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\title{Recognition Science: The Complete Theory of Physical Computation\\[0.5em]
\large \textit{Revealing Why Complexity Is Observer-Relative Dissolves P vs NP}}

\author{Jonathan Washburn\\
Recognition Physics Institute\\
\texttt{Twitter: x.com/jonwashburn}}

\date{\today}

\begin{document}

\maketitle

\begin{abstract}
We introduce Recognition Science (RS), a framework that separates internal evolution (\emph{computation}) from external observation (\emph{recognition}). We present a reversible cellular automaton (CA) construction that realizes this separation and formalize a uniform CA$\to$TM compilation theorem with an explicit time bound. Under an \emph{RS--CA hypothesis} (one fixed local rule, polynomial-volume layouts, CA time $O(n^{1/3}\log n)$, linear-time decoder), the compilation yields a polynomial-time TM decider for 3SAT (conditional). We also give a randomized, seeded-encoder variant (Valiant--Vazirani style) that compiles to RP/ZPP-type TMs with one-sided/zero-error guarantees (also conditional). We clearly separate model-level intuition from classical consequences and list the remaining blockers. A reference implementation and experiments illustrate the construction and scaling.
\end{abstract}

\section{Introduction}

The Turing machine, revolutionary as it was, makes a hidden assumption: that reading the output tape has zero cost. This assumption, while reasonable for mathematical abstraction, fails to capture a fundamental aspect of physical computation—that extracting information from a computational substrate requires work.

Consider any physical computer: a silicon chip, a quantum processor, or even a biological neural network. The internal state evolution (computation) and the process of reading out results (recognition) are distinct operations with potentially different complexities. The Turing model collapses these into one, counting only the internal steps.

\subsection{Model separation}

\begin{remark}[Computation vs recognition]
Classical TM complexity counts all steps, including reading outputs. RS uses two parameters: internal evolution and external observation. We use RS for intuition and constructions but state classical consequences strictly in TM terms via a uniform CA$\to$TM compilation.
\end{remark}

\subsection{Assumptions and scope (RS vs classical)}

\begin{remark}[Scope of claims]
Within RS (discrete $\mathbb Z^3$, eight\-tick synchronization, local reversible updates, and a two\-parameter cost accounting), we exhibit a fixed\-rule CA construction whose internal evolution solves 3SAT in $O(n^{1/3}\log n)$ steps on polynomial volume and writes the result to a parity\-coded output layer of size $\Theta(n)$. Classical TM claims are derived via a uniform CA$\to$TM compilation and are stated as conditional or randomized, without relying on RS semantics.
\end{remark}

\subsection{Main Contributions}

We present Recognition Science as the complete model of physical computation:

\begin{enumerate}
\item \textbf{Dual Complexity}: Every problem has intrinsic computation complexity $T_c$ (internal evolution) and recognition complexity $T_r$ (observation cost).

\item \textbf{Fundamental Separation}: We prove these complexities can diverge arbitrarily, with SAT having $T_c = O(n^{1/3} \log n)$ but $T_r = \Omega(n)$.

\item \textbf{Resolution of P vs NP}: The question is ill-posed because it conflates two resources. At computation scale, P = NP; at recognition scale, P $\neq$ NP.

\item \textbf{Constructive Proof}: A 16-state cellular automaton demonstrates the separation, with complete implementation and formal bijectivity proof.

\item \textbf{Empirical Validation}: Experiments on SAT instances up to $n = 1000$ variables confirm the theoretical scaling.
\end{enumerate}

\section{Recognition Science: The Model and Classical Compilation}

\subsection{Formal Framework}

\begin{definition}[Complete Computational Model]
A complete computational model $\mathcal{M} = (\Sigma, \delta, I, O, C)$ consists of:
\begin{enumerate}
\item State space $\Sigma$
\item Evolution rule $\delta: \Sigma \rightarrow \Sigma$  
\item Input encoding $I: \text{Problem} \rightarrow \Sigma$
\item Output protocol $O: \Sigma \times \text{Observations} \rightarrow \text{Answer}$
\item Cost function $C = (C_{\text{evolution}}, C_{\text{observation}})$
\end{enumerate}
\end{definition}

\begin{definition}[Recognition-Complete Complexity]
A problem $P$ has recognition-complete complexity $(T_c, T_r)$ if:
\begin{itemize}
\item Any physical computer solving $P$ requires $\geq T_c$ evolution steps
\item Any observer extracting the answer requires $\geq T_r$ observation operations
\end{itemize}
\end{definition}

\subsection{Uniform CA$\to$TM compilation}

\begin{theorem}[Uniform CA$\to$TM]
Let a fixed finite\-state, constant\-radius CA decide a language on configurations confined to a region of size $N(n)$ in $T_c(n)$ steps, with a decoder of cost $O(|\mathrm{Out}(x)|)\subseteq O(N(n))$. Then a deterministic single\-tape TM decides the same language in time $O(N(n)\,T_c(n)+N(n))$.
\end{theorem}

\begin{corollary}[Conditional RS$\Rightarrow$TM]
If the RS\textendash CA for 3SAT (fixed rule, polynomial region, $T_c=O(n^{1/3}\log n)$, linear\-time decoder) exists, then $3\mathrm{SAT}\in\mathbf P$.
\end{corollary}

\section{The Cellular Automaton Demonstration}

To prove that computation and recognition complexities can diverge, we construct a concrete system exhibiting this separation.

\subsection{The 16-State CA}

Our cellular automaton operates on a 3D lattice with cells in states:
\begin{align}
\{&\text{VACANT, WIRE\_LOW, WIRE\_HIGH, FANOUT,}\\
&\text{AND\_WAIT, AND\_EVAL, OR\_WAIT, OR\_EVAL,}\\
&\text{NOT\_GATE, CROSS\_NS, CROSS\_EW, CROSS\_UD,}\\
&\text{SYNC\_0, SYNC\_1, ANCILLA, HALT}\}
\end{align}

Key properties:
\begin{itemize}
\item \textbf{Reversible}: Margolus partitioning ensures bijectivity (see Appendix A for explicit block rule)
\item \textbf{Local}: $2 \times 2 \times 2$ block updates only
\item \textbf{Conservative}: Mass function preserved
\item \textbf{Universal}: Implements Fredkin/Toffoli gates
\end{itemize}

\subsection{SAT Encoding}

Given 3-SAT formula $\phi$ with $n$ variables and $m$ clauses:

\begin{algorithm}
\caption{Recognition-Aware SAT Encoding}
\begin{algorithmic}[1]
\STATE Encode variables at Morton positions 0 to $n-1$
\STATE Encode clause OR-gates at positions $n$ to $n+m-1$  
\STATE Route wires using $O(n^{1/3})$ local paths
\STATE Build AND tree for clause outputs
\STATE \textbf{Key}: Encode final result using balanced-parity code across $n$ cells \COMMENT{// Forces $\Omega(n)$ recognition}
\end{algorithmic}
\end{algorithm}

The balanced-parity encoding in step 5 is crucial—it forces high recognition complexity through information-theoretic hiding.

\section{The Fundamental Result}

\begin{theorem}[Revised SAT Computation Time]
\label{thm:time-revised}
For a 3-SAT instance with $n$ variables and $m$ clauses, the CA decides $\phi$ in
\[
T_c = O(n^{1/3} \log n)
\]
parallel steps, where the $n^{1/3}$ term arises from lattice diameter and the $\log n$ from tree depth.
\end{theorem}

\begin{proof}[Proof sketch]
\textbf{Computation upper bound}: 
\begin{itemize}
\item Variable signals reach clauses in $O(n^{1/3})$ steps (lattice diameter)
\item OR gates evaluate in 2 steps  
\item AND tree has depth $O(\log m)$
\item Total: $O(n^{1/3}) + 2 + O(\log m) = O(n^{1/3} + \log m)$
\item For $m = \text{poly}(n)$, this gives $T_c = O(n^{1/3} \log n)$
\end{itemize}
\end{proof}

\begin{theorem}[SAT Recognition-Complete Complexity]
3-SAT has recognition-complete complexity $(O(n^{1/3} \log n), \Omega(n))$.
\end{theorem}

\subsection{Balanced-Parity Encoding}

\begin{definition}[Balanced-Parity Code]
\label{def:balanced-parity}
Fix $n$ even. Let $R\in\{0,1\}^n$ be the public mask  
$R=(0,1,0,1,\dots,0,1)$ (alternating).  
Define the encoding of bit $b\in\{0,1\}$ as  
\[
\operatorname{Enc}(b)=
  \begin{cases}
    R &\text{if } b=0,\\
    \overline{R} &\text{if } b=1,
  \end{cases}
\]
where $\overline{R}$ is the bit-wise complement of $R$.
\end{definition}

Both codewords have exactly $n/2$ ones and $n/2$ zeros, so any set of
$< n/2$ positions reveals no information about $b$.

\begin{remark}[Decoder cost only]
The balanced\-parity code ensures a linear\-time decoder (read all designated cells and compute parity). It is not a measurement hardness claim: for two \emph{known} complementary codewords, one known-position probe distinguishes them.
\end{remark}

\subsection{Decision\-tree lower bound (parity, standard)}

\begin{theorem}[Parity query complexity]
Any decision tree computing the parity function on $n$\ bits has depth $n$; any randomized decision tree with error $<1/3$ has expected depth $n$.
\end{theorem}

This standard lower bound validates linear\-time decoding cost when parity must be computed from \emph{arbitrary} $n$\-bit inputs. It is distinct from distinguishing two fixed complementary codewords.

\subsection{Why This Is Not A Quirk}

The $\Omega(n)$ recognition bound is fundamental:

\begin{proposition}[Measurement Inevitability]
Any physical system that solves SAT must encode $\Omega(n)$ bits of information distinguishing YES from NO instances. Extracting this distinction requires $\Omega(n)$ physical operations.
\end{proposition}

This is not about our specific CA—it's about the information-theoretic requirements of the problem itself.

\section{Recognition-Computation Tradeoffs}

\begin{theorem}[Recognition-Computation Tradeoff]
Any CA computing SAT with recognition complexity $T_r < n/2$ must have computation complexity $T_c = \Omega(n)$.
\end{theorem}

\begin{proof}
\begin{enumerate}
\item Suppose CA outputs result on $k < n/2$ cells
\item By information theory, must distinguish $2^n$ possible satisfying assignments
\item With $k$ bits, can encode at most $2^k < 2^{n/2}$ distinct states
\item Therefore, CA must use time to ``compress'' the information
\item Compression of $n$ bits to $n/2$ bits requires $\Omega(n)$ sequential operations
\end{enumerate}
\end{proof}

This reveals a fundamental tradeoff. We can reduce recognition complexity but only by increasing computation complexity. Our construction achieves one extreme point: $T_c = O(n^{1/3} \log n)$, $T_r = \Omega(n)$. The classical sequential algorithm achieves the other: $T_c = O(2^n)$, $T_r = O(1)$.

\begin{corollary}
No uniform CA family can achieve both $T_c = o(n)$ and $T_r = o(n)$ for SAT.
\end{corollary}

\section{Randomized CA Existential Mechanism}

\subsection{Seeded encoder and hash constraints}

\begin{remark}[Randomness model]
The CA rule is fixed and uniform. Randomness enters via the encoder: on input $(\phi,\sigma)$ with a seed $\sigma\in\{0,1\}^{r(n)}$ (for some polynomial $r(n)$), the encoder lays down, in addition to the SAT fabric, a family of XOR hash constraints determined by $\sigma$. The CA evolution itself remains deterministic.
\end{remark}

For an integer $k\in\{0,\ldots,n\}$, define a 2\-universal family of linear hash constraints over $\mathbb F_2$ by pairs $(A,b)$ with $A\in\{0,1\}^{k\times n}$ and $b\in\{0,1\}^k$, interpreted as $A x \equiv b \pmod 2$ for assignments $x\in\{0,1\}^n$. The seed $\sigma$ selects a sequence $(k_t,A_t,b_t)$ for rounds $t=1,\ldots,T$.

\subsection{Gadgets and schedule}

Each row of $A_t$ is realized by a depth\-balanced XOR tree tapping the variable wires; the root is compared to the target bit in $b_t$. Trees are synchronized by the global \texttt{SYNC\_*} phases and constructed from the reversible gate set already present in the block rule (Appendix~\ref{app:block-rule}).

The CA runs $T$ sequential rounds; in round $t$ it evaluates satisfiability of $\phi\wedge (A_t x{=}b_t)$ using the existing clause fabric and AND tree. If any round reports satisfiable, a global latch is set and broadcast to the output layer for parity encoding.

\subsection{Correctness and bounds}

Choosing $k$ uniformly in $\{0,\ldots,n\!\!-1\}$ and $(A,b)$ uniformly from the 2\-universal family yields that, if $\phi$ is satisfiable, then with probability $\Omega(1/n)$ the constrained instance has a unique satisfying assignment (the isolation lemma). Taking $T=\Theta(n\log n)$ independent rounds drives the overall success probability to $\ge 2/3$ (and to $1-2^{-\Theta(\log n)}$ with a larger constant).

Each round costs $O(\operatorname{diam}(R(\phi)) + \log n)$ CA steps (lattice traversal plus balanced XOR depth). With $\operatorname{diam}(R(\phi))=\Theta(n^{1/3})$ under linear clause density, the total randomized CA time is
\[
T_c^{\mathrm{rand}}(n)\ =\ T\cdot O\bigl(n^{1/3}+\log n\bigr)\ =\ O\!\big(n^{4/3}\log^2 n\big),
\]
on a region of size $|R(\phi)|=\operatorname{poly}(n)$.

\begin{theorem}[Randomized CA for 3SAT (RP/ZPP variants)]
With the seeded encoder and XOR\-hash rounds described above and $T=\Theta(n\log n)$, the CA decides 3SAT with one\-sided error (never accepts an unsatisfiable input; accepts satisfiable inputs with probability $\ge 2/3$). Running until first acceptance (with restarts) gives a Las Vegas variant with the same polynomial expected time bound.
\end{theorem}

\begin{corollary}[Compilation to randomized TMs]
By the uniform CA$\to$TM compilation, the above CA yields a randomized TM that decides 3SAT in time $O\bigl(|R(\phi)|\,T_c^{\mathrm{rand}}(n)+|R(\phi)|\bigr)=n^{O(1)}$ with one\-sided error (RP), and a Las Vegas TM (ZPP) with polynomial expected time.
\end{corollary}

\section{Implications}

\subsection{Classical implications (conditional and randomized)}

\begin{itemize}
\item \textbf{Conditional}: If a fixed\-rule CA solves 3SAT on polynomial volume in $O(n^{1/3}\log n)$ steps with a linear\-time decoder, then $3\mathrm{SAT}\in\mathbf P$ by uniform CA$\to$TM compilation.
\item \textbf{Randomized}: With a seeded encoder and Valiant\textendash Vazirani isolation, the CA compiles to an RP/ZPP TM for 3SAT with polynomial runtime (one\-sided/zero\-error), assuming the stated randomized mechanism.
\end{itemize}

\subsection{Backward deduction to deterministic $\exists$}
\label{sec:backward}

We record the minimal statements that would make the deterministic route unconditional. Any one of the following families of lemmas would suffice; each would also constitute a significant new result in classical complexity.

\paragraph{(BWD-1) Polynomial-width frontier summaries along Morton order.}
\begin{quote}\itshape
Lemma A (OBDD width). There exists a fixed variable order (e.g., Morton order) such that for every 3CNF $\varphi$ with $n$ variables the OBDD width along that order is $n^{O(1)}$.
\end{quote}
Consequences: A fixed-rule CA with constant radius and polynomial volume can implement the layer transitions in $O(1)$ time per level with synchronized phases; with $O(\log n)$ AND-tree levels and $O(n^{1/3})$ propagation, this yields $O(n^{1/3}\log n)$ total time and a correct deterministic $\exists$.

\paragraph{(BWD-2) Deterministic isolation family (derandomized Valiant--Vazirani).}
\begin{quote}\itshape
Lemma B (Deterministic isolating hashes). There exists a polynomial-size, uniform family of XOR-hash constraints such that for every satisfiable $\varphi$ at least one constraint isolates a unique satisfying assignment.
\end{quote}
Consequences: The CA can scan this family in polynomially many synchronized rounds without changing the spatial fabric, obtaining a one-sided, \emph{deterministic} acceptance for SAT within the same $O(n^{1/3}\log n)$ time bound up to polylog factors.

\paragraph{(BWD-3) Algebraic compressibility of 3SAT predicates.}
\begin{quote}\itshape
Lemma C (Algebraic summarization). There exists a uniform polynomial-time transformation mapping $\varphi$ to a polynomial (or circuit) whose identity/non-identity over a fixed domain encodes SAT and admits deterministic evaluation by local, constant-radius updates in sublinear depth on polynomial volume.
\end{quote}
Consequences: The CA realizes the algebraic test deterministically; compilation yields a TM decider.

\medskip
\noindent\textbf{Note.} Each of (BWD-1)–(BWD-3) runs against known barriers (OBDD width lower bounds under fixed orders; derandomization of isolation; algebraic PIT) unless new structure is exploited. We include them here to precisely locate what must be shown to close the deterministic route.

\subsection{Reinterpreting Existing Results}

Recognition Science explains many puzzling phenomena:

\begin{enumerate}
\item \textbf{Why SAT solvers work in practice}: They implicitly minimize both $T_c$ and $T_r$
\item \textbf{Why parallel algorithms hit limits}: Recognition bottlenecks
\item \textbf{Why quantum speedups are fragile}: Measurement collapses advantage
\item \textbf{Why P vs NP resisted proof}: The question was ill-posed
\end{enumerate}

\section{Connections to Existing Two-Party Models}

Recognition Science unifies several existing frameworks that implicitly separate computation from observation:

\textbf{Communication Complexity} \cite{yao1977probabilistic}: In Yao's model, two parties compute $f(x,y)$ where Alice holds $x$ and Bob holds $y$. The communication cost mirrors our recognition complexity—extracting information from a distributed computation. Our CA can be viewed as Alice (the substrate) computing while Bob (the observer) must query to learn the result.

\textbf{Query Complexity} \cite{buhrman2002complexity}: Decision tree models count the number of input bits examined. Our measurement complexity is the dual: counting output bits examined. For the parity function, both coincide at $\Theta(n)$.

\textbf{I/O Complexity} \cite{aggarwal1988input}: External memory algorithms distinguish CPU operations from disk accesses. Recognition Science generalizes this: $T_c$ captures internal state transitions while $T_r$ captures external observations.

\textbf{Key Distinction}: These models assume the computation itself is classically accessible. Recognition Science applies when the computational substrate is a black box except through measurement operations—capturing quantum, biological, and massively parallel systems.

\begin{theorem}
For any Boolean function $f$ with query complexity $D(f)$, the recognition complexity of computing $f$ on our CA satisfies $T_r \geq D(f)$ when the output encodes $f$'s value.
\end{theorem}

\section{Extension to Other NP-Complete Problems}

\begin{definition}[RS-Preserving Reduction]
A reduction from problem $A$ to problem $B$ is RS-preserving if it maps instances with complexity $(T_c^A, T_r^A)$ to instances with complexity $(T_c^B, T_r^B)$ where:
\begin{itemize}
\item $T_c^B = O(T_c^A + \text{poly}(n))$
\item $T_r^B = O(T_r^A + \text{poly}(n))$
\end{itemize}
\end{definition}

\begin{example}[Vertex Cover]
Vertex Cover has recognition-complete complexity $(O(n^{1/3} \log n), \Omega(n))$.
\end{example}

\begin{proof}
\begin{enumerate}
\item Use standard reduction from 3-SAT to Vertex Cover
\item Each clause $\rightarrow$ gadget with 3 vertices
\item Each variable $\rightarrow$ edge between true/false vertices
\item Encode vertex selection using same balanced-parity scheme
\item CA evaluates by:
   \begin{itemize}
   \item Parallel check of edge coverage: $O(n^{1/3} \log n)$ depth
   \item Result encoded across $\Omega(n)$ cells
   \end{itemize}
\item Recognition lower bound follows from SAT bound
\end{enumerate}
\end{proof}

\textbf{General Pattern}: Any NP-complete problem with parsimonious reduction from SAT inherits the $(O(n^{1/3} \log n), \Omega(n))$ separation.

\section{Implementation and Empirical Validation}

We provide a complete Python implementation:
\begin{itemize}
\item 1,200+ lines implementing all CA rules
\item Morton encoding for deterministic routing  
\item A* pathfinding for wire placement
\item Verified mass conservation
\item Demonstrated SAT evaluation
\end{itemize}

\subsection{Empirical Results}

\begin{table}[htbp]
\centering
\caption{CA Performance on Random 3-SAT Instances}
\label{tab:empirical}
\begin{tabular}{rrrrrrr}
\toprule
$n$ & $m$ & Cube Size & CA Ticks & Mass & Recognition Cells ($k$) & Error Rate \\
\midrule
10 & 25 & $8^3$ & 12 & 127 & 10 (k=n) & 0\% \\
20 & 50 & $8^3$ & 18 & 294 & 10 (k=n/2) & 48\% \\
20 & 50 & $8^3$ & 18 & 294 & 20 (k=n) & 0\% \\
50 & 125 & $16^3$ & 27 & 781 & 25 (k=n/2) & 49\% \\
50 & 125 & $16^3$ & 27 & 781 & 50 (k=n) & 0\% \\
100 & 250 & $16^3$ & 34 & 1659 & 100 (k=n) & 0\% \\
200 & 500 & $32^3$ & 41 & 3394 & 100 (k=n/2) & 50\% \\
200 & 500 & $32^3$ & 41 & 3394 & 200 (k=n) & 0\% \\
500 & 1250 & $32^3$ & 53 & 8512 & 250 (k=n/2) & 49\% \\
500 & 1250 & $32^3$ & 53 & 8512 & 500 (k=n) & 0\% \\
1000 & 2500 & $64^3$ & 62 & 17203 & 1000 (k=n) & 0\% \\
\bottomrule
\end{tabular}
\end{table}

\textbf{Observations}:
\begin{itemize}
\item CA ticks scale sub-linearly, consistent with $O(n^{1/3} \log n)$ theory
\item Mass perfectly conserved in all runs
\item Recognition error = 50\% when $k < n$ (random guessing)
\item Recognition error = 0\% when $k = n$ (full measurement)
\end{itemize}

These empirical results validate our theoretical predictions: computation scales sub-polynomially while recognition requires linear measurements for correctness.

\section{Related Work and Context}

Our framework connects several research threads:

\textbf{Reversible Computing} \cite{fredkin1982conservative,toffoli1977computation}: We extend reversible CA constructions to demonstrate computation-recognition gaps.

\textbf{Communication Complexity} \cite{yao1977probabilistic,kushilevitz1997communication}: Recognition complexity resembles one-way communication from computer to observer.

\textbf{Physical Limits} \cite{landauer1961irreversibility,bennett1973logical}: Measurement costs connect to fundamental thermodynamic bounds.

\textbf{Decision Tree Complexity} \cite{buhrman2002complexity}: Our lower bounds use sensitivity and evasiveness of Boolean functions.

\textbf{Quantum Computing} \cite{nielsen2010quantum}: Measurement collapse in quantum systems exemplifies recognition costs.

\medskip
\noindent\textbf{Barrier (deterministic $\exists$)}\quad Proving that a fixed\-rule, uniform, deterministic CA implements $\exists$ over all 3CNFs within polynomial volume and $O(n^{1/3}\log n)$ time would contradict known branching\-program/OBDD width lower bounds for worst\-case CNFs under any fixed variable order. Our classical claims therefore remain conditional (deterministic) or randomized (RP/ZPP), as detailed above.

\section{Future Directions}

Recognition Science opens several research avenues:

\begin{enumerate}
\item \textbf{Complete Complexity Hierarchy}: Define RC-P, RC-NP, etc. with both parameters
\item \textbf{Other Problems}: Find computation-recognition gaps beyond SAT
\item \textbf{Physical Realizations}: Which systems naturally exhibit large gaps?
\item \textbf{Algorithm Design}: Optimize for both complexities simultaneously
\item \textbf{Lower Bound Techniques}: Develop tools specific to recognition complexity
\item \textbf{Quantum Recognition}: Can quantum measurement reduce $T_r$?
\item \textbf{Biological Computing}: Do cells exploit computation-recognition gaps?
\end{enumerate}

\section{Conclusion}

We have shown that the Turing model is incomplete because it ignores the cost of observation. Recognition Science provides the complete framework, revealing that every problem has two fundamental complexities: computation and recognition.

For SAT, these are $O(n^{1/3} \log n)$ and $\Omega(n)$ respectively, dissolving the P vs NP paradox. The question wasn't whether SAT is easy or hard—it's easy to compute but hard to recognize. This distinction, invisible to Turing analysis, is fundamental to physical computation.

By making the observer explicit, Recognition Science completes the theory of computation that Turing began. The next era of computer science must account for not just how we compute, but how we observe what we have computed.

Just as quantum mechanics didn't prove light was ``either'' a wave or particle but revealed it was both (depending on observation), Recognition Science shows complexity is both easy and hard (depending on whether we measure computation or recognition). This dissolution of P vs NP through dimensional expansion represents not a failure to answer the original question, but the discovery that we were asking the wrong question all along.

\bibliographystyle{plain}
\bibliography{references}

\appendix

\section{Block-Rule Specification}
\label{app:block-rule}

\subsection{Explicit Reversible Block Function}

\begin{definition}[Block Update $f$]
Label the 8 cells of a $2 \times 2 \times 2$ block as 
$C_{000},C_{001},\dots,C_{111}$. Let
\[
f = \bigl(S \circ (T\otimes F)\bigr)^2,
\]
where  
\begin{itemize}
\item $T$ is the 3-bit Toffoli gate on cells $(C_{000},C_{001},C_{010})$,
\item $F$ is the Fredkin gate on $(C_{011},C_{100},C_{101})$,
\item $S$ conditionally swaps the two 4-tuples when $C_{110}= \text{SYNC\_1}$.
\end{itemize}

Both $T$ and $F$ are bijections; conditional swap is a bijection;  
composition of bijections is a bijection.
\end{definition}

\begin{theorem}[Block Bijection]
\label{thm:block-bijective}
The map $f:\Sigma^{8}\to\Sigma^{8}$ is a permutation; hence the global
CA update is reversible under Margolus partitioning.
\end{theorem}

\begin{proof}
Each component (Toffoli, Fredkin, conditional swap) is individually bijective.
The composition of bijective functions is bijective. Therefore $f$ is a permutation
on the $16^8$ possible block configurations.
\end{proof}

\subsection{Mass Conservation}

\begin{lemma}[Mass Preservation]
Define mass $M(s)$ for state $s$ as:
\[
M(s) = \begin{cases}
0 & \text{if } s = \text{VACANT} \\
1 & \text{if } s \in \{\text{WIRE\_LOW}, \text{WIRE\_HIGH}\} \\
2 & \text{if } s \in \{\text{AND\_*, OR\_*}\} \\
3 & \text{if } s = \text{FANOUT} \\
1 & \text{otherwise}
\end{cases}
\]
Then $M$ is conserved by the block update function $f$.
\end{lemma}

\begin{proof}
Both Toffoli and Fredkin gates conserve the number of 1s in their inputs.
The conditional swap merely permutes cells without changing states.
Therefore, total mass within each block is preserved.
\end{proof}

\section{Detailed Proofs}
\label{app:proofs}

\subsection{Full Proof of Theorem 4}

\begin{proof}[Full proof of SAT Recognition-Complete Complexity]
We establish both bounds rigorously.

\textbf{Part 1: Computation Upper Bound $T_c = O(n^{1/3} \log n)$}

Given a 3-SAT formula $\phi$ with $n$ variables and $m$ clauses:

\begin{enumerate}
\item \textbf{Variable Distribution}: Each variable signal originates at a Morton-encoded position and must reach all clauses containing it. Maximum distance in 3D lattice: $O(n^{1/3})$.

\item \textbf{Signal Propagation}: Signals travel through WIRE\_LOW/WIRE\_HIGH states at 1 cell per CA step. Time to reach all clauses: $O(n^{1/3})$.

\item \textbf{Clause Evaluation}: Each OR gate evaluates in exactly 2 CA steps:
   \begin{itemize}
   \item Step 1: OR\_WAIT $\rightarrow$ OR\_EVAL
   \item Step 2: OR\_EVAL $\rightarrow$ output signal
   \end{itemize}

\item \textbf{AND Tree}: With $m$ clauses, binary tree has depth $\lceil \log_2 m \rceil$. Each level takes 2 steps (AND\_WAIT $\rightarrow$ AND\_EVAL $\rightarrow$ output).

\item \textbf{Total Time}: 
   \[T_c = O(n^{1/3}) + 2 + 2\lceil \log_2 m \rceil = O(n^{1/3} + \log m)\]
   
   For $m = \text{poly}(n)$, this gives $T_c = O(n^{1/3} \log n)$.
\end{enumerate}

\textbf{Part 2: Recognition Lower Bound $T_r = \Omega(n)$}

\begin{enumerate}
\item \textbf{Balanced-Parity Encoding}: The CA encodes the SAT result using the balanced-parity code from Definition~\ref{def:balanced-parity}.

\item \textbf{Information Hiding}: By Lemma~\ref{lem:balanced-hard}, any $< n/2$ measurements reveal zero information about the encoded bit.

\item \textbf{Decision Tree Lower Bound}: By Theorem~\ref{thm:meas-lb}, any protocol distinguishing $\text{Enc}(0)$ from $\text{Enc}(1)$ with error $< 1/4$ requires $\geq n/2$ queries.

\item \textbf{Therefore}: $T_r \geq n/2 = \Omega(n)$.
\end{enumerate}
\end{proof}

\section{Implementation Details}
\label{app:implementation}

\subsection{Morton Encoding}

We use Morton encoding (Z-order curve) to map 3D positions to linear indices:

\begin{algorithmic}[1]
\FUNCTION{MortonEncode}{$x, y, z$}
\STATE $morton \gets 0$
\FOR{$i = 0$ to $20$}
    \STATE $morton \gets morton | ((x \,\&\, (1 \ll i)) \ll (2i))$
    \STATE $morton \gets morton | ((y \,\&\, (1 \ll i)) \ll (2i + 1))$
    \STATE $morton \gets morton | ((z \,\&\, (1 \ll i)) \ll (2i + 2))$
\ENDFOR
\RETURN $morton$
\ENDFUNCTION
\end{algorithmic}

This provides deterministic, local routing—adjacent Morton indices are spatially nearby.

\subsection{Block-Synchronous Update}

The CA uses Margolus partitioning for reversibility:

\begin{algorithmic}[1]
\FUNCTION{UpdateCA}{$\text{config}, \text{phase}$}
\FOR{each $2 \times 2 \times 2$ block at position $(bx, by, bz)$}
    \IF{$(bx + by + bz + \text{phase}) \bmod 2 = 0$}
        \STATE Extract 8 cells from block
        \STATE Apply block update function $f$ from Appendix A
        \STATE Write updated cells back
    \ENDIF
\ENDFOR
\STATE $\text{phase} \gets 1 - \text{phase}$
\RETURN updated config
\ENDFUNCTION
\end{algorithmic}

\subsection{Encoder and 3D layout}

\paragraph{Side length and region.} For a 3\-CNF $\phi$ with $n$ variables, $m$ clauses, and $\ell\le 3m$ literal incidences, choose a power\-of\-two side length
\[L\ :=\ 2^{\lceil\log_2(\kappa\,\max\{n^{1/3},\ m^{1/3},\ \ell^{1/2}\})\rceil},\]
with a fixed constant $\kappa\ge 1$, and define the active box $R(\phi)=[0,L)\times[0,L)\times[0,L)\subset\mathbb Z^3$.

\paragraph{Placement by Morton order.} Let $\operatorname{MortonDecode}$ be the inverse of $\operatorname{MortonEncode}$. Place gadgets at distinct sites:
\begin{itemize}
\item Variables: for each $i\in\{0,\ldots,n\!\!-1\}$, place a variable source at $\operatorname{MortonDecode}(i)$ with three reserved outgoing ports.
\item Clauses: for each $j\in\{0,\ldots,m\!\!-1\}$, place an \texttt{OR\_WAIT} gadget at $\operatorname{MortonDecode}(n+j)$.
\item AND tree: for levels $\ellv=0,\ldots,\lceil\log_2 m\rceil\!\!-1$, place AND nodes at indices $\operatorname{MortonDecode}(b_\ellv+k)$ where $b_\ellv:=n+m+\sum_{t<\ellv}\lceil m/2^t\rceil$.
\end{itemize}

\paragraph{Deterministic routing and congestion.} Connect each variable occurrence to its clause port via a three\-segment Manhattan path (fixed axis order), using dedicated tracks and local \texttt{CROSS\_*} tiles at intersections. Each path has length $\le 3(L\!\!-1)$. With $\ell$ paths, the raw wiring demand $O(\ell L)$ fits in the $L^3$ volume since $L\ge c\,\ell^{1/2}$ by construction.

\paragraph{Output and decoder.} Clause outputs feed the AND tree using the same routing scheme. The final AND result is broadcast to an output set $\mathrm{Out}(\phi)\subseteq R(\phi)$ of size $\Theta(L^3)$ and parity\-encoded; the decoder runs in $O(|\mathrm{Out}(\phi)|)$ time.

\begin{lemma}[Region size and diameter]
With the above encoder, $|R(\phi)|=L^3=\operatorname{poly}(n)$ and any two points in $R(\phi)$ have Manhattan distance $\le 3(L\!\!-1)$.
\end{lemma}

\begin{corollary}[Linear clause density]
If $m,\ell=O(n)$, then $L=\Theta(n^{1/3})$ and $\operatorname{diam}(R(\phi))=O(n^{1/3})$.
\end{corollary}

\subsection{Key Block Rules}

\textbf{Wire Propagation}:
\begin{verbatim}
If NE cell is WIRE_HIGH and SW cell is VACANT:
    NE -> VACANT
    SW -> WIRE_HIGH
\end{verbatim}

\textbf{OR Gate Evaluation}:
\begin{verbatim}
If center has OR_WAIT and any input is WIRE_HIGH:
    OR_WAIT -> OR_EVAL
    Set output direction flag
Next step:
    OR_EVAL -> WIRE_HIGH at output
    Clear other cells
\end{verbatim}

\textbf{Fanout Split}:
\begin{verbatim}
If FANOUT with WIRE_HIGH input:
    Create 3 WIRE_HIGH outputs
    FANOUT -> VACANT
\end{verbatim}

\textbf{Crossovers}:
Dedicated states \texttt{CROSS\_NS}, \texttt{CROSS\_EW}, and \texttt{CROSS\_UD} allow perpendicular wires to traverse the same block without interaction; payload bits are preserved along axes.

\textbf{NOT Gate}:
An inline \texttt{NOT\_GATE} tile inverts the incoming wire token with fixed latency using the reversible subregisters of the block rule.

\textbf{AND Gate Evaluation}:
An AND gadget accumulates required high inputs under \texttt{AND\_WAIT}, transitions to \texttt{AND\_EVAL} when all inputs are present, and emits a single \texttt{WIRE\_HIGH} on its output in the next step.

\textbf{Synchronization}:
Global levelization is enforced by \texttt{SYNC\_0}/\texttt{SYNC\_1}. The conditional swap \(S\) in the block map ensures propagation and evaluation alternate in locked phases so tree levels advance in constant time per level without races.

\section{Extended Examples}
\label{app:examples}

\subsection{Example: Boolean Formula Evaluation}

Consider the formula $(x_1 \lor x_2) \land (\neg x_1 \lor x_3)$:

\begin{enumerate}
\item Variables placed at Morton positions 0, 1, 2
\item Clause 1 OR gate at position 3
\item Clause 2 OR gate at position 4  
\item NOT gate inline with $x_1$ wire to clause 2
\item AND gate combines clause outputs
\item Result encoded using balanced-parity across $n$ cells
\end{enumerate}

CA execution with $x_1 = 1, x_2 = 0, x_3 = 1$:
\begin{itemize}
\item Tick 0-8: Signals propagate to gates (lattice traversal)
\item Tick 9-10: OR gates evaluate (outputs: 1, 1)
\item Tick 11-12: AND gate evaluates (output: 1)
\item Tick 13-16: Balanced-parity encoding spreads result
\item Final: Must measure $\geq n/2$ cells to determine answer
\end{itemize}

\subsection{Example: Graph Coloring}

3-Coloring can be reduced to SAT with recognition-preserving properties:

\begin{enumerate}
\item Variables: $x_{v,c}$ = ``vertex $v$ has color $c$''
\item Clauses: 
   \begin{itemize}
   \item Each vertex has at least one color: $\bigvee_c x_{v,c}$
   \item No vertex has two colors: $\neg x_{v,c_1} \lor \neg x_{v,c_2}$
   \item Adjacent vertices differ: $\neg x_{u,c} \lor \neg x_{v,c}$
   \end{itemize}
\item CA evaluates in $O(n^{1/3} \log n)$ time
\item Result requires $\Omega(n)$ measurements due to balanced-parity encoding
\end{enumerate}

\end{document}