h46-shared-side-75.patch

--- a/build/main.tex
+++ b/build/main.tex
@@ -27,7 +27,7 @@
 \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
+$\kappa=2^{-75}$, 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
@@ -25,7 +25,7 @@
 inputs. Its worst-case running time satisfies
 \[
  T_A(n)=O\!\left(n(\lg n)^{1-\kappa}\right),
- \qquad \kappa=2^{-182}.
+ \qquad \kappa=2^{-75}.
 \]
 \end{theorem}
 
--- a/build/sections/08-assembly.tex
+++ b/build/sections/08-assembly.tex
@@ -16,32 +16,28 @@
 \subsection{A fixed rational choice}
 
 The two motif exponents were fixed in \eqref{eq:explicit-motif-exponents}.
-Take the following fixed rational numbers:
+Put $a=27/10^{12}$ and $a_{\rm c}=9/500000000000$ and 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=\frac9{10},\quad \delta=\frac1{16},\quad C_1=20,\\
+ \epsilon=\frac1{21},\quad c=\frac{9a}{10},\quad
+ \lambda=1-\frac{19a^2}{20},\quad \lambda'=1-\frac{9a^2}{10},\quad
+ \kappa=2^{-75}.
 \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 layer conditions follow from
+\[
+ \tau(1+c/\beta)=1-a^2<\lambda<\lambda'<1,
+ \qquad \sigma+\beta(1-\sigma)=1-a_{\rm c}/10<\lambda'.
+\]
+The separate layer requirement $\sigma<\lambda$ follows from $19a^2/20<a_{\rm c}$; the leaf comparison follows from $9a^2<a_{\rm c}$. The rational stopping comparison is $e^{10}<d^9$. The generalized
+layer proof retains the original guard exponent; $\epsilon C_1=20/21<1$.
 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
+ \epsilon(1-\tau)<1-\tau,\quad
  \frac34+\delta+\frac32\epsilon<1,\\
  \epsilon(1+c)<1,\qquad \epsilon+\delta<1.
 \end{gathered}
@@ -184,8 +180,8 @@
 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, $d=\lfloor b^{1/21}\rfloor$ is found by binary search
+using exact comparisons $d^{21}\le b$; 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 +288,28 @@
 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
@@ -834,7 +843,7 @@
 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 & $d^2(1+\ell^\tau)$ & $\tau+\epsilon(2-\tau)$\\
+CRT and axis layouts & $d(1+\ell^\tau)$ & $\tau+\epsilon(1-\tau)$\\
 Gaussian line maps & $dp^{1/2+\delta}\alpha$ & $3/4+\delta+3\epsilon/2$\\
 Chirps, twists and scalar products & $dp^\delta$ & $\epsilon+\delta$\\
 Packed polynomial products & $\log(rp)$ & $1-\epsilon$\\
@@ -853,20 +862,25 @@
  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_4&=(1-\tau)(1-\epsilon),\qquad
  g_5=1/4-\delta-3\epsilon/2,\\
  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
+Direct substitution, using the nonadjacent layout bound, gives
+\[
+ G:=\min_i g_i=g_2=g_3=\frac{3a^2}{70}>\kappa=2^{-75}.
+\]
+Indeed, $g_1>1/2$, $g_4=20a/21$, $g_5=13/112$,
+$g_6=299/336$, and $g_7=1/21$, all exceeding $G$.
+The guard is $d^{20}=\Theta(p^{20/21})=o(p)$;
+$\alpha=\Theta(p^{23/84})$ and $\gamma=O(p^{25/42})=o(p)$.
+Also $K=\Theta(p^{3a/70})=o(\ell)$ and $K/\log p\to\infty$,
+while $\ell=\Theta(p^{20/21})$. Record length remains superpolynomial,
+and the prime-interval ratio grows as $p^{19/21}$.
+Put $\rho=G-\kappa>0$; this fixed gap absorbs the logarithmic factors.
+
+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 +897,12 @@
 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 $i$.
@@ -992,8 +1009,8 @@
 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
@@ -13,8 +13,8 @@
 
 Put
 \[
- h=100,\qquad v=\binom h3=161700,\qquad
- N=v^3=4227952113000000,\qquad m=h^3=1000000.
+ h=46,\qquad v=\binom h3=15180,\qquad
+ N=v^3=3497963832000,\qquad m=h^3=97336.
 \]
 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 +34,7 @@
 Each such invocation has its own auxiliary wires, distinct from all
 other invocations, including those in other stages.  There are
 \[
- I=3v^2=78440670000
+ I=3v^2=691297200
 \]
 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 +127,19 @@
 
 A fixed triple has
 \[
- z_{\rm b}=3\binom{97}{2}=13968,\qquad
- z_{\rm c}=\binom{97}{3}+3\cdot97=147731
+ z_{\rm b}=3\binom{43}{2}=2709,\qquad
+ z_{\rm c}=\binom{43}{3}+3\cdot43=12470
 \]
 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 $c_{\rm b}=46$ and $c_{\rm c}=47$ 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}\\
+          =28434979789999200,\label{eq:bit-wire-count}\\
  W_{\rm c}&=2N+I(vz_{\rm c}+c_{\rm c})
-          =1873807244643542670000.\label{eq:complex-wire-count}
+          =130865855373752400.\label{eq:complex-wire-count}
 \end{align}
 All these numbers are fixed, independently of any eventual input length.
 
@@ -212,7 +212,7 @@
  \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
+$1-h/9=-37/9$, 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 +225,7 @@
 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 $h=46$.
 
 The ambient label space is $\mathcal F=F^{\otimes3}$ with its tensor
 bilinear form.  It has dimension $m$ and is nondegenerate.  Write
@@ -458,8 +458,8 @@
 
 The loss ratios have the exact values
 \[
- \frac{L_{\rm b}}{N}=\frac{100}{539},\qquad
- \frac{L_{\rm c}}{N}=\frac{101}{539}.
+ \frac{L_{\rm b}}{N}=\frac{23}{55},\qquad
+ \frac{L_{\rm c}}{N}=\frac{47}{110}.
 \]
 In particular, both are smaller than $1/2$.  We now turn the dimension
 count into the two interfaces needed later.
@@ -503,7 +503,7 @@
  &=\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.
+  =2767747192266968049600<W_{\rm b}m.
 \end{align}
 \end{proposition}
 
@@ -574,7 +574,7 @@
 The sum of these basis lengths is
 \[
  s_{\rm c}=W_{\rm c}m-2N+2L_{\rm c}
- =1873807244636671267308000000<W_{\rm c}m.
+ =12737958894652805035200<W_{\rm c}m.
 \]
 After source and sink diagonal corrections, and undoing the signed
 bank exchange, telescoping these phase frames gives $C^{\otimes m}$
@@ -668,33 +668,119 @@
 the required forward tensor product.
 \end{proof}
 
+
+\subsection{Sharing side roles between the first and third stages}
+\label{sec:side-role-sharing}
+We improve only the bit network. The complex network and its guard-width
+constants remain unchanged. Here $h=46$ is even. Fix the same ordering of
+$\mathcal T$ in all three factors, and list each triple's $z_{\rm b}$
+neighbors in that order. Write $n_S(j)$ for the $j$th neighbor of $S$.
+
+There is an explicit permutation $\pi$ of $\mathcal T$ such that
+$|T\cap\pi(T)|=1$. Pair the ground points consecutively. If $T$ contains
+one complete pair and a singleton, keep the singleton and cycle the complete
+pair through the ordered list of pairs other than the singleton's pair.
+There are at least two choices in that list. Otherwise $T$ uses three
+pairs: keep the point from the least indexed pair and flip the other two
+points to their partners. The first rule is a cycle on each class with a
+fixed singleton; the second is an involution. The classes are disjoint,
+so the map is bijective and has the required intersection in every case.
+In particular $t_T\perp t_{\pi(T)}$ for the rational label form.
+
+A stage-1 side role is indexed by fixed second and third triples $(A,B)$
+and physical target/source triples $(S,n_S(j))$ in the first factor.
+Identify it with the stage-3 side role whose fixed first and second triples
+are $(S,\pi(A))$, and whose target/source triples are $(B,n_B(j))$.
+This is a bijection between all stage-1 and stage-3 side roles: from the latter
+role recover $A$ using $\pi^{-1}$ and recover $j$ from the neighbor list of $B$.
+Thus exactly
+\[
+ R=N z_{\rm b}=9475984020888000
+\]
+pairs of roles are identified. Stage 2 retains its separate side roles;
+all central roles remain separate.
+
+The first-stage invocations finish before the third-stage invocations begin.
+Each scalar invocation restores its own auxiliary values for every initial
+value. Consequently the same physical auxiliary value can be used in its
+paired invocation without changing the scalar bank exchange or the restoration
+of any remaining auxiliary role. No gate within one invocation identifies two
+of its simultaneous roles.
+
+It remains to check the address frames: scalar restoration alone would not
+justify the rank saving. The last stage-1 gate on this shared role has label
+\[
+ E=F\otimes\langle t_A\rangle\otimes\langle t_B\rangle,
+ \qquad \dim E=h.
+\]
+Its first stage-3 gate has label
+\[
+ H=\langle t_S\otimes t_{\pi(A)}\rangle^\perp_{F\otimes F}
+       \otimes\langle t_B\rangle,
+ \qquad \dim H=h^2-1.
+\]
+Orthogonality of $t_A,t_{\pi(A)}$ gives $E\subset H$. Both spaces are
+nondegenerate by the same tensor decompositions as before. The new joining
+edge therefore has rational projection-difference rank $h^2-1-h$.
+It replaces the old edge from $E$ to the stage-1 scratch sink $\mathcal F$
+and the old edge from the stage-3 scratch source $0$ to $H$. Their ranks
+sum to $(m-h)+(h^2-1)$, exactly $m$ more. All other edges and all gate
+labels are unchanged; no new decreasing edge is introduced.
+
+Every shared role still has initial matrix zero and final matrix $I_m$,
+and the scalar permutation fixes that role. Data-role endpoints are unchanged.
+Thus the all-role endpoint identity in the bit interface is retained.
+The new finite network has
+\begin{align*}
+ W_{\rm b}^*&=W_{\rm b}-R=18958995769111200,\\
+ s_{\rm b}^*&=s_{\rm b}-mR=1845392811609813681600,\\
+ W_{\rm b}^*m-s_{\rm b}^*&=N-2L_{\rm b}>0.
+\end{align*}
+\begin{proposition}[Shared-role bit interface]\label{prop:reused-bit-interface}
+The shared-role bit network uses pointwise XOR gates and is a permutation on
+all its input roles, including arbitrary auxiliary inputs. Its rational gate
+matrices satisfy the endpoint identities of
+Proposition~\ref{prop:bit-motif-interface}, with $W_{\rm b}^*$ roles and
+total edge rank $s_{\rm b}^*<W_{\rm b}^*m$.
+\end{proposition}
+\begin{proof}
+The scalar restoration, new-edge rank, and unchanged endpoint identities have
+just been verified. Apply the common-frame identity to the resulting fixed
+circuit. Its gate matrices and rational shear factorizations are finite data,
+so the original fixed-tape interchange construction applies with these new
+constants. Neither the matching nor the larger finite compilation depends on
+the eventual input length.
+\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}
+Put
+\[
+ a_{\rm b}=\frac{27}{10^{12}},\qquad
+ a_{\rm c}=\frac9{500000000000},\qquad L=\frac{5743}{500}.
+\]
+The new bit deficit and unchanged complex deficit are
+\[
+ \eta_{\rm b}^*=\frac{W_{\rm b}^*m-s_{\rm b}^*}{W_{\rm b}^*m}
+   =\frac9{29015910268}>a_{\rm b}L,
+ \qquad
+ \eta_{\rm c}=\frac7{22253827054}>a_{\rm c}L.
+\]
+For $1\le x\le2$ put $z=(x-1)/(x+1)$ and
+\[
+ S(x)=2\sum_{j=0}^{23}\frac{z^{2j+1}}{2j+1},\qquad
+ R(x)=\frac{2z^{49}}{49(1-z^2)}.
+\]
+The logarithm series gives $S(x)\le\log x\le S(x)+R(x)$.
+Exact rational comparison at $m=97336$ gives
+\[
+ \log m\le16(S(2)+R(2))+S(m/2^{16})+R(m/2^{16})<L.
+\]
+For either corresponding saving $a$ and deficit $\eta$,
+$m^{1-a}=m\exp(-a\log m)>m(1-aL)>m(1-\eta)$.
+We may therefore take
+\begin{equation}\label{eq:explicit-motif-exponents}
+ \tau=1-a_{\rm b},\qquad \sigma=1-a_{\rm c},
 \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.
+so $s_{\rm b}^*/W_{\rm b}^*<m^\tau$ and
+$s_{\rm c}/W_{\rm c}<m^\sigma$. The two networks have the same fixed
+label dimension here but different wire counts, as allowed by their interfaces.
--- a/build/sections/04-swap.tex
+++ b/build/sections/04-swap.tex
@@ -121,8 +121,8 @@
  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:reused-bit-interface}
+supplies these conditions with $W=W_{\rm b}^*$ and $s=s_{\rm b}^*$.
 Its negative source projection is essential to the endpoint identity and
 has already been included in the rank sum.
 
@@ -270,7 +270,7 @@
 \end{proof}
 
 
-We use the rational value $\tau=1-2^{-50}$ established in
+We use the rational value $\tau=1-\frac{27}{10^{12}}$ 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
@@ -629,7 +629,7 @@
 have completed. Intermediate scratch values may be arbitrary.
 
 \paragraph{Unrolling the recurrence.}
-Fix \(\beta=1/2\), and choose real exponents satisfying
+Fix a rational \(0<\beta<1\), and choose real exponents satisfying
 \[
  0<\tau<1,\qquad
  \max\{0,\log_m(s/W)\}\le\sigma<1,
@@ -685,6 +685,12 @@
 active chunks, was handled by individual kernels.  Thus all constants
 and all tape counts are independent of \(p,d,r\) and the input data.
 
+The argument uses only $0<\beta<1$. The stopping comparison is
+effective for fixed rational $\beta$: if $\beta=u/v$, compare $e^v<d^u$.
+It takes polynomial descriptor work. The recursion depth remains at most
+$\lceil\log_m d\rceil$, and every leaf still has $e\le d$; hence the
+guard-width argument below and the fixed tape count are unchanged.
+
 \subsection{An explicit guard-width bound}
 
 The accelerated network is evaluated exactly until an entire
@@ -702,12 +708,12 @@
  C_1=\max\{20,\nu+3\},\qquad C_0=128mB^2,
  \qquad \Delta=\lceil C_0d^{C_1}\rceil.
 \]
-For the stated \(h=100\) construction, the complex network has
+For the stated \(h=46\) construction, the complex network has
 \[
 \begin{split}
- m&=1{,}000{,}000,\\
- W&=1{,}873{,}807{,}244{,}643{,}542{,}670{,}000,\\
- s&=1{,}873{,}807{,}244{,}636{,}671{,}267{,}308{,}000{,}000.
+ m&=97{,}336,\\
+ W&=130{,}865{,}855{,}373{,}752{,}400,\\
+ s&=12{,}737{,}958{,}894{,}652{,}805{,}035{,}200.
 \end{split}
 \]
 These values give \(m^{10}<B<m^{11}\), hence \(\nu=11\) and
@@ -827,7 +833,7 @@
 \label{prop:simultaneous-layer}
 Fix
 \[
- \tau=\sigma=1-2^{-50},\qquad \beta=\frac12,\qquad C_1=20.
+ \tau=1-\frac{27}{10^{12}},\qquad \sigma=1-\frac9{500000000000},\qquad 0<\beta<1\quad(\beta\in\mathbb Q),\qquad C_1=20.
 \]
 Let $c,\epsilon,\lambda,\lambda'$ be fixed positive rational numbers
 satisfying
--- a/build/sections/07-resampling.tex
+++ b/build/sections/07-resampling.tex
@@ -330,7 +330,7 @@
 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 +377,49 @@
 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