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The worst-case distance to stationarity isThe Path coupling theorem says that if a coupling contracts an integer-valued path metric by a factor $\alpha <1$ for every adjacent pair, then

Couple two configurations differing at one vertex $u$ by choosing the same update vertex and the same uniform random number for the heat-bath update. Updating $u$ removes the disagreement. Updating a nonneighbor of $u$ cannot create one. At a neighbor $v$, the two conditional plus-spin probabilities differ by at most $\tanh \beta$, using the supplied identity. Since $u$ has at most $\Delta$ neighbors, the expected Hamming distance after one step is at mostThe Hamming diameter is $n$, so the Path coupling theorem gives

Writing $a=|A\setminus B|$ and $b=|B\setminus A|$ givesMake two states adjacent when one can pair every disagreement except at most one, equivalently when $\rho =1$, and give every such edge length one. Pairing a deletion with an insertion as a swap and then handling the excess disagreements constructs a path of length $\max (a,b)$; every edge changes $\rho$ by at most one, so this is the corresponding path metric.

For adjacent states, couple the lazy coin and coordinate choices so that a distinguished disagreement is removed whenever its coordinate is selected, matching a compensating coordinate in the swap case. A direct check of the nested and equal-cardinality cases givesThe Path coupling theorem extends this to all pairs. Since $\mathrm{diam}(V)=k$, part (a), with $\alpha \ge 1/(2n)$ up to an absolute constant, gives

Give the state graph the unit-edge path metric. The construction in part (i) shows that its diameter is at most $2s$. It remains to couple one step from neighbouring states $X=A\cup \{x\}$ and $Y=A\cup \{y\}$.

Pair the choice $x$ in the first chain with $y$ in the second, and pair every $a\in A$ with itself; use the same proposed vertex $v$ in both chains. When the pair $(x,y)$ is removed, the chains coalesce whenever $v$ is not in the closed neighbourhood of $A$. This has probability at leastWhen a common $a\in A$ is removed, the distance can increase from one to at most two only if $v$ lies in one of the closed neighbourhoods of $x$ and $y$, an event of probability at mostThereforewhere the last inequality uses $n\ge 3s(\Delta +1)$. The Path coupling theorem extends this contraction to arbitrary starting distributions. Since $1-1/(sn)\le e^{-1/(sn)}$, part (a) with diameter at most $2s$ yields