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Fixed database typo and removed unnecessary class identifier.

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Batuhan Berk Başoğlu 2020-10-14 10:10:37 -04:00
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"""
Routines for removing redundant (linearly dependent) equations from linear
programming equality constraints.
"""
# Author: Matt Haberland
import numpy as np
from scipy.linalg import svd
import scipy
from scipy.linalg.blas import dtrsm
def _row_count(A):
"""
Counts the number of nonzeros in each row of input array A.
Nonzeros are defined as any element with absolute value greater than
tol = 1e-13. This value should probably be an input to the function.
Parameters
----------
A : 2-D array
An array representing a matrix
Returns
-------
rowcount : 1-D array
Number of nonzeros in each row of A
"""
tol = 1e-13
return np.array((abs(A) > tol).sum(axis=1)).flatten()
def _get_densest(A, eligibleRows):
"""
Returns the index of the densest row of A. Ignores rows that are not
eligible for consideration.
Parameters
----------
A : 2-D array
An array representing a matrix
eligibleRows : 1-D logical array
Values indicate whether the corresponding row of A is eligible
to be considered
Returns
-------
i_densest : int
Index of the densest row in A eligible for consideration
"""
rowCounts = _row_count(A)
return np.argmax(rowCounts * eligibleRows)
def _remove_zero_rows(A, b):
"""
Eliminates trivial equations from system of equations defined by Ax = b
and identifies trivial infeasibilities
Parameters
----------
A : 2-D array
An array representing the left-hand side of a system of equations
b : 1-D array
An array representing the right-hand side of a system of equations
Returns
-------
A : 2-D array
An array representing the left-hand side of a system of equations
b : 1-D array
An array representing the right-hand side of a system of equations
status: int
An integer indicating the status of the removal operation
0: No infeasibility identified
2: Trivially infeasible
message : str
A string descriptor of the exit status of the optimization.
"""
status = 0
message = ""
i_zero = _row_count(A) == 0
A = A[np.logical_not(i_zero), :]
if not(np.allclose(b[i_zero], 0)):
status = 2
message = "There is a zero row in A_eq with a nonzero corresponding " \
"entry in b_eq. The problem is infeasible."
b = b[np.logical_not(i_zero)]
return A, b, status, message
def bg_update_dense(plu, perm_r, v, j):
LU, p = plu
vperm = v[perm_r]
u = dtrsm(1, LU, vperm, lower=1, diag=1)
LU[:j+1, j] = u[:j+1]
l = u[j+1:]
piv = LU[j, j]
LU[j+1:, j] += (l/piv)
return LU, p
def _remove_redundancy_dense(A, rhs, true_rank=None):
"""
Eliminates redundant equations from system of equations defined by Ax = b
and identifies infeasibilities.
Parameters
----------
A : 2-D sparse matrix
An matrix representing the left-hand side of a system of equations
rhs : 1-D array
An array representing the right-hand side of a system of equations
Returns
----------
A : 2-D sparse matrix
A matrix representing the left-hand side of a system of equations
rhs : 1-D array
An array representing the right-hand side of a system of equations
status: int
An integer indicating the status of the system
0: No infeasibility identified
2: Trivially infeasible
message : str
A string descriptor of the exit status of the optimization.
References
----------
.. [2] Andersen, Erling D. "Finding all linearly dependent rows in
large-scale linear programming." Optimization Methods and Software
6.3 (1995): 219-227.
"""
tolapiv = 1e-8
tolprimal = 1e-8
status = 0
message = ""
inconsistent = ("There is a linear combination of rows of A_eq that "
"results in zero, suggesting a redundant constraint. "
"However the same linear combination of b_eq is "
"nonzero, suggesting that the constraints conflict "
"and the problem is infeasible.")
A, rhs, status, message = _remove_zero_rows(A, rhs)
if status != 0:
return A, rhs, status, message
m, n = A.shape
v = list(range(m)) # Artificial column indices.
b = list(v) # Basis column indices.
# This is better as a list than a set because column order of basis matrix
# needs to be consistent.
d = [] # Indices of dependent rows
perm_r = None
A_orig = A
A = np.zeros((m, m + n), order='F')
np.fill_diagonal(A, 1)
A[:, m:] = A_orig
e = np.zeros(m)
js_candidates = np.arange(m, m+n, dtype=int) # candidate columns for basis
# manual masking was faster than masked array
js_mask = np.ones(js_candidates.shape, dtype=bool)
# Implements basic algorithm from [2]
# Uses some of the suggested improvements (removing zero rows and
# Bartels-Golub update idea).
# Removing column singletons would be easy, but it is not as important
# because the procedure is performed only on the equality constraint
# matrix from the original problem - not on the canonical form matrix,
# which would have many more column singletons due to slack variables
# from the inequality constraints.
# The thoughts on "crashing" the initial basis are only really useful if
# the matrix is sparse.
lu = np.eye(m, order='F'), np.arange(m) # initial LU is trivial
perm_r = lu[1]
for i in v:
e[i] = 1
if i > 0:
e[i-1] = 0
try: # fails for i==0 and any time it gets ill-conditioned
j = b[i-1]
lu = bg_update_dense(lu, perm_r, A[:, j], i-1)
except Exception:
lu = scipy.linalg.lu_factor(A[:, b])
LU, p = lu
perm_r = list(range(m))
for i1, i2 in enumerate(p):
perm_r[i1], perm_r[i2] = perm_r[i2], perm_r[i1]
pi = scipy.linalg.lu_solve(lu, e, trans=1)
js = js_candidates[js_mask]
batch = 50
# This is a tiny bit faster than looping over columns indivually,
# like for j in js: if abs(A[:,j].transpose().dot(pi)) > tolapiv:
for j_index in range(0, len(js), batch):
j_indices = js[j_index: min(j_index+batch, len(js))]
c = abs(A[:, j_indices].transpose().dot(pi))
if (c > tolapiv).any():
j = js[j_index + np.argmax(c)] # very independent column
b[i] = j
js_mask[j-m] = False
break
else:
bibar = pi.T.dot(rhs.reshape(-1, 1))
bnorm = np.linalg.norm(rhs)
if abs(bibar)/(1+bnorm) > tolprimal: # inconsistent
status = 2
message = inconsistent
return A_orig, rhs, status, message
else: # dependent
d.append(i)
if true_rank is not None and len(d) == m - true_rank:
break # found all redundancies
keep = set(range(m))
keep = list(keep - set(d))
return A_orig[keep, :], rhs[keep], status, message
def _remove_redundancy_sparse(A, rhs):
"""
Eliminates redundant equations from system of equations defined by Ax = b
and identifies infeasibilities.
Parameters
----------
A : 2-D sparse matrix
An matrix representing the left-hand side of a system of equations
rhs : 1-D array
An array representing the right-hand side of a system of equations
Returns
-------
A : 2-D sparse matrix
A matrix representing the left-hand side of a system of equations
rhs : 1-D array
An array representing the right-hand side of a system of equations
status: int
An integer indicating the status of the system
0: No infeasibility identified
2: Trivially infeasible
message : str
A string descriptor of the exit status of the optimization.
References
----------
.. [2] Andersen, Erling D. "Finding all linearly dependent rows in
large-scale linear programming." Optimization Methods and Software
6.3 (1995): 219-227.
"""
tolapiv = 1e-8
tolprimal = 1e-8
status = 0
message = ""
inconsistent = ("There is a linear combination of rows of A_eq that "
"results in zero, suggesting a redundant constraint. "
"However the same linear combination of b_eq is "
"nonzero, suggesting that the constraints conflict "
"and the problem is infeasible.")
A, rhs, status, message = _remove_zero_rows(A, rhs)
if status != 0:
return A, rhs, status, message
m, n = A.shape
v = list(range(m)) # Artificial column indices.
b = list(v) # Basis column indices.
# This is better as a list than a set because column order of basis matrix
# needs to be consistent.
k = set(range(m, m+n)) # Structural column indices.
d = [] # Indices of dependent rows
A_orig = A
A = scipy.sparse.hstack((scipy.sparse.eye(m), A)).tocsc()
e = np.zeros(m)
# Implements basic algorithm from [2]
# Uses only one of the suggested improvements (removing zero rows).
# Removing column singletons would be easy, but it is not as important
# because the procedure is performed only on the equality constraint
# matrix from the original problem - not on the canonical form matrix,
# which would have many more column singletons due to slack variables
# from the inequality constraints.
# The thoughts on "crashing" the initial basis sound useful, but the
# description of the procedure seems to assume a lot of familiarity with
# the subject; it is not very explicit. I already went through enough
# trouble getting the basic algorithm working, so I was not interested in
# trying to decipher this, too. (Overall, the paper is fraught with
# mistakes and ambiguities - which is strange, because the rest of
# Andersen's papers are quite good.)
# I tried and tried and tried to improve performance using the
# Bartels-Golub update. It works, but it's only practical if the LU
# factorization can be specialized as described, and that is not possible
# until the SciPy SuperLU interface permits control over column
# permutation - see issue #7700.
for i in v:
B = A[:, b]
e[i] = 1
if i > 0:
e[i-1] = 0
pi = scipy.sparse.linalg.spsolve(B.transpose(), e).reshape(-1, 1)
js = list(k-set(b)) # not efficient, but this is not the time sink...
# Due to overhead, it tends to be faster (for problems tested) to
# compute the full matrix-vector product rather than individual
# vector-vector products (with the chance of terminating as soon
# as any are nonzero). For very large matrices, it might be worth
# it to compute, say, 100 or 1000 at a time and stop when a nonzero
# is found.
c = (np.abs(A[:, js].transpose().dot(pi)) > tolapiv).nonzero()[0]
if len(c) > 0: # independent
j = js[c[0]]
# in a previous commit, the previous line was changed to choose
# index j corresponding with the maximum dot product.
# While this avoided issues with almost
# singular matrices, it slowed the routine in most NETLIB tests.
# I think this is because these columns were denser than the
# first column with nonzero dot product (c[0]).
# It would be nice to have a heuristic that balances sparsity with
# high dot product, but I don't think it's worth the time to
# develop one right now. Bartels-Golub update is a much higher
# priority.
b[i] = j # replace artificial column
else:
bibar = pi.T.dot(rhs.reshape(-1, 1))
bnorm = np.linalg.norm(rhs)
if abs(bibar)/(1 + bnorm) > tolprimal:
status = 2
message = inconsistent
return A_orig, rhs, status, message
else: # dependent
d.append(i)
keep = set(range(m))
keep = list(keep - set(d))
return A_orig[keep, :], rhs[keep], status, message
def _remove_redundancy(A, b):
"""
Eliminates redundant equations from system of equations defined by Ax = b
and identifies infeasibilities.
Parameters
----------
A : 2-D array
An array representing the left-hand side of a system of equations
b : 1-D array
An array representing the right-hand side of a system of equations
Returns
-------
A : 2-D array
An array representing the left-hand side of a system of equations
b : 1-D array
An array representing the right-hand side of a system of equations
status: int
An integer indicating the status of the system
0: No infeasibility identified
2: Trivially infeasible
message : str
A string descriptor of the exit status of the optimization.
References
----------
.. [2] Andersen, Erling D. "Finding all linearly dependent rows in
large-scale linear programming." Optimization Methods and Software
6.3 (1995): 219-227.
"""
A, b, status, message = _remove_zero_rows(A, b)
if status != 0:
return A, b, status, message
U, s, Vh = svd(A)
eps = np.finfo(float).eps
tol = s.max() * max(A.shape) * eps
m, n = A.shape
s_min = s[-1] if m <= n else 0
# this algorithm is faster than that of [2] when the nullspace is small
# but it could probably be improvement by randomized algorithms and with
# a sparse implementation.
# it relies on repeated singular value decomposition to find linearly
# dependent rows (as identified by columns of U that correspond with zero
# singular values). Unfortunately, only one row can be removed per
# decomposition (I tried otherwise; doing so can cause problems.)
# It would be nice if we could do truncated SVD like sp.sparse.linalg.svds
# but that function is unreliable at finding singular values near zero.
# Finding max eigenvalue L of A A^T, then largest eigenvalue (and
# associated eigenvector) of -A A^T + L I (I is identity) via power
# iteration would also work in theory, but is only efficient if the
# smallest nonzero eigenvalue of A A^T is close to the largest nonzero
# eigenvalue.
while abs(s_min) < tol:
v = U[:, -1] # TODO: return these so user can eliminate from problem?
# rows need to be represented in significant amount
eligibleRows = np.abs(v) > tol * 10e6
if not np.any(eligibleRows) or np.any(np.abs(v.dot(A)) > tol):
status = 4
message = ("Due to numerical issues, redundant equality "
"constraints could not be removed automatically. "
"Try providing your constraint matrices as sparse "
"matrices to activate sparse presolve, try turning "
"off redundancy removal, or try turning off presolve "
"altogether.")
break
if np.any(np.abs(v.dot(b)) > tol * 100): # factor of 100 to fix 10038 and 10349
status = 2
message = ("There is a linear combination of rows of A_eq that "
"results in zero, suggesting a redundant constraint. "
"However the same linear combination of b_eq is "
"nonzero, suggesting that the constraints conflict "
"and the problem is infeasible.")
break
i_remove = _get_densest(A, eligibleRows)
A = np.delete(A, i_remove, axis=0)
b = np.delete(b, i_remove)
U, s, Vh = svd(A)
m, n = A.shape
s_min = s[-1] if m <= n else 0
return A, b, status, message