Intersection Cuts for Polynomial Optimization

Abstract

We consider dynamically generating linear constraints (cutting planes) to tighten relaxations for polynomial optimization problems. Many optimization problems have feasible set of the form $S \cap P$, where S is a closed set and P is a polyhedron. Integer programs are in this class and one can construct intersection cuts using convex “forbidden” regions, or S-free sets. Here, we observe that polynomial optimization problems can also be represented as a problem with linear objective function over such a feasible set, where S is the set of real, symmetric matrices representable as outer-products of the form $xx^T$. Accordingly, we study outer-product-free sets and develop a thorough characterization of several (inclusion-wise) maximal intersection cut families. In addition, we present a cutting plane approach that guarantees polynomial-time separation of an extreme point in $P\setminus S$ using our outer-product-free sets. Computational experiments demonstrate the promise of our approach from the point of view of strength and speed.

Appendix

Proof

(Theorem 4). The OPF property is given by Lemma 2, so maximality remains. Let C be a set described by (5a) or (5b). It suffices to construct, for every symmetric matrix $\bar{X} \not \in C$, such that $Z-\bar{X}\in \text {int}(C)$. This implies $Z \in \text {int} ( \text {conv}(C\cup \bar{X}))$. Denote the submatrices of $\bar{X}, Z$:

Furthermore, for convenience let us define the following:

Construction for (5a): Suppose $\bar{X}$ violates (5a). We propose the following:

$$\begin{aligned} a_Z = \bar{q} + \lambda _1\Vert \bar{q}, \bar{r}\Vert _2,\, b_Z = \bar{r}+ \lambda _2\Vert \bar{q},\bar{r}\Vert _2,&\, c_Z = \bar{r} - \lambda _2\Vert \bar{q},\bar{r}\Vert _2,\, d_Z = -\bar{q} + \lambda _1\Vert \bar{q},\bar{r}\Vert _2 \\ \Longrightarrow \, \lambda _1(\bar{a}+\bar{d})/2+\lambda _2(\bar{b}-\bar{c})/2 \,&< \, \Vert (\bar{b}+\bar{c})/2,(\bar{a}-\bar{d})/2\Vert _2 \nonumber \\&= \, \lambda _1(a_Z+ d_Z)/2+\lambda _2(b_Z-c_Z)/2 \nonumber \end{aligned}$$

(6)

where the last equality follows from $\lambda _1^2 + \lambda _2^2 = 1$. This implies

$$\begin{aligned} \lambda _1((a_Z-\bar{a})+ (d_Z- \bar{d}))/2 +\lambda _2((b_Z-\bar{b})-(c_Z-\bar{c}))/2 > 0 \end{aligned}$$

and since $\Vert ((b_Z-\bar{b})+(c_Z-\bar{c}))/2,((a_Z-\bar{a})-(d_Z-\bar{d}))/2\Vert _2 = 0$, we conclude $Z-\bar{X} \in \text {int}(C)$.

Construction for (5b): If $\bar{X}$ violates (5b), we use the following construction:

$$\begin{aligned} a_Z = \bar{p} + \lambda _2\Vert \bar{p},\bar{s}\Vert _2, b_Z = \bar{s}+ \lambda _1\Vert \bar{p},\bar{s}\Vert _2,&\, c_Z = -\bar{s} + \lambda _1\Vert \bar{p},\bar{s}\Vert _2, d_Z = \bar{p} - \lambda _2\Vert \bar{p},\bar{s}\Vert _2. \\ \Longrightarrow \lambda _1(\bar{b}+\bar{c})/2+\lambda _2(\bar{a}-\bar{d})/2 \,&< \, \Vert (\bar{a}+\bar{d})/2,(\bar{b}-\bar{c})/2\Vert _2 \\ {}&= \, \lambda _1(b_Z+ c_Z)/2+\lambda _2(a_Z-d_Z)/2, \\ \Longrightarrow \, \lambda _1((b_Z-\bar{b})+(c_Z-\bar{c}))/2&+\lambda _2((a_Z-\bar{a})- (d_Z- \bar{d}))/2 > 0. \end{aligned}$$

We conclude $Z-\bar{X} \in \text {int}(C)$ as before, since $\Vert ((a_Z-\bar{a})+(d_Z-\bar{d}))/2,((b_Z-\bar{b})-(c_Z-\bar{c}))/2\Vert _2 = 0.$ It remains to set the other entries of Z and to show it is an outer product.

Claim

For each condition (i)–(viii), $a_Zd_Z=b_Zc_Z$ and all diagonal elements among $a_Z,b_Z,c_Z,d_Z$ are nonnegative.

Proof: First consider conditions (i)–(iv). By construction of (6):

$$\begin{aligned}&a_Zd_Z\, =\, -\bar{q}^2 + \lambda _1^2\Vert \bar{q},\bar{r}\Vert ^2_2 \, = \, \bar{r}^2-\lambda _2^2\Vert \bar{q},\bar{r}\Vert _2^2\, = \, b_Zc_Z. \end{aligned}$$

The second equality is derived from the following identity:

$$\begin{aligned}&\Vert \bar{q},\bar{r}\Vert _2^2 = \bar{q}^2+\bar{r}^2 \iff \, -\bar{q}^2 + \lambda _1^2\Vert \bar{q},\bar{r}\Vert _2^2 = \bar{r}^2-\lambda _2^2\Vert \bar{q},\bar{r}\Vert _2^2. \end{aligned}$$

Nonnegativity of diagonal elements follows from $\Vert \bar{q}, \bar{r}\Vert _2 \ge \max \{|\bar{q}|,|\bar{r}|\}$. In case (i) only $a_Z$ or $d_Z$ can be diagonal elements, and they are both nonnegative. The other cases can be directly verified. Similarly, for conditions (v)–(viii):

$$\begin{aligned}&a_Zd_Z = \bar{p}^2 - \lambda _2^2\Vert \bar{p},\bar{s}\Vert ^2_2 \, = \, -\bar{s}^2+\lambda _1^2\Vert \bar{p},\bar{s}\Vert _2^2 \, = \, b_Zc_Z. \end{aligned}$$

The second equality is derived from the following identity:

$$\begin{aligned} \Vert \bar{p},\bar{s}\Vert _2^2 = \bar{p}^2+\bar{s}^2&\iff \, -\bar{s}^2 + \lambda _1^2\Vert \bar{p},\bar{s}\Vert _2^2 = \bar{p}^2-\lambda _2^2\Vert \bar{p},\bar{s}\Vert _2^2. \end{aligned}$$

Nonnegativity of diagonal elements follows from the same argument as before, by using the fact that $\Vert \bar{p}, \bar{s}\Vert _2 \ge \max \{|\bar{p}|,|\bar{s}|\}$.    $\blacksquare$

To maintain symmetry we set $Z_{i_1,j_1}=Z_{j_1,i_1}$, $Z_{i_1,j_2}=Z_{j_2,i_1}, Z_{i_2,j_1}=Z_{j_1,i_2}$, $Z_{i_2,j_2}=Z_{j_2,i_2}$. Now denote $\ell = [i_1,i_2,j_1,j_2]$. If $a_Z=b_Z=c_Z=d_Z=0$, then we simply set all other entries of Z equal to zero and so Z is the outer-product of the vector of zeroes. Otherwise, consider the following cases.

Case 1: $\ell$ has 4 unique entries. Suppose w.l.o.g we have an upper-triangular entry $(i_1<i_2<j_1<j_2)$ and furthermore suppose that $b_Z$ is nonzero. Then set

and set all remaining entries of Z to zero. Other orderings of indices or the use of a different nonzero entry is handled by relabeling/rearranging column/row order.

Case 2: $\ell$ has three unique entries. Then, exactly one of $a_Z,b_Z,c_Z,d_Z$ is a diagonal entry, and so cases (i)–(iii), (v)–(vii) apply. If in any of these cases $a_Z$ or $d_Z$ is on the diagonal, by construction $|b_Z|=|c_Z|$. As $a_Zd_Z=b_Zc_Z$, we have $b_Z=c_Z=0$ iff exactly one of $a_Z$ or $d_Z$ is zero. Likewise, if $b_Z$ or $c_Z$ is a diagonal element, then $|a_Z|=|d_Z|$ and so $a_Z=d_Z=0$ iff exactly one of $b_Z$ or $c_Z$ are zero.

Suppose $a_Z$ is a nonzero diagonal entry. We propose:

$$Z_{\ell '}= \left[ \begin{array}{ccc} a_Z &{} b_Z &{} c_Z \\ b_Z&{} b_Z^2/a_Z&{} d_Z\\ c_Z&{} d_Z&{} c_Z^2/a_Z \end{array}\right] $$

where $\ell '$ are the unique entries of $\ell$. If $a_Z=0$ and on the diagonal, then we replace $b_Z^2/a_Z$ and $c_Z^2/a_Z$ with $|d_Z|$. If $b_Z,c_Z$ or $d_Z$ is on the diagonal, we use the same construction but with relabeling/rearranging column/row order.

Case 3: $\ell$ has two unique entries. All remaining entries of Z are set to zero.

For all cases, our construction ensures that all diagonal entries of Z are nonnegative, and all $2\times 2$ minors are zero; by Proposition 1, Z is an outer-product.    $\square$