In Golub and Van Loan's book, Matrix Computations, page 606, it is stated that:

With standard discretizations, 2-dimensional problems can be solved with $O(n^{3/2})$ work and $O(n \log{n})$ fill-in. For 3-dimensional problems, the typical costs are $O(n^2)$ work and $O(n^{4/3})$ fill-in.

Here, they refer to elliptic partial differential equation problems solved on a grid and $n$ refers to the number of unknowns. I guess it is trivial to calculate the cost (probably linear) for one-dimensional problems, but I don't know how to do that. I would appreciate any help for that.


1 Answer 1


It may help to define $N$, the number of discretization points along a 1D edge, and relate it to $n$, the number of unknowns in the system. In 2D on a square grid of points, $n = O(N^2)$. Nested dissection efficiently reduces the sparse system you usually get from discretizations by eliminating levels of "interior" points. The result is a more dense system the size of a few lines of "separator" points, which is of size $O(n^{1/2})$ and can be solved directly in $O(n^{3/2})$ time. The fill-in costs come from the fact that eliminating interior points at each level couples more points together (creating new connections in the matrix, hence fill-in).

The 3D costs come about similarly, except now $n = N^3$, and the separators are now planes with $O(N^2) = O(n^{2/3})$ points. Solving systems involving these planes then gives you the $O(n^2)$ cost).

However, for 1D problems, $n=N$, and you can order unknowns from left to right to get a tridiagonal system, which can be solved in $O(n)$ time with no fill-in with the Thomas algorithm.

Edit: Thanks to Bill for pointing out that this is for standard low-order finite difference/element schemes - for example, higher order FD lead to increased bandwidth, high order FEM leads to a block structure. The Thomas algo can be modified to give a similar orders of work/storage, though the constants involved increase.

  • $\begingroup$ Thank you @JLC. I need to learn graph theory to digest all but could you give me a hint why it becomes tridiagonal for 1D systems? And I am curious if this is true for graphs that are approximately 1D (long and thin). $\endgroup$
    – Armut
    Commented Dec 11, 2014 at 7:52
  • $\begingroup$ Sure - each point represents a column, and the rows of the column are nonzero for points that connect to it. The diagonal is nonzero since a point connects to itself, and since you can order one point after another in 1D, each point connects to at most the point before/after, which leads to only nonzeros in the entries above and below each diagonal. $\endgroup$
    – Jesse Chan
    Commented Dec 11, 2014 at 13:18
  • 1
    $\begingroup$ @JLC, this is only true in 1D for the 3-point finite difference operator. A 5-point 1D stencil for the Laplacian would have 2 more diagonals. Etc. $\endgroup$
    – Bill Barth
    Commented Dec 11, 2014 at 13:32
  • 1
    $\begingroup$ Yes, agreed, and same w/splines or less-compactly supported discretizations. I'll edit that in. $\endgroup$
    – Jesse Chan
    Commented Dec 11, 2014 at 13:44

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