Re: Regularized lsqr
- From: fas <faisalmufti@xxxxxxxxx>
- Date: Sat, 07 Jul 2007 05:19:30 -0700
On Jul 7, 8:55 pm, "John D'Errico" <woodch...@xxxxxxxxxxxxxxxx> wrote:
fas wrote:
Thanks John for your reply, but I would like to know few things.
I was required to solve the following cost function:
|| A_k *x - b_k || + Sum_k[norm(d)* (x'A_k' A_k x)] + lambda^2 ||
L*x ||
The first problem is that the sum of these norms
CANNOT be written in a form that lsqr can solve.
lsqr will minimize a sum of squares. A vector
2-norm (Euclidean norm) is a sqrt of a sum of
squares. So the sum of several normed terms will
not be truly minimized using lsqr.
If you are willing to transform your problem into
a (weighted) sum of the SQUARES of these norms,
THEN you can apply lsqr to the problem.
My guess is that you were really given the problem
of solving the sum of the squares of norms anyway.
The hint is the lambda^2 term. This typically arrives
from squaring a norm.
On the other hand, you have not squared the norm(d)
term. So I'm not entirely sure what is meant.
Next...
|| A_k *x - b_k || + Sum_k[norm(d)* (x'A_k' A_k x)] + lambda^2 ||
L*x ||
Your first normed term here suggests that A is an
nxp array, x is a px1 column vector. So if A_k refers
to the k'th row of A, then (A_k*x - b_k) is the residual
from an approximation problem. Should there be a
sum around this? Do you really wish to form this:
|| A*x - b ||
Or something else?
Please be more accurate with your mathematics.
John
Hi
I am sorry I made a mistake, here is the correct version
Sum_k [ || A_k *x - b_k || ] + Sum_k[w* (x'A_k' A_k x)] + lambda^2 ||
L*x ||
we have k copies of A (matrices) and b(vectors) but not only one copy
of 'L' operator. So I have transformed all A_k as A (stacked copies of
A_k) and same goes for b_k. Then I think we can write it as,
|| A*x - b || + w* || A*x || + lambda^2 || L*x ||
where w is the weighted scalar for each k
A = rand(20000,1000);
b = rand(20000,1);
lambda = 0.1;
A = [A;sqr(w)*A;lambda*eye(1000)];
b = [b;zeros(21000,1)];
x = lsqr(A,b);
So now what is your opinion ? Do I have to minimize in two steps i.e
once I have minimized the first two terms and then minimized the
result with the third term or is it possible to minimize all together.
Thanks,
.
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