"""
Copyright (C) 2010 David Fong and Michael Saunders

LSMR uses an iterative method.

07 Jun 2010: Documentation updated
03 Jun 2010: First release version in Python

David Chin-lung Fong            clfong@stanford.edu
Institute for Computational and Mathematical Engineering
Stanford University

Michael Saunders                saunders@stanford.edu
Systems Optimization Laboratory
Dept of MS&E, Stanford University.

"""

from __future__ import division, print_function, absolute_import

__all__ = ['lsmr']

from numpy import zeros, infty, atleast_1d
from numpy.linalg import norm
from math import sqrt
from scipy.sparse.linalg.interface import aslinearoperator

from .lsqr import _sym_ortho


def lsmr(A, b, damp=0.0, atol=1e-6, btol=1e-6, conlim=1e8,
         maxiter=None, show=False, x0=None):
    """Iterative solver for least-squares problems.

    lsmr solves the system of linear equations ``Ax = b``. If the system
    is inconsistent, it solves the least-squares problem ``min ||b - Ax||_2``.
    A is a rectangular matrix of dimension m-by-n, where all cases are
    allowed: m = n, m > n, or m < n. B is a vector of length m.
    The matrix A may be dense or sparse (usually sparse).

    Parameters
    ----------
    A : {matrix, sparse matrix, ndarray, LinearOperator}
        Matrix A in the linear system.
        Alternatively, ``A`` can be a linear operator which can
        produce ``Ax`` and ``A^T x`` using, e.g.,
        ``scipy.sparse.linalg.LinearOperator``.
    b : array_like, shape (m,)
        Vector b in the linear system.
    damp : float
        Damping factor for regularized least-squares. `lsmr` solves
        the regularized least-squares problem::

         min ||(b) - (  A   )x||
             ||(0)   (damp*I) ||_2

        where damp is a scalar.  If damp is None or 0, the system
        is solved without regularization.
    atol, btol : float, optional
        Stopping tolerances. `lsmr` continues iterations until a
        certain backward error estimate is smaller than some quantity
        depending on atol and btol.  Let ``r = b - Ax`` be the
        residual vector for the current approximate solution ``x``.
        If ``Ax = b`` seems to be consistent, ``lsmr`` terminates
        when ``norm(r) <= atol * norm(A) * norm(x) + btol * norm(b)``.
        Otherwise, lsmr terminates when ``norm(A^{T} r) <=
        atol * norm(A) * norm(r)``.  If both tolerances are 1.0e-6 (say),
        the final ``norm(r)`` should be accurate to about 6
        digits. (The final x will usually have fewer correct digits,
        depending on ``cond(A)`` and the size of LAMBDA.)  If `atol`
        or `btol` is None, a default value of 1.0e-6 will be used.
        Ideally, they should be estimates of the relative error in the
        entries of A and B respectively.  For example, if the entries
        of `A` have 7 correct digits, set atol = 1e-7. This prevents
        the algorithm from doing unnecessary work beyond the
        uncertainty of the input data.
    conlim : float, optional
        `lsmr` terminates if an estimate of ``cond(A)`` exceeds
        `conlim`.  For compatible systems ``Ax = b``, conlim could be
        as large as 1.0e+12 (say).  For least-squares problems,
        `conlim` should be less than 1.0e+8. If `conlim` is None, the
        default value is 1e+8.  Maximum precision can be obtained by
        setting ``atol = btol = conlim = 0``, but the number of
        iterations may then be excessive.
    maxiter : int, optional
        `lsmr` terminates if the number of iterations reaches
        `maxiter`.  The default is ``maxiter = min(m, n)``.  For
        ill-conditioned systems, a larger value of `maxiter` may be
        needed.
    show : bool, optional
        Print iterations logs if ``show=True``.
    x0 : array_like, shape (n,), optional
        Initial guess of x, if None zeros are used.

        .. versionadded:: 1.0.0
    Returns
    -------
    x : ndarray of float
        Least-square solution returned.
    istop : int
        istop gives the reason for stopping::

          istop   = 0 means x=0 is a solution.  If x0 was given, then x=x0 is a
                      solution.
                  = 1 means x is an approximate solution to A*x = B,
                      according to atol and btol.
                  = 2 means x approximately solves the least-squares problem
                      according to atol.
                  = 3 means COND(A) seems to be greater than CONLIM.
                  = 4 is the same as 1 with atol = btol = eps (machine
                      precision)
                  = 5 is the same as 2 with atol = eps.
                  = 6 is the same as 3 with CONLIM = 1/eps.
                  = 7 means ITN reached maxiter before the other stopping
                      conditions were satisfied.

    itn : int
        Number of iterations used.
    normr : float
        ``norm(b-Ax)``
    normar : float
        ``norm(A^T (b - Ax))``
    norma : float
        ``norm(A)``
    conda : float
        Condition number of A.
    normx : float
        ``norm(x)``

    Notes
    -----

    .. versionadded:: 0.11.0

    References
    ----------
    .. [1] D. C.-L. Fong and M. A. Saunders,
           "LSMR: An iterative algorithm for sparse least-squares problems",
           SIAM J. Sci. Comput., vol. 33, pp. 2950-2971, 2011.
           https://arxiv.org/abs/1006.0758
    .. [2] LSMR Software, https://web.stanford.edu/group/SOL/software/lsmr/

    Examples
    --------
    >>> from scipy.sparse import csc_matrix
    >>> from scipy.sparse.linalg import lsmr
    >>> A = csc_matrix([[1., 0.], [1., 1.], [0., 1.]], dtype=float)

    The first example has the trivial solution `[0, 0]`

    >>> b = np.array([0., 0., 0.], dtype=float)
    >>> x, istop, itn, normr = lsmr(A, b)[:4]
    >>> istop
    0
    >>> x
    array([ 0.,  0.])

    The stopping code `istop=0` returned indicates that a vector of zeros was
    found as a solution. The returned solution `x` indeed contains `[0., 0.]`.
    The next example has a non-trivial solution:

    >>> b = np.array([1., 0., -1.], dtype=float)
    >>> x, istop, itn, normr = lsmr(A, b)[:4]
    >>> istop
    1
    >>> x
    array([ 1., -1.])
    >>> itn
    1
    >>> normr
    4.440892098500627e-16

    As indicated by `istop=1`, `lsmr` found a solution obeying the tolerance
    limits. The given solution `[1., -1.]` obviously solves the equation. The
    remaining return values include information about the number of iterations
    (`itn=1`) and the remaining difference of left and right side of the solved
    equation.
    The final example demonstrates the behavior in the case where there is no
    solution for the equation:

    >>> b = np.array([1., 0.01, -1.], dtype=float)
    >>> x, istop, itn, normr = lsmr(A, b)[:4]
    >>> istop
    2
    >>> x
    array([ 1.00333333, -0.99666667])
    >>> A.dot(x)-b
    array([ 0.00333333, -0.00333333,  0.00333333])
    >>> normr
    0.005773502691896255

    `istop` indicates that the system is inconsistent and thus `x` is rather an
    approximate solution to the corresponding least-squares problem. `normr`
    contains the minimal distance that was found.
    """

    A = aslinearoperator(A)
    b = atleast_1d(b)
    if b.ndim > 1:
        b = b.squeeze()

    msg = ('The exact solution is x = 0, or x = x0, if x0 was given  ',
         'Ax - b is small enough, given atol, btol                  ',
         'The least-squares solution is good enough, given atol     ',
         'The estimate of cond(Abar) has exceeded conlim            ',
         'Ax - b is small enough for this machine                   ',
         'The least-squares solution is good enough for this machine',
         'Cond(Abar) seems to be too large for this machine         ',
         'The iteration limit has been reached                      ')

    hdg1 = '   itn      x(1)       norm r    norm A''r'
    hdg2 = ' compatible   LS      norm A   cond A'
    pfreq = 20   # print frequency (for repeating the heading)
    pcount = 0   # print counter

    m, n = A.shape

    # stores the num of singular values
    minDim = min([m, n])

    if maxiter is None:
        maxiter = minDim

    if show:
        print(' ')
        print('LSMR            Least-squares solution of  Ax = b\n')
        print('The matrix A has %8g rows  and %8g cols' % (m, n))
        print('damp = %20.14e\n' % (damp))
        print('atol = %8.2e                 conlim = %8.2e\n' % (atol, conlim))
        print('btol = %8.2e             maxiter = %8g\n' % (btol, maxiter))

    u = b
    normb = norm(b)
    if x0 is None:
        x = zeros(n)
        beta = normb.copy()
    else:
        x = atleast_1d(x0)
        u = u - A.matvec(x)
        beta = norm(u)

    if beta > 0:
        u = (1 / beta) * u
        v = A.rmatvec(u)
        alpha = norm(v)
    else:
        v = zeros(n)
        alpha = 0

    if alpha > 0:
        v = (1 / alpha) * v

    # Initialize variables for 1st iteration.

    itn = 0
    zetabar = alpha * beta
    alphabar = alpha
    rho = 1
    rhobar = 1
    cbar = 1
    sbar = 0

    h = v.copy()
    hbar = zeros(n)

    # Initialize variables for estimation of ||r||.

    betadd = beta
    betad = 0
    rhodold = 1
    tautildeold = 0
    thetatilde = 0
    zeta = 0
    d = 0

    # Initialize variables for estimation of ||A|| and cond(A)

    normA2 = alpha * alpha
    maxrbar = 0
    minrbar = 1e+100
    normA = sqrt(normA2)
    condA = 1
    normx = 0

    # Items for use in stopping rules, normb set earlier
    istop = 0
    ctol = 0
    if conlim > 0:
        ctol = 1 / conlim
    normr = beta

    # Reverse the order here from the original matlab code because
    # there was an error on return when arnorm==0
    normar = alpha * beta
    if normar == 0:
        if show:
            print(msg[0])
        return x, istop, itn, normr, normar, normA, condA, normx

    if show:
        print(' ')
        print(hdg1, hdg2)
        test1 = 1
        test2 = alpha / beta
        str1 = '%6g %12.5e' % (itn, x[0])
        str2 = ' %10.3e %10.3e' % (normr, normar)
        str3 = '  %8.1e %8.1e' % (test1, test2)
        print(''.join([str1, str2, str3]))

    # Main iteration loop.
    while itn < maxiter:
        itn = itn + 1

        # Perform the next step of the bidiagonalization to obtain the
        # next  beta, u, alpha, v.  These satisfy the relations
        #         beta*u  =  a*v   -  alpha*u,
        #        alpha*v  =  A'*u  -  beta*v.

        u *= -alpha
        u += A.matvec(v)
        beta = norm(u)

        if beta > 0:
            u *= (1 / beta)
            v *= -beta
            v += A.rmatvec(u)
            alpha = norm(v)
            if alpha > 0:
                v *= (1 / alpha)

        # At this point, beta = beta_{k+1}, alpha = alpha_{k+1}.

        # Construct rotation Qhat_{k,2k+1}.

        chat, shat, alphahat = _sym_ortho(alphabar, damp)

        # Use a plane rotation (Q_i) to turn B_i to R_i

        rhoold = rho
        c, s, rho = _sym_ortho(alphahat, beta)
        thetanew = s*alpha
        alphabar = c*alpha

        # Use a plane rotation (Qbar_i) to turn R_i^T to R_i^bar