\documentclass[11pt,a4paper]{article} \usepackage[margin=1in]{geometry} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{lmodern,amsmath,amssymb,amsthm,mathtools,booktabs,array,microtype} \usepackage[hidelinks,pdfusetitle]{hyperref} \usepackage{enumitem} \setlist[enumerate]{itemsep=3pt,topsep=5pt} \setlist[itemize]{itemsep=3pt,topsep=5pt} \setlength{\emergencystretch}{2em} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition}\newtheorem{definition}[theorem]{Definition} \theoremstyle{remark}\newtheorem{remark}[theorem]{Remark} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\BSC}{\operatorname{BSC}} \newcommand{\Id}{\mathrm{I}} \newcommand{\Not}{\mathrm{X}} \newcommand{\ezz}{\boldsymbol{e}_z} \newcommand{\dd}{\boldsymbol{d}} \newcommand{\DD}{\boldsymbol{D}} \newcommand{\nn}{\boldsymbol{n}} \newcommand{\vv}{\boldsymbol{v}} \newcommand{\bsigma}{\boldsymbol{\sigma}} \newcommand{\SC}{S_M^{\mathrm{cl}}} \newcommand{\SQ}{S_M^{\mathrm{qu}}} \title{Finite Alphabet Communication with Noisy Forward Memory in a Normalized Loop} \author{Daniel A. Pyle\\\small Independent Researcher} \date{30 October 2026} \hypersetup{pdfauthor={Daniel A. Pyle},pdftitle={Finite Alphabet Communication with Noisy Forward Memory in a Normalized Loop},pdfsubject={Exact finite-alphabet optimization in a stipulated normalized-loop communication model}} \begin{document} \maketitle \begin{abstract} We study one-use communication of a finite classical message through a stipulated normalized loop containing a backward binary symmetric channel and one noisy forward memory carrier. The carrier is either a classical bit or a qubit; no shared random seed, retained side register, or entanglement is supplied. For every finite message count, we characterize the qubit optimum by a constructive scalar feasibility test. For rational noise parameters and a rational target success, the test uses $O(M^2)$ constant-size rational arithmetic checks. The proof combines an exact decoder reduction, minimum-trace allocation for constant encoders, and a concave geometric cap with a matching polygon construction. For three messages, the qubit and classical-bit optima coincide throughout the allowed noise range; an explicit Bernstein certificate proves the remaining polynomial inequality. For four messages, an exact rational qubit protocol exceeds a bound valid for every classical-bit protocol. These are conditional mathematical results for the declared resource model, not a construction of a physical backward channel. Independent human review and originality assessment remain pending. \end{abstract} \section{Introduction} Normalized-loop communication gives a finite-dimensional optimization problem in which both a correct-output weight and its normalization depend on the message. A useful resource question is what changes when a single noisy forward classical bit is replaced by a noisy qubit, while the backward channel, objective, and available side information are kept fixed. This paper gives a complete scalar characterization of that one-use qubit problem for every finite alphabet. It also proves that three messages do not distinguish the two memory carriers, whereas four messages can. The decoder--memory--encoder architecture is already present in Ji, Lloyd and Wilde~\cite{ji2026}. Their discussion of noisy forward memory motivates the restriction considered here, but does not specify all of its operational conventions. The conditional worst-message objective is also established prior art: Ji, Regula and Wilde~\cite{jiregula}, Eq.~(20), use the same type of message-dependent normalization. The present calculations concern a particular feasible resource set, not a new error criterion or a solution of a general capacity problem. Dimension-matched classical and quantum separations have a substantial earlier literature, including prepare-and-measure dimension witnesses~\cite{gallego}. Independent preparation and measurement devices with local but not shared randomness are studied explicitly by Bowles, Quintino and Brunner~\cite{bowles}. These precedents do not, by themselves, optimize the fixed channel-weighted, normalized objective below. No equivalence with their witnesses is asserted, and no priority claim is made for the exact formulas or characterization in this paper. The main theorem is stated in Section~\ref{sec:finite} and proved using the geometric lemma of Section~\ref{sec:cap}. Section~\ref{sec:m3} proves three-message equality; the complete classical upper bound and the Bernstein coefficient certificate are included in the appendices. Section~\ref{sec:m4} gives a rational four-message separation. All results concern a single use of each specified channel and dimension-two forward memory. They do not assert physical time travel or extend to higher-dimensional memory, additional retained registers, shared randomness, entanglement assistance, or multiple channel uses. \section{The normalized loop model}\label{sec:model} Let $M\ge1$ and let $m,y\in\{0,\ldots,M-1\}$ denote the transmitted and recovered classical messages. The backward input and output are bits $a,b$. We \emph{assume}, as part of the mathematical model, a backward binary symmetric channel \begin{equation}\label{eq:backward} N_p(b\mid a)=\begin{cases}1-p,&b=a,\\p,&b\ne a,\end{cases} \qquad 00$, and set both states to zero for $T_m=0$. Positivity follows from $|\dd_m|\le T_m$. The two row traces sum to one because $\sum T_m=2$ and $\sum u_m=0$, and the construction reproduces $\dd_m$. \end{proof} \begin{lemma}[Extreme effects suffice]\label{lem:extreme} Every protocol can be replaced, without lowering any message success, by one whose $E_0^m$ is $I$, $0$, or a rank-one projector for each $m$. \end{lemma} \begin{proof} The extreme points of the effect interval $0\le E\le I$ are orthogonal projectors. To see this, an eigenvalue strictly between zero and one can be perturbed in both directions, so such an effect is not extreme. Conversely a projector cannot be a nontrivial average of two effects: positivity forces both effects to vanish on its kernel, and the upper bound by $I$ forces both to equal $I$ on its range. In dimension two the projectors are $0$, $I$, and rank-one projectors. For a fixed decoder, success for message $m$ is a ratio of affine functions of $E_0^m$, with positive denominator. If an effect is expressed as a convex combination with coefficients $\gamma_j$, correct weights $v_j$, and denominators $z_j>0$, then \[ \frac{\sum_j\gamma_jv_j}{\sum_j\gamma_jz_j} =\sum_j\frac{\gamma_jz_j}{\sum_k\gamma_kz_k}\frac{v_j}{z_j}. \] Some extreme component therefore has at least the original success. The effect for one message does not change another message's weights, so these replacements can be made independently. Joint convexity of the decoder--encoder problem is not assumed. \end{proof} Call $E_0^m=I$ a constant-$0$ encoder and $E_0^m=0$ a constant-$1$ encoder. Their denominators are one and their successes are, respectively, \begin{equation}\label{eq:constantsuccess} A t_{m0}+p t_{m1},\qquad p t_{m0}+A t_{m1}. \end{equation} For a projective encoder $E_0^m=(I+\nn_m\cdot\bsigma)/2$, $|\nn_m|=1$, direct contraction gives \begin{equation}\label{eq:projectiveratio} S_m=\frac{T_m/2+h\nn_m\cdot\dd_m} {1+h\nn_m\cdot\DD}. \end{equation} The denominator is positive. Consequently a projective label can reach a target $t$ if and only if \begin{equation}\label{eq:threshold} \frac{T_m}{2}+h|\dd_m-t\DD|\ge t. \end{equation} For $h>0$, a maximizing threshold direction is along $\dd_m-t\DD$ whenever this vector is nonzero; otherwise any unit vector suffices. The same threshold condition is valid when $h=0$. \begin{proposition}[A universal bound]\label{prop:universal} For every $M\ge1$, \begin{equation}\label{eq:universal} \SQ(p,\epsilon)\le\frac{A}{A+(M-1)p}. \end{equation} In particular $S_1^{\mathrm{qu}}=1$ and $S_2^{\mathrm{qu}}=S_2^{\mathrm{cl}}=A$ at matched forward noise. \end{proposition} \begin{proof} For a label with trace $T_m$, the correct weight is at most $AT_m$ and the incorrect weight is at least $p(2-T_m)$, by the argument of Lemma~\ref{lem:positive}. Thus \[ S_m\le\frac{AT_m}{AT_m+p(2-T_m)}. \] This expression is increasing in $T_m$. Some label has $T_m\le2/M$, proving \eqref{eq:universal}. A single message is recovered perfectly. For two messages, direct encoding through the backward BSC, ignoring memory, attains $A$. \end{proof} \section{The geometric cap}\label{sec:cap} The next lemma controls the joint effect of trace allocation and the normalization vector $\DD$. It also supplies the attaining geometry used in the finite-alphabet theorem. \begin{lemma}[Exact longitudinal cap]\label{lem:cap} Fix $0A$ and $d_1\le A$, then feasibility is equivalent to \[ C=1+\frac{d_0-A}{p}+\frac{d_1}{A}\le2. \] The minimum is attained by $x_0=1$, $x_1=0$, $y_0=(d_0-A)/p$, and $y_1=d_1/A$. \item If $d_1>A$ and $d_0\le A$, exchange the two groups and the two rows in \textup{(ii)}. \item If both $d_0>A$ and $d_1>A$, the class is infeasible. \end{enumerate} Each nonempty group's minimizing row masses may be split equally among its labels, making every constant-label success exactly $t$. \end{lemma} \begin{proof} Every coefficient in \eqref{eq:constdemands} is at most $A$, so total trace is at least $(d_0+d_1)/A$. This proves optimality in case (i). In case (ii), let $C'=x_0+x_1+y_0+y_1$, $s_0=Ax_0+py_0$, and $s_1=px_1+Ay_1$. Then \begin{align*} C'-s_0/p-s_1/A+A/p-1 &= (A/p-1)(1-x_0)+(1-p/A)x_1\ge0. \end{align*} Consequently $C'\ge d_0/p+d_1/A+1-A/p=C$. The displayed minimizing allocation obeys its second-row capacity exactly when $C\le2$; its other constraints hold automatically. If $C>2$, the lower bound contradicts the total row capacity. Case (iii) follows by relabeling. In case (iv), summed demands exceed $2A$, but summed successes are at most $AC'\le2A$. Finally the group successes of each minimizing allocation equal $d_i=n_it$, so equal splitting gives the assertion about individual labels. \end{proof} \subsection{The exact test} \begin{theorem}[Finite-alphabet scalar characterization]\label{thm:finite} Fix $M\ge1$, $00$, and $\alpha T0$. Rotate coordinates so $\DD=r\ezz$ and apply Lemma~\ref{lem:cap}. Each projective trace is at least $L$ and each projective longitudinal coordinate is at most $g(T_j)$. Constant labels contribute at most $C'$ to the longitudinal sum. Jensen's inequality gives \begin{equation}\label{eq:jensen} r\le C'+\sum_{j=1}^Ng(T_j) \le C'+Ng\bigl((2-C')/N\bigr) \le C+Ng(T). \end{equation} For the last inequality, increasing the average projective trace from $(2-C')/N$ to $T$ increases $Ng$ by at least $C'-C$, since $g$ has slope at least one. Both averages are in its domain. Because $\alpha T0$, distribute $B\ezz$ among constant labels in proportion to their traces; if $C=0$, take $B=0$. Give the projective differences transverse components of common length $\sqrt{T^2-z^2}$ arranged as a regular $N$-gon. Their transverse sum is zero, so their total together with the constants is $\DD=r\ezz$. Each has norm $T$ and longitudinal coordinate at most $g(T)$, hence satisfies \eqref{eq:threshold}. Select its encoder direction accordingly. Lemma~\ref{lem:decoder} gives valid states with the prescribed row traces; the constants still succeed with probability $t$. No shared randomization is used by this fixed geometric construction. The count triples exhaust all extreme encoders, completing both directions. All formulas and possible quadratic vertices are rational when the inputs are rational. \end{proof} \begin{remark}[Meaning of the arithmetic bound] The theorem is a complete characterization and decision procedure, not a single elementary expression for every optimum. The $O(M^2)$ count concerns rational arithmetic tests; it is not a bit-complexity claim polynomial in $\log M$. Thresholds $t\le0$ are trivially feasible and $t>1$ are impossible. Downward closure and attainment imply that exact rational bisection gives arbitrarily narrow enclosures of the optimum for rational channel parameters, using $0$ and any rational value greater than one as initial bounds. No linear-programming or nonlinear-optimization solver is required. \end{remark} \begin{corollary}[Complete erasure]\label{cor:erasure} For $\epsilon=1$, \begin{equation}\label{eq:erasure} S_{2k}^{\mathrm{qu}}(p,1)=\frac{A}{k}\quad(k\ge1),\qquad S_{2k+1}^{\mathrm{qu}}(p,1)=\frac{A}{k+A}\quad(k\ge0). \end{equation} These values also equal the classical optima with erased forward memory. \end{corollary} \begin{proof} After complete depolarization, an effect acts only through the scalar $\Tr(E_0)/2$. Thus every encoder is a mixture of the two constant encoders, and the positive-denominator extremality argument reduces to constants alone. Suppose their multiplicities are $j\ge\ell$ with $j+\ell=M$. If $j=\ell$, Lemma~\ref{lem:constantallocation} gives maximum target $A/j$. If $j>\ell$, the best target lies in the spillover case and is \[ t=\frac{A}{jA+\ell p}. \] Indeed, $jt>A$, $\ell t\le A$, and the spillover formula equals two at this value; larger targets violate it. This also covers $\ell=0$, giving $t=1/j$. The denominator $jA+\ell p=Mp+j(A-p)$ is minimized by the most balanced integer split. Substitution gives \eqref{eq:erasure}. \end{proof} \section{Three messages give equal classical and quantum optima}\label{sec:m3} For three messages the general scalar test simplifies to a closed expression. \begin{theorem}[Three-message equality]\label{thm:m3} For every $00$, so $1/2<\alpha\le A<1$. With at most one projective encoder, all effects commute; dephasing reduces the protocol exactly to the classical model, whose optimum is $K$. It remains to bound the two-projector and three-projector classes. First record the commuting lower bound \begin{equation}\label{eq:lowerL} K\ge L_3:=\frac{2A\alpha}{2A+\alpha}\ge\frac{2\alpha}{3}. \end{equation} For a direct construction, give one constant-$0$ label trace $C=L_3/A$, imbalance $u=C$, and $\dd=0$. Give two projective labels trace $T=1-C/2$, imbalance $u=-C/2$, and opposite Bloch differences $\pm T\ezz$, with matching encoder directions. Then $\DD=0$, the constant succeeds with $AC=L_3$, and each projector succeeds with $\alpha T=L_3$. The trace imbalances are feasible since $C<1$. All effects commute, so this is a classical protocol. The second inequality in \eqref{eq:lowerL} follows from $\alpha\le A$. Suppose a protocol succeeds strictly above $K$. Choose a threshold $t>K$ below its success and with $t2\alpha/3$, \[ P_3(0,t)=3T^2-3(t-T/2)^2/h^2<0. \] If $t\le2/3$, the coefficient of $r^2$ is nonpositive and the maximum for $r\ge0$ occurs at zero. If $t>2/3$, the quadratic is convex, so its maximum on $[0,2]$ occurs at an endpoint. The other endpoint is also strictly negative, by the identity \begin{equation}\label{eq:endpointidentity} P_N(2,t)=N(4-h^{-2})(t-T/2)^2, \qquad C+NT=2. \end{equation} This identity follows by expanding \eqref{eq:P}; its sign is strict because $h<1/2$ and $t>T/2$. Thus the three-projector class cannot improve on $K$. \paragraph{Two projective labels.} Evaluate \eqref{eq:P} at $t=K$, with $C=t/A$ and $T=1-t/(2A)$. Its leading coefficient is $2t(t-1)<0$. Its unconstrained maximum $Q$ and vertex are \begin{equation}\label{eq:Q} Q=2\left[T^2-\frac{(t-T/2)^2}{h^2}\right] +\frac{tC^2}{2(1-t)},\qquad r_{\mathrm v}=\frac{C}{2(1-t)}. \end{equation} Introduce \begin{equation}\label{eq:vw} v=\frac{t}{A(1-t)},\qquad w=pv(v+1). \end{equation} The root equation \eqref{eq:Kroot} gives \begin{equation}\label{eq:substitution} t=\frac{Av}{1+Av},\qquad \alpha=\frac{Av^2}{v^2+1-w},\qquad r_{\mathrm v}=v/2. \end{equation} The strict lower bound $K>A/(A+1)$ gives $v>1$, while $p>0$ and $\alpha\le A$ give $04$, the vertex lies to the right of $[0,2]$. The constrained maximum is then the endpoint $P_2(2,K)<0$ from \eqref{eq:endpointidentity}. Here $K>T/2$ follows from $K\ge L_3$ and $\alpha>1/2$. For $1K$, contradicting its strictly negative maximum at $K$. This excludes the final encoder class and proves the theorem. \end{proof} \section{Four messages admit strict quantum advantage}\label{sec:m4} A short classical ceiling and an explicit rational qubit protocol suffice for a dimension-matched separation. No claim that the following witness is optimal is needed. \begin{lemma}[A classical four-message ceiling]\label{lem:m4classical} For all $0\tfrac12 \ge S_4^{\mathrm{cl}}(1/10,\epsilon/2). \end{equation} At $\epsilon=0$, a rational protocol attains worst-message success $259/512$. \end{theorem} \begin{proof} Use labels $0,1,+,-$ and set $x=9/16$. Define the decoder states by \begin{align} \tau_{0,0}&=x\lvert1\rangle\langle1\rvert,& \tau_{1,1}&=x\lvert0\rangle\langle0\rvert,& \tau_{0,1}&=\tau_{1,0}=0,\label{eq:witnessconstant}\\ \tau_{\pm,b}&=\frac{1-x}{4}\left[I\pm(-1)^b\sigma_x\right].&&&\label{eq:witnessprojectors} \end{align} Each state is positive semidefinite, and each row has trace $x+2(1-x)/2=1$. The effects are \begin{equation}\label{eq:witnesseffects} E_0^0=I,\qquad E_0^1=0,\qquad E_0^\pm=\frac{I\pm(3/5)\sigma_x+(4/5)\sigma_z}{2}, \qquad E_1^m=I-E_0^m. \end{equation} Since $(3/5)^2+(4/5)^2=1$, the nonconstant effects are rank-one projectors. All displayed matrix entries are rational. Let $k=(1-2p)(1-\epsilon)$. The reduced variables of the constants are $(T_0,u_0,\dd_0)=(x,x,-x\ezz)$ and $(T_1,u_1,\dd_1)=(x,-x,-x\ezz)$. The other two labels have $T_\pm=1-x$, $u_\pm=0$, and $\dd_\pm=\pm(1-x)\boldsymbol e_x$. Hence $\DD=-2x\ezz$. Equations~\eqref{eq:constantsuccess} and \eqref{eq:projectiveratio} give \begin{equation}\label{eq:familysuccess} S_0=S_1=Ax,\qquad Z_\pm=1-\frac45kx,\qquad S_\pm=\frac{(1-x)(1+3k/5)}{2(1-4kx/5)}. \end{equation} At $p=1/10$ and $x=9/16$, these simplify exactly to \begin{equation}\label{eq:witnessnoisy} S_0=S_1=\frac{81}{160},\qquad Z_\pm=\frac{16+9\epsilon}{25},\qquad S_\pm=\frac{7(37-12\epsilon)}{32(16+9\epsilon)}. \end{equation} In particular, \begin{equation}\label{eq:witnessgap} S_\pm-\frac12=\frac{3(1-76\epsilon)}{32(16+9\epsilon)}. \end{equation} Both constant successes exceed $1/2$, and \eqref{eq:witnessgap} is positive precisely on the stated interval. Lemma~\ref{lem:m4classical} proves the matched classical comparison. At $\epsilon=0$, the four successes are \[ \left(\frac{81}{160},\frac{81}{160}, \frac{259}{512},\frac{259}{512}\right), \] whose minimum is $259/512$. The protocol uses only the specified qubit: $b$ and the recovered label are not supplied to the encoder as side information. The two projective effects do not commute, so the common-basis classical reduction does not apply. \end{proof} \begin{corollary}[The smallest separating alphabet] Within the stated one-use dimension-two model, four is the smallest number of messages for which a qubit can outperform a classical bit at matched depolarizing/BSC noise. \end{corollary} \begin{proof} One and two messages have equal optima by Proposition~\ref{prop:universal}; three have equal optima by Theorem~\ref{thm:m3}; Theorem~\ref{thm:m4} supplies a four-message separation. \end{proof} \section{Scope and verification status} The scalar characterization is global over the stated local operations and finite alphabet, and includes constructive sufficiency. It does not assume a shared seed to convexify the feasible protocol set. The classical upper bounds likewise permit all allowed local stochastic encoders and decoders. The four-message witness proves separation on a positive-noise interval but is not presented as a global optimum at those parameters. Exact symbolic identities, rational arithmetic checks, and direct matrix contractions were used during development. These checks supplement the proofs printed here. Their independence is limited to separate derivations within an AI-assisted workflow; they are not independent human review or proof-assistant certification. Independent human technical review and a fuller originality assessment remain pending. The conditional normalization is part of an assumed backward-channel model and does not supply a physical mechanism for backward communication. \paragraph{Author and AI disclosure.} Daniel A. Pyle is listed as Independent Researcher. Lyra, an OpenAI AI assistant, carried out the technical derivations, code-based verification, literature checking, and manuscript drafting in the research workflow. Lyra is disclosed as an AI tool and is not listed as an author. Independent human verification of the proofs and a fuller originality assessment remain pending. \paragraph{Data and code availability.} The research materials are available at \url{https://github.com/dapyle11/normalized-loop-memory}. The paper is self-contained: the proofs and the complete finite Bernstein coefficient certificate do not require external scripts or data files. \paragraph{Classification.} Primary subject: Information Theory (math.IT). Secondary subject: Quantum Physics (quant-ph). Mathematics Subject Classification: 94A40 (primary); 81P47, 81P45 (secondary). \appendix \section{Complete classical three-message comparison}\label{app:classical} This appendix proves the classical ingredient in Theorem~\ref{thm:m3} for arbitrary $00$, $F(A)=-\beta A^2<0$, and on $[0,A]$, \[ F'(t)=-\alpha(A+1)+2(\alpha-\beta)t \le-\alpha(A+1)+2\alpha A<0. \] Thus there is exactly one root in $(0,A)$. At $t_0=A/(A+1)$, \[ F(t_0)=(\alpha-\beta)\frac{A^2}{(A+1)^2}\ge0, \] with equality exactly when $\alpha=\beta=1/2$. Hence $K\ge A/(A+1)>A/2$, with strict first inequality when $q<1/2$. Expanding the root equation gives $(2\alpha-1)K^2-\alpha(A+1)K+A\alpha=0$. Rationalizing its smaller root gives \eqref{eq:Kstable}, also valid when $\alpha=1/2$. \subsection{Three distinct encoders} Up to the stated symmetries there are two possibilities. \paragraph{Both constants and one bijection.} Take $\{C_0,\Id,C_1\}$. At $t=K$, choose \[ \lambda_{C_0}=\lambda_{C_1}=u,\quad \lambda_\Id=1, \qquad u=\frac{\beta K}{A-K}=\frac{\alpha(1-K)}K>0. \] The equality of the two expressions for $u$ is the root equation. In row $b=0$, the largest $C_0$ column in \eqref{eq:dualcolumn} is $u(A-K)-\beta K=0$, and the largest identity column is $\alpha(1-K)-uK=0$. Every $C_1$ column is no larger than the corresponding $C_0$ column because $p\le A$. Row $b=1$ is symmetric. Thus $B_0=B_1=0$ proves the upper bound $K$. \paragraph{One constant and both bijections.} Take $\{C_0,\Id,\Not\}$. Its exact optimum is \begin{equation}\label{eq:Uclass} U=\frac{2A\alpha}{2A+\alpha}. \end{equation} For the upper bound use weights $\lambda_{C_0}=z=\alpha/A$ and $\lambda_\Id=\lambda_\Not=1$. Write $N_0=A$, $N_1=p$. The maximum column in row $b$ is \[ B_b=\max\{zN_b(1-t)-t,\ \alpha-t-zN_bt\}. \] The two expressions tie in row zero; the second dominates in row one. Hence $B_0+B_1=2\alpha-t(2+\alpha/A)$, which vanishes at $t=U$. To attain $U$, let $y=2\alpha/(2A+\alpha)$. In row zero, put probability $y/2$ on each of $(C_0,0),(C_0,1)$ and probability $(1-y)/2$ on each of $(\Id,0),(\Not,1)$. In row one, put probability $1/2$ on each of $(\Id,1),(\Not,0)$. All denominators are one, and the successes are $Ay$ and $\alpha(2-y)/2$, all equal to $U$. Finally $U\le K$. Direct expansion gives \begin{align} F(U)&=\frac{A\alpha}{(2A+\alpha)^2}\,J,\label{eq:FU}\\ J&=\alpha^2(2\alpha-1) +2\alpha(A-\alpha)(2-\alpha) +4(A-\alpha)^2(1-\alpha)\ge0.\nonumber \end{align} Every displayed summand is nonnegative, and $0A/2$ requires $X>1$. At fixed $X>1$, repeated-constant success is at most $[A+p(X-1)]/2$, because $x_0\le1$. The singleton success is at most $v(X)$. By the root identity, \begin{equation}\label{eq:vthreshold} v(X)\ge K\quad\Longrightarrow\quad X\le\frac{2\alpha(1-K)}{\alpha(1-K)+\beta K} =\frac{2K}{A}. \end{equation} But the repeated messages require \[ X\ge1+\frac{2K-A}{p}>\frac{2K}{A}, \] where the strict inequality follows because the difference, multiplied by $Ap$, is $(A-p)(2K-A)>0$. This class cannot attain $K$. \paragraph{Repeated bijection and singleton constant.} The repeated messages are bounded by $u(X)$. For $X\le1$ this is at most $\alpha/2\le A/21$, the singleton constant is at most $A(2-X)$. If $K>1/2$, the universal bound $u(X)\le1/2$ already excludes this class. Otherwise singleton success at least $K$ requires $X\le2-K/A$, and \[ u(2-K/A)=\frac{\alpha(2A-K)}{2[\alpha(2A-K)+\beta K]}0, \] using \eqref{eq:Kroot}. This excludes the class. \paragraph{Repeated bijection and singleton opposite bijection.} The bounds are $u(X)$ and $v(X)$. Again $K>1/2$ is excluded immediately. For $K\le1/2$, equation~\eqref{eq:vthreshold} forces $X\le2K/A$. At this endpoint, $u(2K/A)0. \] Indeed, \begin{align*} \Delta &=2K-1+\beta(2A+1-4K)\\ &\ge2K-1+p(2A+1-4K)\\ &=(2A-1)(2K-A)>0. \end{align*} The inequality uses $\beta\ge p$ and $2A+1-4K\ge2A-1>0$. This excludes the last repeated class. These cases exhaust every deterministic triple, hence every stochastic encoder triple. Together with the attaining decoder they prove $S_3^{\mathrm{cl}}(p,q)=K$ for the full declared classical model. \section{An explicit Bernstein sign certificate}\label{app:bernstein} This appendix supplies all coefficients needed to prove the strict polynomial sign used in \eqref{eq:QR}. Define \begin{align} R(v,w)={}&v^9-5v^8-2v^7w-7v^7+14v^6w-v^6 +v^5w^2+24v^5w-9v^5\nonumber\\ &-9v^4w^2+12v^4w-19v^4 -25v^3w^2+50v^3w-25v^3\nonumber\\ &-39v^2w^2+46v^2w-7v^2 +16vw^3-8vw^2-32vw+24v\nonumber\\ &+16w^2-32w+16.\label{eq:Rpoly} \end{align} Substituting \eqref{eq:substitution}, $p=w/[v(v+1)]$, $A=1-p$, $h=\alpha-1/2$, $C=t/A$, and $T=1-C/2$ into \eqref{eq:Q}, and clearing denominators, gives \eqref{eq:QR} with \eqref{eq:Rpoly}. Thus the displayed polynomial and rational identity can be checked by finite polynomial multiplication, without a sign oracle or a numerical approximation. For $n\ge0$, write the Bernstein basis as \[ B_i^n(x)=\binom ni x^i(1-x)^{n-i},\qquad 0\le i\le n. \] The exact degree-$(9,3)$ expansion is \begin{equation}\label{eq:bernstein} R(1+3x,w)=\sum_{i=0}^9\sum_{j=0}^3b_{ij}B_i^9(x)B_j^3(w), \end{equation} where the full coefficient table is \begin{center} \renewcommand{\arraystretch}{1.3} \begin{tabular}{@{}crrrr@{}} \toprule $i$ & $b_{i0}$ & $b_{i1}$ & $b_{i2}$ & $b_{i3}$\\ \midrule 0 & $-32$ & $-16/3$ & $0$ & $0$\\ 1 & $-368/3$ & $-416/9$ & $-112/9$ & $0$\\ 2 & $-1054/3$ & $-1519/9$ & $-1141/18$ & $-17/2$\\ 3 & $-6493/7$ & $-3667/7$ & $-21037/84$ & $-2109/28$\\ 4 & $-50678/21$ & $-98048/63$ & $-8170/9$ & $-3012/7$\\ 5 & $-132970/21$ & $-288037/63$ & $-197195/63$ & $-13744/7$\\ 6 & $-233775/14$ & $-555439/42$ & $-214544/21$ & $-53376/7$\\ 7 & $-517625/12$ & $-1318745/36$ & $-277048/9$ & $-25536$\\ 8 & $-310000/3$ & $-827200/9$ & $-732416/9$ & $-71680$\\ 9 & $-200000$ & $-544000/3$ & $-163840$ & $-147456$\\ \bottomrule \end{tabular} \end{center} To verify the coefficient conversion directly, if $R(1+3x,w)=\sum_{a=0}^9\sum_{b=0}^3c_{ab}x^aw^b$, then \begin{equation}\label{eq:bernsteinconvert} b_{ij}=\sum_{a=0}^i\sum_{b=0}^jc_{ab} \frac{\binom ia}{\binom9a}\frac{\binom jb}{\binom3b}. \end{equation} This follows from the identity \[ x^a=\sum_{i=a}^n\frac{\binom ia}{\binom na}B_i^n(x), \] which is obtained using $\binom ni\binom ia=\binom na\binom{n-a}{i-a}$ and the binomial theorem. Thus \eqref{eq:Rpoly}, \eqref{eq:bernsteinconvert}, and the table form a self-contained finite certificate of \eqref{eq:bernstein}. Each $B_i^n$ is nonnegative on $[0,1]$, and the basis sums to one. All $b_{ij}$ are nonpositive. Moreover, for $x>0$ the final row contributes \[ x^9\sum_{j=0}^3b_{9j}B_j^3(w)<0, \] because all four coefficients in that row are strictly negative. The remaining rows contribute nonpositive quantities. Therefore \begin{equation}\label{eq:Rnegative} R(1+3x,w)<0\qquad(0