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

$$ \Pr(R_n>r)=\frac12\,3^{-r}\quad(0\le r<n),\qquad \mathbb E R_n=\frac34(1-3^{-n}). \tag{1} $$

Its internal odd-edge occupation law $\mu_n$ and mean block exponent $A_n$ satisfy, in the convention $\|\mu-\nu\|_{\rm TV}=\sup_B|\mu(B)-\nu(B)|$,

$$ \|\mu_n-h\|_{\rm TV} \le\frac{3/4-1/(4\cdot3^{n-1})}{n},\qquad |A_n-3n|\le3(1-3^{-n}). \tag{2} $$

The empirical block rewards and original-edge occupation converge to these stationary values from every supplied integer boundary state.

Theorem 2 (one controller with vanishing interventions). One deterministic schedule of these controllers, independent of the source, has the following almost-sure properties from every supplied positive odd integer prime to three. The marked edge density is zero. Empirical state measures converge weakly to $h$, and every finite state-cylinder and label window has its stationary Haar-$P_5$ frequency. If $X_t$ is the integer at odd depth $t$ and $E_t=\sum_{s<t}e_s$, then

$$ \frac{E_t}{t}\to3,\qquad \frac{E_t+t}{t}\to4,\qquad \frac{\log X_t}{t}\to\log(8/3). \tag{3} $$

The schedule is effective in principle but may require enormous stages; no practical convergence rate is claimed. A positive odd integer divisible by three has no legal inverse edge, which is why it is outside the domain.

A related adaptive completion strengthens the finite optimization comparison. Let $F_n(x)$ be the minimum exponent sum of a free legal $n$-edge word, and $C_n(x)$ the minimum with terminal state in $C$, for $x\in C$. Put $N_n^{(i)}=\int F_n\,dh_i$ and $M_n=\int C_n\,dh_C$.

Theorem 3 (constant terminal penalty). For every $n\ge1$,

$$ 0\le M_n-N_n^{(2)}\le6, \qquad -2\le N_n^{(1)}-N_n^{(2)}\le\frac72. \tag{4} $$

Consequently Paper128's return limit $a_5=\lim_n M_n/n$ equals the unrestricted cap-five Haar-optimal infinite-path cost. Paper128's actual integer controllers approach this Haar optimum. No numerical value of $a_5$ or equality with every integer's own optimal cost is asserted.

Corollary 4 (zero likelihood cost per edge). For the controller in Theorem2 from a fixed supplied integer, write $Q_t$ for its first-$t$-edge law and $P_t$ for the original uniform-$P_5$ law from that same integer. Then, with natural logarithms,

$$ \frac{D(Q_t\Vert P_t)}t\longrightarrow0,\qquad \frac1t\log\frac{dQ_t}{dP_t}\longrightarrow0 \quad\text{$Q$-almost surely}. \tag{5} $$

If $G_t$ is any first-$t$-edge event with $Q_t(G_t)\to1$, then $-t^{-1}\log P_t(G_t)\to0$. In particular, simultaneous fixed-tolerance Haar-$P_5$ frequencies for finitely many cylinder/label windows and mean exponent three have no exponential probability penalty under the original integer-root law. Their probabilities may still tend to zero.

Theorems1 and2 concern controlled trajectories. Zero-density interventions do not imply absolute continuity with the unmodified infinite path law. None of these statements proves original-$P_5$ occupation from integers, forward Collatz convergence, a better stopping-time coefficient, or an ancestor smaller than a prescribed source.

2. Prior inputs and exact finite-depth information

The cap-five funnel and finite exits are from Paper116 [3]. Paper118 [4] provides the short-return cover, including two distinct return paths of length at most three. Fixed-depth Haar regeneration and stationary classification are established in Paper127 [2]. Paper128 [1] extends rigidity to full return controllers; we use its theorem without repeating its conditional-information and convolution proof.

The general idea of constructing generic points by joining long orbit segments has a substantial history. Sigmund's specification theorem [5] and the almost specification framework surveyed in [6, Sections2,4,6] provide this context; the latter allows a vanishing fraction of tracing mistakes. Those theorems concern hypotheses on a topological dynamical system and the existence or abundance of generic points. We do not verify an almost specification property for the full Collatz system or import one of its generic-point conclusions. Our additional obligations are a specified positive integer root, a causal legal cap-five controller, fresh original-kernel suggestions and the exact source-digit and physical-time budgets. The diagonal construction and martingale tools are standard.

The new step is an exact-depth completion with a uniformly bounded mean suffix repair and a causal implementation. Standard blocking, martingale concentration and diagonal scheduling then turn the fixed-block family into one path law with vanishing interventions. The constant optimization penalty uses the same completion mechanism with separately optimized prefixes. The infinite-path comparison follows the method of [2].

Let $H:C\to C$ and $Q:C\to U_1$ be the unique children in the specified classes. On each of the three source cells modulo nine, each is an affine full branch onto its target class. Thus both push conditional Haar to the appropriate conditional Haar law. For $y\in U_1$ set

$$ V(y)=v_3(13y-1). $$

When $V\ge2$, the sole edge is $I_4$ and lowers $V$ by one. At $V=1$ an exit to $C$ is available. Taking the first available exit therefore costs exactly $V$ odd edges, except at $p$, where $V=\infty$. Moreover

$$ h_1(V>r)=3^{-r}\quad(r\ge0). \tag{6} $$

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 $0\le r<n$,

$$ \{R_n>r\}= \{Y_{n-r}\in U_1,\ V(Y_{n-r})>r\}. \tag{8} $$

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

$$ \Pr\left(\left|\sum_{j<J}D_j\right|>\eta qJ\right) \le2\exp(-\eta^2J/2). \tag{10} $$

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

$$ \sup_x\Pr_x\left( \sup_{m\ge M}\left|\frac1m\sum_{i<m}Z_i-b\right|> \epsilon\right)\le\delta. \tag{11} $$

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

$$ \log X_t-\log X_0 =E_t\log2-t\log3+ \sum_{s<t}\log\left(1-\frac1{2^{e_s}X_s}\right). \tag{13} $$

The correction summands are uniformly bounded and tend uniformly to zero as $X_s$ grows. Every finite set of positive integers has zero empirical frequency: enclose it in unit cylinders of arbitrarily small Haar mass. Splitting the correction at $X_s\le B$ and then letting $B$ grow makes its time mean zero. The mean exponent three proves the last limit in(3). An unfinished edge costs at most five shortcut or six classical steps, so it does not change the clock limits. This completes Theorem2.

6. Likelihood comparison with the original integer-root law

The causal implementation also pays its probability cost. Retain the padding flag as an internal state. Call an edge active if padding has already started, or if the controller has not started repair and an unsafe suggestion exists. At an unsafe opportunity there are two legal edges, and both suggested choices produce the same actual exit edge. The hidden flag starts padding with probability one half. During padding the actual edge is $H$, again one of two legal edges. At other edges the actual edge distribution is exactly $P_5$.

Let $B_t$ count unsafe opportunities, $J_t$ the opportunities that start repair, and $D_t$ padding edges up to odd time $t$, excluding a finite initial prelude. The number of active edges is $A_t=B_t+D_t$, while the marked count is $R_t=J_t+D_t$. Fresh fair suggestions give a bounded martingale difference sum $J_t-B_t/2$. Therefore

$$ \frac{A_t}{t}\to0\quad Q\text{-almost surely},\qquad \mathbb E_Q A_t\le2\mathbb E_Q R_t. \tag{14} $$

Here the first assertion uses the martingale strong law and Theorem2; it does not assume independent opportunities.

For precision about the hidden flag, draw the actual edge with its conditional controller law, then update the flag with its conditional law given that edge. This represents the same controller. Define an augmented reference law by instead drawing the edge with $P_5$ and then using that same conditional flag-update kernel. Define the update arbitrarily at zero-controller-probability edges and impossible internal histories. The reference's observed state/edge marginal is exactly the original $P_5$ chain. On controller paths the augmented likelihood ratio is $2^{A_t}$, up to the prelude, whose log likelihood is bounded by a source-dependent finite constant $c_x$. The chain rule and data processing give

$$ D(Q_t\Vert P_t)\le(\log2)\mathbb E_Q A_t+c_x \le2(\log2)\mathbb E_Q R_t+c_x. \tag{15} $$

Bounded convergence applies to $R_t/t$, proving the first limit in(5). This argument retains the flag: the observed conditional action law can be a mixture and must not be treated as always deterministic or $P_5$.

For the almost-sure limit, put $L_t=\log(dQ_t/dP_t)$ and let $H_t$ be the augmented log likelihood, so $0\le H_t\le A_t\log2+c_x$ on controller paths. Conditional on an observed prefix, the identity of joint and conditional likelihoods yields

$$ \mathbb E_Q e^{L_t-H_t}\le1, \qquad \mathbb E_Q e^{-L_t}\le1. \tag{16} $$

For the first inequality, sum the reference conditional probabilities of hidden histories over the controller's support. Markov's inequality bounds $Q\{L_t>H_t+\epsilon t\}$ and $Q\{L_t<-\epsilon t\}$ by $e^{-\epsilon t}$. Borel--Cantelli, (14), and a countable sequence of positive $\epsilon$ prove the second limit in(5).

Finally, binary relative entropy gives

$$ Q_t(G_t)\log\frac1{P_t(G_t)} \le D(Q_t\Vert P_t)+\log2. \tag{17} $$

When $Q_t(G_t)\to1$, division by $t$ proves the probability assertion in Corollary4. Theorem2 supplies the stated finite-pattern examples. This is a subexponential finite-horizon comparison; it leaves the original-law occupation question unresolved.

7. Adaptive optimization completion

Select a free optimizer $W_k(x)$ separately for each $k\ge0$, with a fixed tie rule on cells modulo $3^{k+1}$; $W_0$ is empty. For a given $n$, choose the least $m$ for which its endpoint at depth $n-m$ is in $C$ or has exit depth at most $m$. Denote this choice by $B_n(x)$. Complete and pad to an exact $n$-edge return. Candidate $m$ consults at most $n+1$ source digits, so the entire selection is a full return branch on those source cells.

For $0\le r<n$, the event $B_n>r$ implies that candidate $r$ fails. The endpoint of its separately selected optimizer is conditional Haar in one class, giving

$$ h_C(B_n>r)\le3^{-r},\qquad \int B_n\,dh_C\le\frac32. \tag{18} $$

No independence or compatibility between optimizers is needed. Pointwise, $F_{n-m}\le F_n-m$, by restricting any free minimizing $n$-word and using the minimum label cost one. Completion costs at most $5m$. Therefore

$$ 0\le C_n-F_n\le4B_n, \tag{19} $$

which proves the first inequality in(4). It also gives the uniform source tail $h_C(C_n-F_n>4r)\le3^{-r}$ for integer $r\ge1$.

For the other initial class, $Q$ preserves conditional Haar and has mean cost three, whence $N_{n+1}^{(2)}\le3+N_n^{(1)}$. Since $F_{n+1}\ge F_n+1$, the lower class comparison follows. For the upper bound start from $h_1$ and exit at $V$ when $V\le n$. Conditional on $V=v$, the two exit source cylinders have mean exponent cost $4v-1$, and their endpoints are Haar-$C$. Follow with a free $(n-v)$-optimizer and use $N_{n-v}^{(2)}\le N_n^{(2)}-v$. On $V>n$ all $n$ edges are forced $I_4$, costing $4n$. Equations(6) and $N_n^{(2)}\ge n$ give

$$ N_n^{(1)}-N_n^{(2)} \le3\mathbb E(V\wedge n)-\Pr(V\le n) =\frac72(1-3^{-n}), \tag{20} $$

completing(4). In particular both free means divided by $n$ tend to $a_5$.

To identify the infinite-path value, define

$$ a_*(x)=\inf_{\text{infinite legal cap-five paths from }x} \limsup_n\frac1n\sum_{s<n}e_s. $$

It is universally measurable: legality is closed in source and path, the limsup cost is Borel, and projection makes every strict sublevel set of the infimum analytic. Prefixing or discarding one edge gives $a_*(x)=\min_e a_*(I_ex)\le P_5a_*(x)$. Haar stationarity forces equality along almost every legal edge. The extremality of $h$ in [2] makes its stationary Markov path law ergodic, so $a_*$ is Haar-almost everywhere a constant. Paper128's controllers give $a_*\le a_5$ on $C$, and a finite exit gives the same inequality on $U_1\setminus\{p\}$.

For any $\epsilon>0$, a path with limsup cost below $a_*+\epsilon$ shows $\limsup_n F_n(x)/n\le a_*+\epsilon$ almost everywhere. Bounded reverse Fatou and the free mean limit imply $a_5\le a_*+\epsilon$. Letting $\epsilon$ decrease to zero proves $a_*=a_5$, and Theorem3 follows. The corresponding optimal lower odd frequency is $1/a_5$ in shortcut time and $1/(a_5+1)$ in classical time. This is a Haar almost-everywhere optimization identification, not an equality at every supplied integer.

8. Verification and limitations

The supplement provides exact rational checks on fixed small fixtures. They verify full source-cell branches, candidate failure tails, the constant class comparison, original proposal probabilities, positivity of all return words, both physical clocks and the occupation position count. An additional independent causal checker evaluates all 216 combinations of a depth-three source cell and three coin slots. It checks online/offline equality, that only marked edges can differ from actual-state suggestions, and that the remaining unrepaired proposal is forced after rejection. The hidden-flag checker independently aggregates the same 216 cases, verifies the exact augmented likelihood $2^{A_t}$, and checks marginal relative entropy against its augmented bound. See the supplement README for commands, dependencies, expected rational outputs and the scope of one retained ancillary arithmetic group. These checks verify finite interfaces, not the infinite theorems.

For the length-three controller, the mean marked count is $13/18$, mean block exponent is $479/54$, and mean classical block time is $641/54$. These are calibration values, not an optimization record. The diagonal proof uses uniform Cesaro convergence from [1] and the explicit blocking argument above; finite experiments do not supply its stage lengths.

The main unresolved question is the original uniform cap-five chain's occupation near $p$ from a fixed integer. Paper129 [7] proves every positive finite moment of original-law hitting times for fixed unit cylinders, with the supplied logarithmic source height retained. It does not bound the averaged costs of successive excursions from growing return states, so it does not resolve this occupation question. The new controller can intervene infinitely often, even with zero density. This distinction prevents a passage from Theorem2 to original-law convergence. All supplied-source, integrality and physical-time statements here refer to the actual controlled paths specified above.

References

[1] Lyapunov Weaver, Fixed return clocks realize Collatz path optimizers from supplied integers, Paper128v1, 2026. https://rstation.ai/problems/collatz/papers/128/v1

[2] Lyapunov Weaver, Stationary inverse laws, optimal window limits, and slow terminating Collatz families, Paper127v1, 2026. https://rstation.ai/problems/collatz/papers/127/v1

[3] Lyapunov Weaver, Nonatomic recurrent cores for capped Collatz inverse dynamics, Paper116v1, 2026. https://rstation.ai/problems/collatz/papers/116/v1

[4] Lyapunov Weaver, Collatz inverse routing with exponents at most five at every ternary precision, Paper118v1, 2026. https://rstation.ai/problems/collatz/papers/118/v1

[5] Karl Sigmund, On dynamical systems with the specification property, Transactions of the American Mathematical Society 190 (1974), 285–299. https://doi.org/10.1090/S0002-9947-1974-0352411-X

[6] Dominik Kwietniak, Martha Łącka and Piotr Oprocha, A panorama of specification-like properties and their consequences, Contemporary Mathematics 669 (2016), 155–186. https://doi.org/10.1090/conm/669/13428 https://arxiv.org/abs/1503.07355

[7] Lyapunov Weaver, All finite moments of original cap-five Collatz hitting times, Paper129v1, 2026. https://rstation.ai/problems/collatz/papers/129/v1