LAPACK: Are operations on packed storage matrices faster?

问题

I want to tridiagonalize a real symmetric matrix using Fortran and LAPACK. LAPACK basically provides two routines, one operating on the full matrix, the other on the matrix in packed storage. While the latter surely uses less memory, I was wondering if anything can be said about the speed difference?

回答1:

It's an empirical question, of course: but in general, nothing comes for free, and less memory/more runtime is a pretty common tradeoff.

In this case, the indexing for the data is more complex for the packed case, so as you traverse the matrix, the cost of getting your data is a little higher. (Complicating this picture is that for symmetric matrices, the lapack routines also assume a certain kind of packing - that you only have the upper or lower component of the matrix available).

I was messing around with an eigenproblem earlier today, so I'll use that as a measurement benchmark; trying with a simple symmetric test case (The Herdon matrix, from http://people.sc.fsu.edu/~jburkardt/m_src/test_mat/test_mat.html ), and comparing ssyevd with sspevd

$ ./eigen2 500
 Generating a Herdon matrix: 
 Unpacked array:
 Eigenvalues L_infty err =   1.7881393E-06
 Packed array:
 Eigenvalues L_infty err =   3.0994415E-06
 Packed time:   2.800000086426735E-002
 Unpacked time:   2.500000037252903E-002

$ ./eigen2 1000
 Generating a Herdon matrix: 
 Unpacked array:
 Eigenvalues L_infty err =   4.5299530E-06
 Packed array:
 Eigenvalues L_infty err =   5.8412552E-06
 Packed time:   0.193900004029274     
 Unpacked time:   0.165000006556511  

$ ./eigen2 2500
 Generating a Herdon matrix: 
 Unpacked array:
 Eigenvalues L_infty err =   6.1988831E-06
 Packed array:
 Eigenvalues L_infty err =   8.4638596E-06
 Packed time:    3.21040010452271     
 Unpacked time:    2.70149993896484

There's about an 18% difference, which I must admit is larger than I expected (also with a slightly larger error for the packed case?). This is with intel's MKL. The performance difference will depend on your matrix in general, of course, as eriktous points out, and on the problem you're doing; the more random access to the matrix you have to do, the worse the overhead would be. The code I used is as follows:

program eigens
      implicit none

      integer :: nargs,n  ! problem size 
      real, dimension(:,:), allocatable :: A, B, Z
      real, dimension(:), allocatable :: PA
      real, dimension(:), allocatable :: work
      integer, dimension(:), allocatable :: iwork
      real, dimension(:), allocatable :: eigenvals, expected
      real :: c, p
      integer :: worksize, iworksize
      character(len=100) :: nstr
      integer :: unpackedclock, packedclock 
      double precision :: unpackedtime, packedtime
      integer :: i,j,info

! get filename
      nargs = command_argument_count()
      if (nargs /= 1) then
          print *,'Usage: eigen2 n'
          print *,'       Where n = size of array'
          stop
      endif
      call get_command_argument(1, nstr)
      read(nstr,'(I)') n
      if (n < 4 .or. n > 25000) then
          print *, 'Invalid n ', nstr
          stop
      endif


! Initialize local arrays    

      allocate(A(n,n),B(n,n))
      allocate(eigenvals(n)) 

! calculate the matrix - unpacked

      print *, 'Generating a Herdon matrix: '

      A = 0.
      c = (1.*n * (1.*n + 1.) * (2.*n - 5.))/6.
      forall (i=1:n-1,j=1:n-1)
        A(i,j) = -1.*i*j/c
      endforall
      forall (i=1:n-1)
        A(i,i) = (c - 1.*i*i)/c
        A(i,n) = 1.*i/c
      endforall
      forall (j=1:n-1)
        A(n,j) = 1.*j/c
      endforall
      A(n,n) = -1./c
      B = A

      ! expected eigenvalues
      allocate(expected(n))
      p = 3. + sqrt((4. * n - 3.) * (n - 1.)*3./(n+1.))
      expected(1) = p/(n*(5.-2.*n))
      expected(2) = 6./(p*(n+1.))
      expected(3:n) = 1.

      print *, 'Unpacked array:'
      allocate(work(1),iwork(1))
      call ssyevd('N','U',n,A,n,eigenvals,work,-1,iwork,-1,info)
      worksize = int(work(1))
      iworksize = int(work(1))
      deallocate(work,iwork)
      allocate(work(worksize),iwork(iworksize))

      call tick(unpackedclock)
      call ssyevd('N','U',n,A,n,eigenvals,work,worksize,iwork,iworksize,info)
      unpackedtime = tock(unpackedclock)
      deallocate(work,iwork)

      if (info /= 0) then
           print *, 'Error -- info = ', info
      endif
      print *,'Eigenvalues L_infty err = ', maxval(eigenvals-expected)


      ! pack array

      print *, 'Packed array:'
      allocate(PA(n*(n+1)/2))
      allocate(Z(n,n))
      do i=1,n 
        do j=i,n
           PA(i+(j-1)*j/2) = B(i,j)
        enddo
      enddo

      allocate(work(1),iwork(1))
      call sspevd('N','U',n,PA,eigenvals,Z,n,work,-1,iwork,-1,info)
      worksize = int(work(1))
      iworksize = iwork(1)
      deallocate(work,iwork)
      allocate(work(worksize),iwork(iworksize))

      call tick(packedclock)
      call sspevd('N','U',n,PA,eigenvals,Z,n,work,worksize,iwork,iworksize,info)
      packedtime = tock(packedclock)
      deallocate(work,iwork)
      deallocate(Z,A,B,PA)

      if (info /= 0) then
           print *, 'Error -- info = ', info
      endif
      print *,'Eigenvalues L_infty err = ', &
      maxval(eigenvals-expected)

      deallocate(eigenvals, expected)


      print *,'Packed time: ', packedtime
      print *,'Unpacked time: ', unpackedtime


contains
    subroutine tick(t)
        integer, intent(OUT) :: t

        call system_clock(t)
    end subroutine tick

    ! returns time in seconds from now to time described by t
    real function tock(t)
        integer, intent(in) :: t
        integer :: now, clock_rate

        call system_clock(now,clock_rate)

        tock = real(now - t)/real(clock_rate)
    end function tock

end program eigens

来源：https://stackoverflow.com/questions/8941678/lapack-are-operations-on-packed-storage-matrices-faster

标签

fortran

lapack