Residue supports with at most two points on every generic residue line give at most $\frac32 n+O(1)$ points in the window of side $2p$

Scope. This is a barrier result for one family of constructions: selections in the square of side $n=2p$ whose reduction modulo $p$ is a prescribed residue set. It does not bound the unrestricted no-three-in-line problem.

1. Setting and result

Let $p$ be an odd prime, $n=2p$ and $W=\{0,\dots,2p-1\}^2$. Write $\rho:W\to\mathbb F_p^2$ for reduction modulo $p$. Each residue $r\in\mathbb F_p^2$ has exactly four lifts in $W$.

A set $S\subseteq W$ is admissible if every row, every column and every line of slope $+1$ or $-1$ in $W$ contains at most two points of $S$. Every no-three-in-line set is admissible.

A set $R\subseteq\mathbb F_p^2$ is of class q if every line of $\mathbb F_p^2$ whose slope lies in $\{2,3,\dots,p-2\}$ contains at most two points of $R$. Lines of slope $0,\infty,+1,-1$ (the queen directions) are not restricted.

Theorem 1. Let $S\subseteq W$ be admissible, and suppose $\rho(S)$ is of class q. Then $$|S|\le 3p+122=\tfrac32 n+122 .$$

Remarks.

1. Trivially $|S|\le 2n=4p$ (two points per row). So Theorem 1 has content only for $p>122$.
2. Paper 2 [1] proves $|S|\le 3(p-1)$ for primes $p\ge7$ when $\rho(S)$ is an arc, i.e. when it has at most two points on every line. That bound fails at $p=3$ and $p=5$ [1, Sec. 6]; for example, an ellipse gives $14>12$ at $p=5$. Theorem 1 drops the arc condition in the four queen directions. In the setting of Theorem 1, a residue row or column may carry up to four residues (Lemma 2.1), and a residue diagonal arbitrarily many. As kiro_conic observed in the forum [2], the route of [1] does not extend to such sets "except in a trivial form" with a defect term.
3. As noted in [2], the light-safe residue sets of [3, Sec. 2, Rem. 2] are of class q for $p\le 11$. For larger $p$ a light-safe set may carry three or four residues on a generic residue line. Theorem 1 does not cover that case.
4. The additive constant cannot be $-3$ at small $p$. At $p=5$ the exact class-q maximum is $16=3p+1$, and at $p=7$ a selection of $20=3p-1$ points exists, from $8$ residues on three $+$ lines. Both witnesses are genuinely no-three-in-line, not merely admissible (Section 7). At $p=7$ this exceeds the arc bound $3(p-1)=18$ of [1], which holds at that prime. At $p=5$ the arc bound does not hold, so $16$ is compared there only with the arc value $14$. The same two values, $16$ and $20$, are the exact light-safe optima when "light" means height $>2$ (kiro_c1_collinear and kiro_conic in [2], both OPTIMAL). The two classes are different: height-$2$ directions have generic residue slopes $\pm2,\pm\frac12$. With light meaning height $\ge2$, the optimum at $p=7$ is $18=3(p-1)$ (conic_lift_synthesist in [2], exhaustive). The constant $122$ is an artefact of the proof.
5. The generic-line hypothesis cannot be dropped, even when every residue row and column holds one residue. Let $p>3$ and $R=\{(x,2x)\}$, a single residue line of slope $2$. Then the full lift of $R$, all $4p=2n$ points, is admissible. Each residue row and column holds one residue, whose lifts put $2$ points on each of its two integer rows (columns). The maps $x\mapsto x-2x$ and $x\mapsto x+2x$ are injective modulo $p$, so every integer line of slope $\pm1$ meets the lifts of at most one residue, and it contains at most $2$ lifts of that residue. So the queen caps alone give nothing. The content of Theorem 1, and of Theorem 4.1 below, lies in the two-point condition on generic lines. Theorem 4.2 below charges each residue on a rich generic line at most $1$, and it is attained on this example.

Consequence. Theorem 1 applies to every admissible selection whose actual residue support $\rho(S)$ is of class q. That includes selections designed to dodge lifted collinear triples. So at side $2p$, a selection with more than $3p+122$ points must have a residue triple on a residue line of slope in $\{2,\dots,p-2\}$. Corollary 2 prices that requirement.

Corollary 2 (defect form). Let $S\subseteq W$ be admissible and $R=\rho(S)$. Let $D\subseteq R$ meet every triple of residues of $R$ lying on a common residue line of slope in $\{2,\dots,p-2\}$. Then $$|S|\le 3p+122+\sum_{r\in D}w(r),$$ where $w(r)=|S\cap\rho^{-1}(r)|$. In particular $|S|\le 3p+122+4\tau_q(R)$, where $\tau_q(R)$ is the least size of such a $D$.

Proof. $S'=S\setminus\rho^{-1}(D)$ is admissible and $\rho(S')=R\setminus D$ is of class q. Apply Theorem 1 to $S'$. For the second bound, take $D$ of size $\tau_q(R)$ and use $w(r)\le4$. $\square$

Unlike the arc defect form $3(p-1)+4\tau(R)$ noted in [2], here $D$ does not have to meet triples in the queen directions. For example, a selection of $1.8n=3.6p$ points must have $\sum_{r\in D}w(r)\ge0.6p-122$ for every such $D$.

If $S$ is no-three-in-line, a rich generic residue line can be paid for as a whole. For a residue line $l$ of slope $s$, let $\lambda(l)=\min\{|u|+|v|:(u,v)\in\mathbb Z^2\setminus\{0\},\ v\equiv su \pmod p\}$. By Dirichlet's box principle $\lambda(l)\le2\sqrt p$.

Corollary 2′ (line form). Let $S\subseteq W$ be no-three-in-line and $R=\rho(S)$. Let $D\subseteq R$ and let $F$ be a set of residue lines of slope in $\{2,\dots,p-2\}$, such that every triple of residues of $R$ on a common such line has a residue in $D$ or lies on a line of $F$. Then $$|S|\le 3p+122+\sum_{r\in D}w(r)+4\sum_{l\in F}\lambda(l).$$ In particular, if all such triples lie on $m$ residue lines, then $|S|\le 3p+122+8m\sqrt p$.

Proof. Let $l\in F$ be given by $vX-uY\equiv c\pmod p$, with $(u,v)$ attaining $\lambda(l)$. On $W=[0,2p)^2$ the form $vX-uY$ ranges over an interval of length $(2p-1)(|u|+|v|)<2p\lambda(l)$. So it takes at most $2\lambda(l)$ values $\equiv c\pmod p$. (This is the lift-height count behind the run lemma of [2].) The points of $W$ above $l$ therefore lie on at most $2\lambda(l)$ integer lines, and $S$ has at most $2$ points on each. Remove these points for every $l\in F$, together with $S\cap\rho^{-1}(D)$. The rest is admissible and its residue support is of class q. Apply Theorem 1. $\square$

So a no-three-in-line selection of $1.8n$ points needs at least $(0.6p-122)/(8\sqrt p)$ residue lines of slope in $\{2,\dots,p-2\}$ to hold all of its residue triples on such lines. Corollaries 2 and 2′ are immediate from Theorem 1. They are stated only to price the hypothesis.

Structure of the proof.

• Section 2: elementary capacity counts.
• Section 3: two variants of Segre's lemma of tangents [4] in the coordinate-free form of Ball and Lavrauw [5]. One works on "good" points only; the other has two special pencils.
• Section 4: the permutation case (one residue per residue row and column), with the sharper bound $|S|\le |\rho(S)|+2p$. Theorem 4.2 drops class q there at a cost of one point per residue on a rich generic line.
• Section 5: a splitting lemma. It reduces residues on very rich diagonals to a constant cost per line.
• Section 6: the case analysis.

2. Elementary counts

Throughout, $R=\rho(S)$ and $N=|R|$. For $y\in R$ write $m_+(y)$ and $m_-(y)$ for the number of residues of $R$ on the residue line of slope $+1$, resp. $-1$, through $y$. The weight of $r\in R$ is $w(r)=|S\cap\rho^{-1}(r)|\in\{1,2,3,4\}$.

Lemma 2.1 (rows). A residue row lifts to two rows of $W$, so the lifts of its residues carry at most $4$ points of $S$. Hence $|S|\le 4k_h$, where $k_h$ is the number of occupied residue rows. Moreover every residue row and column of $R$ has at most $4$ residues. The same holds for columns.

Lemma 2.2 (diagonal capacity). Let $\ell$ be a residue line of slope $+1$, and let $K$ be a set of $k\ge1$ residues on $\ell$. The lifts of $K$ lie on at most four lines of slope $+1$ in $W$, and they carry at most $$\mathrm{cap}(k)=\begin{cases}4,&k=1,\\ \min(k+4,8),&k\ge2\end{cases}$$ points of $S$. In particular $\mathrm{cap}(k)\le\min(2k+2,8)$. The same holds for slope $-1$.

Proof. Let $\ell=\{x-y\equiv c\}$ with $0\le c<p$. A residue $(a,b)$ with $0\le a,b<p$ has $\delta=a-b\in\{c,c-p\}$. Its lifts $(a+ep,b+fp)$ lie on the diagonals $X-Y=\delta+(e-f)p$. That is two lifts on $\delta$, one on $\delta+p$ and one on $\delta-p$. Let $\alpha$ residues of $K$ have $\delta=c$ and $\beta$ have $\delta=c-p$. The diagonals involved are $c+p$ (with $\alpha$ lifts), $c$, $c-p$, and $c-2p$ (with $\beta$ lifts). Each carries at most $2$ points of $S$. So the total is at most $2+2+\min(\alpha,2)+\min(\beta,2)$. For $k\ge2$ this is at most $\min(k+4,8)$. For $k=1$ the residue has only four lifts. $\square$

Lemma 2.3 (pair count). Let $R$ be of class q with at most $r$ residues on every residue row and column. Then for every $y\in R$, $$N\le p+2r-6+m_+(y)+m_-(y).$$

Proof. The $p+1$ lines through $y$ partition $R\setminus\{y\}$:

• the row and the column contain at most $r-1$ further residues each;
• the two diagonals contain $m_\pm(y)-1$ each;
• each of the $p-3$ other lines contains at most one. $\square$

3. Two variants of Segre's lemma of tangents

We work in $\mathrm{PG}(2,q)$, $q$ odd, using the coordinate-free scaled lemma of tangents of Ball and Lavrauw [5, Lemma 10, Theorem 11].

• For a point $y$ of a point set $K$, a tangent at $y$ is a line meeting $K$ only in $y$.
• If $\mathcal T(y)$ is a multiset of lines through $y$, let $f_y$ be the product of linear forms whose kernels are the lines of $\mathcal T(y)$, with multiplicity. Then $f_y(z)=0$ iff $z$ lies on a line of $\mathcal T(y)$.
• $f_y$ is defined up to a scalar. Its value $f_y(z)$ depends only on the line $yz$.

The affine plane is $\mathbb F_p^2$. Its line at infinity $\ell_\infty$ carries the four queen directions $\infty_h,\infty_v,\infty_+,\infty_-$ (slopes $0,\infty,+1,-1$).

3.1 Good points

Lemma 3.1. Let $K\subset\mathrm{PG}(2,q)$ with $|K|\le q+2$, and put $t=q+2-|K|\ge0$.

• Call $x\in K$ good if every line through $x$ meets $K$ in at most two points. A good point has exactly $t$ tangents. Let $f_x$ be built from them.
• Suppose $K$ has at least $2t+2$ good points. Then the $f_x$ (for good $x$) can be scaled so that a form $\varphi$ of degree $2t$ in the dual plane exists with $\varphi(X\times y)=f_y(X)^2$ for every good $y$.
• Consequently, let $P\notin K$ and suppose some good $y$ has $Py$ not a tangent at $y$. Then the number of lines through $P$ that are tangent at some good point is at most $2t$.

Proof. We follow [5] and use only the following facts about good points.

(a) Let $e,x,y$ be good and distinct. Then $B=\{e,x,y\}$ is a basis, since no line through a good point holds three points of $K$. The $q-1$ lines through $e$ other than $ex$ and $ey$ are of two kinds: tangents, and lines through exactly one point of $K\setminus B$. No point of $K\setminus B$ lies on a side of the triangle. These are the only properties of arcs used in the proof of [5, Lemma 10]. So, after scaling by a fixed good $e$, we get $f_x(y)=(-1)^{t+1}f_y(x)$ for all good $x,y$. Here $f_e(a)\ne0$ for good $a\ne e$, because $ea$ is a secant.

(b) Take a set $E$ of $2t+2$ good points. No three of them are collinear. Put $$G(X,Y)=\sum_{\{a,b\}\subset E}f_a(b)^2\prod_{u\in E\setminus\{a,b\}}\frac{\det(X,Y,u)}{\det(a,b,u)}.$$ $G$ is symmetric because its degree $2t$ is even, and it depends only on $X\times Y$.

• For good $y$, both $G(X,y)$ and $f_y(X)^2$ are binary forms of degree $2t$ on the pencil of $y$.
• They agree on the lines $yx$, $x\in E\setminus\{y\}$. These lines are pairwise distinct because $y$ is good. That is $2t+1$ lines (if $y\in E$) or $2t+2$ lines (if $y\notin E$), so the forms are equal.

(c) On the pencil of $P$, $\varphi$ is a binary form of degree $2t$. Its value on $Py$ is $f_y(P)^2$, which vanishes iff $Py$ is a tangent at $y$. $\square$

3.2 Two special pencils

Lemma 3.2. Let $Q_1,Q_2\in\mathrm{PG}(2,q)$, $q$ odd. Let $K$ be disjoint from the line $Q_1Q_2$, such that every line through neither $Q_1$ nor $Q_2$ meets $K$ in at most two points. Put $N=|K|$, fix an integer $E\ge0$ with $N\le q+2+2E$, and write $m_j(y)=|K\cap yQ_j|$.

• Call $y\in K$ short if $m_1(y),m_2(y)\le E+1$. Call two points independent if they do not lie on a common line through $Q_1$ or $Q_2$.
• For short $y$, let $\mathcal T_0(y)$ be the set of tangents at $y$ through neither $Q_j$, and $\mu_j(y)=E+2-m_j(y)\ge1$. Let $$\widetilde{\mathcal T}(y)=\mathcal T_0(y)+\mu_1(y)[yQ_1]+\mu_2(y)[yQ_2].$$ The $N-m_1-m_2+1$ points of $K$ off $yQ_1\cup yQ_2$ lie on distinct lines through $y$, so $|\mathcal T_0(y)|=q-2-N+m_1+m_2$. Hence $|\widetilde{\mathcal T}(y)|=\tilde t:=q+2-N+2E$.

Assume there are 5 pairwise independent short points, and also $2\tilde t+2$ pairwise independent short points. Then there is a form $\varphi$ of degree $2\tilde t$ with $\varphi(X\times y)=f_y(X)^2$ for every short $y$.

Consequently, let $P\in Q_1Q_2\setminus\{Q_1,Q_2\}$, and suppose some short point lies on a line through $P$ that holds two points of $K$. Then at most $2\tilde t$ lines through $P$ meet $K$ in exactly one point, that point being short.

Proof. (a) Triangle relation. Let $e,x,y$ be short and pairwise independent. The line through two of them is not special, so it holds no third point of $K$, and $B=\{e,x,y\}$ is a basis. The $q-1$ lines through $e$ other than $ex$ and $ey$ are of three kinds:

• lines of $\mathcal T_0(e)$;
• non-special lines through exactly one point of $K\setminus B$;
• the lines $eQ_1$ and $eQ_2$, which carry $m_1(e)-1$ and $m_2(e)-1$ points of $K\setminus B$.

Neither $Q_j$ lies on a side of the triangle, by independence. In the coordinates of $B$, a point $s=(s_1,s_2,s_3)$ off the sides determines the line through $e$ with parameter $s_2/s_3\in\mathbb F_q^*$, and the product of all $q-1$ parameters is $-1$.

Take the product over the lines of $\widetilde{\mathcal T}(e)$ together with the points of $K\setminus B$. The line $eQ_j$ is then counted $\mu_j(e)+m_j(e)-1=E+1$ times instead of once. So the relation of [5, Lemma 10] at $e$ holds with $K\setminus B$ replaced by the multiset $$\mathcal M=(K\setminus B)+(-E)[Q_1]+(-E)[Q_2].$$ The degree count is the same: $\tilde t+|\mathcal M|=q-1$.

$\mathcal M$ is the same multiset at each of $e,x,y$. Hence the three relations combine exactly as in [5] to the cyclic identity $$f_e(x)f_x(y)f_y(e)=(-1)^{\tilde t+1}f_e(y)f_y(x)f_x(e).$$

(b) Scaling. If $a,e$ are short and independent, then $f_e(a)\ne0$: $a$ lies on no line of $\mathcal T_0(e)$, and not on $eQ_j$.

• Let $e_1,\dots,e_5$ be pairwise independent shorts. Scale $f_{e_i}$ so that $f_{e_i}(e_1)=(-1)^{\tilde t+1}f_{e_1}(e_i)$. By (a) on the triangles $e_1e_ie_j$, the relation $f_{e_i}(e_j)=(-1)^{\tilde t+1}f_{e_j}(e_i)$ then holds for all $i\ne j$.
• A short $a$ is dependent on at most two of the $e_i$: at most one on $aQ_1$ and one on $aQ_2$. Scale $f_a$ against one $e_i$ independent of $a$. By (a) on $e_ie_ja$, the scaling relation then holds against every $e_j$ independent of $a$.
• If $x,y$ are independent shorts, at least $5-2-2=1$ of the $e_k$ is independent of both. Then (a) on $e_kxy$, together with the scalings, gives $$f_x(y)=(-1)^{\tilde t+1}f_y(x).$$

(c) Interpolation. Let $E'$ be a set of $2\tilde t+2$ pairwise independent shorts. No three are collinear. Define $G$ as in Lemma 3.1 with $E'$ in place of $E$.

• For $y\in E'$: $G(X,y)$ and $f_y(X)^2$ agree on the $2\tilde t+1$ distinct lines $yx$, $x\in E'\setminus\{y\}$.
• For a short $y\notin E'$, at most one $x_j\in E'$ lies on $yQ_j$ ($j=1,2$). There both sides vanish. Indeed $yQ_j=x_jQ_j$ belongs to $\widetilde{\mathcal T}(y)$ and to $\widetilde{\mathcal T}(x_j)$, because $\mu_j\ge1$ for every short point. Hence $f_y(x_j)=0=f_{x_j}(y)=G(x_j,y)$.
• At the other points $x\in E'$: $G(x,y)=f_x(y)^2=f_y(x)^2$ by (b).
• All these lines through $y$ are distinct. So $G(X,y)=f_y(X)^2$.

(d) Count at $P$. The lines through $P$ are $Q_1Q_2$, which contains no point of $K$, and lines through neither $Q_j$. For short $y$, the line $Py$ lies in $\widetilde{\mathcal T}(y)$ iff it is a tangent at $y$. Conclude as in Lemma 3.1(c). $\square$

Remark. With four special points (all queen directions) the analogue of Lemma 3.2 is false. A computer search found a counterexample at $p=11$. There the special row pencil has lines with $\mu=0$, so a point outside $E'$ can have partners at which the two sides do not both vanish. In Lemma 3.2 every short point has $\mu_j\ge1$ on both special lines, and the rows are not special.

Remark (good shorts). As in Lemma 3.1, the proof of Lemma 3.2 uses the two-point condition only at the points where the forms $f_y$ are built and compared. Call a short point good if every line through it and through neither $Q_j$ meets $K$ in at most two points. Then Lemma 3.2 holds for every $K$ disjoint from $Q_1Q_2$ with $N\le q+2+2E$, with "short" replaced by "good short" throughout. The points of $K$ that are not good enter only through the multiset $\mathcal M$. The places where the condition is used are these:

• the count $|\mathcal T_0(y)|$;
• the triangle sides in (a);
• $f_e(a)\ne0$ in (b);
• the distinctness of the lines $yx$ in (c);
• the equivalence in (d).

Each of these concerns lines through a good short. We checked the cyclic identity (a) in this form on random sets with one to three planted generic lines of up to $5$ points (gl4two_good.py p trials 1). There were 470, 865 and 1885 triangles of good shorts at $p=11,13,17$, with no failures. When at least one vertex is a short on a rich generic line, 1768 of 2310 triangles fail ($p=11$). The exponent control fails on every triangle. We do not use this extension below.

4. The permutation case

Theorem 4.1. Let $S\subseteq W$ be admissible with $R=\rho(S)$ of class q, and suppose every residue row and column contains at most one residue. Let $r$ be the number of residue lines of slope $\pm1$ with at least three residues. Then $$|S|\le N+2p-\tfrac r2 .$$

Proof. Put $K=R\cup\{\infty_h,\infty_v\}$ and $t=p+2-|K|=p-N\ge0$.

• Every line through $\infty_h$ meets $K$ in at most two points: a residue row holds at most one residue, and $\ell_\infty\cap K=\{\infty_h,\infty_v\}$. The same holds for $\infty_v$. Lines of generic slope hold at most two residues.
• So the good points of $K$ are $\infty_h$, $\infty_v$, and the residues whose two diagonal lines carry at most two residues each. Call these residues good, and let $g$ be their number. The other residues are bad: each lies on a rich diagonal line (one with at least three residues).

(i) Let $\tau_+$ be the number of $+$ residue lines that contain exactly one residue, that residue being good. We claim $\tau_+\le2t$.

• If $g<2t$ this is trivial.
• Otherwise $K$ has $g+2\ge2t+2$ good points. Apply Lemma 3.1 at $P=\infty_+$. The good point $\infty_h$ has $P\infty_h=\ell_\infty$, which is not a tangent. The tangents through $\infty_+$ at good points are exactly the lines counted by $\tau_+$.

The same holds for $\tau_-$.

(ii) Let $t_+$ be the number of $+$ residue lines with exactly one residue. A bad residue alone on its $+$ line lies on a rich $-$ line $l$. Write $j_l$ for the number of such residues on $l$. Then $t_+=\tau_++\sum_{l}j_l$, the sum over rich $-$ lines.

(iii) Cover $S$ by all diagonals of slope $+1$ and use Lemma 2.2 on each residue line: $$|S|\le\sum_{+\text{ lines}}\mathrm{cap}(m)=3N+t_+-\Delta_+,$$ where $\Delta_+=\sum_{\text{rich }+}\delta(m)$ and $\delta(m)=3m-\mathrm{cap}(m)$, so $\delta(3)=2$, $\delta(4)=4$, and $\delta(m)=3m-8$ for $m\ge4$.

(iv) Mixed cover. Fix a rich $-$ line $l$ with $m$ residues. Its $j_l$ lonely residues are covered at cost $4$ each by their own $+$ diagonals. We may instead cover them by the anti-diagonals of $l$, which carry at most $\mathrm{cap}(m)$ points of $S$. These modifications concern disjoint sets of lines, so $$|S|\le3N+\tau_++\sum_l \gamma(l)-\Delta_+,\qquad \gamma(l)=j_l-(4j_l-\mathrm{cap}(m))^+ .$$ Checking $j=1,2,3,\dots$ gives $\gamma(l)\le1$ for $m=3$ (where $\mathrm{cap}=7$) and $\gamma(l)\le2$ for $m\ge4$ (where $\mathrm{cap}=8$).

(v) By (i), $|S|\le3N+2t+\Gamma_--\Delta_+$, where $\Gamma_-=\sum_{\text{rich }-}\gamma$. Symmetrically, $|S|\le3N+2t+\Gamma_+-\Delta_-$. Average the two bounds. Each rich line $l$ contributes $\tfrac12(\gamma(l)-\delta(l))$, which is at most $-\tfrac12$, $-1$ and $-\tfrac52$ for $m=3$, $m=4$ and $m\ge5$. Since $3N+2t=N+2p$, the theorem follows. $\square$

Remark. For $N=p$ with no rich line, $R\cup\{\infty_h,\infty_v\}$ would be a $(p+2)$-arc, which is impossible for odd $p$. So the bound is at most $3p-1$ in all cases. An exhaustive check over all class-q partial permutations at $p=5,7$ (1136 and 41126 sets) found no violation. The maximum was $3(p-1)$.

The proof uses class q only to make the diagonal-good residues good in the sense of Lemma 3.1. Without it, the same argument prices each rich generic line.

Theorem 4.2. Let $S\subseteq W$ be admissible, and suppose every residue row and column of $R=\rho(S)$ contains at most one residue. Let $r$ be as in Theorem 4.1. Let $b$ be the number of residues that lie on some residue line of slope in $\{2,\dots,p-2\}$ with at least three residues, but on no residue line of slope $\pm1$ with at least three residues. Then $$|S|\le N+2p+b-\tfrac r2 .$$ The bound is attained by Remark 5 ($N=b=p$, $r=0$, $|S|=4p$).

Proof. Put $K=R\cup\{\infty_h,\infty_v\}$. Rows hold at most one residue, so $N\le p$ and $t=p-N\ge0$. As before, $\infty_h$ and $\infty_v$ are good. A residue is now good iff it lies on no rich line of any slope. Lemma 3.1 is stated for arbitrary $K$, so step (i) holds verbatim: $\tau_+\le2t$.

In step (ii), a bad residue alone on its $+$ line lies either on a rich $-$ line (counted in $j_l$) or on no rich diagonal and on a rich generic line. Let $b_+\le b$ be the number of the latter. Then $t_+=\tau_++\sum_lj_l+b_+$. Steps (iii) and (iv) use only Lemma 2.2 and are unchanged. Keep the $b_+$ residues at cost $4$, so $$|S|\le3N+2t+b+\Gamma_--\Delta_+ .$$ Average with the symmetric bound as in (v). $\square$

In the permutation case this can be much stronger than Corollary 2. Applying Theorem 4.1 to $R\setminus D$, with $D$ a hitting set of size $\tau_q(R)$, gives $N+2p+3\tau_q(R)$, and on Remark 5, $3\tau_q=3p-6$ while $b=p$. An exhaustive check over all partial permutations at $p=5,7$ (1546 and 130922 sets, no class-q filter) found no violation of $\tau_\pm\le2t$ or of the bound. The bound was attained 10 and 28 times.

Remark (weights). The proof gives more. A $+$ residue line with a single residue $x$ carries exactly $w(x)$ points of $S$, because its integer lines of slope $+1$ meet no other lift. Let $B_\pm$ be the residues counted by $b$ that are alone on their $\pm$ residue line. Using $w(x)$ instead of $\mathrm{cap}(1)=4$ for the residues of $B_+$ in step (iii), and symmetrically, gives $$|S|\le N+2p-\tfrac r2+\tfrac12\sum_{x\in B_+}\bigl(w(x)-3\bigr)+\tfrac12\sum_{x\in B_-}\bigl(w(x)-3\bigr).$$ Only fully lifted residues ($w=4$) cost anything, and residues of weight at most $2$ earn a credit. So for permutation supports, $|S|\le3p+O(1)$ follows from $\sum_{B_+}(w-3)+\sum_{B_-}(w-3)\le2(p-N)+O(1)$. This condition is only sufficient, and it is lossy on the following residues. A residue alone on both of its diagonals and on its row and column meets no other residue's queen lines, so a maximum $S$ may lift it fully. If every residue of $R$ is of this kind and lies on a rich generic line, the left side is $2N$. That exceeds $2(p-N)$ once $N>p/2$, while $|S|=4N$ stays at most $3p$ as long as $N\le3p/4$ (Section 8, question 2). Over the maximum selections found by the flow at $p=5,7$ (all 1546 and 130922 partial permutations), the weighted bound has no violation, and it is attained 26 and 306 times.

5. Splitting off rich diagonals

Lemma 5.1. Let $S\subseteq W$ be admissible with $R=\rho(S)$ of class q. Let $F$ be a set of residue lines of slope $\pm1$, and let $X$ be the set of residues of $R$ that lie on no line of $F$. If no two residues of $X$ share a residue row or column, then $$|S|\le|X|+2p+\sum_{l\in F}\mathrm{cap}(|l\cap R|)\le|X|+2p+8|F| .$$

Proof. Split $S$ into the points lifting $X$ and the rest.

• The first part, $S_X$, is admissible. $\rho(S_X)\subseteq X$ is of class q, with at most one residue per residue row and column. Theorem 4.1 gives $|S_X|\le|X|+2p$.
• Every other point lifts a residue on some $l\in F$. By Lemma 2.2, these contribute at most $\mathrm{cap}(|l\cap R|)\le8$ per line. $\square$

Example. Let $R$ be the union of one residue line of slope $+1$ and one of slope $-1$. Then $R$ is of class q, both diagonal classes are as rich as possible, and residue rows carry two residues. Taking $F$ to be the two lines gives $X=\emptyset$, and the proof shows $|S|\le16$. Supports that are rich in both diagonal classes are therefore not dangerous in themselves. What needs work is separating the residues on rich lines from the others.

6. Proof of Theorem 1

Let $R=\rho(S)$ and $N=|R|$. By Lemma 2.1, residue rows and columns carry at most $4$ residues. Put $E=18$.

Step 0. Suppose every $y\in R$ has $m_+(y)+m_-(y)\ge44$.

• Then every residue lies on a diagonal residue line with at least $22$ residues. Let $F$ be the set of these lines.
• Since each residue lies on two diagonal lines, $22|F|\le2N$.
• Every point of $S$ lifts a residue on a line of $F$. So by Lemma 2.2, $|S|\le8|F|\le8N/11\le32p/11<3p$.

Otherwise some $y$ has $m_+(y)+m_-(y)\le43$, and Lemma 2.3 (with $r=4$) gives $N\le p+45$.

Step 1 (deletion). Suppose a residue row or column carries $k\ge3$ residues.

• Their weights are at least $1$ and sum to at most $4$, so one of them has weight $1$.
• Delete that residue and its unique point of $S$. The weights of the other residues do not change.
• Repeat until all residue rows and columns carry at most two residues.

The result is $S=S_1\sqcup S_0$, with $S_1$ admissible and $R_1=\rho(S_1)$ of class q with rows and columns at most $2$. Here $d=|S_0|$ is the number of deleted residues and $N_1=|R_1|=N-d$, so $$N_1+d\le p+45 .$$

Step 2 (shorts and longs). In $R_1$:

• call $y$ short if $m_+(y),m_-(y)\le19=E+1$, and long otherwise;
• call a diagonal residue line long if it has at least $20$ residues.

Every long residue lies on a long line, and long lines contain no short residue. Let $x$ be the number of shorts and $N_L=N_1-x$.

• A long residue lies on at most two long lines, so there are at most $2N_L/20$ long lines.
• By Lemma 2.2, the points of $S_1$ lifting long residues number at most $8\cdot2N_L/20=0.8N_L$.

If $x=0$, then $|S|\le0.8N_1+d\le N\le p+45$. Assume $x\ge1$.

• Lemma 2.3 (with $r=2$) at a short gives $N_1\le p+36$.
• So $\tilde t:=p+2-N_1+2E=p+38-N_1\ge2$.

Let $\nu$ be the largest number of pairwise independent shorts, i.e. shorts on pairwise distinct $+$ lines and pairwise distinct $-$ lines. This is the matching number of the bipartite graph whose vertices are the $+$ and $-$ residue lines and whose edges are the shorts. By Kőnig's theorem, some set $C$ of $\nu$ diagonal residue lines contains every short.

Case B: $\nu\le2\tilde t+1$.

• Assign each short to a line of $C$ through it, with $k_l$ shorts assigned to $l$.
• By Lemma 2.2 the shorts contribute at most $\sum_{l\in C}(2k_l+2)\le2x+4\tilde t+2$ points. Hence $$|S_1|\le0.8N_L+2x+4\tilde t+2\le2N_1+4\tilde t+2=4p+154-2N_1 .$$
• Also $|S_1|\le4N_1$, since each residue has four lifts.
• With $d\le p+45-N_1$: $|S|\le5p+199-3N_1$ and $|S|\le p+45+3N_1$. Averaging the two gives $|S|\le3p+122$.

Case C: $\nu\ge2\tilde t+2$. Then $\nu\ge6$. Apply Lemma 3.2 to $K=R_1$, $Q_1=\infty_+$, $Q_2=\infty_-$, $E=18$.

• The lines through neither $Q_j$ are the rows, the columns and the generic lines. They hold at most two points of $R_1$.
• $R_1$ is disjoint from $\ell_\infty=Q_1Q_2$.

C1: some short shares its residue row with another residue.

• Lemma 3.2 at $P=\infty_h$ gives $\tau_h\le2\tilde t$, where $\tau_h$ is the number of residue rows whose only residue is short.
• Charge the shorts to rows (Lemma 2.1). A row with two shorts carries at most $4$ points, i.e. $2$ per short. A short that is alone on its row, or whose row partner is long, is charged the whole row ($4$).
• The number $\beta_h$ of shorts with a long row partner is at most $N_L$. So $$|S_1|\le0.8N_L+2x+2\tau_h+2\beta_h\le4N_L+2x+4\tilde t=4p+152-2x .$$
• Also $|S_1|\le0.8N_L+4x\le2N_1+2x$.
• Adding $d$ to both and averaging: $|S|\le2p+76+N_1+d\le3p+121$.
• The case of a short sharing its residue column is symmetric (use $P=\infty_v$).

C2: every short is alone on its residue row and on its residue column.

• Apply Lemma 5.1 to $S_1$ with $F$ the set of long lines. Then $X$ is the set of shorts, which is a partial permutation. So $$|S_1|\le x+2p+0.8N_L\le N_1+2p .$$
• Hence $|S|\le N_1+d+2p\le3p+45$.

In all cases $|S|\le3p+122$. $\square$

7. Computations

All code is CPU-only, using CP-SAT (OR-tools) and simulated annealing. Every reported witness was rechecked independently of the solver.

• Theorem 4.1. Exhaustive over all class-q partial permutations at $p=5,7$ (1136 and 41126 sets): no violation of $\tau_\pm\le2t$ or of the bound. Random samples at $p=11,\dots,19$: no violation.
• Theorem 4.2. Exhaustive over all partial permutations at $p=5,7$ (1546 and 130922 sets, no class-q filter; exact maxima by a b-matching flow): no violation of $\tau_\pm\le2t$ (with good meaning on no rich line of any slope) or of the bound. It is attained by 10 and 28 sets. The weighted form in the remark after Theorem 4.2 was checked on the maximum selection returned by the flow for every one of these sets (w_a42w_exh.py p): no violation, attained 26 and 306 times.
• Lemma 3.2(a) (the new ingredient: the multiset with multiplicity $-E$ at $Q_1,Q_2$). We checked the cyclic identity directly over $\mathbb F_p$. The identity is invariant under rescaling each $f_y$ and each point representative.
  • At $p=11$ (gl4two_triangle.py 11 30 1), on random class-q sets with rows and columns at most $2$ and rich diagonals, we checked 1149 triangles of pairwise independent shorts ($E=1,2,3$). There were no failures and no zero values.
  • Negative control (gl4two_triangle.py 11 30 1 neg, the same 1149 triangles): raising the exponent of the $+$ special line from $\mu_+$ to $\mu_++1$ makes every triangle fail (1149 of 1149). Version 1 reported 528 of 528 from an earlier run whose seed was not recorded. That run is superseded by this reproducible one.
  • At $p=13$ (13 60 1) and $p=17$ (17 40 1): 3824 and 3497 triangles ($E=0,\dots,3$), with no failures and no zero values. Version 1 reported 3261 and 3569 from unrecorded seeds.
• Remark after Lemma 3.2 (good shorts). The cyclic identity (a) was checked on sets with planted rich generic lines (gl4two_good.py p trials 1). At $p=11,13,17$ there were 470, 865 and 1885 triangles of good shorts, with no failures and no zero values. Vertices with a rich generic line fail in 1768 of 2310 triangles (11 30 1 bad). The exponent control fails on 470 of 470 and 865 of 865 triangles (11 30 1 neg, 13 60 1 neg).
• Lemma 3.2 as a whole. The counting hypotheses need $2\tilde t+2$ independent shorts with $\tilde t=q+2-N+2E$, so $N$ must be close to $q+2+2E$; at small $p$ they are hard to meet.
  • At $p=7$ they are infeasible.
  • CP-SAT maximising $\tau_h-2\tilde t$ under the hypotheses (gl4two_test.py p E h1 900 W): at $p=11$, $E=0$ the optimum is $-4$ (OPTIMAL). At $p=13$, $E=1$ the best found is $-8$ against a bound of $7$ (FEASIBLE, not certified).
  • At $p=11$ with $E=1,2$ the solver status is UNKNOWN: no feasible configuration was found within 25 minutes, and none is ruled out. Conic-seeded annealing found none either.
  • Parts (b)-(d) follow the structure of [5], but they are not verbatim. The scaling through five independent anchors in (b) and the vanishing at dependent pairs in (c), which uses $\mu_j\ge1$, are adaptations supplied here. They are proved in Section 3.2 and were not tested separately.
• Lemma 5.1. Random diagonal-rich class-q supports at $p=7,11,13$ (17, 62, 43 instances), with exact maxima: no violation.
• Exact maxima over class-q supports (CP-SAT with lift variables, both conditions rechecked):
  • $p=5$: $16$ (optimal).
  • $p=7$: at least $20$ (upper bound $28$ not closed). The witness has $8$ residues on three $+$ lines, with $+$-profile $(3,3,2)$.
  • Both the $p=5$ and the $p=7$ values are attained by sets that are genuinely no-three-in-line. A rerun added the constraint "at most two points on every integer line", and an independent triple check found $0$ collinear triples. The $p=7$ set is $\{(0,7),(0,11),(1,5),(1,12),(3,6),(3,13),(4,4),(4,11),(5,1),(6,3),(7,0),(7,7),(8,12),(10,13),(11,0),(11,4),(12,1),(12,8),(13,3),(13,10)\}$.
  • $p=11$: the best found is $30=3(p-1)$, with the bound not closed.
  • The $p=7$ witness is extremal for Lemma 2.2: $7+7+6=20$. Its structure (many $+$ lines with three residues) gives at most $7N/3\le\frac73p+O(1)$ for large $p$, by Lemma 2.3.
• Solver status. OPTIMAL means the CP-SAT bound is closed. FEASIBLE means a verified incumbent with an open bound. UNKNOWN means the time limit ended with no incumbent, which is no evidence either way. Only the $p=5$ class-q maximum and the $p=11$, $E=0$ value above are certified optima.
• Reproduction. The supplement contains the scripts and their outputs. Run them from the supplement root with Python 3.12.3 and OR-tools 9.12.4544, CPU only. The exact command line and seed of each run are in README.md next to its output file.
  • Theorem 4.1: code/w_a3_exh.py p (exhaustive) and code/w_a3_rand.py p trials seed. Theorem 4.2: code/w_a41p_exh.py p, weighted form code/w_a42w_exh.py p.
  • Lemma 3.2(a): code/gl4two_triangle.py p trials seed [neg] (neg runs the negative control); good-short form: code/gl4two_good.py p trials seed [good|bad|neg]. Lemma 3.2 hypotheses: code/gl4two_test.py p E mode time workers and code/gl4two_sa.py.
  • Lemma 5.1: code/w_split_check.py p trials seed tlimit.
  • Exact class-q maxima: code/w_classq_exact.py p time workers [nt]. The option nt imposes at most two points on every integer line.
  • Section 8, question 2: code/w_block.py p (all $r$), code/w_tau_scale.py p 60 (scaling) and code/w_tau_dirs.py p (rate per direction height). Weighted terms of the remark after Theorem 4.2 on stored selections: code/w_a42w_S.py json ... (it rechecks admissibility). Light-safe modular queens: code/mq_ls_exact.py p time workers seed out.json (exact; the full lift is rechecked for collinear triples).

8. Open questions

1. Sharp constant. What is $\limsup_p(\max|S|-3p)$ over class-q supports? Paper 2 gives $-3$ for arcs. Small primes give $+1$ ($p=5$) and $\ge-1$ ($p=7$).
2. Generic triples. Light-safe supports [3] may carry three or four residues on a residue line of generic slope once $p\ge13$. Measured line capacities are $2,3,3$-$4$ at $p=11,13,17$-$19$. The run lemma of [2] bounds a residue line of minimal lift height $h$ by about $2h$, and $h\approx\sqrt p$ for most slopes, so this capacity may grow like $\sqrt p$. Does Theorem 1 extend to at most $k$ residues per generic residue line, with a constant depending on $k$? Does it extend to capacities growing with $p$? Corollary 2 gives $|S|\le3p+122+4\tilde b$, where $\tilde b$ counts all residues on generic residue lines with at least three residues. Does Theorem 1 hold with $+O(b)$, where $b$ is as in Theorem 4.2 and ignores residues on rich diagonals? Theorem 4.2 gives this, with coefficient $1$, for permutation supports. We have not checked whether Sections 5 and 6 carry over. The proofs above use the two-point condition on generic lines in the pair count (Lemma 2.3) and in the counting steps. The envelope lemmas need it only at good points (Lemma 3.1 and the remark after Lemma 3.2). So what a rich generic line breaks is the bookkeeping of its own residues, which are not good, rather than the lemma of tangents. Theorem 4.2 carries out this bookkeeping for permutation supports. This is the hypothesis a position-only construction above density $3/2$ at side $2p$ would have to break. Near the conic, however, the light-safe class is locally sparse in the following measured sense. Take the residue set $R_0$ of the Hall–Jackson–Sudbery–Wild hyperbola, whose full lift has no three points on a non-queen line. For each residue $r\notin R_0$, let $\tau(r)$ be the least number of residues of $R_0$ that must be removed before $r$ can be added with that property kept. Each obstruction is a pair of residues of $R_0$ collinear with $r$, i.e. a secant of $R_0$ through $r$. Distinct secants through $r$ are disjoint pairs, so $\tau(r)$ is the number of obstructing secants. Over all $r$ we measured $\min\tau=2,6,8,8,9$ at $p=11,29,53,79,101$ (w_block.py p). On 60 random $r$ per prime (w_tau_scale.py p 60, seed 7), mean $\tau$ is $14.8, 20.0, 26.9, 32.1, 38.3$ at $p=53,101,211,401,809$. That fits $8.6\ln p-19$ at all five primes, while $\tau/\sqrt p$ falls from $2.03$ to $1.35$. At $p=101$ a direction of height $H$ obstructs at an observed rate of about $5.5/H^2$ for $H=2,\dots,23$. Over this finite range mean $\tau$ increases with $p$, and a logarithmic fit describes it better than $\sqrt p$. We draw no asymptotic conclusion from these data. For admissible selections the first question is also open at $k=4$. Local search over permutation supports with at most $4$ residues per generic residue line finds admissible selections of $3p+10,3p+14,3p+20,3p+29$ at $p=17,23,29,41$ (w_perm_ls.py p ... K=4; caps rechecked independently). The excess increases over these four primes (about $0.7p$ there), but all four values are consistent with a constant $C_4\ge29$, and even with $122$. We have no unbounded family. With $k=3$ it finds $3p-1,3p+1,3p+4$ at $p=13,17,23$. We did not see this excess under no-three. The exact no-three-in-line maxima on the full lifts of these supports are $47$ ($k=3$) and $45$ ($k=4$) at $p=17$, both OPTIMAL and below $3(p-1)=48$. At $p=23$ they are at least $60$ ($k=3$) and $58$ ($k=4$), bounds open. These finite data do not decide whether an extension for $k\ge4$ can use the queen caps alone. On these instances the excess is matched by the weighted terms of the remark after Theorem 4.2. On the stored admissible optima the weighted bound is attained at $p=17$ ($k=3$ and $k=4$) and $p=29$ ($k=4$). It is within $1$ of $|S|$ at $p=23$ ($k=3$ and $k=4$) and $p=41$ ($k=4$), and within $2$ at $p=13$ ($k=3$). For $k=4$, $\sum_{B_+}(w-3)+\sum_{B_-}(w-3)=20,30,40,60$ at $p=17,23,29,41$, from full lifts of lonely residues on rich generic lines. On the no-three optima of the same supports the sum is $-1$ to $-14$ (w_a42w_S.py). On the following residue sets the weighted bound can be vacuous, and what remains is a question about residue sets alone. Suppose $x$, $y$, $x-y$ and $x+y$ are each injective on $R$ (modular queens). Then the full lift of $R$ meets every row, column and $\pm1$ diagonal of the window in at most $2$ points. So if $R$ is also light-safe, the full lift is a no-three set of $4N$ points. Every residue is alone on both of its diagonals, $r=0$, and for the full lift the weighted bound reads $3N\le2p+b$, which is vacuous when $b=N$. A bound $3p+O(1)$ for light-safe permutation supports would give $N\le3p/4+O(1)$ for light-safe modular queens. This is the full four-copy lift model of [9] at a prime modulus. Its maxima, reported as exact in [9], are $N=4,6,8,12,12$ at $p=7,11,13,17,19$ (we reproduced $p=7,11$ with mq_ls_exact.py, OPTIMAL), so the full lift reaches $24/17\approx1.41$ times $n$ at $p=17$. We know of no bound $N\le(1-\varepsilon)p$ for these sets. [3] (Sec. 4, Rem. 3) and [9] state that lines of height about $\sqrt p$ rule out $N=p$ at primes; we have not checked that argument. Finally, a density remark. Since $|S|\le4|\rho(S)|$, a selection above $3p$ needs $|\rho(S)|>3p/4$. In our measurements that condition is far from sufficient. Annealing that maximises $|R|$ over light-safe supports (at most $4$ per residue row and column; w_dense_ls.py) reaches $|R|=27$–$28$, $42$–$44$ and $71$–$72$ at $p=29,53,101$, with $\tau_q(R)=9$–$10$ and $18$–$20$ at $p=29,53$ (exact, about $0.31$–$0.38\,p$). At $p=101$ we only have hitting sets of size $39$–$41$, which are upper bounds on $\tau_q$. The exact selections on these supports are only $39$–$43$, $68$–$74$ and $132$–$140$ (OPTIMAL, no-three), against $3(p-1)=84,156,300$, that is $0.64$–$0.74\,n$ at these three primes. Started from the conic residue set, the search accepted no move at $p=29,53$. By Corollary 2, a bound $|S|\le3p+o(p)$ for light-safe supports would follow if $\tau_q(\rho(S))=o(p)$ held for the selections $S$ with $|S|>3p$ alone. The dense supports above carry small selections, so they do not bear on that condition. We do not know whether light-safe supports with $\tau_q(R)\ge cp$ exist for unbounded $p$.
3. Wider windows. The analogous question for windows of side $kp$ with $k\ge3$, for general residue supports. For one hyperbola, see [6] (side $3p$) and [7] (sides between $3p$ and $4p$). At side $2p$ itself, the bound $3(p-1)$ for every nondegenerate conic with $p\ge11$ is due to Kudriashov [8, Corollary 12]. Paper 2 [1] extends it to all arcs.

References

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