text: """Sparse Equations and Least Squares.
  
  The original Fortran code was written by C. C. Paige and M. A. Saunders as
  described in
  
  C. C. Paige and M. A. Saunders, LSQR: An algorithm for sparse linear
  equations and sparse least squares, TOMS 8(1), 43--71 (1982).
  
  C. C. Paige and M. A. Saunders, Algorithm 583; LSQR: Sparse linear
  equations and least-squares problems, TOMS 8(2), 195--209 (1982).
  
  It is licensed under the following BSD license:
  
  Copyright (c) 2006, Systems Optimization Laboratory
  All rights reserved.
  
  Redistribution and use in source and binary forms, with or without
  modification, are permitted provided that the following conditions are
  met:
  
      * Redistributions of source code must retain the above copyright
        notice, this list of conditions and the following disclaimer.
  
      * Redistributions in binary form must reproduce the above
        copyright notice, this list of conditions and the following
        disclaimer in the documentation and/or other materials provided
        with the distribution.
  
      * Neither the name of Stanford University nor the names of its
        contributors may be used to endorse or promote products derived
        from this software without specific prior written permission.
  
  THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS
  "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT
  LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR
  A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT
  OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
  SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT
  LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
  DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
  THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
  (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
  OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
  
  The Fortran code was translated to Python for use in CVXOPT by Jeffery
  Kline with contributions by Mridul Aanjaneya and Bob Myhill.
  
  Adapted for SciPy by Stefan van der Walt.
  
  """
  
  from __future__ import division, print_function, absolute_import
  
  __all__ = ['lsqr']
  
  import numpy as np
  from math import sqrt
  from scipy.sparse.linalg.interface import aslinearoperator
  
  eps = np.finfo(np.float64).eps
  
  
  def _sym_ortho(a, b):
      """
      Stable implementation of Givens rotation.
  
      Notes
      -----
      The routine 'SymOrtho' was added for numerical stability. This is
      recommended by S.-C. Choi in [1]_.  It removes the unpleasant potential of
      ``1/eps`` in some important places (see, for example text following
      "Compute the next plane rotation Qk" in minres.py).
  
      References
      ----------
      .. [1] S.-C. Choi, "Iterative Methods for Singular Linear Equations
             and Least-Squares Problems", Dissertation,
             http://www.stanford.edu/group/SOL/dissertations/sou-cheng-choi-thesis.pdf
  
      """
      if b == 0:
          return np.sign(a), 0, abs(a)
      elif a == 0:
          return 0, np.sign(b), abs(b)
      elif abs(b) > abs(a):
          tau = a / b
          s = np.sign(b) / sqrt(1 + tau * tau)
          c = s * tau
          r = b / s
      else:
          tau = b / a
          c = np.sign(a) / sqrt(1+tau*tau)
          s = c * tau
          r = a / c
      return c, s, r
  
  
  def lsqr(A, b, damp=0.0, atol=1e-8, btol=1e-8, conlim=1e8,
           iter_lim=None, show=False, calc_var=False):
      """Find the least-squares solution to a large, sparse, linear system
      of equations.
  
      The function solves ``Ax = b``  or  ``min ||b - Ax||^2`` or
      ``min ||Ax - b||^2 + d^2 ||x||^2``.
  
      The matrix A may be square or rectangular (over-determined or
      under-determined), and may have any rank.
  
      ::
  
        1. Unsymmetric equations --    solve  A*x = b
  
        2. Linear least squares  --    solve  A*x = b
                                       in the least-squares sense
  
        3. Damped least squares  --    solve  (   A    )*x = ( b )
                                              ( damp*I )     ( 0 )
                                       in the least-squares sense
  
      Parameters
      ----------
      A : {sparse matrix, ndarray, LinearOperator}
          Representation of an m-by-n matrix.  It is required that
          the linear operator can produce ``Ax`` and ``A^T x``.
      b : (m,) ndarray
          Right-hand side vector ``b``.
      damp : float
          Damping coefficient.
      atol, btol : float
          Stopping tolerances. If both are 1.0e-9 (say), the final
          residual norm should be accurate to about 9 digits.  (The
          final x will usually have fewer correct digits, depending on
          cond(A) and the size of damp.)
      conlim : float
          Another stopping tolerance.  lsqr 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.
          Maximum precision can be obtained by setting ``atol = btol =
          conlim = zero``, but the number of iterations may then be
          excessive.
      iter_lim : int
          Explicit limitation on number of iterations (for safety).
      show : bool
          Display an iteration log.
      calc_var : bool
          Whether to estimate diagonals of ``(A'A + damp^2*I)^{-1}``.
  
      Returns
      -------
      x : ndarray of float
          The final solution.
      istop : int
          Gives the reason for termination.
          1 means x is an approximate solution to Ax = b.
          2 means x approximately solves the least-squares problem.
      itn : int
          Iteration number upon termination.
      r1norm : float
          ``norm(r)``, where ``r = b - Ax``.
      r2norm : float
          ``sqrt( norm(r)^2  +  damp^2 * norm(x)^2 )``.  Equal to `r1norm` if
          ``damp == 0``.
      anorm : float
          Estimate of Frobenius norm of ``Abar = [[A]; [damp*I]]``.
      acond : float
          Estimate of ``cond(Abar)``.
      arnorm : float
          Estimate of ``norm(A'*r - damp^2*x)``.
      xnorm : float
          ``norm(x)``
      var : ndarray of float
          If ``calc_var`` is True, estimates all diagonals of
          ``(A'A)^{-1}`` (if ``damp == 0``) or more generally ``(A'A +
          damp^2*I)^{-1}``.  This is well defined if A has full column
          rank or ``damp > 0``.  (Not sure what var means if ``rank(A)
          < n`` and ``damp = 0.``)
  
      Notes
      -----
      LSQR uses an iterative method to approximate the solution.  The
      number of iterations required to reach a certain accuracy depends
      strongly on the scaling of the problem.  Poor scaling of the rows
      or columns of A should therefore be avoided where possible.
  
      For example, in problem 1 the solution is unaltered by
      row-scaling.  If a row of A is very small or large compared to
      the other rows of A, the corresponding row of ( A  b ) should be
      scaled up or down.
  
      In problems 1 and 2, the solution x is easily recovered
      following column-scaling.  Unless better information is known,
      the nonzero columns of A should be scaled so that they all have
      the same Euclidean norm (e.g., 1.0).
  
      In problem 3, there is no freedom to re-scale if damp is
      nonzero.  However, the value of damp should be assigned only
      after attention has been paid to the scaling of A.
  
      The parameter damp is intended to help regularize
      ill-conditioned systems, by preventing the true solution from
      being very large.  Another aid to regularization is provided by
      the parameter acond, which may be used to terminate iterations
      before the computed solution becomes very large.
  
      If some initial estimate ``x0`` is known and if ``damp == 0``,
      one could proceed as follows:
  
        1. Compute a residual vector ``r0 = b - A*x0``.
        2. Use LSQR to solve the system  ``A*dx = r0``.
        3. Add the correction dx to obtain a final solution ``x = x0 + dx``.
  
      This requires that ``x0`` be available before and after the call
      to LSQR.  To judge the benefits, suppose LSQR takes k1 iterations
      to solve A*x = b and k2 iterations to solve A*dx = r0.
      If x0 is "good", norm(r0) will be smaller than norm(b).
      If the same stopping tolerances atol and btol are used for each
      system, k1 and k2 will be similar, but the final solution x0 + dx
      should be more accurate.  The only way to reduce the total work
      is to use a larger stopping tolerance for the second system.
      If some value btol is suitable for A*x = b, the larger value
      btol*norm(b)/norm(r0)  should be suitable for A*dx = r0.
  
      Preconditioning is another way to reduce the number of iterations.
      If it is possible to solve a related system ``M*x = b``
      efficiently, where M approximates A in some helpful way (e.g. M -
      A has low rank or its elements are small relative to those of A),
      LSQR may converge more rapidly on the system ``A*M(inverse)*z =
      b``, after which x can be recovered by solving M*x = z.
  
      If A is symmetric, LSQR should not be used!
  
      Alternatives are the symmetric conjugate-gradient method (cg)
      and/or SYMMLQ.  SYMMLQ is an implementation of symmetric cg that
      applies to any symmetric A and will converge more rapidly than
      LSQR.  If A is positive definite, there are other implementations
      of symmetric cg that require slightly less work per iteration than
      SYMMLQ (but will take the same number of iterations).
  
      References
      ----------
      .. [1] C. C. Paige and M. A. Saunders (1982a).
             "LSQR: An algorithm for sparse linear equations and
             sparse least squares", ACM TOMS 8(1), 43-71.
      .. [2] C. C. Paige and M. A. Saunders (1982b).
             "Algorithm 583.  LSQR: Sparse linear equations and least
             squares problems", ACM TOMS 8(2), 195-209.
      .. [3] M. A. Saunders (1995).  "Solution of sparse rectangular
             systems using LSQR and CRAIG", BIT 35, 588-604.
  
      """
      A = aslinearoperator(A)
      if len(b.shape) > 1:
          b = b.squeeze()
  
      m, n = A.shape
      if iter_lim is None:
          iter_lim = 2 * n
      var = np.zeros(n)
  
      msg = ('The exact solution is  x = 0                              ',
           '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                      ')
  
      if show:
          print(' ')
          print('LSQR            Least-squares solution of  Ax = b')
          str1 = 'The matrix A has %8g rows  and %8g cols' % (m, n)
          str2 = 'damp = %20.14e   calc_var = %8g' % (damp, calc_var)
          str3 = 'atol = %8.2e                 conlim = %8.2e' % (atol, conlim)
          str4 = 'btol = %8.2e               iter_lim = %8g' % (btol, iter_lim)
          print(str1)
          print(str2)
          print(str3)
          print(str4)
  
      itn = 0
      istop = 0
      ctol = 0
      if conlim > 0:
          ctol = 1/conlim
      anorm = 0
      acond = 0
      dampsq = damp**2
      ddnorm = 0
      res2 = 0
      xnorm = 0
      xxnorm = 0
      z = 0
      cs2 = -1
      sn2 = 0
  
      """
      Set up the first vectors u and v for the bidiagonalization.
      These satisfy  beta*u = b,  alfa*v = A'u.
      """
      v = np.zeros(n)
      u = b
      x = np.zeros(n)
      alfa = 0
      beta = np.linalg.norm(u)
      w = np.zeros(n)
  
      if beta > 0:
          u = (1/beta) * u
          v = A.rmatvec(u)
          alfa = np.linalg.norm(v)
  
      if alfa > 0:
          v = (1/alfa) * v
          w = v.copy()
  
      rhobar = alfa
      phibar = beta
      bnorm = beta
      rnorm = beta
      r1norm = rnorm
      r2norm = rnorm
  
      # Reverse the order here from the original matlab code because
      # there was an error on return when arnorm==0
      arnorm = alfa * beta
      if arnorm == 0:
          print(msg[0])
          return x, istop, itn, r1norm, r2norm, anorm, acond, arnorm, xnorm, var
  
      head1 = '   Itn      x[0]       r1norm     r2norm '
      head2 = ' Compatible    LS      Norm A   Cond A'
  
      if show:
          print(' ')
          print(head1, head2)
          test1 = 1
          test2 = alfa / beta
          str1 = '%6g %12.5e' % (itn, x[0])
          str2 = ' %10.3e %10.3e' % (r1norm, r2norm)
          str3 = '  %8.1e %8.1e' % (test1, test2)
          print(str1, str2, str3)
  
      # Main iteration loop.
      while itn < iter_lim:
          itn = itn + 1
          """
          %     Perform the next step of the bidiagonalization to obtain the
          %     next  beta, u, alfa, v.  These satisfy the relations
          %                beta*u  =  a*v   -  alfa*u,
          %                alfa*v  =  A'*u  -  beta*v.
          """
          u = A.matvec(v) - alfa * u
          beta = np.linalg.norm(u)
  
          if beta > 0:
              u = (1/beta) * u
              anorm = sqrt(anorm**2 + alfa**2 + beta**2 + damp**2)
              v = A.rmatvec(u) - beta * v
              alfa = np.linalg.norm(v)
              if alfa > 0:
                  v = (1 / alfa) * v
  
          # Use a plane rotation to eliminate the damping parameter.
          # This alters the diagonal (rhobar) of the lower-bidiagonal matrix.
          rhobar1 = sqrt(rhobar**2 + damp**2)
          cs1 = rhobar / rhobar1
          sn1 = damp / rhobar1
          psi = sn1 * phibar
          phibar = cs1 * phibar
  
          # Use a plane rotation to eliminate the subdiagonal element (beta)
          # of the lower-bidiagonal matrix, giving an upper-bidiagonal matrix.
          cs, sn, rho = _sym_ortho(rhobar1, beta)
  
          theta = sn * alfa
          rhobar = -cs * alfa
          phi = cs * phibar
          phibar = sn * phibar
          tau = sn * phi
  
          # Update x and w.
          t1 = phi / rho
          t2 = -theta / rho
          dk = (1 / rho) * w
  
          x = x + t1 * w
          w = v + t2 * w
          ddnorm = ddnorm + np.linalg.norm(dk)**2
  
          if calc_var:
              var = var + dk**2
  
          # Use a plane rotation on the right to eliminate the
          # super-diagonal element (theta) of the upper-bidiagonal matrix.
          # Then use the result to estimate norm(x).
          delta = sn2 * rho
          gambar = -cs2 * rho
          rhs = phi - delta * z
          zbar = rhs / gambar
          xnorm = sqrt(xxnorm + zbar**2)
          gamma = sqrt(gambar**2 + theta**2)
          cs2 = gambar / gamma
          sn2 = theta / gamma
          z = rhs / gamma
          xxnorm = xxnorm + z**2
  
          # Test for convergence.
          # First, estimate the condition of the matrix  Abar,
          # and the norms of  rbar  and  Abar'rbar.
          acond = anorm * sqrt(ddnorm)
          res1 = phibar**2
          res2 = res2 + psi**2
          rnorm = sqrt(res1 + res2)
          arnorm = alfa * abs(tau)
  
          # Distinguish between
          #    r1norm = ||b - Ax|| and
          #    r2norm = rnorm in current code
          #           = sqrt(r1norm^2 + damp^2*||x||^2).
          #    Estimate r1norm from
          #    r1norm = sqrt(r2norm^2 - damp^2*||x||^2).
          # Although there is cancellation, it might be accurate enough.
          r1sq = rnorm**2 - dampsq * xxnorm
          r1norm = sqrt(abs(r1sq))
          if r1sq < 0:
              r1norm = -r1norm
          r2norm = rnorm
  
          # Now use these norms to estimate certain other quantities,
          # some of which will be small near a solution.
          test1 = rnorm / bnorm
          test2 = arnorm / (anorm * rnorm + eps)
          test3 = 1 / (acond + eps)
          t1 = test1 / (1 + anorm * xnorm / bnorm)
          rtol = btol + atol * anorm * xnorm / bnorm
  
          # The following tests guard against extremely small values of
          # atol, btol  or  ctol.  (The user may have set any or all of
          # the parameters  atol, btol, conlim  to 0.)
          # The effect is equivalent to the normal tests using
          # atol = eps,  btol = eps,  conlim = 1/eps.
          if itn >= iter_lim:
              istop = 7
          if 1 + test3 <= 1:
              istop = 6
          if 1 + test2 <= 1:
              istop = 5
          if 1 + t1 <= 1:
              istop = 4
  
          # Allow for tolerances set by the user.
          if test3 <= ctol:
              istop = 3
          if test2 <= atol:
              istop = 2
          if test1 <= rtol:
              istop = 1
  
          # See if it is time to print something.
          prnt = False
          if n <= 40:
              prnt = True
          if itn <= 10:
              prnt = True
          if itn >= iter_lim-10:
              prnt = True
          # if itn%10 == 0: prnt = True
          if test3 <= 2*ctol:
              prnt = True
          if test2 <= 10*atol:
              prnt = True
          if test1 <= 10*rtol:
              prnt = True
          if istop != 0:
              prnt = True
  
          if prnt:
              if show:
                  str1 = '%6g %12.5e' % (itn, x[0])
                  str2 = ' %10.3e %10.3e' % (r1norm, r2norm)
                  str3 = '  %8.1e %8.1e' % (test1, test2)
                  str4 = ' %8.1e %8.1e' % (anorm, acond)
                  print(str1, str2, str3, str4)
  
          if istop != 0:
              break
  
      # End of iteration loop.
      # Print the stopping condition.
      if show:
          print(' ')
          print('LSQR finished')
          print(msg[istop])
          print(' ')
          str1 = 'istop =%8g   r1norm =%8.1e' % (istop, r1norm)
          str2 = 'anorm =%8.1e   arnorm =%8.1e' % (anorm, arnorm)
          str3 = 'itn   =%8g   r2norm =%8.1e' % (itn, r2norm)
          str4 = 'acond =%8.1e   xnorm  =%8.1e' % (acond, xnorm)
          print(str1 + '   ' + str2)
          print(str3 + '   ' + str4)
          print(' ')
  
      return x, istop, itn, r1norm, r2norm, anorm, acond, arnorm, xnorm, var
  