def cvxopt_solver(G, h, A, b, c, n):
# cvxopt doesn't allow redundant constraints in the linear program Ax = b,
# so we need to do some preprocessing to find and remove any linearly
# dependent rows in the augmented matrix [A | b].
#
# First we do Gaussian elimination to put the augmented matrix into row
# echelon (reduced row echelon form is not necessary). Since b comes as a
# numpy array (that is, a row vector), we need to convert it to a numpy
# matrix before transposing it (that is, to a column vector).
b = np.mat(b).T
A_b = np.hstack((A, b))
A_b, permutation = to_row_echelon(np.hstack((A, b)))
# Next, we apply the inverse of the permutation applied to compute the row
# echelon form.
P = dictionary_to_permutation(permutation)
A_b = P.I * A_b
# Trim any rows that are all zeros. The call to np.any returns an array of
# Booleans that correspond to whether a row in A_b is all zeros. Indexing
# A_b by an array of Booleans acts as a selector. We need to use np.asarray
# in order for indexing to work, since it expects a row vector instead of a
# column vector.
A_b = A_b[np.any(np.asarray(A_b) != 0, axis=1)]
# Split the augmented matrix back into a matrix and a vector.
A, b = A_b[:, :-1], A_b[:, -1]
# Apply the linear programming solver; cvxopt requires that these are all
# of a special type of cvx-specific matrix.
G, h, A, b, c = (cvx_matrix(M) for M in (G, h, A, b, c))
solution = cvx_solvers.lp(c, G, h, A, b)
if solution['status'] == 'optimal':
return True, solution['x']
# TODO status could be 'unknown' here, but we're currently ignoring that
return False, None
def cvxopt_solver(G, h, A, b, c, n):
# cvxopt doesn't allow redundant constraints in the linear program Ax = b,
# so we need to do some preprocessing to find and remove any linearly
# dependent rows in the augmented matrix [A | b].
#
# First we do Gaussian elimination to put the augmented matrix into row
# echelon (reduced row echelon form is not necessary). Since b comes as a
# numpy array (that is, a row vector), we need to convert it to a numpy
# matrix before transposing it (that is, to a column vector).
b = np.mat(b).T
A_b = np.hstack((A, b))
A_b, permutation = to_row_echelon(np.hstack((A, b)))
# Next, we apply the inverse of the permutation applied to compute the row
# echelon form.
P = dictionary_to_permutation(permutation)
A_b = P.I * A_b
# Trim any rows that are all zeros. The call to np.any returns an array of
# Booleans that correspond to whether a row in A_b is all zeros. Indexing
# A_b by an array of Booleans acts as a selector. We need to use np.asarray
# in order for indexing to work, since it expects a row vector instead of a
# column vector.
A_b = A_b[np.any(np.asarray(A_b) != 0, axis=1)]
# Split the augmented matrix back into a matrix and a vector.
A, b = A_b[:, :-1], A_b[:, -1]
# Apply the linear programming solver; cvxopt requires that these are all
# of a special type of cvx-specific matrix.
G, h, A, b, c = (cvx_matrix(M) for M in (G, h, A, b, c))
solution = cvx_solvers.lp(c, G, h, A, b)
if solution['status'] == 'optimal':
return True, solution['x']
# TODO status could be 'unknown' here, but we're currently ignoring that
return False, None
def test_dictionary_to_permutation():
permutation = {0: 1, 1: 2, 2: 0}
expected = np.mat([[0, 1, 0], [0, 0, 1], [1, 0, 0]])
actual = dictionary_to_permutation(permutation)
assert np.all(expected == actual)
def test_dictionary_to_permutation():
permutation = {0: 1, 1: 2, 2: 0}
expected = np.mat([[0, 1, 0], [0, 0, 1], [1, 0, 0]])
actual = dictionary_to_permutation(permutation)
assert np.all(expected == actual)
