Matrix Text rain effect in Processing 3.3 - processing

I'm working on making a matrix text rain effect in Processing 3.3 as a simple starter project for learning the processing library and Java. My code so far:
class Symbol {
int x, y;
int switchInterval = round(random(2, 50));
float speed;
char value;
Symbol(int x, int y, float speed) {
this.x = x;
this.y = y;
this.speed = speed;
}
//Sets to random symbol based on the Katakana Unicode block
void setToRandomSymbol() {
if(frameCount % switchInterval == 0) {
value = char((int) random(0x30A0, 0x3100));
}
}
//rains the characters down the screen and loops them to the top when they
// reach the bottom of the screen
void rain() {
if(y <= height) {
y += speed;
}else {
y = 0;
}
}
}
Symbol symbol;
class Stream {
int totalSymbols = round(random(5, 30));
Symbol[] symbols = new Symbol[500];
float speed = random(5, 20);
//generates the symbols and adds them to the array, each symbol one symbol
//height above the one previous
void generateSymbols() {
int y = 0;
int x = width / 2;
for (int i = 0; i <= totalSymbols; i++) {
symbols[i] = new Symbol(x, y, speed);
symbols[i].setToRandomSymbol();
y -= symbolSize;
}
}
void render() {
for(Symbol s : symbols) {
fill(0, 255, 70);
s.setToRandomSymbol();
text(s.value, s.x, s.y);
s.rain();
}
}
}
Ok, so that was a lot of code, Let me explain my dilemma. The issue I'm having is that when I run the code I get a NullpointerException at the s.setToRandomSymbol(); method call in the for each loop in the render function. The weird part about this NullPointerException error and the part I'm not understanding is that it's being thrown on a method that doesn't take in any arguments that could be coming back empty, and the method itself is void, so it shouldn't be returning anything, right? Why is this returning Null and what did I do wrong to have it return this way?

First you come up with a random number betwen 5 and 30:
int totalSymbols = round(random(5, 30));
Then you create an array that holds 500 instances of your Symbol class:
Symbol[] symbols = new Symbol[500];
Note that this array holds 500 null values at this point.
Then you add a maximum of 30 instances of Symbol to your array:
for (int i = 0; i <= totalSymbols; i++) {
symbols[i] = new Symbol(x, y, speed);
Note that this array now holds at least 470 null values at this point.
Then you iterate over all 500 indexes:
for(Symbol s : symbols) {
s.setToRandomSymbol();
But remember that at least 470 of these indexes are null, which is why you're getting a NullPointerException.
Some basic debugging would have told you all of this. I would have started by adding a basic println() statement just before you get the error:
for(Symbol s : symbols) {
println("s: " + s);
s.setToRandomSymbol();
This would have showed you that you're iterating over null values.
Anyway, to fix your problem you need to stop iterating over your entire array, or you need to stop making room for indexes you never use.
In the future, please try to narrow your problem down to a MCVE before posting. Note that this much smaller example program shows your error:
String[] array = new String[10];
array[0] = "test";
for(String s : array){
println(s.length());
}

Related

Paper cut algorithm

I want to create a function to determine the most number of pieces of paper on a parent paper size
The formula above is still not optimal. If using the above formula will only produce at most 32 cut/sheet.
I want it like below.
This seems to be a very difficult problem to solve optimally. See http://lagrange.ime.usp.br/~lobato/packing/ for a discussion of a 2008 paper claiming that the problem is believed (but not proven) to be NP-hard. The researchers found some approximation algorithms and implemented them on that website.
The following solution uses Top-Down Dynamic Programming to find optimal solutions to this problem. I am providing this solution in C#, which shouldn't be too hard to convert into the language of your choice (or whatever style of pseudocode you prefer). I have tested this solution on your specific example and it completes in less than a second (I'm not sure how much less than a second).
It should be noted that this solution assumes that only guillotine cuts are allowed. This is a common restriction for real-world 2D Stock-Cutting applications and it greatly simplifies the solution complexity. However, CS, Math and other programming problems often allow all types of cutting, so in that case this solution would not necessarily find the optimal solution (but it would still provide a better heuristic answer than your current formula).
First, we need a value-structure to represent the size of the starting stock, the desired rectangle(s) and of the pieces cut from the stock (this needs to be a value-type because it will be used as the key to our memoization cache and other collections, and we need to to compare the actual values rather than an object reference address):
public struct Vector2D
{
public int X;
public int Y;
public Vector2D(int x, int y)
{
X = x;
Y = y;
}
}
Here is the main method to be called. Note that all values need to be in integers, for the specific case above this just means multiplying everything by 100. These methods here require integers, but are otherwise are scale-invariant so multiplying by 100 or 1000 or whatever won't affect performance (just make sure that the values don't overflow an int).
public int SolveMaxCount1R(Vector2D Parent, Vector2D Item)
{
// make a list to hold both the item size and its rotation
List<Vector2D> itemSizes = new List<Vector2D>();
itemSizes.Add(Item);
if (Item.X != Item.Y)
{
itemSizes.Add(new Vector2D(Item.Y, Item.X));
}
int solution = SolveGeneralMaxCount(Parent, itemSizes.ToArray());
return solution;
}
Here is an example of how you would call this method with your parameter values. In this case I have assumed that all of the solution methods are part of a class called SolverClass:
SolverClass solver = new SolverClass();
int count = solver.SolveMaxCount1R(new Vector2D(2500, 3800), new Vector2D(425, 550));
//(all units are in tenths of a millimeter to make everything integers)
The main method calls a general solver method for this type of problem (that is not restricted to just one size rectangle and its rotation):
public int SolveGeneralMaxCount(Vector2D Parent, Vector2D[] ItemSizes)
{
// determine the maximum x and y scaling factors using GCDs (Greastest
// Common Divisor)
List<int> xValues = new List<int>();
List<int> yValues = new List<int>();
foreach (Vector2D size in ItemSizes)
{
xValues.Add(size.X);
yValues.Add(size.Y);
}
xValues.Add(Parent.X);
yValues.Add(Parent.Y);
int xScale = NaturalNumbers.GCD(xValues);
int yScale = NaturalNumbers.GCD(yValues);
// rescale our parameters
Vector2D parent = new Vector2D(Parent.X / xScale, Parent.Y / yScale);
var baseShapes = new Dictionary<Vector2D, Vector2D>();
foreach (var size in ItemSizes)
{
var reducedSize = new Vector2D(size.X / xScale, size.Y / yScale);
baseShapes.Add(reducedSize, reducedSize);
}
//determine the minimum values that an allowed item shape can fit into
_xMin = int.MaxValue;
_yMin = int.MaxValue;
foreach (var size in baseShapes.Keys)
{
if (size.X < _xMin) _xMin = size.X;
if (size.Y < _yMin) _yMin = size.Y;
}
// create the memoization cache for shapes
Dictionary<Vector2D, SizeCount> shapesCache = new Dictionary<Vector2D, SizeCount>();
// find the solution pattern with the most finished items
int best = solveGMC(shapesCache, baseShapes, parent);
return best;
}
private int _xMin;
private int _yMin;
The general solution method calls a recursive worker method that does most of the actual work.
private int solveGMC(
Dictionary<Vector2D, SizeCount> shapeCache,
Dictionary<Vector2D, Vector2D> baseShapes,
Vector2D sheet )
{
// have we already solved this size?
if (shapeCache.ContainsKey(sheet)) return shapeCache[sheet].ItemCount;
SizeCount item = new SizeCount(sheet, 0);
if ((sheet.X < _xMin) || (sheet.Y < _yMin))
{
// if it's too small in either dimension then this is a scrap piece
item.ItemCount = 0;
}
else // try every way of cutting this sheet (guillotine cuts only)
{
int child0;
int child1;
// try every size of horizontal guillotine cut
for (int c = sheet.X / 2; c > 0; c--)
{
child0 = solveGMC(shapeCache, baseShapes, new Vector2D(c, sheet.Y));
child1 = solveGMC(shapeCache, baseShapes, new Vector2D(sheet.X - c, sheet.Y));
if (child0 + child1 > item.ItemCount)
{
item.ItemCount = child0 + child1;
}
}
// try every size of vertical guillotine cut
for (int c = sheet.Y / 2; c > 0; c--)
{
child0 = solveGMC(shapeCache, baseShapes, new Vector2D(sheet.X, c));
child1 = solveGMC(shapeCache, baseShapes, new Vector2D(sheet.X, sheet.Y - c));
if (child0 + child1 > item.ItemCount)
{
item.ItemCount = child0 + child1;
}
}
// if no children returned finished items, then the sheet is
// either scrap or a finished item itself
if (item.ItemCount == 0)
{
if (baseShapes.ContainsKey(item.Size))
{
item.ItemCount = 1;
}
else
{
item.ItemCount = 0;
}
}
}
// add the item to the cache before we return it
shapeCache.Add(item.Size, item);
return item.ItemCount;
}
Finally, the general solution method uses a GCD function to rescale the dimensions to achieve scale-invariance. This is implemented in a static class called NaturalNumbers. I have included the rlevant parts of this class below:
static class NaturalNumbers
{
/// <summary>
/// Returns the Greatest Common Divisor of two natural numbers.
/// Returns Zero if either number is Zero,
/// Returns One if either number is One and both numbers are >Zero
/// </summary>
public static int GCD(int a, int b)
{
if ((a == 0) || (b == 0)) return 0;
if (a >= b)
return gcd_(a, b);
else
return gcd_(b, a);
}
/// <summary>
/// Returns the Greatest Common Divisor of a list of natural numbers.
/// (Note: will run fastest if the list is in ascending order)
/// </summary>
public static int GCD(IEnumerable<int> numbers)
{
// parameter checks
if (numbers == null || numbers.Count() == 0) return 0;
int first = numbers.First();
if (first <= 1) return 0;
int g = (int)first;
if (g <= 1) return g;
int i = 0;
foreach (int n in numbers)
{
if (i == 0)
g = n;
else
g = GCD(n, g);
if (g <= 1) return g;
i++;
}
return g;
}
// Euclidian method with Euclidian Division,
// From: https://en.wikipedia.org/wiki/Euclidean_algorithm
private static int gcd_(int a, int b)
{
while (b != 0)
{
int t = b;
b = (a % b);
a = t;
}
return a;
}
}
Please let me know of any problems or questions you might have with this solution.
Oops, forgot that I was also using this class:
public class SizeCount
{
public Vector2D Size;
public int ItemCount;
public SizeCount(Vector2D itemSize, int itemCount)
{
Size = itemSize;
ItemCount = itemCount;
}
}
As I mentioned in the comments, it would actually be pretty easy to factor this class out of the code, but it's still in there right now.

I cannot get operator+ function to compile, have I maybe made a mistake with one of the constructors?

This operator will receive a string with the format "x,y."
x represents an x coordinate and y represents a y coordinate. If the
coordinates are valid (that is within the valid range of the chess
board) then the appropriate variables should be set with these
variables. If the coordinates are invalid,
then nothing should happen.
operator<<: It also prints out (with a new line at the end), a message in the
following format:
b rook at [0,0]
The side is printed first, then the piece type followed by the coordinates at the end.
``This message must be sent to the output variable.
Piece::Piece(){
}
Piece::Piece(Piece *newPiece){
*newPiece = Piece;
}
Piece::Piece(string pType, char side, int x, int y){
//string pT = pType;
//char s = side;
}
Piece::~Piece(){
}
char Piece::getSide(){
return side;
}
string Piece::getPieceType(){
return PieceType;
}
int Piece::getX(){
return xPos;
}
int Piece::getY(){
return yPos;
}
void Piece::setX(int x){
xPos = x;
}
void Piece::setY(int y){
yPos = y;
}
void Piece::operator[](int pos){
cin>>pos;
if(pos != 0 || pos != 1){
cout<<"Invalid Index"<<endl;
}
else{
if(pos == 0 ){
cout<<"x coord"<<xPos;
}
if(pos == 1){
cout<<"y cord"<<yPos;
}
}
}
//Everything above compiled
Piece& Piece::operator+(string move){
if(xPos<=side && yPos<=side){
move = xPos;",";yPos;
return move;
}
return 0;
}
ostream& Piece::operator<<(ostream& output,const Piece& t ){
t = Piece;
output = side + PieceType + pos;
cout<<output;
}
There is no reason to call new inside the constructor.
Since you're calling new Piece in the constructor of Piece, you will end up in an infinite loop. The expression new Piece again invokes the constructor of Piece, which again calls new Piece, which invokes the constructor, ...
Also, think a second about what's happening here: *newPiece = new Piece; is constructing a new Piece object and assigning the result to the local pointer value newPiece. After the constructor ends (which in your case, it never does, because of the infinite loop), that local variable goes out of scope and your new Piece is lost.

How to set up if statements so that loop goes forward and then in reverse? Processing

int x = 31;
int y = 31;
int x_dir = 4;
int y_dir = 0;
void setup ()
{
size (800, 800);
}
void draw ()
{
background (150);
ellipse (x,y,60, 60);
if (x+30>=width)
{
x_dir =-4;
y_dir = 4;
}
if (y+30>=height)
{
x_dir=4;
y_dir = 0;
}
if (x+30>=width)
{
x_dir = -4;
}
x+=x_dir;
y+=y_dir;
println(x,y);
}
Hi,
I have to create this program in processing which produces an animation of a ball going in a Z pattern (top left to top right, diagonal top right to bottom left, and then straight from bottom left to bottom right) which then goes backwards along the same path it came.
While I have the code written out for the forward direction, I don't know what 2 if or else statements I need to write for the program so that based on one condition it goes forwards, and based on another condition it will go backwards, and it will continue doing so until it terminates.
If I am able to figure out which two if statements I need to write, all I need to do is copy and reverse the x_dir and y_dir signs on the forward loop.
There are a ton of different ways you can do this.
One approach is to keep track of which "mode" you're in. You could do this using an int variable that's 0 when you're on the first part of the path, 1 when you're on the second part of the path, etc. Then just use an if statement to decide what to do, how to move the ball, etc.
Here's an example:
int x = 31;
int y = 31;
int mode = 0;
void setup ()
{
size (800, 800);
}
void draw ()
{
background (150);
ellipse (x, y, 60, 60);
if (mode == 0) {
x = x + 4;
if (x+30>=width) {
mode = 1;
}
} else if (mode == 1) {
x = x - 4;
y = y + 4;
if (y+30>=height) {
mode = 2;
}
} else if (mode == 2) {
x = x + 4;
if (x+30>=width) {
mode = 3;
}
} else if (mode == 3) {
x = x - 4;
y = y - 4;
if (y-30 < 0) {
mode = 2;
}
}
}
Like I said, this is only one way to approach the problem, and there are some obvious improvements you could make. For example, you could store the movement speeds and the conditions that change the mode in an array (or better yet, in objects) and get rid of all of the if statements.

Simple image processing algorithm causes Processing to freeze

I've written an algorithm in Processing to do the following:
1. Instantiate a 94 x 2 int array
2. Load a jpg image of dimensions 500 x 500 pixels
3. Iterate over every pixel in the image and determine whether it is black or white then change a variable related to the array
4. Print the contents of the array
For some reason this algorithm freezes immediately. I've put print statements in that show me that it freezes before even attempting to load the image. This is especially confusing to me in light of the fact that I have written another very similar algorithm that executes without complications. The other algorithm reads an image, averages the color of each tile of whatever size is specified, and then prints rectangles over the region that was averaged with the average color, effectively pixelating the image. Both algorithms load an image and examine each of its pixels. The one in question is mostly different in that it doesn't draw anything. I was going to say that it was different for having an array but the pixelation algorithm holds all of the colors in a color array which should take up far more space than the int array.
From looking in my mac's console.app I see that there was originally this error: "java.lang.OutOfMemoryError: GC overhead limit exceeded". From other suggestions/sources on the web I tried bumping the memory allocation from 256mb to 4000mb (doing this felt meaningless because my analysis of the algorithms showed they should be the same complexity but I tried anyways). This did not stop freezing but changed the error to a combination of "JavaNativeFoundation error occurred obtaining Java exception description" and "java.lang.OutOfMemoryError: Java heap space".
Then I tried pointing processing to my local jdk with the hope of utilizing the 64 bit jdk over processing's built in 32 bit jdk. From within Processing.app/Contents I executed the following commands:
mv Java java-old
ln -s /Library/Java/JavaVirtualMachines/jdk1.7.0_79.jdk Java
Processing would not start after this attempt with the following error populating my console:
"com.apple.xpc.launchd[1]: (org.processing.app.160672[13559]) Service exited with abnormal code: 1"
Below is my code:
First the noncompliant algorithm
int squareSize=50;
int numRows = 10;
int numCols = 10;
PFont myFont;
PImage img;
//33-126
void setup(){
size(500,500);
count();
}
void count(){
ellipseMode(RADIUS);
int[][] asciiArea = new int[94][2];
println("hello?");
img=loadImage("countingPicture.jpg");
println("image loaded");
for(int i=0; i<(500/squareSize); i++){
for(int j=0; j<(500/squareSize); j++){
int currentValue=i+j*numCols;
if(currentValue+33>126){
break;
}
println(i+", "+j);
asciiArea[currentValue][0]=currentValue+33;
asciiArea[currentValue][1]=determineTextArea(i,j,squareSize);
//fill(color(255,0,0));
//ellipse(i*squareSize,j*squareSize,3,3);
}
}
println("done calculating");
displayArrayContents(asciiArea);
}
int determineTextArea(int i, int j, int squareSize){
int textArea = 0;
double n=0.0;
while(n < squareSize*squareSize){
n+=1.0;
int xOffset = (int)(n%((double)squareSize));
int yOffset = (int)(n/((double)squareSize));
color c = img.get(i*squareSize+xOffset, j*squareSize+yOffset);
if(red(c)!=255 || green(c)!=255 || blue(c)!=255){
println(red(c)+" "+green(c)+" "+blue(c));
textArea++;
}
}
return textArea;
}
void displayArrayContents(int[][] arr){
int i=0;
println("\n now arrays");
while(i<94){
println(arr[i][0]+" "+arr[i][1]);
}
}
The pixelation algorithm that works:
PImage img;
int direction = 1;
float signal;
int squareSize = 5;
int wideness = 500;
int highness = 420;
int xDimension = wideness/squareSize;
int yDimension= highness/squareSize;
void setup() {
size(1500, 420);
noFill();
stroke(255);
frameRate(30);
img = loadImage("imageIn.jpg");
color[][] colors = new color[xDimension][yDimension];
for(int drawingNo=0; drawingNo < 3; drawingNo++){
for(int i=0; i<xDimension; i++){
for(int j=0; j<yDimension; j++){
double average = 0;
double n=0.0;
while(n < squareSize*squareSize){
n+=1.0;
int xOffset = (int)(n%((double)squareSize));
int yOffset = (int)(n/((double)squareSize));
color c = img.get(i*squareSize+xOffset, j*squareSize+yOffset);
float cube = red(c)*red(c) + green(c)*green(c) + blue(c)*blue(c);
double grayValue = (int)(sqrt(cube)*(255.0/441.0));
double nAsDouble = (double)n;
average=(grayValue + (n-1.0)*average)/n;
average=(grayValue/n)+((n-1.0)/(n))*average;
}
//average=discretize(average);
println(i+" "+j+" "+average);
colors[i][j]=color((int)average);
fill(colors[i][j]);
if(drawingNo==0){ //stroke(colors[i][j]); }
stroke(210);}
if(drawingNo==1){ stroke(150); }
if(drawingNo==2){ stroke(90); }
//stroke(colors[i][j]);
rect(drawingNo*wideness+i*squareSize,j*squareSize,squareSize,squareSize);
}
}
}
save("imageOut.jpg");
}
You're entering an infinite loop, which makes the println() statements unreliable. Fix the infinite loop, and your print statements will work again.
Look at this while loop:
while(i<94){
println(arr[i][0]+" "+arr[i][1]);
}
When will i ever become >= 94?
You never increment i, so its value is always 0. You can prove this by adding a println() statement inside the while loop:
while(i<94){
println("i: " + i);
println(arr[i][0]+" "+arr[i][1]);
}
You probably wanted to increment i inside the while loop. Or just use a for loop instead.

Cannot convert from void to PVector in Processing

I'm learning processing and am trying generate fractals using Pascal's triangle. This requires arrays of PVectors. I have run into an issue that I don't understand—I'm getting cannot convert from void to PVector
This is an excerpt, not every variable is defined in the excerpt, but I figure the issue is probably something such that I won't need to reveal more code than this—more code shown might just obfuscate the problem.
arrayCopy(points,old_points);
points = new PVector[int(pow(2, j))];
if (j == 1)
{
points[0] = new PVector(0,0);
}
if (j == 2)
{
points[0] = new PVector(1,0);
points[1] = new PVector(-1,0);
}
else
{
//j will be 3 for the THIRD term in the series
int number_of_terms_to_fill = int(pow(j - 1, 2));
int[] pasc = PascalTriangle(j - 1);
float real = findReal(pasc, x, y, number_of_terms_to_fill);
float imagi = findImagi(pasc, x, y, number_of_terms_to_fill);
PVector v = new PVector(real, imagi);
for (int k = 0; k < number_of_terms_to_fill; k = k + 2)
{
points[k] = old_points[k].add(v); //!!***PROBLEM LINE***!!!
points[k+1] = old_points[k].sub(v);
}
}
My other functions, such as findReal and findImagi, I believe are correct. The addition on the problem line should be an addition between two PVectors—a legal operation. Instead something is void? Perhaps arrayCopy isn't a deep copy like I'd want?
Not sure what is going on.
The method .add() does not return a PVector (it returns nothing, thus void) so you can't do
points[k] = old_points[k].add(v);
I suppose what you are trying to do is add v to old_points[k] and pass it to points[k] which will not work like that... You have to do it like this:
old_points[k].add(v);
points[k] = old_points[k];
old_points[k].sub(v);
points[k+1] = old_points[k];
In order for what you wrote to be valid, the add function should have been like this:
PVector add(PVector v) {
this.x += v.x;
this.y += v.y;
this.z += v.z;
return this;
}
Instead it looks sort of like this:
void add(PVector v) {
this.x += v.x;
this.y += v.y;
this.z += v.z;
}
Edit: After Ryan's comment below, I provide here another way, using the static .add() method which does return a PVector...
points[k] = PVector.add(old_points[k],v);
points[k+1] = PVector.sub(old_points[k],v);

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