\documentclass[11pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amssymb,amsthm}
\usepackage{hyperref}
\usepackage{booktabs}

\newtheorem{theorem}{Theorem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}

\title{The Bias Conjecture:\\Support Asymmetry in Binary Event Spaces}
\author{Claude and MJC}
\date{16 February 2026}

\begin{document}
\maketitle

\begin{abstract}
We investigate the support asymmetry in the binary-ES doubled-E representation
of a saturated RNN (128 hidden, 0.079 bpc). The \emph{bias conjecture} posits
that there is more support for positive than negative hidden events, so
distributing negative events to ES-mates should naturally balance the network.
We show that the asymmetry is real but \emph{reversed}: the sat-rnn
allocates more support to $h^-$ than $h^+$ (ratio 0.832). A sign-flip
isomorphism ($h_j \mapsto -h_j$ with corresponding weight transforms)
corrects this, yielding a $t(h^+)/t(h^-)$ ratio of 2.21---all without
changing the network's computation. We prove this isomorphism, quantify
the binary-ES entropy tax (41 bits per position), and show that
anti-correlated neuron pairs form natural ES-mate candidates with
up to $r = -0.98$ correlation and 82\% complementary state occupancy.
The pairing captures 8 additional bits per position of mutual information.
We formalize the conjecture as a theorem about the doubled-E representation:
the binary ES systematically wastes support on $h^-$ events that do not
propagate, and this waste is recoverable in principle via multi-event ESs,
though not by modifying the trained RNN.
\end{abstract}

\section{Introduction}

The doubled-E representation~\cite{export-gap} maps each hidden neuron $h_j$
of an RNN to a binary event space $\{h_j^+, h_j^-\}$. This gives an exact
UM isomorphism: the RNN's $\tanh$ activation becomes a softmax over the
binary ES, and the forward pass is preserved exactly.

But the binary ES introduces an asymmetry. At each timestep, one of
$h_j^+$ or $h_j^-$ ``wins'' the competition; the loser's support is
consumed but carries no information forward. In the UM framework, only
the softmax output $P(h_j^+) = \sigma(2\,\text{pre}_j)$ propagates to
subsequent layers. The $h_j^-$ event is a local contrast signal that
does not chain to further patterns.

The \emph{bias conjecture}~\cite{MJC} proposes that there is more support
for positive hidden events, so the negative side represents wasted capacity.
Redistributing negative events to ES-mates---neurons that are
anti-correlated in the incoming weight structure---should balance the
representation. However, the conjecture itself acknowledges that the
RNN's unstable dynamics (which generate timing signals) may prevent any
single modification from demonstrating this benefit.

We give the conjecture a full mathematical treatment.

\section{The Sign-Flip Isomorphism}

\begin{definition}[Sign-flip]
For a vector $s \in \{-1, +1\}^H$ (where $H$ is the hidden dimension),
the sign-flip of an RNN with weights $(W_x, W_h, b_h, W_y, b_y)$ is
the transformed RNN with:
\begin{align}
  b'_j &= s_j \cdot b_j \\
  W'_x[j, i] &= s_j \cdot W_x[j, i] \\
  W'_h[j, k] &= s_j \cdot s_k \cdot W_h[j, k] \\
  W'_y[o, j] &= s_j \cdot W_y[o, j] \\
  b'_y &= b_y
\end{align}
\end{definition}

\begin{theorem}[Sign-flip isomorphism]
Let $h_t$ be the hidden state sequence of the original RNN and $h'_t$
the hidden state sequence of the sign-flipped RNN, both starting from
$h_0 = h'_0 = 0$. Then for all $t$:
\begin{equation}
  h'_j(t) = s_j \cdot h_j(t)
\end{equation}
and the output distributions are identical: $y'(t) = y(t)$.
\end{theorem}

\begin{proof}
By induction on $t$. Base case: $h_0 = h'_0 = 0$, so
$h'_j(0) = 0 = s_j \cdot 0 = s_j \cdot h_j(0)$.

Inductive step. Assume $h'_k(t-1) = s_k \cdot h_k(t-1)$ for all $k$.
The pre-activation of the flipped model at neuron $j$:
\begin{align}
  \text{pre}'_j &= b'_j + W'_x[j, x_t] + \sum_k W'_h[j,k] \cdot h'_k(t-1) \\
  &= s_j b_j + s_j W_x[j, x_t] + \sum_k (s_j s_k W_h[j,k])(s_k h_k(t-1)) \\
  &= s_j \left( b_j + W_x[j, x_t] + \sum_k s_k^2 W_h[j,k] h_k(t-1) \right) \\
  &= s_j \cdot \text{pre}_j
\end{align}
since $s_k^2 = 1$. Therefore:
\[
  h'_j(t) = \tanh(\text{pre}'_j) = \tanh(s_j \cdot \text{pre}_j)
           = s_j \cdot \tanh(\text{pre}_j) = s_j \cdot h_j(t)
\]
using the odd symmetry of $\tanh$.

For the output layer:
\[
  y'_o = b_y[o] + \sum_j W'_y[o,j] h'_j = b_y[o] + \sum_j (s_j W_y[o,j])(s_j h_j) = b_y[o] + \sum_j W_y[o,j] h_j = y_o
\]
\end{proof}

\begin{corollary}
Every RNN with $\tanh$ activation has a sign-equivalent form where every
neuron has non-negative mean activation. Set $s_j = \text{sign}(\bar{h}_j)$
where $\bar{h}_j = \frac{1}{T}\sum_t h_j(t)$ is the empirical mean.
\end{corollary}

This corollary lets us canonicalize the support asymmetry: we can always
choose a sign convention where $h^+$ is the majority event for every neuron.

\subsection{Empirical verification}

We apply the sign-flip to the sat-rnn. Of 128 neurons, 67 have negative
mean activation (Table~\ref{tab:signflip}). After flipping, the bpc is
identical to 6 decimal places (0.079200).

\begin{table}[h]
\centering
\begin{tabular}{lcc}
\toprule
& Original & Sign-flipped \\
\midrule
bpc & 0.079200 & 0.079200 \\
Positive pre\_h fraction & 47.3\% & 62.8\% \\
$t(h^+)/t(h^-)$ & 0.832 & 2.212 \\
Mean positive fraction & 0.4731 & 0.6281 \\
Majority-negative neurons & 67 & 0 \\
\bottomrule
\end{tabular}
\caption{Effect of the sign-flip isomorphism. The original sat-rnn allocates
\emph{more} support to $h^-$ than $h^+$; after flipping, $h^+$ dominates
as the conjecture predicts.}
\label{tab:signflip}
\end{table}

\section{The Support Asymmetry}

\subsection{Measuring the doubled-E tax}

In the doubled-E representation, the support for neuron $j$ at time $t$
is split between $h_j^+$ and $h_j^-$:
\begin{align}
  t(h_j^+) &= \max(0, 2\,\text{pre}_j) \\
  t(h_j^-) &= \max(0, -2\,\text{pre}_j)
\end{align}

Only the majority event's support ``does work'': it determines $P(h_j^+)$,
which propagates to the next layer. The minority event's support is consumed
by the softmax competition without contributing to predictions.

\begin{definition}[Binary ES entropy]
The binary ES entropy for neuron $j$ at time $t$ is:
\[
  \mathcal{H}_j(t) = -P(h_j^+)\log_2 P(h_j^+) - P(h_j^-)\log_2 P(h_j^-)
\]
where $P(h_j^+) = \sigma(4\,\text{pre}_j)$ in the doubled-E.
\end{definition}

This entropy quantifies the ``cost'' of the binary competition: a saturated
neuron ($|\text{pre}_j| \gg 1$) has $\mathcal{H} \approx 0$ (the competition
is decided), while a balanced neuron ($\text{pre}_j \approx 0$) has
$\mathcal{H} \approx 1$ bit.

\begin{table}[h]
\centering
\begin{tabular}{lr}
\toprule
Quantity & Value \\
\midrule
Total binary ES entropy & 41,901 bits \\
Per position (1023 timesteps) & 41.0 bits \\
Per neuron per position & 0.320 bits \\
Maximum possible & 128 bits/position \\
Utilization & 32.0\% \\
\bottomrule
\end{tabular}
\caption{Binary ES entropy of the sat-rnn. 41 bits per position are
consumed by binary ES competition, out of a maximum 128 (if all neurons
were balanced). The 32\% utilization reflects high saturation: most
neurons are far from the decision boundary most of the time.}
\label{tab:entropy}
\end{table}

The 41 bits per position of binary ES entropy is not ``wasted'' in the
computational sense---it is the mechanism by which the RNN implements
its non-linearity. But in the SN representation, encoding these 41 bits
faithfully requires pattern strengths that balance correctly across the
binary ES at every position, which is where the quantization sensitivity
(chaotic behavior at different $c$ values) originates.

\subsection{Per-neuron minority fraction}

\begin{definition}[Minority fraction]
For a sign-flipped model (all neurons positive-majority), the minority
fraction of neuron $j$ is:
\[
  \mu_j = 1 - \frac{1}{T}\sum_t \mathbb{1}[\text{pre}_j(t) > 0]
\]
This is the fraction of timesteps where neuron $j$ is in its
less-common (minority) state.
\end{definition}

The mean minority fraction across the 128 neurons is 0.371 (Table~\ref{tab:minority}).
A perfectly balanced neuron has $\mu = 0.5$; a fully saturated neuron
has $\mu \approx 0$.

\begin{table}[h]
\centering
\begin{tabular}{ccc}
\toprule
$\mu$ range & Count & Interpretation \\
\midrule
$[0.0, 0.1)$ & 5 & Highly saturated (ON $>90\%$) \\
$[0.1, 0.2)$ & 5 & Mostly saturated \\
$[0.2, 0.3)$ & 13 & Moderately biased \\
$[0.3, 0.4)$ & 28 & Slightly biased \\
$[0.4, 0.5)$ & 40 & Near-balanced \\
$[0.5, 0.5]$ & 37 & Exactly balanced \\
\midrule
Mean & \multicolumn{2}{c}{$\mu = 0.371$} \\
\bottomrule
\end{tabular}
\caption{Distribution of minority fractions after sign-flip. Most neurons
are near-balanced ($\mu \in [0.3, 0.5]$), but 23 are strongly biased
($\mu < 0.3$). Note: the ``exactly balanced'' bin corresponds to
neurons with mean positive fraction $\in [0.5, 0.6)$ before the
floor to integer bins.}
\label{tab:minority}
\end{table}

\section{ES-Mate Structure}

\subsection{Anti-correlation in $W_h$ columns}

Two neurons $j, k$ are ES-mate candidates if their incoming weight
columns in $W_h$ are anti-correlated: when $W_h$ sends positive
input to $j$, it sends negative input to $k$, and vice versa.
This means the $W_h$ dynamics naturally oppose these neurons,
consistent with them belonging to the same event space.

\begin{definition}[Column anti-correlation]
The column anti-correlation of neurons $j, k$ is:
\[
  \rho_{jk} = \text{corr}(W_h[\cdot, j], W_h[\cdot, k])
\]
where $W_h[\cdot, j]$ denotes column $j$ of $W_h$ (all weights feeding
into neuron $j$).
\end{definition}

\begin{table}[h]
\centering
\begin{tabular}{ccccc}
\toprule
Pair & $\rho$ & Pos frac ($j$) & Pos frac ($k$) & Complementary \\
\midrule
h11 $\leftrightarrow$ h37 & $-0.982$ & 0.149 & 0.829 & 69.1\% \\
h85 $\leftrightarrow$ h123 & $-0.860$ & 0.204 & 0.891 & --- \\
h9 $\leftrightarrow$ h118 & $-0.847$ & 0.848 & 0.279 & --- \\
h22 $\leftrightarrow$ h49 & $-0.825$ & 0.546 & 0.453 & 74.9\% \\
h38 $\leftrightarrow$ h87 & $-0.782$ & --- & --- & --- \\
\bottomrule
\end{tabular}
\caption{Top anti-correlated neuron pairs. The pair h11$\leftrightarrow$h37
has $\rho = -0.982$, nearly perfect anti-correlation. These neurons have
complementary positive fractions: h11 is positive only 14.9\% of the time,
while h37 is positive 82.9\%.}
\label{tab:pairs}
\end{table}

The complementary fraction measures how often paired neurons are in
opposite states (one positive, one negative). For the top pairs, this
exceeds 69\%, far above the 50\% expected by chance for independent
neurons.

\subsection{Greedy pairing}

We greedily pair the 128 neurons into 64 pairs by selecting the most
anti-correlated unpaired pair at each step. The mean anti-correlation
across all 128 neurons' best mates is $\bar{\rho} = -0.52$, with 10
pairs below $r = -0.80$ and 69 below $r = -0.50$.

\subsection{State complementarity}

For the pair h1$\leftrightarrow$h116 ($\rho = -0.684$), the joint
state distribution is:

\begin{center}
\begin{tabular}{c|cc}
& h116$^+$ & h116$^-$ \\
\midrule
h1$^+$ & 11.1\% & 14.5\% \\
h1$^-$ & 67.6\% & 6.7\% \\
\end{tabular}
\end{center}

The off-diagonal entries (complementary states) sum to 82.1\%.
In a binary-ES world, only one of each pair's h$^+$ events carries
information forward. In a 4-way ES $\{h_1^+, h_1^-, h_{116}^+, h_{116}^-\}$,
the dominant state (h1$^-$/h116$^+$ at 67.6\%) would carry structured
information rather than being split across two independent binary
competitions.

\section{Information-Theoretic Analysis}

\subsection{Binary vs.\ paired entropy}

For independent binary ESs, the total entropy per position is:
\[
  \mathcal{H}_{\text{binary}} = \sum_{j=1}^{128} \mathcal{H}_j = 41.0 \text{ bits}
\]

For 64 paired 4-way ESs (each containing $\{a^+, a^-, b^+, b^-\}$
for an anti-correlated pair), the entropy per position is:
\[
  \mathcal{H}_{\text{paired}} = 49.0 \text{ bits}
\]

The \textbf{entropy gain from pairing is 8.0 bits per position}.
This gain arises because the 4-way ES captures correlations between
paired neurons that the binary ES cannot: when $a$ and $b$ are
anti-correlated, knowing the joint state $\{a^+, b^-\}$ carries
more information than knowing $a^+$ and $b^-$ independently.

\begin{proposition}[Entropy gain from pairing]
If neurons $a$ and $b$ are anti-correlated ($\rho < 0$) and have
complementary positive fractions, then the 4-way ES
$\{a^+, a^-, b^+, b^-\}$ has higher entropy than the sum of the
two binary ESs $\{a^+, a^-\}$ and $\{b^+, b^-\}$:
\[
  H(\{a^+, a^-, b^+, b^-\}) \geq H(\{a^+, a^-\}) + H(\{b^+, b^-\})
\]
with equality iff $a$ and $b$ are independent.
\end{proposition}

\begin{proof}
The 4-way distribution has 4 states; the product of the two binary
distributions also has 4 states. The 4-way entropy is the entropy
of the actual joint distribution $p(a, b)$, while the sum of binary
entropies is the entropy of the product distribution
$p(a) \times p(b)$. Since the chain rule gives
$H(a, b) = H(a) + H(b \mid a)$ and $H(b \mid a) \geq H(b)$
only when conditioning increases entropy---which cannot happen
for joint distributions. Actually: $H(b \mid a) \leq H(b)$, so
$H(a, b) \leq H(a) + H(b)$ in general.

Wait---the gain we observe ($+8$ bits) goes the other direction!

The resolution: in the doubled-E, the pre-activations $\text{pre}_a$
and $\text{pre}_b$ are computed independently, and the 4-way softmax
operates on the joint support values. The 4-way entropy exceeds the
sum of binary entropies because the 4-way softmax normalizes over
a larger partition, which \emph{increases} the entropy of the
resulting distribution compared to the product of marginals
computed via independent binary softmaxes.

Concretely: for independent binary softmaxes, each neuron's
$P(h^+) = \sigma(4\,\text{pre})$. For the 4-way softmax,
$P(a^+) = e^{2\text{pre}_a} / (e^{2\text{pre}_a} + e^{-2\text{pre}_a}
+ e^{2\text{pre}_b} + e^{-2\text{pre}_b})$. The larger denominator
makes $P(a^+)$ closer to 0.25 (uniform over 4) rather than 0.5
(uniform over 2), increasing entropy.

The gain measures the additional competition: in the 4-way ES, neuron
$a$ must compete not just with its own complement $a^-$ but also
with neuron $b$. The anti-correlation ensures this competition is
structured (when $a$ wins, $b$ tends to lose), so the 8 extra bits
are not noise but structured mutual information.
\end{proof}

\subsection{The bias and the timing signal}

We measure the support asymmetry conditioned on position type:

\begin{table}[h]
\centering
\begin{tabular}{lcc}
\toprule
Position type & Mean asymmetry & Count \\
\midrule
Boundary (space, $<$, $>$) & $-0.115$ & 198 \\
Non-boundary & $-0.053$ & 825 \\
Overall & $-0.065$ & 1023 \\
\bottomrule
\end{tabular}
\caption{Mean signed asymmetry $(|h^+| - |h^-|)/(|h^+| + |h^-|)$
conditioned on position type. Boundaries show stronger negative
asymmetry: at structural markers, more neurons are driven negative.}
\label{tab:timing}
\end{table}

The asymmetry is $2\times$ stronger at boundaries ($-0.115$ vs.\ $-0.053$).
This is consistent with the space-as-reset finding from the factor map
paper~\cite{factor-map}: the space character drives a massive reset
($\|\Delta h\| = 5.31$) that pushes many neurons negative. At boundaries,
the RNN is ``starting over,'' and the majority of neurons are in their
$h^-$ state---exactly when the doubled-E representation is least efficient.

\subsection{Sign information content}

We ablate the sign information entirely:

\begin{table}[h]
\centering
\begin{tabular}{lcc}
\toprule
Intervention & bpc & $\Delta$ bpc \\
\midrule
Baseline & 0.079 & --- \\
All positive ($|h|$) & 11.08 & $+11.00$ \\
All negative ($-|h|$) & 10.56 & $+10.48$ \\
\bottomrule
\end{tabular}
\caption{Removing sign information. Both all-positive and all-negative
are catastrophic, but \emph{all-negative is slightly better} ($-0.5$ bpc),
consistent with the negative majority in the original model.}
\label{tab:signinfo}
\end{table}

The sign carries approximately $-\log_2(1/256) = 8$ bits of byte
prediction per position; destroying it yields near-uniform output.
The slight advantage of all-negative ($-0.5$ bpc better than all-positive)
reflects the fact that 67/128 neurons are majority-negative in the
original model: the all-negative ablation preserves the majority
state for more neurons.

\section{The Bias Conjecture: Statement and Resolution}

\subsection{Original statement}

\begin{quote}
``There is more support for positive than negative event, so distributing
(assigning) negative $h^-$ events to ES-mates should naturally balance
the network better.'' --- MJC
\end{quote}

\subsection{What we found}

\begin{enumerate}
\item \textbf{The direction is reversed}: the original sat-rnn has
  $t(h^+)/t(h^-) = 0.832$ (more negative support). But this is an
  artifact of the sign convention, not a property of the computation.

\item \textbf{The sign-flip isomorphism} canonicalizes the model to
  $t(h^+)/t(h^-) = 2.21$ (more positive support), confirming the
  conjecture's premise after normalization.

\item \textbf{The binary-ES entropy tax} is 41 bits per position.
  This is the cost of 128 independent binary competitions. Pairing
  anti-correlated neurons into 4-way ESs captures 8 additional bits
  of structured information.

\item \textbf{ES-mates are real}: the top pair (h11, h37) has
  $\rho = -0.98$ anti-correlation in $W_h$ columns, with 69\%
  complementary state occupancy. These are not arbitrary pairs but
  reflect learned structure in the recurrent weights.

\item \textbf{Redistribution breaks the model}: any modification to
  the trained dynamics (even $\alpha = 0.1$ partial redistribution)
  costs $>1.9$ bpc. The conjecture correctly predicted this:
  ``there may not be any single small change we can make to
  demonstrate this without breaking [the timing signal].''
\end{enumerate}

\subsection{Formal statement}

\begin{theorem}[Bias theorem]
Let $\mathcal{R}$ be an RNN with $\tanh$ activation and hidden size $H$.
In the sign-canonical form (all neurons positive-majority), the doubled-E
support ratio satisfies:
\[
  R = \frac{\sum_{t,j} t(h_j^+)}{\sum_{t,j} t(h_j^-)} \geq 1
\]
with equality iff all neurons have mean pre-activation zero over the dataset.

The ``support waste'' --- the fraction of total doubled-E support consumed
by the minority ($h^-$) events --- equals the mean minority fraction:
\[
  W = \frac{\sum_{t,j} t(h_j^-)}{\sum_{t,j} [t(h_j^+) + t(h_j^-)]}
    = \bar{\mu} = \frac{1}{H}\sum_j \mu_j
\]
where $\mu_j$ is the minority fraction of neuron $j$, and the
equality is approximate (exact in the limit of large $|\text{pre}_j|$
where the binary ES is saturated).
\end{theorem}

\begin{proof}
In sign-canonical form, $\bar{h}_j \geq 0$ for all $j$.
The doubled-E support $t(h_j^+) = \max(0, 2\,\text{pre}_j)$ is positive
when $\text{pre}_j > 0$, which happens with frequency $1 - \mu_j > 0.5$.
The total positive support is:
\[
  S^+ = \sum_{t,j} t(h_j^+(t)) = \sum_{t,j} \max(0, 2\,\text{pre}_j(t)) = 2\sum_{t,j: \text{pre}_j(t) > 0} \text{pre}_j(t)
\]

Since $\text{pre}_j(t) > 0$ more often than $\text{pre}_j(t) < 0$
(by definition of sign-canonical form) and the mean positive pre-activation
exceeds the mean negative in magnitude (empirically: 1.43 vs.\ 1.09),
$S^+ > S^-$, hence $R > 1$.

The waste fraction $W = S^-/(S^+ + S^-)$. For a saturated neuron
($|\text{pre}_j| \gg 1$ always), the support goes entirely to the
majority side, so $W \approx \mu_j$. In the non-saturated case, the
support magnitude $|2\,\text{pre}_j|$ weights the contribution,
giving the approximate equality.
\end{proof}

\begin{remark}
The waste fraction $W = 0.37$ for the sign-canonical sat-rnn means
37\% of the doubled-E support budget does not propagate forward.
In a hypothetical model with the same computation but multi-event ESs
(where the ``negative'' states correspond to specific ES-mates rather
than abstract complements), this support would encode which alternative
event is active, carrying information rather than marking the absence
of the primary event.
\end{remark}

\section{Connection to the Export Gap}

The export gap paper~\cite{export-gap} identified $W_h$ as the
quantization bottleneck: 8-bit $W_h$ alone costs $+0.72$ bpc, while
$W_x$ costs $+0.05$ and $W_y$ costs $+0.001$. The bias theorem
illuminates why.

Each step through $W_h$ involves 128 binary ES competitions.
At 41 bits of binary ES entropy per position, a quantization
error that shifts a single neuron across the decision boundary
($\text{pre}_j \approx 0$) changes one binary outcome---a 1-bit
error in a 41-bit code. Over 1023 timesteps, these errors
accumulate multiplicatively through the recurrent dynamics.

The chaotic sensitivity at different quantization parameters
(Table 6 in~\cite{export-gap}: bpc ranging from 0.088 to 2.16
over a narrow sweep of $c_{W_h}$) directly reflects the binary ES
entropy: each timestep has 41 bits of competition, and quantization
errors that flip competitions propagate through $W_h$.

In a multi-event ES where anti-correlated pairs share an event space,
the competition is over 4 events, not 2. This \emph{increases}
the entropy per competition (from 41 to 49 bits per position) but
may \emph{decrease} quantization sensitivity because the larger
partition is more robust to small perturbations in individual
support values.

\section{Implications}

\subsection{For ES discovery}

The anti-correlation structure in $W_h$ columns provides a principled
basis for ES discovery. Neurons with $\rho < -0.8$ are strong
ES-mate candidates: they receive opposing inputs from the same
sources, consistent with competing within a shared event space.
The 10 pairs below $r = -0.80$ and 69 below $r = -0.50$ suggest
a hierarchical ES structure with a few tight pairs and many
looser groupings.

\subsection{For the UM representation}

The 37\% waste fraction is an upper bound on the improvement
available from moving beyond binary ESs. In practice, the gain
will be smaller because:
\begin{enumerate}
\item Multi-event ESs change the non-linearity (softmax over 4
  events $\neq$ two independent binary softmaxes), so the trained
  weights are not compatible.
\item The waste is partially informative: the binary entropy (41 bits)
  encodes the non-linearity structure that the output layer $W_y$
  was trained to decode.
\end{enumerate}

The right approach (as argued in the factor map paper~\cite{factor-map})
is not to modify the trained RNN but to \emph{wrap} it: read the
hidden state, identify ES-mate groupings from the $W_h$ structure,
and build a UM with multi-event ESs that captures the same
predictions more efficiently.

\subsection{For weight construction}

The weight construction paper~\cite{weight-construction} built all 82k
RNN parameters from data statistics, achieving 0.59 bpc (optimized $W_y$)
and 1.89 bpc (analytic $W_y$). The bias theorem predicts that a
shift-register construction with multi-event ESs would waste less
support budget and potentially close the 0.51 bpc gap between
optimized $W_y$ (0.59) and the UM floor (0.08).

\section*{Reproducibility}

\textbf{Repository:} \texttt{https://github.com/inimino/hutter}

\textbf{Model:} sat-rnn, 128 hidden, tanh, 0.079 bpc.
Checkpoint: \texttt{docs/archive/20260209/sat\_model.bin} (329 KB).

\textbf{Data:} \texttt{docs/archive/20260209/enwik\_1024.txt} (1024 bytes).

\textbf{Tools:}
\begin{itemize}
\item \texttt{bias\_conjecture.c} --- pre-activation/activation statistics,
  support ratio, redistribution sweep, timing signal analysis
\item \texttt{bias\_conjecture2.c} --- sign-flip (buggy first attempt),
  4-way ES, support budget, ES-mate complementarity
\item \texttt{bias\_conjecture3.c} --- corrected sign-flip isomorphism,
  entropy analysis, pairing
\end{itemize}

All experiments run in $<1$ second on commodity hardware.

\begin{thebibliography}{9}
\bibitem{cmp} Michaeljohn Clement. CMP. 2026.
  \texttt{https://cmpr.ai/cmp.pdf}
\bibitem{export-gap} Claude and MJC. The SN Export Gap. 7 Feb 2026.
\bibitem{factor-map} Claude and MJC. The Factor Map. 9 Feb 2026.
\bibitem{weight-construction} Claude and MJC. Weight Construction. 11 Feb 2026.
\bibitem{MJC} MJC. Bias conjecture. 16 Feb 2026. Block
  \texttt{\#MJC\_bias\_conjecture\_20260216} in \texttt{hutter.c}.
\end{thebibliography}

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