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Here $\xi ⪯\zeta$ means each block of $\xi$ lies in a block of $\zeta$, and the measure order is stochastic domination of probability measures. The random-cluster single-edge conditional probability is increasing in both exterior open edges and wiring for $q\ge 1$. Couple two heat-bath Markov chains by identical update edges and uniform random variables, starting from ordered states. Order persists, and convergence of the finite chains to their stationary laws proves the displayed inequality. At $q=1$, the law does not depend on the boundary partition.

Let $m=|V|\ge 1$ and put $H(\sigma )={\sum }_{\{x,y\}\in E}{\sigma }_x{\sigma }_y$, counting each edge once. If only the spin at $x$ changes from $-1$ to $+1$, the change in $H$ is $2N(x)$, soThe given opposite transition probabilities have the same ratio. More explicitly,For pairs differing at more than one site both sides vanish, and for equal configurations balance is automatic. This proves detailed balance, hence the chain is reversible with stationary law $\pi$. All single-spin changes have positive probability, so the finite chain is irreducible; it also has positive holding probability and is aperiodic.

Its random-map version is a heat-bath Markov chain. Choose a vertex $I$ uniformly and a uniform $U\in [0,1]$, independently, and set its spin to $+1$ whenand to $-1$ otherwise, leaving all other spins unchanged. This also specifies the holding probabilities omitted from the off-diagonal formulas. The conditional Ising law at $I$ is exactly this Bernoulli choice.

Order spin configurations coordinatewise with $-1<+1$. Since both the neighbour sum and $h$ are increasing, every map defined by a fixed $(I,U)$ preserves this order. For monotone coupling from the past, generate one two-sided independent sequence of such maps $F_n$, $n\in \mathbb{Z}$, and, for a horizon $T$, apply $F_{-T},\dots ,F_{-1}$ to the all-minus and all-plus states at time $-T$. If their time-zero images agree, every initial state has that same image by monotonicity. Return it. Otherwise enlarge $T$, for example by doubling it, retaining all maps already generated at existing negative times. Any earlier starting configuration is also forced to the same output once a horizon coalesces, so the result is consistent under further backward extension.

To prove termination, let $D={\max }_x\mathrm{deg}(x)$ andList the vertices $v_1,\dots ,v_m$. A block of $m$ updates choosing them in that order and having every uniform less than $a$ sets every vertex to $+1$, regardless of the starting configuration. Each choice has probability $1/m$, and every uniform condition has probability $a$. Thus a block is a universal reset with probabilityIndependent disjoint blocks in the past giveOnce such a block occurs, later maps preserve agreement. Therefore the backward search terminates in finite time almost surely, including the doubling implementation.

Finally we prove that the common output $Y$ has law $\pi$. For a deterministic horizon $T$, independently start another chain at time $-T$ with initial law $\pi$ and use the same maps. Its time-zero state $Z_0^{(T)}$ has law $\pi$ by stationarity. On the event that this deterministic horizon coalesces, $Z_0^{(T)}=Y$, so for every event $B\subseteq \Sigma$,Letting deterministic $T\to \infty$ provesThis argument justifies exact sampling without treating a random, data-dependent horizon as a deterministic stationary run.

For a finite graph $G=(V,E)$, let $o(\omega )={\sum }_{e\in E}\omega (e)$, and let $k_G(\omega )$ count connected components of the spanning open subgraph, including isolated graph vertices. The random-cluster model with $p\in [0,1]$ and $q>0$ isThe partition function $Z_{G,p,q}$ is the sum of these weights over all configurations, making the expression a probability measure.

For the lattice box, take all nearest-neighbor edges with both endpoints in ${\Lambda }_n$. A random-cluster boundary condition $\xi$ is a partition of the boundary graph vertices. Vertices in one block are identified, or wired together, before counting components; the identifications do not add random edges. Let $k_{\xi }(\omega )$ be the number of components of the resulting quotient open graph. ThenThe free random-cluster boundary condition has singleton blocks, while the wired random-cluster boundary condition has one boundary block. Arbitrary partitions are allowed; no assumption that the partition itself has a planar realization is needed for the monotonicity statement.

Write $\xi ⪯\zeta$ if every block of $\xi$ is contained in a block of $\zeta$, so $\zeta$ makes at least as many identifications. The precise boundary monotonicity of the random-cluster measure isEquivalently, the expectation of every real-valued order-preserving function is larger under the more wired measure. This is stochastic domination of probability measures on the coordinatewise configuration order.

To prove it, condition on every edge except $e=\{u,v\}$. If $u,v$ are already connected using those open edges and the boundary wiring, opening $e$ does not change $k_{\xi }$, so its conditional open probability is $p$. If they are not connected, opening it reduces $k_{\xi }$ by one. The ratio of the open weight to the closed weight is then $p/[q(1-p)]$. Thus the random-cluster single-edge conditional probability isFor $q\ge 1$ the second number is at most the first. Adding open edges or making the boundary partition coarser can only turn a disconnected pair into a connected one. Therefore these conditional open probabilities are increasing in both the exterior configuration and the amount of wiring.

Run a heat-bath Markov chain for each boundary partition. At every step choose the same uniformly sampled edge in both chains, sample the same independent uniform variable $U$, and set that edge open if $U$ is below its conditional open probability. Start both chains at the all-closed configuration. By the preceding inequality, the two configurations remain ordered at every update. Each marginal chain has its corresponding random-cluster measure as its stationary distribution: resampling one coordinate from its conditional distribution preserves that law. Because $0<p<1$ and $q>0$, every update gives both possible states positive probability; the finite chain is an irreducible Markov chain with aperiodicity, and hence converges to its unique stationary distribution. Taking expectations of any order-preserving function and passing to the limit proves the claimed stochastic domination of probability measures. At $q=1$ the two conditional probabilities agree, and the boundary condition has no effect on the independent bond percolation law.

For the final identity fix a plane embedding of the finite planar graph, and include the unbounded face in its planar dual graph. In the dual configuration ${\bar{\omega }}^{\ast }$, a dual edge is open exactly when its crossed primal edge is closed. Let $m=o(\omega )$. The spanning open primal subgraph has $|V|$ graph vertices, $m$ edges, and $k(\omega )$ components. The Euler formula for a connected planar graph, applied componentwise with the common exterior face accounted for, gives its number of faces asDeleting a closed primal edge merges its incident original faces precisely when its dual edge joins distinct dual components. Consequently the faces of the open primal subgraph correspond exactly to components of the open complementary dual subgraph. This remains true for bridges and loops, with their dual loops and bridges, and for a disconnected primal graph. Thus the planar cluster-count identity isAt the self-dual parameter of the random-cluster model, put $s=\sqrt{q}$ andFor fixed $G$ the factor $(1-p_{\mathrm{sd}}{)}^{|E|}$ is independent of $\omega$. The weight is therefore proportional toAbsorbing the remaining graph-dependent factor into the normalization gives the symmetric cluster weight at the self-dual parameter:Counting the outer dual face and all isolated primal graph vertices is essential for the constant in the Euler identity. No assumption that $G$ is self-dual is required for this proportionality.

Use the single-edge heat-bath Markov chain: choose an edge uniformly and resample it from its conditional random-cluster law. If its endpoints are already connected without that edge, its conditional open probability is $p$; otherwise opening it merges two components and the probability isDetailed balance makes ${\varphi }_{p,2}$ stationary. Driving this chain and Bernoulli heat-bath chains by the same update edges and uniforms gives the stochastic domination