Randomized Algorithms (Spring 2010)/Approximate counting, linear programming

Approximate Counting

Let us consider the following abstract problem.

Let [math]\displaystyle{ U }[/math] be a finite set of known size, and let [math]\displaystyle{ G\subseteq U }[/math]. We want to compute the size of [math]\displaystyle{ G }[/math], namely [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 \alpha}\ln\frac{2}{\delta} }[/math].

Proof: Use the Chernoff bound.

[math]\displaystyle{ \square }[/math]

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

• Implement the membership oracle [math]\displaystyle{ \mathcal{O} }[/math]. This is usually straightforward, or assumed by the model.
• 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.
• 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}\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.

Permanents and perfect matchings

Let [math]\displaystyle{ U=\{u_1,u_2,\ldots,u_n\} }[/math] and [math]\displaystyle{ V=\{v_1,v_2,\ldots,v_n\} }[/math]. Consider a bipartite graph [math]\displaystyle{ G(U,V,E) }[/math]. An [math]\displaystyle{ M\subseteq E }[/math] is a perfect matching of [math]\displaystyle{ G }[/math] if every vertex of [math]\displaystyle{ G }[/math] has exactly one edge in [math]\displaystyle{ M }[/math] adjacent to it.

Given a bipartite graph [math]\displaystyle{ G(U,V,E) }[/math] with [math]\displaystyle{ |U|=|V|=n }[/math], we want to count the number of perfect matchings of [math]\displaystyle{ G }[/math]. This problem can be reduced to computing the permanent of a square matrix.

Definition (permanent) Let [math]\displaystyle{ Q }[/math] be an [math]\displaystyle{ n\times n }[/math] matrix. The permanent of the matrix is defined as [math]\displaystyle{ \mathrm{per}(Q)=\sum_{\pi\in\mathbb{S}_n}\prod_{i=1}^n Q_{i,\pi(i)} }[/math], where [math]\displaystyle{ \mathbb{S}_n }[/math] is the symmetric group of permutation of size [math]\displaystyle{ n }[/math].

If we multiply each term of the sum the sign of the permutation, then it gives us the determinant of the matrix,

[math]\displaystyle{ \det(Q)=\sum_{\pi\in\mathbb{S}_n}\sgn(\pi)\prod_{i=1}^n Q_{i,\pi(i)} }[/math],

where [math]\displaystyle{ \sgn(\pi) }[/math], the sign of a permutation, is either [math]\displaystyle{ -1 }[/math] or [math]\displaystyle{ +1 }[/math], according to whether the minimum number of pair-wise interchanges to achieve [math]\displaystyle{ \pi }[/math] from [math]\displaystyle{ (1,2,\ldots,n) }[/math] is odd or even.

Unlike the determinants, which are computable in poly-time, permanents are hard to compute, as permanents can be used to count the number of perfect matchings in a bipartite graph, which is #P-complete.

A bipartite graph [math]\displaystyle{ G(U,V,E) }[/math] with [math]\displaystyle{ |U|=|V|=n }[/math] can be represented by an [math]\displaystyle{ n\times n }[/math] matrix [math]\displaystyle{ Q }[/math] with 0-1 entries as follows:

• Each row of [math]\displaystyle{ Q }[/math] corresponds to a vertex in [math]\displaystyle{ U }[/math] and each column of [math]\displaystyle{ Q }[/math] corresponds to a vertex in [math]\displaystyle{ V }[/math].

[math]\displaystyle{ Q_{ij}=\begin{cases} 1 & \mbox{if }i\sim j,\\ 0 & \mbox{otherwise}. \end{cases} }[/math]

Note the subtle difference between the definition of [math]\displaystyle{ Q }[/math] and the adjacency matrix.

Each perfect matching corresponds to a permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ U }[/math] with [math]\displaystyle{ (u,\pi(u))\in E }[/math] for every [math]\displaystyle{ u\in U }[/math], which corresponds to a permutation [math]\displaystyle{ \pi\in\mathbb{S}_n }[/math] with [math]\displaystyle{ \prod_{i=1}^n Q_{i,\pi(i)}=1 }[/math]. It is than easy to see that [math]\displaystyle{ \mathrm{per}(Q) }[/math] gives the number of perfect matchings in the bipartite graph [math]\displaystyle{ G }[/math].

It is known that counting the number of perfect matchings in a bipartite graph is #P-hard. Since this problem can be reduced to computing the permanent, thus the problem of computing the permanents is also #P-hard.

Now we show that with randomization, we can approximate the number of perfect matchings in a bipartite graph. In particular, we will give an FPRAS for counting the perfect matchings in a dense bipartite graph.

The Jerrum-Sinclair algorithm

Fix a bipartite graph [math]\displaystyle{ G(U,V,E) }[/math] with [math]\displaystyle{ |U|=|V|=n }[/math]. Let [math]\displaystyle{ \mathcal{M}_k }[/math] be the set of matchings of size [math]\displaystyle{ k }[/math] in [math]\displaystyle{ G }[/math], and let [math]\displaystyle{ m_k=|\mathcal{M}_k| }[/math]. Thus, [math]\displaystyle{ \mathcal{M}_k }[/math] is the set of perfect matchings in [math]\displaystyle{ G }[/math], and our goal is to compute [math]\displaystyle{ m_n }[/math].

Let [math]\displaystyle{ r_k=\frac{m_k}{m_{k-1}} }[/math] for [math]\displaystyle{ 1\lt k\le n }[/math]. Then

[math]\displaystyle{ m_k=m_{k-1}r_k }[/math],

which gives us a recursion to compute the [math]\displaystyle{ m_n }[/math], as

[math]\displaystyle{ m_n=m_{1}\frac{m_2}{m_1}\cdot\frac{m_3}{m_2}\cdots\frac{m_n}{m_{n-1}}=m_1\prod_{k=2}^n r_k }[/math],

where [math]\displaystyle{ m_1=|\mathcal{M}_1| }[/math] is the number of matchings of size 1 in the bipartite graph [math]\displaystyle{ G }[/math], which is just the number of edges in [math]\displaystyle{ G }[/math]. Therefore, [math]\displaystyle{ m_n }[/math] can be computed once we know [math]\displaystyle{ r_k }[/math] for [math]\displaystyle{ 1\lt k\le n }[/math].

Each [math]\displaystyle{ r_k }[/math] can be estimated by sampling uniformly from the set [math]\displaystyle{ \mathcal{M}_k\cup\mathcal{M}_{k-1} }[/math]. The algorithm for estimating [math]\displaystyle{ m_n }[/math] is outlined as:

1. For each [math]\displaystyle{ 1\lt k\le n }[/math], have an FPRAS for computing the [math]\displaystyle{ r_k }[/math] by uniform sampling sufficiently many members from [math]\displaystyle{ \mathcal{M}_k\cup\mathcal{M}_{k-1} }[/math] as

• uniformly sample [math]\displaystyle{ N }[/math] matching from [math]\displaystyle{ \mathcal{M}_k\cup\mathcal{M}_{k-1} }[/math], for some polynomially large [math]\displaystyle{ N }[/math];
• assuming that there are [math]\displaystyle{ X }[/math] sampled matchings of size [math]\displaystyle{ k }[/math], return [math]\displaystyle{ r_k }[/math] as [math]\displaystyle{ r_k=\frac{X}{N-X} }[/math].

2. Compute [math]\displaystyle{ m_n }[/math] as [math]\displaystyle{ m_n=m_1\prod_{k=2}^n r_k }[/math], where [math]\displaystyle{ m_1=|E| }[/math].

There are several issues that we have to deal with in order to have a fully functional FPRAS for counting perfect matchings.

• By taking the product of [math]\displaystyle{ r_k }[/math]'s, the errors for individual [math]\displaystyle{ r_k }[/math]'s add up.
• In order to accurately estimate [math]\displaystyle{ r_k }[/math] by sampling from [math]\displaystyle{ \mathcal{M}_k\cup\mathcal{M}_{k-1} }[/math], the ratio [math]\displaystyle{ r_k }[/math] should be within the range [math]\displaystyle{ \left[\frac{1}{\alpha},\alpha\right] }[/math] for some [math]\displaystyle{ \alpha }[/math] within polynomial of [math]\displaystyle{ n }[/math].
• Implement the uniform sampling from [math]\displaystyle{ \mathcal{M}_k\cup\mathcal{M}_{k-1} }[/math].

Estimator for each [math]\displaystyle{ r_k }[/math]

Accumulation of errors

Near-uniform sampling from [math]\displaystyle{ \mathcal{M}_k\cup\mathcal{M}_{k-1} }[/math]

In the last lecture, we have shown that by random walk, we can sample a near-uniform member of [math]\displaystyle{ \mathcal{M}_n\cup\mathcal{M}_{n-1} }[/math] in poly-time.

Volume estimation

We consider the problem of computing the volume of a given convex body [math]\displaystyle{ K }[/math] in [math]\displaystyle{ n }[/math] dimensions.

We use [math]\displaystyle{ \Upsilon(K)\, }[/math] to denote the volume of the convex body [math]\displaystyle{ K }[/math]. Abstractly, the problem is that given as input a convex body [math]\displaystyle{ K }[/math] in [math]\displaystyle{ n }[/math] dimensions, return the [math]\displaystyle{ \Upsilon(K)\, }[/math].

We should be more specific about the input model. Since we allow an arbitrary convex body as input, it is not even clear how to describe the body. We assume that [math]\displaystyle{ K }[/math] is described by means of a membership oracle [math]\displaystyle{ \mathcal{O} }[/math], such that for a query of an [math]\displaystyle{ n }[/math]-dimensional point [math]\displaystyle{ x }[/math], [math]\displaystyle{ \mathcal{O}(x) }[/math] indicates whether [math]\displaystyle{ x\in K }[/math].

For example, the [math]\displaystyle{ n }[/math]-dimensional convex body defined by the intersection of [math]\displaystyle{ m }[/math] half-spaces, which is the set of feasible solutions to a system of [math]\displaystyle{ m }[/math] linear constraints, can be described as

[math]\displaystyle{ A x\le \boldsymbol{b} }[/math],

for some [math]\displaystyle{ m\times n }[/math] matrix [math]\displaystyle{ A }[/math], and [math]\displaystyle{ m }[/math]-dimensional vector [math]\displaystyle{ b }[/math]. For a query of an [math]\displaystyle{ x }[/math], the membership oracle [math]\displaystyle{ \mathcal{O} }[/math] just check whether [math]\displaystyle{ Ax\le b }[/math].

For deterministic algorithms, there are negative news for this problem.

Theorem (Bárány-Füredi 1987) Suppose that a deterministic poly-time algorithm uses the membership oracle for a convex body [math]\displaystyle{ K }[/math] in [math]\displaystyle{ n }[/math] dimensions, and generates an upper bound [math]\displaystyle{ \Upsilon_u\, }[/math] and a lower bound [math]\displaystyle{ \Upsilon_\ell\, }[/math] on the volume [math]\displaystyle{ \Upsilon(K)\, }[/math] of [math]\displaystyle{ K }[/math]. Then, there is a convex body [math]\displaystyle{ K }[/math] and a constant [math]\displaystyle{ c }[/math] such that [math]\displaystyle{ \frac{\Upsilon_u}{\Upsilon_\ell}\ge c\left(\frac{n}{\log n}\right)^n }[/math].

That is said, with deterministic algorithms, we cannot even approximate the volume within a wildly loose range.

Dyer-Frieze-Kannan come up with an idea of estimating the volume of [math]\displaystyle{ K }[/math] by sampling near-uniformly from convex sets. They reduce the problem of computing approximately the volume of convex bodies to this sampling problem and thus give the first FPRAS for the volume of convex bodies.

For any convex body [math]\displaystyle{ K }[/math], it encloses some [math]\displaystyle{ n }[/math]-dimensional ball and is also enclosed by another ball with larger radius. We assume that the convex body [math]\displaystyle{ K }[/math] encloses the unit ball around the origin, and is enclosed by a larger ball round the origin with polynomially large radius. Formally, we assume that

[math]\displaystyle{ B(0,1)\subseteq K\subseteq B(0,n^c) }[/math] for some constant [math]\displaystyle{ c }[/math],

where [math]\displaystyle{ B(p,r) }[/math] denotes a ball of radius [math]\displaystyle{ r }[/math] with [math]\displaystyle{ p }[/math] as center, i.e. [math]\displaystyle{ B(p,r)=\{x\mid \|x-p\|\le r \} }[/math]. We can make this assumption because it is known that for any convex body [math]\displaystyle{ K }[/math], there exists a linear transformation [math]\displaystyle{ \tau }[/math] which can be found within poly-time such that [math]\displaystyle{ \tau K }[/math] transforms the [math]\displaystyle{ K }[/math] to a convex body that satisfies the assumption, and preserves the ratio of the volumes of the balls to the convex body.

The volumes of the balls are easy to compute. If only the ratio between the convex body and the ball which encloses it, is sufficiently large, then we can apply the Monte Carlo method to estimate [math]\displaystyle{ \Upsilon(K)\, }[/math] by uniformly sampling from the ball.

However, in high-dimension, the ratio between the volume of a convex body and the volume of ball which encloses it, can be exponentially small. This is caused by the so called the "curse of dimensionality".

Instead of having an outer ball and an inner ball, we define a sequence of balls:

[math]\displaystyle{ B_0=B(0,\lambda^0), B_1=B(0,\lambda^1), B_2=B(0,\lambda^2),\ldots, B_m=B(0,\lambda^m) }[/math],

where [math]\displaystyle{ \lambda=(1+\frac{1}{n}) }[/math] and [math]\displaystyle{ m }[/math] is the smallest positive integer such that [math]\displaystyle{ \lambda^m\ge n^c }[/math]. Therefore, the inner ball [math]\displaystyle{ B(0,1)=B_0 }[/math], the outer ball [math]\displaystyle{ B(0,n^c)\subseteq B_m }[/math], and [math]\displaystyle{ m }[/math] is within polynomial of [math]\displaystyle{ n }[/math]. In fact, [math]\displaystyle{ m\approx cn\ln n }[/math].

This sequence of balls naturally defines a sequence of convex bodies by intersections as [math]\displaystyle{ K_i=B_i\cap K }[/math]. It is obvious that

[math]\displaystyle{ B(0,1)=K_0\subseteq K_1\subseteq K_2\subseteq\cdots\subseteq K_m=K }[/math].

Balls are convex, and since the intersection of convex bodies is still convex, the sequence of [math]\displaystyle{ K_i }[/math] is a sequence of convex bodies.

We have the telescopic product:

[math]\displaystyle{ \frac{\Upsilon(K_0)}{\Upsilon(K_1)}\cdot\frac{\Upsilon(K_1)}{\Upsilon(K_2)}\cdots\frac{\Upsilon(K_{m-1})}{\Upsilon(K_m)}=\frac{\Upsilon(K_0)}{\Upsilon(K_m)}=\frac{\Upsilon(B(0,1))}{\Upsilon(K)}. }[/math]

Therefore, the volume [math]\displaystyle{ \Upsilon(K)\, }[/math] can be computed as

[math]\displaystyle{ \Upsilon(K)=\Upsilon(B(0,1))\cdot\prod_{i=1}^{m}\frac{\Upsilon(K_{i})}{\Upsilon(K_{i-1})} }[/math].

The volume of unite ball [math]\displaystyle{ \Upsilon(B(0,1))\, }[/math] can be precisely computed in poly-time. Each [math]\displaystyle{ \frac{\Upsilon(K_{i})}{\Upsilon(K_{i-1})} }[/math] is computed by near-uniform sampling from [math]\displaystyle{ K_{i} }[/math], which encloses [math]\displaystyle{ K_{i-1} }[/math].

Another observation is that the ratio [math]\displaystyle{ \frac{\Upsilon(K_{i})}{\Upsilon(K_{i-1})} }[/math] is well-bounded. Recall that [math]\displaystyle{ K_i=B(0,\lambda^i)\cap K }[/math] where [math]\displaystyle{ \lambda=(1+\frac{1}{n}) }[/math]. It can be proved that in [math]\displaystyle{ n }[/math] dimensions, the volume of [math]\displaystyle{ K_i }[/math] is at most [math]\displaystyle{ \lambda^n }[/math] times the [math]\displaystyle{ K_{i-1} }[/math], thus the ratio [math]\displaystyle{ \frac{\Upsilon(K_{i})}{\Upsilon(K_{i-1})} }[/math] is at most [math]\displaystyle{ \lambda^n=O(1) }[/math]. By the estimator theorem, we can have an FPRAS for the ratio [math]\displaystyle{ \frac{\Upsilon(K_{i})}{\Upsilon(K_{i-1})} }[/math] if we can uniformly sample from [math]\displaystyle{ K_i }[/math].

Uniformly sampling from an arbitrary convex body is replaced by near-uniform sampling achieved by random walks. In the original walk of Dyer-Frieze-Kannan, they consider the random walk over [math]\displaystyle{ n }[/math]-dimensional discrete grid points enclosed by [math]\displaystyle{ K }[/math], and prove that the walk is rapid mixing. This gives us the first FPRAS for volume estimation which runs in [math]\displaystyle{ \tilde{O}(n^{23}) }[/math], where [math]\displaystyle{ \tilde{O}(\cdot) }[/math] ignores the polylogarithmic factors.

The time bound was later improved by a series of works, each introducing some new ideas.

	complexity	new ingredient(s)
Dyer-Frieze-Kannan 1991	[math]\displaystyle{ n^{23} }[/math]	everything
Lovász-Simonovits 1990	[math]\displaystyle{ n^{16} }[/math]	localization lemma
Applegate-Kannan 1990	[math]\displaystyle{ n^{10} }[/math]	logconcave sampling
Lovász 1990	[math]\displaystyle{ n^{10} }[/math]	ball walk
Dyer-Frieze 1991	[math]\displaystyle{ n^8 }[/math]	better error analysis
Lovász-Simonovits 1993	[math]\displaystyle{ n^7 }[/math]	many improvements
Kannan-Lovász-Simonovits 1997	[math]\displaystyle{ n^5 }[/math]	isotropy, speedy walk
Lovász-Vempala 2003	[math]\displaystyle{ n^4 }[/math]	simulated annealing, hit-and-run

(cited from "Geometric Random Walks: A Survey" by Santosh Vempala.)

The current best upper bound is [math]\displaystyle{ \tilde{O}(n^4) }[/math] due to Lovász and Vempala in 2003. It is conjectured that the optimal bound is [math]\displaystyle{ \Theta(n^3) }[/math].

Linear Programming

Given a function [math]\displaystyle{ f:\mathbb{R}^n\rightarrow\mathbb{R} }[/math] and a set [math]\displaystyle{ \mathcal{F}\subseteq\mathbb{R}^n }[/math], an optimization problem is the problem of finding an [math]\displaystyle{ x\in\mathcal{F} }[/math] with the optimal (minimum) [math]\displaystyle{ f(x) }[/math]. Formally, the problem can be expressed as:

[math]\displaystyle{ \begin{align} \mbox{minimize} & \quad f(x)\\ \mbox{subject to} &\quad x\in\mathcal{F} \end{align} }[/math]

where [math]\displaystyle{ x\in\mathcal{R}^n }[/math] is a vector of [math]\displaystyle{ n }[/math] variables. For the problem of maximizing a function [math]\displaystyle{ f }[/math], we just minimizes [math]\displaystyle{ -f }[/math] instead.

We call the function [math]\displaystyle{ f }[/math] the objective function and call any [math]\displaystyle{ x\in\mathcal{F} }[/math] a feasible solution of the problem. A feasible solution that minimizes [math]\displaystyle{ f(x) }[/math] is called an optimal solution. Our task is to find an optimal solution.

The feasible set [math]\displaystyle{ \mathcal{F} }[/math] is usually given by a number of constraints [math]\displaystyle{ P_1,P_2,\ldots,P_m }[/math], which are predicates defined on [math]\displaystyle{ x }[/math]. An [math]\displaystyle{ x }[/math] is a feasible solution if it satisfies all the constraints.

A linear programming (LP) problem is an optimization problem with a linear objective function subject to a number of linear constraints. Formally, an LP is a problem that can be expressed in the following form:

[math]\displaystyle{ \begin{align} \mbox{minimize} & \quad c_1x_1+c_2x_2+\cdots+c_nx_n\\ \mbox{subject to} & \quad a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n\le b_1\\ & \quad a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n\le b_2\\ & \qquad\qquad \vdots\\ & \quad a_{m1}x_1+a_{m2}x_2+\cdots+a_{mn}x_n\le b_m\\ \end{align} }[/math]

where [math]\displaystyle{ a_{ij} }[/math], [math]\displaystyle{ b_i }[/math] and [math]\displaystyle{ c_j }[/math] are constants and [math]\displaystyle{ x_j }[/math] are variables. For maximization problem, or constraints given by "[math]\displaystyle{ \ge }[/math]", we can multiply the coefficients by [math]\displaystyle{ -1 }[/math] and still write the LP in the above form.

We can describe the programming in the vector form. Let [math]\displaystyle{ x }[/math] be a vector of [math]\displaystyle{ n }[/math] variables. Let [math]\displaystyle{ A=(a_{ij}) }[/math] be an [math]\displaystyle{ m\times n }[/math] matrix of constant entries, [math]\displaystyle{ b=(b_1,\ldots,b_m) }[/math] be a vector of [math]\displaystyle{ m }[/math] constant entries, and [math]\displaystyle{ c=(c_1,\ldots,c_n) }[/math] be a vector of [math]\displaystyle{ n }[/math] constant entries. Then an LP is expressed as:

[math]\displaystyle{ \begin{align} \mbox{minimize} & \quad c^T x\\ \mbox{subject to} & \quad Ax\le b \end{align} }[/math]

We call it the canonical form of linear programming.