随机算法 (Spring 2014)/The Monte Carlo Method

Parameter Estimation

Let us consider the following abstract problem of parameter estimation.

Let [math]\displaystyle{ U }[/math] be a finite set of known size, and let [math]\displaystyle{ G\subseteq U }[/math]. We want to estimate the parameter [math]\displaystyle{ |G| }[/math], i.e. the size of [math]\displaystyle{ G }[/math].

We assume two devices:

• A uniform sampler [math]\displaystyle{ \mathcal{U} }[/math], which uniformly and independently samples a member of [math]\displaystyle{ U }[/math] upon each calling.
• A membership oracle of [math]\displaystyle{ G }[/math], denoted [math]\displaystyle{ \mathcal{O} }[/math]. Given as the input an [math]\displaystyle{ x\in U }[/math], [math]\displaystyle{ \mathcal{O}(x) }[/math] indicates whether or not [math]\displaystyle{ x }[/math] is a member of [math]\displaystyle{ G }[/math].

Equipped by [math]\displaystyle{ \mathcal{U} }[/math] and [math]\displaystyle{ \mathcal{O} }[/math], we can have the following Monte Carlo algorithm:

• Choose [math]\displaystyle{ N }[/math] independent samples from [math]\displaystyle{ U }[/math] by the uniform sampler [math]\displaystyle{ \mathcal{U} }[/math], represented by the random variables [math]\displaystyle{ X_1,X_2,\ldots, X_N }[/math].
• Let [math]\displaystyle{ Y_i }[/math] be the indicator random variable defined as [math]\displaystyle{ Y_i=\mathcal{O}(X_i) }[/math], namely, [math]\displaystyle{ Y_i }[/math] indicates whether [math]\displaystyle{ X_i\in G }[/math].
• Define the estimator random variable

[math]\displaystyle{ Z=\frac{|U|}{N}\sum_{i=1}^N Y_i. }[/math]

It is easy to see that [math]\displaystyle{ \mathbf{E}[Z]=|G| }[/math] and we might hope that with high probability the value of [math]\displaystyle{ Z }[/math] is close to [math]\displaystyle{ |G| }[/math]. Formally, [math]\displaystyle{ Z }[/math] is called an [math]\displaystyle{ \epsilon }[/math]-approximation of [math]\displaystyle{ |G| }[/math] if

[math]\displaystyle{ (1-\epsilon)|G|\le Z\le (1+\epsilon)|G|. }[/math]

The following theorem states that the probabilistic accuracy of the estimation depends on the number of samples and the ratio between [math]\displaystyle{ |G| }[/math] and [math]\displaystyle{ |U| }[/math]

Theorem (estimator theorem)

Let [math]\displaystyle{ \alpha=\frac{|G|}{|U|} }[/math]. Then the Monte Carlo method yields an [math]\displaystyle{ \epsilon }[/math]-approximation to [math]\displaystyle{ |G| }[/math] with probability at least [math]\displaystyle{ 1-\delta }[/math] provided [math]\displaystyle{ N\ge\frac{4}{\epsilon^2 \alpha}\ln\frac{2}{\delta} }[/math].

Proof. Recall that [math]\displaystyle{ Y_i }[/math] indicates whether the [math]\displaystyle{ i }[/math]-th sample is in [math]\displaystyle{ G }[/math]. Let [math]\displaystyle{ Y=\sum_{i=1}^NY_i }[/math]. Then we have [math]\displaystyle{ Z=\frac{|U|}{N}Y }[/math], and hence the event [math]\displaystyle{ (1-\epsilon)|G|\le Z\le (1+\epsilon)|G| }[/math] is equivalent to that [math]\displaystyle{ (1-\epsilon)\frac{|G|}{|U|}N\le Y\le (1+\epsilon)\frac{|G|}{|U|}N }[/math]. Note that each [math]\displaystyle{ Y_i }[/math] is a Bernoulli trial that succeeds with probability [math]\displaystyle{ \frac{|G|}{|U|} }[/math], thus [math]\displaystyle{ \mathbb{E}[Y]=\frac{|G|}{|U|}N }[/math]. Then the rest is due to Chernoff bound. [math]\displaystyle{ \square }[/math]

A counting algorithm for the set [math]\displaystyle{ G }[/math] has to deal with the following three issues:

1. Implement the membership oracle [math]\displaystyle{ \mathcal{O} }[/math]. This is usually straightforward, or assumed by the model.
2. Implement the uniform sampler [math]\displaystyle{ \mathcal{U} }[/math]. As we have seen, this is usually approximated by random walks. How to design the random walk and bound its mixing rate is usually technical challenging, if possible at all.
3. Deal with exponentially small [math]\displaystyle{ \alpha=\frac{|G|}{|U|} }[/math]. This requires us to cleverly choose the universe [math]\displaystyle{ U }[/math]. Sometimes this needs some nontrivial ideas.

Counting DNFs

A disjunctive normal form (DNF) formular is a disjunction (OR) of clauses, where each clause is a conjunction (AND) of literals. For example:

[math]\displaystyle{ (x_1\wedge \overline{x_2}\wedge x_3)\vee(x_2\wedge x_4)\vee(\overline{x_1}\wedge x_3\wedge x_4) }[/math].

Note the difference from the conjunctive normal forms (CNF).

Given a DNF formular [math]\displaystyle{ \phi }[/math] as the input, the problem is to count the number of satisfying assignments of [math]\displaystyle{ \phi }[/math]. This problem is #P-complete.

Naively applying the Monte Carlo method will not give a good answer. Suppose that there are [math]\displaystyle{ n }[/math] variables. Let [math]\displaystyle{ U=\{\mathrm{true},\mathrm{false}\}^n }[/math] be the set of all truth assignments of the [math]\displaystyle{ n }[/math] variables. Let [math]\displaystyle{ G=\{x\in U\mid \phi(x)=\mathrm{true}\} }[/math] be the set of satisfying assignments for [math]\displaystyle{ \phi }[/math]. The straightforward use of Monte Carlo method samples [math]\displaystyle{ N }[/math] assignments from [math]\displaystyle{ U }[/math] and check how many of them satisfy [math]\displaystyle{ \phi }[/math]. This algorithm fails when [math]\displaystyle{ |G|/|U| }[/math] is exponentially small, namely, when exponentially small fraction of the assignments satisfy the input DNF formula.

The union of sets problem

We reformulate the DNF counting problem in a more abstract framework, called the union of sets problem.

Let [math]\displaystyle{ V }[/math] be a finite universe. We are given [math]\displaystyle{ m }[/math] subsets [math]\displaystyle{ H_1,H_2,\ldots,H_m\subseteq V }[/math]. The following assumptions hold:

• For all [math]\displaystyle{ i }[/math], [math]\displaystyle{ |H_i| }[/math] is computable in poly-time.
• It is possible to sample uniformly from each individual [math]\displaystyle{ H_i }[/math].
• For any [math]\displaystyle{ x\in V }[/math], it can be determined in poly-time whether [math]\displaystyle{ x\in H_i }[/math].

The goal is to compute the size of [math]\displaystyle{ H=\bigcup_{i=1}^m H_i }[/math].

DNF counting can be interpreted in this general framework as follows. Suppose that the DNF formula [math]\displaystyle{ \phi }[/math] is defined on [math]\displaystyle{ n }[/math] variables, and [math]\displaystyle{ \phi }[/math] contains [math]\displaystyle{ m }[/math] clauses [math]\displaystyle{ C_1,C_2,\ldots,C_m }[/math], where clause [math]\displaystyle{ C_i }[/math] has [math]\displaystyle{ k_i }[/math] literals. Without loss of generality, we assume that in each clause, each variable appears at most once.

• [math]\displaystyle{ V }[/math] is the set of all assignments.
• Each [math]\displaystyle{ H_i }[/math] is the set of satisfying assignments for the [math]\displaystyle{ i }[/math]-th clause [math]\displaystyle{ C_i }[/math] of the DNF formular [math]\displaystyle{ \phi }[/math]. Then the union of sets [math]\displaystyle{ H=\bigcup_i H_i }[/math] gives the set of satisfying assignments for [math]\displaystyle{ \phi }[/math].
• Each clause [math]\displaystyle{ C_i }[/math] is a conjunction (AND) of literals. It is not hard to see that [math]\displaystyle{ |H_i|=2^{n-k_i} }[/math], which is efficiently computable.
• Sampling from an [math]\displaystyle{ H_i }[/math] is simple: we just fix the assignments of the [math]\displaystyle{ k_i }[/math] literals of that clause, and sample uniformly and independently the rest [math]\displaystyle{ (n-k_i) }[/math] variable assignments.
• For each assignment [math]\displaystyle{ x }[/math], it is easy to check whether it satisfies a clause [math]\displaystyle{ C_i }[/math], thus it is easy to determine whether [math]\displaystyle{ x\in H_i }[/math].

The coverage algorithm

We now introduce the coverage algorithm for the union of sets problem.

Consider the multiset [math]\displaystyle{ U }[/math] defined by

[math]\displaystyle{ U=H_1\uplus H_2\uplus\cdots \uplus H_m }[/math],

where [math]\displaystyle{ \uplus }[/math] denotes the multiset union. It is more convenient to define [math]\displaystyle{ U }[/math] as the set

[math]\displaystyle{ U=\{(x,i)\mid x\in H_i\} }[/math].

For each [math]\displaystyle{ x\in H }[/math], there may be more than one instances of [math]\displaystyle{ (x,i)\in U }[/math]. We can choose a unique representative among the multiple instances [math]\displaystyle{ (x,i)\in U }[/math] for the same [math]\displaystyle{ x\in H }[/math], by choosing the [math]\displaystyle{ (x,i) }[/math] with the minimum [math]\displaystyle{ i }[/math], and form a set [math]\displaystyle{ G }[/math].

Formally, [math]\displaystyle{ G=\{(x,i)\in U\mid \forall (x,j)\in U, j\le i\} }[/math]. Every [math]\displaystyle{ x\in H }[/math] corresponds to a unique [math]\displaystyle{ (x,i)\in G }[/math] where [math]\displaystyle{ i }[/math] is the smallest among [math]\displaystyle{ x\in H_i }[/math].

It is obvious that [math]\displaystyle{ G\subseteq U }[/math] and

[math]\displaystyle{ |G|=|H| }[/math].

Therefore, estimation of [math]\displaystyle{ |H| }[/math] is reduced to estimation of [math]\displaystyle{ |G| }[/math] with [math]\displaystyle{ G\subseteq U }[/math]. Then [math]\displaystyle{ |G| }[/math] can have an [math]\displaystyle{ \epsilon }[/math]-approximation with probability [math]\displaystyle{ (1-\delta) }[/math] in poly-time, if we can uniformly sample from [math]\displaystyle{ U }[/math] and [math]\displaystyle{ |G|/|U| }[/math] is suitably small.

An uniform sample from [math]\displaystyle{ U }[/math] can be implemented as follows:

• generate an [math]\displaystyle{ i\in\{1,2,\ldots,m\} }[/math] with probability [math]\displaystyle{ \frac{|H_i|}{\sum_{i=1}^m|H_i|} }[/math];
• uniformly sample an [math]\displaystyle{ x\in H_i }[/math], and return [math]\displaystyle{ (x,i) }[/math].

It is easy to see that this gives a uniform member of [math]\displaystyle{ U }[/math]. The above sampling procedure is poly-time because each [math]\displaystyle{ |H_i| }[/math] can be computed in poly-time, and sampling uniformly from each [math]\displaystyle{ H_i }[/math] is poly-time.

We now only need to lower bound the ratio

[math]\displaystyle{ \alpha=\frac{|G|}{|U|} }[/math].

We claim that

[math]\displaystyle{ \alpha\ge\frac{1}{m} }[/math].

It is easy to see this, because each [math]\displaystyle{ x\in H }[/math] has at most [math]\displaystyle{ m }[/math] instances of [math]\displaystyle{ (x,i) }[/math] in [math]\displaystyle{ U }[/math], and we already know that [math]\displaystyle{ |G|=|H| }[/math].

Due to the estimator theorem, this needs [math]\displaystyle{ \frac{4m}{\epsilon^2}\ln\frac{2}{\delta} }[/math] uniform random samples from [math]\displaystyle{ U }[/math].

This gives the coverage algorithm for the abstract problem of the union of sets. The DNF counting is a special case of it.