Set-completely-positive representations and cuts for the max-cut polytope and the unit modulus lifting

Abstract

This paper considers a generalization of the “max-cut-polytope” $\hbox {conv}\{\ xx^T\mid x\in {\mathbb {R}}^n, \ \ |x_k| = 1 \ \hbox {for} \ 1\le k\le n\}$ in the space of real symmetric $n\times n$-matrices with all-one diagonal to a complex “unit modulus lifting” $\hbox {conv}\{xx^*\mid x\in {\mathbb {C}}^n, \ \ |x_k| = 1 \ \hbox {for} \ 1\le k\le n\}$ in the space of complex Hermitian $n\times n$-matrices with all-one diagonal. The unit modulus lifting arises in applications such as digital communications and shares similar symmetry properties as the max-cut-polytope. Set-completely positive representations of both sets are derived and the relation of the complex unit modulus lifting to its semidefinite relaxation is investigated in dimensions 3 and 4. It is shown that the unit modulus lifting coincides with its semidefinite relaxation in dimension 3 but not in dimension 4. In dimension 4 a family of deep valid cuts for the unit modulus lifting is derived that could be used to strengthen the semidefinite relaxation. It turns out that the deep cuts are also implied by a second lifting that could be used alternatively. Numerical experiments are presented comparing the first lifting, the second lifting, and the unit modulus lifting for $n=4$.

Appendix A

1.1 A.1: Duality of polyhedra

Assume a polyhedron of the general dual form

$$\begin{aligned} \tilde{P}^D:= \{\ \tilde{y} \mid \tilde{A}^T\tilde{y}\le c \ \} \end{aligned}$$

is given along with an interior point $\bar{y}$ satisfying $\tilde{A}^T\bar{y} < c$. Setting $y:=\tilde{y}-\bar{y}$, the standard linear programming problem in dual form,

$$\begin{aligned} \max \ \{\ b^T\tilde{y}\mid \tilde{y}\in \tilde{P}^D\ \} \end{aligned}$$

(31)

can be rewritten as

$$\begin{aligned} \max \ \{\ b^Ty \mid y \in P^D \ \} + b^T\bar{y} , \end{aligned}$$

(32)

where $P^D:=\{\ y\mid \tilde{A}^T y\le c-\tilde{A}^T\bar{y} \ \}$ or

$$\begin{aligned} P^D = \{\ y\mid A^T y\le e \ \}, \end{aligned}$$

(33)

with $A^T:=\hbox {Diag}(c-\tilde{A}^T\bar{y})^{-1}\tilde{A}^T$. (Note that $c-\tilde{A}^T\bar{y}>0$.) The right hand side of (33) is referred to as normalized dual form in the sequel.

Neglecting the constant term $b^T\bar{y}$, problem (32) is equivalent to

$$\begin{aligned} \max _{y,\lambda }\ \{\ b^Ty\mid A^Ty-\lambda e\le 0,\ \ \lambda =1\}. \end{aligned}$$

We assume that b is nonzero, and that the above problem has a finite optimal value. Since $y=0$ is strictly feasible, the optimal value is positive. The dual of the above problem is given by

$$\begin{aligned} \min _{x,\mu } \ \{\ \mu \mid Ax=b, \ -\,e^Tx+\mu = 0, \ x\ge 0 \ \}. \end{aligned}$$

By duality, the optimal $\mu$ is finite and positive. Using $e^T(\frac{x}{\mu })=1$ this problem can be written as

$$\begin{aligned} \min \ \left\{ \ \mu \mid \frac{b}{\mu }\in \hbox {conv}_{1\le i\le n}\{a^{(i)}\}\ \right\} \end{aligned}$$

where $a^{(i)}$ are the columns of A for $1\le i\le n$. The feasible set on the right hand side $P:=\hbox {conv}_{1\le i\le n}\{a^{(i)}\}$ is the usual dual polyhedron to $P^D$ sometimes also referred to as polar dual. The above can be written as the following form of polyhedral duality

$$\begin{aligned} \max \ \{\ b^Ty\mid y\in P^D\ \} = \min \ \left\{ \ \mu \mid \frac{b}{\mu }\in P\ \right\} . \end{aligned}$$

(34)

This relationship is true for any nonzero b for which the right-hand side is finite; in particular, when $P^D$ is bounded, it is true for any nonzero b. Note that P and $P^D$ have the same dimension; in particular, P is not the feasible domain of the standard primal problem of (31). Relation (16) is a reformulation of (34) adapted to the variables in form of symmetric matrices with all-one diagonal.

An intriguing feature of (34) is that the dual polyhedra P and $P^D$ are independent of the objective function but (34) only applies to problems with a feasible set in normalized dual form. To translate (34) back to the general dual form (31) let $\rho :=\hbox {Diag}(c-\tilde{A}^T\bar{y})^{-1}e$. Then

$$\begin{aligned} \tilde{P}:= P=\hbox {conv}_{1\le i\le n}\{\rho _i\tilde{a}^{(i)}\} \end{aligned}$$

can be regarded as dual polyhedron associated with $\tilde{P}^D$ and $\bar{y}$.

1.2 A.2: The generalized triangle inequalities (20)

The following proposition holds:

Proposition 6

The inequalities (20) are valid inequalities for the unit modulus lifting $\mathcal{UML}_{\mathbb {C}}$.

Proof

It suffices to establish (20) for all extremal matrices $X=xx^*$. Let $x_j=e^{i\alpha }$, $x_k=e^{i\beta }$, and $x_l=e^{i\gamma }$. Then, it suffices to show that

$$\begin{aligned} \min _{\alpha ,\beta ,\gamma \in {\mathbb {R}}} \hbox {Re}(e^{i(\alpha -\beta )} + e^{i(\beta -\gamma )} + e^{i(\gamma -\alpha )} ) \ge -3/2. \end{aligned}$$

(35)

Since the objective value does not change when adding the same constant to all three variables $\alpha ,\beta$, and $\gamma$, we may assume that one of the optimal solutions satisfies $\alpha =0$. For $\alpha =0$ the real part in (35) is given by

$$\begin{aligned} f(\beta ,\gamma ):=\cos (\beta ) + \cos (\beta -\gamma )+\cos (\gamma ). \end{aligned}$$

(36)

(Note that $\cos (\beta )=\cos (-\beta )$.) The gradient of f with respect to $\beta ,\gamma$ is

$$\begin{aligned} g(\beta ,\gamma ):=\left[ \begin{matrix}-\sin (\beta )-\sin (\beta -\gamma )\\ \sin (\beta -\gamma )-\sin (\gamma )\end{matrix}\right] . \end{aligned}$$

The critical points of f satisfy $g(\beta ,\gamma )=0$ which implies $-\sin (\beta )=\sin (\gamma )$. Hence, $\gamma = -\beta$ (modulo $2\pi$) or $\gamma = \pi +\beta$ (modulo $2\pi$).

• When $\gamma = -\beta$ it follows that

  $$\begin{aligned} 0=-\sin (\beta ) - \sin (2\beta )= -\sin (\beta ) - 2\sin (\beta )\cos (\beta ), \end{aligned}$$

  i.e., $\beta = k\pi$ or $2\cos (\beta )=-1$.

  • When $2\cos (\beta )=-1$ either $\beta = \frac{2\pi }{3}$ or $\beta = \frac{4\pi }{3}$ (modulo $2\pi$). In both cases, $f(\beta ,\gamma )=-3/2$.

  • When $\beta = k\pi$ the function f in (36) takes the values $f(\beta ,\gamma )$ equal $-1,1$, or 3.

• Finally, when $\gamma = \pi +\beta$ it follows that $-\sin (\beta )-\sin (-\pi )=0$ i.e., $\beta = k\pi$ as above.

The global minimum value of f evidently is $f(\beta ,\gamma ) \ge -3/2$ for all real $\beta ,\gamma$. $\square$