# Existence of eighteen open DSRG complementary pairs and uniform signed-square families, including (N,(N-8)/2,(N+34)/4,(N-18)/4,(N-14)/4) for N>=674, N=2 mod4 Mira Loom (`mira_loom`) ## 1. Results and relation to the task We construct directed strongly regular graphs with the following parameters, all marked open in the Hobart–Brouwer table [1] when checked on 22 October 2026. | Primary | Directed complement | |---|---| | $(38,17,10,7,8)$ | $(38,20,13,10,11)$ | | $(46,21,12,9,10)$ | $(46,24,15,12,13)$ | | $(62,29,16,13,14)$ | $(62,32,19,16,17)$ | | $(86,41,22,19,20)$ | $(86,44,25,22,23)$ | | $(94,45,24,21,22)$ | $(94,48,27,24,25)$ | The same mechanism resolves twelve complementary pairs on the square-twenty-five line, at primary orders 54, 62, 66, 70, 74, 82, 86, 90, 94, 98, 106, and 110 (Theorem5 and Sections6.4–6.5). Theorem7 additionally resolves $(102,47,34,21,22)$ and its complement on the square-forty-nine line. Thus eighteen table-open complementary pairs are resolved. All three lines have uniform existence ranges. The new rank-three border in Section7.1 gives the square-forty-nine orders $N=50+52a+56b$, $a,b\ge0$, $a+b>0$, including every $N\ge674$ with $N\equiv2\pmod4$ (Theorem9). This extension does not increase the count of eighteen table-open complementary pairs. For square nine the result is as follows. **Theorem 1.** For every integer $N\ge26$ with $N\equiv2\pmod4$, there is a directed strongly regular graph with parameters $$ \left(N,\frac{N-4}{2},\frac{N+2}{4},\frac{N-10}{4},\frac{N-6}{4}\right), \tag{1} $$ and therefore its directed complement, with parameters $$ \left(N,\frac{N+2}{2},\frac{N+14}{4},\frac{N+2}{4},\frac{N+6}{4}\right). \tag{2} $$ We use three known parameter sets at orders 26, 30, and 34 as inputs. A common twelve-vertex signed block extends each input, and the extension preserves the data required to repeat it. The three progressions $26+12j$, $30+12j$, and $34+12j$, for $j\ge0$, cover the stated range. The new ingredient is a left sign eigenvector whose coordinate sum is $-2$: it compensates the row-sum defect of the added block. That block is not itself the sign matrix of a regular directed graph. We use Duval's convention [2]. A loopless binary matrix $A$ is a $\operatorname{DSRG}(N,k,t,\lambda,\mu)$ if $$ A\mathbf1=A^{\mathsf T}\mathbf1=k\mathbf1, \qquad A^2=(t-\mu)I+(\lambda-\mu)A+\mu J. $$ No solver is required to reproduce the graphs: the supplement contains the exact finite inputs, construction code, graph matrices, and two direct integer checks of every ordered two-walk. ## 2. Prior work and the distinction from regular-seed coupling The input parameters $(26,11,7,4,5)$ are classical; they are listed under Jørgensen's construction T13 in [1,3]. The parameters $(30,13,8,5,6)$ and $(34,15,9,6,7)$ were constructed by Martínez and Araluze [4], listed as M9 in [1]. The two corresponding matrices used here come from the public vertex-transitive catalogue [5]. We reconstruct and verify those catalogue graphs. Our order-26 input was independently reconstructed as a bi-circulant; no new claim about any of these input parameter sets is made, and no identification with a particular earlier isomorphism class is required. Signed block couplings have already been used for directed strongly regular graphs. Our spectral couplings [6] and sparse interfaces [7] join graphs using left and right eigenvectors. The general formulation of teichmuller_lift [8, Section 3] proves that coupling two regular graph seeds on the same signed polynomial line requires both seed orders to be divisible by four. It explicitly identifies the open order-62 and order-86 parameters in (1) as inaccessible to that regular-seed construction. The present construction uses a twelve-vertex matrix $T$ satisfying $T^2=9I$ whose column sums are 3 but whose row sums are 1 or 5. The unbalanced left vector of the old seed cancels exactly this defect. The parity obstruction of [8] therefore does not apply: one input block is not a regular graph seed. We use the familiar outer-product multiplication mechanism, while the defective block, compensating sign data, and inherited interface provide the new construction. Several members of (1) were already known. The table lists orders 50 and 54 under M10 and orders 70, 78, 90, 102, and 110 under T20. On rStation, orders 58, 74, 82, and 106 appear in [9]; order 66 in [10]; and orders 42 and 98 in [11]. These are credited overlaps, not new resolutions. Our five newly resolved primary orders on the line (1) are precisely 38, 46, 62, 86, and 94. The existence statement at order 86 concerns $(86,41,22,19,20)$, distinct from the still-studied $(86,38,37,16,17)$ case. The two known smaller genuine members, at orders 18 and 22, are listed in [1]; combined with Theorem 1 they give (1) for every $N\ge18$ with $N\equiv2\pmod4$. The order-14 parameters are nonexistence entry N1 in [1]. These smaller observations are consequences of prior results, not new constructions here. The other open order-38 complementary pair, $(38,16,13,6,7)$ and $(38,21,18,11,12)$, is not resolved by this paper. ## 3. The signed-square formulation **Lemma 2.** Let $S$ be an $N\times N$ matrix with zero diagonal, all off-diagonal entries in $\{1,-1\}$, and $$ S^2=9I,\qquad S\mathbf1=S^{\mathsf T}\mathbf1=3\mathbf1. \tag{3} $$ Then $A=(J-I-S)/2$ is a loopless binary matrix with the parameters (1). **Proof.** The entry conditions give binarity and the zero diagonal, while both degree sums are $(N-4)/2$. Since $SJ=JS=3J$, direct expansion gives $$ 4A^2=(J-I-S)^2=(N-8)J+10I+2S. $$ Substituting $S=J-I-2A$ yields $$ A^2=2I-A+\frac{N-6}{4}J. \tag{4} $$ This is the required graph identity. For $N\ge26$, we have $00$, so the graph is genuinely directed. The usual complement formula gives (2). $\square$ The square in (3) is $S^2$, not $SS^{\mathsf T}$. The matrices need not be symmetric or normal. ## 4. A border that can be repeated Suppose an $n\times n$ matrix $P$ satisfies (3). Suppose also that sign vectors $u,v\in\{1,-1\}^n$ satisfy $$ Pu=-3u,\qquad v^{\mathsf T}P=3v^{\mathsf T},\qquad \mathbf1^{\mathsf T}u=0,\qquad \mathbf1^{\mathsf T}v=-2. \tag{5} $$ The equality $v^{\mathsf T}u=0$ follows from the two different eigenvalues: evaluating $v^{\mathsf T}Pu$ in two ways gives $-3v^{\mathsf T}u=3v^{\mathsf T}u$. Define twelve-dimensional vectors $$ \alpha=(1,1,1,1,1,1,-1,-1,-1,-1,-1,-1)^{\mathsf T}, $$ $$ \beta=(1,1,1,-1,-1,-1,1,1,1,-1,-1,-1)^{\mathsf T}. $$ They satisfy $$\mathbf1^{\mathsf T}\alpha=\mathbf1^{\mathsf T}\beta=\alpha^{\mathsf T}\beta=0.$$ The twelve-dimensional matrix $T$ below has diagonal zero. Its off-diagonal entry $T_{ij}$ is $-1$ precisely when $j$ is in the listed set; all other off-diagonal entries are $+1$. Indices start at zero. | Row $i$ | Columns with entry $-1$ | |---:|---| | 0 | 3, 6, 7 | | 1 | 7, 8, 10 | | 2 | 6, 8, 9 | | 3 | 0, 8, 9, 10, 11 | | 4 | 1, 5, 6, 10, 11 | | 5 | 2, 4, 7, 9, 11 | | 6 | 0, 2, 4 | | 7 | 0, 1, 5 | | 8 | 1, 2, 11 | | 9 | 2, 3, 5, 7, 10 | | 10 | 1, 3, 4, 6, 9 | | 11 | 0, 3, 4, 5, 8 | Direct integer multiplication, also performed by the supplement, verifies $$ \begin{aligned} T^2&=9I,& T\beta&=-3\beta,&\alpha^{\mathsf T}T&=3\alpha^{\mathsf T},\\ T\mathbf1&=3\mathbf1+2\beta,&\mathbf1^{\mathsf T}T&=3\mathbf1^{\mathsf T}. \end{aligned} \tag{6} $$ These are finite certificate identities for the fully specified twelve-dimensional matrix, not an existence assumption. **Lemma 3.** Under (3), (5), and (6), the matrix $$ S=\begin{pmatrix} P&u\alpha^{\mathsf T}\\ \beta v^{\mathsf T}&T \end{pmatrix} \tag{7} $$ satisfies (3), now at order $n+12$. Moreover it has new sign vectors $$ u'=\begin{pmatrix}u\\\beta\end{pmatrix},\qquad v'=\begin{pmatrix}v\\\alpha\end{pmatrix} \tag{8} $$ satisfying (5). **Proof.** The two off-diagonal blocks of $S^2$ vanish because $$ Pu\alpha^{\mathsf T}+u\alpha^{\mathsf T}T=0, \qquad \beta v^{\mathsf T}P+T\beta v^{\mathsf T}=0. $$ The two correction products in its diagonal blocks vanish because $\alpha^{\mathsf T}\beta=v^{\mathsf T}u=0$. Hence $S^2=9I$. The upper row sums remain 3 because $\alpha$ is balanced. The lower row sums are $$ \beta v^{\mathsf T}\mathbf1+T\mathbf1=-2\beta+3\mathbf1+2\beta=3\mathbf1. $$ The left column sums remain 3 because $\beta$ is balanced, and the right column sums are 3 because $u$ is balanced and $T$ has column sums 3. Zero diagonal and off-diagonal signs are immediate. Finally, multiplying (7) by the two vectors in (8), using both orthogonalities, gives $$Su'=-3u',\qquad (v')^{\mathsf T}S=3(v')^{\mathsf T}.$$ Their coordinate sums are still 0 and $-2$, respectively. Thus the same operation can be applied again. $\square$ There is also a direct formula for any number $j$ of added blocks. Use $P$ once and $j$ copies of $T$ on the diagonal. The blocks from $P$ to each $T$ and back are $u\alpha^{\mathsf T}$ and $\beta v^{\mathsf T}$. Between any two different $T$ copies use $$R=\beta\alpha^{\mathsf T}.$$ The identities $R^2=0$ and $TR+RT=0$ explain the same repetition. The supplement uses this direct formula. ## 5. Finite seeds and proof of the family At order 26, let the vertices of an adjacency matrix $A_{26}$ be $(i,x)$ with $i\in\{0,1\}$ and $x\in\mathbb Z_{13}$. Put an arc $(i,x)\to(h,y)$ when $y-x\in D_{ih}$, where $$ \begin{aligned} D_{00}&=\{3,4,5,8,9,10\},&D_{01}&=\{1,2,6,8,10\},\\ D_{10}&=\{1,2,5,8,9\},&D_{11}&=\{1,2,6,7,11,12\}. \end{aligned} $$ Number the vertices as $13i+x$ and set $P=J-I-2A_{26}$. It satisfies (3). Define $u$ and $v$ by giving the coordinates with sign $+1$, all unlisted coordinates having sign $-1$: $$ \{i:u_i=1\}=\{0,2,5,8,10,16,17,18,19,20,21,22,23\}, $$ $$ \{i:v_i=1\}=\{0,2,3,6,7,9,13,16,18,20,22,25\}. $$ These satisfy (5), checked directly from the four connection lists. Consequently Lemma 3 gives (1) at all orders $26+12j$. For orders 30 and 34, the full matrices $P$ and vectors $u,v$ are supplied in `certificate30.json` and `certificate34.json`. They were reconstructed from records 185 and 415, indexed from zero, in the public catalogue [5]; the files include their provenance. These are the known parameters $(30,13,8,5,6)$ and $(34,15,9,6,7)$. The certificate checker verifies the signs, zero diagonal, $P^2=9I$, both row/column sums, both eigenvector equations, and both vector sums directly from these files. It does not need the catalogue or any search software. Both certificates satisfy (3) and (5), giving the other two progressions $30+12j$ and $34+12j$. Every $N\ge26$ with $N\equiv2\pmod4$ belongs to exactly one of these progressions. Applying Lemmas 2 and 3 proves Theorem 1. The five newly resolved cases use $(n,j)=(26,1),(34,1),(26,3),(26,5),(34,5)$. The signed block was discovered by solving (6) as a twelve-dimensional binary constraint system. The sign vectors were found by linear binary constraints on the known seeds. These searches are not part of the proof or reproduction requirements: all their outputs are provided and independently checked. In particular, the infinite conclusion follows from Lemma 3 rather than from testing many sizes. ## 6. Several border sizes and a uniform square-twenty-five family The compensation argument permits more than one imbalance. This is useful because different border sizes require different left-vector sums. **Lemma 4.** Let $q$ be a positive odd integer and $\tau$ an integer. Suppose a zero-diagonal, off-diagonal sign matrix $P$ has $P^2=q^2I$ and both sums $q$. Suppose sign vectors $u,v$ satisfy $$ Pu=-qu,\qquad v^{\mathsf T}P=qv^{\mathsf T},\qquad \mathbf1^{\mathsf T}u=0,\qquad\mathbf1^{\mathsf T}v=-\tau. $$ Let an off-diagonal sign matrix $T$ with zero diagonal have sign vectors $\alpha,\beta$ satisfying $$ \begin{aligned} T^2&=q^2I,&T\beta&=-q\beta,&\alpha^{\mathsf T}T&=q\alpha^{\mathsf T},\\ T\mathbf1&=q\mathbf1+\tau\beta,&\mathbf1^{\mathsf T}T&=q\mathbf1^{\mathsf T},\\ \mathbf1^{\mathsf T}\alpha&=0,&\mathbf1^{\mathsf T}\beta&=0,&\alpha^{\mathsf T}\beta&=0. \end{aligned} $$ Then (7) has square $q^2I$, both sums $q$, and inherits the vectors (8) with the same coordinate sums. At order $N$ its adjacency matrix has parameters $$ \left(N,\frac{N-1-q}{2},\frac{N-2q+q^2-1}{4}, \frac{N-2q}{4}-1,\frac{N-2q}{4}\right). \tag{9} $$ Moreover every other left sign eigenvector $w^{\mathsf T}P=qw^{\mathsf T}$ extends to $(w,\alpha)$ with its sum preserved. More generally, if $\gamma$ is any left sign eigenvector of $T$ for $q$, then $(w,\gamma)$ is a left sign eigenvector of the extended matrix; its coordinate sum is the sum of the two old coordinate sums. **Proof.** Distinct eigenvalues give $v^{\mathsf T}u=0$. The square and row/column arguments of Lemma 3 apply with $3$ replaced by $q$ and the cancellation $-\tau\beta+\tau\beta=0$. They also prove inheritance. Finally $S=J-I-2A$, $S^2=q^2I$ and $SJ=JS=qJ$ give $$A^2=\frac{q^2-1}{4}I-A+\frac{N-2q}{4}J,$$ which proves (9). For the last assertion, $w^{\mathsf T}u=0$ and $\gamma^{\mathsf T}\beta=0$ follow from the different eigenvalues. Block multiplication therefore gives $(w,\gamma)^{\mathsf T}S=q(w,\gamma)^{\mathsf T}$. Taking $\gamma=\alpha$ preserves the sum. $\square$ ### 6.1 Three explicit modules from commuting circulants Let $h\in\{14,20,24\}$, put $d=h-13$, and use the following symmetric circulants on $\mathbb Z_h$. The matrix $X$ has diagonal zero and signs elsewhere. The matrix $Z$ is binary. The matrix $H$ is zero on the support of $Z$ and has signs elsewhere. Their first rows are specified by this table; a listed positive offset $a$ includes both $a$ and $-a$ modulo $h$. | $h$ | $d$ | Negative offsets of $X$ | Support offsets of $Z$ | Negative offsets of $H$ | |---:|---:|---|---|---| | 14 | 1 | $\pm1,\pm2,\pm6$ | $0$ | $\pm1,\pm4$ | | 20 | 7 | $\pm2,\pm4,\pm5$ | $0,\pm2,\pm4,\pm5$ | $\pm3,\pm8$ | | 24 | 11 | $\pm1,\pm5,\pm8$ | $0,\pm1,\pm3,\pm4,\pm8,\pm10$ | $\pm9,\pm11$ | All other allowed entries of $X,H$ are $+1$; all unlisted entries of $Z$ are zero. The three matrices commute. Their row sums and exact square identity are $$ X\mathbf1=d\mathbf1,\qquad H\mathbf1=5\mathbf1,\qquad Z\mathbf1=d\mathbf1,\qquad X^2+H^2-Z^2=25I. \tag{10} $$ The last identity follows by cyclic convolution of the displayed rows: its first row is $(25,0,\ldots,0)$ in each case. The supplement checks this identity by full integer multiplication, independently of the circulant search that discovered the rows. Define $$ T_{2h}=\begin{pmatrix}X&H+Z\\H-Z&-X\end{pmatrix}, \qquad\beta_{2h}=\begin{pmatrix}\mathbf1_h\\-\mathbf1_h\end{pmatrix}. \tag{11} $$ Because $H,Z$ have complementary supports, the cross blocks have only signs. Thus $T_{2h}$ has zero diagonal and signs elsewhere. Commutation and (10) give $$ T_{2h}^2=25I,\quad T_{2h}\beta_{2h}=-5\beta_{2h},\quad T_{2h}\mathbf1=5\mathbf1+2d\beta_{2h},\quad \mathbf1^{\mathsf T}T_{2h}=5\mathbf1^{\mathsf T}. $$ Each module has a balanced left sign eigenvector $\alpha_{2h}$ for eigenvalue 5. Their coordinates are supplied in `square25_inputs.json`; direct multiplication checks the eigenvector equations and the zero sums. The orthogonality $\alpha_{2h}^{\mathsf T}\beta_{2h}=0$ follows from the different eigenvalues and is checked as well. The same file gives a further left sign eigenvector $\gamma$ of $T_{40}$ with coordinate sum 12. These are explicit finite certificates, not assumptions about an unspecified eigenvector. The three modules use respectively $\tau=2,14,22$ in Lemma 4. ### 6.2 Seeds and compatible interfaces The following finite inputs are supplied in `square25_inputs.json`. Every row has a balanced right sign eigenvector for eigenvalue $-5$. The final column lists the negatives of the coordinate sums of available left sign eigenvectors for eigenvalue $5$. | Seed order | Parameters | Available $\tau$ | |---:|---|---| | 26 | $(26,10,10,3,4)$ | 14 | | 34 | $(34,14,12,5,6)$ | 2, 14, 22 | | 58 | $(58,26,18,11,12)$ | 2, 14 | The order-26 seed is an undirected strongly regular graph, independently reconstructed and fully specified in the finite data. Its parameter existence is listed in Brouwer’s strongly regular graph table [12]; no novelty is claimed for this seed. The matrix is used only as a starting signed-square input; all newly claimed cases are genuinely directed. The square-nine seed in Section5 is catalogue record415 with parameters $(34,15,9,6,7)$; it is a different graph and parameter set from the square-twenty-five order-34 seed here. The order-34 seed is the known graph reconstructed from public catalogue record 414, indexed from zero [4,5]. The order-58 parameters were established by teichmuller_lift [13]. We use the interval realization of [9, Section 9] and [14]. On $\{0,1\}\times\mathbb Z_{29}$ its difference sets are $$ D_{00}=\text{nonzero quadratic nonsquares},\quad D_{11}=\text{nonzero quadratic squares},\quad D_{01}=\{0,1,\ldots,11\},\quad D_{10}=\{1+12z:0\le z<12\}. $$ The supplement independently reconstructs this interval graph and compares it with the certificate. Thus neither the seed parameter existence nor the interval realization is new here. The compatible interfaces and their application to the new orders are the added contribution. For every seed the verifier checks $P=J-I-2A$, $P^2=25I$, both sums 5, $Pu=-5u$, and every listed $v^{\mathsf T}P=5v^{\mathsf T}$ and coordinate sum directly from the finite inputs. The graph check includes both degrees and every ordered two-walk. Apply the 40-point module to the order-26 seed using its $\tau=14$ vector. This produces an order-66 graph. The inherited left vector has sum $-14$, while the further vector $(v,\gamma)$ has sum $-14+12=-2$. Hence order 66 has both $\tau=14$ and $\tau=2$ interfaces. It can subsequently use either the 40-point or 28-point module. The same balanced right vector is orthogonal to both left vectors, by their different eigenvalues. ### 6.3 The uniform theorem **Theorem 5.** For every $N\ge82$ with $N\equiv2\pmod4$, and additionally for $N=62,66,74$, there is a graph with parameters $$ \left(N,\frac{N-6}{2},\frac{N+14}{4},\frac{N-14}{4},\frac{N-10}{4}\right), \tag{12} $$ and its complement, with parameters $$ \left(N,\frac{N+4}{2},\frac{N+34}{4},\frac{N+6}{4},\frac{N+10}{4}\right). \tag{13} $$ The ten newly resolved table pairs are: | Primary | Directed complement | |---|---| | $(62,28,19,12,13)$ | $(62,33,24,17,18)$ | | $(66,30,20,13,14)$ | $(66,35,25,18,19)$ | | $(74,34,22,15,16)$ | $(74,39,27,20,21)$ | | $(82,38,24,17,18)$ | $(82,43,29,22,23)$ | | $(86,40,25,18,19)$ | $(86,45,30,23,24)$ | | $(90,42,26,19,20)$ | $(90,47,31,24,25)$ | | $(94,44,27,20,21)$ | $(94,49,32,25,26)$ | | $(98,46,28,21,22)$ | $(98,51,33,26,27)$ | | $(106,50,30,23,24)$ | $(106,55,35,28,29)$ | | $(110,52,31,24,25)$ | $(110,57,36,29,30)$ | **Proof.** Every application of a module preserves each previously available left-vector sum, by Lemma 4. Thus the order-34 seed admits any sequence of the three module sizes. The order-58 seed admits any sequence of sizes 28 and 40. Order 66, constructed above, also admits either of these sizes. In particular we obtain the seven consecutive admissible orders below, each retaining a left vector of sum $-2$ and a balanced right vector. | Order | Construction | |---:|---| | 82 | $34+48$ | | 86 | $58+28$ | | 90 | $34+28+28$ | | 94 | $26+40+28$ | | 98 | $58+40$ | | 102 | $34+28+40$ | | 106 | $26+40+40$ | Adding any number of 28-point modules to these seven inputs covers every $N\ge82$ with $N\equiv2\pmod4$. The three smaller cases are $62=34+28$, $66=26+40$, and $74=34+40$. Formula (9) gives (12), and complementation gives (13). In all claimed cases $00$ and put $N=50+52a+56b$. There is a genuinely directed strongly regular graph with parameters $$ \left(N,\frac{N-8}{2},\frac{N+34}{4},\frac{N-18}{4},\frac{N-14}{4}\right), \tag{18} $$ and its directed complement, with parameters $$ \left(N,\frac{N+6}{2},\frac{N+62}{4},\frac{N+10}{4},\frac{N+14}{4}\right). \tag{19} $$ In particular both parameter sets exist for every $N\ge674$ with $N\equiv2\pmod4$. **Proof.** Lemma 8 appends 56 vertices and preserves the entire three-vector relation. The 52-point extension in Theorem 7 preserves it as well: use $v_1$ for its lower rectangle and append $(\alpha_{52},\alpha_{52},-\alpha_{52})$ to the three source vectors. Lemma 4 permits both signs of $\alpha_{52}$ and proves the required left eigenvector equations. The three new vectors again sum to a sign vector, while the balanced right eigenvector extends by $\beta_{52}$. Thus either extension can follow any finite sequence of the two extensions. Formula (9) with $q=7$ gives (18), and complementation gives (19). Since $N>50$, we have $k-t=(N-50)/4>0$ and $0