Comparing LS and TLS
- From: spasmous <spasmous@xxxxxxxxx>
- Date: Tue, 25 Mar 2008 12:38:27 -0700 (PDT)
I wrote some code to test the total least squares method. Results were
disappointing: for 10000 trials with different added Gaussian noise
the least squares produces the smallest error. When there are multiple
right hand side vectors there is a specific algorithm for TLS, which I
also compared (Ref Golub & van Loan). Results of this were worse than
LS and randomly better or worse than single RHS TLS.
I'm curious why is the special purpose TLS for multiple RHS not the
best? And why LS is superior to either TLS approach?!?
% test what is best: ls, tls or tls with multiple RHS
function test_tls()
m = 10; % rows
n = 4; % cols
k = 2; % RHS vectors
sd = 0.01; % added noise std. dev.
trials = 10000; % number of trials
% matrices (condition 100)
Atrue = full(sprand(m,n,0.5,1/100));
xtrue = ones(n,k);
btrue = Atrue*xtrue;
err = zeros(n,k,trials,3);
% loop here
for i = 1:trials
% add noise
A = Atrue + sd*randn(m,n);
b = btrue + sd*randn(m,k);
% least squares
err(:,:,i,1) = pinv(A)*b - xtrue;
% tls separate RHS
for j = 1:k
err(:,j,i,2) = tls(A,b(:,j)) - xtrue(:,j);
end
% tls multiple RHS
err(:,:,i,3) = tls(A,b) - xtrue;
end
% analysis
err = reshape(err,n*k*trials,3);
disp(['LS error = ' num2str(norm(err(:,1)))])
disp(['TLS1 error = ' num2str(norm(err(:,2)))])
disp(['TLS2 error = ' num2str(norm(err(:,3)))])
% total least squares function
function X = tls(A,B)
% Solves the linear equation AX=B using
% total least squares.
% Ref: Golub & van Loan 3rd ed (p 596)
%
% A is the matrix (m x n)
% B is the RHS matrix (m x k)
[m n] = size(A);
if size(B,1)~=m
error('A, B size mis-match')
end
% augmented matrix
C = [A B];
[U S V] = svd(C,0);
% solve
V12 = V(1:n,1+n:end);
V22 = V(1+n:end,1+n:end);
X = -V12/V22;
.
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