\section{Proofs for Section~\ref{sec:amplification}}\label{app:amplification}

\subsection{Scales, the starting map, one step and the iteration}\label{app:amp-routine}

\begin{proof}[Proof of Lemma~\ref{lem:scales}]
Bounds~(a)--(c) are~\cite[Lemma~4.1]{OAINU} with site clusters, and its proof applies verbatim:
$s_o=0$ when $o$ is closed, which contributes nothing to $[\E s_o^\alpha]$ or to any bin
$2^j\le s_o<2^{j+1}$ with $j\ge0$, and the transport is uniform within each finite cluster in the bin.
For~(d), put $\pi_j=[\Pp_{p_c}(2^j\le s_o<2^{j+1})]$; at a record index $j$ of $2^{\beta j}\pi_j$,
$2^{\beta i}\pi_i<2^{\beta j}\pi_j$ for every $i<j$, and~(b) is the case $i=j-1$. Passing
to a subsequence of record indices, as in the type selection of~\cite[Lemma~4.1]{OAINU}, keeps this.
Then for $x\in O$, $\Pp(2^i\le s_x<2^{i+1})\le\pi_i/\sigma_O$, and
\[
\E[s_x^2;\ 1\le s_x<2n]\le\sum_{i\le j}4^{i+1}\frac{\pi_i}{\sigma_O}\le\frac4{\sigma_O}\,\pi
\sum_{i\le j}4^i2^{\beta(j-i)}\le C_O4^j\pi.\qedhere
\]
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lem:initial}]
From $x\in O$ sample a length-$\ell$ walk with kernel $P$, independently of percolation. Require its
endpoint to lie in $O$ and every distinct vertex visited by the walk, \emph{including $x$}, to be open.
If $e$ is the number of distinct visited vertices ($1\le e\le\ell+1$; holding steps add none), accept also
an independent coin of probability $p_c^{\ell+1-e}$. Each walk path then succeeds with probability
exactly $p_c^{\ell+1}$, so before any cap the endpoint mass kernel is $p_c^{\ell+1}P^\ell|_{O\times O}$.
The output lies in $C_x\cap O$, being joined to $x$ by an open walk. By Lemma~\ref{lem:walk} its row sum
is at least $c_*p_c^{\ell+1}$ for $\ell\ge l_0$, and its norm is at most $p_c^{\ell+1}\rho^\ell$. Fix
$\ell$ with $2\rho^\ell/c_*<e^{-H_0}$. Critical finiteness (Lemma~\ref{lem:crit-finite}) and
Proposition~\ref{prop:good} allow any sufficiently large $M$, rejecting unless $s_x\le M$ and $C_x$ is
good for cap $M$, at a loss of at most half the row sum, uniformly on $O$ by invariance. Entrywise
domination gives $r\le2\rho^\ell/c_*<e^{-H_0}$. The capped cluster and its goodness are determined by the
site bits in a deterministic finite ball (Proposition~\ref{prop:good}), and the walk visits only the
ball of radius $\ell$. Covariance holds because $P$ is defined from the graph.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lem:step}]
Put $b=C_0\cQ(k,M)u^kc_0^{-k}$ with a fixed large $C_0$. By Lemmas~\ref{lem:mean}
and~\ref{lem:second-moment}, $\E[L;\ L>b]\le\E L^2/b\le\pi u^kc_0^k/C_0\le b_0/4$ once $C_0\ge4/c$. On the
remaining fields discard also those whose root cluster is not good for cap $M'$; since $L$ is supported
on $s_x\le M'$ by~\eqref{eq:support}, this loses at most $be^{-(M')^\eps}$ by Proposition~\ref{prop:good}
at cap $M'$, which relative to $b_0\ge cc_0^k\pi u^k$ is at most
$C\cQ(k,M)c_0^{-2k}n^\beta e^{-(M')^\eps}\to0$, because $c_0^{-2k}=n^{O(1)}$ for $k=O(\log n)$ and
$M'\ge2n$. For large $n$ it is at most $b_0/4$. On a retained field output $z$ with probability
$L_x(z,\omega')/b$ and fail otherwise, which is legitimate as $L\le b$. The rule is local and covariant, and
every successful input has $z\in C_x(\omega')\cap O$, $s_x(\omega')\le M'$ and $C_x(\omega')$ good for cap $M'$. The
retained unnormalized kernel $F'$ satisfies $0\le F'\le F$ entrywise and has row sum $b_0'\ge b_0/2$.
Hence $u'=b_0'/b\ge cc_0^k\pi u^k/(2C_0\cQ u^kc_0^{-k})$ and $r'=\norm{F'/b_0'}\le2\norm F/b_0\le
2c_0^{-k}r^k$. The probabilities $L_x(z,\omega')/b$ are fixed numbers computed at $p_c$, so the rule is a fixed
kernel from the read site bits to outputs; it can be applied at another parameter unchanged, and its
success conditions still hold.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:efficient}]
We fix the parameters in the following order.
\begin{enumerate}[label=\textup{(\arabic*)},leftmargin=2.2em]
\item Choose $\xi>0$ with $c_\xi=(\beta+\xi+(\tau+\xi)(1+\xi))/(1-\xi)<1$, possible as $\beta+\tau<1$, as
in~\cite[Section~4]{OAINU}.
\item Put $H_*=\max\{1,2\log(1/c_0)/\xi\}$ and $T_*=4\log(1/c_0)/\xi$.
\item By Lemma~\ref{lem:initial}, start with $H>H_*$ and $M$ so large that $(\tau+\xi)\log M\ge T_*$, that
the estimate in~(4) giving $\log a\le t$ holds, and that Proposition~\ref{prop:good} applies at every cap
at least $M$.
\item For a fixed old map put $t=U+(\tau+\xi)m$ and $a=\lceil CM^\tau\log M/u\rceil$ with $C$ large enough that
$a\ge\max\{2,W/u\}$ (Lemma~\ref{lem:coding}). For large $M$, uniformly in $0<u\le1$,
$\log a\le U+\tau m+\log\log M+O(1)\le t$.
\item Only then choose $n\in\cN$ large and $k=\lfloor\log n/\log a\rfloor$. Then $k\to\infty$,
$k=O(\log n)$, $n\ge a^k\ge(W/u)^k$, $kM\le n/2$ and every goodness and truncation threshold hold
eventually, and $k\log a\le\log n<(k+1)t$.
\item For $n$ large and the old map fixed, Lemma~\ref{lem:step} gives
$U'\le\beta\log n+\log \cQ(k,M)+2k\log(1/c_0)+O(1)$. Since $t\ge(\tau+\xi)m\ge T_*$,
$2k\log(1/c_0)\le k\xi t/2$; also $\beta\log n<\beta kt+\beta t$ and $\log \cQ(k,M)=O_M(\log(k+1))$.
Hence $U'\le k(\beta+\xi)t$. Next $M'\le(5/2)n$ gives $m'\le k(1+\xi)t$. From $r'\le2(r/c_0)^k$,
$H'\ge k(H-\log(1/c_0))-\log2\ge k(1-\xi/2)H-\log2\ge k(1-\xi)H$, using $\log(1/c_0)\le\xi H/2$ (as
$H\ge H_*$) and $k\ge2\log2/(\xi H)$. Consequently, with $t'=U'+(\tau+\xi)m'$, $t'/H'\le c_\xi t/H$.
\item Choose $n$ larger if necessary so that $H'\ge2H$, which holds once $k(1-\xi)\ge2$. The invariants
persist: $H'\ge2H>H_*$ and $M'\ge M$, so $t'\ge(\tau+\xi)\log M'\ge T_*$ and~(4) applies to the new map.
\item Iterate with the same $O$ and $\cN$. Then $t/H\to0$ geometrically and $H\to\infty$. Since
$U+m\le\max\{1,(\tau+\xi)^{-1}\}t$, eventually $(U+m)/H<\eta$, with $M\ge2$ and $H>1$. Each map has a finite
radius.
\end{enumerate}
The maps are covariant under $\Aut(G)$, since every rule (the walk $P$, the switching rules, rejection
sets, masks, the load, truncation and the goodness test) is defined from $G$ and seeds whose joint law is
invariant. On success the output satisfies $y\in C_x\cap O$, $s_x\le M$ and goodness for every
configuration agreeing on the read ball, and the auxiliary randomness (the walk, a coin of fixed
probability $p_c^{\ell+1-e}$, or one uniform for a categorical draw with fixed probabilities
$L_x(z,\omega')/b$) does not depend on $p$.
\end{proof}

\subsection{The switching procedure, the inverse codes and the mean load}\label{app:amp-switch}

\begin{proof}[Proof of Lemma~\ref{lem:splicing}]
\emph{Existence of a selection.} Initially $C_1\ni y_0\in A$, so index $1$ is eligible with
$C_1\cap\Nb[A]\ne\varnothing$. After a selection at $i<k$ the new $S$ contains $y_i$ (Step~(5), and
$y_i\in Y_i$ in the standard case), and $y_i\in C_{i+1}$, so the eligible index $i+1$ meets $\Nb[S]$
while $z\notin S$. Hence the procedure runs until $z\in S$, the indices increase, and there are at most
$k$ steps.

\emph{Position of $y_i$.} Let $i$ be selected with current $S$. If $i<k$, the index $i+1$ is eligible
and larger, so by maximality $C_{i+1}\cap\Nb[S]=\varnothing$; as $y_i\in C_{i+1}$, $y_i\notin\Nb[S]$,
which is the standard case. If $i=k$, then $y_k=z\notin S$ by the stopping rule, so either
$z\notin\Nb[S]$ (standard) or $z\in\partial S$ (degenerate). In the degenerate case $z$ is $E_k$-open,
since success of the old step at $z$ gives $y_{k-1}\in C_z(E_k)$, which is nonempty only if $z$ is open.

\emph{$K_i\ge1$ in the standard case.} $C_i$ is connected, contains $y_i\notin\Nb[S]$ and meets
$\Nb[S]$. Along a path in $C_i$ from $y_i$ to $\Nb[S]$, the vertices before the first vertex
$u\in\Nb[S]$ lie in $C_i\setminus\Nb[S]$ and are connected to $y_i$, hence lie in $Y_i$; $u$ is adjacent
to a vertex outside $\Nb[S]$, so $u\notin S$, that is, $u\in\partial S$; and $u$ is $E_i$-open. So
$u\in W_i$ and $K_i\ge1$.

\emph{Induction hypothesis.} Before each step, $S=C_x(\text{current }\omega)$; in particular all of
$\partial S$ is closed in the current $\omega$. This holds for $S=A$. Each field $E_i$ is exchanged at most
once, at its own selection, so it is unmodified before index $i$ is selected.

\emph{One step.} Let the rejection test pass. After the exchange on $\mathrm{Mask}_i$:
\begin{itemize}
\item $S$ is unchanged and open, since $\mathrm{Mask}_i\cap S=\varnothing$ ($w_i\notin S$, and
$\Nb[Y_i]\setminus\Nb[S]$ misses $S$);
\item $w_i$ and $Y_i\subseteq\Nb[Y_i]\setminus\Nb[S]$ are open, being $E_i$-open;
\item $\partial Y_i\setminus\Nb[S]$ is closed: a vertex there is $E_i$-closed, since otherwise it would
be an $E_i$-open vertex of $V\setminus\Nb[S]$ adjacent to $Y_i$, hence in $Y_i$;
\item $\partial S\setminus\{w_i\}$ is closed: it was closed, and $\mathrm{Mask}_i\cap\Nb[S]=\{w_i\}$;
\item $R_i$ is closed, by the rejection test and $R_i\cap\mathrm{Mask}_i=\varnothing$.
\end{itemize}
The set $S\cup\{w_i\}\cup Y_i$ is connected ($w_i\in\partial S$, $w_i\sim b_i\in Y_i$ and $Y_i$ is
connected; in the degenerate case $w_k\in\partial S$). Its outer boundary is contained in
$(\partial S\setminus\{w_i\})\cup(\partial Y_i\setminus\Nb[S])\cup R_i$: a vertex of $\partial Y_i$ is not in
$S$, as $Y_i\cap\Nb[S]=\varnothing$, so it lies in $\partial S$ or in $\partial Y_i\setminus\Nb[S]$; and a
neighbour $v\notin S\cup Y_i\cup\{w_i\}$ of $w_i$ lies in $\partial S\setminus\{w_i\}$ if $v\in\Nb[S]$, in
$\partial Y_i\setminus\Nb[S]$ if $v\in\Nb[Y_i]\setminus\Nb[S]$, and in $R_i$ otherwise. All of these are
closed, so the new root cluster is exactly $S\cup\{w_i\}\cup Y_i$, which re-establishes the induction
hypothesis. No vertex of $Y_i$ is adjacent to $S$, so $S$ and $Y_i$ lie in different components after
deleting $w_i$.

\emph{Freshness of the rejection bits.} $R_i\subseteq V\setminus\Nb[S]$. Every earlier mask and
rejection set lies in $\Nb[S_{j+1}]\subseteq\Nb[S]$, because
$\mathrm{Mask}_{i_j}\subseteq\Nb[P_{i_j}]\subseteq\Nb[S_{j+1}]$ and
$R_{i_j}\subseteq N(w_{i_j})\subseteq\Nb[S_{j+1}]$; and the bin test reads only $\Nb[A]$ on success. So
the current $\omega$ on $R_i$ is the original $\omega$ there, and no $\omega$-bit of $V\setminus\Nb[S]$ has been read.

\emph{Chain structure.} Let $j<j'$. At the selection of $i_{j'-1}$, every larger eligible index, in
particular $i_{j'}$, has $C_{i_{j'}}\cap\Nb[S_{j'-1}]=\varnothing$. Since $P_{i_{j'}}\subseteq C_{i_{j'}}$
($w_{i_{j'}}\in W_{i_{j'}}\subseteq C_{i_{j'}}$, being $E$-open and adjacent to
$Y_{i_{j'}}\subseteq C_{i_{j'}}$; in the degenerate case $w=z\in C_k$) and $S_j\subseteq S_{j'-1}$, we get
$P_{i_{j'}}\cap\Nb[S_j]=\varnothing$. By definition $Y_{i_j}\cap\Nb[S_j]=\varnothing$. So in
$C'=S_j\cup P_{i_j}\cup\dots\cup P_{i_m}$ the only vertex outside $S_j$ adjacent to $S_j$ is $w_{i_j}$;
as $S_j$ is connected and contains $x$, it is the component of $x$ in $C'\setminus\{w_{i_j}\}$. For
$j<m$, $z\notin S_j$ (no stop) and $z\ne w_{i_j}$ ($z=y_k\in C_k$ with $k$ eligible at step $j$, so
$C_k\cap\Nb[S_j]=\varnothing$, while $w_{i_j}\in\partial S_j$), so $w_{i_j}$ separates $x$ from $z$. For
$j=m$ in the standard case, $z\in Y_{i_m}$, since the stop occurs when $z\in S$ and
$z\notin S_m\cup\{w_{i_m}\}$; again $w_{i_m}$ separates.

\emph{Size.} $|A|<2n$, and there are at most $k$ pieces, each inside some $C_i$, of size at most $M$.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lem:coding}]
\emph{Step 1 (order).} If $v\in \Hang(c)$, every $x$--$v$ path in $C'$ meets $c$. Hanging sets are nested
or disjoint: if $c'\in \Hang(c)$ then $\Hang(c')\subseteq \Hang(c)$, since an $x$--$v$ path meets $c'$ and its
initial segment to $c'$ meets $c$; and if neither $c'\in \Hang(c)$ nor $c\in \Hang(c')$, then $\Hang(c)\cap \Hang(c')=
\varnothing$, for if $v$ lay in both, a simple $x$--$v$ path would meet both, say $c$ first, its segment
from $c'$ to $v$ would avoid $c$, and if some $x$--$c'$ path avoided $c$ the concatenation would be an
$x$--$v$ walk avoiding $c$, so $c'\in \Hang(c)$, a contradiction. Every entry of $\Lambda$ has $z$ in its
hanging set, so the hanging sets of the entries of $\Lambda$ are strictly nested, in the order of
$\Lambda$: $\lambda_a\in \Hang(\lambda_b)$ if and only if $a\le b$.

\emph{Step 2 (the code).} By Lemma~\ref{lem:splicing}, $\Hang(w_{i_j})=C'\setminus S_j$. Hence
$w_{i_{j+1}}\in \Hang(w_{i_j})$, and the seams appear in $\Lambda$ in the order $w_{i_m},\dots,w_{i_1}$, at
positions $p_m<p_{m-1}<\dots<p_1$, with $p_m=0$ exactly in the degenerate case ($w_{i_m}=z=\lambda_0$;
in the standard case $w_{i_m}$ is a separating cut vertex, so $p_m\ge1$). The code is the index set
$\{i_1<\dots<i_m\}$ together with the gaps $l_m=p_m+1$ and $l_j=p_j-p_{j+1}$ for $j<m$, all at least
one.

\emph{Step 3 (at most one input per code).} From the code read off the seams; put $S_j=X(w_{i_j})$,
$S_{m+1}=C'$, $Y_{i_j}=S_{j+1}\setminus(S_j\cup\{w_{i_j}\})$ (Lemma~\ref{lem:splicing}), and
$\mathrm{Mask}_{i_j}=\{w_{i_j}\}\cup(\Nb[Y_{i_j}]\setminus\Nb[S_j])$. Undo the exchanges in reverse
chronological order, $j=m,\dots,1$, exchanging the current distinguished field with $E_{i_j}$ on
$\mathrm{Mask}_{i_j}$. The candidate input tuple is now fixed, and no $V_i$ has been used. Replay the
old rules from $z$ with the restored fields and seeds (successes, endpoints $y_i$, clusters $C_i$), the
bin test, $y_0\in A$ and the switching procedure, and discard the code if any test fails or the
procedure does not reproduce the given output and code. The degenerate or standard case is read off
from $p_m=0$.

\emph{Step 4 (gap bounds; goodness enters only here).} Consider a standard step $j$. Apply
Definition~\ref{def:good} to the old cluster $C=C_{i_j}$ with $A_{\mathrm{set}}=C_{i_j}\cap\Nb[S_j]$.
The component of $y_{i_j}$ in $C_{i_j}\setminus A_{\mathrm{set}}$ is $Y_{i_j}$, since an $E$-open path
from $y_{i_j}$ avoiding $\Nb[S_j]$ stays in the full cluster $C_{i_j}$. The neighbours of $Y_{i_j}$ in
$\Nb[S_j]$ lie in $\partial S_j$, as $Y_{i_j}\cap\Nb[S_j]=\varnothing$, and those in $C_{i_j}$ are exactly
the $E$-open ones, that is, $W_{i_j}$. So the number of graph edges between $Y_{i_j}$ and
$A_{\mathrm{set}}$ is $K_{i_j}\ge1$, and goodness gives
\[
\dpiv(Y_{i_j})\le K_{i_j}M^\tau.
\]
Fix $a\in N(w_{i_j})\cap Y_{i_j}$, for example $b_{i_j}$.
\begin{itemize}
\item \emph{$j<m$.} The entries of $\Lambda$ strictly between $w_{i_{j+1}}$ and $w_{i_j}$ lie in
$\Hang(w_{i_j})\setminus\{w_{i_j}\}$ and outside $\Hang(w_{i_{j+1}})$, that is, in
$(C'\setminus S_j\setminus\{w_{i_j}\})\cap S_{j+1}=Y_{i_j}$. By Lemma~\ref{lem:splicing}, $w_{i_{j+1}}$ is
adjacent to $P_{i_j}$ and not to $S_j$. If it has no neighbour in $Y_{i_j}$, it is adjacent to
$w_{i_j}$; then a path from $x$ through $S_j\cup\{w_{i_j}\}$ to $w_{i_{j+1}}$, continued through the
connected set $P_{i_{j+1}}\cup\dots\cup P_{i_m}$ to $z$, avoids $Y_{i_j}$, so no entry lies strictly
between and $l_j=1$. Otherwise fix $b\in N(w_{i_{j+1}})\cap Y_{i_j}$. For every $a$--$b$ path $\pi$ in
$Y_{i_j}$, the concatenation $x\leadsto w_{i_j}\to a\leadsto_\pi b\to w_{i_{j+1}}\leadsto z$ is an
$x$--$z$ path in $C'$, so every separating cut vertex in $Y_{i_j}$ lies on every such $\pi$: it is $a$,
$b$, or a cut vertex of $Y_{i_j}$ separating $a$ from $b$. Hence
$l_j-1\le2+\piv_{Y_{i_j}}(a,b)\le2+K_{i_j}M^\tau$.
\item \emph{$j=m$, standard.} Then $z\in Y_{i_m}$, and the entries of $\Lambda$ before $w_{i_m}$ are $z$
and separating cut vertices in $\Hang(w_{i_m})\setminus\{w_{i_m}\}=Y_{i_m}$; each of the latter lies on
every $a$--$z$ path in $Y_{i_m}$, so it is $a$ or separates $a$ from $z$ in $Y_{i_m}$. Hence
$p_m\le1+(1+K_{i_m}M^\tau)$, that is, $l_m\le3+K_{i_m}M^\tau$.
\end{itemize}
In both cases $l_j\le3+K_{i_j}M^\tau$, and $l_j\le|P_{i_j}|\le M$, since the counted entries (and $z$ when
$j=m$) are distinct vertices of $P_{i_j}\subseteq C_{i_j}$. A degenerate step has $l_m=1\le M$.

\emph{Step 5 (contact probability).} In the reconstructed input each contact list and its priority
order are fixed without any $V_i$. If the prescribed seam $w_{i_j}$ is an endpoint of
$\kappa_j\le\min\{d,K_{i_j}\}$ contact edges, the uniform $V_{i_j}$ must lie in the union of their rank
intervals $((a-1)/K_{i_j},a/K_{i_j}]$, of Lebesgue measure $\kappa_j/K_{i_j}\le\min\{1,d/K_{i_j}\}$. A
degenerate step imposes no condition. These sets are determined without the uniforms, so the measure of
valid $(V_i)$ is at most $\prod_j\min\{1,d/K_{i_j}\}$, and further validity restrictions only decrease
it. From $l\le3+KM^\tau$: if $l\le3$, then $\min\{1,d/K\}\le1=\min\{1,4dM^\tau/l\}$; if $l\ge4$, then
$K\ge(l-3)/M^\tau\ge l/(4M^\tau)$, so $d/K\le4dM^\tau/l$. Thus each selected index contributes at most
$\min\{1,4dM^\tau/l_j\}$. The priorities are integrated once under their law, not counted as inverse
choices.

\emph{Step 6 (sum).} Summing over a skipped index or a gap $1\le l\le M$ at each of the $k$ indices
gives $W^k$, and $W\le1+4dM^\tau(1+\log M)\le CM^\tau\log M$ for $M\ge2$.
\end{proof}

\paragraph{The finite bit space and the load.} All comparisons are made on finite bit spaces, as in~\cite[Section~4.3]{OAINU}. In a successful
experiment the old clusters $C_k\ni y_{k-1}\in C_{k-1}\ni\dots\ni y_0\in A$ join $z$ to $A$ with
$\operatorname{dist}(y_i,y_{i-1})<M$, and $A$ is connected with $|A|<2n$, so
$\operatorname{dist}(x,z)\le2n+kM$. A capped whole-cluster site test explores the cluster and its
closed outer vertex boundary, or stops as soon as the cap is exceeded; the old rule has a fixed query
radius and its successful steps move at most $M$; and every mask and rejection set lies in $\Nb[C']$.
So a deterministic larger ball about $x$ contains every query, mask and rejection set of every
successful experiment at $(x,z)$; there are finitely many endpoints and codes; and checking an inverse
stops at the first failed test. Condition on all bits outside this ball.


At fixed seeds an exchange swaps bits between equally distributed i.i.d.\ fields on a common mask, and
rejection changes no bit; so the number of open bits is preserved, and each successful input tuple and
its output have the same product probability. With the injective inverse at a fixed code
(Lemma~\ref{lem:coding}), define, for a single reference i.i.d.\ distinguished site field $\omega'$,
$L_x(z,\omega')$ as $1/n$ times the sum over codes of the valid-code indicators, integrated over reference
i.i.d.\ auxiliary output fields and all seeds, including the $V_i$. This is the density of the switched
mass at $z$ relative to the law of $\omega'$, and by~\eqref{eq:W}, $0\le L_x(z,\omega')\le W^k/n$. The function
$L_x(z,\cdot)$ vanishes unless~\eqref{eq:support} holds, and on that event it is a function of $\omega'$ on
$\Nb[C_x(\omega')]$, since the cluster, masks, rejection sets, contacts and the restored bin test all lie
there; hence it is a function of $\omega'$ in the ball of radius $M'$ about $x$, and it is covariant. Put
$L=\sum_{z\in O}L_x(z,\omega')$ and $b_0=\E L$. As in~\cite{OAINU}, the reference product law is not a claim
about the law of the actual output of a switch.


\begin{proof}[Proof of Lemma~\ref{lem:mean}]
The density comparison preserves total mass, so $nF(x,z)$ is the probability that the experiment at
$(x,z)$ succeeds. Without the rejection tests the success probability is that of~\cite{OAINU}: the
$k$ independent old steps have unconditional endpoint kernel $u^kK^k$. Dropping the tests gives
$F\le F_0$.

\emph{Lower bound.} Condition on $\{C_x(\omega)=A\}$ for a fixed finite $A$ in the bin, an event measurable
with respect to $\omega$ on $\Nb[A]$, on all $E_i$ and on all seeds, and assume that the chain succeeds with
$y_0\in A$. On the event that the tests of steps $1,\dots,j-1$ pass, the $j$-th selection, piece, seam
and $R_{i_j}$ are functions of the conditioning data only: apart from the rejection tests the
procedure reads $\omega$ only through $S$, which on passing equals $A\cup P_{i_1}\cup\dots\cup P_{i_{j-1}}$,
and contact sets are read in $E_i$, not in $\omega$. The sets $R_{i_j}$ miss all earlier masks, by the
freshness part of Lemma~\ref{lem:splicing}, so each test reads original $\omega$-bits. So there are sets
$R_{i_1},\dots,R_{i_m}$, determined by the conditioning, with
$\{\text{all tests pass}\}=\{\omega\text{ closed on }R_{i_1}\cup\dots\cup R_{i_m}\}$. They are disjoint from
$\Nb[A]$, as $R_{i_j}\subseteq V\setminus\Nb[S_j]$, so under the conditioning $\omega$ is i.i.d.\
Bernoulli($p_c$) on them, and
\[
\Pp(\text{all tests pass}\mid\text{conditioning})=(1-p_c)^{|\bigcup_jR_{i_j}|}\ge(1-p_c)^{(d-1)k}=c_0^k,
\]
since $|R_i|\le\deg(w_i)-1$ ($w_i$ has a neighbour in $S$). Integrating gives $F\ge c_0^kF_0$.

\emph{Row sums and norm.} As in~\cite[Lemma~4.7]{OAINU}: $B_n$ is symmetric (on $y\in C_x$ the
clusters coincide, and both are empty unless $x$ and $y$ are open), its $O$-row sum $b_n$ satisfies
$cn\pi\le b_n\le2n\pi/\sigma_O$ by~\eqref{eq:scales}, and $K^k$ has column sums one, so the row sum of
$F_0$ is $b_0^{(0)}=u^kb_n/n$ and $F_0/b_0^{(0)}=(B_n/b_n)(K^k)^*$ has norm at most $r^k$. Then
$c_0^kb_0^{(0)}\le b_0\le b_0^{(0)}$, and since $0\le F\le F_0$ entrywise,
$\norm F\le\norm{F_0}\le b_0^{(0)}r^k\le c_0^{-k}b_0r^k$.
\end{proof}

\subsection{Proof of Lemma~\ref{lem:second-moment}}\label{app:amp-second}

\begin{proof}[Proof of Lemma~\ref{lem:second-moment}]
Expand $L^2$ with a common reference field $\omega'$ and two independent families of auxiliary output
fields and seeds; a summand is a pair of valid coded inverses with endpoints $z_1,z_2$. For a switching
inverse put $B=\Hang(w_{i_1})=C'\setminus A=P_{i_1}\cup\dots\cup P_{i_m}$ (Lemma~\ref{lem:splicing}); it
contains its endpoint and has $|B|\le \nu$. There are three classes.

\emph{Class 1: at least one inverse makes no switch.} This is verbatim from~\cite[Lemma~4.8]{OAINU}. A
no-switch inverse at $z$ is valid only if $A=C_x(\omega')$ is in the bin, $z\in A$, and the chain from $z$
succeeds with $y_0\in A$; so its integrated part satisfies $L_0\le(u^k/n)\1\{n\le|A|<2n\}
\sum_{z\in A\cap O}\sum_{y\in A\cap O}K^k(z,y)\le2u^k$ by the column sums. These summands contribute at
most $2\E[L_0L]\le4u^kb_0\le C\pi u^{2k}$ by Lemma~\ref{lem:mean}.

\emph{Class 2: both switch, and $B^1,B^2$ are nested or at graph distance at most $2$.} Hanging sets are
nested or disjoint (Lemma~\ref{lem:coding}, Step~1). Fix $\omega'$ and the first inverse, hence
$B^1=\Hang(c^1)$ and $z_1$. The second inverse has $B^2=\Hang(c^2)$, where $c^2$ is $z_2$ or a cut vertex
separating $x$ from $z_2$, with $z_2\in B^2$ and $|B^2|\le \nu$. The number of possible $z_2$ is at most
$\nu+\nu^2+d^2\nu^3$:
\begin{itemize}
\item if $B^2\subseteq B^1$, then $z_2\in B^1$: at most $\nu$;
\item if $B^1\subsetneq B^2$, then $c^1\in \Hang(c^2)$ and $c^2\ne c^1$, so $c^2$ separates $x$ from $c^1$;
the cut vertices separating $x$ from $c^1$, followed by $c^1$, are linearly ordered by nesting of
hanging sets, and those from $c^2$ onward lie in $\Hang(c^2)$, of size at most $\nu$; so $c^2$ is one of the
last $\nu$ entries, and each determines $B^2$ and at most $\nu$ endpoints: at most $\nu^2$;
\item if they are disjoint with $1\le\operatorname{dist}(B^1,B^2)\le2$, choose $b\in B^2$ within distance
$2$ of $B^1$, in at most $d^2\nu$ ways ($d+d(d-1)=d^2$ vertices at distance $1$ or $2$ from a vertex);
$c^2$ is $b$ or a cut vertex separating $x$ from $b$, and lies among the last $\nu$ entries of that ordered
list, since all later entries and $b$ lie in $\Hang(c^2)$; then at most $\nu$ endpoints: at most $d^2\nu^3$.
\end{itemize}
For each endpoint, Lemma~\ref{lem:coding} bounds the integrated second-inverse mass by $W^k/n$.
Integrating the first inverse, this class is at most $b_0(\nu+\nu^2+d^2\nu^3)W^k/n\le C(\nu+\nu^2+d^2\nu^3)\pi u^{2k}$, by
Lemma~\ref{lem:mean} and $n\ge(W/u)^k$.

\emph{Class 3: both switch, and $\operatorname{dist}(B^1,B^2)\ge3$, that is, $\Nb[B^1]\cap\Nb[B^2]=
\varnothing$.}

Locality of each inverse, the joint restoration and the claims~(J1)--(J3) are as in
Section~\ref{sec:second-moment}.

\emph{Injectivity with the extra datum.} Fix $z_1,z_2$ and all seeds. The restored auxiliary fields and
old-map seeds recover both old chains by replay from their endpoints, and $\xi$ gives $A_{00}$. Run the
switching procedure of Section~\ref{sec:switching} virtually from the \emph{set} $S_0=A_{00}$ with field
$\omega_{00}$ and the restored first chain, ignoring only the bin test and the test $y_0\in A$; the rules use
$S$ only as a set, and connectivity of $S_0$ is never used. Let $S$ denote the current set of the
original first forward run, from $A^1$ with $\omega^1$. Throughout, $B^2\subseteq A^1\subseteq S$, and
initially $S_0=A_{00}=A^1\setminus B^2$. The key fact is that for $U\subseteq\Nb[B^1]$,
\[
U\cap S=U\cap S_0,\qquad U\cap\Nb[S]=U\cap\Nb[S_0],\qquad U\cap\partial S=U\cap\partial S_0
\qquad\text{whenever }S_0=S\setminus B^2,
\]
since a vertex of $\Nb[B^1]$ is neither in $B^2$ nor adjacent to it. We show by induction that
$S_0=S\setminus B^2$, that the current distinguished fields of the two runs agree off $\mathrm{Mask}^2$
and their auxiliary fields agree, and that each step selects the same index, piece, contacts, seam,
rejection set and mask.
\begin{enumerate}[label=\textup{(\arabic*)},leftmargin=2.2em]
\item \emph{Index.} The original seam satisfies $w_i\in C_i\cap\partial S$ and $w_i\in B^1$, so $w_i\in
C_i\cap\Nb[S_0]$. Every larger eligible index misses $\Nb[S]\supseteq\Nb[S_0]$. Maximality selects the
same $i$, also when unused clusters meet $B^2$ or $y_0\in B^2$. The case, standard or degenerate, agrees:
$y_i\in B^1$, and the key fact applies.
\item \emph{Piece.} $Y_i^0\supseteq Y_i$, since $\Nb[S_0]\subseteq\Nb[S]$. If it were larger, some
$E_i$-open $u\notin Y_i\cup\Nb[S_0]$ would be adjacent to $Y_i$; then $u\in\Nb[B^1]$, so $u\notin\Nb[S]$
by the key fact, and $u$ would belong to $Y_i$. So $Y_i^0=Y_i$. This is where distance $3$ is needed: at
distance $2$ the slit domain could grow through a vertex of $\partial S\cap\partial B^2$ adjacent to
$Y_i$.
\item \emph{Contacts and seam.} $W_i^0=W_i$, by the key fact on $N(Y_i)$, so the contact list, its
priority order and, with the same $V_i$, the seam agree.
\item \emph{Rejection.} $R_i^0=N(w_i)\setminus(\Nb[S_0]\cup\Nb[Y_i])=R_i$, by the key fact on $N(w_i)$;
the two current fields agree on $R_i\subseteq\Nb[B^1]$, which is disjoint from $\mathrm{Mask}^2$, so the
test passes in the virtual run.
\item \emph{Mask and update.} $\mathrm{Mask}_i^0=\mathrm{Mask}_i$, the fields agree there, so after the
exchange they still agree off $\mathrm{Mask}^2$, and $S_0\cup P_i=(S\cup P_i)\setminus B^2$.
\item \emph{Stopping.} $z_1\in S_0$ if and only if $z_1\in S$, as $z_1\in B^1$.
\end{enumerate}
The same holds for the second virtual run. Running both separately from the jointly restored tuple and
combining their disjoint distinguished-field changes reconstructs $\omega'$, both auxiliary output families
and all seams, hence both codes. Thus, at fixed $z_1,z_2$ and seeds,
$(\omega',E'^1,E'^2,\mathrm{code}^1,\mathrm{code}^2)\mapsto(\omega_{00},E^1,E^2,\xi)$ is injective, so
$(\omega',E'^1,E'^2,\mathrm{code}^1,\mathrm{code}^2)\mapsto(\omega_{00},E^1,E^2)$ is at most $4^d\nu^2$-to-one, by
(J3). By (J2), each restored successful trajectory, the union of its old-map whole clusters, contains
$c^a\in\partial A_x$, so it hits $\Nb[C_x(\omega_{00})]$.

\emph{Mass.} As in~\cite[Lemma~4.8]{OAINU}, the joint undo preserves the weight
$p_c^{a}(1-p_c)^{N-a}$ of the full finite bit tuple, with $a$ open bits among $N$. With the
multiplicity bound, summing weights gives at most $4^d\nu^2$ times the mass of the image in an
unrestricted reference product experiment, weighted by $1/n^2$, with an i.i.d.\ root field and two
independent old-chain families; integrate the seeds by Tonelli's theorem. Retain only
$s_x=|C_x|<2n$ (from $A_x\subseteq A_{00}$), success of both chains, and that each trajectory hits
$\Nb[C_x]$. The lower size window of~\cite[Lemma~4.8]{OAINU} is not available, since $A_x$ can be much
smaller than $A_{00}$; it is replaced by the record-scale bound~(e) of~\eqref{eq:scales}.

\emph{Trajectory hits}, as in~\cite[Lemma~4.8]{OAINU} with site clusters. For finite $A$, let $h_z(A)$ be
the probability that a successful $k$-step old trajectory from $z\in O$ hits $A$. Then
$\sum_{z\in O}h_z(A)\le c_\sigma|A|\nu u^k$, by transporting unit mass from $z$ to each vertex of its
trajectory union on success (expected outgoing mass at most $\nu u^k$) and Lemma~\ref{lem:MTP}. Apply it
to $\Nb[A]$, with $|\Nb[A]|\le(d+1)|A|$. Conditional on the reference root cluster the two chain
families are independent, so this class is at most
\[
4^d\nu^2\cdot\frac{C\nu^2u^{2k}}{n^2}\E_{p_c}[s_x^2;\ 1\le s_x<2n]\le C4^d\nu^4\pi u^{2k},
\]
by the record-scale bound $\E_{p_c}[s_x^2;\ 1\le s_x<2n]\le C_On^2\pi$ for $x\in O$
(Lemma~\ref{lem:scales}). Summing the three classes proves~\eqref{eq:second-moment} with
$\cQ(k,M)=C(1+\nu+\nu^2+d^2\nu^3+4^d\nu^4)$.
\end{proof}