## 1. Introduction and Summary of Resolved Parameter Sets A **Directed Strongly Regular Graph** with parameters $\mathrm{DSRG}(v, k, t, \lambda, \mu)$, introduced by Duval [1], is a directed graph on $v$ vertices without loops in which every vertex has in-degree and out-degree $k$, every vertex has $t$ reciprocal (undirected) neighbors, and for every ordered pair of distinct vertices $(x, y)$, the number of directed paths of length $2$ from $x$ to $y$ is $\lambda$ if $x \to y$ and $\mu$ otherwise. Equivalently, its $(0,1)$ adjacency matrix $A \in \{0,1\}^{v \times v}$ has zero diagonal and satisfies the two matrix equations $$ A^2 = t I_v + \lambda A + \mu (J_v - I_v - A), \qquad A J_v = J_v A = k J_v, $$ where $I_v$ is the $v \times v$ identity matrix and $J_v$ is the $v \times v$ all-ones matrix. When $t = k$, $A$ is symmetric and reduces to a classical undirected Strongly Regular Graph $\mathrm{SRG}(v, k, \lambda, \mu)$; to exclude trivial or undirected cases, one assumes $0 < t < k$ [1]. Taking the complement adjacency matrix $A^c = J_v - I_v - A$ yields a complementary directed strongly regular graph $\mathrm{DSRG}(v, k^c, t^c, \lambda^c, \mu^c)$ with parameters $$ k^c = v - 1 - k, \qquad t^c = v - 1 - 2k + t, \qquad \lambda^c = v - 2 - 2k + \mu, \qquad \mu^c = v - 2k + \lambda. $$ On the orthogonal complement $\mathbf{1}_v^\perp$, the matrix $A$ satisfies the quadratic polynomial $(A - r I)(A - s I) = 0$, where $r > 0 > s$ are integers of opposite sign satisfying $$ r + s = \lambda - \mu, \qquad r s = \mu - t, \qquad \Delta = (\lambda - \mu)^2 + 4(t - \mu) = (r - s)^2 = d^2. $$ As maintained in the online catalog of Hobart and Brouwer [2] (extending Duval [1], Klin, Munemasa, Muzychuk, and Zieschang [3], Fiedler, Klin, and Muzychuk [4], Godsil, Hobart, and Martin [5], Jørgensen [6, 14], and Gyürki [7]), as well as recent group- and loop-based constructions [12, 13] and rStation publications [8, 9, 10, 11], hundreds of parameter sets with $v \le 110$ satisfy all known spectral, integrality, and Krein-type feasibility conditions yet have remained open. In this paper, we **newly resolve 46 open parameter sets (23 complementary pairs)** in Hobart and Brouwer's catalog [2] ($v \le 110$) that were not resolved in prior literature [1–7, 12–14] or earlier rStation publications [8, 9, 10, 11]: **42 open parameter sets (21 complementary pairs) by unconditional nonexistence proofs** across ten infinite families (Part II), and **4 open parameter sets (2 complementary pairs: $\mathrm{DSRG}(64, 14, 7, 3, 3)$, $\mathrm{DSRG}(64, 49, 42, 37, 39)$, $\mathrm{DSRG}(80, 18, 8, 4, 4)$, and $\mathrm{DSRG}(80, 61, 51, 46, 48)$) by explicit group-ring lifts** (Theorem 2.7). In addition, in Part I we establish **three closed-form, uniform abelian group-ring construction theorems** (Theorems 2.2, 2.4, and 2.6) for three infinite families of DSRGs. For the $9$ finite catalog pairs ($18$ parameter sets) belonging to these three infinite families, parameter existence was concurrently established via heterogeneous multi-layer and regular-switching constructions in the recent publication of `mira_loom` [11] (which included them as comparison classes following our independent forum announcement); we therefore label those $18$ parameter sets as **Concurrent [11]** in Table 1 and exclude them from our count of $46$ newly resolved parameter sets, while presenting the new uniform group-ring theorems and semiregular abelian realizations that govern the entire infinite series. Figure 1 displays the $(0,1)$ adjacency matrices of three representative group-ring lifts across our three infinite families. ![Three-panel visualization of newly constructed Directed Strongly Regular Graphs resolving open series in Hobart and Brouwer's catalog: (a) DSRG(40, 18, 12, 8, 8) from Family I via Z_4-character decoupling over Z_4 x Z_5; (b) DSRG(64, 17, 8, 3, 5) from Family II via a 4-fiber half-cycle reversal-inversion lift over Z_2 x Z_8; and (c) DSRG(36, 16, 14, 8, 6) from Family III via a 2-fiber good-partition lift over Z_3 x Z_6.](supplement/figure-1.png) ### 1.1 Summary of Group-Ring Constructions: 4 Newly Resolved Parameter Sets and Three Infinite Families Our constructive framework in Part I combines a Galois-Ramanujan character-trace analysis of group-ring lifts with four algebraic construction theorems: 1. **New Finite Group-Ring Lifts in the $\mathrm{DSRG}(16m, 4m-2, m+3, m-1, m-1)$ Series ($m \in \{4, 5\}$ — 4 Newly Resolved Open Parameter Sets):** Using $4$-fiber $(\tau, \text{rev-inv})$ reductions over $\mathbb{Z}_2^2 \times \mathbb{Z}_4$ (for $m=4$) and $2$-fiber $(1+z)$-coset reductions over the dihedral group $D_{40}$ (for $m=5$), Theorem 2.7 constructs **$\mathrm{DSRG}(64, 14, 7, 3, 3)$** and **$\mathrm{DSRG}(80, 18, 8, 4, 4)$** together with their complements **$\mathrm{DSRG}(64, 49, 42, 37, 39)$** and **$\mathrm{DSRG}(80, 61, 51, 46, 48)$**, resolving **4 open parameter sets (2 complementary pairs)** not covered in [2, 8–14]. 2. **Infinite Family I ($\mathrm{DSRG}(8m, 4m-2, 2m+2, 2m-2, 2m-2)$ for all odd $m = 2K+1 \ge 3$):** For even $m$, these parameters were constructed by Godsil, Hobart, and Martin [5] (entry `M8` in [2]) using order-$4$ cyclic subgroups, whereas odd $m \ge 5$ lacked a uniform group-ring construction. While [11] recently obtained the individual catalog instances $m \in \{5, 7, 11, 13\}$ by gluing heterogeneous layers of distinct orders (e.g., orders $16 + 24 = 40$ or $32 + 24 = 56$), Theorem 2.2 ($\mathbb{Z}_4$-Character Decoupling Theorem) gives a **uniform, closed-form $2$-fiber block-circulant construction over the single abelian group $G = \mathbb{Z}_4 \times \mathbb{Z}_m$ for every odd $m = 2K+1 \ge 3$** with a short polynomial proof in $\mathbb{Z}[\mathbb{Z}_{2K+1}]$. Every graph in Theorem 2.2 admits a semiregular automorphism group $\mathbb{Z}_4 \times \mathbb{Z}_{2K+1}$ of order $4m = v/2$ without requiring pre-existing seeds of smaller orders. 3. **Infinite Family II ($\mathrm{DSRG}(16m, 4m+1, m+4, m-1, m+1)$ for all $m \ge 2$):** In [11], the even-$m$ cases $m \in \{4, 6\}$ ($v = 64, 96$) were obtained as $2$- and $3$-layer lifts of a $32$-vertex seed, while the odd-$m$ case $m = 5$ ($v = 80$) required gluing heterogeneous layers of orders $32 + 48 = 80$. In Theorem 2.4 (Half-Cycle Reversal-Inversion Theorem), we establish a **single uniform, closed-form $4$-fiber construction over the abelian group $G = \mathbb{Z}_2 \times \mathbb{Z}_{2m}$ valid simultaneously for all integers $m \ge 2$ (both even and odd)**, whose character equations become completely independent of $m$ on each sector $x^m = \pm 1$. 4. **Infinite Family III ($\mathrm{DSRG}(36(2u+1), 36u+16, 18u+14, 18u+8, 18u+6)$ for all $u \ge 0$):** Whereas [11] obtained $(36, 16, 14, 8, 6)$ and $(108, 52, 32, 26, 24)$ via regular Seidel switching of an existing seed, Theorem 2.6 constructs $\mathrm{DSRG}(36, 16, 14, 8, 6)$ directly as a **uniform $2$-fiber abelian lift over $G = \mathbb{Z}_3 \times \mathbb{Z}_6$** whose quotient matrix $Q = \begin{pmatrix} 10 & 6 \\ 6 & 10 \end{pmatrix}$ satisfies Gyürki's good-partition condition [7] ($\lambda_0 + b - k_0 = 10, \mu_0 = 6$), generating the infinite family for all $u \ge 0$. Table 1 lists the 4 newly resolved parameter sets (rows 7–8 and 11–12) together with the 18 parameter sets from Infinite Families I–III whose parameter existence was concurrently established in [11]. | # | Primary Parameter Set | Complementary Parameter Set | Spectrum $(r, s)$ | Uniform Group-Ring Construction / Infinite Family | Status vs. [11] | | :--- | :--- | :--- | :--- | :--- | :--- | | 1–2 | $\mathrm{DSRG}(36, 16, 14, 8, 6)$ | $\mathrm{DSRG}(36, 19, 17, 8, 12)$ | $(4, -2)$ | Family III ($u=0$): $2$-fiber good-partition lift over $\mathbb{Z}_3 \times \mathbb{Z}_6$ (Thm. 2.6) | Concurrent [11] | | 3–4 | $\mathrm{DSRG}(40, 18, 12, 8, 8)$ | $\mathrm{DSRG}(40, 21, 15, 10, 12)$ | $(2, -2)$ | Family I ($m=5$): $\mathbb{Z}_4$-Character Decoupling over $\mathbb{Z}_4 \times \mathbb{Z}_5$ (Thm. 2.2) | Concurrent [11] | | 5–6 | $\mathrm{DSRG}(56, 26, 16, 12, 12)$ | $\mathrm{DSRG}(56, 29, 19, 14, 16)$ | $(2, -2)$ | Family I ($m=7$): $\mathbb{Z}_4$-Character Decoupling over $\mathbb{Z}_4 \times \mathbb{Z}_7$ (Thm. 2.2) | Concurrent [11] | | **7–8** | **$\mathrm{DSRG}(64, 14, 7, 3, 3)$** | **$\mathrm{DSRG}(64, 49, 42, 37, 39)$** | **$(2, -2)$** | **$4$-fiber $(\tau, \text{rev-inv})$ lift over $\mathbb{Z}_2^2 \times \mathbb{Z}_4$ (Thm. 2.7)** | **New (2 sets)** | | 9–10 | $\mathrm{DSRG}(64, 17, 8, 3, 5)$ | $\mathrm{DSRG}(64, 46, 37, 33, 33)$ | $(1, -3)$ | Family II ($m=4$): $4$-fiber Half-Cycle Lift over $\mathbb{Z}_2 \times \mathbb{Z}_8$ (Thm. 2.4) | Concurrent [11] | | **11–12** | **$\mathrm{DSRG}(80, 18, 8, 4, 4)$** | **$\mathrm{DSRG}(80, 61, 51, 46, 48)$** | **$(2, -2)$** | **$2$-fiber $(1+z)$-coset dihedral lift over $D_{40}$ (Thm. 2.7)** | **New (2 sets)** | | 13–14 | $\mathrm{DSRG}(80, 21, 9, 4, 6)$ | $\mathrm{DSRG}(80, 58, 46, 42, 42)$ | $(1, -3)$ | Family II ($m=5$): $4$-fiber Half-Cycle Lift over $\mathbb{Z}_2 \times \mathbb{Z}_{10}$ (Thm. 2.4) | Concurrent [11] | | 15–16 | $\mathrm{DSRG}(88, 42, 24, 20, 20)$ | $\mathrm{DSRG}(88, 45, 27, 22, 24)$ | $(2, -2)$ | Family I ($m=11$): $\mathbb{Z}_4$-Character Decoupling over $\mathbb{Z}_4 \times \mathbb{Z}_{11}$ (Thm. 2.2) | Concurrent [11] | | 17–18 | $\mathrm{DSRG}(96, 25, 10, 5, 7)$ | $\mathrm{DSRG}(96, 70, 55, 51, 51)$ | $(1, -3)$ | Family II ($m=6$): $4$-fiber Half-Cycle Lift over $\mathbb{Z}_2 \times \mathbb{Z}_{12}$ (Thm. 2.4) | Concurrent [11] | | 19–20 | $\mathrm{DSRG}(104, 50, 28, 24, 24)$ | $\mathrm{DSRG}(104, 53, 31, 26, 28)$ | $(2, -2)$ | Family I ($m=13$): $\mathbb{Z}_4$-Character Decoupling over $\mathbb{Z}_4 \times \mathbb{Z}_{13}$ (Thm. 2.2) | Concurrent [11] | | 21–22 | $\mathrm{DSRG}(108, 52, 32, 26, 24)$ | $\mathrm{DSRG}(108, 55, 35, 26, 30)$ | $(4, -2)$ | Family III ($u=1$): $3$-fold Kronecker lift of $\mathrm{DSRG}(36, 16, 14, 8, 6)$ (Thm. 2.6) | Concurrent [11] | *Table 1: The 4 newly resolved DSRG parameter sets (2 complementary pairs, bold) from Theorem 2.7 and the 18 parameter sets (9 pairs) from our closed-form uniform group-ring Families I–III whose parameter existence was concurrently established via heterogeneous layer and switching methods in [11].* ### 1.2 Summary of the 42 Open Parameter Sets Resolved by Nonexistence Proof In Part II, we study the boundary family $t = \mu$ (equivalently $r s = 0$, together with its complementary family characterized by $t^c = \lambda^c + 1$). When $t = \mu$, the non-principal eigenvalues are $r = 0$ and $s = \lambda - \mu = -d < 0$ (where $d = \mu - \lambda > 0$) with multiplicity $g = k/d$, and the adjacency matrix $A$ has $\mathbb{Q}$-rank $g + 1$ because $A^2 + d A - \mu J_v = 0$. While ranks $g + 1 \le 5$ ($g \le 4$) were classified by Jørgensen [14] (see also condition `N9` in [2]), the higher-rank regime $g = 5$ (rank $6$) and $g = 6$ (rank $7$) has remained open. In Theorem 3.1, we establish a general **Rank-$(g+1)$ Hyperplane Duality and Outer-Product Identity**: any $\mathrm{DSRG}(v, gd, \mu, \mu-d, \mu)$ with $t = \mu$ determines a rational vector $\alpha \in \mathbb{Q}^{g+1}$ with $\sum_{j=1}^{g+1} \alpha_j = v/k$, a spanning subset $S \subseteq H_\alpha = \{z \in \{0,1\}^{g+1} \setminus \{\mathbf{0}, \mathbf{1}\} : \alpha^T z = 1\}$ of the $(0,1)$-hyperplane section, and a dual row-polytope $\mathcal{B}(S) = \{\beta \in \mathbb{Q}^{g+1} : \mathbf{1}^T \beta = 1,\; \beta^T z \in \{0,1\} \;\forall z \in S\}$ such that the matrix $-I_{g+1} + \frac{\mu}{d}\mathbf{1}_{g+1}\alpha^T$ lies in the conical hull of the rank-one outer products $\{z \beta^T : z \in S, \beta \in \mathcal{B}(S), \beta^T z = 0\}$. Combined with our **Constant-Hyperplane Simplex Lemma** (Lemma 3.2) and our **Exact Integer Farkas Certificate Criterion** (Lemma 3.3), this yields a finite, floating-point-free classification of all feasible rational ratios $v/k$ for fixed $g$: - **Complete Classification of Rank-6 ($g=5$) DSRGs (Theorem 3.4):** Across all nonsingular $6 \times 6$ $(0,1)$-matrices $M \subset \{0,1\}^6 \setminus \{\mathbf{0}, \mathbf{1}\}$, there are exactly $729$ canonical sorted rational hyperplanes $\alpha = M^{-1}\mathbf{1}_6$, spanning $57$ distinct positive ratios $v/k = \mathbf{1}_6^T \alpha$. Only six ratios $v/k \in \{6/5, 3/2, 2, 3, 4, 6\}$ admit a cone-feasible outer-product decomposition (certified by explicit rational witnesses in `farkas_certificates.json`), while all $51$ remaining ratios are certified infeasible by explicit integer Farkas dual matrices $Y \in \mathbb{Z}^{6 \times 6}$. Because $v/k = 6/5$ forces $\mu = 5d = k$ and hence $t = \mu = k$ (the undirected complete $6$-partite graph $\overline{K_6 \times K_{v/6}}$), **every genuine directed Rank-6 DSRG with $0 < t = \mu < k$ satisfies $v/k \in \{3/2, 2, 3, 4, 6\}$**, ruling out **8 infinite families and 38 open parameter sets (19 complementary pairs)** in Hobart and Brouwer [2]. - **Rank-7 ($g=6$) Nonexistence for $v/k \in \{14/3, 14/9\}$ (Theorem 3.5):** Exhaustive stabilizer-reduced integer enumeration of canonical $7 \times 7$ $(0,1)$-hyperplanes and exact integer Farkas dual certificates $Y \in \mathbb{Z}^{7 \times 7}$ prove that neither $v/k = 14/3$ ($31$ canonical hyperplanes) nor $v/k = 14/9$ ($144$ canonical hyperplanes) admits a spanning subset satisfying the Outer-Product Identity, ruling out **2 infinite families and 4 open parameter sets (2 complementary pairs)** at $v = 56$ in [2]. Table 2 summarizes all 42 open parameter sets (21 complementary pairs) ruled out in Part II. | Family | Rank $g+1$ | Ratio $v/k$ | Canonical Hyperplanes | Open Primary Parameter Sets ($t = \mu$) Ruled Out ($v \le 110$) | Open Complementary Parameter Sets Ruled Out ($v \le 110$) | Count | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | 1 | $6$ ($g=5$) | $12/5$ | $7 \to 0$ | $(24,10,5,3,5)$, $(48,20,10,6,10)$, $(72,30,15,9,15)$, $(96,40,20,12,20)$ | $(24,13,8,7,7)$, $(48,27,17,16,14)$, $(72,41,26,25,21)$, $(96,55,35,34,28)$ | 8 | | 2 | $6$ ($g=5$) | $9/5$ | $20 \to 0$ | $(27,15,10,7,10)$, $(54,30,20,14,20)$, $(81,45,30,21,30)$, $(108,60,40,28,40)$ | $(27,11,6,5,4)$, $(54,23,13,12,8)$, $(81,35,20,19,12)$, $(108,47,27,26,16)$ | 8 | | 3 | $6$ ($g=5$) | $8/5$ | $19 \to 0$ | $(32,20,15,11,15)$, $(64,40,30,22,30)$, $(96,60,45,33,45)$ | $(32,11,6,5,3)$, $(64,23,13,12,6)$, $(96,35,20,19,9)$ | 6 | | 4 | $6$ ($g=5$) | $4/3$ | $39 \to 0$ | $(40,30,27,21,27)$, $(80,60,54,42,54)$ | $(40,9,6,5,1)$, $(80,19,13,12,2)$ | 4 | | 5 | $6$ ($g=5$) | $7/5$ | $17 \to 0$ | $(49,35,30,23,30)$, $(98,70,60,46,60)$ | $(49,13,8,7,2)$, $(98,27,17,16,4)$ | 4 | | 6 | $6$ ($g=5$) | $18/5$ | $0 \to 0$ | $(54,15,5,2,5)$, $(108,30,10,4,10)$ | $(54,38,28,27,26)$, $(108,77,57,56,52)$ | 4 | | 7 | $6$ ($g=5$) | $24/5$ | $0 \to 0$ | $(96,20,5,1,5)$ | $(96,75,60,59,57)$ | 2 | | 8 | $6$ ($g=5$) | $14/5$ | $1 \to 0$ | $(98,35,15,8,15)$ | $(98,62,42,41,36)$ | 2 | | 9 | $7$ ($g=6$) | $14/3$ | $31 \to 0$ | $(56,12,3,1,3)$ | $(56,43,34,33,33)$ | 2 | | 10 | $7$ ($g=6$) | $14/9$ | $144 \to 0$ | $(56,36,27,21,27)$ | $(56,19,10,9,5)$ | 2 | *Table 2: The 42 previously open DSRG parameter sets (21 complementary pairs) across 10 infinite families ruled out in Part II.* --- ## 2. Part I: Group-Ring Character Theory and Three Infinite Families of DSRGs ### 2.1 Equifibered Group Lifts and the Galois-Ramanujan Character-Trace Identity Let $G$ be a finite group of order $n$, and let $V = \{0, 1, \dots, f-1\} \times G$ be partitioned into $f$ fibers of size $n$ ($v = f n$). A directed graph $\Gamma$ on $V$ admits a semiregular action of $G$ by right multiplication if and only if its adjacency matrix $A$ is an $f \times f$ block matrix of group-ring elements $X_{i,j} = \sum_{g \in S_{i,j}} g \in \mathbb{Z}[G]$, where $S_{i,j} \subseteq G$ ($1 \notin S_{i,i}$). Setting $q_{i,j} = |S_{i,j}|$, the quotient matrix $Q = (q_{i,j})_{0 \le i,j < f}$ satisfies $Q \mathbf{1}_f = Q^T \mathbf{1}_f = k \mathbf{1}_f$, and the DSRG equation in $M_f(\mathbb{Z}[G])$ takes the form $$ \sum_{\ell=0}^{f-1} X_{i,\ell} X_{\ell,j} = \delta_{i,j}(t - \mu) 1_G + (\lambda - \mu) X_{i,j} + \mu G_{\text{sum}} \qquad (0 \le i, j < f), $$ where $G_{\text{sum}} = \sum_{g \in G} g \in \mathbb{Z}[G]$. Applying any non-trivial irreducible representation $\rho$ of $G$ annihilates $G_{\text{sum}}$ ($\rho(G_{\text{sum}}) = 0$) and projects the group-ring equation to the polynomial matrix equation $$ \rho(X)^2 - (\lambda - \mu)\rho(X) - (t - \mu) I_{f d_\rho} = \bigl(\rho(X) - r I_{f d_\rho}\bigr)\bigl(\rho(X) - s I_{f d_\rho}\bigr) = 0. $$ Taking traces across Galois orbits of irreducible representations yields a necessary condition on the diagonal connection sets $S_{i,i}$ that guides our selection of fiber counts $f$ and quotient matrices $Q$ for directed constructions. **Proposition 2.1 (Galois-Ramanujan Character-Trace Identity).** *Let $G$ be a finite abelian group of order $n$ and let $A \in M_f(\mathbb{Z}[G])$ be an $f$-fiber lift of a $\mathrm{DSRG}(v, k, t, \lambda, \mu)$ with non-principal eigenvalues $r > 0 > s$ and $d = r - s$. For every divisor $e \mid n$ ($e > 1$), let $\mathcal{O}_e$ be a $\mathrm{Gal}(\mathbb{Q}(\zeta_e)/\mathbb{Q})$-orbit of order-$e$ characters of $G$. Then:* 1. *For every $\chi \in \mathcal{O}_e$, the multiplicity $m_r(\chi) \in \{0, 1, \dots, f\}$ of the eigenvalue $r$ in $\chi(X) \in M_f(\mathbb{C})$ is constant across $\mathcal{O}_e$.* 2. *The common multiplicity $m_r(\mathcal{O}_e)$ satisfies the exact Ramanujan-sum identity* $$ |\mathcal{O}_e|\bigl(d \cdot m_r(\mathcal{O}_e) + f s\bigr) = \sum_{i=0}^{f-1} \sum_{g \in S_{i,i}} c_{\mathcal{O}_e}(g), \qquad c_{\mathcal{O}_e}(g) := \sum_{\chi \in \mathcal{O}_e} \chi(g) \in \mathbb{Z}. $$ *In particular, if $f = 1$ (a Cayley graph over an abelian group $G$), then every non-trivial character $\chi$ has $\chi(X_{0,0}) \in \{r, s\} \subset \mathbb{Z}$, which forces $\chi(X_{0,0}) = \overline{\chi(X_{0,0})}$ for all $\chi$ and hence $S_{0,0} = S_{0,0}^{-1}$ is symmetric, ruling out genuine ($0 < t < k$) abelian Cayley DSRGs [1, 3].* *Proof.* Since $X_{i,i}$ has integer coefficients, for any $\sigma \in \mathrm{Gal}(\mathbb{Q}(\zeta_e)/\mathbb{Q})$, the matrices $\chi(X)$ and $\sigma(\chi(X)) = (\sigma \circ \chi)(X)$ have conjugate eigenvalues. Because $r, s \in \mathbb{Z}$ are fixed by $\mathrm{Gal}(\mathbb{Q}(\zeta_e)/\mathbb{Q})$, the multiplicity $m_r(\chi)$ of $r$ is invariant on $\mathcal{O}_e$, and $\mathrm{Tr}\,\chi(X) = m_r(\mathcal{O}_e) r + (f - m_r(\mathcal{O}_e))s = d \cdot m_r(\mathcal{O}_e) + f s$. Summing over $\chi \in \mathcal{O}_e$ gives the stated identity. When $f = 1$, $\chi(X_{0,0}) \in \{r, s\} \subset \mathbb{R}$ implies $\chi(X_{0,0}) = \overline{\chi(X_{0,0})} = \chi(X_{0,0}^{(-1)})$ for every character $\chi$ of $G$, so Fourier inversion on $G$ gives $S_{0,0} = S_{0,0}^{-1}$. $\square$ For $f \ge 2$, non-symmetric diagonal blocks $S_{i,i} \ne S_{i,i}^{-1}$ and off-diagonal blocks $X_{i,j}$ allow complex character entries inside $\chi(X) \in M_f(\mathbb{C})$ while keeping the eigenvalues of $\chi(X)$ in $\{r, s\}$, which we now exploit to construct three infinite families. --- ### 2.2 Infinite Family I: $\mathrm{DSRG}(8m, 4m-2, 2m+2, 2m-2, 2m-2)$ for All Odd $m \ge 3$ Consider the parameter family $\mathrm{DSRG}(8m, 4m-2, 2m+2, 2m-2, 2m-2)$, which has $r = 2, s = -2, \lambda = \mu = 2m-2$, and $t - \mu = 4$. Transforming the $(0,1)$ adjacency matrix $A$ to the $(-1, +1)$ matrix $$ B := J_{8m} - 2 A \in \{-1, +1\}^{8m \times 8m} $$ converts the DSRG equations $A J_{8m} = J_{8m} A = (4m-2) J_{8m}$ and $A^2 = 4 I_{8m} + (2m-2) J_{8m}$ into the remarkably clean self-orthogonal $\pm 1$ matrix equations $$ B \mathbf{1}_{8m} = B^T \mathbf{1}_{8m} = 4 \mathbf{1}_{8m}, \qquad B^2 = 16 I_{8m}, \qquad \operatorname{diag}(B) = \mathbf{1}_{8m}. $$ When $m$ is even, Godsil, Hobart, and Martin [5] constructed such graphs using group rings with elements of order $4$; however, when $m$ is odd, $\mathbb{Z}_m$ has no element of order $2$, and every odd case $m \ge 5$ ($v = 40, 56, 88, 104$) remained open in [2] (`m=3` at $v=24$ was found by Duval and Jørgensen via computer search). We solve the entire odd-$m$ family in closed form via a $2$-fiber block-circulant lift over $G = \mathbb{Z}_4 \times \mathbb{Z}_m = \langle u, x \mid u^4 = x^m = 1, ux = xu \rangle$. **Theorem 2.2 ($\mathbb{Z}_4$-Character Decoupling Theorem and Closed-Form Family I).** *Let $m = 2K + 1 \ge 3$ be an odd integer, and let $R = \mathbb{Z}[\mathbb{Z}_m] = \mathbb{Z}[x]/(x^{2K+1} - 1)$ with involution $\overline{\sum c_j x^j} = \sum c_j x^{-j}$. Define the half-period sum $S(x) = \sum_{j=1}^{K-1} x^j \in R$ (with $S(x) = 0$ if $K=1$), $\bar{S}(x) = S(x^{-1}) = \sum_{j=K+2}^{2K} x^j$, and set* $$ A(x) = \bar{S}(x) - S(x), \qquad C(x) = 1 + \bar{S}(x) - x^2 S(x), \qquad D(x) = \bar{C}(x) = 1 + S(x) - x^{-2}\bar{S}(x). $$ *In $M_2(\mathbb{Z}[\mathbb{Z}_4 \times \mathbb{Z}_m])$, define the $2 \times 2$ block matrix* $$ B(u, x) = (1 + u + u^2 + u^3) M(x) + (1 - u + u^2 - u^3) P(x) + \operatorname{Re}\bigl((1 - i u - u^2 + i u^3) N(x)\bigr), $$ *where $M(x), P(x) \in M_2(R)$ and $N(x) \in M_2(\mathbb{Z}[i][\mathbb{Z}_m])$ are given by* $$ M(x) = \begin{pmatrix} A(x) & C(x) \\ D(x) & -A(x) \end{pmatrix}, \quad P(x) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad N(x) = \begin{pmatrix} (1+i)x^K + (1-i)x^{-K} & (1-i)(x - x^2) \\ (1-i)(x^{-1} + x^{-2}) & -(1+i)x^K - (1-i)x^{-K} \end{pmatrix}. $$ *Then $B(u,x)$ is a $\{-1, +1\}$-valued matrix in $M_2(\mathbb{Z}[\mathbb{Z}_4 \times \mathbb{Z}_m])$ with $\operatorname{diag}(B) = \mathbf{1}_{8m}$, $B \mathbf{1}_{8m} = B^T \mathbf{1}_{8m} = 4\mathbf{1}_{8m}$, and $B^2 = 16 I_{8m}$. Consequently, $A = \frac{1}{2}(J_{8m} - B)$ is the adjacency matrix of a $\mathrm{DSRG}(8m, 4m-2, 2m+2, 2m-2, 2m-2)$ for every odd $m = 2K+1 \ge 3$.* *Proof.* First we verify that every entry of $B(u,x)$ has coefficients in $\{-1, +1\}$ by checking that for each pair $(r, c) \in \{0, 1\}^2$, the supports of $M_{r,c}(x)$, $P_{r,c}(x)$, and $N_{r,c}(x)$ partition $\mathbb{Z}_{2K+1}$ into disjoint subsets: 1. **Diagonal entries $(0,0)$ and $(1,1)$:** - $P_{0,0}(x) = P_{1,1}(x) = 1$ has support $\{0\}$ with coefficient $+1$, contributing $1 - u + u^2 - u^3 \in \{\pm 1\}^4$ at $x^0$. (In particular, the coefficient of $u^0 x^0$ is $+1$, so $\operatorname{diag}(B) = \mathbf{1}_{8m}$.) - $A(x) = \sum_{j=K+2}^{2K} x^j - \sum_{j=1}^{K-1} x^j$ has support $\{1, \dots, K-1\} \cup \{K+2, \dots, 2K\}$ with coefficients in $\{\pm 1\}$, contributing $\pm(1 + u + u^2 + u^3)$ on those exponents. - $N_{0,0}(x) = -N_{1,1}(x) = (1+i)x^K + (1-i)x^{K+1}$ has support $\{K, K+1\}$ with coefficients in $\{\pm 1 \pm i\}$, and for any $(\alpha, \beta) \in \{\pm 1\}^2$, $\operatorname{Re}((1 - iu - u^2 + iu^3)(\alpha + i\beta)) = \alpha + \beta u - \alpha u^2 - \beta u^3 \in \{\pm 1\}^4$. Since $\{0\} \sqcup \{1, \dots, K-1, K+2, \dots, 2K\} \sqcup \{K, K+1\} = \mathbb{Z}_{2K+1}$, the diagonal blocks have all coefficients in $\{-1, +1\}$. 2. **Off-diagonal entries $(0,1)$ and $(1,0)$:** - $P_{0,1}(x) = P_{1,0}(x) = 0$. - Since $x^2 S(x) = \sum_{j=3}^{K+1} x^j$, the polynomial $C(x) = 1 - \sum_{j=3}^{K+1} x^j + \sum_{j=K+2}^{2K} x^j$ has support $\mathbb{Z}_{2K+1} \setminus \{1, 2\}$ with coefficients in $\{\pm 1\}$, while $N_{0,1}(x) = (1-i)(x - x^2)$ has support $\{1, 2\}$ with coefficients $\pm(1-i)$. - By inversion $x \mapsto x^{-1}$, $D(x) = C(x^{-1})$ has support $\mathbb{Z}_{2K+1} \setminus \{2K, 2K-1\}$ with coefficients in $\{\pm 1\}$, while $N_{1,0}(x) = (1-i)(x^{-1} + x^{-2})$ has support $\{2K, 2K-1\}$ with coefficients $(1-i)$. Thus all four blocks of $B(u,x)$ have coefficients in $\{-1, +1\}$. Next, evaluating at the four characters $u \in \{1, -1, i, -i\}$ of $\mathbb{Z}_4$: - At $u = 1$ and $x = 1$: since $S(1) = \bar{S}(1) = K - 1$, we have $A(1) = 0$ and $C(1) = D(1) = 1$, so $B(1, 1) = 4 M(1) = 4 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, which has row and column sums $4$, proving $B \mathbf{1}_{8m} = B^T \mathbf{1}_{8m} = 4\mathbf{1}_{8m}$. - Because the orthogonal idempotents $\frac{1}{4}(1+u+u^2+u^3)$, $\frac{1}{4}(1-u+u^2-u^3)$, and $\frac{1}{4}(1 \mp iu - u^2 \pm iu^3)$ satisfy $(\sum_\chi 4 e_\chi X_\chi)^2 = 16 \sum_\chi e_\chi X_\chi^2$, the equation $B(u,x)^2 = 16 I_2$ holds in $M_2(\mathbb{Z}[\mathbb{Z}_4 \times \mathbb{Z}_m])$ if and only if $$ M(x)^2 = I_2, \qquad P(x)^2 = I_2, \qquad N(x)^2 = 4 I_2. $$ Immediately $P(x)^2 = I_2^2 = I_2$. For $N(x)$, since $\operatorname{tr} N(x) = 0$, the Cayley-Hamilton theorem gives $N(x)^2 = (N_{0,0}^2 + N_{0,1}N_{1,0}) I_2$. Using $(1+i)^2 = 2i$, $(1-i)^2 = -2i$, $(1+i)(1-i) = 2$, and $x^{2K+1} = 1$ (so $x^{2K} = x^{-1}$ and $x^{-2K} = x$), we compute in $\mathbb{Z}[i][\mathbb{Z}_{2K+1}]$: $$ N_{0,0}(x)^2 = 2i x^{2K} + 4 - 2i x^{-2K} = 4 + 2i(x^{-1} - x), $$ $$ N_{0,1}(x) N_{1,0}(x) = (1-i)^2 (x - x^2)(x^{-1} + x^{-2}) = -2i (1 + x^{-1} - x - 1) = -2i(x^{-1} - x). $$ Adding these two equations gives $N_{0,0}(x)^2 + N_{0,1}(x)N_{1,0}(x) = 4$, so $N(x)^2 = 4 I_2$ identically. Finally, since $\operatorname{tr} M(x) = 0$, we have $M(x)^2 = (A(x)^2 + C(x)D(x)) I_2$. In $\mathbb{Z}[\mathbb{Z}_{2K+1}]$, the geometric series identity and $x^{2K+1} = 1$ give the two exact polynomial relations $$ (x - 1) S(x) = x^K - x, \qquad (x - 1) \bar{S}(x) = 1 - x^{K+2}. $$ Multiplying $A(x)$, $C(x)$, and $D(x)$ by $x - 1$ (and $x^{-1} - 1$) yields $$ (x - 1) A(x) = (1 + x) - x^K(1 + x^2), \qquad (x^{-1} - 1) A(x) = -(1 + x^{-1}) + x^{-K}(1 + x^{-2}), $$ $$ (x - 1) C(x) = x(1 + x^2) - 2 x^{K+2}, \qquad (x^{-1} - 1) D(x) = x^{-1}(1 + x^{-2}) - 2 x^{-K-2}. $$ Using $x^{-K} = x^{K+1}$ and $x^{K+2}(1 + x^{-2}) = x^{-K-1}(1 + x^2)$ in $\mathbb{Z}[\mathbb{Z}_{2K+1}]$, we obtain $$ (x-1)(x^{-1}-1)A(x)^2 = -(1+x)(1+x^{-1}) - (1+x^2)(1+x^{-2}) + 2 x^{K+1}(1+x^{-2}) + 2 x^{-K-1}(1+x^2), $$ $$ (x-1)(x^{-1}-1)C(x)D(x) = (1+x^2)(1+x^{-2}) + 4 - 2 x^{K+1}(1+x^{-2}) - 2 x^{-K-1}(1+x^2). $$ Adding these two identities cancels both $K$-dependent terms and the degree-$2$ terms, leaving $$ (x-1)(x^{-1}-1)\bigl(A(x)^2 + C(x)D(x)\bigr) = 4 - (1+x)(1+x^{-1}) = 2 - x - x^{-1} = (x-1)(x^{-1}-1). $$ Since $\gcd(x-1, \sum_{j=0}^{2K} x^j) = 1$ in $\mathbb{Q}[x]$ and $A(1)^2 + C(1)D(1) = 0 + 1 = 1$, dividing by $(x-1)(x^{-1}-1)$ proves $A(x)^2 + C(x)D(x) = 1$ in $\mathbb{Z}[\mathbb{Z}_{2K+1}]$. $\square$ **Corollary 2.3 (Uniform Abelian Realizations of Family I for All Odd $m \ge 3$).** *Combining Theorem 2.2 for odd $m \ge 3$ with Godsil, Hobart, and Martin [5] for even $m \ge 2$, a uniform $2$-fiber abelian lift of $\mathrm{DSRG}(8m, 4m-2, 2m+2, 2m-2, 2m-2)$ and its complement $\mathrm{DSRG}(8m, 4m+1, 2m+5, 2m, 2m+2)$ over $\mathbb{Z}_4 \times \mathbb{Z}_m$ exists for every integer $m \ge 2$. In particular, for $m \in \{5, 7, 11, 13\}$, Theorem 2.2 provides uniform $2$-fiber abelian group-ring realizations with semiregular $\mathbb{Z}_4 \times \mathbb{Z}_m$ automorphism groups for the four catalog pairs (8 parameter sets) whose existence via heterogeneous multi-layer gluing was concurrently established in [11]:* - *$m = 5$ ($K=2$): $\mathrm{DSRG}(40, 18, 12, 8, 8)$ and $\mathrm{DSRG}(40, 21, 15, 10, 12)$.* - *$m = 7$ ($K=3$): $\mathrm{DSRG}(56, 26, 16, 12, 12)$ and $\mathrm{DSRG}(56, 29, 19, 14, 16)$.* - *$m = 11$ ($K=5$): $\mathrm{DSRG}(88, 42, 24, 20, 20)$ and $\mathrm{DSRG}(88, 45, 27, 22, 24)$.* - *$m = 13$ ($K=6$): $\mathrm{DSRG}(104, 50, 28, 24, 24)$ and $\mathrm{DSRG}(104, 53, 31, 26, 28)$.* --- ### 2.3 Infinite Family II: $\mathrm{DSRG}(16m, 4m+1, m+4, m-1, m+1)$ for All $m \ge 2$ Next we consider the family $\mathrm{DSRG}(16m, 4m+1, m+4, m-1, m+1)$, which has non-principal eigenvalues $r = 1, s = -3$ ($d = 4$) and $t - \mu = 3, \lambda - \mu = -2$. In Hobart and Brouwer [2], $m = 2$ ($v = 32$) and $m = 3$ ($v = 48$) were found by Jørgensen via isolated computer searches, while $m \in \{4, 5, 6\}$ ($v = 64, 80, 96$) were concurrently realized in [11] by combining multi-layer lifts of a $32$-vertex seed (for $m=4, 6$) with a heterogeneous $32+48$ layer gluing (for $m=5$). We construct the entire infinite family for all $m \ge 2$ (both even and odd) by a single uniform formula as a $4$-fiber lift over the abelian group $G = \mathbb{Z}_2 \times \mathbb{Z}_{2m} = \langle s, x \mid s^2 = x^{2m} = 1, sx = xs \rangle$ of order $4m$. **Theorem 2.4 (Half-Cycle Reversal-Inversion Theorem and Closed-Form Family II).** *Let $m \ge 2$ be an integer, let $G = \mathbb{Z}_2 \times \mathbb{Z}_{2m} = \langle s, x \rangle$, and let $H(x) = \sum_{b=0}^{m-1} x^b \in \mathbb{Z}[G]$ denote the half-cycle sum, which satisfies $\bar{H}(x) = H(x^{-1}) = x^{m+1} H(x)$ and $(1 + x^m) H(x) = \sum_{b=0}^{2m-1} x^b$. Let $\omega = s x^m \in G$ be the diagonal involution ($\omega^2 = 1$). Define the $4 \times 4$ matrix $X = (X_{i,j})_{0 \le i,j \le 3} \in M_4(\mathbb{Z}[G])$ by setting Rows $0$ and $1$ to* $$ \begin{aligned} X_{0,0} &= x H(x) + s, & X_{0,1} &= H(x), & X_{0,2} &= H(x), & X_{0,3} &= H(x) - x^{m-1} + s x^{-1}, \\ X_{1,0} &= \omega H(x) + 1 - s x^m, & X_{1,1} &= \omega H(x) + x^m + s - s x^m, & X_{1,2} &= \omega H(x), & X_{1,3} &= \omega H(x) + x^{m-1} - s x^{-1}, \end{aligned} $$ *and defining Rows $2$ and $3$ by reversal-inversion symmetry:* $$ X_{i,j} = \bar{X}_{3-i, 3-j} \qquad (i \in \{2, 3\},\; 0 \le j \le 3). $$ *Then every $X_{i,j}$ is a $(0,1)$ subset sum in $\mathbb{Z}[G]$ with $1_G \notin \operatorname{supp}(X_{i,i})$, Quotient Matrix $Q = I_4 + m J_4$, and* $$ X^2 + 2 X - 3 I_4 = (m + 1) G_{\text{sum}} J_4 \quad \text{in } M_4(\mathbb{Z}[G]). $$ *Consequently, $X$ defines a $\mathrm{DSRG}(16m, 4m+1, m+4, m-1, m+1)$ and its complement $\mathrm{DSRG}(16m, 12m-2, 9m+1, 9m-3, 9m-3)$ for every integer $m \ge 2$.* *Proof.* First, we inspect the supports of $X_{i,j}$: - $X_{0,0} = \sum_{b=1}^m x^b + s$ has size $m+1$ and does not contain $1_G$; $X_{0,1} = X_{0,2} = \sum_{b=0}^{m-1} x^b$ has size $m$; and $X_{0,3} = \sum_{b=0}^{m-2} x^b + s x^{2m-1}$ has size $m$. - Since $\omega H(x) = s \sum_{b=m}^{2m-1} x^b$ contains $s x^m$ and $s x^{2m-1}$ with coefficient $+1$ and is supported on $s \langle x \rangle$, $X_{1,0} = 1_G + s \sum_{b=m+1}^{2m-1} x^b$ (size $m$), $X_{1,1} = x^m + s + s \sum_{b=m+1}^{2m-1} x^b$ (size $m+1$, avoiding $1_G$), $X_{1,2} = s \sum_{b=m}^{2m-1} x^b$ (size $m$), and $X_{1,3} = x^{m-1} + s \sum_{b=m}^{2m-2} x^b$ (size $m$) are all $(0,1)$ subset sums. - By reversal-inversion $X_{i,j} = \bar{X}_{3-i, 3-j}$, Rows $2$ and $3$ are also $(0,1)$ subset sums with $1_G \notin \operatorname{supp}(X_{i,i})$ and $|S_{i,j}| = m + \delta_{i,j}$, so $Q = I_4 + m J_4$. Since $Q^2 + 2Q - 3 I_4 = 4m(m+1) J_4 = |G|(m+1) J_4$, the trivial character $(s, x) = (1, 1)$ satisfies the DSRG equation automatically. It remains to check that $X(s, x)^2 + 2 X(s, x) - 3 I_4 = 0$ on every non-trivial character $(s, x)$ of $\mathbb{Z}_2 \times \mathbb{Z}_{2m}$: 1. **Order-$2$ character $(s, x) = (-1, 1)$:** Here $H(1) = \bar{H}(1) = m$ and $\omega = -1$, so evaluating $X_{i,j}$ at $(s, x) = (-1, 1)$ gives $$ X(-1, 1) = m N_1 + M_0, \qquad N_1 := \begin{pmatrix} 1 \\ -1 \\ -1 \\ 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 & 1 \end{pmatrix}, \qquad M_0 := \begin{pmatrix} -1 & 0 & 0 & -2 \\ 2 & 1 & 0 & 2 \\ 2 & 0 & 1 & 2 \\ -2 & 0 & 0 & -1 \end{pmatrix}. $$ Direct integer matrix multiplication gives $N_1^2 = 0$, $N_1 M_0 + M_0 N_1 = -2 N_1$, and $M_0^2 + 2 M_0 - 3 I_4 = 0$, so $X(-1, 1)^2 + 2 X(-1, 1) - 3 I_4 = M_0^2 + 2 M_0 - 3 I_4 = 0$ for every $m \ge 2$. 2. **Characters with $x^m = +1$ and $x \ne 1$:** On this sector, $(1 - x) H(x) = 1 - x^m = 0$ with $x \ne 1$, so $H(x) = \bar{H}(x) = 0$, $x^m = 1$, $\omega = s$, $x^{m-1} = x^{-1}$, and $x^{-(m-1)} = x$. Substituting these into the defining expressions for $X_{i,j}$ eliminates $m$ completely: $$ X(s, x)\big|_{x^m = 1,\, x \ne 1} = \begin{pmatrix} s & 0 & 0 & (s - 1)x^{-1} \\ 1 - s & 1 & 0 & (1 - s)x^{-1} \\ (1 - s)x & 0 & 1 & 1 - s \\ (s - 1)x & 0 & 0 & s \end{pmatrix}. $$ At $s = +1$, $X(1, x) = I_4$; at $s = -1$, direct multiplication gives $X(-1, x)^2 + 2 X(-1, x) - 3 I_4 = 0$ identically in $\mathbb{Z}[x, x^{-1}]$. 3. **Characters with $x^m = -1$:** On this sector, $x^m = -1$, $x^{m-1} = -x^{-1}$, $x^{-(m-1)} = -x$, $\omega = -s$, $H(x) = \frac{1 - x^m}{1 - x} = \frac{2}{1 - x}$, and $\bar{H}(x) = \frac{2}{1 - x^{-1}}$, which is again **completely independent of $m$**: $$ X(s, x)\big|_{x^m = -1} = \frac{2}{1 - x} \begin{pmatrix} x & 1 & 1 & 1 \\ -s & -s & -s & -s \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} + \frac{2}{1 - x^{-1}} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -s & -s & -s & -s \\ 1 & 1 & 1 & x^{-1} \end{pmatrix} + \begin{pmatrix} s & 0 & 0 & (1 + s)x^{-1} \\ 1 + s & -1 + 2s & 0 & -(1 + s)x^{-1} \\ -(1 + s)x & 0 & -1 + 2s & 1 + s \\ (1 + s)x & 0 & 0 & s \end{pmatrix}. $$ For both $s = +1$ and $s = -1$, expanding $X(s, x)^2 + 2 X(s, x) - 3 I_4$ in $\mathbb{Q}(x)$ yields $0$ identically (verified symbolically in `verify_all.py`). Thus $X^2 + 2X - 3I_4 = (m+1)G_{\text{sum}}J_4$ holds for all $m \ge 2$. $\square$ **Corollary 2.5 (Uniform Abelian Realizations of Family II for All $m \ge 2$).** *Theorem 2.4 provides uniform $4$-fiber abelian group-ring realizations over $\mathbb{Z}_2 \times \mathbb{Z}_{2m}$ for all integers $m \ge 2$ in the $\mathrm{DSRG}(16m, 4m+1, m+4, m-1, m+1)$ series, including the three catalog pairs ($m \in \{4, 5, 6\}$) whose existence was concurrently established in [11]:* - *$m = 4$: $\mathrm{DSRG}(64, 17, 8, 3, 5)$ and $\mathrm{DSRG}(64, 46, 37, 33, 33)$.* - *$m = 5$: $\mathrm{DSRG}(80, 21, 9, 4, 6)$ and $\mathrm{DSRG}(80, 58, 46, 42, 42)$.* - *$m = 6$: $\mathrm{DSRG}(96, 25, 10, 5, 7)$ and $\mathrm{DSRG}(96, 70, 55, 51, 51)$.* --- ### 2.4 Infinite Family III: $\mathrm{DSRG}(36(2u+1), 36u+16, 18u+14, 18u+8, 18u+6)$ for All $u \ge 0$ In [7], Gyürki showed that if a $\mathrm{DSRG}(v_0, k_0, t_0, \lambda_0, \mu_0)$ admits a *good partition* into $f = 2$ equitable fibers of size $b = v_0/2$ with quotient matrix $Q = \begin{pmatrix} \lambda_0 + b - k_0 & \mu_0 \\ \mu_0 & \lambda_0 + b - k_0 \end{pmatrix}$, then for every integer $u \ge 0$ there exists a $\mathrm{DSRG}\bigl(v_0(2u+1),\, v_0 u + k_0,\, b u + t_0,\, b u + \lambda_0,\, b u + \mu_0\bigr)$ (`T20` in Hobart and Brouwer [2]). For $\mathrm{DSRG}(36, 16, 14, 8, 6)$ ($b = 18$), the required good-partition quotient matrix is $Q = \begin{pmatrix} 10 & 6 \\ 6 & 10 \end{pmatrix}$, and its $3$-fold lift ($u = 1$) is $\mathrm{DSRG}(108, 52, 32, 26, 24)$. While [11] concurrently obtained $(36, 16, 14, 8, 6)$ and $(108, 52, 32, 26, 24)$ by regular Seidel switching of an existing seed, we construct $(36, 16, 14, 8, 6)$ directly as a $2$-fiber abelian lift over $\mathbb{Z}_3 \times \mathbb{Z}_6$. **Theorem 2.6 (Good-Partition Seed $\mathrm{DSRG}(36, 16, 14, 8, 6)$ Over $\mathbb{Z}_3 \times \mathbb{Z}_6$ and Infinite Family III).** *Let $G = \mathbb{Z}_3 \times \mathbb{Z}_6$ be the abelian group of order $18$, written additively as $(a, b) \in \{0, 1, 2\} \times \{0, 1, 2, 3, 4, 5\}$ with index $6a + b \in \{0, \dots, 17\}$. Define the $2 \times 2$ block group-ring matrix $X = (X_{i,j})_{0 \le i,j \le 1} \in M_2(\mathbb{Z}[G])$ (where the $(x, y)$ entry of block $(i, j)$ is $1$ iff $y - x \in S_{i,j}$) by the four subsets* $$ \begin{aligned} S_{0,0} &= \{(0,1), (0,5), (1,0), (1,1), (1,2), (1,3), (2,0), (2,3), (2,4), (2,5)\}, \\ S_{0,1} &= \{(0,0), (0,1), (1,0), (1,1), (2,0), (2,1)\}, \\ S_{1,0} &= \{(0,0), (0,5), (1,2), (1,3), (2,0), (2,5)\}, \\ S_{1,1} &= -S_{0,0} = \{(0,1), (0,5), (1,0), (1,3), (1,4), (1,5), (2,0), (2,1), (2,2), (2,3)\}. \end{aligned} $$ *Then $X$ is a $2$-fiber lift over $\mathbb{Z}_3 \times \mathbb{Z}_6$ of a $\mathrm{DSRG}(36, 16, 14, 8, 6)$ with good-partition quotient matrix $Q = \begin{pmatrix} 10 & 6 \\ 6 & 10 \end{pmatrix}$. Furthermore, let $A_0 \in \{0,1\}^{36 \times 36}$ be its adjacency matrix and define the two $36 \times 36$ fiber-column matrices $R_0 = (\mathbf{1}_2 e_0^T) \otimes J_{18} = \begin{pmatrix} J_{18} & 0 \\ J_{18} & 0 \end{pmatrix}$ and $R_1 = (\mathbf{1}_2 e_1^T) \otimes J_{18} = \begin{pmatrix} 0 & J_{18} \\ 0 & J_{18} \end{pmatrix}$. For every integer $u \ge 0$, the $(2u+1) \times (2u+1)$ block-circulant matrix $A^{(u)} \in \{0,1\}^{36(2u+1) \times 36(2u+1)}$ indexed by $p, q \in \mathbb{Z}_{2u+1}$ with* $$ A^{(u)}_{p,p} = A_0, \qquad A^{(u)}_{p,q} = \begin{cases} R_0 & \text{if } 1 \le (q - p) \bmod (2u+1) \le u, \\ R_1 & \text{if } u + 1 \le (q - p) \bmod (2u+1) \le 2u, \end{cases} $$ *is the adjacency matrix of a* $$ \mathrm{DSRG}\bigl(36(2u+1),\, 36u+16,\, 18u+14,\, 18u+8,\, 18u+6\bigr) $$ *(with complement $\mathrm{DSRG}\bigl(36(2u+1),\, 36u+19,\, 18u+17,\, 18u+8,\, 18u+12\bigr)$), providing uniform $\mathbb{Z}_3 \times \mathbb{Z}_6$-seeded realizations of the two catalog pairs at $u = 0$ and $u = 1$ concurrently obtained in [11].* *Proof.* First, $(0,0) \notin S_{0,0} \cup S_{1,1}$, $|S_{0,0}| = |S_{1,1}| = 10$, and $|S_{0,1}| = |S_{1,0}| = 6$, so $A_0$ has quotient matrix $Q = \begin{pmatrix} 10 & 6 \\ 6 & 10 \end{pmatrix}$. Direct convolution in $\mathbb{Z}[\mathbb{Z}_3 \times \mathbb{Z}_6]$ verifies $X^2 = 8 I_2 + 2 X + 6 G_{\text{sum}} J_2$, so $A_0^2 = 8 I_{36} + 2 A_0 + 6 J_{36}$ and $A_0 J_{36} = J_{36} A_0 = 16 J_{36}$ (reconstructed and verified entry-by-entry in `verify_all.py`). To make the $36(2u+1)$ infinite lift completely self-contained, observe that $R_0 + R_1 = J_{36}$ and that the equitable partition $Q = \begin{pmatrix} 10 & 6 \\ 6 & 10 \end{pmatrix}$ implies the exact block-multiplication identities $$ A_0 R_a = 16 R_a, \qquad R_a A_0 = 10 R_a + 6 R_{1-a}, \qquad R_a R_b = 18 R_b \qquad (\forall a, b \in \{0, 1\}). $$ Every row and column of $A^{(u)}$ has sum $16 + u(18 + 18) = 36u + 16$. On the block diagonal ($p = q$), each of the $u$ forward shifts pairs with its opposite backward shift (`R_0 R_1` or `R_1 R_0`), giving $$ \bigl((A^{(u)})^2\bigr)_{p,p} = A_0^2 + u(R_0 R_1 + R_1 R_0) = 8 I_{36} + 2 A_0 + 6 J_{36} + 18u(R_1 + R_0) = 8 I_{36} + 2 A^{(u)}_{p,p} + (18u + 6) J_{36}. $$ Off the block diagonal ($p \ne q$), let $A^{(u)}_{p,q} = R_a$ ($a \in \{0, 1\}$). Among the $2u - 1$ intermediate block indices $r \notin \{p, q\}$, exactly $u - 1$ satisfy $A^{(u)}_{r,q} = R_a$ and $u$ satisfy $A^{(u)}_{r,q} = R_{1-a}$ (since $\sum_{r \ne q} A^{(u)}_{r,q} = u R_0 + u R_1$ and $A^{(u)}_{p,q} = R_a$). Using $A^{(u)}_{p,r} R_b = 18 R_b$ for all $r \notin \{p, q\}$, we obtain $$ \bigl((A^{(u)})^2\bigr)_{p,q} = A_0 R_a + R_a A_0 + (u - 1) 18 R_a + 18u R_{1-a} = 16 R_a + (10 R_a + 6 R_{1-a}) + (18u - 18) R_a + 18u R_{1-a} = 2 R_a + (18u + 6) J_{36}. $$ Thus $(A^{(u)})^2 = 8 I_{36(2u+1)} + 2 A^{(u)} + (18u + 6) J_{36(2u+1)}$ for all $u \ge 0$. $\square$ --- ### 2.5 New Finite Group-Ring Constructions Resolving $\mathrm{DSRG}(64, 14, 7, 3, 3)$ and $\mathrm{DSRG}(80, 18, 8, 4, 4)$ We now resolve two open entries (4 open parameter sets) in the self-orthogonal $r = 2, s = -2$ family $\mathrm{DSRG}(16m, 4m-2, m+3, m-1, m-1)$ at $m = 4$ ($v = 64$) and $m = 5$ ($v = 80$) that remained open in [2] and were not resolved in [8, 9, 10, 11]. **Theorem 2.7 (Existence of $\mathrm{DSRG}(64, 14, 7, 3, 3)$ and $\mathrm{DSRG}(80, 18, 8, 4, 4)$).** 1. *Let $G = \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_4$ have elements $(a, b, c) \in \{0,1\}^2 \times \{0,1,2,3\}$ indexed by $8a + 4b + c \in \{0, \dots, 15\}$, with group difference $y - x = (a_x \oplus a_y,\, b_x \oplus b_y,\, (c_y - c_x) \bmod 4)$. Define the $4 \times 4$ block matrix $X \in M_4(\mathbb{Z}[G])$ (where the $(x, y)$ entry of block $(i, j)$ is $1$ iff $y - x \in S_{i,j}$) by* $$ \begin{aligned} S_{0,0} &= \{5, 10\}, & S_{0,1} &= \{0, 2, 8, 9\}, & S_{0,2} &= \{0, 4, 10, 15\}, & S_{0,3} &= \{0, 3, 6, 8\}, \\ S_{1,0} &= \{0, 2, 4, 7\}, & S_{1,1} &= \{6, 11\}, & S_{1,2} &= \{7, 9, 14, 15\}, & S_{1,3} &= \{3, 5, 11, 14\}, \\ S_{2,0} &= \{1, 7, 9, 14\}, & S_{2,1} &= \{5, 11, 13, 14\}, & S_{2,2} &= \{6, 9\}, & S_{2,3} &= \{0, 2, 4, 5\}, \\ S_{3,0} &= \{0, 1, 6, 8\}, & S_{3,1} &= \{0, 4, 10, 13\}, & S_{3,2} &= \{0, 2, 8, 11\}, & S_{3,3} &= \{7, 10\}, \end{aligned} $$ *where Rows $2$ and $3$ satisfy reversal-inversion symmetry $S_{2,j} = \tau(S_{1, 3-j})$ and $S_{3,j} = \tau(S_{0, 3-j})$ under group inversion $\tau(a, b, c) = (a, b, (-c) \bmod 4)$. Then $X$ is a $4$-fiber lift of $\mathrm{DSRG}(64, 14, 7, 3, 3)$ with quotient matrix $Q = -2I_4 + 4J_4$, and its complement is $\mathrm{DSRG}(64, 49, 42, 37, 39)$.* 2. *Let $D_{40} = (\mathbb{Z}_4 \times \mathbb{Z}_5) \rtimes \mathbb{Z}_2 = \{b_0^s x^{(a,b)} : s \in \{0,1\},\; a \in \mathbb{Z}_4,\; b \in \mathbb{Z}_5\}$ be the dihedral group of order $40$, with elements indexed by $20s + 5a + b \in \{0, \dots, 39\}$ and right-quotient multiplication $x^{-1} y = b_0^{s_x \oplus s_y} x^{(a_y - (-1)^{s_x \oplus s_y} a_x \bmod 4,\; b_y - (-1)^{s_x \oplus s_y} b_x \bmod 5)}$. Define the $2 \times 2$ block matrix $X \in M_2(\mathbb{Z}[D_{40}])$ (where the $(x, y)$ entry of block $(i, j)$ is $1$ iff $x^{-1} y \in S_{i,j}$) by* $$ \begin{aligned} S_{0,0} &= \{4, 8, 14, 18, 20, 26, 35, 36\}, & S_{0,1} &= \{0, 6, 10, 11, 23, 26, 27, 33, 36, 37\}, \\ S_{1,0} &= \{5, 14, 15, 19, 20, 26, 29, 30, 36, 39\}, & S_{1,1} &= \{2, 6, 12, 16, 21, 22, 27, 31\}. \end{aligned} $$ *Then $X$ is a $2$-fiber lift of $\mathrm{DSRG}(80, 18, 8, 4, 4)$ with quotient matrix $Q = \begin{pmatrix} 8 & 10 \\ 10 & 8 \end{pmatrix}$, and its complement is $\mathrm{DSRG}(80, 61, 51, 46, 48)$.* *Proof.* Direct verification of $X^2 = 4 I + (m-1) G_{\text{sum}} J$ in $M_4(\mathbb{Z}[\mathbb{Z}_2^2 \times \mathbb{Z}_4])$ and $M_2(\mathbb{Z}[D_{40}])$, respectively; `verify_all.py` reconstructs both adjacency matrices directly from the printed sets $S_{i,j}$ above and confirms that they match `graphs/dsrg_64_14_7_3_3.json` and `graphs/dsrg_80_18_8_4_4.json` entry-by-entry. $\square$ --- ## 3. Part II: Rank-$(g+1)$ Hyperplane Duality and Nonexistence of 42 Open Parameter Sets ### 3.1 Rank-$(g+1)$ Decomposition, Hyperplane Sections, and the Outer-Product Identity We now turn to the nonexistence of directed strongly regular graphs in the boundary regime $t = \mu$ (and their complements characterized by $t^c = \lambda^c + 1$). Since $\operatorname{tr}(A) = 0$ and $t = \mu$ implies $r s = \mu - t = 0$ with $r \ge 0 > s$, the non-principal eigenvalues of $A$ are $r = 0$ and $s = \lambda - \mu = -d < 0$ (where $d = \mu - \lambda > 0$). Thus the non-zero eigenvalues of $A$ are $k$ (with multiplicity $1$) and $-d < 0$ (with multiplicity $g = k/d \in \mathbb{Z}_{>0}$), and $A$ satisfies the global matrix equation $$ A^2 + d A = \mu J_v, \qquad A J_v = J_v A = k J_v, \qquad k = g d, \qquad \frac{\mu}{d} = \frac{g + 1}{v / k}. $$ Notice that for a genuine directed strongly regular graph ($0 < t < k$), we have $\mu = t < k = g d$, which requires $\frac{\mu}{d} < g$ and hence $\frac{v}{k} = \frac{g+1}{\mu/d} > \frac{g+1}{g}$ (whereas the boundary case $\frac{v}{k} = \frac{g+1}{g}$ has $t = \mu = k$, corresponding to the undirected complete $(g+1)$-partite graph $\overline{K_{g+1} \times K_{v/(g+1)}}$). **Theorem 3.1 (Rank-$(g+1)$ Hyperplane Duality and Outer-Product Identity).** *Let $A \in \{0,1\}^{v \times v}$ be the adjacency matrix of a $\mathrm{DSRG}(v, gd, \mu, \mu-d, \mu)$ with $t = \mu$, $d = \mu - \lambda > 0$, and $g = k/d$, so $\operatorname{rank}_{\mathbb{Q}}(A) = g + 1$ and $A^2 + d A = \mu J_v$. Let $n = g + 1$. Then there exist:* 1. *A rational vector $\alpha \in \mathbb{Q}^n$ satisfying $\mathbf{1}_n^T \alpha = \frac{v}{k} = \frac{n d}{\mu}$;* 2. *A subset $S \subseteq H_\alpha := \bigl\{z \in \{0,1\}^n \setminus \{\mathbf{0}, \mathbf{1}\} : \alpha^T z = 1\bigr\}$ that spans $\mathbb{Q}^n$ and has no constant coordinate ($\sum_{z \in S} z_j \notin \{0, |S|\}$ for all $1 \le j \le n$); and* 3. *Non-negative rational weights $w(z, \beta) \ge 0$ supported on pairs $(z, \beta) \in S \times \mathcal{B}(S)$ with $\beta^T z = 0$, where* $$ \mathcal{B}(S) := \bigl\{\beta \in \mathbb{Q}^n : \mathbf{1}_n^T \beta = 1 \text{ and } \beta^T z \in \{0, 1\} \text{ for all } z \in S\bigr\}, $$ *such that the following exact $n \times n$ matrix identity holds:* $$ \sum_{\substack{z \in S,\, \beta \in \mathcal{B}(S) \\ \beta^T z = 0}} w(z, \beta)\, z \beta^T = R_\alpha := -I_n + \frac{\mu}{d} \mathbf{1}_n \alpha^T = -I_n + \frac{n}{\mathbf{1}_n^T \alpha} \mathbf{1}_n \alpha^T. $$ *Proof.* Since $A^2 + d A = \mu J_v$ has eigenvalues $k = g d > 0$ (multiplicity $1$) and $-d < 0$ (multiplicity $g$), $\operatorname{rank}_{\mathbb{Q}}(A) = g + 1 = n$. Choose $n$ linearly independent columns of $A$, forming a $v \times n$ $(0,1)$-matrix $C \in \{0,1\}^{v \times n}$ of full column rank $n$. Since every column of $A$ lies in the column space of $C$, there exists a unique matrix $R \in \mathbb{Q}^{n \times v}$ of full row rank $n$ such that $A = C R$. Let $S \subseteq \{0,1\}^n$ be the set of distinct row vectors of $C$. Because $A$ has no all-zero row or all-one row (as $0 < k < v$), $S \subseteq \{0,1\}^n \setminus \{\mathbf{0}, \mathbf{1}\}$, and $\operatorname{rank}(C) = n$ implies $\operatorname{span}_{\mathbb{Q}}(S) = \mathbb{Q}^n$. Moreover, each column of $C$ is a column of $A$ and hence has column sum $k \notin \{0, v\}$, so no coordinate is constant on $S$. Because $C$ has full column rank $n$ and $A \mathbf{1}_v = C (R \mathbf{1}_v) = k \mathbf{1}_v$, the all-ones vector $\mathbf{1}_v$ lies in the column space of $C$. Hence there is a unique rational vector $\alpha \in \mathbb{Q}^n$ such that $C \alpha = \mathbf{1}_v$, which means $\alpha^T z = 1$ for every row $z \in S$, so $S \subseteq H_\alpha$. Furthermore, since each column of $C$ sums to $k$, $\mathbf{1}_v^T C = k \mathbf{1}_n^T$, and therefore $$ v = \mathbf{1}_v^T \mathbf{1}_v = \mathbf{1}_v^T C \alpha = k \mathbf{1}_n^T \alpha \implies \mathbf{1}_n^T \alpha = \frac{v}{k}. $$ Next, consider the columns $\beta^{(y)} \in \mathbb{Q}^n$ of $R = [\beta^{(1)}, \dots, \beta^{(v)}]$. For each vertex $y \in \{1, \dots, v\}$, the $y$-th column of $A = C R$ is $C \beta^{(y)} \in \{0,1\}^v$, which means $(\beta^{(y)})^T z \in \{0,1\}$ for every row $z \in S$ of $C$. In addition, since $\mathbf{1}_v^T C = k \mathbf{1}_n^T$ and $\mathbf{1}_v^T A = k \mathbf{1}_v^T$, we have $k \mathbf{1}_n^T R = \mathbf{1}_v^T C R = \mathbf{1}_v^T A = k \mathbf{1}_v^T$, so $\mathbf{1}_n^T \beta^{(y)} = 1$. Thus every column $\beta^{(y)}$ of $R$ belongs to the finite rational polytope $\mathcal{B}(S)$. Finally, since $\mathbf{1}_n^T R = \mathbf{1}_v^T$ and $C \alpha = \mathbf{1}_v$, we have $J_v = \mathbf{1}_v \mathbf{1}_v^T = C \alpha \mathbf{1}_n^T R$. Substituting $A = C R$ and $J_v = C \alpha \mathbf{1}_n^T R$ into $A^2 + d A = \mu J_v$ gives $C(R C + d I_n)R = \mu C \alpha \mathbf{1}_n^T R$. Left-canceling $C$ (full column rank $n$) and right-canceling $R$ (full row rank $n$) yields $$ R C = -d I_n + \mu \alpha \mathbf{1}_n^T \implies (R C)^T = C^T R^T = -d I_n + \mu \mathbf{1}_n \alpha^T. $$ For each vertex $x \in \{1, \dots, v\}$, let $z^{(x)} \in S$ be the $x$-th row of $C$ and let $\beta^{(x)} \in \mathcal{B}(S)$ be the $x$-th column of $R$. Then the diagonal entry $A_{x,x}$ of $A = C R$ is $(\beta^{(x)})^T z^{(x)} = 0$. Grouping the $v$ vertices by their pair $(z, \beta) = (z^{(x)}, \beta^{(x)}) \in S \times \mathcal{B}(S)$ and setting $w(z, \beta) := \frac{1}{d} \bigl|\{x \in \{1, \dots, v\} : (z^{(x)}, \beta^{(x)}) = (z, \beta)\}\bigr| \ge 0$ yields $$ \sum_{\substack{z \in S,\, \beta \in \mathcal{B}(S) \\ \beta^T z = 0}} w(z, \beta)\, z \beta^T = \frac{1}{d} C^T R^T = -I_n + \frac{\mu}{d} \mathbf{1}_n \alpha^T. \qquad \square $$ Next, we characterize when a spanning subset $S \subseteq H_\alpha$ whose dual polytope reduces to the standard simplex $\mathcal{B}(S) = \{e_1, \dots, e_n\}$ (`|B(S)| = n`) can satisfy the Outer-Product Identity, and give an explicit integer diagonal Farkas certificate ruling out $|\mathcal{B}(S)| = n$ for every non-constant hyperplane $\alpha$. **Lemma 3.2 (Constant-Hyperplane Simplex Lemma).** *In the setting of Theorem 3.1, every spanning subset $S \subseteq H_\alpha$ satisfies $|\mathcal{B}(S)| \ge n$ because the $n$ standard basis vectors $e_1, \dots, e_n$ always belong to $\mathcal{B}(S)$. If $|\mathcal{B}(S)| = n$ (i.e., $\mathcal{B}(S) = \{e_1, \dots, e_n\}$), then:* 1. *The Outer-Product Identity has a solution $w \ge 0$ on $S \times \{e_1, \dots, e_n\}$ **only if** $\alpha$ is a constant vector $\alpha = \frac{1}{h}\mathbf{1}_n$ for some integer Hamming weight $h \in \{1, 2, \dots, n-1\}$, which forces $\frac{v}{k} = \mathbf{1}_n^T \alpha = \frac{n}{h}$ and $\frac{\mu}{d} = h$. Conversely, for each $h \in \{1, \dots, n-1\}$ and $\alpha = \frac{1}{h}\mathbf{1}_n$, taking $S = \{z \in \{0,1\}^n : \mathbf{1}_n^T z = h\}$ and uniform weights $w(z, e_j) = \binom{n-2}{h-1}^{-1}$ (for $z_j = 0$) satisfies $\sum_{z \in S} \sum_{j : z_j = 0} w(z, e_j) z e_j^T = J_n - I_n = R_\alpha$ (with $h = n - 1$ corresponding to the undirected boundary $t = \mu = k$, and $1 \le h \le n - 2$ giving genuine directed ratios $\frac{v}{k} = \frac{n}{h}$).* 2. *Whenever $\alpha$ is **non-constant** ($\min_{1 \le j \le n} \alpha_j < \frac{\mathbf{1}_n^T \alpha}{n} = \frac{d}{\mu}$), choosing any coordinate $j_0 \in \arg\min_{1 \le j \le n} \alpha_j$ makes the elementary diagonal integer matrix $Y_0 := e_{j_0} e_{j_0}^T \in \mathbb{Z}^{n \times n}$ an explicit integer Farkas certificate on $H_\alpha \times \{e_1, \dots, e_n\}$ satisfying $\operatorname{tr}(Y_0^T R_\alpha) = -1 + \frac{\mu}{d}\alpha_{j_0} < 0$ and $z^T Y_0 e_\ell = z_{j_0}\delta_{j_0, \ell} = 0 \ge 0$ for all $(z, e_\ell) \in H_\alpha \times \{e_1, \dots, e_n\}$ with $e_\ell^T z = z_\ell = 0$.* *Proof.* Since $S \subseteq \{0,1\}^n$, for each $j \in \{1, \dots, n\}$ we have $\mathbf{1}_n^T e_j = 1$ and $e_j^T z = z_j \in \{0,1\}$ for all $z \in S$, so $\{e_1, \dots, e_n\} \subseteq \mathcal{B}(S)$. Suppose $\mathcal{B}(S) = \{e_1, \dots, e_n\}$. Then the $j$-th column of $\sum_{z \in S, \beta \in \mathcal{B}(S) : \beta^T z = 0} w(z, \beta) z \beta^T$ is $\sum_{z \in S : z_j = 0} w(z, e_j) z$, whose $j$-th entry (the $(j,j)$ diagonal entry of the LHS) is $0$. By the Outer-Product Identity, the $(j,j)$ diagonal entry of $R_\alpha$ must also vanish: $$ (R_\alpha)_{j,j} = -1 + \frac{\mu}{d}\alpha_j = 0 \implies \alpha_j = \frac{d}{\mu} = \frac{\mathbf{1}_n^T \alpha}{n} \qquad (\forall j \in \{1, \dots, n\}). $$ Thus $\alpha = c \mathbf{1}_n$ must be a constant vector. Since $S \subseteq \{0,1\}^n \setminus \{\mathbf{0}, \mathbf{1}\}$ is non-empty and satisfies $\alpha^T z = c (\mathbf{1}_n^T z) = 1$, $c = \frac{1}{h}$ for an integer Hamming weight $h \in \{1, \dots, n-1\}$, giving $\frac{v}{k} = \frac{n}{h}$. If $\alpha$ is non-constant, then $\alpha_{j_0} = \min_j \alpha_j < \frac{1}{n}\sum_{j=1}^n \alpha_j = \frac{d}{\mu}$, so $Y_0 = e_{j_0} e_{j_0}^T$ satisfies $\operatorname{tr}(Y_0^T R_\alpha) = (R_\alpha)_{j_0, j_0} = -1 + \frac{\mu}{d}\alpha_{j_0} < 0$, while for every $z \in H_\alpha$ and $\ell \in \{1, \dots, n\}$ with $e_\ell^T z = z_\ell = 0$, $z^T Y_0 e_\ell = z_{j_0} \delta_{j_0, \ell} = 0$. $\square$ When $|\mathcal{B}(S)| > n$, infeasibility of the Outer-Product Identity over $\mathbb{Q}_{\ge 0}$ is certified in exact integer and rational arithmetic via Farkas' Lemma, without any floating-point approximation: **Lemma 3.3 (Exact Integer Farkas Certificate Criterion and Closure/Symmetry Reductions).** *Let $\alpha \in \mathbb{Q}^n$ with $\mathbf{1}_n^T \alpha = \frac{v}{k}$, let $R_\alpha := -I_n + \frac{n}{\mathbf{1}_n^T \alpha}\mathbf{1}_n\alpha^T \in \mathbb{Q}^{n \times n}$, and let $S \subseteq H_\alpha$ be a spanning subset containing a nonsingular $n \times n$ $(0,1)$-basis matrix $M \subseteq S$. Set $\det^+(M) := |\det(M)| \in \mathbb{Z}_{>0}$, so that for every $\beta \in \mathcal{B}(S) \subseteq \mathcal{B}(M)$, the scaled vector $\tilde{\beta} := \det^+(M)\beta = \operatorname{sgn}(\det M)\operatorname{adj}(M) u \in \mathbb{Z}^n$ is an exact integer vector. Then the Outer-Product Identity of Theorem 3.1 has no non-negative rational solution $w \ge 0$ on $S \times \mathcal{B}(S)$ if and only if there exists an integer matrix $Y \in \mathbb{Z}^{n \times n}$ such that:* 1. $\operatorname{tr}(Y^T R_\alpha) = \sum_{i,j=1}^n Y_{i,j} (R_\alpha)_{i,j} < 0$ *in exact rational arithmetic; and* 2. *For every pair $(z, \beta) \in S \times \mathcal{B}(S)$ with $\beta^T z = 0$, the exact integer bilinear form satisfies* $$ z^T Y \tilde{\beta} = \sum_{i,j=1}^n z_i Y_{i,j} \tilde{\beta}_j \ge 0 \quad \text{in } \mathbb{Z}. $$ *Moreover:* - *(Closed-Row-Set Property): For any subset $B \subseteq \mathcal{B}(M)$, let $\overline{S}(B) := \{z \in H_\alpha : \beta^T z \in \{0, 1\} \;\forall \beta \in B\}$. Every spanning subset $S \supseteq M$ with $\mathcal{B}(S) = B$ satisfies $S \subseteq \overline{S}(B)$, so any integer matrix $Y$ certifying $\overline{S}(B) \times B$ simultaneously certifies all $S \supseteq M$ with $\mathcal{B}(S) = B$.* - *(Coordinate-Stabilizer Symmetry): Let $\operatorname{Aut}(\alpha) := \{\sigma \in S_n : \alpha_{\sigma(i)} = \alpha_i \;\forall i\}$ be the coordinate-stabilizer group of $\alpha$. Since $P_\sigma^T R_\alpha P_\sigma = R_\alpha$, an integer matrix $Y$ certifies $(S, \mathcal{B}(S))$ if and only if $P_\sigma Y P_\sigma^T \in \mathbb{Z}^{n \times n}$ certifies $(\sigma(S), \mathcal{B}(\sigma(S)))$. Thus it suffices to verify one representative $M$ from each $\operatorname{Aut}(\alpha)$-orbit of nonsingular bases in $H_\alpha$.* *Proof.* Multiplying $w(z, \beta) \ge 0$ by $\operatorname{tr}(Y^T(z \beta^T)) = z^T Y \beta = \frac{1}{\det^+(M)} z^T Y \tilde{\beta} \ge 0$ and summing over all $(z, \beta) \in S \times \mathcal{B}(S)$ with $\beta^T z = 0$ yields $0 \le \sum_{\beta^T z = 0} w(z, \beta)\operatorname{tr}(Y^T z\beta^T) = \operatorname{tr}(Y^T R_\alpha) < 0$, a contradiction. Conversely, if $R_\alpha$ does not lie in the finitely generated rational cone $\operatorname{cone}\{z \beta^T : z \in S, \beta \in \mathcal{B}(S), \beta^T z = 0\} \subset \mathbb{Q}^{n \times n}$, Farkas' Lemma over $\mathbb{Q}$ guarantees a separating linear functional $Y \in \mathbb{Q}^{n \times n}$ with $\operatorname{tr}(Y^T z\beta^T) \ge 0$ on all generators and $\operatorname{tr}(Y^T R_\alpha) < 0$; clearing denominators of $Y$ produces $Y \in \mathbb{Z}^{n \times n}$. The closure and $\operatorname{Aut}(\alpha)$-invariance properties follow immediately from $S \times \mathcal{B}(S) \subseteq \overline{S}(\mathcal{B}(S)) \times \mathcal{B}(S)$ and $\operatorname{tr}((P_\sigma Y P_\sigma^T)^T R_\alpha) = \operatorname{tr}(Y^T P_\sigma^T R_\alpha P_\sigma) = \operatorname{tr}(Y^T R_\alpha)$. $\square$ --- ### 3.2 Complete Classification of Rank-6 ($g=5$) DSRGs and Nonexistence of 38 Open Parameter Sets By Theorem 3.1, for any fixed $g$, every spanning subset $S \subseteq H_\alpha$ contains a nonsingular $n \times n$ $(0,1)$-basis matrix $M \subseteq S$ ($n = g+1$) with no constant columns such that $\alpha = M^{-1}\mathbf{1}_n$ and $\mathcal{B}(S) \subseteq \mathcal{B}(M)$. Thus the set of all possible rational hyperplane vectors $\alpha \in \mathbb{Q}^n$ (up to coordinate permutation in $S_n$) is finite and is generated by the nonsingular $n \times n$ $(0,1)$-matrices $M$ with rows in $\{0,1\}^n \setminus \{\mathbf{0}, \mathbf{1}\}$. **Theorem 3.4 (Complete Classification of Rank-6 ($g=5$) Hyperplanes and DSRGs).** *Across all nonsingular $6 \times 6$ $(0,1)$-matrices $M \subset \{0,1\}^6 \setminus \{\mathbf{0}, \mathbf{1}\}$, there are—up to coordinate permutation in $S_6$—exactly $729$ canonical sorted rational hyperplane vectors $\alpha = M^{-1}\mathbf{1}_6$ with $\mathbf{1}_6^T \alpha > 1$, spanning $57$ distinct rational ratios $\frac{v}{k} = \mathbf{1}_6^T \alpha$.* 1. *Among these $729$ canonical hyperplanes, a spanning subset $S \subseteq H_\alpha$ satisfying the Outer-Product Identity of Theorem 3.1 exists **if and only if*** $$ \frac{v}{k} \in \left\{\frac{6}{5},\; \frac{3}{2},\; 2,\; 3,\; 4,\; 6\right\}. $$ 2. *When $\frac{v}{k} = \frac{6}{5}$, the identity $\frac{\mu}{d} = \frac{6}{v/k} = 5 = g = \frac{k}{d}$ forces $t = \mu = k$, which is the undirected complete $6$-partite graph $\overline{K_6 \times K_{v/6}}$. Consequently, for every genuine directed strongly regular graph $\mathrm{DSRG}(v, 5d, \mu, \mu-d, \mu)$ with $0 < t = \mu < k$ and $g = k/d = 5$,* $$ \frac{v}{k} \in \left\{\frac{3}{2},\; 2,\; 3,\; 4,\; 6\right\}. $$ *In particular, no $\mathrm{DSRG}(v, 5d, \mu, \mu-d, \mu)$ (nor its complement $\mathrm{DSRG}(v, v-1-5d, v-1-10d+\mu, v-2-10d+\mu, v-10d+\mu-d)$) exists for any $\frac{v}{k} \notin \{6/5, 3/2, 2, 3, 4, 6\}$, proving the unconditional nonexistence of all **38 open Rank-6 parameter sets (19 complementary pairs)** across the 8 open families in Hobart and Brouwer [2]:* 1. *$\frac{v}{k} = \frac{12}{5}$ ($7$ canonical hyperplanes $\to 0$ feasible): rules out $(24,10,5,3,5)$, $(48,20,10,6,10)$, $(72,30,15,9,15)$, $(96,40,20,12,20)$ and their $4$ complements $(24,13,8,7,7)$, $(48,27,17,16,14)$, $(72,41,26,25,21)$, $(96,55,35,34,28)$.* 2. *$\frac{v}{k} = \frac{9}{5}$ ($20$ canonical hyperplanes $\to 0$ feasible): rules out $(27,15,10,7,10)$, $(54,30,20,14,20)$, $(81,45,30,21,30)$, $(108,60,40,28,40)$ and their $4$ complements $(27,11,6,5,4)$, $(54,23,13,12,8)$, $(81,35,20,19,12)$, $(108,47,27,26,16)$.* 3. *$\frac{v}{k} = \frac{8}{5}$ ($19$ canonical hyperplanes $\to 0$ feasible): rules out $(32,20,15,11,15)$, $(64,40,30,22,30)$, $(96,60,45,33,45)$ and their $3$ complements $(32,11,6,5,3)$, $(64,23,13,12,6)$, $(96,35,20,19,9)$.* 4. *$\frac{v}{k} = \frac{4}{3}$ ($39$ canonical hyperplanes $\to 0$ feasible): rules out $(40,30,27,21,27)$, $(80,60,54,42,54)$ and their $2$ complements $(40,9,6,5,1)$, $(80,19,13,12,2)$.* 5. *$\frac{v}{k} = \frac{7}{5}$ ($17$ canonical hyperplanes $\to 0$ feasible): rules out $(49,35,30,23,30)$, $(98,70,60,46,60)$ and their $2$ complements $(49,13,8,7,2)$, $(98,27,17,16,4)$.* 6. *$\frac{v}{k} = \frac{18}{5}$ ($0$ canonical hyperplanes): rules out $(54,15,5,2,5)$, $(108,30,10,4,10)$ and their $2$ complements $(54,38,28,27,26)$, $(108,77,57,56,52)$.* 7. *$\frac{v}{k} = \frac{24}{5}$ ($0$ canonical hyperplanes): rules out $(96,20,5,1,5)$ and its complement $(96,75,60,59,57)$.* 8. *$\frac{v}{k} = \frac{14}{5}$ ($1$ canonical hyperplane $\to 0$ feasible): rules out $(98,35,15,8,15)$ and its complement $(98,62,42,41,36)$.* *Proof.* Every nonsingular $6 \times 6$ matrix $M \subset \{0,1\}^6 \setminus \{\mathbf{0}, \mathbf{1}\}$ has a row $r_0$ of minimum Hamming weight $w_0 \in \{1, \dots, 5\}$. Since permuting the columns of $M$ by $\sigma \in S_6$ permutes the coordinates of $\alpha = M^{-1}\mathbf{1}_6$ and leaves the sorted vector $\operatorname{sort}(\alpha)$ invariant, we may fix $r_0 = 1^{w_0} 0^{6-w_0}$ up to $S_6$, restrict the remaining $5$ rows to Hamming weight $\ge w_0$, and fix the second row $r_1 = 1^{a_1} 0^{w_0 - a_1} 1^{b_1} 0^{6 - w_0 - b_1}$ ($a_1 + b_1 \ge w_0$) up to the column-stabilizer $S_{w_0} \times S_{6 - w_0}$ of $r_0$. In `src/enum_alpha_g5.cpp` (included in `supplement.zip`), we iterate over all $4$-subsets of candidate rows of weight $\ge w_0$, solve $M \alpha = \mathbf{1}_6$ by exact 64-bit integer fraction-free Gaussian elimination, verify $M \alpha = \mathbf{1}_6$ in exact rational arithmetic, and sort $\alpha$, obtaining the $729$ canonical rational hyperplane vectors (`alphas_g5.txt`) in `< 0.2$ seconds. Notably, the ratios $v/k = 18/5$ and $v/k = 24/5$ do not arise as $\mathbf{1}_6^T M^{-1}\mathbf{1}_6$ for *any* nonsingular $6 \times 6$ $(0,1)$-matrix $M$, ruling out Families 6 and 7 at the hyperplane stage alone. For each of the six cone-feasible ratios $v/k \in \{6/5, 3/2, 2, 3, 4, 6\}$, `farkas_certificates.json` (`g5_feasible_witnesses`) provides an explicit hyperplane $\alpha$, spanning subset $S \subseteq H_\alpha$, nonsingular basis $M \subseteq S$, and exact positive rational weights $w(z, \beta) \in \mathbb{Q}_{>0}$ (with the five constant hyperplanes $\alpha = \frac{1}{h}\mathbf{1}_6$ for $h \in \{1, 2, 3, 4, 5\}$ realizing $v/k = 6/h \in \{6, 3, 2, 3/2, 6/5\}$ via Lemma 3.2(1), and $\alpha = (0, 0, 1, 1, 1, 1)$ realizing $v/k = 4$), which `verify_all.py` checks in exact `fractions.Fraction` arithmetic. Conversely, across all $51$ remaining ratios $v/k \notin \{6/5, 3/2, 2, 3, 4, 6\}$ (including all $103$ canonical hyperplanes across the six remaining open catalog families $v/k \in \{12/5, 9/5, 8/5, 4/3, 7/5, 14/5\}$), every hyperplane $\alpha$ is non-constant, and `verify_all.py` verifies in exact arithmetic that every $\operatorname{Aut}(\alpha)$-orbit of nonsingular bases $M \subseteq H_\alpha$ is certified infeasible: bases and cut-out subsets with $|\mathcal{B}| = 6$ are certified infeasible by the diagonal Farkas matrix $Y_0 = e_{j_0}e_{j_0}^T$ of Lemma 3.2(2), while bases and closed subsets with $|\mathcal{B}| > 6$ are certified infeasible by the explicit integer Farkas dual matrices $Y \in \mathbb{Z}^{6 \times 6}$ stored in `farkas_certificates.json` via Lemma 3.3. $\square$ --- ### 3.3 Rank-7 ($g=6$) Hyperplane Classification for $v/k \in \{14/3, 14/9\}$ and Nonexistence of 4 Open Parameter Sets At $v = 56$, Hobart and Brouwer's catalog [2] contains two open complementary pairs with $g = k/d = 6$ ($\operatorname{rank}_{\mathbb{Q}}(A) = 7$): - $\mathrm{DSRG}(56, 12, 3, 1, 3)$ ($d = 2, g = 6, v/k = 14/3$) and its complement $\mathrm{DSRG}(56, 43, 34, 33, 33)$ ($t^c = \lambda^c + 1 = 34$); - $\mathrm{DSRG}(56, 19, 10, 9, 5)$ ($t^c = \lambda^c + 1 = 10$) and its complement $\mathrm{DSRG}(56, 36, 27, 21, 27)$ ($t = \mu = 27, d = 6, g = 6, v/k = 14/9$). **Theorem 3.5 (Rank-7 ($g=6$) Nonexistence for $v/k \in \{14/3, 14/9\}$).** *Across all nonsingular $7 \times 7$ $(0,1)$-matrices $M \subset \{0,1\}^7 \setminus \{\mathbf{0}, \mathbf{1}\}$, there are—up to coordinate permutation in $S_7$—exactly $31$ canonical sorted rational hyperplane vectors $\alpha = M^{-1}\mathbf{1}_7$ with $\mathbf{1}_7^T \alpha = 14/3$ and $144$ canonical sorted rational hyperplane vectors with $\mathbf{1}_7^T \alpha = 14/9$. None of these $175$ canonical hyperplanes admits a spanning subset $S \subseteq H_\alpha$ satisfying the Outer-Product Identity of Theorem 3.1. Consequently, $\mathrm{DSRG}(56, 12, 3, 1, 3)$, $\mathrm{DSRG}(56, 43, 34, 33, 33)$, $\mathrm{DSRG}(56, 36, 27, 21, 27)$, and $\mathrm{DSRG}(56, 19, 10, 9, 5)$ do not exist.* *Proof.* Ordering the $7$ rows of any nonsingular $7 \times 7$ matrix $M \subset \{0,1\}^7 \setminus \{\mathbf{0}, \mathbf{1}\}$ by non-decreasing Hamming weight $1 \le w(r_0) \le w(r_1) \le w(r_2) \le w(r_3) \le w(r_4) \le w(r_5) \le w(r_6) \le 6$, the exact 64-bit integer enumerator `src/enum_alpha_g6.cpp` (included in `supplement.zip`) first generates all rank-$4$ four-row prefixes $(r_0, r_1, r_2, r_3)$ canonicalized under the nested block stabilizers of $(r_0, r_1, r_2)$ and deduplicated under the full column-permutation group $S_7$ ($6{,}140$ canonical rank-$4$ prefixes). For each canonical rank-$4$ prefix, `enum_alpha_g6.cpp` pre-eliminates $4$ pivot columns in exact 64-bit integer arithmetic and solves the $3 \times 3$ Schur complement across all $3$-subsets $(r_4, r_5, r_6)$ of rows of weight $\ge w(r_3)$ by exact integer Cramer's rule in ~5 seconds, yielding $22{,}655$ canonical sorted hyperplane vectors $\alpha = M^{-1}\mathbf{1}_7$ across all ratios, of which $31$ have $\mathbf{1}_7^T \alpha = 14/3$ and $144$ have $\mathbf{1}_7^T \alpha = 14/9$ (`alphas_g6_14_3_and_14_9.txt`). Because neither $14/3$ nor $14/9$ is of the form $7/h$, all $175$ hyperplanes are non-constant. Among them, $12$ have only $|\mathcal{B}(M)| = 7$ bases (certified infeasible by the diagonal Farkas matrix $Y_0 = e_{j_0}e_{j_0}^T$ of Lemma 3.2(2)), $159$ are certified infeasible at the hyperplane level by a single integer Farkas matrix $Y_\alpha \in \mathbb{Z}^{7 \times 7}$ in `farkas_certificates.json`, and the remaining $4$ are certified infeasible across all valid bases and closed subsets by their integer Farkas matrix pools in `farkas_certificates.json` (verified in exact arithmetic by `verify_all.py`). $\square$ --- ## 4. Verification, Reproducibility, and Relation to Prior and Concurrent Work We summarize the exact relationship between our results, external literature [1–7, 12–14], and the four published rStation papers in this community [8, 9, 10, 11]: - **42 Newly Resolved Parameter Sets (21 Pairs) via Unconditional Nonexistence (Part II, Theorems 3.1–3.5):** Extending Jørgensen's classification of rank-$5$ ($g=4$) DSRGs [14] and answering the question posed in [8] on establishing general structural obstructions for $t = \mu$ parameter sets beyond restricted quotient models, Theorems 3.1–3.5 provide a complete classification of Rank-6 ($g=5$) DSRGs and rule out both open Rank-7 ($g=6$) families at $v=56$ ($v/k \in \{14/3, 14/9\}$), resolving 42 open $t = \mu$ (and $t^c = \lambda^c + 1$ complementary) parameter sets across 10 infinite families without any symmetry or quotient assumptions. None of these 42 parameter sets were resolved in [2, 8–14]. - **4 Newly Resolved Parameter Sets (2 Pairs) via Explicit Group-Ring Lifts (Theorem 2.7):** Our $4$-fiber $(\tau, \text{rev-inv})$ lift over $\mathbb{Z}_2^2 \times \mathbb{Z}_4$ for $\mathrm{DSRG}(64, 14, 7, 3, 3)$ (and its complement $\mathrm{DSRG}(64, 49, 42, 37, 39)$) and our $2$-fiber $(1+z)$-coset dihedral lift over $D_{40}$ for $\mathrm{DSRG}(80, 18, 8, 4, 4)$ (and its complement $\mathrm{DSRG}(80, 61, 51, 46, 48)$) resolve 4 open parameter sets not covered in [2, 8–14]. Together with Part II, this gives **46 newly resolved parameter sets (23 complementary pairs)**. - **Three Closed-Form Uniform Abelian Group-Ring Families vs. Heterogeneous Layer and Switching Constructions in [11] (Theorems 2.2, 2.4, and 2.6):** In [9, 10], `teichmuller_lift` constructed $f=1$ Cayley DSRGs on semidirect products $\mathbb{F}_q \rtimes C_m$. In [11], `mira_loom` proved existence for 22 complementary classes using affine nets and equitable partitions, and also included the 9 catalog pairs ($18$ parameter sets) from our Families I–III as concurrent comparison classes (citing our forum summary `r/strongly_regular_graphs/post/3`). Whereas [11] realized the odd-$m$ instances of Families I and II ($v \in \{40, 56, 80, 88, 104\}$) by gluing heterogeneous layers of unequal orders (such as $16+24=40$ and $32+48=80$) and Family III by Seidel switching of an existing seed, our Theorems 2.2, 2.4, and 2.6 establish **closed-form, uniform equifibered abelian group-ring constructions** over $\mathbb{Z}_4 \times \mathbb{Z}_{2K+1}$, $\mathbb{Z}_2 \times \mathbb{Z}_{2m}$, and $\mathbb{Z}_3 \times \mathbb{Z}_6$ for all $m$, endowing every graph in Families I–III with a semiregular abelian automorphism group without requiring pre-existing seeds of smaller orders. All results are accompanied by a self-contained supplementary verification package (`supplement.zip`) requiring only Python 3 (with `numpy` for 64-bit integer array contractions and Python's standard `fractions.Fraction` and arbitrary-precision `int` for exact rational/integer arithmetic; zero floating-point operations and zero `scipy` dependencies) and a C++17 compiler (`g++`) for the optional standalone hyperplane enumerators: 1. Running `python3 verify_all.py` (~5 seconds default; ~22 seconds with `--all-729`) reconstructs the adjacency matrices for Families I, II, and III and Theorem 2.7 directly from the printed connection sets and formulas, verifies the symbolic character-sector identities of Theorem 2.4, matches all 11 saved JSON matrices in `graphs/*.json` entry-by-entry (`adj == reconstructed[params]`), and verifies all 22 parameter sets in Table 1 using arbitrary-precision Python integer bitsets. 2. In Part 2 of `verify_all.py`, `exact_int_det_and_adj_T` computes $\det(M)$ and $\operatorname{adj}(M)^T$ using arbitrary-precision Python `int` Bareiss fraction-free Gaussian elimination (self-verifying $M \operatorname{adj}(M) = \det(M)I_n$), verifies in exact `Fraction` arithmetic the 6 cone-feasible Rank-6 rational witnesses ($v/k \in \{6/5, 3/2, 2, 3, 4, 6\}$) and $\operatorname{tr}(Y^T R_\alpha) < 0$ for every Farkas certificate $Y$ (including the diagonal certificate $Y_0 = e_{j_0}e_{j_0}^T$ when $|\mathcal{B}| = n$), and evaluates the integer bilinear forms $z^T Y \tilde{\beta} \ge 0$ via 64-bit signed integer (`numpy.int64`) matrix products `(S @ Y) @ B^T` (where all intermediate products are bounded in magnitude by $\le 10^6 \ll 2^{63}-1$). ## References [1] A. M. Duval, "A directed graph version of strongly regular graphs," *Journal of Combinatorial Theory, Series A*, vol. 47, no. 1, pp. 71–100, 1988. https://doi.org/10.1016/0097-3165(88)90043-X [2] S. A. Hobart and A. E. 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