Separation routine and extended formulations for the stable set problem in claw-free graphs

Abstract

The maximum weighted stable set problem in claw-free graphs is a well-known generalization of the maximum weighted matching problem, and a classical problem in combinatorial optimization. In spite of the recent development of fast(er) combinatorial algorithms and some progresses in the characterization of the corresponding stable set polytope, the problem of “providing a decent linear description” for this polytope (Grötschel et al. in Geometric algorithms and combinatorial optimization, Springer, New York, 1988) is still open. The main contribution of this paper is to propose an algorithmic answer to that question by providing a polynomial-time and computationally attractive separation routine for the stable set polytope of claw-free graphs, that only requires a combinatorial decomposition algorithm, the solution of (moderate sized) compact linear programs, and Padberg and Rao’s algorithm for separating over the matching polytope. In particular, it is a generalization of the latter and avoids the heavy computational burden of resorting to the ellipsoid method, on which the only poly-time separation routine known so far relied. Besides, our separation routine comes with a ‘small’ (but not polynomial) extended linear programming formulation and a procedure to derive a linear description of the stable set polytope of claw-free graphs in the original space.

A Proof of Lemma 6

Proof

Claim Let $x^* \in \hbox {STAB}(G)$ with $x^*_i=x^*[G_i]\in \hbox {STAB}(G_i)$. Consider two vectors $\underline{s},\overline{s}\in [0,1]^k$ such that $\underline{t}\le \underline{s}\le \overline{s}\le \overline{t}$ and we define $y=f(x^*,\underline{s},\overline{s}) \in \mathbb {R}^{V(H(G))}$. We denote by $y_i{:}{=}y[gg_i]$. Let $(G_i,\mathcal{A}_i)$ be a strip of G with $\mathcal{A}_i=\{A_i,B_i\}$. Then, for all $i\in [k]$, $\underline{s}_i=\min \{t_i: y_i\in \hbox {STAB}(gg_i,t_i,\{u_i\},\{v_i\})$ and $\overline{s}_i=\max \{t_i: y_i \in \hbox {STAB}(gg_i,t_i,\{u_i\},\{v_i\})$.

Proof

Fix $i\in [k]$ and $t_i \in [0,1]$. We will show that $y_i \in \hbox {STAB}(gg_i,t_i,\{u_i\},\{v_i\})$ if and only if $\underline{s_i}\le t_i \le \overline{s_i}$. In order to prove the latter fact, we take a small detour. We define the graph $h_i$ obtained from $gg_i$ by adding the edge $\{u_i,v_i\}$. We will show that:

• (i) $y_i$ is in $\hbox {STAB}(gg_i,t_i,\{u_i\},\{v_i\})$ if and only if $y_i' = y_i - t_i \cdot \chi ^{\{u_i,v_i\}}$ is in $(1-t_i) \cdot \hbox {STAB}(h_i)$ (a point y is in $\rho \cdot Q$ with $Q=\{x:Ax\le b\}$ if $A y \le \rho \cdot b$);

• (ii) $y_i'\in (1-t_i) \cdot \hbox {STAB}(h_i)$ if and only if $\underline{s_i}\le t_i \le \overline{s_i}$.

Clearly, (i) and (ii) are enough to prove that $y_i$ is in $\hbox {STAB}(gg_i,t_i,\{u_i\},\{v_i\})$ if and only if $\underline{s_i}\le t_i \le \overline{s_i}$, and therefore the claim.

• (i) If $y_i \in \hbox {STAB}(gg_i,t_i,\{u_i\},\{v_i\})$, then there exist $\lambda \in \mathbb {R}^{\mathcal{S}(gg_i)}_+ : y_i=\sum _{S \in \mathcal{S}(gg_i)} \lambda _S \chi ^S$, $\sum _{S \in \mathcal{S}(gg_i)} \lambda _S =1$, with $\lambda _{\{u_i,v_i\}} =t_i$, since a stable set of $gg_i$ containing $u_i$ and $v_i$ does not contain any other node. If we remove the stable set $\{u_i,v_i\}$ from the combination we get $y_i'$, which is then expressed as a non-negative combination of stable sets of $h_i$ with coefficients summing to $1-t_i$. Hence $y_i' \in (1-t_i) \cdot \hbox {STAB}(h_i)$; by reversing the operations above we deduce that the converse also holds and the claim follows.

• (ii) Since $h_i$ is a perfect graph, $\hbox {STAB}(h_i)$ is described by clique inequalities and non-negativity constraints. The clique inequalities in $(1-t_i) \cdot \hbox {STAB}(h_i)$ read: $y'(u_i)+ y'({w^1_i}) + y'(w_i) \le 1-t_i, y'(w_i)+y'({w^1_i})+y'({w^2_i}) \le 1-t_i, y'({w_i})+y'({w^2_i})+y'({v_i}) \le 1-t_i$ and $y'({u_i}) + y'({v_i}) + y'(w_i) \le 1-t_i$. The non-negativity constraints read: $y'({u_i}) , y'({v_i}) , y'(w_i),y'({w^1_i}),y'({w^2_i}) \ge 0$. Substituting $y_i'$ by its expression in term of $x^*_i$ and $t_i$, performing simple calculations and using the bounds from Lemma 5, it follows that $y_i'\in (1-t_i) \cdot \hbox {STAB}(h_i)$ if and only if $\underline{s_i}\le t_i \le \overline{s_i}$, as required.

$\square$

We need one more technical claim, where we assume that G is the gluing of a strip $(G_i, \{A_i, B_i\})$ with a strip $(G',\{A',B'\})$ and show how to reduce the membership problem for $\hbox {STAB}(G)$ to the membership problems for two suitable sets that are derived from $\hbox {STAB}(G_i)$ and $\hbox {STAB}(G')$. (Note that the second strip is denoted differently since, in general, it will not be a strip from our original strip decomposition.) In the following, we let $\overline{G'}$ be the graph obtained from $G'$ by adding two non-adjacent vertices $z^1, z^2$ that are complete respectively to $A'$ and $B'$, and anti-complete to the other vertices of $G'$ (i.e., $G'$ is the gluing of $(G',\{A',B'\})$ with $(L,\{\{z^1\}, \{z^2\}\})$, where L is the graph with vertices $z^1$, $z^2$ and no edges). Note that $(\overline{G'}, \{\{z^1\},\{z^2\}\})$ is a 2-strip.

Claim 1

Let G be the gluing of $(G_i,\{A_i,B_i\})$ and $(G',\{A',B'\})$, and let $x^*\in \mathbb {R}^{V(G)}_+$. We extend $x^*[G']$ to vertices of $\overline{G'}$ by setting $x^*(z^1)=x^*(A_i)$ and $x^*(z^2)=x^*(B_i)$, and denote by $x^*[\overline{G'}]$ the resulting vector. Let $t_i$ such that $\underline{t_i}\le t_i \le \overline{t_i}$. Then $x^* \in \hbox {STAB}(G,t_i,A_i,B_i)$ if and only if $x^*[\overline{G'}] \in \hbox {STAB}(\overline{G'}, t_i, \{z^1\},\{z^2\})$.

Proof

Necessity. Let $\lambda \in \mathbb {R}^{\mathcal{S}(G)}_+$ such that $e\sum _{S \in \mathcal{S}(G)} \lambda _S=1$, $x^*=\sum _{S \in \mathcal{S}(G)} \lambda _S \chi ^S$ with $t_i=\sum _{S \in \mathcal{S}_{A_i,B_i}(G)} \lambda _S$ and $\underline{t_i}\le t_i \le \overline{t_i}$. We associate to each $S \in \mathcal{S}(G)$ a stable set $S[\overline{G'}]\in \mathcal{S}(\overline{G'})$ by first taking the restriction of S to $V(G')$, and then adding $z^1$ and/or $z^2$ whenever respectively $S\cap A_i\ne \emptyset$ and/or $S\cap B_i\ne \emptyset$. Then $x^*[\overline{G'}]=\sum _{S \in \mathcal{S}(G)} \lambda _S \chi ^{S[\overline{G'}]}$, and therefore $x^*[\overline{G'}]\in \hbox {STAB}(\overline{G'}, t, \{z^1\},\{z^2\})$.

Sufficiency. $t_i\in [\underline{t_i},\overline{t_i}]$ implies that $x^*[G_i] \in \hbox {STAB}(G_i,t_i,A_i,B_i)$. In particular there exist multipliers $\{\lambda _S\}_{S \in \mathcal{S}(G_i)} \ge 0$ such that $\sum _{S \in \mathcal{S}(G_i)} \lambda _S=1$, $x^*[G_i]=\sum _{S \in \mathcal{S}(G_i)} \lambda _S \chi ^S$, and $\sum _{\mathcal{S}_{A_i,B_i}(G)} \lambda _S=t_i$. Now $x^*[\overline{G'}] \in \hbox {STAB}(\overline{G'}, t_i, \{z^1\},\{z^2\})$ implies that there exists multipliers $\{\mu _W\}_{W \in \mathcal{S}(\overline{G'})} \ge 0$ such that $\sum _{W \in \mathcal{S}(\overline{G'})} \mu _W=1$, $x^*[\overline{G'}]=\sum _{W \in \mathcal{S}(\overline{G'})} \mu _W \chi ^W$, and $\sum _{W \in \mathcal{S}(\overline{G'}) : z^1, z^2 \in W} \mu _W=t_i$. We want to show that $x^* \in \hbox {STAB}(G,t_i,A_i,B_i)$. It is easy to associate stable sets from $\mathcal{S}(G_i)$ with stable sets from $\mathcal{S}(\overline{G'})$ into stable sets of G as $x^*(z^1)=x^*(A_i)$ and $x^*(z^2)=x^*(B_i)$. Indeed, consider the union of the four complete bipartite networks with vertex set $\mathcal{S}_{a,b}(G_i) \cup \mathcal{S}_{c,d}(\bar{G}')$ for $a \in \{A_i,\lnot A_i\},b \in \{B_i,\lnot B_i\}$, $c=\{z^1\}$ if $a=A_i$ and $\lnot \{z^1\}$ otherwise, and $d=\{z^2\}$ if $b=B_i$ and $\lnot \{z^2\}$ otherwise. The offer in vertex $S \in \mathcal{S}(G_i)$ is $\lambda _S$ and the demand in vertex $W \in \mathcal{S}(\bar{G}')$ is $\mu _W$. Arcs have no capacities. It is easy to see that any feasible flow (i.e., a flow meeting offers and demands), and there is a least one, in this network will correspond to a certificate of feasibility for x, since we can interprete f(S, W) as the multiplier for stable set $S \cup W{\setminus } \{z^1,z^2\}$. The sum over all multipliers f(S, W) intersecting both $A_i$ and $B_i$ is by construction $t_i$. $\square$

Now express G as the gluing of $(G_1,A_1,B_1)$ and the appropriate strip $(G',A',B')$. Let $x^*\in \mathbb {R}^{V(G)}$ such that $x^*_i=x^*[G_i]\in STAB(G_i)$ for $i \in [k]$, and let $\underline{s}_1, \overline{s}_1$ such that $\underline{t}_1 \le \underline{s}_1 \le \overline{s}_1 \le \overline{t}_1$.

We next show that $x^* \in \hbox {STAB}(G, t^*,A_1,B_1)$ for some $t^* \in [\underline{s}_1,\overline{s}_1]$ if and only if $\widetilde{x}=\left( \begin{array}{c}x^*[G']\\ y_1\\ \end{array}\right) \in \hbox {STAB}(G'+gg_1)$, where $y_1$ is the restriction of $f_1(x^*_1,\underline{s}_1,\overline{s}_1)$ to $\mathbb {R}^{V(gg_i)}$. Then we can iterate the argument replacing at each step one strip $(G_i,\mathcal{A}_i)$ with the corresponding strip $(gg_i,\mathcal{A}^{gg}_i)$: we will finally obtain the graph H(G) and conclude that $\exists t$ such that $\underline{s}\le t \le \overline{s}$ and $x^*\in \hbox {STAB}(G, \{ t_i,A_i,B_i\})$ if and only if $y=f(x^*,\underline{s},\overline{s}) \in \hbox {STAB}(H(G))$. This proves the equivalence of (i) and (ii).

We observe first that we have (*) $\underline{s}_1=\min \{t^*: y_1\in \hbox {STAB}(gg_1,t^*,\{u_1\},\{v_1\})$ and $\overline{s}_1=\max \{t^*: y_1 \in \hbox {STAB}(gg_1,t^*,\{u_1\},\{v_1\})$ by Claim A. Then:

$$\begin{aligned} \begin{array}{cl} \exists t^* \in [\underline{s}_1, \overline{s}_1] : x^* \in \hbox {STAB}(G, t^*,A_1,B_1)\\ \Leftrightarrow &{} \hbox {(by Claim A)}\\ \exists t^* \in [\underline{s}_1, \overline{s}_1]: x^*[\overline{G'}]\in \hbox {STAB}(\overline{G'}, t^*, \{z^1\},\{z^2\}) \\ \Leftrightarrow &{} \hbox {(by Claim A, with } gg1\hbox { playing the role of } G_1\\ &{} \hbox { and because of (*))}\\ \exists t^* \in [\underline{s}_1, \overline{s}_1]: \widetilde{x} \in \hbox {STAB}(G'+gg_1,t^*,\{u_1\},\{v_1\}) \\ \Leftrightarrow &{} (\text{ because } \text{ of } \text{(*) }) \\ \widetilde{x} \in \hbox {STAB}(G'+gg_1) \end{array} \end{aligned}$$

Now we need to prove equivalence of (iii) with (i) and (ii). It’s enough to observe that we can apply equivalence between (i) and (ii), by choosing $\underline{s}=\overline{s}=t$, to derive equivalence between (i) and (iii).