Maximal Quadratic-Free Sets

Abstract

The intersection cut paradigm is a powerful framework that facilitates the generation of valid linear inequalities, or cutting planes, for a potentially complex set S. The key ingredients in this construction are a simplicial conic relaxation of S and an S-free set: a convex zone whose interior does not intersect S. Ideally, such S-free set would be maximal inclusion-wise, as it would generate a deeper cutting plane. However, maximality can be a challenging goal in general. In this work, we show how to construct maximal S-free sets when S is defined as a general quadratic inequality. Our maximal S-free sets are such that efficient separation of a vertex in LP-based approaches to quadratically constrained problems is guaranteed. To the best of our knowledge, this work is the first to provide maximal quadratic-free sets.

Appendix

Example 1 (Homogeneous case)

Consider the set $S_{\le 0}$, defined as

$$\begin{aligned} S_{\le 0}= \{(x_1, x_2 , y) \in \mathbb {R}^3 \,:\,\Vert x \Vert \le |y |, \quad a^\mathsf {T}x + d y \le 0 \} \end{aligned}$$

with $a = (-1/\sqrt{2},1/\sqrt{2})^\mathsf {T}$ and $d=1/\sqrt{2}$, and $(\bar{x},\bar{y})=(-1, -1 ,0)^\mathsf {T}$. This point satisfies the linear inequality in $S_{\le 0}$, but it is not in $S_{\le 0}$. It is not too hard to check that $G(\lambda ) = \{-1\}$. In Fig. 1 we show $S_{\le 0}$, the $S_{\le 0}$-free set given by $C_{\lambda }$ and the set $C_{G(\lambda )}$. In this case $\Vert a\Vert = 1 > 1/\sqrt{2} = | d |$, so we have no guarantee on the $S_{\le 0}$-freeness of $C_{G(\lambda )}$. Even more, it is not $S_{\le 0}$-free.

Moving forward, since $\lambda ^\mathsf {T}a = 0$ we have

$$\begin{aligned} \lambda ^\mathsf {T}a \Vert y\Vert + d^\mathsf {T}y \le 0 \Longleftrightarrow y\le 0. \end{aligned}$$

A simple calculation using (8) yields

$$\phi _{\lambda }(y) = {\left\{ \begin{array}{ll} - y , &{}\text { if } y \le 0 \\ \frac{y}{\sqrt{2}} &{}\text { if } y > 0 \end{array}\right. } $$

In Fig. 4 we show the set $C_{\phi _{\lambda }}$, which is maximal $S_{\le 0}$-free.

Fig. 4.

$S_{\le 0}$ in Example 1 (orange) and $C_{\phi _{\lambda }}$ set (blue). The latter is maximal $S_{\le 0}$-free. (Color figure online)

Fig. 5.

Plots of $S_{\le 0}$, H and $C_1$ as defined in Example 2 showing that $C_1$ is maximal $S_{\le 0}$-free with respect to H.

Example 2 (Non-homogeneous case)

We continue with $S_{\le 0}$ defined in Example 1, but we now look for maximality with respect to

$$\begin{aligned} H= \{ (x,y) \in \mathbb {R}^{n+m} \,:\,a^\mathsf {T}x + d^\mathsf {T}y = -1 \}. \end{aligned}$$

In this case, $C_{\phi _{\lambda }}\cap H$ is not maximal $S_{\le 0}\cap H$-free, as shown in Fig. 2. Since $\lambda = \frac{1}{\sqrt{2}}(-1,-1)^\mathsf {T}$, we see that

$$\begin{aligned} C_{\phi _{\lambda }}= \{ (x,y) \,:\,&\frac{1}{\sqrt{2}}(x_1 + x_2) - y \le 0, \end{aligned}$$

(12a)

$$\begin{aligned}&\frac{1}{\sqrt{2}}(x_1 + x_2) + \frac{1}{\sqrt{2}}y \le 0 \}. \end{aligned}$$

(12b)

It is not hard to check that $-(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}},\sqrt{2}) \in S_{\le 0}\cap H \cap C_{\phi _{\lambda }}$ exposes inequality (12a). This is the tangent point in Fig. 2b. On the other hand, (12b), which is obtained from $\beta =1$, does not have an exposing point in $S_{\le 0}\cap H \cap C_{\phi _{\lambda }}$, and corresponds to an inequality we should relax as per our discussion. This inequality, however, is exposed by $({x_\beta }, \beta ) =(0,-1, 1) \in S_{\le 0}\cap C_{\phi _{\lambda }}$. Consider now the sequence defined as

$$\begin{aligned} (x_n,\beta ) = \left( \frac{1}{\sqrt{n^2+1}} , -\frac{n}{\sqrt{n^2+1}}, 1\right) \in S_{\le 0}. \end{aligned}$$

Clearly the limit of this sequence is $(0,-1, 1)$ and

$$\begin{aligned} a^\mathsf {T}x_n + d^\mathsf {T}\beta = \frac{1}{\sqrt{2}}\left( -\frac{1}{\sqrt{n^2+1}} -\frac{n}{\sqrt{n^2+1}} + 1\right) < 0. \end{aligned}$$

Now we let

$$\begin{aligned} z_n = -\frac{(x_n, \beta )}{a^\mathsf {T}x_n + d^\mathsf {T}\beta } \in S_{\le 0}\cap H. \end{aligned}$$

which diverges. In Fig. 3, we plot the first two components of the sequence $(z_n)_{n\in \mathbb {N}}$ along with $S_{\le 0}\cap H$ and $C_{\phi _{\lambda }}\cap H$. The sequence $(z_n)_{n\in \mathbb {N}}$ moves along the boundary of $S_{\le 0}\cap H$ towards an “asymptote” from where we deduce $r(\beta )$.

In this case $r(-1) = 0$, and it can be checked that $r(1) = 1$ using its formula. Now, let

$$\begin{aligned} C_1&= \{ (x,y) \,:\,-\lambda ^\mathsf {T}x + \nabla \phi _{\lambda }(\beta )^\mathsf {T}y \le r(\beta ), \text { for all } \beta \in D_1(0) \}\\&= \{ (x,y) \,:\,\frac{1}{\sqrt{2}}(x_1 + x_2) - y \le 0, \frac{1}{\sqrt{2}}(x_1 + x_2) + \frac{1}{\sqrt{2}}y \le 1 \}. \end{aligned}$$

Figure 5 shows the same plots as Fig. 2 with $C_1$ instead of $C_{\phi _{\lambda }}$.