In the paper "An Approach Based on Hybrid Genetic Algorithm Applied to Image Denoising Problem", I encounter this fitness function that by minimizing it, a denoised image of a noisy image is obtained.

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It is called Discretized Beltrami energy. According to the paper, $I$ is the image being evaluated, $I_0$ is noisy image, $\nabla I$ is total variation (TV) regularizing term, $\beta$ and $\lambda$ are balancing parameters. $\Omega$ is the set of all points of the image.

What is $\nabla I$? Determinant of $\nabla I$ is a constant, why is there a sum over $\Omega$?


In this particular case, $\nabla I$ is the gradient because the image $I$ is a scalar field. The summation over $\Omega$ would make more sense if $I$ was a vector field. Since the image pixels are samples over fixed locations in space, the $\Omega$ can be dropped.

To understand why this is used as a fitness function, you need to understand Total Variation Minimisation. The assumption there is that a signal has some broad structure that is disturbed by noise. Noise is seen here as large deviations around a central value. But this view would include edges which do not belong to the "noise" part of an image but to what gives it "structure". So, what you are trying to do with total variation regularisation is to estimate the total length of the signal "curve" which largely depends on its structure and then try to minimise the differences between the signal values in areas that it does not "cost" much to do so. If the denoising algorithm tried to suppress an image edge, that would mean a huge change in the length of the curve. But if the denoising algorithm tried to suppress some spurious spike (as low pass filtering would do), then that would not present a large error.

What the authors are trying to do is to use the outputs of three denoising algorithms to descend to the lowest error of this fitness function as quickly as possible by selecting the "best" areas from each denoised image and blend them together with low pass filtering.

Hope this helps.

  • $\begingroup$ Thank you, It helped me to understand what's going on. $\endgroup$ – Farhad Bat Dec 17 '18 at 14:18
  • $\begingroup$ @FarhadBat You are welcome, good luck with your project. $\endgroup$ – A_A Dec 17 '18 at 14:41

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