Compute the restoration bound along the orbit from each row's margin and push direction, with its Table 7 column and whole-run figures

paper/universal_construction_ternary.pdf

@@ -1,3 +1,3 @@
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paper/universal_construction_ternary.tex

@@ -1539,60 +1539,107 @@ second order in the probability of a flip.
 
 \begin{proposition}[Restoration]\label{prop:restore}
 Let the perturbations of $N_3$ be independent centred Gaussians of standard
-deviation $s$ on every pre-activation, and put $p=Q(3/(2s))$ and
-$\sigma=Q(1/(2s))$. Unroll $\Lev(\Nhost)$ over the steps, the rows of layer~$1$ of
-each step reading the rows of layer~$L$ of the step before, and write $u\to v$
-when row $v$ reads row $u$. For values $x$ on the rows of a layer, every copy of a
-row $w$ carrying $x_w$, let $\pi(x)$ be the sum of $x_wx_{w'}$ over the unordered
-pairs of distinct copies that some row of the next layer reads, together with
-the pairs of copies of one row when the layer is the last. Let $a$ be the least
-solution of $a_v=p+3\sigma\sum_{u\to v}a_u$; for a row $u$ let $V^u$ be the least
-solution of $V^u_v=\sigma[u\to v]+3\sigma\sum_{w\to v}V^u_w$, let $G_u$ be the sum of
-$\pi(V^u)$ over the layers after $u$, and let $X_u=G_u-3\sigma^2\sum_{u\to v}G_v$.
-Then the probability that the state of $N_3$ read by majority differs from the
-orbit at some step $t\le T$ is at most
+deviation $s$ on every pre-activation, and unroll $\Lev(\Nhost)$ along the orbit
+of $x_0$, the rows of layer~$1$ of each step reading the rows of layer~$L$ of the
+step before. For a row $v$ of a step let $z_v$ be its pre-activation and $y_v$
+its value on the orbit, $D_v=3|z_v+\tfrac12|$, $q_j(v)=Q\bigl((D_v-j)/s\bigr)$ for
+$j=0,1,2,3$ and $\Delta_v=q_1(v)-q_0(v)$, and let $T(v)$ be the set of rows $u$
+that $v$ reads with a weight $w$ satisfying $w(1-2y_u)(z_v+\tfrac12)<0$. Define,
+layer by layer,
 \[
-T\Bigl(\sum_{\ell}\pi(a)+3\sum_{u}a_u\min(1,X_u)\Bigr),
+\begin{gathered}
+P_v=3q_0(v)+3\Delta_vX_v,\qquad
+P^{(2)}_v=3q_0(v)^2+3\bigl(q_1(v)^2-q_0(v)^2\bigr)X_v,\\
+P^{(3)}_v=q_0(v)^3+\bigl(q_1(v)^3-q_0(v)^3\bigr)X_v,
+\end{gathered}
 \]
-the sums over the layers and the rows of one step.
+where $X_v=\sum_{u\in T(v)}P_u$, and for distinct rows $v_1,v_2$ of one layer
+\[
+K_{v_1v_2}=9\Delta_{v_1}\Delta_{v_2}\Bigl(\sum_{u\in T(v_1)\cap T(v_2)}P_u
++\sum_{\substack{u_1\in T(v_1),\ u_2\in T(v_2)\\ u_1\neq u_2}}K_{u_1u_2}\Bigr),
+\]
+with $P$ and $K$ zero on the state of step~$0$ and on the device cells of every
+state, and $P^{(2)}$ and $P^{(3)}$ zero on every state. Let
+\begin{multline*}
+b(t)=\sum_{w}\Bigl(\min\bigl(1,3q_2(w)\bigr)\sum_{u\in T(w)}P^{(2)}_u
++\min\bigl(1,3q_3(w)\bigr)\sum_{u\in T(w)}P^{(3)}_u\\
++\sum_{\substack{u_1,u_2\in T(w)\\ u_1<u_2}}\bigl(P_{u_1}P_{u_2}+K_{u_1u_2}\bigr)\Bigr)
++\sum_{v}P^{(2)}_v,
+\end{multline*}
+the first sum over the rows of the layers of step~$t$ and the last over the rows
+of its last layer. Then the probability that the state of $N_3$ read by majority
+differs from the orbit after one of the first $T$ steps is at most
+$\sum_{t<T}b(t)$.
 \end{proposition}
 
 \begin{proof}
-Call a copy wrong when its output differs from the replicated orbit. By
-Proposition~\ref{prop:amplify}(ii) and~(iii), a copy that reads no wrong copy is
-wrong only if its perturbation carries it across its distance
-$3|z+\tfrac12|\ge\tfrac32$ from the comparator, and a copy that reads one wrong
-copy only if its perturbation carries it across a distance of at least $\tfrac12$;
-the first event has probability at most $p$, the second at most $\sigma$, and each
-depends on the perturbation of that copy alone. Call two wrong copies that some
-row reads, or two wrong copies of one row of the last layer, a \emph{collision};
-the device reads and writes every register by majority, so the state read by
-majority is wrong only after a collision. Before the first collision every row
-reads at most one wrong copy, so a wrong copy either reads none, and its first
-event occurred, or reads exactly one, its parent, and its second event occurred;
-following parents gives a chain from a copy of the first kind. The two copies of
-the first collision therefore end two chains that are disjoint or share an
-initial segment up to a last common copy $z$, after which they continue as two
-paths through distinct copies. Distinct copies have independent perturbations,
-so such a configuration occurs with probability at most the product of $p$ for
-the first copy of each chain and $\sigma$ for every other copy.
-
-The sum of these products over the chains ending at a copy of $v$ is $a_v$, and
-over the paths from one copy of $u$ to a copy of $v$ it is $V^u_v$, every copy of a
-row reading all three copies of its inputs. Pairs of disjoint chains whose ends
-collide in one step therefore contribute at most $\sum_\ell\pi(a)$. For a
-configuration parting at a copy $z$ of $u$, the chain to $z$ and the paths after
-it involve disjoint copies, so its probability is at most $a_u$ times the
-probability that two paths leaving $z$ through distinct copies reach a collision.
-The pairs of paths from $z$ that reach a collision number $G_u$ in expectation,
-and those that leave $z$ through the same copy of a consumer $v$ number
-$\sigma^2G_v$ for each of its three copies, so the pairs that part at $z$ number at
-most $X_u$ in expectation and occur with probability at most $\min(1,X_u)$.
-Summing over the three copies of every row and over the $T$ steps at which the
-first collision can occur gives the bound.
+Call a copy wrong when its output differs from the replicated orbit. Every copy
+of a row $v$ reads the same copies, and a wrong copy of an input $u$ moves the
+pre-activation of a copy of $v$ by $w(1-2y_u)=\pm1$, toward the comparator
+exactly when $u\in T(v)$. Let $S_v$ be the set of wrong copies of the rows of
+$T(v)$. By Proposition~\ref{prop:amplify}(ii) a copy of $v$ is wrong only if its
+perturbation, on the far side of the comparator, has magnitude at least
+$D_v-|S_v|$; given the perturbations of the earlier layers the three copies of
+$v$ therefore fail independently, each with probability at most $q_{|S_v|}(v)$.
+Call a layer a \emph{stop} when it holds a wrong copy of a row $v$ with
+$|S_v|\ge2$, or when it is the last layer of a step and one of its rows has two
+wrong copies. The state read by majority is wrong after a step only if the last
+layer of that step is a stop, and while no layer is a stop the device reads
+every cell correctly and writes every copy of it, so the probability to be
+bounded is at most the sum, over the layers of the first $T$ steps, of the
+probability that the first stop is that layer.
+
+For a row $v$ let $W_v$ be the event that $v$ has a wrong copy and neither its
+layer nor an earlier one is a stop. On $W_v$, $|S_v|\le1$, and $|S_v|=1$ only on
+the union of the $W_u$ with $u\in T(v)$, so
+$\Pr(W_v)\le3q_0(v)+3\Delta_v\sum_{u\in T(v)}\Pr(W_u)$ and, by induction over the
+layers, $\Pr(W_v)\le P_v$; on the device cells of a state $W_v$ is empty, the
+device having written every copy. Two of three independent copies fail with
+probability at most $3q_j^2$ and all three with probability at most $q_j^3$, so
+the probability that no earlier layer is a stop, $|S_v|\le1$, and $v$ has two or
+three wrong copies is at most $P^{(2)}_v$ or $P^{(3)}_v$; for a row of a state it
+is zero as far as the layers after it are concerned, since two wrong copies of a
+state row make the last layer of the step before a stop. For distinct rows
+$v_1,v_2$ of one layer the copies of $v_1$ and of $v_2$ fail independently given
+the earlier layers, so
+\[
+\Pr(W_{v_1}\cap W_{v_2})\le9\,\mathbb E\Bigl[\bigl(q_0(v_1)+\Delta_{v_1}\mathbf 1_{|S_{v_1}|=1}\bigr)\bigl(q_0(v_2)+\Delta_{v_2}\mathbf 1_{|S_{v_2}|=1}\bigr)\Bigr],
+\]
+the expectation taken on the event that no earlier layer is a stop. Both
+indicators are $1$ only on the union of the $W_u$ with $u\in T(v_1)\cap T(v_2)$
+and of the $W_{u_1}\cap W_{u_2}$ with $u_1\in T(v_1)$, $u_2\in T(v_2)$,
+$u_1\neq u_2$; expanding the product and bounding the latter events by induction
+gives $\Pr(W_{v_1}\cap W_{v_2})\le P_{v_1}P_{v_2}+K_{v_1v_2}$, since
+$\sum_{u_1\neq u_2}P_{u_1}P_{u_2}\le X_{v_1}X_{v_2}$.
+
+If the first stop holds a wrong copy of a row $w$ with $|S_w|\ge2$, then either
+$S_w$ consists of two or of three copies of one row $u\in T(w)$, events of
+probability at most $P^{(2)}_u$ and $P^{(3)}_u$ after which a copy of $w$ fails
+with probability at most $\min(1,3q_2(w))$ or $\min(1,3q_3(w))$, or two distinct
+rows $u_1,u_2\in T(w)$ have wrong copies, with probability at most
+$P_{u_1}P_{u_2}+K_{u_1u_2}$. Otherwise the first stop is the last layer of a step
+and one of its rows $v$ has two wrong copies with $|S_v|\le1$, with probability
+at most $P^{(2)}_v$. Summing over the rows of the layers of the first $T$ steps
+gives $\sum_{t<T}b(t)$.
 \end{proof}
 
-For the self-reproducing run, $T=1{,}664{,}939$, the bound of Proposition~\ref{prop:restore} is $3.80\times10^{-6}$ at $s=0.20$, where Theorem~\ref{thm:amplify}(ii) gives $1.34\times10^{-2}$, and $2.98\times10^{-3}$ at $s=0.22$, where that bound exceeds one; from $s=0.25$ on it exceeds one as well, and from $s=0.33$ its sums diverge (Table~\ref{tab:bound3}). Along the orbit, over $64$ noisy copies, the state read by majority failed at $4.04\times10^{-6}$ per step at $s=0.27$ and $2.13\times10^{-4}$ at $s=0.30$, below the bound by factors of $23$ and $53$, and $21$ and $12$ times less often than the replicated state left its orbit.
+The sums are taken along the exact orbit, from the pre-activation of every row at
+every step. $K$ is held sparse: entries below $10^{-20}\cdot3Q(3/(2s))$, sources
+$u$ with $P_u$ below that level, and pairs reached through a row with
+$\Delta_v<10^{-14}$ are carried by rank-one majorants, which pass through the
+recursion exactly and contribute less than $10^{-12}$ of $b(t)$. Every quantity
+of the recursion bounds a probability, so it may be restarted at any step from
+$P=1$ and $K=0$ on the state; restarted twenty steps before its first counted
+step, a chunk reproduces the uninterrupted recursion to within $10^{-14}$, and
+the sums over the self-reproducing run are computed in $36$ chunks in parallel.
+Along that run, $T=1{,}664{,}939$, $\sum_{t<T}b(t)$ is $6.13\times10^{-7}$ at
+$s=0.20$, where Theorem~\ref{thm:amplify}(ii) gives $1.34\times10^{-2}$,
+$3.11\times10^{-4}$ at $s=0.22$ and $0.271$ at $s=0.25$, where that bound exceeds
+one, and it exceeds one from $s=0.27$. Along the first $20{,}000$ steps of the
+self-describing instance, weighted by the steps each of the $64$ noisy copies
+ran, $b(t)$ lies $1.06$ to $1.47$ times above the measured failure rates
+(Table~\ref{tab:bound3}), which are $21$ and $12$ times below the rates at which
+the replicated state left its orbit at $s=0.27$ and $0.30$.
 
 
 % =============================================================================
@@ -2420,10 +2467,9 @@ copies of $N_3$ were stepped together from the initial state of the
 self-describing instance for up to $20{,}000$ steps, the state of each noisy copy
 read by majority and compared with the exact state after every step, and each
 noisy copy stopped at its first wrong majority-read state. The restoration bound
-of Proposition~\ref{prop:restore} is computed with the worst margins, every row
-at distance $\tfrac12$ from the comparator, so it holds along every orbit; along
-this one most rows of the decoders sit at distance at least $\tfrac32$, which is
-why it lies one to two orders of magnitude above the measured rate.
+of Proposition~\ref{prop:restore} is computed along the same orbit; at $s=0.27$
+the probability it gives that a copy fails within the $20{,}000$ steps is $0.091$,
+against $5$ of the $64$ copies.
 
 \begin{table}[ht]
 \centering\small
@@ -2431,18 +2477,18 @@ why it lies one to two orders of magnitude above the measured rate.
 \toprule
 $s$ & Failed & Exposure & Measured rate & Restoration bound & $rM\erfc(r/2\sqrt2s)$\\
 \midrule
-$0.10$ & --- & --- & --- & $2.13\times10^{-57}$ & $9.24\times10^{-46}$\\
-$0.15$ & --- & --- & --- & $9.96\times10^{-24}$ & $1.92\times10^{-18}$\\
-$0.20$ & --- & --- & --- & $2.28\times10^{-12}$ & $8.03\times10^{-9}$\\
-$0.22$ & --- & --- & --- & $1.79\times10^{-9}$ & $1.16\times10^{-6}$\\
-$0.25$ & --- & --- & --- & $2.10\times10^{-6}$ & $2.48\times10^{-4}$\\
-$0.27$ & 5/64 & 1{,}237{,}505 & $4.04\times10^{-6}$ & $9.28\times10^{-5}$ & $3.48\times10^{-3}$\\
-$0.30$ & 63/64 & 295{,}163 & $2.13\times10^{-4}$ & $1.14\times10^{-2}$ & $7.22\times10^{-2}$\\
-$0.33$ & 64/64 & 17{,}472 & $3.66\times10^{-3}$ & diverges & $6.90\times10^{-1}$\\
-$0.36$ & 64/64 & 2{,}188 & $2.93\times10^{-2}$ & diverges & $3.89$\\
+$0.10$ & --- & --- & --- & $8.62\times10^{-59}$ & $9.24\times10^{-46}$\\
+$0.15$ & --- & --- & --- & $4.02\times10^{-25}$ & $1.92\times10^{-18}$\\
+$0.20$ & --- & --- & --- & $3.59\times10^{-13}$ & $8.03\times10^{-9}$\\
+$0.22$ & --- & --- & --- & $1.82\times10^{-10}$ & $1.16\times10^{-6}$\\
+$0.25$ & --- & --- & --- & $1.58\times10^{-7}$ & $2.48\times10^{-4}$\\
+$0.27$ & 5/64 & 1{,}237{,}505 & $4.04\times10^{-6}$ & $4.57\times10^{-6}$ & $3.48\times10^{-3}$\\
+$0.30$ & 63/64 & 295{,}163 & $2.13\times10^{-4}$ & $2.26\times10^{-4}$ & $7.22\times10^{-2}$\\
+$0.33$ & 64/64 & 17{,}472 & $3.66\times10^{-3}$ & $4.34\times10^{-3}$ & $6.90\times10^{-1}$\\
+$0.36$ & 64/64 & 2{,}188 & $2.93\times10^{-2}$ & $4.30\times10^{-2}$ & $3.89$\\
 \bottomrule
 \end{tabular}
-\caption{The per-step probability that the state of $N_3$ read by majority leaves the orbit under read noise of standard deviation $s$: $64$ noisy copies stepped beside an exact one from the initial state of the self-describing instance for up to $20{,}000$ steps, each stopped at its first wrong majority-read state, against the bound of Proposition~\ref{prop:restore} per step and the first-order bound of Theorem~\ref{thm:amplify}(ii). From $s=0.33$ the sums defining the restoration bound diverge.}
+\caption{The per-step probability that the state of $N_3$ read by majority leaves the orbit under read noise of standard deviation $s$: $64$ noisy copies stepped beside an exact one from the initial state of the self-describing instance for up to $20{,}000$ steps, each stopped at its first wrong majority-read state, against the bound $b(t)$ of Proposition~\ref{prop:restore} along the same orbit, weighted by the steps each copy ran where copies ran and averaged over the $20{,}000$ steps elsewhere, and the first-order bound of Theorem~\ref{thm:amplify}(ii).}
 \label{tab:bound3}
 \end{table}