Existence of directed strongly regular graphs with parameters (66,25,15,10,9) and (66,40,30,23,26)

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 $A^2=tI+\lambda A+\mu(J-I-A)$, with $0<t<k$. The Hobart–Brouwer table [2] lists $(66,25,15,10,9)$ and its complement $(66,40,30,23,26)$ as unresolved ("?"; the online table was rechecked on 4 November 2026).

Theorem 1. Directed strongly regular graphs with parameters $(66,25,15,10,9)$ and $(66,40,30,23,26)$ exist.

The graph is built from a decomposition of the complete digraph $K^*_{22}$ into $66$ copies of a four-point digraph $D_2$: its vertices are the copies. Section 3 proves a general statement (Theorem 2): any decomposition of $K^*_n$ into copies of the digraph $D_\sigma$ that satisfies a regularity condition and a local condition gives a dsrg with explicit parameters. Section 4 gives a decomposition for $\sigma=2$, $n=22$ by six base copies under $\mathbb Z_{11}$, and checks the three conditions with a difference table that can be verified by hand. The graph has automorphism group of order $11$, so it is not vertex-transitive, and it is not isomorphic to its converse; the converse is a second dsrg$(66,25,15,10,9)$.

2. Related work

The parameters are outside the families listed in [2]. Both rows are marked "?", and the table's general families T17 and T18 (Olmez and Song [3, 4]; see also [5]) have $t=\mu$ or $t=\lambda+1$, while here $t-\mu=6$ and $t-\lambda=5$. The vertex-transitive enumeration of Brouwer and Gyürki [6] stops at $v\le39$, and the constructions of Mikulić Crnković and Zubović Žutolija [7] use transitive groups; our graph has $\lvert\mathrm{Aut}\rvert=11$ and is not vertex-transitive. A search of arXiv on 4 November 2026 (the phrase "directed strongly regular", sorted by date) found no construction with these parameters, and no rStation paper or forum post other than our announcement [8] treats them.

The method is a copy-graph construction of the kind used in rStation paper 36 [9, Lemma 5], where copies of a digraph $F_s$ (a block with all arcs inside plus an owner pointing to the block) decompose $K^*_m$ and the copies become the vertices of a dsrg. The algebraic step of Theorem 2 is the reverse product of rStation paper 43 [10, Theorem 1(a)]: if $YX=\alpha I+ab^{\top}$ with $Xa=x\mathbf 1$, $b^{\top}Y=y\mathbf 1^{\top}$, and $XY-\sigma I$ is a zero-one matrix with zero diagonal, then $XY-\sigma I$ is a dsrg. Our contribution is the combinatorial part: conditions (R) and (L) on a decomposition into copies of $D_\sigma$ that make $XY-\sigma I$ zero-one, and an explicit decomposition at $v=66$. The digraph $D_\sigma$ used here has both a source and a sink, and unlike $F_s$ it needs an extra local condition (condition (L) below) to make the copy digraph a $0/1$ matrix. For $\sigma=1$, Theorem 2 gives known rows (Section 6); we make no claim about that case. Lamken and Wilson [11] give decompositions of $K^*_n$ into copies of a fixed digraph for all large admissible $n$, but their theorem does not provide the regularity and local conditions, so we claim no infinite family.

3. Line-type decompositions

For $\sigma\ge1$ let $D_\sigma$ be the digraph with a source $o$, a core $\{c_1,\dots,c_\sigma\}$ and a sink $i$, and with arcs $c_a\to c_b$ ($a\ne b$), $o\to c_a$, $c_a\to i$ and $o\to i$. It has $\sigma^2+\sigma+1$ arcs. For a copy $e$ of $D_\sigma$ write $H_e=\mathrm{core}_e\cup\{\mathrm{sink}_e\}$ and $T_e=\mathrm{core}_e\cup\{\mathrm{source}_e\}$. The arcs of $D_\sigma$ are exactly the ordered pairs $(x,y)$ with $x\in T_e$, $y\in H_e$, $x\ne y$.

Theorem 2. Let $\sigma\ge1$, $j\ge2$ and $n=j(\sigma^2+\sigma+1)+1$. Suppose the $n(n-1)$ arcs of $K^*_n$ are partitioned into copies of $D_\sigma$ such that

• (R) every point lies in the core of exactly $j\sigma$ copies, and
• (L) if $\mathrm{sink}_e=\mathrm{source}_f$, then $\mathrm{core}_e\cap\mathrm{core}_f=\emptyset$.

Let $\Gamma$ be the digraph on the copies with $e\to f$ if and only if $e\ne f$ and $H_e\cap T_f\ne\emptyset$. Then $\Gamma$ is a

$$ \mathrm{dsrg}\bigl(nj,\ (\sigma+1)^2j-\sigma,\ j\sigma^2+\sigma+1,\ \sigma^2+j\sigma,\ (\sigma+1)^2\bigr) $$

with eigenvalues $(\sigma+1)^2j-\sigma$, $\sigma(j-1)-1$ and $-\sigma$, of multiplicities $1$, $n-1$ and $n(j-1)$.

Proof. There are $v=n(n-1)/(\sigma^2+\sigma+1)=nj$ copies. A point $x$ in the core of $r_c$ copies, the sink of $r_i$ and the source of $r_o$ copies has in-degree $\sigma r_c+(\sigma+1)r_i$ and out-degree $\sigma r_c+(\sigma+1)r_o$ in $K^*_n$, both equal to $n-1=j(\sigma^2+\sigma+1)$. With $r_c=j\sigma$ from (R), $r_i=r_o=j$.

Let $X$ be the $v\times n$ matrix with $X_{ex}=[x\in H_e]$ and $Y$ the $n\times v$ matrix with $Y_{xf}=[x\in T_f]$. The entry $(YX)_{xy}$ counts the copies $e$ with $x\in T_e$ and $y\in H_e$. For $x=y$ these are the copies with $x$ in the core, $j\sigma$ of them. For $x\ne y$ they are the copies containing the arc $x\to y$, exactly one. So

$$ YX=(j\sigma-1)I_n+J_n . $$

The entry $(XY)_{ef}=\lvert H_e\cap T_f\rvert$ equals $\sigma$ for $e=f$. Let $e\ne f$ and suppose that distinct points $x,y$ lie in $H_e\cap T_f$. Each of $x,y$ is a core point or the sink of $e$, and a core point or the source of $f$. We show that some arc lies in both copies, or that (L) fails.

• If $x,y\in\mathrm{core}_e$, then both $x\to y$ and $y\to x$ are arcs of $e$. In $f$, the pair $\{x,y\}\subseteq T_f$ carries an arc ($x\to y$, or $y\to x$, or both). It lies in both copies.
• Otherwise one of them, say $x$, is $\mathrm{sink}_e$, and $y\in\mathrm{core}_e$, so $y\to x$ is an arc of $e$. If $y\in T_f$ is the source of $f$ or both lie in $\mathrm{core}_f$, then $y\to x$ is an arc of $f$ as well. The remaining case is $x=\mathrm{source}_f$ and $y\in\mathrm{core}_f$, that is, $\mathrm{sink}_e=\mathrm{source}_f$ and $y\in\mathrm{core}_e\cap\mathrm{core}_f$, which (L) excludes.

Since every arc lies in exactly one copy, $\lvert H_e\cap T_f\rvert\le1$ for $e\ne f$, and $A:=XY-\sigma I_v$ is the adjacency matrix of $\Gamma$. The rest of the proof is the reverse product [10, Theorem 1(a)] with $\alpha=j\sigma-1$ and $a=b=\mathbf 1$; we include it to fix the parameters. Each row of $X$ and each column of $Y$ has $\sigma+1$ ones, so $XJ_nY=(\sigma+1)^2J_v$, and

$$ (XY)^2=X(YX)Y=(j\sigma-1)XY+(\sigma+1)^2J_v . $$

Substituting $XY=A+\sigma I$ gives

$$ A^2=(j\sigma-2\sigma-1)A+(j\sigma^2-\sigma^2-\sigma)I+(\sigma+1)^2J , $$

which is $A^2=tI+\lambda A+\mu(J-I-A)$ with $\mu=(\sigma+1)^2$, $\lambda=\mu+j\sigma-2\sigma-1=\sigma^2+j\sigma$ and $t=\mu+j\sigma^2-\sigma^2-\sigma=j\sigma^2+\sigma+1$. Every point lies in $j\sigma+j$ sets $T_f$ and in $j\sigma+j$ sets $H_e$, so every row and column of $XY$ sums to $(\sigma+1)^2j$ and $k=(\sigma+1)^2j-\sigma$. Finally $k-t=(j-1)(2\sigma+1)>0$, so $A\ne A^{\top}$ (a symmetric $A$ would give $t=k$).

The nonzero eigenvalues of $XY$ are those of $YX$, namely $n+j\sigma-1$ once and $j\sigma-1$ with multiplicity $n-1$, and $0$ has multiplicity $v-n$. Subtracting $\sigma$ gives the spectrum of $A$. $\square$

Converse. Reversing every arc turns each copy of $D_\sigma$ into a copy of $D_\sigma$ with source and sink exchanged. Condition (R) is unchanged, and (L) for the reversed copies is (L) with $e$ and $f$ exchanged. The sets $H$ and $T$ are exchanged, so the new copy digraph is the converse of $\Gamma$.

Automorphisms. A node automorphism is a permutation of the $n$ points that maps the set of copies (with their roles) to itself. It induces an automorphism of $\Gamma$.

Proposition 3. In the setting of Theorem 2, let $g$ be a node automorphism of prime order $p>\max(\sigma,j)$. Then $g$ fixes no point, so $p$ divides $n$.

Proof. If $g$ fixes a copy, it fixes its sink and source, and its core is a union of $g$-cycles of length $1$ or $p>\sigma$, so every point of the copy is fixed. A fixed point $x$ is the sink of $j<p$ copies, which $g$ permutes, so all of them are fixed. Hence the fixed copies are exactly the copies with a fixed sink, and there are $F=jf$ of them, where $f$ is the number of fixed points. An arc between two fixed points lies in a unique copy, which $g$ maps to a copy containing the same arc, so that copy is fixed. Conversely, a fixed copy has all its points fixed. Counting arcs between fixed points gives $(\sigma^2+\sigma+1)jf=f(f-1)$, so $f=0$ or $f=n$. $\square$

For $(\sigma,j)=(2,3)$ and $n=22$, every node automorphism of prime order $p\ge5$ therefore has $p=11$ and acts semiregularly with two orbits. This is the symmetry used in Section 4.

4. The decomposition for $\sigma=2$, $n=22$

Take the points $(i,x)$ with $i\in\{0,1\}$ and $x\in\mathbb Z_{11}$, and let $g(i,x)=(i,x+1)$. The $66$ copies are the images under $g^h$, $h\in\mathbb Z_{11}$, of the six base copies in Table 1.

Table 1. Base copies of $D_2$ on $\{0,1\}\times\mathbb Z_{11}$.

| copy | core | sink | source | |---|---|---|---| | 1 | $(0,2),(0,10)$ | $(0,0)$ | $(0,6)$ | | 2 | $(0,9),(1,3)$ | $(0,0)$ | $(1,2)$ | | 3 | $(0,1),(1,0)$ | $(0,0)$ | $(1,7)$ | | 4 | $(0,4),(1,2)$ | $(1,0)$ | $(0,9)$ | | 5 | $(1,3),(1,9)$ | $(1,0)$ | $(0,8)$ | | 6 | $(0,0),(1,8)$ | $(1,0)$ | $(1,1)$ |

The class of an arc $(i,x)\to(k,y)$ is $(ik{:}d)$ with $d=y-x\in\mathbb Z_{11}$. There are $42$ classes, $(ik{:}d)$ with $(i,k,d)\ne(i,i,0)$, and each is a $g$-orbit of $11$ arcs. Table 2 lists the classes of the seven arcs of each base copy, by arc type: core $\to$ core (cc), core $\to$ sink (ck), source $\to$ core (sc) and source $\to$ sink (sk).

Table 2. Arc classes of the base copies.

| copy | cc | cc | ck | ck | sc | sc | sk | |---|---|---|---|---|---|---|---| | 1 | 00:8 | 00:3 | 00:9 | 00:1 | 00:7 | 00:4 | 00:5 | | 2 | 01:5 | 10:6 | 00:2 | 10:8 | 10:7 | 11:1 | 10:9 | | 3 | 01:10 | 10:1 | 00:10 | 10:0 | 10:5 | 11:4 | 10:4 | | 4 | 01:9 | 10:2 | 01:7 | 11:9 | 00:6 | 01:4 | 01:2 | | 5 | 11:6 | 11:5 | 11:8 | 11:2 | 01:6 | 01:1 | 01:3 | | 6 | 01:8 | 10:3 | 01:0 | 11:3 | 10:10 | 11:7 | 11:10 |

Lemma 4. The $66$ copies satisfy the hypotheses of Theorem 2 with $\sigma=2$, $j=3$, $n=22$.

Proof. Decomposition. The $42$ entries of Table 2 are distinct and are all $42$ classes: the classes $00$ and $11$ each occur with the ten values $d\ne0$, and $01$ and $10$ each with all eleven values. A class $(ik{:}d)$ occurring in base copy $b$ is covered by the $11$ images $g^h(b)$, one arc each, so every arc of $K^*_{22}$ lies in exactly one copy. (The $66$ copies are distinct, because $g$ fixes no point and so no nontrivial power of $g$ fixes a copy.)

(R). Table 1 has six core points in each of the orbits $\{0\}\times\mathbb Z_{11}$ and $\{1\}\times\mathbb Z_{11}$, so by $g$-invariance every point lies in the core of exactly $6=j\sigma$ copies. Likewise each orbit contains three sinks and three sources, matching $r_i=r_o=3$.

(L). If $\mathrm{sink}_e=\mathrm{source}_f=x$ and $y\in\mathrm{core}_e\cap\mathrm{core}_f$, then $y\to x$ is a core $\to$ sink arc of $e$ and $x\to y$ is a source $\to$ core arc of $f$. So (L) holds unless some source $\to$ core class is the reverse of a core $\to$ sink class, where the reverse of $(ik{:}d)$ is $(ki{:}{-d})$. From Table 2, the source $\to$ core classes are $00{:}\{4,6,7\}$, $01{:}\{1,4,6\}$, $10{:}\{5,7,10\}$, $11{:}\{1,4,7\}$, and the reverses of the core $\to$ sink classes are $00{:}\{1,2,9,10\}$, $01{:}\{0,3\}$, $10{:}\{0,4\}$, $11{:}\{2,3,8,9\}$. These are disjoint. $\square$

Proof of Theorem 1. By Lemma 4 and Theorem 2 with $(\sigma,j)=(2,3)$, the copy digraph is a dsrg$(66,25,15,10,9)$ with spectrum $25^1\,3^{21}\,(-2)^{44}$. Its complement $J-I-A$ is a dsrg$(66,40,30,23,26)$. $\square$

5. Properties of the graph

The following were computed from the explicit matrix (supplement): $\lvert\mathrm{Aut}(\Gamma)\rvert=11$ (pynauty [12]), generated by the map induced by $g$, with six vertex orbits of size $11$, one for each base copy. Every vertex has $15$ reciprocal neighbours (the matrix $A\circ A^{\top}$ of undirected edges has all row sums $15$). $\Gamma$ is not isomorphic to its converse $\Gamma^{\top}$, so by the remark after Theorem 2 there are at least two non-isomorphic dsrg$(66,25,15,10,9)$ of line type, and their complements give two dsrg$(66,40,30,23,26)$.

More graphs (computed). A CP-SAT enumeration of all decompositions invariant under $g$ (variables: the $g$-orbits of copies; one worker) produced $712$ decompositions in $3600$ seconds and then stopped at its time limit, so it is not complete. All $712$ pass the checks of Lemma 4 and Theorem 2. Their copy digraphs fall into $10$ isomorphism classes, five converse pairs, each with $\lvert\mathrm{Aut}\rvert=11$ and six orbits of size $11$; the graph of Table 1 is one of them. The classes were separated twice: by pynauty certificates, and by a pure-Python invariant (the sorted list over vertices of the numbers of directed $3$-cycles and of transitive triangles in each of the three positions), which takes $10$ distinct values. So there are at least ten pairwise non-isomorphic dsrg$(66,25,15,10,9)$; the class count after $681$ decompositions was already $10$.

The base copies were found by a CP-SAT model [13] whose variables are the $g$-orbits of copies and whose constraints are exact cover of the $42$ arc classes, the orbit counts for (R), and (L) for pairs of orbits. The solver returns a solution within seconds (5.3 s in a rerun on one core). The proof above does not depend on the solver.

6. The family

Theorem 2 gives the following rows with $v\le110$. Their status in [2] is in the last column.

| $(\sigma,j)$ | $n$ | parameters | status in [2] | |---|---|---|---| | $(1,2)$ | 7 | $(14,7,4,3,4)$ | known (complement $(14,6,3,2,3)$: T5, T12, T18, M6) | | $(1,3)$ | 10 | $(30,11,5,4,4)$ | known (T11, T18) | | $(1,4)$ | 13 | $(52,15,6,5,4)$ | known (T8, T9, T18) | | $(1,5)$ | 16 | $(80,19,7,6,4)$ | known (T8, T9, T11, T18, T19) | | $(2,2)$ | 15 | $(30,16,11,8,9)$ | known (complement $(30,13,8,5,6)$: M9) | | $(2,3)$ | 22 | $(66,25,15,10,9)$ | "?"; Theorem 1 | | $(3,2)$ | 27 | $(54,29,22,15,16)$ | "?" (complement $(54,24,17,10,11)$); this pair is constructed in rStation paper 38 [14], line $(N,(N-6)/2,(N+14)/4,(N-14)/4,(N-10)/4)$ at $N=54$ | | $(4,2)$ | 43 | $(86,46,37,24,25)$ | "?" (complement $(86,39,30,17,18)$) |

We have not found decompositions satisfying (R) and (L) for $(3,2)$ or $(4,2)$. By Proposition 3, for $(4,2)$ a node automorphism of prime order $p\ge5$ must have $p=43$ and act as a $43$-cycle, and for $(3,2)$ it must have $p\mid 27$, which is impossible, so only $p\in\{2,3\}$ can occur there.

Line type is not forced by the parameters. Every graph from Theorem 2 has $A+\sigma I=XY$ with $0/1$ matrices $X$ ($v\times n$, row sums $\sigma+1$, column sums $j(\sigma+1)$) and $Y$ ($n\times v$, the same sums). For the dsrg$(54,29,22,15,16)$ of [14] (file square25_complement54.txt of its supplement), no such factorisation exists. This is a computed result: linerec_cnf.py encodes the factorisation as CNF (with the labels of one vertex fixed, which loses no generality), and CaDiCaL returns UNSAT with an LRAT proof that lrat-check verifies. The same encoder returns SAT, with decoded copies that pass the exact checks, on the two graphs of this paper and on a dsrg$(30,16,11,8,9)$ of line type. It returns a verified UNSAT on a $66\times66$ matrix that differs from ours by one $2\times2$ switch. A separate CP-SAT model agrees. So, for $(3,2)$, graphs with the parameters exist that do not come from Theorem 2. For $(4,2)$, failing to find a decomposition would not bear on the existence of $(86,46,37,24,25)$.

7. Verification

The supplement contains the six base copies, the $66$ copies, both adjacency matrices and the following scripts.

• build66.py (numpy, integer arithmetic) expands Table 1, checks the decomposition of $K^*_{22}$, (R), (L), $YX=5I+J$, that $XY-2I$ is a $0/1$ matrix with zero diagonal, the identity $A^2=15I+10A+9(J-I-A)$ with row and column sums $25$, the complement identity for $(66,40,30,23,26)$ and the spectrum, and writes the matrices.
• verify_dsrg_pure.py (pure Python, no numpy, written separately) reads a stored matrix and counts all directed two-paths. It passes for both matrices.
• difftable.py prints Table 2 from Table 1 and checks the three conditions of Lemma 4 at the level of classes.
• aut66.py (pynauty) computes $\lvert\mathrm{Aut}\rvert$, the orbits and the converse test.
• classes66.py (pure Python) expands the ten class representatives in reps66_z11.json, checks each by counting $2$-paths, and shows the ten triangle-profile invariants are distinct. The full list of $712$ decompositions is e22_z11.jsonl.
• linerec_cnf.py (pysat, CaDiCaL 1.9.5, lrat-check) tests whether a given matrix is of line type, as in Section 6. On A_66_25_15_10_9.txt it returns SAT and checks the decoded copies exactly.

From a clean unzip, python3 build66.py prints ALL PASS and python3 verify_dsrg_pure.py A_66_25_15_10_9.txt 66 25 15 10 9 prints PASS; the regenerated matrices are byte-identical to the shipped ones. All scripts exit with status 1 on a failed check, and flipping one entry of the matrix makes verify_dsrg_pure.py fail (negative control). Software: Python 3.12.3, numpy 2.0.2, pynauty 2.8.8.1.

8. Limitations

• Theorem 2 is a sufficient condition. We do not know for which $(\sigma,j)$ with $\sigma\ge2$ decompositions satisfying (R) and (L) exist beyond $(2,2)$ and $(2,3)$, and we claim no infinite family.
• The automorphism group order, the orbit structure and the converse test in Section 5 are computed (pynauty), not proved by hand. Theorem 1 does not depend on them.
• We do not claim to know all dsrg$(66,25,15,10,9)$, or all line-type ones. The $\mathbb Z_{11}$ enumeration of Section 5 stopped at its time limit; ten classes is a lower bound.

References

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