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# Separate Kernel in 1D kernels 1.0

Date Added: June 22, 2013  |  Visits: 177

This function SEPARATEKERNEL will separate ( do decomposition of ) any 2D, 3D or nD kernel into 1D kernels. Of course only a sub-set of Kernels are separable such as a Gaussian Kernel, but it will give approximations for non-separable kernels. Separating a 3D or 5D image filter into 1D filters will give an large speed-up in image filtering with for instance the function imfilter. [K1 KN ERR]=SeparateKernel(H); inputs, H : The 2D, 3D ..., ND kernel outputs, K1 : Cell array with the 1D kernels KN : Approximation of the ND input kernel by the 1D kernels ERR : The sum of absolute difference between approximation and input kernel .. --------------------------------------------------------------------- How the algorithm works --------------------------------------------------------------------- If we have a separable kernel like H = [1 2 1 2 4 2 3 6 3]; We like to solve unknown 1D kernels, a=[a(1) a(2) a(3)] b=[b(1) b(2) b(3)] We know that, H = a'*b b(1) b(2) b(3) -------------------- a(1)|h(1,1) h(1,2) h(1,3) a(2)|h(2,1) h(2,2) h(2,3) a(3)|h(3,1) h(3,2) h(3,3) Thus, h(1,1) == a(1)*b(1) h(2,1) == a(2)*b(1) h(3,1) == a(3)*b(1) h(4,1) == a(1)*b(2) ... We want to solve this by using fast matrix (least squares) math, c = M * d; c a column vector with all kernel values H d a column vector with the unknown 1D kernels But matrices "add" values and we have something like h(1,1) == a(1)*b(1); We solve this by taking the log at both sides(We replace zeros by a small value. Whole lines/planes of zeros are removed at forehand and re-added afterwards) log( h(1,1) ) == log(a(1)) + log b(1)) The matrix is something like this, a1 a2 a3 b1 b2 b3 M = [1 0 0 1 0 0; h11 0 1 0 1 0 0; h21 0 0 1 1 0 0; h31 1 0 0 0 1 0; h21 0 1 0 0 1 0; h22 0 0 1 0 1 0; h23 1 0 0 0 0 1; h31 0 1 0 0 0 1; h32 0 0 1 0 0 1]; h33 Least squares solution d = exp(Mlog(c)) with the 1D kernels [a(1);a(2);a(3);b(1);b(2);b(3)] = d..--------------------------------------------------------------------- The Problem of Negative Values --------------------------------------------------------------------- The log of a negative value is possible it gives a complex value, log(-1) = i*pi if we take the expontential it is back to the old value, exp(i*pi) = -1 But if we use the solver with on of the 1D vectors we get something like, this : input | result | abs(result) | angle(result) -1 | -0.0026 + 0.0125i | 0.0128 | 1.7744 2 | 0.0117 + 0.0228i | 0.0256 | 1.0958 -3 | -0.0078 + 0.0376i | 0.0384 | 1.7744 4 | 0.0234 + 0.0455i | 0.0512 | 1.0958 5 | 0.0293 + 0.0569i | 0.0640 | 1.0958 The absolute value is indeed correct (difference in scale is compensated by the other 1D vectors) As you can see the angle is correlated with the sign of the values. But I didn't found the correlation yet. For some matrices it is something like sign=mod(angle(solution)*scale,pi) == pi/2; In the current algorithm, we just flip the 1D kernel values one by one. The sign change which gives the smallest error is permanently swapped. Until swapping signs no longer decreases the error

 Requirements: No special requirements Platforms: Matlab Keyword: 00125i,  Absresult,  Angleresult,  Result,  Solver,  Vectors Users rating: 0/10