def cond(A, norm=lambda x: mnorm(x,1)):
"""
Calculate the condition number of a matrix using a specified matrix norm.
The condition number estimates the sensitivity of a matrix to errors.
Example: small input errors for ill-conditionded coefficient matrices
alter the solution of the system dramatically.
For ill-conditioned matrices it's recommended to use qr_solve() instead
of lu_solve(). This does not help with input errors however, it just avoids
to add additional errors.
Definition: cond(A) = ||A|| * ||A**-1||
"""
return norm(A) * norm(inverse(A))
def cond(A, norm=lambda x: mnorm(x,1)):
"""
Calculate the condition number of a matrix using a specified matrix norm.
The condition number estimates the sensitivity of a matrix to errors.
Example: small input errors for ill-conditioned coefficient matrices
alter the solution of the system dramatically.
For ill-conditioned matrices it's recommended to use qr_solve() instead
of lu_solve(). This does not help with input errors however, it just avoids
to add additional errors.
Definition: cond(A) = ||A|| * ||A**-1||
"""
return norm(A) * norm(inverse(A))
def LU_decomp(A, overwrite=False, use_cache=True):
"""
LU-factorization of a n*n matrix using the Gauss algorithm.
Returns L and U in one matrix and the pivot indices.
Use overwrite to specify whether A will be overwritten with L and U.
"""
if not A.rows == A.cols:
raise ValueError('need n*n matrix')
# get from cache if possible
if use_cache and isinstance(A, matrix) and A._LU:
return A._LU
if not overwrite:
orig = A
A = A.copy()
tol = absmin(mnorm(A,1) * eps) # each pivot element has to be bigger
n = A.rows
p = [None]*(n - 1)
for j in xrange(n - 1):
# pivoting, choose max(abs(reciprocal row sum)*abs(pivot element))
biggest = 0
for k in xrange(j, n):
s = fsum([absmin(A[k,l]) for l in xrange(j, n)])
if absmin(s) <= tol:
raise ZeroDivisionError('matrix is numerically singular')
current = 1/s * absmin(A[k,j])
if current > biggest: # TODO: what if equal?
biggest = current
p[j] = k
# swap rows according to p
swap_row(A, j, p[j])
if absmin(A[j,j]) <= tol:
raise ZeroDivisionError('matrix is numerically singular')
# calculate elimination factors and add rows
for i in xrange(j + 1, n):
A[i,j] /= A[j,j]
for k in xrange(j + 1, n):
A[i,k] -= A[i,j]*A[j,k]
if absmin(A[n - 1,n - 1]) <= tol:
raise ZeroDivisionError('matrix is numerically singular')
# cache decomposition
if not overwrite and isinstance(orig, matrix):
orig._LU = (A, p)
return A, p
def LU_decomp(A, overwrite=False, use_cache=True):
"""
LU-factorization of a n*n matrix using the Gauss algorithm.
Returns L and U in one matrix and the pivot indices.
Use overwrite to specify whether A will be overwritten with L and U.
"""
if not A.rows == A.cols:
raise ValueError('need n*n matrix')
# get from cache if possible
if use_cache and isinstance(A, matrix) and A._LU:
return A._LU
if not overwrite:
orig = A
A = A.copy()
tol = absmin(mnorm(A,1) * eps) # each pivot element has to be bigger
n = A.rows
p = [None]*(n - 1)
for j in xrange(n - 1):
# pivoting, choose max(abs(reciprocal row sum)*abs(pivot element))
biggest = 0
for k in xrange(j, n):
s = fsum([absmin(A[k,l]) for l in xrange(j, n)])
if absmin(s) <= tol:
raise ZeroDivisionError('matrix is numerically singular')
current = 1/s * absmin(A[k,j])
if current > biggest: # TODO: what if equal?
biggest = current
p[j] = k
# swap rows according to p
swap_row(A, j, p[j])
if absmin(A[j,j]) <= tol:
raise ZeroDivisionError('matrix is numerically singular')
# calculate elimination factors and add rows
for i in xrange(j + 1, n):
A[i,j] /= A[j,j]
for k in xrange(j + 1, n):
A[i,k] -= A[i,j]*A[j,k]
if absmin(A[n - 1,n - 1]) <= tol:
raise ZeroDivisionError('matrix is numerically singular')
# cache decomposition
if not overwrite and isinstance(orig, matrix):
orig._LU = (A, p)
return A, p
def exp_pade(a):
"""Exponential of a matrix using Pade approximants.
See G. H. Golub, C. F. van Loan 'Matrix Computations',
third Ed., page 572
TODO:
- find a good estimate for q
- reduce the number of matrix multiplications to improve
performance
"""
def eps_pade(p):
return mpf(2)**(3-2*p) * factorial(p)**2/(factorial(2*p)**2 * (2*p + 1))
q = 4
extraq = 8
while 1:
if eps_pade(q) < eps:
break
q += 1
q += extraq
j = max(1, int(log(mnorm(a,'inf'),2)))
extra = q
mp.dps += extra
try:
a = a/2**j
na = a.rows
den = eye(na)
num = eye(na)
x = eye(na)
c = mpf(1)
for k in range(1, q+1):
c *= mpf(q - k + 1)/((2*q - k + 1) * k)
x = a*x
cx = c*x
num += cx
den += (-1)**k * cx
f = lu_solve_mat(den, num)
for k in range(j):
f = f*f
finally:
mp.dps -= extra
return f
def exp_pade(a):
"""Exponential of a matrix using Pade approximants.
See G. H. Golub, C. F. van Loan 'Matrix Computations',
third Ed., page 572
TODO:
- find a good estimate for q
- reduce the number of matrix multiplications to improve
performance
"""
def eps_pade(p):
return mpf(2)**(3-2*p) * factorial(p)**2/(factorial(2*p)**2 * (2*p + 1))
q = 4
extraq = 8
while 1:
if eps_pade(q) < eps:
break
q += 1
q += extraq
j = max(1, int(log(mnorm(a,'inf'),2)))
extra = q
mp.dps += extra
try:
a = a/2**j
na = a.rows
den = eye(na)
num = eye(na)
x = eye(na)
c = mpf(1)
for k in range(1, q+1):
c *= mpf(q - k + 1)/((2*q - k + 1) * k)
x = a*x
cx = c*x
num += cx
den += (-1)**k * cx
f = lu_solve_mat(den, num)
for k in range(j):
f = f*f
finally:
mp.dps -= extra
return f
