Rank-one nonnegative matrix factorization
=========================================
The DGP atom library has several functions of positive matrices,
including the trace, (matrix) product, sum, Perron-Frobenius eigenvalue,
and :math:`(I - X)^{-1}` (eye-minus-inverse). In this notebook, we use
some of these atoms to approximate a partially known elementwise
positive matrix as the outer product of two positive vectors.
We would like to approximate :math:`A` as the outer product of two
positive vectors :math:`x` and :math:`y`, with :math:`x` normalized so
that the product of its entries equals :math:`1`. Our criterion is the
average relative deviation between the entries of :math:`A` and
:math:`xy^T`, that is,
.. math::
\frac{1}{mn} \sum_{i=1}^{m} \sum_{j=1}^{n} R(A_{ij}, x_iy_j),
where :math:`R` is the relative deviation of two positive numbers,
defined as
.. math::
R(a, b) = \max\{a/b, b/a\} - 1.
The corresponding optimization problem is
.. math::
\begin{equation}
\begin{array}{ll}
\mbox{minimize} & \frac{1}{mn} \sum_{i=1}^{m} \sum_{j=1}^{n} R(X_{ij}, x_iy_j)
\\
\mbox{subject to} & x_1x_2 \cdots x_m = 1 \\
& X_{ij} = A_{ij}, \quad \text{for } (i, j) \in \Omega,
\end{array}
\end{equation}
with variables :math:`X \in \mathbf{R}^{m \times n}_{++}`,
:math:`x \in \mathbf{R}^{m}_{++}`, and
:math:`y \in \mathbf{R}^{n}_{++}`. We can cast this problem as an
equivalent generalized geometric program by discarding the :math:`-1`
from the relative deviations.
The below code constructs and solves this optimization problem, with
specific problem data
.. math::
A = \begin{bmatrix}
1.0 & ? & 1.9 \\
? & 0.8 & ? \\
3.2 & 5.9& ?
\end{bmatrix},
.. code:: python
import cvxpy as cp
m = 3
n = 3
X = cp.Variable((m, n), pos=True)
x = cp.Variable((m,), pos=True)
y = cp.Variable((n,), pos=True)
outer_product = cp.vstack([x[i] * y for i in range(m)])
relative_deviations = cp.maximum(
cp.multiply(X, outer_product ** -1),
cp.multiply(X ** -1, outer_product))
objective = cp.sum(relative_deviations)
constraints = [
X[0, 0] == 1.0,
X[0, 2] == 1.9,
X[1, 1] == 0.8,
X[2, 0] == 3.2,
X[2, 1] == 5.9,
x[0] * x[1] * x[2] == 1.0,
]
problem = cp.Problem(cp.Minimize(objective), constraints)
problem.solve(gp=True)
print("Optimal value:\n", 1.0/(m * n) * (problem.value - m * n), "\n")
print("Outer product approximation\n", outer_product.value, "\n")
print("x: ", x.value)
print("y: ", y.value)
.. parsed-literal::
Optimal value:
1.7763568394002505e-14
Outer product approximation
[[1. 1.84375 1.9 ]
[0.43389831 0.8 0.82440678]
[3.2 5.89999999 6.07999999]]
x: [0.89637009 0.38893346 2.86838428]
y: [1.11561063 2.0569071 2.1196602 ]