def test_vec_to_upper_tri(self) -> None:
from cvxpy.atoms.affine.upper_tri import vec_to_upper_tri
x = Variable(shape=(3, ))
X = vec_to_upper_tri(x)
x.value = np.array([1, 2, 3])
actual = X.value
expect = np.array([[1, 2], [0, 3]])
assert np.allclose(actual, expect)
y = Variable(shape=(1, ))
y.value = np.array([4])
Y = vec_to_upper_tri(y, strict=True)
actual = Y.value
expect = np.array([[0, 4], [0, 0]])
assert np.allclose(actual, expect)
A_expect = np.array([[0, 11, 12, 13], [0, 0, 16, 17], [0, 0, 0, 21],
[0, 0, 0, 0]])
a = np.array([11, 12, 13, 16, 17, 21])
A_actual = vec_to_upper_tri(a, strict=True).value
assert np.allclose(A_actual, A_expect)
def log_det_canon(expr, args):
"""Reduces the atom to an affine expression and list of constraints.
Creates the equivalent problem::
maximize sum(log(D[i, i]))
subject to: D diagonal
diag(D) = diag(Z)
Z is upper triangular.
[D Z; Z.T A] is positive semidefinite
The problem computes the LDL factorization:
.. math::
A = (Z^TD^{-1})D(D^{-1}Z)
This follows from the inequality:
.. math::
\\det(A) >= \\det(D) + \\det([D, Z; Z^T, A])/\\det(D)
>= \\det(D)
because (Z^TD^{-1})D(D^{-1}Z) is a feasible D, Z that achieves
det(A) = det(D) and the objective maximizes det(D).
Parameters
----------
expr : log_det
args : list
The arguments for the expression
Returns
-------
tuple
(Variable for objective, list of constraints)
"""
A = args[0] # n by n matrix.
n, _ = A.shape
z = Variable(shape=(n * (n + 1) // 2, ))
Z = vec_to_upper_tri(z, strict=False)
d = diag_mat(Z) # a vector
D = diag_vec(d) # a matrix
X = bmat([[D, Z], [Z.T, A]])
constraints = [PSD(X)]
log_expr = log(d)
obj, constr = log_canon(log_expr, log_expr.args)
constraints += constr
return sum(obj), constraints