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\title{\LARGE The Shared Randomness Cost of Optimal\\ One Use Loop Communication}
\author{Daniel A. Pyle\\\normalsize Independent Researcher\\\small\href{https://orcid.org/0009-0005-6456-1370}{ORCID 0009-0005-6456-1370}}
\date{Preprint\\31 October 2026}
\hypersetup{pdftitle={The Shared Randomness Cost of Optimal One Use Loop Communication},pdfauthor={Daniel A. Pyle},pdfsubject={Exact raw seed support and entropy in a stipulated normalized-loop model}}
\begin{document}
\maketitle
\begin{abstract}
We determine the minimum support and Shannon entropy of a shared classical seed needed to attain optimal worst-message success in a specified one-use normalized-loop model. The loop contains a backward binary symmetric channel with crossover $0
0\}|$, and its Shannon entropy in bits is $H(\pi)=-\sum_{s:\pi_s>0}\pi_s\log_2\pi_s$. Let $K_*$ and $H_*$ be their respective minima among protocols attaining $t_*$ in \eqref{eq:rawopt}.
\end{definition}
Seed support counts positive seed probabilities, not decoder outcomes or local random choices. A seed with support $K$ can be stored using $\lceil\log_2K\rceil$ binary memory cells. Its entropy is not an exact single-trial fair-bit sampling cost; exact sampling may require a variable number of fair bits. No claim equating these different resources is made.
\begin{theorem}[Minimum support and entropy]\label{thm:cost}
Let $M\ge3$, $0
2$ and $k>0$, \eqref{eq:R2} is strictly increasing in $d$, forcing $d=\delta$. Also $00$. Associate each seed with an edge between its two active labels and give it weight $\pi_s$. Parallel edges can be merged: a standard diameter-pair construction with their summed probability has the same pattern \eqref{eq:pairpattern}. Merging cannot increase support or entropy. If $a_m$ is the total probability of edges incident with message $m$, then
\begin{equation}
P(m\mid m)=\frac{Aa_m}{\ell+(1-\ell)a_m}.\label{eq:degree}
\end{equation}
This is strictly increasing in $a_m$, and $\sum_ma_m=2$. Equality with $t_*$ for every message is equivalent to
\begin{equation}
a_m=\frac2M\quad\text{for every }m.\label{eq:degrees}
\end{equation}
Conversely, any weighted graph obeying \eqref{eq:degrees} gives an optimum by using the same diameter-attaining pair on each edge. Rescale $x_e=M\pi_e/2$. The entire seed-resource problem is now
\begin{equation}
x_e\ge0,\qquad \sum_{e\ni m}x_e=1\quad(m\in\mathbb Z_M),\label{eq:fpm}
\end{equation}
the loopless fractional perfect matching polytope. Its graph-theoretic structure is standard \cite{Behrend}; we include the needed special-case proof.
\subsection{Minimum support in the interior}
Every message must lie on an edge, so at least $M/2$ edges are required. For even $M$, disjoint pairs of weight $2/M$ attain that bound. For odd $M$, a degree-one vertex forces its unique edge to have probability $2/M$, which saturates its neighbor's degree too. Such vertices therefore form isolated pairs. Remove all such pairs. The remaining graph has minimum degree at least two and, because the original number of vertices was odd, contains at least three vertices. If it has $L$ vertices, it has at least $L$ edges, while the removed pairs contribute $(M-L)/2$. Thus
\begin{equation}
K\ge\frac{M+L}{2}\ge\frac{M+3}{2}.\label{eq:Kodd}
\end{equation}
A triangle whose three edges each have probability $1/M$, together with $(M-3)/2$ disjoint edges of probability $2/M$, attains this lower bound.
\subsection{Minimum entropy in the interior}
Entropy is continuous and concave on the compact polytope \eqref{eq:fpm}, so a minimum is attained at an extreme point. Its support components are isolated edges with $x_e=1$ and odd cycles with $x_e=1/2$. To verify this structure, first remove leaf components, which as above are isolated pairs. Any remaining connected component has minimum degree two. If it had more edges than vertices, its unsigned incidence columns would be linearly dependent, giving a nonzero signed perturbation that preserves the constraints and, for sufficiently small magnitude, nonnegativity. It would not be extreme. Hence it is a cycle. An even cycle admits an alternating perturbation; an odd cycle has the unique solution $x_e=1/2$. Conversely, these isolated-edge and odd-cycle components each have a unique supported solution.
If $L$ vertices lie on odd cycles, the seed probabilities are $(M-L)/2$ entries of $2/M$ and $L$ entries of $1/M$. Their entropy is
\begin{equation}
H=\log_2M-1+\frac LM.\label{eq:entropyL}
\end{equation}
For even $M$, $L=0$ is possible and optimal. For odd $M$, the minimum is $L=3$. The constructions minimizing support thus minimize entropy too. This proves the interior rows of Theorem~\ref{thm:cost}.
\section{The two diameter endpoints}\label{sec:endpoints}
The interior equality argument uses both $\delta>0$ and $\delta<1$. Their failures correspond to different feasible decoder structures, not failures of the success formula.
\subsection{Perfect distinguishability}\label{sec:perfect}
Suppose $\delta=1$ and the output is a qubit. The common-label bound \eqref{eq:R1} may now attain $t_*$. Any attaining extreme component uses outputs at distance one, which are orthogonal pure qubit states. Let $H_+,H_-$ be their complementary rank-one projectors, with $\Tr(H_\pm\Delta)=\pm1$.
A distinct-label component still uses $\id,0$ at its two active labels and $H_-$ elsewhere. A common-label component at $u$ uses $H_+$ at $u$ and $H_-$ at every other message. No two different optimal label patterns can coexist in one seed's extreme decomposition. Two distinct pair patterns conflict as in Lemma~\ref{lem:equality}. A pair and a singleton conflict because the pair has constant active effects but the singleton has only rank-one effects. Two singleton patterns at different labels also conflict: because $M\ge3$, a third label is inactive in both and fixes their $H_-$ to be the same; the first active label would then need to have both complementary projectors. Thus each supported seed is a single pair or a single singleton.
Represent a singleton at $m$ by a loop and count that loop twice in its weighted degree. This loop convention is stated explicitly; it differs from the once-counted convention used in \cite{Behrend}. If $a_m$ is the weighted degree, the averaged correct weight and normalizer are
\begin{equation}
\overline w_m(m)=Aa_m,\qquad
\overline Z_m=(1-k)+ka_m.\label{eq:perfectdegrees}
\end{equation}
A singleton of weight $\pi$ contributes correct weight $\pi(1+k)=2A\pi$ and increases the normalizer above its inactive value by $2k\pi$, which explains the double counting. Optimality again requires $a_m=2/M$.
At least $\lceil M/2\rceil$ seeds are needed to cover all messages. Disjoint pairs, with one singleton when $M$ is odd, attain this count. Every pair probability is at most $2/M$ and every singleton probability at most $1/M$. Among all probability vectors with atom cap $2/M$, entropy is minimized by filling as many atoms to that cap as possible and placing any residual in one atom. This follows, for example, by transferring mass from a smaller nonzero atom to a larger unsaturated atom: concavity of $-x\log_2x$ makes entropy nonincreasing. The resulting vector is $M/2$ entries of $2/M$ for even $M$, or $(M-1)/2$ entries of $2/M$ and one of $1/M$ for odd $M$. The matching-plus-singleton construction realizes it, proving the unit-diameter rows of Theorem~\ref{thm:cost}.
\subsection{Complete erasure of distinguishability}\label{sec:zero}
If $\delta=0$, the channel is a replacer: for a fixed state $\sigma$, $\Phi(\tau)=\Tr(\tau)\sigma$. Put $q_{yb}=\Tr\tau_{yb}$ and $e_m=\Tr(E_m\sigma)$. Every normalizer is one, and
\begin{equation}
w_m(m)=(p+ke_m)q_{m0}+(A-ke_m)q_{m1}.\label{eq:replacer}
\end{equation}
The sum of correct weights is at most $2A$. Equality forces $e_m=1$ whenever $q_{m0}>0$ and $e_m=0$ whenever $q_{m1}>0$. Thus the two row label distributions have nonempty disjoint supports. Conversely, any two such distributions with these constant effects attain aggregate correct weight $2A$.
At the unrestricted optimum $2A/M$, every supported seed must attain that aggregate bound. Consider a seed of probability $\pi$. Its row supports have sizes $r_0,r_1$ with $r_0+r_1\le M$. One support has at most $\lfloor M/2\rfloor$ labels and receives total correct weight $\pi A$ from this seed. Its total correct weight from all seeds is $2A$ times its size divided by $M$. Hence
\begin{equation}
\pi\le\frac{2\lfloor M/2\rfloor}{M}.\label{eq:zerocap}
\end{equation}
For even $M$, one seed suffices: split the messages into equal groups and make each row uniform on its group. For odd $M$, \eqref{eq:zerocap} gives the cap $(M-1)/M<1$, so at least two seeds are required and entropy is at least $\hbin(1/M)$ by the same mass-transfer argument.
For an explicit attaining odd construction, write $M=2r+1$. Partition the labels into $U$ of size $r$ and $V$ of size $r+1$, and choose distinct $v_1,v_2\in V$. In a dominant seed of probability $2r/M$, let $q_{\cdot0}$ be uniform on $U$. Let $q_{\cdot1}$ assign probability $1/(2r)$ to each of $v_1,v_2$ and $1/r$ to each other member of $V$. In a second seed of probability $1/M$, concentrate row zero at $v_1$ and row one at $v_2$. Each seed has disjoint row supports. Each message's averaged $q_{m0}+q_{m1}$ equals $2/M$, and its success is $2A/M$. This attains both lower bounds. No dimensional assumption was used, completing Theorem~\ref{thm:cost}.
\section{Why output dimension matters}\label{sec:dimension}
The full-joint-support extension in Theorem~\ref{thm:cost} is sufficient, not asserted necessary. Without an appropriate support condition, the forward output can retain information about the decoder's private random choice. Then several extreme label patterns may coexist within one seed without forcing contradictory encoder effects.
\begin{proposition}[Same diameter and different seed cost]\label{prop:dimension}
For every $0