Sequential Monte Carlo for counting vertex covers in general graphs

Abstract

In this paper we describe a sequential importance sampling (SIS) procedure for counting the number of vertex covers in general graphs. The optimal SIS proposal distribution is the uniform over a suitably restricted set, but is not implementable. We will consider two proposal distributions as approximations to the optimal. Both proposals are based on randomization techniques. The first randomization is the classic probability model of random graphs, and in fact, the resulting SIS algorithm shows polynomial complexity for random graphs. The second randomization introduces a probabilistic relaxation technique that uses Dynamic Programming. The numerical experiments show that the resulting SIS algorithm enjoys excellent practical performance in comparison with existing methods. In particular the method is compared with cachet—an exact model counter, and the state of the art SampleSearch, which is based on Belief Networks and importance sampling.

Appendices

Appendix 1: Rasmusen Algorithm for counting the number of cliques

Appendix 2: Proofs

Lemma 7.1

Given a simple graph $G = G(V,E)$ with $|V|=n$, and let $\overline{G}=G(V,\overline{E})$ be its complement. Let $\varvec{x}\in \{0,1\}^n$ be a binary $n$-tuple such that its associated vertex set is a vertex cover in $G$; i.e., $V(\varvec{x})\in \fancyscript{G}^\mathrm{vc}$, and denote its complement by $\overline{\varvec{x}}$. Furthermore, $f(\cdot )$ and $g(\cdot )$ are PMF’s on binary $n$-tuples given by (16) and (17), respectively. Then $f(\varvec{x})=g(\overline{\varvec{x}})$.

Proof

Note that $V(\overline{\varvec{x}})$ is a clique in the complement graph $\overline{G}$.

First we show that $f(1)=g(0)$. This follows directly from their definitions:

$$\begin{aligned} f_1(1)=\frac{E_{k_{+}}}{E_{k_{+}}+E_{k_{-}}}, \end{aligned}$$

where $k_{+}=n-1$, and $k_{-}=d_1$ (degree of vertex $v_1$ in $G$), see Definition 2.1. And

$$\begin{aligned} g_1(0)=\frac{E_{m_{-}}}{E_{m_{-}}+E_{m_{+}}}, \end{aligned}$$

where $m_{-}=n-1$ and $m_{+}=d_1$, see Definition 3.2.

Consider the vertex $v_{i}$. The following cases are of interest.

• 1. Suppose that in graph $G$ there exists $v_j, j<i$, such that $x_j=0$ and $(v_j,v_i)\in E$. This means that $v_i$ is adjacent to a certain vertex that is not part of the vertex cover. Thus, $v_i$ must be chosen with probability 1; i.e., $f_i(1|x_1,\ldots ,x_{i-1})=1$. Then, in the complement graph $\overline{x}_j=1$ and $(v_j,v_i)\not \in \overline{E}$. This means $v_i$ is not adjacent to a certain vertex that is part of the clique. Thus, $v_i$ is not chosen with probability 1; i.e., $g_i(0|\overline{x}_1,\ldots , \overline{x}_{i-1})=1$.

• 2. Suppose that in graph $G$ all vertices $v_j, j<i$ that are part of the vertex cover, are adjacent to $v_i$. Mathematically:

  $$\begin{aligned} x_j=1\; (j\le i-1) \;\;\Rightarrow \;\; (v_j,v_i)\in E. \end{aligned}$$

  Thus $v_i$ may or may be not part of the vertex cover in $G$. Let

  $$\begin{aligned} A \!&=\! \{v_k | k\ge i+1; \,\exists \, j\le i-1 \; x_j=0\;\text {and}\; (v_j,v_k)\!\in \! E\},\\ B&= \{v_k | k\ge i+1; v_k\not \in A;\text {and}\; (v_i,v_k)\in E\}. \end{aligned}$$

  Then $v_i$ is chosen in the vertex cover with probability

  $$\begin{aligned} f_i(1|x_1,\ldots ,x_{i-1})=\frac{E_{k_{+}}}{E_{k_{+}}+E_{k_{-}}}, \end{aligned}$$

  with, according to Definition 2.1:

  $$\begin{aligned} k_{+}=n-i-|A|;\quad k_{-}=n-i-|A|-|B|. \end{aligned}$$

  In the complement graph $\overline{G}$ it holds that

  $$\begin{aligned} \overline{x}_j=0\; (j\le i-1) \;\;\Rightarrow \;\; (v_j,v_i)\not \in \overline{E}. \end{aligned}$$

  Thus $v_i$ may or may be not part of the clique in $\overline{G}$. Let

  $$\begin{aligned} \overline{A}&= \{v_k | k\ge i+1; \,\exists \, j\le i-1 \; \overline{x}_j=1 \;\text {and}\; (v_j,v_k)\not \in \overline{E}\},\\ \overline{B}&= \{v_k | k\ge i+1; v_k\not \in \overline{A} \;\text {and}\; (v_i,v_k)\in \overline{E}\}. \end{aligned}$$

  Then $v_i$ is not chosen in the clique with probability

  $$\begin{aligned} g_i(0|\overline{x}_1,\ldots ,\overline{x}_{i-1})= \frac{E_{m_{-}}}{E_{m_{-}}+E_{m_{+}}}, \end{aligned}$$

  with, according to Definition 3.2:

  $$\begin{aligned} m_{-}=n-i-|\overline{A}|;\quad m_{+}=|\overline{B}|. \end{aligned}$$

  We see immediately that the set $A$ and $\overline{A}$ are the same, and that $\overline{B}=\{v_{i+1},\ldots ,v_n\}{\setminus }(\overline{A}\cup B)$. In other words, $k_{+}=m_{-}$ and $k_{-}=m_{+}$. $\square$

Lemma 7.2

Given a simple graph $G = G(V,E)$ with $|V|=n$, and let $\overline{G}=G(V,\overline{E})$ be its complement. Let $\varvec{x}\in \{0,1\}^n$ be a binary $n$-tuple such that its associated vertex set is a clique in $G$; i.e., $V(\varvec{x})\in \fancyscript{G}^\mathrm{cl}$, and denote its complement by $\overline{\varvec{x}}$. Furthermore, $f(\cdot )$ and $g(\cdot )$ are PMF’s on binary $n$-tuples given by (16) and (17), respectively. Then $g(\varvec{x})=f(\overline{\varvec{x}})$.

The proof goes similarly as the proof of the previous Lemma, and therefore it is omitted.