# Pairwise quarter-turn compatibility has an unbounded gap for no-three-in-line ## 1. Contribution and relation to the target For a set of grid points invariant under a quarter-turn, one can treat each four-point rotation orbit as a single choice. Two orbits are incompatible if their union contains a collinear triple. This produces a conflict graph, to which exact row and column capacities can be added. The remaining no-three constraints involve three distinct orbits. We prove that the conflict graph and axis capacities alone have optimum $2n$ on every even $n\times n$ board. More strongly, an explicit candidate set $S_n$ attains that relaxed optimum, while the largest actual no-three subset of $S_n$ has exactly 12 points for every even $n\ge6$. Thus the ratio of these two optima on this candidate set is $n/6$. The example also has a connected cycle in the reflected-coordinate graph: its failure cannot be fixed merely by insisting on a single cycle instead of a disconnected degree-two configuration. This is a limitation of an otherwise natural graph reduction, not an upper bound on the original no-three-in-line maximum. We give no unbounded family satisfying the actual no-three condition with density above $1.8$ or above the classical $1.5$. ## 2. Context and distinction from prior work The Hall--Jackson--Sudbery--Wild construction gives asymptotic density $3/2$ [1]; a recent account and generalizations appear in [2]. The reviewed rStation arc-lift barrier [3] concerns modular arcs in a two-period window. Our construction is not a modular arc construction and gives no improvement to the asymptotic lower bound. It concerns the exact logical strength of a symmetry-based relaxation. Quarter-turn symmetry has long been central to finite searches. Flammenkamp documents the symmetry classes, extensive finite configurations, and a heuristic explanation of why rotationally symmetric choices have much less severe immediate blocking than reflection-symmetric choices [4,5]. We do not claim that orbit-based search, the finite success of quarter-turn symmetry, or the qualitative distinction between these symmetries is new. Our result is the explicit all-even-size saturation theorem and its exact constant no-three optimum on the same support. Unlike the density-profile heuristic of [6], this statement is a finite deterministic theorem about a relaxation, not a probabilistic estimate for actual no-three sets. Prellberg's four-corner permutation note [7] studies the stronger necessary condition that genuine maximum configurations cannot have permutation matrices in all four quadrants for large sizes, conjecturally. With the $A_N$ wrap our construction does have such permutation quadrants, but is not a genuine no-three set. It neither proves nor contradicts that conjecture. It shows that the pair relaxation alone cannot establish it, even on a connected cycle. The exceptional $B_N$ choice is used only to retain pair compatibility at the specified quadratic sequence of sizes. ## 3. Coordinates, orbits, and the relaxation Write $n=2m$, $N=n-1$, and use doubled centered coordinates $$ D_N=\{-N,-N+2,\ldots,N\}^2. $$ The affine map $(X,Y)\mapsto((X+N)/2,(Y+N)/2)$ takes this set onto $\{0,\ldots,n-1\}^2$ and preserves collinearity. Let $$ R(X,Y)=(-Y,X),\qquad O(X,Y)=\{(X,Y),R(X,Y),R^2(X,Y),R^3(X,Y)\}. $$ Every orbit has four distinct points and is a square, so it contains no collinear triple. Say that two distinct orbits are *compatible* when their union contains no collinear triple. The pair relaxation selects compatible orbits with at most two selected points in each row and column. Its objective is the number of selected points. Actual no-three constraints imply all these requirements, but may involve three distinct orbits. A useful exact representation pairs each coordinate with its negative. The $m$ positive odd coordinate magnitudes are vertices. An orbit with distinct magnitudes $u,v$ is an edge joining $u$ and $v$; there are two possible orbits for such an unordered pair, represented by $(u,v)$ and $(v,u)$. An orbit with equal magnitudes is a loop. An edge contributes one point to each row of either endpoint magnitude; a loop contributes two to each row of its magnitude. Consequently exact degree two in this multigraph is equivalent to exact row degree two, and rotation supplies column degree two as well. This is an elementary encoding, not a separate novelty claim. ## 4. An explicit family saturating the pair relaxation **Theorem 1.** For every even $n\ge6$ there is an explicit set $S_n\subseteq D_N$ with $2n$ points such that: 1. $S_n$ is quarter-turn invariant and has exactly two points in every row and column; 2. every pair of its rotation orbits is compatible; 3. its reflected-coordinate graph is a single cycle. Define the path orbits $$ P_u=O(u,u+2),\qquad u=1,3,\ldots,N-2. $$ There are $m-1$ of them. Close the path by one of the two wrap orbits $$ A_N=O(N,1),\qquad B_N=O(1,N). $$ Use $B_N$ if $N=t^2+t+1$ for an even positive integer $t$; otherwise use $A_N$. Let $S_n$ be the union of the path orbits and the chosen wrap. *Proof of size and axis capacities.* The underlying cycle is $1,3,\ldots,N,1$. For $m\ge3$ all its edges are distinct, and exactly two edges meet each vertex. Thus the $m$ orbits are disjoint, yield $4m=2n$ points, and meet each row and column exactly twice. It remains to check compatibility. $\square$ ### 4.1 Reducing a pair check to two lines Two vertices of a square orbit are adjacent or opposite. Up to applying a quarter-turn, the line through them is therefore either the diameter through $(x,y)$ and the origin, or the side line $$ (x-y)X+(x+y)Y=x^2+y^2. \tag{1} $$ Thus a collinear triple using two points of one orbit and one of another exists precisely when some point of the second orbit lies on a diameter or side line of the first. Both directions of this test must be checked. ### 4.2 Compatibility of path orbits The positive slope of a diameter of $P_u$ is $(u+2)/u$. These slopes are distinct for distinct positive $u$, and the other diameter is perpendicular. Hence distinct path orbits have no shared diameter. After dividing (1) by two, a representative side of $P_u$ is $$ -X+(u+1)Y=(u+1)^2+1. \tag{2} $$ At the four points of $P_v$, the left side takes the values $$ uv+2u+2,\quad (u+2)v+2,\quad -uv-2u-2,\quad -(u+2)v-2. $$ The first two equal the right side only when $v=u$; the last two are negative. This calculation works in both directions, proving pairwise compatibility of all path orbits. ### 4.3 Compatibility with the wrap Neither wrap shares a diameter with a path orbit when $N\ge5$: a positive wrap slope is $N$ or $1/N$, whereas $(u+2)/u$ lies in $(1,3]$. Substitute the four points of $A_N$ into (2). The four left sides are $$ u+1-N,\quad 1+(u+1)N,\quad N-u-1,\quad -1-(u+1)N. $$ The first and fourth are negative; the second would force $N=u+1$, contrary to parity. The third equals the right side precisely when $$ N=u^2+3u+3. \tag{3} $$ For $B_N$, the corresponding values are $$ -1+(u+1)N,\quad N+u+1,\quad 1-(u+1)N,\quad -N-u-1. $$ The first would require $(u+1)(N-u-1)=2$, which for positive odd $u$ gives only $u=1,N=3$, outside our range. The last two are negative. The second equals the right side precisely when $$ N=u^2+u+1. \tag{4} $$ We must also exclude a point of $P_u$ on a side of the wrap. For $A_N$, the side equation is $$ (N-1)X+(N+1)Y=N^2+1. $$ Substitution of the path orbit gives $\pm(2N(u+1)+2)$ and $\pm2(u-N+1)$. The smaller absolute values are at most $2N-4