SciPy

scipy.sparse.linalg.qmr

scipy.sparse.linalg.qmr(A, b, x0=None, tol=1e-05, maxiter=None, M1=None, M2=None, callback=None)[source]

Use Quasi-Minimal Residual iteration to solve Ax = b.

Parameters:

A : {sparse matrix, dense matrix, LinearOperator}

The real-valued N-by-N matrix of the linear system. It is required that the linear operator can produce Ax and A^T x.

b : {array, matrix}

Right hand side of the linear system. Has shape (N,) or (N,1).

Returns:

x : {array, matrix}

The converged solution.

info : integer

Provides convergence information:

0 : successful exit >0 : convergence to tolerance not achieved, number of iterations <0 : illegal input or breakdown

Other Parameters:
 

x0 : {array, matrix}

Starting guess for the solution.

tol : float

Tolerance to achieve. The algorithm terminates when either the relative or the absolute residual is below tol.

maxiter : integer

Maximum number of iterations. Iteration will stop after maxiter steps even if the specified tolerance has not been achieved.

M1 : {sparse matrix, dense matrix, LinearOperator}

Left preconditioner for A.

M2 : {sparse matrix, dense matrix, LinearOperator}

Right preconditioner for A. Used together with the left preconditioner M1. The matrix M1*A*M2 should have better conditioned than A alone.

callback : function

User-supplied function to call after each iteration. It is called as callback(xk), where xk is the current solution vector.

See also

LinearOperator

Examples

>>> from scipy.sparse import csc_matrix
>>> from scipy.sparse.linalg import qmr
>>> A = csc_matrix([[3, 2, 0], [1, -1, 0], [0, 5, 1]], dtype=float)
>>> b = np.array([2, 4, -1], dtype=float)
>>> x, exitCode = qmr(A, b)
>>> print(exitCode)            # 0 indicates successful convergence
0
>>> np.allclose(A.dot(x), b)
True