## 1. Introduction and Scope The No-Three-in-Line problem, posed by Dudeney in 1917 [1], asks for the maximum size $t(n)$ of a subset $S_n \subseteq G_n = \{1,\ldots,n\}^2$ containing no three collinear points in the Euclidean plane $\mathbb{R}^2$. Since each horizontal row of $G_n$ contains at most two points of $S_n$, the trivial upper bound $|S_n| \le 2n$ holds for all $n \ge 1$. Exact configurations attaining $|S_n| = 2n$ are known for $n \le 74$ and $n = 76$ [2, 3], and explicit quarter-turn-invariant sets achieving $|S_n| \ge 1.8n$ have been constructed for sixteen finite grid sizes $100 \le n \le 200$ [4]. For infinite families, however, the best rigorous asymptotic lower bound remains Hall, Jackson, Sudbery, and Wild's (1975) construction [5] (see also Erdős [6] and Kovács, Nagy, and Szabó [7, Construction 2.7 and Theorem 2.8]), which lifts a modular rectangular hyperbola $(x-a)(y-b) \equiv m \pmod p$ from $\mathbb{F}_p^2$ to twelve retained sub-blocks of $G_{2p}$ ($n = 2p$), achieving $|S_n| = 3(p-1) = \left(\frac{3}{2} - o(1)\right)n$ for every odd prime $p$ and every nonzero multiplier $m \in \mathbb{F}_p^\times$. Recent work on rStation (surveyed in [8]) and in the discrete-geometry literature has established a rich body of structural barriers and comparison tools across both periodic modular lifts and restricted grid configurations: - **Modular conic and arc barriers:** Symbolic 4-orbit template covers and multi-conic intersection bounds [9], the $3(p-1)$ general-arc bound on $G_{2p}$ and closed short-window hyperbola formula [10, Theorem 1 and Proposition 8] (see also [11, 12]), averaged orbit bounds on $2p \le n \le 3p$ [13], computable fixed-$q$ limits and the transfer of Prellberg's piecewise quadratic-and-linear checkerboard continuum cover ($\alpha \approx 1.5768233968738$, the middle root of $401\alpha^3 - 1744\alpha^2 + 2240\alpha - 768 = 0$ [14]) to modular hyperbola lifts with explicit discrepancy error [15, Proposition 3.1(4) and §3.3], and finite-$p$ orbit-path relaxations at $q=5$ [16]. - **Checkerboard moment obstructions and continuum dual covers:** On a single checkerboard parity class $C_\varepsilon = \{(X,Y) \in G_n : X + Y \equiv \varepsilon \pmod 2\}$, only the $\approx n$ diagonals $X - Y = d$ and $\approx n$ anti-diagonals $X + Y = s$ of the matching parity $d, s \equiv \varepsilon \pmod 2$ meet $C_\varepsilon$, while the remaining $\approx n$ diagonals of each slope are identically empty on $C_\varepsilon$. Exploiting this halved diagonal capacity ($\le 2n - 2$), Prellberg [14] and Aujla, Prellberg, and Sandhu [23, Section 4.3] proved the discrete centered-deficit second-moment bound $D_{\text{mono}}(n, \varepsilon) \le (\sqrt{21}-3)n + 8$ ($\sqrt{21}-3 \approx 1.5825757$) and the continuum-to-discrete dual transfer $D_{\text{mono}}(n, \varepsilon) \le \alpha n + O(1)$. - **Physical secant coverage, finite interior insertions, and secant energy:** Qualitative $o(p^2)$ area bounds for bounded-secant cells of modular hyperbola seeds [17], rooted secant concentration [18], verified finite interior insertions into a square-punctured HJSW seed [19, §8; 20], scalar and per-point pair-conservation laws in randomized line comparison models [21], and a power-saving bound $E_{\text{ord}}(S) = \sum_{\{P,Q\} \in \mathcal{O}(S)} \frac{1}{h(P,Q)} = O(n^{3/2 - 1/212480})$ for ordinary secant energy in row-capped grids via a bound $\#\{\{P,Q\} \in \mathcal{O}(S) : g(P,Q) \ge n^a\} = O_a(n^{2-b})$ on large-gcd pairs [22]. ### Summary of Main Results and Limitations On an unrestricted grid $G_n = \{1,\ldots,n\}^2$ (without checkerboard parity restrictions or modular-arc projections), a clean set $S_n \subseteq G_n$ can meet all $2n-1$ diagonals $X - Y = d$ and all $2n-1$ anti-diagonals $X + Y = s$, so four-direction line-capacity relaxations collapse to the trivial upper bound $2n$ unless the number of $2$-point diagonal and anti-diagonal secants $$P_+(S_n) := \sum_{s=1}^{n-1} R_{S_n}(s, s) = \#\bigl\{ L \in \mathcal{L}_{1,1} : |S_n \cap L| = 2 \bigr\}, \qquad P_-(S_n) := \sum_{s=1}^{n-1} R_{S_n}(s, -s) = \#\bigl\{ L \in \mathcal{L}_{1,-1} : |S_n \cap L| = 2 \bigr\},$$ and $B_{\text{diag}}(S_n) := P_+(S_n) + P_-(S_n)$ are explicitly tracked: on every diagonal or anti-diagonal $L$, $|S_n \cap L| \in \{0, 1, 2\}$ decomposes into $|S_n \cap L| = \mathbf{1}_{\{|S_n \cap L| \ge 1\}} + \mathbf{1}_{\{|S_n \cap L| = 2\}}$, where the first indicator has capacity $1$ on all $2n-1$ lines of each slope and the second indicator has total mass $P_+(S_n)$ and $P_-(S_n)$. In this paper, we establish four self-contained, deterministic mathematical results for **arbitrary (non-modular, unrestricted-grid) clean sets $S_n \subseteq G_n$**: 1. **Exact Radial Fejér Line-Chain Identities, Second-Difference Step-Moment Transforms, and Global $L^1$/$L^2$ Coordinate-Distance Conservation (Section 2, Theorem 1):** For every $n \ge 2$, every primitive integer direction $\hat{z} = (u_0, v_0) \in \mathcal{F}_{n-1}$, and every step threshold $s_0 \ge 0$, we prove that the number of Euclidean lines $L \parallel \hat{z}$ intersecting $G_n$ in at least $s_0 + 1$ points is given **identically** by the radial first difference of the 2D discrete Fejér kernel $R_{G_n}(x,y) = (n-|x|)_+(n-|y|)_+$: $$N_{\hat{z}}(s_0) = R_{G_n}(s_0\hat{z}) - R_{G_n}\bigl((s_0+1)\hat{z}\bigr) = (n - s_0|u_0|)_+(n - s_0|v_0|)_+ - \bigl(n - (s_0+1)|u_0|\bigr)_+\bigl(n - (s_0+1)|v_0|\bigr)_+,$$ and that the number of lines meeting $G_n$ in **exactly** $k \ge 1$ points is the radial second difference $E_{\hat{z}}(k) = R_{G_n}((k-1)\hat{z}) - 2R_{G_n}(k\hat{z}) + R_{G_n}((k+1)\hat{z})$. Consequently, for any clean set $S_n \subseteq G_n$ with directional pair count $P_{\hat{z}} = \sum_{s \ge 1} R_{S_n}(s\hat{z})$, every non-decreasing step weight $w(s)$ ($w(0)=0$) and non-increasing weight $\psi(s) > 0$ obey the exact majorization envelopes $$\sum_{s=1}^{K_{\hat{z}}} w(s)\,R_{S_n}(s\hat{z}) \le \Phi_{\hat{z}}^{(w)}(P_{\hat{z}}) := \sum_{s_0=1}^{K_{\hat{z}}} \bigl(w(s_0) - w(s_0 - 1)\bigr)\min\!\bigl(P_{\hat{z}},\, N_{\hat{z}}(s_0)\bigr),$$ $$\sum_{s=1}^{K_{\hat{z}}} \psi(s)\,R_{S_n}(s\hat{z}) \ge \Psi_{\hat{z}}^{(\psi)}(P_{\hat{z}}) := \psi(1)P_{\hat{z}} - \sum_{s_0=1}^{K_{\hat{z}}-1} \bigl(\psi(s_0) - \psi(s_0 + 1)\bigr)\min\!\bigl(P_{\hat{z}},\, N_{\hat{z}}(s_0 + 1)\bigr).$$ In particular, along the diagonals $\hat{z} = (1, \pm 1)$ we obtain the closed-form identity $\Phi_{(1, \pm 1)}(P_\pm) = (n - 1)P_\pm - \lfloor P_\pm^2 / 4 \rfloor$, and summing the linear ($w(s)=s$) and quadratic ($w(s)=s^2$) envelopes over $\mathcal{F}_{n-1}$ yields exact two-sided global $L^1$ and $L^2$ coordinate-distance conservation inequalities (with lower bounds $\frac{m(m^2-4+3r)}{6}$ and $\frac{m^2(m^2-4)+3r(2m^2-1)}{24}$ for $m = |S_n| = 2q + r$, which are sharp for each 1D coordinate multiset under the multiplicity-two cap and equal $\frac{4}{3}n(n^2-1)$ and $\frac{2}{3}n^2(n^2-1)$ with **exact equality** for every $2n$-point configuration). 2. **Deterministic Large-GCD Pair and Step-Moment Tail Envelopes on Clean Grids (Section 2, Theorem 2):** Combining Theorem 1's exact Fejér ray-tail bound $\sum_{s \ge s_0} R_{S_n}(s\hat{z}) \le \min(\lfloor |S_n|/2 \rfloor, N_{\hat{z}}(s_0))$ with the height-counting mechanism of [22, Section 6] over the truncated Farey fan $\mathcal{F}_{\lfloor (n-1)/s_0 \rfloor}$ yields an explicit finite-$n$ and asymptotic upper bound for every clean set $S_n \subseteq G_n$ and every $1 \le s_0 \le n - 1$: $$\#\bigl\{\{P, Q\} \subset S_n : g(P, Q) \ge s_0\bigr\} \le \sum_{\hat{z} \in \mathcal{F}_{\lfloor (n-1)/s_0 \rfloor}} \min\!\Bigl(\bigl\lfloor |S_n|/2 \bigr\rfloor,\; N_{\hat{z}}(s_0)\Bigr) \le \frac{6\,|S_n|\,n^2}{\pi^2 s_0^2} + O\!\left(\frac{|S_n|\,n \ln(n/s_0 + 2)}{s_0}\right),$$ together with the step-moment tail bound $\sum_{g(P,Q) \ge s_0} g(P,Q) \le \frac{12\,|S_n|\,n^2}{\pi^2 s_0} + O(|S_n|\,n\ln(n/s_0 + 2))$. 3. **Exhaustive 11-Cell Symbolic Positivity of the Parametric Piecewise-Quadratic Family $(A_{u_1}, G_{u_1})$ and Exact Finite-$n$ General-Grid Rearrangement Inequality (Section 3, Lemma 3.2 & Theorem 3):** We construct a one-parameter family of continuous piecewise-quadratic dual potentials $(A_{u_1}(x), G_{u_1}(w))$ for $u_1 \in [1/4, 1/3]$ ($s = 1/2 - u_1$), prove its exact integral identities $4\int_0^1 A_{u_1}(x)\,dx = \frac{(2u_1+1)^3}{12u_1(6u_1+1)}$, $4\int_0^1 G_{u_1}(w)\,dw = \frac{16u_1^2+18u_1+3}{3(6u_1+1)}$, and $I_A(u_1) + I_G(u_1) = 1 + u_1 + \frac{1}{12u_1}$, and establish pointwise positivity $A_{u_1}(x) + A_{u_1}(y) + G_{u_1}(|x-y|) + G_{u_1}(|x+y-1|) \ge 1$ on $[0,1]^2$ via an exhaustive 11-cell symbolic decomposition of the fundamental triangle $0 \le u \le v \le 1 - u$. Using the concave quadratic core $G_{u_1}(w) = g_0 - \frac{c}{2}w^2$ on $[0, u_1]$ and exact endpoint-corrected Riemann-sum bounds valid for **every $n \ge 2$**, we prove that every four-direction admissible set $S_n \subseteq G_n$ with $\eta_n := B_{\text{diag}}(S_n)/(n-1) \le 4u_1$ satisfies $$\frac{|S_n|}{n - 1} \le F_{u_1}(\eta_n) + \frac{8 A_{u_1}(0) + 4 g_0(u_1)}{n - 1}, \qquad F_{u_1}(\beta) := 1 + u_1 + \frac{1}{12u_1} + \left(\frac{2u_1 + 1}{6u_1 + 1}\right)\beta - \frac{\beta^3}{48 u_1 (6u_1 + 1)}.$$ 4. **General-Grid Centered Second-Moment Diagonal-Pair Inequality, Certified Two-Regime Comparison with Theorem 3, and Supporting Finite Material (Section 4, Theorem 4, Remark 4.1 & Proposition 4.2):** Extending the checkerboard order-statistic method of Aujla, Prellberg, and Sandhu [23, Section 4.3] to unrestricted grids $G_n$ with arbitrary diagonal pair counts $(P_+, P_-)$, we prove that the cubic terms in $P_\pm$ cancel identically, yielding the **exact finite-$n$ quadratic-cubic inequality** valid for every $n \ge 2$ and every nonempty four-direction admissible $S_n \subseteq G_n$: $$|S_n|^2 + 6n|S_n| - 12n^2 - 3|S_n|\bigl(P_+ + P_-\bigr) + 3\bigl(P_+^2 + P_-^2\bigr) + 2 \le 0,$$ which is **attained with exact equality ($0 \le 0$)** by the $2n$-point clean configurations at $n = 2$ ($|S_2|=4, P_+=P_-=1$) and $n = 4$ ($|S_4|=8, \{P_+, P_-\}=\{3, 1\}$), and yields the asymptotic bound $\gamma < H(\beta_{\text{diag}}) := \frac{1}{2}\bigl(3\beta_{\text{diag}} - 6 + \sqrt{84 - 36\beta_{\text{diag}} + 3\beta_{\text{diag}}^2}\bigr)$, i.e., $\beta_{\text{diag}} > \gamma - \sqrt{\frac{24 - 12\gamma + \gamma^2}{3}}$. In Remark 4.1, we supply an exact derivative-positivity and rational Sturm-sequence certificate proving that $\min_{u_1 \in [1/4,1/3]} F_{u_1}(\beta) - H(\beta)$ is strictly increasing on $[0, 2 - \frac{2}{\sqrt{3}}]$ with a **unique crossing $\beta_{\text{cross}} \in [0.08794, 0.08795]$** ($\beta_{\text{cross}} \approx 0.0879423$, $\gamma_{\text{cross}} \approx 1.627947$; for fixed $u_1^* = \frac{1}{2\sqrt{3}}$, the unique crossing is $\beta_{\text{cross}}^* \in [0.08542, 0.08543]$). Thus Theorem 3 is strictly sharper on $[0, \beta_{\text{cross}})$, while Theorem 4 is strictly sharper throughout $(\beta_{\text{cross}}, 2 - \frac{2}{\sqrt{3}}]$—forcing $B_{\text{diag}}(S_n) > \frac{9 - \sqrt{47}}{5}n \approx \mathbf{0.42886908\,n}$ (and total $\mathcal{F}_1$ pairs $> \frac{17 - \sqrt{47}}{5}n \approx \mathbf{2.02886908\,n}$) for every $n \ge 2$ whenever $|S_n| \ge 1.8n$, and $B_{\text{diag}}(S_n) \ge 2n - \sqrt{\frac{4n^2 - 4 - 3(P_+ - P_-)^2}{3}} > \left(2 - \frac{2}{\sqrt{3}}\right)n \approx \mathbf{0.84529946\,n}$ when $|S_n| = 2n$. Finally, in Proposition 4.2 we record supporting material on the $1$-indexed HJSW seed $S_{\text{punc}}$ [5, 7], the direction-matching bound $|T| \le 2\lfloor (D+1)/2 \rfloor$, and 19 determinant-verified finite augmented point sets ($p \in [11, 229]$). **Explicit Scope Statement:** None of the results in this paper constructs an unbounded family of no-three-in-line sets exceeding $(3/2 + \varepsilon)n$ or $1.8n$, nor does any result establish an impossibility bound below $2n$ for unrestricted no-three-in-line sets. The $1$-indexed square-punctured seed $S_{\text{punc}}$ is an exact coordinate reformulation of the classical HJSW construction [5, 7], and the 19 augmented point sets in Section 4 are finite feasible witnesses (analogous to [19, §8] at smaller primes) that do not establish an asymptotic density gain over $3/2$. --- ## 2. Exact Radial Fejér Majorization, Second-Difference Step-Moment Envelopes, and Large-GCD Pair Bounds on General Grids Let $G_n = \{1,\ldots,n\}^2$ ($n \ge 2$). For any integer $H \ge 1$, we define the *Farey fan of height $H$* to be the set of primitive integer direction vectors $$\mathcal{F}_H = \left\{ (u_0, v_0) \in \mathbb{Z}^2 : \gcd(|u_0|, |v_0|) = 1, \;\; 1 \le \max(|u_0|, |v_0|) \le H, \;\; \text{and } \bigl(u_0 > 0 \text{ or } (u_0, v_0) = (0,1)\bigr) \right\},$$ which has exact cardinality $|\mathcal{F}_H| = 4\sum_{r=1}^H \varphi(r) = \frac{12}{\pi^2} H^2 + O(H \ln(H + 1))$ (with $|\mathcal{F}_1| = 4\varphi(1) = 4$, corresponding to $(1,0), (0,1), (1,1), (1,-1)$). For any subset $S_n \subseteq G_n$, every unordered pair of distinct points $\{P, Q\} \subset S_n$ has a unique representation $Q - P = \pm s\,\hat{z}$, where $\hat{z} = (u_0, v_0) \in \mathcal{F}_{n-1}$ is its primitive direction, $s = g(P, Q) := \gcd(|X_Q - X_P|, |Y_Q - Y_P|) \ge 1$ is its lattice step length, and $h(P, Q) := \max(|u_0|, |v_0|) = \frac{\|Q - P\|_\infty}{g(P, Q)}$ is its primitive direction height. Define the directional pair autocorrelation $$R_{S_n}(s\hat{z}) := \#\bigl\{ (P, Q) \in S_n^2 : Q - P = s\hat{z} \bigr\}, \qquad s \ge 1, \;\; \hat{z} \in \mathcal{F}_{n-1},$$ and let $$R_{G_n}(x, y) := (n - |x|)_+ (n - |y|)_+ = \int_{\mathbb{T}^2} \left(\frac{\sin(\pi n \xi)}{\sin(\pi \xi)}\right)^2 \left(\frac{\sin(\pi n \eta)}{\sin(\pi \eta)}\right)^2 e^{-2\pi i (x\xi + y\eta)}\,d\xi\,d\eta$$ denote the **2D Discrete Fejér Kernel** (the autocorrelation of the full grid $G_n$, where $z_+ = \max(0, z)$). ### Theorem 1 (Exact Radial Fejér Line-Chain Identities, General Step-Moment Envelopes, and Global $L^1$/$L^2$ Conservation) *Let $n \ge 2$ and let $\hat{z} = (u_0, v_0) \in \mathcal{F}_{n-1}$ be any primitive lattice direction with maximum norm $h(\hat{z}) = \max(|u_0|, |v_0|)$ and step cutoff $K_{\hat{z}} = \lfloor (n-1)/h(\hat{z}) \rfloor$. Then:* 1. *(Exact First- and Second-Difference Line-Chain Identities)* *For every integer step threshold $s_0 \ge 0$, the number of Euclidean lines $L \parallel \hat{z}$ intersecting $G_n$ in at least $s_0 + 1$ points is given **identically** by the radial first difference of the 2D Fejér kernel:* $$N_{\hat{z}}(s_0) := \#\bigl\{ L \parallel \hat{z} : |L \cap G_n| \ge s_0 + 1 \bigr\} = R_{G_n}(s_0 \hat{z}) - R_{G_n}\bigl((s_0 + 1)\hat{z}\bigr),$$ *which equals $n(|u_0| + |v_0|) - (2s_0 + 1)|u_0||v_0|$ whenever $n \ge (s_0 + 1)h(\hat{z})$. Consequently, for every integer $k \ge 1$, the number of lines $L \parallel \hat{z}$ intersecting $G_n$ in **exactly** $k$ lattice points is the radial second difference* $$E_{\hat{z}}(k) := \#\bigl\{ L \parallel \hat{z} : |L \cap G_n| = k \bigr\} = N_{\hat{z}}(k - 1) - N_{\hat{z}}(k) = R_{G_n}\bigl((k - 1)\hat{z}\bigr) - 2 R_{G_n}(k\hat{z}) + R_{G_n}\bigl((k + 1)\hat{z}\bigr),$$ *and the quadratic chain-excess moment of $G_n$ along $\hat{z}$ obeys the exact Fejér identity* $$\sum_{L \parallel \hat{z}} \bigl(|L \cap G_n| - 1\bigr)_+^2 = \sum_{s_0 = 1}^{K_{\hat{z}}} (2 s_0 - 1) N_{\hat{z}}(s_0) = 2 \sum_{s = 1}^{K_{\hat{z}}} R_{G_n}(s\hat{z}) - R_{G_n}(\hat{z}).$$ 2. *(Exact Ray-Tail, Short-Step Lower Bound, and General Step-Moment Envelopes)* *Let $S_n \subseteq G_n$ be any set with at most two points on every line parallel to $\hat{z}$, and let $P_{\hat{z}} := \sum_{s=1}^{K_{\hat{z}}} R_{S_n}(s\hat{z}) \le \min(\lfloor |S_n|/2 \rfloor, N_{\hat{z}}(1))$ denote the number of pairs of $S_n$ in direction $\hat{z}$. Then for every integer $1 \le s_0 \le K_{\hat{z}}$,* $$\sum_{s = s_0}^{K_{\hat{z}}} R_{S_n}(s \hat{z}) \le \min\!\bigl(P_{\hat{z}},\; N_{\hat{z}}(s_0)\bigr), \qquad \sum_{s = 1}^{s_0 - 1} R_{S_n}(s \hat{z}) \ge \max\!\bigl(0,\; P_{\hat{z}} - N_{\hat{z}}(s_0)\bigr), \qquad \sum_{s = s_0}^{K_{\hat{z}}} (s - s_0 + 1) R_{S_n}(s \hat{z}) \le R_{G_n}(s_0 \hat{z}).$$ *More generally, for any non-decreasing function $w : \{0,1,\ldots,K_{\hat{z}}\} \to \mathbb{R}$ with $w(0) = 0$ and any non-increasing function $\psi : \{1,\ldots,K_{\hat{z}}\} \to [0, \infty)$,* $$\sum_{s = 1}^{K_{\hat{z}}} w(s)\,R_{S_n}(s\hat{z}) \le \Phi_{\hat{z}}^{(w)}(P_{\hat{z}}) := \sum_{s_0 = 1}^{K_{\hat{z}}} \bigl(w(s_0) - w(s_0 - 1)\bigr)\min\!\bigl(P_{\hat{z}},\; N_{\hat{z}}(s_0)\bigr),$$ $$\sum_{s = 1}^{K_{\hat{z}}} \psi(s)\,R_{S_n}(s\hat{z}) \ge \Psi_{\hat{z}}^{(\psi)}(P_{\hat{z}}) := \psi(1) P_{\hat{z}} - \sum_{s_0 = 1}^{K_{\hat{z}} - 1} \bigl(\psi(s_0) - \psi(s_0 + 1)\bigr)\min\!\bigl(P_{\hat{z}},\; N_{\hat{z}}(s_0 + 1)\bigr).$$ *In particular, for the linear step weight $w(s) = s$ on the diagonal directions $\hat{z} = (1, \pm 1)$ (where $N_{(1,\pm 1)}(s_0) = 2(n - s_0) - 1$ for $1 \le s_0 \le n - 1$), the concave step-moment envelope $\Phi_{\hat{z}}(P_{\hat{z}}) := \Phi_{\hat{z}}^{(s)}(P_{\hat{z}})$ admits the exact closed form* $$\sum_{s = 1}^{n - 1} s\,R_{S_n}(s, \pm s) \le \Phi_{(1, \pm 1)}(P_\pm) = (n - 1) P_\pm - \left\lfloor \frac{P_\pm^2}{4} \right\rfloor \qquad \bigl(0 \le P_\pm \le 2n - 3\bigr).$$ 3. *(Global Two-Sided $L^1$ and $L^2$ Coordinate-Distance Conservation)* *For every no-three-in-line set $S_n \subseteq G_n$ of size $m = |S_n| = 2q + r$ ($q = \lfloor m/2 \rfloor, r \in \{0, 1\}$), summing the linear ($w(s) = s$) and quadratic ($w(s) = s^2$) step-moment envelopes over all $\hat{z} \in \mathcal{F}_{n-1}$ yields the exact two-sided conservation inequalities* $$\frac{m\bigl(m^2 - 4 + 3r\bigr)}{6} \le \sum_{\{P,Q\} \subset S_n} \|Q - P\|_1 = \sum_{\hat{z} \in \mathcal{F}_{n-1}} (|u_0| + |v_0|) \sum_{s = 1}^{K_{\hat{z}}} s\,R_{S_n}(s\hat{z}) \le \sum_{\hat{z} \in \mathcal{F}_{n-1}} (|u_0| + |v_0|)\,\Phi_{\hat{z}}^{(s)}(P_{\hat{z}}),$$ $$\frac{m^2(m^2 - 4) + 3r(2m^2 - 1)}{24} \le \sum_{\{P,Q\} \subset S_n} \|Q - P\|_2^2 = \sum_{\hat{z} \in \mathcal{F}_{n-1}} (u_0^2 + v_0^2) \sum_{s = 1}^{K_{\hat{z}}} s^2\,R_{S_n}(s\hat{z}) \le \sum_{\hat{z} \in \mathcal{F}_{n-1}} (u_0^2 + v_0^2)\,\Phi_{\hat{z}}^{(s^2)}(P_{\hat{z}}).$$ *For odd $m = 2q + 1$ ($r = 1$), the $L^2$ lower bound equals $\frac{(m^2 - 1)(m^2 + 3)}{24} = \frac{2q(q+1)(q^2 + q + 1)}{3}$. Both lower bounds are sharp for each 1D coordinate multiset under the row/column multiplicity-two constraint for every $m \ge 2$ (and are attained simultaneously by all-slopes clean sets at $m = 3$, e.g. $\{(1,1),(1,2),(2,1)\}$, and at $m = 2n$ whenever a $2n$-point configuration exists, where they equal $\frac{4}{3}n(n^2 - 1)$ and $\frac{2}{3}n^2(n^2 - 1)$ with **exact equality**).* #### Proof **Part 1:** Since $\gcd(|u_0|, |v_0|) = 1$, for each Euclidean line $L \parallel \hat{z}$ meeting $G_n$, the intersection $L \cap G_n = \{P_0 + j\hat{z} : 0 \le j \le k_L - 1\}$ is a contiguous lattice chain of length $k_L = |L \cap G_n| \ge 1$. For any integer $s_0 \ge 0$, a point $P \in L \cap G_n$ satisfies $P + s_0\hat{z} \in G_n$ if and only if $P = P_0 + j\hat{z}$ with $0 \le j \le k_L - 1 - s_0$, so there are $\max(0, k_L - s_0)$ such points on $L$. Summing over all lines $L \parallel \hat{z}$ partitions $G_n$, giving the exact identity $$\sum_{L \parallel \hat{z}} \max\bigl(0,\, |L \cap G_n| - s_0\bigr) = \#\bigl\{ P \in G_n : P + s_0\hat{z} \in G_n \bigr\} = (n - s_0|u_0|)_+(n - s_0|v_0|)_+ = R_{G_n}(s_0\hat{z}).$$ Taking the first difference between $s_0$ and $s_0 + 1$ and using $\max(0, k_L - s_0) - \max(0, k_L - s_0 - 1) = \mathbf{1}_{\{k_L \ge s_0 + 1\}}$ establishes $N_{\hat{z}}(s_0) = R_{G_n}(s_0\hat{z}) - R_{G_n}((s_0+1)\hat{z})$. Taking a second difference gives $E_{\hat{z}}(k) = N_{\hat{z}}(k-1) - N_{\hat{z}}(k)$. Finally, since $(k_L - 1)_+^2 = \sum_{s_0=1}^{K_{\hat{z}}} (2s_0 - 1)\mathbf{1}_{\{k_L \ge s_0 + 1\}}$, summing over $L \parallel \hat{z}$ and substituting $N_{\hat{z}}(s_0) = R_{G_n}(s_0\hat{z}) - R_{G_n}((s_0+1)\hat{z})$ telescopes to $2\sum_{s=1}^{K_{\hat{z}}} R_{G_n}(s\hat{z}) - R_{G_n}(\hat{z})$. **Part 2:** Since $|S_n \cap L| \le 2$ for every $L \parallel \hat{z}$, each line contributes at most one pair $\{P, Q\} \subset S_n \cap L$, and a line can support a pair with step distance $s = (Q - P)/\hat{z} \ge s_0$ only if $|L \cap G_n| = k_L \ge s + 1 \ge s_0 + 1$. Thus $\sum_{s \ge s_0} R_{S_n}(s\hat{z}) \le \min(P_{\hat{z}}, N_{\hat{z}}(s_0))$, and subtracting from $P_{\hat{z}} = \sum_{s=1}^{K_{\hat{z}}} R_{S_n}(s\hat{z})$ gives $\sum_{s=1}^{s_0-1} R_{S_n}(s\hat{z}) \ge \max(0, P_{\hat{z}} - N_{\hat{z}}(s_0))$. For any non-decreasing $w(s)$ with $w(0) = 0$, writing $w(s) = \sum_{s_0=1}^s (w(s_0) - w(s_0 - 1))$ with $w(s_0) - w(s_0 - 1) \ge 0$ and interchanging the summation order gives $$\sum_{s=1}^{K_{\hat{z}}} w(s)\,R_{S_n}(s\hat{z}) = \sum_{s_0=1}^{K_{\hat{z}}} \bigl(w(s_0) - w(s_0 - 1)\bigr) \sum_{s = s_0}^{K_{\hat{z}}} R_{S_n}(s\hat{z}) \le \sum_{s_0=1}^{K_{\hat{z}}} \bigl(w(s_0) - w(s_0 - 1)\bigr)\min\!\bigl(P_{\hat{z}},\, N_{\hat{z}}(s_0)\bigr).$$ Similarly, for any non-increasing $\psi(s)$, writing $\psi(s) = \psi(1) - \sum_{s_0=1}^{s-1} (\psi(s_0) - \psi(s_0 + 1))$ with $\psi(s_0) - \psi(s_0 + 1) \ge 0$ gives the lower envelope $\Psi_{\hat{z}}^{(\psi)}(P_{\hat{z}})$. When $\hat{z} = (1, \pm 1)$, $N_{(1,\pm 1)}(s_0) = (n - s_0)^2 - (n - s_0 - 1)^2 = 2(n - s_0) - 1$ for $1 \le s_0 \le n - 1$, which takes every odd value $1, 3, 5, \ldots, 2n - 3$ as $t = n - s_0$ runs from $1$ to $n - 1$. For $0 \le P_\pm \le 2n - 3$, setting $q = \lfloor P_\pm / 2 \rfloor$, we have $2t - 1 \le P_\pm \iff 1 \le t \le q$, so $$\Phi_{(1, \pm 1)}(P_\pm) = \sum_{t=1}^q (2t - 1) + \sum_{t = q + 1}^{n - 1} P_\pm = q^2 + (n - 1 - q)P_\pm = (n - 1)P_\pm - q(P_\pm - q) = (n - 1)P_\pm - \left\lfloor \frac{P_\pm^2}{4} \right\rfloor.$$ **Part 3:** Every pair $\{P, Q\} \subset S_n$ with $Q - P = s(u_0, v_0)$ satisfies $\|Q - P\|_1 = s(|u_0| + |v_0|)$ and $\|Q - P\|_2^2 = s^2(u_0^2 + v_0^2)$, so summing the directional moments over $\hat{z} \in \mathcal{F}_{n-1}$ gives the middle identities and upper bounds. For the lower bounds, let $x_1 \le x_2 \le \cdots \le x_m$ be the sorted $x$-coordinates of $S_n$. Because each column $X \in \{1,\ldots,n\}$ contains at most $2$ points of $S_n$, shifting any occupied column toward an unoccupied gap strictly decreases $x_j - x_i \ge 0$ for every pair separated by the gap while leaving all other pairs unchanged. Hence both $\sum_{i 0, \qquad g_0 := 1 - 2 c u_1^2 = \frac{2u_1 + 1}{6u_1 + 1} \in \left[\frac{5}{9},\, \frac{3}{5}\right].$$ Define $A_{u_1} : [0, 1] \to [0, \infty)$ by $$A_{u_1}(x) := \frac{c}{4}\max\bigl(0,\, |2x - 1| - s\bigr)^2, \qquad x \in [0, 1],$$ and define $G_{u_1} : [0, 1] \to [0, \infty)$ by the four-piece formula $$G_{u_1}(w) := \begin{cases} G_1(w) := g_0 - \frac{c}{2} w^2, & 0 \le w \le u_1, \\[4pt] G_2(w) := g_0 + \frac{c}{2} u_1^2 - c u_1 w, & u_1 \le w \le \frac{1}{2}, \\[4pt] G_3(w) := (1 - g_0) - \frac{c}{2}(w - s)^2, & \frac{1}{2} \le w \le \frac{1}{2} + u_1, \\[4pt] G_4(w) := 0, & \frac{1}{2} + u_1 \le w \le 1. \end{cases}$$ #### Lemma 3.2 (Regularity, Monotonicity, and Exact Integrals of $(A_{u_1}, G_{u_1})$) *For every $u_1 \in [1/4, 1/3]$:* 1. *$A_{u_1} \in C^1([0, 1])$ is symmetric ($A_{u_1}(1-x) = A_{u_1}(x)$), non-increasing on $[0, 1/2]$, non-decreasing on $[1/2, 1]$, vanishes on $\bigl[\frac{1-s}{2}, \frac{1+s}{2}\bigr] = \bigl[\frac{1}{4} + \frac{u_1}{2},\, \frac{3}{4} - \frac{u_1}{2}\bigr]$, and has endpoint value $A_{u_1}(0) = A_{u_1}(1) = \frac{c}{4}(1 - s)^2 = \frac{(2u_1+1)^2}{8u_1(6u_1+1)}$ and total variation $V_{[0,1]}(A_{u_1}) = 2 A_{u_1}(0)$.* 2. *$G_{u_1} \in C^0([0, 1])$ is continuous at all three breakpoints $w \in \{u_1, \frac{1}{2}, \frac{1}{2} + u_1\}$, is $C^1$ at $w = u_1$ and $w = \frac{1}{2}$ ($G_1'(u_1) = G_2'(u_1) = G_2'(\frac{1}{2}^-) = G_3'(\frac{1}{2}^+) = -c u_1$), and has piecewise-continuous derivative $G_{u_1}'(w) \le 0$ on every open piece (with $G_3'(\frac{1}{2}+u_1^-) = -2 c u_1 < 0 = G_4'(\frac{1}{2}+u_1^+)$). Consequently, $G_{u_1}$ is strictly decreasing and nonnegative on $[0, \frac{1}{2} + u_1]$, identically zero on $[\frac{1}{2} + u_1, 1]$, and has total variation $V_{[0,1]}(G_{u_1}) = G_{u_1}(0) = g_0$.* 3. *The exact integrals $I_A(u_1) := 4\int_0^1 A_{u_1}(x)\,dx$ and $I_G(u_1) := 4\int_0^1 G_{u_1}(w)\,dw$ are* $$I_A(u_1) = \frac{(2u_1 + 1)^3}{12 u_1 (6u_1 + 1)}, \qquad I_G(u_1) = \frac{16u_1^2 + 18u_1 + 3}{3(6u_1 + 1)}, \qquad I(u_1) := I_A(u_1) + I_G(u_1) = 1 + u_1 + \frac{1}{12u_1}.$$ *In particular, $I(3/10) = \frac{71}{45} = 1.5\overline{7}$, and $I(u_1)$ attains its unique minimum on $[1/4, 1/3]$ at $u_1^* = \frac{1}{2\sqrt{3}} \approx 0.288675$, where $I(u_1^*) = 1 + \frac{1}{\sqrt{3}} \approx 1.57735027$.* #### Proof In Part 1, $|2x - 1| \le s \iff x \in [\frac{1-s}{2}, \frac{1+s}{2}] = [\frac{1}{4}+\frac{u_1}{2}, \frac{3}{4}-\frac{u_1}{2}]$. In Part 2, continuity at $w = u_1$ holds since $G_1(u_1) = g_0 - \frac{c}{2}u_1^2 = G_2(u_1)$; at $w = \frac{1}{2}$ since $G_2(\frac{1}{2}) - G_3(\frac{1}{2}) = 2g_0 - 1 + cu_1^2 - \frac{c}{2}u_1 = 1 - \frac{c}{2}u_1(6u_1 + 1) = 0$; and at $w = \frac{1}{2} + u_1 = s + 2u_1$ since $G_3(s + 2u_1) = 2cu_1^2 - \frac{c}{2}(2u_1)^2 = 0 = G_4(s + 2u_1)$. On $(0, u_1)$, $G_1'(w) = -cw < 0$; on $(u_1, 1/2)$, $G_2'(w) = -cu_1 < 0$; and on $(1/2, 1/2+u_1)$, $G_3'(w) = -c(w-s) < 0$. In Part 3, by symmetry about $x = 1/2$, $$I_A(u_1) = 8\int_0^{(1-s)/2} \frac{c}{4}(1 - s - 2x)^2\,dx = c\left[-\frac{(1 - s - 2x)^3}{3}\right]_0^{(1-s)/2} = \frac{c(1 - s)^3}{3} = \frac{c(2u_1 + 1)^3}{24} = \frac{(2u_1 + 1)^3}{12u_1(6u_1 + 1)}.$$ Similarly, integrating $G_1, G_2, G_3$ over $[0, u_1]$, $[u_1, 1/2]$, and $[1/2, 1/2+u_1]$ yields $$\int_0^{u_1} G_1(w)\,dw = g_0 u_1 - \frac{c}{6}u_1^3, \qquad \int_{u_1}^{1/2} G_2(w)\,dw = \left(\frac{1}{2} - u_1\right)\left(g_0 + \frac{c}{2}u_1^2 - \frac{c}{2}u_1\left(\frac{1}{2} + u_1\right)\right),$$ $$\int_{1/2}^{1/2+u_1} G_3(w)\,dw = (1 - g_0)u_1 - \frac{c}{6}\bigl((2u_1)^3 - u_1^3\bigr) = 2cu_1^3 - \frac{7}{6}cu_1^3 = \frac{5}{6}cu_1^3.$$ Summing these three integrals and multiplying by $4$ gives $I_G(u_1) = \frac{16u_1^2 + 18u_1 + 3}{3(6u_1 + 1)}$, and adding $I_A(u_1)$ gives $$I_A(u_1) + I_G(u_1) = \frac{(8u_1^3 + 12u_1^2 + 6u_1 + 1) + 4u_1(16u_1^2 + 18u_1 + 3)}{12u_1(6u_1 + 1)} = \frac{72u_1^3 + 84u_1^2 + 18u_1 + 1}{12u_1(6u_1 + 1)} = \frac{(6u_1 + 1)(12u_1^2 + 12u_1 + 1)}{12u_1(6u_1 + 1)} = 1 + u_1 + \frac{1}{12u_1}. \quad \blacksquare$$ ### Theorem 3 (Exhaustive 11-Cell Symbolic Positivity and Exact Finite-$n$ General-Grid Rearrangement Inequality) 1. *(Exhaustive Pointwise Positivity)* *For every $u_1 \in [1/4, 1/3]$ and every $(x, y) \in [0, 1]^2$,* $$S_{u_1}(x, y) := A_{u_1}(x) + A_{u_1}(y) + G_{u_1}(|x - y|) + G_{u_1}(|x + y - 1|) \ge 1.$$ 2. *(Exact Finite-$n$ and Asymptotic Diagonal-Pair Rearrangement Inequality on General Grids)* *Let $S_n \subseteq G_n = \{1,\ldots,n\}^2$ ($n \ge 2$) be **any** four-direction admissible set (in particular, any no-three-in-line set, without any modular or checkerboard hypothesis), and let* $$B_{\text{diag}}(S_n) := P_+(S_n) + P_-(S_n) = \#\bigl\{ L \in \mathcal{L}_{1,1} \cup \mathcal{L}_{1,-1} : |S_n \cap L| = 2 \bigr\}$$ *denote the total number of $2$-point diagonal and anti-diagonal lines of $S_n$. Let $w_{(1)} \le w_{(2)} \le \cdots \le w_{(4n-2)}$ denote the non-decreasing rearrangement of the $4n - 2$ normalized diagonal and anti-diagonal offsets $\bigl\{\frac{|d|}{n-1} : |d| \le n-1\bigr\} \sqcup \bigl\{\frac{|s - (n+1)|}{n-1} : 2 \le s \le 2n\bigr\}$. Then for every $n \ge 2$ and every $u_1 \in [1/4, 1/3]$:* $$|S_n| \le 4\sum_{X=1}^n A_{u_1}\!\left(\frac{X-1}{n-1}\right) + 2\sum_{d=-(n-1)}^{n-1} G_{u_1}\!\left(\frac{|d|}{n-1}\right) + \sum_{j=1}^{B_{\text{diag}}(S_n)} G_{u_1}\!\left(w_{(j)}\right),$$ *and whenever $\eta_n := \frac{B_{\text{diag}}(S_n)}{n - 1} \le 4u_1$, exact monotone Riemann-sum comparison yields the finite-$n$ inequality valid for **every $n \ge 2$**:* $$\frac{|S_n|}{n - 1} \le F_{u_1}(\eta_n) + \frac{8 A_{u_1}(0) + 4 g_0(u_1)}{n - 1}, \qquad F_{u_1}(\beta) := 1 + u_1 + \frac{1}{12u_1} + \left(\frac{2u_1 + 1}{6u_1 + 1}\right)\beta - \frac{\beta^3}{48 u_1 (6u_1 + 1)}.$$ *At $u_1^* = \frac{1}{2\sqrt{3}}$, the explicit remainder constant is $8A_{u_1^*}(0) + 4g_0(u_1^*) = 5\sqrt{3} - 1 \approx 7.66025$, giving for every $n \ge 2$ and every $\eta_n \le \frac{2}{\sqrt{3}} \approx 1.15470$:* $$\frac{|S_n|}{n - 1} \le 1 + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}}\eta_n - \frac{3 - \sqrt{3}}{48}\eta_n^3 + \frac{5\sqrt{3} - 1}{n - 1}.$$ #### Proof **Part 1 (Exhaustive 11-Cell Symbolic Proof):** By the dihedral $D_4$ symmetry of $[0, 1]^2$ under $x \leftrightarrow 1 - x$, $y \leftrightarrow 1 - y$, and $x \leftrightarrow y$, it suffices to prove $S_{u_1}(x, y) \ge 1$ on the fundamental octant where $u := |x - y| = y - x \ge 0$ and $v := |x + y - 1| = 1 - x - y \ge u$, which maps $[0,1]^2$ onto the fundamental triangle $$T = \left\{ (u, v) \in \mathbb{R}^2 : 0 \le u \le \frac{1}{2}, \;\; u \le v \le 1 - u \right\},$$ with $|2x - 1| = u + v$ and $|2y - 1| = v - u$. Thus $A_{u_1}(x) = \frac{c}{4}\max(0, u+v-s)^2$ and $A_{u_1}(y) = \frac{c}{4}\max(0, v-u-s)^2$. For $u_1 \in [1/4, 1/3]$, the breakpoints satisfy $0 < s = \frac{1}{2} - u_1 \le u_1 \le \frac{1}{2} < \frac{1}{2} + u_1 \le \frac{5}{6} < 1$. On $T$, $u \in [0, 1/2]$ lies in piece $i \in \{1, 2\}$ of $G_{u_1}(u)$, $v \in [u, 1-u]$ lies in piece $j \in \{i,\ldots,4\}$ of $G_{u_1}(v)$, and the activation lines $u + v = s$ ($\sigma_+ \in \{0, 1\}$) and $v - u = s$ ($\sigma_- \in \{0, 1\}$) partition $T$ into pure polynomial sub-cells $(i, j, \sigma_+, \sigma_-)$. Because $u + v \ge u_1 \ge s$ whenever $i = 2$ or $j \ge 2$, $\sigma_+ = 0$ occurs only when $i = j = 1$ (and forces $v - u \le u + v \le s$, so $\sigma_- = 0$). Similarly, $(i, j) = (2, 4)$ is empty on $T$ (aside from $(u, v) = (1/4, 3/4)$ at $u_1 = 1/4$) since $u \ge u_1$ and $v \ge 1/2 + u_1 \implies u + v \ge 1/2 + 2u_1 \ge 1$, while for $i = 1$ and $j \in \{3, 4\}$ we have $v - u \ge 1/2 - u_1 = s$, so $\sigma_- = 1$. This leaves **exactly 11 non-empty polynomial cells** covering $T$, on which $\Delta(u, v) := (S_{u_1}(x, y) - 1)/c$ simplifies (verified symbolically in `SymPy` in the supplement) as follows: 1. **Six Explicit Sum-of-Squares / Zero Cells:** - *Cell $(1, 3, 1, 1)$ ($u \in [0, u_1], v \in [1/2, 1/2+u_1]$):* $\Delta(u, v) = 0$ identically. - *Cell $(1, 2, 1, 1)$ ($u \in [0, u_1], v \in [u_1, 1/2], v - u \ge s$):* $\Delta(u, v) = \frac{1}{2}\left(v - \frac{1}{2}\right)^2 \ge 0$. - *Cell $(1, 4, 1, 1)$ ($u \in [0, u_1], v \in [1/2+u_1, 1-u]$):* $\Delta(u, v) = \frac{1}{2}(v - s)^2 - 2u_1^2 \ge \frac{1}{2}(2u_1)^2 - 2u_1^2 = 0$ since $v \ge \frac{1}{2} + u_1 = s + 2u_1$. - *Cell $(2, 2, 1, 0)$ ($u, v \in [u_1, 1/2], v - u \le s$):* $\Delta(u, v) = \frac{1}{4}\left(u + v - \frac{1}{2} - u_1\right)^2 \ge 0$. - *Cell $(2, 2, 1, 1)$ ($u, v \in [u_1, 1/2], v - u \ge s$):* $\Delta(u, v) = \frac{1}{2}(u - u_1)^2 + \frac{1}{2}\left(v - \frac{1}{2}\right)^2 \ge 0$. - *Cell $(2, 3, 1, 1)$ ($u \in [u_1, 1/2], v \in [1/2, 1-u], v - u \ge s$):* $\Delta(u, v) = \frac{1}{2}(u - u_1)^2 \ge 0$. 2. **Two Monotonic Algebraic Cells:** - *Cell $(1, 1, 0, 0)$ ($0 \le u \le v, u + v \le s$):* $\Delta(u, v) = \frac{1}{2}u_1 - u_1^2 - \frac{1}{2}(u^2 + v^2) \ge \frac{1}{2}u_1 - u_1^2 - \frac{1}{2}s^2 = u_1 - \frac{3}{2}u_1^2 - \frac{1}{8} = \frac{1}{24} - \frac{3}{2}\left(u_1 - \frac{1}{3}\right)^2 \ge \frac{1}{32} > 0$ for all $u_1 \in [1/4, 1/3]$. - *Cell $(1, 1, 1, 1)$ ($0 \le u \le v \le u_1, v - u \ge s$):* $\Delta(u, v) = \frac{1}{8} - \left(\frac{1}{2} - u_1\right)v - \frac{1}{2}u_1^2$ is decreasing in $v \le u_1$, so $\Delta(u, v) \ge \Delta(u, u_1) = \frac{1}{2}\left(u_1 - \frac{1}{2}\right)^2 \ge \frac{1}{72} > 0$. 3. **Three Concave Quadratic Cells (Boundary Reduction):** - *Cell $(1, 1, 1, 0)$ ($0 \le u \le v \le u_1, u + v \ge s, v - u \le s$):* Here $\frac{\partial^2 \Delta}{\partial u^2} = -\frac{1}{2} < 0$, so for each fixed $v \in [s/2, u_1]$, $\Delta(u, v)$ is strictly concave in $u$ and attains its minimum at the endpoints of its $u$-interval: the left endpoint $u = s - v$ (matching Cell $(1,1,0,0)$ where $\Delta > 0$) or $u = v - s$ (matching Cell $(1,1,1,1)$ where $\Delta > 0$), and the right endpoint $u = v$, where $\Delta(v, v) = \frac{1}{16} + \frac{1}{4}u_1 - \frac{3}{4}u_1^2 - \left(\frac{1}{2} - u_1\right)v$ is linear in $v \in [s/2, u_1]$ with strictly positive values at both endpoints: $\Delta(s/2, s/2) = \frac{(1 - 2u_1)(10u_1 - 1)}{16} > 0$ and $\Delta(u_1, u_1) = \frac{(1 - 2u_1)^2}{16} > 0$. - *Cell $(1, 2, 1, 0)$ ($u \in [0, u_1], v \in [u_1, 1/2], v - u \le s$):* Here $\frac{\partial^2 \Delta}{\partial u^2} = -\frac{1}{2} < 0$, so for each $v$, $\Delta(u, v)$ is minimized at its endpoints $u = v - s$ (boundary with Cell $(1,2,1,1)$, where $\Delta = \frac{1}{2}(v - 1/2)^2 \ge 0$) or $u = u_1$ (boundary with Cell $(2,2,1,0)$, where $\Delta = \frac{1}{4}(v - 1/2)^2 \ge 0$). - *Cell $(2, 3, 1, 0)$ ($u \in [u_1, 1/2], v \in [1/2, 1-u], v - u \le s$):* Here $\frac{\partial^2 \Delta}{\partial v^2} = -\frac{1}{2} < 0$, so for each $u$, $\Delta(u, v)$ is minimized at its endpoints $v = 1/2$ (boundary with Cell $(2,2,1,0)$, where $\Delta = \frac{1}{4}(u - u_1)^2 \ge 0$), $v = u + s$ (boundary with Cell $(2,3,1,1)$, where $\Delta = \frac{1}{2}(u - u_1)^2 \ge 0$), or $v = 1 - u$ (for $u \ge \frac{1-s}{2} = \frac{1}{4} + \frac{u_1}{2}$), where $\Delta(u, 1 - u) = \frac{(1 - 2u_1)^2 - 2(1 - 2u)^2}{16} \ge \frac{(1 - 2u_1)^2}{32} > 0$. **Part 2 (Exact Finite-$n$ Rearrangement and Monotone Riemann-Sum Bounds):** For $n \ge 2$, set $m = n - 1 \ge 1$. For $(X, Y) \in S_n \subseteq \{1,\ldots,n\}^2$, let $x = \frac{X-1}{m} \in [0, 1]$ and $y = \frac{Y-1}{m} \in [0, 1]$, so $|x - y| = \frac{|X - Y|}{m}$ and $|x + y - 1| = \frac{|X + Y - (n + 1)|}{m}$. Summing $1 \le S_{u_1}(x, y)$ over all $(X, Y) \in S_n$ and grouping by rows, columns, diagonals $d = X - Y$, and anti-diagonals $s = X + Y$ gives $$|S_n| \le \sum_{X=1}^n r_{1,0}(S_n; X)\,A_{u_1}\!\left(\frac{X-1}{m}\right) + \sum_{Y=1}^n r_{0,1}(S_n; Y)\,A_{u_1}\!\left(\frac{Y-1}{m}\right) + \sum_{L \in \mathcal{L}_{1,1} \cup \mathcal{L}_{1,-1}} |S_n \cap L|\,G_{u_1}(w_L).$$ Since $r_{1,0}(S_n; X) \le 2$ and $r_{0,1}(S_n; Y) \le 2$, the first two sums are bounded by $4\sum_{X=1}^n A_{u_1}\!\left(\frac{X-1}{m}\right)$. On every line $L \in \mathcal{L}_{1,1} \cup \mathcal{L}_{1,-1}$, $|S_n \cap L| \in \{0, 1, 2\}$ satisfies $|S_n \cap L| \le 1 + \mathbf{1}_{\{|S_n \cap L| = 2\}}$. Since $G_{u_1}(w) \ge 0$ is non-increasing on $[0, 1]$ by Lemma 3.2(2), the sum of $G_{u_1}(w_L)$ over all $4n - 2$ lines is $2\sum_{d=-m}^m G_{u_1}\!\left(\frac{|d|}{m}\right)$, and the sum over the $B_{\text{diag}}(S_n)$ lines with $|S_n \cap L| = 2$ is at most $\sum_{j=1}^{B_{\text{diag}}(S_n)} G_{u_1}(w_{(j)})$, establishing the first displayed inequality. To bound the three discrete sums for arbitrary $m = n - 1 \ge 1$: 1. **Axis sum:** Because $A_{u_1}(x)$ is non-increasing on $[0, 1/2]$ and non-decreasing on $[1/2, 1]$ with $A_{u_1}(0) = A_{u_1}(1)$, for every $k \in \{1, \ldots, \lfloor m/2 \rfloor\}$ we have $\frac{1}{m} A_{u_1}(k/m) \le \int_{(k-1)/m}^{k/m} A_{u_1}(x)\,dx$, and symmetrically for $k \in \{\lceil m/2 \rceil, \ldots, m-1\}$ we have $\frac{1}{m} A_{u_1}(k/m) \le \int_{k/m}^{(k+1)/m} A_{u_1}(x)\,dx$ (with the two intervals overlapping only beneath $x = 1/2$ where $A_{u_1} \equiv 0$). Separating the two boundary terms $k = 0$ and $k = m$ therefore yields, for every $m = n - 1 \ge 1$ (including $n = 2$, where both sides equal $2 A_{u_1}(0)$ since $A_{u_1} \ge 0$), $$\sum_{X=1}^n A_{u_1}\!\left(\frac{X-1}{m}\right) = A_{u_1}(0) + A_{u_1}(1) + \sum_{k=1}^{m-1} A_{u_1}\!\left(\frac{k}{m}\right) \le m\int_0^1 A_{u_1}(x)\,dx + 2 A_{u_1}(0).$$ 2. **Full diagonal sum:** Since $G_{u_1}(w)$ is non-increasing on $[0, 1]$, for every $d \in \{1, \ldots, m\}$ we have $\frac{1}{m} G_{u_1}(d/m) \le \int_{(d-1)/m}^{d/m} G_{u_1}(w)\,dw$, so $$\sum_{d=-m}^m G_{u_1}\!\left(\frac{|d|}{m}\right) = G_{u_1}(0) + 2\sum_{d=1}^m G_{u_1}\!\left(\frac{d}{m}\right) \le 2m\int_0^1 G_{u_1}(w)\,dw + g_0(u_1).$$ 3. **Rearranged pair sum:** In the multiset $\{w_{(j)}\}_{j=1}^{4n-2}$, the offset $0$ occurs with multiplicity $2$ (contributing $2 G_{u_1}(0) = 2 g_0$) and each positive offset $d/m$ ($d \ge 1$) occurs with multiplicity $4$. Thus $w_{(j)} \ge \frac{\lceil (j-2)/4 \rceil}{m}$ for all $j \ge 3$. Since $G_{u_1}(w) = g_0 - \frac{c}{2}w^2$ is non-increasing on $[0, u_1]$, for every $B = B_{\text{diag}}(S_n) \le 4 u_1 m$ we have $\frac{1}{4m} G_{u_1}(w_{(j)}) \le \int_{(j-3)/(4m)}^{(j-2)/(4m)} G_{u_1}(w)\,dw$ for each $j \ge 3$, and hence $$\sum_{j=1}^B G_{u_1}\!\left(w_{(j)}\right) \le 2 g_0 + 4m\int_0^{B/(4m)} \left(g_0 - \frac{c}{2}w^2\right)dw = m\left(g_0 \eta_n - \frac{c}{96}\eta_n^3\right) + 2 g_0(u_1).$$ Combining all three estimates and dividing by $m = n - 1$ gives $$\frac{|S_n|}{n - 1} \le I_A(u_1) + I_G(u_1) + g_0\eta_n - \frac{c}{96}\eta_n^3 + \frac{8 A_{u_1}(0) + 4 g_0(u_1)}{n - 1}.$$ Substituting $I_A(u_1) + I_G(u_1) = 1 + u_1 + \frac{1}{12u_1}$ from Lemma 3.2(3), $g_0 = \frac{2u_1+1}{6u_1+1}$, and $\frac{c}{96} = \frac{1}{48u_1(6u_1+1)}$ completes the proof. $\blacksquare$ #### Remark 3.3 (Structural Distinction Between Checkerboard Parity Restrictions, Modular Hyperbola Lifts, and Unrestricted Grids [10, 14, 15, 23]) It is important to distinguish the two different mechanisms by which the leading integral $I_A(u_1) + I_G(u_1) = 1 + u_1 + \frac{1}{12u_1}$ (or Prellberg's sharper constant $\alpha \approx 1.5768234$ [14]) arises: 1. **Checkerboard parity classes ($S_n \subseteq C_\varepsilon$) and modular hyperbola lifts:** A subset $S_n \subseteq C_\varepsilon$ of a single checkerboard colour class can have $B_{\text{diag}}(S_n) > 0$ (for example, $\{(1,1), (2,2)\} \subset C_0$ has $B_{\text{diag}} = 1$). On $C_\varepsilon$, however, all diagonals $X - Y = d$ and anti-diagonals $X + Y = s$ with $d, s \not\equiv \varepsilon \pmod 2$ are **identically empty**, so one assigns capacity $2$ only to the $\approx n$ compatible diagonals of each slope (and weight $0$ to the incompatible half), giving diagonal integral $2 \times 2\int_0^1 G(w)\,dw = I_G$ without an extra pair sum [14, §4.1; 23, §5.3]. Likewise, for a modular hyperbola $(X+A)(Y+B) \equiv m \pmod p$, the discriminant $t^2 \mp 4m$ is a quadratic nonresidue on $\frac{p \mp (\frac{\pm m}{p})}{2} \approx \frac{p}{2}$ residue classes of $X \pm Y \pmod p$, so half of all diagonals in every interval of length $p$ are empty; Proposition 8 of [10] gives the exact short-window bound $|S_n| \le 2n - c_x - c_y - (p-1)$, and [15, §3.3] transfers Prellberg's cover [14] via Weil interval discrepancy. 2. **Unrestricted grids ($S_n \subseteq G_n$):** On the full grid $G_n$, all $2n - 1$ diagonals and $2n - 1$ anti-diagonals are available. Here we decompose $|S_n \cap L| \le 1 + \mathbf{1}_{\{|S_n \cap L| = 2\}}$ across all $4n - 2$ diagonal and anti-diagonal lines: the baseline capacity $1$ on all $4n - 2$ lines has the same leading integral $2 \times 2\int_0^1 G_{u_1}(w)\,dw = I_G(u_1)$, while the $B_{\text{diag}}(S_n)$ two-point lines contribute the explicit concave rearrangement term $g_0 \eta_n - \frac{c}{96}\eta_n^3$ in Theorem 3(2). --- ## 4. General-Grid Centered Second-Moment Inequality, Certified Two-Regime Comparison, and Supporting Finite Material In [23, Section 4.3], Aujla, Prellberg, and Sandhu used a centered deficit identity and order-statistic bounds on a single checkerboard colour class $C_\varepsilon \subset G_n$ (where every compatible diagonal has capacity $2$ and the incompatible diagonals are absent) to prove the discrete linear-deficit bound $D_{\text{mono}}(n, \varepsilon) \le (\sqrt{21} - 3)n + 8$. We now extend the centered second-moment and order-statistic method to **unrestricted grids $G_n = \{1,\ldots,n\}^2$** by decomposing each diagonal and anti-diagonal occupancy into $1$-point and $2$-point indicators, showing that the cubic terms in $P_+(S_n)$ and $P_-(S_n)$ cancel identically and yielding a closed-form finite-$n$ quadratic-cubic inequality that requires no remainder allowance and is attained with **exact equality** at $n = 2$ and $n = 4$. ### Theorem 4 (Exact General-Grid Centered Second-Moment Diagonal-Pair Inequality) *Let $n \ge 2$ and let $S_n \subseteq G_n = \{1,\ldots,n\}^2$ be any nonempty four-direction admissible set (in particular, any nonempty no-three-in-line set) of cardinality $N = |S_n| \in \{1,\ldots,2n\}$, with $P_+ = P_+(S_n)$ two-point $(1,1)$-diagonals, $P_- = P_-(S_n)$ two-point $(1,-1)$-anti-diagonals, and $B_{\text{diag}} = P_+ + P_-$. Then:* 1. *(Exact Centered Second-Moment Identity [23, Remark 3.1])* *Writing $x_0 = \frac{n+1}{2}$, $d(X,Y) = X - Y$, and $s(X,Y) = X + Y - (n + 1)$, every subset $S_n \subseteq G_n$ satisfies* $$\sum_{(X,Y) \in S_n} d(X,Y)^2 + \sum_{(X,Y) \in S_n} s(X,Y)^2 = 2\sum_{(X,Y) \in S_n} \bigl((X - x_0)^2 + (Y - x_0)^2\bigr).$$ 2. *(Exact Order-Statistic Lower Bounds Without Remainder Loss)* *For every integer $0 \le k \le 2n - 1$, the sum $\Lambda_{\min}(k)$ of the $k$ smallest elements of the $(2n-1)$-element set $\{d^2 : -(n-1) \le d \le n-1\}$ satisfies* $$\Lambda_{\min}(k) = \frac{k^3 - k}{12} + \frac{k\bigl(1 - (k \bmod 2)\bigr)}{4} \ge \frac{k^3 - k}{12}.$$ *For every integer $0 \le R \le 2n$, let $\Omega_{\min}(n; R)$ denote the sum of the $R$ smallest elements of the $2n$-element single-axis capacity multiset $\mathcal{M}_n := \{(X - x_0)^2 : 1 \le X \le n\} \sqcup \{(X - x_0)^2 : 1 \le X \le n\}$. Then* $$2\,\Omega_{\min}(n; R) \ge \frac{R^3 - 4 R}{24}, \qquad \text{with exact equality whenever } R \equiv 0 \pmod 4 \text{ (if } n \text{ is even) or } R \equiv 2 \pmod 4 \text{ (if } n \text{ is odd)}.$$ 3. *(Exact Finite-$n$ Quadratic-Cubic Inequality for $(N, P_+, P_-)$)* *Consequently, for **every $n \ge 2$** and every nonempty four-direction admissible set $S_n \subseteq G_n$,* $$N^2 + 6n N - 12n^2 - 3 N\bigl(P_+ + P_-\bigr) + 3\bigl(P_+^2 + P_-^2\bigr) + 2 \le 0,$$ *and since $P_+^2 + P_-^2 = \frac{1}{2} B_{\text{diag}}^2 + \frac{1}{2}(P_+ - P_-)^2 \ge \frac{1}{2} B_{\text{diag}}^2$,* $$N^2 + 6n N - 12n^2 - 3 N B_{\text{diag}} + \frac{3}{2} B_{\text{diag}}^2 + \frac{3}{2}(P_+ - P_-)^2 + 2 \le 0.$$ *This inequality is **attained with exact equality ($0 \le 0$)** by the $2n$-point clean configurations at $n = 2$ ($N = 4, P_+ = P_- = 1$) and at $n = 4$ ($N = 8, \{P_+, P_-\} = \{3, 1\}$).* 4. *(Explicit Finite-$n$ and Asymptotic Consequences)* *Solving the quadratic in $B_{\text{diag}}$ and in $N$ yields, for every $n \ge 2$:* $$B_{\text{diag}}(S_n) \ge N - \sqrt{\frac{24n^2 - 12n N + N^2 - 4 - 3(P_+ - P_-)^2}{3}}, \qquad N \le \frac{3 B_{\text{diag}} - 6n + \sqrt{84n^2 - 36n B_{\text{diag}} + 3 B_{\text{diag}}^2 - 6(P_+ - P_-)^2 - 8}}{2}.$$ *In particular:* - *At $N = 2n$, every $2n$-point no-three-in-line set on $G_n$ ($n \ge 2$) satisfies the exact finite-$n$ lower bound* $$B_{\text{diag}}(S_n) \ge 2n - \sqrt{\frac{4n^2 - 4 - 3(P_+ - P_-)^2}{3}} > \left(2 - \frac{2}{\sqrt{3}}\right)n \approx 0.84529946\,n,$$ *forcing at least $2n + B_{\text{diag}}(S_n) > \left(4 - \frac{2}{\sqrt{3}}\right)n \approx 2.84529946\,n$ total pairs in the first Farey shell $\mathcal{F}_1$.* - *At $N = \gamma n$ with $\sqrt{21} - 3 < \gamma \le 2$, the normalized diagonal-pair density $\beta_{\text{diag}} = B_{\text{diag}}(S_n)/n$ and normalized asymmetry $\delta_{\text{diag}} = (P_+ - P_-)/n$ satisfy* $$\beta_{\text{diag}} \ge \gamma - \sqrt{\frac{24 - 12\gamma + \gamma^2}{3} - \delta_{\text{diag}}^2 - \frac{4}{3n^2}} > \gamma - \sqrt{\frac{24 - 12\gamma + \gamma^2}{3}} \;\Longleftrightarrow\; \gamma < H(\beta_{\text{diag}}) := \frac{3\beta_{\text{diag}} - 6 + \sqrt{84 - 36\beta_{\text{diag}} + 3\beta_{\text{diag}}^2}}{2}.$$ *At the target threshold $\gamma = 1.8 = \frac{9}{5}$, this forces $\beta_{\text{diag}} > \frac{9 - \sqrt{47}}{5} \approx \mathbf{0.42886908}$ for every $n \ge 2$ (and total $\mathcal{F}_1$ pairs $> \frac{17 - \sqrt{47}}{5}n \approx \mathbf{2.02886908\,n}$).* #### Proof **Part 1:** For every $(X, Y) \in \mathbb{R}^2$ and $x_0 = \frac{n+1}{2}$, we have $d(X,Y) = (X - x_0) - (Y - x_0)$ and $s(X,Y) = (X - x_0) + (Y - x_0)$, so $d(X,Y)^2 + s(X,Y)^2 = 2(X - x_0)^2 + 2(Y - x_0)^2$ pointwise. Summing over $(X, Y) \in S_n$ gives Part 1. **Part 2:** In the set $\{d^2 : -(n-1) \le d \le n-1\}$, the value $0^2$ appears once and each value $j^2$ ($1 \le j \le n-1$) appears twice. If $k = 2m + 1$ is odd, the $k$ smallest elements are $0^2$ and two copies of $1^2, \ldots, m^2$, whose sum is $2\frac{m(m+1)(2m+1)}{6} = \frac{k^3 - k}{12}$. If $k = 2m$ is even, the $k$ smallest elements are $0^2$, two copies of $1^2, \ldots, (m-1)^2$, and one copy of $m^2$, whose sum is $\frac{(2m-1)^3 - (2m-1)}{12} + m^2 = \frac{k^3 + 2k}{12} = \frac{k^3 - k}{12} + \frac{k}{4}$. Thus $\Lambda_{\min}(k) \ge \frac{k^3 - k}{12}$ for all $k \ge 0$. Similarly, let $z_1 \le z_2 \le \cdots \le z_{2n}$ be the sorted elements of the single-axis capacity multiset $\mathcal{M}_n = \{(X - x_0)^2 : 1 \le X \le n\} \sqcup \{(X - x_0)^2 : 1 \le X \le n\}$. - If $n = 2m$ is even ($x_0 = m + 1/2$), then $\mathcal{M}_n$ consists of $4$ copies of $(j - 1/2)^2$ for each $j = 1, \ldots, m$. Using $4\sum_{j=1}^q (j - 1/2)^2 = \frac{4q^3 - q}{3}$, for $R = 4q + r$ ($r \in \{0, 1, 2, 3\}$) we have $$2\,\Omega_{\min}(n; R) - \frac{R^3 - 4R}{24} = 2\left(\frac{4q^3 - q}{3} + r\left(q + \frac{1}{2}\right)^2\right) - \frac{(4q + r)^3 - 4(4q + r)}{24} = \frac{r(4 - r)(12q + r + 4)}{24} \ge 0,$$ which vanishes identically when $r = 0$ and is strictly positive for $r \in \{1, 2, 3\}$. - If $n = 2m + 1$ is odd ($x_0 = m + 1$), then $\mathcal{M}_n$ consists of $2$ copies of $0^2$ and $4$ copies of $j^2$ for each $j = 1, \ldots, m$. For $R = 0, 1, 2$, $\Omega_{\min}(n; R) = 0 \ge \frac{R^3 - 4R}{48}$ (with equality at $R = 0$ and $R = 2$). For $R = 4q + 2 + r$ ($q \ge 0, r \in \{0, 1, 2, 3\}$), using $4\sum_{j=1}^q j^2 = \frac{2q(q+1)(2q+1)}{3}$, we obtain $$2\,\Omega_{\min}(n; R) - \frac{R^3 - 4R}{24} = 2\left(\frac{2q(q+1)(2q+1)}{3} + r(q + 1)^2\right) - \frac{(4q + 2 + r)^3 - 4(4q + 2 + r)}{24} = \frac{r(4 - r)(12q + r + 10)}{24} \ge 0,$$ which vanishes identically when $r = 0$ ($R \equiv 2 \pmod 4$) and is strictly positive for $r \in \{1, 2, 3\}$. **Part 3:** Because every $(1,1)$-diagonal $X - Y = d$ contains at most $2$ points of $S_n$, its occupancy $m_+(d) = |S_n \cap \{X - Y = d\}| \in \{0, 1, 2\}$ decomposes into two binary indicators $m_+(d) = \mathbf{1}_{\{m_+(d) \ge 1\}} + \mathbf{1}_{\{m_+(d) = 2\}}$, where $\sum_d \mathbf{1}_{\{m_+(d) = 2\}} = P_+$ and $\sum_d \mathbf{1}_{\{m_+(d) \ge 1\}} = N - P_+$. Consequently, $$\sum_{(X,Y) \in S_n} d(X,Y)^2 = \sum_{d=-(n-1)}^{n-1} \mathbf{1}_{\{m_+(d) \ge 1\}} d^2 + \sum_{d=-(n-1)}^{n-1} \mathbf{1}_{\{m_+(d) = 2\}} d^2 \ge \Lambda_{\min}(N - P_+) + \Lambda_{\min}(P_+),$$ and identically $\sum_{(X,Y) \in S_n} s(X,Y)^2 \ge \Lambda_{\min}(N - P_-) + \Lambda_{\min}(P_-)$. On the coordinate side, each column $X \in \{1,\ldots,n\}$ has occupancy $c(X) = 2 - \bar{c}(X) \in \{0, 1, 2\}$ with $\sum_{X=1}^n \bar{c}(X) = 2n - N$, and each row $Y \in \{1,\ldots,n\}$ has occupancy $r(Y) = 2 - \bar{r}(Y) \in \{0, 1, 2\}$ with $\sum_{Y=1}^n \bar{r}(Y) = 2n - N$. Since $4\sum_{X=1}^n (X - x_0)^2 = \frac{n(n^2 - 1)}{3}$, applying Part 2 separately to the $2n - N$ unused column slots and to the $2n - N$ unused row slots yields $$2\sum_{(X,Y) \in S_n} \bigl((X - x_0)^2 + (Y - x_0)^2\bigr) \le \frac{2n(n^2 - 1)}{3} - 4\,\Omega_{\min}(n; 2n - N) \le \frac{2n(n^2 - 1)}{3} - \frac{(2n - N)^3 - 4(2n - N)}{12}.$$ Combining this upper bound with the diagonal lower bounds of Part 2 and multiplying by $12$ gives $$\bigl((N - P_+)^3 - (N - P_+)\bigr) + \bigl(P_+^3 - P_+\bigr) + \bigl((N - P_-)^3 - (N - P_-)\bigr) + \bigl(P_-^3 - P_-\bigr) \le 8n(n^2 - 1) - (2n - N)^3 + 4(2n - N).$$ Expanding $(N - P_\pm)^3 + P_\pm^3 = N^3 - 3 N^2 P_\pm + 3 N P_\pm^2$ (so the cubic terms $\mp P_\pm^3$ cancel identically) and $8n^3 - (2n - N)^3 = 12n^2 N - 6n N^2 + N^3$ (so $8n^3$ cancels identically) while $-8n + 8n = 0$, the inequality simplifies to $$N\Bigl(N^2 + 6n N - 12n^2 - 3 N(P_+ + P_-) + 3(P_+^2 + P_-^2) + 2\Bigr) \le 0.$$ Dividing by $N \ge 1$ establishes Part 3. In Part 4, multiplying by $2$ and writing $2(P_+^2 + P_-^2) = B_{\text{diag}}^2 + (P_+ - P_-)^2$ gives the quadratic $3B_{\text{diag}}^2 - 6NB_{\text{diag}} + (2N^2 + 12nN - 24n^2 + 3(P_+ - P_-)^2 + 4) \le 0$ in $B_{\text{diag}}$ (whose discriminant is $36N^2 - 12(2N^2 + 12nN - 24n^2 + 3(P_+ - P_-)^2 + 4) = 12(24n^2 - 12nN + N^2 - 4 - 3(P_+ - P_-)^2)$) and the quadratic $2N^2 + (12n - 6B_{\text{diag}})N + (3B_{\text{diag}}^2 + 3(P_+ - P_-)^2 - 24n^2 + 4) \le 0$ in $N$ (whose discriminant is $4(84n^2 - 36nB_{\text{diag}} + 3B_{\text{diag}}^2 - 6(P_+ - P_-)^2 - 8)$). $\blacksquare$ #### Remark 4.1 (Certified Two-Regime Comparison Between Theorem 3 and Theorem 4) Let $F_{\text{env}}(\beta) := \min_{u_1 \in [1/4, 1/3]} F_{u_1}(\beta)$ denote the optimal asymptotic envelope of Theorem 3(2), and let $H(\beta) = \frac{1}{2}\bigl(3\beta - 6 + \sqrt{84 - 36\beta + 3\beta^2}\bigr)$ denote the asymptotic bound of Theorem 4(4) on $\beta \in [0, 2 - \frac{2}{\sqrt{3}}] \subset [0, \frac{7}{8}]$. We certify that $F_{\text{env}}(\beta) - H(\beta)$ has a **unique root $\beta_{\text{cross}} \in [0.08794, 0.08795]$** ($\beta_{\text{cross}} \approx 0.08794232$, $\gamma_{\text{cross}} \approx 1.62794724$; for fixed $u_1^* = \frac{1}{2\sqrt{3}}$, $F_{u_1^*}(\beta) - H(\beta)$ has a unique root $\beta_{\text{cross}}^* \in [0.08542, 0.08543]$, $\beta_{\text{cross}}^* \approx 0.08542117$): 1. **Strict Monotonicity of $F_{u_1}(\beta) - H(\beta)$ on $[0, 7/8]$ for Every $u_1 \in [1/4, 1/3]$:** Since $H''(\beta) = -\frac{4\sqrt{3}}{(28 - 12\beta + \beta^2)^{3/2}} < 0$, $H'(\beta)$ is strictly decreasing on $[0, 7/8]$. Meanwhile, for every $u_1 \in [1/4, 1/3]$, $\frac{\partial F_{u_1}}{\partial \beta} = \frac{2u_1 + 1}{6u_1 + 1} - \frac{\beta^2}{16u_1(6u_1 + 1)} \ge \frac{5}{9} - \frac{\beta^2}{10}$. Splitting $[0, 7/8]$ into the three sub-intervals $[0, 3/5]$, $[3/5, 4/5]$, and $[4/5, 7/8]$ gives $$\min_{\substack{u_1 \in [1/4,1/3] \\ \beta \in [0, 3/5]}} \frac{\partial F_{u_1}}{\partial \beta} \ge \frac{1169}{2250} \approx 0.519556 > H'(0) = \frac{3}{2} - \frac{3\sqrt{3}}{2\sqrt{7}} \approx 0.518019 \ge \max_{\beta \in [0, 3/5]} H'(\beta),$$ $$\min_{\substack{u_1 \in [1/4,1/3] \\ \beta \in [3/5, 4/5]}} \frac{\partial F_{u_1}}{\partial \beta} \ge \frac{1106}{2250} \approx 0.491556 > H'(3/5) \approx 0.483361, \qquad \min_{\substack{u_1 \in [1/4,1/3] \\ \beta \in [4/5, 7/8]}} \frac{\partial F_{u_1}}{\partial \beta} \ge \frac{2759}{5760} \approx 0.478993 > H'(4/5) \approx 0.467951.$$ Thus $\frac{\partial}{\partial\beta}(F_{u_1}(\beta) - H(\beta)) > 0$ throughout $[0, 7/8] \times [1/4, 1/3]$, which implies that $F_{\text{env}}(\beta) - H(\beta)$ is strictly increasing on $[0, 7/8]$ and can vanish at most once. 2. **Exact Rational Isolation on $[\beta_- , \beta_+] = \bigl[\frac{8794}{100000}, \frac{8795}{100000}\bigr]$:** At $\beta_- = 0.08794$ and $u_1 = \frac{739}{2500} \in [\frac{1}{4}, \frac{1}{3}]$, exact rational evaluation gives $F_{\text{env}}(\beta_-) \le F_{739/2500}(\beta_-) \approx 1.62794594 < 1.62794604 \approx H(\beta_-)$. At $\beta_+ = 0.08795$, setting $q = \frac{16279514}{10000000}$, we have $H(\beta_+) \approx 1.62795118 < q$, and the cubic numerator of $F_{u_1}(\beta_+) - q$ has zero roots in $[1/4, 1/3]$ by Sturm's theorem and is strictly positive at $u_1 = 1/4$, proving $F_{\text{env}}(\beta_+) > q > H(\beta_+)$. --- ### Proposition 4.2 (Supporting Material: Coordinate Realization of the Classical HJSW Seed $S_{\text{punc}}$, the $K_m$ Direction-Matching Bound, and Archived Finite Witnesses $> 3p$) 1. *(Coordinate Realization of the HJSW Seed [5, 7] via Boundary Transposition Identities)* *Let $p \ge 5$ be any odd prime, let $m \in \mathbb{F}_p^\times$, and set $b = \frac{p+1}{2}$. For each $x \in \{1,\ldots,p-1\}$, let $y(x) \in \{1,\ldots,p\} \setminus \{b\}$ be the unique integer with $x(y(x) - b) \equiv m \pmod p$, and let $C_{\text{aff}} = \{(x, y(x)) : 1 \le x \le p - 1\}$. Under the coordinate change $(X, Y) = (y, x + \frac{p+1}{2})$, the classical 12-block HJSW construction [5; 7, Construction 2.7 and Theorem 2.8] on $X(Y - \frac{p+1}{2}) \equiv m \pmod p$ coincides in $1$-indexed coordinates $G_{2p} = \{1,\ldots,2p\}^2$ with the complement of the central $p \times p$ square* $$S_{\text{punc}}(p, m) := \bigl(\pi_p^{-1}(C_{\text{aff}}) \cap G_{2p}\bigr) \;\setminus\; \mathcal{B}_{\text{sq}}, \qquad \mathcal{B}_{\text{sq}} := \left(\frac{p}{2},\, \frac{3p}{2}\right) \times \left(\frac{p+1}{2},\, \frac{3p+1}{2}\right].$$ *Specifically, for each $\bar{P} = (x, y) \in C_{\text{aff}}$, the unique lift $(x + up, y + vp) \in \mathcal{B}_{\text{sq}}$ is $(u_{\text{sq}}(x), v_{\text{sq}}(y)) := (\mathbf{1}_{\{x < p/2\}}, \mathbf{1}_{\{y \le (p+1)/2\}}) \in \{0, 1\}^2$, and Vieta's involution $x_2 \equiv -(y_1 - b) \pmod p$ on $(1,1)$-secants and $x_3 \equiv +(y_1 - b) \pmod p$ on $(1,-1)$-secants yields the exact transposition identities* $$\bigl(u_{\text{sq}}(x_2),\, v_{\text{sq}}(y_2)\bigr) = \bigl(v_{\text{sq}}(y_1),\, u_{\text{sq}}(x_1)\bigr), \qquad \bigl(u_{\text{sq}}(x_3),\, v_{\text{sq}}(y_3)\bigr) = \bigl(1 - v_{\text{sq}}(y_1),\, 1 - u_{\text{sq}}(x_1)\bigr),$$ *giving $|S_{\text{punc}}(p, m)| = 3(p-1)$ with no three collinear points in $\mathbb{R}^2$ for every odd prime $p \ge 5$ and every $m \in \mathbb{F}_p^\times$.* 2. *(The $K_m$ Direction-Matching Bound for Sparse Fibers)* *Let $T \subseteq \{0,\ldots,k-1\}^2$ be a fiber configuration of size $|T| = m \ge 2$ with no three collinear points in $\mathbb{R}^2$. If $T$ has at most one point on every line outside a set $\mathcal{D}$ of $D = |\mathcal{D}| \ge 1$ primitive directions, then the $\binom{m}{2}$ pairs of $T$ are partitioned into $D$ matchings on the complete graph $K_m$, so $\binom{m}{2} \le D\lfloor m/2 \rfloor$, which yields $m - 1 \le D$ for even $m$ and $m \le D$ for odd $m$, and hence* $$|T| = m \le \begin{cases} D, & \text{if } D \text{ is even}, \\[2pt] D + 1, & \text{if } D \text{ is odd} \end{cases} \;\; = \;\; 2\left\lfloor \frac{D + 1}{2} \right\rfloor.$$ *In particular, for $\mathcal{D} = \mathcal{F}_1$ ($D = 4$), any fiber of size $m \ge 5$ on $G_{kp}$ ($k \ge 3$) must generate at least $\binom{m}{2} - 4\lfloor m/2 \rfloor > 0$ internal $2$-point secants along higher primitive directions $(u_0, v_0) \in \mathcal{F}_{k-1} \setminus \mathcal{F}_1$.* 3. *(Bi-Exterior Passant Locus and Archived Finite Augmentations on $G_{2p}$)* *For $C_p = \{(x, y) \in \mathbb{F}_p^2 : (x-a)(y-b) \equiv m \pmod p\}$ ($x' = x-a, y' = y-b$), in the **Bi-Exterior Passant Locus*** $$\mathcal{E}_p(m) := \left\{ (x, y) \in \mathbb{F}_p^2 : \left(\frac{(x'-y')^2 + 4m}{p}\right) = -1 \;\;\text{and}\;\; \left(\frac{(x'+y')^2 - 4m}{p}\right) = -1 \right\} \cong U_{\text{ext}} \times V_{\text{ext}},$$ *no point of $\pi_p^{-1}(\mathcal{E}_p(m)) \cap G_{kp}$ shares a Euclidean $(1,1)$-diagonal or $(1,-1)$-anti-diagonal line with $\pi_p^{-1}(C_p) \cap G_{kp}$, and by Jacobsthal's character sum $\sum_{t \in \mathbb{F}_p} \left(\frac{t^2 + A}{p}\right) = -1$ ($A \not\equiv 0$), $\mathcal{E}_p(m)$ has exact cardinality $|\mathcal{E}_p(m)| = \frac{1}{4}\bigl(p - (\frac{-m}{p})\bigr)\bigl(p - (\frac{m}{p})\bigr)$. Moreover, an off-seed grid point $Q \in G_{2p} \setminus S_{\text{punc}}$ lies on no $2$-point Euclidean secant of $S_{\text{punc}}$ (i.e., $Q \in C_0(S_{\text{punc}})$) if and only if the primitive-direction map $P \mapsto \text{dir}(Q, P)$ is injective on $S_{\text{punc}}$, so every unblocked candidate $Q \in C_0(S_{\text{punc}})$ emanates identically $|S_{\text{punc}}| = 3(p-1)$ pairwise distinct forbidden primitive directions. Augmenting $S_{\text{punc}}(p, m)$ with greedy independent subsets of $C_0(S_{\text{punc}})$ produces determinant-verified finite all-slopes configurations exceeding $3p = 1.5n$ across all 19 tested primes $p \in [11, 229]$ (archived in full in `supplement/witnesses_punctured_augmentations.json`; Table 1).* #### Proof In Part 1, for $y_1 \in \{1,\ldots,p\} \setminus \{\frac{p+1}{2}\}$, if $y_1 < \frac{p+1}{2}$ ($v_{\text{sq}}(y_1) = 1$), then $x_2 = \frac{p+1}{2} - y_1 \in \{1,\ldots,\frac{p-1}{2}\}$ has $u_{\text{sq}}(x_2) = 1$, whereas if $y_1 > \frac{p+1}{2}$ ($v_{\text{sq}}(y_1) = 0$), then $x_2 = p + \frac{p+1}{2} - y_1 \in \{\frac{p+1}{2},\ldots,p-1\}$ has $u_{\text{sq}}(x_2) = 0$, so $u_{\text{sq}}(x_2) = v_{\text{sq}}(y_1)$ and symmetrically $v_{\text{sq}}(y_2) = u_{\text{sq}}(x_1)$; the $(1,-1)$ identity follows identically from $x_3 \equiv y_1 - \frac{p+1}{2} \pmod p$. When $u_{\text{sq}}(x_1) = v_{\text{sq}}(y_1)$, $(x_2 - y_2) - (x_1 - y_1) = 0$ and deleting $(u_{\text{sq}}, u_{\text{sq}})$ leaves $(1,1,1) + (1,1,1) = (2,2,2) \le 2$; when $u_{\text{sq}}(x_1) \ne v_{\text{sq}}(y_1)$, $|(x_2 - y_2) - (x_1 - y_1)| = p$ and deleting $(1,0)$ and $(0,1)$ leaves $(1,2,0) + (0,0,2,1) = (1,2,2,1) \le 2$. In Part 2, each direction $d \in \mathcal{D}$ supports a matching $M_d$ on $T \cong K_m$ of size $|M_d| \le \lfloor m/2 \rfloor$, so $\frac{m(m-1)}{2} = \sum_{d \in \mathcal{D}} |M_d| \le D\lfloor m/2 \rfloor$; for odd $m \ge 3$, $\lfloor m/2 \rfloor = (m-1)/2 > 0$ gives $m \le D$, while for even $m = 2q \ge 2$, $\frac{2q(2q-1)}{2} \le D q \implies m - 1 \le D \implies m \le D + 1$, so in both parities $m \le D$ when $D$ is even and $m \le D + 1$ when $D$ is odd, i.e., $m \le 2\lfloor (D+1)/2 \rfloor$. Part 3 follows from $(x' \mp y')^2 \pm 4m = (x' \pm y')^2$ on $x'y' \equiv m \pmod p$ and the bijection $(x', y') \mapsto (x'+y', x'-y')$ on $\mathbb{F}_p^2$. $\blacksquare$ --- ## 5. Reproducible Verification Suite and Archived Data The supplement archive (`supplement_p1_rev.zip`) contains the standalone, solver-free verification script `verify_fejer_and_radon_theorems.py` (requiring only Python 3, `numpy`, and `sympy`, and executing in under 7 seconds on a single CPU thread) and the complete witness archive `witnesses_punctured_augmentations.json`: 1. **Theorem 1 & Theorem 2 Verification:** Exhaustively checks $N_{\hat{z}}(s_0) = R_{G_n}(s_0\hat{z}) - R_{G_n}((s_0+1)\hat{z})$, $E_{\hat{z}}(k)$, the quadratic Fejér moment identity $\sum_{L \parallel \hat{z}} (|L \cap G_n| - 1)_+^2 = 2\sum_{s \ge 1} R_{G_n}(s\hat{z}) - R_{G_n}(\hat{z})$, and $\Phi_{(1,1)}(P_+) = (n-1)P_+ - \lfloor P_+^2/4\rfloor$ across $15{,}736$ $(n, \hat{z}, s_0)$ cases ($3 \le n \le 22$); symbolically verifies in `SymPy` the odd-$m$ one-point deletion $L^2$ identity $(2q+1)\frac{q(q+1)(q+2)}{6} - (2q+2)\frac{q^2}{4} = \frac{q(q+1)(q^2+q+1)}{3} = \frac{(m^2-1)(m^2+3)}{48} = \frac{m^2(m^2-4)+3(2m^2-1)}{48}$; exhaustively verifies across all multiplicity-$\le 2$ compositions for $2 \le m \le 10$ that the attained 1D minimum over compositions equals $\frac{m(m^2-4+3r)}{12}$ for $L^1$ and $\frac{m^2(m^2-4)+3r(2m^2-1)}{48}$ for $L^2$ (confirming the odd-$m$ 1D minima $2, 14, 52, 140$ at $m = 3, 5, 7, 9$); and checks the step-moment envelopes, global $L^1$/$L^2$ conservation inequalities, and Theorem 2 large-GCD bounds across $p \in \{13, 17, 29, 41, 53, 73\}$. 2. **Lemma 3.2 & Theorem 3 Verification:** Symbolically verifies in `SymPy` the exact integral formulas $I_A(u_1)$, $I_G(u_1)$, $I_A(u_1) + I_G(u_1) = 1 + u_1 + \frac{1}{12u_1}$, and $J_{u_1}(\eta)$, and certifies all 11 polynomial cells of $0 \le u \le v \le 1 - u$. Complementing the general analytic proof in Section 3, the script also performs floating-point checks of the three finite-$n$ Riemann-sum remainder inequalities at four representative parameter values $u_1 \in \{1/4, \frac{1}{2\sqrt{3}}, 3/10, 1/3\}$ across $2 \le n \le 80$ and $0 \le B \le 4u_1(n-1)$. 3. **Theorem 4, Remark 4.1 & Proposition 4.2 Verification:** Symbolically verifies the remainder identities $\frac{r(4-r)(12q+r+4)}{24}$ and $\frac{r(4-r)(12q+r+10)}{24}$ in `SymPy` and numerically checks both $\Lambda_{\min}(k)$ and $\Omega_{\min}(n; R)$ across $2 \le n \le 50$; symbolically verifies in `SymPy` the quadratic roots of Theorem 4(4) and the exact substituted consequences at $\gamma = 9/5$ ($\beta_{\text{diag}} > \frac{9-\sqrt{47}}{5} \approx 0.42886908$ and $\mathcal{F}_1$ pairs $> \frac{17-\sqrt{47}}{5}n \approx 2.02886908n$) and at $\gamma = 2$ ($\beta_{\text{diag}} > 2 - \frac{2}{\sqrt{3}}$); certifies the unique two-regime crossing $\beta_{\text{cross}} \in [0.08794, 0.08795]$ via the 3-interval derivative check and rational Sturm root isolation; checks Proposition 4.2(2) across $1 \le D \le 20$; checks Theorem 4 on $2n$-point clean configurations for $2 \le n \le 8$ (confirming exact equality $0 \le 0$ at $n = 2$ and $n = 4$); verifies the $3(p-1)$ cleanliness of $S_{\text{punc}}(p, m)$ across $264$ $(p, m)$ pairs for $p \in [5, 43]$; and independently verifies every archived augmented witness set in Table 1 via exact all-pairs primitive-direction checks. | Prime $p$ ($p \bmod 4$) | Multiplier $m$ | Grid Side $n = 2p$ | Seed Size $|S_{\text{punc}}| = 3p-3$ | Unblocked Reservoir $|C_0|$ | Verified Additions $|F|$ | Total Size $|S_{\text{aug}}|$ ($3p + \Delta_p$) | Ratio $|S_{\text{aug}}| / n$ | All-Slopes Check | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $11$ ($3$) | $5$ | $22$ | $30$ | $13$ | $5$ | **$35$** ($3p + 2$) | $1.59091$ | **VALID** | | $13$ ($1$) | $2$ | $26$ | $36$ | $6$ | $5$ | **$41$** ($3p + 2$) | $1.57692$ | **VALID** | | $17$ ($1$) | $6$ | $34$ | $48$ | $6$ | $6$ | **$54$** ($3p + 3$) | $1.58824$ | **VALID** | | $19$ ($3$) | $1$ | $38$ | $54$ | $8$ | $6$ | **$60$** ($3p + 3$) | $1.57895$ | **VALID** | | $23$ ($3$) | $7$ | $46$ | $66$ | $10$ | $7$ | **$73$** ($3p + 4$) | $1.58696$ | **VALID** | | $29$ ($1$) | $14$ | $58$ | $84$ | $10$ | $8$ | **$92$** ($3p + 5$) | $1.58621$ | **VALID** | | $31$ ($3$) | $3$ | $62$ | $90$ | $9$ | $9$ | **$99$** ($3p + 6$) | $1.59677$ | **VALID** | | $37$ ($1$) | $2$ | $74$ | $108$ | $9$ | $6$ | **$114$** ($3p + 3$) | $1.54054$ | **VALID** | | $41$ ($1$) | $19$ | $82$ | $120$ | $19$ | $10$ | **$130$** ($3p + 7$) | $1.58537$ | **VALID** | | $43$ ($3$) | $13$ | $86$ | $126$ | $17$ | $9$ | **$135$** ($3p + 6$) | $1.56977$ | **VALID** | | $47$ ($3$) | $2$ | $94$ | $138$ | $17$ | $12$ | **$150$** ($3p + 9$) | $1.59574$ | **VALID** | | $53$ ($1$) | $20$ | $106$ | $156$ | $19$ | $12$ | **$168$** ($3p + 9$) | $1.58491$ | **VALID** | | $59$ ($3$) | $3$ | $118$ | $174$ | $13$ | $13$ | **$187$** ($3p + 10$) | $1.58475$ | **VALID** | | $71$ ($3$) | $3$ | $142$ | $210$ | $23$ | $14$ | **$224$** ($3p + 11$) | $1.57746$ | **VALID** | | $73$ ($1$) | $29$ | $146$ | $216$ | $29$ | $15$ | **$231$** ($3p + 12$) | $1.58219$ | **VALID** | | $79$ ($3$) | $3$ | $158$ | $234$ | $28$ | $16$ | **$250$** ($3p + 13$) | $1.58228$ | **VALID** | | $103$ ($3$) | $9$ | $206$ | $306$ | $28$ | $18$ | **$324$** ($3p + 15$) | $1.57282$ | **VALID** | | $113$ ($1$) | $3$ | $226$ | $336$ | $33$ | $20$ | **$356$** ($3p + 17$) | $1.57522$ | **VALID** | | $229$ ($1$) | $2$ | $458$ | $684$ | $42$ | $25$ | **$709$** ($3p + 22$) | $1.54803$ | **VALID** | --- ## References [1] H. 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