This function does the decomposition of a separable nD kernel into
its 1D components, such that a convolution with each of these
components yields the same result as a convolution with the full nD
kernel, at a drastic reduction in computational cost.
SYNTAX:
=======
[K1,KN,ERR] = DECOMPOSE_KERNEL(H)
computes a set of 1D kernels K1{1}, K1{2}, ... K1{N} such that the
convolution of an image with all of these in turn yields the same
result as a convolution with the N-dimensional input kernel H:
RES1 = CONVN(IMG,H);
RES2 = IMG;
FOR II=1:LENGTH(K1)
RES2 = CONVN(RES2,K1{II});
END
KN is the reconstruction of the original kernel H from the 1D
kernels K1, and ERR is the sum of absolute differences between H
and KN.
The syntax mimics Dirk-Jan Kroon's submission to the FileExchange
(see below).
EXPLANATION:
============
In general, for a 2D kernel H, the convolution with 2D image F:
G = F * H
is identical to the convolution of the image with column vector H1
and convolution of the result with row vector H2:
G = ( F * H1 ) * H2 .
In MATLAB speak, this means that
> CONV2(F,H) == CONV2(CONV2(F,H1),H2)
Because of the properties of the convolution,
( F * H1 ) * H2 = F * ( H1 * H2 ) ,
meaning that the convolution of the two 1D filters with each other
results in the original filter H. And because H1 is a column vector
and H2 a row vector,
H = H1 * H2 = H1 H2 .
Thus, we need to find two vectors whose product yields the matrix H.
In MATLAB speak we need to solve the equation
> H1*H2 == H
The function in the standard MATLAB toolbox, FILTER2, does just
this, and it does it using singular value decomposition:
U S V' = H ,
H1(i) = U(i,1) S(1,1)^0.5 ,
H2(i) = V(i,1)* S(1,1)^0.5 . (the * here is the conjugate!)
Note that, if the kernel H is separable, all values of S are zero
except S(1,1). Also note that this is an under-determined problem,
in the sense that
H = H1 H2 = ( a H1 ) ( 1/a H2 ) ;
that is, it is possible to multiply one 1D kernel with any value
and compensate by dividing the other kernel with the same value.
Our solution will, in effect, just choose one of the infinite number
of (equivalent) solutions.
To extend this concept to nD, what we need to understand is that it
is possible to collapse all dimensions except one, obtaining a 2D
matrix, and solve the above equation. This results in a 1D kernel
and an (n-1)D kernel. Repeat the process until all you have is a
set of 1D kernels and you're done!
This function is inspired by a solution to this problem that
Dirk-Jan Kroon posted on the File Exchange recently:
http://www.mathworks.com/matlabcentral/fileexchange/28218-separate-kernel-in-1d-kernels
His solution does the whole decomposition in one go, by setting up
one big set of equations. He noted a problem with negative values,
which produce complex 1D kernels. The magnitude of the result is
correct, but the sign is lost. He needs to resort to some heuristic
to determine the sign of each element. What he didn't notice (or
didn't mention) is the problem that his solution has with 0 values.
The SVD solution doesn't have this problem, although it sometimes
does produce a slightly worse solution. For example, in the first
example below, Dirk-Jan Kroon's solution is exact, whereas this one
produces a very small error. Where Dirk-Jan Kroon's solution cannot
find the exact solution, this algorithm generally does better.
EXAMPLES:
=========
Simplest 5D example:
H = ones(5,7,4,1,5);
[K1,~,err] = SeparateKernel(H); % D.Kroon's submission to FileEx.
err
[k1,~,err] = decompose_kernel(H);
err
2D example taken from Dirk-Jan Kroon's submission:
a = permute(rand(4,1),[1 2 3])-0.5;
b = permute(rand(4,1),[2 1 3])-0.5;
H = repmat(a,[1 4]).*repmat(b,[4 1]);
[K1,~,err] = SeparateKernel(H);
err
[k1,~,err] = decompose_kernel(H);
err
2D example for which Dirk-Jan Kroon's solution has problems:
H = [1,2,3,2,1]'*[1,1,3,0,3,1,1];
[K1,~,err] = SeparateKernel(H);
err
[k1,~,err] = decompose_kernel(H);
err
3D example that's not separable:
H = rand(5,5,3);
[K1,~,err] = SeparateKernel(H);
err
[k1,~,err] = decompose_kernel(H);
Example to apply a convolution using the decomposed kernel:
img = randn(50,50,50);
h = ones(7,7,7);
tic;
res1 = convn(img,h);
toc
k1 = decompose_kernel(h);
tic;
res2 = img;
for ii=1:length(k1)
res2 = convn(res2,k1{ii});
end
toc
rms_diff = sqrt(mean((res1(:)-res2(:)).^2))
Cris Luengo (2020). Kernel decomposition (https://www.mathworks.com/matlabcentral/fileexchange/28238-kernel-decomposition), MATLAB Central File Exchange. Retrieved .
Inspired by: Separate Kernel in 1D kernels
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