# pylops.optimization.cls_basic.CGLS¶

class pylops.optimization.cls_basic.CGLS(Op, callbacks=None)[source]

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

Parameters: Op : pylops.LinearOperator Operator to invert of size $$[N \times M]$$

Notes

Minimize the following functional using conjugate gradient iterations:

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

where $$\epsilon$$ is the damping coefficient.

Methods

 __init__(Op[, callbacks]) Initialize self. callback(x, *args, **kwargs) Callback routine finalize([show]) Finalize solver run(x[, niter, show, itershow]) Run solver setup(y[, x0, niter, damp, tol, show]) Setup solver solve(y[, x0, niter, damp, tol, show, itershow]) Run entire solver step(x[, show]) Run one step of solver
setup(y, x0=None, niter=None, damp=0.0, tol=0.0001, show=False)[source]

Setup solver

Parameters: y : np.ndarray Data of size $$[N \times 1]$$ x0 : np.ndarray, optional Initial guess of size $$[M \times 1]$$. If None, initialize internally as zero vector niter : int, optional Number of iterations (default to None in case a user wants to manually step over the solver) damp : float, optional Damping coefficient tol : float, optional Tolerance on residual norm show : bool, optional Display setup log x : np.ndarray Initial guess of size $$[N \times 1]$$
step(x, show=False)[source]

Run one step of solver

Parameters: x : np.ndarray Current model vector to be updated by a step of CG show : bool, optional Display iteration log
run(x, niter=None, show=False, itershow=[10, 10, 10])[source]

Run solver

Parameters: x : np.ndarray Current model vector to be updated by multiple steps of CGLS niter : int, optional Number of iterations. Can be set to None if already provided in the setup call show : bool, optional Display iterations log itershow : list, optional Display set log for the first N1 steps, last N2 steps, and every N3 steps in between where N1, N2, N3 are the three element of the list. x : np.ndarray Estimated model of size $$[M \times 1]$$
finalize(show=False)[source]

Finalize solver

Parameters: show : bool, optional Display finalize log
solve(y, x0=None, niter=10, damp=0.0, tol=0.0001, show=False, itershow=[10, 10, 10])[source]

Run entire solver

Parameters: y : np.ndarray Data of size $$[N \times 1]$$ x0 : np.ndarray Initial guess of size $$[M \times 1]$$. If None, initialize internally as zero vector niter : int, optional Number of iterations (default to None in case a user wants to manually step over the solver) damp : float, optional Damping coefficient tol : float, optional Tolerance on residual norm show : bool, optional Display logs itershow : list, optional Display set log for the first N1 steps, last N2 steps, and every N3 steps in between where N1, N2, N3 are the three element of the list. x : np.ndarray Estimated model of size $$[M \times 1]$$ istop : int Gives the reason for termination 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 iit : int Iteration number upon termination r1norm : float $$||\mathbf{r}||_2$$, 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 $$\epsilon=0$$ cost : numpy.ndarray, optional History of r1norm through iterations