why tis matrix is sparse?

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Stefanos Petsios
Stefanos Petsios on 17 Jan 2019
Edited: John D'Errico on 17 Jan 2019
Hello everyone, why is this matrix is classyfied as sparse although its picture dont seem like sparse?

Accepted Answer

John D'Errico
John D'Errico on 17 Jan 2019
Edited: John D'Errico on 17 Jan 2019
You can define anything you want to be sparse if you so desire. So sparse(ones(1000)) will produce a sparse matrix. ;-)
But seriously, this matrix really is reasonably sparse. Just read what is shown on the page you direct to!
We see that it is shown to be an 1899x1899 square matrix. There are 20296 non-zeros in the matrix.
20296/1899
ans =
10.688
So on average, roughly 10.7 non-zeros per row of a matrix, where each row has 1899 elements.
20296/1899/1899
ans =
0.0056281
Total density of around 0.6% non-zeros. Is that matrix sparse? Well, yes, it is. Not massively so, but it is sparse. Since I don't have that matrix, I'll do a little memory comparison on a random one with the same density.
As = sprand(1899,1899,0.0056);
Af = full(As);
whos As Af
Name Size Bytes Class Attributes
Af 1899x1899 28849608 double
As 1899x1899 337568 double sparse
So we see that storing the matrix as sparse gives me a decrese in memory of almost a factor of 100-1. (Sparse matrix storage also requires we store the location of those elements, so it is not quite as efficient as we might like.)
b = rand(1899,1);
timeit(@() Af*b)
ans =
0.0023482
timeit(@() As*b)
ans =
6.1858e-05
And working with the matrix with a matrix multiply does give a significant savings.
Much of the time, when you work with a sparse matrix, you will perform a decomposition of some sort. Odds are that will not produce a speed increase here, because the matrix itself is not a nicely tightly banded matrix. I think much of the time, people think of a sparse matrix in the form of something like a tridiagonal matrix. A matrix factorization of a tridiagonal matrix will produce no fill-in at all. On the matrix you show, if we tried to do a Cholesky or LU factoriation, for example, the result will be a virtually completely full triangular matrix.
tic,[L,U] = lu(Af);toc
Elapsed time is 0.205769 seconds.
tic,[L,U] = lu(As);toc
Elapsed time is 3.032499 seconds.
As you see, not all computations on such a matrix will see a gain in speed. Here, we see the result ends up as a nearly full triangular matrix for the factors. So treating that matrix as sparse, IF you intend to do a decomposition of the matrix will be a time loss, because the overhead from the sparse algorithms is too costly.
nnz(L)
ans =
1177842
nnz(U)
ans =
1208969
numel(L)
ans =
3606201
That is to be expected on this matrix. It does not mean the original was not sparse. It is indeed sparse. Just perhaps not as massively sparse as you may want a matrix to be. Sometimes sparse must be measured by the eye of the beholder, in proper context.
spy(As)

More Answers (1)

Torsten
Torsten on 17 Jan 2019
Calculate
number_of_nonzeros/(number_of_rows*number_of_columns)
and you'll see that the matrix is sparse.

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