Vanishing return centers in a twelve-orbit construction for DSRG(108,15,11,2,2)

1. The construction question and the result

An adjacency matrix for a directed strongly regular graph with parameters $(108,15,11,2,2)$ must satisfy

$$ A^2=9I_{108}+2J_{108},\qquad A\mathbf1=A^T\mathbf1=15\mathbf1. \tag{1} $$

This parameter set remains unresolved in the Hobart–Brouwer table [1]. We study a particular construction with twelve regular orbits of the translation group $G=\mathbb F_3^2$. The paper proves a global restriction on this construction, not unrestricted nonexistence.

Label the orbits by $i=1,\ldots,12$ and write $U_{ij}\subseteq G$ for the connection set: $(i,x)$ points to $(j,y)$ when $y-x\in U_{ij}$. The principal quotient is $Q_{ij}=|U_{ij}|$. If $Q$ is symmetric and has zero diagonal, (1) requires

$$ Q\mathbf1=15\mathbf1,\qquad Q^2=9I_{12}+18J_{12}. \tag{2} $$

We first classify these quotients. There are exactly two permutation types. The second has a convenient cube description. Let

$$ v_0=(1,1,1),\ v_1=(1,-1,-1),\ v_2=(-1,1,-1),\ v_3=(-1,-1,1), $$

and let $S$ have rows consisting of two copies of each $v_a$, followed by one copy of each $-v_a$. The cube quotient is

$$ Q=\frac{3J_{12}+SS^T}{2}-3I_{12}. \tag{3} $$

We call its first eight cells positive and the last four negative. The positive cells of type $a$ are denoted $(a,0),(a,1)$; the negative cell of type $a$ is denoted $a^-$. The physical assumptions used below are:

1. $|U_{ij}|=Q_{ij}$, with $Q$ given by (3).
2. Whenever $Q_{ij}=1$, $U_{ji}=-U_{ij}$.
3. Each opposite degree-three pair consists of affine lines with distinct directions in $\mathbb F_3^2$.

All two-element connection sets are unrestricted. Condition 3 makes the product of the opposite degree-three sets the group sum. It imposes no common direction across different pairs.

For a return product $P_{ij}=U_{ij}U_{ji}\in\mathbb Z[G]$, define its center by

$$ c(P_{ij})=\sum_{g\in G}P_{ij}(g)g\in\mathbb F_3^2. \tag{4} $$

The total coefficient of this product is four, so (4) is also its geometric center: the coefficient array is centrally symmetric and $4=1$ in $\mathbb F_3$.

Theorem 1 (vanishing centers). Under assumptions 1–3, every graph satisfying (1) has $c(P_{ij})=0$ on every degree-two pair. At every positive cell the three return products therefore have identity coefficients $(0,2,2)$, in some order. In particular, the locally feasible $(1,1,2)$ return pattern cannot occur in a complete graph in this class.

The proof connects three different levels of the construction. First moments force either zero centers or a single nonzero circulation around a four-cycle of the tetrahedron types. The nonzero circulation forces an actual symmetric graph on 36 vertices after quotienting by its common center direction. We describe all these 36-vertex bases by Hermitian cube-root weighing matrices and prove that none admits the required cyclic three-cover. The latter exclusion allows arbitrary phases on the 36-vertex base; it does not require that its additional order-three automorphism lift to the cover.

Thus the conclusion is stronger than a rejection of one chosen MUB base or a list of trial phases. It identifies the integral return condition that every base in the forced family fails. The remaining zero-center mode, different local assumptions, asymmetric quotients, and arbitrary graphs at (1) remain open.

2. Relation to earlier work

The connection-set and character-matrix language is standard partial-sum-family theory. Martínez [2], particularly Lemma 5.1, gives the character framework for strongly regular $m$-Cayley digraphs. We do not claim a new Fourier formalism or character trace formula.

Unit weighing matrices and their relation to mutually unbiased bases are established objects; Best, Kharaghani, and Ramp [3] classify small weights and explain the general framework. Our weighing matrices have weight nine, entries restricted to third roots of unity, Hermitian symmetry, and a prescribed four-partite support. The finite calculation here concerns precisely that additional support and symmetry, and its role in the directed graph construction.

The undirected graphs with parameters $(36,15,6,6)$ are already classified by McKay and Spence [4]. No novelty is claimed for any 36-vertex graph in this paper. We regenerate the much smaller family forced by (3), keeping the action and markings needed for the proof. The graph-classification literature alone does not decide whether these graphs admit directed cyclic three-covers with parameters (1).

Our Paper 82 [5] supplies the characteristic-three phase method used in the final step. It already proves uniform-sign cover exclusions for the Latin-square graph of $\mathbb Z_6$. We reuse its algebra and certificate principle. The new contribution is the chain from the cube's physical return products through a conserved center circulation to a complete forced base family, together with the exclusion of that family. Neither the Latin-square exclusion nor the standard MUB construction supplies this implication. The principal-quotient and first-moment reductions were previously shared in our public discussion [6]; this paper adds the complete nonzero-mode obstruction and presents the connected argument.

3. The two symmetric principal quotients

Proposition 2. Up to simultaneous row and column permutation, the symmetric nonnegative integral matrices satisfying (2) and $\operatorname{diag}Q=0$ are (3) and

$$ Q=(2I_4+J_4)\otimes J_3-3I_{12}. \tag{5} $$

Proof. The eigenvalues off the all-ones space are $3$ and $-3$. The trace condition gives their multiplicities as three and eight. Consequently

$$ H=Q+3I-\frac32J $$

is positive semidefinite of rank three, $H^2=6H$, and has diagonal $3/2$. Write $H=(3/2)XX^T$ with unit row vectors $x_i\in\mathbb R^3$. Then $X^TX=4I$ and $\sum_i x_i=0$. Positivity of $Q+3I$ gives $0\leq Q_{ij}\leq3$, so for distinct rows

$$ x_i\mathbin{\cdot}x_j=\frac{2Q_{ij}-3}{3} \in\{-1,-1/3,1/3,1\}. $$

If the unoriented line through $x_i$ contains $N$ rows, the tight-frame identity gives

$$ 4=\sum_j(x_i\mathbin{\cdot}x_j)^2=N+\frac{12-N}{9}. $$

Thus $N=3$, and there are four lines. The Gram matrix of one representative from each line has the form $(4/3)(I-zz^T)$ with $\|z\|=1$. Its diagonal forces $z_i^2=1/4$. Reorienting the representatives gives $z=(1,1,1,1)/2$, so these representatives are the vertices of a regular tetrahedron. Their only relation is their sum being zero.

On each line, the difference between the numbers of the two orientations is one of $-3,-1,1,3$. The equation $\sum_i x_i=0$ makes these four differences equal. Up to reversing all vectors, all are three or all are one. These give (5) and (3). Their row profiles differ, so they are inequivalent. $\square$

4. Local returns and the global circulation

Write $e$ for the identity of $\mathbb Z[G]$ and $\mathcal G$ for the sum of all nine group elements. At a positive cell, the six singleton returns contribute $6e$ and the opposite line pair contributes $\mathcal G$. Its three two-set products must therefore satisfy

$$ P_1+P_2+P_3=3e+\mathcal G. \tag{6} $$

At a negative cell, its six products satisfy

$$ P_1+\cdots+P_6=6e+2\mathcal G. \tag{7} $$

Every two-set is a translate of $\{u,-u\}$. The product of two such sets is either an affine four-cap with coefficients one or a line product with coefficients $(1,2,1)$. In (6), a coefficient two must be at the identity. Enumerating the four possible line directions gives exactly the following two local types:

• Two distinct centered line products and their complementary cap. Their identity coefficients are $(0,2,2)$ and all three centers are zero. There are six unordered product triples.
• Two opposite caps through zero and the remaining centered line product. Representatives are $\{0,a,b,a+b\}$, its negative, and $2e+\delta_{a-b}+\delta_{b-a}$ for independent $a,b$. Their identity coefficients are $(1,1,2)$, with centers $c,-c,0$ and $c\ne0$. There are twelve unordered triples.

These claims also have a short independent exhaustive verification: enumerate all $\binom92^2$ ordered pairs of two-sets, convolve their indicator arrays, and select the triples summing to (6). The supplied standard-library checker does this literally. It obtains 90 distinct products, 34 occurring in the 18 allowed triples. At (7), these products give 195 multisets, including 24 that are not unions of two positive triples. These 24 diagonal-level possibilities will disappear after the global equations are imposed.

For a connection set define $m(U)=\sum_{u\in U}u$. The first moment of a convolution is $|U|m(V)+|V|m(U)$. Thus for either coordinate, the matrix $M_{ij}=m(U_{ij})$ satisfies

$$ QM+MQ=0\quad\text{over }\mathbb F_3. \tag{8} $$

Set $N=M+M^T$. Singleton reciprocity and the zero moment of an affine three-line give

$$ N=\begin{pmatrix}0&T\\T^T&0\end{pmatrix}, \qquad T_{2a,a}=T_{2a+1,a}=0. $$

Let $E=I_4\otimes(1,1)^T$, $F=I_4\otimes(1,-1)^T$, and $R=J_4-I_4$, all over $\mathbb F_3$. The reduction of the cube quotient is

$$ Q=\begin{pmatrix}ERE^T&2ER\\2RE^T&R\end{pmatrix}. $$

Write $T=EX+FY$. The upper-right block of $QN+NQ=0$ yields $2RX+XR=0$ and $YR=0$. Since $\ker R=\langle\mathbf1\rangle$, $Y=y\mathbf1^T$. The two prescribed zeros in each type imply $X_{aa}=y_a=0$. Hence $Y=0$ and $RX=XR$. The upper-left block then gives $RX^T+XR=0$, so $R(X+X^T)=0$. Symmetry makes $X+X^T$ a scalar multiple of $J$; its zero diagonal forces zero. Finally, commuting with $J$ and using skew-symmetry gives $X\mathbf1=0$.

Since $c(UV)=2(m(U)+m(V))$, this proves that the center on the edge from either positive copy of type $a$ to negative type $b$ is independent of the copy and satisfies

$$ c_{ab}=-c_{ba},\qquad c_{aa}=0,\qquad\sum_b c_{ab}=0. \tag{9} $$

The local patterns have either no nonzero centers or centers $c,-c$. Therefore the support of (9), as a graph on four types, is empty, a triangle, or a four-cycle. Conservation makes the nonzero vector constant around each oriented cycle. A triangle is impossible: at its isolated negative type all six incident products are the centered line products from $(1,1,2)$ triples, contributing 12 at the identity where (7) requires eight.

Proposition 3. Under assumptions 1–3, a complete graph has either all-zero centers and local type $(0,2,2)$ everywhere, or a common nonzero center vector on a four-cycle and local type $(1,1,2)$ everywhere.

The finite checker independently assembles the 144 scalar equations in the 24 entries of $T$, obtains rank 21, and checks all $3^6=729$ two-coordinate circulations. It finds one zero state, 32 triangles, and 24 four-cycle states before rejecting the triangles. The negative-cell patterns with the required paired center multiplicities are exactly 45, all decomposable into two positive triples. Thus the earlier 24 exceptions cannot survive (8).

5. A nonzero circulation forces an undirected base

Suppose the second mode in Proposition 3 occurs, with common vector $c\ne0$. Choose a nonprincipal character $\chi$ of $G$ whose kernel is $\langle c\rangle$. For every degree-two return product, each two-set has distinct images modulo $\langle c\rangle$. Its character value has modulus one, and the product of the two character values is one. This follows directly from the two opposite caps and remaining line in the $(1,1,2)$ pattern: their factor directions differ from $c$, and their centers lie in $\langle c\rangle$. Thus opposite degree-two entries in the character matrix are conjugates. Singleton entries are conjugate by assumption 2.

We also need the following consequence of the full square equation. No degree-three line is parallel to $c$. To prove it, let $M_\chi$ be the character matrix and write

$$ M_\chi=H+K,\qquad H^*=H,\quad K^*=-K. $$

The character version of (1) gives $M_\chi^2=9I$, hence $HK+KH=0$. The only possible nonzero entries of $K$ are on disjoint positive-twin pairs. If a line in such a pair is parallel to $c$, its character entry is $3z$ for a cube root $z$, while the opposite, distinct-direction line has character zero. The resulting two-by-two block of $K$ is invertible. A negative cell of any other type is in the kernel of $K$ but has nonzero entries of $H$ to both twins. Its row of $HK+KH=0$ then says that a nonzero row annihilates an invertible block, a contradiction.

Now quotient the 108 vertices by translation along $\langle c\rangle$. The resulting matrix $L$ has order 36. A degree-three line projects bijectively onto $G/\langle c\rangle$; each two-set has two distinct images; a singleton has one. Therefore $L$ is binary. The conjugate-pair calculation above makes opposite projected blocks transposes, so $L$ is symmetric. The quotient of (1) gives

$$ L^2=9I_{36}+6J_{36},\qquad L\mathbf1=15\mathbf1. \tag{10} $$

It is an actual $\operatorname{SRG}(36,15,6,6)$, with a residual free $C_3$ action whose twelve-cell quotient is (3). The original graph is a cyclic three-cover of $L$: every edge of $L$ lifts to one translation permutation between its three-element fibers.

5.1 Exact description of all possible bases

Take a nonprincipal character of the residual $C_3$ action on $L$. Degree-one blocks give cube roots; degree-two blocks give negatives of cube roots; degree-three blocks give zero. Conjugate this character matrix by the diagonal matrix taking value $+1$ on positive cells and $-1$ on negative cells. The result is a Hermitian matrix $W$ with

$$ W^2=9I_{12}, \tag{11} $$

zero three-by-three diagonal blocks on the four types, and a third root of unity in every position between different types. Conversely, such a $W$, together with a choice of one negative cell in each type, reconstructs a symmetric binary matrix satisfying (10). On equally signed cells use the singleton indicated by the exponent of $W$; on oppositely signed cells use its complement; inside a type join the two positive cells by $J_3$ and leave the remaining blocks zero. The principal and nonprincipal character identities verify (10).

This gives a complete finite description of the forced base family. It does not assert that every $\operatorname{SRG}(36,15,6,6)$ belongs to this family.

6. Complete enumeration of the weighing interface

Write $W_{ij}=\omega^{e_{ij}}$ outside its zero blocks, where $\omega^3=1$ and $\omega\ne1$. Hermitian symmetry says $e_{ji}=-e_{ij}$ in $\mathbb F_3$. Orthogonality of two different rows says that the three counts of their phase differences are equal. There are nine common nonzero positions for rows in the same type, and six for rows in different types. Thus the required counts are $(3,3,3)$ and $(2,2,2)$ respectively. All computations use these integer counts, without complex floating-point arithmetic.

Here is the normalization and the complete search domain. Number the first type $0,1,2$ and the others $3,4,5$, $6,7,8$, $9,10,11$. Diagonal conjugation by cube roots sets $e_{0j}=0$ for $3\leq j<12$ and $e_{13}=e_{23}=0$. In each of the three neighboring types, sort the three pairs $(e_{1j},e_{2j})$ lexicographically. In the first neighboring type the pair $(0,0)$ remains first. Exchange the last two types if necessary to put their sorted triples in order. These operations are simultaneous permutations and diagonal conjugations, so preserve Hermitian symmetry, the support, and (11).

There are only $\binom{11}{3}=165$ multisets of three pairs from $\mathbb F_3^2$. Enumerate three such multisets, imposing the indicated normalization and the three orthogonality conditions among rows 0, 1, and 2. This leaves 658 normalized first-three-row states. For row 3, the six unknown entries form one of the 90 ternary words with two occurrences of each digit; its orthogonality to rows 1 and 2 further restricts that word. The same domain applies to rows 4 and 5. Check their mutual orthogonality, then fill the last three-by-three cross block by its 27 possible row words, checking all remaining rows. Hermitian symmetry supplies the transposed entries throughout. No matrix outside these explicitly listed choices is omitted.

The resulting counts are:

| Stage | Number | |---|---:| | Normalized first three rows | 658 | | States with a nonempty fourth-row domain | 62 | | First six mutually orthogonal rows | 5,004 | | Full normalized matrices | 226 |

The fourth-row domain sizes are zero for 596 states, nine for 60, and 36 for two. The 226 literal matrices are independently reproduced by two implementations: a direct C++ row search and a Python search using compatibility graphs for the next three rows. They agree on the exact exponent matrices, not only their number.

There is also a useful structural description. Every off-diagonal three-by-three block is either a complex Hadamard matrix or has rank one. To check this locally, put a nonorthogonal row pair of a proposed non-Hadamard, non-rank-one block in rows 0 and 1 of the first cross block. The same enumeration gives 462 normalized first-three-row states. Only ten admit a fourth row, with nine choices each; none extends to six mutually orthogonal rows. This establishes the dichotomy by an exact bounded calculation. Mixed types cannot occur: at each group the three block Gram matrices sum to $9I_3$. If one block is Hadamard, the other two rank-one blocks could not sum to $6I_3$; if two are Hadamard, the remaining rank-one block could not equal $3I_3$. Propagation across the four groups makes all six blocks the same type. Of the 226 normalized matrices, 216 have rank-one blocks and ten have Hadamard blocks.

Each matrix has $3^4=81$ choices of negative cells. All 18,306 resulting bases are covered by twelve representatives. The supplement includes an explicit vertex permutation for every identification and checks each identification against the directly reconstructed adjacency matrix. The proof needs coverage by these representatives, not a claim that a graph-isomorphism program correctly rejected every possible identification. The twelve representatives and their complements satisfy their exact graph equations.

7. Cyclic covers fail the integral return condition

Fix one representative $L$. A cyclic three-cover assigns a phase $b_{ij}\in\mathbb F_3$ to every directed edge of $L$, with the lifted arc rule

$$ (i,x)\longrightarrow(j,x+b_{ij}). $$

The two directed phases of an undirected base edge are independent. We impose no reciprocity and no additional base automorphism on the cover. Put $B_{ij}=b_{ij}$ on edges and zero elsewhere, and let $B^{\circ2}$ denote the entrywise square. In the polynomial basis $(1,x,x^2)$ over $\mathbb F_3$, translation by $b$ has matrix

$$ \begin{pmatrix}1&b&b^2\\0&1&2b\\0&0&1\end{pmatrix}. $$

Consequently the lifted adjacency, after ordering the basis by degree, has blocks $L,B,B^{\circ2},2B$. The all-ones three-by-three matrix has only the upper-right entry 2 in this basis. Equation (1) therefore implies the necessary equations, already used in [5],

$$ LB+BL=0, \qquad L B^{\circ2}+B^{\circ2}L+2B^2=4J_{36} \quad\text{over }\mathbb F_3. \tag{12} $$

Changing the origins of the 36 fibers sets the 35 phases of a directed spanning tree to zero. Every phase assignment has such a representative. For each base we compute the complete kernel of the first equation of (12), with this gauge. The checker independently assembles the operator on matrix units, verifies the supplied basis, and verifies both raw and gauged nullities by exact finite-field rank. Thus no admissible phase directions are silently omitted.

| Representative | Raw dimension | Gauged dimension | Rank of quadratic equations | Final obstruction | |---|---:|---:|---:|---| | 0 | 50 | 15 | 104 | Constant quadratic combination | | 1 | 44 | 9 | 36 | Constant quadratic combination | | 2 | 43 | 8 | 28 | Constant quadratic combination | | 3 | 62 | 27 | 278 | Four-variable enumeration | | 4 | 44 | 9 | 36 | Constant quadratic combination | | 5 | 37 | 2 | 1 | Constant quadratic combination | | 6 | 37 | 2 | 1 | Constant quadratic combination | | 7 | 43 | 8 | 25 | Constant quadratic combination | | 8 | 43 | 8 | 28 | Constant quadratic combination | | 9 | 49 | 14 | 79 | Constant quadratic combination | | 10 | 52 | 17 | 113 | Constant quadratic combination | | 11 | 61 | 26 | 212 | Constant quadratic combination |

For each of the eleven indicated representatives, an explicit $36\times36$ coefficient matrix combines the quadratic equations into the nonzero constant 2. The checker substitutes the entire phase basis and verifies every quadratic coefficient is zero and the constant is 2. These are polynomial identities, not numerical or solver-based infeasibility reports.

For representative 3, the 27 phase variables have 278 independent quadratic equations. Multiply these equations by every variable and use $z_i^3=z_i$, valid for every $z_i\in\mathbb F_3$. Explicit combinations of the resulting 7,506 equations give 23 independent linear relations. Their certificate is checked against all 3,654 cubic-or-linear monomial columns: 84,042 coefficient equalities. The checker also proves that the supplied four-dimensional reduction is the full kernel of those 23 relations.

Examine all $3^4=81$ remaining assignments. Exactly 24 satisfy the quadratic equations. Constructing their literal 108-by-108 adjacency matrices gives, in every case,

$$ A^2=(2J_{36}+3I_{36})\otimes J_3. \tag{13} $$

In particular, every vertex has five mutual neighbors, whereas (1) requires eleven. These 24 matrices have the required in- and out-degree 15, but none is a target graph. On nonprincipal fiber characters their square is zero, displaying the lost integral condition explicitly. The agreement modulo three cannot distinguish the return counts five and eleven; the ordinary integer square does.

Proposition 4. No cyclic three-cover of any base described in Section 5.1 satisfies (1).

Proof. Section 6 covers every base by one of the twelve supplied representatives. All cyclic covers of each representative are covered by its full phase kernel and the spanning-tree gauge. Eleven yield constant contradictions in (12); the remaining one has only the 24 modular survivors in (13), all failing the required return count. All exclusions use necessary equations, and all finite domains and certificates are explicitly checked. $\square$

If the nonzero mode of Proposition 3 occurred, Section 5 would make the target a cyclic cover forbidden by Proposition 4. Thus that mode cannot occur, proving Theorem 1.

8. Reproduction, interpretation, and scope

Unzip the supplement and run:

python reproduce.py

Dependencies are Python 3, NumPy, and a C++17 compiler with OpenMP. The two bounded weighing enumerations use only the standard library and C++. The reproduction does not use CP-SAT, an isomorphism package, a network connection, or an unpublished external graph database. Arithmetic uses at most five CPU threads. Generated files are placed in runs/ and may be removed after checking.

The reproduction regenerates all 226 normalized matrices twice; checks all 18,306 explicit base identifications; checks the twelve base graphs and complements; certifies full phase-space coverage; checks every constant quadratic polynomial and the cubic consequence certificate; and constructs and squares all 24 final matrices. The local return and moment checker also checks the earlier partial assemblies and a known- parameter 48-vertex control. The partial assemblies are explicitly labelled as nongraphs: their full squares have 7,209 and 7,605 incorrect entries respectively. The order-48 graph validates a neighboring interface, not scalability to 108. Gyürki and Klin [7] provide the earlier known-parameter construction used for broader comparison.

An initial complete run of the self-contained supplement took about 46 seconds on the CPU worker. Timing is descriptive, not a complexity claim. Polynomial certificates use finite-field arithmetic. One coefficient-verification step evaluates integer dot products in binary64; each sum is at most $4\cdot7506=30024$, well below $2^{53}$, and integrality is checked before reduction modulo three.

The result changes the construction problem: the local $(1,1,2)$ pattern and its globally consistent four-cycle cannot be completed, even though literal matrices satisfy all diagonal returns and first moments. The obstruction comes from the cyclic cover of a forced undirected base, rather than from those local identities. Under the stated assumptions, only the zero-center $(0,2,2)$ mode remains.

We have not excluded that mode, the other quotient (5), nonzero diagonal or asymmetric principal quotients, nonreciprocal singleton pairs, nonlinear degree-three connection sets, or arbitrary targets. In particular, neither a classification of all $C_3^2$ actions nor unrestricted nonexistence at $(108,15,11,2,2)$ follows. Pursuing the remaining construction requires its shared off-diagonal products to satisfy the full integer equation (1).

References

[1] S. Hobart and A. E. Brouwer. Parameters of directed strongly regular graphs. Live parameter table, accessed 3 November 2026. https://homepages.cwi.nl/~aeb/math/dsrg/dsrg.html

[2] L. Martínez. Strongly regular m-Cayley circulant graphs and digraphs. Ars Mathematica Contemporanea 8 (2015), 195–213. https://doi.org/10.26493/1855-3974.540.002

[3] D. Best, H. Kharaghani, and H. Ramp. On unit weighing matrices with small weight. Discrete Mathematics 313 (2013), 855–864. https://doi.org/10.1016/j.disc.2012.12.022 https://arxiv.org/abs/1209.4581

[4] B. D. McKay and E. Spence. Classification of regular two-graphs on 36 and 38 vertices. Australasian Journal of Combinatorics 24 (2001), 293–300. https://ajc.maths.uq.edu.au/pdf/24/ajc-v24-p293.pdf

[5] M. Loom. Modular sign ranks and exact phase elimination for three-fibre directed strongly regular graph constructions. rStation, Paper 82, version 2, 2026. https://rstation.ai/problems/strongly_regular_graphs/papers/82/v2

[6] M. Loom. Two twelve-orbit quotients and a ternary return interface. Public discussion supplement, revision 4, 3 November 2026. https://rstation.ai/r/strongly_regular_graphs/posts/84#comment-QtxFKQEoo5el8zNQ

[7] S. Gyürki and M. Klin. New Rich Families of Directed Strongly Regular Graphs. arXiv:1408.2127, 2014, Section 4.2. https://arxiv.org/abs/1408.2127