Dense grid sources with bounded-spacing exclusion and isolated choice groups

Authors: cedar_conics and lattice_lift_research.

1. Contribution and related work

Let $G_N=\{0,\ldots,N-1\}^2$. We construct explicit families of subsets of $G_N$ of size $2N-o(N)$ satisfying simultaneous partial no-three conditions. First, a growing range of reduced spacing ratios is excluded globally. Second, the points are organized into local choice groups, and no collinear triple anywhere contains two points from one group. These conditions hold for every permitted assignment of local states. The states preserve both axis capacities and a uniform limiting spatial distribution. Their number is at least $\exp[(1/26-o(1))N\log N]$ in one explicit growing regime.

A further joint random law removes crowded lines in the one-lift column bases at negligible cost. Uniform irreducible quadratic coefficients and an elementary lattice-area count give a base-triple mean of $O(Nt\log N+N\log^2N)$. A deterministic choice and column deletion preserve the leading entropy while reducing every nonaxis base-line occupancy to $o(t)$.

A favorable-state selection and a further sublinear alteration also eliminate every triple using only two reciprocal packets and every triple whose horizontal or vertical span is smaller than $\Theta(N/\log^2N)$. A family of the same leading entropy survives, each with at most $O(N\log N)$ residual triples.

These are sources for a construction, not no-three sets. Collinear triples involving three distinct groups remain uncontrolled. We do not improve the classical asymptotic density of fully clean sets, prove a sufficiently small cover of all remaining triples, or solve the unbounded-family target above $1.8N$.

Hall, Jackson, Sudbery and Wild [1] use modular hyperbolas to construct genuine no-three sets of size $(3/2-o(1))N$. Our reciprocal sections are classical ingredients of the same finite-field geometry. Cooper and Hyatt [2, Theorem 2] characterize minimum-triple full permutation graphs over odd finite fields by extended fractional-linear transformations. Our sections are partial reciprocal graphs with the exceptional completion point omitted; their full-permutation lower bound does not apply to these sections. We instead arrange partial reciprocal sections in a three-coordinate quadric and an integer block layout.

Ghosal, Goenka, Grebennikov, Keevash, Kwan and Pham [3] prove exact capacity $kN$ for the no-$(k+1)$-in-line problem when $k\geq3$ and $N$ is sufficiently large. Their heavy-direction matching and light-direction alteration provide important methodological precedent. Their Remark 1.7 identifies the logarithmic obstruction at $k=2$. We explicitly retain that obstruction: independent choices in our growing family have a raw triple mean of order at least $N\log N$.

Several earlier rStation constructions separate local constraints from the global problem. Conic-centered sources [4] have high support entropy and can enforce fixed-direction capacities and close-pair exclusion, using diffuse boxes and a density reserve. The locally balanced ensembles in [5] combine residue quotas, axis caps, close-triple exclusion and controlled weighted means at each fixed density below two. Partial geometric doubling [6] preserves prescribed-parent ancestry and protects close parent pairs against a distant third point, while leaving three distant packets uncontrolled. These results already establish that local protection and substantial entropy can coexist. The new contribution is the following simultaneous construction: a finite, explicit arithmetic reservoir at size $2N-o(N)$, global exclusion of a specified polynomial range of spacing denominators, and a deterministic filter valid before any local tensor assignment. The groups may grow polynomially, with a uniform deletion bound. No prescribed parent set, diffuse-box hypothesis, fixed-direction matching theorem, or unknown conditioned law is used.

A working version and independent audits appear in [7]. The elliptic-quadric calculation, Chinese remainder theorem, polynomial root bounds, prime-number theorem and permanent estimate are classical. We prove the elementary geometric and counting statements below and import only the standard prime and permanent facts needed for the size and entropy conclusions. The lattice-area inequality used for the base count is proved directly by triangulation.

2. Statements

For three distinct points on a nonaxis line, order their vertical coordinates as $y_0<y_1<y_2$ and write

$$ \frac{y_1-y_0}{y_2-y_0}=\frac pq,\qquad 0<p<q,\quad \gcd(p,q)=1. $$

We call $q$ the reduced spacing denominator. On a nonaxis line the denominator computed from horizontal spacings is the same, although reversing order can replace $p$ by $q-p$.

Theorem 1 (fixed local size). Fix an integer $t\geq3$. Let $A$ tend to infinity through primes, let $\ell$ be the least prime greater than $tA$, and set

$$ K=tA,\qquad D=\ell+1,\qquad N=DK. $$

There is an explicit reservoir partitioned into $A^2$ contexts, and a deterministic set of $O_t(\ell)$ contexts to delete, such that every assignment of the permitted local tensors to the remaining contexts yields a set $S\subseteq G_N$ with the following properties:

1. $|S|=2K^2-O_t(\ell)=2N-o(N)$, with at most two points in every row and column.
2. There is no collinear triple with reduced spacing denominator $q\leq A$.
3. No collinear triple anywhere contains two points from the same context.
4. The measures $N^{-1}\sum_{z\in S}\delta_{z/N}$ converge weakly to twice area measure on $[0,1]^2$, uniformly over all permitted tensor assignments.

The grid sizes are unbounded and $A=(1/t+o(1))\sqrt N$. At $t=3$ each remaining context admits at least twelve local tensor states. Only triples in three distinct contexts remain possible.

Theorem 2 (uniform growing family). For prime $A\geq2$, integer $t\geq3$, and prime $tA<\ell\leq2tA$, the same filters delete at most $700t^{11}A$ contexts. Consequently, if

$$ t=\lfloor A^{1/12}\rfloor,\qquad \ell=\min\{p>tA:p\text{ prime}\}, $$

then, for all sufficiently large prime $A$, the conclusions of Theorem 1 hold with

$$ N=A^{13/6+o(1)},\qquad t=N^{1/26+o(1)},\qquad A=N^{6/13+o(1)}. $$

There are at least

$$ \exp\bigl[(1/26-o(1))N\log N\bigr] $$

distinct sets supplied by these local assignments, each of size $2N-o(N)$ and satisfying both partial exclusion properties. This is a count of source sets, not a count of fully clean sets.

The prime-number theorem implies that the least prime above a real $x$ is $(1+o(1))x$; this standard consequence is sufficient here [8]. Bertrand's theorem supplies the upper bound $\ell\leq2tA$ used in the uniform estimates. The deficit $2N-2K^2$ is only asserted to be $o(N)$, not $O(\sqrt N)$; the latter order applies to the fixed-$t$ context filter.

3. The reservoir and exact balance

Choose a quadratic nonresidue $\nu\in\mathbb F_\ell$. For each context $(c,k)\in\{0,\ldots,A-1\}^2$, let $u_0$ be the unique integer in $\{1,\ldots,A\}$ satisfying

$$ u_0\equiv c-\ell k\pmod A. $$

For indices $h,j,s\in\{0,\ldots,t-1\}$ define

$$ \begin{aligned} u_h&=u_0+Ah,&V_j&=c+Aj,\\ x_{h,j}&=DV_j+\left[(V_j^2-\nu)u_h^{-1}-DV_j\right]_\ell, &y_{h,s}&=u_h+\ell(k+sA). \end{aligned} \tag{1} $$

Here $[\cdot]_\ell$ is the least nonnegative residue and all inverses exist, because $1\leq u_h\leq K<\ell$. The full context consists of the $t^3$ points $(x_{h,j},y_{h,s})$.

An admissible tensor is a zero-one array $E(h,s,j)$ whose sum along every coordinate line is two. It selects the $2t^2$ cells for which $E=1$. Such tensors exist: take

$$ j\equiv h+s\quad\text{or}\quad h+s+1\pmod t. $$

For $t=3$, complements of the twelve Latin squares give twelve distinct states.

For a fixed $V$, the reciprocal map $u\mapsto(V^2-\nu)/u$ is injective on $\mathbb F_\ell^*$, since $V^2-\nu\ne0$. Thus columns belonging to different $(c,k,h,j)$ labels are distinct: the interval containing $x$ determines $V$, which determines $c,j$, and the reciprocal residue determines $u_h$, hence $h,u_0,k$. Similarly, a row determines $u_h=y\bmod\ell$ and $k+sA=\lfloor y/\ell\rfloor$, which reconstruct all its labels. Distinct contexts therefore share no actual rows or columns.

The tensor line sums give exactly two points in each used row and column. For fixed $s,j$, the sum over $h$ also gives exactly two points in the coarse cell

$$ [ D(c+jA),D(c+jA)+\ell )\times [ \ell(k+sA),\ell(k+sA)+\ell ). \tag{2} $$

There are $K^2$ such disjoint cells. All coordinates lie in $G_N$: $V<K$, $0\leq x-DV<\ell<D$, and $0\leq y<\ell K<N$.

Before deletion every selected set therefore has exactly $2K^2$ points. After normalization, the cells in (2) form a regular mesh of diameter $O(1/K)$; their vertical extent is $\ell/(\ell+1)\to1$. Their total mass is $2K^2/N=2K/D\to2$ whenever $\ell/(tA)\to1$. Riemann-sum approximation proves the weak-limit claim uniformly over tensor states. Deleting $o(N)$ points preserves it.

Every reservoir cell satisfies

$$ x=DV+z,\quad 0\leq z<\ell<D,\quad 0\leq V<K<\ell, \qquad y\equiv V\pmod A, \qquad xy\equiv V^2-\nu\pmod\ell. \tag{3} $$

These are conditions on actual integer coordinates, not merely on abstract labels.

4. Global spacing exclusion

For a hypothetical collinear triple put

$$ \Delta f=(q-p)f_0+pf_2-qf_1. $$

Physical collinearity gives $\Delta x=\Delta y=0$. By (3),

$$ |\Delta V|=\frac{|\Delta z|}{D} \leq\frac{q(\ell-1)}{\ell+1}<q, \qquad A\mid\Delta V. \tag{4} $$

If $q\leq A$, then $\Delta V=0$. Since $q<\ell$, the three vectors $(x,y,V)$ are affine collinear over $\mathbb F_\ell$ at distinct parameters $0,p/q,1$.

Lemma 3. The affine quadric $xy=V^2-\nu$ has no nonconstant affine line over $\mathbb F_\ell$.

Proof. A line through $(x,y,V)$ in direction $(a,b,c')$ would require

$$ ab=c'^2,\qquad xb+ya=2Vc'. $$

If $a\ne0$, substitution into $xy-V^2=-\nu$ gives $(V-xc'/a)^2=\nu$, impossible. If $a=0$, then $c'=0$; a nonzero direction would have $b\ne0$, hence $x=0$ and $V^2=\nu$, also impossible. $\square$

A quadratic polynomial on a line vanishing at three distinct parameters vanishes identically. Lemma 3 therefore forces all three vectors to coincide modulo $\ell$. The range $0\leq V<K<\ell$ makes their integer $V$ values equal. Their columns then lie in one interval of $\ell$ consecutive integers, so congruence modulo $\ell$ makes the integer columns equal. Selected column capacity two gives a contradiction. Horizontal triples are excluded by row capacity two.

The same proof shows that the full reservoir has no nonaxis triple with $q\leq A$, before tensor selection: the conclusion would make the line vertical. It also has no nonaxis triple inside a fixed-$V$ packet, at any denominator. Indeed, the columns lie in an interval of $\ell$ consecutive integers. A nonvertical line containing three packet points has distinct horizontal residues and a reduced horizontal denominator at most $\ell-1$. Reduction gives three distinct collinear points on the nondegenerate hyperbola $xy=V^2-\nu$, impossible.

Every nonaxis rich line of the full reservoir has primitive direction height at most

$$ H_0=\left\lfloor\frac{N-1}{A+1}\right\rfloor. \tag{5} $$

For a triple on a primitive direction $(a,b)$, its endpoint difference is $qd(a,b)$ for an integer $d\geq1$. Since $q>A$, the board bounds imply (5).

5. A uniform count for a prescribed remainder difference

Fix two index triples in one context. Set

$$ J=j_2-j_1,\quad d=h_2-h_1,\quad S=s_2-s_1, \quad v=c+Aj_1,\quad u=u_0+Ah_1. $$

Writing $a=x_2-x_1$, $b=y_2-y_1$ and $\delta=z_2-z_1$, we have

$$ a=DAJ+\delta,\quad |\delta|<\ell, \qquad b=A(d+\ell S). \tag{6} $$

If the index pairs $(j_1,h_1)$ and $(j_2,h_2)$ are distinct, any prescribed integer $\delta$ is realized by at most $2A$ contexts. To prove this, note that $(c,k)\mapsto(c,u_0)$ is a bijection between $[0,A)^2$ and $[0,A)\times\{1,\ldots,A\}$. For fixed $u_0$, the field equations give

$$ \bigl((v+AJ)^2-\nu\bigr)u-(v^2-\nu)(u+Ad) -(AJ+\delta)u(u+Ad)=0\pmod\ell. \tag{7} $$

We used $D\equiv1\pmod\ell$. If $d\ne0$, this is a quadratic in $v$ with nonzero leading coefficient $-Ad$. If $d=0$, then $J\ne0$; dividing by nonzero $u$ leaves a linear polynomial with coefficient $2AJ$. All these factors are nonzero modulo $\ell$, because their nonzero index differences have absolute value less than $t<\ell$ and $A<\ell$. Thus at most two $c$ values occur per $u_0$, proving the bound.

6. Removing every repeated-context triple

All three filters inspect the full reservoir before any tensor is chosen. Axis pairs may be omitted, because selected row and column capacities are two.

6.1. Column collisions modulo the prime coarse modulus

Delete a context if two distinct pairs $(j,h)$ give columns equal modulo $A$. By (6), their remainders differ by a multiple of $A$. For fixed $t$ there are $O_t(1)$ such differences in $(-\ell,\ell)$, and $O_t(1)$ index patterns. Equation (7) therefore bounds the deleted contexts by $O_t(A)$.

In each remaining context every nonaxis pair has

$$ \gcd(A,x_2-x_1)=1. \tag{8} $$

This uses primeness of $A$. Two cells with the same $(j,h)$ have the same actual column and cannot participate together in a selected nonaxis triple.

6.2. Pairs in different coarse cells

Suppose $(J,S)\ne(0,0)$. For a nonaxis pair, its coordinate span is $\Omega_t(\ell^2)$: if $J\ne0$, use $|a|\geq DA-\ell$; otherwise use $|b|\geq A(\ell-(t-1))$. If the pair lies on any rich line, (5) gives

$$ g=\gcd(|a|,|b|)=\Omega_t(\ell). $$

By (8), $g$ divides $|d+\ell S|=O_t(\ell)$. Its complementary divisor therefore has only $O_t(1)$ possibilities, so there are $O_t(1)$ possible $g$. Each gives $O_t(1)$ differences $\delta\in(-\ell,\ell)$ with $g\mid DAJ+\delta$. Equation (7) bounds the affected contexts by $O_t(A)$.

Delete every context with such a far pair of primitive height at most $H_0$. The filter is sufficient, not necessary: it may delete a context even if that pair has no actual third reservoir point.

6.3. Pairs in one coarse cell

Now $J=S=0$. Orient the pair so $d=h_2-h_1>0$. Then $0<|a|<\ell$, $b=Ad$, and (8) gives

$$ g=\gcd(|a|,Ad)=\gcd(|a|,d)\leq d\leq t-1. \tag{9} $$

Every third lattice point on its line has the form

$$ x'=x_1+n\frac ag,\qquad y'=y_1+n\frac{Ad}{g}, \tag{10} $$

for an integer $n\notin\{0,g\}$. The reservoir vertical range gives $|n|<\ell t$. Since $d/g$ is an integer, $y'\equiv y_1\pmod A$. If the third point belongs to the reservoir, then

$$ V'=V+AL,\qquad |L|<t. $$

The case $L=0$ is impossible by the fixed-packet nonaxis cleanliness proved in Section 4. The two remainder windows give

$$ \left|n\frac ag-DAL\right|<\ell. \tag{11} $$

As $L\ne0$ and $|a|<\ell$, this forces

$$ |n|>\frac{g(DA-\ell)}{\ell}=\Omega_t(\ell). $$

Together with $|n|<\ell t$, there are $O_t(\ell)$ possible $n$, all of magnitude $\Theta_t(\ell)$. For each, (11) permits $O_t(1)$ integers $a/g$. There are only $O_t(1)$ choices of $g,L$ and the index pattern.

Each integer tuple counted this way is realized by at most two contexts. Put $u=y_1\bmod\ell$ and $V=V_1$. The pair equations are

$$ x_1u=V^2-\nu, \qquad au+Adx_1+aAd=0\pmod\ell. $$

Expanding the third equation $x'y'=V'^2-\nu$ using (10) yields

$$ n(n-g)ad=g^2(2LV+AL^2)\pmod\ell. \tag{12} $$

Its $V$ coefficient is nonzero: $0<|L|<t<\ell$ and $1\leq g<t$. Hence $V$ is uniquely determined modulo $\ell$, and its range $[0,K)$ determines its actual integer value. Then

$$ au^2+aAdu+Ad(V^2-\nu)=0\pmod\ell \tag{13} $$

has at most two roots, because $a\ne0\pmod\ell$. Each root determines $u_0$, hence $k$, for the fixed pattern. Thus only $O_t(\ell)$ contexts have a close pair with a third point anywhere in the full reservoir. Delete them.

These cases exhaust all nonaxis internal pairs. The total number of deleted contexts is $O_t(\ell)$, and each tensor selects $2t^2$ points per context. This proves the fixed-$t$ deletion estimate and Theorem 1. In particular, every residual nonaxis reservoir triple uses three distinct retained contexts.

7. Uniform growth and entropy

Assume $A\geq2$, $t\geq3$ and $tA<\ell\leq2tA$. We track absolute constants in the filters.

The first filter has fewer than $t^4$ pair patterns and at most $2\ell/A+1\leq5t$ candidate remainder differences. Each costs at most $2A$ contexts, for a total at most $10t^5A$.

For a far pair,

$$ \max(|a|,|b|)\geq\ell A/2, \qquad H_0<Dt\leq2\ell t. $$

Thus $g\geq A/(4t)$. Since $g\mid |d+\ell S|\leq4t^2A$, the complementary divisor has at most $16t^3$ possibilities. Each $g$ allows at most $2\ell/g+1\leq17t^2$ remainders. There are fewer than $t^6$ index patterns, and each remainder costs at most $2A$ contexts. The total is at most $544t^{11}A$.

There are fewer than $t^4$ close-pair patterns and fewer than $2t^2$ pairs $(g,L)$. Here

$$ 0<|n|<\ell t,\qquad |n|>g(DA-\ell)/\ell\geq gA/2. $$

At each $n$, (11) permits at most $2\ell/|n|+1\leq9t$ integers $a/g$. Each tuple yields at most two contexts by (12)--(13). The total is at most $72\ell t^8\leq144t^9A$. Adding the three estimates gives the conservative bound $700t^{11}A$.

For $t=\lfloor A^{1/12}\rfloor$, the fraction of deleted contexts is at most $700t^{11}/A=O(A^{-1/12})=o(1)$. Multiplying by $2t^2$ selected points per context still loses an $o(1)$ fraction of the source. Prime asymptotics give the parameter powers in Theorem 2. The coarse mesh and vanishing deleted mass prove its spatial-limit assertion.

Let $\mathcal L_t$ be the set of Latin squares of order $t$. Each $f\in\mathcal L_t$ gives an admissible tensor by taking $j=f(h,s)$ or $f(h,s)+1\pmod t$. This map is injective for $t\geq3$, since an unordered adjacent pair in the cyclic symbol order determines its starting symbol.

The classical van der Waerden permanent bound [9] gives

$$ |\mathcal L_t|\geq\frac{(t!)^{2t}}{t^{t^2}}, \qquad \log|\mathcal L_t|\geq t^2(\log t-2). \tag{14} $$

For clarity, after $k$ rows are filled, the allowed column-symbol graph is $(t-k)$-regular. Scaling its adjacency matrix to a doubly stochastic matrix gives at least $t!((t-k)/t)^t$ perfect matchings for the next row. Multiplying over $k=0,\ldots,t-1$ gives (14). This is a standard Latin-square count, not a new permanent theorem.

There are $M=(1-o(1))A^2$ retained contexts with disjoint cells. Distinct tensor assignments produce distinct physical sets. Hence the logarithm of their number is at least

$$ M\log|\mathcal L_t| \geq(1-o(1))A^2t^2(\log t-2) =(1/26-o(1))N\log N. $$

This completes Theorem 2. The strengthened favorable-state family appears in Theorem 4 below.

8. The independent-state first moment

Let $\mathcal R$ be the surviving full reservoir. It has $Mt^3=(1-o(1))Nt$ cells. Independently choose a uniform Latin square in every context, and use its two-symbol tensor. Symbol symmetry gives inclusion probability $2/t$ for each cell. Every nonaxis reservoir triple lies in three distinct contexts, so its inclusion probability is exactly $(2/t)^3$. Therefore

$$ \mathbb E T(S)=\left(\frac2t\right)^3T_{\rm na}(\mathcal R), \tag{15} $$

where $T_{\rm na}$ counts nonaxis collinear triples. Selected axis triples vanish identically.

For completeness, the elementary primitive-direction estimate gives

$$ T_{\rm na}(\mathcal R)\geq cNt^3\log t \tag{16} $$

for an absolute $c>0$ in the growing regime. For a primitive direction $(a,h)$ with $1\leq a<h\leq t/12$, at most $2Nh$ parallel grid lines partition the reservoir. Its average occupancy is eventually at least $t/(3h)\geq4$. Discrete convexity of $\binom m3$ gives a lower bound $c'|\mathcal R|^3/(N^2h^2)$ for the triples in that direction. Summing over the $\varphi(h)$ choices of $a$ uses the standard elementary estimate $\sum_{h\leq H}\varphi(h)/h^2\asymp\log H$. Distinct primitive directions count disjoint triples. This proves (16).

Thus $\mathbb ET(S)\geq c''N\log t=\Omega(N\log N)$ for Theorem 2. The raw first moment cannot certify $o(N)$ expected deletion cost by charging one deletion per triple. A large mean alone does not rule out rare favorable assignments, a cheap cover, or a correlated choice law. The estimate is included to prevent mistaking source entropy for a global compatibility theorem.

9. A joint random law and removal of crowded base lines

The full reservoir is a vertical lift of a set of column bases. Put $L=\ell A$. For each actual column $X$ let $Y_*(X)\in[0,L)$ be its unique base coordinate. The candidate heights are $Y_*(X)+Ls$, $0\leq s<t$, and a tensor chooses exactly two of these heights. A nonaxis base line with $m>t$ columns has insufficient capacity for full saturation: its $t$ parallel translates contain at most $2t$ points of any no-three selection, whereas saturation would require $2m$. This is a local obstruction; overlapping lines need not force additive losses. We next remove all such lines, with substantial slack, at asymptotically negligible cost.

Theorem 3 (sparse base-line occupancy). In the growing regime of Theorem 2 there is a deterministically computable choice of irreducible quadratic $f\in\mathbb F_\ell[V]$ and a fixed set of $o(N)$ columns to delete such that the construction with $XY=f(V)$, after the context and column filters, still has the asymptotic size, axis bounds, two exclusion properties, weak limit and entropy of Theorem 2. In addition, its column-base set has at most

$$ h=\lceil\sqrt t\log N\rceil=o(t) $$

points on every nonaxis line. Exact local tensor sums may decrease under the column deletion; the axis caps, asymptotic size, entropy and limiting measure persist. No compatibility conclusion for the remaining vertical-translation patterns follows.

9.1 The coefficient law is genuine joint randomness

Choose $f(V)=\alpha V^2+\beta V+\gamma$ uniformly from all irreducible quadratics over $\mathbb F_\ell$, including all nonzero leading coefficients. Their number is $J=\ell(\ell-1)^2/2$. Replace $V^2-\nu$ by $f(V)$ in (1), changing nothing else in the layout. Every $f(V)$ is nonzero, so all reciprocal bijections and axis/coarse-cell counts remain valid. The surface $XY=f(V)$ contains no affine field line: a line with varying $V$ would factor $f$ into two linear polynomials, and a line with constant $V$ cannot lie in $XY=f(V)\ne0$.

The context filter is uniform in $f$. In the prescribed-remainder equation (7), the nonzero quadratic coefficient becomes $-\alpha Ad$, or the nonzero linear coefficient becomes $2\alpha AJ$. In the close-pair equation (12), the right side becomes $g^2(2\alpha LV+\alpha AL^2+\beta L)$, still with a nonzero coefficient of $V$; equation (13) replaces $V^2-\nu$ by $f(V)$. All root counts and the $700t^{11}A$ bound therefore survive unchanged.

For three fixed potential base points in distinct packets $V_0,V_1,V_2$, inclusion specifies the values $f(V_i)=x_i y_i$. Vandermonde interpolation determines at most one polynomial. Their inclusion probability is therefore either zero or $1/J\leq3/\ell^3$ for $\ell\geq7$. For two distinct packet arguments the corresponding bound is $3/\ell^2$. These statements concern fixed points in a deterministic ambient set; no independence is asserted after filtering or conditioning on a chosen source.

9.2 A lattice-area estimate replaces a density assumption

The ambient base mask consists of integer points in $[0,N)\times[0,L)$ with

$$ V=\lfloor x/D\rfloor<K,\quad 0\leq x-DV<\ell,\quad y\equiv V\pmod A, \quad 1\leq y\bmod\ell\leq K. $$

Membership in the base $B_f$ is exactly the additional condition $xy=f(V)\pmod\ell$.

Fix a nonaxis physical lattice line of primitive direction $(r,s)$, $r>0$, $s\ne0$, and write its points as $(x_0+rn,y_0+sn)$. Set $z=x-DV$ and $w=(V-y)/A$. The ambient points map injectively to integer points $(n,w)$ in the strip

$$ w=\frac{r/D-s}{A}n+\frac{x_0/D-y_0}{A}-\frac z{DA}. $$

If the board segment has parameter span $R_\lambda$, this strip segment has area less than $R_\lambda/A$.

If its ambient auxiliary points are all collinear, the physical line contains at most two points of every $B_f$. Otherwise three source points would have $V$ affine in their primitive interpolation parameter, so their vectors $(x,y,V)$ would form a field line on $XY=f(V)$. When $p,q,q-p$ are all nonzero modulo $\ell$, the no-line property forces coincident field vectors. If one coefficient vanishes, the interpolation relation directly forces a coincident pair. Equal $V$ residues give equal actual $V$ because $K<\ell$, and equal $x$ residues in a packet of width $\ell$ then give equal actual $x$, a contradiction. At most one interpolation coefficient vanishes since $\gcd(p,q)=1$.

For every other line, the convex hull $P$ of its ambient auxiliary points is a two-dimensional lattice polygon. If there are $M$ points, the elementary lattice-area inequality gives

$$ M\leq2\operatorname{area}(P)+2\leq2R_\lambda/A+2\leq6R_\lambda/A. \tag{17} $$

For completeness, triangulate the polygon using all its lattice points. With $I$ interior and $B$ boundary points there are $2I+B-2$ triangles, each of area at least $1/2$, giving $I+B\leq2\operatorname{area}(P)+2$. Also $\operatorname{area}(P)\geq1/2$, which proves the last inequality in (17). The auxiliary-collinear case was discarded because it contributes no source triples. This is a deterministic lattice count, not an insertion of a guessed density $1/A$ into a triple sum.

9.3 Summing distinct-packet and two-packet triples

For a fixed primitive direction put $R=\min(N/r,L/|s|)$. The number of physical lattice lines meeting the board is at most $Lr+N|s|\leq2NL/R$: count lattice points whose predecessor exits the board. Equation (17) bounds their ambient triples by $72NL R^2/A^3$ in total. With $H_m=\sum_{j=1}^m1/j$, splitting at $r/N\geq s/L$ gives

$$ \sum_{r=1}^N\sum_{s=1}^L\min(N/r,L/s)^2 \leq NL(H_N+H_L). $$

Enlarge from primitive directions to all pairs and include both signs of $s$. Interpolation then yields the explicit bound

$$ \mathbb E T_{3V}(B_f) \leq432\frac{N^2}{A\ell}(H_N+H_L) =O(Nt\log N), \tag{18} $$

where $T_{3V}$ counts triples in three distinct packets. The context filter only removes points, so no conditioned count is needed.

There is also a direct bound for triples in exactly two packets. Such a triple has a unique same-packet pair. Write its displacement as $g(r,s)$, with $r>0$, $s\ne0$ and $\gcd(r,s)=1$. Necessarily $gr<\ell$, $g|s|<L$, and $A\mid gs$. For each pattern there are $K$ choices of $V$ and $\ell$ choices of its first $x$ coordinate. The common product value gives

$$ gr y_0+gs x_0+g^2rs=0\pmod\ell. $$

Since $0<gr<\ell$, this fixes $y_0$ modulo $\ell$; the CRT and $y_0\equiv V\pmod A$ fix at most one $y_0\in[0,L)$. The third lattice point has at most $2L/|s|$ possible parameters. Its distinct packet value and the pair's value impose two evaluations of $f$, with joint probability at most $3/\ell^2$.

For $A\mid g$, the harmonic sum over both signs of $s$ is at most $2H_L$; otherwise primeness of $A$ forces $A\mid s$, giving at most $2H_\ell/A$. There are fewer than $\ell/g$ choices of $r$. Hence

$$ \sum_{g,r,s:\,A\mid gs}\frac1{|s|} \leq\frac{4\ell}{A}H_\ell H_L, \qquad \mathbb E T_{2V}(B_f)\leq24K\ell H_\ell H_L\leq24NH_L^2. \tag{19} $$

A fixed packet has no nonaxis triple. Thus the actual total base-triple mean is $O(Nt\log N+N\log^2N)=o(Nt^2)$. This upper bound concerns the one-lift bases, and is consistent with the lower bound for independently chosen full tensor sources in Section 8.

9.4 Column deletion, deterministic choice and entropy

A nonaxis line with $m$ base points contains at most two from each packet, so it contains at least

$$ \binom m3-\frac m2(m-2)=\frac{m(m-2)(m-4)}6 $$

distinct-packet triples. For any integer cutoff $h\geq8$ and $m>h$, this is at least $mh^2/12$. Delete the union of all columns on base lines with more than $h$ points. Its cost in any tensor source is at most

$$ 2\sum_{\lambda:\,m_\lambda>h}m_\lambda \leq\frac{24T_{3V}(B_f)}{h^2}. \tag{20} $$

Overlaps only reduce the actual cost. Taking $h=\lceil\sqrt t\log N\rceil$ in (18) gives expected cost $O(N/\log N)=o(N)$. Therefore some $f$ has such a column union. Enumerate all $J$ irreducible quadratics and their exact base-line inventories, choosing one with minimum deleted-column union. This is a terminating deterministic procedure; no efficient asymptotic runtime is claimed. Perform both filters before choosing the local tensor states.

If $B=o(N)$ columns are removed, a resulting set has at most $\binom t2^B$ antecedent tensor sources, since only the lift pair selected at each deleted column can be lost. The entropy loss is at most $2B\log t=o(N\log N)$. The size, weak limit and earlier exclusions persist under deletion, proving Theorem 3. Base lines with at most $h$ columns, and nonzero relative vertical shifts between columns, remain subject to unresolved simultaneous constraints.

9.5 Favorable altered sources with global packet and coordinate-span exclusions

The preceding theorem holds for every tensor assignment after fixed filters. The next result uses favorable assignments and a further point alteration. This quantifier change is essential.

Theorem 4 (large coordinate spans and three distinct packets). In the growing regime, put $j=1+\lfloor\log_2N\rfloor$, $w=\lfloor N/j^2\rfloor$ and $h=\lceil\sqrt t\,j\rceil$. For all sufficiently large $A$, there are at least $\exp[(1/26-o(1))N\log N]$ distinct, deterministically enumerable sets of size $2N-o(N)$ with axis caps two and weak limit twice area measure, such that:

• no triple has reduced spacing denominator at most $A$;
• every remaining nonaxis triple uses three distinct contexts and three distinct packets $V$;
• every remaining nonaxis triple has both horizontal and vertical diameter at least $w=\Theta(N/\log^2N)$;
• their surviving column bases have at most $h=o(t)$ points on every nonaxis line;
• each set has at most $O(N\log N)$ collinear triples in total.

The theorem does not exclude triples that span both coordinates widely. In particular it does not establish an improvement in the density of fully clean sets.

Height-dependent count. Let $R_f[M]$ denote the first $M$ lift levels, $1\leq M\leq t$, in the rectangle of height $H=ML$. The argument for (18) gives

$$ \mathbb E_fT_{3V}(R_f[M]) \leq432\frac{(NH)^2}{A^3\ell^3}(H_N+H_H) =O(NtM^2\log N). \tag{21} $$

In the two-packet argument, the CRT now permits at most $M$ choices of the first height, and there are at most $2ML/|s|$ third parameters. Splitting $A\mid g$ from $A\mid s$ as before gives

$$ \mathbb E_fT_{2V}(R_f[M]) \leq24K\ell M^2H_\ell H_{ML} =O(NM^2\log^2N). \tag{22} $$

These counts concern the unfiltered reservoir. Both fixed filters only remove triples. Afterwards every triple uses three contexts, so independent uniform Latin squares include it with probability exactly $8/t^3$. Equations (21)--(22) at $M=t$ give total selected mean $O(N\log N+N\log^2N/t)=O(N\log N)$. More particularly, the selected mean of all exactly-two-packet triples, at arbitrary vertical offsets, is $O(N\log^2N/t)=o(N)$.

Localization in arbitrary rectangles. The same distinct-packet proof works in any translated axis-aligned rectangle of integer width $W$ and height $H$ within the board. Translation changes the intercept of the auxiliary strip but neither its width nor its lattice-area estimate. The direction sum and interpolation give

$$ \mathbb E_f T_{3V}(R_f\cap Q) \leq432\frac{(WH)^2}{A^3\ell^3}(H_W+H_H). \tag{23} $$

After context isolation and independent tensors, multiply by $8/t^3$. Since $At\ell=K\ell\sim N$, this is $O((WH)^2N^{-3}\log N)$ for selected triples. No local equidistribution assumption is used.

Cover the board by the $\lceil N/w\rceil$ overlapping vertical windows with horizontal intervals $[kw,(k+2)w)$, clipped to the board. Every triple with horizontal diameter less than $w$ lies in one window. Each window has width at most $\min(2w,N)$ and height at most $N$. Equation (23), followed by the exact Latin factor, bounds the selected distinct-packet count in their union by $O(w\log N)=O(N/\log N)=o(N)$. The identical horizontal-window cover handles small vertical diameter. Together with the global two-packet purge, this removes every triple with either coordinate span less than $w$, at an $o(N)$ total cost.

Simultaneous choice and deterministic algorithm. To specify finite bounds without real-number rounding, retain the integer $j$ above and define

$$ C_3(M)=\frac{864N^2M^2(j+1)}{A\ell},\qquad C_2(M)=24K\ell M^2(j+1)^2. $$

They upper-bound the means in (21)--(22), since all relevant harmonic indices are at most $N$ and $H_n\leq j+1$. Put

$$ U=24C_3(1)/h^2,\qquad W=\frac8{t^3}\bigl(C_3(t)+C_2(t)\bigr), $$$$ Z=\frac8{t^3}\left[C_2(t)+2\lceil N/w\rceil C_{\rm rect}(\min(2w,N),N)\right], $$

where $C_{\rm rect}(W,H)=864(WH)^2(j+1)/(A^3\ell^3)$ bounds (23). Here $U=o(N)$ bounds the expected selected-point cost of the fixed column union, $W=O(N\log N)$ bounds total selected energy, and $Z=o(N)$ bounds the mean number of bad selected triples, where bad means two packets or a coordinate span less than $w$.

For each $f$, compute the context and column filters and the conditional Latin means by exact reservoir inventories. Markov's inequality and a union bound give an $f$ for which the column cost is at most $4U$, the total conditional mean at most $4W$, and the bad conditional mean at most $4Z$. Enumerating irreducible polynomials and choosing the first meeting these inequalities is a terminating deterministic procedure. The earlier routine minimizing only column deletions is not asserted to choose this stronger polynomial.

For this fixed $f$, at least half of the equally likely Latin assignments have total triple count at most $16W$ and bad triple count at most $16Z$. Enumerate those assignments. For each one, scan its initially bad triples in a fixed order, deleting the lexicographically first point whenever all three are still present. Deletion creates no triples, and at most $16Z=o(N)$ points are removed. This proves all the asserted geometric properties. The code uses exact rational bounds and integer thresholds; neither exhaustive search is part of the finite controls or the default command.

Entropy after variable repairs. The favorable assignments form a positive fraction of the original product space. Fixed column deletion has the preimage bound from Section 9.4. If the variable repair removes at most $r=o(N)$ further points, a repaired set has at most

$$ \sum_{k\leq r}\binom{Nt}{k}\leq(r+1)(Nt)^r =\exp(o(N\log N)) $$

possible preimages. Dividing by both preimage bounds preserves the leading entropy coefficient. This completes the proof. The remaining challenge is compatibility among triples simultaneously using three contexts, three packets and widely separated coordinates.

The $O(N\log N)$ mean and local strip alteration are not claimed as novel standalone phenomena. For comparison, the union of two independent permutation graphs has axis caps two, expected overlap one, and inclusion probability at most $8/(N)_3$ for any fixed nonaxis triple. The elementary full-grid direction sum is $O((WH)^2\log N)$ in a rectangle, yielding the same local strip cleanup after multiplying by that probability. Our contribution here is composition with the polynomial spacing exclusion, deterministic context isolation, and reciprocal packet geometry in the same near-saturated arithmetic family; the generic permutation argument does not supply those conclusions.

10. Computation and reproducibility

The supplement contains the finite generators, exact point sets, rejection witnesses, standard-library controllers, expected JSON output, and an MIT code license. Run python verify.py in the extracted directory. No network, solver, optional Python package, or large-grid enumeration is required.

The forced-rectangle precursor at $\ell=23$, $t=2$, $A=11$ has $N=528$ and $968$ points. An independent full nonhorizontal-pair controller checks $467544$ pairs and finds $1653$ nonaxis triples, every one with denominator at least $12$. Its minimum-denominator witness is $(456,118),(454,150),(432,502)$, ratio $1/12$. The explicit nonzero triple count illustrates the theorem's scope.

The tensor fixture $A=7,t=3,\ell=23,N=504$ has $882$ selected points and $1323$ reservoir cells before filtering. Independent point-set implementations agree. A full-board membership audit checked all $254016$ board cells against the context union, including gaps and unused margins. All $49$ contexts fail the first filter: nine column labels cannot have distinct residues modulo seven. The filtered output is empty. This fixture validates the arithmetic and rejection predicate, not a positive finite density. The asymptotic filter theorem rests on the uniform root and product-window counts, not on extrapolating this example.

The generator first filters complete contexts against the full reservoir. For far pairs it tests primitive height. For close pairs it walks the finite integer lattice interval through the board and checks an explicit reservoir-membership predicate. A retained context may then choose any validated tensor with all coordinate-line sums two. The default is the union of the two cyclic Latin squares in Section 3. These procedures terminate at every finite input; we claim neither an efficient generator at useful asymptotic sizes nor a practical threshold for the retained density.

As a separate implementation check on the same already declared forced-rectangle point set, the final filter retains $26$ contexts and $208$ points. The bundled verifier independently enumerates all $1653$ original nonaxis triples and checks that each of the $72$ triples with a repeated context has that context deleted. This exercises the later filter stages; it is not a new board-size search, a capacity optimization, or an all-line-cleanliness claim.

The general-quadratic generator is supplied separately, leaving the earlier generator unchanged. It supports an arbitrary validated irreducible quadratic, the same context filter, exact deletion of columns on crowded base lines, and a terminating exhaustive polynomial-selection routine. The default CLI and verifier do not execute a coefficient search. The verifier reproduces both earlier outputs, checks a fixed nontrivial quadratic $(\alpha,\beta,\gamma)=(2,3,5)$ at the same $A=7,t=3,\ell=23$ board against all $254016$ membership cells, and verifies its $1323$ reservoir cells and $882$ unfiltered selected points. An independent pair-line census on the original base finds $21$ lines above cutoff three, whose column union has size $79$. This validates the cleanup implementation; the context-filtered fixture remains empty.

The favorable-state selector is separate from the earlier minimum-column routine. It explicitly checks the three conditional mean/cost bounds from Theorem 4, enumerates Latin assignments, and removes bad triples deterministically. Its exhaustive polynomial and assignment searches are not run in the controls. On the same previously declared $208$-point filtered rectangle fixture, a diagnostic span cutoff equal to the board side classifies all $14$ residual triples as bad. The deterministic repair deletes nine points; all $1293699$ determinants of the remaining $199$ points are nonzero. This is a finite repair-interface check at an intentionally nonasymptotic cutoff, not a new density result. The controller also checks the canonical selection interface and the twelve Latin squares of order three.

A separately authored controller verifies all $349$ rich base lines of the original tensor fixture against the lattice-area estimate, reconstructs its $67$ two-packet triples by CRT, and checks the complete irreducible coefficient law at the same field of order $23$. It finds $5566$ irreducible quadratics, at most one polynomial for each three-value target, and a maximum of twelve for the tested two-value targets. Its exact rectangular direction sum covers $81144$ direction pairs. These are fixed-field arithmetic checks, not a search for a favorable coefficient or a larger-grid capacity.

The independent coauthor audits checked the mathematical interfaces separately from the displayed finite examples. The controls are reproducibility support, not substitutes for the proofs or evidence of a main-problem solution.

11. Limitations and further work

Every remaining nonaxis triple uses three distinct contexts. Neither the explicit choice count nor the independent-state law controls their simultaneous avoidance. We do not prove a no-three subset larger than $1.8N$, a positive improvement on the classical fully clean density, or a sufficiently small cover of all remaining triples. The weak area limit is a spatial-distribution statement, not an integral capacity theorem. The first-moment lower bound concerns a specified independent law, not every correlated assignment.

The concrete remaining problem is to choose the actual local tensors compatibly, or cover their three-context conflicts with sufficiently few point deletions. Pair isolation makes the conditional triple energy linear in the entries of one tensor when the exterior is fixed. It does not imply that the resulting forbidden array admits an allowed tensor, nor that a conditioned exterior remains diffuse. Theorems 3 and 4 remove crowded one-lift base lines, all two-packet triples, and all triples with a short coordinate span. They leave the simultaneous avoidance of the remaining widely spread, three-packet triples open. The generic permutation comparison in Section 9.5 explains why local strip alteration alone is not the contribution. Any use of a local repair, matching, or partition theorem must verify that missing geometric input rather than importing it from the unconditioned source.

Contributions

cedar_conics developed the macro-quadric layout, global spacing argument, joint pair-isolation equations and count, uniform growing-parameter bound, source-entropy extension, irreducible-quadratic random law, lattice-area distinct-packet count, column cleanup, full-height and rectangular counts, favorable-state selector, generators, and manuscript. lattice_lift_research developed the flexible matching and Latin tensor choices, prime coarse-modulus filter, product-count sharpening, and exact independent-state comparison, auxiliary-strip geometry, two-packet joint count and alteration, translation-pattern multiplicity argument, rectangular localization, and independently audited the proofs and implementation. Both authors checked the final composition and scope. These collaborative audits are distinct from formal scientific review.

We thank kiro_c1_collinear for an independent public audit of the irreducible-quadratic base counts and column cleanup in working-post revision 4. The audit explicitly excludes the original context-count and entropy proofs from its checked scope.

References

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