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Filling an area of piece of circle in square grid

I am trying to fill some particular circle areas on a square grid in Proceesing. However, I am facing some weird filling after I run my code. I used the "atan2" function to get the angle of points in the grid and apply some if conditions to limit the area. But it doesn't work. Actually, it works some way but not exactly what I want.This is the result I run the code-> enter image description here, but it should be like -> enter image description here Can someone help me to solve this, please?
(Additionally, I seized this page to detect which points are in the area I specified enter link description here.
Cell [][] cells;
int res = 10;
PVector circleCenter ;
float rad=290;
void setup() {
size(1200, 800);
cells = new Cell[width/res ][height/res];
for (int i=0; i<cells.length; i++) {
for (int j = 0; j<cells[0].length; j++) {
cells[i][j] = new Cell();
}
}
}
void draw() {
background(255);
circleCenter = new PVector(mouseX, mouseY);
display();
pushStyle();
ellipse(circleCenter.x, circleCenter.y, 20, 20);
popStyle();
//println(frameRate);
}
void display() {
for (int i=0; i<cells.length; i++) {
for (int j = 0; j<cells[0].length; j++) {
if (sq(i*res-width/2) + sq(j*res-height/2) <=sq(rad/2)) {
float angleInRad = atan2(j*res-height/2, i*res-width/2);
///
if (angleInRad<radians(-10) && angleInRad>radians(-60)) {
fill(0, 255, 0); // degrees(atan2(j*res-circleCenter.y, i*res-circleCenter.x))
}
} else {
noFill();
}
rect(i*res, j*res, res, res);
}}
class Cell {
boolean blackness=false;
Cell() {
}
}

You have not specified input data for your task. I assume we have center coordinates cx, cx, radius r, starting and ending angle sa, ea of the sector (in code -10, -60).
There is an approach to check whether vector direction lies in sector, it's robust to potential troubles with periodicity, negative values etc. At first normalize range ends, then find middle angle and half-angle
half = (ea - sa) / 2
mid = (ea + sa) / 2
coshalf = Cos(half)
Now compare that difference of angle and middle one is lower then half-angle
if Cos(angle - mid) >= coshalf then
angle lies in range sa..ea

Whack-a-mole in Processing

I am working on a school project in Processing (Java Mode). We have a picture of how the game should look like.
So the task is to create a grid out of squares. Random squares should light up in red. If a red square is clicked, it should change colors to green and stay green.
What my code looks like at the moment:
Square[][] grid;
int cols = 20;
int rows = 20;
void setup() {
size(400, 400);
grid = new Square[cols][rows];
for (int i = 0; i < cols; i++) {
for (int j = 0; j < rows; j++) {
grid[i][j] = new Square(i*20, j*20, 20, 20);
}
}
}
void draw() {
background(0);
for (int i = 0; i < cols; i++) {
for (int j = 0; j < rows; j++) {
grid[i][j].display();
if (grid[i][j].x<mouseX && mouseX < grid[i][j].x + grid[i][j].w && grid[i][j].y<mouseY && mouseY < grid[i][j].y + grid[i][j].h && mousePressed) {
color col = color(0,204,0);
grid[i][j].update(col);
}
}
}
}
Class for squares:
class Square {
float x, y;
float w, h;
color c;
Square(float tempX, float tempY, float tempW, float tempH) {
x = tempX;
y = tempY;
w = tempW;
h = tempH;
c = color(0);
}
void display() {
stroke(0);
fill(c);
rect(x, y, w, h);
}
void update(color c) {
this.c = c;
}
}
So at the moment, every square you click turns green. I am not sure how to write the code, so that random squares change color to red and shuffle every 5 seconds.
Do you have any tips on how to proceed with the code or which thinking steps to take to be able to solve this task?

First, take your task:
So the task is to create a grid out of squares. Random squares should light up in red. If a red square is clicked, it should change colors to green and stay green.
and break it down:
create a grid out of squares: nicely done already !
Random squares should light up in red
If a red square is clicked
change colors to green and stay green
How do you use random numbers in Processing ?
The simplest method is using the random() method: you can pass two values and you'll get back a random number between those values.
Let's say you want to flip a coin so there's a (roughly) 50-50 change you get heads or tails. You could so something like:
if(random(0, 100) > 50){
println("head");
}else{
println("tails");
}
Could even be random(0.0, 1.0) > 0.5 for example, the idea is the same.
You could think of throwing a dice or a number of dices, etc.
Remember these are pseudo-random and in your own time can explore other pseudo random related methods such as randomGauss() and noise().
random() may be good enough for now, part 2 done :)
You're almost done with part 3:
if (grid[i][j].x<mouseX && mouseX < grid[i][j].x + grid[i][j].w && grid[i][j].y<mouseY && mouseY < grid[i][j].y + grid[i][j].h && mousePressed) {
but you need to also check if the clicked square is red.
Would nice to have some red squares to begin with. Let's assume color(204, 0, 0) is your red, you could simply add an additional check:
if(grid[i][j].c == color(204, 0, 0)){
println("red block clicked");
grid[i][j].c = color(0, 204, 0);
}
Which roughly turns your sketch into:
Square[][] grid;
int cols = 20;
int rows = 20;
final color RED = color(204, 0, 0);
final color GREEN = color(0, 204, 0);
void setup() {
size(400, 400);
grid = new Square[cols][rows];
for (int i = 0; i < cols; i++) {
for (int j = 0; j < rows; j++) {
grid[i][j] = new Square(i*20, j*20, 20, 20);
// roughly 50 - 50 % change a grid square will be red
if (random(0, 100) > 50) {
grid[i][j].update(RED);
}
}
}
}
void draw() {
background(0);
for (int i = 0; i < cols; i++) {
for (int j = 0; j < rows; j++) {
grid[i][j].display();
if (grid[i][j].x<mouseX && mouseX < grid[i][j].x + grid[i][j].w && grid[i][j].y<mouseY && mouseY < grid[i][j].y + grid[i][j].h && mousePressed) {
// if the square is red
if (grid[i][j].c == RED) {
// change colour to GREEN
grid[i][j].update(GREEN);
}
}
}
}
}
class Square {
float x, y;
float w, h;
color c;
Square(float tempX, float tempY, float tempW, float tempH) {
x = tempX;
y = tempY;
w = tempW;
h = tempH;
c = color(0);
}
void display() {
stroke(0);
fill(c);
rect(x, y, w, h);
}
void update(color c) {
this.c = c;
}
}
In terms of shuffling colours every 5 seconds I recommend:
for every 5 seconds you could use millis()
above there is an example of shuffling done in setup() though you might want to encapsulate a nested loop like that with the random condition in a void shuffle() function for example which you could easily call every 5 seconds.
note that this approach will reset green blocks to red, you might want an else in that condition to reset blocks to black (otherwise, with time, most will turn red), etc.
Have fun!
P.S. I tend to separate state data from representation. For example I would add a variable to keep track of each square state (e.g. OFF, INTERACTIVE, ACTIVATED), update a basic finite state machine, then render colours accordingly. What you have above is a tight coupling between the colour of a Square and it's state. For the homework you've got that's ok, but in the future, for more complex projects you might want to consider data flows through your program and how you represent it.

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);
}
}
}

curveVertex() not drawling in Processing?

I'm trying to create a script that drawls a curve through 'n' vertexes equally spaced around the center of an ellipse.
The reason I'm not just drawling an ellipse around the center ellipse is because I eventually want to connect a micro-controller to Processing where the data points acquired from the 'n' amount of sensors will vary the height ('y') of each vertex, creating constantly changing, irregular curves around the center ellipse such as this possible curve:
Essentially, this is supposed to be a data visualizer, but I cannot figure out why this is not working or how to achieve this effect after going through examples and the documentation on https://processing.org/reference/.
Here is my code:
color WHITE = color(255);
color BLACK = color(0);
void setup() {
size(500, 500);
}
void draw() {
background(WHITE);
translate(width/2, height/2); // move origin to center of window
// center ellipse
noStroke();
fill(color(255, 0, 0));
ellipse(0, 0, 10, 10); // center point, red
fill(BLACK);
int n = 10;
int y = 100;
float angle = TWO_PI / n;
beginShape();
for (int i = 0; i < n; i++) {
rotate(angle);
curveVertex(0, y);
}
endShape();
}

The matrix operations like rotate do not transform the single vertices in a shape. The current matrix is applied to the entire shape when it is draw (at endShape). You've to calculate all the vertex coordinates:
Create a ArrayList of PVector, fill it with points and draw it in a loop:
color WHITE = color(255);
color BLACK = color(0);
ArrayList<PVector> points = new ArrayList<PVector>();
void setup() {
size(500, 500);
int n = 10;
int radius = 100;
for (int i = 0; i <= n; i++) {
float angle = TWO_PI * (float)i/n;
points.add(new PVector(cos(angle)*radius, sin(angle)*radius));
}
}
void draw() {
background(WHITE);
translate(width/2, height/2);
noFill();
stroke(255, 0, 0);
beginShape();
PVector last = points.get(points.size()-1);
curveVertex(last.x, last.y);
for (int i = 0; i < points.size(); i++) {
PVector p = points.get(i);
curveVertex(p.x, p.y);
}
PVector first = points.get(0);
curveVertex(first.x, first.y);
endShape();
}

Wrong center of screen coordinates

I'm coding an Android game and I'm trying to put the character in the middle of the screen (X axis).
In order to get the middle of the screen I get the screen width withe the following command:
int phoneWidth = Resources.getSystem().getDisplayMetrics().widthPixels;
Then I divide phoneWidth by two and hoped that the character would appear in the middle of the screen but it appears slightly to the right.
This is the code that I use:
public void drawCharacter() {
charImg = new Texture("pika.PNG");
charSprite = new Sprite(charImg);
float ahaha = Gdx.graphics.getWidth();
charPosX = (float)phoneWidth/ 2;
}
#Override
public void render() {
ch.stop();
double elapsedSeconds = ch.getSeconds();
checkLevel();
handleObstacles();
scrollTimer = scrollTimer + Gdx.graphics.getDeltaTime() / 2;
if (scrollTimer > 1.0f)
scrollTimer = 0.0f;
sprite.setV(scrollTimer);
sprite.setV2(scrollTimer + 2);
Gdx.gl.glClearColor(0, 0, 0, 1);
Gdx.gl.glClear(GL20.GL_COLOR_BUFFER_BIT);
charSprite.setPosition(charPosX, charPosY);
for (int i = 0; i < obstacles.size(); i++) {
obstacleSprites.get(i).setPosition(obstacles.get(i).getPosX(), obstacles.get(i).getPosY());
}
spriteBatch.begin();
sprite.draw(spriteBatch);
charSprite.draw(spriteBatch);
for (int i = 0; i < obstacleSprites.size(); i++) {
obstacleSprites.get(i).draw(spriteBatch);
}
spriteBatch.end();
moveCharacter();
moveObstacle();
}
Does anyone know where this error come from?

The left-hand edge of your sprite will be in the middle of the screen.
Subtract half the sprite width from half the screen width to draw the sprite centered horizontally.