Sum-of-squares hierarchy lower bounds for symmetric formulations

Abstract

We introduce a method for proving Sum-of-Squares (SoS)/Lasserre hierarchy lower bounds when the initial problem formulation exhibits a high degree of symmetry. Our main technical theorem allows us to reduce the study of the positive semidefiniteness to the analysis of “well-behaved” univariate polynomial inequalities. We illustrate the technique on two problems, one unconstrained and the other with constraints. More precisely, we give a short elementary proof of Grigoriev/Laurent lower bound for finding the integer cut polytope of the complete graph. We also show that the SoS hierarchy requires a non-constant number of rounds to improve the initial integrality gap of 2 for the Min-Knapsack linear program strengthened with cover inequalities.

Appendices

Appendix

Omitted calculations for Max-Cut

Grigoriev [20] and Laurent [30] proved that the following solution is feasible for any $\omega \le n/2$ up to round $t\le \lfloor \omega \rfloor$ for the SoS hierarchy given in Definition 1:

$$\begin{aligned} y_{I} = \frac{\left( {\begin{array}{c}\omega \\ |I|\end{array}}\right) }{\left( {\begin{array}{c}n\\ |I|\end{array}}\right) } \qquad \forall I\subseteq N:|I|\le 2t. \end{aligned}$$

(36)

For a graph $G = (V, E)$, the objective function of the Max-Cut problem in the usual formulation for 0 / 1 variables is $\sum _{(i,j) \in E} (x_i-x_j)^2$. The cut value of the SoS relaxation of Definition 1 is then $\sum _{(i,j) \in E} y_{\left\{ i \right\} }+y_{\left\{ j \right\} }-2y_{\left\{ i,j \right\} }$. When G is the complete graph and the solution to the SoS relaxation is taken to be (36), the cut value of the SoS relaxation is

$$\begin{aligned} \sum _{i=1}^{n-1} \sum _{j=i+1}^n y_{\left\{ i \right\} }+y_{\left\{ j \right\} }-2y_{\left\{ i,j \right\} }=2 \left( {\begin{array}{c}n\\ 2\end{array}}\right) \left( \frac{\left( {\begin{array}{c}\omega \\ 1\end{array}}\right) }{\left( {\begin{array}{c}n\\ 1\end{array}}\right) }-\frac{\left( {\begin{array}{c}\omega \\ 2\end{array}}\right) }{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }\right) = \omega (n-\omega ). \end{aligned}$$

It is straightforward to check that for odd n and $\omega = n/2$ this value is larger than the maximum cut in the complete graph of n vertices.

Change of basis. Using the change of basis of Lemma 1, solution $\{y_I\}$ is equivalent to solution $\{y^N_I\}$:

$$\begin{aligned} y^N_I&= \sum _{H\subseteq N{\setminus } I} (-1)^{|H|} y_{I\cup H} = \sum _{h=0}^{n-|I|} \left( {\begin{array}{c}n-|I|\\ h\end{array}}\right) (-1)^{h} \frac{\left( {\begin{array}{c}\omega \\ |I|+h\end{array}}\right) }{\left( {\begin{array}{c}n\\ |I|+h\end{array}}\right) } \nonumber \\&= y_I \left( {\begin{array}{c}\omega -|I|-1\\ n-|I|\end{array}}\right) (-1)^{n-|I|}= (n+1) \left( {\begin{array}{c}{\omega }\\ n+1\end{array}}\right) \frac{(-1)^{n-|I|}}{{\omega }-|I|} \end{aligned}$$

(37)

where we use the identity $\sum _{\omega =0}^m (-1)^{\omega } \left( {\begin{array}{c}n\\ {\omega }\end{array}}\right) =(-1)^m\left( {\begin{array}{c}n-1\\ m\end{array}}\right)$.