pylops.optimization.solver.lsqr

pylops.optimization.solver.lsqr(Op, y, x0, damp=0.0, atol=1e-08, btol=1e-08, conlim=100000000.0, niter=10, calc_var=True, show=False, callback=None)[source]

LSQR

Solve an overdetermined system of equations given an operator Op and data y using LSQR iterations.

\[\DeclareMathOperator{\cond}{cond}\]
Parameters:
Op : pylops.LinearOperator

Operator to invert of size \([N \times M]\)

y : np.ndarray

Data of size \([N \times 1]\)

x0 : np.ndarray, optional

Initial guess

damp : float, optional

Damping coefficient

atol, btol : float, optional

Stopping tolerances. If both are 1.0e-9, the final residual norm should be accurate to about 9 digits. (The solution will usually have fewer correct digits, depending on \(\cond(\mathbf{Op})\) and the size of damp.)

conlim : float, optional

Stopping tolerance on \(\cond(\mathbf{Op})\) exceeds conlim. For square, conlim could be as large as 1.0e+12. For least-squares problems, conlim should be less than 1.0e+8. Maximum precision can be obtained by setting atol = btol = conlim = 0, but the number of iterations may then be excessive.

niter : int, optional

Number of iterations

calc_var : bool, optional

Estimate diagonals of \((\mathbf{Op}^H\mathbf{Op} + \epsilon^2\mathbf{I})^{-1}\).

show : bool, optional

Display iterations log

callback : callable, optional

Function with signature (callback(x)) to call after each iteration where x is the current model vector
Returns:
x : np.ndarray

Estimated model of size \([M \times 1]\)

istop : int

Gives the reason for termination

0 means the exact solution is \(\mathbf{x}=0\)

1 means \(\mathbf{x}\) is an approximate solution to \(\mathbf{y} = \mathbf{Op}\,\mathbf{x}\)

2 means \(\mathbf{x}\) approximately solves the least-squares problem

3 means the estimate of \(\cond(\overline{\mathbf{Op}})\) has exceeded conlim

4 means \(\mathbf{y} - \mathbf{Op}\,\mathbf{x}\) is small enough for this machine

5 means the least-squares solution is good enough for this machine

6 means \(\cond(\overline{\mathbf{Op}})\) seems to be too large for this machine

7 means the iteration limit has been reached

r1norm : float

\(||\mathbf{r}||_2^2\), where \(\mathbf{r} = \mathbf{y} - \mathbf{Op}\,\mathbf{x}\)

r2norm : float

\(\sqrt{\mathbf{r}^T\mathbf{r} + \epsilon^2 \mathbf{x}^T\mathbf{x}}\). Equal to r1norm if \(\epsilon=0\)

anorm : float

Estimate of Frobenius norm of \(\overline{\mathbf{Op}} = [\mathbf{Op} \; \epsilon \mathbf{I}]\)

acond : float

Estimate of \(\cond(\overline{\mathbf{Op}})\)

arnorm : float

Estimate of norm of \(\cond(\mathbf{Op}^H\mathbf{r}- \epsilon^2\mathbf{x})\)

var : float

Diagonals of \((\mathbf{Op}^H\mathbf{Op})^{-1}\) (if damp=0) or more generally \((\mathbf{Op}^H\mathbf{Op} + \epsilon^2\mathbf{I})^{-1}\).

cost : numpy.ndarray, optional

History of r1norm through iterations

Notes

Minimize the following functional using LSQR iterations [1]:

\[J = || \mathbf{y} - \mathbf{Op}\,\mathbf{x} ||_2^2 + \epsilon^2 || \mathbf{x} ||_2^2\]

where \(\epsilon\) is the damping coefficient.

[1]Paige, C. C., and Saunders, M. A. “LSQR: An algorithm for sparse linear equations and sparse least squares”, ACM TOMS, vol. 8, pp. 43-71, 1982.

Examples using pylops.optimization.solver.lsqr