code: def lsqr(
      A,
      b,
      damp=0.0,
      atol=1e-8,
      btol=1e-8,
      conlim=1e8,
      iter_lim=None,
      show=False,
      calc_var=False,
      x0=None,
  ):
      """Find the least-squares solution to a large, sparse, linear system
      of equations.
  
      The function solves ``Ax = b``  or  ``min ||Ax - b||^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.
          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,)
          Right-hand side vector ``b``.
      damp : float
          Damping coefficient.
      atol, btol : float, optional
          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, optional
          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, optional
          Explicit limitation on number of iterations (for safety).
      show : bool, optional
          Display an iteration log.
      calc_var : bool, optional
          Whether to estimate diagonals of ``(A'A + damp^2*I)^{-1}``.
      x0 : array_like, shape (n,), optional
          Initial guess of x, if None zeros are used.
  
          .. versionadded:: 1.0.0
  
      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.
  
      Examples
      --------
      >>> from scipy.sparse import csc_matrix
      >>> from scipy.sparse.linalg import lsqr
      >>> 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 = lsqr(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, r1norm = lsqr(A, b)[:4]
      >>> istop
      1
      >>> x
      array([ 1., -1.])
      >>> itn
      1
      >>> r1norm
      4.440892098500627e-16
  
      As indicated by `istop=1`, `lsqr` 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, r1norm = lsqr(A, b)[:4]
      >>> istop
      2
      >>> x
      array([ 1.00333333, -0.99666667])
      >>> A.dot(x)-b
      array([ 0.00333333, -0.00333333,  0.00333333])
      >>> r1norm
      0.005773502691896255
  
      `istop` indicates that the system is inconsistent and thus `x` is rather an
      approximate solution to the corresponding least-squares problem. `r1norm`
      contains the norm of the minimal residual that was found.
      """
      A = aslinearoperator(A)
      b = np.atleast_1d(b)
      if b.ndim > 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 = f"The matrix A has {m} rows and {n} columns"
          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 - A*x,  alfa*v = A'*u.
      u = b
      bnorm = np.linalg.norm(b)
      if x0 is None:
          x = np.zeros(n)
          beta = bnorm.copy()
      else:
          x = np.asarray(x0)
          u = u - A.matvec(x)
          beta = np.linalg.norm(u)
  
      if beta > 0:
          u = (1 / beta) * u
          v = A.rmatvec(u)
          alfa = np.linalg.norm(v)
      else:
          v = x.copy()
          alfa = 0
  
      if alfa > 0:
          v = (1 / alfa) * v
      w = v.copy()
  
      rhobar = alfa
      phibar = 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:
          if show:
              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 + dampsq)
              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.
          if damp > 0:
              rhobar1 = sqrt(rhobar**2 + dampsq)
              cs1 = rhobar / rhobar1
              sn1 = damp / rhobar1
              psi = sn1 * phibar
              phibar = cs1 * phibar
          else:
              # cs1 = 1 and sn1 = 0
              rhobar1 = rhobar
              psi = 0.0
  
          # 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.
          if damp > 0:
              r1sq = rnorm**2 - dampsq * xxnorm
              r1norm = sqrt(abs(r1sq))
              if r1sq < 0:
                  r1norm = -r1norm
          else:
              r1norm = rnorm
          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
  
          if show:
              # 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:
                  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,
      )
  
imports: import numpy
  from math import sqrt
  from scipy.sparse.linalg.interface import aslinearoperator
label_desc: Use image node_id 2 for calling a global function with example usage: lsqr(A, b, damp, atol, btol, conlim, iter_lim, show, calc_var, x0) and returns: x,  istop,  itn,  r1norm,  r2norm,  anorm,  acond,  arnorm,  xnorm,  var, x,  istop,  itn,  r1norm,  r2norm,  anorm,  acond,  arnorm,  xnorm,  var