# PythonOptimizers/pykrylov

Switch branches/tags
Nothing to show
Fetching contributors…
Cannot retrieve contributors at this time
473 lines (389 sloc) 17.3 KB
 """ 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 with several Pythonic enhancements. Michael P. Friedlander, University of British Columbia Dominique Orban, Ecole Polytechnique de Montreal """ from pykrylov.generic import KrylovMethod from numpy import zeros, dot, inf from numpy.linalg import norm from math import sqrt __docformat__ = 'restructuredtext' # Simple shortcuts---linalg.norm is too slow for small vectors def normof2(x,y): return sqrt(x*x + y*y) def normof4(x1,x2,x3,x4): return sqrt(x1*x1 + x2*x2 + x3*x3 + x4*x4) class LSQRFramework(KrylovMethod): r""" LSQR solves `Ax = b` or `minimize |b - Ax|` in Euclidian norm if `damp = 0`, or `minimize |b - Ax| + damp * |x|` in Euclidian norm if `damp > 0`. `A` is an (m x n) linear operator defined by `y = A * x` (or `y = A(x)`), where `y` is the result of applying the linear operator to `x`. Application of transpose linear operator must be accessible via `u = A.T * x` (or `u = A.T(x)`). The shape of the linear operator `A` must be accessible via `A.shape`. A convenient way to achieve this is to make sure that `A` is a `LinearOperator` instance. 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. """ def __init__(self, A, **kwargs): # Initialize. KrylovMethod.__init__(self, A, **kwargs) self.name = 'Least-Squares QR' self.acronym = 'LSQR' self.prefix = self.acronym + ': ' self.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 ', 'The truncated direct error is small enough, given etol '] self.A = A self.x = None ; self.var = None self.itn = 0; self.istop = 0 self.Anorm = 0.; self.Acond = 0. ; self.Arnorm = 0. self.xnorm = 0.; self.r1norm = 0.; self.r2norm = 0. self.optimal = False self.resids = [] # Least-squares objective function values. self.normal_eqns_resids = [] # Residuals of normal equations. self.dir_errors_window = [] # Direct error estimates. self.error_upper_bound = [] # Upper bound on direct error. self.iterates = [] return def solve(self, rhs, itnlim=0, damp=0.0, M=None, N=None, atol=1.0e-9, btol=1.0e-9, conlim=1.0e+8, show=False, wantvar=False, **kwargs): """ Solve the linear system, linear least-squares problem or regularized linear least-squares problem with specified parameters. All return values below are stored in members of the same name. :parameters: :rhs: right-hand side vector. :itnlim: is an explicit limit on iterations (for safety). :damp: damping/regularization parameter. :keywords: :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`.) :etol: stopping tolerance based on direct error (default 1.0e-6). :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. :show: if set to `True`, gives an iteration log. If set to `False`, suppresses output. :store_resids: Store full residual norm history (default: False). :window: Number of consecutive iterations over which the director error should be measured (default: 5). :return: :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 (regularized) A. :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.) """ etol = kwargs.get('etol', 1.0e-6) store_resids = kwargs.get('store_resids', False) store_iterates = kwargs.get('store_iterates', False) window = kwargs.get('window', 5) self.resids = [] # Least-squares objective function values. self.normal_eqns_resids = [] # Residuals of normal equations. self.dir_errors_window = [] # Direct error estimates. self.iterates = [] A = self.A m, n = A.shape if itnlim == 0: itnlim = 3*n if wantvar: var = zeros(n,1) else: var = None dampsq = damp*damp; itn = istop = 0 ctol = 0.0 if conlim > 0.0: self.ctol = 1.0/conlim Anorm = Acond = 0. z = xnorm = xxnorm = ddnorm = res2 = 0. cs2 = -1. ; sn2 = 0. if show: print ' ' print 'LSQR Least-squares solution of Ax = b' str1='The matrix A has %8d rows and %8d 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; # Set up the first vectors u and v for the bidiagonalization. # These satisfy beta*M*u = b, alpha*N*v = A'u. x = zeros(n) xNrgNorm2 = 0.0 # Squared energy norm of final solution. dErr = zeros(window) # Truncated direct error terms. trncDirErr = 0 # Truncated direct error. if store_iterates: self.iterates.append(x.copy()) Mu = rhs[:m].copy() if M is not None: u = M(Mu) else: u = Mu alpha = 0. beta = sqrt(dot(u,Mu)) # norm(u) if beta > 0: u /= beta if M is not None: Mu /= beta Nv = A.T * u if N is not None: v = N(Nv) else: v = Nv alpha = sqrt(dot(v,Nv)) # norm(v) if alpha > 0: v /= alpha if N is not None: Nv /= alpha w = v.copy() # Should this be Nv ??? x_is_zero = False # Is x=0 the solution to the least-squares prob? Arnorm = alpha * beta if Arnorm == 0.0: if show: print self.msg[0] x_is_zero = True istop = 0 rhobar = alpha ; 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 = alpha / beta if not x_is_zero else 1.0 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 if store_resids: self.resids.append(r2norm) self.normal_eqns_resids.append(Arnorm) # ------------------------------------------------------------------ # Main iteration loop. # ------------------------------------------------------------------ while itn < itnlim and not x_is_zero: itn = itn + 1 # Perform the next step of the bidiagonalization to obtain the # next beta, u, alpha, v. These satisfy the relations # beta*M*u = A*v - alpha*M*u, # alpha*N*v = A'*u - beta*N*v. Mu = A*v - alpha*Mu if M is not None: u = M(Mu) else: u = Mu beta = sqrt(dot(u,Mu)) # norm(u) if beta > 0: u /= beta if M is not None: Mu /= beta Anorm = normof4(Anorm, alpha, beta, damp) Nv = A.T*u - beta*Nv if N is not None: v = N(Nv) else: v = Nv alpha = sqrt(dot(v,Nv)) # norm(v) if alpha > 0: v /= alpha if N is not None: Nv /= alpha # 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 * alpha rhobar = - cs * alpha 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 += t1*w w *= t2 ; w += v ddnorm = ddnorm + norm(dk)**2 if wantvar: var += dk*dk if store_iterates: self.iterates.append(x.copy()) # Update energy norm of x. xNrgNorm2 += phi*phi dErr[itn % window] = phi if itn > window: trncDirErr = norm(dErr) xNrgNorm = sqrt(xNrgNorm2) self.dir_errors_window.append(trncDirErr / xNrgNorm) if trncDirErr < etol * xNrgNorm: istop = 8 # 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 += z*z # 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 = alpha * 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 if Anorm == 0. or rnorm == 0.: test2 = inf else: test2 = Arnorm/(Anorm * rnorm) if Acond == 0.0: test3 = inf else: test3 = 1.0 / Acond t1 = test1 / (1 + Anorm * xnorm / bnorm) rtol = btol + atol * Anorm * xnorm / bnorm if store_resids: self.resids.append(r2norm) self.normal_eqns_resids.append(Arnorm) # 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 # End of iteration loop. # Print the stopping condition. if show: print ' ' print 'LSQR finished' print self.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 str6 = 'xNrgNorm2 = %7.1e trnDirErr = %7.1e' % \ (xNrgNorm2, trncDirErr) print str1 + ' ' + str2 print str3 + ' ' + str4 print str5 print str6 print ' ' if istop == 0: self.status = 'solution is zero' if istop in [1,2,4,5]: self.status = 'residual small' if istop in [3,6]: self.status = 'ill-conditioned operator' if istop == 7: self.status = 'max iterations' if istop == 8: self.status = 'direct error small' self.optimal = istop in [1,2,4,5,8] self.x = self.bestSolution = x self.istop = istop self.itn = itn self.nMatvec = 2*itn self.r1norm = r1norm self.r2norm = r2norm self.residNorm = r2norm self.Anorm = Anorm self.Acond = Acond self.Arnorm = Arnorm self.xnorm = xnorm self.var = var return if __name__ == '__main__': # Solve the SQD system # [ 2 1 ] [x] = [2] # [ 1 -3 ] [y] [0] from pykrylov.linop import LinearOperator import numpy as np A = LinearOperator(1, 1, matvec=lambda u: u/2, symmetric=True) C = LinearOperator(1, 1, matvec=lambda v: v/3, symmetric=True) B = LinearOperator(1, 1, matvec=lambda x: x.copy(), symmetric=True) rhs = np.array([2.0]) lsqr = LSQRFramework(B) lsqr.solve(rhs, M=A, N=C, damp=1.0, show=True) print 'Solution: ', lsqr.x