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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:

Sample Signal

which should evaluate to the following real and imaginary component respectively

FFT real

FFT imaginary

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.