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I am developing a finite element software that minimizes the energy of a mechanical structure. Using octave and its optim package, I run into a strange issue: The lm_feasible algorithm doesn't calculate at all when I use more than 300 degrees of freedom (DoF). Another algorithm (sqp) performs the calculation but doesn't work well when I complexify the structure and are out of my test case.
Is there a limit in the number of DoF with lm_feasible algorithm?
If so, how many DoF are maximally possible?
To give an overview and general idea of how the code works:
[x,y] = geometryGenerator()
U = zeros(lenght(x)*2,1);
U(1:2:end-1) = x;
U(2:2:end) = y;
%Non geometric argument are not optimised, and fixed during calculation
fct =#(U)complexFunctionOfEnergyIWrap(U(1:2:end-1),U(2:2:end), variousMaterialPropertiesAndOtherArgs)
para = optimset("f_equc_idx",contEq,"lb",lb,"ub",ub,"objf_grad",dEne,"objf_hessian",d2Ene,"MaxIter",1000);
[U,eneFinale,cvg,outp] = nonlin_min(fct,U,para)
Full example:
clear
pkg load optim
function [x,y] = geometryGenerator(r,elts = 100)
teta = linspace(0,pi,elts = 100);
x = r * cos(teta);
y = r * sin(teta);
endfunction
function ene = complexFunctionOfEnergyIWrap (x,y,E,P, X,Y)
ene = 0;
for i = 1:length(x)-1
ene += E*(x(i)/X(i))^4+ E*(y(i)/Y(i))^4- P *(x(i)^2+(x(i+1)^2)-x(i)*x(i+1))*abs(y(i)-y(i+1));
endfor
endfunction
[x,y] = geometryGenerator(5,100)
%Little distance from axis to avoid division by zero
x +=1e-6;
y +=1e-6;
%Saving initial geometry
X = x;
Y = y;
%Vectorisation of the function
%% Initial vector
U = zeros(length(x)*2,1);
U(1:2:end-1) = linspace(min(x),max(x),length(x));
U(2:2:end) = linspace(min(y),max(y),length(y));
%%Constraints
Aeq = zeros(3,length(U));
%%% Blocked bottom
Aeq(1,1) = 1;
Aeq(2,2) = 1;
%%% Sliding top
Aeq(3,end-1) = 1;
%%%Initial condition
beq = zeros(3,1);
beq(1) = U(1);
beq(2) = U(2);
beq(3) = U(end-1);
contEq = #(U) Aeq * U - beq;
%Parameter
Mat = 0.2e9;
pressure = 50;
%% Vectorized function. Non geometric argument are not optimised, and fixed during calculation
fct =#(U)complexFunctionOfEnergyIWrap(U(1:2:end-1),U(2:2:end), Mat, pressure, X, Y)
para = optimset("Algorithm","lm_feasible","f_equc_idx",contEq,"MaxIter",1000);
[U,eneFinale,cvg,outp] = nonlin_min(fct,U,para)
xFinal = U(1:2:end-1);
yFinal = U(2:2:end);
plot(x,y,';Initial geo;',xFinal,yFinal,'--x;Final geo;')
Finite Element Method is typically formulated as the optimal criteria for the minimization problem, which is equivalent to the Virtual Work Principle (see books like Hughes of Bathe). The Virtual Work, represents a set of linear (or nonlinear) equations which can be solved more efficiently (with fsolve).
If for some motive you must solve the problem as an optimization problem, then, if you are considering linear elasticity, your strain energy is quadratic, thus you could use the qp Octave function.
To use sparse matrices could also be helpful.
I am doing a simple task on ocatve.
I have to administer drug dose at day 1,7,21 and 28..
i wrote a function like that:
function xdot = f (x,t)
a=[1;7;21;28]
drug=0; ### initially drug is zero
if (t==a)
drug=drug+57.947;
else
drug=drug+0;
endif
xdot=(-0.4077)*(x)+ drug; passing the value of drug to differential equation
endfunction
In the main file i called this function in lsode:
t=linspace(0,30,30);
x0=0;
y=lsode(#ex,x0,t); ### ex is the file name where function is written
plot(t,y,'o')
This program doesn't work.. it displays all the time zero value for drug. Can anybody help me that how to administer dug dose with certain time step by manipulating linspace function.
It looks like you have a simple clearance model, and at each time in a, you want the dose to be delivered instantaneously. That is, at each time in a, the amount of drug in the subject increases by 57.947.
If that is the model that you have in mind, implementing it in the formula for xdot will not work well. You would actually need to implement it as a "delta function", and lsode won't work with that.
Instead, I recommend solving the problem in stages, corresponding to the time intervals [0, 1], [1, 7], [7, 21], and [21, 28]. During one stage, the differental equation is simply xdot = -0.4077*x. In the first stage, the initial condition is 0. In the next stage, the initial condition is the final value of the previous stage plus the dosage amount 57.947.
Here's a script:
dose = 57.947;
a = [1 7 21 28 30];
x0 = 0.0;
t0 = 0;
t = [];
y = [];
for t1 = a
tinterval = linspace(t0, t1, 2*(t1 - t0) + 1);
yinterval = lsode(#(x, t) -0.4077*x, x0, tinterval);
t = [t tinterval];
y = [y yinterval'];
t0 = t1;
x0 = yinterval(end) + dose;
end
plot(t, y, "linewidth", 2);
The script creates this plot:
Note that the differential equation xdot = -k*x has the solution x0*exp(-k*(t-t0)), so the call to lsode could be replaced with
yinterval = x0*exp(-0.4077*(tinterval - t0));
If you do this, also remove the transpose from yinterval two lines below:
y = [y yinterval];
If you want to keep the implementation of the administration of the drug dose within the formula for xdot, you'll need to distribute it over a small time interval. It could be implemented as a short rectangular pulse of width w and height 57.974/w. You'll also need to ensure that lsode takes sufficiently small internal time steps (less than w) so that it "sees" the drug dose.
You probably want replace t==a by ismember (t, a). Also, you should erase the else clause as it has no effect on the answer.
UPDATE
Consider rewriting your function as:
function xdot = f (x,t)
xdot = -0.4077*x + 57.947*ismember (t, [1 7 21 28])
endfunction
I'm trying to compute the Fourier coefficients for a waveform using MATLAB. The coefficients can be computed using the following formulas:
T is chosen to be 1 which gives omega = 2pi.
However I'm having issues performing the integrals. The functions are are triangle wave (Which can be generated using sawtooth(t,0.5) if I'm not mistaking) as well as a square wave.
I've tried with the following code (For the triangle wave):
function [ a0,am,bm ] = test( numTerms )
b_m = zeros(1,numTerms);
w=2*pi;
for i = 1:numTerms
f1 = #(t) sawtooth(t,0.5).*cos(i*w*t);
f2 = #(t) sawtooth(t,0.5).*sin(i*w*t);
am(i) = 2*quad(f1,0,1);
bm(i) = 2*quad(f2,0,1);
end
end
However it's not getting anywhere near the values I need. The b_m coefficients are given for a
triangle wave and are supposed to be 1/m^2 and -1/m^2 when m is odd alternating beginning with the positive term.
The major issue for me is that I don't quite understand how integrals work in MATLAB and I'm not sure whether or not the approach I've chosen works.
Edit:
To clairify, this is the form that I'm looking to write the function on when the coefficients have been determined:
Here's an attempt using fft:
function [ a0,am,bm ] = test( numTerms )
T=2*pi;
w=1;
t = [0:0.1:2];
f = fft(sawtooth(t,0.5));
am = real(f);
bm = imag(f);
func = num2str(f(1));
for i = 1:numTerms
func = strcat(func,'+',num2str(am(i)),'*cos(',num2str(i*w),'*t)','+',num2str(bm(i)),'*sin(',num2str(i*w),'*t)');
end
y = inline(func);
plot(t,y(t));
end
Looks to me that your problem is what sawtooth returns the mathworks documentation says that:
sawtooth(t,width) generates a modified triangle wave where width, a scalar parameter between 0 and 1, determines the point between 0 and 2π at which the maximum occurs. The function increases from -1 to 1 on the interval 0 to 2πwidth, then decreases linearly from 1 to -1 on the interval 2πwidth to 2π. Thus a parameter of 0.5 specifies a standard triangle wave, symmetric about time instant π with peak-to-peak amplitude of 1. sawtooth(t,1) is equivalent to sawtooth(t).
So I'm guessing that's part of your problem.
After you responded I looked into it some more. Looks to me like it's the quad function; not very accurate! I recast the problem like this:
function [ a0,am,bm ] = sotest( t, numTerms )
bm = zeros(1,numTerms);
am = zeros(1,numTerms);
% 2L = 1
L = 0.5;
for ii = 1:numTerms
am(ii) = (1/L)*quadl(#(x) aCos(x,ii,L),0,2*L);
bm(ii) = (1/L)*quadl(#(x) aSin(x,ii,L),0,2*L);
end
ii = 0;
a0 = (1/L)*trapz( t, t.*cos((ii*pi*t)/L) );
% now let's test it
y = ones(size(t))*(a0/2);
for ii=1:numTerms
y = y + am(ii)*cos(ii*2*pi*t);
y = y + bm(ii)*sin(ii*2*pi*t);
end
figure; plot( t, y);
end
function a = aCos(t,n,L)
a = t.*cos((n*pi*t)/L);
end
function b = aSin(t,n,L)
b = t.*sin((n*pi*t)/L);
end
And then I called it like:
[ a0,am,bm ] = sotest( t, 100 );
and I got:
Sweetness!!!
All I really changed was from quad to quadl. I figured that out by using trapz which worked great until the time vector I was using didn't have enough resolution, which led me to believe it was a numerical issue rather than something fundamental. Hope this helps!
To troubleshoot your code I would plot the functions you are using and investigate, how the quad function samples them. You might be undersampling them, so make sure your minimum step size is smaller than the period of the function by at least factor 10.
I would suggest using the FFTs that are built-in to Matlab. Not only is the FFT the most efficient method to compute a spectrum (it is n*log(n) dependent on the length n of the array, whereas the integral in n^2 dependent), it will also give you automatically the frequency points that are supported by your (equally spaced) time data. If you compute the integral yourself (might be needed if datapoints are not equally spaced), you might calculate frequency data that are not resolved (closer spacing than 1/over the spacing in time, i.e. beyond the 'Fourier limit').
I'm trying to get the skew values out of a transformation matrix in a flash movie clip. The transformation matrix is represented by
a b tx
c d ty
0 0 1
I have no information on what kind of transformation is performed and which comes first. I do know that in flash, you may only rotate OR skew a movie clip (correct me if I am wrong). I can get scale values from scaleX and scaleY properties of the movie clip. I believe translation does not quite matter i can just equate tx and ty to zero.
so my question has 2 parts. How do I determine if a skew or a rotation had been applied, and how do I get the respective values?
The 2D rotation matrix is
cos(theta) -sin(theta)
sin(theta) cos(theta)
so if you have no scaling or shear applied,
a = d
and
c = -b
and the angle of rotation is
theta = asin(c) = acos(a)
If you've got scaling applied and can recover the scaling factors sx and sy, just divide the first row by sx and the second by sy in your original transformation matrix and then recover the rotation angle as above.
If you've got a shear (skew) applied anywhere in there, I'm with the previous commenters, it might not be possible except in very limited cases (such as shear in only one known direction at a time and in a known order).
You need to do a polar decomposition. This Wikipedia article explains how it works:
http://en.wikipedia.org/wiki/Polar_decomposition
Here is the code I wrote for my own program using the OpenCV library.
const double PI = 3.141592653;
cv::Mat rotationOutput = cv::Mat::zeros(warp00.size(),CV_64F);
cv::Mat_<double>::iterator rotIter = rotationOutput.begin<double>();
cv::Mat_<double>::iterator warp00Iter = warp00.begin<double>();
cv::Mat_<double>::iterator warp01Iter = warp01.begin<double>();
cv::Mat_<double>::iterator warp10Iter = warp10.begin<double>();
cv::Mat_<double>::iterator warp11Iter = warp11.begin<double>();
for(; warp00Iter != warp00.end<double>(); ++warp00Iter, ++warp01Iter, ++warp10Iter,
++warp11Iter, ++rotIter){
cv::Matx22d fMatrix(*warp00Iter,*warp01Iter, *warp10Iter, *warp11Iter);
cv::Matx22d cMatrix;
cv::Matx22d cMatSqrt(0.,0.,0.,0.);
cv::mulTransposed(fMatrix, cMatrix, true);
cv::Matx21d eigenVals;
cv::Matx22d eigenVecs;
if((cMatrix(0,0) !=0.) && (cMatrix(1,1) !=0.)){
if(cv::eigen(cMatrix,true,eigenVals,eigenVecs)){
cMatSqrt = eigenVecs.t()*
cv::Matx22d(sqrt(eigenVals(0,0)),0.,0.,sqrt(eigenVals(1,0)))*eigenVecs;
}
}
cv::Matx22d rMat = fMatrix*cMatSqrt.inv();
*rotIter = atan(rMat(1,0)/rMat(0,0));
}
warp00, warp01, warp10 and warp11 contains the first 4 params of the affine transform (translation params warp02 and warp12 are not needed). IN your case it would be a,b,c,d.
You'll notice in the wikipedia article that you need to compute the square root of a matrix. The only way to do so is by computing the eigen values, then compute their square roots and rotate the diagonal matrix back to the original coordinate system.
It's complicated, but it is the only way to compute the rotations when you have an affine transform.
In my case, I only cared about the rotations, so my code won't give you the skew.
The term for this is matrix decomposition. Here is a solution that includes skew as described by Frédéric Wang.
Works when transforms are applied in this order: skew, scale, rotate, translate.
function decompose_2d_matrix(mat) {
var a = mat[0];
var b = mat[1];
var c = mat[2];
var d = mat[3];
var e = mat[4];
var f = mat[5];
var delta = a * d - b * c;
let result = {
translation: [e, f],
rotation: 0,
scale: [0, 0],
skew: [0, 0],
};
// Apply the QR-like decomposition.
if (a != 0 || b != 0) {
var r = Math.sqrt(a * a + b * b);
result.rotation = b > 0 ? Math.acos(a / r) : -Math.acos(a / r);
result.scale = [r, delta / r];
result.skew = [Math.atan((a * c + b * d) / (r * r)), 0];
} else if (c != 0 || d != 0) {
var s = Math.sqrt(c * c + d * d);
result.rotation =
Math.PI / 2 - (d > 0 ? Math.acos(-c / s) : -Math.acos(c / s));
result.scale = [delta / s, s];
result.skew = [0, Math.atan((a * c + b * d) / (s * s))];
} else {
// a = b = c = d = 0
}
return result;
}
First, you can do both skew and rotate, but you have to select the order first. A skew matrix is explained here, to add a skew matrix to a transformation you create a new matrix and do yourTransformMatrix.concat(skewMatrix);
I can't currently say if you can retrieve values for transformation in terms of "rotation angle", "skew_X angle", "skew_Y angle", "translation_X","translation_Y", this in general is a nonlinear equation system which might not have a solution for a specific matrix.
I've been writing an image processing program which applies effects through HTML5 canvas pixel processing. I've achieved Thresholding, Vintaging, and ColorGradient pixel manipulations but unbelievably I cannot change the contrast of the image!
I've tried multiple solutions but I always get too much brightness in the picture and less of a contrast effect and I'm not planning to use any Javascript libraries since I'm trying to achieve these effects natively.
The basic pixel manipulation code:
var data = imageData.data;
for (var i = 0; i < data.length; i += 4) {
//Note: data[i], data[i+1], data[i+2] represent RGB respectively
data[i] = data[i];
data[i+1] = data[i+1];
data[i+2] = data[i+2];
}
Pixel manipulation example
Values are in RGB mode which means data[i] is the Red color. So if data[i] = data[i] * 2; the brightness will be increased to twice for the Red channel of that pixel. Example:
var data = imageData.data;
for (var i = 0; i < data.length; i += 4) {
//Note: data[i], data[i+1], data[i+2] represent RGB respectively
//Increases brightness of RGB channel by 2
data[i] = data[i]*2;
data[i+1] = data[i+1]*2;
data[i+2] = data[i+2]*2;
}
*Note: I'm not asking you guys to complete the code! That would just be a favor! I'm asking for an algorithm (even Pseudo code) that shows how Contrast in pixel manipulation is possible!
I would be glad if someone can provide a good algorithm for Image Contrast in HTML5 canvas.
A faster option (based on Escher's approach) is:
function contrastImage(imgData, contrast){ //input range [-100..100]
var d = imgData.data;
contrast = (contrast/100) + 1; //convert to decimal & shift range: [0..2]
var intercept = 128 * (1 - contrast);
for(var i=0;i<d.length;i+=4){ //r,g,b,a
d[i] = d[i]*contrast + intercept;
d[i+1] = d[i+1]*contrast + intercept;
d[i+2] = d[i+2]*contrast + intercept;
}
return imgData;
}
Derivation similar to the below; this version is mathematically the same, but runs much faster.
Original answer
Here is a simplified version with explanation of an approach already discussed (which was based on this article):
function contrastImage(imageData, contrast) { // contrast as an integer percent
var data = imageData.data; // original array modified, but canvas not updated
contrast *= 2.55; // or *= 255 / 100; scale integer percent to full range
var factor = (255 + contrast) / (255.01 - contrast); //add .1 to avoid /0 error
for(var i=0;i<data.length;i+=4) //pixel values in 4-byte blocks (r,g,b,a)
{
data[i] = factor * (data[i] - 128) + 128; //r value
data[i+1] = factor * (data[i+1] - 128) + 128; //g value
data[i+2] = factor * (data[i+2] - 128) + 128; //b value
}
return imageData; //optional (e.g. for filter function chaining)
}
Notes
I have chosen to use a contrast range of +/- 100 instead of the original +/- 255. A percentage value seems more intuitive for users, or programmers who don't understand the underlying concepts. Also, my usage is always tied to UI controls; a range from -100% to +100% allows me to label and bind the control value directly instead of adjusting or explaining it.
This algorithm doesn't include range checking, even though the calculated values can far exceed the allowable range - this is because the array underlying the ImageData object is a Uint8ClampedArray. As MSDN explains, with a Uint8ClampedArray the range checking is handled for you:
"if you specified a value that is out of the range of [0,255], 0 or 255 will be set instead."
Usage
Note that while the underlying formula is fairly symmetric (allows round-tripping), data is lost at high levels of filtering because pixels only allow integer values. For example, by the time you de-saturate an image to extreme levels (>95% or so), all the pixels are basically a uniform medium gray (within a few digits of the average possible value of 128). Turning the contrast back up again results in a flattened color range.
Also, order of operations is important when applying multiple contrast adjustments - saturated values "blow out" (exceed the clamped max value of 255) quickly, meaning highly saturating and then de-saturating will result in a darker image overall. De-saturating and then saturating however doesn't have as much data loss, because the highlight and shadow values get muted, instead of clipped (see explanation below).
Generally speaking, when applying multiple filters it's better to start each operation with the original data and re-apply each adjustment in turn, rather than trying to reverse a previous change - at least for image quality. Performance speed or other demands may dictate differently for each situation.
Code Example:
function contrastImage(imageData, contrast) { // contrast input as percent; range [-1..1]
var data = imageData.data; // Note: original dataset modified directly!
contrast *= 255;
var factor = (contrast + 255) / (255.01 - contrast); //add .1 to avoid /0 error.
for(var i=0;i<data.length;i+=4)
{
data[i] = factor * (data[i] - 128) + 128;
data[i+1] = factor * (data[i+1] - 128) + 128;
data[i+2] = factor * (data[i+2] - 128) + 128;
}
return imageData; //optional (e.g. for filter function chaining)
}
$(document).ready(function(){
var ctxOrigMinus100 = document.getElementById('canvOrigMinus100').getContext("2d");
var ctxOrigMinus50 = document.getElementById('canvOrigMinus50').getContext("2d");
var ctxOrig = document.getElementById('canvOrig').getContext("2d");
var ctxOrigPlus50 = document.getElementById('canvOrigPlus50').getContext("2d");
var ctxOrigPlus100 = document.getElementById('canvOrigPlus100').getContext("2d");
var ctxRoundMinus90 = document.getElementById('canvRoundMinus90').getContext("2d");
var ctxRoundMinus50 = document.getElementById('canvRoundMinus50').getContext("2d");
var ctxRound0 = document.getElementById('canvRound0').getContext("2d");
var ctxRoundPlus50 = document.getElementById('canvRoundPlus50').getContext("2d");
var ctxRoundPlus90 = document.getElementById('canvRoundPlus90').getContext("2d");
var img = new Image();
img.onload = function() {
//draw orig
ctxOrig.drawImage(img, 0, 0, img.width, img.height, 0, 0, 100, 100); //100 = canvas width, height
//reduce contrast
var origBits = ctxOrig.getImageData(0, 0, 100, 100);
contrastImage(origBits, -.98);
ctxOrigMinus100.putImageData(origBits, 0, 0);
var origBits = ctxOrig.getImageData(0, 0, 100, 100);
contrastImage(origBits, -.5);
ctxOrigMinus50.putImageData(origBits, 0, 0);
// add contrast
var origBits = ctxOrig.getImageData(0, 0, 100, 100);
contrastImage(origBits, .5);
ctxOrigPlus50.putImageData(origBits, 0, 0);
var origBits = ctxOrig.getImageData(0, 0, 100, 100);
contrastImage(origBits, .98);
ctxOrigPlus100.putImageData(origBits, 0, 0);
//round-trip, de-saturate first
origBits = ctxOrig.getImageData(0, 0, 100, 100);
contrastImage(origBits, -.98);
contrastImage(origBits, .98);
ctxRoundMinus90.putImageData(origBits, 0, 0);
origBits = ctxOrig.getImageData(0, 0, 100, 100);
contrastImage(origBits, -.5);
contrastImage(origBits, .5);
ctxRoundMinus50.putImageData(origBits, 0, 0);
//do nothing 100 times
origBits = ctxOrig.getImageData(0, 0, 100, 100);
for(i=0;i<100;i++){
contrastImage(origBits, 0);
}
ctxRound0.putImageData(origBits, 0, 0);
//round-trip, saturate first
origBits = ctxOrig.getImageData(0, 0, 100, 100);
contrastImage(origBits, .5);
contrastImage(origBits, -.5);
ctxRoundPlus50.putImageData(origBits, 0, 0);
origBits = ctxOrig.getImageData(0, 0, 100, 100);
contrastImage(origBits, .98);
contrastImage(origBits, -.98);
ctxRoundPlus90.putImageData(origBits, 0, 0);
};
img.src = 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";
});
canvas {width: 100px; height: 100px}
div {text-align:center; width:120px; float:left}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div>
<canvas id="canvOrigMinus100" width="100" height="100"></canvas>
-98%
</div>
<div>
<canvas id="canvOrigMinus50" width="100" height="100"></canvas>
-50%
</div>
<div>
<canvas id="canvOrig" width="100" height="100"></canvas>
Original
</div>
<div>
<canvas id="canvOrigPlus50" width="100" height="100"></canvas>
+50%
</div>
<div>
<canvas id="canvOrigPlus100" width="100" height="100"></canvas>
+98%
</div>
<hr/>
<div style="clear:left">
<canvas id="canvRoundMinus90" width="100" height="100"></canvas>
Round-trip <br/> (-98%, +98%)
</div>
<div>
<canvas id="canvRoundMinus50" width="100" height="100"></canvas>
Round-trip <br/> (-50%, +50%)
</div>
<div>
<canvas id="canvRound0" width="100" height="100"></canvas>
Round-trip <br/> (0% 100x)
</div>
<div>
<canvas id="canvRoundPlus50" width="100" height="100"></canvas>
Round-trip <br/> (+50%, -50%)
</div>
<div>
<canvas id="canvRoundPlus90" width="100" height="100"></canvas>
Round-trip <br/> (+98%, -98%)
</div>
Explanation
(Disclaimer - I am not an image specialist or a mathematician. I am trying to provide a common-sense explanation with minimal technical details. Some hand-waving below, e.g. 255=256 to avoid indexing issues, and 127.5=128, for simplifying the numbers.)
Since, for a given pixel, the possible number of non-zero values for a color channel is 255, the "no-contrast", average value of a pixel is 128 (or 127, or 127.5 if you want argue, but the difference is negligible). For purposed of this explanation, the amount of "contrast" is the distance from the current value to the average value (128). Adjusting the contrast means increasing or decreasing the difference between the current value and the average value.
The problem the algorithm solves then is to:
Chose a constant factor to adjust contrast by
For each color channel of each pixel, scale "contrast" (distance from average) by that constant factor
Or, as hinted at in the CSS spec, simply choosing the slope and intercept of a line:
<feFuncR type="linear" slope="[amount]" intercept="-(0.5 * [amount]) + 0.5"/>
Note the term type='linear'; we are doing linear contrast adjustment in RGB color space, as opposed to a quadratic scaling function, luminence-based adjustment, or histogram matching.
If you recall from geometry class, the formula for a line is y=mx+b. y is the final value we are after, the slope m is the contrast (or factor), x is the initial pixel value, and b is the intercept of the y-axis (x=0), which shifts the line vertically. Recall also that since the y-intercept is not at the origin (0,0), the formula can also be represented as y=m(x-a)+b, where a is the x-offset shifting the line horizontally.
For our purposes, this graph represents the input value (x-axis) and the result (y-axis). We already know that b, the y-intercept (for m=0, no contrast) must be 128 (which we can check against the 0.5 from the spec - 0.5 * the full range of 256 = 128). x is our original value, so all we need is to figure out the slope m and x-offset a.
First, the slope m is "rise over run", or (y2-y1)/(x2-x1) - so we need 2 points known to be on the desired line. Finding these points requires bringing a few things together:
Our function takes the shape of a line-intercept graph
The y-intercept is at b = 128 - regardless of the slope (contrast).
The maximum expected 'y' value is 255, and the minimum is 0
The range of possible 'x' values is 256
A neutral value should always stay neutral: 128 => 128 regardless of slope
A contrast adjustment of 0 should result in no change between input and output; that is, a 1:1 slope.
Taking all these together, we can deduce that regardless of the contrast (slope) applied, our resulting line will be centered at (and pivot around) 128,128. Since our y-intercept is non-zero, the x-intercept is also non-zero; we know the x-range is 256 wide and is centered in the middle, so it must be offset by half of the possible range: 256 / 2 = 128.
So now for y=m(x-a)+b, we know everything except m. Recall two more important points from geometry class:
Lines have the same slope even if their location changes; that is, m stays the same regardless of the values of a and b.
The slope of a line can be found using any 2 points on the line
To simplify the slope discussion, let's move the coordinate origin to the x-intercept (-128) and ignore a and b for a moment. Our original line will now pivot through (0,0), and we know a second point on the line lies away the full range of both x (input) and y (output) at (255,255).
We'll let the new line pivot at (0,0), so we can use that as one of the points on the new line that will follow our final contrast slope m. The second point can be determined by moving the current end at (255,255) by some amount; since we are limited to a single input (contrast) and using a linear function, this second point will be moved equally in the x and y directions on our graph.
The (x,y) coordinates of the 4 possible new points will be 255 +/- contrast. Since increasing or decreasing both x and y would keep us on the original 1:1 line, let's just look at +x, -y and -x, +y as shown.
The steeper line (-x, +y) is associated with a positive contrast adjustment; it's (x,y) coordinates are (255 - contrast,255 + contrast). The coordinates of the shallower line (negative contrast) are found the same way. Notice that the biggest meaningful value of contrast will be 255 - the most that the initial point of (255,255) can be translated before resulting in a vertical line (full contrast, all black or white) or a horizontal line (no contrast, all gray).
So now we have the coordinates of two points on our new line - (0,0) and (255 - contrast,255 + contrast). We plug this into the slope equation, and then plug that into the full line equation, using all the parts from before:
y = m(x-a) + b
m = (y2-y1)/(x2-x1) =>
((255 + contrast) - 0)/((255 - contrast) - 0) =>
(255 + contrast)/(255 - contrast)
a = 128
b = 128
y = (255 + contrast)/(255 - contrast) * (x - 128) + 128 QED
The math-minded will notice that the resulting m or factor is a scalar (unitless) value; you can use any range you want for contrast as long as it matches the constant (255) in the factor calculation. For example, a contrast range of +/-100 and factor = (100 + contrast)/(100.01 - contrast), which is was I really use to eliminate the step of scaling to 255; I just left 255 in the code at the top to simplify the explanation.
Note about the "magic" 259
The source article uses a "magic" 259, although the author admits he doesn't remember why:
"I can’t remember if I had calculated this myself or if I’ve read it in a book or online.".
259 should really be 255 or perhaps 256 - the number of possible non-zero values for each channel of each pixel. Note that in the original factor calculation, 259/255 cancels out - technically 1.01, but final values are whole integers so 1 for all practical purposes. So this outer term can be discarded. Actually using 255 for the constant in the denominator, though, introduces the possibility of a Divide By Zero error in the formula; adjusting to a slightly larger value (say, 259) avoids this issue without introducing significant error to the results. I chose to use 255.01 instead as the error is lower and it (hopefully) seems less "magic" to a newcomer.
As far as I can tell though, it doesn't make much difference which you use - you get identical values except for minor, symmetric differences in a narrow band of low contrast values with a low positive contrast increase. I'd be curious to round-trip both versions repeatedly and compare to the original data, but this answer already took way too long. :)
After trying the answer by Schahriar SaffarShargh, it wasn't behaving like contrast should behave. I finally came across this algorithm, and it works like a charm!
For additional information about the algorithm, read this article and it's comments section.
function contrastImage(imageData, contrast) {
var data = imageData.data;
var factor = (259 * (contrast + 255)) / (255 * (259 - contrast));
for(var i=0;i<data.length;i+=4)
{
data[i] = factor * (data[i] - 128) + 128;
data[i+1] = factor * (data[i+1] - 128) + 128;
data[i+2] = factor * (data[i+2] - 128) + 128;
}
return imageData;
}
Usage:
var newImageData = contrastImage(imageData, 30);
Hopefully this will be a time-saver for someone. Cheers!
This javascript implementation complies with the SVG/CSS3 definition of "contrast" (and the following code will render your canvas image identically):
/*contrast filter function*/
//See definition at https://drafts.fxtf.org/filters/#contrastEquivalent
//pixels come from your getImageData() function call on your canvas image
contrast = function(pixels, value){
var d = pixels.data;
var intercept = 255*(-value/2 + 0.5);
for(var i=0;i<d.length;i+=4){
d[i] = d[i]*value + intercept;
d[i+1] = d[i+1]*value + intercept;
d[i+2] = d[i+2]*value + intercept;
//implement clamping in a separate function if using in production
if(d[i] > 255) d[i] = 255;
if(d[i+1] > 255) d[i+1] = 255;
if(d[i+2] > 255) d[i+2] = 255;
if(d[i] < 0) d[i] = 0;
if(d[i+1] < 0) d[i+1] = 0;
if(d[i+2] < 0) d[i+2] = 0;
}
return pixels;
}
I found out that you have to use the effect by separating the darks and lights or technically anything that is less than 127 (average of R+G+B / 3) in rgb scale is a black and more than 127 is a white, therefore by your level of contrast you minus a value say 10 contrast from the blacks and add the same value to the whites!
Here is an example:
I have two pixels with RGB colors, [105,40,200] | [255,200,150]
So I know that for my first pixel 105 + 40 + 200 = 345, 345/3 = 115
and 115 is less than my half of 255 which is 127 so I consider the pixel closer to [0,0,0] therefore if I want to minus 10 contrast then I take away 10 from each color on it's average
Thus I have to divide each color's value by the total's average which was 115 for this case and times it by my contrast and minus out the final value from that specific color:
For example I'll take 105 (red) from my pixel, so I divide it by total RGB's avg. which is 115 and times it by my contrast value of 10, (105/115)*10 which gives you something around 9 (you have to round it up!) and then take that 9 away from 105 so that color becomes 96 so my red after having a 10 contrast on a dark pixel.
So if I go on my pixel's values become [96,37,183]! (note: the scale of contrast is up to you! but my in the end you should convert it to some scale like from 1 to 255)
For the lighter pixels I also do the same except instead of subtracting the contrast value I add it! and if you reach the limit of 255 or 0 then you stop your addition and subtraction for that specific color! therefore my second pixel which is a lighter pixel becomes [255,210,157]
As you add more contrast it will lighten the lighter colors and darken the darker and therefore adds contrast to your picture!
Here is a sample Javascript code ( I haven't tried it yet ) :
var data = imageData.data;
for (var i = 0; i < data.length; i += 4) {
var contrast = 10;
var average = Math.round( ( data[i] + data[i+1] + data[i+2] ) / 3 );
if (average > 127){
data[i] += ( data[i]/average ) * contrast;
data[i+1] += ( data[i+1]/average ) * contrast;
data[i+2] += ( data[i+2]/average ) * contrast;
}else{
data[i] -= ( data[i]/average ) * contrast;
data[i+1] -= ( data[i+1]/average ) * contrast;
data[i+2] -= ( data[i+2]/average ) * contrast;
}
}
You can take a look at the OpenCV docs to see how you could accomplish this: Brightness and contrast adjustments.
Then there's the demo code:
double alpha; // Simple contrast control: value [1.0-3.0]
int beta; // Simple brightness control: value [0-100]
for( int y = 0; y < image.rows; y++ )
{
for( int x = 0; x < image.cols; x++ )
{
for( int c = 0; c < 3; c++ )
{
new_image.at<Vec3b>(y,x)[c] = saturate_cast<uchar>( alpha*( image.at<Vec3b>(y,x)[c] ) + beta );
}
}
}
which I imagine you are capable of translating to javascript.
By vintaging I assume your trying to apply LUTS..Recently I have been trying to add color treatments to canvas windows. If you want to actually apply "LUTS" to the canvas window I believe you need to actually map the array that imageData returns to the RGB array of the LUT.
(From Light illusion)
As an example the start of a 1D LUT could look something like this:
Note: strictly speaking this is 3x 1D LUTs, as each colour (R,G,B) is a 1D LUT
R, G, B
3, 0, 0
5, 2, 1
7, 5, 3
9, 9, 9
Which means that:
For an input value of 0 for R, G, and B, the output is R=3, G=0, B=0
For an input value of 1 for R, G, and B, the output is R=5, G=2, B=1
For an input value of 2 for R, G, and B, the output is R=7, G=5, B=3
For an input value of 3 for R, G, and B, the output is R=9, G=9, B=9
Which is a weird LUT, but you see that for a given value of R, G, or B input, there is a given value of R, G, and B output.
So, if a pixel had an input value of 3, 1, 0 for RGB, the output pixel would be 9, 2, 0.
During this I also realized after playing with imageData that it returns a Uint8Array and that the values in that array are decimal. Most 3D LUTS are Hex. So you first have to do some type of hex to dec conversion on the entire array before all this mapping.
This is the formula you are looking for ...
var data = imageData.data;
if (contrast > 0) {
for(var i = 0; i < data.length; i += 4) {
data[i] += (255 - data[i]) * contrast / 255; // red
data[i + 1] += (255 - data[i + 1]) * contrast / 255; // green
data[i + 2] += (255 - data[i + 2]) * contrast / 255; // blue
}
} else if (contrast < 0) {
for (var i = 0; i < data.length; i += 4) {
data[i] += data[i] * (contrast) / 255; // red
data[i + 1] += data[i + 1] * (contrast) / 255; // green
data[i + 2] += data[i + 2] * (contrast) / 255; // blue
}
}
Hope it helps!