Counting Girth Cycles in the Graphs D(4, q)

Abstract

A girth cycle refers to a cycle that has the minimum length within a graph. The graphs $D(k,q)$ form an important family of algebraically defined bipartite graphs over finite fields, and their short-cycle structure is closely related to questions in extremal graph theory and finite geometry. Motivated by the problem of determining the edge-girth-regular parameter of $D(k,q)$, we determine the exact number of girth cycles in $D(4,q)$ for every prime power q. Our proof uses a convenient isomorphic model $\Gamma (4,q)$ and edge-transitivity to reduce the global enumeration to counting the girth cycles containing one fixed edge. Specifically, we prove that when q is odd with $q>3$, the number of girth cycles in $D(4,q)$ is $q^5{(q-1)}^2(q-3)/8$. Moreover, when q is even, the number of girth cycles in $D(4,q)$ is $q^5{(q-1)}^2(2q-3)/8$. When $q=3$, the girth is 12 and the number of girth cycles is 729. Together, these results resolve the case $k=4$ of the problem posed in our earlier work.

1. Introduction

Let $G=\left(V\right(G),E(G\left)\right)$ be a finite simple undirected graph. We write $u\sim v$ when u and v are adjacent. The girth $g\left(G\right)$ is the length of a shortest cycle in G, and a cycle of length $g\left(G\right)$ is called a girth cycle. A graph is edge-transitive if its automorphism group acts transitively on its edge set. Standard graph-theoretic terminology can be found in [1,2].

The graphs $D(k,q)$ were introduced by Lazebnik and Ustimenko in [3], following their earlier algebraic constructions of graphs with a large girth [4]. Since then, their girth, connectivity, symmetry, and spectra have been studied extensively; see, for example, Refs. [3,5,6,7,8,9,10,11,12] and the recent survey conducted in [13] (shown in Section 6).

We recall the definition of $D(k,q)$ here because the coordinate recurrences are used throughout the paper. Let $k\ge 2$ and let q be a prime power. The two partite sets of $D(k,q)$ are copies of ${\mathbb{F}}_q^k$. We write points as $\left(p\right)=(p_1,p_2,\dots ,p_k)$ and lines as $\left[\mathsf{\ell }\right]=[{\mathsf{\ell }}_1,{\mathsf{\ell }}_2,\dots ,{\mathsf{\ell }}_k]$. A point $\left(p\right)$ and a line $\left[\mathsf{\ell }\right]$ are adjacent when

$$p_2+{\mathsf{\ell }}_2=p_1{\mathsf{\ell }}_1,p_3+{\mathsf{\ell }}_3=p_1{\mathsf{\ell }}_2,\mathrm{and}{p}_i+{\mathsf{\ell }}_i=\left\{\begin{array}{cc}{p}_1{\mathsf{\ell }}_{i-2},\hfill & \mathrm{if}i\equiv 2,3\left(\mathrm{mod}4\right),\hfill \\ p_{i-2}{\mathsf{\ell }}_1,\hfill & \mathrm{if}i\equiv 0,1\left(\mathrm{mod}4\right).\hfill \end{array}\right.$$

The survey conducted by Lazebnik and Wang first recalls an earlier sign convention for $D(k,q)$ and then, as shown in Section 6.1, passes via Proposition 38 to the equivalent coordinate representation used above (see Section 6.1, Proposition 38 from [13]). We use this equivalent representation throughout and keep its definition in the Introduction because the recurrences are used explicitly in the proof.

Girth cycles encode finer local information than the girth alone. Their enumeration is relevant to extremal questions and to local regularity conditions on graphs; see, for example, Refs. [11,14,15,16]. In particular, Solymosi and Wong [16] obtain optimal asymptotic lower bounds for the number of even cycles of any fixed length in sufficiently large graphs of prescribed girth, while Gerbner, Győri, Methuku and Vizer [14] place related cycle-counting questions in the generalized Turán framework. Xu, Cheng and Tang [11] determine the girth cycles of several small members of the family $D(k,q)$. Thus, once the girth of a structured graph family is understood, it is natural to ask how many shortest cycles occur and how uniformly they are distributed over the edges.

We use the following formulation, which appears as Conjecture 46 in the published survey of Lazebnik and Wang (Conjecture 46 from [13]). As explained immediately before that conjecture, the statement was given in [17] for prime powers $q\ge 5$, while the survey extends the stated range to include $q=4$.

Conjecture 1.

For all prime powers $q\ge 4$, the graph $D(k,q)$ has girth $k+5$ when k is odd and $k+4$ when k is even.

The conjecture remains open in general; a summary of the known cases and recent progress is given in Section 6.3 (from [13]).

An appropriate framework for counting shortest cycles is edge-girth-regularity. Following Jajcay, Kiss and Miklavič [15], an edge-girth-regular graph with parameters $(v,c,g,\lambda )$, denoted by $egr(v,c,g,\lambda )$, is a c-regular graph on v vertices with girth g such that every edge lies in exactly $\lambda$ girth cycles. Every finite regular edge-transitive graph of finite girth is edge-girth-regular, although the converse need not hold. The parameter $\lambda$ measures a local multiplicity that is invisible from the usual parameters v, c, and g. For example, the complete graph $K_t$ is an $egr(t,t-1,3,t-2)$-graph, while the Heawood graph is an $egr(14,3,6,8)$-graph [15]. Edge-girth-regularity is therefore a natural refinement when one wants to distinguish graphs with the same order, degree, and girth but different distributions of shortest cycles. Jajcay et al. [15] also showed that the class is broader than the edge-transitive setting, so determining $\lambda$ is of independent structural interest rather than merely a reformulation of symmetry.

Subsequent work has developed both the extremal and constructive aspects of edge-girth-regularity. Araujo-Pardo and Leemans [18] constructed further infinite families from incidence geometry and Suzuki groups; in particular, they obtained an extremal family with parameters $egr(2q^2,q,6,{(q-1)}^2(q-2))$ for prime powers $q\ge 3$. More recently, Goedgebeur and Jooken [19] developed exhaustive-generation methods for edge-girth-regular graphs. These results illustrate that the parameter $\lambda$ is useful not only for describing highly symmetric examples but also in extremal and computational questions. The family $D(4,q)$ considered here has a different parameter regime, with degree growing with q and a girth of 8, except for $q=3$, where the girth is 12. An explicit determination of $\lambda$ for this family therefore supplies concrete algebraic examples to the broader edge-girth-regular framework.

A closely related local regularity notion has recently been studied systematically by Jajcay, Jooken and Porupsánszki [20]. A c-regular graph of order v and girth g is called vertex-girth-regular, denoted by $vgr(v,c,g,\mu )$, if every vertex lies on exactly $\mu$ girth cycles. Every edge-girth-regular graph is vertex-girth-regular: indeed, if G is an $egr(v,c,g,\lambda )$-graph, then each of the c edges incident with a fixed vertex lies on $\lambda$ girth cycles and each such cycle uses exactly two incident edges, so G is a $vgr(v,c,g,c\lambda /2)$-graph (see Observation 6 from [20]). This recent vertex-based framework will allow us to interpret our fixed-edge count from a second local viewpoint in Section 4.

There is also an intermediate local notion, namely girth-regularity, introduced by Potočnik and Vidali [21]. For a vertex u, one records the number of girth cycles containing the individual edges incident with u and orders these numbers into a signature; a graph is girth-regular when the same signature occurs at every vertex. Stability phenomena for girth-regular graphs of even girth were subsequently studied by Kiss, Miklavič and Szőnyi [22]. An edge-girth-regular graph has the constant signature $(\lambda ,\dots ,\lambda )$ and is therefore automatically girth-regular as well as vertex-girth-regular. Thus, the fixed-edge parameter computed in this paper simultaneously determines three levels of local information for $D(4,q)$: the number of shortest cycles through an edge, the resulting signature at a vertex, and the total number of shortest cycles through a vertex. This observation is particularly natural here because edge-transitivity makes the edge parameter independent of the chosen edge.

In our previous work (Problem 5.4 from [23]), we asked for the edge-girth-regular parameter of both the Lie graph $L(M_3,6,q)$ and the graphs $D(k,q)$ for $k\ge 4$. The part of that problem concerning $D(k,q)$ is as follows:

Problem 1.

Determine the edge-girth-regular parameter λ of $D(k,q)$ for $k\ge 4$.

The present paper resolves Problem 1 for $k=4$. For $q\ne 3$, we determine the precise multiplicity of shortest cycles through each edge and then the global count. The formulas reveal a genuine characteristic-dependent phenomenon: the local number of 8-cycles through an edge is ${(q-1)}^2(q-3)$ in odd characteristic but ${(q-1)}^2(2q-3)$ in characteristic two. The proof is constructive. We introduce an isomorphic model $\Gamma (4,q)$ whose adjacency equations make an 8-cycle through a fixed edge amenable to explicit parametrization. The resulting three equations are then solved completely, with all non-backtracking conditions incorporated into the count. The remaining case $q=3$, whose girth is 12, is treated separately in Section 4.

This paper is organized as follows: Section 2 illustrates the defining recurrences, records the symmetry and counting facts used later, and proves $D(4,q)\cong \Gamma (4,q)$. Section 3 derives the fixed-edge parametrization and proves the main counting theorem in detail. Section 4 records consequences for edge- and vertex-girth-regularity, treats the exceptional case $q=3$, and gives small-parameter computational checks. Section 5 summarizes the results and indicates natural directions for further work. The 12-cycle types for $D(4,3)$ were also studied by Xu, Cheng and Tang [11]; in Section 4, we give a direct fixed-edge count adapted to the present edge-girth-regular setting.

2. Preliminaries and an Isomorphic Model of $\mathit{D}(\mathbf{4},\mathit{q})$

We begin with a concrete illustration of the recurrences defined in the introduction. Recall that the graph $D(4,q)$ is a bipartite graph defined over the finite field ${\mathbb{F}}_q$. The adjacency relationship between a point $\left(p\right)$ and a line $\left[\mathsf{\ell }\right]$ is as follows:

$$p_2+{\mathsf{\ell }}_2=p_1{\mathsf{\ell }}_1,p_3+{\mathsf{\ell }}_3=p_1{\mathsf{\ell }}_2,p_4+{\mathsf{\ell }}_4=p_2{\mathsf{\ell }}_1.$$

This example explains why the successive first coordinates are the natural parameters for the cycle computation.

Example 1.

The defining equations can be read recursively. Fix a point $(p_1,p_2,p_3,p_4)$ of $D(4,q)$ and prescribe the first coordinate ${\mathsf{\ell }}_1=s$ of a neighboring line. The remaining coordinates are then uniquely determined by ${\mathsf{\ell }}_2=p_{1}s-p_2$, ${\mathsf{\ell }}_3=p_1{\mathsf{\ell }}_2-p_3$, and ${\mathsf{\ell }}_4=p_{2}s-p_4$. For instance, in $D(4,5)$, if $\left(p\right)=(1,2,3,4)$ and $s=2$, then ${\mathsf{\ell }}_2=0$, ${\mathsf{\ell }}_3=2$, and ${\mathsf{\ell }}_4=0$ in ${\mathbb{F}}_5$, so $(1,2,3,4)\sim [2,0,2,0]$. Since s can be chosen arbitrarily in ${\mathbb{F}}_q$, every point has exactly q neighbors. The same argument starting from a line shows that every line also has degree q. Hence, $D(4,q)$ is q-regular.

The preceding example also explains why the first coordinate is the natural parameter in later walks. If a point $\left(p\right)$ and the first coordinate s of a neighboring line are fixed, there is no remaining choice: the other three line coordinates are determined in order. Conversely, if a line $\left[\mathsf{\ell }\right]$ and the first coordinate r of a neighboring point are fixed, then $p_2=r{\mathsf{\ell }}_1-{\mathsf{\ell }}_2$, $p_3=r{\mathsf{\ell }}_2-{\mathsf{\ell }}_3$, and $p_4=p_2{\mathsf{\ell }}_1-{\mathsf{\ell }}_4$. Thus, a walk can be specified by recording only the successive first coordinates, while all other coordinates are recovered recursively. This observation is elementary, but it is the mechanism behind the six-parameter description in Section 3. Thus, $D(4,q)$ has $2q^4$ vertices and $q^5$ edges.

The symmetry recorded next is the reason why the later cycle enumeration can be carried out locally. Once edge-transitivity is known, the number of girth cycles through an edge is constant on the whole edge set. Therefore, one may choose the edge whose coordinates are most convenient, compute the corresponding local multiplicity once, and recover the total number of girth cycles by double counting. In the present paper we choose the zero edge $\left(0\right)\sim \left[0\right]$. The isomorphism with $\Gamma (4,q)$ is useful precisely because the triangular equations in that model allow all remaining coordinates along a walk through this fixed edge to be recovered from first-coordinate data. The same local-to-global principle is used twice: for 8-cycles when $q\ne 3$ and for 12-cycles when $q=3$. In this way, the algebraic parametrization and edge-transitivity play complementary roles rather than independent ones.

An important observation is that every finite regular edge-transitive graph of finite girth is edge-girth-regular. Since $D(k,q)$ is q-regular, the following lemma, together with this observation, demonstrates the edge-girth-regularity of $D(k,q)$.

Lemma 1

(Theorem 3.2 from [3]). The graph $D(k,q)$ is edge-transitive for all k and q.

Let $\mathcal{C}(\Gamma )$ denote the number of girth cycles of $\Gamma$. The subsequent lemma shows that $\mathcal{C}(\Gamma )$ can be obtained using the parameters associated with the edge-girth-regularity of $\Gamma$.

Lemma 2

(Proposition 2.3 (iii) from [15]). Let Γ be an edge-girth-regular graph with parameters $(v,c,g,\lambda )$. Then $\mathcal{C}(\Gamma )={\displaystyle \frac{vc\lambda }{2g}}$.

For completeness, let $\mathcal{I}=\left\{\right(e,C):e\in E(\Gamma ),C\mathrm{is}\mathrm{a}\mathrm{girth}\mathrm{cycle}\mathrm{containing}e\}$. Counting first by edges gives $\left|\mathcal{I}\right|=\left|E\right(\Gamma \left)\right|\lambda =vc\lambda /2$, whereas counting first by girth cycles gives $\left|\mathcal{I}\right|=g\mathcal{C}(\Gamma )$, because every girth cycle contains exactly g edges. Equating the two expressions gives the displayed formula. This simple double count is the final step that turns our fixed-edge enumeration into the global count.

For the counting argument it is convenient to replace $D(4,q)$ by a model whose last two adjacency equations depend only on the first coordinates. Let $\Gamma (4,q)$ be the bipartite graph with the same two partite sets ${\mathbb{F}}_q^4$, where $\left(p\right)$ and $\left[\mathsf{\ell }\right]$ are adjacent if

$$p_2+{\mathsf{\ell }}_2=p_1{\mathsf{\ell }}_1,p_3+{\mathsf{\ell }}_3=p_1{\mathsf{\ell }}_1^2,p_4+{\mathsf{\ell }}_4=p_1^2{\mathsf{\ell }}_1.$$

(1)

It should be noted that the graph $D(4,q)$ exhibits q-regularity (every vertex has exactly q neighbors). The reason is that for any vertex $a\in V\left(D\right(4,q\left)\right)$ and $x\in {\mathbb{F}}_q$, there exists a unique vertex b such that a is adjacent to b and the first component of b is x. Similarly, $\Gamma (4,q)$ is also a q-regular graph.

Next, we recall the definition of graph isomorphism between two graphs. Let $G_1$ and $G_2$ be two graphs. A graph isomorphism from $G_1$ to $G_2$ is a bijective map $\psi :V\left(G_1\right)\to V\left(G_2\right)$ such that for any two vertices $u,v\in V\left(G_1\right)$, $u\sim v\in E\left(G_1\right)$ if and only if $\psi \left(u\right)\sim \psi \left(v\right)\in E\left(G_2\right)$. We say that $G_1$ and $G_2$ are isomorphic, denoted as $G_1\cong G_2$, if there exists an isomorphism between them.

Lemma 3.

For every prime power q, $D(4,q)\cong \Gamma (4,q)$.

Proof. 

We define the assignments $\sigma :V\left(D\right(4,q\left)\right)\to V(\Gamma (4,q\left)\right)$ as follows:

$$\sigma \left((p_1,p_2,p_3,p_4)\right):=(p_1,p_2,p_4,p_3+p_1{p}_2),\mathrm{and}\sigma \left([{\mathsf{\ell }}_1,{\mathsf{\ell }}_2,{\mathsf{\ell }}_3,{\mathsf{\ell }}_4]\right):=[{\mathsf{\ell }}_1,{\mathsf{\ell }}_2,{\mathsf{\ell }}_4+{\mathsf{\ell }}_1{\mathsf{\ell }}_2,{\mathsf{\ell }}_3].$$

To prove that $\sigma$ is a bijection, it suffices to show that $\sigma$ is an injection since $V\left(D\right(4,q\left)\right)$ is finite. Let

$$\left(p\right)=(p_1,p_2,p_3,p_4),\left(p^{\ast }\right)=(p_1^{\ast },p_2^{\ast },p_3^{\ast },p_4^{\ast }),\left[\mathsf{\ell }\right]=[{\mathsf{\ell }}_1,{\mathsf{\ell }}_2,{\mathsf{\ell }}_3,{\mathsf{\ell }}_4],\mathrm{and}\left[{\mathsf{\ell }}^{\ast }\right]=[{\mathsf{\ell }}_1^{\ast },{\mathsf{\ell }}_2^{\ast },{\mathsf{\ell }}_3^{\ast },{\mathsf{\ell }}_4^{\ast }].$$

If $\sigma \left(\left(p\right)\right)=\sigma \left(\left(p^{\ast }\right)\right)$ and $\sigma \left(\left[\mathsf{\ell }\right]\right)=\sigma \left(\left[{\mathsf{\ell }}^{\ast }\right]\right)$, then $\left(p\right)=\left(p^{\ast }\right)$ and $\left[\mathsf{\ell }\right]=\left[{\mathsf{\ell }}^{\ast }\right]$ by a straightforward check. Therefore, $\sigma$ is an injection. More explicitly, the inverse maps are

$$(a_1,a_2,a_3,a_4)⟼(a_1,a_2,a_4-a_1{a}_2,a_3)$$

on points and

$$[b_1,b_2,b_3,b_4]⟼[b_1,b_2,b_4,b_3-b_1{b}_2]$$

on lines.

Next, we will verify that $\sigma$ preserves the adjacency relation between any two vertices, that is, $\sigma \left(\right(p\left)\right)\sim \sigma \left(\right[\mathsf{\ell }\left]\right)\in E(\Gamma (4,q\left)\right)$ if $\left(p\right)\sim \left[\mathsf{\ell }\right]\in E\left(D\right(4,q\left)\right)$. Let $\sigma \left(\left(p\right)\right)=\left(p^{\ast }\right)$ and $\sigma \left(\left[\mathsf{\ell }\right]\right)=\left[{\mathsf{\ell }}^{\ast }\right]$. The following equations show the detailed verification process:

$$\begin{array}{cc}\hfill p_2^{\ast }+{\mathsf{\ell }}_2^{\ast }& =p_2+{\mathsf{\ell }}_2=p_1{\mathsf{\ell }}_1=p_1^{\ast }{\mathsf{\ell }}_1^{\ast },\hfill \\ \hfill p_3^{\ast }+{\mathsf{\ell }}_3^{\ast }& =p_4+{\mathsf{\ell }}_4+{\mathsf{\ell }}_1{\mathsf{\ell }}_2=p_2{\mathsf{\ell }}_1+{\mathsf{\ell }}_1{\mathsf{\ell }}_2=(p_2+{\mathsf{\ell }}_2){\mathsf{\ell }}_1=p_1{\left({\mathsf{\ell }}_1\right)}^2=p_1^{\ast }{\left({\mathsf{\ell }}_1^{\ast }\right)}^2,\hfill \\ \hfill p_4^{\ast }+{\mathsf{\ell }}_4^{\ast }& =p_3+{\mathsf{\ell }}_3+p_1{p}_2=p_1{\mathsf{\ell }}_2+p_1{p}_2=p_1({\mathsf{\ell }}_2+p_2)={\left(p_1\right)}^2{\mathsf{\ell }}_1={\left(p_1^{\ast }\right)}^2{\mathsf{\ell }}_1^{\ast }.\hfill \end{array}$$

The above computation shows that

$$\sigma \left(E\left(D\right(4,q\left)\right)\right)\subseteq E(\Gamma (4,q)).$$

Since $\sigma$ is a bijection and both graphs are q-regular on $2q^4$ vertices, each graph has exactly $q^5$ edges. Hence, the above inclusion is an equality. Therefore, for every point $\left(p\right)$ and line $\left[\mathsf{\ell }\right]$,

$$\left(p\right)\sim \left[\mathsf{\ell }\right]\mathrm{in}D(4,q)⟺\sigma \left(\right(p\left)\right)\sim \sigma \left(\right[\mathsf{\ell }\left]\right)\mathrm{in}\Gamma (4,q),$$

and $\sigma$ is a graph isomorphism. □

The advantage of $\Gamma (4,q)$ is visible in (1): once the first coordinates of a point and a neighboring line are known, each of the remaining coordinates is obtained immediately. This triangular form will allow us to write every 8-cycle through a fixed edge explicitly and to isolate the only nonautomatic adjacency condition.

3. Determining the Exact Value of $\mathcal{C}\left(\mathit{D}\right(\mathbf{4},\mathit{q}\left)\right)$

It should be noted that the girth of the graph $D(4,q)$ is 8 for $q>3$ (see Theorem 2 (e) in [11]); the case $q=2$ is recorded in Section 6.3 (from [13]). Moreover, the girth of $D(4,3)$ is 12 (see Table 4 in [13]). Therefore, for $q\ne 3$, given that the order, degree, and girth of $D(4,q)$ have been established, it remains to determine the number of girth cycles containing a specific edge. The theorem below states the complete answer for all prime powers q; the $q\ne 3$ cases are proved in this section, while the case $q=3$ is established separately in Proposition 2 of Section 4.

Main Theorem 1.

Let $\mathcal{C}\left(D\right(4,q\left)\right)$ denote the number of girth cycles in $D(4,q)$. Then

$$\mathcal{C}\left(D(4,q)\right)=\left\{\begin{array}{cc}{\displaystyle \frac{q^5{(q-1)}^2(q-3)}{8}},\hfill & q>3\mathit{odd},\hfill \\ 729,\hfill & q=3,\hfill \\ {\displaystyle \frac{q^5{(q-1)}^2(2q-3)}{8}},\hfill & q\mathit{even}.\hfill \end{array}\right.$$

Proof. 

By Lemma 3, we may work in $\Gamma (4,q)$. Fix the edge $(0,0,0,0)\sim [0,0,0,0]$, abbreviated to $\left(0\right)\sim \left[0\right]$.

We first explain in detail how an 8-cycle through this edge is parametrized. This is the point in the proof where the special form of (1) is used most strongly. Start from the oriented edge $\left(0\right)\to \left[0\right]$. Let x be the first coordinate of the next point $P_1$. Because $P_1\sim \left[0\right]$, the three adjacency equations give its remaining coordinates as zero, so $P_1=(x,0,0,0)$. Next, let y be the first coordinate of the line $L_1$ following $P_1$. From $P_1\sim L_1$ we successively obtain ${\mathsf{\ell }}_2=xy$, ${\mathsf{\ell }}_3=x{y}^2$, and ${\mathsf{\ell }}_4=x^{2}y$; hence, $L_1=[y,xy,x{y}^2,x^{2}y]$.

Now prescribe the first coordinate $x_1$ of the next point $P_2$. Since $P_2\sim L_1$, Equation (1) gives $p_2=x_{1}y-xy=(x_1-x)y$, $p_3=x_1{y}^2-x{y}^2=(x_1-x)y^2$, and $p_4=x_1^{2}y-x^{2}y=(x_1^2-x^2)y$. Thus, $P_2=(x_1,(x_1-x)y,(x_1-x)y^2,(x_1^2-x^2)y)$. If $y_1$ is the first coordinate of the next line $L_2$, the same three equations force its other coordinates to be $x_1{y}_1-(x_1-x)y$, $x_1{y}_1^2-(x_1-x)y^2$, and $x_1^2{y}_1-(x_1^2-x^2)y$, respectively.

It is useful to build the last two vertices from the opposite end of the fixed edge. A line $L_3$ adjacent to $\left(0\right)$ and having first coordinate t must satisfy ${\mathsf{\ell }}_2={\mathsf{\ell }}_3={\mathsf{\ell }}_4=0$, so $L_3=[t,0,0,0]$. If $P_3$ is the preceding point and its first coordinate is $x_2$, then $P_3\sim L_3$ forces $P_3=(x_2,x_{2}t,x_2{t}^2,x_2^{2}t)$. After these choices, every required adjacency in the closed walk has already been imposed except $P_3\sim L_2$. Consequently, the entire candidate walk is determined by the six first-coordinate parameters $(x,y,x_1,y_1,x_2,t)$.

$$\begin{array}{cc}\hfill W_8:& (0,0,0,0)\sim [0,0,0,0]\sim (x,0,0,0)\sim [y,xy,x{y}^2,x^{2}y]\hfill \\ \hfill & \sim (x_1,(x_1-x)y,(x_1-x)y^2,(x_1^2-x^2)y)\hfill \\ \hfill & \sim [y_1,x_1{y}_1-(x_1-x)y,x_1{y}_1^2-(x_1-x)y^2,x_1^2{y}_1-(x_1^2-x^2)y]\hfill \\ \hfill & \sim (x_2,x_{2}t,x_2{t}^2,x_2^{2}t)\sim [t,0,0,0]\sim (0,0,0,0).\hfill \end{array}$$

All adjacencies displayed above are automatic from the construction except $P_3\sim L_2$. We spell out this last condition coordinate by coordinate. The second-coordinate equation is $x_{2}t+x_1{y}_1-(x_1-x)y=x_2{y}_1$, which is equivalent to (2). The third-coordinate equation is $x_2{t}^2+x_1{y}_1^2-(x_1-x)y^2=x_2{y}_1^2$; after moving terms and using $y_1^2-t^2=(y_1-t)(y_1+t)$, this is (3). Finally, the fourth-coordinate equation is $x_2^{2}t+x_1^2{y}_1-(x_1^2-x^2)y=x_2^2{y}_1$, which is exactly (4). Thus, the following three equations are neither auxiliary assumptions nor an ad hoc system: they are precisely the three coordinates of the single missing adjacency.

$$\begin{array}{cc}\hfill x_2(y_1-t)-x_1{y}_1+(x_1-x)y& =0,\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill x_2(y_1-t)(y_1+t)-x_1{y}_1^2+(x_1-x)y^2& =0,\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill x_2^2(y_1-t)-x_1^2{y}_1+(x_1^2-x^2)y& =0.\hfill \end{array}$$

(4)

Claim. 

A tuple $(x,y,x_1,y_1,x_2,t)$ satisfying (2)–(4) determines an 8-cycle through $\left(0\right)\sim \left[0\right]$ if and only if

$$x\ne 0,x\ne x_1,x_1\ne x_2,x_2\ne 0,\mathrm{and}y\ne 0,y\ne y_1,y_1\ne t,t\ne 0.$$

Moreover, after the initial oriented edge $\left(0\right)\to \left[0\right]$ is fixed, different admissible tuples determine different 8-cycles.

Proof of the Claim. 

Each inequality rules out one possible immediate reversal. For example, $x=0$ would make the third vertex equal to the initial point, $y=0$ would make the next line equal to $\left[0\right]$, $x_1=x$ would make $P_2=P_1$, and $y_1=y$ would make $L_2=L_1$. Continuing around the walk gives the remaining four inequalities $x_1\ne x_2$, $y_1\ne t$, $x_2\ne 0$, and $t\ne 0$. Hence, the displayed conditions are necessary and sufficient for the closed walk to have no immediate backtracking.

It remains to exclude a repetition at nonconsecutive positions. Suppose that such a repetition occurs and choose two equal vertices whose cyclic separation along $W_8$ is minimal. The vertices strictly between these two occurrences are then pairwise distinct. A separation of two would be an immediate reversal and has already been excluded, while a separation of one would be a loop, which is impossible because the graph is simple and bipartite. Therefore, the segment between the two occurrences is a cycle with a length of at least four and strictly less than eight. This contradicts $g(\Gamma (4,q\left)\right)=8$. Hence, a closed walk satisfying the eight inequalities has eight distinct vertices and is an 8-cycle. Conversely, every 8-cycle clearly satisfies the inequalities. Finally, an undirected 8-cycle containing the fixed edge has exactly one traversal beginning with the oriented edge $\left(0\right)\to \left[0\right]$. Reading the successive first coordinates from this traversal recovers $(x,y,x_1,y_1,x_2,t)$ uniquely, so different admissible tuples cannot represent the same cycle. □

Moreover, after fixing the oriented initial edge $\left(0\right)\to \left[0\right]$, every 8-cycle containing this edge determines a unique tuple $(x,y,x_1,y_1,x_2,t)$, and every admissible tuple satisfying Equations (2)–(4) determines one such cycle. Thus, counting admissible tuples counts the desired cycles without any additional factor.

Suppose that q is a power of an odd prime and $q\ne 3$. Let ${\mathcal{C}}^{\prime }\left(e\right)$ denote the number of 8-cycles containing an edge e. We now determine ${\mathcal{C}}^{\prime }(\left(0\right)\sim \left[0\right])$ by considering the four possibilities according to whether $x_1$ and $y_1$ are zero.

Case A. Suppose that $x_1=0$ and $y_1=0$. Equation (2) becomes $x_{2}t=-xy$, while (3) becomes $x_2{t}^2=-x{y}^2$. Because $x,y,x_2,t$ are nonzero by the claim above, division of the second equality by the first gives $t=y$, and then the first equality gives $x_2=-x$. With $x_1=y_1=0$, Equation (4) is

$$-x_2^{2}t-x^{2}y=0.$$

Substituting $t=y$ and $x_2=-x$ yields $-x^{2}y-x^{2}y=-2x^{2}y=0$, or equivalently $2x^{2}y=0$. Since the characteristic is odd and $x,y\ne 0$, this is impossible.

Case B. Suppose that $x_1=0$ and $y_1\ne 0$. Equations (2) and (3) give

$$t=y-y_1,x_2(2y_1-y)=xy.$$

Equation (4) gives $x_2^2(2y_1-y)=x^{2}y$. Since $x,y,x_2\ne 0$, comparison with the preceding equality yields $x_2=x$, and then $y_1=y$, contradicting $y\ne y_1$.

Case C. Suppose that $x_1\ne 0$ and $y_1=0$. Equations (2) and (3) read $x_{2}t=(x_1-x)y$ and $x_2{t}^2=(x_1-x)y^2$. Since $x_1\ne x$ and $x_2,t,y\ne 0$, division of the second equality by the first gives $t=y$, and the first equality then gives $x_2=x_1-x$. With $y_1=0$, Equation (4) becomes

$$-x_2^{2}t+(x_1^2-x^2)y=0.$$

After substituting $t=y$ and $x_2=x_1-x$, its left-hand side is

$$-{(x_1-x)}^{2}y+(x_1^2-x^2)y=2x(x_1-x)y.$$

Thus, $2x(x_1-x)y=0$. Here, $x\ne 0$, $y\ne 0$, and $x_1\ne x$ by the claim, so this is impossible in odd characteristic.

Case D. Suppose that $x_1\ne 0$ and $y_1\ne 0$. Put

$$A=x_1{y}_1-(x_1-x)y.$$

By Equation (2),

$$A=x_2(y_1-t).$$

The cycle conditions give $x_2\ne 0$ and $y_1\ne t$, so $A\ne 0$. Using this identity in Equation (3), we obtain

$$A(y_1+t)=x_1{y}_1^2-(x_1-x)y^2.$$

Since $A{y}_1=x_1{y}_1^2-(x_1-x)y{y}_1$, subtraction yields

$$At=(x_1-x)y(y_1-y),t={\displaystyle \frac{(x_1-x)y(y_1-y)}{A}}.$$

Similarly, Equation (4) and $A=x_2(y_1-t)$ give

$$x_{2}A=x_1^2{y}_1-(x_1^2-x^2)y,x_2={\displaystyle \frac{x_1^2{y}_1-(x_1^2-x^2)y}{A}}.$$

Substituting these two expressions back into $A=x_2(y_1-t)$ and clearing the nonzero denominator A gives

$$A^3=\left(x_1^2{y}_1-(x_1^2-x^2)y\right)\left(A{y}_1-(x_1-x)y(y_1-y)\right).$$

Expanding the two sides and collecting factors yields the polynomial identity

$$xy(x-x_1)(y-y_1)\left(x_1{y}_1-2(x_1-x)y\right)=0.$$

(5)

Because $x,y\ne 0$, $x\ne x_1$, and $y\ne y_1$, we obtain

$$x_1{y}_1=2(x_1-x)y,y_1={\displaystyle \frac{2(x_1-x)y}{x_1}}.$$

Consequently,

$$A=(x_1-x)y,x_2=x_1-x,t=y_1-y={\displaystyle \frac{x_1-2x}{x_1}}y.$$

Thus, once x, y, and $x_1$ are chosen, the remaining parameters are uniquely determined. It remains only to impose the cycle conditions. They are equivalent to

$$x\ne 0,y\ne 0,x_1\notin \{0,x,2x\}.$$

Indeed, $x_2\ne 0$ is equivalent to $x_1\ne x$; $x_1\ne x_2$ follows from $x\ne 0$ and $y_1-t=y\ne 0$; and both $t\ne 0$ and $y_1\ne y$ are equivalent to $x_1\ne 2x$. Since the characteristic is odd and $x\ne 0$, the elements $0,x,2x$ are distinct.

It follows that x, y, and $x_1$ have, respectively, $q-1$, $q-1$, and $q-3$ admissible choices. For each such choice, the values of $y_1$, $x_2$, and t are uniquely determined. Therefore,

$${\mathcal{C}}^{\prime }(\left(0\right)\sim \left[0\right])={(q-1)}^2(q-3).$$

Thus, in odd characteristic, there are exactly ${(q-1)}^2(q-3)$ 8-cycles containing the edge $\left(0\right)\sim \left[0\right]$.

Next, suppose that $q=2^e$ and $e\ge 1$. We again distinguish four cases according to whether $x_1$ and $y_1$ are zero. The equations are the same, but the relation $2=0$ changes which cases are admissible.

Case i. Suppose that $x_1=0$ and $y_1=0$. The first two equations, Equations (2) and (3), become $x_{2}t=xy$ and $x_2{t}^2=x{y}^2$ because minus and plus coincide in characteristic two. All four variables are nonzero in a cycle, so division gives $t=y$ and then $x_2=x$. Equation (4) becomes $x^{2}y+x^{2}y=0$ and is automatic. The eight inequalities in the claim above now reduce simply to $x\ne 0$ and $y\ne 0$, because $x_1=y_1=0$, $x_2=x$, and $t=y$. Conversely, for every pair $x,y\in {\mathbb{F}}_q^{\ast }$, the tuple $(x,y,0,0,x,y)$ satisfies (2)–(4) and all eight inequalities. Therefore, this case contributes exactly ${(q-1)}^2$ cycles.

Case ii. Suppose that $x_1=0$ and $y_1\ne 0$. Equations (2) and (3) become $x_2(y_1+t)=xy$ and $x_2{(y_1+t)}^2=x{y}^2$. The first equality shows $y_1+t\ne 0$, so division gives $y_1+t=y$; hence, $t=y+y_1$, and the first equality then gives $x_2=x$. Equation (4) is automatically satisfied. The cycle conditions require $x\ne 0$, $y\ne 0$, $y_1\ne 0$, and $y_1\ne y$; indeed, $t\ne 0$ is exactly $y_1\ne y$, while $y_1\ne t$ follows from $y\ne 0$. Conversely, choose arbitrary $x,y,y_1\in {\mathbb{F}}_q^{\ast }$ with $y_1\ne y$, and put $x_2=x$ and $t=y+y_1$. Direct substitution gives $x_2(y_1+t)=xy$, $x_2{(y_1+t)}^2=x{y}^2$, and (4) = 0, and the stated restrictions give all eight inequalities. Thus, x and y have $q-1$ choices each and, after y is fixed, $y_1$ has $q-2$ choices. The contribution is exactly ${(q-1)}^2(q-2)$.

Case iii. Suppose that $x_1\ne 0$ and $y_1=0$. The first two equations become $x_{2}t=(x_1+x)y$ and $x_2{t}^2=(x_1+x)y^2$. Because an admissible cycle has $x_1\ne x$, the factor $x_1+x$ is nonzero. Division therefore gives $t=y$ and then $x_2=x_1+x$. Substituting these values into (4) gives zero, so the third equation is automatic. The cycle conditions reduce to $x\ne 0$, $y\ne 0$, $x_1\ne 0$, and $x_1\ne x$: the condition $x_2\ne 0$ is precisely $x_1\ne x$, while $x_1\ne x_2$ follows from $x\ne 0$. Conversely, every choice $x,y,x_1\in {\mathbb{F}}_q^{\ast }$ with $x_1\ne x$, together with $x_2=x_1+x$ and $t=y$, satisfies all three equations and all eight inequalities. Hence, the contribution is exactly ${(q-1)}^2(q-2)$.

Case iv. Suppose that $x_1\ne 0$ and $y_1\ne 0$. For an admissible cycle we again have $A=x_1{y}_1-(x_1-x)y=x_2(y_1-t)\ne 0$. Every step used to derive (5) consists only of field operations and division by the nonzero element A; in particular, that derivation did not divide by 2. Hence, (5) is valid in characteristic two as well. Since $2=0$, its last factor is simply $x_1{y}_1$, and we obtain

$$xy(x-x_1)(y-y_1)x_1{y}_1=0.$$

Every factor is nonzero by the assumptions of Case iv and by the claim, a contradiction. Therefore, Case iv contains no admissible 8-cycle.

Combining Cases (i)–(iv), when q is an even number, we conclude that ${\mathcal{C}}^{\prime }(\left(0\right)\sim \left[0\right])$ is equal to

$${(q-1)}^2+2{(q-1)}^2(q-2)={(q-1)}^2(2q-3).$$

Finally, we count all 8-cycles in $D(4,q)$. The graph $D(4,q)$ is q-regular, has $2q^4$ vertices, and is edge-transitive by Lemma 1; therefore it is edge-girth-regular. Applying Lemma 2 with $g=8$ and with the above values of $\lambda ={\mathcal{C}}^{\prime }(\left(0\right)\sim \left[0\right])$, we obtain

$$\mathcal{C}\left(D(4,q)\right)=\left\{\begin{array}{cc}{\displaystyle \frac{q^5{(q-1)}^2(q-3)}{8}},\hfill & q>3\mathrm{odd},\hfill \\ {\displaystyle \frac{q^5{(q-1)}^2(2q-3)}{8}},\hfill & q\mathrm{even}.\hfill \end{array}\right.$$

The remaining case $q=3$ is proven in Proposition 2 of Section 4. This completes the proof. □

Remark 1.

The case $q=3$ in Main Theorem 1 is exceptional because its girth is 12, not 8 (Table 4 from [13]). Hence, the factor $q-3$ in the odd-characteristic 8-cycle formula should not be interpreted as saying that $D(4,3)$ has no girth cycles. The fixed-edge parametrization itself explains why the odd formula vanishes at $q=3$: for every nonzero $x\in {\mathbb{F}}_3$, the three elements 0, x, and $2x$ exhaust the whole field, so there is no admissible choice of $x_1$ in Case D. Thus there are no 8-cycles, consistently with the known girth of 12. We count the actual 12-cycles separately in Section 4.

4. Consequences and Applications

The fixed-edge count has immediate consequences for edge- and vertex-girth-regularity. We also record the fixed-edge parametrization obtained in the proof and give computational checks for several small finite fields.

It is useful to distinguish the three counting parameters that appear below. The quantity $\lambda$ is local to an edge, the quantity $\mu$ is local to a vertex, and $\mathcal{C}\left(D\right(4,q\left)\right)$ is global. For a c-regular edge-girth-regular graph, these quantities are not independent: every girth cycle through a fixed vertex uses exactly two incident edges, so $\mu =c\lambda /2$, while counting edge–cycle incidences gives $\mathcal{C}=vc\lambda /\left(2g\right)$. Consequently, once the fixed-edge calculation is complete, both the vertex parameter and the global number follow without a second cycle enumeration. This is why the main computational burden of the paper is concentrated on determining $\lambda$ for a single edge. The corollaries below make these consequences explicit, and Proposition 2 shows that exactly the same strategy continues to work in the exceptional field ${\mathbb{F}}_3$, although the relevant girth changes from 8 to 12.

Corollary 1.

For $q\ne 3$, the graph $D(4,q)$ is edge-girth-regular with parameters

$$(2q^4,q,8,{\lambda }_q),{\lambda }_q=\left\{\begin{array}{cc}{(q-1)}^2(q-3),\hfill & q>3\mathit{odd},\hfill \\ {(q-1)}^2(2q-3),\hfill & q\mathit{even}.\hfill \end{array}\right.$$

Proof. 

This is exactly the fixed-edge count in the proof of Main Theorem 1, together with the edge-transitivity of $D(4,q)$. □

The next consequence places $D(4,q)$ in the vertex-girth-regular framework of Jajcay, Jooken and Porupsánszki [20].

Corollary 2.

For every prime power $q\ne 3$, the graph $D(4,q)$ is vertex-girth-regular. More precisely,

$$D(4,q)\mathit{is}vgr(2q^4,q,8,{\mu }_q),$$

where

$${\mu }_q={\displaystyle \frac{q{\lambda }_q}{2}}=\left\{\begin{array}{cc}{\displaystyle \frac{q{(q-1)}^2(q-3)}{2}},\hfill & q>3\mathit{odd},\hfill \\ {\displaystyle \frac{q{(q-1)}^2(2q-3)}{2}},\hfill & q\mathit{even}.\hfill \end{array}\right.$$

Proof. 

Observation 6 of [20] states that every $egr(v,c,g,\lambda )$-graph is a $vgr(v,c,g,c\lambda /2)$-graph. Applying it to Corollary 1 with $c=q$ gives the stated value of ${\mu }_q$. This relation can also be seen directly: the q edges incident with a fixed vertex contribute $q{\lambda }_q$ edge–cycle incidences, while every girth cycle through that vertex uses exactly two of its incident edges. □

The earlier description in [11] and the present fixed-edge parametrization address the same shortest-cycle structure from different viewpoints. Here, we only need the latter in the form produced by the proof above: after fixing $\left(0\right)\sim \left[0\right]$, the admissible tuples can be listed explicitly, and their cardinalities are exactly the edge-girth-regular parameters. Recording these tuples is also useful for checking that the case analysis has neither omitted nor duplicated an admissible 8-cycle. No additional structural assertion about $D(4,q)$ is needed for the applications below. The following proposition records the fixed-edge parametrization that arises from the proof of Main Theorem 1 and is used here to calculate ${\lambda }_q$.

Proposition 1.

The proof of Main Theorem 1 yields the following complete fixed-edge parametrization. If $q>3$ is odd, every 8-cycle through $\left(0\right)\sim \left[0\right]$ in $\Gamma (4,q)$ corresponds uniquely to

$$\left(x,y,u,{\displaystyle \frac{2(u-x)y}{u}},u-x,{\displaystyle \frac{(u-2x)y}{u}}\right),$$

where $x,y\in {\mathbb{F}}_q^{\ast }$ and $u\in {\mathbb{F}}_q\setminus \{0,x,2x\}$. If q is even, every such cycle belongs to exactly one of the three families

$$\begin{array}{cccc}\hfill \mathrm{I}:& (x,y,0,0,x,y),\hfill & & x,y\in {\mathbb{F}}_q^{\ast },\hfill \\ \hfill \mathrm{II}:& (x,y,0,v,x,y+v),\hfill & & x,y,v\in {\mathbb{F}}_q^{\ast },v\ne y,\hfill \\ \hfill \mathrm{III}:& (x,y,u,0,u+x,y),\hfill & & x,y,u\in {\mathbb{F}}_q^{\ast },u\ne x.\hfill \end{array}$$

There are no additional admissible tuples.

Proof. 

In odd characteristic, Cases A–C in the proof of Main Theorem 1 have no solutions, while Case D gives $y_1=2(x_1-x)y/x_1$, $x_2=x_1-x$, and $t=(x_1-2x)y/x_1$, with $x,y\ne 0$ and $x_1\notin \{0,x,2x\}$. Renaming $x_1$ as u gives the first family. In characteristic two, Cases i, ii, and iii give Families I, II, and III, respectively, while Case iv has no solutions. The restrictions displayed above are precisely the non-backtracking conditions established in the claim inside the proof of Main Theorem 1. □

For odd $q>3$, the family in Proposition 1 has ${(q-1)}^2(q-3)$ members. In even characteristic, Families I–III have ${(q-1)}^2$, ${(q-1)}^2(q-2)$, and ${(q-1)}^2(q-2)$ members, respectively, and hence, their total number is ${(q-1)}^2(2q-3)$. The even-characteristic statement also includes $q=2$.

We now treat the exceptional field ${\mathbb{F}}_3$ directly. The case $q=3$ requires a separate count only because the girth changes from 8 to 12; the underlying method is the same as in the proof of Main Theorem 1. We again pass to the isomorphic graph $\Gamma (4,3)$, fix the edge $\left(0\right)\sim \left[0\right]$, and recover all remaining coordinates recursively from successive first coordinates. For an 8-cycle, this led to six first-coordinate parameters. For a 12-cycle, it is more convenient to record the successive differences of the first coordinates on the two partite sets. Since every nonzero element of ${\mathbb{F}}_3$ is 1 or $-1$, the non-backtracking condition forces each of these differences to be a sign. The three remaining closing conditions are then obtained by telescoping exactly the same triangular adjacency relations used in Section 3.

Thus, the argument below is not a separate computational device: it is the fixed-edge parametrization of the main proof adapted to a longer girth. The small field has an additional advantage, namely that the possible point-increment sequences fall into only four symmetry classes. Counting the compatible line-increment sequences in these four classes gives the number of 12-cycles through the fixed edge; edge-transitivity and the same double-counting lemma then give the global number of girth cycles. For context, the 12-cycle types of $D(4,3)$ were also considered by Xu, Cheng and Tang [11]; the proof below is self-contained and is formulated directly for the edge-girth-regular parameter needed here.

The increment notation also separates the two requirements that were intertwined in the six-parameter calculation for 8-cycles. Non-backtracking is expressed simply by $u_i,v_i\ne 0$, while closure of the first coordinates is given by $\sum u_i=\sum v_i=0$. The other three coordinates are handled by telescoping the adjacency equations. This makes the $q=3$ calculation short enough to be carried out by hand and shows explicitly why the same local-to-global strategy continues to work when the girth increases from 8 to 12.

Proposition 2.

The graph $D(4,3)$ is an $egr(162,3,12,36)$-graph and a $vgr(162,3,12,54)$-graph. Moreover, $\mathcal{C}\left(D\right(4,3\left)\right)=729$.

Proof. 

Following the same fixed-edge method used in the proof of Main Theorem 1, by Lemma 3 it is enough to work in $\Gamma (4,3)$. Fix the oriented edge $P_0=\left(0\right)\to L_0=\left[0\right]$ and consider a backtrackless closed walk of length 12 through this edge,

$$P_0,L_0,P_1,L_1,\dots ,P_5,L_5,P_6=P_0.$$

Let $p_i$ and ${\mathsf{\ell }}_i$ denote the first coordinates of $P_i$ and $L_i$, respectively, and put $p_0=p_6={\mathsf{\ell }}_0={\mathsf{\ell }}_6=0$. Define

$$u_i=p_i-p_{i-1},v_i={\mathsf{\ell }}_i-{\mathsf{\ell }}_{i-1}(1\le i\le 6).$$

Since the walk has no immediate backtracking, $u_i,v_i\in {\mathbb{F}}_3^{\ast }=\{1,-1\}$ for every i. Conversely, these nonzero differences are exactly the first-coordinate non-backtracking conditions. The closing conditions on the first coordinates are ${\sum }_{i=1}^6{u}_i=0$ and ${\sum }_{i=1}^6{v}_i=0$.

We next record the remaining three closing conditions. The points $P_{i-1}$ and $P_i$ have the common neighbor $L_{i-1}$. Subtracting the three adjacency equations in (1) for these two points gives

$$p_{i,2}-p_{i-1,2}=u_i{\mathsf{\ell }}_{i-1},p_{i,3}-p_{i-1,3}=u_i{\mathsf{\ell }}_{i-1}^2,p_{i,4}-p_{i-1,4}=(p_i^2-p_{i-1}^2){\mathsf{\ell }}_{i-1}.$$

Summing from $i=1$ to 6, the condition $P_6=P_0$ is therefore equivalent to

$$\begin{array}{cc}\hfill {\displaystyle \sum _{i=1}^6}{u}_i{\mathsf{\ell }}_{i-1}& =0,\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill {\displaystyle \sum _{i=1}^6}{u}_i{\mathsf{\ell }}_{i-1}^2& =0,\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill {\displaystyle \sum _{i=1}^6}(p_i^2-p_{i-1}^2){\mathsf{\ell }}_{i-1}& =0.\hfill \end{array}$$

(8)

Conversely, the triangular adjacency relations of $\Gamma (4,3)$ uniquely recover the successive vertices from the successive first coordinates. Hence, any pair of sign sequences satisfying the two first-coordinate closing conditions and Equations (6)–(8) reconstructs a unique backtrackless closed walk of length 12 through the fixed oriented edge. Thus, the problem is reduced to counting such pairs of sign sequences.

Because each $u_i$ is 1 or $-1$ and $\sum u_i=0$ in ${\mathbb{F}}_3$, the number of entries equal to 1 is 0, 3, or 6. Hence, there are $2+(\genfrac{}{}{0pt}{}{6}{3})=22$ possible u-sequences. Cyclic reindexing preserves the system (6)–(8). Indeed, denote the three left-hand sides by

$$A={\displaystyle \sum _{i=1}^6}{u}_i{\mathsf{\ell }}_{i-1},B={\displaystyle \sum _{i=1}^6}{u}_i{\mathsf{\ell }}_{i-1}^2,C={\displaystyle \sum _{i=1}^6}(p_i^2-p_{i-1}^2){\mathsf{\ell }}_{i-1}.$$

Fix $r\in \{0,\dots ,5\}$ and cyclically shift the increments by r, so that

$$u_i^{\prime }=u_{r+i},v_i^{\prime }=v_{r+i},$$

where the indices are taken modulo 6. Since ${\sum }_{i=1}^6{u}_i={\sum }_{i=1}^6{v}_i=0$, we may extend the partial sums periodically, so that $p_{i+6}=p_i$ and ${\mathsf{\ell }}_{i+6}={\mathsf{\ell }}_i$. By definition of the new partial sums,

$$p_i^{\prime }={\displaystyle \sum _{j=1}^i}{u}_j^{\prime }={\displaystyle \sum _{j=1}^i}{u}_{r+j}=p_{r+i}-p_r,$$

and similarly,

$${\mathsf{\ell }}_i^{\prime }={\displaystyle \sum _{j=1}^i}{v}_j^{\prime }={\displaystyle \sum _{j=1}^i}{v}_{r+j}={\mathsf{\ell }}_{r+i}-{\mathsf{\ell }}_r.$$

Thus, the cyclic shift amounts to taking the old rth position as the new origin. Hence,

$$A^{\prime }={\displaystyle \sum _{i=1}^6}{u}_i^{\prime }{\mathsf{\ell }}_{i-1}^{\prime }={\displaystyle \sum _{i=1}^6}{u}_{r+i}\left({\mathsf{\ell }}_{r+i-1}-{\mathsf{\ell }}_r\right)={\displaystyle \sum _{i=1}^6}{u}_i({\mathsf{\ell }}_{i-1}-{\mathsf{\ell }}_r)=A,$$

because ${\sum }_i{u}_i=0$. Similarly,

$$B^{\prime }={\displaystyle \sum _{i=1}^6}{u}_i^{\prime }{\left({\mathsf{\ell }}_{i-1}^{\prime }\right)}^2={\displaystyle \sum _{i=1}^6}{u}_i{({\mathsf{\ell }}_{i-1}-{\mathsf{\ell }}_r)}^2=B-2{\mathsf{\ell }}_{r}A.$$

For the third expression, since ${(p_i-p_r)}^2-{(p_{i-1}-p_r)}^2=(p_i^2-p_{i-1}^2)-2p_r{u}_i$, we obtain

$$C^{\prime }={\displaystyle \sum _{i=1}^6}\left((p_i^2-p_{i-1}^2)-2p_r{u}_i\right)({\mathsf{\ell }}_{i-1}-{\mathsf{\ell }}_r)=C-2p_{r}A,$$

where we have used

$${\displaystyle \sum _{i=1}^6}(p_i^2-p_{i-1}^2)=p_6^2-p_0^2=0\mathrm{and}{\displaystyle \sum _{i=1}^6}{u}_i=0.$$

Therefore, a cyclic shift sends $(A,B,C)⟼\left(A,B-2{\mathsf{\ell }}_{r}A,C-2p_{r}A\right)$, and hence preserves the condition $A=B=C=0$. Moreover, replacing every $u_i$ with $-u_i$ replaces $p_i$ with $-p_i$ and sends $(A,B,C)⟼(-A,-B,C)$. Hence, it also preserves the number of admissible v-sequences. Consequently, the 22 possible u-sequences fall into the following four classes under cyclic shifts and global sign reversal, with representatives as follows:

$$\begin{array}{ccc}\mathrm{class}& \mathrm{representative}& \mathrm{number}\mathrm{of}u\text{-}\mathrm{sequences}\\ \mathrm{I}& ++++++& 2\\ \mathrm{II}& +++---& 6\\ \mathrm{III}& +-+-+-& 2\\ \mathrm{IV}& ++-+--& 12.\end{array}$$

Here, + and − denote 1 and $-1$, respectively. We count the admissible v-sequences for one representative of each class. Write ${\mathsf{\ell }}_i=v_1+\cdots +v_i$; then, ${\mathsf{\ell }}_0={\mathsf{\ell }}_6=0$, ${\mathsf{\ell }}_1,{\mathsf{\ell }}_5\ne 0$, and consecutive ${\mathsf{\ell }}_i$ are distinct.

For Class I, Equations (6)–(8) become

$${\mathsf{\ell }}_1+{\mathsf{\ell }}_2+{\mathsf{\ell }}_3+{\mathsf{\ell }}_4+{\mathsf{\ell }}_5=0,{\mathsf{\ell }}_1^2+{\mathsf{\ell }}_2^2+{\mathsf{\ell }}_3^2+{\mathsf{\ell }}_4^2+{\mathsf{\ell }}_5^2=0,-{\mathsf{\ell }}_2+{\mathsf{\ell }}_3-{\mathsf{\ell }}_5=0.$$

In ${\mathbb{F}}_3$, a nonzero square equals 1. Since ${\mathsf{\ell }}_1$ and ${\mathsf{\ell }}_5$ are nonzero, the middle equation implies that exactly three of ${\mathsf{\ell }}_1,\dots ,{\mathsf{\ell }}_5$ are nonzero. The two zero entries therefore have to be ${\mathsf{\ell }}_2={\mathsf{\ell }}_4=0$. The other two equations then give ${\mathsf{\ell }}_1={\mathsf{\ell }}_3={\mathsf{\ell }}_5=a$ with $a\in {\mathbb{F}}_3^{\ast }$. Hence,

$$(v_1,\dots ,v_6)=(a,-a,a,-a,a,-a),$$

so Class I has exactly two admissible v-sequences.

For Class II, the three equations are

$${\mathsf{\ell }}_1+{\mathsf{\ell }}_2-{\mathsf{\ell }}_3-{\mathsf{\ell }}_4-{\mathsf{\ell }}_5=0,{\mathsf{\ell }}_1^2+{\mathsf{\ell }}_2^2-{\mathsf{\ell }}_3^2-{\mathsf{\ell }}_4^2-{\mathsf{\ell }}_5^2=0,-{\mathsf{\ell }}_2+{\mathsf{\ell }}_3-{\mathsf{\ell }}_5=0.$$

Because ${\mathsf{\ell }}_1\ne 0$ and ${\mathsf{\ell }}_2\ne {\mathsf{\ell }}_1$, either ${\mathsf{\ell }}_2=0$ or ${\mathsf{\ell }}_2=-{\mathsf{\ell }}_1$. If ${\mathsf{\ell }}_2=0$, the last equation gives ${\mathsf{\ell }}_3={\mathsf{\ell }}_5$, and the first gives ${\mathsf{\ell }}_4={\mathsf{\ell }}_1+{\mathsf{\ell }}_5$. Substitution into the square equation would give ${\mathsf{\ell }}_4^2=2$, impossible in ${\mathbb{F}}_3$. Thus, ${\mathsf{\ell }}_2=-{\mathsf{\ell }}_1$. Put $a={\mathsf{\ell }}_1$ and $b={\mathsf{\ell }}_5$, where $a,b\in {\mathbb{F}}_3^{\ast }$. The first and third equations then give

$${\mathsf{\ell }}_3=b-a,{\mathsf{\ell }}_4=a+b.$$

The square equation is automatic, since

$${(b-a)}^2+{(a+b)}^2=2(a^2+b^2)=1$$

in ${\mathbb{F}}_3$. All four choices of $(a,b)$ satisfy the non-backtracking conditions. Hence, Class II has exactly 4 admissible v-sequences.

For Class III, Equation (6) is

$$-{\mathsf{\ell }}_1+{\mathsf{\ell }}_2-{\mathsf{\ell }}_3+{\mathsf{\ell }}_4-{\mathsf{\ell }}_5=0.$$

In terms of the increments this is $v_1+v_3+v_5=0$. Three elements of $\{1,-1\}$ sum to zero in ${\mathbb{F}}_3$ if and only if they are equal, so $v_1=v_3=v_5=a$. Since $\sum v_i=0$, the same argument gives $v_2=v_4=v_6=b$, where $a,b\in {\mathbb{F}}_3^{\ast }$. Thus

$$(v_1,\dots ,v_6)=(a,b,a,b,a,b).$$

Equation (8) coincides with (6) for this u-sequence. Moreover, ${\mathsf{\ell }}_1=a$, ${\mathsf{\ell }}_2=a+b$, ${\mathsf{\ell }}_3=b-a$, ${\mathsf{\ell }}_4=-(a+b)$, and ${\mathsf{\ell }}_5=-b$, so direct substitution into (7) gives zero. Therefore, all four choices of $(a,b)$ are admissible, and Class III contributes four v-sequences.

For Class IV, the partial sums of the representative $u=(+,+,-,+,-,-)$ are

$$(p_0,p_1,\dots ,p_6)=(0,1,-1,1,-1,1,0).$$

Hence, Equation (8) reduces to $-{\mathsf{\ell }}_5=0$. This is impossible because $v_6={\mathsf{\ell }}_6-{\mathsf{\ell }}_5=-{\mathsf{\ell }}_5$ must be nonzero. Thus, Class IV contributes no solutions.

It follows that the number of backtrackless closed walks of length 12 through the fixed oriented edge is

$$2·2+6·4+2·4+12·0=36.$$

Since $g(\Gamma (4,3\left)\right)=12$, such a backtrackless closed walk cannot contain a repeated nonconsecutive vertex; otherwise, it would contain a cycle of length smaller than 12. Hence, these 36 walks are precisely the 36 girth cycles containing the fixed edge. Fixing the orientation $\left(0\right)\to \left[0\right]$ introduces no extra factor, because each undirected cycle through this edge has a unique traversal beginning with that oriented edge.

Therefore, every edge of $D(4,3)$ lies in 36 girth cycles, so $D(4,3)$ is $egr(162,3,12,36)$. The corresponding vertex parameter is $\frac{3·36}{2}=54$, and Lemma 2 gives

$$\mathcal{C}\left(D(4,3)\right)={\displaystyle \frac{162·3·36}{2·12}}=729.$$

This completes the proof. □

For computational verification, we note that the algorithmic framework of Goedgebeur and Jooken (Section 2 from [19]) computes the number of girth cycles through a prescribed edge. Since $D(4,q)$ is edge-transitive, a computation for one edge is sufficient.

Table 1. Small-parameter computational verification of the closed formulas.

q	Enumerated ${\mathit{\lambda }}_{\mathit{q}}$	Formula ${\mathit{\lambda }}_{\mathit{q}}$	${\mathit{\mu }}_{\mathit{q}}=\mathit{q}{\mathit{\lambda }}_{\mathit{q}}/2$	$\mathcal{C}\left(\mathit{D}\right(4,\mathit{q}\left)\right)$
2	1	1	1	4
3	36	36	54	729
4	45	45	90	5760
5	32	32	80	12,500
7	144	144	504	302,526
8	637	637	2548	2,609,152
9	384	384	1728	2,834,352

lists small-parameter computations and compares them with the values proven above. The cases $q=4$, 8, and 9 also provide checks over non-prime finite fields.

These computations are included only as an independent verification of the closed formulas; none of the proofs depends on the numerical output. In particular, the $q=3$ entry checks the separate 12-cycle calculation in Proposition 2, while the entries for $q=4,8,9$ test the formulas in genuine extension fields rather than only over prime fields. This provides a useful consistency check on both the characteristic-two and odd-characteristic case distinctions.

5. Concluding Remarks

We have determined the exact number of girth cycles in $D(4,q)$ for every prime power q. For $q\ne 3$, the main step is a fixed-edge enumeration in the isomorphic graph $\Gamma (4,q)$: the triangular adjacency equations produce a six-parameter description of all candidate 8-cycles, and the remaining adjacency condition reduces to three polynomial equations. Solving these equations together with the non-backtracking conditions yields the edge-girth-regular parameter $\lambda ={(q-1)}^2(q-3)$ for odd $q>3$ and $\lambda ={(q-1)}^2(2q-3)$ for even q. Double counting then gives the total numbers stated in Main Theorem 1.

From the viewpoint of Problem 1, Corollary 1 together with Proposition 2 gives a complete solution for $k=4$. The results also place all graphs $D(4,q)$ in the vertex-girth-regular framework: for $q\ne 3$ the vertex parameter is $q{\lambda }_q/2$, while Proposition 2 gives the parameter 54 when $q=3$. The computational values in Table 1 agree with the closed formulas and provide independent checks for several small prime powers. For the exceptional graph $D(4,3)$, Proposition 2 gives the corresponding parameters and the exact number of girth cycles, 729. Natural next questions are to determine the corresponding parameters for $D(k,q)$ with $k\ge 5$ and to understand whether similarly explicit parametrizations can be organized efficiently as the girth increases.