Quasi-thin weakly distance-regular digraphs

Abstract

A weakly distance-regular digraph is quasi-thin if the maximum value of its intersection numbers is 2. In this paper, we focus on commutative quasi-thin weakly distance-regular digraphs, and classify such digraphs with valency more than 3. As a result, this family of digraphs is completely determined.

1 Introduction

Throughout this paper, $\Gamma$ always denotes a finite simple digraph. We write $V\Gamma$ and $A\Gamma$ for the vertex set and arc set of $\Gamma$, respectively. A path of length r from x to y is a sequence of vertices $(x=w_{0},w_{1},\ldots ,w_{r}=y)$ such that $(w_{t-1}, w_{t})\in A\Gamma$ for $t=1,2,\ldots ,r$. A digraph is said to be strongly connected if, for any two distinct vertices x and y, there is a path from x to y. The length of a shortest path from x to y is called the distance from x to y in $\Gamma$, denoted by $\partial _\Gamma (x,y)$. Let $\tilde{\partial }_\Gamma (x,y)=(\partial _\Gamma (x,y),\partial _\Gamma (y,x))$ and $\tilde{\partial }(\Gamma )=\{\tilde{\partial }_\Gamma (x,y)\mid x,y\in V\Gamma \}$. We call $\tilde{\partial }_\Gamma (x,y)$ the two-way distance from x to y in $\Gamma$. If no confusion occurs, we write $\partial (x,y)$ (resp. $\tilde{\partial }(x,y)$) instead of $\partial _\Gamma (x,y)$ (resp. $\tilde{\partial }_\Gamma (x,y)$). An arc (u, v) of $\Gamma$ is of type (1, r) if $\partial (v,u)=r$. A path $(w_{0},w_{1},\ldots ,w_{r-1})$ is said to be a circuit of length r if $\partial (w_{r-1},w_{0})=1$. A circuit is undirected if each of its arcs is of type (1, 1). Let $C_{r}$ denote the undirected circuit of length r.

A strongly connected digraph $\Gamma$ is said to be weakly distance-regular if, for any $\tilde{h}$, $\tilde{i}$, $\tilde{j}\in \tilde{\partial }(\Gamma )$, the cardinality of the set

$$\begin{aligned} P_{\tilde{i},\tilde{j}}(x,y):=\{z\in V\Gamma \mid \tilde{\partial }(x,z)= \tilde{i}~\text {and}~\tilde{\partial }(z,y)=\tilde{j}\} \end{aligned}$$

is constant whenever $\tilde{\partial }(x,y)=\tilde{h}$. This constant is denoted by $p_{\tilde{i},\tilde{j}}^{\tilde{h}}$. The integers $p_{\tilde{i},\tilde{j}}^{\tilde{h}}$ are called the intersection numbers. We say that $\Gamma$ is commutative if $p_{\tilde{i},\tilde{j}}^{\tilde{h}}=p_{\tilde{j},\tilde{i}}^{\tilde{h}}$ for all $\tilde{i}$, $\tilde{j}$, $\tilde{h}\in \tilde{\partial }(\Gamma )$. A weakly distance-regular digraph is quasi-thin (resp. thin) if the maximum value of its intersection numbers is 2 (resp. 1). The size of $\Gamma _{\tilde{i}}(x):=\{y\in V\Gamma \mid \tilde{\partial }(x,y)=\tilde{i}\}$ depends only on $\tilde{i}$, denoted by $k_{\tilde{i}}$. The integer $k:=\sum _{(1,j)\in \tilde{\partial }(\Gamma )}k_{1,j}$ is called the valency of $\Gamma$, which is often called the out-degree of $\Gamma$.

Some special families of weakly distance-regular digraphs were classified. See [7, 8] for valency 2, [9,10,11] for valency 3 and [7] for thin case. In this paper, we classify commutative quasi-thin weakly distance-regular digraphs of valency more than 3, and obtain the following main result.

Theorem 1.1

If $\Gamma$ is a commutative quasi-thin weakly distance-regular digraph of valency more than 3, then $\Gamma$ is isomorphic to one of the following Cayley digraphs:

• (i) $\mathrm{Cay}(\mathbb {Z}_{8},\{1,2,3,6\})$.

• (ii) $\mathrm{Cay}(\mathbb {Z}_{4p},\{1,2,2p+i,2p+1,2p+2\})$, $p\ne 2-i$.

• (iii) $\mathrm{Cay}(\mathbb {Z}_{4}\times \mathbb {Z}_{4},\{(0,1),(1,0),(2,0),(0,2)\})$.

• (iv) $\mathrm{Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{4},\{(0,1),(1,0),(1,2),(0,2+i)\})$, $q\ne 3+i$.

• (v) $\mathrm{Cay}(\mathbb {Z}_{2q}\times \mathbb {Z}_{2},\{(0,1),(1,0),(2,0),(1,1)\})$.

• (vi) $\mathrm{Cay}(\mathbb {Z}_{4q}\times \mathbb {Z}_{2},\{(0,1),(1,0),(2,0),(2q+1,0),(2q+2,0),(2qi,1)\})$, $q\notin \{3,3+i\}$.

• (vii) $\mathrm{Cay}(\mathbb {Z}_{2q}\times \mathbb {Z}_{4},\{(0,1),(1,0),(1,2),(0,2-i),(2,0),(2,2)\})$, $q\notin \{3,3+i\}$.

• (viii) $\mathrm{Cay}(\mathbb {Z}_{2q}\times \mathbb {Z}_{n},\{(0,1),(1,0),(2,0),(0,-1)\})$.

• (ix) $\mathrm{Cay}(\mathbb {Z}_{2q}\times \mathbb {Z}_{n},\{(0,1),(1,(c+1)/2),(1,(c-1)/2),(2,c),(0,-1)\})$.

• (x) $\mathrm{Cay}(\mathbb {Z}_{2n}\times \mathbb {Z}_{q},\{(0,1),(1,(t+1)/2),(-1,(1-t)/2),(2,t),(-2,-t)\})$.

Here, $i\in \{0,1\}$, $2\le p$, $3\le q$, $3\le n\le q-(1+(-1)^{q})/2$, $c=n/\mathrm{gcd}(q,n)$, $t=q/\mathrm{gcd}(q,n)$ and c, t are both odd.

Routinely, all digraphs in above theorem are commutative quasi-thin weakly distance-regular. For the last seven families of Cayley digraphs, in Table 1, we list the two-way distance from the identity element to any other element of the corresponding group.

Table 1 Two-way distance of digraphs in Theorem 1.1

In order to give a high-level description of our proof of Theorem 1.1, we need additional notations and terminologies. Let $\Gamma$ be a weakly distance-regular digraph and $R=\{\Gamma _{\tilde{i}}\mid \tilde{i}\in \tilde{\partial }(\Gamma )\}$, where $\Gamma _{\tilde{i}}=\{(x,y)\in V\Gamma \times V\Gamma \mid \tilde{\partial }(x,y)=\tilde{i}\}$. Then $(V\Gamma ,R)$ is an association scheme ([2, 12, 13]). Moreover, if $\Gamma$ is quasi-thin, then $(V\Gamma ,R)$ is quasi-thin. About this special scheme, see [4,5,6]. For two nonempty subsets E and F of R, define

$$\begin{aligned} EF:= & {} \left\{ \Gamma _{\tilde{h}}\Big |\sum _{\Gamma _{\tilde{i}}\in E}\sum _{\Gamma _{\tilde{j}}\in F}p_{\tilde{i},\tilde{j}}^{\tilde{h}}\ne 0\right\} , \end{aligned}$$

and write $\Gamma _{\tilde{i}}\Gamma _{\tilde{j}}$ instead of $\{\Gamma _{\tilde{i}}\}\{\Gamma _{\tilde{j}}\}$. For any $(a,b)\in \tilde{\partial }(\Gamma )$, we usually write $k_{a,b}$ (resp. $\Gamma _{a,b}$) instead of $k_{(a,b)}$ (resp. $\Gamma _{(a,b)}$). Now we list basic properties of intersection numbers which are used frequently in this paper.

Lemma 1.2

([2, Chapter II, Proposition 2.2] and [1, Proposition 5.1]) For each $\tilde{i}:=(a,b)\in \tilde{\partial }(\Gamma )$, define $\tilde{i}^{*}=(b,a)$. The following hold:

• (i) $k_{\tilde{d}}\;k_{\tilde{e}}=\sum _{\tilde{f}\in \tilde{\partial }(\Gamma )}p_{\tilde{d},\tilde{e}}^{\tilde{f}}\;k_{\tilde{f}}$.

• (ii) $p_{\tilde{d},\tilde{e}}^{\tilde{f}}\;k_{\tilde{f}}=p_{\tilde{f},\tilde{e}^{*}}^{\tilde{d}}\;k_{\tilde{d}}=p_{\tilde{d}^{*},\tilde{f}}^{\tilde{e}}\;k_{\tilde{e}}$.

• (iii) $|\Gamma _{\tilde{d}}\;\Gamma _{\tilde{e}}|\le \mathrm{gcd}(k_{\tilde{d}},k_{\tilde{e}})$.

• (iv) $\sum _{\tilde{e}\in \tilde{\partial }(\Gamma )}p_{\tilde{d},\tilde{e}}^{\tilde{f}}=k_{\tilde{d}}$.

• (v) $\mathrm{lcm}(k_{\tilde{d}},k_{\tilde{e}})\mid p_{\tilde{d},\tilde{e}}^{\tilde{f}}\;k_{\tilde{f}}$.

• (vi) $\sum _{\tilde{f}\in \tilde{\partial }(\Gamma )}p_{\tilde{d},\tilde{e}}^{\tilde{f}}\;p_{\tilde{g},\tilde{f}}^{\tilde{h}}=\sum _{\tilde{l}\in \tilde{\partial }(\Gamma )}p_{\tilde{g},\tilde{d}}^{\tilde{l}}\;p_{\tilde{l},\tilde{e}}^{\tilde{h}}$.

We now introduce the concepts about arcs. An arc of type $(1,q-1)$ is said to be pure, if every circuit of length q containing it consists of arcs of type $(1,q-1)$; otherwise, this arc is said to be mixed. We say that $(1,q-1)$ is pure if any arc of type $(1,q-1)$ is pure; otherwise, we say that $(1,q-1)$ is mixed. The concepts of pure arc and mixed arc are inspired by Suzuki in [7].

Another concept we need is a configuration. Let h and q be distinct integers more than 2. If $(\Gamma _{1,q-1})^{2}=\{\Gamma _{2,q-2}\}$ and $(\Gamma _{1,h-1})^{2}\subseteq \Gamma _{1,q-1}\Gamma _{q-1,1}$, we say that the configuration $C_{q,h}$ exists.

For fixed $x\in V\Gamma$, let $\Delta _{q_{1},q_{2},\ldots ,q_{l}}(x)$ be the connected component of digraph $(V\Gamma ,\cup _{i=1}^{l}\Gamma _{1,q_{i}-1})$ containing vertex x. Note that $\Delta _{q_{1},q_{2},\ldots ,q_{l}}(x)$ does not depend on the choice of vertex x up to isomorphism. If no confusion occurs, we write $\Delta _{q_{1},q_{2},\ldots ,q_{l}}$ instead of $\Delta _{q_{1},q_{2},\ldots ,q_{l}}(x)$.

Let $\Gamma$ be a commutative quasi-thin weakly distance-regular digraph of valency more than 3 in the remaining of this paper. We are now ready to give a high-level description of our proof of Theorem 1.1.

Outline of the proof of Theorem 1.1.

In Sect. 2, we give a characterization of mixed arcs, i.e., we show that $(1,q-1)$ is mixed if and only if $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$ and $(1,q-2)$ is pure.

In Sect. 3, we discuss the basic properties about the configuration $C_{q,h}$. In particular, we show that, if $C_{q,h}$ exists, then $(1,q-1)$ is pure, h is a constant and $h\in \{3,4\}$.

In Sect. 4, applying the results in Sects. 2 and 3, we prove the following result.

Proposition 1.3

Let $K=\{(1,r)\mid (1,r)\in \tilde{\partial }(\Gamma )\}$. Then one of the following holds:

• C1) $K=\{(1,1),(1,2),(1,q-1)\}$, where $C_{q,3}$ exists.

• C2) $K=\{(1,3),(1,q-1),(1,q)\}$, where $C_{q,4}$ exists and (1, q) is mixed.

• C3) $K=\{(1,1),(1,2),(1,q-1),(1,q)\}$, where $C_{q,3}$ exists and (1, q) is mixed.

• C4) $K=\{(1,1),(1,q-1)\}$, where $(1,q-1)$ is pure.

• C5) $K=\{(1,q-1),(1,q)\}$, where (1, q) is mixed.

• C6) $K=\{(1,1),(1,q-1),(1,q)\}$, where (1, q) is mixed.

In Sect. 5, we determine the subdigraphs $\Delta _{q,3}$ for the cases C1 and C3, the subdigraphs $\Delta _{q,4}$ for case C2, the subdigraphs $\Delta _{2,q}$ for cases C4 and C6, and the subdigraphs $\Delta _{q,q+1}$ for cases C5 and C6.

In Sect. 6, we give a proof of Theorem 1.1. For the cases C1, C2 and C3, we determine $\Gamma$ based on the subdigraphs $\Delta _{q,3}$ and $\Delta _{q,4}$. For the cases C4, C5 and C6, we determine $\Gamma$ based on the subdigraphs $\Delta _{2,q}$ and $\Delta _{q,q+1}$.

2 Characterization of mixed arcs

The main result of this section is the following important result which characterizes mixed arcs.

Theorem 2.1

Let $q\ge 3$ and $(1,q-1)\in \tilde{\partial }(\Gamma )$.

• (i) If $p_{(1,s-1),(1,t-1)}^{(1,q-1)}\ne 0$ with $s<t$, then $s=2$ and $t=q$.

• (ii) The following are equivalent:

  (a) $(1,q-1)$ is mixed; (b) $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$ and $(1,q-2)$ is pure; (c) $p_{(1,q-1),(1,q-1)}^{(1,s-1)}\ne 0$ for some s.

• (iii) If $p_{(1,q-1),(1,q-1)}^{(1,s-1)}\ne 0$, then $s=q-1$.

In the proof of Theorem 2.1, we use the following auxiliary lemmas.

Lemma 2.2

Suppose $\tilde{d},\tilde{h},\tilde{l}\in \tilde{\partial }(\Gamma )$ and $k_{\tilde{d}}=2$. The following hold:

• (i) $k_{\tilde{h}}=k_{\tilde{h}^{*}}\le 2$.

• (ii) $|\Gamma _{\tilde{h}}\Gamma _{\tilde{l}}|\le 2$ and equality holds only if $k_{\tilde{h}}=k_{\tilde{l}}=2$.

• (iii) $p_{\tilde{d},\tilde{d}}^{\tilde{e}}=2$ for some $\tilde{e}\in \tilde{\partial }(\Gamma )$.

• (iv) $\Gamma _{\tilde{d}}\Gamma _{\tilde{d}^{*}}=\{\Gamma _{0,0},\Gamma _{e,e}\}$. In particular, if $p_{\tilde{d},\tilde{d}^{*}}^{\tilde{e}}\ne 0$, then $\tilde{e}=\tilde{e}^{*}.$

Proof

Since $k_{\tilde{h}^{*}}=k_{\tilde{h}}=p_{\tilde{h},\tilde{h}^{*}}^{(0,0)}$ by Lemma 1.2 (ii), (i) is valid. (ii) follows from (i) and Lemma 1.2 (iii). By the commutativity of $\Gamma$, (iii) holds. In view of (ii) and Lemma 1.2 (i), (iv) is valid. $\square$

The commutativity of $\Gamma$ will be used frequently in the sequel, so we no longer refer to it for the sake of simplicity.

Lemma 2.3

If $(x_{0},x_{1},\ldots ,x_{n-1})$ is an undirected circuit in $\Gamma$, then $\partial (x_{0},x_{i})=\partial (x_{i},x_{0})=\partial (x_{0},x_{n-i})$ for $1\le i\le n-1$.

Proof

It is routine by induction. $\square$

Lemma 2.4

Let $q\ge 3$. Suppose that $(1,q-1)$ is pure and $k_{1,q-1}=2$. Then one of the following holds:

• (i) $p_{(1,q-1),(1,q-1)}^{(2,q-2)}=2$, $\Delta _{q}\simeq \mathrm{Cay}(\mathbb {Z}_{2q},\{1,q+1\})$, $(\Gamma _{1,q-1})^{i}=\{\Gamma _{i,q-i}\}$ for $2\le i\le q-1$.

• (ii) $p_{(1,q-1),(1,q-1)}^{(2,q-2)}=1$, $\Delta _{q}\simeq \mathrm{Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{q},\{(1,0),(0,1)\})$, $|(\Gamma _{1,q-1})^{2}|=2$.

Proof

Similar to the proofs of Lemma 12 in [10] and Proposition 4.3 in [8]. $\square$

Lemma 2.5

Let $q\ge 3$. Suppose that $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$ and $(1,q-2)$ is pure. Then the following hold:

• (i) $p_{(1,q-1),(1,q-2)}^{(2,q-2)}\ne 0$ and $p_{(1,q-1),(1,q-1)}^{(2,q-2)}=0$.

• (ii) Any circuit of length q containing an arc of type $(1,q-1)$ consists of arcs of types $(1,q-1)$ and $(1,q-2)$.

• (iii) If $|(\Gamma _{1,q-1})^{2}|=2$ and $k_{1,q-2}=1$, then $p_{(1,q-1),(1,q-1)}^{(2,q-1)}\ne 0$.

Proof

(i) Let $(z,z_{0})$ be an arc of type $(1,q-1)$. By $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$ and Lemma 1.2 (ii), there exists a vertex $z_{q-2}\in P_{(q-1,1),(1,q-2)}(z,z_{0})$. Since $(1,q-2)$ is pure, we assume that $(z_{0},z_{1},\ldots ,z_{q-2})$ is a circuit consisting of arcs of the same type. Hence, $\tilde{\partial }(z,z_{1})=(2,q-2)$. The fact that $\tilde{\partial }(z,z_{0})=(1,q-1)$ and $\tilde{\partial }(z_{0},z_{1})=(1,q-2)$ imply $p_{(1,q-1),(1,q-2)}^{(2,q-2)}\ne 0$.

Suppose $p_{(1,q-1),(1,q-1)}^{(2,q-2)}\ne 0$. Let $(y_{0},y_{1})$ and $(y_{1},y_{2})$ be arcs of type $(1,q-1)$ such that $\tilde{\partial }(y_{0},y_{2})=(2,q-2)$. Since $p_{(1,q-1),(1,q-2)}^{(2,q-2)}\ne 0$, there exists a vertex $y_{1}'\in P_{(1,q-1),(1,q-2)}(y_{0},y_{2})$. By Lemma 2.2 (i), one has $k_{1,q-1}=2$ and $p_{(1,q-1),(1,q-1)}^{(2,q-2)}=1$. Lemma 2.2 (ii) and (iii) imply that $p_{(1,q-1),(1,q-1)}^{(1,q-2)}=2$ and $(y_{1}',y_{1})$ is an arc of type $(1,q-1)$. Since $y_{0}\in P_{(q-1,1),(1,q-1)}(y_{1},y_{1}')$, from Lemma 2.2 (iv), we get $q=2$, a contradiction. Thus, (i) holds

(ii) Let $(x_{0},x_{1},\ldots ,x_{q-1})$ be a circuit such that $\tilde{\partial }(x_{q-1},x_{0})=(1,q-1)$. Suppose $\tilde{\partial }(x_{0},x_{1})=(1,p-1)$ with $p\notin \{q,q-1\}$. It follows that $q>3$ and $\partial (x_{1},x_{q-1})=q-2$.

Case 1. $\partial (x_{q-1},x_{1})=1$.

Since $x_{0}\in P_{(1,q-1),(1,p-1)}(x_{q-1},x_{1})$, there exists $y\in P_{(1,p-1),(1,q-1)}(x_{q-1},x_{1})$. By $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$, we can pick a vertex $z\in P_{(1,q-1),(1,q-1)}(x_{q-1},x_{1})$. Note that $|\{x_{0},y,z\}|=3$. Since $(1,q-2)$ is pure, one gets $\tilde{\partial }(x_{i},x_{i+1})=(1,q-2)$ for $1\le i\le q-2$, which implies $\{x_{0},y,z\}\subseteq \Gamma _{2,q-2}(x_{q-2})$, contrary to Lemma 2.2 (i).

Case 2. $\partial (x_{q-1},x_{1})=2$.

Let $(z_{0}',z_{1}')$ and $(z_{1}',z_{2}')$ be arcs of type $(1,q-2)$ such that $\tilde{\partial }(z_{0}',z_{2}')=(2,q-3)$. By $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$, there exists a vertex $z'\in P_{(1,q-1),(1,q-1)}(z_{0}',z_{1}')$. Since $(1,q-2)$ is pure, one gets $\tilde{\partial }(z',z_{2}')=(2,q-2)$. By $x_{0}\in P_{(1,q-1),(1,p-1)}(x_{q-1},x_{1})$, there exists a vertex $w\in P_{(1,q-1),(1,p-1)}(z',z_{2}')$, which implies $\tilde{\partial }(z_{0}',w)=(2,q-2)$. Since $\tilde{\partial }(z_{0}',z')=\tilde{\partial }(z',w)=(1,q-1)$, we have $p_{(1,q-1),(1,q-1)}^{(2,q-2)}\ne 0$, contrary to (i).

Note that $\tilde{\partial }(x_{i},x_{i+1})=(1,q-1)$ or $(1,q-2)$ for $0\le i\le q-2$. If $\tilde{\partial }(x_{i},x_{i+1})=(1,q-1)$ for each i, by $q\ge 3$, then $\tilde{\partial }(x_{0},x_{2})=(2,q-2)$, contrary to $p_{(1,q-1),(1,q-1)}^{(2,q-2)}=0$. Thus, (ii) holds.

(iii) By Lemma 2.2 (ii), $k_{1,q-1}=2$. Since $k_{1,q-2}=1$, we get $p_{(1,q-1),(1,q-1)}^{(1,q-2)}=2$ from Lemma 1.2 (v). Let $(w_{0}=w_{q-1},w_{1},\ldots ,w_{q-2})$ be a circuit consisting of arcs of type $(1,q-2)$. Pick vertices $w'\in P_{(1,q-1),(1,q-1)}(w_{0},w_{1})$ and $w''\in P_{(1,q-1),(1,q-1)}(w_{1},w_{2})$ such that $\tilde{\partial }(w',w'')\ne (1,q-2)$. Note that $(w'',w_{0})=q-2$. By (i), one has $q-2<\partial (w'',w')\le 1+\partial (w'',w_{0})=q-1$. Since $q\ge 3$, we obtain $\partial (w',w'')=2$ from Lemma 2.2 (iv). The desired result follows. $\square$

Lemma 2.6

Let $(1,h-1),(1,l-1)\in \tilde{\partial }(\Gamma )$ and $v=\min \{j\mid p_{(1,h-1),(1,l-1)}^{(i,j)}\ne 0\}$ with $h,l>2$. Suppose that $(1,l-1)$ is pure, or $p_{(1,l-1),(1,l-1)}^{(1,l-2)}\ne 0$ and $(1,l-2)$ is pure. If $(\Gamma _{1,l-1})^{v}\cap \Gamma _{l-1,1}\Gamma _{h-1,1}\ne \emptyset$, then $h=l$ or $p_{(1,h-1),(1,h-1)}^{(1,l-2)}\ne 0$.

Proof

Let $(x_{0},x_{1},\ldots ,x_{v+1})$ be a circuit of length $v+2$ such that $\tilde{\partial }(x_{v+1},x_{0})=(1,h-1)$ and $\tilde{\partial }(x_{i},x_{i+1})=(1,l-1)$ for $0\le i\le v$. Suppose that $h\ne l$.

Case 1. $(1,l-1)$ is pure.

Note that $v+2>l$. Since $x_{0}\ne x_{l}$, by Lemma 2.2 (i), we have $k_{1,l-1}=2$. In view of $l>2$ and Lemma 2.4, we get $\Delta _{l}\simeq \mathrm{Cay}(\mathbb {Z}_{2l},\{1,l+1\})$ or $\mathrm{Cay}(\mathbb {Z}_{l}\times \mathbb {Z}_{l},\{(1,0),(0,1)\})$.

Case 1.1.$\Delta _{l}\simeq \mathrm{Cay}(\mathbb {Z}_{2l},\{1,l+1\})$.

In view of Lemma 2.4 (i), we obtain $(\Gamma _{1,l-1})^{i}=\{\Gamma _{i,l-i}\}$ for $2\le i\le l-1$. Then $\tilde{\partial }(x_{0},x_{l-1})=(l-1,1)$. If $v\ge l$, by Lemma 2.2 (iii), then $\tilde{\partial }(x_{0},x_{l+1})=(1,l-1)$, contrary to the minimality of v; if $v=l-1$, by $x_{l-1}\in P_{(l-1,1),(1,l-1)}(x_{l},x_{0})$ and Lemma 2.2 (iv), then $h=2$, a contradiction.

Case 1.2.$\Delta _{l}\simeq \mathrm{Cay}(\mathbb {Z}_{l}\times \mathbb {Z}_{l},\{(1,0),(0,1)\})$.

Let $\tau$ be an isomorphism from $\mathrm{Cay}(\mathbb {Z}_{l}\times \mathbb {Z}_{l},\{(1,0),(0,1)\})$ to $\Delta _{l}$. Pick $\tau (a,b)\in \Gamma _{1,h-1}(\tau (0,0))$. Then $0\notin \{a,b\}$. Since $\tau (a,b)\in P_{(1,h-1),(l-\hat{a},\hat{a})}(\tau (0,0),\tau (0,b))$, we get $\tau (e+a,f+b)\in \Gamma _{1,h-1}(\tau (e,f))\cap \Gamma _{\hat{a},l-\hat{a}}(\tau (e,f+b))$ and $\tau (e+b,f+a)\in \Gamma _{1,h-1}(\tau (e,f))\cap \Gamma _{\hat{a},l-\hat{a}}(\tau (e+b,f))$ for each (e, f). By $h\ne 2$, one has $\hat{a}+\hat{b}\ne l$.

Suppose $a=-1$. Since $(\tau (0,0),\tau (1,0),\tau (1+a,b)=\tau (0,b),\tau (0,b+1),\ldots ,\tau (0,l-1))$ is a circuit of length $l-\hat{b}+2$ containing arcs of types $(1,h-1)$ and $(1,l-1)$, we get $b=1$, contrary to $\hat{a}+\hat{b}\ne l$. Hence, $a\ne -1$. Similarly, $b\ne -1$. By $(\Gamma _{1,l-1})^{v}\cap \Gamma _{l-1,1}\Gamma _{h-1,1}\ne \emptyset$ and the minimality of v, one gets $v=\partial _{\Gamma }(\tau (a+1,b),\tau (0,0))=2l-\hat{a}-\hat{b}-1$. By $l-1\le v$, we obtain $\hat{a}+\hat{b}<l$. Note that $(\tau (a+1,b),\tau (a+b+1,a+b),\tau (a+b+2,a+b),\ldots ,\tau (0,a+b),\tau (0,a+b+1),\tau (0,a+b+2),\ldots ,\tau (0,0))$ is a path. If $a+b=-1$, then $l+1-\hat{a}-\hat{b}\ge v$, contrary to $l>2$; if $a+b\ne -1$, then $2l-2\hat{a}-2\hat{b}\ge v$, contrary to $\hat{a}+\hat{b}>1$.

Case 2. $p_{(1,l-1),(1,l-1)}^{(1,l-2)}\ne 0$ and $(1,l-2)$ is pure.

Since $h\ne l$ and $h,l>2$, one has $v\ge 2$. By the minimality of v, we obtain $\partial (x_{j},x_{j+2})=2$ for $0\le j\le v-1$. Lemma 2.2 (ii) implies $|(\Gamma _{1,l-1})^{2}|=2$ and $k_{1,l-1}=2$. If $|P_{(1,l-1),(1,l-1)}(x_{0},x_{2})|=2$, then there exists a vertex $x_{1}'\in P_{(1,l-1),(1,l-1)}(x_{0},x_{2})$ such that $\tilde{\partial }(x_{1}',x_{3})=(1,l-2)$, contrary to the minimality of v. Then $|P_{(1,l-1),(1,l-1)}(x_{0},x_{2})|=1$. By Lemma 2.2 (iii), $p_{(1,l-1),(1,l-1)}^{(1,l-2)}=2$. It follows from Lemma 1.2 (i) and (v) that $k_{1,l-2}=1$. In view of Lemma 2.5 (iii), we have $\tilde{\partial }(x_{j},x_{j+2})=(2,l-1)$ for $0\le j\le v-1$. Hence, $v\ge l-1$. By Lemma 2.5 (i) and Lemma 1.2 (iii), we obtain $h\ne l-1$.

Let $(y_{0},y_{1},\ldots ,y_{v+1})$ be a path consisting of arcs of type $(1,l-1)$ such that $\tilde{\partial }(y_{j},y_{j+2})=(2,l-1)$ for $0\le j\le v-1$. Pick $x_{v+1}'$ and $y_{v+1}'$ such that $\Gamma _{1,l-1}(x_{v})=\{x_{v+1},x_{v+1}'\}$ and $\Gamma _{1,l-1}(y_{v})=\{y_{v+1},y_{v+1}'\}$. Then $\tilde{\partial }(x_{v-1},x_{v+1}')=\tilde{\partial }(y_{v-1},y_{v+1}')=(1,l-2)$. Since $k_{1,l-2}=1$, by Lemma 1.2 (iii) and the inductive hypothesis, we have $\tilde{\partial }(x_{0},x_{v+1}')=\tilde{\partial }(y_{0},y_{v+1}')$, which implies $\tilde{\partial }(x_{0},x_{v+1})=\tilde{\partial }(y_{0},y_{v+1})$. Thus, $\tilde{\partial }(x_{0},x_{v+1})$ only depends on v.

Since $(1,l-2)$ is pure and $k_{1,l-2}=1$, each $\Delta _{l-1}(x_{i})$ is a circuit of length $l-1$, denoted by $(x_{i}=x_{0,i},x_{1,i},\ldots ,x_{l-2,i})$, where the first subscription of x is taken modulo $l-1$. The fact that $p_{(1,l-1),(1,l-1)}^{(1,l-2)}=2$ implies that $\tilde{\partial }(x_{a,b},x_{a,b+1})=\tilde{\partial }(x_{a,b+1},x_{a+1,b})=(1,l-1)$ for any $a\in \{0,1,\ldots ,l-2\}$ and $b\in \{0,1,\ldots ,v\}$. By $k_{1,l-2}=1$, one gets $\tilde{\partial }(x_{j,v-j+1},x_{j+2,v-j-1})=(2,l-1)$ for $0\le j\le v-1$. Since $\tilde{\partial }(x_{0},x_{v+1})$ only depends on v, we obtain $\tilde{\partial }(x_{0,v+1},x_{v+1,0})=\tilde{\partial }(x_{0},x_{v+1})=(h-1,1)$. Let r be the minimal nonnegative integer such that $r\equiv v+1$ (mod $l-1$). It suffices to show that $r=l-2$. Note that $(x_{0,0},x_{1,0},\ldots ,x_{r,0}=x_{v+1,0},x_{0,v+1})$ is a circuit. By $h\ne 2$, $r\ne 0$. Since $h\ne l-1$ and $(1,l-2)$ is pure, one gets $r=l-2$.

This completes the proof of Lemma 2.6. $\square$

Lemma 2.7

Let $q\ge 3$. If $(1,q-1)$ is pure, or $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$ and $(1,q-2)$ is pure, then (i) in Theorem 2.1 is valid.

Proof

Let $x_{0},x,x_{1}$ be vertices such that $\tilde{\partial }(x_{0},x)=(1,s-1)$, $\tilde{\partial }(x,x_{1})=(1,t-1)$ and $\tilde{\partial }(x_{0},x_{1})=(1,q-1)$. By Lemma 2.2 (iv) and $s<t$, we have $s\ne q$. Suppose $t\ne q$. Observe that $p_{(1,q-1),(1,q-1)}^{(2,q-2)}\ne 0$ or $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$. Pick $x_{2}$ such that $\tilde{\partial }(x_{1},x_{2})=(1,q-1)$ and $\partial (x_{2},x_{0})=q-2$.

We claim that $\tilde{\partial }(x,x_{2})=(2,q-1)$. If $(1,q-1)$ is pure, by $q\notin \{s,t\}$, then our claim is valid. Suppose that $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$ and $(1,q-2)$ is pure. It follows from Lemma 2.5 (i) that $\tilde{\partial }(x_{0},x_{2})=(1,q-2)$.

Suppose $s=q-1$. Since $x_{0}\in P_{(q-2,1),(1,q-2)}(x,x_{2})$ and $x\ne x_{2}$, by Lemma 2.2 (iv), we get $\tilde{\partial }(x,x_{2})=(1,1)$ or (2, 2). In view of Lemma 2.3 and $t\ne q$, $\tilde{\partial }(x,x_{2})=(2,2)$. Since $t\ne q-1$, from Lemma 2.5 (ii), one has $\partial (x_{2},x)>q-2$. Hence, $q=3$, $s=2$ and $t=4$. By $x_{1}\in P_{(1,3),(1,2)}(x,x_{2})$, there exists a vertex $x_{1}'\in P_{(1,2),(1,3)}(x_{2},x)$, which implies $\partial (x_{1},x_{1}')=2$. It follows from Lemma 2.5 (i) that $\partial (x_{1}',x_{1})=2$. By Lemma 2.2 (ii), we obtain $(\Gamma _{1,2})^{2}=\{\Gamma _{1,1},\Gamma _{2,2}\}$. Since $x\ne x_{2}$, from Lemma 2.2 (i), one has $k_{1,1}=2$. In view of Lemma 1.2 (i) and (v), we get $p_{(1,2),(1,2)}^{(1,1)}=1$. By Lemma 2.2 (ii), $p_{(1,2),(1,2)}^{(2,2)}=2$. Hence, $\tilde{\partial }(x,x_{1})=(1,2)$, a contradiction. Thus, $s\ne q-1$. Similarly, $t\ne q-1$.

Since $t\notin \{q-1,q\}$, by Lemma 2.5 (ii), we get $q-1\le \partial (x_{2},x)\le 1+\partial (x_{2},x_{0})=q-1$. The fact that $s\notin \{q-1,q\}$ and $\partial (x_{2},x_{0})=q-2$ imply $\partial (x,x_{2})=2$. Therefore, our claim is valid.

Since $x\in P_{(1,s-1),(1,t-1)}(x_{0},x_{1})$, there exists a vertex $x'\in P_{(1,t-1),(1,s-1)}(x_{0},x_{1})$. Similarly, $\tilde{\partial }(x',x_{2})=(2,q-1)$. Since $x_{1}\in \Gamma _{1,t-1}(x)\cap \Gamma _{1,s-1}(x')\cap \Gamma _{q-1,1}(x_{2})$, there exist vertices $y_{1}'\in P_{(1,q-1),(1,t-1)}(x,x_{2})$ and $y_{1}''\in P_{(1,q-1),(1,s-1)}(x,x_{2})$. It follows from Lemma 2.2 (i) that $k_{1,q-1}=2$. Similarly, $y_{1}',y_{1}''\in \Gamma _{2,q-1}(x_{0})$. Then $\Gamma _{1,s-1}\Gamma _{1,q-1}=\{\Gamma _{2,q-1}\}$. By Lemma 1.2 (i), we have $k_{1,s-1}=p_{(1,s-1),(1,q-1)}^{(2,q-1)}$. Since $x_{1}\in P_{(1,t-1),(1,q-1)}(x,x_{2})$, from Lemma 1.2 (iv), one gets $k_{1,s-1}=1$. Similarly, $k_{1,t-1}=1$. In view of $p_{(1,s-1),(1,t-1)}^{(1,q-1)}\ne 0$, we obtain $k_{1,q-1}=1$, a contradiction. Thus, $t=q$. By Lemma 2.2 (iv), one has $s=2$. $\square$

Proof of Theorem 2.1

(ii) (a)$\Rightarrow$(b): By way of contradiction, we may assume that q is the minimal integer such that $(1,q-1)$ is mixed and (b) does not hold. Since (1, 1) is pure, $q\ge 3$. Pick a circuit $(x_{0},x_{1},\ldots ,x_{q-1})$ such that $\tilde{\partial }(x_{q-1},x_{0})=(1,q-1)$ and $\tilde{\partial }(x_{0},x_{1})=(1,c-1)$ with $c<q$.

Suppose $\tilde{\partial }(x_{i},x_{i+1})=(1,d-1)$ for some $i\in \{1,2,\ldots ,q-2\}$ and $d\notin \{q,c\}$. Without loss of generality, we may assume $i=q-2$. Lemmas 2.3, 2.7 and the minimality of q imply $\tilde{\partial }(x_{q-2},x_{0})=\tilde{\partial }(x_{q-1},x_{1})=(2,q-2)$. Since $x_{0}\in P_{(1,q-1),(1,c-1)}(x_{q-1},x_{1})$, there exist vertices $z_{0}\in P_{(1,c-1),(1,q-1)}(x_{q-1},x_{1})$ and $z_{q-1}\in P_{(1,c-1),(1,q-1)}(x_{q-2},x_{0})$. In view of Lemma 2.2 (i), $k_{1,q-1}=2$. By Lemmas 2.3, 2.7 and the minimality of q, we get $\tilde{\partial }(z_{q-1},x_{1})=(2,q-2)$ and $\Gamma _{1,q-1}\Gamma _{1,c-1}=\{\Gamma _{2,q-2}\}$. It follows from Lemma 1.2 (i) that $k_{1,c-1}=p_{(1,q-1),(1,c-1)}^{(2,q-2)}$. Since $x_{q-1}\in P_{(1,d-1),(1,q-1)}(x_{q-2},x_{0})$, by Lemma 1.2 (iv), we obtain $k_{1,c-1}=1$. Similarly, $k_{1,d-1}=1$.

Since $k_{1,q-1}=2$, by Lemma 1.2 (i) and Lemma 2.2 (i), one gets $\tilde{\partial }(x_{j},x_{j+1})=(1,q'-1)$ for some $j\in \{1,2,\ldots ,q-3\}$, and $k_{1,q'-1}=2$. Without loss of generality, we may assume $j=1$. It follows from Lemmas 2.3, 2.7 and the minimality of q that $\tilde{\partial }(z_{0},x_{2})=(2,q-2)$. Since $x_{1}\in P_{(1,q-1),(1,q'-1)}(z_{0},x_{2})$, we have $x_{q-1}$ or $z_{q-1}\in P_{(1,q'-1),(1,q-1)}(x_{q-2},x_{0})$, a contradiction. Hence, $\tilde{\partial }(x_{i},x_{i+1})=(1,q-1)$ or $(1,c-1)$ for each i.

Since $c<q$, by Lemmas 2.3 and 2.6, we have $\tilde{\partial }(x_{i},x_{i+1})=(1,q-1)$ for some $i\in \{1,2,\ldots ,q-2\}$. Without loss of generality, we may assume $i=q-2$. Suppose $\partial (x_{q-2},x_{0})=2$. Then $\tilde{\partial }(x_{q-2},x_{0})=\tilde{\partial }(x_{q-1},x_{1})=(2,q-2)$. Since $x_{q-1}\in P_{(1,q-1),(1,q-1)}(x_{q-2},x_{0})$, there exists a vertex $x_{0}'\in P_{(1,q-1),(1,q-1)}(x_{q-1},x_{1})$, which implies $\tilde{\partial }(x_{q-2},x_{0}')=(2,q-2)$ and $k_{1,q-1}=2$ from Lemma 2.2 (i). Hence, $(\Gamma _{1,q-1})^{2}=\{\Gamma _{2,q-2}\}$. By Lemma 2.2 (iii), we get $p_{(1,q-1),(1,q-1)}^{(2,q-2)}=2$ and $\tilde{\partial }(x_{0},x_{1})=(1,q-1)$, a contradiction. Thus, $\tilde{\partial }(x_{q-2},x_{0})=(1,q-2)$ and $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$.

Note that $(1,q-2)$ is mixed. By the minimality of q, $p_{(1,q-2),(1,q-2)}^{(1,q-3)}\ne 0$ and $(1,q-3)$ is pure. It follows from Lemma 2.5 (ii) that the path $(x_{0},x_{1},\ldots ,x_{q-2})$ contains an arc of type $(1,q-3)$. Hence, $c=q-2$ and $\tilde{\partial }(x_{j},x_{j+1})=(1,q-3)$ for $0\le j\le q-3$. By Lemma 2.3, we get $q>4$. In view of Lemma 2.6, we obtain $p_{(1,q-2),(1,q-2)}^{(1,q-4)}\ne 0$, a contradiction. Thus, our desired result holds.

(b)$\Rightarrow$(c): It is obvious.

(c)$\Rightarrow$(a): Suppose for the contrary that $(1,q-1)$ is pure. By Lemma 2.2 (ii), we have $|(\Gamma _{1,q-1})^{2}|=2$ and $k_{1,q-1}=2$. Lemma 2.4 implies that $p_{(1,q-1),(1,q-1)}^{(2,q-2)}=1$ and there exists an isomorphism $\tau$ from $\mathrm{Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{q},\{(1,0),(0,1)\})$ to $\Delta _{q}$. It follows from Lemma 2.2 (iii) and Lemma 1.2 (i), (v) that $k_{1,s-1}=1$. Observe that $(\tau (0,0),\tau (1,1),\ldots ,\tau (-1,-1))$ is a circuit consisting of arcs of type $(1,s-1)$. Since $s\ne q$ from Lemma 2.2 (iv), $(1,s-1)$ is mixed. Then $p_{(1,s-1),(1,s-1)}^{(1,s-2)}\ne 0$. By Lemma 1.2 (i), we get $(\tau (1,1),\tau (3,3))\in \Gamma _{1,s-2}$ and $k_{1,s-2}=1$. Note that $(\tau (0,0),\tau (1,0),\tau (1,1),\tau (3,3),\tau (4,4),\ldots ,\tau (-1,-1))$ is a circuit of length q containing arcs of types $(1,q-1)$ and $(1,s-2)$, contrary to the fact that $(1,q-1)$ is pure. Thus, we have the assertion.

(i) follows by (ii) and Lemma 2.7.

(iii) By Lemma 2.2 (iv), $s\ne q$. (ii) implies that $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$ and $(1,q-2)$ is pure. In view of Lemma 2.2 (ii) and Lemma 2.5 (iii), we only need to consider the case that $|(\Gamma _{1,q-1})^{2}|=2$ and $k_{1,q-2}=2$. Then $k_{1,q-1}=2$. By Lemma 1.2 (i) and (v), we have $p_{(1,q-1),(1,q-1)}^{(1,q-2)}=1$. Suppose $s\ne q-1$. In view of Lemma 2.2 (iii), one gets $p_{(1,q-1),(1,q-1)}^{(1,s-1)}=2$ and $k_{1,s-1}=1$. Let $(x_{0},x),(x,x_{1})$ and $(x,x_{1}')$ be arcs of type $(1,q-1)$ such that $\tilde{\partial }(x_{0},x_{1})=(1,s-1)$ and $\tilde{\partial }(x_{0},x_{1}')=(1,q-2)$. Pick vertices $x_{2},z$ such that $\tilde{\partial }(x_{1},x_{2})=\tilde{\partial }(x,z)=(1,s-1)$. Since $p_{(1,q-1),(1,q-1)}^{(1,s-1)}=2$, we obtain $\tilde{\partial }(x_{1},z)=\tilde{\partial }(x_{1}',z)=\tilde{\partial }(z,x_{2})=(1,q-1)$ and $\tilde{\partial }(x_{1}',x_{2})=(1,q-2)$. The fact that $x_{1}'\in P_{(1,q-2),(1,q-2)}(x_{0},x_{2})$ and $k_{1,s-1}=1$ imply that $(1,s-1)$ is mixed. It follows from (ii) that $\tilde{\partial }(x_{0},x_{2})=(1,s-2)$ and $(1,q-2)$ is mixed, contrary to the fact that $(1,q-2)$ is pure. $\square$

3 Configuration $C_{q,h}$

In this section, we will discuss some useful properties of the configuration $C_{q,h}$.

Lemma 3.1

Suppose that $C_{q,h}$ exists. Then $k_{1,h-1}=1$, $k_{1,q-1}=2$, $(1,q-1)$ is pure and $\Delta _{q}\simeq \mathrm{Cay}(\mathbb {Z}_{2q},\{1,q+1\})$. Moreover, if (1, q) is mixed, then $k_{1,q}=2$.

Proof

Pick four distinct vertices x, y, z, w such that $\tilde{\partial }(x,w)=\tilde{\partial }(y,w)=(1,q-1)$ and $\tilde{\partial }(x,z)=\tilde{\partial }(z,y)=(1,h-1)$. By Lemma 2.2 (i), $k_{1,q-1}=2$. In view of $h>2$ and Lemma 2.2 (iv), we have $|(\Gamma _{1,h-1})^{2}|=1$. Since $(\Gamma _{1,q-1})^{2}=\{\Gamma _{2,q-2}\}$, from Theorem 2.1 (ii), $(1,q-1)$ is pure. Lemma 2.4 implies $\Delta _{q}\simeq \mathrm{Cay}(\mathbb {Z}_{2q},\{1,q+1\})$. So that there exists a vertex $w'\in P_{(1,q-1),(q-1,1)}(x,y)$ with $w'\ne w$. Write $\tilde{\partial }(x,y)=\tilde{f}$. By Lemma 1.2 (i) and (v), one has $k_{\tilde{f}}=1$. Since $(\Gamma _{1,h-1})^{2}=\{\Gamma _{\tilde{f}}\}$, we get $k_{1,h-1}=1$. If (1, q) is mixed, then $k_{1,q}=2$ from Theorem 2.1 (ii) and Lemma 1.2 (i). $\square$

Lemma 3.2

Suppose that $C_{q,h}$ exists. The following hold:

• (i) If $(1,h-1)$ is pure, then $h=4$.

• (ii) If $(1,h-1)$ is mixed, then $h=3$.

Proof

Let (x, z), (z, y) be two arcs of type $(1,h-1)$. Observe $P_{(1,q-1),(q-1,1)}(x,y)\ne \emptyset$. It follows from Lemma 2.2 (iv) that $\partial (x,y)=\partial (y,x)$. In view of Lemma 3.1, one has $k_{1,h-1}=1$. If $(1,h-1)$ is pure, by Lemma 2.2 (ii), then $\tilde{\partial }(x,y)=(2,2)$ and $h=4$; if $(1,h-1)$ is mixed, by Theorem 2.1 (ii), then $\tilde{\partial }(x,y)=(1,1)$ and $h=3$. $\square$

Lemma 3.3

If $C_{q,h}$ exists, then $\Gamma _{1,q-1}\Gamma _{1,h-1}=\{\Gamma _{2,q}\}$ and $\Gamma _{q,2}\in \Gamma _{1,h-1}(\Gamma _{1,q-1})^{q-1}$.

Proof

Pick four distinct vertices x, y, z, w such that $\tilde{\partial }(x,y)=\tilde{\partial }(x,w)=(1,q-1)$ and $\tilde{\partial }(y,z)=\tilde{\partial }(z,w)=(1,h-1)$. By Lemma 3.1, $(1,q-1)$ is pure and $k_{1,h-1}=1$. In view of Lemma 2.2 (ii), we have $|\Gamma _{1,q-1}\Gamma _{1,h-1}|=1$. It follows from Theorem 2.1 (i) that $\partial (x,z)=2$. Note that $q-1\le \partial (z,x)\le 1+\partial (w,x)=q$. It suffices to show that $\partial (z,x)=q$.

Assume the contrary, namely there exists a path $(z=x_{0},x_{1},\ldots ,x_{q-1}=x)$. Suppose that $\tilde{\partial }(x_{i},x_{i+1})=(1,p-1)$ for some $i\in \{0,1,\ldots ,q-2\}$ and $p\ne q$. Since $k_{1,h-1}=1$, we obtain $\tilde{\partial }(z,x_{1})\ne (1,h-1)$. Hence, $p\ne h$. Without loss of generality, we may assume $i=q-2$. Since $(1,q-1)$ is pure, one has $\partial (y,x_{q-2})=q-1$, which implies $\partial (x_{q-2},y)=2$. By $x\in P_{(1,p-1),(1,q-1)}(x_{q-2},y)$ and Lemma 2.2 (i), we get $\tilde{\partial }(w,z)=(1,p-1)$, contrary to $h\ge 3$. Hence, $\tilde{\partial }(x_{i},x_{i+1})=(1,q-1)$ for each i. It follows from Lemma 3.1 and Lemma 2.4 (i) that $\tilde{\partial }(z,x)=(q-1,1)$, contrary to Lemma 2.2 (i). $\square$

Lemma 3.4

If $(1,q-1)$ is mixed and $C_{q-1,h}$ exists, then $\Gamma _{1,q-1}\Gamma _{1,h-1}=\{\Gamma _{2,q}\}$ and $\Gamma _{q,2}\in \Gamma _{1,h-1}(\Gamma _{1,q-2})^{q-2}\Gamma _{1,q-1}$.

Proof

Let x, y, z be vertices such that $\tilde{\partial }(x,y)=(1,q-1)$ and $\tilde{\partial }(y,z)=(1,h-1)$. By Theorem 2.1 (ii), we have $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$. Pick a vertex $w\in P_{(q-1,1),(1,q-2)}(x,y)$. It follows from Theorem 2.1 (i) that $\partial (x,z)=2$. Since $h\notin \{q,q-1\}$ from Lemma 3.1, by Lemma 2.5 (ii), one obtains $\partial (z,x)\ge q-1$. In view of Lemma 3.3, we get $\tilde{\partial }(w,z)=(2,q-1)$, which implies $\tilde{\partial }(x,z)=(2,q)$ from Lemma 2.2 (iv). The desired results follow by Lemma 3.3. $\square$

Lemma 3.5

Suppose $q\ge 3$ and $p_{(1,q-1),(1,1)}^{(2,s)}\ne 0$. The following hold:

• (i) If $(1,q-1)$ is pure, then $s=q$ and $\Gamma _{q,2}\in \Gamma _{1,1}(\Gamma _{1,q-1})^{q-1}$.

• (ii) If $(1,q-1)$ is mixed and $s=q$, then $\Gamma _{q,2}\in \Gamma _{1,1}\Gamma _{1,q-1}(\Gamma _{1,q-2})^{q-2}.$

Proof

(i) Note that $s=q-1$ or q. Suppose for the contrary that $s=q-1$. Let $x_{q-1},x_{q},x_{0}$ be three vertices such that $\tilde{\partial }(x_{q-1},x_{q})=(1,q-1)$, $\tilde{\partial }(x_{q},x_{0})=(1,1)$ and $\tilde{\partial }(x_{q-1},x_{0})=(2,q-1)$. Pick a path $(x_{0},x_{1},\ldots ,x_{q-1})$.

Case 1.$\partial (x_{i+1},x_{i})\notin \{1,q-1\}$ for some $i\in \{0,1,\ldots ,q-2\}$.

Without loss of generality, we may assume $\tilde{\partial }(x_{q-2},x_{q-1})=(1,p-1)$ with $p\notin \{2,q\}$. Since $(1,q-1)$ is pure, we get $\tilde{\partial }(x_{q-2},x_{q})=(2,q-1)$ from Theorem 2.1 (i). In view of $x_{q}\in P_{(1,q-1),(1,1)}(x_{q-1},x_{0})$, there exists a vertex $x'\in P_{(1,1),(1,q-1)}(x_{q-2},x_{q})$, which implies $k_{1,q-1}=2$ by Lemma 2.2 (i). Since $(1,q-1)$ is pure, we have $\tilde{\partial }(x',x_{0})=(2,q-1)$ and $\Gamma _{1,q-1}\Gamma _{1,1}=\{\Gamma _{2,q-1}\}$. It follows from Lemma 1.2 (i) that $k_{1,1}=p_{(1,q-1),(1,1)}^{(2,q-1)}$. In view of $x_{q-1}\in P_{(1,p-1),(1,q-1)}(x_{q-2},x_{q})$ and Lemma 1.2 (iv), we obtain $k_{1,1}=1$. By $x_{q}\in P_{(1,q-1),(q-1,1)}(x',x_{q-1})$ and Lemma 2.2 (iv), one gets $\partial (x',x_{q-1})=\partial (x_{q-1},x')$. Since $x_{q-2}\in P_{(1,1),(1,p-1)}(x',x_{q-1})$, we obtain $\tilde{\partial }(x_{q-1},x_{q-2})=(1,p-1)$, contrary to $p\ne 2$.

Case 2.$\partial (x_{i+1},x_{i})\in \{1,q-1\}$ for $0\le i\le q-2$.

Let $r-1$ be the number of arcs of type $(1,q-1)$ in the path $(x_{0},x_{1},\ldots ,x_{q-1})$. Lemma 2.3 implies $r>1$. Without loss of generality, we may assume $\tilde{\partial }(x_{j},x_{j+1})=(1,q-1)$ with $q-r\le j\le q-2$. It follows from Theorem 2.1 (ii) that $\tilde{\partial }(x_{j},x_{j+2})=(2,q-2)$ or $(2,q-1)$ for each j.

Suppose $\tilde{\partial }(x_{j},x_{j+2})=(2,q-1)$ for some j. It follows from Lemma 2.2 (ii) that $(\Gamma _{1,q-1})^{2}=\{\Gamma _{2,q-2},\Gamma _{2,q-1}\}$ and $k_{1,q-1}=2$. Lemma 2.4 implies $p_{(1,q-1),(1,q-1)}^{(2,q-2)}=1$. By Lemma 2.2 (iii), $p_{(1,q-1),(1,q-1)}^{(2,q-1)}=2$. Hence, $\tilde{\partial }(x_{q},x_{0})=(1,q-1)$, a contradiction.

Suppose $\tilde{\partial }(x_{j},x_{j+2})=(2,q-2)$ for each j. Since $\tilde{\partial }(x_{q-1},x_{0})\ne (1,q-1)$, we have $r<q$ from Lemma 2.4. Hence, $\tilde{\partial }(x_{q-r},x_{q})=(r,q-r)$. By Lemma 2.3, $r=\frac{q}{2}$. Since $\tilde{\partial }(x_{0},x_{r})=(\frac{q}{2},\frac{q}{2})$, there exists a path $(y_{r}=x_{r},y_{r+1},\ldots ,y_{q}=x_{0})$ consisting of arcs of type $(1,q-1)$. Then $(x_{0},x_{1},\ldots ,x_{r}=y_{r},y_{r+1},\ldots ,y_{q-1})$ is a circuit of length q containing arcs of types $(1,q-1)$ and (1,1), a contradiction.

(ii) It is an immediate consequence of Theorem 2.1 (ii). $\square$

Let $A_{i,j}$ denote a matrix with rows and columns indexed by $V\Gamma$ such that $(A_{i,j})_{x,y}=1$ if $\tilde{\partial }(x,y)=(i,j)$, and $(A_{i,j})_{x,y}=0$ otherwise.

Lemma 3.6

Suppose that $q>2$, $(1,q-1)$ is pure and $\Delta _{q}\simeq \mathrm{Cay}(\mathbb {Z}_{2q},\{1,q+1\})$. The following hold:

• (i) If $(1,1)\in \tilde{\partial }(\Gamma )$, then $A_{1,q-1}A_{1,1}=A_{1,q-1}$ or $A_{1,q-1}A_{1,1}=k_{1,1}A_{2,q}$.

• (ii) If (1, q) is mixed, then $A_{1,q-1}A_{1,q}=2A_{2,q-1}$ and $(A_{1,q})^{2}=2A_{1,q-1}$.

• (iii) If (1, q) is mixed and $A_{1,q-1}A_{1,1}=A_{1,q-1}$, then $A_{1,q}A_{1,1}=A_{1,q}$.

• (iv) If (1, q) is mixed and $A_{1,q-1}A_{1,1}=k_{1,1}A_{2,q}$, then $A_{1,q}A_{1,1}=k_{1,1}A_{2,q+1}$.

Proof

(i) Suppose $p_{(1,q-1),(1,1)}^{(1,q-1)}\ne 0$. Since $\Delta _{q}\simeq \mathrm{Cay}(\mathbb {Z}_{2q},\{1,q+1\})$, we obtain $p_{(1,q-1),(q-1,1)}^{(1,1)}=2$. By Lemma 1.2 (i) and (v), we get $k_{1,1}=1$, which implies $A_{1,q-1}A_{1,1}=A_{1,q-1}$. Suppose $p_{(1,q-1),(1,1)}^{(1,q-1)}=0$. By Theorem 2.1 (i), Lemma 2.3 and Lemma 3.5 (i), we have $\Gamma _{1,q-1}\Gamma _{1,1}=\{\Gamma _{2,q}\}$, which implies $A_{1,q-1}A_{1,1}=k_{1,1}A_{2,q}$ from Lemma 1.2 (i).

(ii) By Theorem 2.1 (ii), we get $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$. Since $k_{1,q-1}=2$, from Lemma 1.2 (i) and Lemma 2.2 (i), we have $k_{1,q}=2$.

Let x, y, z, w be vertices such that $\tilde{\partial }(x,y)=(1,q-1)$, $\tilde{\partial }(y,z)=(1,q)$ and $w\in P_{(1,q-1),(q,1)}(y,z)$. By Lemma 2.4 (i), we have $\tilde{\partial }(x,w)=(2,q-2)$. In view of Theorem 2.1 (i), one gets $\tilde{\partial }(x,z)=(2,q-1)$, which implies $A_{1,q-1}A_{1,q}=2A_{2,q-1}$ from Lemma 1.2 (i) and Lemma 2.2 (i).

By Lemma 2.4 (i), there exists a vertex $y'\in P_{(1,q-1),(1,q-1)}(x,w)$ with $y\ne y'$. Since $p_{(1,q-1),(1,q)}^{(2,q-1)}=2$, one has $\tilde{\partial }(y',z)=(1,q)$, which implies $(A_{1,q})^{2}=2A_{1,q-1}$ from Lemma 1.2 (i).

(iii) By Lemma 1.2 (i), we have $k_{1,1}=1$. Let $x_{0},x_{1},x_{2},x_{3}$ be vertices such that $\tilde{\partial }(x_{0},x_{2})=(1,q-1)$, $x_{1}\in P_{(1,q-1),(1,1)}(x_{0},x_{2})$ and $x_{3}\in P_{(1,q),(1,q)}(x_{0},x_{2})$. It follows from (ii) that $\tilde{\partial }(x_{3},x_{1})=(1,q)$. Since $x_{1}\in P_{(1,q),(1,1)}(x_{3},x_{2})$, by Lemma 1.2 (i), we get $A_{1,q}A_{1,1}=A_{1,q}$.

(iv) Let $z_{0},z_{1},z_{2},z_{0}'$ be vertices such that $\tilde{\partial }(z_{0},z_{1})=(1,q)$, $\tilde{\partial }(z_{1},z_{2})=(1,1)$ and $z_{0}'\in P_{(q,1),(1,q-1)}(z_{0},z_{1})$. Since $A_{1,q-1}A_{1,1}=k_{1,1}A_{2,q}$, $\tilde{\partial }(z_{0}',z_{2})=(2,q)$. In view of (ii), one has $\tilde{\partial }(z_{0},z_{2})\ne (1,q)$, which implies $\partial (z_{0},z_{2})=2$ from Lemma 2.3 and Theorem 2.1 (i). Since $\tilde{\partial }(z_{0}',z_{2})=(2,q)$, by Lemma 2.2 (iv), we get $\partial (z_{2},z_{0})\ne q$. It follows from Lemma 2.5 (ii) that $\partial (z_{2},z_{0})=q+1$. The desired result holds by Lemma 1.2 (i) and Lemma 2.2 (i). $\square$

Lemma 3.7

Suppose that $(1,1)\in \tilde{\partial }(\Gamma )$ and $C_{q,h}$ exists. The following hold:

• (i) $h=3$, $k_{1,1}=1$ and $A_{1,q-1}A_{1,1}=A_{1,q-1}$.

• (ii) If (1, q) is mixed, then $A_{1,q}A_{1,1}=A_{1,q}$.

Proof

(i) Let x, y, z be vertices such that $\tilde{\partial }(x,y)=(1,q-1)$ and $\tilde{\partial }(y,z)=(1,1)$. Suppose $\tilde{\partial }(x,z)=(2,q)$. Since $C_{q,h}$ exists, by Lemma 3.3, there exists a vertex $w\in P_{(1,q-1),(1,h-1)}(x,z)$, which implies $\tilde{\partial }(z,y)=(1,h-1)$, contrary to $h\ge 3$. It follows from Lemma 3.1 and Lemma 3.6 (i) that $A_{1,q-1}A_{1,1}=A_{1,q-1}$ and $\tilde{\partial }(x,z)=(1,q-1)$. Then $(\Gamma _{1,h-1})^{2}=\{\Gamma _{1,1}\}$ and $h=3$. By Lemma 1.2 (i), $k_{1,1}=1$.

(ii) It is an immediate consequence of Lemma 3.1 and Lemma 3.6 (iii). $\square$

Proposition 3.8

If $C_{q,h}$ and $C_{q',h'}$ both exist, then $h=h'$.

Proof

If $(1,1)\in \tilde{\partial }(\Gamma )$, by Lemma 3.7 (i), then $h=h'=3$; if $(1,1)\notin \tilde{\partial }(\Gamma )$, by Theorem 2.1 (ii) and Lemma 3.2, then $h=h'=4$. $\square$

4 Proof of Proposition 1.3

We shall prove Proposition 1.3 by contradiction. Suppose that C1–C6 do not hold. Let $\mathscr {B}$ be the set consisting of $(p,p-1)$ and $(p-1,p)$ where $(1,p-1)$ is mixed, $\mathscr {C}=\{(p,q)\mid C_{p,q}~\text {or}~C_{q,p}~\text {exists}\}$ and $\mathscr {D}=\{(p,q)\mid (p,p-1)\in \mathscr {B}~\text {and}~C_{p-1,q}~\text {exists,}~\text {or}~(q,q-1)\in \mathscr {B}~\text {and}~C_{q-1,p}~\text {exists}\}$.

Suppose that $C_{q,h}$ exists for some q and h. In view of Lemma 3.1, $(1,q-1)$ is pure. If $(1,1)\in K$, from Lemma 3.7 (i), then $h=3$; if $(1,1)\notin K$, from Lemma 3.2 and Theorem 2.1 (ii), then $h=4$. Since C1, C2 and C3 do not hold, by Proposition 3.8, there exists $(1,p-1)\in K$ such that $p\ne 2$ and $(q,p)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$. Suppose that $C_{t,h}$ does not exist for any t and h. Since the valency of $\Gamma$ is more than 3, we may assume that $(1,q-1)\in K$ with $q\ne 2$. Since C4, C5 and C6 do not hold, from Theorem 2.1 (ii), there exists $(1,p-1)\in K$ such that $p\ne 2$ and $(q,p)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$.

We set

$$\begin{aligned} l=\min \left\{ r\mid p_{(1,i-1),(1,j-1)}^{(2,r)}\ne 0,~i\ne j,~i,j\ge 3,~(i,j)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}\right\} . \end{aligned}$$

Without loss of generality, we may assume $p_{(1,q-1),(1,p-1)}^{(2,l)}\ne 0$. By Lemma 2.5 (ii) and Theorem 2.1 (ii), one has $l\ge 3$.

Choose vertices x, y and z with $\tilde{\partial }(x,y)=(1,q-1)$, $\tilde{\partial }(y,z)=(1,p-1)$ and $\tilde{\partial }(x,z)=(2,l)$. Then there exists $y'$ such that $\tilde{\partial }(x,y')=(1,p-1)$ and $\tilde{\partial }(y',z)=(1,q-1)$.

The minimality of l will be used many times in the sequel, so we will not refer to it every time for the sake of simplicity. We will reach a contradiction under the following two separate cases:

• A) There exists a shortest path from z to x containing an arc of type $(1,h-1)$ with $h\notin \{2,q,p\}$.

• B) Each arc of any shortest path from z to x is of type (1, 1), $(1,q-1)$ or $(1,p-1)$.

4.1 The case A

Without loss of generality, we may assume that $(z=x_{0},x_{1},\ldots ,x_{l}=x)$ is a path such that $\tilde{\partial }(x_{0},x_{1})=(1,h-1)$. For each i, write $h_{i}=\partial (x_{i+1},x_{i})+1$.

Step 1

Show that $C_{t,h}$ exists for some t.

Assume the contrary, namely $C_{t,h}$ does not exist for any t. Suppose that $(h,q),(h,p)\in \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$. Observe $\{(h,q),(h,p)\}\nsubseteq \mathscr {B}$. It follows from Proposition 3.8 that $\{(h,q),(h,p)\}\nsubseteq \mathscr {C}\cup \mathscr {D}$. Without loss of generality, we may assume $(h,q)\in \mathscr {B}$ and $(h,p)\in \mathscr {C}\cup \mathscr {D}$. If $C_{h,p}$ exists, by Lemma 3.1, then $q=h+1$ and $(1,q-1)$ is mixed, contrary to $(q,p)\notin \mathscr {D}$. If $(1,h-1)$ is mixed and $C_{h-1,p}$ exists, then $C_{q,p}$ exists, a contradiction. Thus, (h, q) or $(h,p)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$.

Without loss of generality, we may assume that $(h,p)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$. Theorem 2.1 (i) implies $\tilde{\partial }(y,x_{1})=(2,l)$. Since $z\in P_{(1,p-1),(1,h-1)}(y,x_{1})$, there exists a vertex $y''\in P_{(1,h-1),(1,p-1)}(x,z)$, which implies $k_{1,p-1}=2$ from Lemma 2.2 (i). By Theorem 2.1 (i) again, we get $\partial (y',x_{1})=2$ and $\tilde{\partial }(y'',x_{1})=(2,l)$. Then $\Gamma _{1,p-1}\Gamma _{1,h-1}=\{\Gamma _{2,l}\}$ and $k_{2,l}=2$. Since $p_{(1,q-1),(1,p-1)}^{(2,l)}\ne 0$, from Lemma 1.2 (i) and (iv), we obtain $k_{1,h-1}=p_{(1,p-1),(1,h-1)}^{(2,l)}$=1. By $k_{2,l}=2$, $\partial (x_{1},y')<l$. Hence, $(h,q)\in \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$. Since $k_{1,h-1}=1$, one gets $(h,q)\in \mathscr {B}$ from Lemma 3.1. Suppose that $(1,q-1)$ is mixed. By Theorem 2.1 (ii) and Lemma 1.2 (ii), one has $p_{(1,q-1),(1,q-1)}^{(1,h-1)}=k_{1,q-1}$, which implies $\tilde{\partial }(y,y'')=(1,q-1)$. Since $z\in P_{(1,p-1),(p-1,1)}(y'',y)$, from Lemma 2.2 (iv), one obtains $q=2$, a contradiction. Now suppose that $(1,h-1)$ is mixed. By Theorem 2.1 (ii) again, $\tilde{\partial }(y'',y)=(1,h-1)$. In view of Lemma 2.2 (iv), $h=2$, a contradiction. Thus, $C_{t,h}$ exists for some t.

Step 2

Show that $\{(q,h),(p,h)\}\nsubseteq \mathscr {C}\cup \mathscr {D}$.

Suppose for the contrary that $\{(q,h),(p,h)\}\subseteq \mathscr {C}\cup \mathscr {D}$. By Step 1 and Lemma 3.1, we have $k_{1,h-1}=1$. We conclude that $(x_{1},x_{2},\ldots ,x_{l})$ consists of arcs of types $(1,q-1)$ and $(1,p-1)$.

By Proposition 3.8, $C_{q,h}$ exists, or $C_{q-1,h}$ exists and $(1,q-1)$ is mixed. Suppose $h_{l-1}=2$. By Lemma 3.7 (i) or (ii), we have $\tilde{\partial }(x_{l-1},y)=(1,q-1)$. Theorem 2.1 (i) implies $\partial (x_{l-1},z)=2$, contrary to $\partial (z,x_{l-1})<l$. Then $h_{j}\ne 2$ for $1\le j\le l-1$. Step 1 and Proposition 3.8 imply that $h_{j}\in \{q,p,h\}$ for any j. Since $h\notin \{q,p\}$, one gets $l\ge 4$ from Lemma 2.5 (ii) and Theorem 2.1 (ii). If $h_{j}=h$ for any j, by Lemma 3.2 and $k_{1,h-1}=1$, then $(1,h-1)$ is pure and $h=4$, which imply $z=x_{4}$, a contradiction. In the path $(x_{1},x_{2},\ldots ,x_{l})$, without loss of generality, we may assume that the number of arcs of type $(1,p-1)$ is not less than the number of arcs of type $(1,q-1)$.

Without loss of generality, we may assume $h_{l-1}=p$. By Proposition 3.8 again, $C_{p,h}$ exists, or $C_{p-1,h}$ exists and $(1,p-1)$ is mixed. Suppose that $C_{p-1,h}$ exists and $(1,p-1)$ is mixed. Then $p\ge 4$. In view of Lemma 3.1 and Proposition 3.8, we obtain $(q,p-1)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$. Lemma 3.6 (ii) implies $\tilde{\partial }(x_{l-1},y')=(1,p-2)$. It follows from Theorem 2.1 (i) that $\partial (x_{l-1},z)=2$, contrary to $\partial (z,x_{l-1})<l$. Hence, $C_{p,h}$ exists. Suppose $h_{1}=h$. Then $\tilde{\partial }(y,x_{2})=(1,p-1)$. By Theorem 2.1 (i) again, one has $\partial (x,x_{2})=2$, contrary to $\partial (x_{2},x)<l$. Thus, $h_{j}\in \{q,p\}$ for $1\le j\le l-1$.

Suppose $h_{j}=p$ for each j. Since $C_{p,h}$ exists, by Lemma 3.3, we get $\partial (x_{1},y)=p$, and so $l\ge p$. In view of Lemma 3.1 and Lemma 2.4 (i), one has $\tilde{\partial }(x_{1},x_{p})=(p-1,1)$. Since $\tilde{\partial }(y',z)=(1,q-1)$, we obtain $\tilde{\partial }(x,x_{1})\ne (1,p-1)$ and $l>p$. Let $x_{l+1}=y'$. By Lemma 2.4 (i), one gets $\tilde{\partial }(x_{1},x_{p+2})=(1,p-1)$. Then $x_{p+2}=y'$ and $(y',z,x_{1})$ is a circuit, a contradiction. Therefore, our conclusion is valid.

Without loss of generality, we may assume $h_{l-3}=q$ and $h_{l-2}=p$. Observe that $C_{p,h}$ exists and $(q,h)\in \mathscr {C}\cup \mathscr {D}$. From Lemma 3.1 and Proposition 3.8, we get $k_{1,q-1}=k_{1,p-1}=2$ and there exists a vertex $x_{l-1}'\in P_{(1,p-1),(1,p-1)}(x_{l-2},x)$ with $x_{l-1}'\ne x_{l-1}$. Hence, $x_{l-1},x_{l-1}'\in \Gamma _{2,l}(x_{l-3})$. In view of Lemma 1.2 (i) and Lemma 2.2 (i), we obtain $A_{1,q-1}A_{1,p-1}=2A_{2,l}$. Since $\tilde{\partial }(x_{l-1},z)=\tilde{\partial }(x,x_{1})=(3,l-1)$ and $x\in P_{(1,p-1),(2,l)}(x_{l-1},z)$, there exists a vertex $z''\in P_{(2,l),(1,p-1)}(x,x_{1})$, which implies $\tilde{\partial }(y',z'')=(1,q-1)$. Then $\tilde{\partial }(y',x_{1})=(2,l)$ and $\tilde{\partial }(z,x_{1})=(1,p-1)$, contrary to $h\ne p$. The desired result follows.

In the following, we reach a contradiction based on the above discussion.

By Step 2, we may assume $(p,h)\notin \mathscr {C}\cup \mathscr {D}$. It follows from Step 1 and Lemma 3.1 that $k_{1,h-1}=1$. In view of Theorem 2.1 (i), we have $\partial (y,x_{1})=\partial (y',x_{1})=2$.

Case 1. $\partial (x_{1},y)=l$.

Since $y'\in P_{(1,p-1),(1,q-1)}(x,z)$, there exists a vertex $z'\in P_{(1,p-1),(1,q-1)}(y,x_{1})$. It follows from Lemma 2.2 (i) that $k_{1,p-1}=2$. By Theorem 2.1 (i), we have $\tilde{\partial }(x,z')=(2,l)$, which implies $\Gamma _{1,q-1}\Gamma _{1,p-1}=\{\Gamma _{2,l}\}$. In view of Lemma 1.2 (i), $k_{1,q-1}=p_{(1,q-1),(1,p-1)}^{(2,l)}$. Observe $z\in P_{(1,p-1),(1,h-1)}(y,x_{1})$. Lemma 1.2 (iv) implies $k_{1,q-1}=1$. Since $k_{1,h-1}=1$, we obtain $\partial (x_{1},y')<l$. Note that $(q,h)\in \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$. From Lemma 3.1, one has $(q,h)\in \mathscr {B}$. By Theorem 2.1 (ii), we get $\tilde{\partial }(z,z')=(1,q-1)$ or $\tilde{\partial }(z',z)=(1,h-1)$, contrary to Lemma 2.2 (iv).

Case 2. $\partial (x_{1},y)<l$.

Note that $(p,h)\in \mathscr {B}$. By Lemma 3.2, $(1,p-1)$ is mixed and $p=h+1=5$. Since $k_{1,3}=1$, one gets $k_{1,4}=p_{(1,4),(1,4)}^{(1,3)}$ from Theorem 2.1 (ii) and Lemma 1.2 (ii). If $\partial (x_{1},y')=l$, then there exists a vertex $w\in P_{(1,q-1),(1,4)}(y',x_{1})$, which implies $\tilde{\partial }(z,w)=(1,4)$, contrary to Lemma 2.2 (iv). Hence, $\partial (x_{1},y')<l$.

Pick a vertex $w'\in \Gamma _{1,3}(y)$. Since $k_{1,4}=p_{(1,4),(1,4)}^{(1,3)}$, one has $\tilde{\partial }(z,w')=\tilde{\partial }(w',x_{1})=(1,4)$. The fact that $\partial (x_{1},y')<l$ implies $(q,4)\in \mathscr {C}\cup \mathscr {D}$. By Proposition 3.8, $C_{q,4}$ exists, or $C_{q-1,4}$ exists and $(1,q-1)$ is mixed. In view of Lemma 3.3 or 3.4, we get $q=\partial (x_{1},y')<\partial (z,x)\le 1+\partial (w',x)=q+1$. Thus, $l=q+1$.

Suppose that $C_{q,4}$ exists. Pick a vertex $x_{2}'\in P_{(1,3),(4,1)}(w',x_{1})$. Then $\tilde{\partial }(x,x_{2}')=(1,q-1)$. By Lemma 3.1, there exists a circuit $(x_{2}',x_{3}',\ldots ,x_{l}'=x)$ consisting of arcs of type $(1,q-1)$. Since $(z,x_{1},x_{2}',x_{3}',\ldots ,x_{l}')$ is a shortest path, $\tilde{\partial }(x_{1},x_{3}')=(2,l)$. It follows that $\tilde{\partial }(x_{l-1}',z)=\tilde{\partial }(x,x_{1})=(3,l-1)$. The fact that $x\in P_{(1,q-1),(2,l)}(x_{l-1}',z)$ and $\partial (x_{1},y)<l$ imply $\tilde{\partial }(x_{2}',x_{1})=(2,l)$, a contradiction.

Suppose that $C_{q-1,4}$ exists and $(1,q-1)$ is mixed. Since (1, 4) is mixed, by Lemma 3.1, we obtain $q\ge 7$. It follows from Lemma 3.4 that there exists a vertex $y_{1}$ such that $\tilde{\partial }(w',y_{1})=(1,q-2)$ and $\partial (y_{1},x)=q-1$. By Proposition 3.8, we have $(5,q-1)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$. Theorem 2.1 (i) implies $\tilde{\partial }(z,y_{1})=(2,l)=(2,q+1)$. In view of $w'\in P_{(1,4),(1,q-2)}(z,y_{1})$, there exists a vertex $y''\in P_{(1,4),(1,q-2)}(x,z)$. By Lemma 3.1 and Lemma 3.6 (ii), we get $\tilde{\partial }(y'',y')=(1,q-1)$, contrary to Lemma 2.2 (iv).

By the above discussion, we finish the proof of Proposition 1.3 for the case A.

4.2 The case B

Let $(z=x_{0},x_{1},\ldots ,x_{l}=x)$ be a path. For each i, write $h_{i}=\partial (x_{i+1},x_{i})+1$. Note that $h_{i}\in \{2,q,p\}$.

Step 1

Show that $|\{i\mid h_{i}\ne 2,~0\le i\le l-1\}|\ge 2$ and $|\{h_{i}\mid 0\le i\le l-1\}|\ge 2$.

Suppose that $h_{j}=2$ for $0\le j\le l-2$. It follows from Lemma 2.3 that $l=3$ or 4. In view of Lemma 2.2 (i), we obtain $k_{1,1}=2$. By Lemma 2.3, $\tilde{\partial }(x_{0},x_{2})=(2,2)$ and $\Delta _{2}$ is not isomorphic to $C_{3}$. If (1, 3) is pure, then there exists a vertex $x_{1}'\in P_{(1,3),(1,3)}(x_{2},x_{0})$, a contradiction. Then $(1,3)\notin \tilde{\partial }(\Gamma )$ or (1, 3) is mixed. Since $(q,p)\notin \mathscr {B}$, we get $\{q,p\}\ne \{3,4\}$ from Theorem 2.1 (ii). By Lemma 2.5 (ii), one has $l=4$ and $5\in \{q,p\}$. Lemma 2.3 and Theorem 2.1 (i) imply $\partial (y,x_{1})=\partial (y',x_{1})=2$. Since $\partial (x_{1},y)\le 4$ and $\partial (x_{1},y')\le 4$, (1, 4) is mixed from Lemma 3.5 (i). By Theorem 2.1 (ii), (1, 3) is pure, a contradiction. Therefore, the first statement is valid. The second statement follows from Lemma 2.6 and Theorem 2.1 (iii).

Step 2

Show that $k_{1,q-2}=1$ if $|\{i\mid h_{i}=q\}|\ge 2$ and $(1,q-1)$ is mixed.

Without loss of generality, we may assume that $h_{l-2}=h_{l-1}=q$. Note that $\partial (x_{l-2},x)=2$. By Theorem 2.1 (ii) and Lemma 2.2 (ii), we have $|(\Gamma _{1,q-1})^{2}|=2$ and $k_{1,q-1}=2$. Suppose $p_{(1,q-1),(1,q-1)}^{(1,q-2)}=1$. It follows from Lemma 2.2 (iii) that there exists a vertex $x_{l-1}'\in P_{(1,q-1),(1,q-1)}(x_{l-2},x)$ such that $\tilde{\partial }(x_{l-1}',y)=(1,q-2)$. Pick a vertex $x'\in P_{(1,q-2),(1,q-1)}(x_{l-2},y)$. By Theorem 2.1 (i), one gets $\partial (x',z)=2$, contrary to $\partial (z,x')<l$. Hence, $p_{(1,q-1),(1,q-1)}^{(1,q-2)}=2$. In view of Lemma 1.2 (i) and (v), we obtain $k_{1,q-2}=1$.

Step 3

Show that $\partial (x_{l-1},z)\ge 2$.

Suppose for the contrary that $\partial (x_{l-1},z)=1$.

Case 1.$(1,l-1)$ is mixed.

By Theorem 2.1 (ii), $(1,l-2)$ is pure and there exists a vertex $x_{l-2}'$ such that $\tilde{\partial }(x_{l-2}',x_{l-1})=(1,l-1)$ and $\tilde{\partial }(x_{l-2}',z)=(1,l-2)$. Observe that $l-1,l\in \{2,q,p\}$. Since $(q,p)\notin \mathscr {B}$, $l=3$. From Lemma 2.5 (ii), $\{q,p\}=\{3,4\}$. Without loss of generality, we may assume $p=4$. By Theorem 2.1 (ii), (1, 3) is pure. In view of $\partial (x_{l-2}',y)\le 3$ and Lemma 3.5 (i), we get $\partial (y,x_{l-2}')=1$. It follows from Lemma 2.3 and Theorem 2.1 (i) that $\tilde{\partial }(y,x_{l-2}')=(1,3)$ and $\partial (x,x_{l-2}')=2$, contrary to $\partial (x_{l-2}',x)<l$.

Case 2.$(1,l-1)$ is pure.

Observe that $l\in \{q,p\}$.

Case 2.1.$h_{l-1}\ne 2$.

Without loss of generality, we may assume $h_{l-1}=q$. By $l\ge 3$ and Lemma 2.2 (iv), one has $l=p$, which implies $h_{j}=p$ for $0\le j\le l-2$. In view of Lemma 2.2 (i), $k_{1,p-1}=2$.

We claim that $k_{1,q-1}=2$ and there exists $z'\in P_{(1,p-1),(1,p-1)}(y,x_{1})\setminus \{z\}$. Lemma 2.4 implies $\Delta _{p}\simeq \mathrm{Cay}(\mathbb {Z}_{2p},\{1,p+1\})$ or $\mathrm{Cay}(\mathbb {Z}_{p}\times \mathbb {Z}_{p},\{(1,0),(0,1)\})$. Suppose $\Delta _{p}\simeq \mathrm{Cay}(\mathbb {Z}_{2p},\{1,p+1\})$. Since $C_{p,q}$ does not exist, from Lemma 2.2 (ii), we have $|(\Gamma _{1,q-1})^{2}|=2$ and $k_{1,q-1}=2$. It follows from Lemma 2.4 (i) that the claim is valid. Suppose $\Delta _{p}\simeq \mathrm{Cay}(\mathbb {Z}_{p}\times \mathbb {Z}_{p},\{(1,0),(0,1)\})$. Then $|P_{(1,p-1),(p-1,1)}(x_{l-1},y)|=1$. Lemma 1.2 (v) and Lemma 2.2 (i) imply $k_{\tilde{\partial }(x_{l-1},y)}=2$. By Lemma 1.2 (i), $k_{1,q-1}=2$. Since $\tilde{\partial }(x_{l-1},x_{1})=(2,p-2)$, one has $\tilde{\partial }(y,x_{1})\ne (2,p-2)$ from Lemma 2.4 (ii). In view of Lemma 2.2 (iii), the claim is valid.

By Theorem 2.1 (i), $\tilde{\partial }(x,z')=(2,l)$. In view of Lemma 1.2 (i) and Lemma 2.2 (i), one gets $A_{1,q-1}A_{1,p-1}=2A_{2,l}$, which implies that $\tilde{\partial }(x,x_{l-1})=(1,q-1)$, contrary to $q\ne 2$.

Case 2.2.$h_{l-1}=2$.

Without loss of generality, we may assume $l=q$. By $x_{l-1}\ne y'$ and Lemma 2.2 (i), one gets $k_{1,q-1}=2$. Since $z\in P_{(1,q-1),(q-1,1)}(x_{l-1},y')$, we have $\tilde{\partial }(y',x_{l-1})=(2,2)$ from Lemma 2.2 (iv) and Lemma 2.3. In view of $x\in P_{(1,1),(1,p-1)}(x_{l-1},y')$, there exists a vertex $x''\in P_{(1,p-1),(1,1)}(y',x_{l-1})$. Then $x''\ne x$ and $k_{1,1}=2$. The fact that $\tilde{\partial }(x_{l-1},y')=(2,2)$ implies $p_{(1,q-1),(q-1,1)}^{(1,1)}=p_{(1,1),(1,q-1)}^{(1,q-1)}=0$. Since $(1,q-1)$ is pure, by Lemma 2.3, Theorem 2.1 (i) and Lemma 3.5 (i), we obtain $A_{1,1}A_{1,q-1}=2A_{2,q}$. Hence, $y'\in P_{(1,1),(1,q-1)}(x'',z)$, a contradiction.

Step 4

Show that $p_{(1,s-1),(1,1)}^{(2,l-1)}\ne 0$ for some $s>2$ if $\partial (x_{l-1},z)=2$.

Pick a path $(x_{l-1},w,z)$ such that $\tilde{\partial }(x_{l-1},w)=(1,s-1)$, $\tilde{\partial }(w,z)=(1,t-1)$ and $s\ge t$. By Step 1, we may assume $h_{0}=q$. If $t=2$, then $s>2$ since $l\ge 3$, and the desired result holds. Suppose $t\ne 2$. Since $(q,p)\notin \mathscr {C}\cup \mathscr {D}$, from Lemmas 3.3 and 3.4, we have $t=s$ or $(s,t)\in \mathscr {B}$.

Case 1.$s=t$.

Since $\partial (x_{1},w)<l$, by Theorem 2.1 (i), one has $t=q$ or $(t,q)\in \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$.

Case 1.1.$t=q$.

Suppose $\tilde{\partial }(x_{l-1},y')=(1,q-1)$. By Theorem 2.1 (i), $h_{l-1}=p$. It follows from Theorem 2.1 (ii) and (iii) that $(1,p-1)$ is mixed and $q=p-1$, contrary to $(q,p)\notin \mathscr {B}$. Since $w\ne y'$, by Lemma 2.2 (i), one gets $k_{1,q-1}=2$ and $p_{(1,q-1),(1,q-1)}^{(2,l-1)}=1$.

Case 1.1.1.$(1,q-1)$ is pure.

By Lemma 2.4, one has $|(\Gamma _{1,q-1})^{2}|=2$ and $p_{(1,q-1),(1,q-1)}^{(2,q-2)}=1$. In view of Lemma 2.2 (iii), we get $\tilde{\partial }(x_{l-1},z)=(2,q-2)$, which implies $l=q-1$ and $\tilde{\partial }(w,x_{1})=(2,q-2)$. It follows from Theorem 2.1 (ii) that $\tilde{\partial }(y',x_{1})=(2,q-1)$. By Lemma 2.2 (iii) again, we obtain $p_{(1,q-1),(1,q-1)}^{(2,q-1)}=2$. Then $\tilde{\partial }(x,y')=(1,q-1)$, a contradiction.

Case 1.1.2.$(1,q-1)$ is mixed.

By Theorem 2.1 (ii), $(1,q-2)$ is pure. Since $p_{(1,q-1),(1,q-1)}^{(2,l-1)}=1$, one gets $p_{(1,q-1),(1,q-1)}^{(1,q-2)}=2$ from Lemma 2.2 (ii) and (iii). In view of Lemma 1.2 (i) and (v), one has $k_{1,q-2}=1$. Lemma 2.5 (iii) implies $\tilde{\partial }(x_{l-1},z)=(2,q-1)$ and $l=q$. Pick vertices $x_{l-2}'\in P_{(q-1,1),(1,q-2)}(x_{l-1},w)$ and $x_{1}'\in P_{(1,q-2),(q-1,1)}(w,z)$. Note that $\tilde{\partial }(x_{l-2}',x_{1}')\ne (2,q-3)$. By $k_{1,q-2}=1$, we obtain $l=q=3$. In view of Lemma 2.5 (ii) and Theorem 2.1 (ii), we get $p=4$, which implies that (1, 3) is pure. Since $\tilde{\partial }(x_{l-1},z)=(2,2)$, there exists a vertex $x'\in P_{(1,3),(1,3)}(z,x_{l-1})$. Then $(x_{l-1},w,z,x')$ is a circuit containing arcs of types (1, 2) and (1, 3), a contradiction.

Case 1.2.$(t,q)\in \mathscr {B}$ and $t=q-1$.

Note that $q>3$. Theorem 2.1 (ii) implies that $(1,q-2)$ is pure. Since $q-1\notin \{2,q,p\}$, we have $\tilde{\partial }(x_{l-1},z)\ne (2,q-3)$. In view of Lemma 2.2 (ii), we get $|(\Gamma _{1,q-2})^{2}|=2$ and $k_{1,q-2}=2$, which imply $p_{(1,q-2),(1,q-2)}^{(2,q-3)}=1$ from Lemma 2.4. By Lemma 2.2 (iii), we obtain $p_{(1,q-2),(1,q-2)}^{(2,l-1)}=2$. In view of Lemma 1.2 (i) and (v), one gets $k_{2,l-1}=1$.

Since $k_{1,q-2}=2$ and $(q,p)\notin \mathscr {D}$, by Lemma 3.1, we have $(q-1,p)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$. It follows from $\partial (w,x_{l-2})<l$ and Theorem 2.1 (i) that $h_{l-2}\ne p$. Hence, $h_{i}\ne p$ for $0\le i\le l-2$. In view of Step 2, we get $|\{j\mid h_{j}=q\}|<2$. By Step 1, one has $h_{l-1}=p$.

Since $\partial (x_{1},x)<l$ and $(q-1,p)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$, we have $\tilde{\partial }(y',x_{1})\ne (1,q-2)$ from Theorem 2.1 (i). In view of Theorem 2.1 (ii), we get $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$, which implies $|(\Gamma _{1,q-1})^{2}|=2$ and $k_{1,q-1}=2$ from Lemma 2.2 (ii). Since $k_{1,q-2}=2$, by Lemma 1.2 (i) and (v), we get $p_{(1,q-1),(1,q-1)}^{(1,q-2)}=1$. It follows from Lemma 2.2 (iii) that there exists a vertex $z'\in P_{(1,q-1),(1,q-1)}(y',x_{1})\setminus \{z\}$. In view of Theorem 2.1 (i), we obtain $z'\in \Gamma _{2,l}(x)$ and $\Gamma _{1,p-1}\Gamma _{1,q-1}=\{\Gamma _{2,l}\}$. Since $x\in P_{(1,p-1),(2,l)}(x_{l-1},z)$ and $k_{2,l-1}=1$, we obtain $k_{1,p-1}=2$ from Lemma 1.2 (i) and Lemma 2.2 (i). Hence, $p_{(1,p-1),(1,q-1)}^{(2,l)}=2$ and there exists a vertex $y''\in P_{(1,p-1),(1,q-1)}(x,z)$ such that $\tilde{\partial }(y'',x_{1})=(1,q-2)$. By Theorem 2.1 (i), one has $\partial (x,x_{1})=2$, contrary to $\partial (x_{1},x)<l$.

Case 1.3.$(t,q)\in \mathscr {B}$ and $t=q+1$.

Since (1, q) is mixed, $(1,q-1)$ is pure and $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$ from Theorem 2.1 (ii). By $\partial (x_{l-1},z)=2$ and Lemma 2.2 (ii), we have $|(\Gamma _{1,q})^{2}|=2$ and $k_{1,q}=2$. If $p_{(1,q),(1,q)}^{(2,l-1)}=2$, then there exists a vertex $w'\in P_{(1,q),(1,q)}(x_{l-1},z)$ such that $\tilde{\partial }(y',w')=(1,q)$; if $p_{(1,q),(1,q)}^{(2,l-1)}=1$, by Lemma 2.2 (iii), then $p_{(1,q),(1,q)}^{(1,q-1)}=2$ and $\tilde{\partial }(y',w)=(1,q)$. Without loss of generality, we may assume $\tilde{\partial }(y',w)=(1,q)$.

By Theorem 2.1 (i), we have $\partial (x,w)=2$. Since $k_{1,q}=2$ and $C_{q,p}$ does not exist, we obtain $(q+1,p)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$ from Lemma 3.1. It follows that $l\le \partial (w,x)\le \partial (w,x_{l-1})+1=q+1$. In view of Lemma 2.5 (i), one gets $l-1=\partial (z,x_{l-1})>q-1$. Then $\partial (w,x)=l$. Since $x\in P_{(l,2),(2,l)}(w,z)$, by Lemma 2.2 (iv), one has $q=1$, a contradiction.

Case 1.4.$(t,q)\in \mathscr {C}\cup \mathscr {D}$.

Suppose that $(1,t-1)$ is pure. By Lemma 3.1, $k_{1,t-1}=1$ or $\Delta _{t}\simeq \mathrm{Cay}(\mathbb {Z}_{2t},\{1,t+1\})$, which implies $\tilde{\partial }(x_{l-1},z)=(2,t-2)$. Hence, $t=p$, a contradiction.

Suppose that $(1,t-1)$ is mixed. It follows from Lemma 3.1 that $k_{1,t-1}=1$ or $C_{t-1,q}$ exists. If $k_{1,t-1}=1$, by Theorem 2.1 (ii), then $\partial (x_{l-1},z)=1$, a contradiction; if $C_{t-1,q}$ exists, by Lemma 3.1 and Lemma 3.6 (ii), then $\partial (x_{l-1},z)=1$, a contradiction.

Case 2.$(s,t)\in \mathscr {B}$.

Note that $(1,s-1)$ is mixed and $s=t+1$. By Theorem 2.1 (ii), $(1,t-1)$ is pure. Since $\partial (x_{1},w)<l$, from Theorem 2.1 (i), one has $t=q$ or $(t,q)\in \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$.

Case 2.1.$t=q$.

Note that $s=q+1$ and $l-1=\partial (z,x_{l-1})\ge q-1$. Since (1, q) is mixed, by Theorem 2.1 (ii), one has $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$. Pick a vertex $x_{l-2}'\in P_{(q,1),(1,q-1)}(x_{l-1},w)$. The fact that $q+1\notin \{2,q,p\}$ implies $\tilde{\partial }(x_{l-2}',z)\ne (2,q-2)$. By Lemma 2.2 (ii), we get $|(\Gamma _{1,q-1})^{2}|=2$ and $k_{1,q-1}=2$. In view of Lemma 2.4, we get $p_{(1,q-1),(1,q-1)}^{(2,q-2)}=1$. Since $\tilde{\partial }(x_{l-2}',z)\ne (2,q-2)$, we obtain $\tilde{\partial }(x_{l-2}',y')=(1,q-1)$ from Lemma 2.2 (iii).

Note that $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$ and $k_{1,q-1}=2$. By Lemma 1.2 (i) and Lemma 2.2 (i), we obtain $k_{1,q}=2$. Since $C_{q,p}$ does not exist, $(q+1,p)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$ from Lemma 3.1. Since $x\in P_{(1,h_{l-1}-1),(1,p-1)}(x_{l-1},y')$, there exists a vertex $x'\in P_{(1,p-1),(1,h_{l-1}-1)}(x_{l-1},y')$. In view of Theorem 2.1 (i), we have $\partial (x_{l-2}',x')=2$. Since $q\le l$ and $l\le \partial (x',x_{l-2}')\le 1+\partial (y',x_{l-2}')=q$, we get $\tilde{\partial }(x_{l-2}',x')=(2,l)$.

Suppose $w=y'$. Since $x\in P_{(1,h_{l-1}-1),(1,p-1)}(x_{l-1},w)$ and $p\ne q+1$, by Theorem 2.1 (i), we have $h_{l-1}=p$. Theorem 2.1 (ii) and (iii) imply that (1, q) is pure, a contradiction. Suppose $w\ne y'$. By $p_{(1,q-1),(1,p-1)}^{(2,l)}\ne 0$ and $p>2$, we get $w\in P_{(1,q-1),(1,p-1)}(x_{l-2}',x')$. Since $x'\in P_{(1,p-1),(p-1,1)}(x_{l-1},w)$, from Lemma 2.2 (iv), we obtain $q=1$, a contradiction.

Case 2.2.$(t,q)\in \mathscr {B}$.

Note that $t=q-1$ and $s=q$. By Theorem 2.1 (ii), $p_{(1,q-1),(1,q-1)}^{(1,q-2)}\ne 0$ and $(1,q-2)$ is pure. Lemma 2.2 (i) implies $k_{1,q-2}=1$ or 2.

Case 2.2.1.$k_{1,q-2}=1$.

Since $\partial (z,x_{l-1})\ge q-2$, $l\ge q-1$. Pick a vertex $x_{l-2}'\in P_{(q-1,1),(1,q-2)}(x_{l-1},w)$. The fact that $(1,q-2)$ is pure implies that $\tilde{\partial }(x_{l-2}',z)=(2,q-3)$ and $l=q-1$, contrary to $q-1\notin \{2,q,p\}$.

Case 2.2.2.$k_{1,q-2}=2$.

Since $\partial (w,x_{l-2})<l$, one gets $h_{l-2}\ne p$ from Theorem 2.1 (i). Then $h_{j}\ne p$ for $0\le j\le l-2$. Step 2 implies $|\{i\mid h_{i}=q\}|<2$. It follows from Step 1 that $h_{l-1}=p$ and $h_{j}=2$ for $1\le j\le l-2$.

Since $(q,p)\notin \mathscr {D}$ and $(1,q-2)$ is pure, by Lemma 3.1, one has $(q-1,p)\notin \mathscr {B}\cup \mathscr {C}\cup \mathscr {D}$. In view of $w\in P_{(1,q-1),(1,q-2)}(x_{l-1},z)$, there exists a vertex $w'\in P_{(1,q-2),(1,q-1)}(x_{l-1},z)$. It follows from Theorem 2.1 (iii) that $w'\ne y'$. Observe $\partial (x_{1},x)<l$. Theorem 2.1 (i) and Lemma 2.2 (i) imply $\tilde{\partial }(y',x_{1})\ne (1,q-2)$ and $\tilde{\partial }(w',x_{1})=(1,q-2)$. Since $(1,q-2)$ is pure, by Lemma 2.3, we have $\tilde{\partial }(x_{l-1},x_{1})=(2,2)$ and $l=q=4$. In view of $\partial (x_{2},w')\le 2$ and Lemma 3.5 (i), we obtain $\partial (w',x_{2})=1$. Then $(x_{2},x_{3},w')$ is a circuit containing arcs of types (1, 1) and (1, 2), a contradiction.

Case 2.3.$(t,q)\in \mathscr {C}\cup \mathscr {D}$.

Observe that $(1,t-1)$ is pure and (1, t) is mixed. By Lemma 3.1, we have $k_{1,t-1}=1$ or $\Delta _{t}\simeq \mathrm{Cay}(\mathbb {Z}_{2t},\{1,t+1\})$. It follows from Theorem 2.1 (ii) that $p_{(1,t),(1,t)}^{(1,t-1)}\ne 0$. Pick a vertex $x_{l-2}'\in P_{(t,1),(1,t-1)}(x_{l-1},w)$. In view of $k_{1,t-1}=1$ or Lemma 2.4 (i), one gets $\tilde{\partial }(x_{l-2}',z)=(2,t-2)$. Note that $l-1=\partial (z,x_{l-1})\ge t-1$. Hence, $l=t$. Since $(1,t-1)$ is pure, we obtain $t=p$, contrary to $(q,p)\notin \mathscr {C}\cup \mathscr {D}$.

We complete the proof of Step 4.

Step 5

Show that $(1,s-1)$ is pure if $\partial (x_{l-1},z)=2$.

Suppose for the contrary that $(1,s-1)$ is mixed. Theorem 2.1 (ii) implies that $p_{(1,s-1),(1,s-1)}^{(1,s-2)}\ne 0$ and $(1,s-2)$ is pure. Pick vertices $w\in P_{(1,s-1),(1,1)}(x_{l-1},z)$ and $x_{l-2}'\in P_{(s-1,1),(1,s-2)}(x_{l-1},w)$.

Case 1.$s>3$.

By Lemma 2.5 (ii), we have $\partial (z,x_{l-1})\ne s-2$. Since $(q,p)\notin \mathscr {B}$, one gets $\{s-1,s\}\ne \{q,p\}$. It follows from Lemma 3.5 (ii) that $\tilde{\partial }(x_{l-1},z)=(2,s-1)$ and $l=s$. Since $(1,s-2)$ is pure, by Lemma 2.3 and Theorem 2.1 (i), we obtain $\partial (x_{l-2}',z)=2$. From Lemma 3.5 (i), we obtain $\tilde{\partial }(x_{l-2}',z)=(2,s-1)$. Since $z\in P_{(2,s-1),(s-1,2)}(x_{l-2}',x_{l-1})$, by Lemma 2.2 (iv), one has $s=2$, a contradiction.

Case 2.$s=3$.

Note that $l\le 4$ and $x_{l-2}'\ne z$. Lemma 2.2 (i) implies $k_{1,1}=2$. Without loss of generality, we may assume $p\ne 3$. By Lemma 2.5 (ii) and Theorem 2.1 (ii), we have $p=4$ or 5.

Suppose $p=4$. Theorem 2.1 (ii) implies that (1, 3) is pure. If $\tilde{\partial }(x_{l-2}',z)=(2,2)$, by $p_{(1,3),(1,3)}^{(2,2)}\ne 0$, then there exists a vertex $w'\in P_{(1,3),(1,3)}(x_{l-2}',z)$, which implies that $(x_{l-2}',w',z,w)$ is a circuit containing arcs of types (1, 1) and (1, 3), a contradiction. It follows from Lemma 2.3 that $\tilde{\partial }(x_{l-2}',z)=(1,1)$. Since $\partial (x_{l-2}',y)\le 3$, by Theorem 2.1 (i) and Lemma 3.5 (i), we get $\tilde{\partial }(y,x_{l-2}')=(1,3)$. Then $\partial (x,x_{l-2}')=2$, contrary to $l\ge 3$.

Suppose $p=5$. By Lemma 2.5 (ii) and Theorem 2.1 (ii), one has $l=4$. Since $\partial (w,x)\le 3$, we obtain $\partial (y,w)=2$ from Lemma 2.3 and Theorem 2.1 (i). In view of $\partial (w,y)\le 4$ and Lemma 3.5 (i), (1, 4) is mixed. It follows from Theorem 2.1 (ii) that (1, 3) is pure. By Lemma 2.3, we get $\tilde{\partial }(x_{l-2}',z)=(2,2)$, which implies that there exists a vertex $w'\in P_{(1,3),(1,3)}(x_{l-2}',z)$. Hence, $(x_{l-2}',w',z,w)$ is a circuit containing arcs of types (1, 1) and (1, 3), a contradiction.

Step 6

Show that $\{h_{l-1},s\}=\{q,p\}$ if $\partial (x_{l-1},z)=2$.

By Step 5, $(1,s-1)$ is pure. From Step 4 and Lemma 3.5 (i), we get $\tilde{\partial }(x_{l-1},z)=(2,s)$ and $l=s+1$, which imply $s\in \{q,p\}$. Pick a vertex $w\in P_{(1,s-1),(1,1)}(x_{l-1},z)$. Without loss of generality, we may assume $s=q$.

Suppose $h_{l-1}=2$. Observe that $\tilde{\partial }(x_{l-1},z)=(2,q)=(2,l-1)$. By Lemma 2.3 and Theorem 2.1 (i), we get $\partial (x_{l-1},y)=2$. In view of Lemma 3.5 (i), one has $\tilde{\partial }(x_{l-1},y)=(2,q)$. It follows from Lemma 2.2 (iv) that $p=2$, a contradiction. Hence, $h_{l-1}\in \{q,p\}$.

Suppose $h_{l-1}=s$. Since $l\ge 3$, one gets $w\ne x$ and $k_{1,q-1}=2$ from Lemma 2.2 (i). By Lemma 2.2 (iv) and $x_{l-1}\in P_{(q-1,1),(1,q-1)}(x,w)$, we have $\partial (x,w)=\partial (w,x)$. In view of $z\in P_{(2,l),(1,1)}(x,w)$, there exists $z'\in P_{(1,1),(2,l)}(w,x)$. Since $l\ge 3$, we get $z'\ne z$ and $k_{1,1}=2$. By $\tilde{\partial }(w,x)\ne (1,1)$, we obtain $p_{(1,q-1),(q-1,1)}^{(1,1)}=p_{(1,q-1),(1,1)}^{(1,q-1)}=0$. In view of Lemma 2.3, Theorem 2.1 (i) and Lemma 3.5 (i), one has $\Gamma _{1,q-1}\Gamma _{1,1}=\{\Gamma _{2,q}\}$. Since $p_{(1,q-1),(1,1)}^{(2,q)}=2$ from Lemma 1.2 (i), one obtains $\tilde{\partial }(x,z)=(1,1)$, a contradiction. Thus, $h_{l-1}=p$.

Step 7

For $a,b\in \{2,q,p\}$ and $a<b$, show that $p_{(1,b-1),(2,l)}^{(3,l-1)}\ne 0$ if $p_{(1,a-1),(2,l)}^{(3,l-1)}=0$.

Without loss of generality, we may assume $b=q$. We claim that $h_{i}=q$ for some $i\in \{0,1,\ldots ,l-1\}$. Assume the contrary, namely $h_{i}\ne q$ for each i. Suppose $a=2$. By Step 1, we may assume $h_{l-1}=2$. It follows from Steps 3, 4 and 6 that $\tilde{\partial }(x_{l-1},z)=(3,l-1)$, contrary to $p_{(1,1),(2,l)}^{(3,l-1)}=0$. Suppose $a=p$. By Step 1, we may assume $h_{l-2}=h_{l-1}=p$. It follows from Steps 3, 4 and 6 that $\partial (x_{l-1},z)=2$ and there exists a vertex $w\in P_{(1,q-1),(1,1)}(x_{l-1},z)$. Theorem 2.1 (i) implies $\partial (x_{l-2},w)=2$, contrary to $\partial (w,x_{l-2})<l$. So our claim is valid.

Without loss of generality, we may assume $h_{l-1}=q$. It suffices to show that $\partial (x_{l-1},z)=3$.

Suppose $\partial (x_{l-1},z)=2$. It follows from Steps 4–6 that $(1,p-1)$ is pure and there exists a vertex $w'\in P_{(1,1),(1,p-1)}(x_{l-1},z)$. By Lemma 3.5 (i), we have $\tilde{\partial }(x_{l-1},z)=(2,p)$ and $l=p+1$. Let $(y_{0}=z,y_{1},\ldots ,y_{l-2}=w')$ be a path consisting of arcs of type $(1,p-1)$. Since $x_{l-1}\in P_{(1,1),(1,q-1)}(w',x)$, there exists $x_{l-1}'\in P_{(1,q-1),(1,1)}(w',x)$. Note that $(z=y_{0},y_{1},\ldots ,y_{l-2},x_{l-1}',x)$ is a shortest path. Then $\tilde{\partial }(y_{l-3},x_{l-1}')=(2,l)$. Hence, $\tilde{\partial }(x,y_{1})=\tilde{\partial }(x_{l-1}',z)=(3,l-1)$, contrary to $a\in \{2,p\}$. By Step 3, we obtain $\partial (x_{l-1},z)=3$, as desired.

Based on the above discussion, we consider two cases, and reach a contradiction, respectively.

Case 1. $p_{(2,l),(1,1)}^{(3,l-1)}\ne 0$.

Pick a vertex $y_{1}\in P_{(3,l-1),(1,1)}(x,z)$. By Step 7, we may assume $p_{(1,q-1),(2,l)}^{(3,l-1)}\ne 0$. Then there exist vertices $z'\in P_{(2,l),(1,q-1)}(x,y_{1})$ and $y_{1}'\in P_{(1,p-1),(1,q-1)}(x,z')$. It follows from Lemma 2.2 (i) that $k_{2,l}=2$. Observe that $x\in P_{(l,2),(2,l)}(z,z')$. By Lemma 2.2 (iv) and Lemma 2.3, we get $\tilde{\partial }(z',z)=(2,2)$. Lemma 2.5 (ii) and Theorem 2.1 (ii) imply $q=3$.

By $\tilde{\partial }(z',z)=(2,2)$ and Lemma 3.5 (i), (1, 2) is mixed, which implies $p_{(1,2),(1,2)}^{(1,1)}\ne 0$ from Theorem 2.1 (ii). Since $\partial (y_{1}',y_{1})=2$, by Lemma 2.2 (ii), we have $|(\Gamma _{1,2})^{2}|=2$ and $k_{1,2}=2$. In view of $\tilde{\partial }(z,z')=(2,2)$, $p_{(1,2),(1,2)}^{(1,1)}=1$. It follows from Lemma 2.2 (iii) that there exists a vertex $z''\in P_{(1,2),(1,2)}(y_{1}',y_{1})\setminus \{z'\}$. In view of Theorem 2.1 (i), we get $z''\in \Gamma _{2,l}(x)$. Since $k_{2,l}=2$, we obtain $z''=z$, a contradiction.

Case 2. $p_{(2,l),(1,1)}^{(3,l-1)}=0$.

We claim that any shortest path from z to x does not contain an edge. Suppose for the contrary that $h_{l-1}=2$. It follows from Steps 3, 4 and 6 that $\partial (x_{l-1},z)=3$, contrary to $p_{(2,l),(1,1)}^{(3,l-1)}=0$. Thus, the claim is valid. By Step 7, we have $p_{(1,q-1),(2,l)}^{(3,l-1)}\ne 0$ and $p_{(1,p-1),(2,l)}^{(3,l-1)}\ne 0$. Pick a vertex $y_{l-1}\in P_{(q-1,1),(3,l-1)}(x,z)$. It follows that there exist vertices $x'\in P_{(1,p-1),(2,l)}(y_{l-1},z)$ and $y''\in P_{(1,p-1),(1,q-1)}(x',z)$. By Lemma 2.2 (i), $k_{2,l}=2$. In view of Lemma 1.2 (i), one obtains $k_{1,q-1}=2$ or $k_{1,p-1}=2$.

Case 2.1. $k_{1,q-1}=2$ and $k_{1,p-1}=2$.

In view of the claim and Step 1, there exists a vertex $z_{1}$ such that $\tilde{\partial }(z,z_{1})=(1,p-1)$ and $\partial (z_{1},y_{l-1})=l-2$. By Theorem 2.1 (i), if $y'=y''$, then $x,x'\in \Gamma _{l,2}(z)$; if $y'\ne y''$, then $y',y''\in \Gamma _{l,2}(z_{1})$. In view of Lemma 1.2 (i), we have $A_{1,q-1}A_{1,p-1}=2A_{2,l}$, which implies $\tilde{\partial }(x',y)=(1,q-1)$. Thus, $\tilde{\partial }(y_{l-1},y)=(2,l)$ and $\tilde{\partial }(x,y)=(1,p-1)$, a contradiction.

Case 2.2. $k_{1,q-1}=1$ or $k_{1,p-1}=1$.

Without loss of generality, we may assume $k_{1,q-1}=1$. Then $y'=y''$ and $k_{1,p-1}=2$. It follows from Theorem 2.1 (i) that $\tilde{\partial }(y_{l-1},y')=(2,l)$. In view of $x'\in P_{(1,p-1),(1,p-1)}(y_{l-1},y')$, one gets $p_{(1,p-1),(1,p-1)}^{(2,l)}=1$ and there exists a vertex $y_{0}\in P_{(1,p-1),(1,p-1)}(x,z)$. Since $k_{2,l}=2$, by Lemma 1.2 (i) and Lemma 2.2 (ii), we have $|(\Gamma _{1,p-1})^{2}|=2$. In view of Theorem 2.1 (i), we get $y_{0}\in \Gamma _{2,l}(y_{l-1})$. Then $\tilde{\partial }(x',y_{0})\ne (1,p-1)$. Since $p_{(1,p-1),(1,p-1)}^{(2,l)}=1$, we obtain $y\in P_{(1,p-1),(1,p-1)}(x',z)$ and $\tilde{\partial }(y_{l-1},y)\ne (2,l)$. By $\partial (y_{l-1},y)=2$ and Theorem 2.1 (ii), $(1,p-1)$ is pure, which implies $l>p-2$. Then $\tilde{\partial }(y_{l-1},y)=(2,p-2)$, contrary to $x\in P_{(1,q-1),(1,q-1)}(y_{l-1},y)$.

Thus, we finish the proof of Proposition 1.3 for the case B.

5 Subdigraphs

In this section, we focus on the existence of some special subdigraphs of commutative quasi-thin weakly distance-regular digraphs.

Let F be a nonempty subset of R and $x\in V\Gamma$. Set $F(x):=\{y\in V\Gamma \mid (x,y)\in \cup _{f\in F}f\}$, and $F_{q_{1},q_{2},\ldots ,q_{l}}(x)$ is a collection of vertices y satisfying each arc in one of the paths from x to y is of type $(1,q_{1}-1),(1,q_{2}-1),\ldots ,(1,q_{l-1}-1)$ or $(1,q_{l}-1)$. If $\Gamma _{\tilde{i}^{*}}\Gamma _{\tilde{j}}\subseteq F$ for any $\Gamma _{\tilde{i}},\Gamma _{\tilde{j}}\in F$, we say that F is closed. Let $\langle F\rangle$ be the minimum closed subset containing F. We write $\langle \Gamma _{1,q-1}\rangle$ instead of $\langle \{\Gamma _{1,q-1}\}\rangle$.

Proposition 5.1

If $C_{q,h}$ exists, then $\Delta _{q,h}\simeq \mathrm{Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{4},\{(1,0),(0,1),(1,2)\})$ for $q>2$ and $q\ne h$.

Proof

For fixed $x\in V\Gamma$, by Lemma 3.1, there exists an isomorphism $\tau$ from $\mathrm{Cay}(\mathbb {Z}_{2q},\{1,q+1\})$ to $\Delta _{q}(x)$. Write $\tau (a)=(a,0)$ for each $a\in \mathbb {Z}_{2q}$. Suppose that there exists a vertex $(s,0)\in \Gamma _{1,h-1}(0,0)$. From Lemma 2.4 (i), we have $s=q$. Since $(1,0)\in P_{(1,q-1),(q-1,1)}((0,0),(q,0))$ , by Lemma 2.2 (iv), we get $h=2$, contrary to $h\ge 3$. Hence, $\Gamma _{1,h-1}\notin \langle \Gamma _{1,q-1}\rangle$. In view of Lemma 3.1, one obtains $k_{1,h-1}=1$. Since $C_{q,h}$ exists, $V\Delta _{q,h}(x)$ has a partition $F_{q}(x)\dot{\cup }F_{q}(x')$. It follows that $\sigma :F_{q}(x)\rightarrow F_{q}(x')$, $y\mapsto y'$ is an isomorphism from $\Delta _{q}(x)$ to $\Delta _{q}(x')$, where $y'\in \Gamma _{1,h-1}(y)$. Write $\sigma (a,0)=(a,1)$ for each a. Since $C_{q,h}$ exists again, $((a,1),(a+q,0))\in \Gamma _{1,h-1}$. The desired result holds. $\square$

Proposition 5.2

Let $q\ge 3$. If $k_{1,q-1}=2$ and $(1,q-1)$ is pure, then $\Delta _{q}\simeq \mathrm{Cay}(\mathbb {Z}_{2q},\{1,q+1\})$.

Proof

Suppose not. By Lemma 2.4, there exists an isomorphism $\tau$ from $\mathrm{Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{q},\{(1,0),(0,1)\})$ to $\Delta _{q}$.

By Lemma 3.1 and Proposition 1.3, C4, C5 or C6 holds, which implies that $K\subseteq \{(1,1),(1,q-1),(1,q)\}$. If $(1,q)\in \tilde{\partial }(\Gamma )$, then (1, q) is mixed, which implies $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$ and $k_{1,q}=2$ by Lemma 1.2 (i), Lemma 2.2 (i) and Theorem 2.1 (ii).

Step 1

Show that $\Gamma _{1,q-1}\Gamma _{1,1}=\{\Gamma _{2,q}\}$ if $(1,1)\in \tilde{\partial }(\Gamma )$.

Suppose $p_{(1,q-1),(q-1,1)}^{(1,1)}\ne 0$. Note that $\tilde{\partial }_{\Gamma }(\tau (a,b),\tau (a+1,b-1))=(1,1)$. By Lemma 2.2 (i), $k_{1,1}=2$. Observe that $\tau (1,0)\in P_{(1,1),(1,q-1)}(\tau (0,1),\tau (2,0))$ and $(\tau (0,1),\tau (2,0))\notin \Gamma _{1,q-1}\cup \Gamma _{1,1}$. In view of Theorem 2.1 (i) and Lemma 3.5 (i), we get $(\tau (0,1),\tau (2,0))\in \Gamma _{2,q}$, contrary to the fact that $(\tau (2,0),\tau (3,0),\ldots ,\tau (0,0),\tau (0,1))$ is a path of length $q-1$. Thus, $p_{(1,q-1),(q-1,1)}^{(1,1)}=0$. It follows that $\Gamma _{1,q-1}\Gamma _{1,1}=\{\Gamma _{2,q}\}$.

Step 2

Show that $\Gamma _{1,q}\Gamma _{1,1}=\{\Gamma _{2,q+1}\}$ if $(1,1),(1,q)\in \tilde{\partial }(\Gamma )$.

Let x, y, z, w be vertices such that $\tilde{\partial }(x,y)=(1,q)$, $\tilde{\partial }(y,z)=(1,1)$ and $w\in P_{(q,1),(1,q-1)}(x,y)$. By Step 1, $\tilde{\partial }(w,z)=(2,q)$. Since $k_{1,q-1}=2$, from Lemma 1.2 (i) and Lemma 2.2 (i), we obtain $k_{2,q}=2$. Suppose $\partial (x,z)=1$. In view of Lemma 2.3 and Theorem 2.1 (i), one has $\tilde{\partial }(x,z)=(1,q)$. Note that $x\in P_{(1,q),(1,q)}(w,z)$ and $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$. By Lemma 2.2 (ii), we get $(\Gamma _{1,q})^{2}=\{\Gamma _{1,q-1},\Gamma _{2,q}\}$. Since $k_{1,q}=2$, from Lemma 1.2 (i) and (v), we obtain $p_{(1,q),(1,q)}^{(1,q-1)}=1$. In view of Lemma 2.2 (iii), we have $p_{(1,q),(1,q)}^{(2,q)}=2$, which implies $k_{2,q}=1$, a contradiction. Then $\partial (x,z)=2$. Since $\tilde{\partial }(w,z)=(2,q)$, by Lemma 2.2 (iv), we have $\partial (z,x)\ne q$. In view of Lemma 2.5 (ii), $\partial (z,x)=q+1$. Thus, $\Gamma _{1,q}\Gamma _{1,1}=\{\Gamma _{2,q+1}\}$.

Step 3

Show that $(A_{1,q-1})^{2}=A_{2,q-2}+2A_{2,2q-2}$.

In view of Lemma 2.4 (ii) and Theorem 2.1 (ii), we have $(A_{1,q-1})^{2}=A_{2,q-2}+p_{(1,q-1),(1,q-1)}^{(2,t)}A_{2,t}$ with $t\ne q-2$. By Lemma 2.2 (iii), one gets $p_{(1,q-1),(1,q-1)}^{(2,t)}=2$, which implies $k_{2,t}=1$ from Lemma 1.2 (i) and (v). Let $x,y,y',z$ be vertices such that $\tilde{\partial }(x,z)=(2,t)$ and $P_{(1,q-1),(1,q-1)}(x,z)=\{y,y'\}$.

We claim that $\partial (x,x_{1})=3$ for any path $(z=x_{0},x_{1},\ldots ,x_{t}=x)$. Assume the contrary, namely $\partial (x,x_{1})=1$ or 2.

Case 1.$\partial (x,x_{1})=1$.

Since $x_{1}\notin \{y,y'\}$, we have $\tilde{\partial }(x,x_{1})=(1,1)$ or (1, q). If $\tilde{\partial }(x,x_{1})=(1,1)$, by Step 1, then $\tilde{\partial }(x_{1},y)=(2,q)$, contrary to $q>2$; if $\tilde{\partial }(x,x_{1})=(1,q)$, by $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$, then y or $y'\in \Gamma _{1,q}(x_{1})$, which implies that $(y,z,x_{1})$ or $(y',z,x_{1})$ is a circuit, contrary to $q>2$.

Case 2.$\partial (x,x_{1})=2$.

Pick a vertex $w\in P_{(1,h-1),(1,l-1)}(x,x_{1})$. Suppose $h=q$. Then $w\in \{y,y'\}$. Since $(1,q-1)$ is pure, $\tilde{\partial }(w,x_{1})\ne (1,1)$. In view of Theorem 2.1 (i) and (ii), we have $\tilde{\partial }(z,x_{1})=(1,1)$, and y or $y'\in \Gamma _{q-1,1}(x_{1})$, which imply $p_{(1,q-1),(1,1)}^{(1,q-1)}\ne 0$, contrary to Step 1. Thus, $h\ne q$ and $l\ne q$.

Suppose $h=l=2$. Lemma 2.2 (i) implies $k_{1,1}=2$. By Step 1, $y,y'\in \Gamma _{2,q}(w)$. It follows from Lemma 1.2 (i) that $p_{(1,1),(1,q-1)}^{(2,q)}=2$ and $y,y'\in \Gamma _{1,q-1}(x_{1})$. Since $(1,q-1)$ is pure, we get $q=3$ and $\tilde{\partial }(z,x_{1})=(1,2)$. Observe that $y,y'\in P_{(1,2),(1,2)}(x_{1},z)$, contrary to $p_{(1,2),(1,2)}^{(2,1)}=1$.

Suppose $h=q+1$ or $l=q+1$. By $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$, we may assume that $h=q+1$ and $\tilde{\partial }(w,y)=(1,q)$. Since $\partial (y,x_{1})\le 2$, one gets $l=q+1$ from Step 2. In view of $\tilde{\partial }(x,x_{1})=(2,t-1)$ and Lemma 2.2 (ii), one has $(\Gamma _{1,q})^{2}=\{\Gamma _{1,q-1},\Gamma _{2,t-1}\}$. Since $k_{1,q-1}=2$, by Lemma 1.2 (i) and (v), we obtain $p_{(1,q),(1,q)}^{(1,q-1)}=1$. By Lemma 2.2 (iii), we get $p_{(1,q),(1,q)}^{(2,t-1)}=2$, which implies $k_{2,t-1}=1$. Since $k_{1,q-1}=k_{1,q}=2$ and $k_{2,t}=1$, from Lemma 1.2 (i), one has $\tilde{\partial }(z,x_{1})=(1,1)$. In view of Step 1, $\tilde{\partial }(y,x_{1})=(2,q)$. Since $w\in P_{(q,1),(1,q)}(y,x_{1})$, from Lemma 2.2 (iv), we get $q=2$, a contradiction.

Thus, our claim is valid.

Suppose that the path $(x_{0},x_{1},\ldots ,x_{t})$ contains arcs of different types. Without loss of generality, we may assume $\tilde{\partial }(z,x_{1})=(1,u-1)$ and $\tilde{\partial }(x_{1},x_{2})=(1,v-1)$ with $u\ne v$. Pick a vertex $x_{1}'\in P_{(1,v-1),(1,u-1)}(z,x_{2})$. By the claim, we get $\tilde{\partial }(x,x_{1})=\tilde{\partial }(x,x_{1}')=(3,t-1)$. It follows from Lemma 1.2 (iv) that $k_{2,t}\ge 2$, a contradiction. Then the path $(x_{0},x_{1},\ldots ,x_{t})$ consists of arcs of the same type.

Suppose $\tilde{\partial }(x_{i},x_{i+1})=(1,1)$ for $0\le i\le t-1$. By Lemma 2.3, $t=2$. In view of Step 1, we get $\tilde{\partial }(y,x_{1})=(2,q)$. Since $(x_{1},x_{2}=x,y)$ is a path, one has $q\le 2$, a contradiction.

Suppose $\tilde{\partial }(x_{i},x_{i+1})=(1,q)$ for $0\le i\le t-1$. Then $\partial (z,x_{2})=2$. In view of $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$ and Lemma 2.5 (i), we have $\partial (x_{2},z)>q-1$, which implies $t\ge 3$. Since $k_{1,q-1}=2$ and $|(\Gamma _{1,q})^{2}|=2$ from Lemma 2.2 (ii), one gets $p_{(1,q),(1,q)}^{(1,q-1)}=1$ by Lemma 1.2 (i) and (v). In view of Lemma 2.2 (iii), there exists a vertex $x_{1}''\in P_{(1,q),(1,q)}(z,x_{2})$ such that $\tilde{\partial }(x_{1}'',x_{3})=(1,q-1)$, a contradiction.

Hence, $\tilde{\partial }(x_{i},x_{i+1})=(1,q-1)$ for $0\le i\le t-1$. Since $\Delta _{q}\simeq \mathrm{Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{q},\{(1,0),(0,1)\})$, we have $t=2q-2$.

In the following, we reach a contradiction based on the above discussion.

Suppose $q>3$. Note that $\tilde{\partial }_{\Gamma }(\tau (a,b),\tau (a+1,b+1))=(2,2q-2)$. Since

$$\begin{aligned}&(\tau (1,1),\tau (2,1),\tau (3,1),\ldots ,\tau (-1,1),\tau (0,1),\tau (0,2),\ldots ,\tau (0,0)),\\&(\tau (1,1),\tau (2,1),\tau (2,2),\ldots ,\tau (2,-1),\tau (2,0),\tau (3,0),\ldots ,\tau (0,0)) \end{aligned}$$

are two shortest paths, we get $\tau (3,1),\tau (2,2)\in \Gamma _{4,2q-4}(\tau (0,0))$. But $\tau (1,1)\in P_{(2,2q-2),(2,q-2)}(\tau (0,0),\tau (3,1))$ and $P_{(2,2q-2),(2,q-2)}(\tau (0,0),\tau (2,2))=\emptyset$, a contradiction. In the following, we consider $q=3$.

Case 1. $(1,1)\in \tilde{\partial }(\Gamma )$.

By Step 3 and Lemma 1.2 (i), we have $k_{2,4}=1$. From Step 2, one gets $(1,3)\notin \tilde{\partial }(\Gamma )$. Since the valency of $\Gamma$ is more than 3, by Lemma 2.2 (i), one has $k_{1,1}=2$. Let $x,y,z,z'$ be distinct vertices such that $\tilde{\partial }(x,y)=(1,2)$ and $\tilde{\partial }(y,z)=\tilde{\partial }(y,z')=(1,1)$. By Step 1, we obtain $z,z'\in \Gamma _{2,3}(x)$. In view of Lemma 1.2 (i), one has $p_{(1,2),(1,1)}^{(2,3)}=2$, which implies that there exists a vertex $y'$ such that $\tilde{\partial }(x,y')=(1,2)$ and $\tilde{\partial }(y',z)=\tilde{\partial }(y',z')=(1,1)$ with $y'\ne y$. Hence, $(y,z,y',z')$ is an undirected circuit of length 4. By Lemma 2.3, we get $\tilde{\partial }(y,y')=(2,2)$ and $p_{(1,1),(1,1)}^{(2,2)}=2$. From Lemma 1.2 (i) and (v), $k_{2,2}=1$. Since $x\in P_{(2,1),(1,2)}(y,y')$, we have $p_{(2,1),(1,2)}^{(2,2)}=2$, contrary to $\Delta _{3}\simeq \mathrm{Cay}(\mathbb {Z}_{3}\times \mathbb {Z}_{3},\{(1,0),(0,1)\})$.

Case 2. $(1,1)\notin \tilde{\partial }(\Gamma )$.

Note that $(1,3)\in \tilde{\partial }(\Gamma )$. Pick a vertex $w\in P_{(1,3),(1,3)}(\tau (0,0),\tau (0,1))$. By Lemma 2.2 (ii), we have $|(\Gamma _{1,3})^{2}|=1$ or 2.

Case 2.1. $|(\Gamma _{1,3})^{2}|=1$.

Since $k_{1,2}=2$, by Lemma 1.2 (i), we have $p_{(1,3),(1,3)}^{(1,2)}=2$ and $\tilde{\partial }_{\Gamma }(w,\tau (1,0))=(1,3)$. Pick a vertex $x'\in P_{(1,3),(1,3)}(\tau (0,0),\tau (1,0))$ with $x'\ne w$. Observe $x'\in P_{(1,3),(1,3)}(\tau (0,0),\tau (0,1))$. Since $w,x'\in P_{(3,1),(1,3)}(\tau (0,1),\tau (1,0))$, from Lemma 1.2 (i) and (v), we obtain $k_{\tilde{\partial }_{\Gamma }(\tau (0,1),\tau (1,0))}=1$ and $|P_{(1,2),(2,1)}(\tau (0,1),\tau (1,0))|=2$, contrary to $\Delta _{3}\simeq \mathrm{Cay}(\mathbb {Z}_{3}\times \mathbb {Z}_{3},\{(1,0),(0,1)\})$.

Case 2.2. $|(\Gamma _{1,3})^{2}|=2$.

Since $|(\Gamma _{1,3})^{2}|=2$, $(w,\tau (1,0))\notin \Gamma _{1,3}$. It follows that $P_{(1,2),(1,3)}(w,\tau (1,1))=\{\tau (0,1)\}$. By Lemma 1.2 (i) and Lemma 2.2 (i),(ii), we have $|\Gamma _{1,2}\Gamma _{1,3}|=2$. In view of Theorem 2.1 (i), one obtains $\tilde{\partial }_{\Gamma }(w,\tau (0,2))=(2,2)$ and $\partial _{\Gamma }(w,\tau (1,1))=2$. By Step 3, we get $p_{(1,2),(1,2)}^{(2,4)}=2$. Hence, $\partial _{\Gamma }(\tau (1,1),w)=3$ or 5.

Case 2.2.1. $\partial _{\Gamma }(\tau (1,1),w)=3$.

Pick a path $(\tau (1,1),z_{1},z_{2},w)$. Suppose that $(z_{2},w)\in \Gamma _{1,3}$. The fact that $\partial _{\Gamma }(\tau (1,1),\tau (0,0))=4$ implies $z_{2}\ne \tau (0,0)$. Since $|(\Gamma _{1,3})^{2}|=2$, from Theorem 2.1 (iii) and Lemma 2.5 (i), we get $(z_{2},\tau (0,1))\in \Gamma _{2,3}$, which implies $(\Gamma _{1,3})^{2}=\{\Gamma _{1,2},\Gamma _{2,3}\}$. Since $k_{1,2}=2$, by Lemma 1.2 (i) and (v), we obtain $p_{(1,3),(1,3)}^{(1,2)}=1$. In view of Lemma 2.2 (iii), one has $p_{(1,3),(1,3)}^{(2,3)}=2$ and $\tilde{\partial }_{\Gamma }(\tau (0,1),\tau (1,1))=(1,3)$, a contradiction.

Observe that the path $(\tau (1,1),z_{1},z_{2},w)$ consists of arcs of type (1, 2). Since $(\tau (0,1),\tau (1,2)),(\tau (1,1),\tau (2,2))\in \Gamma _{2,4}$, we have $z_{1}=\tau (2,1)$ and $z_{2}=\tau (0,1)$, a contradiction.

Case 2.2.2. $\partial _{\Gamma }(\tau (1,1),w)=5$.

By $\tilde{\partial }_{\Gamma }(w,\tau (0,2))=(2,2)$ and Lemma 2.2 (ii), $\Gamma _{1,2}\Gamma _{1,3}=\{\Gamma _{2,2},\Gamma _{2,5}\}$. Then $\tau (2,0)\in P_{(2,1),(2,5)}(\tau (0,0),w)$. Since $(\tau (1,1),\tau (1,2),\tau (2,2),\tau (2,0),\tau (0,0),w)$ and $(\tau (1,1),\tau (2,1),\tau (2,2),\tau (2,0),\tau (0,0),w)$ are two shortest paths, $\tilde{\partial }_{\Gamma }(w,\tau (1,2))=\tilde{\partial }_{\Gamma }(w,\tau (2,1))=(3,4)$. It follows from Step 3 and Lemma 1.2 (i) that $k_{2,4}=1$. Since $\tau (0,1)\in P_{(1,3),(2,4)}(w,\tau (1,2))$, we obtain $(w,\tau (1,0))\in \Gamma _{1,3}$, contrary to $|(\Gamma _{1,3})^{2}|=2$.

This completes the proof of the proposition.$\square$

Proposition 5.3

Let $q>2$, $k_{1,q-1}=2$ and $(1,q-1)$ be pure. The following hold:

• (i) If (1, q) is mixed, then $\Delta _{q,q+1}\simeq \mathrm{Cay}(\mathbb {Z}_{4q},\{1,2,2q+1,2q+2\})$.

• (ii) If $k_{1,1}=2$, then $\Delta _{2,q}\simeq \mathrm{Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{4},\{(1,0),(1,2),(0,1),(0,3)\})$ for $q\ne 4$.

Proof

Assume that $l=q+1$ and (1, q) is mixed, or $l=2$ and $k_{1,1}=2$. In view of Theorem 2.1 (ii), Lemma 1.2 (i) and Lemma 2.2 (i), we have $k_{1,l-1}=2$. By Proposition 5.2, there exists an isomorphism $\tau$ from $\mathrm{Cay}(\mathbb {Z}_{2q},\{1,q+1\})$ to $\Delta _{q}(x)$ for fixed $x\in V\Gamma$. Write $\tau (a)=(a,0)$ for any a. Suppose that there exists a vertex $(s,0)\in \Gamma _{1,l-1}(0,0)$. By Lemma 2.4 (i), we have $s=q$. Since $(1,0),(q+1,0)\in P_{(1,q-1),(q-1,1)}((0,0),(q,0))$, from Lemma 2.2 (iv), one gets $l=2$. In view of Lemma 1.2 (i) and (v), we obtain $k_{1,1}=1$, a contradiction. Hence, $\Gamma _{1,l-1}\notin \langle \Gamma _{1,q-1}\rangle$.

If $l=q+1$, by Lemma 3.6 (ii), then $(A_{1,q})^{2}=2A_{1,q-1}$; if $l=2$, by Lemma 3.6 (i) and Lemma 1.2 (i), then $A_{1,q-1}A_{1,1}=2A_{2,q}$. Then $V\Delta _{l,q}(x)$ has a partition $F_{q}(x)\dot{\cup }F_{q}(x')$. Let $\sigma$ be an isomorphism from $\Delta _{q}(x)$ to $\Delta _{q}(x')$ such that $\sigma (0,0)\in \Gamma _{1,l-1}(0,0)$. Write $\sigma (a,0)=(a,1)$ for each a. Suppose $l=q+1$. Since $(A_{1,q})^{2}=2A_{1,q-1}$, we have $(a,1),(a+q,1)\in \Gamma _{1,q}(a,0)$ and $(a+1,0),(a+q+1,0)\in \Gamma _{1,q}(a,1)$, which imply that (i) holds. Suppose $l=2$. Since $A_{1,q-1}A_{1,1}=2A_{2,q}$, one gets $(a,1),(a+q,1)\in \Gamma _{1,1}(a,0)$. If $q=4$, by Lemma 2.3, then $(4,0),(2,0),(6,0)\in \Gamma _{2,2}(0,0)$ since (1, 3) is pure, contrary to Lemma 2.2 (i). Thus, (ii) holds. $\square$

Proposition 5.4

Suppose that C6 holds. If $k_{1,1}=2$ and $k_{1,q-1}=1$, then $\Gamma _{1,q}\notin \langle \{\Gamma _{1,1},\Gamma _{1,q-1}\}\rangle$ and $\Delta _{2,q}\simeq \mathrm{Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{n},\{(1,0),(0,1),(0,-1)\}$ with $n\le q-(1+(-1)^{q})/2$.

Proof

Since (1, q) is mixed, from Theorem 2.1 (ii), $(1,q-1)$ is pure. By Lemma 2.3, we get $\Gamma _{1,q-1}\notin \langle \Gamma _{1,1}\rangle$. For fixed $x_{0}\in V\Gamma$, $V\Delta _{2,q}(x_{0})$ has a partition $\dot{\cup }_{i=0}^{m-1}F_{2}(x_{i})$ with $m>1$. Let $\tau$ be an isomorphism from $\text {Cay}(\mathbb {Z}_{n},\{1,n-1\})$ to $\Delta _{2}(x_{0})$. Write $\tau (a)=(0,a)$ for each a. Since $k_{1,q-1}=1$, $\sigma _{j}:F_{2}(x_{j})\rightarrow F_{2}(x_{j+1})$, $y_{j}\mapsto y_{j+1}$ is an isomorphism from $\Delta _{2}(x_{j})$ to $\Delta _{2}(x_{j+1})$, where $y_{j+1}\in \Gamma _{1,q-1}(y_{j})$ for $0\le j\le m-2$. Write $\sigma _{j}(j,a)=(j+1,a)$.

Assume that $(s,t)\in \Gamma _{1,q-1}(m-1,0)$. Since $k_{1,q-1}=1$, we have $s=0$. It follows from Lemma 1.2 (i) and Lemma 2.3 that $t=0$, or $2\mid n$ and $t=n/2$.

Suppose $2\mid n$ and $t=n/2$. Since $(1,q-1)$ is pure and $k_{1,q-1}=1$, from Lemma 1.2 (i), we get $\tilde{\partial }_{\Gamma }((0,0),(0,n/2))=(m,q-m)$, which implies $q=2m$ from Lemma 2.3. Hence, $q\le n$ and $(0,0)\in \Gamma _{1,q-1}(m-1,n/2)$. Since $\{i\mid (1,i-1)\in \tilde{\partial }(\Gamma )\}=\{2,q,q+1\}$, one has $(0,m),(0,-m)\in \Gamma _{m,m}(0,0)$. Since $k_{m,m}\le 2$ by Lemma 2.2 (i), we obtain $m=n/2$ and $n=q$. Hence, $((0,0),(1,0),\ldots ,(m-1,0),(0,n/2),(0,n/2-1),\ldots ,(0,1))$ is a circuit of length q containing arcs of types (1, 1) and $(1,q-1)$, contrary to the fact that $(1,q-1)$ is pure. Then $t=0$ and $m=q$. Since $(1,q-1)$ is pure and $k_{1,q-1}=1$, one has $((m-1,a),(0,a))\in \Gamma _{1,q-1}$ for each a. Thus, $\Delta _{2,q}\simeq \mathrm{Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{n},\{(1,0),(0,1),(0,-1)\}$.

Since (1, q) is mixed, we have $p_{(1,q),(1,q)}^{(1,q-1)}=k_{1,q}$ from Theorem 2.1 (ii) and Lemma 1.2 (ii). We prove $n\le q-(1+(-1)^{q})/2$ by the way of contradiction. Assume that $n>q-(1+(-1)^{q})/2$. Suppose that q is even. Since $(1,q-1)$ is pure and $k_{1,q-1}=1$, by Lemma 1.2 (i), we get $\tilde{\partial }_{\Gamma }((0,0),(q/2,0))=(q/2,q/2)$ and $k_{q/2,q/2}=1$. Observe $\tilde{\partial }_{\Gamma }((0,0),(0,q/2))=(q/2,q/2)$, a contradiction. Suppose that q is odd. Pick a vertex $x\in P_{(1,q),(1,q)}(((q-1)/2,0),((q+1)/2,0))$. Note that $x,(0,(q+1)/2)\in \Gamma _{(q+1)/2,(q+1)/2}(0,0)$. Since $(x,((q+1)/2,0),((q+3)/2,0),\ldots ,(0,0))$ is a path containing arcs of types $(1,q-1)$ and (1, q), there exists a path $((0,(q+1)/2)=x_{0},x_{1},\ldots ,x_{(q+1)/2}=(0,0))$ containing arcs of types $(1,q-1)$ and (1, q). Then $((0,0),(0,1),\ldots ,(0,(q+1)/2)=x_{0},x_{1},\ldots ,x_{(q-1)/2})$ is a circuit of length $q+1$ containing arcs of types $(1,1),(1,q-1)$ and (1, q), contrary to Lemma 2.5 (ii).

Suppose that $(h,l)\in \Gamma _{1,q}(0,0)$ for some $h\in \{0,1,\ldots ,q-1\}$ and $l\in \mathbb {Z}_{n}$. By Lemma 2.3, $h\ne 0$. Without loss of generality, we may assume $2\hat{l}\le n$. The fact that $p_{(1,q),(1,q)}^{(1,q-1)}=k_{1,q}$ implies $\tilde{\partial }_{\Gamma }((h,l),(1,0))=(1,q)$. Since $((0,0),(h,l),(h+1,l),\ldots ,(0,l),(0,l-1),\ldots ,(0,1))$ and $((1,0),(2,0),\ldots ,(h,0),(h,1),\ldots ,(h,l))$ are two circuits, one has $q-h+\hat{l}+1\ge q+1$ and $h+\hat{l}\ge q+1$. Hence, $q+1\le 2\hat{l}\le n$, contrary to $n\le q-(1+(-1)^{q})/2$. Thus, $\Gamma _{1,q}\notin \langle \{\Gamma _{1,1},\Gamma _{1,q-1}\}\rangle$. $\square$

6 Proof of Theorem 1.1

For any nonempty subset F of R with $F=\langle F\rangle$, let

$$\begin{aligned} V\Gamma /F:=\{F(x)\mid x\in V\Gamma \}\quad \text {and}\quad \Gamma _{\tilde{i}}^{F}:=\{(F(x),F(y))\mid y\in F\Gamma _{\tilde{i}}F(x)\}. \end{aligned}$$

The digraph $(V\Gamma /F,\ \cup _{(1,s)\in \tilde{\partial }(\Gamma )}\Gamma _{1,s}^{F})$ is said to be the quotient digraph of $\Gamma$ over F, denoted by $\Gamma /F$.

In the following, we divide the proof of Theorem 1.1 into four subsections according to separate assumptions based on Proposition 1.3.

6.1 The cases C1, C2 and C3

By Lemma 3.1, $k_{1,q-1}=2$. If $(1,1)\in \tilde{\partial }(\Gamma )$, by Lemma 3.7 (i), then $k_{1,1}=1$; if $(1,q)\in \tilde{\partial }(\Gamma )$, then (1, q) is mixed, which imply $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$ and $k_{1,q}=2$ from Theorem 2.1 (ii) and Lemma 3.1.

Case 1.$(1,q)\notin \tilde{\partial }(\Gamma )$.

Note that C1 holds. Since $C_{q,3}$ exists, from Lemma 3.2, (1, 2) is mixed. By Lemma 3.1, we have $k_{1,2}=1$, which implies $p_{(1,2),(1,2)}^{(1,1)}=1$ from Theorem 2.1 (ii). In view of Proposition 5.1, $\Gamma$ is isomorphic to one of the digraphs in Theorem 1.1 (iv) for $i=0$.

Case 2.$(1,q)\in \tilde{\partial }(\Gamma )$.

Note that C2 or C3 holds. Assume that $h=4$ or 3. Since $C_{q,h}$ exists, by Proposition 5.1, there exists an isomorphism $\tau$ from $\text {Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{4},\{(1,0),(0,1),(1,2)\})$ to $\Delta _{q,h}(x)$ for fixed $x\in V\Gamma$. Write $\tau (a,b)=(a,b,0)$ for each (a, b). Suppose that there exists (c, d, 0) such that $\tilde{\partial }_{\Gamma }((0,0,0),(c,d,0))=(1,q)$. Since $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$ and $(1,q-1)$ is pure from Lemma 3.1, we get $d\in \{1,3\}$ and $c\ne 0$. Observe that $((0,0,0),(c,d,0),(c+1,3,0),(c+2,3,0),\ldots ,(0,3,0))$ is a circuit of length $q-\hat{c}+2$ containing arcs of types (1, q) and $(1,h-1)$, contrary to Lemma 2.5 (ii). Hence, $\Gamma _{1,q}\notin \langle \{\Gamma _{1,q-1},\Gamma _{1,h-1}\}\rangle$.

By Lemma 3.1 and Lemma 3.6 (ii), we have $(A_{1,q})^{2}=2A_{1,q-1}$, which implies that $V\Delta _{q,q+1,h}$ has a partition $F_{q,h}(x)\dot{\cup }F_{q,h}(x')$. Let $\sigma$ be an isomorphism from $\Delta _{q,h}(x)$ to $\Delta _{q,h}(x')$ such that $\sigma (0,0,0)\in \Gamma _{1,q}(0,0,0)$. Write $\sigma (a,b,0)=(a,b,1)$ for each (a, b). Since $(0,0,1)\in P_{(1,q),(1,h-1)}((0,0,0),(0,1,1))$ and $k_{1,h-1}=1$, we get $(0,1,1)\in \Gamma _{1,q}(0,1,0)$. Similarly, $(0,b,1)\in \Gamma _{1,q}(0,b,0)$ for each b. The fact that $(A_{1,q})^{2}=2A_{1,q-1}$ implies that $(a,b,1),(a,b+2,1)\in \Gamma _{1,q}(a,b,0)$ and $(a+1,b,0),(a+1,b+2,0)\in \Gamma _{1,q}(a,b,1)$ for each (a, b). Thus, $\Delta _{q,q+1,h}\simeq \mathrm{Cay}(\mathbb {Z}_{2q}\times \mathbb {Z}_{4},\{(2,0),(2,2),(1,0),(1,2),(0,1)\})$.

If C2 holds, then $\Gamma$ is isomorphic to one of the digraphs in Theorem 1.1 (vii) for $i=1$. Suppose that C3 holds. Since $C_{q,3}$ exists, from Lemma 3.2, (1, 2) is mixed. By Lemma 3.1, we have $k_{1,2}=1$, which implies $p_{(1,2),(1,2)}^{(1,1)}=1$ from Theorem 2.1 (ii). Hence, $\Gamma$ is isomorphic to one of the digraphs in Theorem 1.1 (vii) for $i=0$.

We complete the proof of the main theorem for the cases C1, C2 and C3.

6.2 The case C4

Since the valency of $\Gamma$ is more than 3, from Lemma 2.2 (i), we have $k_{1,1}=k_{1,q-1}=2$. By Proposition 5.3 (ii), $\Gamma$ is isomorphic to one of the digraphs in Theorem (iv) for $i=1$. We complete the proof of the main theorem for the case C4.

6.3 The case C5

Since the valency of $\Gamma$ is more than 3, from Lemma 2.2 (i), we have $k_{1,q-1}=k_{1,q}=2$. Note that (1, q) is mixed. By Theorem 2.1, $(1,q-1)$ is pure. If $q>2$, from Proposition 5.3 (i), then $\Gamma \simeq \mathrm{Cay}(\mathbb {Z}_{4q},\{1,2,2q+1,2q+2\})$. We consider $q=2$ in the following.

By Theorem 2.1 (ii), $p_{(1,2),(1,2)}^{(1,1)}\ne 0$. It follows from Lemma 2.3 that $\Gamma _{1,2}\notin \langle \Gamma _{1,1}\rangle$. Suppose $\tilde{\partial }(x_{0},x_{1})=(1,2)$ for $x_{0},x_{1}\in V\Gamma$. Then $\partial (F_{2}(x_{0}),F_{2}(x_{1}))=1$ in $\Gamma /\langle \Gamma _{1,1}\rangle$. Since $p_{(1,2),(1,2)}^{(1,1)}\ne 0$, we get $\Gamma _{1,1}(x_{0})\cap \Gamma _{1,2}(x_{1})\ne \emptyset$, which implies $\partial (F_{2}(x_{1}),F_{2}(x_{0}))=1$. Hence, $\Gamma /\langle \Gamma _{1,1}\rangle$ is a connected undirected graph. By $k_{1,2}=2$, $\Gamma /\langle \Gamma _{1,1}\rangle \simeq C_{l}$.

Let $(F_{2}(x_{0}),F_{2}(x_{1}),\ldots ,F_{2}(x_{l-1}))$ be an undirected circuit. Suppose $l\ne 2$. Without loss of generality, we may assume that $(x_{0},x_{1}),(x_{1},x_{2}),(x_{3},x_{2})\in \Gamma _{1,2}$. Then $x_{1}\ne x_{3}$. In view of $\tilde{\partial }(x_{0},x_{2})\ne (1,1)$ and Lemma 2.2 (ii), one gets $|(\Gamma _{1,2})^{2}|=2$. Since $k_{1,1}=2$, by Lemma 1.2 (i) and (v), we have $p_{(1,2),(1,2)}^{(1,1)}=1$, which implies $x_{3}\in P_{(1,2),(1,2)}(x_{0},x_{2})$ from Lemma 2.2 (iii). Hence, $\partial (F_{2}(x_{0}),F_{2}(x_{3}))=1$ and $l=4$. Thus, $l=2$ or 4.

Case 1.$\Gamma /\langle \Gamma _{1,1}\rangle \simeq C_{2}$.

Note that $V\Gamma =F_{2}(x_{0})\dot{\cup }F_{2}(x_{1})$. Let $\tau _{i}$ be an isomorphism from $\text {Cay}(\mathbb {Z}_{n},\{1,n-1\})$ to $\Delta _{2}(x_{i})$. Write $\tau _{i}(a)=(a,i)$ for each a. Without loss of generality, we may assume $\tilde{\partial }_{\Gamma }((0,0),(0,1))=(1,2)$. By Lemma 2.2 (ii), we get $|(\Gamma _{1,2})^{2}|=1$ or 2.

Case 1.1.$(\Gamma _{1,2})^{2}=\{\Gamma _{1,1}\}$.

By Lemma 1.2 (i), one has $p_{(1,2),(1,2)}^{(1,1)}=2$, which implies $(1,0),(-1,0)\in \Gamma _{1,2}(0,1)$. It follows from Lemma 2.3 that $\tilde{\partial }_{\Gamma }((1,0),(-1,0))=(2,2)$. In view of Lemma 1.2 (ii) and (vi), we get $p_{(1,2),(1,2)}^{(1,1)}p_{(2,1),(1,1)}^{(1,2)}=2+p_{(2,1),(1,2)}^{(2,2)}=4$. By Lemma 1.2 (i) and (v), we obtain $k_{2,2}=1$. It follows from Lemma 2.3 that $n=4$ and $|V\Gamma |=8$. Since $(\Gamma _{1,2})^{2}=\{\Gamma _{1,1}\}$, by [3], we obtain $\Gamma \simeq \text {Cay}(\mathbb {Z}_{8},\{1,2,5,6\})$.

Case 1.2.$|(\Gamma _{1,2})^{2}|=2$.

Assume that $((0,1),(t,0))\in \Gamma _{1,2}$ and $((0,0),(t,0))\notin \Gamma _{1,1}$. By Theorem 2.1 (iii) and Lemma 2.3, we have $\tilde{\partial }_{\Gamma }((0,0),(t,0))=(2,2)$. Hence, $n>3$. Since $k_{1,1}=2$, we get $p_{(1,2),(1,2)}^{(1,1)}=1$ from Lemma 1.2 (i) and (v). In view of Lemma 2.2 (iii), we obtain $p_{(1,2),(1,2)}^{(2,2)}=2$ and $k_{2,2}=1$. By Lemma 2.3, one has $(2,0),(-2,0)\in \Gamma _{2,2}(0,0)$, which implies $n=4$ and $|V\Gamma |=8$. Since $|(\Gamma _{1,2})^{2}|=2$, from [3], $\Gamma \simeq \text {Cay}(\mathbb {Z}_{8},\{1,2,3,6\})$.

Case 2.$\Gamma /\langle \Gamma _{1,1}\rangle \simeq C_{4}$.

Note that $V\Gamma =F_{2}(x_{0})\dot{\cup }F_{2}(x_{1})\dot{\cup }F_{2}(x_{2})\dot{\cup }F_{2}(x_{3})$. Let $\sigma _{i}$ be an isomorphism from $\text {Cay}(\mathbb {Z}_{n},\{1,n-1\})$ to $\Delta _{2}(x_{i})$ for each i. Write $\tau _{i}(a)=(a,i)$ for any a. Without loss of generality, we may assume $\tilde{\partial }_{\Gamma }((0,j),(0,j+1))=(1,2)$ for $j=0,1,2$.

Since $(0,j+1)\in P_{(1,2),(1,1)}((0,j),(1,j+1))$, we have (1, j) or $(-1,j)\in \Gamma _{1,2}(1,j+1)$. Without loss of generality, we may assume that $\tilde{\partial }_{\Gamma }((1,j),(1,j+1))=(1,2)$. Since $(1,j+1)\in P_{(1,2),(1,1)}((1,j),(2,j+1))$ and $\Gamma /\langle \Gamma _{1,1}\rangle \simeq C_{4}$, one gets $\tilde{\partial }_{\Gamma }((2,j),(2,j+1))=(1,2)$. Similarly, $\tilde{\partial }_{\Gamma }((a,j),(a,j+1))=(1,2)$ for each $a\in \mathbb {Z}_{n}$ and $j\in \{0,1,2\}$.

By $p_{(1,2),(1,2)}^{(1,1)}\ne 0$, we may assume $\tilde{\partial }_{\Gamma }((0,1),(1,0))=(1,2)$. Since $(1,0)\in P_{(1,2),(1,1)}((0,1),(2,0))$, we get (1, 1) or $(-1,1)\in \Gamma _{2,1}(2,0)$.

Case 2.1.$\tilde{\partial }_{\Gamma }((1,1),(2,0))=(1,2)$.

Since $(2,0)\in P_{(1,2),(1,1)}((1,1),(3,0))$ and $\Gamma /\langle \Gamma _{1,1}\rangle \simeq C_{4}$, $\tilde{\partial }_{\Gamma }((2,1),(3,0))=(1,2)$. Similarly, $\tilde{\partial }_{\Gamma }((a,1),(a+1,0))=(1,2)$ for each $a\in \mathbb {Z}_{n}$. The fact that $p_{(1,2),(1,2)}^{(1,1)}\ne 0$ and $\Gamma /\langle \Gamma _{1,1}\rangle \simeq C_{4}$ imply that $(0,2)\in P_{(1,2),(1,2)}((0,1),(-1,1))$. Hence, $((0,0),(0,1),(0,2),(-1,1))$ is a circuit consisting of arcs of type (1, 2). In view of Theorem 2.1 (iii), one gets $\tilde{\partial }_{\Gamma }((0,0),(0,2))=(2,2)$, which implies $(\Gamma _{1,2})^{2}=\{\Gamma _{1,1},\Gamma _{2,2}\}$ by Lemma 2.2 (ii). Since $k_{1,1}=2$, from Lemma 1.2 (i) and (v), we obtain $p_{(1,2),(1,2)}^{(1,1)}=1$. In view of Lemma 2.2 (iii), one has $p_{(1,2),(1,2)}^{(2,2)}=1$ and $k_{2,2}=1$. By Lemma 2.3, we get $\tilde{\partial }_{\Gamma }((0,0),(2,0))=(1,1)$. Since $\Gamma /\langle \Gamma _{1,1}\rangle \simeq C_{4}$, from Theorem 2.1 (i), one obtains $\tilde{\partial }_{\Gamma }((0,0),(1,1))=(2,2)$, a contradiction.

Case 2.2.$\tilde{\partial }_{\Gamma }((-1,1),(2,0))=(1,2)$.

Since $p_{(1,2),(1,2)}^{(1,1)}\ne 0$ and $((-1,0),(-1,2))\notin \Gamma _{1,1}$, we have $\tilde{\partial }_{\Gamma }((-1,0),(2,0))=(1,1)$, $n=4$ and $|V\Gamma |=16$. By [3], $\Gamma \simeq \mathrm{Cay}(\mathbb {Z}_{4~}\times \mathbb {Z}_{4},\{(0,1),(1,0),(2,0),(0,2)\})$.

We complete the proof of the main theorem for the case C5.

6.4 The case C6

By Theorem 2.1 (ii), $p_{(1,q),(1,q)}^{(1,q-1)}\ne 0$ and $(1,q-1)$ is pure. In view of Lemma 2.2 (i), we have $k_{1,1},k_{1,q-1},k_{1,q}\in \{1,2\}$.

Case 1.$k_{1,q-1}=1$.

By Lemma 1.2 (ii), we have $p_{(1,q),(1,q)}^{(1,q-1)}=k_{1,q}$.

Case 1.1.$k_{1,q}=1$.

Since the valency of $\Gamma$ is more than 3, one has $k_{1,1}=2$. In view of Proposition 5.4 and $p_{(1,q),(1,q)}^{(1,q-1)}=1$, $V\Gamma$ has a partition $F_{2,q}(x_{0})\dot{\cup }F_{2,q}(x_{1})$ and there exists an isomorphism $\tau$ from $\text {Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{n},\{(1,0),(0,1),(0,-1)\}$ to $\Delta _{2,q}(x_{0})$ for $n\le q-(1+(-1)^{q})/2$. Write $\tau (a,b)=(a,b,0)$ for each (a, b). Since $k_{1,q}=1$, $\sigma :F_{2,q}(x_{0})\rightarrow F_{2,q}(x_{1})$, $x\mapsto x'$ is an isomorphism from $\Delta _{2,q}(x_{0})$ to $\Delta _{2,q}(x_{1})$, where $x'\in \Gamma _{1,q}(x)$. Write $\sigma (a,b,0)=(a,b,1)$ for each (a, b). The fact that $p_{(1,q),(1,q)}^{(1,q-1)}=1$ implies $\tilde{\partial }_{\Gamma }((a,b,1),(a+1,b,0))=(1,q)$. Thus, $\Gamma$ is isomorphic to one of the digraphs in Theorem 1.1 (viii).

Case 1.2.$k_{1,q}=2$.

We claim that $p_{(1,q),(q,1)}^{(1,1)}\ne 0$. Let x, y, z be vertices such that $\tilde{\partial }(x,y)=(1,1)$ and $\tilde{\partial }(y,z)=(1,q-1)$. Since $k_{1,q-1}=1$, by Lemma 2.3, Theorem 2.1 (i) and Lemma 3.5 (i), one gets $\tilde{\partial }(x,z)=(2,q)$. It follows from Lemma 1.2 (i) and Lemma 2.2 (ii) that $|(\Gamma _{1,q})^{2}|=2$. In view of Lemma 2.5 (iii), one gets $p_{(1,q),(1,q)}^{(2,q)}\ne 0$, which implies that there exists a vertex $y'\in P_{(1,q),(1,q)}(x,z)$. Since $p_{(1,q),(1,q)}^{(1,q-1)}=2$, we obtain $\tilde{\partial }(y,y')=(1,q)$. Thus, our claim is valid.

Case 1.2.1.$k_{1,1}=1$.

Since $k_{1,q-1}=1$, we have $\Gamma _{1,1}\notin \langle \Gamma _{1,q-1}\rangle$. Let $\varphi$ be an isomorphism from $\text {Cay}(\mathbb {Z}_{q},\{1\})$ to $\Delta _{q}(x_{0})$ for fixed $x_{0}\in V\Gamma$. Write $\varphi (a)=(a,0,0)$ for any $a\in \mathbb {Z}_{q}$. Since $k_{1,1}=1$, $V\Delta _{2,q}(x_{0})$ has a partition $F_{q}(x_{0})\dot{\cup }F_{q}(x_{1})$. It follows that $\sigma :F_{q}(x_{0})\rightarrow F_{q}(x_{1})$, $x\mapsto x'$ is an isomorphism from $\Delta _{q}(x_{0})$ to $\Delta _{q}(x_{1})$, where $x'\in \Gamma _{1,1}(x)$. Write $\sigma (a,0,0)=(a,1,0)$ for each a.

Suppose that there exists (c, d, 0) such that $\tilde{\partial }_{\Gamma }((0,0,0),(c,d,0))=(1,q)$. The fact that $(1,q-1)$ is pure and $k_{1,q-1}=1$ imply $d=1$ and $c\ne 0$. Since $k_{1,1}=1$, by the claim and Lemma 1.2 (v), we get $p_{(1,q),(q,1)}^{(1,1)}=2$, which implies $(c,1,0)\in P_{(1,q),(q,1)}((0,0,0),(0,1,0))$. Then $((0,1,0),(c,1,0),(c+1,1,0),\ldots ,(-1,1,0))$ is a circuit of length $q-\hat{c}+1$, a contradiction. Hence, $\Gamma _{1,q}\notin \langle \{\Gamma _{1,1},\Gamma _{1,q-1}\}\rangle$.

Since $p_{(1,q),(1,q)}^{(1,q-1)}=2$, $V\Gamma$ has a partition $F_{2,q}(x_{0})\dot{\cup }F_{2,q}(x_{0}')$. Let $\psi$ be an isomorphism from $\Delta _{2,q}(x_{0})$ to $\Delta _{2,q}(x_{0}')$ such that $\psi (0,0,0)\in \Gamma _{1,q}(0,0,0)$. Write $\psi (a,b,0)=(a,b,1)$ for each $a\in \mathbb {Z}_{q}$ and $b\in \{0,1\}$. Since $p_{(1,q),(1,q)}^{(1,q-1)}=p_{(1,q),(q,1)}^{(1,1)}=2$, we obtain $(a,0,1),(a,1,1)\in \Gamma _{1,q}(a,b,0)$ and $(a+1,0,0),(a+1,1,0)\in \Gamma _{1,q}(a,b,1)$. Then $\Gamma$ is isomorphic to one of the digraphs in Theorem 1.1 (v).

Case 1.2.1.$k_{1,1}=2$.

Since $p_{(1,q),(1,q)}^{(1,q-1)}=2$, from Proposition 5.4, $V\Gamma$ has a partition $F_{2,q}(x_{0})\dot{\cup }F_{2,q}(x_{1})$ and there exists an isomorphism $\tau _{i}$ from $\text {Cay}(\mathbb {Z}_{q}\times \mathbb {Z}_{n},\{(1,0),(0,1),(0,-1)\})$ to $\Delta _{2,q}(x_{i})$ for $i=0,1$, where $n\le q-(1+(-1)^{q})/2$. Write $\tau _{i}(a,b)=(a,b,i)$ for each $(a,b)\in \mathbb {Z}_{q}\times \mathbb {Z}_{n}$.

By the claim, we have $p_{(1,q),(q,1)}^{(1,1)}\ne 0$. Without loss of generality, we may assume $(0,0,1),(0,-1,1)\in \Gamma _{1,q}(0,0,0)$. In view of $(0,-1,1)\in P_{(1,q),(1,1)}((0,0,0),(0,0,1))$, we may assume $(0,0,1)\in \Gamma _{1,q}(0,1,0)$. Since $(0,0,0)\notin P_{(q,1),(1,q)} ((0,0,1),(0,1,1))$ and $p_{(1,q),(q,1)}^{(1,1)}\ne 0$, we get $((0,1,0),(0,1,1))\in \Gamma _{1,q}$. Similarly, $(0,b,1),(0,b-1,1)\in \Gamma _{1,q}(0,b,0)$ for each b. In view of $p_{(1,q),(1,q)}^{(1,q-1)}=2$. we obtain $(a,b,1),(a,b-1,1)\in \Gamma _{1,q}(a,b,0)$ and $(a+1,b,0),(a+1,b+1,0)\in \Gamma _{1,q}(a,b,1)$ for any $(a,b)\in \mathbb {Z}_{q}\times \mathbb {Z}_{n}$.

Suppose that $c=n/\mathrm{gcd}(q,n)$ and c is odd. Let $\varphi$ be the mapping from $\Gamma$ to the corresponding digraph in Theorem 1.1 (ix) satisfying $\varphi (a,b,i)=(2\hat{a}+i,(2\hat{a}c+ic+i)/2+\hat{b})$. Routinely, $\varphi$ is an isomorphism.

Suppose that $t=q/\mathrm{gcd}(q,n)$ and t is odd. Let $\psi$ be the mapping from $\Gamma$ to the corresponding digraph in Theorem 1.1 (x) such that $\psi (a,b,i)=(2\hat{b}+i,\hat{a}+\hat{b}t+i(1+t)/2)$. Note that $\psi$ is well defined. Assume that $\psi (a,b,i)=\psi (x,y,j)$ for some (a, b, i) and (x, y, j). Since $2\hat{b}+i\equiv 2\hat{y}+j~(\text {mod}~2n)$, we have $i=j$ and $b=y$. By $\hat{a}+\hat{b}t+i(1+t)/2\equiv \hat{x}+\hat{y}t+j(1+t)/2~(\text {mod}~q)$, one gets $a=x$. Therefore, $\psi$ is a bijection. One can verify that $((x_{1},y_{1},i_{1}),(x_{2},y_{2},i_{2}))$ is an arc if and only if $(\psi (x_{1},y_{1},i_{1}),\psi (x_{2},y_{2},i_{2}))$ is an arc. Hence, $\psi$ is an isomorphism.

Case 2.$k_{1,q-1}=2$.

If $\Gamma _{1,1}\in \Gamma _{1,q-1}\Gamma _{q-1,1}$, by Proposition 5.3 (i), then $\Gamma \simeq \mathrm{Cay}(\mathbb {Z}_{4q},\{1,2,2q,2q+1,2q+2\})$ for $q\ge 3$. In the following, we consider the case that $\Gamma _{1,1}\notin \Gamma _{1,q-1}\Gamma _{q-1,1}$.

By Proposition 5.3 (i), there exists an isomorphism $\tau$ from $\text {Cay}(\mathbb {Z}_{4q},\{1,2,2q+1,2q+2\})$ to $\Delta _{q,q+1}(x)$ for fixed $x\in V\Gamma$. Write $\tau (a)=(a,0)$ for each $a\in \mathbb {Z}_{4q}$. Observe that $\partial _{\Gamma }((0,0),(b,0))+\partial _{\Gamma }((b,0),(0,0))=q+(1+(-1)^{\hat{b}+1})/2$ for $b\notin \{0,2q\}$. Since $\Gamma _{1,1}\notin \Gamma _{1,q-1}\Gamma _{q-1,1}$, we have $\Gamma _{1,1}\notin \langle \{\Gamma _{1,q-1},\Gamma _{1,q}\}\rangle$.

Case 2.1.$k_{1,1}=1$.

Observe that $V\Gamma$ has a partition $F_{q,q+1}(x)\dot{\cup }F_{q,q+1}(x')$. Note that $\sigma :F_{q,q+1}(x)\rightarrow F_{q,q+1}(x')$, $y\mapsto y'$ is an isomorphism from $\Delta _{q,q+1}(x)$ to $\Delta _{q,q+1}(x')$, where $y'\in \Gamma _{1,1}(y)$. Write $\sigma (a,0)=(a,1)$ for each a. If $q=3$, then $(6,0),(3,1),(9,1)\in \Gamma _{3,3}(0,0)$, a contradiction. Hence, $\Gamma$ is isomorphic to one of the digraphs in Theorem 1.1 (vi) for $i=0$.

Case 2.2.$k_{1,1}=2$.

By Proposition 5.2 and Lemma 3.6 (i), (iv), one gets $A_{1,q-1}A_{1,1}=2A_{2,q}$ and $A_{1,q}A_{1,1}=2A_{2,q+1}$. Hence, $V\Gamma$ has a partition $F_{q,q+1}(x)\dot{\cup }F_{q,q+1}(x')$. Let $\varphi$ be an isomorphism from $\Delta _{q,q+1}(x)$ to $\Delta _{q,q+1}(x')$ such that $\varphi (0,0)\in \Gamma _{1,1}(0,0)$. Write $\varphi (a,0)=(a,1)$ for each a. Since $A_{1,q-1}A_{1,1}=2A_{2,q}$ and $A_{1,q}A_{1,1}=2A_{2,q+1}$, we have $(a,1),(a+2q,1)\in \Gamma _{1,1}(a,0)$ for each a. If $2<q<5$, then $(2q,0),(q,0),(3q,0)\in \Gamma _{2,2}(0,0)$, contrary to Lemma 2.2 (i). Therefore, $\Gamma$ is isomorphic to one of the digraphs in Theorem 1.1 (vi) for $i=1$.

We complete the proof of the main theorem for the case C6.