MATLAB: Expression representation in an Iteration

optimizationOptimization Toolbox

I'd love to know how i can express the expression

in a loop. n is the iteration number. I know it may seem trivial but i've been cracking my skull over it. My approach, which I figure is wrong anyways, was to index the second term with n. But the iteration number is greater than the the matrix elements and i get an error. Here is my attempt at the issue.

w=[randn(2,1)+ij*randn(2,1)];
W_k=w*ctranspose(w);
for n=1:N;
    w=[randn(2,1)+ij*randn(2,1)];
    W_k1=w*ctranspose(w);
    t=W_k - W_k1;
end

Is this doing what the expression requires?

Best Answer

W_k = cell(N,1);
w=[randn(2,1)+ij*randn(2,1)];
W_k{1} = w*ctranspose(w);
for n=1:N;
    w = randn(2,1)+ij*randn(2,1);
    W_k{n+1} = w*ctranspose(w);
    t = W_k{n} - W_k1{n+1};
    %do something with t
end

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There has been a lot of research on how to deal with floating-point error in the Lanczos algorithm, see the Wikipedia page for some discussion.

However, even in exact arithmetic, you should only expect beta(j) to become zero reliably after having performed n iterations (where the matrix A is of size n-by-n). Now usually, the goal of the Lanczos algorithm is to converge in significantly fewer iterations than that, so this is not usually a good convergence criterion.

It's still a good idea to check if beta(j) has become zero, because this would be a breakdown condition - if beta(j) becomes zero, it means that a complete eigenspace of A has been constructed. Either all information you need is already contained in this space, or you need to restart another Lanczos iteration with a new random vector.

MATLAB: How to speed up a parfor loop with large broadcast variables

It seems from your code that you should be able to slice your large variables idx and W. This would mean that each element of those arrays is transmitted only to the worker that needs it, rather than to all workers. Something like this:

% Generate dummy data as cell arrays
idx_vals = arrayfun(@rand, 1:10, 'UniformOutput', false);
W = arrayfun(@rand, 1:10, 'UniformOutput', false);
n = length(idx_vals);
val = zeros(1, n);
parfor k = 1:n
    idx_k = idx_vals{k};
    W_k = W{k};
    val(k) = max(abs(eig(idx_k * W_k)));
end

Best Answer

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