## Tag Archives: rgb2gray

Note: My thesis is done. However, there are still some things which I find of interest to address here. Due to how our thesis was (mis)handled, I’m only finding the time to post about these things now.

Last time, I told you how we are porting all our Matlab/Octave code to Java. I wrote about a huge difference between Matlab and Octave’s implementation of rgb2gray. However, as it became apparent to me, my issues with rgb2gray are not quite done yet.

First a little background. Our project relies heavily on the Fast Fourier Transform, a numerical algorithm whose products I bet you encounter everyday though you probably are not aware of it. The 2D FFT is built-in with Octave, as expected for a language meant specifically for mathematical processing. The same, however, cannot be said for Java.

It’d be crazy for me to attempt to code it from scratch due to a number of reasons not the least of which is the fact that I don’t really have a solid training with numerical algorithms. So, after browsing through various libraries, we decided to use JTransforms, largely because we can see the source code for ourselves and it is natively in Java, in contrast to, say, FFTW, which has Java bindings but is natively in C. Plus, it does not have power-of-2 constraints, a common trait among various implementations of the FFT.

Now, the thing is, there seems to be quite a considerable offset between the FFT results of Octave1 and JTransforms; the kind of offset which you’d be so tempted to chalk up to floating-point handling differences though some part of your reason tells you it is too large to be so. The offset I was seeing ranged from 2 to 12 integer values.

As I admitted earlier, I don’t have any solid grounding with numerical algorithms. The way I see it, there are a number of possible culprits on why I’m not getting the results I should get. The FFT, after all, is a blanket term for a family of algorithms which all do the same thing, namely, apply the Fourier Transform to a discrete collection of values (i.e., vectors or matrices).

What I’ve Tried (Or, the scientific probable reason on why there is this offset, as discussed by my adviser)

Warning: Some equations ahead. I wouldn’t be delving too much into the equations and you should be able to understand my main point with just some basic algebra. But just so you know.

Consider this tutorial on the FFT.

It defines the 2D FFT as:

$F[k,l] = \frac{1}{\sqrt{MN}}\sum_{n=0}^{N-1}\sum_{m=0}^{M-1}f[m,n]e^{-2\pi i(\frac{mk}{M}+\frac{nl}{N})}$ $f[m,n]= \frac{1}{\sqrt{MN}}\sum_{l=0}^{N-1}\sum_{k=0}^{M-1}F[k,l]e^{2\pi i(\frac{mk}{M}+\frac{nl}{N})}$ $0 \leq m,k \leq M-1, 0 \leq n,l \leq N-1$

where M and N is the number of samples in the x and y direction (or, for our particular problem, the dimensions of the image).

While it may seem interesting to find mutual recursion here, what my adviser told me to note is the constant at the beginning of the equations, $\frac{1}{\sqrt{MN}}$. According to him, depending on the implementation of the FFT we are using (remember what I told you above that Fast Fourier Transform is just a blanket term for a family of algorithms), the constant varies to either $\frac{1}{\sqrt{MN}}$ or $\frac{1}{MN}$. It’s now just a matter of scaling the samples (the pixel values) by the appropriate constant to get the desired results.

The tutorial I linked to presents the following sample signal to demonstrate the FFT:

which should evaluate to the following real and imaginary component respectively

The following code listing directly translates the above example to Octave code:

 x_real = zeros(8, 8); x_real(2,3) = 70; x_real(2,4) = 80; x_real(2,5) = 90; x_real(3,3) = 90; x_real(3,4) = 100; x_real(3,5) = 110; x_real(4,3) = 110; x_real(4,4) = 120; x_real(4,5) = 130; x_real(5,3) = 130; x_real(5,4) = 140; x_real(5,5) = 150 x_fft = fft2(x_real); x_r = real(x_fft) x_i = imag(x_fft)

Which results to the following output:

 x_real =   0 0 0 0 0 0 0 0 0 0 70 80 90 0 0 0 0 0 90 100 110 0 0 0 0 0 110 120 130 0 0 0 0 0 130 140 150 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0   x_r =   1.3200e+03 -7.9113e+02 8.0000e+01 -1.6887e+02 4.4000e+02 -1.6887e+02 8.0000e+01 -7.9113e+02 -5.0485e+02 -9.0711e+01 2.2142e+02 1.0586e+02 -1.6828e+02 1.2721e+01 -2.6142e+02 6.8527e+02 1.2000e+02 0.0000e+00 -4.0000e+01 -2.3431e+01 4.0000e+01 0.0000e+00 4.0000e+01 -1.3657e+02 -3.3515e+02 1.3414e+02 2.1421e+01 5.0711e+01 -1.1172e+02 3.4731e+01 -6.1421e+01 2.6728e+02 1.2000e+02 -6.8284e+01 0.0000e+00 -1.1716e+01 4.0000e+01 -1.1716e+01 0.0000e+00 -6.8284e+01 -3.3515e+02 2.6728e+02 -6.1421e+01 3.4731e+01 -1.1172e+02 5.0711e+01 2.1421e+01 1.3414e+02 1.2000e+02 -1.3657e+02 4.0000e+01 0.0000e+00 4.0000e+01 -2.3431e+01 -4.0000e+01 0.0000e+00 -5.0485e+02 6.8527e+02 -2.6142e+02 1.2721e+01 -1.6828e+02 1.0586e+02 2.2142e+02 -9.0711e+01   x_i =   0.00000 -711.12698 440.00000 88.87302 0.00000 -88.87302 -440.00000 711.12698 -724.26407 713.55339 -216.56854 55.56349 -241.42136 134.14214 120.00000 158.99495 120.00000 -136.56854 40.00000 0.00000 40.00000 -23.43146 -40.00000 0.00000 -124.26407 255.56349 -120.00000 -6.44661 -41.42136 38.99495 103.43146 -105.85786 0.00000 -68.28427 40.00000 11.71573 0.00000 -11.71573 -40.00000 68.28427 124.26407 105.85786 -103.43146 -38.99495 41.42136 6.44661 120.00000 -255.56349 -120.00000 -0.00000 40.00000 23.43146 -40.00000 -0.00000 -40.00000 136.56854 724.26407 -158.99495 -120.00000 -134.14214 241.42136 -55.56349 216.56854 -713.55339

As evident, the results are very far from what our source says. However, dividing the sample (matrix x_r) by 8 (cf sample size) produces the results as given by our source.

Add the following line before calling fft2

 x_real /= 8;

And we get,

 x_r =   165.00000 -98.89087 10.00000 -21.10913 55.00000 -21.10913 10.00000 -98.89087 -63.10660 -11.33883 27.67767 13.23223 -21.03553 1.59010 -32.67767 85.65864 15.00000 0.00000 -5.00000 -2.92893 5.00000 0.00000 5.00000 -17.07107 -41.89340 16.76777 2.67767 6.33883 -13.96447 4.34136 -7.67767 33.40990 15.00000 -8.53553 0.00000 -1.46447 5.00000 -1.46447 0.00000 -8.53553 -41.89340 33.40990 -7.67767 4.34136 -13.96447 6.33883 2.67767 16.76777 15.00000 -17.07107 5.00000 0.00000 5.00000 -2.92893 -5.00000 0.00000 -63.10660 85.65864 -32.67767 1.59010 -21.03553 13.23223 27.67767 -11.33883   x_i =   0.00000 -88.89087 55.00000 11.10913 0.00000 -11.10913 -55.00000 88.89087 -90.53301 89.19417 -27.07107 6.94544 -30.17767 16.76777 15.00000 19.87437 15.00000 -17.07107 5.00000 0.00000 5.00000 -2.92893 -5.00000 0.00000 -15.53301 31.94544 -15.00000 -0.80583 -5.17767 4.87437 12.92893 -13.23223 0.00000 -8.53553 5.00000 1.46447 0.00000 -1.46447 -5.00000 8.53553 15.53301 13.23223 -12.92893 -4.87437 5.17767 0.80583 15.00000 -31.94544 -15.00000 -0.00000 5.00000 2.92893 -5.00000 -0.00000 -5.00000 17.07107 90.53301 -19.87437 -15.00000 -16.76777 30.17767 -6.94544 27.07107 -89.19417

Which, at last, agrees with our source.

The Blooper (or, the actual and embarassing reason why JTransforms FFT is not coinciding with Octave FFT)

In my first post about rgb2gray, I mentioned that I was taking the floor of the average of the RGB components of a pixel (or, in Java code, just perform integer division as the resulting precision loss is tantamount to floor). Big mistake. Octave does not merely floor values. It rounds off values.

In code, so there be no misunderstandings,

 public static int[][] rgb2gray(BufferedImage bi) { int heightLimit = bi.getHeight(); int widthLimit = bi.getWidth(); int[][] converted = new int[heightLimit][widthLimit];   for (int height = 0; height < heightLimit; height++) { for (int width = 0; width < widthLimit; width++) { Color c = new Color(bi.getRGB(width, height) & 0x00ffffff); float rgbSum = c.getRed() + c.getGreen() + c.getBlue(); float ave = (rgbSum / 3f); float diff = ave % 1f; int iave = diff < 0.5f ? (int) ave : (int) (ave + 1); converted[height][width] = iave; } }   return converted; }

The moral of this long story? If two supposedly-similar functions are returning different outputs, check that the inputs are exactly the same before jumping to any wild conclusions.

1. I’d like to note here that, according to the help file, Octave’s FFT is based on FFTW []

Due to the nature of the algorithms we are testing for our thesis, we had to “prototype” the procedures in Matlab so that we can easily modify parameters and test variables. However, since Matlab is expensive and we are a university that does not tolerate piracy ;), we used GNU Octave, a FOSS equivalent of Matlab (think Mono for C#).

We are done with the algorithm-prototyping part and we are now porting our Matlab code to Java, since this thesis is meant to be used by scientists, with a GUI and all that comes with standard software.  A big part of this task is in coding the functions that are built-in in Matlab; remember that Matlab is meant specially for mathematical purposes (it is a portmanteau of Matrix Laboratory) while Java is more general purpose, closer to metal, if you will.

For the past few days, I’ve been trying to implement the Matlab function rgb2gray which, as the name suggests, converts an RGB-colored image to grayscale. Now, there are a lot of ways to convert an image to grayscale but getting a grayscale isn’t the main point here. Getting it the way Matlab/Octave does is essential so that we can recreate in Java the recognition accuracy we achieved in Octave. We will be manipulating these pixel values after all.

So, I looked into Matlab’s documentation of rgb2gray and found that, for a given RGB pixel, it gets the the corresponding grayscale value by the following weighted sum:

0.2989 * R + 0.5870 * G + 0.1140 * B

(Or something close to those constants/giving the same priority over the RGB components. That is, green most weighted, followed by red, and then blue. This priority reflects the sensitivity of the human eye to these colors. See Luma (video).)

I then ran some tests on Octave to verify the docs:

 octave3.2:1> four = imread("four.JPG"); octave3.2:2> four(1,1,1) # The red component of the first pixel of four.JPG ans = 159 octave3.2:3> four(1,1,2) # The green component of the first pixel of four.JPG ans = 125 octave3.2:4> four(1,1,3) # The blue component of the first pixel of four.JPG ans = 64 octave3.2:5> grayval = 0.2989 * 159 + 0.5870 * 125 + 0.1140 * 64 grayval = 128.20

So, the grayscale equivalent of the first pixel of four.JPG will have the value floor(128.20)=128. Sure enough, when I encoded the procedure in Java, the first pixel of the grayscaled four.JPG has the value 127—close enough taking into account the possible differences in how Java and Octave handle floating point computation.

But wait, there’s more…

 octave3.2:6> fourgray = rgb2gray(four); octave3.2:7> fourgray(1,1) ans = 116

The value of the first pixel of four.JPG after rgb2gray is 116! Now that’s something no amount of discrepancy in floating-point handling can account for. Besides, hasn’t Octave itself computed a value close to Java’s 127 when done manually?

That’s when I realized that Octave may not be an exact port/clone of Matlab after all. I decided to Google “rgb2gray octave” and, sure enough, the documentation of Octave at SourceForge points to a departure from Matlab’s implementation:

Function File: gray = rgb2gray (rgb)

If the input is an RGB image, the conversion to a gray image is computed as the mean value of the color channels.

And verifying the docs…

 octave3.2:8> floor((159 + 125 + 64)/3) ans = 116

Problem solved.

I’m pretty sure that this isn’t the only difference between Matlab and Octave. The next time I encounter another one, I’ll try to document it here, time permitting.

BONUS: My encounters with Octave so far gives credence to this but I have yet to verify this thoroughly. It seems that Matlab/Octave loops through matrices (arrays of at least two dimensions, in Java/C-speak) column-major order. This isn’t exactly difficult to do in Java/C but Java/C programmers are more used to traversing multidimensional arrays in row-major order, since this should result to less page faults and therefore faster code. Still, for some computations, the order with which you traverse a matrix matters a lot. Be careful there.