Particle Swarm Optimization to solve matrix inversion
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Nurulhuda Ismail
on 7 Jan 2019
Commented: Walter Roberson
on 8 Jan 2019
Hi everyone,
I am Huda, and I am working with matrix inversion. I am really new with PSO and I would like to know is it possible to use PSO for a large size matrix inversion (eg., 16x16, or 64x64)? From a linear equation, Ax=b, we can find x by several techniques such as Gauss-Jordan elimination, and we know that AA^(-1) =I. Thus, I wonder if PSO can solve this problem by computing A^(-1) iteratively.
If yes, can you share with me any material that can guide me to start my work. But, If matrix inversion with PSO will give bad result, can you advise any technique that suitable to use?
Thank you for your response.
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Accepted Answer
Walter Roberson
on 8 Jan 2019
N = 8;
rng(12345);
M = randi([-50 50], N, N);
obj = @(v) sum(sum((M * reshape(v,N,N) - eye(N)).^2));
N2 = N.^2;
LB = []; UB = [];
options = optimoptions(@particleswarm, 'Display', 'final', 'MaxIterations', 500*N2, 'PlotFcn', @pswplotbestf );
tic
[bestinv, fval] = particleswarm(obj, N2, LB, UB, options);
time_needed_ps = toc;
Minvps = reshape(bestinv, N, N);
tic
Minv = inv(M);
time_needed_numeric = toc;
residue = sum((Minvps(:) - Minv(:)).^2);
fprintf('\nResidue between particle swarm inverse and numeric inverse is: %.6g\n', residue);
fprintf('\nDifference between particle swarm inverse and numeric inverse is: \n\n')
disp(Minvps - Minv)
fprintf('\ntime required for particle swarm partial inverse: %.6gs\n', time_needed_ps);
fprintf('time required for numeric inverse: %.6gs\n', time_needed_numeric);
Approximate time multiplication for the particle swarm approach: 620000 times more expensive. The calculated inverse will not necessarily be especially close -- it is not bad for 5 x 5, but by 8 x 8 the residue is still about a million.
3 Comments
Walter Roberson
on 8 Jan 2019
On my system inv() of a 64x64 array takes less than 1e-4 seconds. Any optimization routine would have to finish faster than that to be worth bothering with.
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