I found a way to increase the gamma, but no way to decrease it! This article states a formula for increasing the gamma. The formula works for increasing the gamma but not for decreasing, even if I apply the reduction on a new instance of the canvas. I tried redrawing the canvas and using a negative value for gamma calculation, but I don't get my original canvas back.
//For increasing, I tried
gamma = 0.5;
gammacorrection = 1/gamma;
r = Math.pow(255 * (r / 255), gammacorrection);
g = ...
b = ...
//For decreasing
gamma = -0.5;
gammacorrection = 1/gamma;
r = Math.pow(255 * (r / 255), gammacorrection);
g = ...
b = ...
First part works. Second doesn't.
For sake of completeness here's a working piece of code
async function adjustGamma(gamma) {
const gammaCorrection = 1 / gamma;
const canvas = document.getElementById('canvasOutput');
const ctx = canvas.getContext('2d');
const imageData = ctx.getImageData(0.0, 0.0, canvas.width, canvas.height);
const data = imageData.data;
for (var i = 0; i < data.length; i += 4) {
data[i] = 255 * Math.pow((data[i] / 255), gammaCorrection);
data[i+1] = 255 * Math.pow((data[i+1] / 255), gammaCorrection);
data[i+2] = 255 * Math.pow((data[i+2] / 255), gammaCorrection);
}
ctx.putImageData(imageData, 0, 0);
}
Here the function adjusts the gamma based on the formula in the Article linked by OP on the Canvas with id "canvasOutput"
There is no negative gamma correction. You should save the original values and use them when making gamma changes, and set gamma to 1.0 to revert back to the original.
Also note that you have the wrong order of operations (exponents come before multiplication).
var originals = { r: r, g: g, b: b };
// increase
gamma = 0.5;
gammacorrection = 1/gamma;
r = 255 * Math.pow(( originals.r / 255), gammacorrection);
g = ...
b = ...
// revert to original
gamma = 1;
gammacorrection = 1/gamma;
r = 255 * Math.pow(( originals.r / 255), gammacorrection);
g = ...
b = ...
There is no negative value for gamma. Ideally this value will range between 0.01 and 7.99. So reverting back the gamma to the original value should be possible either by creating a new canvas instance with the original values of the image, then instantiating it, or either by creating a pool of pixels with the original image and reverting back to it.
I wrote a script how would i construct the algorithm for gamma reduction.
var gamma = 0.5;
var gammaCorrection = 1 / gamma;
var canvas = document.getElementById('canvas');
var ctx = canvas.getContext('2d');
var imageData = ctx.getImageData(0.0, canvas.width, canvas.height);
function GetPixelColor(x, y) {
var index = parseInt(x + canvas.width * y) * 4;
var rgb = {
r : imageData.data[index + 0],
g : imageData.data[index + 1],
b : imageData.data[index + 2]
};
return rgb;
}
function SetPixelColor(x, y, color) {
var index = parseInt(x + this.width * y) * 4;
var data = imageData.data;
data[index+0] = color.r;
data[index+1] = color.g;
data[index+2] = color.b;
};
for (y = 0; y < canvas.height; y++) {
for (x = 0; x < canvas.width; x++) {
var color = GetPixelColor(x, y)
var newRed = Math.pow(255 * (color.r / 255), gammaCorrection);
var newGreen = Math.pow(255 * (color.g / 255), gammaCorrection);
var newBlue = Math.pow(255 * (color.b / 255), gammaCorrection);
var color = {
r: newRed,
g: newGreen,
b: newBlue
}
SetPixelColor(x, y, color);
}
}
I don't know how the application should adjust the gamma value, but i suppose it's done with a value adjuster. If so you should adjust the gamma value dynamically giving the min and max range. I didn't tested the code, this wasn't my scope, but the idea is hopefully clear.
EDIT:
To understand the principle of gamma correction first how about to define the gamma instead.
Gamma is the monitor particularity altering the pixels input. Gamma correction is the act of inverting that process for linear RGB values so that the final output remains linear. For example, if you calculated the light intensity of an object is 0.5, you don't store the result as 0.5 in the pixel. Store it as pow(0.5, 1.0/2.2) = 0.73. When you send 0.73 to the monitor, it will apply a gamma on the value and produce pow(0.73, 2.2) = 0.5, which is what you want. To do this, you apply the inverse gamma function.
o=pow(i, 1.0/gamma)
Where
o is the output value.
i is the input value.
gamma is the gamma value used by your monitor.
So the gamma correction is nothing else than the rise of input value to the power of inverse of gamma. So to restore the gamma to the original value you apply the formula before the gamma correction has been applied.
The blue line represents the inverse gamma curve you need to apply to your pixels before they're sent to the monitor. When your monitor applies its gamma curve (red line) to the pixels, the result is a linear line (green line) that represents your intended RGB pixel values.
Related
How can I curve a sheet (cube)? I'd like to control the angle of the bend/curve.
e.g.
cube([50,50,2]);
You can rotate_extrude() an rectangle with the parameter angle. This requires the openscad version 2016.xx or newer, see documentation.
It is necessary to install a development snapshot, see download openscad
$fn= 360;
width = 10; // width of rectangle
height = 2; // height of rectangle
r = 50; // radius of the curve
a = 30; // angle of the curve
rotate_extrude(angle = a) translate([r, 0, 0]) square(size = [height, width], center = true);
looks like this:
The curve is defined by radius and angle. I think it is more realistic, to use other dimensions like length or dh in this sketch
and calculate radius and angle
$fn= 360;
w = 10; // width of rectangle
h = 2; // height of rectangle
l = 25; // length of chord of the curve
dh = 2; // delta height of the curve
module curve(width, height, length, dh) {
// calculate radius and angle
r = ((length/2)*(length/2) - dh*dh)/(2*dh);
a = asin((length/2)/r);
rotate_extrude(angle = a) translate([r, 0, 0]) square(size = [height, width], center = true);
}
curve(w, h, l, dh);
Edit 30.09.2019:
considering comment of Cfreitas, additionally moved the resulting shape to origin, so dimensions can be seen on axes of coordinates
$fn= 360;
w = 10; // width of rectangle
h = 2; // height of rectangle
l = 30; // length of chord of the curve
dh = 4; // delta height of the curve
module curve(width, height, length, dh) {
r = (pow(length/2, 2) + pow(dh, 2))/(2*dh);
a = 2*asin((length/2)/r);
translate([-(r -dh), 0, -width/2]) rotate([0, 0, -a/2]) rotate_extrude(angle = a) translate([r, 0, 0]) square(size = [height, width], center = true);
}
curve(w, h, l, dh);
and the result:
Edit 19.09.2020: There was a typo in the last edit: In the first 'translate' the local 'width' should be used instead of 'w'. Corrected it in the code above.
I can do it this way but it would be better if you could specify the bend/curve in #degrees as an argument to the function:
$fn=300;
module oval(w, h, height, center = false) {
scale([1, h/w, 1]) cylinder(h=height, r=w, center=center);
}
module curved(w,l,h) {
difference() {
oval(w,l,h);
translate([0.5,-1,-1]) color("red") oval(w,l+2,h+2);
}
}
curved(10,20,30);
Using the concept used by a_manthey_67, corrected the math and centered (aligned the chord with y axis) the resulting object:
module bentCube(width, height, length, dh) {
// calculate radius and angle
r = (length*length + 4*dh*dh)/(8*dh);
a = 2*asin(length/(2*r));
translate([-r,0,0]) rotate([0,0,-a/2])
rotate_extrude(angle = a) translate([r, 0, 0]) square(size = [height, width], center = true);}
Or, if you just want something with a fixed length, and a certain bent angle do this:
module curve(width, height, length, a) {
if( a > 0 ) {
r = (360 * (length/a)) / (2 * pi);
translate( [-r-height/2,0,0] )
rotate_extrude(angle = a)
translate([r, 0, 0])
square(size = [height, width], center = false);
} else {
translate( [-height/2,0,width] )
rotate( a=270, v=[1,0,0] )
linear_extrude( height = length )
square(size = [height, width], center = false);
}
}
The if (a > 0) statement is needed to make an exception when the bending angle is 0 (which, if drawing a curved surface, would result in an infinite radius).
Animated GIF here
GPU.js converts a JS func into a shader. The following function knows this.thread.x as the current index being operated on, but it is ultimately working as a WebGL shader.
export default function(sprite, w, h, scale) {
var bufferWidth = w * 4;
var channel = this.thread.x % 4;
var thread = this.thread.x - channel;
var y = Math.round(this.thread.x / bufferWidth);
var x = (thread % bufferWidth) / 4;
var upscale = scale * 10;
var upscaleY = y * 10;
var upscaleX = x * 10;
var scaledY = Math.round(upscaleY / upscale);
var scaledX = Math.round(upscaleX / upscale);
var newIndex = scaledY * bufferWidth + scaledX * 4;
if (x <= w * scale && y <= h * scale) {
return sprite[newIndex + channel];
} else {
return 0;
}
}
This almost works, but rows become skipped completely, actually making the result shorter than it should, and lines where those missing rows travel up and down and left to right on the image as it's scaled over time.
You can see this effect here: https://enviziion.github.io/lost-worlds/
What's wrong with my algo? Ive tried tweaking rounding and all sorts of stuff but no luck.
Use Math.floor when computing y:
var y = Math.floor(thread / bufferWidth);
If you use Math.round then it will start rounding up to the next row halfway across the buffer, which will produce a weird discontinuity.
Mathematically, you should be able to get back thread.x from y * bufferWidth + x * 4, which works for floor but not round.
Consider this binary image:
A normal edge detection algorithm (Like Canny) takes the binary image as input and results into the contour shown in red. I need another algorithm that takes a point "P" as a second piece of input data. "P" is the black point in the previous image. This algorithm should result into the blue contour. The blue contours represents the point "P" lines-of-sight edge of the binary image.
I searched a lot of an image processing algorithm that achieve this, but didn't find any. I also tried to think about a new one, but I still have a lot of difficulties.
Since you've got a bitmap, you could use a bitmap algorithm.
Here's a working example (in JSFiddle or see below). (Firefox, Chrome, but not IE)
Pseudocode:
// part 1: occlusion
mark all pixels as 'outside'
for each pixel on the edge of the image
draw a line from the source pixel to the edge pixel and
for each pixel on the line starting from the source and ending with the edge
if the pixel is gray mark it as 'inside'
otherwise stop drawing this line
// part 2: edge finding
for each pixel in the image
if pixel is not marked 'inside' skip this pixel
if pixel has a neighbor that is outside mark this pixel 'edge'
// part 3: draw the edges
highlight all the edges
At first this sounds pretty terrible... But really, it's O(p) where p is the number of pixels in your image.
Full code here, works best full page:
var c = document.getElementById('c');
c.width = c.height = 500;
var x = c.getContext("2d");
//////////// Draw some "interesting" stuff ////////////
function DrawScene() {
x.beginPath();
x.rect(0, 0, c.width, c.height);
x.fillStyle = '#fff';
x.fill();
x.beginPath();
x.rect(c.width * 0.1, c.height * 0.1, c.width * 0.8, c.height * 0.8);
x.fillStyle = '#000';
x.fill();
x.beginPath();
x.rect(c.width * 0.25, c.height * 0.02 , c.width * 0.5, c.height * 0.05);
x.fillStyle = '#000';
x.fill();
x.beginPath();
x.rect(c.width * 0.3, c.height * 0.2, c.width * 0.03, c.height * 0.4);
x.fillStyle = '#fff';
x.fill();
x.beginPath();
var maxAng = 2.0;
function sc(t) { return t * 0.3 + 0.5; }
function sc2(t) { return t * 0.35 + 0.5; }
for (var i = 0; i < maxAng; i += 0.1)
x.lineTo(sc(Math.cos(i)) * c.width, sc(Math.sin(i)) * c.height);
for (var i = maxAng; i >= 0; i -= 0.1)
x.lineTo(sc2(Math.cos(i)) * c.width, sc2(Math.sin(i)) * c.height);
x.closePath();
x.fill();
x.beginPath();
x.moveTo(0.2 * c.width, 0.03 * c.height);
x.lineTo(c.width * 0.9, c.height * 0.8);
x.lineTo(c.width * 0.8, c.height * 0.8);
x.lineTo(c.width * 0.1, 0.03 * c.height);
x.closePath();
x.fillStyle = '#000';
x.fill();
}
//////////// Pick a point to start our operations: ////////////
var v_x = Math.round(c.width * 0.5);
var v_y = Math.round(c.height * 0.5);
function Update() {
if (navigator.appName == 'Microsoft Internet Explorer'
|| !!(navigator.userAgent.match(/Trident/)
|| navigator.userAgent.match(/rv 11/))
|| $.browser.msie == 1)
{
document.getElementById("d").innerHTML = "Does not work in IE.";
return;
}
DrawScene();
//////////// Make our image binary (white and gray) ////////////
var id = x.getImageData(0, 0, c.width, c.height);
for (var i = 0; i < id.width * id.height * 4; i += 4) {
id.data[i + 0] = id.data[i + 0] > 128 ? 255 : 64;
id.data[i + 1] = id.data[i + 1] > 128 ? 255 : 64;
id.data[i + 2] = id.data[i + 2] > 128 ? 255 : 64;
}
// Adapted from http://rosettacode.org/wiki/Bitmap/Bresenham's_line_algorithm#JavaScript
function line(x1, y1) {
var x0 = v_x;
var y0 = v_y;
var dx = Math.abs(x1 - x0), sx = x0 < x1 ? 1 : -1;
var dy = Math.abs(y1 - y0), sy = y0 < y1 ? 1 : -1;
var err = (dx>dy ? dx : -dy)/2;
while (true) {
var d = (y0 * c.height + x0) * 4;
if (id.data[d] === 255) break;
id.data[d] = 128;
id.data[d + 1] = 128;
id.data[d + 2] = 128;
if (x0 === x1 && y0 === y1) break;
var e2 = err;
if (e2 > -dx) { err -= dy; x0 += sx; }
if (e2 < dy) { err += dx; y0 += sy; }
}
}
for (var i = 0; i < c.width; i++) line(i, 0);
for (var i = 0; i < c.width; i++) line(i, c.height - 1);
for (var i = 0; i < c.height; i++) line(0, i);
for (var i = 0; i < c.height; i++) line(c.width - 1, i);
// Outline-finding algorithm
function gb(x, y) {
var v = id.data[(y * id.height + x) * 4];
return v !== 128 && v !== 0;
}
for (var y = 0; y < id.height; y++) {
var py = Math.max(y - 1, 0);
var ny = Math.min(y + 1, id.height - 1);
console.log(y);
for (var z = 0; z < id.width; z++) {
var d = (y * id.height + z) * 4;
if (id.data[d] !== 128) continue;
var pz = Math.max(z - 1, 0);
var nz = Math.min(z + 1, id.width - 1);
if (gb(pz, py) || gb(z, py) || gb(nz, py) ||
gb(pz, y) || gb(z, y) || gb(nz, y) ||
gb(pz, ny) || gb(z, ny) || gb(nz, ny)) {
id.data[d + 0] = 0;
id.data[d + 1] = 0;
id.data[d + 2] = 255;
}
}
}
x.putImageData(id, 0, 0);
// Draw the starting point
x.beginPath();
x.arc(v_x, v_y, c.width * 0.01, 0, 2 * Math.PI, false);
x.fillStyle = '#800';
x.fill();
}
Update();
c.addEventListener('click', function(evt) {
var x = evt.pageX - c.offsetLeft,
y = evt.pageY - c.offsetTop;
v_x = x;
v_y = y;
Update();
}, false);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.2.3/jquery.min.js"></script>
<center><div id="d">Click on image to change point</div>
<canvas id="c"></canvas></center>
I would just estimate P's line of sight contour with ray collisions.
RESOLUTION = PI / 720;
For rad = 0 To PI * 2 Step RESOLUTION
ray = CreateRay(P, rad)
hits = Intersect(ray, contours)
If Len(hits) > 0
Add(hits[0], lineOfSightContour)
https://en.wikipedia.org/wiki/Hidden_surface_determination with e.g. a Z-Buffer is relatively easy. Edge detection looks a lot trickier and probably needs a bit of tuning. Why not take an existing edge detection algorithm from a library that somebody else has tuned, and then stick in some Z-buffering code to compute the blue contour from the red?
First approach
Main idea
Run an edge detection algorithm (Canny should do it just fine).
For each contour point C compute the triplet (slope, dir, dist), where:
slope is the slope of the line that passes through P and C
dir is a bit which is set if C is to the right of P (on the x axis) and reset if it is to the left; it used in order to distinguish in between points having the same slope, but on opposite sides of P
dist is the distance in between P and C.
Classify the set of contour points such that a class contains the points with the same key (slope, dir) and keep the one point from each such class having the minimum dist. Let S be the set of these closest points.
Sort S in clockwise order.
Iterate once more through the sorted set and, whenever two consecutive points are too far apart, draw a segment in between them, otherwise just draw the points.
Notes
You do not really need to compute the real distance in between P and C since you only use dist to determine the closest point to P at step 3. Instead you can keep C.x - P.x in dist. This piece of information should also tell you which of two points with the same slope is closest to P. Also, C.x - P.x swallows the dir parameter (in the sign bit). So you do not really need dir either.
The classification in step 3 can ideally be done by hashing (thus, in linear number of steps), but since doubles/floats are subject to rounding, you might need to allow small errors to occur by rounding the values of the slopes.
Second approach
Main idea
You can perform a sort of BFS starting from P, like when trying to determine the country/zone that P resides in. For each pixel, look at the pixels around it that were already visited by BFS (called neighbors). Depending on the distribution of the neighbor pixels that are in the line of sight, determine if the currently visited pixel is in the line of sight too or not. You can probably apply a sort of convolution operator on the neighbor pixels (like with any other filter). Also, you do not really need to decide right away if a pixel is for sure in the line of sight. You could instead compute some probability of that to be true.
Notes
Due to the fact that your graph is a 2D image, BFS should be pretty fast (since the number of edges is linear in the number of vertices).
This second approach eliminates the need to run an edge detection algorithm. Also, if the country/zone P resides in is considerably smaller than the image the overall performance should be better than running an edge detection algorithm solely.
Is the intent of the TrackballControl to have a "border" outside the trackball that induces roll? I personally dislike it. It is a bit discontinuous, and does't really have a lot of purpose (imho).
If not, the function getMouseProjectionOnBall can be changed similar to the following. This does two things (not necessarily "correctly"):
Normalize the radius to fill both axis
Map z values outside of the ball (ie where z was previously 0)
I find this a lot more natural, personally.
Thoughts?
this.getMouseProjectionOnBall = function(clientX, clientY) {
var xnormalized = (clientX - _this.screen.width * 0.5 - _this.screen.offsetLeft) / (_this.screen.width / 2.0);
var ynormalized = (_this.screen.height * 0.5 + _this.screen.offsetTop - clientY) / (_this.screen.height / 2.0);
var mouseOnBall = new THREE.Vector3(
xnormalized,
ynormalized,
0.0
);
var length = mouseOnBall.length();
var ballRadius = 1.0; // As a fraction of the screen
if (length > ballRadius * 0.70710678118654752440) {
var temp = ballRadius / 1.41421356237309504880;
mouseOnBall.z = temp * temp / length;
// Remove old method.
// This Left z = 0, which meant rotation axis
// becomes z, which is a roll
//mouseOnBall.normalize();
} else {
mouseOnBall.z = Math.sqrt(1.0 - length * length);
}
_eye.copy(_this.object.position).sub(_this.target);
var projection = _this.object.up.clone().setLength(mouseOnBall.y);
projection.add(_this.object.up.clone().cross(_eye).setLength(mouseOnBall.x));
projection.add(_eye.setLength(mouseOnBall.z));
return projection;
};
I'm currently writing a program to generate really enormous (65536x65536 pixels and above) Mandelbrot images, and I'd like to devise a spectrum and coloring scheme that does them justice. The wikipedia featured mandelbrot image seems like an excellent example, especially how the palette remains varied at all zoom levels of the sequence. I'm not sure if it's rotating the palette or doing some other trick to achieve this, though.
I'm familiar with the smooth coloring algorithm for the mandelbrot set, so I can avoid banding, but I still need a way to assign colors to output values from this algorithm.
The images I'm generating are pyramidal (eg, a series of images, each of which has half the dimensions of the previous one), so I can use a rotating palette of some sort, as long as the change in the palette between subsequent zoom levels isn't too obvious.
This is the smooth color algorithm:
Lets say you start with the complex number z0 and iterate n times until it escapes. Let the end point be zn.
A smooth value would be
nsmooth := n + 1 - Math.log(Math.log(zn.abs()))/Math.log(2)
This only works for mandelbrot, if you want to compute a smooth function for julia sets, then use
Complex z = new Complex(x,y);
double smoothcolor = Math.exp(-z.abs());
for(i=0;i<max_iter && z.abs() < 30;i++) {
z = f(z);
smoothcolor += Math.exp(-z.abs());
}
Then smoothcolor is in the interval (0,max_iter).
Divide smoothcolor with max_iter to get a value between 0 and 1.
To get a smooth color from the value:
This can be called, for example (in Java):
Color.HSBtoRGB(0.95f + 10 * smoothcolor ,0.6f,1.0f);
since the first value in HSB color parameters is used to define the color from the color circle.
Use the smooth coloring algorithm to calculate all of the values within the viewport, then map your palette from the lowest to highest value. Thus, as you zoom in and the higher values are no longer visible, the palette will scale down as well. With the same constants for n and B you will end up with a range of 0.0 to 1.0 for a fully zoomed out set, but at deeper zooms the dynamic range will shrink, to say 0.0 to 0.1 at 200% zoom, 0.0 to 0.0001 at 20000% zoom, etc.
Here is a typical inner loop for a naive Mandelbrot generator. To get a smooth colour you want to pass in the real and complex "lengths" and the iteration you bailed out at. I've included the Mandelbrot code so you can see which vars to use to calculate the colour.
for (ix = 0; ix < panelMain.Width; ix++)
{
cx = cxMin + (double )ix * pixelWidth;
// init this go
zx = 0.0;
zy = 0.0;
zx2 = 0.0;
zy2 = 0.0;
for (i = 0; i < iterationMax && ((zx2 + zy2) < er2); i++)
{
zy = zx * zy * 2.0 + cy;
zx = zx2 - zy2 + cx;
zx2 = zx * zx;
zy2 = zy * zy;
}
if (i == iterationMax)
{
// interior, part of set, black
// set colour to black
g.FillRectangle(sbBlack, ix, iy, 1, 1);
}
else
{
// outside, set colour proportional to time/distance it took to converge
// set colour not black
SolidBrush sbNeato = new SolidBrush(MapColor(i, zx2, zy2));
g.FillRectangle(sbNeato, ix, iy, 1, 1);
}
and MapColor below: (see this link to get the ColorFromHSV function)
private Color MapColor(int i, double r, double c)
{
double di=(double )i;
double zn;
double hue;
zn = Math.Sqrt(r + c);
hue = di + 1.0 - Math.Log(Math.Log(Math.Abs(zn))) / Math.Log(2.0); // 2 is escape radius
hue = 0.95 + 20.0 * hue; // adjust to make it prettier
// the hsv function expects values from 0 to 360
while (hue > 360.0)
hue -= 360.0;
while (hue < 0.0)
hue += 360.0;
return ColorFromHSV(hue, 0.8, 1.0);
}
MapColour is "smoothing" the bailout values from 0 to 1 which then can be used to map a colour without horrible banding. Playing with MapColour and/or the hsv function lets you alter what colours are used.
Seems simple to do by trial and error. Assume you can define HSV1 and HSV2 (hue, saturation, value) of the endpoint colors you wish to use (black and white; blue and yellow; dark red and light green; etc.), and assume you have an algorithm to assign a value P between 0.0 and 1.0 to each of your pixels. Then that pixel's color becomes
(H2 - H1) * P + H1 = HP
(S2 - S1) * P + S1 = SP
(V2 - V1) * P + V1 = VP
With that done, just observe the results and see how you like them. If the algorithm to assign P is continuous, then the gradient should be smooth as well.
My eventual solution was to create a nice looking (and fairly large) palette and store it as a constant array in the source, then interpolate between indexes in it using the smooth coloring algorithm. The palette wraps (and is designed to be continuous), but this doesn't appear to matter much.
What's going on with the color mapping in that image is that it's using a 'log transfer function' on the index (according to documentation). How exactly it's doing it I still haven't figured out yet. The program that produced it uses a palette of 400 colors, so index ranges [0,399), wrapping around if needed. I've managed to get pretty close to matching it's behavior. I use an index range of [0,1) and map it like so:
double value = Math.log(0.021 * (iteration + delta + 60)) + 0.72;
value = value - Math.floor(value);
It's kind of odd that I have to use these special constants in there to get my results to match, since I doubt they do any of that. But whatever works in the end, right?
here you can find a version with javascript
usage :
var rgbcol = [] ;
var rgbcol = MapColor ( Iteration , Zy2,Zx2 ) ;
point ( ctx , iX, iY ,rgbcol[0],rgbcol[1],rgbcol[2] );
function
/*
* The Mandelbrot Set, in HTML5 canvas and javascript.
* https://github.com/cslarsen/mandelbrot-js
*
* Copyright (C) 2012 Christian Stigen Larsen
*/
/*
* Convert hue-saturation-value/luminosity to RGB.
*
* Input ranges:
* H = [0, 360] (integer degrees)
* S = [0.0, 1.0] (float)
* V = [0.0, 1.0] (float)
*/
function hsv_to_rgb(h, s, v)
{
if ( v > 1.0 ) v = 1.0;
var hp = h/60.0;
var c = v * s;
var x = c*(1 - Math.abs((hp % 2) - 1));
var rgb = [0,0,0];
if ( 0<=hp && hp<1 ) rgb = [c, x, 0];
if ( 1<=hp && hp<2 ) rgb = [x, c, 0];
if ( 2<=hp && hp<3 ) rgb = [0, c, x];
if ( 3<=hp && hp<4 ) rgb = [0, x, c];
if ( 4<=hp && hp<5 ) rgb = [x, 0, c];
if ( 5<=hp && hp<6 ) rgb = [c, 0, x];
var m = v - c;
rgb[0] += m;
rgb[1] += m;
rgb[2] += m;
rgb[0] *= 255;
rgb[1] *= 255;
rgb[2] *= 255;
rgb[0] = parseInt ( rgb[0] );
rgb[1] = parseInt ( rgb[1] );
rgb[2] = parseInt ( rgb[2] );
return rgb;
}
// http://stackoverflow.com/questions/369438/smooth-spectrum-for-mandelbrot-set-rendering
// alex russel : http://stackoverflow.com/users/2146829/alex-russell
function MapColor(i,r,c)
{
var di= i;
var zn;
var hue;
zn = Math.sqrt(r + c);
hue = di + 1.0 - Math.log(Math.log(Math.abs(zn))) / Math.log(2.0); // 2 is escape radius
hue = 0.95 + 20.0 * hue; // adjust to make it prettier
// the hsv function expects values from 0 to 360
while (hue > 360.0)
hue -= 360.0;
while (hue < 0.0)
hue += 360.0;
return hsv_to_rgb(hue, 0.8, 1.0);
}