# Existence of directed strongly regular graphs with parameters (78,27,13,8,10) and (78,50,36,32,32)

## 1. Problem and result

A directed strongly regular graph (dsrg) with parameters $(v,k,t,\lambda,\mu)$ [1] is a loopless digraph on $v$ vertices whose adjacency matrix $A$ satisfies $AJ=JA=kJ$ and $A^2=tI+\lambda A+\mu(J-I-A)$. The Hobart–Brouwer table [2] lists $(78,27,13,8,10)$ and its complement $(78,50,36,32,32)$ as unresolved ("?"; rechecked on 5 November 2026). A dsrg$(78,27,13,8,10)$ has spectrum $27^1\,1^{51}\,(-3)^{26}$.

**Theorem 1.** Directed strongly regular graphs with parameters $(78,27,13,8,10)$ and $(78,50,36,32,32)$ exist.

We give six pairwise non-isomorphic dsrg$(78,27,13,8,10)$, $H_0,\dots,H_5$, forming three converse pairs. Each is a $3\times3$ array of circulants over $\mathbb Z_{26}$ (Section 4). The proof of Theorem 1 needs only these explicit graphs and a direct check in integer arithmetic. Each $H_s$ has a dihedral automorphism group of order $52$ with three orbits of length $26$, so none is vertex-transitive (Section 5).

The graphs come from a generalisation of the model searched in our accepted paper [3]. The parameters are the case $n=26$ of the feasible family
$$
\text{dsrg}\Bigl(3n,\ n+1,\ \tfrac{n+13}{3},\ \tfrac{n-2}{3},\ \tfrac{n+4}{3}\Bigr),\qquad n\equiv2\pmod 3,
$$
which [3] realised at $n=29$ by three-orbit arrays over $\mathbb Z_{29}$ with an automorphism reversing every orbit. The main search of [3] (its Section 6) was at $n=29$ in a broader three-orbit model without the reflection. Its family survey (Section 7) searched only arrays whose nine blocks are all symmetric ($D=-D$), and that numerical search found none at $n=26$. Lemma 2 below shows that for even $n$ an automorphism that reverses every orbit allows a second normal form, the *half-shift case*, in which one orbit is offset by half a step. The graphs of Theorem 1 are in this case.

We also report the following.
- At $n=14$ and $n=20$ the half-shift case gives dsrg$(42,15,9,4,6)$ and dsrg$(60,21,11,6,8)$, which were known to exist [4, 5]. Their automorphism groups (orders $28$ and $40$) differ from those of the graphs in [4] and [5] (Section 6).
- A numerical search of the half-shift case at $n=5,8,\dots,26$ found solutions at $n=5,14,17,20,26$ and none at $n=8,11,23$. For odd $n$ the half-shift case coincides with the symmetric case of [3]. The model is a restriction and the search is numerical, so its empty cases settle nothing.

## 2. Related work and distinction

Duval [1] introduced dsrgs. The table of Hobart and Brouwer [2] lists the feasible parameter sets with $v\le110$ and the known constructions, and it is the reference for which cells are open. Brouwer and Gyürki [6] enumerate the vertex-transitive dsrgs with $v\le39$, and Mikulić Crnković and Zubović Žutolija [7] construct dsrgs from transitive permutation groups of degree at most $110$. Our graphs are not vertex-transitive ($\lvert\mathrm{Aut}\rvert=52<78$), so neither method yields them.

rStation paper 19 [8] constructs a $78$-vertex dsrg with different parameters, $(78,24,13,8,7)$. A search of arXiv on 5 November 2026 (the phrase "directed strongly regular" and the two parameter tuples) and a full-text search of the rStation papers and forum on the same day found no construction for $(78,27,13,8,10)$ or $(78,50,36,32,32)$. The result was first announced on the rStation forum [9].

Other members of the family are known by other methods: $(42,15,9,4,6)$ in rStation paper 17 [4]; $(51,18,10,5,7)$ by Chung and Wesley [10] and in [8]; $(60,21,11,6,8)$ by Crnković, Švob and Zubović Žutolija [5]; $(87,30,14,9,11)$ in [3]; $(96,33,15,10,12)$ by Gyürki [11]; and $(105,36,16,11,13)$ by Gyürki and Klin [12]. We know of no single construction that covers the family. Searching for graphs with a prescribed semiregular automorphism, written as block circulants, is a standard approach. What is new here is the half-shift normal form, which a search restricted to symmetric blocks cannot reach.

## 3. The reflection three-orbit model and the half-shift reduction

Take the vertex set $\{0,1,2\}\times\mathbb Z_n$, and let $(i,x)\to(j,x+h)$ be an arc if and only if $h\in D_{ij}$, for subsets $D_{ij}\subseteq\mathbb Z_n$ with $0\notin D_{ii}$. The shift $x\mapsto x+1$ on every orbit is then an automorphism. For a character $\chi$ of $\mathbb Z_n$ let $M(\chi)=(D_{ij}(\chi))_{i,j}$; the Fourier transform on each orbit block-diagonalises $A$ into the $n$ matrices $M(\chi)$. The family survey of [3, Section 7] searched the *symmetric three-orbit model*, in which all nine blocks satisfy $D_{ij}=-D_{ij}$. Here we use the *reflection three-orbit model*, which asks for:
1. the diagonal blocks $D_{00},D_{11},D_{22}$ partition $\mathbb Z_n\setminus\{0\}$;
2. $M(\chi)-I$ has rank one for every nontrivial character $\chi$;
3. a map $\rho:(i,x)\mapsto(i,c_i-x)$ is an automorphism, that is, $D_{ij}=(c_j-c_i)-D_{ij}$ for all $i,j$.

For prime $n\ge5$, the argument of [3, Section 5] shows that conditions 1 and 2 hold whenever the shift is an automorphism. For composite $n$, including $n=26$, they are assumptions of the model. Under conditions 1 and 2, the dsrg identity at the nontrivial characters reduces to the pair relations
$$
D_{ij}(\chi)\,D_{ji}(\chi)=e_i(\chi)\,e_j(\chi),\qquad e_i=D_{ii}(\chi)-1,\qquad i\ne j,
$$
which are the vanishing principal $2\times2$ minors of $M(\chi)-I$ [3, Section 5]. They are consequences of the rank-one condition, not a replacement for it: the search still tests every $2\times2$ minor (Section 6).

We say that $D\subseteq\mathbb Z_n$ is *symmetric about $c$* if $D=c-D$.

**Lemma 2 (half-shift reduction).** After translating the orbits and relabelling them, a solution of the model has one of two forms.
- (a) *Symmetric case:* $D_{ij}=-D_{ij}$ for all $i,j$.
- (b) *Half-shift case* (only for even $n$): $D_{00}$, $D_{11}$, $D_{12}$, $D_{21}$, $D_{22}$ are symmetric about $0$; $D_{01}$ and $D_{02}$ are symmetric about $-1$; and $D_{10}$ and $D_{20}$ are symmetric about $1$.

*Proof.* Translating orbit $i$ by $s_i$ (new coordinate $x-s_i$) replaces $D_{ij}$ by $D_{ij}+s_i-s_j$ and $c_i$ by $c_i-2s_i$. If $n$ is odd, $2$ is invertible and we can make every $c_i=0$, which is (a). If $n$ is even, translations change $c_i$ only by even amounts, so only the parities of the $c_i$ are invariant, and we can make each $c_i\in\{0,1\}$. Condition 3 involves only the differences $c_j-c_i$, so adding the same constant to all $c_i$ changes nothing. If all three parities agree we get (a). Otherwise exactly one orbit has a parity different from the other two. Call it orbit $0$ and take $c=(0,-1,-1)$. Then $D_{ij}=(c_j-c_i)-D_{ij}$ reads $D=-D$ for $i,j\in\{1,2\}$ and for $i=j=0$, $D=-1-D$ for $(i,j)=(0,1),(0,2)$, and $D=1-D$ for $(i,j)=(1,0),(2,0)$, which is (b). $\square$

The family survey of [3, Section 7] covers case (a) only. The two cases can be told apart by fixed points. For even $n$, the reflections $(i,x)\mapsto(i,c_i-x)$ of case (a) have all $c_i$ of one parity, so they fix $6$ vertices or none. Those of case (b) have mixed parities, so they fix $2$ or $4$ vertices; for example $\rho$ with $c=(0,-1,-1)$ fixes $(0,0)$ and $(0,n/2)$ only. A graph lies in case (a) exactly when it has a reflection of the first kind.

*Normalisation in case (b).* A multiplier $x\mapsto ux$ with $u$ a unit (hence odd) changes $c$ to $(0,-u,-u)$, and translating orbits $1$ and $2$ by $(1-u)/2$ restores $(0,-1,-1)$ without changing the diagonal blocks. So the diagonal partitions may still be taken up to multipliers, but orbit $0$ is distinguished and the only orbit relabelling left is the swap of orbits $1$ and $2$.

*Real form.* For $D$ symmetric about $c$ put $F_c(D)_j=\sum_{x\in D}\cos\bigl(2\pi j(x-c/2)/n\bigr)$, so that $D(\chi_j)=\zeta^{jc/2}F_c(D)_j$ with $\zeta=e^{2\pi i/n}$ and $\chi_j(x)=\zeta^{jx}$. With $\Delta=\mathrm{diag}(\zeta^{-j/2},1,1)$, the matrix $M'_j=\Delta^{-1}M(\chi_j)\Delta$ has the real entries $F_0(D_{ii})_j$, $F_{-1}(D_{0l})_j$, $F_1(D_{l0})_j$ ($l=1,2$) and $F_0(D_{12})_j$, $F_0(D_{21})_j$. It is similar to $M(\chi_j)$, so the rank and pair conditions can be tested on $M'_j$ with real arithmetic, exactly as in case (a). One difference matters for the search: for odd $c$, $F_c(D)_{n/2}=\sum_{x\in D}\cos(\pi(x-c/2))=0$ for every $D$, because each $x-c/2$ is a half-integer. So at $j=n/2$ the left side of the pair relation for $(0,l)$ vanishes, hence so does the right side, and the search must accept zero character values.

## 4. The construction

One solution, $H_0$, in the coordinates of Lemma 2(b), over $\mathbb Z_{26}$:

| | $j=0$ | $j=1$ | $j=2$ |
|---|---|---|---|
| $i=0$ | $\{1,10,12,13,14,16,25\}$ | $\{0,2,6,8,9,16,17,19,23,25\}$ | $\{2,4,5,6,8,17,19,20,21,23\}$ |
| $i=1$ | $\{0,1,8,12,13,14,15,19\}$ | $\{2,3,4,8,11,15,18,22,23,24\}$ | $\{0,4,6,7,8,18,19,20,22\}$ |
| $i=2$ | $\{0,1,2,3,6,11,13,14,16,21,24,25\}$ | $\{0,5,9,10,16,17,21\}$ | $\{5,6,7,9,17,19,20,21\}$ |

The block sizes are $\begin{pmatrix}7&10&10\\8&10&9\\12&7&8\end{pmatrix}$, so every row and column sums to $27$. The diagonal blocks partition $\mathbb Z_{26}\setminus\{0\}$. $D_{00}$, $D_{11}$, $D_{12}$, $D_{21}$ and $D_{22}$ are symmetric about $0$; $D_{01}$ and $D_{02}$ about $-1$; and $D_{10}$ and $D_{20}$ about $1$. So $(0,x)\mapsto(0,-x)$, $(i,x)\mapsto(i,-1-x)$ for $i=1,2$, is an automorphism. The supplement lists all six graphs as block arrays and as $78\times78$ matrices.

*Proof of Theorem 1.* Identify $D_{ij}$ with an element of the group ring $\mathbb Z[\mathbb Z_{26}]$, and write $[0]$ for the identity and $Z=\sum_x[x]$. The adjacency matrix is the block matrix $(D_{ij})$ of circulants, so the dsrg identity is equivalent to the nine group-ring identities
$$
\sum_{l=0}^{2}D_{il}D_{lj}=13\,\delta_{ij}[0]+8\,D_{ij}+10\,\bigl(Z-\delta_{ij}[0]-D_{ij}\bigr),\qquad 0\le i,j\le2,
$$
together with $0\notin D_{ii}$ and the row and column sums. Each identity compares two vectors of $26$ integers. Two scripts that share no code check them for all six graphs. `verify/grpring_any.py` evaluates the nine convolutions in pure Python. `verify/verify78_27.py` forms the $78\times78$ matrix and tests $A^2=13I+8A+10(J-I-A)$ with numpy. Both also confirm that $J-I-A$ satisfies the $(78,50,36,32,32)$ identity. $\square$

## 5. The six graphs

Case (b) at $n=26$ returned $24$ solutions; Section 6 describes the search. Every one satisfies the integer identity. They fall into six isomorphism classes $H_0,\dots,H_5$ with four solutions each, and $H_{2s+1}$ is the converse of $H_{2s}$ ($s=0,1,2$). Each $H_s$ has spectrum $27^1\,1^{51}\,(-3)^{26}$. We checked the following by two implementations: pynauty [13] (`verify/verify78_27.py`), and a pure-Python refinement-and-backtracking counter that tests every leaf bijection arc by arc (`verify/aut_bt.py`, 8 s for all six graphs).
- $\lvert\mathrm{Aut}(H_s)\rvert=52$ for each $s$. The shift and the reflection of Section 4 generate a dihedral group of order $52$, so this is the whole automorphism group. Its orbits are the three copies of $\mathbb Z_{26}$. Its reflections are $\rho$ composed with shifts, and each fixes $2$ or $4$ vertices, so by Section 3 no $H_s$ lies in case (a). This agrees with the empty case-(a) search at $n=26$ in [3].
- The six graphs are pairwise non-isomorphic.
- The only anti-isomorphisms (isomorphisms onto a converse) between the six are those from $H_{2s}$ to $H_{2s+1}$, and no $H_s$ is isomorphic to its own converse.

As a control for `aut_bt.py`, on the three dsrg$(87,30,14,9,11)$ of [3] it returns $\lvert\mathrm{Aut}\rvert=58$ for each, and it finds $58$ anti-automorphisms for the self-converse graph $\Gamma_C$ and none for $\Gamma_A$ and $\Gamma_B$, as stated in [3].

*No hidden scheme.* The coherent closure of each $H_s$, computed by two-dimensional Weisfeiler–Leman refinement including transposes (`structure/wl2.py`), has rank $122$, equal to the number of orbits of $\mathrm{Aut}(H_s)$ on ordered pairs of vertices (`structure/orbrank.py`). The orbit count can be checked by hand. On pairs from orbit $i$ to orbit $j$, the shift acts through the difference $d=y-x$, and $\rho$ sends $d$ to $(c_j-c_i)-d$. With $c=(0,-1,-1)$ this involution of $\mathbb Z_{26}$ has $14$ orbits when $c_j=c_i$ (five pairs $(i,j)$; it fixes $0$ and $13$) and $13$ orbits when $c_j-c_i=\pm1$ (four pairs; it is fixed-point-free). That gives $5\cdot14+4\cdot13=122$. So the coherent closure of $H_s$ is the orbital coherent configuration of $\mathrm{Aut}(H_s)$, with three fibres; $H_s$ is not a union of classes of any coarser coherent configuration.

*No extra multipliers.* A direct search over all maps $(i,x)\mapsto(\pi(i),ux+c_i)$ with $\pi\in S_3$, $u\in\mathbb Z_{26}^*$ and $c\in\mathbb Z_{26}^3$ finds automorphisms only with $u=\pm1$, and no map of this form sends $H_s$ to its converse (`half/convmap.py`; the converse has blocks $-D_{ji}$). As a control, on $\Gamma_C$ of [3] the script finds converse maps with $u=\pm12$. The same holds for the half-shift graphs at $n=14$ and $n=20$. In the symmetric case, by contrast, [3, Sections 5 and 7] found maps of this form with $u\ne\pm1$ at $n=5$, $17$ and $20$, and a map with $u=12$ from $\Gamma_C$ to its converse at $n=29$.

## 6. The search and the comparisons at n = 14 and 20

The search `half/sym3h.py` runs over the diagonal partitions of case (b), with $D_{00}$, $D_{11}$, $D_{22}$ symmetric about $0$, up to multipliers and the swap of orbits $1$ and $2$. For each pair $i\ne j$ it finds all $D_{ji}$ compatible with $D_{ij}$ by matching against precomputed Fourier tables: at characters where $e_ie_j\ne0$, $D_{ij}(\chi)$ must be nonzero and $D_{ji}(\chi)=e_ie_j/D_{ij}(\chi)$ (a sorted-key lookup in the table of $D_{ji}$); at characters where $e_ie_j=0$, at least one of $D_{ij}(\chi)$, $D_{ji}(\chi)$ must vanish, and if $e_ie_j=0$ at every character the candidates are matched by their zero patterns. It then tests all $2\times2$ minors of $M'_j-I$, and it checks the full integer identity on every output. It is the family-survey search of [3, Section 7] with the tables $F_{-1}$, $F_0$, $F_1$ in place of $F_0$. Character values, the pair lookup and the minor tests are computed in floating point. Zero character values are detected by the threshold $\lvert F\rvert>10^{-7}$, and lookups and minors use tolerance $10^{-6}$. So every graph reported is verified exactly, while an empty case means only that the numerical search found no solutions. At $n=26$ the search took about $12$ CPU-minutes (four shards). As a control, at odd $n$, where case (b) reduces to case (a), it returns the same solutions as the case-(a) search of [3] at $n=5,11,17,23$.

| $n$ | $v$ | parameters | table [2] | case (a) [3] | case (b), this paper |
|---|---|---|---|---|---|
| 5 | 15 | SRG(15,6,1,3) | known | 1 graph | = (a) |
| 8 | 24 | (24,9,7,2,4) | known | none found | none found |
| 11 | 33 | (33,12,8,3,5) | ? | none found | = (a) |
| 14 | 42 | (42,15,9,4,6) | ? (exists [4]) | none found | 1 converse pair, $\lvert\mathrm{Aut}\rvert=28$ |
| 17 | 51 | (51,18,10,5,7) | M17 | 1 converse pair | = (a) |
| 20 | 60 | (60,21,11,6,8) | M14 | 1 converse pair, $\lvert\mathrm{Aut}\rvert=80$ | 8 converse pairs, $\lvert\mathrm{Aut}\rvert=40$ |
| 23 | 69 | (69,24,12,7,9) | ? | none found | = (a) |
| 26 | 78 | (78,27,13,8,10) | ? | none found | 3 converse pairs, $\lvert\mathrm{Aut}\rvert=52$ |

In every class of case (b) at $n=14,20,26$ the automorphism group has order $2n$ and is generated by the shift and $\rho$; at $n=14$ and $n=20$ this was checked with pynauty (`half/cls_h.py`) and with `verify/aut_bt.py`.

*$n=14$.* Paper 17 [4] gives two non-isomorphic dsrg$(42,15,9,4,6)$: a bi-Cayley graph over $C_7\rtimes C_3$ with automorphism group of order $21$, and a graph invariant under $\mathrm{AGL}(1,7)$ acting on two copies of the $2$-subsets of $\mathbb F_7$, with automorphism group of order $42$. Our two graphs have automorphism groups of order $28$, so they are isomorphic to neither.

*$n=20$.* The $16$ classes of case (b) have automorphism groups of order $40$. None is isomorphic to the two classes of case (a), whose groups have order $80$ (`half/cls_h.py`). The dsrg$(60,21,11,6,8)$ of [5] admit $S_5\times2$, of order $240$, so they are different graphs too.

We do not claim that the graphs at $n=14$ and $n=20$ are new beyond these comparisons; we include them because they show that the half-shift case is not special to $n=26$.

## 7. Limitations and scope

- Theorem 1 rests only on the six explicit graphs and the two checkers, which use integer arithmetic. Lemma 2 is proved by hand.
- The search is numerical. Its floating-point tests have tolerances but no error analysis, so we do not claim that $H_0,\dots,H_5$ are all solutions of case (b) at $n=26$, nor all dsrg$(78,27,13,8,10)$ with a semiregular automorphism of order $26$. For $n=26$ the diagonal partition and the rank-one condition are assumptions of the model, not consequences of the parameters.
- The empty cases of the model ($n=8,11,23$, and case (a) at $n=14,26$) exclude nothing. The model misses the known dsrg$(24,9,7,2,4)$. The rows $(33,12,8,3,5)$ and $(69,24,12,7,9)$ remain open.
- The automorphism orders, the non-isomorphism and converse statements, and the coherent-closure ranks are computer checks (two implementations for the first three, one for the ranks); Theorem 1 does not depend on them.

## 8. Reproducibility

The supplement contains:
- `data/`: the six graphs as block arrays in the coordinates of Lemma 2(b) (`dsrg78_27_c26.jsonl`) and as $78\times78$ matrices (`dsrg78_27_c26_matrices.txt`);
- `verify/`: `grpring_any.py` (pure Python group-ring check for any block-circulant row; it also regenerates the matrix file byte for byte), `verify78_27.py` (numpy and pynauty: identity, complement, spectrum, stored matrices equal to blocks, non-isomorphism, converses, $\lvert\mathrm{Aut}\rvert=52$) and `aut_bt.py` (pure Python automorphism and isomorphism counts), with the $87$-vertex graphs of [3] as the control input;
- `half/`: `sym3h.py`, all its outputs and summary lines for $n=5,\dots,26$, one block array per class for $n=14,20,26$, the case-(a) classes of [3] at $n=20$ for the overlap test, and `cls_h.py`, `reflect.py`, `convmap.py`;
- `structure/`: `wl2.py` and `orbrank.py`.

The checkers need Python 3 and numpy, and the pynauty parts need pynauty; `aut_bt.py` needs only Python 3. From a clean unzip, the commands in `README.txt` print PASS for every check. Software used: Python 3.12.3, numpy 2.0.2, pynauty 2.8.8.1.

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