;
part2[0] /= (10935 * cbrt(2));
toRect(part1, part2);
double looseparts = 4.0 / 9 - 4.0 / 9 * pow(x, 2) - 4.0 / 9 * pow(y, 2);
double real_part = looseparts + part1[0] + part2[0];
double imag_part = part1[1] + part2[1];
double[] result = sqrt_complex(real_part, imag_part);
toRect(result);
// theoretically, result[1] == 0 should work, but floating point says otherwise
if (Math.abs(result[1]) < 1e-5) {
return result[0];
}
return -1;
}
/**
* returns a specific cuberoot of this complex number, in polar form
*/
public static double[] cbrt_complex(double a, double b) {
double r = Math.hypot(a, b);
double theta = Math.atan2(b, a);
double cbrt_r = cbrt(r);
double cbrt_theta = 1.0 / 3 * (2 * PI * Math.floor((PI - theta) / (2 * PI)) + theta);
return new double[]{cbrt_r, cbrt_theta};
}
/**
* returns a specific squareroot of this complex number, in polar form
*/
public static double[] sqrt_complex(double a, double b) {
double r = Math.hypot(a, b);
double theta = Math.atan2(b, a);
double sqrt_r = Math.sqrt(r);
double sqrt_theta = 1.0 / 2 * (2 * PI * Math.floor((PI - theta) / (2 * PI)) + theta);
return new double[]{sqrt_r, sqrt_theta};
}
public static double[] polar_reciprocal(double[] polar) {
return new double[]{1 / polar[0], -polar[1]};
}
public static void toRect(double[]... polars) {
for (double[] polar: polars) {
double a = Math.cos(polar[1]) * polar[0];
double b = Math.sin(polar[1]) * polar[0];
polar[0] = a;
polar[1] = b;
}
}
To join this with your program, simply change your function to reflect this:
if (temp1 < 0) {
return topOfHeart(x, y, temp, temp1);
}
And running it, we get the desired result:
It should be pretty clear that this new function implements exactly the same formula. But how does each part work?
double[] temp3 = cbrt_complex(temp, sqrt(-temp1));
cbrt_complex takes a complex number in the form of a + b i. That's why the second argument is simply sqrt(-temp1) (notice that temp1 < 0, so I use - instead of Math.abs; Math.abs is probably a better idea). cbrt_complex returns the cube root of the complex number, in polar form: r eiθ. We can see from wolframalpha that with positive r and θ, we can write an n-th root of a complex numbers as follows:
And that's exactly how the code for the cbrt_complex and sqrt_complex work. Note that both take a complex number in rectangular coordinates (a + b i) and return a complex number in polar coordinates (r eiθ)
double[] part1 = polar_reciprocal(temp3);
It is easier to take the reciprocal of a polar complex number than a rectangular complex number. If we have r eiθ, its reciprocal (this follows standard power rules, luckily) is simply 1/r e-iθ. This is actually why we are staying in polar form; polar form makes multiplication-type operations easier, and addition type operations harder, while rectangular form does the opposite.
Notice that if we have a polar complex number r eiθ and we want to multiply by a real number d, the answer is as simple as d r eiθ.
The toRect function does exactly what it seems like it does: it converts polar coordinate complex numbers to rectangular coordinate complex numbers.
You may have noticed that the if statement doesn't check that there is no imaginary part, but only if the imaginary part is really small. This is because we are using floating point numbers, so checking result[1] == 0 will likely fail.
And there you are! Notice that we could actually implement the entire heart function with this complex number arithmetic, but it's probably faster to avoid this.

How can you find the point on an ellipse that sweeps a given area?

I am working on the problem of dividing an ellipse into equal sized segments. This question has been asked but the answers suggested numerical integration so that I what I'm attempting. This code short-circuits the sectors so the integration itself should never cover more than 90 degrees. The integration itself is being done by totaling the area of intermediate triangles. Below is the code I have tried, but it is sweeping more than 90 degrees in some cases.
public class EllipseModel {
protected double r_x;
protected double r_y;
private double a,a2;
private double b,b2;
boolean flip;
double area;
double sector_area;
double radstep;
double rot;
int xp,yp;
double deviation;
public EllipseModel(double r_x, double r_y, double deviation)
{
this.r_x = r_x;
this.r_y = r_y;
this.deviation = deviation;
if (r_x < r_y) {
flip = true;
a = r_y;
b = r_x;
xp = 1;
yp = 0;
rot = Math.PI/2d;
} else {
flip = false;
xp = 0;
yp = 1;
a = r_x;
b = r_y;
rot = 0d;
}
a2 = a * a;
b2 = b * b;
area = Math.PI * r_x * r_y;
sector_area = area / 4d;
radstep = (2d * deviation) / a;
}
public double getArea() {
return area;
}
public double[] getSweep(double sweep_area)
{
System.out.println(String.format("getSweep(%f) a = %f b = %f deviation = %f",sweep_area,a,b,deviation));
double[] ret = new double[2];
double[] next = new double[2];
double t_base, t_height, swept,x_mid,y_mid;
double t_area;
sweep_area = sweep_area % area;
if (sweep_area < 0d) {
sweep_area = area + sweep_area;
}
if (sweep_area == 0d) {
ret[0] = r_x;
ret[1] = 0d;
return ret;
}
double sector = Math.floor(sweep_area/sector_area);
double theta = Math.PI * sector/2d;
double theta_last = theta;
System.out.println(String.format("- Theta start = %f",Math.toDegrees(theta)));
ret[xp] = a * Math.cos(theta + rot);
ret[yp] = (1 + (((theta / Math.PI) % 2d) * -2d)) * Math.sqrt((1 - ( (ret[xp] * ret[xp])/a2)) * b2);
next[0] = ret[0];
next[1] = ret[1];
swept = sector * sector_area;
System.out.println(String.format("- Sweeping for %f sector_area=%f",sweep_area-swept,sector_area));
int c = 0;
while(swept < sweep_area) {
c++;
ret[0] = next[0];
ret[1] = next[1];
theta_last = theta;
theta += radstep;
// calculate next point
next[xp] = a * Math.cos(theta + rot);
next[yp] = (1 + (((theta / Math.PI) % 2d) * -2d)) * // selects +/- sqrt
Math.sqrt((1 - ( (ret[xp] * ret[xp])/a2)) * b2);
// calculate midpoint
x_mid = (ret[xp] + next[xp]) / 2d;
y_mid = (ret[yp] + next[yp]) / 2d;
// calculate triangle metrics
t_base = Math.sqrt( ( (ret[0] - next[0]) * (ret[0] - next[0]) ) + ( (ret[1] - next[1]) * (ret[1] - next[1])));
t_height = Math.sqrt((x_mid * x_mid) + (y_mid * y_mid));
// add triangle area to swept
t_area = 0.5d * t_base * t_height;
swept += t_area;
}
System.out.println(String.format("- Theta end = %f (%d)",Math.toDegrees(theta_last),c));
return ret;
}
}
In the output I see the following case where it sweeps over 116 degrees.
getSweep(40840.704497) a = 325.000000 b = 200.000000 deviation = 0.166667
- Theta start = 0.000000
- Sweeping for 40840.704497 sector_area=51050.880621
- Theta end = 116.354506 (1981)
Is there any way to fix the integration formula to create a function that returns the point on an ellipse that has swept a given area? The application that is using this code divides the total area by the number of segments needed, and then uses this code to determine the angle where each segment starts and ends. Unfortunately it doesn't work as intended.
* edit *
I believe the above integration failed because the base and height formula's aren't correct.

No transformation needed use parametric equations for ellipse ...
x=x0+rx*cos(a)
y=y0+ry*sin(a)
where a = < 0 , 2.0*M_PI >
if you divide ellipse by lines from center to x,y from above equation
and angle a is evenly encreased
then the segments will have the same size
btw. if you apply affine transform you will get the same result (even the same equation)
This code will divide ellipse to evenly sized chunks:
double a,da,x,y,x0=0,y0=0,rx=50,ry=20; // ellipse x0,y0,rx,ry
int i,N=32; // divided to N = segments
da=2.0*M_PI/double(N);
for (a=0.0,i=0;i<N;i++,a+=da)
{
x=x0+(rx*cos(a));
y=y0+(ry*sin(a));
// draw_line(x0,y0,x,y);
}
This is what it looks like for N=5
[edit1]
I do not understood from your comment what exactly you want to achieve now
sorry but my English skills are horrible
ok I assume these two possibilities (if you need something different please specify closer)
0.but first some global or member stuff needed
double x0,y0,rx,ry; // ellipse parameters
// [Edit2] sorry forgot to add these constants but they are I thin straight forward
const double pi=M_PI;
const double pi2=2.0*M_PI;
// [/Edit2]
double atanxy(double x,double y) // atan2 return < 0 , 2.0*M_PI >
{
int sx,sy;
double a;
const double _zero=1.0e-30;
sx=0; if (x<-_zero) sx=-1; if (x>+_zero) sx=+1;
sy=0; if (y<-_zero) sy=-1; if (y>+_zero) sy=+1;
if ((sy==0)&&(sx==0)) return 0;
if ((sx==0)&&(sy> 0)) return 0.5*pi;
if ((sx==0)&&(sy< 0)) return 1.5*pi;
if ((sy==0)&&(sx> 0)) return 0;
if ((sy==0)&&(sx< 0)) return pi;
a=y/x; if (a<0) a=-a;
a=atan(a);
if ((x>0)&&(y>0)) a=a;
if ((x<0)&&(y>0)) a=pi-a;
if ((x<0)&&(y<0)) a=pi+a;
if ((x>0)&&(y<0)) a=pi2-a;
return a;
}
1.is point inside segment ?
bool is_pnt_in_segment(double x,double y,int segment,int segments)
{
double a;
a=atanxy(x-x0,y-y0); // get sweep angle
a/=2.0*M_PI; // convert angle to a = <0,1>
if (a>=1.0) a=0.0; // handle extreme case where a was = 2 Pi
a*=segments; // convert to segment index a = <0,segments)
a-=double(segment );
// return floor(a); // this is how to change this function to return points segment id
// of course header should be slightly different: int get_pnt_segment_id(double x,double y,int segments)
if (a< 0.0) return false; // is lower then segment
if (a>=1.0) return false; // is higher then segment
return true;
}
2.get edge point of segment area
void get_edge_pnt(double &x,double &y,int segment,int segments)
{
double a;
a=2.0*M_PI/double(segments);
a*=double(segment); // this is segments start edge point
//a*=double(segment+1); // this is segments end edge point
x=x0+(rx*cos(a));
y=y0+(ry*sin(a));
}
for booth:
x,y is point
segments number of division segments.
segment is sweep-ed area < 0,segments )

Apply an affine transformation to turn your ellipse into a circle, preferrably the unit circle. Then split that into equal sized segments, before you apply the inverse transform. The transformation will scale all areas (as opposed to lengths) by the same factor, so equal area translates to equal area.

Detect black circles(not just pixels) in a image using JUI

I have a image with blackened circles.
The image is a scanned copy of an survey sheet pretty much like an OMR questionnaire sheet.
I want to detect the circles that have been blackened using the JUI(if any other api required)
I have a few examples while searching, but they dont give me accurate result.
I tried..UDAI,Moodle...etc...
Then I decided to make my own. I am able to detect the black pixels but as follows.
BufferedImage mapa = BMPDecoder.read(new File("testjui.bmp"));
final int xmin = mapa.getMinX();
final int ymin = mapa.getMinY();
final int ymax = ymin + mapa.getHeight();
final int xmax = xmin + mapa.getWidth();
for (int i = xmin;i<xmax;i++)
{
for (int j = ymin;j<ymax;j++)
{
int pixel = mapa.getRGB(i, j);
if ((pixel & 0x00FFFFFF) == 0)
{
System.out.println("("+i+","+j+")");
}
}
}
This gives me the co-ordinates of all the black pixels but i cannot make out if its a circle or not.
How can I identify if its a circle.
2] Also I want to know if the image scanned is tilted....I know that the Udai api takes care of that, but for some reason I am not able to get my survey template to run with that code.

So if I understood correctly, you have code that picks out the black pixels so now you have the coordinates of all black pixels and you want to determine all of those that fall on a circle.
The way I would approach this is in 2 steps.
1) Cluster the pixels. Create a class called Cluster, that contains a list of points and use your clustering algorithm to put all the points in the right cluster.
2) Determine which clusters are circles. To do this find the midpoint of all of the points in each cluster (just take the mean of all the points). Then find the minimum and maximum distances from the center, The difference between these should be less than the maximum thickness for a circle in your file. These will give you the radii for the innermost and outermost circles contained within the circle. Now use the equation of a circle x^2 + y^2 = radius, with the radius set to a value between the maximum and minimum found previously to find the points that your cluster should contain. If your cluster contains these it is a circle.
Of course other considerations to consider is whether the shapes you have approximate ellipses rather than circles, in which case you should use the equation of an ellipse. Furthermore, if your file contains circle-like shapes you will need to write additional code to exclude these. On the other hand if all of your circles are exactly the same size you can cut the work that needs to be done by having your algorithm search for circles of that size only.
I hope I could be of some help, good luck!

To answer your first question, I created a class that checks weather an image contains a single non black filled black outlined circle.
This class is experimental, it does not provide exact results all the time, feel free to edit it and to correct the bugs you might encounter.
The setters do not check for nulls or out of range values.
import java.awt.Point;
import java.awt.image.BufferedImage;
import java.io.File;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import javax.imageio.ImageIO;
/**
* Checks weather an image contains a single non black filled black outlined circle<br />
* This class is experimental, it does not provide exact results all the time, feel free to edit it and to correct
* the bugs you might encounter.
* #author Ahmed KRAIEM
* #version 0.9 alpha
* #since 2013-04-03
*/
public class CircleChecker {
private BufferedImage image;
/**
* Points that are equal to the calculated radius±<code>radiusesErrorMargin%</code> are not considered rogue points.<br />
* <code>radiusesErrorMargin</code> must be <code>>0 && <1</code>
*/
private double radiusesErrorMargin = 0.2;
/**
* A shape that has fewer than roguePointSensitivity% of rogue points is considered a circle.<br />
* <code>roguePointSensitivity</code> must be <code>>0 && <1</code>
*/
private double roguePointSensitivity = 0.05;
/**
* The presumed circle is divided into <code>angleCompartimentPrecision</code> parts,<br />
* each part must have <code>minPointsPerCompartiment</code> points
* <code>angleCompartimentPrecision</code> must be <code>> 0</code>
*/
private int angleCompartimentPrecision = 50;
/**
* The minimum number of points requiered to declare a part valid.<br />
* <code>minPointsPerCompartiment</code> must be <code>> 0</code>
*/
private int minPointsPerCompartiment = 20;
public CircleChecker(BufferedImage image) {
super();
this.image = image;
}
public CircleChecker(BufferedImage image, double radiusesErrorMargin,
int minPointsPerCompartiment, double roguePointSensitivity,
int angleCompartimentPrecision) {
this(image);
this.radiusesErrorMargin = radiusesErrorMargin;
this.minPointsPerCompartiment = minPointsPerCompartiment;
this.roguePointSensitivity = roguePointSensitivity;
this.angleCompartimentPrecision = angleCompartimentPrecision;
}
public BufferedImage getImage() {
return image;
}
public void setImage(BufferedImage image) {
this.image = image;
}
public double getRadiusesErrorMargin() {
return radiusesErrorMargin;
}
public void setRadiusesErrorMargin(double radiusesErrorMargin) {
this.radiusesErrorMargin = radiusesErrorMargin;
}
public double getMinPointsPerCompartiment() {
return minPointsPerCompartiment;
}
public void setMinPointsPerCompartiment(int minPointsPerCompartiment) {
this.minPointsPerCompartiment = minPointsPerCompartiment;
}
public double getRoguePointSensitivity() {
return roguePointSensitivity;
}
public void setRoguePointSensitivity(double roguePointSensitivity) {
this.roguePointSensitivity = roguePointSensitivity;
}
public int getAngleCompartimentPrecision() {
return angleCompartimentPrecision;
}
public void setAngleCompartimentPrecision(int angleCompartimentPrecision) {
this.angleCompartimentPrecision = angleCompartimentPrecision;
}
/**
*
* #return true if the image contains no more than <code>roguePointSensitivity%</code> rogue points
* and all the parts contain at least <code>minPointsPerCompartiment</code> points.
*/
public boolean isCircle() {
List<Point> list = new ArrayList<>();
final int xmin = image.getMinX();
final int ymin = image.getMinY();
final int ymax = ymin + image.getHeight();
final int xmax = xmin + image.getWidth();
for (int i = xmin; i < xmax; i++) {
for (int j = ymin; j < ymax; j++) {
int pixel = image.getRGB(i, j);
if ((pixel & 0x00FFFFFF) == 0) {
list.add(new Point(i, j));
}
}
}
if (list.size() == 0)
return false;
double diameter = -1;
Point p1 = list.get(0);
Point across = null;
for (Point p2 : list) {
double d = distance(p1, p2);
if (d > diameter) {
diameter = d;
across = p2;
}
}
double radius = diameter / 2;
Point center = center(p1, across);
int diffs = 0;
int diffsUntilError = (int) (list.size() * roguePointSensitivity);
double minRadius = radius - radius * radiusesErrorMargin;
double maxRadius = radius + radius * radiusesErrorMargin;
int[] compartiments = new int[angleCompartimentPrecision];
for (int i=0; i<list.size(); i++) {
Point p = list.get(i);
double calRadius = distance(p, center);
if (calRadius>maxRadius || calRadius < minRadius)
diffs++;
else{
//Angle
double angle = Math.atan2(p.y -center.y,p.x-center.x);
//angle is between -pi and pi
int index = (int) ((angle + Math.PI)/(Math.PI * 2 / angleCompartimentPrecision));
compartiments[index]++;
}
if (diffs >= diffsUntilError){
return false;
}
}
int sumCompartiments = list.size() - diffs;
for(int comp : compartiments){
if (comp < minPointsPerCompartiment){
return false;
}
}
return true;
}
private double distance(Point p1, Point p2) {
return Math.sqrt(Math.pow(p1.x - p2.x, 2) + Math.pow(p1.y - p2.y, 2));
}
private Point center(Point p1, Point p2) {
return new Point((p1.x + p2.x) / 2, (p1.y + p2.y) / 2);
}
public static void main(String[] args) throws IOException {
BufferedImage image = ImageIO.read(new File("image.bmp"));
CircleChecker cc = new CircleChecker(image);
System.out.println(cc.isCircle());
}
}

You'll need to program in a template of what a circle would look like, and then make it scalable to suit the different circle sizes.
For example circle of radius 3 would be:
o
ooo
o
This assumes you have a finite set of circles you need to find, maybe up to 5x5 or 6x6 this would be feasible.
or you could use: Midpoint circle algorithm
This would involve finding all black pixel groups and then selecting the middle pixel for each one.
Apply this algorithm using the outer pixels as a guid to how big the circle could be.
Finding the difference between black /expected black pixels.
If the black to expected black ratio is high enough, its a black circle and you can delete / whiten it.

Finding Points contained in a Path in Android

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();
}

Polygon Intersection fails, collision "size" too big

OK, so I'm trying to make a simple asteroids clone. Everything works fine, except for the collision detection.
I have two different versions, the first one uses java.awt.geom.Area:
// polygon is a java.awt.Polygon and p is the other one
final Area intersect = new Area();
intersect.add(new Area(polygon));
intersect.intersect(new Area(p.polygon));
return !intersect.isEmpty();
This works like a charm... if you don't care about 40% CPU for only 120 asteroids :(
So I searched the net for the famous separating axis theorem, since I'm not thaaaaaat good a the math I took the implementation from here and converted it to fit my Java needs:
public double dotProduct(double x, double y, double dx, double dy) {
return x * dx + y * dy;
}
public double IntervalDistance(double minA, double maxA, double minB,
double maxB) {
if (minA < minB) {
return minB - maxA;
} else {
return minA - maxB;
}
}
public double[] ProjectPolygon(double ax, double ay, int p, int[] x, int[] y) {
double dotProduct = dotProduct(ax, ay, x[0], y[0]);
double min = dotProduct;
double max = dotProduct;
for (int i = 0; i < p; i++) {
dotProduct = dotProduct(x[i], y[i], ax, ay);
if (dotProduct < min) {
min = dotProduct;
} else if (dotProduct > max) {
max = dotProduct;
}
}
return new double[] { min, max };
}
public boolean PolygonCollision(Asteroid ast) {
int edgeCountA = points;
int edgeCountB = ast.points;
double edgeX;
double edgeY;
for (int edgeIndex = 0; edgeIndex < edgeCountA + edgeCountB; edgeIndex++) {
if (edgeIndex < edgeCountA) {
edgeX = xp[edgeIndex] * 0.9;
edgeY = yp[edgeIndex] * 0.9;
} else {
edgeX = ast.xp[edgeIndex - edgeCountA] * 0.9;
edgeY = ast.yp[edgeIndex - edgeCountA] * 0.9;
}
final double x = -edgeY;
final double y = edgeX;
final double len = Math.sqrt(x * x + y * y);
final double axisX = x / len;
final double axisY = y / len;
final double[] minMaxA = ProjectPolygon(axisX, axisY, points, xp,
yp);
final double[] minMaxB = ProjectPolygon(axisX, axisY, ast.points,
ast.xp, ast.yp);
if (IntervalDistance(minMaxA[0], minMaxA[1], minMaxB[0], minMaxB[1]) > 0) {
return false;
}
}
return true;
}
It works... kinda. Actually it seems that the "collision hull" of the asteroids is too big when using this code, it's like 1.2 times the size of the asteroid. And I don't have any clue why.
Here are two pictures for comparison:
http://www.spielecast.de/stuff/asteroids1.png
http://www.spielecast.de/stuff/asteroids2.png
As you can hopefully see, the asteroids in picture one are much denser than the ones in picture 2 where is use the SAT code.
So any ideas? Or does anyone knows a Polygon implementation for Java featuring intersection tests that I could use?

It looks like your second result is doing collision detection as if the polygons were circles with their radius set to the most distant point of the polygon from the center. Most collision detection stuff I've seen creates a simple bounding box (either a circle or rectangle) into which the polygon can fit. Only if two bounding boxes intersect (a far simpler calculation) do you continue on to the more detailed detection. Perhaps the appropriated algorithm is only intended as a bounding box calculator?
EDIT:
Also, from wikipedia
The theorem does not apply if one of the bodies is not convex.
Many of the asteroids in your image have concave surfaces.