def test_onenormest_table_6_t_1(self):
#TODO this test seems to give estimates that match the table,
#TODO even though no attempt has been made to deal with
#TODO complex numbers in the one-norm estimation.
# This will take multiple seconds if your computer is slow like mine.
# It is stochastic, so the tolerance could be too strict.
np.random.seed(1234)
t = 1
n = 100
itmax = 5
nsamples = 5000
observed = []
expected = []
nmult_list = []
nresample_list = []
for i in range(nsamples):
A_inv = np.random.rand(n, n) + 1j * np.random.rand(n, n)
A = scipy.linalg.inv(A_inv)
est, v, w, nmults, nresamples = _onenormest_core(A, A.T, t, itmax)
observed.append(est)
expected.append(scipy.linalg.norm(A, 1))
nmult_list.append(nmults)
nresample_list.append(nresamples)
observed = np.array(observed, dtype=float)
expected = np.array(expected, dtype=float)
relative_errors = np.abs(observed - expected) / expected
# check the mean underestimation ratio
underestimation_ratio = observed / expected
underestimation_ratio_mean = np.mean(underestimation_ratio)
assert_(0.90 < underestimation_ratio_mean < 0.99)
# check the required column resamples
max_nresamples = np.max(nresample_list)
assert_equal(max_nresamples, 0)
# check the proportion of norms computed exactly correctly
nexact = _count_nonzero(relative_errors < 1e-14)
proportion_exact = nexact / float(nsamples)
assert_(0.7 < proportion_exact < 0.8)
# check the average number of matrix*vector multiplications
mean_nmult = np.mean(nmult_list)
assert_(4 < mean_nmult < 5)
def test_onenormest_table_6_t_1(self):
#TODO this test seems to give estimates that match the table,
#TODO even though no attempt has been made to deal with
#TODO complex numbers in the one-norm estimation.
# This will take multiple seconds if your computer is slow like mine.
# It is stochastic, so the tolerance could be too strict.
np.random.seed(1234)
t = 1
n = 100
itmax = 5
nsamples = 5000
observed = []
expected = []
nmult_list = []
nresample_list = []
for i in range(nsamples):
A_inv = np.random.rand(n, n) + 1j * np.random.rand(n, n)
A = scipy.linalg.inv(A_inv)
est, v, w, nmults, nresamples = _onenormest_core(A, A.T, t, itmax)
observed.append(est)
expected.append(scipy.linalg.norm(A, 1))
nmult_list.append(nmults)
nresample_list.append(nresamples)
observed = np.array(observed, dtype=float)
expected = np.array(expected, dtype=float)
relative_errors = np.abs(observed - expected) / expected
# check the mean underestimation ratio
underestimation_ratio = observed / expected
underestimation_ratio_mean = np.mean(underestimation_ratio)
assert_(0.90 < underestimation_ratio_mean < 0.99)
# check the required column resamples
max_nresamples = np.max(nresample_list)
assert_equal(max_nresamples, 0)
# check the proportion of norms computed exactly correctly
nexact = _count_nonzero(relative_errors < 1e-14)
proportion_exact = nexact / float(nsamples)
assert_(0.7 < proportion_exact < 0.8)
# check the average number of matrix*vector multiplications
mean_nmult = np.mean(nmult_list)
assert_(4 < mean_nmult < 5)
def test_onenormest_table_3_t_2(self):
# This will take multiple seconds if your computer is slow like mine.
# It is stochastic, so the tolerance could be too strict.
np.random.seed(1234)
t = 2
n = 100
itmax = 5
nsamples = 5000
observed = []
expected = []
nmult_list = []
nresample_list = []
for i in range(nsamples):
A = scipy.linalg.inv(np.random.randn(n, n))
est, v, w, nmults, nresamples = _onenormest_core(A, A.T, t, itmax)
observed.append(est)
expected.append(scipy.linalg.norm(A, 1))
nmult_list.append(nmults)
nresample_list.append(nresamples)
observed = np.array(observed, dtype=float)
expected = np.array(expected, dtype=float)
relative_errors = np.abs(observed - expected) / expected
# check the mean underestimation ratio
underestimation_ratio = observed / expected
assert_(0.99 < np.mean(underestimation_ratio) < 1.0)
# check the max and mean required column resamples
assert_equal(np.max(nresample_list), 2)
assert_(0.05 < np.mean(nresample_list) < 0.2)
# check the proportion of norms computed exactly correctly
nexact = _count_nonzero(relative_errors < 1e-14)
proportion_exact = nexact / float(nsamples)
assert_(0.9 < proportion_exact < 0.95)
# check the average number of matrix*vector multiplications
assert_(3.5 < np.mean(nmult_list) < 4.5)
def test_onenormest_table_3_t_2(self):
# This will take multiple seconds if your computer is slow like mine.
# It is stochastic, so the tolerance could be too strict.
np.random.seed(1234)
t = 2
n = 100
itmax = 5
nsamples = 5000
observed = []
expected = []
nmult_list = []
nresample_list = []
for i in range(nsamples):
A = scipy.linalg.inv(np.random.randn(n, n))
est, v, w, nmults, nresamples = _onenormest_core(A, A.T, t, itmax)
observed.append(est)
expected.append(scipy.linalg.norm(A, 1))
nmult_list.append(nmults)
nresample_list.append(nresamples)
observed = np.array(observed, dtype=float)
expected = np.array(expected, dtype=float)
relative_errors = np.abs(observed - expected) / expected
# check the mean underestimation ratio
underestimation_ratio = observed / expected
assert_(0.99 < np.mean(underestimation_ratio) < 1.0)
# check the max and mean required column resamples
assert_equal(np.max(nresample_list), 2)
assert_(0.05 < np.mean(nresample_list) < 0.2)
# check the proportion of norms computed exactly correctly
nexact = _count_nonzero(relative_errors < 1e-14)
proportion_exact = nexact / float(nsamples)
assert_(0.9 < proportion_exact < 0.95)
# check the average number of matrix*vector multiplications
assert_(3.5 < np.mean(nmult_list) < 4.5)
def _linprog_simplex(c,
A_ub=None,
b_ub=None,
A_eq=None,
b_eq=None,
bounds=None,
maxiter=1000,
disp=False,
callback=None,
tol=1.0E-12,
bland=False,
**unknown_options):
"""
Solve the following linear programming problem via a two-phase
simplex algorithm.
maximize: c^T * x
subject to: A_ub * x <= b_ub
A_eq * x == b_eq
Parameters
----------
c : array_like
Coefficients of the linear objective function to be maximized.
A_ub :
2-D array which, when matrix-multiplied by x, gives the values of the
upper-bound inequality constraints at x.
b_ub : array_like
1-D array of values representing the upper-bound of each inequality
constraint (row) in A_ub.
A_eq : array_like
2-D array which, when matrix-multiplied by x, gives the values of the
equality constraints at x.
b_eq : array_like
1-D array of values representing the RHS of each equality constraint
(row) in A_eq.
bounds : array_like
The bounds for each independent variable in the solution, which can take
one of three forms::
None : The default bounds, all variables are non-negative.
(lb, ub) : If a 2-element sequence is provided, the same
lower bound (lb) and upper bound (ub) will be applied
to all variables.
[(lb_0, ub_0), (lb_1, ub_1), ...] : If an n x 2 sequence is provided,
each variable x_i will be bounded by lb[i] and ub[i].
Infinite bounds are specified using -np.inf (negative)
or np.inf (positive).
maxiter : int
The maximum number of iterations to perform.
disp : bool
If True, print exit status message to sys.stdout
callback : callable
If a callback function is provide, it will be called within each
iteration of the simplex algorithm. The callback must have the
signature `callback(xk, **kwargs)` where xk is the current solution
vector and kwargs is a dictionary containing the following::
"tableau" : The current Simplex algorithm tableau
"nit" : The current iteration.
"pivot" : The pivot (row, column) used for the next iteration.
"phase" : Whether the algorithm is in Phase 1 or Phase 2.
"bv" : A structured array containing a string representation of each
basic variable and its current value.
tol : float
The tolerance which determines when a solution is "close enough" to zero
in Phase 1 to be considered a basic feasible solution or close enough
to positive to to serve as an optimal solution.
bland : bool
If True, use Bland's anti-cycling rule [3] to choose pivots to
prevent cycling. If False, choose pivots which should lead to a
converged solution more quickly. The latter method is subject to
cycling (non-convergence) in rare instances.
Returns
-------
A scipy.optimize.OptimizeResult consisting of the following fields::
x : ndarray
The independent variable vector which optimizes the linear
programming problem.
slack : ndarray
The values of the slack variables. Each slack variable corresponds
to an inequality constraint. If the slack is zero, then the
corresponding constraint is active.
success : bool
Returns True if the algorithm succeeded in finding an optimal
solution.
status : int
An integer representing the exit status of the optimization::
0 : Optimization terminated successfully
1 : Iteration limit reached
2 : Problem appears to be infeasible
3 : Problem appears to be unbounded
nit : int
The number of iterations performed.
message : str
A string descriptor of the exit status of the optimization.
Examples
--------
Consider the following problem:
Minimize: f = -1*x[0] + 4*x[1]
Subject to: -3*x[0] + 1*x[1] <= 6
1*x[0] + 2*x[1] <= 4
x[1] >= -3
where: -inf <= x[0] <= inf
This problem deviates from the standard linear programming problem. In
standard form, linear programming problems assume the variables x are
non-negative. Since the variables don't have standard bounds where
0 <= x <= inf, the bounds of the variables must be explicitly set.
There are two upper-bound constraints, which can be expressed as
dot(A_ub, x) <= b_ub
The input for this problem is as follows:
>>> c = [-1, 4]
>>> A = [[-3, 1], [1, 2]]
>>> b = [6, 4]
>>> x0_bnds = (None, None)
>>> x1_bnds = (-3, None)
>>> res = linprog(c, A, b, bounds=(x0_bnds, x1_bnds), options={"disp":True})
>>> print(res)
Optimization terminated successfully.
Current function value: 11.428571
Iterations: 2
status: 0
success: True
fun: 11.428571428571429
x: array([-1.14285714, 2.57142857])
slack: array([], dtype=np.float64)
message: 'Optimization terminated successfully.'
nit: 2
References
----------
.. [1] Dantzig, George B., Linear programming and extensions. Rand
Corporation Research Study Princeton Univ. Press, Princeton, NJ, 1963
.. [2] Hillier, S.H. and Lieberman, G.J. (1995), "Introduction to
Mathematical Programming", McGraw-Hill, Chapter 4.
.. [3] Bland, Robert G. New finite pivoting rules for the simplex method.
Mathematics of Operations Research (2), 1977: pp. 103-107.
"""
_check_unknown_options(unknown_options)
status = 0
messages = {
0: "Optimization terminated successfully.",
1: "Iteration limit reached.",
2: "Optimzation failed. Unable to find a feasible"
" starting point.",
3: "Optimization failed. The problem appears to be unbounded.",
4: "Optimization failed. Singular matrix encountered."
}
have_floor_variable = False
cc = np.asarray(c)
# The initial value of the objective function element in the tableau
f0 = 0
# The number of variables as given by c
n = len(c)
# Convert the input arguments to arrays (sized to zero if not provided)
Aeq = np.asarray(A_eq) if A_eq is not None else np.empty([0, len(cc)])
Aub = np.asarray(A_ub) if A_ub is not None else np.empty([0, len(cc)])
beq = np.ravel(np.asarray(b_eq)) if b_eq is not None else np.empty([0])
bub = np.ravel(np.asarray(b_ub)) if b_ub is not None else np.empty([0])
# Analyze the bounds and determine what modifications to me made to
# the constraints in order to accommodate them.
L = np.zeros(n, dtype=np.float64)
U = np.ones(n, dtype=np.float64) * np.inf
if bounds is None or len(bounds) == 0:
pass
elif len(bounds) == 2 and not hasattr(bounds[0], '__len__'):
# All bounds are the same
L = np.asarray(n * [bounds[0]], dtype=np.float64)
U = np.asarray(n * [bounds[1]], dtype=np.float64)
else:
if len(bounds) != n:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Length of bounds is inconsistent with the length of c")
else:
try:
for i in range(n):
if len(bounds[i]) != 2:
raise IndexError()
L[i] = bounds[i][0] if bounds[i][0] is not None else -np.inf
U[i] = bounds[i][1] if bounds[i][1] is not None else np.inf
except IndexError:
status = -1
message = ("Invalid input for linprog with "
"method = 'simplex'. bounds must be a n x 2 "
"sequence/array where n = len(c).")
if np.any(L == -np.inf):
# If any lower-bound constraint is a free variable
# add the first column variable as the "floor" variable which
# accommodates the most negative variable in the problem.
n = n + 1
L = np.concatenate([np.array([0]), L])
U = np.concatenate([np.array([np.inf]), U])
cc = np.concatenate([np.array([0]), cc])
Aeq = np.hstack([np.zeros([Aeq.shape[0], 1]), Aeq])
Aub = np.hstack([np.zeros([Aub.shape[0], 1]), Aub])
have_floor_variable = True
# Now before we deal with any variables with lower bounds < 0,
# deal with finite bounds which can be simply added as new constraints.
# Also validate bounds inputs here.
for i in range(n):
if (L[i] > U[i]):
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Lower bound %d is greater than upper bound %d" %
(i, i))
if np.isinf(L[i]) and L[i] > 0:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Lower bound may not be +infinity")
if np.isinf(U[i]) and U[i] < 0:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Upper bound may not be -infinity")
if np.isfinite(L[i]) and L[i] > 0:
# Add a new lower-bound (negative upper-bound) constraint
Aub = np.vstack([Aub, np.zeros(n)])
Aub[-1, i] = -1
bub = np.concatenate([bub, np.array([-L[i]])])
L[i] = 0
if np.isfinite(U[i]):
# Add a new upper-bound constraint
Aub = np.vstack([Aub, np.zeros(n)])
Aub[-1, i] = 1
bub = np.concatenate([bub, np.array([U[i]])])
U[i] = np.inf
# Now find negative lower bounds (finite or infinite) which require a
# change of variables or free variables and handle them appropriately
for i in range(0, n):
if L[i] < 0:
if np.isfinite(L[i]) and L[i] < 0:
# Add a change of variables for x[i]
# For each row in the constraint matrices, we take the
# coefficient from column i in A,
# and subtract the product of that and L[i] to the RHS b
beq[:] = beq[:] - Aeq[:, i] * L[i]
bub[:] = bub[:] - Aub[:, i] * L[i]
# We now have a nonzero initial value for the objective
# function as well.
f0 = f0 - cc[i] * L[i]
else:
# This is an unrestricted variable, let x[i] = u[i] - v[0]
# where v is the first column in all matrices.
Aeq[:, 0] = Aeq[:, 0] - Aeq[:, i]
Aub[:, 0] = Aub[:, 0] - Aub[:, i]
cc[0] = cc[0] - cc[i]
if np.isinf(U[i]):
if U[i] < 0:
status = -1
message = ("Invalid input for linprog with "
"method = 'simplex'. Upper bound may not be -inf.")
# The number of upper bound constraints (rows in A_ub and elements in b_ub)
mub = len(bub)
# The number of equality constraints (rows in A_eq and elements in b_eq)
meq = len(beq)
# The total number of constraints
m = mub + meq
# The number of slack variables (one for each of the upper-bound constraints)
n_slack = mub
# The number of artificial variables (one for each lower-bound and equality
# constraint)
n_artificial = meq + _count_nonzero(bub < 0)
try:
Aub_rows, Aub_cols = Aub.shape
except ValueError:
raise ValueError("Invalid input. A_ub must be two-dimensional")
try:
Aeq_rows, Aeq_cols = Aeq.shape
except ValueError:
raise ValueError("Invalid input. A_eq must be two-dimensional")
if Aeq_rows != meq:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"The number of rows in A_eq must be equal "
"to the number of values in b_eq")
if Aub_rows != mub:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"The number of rows in A_ub must be equal "
"to the number of values in b_ub")
if Aeq_cols > 0 and Aeq_cols != n:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Number of columns in A_eq must be equal "
"to the size of c")
if Aub_cols > 0 and Aub_cols != n:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Number of columns in A_ub must be equal to the size of c")
if status != 0:
# Invalid inputs provided
raise ValueError(message)
# Create the tableau
T = np.zeros([m + 2, n + n_slack + n_artificial + 1])
# Insert objective into tableau
T[-2, :n] = cc
T[-2, -1] = f0
b = T[:-2, -1]
if meq > 0:
# Add Aeq to the tableau
T[:meq, :n] = Aeq
# Add beq to the tableau
b[:meq] = beq
if mub > 0:
# Add Aub to the tableau
T[meq:meq + mub, :n] = Aub
# At bub to the tableau
b[meq:meq + mub] = bub
# Add the slack variables to the tableau
np.fill_diagonal(T[meq:m, n:n + n_slack], 1)
# Further setup the tableau
# If a row corresponds to an equality constraint or a negative b (a lower
# bound constraint), then an artificial variable is added for that row.
# Also, if b is negative, first flip the signs in that constraint.
slcount = 0
avcount = 0
basis = np.zeros(m, dtype=int)
r_artificial = np.zeros(n_artificial, dtype=int)
for i in range(m):
if i < meq or b[i] < 0:
# basic variable i is in column n+n_slack+avcount
basis[i] = n + n_slack + avcount
r_artificial[avcount] = i
avcount += 1
if b[i] < 0:
b[i] *= -1
T[i, :-1] *= -1
T[i, basis[i]] = 1
T[-1, basis[i]] = 1
else:
# basic variable i is in column n+slcount
basis[i] = n + slcount
slcount += 1
# Make the artificial variables basic feasible variables by subtracting
# each row with an artificial variable from the Phase 1 objective
for r in r_artificial:
T[-1, :] = T[-1, :] - T[r, :]
nit1, status = _solve_simplex(T,
n,
basis,
phase=1,
callback=callback,
maxiter=maxiter,
tol=tol,
bland=bland)
# if pseudo objective is zero, remove the last row from the tableau and
# proceed to phase 2
if abs(T[-1, -1]) < tol:
# Remove the pseudo-objective row from the tableau
T = T[:-1, :]
# Remove the artificial variable columns from the tableau
T = np.delete(T, np.s_[n + n_slack:n + n_slack + n_artificial], 1)
else:
# Failure to find a feasible starting point
status = 2
if status != 0:
message = messages[status]
if disp:
print(message)
return OptimizeResult(x=np.nan,
fun=-T[-1, -1],
nit=nit1,
status=status,
message=message,
success=False)
# Phase 2
nit2, status = _solve_simplex(T,
n,
basis,
maxiter=maxiter - nit1,
phase=2,
callback=callback,
tol=tol,
nit0=nit1,
bland=bland)
solution = np.zeros(n + n_slack + n_artificial)
solution[basis[:m]] = T[:m, -1]
x = solution[:n]
slack = solution[n:n + n_slack]
# For those variables with finite negative lower bounds,
# reverse the change of variables
masked_L = np.ma.array(L, mask=np.isinf(L), fill_value=0.0).filled()
x = x + masked_L
# For those variables with infinite negative lower bounds,
# take x[i] as the difference between x[i] and the floor variable.
if have_floor_variable:
for i in range(1, n):
if np.isinf(L[i]):
x[i] -= x[0]
x = x[1:]
# Optimization complete at this point
obj = -T[-1, -1]
if status in (0, 1):
if disp:
print(messages[status])
print(" Current function value: {: <12.6f}".format(obj))
print(" Iterations: {:d}".format(nit2))
else:
if disp:
print(messages[status])
print(" Iterations: {:d}".format(nit2))
return OptimizeResult(x=x,
fun=obj,
nit=int(nit2),
status=status,
slack=slack,
message=messages[status],
success=(status == 0))
def _remainder_matrix_power(A, t):
"""
Compute the fractional power of a matrix, for fractions -1 < t < 1.
This uses algorithm (3.1) of [1]_.
The Pade approximation itself uses algorithm (4.1) of [2]_.
Parameters
----------
A : (N, N) array_like
Matrix whose fractional power to evaluate.
t : float
Fractional power between -1 and 1 exclusive.
Returns
-------
X : (N, N) array_like
The fractional power of the matrix.
References
----------
.. [1] Nicholas J. Higham and Lijing Lin (2013)
"An Improved Schur-Pade Algorithm for Fractional Powers
of a Matrix and their Frechet Derivatives."
.. [2] Nicholas J. Higham and Lijing lin (2011)
"A Schur-Pade Algorithm for Fractional Powers of a Matrix."
SIAM Journal on Matrix Analysis and Applications,
32 (3). pp. 1056-1078. ISSN 0895-4798
"""
# This code block is copied from numpy.matrix_power().
A = np.asarray(A)
if len(A.shape) != 2 or A.shape[0] != A.shape[1]:
raise ValueError('input must be a square array')
# Get the number of rows and columns.
n, n = A.shape
# Triangularize the matrix if necessary,
# attempting to preserve dtype if possible.
if np.array_equal(A, np.triu(A)):
Z = None
T = A
else:
if np.isrealobj(A):
T, Z = schur(A)
if not np.array_equal(T, np.triu(T)):
T, Z = rsf2csf(T, Z)
else:
T, Z = schur(A, output='complex')
# Zeros on the diagonal of the triangular matrix are forbidden,
# because the inverse scaling and squaring cannot deal with it.
T_diag = np.diag(T)
if _count_nonzero(T_diag) != n:
raise FractionalMatrixPowerError(
'cannot use inverse scaling and squaring to find '
'the fractional matrix power of a singular matrix')
# If the triangular matrix is real and has a negative
# entry on the diagonal, then force the matrix to be complex.
if np.isrealobj(T) and np.min(T_diag) < 0:
T = T.astype(complex)
# Get the fractional power of the triangular matrix,
# and de-triangularize it if necessary.
U = _remainder_matrix_power_triu(T, t)
if Z is not None:
ZH = np.conjugate(Z).T
return Z.dot(U).dot(ZH)
else:
return U
def _inverse_squaring_helper(T0, theta):
"""
A helper function for inverse scaling and squaring for Pade approximation.
Parameters
----------
T0 : (N, N) array_like upper triangular
Matrix involved in inverse scaling and squaring.
theta : indexable
The values theta[1] .. theta[7] must be available.
They represent bounds related to Pade approximation, and they depend
on the matrix function which is being computed.
For example, different values of theta are required for
matrix logarithm than for fractional matrix power.
Returns
-------
R : (N, N) array_like upper triangular
Composition of zero or more matrix square roots of T0, minus I.
s : non-negative integer
Number of square roots taken.
m : positive integer
The degree of the Pade approximation.
Notes
-----
This subroutine appears as a chunk of lines within
a couple of published algorithms; for example it appears
as lines 4--35 in algorithm (3.1) of [1]_, and
as lines 3--34 in algorithm (4.1) of [2]_.
The instances of 'goto line 38' in algorithm (3.1) of [1]_
probably mean 'goto line 36' and have been intepreted accordingly.
References
----------
.. [1] Nicholas J. Higham and Lijing Lin (2013)
"An Improved Schur-Pade Algorithm for Fractional Powers
of a Matrix and their Frechet Derivatives."
.. [2] Awad H. Al-Mohy and Nicholas J. Higham (2012)
"Improved Inverse Scaling and Squaring Algorithms
for the Matrix Logarithm."
SIAM Journal on Scientific Computing, 34 (4). C152-C169.
ISSN 1095-7197
"""
if len(T0.shape) != 2 or T0.shape[0] != T0.shape[1]:
raise ValueError('expected an upper triangular square matrix')
n, n = T0.shape
T = T0
# Find s0, the smallest s such that the spectral radius
# of a certain diagonal matrix is at most theta[7].
# Note that because theta[7] < 1,
# this search will not terminate if any diagonal entry of T is zero.
s0 = 0
tmp_diag = np.diag(T)
if _count_nonzero(tmp_diag) != n:
raise Exception('internal inconsistency')
while np.max(np.absolute(tmp_diag - 1)) > theta[7]:
tmp_diag = np.sqrt(tmp_diag)
s0 += 1
# Take matrix square roots of T.
for i in range(s0):
T = _sqrtm_triu(T)
# Flow control in this section is a little odd.
# This is because I am translating algorithm descriptions
# which have GOTOs in the publication.
s = s0
k = 0
d2 = _onenormest_m1_power(T, 2)**(1 / 2)
d3 = _onenormest_m1_power(T, 3)**(1 / 3)
a2 = max(d2, d3)
m = None
for i in (1, 2):
if a2 <= theta[i]:
m = i
break
while m is None:
if s > s0:
d3 = _onenormest_m1_power(T, 3)**(1 / 3)
d4 = _onenormest_m1_power(T, 4)**(1 / 4)
a3 = max(d3, d4)
if a3 <= theta[7]:
j1 = min(i for i in (3, 4, 5, 6, 7) if a3 <= theta[i])
if j1 <= 6:
m = j1
break
elif a3 / 2 <= theta[5] and k < 2:
k += 1
T = _sqrtm_triu(T)
s += 1
continue
d5 = _onenormest_m1_power(T, 5)**(1 / 5)
a4 = max(d4, d5)
eta = min(a3, a4)
for i in (6, 7):
if eta <= theta[i]:
m = i
break
if m is not None:
break
T = _sqrtm_triu(T)
s += 1
# The subtraction of the identity is redundant here,
# because the diagonal will be replaced for improved numerical accuracy,
# but this formulation should help clarify the meaning of R.
R = T - np.identity(n)
# Replace the diagonal and first superdiagonal of T0^(1/(2^s)) - I
# using formulas that have less subtractive cancellation.
# Skip this step if the principal branch
# does not exist at T0; this happens when a diagonal entry of T0
# is negative with imaginary part 0.
has_principal_branch = all(x.real > 0 or x.imag != 0 for x in np.diag(T0))
if has_principal_branch:
for j in range(n):
a = T0[j, j]
r = _briggs_helper_function(a, s)
R[j, j] = r
p = np.exp2(-s)
for j in range(n - 1):
l1 = T0[j, j]
l2 = T0[j + 1, j + 1]
t12 = T0[j, j + 1]
f12 = _fractional_power_superdiag_entry(l1, l2, t12, p)
R[j, j + 1] = f12
# Return the T-I matrix, the number of square roots, and the Pade degree.
if not np.array_equal(R, np.triu(R)):
raise Exception('internal inconsistency')
return R, s, m
def _linprog_simplex(c, A_ub=None, b_ub=None, A_eq=None, b_eq=None,
bounds=None, maxiter=1000, disp=False, callback=None,
tol=1.0E-12, bland=False, **unknown_options):
"""
Solve the following linear programming problem via a two-phase
simplex algorithm.
maximize: c^T * x
subject to: A_ub * x <= b_ub
A_eq * x == b_eq
Parameters
----------
c : array_like
Coefficients of the linear objective function to be maximized.
A_ub :
2-D array which, when matrix-multiplied by x, gives the values of the
upper-bound inequality constraints at x.
b_ub : array_like
1-D array of values representing the upper-bound of each inequality
constraint (row) in A_ub.
A_eq : array_like
2-D array which, when matrix-multiplied by x, gives the values of the
equality constraints at x.
b_eq : array_like
1-D array of values representing the RHS of each equality constraint
(row) in A_eq.
bounds : array_like
The bounds for each independent variable in the solution, which can take
one of three forms::
None : The default bounds, all variables are non-negative.
(lb, ub) : If a 2-element sequence is provided, the same
lower bound (lb) and upper bound (ub) will be applied
to all variables.
[(lb_0, ub_0), (lb_1, ub_1), ...] : If an n x 2 sequence is provided,
each variable x_i will be bounded by lb[i] and ub[i].
Infinite bounds are specified using -np.inf (negative)
or np.inf (positive).
maxiter : int
The maximum number of iterations to perform.
disp : bool
If True, print exit status message to sys.stdout
callback : callable
If a callback function is provide, it will be called within each
iteration of the simplex algorithm. The callback must have the
signature `callback(xk, **kwargs)` where xk is the current solution
vector and kwargs is a dictionary containing the following::
"tableau" : The current Simplex algorithm tableau
"nit" : The current iteration.
"pivot" : The pivot (row, column) used for the next iteration.
"phase" : Whether the algorithm is in Phase 1 or Phase 2.
"bv" : A structured array containing a string representation of each
basic variable and its current value.
tol : float
The tolerance which determines when a solution is "close enough" to zero
in Phase 1 to be considered a basic feasible solution or close enough
to positive to to serve as an optimal solution.
bland : bool
If True, use Bland's anti-cycling rule [3] to choose pivots to
prevent cycling. If False, choose pivots which should lead to a
converged solution more quickly. The latter method is subject to
cycling (non-convergence) in rare instances.
Returns
-------
A scipy.optimize.OptimizeResult consisting of the following fields::
x : ndarray
The independent variable vector which optimizes the linear
programming problem.
slack : ndarray
The values of the slack variables. Each slack variable corresponds
to an inequality constraint. If the slack is zero, then the
corresponding constraint is active.
success : bool
Returns True if the algorithm succeeded in finding an optimal
solution.
status : int
An integer representing the exit status of the optimization::
0 : Optimization terminated successfully
1 : Iteration limit reached
2 : Problem appears to be infeasible
3 : Problem appears to be unbounded
nit : int
The number of iterations performed.
message : str
A string descriptor of the exit status of the optimization.
Examples
--------
Consider the following problem:
Minimize: f = -1*x[0] + 4*x[1]
Subject to: -3*x[0] + 1*x[1] <= 6
1*x[0] + 2*x[1] <= 4
x[1] >= -3
where: -inf <= x[0] <= inf
This problem deviates from the standard linear programming problem. In
standard form, linear programming problems assume the variables x are
non-negative. Since the variables don't have standard bounds where
0 <= x <= inf, the bounds of the variables must be explicitly set.
There are two upper-bound constraints, which can be expressed as
dot(A_ub, x) <= b_ub
The input for this problem is as follows:
>>> c = [-1, 4]
>>> A = [[-3, 1], [1, 2]]
>>> b = [6, 4]
>>> x0_bnds = (None, None)
>>> x1_bnds = (-3, None)
>>> res = linprog(c, A, b, bounds=(x0_bnds, x1_bnds), options={"disp":True})
>>> print(res)
Optimization terminated successfully.
Current function value: 11.428571
Iterations: 2
status: 0
success: True
fun: 11.428571428571429
x: array([-1.14285714, 2.57142857])
slack: array([], dtype=np.float64)
message: 'Optimization terminated successfully.'
nit: 2
References
----------
.. [1] Dantzig, George B., Linear programming and extensions. Rand
Corporation Research Study Princeton Univ. Press, Princeton, NJ, 1963
.. [2] Hillier, S.H. and Lieberman, G.J. (1995), "Introduction to
Mathematical Programming", McGraw-Hill, Chapter 4.
.. [3] Bland, Robert G. New finite pivoting rules for the simplex method.
Mathematics of Operations Research (2), 1977: pp. 103-107.
"""
_check_unknown_options(unknown_options)
status = 0
messages = {0: "Optimization terminated successfully.",
1: "Iteration limit reached.",
2: "Optimzation failed. Unable to find a feasible"
" starting point.",
3: "Optimization failed. The problem appears to be unbounded.",
4: "Optimization failed. Singular matrix encountered."}
have_floor_variable = False
cc = np.asarray(c)
# The initial value of the objective function element in the tableau
f0 = 0
# The number of variables as given by c
n = len(c)
# Convert the input arguments to arrays (sized to zero if not provided)
Aeq = np.asarray(A_eq) if A_eq is not None else np.empty([0, len(cc)])
Aub = np.asarray(A_ub) if A_ub is not None else np.empty([0, len(cc)])
beq = np.ravel(np.asarray(b_eq)) if b_eq is not None else np.empty([0])
bub = np.ravel(np.asarray(b_ub)) if b_ub is not None else np.empty([0])
# Analyze the bounds and determine what modifications to me made to
# the constraints in order to accommodate them.
L = np.zeros(n, dtype=np.float64)
U = np.ones(n, dtype=np.float64)*np.inf
if bounds is None or len(bounds) == 0:
pass
elif len(bounds) == 2 and not hasattr(bounds[0], '__len__'):
# All bounds are the same
L = np.asarray(n*[bounds[0]], dtype=np.float64)
U = np.asarray(n*[bounds[1]], dtype=np.float64)
else:
if len(bounds) != n:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Length of bounds is inconsistent with the length of c")
else:
try:
for i in range(n):
if len(bounds[i]) != 2:
raise IndexError()
L[i] = bounds[i][0] if bounds[i][0] is not None else -np.inf
U[i] = bounds[i][1] if bounds[i][1] is not None else np.inf
except IndexError:
status = -1
message = ("Invalid input for linprog with "
"method = 'simplex'. bounds must be a n x 2 "
"sequence/array where n = len(c).")
if np.any(L == -np.inf):
# If any lower-bound constraint is a free variable
# add the first column variable as the "floor" variable which
# accommodates the most negative variable in the problem.
n = n + 1
L = np.concatenate([np.array([0]), L])
U = np.concatenate([np.array([np.inf]), U])
cc = np.concatenate([np.array([0]), cc])
Aeq = np.hstack([np.zeros([Aeq.shape[0], 1]), Aeq])
Aub = np.hstack([np.zeros([Aub.shape[0], 1]), Aub])
have_floor_variable = True
# Now before we deal with any variables with lower bounds < 0,
# deal with finite bounds which can be simply added as new constraints.
# Also validate bounds inputs here.
for i in range(n):
if(L[i] > U[i]):
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Lower bound %d is greater than upper bound %d" % (i, i))
if np.isinf(L[i]) and L[i] > 0:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Lower bound may not be +infinity")
if np.isinf(U[i]) and U[i] < 0:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Upper bound may not be -infinity")
if np.isfinite(L[i]) and L[i] > 0:
# Add a new lower-bound (negative upper-bound) constraint
Aub = np.vstack([Aub, np.zeros(n)])
Aub[-1, i] = -1
bub = np.concatenate([bub, np.array([-L[i]])])
L[i] = 0
if np.isfinite(U[i]):
# Add a new upper-bound constraint
Aub = np.vstack([Aub, np.zeros(n)])
Aub[-1, i] = 1
bub = np.concatenate([bub, np.array([U[i]])])
U[i] = np.inf
# Now find negative lower bounds (finite or infinite) which require a
# change of variables or free variables and handle them appropriately
for i in range(0, n):
if L[i] < 0:
if np.isfinite(L[i]) and L[i] < 0:
# Add a change of variables for x[i]
# For each row in the constraint matrices, we take the
# coefficient from column i in A,
# and subtract the product of that and L[i] to the RHS b
beq[:] = beq[:] - Aeq[:, i] * L[i]
bub[:] = bub[:] - Aub[:, i] * L[i]
# We now have a nonzero initial value for the objective
# function as well.
f0 = f0 - cc[i] * L[i]
else:
# This is an unrestricted variable, let x[i] = u[i] - v[0]
# where v is the first column in all matrices.
Aeq[:, 0] = Aeq[:, 0] - Aeq[:, i]
Aub[:, 0] = Aub[:, 0] - Aub[:, i]
cc[0] = cc[0] - cc[i]
if np.isinf(U[i]):
if U[i] < 0:
status = -1
message = ("Invalid input for linprog with "
"method = 'simplex'. Upper bound may not be -inf.")
# The number of upper bound constraints (rows in A_ub and elements in b_ub)
mub = len(bub)
# The number of equality constraints (rows in A_eq and elements in b_eq)
meq = len(beq)
# The total number of constraints
m = mub+meq
# The number of slack variables (one for each of the upper-bound constraints)
n_slack = mub
# The number of artificial variables (one for each lower-bound and equality
# constraint)
n_artificial = meq + _count_nonzero(bub < 0)
try:
Aub_rows, Aub_cols = Aub.shape
except ValueError:
raise ValueError("Invalid input. A_ub must be two-dimensional")
try:
Aeq_rows, Aeq_cols = Aeq.shape
except ValueError:
raise ValueError("Invalid input. A_eq must be two-dimensional")
if Aeq_rows != meq:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"The number of rows in A_eq must be equal "
"to the number of values in b_eq")
if Aub_rows != mub:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"The number of rows in A_ub must be equal "
"to the number of values in b_ub")
if Aeq_cols > 0 and Aeq_cols != n:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Number of columns in A_eq must be equal "
"to the size of c")
if Aub_cols > 0 and Aub_cols != n:
status = -1
message = ("Invalid input for linprog with method = 'simplex'. "
"Number of columns in A_ub must be equal to the size of c")
if status != 0:
# Invalid inputs provided
raise ValueError(message)
# Create the tableau
T = np.zeros([m+2, n+n_slack+n_artificial+1])
# Insert objective into tableau
T[-2, :n] = cc
T[-2, -1] = f0
b = T[:-2, -1]
if meq > 0:
# Add Aeq to the tableau
T[:meq, :n] = Aeq
# Add beq to the tableau
b[:meq] = beq
if mub > 0:
# Add Aub to the tableau
T[meq:meq+mub, :n] = Aub
# At bub to the tableau
b[meq:meq+mub] = bub
# Add the slack variables to the tableau
np.fill_diagonal(T[meq:m, n:n+n_slack], 1)
# Further setup the tableau
# If a row corresponds to an equality constraint or a negative b (a lower
# bound constraint), then an artificial variable is added for that row.
# Also, if b is negative, first flip the signs in that constraint.
slcount = 0
avcount = 0
basis = np.zeros(m, dtype=int)
r_artificial = np.zeros(n_artificial, dtype=int)
for i in range(m):
if i < meq or b[i] < 0:
# basic variable i is in column n+n_slack+avcount
basis[i] = n+n_slack+avcount
r_artificial[avcount] = i
avcount += 1
if b[i] < 0:
b[i] *= -1
T[i, :-1] *= -1
T[i, basis[i]] = 1
T[-1, basis[i]] = 1
else:
# basic variable i is in column n+slcount
basis[i] = n+slcount
slcount += 1
# Make the artificial variables basic feasible variables by subtracting
# each row with an artificial variable from the Phase 1 objective
for r in r_artificial:
T[-1, :] = T[-1, :] - T[r, :]
nit1, status = _solve_simplex(T, n, basis, phase=1, callback=callback,
maxiter=maxiter, tol=tol, bland=bland)
# if pseudo objective is zero, remove the last row from the tableau and
# proceed to phase 2
if abs(T[-1, -1]) < tol:
# Remove the pseudo-objective row from the tableau
T = T[:-1, :]
# Remove the artificial variable columns from the tableau
T = np.delete(T, np.s_[n+n_slack:n+n_slack+n_artificial], 1)
else:
# Failure to find a feasible starting point
status = 2
if status != 0:
message = messages[status]
if disp:
print(message)
return OptimizeResult(x=np.nan, fun=-T[-1, -1], nit=nit1, status=status,
message=message, success=False)
# Phase 2
nit2, status = _solve_simplex(T, n, basis, maxiter=maxiter-nit1, phase=2,
callback=callback, tol=tol, nit0=nit1,
bland=bland)
solution = np.zeros(n+n_slack+n_artificial)
solution[basis[:m]] = T[:m, -1]
x = solution[:n]
slack = solution[n:n+n_slack]
# For those variables with finite negative lower bounds,
# reverse the change of variables
masked_L = np.ma.array(L, mask=np.isinf(L), fill_value=0.0).filled()
x = x + masked_L
# For those variables with infinite negative lower bounds,
# take x[i] as the difference between x[i] and the floor variable.
if have_floor_variable:
for i in range(1, n):
if np.isinf(L[i]):
x[i] -= x[0]
x = x[1:]
# Optimization complete at this point
obj = -T[-1, -1]
if status in (0, 1):
if disp:
print(messages[status])
print(" Current function value: {: <12.6f}".format(obj))
print(" Iterations: {:d}".format(nit2))
else:
if disp:
print(messages[status])
print(" Iterations: {:d}".format(nit2))
return OptimizeResult(x=x, fun=obj, nit=int(nit2), status=status, slack=slack,
message=messages[status], success=(status == 0))
def _remainder_matrix_power(A, t):
"""
Compute the fractional power of a matrix, for fractions -1 < t < 1.
This uses algorithm (3.1) of [1]_.
The Pade approximation itself uses algorithm (4.1) of [2]_.
Parameters
----------
A : (N, N) array_like
Matrix whose fractional power to evaluate.
t : float
Fractional power between -1 and 1 exclusive.
Returns
-------
X : (N, N) array_like
The fractional power of the matrix.
References
----------
.. [1] Nicholas J. Higham and Lijing Lin (2013)
"An Improved Schur-Pade Algorithm for Fractional Powers
of a Matrix and their Frechet Derivatives."
.. [2] Nicholas J. Higham and Lijing lin (2011)
"A Schur-Pade Algorithm for Fractional Powers of a Matrix."
SIAM Journal on Matrix Analysis and Applications,
32 (3). pp. 1056-1078. ISSN 0895-4798
"""
# This code block is copied from numpy.matrix_power().
A = np.asarray(A)
if len(A.shape) != 2 or A.shape[0] != A.shape[1]:
raise ValueError('input must be a square array')
# Get the number of rows and columns.
n, n = A.shape
# Triangularize the matrix if necessary,
# attempting to preserve dtype if possible.
if np.array_equal(A, np.triu(A)):
Z = None
T = A
else:
if np.isrealobj(A):
T, Z = schur(A)
if not np.array_equal(T, np.triu(T)):
T, Z = rsf2csf(T, Z)
else:
T, Z = schur(A, output='complex')
# Zeros on the diagonal of the triangular matrix are forbidden,
# because the inverse scaling and squaring cannot deal with it.
T_diag = np.diag(T)
if _count_nonzero(T_diag) != n:
raise FractionalMatrixPowerError(
'cannot use inverse scaling and squaring to find '
'the fractional matrix power of a singular matrix')
# If the triangular matrix is real and has a negative
# entry on the diagonal, then force the matrix to be complex.
if np.isrealobj(T) and np.min(T_diag) < 0:
T = T.astype(complex)
# Get the fractional power of the triangular matrix,
# and de-triangularize it if necessary.
U = _remainder_matrix_power_triu(T, t)
if Z is not None:
ZH = np.conjugate(Z).T
return Z.dot(U).dot(ZH)
else:
return U
def _inverse_squaring_helper(T0, theta):
"""
A helper function for inverse scaling and squaring for Pade approximation.
Parameters
----------
T0 : (N, N) array_like upper triangular
Matrix involved in inverse scaling and squaring.
theta : indexable
The values theta[1] .. theta[7] must be available.
They represent bounds related to Pade approximation, and they depend
on the matrix function which is being computed.
For example, different values of theta are required for
matrix logarithm than for fractional matrix power.
Returns
-------
R : (N, N) array_like upper triangular
Composition of zero or more matrix square roots of T0, minus I.
s : non-negative integer
Number of square roots taken.
m : positive integer
The degree of the Pade approximation.
Notes
-----
This subroutine appears as a chunk of lines within
a couple of published algorithms; for example it appears
as lines 4--35 in algorithm (3.1) of [1]_, and
as lines 3--34 in algorithm (4.1) of [2]_.
The instances of 'goto line 38' in algorithm (3.1) of [1]_
probably mean 'goto line 36' and have been intepreted accordingly.
References
----------
.. [1] Nicholas J. Higham and Lijing Lin (2013)
"An Improved Schur-Pade Algorithm for Fractional Powers
of a Matrix and their Frechet Derivatives."
.. [2] Awad H. Al-Mohy and Nicholas J. Higham (2012)
"Improved Inverse Scaling and Squaring Algorithms
for the Matrix Logarithm."
SIAM Journal on Scientific Computing, 34 (4). C152-C169.
ISSN 1095-7197
"""
if len(T0.shape) != 2 or T0.shape[0] != T0.shape[1]:
raise ValueError('expected an upper triangular square matrix')
n, n = T0.shape
T = T0
# Find s0, the smallest s such that the spectral radius
# of a certain diagonal matrix is at most theta[7].
# Note that because theta[7] < 1,
# this search will not terminate if any diagonal entry of T is zero.
s0 = 0
tmp_diag = np.diag(T)
if _count_nonzero(tmp_diag) != n:
raise Exception('internal inconsistency')
while np.max(np.absolute(tmp_diag - 1)) > theta[7]:
tmp_diag = np.sqrt(tmp_diag)
s0 += 1
# Take matrix square roots of T.
for i in range(s0):
T = _sqrtm_triu(T)
# Flow control in this section is a little odd.
# This is because I am translating algorithm descriptions
# which have GOTOs in the publication.
s = s0
k = 0
d2 = _onenormest_m1_power(T, 2) ** (1/2)
d3 = _onenormest_m1_power(T, 3) ** (1/3)
a2 = max(d2, d3)
m = None
for i in (1, 2):
if a2 <= theta[i]:
m = i
break
while m is None:
if s > s0:
d3 = _onenormest_m1_power(T, 3) ** (1/3)
d4 = _onenormest_m1_power(T, 4) ** (1/4)
a3 = max(d3, d4)
if a3 <= theta[7]:
j1 = min(i for i in (3, 4, 5, 6, 7) if a3 <= theta[i])
if j1 <= 6:
m = j1
break
elif a3 / 2 <= theta[5] and k < 2:
k += 1
T = _sqrtm_triu(T)
s += 1
continue
d5 = _onenormest_m1_power(T, 5) ** (1/5)
a4 = max(d4, d5)
eta = min(a3, a4)
for i in (6, 7):
if eta <= theta[i]:
m = i
break
if m is not None:
break
T = _sqrtm_triu(T)
s += 1
# The subtraction of the identity is redundant here,
# because the diagonal will be replaced for improved numerical accuracy,
# but this formulation should help clarify the meaning of R.
R = T - np.identity(n)
# Replace the diagonal and first superdiagonal of T0^(1/(2^s)) - I
# using formulas that have less subtractive cancellation.
# Skip this step if the principal branch
# does not exist at T0; this happens when a diagonal entry of T0
# is negative with imaginary part 0.
has_principal_branch = all(x.real > 0 or x.imag != 0 for x in np.diag(T0))
if has_principal_branch:
for j in range(n):
a = T0[j, j]
r = _briggs_helper_function(a, s)
R[j, j] = r
p = np.exp2(-s)
for j in range(n-1):
l1 = T0[j, j]
l2 = T0[j+1, j+1]
t12 = T0[j, j+1]
f12 = _fractional_power_superdiag_entry(l1, l2, t12, p)
R[j, j+1] = f12
# Return the T-I matrix, the number of square roots, and the Pade degree.
if not np.array_equal(R, np.triu(R)):
raise Exception('internal inconsistency')
return R, s, m