ternary-30.patch

--- a/build/main.tex
+++ b/build/main.tex
@@ -1,3 +1,5 @@
+% Modified by Douglas Colkitt, October 7, 2026.
+% Conditional compact-control, complex-compression and ternary research draft; see main.tex for scope.
 \documentclass[11pt]{article}
 \usepackage[T1]{fontenc}
 \usepackage{lmodern,microtype}
@@ -7,7 +9,7 @@
 \usetikzlibrary{positioning,arrows.meta}
 \usepackage[hyphens]{url}
 \usepackage{hyperref}
-\hypersetup{colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=blue!45!black,pdftitle={Integer multiplication below n log n},pdfauthor={OpenAI}}
+\hypersetup{colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=blue!45!black,pdftitle={Integer multiplication below n log n},pdfauthor={OpenAI (original manuscript); Douglas Colkitt (modifications)}}
 \pdfinfoomitdate=1
 \pdftrailerid{}
 \pdfsuppressptexinfo=15
@@ -20,14 +22,24 @@
 \newtheorem{remark}[theorem]{Remark}
 \newcommand{\norm}[1]{\lVert#1\rVert_\infty}
 \title{Integer multiplication below \texorpdfstring{$n\log n$}{n log n}}
-\author{OpenAI}
-\date{September 23, 2026}
+\author{\small Original manuscript: OpenAI\\
+\small Research modifications: Douglas Colkitt}
+\date{Original: September 23, 2026; modified: October 7, 2026}
 \begin{document}
 \maketitle
+\begin{quote}\small
+\textbf{Modified research draft; not an OpenAI release.}
+This derivative incorporates Douglas Colkitt's compact-control, complex-network and ternary constructions
+and retained project refinements, prepared with assistance from OpenAI Codex.
+The revised bound is conditional on the original manuscript's retained
+algorithmic and analytic interfaces and the new written arguments. These
+modifications have not received independent mathematical review or formal
+verification; no endorsement by OpenAI is implied.
+\end{quote}
 \begin{abstract}
 We give a deterministic algorithm that multiplies two $n$-bit integers
 in $O(n(\lg n)^{1-\kappa})$ worst-case time, with
-$\kappa=2^{-182}$, on one fixed finite-alphabet Turing machine with a
+$\kappa=2^{-30}$, on one fixed finite-alphabet Turing machine with a
 fixed finite number of one-dimensional tapes. The algorithm is exact
 for every input length and disproves the $n\log n$ optimality conjecture
 of Sch\"onhage and Strassen in this model.
--- a/build/sections/00-introduction.tex
+++ b/build/sections/00-introduction.tex
@@ -1,3 +1,5 @@
+% Modified by Douglas Colkitt, October 7, 2026.
+% Conditional compact-control, complex-compression and ternary research draft; see main.tex for scope.
 \section{Introduction}
 
 Sch\"onhage and Strassen proved in 1971 that two $n$-bit integers can
@@ -25,7 +27,7 @@
 inputs. Its worst-case running time satisfies
 \[
  T_A(n)=O\!\left(n(\lg n)^{1-\kappa}\right),
- \qquad \kappa=2^{-182}.
+ \qquad \kappa=2^{-30}.
 \]
 \end{theorem}
 
--- a/build/sections/08-assembly.tex
+++ b/build/sections/08-assembly.tex
@@ -1,3 +1,5 @@
+% Modified by Douglas Colkitt, October 7, 2026.
+% Conditional compact-control, complex-compression and ternary research draft; see main.tex for scope.
 \section{Exact integer multiplication and its cost}
 \label{sec:assembly}
 
@@ -16,35 +18,33 @@
 \subsection{A fixed rational choice}
 
 The two motif exponents were fixed in \eqref{eq:explicit-motif-exponents}.
-Take the following fixed rational numbers:
+Put $a=467/10^{11}$ and $a_{\rm c}=5/10^9$.
+Choose
 \begin{equation}\label{eq:fixed-parameters}
 \begin{gathered}
- \tau=\sigma=1-2^{-50},\qquad \beta=\frac12,\qquad \delta=\frac1{16},\\
- \lambda=1-2^{-52},\qquad c=2^{-56},\qquad
- \lambda'=1-2^{-54},\\
- C_1=20,\qquad \epsilon=2^{-75},\qquad \kappa=2^{-182}.
+ \tau=1-a,\quad\sigma=1-a_{\rm c},\quad
+ \beta=\frac1{100},\quad\zeta=\frac1{1000},\quad
+ C_1=\frac{4961}{1000},\\
+ \epsilon=\frac{1999}{10000},\quad c=1,\quad\delta=\frac1{10^6},\\
+ \lambda=1-\frac{4669}{10^{12}},\quad
+ \lambda'=1-\frac{4668}{10^{12}},\quad
+ \kappa=2^{-30}.
 \end{gathered}
 \end{equation}
-These constants are deliberately conservative; no optimization is claimed.
-
-The layer conditions hold with room to spare:
-\[
- \max\{\tau,\sigma\}<\lambda<1,\qquad
- \tau(1+c/\beta)<1-31\cdot2^{-55}<\lambda,
-\]
-and
-\[
- \sigma+\beta(1-\sigma)=1-2^{-51}<\lambda<\lambda'<1.
-\]
-The strict final inequality absorbs the logarithmic number of groups in
-a base-$m$ partition.  The dimension exponent also satisfies
-\[
-\begin{gathered}
- \epsilon<\frac1{12},\quad \epsilon C_1<1,\quad
- \epsilon(2-\tau)<1-\tau,\quad
- \frac34+\delta+\frac32\epsilon<1,\\
- \epsilon(1+c)<1,\qquad \epsilon+\delta<1.
-\end{gathered}
+The new compact-control recurrence has
+$\chi=\tau$, because $\sigma<\tau$, and exact rational comparisons give
+\[
+ \max\{\tau,\sigma,\chi\}<\lambda<\lambda'<1,\quad
+ \sigma+\beta(1-\sigma)<\lambda',\quad 1-c<\lambda'.
+\]
+The stopping test is $e^{100}<d$. Section~\ref{sec:compact-stopped-guard}
+proves the displayed $C_1$, with $\epsilon C_1<1$.
+The retained Gaussian choice gives the remaining comparisons
+\[
+ \epsilon<\tfrac13,\quad 2\epsilon<1,\quad
+ \epsilon(1-\tau)<1-\tau,\quad
+ \tfrac34+\delta+\tfrac54\epsilon<1,\quad
+ \epsilon(1+c)<1,\quad \epsilon+\delta<1.
 \]
 
 \subsection{Input and transform sizes}
@@ -96,21 +96,22 @@
 These bounds also give that $r$ exceeds every fixed power of $p$,
 $K/\log p\to\infty$, and
 $\ell/K\to\infty$.  There are eventually enough axes for the
-$O(\log d)$ reserved row axes and enough bits for all work blocks.
+$O(\log d+d^{1-c}\log p)$ reserved row and compact-control axes,
+which form $o(d)$ axes, and enough bits for all work blocks.
 
 Set
 \[
- \alpha=\left\lceil(12d^2b)^{1/4}\right\rceil,
+ \alpha=\left\lceil(32db)^{1/4}\right\rceil,
  \qquad \gamma=2d\alpha^2,\qquad \eta=\frac1{4d}.
 \]
 The ceiling does not affect the needed bounds: for $A\ge1$,
 $\lceil A\rceil\le2A$, and hence
 \begin{equation}\label{eq:gamma}
- \alpha\le2(12d^2b)^{1/4},\qquad
- \gamma<28d^2\sqrt b\le28b^{1/2+2\epsilon}\le28b^{2/3}.
+ \alpha\le2(32db)^{1/4},\qquad
+ \gamma<46d^{3/2}\sqrt b\le46b^{1/2+3\epsilon/2}=46b^{15997/20000}.
 \end{equation}
 It follows that $2\le\alpha<\sqrt p$ eventually and $\gamma=o(b)$.
-For example $b\ge2^{24}$ implies $\gamma\le b/4$.
+For example $b\ge2^{40}$ implies $\gamma\le b/4$.
 
 \subsubsection{Prime source lengths}
 
@@ -158,10 +159,10 @@
 Writing $\theta_i=t_i/s_i-1$, the upper bound on $s_i$ gives
 $\theta_i\geq\eta/(1-\eta)>\eta$. Therefore
 \[
- \alpha^4\theta_i>\frac{12d^2b}{4d}=3db\geq6b=p.
+ \alpha^4\theta_i>\frac{32db}{4d}=8b>6b=p.
 \]
 The choice of $\alpha$ gives
-$\alpha=\Theta(p^{1/4+\epsilon/2})$, and \eqref{eq:gamma} already
+$\alpha=\Theta(p^{1/4+\epsilon/4})$, and \eqref{eq:gamma} already
 ensures $2\leq\alpha<\sqrt p$ for large $n$. We also have
 $t_i\leq r\leq T<8n/b<2^{6b}=2^p$.
 Thus every hypothesis of Lemma~\ref{lem:no-sort-resampling} holds for
@@ -184,8 +185,8 @@
 All remaining scalar setup, including comparisons for fixed rational
 powers, modular inverses,
 prime products and shape descriptors, has cost polynomial in $p$.
-For example, $d=\lfloor b^{2^{-75}}\rfloor$ is found by binary search
-using exact comparisons $d^{2^{75}}\le b$; the exponent is fixed.
+For example, $d=\lfloor b^{1999/10000}\rfloor$ is found by binary search
+using exact comparisons $d^{10000}\le b^{1999}$; the exponent is fixed.
 The same procedure computes $K$ and all other fixed rational powers.
 All primes and length parameters are therefore generated by the machine;
 they are not advice.
@@ -292,15 +293,28 @@
 offsets therefore cost $O(dTp^C/r)=o(Tp)$ to prepare, since
 $t_i\ge r/2$ and $r$ exceeds every fixed power of $p$.
 
-The axis widths are $\ell$ or $\ell-1$.  To exchange two adjacent axes
-of unequal width, move the longer axis's leading bit outside their
-equal-width pair, swap that pair by the proved chunk-swap procedure,
-and put the leading bit at its new position.  This adds a fixed number
-of linear passes to the equal-width swap cost.  Reversing all axes
-uses $O(d^2)$ such moves.  Therefore the map $\Phi$, its inverse, and
-their padding operations cost
-\[
- O\bigl(Tp\,d^2(1+\ell^\tau)\bigr).
+The axis widths are $\ell$ or $\ell-1$. Reverse the axis order by
+interchanging physical positions $j$ and $d+1-j$ for
+$j=1,\ldots,\lfloor d/2\rfloor$. Every pair is disjoint, so this uses
+$\lfloor d/2\rfloor$ full-field interchanges, not a sequence of adjacent moves.
+Each is covered by Lemma~\ref{lem:chunk-swap} and its subsequent one-bit-width
+extension, with all intervening fields collapsed into its spectator gap.
+Its cost is $O(Tp(1+\ell^\tau))$. Keep each axis name and width attached
+to that field in the descriptor. The final order is exactly
+$(a_1,\ldots,a_d)$, as required by the triangular updates above. Their target
+order and their requirement that controls precede targets are unchanged.
+The inverse uses the reversed interchange schedule after recovering the digits.
+
+The completed reversal and its inverse preserve complete records and their
+internal bit order. Padding is inserted before the forward reversal and
+removed only after the inverse reversal, so the insertion/deletion enumeration
+is unchanged. Since $d\le\log_2T$, the named-axis list and all collapsed
+lengths have polynomial length in $\log(2Tp)$. Each call descriptor can thus
+be prepared in $\operatorname{poly}(\log(2Tp))=O(Tp)$ time. The $d$ controlled
+rotations and the padding scans cost $O(dTp)$ as above. Therefore $\Phi$,
+its inverse, and their padding operations cost
+\[
+ O\bigl(Tp\,d(1+\ell^\tau)\bigr).
 \]
 This construction describes payload movement from address $a$ to its
 new address.  It is distinct from the operator convention
@@ -558,7 +572,7 @@
 \subsection{Precision and exact recovery}
 
 Proposition~\ref{prop:simultaneous-layer} uses
-$C_1=20$, so $\epsilon C_1<1$ makes its coefficient width $O(p)$.
+$C_1=4961/1000$, so $\epsilon C_1<1$ makes its coefficient width $O(p)$.
 Lemma~\ref{lem:synthetic-transform-cost} therefore applies to every
 synthetic transform here, with error less than
 $\sqrt2\ell\,2^{-p}<L2^{-p}$, where $L=\log_2T$, for all
@@ -670,7 +684,7 @@
 Conjugation preserves the grid and norm, so the opposite procedure
 has the same error and disk guarantee.
 All these margins may be enforced by one fixed cutoff.  For example,
-$b\ge2^{24}$, together with the other eventual construction conditions,
+$b\ge2^{40}$, together with the other eventual construction conditions,
 makes $dp^2h_p$, $8TLh_p$, $E_sh_p$ and
 $28\,2^\gamma T^2Lh_p$ smaller than $10^{-3}$.
 
@@ -834,8 +848,8 @@
 Prefix-slot moves and individual butterfly rounds & $dK$ & $\epsilon(1+c)$\\
 Chunk exchanges for transform layout & $pK^{\tau-1}$ & $1-\epsilon c(1-\tau)$\\
 Simultaneous butterfly rounds & $\ell d^{\lambda'}$ & $1-\epsilon(1-\lambda')$\\
-CRT and axis layouts & $d^2(1+\ell^\tau)$ & $\tau+\epsilon(2-\tau)$\\
-Gaussian line maps & $dp^{1/2+\delta}\alpha$ & $3/4+\delta+3\epsilon/2$\\
+CRT and axis layouts & $d(1+\ell^\tau)$ & $\tau+\epsilon(1-\tau)$\\
+Gaussian line maps & $dp^{1/2+\delta}\alpha$ & $3/4+\delta+5\epsilon/4$\\
 Chirps, twists and scalar products & $dp^\delta$ & $\epsilon+\delta$\\
 Packed polynomial products & $\log(rp)$ & $1-\epsilon$\\
 Final scaling and rounding & $p^\delta$ & absorbed above\\
@@ -853,20 +867,28 @@
  g_1&=1-\epsilon(1+c),\qquad
  g_2=\epsilon c(1-\tau),\qquad
  g_3=\epsilon(1-\lambda'),\\
- g_4&=1-\tau-\epsilon(2-\tau),\qquad
- g_5=1/4-\delta-3\epsilon/2,\\
+ g_4&=(1-\tau)(1-\epsilon),\qquad
+ g_5=1/4-\delta-5\epsilon/4,\\
  g_6&=1-\delta-\epsilon,\qquad g_7=\epsilon.
 \end{split}
 \end{equation}
-Direct substitution gives
-\[
- g_1>\frac12,\quad g_2=2^{-181},\quad g_3=2^{-129},\quad
- g_4>2^{-51},\quad g_5>\frac18,\quad g_6>\frac12,\quad g_7=2^{-75}.
-\]
-Thus $\min_i g_i=2^{-181}=2\kappa$.  This explicit slack will cover
-any remaining fixed power of $\log p$.
-
-Here $d^2$ is absorbed by $d^2\ell^\tau$, and
+Exact substitution in the unchanged seven-term assembly accounting gives
+\[
+ G_*:=\min_i g_i=g_3=\epsilon(1-\lambda')
+       =\frac{2332833}{2500000000000000}>2^{-30}=\kappa.
+\]
+The new reservation, compact movement and repair costs are all included
+in the completed-layer term $g_3$. Put $\rho=G_*-\kappa>0$; this fixed
+strict gap absorbs all remaining fixed powers of $\log p$.
+The guard is $O(d^{4961/1000})=o(p)$, while
+$\alpha=\Theta(p^{11999/40000})$ and
+$\gamma=O(p^{15997/20000})=o(p)$. Also
+$K=\Theta(p^{1999/10000})=o(\ell)$, $K/\log p\to\infty$,
+and $\ell=\Theta(p^{8001/10000})$. The prime-interval ratio grows as
+$p^{3001/5000}$. All construction and precision conditions hold beyond
+a common fixed cutoff.
+
+Here $d$ is absorbed by $d\ell^\tau$, and
 $\log(rp)=O(p^{1-\epsilon}+\log p)=O(p^{1-\epsilon})$.
 The first three rows are the terms in
 \eqref{eq:synthetic-transform-cost}. The prefix-slot moves and
@@ -883,9 +905,12 @@
 deterministic repair, row padding and removal, all base-$m$ groups,
 and fixed-tape cleanup.
 
-Axis exposure and restoration require $O(d^2)$ adjacent axis moves.
-Each costs $O(V(1+\ell^\tau))$: peel a single excess bit when widths
-differ by one, exchange their equal-width parts, and restore the bit.
+Axis exposure and restoration require $O(d)$ nonadjacent full-field
+interchanges by Lemma~\ref{lem:tensor-resampling}; the CRT reversal also
+uses $O(d)$ such interchanges. Each costs $O(V(1+\ell^\tau))$, using the
+arbitrary spectator gap and the one-bit-width extension of
+Lemma~\ref{lem:chunk-swap}. Thus this row has exponent
+$\epsilon+\tau(1-\epsilon)$ rather than $2\epsilon+\tau(1-\epsilon)$.
 The $d$ CRT rotations copy valid and invalid intervals in $O(dV)$ time.
 Their offset setup was bounded above by $O(dTp^C/r)=o(Tp)$,
 using at most $T/t_i$ control prefixes for target $i$.
@@ -898,7 +923,7 @@
  =O(dTp^{3/2+\delta}\alpha).
 \]
 Dividing by $V=Tp$ gives the fifth row.  The factor
-$\alpha=\Theta(p^{1/4+\epsilon/2})$ explains its displayed power of $p$.
+$\alpha=\Theta(p^{1/4+\epsilon/4})$ explains its displayed power of $p$.
 
 Chirp numerators have $O(p)$ bits and can be computed modulo $2r$ as
 sums of $d$ terms $s_i j_i^2(r/t_i)$.  Computing their squares and
@@ -960,29 +985,19 @@
 does scan the axis descriptors at every coefficient; its
 $O(Tdp^{1+\delta})$ cost was included in the chirp row.
 
-For one packed change of basis, let $V_{\rm call}$ be its current
-logical volume.  The bad-address fraction is
-$O(d2^{-K})$, independently of the payload.  Each polynomial record
-has $\Theta(rp)$ payload bits, and an exceptional record adds an
-$O(p)$-bit destination key.  Thus the compact exceptional stream has
-length
-\[
- O\!\left(\frac{V_{\rm call}}{rp}\,d2^{-K}(rp+p)\right)
-   =O(V_{\rm call}d2^{-K}).
-\]
-Stable bit sorting makes $O(p)$ scans of this stream, for one-call
-cost $O(V_{\rm call}dp2^{-K})$.  Extraction and reinsertion use a
-constant number of scans of the full current volume; those scans are
-already in the node overhead.  Since there are at most $p^A$ visits,
-each of volume $O(V)$, the compact sorting costs sum to at most
-\[
- O\!\left(dp2^{-K}\sum_{\rm calls}V_{\rm call}\right)
-   =O(Vp^{A'}2^{-K})=o(V)
-\]
-for another fixed exponent $A'$.  Here $d=O(p^\epsilon)$ and
-$K=\Theta(p^{\epsilon c})$ with $\epsilon c>0$.  Predicate and key
-arithmetic is the polynomial work per record counted above.  This is
-a deterministic worst-case bound.
+For a compact-control change of basis of current volume $V_{\rm call}$,
+the exceptional fraction is at most $5/(128p^3)$ beyond the common cutoff.
+Sorting its keyed exceptional records costs
+\[
+ O\bigl(V_{\rm call}\delta p(1+1/r)\bigr)=O(V_{\rm call}p^{-2}).
+\]
+Extraction and reinsertion cost $O(V_{\rm call})$, and polynomial predicate
+and key work per record is absorbed by the whole polynomial record suffix.
+These are already charged at \emph{each node} in the compact-control
+recurrence. Their sum is bounded by that volume-weighted recurrence and
+therefore by the completed-layer cost in the table. No claim that their
+global total is $o(V)$ is needed, and the old global estimate using an
+exponentially small $2^{-K}$ bad fraction is not used for this new gadget.
 
 Finally, all role, arithmetic, and stack tapes form fixed finite
 families.  A child parks and restores only its parent's streams, with
@@ -992,8 +1007,8 @@
 alphabet symbols.  Global setup and prime search cost $o(V)$, as
 proved earlier.
 
-Each of the seven displayed powers is at most $1-2\kappa$ by
-\eqref{eq:margin-list}.  Any fixed power of $\log p$ is at most $p^\kappa$
+Each of the seven displayed powers is at most $1-G_*$ by
+\eqref{eq:margin-list}. Any fixed power of $\log p$ is at most $p^\rho$
 eventually.  The complete large-input bound is consequently
 \[
  O(Tp\,p^{1-\kappa})=O(n(\log n)^{1-\kappa}).
--- a/build/sections/03-motifs.tex
+++ b/build/sections/03-motifs.tex
@@ -1,3 +1,5 @@
+% Modified by Douglas Colkitt, October 7, 2026.
+% Conditional compact-control, complex-compression and ternary research draft; see main.tex for scope.
 \section{Two finite networks with a rank saving}
 \label{sec:finite-networks}
 
@@ -13,8 +15,8 @@
 
 Put
 \[
- h=100,\qquad v=\binom h3=161700,\qquad
- N=v^3=4227952113000000,\qquad m=h^3=1000000.
+ h=50,\qquad v=\binom h3=19600,\qquad
+ N=v^3=7529536000000,\qquad m=h^3=125000.
 \]
 Let $\mathcal T$ be the set of three-element subsets of $[h]$.
 There are two data banks $X_a,Y_a$, indexed by
@@ -34,7 +36,7 @@
 Each such invocation has its own auxiliary wires, distinct from all
 other invocations, including those in other stages.  There are
 \[
- I=3v^2=78440670000
+ I=3v^2=1152480000
 \]
 invocations.  In a forward invocation, write $x_T$ for the data used as
 controls and $y_S$ for the data to be updated; $S,T\in\mathcal T$ are
@@ -127,19 +129,19 @@
 
 A fixed triple has
 \[
- z_{\rm b}=3\binom{97}{2}=13968,\qquad
- z_{\rm c}=\binom{97}{3}+3\cdot97=147731
+ z_{\rm b}=3\binom{47}{2}=3243,\qquad
+ z_{\rm c}=\binom{47}{3}+3\cdot47=16356
 \]
 neighbors in the two constructions, respectively.  The second count
 separates intersection zero and intersection two.  Ordered pairs must
 be counted: there are $vz$ side wires in each invocation, not $vz/2$.
-With $c_{\rm b}=100$ and $c_{\rm c}=101$ central wires per invocation,
+With $c_{\rm b}=50$ and $c_{\rm c}=51$ central wires per invocation,
 the total wire counts are
 \begin{align}
  W_{\rm b}&=2N+I(vz_{\rm b}+c_{\rm b})
-          =177176569091445000000,\label{eq:bit-wire-count}\\
+          =73269972440000000,\label{eq:bit-wire-count}\\
  W_{\rm c}&=2N+I(vz_{\rm c}+c_{\rm c})
-          =1873807244643542670000.\label{eq:complex-wire-count}
+          =369474390296480000.\label{eq:complex-wire-count}
 \end{align}
 All these numbers are fixed, independently of any eventual input length.
 
@@ -212,7 +214,7 @@
  \langle x,y\rangle=x^{\mathsf T}(I-J/9)y,
 \]
 where $J$ is the all-one matrix.  This form has eigenvalues $1$ and
-$1-h/9=-91/9$, so it is nondegenerate.  For a triple $T$ let $t_T$ be
+$1-h/9=-41/9$, so it is nondegenerate.  For a triple $T$ let $t_T$ be
 its indicator vector.  Then
 \[
  \langle t_S,t_T\rangle=|S\cap T|-1,
@@ -225,7 +227,7 @@
 The scalar domain need not equal the label field.
 These degree-one intersection formulas specialize the construction
 described by Alon~\cite[Section~3, remark after Theorem~1.1]{Alon1998}
-to the prime $2$, with the ground-set size changed to $h=100$.
+to the prime $2$, with the ground-set size changed to $h=50$.
 
 The ambient label space is $\mathcal F=F^{\otimes3}$ with its tensor
 bilinear form.  It has dimension $m$ and is nondegenerate.  Write
@@ -458,8 +460,8 @@
 
 The loss ratios have the exact values
 \[
- \frac{L_{\rm b}}{N}=\frac{100}{539},\qquad
- \frac{L_{\rm c}}{N}=\frac{101}{539}.
+ \frac{L_{\rm b}}{N}=\frac{75}{196},\qquad
+ \frac{L_{\rm c}}{N}=\frac{153}{392}.
 \]
 In particular, both are smaller than $1/2$.  We now turn the dimension
 count into the two interfaces needed later.
@@ -503,7 +505,7 @@
  &=\sum_{e}\operatorname{rank}_{\mathbb Q}
       (M_{\mathrm{head}(e)}-M_{\mathrm{tail}(e)})\\
  &=W_{\rm b}m-N+2L_{\rm b}
-  =177176569088785861287000000<W_{\rm b}m.
+  =9158746553232864000000<W_{\rm b}m.
 \end{align}
 \end{proposition}
 
@@ -574,7 +576,7 @@
 The sum of these basis lengths is
 \[
  s_{\rm c}=W_{\rm c}m-2N+2L_{\rm c}
- =1873807244636671267308000000<W_{\rm c}m.
+ =46184298777878576000000<W_{\rm c}m.
 \]
 After source and sink diagonal corrections, and undoing the signed
 bank exchange, telescoping these phase frames gives $C^{\otimes m}$
@@ -668,33 +670,666 @@
 the required forward tensor product.
 \end{proof}
 
-\subsection{Explicit rational bounds for the two recurrences}
-
-The relative rank deficits are
-\[
- \eta_{\rm b}=\frac{W_{\rm b}m-s_{\rm b}}{W_{\rm b}m}
-       =\frac{339}{22587335000000},\qquad
- \eta_{\rm c}=\frac{W_{\rm c}m-s_{\rm c}}{W_{\rm c}m}
-       =\frac{73}{19906842167500}.
-\]
-For both recurrences we may use the fixed rational exponent
-\begin{equation}
- \tau=\sigma=1-2^{-50}.
- \label{eq:explicit-motif-exponents}
+\subsection{A paired-block circuit for the bit correction}
+\label{sec:paired-bit-construction}
+Use $h=50$, $v=\binom h3=19600$, $N=v^3$, and $m=h^3=125000$.
+The complex network above is retained at this ground size. For the bit
+network, we replace the intersection-one side correction by a smaller circuit.
+Fix a common point $i$. On the other $n=h-1=49$ points, inputs $x_{ab}$
+represent source triples $\{i,a,b\}$. The required partial output is
+\[
+ z_{cd}=\sum_{\{a,b\}\cap\{c,d\}=\varnothing}x_{ab}.
+\]
+All scalar additions are in $\mathbb F_2$, but the computation below always
+adds disjoint formal input supports.
+
+It is useful to solve a weighted version. Give a graph an edge value $e_{uv}$
+and a vertex value $w_u$, with pairwise disjoint supports. Return its total
+value and the values left after deleting each one or two vertices, including
+all their incident edges. At the root, take $w_u=0$ and $e_{uv}=x_{uv}$.
+Group the ordered vertices consecutively into pairs, with a final singleton
+when necessary. A block $A$ becomes a vertex of a smaller graph with weight
+\[
+ \widetilde w_A=\sum_{u\in A}w_u+\sum_{\{u,v\}\subset A}e_{uv},
+ \qquad
+ \widetilde e_{AB}=\sum_{u\in A,\ v\in B}e_{uv}.
+\]
+Recursively compute the smaller graph's total, its value $O_A$ after deleting
+block $A$, and its value $F_{AB}$ after deleting both $A,B$.
+
+For $a\in A$, put
+\[
+ C_a=\sum_{u\in A\setminus\{a\}}w_u,\qquad
+ E_{a,B}=\sum_{u\in A\setminus\{a\},\ v\in B}e_{uv},\qquad
+ H_{a;B}=C_a+\sum_{C\ne A,B}E_{a,C}.
+\]
+Compute all $H_{a;B}$ by the leave-one-out prefix/suffix routine on the
+vector $(C_a,(E_{a,C})_{C\ne A})$, requesting only the outputs that omit
+one of the $E$ terms. The total of this vector is denoted $H_a$.
+If $a,b$ are in different blocks $A,B$, their exclusion output is
+\[
+ F_{AB}+H_{a;B}+H_{b;A}
+   +\sum_{u\in A\setminus\{a\},\ v\in B\setminus\{b\}}e_{uv}.
+\]
+For two vertices in one block it is $O_A$; the single exclusion of $a$ is
+$O_A+H_a$. These are disjoint partitions of the surviving weighted graph.
+This proves the recursion by induction. The weights created by internal edges
+remain inside the recursion, avoiding a separate exclusion circuit for them.
+
+Use direct sums for at most four vertices. Compute other totals by balanced
+binary addition. In the four-term formula, first add $F_{AB}+H_{a;B}$,
+then add the other two terms, then combine those two results. Intern equal
+formal supports and remove unused nodes. At $n=49$, the exact circuit has
+9813 additions and 1176 designated outputs. All 1271256 nonzero coefficients,
+and every required zero, are checked by comparing all output supports with
+their definitions.
+
+Identify equal inputs and equal sums across the $h$ common-point groups.
+A sum shared by groups $i\ne j$ consists entirely of triples $\{i,j,k\}$;
+its fixed pair and variable vertices identify it exactly. Keep the first
+encountered decomposition, in common-point and then local-node order. Retain
+all $q=h\binom{h-1}{2}=58800$ partial outputs, injecting their sum into each
+physical target. Since every neighbor pair has a unique common point, this
+remains exactly the intersection-one matrix. The global graph has
+\[
+ c=450394,\qquad q=58800,\qquad c+q=509194.
+\]
+Here 40256 additions have been merged from the separate local graphs.
+
+\subsection{Reversible roles and frames in both directions}
+For every node, regard its later uses and designated outputs as outgoing edges.
+At an input node, allocate one role per outgoing edge and copy a pivot into
+the others. At an addition node, add the second incoming role into the first,
+reuse the first as one outgoing pivot, and copy it into fresh roles for the
+other outgoing edges. The other input role retires. These are invertible
+pointwise gates. There are $2c+q$ outgoing uses and $c$ pivot identifications,
+so $R=c+q=509194$ roles suffice. Roles are compiled from the merged graph;
+no simultaneously occupied physical slots are identified.
+
+Let $L$ be the product of these gates, $V$ the source copy, $J$ the injection
+of the designated partial outputs, and $G,R_0$ the original central operations.
+The chronological schedule is
+\[
+ L,\ J,\ L^{-1},\ R_0,\ V,\ G,\ R_0,\ L,\ J,\ L^{-1},\ G,\ V.
+\]
+The two injections give $JLz+JL(z+Vx)=JLVx$ for arbitrary initial scratch $z$.
+The cleanup restores $z$ and all central scratch. Since $JLV+R_0G=I$,
+the invocation is $y\gets y+x$. Reverse and invert the schedule for stage 2.
+Forward, inverse, forward invocations therefore exchange the two data banks.
+
+Use the original rational form $I-J/9$ on $F=\mathbb Q^{50}$. At a stage,
+write $A=B\mathbin\perp P$ as before, suppressing the future line factor $Q$.
+Set $D_0=B\otimes F$, $D_1=A\otimes F$, and
+$D_U=(B\otimes F)\mathbin\perp(P\otimes U)$.
+Let $U_z$ be the span of the source indicators reaching node $z$.
+Every retained node is a sum from an original common-point group, so all its
+triples contain some $i$. For a vector $u$ in that span,
+\[
+ \sum_j u_j=3u_i,\qquad
+ \langle u,u\rangle=\sum_{j\ne i}u_j^2>0\quad(u\ne0).
+\]
+Thus $U_z$ is nondegenerate. Along every graph edge $z\to w$,
+$U_z\subset U_w$. Forward middle gates use $D_{U_z}$; the physical roles
+follow paths, so these frames increase. A designated output for target $T$
+has $U_z\subset t_T^\perp$, because every contributing triple is a neighbor.
+
+For the reversed middle computation use $D_{U_z^\perp}$. The ambient form
+is nondegenerate, so all these complements are nondegenerate. Reversing an
+edge reverses inclusion: $U_w^\perp\subset U_z^\perp$.
+The original output begins at the physical $X$ line
+$\langle t_T\rangle\subset U_z^\perp$; an original input ends at exactly
+its physical $Y$ complement $t_S^\perp$. No positivity assumption on the
+reachable-target spans is needed. At every gate all incident roles receive
+the same frame. Early mixers use $D_0$, and late cleanup uses $D_1$.
+Newly active roles enter from $D_0$ and retired roles grow to $D_1$.
+Tensoring by $Q$ gives the original data-stage boundaries in both orientations.
+The only decreases remain the central returns, each of dimension $h$.
+
+Share the complete auxiliary bank of a stage-1 invocation $(A,B)$ with stage 3
+at $(B,\pi(A))$, matching local role indices. Here $\pi$ is a bijection of
+triples with $|T\cap\pi(T)|=1$: pair consecutive ground points; for a triple
+with a full pair and singleton, keep the singleton and cycle the full pair
+among pairs not containing it; otherwise keep the point in the least indexed
+pair and flip the other two to their partners. Each rule is bijective on its
+class. All invocations restore arbitrary scratch, and the joining labels are
+\[
+ E=F\otimes\langle t_A\rangle\otimes\langle t_B\rangle,
+ \qquad H=\langle t_B\otimes t_{\pi(A)}\rangle^\perp\otimes F.
+\]
+They are nondegenerate and $E\subset H$ by the chosen orthogonality.
+Each join removes one auxiliary role and exactly $m$ of edge rank.
+
+\begin{proposition}[Paired-block bit interface]\label{prop:paired-bit-interface}
+The new bit network restores arbitrary scratch, exchanges the data banks,
+and satisfies the original all-role endpoint identity with
+\begin{align*}
+ W_{\rm b}^{\rm pair}&=2N+2v^2(R+h)=406321422080000,\\
+ s_{\rm b}^{\rm pair}&=W_{\rm b}^{\rm pair}m-N+6v^2h^2
+                         =50790175992864000000,\\
+ W_{\rm b}^{\rm pair}m-s_{\rm b}^{\rm pair}&=1767136000000>0.
+\end{align*}
+\end{proposition}
+\begin{proof}
+The scalar schedule and both frame assignments were verified above. The total
+decreasing dimension is $3v^2h^2$. Telescoping gives total absolute label change
+$W_{\rm b}^{\rm pair}m-2N+6v^2h^2$; the unchanged negative source projections
+add $N$ rational rank. Data endpoints are unchanged and every surviving
+auxiliary role has endpoints zero and $I_m$, with its scalar value restored.
+The original common-frame identity and rational-shear transfer apply. All
+circuits, rational bases, and gates are fixed finite data.
+\end{proof}
+
+\subsection{Explicit exponents at ground size fifty}
+Take $a=296/10^{11}$, $a_{\rm c}=1/10^{11}$, and $L_0=11737/1000$.
+The exact deficits satisfy
+\[
+ \eta_{\rm b}^{\rm pair}=\frac{23}{661055000}>aL_0,\qquad
+ \eta_{\rm c}=\frac{239}{1202215191250}>a_{\rm c}L_0.
+\]
+For $1\le x\le2$, put $z=(x-1)/(x+1)$,
+$S(x)=2\sum_{j=0}^{23}z^{2j+1}/(2j+1)$, and
+$E_0(x)=2z^{49}/(49(1-z^2))$. Then
+$S(x)\le\log x\le S(x)+E_0(x)$, and exact rational comparison gives
+\[
+ \log m\le16(S(2)+E_0(2))+S(m/2^{16})+E_0(m/2^{16})<L_0.
+\]
+The inequality $e^{-x}>1-x$ therefore proves the required strict bounds with
+\begin{equation}\label{eq:paired-intermediate-exponents}
+ \tau_{\rm pair}=1-\frac{296}{10^{11}},\qquad \sigma_{50}=1-\frac1{10^{11}}.
 \end{equation}
-Both exact deficits displayed above are greater than $20\cdot2^{-50}$.
-Since $m<2^{20}$ and $\log 2<1$, we have $\log m<20$, whence
-\[
- m^{1-2^{-50}}
- =m\exp(-2^{-50}\log m)
- >m(1-20\cdot2^{-50})>m(1-\eta)
-\]
-for either $\eta=\eta_{\rm b}$ or $\eta=\eta_{\rm c}$.
+The complex network itself is unchanged apart from the ground size $h=50$;
+its construction, binary residual bases, and scalar corrections remain those
+proved above.
+
+\subsection{Weighted side computations with binary phase frames}
+\label{sec:complex-compression}
+We retain the complex motif's data banks, central operations, three tensor
+stages, and endpoint corrections. Scalars lie in $\mathbb Z[i,1/2]$ and
+labels in $F=\mathbb F_2^h$ with its standard dot product. Throughout this
+construction $h\ge8$ is even. Write $\mathcal T=\binom{[h]}3$,
+$v=\binom h3$, $N=v^3$ and $m=h^3$.
+
+For triples $S,T$, the central operation has coefficient
+$(|S\cap T|-1)/2$. Its side correction has coefficient $+1/2$ for
+disjoint triples and $-1/2$ for intersection-two triples. Intersection-one
+pairs need no side correction; the diagonal central coefficient is one.
+We compress these two side relations separately. All coefficients below
+are characteristic-zero coefficients; binary labels do not turn the scalar
+arithmetic into XOR arithmetic.
+
+\subsubsection{A reusable weighted circuit transfer}
+An all-ones rectangle with $a$ sources and $b$ targets uses $a+b-1$ roles.
+Allocate $a$ input slots and $b$ output slots sharing one pivot. Add the
+other inputs into the pivot, then add it into the other outputs. This
+invertible mixer $L$ has an all-ones output-by-input submatrix. Its inverse
+undoes these additions in reverse order. A fixed dyadic rectangle weight
+is placed in the output injection $J$, not in the binary labels.
+
+More generally, a circuit of disjoint-support binary sums with $c$ addition
+nodes and $q$ output uses has an invertible embedding on $c+q$ roles.
+Allocate one outgoing role per use of each input. At an addition, reuse
+its first incoming role as one outgoing pivot, allocate the other outgoing
+roles freshly, add the second input to the pivot, and fan it out. The
+second incoming role retires. There are $2c+q$ uses and $c$ role
+identifications, proving the count. Starting from zero side values,
+disjoint input supports give the exact integer coefficients, not merely
+their reductions modulo two. Equal sums may have shared fanout.
+
+Let $V$ copy data into the input slots, let $J$ inject weighted outputs,
+and retain the central gather $G$ and scatter $R$. The signed chronological
+schedule is
+\[
+ L,\ -J,\ L^{-1},\ -R,\ V,\ G,\ R,\ L,\ J,\ L^{-1},\ -G,\ -V.
+\]
+An arbitrary side vector $z$ contributes $-JLz+JL(z+Vx)=JLVx$.
+The two inverse mixers and final negative copy restore $z$. Similarly the
+centers contribute $RGx$ and return to their original values. Thus the
+invocation is $y\gets y+x$ precisely when $JLV+RG=I$. Reversing and
+inverting the signed operations implements $y\gets y-x$.
+
+For the frame transfer use the original decomposition $A=B\perp P$,
+where $P$ is the earlier tensor-indicator line, and suppress the future
+norm-one tensor line $Q$. Put
+\[
+ D_0=B\otimes F,\quad D_1=A\otimes F,\quad
+ D_U=(B\otimes F)\perp(P\otimes U),
+\]
+\[
+ X_t=D_{\langle t\rangle},\qquad Y_t=D_{t^\perp}.
+\]
+The forward schedule's common gate frames, in the displayed order, are
+\[
+ D_0,D_0,D_0,D_0,X_t,D_1,D_0,D_U,Y_t,D_1,D_1,D_1.
+\]
+Copy and injection gates are grouped by data triple. In a rectangle the
+middle mixer has one common $U$. In a sum circuit it has one $U$ per node.
+For the inverse invocation with logical banks exchanged, the frames are
+\[
+ D_0,D_0,D_0,X_t,D_U,D_1,D_0,Y_t,D_1,D_1,D_1,D_1.
+\]
+The inverse middle mixer uses physical-$X$ target supports reachable from
+each forward node. These grow along reversed edges; forward source supports
+must not be substituted for them.
+
+A valid assignment must make every side-role transition increasing and
+nondegenerate, with each nonzero orthogonal residual nonalternating. Merely
+having nested source sets is insufficient over $\mathbb F_2$. The following
+two constructions explicitly meet this stronger condition in both directions.
+
+\subsubsection{Disjoint triples: coordinate enclosures}
+Partition the ordered disjoint-triple relation into rectangles
+$\mathcal S\times\mathcal U$. Their ground-point unions are disjoint.
+In a forward middle mixer choose its frame $U$ as follows:
+\begin{itemize}
+\item If $\mathcal S=\{S\}$, use $U=\langle t_S\rangle$.
+\item Otherwise, if $\mathcal U=\{T\}$, use $U=t_T^\perp$.
+\item Otherwise, use the coordinate subspace on $\bigcup\mathcal S$.
+\end{itemize}
+For the inverse invocation interchange the two families in this rule.
+In every case $\langle t_S\rangle\subset U\subset t_T^\perp$ for
+every rectangle pair. These spaces are nondegenerate. In the two singleton
+cases the only new nontrivial local complement is that of two orthogonal
+triple lines. It has a norm-one coordinate vector outside their six-point
+union, since $h>6$.
+
+In the third case both unions have at least four points. The complement
+of any source triple line within $U$ contains a coordinate unit outside
+that triple. The complement of $U$ within $t_T^\perp$ contains a coordinate
+unit in the target union outside $T$. The spaces $U$ and $U^\perp$ are
+coordinate spaces. Consequently all nonzero residuals in
+$0\subset\langle t_S\rangle\subset U\subset t_T^\perp\subset F$,
+and all skipped transitions actually used by input-only or output-only
+roles, have norm-one vectors. Every such nondegenerate residual has an
+orthonormal basis by the retained binary-space lemma.
+
+The partition is explicit. For disjoint $a$- and $b$-subsets of $[n]$,
+$0\le a,b\le3$, allow row stars, column stars, and a split into $l$ and
+$n-l$ points. Split cases by the source and target cardinalities in the
+first part, and use the product of the already chosen smaller partitions.
+For a plan let $(C,L,R)$ be its rectangle count and source/target membership
+counts. Products multiply these counts coordinatewise and cases add them.
+An empty relation has zero counts; if one subset size is zero, use the full
+rectangle. Try all split sizes and choose the least $L+R-C$, breaking ties
+by row stars, then column stars, then increasing split size. This finite
+recursion is a construction, not a global rectangle optimality claim.
+
+At $h=26$, the counts are
+\[
+ (C,L,R)=(24870,217255,299571),\qquad R_0=L+R-C=491956.
+\]
+The exact checker verifies that each of the $4604600$ ordered disjoint
+triple pairs appears once. It also checks the frame-case hypotheses for
+every generated rectangle. Every such rectangle has scalar weight $+1/2$.
+
+\subsubsection{Intersection two: an orthonormal sum compiler}
+Fix the unique common pair $P=\{i,j\}$. The source and target triples
+are $P\cup\{a\}$ and $P\cup\{b\}$ with $a\ne b$ outside $P$.
+The binary vectors
+\[
+ t_a=e_i+e_j+e_a\quad(a\notin P)
+\]
+are pairwise orthonormal. Since $h$ is even, adjoining
+\[
+ r_i=e_i+\sum_{a\notin P}e_a,\qquad
+ r_j=e_j+\sum_{a\notin P}e_a
+\]
+gives an orthonormal basis of the entire $F$: each extra vector has odd
+weight $h-1$, their mutual product is $h-2=0$ in $\mathbb F_2$, and both
+are orthogonal to every $t_a$.
+
+Compute each leave-one-out sum with prefix and suffix sums on the $n=h-2$
+inputs. Omit unused totals and intern equal input-support sets. The active
+circuit has $3n-6$ additions and $n$ outputs, hence $4n-6$ roles. Each sum
+is cancellation-free over the integers. Inject each output with weight
+$-1/2$. Distinct triples intersecting in two points have a unique common
+pair, so these circuits give every required coefficient exactly once.
+
+At a forward node let $U$ span its ancestor vectors $t_a$; in the inverse
+middle circuit let $U$ span the reachable output vectors $t_b$. Nested
+support differences are spanned by subsets of an orthonormal family.
+No source can reach its excluded output. At the final transition into
+the relevant $t_b^\perp$, the residual has an orthonormal basis consisting
+of the remaining $t_a$ and the two extra vectors $r_i,r_j$. Cleanup
+complements have the same property. Thus every required residual is
+zero or nonalternating and nondegenerate in both directions.
+
+This argument applies to any disjoint-support circuit for the same relation,
+not only prefix/suffix sums. It is a reusable phase-frame transfer criterion.
+At $h=26$ the role count is
+\[
+ R_2=\binom{26}{2}(4\cdot24-6)=29250,
+ \qquad R_{\rm side}=R_0+R_2=521206.
+\]
+
+\subsection{Whole-network composition and accounting}
+All local frames above are tensored with the original future line $Q$.
+The original data boundaries remain
+$A\otimes\langle t\rangle\otimes Q$ to $D_1\otimes Q$ on $X$ and
+$B\otimes\langle t\rangle\otimes Q$ to $Y_t\otimes Q$ on $Y$.
+Their residuals and the original tensor complements have norm-one witnesses
+as in the retained proof. Tensoring nonalternating nondegenerate factors
+preserves a norm-one witness. All new side transitions are increasing.
+Only the $h+1$ centers decrease once per invocation, by dimension $h$.
+Thus $L=3v^2h(h+1)$ is unchanged.
+
+Pair consecutive ground points and put $\pi(T)=\{x\mathbin\oplus1:x\in T\}$.
+This is an involution of triples. A triple with a full ground-point pair
+intersects its image in two points; otherwise the intersection is empty.
+Hence $t_T\perp t_{\pi(T)}$. Match the complete stage-1 auxiliary bank
+at fixed triples $(A,B)$ to the stage-3 bank at $(B,\pi(A))$, preserving
+each local role index. These banks have equal size, the matching is bijective,
+and no simultaneous roles are identified. Each invocation restores arbitrary
+scratch, so its scalar correctness is preserved.
+
+All auxiliary first frames are $D_0\otimes Q$ and all last frames are
+$D_1\otimes Q$. The joining labels are therefore
+\[
+ E=F\otimes\langle t_A\rangle\otimes\langle t_B\rangle,
+ \qquad H=\langle t_B\otimes t_{\pi(A)}\rangle^\perp\otimes F.
+\]
+They are nondegenerate and $E\subset H$. The residual $H\cap E^\perp$
+contains a norm-one coordinate tensor whose second coordinate lies outside
+$A\cup\pi(A)$; such a coordinate exists because $h>6$. Thus the join
+also has an orthonormal residual basis. Replacing the old edges
+$E\to F^{\otimes3}$ and $0\to H$ by $E\to H$ saves exactly $m$
+residual dimensions per removed role and introduces no decrease.
+
+Consequently the entire network has
+\[
+ W=2N+2v^2(R_{\rm side}+h+1),\qquad s=Wm-2N+2L.
+\]
+Forward, inverse, forward stages send $(X,Y)$ to $(-Y,X)$ and restore every
+surviving auxiliary role. Data endpoint labels and all scratch endpoints
+$0,F^{\otimes3}$ are unchanged. The retained weight-27 phase corrections,
+signed-bank correction and return permutation therefore give the same
+$C^{\otimes m}$ operator on every role, for any number of columns. Each
+orthonormal residual contributes precisely its dimension in forward or
+inverse translation kernels. These are the full complex-interface obligations.
+
+At $h=26$,
+\[
+ m=17576,\quad v=2600,\quad N=17576000000,\quad L=14236560000,
+\]
+\[
+ W=7082222160000,\quad s=124477130005280000,
+ \quad Wm-s=6678880000,\quad \eta=\frac{19}{354111108}.
+\]
+Exact logarithm bounds give $\eta>(5/10^9)\log m$, and hence
+$s/W<m^{1-5/10^9}$. Thus this construction supports
+$a_{\rm c}=1-\sigma=5\cdot10^{-9}$. The exact saving is approximately
+$5.48945\cdot10^{-9}$. The finite even-$h$ screen is not an all-$h$ ceiling.
+
+
+\begin{proposition}\label{prop:compressed-complex-interface}
+The complex motif interface holds with
+$m_{\rm c}=17576$, $W_{\rm c}=7082222160000$ and
+$s_{\rm c}=124477130005280000$, and strict exponent $\sigma=1-5/10^9$.
+Its scalar circuit restores arbitrary auxiliary inputs; after phase-frame
+telescoping and endpoint corrections, the network implements the normalized
+$C^{\otimes m_{\rm c}}$ operator on every role, for any number of columns,
+using $s_{\rm c}$ forward or inverse translation kernels.
+\end{proposition}
+\begin{proof}
+The construction above supplies the signed scalar exchange, nested binary
+frames with orthonormal residual bases, shared-bank joins and unchanged
+endpoint corrections. Its displayed residual accounting and logarithm
+inequality give the claimed constants and strict exponent.
+\end{proof}
+
+\subsection{A ternary five-subset network}
+\label{sec:ternary-five-subsets}
+
+This section changes the finite scalar alphabet for the bit interchange
+network. It retains the rational-frame contract and the compact-control
+movement proof. The complex network and its guard constants are unchanged.
+The result remains conditional on the pinned upstream manuscript and the
+previously identified compact-control and Gaussian extensions.
+
+Zhihao Chen's earlier
+\href{https://github.com/CrocSwap/integer-mult-bounds/pull/7}{PR~\#7}
+introduces the same ternary five-subset motif, with a different producer and
+a stronger claimed witness. The circuit below is a separate implementation;
+no priority claim for the motif or verification of that pending PR is made.
+
+\subsubsection{The two fields and the side map}
+Let $h=29$, let $\mathcal T=\binom{[h]}5$, and put
+$n=\binom h5=118755$, $r=\binom h2=406$. Over $\mathbb F_3$ let $B$ be
+the incidence matrix of five-subsets against pairs. Then
+\[
+ C=BB^{\mathsf T},\qquad C_{ST}=\binom{|S\cap T|}{2}\pmod3.
+\]
+For $t=0,1,2,3,4,5$ the binomial residues are $0,0,1,0,0,1$.
+Consequently $C=I+A_2$, where $A_2$ is the adjacency matrix for
+intersection exactly two. The side map is $-A_2$; thus the central
+and side maps add to the identity. Only the displayed factor size $r$
+is needed, not a claim that it is a minimal factor.
+
+The rational label space is $F=\mathbb Q^h$ with form
+\[
+ H=I-\frac2{25}J.
+\]
+It is nondegenerate, since $1-2h/25=-33/25\ne0$. Write $t_T$ for the
+indicator of $T$. Then
+\[
+ t_S^{\mathsf T}Ht_T=|S\cap T|-2,\qquad
+ t_T^{\mathsf T}Ht_T=3.
+\]
+Let $P_T$ be projection onto $\langle t_T\rangle$ in this form.
+Intersection-two neighbors have mutually annihilating projections.
+The scalar characteristic and rational label field are different;
+the address prime in the matrix-shear procedure is a third, independently
+chosen fixed prime avoiding its rational denominators.
+
+\subsubsection{An aligned side sum circuit}
+For each pair $J\subset[h]$, take all source variables $x_{J\cup S}$
+with $|S|=3$ and $S\cap J=\varnothing$. For each triple $T$ disjoint
+from $J$ compute
+\[
+ z_{J,T}=\sum_{S\cap(J\cup T)=\varnothing,\ |S|=3}x_{J\cup S}.
+\]
+Inject $-z_{J,T}$ into target $J\cup T$. An ordered intersection-two
+pair has the unique common pair $J$, so these partial outputs implement
+$-A_2$ exactly. There are $q=\binom h2\binom{h-2}3=10n$ output uses.
+
+Use the deterministic disjoint-triple sum DAG of the tensor construction
+on all $h$ points. In group $J$, delete formal source variables meeting
+$J$ and retain only output triples disjoint from $J$. A zero in this
+substitution denotes an absent formal input, not a cleared physical
+scratch register. Every retained addition still has disjoint source
+supports. The fixed global partition order aligns intermediate sums
+between different common-pair groups.
+
+For completeness, the tensor sum recipe is a finite recursion on ordered
+ground sets. Let $Z(n,a,b)$ be its uninterned addition count for summing
+$a$-subset inputs over all $b$-subset disjointness queries. If $a+b>n$,
+the outputs are empty sums; if $a=0$, copy the sole input. Both have
+zero cost. If $b=0$, sum the $\binom na$ inputs at cost $\binom na-1$.
+Otherwise start with separate query sums, costing
+$\binom nb(\binom{n-b}a-1)$. For each $1\le l\le\lfloor n/2\rfloor$,
+put $r_0=n-l$ and split the ground set into its first $l$ and last $r_0$
+points. For every feasible input split $i$ and query split $j$, contract
+one factor and then the other, at cost
+\[
+ \min\left\{
+ \binom{r_0}{a-i}Z(l,i,j)+\binom lj Z(r_0,a-i,b-j),\quad
+ \binom li Z(r_0,a-i,b-j)+\binom{r_0}{b-j}Z(l,i,j)
+ \right\}.
+\]
+Feasibility requires the subset sizes to fit their factors and
+$i+j\le l$, $a-i+b-j\le r_0$. For each $j$, add the resulting terms
+for each full query: if there are $t_j>0$ feasible $i$, this costs
+$\binom lj\binom{r_0}{b-j}(t_j-1)$. Choose a split only if its total
+strictly improves the current best, testing $l$ in increasing order.
+For the displayed minimum prefer its first contraction order on a tie.
+Use lexicographic subset order and increasing split indices; sum lists
+by recursively splitting them into two halves. Intern every equal formal
+support at each addition, retaining its first decomposition. The initial
+template is the resulting $(n,a,b)=(h,3,3)$ DAG. This describes a
+cancellation-free integer coefficient computation: disjointness factors
+across the two ground sets, and the input split partitions the sources.
+
+Within a group, intern equal full formal support sets. Across groups,
+let $I$ be the intersection of all full five-subset source labels of a
+nontrivial sum. A sum shared by distinct groups has $|I|\ge3$.
+If $|I|=3$, its exact signature is $I$ and the set of residual pairs
+$S\setminus I$; if $|I|=4$, use $I$ and its residual points. These
+signatures reconstruct the full set of five-subset variables, so there
+are no hash collisions at the mathematical level. A nontrivial sum
+cannot have $|I|=5$. Sums with $|I|=2$ belong only to that common-pair
+group. Equal singleton inputs are identified by their five-subset labels.
+Retain the first decomposition of each sum and prune unused nodes after
+all groups are processed.
+
+The exact integer DAG-array construction at $h=29$ gives
+\[
+ c=19593239,\qquad q=1187550,\qquad S=c+q=20780789.
+\]
+Here $c$ counts active additions after pruning, not allocated nodes.
+The implementation independently expands every support and output row
+at $h=8,9$, checks the signature identifications there, and regenerates
+the full integer DAG counts at $h=29$. The general exact-signature
+argument above justifies the large sharing operation; the small tests
+are implementation controls, not an extrapolated proof.
+
+\subsubsection{Restoration and rational side frames}
+A cancellation-free addition DAG with $c$ additions and $q$ output uses
+has an invertible gather-and-fanout embedding on $S=c+q$ roles: each
+binary addition identifies one input pivot with an output use, while
+all other uses receive their own role. No initial side value is assumed
+zero. Let its invertible scalar operation be $L$, its data copy be $V$,
+and its signed output injection be $J$ (coefficient $-1$ at every partial
+output). With central gather $G=B^{\mathsf T}$ and scatter $R=B$, use
+\[
+ L,\ -J,\ L^{-1},\ -R,\ V,\ G,\ R,\ L,\ J,\ L^{-1},\ -G,\ -V.
+\]
+On arbitrary side input $z$, the two injections contribute
+$-JLz+JL(z+Vx)=JLVx=-A_2x$. The inverse mixers and final negative copy
+restore $z$. Central input cancels in the same way, leaving $RGx=Cx$.
+Thus $y\gets y+x$ and every auxiliary input is restored. The actual
+reverse inverse schedule gives $y\gets y-x$.
+
+Every nonzero vector $u$ in the span of five-subset indicators containing
+one fixed pair $\{a,b\}$ satisfies
+\[
+ u_a=u_b,\quad \sum_i u_i=5u_a,\quad
+ u^{\mathsf T}Hu=\sum_{i\notin\{a,b\}}u_i^2>0.
+\]
+If the last sum vanished, the first two identities would give
+$2u_a=5u_a$, hence $u=0$. Therefore every source span appearing in an
+original group is nondegenerate. Interning an equal formal sum does not
+change its span, so this remains true after cross-group sharing.
+
+For a DAG node $v$ choose the orthogonal projection $E_v$ onto its source
+span. These projections increase along forward edges. Every ancestor
+source is orthogonal to every reachable target, so the final projection
+at a partial output lies below $I-P_T$ for its target $T$. For the
+reverse mixer use $I-E_v$. Orthogonal complementation reverses the
+inclusions and gives increasing reverse edges, including the target
+lines at entry and source complements at exit. Rational nondegeneracy
+suffices; no characteristic-two orthonormal-residual assumption is used.
+
+\subsubsection{Complete tensor circuit and rank ledger}
+Index the data banks by $\mathcal T^3$, with $N=n^3$ and $m=h^3$.
+Use the first-coordinate shears, inverse second-coordinate shears with
+logical banks exchanged, and third-coordinate shears. The scalar action
+is $(X,Y)\mapsto(-Y,X)$. Negate the physical $X$ bank at its final identity
+frame to obtain the unsigned bank exchange. This fixed pointwise gate
+has zero added edge rank and leaves all restored auxiliaries unchanged.
+
+For a stage write $A=B_0\mathbin\perp P$, where $P$ is its preceding
+label tensor line, and suppress the future tensor line $Q$. Put
+\[
+ D_0=B_0\otimes I,\quad D_1=A\otimes I,\quad
+ D(E)=D_0+P\otimes E.
+\]
+The common frames of the forward twelve operations are
+\[
+ D_0,D_0,D_0,D_0,D(P_T),D_1,D_0,D(E_v),D(I-P_T),D_1,D_1,D_1.
+\]
+For the reversed inverse invocation they are
+\[
+ D_0,D_0,D_0,D(P_T),D(I-E_v),D_1,D_0,D(I-P_T),D_1,D_1,D_1,D_1.
+\]
+Copy and injection gates are grouped by their data role, including all
+partial outputs for that target. Central gates retain every data and
+central incidence, including zero factor coefficients. All frames are
+tensored with $Q$. These are the same tensor boundaries as the general
+three-stage compiler. All side changes increase; only the central return
+decreases, by $h$ on each of the $r$ central roles in each invocation.
+Thus the total decreasing dimension is
+\[
+ L_{\mathrm{dec}}=3n^2rh.
+\]
+The signed dimension telescoping gives $Wm-2N+2L_{\mathrm{dec}}$.
+Replace the physical $X$ source projections $P_a$ by $-P_a$; because
+$2\ne0$ in the rational frame field, this adds exactly $N$ to the rank
+sum. Source frames are now $-P_a,0$, sink frames $I,I-P_a$, and each
+restored auxiliary has source $0$, sink $I$. The unsigned scalar exchange
+therefore satisfies every endpoint difference $I_m$.
+
+The bipartite double cover of the intersection-two graph is regular of
+positive degree $10\binom{h-5}3$. Hall's condition follows by edge
+counting, and a fixed deterministic matching algorithm supplies an
+orthogonal label permutation $\pi$. Share stage-one invocation $(A,B)$
+with stage-three invocation $(B,\pi(A))$, keeping local auxiliary indices.
+The boundary projections are
+\[
+ E=I\otimes P_A\otimes P_B,\qquad
+ H_0=(I-P_B\otimes P_{\pi(A)})\otimes I.
+\]
+They satisfy $EH_0=H_0E=E$. The twelve-operation schedule already has
+initial frame $D_0$ on every auxiliary, so the join replaces two boundary
+segments by one and saves exactly $m$ ranks per identified role.
 Consequently
 \[
- \frac{s_{\rm b}}{W_{\rm b}}<m^\tau,
- \qquad \frac{s_{\rm c}}{W_{\rm c}}<m^\sigma,
- \qquad 0<\tau=\sigma<1.
-\]
-Here $\sigma$ is a rational upper bound for the logarithmic exponent
-of the complex recurrence.
+ W=2n^3+2n^2(S+r),\quad
+ \Delta=Wm-s=n^3-6n^2rh.
+\]
+Numerically,
+\[
+\begin{split}
+ m&=24389,\qquad N=1674772079218875,\\
+ W&=589493540769997500,\\
+ L_{\mathrm{dec}}&=498137336383050,\\
+ \Delta&=678497406452775,\\
+ s&=14377157287342062574725.
+\end{split}
+\]
+Writing $\eta=\Delta/(Wm)$, the exact logarithm enclosure verifies
+\[
+ \eta>\frac{467}{10^{11}}\log m.
+\]
+Since $1-\eta<e^{-\eta}$, this proves
+$s/W<m^{1-a_{\rm b}}$ with $a_{\rm b}=467/10^{11}$.
+The finite-alphabet transfer therefore supplies the bit interchange
+exponent $\tau=1-467/10^{11}$.
+
+
+\begin{proposition}\label{prop:ternary-bit-interface}
+There is a fixed pointwise circuit over $\mathbb F_3$ satisfying the
+rational finite-shear contract on every scalar role, with
+\[
+ m_{\rm b}=24389,\quad W_{\rm b}=589493540769997500,\quad
+ s_{\rm b}=14377157287342062574725,
+\]
+and strict exponent $\tau=1-467/10^{11}$. Every auxiliary input is restored. Via
+Section~\ref{sec:finite-alphabet-transfer}, it supplies the bit chunk
+interchange interface with this exponent on the same fixed-tape model.
+\end{proposition}
+\begin{proof}
+The construction above proves the scalar permutation, all rational
+endpoint identities, full edge-rank accounting and the strict logarithm
+comparison. The finite-alphabet extension applies to its ternary symbols
+and returns the original bit payload at completed interchange boundaries.
+\end{proof}
+
+The bit network used below has $m_{\rm b}=24389$ by Proposition~\ref{prop:ternary-bit-interface}. Its swap interface works at
+arbitrary binary widths; the complex recursion uses that interface's
+exponent and fixed tape guarantee, not its arity. The complex recursion's
+branching, role split and depth use $m_{\rm c},W_{\rm c},s_{\rm c}$.
+No equality of bit and complex dimensions is required by the assembly.
+
+The exponents used from now on are
+\begin{equation}\label{eq:explicit-motif-exponents}
+ \tau=1-467/10^{11},\qquad\sigma=1-5/10^9.
+\end{equation}
--- a/build/sections/04-swap.tex
+++ b/build/sections/04-swap.tex
@@ -1,3 +1,5 @@
+% Modified by Douglas Colkitt, October 7, 2026.
+% Conditional compact-control, complex-compression and ternary research draft; see main.tex for scope.
 \section{Faster interchange of address chunks}
 \label{sec:tape-interchange}
 
@@ -104,7 +106,7 @@
 
 Fix integers $m,W\ge2$.  The circuit has $W$ wires, including its
 scratch wires, with separate input and output terminals.  Its fixed,
-pointwise bit gates induce a permutation $\rho$ of the $W$ scalar inputs
+pointwise finite-alphabet gates induce a permutation $\rho$ of the $W$ scalar inputs
 at completion: the input value at role $w$ appears at output role
 $\rho(w)$.  Assign a
 rational $m\times m$ matrix to each terminal and to each gate, the latter
@@ -121,8 +123,8 @@
  M_{\mathrm{out}(\rho(w))}-M_{\mathrm{in}(w)}=I_m
  \quad(1\le w\le W),\qquad s<Wm.
 \end{equation}
-All these data are finite and fixed. Proposition~\ref{prop:bit-motif-interface}
-supplies these conditions with $W=W_{\rm b}$ and $s=s_{\rm b}$.
+All these data are finite and fixed. Proposition~\ref{prop:ternary-bit-interface}
+supplies these conditions over $\mathbb F_3$, with $m=m_{\rm b}$, $W=W_{\rm b}$ and $s=s_{\rm b}$.
 Its negative source projection is essential to the endpoint identity and
 has already been included in the rank sum.
 
@@ -170,7 +172,7 @@
 vertex with matrix $M$ is the address permutation $\Phi_M$, and every
 edge changes it by $\Phi_{M'-M}$. The same invariant proof applies
 because a common address permutation commutes with every pointwise
-bit gate. For the physical input arrays $f_w$, the logical arrays are
+finite-alphabet gate. For the physical input arrays $f_w$, the logical arrays are
 $g_w=\Phi_{-M_{\mathrm{in}(w)}}f_w$; this only specifies their
 interpretation, with no initial tape operation. The logical circuit
 routes $g_w$ to role $\rho(w)$. Thus its physical output satisfies
@@ -270,11 +272,91 @@
 \end{proof}
 
 
-We use the rational value $\tau=1-2^{-50}$ established in
+We use the rational value $\tau=1-\frac{467}{10^{11}}$ established in
 \eqref{eq:explicit-motif-exponents}. Its verification uses fixed integer
 comparisons and the displayed exponential inequality; it needs no
 real-arithmetic oracle.
 
+
+\subsection{A fixed finite scalar alphabet in the interchange transfer}
+\label{sec:finite-alphabet-transfer}
+
+The rational-frame interchange interface does not intrinsically require
+XOR as its pointwise scalar operation. The following extension concerns
+the swap interface only; it does not change the complex phase interface.
+All conclusions remain conditional on the upstream stream procedures.
+
+\begin{proposition}
+Let $\mathcal A$ be a fixed finite alphabet containing distinct symbols
+$0,1$. Suppose a fixed finite circuit on $W$ roles consists of fixed
+pointwise maps on symbols in $\mathcal A$ and has, on every input in
+$\mathcal A^W$, the complete scalar action of a role permutation $\rho$.
+Assign the same rational $m$-by-$m$ matrix to all incidences of each gate.
+If
+\[
+ M_{\mathrm{out}(\rho(w))}-M_{\mathrm{in}(w)}=I_m,
+ \qquad s=\sum_e\operatorname{rank}_{\mathbb Q}
+ (M_{\mathrm{head}(e)}-M_{\mathrm{tail}(e)})<Wm,
+\]
+then the upstream power-width interchange proposition holds with these
+$m,W,s$ and every rational $\tau\in(0,1)$ satisfying $s/W<m^\tau$.
+It acts on bit data as well as arbitrary $\mathcal A$-valued data.
+\end{proposition}
+
+\begin{proof}
+Use the fixed tape alphabet $\mathcal A$ on the data track. Its symbols
+can alternatively be represented by a fixed number of bits per cell;
+this changes every scan cost by a fixed factor. For this induction let
+$V$ count data entries, so a role stream has exactly $V/W$ entries.
+For an original bit array this is its original logical volume. No
+conversion factor enters the recursive branching coefficient.
+
+The elementary stream procedures move whole symbols and apply fixed
+pointwise maps to aligned symbols. The machine model already permits
+fixed finite alphabets and fixed tracks. All counters, descriptors and
+address radices remain as in the upstream proof. The prime used for
+rational address shears is chosen from their finite rational tables and
+is unrelated to any field used to name the data symbols.
+
+Induct simultaneously for all $\mathcal A$-valued arrays. Base-case scans
+permute symbols. Recursive matrix-shear calls therefore remain within
+the induction hypothesis even when a surrounding gate has produced
+symbols other than $0,1$. A common address permutation commutes with
+every fixed pointwise map, without any linearity assumption on that map.
+Consequently the same frame invariant gives
+\[
+ f'_{\rho(w)}=\Phi_{M_{\mathrm{out}(\rho(w))}}
+             \Phi_{-M_{\mathrm{in}(w)}}f_w=\Phi_{I_m}f_w.
+\]
+Merging the role streams undoes $\rho$, including on all scratch-role
+payload. The two other triangular updates convert this shear into the
+desired chunk interchange. The identical recurrence is
+\[
+ T(V,e)\le\sum_{j=1}^s T(V/W,e/m)+O(V),
+\]
+with constants depending on the fixed alphabet and circuit. Restricting
+initial data to $0,1$ is now legitimate: every completed recursive
+interchange restores those data values, while its interior may use other
+symbols. No invalid binary encoding or initialized scratch symbol is
+assumed.
+
+The arbitrary-width reduction is unchanged as well. Its reserved high
+address digits provide at least $W^k$ rows, using only $O(\log e)$ digits
+for the fixed $m,W$. Row padding increases volume by less than two, and
+radix padding by a fixed factor. Use the symbol $0$ in padded entries;
+completed recursive calls are data permutations and therefore preserve
+the padded zero rows at their boundaries. Removing those rows and invalid
+address ranges consequently returns the original symbol array in the
+required order. No bound on the numerical size of $W$ relative to $m$
+is added by this reduction.
+\end{proof}
+
+In particular one may take $\mathcal A=\mathbb F_p$ for any fixed prime
+$p$. A signed exchange $(X,Y)\mapsto(-Y,X)$ is corrected to an unsigned
+exchange by negating the physical $X$ bank at its final identity frame.
+This is a fixed pointwise map with zero added edge rank. The rational
+frames and all rank charges must still be checked; changing the scalar
+alphabet by itself supplies no improvement.
 
 \subsection{Removing width and row restrictions}
 
--- a/build/sections/05-layers.tex
+++ b/build/sections/05-layers.tex
@@ -1,5 +1,13 @@
+% Modified by Douglas Colkitt, October 7, 2026.
+% Conditional compact-control, complex-compression and ternary research draft; see main.tex for scope.
 \section{Fast simultaneous synthetic butterfly layers}
 \label{sec:fast-butterfly-layers}
+Throughout this section, the complex constants are
+$m=m_{\rm c}=17576$, $W=W_{\rm c}=7082222160000$ and
+$s=s_{\rm c}=124477130005280000$ from
+Proposition~\ref{prop:compressed-complex-interface}. The bit-swap primitive
+retains its separate arity $m_{\rm b}=24389$ and exponent
+$\tau=1-467/10^{11}$.
 
 We now apply the complex network to a common butterfly layer on several
 axes. The array has complete address set
@@ -32,10 +40,10 @@
 factors would require $D$ whole-array passes.
 
 We first construct tensor products of the auxiliary two-point kernel
-$C$ from Proposition~\ref{prop:complex-motif-interface}. Its phase
+$C$ from Proposition~\ref{prop:compressed-complex-interface}. Its phase
 frames express a $C$ tensor product using smaller groups of
 selected bits, separated by binary changes of basis. We implement those
-changes by moving whole address chunks. Splitting complete rows among
+changes by moving compact dirty-control fields with exact exceptional repair. Splitting complete rows among
 the network's wire roles then reduces the volume of each recursive
 call. Finally, we bound the intermediate dyadic values and convert the
 completed $C$ layer to the normalized $H_0$ layer. The entire layer is
@@ -72,7 +80,7 @@
 
 Consider an edge from label $U_-$ to label $U_+$, and choose an
 orthonormal basis $v_1,\ldots,v_{\rho_{\rm edge}}$ of its residual.
-Proposition~\ref{prop:complex-motif-interface} gives the frame change
+Proposition~\ref{prop:compressed-complex-interface} gives the frame change
 \[
  \mathcal C_{U_+}\mathcal C_{U_-}^{-1}
  =\prod_{j=1}^{\rho_{\rm edge}}(aI+bX_{v_j})^{\epsilon_j},
@@ -110,7 +118,7 @@
 chunk $j$ of slot $h$, where $1\le h\le m$.  Column $j$ is the vector
 $(x_{1,j},\ldots,x_{m,j})$.  A frame on this group is the tensor
 product of the same phase frame
-$\mathcal C_U$ of Proposition~\ref{prop:complex-motif-interface}
+$\mathcal C_U$ of Proposition~\ref{prop:compressed-complex-interface}
 on these $f$ columns.
 
 To implement \eqref{eq:phase-edge-coordinate} on the group, first apply
@@ -146,7 +154,7 @@
 $i^{27f}$ at each affected sink, turn every endpoint kernel into a
 forward one.  Negating the prescribed physical bank
 and undoing its exchange returns each result to its original role.
-Proposition~\ref{prop:complex-motif-interface} therefore gives
+Proposition~\ref{prop:compressed-complex-interface} therefore gives
 $C^{\otimes e}$ on every role, including roles whose initial scratch
 values were arbitrary.  Formula~\eqref{eq:phase-inverse-child} supplies
 each inverse child from one forward call, and none of the corrections
@@ -179,6 +187,9 @@
 
 \subsection{Packed changes of selected address bits}
 \label{sec:packed-selected-bits}
+This wide-slot implementation remains available for the grouped-rectangle
+appendix. The accelerated layer below instead uses the compact-control
+construction of Section~\ref{sec:compact-control-movement}.
 
 We implement a binary row addition on corresponding selected bits in two
 address chunks. A third chunk supplies temporary address bits; its arbitrary
@@ -468,6 +479,174 @@
 every role stream retains a complete suffix with the same fixed field
 lengths in every row.
 
+\subsection{Compact dirty controls for selected-bit additions}
+\label{sec:compact-control-movement}
+The following replacement for the wide-slot packed operation moves only
+compact control fields. Temporary address fields have arbitrary contents;
+they are complete ranges of existing address bits, not newly allocated
+independent coordinates. Their availability in every recursive call is
+proved separately below.
+
+\begin{lemma}[Controlled rotations with record-paid offset work]
+\label{lem:record-paid-rotation}
+Consider $M$ records of $R$ bits on a complete rectangular address set, of
+volume $V=MR$. Its fixed number of field lengths and controls have at most
+$A=\lceil\log_2(2V)\rceil$ bits each. A binary target field follows all its
+controls. Suppose its rotation offset, independent of that target, is
+computable from the controls and descriptors in $O(A^C)$ tape steps for
+some fixed $C$. Then the rotation can be performed on fixed tapes in
+$O(V+MA^{C'})$ time for a fixed $C'$. In particular, this is $O(V)$ when
+$A=O(p)$ and $R$ exceeds every fixed power of $p$.
+\end{lemma}
+\begin{proof}
+Fix the prefix before the target. If its range is $Q$ and each target block
+has $B$ suffix bits, rotate by $a$ through the two-piece split at $Q-a$ and
+merge in the opposite piece order. This is the controlled-rotation
+construction from the ordered-affine streaming lemma. Each data bit is
+copied a bounded number of times. There are at most $M$ prefix fibers, since
+every fiber contains at least one complete record. At each fiber, compute
+the offset, initialize its two piece lengths, and reset the finite collection
+of counters; recompute the offset at the merge. Ordinary binary arithmetic
+and descriptor copies cost a fixed polynomial in $A$ per fiber. The controls
+can be wider than the target: their lengths are bounded by $A$, not by the
+target width. Local suffix-length copies and ripple counters are charged as
+in the original streaming proof, so no growing descriptor is scanned per
+payload bit. Two piece tapes, output and fixed arithmetic work tapes suffice;
+rewinds and clearing are over the regions just written. This gives the stated
+bound. Finally $MA^{C'}/V=A^{C'}/R=o(1)$ in the specified record regime.
+\end{proof}
+
+Let $x,y$ be disjoint slots of $fK$ bits, with corresponding selected
+positions $j_i=\rho+iK$, $0\le i<f$, $0\le\rho<K$. Put $n=f-1$.
+For $f=1$ use the elementary two-bit XOR. Otherwise fix a positive integer
+$G$, put $B=2^G$, and assume two disjoint complete $nG$-bit fields $u,t$
+precede both slots and a third such field $b$ follows both slots. Arbitrary
+spectator fields may intervene. Each compact field has $n$ radix-$B$ digits.
+The selected-bit operation will preserve these three entire fields.
+
+\paragraph{Earlier source.}
+Suppose $x$ precedes $y$. The unused field $u$ is a spectator. For an integer
+target segment $v$, a bit $z$, and an arbitrary integer temporary $w$, perform
+\[
+ v\gets v+2zw,\qquad w\gets w+(v\bmod2),\qquad
+ v\gets v+z(1-2w),\qquad w\gets w-((v\bmod2)\mathbin\oplus z).
+\]
+Writing the original parity as $a$, the first update leaves it unchanged.
+The final values are $v+z(1-2a)$ and the original $w$. Thus this toggles
+only the parity of $v$ when $z=1$, and restores the dirty temporary.
+
+Apply these four updates simultaneously, with $z_i=x_{j_i}$, to the low
+$n$ selected segments of $y$ and the digits $t_i$ of $t$. A target-slot
+update is one modular rotation of $y$, whose offsets are respectively
+\[
+ \sum_{i<n}2x_{j_i}t_i2^{j_i},\qquad
+ \sum_{i<n}x_{j_i}(1-2t_i)2^{j_i} \pmod{2^{fK}}.
+\]
+Both control fields precede $y$. For a temporary update, swap $t$ with $b$,
+rotate the back field by the packed selected bits of the current $y$, and
+swap back. For unloading, use minus the packed bits
+$y_{j_i}\mathbin\oplus x_{j_i}$. These controls precede the back field.
+There are four width-$nG$ swaps and four rotations. The arbitrary original
+contents of $b$ return after each pair of swaps.
+
+\paragraph{Later source.}
+If $x$ follows $y$, first execute the earlier-source construction with control
+bits $u_i\bmod2$. Load the selected bits of $x$ into $u$ by swapping $u,b$,
+rotating the back field by their packed radix-$B$ representation, and swapping
+back. Repeat the earlier-source construction with the new parities of $u$,
+then unload the selected bits of $x$ from $u$ by the inverse load.
+When the digit loads do not overflow, the two control parities differ by
+exactly $x_{j_i}$. The resulting target update is the required XOR, with
+$u,t,b$ restored. This version uses twelve compact swaps and ten rotations.
+Every offset depends on a fixed number of fields preceding its target.
+
+\paragraph{Guards and deterministic repair.}
+Temporarily omit the highest selected position. The low $n$ segments
+$[j_i,j_i+K)$ are disjoint and fit inside $y$, including when $\rho=K-1$.
+Write each initial segment as $2g_i+a_i$, $a_i\in\{0,1\}$.
+Declare an address bad if a $t$ digit is $B-1$, or a target guard fails
+\[
+ 2B\le g_i<2^{K-1}-2B.
+\]
+For a later source also declare it bad if any $u$ digit is $B-1$.
+On a good address each temporary load stays within its digit, and the target
+displacement is at most $2B$ in either execution of the four-step identity.
+The conservative guard interval therefore prevents every carry or borrow
+between the wide segments. After each completed identity the guard is
+unchanged. Induction through the later-source sequence proves the desired
+map on the good set, including restoration of every temporary and every
+unselected address bit.
+
+Uniform complete ranges give the common bound
+\[
+ \delta=\min\{1,n(2\cdot2^{-G}+8\cdot2^{G-K})\}
+\]
+on the bad fraction; for an earlier source the first factor 2 can be omitted.
+The ideal map $T$ changes only the selected target parities, so it preserves
+the bad set. The actual sequence $S$ is a bijection on the entire rectangle:
+each rotation is independent of its target and all swaps are bijections.
+Since $S=T$ off the bad set, $S$ preserves that set too. Thus $TS^{-1}$ is
+supported there, exactly as in the original exceptional-stream repair.
+
+Extract the bad records, retaining marked holes in the full stream. For each
+extracted address $q$, reverse the swaps and modular rotations to compute
+$S^{-1}q$, then compute $T(S^{-1}q)$. All fields and descriptors have $O(A)$
+bits. A constant number of offset calculations each scan at most $n\le A$
+digits and combine shifted integers of $O(A)$ bits. Integer arithmetic,
+mixed-radix decoding, inverse offsets and the destination key can consequently
+be computed in $O(A^3)$ per record. Sort only the extracted records by their
+$O(A)$-bit destination keys, using stable binary radix passes, then reinsert
+them in their marked holes. Keys are distinct and fill precisely those holes
+because $TS^{-1}$ permutes the bad set. This uses fixed tapes and costs
+\[
+ O(V+MA^3+\delta MA(R+A)).
+\]
+Every inverse offset is evaluated on the current inverse controls; modular
+overflow on bad addresses is handled exactly, rather than treated as an
+integer identity. Finally, supply the omitted highest selected position
+through the elementary two-bit XOR in $O(V)$ time.
+
+\begin{proposition}[Compact-control selected-bit addition]
+\label{prop:compact-selected-addition}
+Under the stated complete-field hypotheses, take
+\[
+ G=4\lceil\log_2 p\rceil+6,\quad f\le p,\quad A=O(p),
+ \quad K/\log p\longrightarrow\infty,
+\]
+and let $R$ exceed every fixed power of $p$. The exact selected-bit row
+addition, with all temporary fields restored, costs
+\[
+ O\bigl(V((f\log p)^\tau+1)\bigr)
+\]
+on a fixed number of tapes, for either order of the two slots. Constants and
+cutoffs are uniform within each fixed parameter family of the layer theorem.
+\end{proposition}
+\begin{proof}
+Use the retained arbitrary-width chunk-swap primitive for the at most twelve
+swaps, each of width $nG$. Its padding, cleanup and all spectator intervals
+are covered by that interface. For every rotation, apply
+Lemma~\ref{lem:record-paid-rotation}; its offset arithmetic is polynomial
+in $A$. No new record volume is introduced by describing the compact fields.
+For $\ell_p=\lceil\log_2 p\rceil$, eventually
+$K\ge G+4\ell_p+10$. Then
+\[
+ \delta\le \frac{5}{128p^3},\qquad
+ \frac{A^3}{R}=o(1),\qquad
+ \delta A(1+A/R)=O(p^{-2}).
+\]
+The repair is therefore $O(V)$ per call, including predicate scans and key
+arithmetic. The elementary highest-bit operation costs $O(V)$. All calls
+reuse their fixed work tapes and finish in the declared physical field order.
+\end{proof}
+
+\paragraph{Application to a fixed binary change of basis.}
+The original phase-frame construction expresses each binary basis change
+as a fixed number of row additions on $m$ slots of $f=e/m$ axes. Use the
+two slots of each addition as $x,y$ and the reserved front/back compact
+fields as controls. The complete basis change and its inverse cost
+$O(V((e\log p)^\tau+1))$. The reservations below guarantee their complete
+ranges, independently of which slots are involved, at every recursive node.
+
 \subsection{Batching a common layer on a fixed number of tapes}
 
 We now turn the finite complex network into an algorithm for a whole
@@ -530,10 +709,11 @@
 is \(V\), binary changes of basis, their inverses, scalar gates,
 endpoint corrections and reassembly of the physical banks cost
 \[
- O\bigl(V((eK)^\tau+1)\bigr)
+ O\bigl(V((e\log p)^\tau+1)\bigr)
 \]
 at this node.  Each slot has \((e/m)K\) address bits, so
-Lemma~\ref{lem:packed-selected-bit-rectangle} gives the stated bound for a row addition.  The
+Proposition~\ref{prop:compact-selected-addition} gives the stated bound for a row addition,
+using the complete compact fields constructed below.  The
 network and all of its \(m\)-dimensional residual bases are fixed,
 so the total number of row additions is fixed as well.  At the same
 guarded width, the scalar gates cost a constant
@@ -541,301 +721,205 @@
 work bounded by any fixed polynomial in \(p\) per record is also
 absorbed: its ratio to the record's \(\Theta(rp)\) bits tends to zero.
 
-\paragraph{Where the rows come from.}
-Put
-\[
- q_0=\lceil\log_2 W\rceil\,
-       \lceil\log_m(2d)\rceil,
- \qquad k_0=\lceil\log_m d\rceil.
-\]
-If \(D\le q_0\), use individual \(C\) kernels, at cost \(O(V\log d)\).
-Otherwise apply \(C\) individually on the selected bits of the first
-\(q_0\) chunks, in \(O(V\log d)\) time.  These chunks remain a
-complete consecutive address block before all the remaining chunks;
-regard their concatenated bits as one row index.  Its range has
-length \(R_{\rm row}=2^{q_0K}\ge W^{k_0}\), since \(K\ge1\) for sufficiently
-large \(p\) and \(q_0K\ge k_0\log_2W\).  Within each preceding prefix, add
-the same number of zero rows to reach the next multiple of \(W^{k_0}\);
-the volume grows
-by a factor at most two.  At any recursive depth \(j\le k_0\), the
-row count is divisible by \(W^{k_0-j}\).
-
-At a node, write a row index as \(u=Wg+w\), with \(0\le w<W\),
-and send that entire row to physical wire role \(w\), indexed there
-by \(g\).  Every role stream has the same remaining address shape.
-Thus a pointwise gate reads the records with the same \(g\) and the
-same remaining address in its touched roles.  The scalar network
-allows these role values to be arbitrary, so the split supplies its
-scratch roles from existing rows.  Each recursive edge call acts on
-one role stream with volume exactly \(1/W\) of its parent's logical
-volume.
-
-Splitting the row index removes no address within a row. Every row in
-every role stream, including a padded row, retains the complete suffix
-\([2^K]^{D-q_0}\times[S]\) in the same field order and with the same
-lengths; inactive chunks remain complete spectator fields. The row
-counts also remain equal across preceding prefixes, decreasing by
-the same factor \(W\) at each split. The network is invoked only on
-these equal rows: their common suffix gives the aligned role addresses
-and exact \(V/W\) child volumes above. For each packed call, collapse
-the preceding prefix and row index into one field, and collapse the
-spectator intervals around its three slots. This gives the rectangle of
-Lemma~\ref{lem:packed-selected-bit-rectangle}, whose guard count and
-deterministic repair apply to every child.
-
-The \(W\) role streams use a fixed number of tapes.  Before entering a
-child, push the other role streams onto a LIFO work tape, copy the
-active stream to a reserved child input area, and clear or reuse the
-role areas.  On return, copy the result back and pop the parked
-streams, reversing each popped stream when necessary to restore its
-order.  The stack head is at its top at every call boundary.  Pushing,
-popping, copying, resetting and clearing at this node cost its own
-volume times a fixed constant; no step traverses an ancestor's parked
-data.  The call descriptor and program counter use a separate stack.
-The number of child calls is the fixed integer \(s\), so these costs
-are included in the node overhead above.  The fixed-tape chunk-swap
-subroutine has its own fixed work areas, reused on every call.
-
-For each packed call, padding has increased the row count by at most
-a factor two and row splitting can only decrease it.  Its active
-address still has \(O(p)\) bits. The four collapsed spectator lengths
-therefore have \(O(p)\)-bit descriptions. Its record-width parameter
-describes a polynomial of \(\Theta(rp)\) bits and has
-\(O(\log(rp))=O(p)\) bits as well. These lengths, together with
-\(K,f,\rho\) and the finite chunk-name permutation, are precisely the
-local descriptor required by
-Lemma~\ref{lem:packed-selected-bit-rectangle}. Both the record count
-and record width have \(O(p)\)-bit lengths, so its bound uses
-\(A_{\rm rect}=\lceil\log_2(2V)\rceil=O(p)\). A selected-position
-mask has at most \(DK=O(p)\) bits.
-
-The binary descriptions of all growing shape parameters have
-polynomial length in \(p\).  Each nonempty child stream still contains
-a complete length-\(r\) polynomial record.  Its volume therefore
-absorbs polynomial descriptor setup at that node, as well as
-polynomial address work per record.  Even if every node is counted
-separately, there are only \(d^{O(1)}\) nodes: the branching factor
-\(s\) is fixed and the depth is at most
-\(\lceil\log_m d\rceil\).  Any additional fixed power of \(p\)
-from this count is still dominated by \(r\).
-
-Every completed network invocation applies \(C^{\otimes e}\)
-separately to each input row.  Its selected bits lie in the
-remaining chunks, disjoint from the bits used by the earlier
-preprocessing.  On the genuine rows, those two completed operations
-therefore commute.
-Each padded zero row is again zero when an invocation finishes, so the
-added rows can be deleted after the invocations on the remaining chunks
-have completed. Intermediate scratch values may be arbitrary.
-
-\paragraph{Unrolling the recurrence.}
-Fix \(\beta=1/2\), and choose real exponents satisfying
-\[
- 0<\tau<1,\qquad
- \max\{0,\log_m(s/W)\}\le\sigma<1,
- \qquad \max\{\tau,\sigma\}<\lambda<1,
- \qquad \tau(1+c/\beta)<\lambda.
-\]
-The exponents can all be chosen rational, with a strict upper bound
-in the condition on \(\sigma\).  Stop recursion when
-\(e<d^\beta\) and apply the \(e\) \(C\) kernels individually.  If
-\(F(e)\) denotes time divided by the current logical stream volume,
-then
-\[
- F(e)\le (s/W)F(e/m)+O((eK)^\tau+1),
- \qquad F(e)=O(e)\quad(e<d^\beta).
-\]
-Indeed, the \(s\) child calls are executed on streams of volume
-\(V/W\), so their total cost is \(s(V/W)F(e/m)\).  Dividing by
-the current volume \(V\) gives the factor \(s/W\).
-At an internal node \(e\ge d^\beta\), hence
-\(K\le d^c\le e^{c/\beta}\).  The overhead is therefore
-\(O(e^{\tau(1+c/\beta)}+1)=O(e^\lambda)\).
-Put \(a=s/W\).  Since \(a\le m^\sigma<m^\lambda\), the internal
-costs sum to at most
-\[
- C e^\lambda\sum_{j\ge0}(a/m^\lambda)^j=O(e^\lambda).
-\]
-If the leaves after \(j\) levels contain \(u=e/m^j<d^\beta\)
-active chunks each, their combined normalized
-contribution is at most
-\[
- O(a^ju)\le O(e^\sigma u^{1-\sigma})
-          \le O(e^\sigma d^{\beta(1-\sigma)}).
-\]
-The first inequality uses
-\(a^j\le m^{j\sigma}=(e/u)^\sigma\).  It also covers \(a<1\)
-by taking \(\sigma\ge0\).
-
-Write \(D-q_0=\sum_j a_jm^j\) in base \(m\), with \(0\le a_j<m\).
-Partition the remaining chunks into \(a_j\) consecutive pieces
-containing \(m^j\) chunks for each \(j\).
-There are at most \((m-1)(1+\lfloor\log_m d\rfloor)\) pieces.
-Adding their costs and the row preprocessing gives
-\[
- O\!\left(\log(2d)
-    \bigl[d^\lambda+d^{\sigma+\beta(1-\sigma)}+1\bigr]\right)
-   =O(d^{\lambda'})
-\]
-for any fixed
-\[
- \max\{\lambda,\sigma+\beta(1-\sigma)\}<\lambda'<1.
-\]
-The case \(D\le q_0\), in which preprocessing would consume all
-active chunks, was handled by individual kernels.  Thus all constants
-and all tape counts are independent of \(p,d,r\) and the input data.
-
-\subsection{An explicit guard-width bound}
-
-The accelerated network is evaluated exactly until an entire
-normalized butterfly layer is complete.  Its intermediate operators
-need not be contractions.  We now give explicit constants that bound
-denominator and magnitude growth within one layer.
-
-For the complex network, let
-\[
- E=64(W+m+1)^3,\qquad B=s+E,
- \qquad \nu=\min\{j\ge1:m^j\ge B\},
-\]
-and set
-\[
- C_1=\max\{20,\nu+3\},\qquad C_0=128mB^2,
+\paragraph{Rows and complete compact fields.}
+Let $m,W,s$ denote the complex network constants and keep the layer's global
+$p,d,K$ at every node. Put
+\[
+ G=4\lceil\log_2p\rceil+6,\quad H=dG,\quad
+ q_F=\left\lceil\frac{2H}{K}\right\rceil,\quad
+ q_B=\left\lceil\frac{H}{K}\right\rceil,
+\]
+\[
+ q_0=\lceil\log_2W\rceil\lceil\log_m(2d)\rceil,
+ \quad k_0=\lceil\log_m d\rceil,\quad Q_0=q_0+q_F+q_B.
+\]
+If $D\le Q_0$, apply all $D$ selected kernels individually. Otherwise,
+process individually the first $q_0+q_F$ and last $q_B$ selected axes.
+This uses existing coordinates only. The remaining active axes are a
+consecutive middle block of length $D-Q_0$.
+
+Regard the first $q_0$ chunks as the row index. Immediately after it, the
+next $q_F$ chunks have at least $2H$ bits. Reinterpret these consecutive
+bits as two $H$-bit front fields and a spectator remainder; a remainder
+of length zero is omitted. The last $q_B$ chunks similarly supply one
+$H$-bit back field and a spectator remainder. These reinterpretations move
+no records and preserve the total address space. For a call with $f$ axes
+per slot, carve $nG=(f-1)G\le H$ consecutive bits from each of these
+fields; collapse their unused bits into spectators. The resulting field
+list has a fixed number of entries. Both front fields precede all active
+slots and the back field follows them, so the compact-control proposition
+applies to every ordered pair of slots. For $f=1$ no compact fields are used.
+
+The original row range has length $R_{\rm row}=2^{q_0K}\ge W^{k_0}$.
+For each original prefix, pad this row range to the next multiple of
+$W^{k_0}$ by adding zero rows. A row includes the entire complete front
+fields, active middle block, back field, all spectator coordinates and
+polynomial record suffix. Padding adds whole such rows, so increases total
+volume by at most two and does not change any within-row range.
+
+At a node write the row index as $u=Wg+w$, $0\le w<W$, and send the entire
+row to role $w$. The index $u$ consists solely of the first row region:
+\emph{no compact-field bit is included in it}. Each role therefore has
+the same complete within-row address set and exactly $1/W$ of the parent
+volume. At depth $j$ the remaining row count is divisible by $W^{k_0-j}$.
+This invariant survives further splits, padding, and reassembly. Pointwise
+gates on different roles have aligned $g$ and every remaining coordinate.
+All compact operations restore the temporary fields before a scalar gate,
+recursive child, or returned result uses that layout.
+
+The fixed-tape scheduler is unchanged: park only the other role streams at
+the active stack top, execute the child on a complete role stream, and restore
+the parked streams on return. Every child has its own current logical volume;
+copying, parking, rewinds and clearing cost a fixed multiple of that volume.
+The constant number of compact operations per edge use fixed work areas.
+Descriptors include the global $p,d,K,G,H$ and the carved field lengths;
+all have $O(p)$ bits and there are a fixed number of fields. The existing
+record suffix absorbs polynomial setup at every node and per record.
+
+Each completed invocation acts as the desired $C$ tensor on its active axes
+and as identity on the reserved and spectator coordinates. It therefore
+commutes with the earlier individual kernels, including those on bits now
+interpreted as compact fields. Restoring the row permutation returns the
+result to each original row separately. A padded zero row is consequently
+zero at completion and can be deleted exactly as before. Neither temporary
+field contents nor intermediate scratch payloads were assumed to be zero.
+
+The number of individually processed axes is at most $Q_0$, including the
+fallback $D\le Q_0$. For large $p$, $K\ge d^c/2$ and
+\[
+ q_F+q_B\le\frac{3dG}{K}+2\le6d^{1-c}G+2.
+\]
+Thus preprocessing, padding and reservation cost
+\[
+ O\bigl(V(\log(2d)+d^{\max\{1-c,0\}}\log p+1)\bigr).
+\]
+All reserved axes lie within the original $D$ axes; in particular their
+total number never exceeds $d$. This fact also bounds their coefficient
+dependency depth by the same $8d$ allowance used in the guard proof.
+
+\paragraph{Unrolling the compact-control recurrence.}
+Fix rational $0<\beta<1$. Choose $0<\tau<1$ and
+$\max\{0,\log_m(s/W)\}<\sigma<1$. Stop when $e<d^\beta$ and apply
+the remaining kernels individually. With $F(e)$ denoting time per current
+logical volume, the exact row split gives
+\[
+ F(e)\le(s/W)F(e/m)+O(G^\tau e^\tau+1),\qquad
+ F(e)=O(e)\quad(e<d^\beta).
+\]
+Here $G$ and the compact reservations are global parameters, not rebuilt
+at each node. This is why no factor $K^\tau$ appears in the overhead.
+Repair costs are already included in the node overhead.
+
+Write $b=s/W$. At level $j$ the internal contribution is bounded by
+$O(G^\tau e^\tau(b/m^\tau)^j)$. Summing through the stopping depth gives
+\[
+ O(G^\tau\log(2d)d^\chi),\qquad
+ \chi=\tau+(1-\beta)\max\{\sigma-\tau,0\}.
+\]
+Indeed if $\sigma\le\tau$ the root bounds every summand; otherwise the
+last internal level costs at most the root bound times a fixed constant
+and $d^{(1-\beta)(\sigma-\tau)}$. The additive $1$ is absorbed by
+$G^\tau e^\tau$ at internal nodes. The leaf contribution is at most
+$O(d^{\sigma+\beta(1-\sigma)})$, by the original volume-weighted argument.
+
+Partition $D-Q_0$ into consecutive powers of $m$, using its base-$m$
+expansion. There are $O(\log(2d))$ pieces. Include their costs and the
+reservation cost to obtain, per original volume,
+\[
+ O\!\left((\log p)^3
+ [d^\chi+d^{\sigma+\beta(1-\sigma)}+d^{\max\{1-c,0\}}+1]\right).
+\]
+We used $\log d=\Theta(\log p)$ and $G=O(\log p)$ for each fixed
+$d=\Theta(p^\epsilon)$ family. Any fixed rational exponents satisfying
+\[
+ \max\{\tau,\sigma,\chi\}<\lambda<\lambda'<1,\qquad
+ \max\{\sigma+\beta(1-\sigma),1-c,0\}<\lambda'
+\]
+therefore give the complete $O(Vd^{\lambda'})$ layer bound. The strict gaps
+absorb the logarithms. The small-$D$ fallback obeys the same bound.
+For fixed rational $\beta=u/v$, implement stopping by the integer test
+$e^v<d^u$; its polynomial descriptor cost is already included.
+
+\paragraph{Node charge for the compressed complex motif.}
+\label{par:compressed-complex-node-charge}
+The new scalar mixers use only additions, subtractions and half scalings.
+For explicit guard accounting let $M_0=R_0-C$ be the number of additions
+in all disjoint-rectangle mixers per invocation. Each pair circuit has
+$2(3n-6)$ mixer additions. Let $M$ be their combined mixer count, $V_0$
+the number of input-copy incidences, and $J_0$ the output incidences.
+The signed schedule uses at most
+\[
+ T_{\rm scalar}=3v^2(4M+2V_0+4J_0+18v)
+\]
+elementary scalar operations, including half scalings. The exact checker
+verifies $4T_{\rm scalar}+8s+8W+8m<64(W+m+1)^3$ and $2\le s<m^5$.
+Thus the retained additive node charge $E=64(W+m+1)^3$ dominates these
+operations, per-kernel scalar corrections, and fixed endpoint corrections.
+Binary basis changes only
+permute coefficient encodings. The generalized stopped-depth proof applies
+with the new $W,m,s$, $C_1=5-4\beta+\zeta$ and its explicit $C_0$.
+Since $\epsilon C_1<1$, coefficient widths remain $O(p)$.
+
+\subsection{A generalized stopped-depth guard}
+\label{sec:compact-stopped-guard}
+Use the complex network's own constants $m,W,s$, independently of the bit
+network used by swaps. Assume $m\ge3$ and $2\le s<m^5$, as checked below.
+For rational $0<\beta<1$ and $\zeta>0$, put
+\[
+ E=64(W+m+1)^3,\quad B=s+E,\quad C_1=5-4\beta+\zeta,
+\]
+\[
+ C_0=\left\lceil\max\{128mB^2,18mB^2(1+1/\zeta)\}\right\rceil,
  \qquad \Delta=\lceil C_0d^{C_1}\rceil.
 \]
-For the stated \(h=100\) construction, the complex network has
-\[
-\begin{split}
- m&=1{,}000{,}000,\\
- W&=1{,}873{,}807{,}244{,}643{,}542{,}670{,}000,\\
- s&=1{,}873{,}807{,}244{,}636{,}671{,}267{,}308{,}000{,}000.
-\end{split}
-\]
-These values give \(m^{10}<B<m^{11}\), hence \(\nu=11\) and
-\(C_1=20\).  In fact the bound below needs only exponent fourteen;
-the choice twenty leaves an explicit additional margin.
-
-We measure arithmetic dependency depth using complex additions,
-subtractions, halvings, sign changes and multiplication by \(i\).
-For a group of \(e\) axes, let \(A(e)\) bound the number of these
-operations on any chain from an input coefficient to an intermediate
-coefficient in its recursive \(C\)-layer.  A chain follows an operand
-into a later arithmetic result.  Work on other coefficients or records
-that do not feed this value lies on other chains, even though the
-fixed-tape machine executes that work sequentially.  To obtain a
-conservative bound, we concatenate all scalar gates and all recursive
-edge calls within one network invocation.
-
-The scalar schedule \eqref{eq:motif-schedule} touches a data wire
-at most four times in each of the three stages.  A scratch wire belongs
-to one invocation and is touched at most four times there.  Thus every
-physical wire occurs in at most twelve scalar gates, and there are
-at most \(12W\) gates.  A gate has at most \(W\) inputs and outputs
-and coefficients in \(\{0,\pm1,\pm1/2\}\).  One output is a sum of
-at most \(W\) terms, each requiring at most one halving and one
-addition or subtraction.  Evaluating all outputs from saved inputs
-therefore takes at most \(2W^2\) elementary operations.
-
-For an inverse child, each \(Z^{\otimes f}\) is one sign chosen by
-the parity of the selected address bits, and the scalar fourth-root
-phase takes at most one sign change and one multiplication by \(i\).
-This gives at most \(4s\) operations for the child corrections.
-The source, sink and bank corrections take at most \(4W+4\) more.
-Calculating the address parities and phase exponents is address or descriptor
-work, not arithmetic on coefficients.  Since \(s<Wm\), the total
-\(24W^3+4s+4W+4\) is less than
-\(E=64(W+m+1)^3\).  Binary address permutations add no coefficient
-arithmetic at their interfaces.
-
-The paired evaluation in \eqref{eq:individual-C-pair} uses at most
-eight elementary operations: form the sum and difference, multiply
-the difference by \(i\), then form the two half-sums.  We may therefore use
-\[
- A(e)\le sA(e/m)+E
- \quad\hbox{at an internal node},\qquad A(e)\le8e
- \quad\hbox{at a leaf}.
-\]
-The depth estimate concatenates the \(s\) calls, so it uses \(s\).
-The factor \(1/W\) in the earlier time recurrence accounted for
-their smaller stream volumes.
-
-There are at most \(k\le\lceil\log_m d\rceil\) internal levels.
-For any bound \(x\ge1\), \(sx+E\le(s+E)x=Bx\).
-Starting with \(8e\le8d\le8dB\) at a leaf and applying this
-inequality through the levels gives the first bound below.
-For the second, use \(B\le m^\nu\) and
-\[
- B^{k+1}\le B^{\log_m d+2}
-          =B^2d^{\log_m B}\le B^2d^\nu.
-\]
-Thus
-\[
- A(e)\le 8dB^{k+1}\le8B^2d^{\nu+1}.
-\]
-For the sufficiently large sizes under consideration, the earlier
-piece count is at most \(md\).  Concatenating their depths contributes
-at most \(8mB^2d^{\nu+2}\).  In either preprocessing branch, at most
-\(d\) individual kernels are used, contributing at most \(8d\).
-Splitting rows, padding, copying and removing padding add no
-coefficient arithmetic.
-
-Each of the two outer diagonals applies a fourth-root phase and costs
-at most two elementary phase operations.  Put
-\(t=\lfloor D/2\rfloor\).  Since \(b^2=-i/2\), the scalar \(b^D\)
-is \((-i)^t2^{-t}b^{D\bmod2}\).  Its phase costs at most two
-operations, and the possible factor \(b\) is computed as
-\((z-iz)/2\) in three operations.  The shift by \(t\) consumes
-\(t\) fractional bits, so we charge it \(t\) units of denominator
-growth even though one tape scan performs the shift.  Charging these
-same units to magnitude growth gives a common upper bound for both:
-\[
-\begin{aligned}
- 8mB^2d^{\nu+2}+8d+t+9
- &\le 8mB^2d^{\nu+2}+18d\\
- &\le 26mB^2d^{\nu+2}\\
- &<128mB^2d^{\nu+3}\le C_0d^{C_1}.
-\end{aligned}
-\]
-Here \(d\ge1\), \(mB^2\ge1\), and \(C_1\ge\nu+3\).
-
-To see how this count controls numerical size, start with disk-valued
-coefficients in \(2^{-p}\mathbb Z[i]\).  A value at depth \(\ell\)
-lies in \(2^{-(p+\ell)}\mathbb Z[i]\) and has modulus at most
-\(2^\ell\).  This follows by induction: addition or subtraction
-can at most double the maximum modulus of its operands; halving adds
-at most one denominator bit; and signs and multiplication by \(i\)
-preserve both bounds.  The charge of \(t\) for the final shift is
-the same as \(t\) successive halvings.  Thus reserving \(p+\Delta\)
-fractional bits and \(\Delta+3\) integer and sign bits per real
-component represents every intermediate coefficient exactly and
-prevents overflow.  Shifts consume the reserved fractional zeros
-and discard no bits.
-
-An address-swap subroutine may temporarily hold XOR combinations of
-the fixed-width encoding bits.  These strings are not used as
-coefficients during the swap.  It returns an exact permutation of
-the encodings before coefficient arithmetic resumes, so these
-internal bit operations cause no numerical rounding or denominator
-growth.
-
-Choose \(\epsilon C_1<1\).  Since \(d=\Theta(p^\epsilon)\),
-\(\Delta=o(p)\), and the full coefficient width is \(O(p)\), as
-required in the time analysis.  Addition, signed shifts, copying and
-phase changes consequently take \(O(p)\) bit operations per
-coefficient.
+As in the retained exact-arithmetic guard proof, a dependency-depth upper
+bound is
+\[
+ A(e)\le sA(e/m)+E\quad\hbox{internally},\qquad A(e)\le8e
+ \quad(e<d^\beta).
+\]
+The fixed node charge $E$ remains additive. All new compact-control,
+padding and repair operations only permute complete coefficient encodings
+and introduce no coefficient arithmetic.
+
+For a root $e\le d$, the number of internal levels is at most
+$(1-\beta)\log_m d+1$. Hence
+\[
+ s^j\le s d^{5(1-\beta)},\qquad
+ A(e)\le s(8+E)d^{5-4\beta}\le9B^2d^{5-4\beta}.
+\]
+A root that is already a leaf also satisfies this bound. There are at most
+$m(1+1/\zeta)d^\zeta$ base-$m$ pieces, using
+$\log d\le d^\zeta/\zeta$ and $\log m>1$. Individually processed
+row, front and back axes are disjoint and number at most $d$, so their
+depth is at most $8d$. The outer phases and $b^D$ conversion add at most
+$\lfloor D/2\rfloor+9$ units, including denominator shifts. Consequently
+\[
+ A_{\rm layer}\le9mB^2(1+1/\zeta)d^{C_1}+18d
+              \le C_0d^{C_1}.
+\]
+The exact integer comparison for the final inequality is included in the
+certificate. The usual induction gives denominator exponent at most
+$p+\Delta$ and magnitude at most $2^\Delta$. Thus $p+\Delta$ fractional
+bits and $\Delta+3$ integer/sign bits suffice per component. The condition
+$\epsilon C_1<1$ makes these widths $O(p)$, uniformly over the retained
+parameter bands. Integer power comparisons implement the rational guard
+ceilings in polynomial descriptor time.
 
 \begin{proposition}[Simultaneous normalized butterfly layer]
 \label{prop:simultaneous-layer}
-Fix
-\[
- \tau=\sigma=1-2^{-50},\qquad \beta=\frac12,\qquad C_1=20.
-\]
-Let $c,\epsilon,\lambda,\lambda'$ be fixed positive rational numbers
-satisfying
-\[
- \max\{\tau,\sigma\}<\lambda<1,\qquad
- \tau(1+c/\beta)<\lambda,\qquad
- \max\{\lambda,\sigma+\beta(1-\sigma)\}<\lambda'<1,
- \qquad \epsilon C_1<1.
+Fix $\tau=1-467/10^{11}$ and $\sigma=1-5/10^9$.
+Let $c,\epsilon,\lambda,\lambda',\beta,\zeta$ be fixed positive rational numbers
+with $0<\beta<1$, and put
+\[
+ C_1=5-4\beta+\zeta,\qquad
+ \chi=\tau+(1-\beta)\max\{\sigma-\tau,0\}.
+\]
+Require
+\[
+ \max\{\tau,\sigma,\chi\}<\lambda<\lambda'<1,\qquad
+ \max\{\sigma+\beta(1-\sigma),1-c,0\}<\lambda',\qquad
+ \epsilon C_1<1.
 \]
 Fix positive constants $a_d\le b_d$ and $a_r\le b_r$, and fix
 positive constants $C_M,C_{\rm desc},C_w$ with $C_w>1$.
@@ -909,10 +993,11 @@
 The procedure evaluates the entire layer exactly before this final
 truncation. Put
 \[
- C_0=128m\bigl(s_{\rm c}+64(W_{\rm c}+m+1)^3\bigr)^2,
+ C_0=\left\lceil\max\{128mB^2,18mB^2(1+1/\zeta)\}\right\rceil,
+ \qquad B=s_{\rm c}+64(W_{\rm c}+m+1)^3,
 \]
 where $m,W_{\rm c},s_{\rm c}$ are the fixed constants of
-Proposition~\ref{prop:complex-motif-interface}. The procedure uses
+Proposition~\ref{prop:compressed-complex-interface}. The procedure uses
 at most $p+\lceil C_0d^{C_1}\rceil$ fractional bits and
 $\lceil C_0d^{C_1}\rceil+3$ integer and sign bits per real component
 internally; these widths are $O(p)$. The time includes address changes,
@@ -928,7 +1013,7 @@
 descriptor work above.
 The phase-frame construction reduces a group of $e$ selected bits to
 $s_{\rm c}$ children on $e/m$ selected bits.  The row split gives
-each child $1/W_{\rm c}$ of the parent's logical volume. The packed
+each child $1/W_{\rm c}$ of the parent's logical volume. The compact-control
 selected-bit procedure supplies the binary basis changes. We verify
 that its cost is uniform over the displayed bands for $d,r$.
 Put $\Delta=\lceil C_0d^{C_1}\rceil$. Since $d\le b_dp^\epsilon$
@@ -951,26 +1036,26 @@
 
 Uniformly for large $p$, the bands give $f\le d\le p$ and
 \[
- K\ge\tfrac12a_d^c p^{\epsilon c},\qquad
+ K\ge\tfrac12a_d^cp^{\epsilon c},\qquad
  r\ge2^{(a_r/b_d)p^{1-\epsilon}}.
 \]
-The first inequality follows from $K=\lfloor d^c\rfloor$ once
-$d^c\ge2$. In particular, $K\ge6$ and $K/\log p\to\infty$, while
-$r$ dominates every fixed polynomial. Substituting the common bound
-$A_*\le C_Ap$ into the two additional normalized
-terms in \eqref{eq:packed-rectangle-time} gives
-\[
-\begin{aligned}
- \frac{A_*^3}{R_{\rm g}}
- &\le\frac{C_A^3p^2}{2r},\\
- 80(f-1)2^{-K}A_*\left(1+\frac{A_*}{R_{\rm g}}\right)
- &\le80C_Ap^2 2^{-K}\left(1+\frac{C_A}{2r}\right).
-\end{aligned}
-\]
-Both bounds tend to zero uniformly over all current allowed $d,r$.
-Thus Lemma~\ref{lem:packed-selected-bit-rectangle} has the required
-$O(M_{\rm child}R_{\rm g}(L^\tau+1))$ cost for every packed call,
-with one common time constant.
+Thus $K/\log p\to\infty$, $r$ dominates every fixed polynomial, and
+$G=4\lceil\log_2p\rceil+6$ eventually obeys
+$K\ge G+4\lceil\log_2p\rceil+10$. Every role stream retains both complete
+front fields and the complete back field outside its row index. The above
+$A_*\le C_Ap$ bound applies to each compact rectangle. For its repair,
+\[
+ \frac{A_*^3}{R_{\rm g}}\le\frac{C_A^3p^2}{2r},\qquad
+ \delta A_*\left(1+\frac{A_*}{R_{\rm g}}\right)
+ \le\frac{5C_A}{128p^2}\left(1+\frac{C_A}{2r}\right).
+\]
+The wider-control rotation bound is also uniform, because every polynomial
+offset charge divided by $R_{\rm g}$ tends to zero. Hence
+Proposition~\ref{prop:compact-selected-addition} costs
+$O(M_{\rm child}R_{\rm g}((f\log p)^\tau+1))$ with a common constant.
+Reservation preprocessing costs
+$O(V(\log(2d)+d^{\max\{1-c,0\}}\log p+1))$ and is included by the
+strict condition $1-c<\lambda'$.
 The row construction and the recurrence analysis
 above give total time $O(Vd^{\lambda'})$ on fixed tapes, including
 all padded and auxiliary streams. Equation~\eqref{eq:phase-butterfly-layer}
--- a/build/sections/07-resampling.tex
+++ b/build/sections/07-resampling.tex
@@ -1,3 +1,5 @@
+% Modified by Douglas Colkitt, October 7, 2026.
+% Conditional compact-control, complex-compression and ternary research draft; see main.tex for scope.
 \section{Resampling with the permutations left in the transform}
 \label{sec:resampling}
 
@@ -330,7 +332,7 @@
 full padded box of $T$ entries, and each entry occupies $O(p)$ bits,
 then exposing and restoring all the lines costs
 \[
- O\bigl(Tp\,d^2(1+\ell^\tau)\bigr)
+ O\bigl(Tp\,d(1+\ell^\tau)\bigr)
 \]
 using the equal-width chunk-swap bound $O(Vu^\tau)$ and the fixed-digit
 movement primitive. Any fixed polynomial in $p$ of setup per line is
@@ -377,21 +379,49 @@
 tapes. Loop indices and the growing shape descriptor reside on tapes;
 they are not extra heads.
 
-The descriptor records which coordinate each physical address field
-holds. Thus moving fields changes their positions, but not the validity
-conditions above. To move an axis past its neighbor, interchange their
-$\ell$-bit or $(\ell-1)$-bit address fields. Equal widths use
-Lemma~\ref{lem:chunk-swap}. For unequal widths, write the longer field as
-a leading bit $e$ followed by $\ell-1$ bits. If it comes first, the order
-is $e\,x\,y$, with $x,y$ of equal width. Interchange $x,y$, then move $e$
-between them, obtaining $y\,e\,x$. If the longer field comes second,
-first move its leading bit left of the shorter field, then interchange
-the two equal-width pieces. The bit move uses
-Lemma~\ref{lem:elementary-streams}. Each adjacent interchange therefore
-costs $O(Tp(1+\ell^\tau))$. Exposing and restoring one axis needs
-$O(d)$ adjacent interchanges, hence all axes need $O(d^2)$. These are
-exact permutations of the padded box, so they change neither norms nor
-errors.
+The descriptor records which named coordinate each physical address field
+holds, together with its width and valid range. Before processing axis $i$,
+interchange its entire address field with the innermost field; after processing
+its lines, undo precisely that interchange. If it is already innermost, no
+interchange is needed. Thus each axis starts and ends in the original field
+order, and all axes together require at most $2(d-1)$ field interchanges.
+
+These interchanges need not be adjacent. Lemma~\ref{lem:chunk-swap} permits
+an arbitrary intervening field $[G]$, obtained by collapsing the intervening
+axes, and a suffix containing the complete coefficient word. For widths that
+differ by one, use the explicit nonadjacent construction following that lemma.
+Suppressing the prefix and suffix, its two cases are
+\[
+ (e,x,G,y)\longmapsto(e,y,G,x)\longmapsto(y,G,e,x),
+\]
+\[
+ (x,G,e,y)\longmapsto(e,x,G,y)\longmapsto(e,y,G,x),
+\]
+where $x,y$ have equal bit width and $e$ is one bit. Each uses one equal-width
+interchange and one single-bit move. Empty equal-width pieces require no swap.
+The intervening axes keep their internal order, and each complete record keeps
+its bit order. The cost of a full field interchange is
+$O(Tp(1+\ell^\tau))$, including workspace cleanup.
+
+At a completed interchange the descriptor moves the coordinate names, widths,
+and validity bounds with their fields. A processed line therefore still runs
+through the same named coordinate in increasing numerical order. Other named
+coordinates retain their values; their physical order is immaterial to the
+validity test. After the line algorithm, the inverse interchange restores the
+original field order. The padded box keeps its full physical size throughout.
+Invalid entries remain zero at the completed axis-map boundaries specified
+above; temporary bit moves and the internal arithmetic of a swap need not
+preserve an intermediate validity mask. The completed permutations and their
+inverses are exact, so the contraction and error arguments do not change.
+
+The list and collapsed descriptors are stored on tapes, not in new heads.
+Here $t_i\ge2$, so $d\le\log_2T$. With $V$ the physical bit volume,
+each collapsed length is at most $V$, and the list of axis names, ranges,
+and widths has length polynomial in $\log(2V)$. Ordinary integer arithmetic
+therefore prepares each call descriptor in $\operatorname{poly}(\log(2V))=O(V)$
+time. Each primitive reuses its fixed work tapes. Thus all exposure and
+restoration costs
+$O(Tp\,d(1+\ell^\tau))$.
 
 Since $t_i\geq r/2$, each tensor map invokes at most
 $\sum_iT/t_i\leq2dT/r$ lines. For any fixed setup degree $c$, work