On Sun, 22 Jun 2003 13:17:49 -0700 Joel Rodriguez <[EMAIL PROTECTED]> wrote:

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> >In practice this means, that I can have two completely different > >images, one that shows a "normal" image and another one that > >completely looks like random noise. But when I use a convolution > >on both images the result can be almost identical. The problem with > >least squares optimization is that this procedure cannot > >distinguish between these two images. > > I almost agree with you, in the sense that the forward modeling Fourier analysis > based > technique might not be sufficient, although the least squares method (as a hole) > might be. > > Might be if there is a way to incorporate such image information that characterizes > it, > in some other way or aside Fourier analysis, such information (invariant measures) > could > be incorporated into a least squares reconstruction task as regularization scheme > technique. > > although the above sentence it is not a fact to my knowledge. To my knowledge it is a fact. Perhaps the following explanation might help. Given a convolution C and a blurred image B, the least squares technique tries to find an image I such that the least squares distance between C(I) and B is minimal. In my previous post I describe how you could construct an inverse of C, let's call it D. Now for all images X it is true that C(D(X)) = X, therefore I=D(B) is the least square solution because C(I)=C(D(B))=B. When the Fourier transform of C contains any zero coeficients, the inverse D is not uniquely determined. In fact, I can choose arbitrary values for the coefficients in the Fourier transform of D that correspond with the zero coeficients in C. This is true because you can describe convolution in the Fourier domain by multiplication of the coeficients. When I multiply a value by zero the end result is always zero. So when the Fourier transform of C contains zero coeficients there are an infinite number of least square solutions, and the least square criterium alone cannot select one of them. Even when the Fourier transform does not contain zero coeficients the least squares solution is generally not a visually good solution because it is extremely sensitive to errors. The problem is that C frequently contains some small coeficients, the corresponding coefficients in D are therefore very large. But this means that D will greatly magnify any errors in the corresponding Fourier coeficients of the blurred image B. So even very small changes in B will have a great impact on the least square solution I=D(B). In practice this means that in general you don't want to use the true least square solution because it is highly unstable. greetings, Ernst Lippe _______________________________________________ Gimp-developer mailing list [EMAIL PROTECTED] http://lists.xcf.berkeley.edu/mailman/listinfo/gimp-developer