Solving $k$-cluster problems to optimality with semidefinite programming

Abstract

This paper deals with the computation of exact solutions of a classical NP-hard problem in combinatorial optimization, the $k$-cluster problem. This problem consists in finding a heaviest subgraph with $k$ nodes in an edge weighted graph. We present a branch-and-bound algorithm that applies a novel bounding procedure, based on recent semidefinite programming techniques. We use new semidefinite bounds that are less tight than the standard semidefinite bounds, but cheaper to get. The experiments show that this approach is competitive with the best existing ones.

Appendix: Formulation of the \(k\)-cluster as a quadratic \(\{-1,1\}\)-problem

The spherical constraint appears more easily on purely quadratic $\{-1,1\}$-optimization problems (see [32]). The second reformulation in Sect. 2 considers the transformation of the $k$-cluster problem, from the natural modeling as a quadratic $\{0,1\}$-problem (with linear and quadratic constraints), to a purely quadratic $\{-1,1\}$-problem. We specify here this transformation, that uses standard techniques.

Lemma 3

With the notation of this section, the $k$-cluster problem (1) is equivalent to the quadratic problem in dimension $n+ 1$

$$\begin{aligned} \left\{ \begin{array}{l} \max \quad x^{\top }Q\,x\\ \quad x^{\top }Q_j\,x = 4k-2n, \ \ j\in \{0,\ldots ,n\}\\ \quad x\in \{-1,1\}^{n+1} \end{array}\right. \end{aligned}$$

(17)

where the symmetric $(n+1)\times (n+1)$-matrices $Q$ and $Q_j\ (j\in \{0,\ldots ,n\})$ are defined by (5). More precisely, this equivalence means that the optimal values of (1) and (17) are the same and that the optimal solutions coincide as follows:

• if $\bar{z}$ is a solution of (1) then $\bar{x} = (1, 2\bar{z}-e)$ is a solution of (1);

• if $\bar{x}$ is a solution of (17), then $\bar{z} = ((\bar{x}_0\bar{x}_1,\ldots , \bar{x}_0\bar{x}_{n}) +e)/2$ is a solution of (1).

Proof

Operate first the change of variable $x=2z-e$, and express the objective and the constraints with respect to $x=(x_1,\ldots ,x_n)\in \mathbb R ^n$. Just develop the objective

$$\begin{aligned} {z}^{\top }Wz = \frac{{(x+e)}^{\top }}{2}W\frac{(x+e)}{2} = \frac{1}{4}({x}^{\top }Wx + 2 {x}^{\top }We + {e}^{\top }We), \end{aligned}$$

and similarly transform ${e}^{\top }z = k$ as ${e}^{\top }x = 2k-n$, and for all $j=1,\ldots ,n$

$$\begin{aligned} {z\!}^{\top }\!C_jz&= 2kz_j \iff {x\!}^{\top }\!C_jx + 2 {x\!}^{\top }\!(C_je - 2ke_j)\\&= 4k - {e\!}^{\top }\!C_je \iff {x\!}^{\top }\!C_jx + 2 {x\!}^{\top }\!\tilde{e}_j = 4k - 2n \end{aligned}$$

the last equivalence coming from ${e}^{\top }C_je= 2n$ and $\tilde{e}_j = C_je - 2ke_j$. So we get quadratic problem in $x\in \{-1,1\}^n$

$$\begin{aligned} \left\{ \begin{array}{l} \max \quad \frac{1}{4}({x}^{\top }Wx + 2 {x}^{\top }We + {e}^{\top }We)\\ \quad {e}^{\top }x = 2k-n\\ \quad {x}^{\top }C_jx + 2 {x}^{\top }\tilde{e}_j= 4k-2n, \ \ j\in \{1,\ldots ,n\}\\ \quad x_i\in \{-1,1\} \ \ i\in \{1,\ldots ,n\}. \end{array}\right. \end{aligned}$$

(18)

The formulation of (3) is equivalent to (18) in the sense that the optimal values are the same and the solutions are in one-to-one correspondance with $x=2z-e$.

We consider now that the following purely quadratic problem with the additional variable $x_0\in \{-1,1\}$

$$\begin{aligned} \left\{ \begin{array}{l} \max \quad \frac{1}{4}({x}^{\top }Wx + 2 {x}^{\top }We\,x_0 + {e}^{\top }We\,{x_0}^2)\\ \quad {e}^{\top }x\, x_0 = 2k-n\\ \quad {x}^{\top }C_jx + 2 {x}^{\top }\tilde{e}_j\,x_0= 4k-2n, \ \ j\in \{1,\ldots ,n\}\\ \quad x_i\in \{-1,1\} \ \ i\in \{0,\ldots ,n\}. \end{array}\right. \end{aligned}$$

(19)

We observe that this problem is equivalent to (17) in view of the definitions of the matrices in (5). So we just have to establish that (18) is equivalent to (19); we do so in two steps. If $(x_1,\ldots ,x_n)$ is feasible in (18), then obviously $(1,x)$ and $(-1,-x)$ are both feasible in (19), with same objective value. It follows that $\mathrm{val}(19) \le \mathrm{val}(18)$. Conversely if $(x_0,\ldots ,x_n)$ is feasible in (19), then $x_0=\pm 1$ and $(x_0x_1,\ldots ,x_0x_n)$ is feasible in (18) with the same objective value. It follows that $\mathrm{val}(18)\le (19)$, so that we have the equality in fact. The relation between the argmins then becomes clear.

$\square$