Sum-of-squares rank upper bounds for matching problems

Abstract

The matching problem is one of the most studied combinatorial optimization problems in the context of extended formulations and convex relaxations. In this paper we provide upper bounds for the rank of the sum-of-squares/Lasserre hierarchy for a family of matching problems. In particular, we show that when the problem formulation is strengthened by incorporating the objective function in the constraints, the hierarchy requires at most $\left\lceil \frac{k}{2} \right\rceil$ levels to refute the existence of a perfect matching in an odd clique of size $2k+1$.

Appendix: Proof of Lemma 8

As it is presented in e.g. Laurent (2003) the 0/1 Lasserre hierarchy of level t can be formulated in terms of matrices. For $0 \le t \le n$, let $\left( {\begin{array}{c}[n]\\ t\end{array}}\right)$, $\left( {\begin{array}{c}[n]\\ \le t\end{array}}\right)$ denote all subsets of size exactly and at most t, respectively. The moment matrix of variables is symmetric, real-valued, set-indexed matrix $M_{t+1} \in \mathbb {S}^{\left( {\begin{array}{c}[n]\\ \le r\end{array}}\right) \times \left( {\begin{array}{c}[n]\\ \le r\end{array}}\right) }$ such that $\left( M_{t+1}\right) _{I,J}=\tilde{\mathbb {E}}{\left( \prod _{i \in I \cup J}x_i \right) }$.

Lemma 9

$\tilde{\mathbb {E}}{\left( u^2(x) \right) } \ge 0, ~ \forall u \in \mathbb {R}[x]_{t+1}$ iff $M_{t+1} \succeq 0$

Proof

$$\begin{aligned} \tilde{\mathbb {E}}{\left( u^2(x) \right) }=\tilde{\mathbb {E}}{\left( \sum _{|I|\le t+1} a_I \prod _{i\in I}x_i \right) }^2 = \sum _{|I|,|J|\le t+1} a_I a_J \tilde{\mathbb {E}}{\left( \prod _{i \in I \cup J}x_i \right) }= a^\top M_{t+1} a \end{aligned}$$

where $a \in \mathbb {R}^{\left( {\begin{array}{c}[n]\\ \le t\end{array}}\right) }$. The claim follows since $a^\top M_{t+1} a \ge 0, ~ \forall a \in \mathbb {R}^{\left( {\begin{array}{c}[n]\\ \le t\end{array}}\right) }$ iff $M_{t+1} \succeq 0$.$\square$

We consider a complete graph $K_5$, with $n=10$ edges, and $t=1$. For a sake of convenience we denote the moment matrix $M_{2}$ by M and, since the considered pseudoexpectation operator is symmetric, for $\tilde{\mathbb {E}}{\left( \prod _{i \in S}x_i \right) } \ne 0$ we denote $\tilde{\mathbb {E}}{\left( \prod _{i \in S}x_i \right) }$ by $\alpha _1$ or $\alpha _2$ for $|S|=1$ or $|S|=2$, respectively. Let M be of the form:

$$\begin{aligned} M=\begin{pmatrix}1 &{} y^{\top } &{}z^{\top } \\ y &{} A&{} B \\ z &{} B^{\top } &{} C\end{pmatrix} \end{aligned}$$

where $y\in \mathbb {R}^{\left( {\begin{array}{c}[n]\\ 1\end{array}}\right) }$, $z\in \mathbb {R}^{\left( {\begin{array}{c}[n]\\ 2\end{array}}\right) }$, $A\in \mathbb {R}^{\left( {\begin{array}{c}[n]\\ 1\end{array}}\right) \times \left( {\begin{array}{c}[n]\\ 1\end{array}}\right) }$, $B\in \mathbb {R}^{\left( {\begin{array}{c}[n]\\ 1\end{array}}\right) \times \left( {\begin{array}{c}[n]\\ 2\end{array}}\right) }$ and $C\in \mathbb {R}^{\left( {\begin{array}{c}[n]\\ 2\end{array}}\right) \times \left( {\begin{array}{c}[n]\\ 2\end{array}}\right) }$.

In the following we cite a Lemma from Kurpisz et al. (2015) which gives necessary and sufficient conditions for a diagonal matrix plus rank one matrix to be PSD.

Lemma 10

(Kurpisz et al. 2015) For a diagonal matrix $D\in \mathbb {R}^{m \times m}$ with diagonal entries $d_1,\ldots , d_{m}$, the matrix $D + d_{m+1} \mathbf 1 \mathbf 1 ^\top$ is positive-semidefinite if and only if either $d_i\ge 0$ for all $i \in [m+1]$, or the following holds

$$\begin{aligned} d_j&< 0, \quad \textit{for exactly one } j \in [m+1],\end{aligned}$$

(6)

$$\begin{aligned} d_k&> 0, \quad \textit{for all } k \ne j \in [m+1], \end{aligned}$$

(7)

$$\begin{aligned} \sum _{i \in [m+1]} \frac{1}{d_i}&\le 0 \end{aligned}$$

(8)

We prove the Lemma 8 by contradiction. Assume that there exists a pseudoexpectation operator $\tilde{\mathbb {E}}{\left( \cdot \right) }$ satisfying (2) for maximum matching problem  (3) with $b=1$ in $K_5$ at level $t=1$ with an objective value greater than 2. Since by assumption there exists such $\tilde{\mathbb {E}}{\left( \cdot \right) }$, the corresponding moment matrix M is PSD.

Lemma 11

For every feasible symmetric pseudoexpectation operator $\tilde{\mathbb {E}}{\left( \cdot \right) }$, $\alpha _2 \le 1/15$.

Proof

For every feasible pseudoexpectation operator, the corresponding moment matrix M must be PSD, and (by Observation 7.1.12 in  Horn and Johnson (2013)) every principal submatrix of M must be PSD. Consider the following principal submatrix $P_1$

$$\begin{aligned} P_1=\begin{pmatrix}1 &{}{z'}^{\top } \\ {z'} &{} C'\end{pmatrix} \end{aligned}$$

where $z'$ and $C'$ are composed of the nonzero rows/columns of z and C, respectively (clearly, by Lemma 5 which holds also for relaxation (3), there are some zero rows). Since, by Lemma 6, maximum matching in $K_5$ is of size 2 and by Lemma 5, the matrix $C'$ is a diagonal matrix. By the Schur complement theorem,

$$\begin{aligned} P_1 \succeq 0 \Leftrightarrow C' - z' {z'}^{\top } \succeq 0 \end{aligned}$$

Now since we consider symmetric pseudoexpectation operator, $z=\alpha _2 \mathbf 1$. We have

$$\begin{aligned} C' - z' {z'}^{\top } = C'-\alpha _2^2\mathbf 1 \mathbf 1 ^\top \end{aligned}$$

where $C'$ is a diagonal matrix with diagonal entries $\alpha _2$. Finally, since there are 15 subsets of cardinality 2 corresponding to matchings of size 2, that produce 15 nonzero row/columns in C by Lemma 10 get that:

$$\begin{aligned} \frac{15}{\alpha _2}\le \frac{1}{\alpha _2^2} \end{aligned}$$

thus, finally $\alpha _2\le 1/15$. $\square$

Lemma 12

For every feasible symmetric pseudoexpectation operator $\tilde{\mathbb {E}}{\left( \cdot \right) }$, $\alpha _1\le 1/5$.

Proof

Consider the following principal submatrix $P_2$ of M

$$\begin{aligned} P_2=\begin{pmatrix}1 &{}{y}^{\top } \\ {y} &{} A\end{pmatrix} \end{aligned}$$

For every feasible pseudoexpectation operator the matrix $P_2$ has to be PSD. By the Schur complement Theorem a matrix

$$\begin{aligned} P_2 \succeq 0 \Leftrightarrow A - y {y}^{\top } \succeq 0 \end{aligned}$$

It remains to show that PSDness of a matrix $A - y {y}^{\top }$ implies that $\alpha _1\le 1/5$. We use the fact that $A - y {y}^{\top }$ belongs to Bose–Mesner algebra of the Johnson Scheme J(5, 2). Indeed, A belongs to J(5, 2) so does $A - y {y}^{\top }$. This implies that the eigenvalues of $A - y {y}^{\top }$ can be explicitly computed using Eberlein polynomials.

More precisely Johnson scheme J(v, k) is defined as a sequence of 0/1 matrices $A_0,\ldots , A_k$ row-column indexed by the k subsets of [v] such that $\left( A_i \right) _{I,J}=1$ iff $|I \triangle J|=2i$ . Thus $A_0=I$ and $\sum _{i=0}^k A_i=J$. A fundamental property of the Johnson scheme is that the distinct eigenvalues of a matrix $Z=\sum _{i=0}^k a_i A_i$ can be expressed by $\lambda _u=\sum _{i=0}^k a_i Q_i(u)$, for $u\in \{0,\ldots ,k\}$ where $Q_i(u)$ is the Eberlein polynomial of the form:

$$\begin{aligned} Q_l(u)=\sum _{j=0}^l (-1)^j \left( {\begin{array}{c}u\\ j\end{array}}\right) \left( {\begin{array}{c}k-u\\ l-j\end{array}}\right) \left( {\begin{array}{c}v-k-u\\ l-j\end{array}}\right) \end{aligned}$$

It is easy to see that the matrix A is a Johnson scheme J(5, 2) which can be express as

$$\begin{aligned} A=\alpha _1 A_0 + \alpha _2 A_2 \end{aligned}$$

Thus $A- y {y}^{\top }$ is a Johnson scheme J(5, 2) which can be express as

$$\begin{aligned} A- y {y}^{\top }=(\alpha _1-\alpha _1^2) A_0 - \alpha _1^2 A_1 + (\alpha _2-\alpha _1^2) A_2 \end{aligned}$$

and distinct eigenvalues of $A- y {y}^{\top }$ can be expressed by

$$\begin{aligned} \lambda _u=(\alpha _1-\alpha _1^2) Q_0(u) -\alpha _1^2 Q_1(u) +(\alpha _2-\alpha _1^2) Q_2(u) \quad u=\{0,1,2\} \end{aligned}$$

First note that:

$$\begin{aligned} Q_0(u)= & {} 1\\ Q_1(u)= & {} 6 + (u-6 ) u\\ Q_2(u)= & {} \frac{1}{4} \left( u(u-6) (u-3)^2+12\right) \end{aligned}$$

and thus we get the following distinct values of $\lambda _u$ for $u=\{0,1,2\}$

$$\begin{aligned} \lambda _0= & {} -10\alpha _1^2 +\alpha _1 +3 \alpha _2\\ \lambda _1= & {} \alpha _1-2 \alpha _2\\ \lambda _2= & {} \alpha _1+\alpha _2 \end{aligned}$$

Since the matrix $A - y {y}^{\top }$ is claimed to be PSD, in particular $\lambda _0 \ge 0$. Note that $\lambda _0$ is increasing in $\alpha _2$ thus, by Lemma 11 we consider $\alpha _2=1/15$, we get

$$\begin{aligned} \lambda _0=-10\alpha _1^2+\alpha _1+\frac{1}{5}\ge 0 \end{aligned}$$

which implies that $\alpha _1 \le 1/5$ $\square$

By Lemma 12 we get that $\alpha _1\le 1/5$, this is clearly a contradiction since the assumed pseudoexpectation operator had an objective value greater than 2.