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-methods1LAPACK:direct solutions for dense matrices
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 A with a unit (1's on the diagonal) lower triangular matrix, [math]\displaystyle{ L_1 }[/math], with non-zeros only in the first column (this case is called an atomic unit lower triangular matrix), 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{ 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{ A'=L_1 A }[/math] by [math]\displaystyle{ L_2 }[/math] (which is atomic unit lower triangular with non-zeros only in the second column):
[math]\displaystyle{ \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]
We note that the result [math]\displaystyle{ L_2 L_1 A = A'' = U }[/math] is upper triangular. Rearranging we get, [math]\displaystyle{ A=L_1^{-1} \cdots L^{-1}_{n-2} L^{-1}_{n-1} U }[/math]. The inverse matrices are very easy to find as the inverse of an atomic unit lower triangular matrix is the same as the original save only for a change of sign in the off diagonal elements. The product is, again, unit lower triangular. Thus we may write:
- [math]\displaystyle{ A=LU }[/math].
Given L and U we can solve [math]\displaystyle{ Ax=b }[/math] 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.
If we look once again at dgesv-example.f90 and set:
logical, parameter :: verbose=.true.
(and recompile and run) we can see the contents of A after the call to dgesv(). It now contains an encoding of the LU decomposition (the unit leading diagonal from L is assumed):
LU decomposition as encoded in matrix A:
3.00000000000000 5.00000000000000 6.00000000000000
0.333333333333333 1.33333333333333 -4.00000000000000
0.666666666666667 0.500000000000000 1.00000000000000
..and the IPIV vector:
2 2 3
This is all well and good, but hang-on a cotton pickin' minute, the matrix U is not as we expected:
- [math]\displaystyle{ \begin{bmatrix} 3 & 5 & 6 \\ 0 & \frac{4}{3} & -4 \\ 0 & 0 & 1 \end{bmatrix} \neq \begin{bmatrix} 1 & 3 & -2 \\ 0 & -4 & 12 \\ 0 & 0 & 1 \end{bmatrix} }[/math]
Why not? The answer lies the in contents of the IPIV. The elements along the leading diagonal are known as pivots. If any pivots are small, our algorithm will become unstable. We may exchange rows (known as partial pivoting) in A without changing the solution. The way to read IPIV is that value j of the ith entry indicates that row i was exchanged with row j. In the case above, row one was exchanged with row 2, but then rows 2 and 3 were kept the same. This upshot of this maneuver is that we moved a 1 from the leading diagonal and replaced it with a 3.
Our second example uses a pre-computed LU decomposition when solving a parabolic equation using the Crank-Nicolson scheme.
cd ../example2
In simple-crank-nicolson.f90 we see that the initial call to dgesv replaces the entries in the matrix A with an encoding of the LU decompostion (the unit diagonal is assumed). For subsequent timesteps we can thus call dgetrs instead:
do ii=2,4
...
call dgetrs('N', N, NRHS, A, LDA, IPIV, B, LDB, INFO)
...
end do
where we have remembered to take acount of the boundary conditions at each time-step too:
do ii=2,4
...
! copy B into B1, so that we can recompute for
! the boundary conditions again.
B1 = B
! Need to recalculate entries for the vector of known
! values--held in B--given the solution at the current
! timestep--held in B1.
! *NB* the loops below are only valid for the case r=1.
! LHS
B(1,1) = 0 + B1(2,1)
! RHS
B(N,1) = 0 + B1(N-1,1)
! middle bit
do jj=2,N-1
B(jj,1) = B1(jj-1,1) + B1(jj+1,1)
end do
...
end do
PetSc: Iterative solutions and sparse matrices
http://www.mcs.anl.gov/petsc/petsc-as/
Iteration:
- [math]\displaystyle{ x^0, x^1, x^2, x^3, \cdots, x^j \rightarrow x }[/math]
- [math]\displaystyle{ A = L + D + U }[/math]
Ax = b, becomes:
- [math]\displaystyle{ Lx + Dx + Ux = b }[/math]
Jacobi (useful when A is diagonally dominant):
- [math]\displaystyle{ Lx^j + Dx^{j+1} + Ux^j = b }[/math]
- [math]\displaystyle{ x^{j+1} = D^{-1}(b - (L + U)x^j) }[/math]
cd ../example3 make ./ex1f
Take a look at 'jacobi-example.f90:
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.