/usr/share/pyshared/openopt/solvers/Standalone/lsqr.py is in python-openopt 0.38+svn1589-1.
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Solve the least-squares problem
minimize ||Ax-b||
using LSQR. This is a line-by-line translation from Matlab code
available at http://www.stanford.edu/~saunders/lsqr.
Michael P. Friedlander, University of British Columbia
Dominique Orban, Ecole Polytechnique de Montreal
$Id$
"""
from numpy import zeros, dot
from openopt.kernel.ooMisc import norm
from math import sqrt
# Simple shortcuts---linalg.norm is too slow for small vectors
def normof2(x,y): return sqrt(x**2 + y**2)
def normof4(x1,x2,x3,x4): return sqrt(x1**2 + x2**2 + x3**2 + x4**2)
def lsqr( m, n, aprod, b, damp, atol, btol, conlim, itnlim, show, wantvar = False, callback = lambda x: None):
"""
[ x, istop, itn, r1norm, r2norm, anorm, acond, arnorm, xnorm, var ]...
= lsqr( m, n, @aprod, b, damp, atol, btol, conlim, itnlim, show );
LSQR solves Ax = b or min ||b - Ax||_2 if damp = 0, or
min || [ b ] - [ A ] x || otherwise.
|| [ 0 ] [ damp I ] ||2
A is an m by n matrix defined by y = aprod(mode, m, n, x),
where aprod refers to a function that performs the matrix-vector operations.
If mode = 1, aprod must return y = Ax without altering x.
If mode = 2, aprod must return y = A'x without altering x.
----------------------------------------------------------------------
LSQR uses an iterative (conjugate-gradient-like) method.
For further information, see
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.
Input parameters:
atol, btol are 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 is also a 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.
itnlim is an explicit limit on iterations (for safety).
show if set to 1, gives an iteration log.
If set to 0, suppresses output.
Output parameters:
x is the final solution.
istop gives the reason for termination.
istop = 1 means x is an approximate solution to Ax = b.
= 2 means x approximately solves the least-squares problem.
r1norm = norm(r), where r = b - Ax.
r2norm = sqrt( norm(r)^2 + damp^2 * norm(x)^2 )
= r1norm if damp = 0.
anorm = estimate of Frobenius norm of Abar = [ A ].
[damp*I]
acond = estimate of cond(Abar).
arnorm = estimate of norm(A'*r - damp^2*x).
xnorm = norm(x).
var (if present) 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.)
----------------------------------------------------------------------
"""
# Initialize.
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 wantvar:
var = zeros(n,1)
else:
var = None
# 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 wantvar = %-5s' % (damp, repr(wantvar))
# str3 = 'atol = %8.2e conlim = %8.2e' % (atol, conlim)
# str4 = 'btol = %8.2e itnlim = %8g' % (btol, itnlim)
# print str1; print str2; print str3; print str4;
itn = 0; istop = 0; nstop = 0
ctol = 0.0
if conlim > 0.0: ctol = 1.0/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.
u = b[:m]; x = zeros(n)
alfa = 0.; beta = norm( u )
if beta > 0:
u = (1.0/beta) * u; v = aprod(2, m, n, u)
alfa = norm( v );
if alfa > 0:
v = (1.0/alfa) * v; w = v.copy();
arnorm = alfa * beta;
if arnorm == 0:
# print(msg[0])
return (x, istop, itn, r1norm, r2norm, anorm, acond, arnorm, xnorm, var)
rhobar = alfa; phibar = beta; bnorm = beta;
rnorm = beta
r1norm = rnorm
r2norm = rnorm
head1 = ' Itn x(1) r1norm r2norm '
head2 = ' Compatible LS Norm A Cond A'
if show:
# print ' '
# print head1+head2
test1 = 1.0; 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 < itnlim:
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 = aprod(1, m, n, v) - alfa*u
beta = norm( u );
if beta > 0:
u = (1.0/beta) * u
anorm = normof4(anorm, alfa, beta, damp)
v = aprod(2, m, n, u) - beta*v
alfa = norm( v )
if alfa > 0: v = (1.0/alfa) * v
# Use a plane rotation to eliminate the damping parameter.
# This alters the diagonal (rhobar) of the lower-bidiagonal matrix.
rhobar1 = normof2(rhobar, damp)
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.
rho = normof2(rhobar1, beta)
cs = rhobar1/ rho
sn = beta / rho
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 = (1.0/rho)*w;
x = x + t1*w
w = v + t2*w
ddnorm = ddnorm + norm(dk)**2
if wantvar: var = var + dk*dk
# 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 = normof2(gambar, theta)
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 )
# 07 Aug 2002:
# 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 )
test3 = 1.0 / acond
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 >= itnlim: 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 >= itnlim-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 and 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
callback(x) # added for OpenOpt kernel
# 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 )
# str5 = ' bnorm =%8.1e' % bnorm
# print str1 + ' ' + str2
# print str3 + ' ' + str4
# print str5
# print ' '
return ( x, istop, itn, r1norm, r2norm, anorm, acond, arnorm, xnorm, var )
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