Randomized Algorithms (Spring 2010)/Random sampling

Random sampling

Rejection 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 or simulation by sampling via random walks. Here we use the name for 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.

We now focus on how to construct a Markov chain that converges to a uniform stationary distribution, and leave the discussion of mixing time to the next section.

Consider the problem of sampling an independent set of a graph [math]\displaystyle{ G(V,E) }[/math]. We consider an enormously large transition graph [math]\displaystyle{ \mathcal{G} }[/math] whose vertex set is the state space [math]\displaystyle{ \mathcal{I} }[/math]. That is, every vertex of [math]\displaystyle{ \mathcal{G} }[/math] is an independent set [math]\displaystyle{ I }[/math] of [math]\displaystyle{ G }[/math].

Irreducibility

Two independent sets [math]\displaystyle{ I_1,I_2\in\mathcal{I} }[/math] are adjacent to each other in the transition graph [math]\displaystyle{ \mathcal{G} }[/math], if they differ from each other in just one vertex, formally, if [math]\displaystyle{ I_1- I_2=\{v\} }[/math] or [math]\displaystyle{ I_2- I_1=\{v\} }[/math] for some vertex [math]\displaystyle{ v }[/math] of [math]\displaystyle{ G }[/math]. It is easy to see that the transition graph [math]\displaystyle{ \mathcal{G} }[/math] is connected, since any independent set is connected to [math]\displaystyle{ \empty\in\mathcal{I} }[/math] by a series of removals of vertices. The connectivity of the transition graph [math]\displaystyle{ \mathcal{G} }[/math] implies the irreducibility of the Markov chain.

Aperiodicity

The Markov chain is aperiodic if it has nonzero loop probabilities, that is, if for some state [math]\displaystyle{ I\in\mathcal{I} }[/math], the loop probability [math]\displaystyle{ P(I,I)\gt 0 }[/math]. Thus, the Markov chain is aperiodic if we make it lazy.

With irreducibility and aperiodicity, the chain converges to its stationary distribution. We now deal with the uniformity.

The Metropolis algorithm

We know that the random walk on a regular graph has uniform distribution as its stationary distribution. We will show in general, how to make the stationary distribution uniform if the transition graph is irregular. This method is called the Metropolis algorithm. (In fact, the Metropolis algorithm transforms a Markov chain to a Markov chain with any required stationary distribution. Here, we only deal with the case that our goal is the uniform distribution.)

Consider a random walk on an undirected graph [math]\displaystyle{ H }[/math], which is not necessarily regular. Let [math]\displaystyle{ d }[/math] be the maximum degree of [math]\displaystyle{ H }[/math].

The Metropolis algorithm Let [math]\displaystyle{ p\le\frac{1}{d} }[/math]. Consider a Markov chain with the transition matrix [math]\displaystyle{ P_{u,v}=\begin{cases} p & \mbox{if }u\neq v\mbox{ and }u\sim v,\\ 0 & \mbox{if }u\neq v\mbox{ and }u\not\sim v,\\ 1-p\cdot d_u & \mbox{if }u=v. \end{cases} }[/math] The Markov chain is time-reversible, and the stationary distribution [math]\displaystyle{ \pi }[/math] is the uniform distribution.

Proof: Because the graph [math]\displaystyle{ H }[/math] is undirected, [math]\displaystyle{ P }[/math] is symmetric. It follows that the uniform distribution is stationary. And [math]\displaystyle{ \pi_uP_{u,v}=\pi_vP_{v,u} }[/math], thus the Markov chain is time-reversible.

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

The only magic that the Metropolis algorithm plays is having every edge has the same probability weight, therefore, in an undirected graph, the transition matrix is symmetric, thus the chain is time-reversible and the stationary distribution is uniform.

Back to our example of sampling an independent set of a fixed graph [math]\displaystyle{ G(V,E) }[/math]. The transition graph [math]\displaystyle{ \mathcal{G} }[/math] is defined on the space [math]\displaystyle{ \mathcal{I} }[/math] of all independent sets of [math]\displaystyle{ G }[/math], and two independent sets are adjacent in [math]\displaystyle{ \mathcal{I} }[/math] if they differ from each other in exact one vertex. This gives us a large undirected graph [math]\displaystyle{ \mathcal{G} }[/math] whose vertices are independent sets of [math]\displaystyle{ G }[/math]. We consider the following random walk on [math]\displaystyle{ \mathcal{G} }[/math].

Given as the input an undirected graph [math]\displaystyle{ G(V,E) }[/math]: [math]\displaystyle{ I_0=\emptyset\, }[/math]. 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].

For any independent sets [math]\displaystyle{ I_1,I_2 }[/math] of [math]\displaystyle{ G(V,E) }[/math], if [math]\displaystyle{ |I_1\triangle I_2|\gt 1 }[/math], where [math]\displaystyle{ \triangle }[/math] is the symmetric difference, then the transition probability is [math]\displaystyle{ P(I_1,I_2)=0 }[/math]; if [math]\displaystyle{ I_1\triangle I_2=\{v\} }[/math] for some [math]\displaystyle{ v\in V }[/math], then with probability [math]\displaystyle{ 1/|V| }[/math], [math]\displaystyle{ v }[/math] is chosen and the transition from [math]\displaystyle{ I_1 }[/math] to [math]\displaystyle{ I_2 }[/math] occurs, thus [math]\displaystyle{ P(I_1,I_2)=1/|V| }[/math]. This corresponds to a random walk over [math]\displaystyle{ \mathcal{G} }[/math] that each undirected edge of [math]\displaystyle{ \mathcal{G} }[/math] has a probability weight of [math]\displaystyle{ 1/|V| }[/math].

Due to the Metropolis algorithm, the Markov chain is time-reversible and has uniform distribution as its stationary distribution. We know this chain is irreducible (because [math]\displaystyle{ \mathcal{G} }[/math]is connected) and it is easy to see it is aperiodic (because it has nonzero self-loop probabilities). Therefore, running the chain for a sufficiently long time, the distribution will be close to the uniform distribution over all independent sets of [math]\displaystyle{ G }[/math].

We still don't know how long we have to run the chain. This is determine be its mixing time. We will discuss how to analyze the mixing time for general random walks.

Conductance

Spectral gap and the mixing time

• We consider a Markov chain with finite space [math]\displaystyle{ \mathcal{S} }[/math], transition matrix [math]\displaystyle{ P }[/math], and stationary distribution [math]\displaystyle{ \pi }[/math]. Let [math]\displaystyle{ N=|\mathcal{S}| }[/math] denote the size of the state space, and let [math]\displaystyle{ \lambda_1\ge\cdots\ge\lambda_N }[/math] be the eigenvalues of [math]\displaystyle{ P }[/math]. For any stochastic matrix [math]\displaystyle{ P }[/math], it holds that [math]\displaystyle{ 1=\lambda_1\ge\cdots\ge\lambda_N\ge -1 }[/math].
• The mixing time [math]\displaystyle{ \tau(\epsilon)\, }[/math] of an irreducible and aperiodic Markov chain is given by the time to be within total variation distance [math]\displaystyle{ \epsilon }[/math] from [math]\displaystyle{ \pi }[/math], starting from a worst-case state. Formally,

[math]\displaystyle{ \tau_i(\epsilon)=\min\left\{t\,\,\bigg|\,\, \frac{1}{2}\sum_{j\in\mathcal{S}}|P^t(i,j)-\pi(i)|\le\epsilon\right\} }[/math] and [math]\displaystyle{ \tau(\epsilon)=\max_{i\in\mathcal{S}}\tau_i(\epsilon) }[/math].

Conditions:

• 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].

For a time-reversible Markov chain, the transition matrix [math]\displaystyle{ P }[/math] can be transformed to a symmetric matrix, thus the orthogonal diagonalization for symmetric matrices can be applied, and the convergence rate is determined by the second largest absolute value of eigenvalues [math]\displaystyle{ \lambda_{\max}=\max(|\lambda_2|,|\lambda_N|) }[/math] as

[math]\displaystyle{ \tau_i(\epsilon)=O\left(\frac{\ln(1/\pi_i)+\ln(1/\epsilon)}{1-\lambda_{\max}}\right) }[/math].

• Lazy random walk: [math]\displaystyle{ P_{i,i}\ge\frac{1}{2} }[/math] for any [math]\displaystyle{ i\in\mathcal{S} }[/math].

A lazy walk is aperiodic. The eigenvalues of a lazy walk are nonnegative, and thus [math]\displaystyle{ \lambda_{\max}=\max(|\lambda_2|,|\lambda_N|)=\lambda_2 }[/math]. The mixing time is determined by the second largest eigenvalue [math]\displaystyle{ \lambda_2 }[/math] as

[math]\displaystyle{ \tau_i(\epsilon)=O\left(\frac{\ln(1/\pi_i)+\ln(1/\epsilon)}{1-\lambda_{2}}\right) }[/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 of a time-reversible lazy random walk with uniform stationary distribution over a state space of size [math]\displaystyle{ N }[/math], is [math]\displaystyle{ \tau(\epsilon)=O\left(\frac{\ln N+\ln(1/\epsilon)}{1-\lambda_2}\right) }[/math].

From the previous section, we see how to construct a time-reversible lazy random walk with uniform stationary distribution. The mixing time is now determined by the spectral gap [math]\displaystyle{ (1-\lambda_2) }[/math]. In order to upper bound the mixing time, we need to lower bound the spectral gap. However, it is difficult to directly bound the eigenvalues of a usually exponentially large transition matrix.

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

The following lemma 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 [math]\displaystyle{ \Phi^2/2 }[/math] is a lower bound of the spectral gap, which in turn gives an upper bound of the mixing time via conductance.

Proposition The mixing time of a time-reversible lazy random walk with uniform stationary distribution over a state space of size [math]\displaystyle{ N }[/math], is [math]\displaystyle{ \tau(\epsilon)=O\left(\frac{\ln N+\ln(1/\epsilon)}{\Phi^2}\right) }[/math].

So analyzing of mixing time is reduced to analyzing of the second largest eigenvalue, which is reduced to analyzing of the conductance. For some simple Markv chain, this can be done directly. For some more complicated Markov chain, this is done by a technique called the canonical paths argument.

Canonical paths

We will first show how to lower bound the expansion ratio of a graph by canonical paths argument, and then we will generalize this technique to the conductance.

As we know, the expansion ratio of an undirected graph [math]\displaystyle{ G(V,E) }[/math],

[math]\displaystyle{ \phi(G)=\min_{\overset{ S\subset V}{|S|\le\frac{|V|}{2}}}\frac{|\partial|}{|S|} }[/math],

is hard to compute, because its definition involves an optimization over exponentially many subsets. We will see that it can be lower bounded by computing the maximum congestion caused by a set of paths.

For every pair of vertices [math]\displaystyle{ x,y\in V }[/math], we construct a simple path [math]\displaystyle{ \gamma_{xy} }[/math] from [math]\displaystyle{ x }[/math] to [math]\displaystyle{ v }[/math]. Each ordered pair [math]\displaystyle{ (x,y) }[/math] of distinct vertices corresponds to exact one path, called the canonical path from [math]\displaystyle{ x }[/math] to [math]\displaystyle{ y }[/math]. Let [math]\displaystyle{ \Gamma=\{\gamma_{xy}\mid x,y\in V, x\neq y\} }[/math] be the collection of canonical paths.

The maximum congestion caused by these path is computed as

[math]\displaystyle{ \sigma=\max_{uv \in E}|\{\gamma_{xy}\in\Gamma\mid uv\in\gamma_{xy}\}|=\max_{uv \in E}\sum_{\gamma_{xy}\ni uv}1 }[/math].

Informally, the congestion of an edge is the number of canonical paths crossing the edge.

For any [math]\displaystyle{ S\subset V }[/math] where [math]\displaystyle{ |S|\le\frac{|V|}{2} }[/math], there are [math]\displaystyle{ |S||\bar{S}| }[/math] ordered pairs [math]\displaystyle{ (x,y) }[/math] of vertices such that [math]\displaystyle{ x\in S }[/math] and [math]\displaystyle{ y\not\in S }[/math], thus there are [math]\displaystyle{ |S||\bar{S}| }[/math] canonical paths going from [math]\displaystyle{ S }[/math] to [math]\displaystyle{ \bar{S} }[/math]. The average congestion of edges crossing from [math]\displaystyle{ S }[/math] and [math]\displaystyle{ \bar{S} }[/math] is [math]\displaystyle{ \frac{|S||\bar{S}|}{|\partial S|} }[/math], which must be no bigger than the maximum congestion. Thus,

[math]\displaystyle{ \sigma\ge\frac{|S||\bar{S}|}{|\partial S|} }[/math]

and

[math]\displaystyle{ \frac{|\partial S|}{|S|}\ge\frac{|\bar{S}|}{\sigma}\ge\frac{|V|}{2\sigma} }[/math].

This inequality holds for any [math]\displaystyle{ S\subset V }[/math] where [math]\displaystyle{ |S|\le\frac{|V|}{2} }[/math], by the definition of expansion ratio,

[math]\displaystyle{ \phi(G)\ge\frac{|V|}{2\sigma} }[/math]

Therefore, for any collection of canonical paths [math]\displaystyle{ \Gamma=\{\gamma_{xy}\} }[/math], the maximum congestion [math]\displaystyle{ \sigma\, }[/math] caused by [math]\displaystyle{ \Gamma }[/math] can be used to lower bound [math]\displaystyle{ \phi(G) }[/math]. By choosing cleverer paths, we can make the bound tighter.

The intuition behind this inequality is that large expansion ratio (the graph is well-connected) is a necessary condition for the existence of a pairwise routing scheme with low-congestion.

---

We now show how to apply the canonical paths argument to the Markov chain and lower bound the conductance of the chain.

For a Markov chain with finite state space [math]\displaystyle{ \mathcal{S} }[/math], and transition matrix [math]\displaystyle{ P }[/math]. The chain corresponds to a transition graph [math]\displaystyle{ \mathcal{G}(\mathcal{S},E) }[/math] on the state space [math]\displaystyle{ \mathcal{S} }[/math]. For any states [math]\displaystyle{ u,v\in\mathcal{S} }[/math], [math]\displaystyle{ uv\in E }[/math] if [math]\displaystyle{ P_{uv}\gt 0 }[/math].

The canonical paths for the chain is a collection of simple paths [math]\displaystyle{ \Gamma=\{\gamma_{xy}\mid x,y\in \mathcal{S}, x\neq y\} }[/math] in the transition graph [math]\displaystyle{ \mathcal{G} }[/math], such that for any ordered pair [math]\displaystyle{ (x,y) }[/math] of distinct states, there is exact one path [math]\displaystyle{ \gamma_{xy} }[/math] going from [math]\displaystyle{ x }[/math] to [math]\displaystyle{ y }[/math] in [math]\displaystyle{ \mathcal{G} }[/math].

Let [math]\displaystyle{ \Gamma=\{\gamma_{xy}\} }[/math] be a collection of canonical paths. The congestion caused by [math]\displaystyle{ \Gamma }[/math] is computed as

[math]\displaystyle{ \rho=\max_{uv \in E}\frac{1}{\pi_u P_{uv}}\sum_{\gamma_{xy}\ni uv}\pi_x\pi_y }[/math].

Note that compared to the example of lower bounding the expansion ratio, in which we directly count the number of paths crossing an edge as the congestion, this time the congestion is defined by the probability "flows". This is analogous to the difference between the definitions of expansion ratio of a graph and the conductance of a Markov chain.

Lemma Let [math]\displaystyle{ \Gamma=\{\gamma_{xy}\} }[/math] be a collection of canonical path. The conductance is lower bounded by [math]\displaystyle{ \Phi\ge\frac{1}{2\rho} }[/math].

Proof:

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

Due to the upper bound of the mixing time via the conductance, we have the following proposition.

Proposition Consider a time-reversible lazy random walk with uniform stationary distribution over a state space of size [math]\displaystyle{ N }[/math]. Let [math]\displaystyle{ \Gamma=\{\gamma_{xy}\} }[/math] be an arbitrary collection of canonical paths. Assuming that the congestion of [math]\displaystyle{ \Gamma }[/math] is [math]\displaystyle{ \rho\, }[/math], the mixing time of the chain is bounded by [math]\displaystyle{ \tau(\epsilon)=O\left(\rho^2\left(\ln N+\ln(1/\epsilon)\right)\right) }[/math].

Example: rapid mixing of the random walk in cube

We then apply the canonical path argument on a simple Markov chain: the random walk on high dimensional cube.

Let the state space be [math]\displaystyle{ \mathcal{S}=[d]^n }[/math]. Consider the random walk:

• Start from an arbitrary [math]\displaystyle{ x\in \mathcal{S} }[/math]
• At each step, assuming the current state is [math]\displaystyle{ x\in \mathcal{S} }[/math], choose an random [math]\displaystyle{ i\in\{1,2,\ldots, n\} }[/math] and a random [math]\displaystyle{ b\in\{-1,1\} }[/math] and move to the state [math]\displaystyle{ y\in\mathcal{S} }[/math] such that [math]\displaystyle{ y_i=(x_i+b)\bmod d }[/math] and [math]\displaystyle{ y_j=x_j }[/math] for [math]\displaystyle{ j\neq i }[/math].

The Markov chain is a random walk on an [math]\displaystyle{ n }[/math]-dimensional [math]\displaystyle{ d }[/math]-ary cube. There are [math]\displaystyle{ d^n }[/math] vertices in the cube. We will show that the walk is rapid mixing by the canonical paths argument.

Consider the canonical paths [math]\displaystyle{ \Gamma=\{\gamma_{xy}\} }[/math] defined by the following "fixing" scheme:

• For each [math]\displaystyle{ x,y \in \mathcal{S} }[/math], [math]\displaystyle{ \gamma_{xy} }[/math] is the path obtained by fixing the 1st, 2nd, ... [math]\displaystyle{ n }[/math]th coordinate of [math]\displaystyle{ x }[/math], by increasing each coordinate by one (modula [math]\displaystyle{ d }[/math]) at a time. The path is described by the following pseudocode.

for(i=1; i<=d; i++){
    while(x[i]!=y[i]){
        x[i]=(x[i]+1)mod d;
    }
}

For any edge [math]\displaystyle{ uv\in E }[/math], i.e. such [math]\displaystyle{ u,v\in[d]^n }[/math] that differ at one coordinate by 1, let

[math]\displaystyle{ \mathrm{cong}(uv)=|\{\gamma_{xy}\in\Gamma\mid uv\in\gamma_{xy}\}| }[/math]

be the congestion of edge [math]\displaystyle{ uv }[/math], which is the number of canonical paths crossing [math]\displaystyle{ uv }[/math].

We claim that for any [math]\displaystyle{ uv\in E }[/math]

[math]\displaystyle{ \mathrm{cong}(uv)\le d^{n+1} }[/math].

To see this, consider an edge from [math]\displaystyle{ u=(u_1,u_2,\ldots,u_n)\, }[/math] to [math]\displaystyle{ v=(u_1,u_2,\ldots,u_{i-1},(u_i+1)\bmod d,u_{i+1},\ldots,u_n)\, }[/math]. The idea of the argument is that the [math]\displaystyle{ \gamma_{xy} }[/math] that cross the edge [math]\displaystyle{ uv }[/math] must satisfy:

• [math]\displaystyle{ y_1=u_1, \ldots,y_{i-1}=u_{i-1}\, }[/math] and
• [math]\displaystyle{ x_{i+1}=u_{i+1},\ldots, x_n=u_n\, }[/math],

because we are modifying the coordinates in order [math]\displaystyle{ 1,2,\ldots, n }[/math]. More precisely, corresponding to any [math]\displaystyle{ \gamma_{xy} }[/math] that cross the edge [math]\displaystyle{ uv }[/math], we may define a complementary point [math]\displaystyle{ w\in[d]^n }[/math] as

[math]\displaystyle{ w=(x_1,x_2,\ldots,x_i,y_{i+1},y_{i+2},\ldots,y_n)\, }[/math].

Then, for any fixed edge [math]\displaystyle{ uv }[/math], given [math]\displaystyle{ w }[/math] and [math]\displaystyle{ y_i }[/math], a pair of [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] whose canonical path cross the [math]\displaystyle{ uv }[/math], can be uniquely determined. In other words, a canonical path [math]\displaystyle{ \gamma_{xy}\in\{\gamma_{xy}\in\Gamma\mid uv\in\gamma_{xy}\} }[/math] can be encoded as a pair of [math]\displaystyle{ w\in[d]^n }[/math] and [math]\displaystyle{ y_i\in [d] }[/math]. Thus,

[math]\displaystyle{ \mathrm{cong}(uv)=|\{\gamma_{xy}\in\Gamma\mid uv\in\gamma_{xy}\}|\le|[d]^n\times [d]|=d^{n+1} }[/math].

Since the cube is a regular graph, the stationary distribution is uniform. The [math]\displaystyle{ \rho\, }[/math] can be easily computed by

[math]\displaystyle{ \rho=\max_{uv \in E}\frac{1}{\pi_u P_{uv}}\sum_{\gamma_{xy}\ni uv}\pi_x\pi_y=\max_{uv\in E}\frac{\mathrm{cong}(uv)}{d^n\cdot P_{uv}}\le 2nd }[/math].

Therefore, the mixing time is bounded by [math]\displaystyle{ O(n^2d^2\cdot \log (d^n))=O(n^3d^2\log d) }[/math], which is logarithmic to the size of the state space [math]\displaystyle{ |\mathcal{S}|=d^n }[/math].