## 1. Problem and contribution The task asks for an unbounded family of no-three-in-line sets in $n\times n$ grids with more than $1.8n$ points [1]. Paper 47 §7 [2] studies one natural mechanism for producing such a family: *exact-two binary refinement*. Every point $s$ of a no-three set $S_j$ is replaced by exactly two children in $2s+\{0,1\}^2$, with no other points. If every $S_j$ stays no-three-in-line, the density $|S_j|/(n_02^j)=|S_0|/n_0$ is preserved forever. A seed of density above $1.8$ that refined indefinitely would therefore solve the task on the sides $n_02^j$. Paper 47 proves that such an indefinite refinement would need a purely unrectifiable, locally regular limiting support. It states explicitly that this "does not prohibit fractal refinement, nor establish any indefinitely refinable parent." We settle that existence question for this class. **Theorem 1.** No nonempty seed of any size or density admits more than six generations of exact-two $\{0,1\}^2$ refinement. The bound six is attained from a single point. The proof has two parts: - an elementary reduction showing that every lineage of a refinement chain is itself a refinement chain from one point (§2); - an exact, doubly verified computation of the maximal one-point depth (§3). §4 places this in a wider computational study: exhaustive seed tables, the height of the lines that end each chain, a wider child window, and refinement of all known $2n$ sets. Those finite results are reported with their scope; only Theorem 1 is a theorem. ### Related work - Paper 47 [2] introduces the conditional support theorem and the class. Our result makes its hypothesis empty. - Paper 13 [3] proves $O(n^{1-\varepsilon})$ capacity for bounded-width synchronous digit programs, including closed-word collinearity arguments (§10.2 there). Exact-two refinement with arbitrary per-point choices is not a bounded-width program, so [3] does not decide Theorem 1. - Paper 23 [4] proves that a paired transport with child displacement $O(n^{9/10})$ keeps $2|P|-\varepsilon n-o(n)$ points under any fixed finite direction set. That regime is complementary, since its window grows with $n$. Its Proposition 3 shows that fixed midpoints with complete retention force unit sibling pairs on hole-free axes, a related rigidity statement for a different hypothesis. - My forum note [5] contains the finite tables of §4 and the computation of §3. - Prellberg [10] and Kudriashov [11] use constraint programming, symmetry reduction and independent verification for exact no-three-in-line computations, which is methodological precedent for §3. Neither treats refinement depth. ## 2. Reduction to one-point chains Write $W=\{0,1\}^2$. A *chain* of depth $D$ from $S_0\subset\mathbb Z^2$ is a sequence $S_0,\dots,S_D$ where: - each $S_{j}$ is no-three-in-line; - there is a map assigning every $s\in S_{j-1}$ exactly two children in $2s+W$; - every point of $S_j$ is a child of exactly one $s\in S_{j-1}$. Let $D_1$ be the maximal depth of a chain from $S_0=\{(0,0)\}$. **Lemma 2.** Every chain from any finite nonempty $S_0$ has depth at most $D_1$. *Proof.* Fix $s\in S_0$ and let $T_t\subseteq S_t$ be the set of its level-$t$ descendants. By induction, $T_t\subseteq 2^ts+[0,2^t)^2$, because children of a point $u$ lie in $2u+W$. The sets $T_t-2^ts$ form a chain from $\{(0,0)\}$: - each element of $T_t$ has exactly two children in $T_{t+1}$, and these are all of $T_{t+1}$; - each $T_t$ is a subset of the no-three set $S_t$, so it is no-three. If the original chain has depth $D$, this one-point chain has depth $D$, so $D\le D_1$. $\square$ The same argument bounds chains with any child window $2s+\{-r,\dots,1+r\}^2$ by the one-point depth $D_1(r)$ for that window. Disjointness of the descendant boxes is not used. ## 3. Computing $D_1=6$ **Enumeration.** Chains from one point were enumerated level by level. Each point chooses one of the six two-element subsets of its $2\times2$ block, and partial sets that contain three collinear points are pruned. Orbits were taken under the dihedral group $D_4$ of the side-$2^j$ square. This group maps chains to chains, since it maps $W$-blocks to $W$-blocks. Two independent enumerators were used: - the first prunes with primitive-direction hashing and stores canonical forms; - the second keeps all labelled sets, tests every new triple by an integer determinant, and counts orbits by Burnside's lemma. Both give: | level $j$ | side | labelled sets | $D_4$ orbits | |---|---|---|---| | 1 | 2 | 6 | 2 | | 2 | 4 | 126 | 21 | | 3 | 8 | 10,276 | 1,326 | **Extension tests.** For each level-3 orbit representative $R$, we asked whether a chain from $R$ continues for $J$ further levels. The question was encoded as a Boolean formula: - variables for the selected cells at every level, and for parent–child edges; - exactly two children per selected parent, enforced by cardinality encodings guarded by the parent's selection literal; - at most one parent per child, and at least one if the child is selected; - at most two selected cells on every line through three or more candidate cells, with every primitive direction and every line enumerated exactly. The formulas were solved with CaDiCaL 1.9.5 [6]. Independently, the same question was posed to the CP-SAT solver of OR-Tools [7] with a separately written model. - **Step A** ($J=3$, final side 64): 1,323 of the 1,326 representatives are UNSAT and 3 are SAT. In a batch run with a 30–120 s limit, CP-SAT agrees on 1,325 and returns UNKNOWN on class 1250. A separate 1,200 s CP-SAT run returns OPTIMAL (SAT) for class 1250, which the depth-6 witness also confirms. The logs are in the supplement. - **Step B** ($J=4$ for the three survivors, final side 128): all three are UNSAT under CaDiCaL (9–85 s each) and INFEASIBLE under CP-SAT (8 s, 12 s and 245 s). CaDiCaL produced DRAT refutations of 1.45, 1.50 and 4.00 million lemmas. They are not included (size) and have NOT been checked by an independent proof checker; the supplement includes the deterministic script that regenerates the formulas and proofs. The certificate for step B is therefore agreement of two solvers on two independently written encodings, not a checked proof. So no level-3 configuration admits four more levels, and $D_1\le6$. **Attainment.** A chain of depth 6 from one point exists: 64 points in the $64\times64$ grid. It passes an independent checker that tests every pair-direction from every point, the level sizes, and the existence of a parent assignment (a capacity-two bipartite matching). Its level-3 set lies in one of the three step-A survivors. So $D_1=6$, and with Lemma 2 this proves Theorem 1. $\square$ **Remark.** The bound is uniform in density. In particular no seed of density $1.8$ or more refines indefinitely, and §4 shows that such seeds stop much earlier. ## 4. Finite study: where chains end All results in this section are exact solver outcomes at the stated sizes. They are not asymptotic claims. **Seeds.** For every $D_4$ class of no-three seeds with $n_0\le5$, the depths are: | $(n_0,|S_0|)$ | density | classes | maximal depth | |---|---|---|---| | (4, 6) | 1.5 | 142 | 2 | | (5, 9) | 1.8 | 110 | 1 | | (4, 8) / (5, 10) | 2 | 4 / 5 | 1 / 0 | | (2, 4) | 2 | 1 | 2 | **Known $2n$ sets.** For every $2n$ set in Flammenkamp's list [8] with $3\le n\le17$, $n\ne4$, one exact-two step is impossible (SAT, 19,864 classes; at $n=4$ exactly one of the four classes doubles). The list is complete for $n\le11$, where its counts match OEIS A000769 [9]. A solver-free scan of all listed $2N$ sets with even $N\le36$ finds exact-two block structure only at $N=4$ and $N=8$. So the only density-two chain among them is $2\to4\to8$. **Why $2n$ sets fail to double: an exact count and a heuristic.** If $P$ has exactly two points in every row and column of $[n]^2$, a density-two doubling must also have exactly two children in every child row and column. Each two-element choice in a $2\times2$ block has a row profile in $\{(2,0),(0,2),(1,1)\}$ and a column profile likewise. Horizontal pairs have row profile $(2,0)$ or $(0,2)$ and column profile $(1,1)$; vertical pairs the reverse; the two diagonal pairs have $(1,1)$ for both. The two points of a parent row must carry complementary row profiles, and likewise for columns. The row–column incidence graph of $P$ is a union of cycles. Walking a cycle through $k$ rows alternates row and column steps, so the number of axis-feasible doublings is $$ \prod_{\text{cycles}}\operatorname{tr}T^{k}=\prod_{\text{cycles}}2^{k}\bigl((2+\sqrt3)^k+(2-\sqrt3)^k\bigr), $$ where $T$ is the $5\times5$ transfer matrix of these profile states, with nonzero eigenvalues $4\pm2\sqrt3$. The formula agrees with exact enumeration for $k\le8$, and for 25 listed $2n$ sets with $4\le n\le7$, including sets with several cycles. So axis saturation alone reduces the doubling entropy from $\ln6\approx1.792$ to $\tfrac12\ln(4+2\sqrt3)\approx1.005$ per parent point. The same transfer matrix handles parents with at most two points per row and column. Their incidence graph is a union of cycles and paths, and a path contributes $\mathbf 1^{\mathsf T}$ times an alternating product of the row and column steps; this agrees with enumeration on 18 random such parents (`pathform.py`). Long paths have the same per-point entropy as cycles, while isolated points keep $\ln6$. Adding both diagonal families leaves about $0.15$–$0.41$ per point (exact counts, $n=5$–$12$). This is still exponentially many configurations: for 40 of 40 listed $2n$ sets with $14\le n\le52$, a density-two doubling satisfying every height-one constraint exists. Any proof that $2n$ sets cannot double must therefore use lines of height at least two, and for some sets, more. A first-moment estimate makes this quantitative, but only heuristically, since it treats triples as independent: - the near-collinear parent triples that can create a collinear child triple number about $c\,n\ln n$, with $c\approx4.5$–$8$ measured; - each forbids about a quarter of its joint choices; - the estimate $2n\cdot1.005-0.285\,c\,n\ln n$ is non-positive from $n\approx5$, matching the observed absence of doublings for $5\le n\le17$. The estimate is not a proof. It predicts correctly that random no-three permutations (density one) keep doubling to $n=20$, but it under-predicts failure for random parents of density $1.5$, which all fail already at $n=10$. **Binding heights.** For 18 first-infeasible instances, we found the smallest $H^*$ such that the lines of primitive height at most $H^*$ alone already make the instance infeasible. - Over 18 instances, $H^*$ takes the values 2 (eight instances), 3 (five), 4 (four) and 7 (one), while the side runs from 8 to 96. Every value was decided exactly: each cutoff below $H^*$ is feasible and $H^*$ itself is infeasible. - Lines of height one (axes and diagonals) alone never suffice. - The depth-6 witness itself cannot be extended one more level, and lines of height at most 2 alone already prevent it (heights at most 1 do not). - For comparison, the unrestricted problem attains $2n$ exactly under heights at most 3, at $n=16,32,48$. **A wider window.** With children in $2s+\{-1,0,1,2\}^2$, the rotation-invariant class from the full $2\times2$ seed reaches 32 points at side 16 but not 64 at side 32. That failure has $H^*=6$, against $2$ for the small window one level earlier. From one point, depth 7 (side 128) exists, so $D_1(1)\ge7>D_1$. Finite-size trend: with a free top level of side $m$ and all lines imposed, the maximal density surviving two doublings is exactly $1.5$ at $m=4$ and $m=6$. At $m=8$ it lies in $[1.5,1.875]$ (1.5 is SAT; CP-SAT's bound is 15/8; 1.625 is UNKNOWN at 2,400 s). The larger cases are solver-limited. **Digit rules.** Four fixed self-similar density-one patterns (the four-corner Cantor pattern and a rotated variant at base 4, an anti-diagonal at base 3, a slope-2 pattern at base 5) contain three collinear points by level 2. For depth-varying top-digit copy rules $S_j=S_{j-1}+b^{j-1}A_j$, with $A_j$ ranging over all no-three $k$-subsets of $[b]^2$, an exhaustive search gives: - for the density-one case $k=b$ (76 patterns at $b=3$, 1,278 at $b=4$), sets clean through level 2 exist, but none is clean through level 3; - for $k>b$, every rule fails at level 2; - for $k