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"""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.
"""
__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, 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]
The exact solution is x = 0
>>> 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:
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