Correction to: Exact semidefinite formulations for a class of (random and non-random) nonconvex quadratic programs

1 Correction to: Mathematical Programming (2020) 181:1–17 https://doi.org/10.1007/s10107-019-01367-2

This note reports an error in Theorem 3 of [1], which was pointed about by Prof. Fatma Kılınç-Karzan and Mr. Alex Wang of Carnegie Mellon University, whom we sincerely thank for bringing this to our attention.

The error can be seen in the proof of Theorem 3, where system (12) is analyzed for given diagonal matrices $\{C, B_i, D_i\}$ and variable diagonal matrices $\{X,Y_i\}$. In this setting, $C, B_i \succ 0$, while each $D_i$ is indefinite. In fact, multiple systems of the form (12) are analyzed. In each, one scalar variable in $\{X, Y_i\}$ is allowed to be free, while all others are nonnegative. Then, within the proof, the goal is to show that, with high probability, all such systems are simultaneously feasible. However, in truth the proof fails because, with this setup, the systems cannot be simultaneously feasible. Said differently, the probability that all systems are feasible is zero.

To see this, we assume all such systems are feasible and derive a contradiction. Fix i and let $\omega > 0$ be minimal such that $D_i + \omega B_i \succeq 0$. In particular, let j be such that $[D_i + \omega B_i]_{jj} = 0$. Then consider the specific system (12) with $[Y_i]_{jj}$ free. Because this system is feasible, the inequality

$$\begin{aligned} \omega \, C \bullet X + \epsilon (D_i + \omega B_i) \bullet Y_i + \epsilon \, \omega \sum _{k \ne i} B_k \bullet Y_k \le -\omega , \end{aligned}$$

is also feasible. Note that $[Y_i]_{jj}$ does not actually appear in this system because its coefficient $\epsilon [D_i + \omega B_i]_{jj}$ equals 0, and so we see that a positive combination of nonnegative variables is negative, thus providing the desired contradiction.

Instead of the purported Theorem 3, we now prove an analogous theorem, except that it only applies to diagonal instances of (1). We first describe a geometric condition on the data $\{C, A_i\}$ and $\{c, a_i\}$ ensuring $r^* = 1$. Define, for each $j = 1,\ldots ,n$,

We remark that, using an epigraph variable t, we may rewrite (1) equivalently as follows:

where $(t,b) \in {\mathbb {R}}^{1+m}$. This formulation ($\ddag$) emphasizes the separable nature of each quadratic function in (1). In addition to ($\dag$), define $K_j := \text {cone}\{ f_k - f_j{:}\,k \ne j \} \subseteq {\mathbb {R}}^{1+m}$ and $L_j := \{ \lambda d_j{:}\,\lambda \in {\mathbb {R}}\} \subseteq {\mathbb {R}}^{1+m}$ for each j. $K_j$ is the cone of feasible directions at $f_j$ in the polytope $Q := \text {conv}\{ f_1,\ldots ,f_n\}$, and $L_j$ is the line spanned by the directions $\pm d_j$.

Lemma

Given an instance of the diagonal QCQP (1), if $\text {int}(K_j) \cap L_j \ne \emptyset$ for all $j = 1,\ldots ,n$, then $r^* = 1$.

Proof

We use Assumptions 1–2, which guarantee that the feasible set of (1) is bounded, to add the redundant constraint $x^T x \le \rho ^2$ for some radius $\rho$. Note that we do not need to know $\rho$ explicitly.

For fixed j, system (5) is certainly feasible if the following more restrictive system is feasible:

$$\begin{aligned}&C \bullet X + c_j x_j< 0 \\&A_i \bullet X + a_{ij} x_j < 0 \ \ \forall \ i = 1, \ldots , m \\&I \bullet X = 0 \\&X \text { diagonal}, \ X_{kk} \ge 0 \ \ \forall \ k \ne j \\&X_{jj} \text { free}, \ x_j \text { free}. \end{aligned}$$

Eliminating the free variable $X_{jj}$ via the equation $I \bullet X = 0$ and using the definitions of $f_j$ and $d_j$ in ($\dag$), we have the equivalent vector-inequality system

$$\begin{aligned}&\sum _{k \ne j} X_{kk} (f_k - f_j) + x_j \, d_j < 0 \\&X_{kk} \ge 0 \ \ \forall \ k \ne j, \ \ x_j \text { free}, \end{aligned}$$

which is feasible if and only if there exists a direction in $K_j$, which equals the sum of a vector in $L_j$ and a strictly negative vector, i.e., if and only if the intersection $K_j \cap (L_j + \text {int}({\mathbb {R}}^{1+m}_-))$ is nonempty. Since $L_j$ is a line, we have $L_j + \text {int}({\mathbb {R}}^{1+m}_-) = \text {int}(L_j + {\mathbb {R}}^{1+m}_-)$.

So, to complete the proof, it suffices to show $K_j \cap \text {int}(L_j + {\mathbb {R}}^{1+m}_-) \ne \emptyset$, which is true by the following argument. Since $\text {int}(K_j) \cap L_j \ne \emptyset$ by assumption, $K_j$ is full-dimensional. In addition, the polyhedron $L_j + {\mathbb {R}}^{1+m}_-$ is full-dimensional with $\emptyset \ne \text {int}(K_j) \cap L_j \subseteq \text {int}(K_j) \cap (L_j + {\mathbb {R}}^{1+m}_-)$, i.e., the interior of the full-dimensional polyhedron $K_j$ intersects the full-dimensional polyhedron $L_j + {\mathbb {R}}^{1+m}_-$. It follows that $K_j$ intersects the interior of $L_j + {\mathbb {R}}^{1+m}_-$, as desired. $\square$

Theorem

Given fixed $m > 0$ and $\theta \in (0,\tfrac{\pi }{2}]$, consider the class of random diagonal QCQPs (1) generated as follow for $n > 0$:

• first, the diagonal entries of C and $A_1,\ldots ,A_m$ are generated randomly such that, for each $j = 1, \ldots , n$ independently, $f_j/\Vert f_j\Vert$ is uniformly distributed on the unit sphere in ${\mathbb {R}}^{1+m}$, where $f_j$ is defined in ($\dag$);

• next, the entries of c and $a_1, \ldots , a_m$ are generated randomly such that, for each $j = 1,\ldots ,n$ independently, the angle between $d_j$ and $f_j$ is outside the interval $\pi /2 \pm \theta$, where $d_j$ is defined by ($\dag$); 

• finally, the entries of b are chosen such that Assumptions 1 and 3 are satisfied, and for some finite $\rho > 0$, the constraint $x^T x \le \rho ^2$ is added to ensure that Assumption 2 is satisfied, while not violating Assumptions 1 and 3.

Then $\text {Prob}(r^* = 1) \rightarrow 1$ as $n \rightarrow \infty$.

Proof

We apply the lemma as $n \rightarrow \infty$. Specifically, we wish to show the probability that $\text {int}(K_j) \cap L_j \ne \emptyset$ for all $j = 1,\ldots ,n$ goes to 1 as $n \rightarrow \infty$, while m and $\theta$ stay fixed. Also define Q as before the lemma.

We first note that, by using the formulation ($\ddag$) along with independent variable substitutions $x_j = {\hat{x}}_j / \sqrt{\Vert f_j\Vert }$ for all $j = 1,\ldots ,n$, (1) is equivalent to

$$\begin{aligned} \min \left\{ t{:}\,\sum _{j=1}^n \left( {\hat{x}}_j^2 \cdot \frac{f_j}{\Vert f_j\Vert } + 2 \, {\hat{x}}_j \cdot \frac{d_j}{\sqrt{\Vert f_j\Vert }} \right) \le {t \atopwithdelims ()b} \right\} . \end{aligned}$$

Note that this corresponds to a new instance of (1) with data ${\hat{f}}_j := f_j / \Vert f_j\Vert$ and ${\hat{d}}_j := d_j / \sqrt{\Vert f_j\Vert }$ such that each ${\hat{f}}_j$ has norm 1 and the angle between ${\hat{d}}_j$ and ${\hat{f}}_j$ equals the angle between $d_j$ and $f_j$. Since the tightness of the Shor SDP relaxation is not affected by simple positive variable scalings, we thus may assume without loss of generality that $\Vert f_j\Vert = 1$.

Since $\{ f_j \}$ follows a uniform distribution on the unit sphere, as $n \rightarrow \infty$, Q approaches the unit ball and the cones $K_j$ approach the halfspaces $\{ v{:}\,f_j^T v \le 0 \}$. Since $L_j$ makes an angle with $f_j$ outside of $\pi /2 \pm \theta$, $L_j$ intersects $\text {int}(K_j)$ for all j with probability going to 1 as $n \rightarrow \infty$. $\square$

Remark

It is well known that, if all of the diagonal entries of $\{C, A_i\}$ are generated i.i.d. standard Gaussian, then $f_j / \Vert f_j\Vert$ is uniform on the unit sphere. In addition, $d_j$ can be generated with entries i.i.d. standard Gaussian along with a simple accept-reject procedure to satisfy the angle condition of the theorem.