\documentclass[11pt]{article} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage[margin=1in]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{microtype} \usepackage{booktabs} \usepackage[hidelinks]{hyperref} \usepackage{url} \urlstyle{same} \linespread{1.035} \setlength{\parskip}{3pt} \setlength{\emergencystretch}{2em} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \DeclareMathOperator{\Tr}{Tr} \DeclareMathOperator{\TD}{TD} \newcommand{\id}{I} \newcommand{\cD}{\mathcal D} \newcommand{\cS}{\mathcal S} \newcommand{\Sraw}{S_{\mathrm{raw}}} \title{\LARGE Shared Randomness and Normalization\\ in One-Use Loop Communication} \author{Daniel A. Pyle\\\normalsize Independent Researcher} \date{Research draft --- 30 October 2026} \hypersetup{pdftitle={Shared Randomness and Normalization in One-Use Loop Communication},pdfauthor={Daniel A. Pyle},pdfsubject={Exact finite-message optimization in a specified normalized-loop model}} \begin{document} \maketitle \begin{abstract} We study a finite-message communication model defined by contracting one backward binary symmetric channel with one fixed, finite-dimensional forward quantum channel and then normalizing the resulting weights. The devices may share a finite classical seed that is independent of the message \emph{before normalization}. The seed branches are mixed at the level of unnormalized weights, followed by a single normalization for each message. For $M\geq2$ messages, backward crossover probability $0
0}\frac{D_s}{Q_s}.
\label{eq:seedbound}
\end{equation}
Thus an upper bound on every unseeded aggregate ratio
\begin{equation}
R=\frac{\sum_m w_m(m)}{\sum_m Z_m}
\label{eq:single}
\end{equation}
also bounds the seeded minimax objective. This step bounds a ratio of sums; it does not replace the objective with an unweighted average of conditional successes.
\begin{lemma}
\label{lem:extreme}
To upper-bound \eqref{eq:single}, it suffices to consider decoders that, for each $b$, choose one deterministic label $y_b$ and prepare one pure state $\rho_b$.
\end{lemma}
\begin{proof}
Fix all encoder effects. For a fixed $b$, the feasible tuples $(\tau_{yb})_y$ in \eqref{eq:decoder} form a compact convex set. Spectrally decomposing each block expresses any such tuple as a convex combination of tuples with exactly one nonzero block, that block being a rank-one unit-trace state. Conversely, each such tuple is extreme: positivity forces any convex decomposition to retain its only nonzero block, and a pure state has no nontrivial state decomposition. The feasible decoder set is the product of the two sets for $b=0,1$. Taking products of the two finite decompositions gives a convex decomposition into the stated decoders.
For fixed effects, the numerator $N$ and denominator $T$ of \eqref{eq:single} are linear in the decoder blocks. If a decoder decomposes with weights $\lambda_j$, then $T_j>0$ by \eqref{eq:positive} and
\[
\frac{N}{T}=\sum_j\frac{\lambda_jT_j}{T}\,\frac{N_j}{T_j}.
\]
At least one component has ratio no smaller than the original. An upper bound for all such components and all encoder effects is therefore an upper bound for every decoder.
\end{proof}
Fix a decoder from Lemma~\ref{lem:extreme}. Write
\begin{equation}
\sigma_b=\Phi(\rho_b),\qquad
\Delta=\sigma_0-\sigma_1,\qquad
d=\tfrac12\|\Delta\|_1\leq\delta(\Phi).
\end{equation}
For each encoder effect,
\begin{equation}
Z_m=1+k\Tr(E_m\Delta),\qquad 1-kd\leq Z_m\leq1+kd.
\label{eq:extremez}
\end{equation}
The endpoint values are attainable by the negative and positive spectral projectors of $\Delta$, respectively. Its zero eigenspace may be assigned arbitrarily.
\subsection{Two different output labels}
Suppose $y_0\ne y_1$ and relabel these messages as $0$ and $1$. Their correct weights are
\begin{equation}
w_0(0)=p+k\Tr(E_0\sigma_0),\qquad
w_1(1)=A-k\Tr(E_1\sigma_1).
\end{equation}
Every other correct weight is zero. For any $t\in[0,1]$,
\begin{align}
w_0(0)-tZ_0
&=p-t+k\Tr\!\left[E_0\bigl((1-t)\sigma_0+t\sigma_1\bigr)\right]
\leq A-t,
\label{eq:activezero}\\
w_1(1)-tZ_1
&=A-t-k\Tr\!\left[E_1\bigl(t\sigma_0+(1-t)\sigma_1\bigr)\right]
\leq A-t.
\label{eq:activeone}
\end{align}
The operators inside both traces are positive and have trace one. The first trace is at most one and the second is nonnegative. The two inequalities hold with equality at $E_0=\id$ and $E_1=0$. For each remaining message, \eqref{eq:extremez} gives
\[
w_m(m)-tZ_m=-tZ_m\leq-t(1-kd).
\]
Summing these inequalities yields
\begin{equation}
\sum_m\bigl(w_m(m)-tZ_m\bigr)
\leq 2(A-t)-t(M-2)(1-kd).
\label{eq:residual}
\end{equation}
Choose
\begin{equation}
R_2(d)=\frac{2A}{2+(M-2)(1-kd)}
=\frac{2A}{M-(M-2)kd}.
\label{eq:r2}
\end{equation}
This value is in $[0,1]$, since its denominator is at least $2$ and $2A<2$. Setting $t=R_2(d)$ makes the right-hand side of \eqref{eq:residual} zero. Dividing by $\sum_mZ_m>0$ proves $R\leq R_2(d)$. Equality is attained by the two constant active effects above and the negative spectral projector of $\Delta$ for every inactive message. Finally, $R_2(d)$ is nondecreasing in $d$, so $R\leq R_2(\delta(\Phi))$.
\subsection{One common output label}
Suppose instead that $y_0=y_1=y_*$. The only nonzero correct weight is $w_{y_*}(y_*)=Z_{y_*}$. The ratio \eqref{eq:single} is increasing in $Z_{y_*}$ and decreasing in every other $Z_m$. These normalizers can be chosen independently through the message-dependent effects. Equation~\eqref{eq:extremez} therefore gives the exact maximum for these preparations as
\begin{equation}
R_1(d)=\frac{1+kd}{(1+kd)+(M-1)(1-kd)}
=\frac{1+kd}{M-(M-2)kd}.
\end{equation}
Because $d\leq1$ and $1+k=2A$, it follows that
\begin{equation}
R_1(d)\leq\frac{1+k}{M-(M-2)kd}=R_2(d)
\leq R_2(\delta(\Phi)).
\end{equation}
Together with Lemma~\ref{lem:extreme} and \eqref{eq:aggregate}--\eqref{eq:seedbound}, this proves the upper bound in Theorem~\ref{thm:main} for every finite seed alphabet.
\subsection{A cyclic construction attaining the bound}
Choose states $\rho_0,\rho_1$ attaining \eqref{eq:diameter}, and choose an effect $H$ such that
\begin{equation}
\Tr\!\left[H\bigl(\Phi(\rho_0)-\Phi(\rho_1)\bigr)\right]
=-\delta(\Phi).
\end{equation}
The negative spectral projector is one choice. Let $s$ be uniform on $\mathbb Z_M$. Given $b=0$, the decoder outputs $s$ and prepares $\rho_0$; given $b=1$, it outputs $s+1$ modulo $M$ and prepares $\rho_1$. Set
\begin{equation}
E^s_m=
\begin{cases}
\id,&m=s,\\
0,&m=s+1,\\
H,&m\notin\{s,s+1\}.
\end{cases}
\label{eq:cyclic}
\end{equation}
For any fixed message $m$, exactly one seed places $m$ in the first active role and exactly one in the second. In either active role, $w^s_m(m)=A$ and $Z^s_m=1$. In each of the remaining $M-2$ roles, $w^s_m(m)=0$ and $Z^s_m=1-k\delta(\Phi)$. Thus
\begin{equation}
\overline w_m(m)=\frac{2A}{M},\qquad
\overline Z_m=\frac{2+(M-2)(1-k\delta(\Phi))}{M}.
\end{equation}
Every message attains the ratio in \eqref{eq:main}, proving equality and attainment. If $M=2$, a single fixed seed with labels $0,1$ and effects $\id,0$ already gives success $A$ for both messages, so no seed is necessary.
The construction uses $M$ seed values, with entropy $\log_2M$. This is a sufficient size, not a minimality result. Storage in binary memory requires $\lceil\log_2M\rceil$ cells. Exact uniform sampling from fair bits may use a variable number of bits when $M$ is not a power of two; storage and random-bit generation are different resource costs.
\section{Why the normalization convention matters}
\label{sec:example}
Take $M=3$, $p=1/10$, and a forward channel with $\delta(\Phi)=1$. For each message, the cyclic construction has three equally likely raw seed branches with pairs
\begin{equation}
\bigl(w^s_m(m),Z^s_m\bigr):
\qquad (9/10,1),\quad(9/10,1),\quad(0,1/5).
\end{equation}
Mixing the weights first gives
\begin{equation}
P_{\mathrm{raw}}(m\mid m)
=\frac{(9/10+9/10+0)/3}{(1+1+1/5)/3}
=\frac9{11}.
\end{equation}
By contrast, externally mixing the same three normalized branches gives
\begin{equation}
P_{\mathrm{external}}(m\mid m)
=\frac13\left(\frac{9/10}{1}+\frac{9/10}{1}+\frac0{1/5}\right)
=\frac35.
\end{equation}
This compares two conventions for the \emph{same construction}; it is not a calculation of the optimum under \eqref{eq:external}. The normalized seed probabilities in the raw model are $5/11,5/11,1/11$, rather than $1/3$ each. Which seed is suppressed depends on the message. Conditional on the inactive seed, success is zero.
There is an elementary postselection interpretation of this bookkeeping. Since $Z^s_m\leq2A$, one may define an abstract joint distribution in which a sampled seed is accepted with probability $Z^s_m/(2A)$ and, upon acceptance, outputs $y$ with probability $w^s_m(y)/Z^s_m$. Conditioning on acceptance gives \eqref{eq:raw} and \eqref{eq:posterior}. This observation explains the change in the seed law. It does \emph{not} implement the stipulated backward channel, establish a causal realization of the loop, or show that the abstract acceptance rule respects a physical communication constraint.
The theorem therefore applies only when the seed is included in the ensemble that is normalized. Requiring the seed distribution to remain message-independent after conditioning, or demanding a guarantee for every realized seed, changes the feasible resource model and requires a separate analysis.
\section{The channel coefficient and concrete comparisons}
\label{sec:diameter}
The reduction to $\delta(\Phi)$ makes the dependence on the forward channel explicit. Its identification with an established contraction coefficient is important for both interpretation and attribution.
\begin{proposition}
\label{prop:coefficient}
For an input space of dimension at least two,
\begin{equation}
\delta(\Phi)=
\sup_{\rho\ne\sigma}
\frac{\TD(\Phi(\rho),\Phi(\sigma))}{\TD(\rho,\sigma)}
=\max_{\langle u,v\rangle=0}
\TD\bigl(\Phi(|u\rangle\langle u|),\Phi(|v\rangle\langle v|)\bigr).
\label{eq:dobrushin}
\end{equation}
Here $u,v$ are unit vectors. For a one-dimensional input space, $\delta(\Phi)=0$.
\end{proposition}
\begin{proof}
For distinct states, write the Jordan decomposition as
\[
\rho-\sigma=t(\omega_+-\omega_-),\qquad
t=\TD(\rho,\sigma)>0,
\]
where $\omega_+$ and $\omega_-$ are states with orthogonal supports. Linearity shows that the ratio in \eqref{eq:dobrushin} is
$\TD(\Phi(\omega_+),\Phi(\omega_-))\leq\delta(\Phi)$. Conversely, if $\delta(\Phi)>0$, a diameter-attaining pair has $0