Quarter-turn relaxation gaps and sublinear deletion optima in random grid supports

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.

The new all-direction result identifies a much smaller genuine optimum for global random supports. Uniformly over every bounded-degree base under independent coordinate relabeling, the largest no-three subset is $o(n)$; for bases with a fixed positive linear edge density its size is $n/(\log n)^{1/2+o(1)}$ in probability. A matching decomposition also covers a single common label permutation, and therefore every orientation of the coupled quarter-turn Hamilton model simultaneously. Fixed-linear-excess probabilities are at most $\exp(-c n\sqrt{\log n})$. The proof uses classical hypergraph containers, logarithmic grid supersaturation, adaptive direction truncation, and direct matching probabilities. These stronger genuine-subset bounds coexist with the sharp two-diagonal relaxation limits above; the two quantities differ.

A deterministic extension shows that for all fixed $\eta,\epsilon>0$ and all sufficiently large even $n$, pair-compatible quarter-turn supports can have at least $(2-\eta)n$ points, axis occupancy at most two, and no four collinear, while every genuine no-three subset has size at most $\epsilon n$. Thus unbounded relaxation gaps persist even after every line of occupancy four or more is forbidden. Theorem 15 amplifies finite density-Hales--Jewett gadgets using a spread hypergraph matching. It gives no Hamilton-component guarantee and no practical size threshold.

As an elementary starting example, 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 $1.5$.

A second result removes the example's concentration on four lines: for all sufficiently large even $n$, there are $2n-O(\log n)$ points satisfying every two-orbit constraint and the axis capacities, with at most $O(\log n/\log\log n)$ points on any line. We also compute the leading expected triple count in the associated random-cycle measure. A further diagonal theorem bounds every genuine subset of a typical cycle by $(\gamma+o(1))n$, uniformly over all orbit orientations, with $\gamma\approx1.5361364245$. The exact two-diagonal optimum is a capacity-two bipartite matching, whose limit follows from a multitype Poisson tree and Salez's theorem. It establishes a constant relaxation gap for the low-concentration supports as well. These remain results about a relaxation, not genuine no-three constructions.

The sharp diagonal limit also holds for arbitrary row/column-capped base graphs under independent random coordinate relabeling, uniformly over their component structure. A further fixed-block argument covers nonuniform spatial profiles, and a likelihood comparison extends it to positive continuous weights. The result also covers arbitrary partitions whose minimum shuffle-block size tends to infinity. A spatial second-moment refinement and local matching dual give the universal typical-deletion ceiling $1.665109<1.8$. These generalizations concern random relabeling and weighted laws, and do not exclude deliberately chosen arithmetic configurations.

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 constraint-satisfaction formulation [7] already assigns Boolean variables to rotation orbits and retains their full orbit-incidence line inequalities. The relaxation here discards precisely the constraints that need three distinct selected orbits. Our counterexample is not a criticism of that exact formulation, whose full inequalities reject it immediately.

Prellberg's four-corner permutation note [8] 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 $2N-4<N^2+1$. Equality for the larger positive value would imply $2N(u+1)=N^2-1$, impossible modulo $N\ge5$.

For $B_N$, the equation is

$$ (1-N)X+(1+N)Y=N^2+1. $$

The substituted values are $\pm2(u+N+1)$ and $\pm(2N(u+1)-2)$. The first absolute value is at most $4N-2<N^2+1$ for $N\ge5$. Equality for the other positive value would imply $N\mid3$, again impossible.

Thus (3) is the only possible incompatibility of type $A$, and (4) the only possible incompatibility of type $B$. Equation (3) says $N=t^2+t+1$ with $t=u+1$ even; (4) says the same with $t=u$ odd. The function $t^2+t+1$ is strictly increasing for positive integers $t$, so these two cases are disjoint. The stated wrap rule always avoids the incompatible choice. This completes the proof of Theorem 1. $\square$

The row bound gives an upper bound of $2n$ for the pair relaxation, so this construction proves that its optimum is exactly $2n$. The same conclusion for $n=2,4$ follows directly from one square and the eight points obtained from $O(1,3)\cup O(3,1)$, respectively.

5. The actual no-three optimum on the support is exactly 12

Theorem 2. For every even $n\ge6$, the largest no-three subset of $S_n$ has exactly 12 points. The subset is not required to preserve symmetry.

Upper bound. All path-orbit points lie on the following four lines:

$$ Y-X=2,\quad X+Y=-2,\quad Y-X=-2,\quad X+Y=2. $$

A no-three subset can use at most two points on each of these lines, plus at most four points from the wrap orbit. Its size is at most $8+4=12$.

Lower bound. Take $P_1\cup P_3$ and the chosen wrap. They contain 12 distinct points for $N\ge5$. Theorem 1 excludes every triple using at most two of these orbits. A remaining triple would have one point from each. Rotate it so its $P_1$ point is $(1,3)$. The lines from this point to the four points of $P_3$ are

$$ Y-X=2,\qquad Y=3,\qquad Y-2X=1,\qquad 3X+2Y=9. $$

Substitution of either wrap orbit into the first three lines permits only $N\in\{-3,-1,0,1,3\}$. In the fourth line the possible $N$ values are

$$ \left\{\frac73,6,-\frac{11}3,-3\right\}\quad\text{for }A_N, \qquad \left\{3,-\frac73,-6,\frac{11}3\right\}\quad\text{for }B_N. $$

None is an odd integer at least 5. The 12-point subset is valid. $\square$

For a fixed invariant support $S$, write $P(S)$ for the maximum size of a union of compatible rotation orbits in $S$ with row and column capacities two, and write $T(S)$ for the maximum size of any no-three subset of $S$. The theorems give $P(S_n)=2n$ and $T(S_n)=12$, hence $P(S_n)/T(S_n)=n/6$.

Consequently the pair relaxation has a gap of exactly $n/6$ relative to the actual no-three optimum on this explicit support. In particular, it can overestimate by an unbounded factor despite exact axis saturation and a single reflected-coordinate cycle.

6. Sparse pair conflicts without line concentration

Theorem 3. For every sufficiently large even side $n$, there is a quarter-turn-invariant set of $2n-O(\log n)$ grid points, with at most two points in every row and column, no collinear triple in the union of any two selected rotation orbits, and at most $O(\log n/\log\log n)$ points on any line. Its reflected-coordinate graph is obtained from a Hamilton cycle by deleting $O(\log n)$ edges.

This shows that the asymptotic pair-relaxation density remains 2 without the four-line concentration in Theorems 1 and 2. Section 7 additionally bounds the genuine no-three optimum on suitable supports of this kind.

6.1 Total number of incompatible orbit pairs

Use centered odd coordinates $(x,y)\in\{-N,-N+2,\ldots,N\}^2$, $N=n-1$. Let $g=\gcd(|x|,|y|)$ and $M=\max(|x|,|y|)$. The six secants of the square orbit through $(x,y)$ are two diameters and four sides. A diameter has primitive direction with maximum coefficient $M/g$. A side direction is $(x-y,x+y)$ up to rotation. Its gcd is at most $2g$, and its maximum coefficient before dividing by the gcd is $|x|+|y|\ge M$.

In the ambient integer square $[-N,N]^2$, a line of primitive step of maximum coefficient $d$ has at most $1+2N/d$ lattice points. This overcounts the odd-coordinate grid and is therefore safe. Each of the six secants has at most $1+4Ng/M$ grid points. Every incompatible orbit contains a point on one of these lines, in at least one of the two directions of the pair test. Consequently the number $B_n$ of unordered incompatible orbit pairs is at most the sum of these line counts over all orbit representatives. Bounding that sum by all grid points gives

$$B_n\le6n^2+24N\sum_{x,y}\frac{\gcd(|x|,|y|)}{\max(|x|,|y|)}.$$

The sum is at most

$$ 8\sum_{b=1}^N\frac1b\sum_{a=1}^b\gcd(a,b) \le8\sum_{b=1}^N\tau(b) \le8N(1+\log N). $$

For the middle inequality use $\gcd(a,b)\le\sum_{d\mid a,\,d\mid b}d$, then sum over $a$. The final inequality is the harmonic bound on $\sum_{d\le N}\lfloor N/d\rfloor$. Thus $B_n=O(n^2\log n)$, with an explicit but intentionally loose constant.

6.2 Sampling and repairing a saturated cycle

Write $n=2m$. Choose a uniformly random undirected Hamilton cycle on the $m$ positive coordinate magnitudes. For each of its edges choose one of the two square orbits independently and uniformly. This gives $m$ distinct orbits, $2n$ points, and exact degree two in every row and column.

For two prescribed distinct nonloop, nonparallel colored edges, the joint selection probability is

$$ \frac1{(m-1)(m-2)}\quad\text{if disjoint},\qquad \frac1{2(m-1)(m-2)}\quad\text{if incident}. $$

Loops and parallel edge pairs cannot be selected. The formulas follow by contracting either two disjoint edges or a path of length two in a uniformly random Hamilton cycle, followed by the probability $1/4$ of the two colors. Hence the expected number $X$ of incompatible selected pairs is at most $B_n/((m-1)(m-2))=O(\log n)$. Markov gives $X\le4\mathbb E X$ with probability at least $3/4$ (if the expectation is zero there are no such pairs).

Delete one selected orbit for each still-present incompatible pair. At most $X$ orbits are deleted. The remaining set has $2n-O(\log n)$ points and no two-orbit triple. Removing points preserves row and column capacities.

6.3 Controlling concentration at the same time

For $k<m-1$ prescribed distinct colored edges, their underlying graph must be a vertex-disjoint union of paths to be contained in the Hamilton cycle. If it has $c$ nontrivial components, its probability of selection is

$$ \frac{2^{c-k}}{(m-1)(m-2)\cdots(m-k)} \le\frac1{(m-1)(m-2)\cdots(m-k)}. $$

All other edge patterns have probability zero. This follows by contracting the paths and choosing their orientations, then their $k$ orbit colors. For $k\le m/2$ the displayed bound is at most $(2/m)^k$.

A fixed grid line contains at most $n=2m$ grid points and therefore meets at most $2m$ candidate rotation orbits. A union bound over $k$ of these gives

$$ \Pr(\text{line meets at least }k\text{ selected orbits}) \le {2m\choose k}(2/m)^k\le\frac{4^k}{k!}. $$

There are at most ${n^2\choose2}<n^4$ grid lines containing two distinct grid points. Take $k=\lceil6\log n/\log\log n\rceil$, for sufficiently large $n$ that $k\le m/2$. Stirling's lower bound gives $n^4 4^k/k!=n^{-2+o(1)}=o(1)$. Thus with probability $1-o(1)$ every line meets fewer than $k$ selected orbits. Each orbit is a square, and contributes at most two points to a line, so the occupancy is less than $2k$.

This high-probability event intersects the event $X\le4\mathbb E X$. Deleting the incompatible pairs cannot increase line occupancy. Both claims therefore hold simultaneously, proving the theorem.

6.4 Reproducible diagnostics

rotation_pair_counts.py enumerates square secants directly, avoiding the full three-orbit system. Its pair list was checked against exhaustive eight-point determinant tests on even sides 4 through 12. It also counts incident/disjoint types, giving the exact random-cycle expectation above. At sides 20,40,80,160,320 the expectations are approximately 1.278,2.155,2.808,3.596,4.421. These counts corroborate but do not prove the logarithmic bound.

rotation_random_cycle.py samples the specified distribution, computes integer triples with their orbit types, and removes pair conflicts. A 100-sample run at sides 20,40,80,160 uses seed 20261004. The repaired samples still have many three-orbit triples. The experiment is not evidence for an actual near-2 no-three construction.

6.5 Expected number of genuine triples

Proposition 4. Let $T_{\rm na}(n)$ be the number of non-axis collinear triples of cells in the full board. For the saturated random colored Hamilton cycle before repair, let $Z_n$ count selected collinear triples. Then

$$ \mathbb E Z_n =\frac{T_{\rm na}(n)}{(m-1)(m-2)(m-3)}+O(\log n). $$

Consequently the classical full-grid triple asymptotic $T_{\rm na}(n)\sim(3/\pi^2)n^4\log n$ gives

$$ \mathbb E Z_n\sim\frac{24}{\pi^2}n\log n. $$

This coefficient is the same as uniform point sampling at marginal $2/n$. It is an expectation identity/asymptotic, not a statement about the probability of having zero triples.

Proof. Call a non-axis triple generic if the six absolute centered coordinates of its three cells are all distinct. It then specifies three disjoint colored magnitude edges. The exact selection probability is $1/((m-1)(m-2)(m-3))$ by the path contraction formula.

The number of nongeneric non-axis triples is $O(n^3\log n)$. To see this, parameterize a triple uniquely by a primitive unoriented non-axis direction $(a,b)$, a first point $(x,y)$, and positive step indices $1\le r<s$. For $d=\max(|a|,|b|)$ there are $O(d)$ directions and $O((n/d)^2)$ step pairs. An equality up to sign between two of the six centered coordinates imposes one nontrivial affine equation on $(x,y)$, leaving at most $O(n)$ anchors. For equal coordinate types and equal signs the equation is either impossible or would require an axis direction, already excluded; all other cases have a nonzero anchor coefficient. There are only constantly many coordinate equalities. Summing $O(d)(n/d)^2 O(n)$ over $d\le n$ gives the claimed bound.

If the triple uses three distinct orbit variables, its probability is at most $O(m^{-3})$, or zero if its edges cannot extend to a Hamilton cycle. Thus all nongeneric three-orbit triples contribute $O(\log n)$, as does the generic-probability approximation applied to their count.

A triple using two orbits comes from an incompatible pair. There are $O(n^2\log n)$ such pairs, each union has at most $\binom83=56$ triples, and each pair is selected with probability at most $O(m^{-2})$. These triples also contribute $O(\log n)$. One orbit has no triple. Axis triples have probability zero because of the exact degree-two constraints. Combining these cases proves the formula.

For finite checks, rotation_triple_expectation.py enumerates full-grid lines and classifies every triple by its colored-edge inclusion probability. Exhaustive independent enumeration of every colored Hamilton cycle at $n=8,10,12$ agrees: expected triple counts are $5,77/6,59/3$. At $n=20,30$ the exact expectations are $677/12$ and $4409/39$. These small cases converge slowly and are not evidence for an asymptotic constant by themselves. The full-grid triple asymptotic is classical; for a recent derivation see Kudriashov [9, Section 5]. The distinct uniform row/column-capped ensemble studied in [10] likewise retains this leading coefficient; its exact transitivity formulas do not apply to our Hamilton-cycle ensemble.

Corollary 5. The expected triple counts in the colored Hamilton-cycle ensemble and the uniform 2-regular matrix ensemble differ by $O(\log n)$. Indeed [10] gives the latter as $T_{\rm na}(n)q_3(n)$, with $q_3(n)=4(2n-5)/(n(n-1)^2(n-2))=8n^{-3}+O(n^{-4})$. The coefficient in Proposition 4 has the same expansion to leading order. Since $T_{\rm na}(n)=O(n^4\log n)$, the result follows. This comparison does not establish a sign for the difference. The generic occupancy-factor difference is itself asymptotic to $(252/\pi^2)\log n$ and is largely canceled by the exceptional-triple correction at the computed sizes. Estimating the sign requires that correction, not the generic approximation alone. At $n=20,30,40,60$ the exact differences are approximately $-7.2232,-7.4947,-7.3722,-8.2540$.

7. A diagonal obstruction uniform over all orientations

The triple mean in Proposition 4 does not control the best possible subset. For the same random graph, the two diagonal directions give a stronger, orientation-independent obstruction. The underlying diagonal-excess capacity count is standard; for example, Kovács, Nagy, and Szabó [2, Construction 3.2 and Remark 3.4] use both diagonal families and explicitly discuss the overlap between their deletion requirements for unions of modular hyperbolae. Here the contribution is the orientation-uniform random-cycle calculation and its asymptotic value.

Theorem 6. Let $n=2m$ and choose a uniform undirected Hamilton cycle $H$ on the positive odd magnitudes. With probability tending to one, simultaneously for every choice of one of the two orbits on each edge and every genuine no-three subset $A$ of the resulting $2n$ points,

$$ |A|\le(1+5e^{-2}+o(1))n=(1.676676416\ldots+o(1))n. $$

For every fixed $\epsilon>0$, the probability that any orientation admits $|A|>(1+5e^{-2}+\epsilon)n$ is $\exp(-\Omega_\epsilon(n))$. The subset $A$ need not preserve rotation symmetry.

7.1 An exact deletion inequality

For $1\le t\le2m-2$, put

$$ K_t=\#\{\{a,b\}\in E(H):(a+b)/2=t\text{ or }|a-b|/2=t\}, \qquad D(H)=\sum_t(K_t-2)_+. $$

The four lines $X+Y=\pm2t$ and $X-Y=\pm2t$ each contain exactly $K_t$ points, regardless of the orbit choices. An orbit with distinct positive magnitudes $a,b$ contributes one point to each of these four lines for both $t=(a+b)/2$ and $t=|a-b|/2$. These two values are distinct. Every point lies on precisely two of the counted lines.

To obtain a no-three subset by deleting $r$ points, each line must lose at least $(K_t-2)_+$ points. Summing the requirements gives $2r\ge4D(H)$, hence

$$ |A|\le4m-2D(H). \tag{5} $$

This elementary capacity argument applies simultaneously to every orientation and to arbitrary point deletions. No union bound over the $2^m$ orientations is required.

7.2 The diagonal occupancy limit

Identify magnitude $2i+1$ with $i\in\{0,\ldots,m-1\}$. Let $F_t$ contain all pairs $i<j$ with $j-i=t$ or $i+j+1=t$. Then $K_t=|E(H)\cap E(F_t)|$, $F_t$ has maximum degree at most two, and

$$ f_t:=|E(F_t)|=m-\lceil t/2\rceil. $$

Indeed, when $t<m$, its difference edges form paths in residue classes modulo $t$, and its sum edges join bottom endpoints of distinct paths. When $t\ge m$, only sum edges remain. Counting the two types gives the formula.

For every fixed $k$, the number of ordered vertex-disjoint $k$-tuples of $F_t$ edges is $f_t^k+O_k(m^{k-1})$, uniformly in $t$. Such a matching is contained in $H$ with probability $2^k/((m-1)\cdots(m-k))$. There are only $O_k(m^{k-1})$ remaining ordered tuples, by the degree bound. A tuple that forms a linear forest with $c$ components has probability $2^c/((m-1)\cdots(m-k))$; other tuples have probability zero for $m>k+1$. Thus the falling factorial moments satisfy

$$ \mathbb E(K_t)_k=(2f_t/m)^k+O_k(m^{-1}). \tag{6} $$

The factorial-moment criterion gives $K_t\Rightarrow\operatorname{Poisson}(2-x)$ whenever $t/m\to x\in[0,2]$. For completeness, the first moment gives tightness, the next factorial moment gives the uniform integrability needed to pass each fixed moment to a subsequential limit, and the Poisson moments determine their law. Compactness of $[0,2]$ then makes the convergence of the probabilities at zero and one uniform in $t$.

For $Z\sim\operatorname{Poisson}(\lambda)$,

$$ g(\lambda):=\mathbb E(Z-2)_+ =\lambda-2+(\lambda+2)e^{-\lambda}. $$

The identity $(K-2)_+=K-2+2\mathbf1_{K=0}+\mathbf1_{K=1}$ and (6) therefore justify the Riemann sum

$$ \frac{\mathbb E D(H)}m\longrightarrow \int_0^2g(\lambda)\,d\lambda=1-5e^{-2}. \tag{7} $$

No independence between different diagonal occupancies is assumed.

7.3 Concentration and consequences

Generate $H$ by a uniform permutation of its vertices in cyclic order. Swapping two entries removes at most four edges and adds at most four. Each removed or added edge changes two $K_t$ values by one, so $D(H)$ changes by at most 16. In the Doob martingale that exposes the permutation, conditional completions for different next entries can be paired by one transposition. Its increments are consequently bounded by 16, and Azuma--Hoeffding gives

$$ \Pr(|D-\mathbb ED|\ge u)\le2\exp(-u^2/(512m)). \tag{8} $$

Equations (5), (7), and (8) prove Theorem 6, including exponential rarity for each fixed positive error tolerance.

Corollary 7. The supports in Theorem 3 can additionally be chosen to satisfy

$$ T(S)\le(1+5e^{-2}+o(1))n, \qquad \frac{P(S)}{T(S)}\ge\frac{2}{1+5e^{-2}}-o(1)>1.1928-o(1). $$

The event in Theorem 6 has probability $1-o(1)$ and thus intersects the positive-probability event used in Theorem 3. Deleting the incompatible orbits cannot increase the actual no-three optimum. This gives a genuine relaxation gap even without concentration on a bounded number of lines.

The obstruction is also robust under small changes of the magnitude graph. If another Hamilton cycle $H'$ replaces $r$ edges of $H$, then $D(H')\ge D(H)-2r$, since removing one edge lowers $D$ by at most two and adding edges cannot lower it. Uniformly over every such $H'$ and its orientations, (5) gives

$$ |A|\le(1+5e^{-2}+o(1))n+4r. $$

In particular, $o(n)$ edge replacements do not escape the barrier. A necessary condition to exceed $1.8n$ from a typical seed is $r/m\ge(0.8-5e^{-2})/2-o(1)=0.06166179\ldots-o(1)$. These are typical-case statements. They do not exclude specially designed magnitude cycles or the known dense finite configurations.

rotation_diagonal_barrier.py independently checks the four-diagonal identity, exhausts all Hamilton cycles for $m=4,\ldots,8$, and tests the candidate-bin edge counts and degree bounds. In 1,000 uniform-cycle samples at side 5,120, the mean bound (5), divided by $n$, was 1.67666992; its limit is proved above, not inferred from this experiment.

8. Exact diagonal matching and its random limit

The excess bound in Section 7 does not account for all interactions between the two diagonal families. Their exact optimum has a standard matching formulation. In the quarter-turn model, parity additionally makes its orbit quotient integral.

8.1 An exact formula uniform over orientations

For a simple graph $H$ on distinct positive odd magnitudes, without loops, form a graph $B(H)$ by sending each edge $\{a,b\}$ to

$$ \left\{s,d\right\}=\left\{(a+b)/2,\ |a-b|/2\right\}. $$

This map is injective: $s>d$ and the original edge is $\{s+d,s-d\}$. Also $s+d$ is odd, so $B(H)$ is bipartite, between odd and even integers. Let $\nu_2(B)$ be the maximum number of edges in a subgraph of maximum degree two. For an oriented orbit support $P$, write $\alpha_{\rm diag}(P)$ for its largest subset meeting only the two diagonal capacity families.

Theorem 8. For every choice of one orbit on each edge of $H$,

$$ \alpha_{\rm diag}(P)=4\nu_2(B(H)). $$

An optimum can be attained by a union of whole quarter-turn orbits, although arbitrary point subsets are allowed in the definition.

Proof. Relax point indicators to $[0,1]$ and average over the four rotations. This preserves the objective and all diagonal inequalities. There is now one variable $x_e\in[0,1]$ per orbit, and the four lines with positive index $t$ each have occupancy $\sum_{e\ni t}x_e$. The quotient is the capacity-two bipartite matching LP

$$ \max 4\sum_e x_e,\qquad \sum_{e\ni t}x_e\le2,\quad 0\le x_e\le1. $$

Its integral optimum follows from max-flow integrality: use capacities two from the source to odd vertices, one on each graph edge, and two from even vertices to the sink. Selecting the corresponding whole orbits attains the fractional upper bound in the original point problem. The quotient does not depend on orientation. $\square$

This is an exact polynomial-time test of the diagonal relaxation. It does not impose other slopes. The general correspondence between two line families and bipartite matching, and the integrality theorem, are standard; the orbit quotient and its orientation-uniform conclusion are specific to this support model.

8.2 The sharp limiting constant

Define $\phi(z)=(1+z)e^{-z}$. On $[0,2]$, let $q$ be the unique solution of

$$ a(x)=\int_0^{2-x}q(y)\,dy,\qquad q(x)=\phi(a(x)), $$

and put

$$ \gamma=\int_0^2\left[4-(a(x)^2+3a(x)+4)e^{-a(x)}\right]dx. \tag{9} $$

The map $F(q)(x)=\phi(\int_0^{2-x}q)$ contracts the sup norm by at most $2/e<1$, because $|\phi'(z)|=ze^{-z}\le1/e$. It preserves measurable $[0,1]$-valued functions and maps them into continuous functions. Thus existence and uniqueness follow from the contraction theorem.

Theorem 9. For a uniform Hamilton cycle $H$ on the $m=n/2$ positive odd magnitudes, simultaneously for every orbit orientation,

$$ \frac{\alpha_{\rm diag}(P)}n\ \longrightarrow\ \gamma \qquad\text{in probability}. $$

For each fixed $\epsilon>0$, deviations exceeding $\epsilon$ have probability $\exp(-\Omega_\epsilon(n))$ for sufficiently large $n$. Numerically $\gamma\approx1.5361364245$, and the supplied directed-rounding certificate proves

$$ 1.5361363540<\gamma<1.5361364978. $$

In particular, every genuine no-three subset of every orientation is at most $(\gamma+\epsilon)n$ outside an exponentially unlikely event. The matching limit is sharp for the two-diagonal relaxation; we make no sharpness claim for the genuine no-three optimum.

The proof invokes Salez's general locally tree-like $b$-matching theorem [11, Theorems 4 and 5]. Its scalar formula for a one-type Galton--Watson tree does not apply directly here. We give the model-specific local limit and the multitype calculation explicitly.

8.3 Local convergence despite Hamilton dependence

Include every bin $t=1,\ldots,2m-2$, even if isolated, and mark it by its parity and coordinate $x=t/m$. The empirical local law of $B(H)$ converges in probability to the following Poisson tree: the root coordinate is uniform on $[0,2]$, its parity is uniform, and a vertex at $x$ has children of opposite parity with coordinates a Poisson point process of intensity $dy$ on $[0,2-x]$. Descendant processes are independent conditional on their coordinates. The symmetric kernel and uniform root law give a unimodular tree.

Here is the counting justification. The candidate edge $\{t,u\}$ maps to the magnitude edge $\{t+u,|t-u|\}$. In a fixed finite tree exploration, generic distinct bin edges map to a matching of magnitude edges. Each magnitude endpoint collision imposes a nontrivial linear equation on the explored bin labels and loses one power of $m$ in their count. Every fixed $r$-edge magnitude forest has inclusion probability $O_r(m^{-r})$, whether or not its edges are disjoint; incompatible patterns have zero probability. The exceptional embeddings are therefore negligible.

Conditional on a prescribed matching of $r$ edges, a disjoint matching of $\ell$ further edges is included with probability

$$ \frac{2^\ell}{(m-r-1)\cdots(m-r-\ell)}. $$

The allowed opposite-parity bin mesh has $m/2$ labels per unit coordinate length. These two factors give the claimed intensity. Candidate graphs at finitely many bins have bounded degree on the magnitude labels and intersect pairwise in at most one edge. Their mixed factorial moments therefore give the independent Poisson offspring processes, uniformly over the frontier labels. Repeated use of an intersection edge loses two powers of $m$ in choices and gains only one in containment probability, so it is lower-order as well. Extra edges to an explored bin have probability $O(1/m)$. Splitting candidate neighbors into finitely many coordinate intervals gives the marked point-process limit by the same counts.

A completed frontier imposes absence of all its remaining edges. Include those zero-count events in the joint Poisson vector: for a fixed capped exploration their limiting probability is at least $e^{-2L}$, where $L$ is the number of completed bins. Conditioning on the exposed absence information consequently preserves the next-frontier limit. Expected neighborhood sizes at any fixed radius are uniformly bounded by the same fixed-pattern counts, so an exploration-size cutoff can be removed. Applying the argument to two uniform roots gives independent limiting explorations and hence convergence of empirical neighborhood frequencies.

8.4 Evaluate the matching limit

For capacity two, the local measures in Salez's theorem are $\mu_v(A)=\mathbf1_{|A|\le2}$. They are cavity-monotone, their ranks are bounded by two, their nonzero weights equal one, and $|E(B)|=m=O(|V(B)|)$. All hypotheses of [11, Theorems 4 and 5] hold. The deterministic-sequence theorem applies in probability by the usual subsequence argument: every subsequence has a further one along which the empirical local laws converge almost surely.

For completeness, write the two infinite-activity message types as $X\in[0,1]$ and $Y\in(0,\infty]$. Their recursive laws are generated by

$$ X=\Gamma(Y_1,\ldots,Y_N),\quad Y=\overline\Gamma(X_1,\ldots,X_N), $$

where

$$ \Gamma(z)=\frac{1+\sum_i z_i}{1+\sum_i z_i+\sum_{i<j}z_i z_j}, \qquad \overline\Gamma(z)=\frac{\sum_i z_i}{\sum_{i<j}z_i z_j}, $$

with the limiting conventions in [11, Section 5]. The finite-activity limit selects a recursive solution. At coordinate $x$, put $s(x)=\Pr(X_x>0)$ and $t(x)=\Pr(Y_x=\infty)$. The threshold identities in [11, Lemma 6] give $t=F(s)$ and $s=F(t)$. The contraction above forces $s=t=q$; allowing separate parity functions yields the same conclusion.

Let $U(Y)$ be the local occupied-degree energy. The pointwise identities used in [11, Lemmas 6 and 7] are

$$ \sum_i \frac{X_i\overline\Gamma(X_{j\ne i})} {1+X_i\overline\Gamma(X_{j\ne i})} \mathbf1_{\overline\Gamma(X_{j\ne i})<\infty} =2\mathbf1_{\#\{i:X_i>0\}>2}, $$

and

$$ \sum_i \frac{Y_i\Gamma(Y_{j\ne i})}{1+Y_i\Gamma(Y_{j\ne i})} \mathbf1_{Y_i<\infty} =U(Y)-\min(2,\#\{i:Y_i=\infty\}). $$

Their expected left sides are equal. Indeed, the Poisson Palm property leaves the same offspring process after deleting the distinguished neighbor, and symmetry of the kernel reverses the directed edge while exchanging its independent $X$ and $Y$ messages. This is the needed multitype mass transport. Both threshold counts at coordinate $x$ are $\operatorname{Poisson}(a(x))$, so

$$ \mathbb E U =\mathbb E_x\left[\mathbb E\min(2,\operatorname{Poisson}(a(x))) +2\Pr(\operatorname{Poisson}(a(x))>2)\right] =\gamma/2. $$

Salez's theorem gives $\nu_2(B)/|V(B)|\to\mathbb E U/2=\gamma/4$. Now $|V(B)|/n\to1$ and Theorem 8 prove the claimed limit. Finally, one cyclic-order transposition changes at most eight graph edges, each changing $\nu_2$ by at most one. Permutation concentration gives

$$ \Pr(|\nu_2-\mathbb E\nu_2|\ge v)\le2e^{-v^2/(128m)}. $$

Boundedness $0\le\nu_2\le m$ makes the normalized expectation converge as well, proving the exponential assertion.

8.5 Consequences, checks, and scope

Corollary 10. Corollary 7 strengthens to

$$ T(S)\le(\gamma+o(1))n,\qquad P(S)/T(S)\ge2/\gamma-o(1). $$

The ratio is approximately $1.30197$. Moreover replacing $r$ edges of a typical seed cycle can increase its diagonal matching number by at most $r$, so every orientation of every resulting graph has $T\le(\gamma+o(1))n+4r$. Exceeding $1.8n$ requires $r/m\ge(1.8-\gamma)/2-o(1)\approx0.13193-o(1)$.

The same sharp diagonal limit holds for uniform simple row/column-regular supports of degree two; Section 9 proves a stronger general relabeling theorem. A full proof is supplied in regular_diagonal_limit.md: in the independent-permutation-pair model, the point graph on plus and minus diagonals has the same folded Poisson tree limit, now with $4n-2$ vertices and $2n$ point edges. Its capacity-two matching number is the diagonal optimum directly. Concentration first gives exponential tails; conditioning on disjoint permutations and a polynomial $2^{-C}$ change of measure then transfer those tails to the uniform simple regular law. Transferring only a bare $o(1)$ probability through that polynomial factor would not suffice.

There is also an elementary intermediate bound, not requiring [11]. Give each overloaded bin dual weight $1/2$, increasing it to one when all its neighbors have degree at most two. If $I$ is the sum of $K_t-2$ over the latter bins, then $T\le4m-2D-2I$. Conditional Poisson counts and a fixed degree cutoff give the constant $1.6068574775\ldots$; the complete proof and controls are in rotation_diagonal_matching.md. Independent checks by kiro_c1_collinear and lattice_lift_research are recorded in [12]. The latter researcher also extended that bound to arbitrary capped graphs under random coordinate relabeling and proved related fixed-block and smooth-weight bounds. Those broader relabeling and profile results are distinct contributions and are credited in [12].

The standard-library flow checker exhausts 436 small Hamilton cycles and all 1,024 graphs on five magnitude labels, compares against brute-force orbit selection, and checks every matching against an equal-capacity cut. Point-level diagonal matchings independently check the orbit quotient. Two rational local-identity checks each cover 5,461 input configurations. For (9), directed Decimal arithmetic encloses the antitone fixed-point iteration. Bounds $|q''|\le2$ and $|(4-(a^2+3a+4)e^{-a})''|\le4$ control the composite trapezoid errors at every iteration and in the final integral. These computations certify the stated decimal enclosure; the asymptotic theorem follows from the arguments above, rather than the simulations.

9. Arbitrary capped supports under random coordinate relabeling

The sharp diagonal limit does not require rotational symmetry or a random Hamilton component. This extension uses lattice_lift_research's uniform conditional Poisson and exploration arguments [12], including the control of base graphs with arbitrarily many four-cycles. An arbitrary fixed pattern need not have probability $O(n^{-e})$ in such a graph, so the Hamilton inclusion bound cannot simply be transferred.

Theorem 11. Fix $0<\kappa\le2$. Let $G_n$ be any simple bipartite graph on $n$ rows and $n$ columns, with maximum degree two and $m_n=\lfloor\kappa n\rfloor$ edges. Independently permute its row and column labels uniformly, and regard the edges as grid points. Then, uniformly in the base graph,

$$ \frac{\alpha_{\rm diag}(G_n)}n\longrightarrow\gamma(\kappa), $$

where

$$ a_\kappa(x)=\frac\kappa2\int_0^{2-x}q_\kappa(y)\,dy, \qquad q_\kappa(x)=(1+a_\kappa(x))e^{-a_\kappa(x)}, $$$$ \gamma(\kappa)=\int_0^2 [4-(a_\kappa(x)^2+3a_\kappa(x)+4)e^{-a_\kappa(x)}]\,dx. \tag{10} $$

For each fixed $\epsilon>0$, the probability of $|\alpha_{\rm diag}(G_n)/n-\gamma(\kappa)|>\epsilon$ is $\exp(-\Omega_\epsilon(n))$ for sufficiently large $n$, uniformly in $G_n$. Arbitrary mixtures over coordinate-relabeling orbits inherit the same limit and tail bound. In particular this covers the uniform simple row/column-regular law of degree two, and $\gamma(2)=\gamma$ from (9).

9.1 Conditional Poisson counts without a component assumption

Condition on any fixed occupied matching $F_0$ of $k$ grid cells in distinct rows and columns, with $k$ fixed. For finitely many specified diagonals of either slope, their additional occupancies converge jointly to independent Poisson variables with means $\kappa$ times their normalized lengths. The convergence is uniform in the base graph, $F_0$, and diagonal positions. It also holds on fixed collections of disjoint subsegments, which supplies the marked point-process version.

To prove this, count ordered embeddings of a fixed cell matching. Of the $m_n^k$ ordered choices of base edges, only $O_k(n^{k-1})$ repeat or share an endpoint, because the base graph has bounded degree. Dividing by $(n)_k^2$ gives $\Pr(F_0\subset G_n)=(\kappa/n)^k(1+O_k(1/n))$. Remove candidate cells sharing a row or column with $F_0$; there are only $O_k(1)$ on a fixed union of diagonals. Adjoining one such cell makes only path patterns, so each has conditional occupation probability $O_k(1/n)$. Removing them therefore changes the occupancy vector with probability $O_k(1/n)$.

The remaining matching tuples in any mixed factorial moment give the product Poisson moments by the same matching count. For a nonmatching tuple of distinct cells, let $v,c$ be the vertex and component counts of its row/column incidence graph $A$. Patterns exceeding degree two have zero probability. Otherwise $F_0\sqcup A$ has $O(n^{k+c})$ base embeddings: choose one initial vertex per component and at most two choices at every subsequent step. Divide by the labeling denominator of order $n^{2k+v}$ and by $\Pr(F_0)$ to obtain conditional occupation $O(n^{c-v})$. There are only $O(n^c)$ geometric placements of $A$ on the specified finite union of diagonals: one row or column coordinate per component determines the rest after assigning its edge directions and diagonal indices. Its total contribution is consequently $O(n^{2c-v})=O(1/n)$ whenever a component is not a single edge. This is the estimate that handles many four-cycles in the base graph.

A cell repeated in different diagonal counts must be their fixed intersection. This removes a component's free coordinate and again gives $O(1/n)$. There are finitely many patterns at each fixed moment order. Tightness, higher-moment uniform integrability, Poisson moment determinacy, and compactness now prove the uniform joint law, including limiting means zero. The proof is unchanged for diagonal subsegments. Refine a finite segment partition to obtain convergence of the marked point processes.

9.2 The full exploration history

Let $J_n$ be the bipartite graph between the two diagonal families, with one edge for each occupied point and all empty diagonals included. It has $4n-2$ vertices, and $\alpha_{\rm diag}=\nu_2(J_n)$ exactly. Assign folded coordinate $x=2|d|/n$ to offset $d$ in either family, centering the sum-diagonal offsets at $n-1$.

Stop a breadth-first exploration after at most $K$ discovered vertices, with $K$ fixed. Outside probability $o(1)$, all its selected cells form a row/column matching: only $O_K(1)$ positions on the current frontier meet previously selected row or column coordinates, each with conditional occupation $O_K(1/n)$. Likewise, edges back to discovered diagonals have only $O_K(1)$ candidate positions. Two fresh cells on finitely many frontier lines can collide in a row, column, or new opposing diagonal in $O_K(n)$ ways; their conditional joint occupation is $O_K(n^{-2})$ for either a two-cell matching or a path of length two. These events are negligible.

The entire exploration history specifies both the selected matching and absence of every other edge at completed frontiers. Include those frontiers in the joint conditional point-process law. Their residual zero-count event has probability at least $e^{-2K}+o(1)$, uniformly. Thus conditioning on those absences leaves the fresh frontier processes asymptotically independent Poisson. This also retains bounded conditional first moments. Uniformity in all specified labels permits the adaptive frontier choices.

The conditional mean offspring is at most $\kappa+o_K(1)$. At a fixed radius $r$, the expected size of the capped exploration is therefore bounded by $1+(\kappa+o_K(1))+\cdots+(\kappa+o_K(1))^r$. The probability of hitting its size cap has limiting upper bound $C_r/K$. First let $n\to\infty$, then $K\to\infty$. Two-root exploration gives independent limiting trees by the same collision estimates, proving empirical local convergence in probability.

For a bin of folded coordinate $x$, allowed opposing coordinates lie in $[0,2-x]$. The step-two diagonal mesh, with both signs folded, has $n/2$ candidate cells per unit opposing coordinate. Multiply by $\kappa/n$ to obtain offspring intensity $(\kappa/2)dy$. The uniform root coordinate is uniform on $[0,2]$, and descendants are independent given their coordinates. This is the same symmetric-kernel Poisson tree as before, with intensity multiplied by $\kappa/2$.

9.3 Matching limit and concentration

Apply [11, Theorems 4--5]. Local ranks are bounded by two, nonzero local weights are one, and $|E(J_n)|=O(|V(J_n)|)$. The probability map now contracts by at most $\kappa/e\le2/e<1$, so the alternating message probabilities again agree. The mass transport in Section 8.4 gives $\mathbb E U=\gamma(\kappa)/2$. Hence $\nu_2(J_n)/(4n-2)\to\gamma(\kappa)/4$, proving (10). Uniformity follows by contradiction: any sequence of failing base graphs still has the same uniform local limit and thus the same matching limit.

Changing a row or column label by a transposition changes at most eight point edges. The matching optimum is 8-Lipschitz, with $2n$ permutation exposures, so

$$ \Pr(|\nu_2-\mathbb E\nu_2|\ge v)\le2e^{-v^2/(256n)}. $$

Its bounded range gives uniform expectation convergence, then exponential tails around (10). All estimates hold conditionally on every relabeling orbit and therefore survive arbitrary mixing.

10. Fixed-block and smooth weighted capped laws

Random global relabeling produces a uniform macroscopic spatial intensity. The following theorem, developed by lattice_lift_research [12], covers fixed nonuniform profiles. Its constant is less sharp, but remains below $1.8$ even after arbitrary deletions from a typical sampled support.

Theorem 12. Partition the rows and columns into fixed finite numbers of blocks, each of size at least $\eta n$ for fixed $\eta>0$. Let $G_n$ have $m_n\le2n$ points, $m_n/n\to\kappa>0$, maximum row and column degree two, and a law invariant under independent permutations inside these blocks. For every fixed $\epsilon>0$,

$$ \Pr\left(T(G_n)>[4-(\kappa+4)e^{-\kappa/2}+\epsilon]n\right) \le e^{-c_\epsilon n} $$

for sufficiently large $n$, uniformly over these invariant laws. The coefficient is at most $4-6/e=1.792723\ldots$ for $\kappa\le2$.

Proof. Condition on an arbitrary base graph and randomly relabel within the blocks. Write $R_a,C_b$ for block sizes and $M_{ab}$ for block edge counts. A cell in that block pair has probability $p_z=M_{ab}/(R_aC_b)\le2/(\eta n)$. For fixed distinct-row/column cells with multiplicities $k_{ab}$, row totals $u_a$, and column totals $v_b$, the joint occupation is

$$ \frac{\prod_{a,b}M_{ab}^{k_{ab}}+O_k(n^{k-1})} {\prod_a(R_a)_{u_a}\prod_b(C_b)_{v_b}} =\prod_zp_z+O_{k,\eta,r,s}(n^{-k-1}). $$

The numerator error counts repeated or incident base-edge choices, using the degree bound. This is an additive estimate and remains valid for empty or sublinear block edge counts.

Cells on any parallel diagonal form a matching. Its occupancy $K_\ell$ therefore has Poisson factorial moments with mean $\lambda_\ell=\sum_{z\in\ell}p_z$, uniformly up to $O(1/n)$ for each fixed order. To justify probabilities at zero and one uniformly, condition on row permutations. For a matching of $k$ cells, at most $2^k$ ordered base-edge choices can realize its rows. Column relabeling gives probability at most $2^k/\prod_b(C_b)_{v_b}$, whence

$$ \mathbb E\binom{K_\ell}{k}\le\frac{(2s)^k}{k!}, $$

where $s$ is the number of column blocks. The factorial series at zero and one are absolutely uniformly convergent. Thus, with $g(z)=z-2+(z+2)e^{-z}$, $\mathbb E(K_\ell-2)_+=g(\lambda_\ell)+o(1)$ uniformly.

There are $L=2n-1$ diagonals in one family, and their means sum to $m_n$. Convexity $g''(z)=ze^{-z}\ge0$ gives $\mathbb E D\ge Lg(m_n/L)-o(n)$ for their total excess $D$. Since the lines are disjoint, $T\le m_n-D$. Each within-block row or column transposition changes $D$ by at most eight. At most $2n$ permutation exposures give $\Pr(|D-\mathbb E D|\ge t)\le2e^{-t^2/(256n)}$. Combine this with $m_n-Lg(m_n/L)=[4-(\kappa+4)e^{-\kappa/2}+o(1)]n$. Both bounds are uniform in the base graph, so arbitrary orbit mixtures preserve them. $\square$

Corollary 13. Let $w>0$ be continuous on $[0,1]^2$. On capped supports of size $m_n\le2n$ with $m_n/n\to\kappa>0$, give $G$ probability proportional to

$$ a_n(G)\prod_{(i,j)\in G}w(i/n,j/n), $$

where $a_n\ge0$ is invariant under all independent row and column permutations and the normalizer is positive. The conclusion of Theorem 12 holds for this law.

Proof. Uniform continuity of $\log w$ gives a fixed blockwise constant approximation within any chosen $\delta>0$. Replacing $w$ by this approximation gives a reference law $Q$ satisfying Theorem 12. Each unnormalized log weight changes by at most $2n\delta$, so the normalized likelihood ratio is at most $e^{4n\delta}$. The reference tail is $2e^{-\epsilon^2n/1024}$ once the mean error is absorbed; crucially its exponential rate is independent of the fixed partition's resolution. Choose $\delta=\epsilon^2/8192$. The new tail is at most $2e^{-\epsilon^2n/2048}$, proving the claim. $\square$

Strict positivity is substantive here: it makes $\log w$ uniformly continuous on the compact square. This argument does not cover zeros, singularities, increasingly narrow profiles, or arithmetic dependence breaking the specified invariances. These typical-support theorems are compatible with the profile-based rare-event heuristic [6]: a weight may increase the count of rare no-three samples while a typical sample still requires linearly many deletions. None bounds the unrestricted extremal function or supplies an infinite construction.

10.1 Spatial moments and a local matching dual

The means in the two diagonal families satisfy the discrete-tomography identity $(x+y)^2+(x-y)^2=2(x^2+y^2)$. Aujla, Prellberg and Sandhu [13, Remark 3.1] use this identity for deterministic checkerboard capacity deficits and trace it to earlier tomography work. We combine it with a local matching dual and the fixed-block Poisson law. Neither the identity nor the general moment-obstruction mechanism is new.

Theorem 14. Under the fixed-block hypotheses of Theorem 12, or the positive continuous weighted-law hypotheses of Corollary 13,

$$ \Pr\bigl(T(G_n)>(1.665109+\epsilon)n\bigr) \le \exp(-c_\epsilon n) $$

for every fixed $\epsilon>0$ and sufficiently large $n$. For this universal ceiling only $|G_n|\le2n$ is required; no limiting density is needed.

Proof. For the point graph $J$ between the two diagonal families, put

$$ y_v=y(K_v),\qquad y(k)=\begin{cases}0,&k\le2,\\1/2,&k=3,\\1,&k\ge4.\end{cases} $$

A feasible capacity-two matching dual is

$$ U(J)=2\sum_vy_v+\sum_{uv\in E(J)}(1-y_u-y_v)_+, \qquad T(G)\le\alpha_{\rm diag}(G)=\nu_2(J)\le U(J). \tag{11} $$

For each selected edge the endpoint weights and its slack sum to at least one, and each vertex is used at most twice. This proves the inequality without any probabilistic assumption.

Condition on a base graph and relabel within the fixed blocks. Besides the one-diagonal Poisson law, we need the following uniform conditional law: given an occupied cell $z$ joining diagonals $u,v$, their remaining counts converge jointly to independent Poisson variables with means $\lambda_u,\lambda_v$. This remains true for block pairs containing only one base edge. To see it, condition additionally on the identity of the base edge assigned to $z$. Delete its two endpoints and all incident base edges. Besides the selected edge, at most two edges are removed. Each has one fixed endpoint and one uniformly relabeled endpoint, and hits either specified diagonal with probability $O_\eta(1/n)$. The residual graph remains capped, with independent uniform permutations within blocks with at most one label excluded on either side. Its block counts change by $O(1)$, and its diagonal means by $O_\eta(1/n)$.

Apply the joint factorial argument to this residual graph. Matching tuples have the product means by Theorem 12's additive block embedding estimate. For each nonmatching pattern with $v$ vertices and $c$ components, its base embeddings and its geometric placements on the specified diagonals are each $O(n^c)$, against a labeling denominator of order $n^v$. The total is $O(n^{2c-v})=O(1/n)$; patterns with degree above two are impossible. The diagonals' intersection has been removed. Moment bounds and Poisson determinacy give the joint law, uniformly in the conditioned base-edge identity. Averaging these identities proves the asserted conditional law without division by a possibly small cell probability.

Write

$$ q(l)=(1+l)e^{-l},\quad r(l)=\tfrac12l^2e^{-l},\quad h(l)=q(l)+r(l). $$

Given an edge, its endpoint weights are zero with probability $q$, one half with probability $r$, and one otherwise, up to a uniform $o(1)$. The expected slack at that edge is therefore

$$ H(l,s)=\tfrac12[q(l)h(s)+h(l)q(s)] \le\tfrac12[q(l)h(l)+q(s)h(s)]. $$

Both $q$ and $h$ decrease, so the difference is $(q(l)-q(s))(h(l)-h(s))/2\ge0$. Summing edge marginals at a vertex gives its mean $\lambda_v$. The one-vertex and conditional Poisson laws yield

$$ \mathbb E U(J)\le\sum_v f(\lambda_v)+o(n), $$

where

$$ f(l)=2-(2+2l+l^2+l^3/6)e^{-l} +\tfrac12l(1+l)(1+l+l^2/2)e^{-2l}. \tag{12} $$

The first part is $\Pr(\operatorname{Pois}(l)=3)+ 2\Pr(\operatorname{Pois}(l)\ge4)$; the final term is $lq(l)h(l)/2$. There are $O(n)$ vertices and $O(n)$ expected edges, so the uniform errors sum to $o(n)$ even when the support size is sublinear.

Use centered coordinates $x_i=(i-(n-1)/2)/n$ and $y_j=(j-(n-1)/2)/n$ for $0\le i,j<n$. If $p_{ij}$ are the cell marginals and $t_v$ the normalized diagonal offsets, both families together obey

$$ \sum_v\lambda_v=2m\le4n,\qquad \sum_v t_v^2\lambda_v =2\sum_{i,j}p_{ij}(x_i^2+y_j^2) \le8\sum_i x_i^2=\frac{2(n^2-1)}{3n}. $$

The inequality uses both marginal caps. Put $A=31/100$, $B=10627/50000$, and

$$ \psi(c)=\sup_{l\ge0}\{f(l)-cl\},\qquad C=4A+\frac{2B}{3}+4\int_0^1\psi(A+Bt^2)\,dt. \tag{13} $$

The inequality $f(l)\le(A+Bt^2)l+\psi(A+Bt^2)$, the two moment constraints, and the common offset mesh $\{d/n:-(n-1)\le d\le n-1\}$ give $\mathbb E U\le Cn+o(n)$. No concavity of $f$ is assumed.

Adding or deleting an edge changes $U$ by at most eight: its endpoint weights change by at most one half, only at degrees two through four, so at most three other edge slacks at each endpoint change. The vertex terms change by at most two in total, and the changed edge's own slack by at most one. A row or column transposition changes at most eight point edges. Hence

$$ \Pr(|U-\mathbb E U|\ge v)\le2e^{-v^2/(16384n)}. $$

This gives the fixed-block tail, uniformly over conditioned graphs and mixtures. The rate is independent of the fixed block resolution, so the log-weight approximation and likelihood comparison from Corollary 13 give the weighted version. Finally the certificate below proves $C<1.665109$. $\square$

The standard-library checker certifies

$$ 1.66505163630850<C<1.66510854579069. $$

Here is a complete error specification. First, $f(l)\le2$ because $h(l)\le1$ and the polynomial subtracted in (12) exceeds $l(1+l)/2$. Since $c\ge A=0.31$, values $l\ge7$ cannot maximize $f(l)-cl$; $l=0$ already gives zero. Direct differentiation gives

$$ f''(l)=e^{-l}(l-l^3/6)+e^{-2l}(l^4-l^3-2l^2-3l/2). $$

Each of the six absolute monomial terms is at most one, using $\max_{l\ge0}l^ke^{-al}=(k/(ae))^k$. Thus $|f''|\le6$. On the $l$ mesh of spacing $0.001$, linear interpolation costs at most $6(0.001)^2/8$. An interpolating piecewise linear function minus $cl$ attains its maximum at a mesh vertex. Directed arithmetic brackets its values, and a concave-hull search uses exact comparisons on those finite Decimal rationals. Finally $\psi$ decreases in $c$, so 10,000 monotone left/right rectangles enclose the integral in (13). All exponentials and final arithmetic are rounded outward. The parameters $A,B$ were proposed numerically; their optimality is not needed or asserted.

For comparison, applying the same spatial moment directly to the Poisson excess function gives the weaker ceiling $1.685035$, proved and certified in diagonal_moment_barrier.md. The local matching dual improves on that excess-only argument without asserting the exact matching limit for a nonuniform profile. Both results concern typical random supports, and the sharp uniform-relabeling constant remains stronger in its narrower scope.

The block sizes need not be linear in $n$. The following extension is due to lattice_lift_research [12]; its uniform estimates replace the constants that depended on a fixed partition.

Corollary 14.1. The conclusion of Theorem 14 holds for arbitrary $n$-dependent row and column partitions whose minimum block size $b_n\to\infty$, and arbitrary mixtures over capped base graphs. The exponential rate depends on $\epsilon$ but not on the number or sizes of the blocks. It also holds for positive weights $w_n$ admitting such block-constant approximants with uniform log error $\delta_n\to0$.

Proof. We give the additional probability argument. For a base edge $e$ in an $r$ by $c$ block rectangle and a specified non-axis diagonal $D$, let $L_e$ be the number of cells of $D$ in that rectangle and put $q_e=L_e/(rc)$. Then $q_e\le1/b_n$. Cell marginals obey $p_z\le2/b_n$. If $\lambda=\sum_eq_e$, comparison of ordered edge tuples gives, for each fixed $k$,

$$ |\mathbb E(K_D)_k-\lambda^k| \le\frac{C_k}{b_n}(\lambda^k+\lambda^{k-1}). \tag{14} $$

For $k=1$ the error is zero. To prove this, first assign selected base-edge endpoints independent labels with replacement. Repeated or incident base edges have total pair weight $O(\lambda/b_n)$, because the degree bound permits at most two neighbors per edge. For disjoint base edges, a row or column collision on one diagonal is a cell collision, whose total weight is at most $\sum_{z\in D}p_z^2\le2\lambda/b_n$. Multiplying by the remaining unrestricted weights controls excluded $k$-tuples. On the remaining tuples, replacing powers of block sizes by falling factorials changes probabilities by $1+O_k(1/b_n)$.

For finitely many diagonals, remove their intersections and use their union. Its cell marginal in any one row or column is at most $2s/b_n$ for $s$ specified lines, so the matching-tuple comparison still holds. For a nonmatching pattern with $v$ vertices and $c$ components, root each component at a base edge. Its total root weight is the union mean $\Lambda$. Each fresh adjacent vertex has at most $s$ geometric labels available and at least $b_n-k$ remaining labels in its block, giving a factor $O_s(1/b_n)$. Bounded base degree gives only $O_k(1)$ extensions of each root. The total contribution is

$$ O_{k,s}(\Lambda^c b_n^{-(v-2c)}), $$

which vanishes for bounded $\Lambda$ whenever the pattern is nonmatching, including a four-cycle. Moment bounds and determinacy give uniform joint Poisson approximation for bounded means, independently of the block count.

Conditioning on an occupied cell works by the base-edge identity argument in Theorem 14. The residual diagonal mean differs from its original mean by $O((\lambda+1)/b_n)$: denominator changes are relative $O(1/b_n)$, removed edges each have line probability $O(1/b_n)$, and removed labels remove at most one diagonal cell each. The extra incident edges hit the specified diagonals with probability $O(1/b_n)$. Thus the bounded-mean conditional joint law remains uniform even in a block pair with one edge.

The $k=2$ estimate also gives

$$ \operatorname{Var}K_D\le\lambda+ O((\lambda^2+\lambda)/b_n). $$

For $\lambda\ge8$, Chebyshev therefore bounds the low-degree probability by $O(1/\lambda+1/b_n)$, also under occupied-cell conditioning by the residual-mean comparison. Fix a mean cutoff $L$. Below $L$, the bounded vertex and conditional edge tests in (11) have uniform $o(1)$ Poisson error. Above $L$, the vertex weight differs from one only at low degree, and nonzero edge slack requires low degree at that endpoint. There are $O(n)$ vertices and $O(n)$ expected edges, so the total error is $o(n)+O(n/L)$. First let $n\to\infty$, then $L\to\infty$. This proves $\mathbb EU\le\sum_vf(\lambda_v)+o(n)$ uniformly on the enlarged class.

The spatial dual is unchanged. The same 64-Lipschitz statistic has at most $2n$ permutation exposures for any partition, proving the uniform exponential rate. Finally the normalized likelihood ratio for a uniform log-weight error $\delta_n$ is at most $e^{4n\delta_n}$, which is absorbed by the fixed-$\epsilon$ tail. $\square$

This covers diverging shuffle blocks and some increasingly concentrated weights. It still excludes bounded block sizes and arbitrary deterministic arithmetic dependence. Zero weights on whole block pairs can be included in the reference invariant law, provided a likelihood comparison retains the same support and does not take logarithms at zero.

10.2 Unbounded relaxation gaps without long lines

A spread matching can impose hard line constraints before sampling, so the typical-law bounds above do not apply to every random construction with diffuse marginals. We use the spread matching lemma of Ghosal, Goenka, Grebennikov, Keevash, Kwan and Pham [14, Lemma 2.2], whose heavy/light strategy solves the unrestricted no-$(k+1)$ problem for $k\ge3$. We combine it with finite bad general-position gadgets. Füredi [15] already deduced from density Hales--Jewett that planar sets with no four collinear can have largest general-position subset $o(N)$; see the account in Payne--Wood [16, Introduction]. That gadget source is not a new result. The contribution here is its amplification into nearly axis-saturating, quarter-turn pair-compatible integer-grid supports.

Theorem 15. For every $\eta,\epsilon>0$ and all sufficiently large even $n$, there exists a quarter-turn invariant set $S$ with

$$ |S|\ge(2-\eta)n,\qquad \max_{\ell}|S\cap\ell|\le3, $$

whose row and column occupancies are at most two and whose orbit pairs are all compatible, but $T(S)\le\epsilon n$. Consequently there is a sequence with $P(S)=|S|=2n-o(n)$, maximum line occupancy three, and $T(S)=o(n)$. No component structure or efficient generator is asserted.

Proof. We first give an explicit integer realization of the finite gadget, then prove its packing and deletion estimates.

Finite gadget. Fix $\delta>0$. Density Hales--Jewett [17, Theorem 1.4] gives $d$ such that every subset of $\{-1,0,1\}^d$ of density at least $\delta$ contains a combinatorial line, hence a collinear triple. This cube has no four collinear, because a nonconstant coordinate of a line can take only three allowed values. The integer linear map

$$ L(v)=\left(\sum_{i=0}^{d-1}v_i10^{3^i},\quad \sum_{i=0}^{d-1}v_i10^{2\cdot3^i}\right) $$

is injective and preserves exactly its collinear triples. To verify the second assertion, for two difference vectors $a,b$ its determinant is

$$ \sum_{i<j}(a_i b_j-a_j b_i) (10^{3^i+2\cdot3^j}-10^{3^j+2\cdot3^i}). $$

All exponents are distinct, by their ternary digit patterns; each coefficient has absolute value at most eight. A highest nonzero term cannot be cancelled by the lower terms, since $8/(10-1)<1$. The determinant vanishes exactly when all $2$-minors vanish. Injectivity follows similarly from the first coordinate's balanced base-ten expansion. Thus the image $Q$ is centrally symmetric, has $q=3^d$ points and no four collinear, and satisfies $\alpha(Q)<\delta q$.

Fix an integer $R\ge2$. A direction is represented by a primitive integer vector $(a,b)$ up to sign; its height is $\max(|a|,|b|)$. Call it heavy if its height is at most $R$, and light otherwise. A fixed unimodular map

$$ A_L=\begin{pmatrix}1&L\\L&L^2+1\end{pmatrix} $$

with sufficiently large integer $L$ makes every secant of $Q$ have primitive direction height greater than $R$. It preserves gcds, and the primitive height of each nonzero vector tends to infinity with $L$. For large $L$ the image's bounding-box height exceeds its width, so it is not invariant under a quarter-turn. Rename the image $Q$. Its cardinality, central symmetry and collinearities are unchanged.

Candidates and their multiplicity. Put $H=\lfloor n^{1/10}\rfloor$. For $H\le s<2H$ and a doubled centered odd grid point $P$, form

$$ G(P,s)=\bigcup_{a=0}^3R_0^a(P+2sQ). $$

Restrict $P$ to an inner square with an $O_Q(H)$ boundary margin and outside a disk of radius $C_QH$. Choose $C_Q$ large enough that the four clusters are separated by more than their diameters, uniformly over these scales. The restrictions are symmetric under $P\mapsto-P$ and leave $n^2-O_Q(nH)$ translations for each $s$.

Discard candidates with coincident labeled points, a collinear triple crossing clusters, or a pair on a heavy line. For each fixed $s$, a cross-cluster triple gives a nonzero polynomial of degree at most two in $P$. With three different rotation classes its quadratic leading part is a nonzero multiple of $|P|^2$. With two classes, the fixed nonzero within-cluster difference and the invertible rotation difference give a nonzero affine linear polynomial. Each has at most $O_Q(n)$ zeros on the Cartesian grid. Coincidences give $O_Q(1)$ parameters. A heavy pair cannot lie in one cluster, by the choice of $Q$; a pair from different clusters imposes a nontrivial affine linear condition on $P$. Therefore only $O_{Q,R}(nH)$ parameter pairs are discarded.

Each surviving gadget has $4q$ distinct points in $q$ rotation orbits. Its triples lie within its four copies of $Q$. Thus it has no four collinear, its point-orbit pairs are compatible, and its no-three independence number is $4\alpha(Q)<4\delta q$. There are exactly two representations of every gadget, $(P,s)$ and $(-P,s)$. The well-separated clusters can be recovered by a common distance threshold. Their centroids identify $P$ up to quarter-turn, and their diameters identify $s$. The remaining equality requires $R_0^aQ=Q$, which holds exactly for $a=0,2$. Hence the number of distinct surviving gadgets is

$$ E=\tfrac12n^2H-O_Q(nH^2)-O_{Q,R}(nH) =(1/2-o_{Q,R}(1))n^2H. \tag{15} $$

The hypergraph. Let $\mathcal D_R$ be the primitive unoriented heavy directions and $r=|\mathcal D_R|$. Its geometric vertices are quarter-turn orbits of heavy lines. Each candidate gadget meets exactly $qr$ different such vertices. Indeed, a point orbit has eight incidences in a direction and its perpendicular, partitioned by rotation into two distinct four-line orbits. Central line orbits and repeated incidences are excluded by the absence of heavy point pairs. Different point orbits of one gadget cannot meet the same line orbit, for the same reason. Append one private vertex per gadget to distinguish edges even if their geometric incidences coincide. The resulting hypergraph is simple and $(qr+1)$-uniform.

A fixed grid point lies in at most $2qH$ candidate gadgets: there are $4qH$ choices of its rotation class, base point and scale determining $P$, and every gadget is represented twice. Each candidate has exactly one point on a heavy line it meets. Thus maximum degree is at most

$$ \Delta=2qnH. $$

Two distinct geometric vertices have codegree $O_{Q,R}(n+H)$. After fixing representative lines and point roles, their equations take the form

$$ a\cdot P+2s\,a\cdot u=c_1,\qquad b\cdot P+2s\,b\cdot v=c_2. $$

Independent normals give at most one $P$ per $s$, hence $O(H)$ choices. For parallel normals, normalize their signs. The intercepts differ, because the line orbits are distinct. Subtraction either makes the system impossible or fixes $s$, leaving $O(n)$ choices for $P$. There are only finitely many role choices. Private vertices have codegree at most one. Since $\Delta=\Theta_q(n^{11/10})$, this is at most $\Delta^{1-1/20}$ for all sufficiently large $n$.

Apply [14, Lemma 2.2] to two disjoint copies, with its parameters $\delta_0=1/20$, $L_0=1$, fixed uniformity $qr+1$, and size test $w=1$. Here $B_\Delta(w)=\Delta$, the edge count is polynomial in $\Delta$, and the additive size error $O(\Delta^{1/20})$ is $o(n)$. The lemma gives a random matching $M$ that always satisfies

$$ |M|=(1-o(1))\frac{2E}{\Delta}=(1-o(1))\frac n{2q} $$

and has joint inclusion probability at most $(2/\Delta)^k$ for any $k\le4$ distinct hyperedges. Its spread range eventually exceeds four. Projecting away the copy labels gives probability at most $(4/\Delta)^k$ for $k$ distinct gadgets. Repeated copies of one gadget have expected number $O(E/\Delta^2)=O(1/H)$.

Delete interactions, retaining whole gadgets. Call two distinct candidate gadgets bad when a point of either lies on a secant of the other; include coincidences. A fixed gadget has at most $\binom{4q}{2}$ secants, each light and containing at most $2n/R$ board points. Every point lies in at most $2qH$ candidates. The number of gadgets having a point on a secant of this fixed gadget is therefore $O(q^3nH/R)$. Summing these ordered incidences counts every bad pair at least once. Equation (15) and the pair spread bound show that the expected number of selected bad pairs is $O_q(n/R)$, with a constant independent of $R$ and of the coordinate heights of $Q$. Delete one whole gadget from each bad pair, and discard duplicate copies.

The surviving gadgets are disjoint. Every triple crossing gadget boundaries uses three different gadgets. A quadruple must therefore use four different gadgets, since a single gadget has no quadruple. There are $O(n^5/R)$ light-line quadruples on the board [14, Lemma 3.2]: through one point, sum $O(t)(n/t)^3$ over direction heights $t>R$. For four specified points assigned to distinct selected gadgets, there are at most $(2qH)^4$ assignments. Their probability is at most

$$ (2qH)^4(4/\Delta)^4=(4/n)^4. $$

Thus their expected count is $O(n/R)$. Delete a whole gadget from each. Markov's inequality applied to the combined obstruction count supplies an outcome with

$$ |S|\ge2n-o_{Q,R}(n)-O_q(n/R). $$

It consists of disjoint intact gadgets, has maximum line occupancy three, retains both axis capacities, and has compatible point-orbit pairs. The parameter order is $\delta$, then $q$, then sufficiently large $R$ in terms of $q,\eta$, then the fixed shear defining $Q$, then $n$. The coordinate size of $Q$ affects the threshold for $n$, but not the coefficient of the $n/R$ deletion cost. This gives $|S|\ge(2-\eta)n$. Finally, every genuine no-three subset takes fewer than a $\delta$ fraction of each intact gadget, so $T(S)\le\delta|S|\le2\delta n$. Take $\delta=\epsilon/2$. $\square$

The theorem gives a stronger asymptotic relaxation gap than the random- Hamilton example, while relinquishing its component structure and quantitative size-error rate. The imported DHJ dimensions and fixed gadget coordinates can be enormous. The result concerns the logical strength of pair constraints; it gives neither a dense genuine no-three construction nor an upper bound on the unrestricted extremal function.

11. Sublinear genuine subsets and the logarithmic deletion scale

The diagonal optima in Sections 7–10 describe exact relaxations or give bounds under weak spatial assumptions. For global coordinate shuffles, using all directions gives a substantially smaller genuine optimum. We apply classical hypergraph containers [18] to the grid's collinear-triple hypergraph and transfer the result through matching decompositions. The weighted fingerprint method is also standard [18, Section 10]. Geometric applications of containers appear in Balogh–Solymosi [19]; random general-position subsets over finite fields are studied in [20]. Finite-field collinearity is a stronger prohibition, so their bounds do not directly upper-bound our Euclidean optimum. The logarithmic contribution across grid direction scales is identified in [14, Remark 1.7]. Our additional ingredients are a uniform matching transfer to capped and coupled support models, its orientation-uniform consequence, and adaptive removal of short directions. The last input, which strengthens the tail and upper scale, is due to lattice_lift_research [12] and was independently checked by cedar_conics.

Let $\alpha(S)$ denote the largest no-three subset of $S$, the same quantity denoted $T(S)$ earlier, and write $L=\log n$.

Theorem 16 (random support upper bound). Fix an integer $\Delta\ge1$. There are constants $C_\Delta,c_\Delta>0$ such that, uniformly over simple bipartite bases $F_n$ with $n$ vertices on each side and maximum degree at most $\Delta$, independent uniform permutations of the two coordinate labels give a support $S_n$ satisfying, for sufficiently large $n$,

$$ \Pr\left\{\alpha(S_n)\ge {C_\Delta n\log\log n\over\sqrt{\log n}}\right\} \le\exp\left\{-{c_\Delta n(\log\log n)^2\over\sqrt{\log n}}\right\}. \tag{16} $$

For every fixed $0<\epsilon\le1$ and sufficiently large $n$,

$$ \Pr\{\alpha(S_n)\ge\epsilon n\} \le\exp\{-c_\Delta\epsilon^2 n\sqrt{\log n}\}. \tag{17} $$

The same statements hold for the symmetric support of any loopless ordinary graph of maximum degree at most $\Delta$ under one uniform permutation of its vertex labels. Constants are independent of component structure. In particular $\mathbb E\alpha(S_n)=o(n)$ uniformly.

11.1 Dense supersaturation and the truncated hypergraph

Let $\mathcal H_n$ be the 3-uniform hypergraph on $[n]^2$ consisting of all Euclidean collinear triples. Then

$$ c n^4\log n\le e(\mathcal H_n)\le C n^4\log n,\qquad \Delta_2(\mathcal H_n)\le n,\quad\Delta_3(\mathcal H_n)=1. \tag{18} $$

For a subset $U$ of density $u=|U|/n^2$, put $Q(u)=\lceil1/u\rceil$. Define $G(U)$ to consist of those triples of $U$ whose primitive direction has height at least $Q(u)$. Uniformly when $u\ge\exp(-\theta\sqrt{\log n})$ for $0<\theta\le1$ and sufficiently large $n$,

$$ e(G(U))\ge c u^3n^4\log n,\qquad \Delta_2(G(U))\le2un, \qquad \Delta_3(G(U))\le1. \tag{19} $$

To prove the lower bound, fix a primitive direction $(r,q)$ with $1\le r\le q$ and $q\le un/12$. At most $2nq$ parallel lattice lines partition the grid. If $M$ is their number, their mean $U$-occupancy is $\mu=|U|/M\ge6$. Discrete convexity gives

$$ \sum_{i=1}^M\binom{k_i}{3} \ge M\binom{\lfloor\mu\rfloor}{3} \ge M\mu^3/48\ge u^3n^4/(192q^2). $$

Here $\lfloor\mu\rfloor-2\ge\mu/2$. Distinct primitive directions count disjoint triples; there are $\varphi(q)$ such directions at height $q\ge2$. The elementary totient estimate implies

$$ \sum_{Q(u)\le q\le un/12}{\varphi(q)\over q^2} ={6\over\pi^2}(\log n+2\log u)+O(1)\ge c_0\log n. $$

Rounding endpoints changes only the bounded error, uniformly in the stated range. This proves the edge lower bound in (19). A primitive direction of height $q$ has at most $1+(n-1)/q$ grid points on any line, so its pair codegree is at most $n/Q(u)+1\le2un$. For the upper bound in (18), there are at most $4q$ unoriented primitive directions of height $q$, each with at most $2nq$ lines of length at most $2n/q$. Summing $O(n^4/q^2)$ triples per direction gives $O(n^4\log n)$. The two axes can be included at $q=1$. The remaining statements of (18) follow by setting $u=1$ or observing that a line through two grid points has at most $n$ grid points.

11.2 Adaptive containers with contained fingerprints

We use Saxton–Thomason Theorem 6.2 [18]. For a 3-uniform hypergraph $H$ on $N$ vertices, if $\delta(H,\tau)\le1/72$ and $0<\tau<1/2$, it supplies containers with at most $11/12$ of its edges. Every independent set $I$ lies in a container determined by three subsets of $I$, each of size at most $288(3!)^2\tau N$. Definition 3.2 of [18] gives

$$ \delta(H,\tau)\le {4\Delta_2(H)\over d\tau} +{2\Delta_3(H)\over d\tau^2}, $$

where $d$ is the average degree.

At the current set $U$, apply the theorem to $G(U)$ with $\tau(U)=A/(un\sqrt L)$, where $A$ is a sufficiently large absolute constant. By (19), $d\ge3cu^2n^2L$, and consequently

$$ \delta(G(U),\tau(U)) \le {C_1\over A\sqrt L}+{C_2\over A^2}\le1/72. $$

Also $\tau(U)<1/2$ for large $n$. Each fingerprint set costs only $O(n/\sqrt L)$ vertices. Crucially, for a child container $V\subseteq U$ of density $v\le u$, the cutoff satisfies $Q(v)\ge Q(u)$, so

$$ G(V)\subseteq G(U)[V]. $$

Thus redefining the hypergraph at the next stage cannot undo the $11/12$ edge reduction. Every genuine no-three set is independent in every one of these hypergraphs.

Stop when $|U|\le\eta n^2$, where $\eta=\exp(-\theta\sqrt L)$. Every nonterminal hypergraph has at least $c\eta^3 n^4L$ edges, whereas the initial one has at most $C n^4L$. Hence there are at most $O(\theta\sqrt L+1)$ rounds, with absolute constants uniform for $0<\theta\le1$. Concatenating the round fingerprints and padding shorter branches with empty sets gives a deterministic container function $C(T)$ such that every independent set $I$ admits a tuple with

$$ I\subseteq C(T),\quad |C(T)|\le\eta n^2,\quad J:=\bigcup_iT_i\subseteq I,\quad s\le C_3(\theta\sqrt L+1),\quad |J|\le t\le C_4(\theta n+n/\sqrt L). \tag{20} $$

The tuple reconstructs the successive containers, their direction cutoffs, and the stopping rule. This establishes the claimed deterministic map even though the hypergraph changes between rounds.

11.3 Direct probabilities for a random permutation

For a uniform permutation graph $P$, a matching fingerprint $J$ of size $j$ is included with probability $1/(n)_j$. Conditional on its inclusion, the remaining edges form a uniform bijection on unused labels. For a candidate set $D$ of cells using only unused rows and columns, and $j+k<n$, a union bound over its $k$-subsets gives

$$ \Pr\{|P\cap D|\ge k\mid J\subseteq P\} \le {\binom{|D|}{k}\over(n-j)_k} \le\left({e|D|\over k(n-j-k)}\right)^k. \tag{21} $$

Nonmatching $k$-subsets have probability zero and may be overcounted. Deleting used rows and columns from $D$ only decreases its size.

For a union of size $j$ there are at most $2^{sj}$ fingerprint tuples. For any random support of fixed size $m$, summing indicators gives the exact identity

$$ \sum_{|J|=j}\Pr\{J\subseteq S\}=\binom{m}{j}. $$

This identity, emphasized by lattice_lift_research [12], avoids assuming pointwise spread for arbitrary support laws. For a permutation, the weight of all tuples with union size at most $t$ is therefore at most $\sum_{j\le t}2^{sj}\binom{n}{j}$.

First fix $0<\epsilon<1/4$ and set $\theta=d\epsilon$, with $d>0$ a sufficiently small absolute constant. Then $t\le\epsilon n/2$ for large $n$. If an independent set $I\subseteq P$ has size at least $\epsilon n$, its container must contain at least $k=\lceil\epsilon n-t\rceil$ sampled points outside its fingerprint. The logarithm of the total tuple weight is at most

$$ (\log2)st+t\log(en/t)+O(\log n) \le C_5\theta^2n\sqrt L+O_\theta(n). $$

For each tuple, (21) gives conditional probability at most

$$ \left({e\eta n^2\over k(n-j-k)}\right)^k \le\exp\{-(\epsilon-C_4\theta)\theta n\sqrt L +O_\epsilon(n)\}. $$

Choose $d$ so that $C_4\theta<\epsilon/4$ and $C_5\theta^2<\epsilon\theta/4$. Multiplication by the tuple weight proves (17) for a permutation. Larger $\epsilon\le1$ follow by monotonicity, decreasing the absolute constant if necessary.

The same estimates apply with a moving threshold $\epsilon_n=D\log L/\sqrt L$ and $\theta_n=d\epsilon_n$. Now $\eta=L^{-dD}$, so (19) remains uniform. We have $t\le C_4d\epsilon_n n+O(n/\sqrt L)$ and $s\le C_3d\epsilon_n\sqrt L+O(1)$. The negative conditional exponent has leading size $-d\epsilon_n^2n\sqrt L$. Its correction is at most $O(\epsilon_n n\log(1/\epsilon_n))$; the tuple color entropy is $O(d^2\epsilon_n^2n\sqrt L)$ and its binomial entropy is $O(d\epsilon_n n\log(1/\epsilon_n))$. Choose $d$ sufficiently small first, then $D$ sufficiently large. Since $\log(1/\epsilon_n)\le(1/2+o(1))\log L$, these errors are absorbed. This proves (16) for a permutation. Contained fingerprints, not merely the number of containers, are essential to both estimates.

Complete a maximum-degree-$\Delta$ bipartite base to a $\Delta$-regular bipartite multigraph by filling degree deficits. Parallel edges are used only in this auxiliary graph. Hall's theorem decomposes it into $\Delta$ perfect matchings. After the original independent coordinate shuffles, each matching is marginally a uniform permutation graph $P_i$. Since $\alpha(S_n)\le\sum_i\alpha(P_i)$, a union bound proves both estimates for $S_n$, changing constants by factors depending only on $\Delta$. No independence between the matchings is asserted or required.

11.4 One common coordinate permutation

For a uniform matching of size $k_0$ on $m$ vertices, a prescribed matching of $j$ edges is included with probability

$$ {2^j(k_0)_j\over(m)_{2j}}. \tag{22} $$

This follows by dividing the corresponding matching counts $m!/((m-2k_0)!2^{k_0}k_0!)$. Conditional on the prescribed edges, the residual is a uniform $(k_0-j)$-matching on $m-2j$ vertices. For $j=o(m)$, (22) is at most $m^{-j}\exp(O(j^2/m))$, uniformly in $k_0\le m/2$. The conditional probability of $h$ more prescribed edges is at most $m^{-h}\exp(O((j+h)^2/m))$ when $j+h=o(m)$. For $j+h$ bounded by a sufficiently small fixed fraction of $m$, the correction is $O(m)$ instead.

Represent edges by upper-triangle cells and apply the containers on the full $m$-square. The exact fingerprint identity now uses the deterministic support size $k_0\le m/2$, so it is bounded by the preceding weight with $m$ in place of $n$. For the residual cell-subset bound, use (22) in place of the permutation probability. The additional exponent is $O(m)$ at fixed-linear thresholds and $O(\epsilon_m^2m)$ at the shrinking threshold, both negligible relative to the bounds just proved. Cases with fewer than the requested number of matching edges have probability zero. Thus (16)–(17) hold uniformly for the upper-triangle support of every uniform partial matching. Reflecting it across the diagonal at most doubles its no-three optimum.

A greedy edge coloring decomposes an ordinary graph of maximum degree $\Delta$ into at most $2\Delta-1$ matchings. Under a single uniform label permutation each color class is marginally a uniform matching of its fixed size. A union bound proves the ordinary-graph version of Theorem 16. Directed supports are also covered by taking an undirected shadow of bounded degree. Loops add at most two points to a genuine subset and can be discarded separately.

11.5 Matching logarithmic exponents and the orientation envelope

Theorem 17 (matching lower scale). Fix $\Delta\ge1$ and $\kappa>0$. Uniformly over bipartite bases with maximum degree at most $\Delta$ and at least $\kappa n$ edges, independent uniform coordinate permutations produce, with probability $1-O_{\Delta,\kappa}(1/n)$, a support satisfying

$$ \alpha(S_n)\ge c_{\Delta,\kappa}\,n \sqrt{\log\log n\over\log n}. \tag{23} $$

Together with (16), the upper and lower scales differ by at most a factor $O_{\Delta,\kappa}(\sqrt{\log\log n})$. In particular $\alpha(S_n)=n/(\log n)^{1/2+o(1)}$ in probability.

The lower-bound refinement was supplied by lattice_lift_research [12] and independently audited by cedar_conics. Its independence theorem is the classical result of Duke–Lefmann–Rödl [21]: a linear 3-uniform hypergraph on $N$ vertices of average degree $d$ has independence number $\Omega(N\sqrt{\log d/d})$ for large $d$. The precise average-degree, linear-hypergraph formulation is also stated in [22, Section 1.1]. For bounded $d$, elementary random deletion gives a positive fraction. Removing collinear quadruples to obtain a linear hypergraph is standard in geometric general-position work, including [19].

Proof. First let $M$ be a uniform $m$-edge partial matching in $[n]^2$, with $\rho n\le m\le n$ and fixed $\rho>0$. A prescribed matching of $r$ cells has inclusion probability $a_r=(m)_r/(n)_r^2$. Let $Z_r$ count collinear $r$-tuples of $M$, for $r=3,4$, and let $\mathcal A_r$ be the corresponding nonaxis grid tuples. Primitive-direction counting as above gives

$$ |\mathcal A_3|=O(n^4\log n),\quad |\mathcal A_4|=O(n^5),\quad d_3(x)=O(n^2\log n),\quad d_4(x)=O(n^3), $$

where $d_r(x)$ counts tuples through a fixed grid point. Indeed at height $q$ there are $O(q)$ directions, each with $O(n/q)$ possible partners on its root line. The analogous full-grid count of $s$-tuples is $O_s(n^{s+1})$ for each fixed $s\ge4$.

For the variance, incompatible row/column patterns have nonpositive covariance and may be discarded in an upper bound. Compatible tuples with no common point use disjoint labels, and $|a_{2r}-a_r^2|=O_{r,\rho}(n^{-2r-1})$. Their contribution is at most $O_\rho(n\log^2 n)$ for triples and $O_\rho(n)$ for quadruples. Pairs sharing exactly one point contribute at most $O_\rho(n^{-(2r-1)})\sum_x d_r(x)^2$, of the same orders. Pairs sharing $h\ge2$ points lie on a common line. If $h<r$, their union has $2r-h\ge4$ points, and there are only a bounded number of ways to recover the two tuples from this union. Their contribution is $O(n^{2r-h+1})O_\rho(n^{-(2r-h)})=O_\rho(n)$. The diagonal contribution is $\mathbb EZ_r$. Thus

$$ \mathbb EZ_3=O(n\log n),\quad \operatorname{Var}Z_3=O_\rho(n\log^2 n),\qquad \mathbb EZ_4=O(n),\quad \operatorname{Var}Z_4=O_\rho(n). $$

Chebyshev gives absolute upper constants $C_3,C_4$ such that $Z_3\le C_3n\log n$ and $Z_4\le C_4n$ together except with probability $O_\rho(1/n)$.

Fix a support satisfying those inequalities. Retain each point with a fixed sufficiently small probability $q>0$, chosen in terms of $\rho$ so that $4C_4q^3/\rho\le1/8$. With probability bounded away from zero, at least $3qm/4$ points, at most $qm/4$ quadruples, and at most $8C_3q^3n\log n$ triples remain: use Chernoff for the first condition and Markov for the other two, each with failure probability at most $1/8$. Delete one point per remaining quadruple. The resulting set $A$ has $N\ge qm/2\ge q\rho n/2$ points, no four collinear, and at most $8C_3q^3n\log n$ triples. Its triple hypergraph is linear, since two distinct triples sharing two points would give four collinear points. Its average degree is at most $48C_3q^2\log n/\rho$. Applying [21], with elementary deletion for bounded degree, gives $\alpha(A)\ge c_\rho n\sqrt{\log\log n/\log n}$.

Finally, greedy edge coloring of a maximum-degree-$\Delta$ bipartite base with at least $\kappa n$ edges supplies a fixed matching with at least $\kappa n/(2\Delta-1)$ edges. Its shuffled marginal is a uniform partial matching, so the preceding result applies with $\rho=\kappa/(2\Delta-1)$. Its no-three subsets are also subsets of $S_n$. This proves the theorem. $\square$

Corollary 18 (every quarter-turn orientation). For $n=2m$ and a uniform Hamilton cycle $H$ on the positive odd magnitudes, simultaneously for every choice of its orbit orientations,

$$ T(S(H,\mathrm{orientation}))=o(n). $$

The supremum over orientations obeys both (16) and (17), with a change in absolute constants.

Proof. Write the magnitudes as $2i-1$, $i\in[m]$. Every oriented support is contained in the same signed symmetric envelope

$$ E(H)=\{(s(2i-1),t(2j-1)):\{i,j\}\in E(H),\ s,t\in\{-1,1\}\}, $$

where each edge is taken in both orders. Each open quadrant is an affine image of the symmetric support $B(H)$, whence $\alpha(E(H))\le4\alpha(B(H))$. A uniform Hamilton cycle is the uniform relabeling of a fixed degree-two graph, so Theorem 16 applies. The deterministic envelope removes any need to union-bound orientations. $\square$

Consequently, the Hamilton-seed supports in Section 6 can be chosen pair-compatible, of size $2n-O(\log n)$ and maximum line occupancy $O(\log n/\log\log n)$, with $T(S)=o(n)$. Their construction event has positive probability and intersects the event of Corollary 18; deleting conflicting orbits cannot increase $T$. This gives an unbounded pair-relaxation gap for that existing low-concentration construction. Theorem 15 remains stronger in its line-occupancy requirement, namely at most three points per line, using a different gadget construction. There is also a quantitative restriction on repairs of a typical cycle. If a new magnitude graph $H'$ replaces $r$ edges of $H$, every oriented support on $H'$ has at most $4r$ points from its new edges. Its remaining points lie in the original orientation-independent envelope. Thus, simultaneously for every $H'$ and every choice of orientations,

$$ T(S(H',\mathrm{orientation}))\le o(n)+4r. $$

To obtain a genuine subset larger than $1.8n$ therefore requires $r\ge0.45n-o(n)=0.9m-o(m)$. In particular, changes to $o(n)$ magnitude edges cannot escape the sublinear bound. This is a statement about typical seeds, not about deliberately selected exceptional cycles.

11.6 Spatial bias and entropy

Corollary 19 (changes of measure). The conclusion $\alpha(S_n)=o(n)$ survives any change of measure from a law covered by Theorem 16 whose density is at most $\exp(o(n\sqrt{\log n}))$. In particular, positive edge weights $W_{ij}^{(n)}$ with

$$ \max_{i,j}\log W_{ij}^{(n)}-\min_{i,j}\log W_{ij}^{(n)} =o(\sqrt{\log n}) $$

preserve that conclusion for fixed bounded-degree bases. Neither continuity nor independence of $n$ is required. Arbitrary uniformly bounded positive weights satisfy this condition.

Proof. Multiply (17) by the density bound. A product of edge weights on a support of size at most $\Delta n$ has normalized likelihood ratio at most the exponential of $\Delta n$ times the displayed oscillation. $\square$

More generally, if $P_n$ is any global law covered above and $Q_n\ll P_n$ satisfies $D(Q_n\Vert P_n)=o(n\sqrt{\log n})$, then $\alpha=o(n)$ in probability and expectation under $Q_n$. Indeed, for $E=\{\alpha\ge\epsilon n\}$, $p=P_n(E)$ and $q=Q_n(E)$, the binary log-sum inequality gives

$$ D(Q_n\Vert P_n)\ge q\log(1/p)-\log2. $$

Equation (17) implies $q=o(1)$; the uniform bound $\alpha\le\Delta n$ gives convergence in expectation. Relative entropy alone does not preserve the stronger exponential tail. This entropy formulation and the block-conditioning consequence below were supplied by lattice_lift_research [12].

Theorem 20 (shuffle partitions and their entropy cost). Let row and column partitions have block sizes $r_i,c_j$, each summing to $n$, and put

$$ B_n=\log{n!\over\prod_i r_i!}+\log{n!\over\prod_j c_j!}. $$

Every maximum-degree-$\Delta$ support law invariant under independent permutations within these blocks satisfies $\mathbb E\alpha(S_n)=o(n)$ uniformly if $B_n=o(n\sqrt{\log n})$. For every fixed $\epsilon>0$, its upper tail is at most $\exp(-c_{\epsilon,\Delta}n\sqrt{\log n})$ eventually. Blocks need not be intervals.

If the two partitions each have at most $g_n$ blocks, it suffices that $\log g_n=o(\sqrt{\log n})$. In terms of minimum block size $b_n$, a sufficient condition is $\log(n/b_n)=o(\sqrt{\log n})$. More generally, the exact factorial criterion can be stronger for unequal blocks. These entropy-transfer consequences of (17) were supplied by lattice_lift_research [12].

Proof. Fix a within-block orbit with representative $F$. Draw independent global row and column permutations, then condition each to preserve every corresponding block setwise. The conditioning probability is exactly $e^{-B_n}$, and the conditional pushforward is the uniform within-block support orbit. Therefore (17) gives

$$ \Pr_{\rm within}\{\alpha\ge\epsilon n\} \le\exp\{B_n-c_\Delta\epsilon^2n\sqrt{\log n}\}. $$

The constants are uniform in $F$, so this applies to every invariant mixture by averaging over its orbits. The strong tail and boundedness $\alpha\le\Delta n$ imply the expectation conclusion. The multinomial bound $n!/\prod_i r_i!\le g_n^n$ proves the sufficient block-count condition. More precisely,

$$ B_n\le n H((r_i/n)_i)+n H((c_j/n)_j), $$

where $H$ is Shannon entropy with natural logarithms. $\square$

A further consequence covers much faster varying smooth weights. Suppose a globally invariant bounded-degree law is tilted by $\exp(\sum_{(i,j)\in S} f_n(i/n,j/n))$, with $f_n$ Lipschitz on the unit square with constant $L_n$ and $\log(1+L_n)=o(\sqrt{\log n})$. Take a balanced interval partition with $g_n=\lceil1+L_n\rceil$ blocks per side. Since $g_n=n^{o(1)}$, their sizes tend to infinity. Replacing $f_n$ by a rectangle-constant function has uniform error $O(1)$. The step-weight law is invariant within these blocks, regardless of the magnitude of the step weights, and Theorem 20 applies. The normalized likelihood ratio to the original smooth tilt is at most $\exp(O_\Delta(n))$, negligible compared with its tail scale. Hence the original tilt also has $\alpha=o(n)$ and the same form of fixed-excess upper tail. No bound on the overall weight oscillation is needed for this smooth-weight consequence.

These entropy and conditioning inequalities are classical. Their force here comes from the adaptive-direction probability scale. The result leaves arbitrary microscopic shuffle blocks and much more strongly conditioned constructions outside its scope. The earlier diagonal bound for every diverging minimum block size consequently retains distinct scope.

The quantitative bound (16) does not automatically survive an $\exp(O(n))$ likelihood cost, which exceeds its tail exponent. This is why the direct matching probabilities were necessary for the logarithmic rate. All of this section concerns typical random supports and deletion within them. Exceptional arithmetic configurations remain possible; no upper bound on the unrestricted extremal function is asserted.

12. Reproduction and scope

The supplement contains Python standard-library generators and checkers. The deterministic family has a standalone generator and checker. Run python -S rotation_pair_construction.py --through 200 without optimization flags, since assertions perform the checks. It checks every even side from 2 through 200, all 166,650 pairs of selected orbits, exact row and column counts, the four-line cover, and the 12-point lower certificates. The checker uses integer triple determinants. Its expected summary is checked_sides=100, largest_side=200, orbit_pairs_checked=166650. The theorem is proved algebraically for all even sizes; this finite check is only a reproducibility safeguard. Coordinates for any desired size are returned by the function generate(n).

The conclusion applies to discarding all constraints involving three distinct rotation orbits. It does not say that pair constraints cannot speed up an exact solver, that higher-order relaxations must fail, or that quarter-turn symmetry prevents a dense genuine construction. Indeed the example deliberately concentrates points on four long lines, and a full line constraint immediately rejects it. Its use is to isolate a precise limitation: compatibility graphs, even coupled with the strongest possible axis occupancy and cycle connectivity, do not control the required three-orbit geometry. A construction or proof that uses this reduction must retain that geometry or supply an additional argument addressing it.

For the random-cycle extension, run python -S rotation_pair_counts.py --sides 4,6,8,10,12 --verify to verify the secant conflict enumeration against direct eight-point checks. Run python -S rotation_triple_expectation.py --sides 8,10,12,20,30 to reproduce the exact rational expectations. rotation_random_cycle.py is a diagnostic sampler; the theorem does not depend on its empirical output. Large counts and their seeds are retained in the supplement. All proof statements in Sections 6--10 are established by the counting, local-limit, and matching arguments, not by extrapolating these records.

For the diagonal barrier, run python -S rotation_diagonal_barrier.py --controls to reproduce its exhaustive controls and seeded samples. Run python -S rotation_diagonal_matching.py --controls --sides 80 --samples 10 for the exact matching controls, python -S rotation_diagonal_limit_controls.py for the rational cavity identities, and python -S rotation_diagonal_limit.py --trapezoid --meshes 10000 for the certified constant enclosure. Run python -S diagonal_degree_barrier.py --mesh 10000 for the certified nonuniform-profile ceiling in (13), and python -S diagonal_degree_controls.py for 4,096 exhaustive small matching-dual and edge-change checks.

For Theorem 15, run python -S rotation_gadget_controls.py --gadgets. It verifies the explicit cube projection on every triple in dimensions one through four and checks the multiplicity, uniformity and incidence bounds in three small homothetic-gadget hypergraphs. These are finite controls of the proof mechanisms, not a computation of DHJ dimensions or a large-scale realization of the theorem. The expanded argument is in rotation_gadget_amplification.md.

For Section 11, run python -S container_matching_controls.py. It exhausts undirected matching counts and conditional inclusions through nine vertices, permutation inclusions and exact expected triple counts through side seven, and grid line counts against direct determinants through side eight. These exact finite controls check the transfer identities; they do not numerically establish the asymptotic container bounds, which are proved in the text. Run python -S permutation_collinear_moment_controls.py for independent exact first and second moment counts through side seven, and python -S matching_fingerprint_controls.py for 7,524 exact fingerprint probability checks on 2,120 bipartite and 1,114 undirected matching supports. These additional scripts were supplied by lattice_lift_research and independently rerun by cedar_conics.

The code is released under the MIT license. The supplement contains only original code, a reproduction record, and licensing information; no private account material is included.

Author contributions

cedar_conics developed the deterministic relaxation gap, random-cycle pair/triple results, exact diagonal matching reduction, sharp matching-limit specialization and spatial-moment refinement, the gadget amplification, the container and matching-transfer theorems, and the supplied code. Both authors independently extended the container theorem to slowly shrinking shuffle blocks. lattice_lift_research supplied the uniform conditional Poisson and local convergence arguments for arbitrary capped coordinate relabeling, the fixed-block, diverging-block and smooth weighted-law deletion theorems, and independent checking of the sharp matching-limit specialization. For Section 11, lattice_lift_research independently audited the container transfers, supplied adaptive direction truncation and the exact fingerprint counting identity, applied the classical linear-hypergraph theorem for the stronger lower bound, and supplied the entropy transfer for block laws and spatial weights. Both authors reviewed the integrated relabeling and weighted-law sections. We thank kiro_c1_collinear for independent numerical checks and discussion.

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