LinAlgebraPacks

Packages for Linear Algebra: Solving your system of equations using (optimised, bug-free) code that someone else has already written!

Introduction

svn co https://svn.ggy.bris.ac.uk/subversion-open/num-methods1 ./num-methods1

LAPACK:direct solutions for dense matrices

http://www.netlib.org/lapack

For example, consider the following system of linear equations:

[math]\displaystyle{ \begin{alignat}{7} x &&\; + \;&& 3y &&\; - \;&& 2z &&\; = \;&& 5 & \\ 3x &&\; + \;&& 5y &&\; + \;&& 6z &&\; = \;&& 7 & \\ 2x &&\; + \;&& 4y &&\; + \;&& 3z &&\; = \;&& 8 & \end{alignat} }[/math]

(From http://en.wikipedia.org/wiki/System_of_linear_equations.)

The following computation shows Gauss-Jordan elimination applied to the matrix version (Ax=b) of the above system:

[math]\displaystyle{ \left[\begin{array}{rrr|r} 1 & 3 & -2 & 5 \\ 3 & 5 & 6 & 7 \\ 2 & 4 & 3 & 8 \end{array}\right] }[/math][math]\displaystyle{ \sim \left[\begin{array}{rrr|r} 1 & 3 & -2 & 5 \\ 0 & -4 & 12 & -8 \\ 2 & 4 & 3 & 8 \end{array}\right] }[/math][math]\displaystyle{ \sim \left[\begin{array}{rrr|r} 1 & 3 & -2 & 5 \\ 0 & -4 & 12 & -8 \\ 0 & -2 & 7 & -2 \end{array}\right] }[/math][math]\displaystyle{ \sim \left[\begin{array}{rrr|r} 1 & 3 & -2 & 5 \\ 0 & 1 & -3 & 2 \\ 0 & -2 & 7 & -2 \end{array}\right] }[/math][math]\displaystyle{ \sim \left[\begin{array}{rrr|r} 1 & 3 & -2 & 5 \\ 0 & 1 & -3 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right] }[/math][math]\displaystyle{ \sim \left[\begin{array}{rrr|r} 1 & 3 & -2 & 5 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 2 \end{array}\right] }[/math][math]\displaystyle{ \sim \left[\begin{array}{rrr|r} 1 & 3 & 0 & 9 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 2 \end{array}\right] }[/math][math]\displaystyle{ \sim \left[\begin{array}{rrr|r} 1 & 0 & 0 & -15 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 2 \end{array}\right]. }[/math]

To solve this using LAPACK take a look inside dgesv-example.f90. First the declarations:

  integer, parameter    :: N = 3
  integer, parameter    :: LDA = N  ! leading dimension of A                                                                        
  integer, parameter    :: LDB = N  ! leading dimension of B                                                                        
  integer, parameter    :: NRHS = 1 ! no. of RHS, i.e columns in b                                                                  
  integer, dimension(N) :: IPIV
  integer               :: INFO
  integer               :: i
  real(kind=8), dimension(LDA,N)    :: A ! LDAxN matrix                                                                             
  real(kind=8), dimension(LDB,NRHS) :: B ! LDBxNRHS matrix

Next, we set up the matrix A:

  ! ( 1  3 -2)                                                                                                                      
  ! ( 3  5  6)                                                                                                                      
  ! ( 2  4 3)               
  A(1,1) =  1.0d+0
  A(1,2) =  3.0d+0
  A(1,3) = -2.0d+0
  A(2,1) =  3.0d+0
  A(2,2) =  5.0d+0
  A(2,3) =  6.0d+0
  A(3,1) =  2.0d+0
  A(3,2) =  4.0d+0
  A(3,3) =  3.0d+0

And the vector B:

  ! ( 5)                                                                                                                            
  ! ( 7)                                                                                                                            
  ! ( 8)                                                                                                                            
  B(1,1) =  5.0d+0
  B(2,1) =  7.0d+0
  B(3,1) =  8.0d+0

Once we have all that in place, all we need to do is call the solver routine. (Note that the solution vector replaces the values in B):

  call dgesv(N, NRHS, A, LDA, IPIV, B, LDB, INFO)

Run the program to satisfy yourself that it works:

cd num-methods1/examples/example1
make
./dgesv-example.exe

Next, take a look at simple-laplace.f90. This program solves the 2D laplace equation for a 4x4 domain, as described on NumMethodsPDEs. This program only differs from the previous one in way in which we initialise the matrix A and vector b. To run the program type:

./simple-laplace.exe

The last program in this first example is mxm-laplace.f90. In this case, we have just added a little more complexity, so that a 2D laplace equation can be solved on a more flexibly sized--mxm--square domain:

./mxm-laplace.exe

Direct Solve using LU Decomposition

In practice, the LAPACK routine dgesv does not perform Gaussian elimination, but rather computes an LU decomposition and then solves the equation using two substitution steps--one forward and one backward. This is because for the same computational cost, we can factor A without knowing about b. We can then find x cheaply, for many different b. This is useful when solving parabolic equations, where x at time-step j forms the basis for b at time-step j+1.

OK, how to we obtain an LU decomposition?

The Guassian elimination process above can be expressed as a matrix multiplication. The elimination in the first column can be achieved by left multiplying [math]\displaystyle{ \mathbf{A} }[/math] with a unit lower triangular matrix, [math]\displaystyle{ \mathbf{L_1} }[/math], with non-zeros only in the first column, such as that schematically shown below:

[math]\displaystyle{ \begin{bmatrix} 1 & 0 & 0 \\ x & 1 & 0 \\ x & 0 & 1 \end{bmatrix} }[/math]

Following our example, [math]\displaystyle{ \mathbf{L_1 A=A'} }[/math] becomes:

[math]\displaystyle{ \begin{bmatrix} 1 & 0 & 0 \\ -3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 3 & -2 \\ 3 & 5 & 6 \\ 2 & 4 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 3 & -2 \\ 0 & -4 & 12 \\ 0 & -2 & 7 \end{bmatrix} }[/math]

The eliminations in the second column are achieved by left multiplying [math]\displaystyle{ \mathbf{A'}=L_1 A\lt \math\gt with \lt math\gt \mathbf{L_2}\lt \math\gt : \lt math\gt \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & \frac{-1}{2} & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 0 \\ -3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 3 & -2 \\ 3 & 5 & 6 \\ 2 & 4 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 3 & -2 \\ 0 & -4 & 12 \\ 0 & 0 & 1 \end{bmatrix} }[/math]

[math]\displaystyle{ A = LU }[/math],

where U is an upper triangular matrix, and L is a unit (1's on the diagonal) lower triangular matrix.

Given L and U we can solve Ax=b by first solving:

[math]\displaystyle{ Ly=b }[/math].

As L is unit lower triangular, we can do this by easily by forward substitution. Once we know y, we can solve:

[math]\displaystyle{ Ux=y }[/math],

easily again, this time by backward substitution.

Computing the LU decomposition of an [math]\displaystyle{ n \times n }[/math] system of equations has a computational cost of [math]\displaystyle{ O(n^3) }[/math] operations, whereas the forward and backward substitution steps have a cost of [math]\displaystyle{ O(n^2) }[/math] operations. Thus the cost of solution is dominated by the LU decomposition.

PetSc: Iterative solutions and sparse matrices

http://www.mcs.anl.gov/petsc/petsc-as/

cd ../example3
make
./ex1f

For further examples see INSTALL_DIR/petsc-3.1-p1/src/ksp/ksp/examples/tutorials.

PLASMA: For Multicore Architectures

http://icl.cs.utk.edu/plasma/index.html

cd ../example4
make
./mxm-laplace-plasma.exe

For further examples, see e.g. INSTALL_DIR/plasma-installer_2.3.1/build/plasma_2.3.1.

MAGMA: To include GPUs in Heterogeneous Systems

http://icl.cs.utk.edu/magma/index.html