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How to draw lines pointing to mouse in Processing

I am trying to make a program where there are lines in a grid pointing towards the mouse like magnets. I am a beginner in Processing, can someone point me towards a tutorial on how to do that or give me some code and explain what it does?
int x1 = 0;
int x2 = 0;
int y1 = 0;
int y2 = 0;
void setup() {
size(200, 200);
}
void draw() {
background(255, 255, 0);
x1 = (mouseX + 100) / 2;
y1 = (mouseY + 100) / 2;
x2 = -1 * x1 + 200;
y2 = -1 * y1 + 200;
line(x1, y1, x2, y2);
}

There's plenty of solutions for this project. One of the easiest is to use Processing's PVector class.
The PVector class can be used for two or three dimensional vectors. A vector is an entity that has both magnitude and direction. The PVector class, however, stores the components of the vector (x,y for 2D, and x,y,z for 3D). The magnitude and direction are calculated from the components and can be accessed via the methods mag() and heading().
A two dimensional vector in Processing is defined through x and y components:
PVector v = new PVector(xComponent, yComponent);
With some mathematical formulae, you can determine magnitude and direction using the x- and y-components. But we don't need to determine these.
Below, I've attached completed solution code. Most of it should make sense to you. But it's worth understanding what is going on with PVector.
A nested for loop within void draw() contains x and y variables that represent the coordinates of each grid vertex.
We first define PVector v as a vector given by an x-component of mouseX - x, or the difference between the x-positions of the mouse and each grid point. Similarly, the y-component given by mouseY - y has the same difference.
Creating a variable PVector u initialized from v.setMag(15) holds a PVector that has the same direction as v, but with a length of just 15.
Now to draw the lines. Vectors represent an offset, not a position (in this case), so drawing a line from a grid point to an offset of a grid point is key.
Hence line(x, y, x + u.x, y + u.y), where u.x and u.y are the x- and y-components of the vector u.
void setup() {
size(600, 600); // Set the size of the canvas to 600x600.
}
void draw() {
background(255);
stroke(200); // Set the stroke color to black
int distVertLine = width / 10; // This variable defines the distance between each subsequent vertical line.
for(int i = 0; i < width; i += distVertLine) {
line(i, 0, i, height); // Draw a line at x=i starting at the top of the canvas (y=0) and going to the bottom (y=height)
}
int distHorizLine = height / 10; // This variable defines the distance between each subsequent vertical line.
for(int i = 0; i < width; i += distHorizLine) {
line(0, i, width, i); // Draw a line at y=i starting at the left of the canvas (x=0) and going to the right (x=width)
}
stroke(0); // Set the stroke to black.
// Use a nested for loop to iterate through all grid vertices.
for(int x = 0; x <= width; x += width/10) {
for(int y = 0; y <= height; y += height/10) {
PVector v = new PVector(mouseX - x, mouseY - y); // Define a vector that points in the direction of the mouse from each grid point.
PVector u = v.setMag(15); // Make the vector have a length of 15 units.
line(x, y, x + u.x, y + u.y); // Draw a line from the grid vertex to the terminal point given by the vector.
}
}
}

The answer already given by Ben Myers is excellent! The code below has a few small modifications:
the two for loops for the grid lines have been combined (since width and height are equal);
the construction of the vector is combined with setting the magnitude;
some minor changes to colors and comments.
Modified code:
void setup() {
// Set the size of the canvas to 600x600 pixels.
size(600, 600);
}
void draw() {
// There are 10x10 grid cells that each have a size of 60x60 pixels.
int gridSize = width / 10;
// Set the background color to anthracite and the stroke color to orange.
background(56, 62, 66);
stroke(235, 113, 52);
// Draw vertical and horizontal grid lines.
for (int lineIndex = 0; lineIndex < gridSize; lineIndex++) {
line(lineIndex * gridSize, 0, lineIndex * gridSize, height);
line(0, lineIndex * gridSize, width, lineIndex * gridSize);
}
// Set the stroke color to blue.
stroke(0, 139, 225);
// Use a nested for loop to iterate through all grid cells.
for (int x = 0; x <= width; x += gridSize) {
for (int y = 0; y <= height; y += gridSize) {
// Define a vector that points in the direction of the mouse from
// each grid point and set the vector length to 15 units.
PVector vector = new PVector(mouseX - x, mouseY - y).setMag(15);
// Draw a line from the grid point to the end point using the vector.
line(x, y, x + vector.x, y + vector.y);
}
}
}

Draw an arc in opengl GL10

I want to draw an arc using center point,starting point,ending point on opengl surfaceview.I have tried this given below code so far. This function draws the expected arc if we give the value for start_line_angle and end_line_angle manually (like start_line_angle=0 and end_line_angle=90) in degree.
But I need to draw an arc with the given co-ordinates(center point,starting point,ending point) and calculating the start_line_angle and end_line_angle programatically.
This given function draws an arc with the given parameters but not giving the desire result. I've wasted my 2 days for this. Thanks in advance.
private void drawArc(GL10 gl, float radius, float cx, float cy, float start_point_x, float start_point_y, float end_point_x, float end_point_y) {
gl.glLineWidth(1);
int start_line_angle;
double sLine = Math.toDegrees(Math.atan((cy - start_point_y) / (cx - start_point_x))); //normal trigonometry slope = tan^-1(y2-y1)/(x2-x1) for line first
double eLine = Math.toDegrees(Math.atan((cy - end_point_y) / (cx - end_point_x))); //normal trigonometry slope = tan^-1(y2-y1)/(x2-x1) for line second
//cast from double to int after round
int start_line_Slope = (int) (sLine + 0.5);
/**
* mapping the tiriogonometric angle system to glsurfaceview angle system
* since angle system in trigonometric system starts in anti clockwise
* but in opengl glsurfaceview angle system starts in clock wise and the starting angle is 90 degree of general trigonometric angle system
**/
if (start_line_Slope <= 90) {
start_line_angle = 90 - start_line_Slope;
} else {
start_line_angle = 360 - start_line_Slope + 90;
}
// int start_line_angle = 270;
// int end_line_angle = 36;
//casting from double to int
int end_line_angle = (int) (eLine + 0.5);
if (start_line_angle > end_line_angle) {
start_line_angle = start_line_angle - 360;
}
int nCount = 0;
float[] stVertexArray = new float[2 * (end_line_angle - start_line_angle)];
float[] newStVertextArray;
FloatBuffer sampleBuffer;
// stVertexArray[0] = cx;
// stVertexArray[1] = cy;
for (int nR = start_line_angle; nR < end_line_angle; nR++) {
float fX = (float) (cx + radius * Math.sin((float) nR * (1 * (Math.PI / 180))));
float fY = (float) (cy + radius * Math.cos((float) nR * (1 * (Math.PI / 180))));
stVertexArray[nCount * 2] = fX;
stVertexArray[nCount * 2 + 1] = fY;
nCount++;
}
//taking making the stVertextArray's data in reverse order
reverseArray = new float[stVertexArray.length];//-2 so that no repeatation occurs of first value and end value
int count = 0;
for (int i = (stVertexArray.length) / 2; i > 0; i--) {
reverseArray[count] = stVertexArray[(i - 1) * 2 + 0];
count++;
reverseArray[count] = stVertexArray[(i - 1) * 2 + 1];
count++;
}
//reseting the counter to initial value
count = 0;
int finalArraySize = stVertexArray.length + reverseArray.length;
newStVertextArray = new float[finalArraySize];
/**Now adding all the values to the single newStVertextArray to draw an arc**/
//adding stVertextArray to newStVertextArray
for (float d : stVertexArray) {
newStVertextArray[count++] = d;
}
//adding reverseArray to newStVertextArray
for (float d : reverseArray) {
newStVertextArray[count++] = d;
}
Log.d("stArray", stVertexArray.length + "");
Log.d("reverseArray", reverseArray.length + "");
Log.d("newStArray", newStVertextArray.length + "");
ByteBuffer bBuff = ByteBuffer.allocateDirect(newStVertextArray.length * 4);
bBuff.order(ByteOrder.nativeOrder());
sampleBuffer = bBuff.asFloatBuffer();
sampleBuffer.put(newStVertextArray);
sampleBuffer.position(0);
gl.glEnableClientState(GL10.GL_VERTEX_ARRAY);
gl.glVertexPointer(2, GL10.GL_FLOAT, 0, sampleBuffer);
gl.glDrawArrays(GL10.GL_LINE_LOOP, 0, nCount * 2);
gl.glLineWidth(1);
}

To begin with the trigonometry you may not simply use the atan to find degrees of the angle. You need to check what quadrant the vector is in and increase or decrease the result you get from atan. Better yet use atan2 which should include both dx and dy and do the job for you.
You seem to create the buffer so that a point is created per degree. This is not the best solution as for large radius that might be too small and for small radius this is way too much. Tessellation should include the radius as well such that number of points N is N = abs((int)(deltaAngle*radius*tessellationFactor)) then use angleFragment = deltaAngle/N but make sure that N is greater then 0 (N = N?N:1). The buffer size is then 2*(N+1) of floats and the iteration if for(int i=0; i<=N; i++) angle = startAngle + angleFragment*i;.
As already pointed out you need to define the radius of the arc. It is quite normal to use an outside source the way you do and simply force it to that value but use the 3 points for center and the two borders. Some other options that usually make sense are:
getting the radius from the start line
getting the radius from the shorter of the two lines
getting the average of the two
interpolate the two to get an elliptic curve (explained below)
To interpolate the radius you need to get the two radiuses startRadius and endRadius. Then you need to find the overall radius which was already used as deltaAngle above (watch out when computing this one, it is more complicated as it seems, for instance drawing from 320 degrees to 10 degrees results in deltaAngle = 50). Anyway the radius for a specific point is then simply radius = startRadius + (endRadius-startRadius)*abs((angleFragment*i)/deltaAngle). This represents a simple linear interpolation in polar coordinate system which is usually used to interpolate vector in matrices and is the core functionality to get nice animations.
There are some other ways of getting the arc points which may be better performance wise but I would not suggest them unless and until you need to optimize your code which should be very late in production. You may simply keep stepping toward the next point and correcting the radius (this is only a concept):
vec2 start, end, center; // input values
float radius; // input value
// making the start and end relative to center
start -= center;
end -= center;
vec2 current = start/length(start) * radius; // current position starts in first vector
vec2 target = end/length(end) * radius; // should be the last point
outputBuffer[0] = current+center; // insert the first point
for(int i=1;; i++) { // "break" will need to exit the loop, we need index only for the buffer
vec2 step = vec2(current.y, -(current.x)); // a tangential vector from current start point according to center
step = step/length(step) / tessellationScale; // normalize and apply tessellation
vec2 next = current + step; // move tangentially
next = next/length(next) * radius; // normalize and set the
if(dot(current-target, next-target) > .0) { // when we passed the target vector
current = next; // set the current point
outputBuffer[i] = current+center; // insert into buffer
}
else {
current = target; // simply use the target now
outputBuffer[i] = current+center; // insert into buffer
break; // exit
}
}

smooth color interpolation along a "bresenham" line

I am trying to interpolate color along a line so that, given two points and their respective RGB values, I can draw a line with a smooth color gradient. Using Bresenham's Line Algorithm, I can now draw lines, but am not sure how to begin interpolating colors between the two end points. The following is part of the drawLine() function that works for all line whose slope are less than 1.
int x_start = p1.x, x_end = p2.x, y_start =p1.y, y_end = p2.y;
int dx = Math.abs(x_end-x_start), dy = Math.abs(y_end-y_start);
int x = x_start, y = y_start;
int step_x = x_start < x_end ? 1:-1;
int step_y = y_start < y_end ? 1:-1;
int rStart = (int)(255.0f * p1.c.r), rEnd = (int)(255.0f * p2.c.r);
int gStart = (int)(255.0f * p1.c.g), gEnd = (int)(255.0f * p2.c.g);
int bStart = (int)(255.0f * p1.c.b), bEnd = (int)(255.0f * p2.c.b);
int xCount = 0;
//for slope < 1
int p = 2*dy-dx;
int twoDy = 2*dy, twoDyMinusDx = 2*(dy-dx);
int xCount = 0;
// draw the first point
Point2D start = new Point2D(x, y, new ColorType(p1.c.r, p1.c.g, p1.c.b));
drawPoint(buff, start);
float pColor = xCount / Math.abs((x_end - x_start));
System.out.println(x_end + " " + x_start);
while(x != x_end){
x+= step_x;
xCount++;
if(p<0){
p+= twoDy;
}
else{
y += step_y;
p += twoDyMinusDx;
}
Point2D draw_line = new Point2D(x, y, new ColorType(p1.c.r*(1-pColor)+p2.c.r*pColor,p1.c.g*(1-pColor)+p2.c.g*pColor,p1.c.b*(1-pColor)+p2.c.b*pColor));
System.out.println(pColor);
drawPoint(buff,draw_line );
}
So what I'm thinking is that, just like drawing lines, I also need some sort of decision parameter p to determine when to change the RGB values. I am thinking of something along lines of as x increments, look at each rgb value and decide if I want to manipualte them or not.
I initialized rStart and rEnd(and so on for g and b) but have no idea where to start. any kind of help or suggestions would be greatly appreciated!
Edit: thanks #Compass for the great suggestion ! Now I've ran into another while trying to implementing that strategy, and I am almost certain it's an easy bug. I just can't see it right now. For some reason my pColor always return 0, I am not sure why. I ran some print statements to make sure xCount is indeed increasing, so I am not sure what else might've made this variable always 0.

I remember figuring this out way back when I was learning GUI! I'll explain the basic concepts for you.
Let's say we have two colors,
RGB(A,B,C)
and
RGB(X,Y,Z)
for simplicity.
If we know the position percentage-wise (we'll call this P, a float 0 for beginning, 1.0 at end) along the line, we can calculate what color should be there using the following:
Resultant Color = RGB(A*(1-P)+X*P,B*(1-P)+Y*P,C*(1-P)+Z*P)
In other words, you average out the individual RGB values along the line.

Actually you will be drawing the line in RGB space as well !
Bresenham lets you compute point coordinates from (X0, Y0) to (X1, Y1).
This is done by a loop on X or Y, with a linear interpolation on the other coordinate.
Just extend the algorithm to draw a line from (X0, Y0, R0, G0, B0) to (X1, Y1, R1, G1, B1), in the same loop on X or Y, with a linear interpolation on the other coordinates.

Fill a shape with points

I try to draw a leaf looking thing on the screen, and try to fill it with a color. It's like drawing a circle, the difference is, that it's only 270 degrees, and the radius starts from 0 to 100. I first draw the left side, and on each degree I fill the inside. At the end I draw the right side.
Here is to code, maybe it's easier to understand:
canvas = new BufferedImage(SIZE, SIZE, BufferedImage.TYPE_INT_ARGB);
Color black = new Color(0,0,0);
Color green = new Color(0,130,0);
double j = 0.0; // radius
double max = 100.0; // max radius
for (int i = 0; i < 135; i++) { // left side (270 degree / 2)
j += max / 135.0;
// x, y coordinate
int x = (int)(Math.cos(Math.toRadians(i)) * j);
int y = (int)(Math.sin(Math.toRadians(i)) * j);
// draw a circle like thing with radius j
for (int l = i; l < 135 + (135 - i); l++) {
int ix = (int)(Math.cos(Math.toRadians(l)) * j);
int iy = (int)(Math.sin(Math.toRadians(l)) * j);
canvas.setRGB(ix + 256, iy + 256, green.getRGB());
}
canvas.setRGB(x + 256, y + 256, black.getRGB());
}
// draw the right side
for (int i = 135; i < 270; i++) {
j -= max / 135.0;
int x = (int)(Math.cos(Math.toRadians(i)) * j);
int y = (int)(Math.sin(Math.toRadians(i)) * j);
canvas.setRGB(x + 256, y + 256, black.getRGB());
}
This is the result:
As you can see, where the radius is bigger, the leaf is not filled completely.
If I change i to 1350, then divide it with 10 where I calculate x, y, then it's filled, but it's much slower. Is there a better way to properly fill my shape?
Later I also would like to fill my shape with a gradient, so from green to a darker green, then back to green. With my method this is easy, but super slow.
Thanks in advance!

I think that for you the best solution is to use a flood fill algorithm, it's easy to implement in Java and efficient in your case, like you have a simple shape.
Here is a wikipedia article that is really complet : http://en.wikipedia.org/wiki/Flood_fill

Here is a simple suggestion: Instead of drawing the leaf, just put the points that create the outline into an array. The array should run from xMin (smallest X coordiate of the leaf outline) to xMax. Each element is two ints: yMin and yMax.
After rendering all the points, you can just draw vertical lines to fill the space between yMin/yMax for each X coordinate.
If you have gaps in the array, fill them by interpolating between the neighboring points.
An alternative would be to sort the points clockwise or counter-clockwise and use them as the outline for a polygon.

Creating a Squircle

I'm a first year programmer. I'm trying to create a squircle. (square with round corners).
So far i have managed to get. I have been given the constants of a,b and r. If anyone could help i would be really thankful. I'm a total noob to this. So be nice :)
package squircle;
import java.awt.*;
import javax.swing.*;
import java.lang.Math;
public class Main extends javax.swing.JApplet {
public void paint(Graphics g){
// (x-a)^4 + (y-b)^4 = r^4
// y = quadroot( r^4 - (x-a)^4 + b)
// x values must fall within a-r < x < a+r
int[] xPoints = new int[200];
int[] yPoints = new int[200];
int[] mypoints = new int[200];
for(int c = 0; c <200; c++){
int a = 100;
int r = 100;
int b = 100;
double x = c ;
double temp = (r*r*r*r);
double temp2 = x-a;
double temp3 = ((temp2)*(temp2)*(temp2)*(temp2));
double temp6 = Math.sqrt(temp-temp3);
double y = (Math.sqrt(temp6) + b );
double z = (y*-1)+300;
mypoints[c]=(int)z;
// if (c>100){
// y = y*1;
// }
// else if(c<100){
// y = y*1;
// }
xPoints[c]=(int)x;
yPoints[c]=(int)y;
// change the equation to find x co-ordinates
// change it to find y co-ordinates.
// r is the minor radius
// (a,b) is the location of the centre
// a = 100
// b = 100
// r = 100
// x value must fall within 0 or 200
}
g.drawPolygon(xPoints, yPoints, xPoints.length);
g.drawPolygon(xPoints, (mypoints), xPoints.length);
}
}

Is it homework or is there some other reason why you're not using Graphics#drawRoundRect()?

If you are submitting this as homework there are some elements of style that may help you. What are the roles of 200, 100 and 300? These are "magic constants" which should be avoided. Are they related or is it just chance that they have these values? Suggest you use symbols such as:
int NPOINTS = 200;
or
double radius = 100.0
That would reveal whether the 300 was actually the value you want. I haven't checked.
Personally I wouldn't write
y*-1
but
-y
as it's too easy to mistype the former.
I would also print out the 200 points as floats and see if you can tell by eye where the error is. It's highly likely that the spurious lines are either drawn at the start or end of the calculation - it's easy to make "end-effect" errors where exactly one point is omitted or calculated twice.
Also it's cheap to experiment. Try iterating c from 0 to 100. or 0 to 10, or 0 to 198 or 1 to 200. Does your spurious line/triangle always occur?
UPDATE Here is what I think is wrong and how to tackle it. You have made a very natural graphics error and a fence-post error (http://en.wikipedia.org/wiki/Off-by-one_error) and it's hard to detect what is wrong because your variable names are poorly chosen.
What is mypoints? I believe it is the bottom half of the squircle - if you had called it bottomHalf then those replying woulod have spotted the problem quicker :-).
Your graphics problem is that you are drawing TWO HALF-squircles. Your are drawing CLOSED curves - when you get to the last point (c==199) the polygon is closed by drawing back to c==0. That makes a D-shape. You have TWO D-shapes, one with the bulge UP and one DOWN. Each has a horizontal line closing the polygon.
Your fence-post error is that you are drawing points from 0 to 199. For the half-squircle you want to draw from 0 to 200. That's 201 points! The loss of one point means that you have a very slightly sloping line. The bottom lines slopes in tghe opposite direction from the top. That gives you a very then wedge shape, which you refer to as a triangle. I'm guessing that your triangle is not actually closed but like a slice from a pie but very then/sharp.
(The code below could be prettier and more compact. However it is often useful to break symmetrical problems into quadrants or octants. It would also be interesting to use an anngle to sweep out the polygon).
You actually want ONE polygon. The code should be something like:
int NQUADRANT = 100;
int NPOINTS = 4*NQUADRANT ; // closed polygon
double[] xpoints = new double[NPOINTS];
double[] ypoints = new double[NPOINTS];
Your squircle is at 100, 100 with radius 100. I have chosen different values here
to emphasize they aren't related. By using symbolic names you can easily vary them.
double xcenter = 500.0;
double ycentre = 200.0;
double radius = 100.;
double deltax = radius/(double) NQUADRANT;
// let's assume squircle is centered on 0,0 and add offsets later
// this code is NOT complete or correct but should show the way
// I might have time later
for (int i = 0; i < NPOINTS; i++) {
if (i < NQUADRANT) {
double x0 = -radius + i* deltax;
double y0 = fourthRoot(radius, x0);
x[i] = x0+xcenter;
y[i] = y0+ycenter;
}else if (i < 2*NQUADRANT) {
double x0 = (i-NQUADRANT)* deltax;
double y0 = fourthRoot(radius, x0);
x[i] = x0+xcenter;
y[i] = y0+ycenter;
}else if (i < 3*NQUADRANT) {
double x0 = (i-2*NQUADRANT)* deltax;
double y0 = -fourthRoot(radius, x0);
x[i] = x0+xcenter;
y[i] = y0+ycenter;
}else {
double x0 = -radius + (i-3*NQUADRANT)* deltax;
double y0 = -fourthRoot(radius, x0);
x[i] = x0+xcenter;
y[i] = y0+ycenter;
}
}
// draw single polygon
private double fourthRoot(double radius, double x) {
return Math.sqrt(Math.sqrt(radius*radius*radius*radius - x*x*x*x));
}

There is a javascript version here. You can view the source and "compare notes" to potentially see what you are doing wrong.

Ok, upon further investigation here is why you are getting the "triangle intersecting it". When you drawPolygon the points are drawn and the last point connects the first point, closing the points and making the polygon. Since you draw one half it is drawn (then connected to itself) and then the same happens for the other side.
As a test of this change your last couple lines to this:
for( int i = 0; i < yPoints.length; i++ ) {
g.drawString( "*", xPoints[ i ], yPoints[ i ] );
}
for( int i = 0; i < mypoints.length; i++ ) {
g.drawString( "*", xPoints[ i ], mypoints[ i ] );
}
// g.drawPolygon( xPoints, yPoints, xPoints.length );
// g.drawPolygon( xPoints, ( mypoints ), xPoints.length );
It is a little crude, but I think you'll get the point. There are lots of solutions out there, personally I would try using an array of the Point class and then sort it when done, but I don't know the specifics of what you can and can not do.

Wow, are you guys overthinking this, or what! Why not just use drawLine() four times to draw the straight parts of the rectangle and then use drawArc() to draw the rounded corners?