Develop Reference

Loop construction in Processing artistic rendition

The code is from here:
t=0
draw=_=>{t||createCanvas(W = 720,W)
t+=.01
B=blendMode
colorMode(HSB)
B(BLEND)
background(0,.1)
B(ADD)
for(y = 0; y < W; y += 7)
for(x = 0; x < W; x+=7)
dist(x, y, H = 360, H) +
!fill(x * y % 360, W, W,
T=tan(noise(x, y) * 9 + t))
< sin(atan2(y - H, x - H) * 2.5 + 84)**8 * 200 + 130?
circle(x, y + 30, 4/T) : 0}
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/1.5.0/p5.js"></script>
I see that t is increased by 0.01 each iteration, but I am unsure whether for each t value the entire canvas is refreshed, or whether there is an endpoint somewhat set by :0}. Is this correct?
It also seems like there is a Boolean call in the middle basically comparing two distances to determine which circles are filled and how. If the < was instead > the circles would form a complementary pattern on the rendition.
The ! is explained here, as a way of saving space and calling a function as soon as it is declared. I presume it determines how points are filled with a certain variable color scheme depending on the Boolean operation result. The +!fill() is unfamiliar, as well as the ? at the end, and I guess that they amount to: if the x, y location is within the boundary of the star the circles are colored (filled) as such, but the negation in '!' is confusing.
Can I get an English explanation of the main structural points on this code (the loop and the Boolean) to match the syntax?
I have so far gathered that the basic loop is
for(y from 0 to the width of the canvas at increments of 7)
for(x from... )
check if the distance from (x , y) to 360 is less than sin()^8 * 200 + 130
If so fill (or not fill with the ! ????) with these colors
otherwise do nothing :0

This is what it might look like if it were written normally
let t = 0;
const W = 720;
// https://p5js.org/reference/#/p5/draw
// `draw` needs to be in the global scope so p5 can use it
draw = () => {
// create a canvas if this is the first frame
if (!t) createCanvas(W, W);
t += 0.01;
// Use HSB and blending to do the fancy effects
// The black circles are because we're ignoring stroke and so using its defaults
// The blending will eventually hide it anyway
colorMode(HSB);
blendMode(BLEND);
background(0, 0.1);
blendMode(ADD);
// iterate over 7px grid
for(y = 0; y < W; y += 7) {
for(x = 0; x < W; x += 7) {
// center
const H = 360;
// distance from center
const D = dist(x, y, H, H);
// pick an alpha based on location and time
const T = tan(noise(x, y) * 9 + t);
// set fill color
fill(x * y % 360, W, W, T);
// magic to calculate the star's boundary
// sine wave in polar coords, I think
const S = sin(atan2(y - H, x - H) * 2.5 + 84)**8 * 200 + 130;
// draw a circle if we're within the star's area
// circle's color, alpha, and radius depend on location and time
if (D < S) circle(x, y + 30, 4/T);
}
}
};
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/1.5.0/p5.js"></script>
Note that there are some hacks in the original code that are solely to help make it a one-liner or to reduce the number of characters. They don't have any meaningful effect on the results.

The main issue is the +! with ? and :0, which is terribly confusing, but clearly it is equivalent to
t=0
draw=_=>{t||createCanvas(W = 720,W)
t+=.01
B=blendMode
colorMode(HSB)
B(BLEND)
background(0,.1)
B(ADD)
for(y = 0; y < W; y += 7)
for(x = 0; x < W; x+=7)
if(dist(x, y, H = 360, H) < sin(atan2(y - H, x - H) * 2.5 + 84)**8 * 200 + 130){
fill(x * y % 360, W, W,
T=tan(noise(x, y) * 9 + t))
circle(x, y + 30, 4/T)}else{0}}
The Boolean in the ? applies to the dist() part (in absolute values the angle has to be less than sin()... + 130.
The + part I still don't understand, and hopefully will be addressed by someone who knows Processing or JS (not me). However, it probably forces the execution to identify and throw out with regards to filling (hence !) values that are too low for the sin(atan2()), which will happen at around 0 and pi.
Because of arctan2() being (here)
the number of spikes in the star will be a multiple of 5 going from 0 to 2 pi. The fact that changing the code to < sin(2 * atan2(...)... doubles the spikes implies that the fill() part is also in Boolean operation of its own.
The result of the Boolean determines whether the colorful fill is applied or not (applied if less than).
The :0 at the end is the else do nothing.

Algorithm for searching a grid radially from a point

I have an object that sits at point 0,0. This object cannot share space with any other object of its type that may appear on top of it, next to it, above it, etc.. There may be more than a few of these objects present overlapping each other and i have no knowledge of where the other ones are placed until i try the collision detection method.
My thinking is that i'll use a collision detection along side a grid search. Along the lines of the photo below.
The object will first try its default best case location. If that doesn't work then it tries to the left, left-above, left-below, etc, until it has searched all the #1 positions. Then it moves onto the #2 positions and so on until it finds a place to drop the element where it won't be overlapping another.
this is the code i'm playing around with right now but it is choosing some very, very random locations for things. I'm pretty sure it isn't following the algorithm i described above.
for (let i = 0; i < 5 && this._hasCollisions(this._tagWrapper); i++) {
/**
* This algorithm explores positions inside nested boxes.
* The move algorithm behaves the following way. It goes,
* down, up, left, down, up, right * 2, repeat.
*
* For example this is how it works given the height of 5 and a width of 7
* numbers are expressed in the offset generated
* 1: 5,0 4: 5,-7 7: 5,7 10: 10,-14
* 2: -5,0 5: -5,-7 8: -5,7 11: -10,-14
* 3: 0,-7 6: 0,7 9: 0,-14
*/
// Calculate which box the collision detector is working in
// This happens every 9 iterations
let multiplier = (i / 9) + 1;
/**
* Get the x offset
*/
if (i % 3 === 0) {
// Clear the height offset on multiples of 3
xOffset = 0;
} else {
// Set the height to the multiplier
xOffset = this._tagWrapper.offsetWidth * multiplier;
}
if (i % 3 === 2) {
// Get the sequence 2, 5, 8, 11, 14, etc..
xOffset *= -1;
}
/**
* Get the y offset
*/
if (i > 2) {
// Set the width to a multiple of the multiplier and assign the existing negativeness
yOffset = this._tagWrapper.offsetHeight * multiplier * (yOffset > 0 ? 1 : -1);
}
if (i % 3 === 0) {
// Flip the sign every 3 numbers
yOffset *= -1;
}
console.log('iteration', i);
this._tagWrapper.style.top = (basePosition.y + yOffset) + 'px';
this._tagWrapper.style.left = (basePosition.x + xOffset) + 'px';
}
What is the best way to go about performing this search? I already hav

Something like this work? (most of the code is just for the visualization)
// just draw a table to visualize
var SIZE = 15;
for (var i = 0; i < SIZE; i++) {
$("#a").append("<tr>");
for (var j = 0; j < SIZE; j++) {
$("#a > tr").last().append("<td>.</td>");
}
}
// where to start searching from
var startX = 8;
var startY = 8;
function loop() {
// tell the world which grid we are on
$("#a > tr:nth-child(" + y + ") > td:nth-child(" + x + ")").css("backgroundColor", "red");
// check if done here!!! - x and y are our positions in the grid
// also do bounds checking here
if (isX) {
x = x + xDirection;
i--;
if (!i) {
// switch dimension
isX = !isX;
i = moveFor;
// switch direction
xDirection *= -1;
}
} else {
y = y + yDirection;
i--;
if (!i) {
// switch dimension
isX = !isX;
// increase the width / height we are spanning
moveFor += 1;
i = moveFor;
// switch direction
yDirection *= -1;
}
}
// jsut so that we have a nice animation
if (x > 0 && y > 0 && x <= SIZE && y <= SIZE) {
setTimeout(loop, 10)
}
}
var x = startX;
var y = startY;
var moveFor = 1;
// our step (down) counter
var i = moveFor;
var xDirection = -1;
var yDirection = -1;
// are we moving along x or y
var isX = true;
loop();
body {
font-family: monospace;
}
td {
height: 20px;
width: 20px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table>
<tbody id="a"></tbody>
</table>

I would suggest using distance solution
point1 has x1 and y1
point2 has x2 and y2
var d = Math.sqrt( (x2-=x1)*x2 + (y2-=y1)*y2 );
link here: Get distance between two points in canvas

Here is an implementation of scanning points in a ring around a center point. You define the center point and the distance you want to sample and it returns the list of points in clock-wise order. It is in Python not JavaScript but it is simple enough that you can translate if needed.

Paper.js: fastest way to draw many iterated shapes over loop

jsfiddle here: http://jsfiddle.net/yw0w18m3/2/
I'm using paper.js to make a background image that looks somthing like this:
Basically, I'm creating a couple thousand triangles over a loop and rotating them on every other iteration.
function Tri(x, y, rotate) {
var tri = new Path([
new Point((x - 42), (y - 48)),
new Point((x - 42), y),
new Point(x, (y - 24)),
new Point((x - 42), (y - 48))
]);
tri.fillColor = {
hue: Math.random() * 360,
saturation: 0,
brightness: ( (( Math.random() ) * .95) + .3 )
};
if(rotate) { tri.rotate(180); }
}
for (var i = 0; i < 2000; i++) {
rotate = false;
if( i % 2 ) {
rotate = true;
}
new Tri(x, y, rotate);
x = x + 42;
if( x > (winWidth + 42) ) {
x = 0 ;
y = y + 24;
}
}
There seems to be a brief 1-2 second pause/freeze though while the shapes are being drawn. Is there a more efficient way to draw all the shapes first (or push to an array) then add that to the canvas all at once?
I based my code off of the example here: http://paperjs.org/examples/candy-crash/ (click "source" in the upper right corner).
Any help is much appreciated.
Thanks!

I would end up creating two triangles, one rotated, so they don't have to be built from new points each time. Then choose the correct triangle based on the rotation variable and clone it, as opposed to create points and a triangle from scratch each time. Finally, just change the position of the cloned triangle.
Last, I would correct the maxTri so it doesn't do more than it needs to. The paren should follow the 48, not the 24. You're doing an order of magnitude more triangles than needed.
Here's a link to the sketch.paperjs.org solution I created based on your code. I find sketch easier to use than jsfiddle for paper examples.
proto1 = new Path([
new Point(0, -24),
new Point(0, 24),
new Point(42, 0)
]);
proto1.closed = true;
proto2 = proto1.clone();
proto2.rotate(180);
function putTriangle(pos, rotate) {
var tri = (rotate ? proto2 : proto1).clone();
tri.position = pos;
tri.position = tri.position.subtract([21, 0])
tri.fillColor = {
hue: Math.random() * 360,
saturation: 0,
brightness: Math.random() * 0.5 + 0.5
}
}
var tris = [],
x = 42,
y = 24,
rotate,
winWidth = paper.view.size.width,
winHeight = paper.view.size.height,
rows = (winHeight + 48) / 24,
cols = (winWidth + 42) / 42,
numTri = rows * cols,
numTriOrig = (winWidth + 42) / 42 * (winHeight + 48 / 24);
//console.log(numTri, numTriOrig);
x = 0;
y = 0;
for (var row = 0; row < rows; row++) {
rowrotate = row % 2;
for (var col = 0; col <= cols; col++) {
rotate = rowrotate ^ col % 2;
putTriangle([x,y], rotate);
x += 42;
}
x = 0;
y = y + 24;
}

Two thoughts:
I see you use rotate to transform you triangles into place. This is an expensive operation. You could replace the rotate with a less geometric & more arithmetic calculation of the triangles orientation.
Also, I see is that the fill color is being changed with each triangle and state changes (like fill) are modestly expensive. You could group all the similarly colored triangles and draw them in a single batch.

hough transform - javascript - node.js

So, i'm trying to implement hough transform, this version is 1-dimensional (its for all dims reduced to 1 dim optimization) version based on the minor properties.
Enclosed is my code, with a sample image... input and output.
Obvious question is what am i doing wrong. I've tripled check my logic and code and it looks good also my parameters. But obviously i'm missing on something.
Notice that the red pixels are supposed to be ellipses centers , while the blue pixels are edges to be removed (belong to the ellipse that conform to the mathematical equations).
also, i'm not interested in openCV / matlab / ocatve / etc.. usage (nothing against them).
Thank you very much!
var fs = require("fs"),
Canvas = require("canvas"),
Image = Canvas.Image;
var LEAST_REQUIRED_DISTANCE = 40, // LEAST required distance between 2 points , lets say smallest ellipse minor
LEAST_REQUIRED_ELLIPSES = 6, // number of found ellipse
arr_accum = [],
arr_edges = [],
edges_canvas,
xy,
x1y1,
x2y2,
x0,
y0,
a,
alpha,
d,
b,
max_votes,
cos_tau,
sin_tau_sqr,
f,
new_x0,
new_y0,
any_minor_dist,
max_minor,
i,
found_minor_in_accum,
arr_edges_len,
hough_file = 'sample_me2.jpg',
edges_canvas = drawImgToCanvasSync(hough_file); // make sure everything is black and white!
arr_edges = getEdgesArr(edges_canvas);
arr_edges_len = arr_edges.length;
var hough_canvas_img_data = edges_canvas.getContext('2d').getImageData(0, 0, edges_canvas.width,edges_canvas.height);
for(x1y1 = 0; x1y1 < arr_edges_len ; x1y1++){
if (arr_edges[x1y1].x === -1) { continue; }
for(x2y2 = 0 ; x2y2 < arr_edges_len; x2y2++){
if ((arr_edges[x2y2].x === -1) ||
(arr_edges[x2y2].x === arr_edges[x1y1].x && arr_edges[x2y2].y === arr_edges[x1y1].y)) { continue; }
if (distance(arr_edges[x1y1],arr_edges[x2y2]) > LEAST_REQUIRED_DISTANCE){
x0 = (arr_edges[x1y1].x + arr_edges[x2y2].x) / 2;
y0 = (arr_edges[x1y1].y + arr_edges[x2y2].y) / 2;
a = Math.sqrt((arr_edges[x1y1].x - arr_edges[x2y2].x) * (arr_edges[x1y1].x - arr_edges[x2y2].x) + (arr_edges[x1y1].y - arr_edges[x2y2].y) * (arr_edges[x1y1].y - arr_edges[x2y2].y)) / 2;
alpha = Math.atan((arr_edges[x2y2].y - arr_edges[x1y1].y) / (arr_edges[x2y2].x - arr_edges[x1y1].x));
for(xy = 0 ; xy < arr_edges_len; xy++){
if ((arr_edges[xy].x === -1) ||
(arr_edges[xy].x === arr_edges[x2y2].x && arr_edges[xy].y === arr_edges[x2y2].y) ||
(arr_edges[xy].x === arr_edges[x1y1].x && arr_edges[xy].y === arr_edges[x1y1].y)) { continue; }
d = distance({x: x0, y: y0},arr_edges[xy]);
if (d > LEAST_REQUIRED_DISTANCE){
f = distance(arr_edges[xy],arr_edges[x2y2]); // focus
cos_tau = (a * a + d * d - f * f) / (2 * a * d);
sin_tau_sqr = (1 - cos_tau * cos_tau);//Math.sqrt(1 - cos_tau * cos_tau); // getting sin out of cos
b = (a * a * d * d * sin_tau_sqr ) / (a * a - d * d * cos_tau * cos_tau);
b = Math.sqrt(b);
b = parseInt(b.toFixed(0));
d = parseInt(d.toFixed(0));
if (b > 0){
found_minor_in_accum = arr_accum.hasOwnProperty(b);
if (!found_minor_in_accum){
arr_accum[b] = {f: f, cos_tau: cos_tau, sin_tau_sqr: sin_tau_sqr, b: b, d: d, xy: xy, xy_point: JSON.stringify(arr_edges[xy]), x0: x0, y0: y0, accum: 0};
}
else{
arr_accum[b].accum++;
}
}// b
}// if2 - LEAST_REQUIRED_DISTANCE
}// for xy
max_votes = getMaxMinor(arr_accum);
// ONE ellipse has been detected
if (max_votes != null &&
(max_votes.max_votes > LEAST_REQUIRED_ELLIPSES)){
// output ellipse details
new_x0 = parseInt(arr_accum[max_votes.index].x0.toFixed(0)),
new_y0 = parseInt(arr_accum[max_votes.index].y0.toFixed(0));
setPixel(hough_canvas_img_data,new_x0,new_y0,255,0,0,255); // Red centers
// remove the pixels on the detected ellipse from edge pixel array
for (i=0; i < arr_edges.length; i++){
any_minor_dist = distance({x:new_x0, y: new_y0}, arr_edges[i]);
any_minor_dist = parseInt(any_minor_dist.toFixed(0));
max_minor = b;//Math.max(b,arr_accum[max_votes.index].d); // between the max and the min
// coloring in blue the edges we don't need
if (any_minor_dist <= max_minor){
setPixel(hough_canvas_img_data,arr_edges[i].x,arr_edges[i].y,0,0,255,255);
arr_edges[i] = {x: -1, y: -1};
}// if
}// for
}// if - LEAST_REQUIRED_ELLIPSES
// clear accumulated array
arr_accum = [];
}// if1 - LEAST_REQUIRED_DISTANCE
}// for x2y2
}// for xy
edges_canvas.getContext('2d').putImageData(hough_canvas_img_data, 0, 0);
writeCanvasToFile(edges_canvas, __dirname + '/hough.jpg', function() {
});
function getMaxMinor(accum_in){
var max_votes = -1,
max_votes_idx,
i,
accum_len = accum_in.length;
for(i in accum_in){
if (accum_in[i].accum > max_votes){
max_votes = accum_in[i].accum;
max_votes_idx = i;
} // if
}
if (max_votes > 0){
return {max_votes: max_votes, index: max_votes_idx};
}
return null;
}
function distance(point_a,point_b){
return Math.sqrt((point_a.x - point_b.x) * (point_a.x - point_b.x) + (point_a.y - point_b.y) * (point_a.y - point_b.y));
}
function getEdgesArr(canvas_in){
var x,
y,
width = canvas_in.width,
height = canvas_in.height,
pixel,
edges = [],
ctx = canvas_in.getContext('2d'),
img_data = ctx.getImageData(0, 0, width, height);
for(x = 0; x < width; x++){
for(y = 0; y < height; y++){
pixel = getPixel(img_data, x,y);
if (pixel.r !== 0 &&
pixel.g !== 0 &&
pixel.b !== 0 ){
edges.push({x: x, y: y});
}
} // for
}// for
return edges
} // getEdgesArr
function drawImgToCanvasSync(file) {
var data = fs.readFileSync(file)
var canvas = dataToCanvas(data);
return canvas;
}
function dataToCanvas(imagedata) {
img = new Canvas.Image();
img.src = new Buffer(imagedata, 'binary');
var canvas = new Canvas(img.width, img.height);
var ctx = canvas.getContext('2d');
ctx.patternQuality = "best";
ctx.drawImage(img, 0, 0, img.width, img.height,
0, 0, img.width, img.height);
return canvas;
}
function writeCanvasToFile(canvas, file, callback) {
var out = fs.createWriteStream(file)
var stream = canvas.createPNGStream();
stream.on('data', function(chunk) {
out.write(chunk);
});
stream.on('end', function() {
callback();
});
}
function setPixel(imageData, x, y, r, g, b, a) {
index = (x + y * imageData.width) * 4;
imageData.data[index+0] = r;
imageData.data[index+1] = g;
imageData.data[index+2] = b;
imageData.data[index+3] = a;
}
function getPixel(imageData, x, y) {
index = (x + y * imageData.width) * 4;
return {
r: imageData.data[index+0],
g: imageData.data[index+1],
b: imageData.data[index+2],
a: imageData.data[index+3]
}
}

It seems you try to implement the algorithm of Yonghong Xie; Qiang Ji (2002). A new efficient ellipse detection method 2. p. 957.
Ellipse removal suffers from several bugs
In your code, you perform the removal of found ellipse (step 12 of the original paper's algorithm) by resetting coordinates to {-1, -1}.
You need to add:
`if (arr_edges[x1y1].x === -1) break;`
at the end of the x2y2 block. Otherwise, the loop will consider -1, -1 as a white point.
More importantly, your algorithm consists in erasing every point which distance to the center is smaller than b. b supposedly is the minor axis half-length (per the original algorithm). But in your code, variable b actually is the latest (and not most frequent) half-length, and you erase points with a distance lower than b (instead of greater, since it's the minor axis). In other words, you clear all points inside a circle with a distance lower than latest computed axis.
Your sample image can actually be processed with a clearing of all points inside a circle with a distance lower than selected major axis with:
max_minor = arr_accum[max_votes.index].d;
Indeed, you don't have overlapping ellipses and they are spread enough. Please consider a better algorithm for overlapping or closer ellipses.
The algorithm mixes major and minor axes
Step 6 of the paper reads:
For each third pixel (x, y), if the distance between (x, y) and (x0,
y0) is greater than the required least distance for a pair of pixels
to be considered then carry out the following steps from (7) to (9).
This clearly is an approximation. If you do so, you will end up considering points further than the minor axis half length, and eventually on the major axis (with axes swapped). You should make sure the distance between the considered point and the tested ellipse center is smaller than currently considered major axis half-length (condition should be d <= a). This will help with the ellipse erasing part of the algorithm.
Also, if you also compare with the least distance for a pair of pixels, as per the original paper, 40 is too large for the smaller ellipse in your picture. The comment in your code is wrong, it should be at maximum half the smallest ellipse minor axis half-length.
LEAST_REQUIRED_ELLIPSES is too small
This parameter is also misnamed. It is the minimum number of votes an ellipse should get to be considered valid. Each vote corresponds to a pixel. So a value of 6 means that only 6+2 pixels make an ellipse. Since pixels coordinates are integers and you have more than 1 ellipse in your picture, the algorithm might detect ellipses that are not, and eventually clear edges (especially when combined with the buggy ellipse erasing algorithm). Based on tests, a value of 100 will find four of the five ellipses of your picture, while 80 will find them all. Smaller values will not find the proper centers of the ellipses.
Sample image is not black & white
Despite the comment, sample image is not exactly black and white. You should convert it or apply some threshold (e.g. RGB values greater than 10 instead of simply different form 0).
Diff of minimum changes to make it work is available here:
https://gist.github.com/pguyot/26149fec29ffa47f0cfb/revisions
Finally, please note that parseInt(x.toFixed(0)) could be rewritten Math.floor(x), and you probably want to not truncate all floats like this, but rather round them, and proceed where needed: the algorithm to erase the ellipse from the picture would benefit from non truncated values for the center coordinates. This code definitely could be improved further, for example it currently computes the distance between points x1y1 and x2y2 twice.

Making a blur algorithm work with canvas image data?

I have the following simple blur algorithm:
for (y = 0; y < height; ++y) {
for (x = 0; x < width; ++x) {
total = 0
for (ky = -radius; ky <= radius; ++ky)
for (kx = -radius; kx <= radius; ++kx)
total += source(x + kx, y + ky)
dest(x, y) = total / (radius * 2 + 1) ^ 2
}
}
And I need to make this work with an array, generated by canvas with getImageData(). The problem is that the array from canvas is one dimentional, and the algorithm needs a two dimentional one, because "kx" and "ky" are distances from the current pixel, and in a two dimentional array you just change one of the indexes in order to move left or right on the grid, but in a one dimentional array (which is trying to represent a two-dim. one) you can't do that.
What do you think, how should I approach this problem? In my opinion, the entire problem is here:
total += source(x + kx, y + ky)
This line needs to get the correct pixels from the one-dimentional array and it should work fine.
P.S. I forgot to say that I'm trying to blur the image one channel at a time, like this:
for (var i=0; i<imgData.data.length; i+=4) {
red[index] = imgData.data[i];
green[index] = imgData.data[i+1];
blue[index] = imgData.data[i+2];
index++;
}
and I'm passing each array (red, green and blue) individually to the algorithm.

In order access a single dimensional array with two dimensional coordinates, you have the following formula:
var pixel = pixels[ ( ( y * width ) + x ) * 4 ];
The * 4 was added to account for the r,g,b,a values.
xResolution is the width if the image.
So source() would be:
function source( imageArray, x, y, imageWidth )
{
var pos = ( ( y * imageWidth ) + x ) * 4;
var val =
{
r : imageArray[ pos ],
g : imageArray[ pos + 1 ],
b : imageArray[ pos + 2 ],
a : imageArray[ pos + 3 ]
};
return val;
}