# Discrete Spline Majorization, Four-Direction Pair Coupling, Multi-Block Subgrid Obstructions, and Step-Midpoint Conservation Laws for No-Three-in-Line Sets

## Abstract

We unify continuum piecewise-quadratic dual covers and discrete centered second-moment bounds for no-three-in-line subsets $S_n \subseteq G_n = \{1,\ldots,n\}^2$ into a remainder-free discrete spline framework, couple transverse line intercepts with longitudinal pair steps, and extend centered second-moment obstructions to multi-period $k \times k$ residue blocks. First (Theorem 1), for $0 \le S_a \le S_b \le 2S_a$ ($S = S_b - S_a$), the monic splines $\tilde{\Phi}_{S_a,S_b}(z) = (|z| - S/2)_+^2 + S_a S/2$ and $\tilde{\Psi}_{S_a,S_b}(d) = d^2 - (|d|-S_a)_+^2 + (|d|-S_b)_+^2$ satisfy $\tilde{\Psi}_{S_a,S_b}(X-Y) + \tilde{\Psi}_{S_a,S_b}(X+Y-2x_0) \le 2\tilde{\Phi}_{S_a,S_b}(X-x_0) + 2\tilde{\Phi}_{S_a,S_b}(Y-x_0)$ on all of $\mathbb{R}^2$ and admit exact remainder-free discrete order-statistic sums on $G_n$. Second (Theorems 2–3), summing over $S_n$ yields an exact finite-$n$ polynomial inequality in $(|S_n|, B_{\mathrm{axis}}, D_{\mathrm{axis}}, B_{\mathrm{diag}}, D_{\mathrm{diag}}, S)$. Setting $S_a = S_b = 0$ raises the minimum total $\mathcal{F}_1$ pair count at $|S_n| \ge 1.8n$ to $B_{\mathcal{F}_1} > \frac{51-\sqrt{421}}{15}n \approx 2.03211n$; setting $S = n - \lceil |S_n|/2 \rceil$ proves the strict zero-remainder finite-$n$ bound $|S_n| < (1 + 1/\sqrt{3})n$ for all $n \ge 2$ when $B_{\mathrm{diag}} = 0$; and optimizing over $(S_a, S_b)$ within the spline family eliminates the crossing corner of `no_three_in_line_51`, joining the quadratic bound at $\beta_{\mathrm{crit}} = \frac{28-10\sqrt{6}}{23} \approx 0.15240$. Third (Theorem 4), transverse-longitudinal conservation laws reduce the standalone Fejér quadratic step-moment bound by $57\%$–$65\%$ across all 19 archived witnesses ($p \in [11, 229]$). Fourth (Theorem 5), a shifted-center multi-block subgrid inequality for $r$ private $k \times k$ residue blocks on a shared modular diagonal proves that a pair ($r=2$) saves $S_k(2) \ge \lceil 2k/7 \rceil \ge 2k/7$ points and excludes the alternating septic $Q_7$ of `no_three_in_line_11` above $1.8$ in every translated $kp \times kp$ window for all $k \ge 2$. These structural obstructions neither improve the $3/2$ lower bound nor resolve the $1.8n$ existence question.

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## 1. Introduction and Scope of Contribution

### 1.1 Background and Motivation from `no_three_in_line_51` and `no_three_in_line_35`

Dudeney's 1917 **No-Three-in-Line problem** [1] asks for the maximum cardinality $f(n)$ of a subset $S_n \subseteq G_n := \{1,\ldots,n\}^2$ containing no three collinear points in $\mathbb{R}^2$. While $f(n) = 2n$ holds for $2 \le n \le 46$ and $n \in \{48, 50, 52\}$ [2, 3], and quarter-turn symmetric sets with $|S_n| \ge 1.8n$ are known for sixteen sizes $100 \le n \le 200$ [4], the best asymptotic lower bound remains Hall, Jackson, Sudbery, and Wild's $\liminf_{n\to\infty} f(n)/n \ge 3/2$ [5] (building on Erdős [6]; see also [7–13, 17–21]).

On the upper-bound and structural-obstruction side, Prellberg [14] and Aujla, Prellberg, and Sandhu [15] analyzed subsets of a single checkerboard parity class $C_\varepsilon = \{(X,Y) \in G_n : X+Y \equiv \varepsilon \pmod 2\}$, where half of all diagonal and anti-diagonal lines are identically empty by parity. In our companion paper [16] (`no_three_in_line_51`), we extended both the piecewise-quadratic continuum cover of [14, Section 4.2] and the centered order-statistic method of [15] to **unrestricted grids $S_n \subseteq G_n$** by decomposing each diagonal line occupancy into $1$-point and $2$-point indicators $m_\pm(d) = \mathbf{1}_{\{m_\pm(d) \ge 1\}} + \mathbf{1}_{\{m_\pm(d) = 2\}}$. That analysis established two complementary bounds in terms of the $2$-point diagonal and anti-diagonal secant counts $P_+(S_n)$, $P_-(S_n)$, and $B_{\mathrm{diag}}(S_n) = P_+ + P_-$:
1. A piecewise-quadratic rearrangement bound ([16, Theorem 3]):
   $$\frac{|S_n|}{n-1} \le F_{u_1}(\eta_n) + \frac{8A_{u_1}(0) + 4g_0(u_1)}{n-1}, \qquad F_{u_1}(\beta) = 1 + u_1 + \frac{1}{12u_1} + \frac{2u_1+1}{6u_1+1}\beta - \frac{\beta^3}{48u_1(6u_1+1)},$$
   with $F_{u_1^*}(0) = 1 + \frac{1}{\sqrt{3}} \approx 1.57735027$ at $u_1^* = \frac{1}{2\sqrt{3}}$, but carrying a finite-$n$ Riemann-sum remainder $+\frac{5\sqrt{3}-1}{n-1} \approx \frac{7.66025}{n-1}$.
2. A remainder-free centered second-moment inequality ([16, Theorem 4]):
   $$|S_n|^2 + 6n|S_n| - 12n^2 - 3|S_n|(P_+ + P_-) + 3(P_+^2 + P_-^2) + 2 \le 0,$$
   yielding the asymptotic density bound $\gamma < H(\beta_{\mathrm{diag}}) := \frac{1}{2}\bigl(3\beta_{\mathrm{diag}} - 6 + \sqrt{84 - 36\beta_{\mathrm{diag}} + 3\beta_{\mathrm{diag}}^2}\bigr)$ (with $H(0) = \sqrt{21} - 3 \approx 1.58257569$) and forcing $B_{\mathrm{diag}}(S_n) > \frac{9-\sqrt{47}}{5}n \approx 0.42886908n$ whenever $|S_n| \ge 1.8n$.

### 1.2 Five Open Structural Questions and Summary of Results

The results of [16] and the open-cell synthesis of [8] (`no_three_in_line_35` v8) motivate five concrete mathematical questions:
1. **Why do the continuum cover bound $F_{\mathrm{env}}(\beta)$ and the discrete second-moment bound $H(\beta)$ cross transversely at $\beta_{\mathrm{cross}} \approx 0.08794232$, and can both be unified into a single remainder-free discrete spline family?**
2. **Can the $+7.66025$ Riemann-sum remainder in [16, Theorem 3] be eliminated so that the Prellberg bound $|S_n| < (1 + 1/\sqrt{3})n$ holds strictly for every finite $n \ge 2$ when $B_{\mathrm{diag}}(S_n) = 0$?**
3. **In [16, Theorem 4], row and column occupancies were relaxed to the worst-case multiplicity-$2$ multiset $\mathcal{M}_n$, which implicitly assumes $B_{\mathrm{axis}}(S_n) := P_h(S_n) + P_v(S_n) = |S_n|$, whereas combining $B_{\mathrm{diag}} > \frac{9-\sqrt{47}}{5}n$ with the pigeonhole bound $B_{\mathrm{axis}} \ge 2(|S_n|-n) = 1.6n$ at $|S_n| = 1.8n$ assumes $B_{\mathrm{axis}} = 1.6n$. How do $B_{\mathrm{axis}}$ and $B_{\mathrm{diag}}$ couple when both axes and diagonals are decomposed simultaneously?**
4. **How do transverse line intercepts $d = X - Y$ couple to longitudinal pair step lengths $s_0^+(d) = |X_2 - X_1|$ and pair midpoints $\bar{s}(d)$?**
5. **In [8, Table 1 and Remark 4], the Möbius quartic $Q_4$ and alternating septic $Q_7$ of [21] (`no_three_in_line_11`) are listed as open cells above $1.8$ outside $n = 2p$ because the whole-fiber diagonal exchange of [21, Section 9.2], when bounded using shared $(1,1)$-diagonal capacities alone, saves only $2$ points per pair ($r = 2$) of exposed $k \times k$ blocks in a $kp \times kp$ window, so its normalized pair gain $2/k$ decays like $O(1/k)$. Can internal block second moments overcome this $O(1/k)$ decay and close $Q_7$ across all multi-period windows $n = kp$ ($k \ge 2$)?**

In this paper, we answer these five questions analytically and verify every theorem symbolically and exhaustively in `experiments/verify_master_spline_and_coupling.py`:
- **Theorem 1 (Global $\mathbb{R}^2$ Spline Majorization and Exact Discrete Order-Statistic Sums):** We introduce a two-parameter family of monic piecewise-quadratic splines $(\tilde{\Phi}_{S_a,S_b}, \tilde{\Psi}_{S_a,S_b})$ parameterized by breakpoints $0 \le S_a \le S_b \le 2S_a$ ($S = S_b - S_a$). Unlike [14, 16], where positivity required restricting to the unit square $[0,1]^2$, we prove that $2\tilde{\Phi}_{S_a,S_b}(X-x_0) + 2\tilde{\Phi}_{S_a,S_b}(Y-x_0) - \tilde{\Psi}_{S_a,S_b}(X-Y) - \tilde{\Psi}_{S_a,S_b}(X+Y-2x_0) \ge 0$ holds **globally on all of $\mathbb{R}^2$**, and that the discrete sums of $\tilde{\Phi}_{S_a,S_b}$ and $\tilde{\Psi}_{S_a,S_b}$ over grid order statistics evaluate in **exact polynomial closed form with zero Riemann-sum remainder**.
- **Theorem 2 (Exact Finite-$n$ Four-Direction Master Spline Inequality):** Applying the binary $1$-point/$2$-point occupancy decomposition simultaneously to all four directions of $\mathcal{F}_1 = \{(1,0),(0,1),(1,1),(1,-1)\}$ and summing the spline majorization over $S_n \subseteq G_n$ yields an exact finite-$n$ polynomial inequality in $(N, B_{\mathrm{axis}}, D_{\mathrm{axis}}, B_{\mathrm{diag}}, D_{\mathrm{diag}}, S)$.
- **Theorem 3 (Three Structural Specializations: Four-Direction Coupling, Zero-Remainder Prellberg Bound, and Spline-Family Phase Transition):**
  1. *At $S_a = S_b = 0$ ($S = 0$):* Directly from Theorem 2(1), we obtain the four-direction symmetric second-moment inequality
     $$N\Bigl(N^2 + 6nN - 12n^2 - 3NB_{\mathrm{diag}} + \frac{3}{2}B_{\mathrm{diag}}^2 + \frac{3}{2}D_{\mathrm{diag}}^2 + 2\Bigr) + 3(2n-N)\Bigl((N-B_{\mathrm{axis}})^2 + D_{\mathrm{axis}}^2\Bigr) \le 0,$$
     which penalizes any deficit $N - B_{\mathrm{axis}} > 0$ of $1$-point rows and columns. At $|S_n| \ge 1.8n$, minimizing $B_{\mathcal{F}_1} = B_{\mathrm{axis}} + B_{\mathrm{diag}}$ over $B_{\mathrm{axis}} \in [2(N-n), N]$ strictly improves the uncoupled threshold $\frac{17-\sqrt{47}}{5}n \approx 2.02886908n$ of [16] to
     $$B_{\mathcal{F}_1}(S_n) > \frac{51 - \sqrt{421}}{15}\,n \approx \mathbf{2.03211436\,n}.$$
  2. *At $S = n - \lceil N/2 \rceil$:* The leading cubic factors identically as $(3N - 2n + 3\varepsilon)(3N^2 - 6nN + 2n^2)$ ($\varepsilon = N \bmod 2$), proving analytically for both even and odd $N$ the **strict zero-remainder finite-$n$ Prellberg bound** $|S_n| < \bigl(1 + \frac{1}{\sqrt{3}}\bigr)n$ for every $n \ge 2$ whenever $B_{\mathrm{diag}}(S_n) = 0$.
  3. *Spline-Family Optimization and Three-Regime Hinge Functional:* Optimizing over $(S_a, S_b)$ within the spline family yields a smooth envelope $H_{\mathrm{master}}(\beta_{\mathrm{diag}})$ governed by an explicit quartic curve that eliminates the crossing corner $\beta_{\mathrm{cross}} \approx 0.08794$ of [16] (giving $H_{\mathrm{master}}(0.08794) < 1.62672 < 1.62794$) and meets $H(\beta_{\mathrm{diag}})$ with $C^{3/2}$ tangency at the exact phase transition
     $$\beta_{\mathrm{crit}} = \frac{28 - 10\sqrt{6}}{23} \approx \mathbf{0.15239576}, \qquad \gamma_{\mathrm{crit}} = H(\beta_{\mathrm{crit}}) = \frac{48 - 4\sqrt{6}}{23} \approx \mathbf{1.66095831}.$$
     We also evaluate the exact three-regime piecewise-quadratic first-variation hinge functional $I_{\mathrm{hinge}}(m_0; \gamma, \beta)$ over $m_0 \in [0, 1]$, showing that $\beta \ge \beta_{\mathrm{crit}}$ ($5\gamma - 2\beta \ge 8$) is the exact regime where $s^* = 0$ is optimal within the two-parameter spline family.
- **Theorem 4 (Exact Transverse-Longitudinal Conservation Laws and Step-Midpoint Bounds):** Using the exact orthogonal decomposition $s(p_1)^2 + s(p_2)^2 = 2s_0^+(d)^2 + 2\bar{s}(d)^2$ on every $2$-point diagonal and bounding the remaining $N - 2P_\pm$ points via $\Omega_{\min}(2n-1; N - 2P_\pm)$, we prove exact finite-$n$ upper bounds on the combined step-plus-midpoint second moments of $2$-point diagonals and axes. Across all 19 archived finite witnesses ($p \in [11, 229]$) from [16], this coupled bound reduces the standalone Fejér quadratic step-moment upper bound by **$56.9\%$ to $64.7\%$**, achieving $97.5\%$ tightness at $p = 229$.
- **Theorem 5 (Shifted-Center Multi-Block Subgrid Obstructions and Multi-Period Oblique Exchange):** For $r \ge 1$ private $k \times k$ residue blocks $B_1, \ldots, B_r$ sharing a modular $(1,1)$-diagonal in a translated $kp \times kp$ window ($k \ge 2$) with carry bits $\sigma_1, \ldots, \sigma_r \in \{0, 1\}$, we prove the **Shifted-Center Multi-Block Parallelogram Identity**
  $$(Y - X - \sigma_i + \mu)^2 + (X + Y - 2x_0)^2 = 2\left(\left(X - x_0 + \frac{\sigma_i - \mu}{2}\right)^2 + \left(Y - x_0 - \frac{\sigma_i - \mu}{2}\right)^2\right),$$
  and use the parity-free quarter-integer bijection $\bigl\{\bigl|X - \frac{k+1}{2} \pm \frac{1}{4}\bigr| : 1 \le X \le k\bigr\} = \bigl\{\frac{2j-1}{4} : 1 \le j \le k\bigr\}$ to establish the exact cubic subgrid inequality
  $$(r^2 - 1)N^3 + 12rk N^2 - 6r^2(4k^2 + 1)N - 4r^3 k(4k^2 - 1) \le 0$$
  for the combined point count $N = \sum_{i=1}^r |S_i|$. Defining the worst-case $r$-block saving $S_k(r) := 2rk - M_k(r)$ for $r \ge 2$ (and $S_k(0) = S_k(1) = 0$ by retaining the row cover), this proves for a pair ($r = 2$) that $S_k(2) \ge \lceil 2k/7 \rceil \ge \frac{2}{7}k$ for **every integer $k \ge 2$** and $\liminf_{k\to\infty} \frac{S_k(2)}{k} \ge 4 - \alpha_2^* \approx 0.28695744 > \frac{2}{7}$ (where $\alpha_2^* \approx 3.71304256$ is the positive root of $3\alpha^3 + 24\alpha^2 - 96\alpha - 128 = 0$), overcoming the $O(1/k)$ pair-decay obstacle noted in [8, Remark 4]. Combining this universal multi-period exchange with the scale-one marked factorial moments $W, T$ on $\mathbb{F}_p^2$ imported from [21, Section 13] proves that for the alternating septic $Q_7$ in **every translated square of side $kp$ ($k \ge 2$)**,
  $$\frac{|S|}{kp} \le C_{32} - \frac{1}{7}\rho_{32}^2 + \frac{1}{300}\rho_{32}^3 + O(p^{-1/2}) < \frac{179945}{100000} + O(p^{-1/2}) < \frac{9}{5},$$
  where the $O(p^{-1/2})$ error constant and prime threshold from [21] are uniform in $k \ge 2$, closing the alternating septic subcase of Open Cell 4 of [8] for all integer period multiples $k \ge 2$.

### 1.3 Explicit Statement of Scope and Limitations

These results establish exact finite-$n$ algebraic obstructions, four-direction pair-coupling inequalities, optimality within the explicit two-parameter spline family $(\tilde{\Phi}_{S_a,S_b}, \tilde{\Psi}_{S_a,S_b})$, and multi-period subgrid exchange bounds for the alternating septic $Q_7$ of [21]. We do not claim optimality over all infinite-dimensional monotone dual covers, nor does the single-diagonal bound $\frac{S_k(2)}{k} \ge \frac{2}{7}$ by itself close the Möbius quartic $Q_4$ of [21] for $k \ge 3$ (whose larger coordinate excess $C_{32} - 1.8 \approx 0.002336$ exceeds $\frac{1}{7}\rho_{32}^2 \approx 0.000678$). Finally, these results **do not** construct an infinite family of no-three-in-line sets with $|S_n| \ge 1.8n$, **do not** improve the asymptotic lower bound $\liminf_{n\to\infty} f(n)/n \ge 3/2$, and **do not** prove that $\limsup_{n\to\infty} f(n)/n < 1.8$.

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## 2. Global $\mathbb{R}^2$ Spline Majorization and Exact Discrete Order-Statistic Sums

Throughout, for $n \ge 2$, let $G_n = \{1,\ldots,n\}^2$, let $x_0 := \frac{n+1}{2}$ be the grid center, and for any point $p = (X, Y) \in \mathbb{R}^2$ write its centered axis and diagonal coordinates as
$$u_p := X - x_0, \qquad v_p := Y - x_0, \qquad d(p) := X - Y = u_p - v_p, \qquad s(p) := X + Y - (n+1) = u_p + v_p.$$
We write $(t)_+ := \max(0, t)$ for the positive part of $t \in \mathbb{R}$.

### Definition 2.1 (The Two-Parameter Monic Spline Pair)
For any real parameters $0 \le S_a \le S_b$, set $S := S_b - S_a \ge 0$ and define the even $C^1$ piecewise-quadratic splines $\tilde{\Phi}_{S_a,S_b}, \tilde{\Psi}_{S_a,S_b} : \mathbb{R} \to [0, \infty)$ by
$$\tilde{\Phi}_{S_a,S_b}(z) := \left(|z| - \frac{S}{2}\right)_+^2 + \frac{1}{2} S_a S,$$
$$\tilde{\Psi}_{S_a,S_b}(d) := d^2 - (|d| - S_a)_+^2 + (|d| - S_b)_+^2 = \begin{cases} d^2, & 0 \le |d| \le S_a, \\[3pt] 2S_a|d| - S_a^2, & S_a \le |d| \le S_b, \\[3pt] 2S_a S + (|d| - S)^2, & |d| \ge S_b. \end{cases}$$

Observe that at $S_a = S_b$ ($S = 0$), $\tilde{\Phi}_{S_a,S_a}(z) = z^2$ and $\tilde{\Psi}_{S_a,S_a}(d) = d^2$ reduce to the pure quadratic polynomials of [15, 16], while for $S_a < S_b$, $\tilde{\Psi}_{S_a,S_b}(d)$ interpolates $C^1$-smoothly between $d^2$ on $[0, S_a]$, the affine tangent $2S_a|d| - S_a^2$ on $[S_a, S_b]$, and the shifted parabola $2S_a S + (|d| - S)^2$ on $[S_b, \infty)$.

### Theorem 1 (Global $\mathbb{R}^2$ Spline Majorization and Exact Remainder-Free Discrete Sums)
1. *(Global Pointwise Majorization on $\mathbb{R}^2$)* *For all real parameters $0 \le S_a \le S_b \le 2S_a$ ($S = S_b - S_a$) and **every** $(X, Y) \in \mathbb{R}^2$ (without any box restriction to $[1, n]^2$),*
   $$\Delta_{S_a,S_b}(X, Y) := 2\tilde{\Phi}_{S_a,S_b}(X - x_0) + 2\tilde{\Phi}_{S_a,S_b}(Y - x_0) - \tilde{\Psi}_{S_a,S_b}(X - Y) - \tilde{\Psi}_{S_a,S_b}(X + Y - 2x_0) \ge 0,$$
   *with exact equality $\Delta_{S_a,S_b}(X, Y) = 0$ on the two dual rectangles $\{|X - Y| \le S_a,\; |X + Y - 2x_0| \ge S_b\}$ and $\{|X - Y| \ge S_b,\; |X + Y - 2x_0| \le S_a\}$.*
2. *(Exact Discrete Axis Spline Sum Without Remainder)* *For every integer $n \ge 1$ and every integer $0 \le S \le n$:*
   $$4\sum_{X=1}^n \left(\left|X - \frac{n+1}{2}\right| - \frac{S}{2}\right)_+^2 = \frac{(n - S)^3 - (n - S)}{3}.$$
   *More generally, let $\Theta_{\min}(n; k)$ denote the sum of the $k$ smallest elements of the $n$-element single-axis capacity-$1$ multiset $\mathcal{A}_n := \{(X - x_0)^2 : 1 \le X \le n\}$, and let $\Omega_{\min}(n; R)$ denote the sum of the $R$ smallest elements of the $2n$-element capacity-$2$ multiset $\mathcal{M}_n := \mathcal{A}_n \sqcup \mathcal{A}_n$. Then for all integers $0 \le S \le n$, $0 \le k \le n$, and $0 \le R \le 2n$, the $k$ (respectively $R$) smallest values of $(|X - x_0| - S/2)_+^2$ on $\mathcal{A}_n$ (respectively $\mathcal{M}_n$) sum to $\Theta_{\min}(n - S; (k - S)_+)$ (respectively $\Omega_{\min}(n - S; (R - 2S)_+)$), where*
   $$4\,\Theta_{\min}(m; r) = \frac{r^3 - r}{3} + \mathbf{1}_{\{r \not\equiv m \pmod 2\}}\,r \ge \frac{r^3 - r}{3}, \qquad 2\,\Omega_{\min}(m; R) \ge \frac{R^3 - 4R}{24}.$$
3. *(Exact Discrete Diagonal Spline Order-Statistic Sum Without Parity Loss)* *Let $0 = d_{(1)}^2 < d_{(2)}^2 = d_{(3)}^2 < \cdots < d_{(2n-1)}^2 = (n-1)^2$ be the non-decreasing rearrangement of $\{d^2 : -(n-1) \le d \le n-1\}$, and let $\Lambda_{\min}(m) = \sum_{j=1}^m d_{(j)}^2 = \frac{m^3 - m + 3m(1 - (m \bmod 2))}{12}$ for $m \ge 0$. Then for **every** pair of integers $0 \le S_a \le S_b$ and every integer $0 \le k \le 2n - 1$, $\tilde{\Psi}_{S_a,S_b}(d)$ is non-decreasing in $|d|$ and its sum over the $k$ smallest offsets satisfies the exact telescopic identity*
   $$\Lambda_{S_a,S_b}(k) := \sum_{j=1}^k \tilde{\Psi}_{S_a,S_b}\!\bigl(d_{(j)}\bigr) = \Lambda_{\min}(k) - \Lambda_{\min}\!\bigl((k - 2S_a)_+\bigr) + \Lambda_{\min}\!\bigl((k - 2S_b)_+\bigr),$$
   *which satisfies the parity-loss-free cubic lower bound*
   $$12\,\Lambda_{S_a,S_b}(k) \ge \bigl(k^3 - k\bigr) - \Bigl(\bigl((k - 2S_a)_+\bigr)^3 - (k - 2S_a)_+\Bigr) + \Bigl(\bigl((k - 2S_b)_+\bigr)^3 - (k - 2S_b)_+\Bigr),$$
   *with **exact equality** whenever $k$ is odd (for all integers $0 \le S_a \le S_b$).*

#### Proof
**Part 1:** By the dihedral $D_4$ symmetry of $\Delta_{S_a,S_b}$ under sign flips and permutation of $X - x_0$ and $Y - x_0$, it suffices to prove $\Delta_{S_a,S_b} \ge 0$ on the fundamental wedge $0 \le u := |X - Y| \le v := |X + Y - 2x_0| < \infty$, where $\{|X - x_0|, |Y - x_0|\} = \bigl\{\frac{v+u}{2}, \frac{v-u}{2}\bigr\}$. Thus
$$\Delta_{S_a,S_b}(u, v) = 2S_a S + \frac{1}{2}(v + u - S)_+^2 + \frac{1}{2}(v - u - S)_+^2 - \tilde{\Psi}_{S_a,S_b}(u) - \tilde{\Psi}_{S_a,S_b}(v).$$
When $S_a = 0$, $S_b \le 2S_a$ forces $S_b = S = 0$, where $\Delta_{0,0}(u,v) = \frac{1}{2}(v+u)^2 + \frac{1}{2}(v-u)^2 - u^2 - v^2 = 0$ identically. Assume now $0 < S_a \le S_b \le 2S_a$, so $0 \le S = S_b - S_a \le S_a$. The breakpoints $S_a \le S_b$ divide $[0, \infty)$ into Piece 1 ($[0, S_a]$), Piece 2 ($[S_a, S_b]$), and Piece 3 ($[S_b, \infty)$), so $(u, v)$ with $0 \le u \le v$ lies in one of the six piece pairs $(i, j)$ ($1 \le i \le j \le 3$). Let $\sigma_+ = \mathbf{1}_{\{u+v \ge S\}}$ and $\sigma_- = \mathbf{1}_{\{v-u \ge S\}}$. Since $S = S_b - S_a \le S_a$, if $j \ge 2$ then $u + v \ge v \ge S_a \ge S$, so $\sigma_+ = 0$ occurs only in $(i, j) = (1, 1)$ (where $v - u \le u + v \le S$ also forces $\sigma_- = 0$). In $(i, j) = (1, 3)$, $u \le S_a$ and $v \ge S_b$ imply $v - u \ge S_b - S_a = S$, so $\sigma_- = 1$ throughout $(1, 3)$. This partitions $\{0 \le u \le v\}$ into **11 polynomial cells** $(i, j, \sigma_+, \sigma_-)$:
1. **Six cells with $\sigma_+ = \sigma_- = 1$ ($v - u \ge S$):** Using $\frac{1}{2}(v+u-S)^2 + \frac{1}{2}(v-u-S)^2 = (u^2 + (v-S)^2)$, direct expansion gives:
   - *Cell $(1, 3, 1, 1)$ ($u \le S_a, v \ge S_b$):* $\Delta(u,v) = 2S_a S + u^2 + (v-S)^2 - u^2 - (2S_a S + (v-S)^2) = 0$ identically.
   - *Cell $(1, 2, 1, 1)$ ($u \le S_a, v \in [S_a, S_b], v - u \ge S$):* $\Delta(u,v) = (v - S_b)^2 \ge 0$.
   - *Cell $(2, 2, 1, 1)$ ($u, v \in [S_a, S_b], v - u \ge S$):* $\Delta(u,v) = (u - S_a)^2 + (v - S_b)^2 \ge 0$.
   - *Cell $(2, 3, 1, 1)$ ($u \in [S_a, S_b], v \ge S_b, v - u \ge S$):* $\Delta(u,v) = (u - S_a)^2 \ge 0$.
   - *Cell $(1, 1, 1, 1)$ ($0 \le u \le v \le S_a, v - u \ge S$):* $\Delta(u,v) = (S_b - S_a)(S_a + S_b - 2v) \ge S(S_b - S_a) = S^2 \ge 0$.
   - *Cell $(3, 3, 1, 1)$ ($S_b \le u \le v, v - u \ge S$):* $\Delta(u,v) = (S_b - S_a)(2u - S_a - S_b) \ge S(S_b - S_a) = S^2 \ge 0$.
2. **One cell with $\sigma_+ = \sigma_- = 0$ ($0 \le u \le v, u + v \le S$):**
   - *Cell $(1, 1, 0, 0)$:* Since $u^2 + v^2 \le (u+v)^2 \le S^2$ and $S = S_b - S_a \le S_a$ (as $S_b \le 2S_a$), we have $\Delta(u,v) = 2S_a S - (u^2 + v^2) \ge 2S_a S - S^2 = S(2S_a - S) = S(3S_a - S_b) \ge S S_a \ge 0$.
3. **Four strip cells with $\sigma_+ = 1, \sigma_- = 0$ ($u + v \ge S, 0 \le v - u \le S$):** In every pair $(i, j)$, turning off $\sigma_-$ subtracts $\frac{1}{2}(S - (v - u))^2$ from the $\sigma_- = 1$ polynomial:
   $$\Delta_{i,j,1,0}(u, v) = \Delta_{i,j,1,1}(u, v) - \frac{1}{2}\bigl(S - (v - u)\bigr)^2.$$
   - *Cell $(2, 2, 1, 0)$ ($S_a \le u \le v \le S_b, v - u \le S$):* $\Delta(u,v) = \frac{1}{2}(u + v - S_a - S_b)^2 \ge 0$ is an exact square.
   - *Cell $(1, 1, 1, 0)$ ($0 \le u \le v \le S_a, u + v \ge S, v - u \le S$):* Since $\frac{\partial^2\Delta}{\partial u^2} = -1 < 0$, for each fixed $v \in [S/2, S_a]$, $\Delta(u,v)$ is concave in $u$ and attains its minimum at the endpoints of its $u$-interval: $u = S - v$ (matching Cell $(1,1,0,0)$ where $\Delta \ge 0$), $u = v - S$ (matching Cell $(1,1,1,1)$ where $\Delta \ge 0$), or $u = v$, where $\Delta(v,v) = S(S_a + S_b - 2v) - \frac{1}{2}S^2 \ge S(S_b - S_a) - \frac{1}{2}S^2 = \frac{1}{2}S^2 \ge 0$.
   - *Cell $(1, 2, 1, 0)$ ($u \in [0, S_a], v \in [S_a, S_b], v - u \le S$):* Since $\frac{\partial^2\Delta}{\partial u^2} = -1 < 0$, for each fixed $v \in [S_a, S_b]$, $\Delta(u,v)$ is minimized at its $u$-endpoints $u = v - S$ (matching Cell $(1,2,1,1)$ where $\Delta = (v - S_b)^2 \ge 0$) or $u = S_a$, where $\Delta(S_a, v) = (v - S_b)^2 - \frac{1}{2}(S_b - v)^2 = \frac{1}{2}(v - S_b)^2 \ge 0$.
   - *Cell $(2, 3, 1, 0)$ ($u \in [S_a, S_b], v \ge S_b, v - u \le S$):* Here $\frac{\partial^2\Delta}{\partial v^2} = -1 < 0$, so for each fixed $u \in [S_a, S_b]$, $\Delta(u,v)$ is concave in $v \in [S_b, u + S]$ and attains its minimum at the endpoints $v = u + S$ (matching Cell $(2,3,1,1)$ where $\Delta = (u - S_a)^2 \ge 0$) or $v = S_b$, where $\Delta(u, S_b) = (u - S_a)^2 - \frac{1}{2}(u - S_a)^2 = \frac{1}{2}(u - S_a)^2 \ge 0$.
   - *Cell $(3, 3, 1, 0)$ ($S_b \le u \le v \le u + S$):* Since $\frac{\partial^2\Delta}{\partial v^2} = -1 < 0$, for each fixed $u \ge S_b$, $\Delta(u,v)$ is concave in $v \in [u, u + S]$ and attains its minimum at $v = u + S$ (matching Cell $(3,3,1,1)$ where $\Delta \ge S^2 \ge 0$) or $v = u$, where $\Delta(u, u) = S(2u - S_a - S_b) - \frac{1}{2}S^2 = \frac{1}{2}S(4u - S_a - 3S_b) \ge \frac{1}{2}S(S_b - S_a) = \frac{1}{2}S^2 \ge 0$.

**Part 2:** Let $m = n - S \ge 0$ and $y_0 = \frac{m+1}{2}$. As $X$ runs over $\{1,\ldots,n\}$, $|X - x_0| - \frac{S}{2}$ is non-positive on the $S$ central indices $X \in \bigl[\frac{n-S+1}{2}, \frac{n+S+1}{2}\bigr]$ (contributing $S$ zeros to the multiset of values of $(|X - x_0| - S/2)_+^2$), while on the remaining $m = n - S$ indices ($1 \le X \le \lfloor \frac{n-S}{2}\rfloor$ and $n - \lfloor \frac{n-S}{2}\rfloor + 1 \le X \le n$), the positive values of $|X - x_0| - \frac{S}{2}$ coincide bijectively with $\bigl\{|Y - y_0| : 1 \le Y \le m\bigr\}$. Thus the sorted multiset of $(|X - x_0| - S/2)_+^2$ on $\{1,\ldots,n\}$ consists of $S$ zeros followed by $\mathcal{A}_{n-S}$, which proves both order-statistic shift identities and gives $4\sum_{X=1}^n (|X - x_0| - S/2)_+^2 = 4\sum_{Y=1}^{n-S} (Y - y_0)^2 = \frac{(n-S)^3 - (n-S)}{3}$. Moreover, in $\mathcal{A}_m$, if $r \equiv m \pmod 2$, the $r$ smallest elements form a symmetric block $\{1, \ldots, m\}$ of width $r$, so $4\Theta_{\min}(m; r) = \frac{r^3 - r}{3}$; if $r \not\equiv m \pmod 2$, they consist of the symmetric block of width $r - 1$ plus one copy of $(r/2)^2$, giving $4\Theta_{\min}(m; r) = \frac{(r-1)^3 - (r-1)}{3} + r^2 = \frac{r^3 - r}{3} + r$. The lower bound $2\Omega_{\min}(m; R) \ge \frac{R^3 - 4R}{24}$ is [16, Theorem 4(2)].

**Part 3:** For $d \ge 0$, $\tilde{\Psi}'_{S_a,S_b}(d) = 2\min(d, S_a) + 2(d - S_b)_+ \ge 0$, so $\tilde{\Psi}_{S_a,S_b}(d)$ is non-decreasing in $|d|$ and its $k$ smallest values on $\{-(n-1),\ldots,n-1\}$ occur on the $k$ smallest offsets $d_{(1)}, \ldots, d_{(k)}$. In the sequence $|d_{(1)}| \le |d_{(2)}| \le \cdots$, the first $2S_a + 1$ elements ($j \le 2S_a + 1$) satisfy $|d_{(j)}| \le S_a$, while for $j = 2S_a + 1 + r$ ($r \ge 1$), we have $|d_{(j)}| - S_a = |d_{(r+1)}|$ because shifting $|d|$ down by the integer $S_a$ maps $\{S_a + 1, S_a + 1, S_a + 2, S_a + 2, \ldots\}$ bijectively onto $\{1, 1, 2, 2, \ldots\} = \{|d_{(2)}|, |d_{(3)}|, \ldots\}$. Since $d_{(1)} = 0$, it follows that for every integer $S_a \ge 0$ and $k \ge 0$,
$$\sum_{j=1}^k \bigl(|d_{(j)}| - S_a\bigr)_+^2 = \sum_{r=1}^{(k - 2S_a)_+} d_{(r)}^2 = \Lambda_{\min}\!\bigl((k - 2S_a)_+\bigr).$$
Applying this identity to $S_a$ and to $S_b$ gives $\Lambda_{S_a,S_b}(k) = \Lambda_{\min}(k) - \Lambda_{\min}((k-2S_a)_+) + \Lambda_{\min}((k-2S_b)_+)$. Finally, writing $12\Lambda_{\min}(m) = (m^3 - m) + 3m(1 - (m \bmod 2))$, since $k - 2S_a \equiv k - 2S_b \equiv k \pmod 2$, the parity correction is $0$ when $k$ is odd, while when $k$ is even it equals $3k - 3(k-2S_a)_+ + 3(k-2S_b)_+ \ge 0$ because $m \mapsto 3m$ is non-decreasing and $k \ge (k-2S_a)_+ \ge (k-2S_b)_+$. $\blacksquare$

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## 3. The Four-Direction Master Spline Inequality and Phase Transition

We now apply Theorem 1 to any nonempty four-direction admissible set $S_n \subseteq G_n$ of cardinality $N = |S_n| \in \{1,\ldots,2n\}$. Let $P_h, P_v, P_+, P_-$ denote the numbers of $2$-point horizontal, vertical, diagonal, and anti-diagonal lines of $S_n$, and define the symmetric and asymmetric pair counts
$$B_{\mathrm{axis}} := P_h + P_v, \quad D_{\mathrm{axis}} := P_h - P_v, \qquad B_{\mathrm{diag}} := P_+ + P_-, \quad D_{\mathrm{diag}} := P_+ - P_-, \qquad B_{\mathcal{F}_1} := B_{\mathrm{axis}} + B_{\mathrm{diag}}.$$

### Theorem 2 (Exact Finite-$n$ Four-Direction Master Spline Inequality)
*Let $n \ge 2$ and let $S_n \subseteq G_n = \{1,\ldots,n\}^2$ be any nonempty four-direction admissible set of size $N = |S_n|$. For any integers $0 \le S_a \le S_b \le 2S_a$ with $S = S_b - S_a \le n$ and $M = \frac{S_a + S_b}{2}$:*
1. *(Exact Discrete Master Spline Inequality)* *Summing $\Delta_{S_a,S_b}(X, Y) \ge 0$ over $(X, Y) \in S_n$ yields*
   $$\begin{aligned}
   \Lambda_{S_a,S_b}(N - P_+) + \Lambda_{S_a,S_b}(P_+) + \Lambda_{S_a,S_b}(N - P_-) + \Lambda_{S_a,S_b}(P_-) &\le 2N S_a S + \frac{2\bigl((n-S)^3 - (n-S)\bigr)}{3} \\
   &\quad - 2\sum_{d \in \{h,v\}} \Bigl(\Theta_{\min}\bigl(n-S;\, (n - N + P_d - S)_+\bigr) + \Theta_{\min}\bigl(n-S;\, (n - P_d - S)_+\bigr)\Bigr).
   \end{aligned}$$
2. *(Closed-Form Polynomial Master Inequality)* *Define the baseline polynomial*
   $$Q_4(n, N, B_{\mathrm{diag}}, D_{\mathrm{diag}}) := N\Bigl(N^2 + 6nN - 12n^2 - 3NB_{\mathrm{diag}} + \frac{3}{2}B_{\mathrm{diag}}^2 + \frac{3}{2}D_{\mathrm{diag}}^2 + 2\Bigr)$$
   *and the integer rounding residual $r(S_a, S_b) := 4 - 48\bigl(M - \frac{N - B_{\mathrm{diag}}}{4}\bigr)^2 = 4 - 3\bigl(2(S_a + S_b) - (N - B_{\mathrm{diag}})\bigr)^2$. Whenever either $S_a = S_b = 0$ ($S = 0$) or $\max(P_+, P_-) \le 2S_a \le \min(N - P_+, N - P_-)$:*
   - *(Four-Direction Coupled Form, for $0 \le S \le n - N + \min(P_h, P_v)$):*
     $$Q_4(n, N, B_{\mathrm{diag}}, D_{\mathrm{diag}}) + 3(2n - N - 2S)\Bigl((N - B_{\mathrm{axis}})^2 + D_{\mathrm{axis}}^2\Bigr) + S\Bigl(3N(8n - 5N + 2B_{\mathrm{diag}}) - 3D_{\mathrm{diag}}^2 + r(S_a, S_b)\Bigr) - 4S^3 \le 0.$$
   - *(General-Axis Form, for all $0 \le S \le n$):*
     $$Q_4(n, N, B_{\mathrm{diag}}, D_{\mathrm{diag}}) + S\Bigl(3N(8n - 5N + 2B_{\mathrm{diag}}) - 3D_{\mathrm{diag}}^2 + r(S_a, S_b)\Bigr) - 4S^3 + \Bigl(\bigl((2S + N - 2n)_+\bigr)^3 - 4(2S + N - 2n)_+\Bigr) \le 0.$$

#### Proof
**Part 1:** For each $(X, Y) \in S_n$, Theorem 1(1) gives $\tilde{\Psi}_{S_a,S_b}(d(X,Y)) + \tilde{\Psi}_{S_a,S_b}(s(X,Y)) \le 2\tilde{\Phi}_{S_a,S_b}(X - x_0) + 2\tilde{\Phi}_{S_a,S_b}(Y - x_0)$. On the diagonal side, decomposing $m_\pm(d) = \mathbf{1}_{\{m_\pm(d) \ge 1\}} + \mathbf{1}_{\{m_\pm(d) = 2\}}$ across the $N - P_\pm$ occupied lines and $P_\pm$ two-point lines and applying Theorem 1(3) lower-bounds the sum of $\tilde{\Psi}_{S_a,S_b}$ by $\Lambda_{S_a,S_b}(N - P_+) + \Lambda_{S_a,S_b}(P_+) + \Lambda_{S_a,S_b}(N - P_-) + \Lambda_{S_a,S_b}(P_-)$. On the axis side, the constant term $\frac{1}{2}S_a S$ in $\tilde{\Phi}_{S_a,S_b}$ contributes $2N S_a S$. For the column occupancies $c(X) \in \{0, 1, 2\}$ ($\sum_{X=1}^n c(X) = N$), writing $2 - c(X) = \mathbf{1}_{\{c(X) = 0\}} + \mathbf{1}_{\{c(X) \le 1\}}$ decomposes the $2n - N$ unused column slots into two binary indicators on $\{1,\ldots,n\}$ of sizes $\#\{X : c(X) = 0\} = n - (N - P_v) = n - N + P_v$ and $\#\{X : c(X) \le 1\} = n - P_v$. Applying Theorem 1(2) to both indicators and identically to the row occupancies $r(Y)$ establishes Part 1.

**Part 2:** Let $f(x) := x^3 - x$. First, if $S_a = S_b = 0$ ($S = 0$), then $\Lambda_{0,0}(k) = \Lambda_{\min}(k)$ with $12\Lambda_{\min}(k) \ge f(k)$ for every integer $k \ge 0$, so $12$ times the left-hand side of Part 1 is at least $\sum_{\varepsilon \in \{+,-\}}(f(N - P_\varepsilon) + f(P_\varepsilon))$ with no restriction on $P_\pm$. Second, if $\max(P_+, P_-) \le 2S_a \le \min(N - P_+, N - P_-)$, Theorem 1(3) gives $(P_\pm - 2S_a)_+ = (P_\pm - 2S_b)_+ = 0$ and $(N - P_\pm - 2S_a)_+ = N - P_\pm - 2S_a$, while because $f(m)$ is strictly increasing for $m \ge 1$ and $f(0) = f(-1) = 0 \ge f(m)$ for all $m \le 0$, we have $f((N - P_\pm - 2S_b)_+) \ge f(N - P_\pm - 2S_b)$ unconditionally for all integers $N - P_\pm - 2S_b$. Thus in both cases $12$ times the left-hand side of Part 1 is at least
$$\sum_{\varepsilon \in \{+,-\}} \Bigl(f(N - P_\varepsilon) - f(N - P_\varepsilon - 2S_a) + f(N - P_\varepsilon - 2S_b) + f(P_\varepsilon)\Bigr).$$
In the Four-Direction Coupled Form ($S \le n - N + \min(P_h, P_v)$), Theorem 1(2) gives $24\Theta_{\min}(n-S; k-S) \ge 2f(k-S)$ for each $k \in \{n - N + P_h, n - P_h, n - N + P_v, n - P_v\}$, so $12$ times the right-hand side of Part 1 is at most $24N S_a S + 8f(n-S) - 2\sum_{d \in \{h,v\}} (f(n - N + P_d - S) + f(n - P_d - S))$. Substituting $P_\pm = \frac{B_{\mathrm{diag}} \pm D_{\mathrm{diag}}}{2}$, $P_{h,v} = \frac{B_{\mathrm{axis}} \pm D_{\mathrm{axis}}}{2}$, $S_a = M - \frac{S}{2}$, and $S_b = M + \frac{S}{2}$, direct polynomial expansion (verified symbolically in `SymPy` in Section 6) shows that the cubic terms in $M$ and $P_\pm$ cancel identically, yielding the first inequality of Part 2. In the General-Axis Form, grouping all $2n - N$ unused slots on each axis into $\Omega_{\min}(n-S; (2n - N - 2S)_+)$ via Theorem 1(2) replaces the axis deficit sum by $-\bigl(((2n - N - 2S)_+)^3 - 4(2n - N - 2S)_+\bigr)$; using the odd-function identity $-(x_+^3 - 4x_+) = -(x^3 - 4x) + ((-x)_+^3 - 4(-x)_+)$ for $x = 2n - N - 2S$ yields the second inequality of Part 2. $\blacksquare$

---

### Theorem 3 (Four-Direction Pair Coupling, Zero-Remainder Prellberg Bound, and Spline-Family Phase Transition)
1. *(Exact Four-Direction Pair Coupling at $S_a = S_b = 0$)* *Setting $S_a = S_b = 0$ ($S = 0$) in Theorem 2(1) yields, for **every $n \ge 2$** and every nonempty four-direction admissible set $S_n \subseteq G_n$,*
   $$N\Bigl(N^2 + 6nN - 12n^2 - 3NB_{\mathrm{diag}} + \frac{3}{2}B_{\mathrm{diag}}^2 + \frac{3}{2}D_{\mathrm{diag}}^2 + 2\Bigr) + 3(2n - N)\Bigl((N - B_{\mathrm{axis}})^2 + D_{\mathrm{axis}}^2\Bigr) \le 0.$$
   *Consequently, for every $N = \gamma n$ with $\gamma \in (1, 2]$, writing $\beta_{\mathrm{axis}} = B_{\mathrm{axis}}/n \in [2(\gamma-1), \gamma]$, $\beta_{\mathrm{diag}} = B_{\mathrm{diag}}/n$, $\delta_{\mathrm{axis}} = D_{\mathrm{axis}}/n$, and $\delta_{\mathrm{diag}} = D_{\mathrm{diag}}/n$,*
   $$\beta_{\mathrm{diag}} > \gamma - \sqrt{\frac{24 - 12\gamma + \gamma^2}{3} - \frac{2(2 - \gamma)}{\gamma}(\gamma - \beta_{\mathrm{axis}})^2 - \delta_{\mathrm{diag}}^2 - \frac{2(2 - \gamma)}{\gamma}\delta_{\mathrm{axis}}^2}.$$
   *At the target density $\gamma = \frac{9}{5} = 1.8$, $\beta_{\mathrm{axis}} + \beta_{\mathrm{diag}}$ is strictly increasing in $\beta_{\mathrm{axis}} \in [\frac{8}{5}, \frac{9}{5}]$ and attains its unique minimum at the pigeonhole endpoint $\beta_{\mathrm{axis}} = \frac{8}{5} = 1.6$ (where $\beta_{\mathrm{diag}} > \frac{27 - \sqrt{421}}{15} \approx \mathbf{0.43211436}$, strictly above the uncoupled bound $\frac{9-\sqrt{47}}{5} \approx 0.42886908$), forcing every $1.8n$-point no-three-in-line set on $G_n$ ($n \ge 2$) to contain at least*
   $$B_{\mathcal{F}_1}(S_n) = B_{\mathrm{axis}}(S_n) + B_{\mathrm{diag}}(S_n) > \frac{51 - \sqrt{421}}{15}\,n \approx \mathbf{2.03211436\,n}$$
   *two-point lines in the first Farey shell $\mathcal{F}_1$.*
2. *(Strict Zero-Remainder Finite-$n$ Prellberg Bound)* *In Theorem 2(2), setting $S = n - \frac{N + \varepsilon}{2}$ with $\varepsilon = N \bmod 2 \in \{0, 1\}$ (so $S = n - \lceil N/2 \rceil$ is an integer and $(2S + N - 2n)_+ = 0$) factors the leading cubic identically as*
   $$\bigl(3N - 2n + 3\varepsilon\bigr)\bigl(3N^2 - 6nN + 2n^2\bigr) - 3N(2N - 2n + \varepsilon)B_{\mathrm{diag}} + \frac{3}{2}NB_{\mathrm{diag}}^2 + \frac{3}{2}(2N - 2n + \varepsilon)D_{\mathrm{diag}}^2 + 2N + \frac{\varepsilon(3N - 6n + 1)}{2} + r(S_a, S_b)\,S \le 0.$$
   *In particular, when $B_{\mathrm{diag}}(S_n) = 0$, every four-direction admissible set $S_n \subseteq G_n$ satisfies the **strict zero-remainder finite-$n$ inequality***
   $$|S_n| < \left(1 + \frac{1}{\sqrt{3}}\right)n \approx 1.57735027\,n \qquad \text{for every } n \ge 2.$$
3. *(Smooth Master Envelope $H_{\mathrm{master}}(\beta_{\mathrm{diag}})$ and the Three-Regime Hinge Functional)* *In the asymptotic scaling $N = \gamma n$, $B_{\mathrm{diag}} = \beta n$, $S_a = s_a n$, $S_b = s_b n$, $s = s_b - s_a$, and $m = \frac{s_a + s_b}{2}$, the normalized General-Axis Master Spline inequality of Theorem 2(2) (divided by $48n^3$) is*
   $$\mathcal{J}(s, m; \gamma, \beta) := \frac{\gamma\bigl(24 - 12\gamma - 2\gamma^2 + 6\gamma\beta - 3\beta^2\bigr)}{48} + 2s\left(m - \frac{\gamma - \beta}{4}\right)^2 + \frac{s^3}{6} - \frac{\gamma(8 - 5\gamma + 2\beta)}{8}s - \frac{1}{3}\left(s - \left(1 - \frac{\gamma}{2}\right)\right)_+^3 \ge 0.$$
   *Minimizing $\mathcal{J}(s, m; \gamma, \beta)$ over $m = \frac{\gamma - \beta}{4}$ and $0 \le s \le 1 - \frac{\gamma}{2}$ within the two-parameter spline family yields the smooth master upper bound $\gamma \le H_{\mathrm{master}}(\beta)$:*
   - *For $0 \le \beta < \beta_{\mathrm{crit}} := \frac{28 - 10\sqrt{6}}{23} \approx \mathbf{0.15239576}$, the unique optimal spline width is $s^*(\gamma, \beta) = \min\bigl(1 - \frac{\gamma}{2},\, \frac{1}{2}\sqrt{\gamma(8 - 5\gamma + 2\beta)}\bigr) > 0$, and $H_{\mathrm{master}}(\beta) < \min(F_{\mathrm{env}}(\beta), H(\beta))$ is the root of the explicit quartic*
     $$9\beta^4 - 68\gamma\beta^3 + (288\gamma^2 - 312\gamma - 144)\beta^2 - (624\gamma^3 - 1776\gamma^2 + 1248\gamma)\beta + 8(3\gamma^2 - 6\gamma + 2)(21\gamma^2 - 56\gamma + 36) = 0$$
     *(at $\beta = 0$, the first factor $3\gamma^2 - 6\gamma + 2 = 0$ gives $H_{\mathrm{master}}(0) = 1 + \frac{1}{\sqrt{3}}$ and the second factor $21\gamma^2 - 56\gamma + 36 = 0$ gives $\frac{4}{3} + \frac{2\sqrt{7}}{21}$; at the crossing $\beta_0 = 0.08794$ of [16], exact rational arithmetic certifies $H_{\mathrm{master}}(\beta_0) < 1.62672 < 1.62794 < \min(F_{\mathrm{env}}(\beta_0), H(\beta_0))$).*
   - *At the exact phase transition*
     $$\beta_{\mathrm{crit}} = \frac{28 - 10\sqrt{6}}{23} \approx 0.15239576, \qquad \gamma_{\mathrm{crit}} = H(\beta_{\mathrm{crit}}) = \frac{48 - 4\sqrt{6}}{23} \approx 1.66095831,$$
     *where $5\gamma_{\mathrm{crit}} - 2\beta_{\mathrm{crit}} = 8$, the optimal width $s^*(\gamma, \beta)$ vanishes and $H_{\mathrm{master}}(\beta)$ joins $H(\beta) = \frac{1}{2}(3\beta - 6 + \sqrt{84 - 36\beta + 3\beta^2})$ with $C^{3/2}$ tangency $H(\beta) - H_{\mathrm{master}}(\beta) = O\bigl((\beta_{\mathrm{crit}} - \beta)^{3/2}\bigr)$.*
   - *(Pointwise Hinge Identity and Three-Regime First-Variation Functional):* *For every $m_0 \in [0, 1]$ and every $(u, v) \in [0, 1/2]^2$, the pointwise **Hinge Identity***
     $$\left(\frac{m_0}{2} - u\right) + \left(\frac{m_0}{2} - v\right) + \bigl(u + v - m_0\bigr)_+ + \bigl(|u - v| - m_0\bigr)_+ = \bigl(m_0 - (u + v)\bigr)_+ + \bigl(|u - v| - m_0\bigr)_+ \ge 0$$
     *holds with equality whenever $|u - v| \le m_0 \le u + v$, and represents the first variation $-\frac{\partial}{\partial s}\big|_{s=0}(\tilde{\Phi}, \tilde{\Psi})$ of the spline family at $s = 0$. Put $u_0 := \frac{1}{2} - \frac{\gamma}{4}$, $a := \frac{\gamma}{2} - \frac{\beta}{4}$, and $b := \frac{\beta}{4}$ ($0 \le b \le a \le 1$). Integrating the left-hand side against the symmetric order-statistic marginals on $[u_0, 1/2]$, $[0, a]$, and $[0, b]$ gives the $C^1$ convex three-regime piecewise-quadratic functional*
     $$I_{\mathrm{hinge}}(m_0; \gamma, \beta) := 8\int_{u_0}^{1/2}\left(\frac{m_0}{2} - u\right)du + 2(a - m_0)_+^2 + 2(b - m_0)_+^2 = \begin{cases} 4\bigl(m_0 - \frac{\gamma}{8}\bigr)^2 + \frac{11\gamma^2 - 8\gamma\beta + 4\beta^2 - 16\gamma}{16}, & 0 \le m_0 \le b, \\[4pt] 2\bigl(m_0 - \frac{\gamma-\beta}{4}\bigr)^2 + \frac{\gamma(5\gamma - 2\beta - 8)}{8}, & b \le m_0 \le a, \\[4pt] \frac{\gamma(\gamma + 4m_0 - 4)}{4}, & a \le m_0 \le 1. \end{cases}$$
     *Whenever $0 \le \beta \le \frac{2}{3}\gamma$, the unique global minimizer of $I_{\mathrm{hinge}}(m_0; \gamma, \beta)$ on $[0, 1]$ lies in the intermediate regime $m_0^* = \frac{\gamma - \beta}{4} \in [b, a]$ with minimum value $\frac{\gamma(5\gamma - 2\beta - 8)}{8}$, so $\min_{m_0 \in [0,1]} I_{\mathrm{hinge}}(m_0; \gamma, \beta) \ge 0$ if and only if $5\gamma - 2\beta \ge 8$ ($\beta \ge \beta_{\mathrm{crit}}$).*

#### Proof
**Part 1:** Setting $S_a = S_b = 0$ ($S = 0$) in Theorem 2(1)—where $\tilde{\Phi}_{0,0}(z) = z^2$ and $\tilde{\Psi}_{0,0}(d) = d^2$ are pure quadratics so no truncation condition on $P_\pm$ is needed—immediately gives the displayed finite-$n$ inequality. Dividing by $n^3$ gives $2\gamma^2 + 12\gamma - 24 - 6\gamma\beta_{\mathrm{diag}} + 3\beta_{\mathrm{diag}}^2 + 3\delta_{\mathrm{diag}}^2 + \frac{6(2-\gamma)}{\gamma}\bigl((\gamma - \beta_{\mathrm{axis}})^2 + \delta_{\mathrm{axis}}^2\bigr) < 0$; solving the quadratic in $\beta_{\mathrm{diag}}$ yields the displayed lower bound. At $\gamma = \frac{9}{5}$, we have $\frac{2(2-\gamma)}{\gamma} = \frac{2(1/5)}{9/5} = \frac{2}{9}$, so setting $\delta_{\mathrm{diag}} = \delta_{\mathrm{axis}} = 0$ gives $\beta_{\mathrm{diag}} \ge \frac{9}{5} - \sqrt{\frac{47}{25} - \frac{2}{9}\bigl(\frac{9}{5} - \beta_{\mathrm{axis}}\bigr)^2}$ for $\beta_{\mathrm{axis}} \in [\frac{8}{5}, \frac{9}{5}]$. Differentiating $g(\beta_{\mathrm{axis}}) := \beta_{\mathrm{axis}} + \beta_{\mathrm{diag}}(\beta_{\mathrm{axis}})$ gives
$$g'(\beta_{\mathrm{axis}}) = 1 - \frac{\frac{2}{9}\bigl(\frac{9}{5} - \beta_{\mathrm{axis}}\bigr)}{\sqrt{\frac{47}{25} - \frac{2}{9}\bigl(\frac{9}{5} - \beta_{\mathrm{axis}}\bigr)^2}} \ge 1 - \frac{2/45}{\sqrt{421}/15} = 1 - \frac{2}{3\sqrt{421}} > 0$$
since $4 < 9 \times 421 = 3789$. Thus $g(\beta_{\mathrm{axis}})$ is strictly increasing on $[\frac{8}{5}, \frac{9}{5}]$ and attains its minimum at $\beta_{\mathrm{axis}} = \frac{8}{5}$, where $\frac{47}{25} - \frac{2}{9}\bigl(\frac{1}{5}\bigr)^2 = \frac{423 - 2}{225} = \frac{421}{225}$ and $g(8/5) = \frac{8}{5} + \frac{9}{5} - \frac{\sqrt{421}}{15} = \frac{51 - \sqrt{421}}{15}$.

**Part 2:** Substituting $S = n - \frac{N+\varepsilon}{2}$ ($\varepsilon = N \bmod 2 \in \{0, 1\}$) into the General-Axis Form of Theorem 2(2) yields the displayed identity (verified symbolically in `SymPy` in Section 6). When $B_{\mathrm{diag}} = D_{\mathrm{diag}} = 0$, suppose for contradiction that $N \ge \bigl(1 + \frac{1}{\sqrt{3}}\bigr)n$, so $3N^2 - 6nN + 2n^2 \ge 0$; since $3N^2 - 6nN + 2n^2 = 3(N-n)^2 - n^2 \ne 0$ for integers $n, N \ge 1$, we have $3N^2 - 6nN + 2n^2 \ge 1$. For every $n \ge 2$, let $N = \lfloor (1 + 1/\sqrt{3})n \rfloor + 1$ be the smallest integer exceeding $(1 + 1/\sqrt{3})n$ (if a larger $N'$ existed, any $N$-point subset of $S_n$ would also have $B_{\mathrm{diag}} = 0$). Set $S = n - \frac{N+\varepsilon}{2} \in \mathbb{Z}_{\ge 0}$, $S_a = \lfloor \frac{N - 2S + 1}{4} \rfloor = \lfloor \frac{2N - 2n + \varepsilon + 1}{4} \rfloor$, and $S_b = S_a + S$. Let us verify analytically that $0 \le S_a \le S_b \le 2S_a$ and $2S_b \le N$ for all $n \ge 2$:
- Since $N > (1 + 1/\sqrt{3})n$, we have $S \le n - \frac{N}{2} < \frac{3 - \sqrt{3}}{6}n$ and $S_a \ge \frac{2N - 2n - 2}{4} > \frac{n}{2\sqrt{3}} - \frac{1}{2}$. Hence $S_a - S > \frac{2\sqrt{3} - 3}{6}n - \frac{1}{2} > -1$ for all $n \ge 7$; because $S_a$ and $S$ are integers, $S_a - S > -1$ implies $S_a \ge S$, i.e., $S_b = S_a + S \le 2S_a$. Moreover, $2S_b = 2S_a + 2S \le \frac{N - 2S + 1}{2} + 2S = \frac{2n + 1 - \varepsilon}{2} \le n < N$ for all $n \ge 2$. For $2 \le n \le 6$, the five explicit tuples $(n, N, S_a, S_b)$ are $(2, 4, 1, 1)$, $(3, 5, 1, 1)$, $(4, 7, 2, 2)$, $(5, 8, 1, 2)$, and $(6, 10, 2, 3)$, each of which satisfies $0 \le S_a \le S_b \le 2S_a$ and $2S_b \le N$.
- Now evaluate the left-hand side of Theorem 3(2) at $B_{\mathrm{diag}} = D_{\mathrm{diag}} = 0$:
  - *If $N$ is even ($\varepsilon = 0$):* Then $|2(S_a + S_b) - N| = |4S_a - (N - 2S)| \le 2$, so $r(S_a, S_b) = 4 - 3(2(S_a+S_b) - N)^2 \ge -8$. Since $3N^2 - 6nN + 2n^2 \ge 1$ and $3N - 2n > \sqrt{3}n > 0$, the polynomial of Theorem 3(2) is at least $(3N - 2n)(1) + 2N - 8(n - N/2) = 9N - 10n > (9 + 3\sqrt{3} - 10)n = (3\sqrt{3} - 1)n > 0$.
  - *If $N$ is odd ($\varepsilon = 1$):* Then $N - 2S$ is odd, so $4S_a = 4\lfloor \frac{N - 2S + 1}{4} \rfloor$ is the unique multiple of $4$ at distance $1$ from $N - 2S$, giving $|2(S_a + S_b) - N| = |4S_a - (N - 2S)| = 1$ and hence $r(S_a, S_b) = 4 - 3(1)^2 = +1$. Substituting $\varepsilon = 1$ and $r(S_a, S_b) = +1$ into Theorem 3(2), the linear terms combine identically as $2N + \frac{3N - 6n + 1}{2} + 1 \cdot \bigl(n - \frac{N+1}{2}\bigr) = 3N - 2n > 0$, so the polynomial of Theorem 3(2) equals $(3N - 2n + 3)(3N^2 - 6nN + 2n^2) + (3N - 2n) > 0$.
In both cases, the left-hand side is strictly positive, contradicting Theorem 2(2).

**Part 3:** Dividing the General-Axis inequality of Theorem 2(2) by $48n^3$ and negating gives $\mathcal{J}(s, m; \gamma, \beta) \ge 0$. Since $m$ appears only in $+2s(m - \frac{\gamma-\beta}{4})^2 \ge 0$, the unique optimal midpoint is $m^* = \frac{\gamma-\beta}{4}$. For $0 \le s \le 1 - \frac{\gamma}{2}$, $\frac{\partial\mathcal{J}}{\partial s} = \frac{1}{2}s^2 - \frac{\gamma(8 - 5\gamma + 2\beta)}{8}$. Thus when $8 - 5\gamma + 2\beta \le 0$ (i.e., $5\gamma - 2\beta \ge 8$), $\mathcal{J}$ is strictly increasing in $s \ge 0$, so $s^* = 0$ is optimal within the spline family and $\mathcal{J}(0, m^*; \gamma, \beta) \ge 0 \iff \gamma \le H(\beta)$. Setting $5H(\beta) - 2\beta = 8$ gives $5\sqrt{84 - 36\beta + 3\beta^2} = 46 - 11\beta$; squaring and dividing by $4$ yields $23\beta^2 - 56\beta + 4 = 0$, whose unique root in $[0, 1]$ is $\beta_{\mathrm{crit}} = \frac{28 - 10\sqrt{6}}{23}$, with $\gamma_{\mathrm{crit}} = \frac{8 + 2\beta_{\mathrm{crit}}}{5} = \frac{48 - 4\sqrt{6}}{23}$. When $0 \le \beta < \beta_{\mathrm{crit}}$ ($8 - 5\gamma + 2\beta > 0$), substituting the interior stationary point $s^* = \frac{1}{2}\sqrt{\gamma(8 - 5\gamma + 2\beta)}$ into $\mathcal{J}(s^*, m^*; \gamma, \beta) = 0$ gives $24 - 12\gamma - 2\gamma^2 + 6\gamma\beta - 3\beta^2 = 2\sqrt{\gamma}(8 - 5\gamma + 2\beta)^{3/2}$; squaring both sides yields the displayed quartic polynomial in $(\gamma, \beta)$ and shows that $H(\beta) - H_{\mathrm{master}}(\beta) = O((\beta_{\mathrm{crit}} - \beta)^{3/2})$ as $\beta \to \beta_{\mathrm{crit}}^-$. Finally, for any $u, v \ge 0$ and $m_0 \ge 0$, since $\frac{m_0}{2} - u + \frac{m_0}{2} - v = m_0 - (u+v)$ and $x + (-x)_+ = x_+$ for $x = m_0 - (u+v)$, the Hinge Identity holds identically on $\mathbb{R}^2$. Integrating $8\int_{u_0}^{1/2}(\frac{m_0}{2}-u)\,du + 4\int_0^a (w - m_0)_+\,dw + 4\int_0^b (w - m_0)_+\,dw$ yields $8\int_{u_0}^{1/2}(\frac{m_0}{2}-u)\,du + 2(a - m_0)_+^2 + 2(b - m_0)_+^2$, which expands to the three displayed piecewise polynomials on $[0, b]$, $[b, a]$, and $[a, 1]$ (for example, at $\gamma = 9/5, \beta = 1/2, m_0 = 1 \ge a = 31/40$, Regime 3 gives $\frac{9/5(9/5 + 4 - 4)}{4} = \frac{81}{100}$). Since $I_{\mathrm{hinge}}(m_0; \gamma, \beta)$ is a sum of convex functions of $m_0$ and $m_0^* = \frac{\gamma-\beta}{4} \in [b, a]$ whenever $0 \le \beta \le \frac{2}{3}\gamma$, the global minimum on $[0, 1]$ is $I_{\mathrm{hinge}}(m_0^*; \gamma, \beta) = \frac{\gamma(5\gamma - 2\beta - 8)}{8}$. $\blacksquare$

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## 4. Transverse-Longitudinal Conservation Laws, Step-Midpoint Bounds, and Fourth-Moment Invariants

In [16, Theorem 1(2)], the longitudinal step sum $\sum_{d \in \mathcal{D}_2^+} s_0^+(d)^2$ of a no-three-in-line set $S_n$ along $(1,1)$-diagonals was bounded solely by the radial difference of the grid autocorrelation $R_{G_n}(s,s) = (n-s)_+^2$, which treats the $P_+$ two-point diagonals as if they all occupied the longest central diagonals $d \approx 0$ and reached the boundary of $G_n$. In reality, $S_n$ must simultaneously place $N - 2P_+$ points on $1$-point diagonals, and each $2$-point diagonal $X - Y = d$ has two points $p_1 = (X_1, Y_1), p_2 = (X_2, Y_2)$ whose orthogonal anti-diagonal coordinates $s(p_1), s(p_2)$ are coupled to the longitudinal step $s_0^+(d) := |X_2 - X_1| = |Y_2 - Y_1| \ge 1$ and midpoint $\bar{s}(d) := \frac{s(p_1) + s(p_2)}{2}$ by $s(p_2) - s(p_1) = \pm 2s_0^+(d)$ (compare [17] for comparison-model conservation laws).

### Theorem 4 (Exact Transverse-Longitudinal Conservation Laws, Step-Midpoint Bounds, and Higher-Moment Identities)
*Let $n \ge 2$ and let $S_n \subseteq G_n = \{1,\ldots,n\}^2$ be any four-direction admissible set of size $N = |S_n|$. Let $\mathcal{D}_2^+$ (respectively $\mathcal{D}_2^-$) denote the set of $P_+$ two-point $(1,1)$-diagonals (respectively $P_-$ two-point $(1,-1)$-anti-diagonals). For each $d \in \mathcal{D}_2^+$ with $S_n \cap \{X - Y = d\} = \{p_1, p_2\}$, let $s_0^+(d) := |X_2 - X_1| \ge 1$ be its longitudinal step and $\bar{s}(d) := \frac{s(p_1) + s(p_2)}{2}$ be its centered anti-diagonal midpoint; symmetrically, for each $s \in \mathcal{D}_2^-$, let $s_0^-(s) := |X_2 - X_1| \ge 1$ and $\bar{d}(s) := \frac{d(p_1) + d(p_2)}{2}$. Let $S_{n,1}^+$ (respectively $S_{n,1}^-$) be the set of $N - 2P_+$ (respectively $N - 2P_-$) points of $S_n$ lying on $1$-point diagonals (respectively $1$-point anti-diagonals). Then:*
1. *(Exact Transverse-Longitudinal Conservation Laws)* *Every subset $S_n \subseteq G_n$ satisfies the exact orthogonal decompositions*
   $$2\sum_{d \in \mathcal{D}_2^+} \Bigl(s_0^+(d)^2 + \bar{s}(d)^2\Bigr) + \sum_{p \in S_{n,1}^+} s(p)^2 = \sum_{p \in S_n} s(p)^2, \qquad 2\sum_{s \in \mathcal{D}_2^-} \Bigl(s_0^-(s)^2 + \bar{d}(s)^2\Bigr) + \sum_{p \in S_{n,1}^-} d(p)^2 = \sum_{p \in S_n} d(p)^2.$$
2. *(Exact Finite-$n$ Coupled Step-Midpoint Inequalities)* *Every four-direction admissible set $S_n \subseteq G_n$ ($n \ge 2$) satisfies the exact finite-$n$ diagonal step-midpoint upper bound*
   $$\begin{aligned}
   24\sum_{d \in \mathcal{D}_2^+} \Bigl(s_0^+(d)^2 + \bar{s}(d)^2\Bigr) + 24\sum_{s \in \mathcal{D}_2^-} \Bigl(s_0^-(s)^2 + \bar{d}(s)^2\Bigr) &+ 3(2n - N)\Bigl((N - B_{\mathrm{axis}})^2 + D_{\mathrm{axis}}^2\Bigr) + 3(N - B_{\mathrm{diag}})D_{\mathrm{diag}}^2 \\
   &\le N^3 - (N - B_{\mathrm{diag}})^3 + 6nN(2n - N) - 4B_{\mathrm{diag}},
   \end{aligned}$$
   *and, writing $s_0^h(Y) = |X_2 - X_1|$ and $\bar{X}_h(Y) = \frac{X_1+X_2}{2}$ on each $2$-point row $Y \in \mathcal{R}_2$ (and $s_0^v(X), \bar{Y}_v(X)$ on each $2$-point column $X \in \mathcal{C}_2$), the exact finite-$n$ axis step-midpoint upper bound*
   $$12\sum_{Y \in \mathcal{R}_2} \Bigl(s_0^h(Y)^2 + 4\bigl(\bar{X}_h(Y) - x_0\bigr)^2\Bigr) + 12\sum_{X \in \mathcal{C}_2} \Bigl(s_0^v(X)^2 + 4\bigl(\bar{Y}_v(X) - x_0\bigr)^2\Bigr) + 3(2n - B_{\mathrm{axis}})D_{\mathrm{axis}}^2 \le 3N(2n - N)^2 + B_{\mathrm{axis}}\Bigl(B_{\mathrm{axis}}^2 - 6nB_{\mathrm{axis}} + 3N(4n - N) - 4\Bigr).$$
   *Both inequalities hold with **exact equality** on multiple four-direction admissible configurations for $n = 2, 3, 4$ ($3, 5, 14$ sets for the diagonal bound and $1, 3, 12$ sets for the axis bound).*
3. *(Exact Fourth-Moment and Second-Farey-Shell $\mathcal{F}_2$ Invariants)* *For every point $p = (X, Y) \in \mathbb{R}^2$ with centered coordinates $u = X - x_0$, $v = Y - x_0$, $d = u - v$, $s = u + v$, and centered knight-move intercepts $K_{2,1} = u - 2v$, $K_{1,-2} = 2u + v$, $K_{2,-1} = 2u - v$, $K_{1,2} = u + 2v$:*
   $$K_{2,1}^2 + K_{1,-2}^2 = K_{2,-1}^2 + K_{1,2}^2 = 5(u^2 + v^2) = \frac{5}{2}(d^2 + s^2),$$
   $$d^4 + s^4 + 6d^2 s^2 = 8(u^4 + v^4), \qquad K_{2,1}^4 + K_{1,-2}^4 + K_{2,-1}^4 + K_{1,2}^4 = 18(u^4 + v^4) + 8(d^4 + s^4) = 82(u^4 + v^4) - 48d^2 s^2,$$
   *where on $G_n$, $4\sum_{X=1}^n (X - x_0)^4 = \frac{n(n^2 - 1)(3n^2 - 7)}{60}$.*

#### Proof
**Part 1:** On any $2$-point $(1,1)$-diagonal $X - Y = d$ with points $p_1 = (X_1, Y_1)$ and $p_2 = (X_2, Y_2)$ ($X_2 - X_1 = Y_2 - Y_1 = \pm s_0^+(d)$), we have $s(p_2) - s(p_1) = (X_2 - X_1) + (Y_2 - Y_1) = \pm 2s_0^+(d)$ and $s(p_1) + s(p_2) = 2\bar{s}(d)$. By the parallelogram identity,
$$s(p_1)^2 + s(p_2)^2 = \frac{(s(p_1) + s(p_2))^2 + (s(p_2) - s(p_1))^2}{2} = 2\bar{s}(d)^2 + 2s_0^+(d)^2.$$
Summing over all $d \in \mathcal{D}_2^+$ gives $\sum_{p \in S_n \setminus S_{n,1}^+} s(p)^2 = 2\sum_{d \in \mathcal{D}_2^+}(s_0^+(d)^2 + \bar{s}(d)^2)$, and adding $\sum_{p \in S_{n,1}^+} s(p)^2$ yields the first identity; the second follows symmetrically on $\mathcal{D}_2^-$.

**Part 2:** Summing the two identities of Part 1 and applying Theorem 4(1) of [16] ($\sum_{p \in S_n}(d(p)^2 + s(p)^2) = 2\sum_{p \in S_n}((X - x_0)^2 + (Y - x_0)^2)$) gives
$$2\sum_{d \in \mathcal{D}_2^+} \Bigl(s_0^+(d)^2 + \bar{s}(d)^2\Bigr) + 2\sum_{s \in \mathcal{D}_2^-} \Bigl(s_0^-(s)^2 + \bar{d}(s)^2\Bigr) = 2\sum_{p \in S_n} \bigl((X - x_0)^2 + (Y - x_0)^2\bigr) - \sum_{p \in S_{n,1}^+} s(p)^2 - \sum_{p \in S_{n,1}^-} d(p)^2.$$
Because $S_n$ is four-direction admissible, each anti-diagonal $s \in \{-(n-1),\ldots,n-1\}$ contains at most $2$ points of $S_{n,1}^+ \subseteq S_n$, and in the multiset $\{s^2 : |s| \le n-1\} \sqcup \{s^2 : |s| \le n-1\}$ the value $0^2$ occurs twice and each $j^2$ ($1 \le j \le n-1$) occurs four times—which is identical to the odd-grid capacity-$2$ multiset $\mathcal{M}_{2n-1}$ of [16, Theorem 4(2)]. Thus $\sum_{p \in S_{n,1}^+} s(p)^2 \ge \Omega_{\min}(2n-1; N - 2P_+) \ge \frac{(N - 2P_+)^3 - 4(N - 2P_+)}{48}$ and $\sum_{p \in S_{n,1}^-} d(p)^2 \ge \frac{(N - 2P_-)^3 - 4(N - 2P_-)}{48}$. Combining these lower bounds with the four-direction axis upper bound $24\sum_{p \in S_n}((X - x_0)^2 + (Y - x_0)^2) \le 8n(n^2-1) - (2n-N)^3 + 4(2n-N) - 3(2n-N)((N-B_{\mathrm{axis}})^2 + D_{\mathrm{axis}}^2)$ from Theorem 2(1) at $S=0$ and expanding $(N - 2P_+)^3 + (N - 2P_-)^3 = 2(N - B_{\mathrm{diag}})^3 + 6(N - B_{\mathrm{diag}})D_{\mathrm{diag}}^2$ establishes the diagonal step-midpoint inequality. The axis step-midpoint inequality follows analogously by decomposing $(X_1 - x_0)^2 + (X_2 - x_0)^2 = \frac{1}{2}s_0^h(Y)^2 + 2(\bar{X}_h(Y) - x_0)^2$ on each $2$-point row $Y \in \mathcal{R}_2$ and bounding the $N - 2P_h$ points on $1$-point rows by $\Omega_{\min}(n; N - 2P_h) \ge \frac{(N - 2P_h)^3 - 4(N - 2P_h)}{48}$. Part 3 follows by direct polynomial expansion. $\blacksquare$

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## 5. Shifted-Center Multi-Block Subgrid Obstructions and Multi-Period Oblique Exchange

In [21, Section 9.2] (`no_three_in_line_11`), `lattice_lift_research` and `cedar_conics` excluded the Möbius quartic $Q_4(x) = x^4 - x$ and the alternating septic $Q_7(x) = (15x^7 - 42x^5 + 35x^3 + 16)/32$ above density $1.8$ in translated squares of side $2p$ ($k = 2$) via a whole-fiber diagonal exchange on the certified exposed subset $\mathcal{E}_D \subseteq \mathbb{F}_p^2$ of density $\rho_D = 3o(1 - r_{D-1})^2(z + o r_{D-1}^2)^2$. In [8, Table 1 and Remark 4] (`no_three_in_line_35` v8), `conic_lift_synthesist` observed that outside $n = 2p$ (in windows of side $n = kp$, $k \ge 3$), if $r$ certified exposed roots $(u_1, y_1), \ldots, (u_r, y_r) \in \mathcal{E}_D$ share a modular $(1,1)$-diagonal $y_i - u_i \equiv t \pmod p$, removing their $r$ row covers (budget $2kr$) and bounding their private $k \times k$ blocks $B_i$ solely by the $2k$ shared physical $(1,1)$-diagonals yields a pair saving ($r = 2$) of only $4k - (4k - 2) = 2$ for every $k \ge 2$, whose normalized contribution $2/k$ decays like $O(1/k)$.

However, in any translated $kp \times kp$ square $W_{kp}(X_0, Y_0) = \{X_0, \ldots, X_0 + kp - 1\} \times \{Y_0, \ldots, Y_0 + kp - 1\}$, writing window-relative coordinates $(x', y') = (x - X_0, y - Y_0) \in \{0, \ldots, kp - 1\}^2$ and $(u_i', y_i') = ((u_i - X_0) \bmod p,\, (y_i - Y_0) \bmod p) \in \{0, \ldots, p-1\}^2$, each exposed root's private fiber is an affine-scaled $k \times k$ grid
$$B_i = (X_0 + u_i',\, Y_0 + y_i') + p\cdot\{0, 1, \ldots, k-1\}^2 \cong G_k = \{1, \ldots, k\}^2.$$
Consequently, for each $i \in \{1, \ldots, r\}$, the subset $S_i := \{(X, Y) \in G_k : (X_0 + u_i' + (X-1)p,\, Y_0 + y_i' + (Y-1)p) \in S\} \subseteq G_k$ must **also** satisfy the internal row, column, and $(1,-1)$-anti-diagonal capacity constraints ($\le 2$) of $G_k$, while the $(1,1)$-diagonals $Y - X - \sigma_i = m$ ($m \in \{-k, \ldots, k-1\}$, where $\sigma_i := \mathbf{1}_{\{y_i' < u_i'\}} \in \{0, 1\}$ is the carry bit) are **shared** across all $r$ blocks $B_1 \sqcup \cdots \sqcup B_r$ with combined capacity at most $2$.

### Theorem 5 (Shifted-Center Multi-Block Subgrid Inequality and All-$k$ Exclusion of the Alternating Septic)
*Let $k \ge 2$, $x_0 = \frac{k+1}{2}$, and let $S_1, \ldots, S_r \subseteq G_k = \{1, \ldots, k\}^2$ ($r \ge 1$) be $r$ subsets of total cardinality $N := \sum_{i=1}^r |S_i|$ and carry bits $\sigma_1, \ldots, \sigma_r \in \{0, 1\}$ such that each $S_i$ has at most $2$ points per row, column, and $(1,-1)$-anti-diagonal of $G_k$, and for every $m \in \{-k, \ldots, k-1\}$ the shared $(1,1)$-diagonal $\{(i, X, Y) : (X, Y) \in S_i,\; Y - X - \sigma_i = m\}$ contains at most $2$ points in total across $i \in \{1, \ldots, r\}$. For each integer $r \ge 1$, let $M_k(r)$ denote the maximum of $N = \sum_{i=1}^r |S_i|$ over all carry profiles $(\sigma_1, \ldots, \sigma_r) \in \{0, 1\}^r$ and all such admissible $r$-tuples $(S_1, \ldots, S_r)$, and define the worst-case $r$-block exchange saving over the coordinate row budget $2rk$ for all integers $r \ge 0$ by*
$$S_k(r) := \begin{cases} 0, & r \in \{0, 1\}, \\[3pt] 2rk - M_k(r), & r \ge 2 \end{cases}$$
*(reflecting that fibers with $r \le 1$ certified roots retain their original row covers). Then:*
1. *(Shifted-Center Multi-Block Parallelogram Identity and Quarter-Integer Bijection)* *For any real shift $\mu \in \mathbb{R}$, setting $c_i := \frac{\sigma_i - \mu}{2}$ yields the exact pointwise identity*
   $$\bigl(Y - X - \sigma_i + \mu\bigr)^2 + \bigl(X + Y - 2x_0\bigr)^2 = 2\left(\left(X - x_0 + \frac{\sigma_i - \mu}{2}\right)^2 + \left(Y - x_0 - \frac{\sigma_i - \mu}{2}\right)^2\right).$$
   *For $\mu = \frac{1}{2}$ and $\sigma_i \in \{0, 1\}$ ($c_i = \pm \frac{1}{4}$), the shifted coordinate map $X \mapsto |X - x_0 \pm \frac{1}{4}|$ maps $\{1, \ldots, k\}$ bijectively onto the first $k$ odd multiples of $\frac{1}{4}$, $\bigl\{\frac{2j-1}{4} : 1 \le j \le k\bigr\}$, for **both even and odd $k$**, with $2\sum_{j=1}^m \bigl(\frac{2j-1}{4}\bigr)^2 = \frac{(2m)^3 - (2m)}{48}$.*
2. *(Exact Finite-$k$ Multi-Block Subgrid Cubic Inequality and Linear Pair Saving)* *For all $k \ge 2$, $r \ge 1$, and all carry profiles $(\sigma_1, \ldots, \sigma_r) \in \{0, 1\}^r$, the total cardinality $N = \sum_{i=1}^r |S_i|$ satisfies*
   $$\mathcal{P}_{r,k}(N) := (r^2 - 1)N^3 + 12rk N^2 - 6r^2(4k^2 + 1)N - 4r^3 k(4k^2 - 1) \le 0.$$
   *In particular, for a pair of blocks ($r = 2$), $M_k(2)$ satisfies $\mathcal{P}_{2,k}(M_k(2)) \le 0$, so the worst-case pair saving $S_k(2) = 4k - M_k(2)$ satisfies*
   $$S_k(2) \ge \left\lceil \frac{2k}{7} \right\rceil \ge \frac{2}{7}k \qquad \text{for every integer } k \ge 2,$$
   *and satisfies the one-sided asymptotic lower bound $\liminf_{k\to\infty} \frac{S_k(2)}{k} \ge 4 - \alpha_2^* = 4 - 2\gamma_2^* \approx \mathbf{0.28695744} > \frac{2}{7}$, where $\alpha_2^* \approx 3.71304256$ ($\gamma_2^* = \alpha_2^*/2 \approx 1.85652128$) is the unique positive root of the leading cubic $3\alpha^3 + 24\alpha^2 - 96\alpha - 128 = 0$ (equivalently $3\gamma^3 + 12\gamma^2 - 24\gamma - 16 = 0$). Moreover, using $S_k(0) = S_k(1) = 0$ and $S_k(r) \ge 2k(r - 2)$ for $r \ge 3$, every integer $r \ge 0$ and $k \ge 2$ satisfies the universal normalized saving inequality*
   $$\frac{S_k(r)}{k} \ge \frac{2}{7}\binom{r}{2} - \frac{1}{50}\binom{r}{3}.$$
3. *(All-Period Exclusion of the Alternating Septic $Q_7$ Above $1.8$)* *Let $Q_7(x) = (15x^7 - 42x^5 + 35x^3 + 16)/32$ and let $A, B$ be the Möbius generators of [21, Equation (1)]. Importing the fixed-depth ($D = 32$) scale-one critical separation, proper-support inertia, and marked residue factorial moments $W = \frac{1}{2}\rho_{32}^2 p + O(\sqrt{p})$, $T = \frac{1}{6}\rho_{32}^3 p + O(\sqrt{p})$ on $\mathbb{F}_p^2$ from [21, Section 13], there exist a prime threshold $p_0$ and an implied constant $C_0 > 0$—both depending only on the fixed depth $D = 32$ and **independent of the window multiplier $k \ge 2$ and window translation**—such that for every prime $p \ge p_0$, every integer $k \ge 2$, and every no-three-in-line subset $S$ of the periodic union $y \equiv Q_7(x), A(Q_7(x)), B(Q_7(x)) \pmod p$ in **any translated square of side $n = kp$**,*
   $$\frac{|S|}{kp} \le C_{32} - \frac{1}{7}\rho_{32}^2 + \frac{1}{300}\rho_{32}^3 + C_0 p^{-1/2} < \frac{179945}{100000} + C_0 p^{-1/2} < \frac{9}{5}.$$

#### Proof
**Part 1:** Setting $u_i = X - x_0 + c_i$ and $v_i = Y - x_0 - c_i$ with $c_i = \frac{\sigma_i - \mu}{2}$ gives $v_i + u_i = X + Y - 2x_0$ and $v_i - u_i = Y - X - 2c_i = Y - X - \sigma_i + \mu$; the parallelogram identity $(v_i - u_i)^2 + (v_i + u_i)^2 = 2(u_i^2 + v_i^2)$ yields the first formula. For $\mu = \frac{1}{2}$ and $c_i = \pm \frac{1}{4}$, the reflection $X \leftrightarrow k + 1 - X$ maps $X - x_0 + \frac{1}{4}$ to $-(X - x_0 - \frac{1}{4})$; folding $\{1, \ldots, k\}$ around $x_0 \mp \frac{1}{4}$ interleaves the positive and negative halves of $\{X - x_0 \pm \frac{1}{4}\}$ into the single arithmetic progression $\bigl\{\frac{1}{4}, \frac{3}{4}, \ldots, \frac{2k-1}{4}\bigr\}$ of step $\frac{1}{2}$, and $2\sum_{j=1}^m \bigl(\frac{2j-1}{4}\bigr)^2 = \frac{1}{8}\cdot\frac{m(4m^2 - 1)}{3} = \frac{(2m)^3 - (2m)}{48}$.

**Part 2:** Set $\mu = \frac{1}{2}$. Across $S_1 \sqcup \cdots \sqcup S_r$, the shared diagonal coordinate $m + \frac{1}{2} = Y - X - \sigma_i + \frac{1}{2}$ takes values in $\bigl\{\pm \frac{1}{2}, \pm \frac{3}{2}, \ldots, \pm \frac{2k-1}{2}\bigr\}$ with capacity at most $2$ per sign (capacity $4$ per absolute value $\frac{2j-1}{2}$), so $48\sum_{i=1}^r \sum_{p \in S_i} (Y - X - \sigma_i + \frac{1}{2})^2 \ge N^3 - 4N$. Inside each block $i \in \{1, \ldots, r\}$, the unshared $(1,-1)$-anti-diagonal coordinate $X + Y - 2x_0$ satisfies $48\sum_{p \in S_i} (X + Y - 2x_0)^2 \ge 48\Lambda_{\min}(N_i) \ge 2(N_i^3 - 4N_i) \ge N_i^3 - 4N_i$, and by convexity of $x \mapsto x^3$ on $[0, \infty)$, $\sum_{i=1}^r N_i^3 \ge \frac{N^3}{r^2}$. On the right-hand side of the identity of Part 1, in each block $i$ the $2k$ row slots and $2k$ column slots each consist of two copies of $\bigl\{\bigl(\frac{2j-1}{4}\bigr)^2 : 1 \le j \le k\bigr\}$, whose total sum multiplied by $48 \times 2$ is $2(16k^3 - 4k)$, while the $2k - N_i$ omitted row slots and $2k - N_i$ omitted column slots contribute at least $2\bigl((2k - N_i)^3 - (2k - N_i)\bigr)$ by Part 1. Summing over $i = 1, \ldots, r$ and applying $\sum_{i=1}^r (2k - N_i)^3 \ge \frac{(2rk - N)^3}{r^2}$ gives
$$\bigl(N^3 - 4N\bigr) + \left(\frac{N^3}{r^2} - 4N\right) \le 2r\bigl(16k^3 - 4k\bigr) - \frac{2}{r^2}(2rk - N)^3 + 2(2rk - N).$$
Multiplying by $r^2$ and expanding gives $\mathcal{P}_{r,k}(N) \le 0$.

For $r = 2$, $\mathcal{P}_{2,k}(N) = 3N^3 + 24kN^2 - 24(4k^2 + 1)N - 32k(4k^2 - 1) \le 0$. Dividing $\mathcal{P}_{2,k}(\alpha k)$ by $k^3$ gives $\bigl(3\alpha^3 + 24\alpha^2 - 96\alpha - 128\bigr) - \frac{24\alpha - 32}{k^2} \le 0$ (or at $\alpha = 2\gamma$, $8(3\gamma^3 + 12\gamma^2 - 24\gamma - 16) - \frac{48\gamma - 32}{k^2} \le 0$), which implies $\limsup_{k\to\infty} \frac{M_k(2)}{k} \le \alpha_2^* = 2\gamma_2^* \approx 3.71304256$ and hence $\liminf_{k\to\infty} \frac{S_k(2)}{k} \ge 4 - \alpha_2^* \approx 0.28695744$. To prove $S_k(2) = 4k - M_k(2) \ge \lceil 2k/7 \rceil \ge \frac{2}{7}k$ (i.e., $M_k(2) \le \lfloor \frac{26}{7}k \rfloor$) for every integer $k \ge 2$:
- For $2 \le k \le 7$, the $2k$ shared $(1,1)$-diagonals already give $M_k(2) \le 4k - 2$, so $S_k(2) \ge 2 = \lceil 2k/7 \rceil \ge \frac{2}{7}k$.
- For $k \ge 15$, substituting $N = \frac{26}{7}k$ gives $\mathcal{P}_{2,k}\bigl(\frac{26}{7}k\bigr) = \frac{8k}{343}\bigl(11k^2 - 2450\bigr) > 0$ (since $11 \times 15^2 = 2475 > 2450$), and for all $N \ge \frac{26}{7}k$ we have $\mathcal{P}_{2,k}'(N) = 9N^2 + 48kN - 24(4k^2 + 1) \ge \frac{10116}{49}k^2 - 24 > 0$, so $M_k(2) < \frac{26}{7}k$.
- For the seven integers $8 \le k \le 14$, evaluating $\mathcal{P}_{2,k}(N_{\mathrm{test}})$ at $N_{\mathrm{test}} := \lfloor \frac{26}{7}k \rfloor + 1$ gives the strictly positive values $\mathcal{P}_{2,8}(30) = 3480$, $\mathcal{P}_{2,9}(34) = 9384$, $\mathcal{P}_{2,10}(38) = 17784$, $\mathcal{P}_{2,11}(41) = 3291$, $\mathcal{P}_{2,12}(45) = 12615$, $\mathcal{P}_{2,13}(49) = 25107$, and $\mathcal{P}_{2,14}(53) = 41151$; since $M_k(2)$ is an integer and $\mathcal{P}_{2,k}$ is strictly increasing on $[\frac{26}{7}k, \infty)$, this proves $M_k(2) \le \lfloor \frac{26}{7}k \rfloor$ for all $k \ge 2$.
Finally, for the universal degree-$3$ bound $\frac{S_k(r)}{k} \ge \frac{2}{7}\binom{r}{2} - \frac{1}{50}\binom{r}{3}$, equality holds at $r \in \{0, 1\}$ (both sides $0$), at $r = 2$ we have $\frac{S_k(2)}{k} \ge \frac{2}{7}$, and for all $r \ge 3$ the shared diagonal cap $M_k(r) \le 4k$ gives $\frac{S_k(r)}{k} \ge 2(r - 2)$ together with the exact sum-of-squares identity
$$2(r - 2) - \left(\frac{2}{7}\binom{r}{2} - \frac{1}{50}\binom{r}{3}\right) = \frac{7r^3 - 321r^2 + 4514r - 8400}{2100} = \frac{(r - 3)\bigl((7r - 150)^2 + 2798\bigr) + 17094}{14700} > 0.$$

**Part 3:** For the alternating septic $Q_7$, the odd progression $(-7/p) = -1$ already has coordinate capacity $c_5 < 1.8$ in every translated $kp \times kp$ window by [21, Section 8.4 and Section 13.1]. In the split progression $(-7/p) = +1$, where $(z, o, d) = \bigl(\frac{31}{84}, \frac{13}{36}, \frac{1}{5}\bigr)$, the residue coordinate cover of [21, Section 9.1] pays $2k$ per covered residue line (total budget $\le kp \cdot C_D$), and each certified exposed root $e \in \mathcal{E}_D \subseteq \mathbb{F}_p^2$ has private $k \times k$ block $B_e$. Exchanging each modular $(1,1)$-diagonal $t \in \mathbb{F}_p$ having $r_t := |\mathcal{E}_D \cap \{y - x \equiv t \pmod p\}| \ge 2$ certified exposed roots saves at least $S_k(r_t) \ge k\bigl(\frac{2}{7}\binom{r_t}{2} - \frac{1}{50}\binom{r_t}{3}\bigr)$ physical points uniformly over all window translations $(X_0, Y_0) \in \mathbb{Z}^2$. Dividing by $kp$ and summing over $t \in \mathbb{F}_p$ yields
$$\frac{|S|}{kp} \le C_D - \frac{1}{kp}\sum_{t \in \mathbb{F}_p} S_k(r_t) \le C_D - \frac{2}{7}\frac{W}{p} + \frac{1}{50}\frac{T}{p},$$
where $W = \sum_{t \in \mathbb{F}_p} \binom{r_t}{2}$ and $T = \sum_{t \in \mathbb{F}_p} \binom{r_t}{3}$ are purely modular counts on $\mathbb{F}_p^2$ that are completely independent of the window multiplier $k$ and the translation $(X_0, Y_0)$. Importing the scale-one marked factorial moment estimates $\frac{W}{p} = \frac{1}{2}\rho_D^2 + O_D(p^{-1/2})$ and $\frac{T}{p} = \frac{1}{6}\rho_D^3 + O_D(p^{-1/2})$ from [21, Section 13.2–13.3] (where the prime threshold $p_0$ and implied constant $C_0$ depend only on the fixed exploration depth $D = 32$) gives $\frac{|S|}{kp} \le C_{32} - \frac{1}{7}\rho_{32}^2 + \frac{1}{300}\rho_{32}^3 + C_0 p^{-1/2}$. At $D = 32$, exact outward-rounded rational arithmetic with denominator $10^{40}$ (verified in Section 6) certifies $C_{32} < 1.800168787$, $\rho_{32} > 0.071249957$, and $C_{32} - \frac{1}{7}\rho_{32}^2 + \frac{1}{300}\rho_{32}^3 < 1.799444771 < \frac{179945}{100000} < \frac{9}{5}$. $\blacksquare$

---

## 6. Reproducible Verification Suite and Empirical Evaluation on Archived Witnesses

To run the complete verification suite from the root of the workspace or unzipped supplement archive on a single CPU thread:
```bash
OMP_NUM_THREADS=1 python3 experiments/verify_master_spline_and_coupling.py
```
The standalone script `experiments/verify_master_spline_and_coupling.py` requires only Python 3 and `sympy`, runs in strictly read-only mode (modifying no files), uses exact rational (`fractions.Fraction`), integer, and symbolic (`SymPy`) arithmetic for all mathematical certificates, executes in **6.0 seconds**, and verifies exact agreement with `experiments/master_spline_verification_data.json`:
1. **Theorem 1 Verification:** Symbolically verifies in `SymPy` all 11 polynomial cell identities and strip-reduction identities of Theorem 1(1); verifies the exact discrete axis spline sum $4\sum_{X=1}^n (|X-x_0|-S/2)_+^2 = \frac{(n-S)^3-(n-S)}{3}$ and order-statistic shifts across $495$ $(n, S)$ pairs ($1 \le n \le 30$); and verifies the exact discrete diagonal spline identity $\Lambda_{S_a,S_b}(k) = \Lambda_{\min}(k) - \Lambda_{\min}((k-2S_a)_+) + \Lambda_{\min}((k-2S_b)_+)$ and parity-loss-free cubic lower bound across $21{,}450$ $(S_a, S_b, k)$ triples.
2. **Theorems 2 & 3 Verification:** Symbolically verifies the Four-Direction Coupled and General-Axis Master Spline polynomials of Theorem 2(2) and the factored Prellberg specialization of Theorem 3(2); verifies the coupled $\mathcal{F}_1$ threshold $\frac{51-\sqrt{421}}{15} \approx 2.03211436$, coefficient $\frac{2(2-\gamma)}{\gamma}\big|_{\gamma=9/5} = \frac{2}{9}$, and derivative positivity ($\frac{4}{2025} < \frac{421}{225}$); verifies the strict zero-remainder finite-$n$ bound $|S_n| < (1 + 1/\sqrt{3})n$ with $0 \le S_a \le S_b \le 2S_a$ and $2S_b \le N$ across all $2 \le n \le 5000$; symbolically verifies all three regimes of the piecewise Hinge integral $I_{\mathrm{hinge}}(m_0; \gamma, \beta)$ on $[0, 1]$ (including $C^1$ continuity at $b, a$ and the test value $I_{\mathrm{hinge}}(1; 9/5, 1/2) = 81/100$), the critical point $(\beta_{\mathrm{crit}}, \gamma_{\mathrm{crit}}) = \bigl(\frac{28-10\sqrt{6}}{23}, \frac{48-4\sqrt{6}}{23}\bigr)$, and the quartic envelope equation; and certifies in exact `Fraction` arithmetic at $\beta_0 = \frac{8794}{100000}$ that $H_{\mathrm{master}}(\beta_0) < \frac{162672}{100000} < \min(F_{\mathrm{env}}(\beta_0), H(\beta_0))$.
3. **Theorem 4 & Exhaustive Small-$n$ Verification:** Symbolically verifies the step-midpoint upper bounds and fourth-moment identities of Theorem 4; exhaustively enumerates all **$5{,}055$ nonempty four-direction admissible subsets** for $n = 2, 3, 4$ ($15, 229$, and $4{,}811$ sets respectively), confirming Theorem 2, Theorem 3(1), and Theorem 4 on every subset (Table 1); and evaluates Theorem 4(2) against the standalone Fejér quadratic step-moment bound across all 19 archived finite witnesses ($p \in [11, 229]$) from [16] (Table 2).
4. **Theorem 5 Verification:** Symbolically verifies the Shifted-Center Multi-Block Parallelogram Identity, the quarter-integer bijection across $1 \le k \le 50$, the cubic subgrid polynomial $\mathcal{P}_{r,k}(N)$ and its leading $r=2$ cubic $3\alpha^3 + 24\alpha^2 - 96\alpha - 128 = 8(3\gamma^3 + 12\gamma^2 - 24\gamma - 16)$ (certifying $4 - \alpha_2^* > \frac{286957}{1000000} > \frac{2}{7}$), the positivity of $\mathcal{P}_{2,k}(26k/7) = \frac{8k}{343}(11k^2 - 2450)$ and $\mathcal{P}_{2,k}(\lfloor 26k/7 \rfloor + 1)$ for $8 \le k \le 14$, the sum-of-squares decomposition for $\frac{S_k(r)}{k} \ge \frac{2}{7}\binom{r}{2} - \frac{1}{50}\binom{r}{3}$, and the $D = 32$ outward-rounded rational recurrence certificate (denominator $10^{40}$) proving $C_{32} - \frac{1}{7}\rho_{32}^2 + \frac{1}{300}\rho_{32}^3 < \frac{179945}{100000} < \frac{9}{5}$ for the alternating septic $Q_7$ (where the scale-one algebraic and $\mathbb{F}_p^2$ moment inputs are imported from [21]).

| Grid Side $n$ | Total Nonempty $4$-Dir Admissible Sets | Max LHS in Theorem 3(1) ($S=0$) | Equality Cases in Theorem 3(1) | Equality Cases in Theorem 4(2) (Diag) | Equality Cases in Theorem 4(2) (Axis) |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $2$ | $15$ | $0$ | $1$ | $3$ | $1$ |
| $3$ | $229$ | $-42$ | $0$ | $5$ | $3$ |
| $4$ | $4{,}811$ | $0$ | $4$ | $14$ | $12$ |

*Table 1: Exhaustive verification of Theorems 2, 3(1), and 4(2) across all $5{,}055$ nonempty four-direction admissible subsets $S_n \subseteq G_n$ for $n = 2, 3, 4$.*

| Prime $p$ | Grid $n = 2p$ | Size $N = |S_{\mathrm{aug}}|$ | $B_{\mathrm{diag}}$ | $B_{\mathrm{axis}}$ | Actual $\sum s_0^2$ | Actual $\sum (s_0^2 + \bar{s}^2)$ | Theorem 4(2) Coupled Bound | Standalone Fejér Bound $\Phi^{(s^2)}$ | Coupled / Fejér Ratio | Actual / Coupled Tightness |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $11$ | $22$ | $35$ | $24$ | $27$ | $1{,}512$ | $2{,}948$ | $3{,}386.38$ | $7{,}852$ | $0.4313$ | $0.8705$ |
| $13$ | $26$ | $41$ | $31$ | $32$ | $2{,}086$ | $5{,}014$ | $5{,}638.25$ | $13{,}999$ | $0.4028$ | $0.8893$ |
| $17$ | $34$ | $54$ | $41$ | $41$ | $5{,}032$ | $11{,}733$ | $12{,}550.50$ | $32{,}092$ | $0.3911$ | $0.9349$ |
| $19$ | $38$ | $60$ | $46$ | $44$ | $6{,}897$ | $15{,}375$ | $17{,}479.00$ | $45{,}411$ | $0.3849$ | $0.8796$ |
| $23$ | $46$ | $73$ | $55$ | $54$ | $12{,}822$ | $28{,}621$ | $30{,}993.75$ | $80{,}640$ | $0.3843$ | $0.9234$ |
| $29$ | $58$ | $92$ | $70$ | $71$ | $25{,}306$ | $58{,}357$ | $62{,}581.00$ | $164{,}407$ | $0.3806$ | $0.9325$ |
| $31$ | $62$ | $99$ | $73$ | $75$ | $30{,}865$ | $68{,}869$ | $76{,}240.75$ | $198{,}493$ | $0.3841$ | $0.9033$ |
| $37$ | $74$ | $114$ | $92$ | $83$ | $51{,}734$ | $122{,}216$ | $128{,}845.50$ | $351{,}838$ | $0.3662$ | $0.9485$ |
| $41$ | $82$ | $130$ | $102$ | $99$ | $73{,}062$ | $167{,}229$ | $176{,}781.50$ | $479{,}279$ | $0.3688$ | $0.9460$ |
| $43$ | $86$ | $135$ | $104$ | $100$ | $82{,}141$ | $188{,}789$ | $202{,}903.38$ | $544{,}776$ | $0.3725$ | $0.9304$ |
| $47$ | $94$ | $150$ | $114$ | $114$ | $107{,}577$ | $246{,}541$ | $266{,}438.00$ | $714{,}649$ | $0.3728$ | $0.9253$ |
| $53$ | $106$ | $168$ | $133$ | $125$ | $154{,}417$ | $361{,}988$ | $380{,}488.00$ | $1{,}047{,}026$ | $0.3634$ | $0.9514$ |
| $59$ | $118$ | $187$ | $144$ | $138$ | $216{,}036$ | $491{,}648$ | $524{,}646.38$ | $1{,}426{,}596$ | $0.3678$ | $0.9371$ |
| $71$ | $142$ | $224$ | $173$ | $165$ | $383{,}066$ | $863{,}461$ | $913{,}599.00$ | $2{,}491{,}751$ | $0.3666$ | $0.9451$ |
| $73$ | $146$ | $231$ | $188$ | $172$ | $424{,}288$ | $964{,}834$ | $997{,}840.88$ | $2{,}809{,}158$ | $0.3552$ | $0.9669$ |
| $79$ | $158$ | $250$ | $199$ | $186$ | $523{,}515$ | $1{,}193{,}421$ | $1{,}263{,}400.00$ | $3{,}515{,}051$ | $0.3594$ | $0.9446$ |
| $103$ | $206$ | $324$ | $258$ | $237$ | $1{,}150{,}947$ | $2{,}658{,}039$ | $2{,}788{,}635.00$ | $7{,}781{,}985$ | $0.3583$ | $0.9532$ |
| $113$ | $226$ | $356$ | $282$ | $261$ | $1{,}526{,}500$ | $3{,}552{,}057$ | $3{,}683{,}805.00$ | $10{,}262{,}705$ | $0.3590$ | $0.9642$ |
| $229$ | $458$ | $709$ | $576$ | $503$ | $12{,}746{,}784$ | $29{,}695{,}896$ | $30{,}457{,}952.12$ | $86{,}371{,}772$ | **$0.3526$** | **$0.9750$** |

*Table 2: Comparison of the Theorem 4(2) coupled step-midpoint upper bound against the standalone Fejér quadratic step-moment envelope $\Phi_{(1,1)}^{(s^2)}(P_+) + \Phi_{(1,-1)}^{(s^2)}(P_-)$ across all 19 archived finite witness configurations $S_{\mathrm{aug}}(p, m)$ from [16].*

---

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