compact-control-34.patch

--- a/build/main.tex
+++ b/build/main.tex
@@ -1,3 +1,5 @@
+% Modified by Douglas Colkitt, October 7, 2026.
+% Conditional compact-control 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 Compact-control 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 construction
+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
+$\kappa=83/10^{12}$, 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 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
 \[
  T_A(n)=O\!\left(n(\lg n)^{1-\kappa}\right),
- \qquad \kappa=2^{-182}.
+ \qquad \kappa=83/10^{12}.
 \]
 \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 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 $a=296/10^{11}$ and $a_{\rm c}=418/10^{12}$.
+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{1000},\quad\zeta=\frac1{10000},\quad
+ C_1=\frac{49961}{10000},\\
+ \epsilon=\frac{1999}{10000},\quad c=\frac15,\quad\delta=\frac1{10^6},\\
+ \lambda=1-\frac{1671}{4\cdot10^{12}},\quad
+ \lambda'=1-\frac{167}{4\cdot10^{11}},\quad
+ \kappa=\frac{83}{10^{12}}>2^{-34}.
 \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
+$\chi=\tau+(1-\beta)(\sigma-\tau)$, 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 $e^{1000}<d$. 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 $O(p)$.
+$C_1=49961/10000$, 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
+\[
+ G_*:=\min_i g_i=g_3=\epsilon(1-\lambda')
+       =\frac{333833}{4\cdot10^{15}}>\frac{83}{10^{12}}=\kappa.
+\]
+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 $O(d^{49961/10000})=o(p)$, while
+$\alpha=\Theta(p^{11999/40000})$ and
+$\gamma=O(p^{15997/20000})=o(p)$. Also
+$K=\Theta(p^{1999/50000})=o(\ell)$, $K/\log p\to\infty$,
+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 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 +670,240 @@
 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=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 $\eta=\eta_{\rm c}$.
-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.
+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{An independently sized complex network}
+\label{sec:independent-complex}
+Keep the paired bit network at $h_{\rm b}=50$, with address-swap exponent
+$\tau=1-296/10^{11}$. Instantiate the original complex construction instead
+at $h_{\rm c}=25$. Its constants, distinguished from those of the bit
+network, are
+\[
+ m_{\rm c}=15625,\quad v_{\rm c}=2300,\quad N_{\rm c}=12167000000,
+\]
+\[
+ W_{\rm c}=58645352620000,\quad L_{\rm c}=10315500000,\quad
+ s_{\rm c}=916333630984500000.
+\]
+Here $W_{\rm c}=2v_{\rm c}^3+3v_{\rm c}^2(v_{\rm c}z_{\rm c}+26)$,
+$z_{\rm c}=\binom{22}{3}+66$, and $L_{\rm c}=3v_{\rm c}^2\cdot25\cdot26$.
+
+\begin{proposition}\label{prop:compact-complex-interface}
+The original complex phase interface, including its signed scalar bank
+exchange, arbitrary-scratch restoration, binary residual bases and endpoint
+corrections, holds with these constants. Its normalized residual saving
+supports the strict exponent $\sigma=1-418/10^{12}$.
+\end{proposition}
+\begin{proof}
+The original construction and residual table apply at this ground size.
+For an intersection of size $j$, the center coefficient is $(j-1)/2$; the
+side correction is its negative for distinct even-intersection neighbors.
+Their sum is zero for $j=0,1,2$ and one for $j=3$. The same eight-step
+schedule therefore restores arbitrary scratch and gives the elementary bank
+shear, and the three stages give the signed exchange.
+
+Binary triple indicators have norm one. The orthogonal decompositions in
+the residual table remain nondegenerate. Every nonzero residual has a
+norm-one vector: choose a coordinate outside the support of a triple or two
+neighboring triples (at most six coordinates), or outside an earlier/future
+tensor indicator (support $3^j<25^j$); tensor with the remaining norm-one
+factors. Thus the original nonalternating-space argument supplies the required
+orthonormal bases. No even-ground-size matching or rational bit-interface
+positivity is needed for this unchanged complex network.
+
+The rank accounting gives
+\[
+ s_{\rm c}=W_{\rm c}m_{\rm c}-2N_{\rm c}+2L_{\rm c},\qquad
+ \eta_{\rm c}=\frac{W_{\rm c}m_{\rm c}-s_{\rm c}}{W_{\rm c}m_{\rm c}}
+             =\frac{14}{3464399375}>0.
+\]
+The needed positivity is $L_{\rm c}<N_{\rm c}$, not
+$2L_{\rm c}<N_{\rm c}$. Endpoint weight-27 phase corrections and the
+signed-exchange correction are unchanged. Exact rational logarithm bounds
+give $\log m_{\rm c}<966/100$ and
+$\eta_{\rm c}>(418/10^{12})(966/100)$. The inequality
+$e^{-x}>1-x$ then yields $s_{\rm c}/W_{\rm c}<m_{\rm c}^{\sigma}$.
+The certificate also checks $2\le s_{\rm c}<m_{\rm c}^5$.
+\end{proof}
+
+The bit and complex dimensions need not agree. The bit interface supplies
+swaps at arbitrary binary widths. The complex recursion uses only that
+interface's exponent $\tau$ and fixed tape guarantee, while its own branching,
+role split and depth use $m_{\rm c},W_{\rm c},s_{\rm c}$. All occurrences of
+$m,W,s$ within the revised layer proof denote these complex constants.
+The retained assembly consumes their time exponents and completed-layer
+contract, so it introduces no equality requirement between the two arities.
+
+The exponents used from now on are
+\begin{equation}\label{eq:explicit-motif-exponents}
+ \tau=1-296/10^{11},\qquad\sigma=1-418/10^{12}.
+\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 research draft; see main.tex for scope.
 \section{Faster interchange of address chunks}
 \label{sec:tape-interchange}
 
@@ -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:paired-bit-interface}
+supplies these conditions with $W=W_{\rm b}^{\rm pair}$ and $s=s_{\rm b}^{\rm pair}$.
 Its negative source projection is essential to the endpoint identity and
 has already been included in the rank sum.
 
@@ -270,7 +272,7 @@
 \end{proof}
 
 
-We use the rational value $\tau=1-2^{-50}$ established in
+We use the rational value $\tau=1-\frac{296}{10^{11}}$ established in
 \eqref{eq:explicit-motif-exponents}. Its verification uses fixed integer
 comparisons and the displayed exponential inequality; it needs no
 real-arithmetic oracle.
--- 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 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
+$m=m_{\rm c}=15625$, $W=W_{\rm c}=58645352620000$ and
+$s=s_{\rm c}=916333630984500000$ from
+Proposition~\ref{prop:compact-complex-interface}. The bit-swap primitive
+retains its separate arity $m_{\rm b}=125000$ and exponent
+$\tau=1-296/10^{11}$.
 
 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:compact-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:compact-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:compact-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:compact-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 +721,185 @@
 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.
+
+\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 $\tau=1-296/10^{11}$ and $\sigma=1-418/10^{12}$.
+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 +973,11 @@
 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:compact-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 +993,7 @@
 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 +1016,26 @@
 
 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 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