2

I have this problem to solve

$$\max_{\mathbf{v} \in \{-1,1\}^n} \mathbf{v}^T A \mathbf{v}$$

where $A$ is a $n$ by $n$ symmetric (real) square matrix where each element $a_{ij} \in [-b,+b]$.

I'd like to determine properties of $A$ where $\mathbf{v}^T A \mathbf{v} < 0$ holds. I know that I can check whether the eigenvalues of $A$ are all negative to see $\mathbf{v}^T A \mathbf{v} < 0$. But I want to determine properties of $A$ that gives $\mathbf{v}^T A \mathbf{v} < 0$ for all $\mathbf{v} \in \{-1,1\}^n$ only. In a sense, I'd like to know the structure of matrix $A$ that is 'negative definite' under only such binary vectors.

I did literature search but couldn't find anything much related to this particular problem. Perhaps someone can shed a light.

Rodrigo de Azevedo
  • 23,223
Adam I.
  • 379

1 Answers1

2

Let $\mathrm Q : = - \mathrm A$. Consider the following unconstrained binary quadratic program

$$\underset{\mathrm x \in \{\pm 1\}^n}{\text{minimize}} \quad \mathrm x^{\top} \mathrm Q \,\mathrm x$$

or, equivalently, the following quadratically constrained quadratic program (QCQP) in $\mathrm x \in \mathbb R^n$

$$\begin{array}{ll} \text{minimize} & \mathrm x^{\top} \mathrm Q \,\mathrm x\\ \text{subject to} & x_i^2 = 1 \quad \forall i \in [n]\end{array}\tag{QCQP}$$

where $[n] := \{1,2,\dots,n\}$. Both optimization problems above are hard (even when $\mathrm Q \succeq \mathrm O_n$).

The QCQP yields the following Lagrangian

$$\mathcal{L} (\mathrm x, \lambda) := \mathrm x^{\top} \mathrm Q \,\mathrm x - \sum_{i=1}^n \lambda_i (x_i^2 - 1) = \mathrm x^{\top} \left( \mathrm Q - \mbox{diag} (\lambda) \right) \mathrm x + 1_n^{\top} \lambda$$

The dual of the QCQP [PP&SL'03, PP&SL'06] is, thus,

$$\begin{array}{ll} \text{maximize} & 1_n^{\top} \lambda\\ \text{subject to} & \mbox{diag} (\lambda) \preceq \mathrm Q\end{array}$$

Hence, we have the following semidefinite program (SDP) in $\Lambda$

$$\begin{array}{ll} \text{maximize} & \mbox{tr} (\Lambda)\\ \text{subject to} & \Lambda_{ij} = 0 \quad \forall i \neq j\\ & \Lambda \preceq \mathrm Q\end{array}$$

which is convex and, thus, easy (unlike the QCQP). This SDP does provide a lower bound on the minimum of the QCQP. If the lower bound is positive, we can immediately conclude that

$$\mathrm x^{\top} \mathrm Q \,\mathrm x > 0$$

for all $\mathrm x \in \{\pm 1\}^n$. If the lower bound is non-positive, we cannot conclude anything, unfortunately.

References

[PP&SL'03] Pablo Parrilo, Sanjay Lall, Quadratically Constrained Quadratic Programming, 2003.

[PP&SL'06] Pablo Parrilo, Sanjay Lall, Quadratically Constrained Quadratic Programming, 2006.

Community
  • 1
Rodrigo de Azevedo
  • 23,223
  • Thank you dearly. Looking at [PP&SL'03], on page 22 there is the 'sandwich' inequality bounding the objective values in QCQP, SDP etc. It suggests the strategy of 'Goemans and Williamson' to obtain an approximate solution for QCQP from the SDP problem. Thus, if one shows that the lower bound is negative, the approximate lower bound is also negative suggesting that $\max_\mathbf{v} \mathbf{v}^TA\mathbf{v}$ is positive, correct? By the use of the sandwich inequality it seems that we can conclude the sign of $\max_\mathbf{v} \mathbf{v}^T A \mathbf{v}$? – Adam I. Feb 04 '17 at 00:10
  • The Goemans and Williamson yields an a solution where an expected bound is derived. But there are other deterministic methods to obtain approximate solutions that yield similar sandwich inequalities. It'll be great if you can comment on this. Thank you. – Adam I. Feb 04 '17 at 00:12
  • @Adam Take a look at chapter 1 of Approximation Algorithms and Semidefinite Programming and at the original G&W paper. Unfortunately, your matrix $\rm A$ has negative entries. If it were a nonnegative matrix, it would be easier. – Rodrigo de Azevedo Feb 04 '17 at 00:18