Suppose we want to generate a uniformly random proper $q$-coloring of a graph $G=(V,E)$. In general, the set $\Omega$ of proper colorings can be exponentially large, and even determining whether $\Omega$ is nonempty is NP-complete. We cannot hope to enumerate $\Omega$ and pick an element at random. Instead, we turn to a local Markov chain: start at an arbitrary valid coloring, repeatedly pick a random vertex and recolor it according to a simple local rule, and run this process until it “forgets” its starting state.

This chapter develops two fundamental perspectives for proving that such a local chain reaches equilibrium quickly. The first is coupling, which tracks two random trajectories simultaneously and gives a concrete, intuitive proof of rapid mixing for graph colorings whenever $q > 2\Delta$. The second is spectral analysis, which studies the transition matrix as a self-adjoint operator on a weighted $L^2$ space; for reversible chains, this yields a clean spectral proof of the fundamental convergence theorem, and connects mixing rates to Dirichlet forms, Poincaré inequalities, and the geometry of bottlenecks via Cheeger’s inequality.

I. Course Content

1.1 From local moves to a sampler

Let $G=(V,E)$ be a finite graph with $n$ vertices and maximum degree $\Delta$. Let $\Omega$ be the set of proper $q$-colorings of $G$, where a coloring $\sigma:V\to[q]$ is proper if $\sigma(u)\ne\sigma(v)$ for all $\{u,v\}\in E$.

The simplest and most direct local random walk on $\Omega$ operates by a propose-and-validate rule—a Metropolis chain:

Given the current proper coloring $\sigma\in\Omega$: 1. Choose a vertex $v\in V$ uniformly at random (with probability $1/n$). 2. Choose a candidate color $c\in[q]$ uniformly at random (with probability $1/q$). 3. Acceptance check: If assigning color $c$ to vertex $v$ yields a valid proper coloring—that is, if $c\notin\{\sigma(u):u\in N(v)\}$—accept the proposal and update $\sigma(v)\leftarrow c$, leaving all other vertices unchanged. Otherwise, reject the proposal and stay at $\sigma$ ($\sigma'\leftarrow\sigma$).

Notice the operational simplicity of this rule: we do not need to compute or count the set of available colors at $v$; we simply draw a uniform pair $(v,c)$ from $V\times[q]$ and perform a purely local validity check against the immediate neighbors $N(v)$.

If the chosen color $c$ happens to match the vertex’s current color $\sigma(v)$, the move is always accepted and leaves the configuration unchanged. Furthermore, whenever $c$ conflicts with a neighbor, the proposal is rejected, which also leaves the configuration unchanged. Thus, the chain naturally possesses a positive probability of staying put (a self-loop), which immediately rules out periodicity issues.

Every step of this procedure requires only $O(\deg(v))=O(\Delta)$ time to execute. But to turn it into a provably sound sampling algorithm, we must resolve two fundamental questions:

1. Correctness: Does this simple propose-and-validate walk converge to the uniform distribution on $\Omega$, or does it subtly favor certain colorings over others?
2. Efficiency: How many steps must we run before the current state is provably close to uniform?

1.2 Discrete Markov chains and equilibrium

We consider discrete-time Markov chains on a finite state space $\Omega$.

Every row of $P$ is a probability distribution: $P(x,y)\ge0$ and $\sum_{y\in\Omega}P(x,y)=1$. If $\mu_t$ denotes the distribution of $X_t$ (represented as a row vector in $\mathbb R^\Omega$), the law of total probability gives $$\mu_{t+1}=\mu_tP, \qquad\text{and hence}\qquad \mu_t=\mu_0P^t.$$ The transition matrix $P$ and the initial distribution $\mu_0$ completely determine the law of the chain at any time $t$.

Three fundamental questions immediately arise: - Existence: Does a stationary distribution always exist? - Uniqueness: When is $\pi$ unique? - Convergence: Under what conditions does $\mu_t=\mu_0P^t$ converge to $\pi$ as $t\to\infty$ from any initial distribution $\mu_0$?

For finite state spaces, Perron–Frobenius ensures that a stationary distribution always exists. It is unique if the chain is irreducible (the transition graph is strongly connected, meaning every state can eventually reach every other state). Convergence from every initial state additionally requires aperiodicity (the chain does not become trapped in a deterministic cycle of length $>1$). In Section 1.12, we will give a complete spectral proof of this convergence for reversible chains, showing exactly how each of these conditions enters.

1.3 Time reversal and detailed balance

Suppose a chain is running at stationarity: $X_t\sim\pi$ for all $t$. The quantity $\pi(x)P(x,y)$ represents the probability that the chain is at $x$ at time $t$ and moves to $y$ at time $t+1$. We call $\pi(x)P(x,y)$ the probability flow from $x$ to $y$ at equilibrium.

If we observe this stationary process running backwards in time, Bayes’ rule gives the transition probability of the reversed process: $$\widehat P(y,x) =\mathbb P(X_t=x\mid X_{t+1}=y) =\frac{\mathbb P(X_t=x,X_{t+1}=y)}{\mathbb P(X_{t+1}=y)} =\frac{\pi(x)P(x,y)}{\pi(y)}.$$

Detailed balance states that at equilibrium, the flow from $x$ to $y$ exactly balances the reverse flow from $y$ to $x$ along every transition. Summing over all $x$ immediately implies that $\pi$ is stationary:

Stationarity only requires that the total flow into $y$ equals the total flow out of $y$ ($\sum_x \pi(x)P(x,y) = \pi(y)$). Detailed balance is a much stronger condition, but it is vastly easier to verify in practice because it is purely local.

A classic example is a simple random walk on a connected undirected graph $G=(V,E)$. If the walk chooses a neighbor of the current vertex uniformly at random, then $P(u,v)=1/\deg(u)$ for every edge $\{u,v\}\in E$. Setting $\pi(u)\propto\deg(u)$ satisfies detailed balance: $$\pi(u)P(u,v)\propto\deg(u)\cdot\frac{1}{\deg(u)}=1=\deg(v)\cdot\frac{1}{\deg(v)}\propto\pi(v)P(v,u).$$ On a regular graph, vertex degrees are equal, so the stationary distribution is uniform.

1.4 The Metropolis algorithm and the coloring chain

Detailed balance ($\pi(x)P(x,y)=\pi(y)P(y,x)$) guarantees that $\pi$ is stationary. Conversely, given a target distribution $\pi$ on a large state space $\Omega$, how can we systematically design a local transition matrix $P$ that satisfies detailed balance?

The Metropolis algorithm (Metropolis, Rosenbluth, Rosenbluth, Teller, and Teller, 1953) provides a general recipe. Suppose we start with a symmetric proposal mechanism $Q$ on $\Omega$: for any pair of states $x,y\in\Omega$, $$Q(x,y)=Q(y,x),$$ where $Q(x,y)$ is the probability of proposing a move to $y$ when currently at $x$. (For instance, $Q$ might choose a uniform neighbor in an undirected regular graph on $\Omega$, or pick a coordinate and a candidate value uniformly at random.)

From the current state $x\in\Omega$, one step of the Metropolis algorithm proceeds as follows: 1. Proposal: Draw a candidate state $y\in\Omega$ according to $Q(x,\cdot)$. 2. Acceptance: Accept the move to $y$ with probability $$\alpha(x,y)=\min\left\{1,\frac{\pi(y)}{\pi(x)}\right\}.$$ If accepted, set $X_{t+1}=y$; if rejected, remain at $X_{t+1}=x$.

For distinct states $x\ne y$, the transition probability is therefore $$P(x,y)=Q(x,y)\alpha(x,y)=Q(x,y)\min\left\{1,\frac{\pi(y)}{\pi(x)}\right\},$$ and the self-loop probability is $P(x,x)=1-\sum_{y\ne x}P(x,y)$.

Crucially, the Metropolis rule never requires the normalizing constant of $\pi$. If $\pi(x)=w(x)/Z$ where $Z=\sum_{z\in\Omega}w(z)$ is an intractable partition function, the acceptance ratio depends only on the ratio of unnormalized weights: $$\frac{\pi(y)}{\pi(x)}=\frac{w(y)/Z}{w(x)/Z}=\frac{w(y)}{w(x)}.$$

Justifying the coloring chain as a Metropolis instance

We can now verify that the propose-and-validate coloring walk from Section 1.1 is an instance of the Metropolis algorithm for the uniform distribution $$\pi(\sigma)=\frac{1}{|\Omega|} \qquad\text{for all }\sigma\in\Omega.$$ Here $\Omega$ is the set of proper $q$-colorings of $G$. An improper assignment $\tau\notin\Omega$ is not a state of the chain; when the algorithm proposes an invalid recoloring and rejects it, we fold that event into the self-proposal $Q(\sigma,\sigma)$ on $\Omega$.

At each step from $\sigma\in\Omega$, picking a vertex $v\in V$ (probability $1/n$) and a color $c\in[q]$ (probability $1/q$) induces a symmetric proposal kernel $Q$ on $\Omega$: - Any two distinct proper colorings $\sigma,\tau\in\Omega$ that differ at a single vertex $v$ have proposal probability $$Q(\sigma,\tau)=\frac{1}{n}\cdot\frac{1}{q}=\frac{1}{nq}=Q(\tau,\sigma),$$ corresponding to picking $(v,\tau(v))$ from $\sigma$ or $(v,\sigma(v))$ from $\tau$. - Proper colorings differing at two or more vertices have $Q(\sigma,\tau)=Q(\tau,\sigma)=0$.

Since $\pi$ is uniform on $\Omega$, every proposed move to a valid proper coloring $\tau=\sigma^{v\to c}\in\Omega$ has acceptance probability $$\alpha(\sigma,\tau)=\min\left\{1,\frac{\pi(\tau)}{\pi(\sigma)}\right\}=1,$$ and every invalid proposal $\sigma^{v\to c}\notin\Omega$ is rejected, leaving the chain at $\sigma$. For any two adjacent proper colorings $\sigma\ne\tau$, we therefore have $$P(\sigma,\tau)=Q(\sigma,\tau)=\frac{1}{nq}=P(\tau,\sigma),$$ so $P$ is symmetric on $\Omega$ and reversible with respect to the uniform distribution $\pi$.

Because proposing the current color $c=\sigma(v)$ always keeps the chain at $\sigma$, every state has a self-loop probability of at least $$P(\sigma,\sigma)\ge\frac{1}{n}\cdot\frac{1}{q}\cdot n=\frac{1}{q}>0,$$ which guarantees that the chain is aperiodic.

What about irreducibility when $q>2\Delta$? Given any two proper colorings $X,Y\in\Omega$, we can transform $X$ into $Y$ through at most $2n$ valid single-vertex recolorings in two sweeps across $V=\{v_1,\ldots,v_n\}$:

1. Sweep 1 (move to safe intermediate colors): Visit $v_1,\ldots,v_n$ in order. At $v_i$, choose a color $c_i\in[q]$ that avoids both the current colors of its neighbors $N(v_i)$ and their target colors $Y(N(v_i))$. Since $|N(v_i)|\le\Delta$, these two constraints forbid at most $2\Delta$ colors, so $q>2\Delta$ guarantees at least one valid choice $c_i$. After Sweep 1, the resulting proper coloring $Z$ satisfies $Z(u)\ne Y(v)$ for every edge $\{u,v\}\in E$.
2. Sweep 2 (move to the target coloring $Y$): Visit $v_1,\ldots,v_n$ again and recolor each $v_i$ to $Y(v_i)$. When $v_i$ is updated, each neighbor $u\in N(v_i)$ carries either its safe color $Z(u)$ from Sweep 1 (which avoids $Y(v_i)$) or its target color $Y(u)$ (which differs from $Y(v_i)$ because $Y$ is proper). Thus every intermediate step remains a proper coloring.

Since every coloring $Y$ is reachable from every coloring $X$, the chain is irreducible and therefore converges to the uniform distribution as $t\to\infty$. We now turn to the efficiency question: how many steps are needed until the coloring is nearly uniform?

1.5 Total variation distance and mixing time

To quantify how close the distribution $\mu_t$ is to stationarity $\pi$, we use total variation distance.

Operationally, $\max_{B}|\mu(B)-\nu(B)|$ is the largest difference in probability that any event $B$ can assign under $\mu$ versus $\nu$. If $d_{\mathrm{TV}}(\mu,\nu)\le0.01$, no statistical test based on a single sample can distinguish $\mu$ from $\nu$ with success probability better than $0.51$.

For a Markov chain with transition matrix $P$ and stationary distribution $\pi$, we measure convergence by the worst-case distance from stationarity over all starting states: $$d(t)=\max_{x\in\Omega}d_{\mathrm{TV}}(P^t(x,\cdot),\pi).$$

When $\varepsilon$ is omitted, one sets $\varepsilon=1/4$ by convention. The constant $1/4$ is convenient because $2\cdot\frac14=\frac12<1$: as shown below using coupling, $d(t)$ satisfies the submultiplicative bound $d(s+t)\le2d(s)d(t)$. Consequently, every additional block of $\tau_{\mathrm{mix}}(1/4)$ steps at least halves the distance to stationarity, and $\lceil\log_2(1/\varepsilon)\rceil\cdot\tau_{\mathrm{mix}}(1/4)$ steps suffice to reduce the error to any $\varepsilon>0$.

1.6 Coupling: make two chains meet

Computing $P^t(x,\cdot)$ directly requires taking powers of an $|\Omega|\times|\Omega|$ matrix, which is intractable when $|\Omega|$ is exponential in $n$. Coupling bounds total variation distance probabilistically without computing matrix powers.

A coupling of two probability distributions $\mu$ and $\nu$ on $\Omega$ is a pair of random variables $(X,Y)$ defined on a common probability space such that $X\sim\mu$ and $Y\sim\nu$.

Warm-up: Total variation distance never increases under a Markov step

Before analyzing concrete chains, let us ask a basic structural question: can a Markov step ever push two distributions further apart? Using the coupling lemma, we can see in a few lines that total variation distance is monotonically non-increasing under any Markov transition matrix $P$.

Iterating this inequality shows that the distance to stationarity never increases as time progresses: $$d_{\mathrm{TV}}(\mu_0 P^t,\pi)\le d_{\mathrm{TV}}(\mu_0 P^{t-1},\pi)\le\cdots\le d_{\mathrm{TV}}(\mu_0,\pi).$$

Motivating Markovian coupling: from non-increasing to coalescence

The proof of Proposition 3 highlights the two ingredients needed to bound mixing times via coupling: 1. Staying together is automatic: Once two trajectories meet ($X_t=Y_t$), feeding them identical random choices keeps them equal forever, ensuring $\{X_{t+1}\ne Y_{t+1}\}\subseteq\{X_t\ne Y_t\}$. 2. Mixing requires bringing distant states together: To make $\mathbb P(X_t\ne Y_t)$ decay to zero, we must replace the independent transitions on $\{X_t\ne Y_t\}$ with a correlated rule that actively drives $X_t$ and $Y_t$ toward coalescence.

Formally, a Markovian coupling is a joint process $(X_t,Y_t)_{t\ge0}$ whose marginals each evolve according to $P$, such that once $X_T=Y_T$ at the coalescence time $T=\inf\{t\ge0:X_t=Y_t\}$, the two chains remain equal for all $t\ge T$. Starting $X_0=x$ and $Y_0\sim\pi$ (so $Y_t\sim\pi$ for all $t$), Theorem 1 gives $$d_{\mathrm{TV}}(P^t(x,\cdot),\pi)\le\mathbb P(X_t\ne Y_t)=\mathbb P(T>t).$$ Bounding the mixing time thus reduces to bounding the tail of the coalescence time $T$.

1.7 First mixing bound: the hypercube walk

Let us see how a simple coupling drives two trajectories to coalesce on the Boolean hypercube.

Although $|\Omega|=2^n$ is exponential, the walk mixes in $O(n\log n)$ steps because its coordinates do not interact: refreshing coordinate $i$ eliminates any disagreement at $i$ and never creates a disagreement elsewhere.

1.8 Rapid mixing of the coloring chain

We now return to the propose-and-validate Metropolis coloring chain from Section 1.4 and track the Hamming distance between two proper colorings $X,Y\in\Omega$: $$d(X,Y)=|\{u\in V:X(u)\ne Y(u)\}|.$$ Unlike the hypercube walk, adjacent vertices in a graph coloring interact. Suppose $X$ and $Y$ currently agree at $v$, and both chains select $v$ and propose the same color $c$. Color $c$ may be legal in $X$ (accepted) but blocked in $Y$ (rejected) because $v$ has a neighbor $w$ where $X(w)\ne Y(w)$. Thus disagreements can propagate across edges.

To prove rapid mixing, we need a one-step coupling under which disagreement healing at disagreeing vertices strictly outpaces disagreement creation at their neighbors.

Warm-up: One disagreement and why naive coupling only reaches $q>3\Delta$

Start with the simplest nontrivial case: $X$ and $Y$ differ at a single vertex $w$, with $X(w)=a$ and $Y(w)=b$ ($a\ne b$), so $d(X,Y)=1$.

Suppose both chains pick the same vertex $v\in V$ (probability $1/n$ each) and also naively propose the same color $c\in[q]$:

1. Healing at $v=w$ (probability $1/n$): Because $X$ and $Y$ agree on $N(w)$, they share the same available color set $A(w)=[q]\setminus X(N(w))$ of size $|A(w)|\ge q-\Delta$.
   • If $c\in A(w)$ (probability $|A(w)|/q\ge(q-\Delta)/q$), both chains accept $c$ and coalesce ($d(X_1,Y_1)=0$).
   • Otherwise, both reject $c$ and $d(X_1,Y_1)=1$.
2. Creation at a neighbor $v\in N(w)$ (at most $\Delta$ vertices, probability $1/n$ each): Here $X(v)=Y(v)$, and all neighbors of $v$ other than $w$ have identical colors in $X$ and $Y$.
   • If $c\notin\{a,b\}$, color $c$ has the same validity in $X$ and $Y$: both chains either accept $c$ or reject $c$, keeping $X_1(v)=Y_1(v)$.
   • If $c=a$, color $a$ is blocked by $X(w)=a$ in $X$ (rejected) but may be accepted in $Y$.
   • Symmetrically, if $c=b$, color $b$ is blocked by $Y(w)=b$ in $Y$ (rejected) but may be accepted in $X$.

The key idea: Matching rejected proposals cuts creation in half

Why did the naive coupling lose $2/q$ at each neighbor $v\in N(w)$? Because $X$ rejected $c=a$ on one draw while $Y$ rejected $c=b$ on a different draw, even though both rejections produce the exact same outcome—staying at the shared color $X(v)=Y(v)$.

A valid coupling only requires each chain’s proposal to be marginally uniform on $[q]$. When updating $v\in N(w)$, couple the proposals via the transposition $\tau=(a\;\;b)$ that swaps $a$ and $b$ and fixes all other colors: draw $c\in[q]$ uniformly, let $X$ propose $c$, and let $Y$ propose $\tau(c)$. On the two exceptional draws $c\in\{a,b\}$:

• When $c=a$ (so $Y$ proposes $\tau(a)=b$): $X$ proposes $a$ (blocked by $X(w)=a$) and $Y$ proposes $b$ (blocked by $Y(w)=b$). Both proposals are rejected, so both chains stay at $X(v)=Y(v)$ and no disagreement is created.
• When $c=b$ (so $Y$ proposes $\tau(b)=a$): $X$ proposes $b$ while $Y$ proposes $a$. This single draw—occurring with probability $1/q$—is the only choice that can cause $X_1(v)\ne Y_1(v)$.

Pairing the two guaranteed rejections cuts disagreement creation at each neighbor from $2/q$ to $1/q$, improving the contraction threshold from $q>3\Delta$ to $$\frac{q-\Delta}{nq}>\frac{\Delta}{nq} \quad\Longleftrightarrow\quad q>2\Delta.$$

Coupling arbitrary colorings via updated-color distributions

How do we extend this rejection-matching idea to arbitrary proper colorings $X,Y\in\Omega$ at Hamming distance $k=d(X,Y)$, where a vertex $v$ may have several disagreeing neighbors at once?

Instead of designing a custom permutation of proposals at each vertex, look directly at the distribution of the color at $v$ after one Metropolis update, where all rejected proposals have already been merged into the single outcome of keeping the current color.

Fix a vertex $v\in V$, and let $$A_X(v)=[q]\setminus\{X(u):u\in N(v)\}$$ be the set of available colors at $v$ in $X$. Because $X$ is proper, $X(v)\in A_X(v)$. Conditioned on selecting $v$ at state $X$, the distribution $K_X^v$ of the updated color $X_1(v)$ is: