cgs

Solve a square linear system using cgs with default settings, and then adjust the tolerance and number of iterations used in the solution process.

Create a random sparse matrix A with 50% density. Also create a random vector for the right-hand side of $\mathrm{Ax}=\mathit{b}$.

rng default
A = sprand(600,600,.5);
A = A'*A + speye(size(A));
b = rand(600,1);

Solve $\mathrm{Ax}=\mathit{b}$ using cgs. The output display includes the value of the relative residual error $\frac{\Vert \mathit{b}-\mathrm{Ax}\Vert }{\Vert \mathit{b}\Vert }$.

x = cgs(A,b);

cgs stopped at iteration 20 without converging to the desired tolerance 1e-06
because the maximum number of iterations was reached.
The iterate returned (number 20) has relative residual 0.068.

By default cgs uses 20 iterations and a tolerance of 1e-6, and the algorithm is unable to converge in those 20 iterations for this matrix. Since the residual is still large, it is a good indicator that more iterations (or a preconditioner matrix) are needed. You also can use a larger tolerance to make it easier for the algorithm to converge.

Solve the system again using a tolerance of 1e-4 and 40 iterations.

x = cgs(A,b,1e-4,40);

cgs stopped at iteration 40 without converging to the desired tolerance 0.0001
because the maximum number of iterations was reached.
The iterate returned (number 40) has relative residual 0.0024.

Even with a looser tolerance and more iterations cgs does not converge. When an iterative algorithm stalls in this manner it is a good indication that a preconditioner matrix is needed.

Calculate the incomplete Cholesky factorization of A, and use the L factor as a preconditioner input to cgs.

L = ichol(A);
x = cgs(A,b,1e-4,40,L);

cgs converged at iteration 13 to a solution with relative residual 7.3e-05.

Using a preconditioner improves the numerical properties of the problem enough that cgs is able to converge.

Examine the effect of using a preconditioner matrix with cgs to solve a linear system.

Load west0479, a real 479-by-479 nonsymmetric sparse matrix.

load west0479
A = west0479;

Define b so that the true solution to $\mathrm{Ax}=\mathit{b}$ is a vector of all ones.

b = sum(A,2);

Set the tolerance and maximum number of iterations.

tol = 1e-12;
maxit = 20;

Use cgs to find a solution at the requested tolerance and number of iterations. Specify five outputs to return information about the solution process:

[x,fl0,rr0,it0,rv0] = cgs(A,b,tol,maxit);
fl0

fl0 = 
1

rr0

rr0 = 
1

it0

it0 = 
0

fl0 is 1 because cgs does not converge to the requested tolerance 1e-12 within the requested 20 iterations. In fact, the behavior of cgs is so poor that the initial guess x0 = zeros(size(A,2),1) is the best solution and is returned, as indicated by it0 = 0.

To aid with the slow convergence, you can specify a preconditioner matrix. Since A is nonsymmetric, use ilu to generate the preconditioner $\mathit{M}=\mathit{L}\mathit{U}$. Specify a drop tolerance to ignore nondiagonal entries with values smaller than 1e-6. Solve the preconditioned system ${\mathit{A}\mathit{M}}^{-1}\mathit{M}\mathit{x}=\mathit{b}$ by specifying L and U as inputs to cgs.

setup = struct('type','ilutp','droptol',1e-6);
[L,U] = ilu(A,setup);
[x1,fl1,rr1,it1,rv1] = cgs(A,b,tol,maxit,L,U);
fl1

fl1 = 
0

rr1

rr1 = 
4.3851e-14

it1

it1 = 
3

The use of an ilu preconditioner produces a relative residual less than the prescribed tolerance of 1e-12 at the third iteration. The output rv1(1) is norm(b), and the output rv1(end) is norm(b-A*x1).

You can follow the progress of cgs by plotting the relative residuals at each iteration. Plot the residual history of each solution with a line for the specified tolerance.

semilogy(0:length(rv0)-1,rv0/norm(b),'-o')
hold on
semilogy(0:length(rv1)-1,rv1/norm(b),'-o')
yline(tol,'r--');
legend('No preconditioner','ILU preconditioner','Tolerance','Location','East')
xlabel('Iteration number')
ylabel('Relative residual')

Examine the effect of supplying cgs with an initial guess of the solution.

Create a tridiagonal sparse matrix. Use the sum of each row as the vector for the right-hand side of $\mathrm{Ax}=\mathit{b}$ so that the expected solution for $\mathit{x}$ is a vector of ones.

n = 900;
e = ones(n,1);
A = spdiags([e 2*e e],-1:1,n,n);
b = sum(A,2);

Use cgs to solve $\mathrm{Ax}=\mathit{b}$ twice: one time with the default initial guess, and one time with a good initial guess of the solution. Use 200 iterations and the default tolerance for both solutions. Specify the initial guess in the second solution as a vector with all elements equal to 0.99.

maxit = 200;
x1 = cgs(A,b,[],maxit);

cgs converged at iteration 17 to a solution with relative residual 8.8e-07.

x0 = 0.99*e;
x2 = cgs(A,b,[],maxit,[],[],x0);

cgs converged at iteration 4 to a solution with relative residual 8e-07.

In this case supplying an initial guess enables cgs to converge more quickly.

x0 = zeros(size(A,2),1);
tol = 1e-8;
maxit = 100;
for k = 1:4
    [x,flag,relres] = cgs(A,b,tol,maxit,[],[],x0);
    X(:,k) = x;
    R(k) = relres;
    x0 = x;
end

Solve a linear system by providing cgs with a function handle that computes A*x in place of the coefficient matrix A.

One of the Wilkinson test matrices generated by gallery is a 21-by-21 tridiagonal matrix. Preview the matrix.

A = gallery('wilk',21)

A = 21×21

    10     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     1     9     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     1     8     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     1     7     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     1     6     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     1     5     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     1     4     1     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     1     3     1     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     1     2     1     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     1     1     1     0     0     0     0     0     0     0     0     0     0
      ⋮

The Wilkinson matrix has a special structure, so you can represent the operation A*x with a function handle. When A multiplies a vector, most of the elements in the resulting vector are zeros. The nonzero elements in the result correspond with the nonzero tridiagonal elements of A. Moreover, only the main diagonal has nonzeros that are not equal to 1.

The expression $\mathrm{Ax}$ becomes:

$\mathrm{Ax}=\left[\begin{array}{cccccccccc}10& 1& 0& \cdots & & \cdots & & \cdots & 0& 0\\ 1& 9& 1& 0& & & & & & 0\\ 0& 1& 8& 1& 0& & & & & ⋮\\ ⋮& 0& 1& 7& 1& 0& & & & \\ & & 0& 1& 6& 1& 0& & & ⋮\\ ⋮& & & 0& 1& 5& 1& 0& & \\ & & & & 0& 1& 4& 1& 0& ⋮\\ ⋮& & & & & 0& 1& 3& \ddots & 0\\ 0& & & & & & 0& \ddots & \ddots & 1\\ 0& 0& \cdots & & \cdots & & \cdots & 0& 1& 10\end{array}\right]\left[\begin{array}{c}{\mathit{x}}_1\\ {\mathit{x}}_2\\ {\mathit{x}}_3\\ {\mathit{x}}_4\\ {\mathit{x}}_5\\ ⋮\\ \\ ⋮\\ \\ {\mathit{x}}_{21}\end{array}\right]=\left[\begin{array}{c}10{\mathit{x}}_1+{\mathit{x}}_2\\ {\mathit{x}}_1+9{\mathit{x}}_2+{\mathit{x}}_3\\ {\mathit{x}}_2+8{\mathit{x}}_3+{\mathit{x}}_4\\ ⋮\\ {\mathit{x}}_{19}+9{\mathit{x}}_{20}+{\mathit{x}}_{21}\\ {\mathit{x}}_{20}+10{\mathit{x}}_{21}\end{array}\right]$.

The resulting vector can be written as the sum of three vectors:

$\mathrm{Ax}=\left[\begin{array}{c}0+10{\mathit{x}}_1+{\mathit{x}}_2\\ {\mathit{x}}_1+9{\mathit{x}}_2+{\mathit{x}}_3\\ {\mathit{x}}_2+8{\mathit{x}}_3+{\mathit{x}}_4\\ ⋮\\ {\mathit{x}}_{19}+9{\mathit{x}}_{20}+{\mathit{x}}_{21}\\ {\mathit{x}}_{20}+10{\mathit{x}}_{21}+0\end{array}\right]$=$\left[\begin{array}{c}0\\ {\mathit{x}}_1\\ ⋮\\ {\mathit{x}}_{20}\end{array}\right]+\left[\begin{array}{c}10{\mathit{x}}_1\\ 9{\mathit{x}}_2\\ ⋮\\ 10{\mathit{x}}_{21}\end{array}\right]+\left[\begin{array}{c}{\mathit{x}}_2\\ ⋮\\ {\mathit{x}}_{21}\\ 0\end{array}\right]$.

In MATLAB®, write a function that creates these vectors and adds them together, thus giving the value of A*x:

function y = afun(x)
y = [0; x(1:20)] + ...
    [(10:-1:0)'; (1:10)'].*x + ...
    [x(2:21); 0];
end

(This function is saved as a local function at the end of the example.)

Now, solve the linear system $\mathrm{Ax}=\mathit{b}$ by providing cgs with the function handle that calculates A*x. Use a tolerance of 1e-12 and 50 iterations.

b = ones(21,1);
tol = 1e-12;  
maxit = 50;
x1 = cgs(@afun,b,tol,maxit)

cgs converged at iteration 11 to a solution with relative residual 1.3e-14.

x1 = 21×1

    0.0910
    0.0899
    0.0999
    0.1109
    0.1241
    0.1443
    0.1544
    0.2383
    0.1309
    0.5000
      ⋮

Check that afun(x1) produces a vector of ones.

afun(x1)

ans = 21×1

    1.0000
    1.0000
    1.0000
    1.0000
    1.0000
    1.0000
    1.0000
    1.0000
    1.0000
    1.0000
      ⋮

function y = afun(x)
y = [0; x(1:20)] + ...
    [(10:-1:0)'; (1:10)'].*x + ...
    [x(2:21); 0];
end