Fastest way to compute J' * J, where J is sparse

I have a sparse rectangular matrix, J, for which I want to compute:

>> H = J' * J;

It's a bit slow (transpose is taking 5s and the matrix multiplication 9s), and given this is a special and very common case of a transpose and multiply, I was wondering if MATLAB had a faster way, e.g. one which avoids an explicit transpose.

[John D'Errico]

I imagine there are tools to do this operation, but they are down at the blas level. I can see as how it would be nice to have a tool that would explicitly compute J'*J, and do so efficiently though, taking advantage of the symmetry of the problem and avoiding the transpose.

[Matt J]

and given this is a special and very common case of a transpose and multiply

It's common to encounter an expression like that, but less common to have to implement it directly. So, it makes me want to ask why you think you have to. For example, I often see it when people make the mistake of solving

    J*x=y

using an explicit pseudo-inverse inv(J'*J)*J'*y as opposed to J\y.

[John D'Errico]

Matt is right, that it is often perhaps a bad idea, IF you are using it for the wrong reason. However, there are cases one can think of, where J'*J is useful, even necessary. Even there however, there might be alternatives. For example, to compute a covariance matrix, you could generate inv(J'*J). But you could also form a QR factorization for J, a Q-less QR in fact. Then

J'*J = R'*R

And since R is triangular, performing that inverse is far more efficiently done.

[Mohammad Abouali]

Between J\y and inv(J'*J)*J'*y, definitely J\y is better. But an alternative is (J'*J) \ (J'*y). This last form actually sometimes help the results, particularly if J is MxN and M>>N and the condition number is not that good. Note that mldivide(A,B) or (A\B) only uses QR-solver if A is not square matrix. However if A is square matrix there are much more options and the inverse solution is selected much more carefully. So sometimes it helps to use (J'*J)\ (J'*y) to deal with an square matrix. Refer to:

http://www.mathworks.com/help/matlab/ref/mldivide.html?refresh=true

go down to the bottom of the page and check the flowchart under the algorithm.

[Oliver Woodford]

Matt, J\y minimizes a different cost to (J'*J)\(J'*y), so when your problem is over constrained (i.e. J is rectangular) it will give a different answer. Note that I did state explicitly that my matrix J was rectangular. If you're doing Gauss-Newton optimization, as I am, then you want to do the latter. I can assure you it's fairly common. Plus the latter approach can be faster. E.g. in my case

>> tic; A = J \ y; toc
Elapsed time is 7.075738 seconds.
>> tic;  J_ = J'; B = (J_ * J) \ (J_ * y); toc
Elapsed time is 1.180669 seconds.
>> max(abs(A-B))
ans = 1.1125e+09

[Torsten]

Maybe J does not have full rank N and there are infinitely many solutions that minimize |J*x-y|_2 .

If this is not the case, I can not find a reason why A=J\y should give a solution different from B=(J'*J)\(J'*y).

Best wishes

Torsten.

[Oliver Woodford]

Torsten, Matt: Sorry, you're both right. Although A=J\y minimizes |J*x-y|_2, and B=(J'*J)\(J'*y) minimizes |(J*J')^0.5*(J*x-y)|_2, the weight matrix (J*J')^0.5 doesn't change the location of the minimum, only the shape of the cost.

In my case J is 3192027 x 3225 with rank 3034, which explains the large difference in solutions. In this case you require the regularization that the weight matrix provides in order to find the correct solution for the Gauss-Newton step. Mohammad's point is relevant here. Plus it's faster.

[Matt J]

This last form actually sometimes help the results, particularly if J is MxN and M >>N and the condition number is not that good.

@Mohammad, did you instead mean to say "if the condition number is not that bad"? The condition number of J'*J is always the square of J, making the conditioning worse.

    >> J=rand(3000,100);

    >> cond(J)

    ans =

       20.9806

    >> cond(J'*J)

    ans =

      440.1848

In this case you require the regularization that the weight matrix provides in order to find the correct solution for the Gauss-Newton step.

@Oliver, No, regularization still doesn't require explicit construction of J'*J. If you are trying to do the regularized mldivide operation,

    (J'*J+beta*speye(N))\ J'*y

then you could equivalently do the following, and it would be better conditioned,

    [J;sqrt(beta)*speye(N)]\[y; zeros(N,1)]

If you are still willing to trade conditioning for speed, there may be further brainstorming to be done, but then please provide more info about J, e.g. the density nnz(J)/numel(J) so that we can perhaps simulate the problem data.

[Oliver Woodford]

Matt, I don't want to overcomplicate things here. I'd like an answer to the question posed, if there is one. If I'm asking the wrong question for my particular work then on my head be it.

[Mohammad Abouali]

@Matt: Good catch. Yes, I meant if the condition number is good or not that bad; but apparently I mixed the two sentences in one, completely changing the meaning.

[Oliver Woodford]

I've just discovered that MATLAB also gives me grief if J is 2^30x1024, even though it's sparse and H will only be 1024x1024. The problem is J' apparently gets explicitly created, and it has too many columns to fit in memory. I'm distinctly unimpressed. Writing a mex file right now to fix this!

[Matt J]

I don't think you ever responded to John's suggestion about using a QR decomposition. For sparse matrices, you can do a Q-less decomposition

    R=qr(J,0);

and then exploit the fact that J'*J=R'*R.

[Oliver Woodford]

Thanks, Matt J. I'll respond now. What is the complexity of qr(J, 0)?

[Matt J]

I don't know the complexity. It will have to be tested. I imagine this could be advantageous only for certain kinds of sparsity structure.

[Oliver Woodford]

For my size of matrix, qr(J,0) is vastly slower (i.e. several orders of magnitude; actually it crashed MATLAB after a minute or so) than a mex file that computes J'*J (a second or so). The mex file is a bit slower than standard MATLAB for smaller sparse matrices, but at least works for very tall matrices (which MATLAB produces an "Matrix too large" error on when transposing).

Answer 1

[Azzi Abdelmalek]

c=sparse(J);
H=full(c*c');

[Oliver Woodford]

The matrix is already stored as a sparse matrix.

[John D'Errico]

Um, J is already assumed to be in sparse form, and one would definitely not want to compute a full result when working with sparse matrices. Finally, you put the transpose on the wrong term, computing J*J', not J'*J.

Answer 2

[Matt J]

I don't think there's anything available to accelerate an exact calculation of J'*J for general J. However, if you know in advance that J'*J happens to be banded to diagonals -k:k for small k (or if it can be approximated as such), then it might help to compute the 2*k+1 non-trivial diagonals individually. You can do so without transposition as below.

    [m,n]=size(J);
    k=2;
    kc=k+1;
    tic;

     B=zeros(n);
     B(:,kc)=sum(J.^2);
     for i=1:k
       tmp=sum(J(:,1:end-i).*J(:,i+1:end));
       B(1:end-i,kc-i)=tmp;
       B(i+1:end,kc+i)=tmp;
     end

      result=spdiags(B,-k:k,n,n);
    toc;

Whether this is actually faster will probably depend on the specifics of J. If nothing else, it spares you the large memory consumption of holding wide sparse matrices such as J' in RAM

>> J=sparse(m,n); Jt=J'; whos J Jt

    Name            Size                    Bytes  Class     Attributes

    J         3192027x3225                  25824  double    sparse    
    Jt           3225x3192027            25536240  double    sparse

Replacing J'*J by a banded approximation is something I haven't tried myself with Gauss-Newton specifically, but the role of J'*J is already as an approximation there, so I think it could work. Other minimization algorithms tend to be robust to small errors in the derivatives.