Randomized Algorithms (Spring 2010)/Random sampling

Random sampling

The Markov Chain Monte Carlo (MCMC) method

The MCMC method provide a very general approach to near uniform sampling. The basic idea of the method is as follows:

• Define a Markov chain whose state space is the sample space, and whose stationary distribution is the uniform distribution.
• Start the chain from an arbitrary state, run the Markov chain for a sufficiently long time, and return the current state.

Usually, the name "MCMC" refers to a class of methods for numerical computation and simulation by sampling via random walks. Here we use the name to refer to the methods of sampling via random walks.

Consider the following problem:

Given an undirected graph [math]\displaystyle{ G(V,E) }[/math] on [math]\displaystyle{ n }[/math] vertices, uniformly sample an independent set of [math]\displaystyle{ G }[/math].

By the MCMC method, we consider a Markov chain whose state space [math]\displaystyle{ \mathcal{I} }[/math] is the space of all independent sets of [math]\displaystyle{ G }[/math]. Our sampling algorithm simply run the Markov chain for a sufficiently long time [math]\displaystyle{ T }[/math], and return the current independent set at time [math]\displaystyle{ T }[/math].

To guarantee that the returned independent set is nearly uniformly distributed, the Markov chain has to meet the following constraints:

• The chain converges (irreducible and aperiodic).
• The stationary distribution is uniform.

The running time of the algorithm is determined by: (1) the time complexity of a transition of a single step, and (2) the total number of steps [math]\displaystyle{ T }[/math]. Thus, to have an efficient algorithm, we have to also guarantee:

• The transition at each step of the chain is efficiently computable.
• The chain is rapid mixing.

Given as the input an undirected graph [math]\displaystyle{ G(V,E) }[/math]:

1. [math]\displaystyle{ I_0=\{v\}\, }[/math] for an arbitrary vertex [math]\displaystyle{ v\in V }[/math].
2. At step [math]\displaystyle{ t }[/math], assuming that the current independent set is [math]\displaystyle{ I_t }[/math], the [math]\displaystyle{ I_{t+1} }[/math] is computed:

• choose a vertex [math]\displaystyle{ v }[/math] uniformly at random from [math]\displaystyle{ V }[/math];
• if [math]\displaystyle{ v\in I_t }[/math] then [math]\displaystyle{ I_{t+1}=I_{t}\setminus \{v\}\, }[/math];
• if [math]\displaystyle{ v\not\in I_t }[/math] and [math]\displaystyle{ \forall u\in I_t, uv\not\in E }[/math], then [math]\displaystyle{ I_{t+1}=I_{t}\cup \{v\} }[/math];
• otherwise, [math]\displaystyle{ I_t=I_{t+1}\, }[/math].

Conductance

Recap

• A Markov chain with finite space [math]\displaystyle{ \mathcal{S} }[/math], where [math]\displaystyle{ N=|\mathcal{S}| }[/math], transition matrix [math]\displaystyle{ P }[/math], whose eigenvalues are [math]\displaystyle{ \lambda_1\ge\cdots\ge\lambda_N }[/math], stationary distribution [math]\displaystyle{ \pi }[/math].
• The mixing time [math]\displaystyle{ \tau(\epsilon)\, }[/math]: time to be close to [math]\displaystyle{ \pi }[/math] within total variation distance [math]\displaystyle{ \epsilon }[/math], starting from a worst-case state.

Conditions:

• Lazy random walk: [math]\displaystyle{ P_{i,i}\ge\frac{1}{2} }[/math] for any [math]\displaystyle{ i\in\mathcal{S} }[/math], so [math]\displaystyle{ 1=\lambda_1\ge\cdots\ge\lambda_N\ge 0 }[/math] and [math]\displaystyle{ \lambda_{\max}=\max(|\lambda_2|,|\lambda_N|)=\lambda_2\, }[/math].
• The Markov chain is time-reversible: [math]\displaystyle{ \pi_{i}P_{i,j}=\pi_{j}P_{j,i} }[/math] for all [math]\displaystyle{ i,j\in\mathcal{S} }[/math].
• The stationary [math]\displaystyle{ \pi }[/math] is the uniform distribution, that is, [math]\displaystyle{ \pi_i=\frac{1}{N} }[/math] for all [math]\displaystyle{ i\in\mathcal{S} }[/math].

Then:

Theorem The mixing time [math]\displaystyle{ \tau(\epsilon)=O\left(\frac{\ln N+\ln(1/\epsilon)}{1-\lambda_2}\right) }[/math]

Conductance and the mixing time

For many problems, such as card shuffling, the state space is exponentially large, so the estimation of [math]\displaystyle{ \lambda_2 }[/math] becomes very difficult. The following technique based on conductance overcomes this issue by considering the conductance of a Markov chain.

Definition (conductance) The conductance of a irreducible Markov chain with finite state space [math]\displaystyle{ \Omega }[/math], transition matrix [math]\displaystyle{ P }[/math], and stationary distribution [math]\displaystyle{ \pi }[/math], is defined as [math]\displaystyle{ \Phi=\min_{\overset{S\subset\Omega}{0\lt \pi(S)\le\frac{1}{2}}} \frac{\sum_{i\in S, j\not\in S}\pi_ip_{ij}}{\pi(S)} }[/math] where [math]\displaystyle{ \pi(S)=\sum_{i\in S}\pi_i }[/math] is the probability density of [math]\displaystyle{ S\subset \Omega }[/math] under the stationary distribution [math]\displaystyle{ \pi }[/math].

The definition of conductance looks quite similar to the expansion ratio of graphs. In fact, for the random walk on a undirected [math]\displaystyle{ d }[/math]-regular graph [math]\displaystyle{ G }[/math], there is a straightforward relation between the conductance [math]\displaystyle{ \Phi }[/math] of the walk and the expansion ratio [math]\displaystyle{ \phi(G) }[/math] of the graph

[math]\displaystyle{ \Phi=\frac{\phi(G)}{d}. }[/math]

Very informally, the conductance can be seen as the weighted normalized version of expansion ratio.

The following theorem states a Cheeger's inequality for the conductance.

Lemma (Jerrum-Sinclair 1988) In a time-reversible Markov chain, [math]\displaystyle{ 1-2\Phi\le\lambda_2\le 1-\frac{\Phi^2}{2} }[/math].

The inequality can be equivalent written for the spectral gap:

[math]\displaystyle{ \frac{\Phi^2}{2}\le1-\lambda_2\le 2\Phi }[/math]

thus a large conductance implies a large spectral gap, which in turn implies the rapid mixing of the random walk.

Proposition For a time-reversible Markov chain with finite state space [math]\displaystyle{ \mathcal{S} }[/math], transition matrix [math]\displaystyle{ P }[/math], and conductance [math]\displaystyle{ \Phi }[/math], if [math]\displaystyle{ P_{i,i}\ge\frac{1}{2} }[/math] for all state [math]\displaystyle{ i\in\mathcal{S} }[/math], and the stationary distribution [math]\displaystyle{ \pi }[/math] is uniform, then the mixing time [math]\displaystyle{ \tau(\epsilon)=O\left(\frac{\ln N+\ln(1/\epsilon)}{\Phi^{2}}\right) }[/math] where [math]\displaystyle{ N=|\mathcal{S}| }[/math].