Image processing: Minimizing function (regularized least square problem)
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Hello,
I'm trying to minimize this function (by A):
argmin A (|L(A)|^2 + a*||A-B||^2*)
where:
- A is a MxN image
- L is the Laplacian Operator
- .|| is the usual norm operator
- a is a weight parameter
- B is a matrix of size (M+2*k)xN where k is an integer parameter.
- * indicates that we just consider the pixels in the boundary (we want to preserve in A the pixels in the boundary of B).
Maybe the problem has a trivial solution, but I'm absolutely blocked.
I will be very grateful for any help provided.
采纳的回答
Below is a direct linear algebraic solution. It works by rewriting the problem as
min || H*A(:)-y ||^2
for an appropriate matrix H and vector y. Whether it will be practical for you depends on the magnitudes of M,N and the power of your computer. Also, you will have to modify sqrt(a)*Imn and B(:) to include only the boundary pixels (just delete the rows corresponding to those pixels).
[M,N]=size(A);
Im=speye(M);
In=speye(N);
Imn=speye(M*N);
Dx=diff(Im,2,1);
Dy=diff(In,2,1);
LA=kron(Dx,In)+kron(Im,Dy); %Laplacian operator
H=[LA; sqrt(a)*Imn];
y=[zeros(size(LA,1),1); B(:) ];
A_solution=reshape( H\y, M,N);
14 个评论
gui_tech
2012-10-1
Matt J
2012-10-1
Not sure how detailed you want me to be. In your original least squares function, the first term has a Laplacian operator applied to A(:). The second term has sqrt(a)*eye(M*N) applied to A(:). The H matrix is just the vertical stacking of these 2 operators.
BTW, I've been assuming that the "usual norm operator" that you're using is the Frobenius norm.
I don't see how do you rewrite the problem. Maybe I have a theorical lack.
Forget about A being a matrix for a moment. Suppose you just had a vector unknown, x, and a two-term lsq cost function like this:
argmin_x norm(P*x-p)^2 +norm(Q*x-q)^2
Can you see that this is equivalent to
argmin_x norm([P;Q]*x - [p;q])^2
and can be solved as [P;Q]\[p;q] ? That's all we're doing here, except in your case the vector is A(:) instead of x.
A-B are just defined in the boundary pixels (only in top and bottom), we want: A(1:k,:)=B(1:k,:) and A((M-k):M,:)=B(M:M+k,:)
Then you would need to modify the code as follows
Amask=false(size(A));
Amask(1:k,M-k:M],:)=true;
Bmask=false(size(B));
Bmask(1:k,M:M+k],:)=true;
H=[LA; sqrt(a)*Imn(Amask,:)]; %definition of H
y=[zeros(size(LA,1),1); B(Bmask) ]; %definition of y
gui_tech
2012-10-1
Matt J
2012-10-1
You need to convert the 8-bit image to double so that MATLAB will use double precision math (not integer math) to do the calculations.
I don't know why you expect the Laplacian operator to be M^2xN^2. The vector A(:) has M*N elements and so the matrix that pre-multiplies it must have M*N columns, not N^2 columns. Similarly, the number of rows should be approximately M*N, since there are M*N pixel locations where the Laplacian gets evaluated. It won't be exactly M*N, because the DIFF command reduces the length of the vector, e.g.
>> diff([1,3,4])
ans =
2 1
One possibility, I edited a line of the code from
y=[zeros(size(LA,1),1); B(Bmask,:) ]; %definition of y
to this
y=[zeros(size(LA,1),1); B(Bmask) ]; %remove colon
Make sure you have the 2nd version.
gui_tech
2012-10-1
ZHUN
2014-1-14
Matt J, kron(Dx,In) and kron(Im,Dy) have different size, how did you add them together please?
OJ27
2017-12-28
if have the following min problem.
I am using young's inequality to say |k|_L1 times(|p|_L2)-||tc||+theta||grad(p)||_L1
I need some help coding this one
OJ27
2017-12-28
in this case, when I create the H, I would replace sqrt(a)*Imn by k, however I don't know what form I have to put k since they are different dimensions. Should I make k sparse? also, since they are different norms, will that change the implementation
Matt J
2017-12-28
OLGA, there is no algebraic solution to the the problem you've written. You will have to use an iterative solver, probably ga(). For ga(), you don't need matrix implementations of the different operators. Just implement the objective function in function form and pass an appropriate function handle to the solver.
更多回答（1 个）
Laurentiu
2014-1-12
0 个投票
gui_tech: I am working on the implementation of the paper as well. Were you able to fix the problem with the low intensities in the solution ? I am also getting these -1*10^3 intensities in A.
Any help would be appreciated.
1 个评论
Possibly, my implementation of the Laplace operator is not what the paper expects. It might be worth trying to use
http://www.mathworks.com/matlabcentral/fileexchange/44669-func2mat-convert-linear-function-to-matrix
to convert whatever routine you normally use to compute L(A) to matrix form.
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