"""Compatible Relaxation"""
__docformat__ = "restructuredtext en"
import numpy
import scipy
from scipy.linalg import norm
from scipy.sparse import isspmatrix, csr_matrix, spdiags
from pyamg.relaxation import gauss_seidel
__all__ = ['CR','binormalize']
[docs]def CR(S, method='habituated',maxiter=20):
"""Use Compatible Relaxation to compute a C/F splitting
Parameters
----------
S : csr_matrix
sparse matrix (n x n) usually matrix A of Ax=b
method : {'habituated','concurrent'}
Method used during relaxation:
- concurrent: GS relaxation on F-points, leaving e_c = 0
- habituated: full relaxation, setting e_c = 0
maxiter : int
maximum number of outer iterations (lambda)
Returns
-------
splitting : array
C/F list of 1's (coarse pt) and 0's (fine pt) (n x 1)
References
----------
.. [1] Livne, O.E., "Coarsening by compatible relaxation."
Numer. Linear Algebra Appl. 11, No. 2-3, 205-227 (2004).
Examples
--------
>>> from pyamg.gallery import poisson
>>> from pyamg.classical.cr import CR
>>> A = poisson((20,20),format='csr')
>>> splitting = CR(A)
"""
# parameters (paper notation)
ntests = 3 # (nu) number of random tests to do per iteration
nrelax = 4 # (eta) number of relaxation sweeps per test
smagic = 1.0 # (s) parameter in [1,5] to account for fill-in
gamma = 1.5 # (gamma) cycle index. use 1.5 for 2d
G = 30 # (G) number of equivalence classes (# of bins)
tdepth = 1 # (t) drop depth on parse of L bins
delta = 0 # (delta) drop threshold on parse of L bins
alphai = 0.25 # (alpha_inc) quota increase
# initializations
alpha = 0.0 # coarsening ratio, quota
beta = numpy.inf # quality criterion
beta1 = numpy.inf # quality criterion, older
beta2 = numpy.inf # quality criterion, oldest
n=S.shape[0] # problem size
nC = 0 # number of current Coarse points
rhs = numpy.zeros((n,1)); # rhs for Ae=0
if not isspmatrix(S): raise TypeError('expecting sparse matrix')
S = binormalize(S)
splitting = numpy.zeros( (S.shape[0],1), dtype='intc' )
# out iterations ---------------
for m in range(0,maxiter):
mu = 0.0 # convergence rate
E = numpy.zeros((n,1)) # slowness measure
# random iterations ---------------
for k in range(0,ntests):
e = 0.5*( 1 + scipy.rand(n,1))
e[splitting>0] = 0
enorm = norm(e)
# relaxation iterations ---------------
for l in range(0,nrelax):
if method == 'habituated':
gauss_seidel(S,e,numpy.zeros((n,1)),iterations=1)
e[splitting>0]=0
elif method == 'concurrent':
raise NotImplementedError, 'not implemented: need an F-smoother'
else:
raise NotImplementedError, 'method not recognized: need habituated or concurrent'
enorm_old = enorm
enorm = norm(e)
if enorm <= 1e-14:
# break out of loops
ntests = k
nrelax = l
maxiter = m
# end relax
# check slowness
E = numpy.where( numpy.abs(e)>E, numpy.abs(e), E )
# update convergence rate
mu = mu + enorm/enorm_old
# end random tests
mu = mu/ntests
# work
alpha = float(nC)/n
W = (1 + (smagic-1)*gamma*alpha)/(1-gamma*alpha)
# quality criterion
beta2 = beta1
beta1 = beta
beta = numpy.power(max([mu, 0.1]), 1.0 / W)
# check if we're doing well
if (beta>beta1 and beta1>beta2) or m==(maxiter-1) or max(E)<1e-13:
return splitting.ravel()
# now add points
#
# update limit on additions to splitting (C)
if alpha < 1e-13:
alpha=0.25
else:
alpha = (1-alphai) * alpha + alphai * (1/gamma)
nCmax = numpy.ceil( alpha * n )
L = numpy.ceil( G * E / E.max() ).ravel()
binid=G
# add whole bins (and t-depth nodes) at a time
u = numpy.zeros((n,1))
# TODO This loop may never halt...
# Perhaps loop over nC < nCmax and binid > 0 ?
while nC < nCmax:
if delta > 0:
raise NotImplementedError
if tdepth != 1:
raise NotImplementedError
(roots,) = numpy.where(L==binid)
for root in roots:
if L[root]>=0:
cols = S[root,:].indices
splitting[root] = 1 # add roots
nC += 1
L[cols]=-1
binid -= 1
#L[troots] = -1 # mark t-rings visited
#u[:]=0.0
#u[roots] = 1.0
#for depth in range(0,tdepth):
# u = numpy.abs(S) * u
#(troots,tmp) = numpy.where(u>0)
return splitting.ravel()
[docs]def binormalize( A, tol=1e-5, maxiter=10):
"""Binormalize matrix A. Attempt to create unit l_1 norm rows.
Parameters
----------
A : csr_matrix
sparse matrix (n x n)
tol : float
tolerance
x : array
guess at the diagonal
maxiter : int
maximum number of iterations to try
Returns
-------
C : csr_matrix
diagonally scaled A, C=DAD
Notes
-----
- Goal: Scale A so that l_1 norm of the rows are equal to 1:
- B = DAD
- want row sum of B = 1
- easily done with tol=0 if B=DA, but this is not symmetric
- algorithm is O(N log (1.0/tol))
Examples
--------
>>> from pyamg.gallery import poisson
>>> from pyamg.classical import binormalize
>>> A = poisson((10,),format='csr')
>>> C = binormalize(A)
References
----------
.. [1] Livne, Golub, "Scaling by Binormalization"
Tech Report SCCM-03-12, SCCM, Stanford, 2003
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.3.1679
"""
if not isspmatrix(A):
raise TypeError('expecting sparse matrix A')
if A.dtype==complex:
raise NotImplementedError('complex A not implemented')
n = A.shape[0]
it = 0
x = numpy.ones((n,1)).ravel()
# 1.
B = A.multiply(A).tocsc() # power(A,2) inconsistent for numpy, scipy.sparse
d=B.diagonal().ravel()
# 2.
beta = B * x
betabar = (1.0/n) * numpy.dot(x,beta)
stdev = rowsum_stdev(x,beta)
#3
while stdev > tol and it < maxiter:
for i in range(0,n):
# solve equation x_i, keeping x_j's fixed
# see equation (12)
c2 = (n-1)*d[i]
c1 = (n-2)*(beta[i] - d[i]*x[i])
c0 = -d[i]*x[i]*x[i] + 2*beta[i]*x[i] - n*betabar
if (-c0 < 1e-14):
print 'warning: A nearly un-binormalizable...'
return A
else:
# see equation (12)
xnew = (2*c0)/(-c1 - numpy.sqrt(c1*c1 - 4*c0*c2))
dx = xnew - x[i]
# here we assume input matrix is symmetric since we grab a row of B
# instead of a column
ii = B.indptr[i]
iii = B.indptr[i+1]
dot_Bcol = numpy.dot(x[B.indices[ii:iii]],B.data[ii:iii])
betabar = betabar + (1.0/n)*dx*(dot_Bcol + beta[i] + d[i]*dx)
beta[B.indices[ii:iii]] += dx*B.data[ii:iii]
x[i] = xnew
stdev = rowsum_stdev(x,beta)
it+=1
# rescale for unit 2-norm
d = numpy.sqrt(x)
D = spdiags( d.ravel(), [0], n,n)
C = D * A * D
C = C.tocsr()
beta = C.multiply(C).sum(axis=1)
scale = numpy.sqrt((1.0/n) * numpy.sum(beta))
return (1/scale)*C
def rowsum_stdev(x,beta):
"""Compute row sum standard deviation
Compute for approximation x, the std dev of the row sums
s(x) = ( 1/n \sum_k (x_k beta_k - betabar)^2 )^(1/2)
with betabar = 1/n dot(beta,x)
Parameters
----------
x : array
beta : array
Returns
-------
s(x)/betabar : float
Notes
-----
equation (7) in Livne/Golub
"""
n=x.size
betabar = (1.0/n) * numpy.dot(x,beta)
stdev = numpy.sqrt((1.0/n)*numpy.sum(numpy.power(numpy.multiply(x,beta) - betabar,2)))
return stdev/betabar