def explainedDemo(N, verbose=True, dense=False):
## view the docstring
#help(openmg.mgSolve)
## set up the problem
threshold = 1e-14
u_true = np.array([np.sin(x / 10.0) for x in range(N)])
A = openmg.operators.poisson(N, sparse=True)
b = openmg.tools.flexibleMmult(A, u_true)
## Use only the coarse solver.
params = {'verbose': False, 'threshold': threshold}
start = time()
soln = openmg.solvers.coarseSolve(A, b)
elapsed = time() - start
if verbose:
print N, "direct", np.linalg.norm(
openmg.tools.getresidual(b, A, soln, N)), elapsed
## Use an iterative solver.
## This Gauss-Seidel solver will be painfully slow for N larger than, say 200.
## It's probably the main bottleneck in this process.
if N <= 200:
start = time()
soln = openmg.smoothToThreshold(A,
b,
np.zeros((N, 1)),
params['threshold'],
verbose=params['verbose'])
elapsed = time() - start
if verbose:
print N, "Gauss-Seidel", np.linalg.norm(
openmg.tools.getresidual(b, A, soln, N)), elapsed
## Use a 3-grid pattern.
params = {
'problemShape': (N, ),
'gridLevels': 3,
'cycles': 0,
'iterations': 2,
'verbose': False,
'dense': dense,
'threshold': threshold,
'giveInfo': True,
'minSize': 30
}
# gridLevels is set to 3, but minSize is set to 30.
# the resulting restriction matrices would have shapes:
# (100, 200), (50, 100), and (25, 50)
# since 25 < 30, the minSize parameter prevails, and only the first two
# operators are generated.
start = time()
soln, info_dict = openmg.mgSolve(A, b, params)
cycles = info_dict['cycle']
elapsed = time() - start
#print params
if verbose: print N, "%i-grid"%params['gridLevels'],\
np.linalg.norm(openmg.tools.getresidual(b, A, soln, N)), elapsed, cycles
def explainedDemo(N, verbose=True, dense=False):
## view the docstring
#help(openmg.mgSolve)
## set up the problem
threshold = 1e-14
u_true = np.array([np.sin(x / 10.0) for x in range(N)])
A = openmg.operators.poisson(N, sparse=True)
b = openmg.tools.flexibleMmult(A, u_true)
## Use only the coarse solver.
params = {'verbose': False, 'threshold': threshold}
start = time()
soln = openmg.solvers.coarseSolve(A, b)
elapsed = time() - start
if verbose: print N, "direct", np.linalg.norm(openmg.tools.getresidual(b, A, soln, N)), elapsed
## Use an iterative solver.
## This Gauss-Seidel solver will be painfully slow for N larger than, say 200.
## It's probably the main bottleneck in this process.
if N <= 200:
start = time()
soln = openmg.smoothToThreshold(A, b, np.zeros((N, 1)),
params['threshold'],
verbose=params['verbose'])
elapsed = time() - start
if verbose: print N, "Gauss-Seidel", np.linalg.norm(openmg.tools.getresidual(b, A, soln, N)), elapsed
## Use a 3-grid pattern.
params = {'problemShape': (N,), 'gridLevels': 3, 'cycles': 0,
'iterations': 2, 'verbose': False, 'dense': dense,
'threshold': threshold, 'giveInfo': True, 'minSize': 30}
# gridLevels is set to 3, but minSize is set to 30.
# the resulting restriction matrices would have shapes:
# (100, 200), (50, 100), and (25, 50)
# since 25 < 30, the minSize parameter prevails, and only the first two
# operators are generated.
start = time()
soln, info_dict = openmg.mgSolve(A, b, params)
cycles = info_dict['cycle']
elapsed = time() - start
#print params
if verbose: print N, "%i-grid"%params['gridLevels'],\
np.linalg.norm(openmg.tools.getresidual(b, A, soln, N)), elapsed, cycles
def simpleDemo(verbose=False):
"""Solve A * u = b, where A is the 1D Poisson matrix,
u=sin(x), and x is in [0, 20)"""
N = 100
u_true = np.array([np.sin(x / 10.0) for x in np.linspace(0, 20, N)])
A = openmg.operators.poisson(N, sparse=True)
b = openmg.tools.flexibleMmult(A, u_true)
params = {
'problemShape': (N, ),
'gridLevels': 3,
'cycles': 10,
'iterations': 2,
'verbose': verbose,
'dense': True,
'threshold': 1e-2,
'giveInfo': True
}
u_mg, infoDict = openmg.mgSolve(A, b, params)
if verbose:
print "info:"
print infoDict
## if verbose==True, output will look something like this:
# Generating restriction matrices; dense=True
# Generating coefficient matrices; dense=True ... made 3 A matrices
# calling amg_cycle at level 0
# calling amg_cycle at level 1
# direct solving at level 2
# Residual norm from cycle 1 is 0.805398.
# cycle 1 < cycles 10
# calling amg_cycle at level 0
# calling amg_cycle at level 1
# direct solving at level 2
# Residual norm from cycle 2 is 0.107866.
# cycle 2 < cycles 10
# calling amg_cycle at level 0
# calling amg_cycle at level 1
# direct solving at level 2
# Residual norm from cycle 3 is 0.018650.
# cycle 3 < cycles 10
# calling amg_cycle at level 0
# calling amg_cycle at level 1
# direct solving at level 2
# Residual norm from cycle 4 is 0.003405.
# Returning mgSolve after 4 cycle(s) with norm 0.003405
# info:
# {'norm': 0.0034051536498270769, 'cycle': 4}
return u_mg
def simpleDemo(verbose=False):
"""Solve A * u = b, where A is the 1D Poisson matrix,
u=sin(x), and x is in [0, 20)"""
N = 100
u_true = np.array([np.sin(x / 10.0) for x in np.linspace(0, 20, N)])
A = openmg.operators.poisson(N, sparse=True)
b = openmg.tools.flexibleMmult(A, u_true)
params = {'problemShape': (N,), 'gridLevels': 3, 'cycles': 10,
'iterations': 2, 'verbose': verbose, 'dense': True,
'threshold': 1e-2, 'giveInfo': True}
u_mg, infoDict = openmg.mgSolve(A, b, params)
if verbose:
print "info:"
print infoDict
## if verbose==True, output will look something like this:
# Generating restriction matrices; dense=True
# Generating coefficient matrices; dense=True ... made 3 A matrices
# calling amg_cycle at level 0
# calling amg_cycle at level 1
# direct solving at level 2
# Residual norm from cycle 1 is 0.805398.
# cycle 1 < cycles 10
# calling amg_cycle at level 0
# calling amg_cycle at level 1
# direct solving at level 2
# Residual norm from cycle 2 is 0.107866.
# cycle 2 < cycles 10
# calling amg_cycle at level 0
# calling amg_cycle at level 1
# direct solving at level 2
# Residual norm from cycle 3 is 0.018650.
# cycle 3 < cycles 10
# calling amg_cycle at level 0
# calling amg_cycle at level 1
# direct solving at level 2
# Residual norm from cycle 4 is 0.003405.
# Returning mgSolve after 4 cycle(s) with norm 0.003405
# info:
# {'norm': 0.0034051536498270769, 'cycle': 4}
return u_mg