Consider the following small example: where
and the vector of right-hand sides Since the matrix is positive definite and symmetric, you can apply the conjugate gradient algorithm to solve the system. Remember that you must specify only the lower-triangular part of the matrix (so row indices must be greater than or equal to the corresponding column indices.)
The code for this example is as follows:
/* value row col */ A = { 3 1 1, 1 2 1, 4 2 2, 1 3 2, 3 4 2, 10 3 3, 3 4 4 }; /* right-hand sides */ b = {1, 1, 1, 1}; /* desired solution tolerance (optional) */ tol = 1e-7; /* maximum number of iterations (optional) */ maxit = 200; /* allocate iteration progress (optional) */ hist = j(50, 1); /* provide an initial guess (optional) */ start = {2, 3, 4, 5}; /* invoke conjugate gradient method */ call itsolver ( x, st, it, /* output parameters */ 'cg', A, b, 'ic', /* input parameters */ tol, /* optional control parameters */ maxit, start, hist ); print x; /* print solution */ print st; /* print solution tolerance */ print it; /* print resultant number of iterations */
Notice that the example used an incomplete Cholesky preconditioner (which is recommended). Here is the program output:
X 0.5882353 -0.764706 0.1764706 1.0980392 ST 1.961E-16 IT 3
The conjugate gradient method converged successfully within three iterations. You can also print out the hist
(iteration progress) array. Different starting points result in different iterative histories.