Intel® oneAPI Math Kernel Library
Ask questions and share information with other developers who use Intel® Math Kernel Library.
6981 Discussions

Efficient storage of sparse matrix with constant non-zero elements

Customer__Intel4
Beginner
311 Views
I have a sparse matrix whose non-zero elements are always ones ( = 1). Is there a storage scheme supported by MKL that will save the memory space required for storing the data array and stores only the indices arrays?
0 Kudos
3 Replies
mecej4
Honored Contributor III
311 Views
While I understand your motivation for asking, I feel that there is little use for such a representation, for the following reason.

If any transformation is performed upon or using the special matrix, it will probably cease to have all nonzero values equal to 1. Therefore, the special representation, if it existed, would only apply to the original matrix. It would also be probably necessary to convert (or provide a routine to convert) from the special representation to the usual (ir,jc,v) representation before calling solver routines.
0 Kudos
Customer__Intel4
Beginner
311 Views
Quoting mecej4
While I understand your motivation for asking, I feel that there is little use for such a representation, for the following reason.

If any transformation is performed upon or using the special matrix, it will probably cease to have all nonzero values equal to 1. Therefore, the special representation, if it existed, would only apply to the original matrix. It would also be probably necessary to convert (or provide a routine to convert) from the special representation to the usual (ir,jc,v) representation before calling solver routines.

I will use this special matrix to multiply it with vectors. My program will not alter it at all!
0 Kudos
mecej4
Honored Contributor III
311 Views
> I will use this special matrix to multiply it with vectors. My program will not alter it at all!

In that case, the matrix-vector product calculation can be expressed quite simply using only the ir() and jc() arrays, with the implied value of 1, using a couple of lines of code. For example, to compute u = A . x with A in coordinate storage format :

u(1:n) = 0
do iz = 1, nnz
c u(ir(iz)) = u(ir(iz)) + val(ir(iz),jc(iz)) * x(jc(iz)) ! general case
u(ir(iz)) = u(ir(iz)) + x(jc(iz)) ! special case
end do

0 Kudos
Reply