Lower Bounds for Cubic Optimization over the Sphere

Abstract

We consider the problem of minimizing a polynomial function of degree three over the boundary of the sphere. If the objective is quadratic instead of cubic, this is the well-studied trust region subproblem, which is known to be tractable. In the cubic case, the problem turns out to be NP-hard. In this paper, we derive and evaluate different approaches for computing lower bounds for the cubic problem. Alternatively to semidefinite programming relaxations proposed in the literature, our approaches do not lift the problem to higher dimensions. The strongest bounds are obtained by Lagrangian decomposition, resulting in a number of parameterized quadratic problems for which the above-mentioned results can be exploited, in particular the existence of a tractable dual problem. In an experimental evaluation, we consider the cubic one-spherical optimization problem, with homogeneous objective function, and compare the bounds generated with the different approaches proposed, for small examples from the literature and for randomly generated instances of varied dimensions.

Appendix

Proof of Theorem 1

We consider each problem on the right-hand side of (2) independently and show that all appearing minima can be obtained in closed form. First, note that

$$\begin{aligned}&\text{ for }\;x_{i}\in (-1,0):\; \displaystyle \min _{\Vert x_{{\hat{\imath }}}\Vert =\sqrt{1-x_{i}^{2}}} x_{i}\displaystyle \sum _{\begin{array}{c} j,k=1\\ j,k \ne i \end{array}}^{n}a_{ijk}x_{j}x_{k} = x_{i}(1-x_{i}^{2})\lambda _{\max }({\tilde{A}}_{i})\; , \\&\text{ for }\;x_{i}\in (0,1):\; \displaystyle \min _{\Vert x_{{\hat{\imath }}}\Vert =\sqrt{1-x_{i}^{2}}} x_{i}\displaystyle \sum _{\begin{array}{c} j,k=1\\ j,k \ne i \end{array}}^{n}a_{ijk}x_{j}x_{k} = x_{i}(1-x_{i}^{2})\lambda _{\min }({\tilde{A}}_{i})\; , \end{aligned}$$

using the notation of Sect. 2. For $x_i\in \{-1,0,1\}$, this expression is zero. In summary, taking the minimum over $x_i\in [-1,1]$, we obtain

$$\begin{aligned}&\displaystyle \min _{x_{i}\in [-1,1]} \Big (\displaystyle \min _{\Vert x_{{\hat{\imath }}}\Vert =\sqrt{1-x_{i}^{2}}}x_{i}\displaystyle \sum _{\begin{array}{c} j,k=1\\ j,k \ne i \end{array}}^{n}a_{ijk}x_{j}x_{k}\Big ) \nonumber \\&\quad =\displaystyle \min \Big \{-\frac{2}{9}\sqrt{3}\lambda _{\max }({\tilde{A}}_{i}),\frac{2}{9}\sqrt{3}\lambda _{\min }({\tilde{A}}_{i})\Big \}\; , \end{aligned}$$

(23)

since this expression is always non-positive. Moreover,

$$\begin{aligned} \min _{\Vert x_{{\hat{\imath }}}\Vert =\sqrt{1-x_{i}^{2}}}2x_{i}^{2}\displaystyle \sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}a_{iij}x_{j} = (-2)\sqrt{\sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}a_{iij}^{2}} \ x_{i}^{2} \sqrt{1-x_{i}^{2}} \end{aligned}$$

and hence

$$\begin{aligned} \displaystyle \min _{x_{i}\in [-1,1]} \Big (\displaystyle \min _{\Vert x_{{\hat{\imath }}}\Vert =\sqrt{1-x_{i}^{2}}}2x_{i}^{2}\displaystyle \sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}a_{iij}x_{j}\Big ) = -\frac{4}{9}\sqrt{3} \sqrt{\sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}a_{iij}^{2}}\;. \end{aligned}$$

(24)

Also,

$$\begin{aligned} \min _{\Vert x_{{\hat{\imath }}}\Vert =\sqrt{1-x_{i}^{2}}}x_{i}\displaystyle \sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}a_{ij}x_{j} = -\sqrt{\sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}a_{ij}^{2}} \ x_{i} \sqrt{1-x_{i}^{2}} \end{aligned}$$

and hence

$$\begin{aligned} \displaystyle \min _{x_{i}\in [-1,1]} \Big (\displaystyle \min _{\Vert x_{{\hat{\imath }}}\Vert =\sqrt{1-x_{i}^{2}}}x_{i}\displaystyle \sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}a_{ij}x_{j}\Big ) = -\frac{1}{2} \sqrt{\sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}a_{ij}^{2}}\;. \end{aligned}$$

(25)

Next, to compute $\min _{x_{i}\in [-1,1]}a_{iii}x_{i}^{3} + a_{ii}x_i^2+a_ix_i$, we must consider the different cases described in the following.

If $a_{iii}\ne 0$, let

$$\begin{aligned} {\hat{x}}_i^-:=\frac{-a_{ii}-\sqrt{a_{ii}^2-3a_{iii}a_i}}{3a_{iii}}, \;\; {\hat{x}}_i^+:=\frac{-a_{ii}+\sqrt{a_{ii}^2-3a_{iii}a_i}}{3a_{iii}}. \end{aligned}$$

If ${\hat{x}}_i^- \in [-1,1]$ and ${\hat{x}}_i^+ \in [-1,1]$, then we define

$$\begin{aligned} \begin{array}{ll} \kappa _i:=&{}\displaystyle \min _{x_{i}\in [-1,1]}a_{iii}x_{i}^{3} + a_{ii}x_i^2+a_ix_i = \min \{a_{ii}-|a_{iii}+a_i|,\\ &{}a_{iii}({\hat{x}}_{i}^-)^{3} + a_{ii}({\hat{x}}_i^-)^2+a_i{\hat{x}}_i^-, a_{iii}({\hat{x}}_{i}^+)^{3} + a_{ii}({\hat{x}}_i^+)^2+a_i{\hat{x}}_i^+\}. \end{array} \end{aligned}$$

(26)

If ${\hat{x}}_i^- \notin [-1,1]$, we should disregard the cubic polynomial in ${\hat{x}}_i^-$ when computing the minimum in (26). Analogously, if ${\hat{x}}_i^+ \notin [-1,1]$, we should disregard the cubic polynomial in ${\hat{x}}_i^+$ when computing the minimum in (26).

If $a_{iii} = 0$, then if $a_{ii}\ne 0$ and $-\frac{a_i}{2a_{ii}}\notin [-1,1]$, we define

$$\begin{aligned} \kappa _i:=\displaystyle \min _{x_{i}\in [-1,1]}a_{iii}x_{i}^{3} + a_{ii}x_i^2+a_ix_i = a_{ii}-|a_i|. \end{aligned}$$

(27)

If $a_{iii} = 0$, then if $a_{ii}\ne 0$ and $-\frac{a_i}{2a_{ii}}\in [-1,1]$, we define

$$\begin{aligned} \kappa _i:=\displaystyle \min _{x_{i}\in [-1,1]}a_{iii}x_{i}^{3} + a_{ii}x_i^2+a_ix_i = \min \{a_{ii}-|a_i|,-\frac{a_i^2}{4a_{ii}}\}. \end{aligned}$$

(28)

Finally, if $a_{iii} = a_{ii}= 0$, we define

$$\begin{aligned} \kappa _i:=\displaystyle \min _{x_{i}\in [-1,1]}a_{iii}x_{i}^{3} + a_{ii}x_i^2+a_ix_i = -|a_i|. \end{aligned}$$

(29)

Adding up (23)–(25) with either (26), (27), (28), or (29), we obtain

$$\begin{aligned}&\displaystyle \min _{\Vert x\Vert =1}\displaystyle \sum _{i,j,k=1}^{n}a_{ijk}x_{i}x_{j}x_{k}+\sum _{i,j=1}^{n}a_{ij}x_{i}x_{j}+\sum _{i=1}^{n}a_{i}x_{i} \\&\quad \ge \displaystyle \sum _{i=1}^{n}\Big (\min \Big \{-\frac{2}{9}\sqrt{3}\lambda _{\max }({\tilde{A}}_{i}),\frac{2}{9}\sqrt{3}\lambda _{\min }({\tilde{A}}_{i})\Big \} -\frac{4}{9}\sqrt{3} \sqrt{\sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}a_{iij}^{2}} \\&\qquad -\,\frac{1}{2} \sqrt{\sum _{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}a_{ij}^{2}} + \kappa _i\Big ). \end{aligned}$$

$\square$