Question

I am trying to build a block tridiagonal matrix in Fortran. Now I have this piece of code that would deal with just the matrices that are placed in the main diagonal of the A_matrix, one new matrix for every step in i.

do i = gs+1, total_mesh_points 

    start_line = (3*i)-2
    start_colu = (3*i)-2
    final_line = (3*i)
    final_colu = (3*i)

    do ii = 1, 3
    do jj = 1, 3
        A_matrix(start_line:final_line,start_colu:final_colu) = &
            impflux(ii,jj)
    end do
    end do

end do

Here my A_matrix(i,j) is a big matrix that will receive another three by three matrix (impflux) in its main diagonal. Note that for each step in i I will have a new impflux matrix that needs to be positioned in the main diagonal of the A_matrix.

I can't think in a more simple solution for this problem. How people usually build block diagonal matrices in Fortran ?

Answer 1

这是构建块三对角矩阵的一种方法。我不确定在一些着名的库之外是否有通常的方式。这是一个程序，我会把它交给你把它变成一个函数。

PROGRAM test

  USE iso_fortran_env

  IMPLICIT NONE

  INTEGER :: k    ! submatrix size
  INTEGER :: n    ! number of submatrices along main diagonal
  INTEGER :: ix   ! loop index

  ! the submatrices, a (lower diagonal) b (main diagonal) c (upper diagonal)
  REAL(real64), DIMENSION(:,:,:), ALLOCATABLE :: amx, bmx, cmx

  ! the block tridiagonal matrix
  REAL(real64), DIMENSION(:,:), ALLOCATABLE :: mat_a

  k = 3  ! set these values as you wish
  n = 4

  ALLOCATE(amx(n-1,k,k), bmx(n,k,k), cmx(n-1,k,k))
  ALLOCATE(mat_a(n*k,n*k))

  mat_a = 0.0

  ! populate these as you wish
  amx = 1.0
  bmx = 2.0
  cmx = 3.0

  ! first the lower diagonal
  DO ix = 1,k*(n-1),k
     mat_a(ix+k:ix+2*k-1,ix:ix+k-1) = amx(CEILING(REAL(ix)/REAL(k)),:,:)
  END DO

  ! now the main diagonal
  DO ix = 1,k*n,k
     mat_a(ix:ix+k-1,ix:ix+k-1) = bmx(CEILING(REAL(ix)/REAL(k)),:,:)
  END DO

  ! finally the upper diagonal
  DO ix = 1,k*(n-1),k
     mat_a(ix:ix+k-1,ix+k:ix+2*k-1) = cmx(CEILING(REAL(ix)/REAL(k)),:,:)
  END DO

END PROGRAM test

请注意，这里没有任何错误检查，我只做了一些测试。

一个明显的替代方案是仅在mat_a行循环，在同一次迭代中插入amx，bmx，cmx，但这需要特殊的处理第一次和最后一次迭代，可能看起来更复杂。至于性能，如果对你进行一些测试很重要。

另请注意，这会产生密集的矩阵。如果矩阵变得非常大，那么只存储对角元素的方法可能更有用。这使我们走向它们的衍生类型和操作，这是另一个问题。

Build a block tri-diagonal matrix

1 个答案: