# Vanishing interventions realize the cap-five Haar law from integer Collatz roots Lyapunov Weaver. 10 November 2026. ## 1. Results and relation to the original chain Let $U=\mathbb Z_3^\times$, $U_i=i+3\mathbb Z_3$, $C=U_2$, and $$ I_e(x)=\frac{2^ex-1}{3},\qquad 1\le e\le5. $$ An edge is legal if its source and target lie in $U$. The kernel $P_5$ chooses uniformly among the legal edges. Write $h$ for normalized Haar probability on $U$, $h_i$ for conditional Haar on $U_i$, and $p=1/13$. The original odd-edge clock counts each application of $I_e$ once. The corresponding shortcut and classical physical clocks cost $e$ and $e+1$. Paper128 [1] proves supplied-state convergence for positive controllers which select full return branches of one fixed length. The original $P_5$ chain presents a different question: its stationary laws are $a\delta_p+(1-a)h$ [2], and the atomic weight for an integer start is still unresolved. We construct a causal controller which retains genuine $P_5$ suggestions except on a set of edges of density zero. **Theorem 1 (bounded mean repair).** For each $n\ge3$, there is a causal cap-five controller $K_n$ whose blocks have exactly $n$ edges and return to $C$. It has unique boundary stationary law $h_C$, and its empirical boundary measures converge to $h_C$ from every supplied $x\in C$. Let $R_n$ be its marked suffix length in a stationary block. Then $$ \Pr(R_n>r)=\frac12\,3^{-r}\quad(0\le rr)=3^{-r}\quad(r\ge0). \tag{6} $$ A length-$k$ word and its probabilities are determined by the source modulo $3^{k+1}$. After selecting a word on such a source cell, its endpoint is conditional Haar in its terminal class: the affine word divides by $3^k$ and multiplies by a ternary unit. A completion test allowing $m$ further edges consults at most $k+m+1$ source digits. This information budget will be retained in every construction below. ## 3. Causal completion and its exact repair law First describe a convenient coupling. From $x\in C$ sample an ordinary $P_5$ proposal of length $n$, with word $W$ and states $Y_0,\ldots,Y_n$. Choose the least $m\in\{0,\ldots,n\}$ such that $$ Y_{n-m}\in C \quad\text{or}\quad Y_{n-m}\in U_1\text{ and }V(Y_{n-m})\le m. \tag{7} $$ At $m=n$ the empty prefix succeeds. Retain the first $n-m$ edges, take the first available exit if needed, and pad by $H$ to exactly $n$ edges. Call the completed word $W'$ and set $R_n=m$. This has an online implementation. With $r$ edges left, sample a genuine $P_5$ suggested edge from the current state $x$. Accept its child $y$ when $y\in C$ or $V(y)\le r-1$. Otherwise reject that edge, take the immediate exit to $C$, and pad the remainder by $H$. Mark the rejected edge and all padding edges. Before rejection the invariant is $x\in C$ or $V(x)\le r$. An unsafe suggestion can occur only at $x\in C$ or $V(x)=1$, so the immediate exit exists. Its unrepaired child has more depth than remaining edges; all remaining proposal edges are then forced $I_4$. Consequently the first rejection is exactly at position $n-R_n$, and the online output is $W'$. No future random coin is inspected. During padding one may still sample and ignore fresh genuine $P_5$ suggestions at the actual states. The whole construction depends only on the source modulo $3^{n+1}$ and finitely many fair coins. Every chosen word maps its source cell onto $C$. Every exact $n$-edge legal return word has positive proposal probability and survives unchanged because $m=0$ succeeds. For $n\ge3$, [4] supplies at least two distinct such words on every source cell, by padding its short non-$H$ return and comparing with $H^n$. Theorem1 of [1] proves unique Haar boundary stationarity and supplied-state empirical convergence. For a proposal started from $h_C$, its first step has law $h$ because $H,Q$ are equiprobable; all later states also have law $h$ by stationarity [2]. For $0\le rr\}= \{Y_{n-r}\in U_1,\ V(Y_{n-r})>r\}. \tag{8} $$ One inclusion follows from the failure of candidate $r$. Conversely, that failure forces the remaining $r$ proposal steps to be $I_4$; their depth decreases by one each time, so every later candidate fails too. Equation(6) now proves(1), without independent-valuation assumptions. Also $$ \mathbb E(R_n-1)_+=\frac14(1-3^{-(n-1)}). \tag{9} $$ Under the coupling, only positions strictly after $n-R_n$ can change. Among the source positions $0,\ldots,n-1$, there are at most $(R_n-1)_+$ such positions. The proposal occupation is $(h_C+(n-1)h)/n$, at total variation distance $1/(2n)$ from $h$. Equations (9) and the coupling inequality prove the first bound in(2). The mean exponent of uniform $P_5$ under either conditional Haar class is three, by its six rows modulo nine. Hence the proposal mean is $3n$. Each of at most $R_n$ changed labels differs by at most four, proving the second bound in(2). The discarded proposal contributes no physical time. Conditional block rewards are bounded continuous functions of the boundary state. Boundary convergence and bounded martingale differences therefore give all claimed supplied-state reward and occupation limits. This proves Theorem1. Half the proposal blocks change, since $\Pr(R_n>0)=1/2$; it is the fraction of edited edges, rather than whole-block total variation, that becomes small. ## 4. Uniform sample averages and a paid diagonal schedule We give the uniform probabilistic step needed for Theorem2. Fix one $K_n$ and a block reward $Z_i\in[0,1]$ with continuous conditional mean $F$ and stationary mean $b=h_C(F)$. The uniform expected Cesaro convergence in [1] implies that, for any $\eta>0$, some $q$ makes the expected reward average over $q$ consecutive blocks lie within $\eta$ of $b$, from every state. Group the process into $q$-block chunks and subtract conditional chunk means. The resulting martingale differences $D_j$ are bounded by $q$. Thus $$ \Pr\left(\left|\sum_{j\eta qJ\right) \le2\exp(-\eta^2J/2). \tag{10} $$ Sum(10) over all $J\ge J_0$ and absorb the at most $q$ remaining terms. Choosing $\eta<\epsilon/4$ and then $J_0$ large gives, for any $\epsilon,\delta>0$, an $M$ such that $$ \sup_x\Pr_x\left( \sup_{m\ge M}\left|\frac1m\sum_{i \epsilon\right)\le\delta. \tag{11} $$ A finite union proves the same simultaneous bound for finitely many rewards. No fixed-time mixing rate or independence of rewards is assumed. Enumerate all unit-cylinder indicators as $f_1,f_2,\ldots$. At stage $j$ use $n_j=j+2$. The normalized rewards are the actual block averages of $f_1,\ldots,f_j$, the marked count divided by $n_j$, and the exponent sum divided by $5n_j$. Apply(11) with tolerance $1/j$ and simultaneous failure probability $2^{-j}$ to obtain $M_j$. Their stationary means tend, for each fixed $f_l$, to $h(f_l)$, zero and $3/5$, by Theorem1. Choose finite deterministic block counts $L_j\ge M_j$ and set $T_j=\sum_{i\le j}n_iL_i$. Enlarge each $L_j$ so that $$ T_j\ge(j+1)n_{j+1}M_{j+1}. \tag{12} $$ All choices can be made independently of the source and future coins. Conditionally on any stage's starting state, its all-prefix failure probability is at most $2^{-j}$. One may couple to an infinite continuation of that fixed kernel when applying(11). Borel--Cantelli shows that all sufficiently late stages are good almost surely. Prefixes of a good stage with at least $M_j$ complete blocks have their specified error bound. A shorter prefix contributes at most $n_jM_j/T_{j-1}\le1/j$ relative to previous elapsed odd time. An unfinished block contributes at most $n_j/T_{j-1}\le1/j$. At stage endpoints, averages are weighted means of complete-stage averages with vanishing errors and target discrepancies. The finitely many early or bad stages have finite length and vanishing eventual weight. This proves state occupation, zero marked density and the exponent and physical-clock limits at every time, not just at stage endpoints. A finite first-exit prelude handles sources in $U_1$. The schedule is effective in principle. For locally constant rational rewards, any finite Cesaro expectation is a rational table on finitely many source cylinders. Search $q$ until its supremum error is smaller than the required strict tolerance. Uniform convergence guarantees termination. Equation(10) then gives an explicit finite $M$. This procedure has no useful complexity bound here. ## 5. Finite windows and real integer growth Use independent fair coin slots at every actual edge, including marked padding edges. Fix a length-$k$ state-cylinder/label observable. From time $t$, simulate a virtual $P_5$ path of length $k$ from $X_t$, using the next $k$ coin slots. If its actual window contains no marked edge, the paths agree. The fraction of windows containing a mark is at most $k$ times the marked density, up to a boundary term, so the discrepancy tends to zero almost surely. The virtual observable has conditional mean $g(X_t)$ for a continuous $P_5$ expectation $g$. Partition $t$ into its $k$ residue classes. Nonoverlapping windows in each class give bounded martingale differences, whose averages vanish. Haar empirical state convergence sends the mean of $g(X_t)$ to $h(g)$, exactly the stationary Haar-$P_5$ window frequency. A countable cylinder and word base gives this conclusion simultaneously for all fixed windows. This comparison uses genuine suggested coins at actual states; it does not identify the controlled and unmodified paths after their first disagreement. For positive integer states the exact real identity is $$ \log X_t-\log X_0 =E_t\log2-t\log3+ \sum_{s