How to create a tri diagonal matrix as 1D - c++

Question

So for my application I would need to create a tri diagonal matrix. This is easy to do with any language, you loop through all rows and columns, then set the main diagonal values, the sub diagonal values and the super diagonal values. Usually, this is performed on a 2d array.
For my application, I need to create a 1d array of "tridiagonal". Otherway to say this is: take the 2d tridiagonal matrix then turn it into 1d. I can just start with 2d then write some functions that convert 2d array to 1d array. This, I can do. I would like to know if we ca go directly to a 1D "tridiagonal"? For example, say the 2D array is 10*10, then my 1D array would be 100 elements long, then I would need to figure out which index is the main, super and sub diagonal.
Is it possible to do this? Please let me know and thank you

Answer 1

The elements on the main diagonal are at indexes (i, i) and there are n of them; the supra- and infra- diagonals at (i, i-1) and (i, i+1) and there are n-1 of them (i starts at 2 and ends at n-1 respectively).
An option is to use three vectors and store the elements at the respective indexes i in those three vectors.
You can also pack all values in a single vector of length 3n (or 3n-2 if you want to spare space). Add n or 2n to the index, depending on the diagonal you want to address. For an element (i, j), the index of the diagonal is given by j-i+2.

Answer 2

You can just look at your 1D array using a 2D array pointer. Fortran:
integer, target :: A(100)
integer, pointer :: B(:,:)
B(1:10,1:10) => A
B = 0
do i = 1, 10
B(i,i) = 1
end do
print '(*(1x,g0))', A
end
> gfortran diag1d.f90
> ./a.out
1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1
In C++ the casting is easy as well.

Answer 3

Please consider also:
#YvesDaoust's answer, because proposes a better storage strategy: instead of storing all the elements of the tridiagonal matrix, store just the non-zero. You could write a derived-type to encapsulate behavior, if it worths.
#VladmirF's answer, because a pointer association maybe a better approach (depending on your use case), than copying over all the array through reshape, if all you want is a temporarily indexing change for convenience while working on data.
Said that, let's go to my answer.
Populating a tridiagonal matrix is not a different problem than constructing any matrix. I don't think it really matters here. Just bear in mind that Fortran stores arrays in Column Major order, 1 based index.
Changing the shape of your data is easy and obvious, if storage is contiuous. You can make a pointer association or transferring it to a new variable with reshape.
Extracting the main, super and sub diagonal is also a trivial problem, and can be done with simple manipulation of the array index triplet. Look:
program tridiagonal
implicit none
integer, parameter :: n = 4
integer :: A(n, n), B(n**2), main(n), sub(n-1), sup(n-1)
A(1,:) = [1, 4, 0, 0]
A(2,:) = [3, 4, 1, 0]
A(3,:) = [0, 2, 3, 4]
A(4,:) = [0, 0, 1, 3]
! Remember, colum major
B = reshape(A, shape(B)) ! 1, 3, 0, 0, 4, 4, 2, 0, 0, 1, 3, 1, 0, 0, 4, 3
main = B( 1:n**2:n+1) ! 1, 4, 3, 3
sub = B( 2:n**2:n+1) ! 3, 2, 1
sup = B(n+1:n**2:n+1) ! 4, 1, 4
end