Extended Formulations from Communication Protocols in Output-Efficient Time

Abstract

Deterministic protocols are well-known tools to obtain extended formulations, with many applications to polytopes arising in combinatorial optimization. Although constructive, those tools are not output-efficient, since the time needed to produce the extended formulation also depends on the number of rows of the slack matrix (hence, of the exact description in the original space). We give general sufficient conditions under which those tools can be implemented as to be output-efficient, showing applications to e.g. Yannakakis’ extended formulation for the stable set polytope of perfect graphs, for which, to the best of our knowledge, an efficient construction was previously not known. For specific classes of polytopes, we give also a direct, efficient construction of those extended formulations. Finally, we deal with extended formulations coming from certain unambiguous non-deterministic protocols.

Appendix

Application of Theorem 5 to (STAB(G), QSTAB(G)). We now sketch how to apply Theorem 5 to the protocol from Theorem 3 in order to obtain an extended formulation for (STAB(G), QSTAB(G)) in time $n^{O(\log (n))}$.

For a detailed description of the protocol from [30] and for full details of the proof, we refer to the journal version. Here we recall that, at the beginning of the protocol, Alice is given a clique C of G as input and Bob a stable set S, and they want to compute the entry of the slack matrix of $\mathrm {STAB}(G)$ corresponding to C, S, i.e. to establish whether C, S intersect or not. To do that, they exchange vertices of their respective sets. The number of vertices exchanged in total is at most $\lceil \log _2 n\rceil$. Hence, the protocol partitions the slack matrix of (STAB(G), QSTAB(G)) in a collection $\mathcal {R}$ of $n^{O(\log n)}$ monochromatic rectangles. Each rectangle $R\in \mathcal {R}$ is univocally identified by the list of vertices sent by Alice and by Bob during the execution of the protocol, which we denote by $C_R$ and $S_R$ respectively. Notice that $|C_R|+|S_R|\le \lceil \log _2 n\rceil$. For any clique C (resp. stable set) whose corresponding row (resp. column) is in R, we write $C\in R$ (resp. $S \in R$). If $C \in R$, then $C_R\subseteq C$ (in particular, $C_R$ is a clique), and similarly $S \in R$ implies $S_R \subseteq S$ (and $S_R$ is a stable set). We let $P_R$ be the convex hull of stable sets $S \in R$ and $Q_R$ the set of clique inequalities corresponding to cliques $C \in R$, together with the unit cube constraints.

To apply Theorem 5 we need to have a description of $\tau ,\ell$ and the extended formulations for $(P_R,Q_R)$, for each $R\in \mathcal {R}$. These are computed as follows.

• 1. $\tau$, $\ell$, and $(C_R,S_R)$ for all $R \in \mathcal{R}$. Enumerate all cliques and stable sets of G of combined size at most $\lceil \log _2 n\rceil$ and run the protocol on each of those input pairs. Each of those inputs gives a path in the tree (with the corresponding $\ell$), terminating in a leaf v, corresponding to a rectangle R. $\tau$ is given by the union of those paths, hence it has size $n^{O(\log n)}$. Moreover, the inclusion-wise minimal input pairs corresponding to the same rectangle R give $C_R,S_R$.

• 2. Extended formulations of $(P_R,Q_R)$, for each $R\in \mathcal {R}$. From the discussion in Sect. 2, it follows that an extended formulation of $(P_R,Q_R)$ is given by

  $$\begin{aligned} {T_R:=}\{x\in \mathbb {R}^d, y_R\in \mathbb {R}: x(C)+y_R= 1 \;\forall \; C\in R, \ y_R\ge 0, 0\le x \le 1 \}.\end{aligned}$$

  (3)

  We now claim that the equations in the description above, which can be exponentially many, are implied by a much smaller system (of size at most n):

  $$\begin{aligned}{T_R=}\{x\in \mathbb {R}^d, y_R\in \mathbb {R}: x(C_R)+y_R= 1\\ x(C_R+v)+y_R=1&\;\forall \; v\in V\setminus C_R: C_R+v\in R\\ y_R\ge 0, \; 0\le x \le 1 \}. \end{aligned}$$

  Let $C \in R$. We only need to show that $x(C)+y_R$ is a linear combination of the left-hand sides of the equations above. For any $v\in C\setminus C_R$, one can show that $C_R+v\in R$. Hence we obtain, as required,

  $$ \sum _{v\in C\setminus C_R} (x(C_R+v)+y_R)- (|C\setminus C_R|-1) (x(C_R)+y_R) = x(C) + y_R.$$

Proof of Theorem 12 (Sketch). The decomposition scheme outlined in the proof of Theorem 11 can be associated to a decomposition tree $\tau =\tau (G)$ on $n^{O(\log n)}$ nodes as follows: at each step, either decompose the current graph H using Lemma 9 (in which case each children is one of the $H_i$’s), or take the complement (in which case there is a single child, associated to $\bar{H}$). We will abuse notation and identify a node of the decomposition tree and the corresponding subgraph. Note, in particular, that this decomposition tree does not depend on the fact that G is perfect, hence it can be applied here as well.

We can assume that, for each leaf L of $\tau$, we are given an extended formulation $T_L$ of $(\mathrm {STAB}(L),\mathrm {QSTAB}(L))$. Consider the extended formulation, which we call $\eta (G)$, obtained by traversing $\tau$ bottom-up and applying the following:

• 1. if a non-leaf node H of $\tau$ has a single child $\bar{H}$, then define $\eta (H)$ to be equal to the extended formulation of $\{x\in \mathbb {R}^{V(H)}: x^Ty\le 1 \;\forall \; y \in \eta (\bar{H})\}$, obtained by applying Lemma 7.

• 2. otherwise if H has children $H_0,\dots ,H_k$, then we define $\eta (H)$ to be the extended formulation obtained from $\eta (H_0),\dots , \eta (H_k)$ using Observation 10.

We only need to prove that $\eta (G)$ is an extended formulation for $(\mathrm {STAB}(G),\mathrm {QSTAB}(G))$, as the efficiency aspects have been discussed in Theorems 5 and 11. We proceed by induction on the height of $\tau$, in particular we prove that for a node H of $\tau$, $\eta (H)$ is an extended formulation of $(\mathrm {STAB}(H),\mathrm {QSTAB}(H))$, assuming this is true for the children of H. If H is a leaf of $\tau$, then there is nothing to prove. Otherwise, we need to analyze two cases:

• 1. H has a single child, labelled $\bar{H}$. Assume that $\mathrm {STAB}(\bar{H})\subseteq \pi (\eta (\bar{H})) \subseteq \mathrm {QSTAB}(\bar{H})$, where $\pi$ is the projection on the appropriate space. Let S be a stable set in H, and $\chi ^S$ the corresponding incidence vector. For any $y\in \pi (\eta (\bar{H})) \subseteq \mathrm {QSTAB}(\bar{H})$, we have $y^T\chi ^S = y(S)\le 1$ as S is a clique in $\bar{H}$, hence since $\eta (H)$ (hence $\pi (\eta (H))$) is clearly convex it follows that $\mathrm {STAB}(H)\subseteq \pi (\eta (H))$. Now, for a clique C of H and $x\in \pi (\eta (H))$, one has $\chi ^C\in \pi (\eta (\bar{H}))$ hence $x(C)=x^T\chi ^C\le 1$, proving $\pi (\eta (H))\subseteq \mathrm {QSTAB}(H)$.

• 2. H has children $H_0,\dots ,H_k$. Assume that for $i=0,\dots ,k$, $\mathrm {STAB}(H_i)\subseteq \pi (\eta (H_i))\subseteq \mathrm {QSTAB}(H_i)$ for $\pi$ as above. Let $\chi ^S$ be the characteristic vector of a stable set S of H, for every $i=0,\dots ,k$ $S\cap V(H_i)$ is a stable set in $H_i$, hence $\chi ^{S}\in \pi (\eta (H_i))\times \mathbb {R}^{V(H)\setminus V(H_i)}$, and by convexity we conclude again that $\mathrm {STAB}(H)\subseteq \pi (\eta (H))$. Finally, let $x\in \pi (\eta (H))$, and let C be a clique of H. It follows from the way we decompose H, that C is contained in $H_i$ for some i: indeed, if $\{v_1,\dots ,v_k\}\cap C\ne \emptyset$, let i be the minimum such that $v_i\in C$, then $C\subseteq N^+(V_i)\setminus \{v_1,\dots ,v_{i-1}\}=V_i$ by definition; if $\{v_1,\dots ,v_k\}\cap C= \emptyset$, then $C\subseteq V_0$. By induction hypothesis, $\pi (\eta (H_i))\subseteq \mathrm {QSTAB}(H_i)$. But then $x(C)\le 1$, and since this holds for all the cliques of H, we have $\pi (\eta (H))\subseteq \mathrm {QSTAB}(H)$.   $\square$