This function SEPARATEKERNEL will separate ( do decomposition of ) any
2D, 3D or nD kernel into 1D kernels. Of course only a subset of Kernels
are separable such as a Gaussian Kernel, but it will give approximations for nonseparable kernels.
Separating a 3D or 5D image filter into 1D filters will give an large
speedup 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 readded 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(M\log(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
.

Alternative SVD Method

Cris Luengo made the same function but then based on the SVD of 2D kernels. His algorithm is faster with large kernels (because SVD is an buildin function), an deals more elegant with negative and zero values.
http://www.mathworks.com/matlabcentral/fileexchange/28238kerneldecomposition
.
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Question

If someone knows the mathematical / analytical solution for finding the signs of the 1D filter kernels, please let me know.
1.2.0.0  Solved zero value problems 

1.1.0.0  Changed text on website 
Inspired: Kernel decomposition
Create scripts with code, output, and formatted text in a single executable document.
dai zhengguo (view profile)
Hi DirkJan. Your code is excellent to be used. However, there is a small bug here, if you use the test code below:
for i = 1:20
v = [1; 2; 1] ;
h = [1 0 1] ;
s = v*h ;
[K,KN,err] = SeparateKernel( s ) ;
err
end
Please check it.
Cris Luengo (view profile)
Hi DirkJan. That's an elegant solution to the zerovalue problem.