Some bounds of the generalized μ-scrambling indices of primitive digraphs with d loops

Abstract

In 2010, Huang and Liu introduced a useful parameter called the generalized μ-scrambling indices of a primitive digraph. In this paper, we give some bounds for μ-scrambling indices of some primitive digraphs with d loops and the digraphs attained the sharp upper bounds are provided.

1 Introduction

For the research on the competition index, m-competition index, the scrambling index and the generalized μ-scrambling index, please refer to [1–3, 5, 6, 8, 9] and [7, 11], respectively. Cho et al. [6] defined the m-step competition graph of a digraph which is an extension of a competition graph. In 2009, Akelbek and Kirkland [2] defined and studied the scrambling index of a primitive digraph and provided an upper bound on the scrambling index of a primitive digraph. The m-competition index of a primitive digraph was introduced by Kim [8]. Kim investigated the m-competition index of a primitive digraph and gave an upper bound for the m-competition indices of primitive digraphs. In 2010, Huang and Liu [7] gave the definition of the generalized μ-scrambling indices for a primitive digraph which are a generalization of the scrambling index and m-competition index and they provided some bounds for the generalized μ-scrambling indices of some primitive digraphs. In this paper, we give some bounds for μ-scrambling indices of some primitive digraphs.

The outline of this paper is as follows: Some notation and notions used throughout this paper are introduced in Sect. 2. In Sect. 3, we study the generalized μ-scrambling indices of the primitive digraphs with d loops.

2 Definitions and terminology

In this section, we introduce some definitions,notations which are needed to use in the presentations and proofs of our main results in this paper.

A digraph D consists of a nonempty set $V=V(D)$ and an arc set $E=E(D)$. In D, loops are permitted but multiple arcs are not. A path $P=x\rightarrow y$ is a sequence of edges $\{(x, v_{1}), (v_{1}, v_{2}), \ldots , (v_{k-1}, y)\}$ in which all vertices are distinct. A cycle C is a closed path with the first and the last vertices coincided. A walk from x to y is a sequence of arcs: $e_{1}, e_{2}, \dots , e_{k}$ such that the terminal vertex of $e_{i}$ is the same as the initial vertex of $e_{i+1}$ for $i =1,2, \dots ,k-1$, denoted by $W=x\rightarrow y$. The length of a walk or cycle is the number of arcs. A walk $W=x\rightarrow y$ of length k is denoted by $x\stackrel{{k }}{\rightarrow }y$. A cycle of length l is denoted by $C_{l}$. The girth of D which has at least one cycle, is the length of a shortest cycle in D.

A digraph D is primitive with a walk of length k from each vertex x to each vertex y (not necessarily distinct). The digraph D is primitive if and only if D is strongly connected and the greatest common divisor of the lengths of its cycles is 1 (see [4]). For a positive integer s, the sth power of D, denoted by $D^{(s)}$, is the digraph on the same vertex set $V(D)$ and with an arc from i to j if and only if $i\stackrel{{s }}{\rightarrow }j$ in D. The scrambling index $k(D)$ of a primitive digraph D is the smallest positive integer k such that, for every pair of vertices u and v, there exists a vertex w such that $u \stackrel{{k}}{\rightarrow }w$ and $v\stackrel{{k}}{\rightarrow }w$ in D (see [2]).

Let D be a digraph with vertex set V and let k be a positive integer. A vertex w of D is a k-step common prey for u and v if $u\stackrel{{k}}{\rightarrow }w$ and $v\stackrel{{k}}{\rightarrow }w$. The k-step m-competition graph of D has the same vertex set of D and an edge between vertices u and v if and only if there are at least m distinct vertices $v_{1},\ldots ,v_{m}$ in D such that $u\stackrel{{k}}{\rightarrow }v_{i}$ and $v\stackrel{{k}}{\rightarrow }v_{i}$ for $i=1,2,\ldots ,m$ (see [6]). The m-competition index $c(D,m)$ of a primitive digraph D is the smallest positive integer k such that, for every pair of vertices u and v, there are m distinct vertices $v_{1},\ldots , v_{m}$ in D such that $u\stackrel{{k}}{\rightarrow }v_{i}$ and $v\stackrel{{k }}{\rightarrow }v_{i}$ for $i=1,2,\ldots , m$ (see [2]). That is to say, the m-competition index of D is the smallest positive integer k such that the k-step m-competition graph is complete.

Let $P_{n}$ denote the set of all primitive digraphs of order n.

Definition 2.1

([7])

Let $D\in P_{n}$, and λ, μ be integers with $1\leq \lambda ,\mu \leq n$. For $X\subseteq V(D)$, let $k^{(\mu )}_{X}$ be the smallest positive integer m such that there exist μ vertices $w_{1},w_{2},\ldots ,w_{\mu }$ of D such that $x\stackrel{{m }}{\rightarrow }w_{i}$ $(i=1,2,\ldots , \mu )$ in D for every vertex x of X. Then

$$\begin{aligned}& h(D, \lambda ,\mu ):=\min \bigl\{ k^{(\mu )}_{X}\mid X \subseteq V(D) \text{ and } \mid X\mid =\lambda \bigr\} \quad \text{and} \\& k(D, \lambda ,\mu ):=\max \bigl\{ k^{(\mu )}_{X}\mid X \subseteq V(D) \text{ and }\mid X\mid =\lambda \bigr\} \end{aligned}$$

are called the λth lower and upper μ-scrambling indices of D, respectively. For convenience, let $k_{X}(D):=k^{(1)}_{X}(D)$, $h(D,\lambda ):= h(D,\lambda ,1)$ and $k(D,\lambda ) := k(D, \lambda , 1)$.

Since $k(D,2)=k(D)$, in [7] Huang and Liu called $h(D, \lambda ,\mu )$ and $k(D, \lambda ,\mu )$ the generalized μ-scrambling indices, $h(D,\lambda )$ and $k(D,\lambda )$ the generalized scrambling indices of D in $P_{n}$. As $k(D,2,m)=c(D,m)$, the generalized μ-scrambling indices are also generalizations of the m-competition index.

3 Generalized μ-scrambling indices

In [7], Huang and Liu investigated generalized scrambling indices of the primitive digraphs with d loops. In this section, we study the generalized μ-scrambling indices of the primitive digraphs with d loops.

For a vertex subset $X \subseteq V(D)$, define $R_{t}^{D}(X)$ to be the set of vertices in D reachable from some vertices in X via a walk of length t.

Let d be an integer with $1\leq d\leq n$ and let $P_{n}(d)$ be the class of primitive digraphs with n vertices and d loops. Let $L_{n,d}$ ($1 \leq d\leq n$) be the digraph with vertex set $V(L_{n,d})=\{1,2,\ldots ,n \}$ and arc set

$$ E(L_{n,d})=\bigl\{ (i,i+1)|1\leq i\leq n-1\bigr\} \cup \bigl\{ (n,1) \bigr\} \cup \bigl\{ (i,i)|n-d+1 \leq i\leq n\bigr\} . $$

Theorem 3.1

Let $D\in P_{n}(d)$ and $1\leq \lambda ,\mu \leq n$.

$$ h(D,\lambda ,\mu ) \leq \textstyle\begin{cases} \lambda +\mu -2,&\lambda +\mu < n+1, \\ n-1,&d\geq \lambda , \lambda +\mu \geq n+1, \\ n-1,&d< \lambda ,n+1 \leq \lambda +\mu \leq n+d, \\ \lambda +\mu -d-1,&d< \lambda , \lambda +\mu > n+d, \end{cases} $$

and the bound can be attained by the digraph $L_{n,d}$.

Proof

Since $D\in P_{n}(d)$, there exists a loop vertex u such that there is a set Y of $\lambda -1$ vertices whose distances to u are at most $\lambda -1$. If $\lambda +\mu < n+1$, let $X=Y\cup \{u\}$. Then $|X|=\lambda$. Since D is strongly connected and u is a loop vertex, the minimum number of vertices that can be reached from u at $(\mu -1)$-step in D is μ. Therefore, $|\bigcap _{x\in X}R^{D}_{\lambda +\mu -2}(\{x\})|\geq \mu$, which implies that $h(D,\lambda ,\mu )\leq \lambda +\mu -2$.

If $d\geq \lambda$ and $\lambda +\mu \geq n+1$, let X be a vertex set which contains λ loop vertices. Since each vertex in X is a loop vertex, we have $R_{n-1}^{D}(X)=V(D)$. Therefore, $|R_{n-1}^{D}(X)|=|V(D)|=n\geq \mu$, which implies that $h(D,\lambda ,\mu )\leq \lambda +\mu -2$.

If $d<\lambda$, let Z be the vertex set of d loop vertices and $X_{i}\subseteq (V(D)\setminus Z)$ be the vertex set of $x_{i}$ vertices whose shortest distance to vertices of Z is i, where $1\leq i\leq \lambda -d$. Assume $\sum^{r}_{i=1}{x_{i}}\leq \lambda -d<\sum^{r+1}_{i=1}{x_{i}}$, where $1\leq r\leq \lambda -d$. Let $X=Z\cup X_{1}\cup \cdots \cup X_{r} \cup \bar{X}_{r+1}$, where $\bar{X}_{r+1}\subseteq X_{r+1}$ contains $\bar{x}_{r+1}$ vertices and $\sum^{r}_{i=1}{x_{i}} +\bar{x}_{r+1}=\lambda -d$. Then $|X|=\lambda$.

If $d<\lambda$ and $n+1 \leq \lambda +\mu \leq n+d$, since $R_{n-1}^{D}(Z)=V(D)$ and $R_{n-1}^{D}(X_{1}\cap \cdots \cap X_{r} \cap \bar{X}_{r+1})$ contains at least $n-\sum^{r}_{i=1}{x_{i}} -\bar{x}_{r+1}=n-\lambda +d$ vertices, we have $|R_{n-1}^{D}(X)|\geq n-\lambda +d\geq \mu$. Therefore, $h(D,\lambda ,\mu )\leq \lambda +\mu -2$.

If $d<\lambda$ and $\lambda +\mu > n+d$, let $k=\lambda +\mu -n-d$. Notice that $R_{\lambda +\mu -d-1}^{D}(X_{i})=V(D)$, for $1\leq i\leq k$. If $k\geq r+1$, then $|R_{\lambda +\mu -d-1}^{D}(X)|=|V(D)|=n\geq \mu$.

If $k< r+1$, then $R_{\lambda +\mu -d-1}^{D}(X)=\bigcap ^{r}_{i={k+1}}[R^{D}_{\lambda + \mu -d-1}(X_{i})]\cap R^{D}_{\lambda +\mu -d-1}(\bar{X}_{r+1})$. Since $\bigcap ^{r}_{i={k+1}}[R^{D}_{\lambda +\mu -d-1}(X_{i})]\cap R^{D}_{ \lambda +\mu -d-1}(\bar{X}_{r+1})$ contains at least $n-(\sum^{r}_{i=k+1}{x_{i}} +\bar{x}_{r+1})$ vertices, we have $|R_{\lambda +\mu -d-1}^{D}(X)|\geq n-(\lambda -d-\sum^{k}_{i=1}{x_{i}}) \geq n-\lambda +d+k=n-\lambda +d+(\lambda +\mu -n-d)=\mu$. We thus arrive at $h(D,\lambda ,\mu )\leq \lambda +\mu -2$.

On the other hand, consider the digraph $L_{n,d}$. Lex X be a vertex set with λ vertices. If $\lambda +\mu < n+1$, since $R_{\lambda +\mu -3}^{L_{n,d}}(i)=\{i,\ldots ,n,\ldots ,\lambda +\mu +i-n-3 \}$, for $n-d+1\leq i\leq n$ and $R_{\lambda +\mu -3}^{L_{n,d}}(i)=\{n-d+1,\ldots ,n,\ldots ,\lambda + \mu +i-n-3\}$, for $1\leq i\leq n-d+1$, we obtain $|\bigcap _{x\in X}R^{L_{n,d}}_{\lambda +\mu -3}(\{x\})|\leq \mu -1$.

Noticing that $R_{n-2}^{L_{n,d}}(i)=\{i,\ldots ,n,\ldots ,i-2\}$, for $n-d+1\leq i\leq n$, and $R_{n-2}^{L_{n,d}}(i)=\{n-d+1,\ldots ,n,\ldots ,i-2\}$, for $1\leq i< n-d+1$. If $d\geq \lambda$ and $\lambda +\mu \geq n+1$, then, for any vertex $x\in X$, $R_{n-2}^{L_{n,d}}(\{x\})$ contains at most $n-2$ vertices. Therefore, $\bigcap _{x\in X}R^{L_{n,d}}_{n-2}(\{x\})$ contains at most $n-\lambda -1$ vertices. As $\lambda +\mu \geq n+1$, we have $|\bigcap _{x\in X}R^{L_{n,d}}_{n-2}(\{x\})|\leq \mu -2$. Consequently, we obtain $h(D,\lambda ,\mu )=\lambda +\mu -2$.

If $d<\lambda$ and $n+1 \leq \lambda +\mu \leq n+d$, we have for any set X, there is at least one vertex $x\in X$ such that $R_{n-2}^{L_{n,d}}(\{x\})$ contains at most $n-3$ vertices. Thus, $\bigcap _{x\in X}R^{L_{n,d}}_{n-2}(\{x\})$ contains at most $n-\lambda -2$ vertices. Since $\lambda +\mu \geq n+1$, we obtain $|\bigcap _{x\in X}R^{L_{n,d}}_{n-2}(\{x\})|\leq \mu -3$. Therefore, $h(D,\lambda ,\mu )=\lambda +\mu -2$.

If $d<\lambda$ and $\lambda +\mu > n+d$, we have $R_{\lambda +\mu -d-2}^{L_{n,d}}(i)=V({L_{n,d}})$, for $2n+2-\lambda -\mu \leq i\leq n$, and $R_{\lambda +\mu -d-2}^{L_{n,d}}(i)=\{n-d+1,\ldots ,n,\ldots , \lambda +\mu +i-n-d-2\}$, for $1\leq i<2n+2-\lambda -\mu$. Thus, for any vertex set X of λ vertices, there is a set $Y\subseteq X$ of at least $n+1-\mu$ vertices, such that, for any vertex $y\in Y$, $R_{\lambda +\mu -d-2}^{L_{n,d}}(\{y\})$ contains at most $\lambda +\mu +y-n-2$ vertices, where $1\leq y<2n+2-\lambda -\mu$. It follows that $\bigcap _{y\in Y}R^{L_{n,d}}_{\lambda +\mu -d-2}(\{y\})$ containing at most $\mu -1$ vertices. Therefore, $|\bigcap _{x\in X}R^{L_{n,d}}_{\lambda +\mu -d-2}(\{x\})|\leq |\bigcap _{y \in Y}R^{L_{n,d}}_{\lambda +\mu -d-2}(\{y\})| \leq \mu -1$. It follows that $h(D,\lambda ,\mu )=\lambda +\mu -2$. This completes the proof. □

Lemma 3.2

([10])

Let $D\in P_{n}(d)$ and $\emptyset \neq X\subseteq V(D)$. Then, for nonnegative integers i, j, t, k, we have $R^{D}_{i}(X)=R^{D}_{i-j}(R^{D}_{j}(X))$ for $i\geq j$, and $|\bigcup ^{k}_{t=0}R^{D}_{t}(X)|\geq \min \{|X|+k,n\}$.

Theorem 3.3

Let $D\in P_{n}(d)$ and $1\leq \lambda ,\mu \leq n$. Then

$$ k(D,\lambda ,\mu ) \leq \textstyle\begin{cases} n-\lceil \frac{ d-\mu +1}{\lambda }\rceil ,&\mu \leq d, \\ n+\mu -d-1,&\mu > d, \end{cases} $$

and the bound can be attained by the digraph $L_{n,d}$.

Proof

Let $X\subseteq V(D)$ be a vertex set of any λ vertices. Set $X=\{v_{1},v_{2},\ldots ,v_{\lambda }\}$.

Case 1. If $\mu \leq d$.

For any vertex $v_{i} \in X$, since $D\in P_{n}(d)\subseteq P_{n}$, by Lemma 3.2,

$$ \Biggl\vert \bigcup ^{n-\lceil \frac{ d-\mu +1}{\lambda }\rceil }_{t=0}R^{D}_{t} \bigl( \{v_{i}\}\bigr) \Biggr\vert \geq n-\biggl\lceil \frac{ d-\mu +1}{\lambda }\biggr\rceil +1, $$

where $i=1,2,\ldots ,\lambda$. Let $\frac{ d-\mu +1}{{\lambda }}=k^{\prime }+a$ where $k^{\prime }$ is a nonnegative integer and $0\leq a<1$. Therefore, if $0< a<1$,

$$ \Biggl\vert \bigcap ^{\lambda }_{i=1}\Biggl[\bigcup ^{n-k^{\prime }-1}_{t=0}R^{D}_{t}\bigl( \{v_{i}\}\bigr)\Biggr] \Biggr\vert \geq \lambda \bigl(n-k^{\prime } \bigr)-n(\lambda -1)\geq n-d+\mu +\lambda a-1. $$

Since $\lambda a\geq 1$,

$$ \Biggl\vert \bigcap ^{\lambda }_{i=1}\Biggl[\bigcup ^{n-k^{\prime }-1}_{t=0}R^{D}_{t}\bigl( \{v_{i}\}\bigr)\Biggr] \Biggr\vert \geq n-d+\mu . $$

If $a=0$,

$$ \Biggl\vert \bigcap ^{\lambda }_{i=1}\Biggl[\bigcup ^{n-k^{\prime }-1}_{t=0}R^{D}_{t}\bigl( \{v_{i}\}\bigr)\Biggr] \Biggr\vert \geq \lambda \bigl(n-k^{\prime }+1 \bigr)-n(\lambda -1)\geq n-d+\mu +\lambda -1. $$

Since $\lambda \geq 1$, we have

$$ \Biggl\vert \bigcap ^{\lambda }_{i=1}\Biggl[\bigcup ^{n-k^{\prime }-1}_{t=0}R^{D}_{t}\bigl( \{v_{i}\}\bigr)\Biggr] \Biggr\vert \geq n-d+\mu . $$

It follows that there are at least μ loop vertices $u_{1},u_{2},\ldots ,u_{\mu }$ such that $u_{i}\in \bigcap ^{\lambda }_{i=1}[\bigcup ^{n-k^{\prime }-1}_{t=0}R^{D}_{t}(\{v_{i} \})]$, where $i=1,2,\ldots ,\mu$. That is to say,

$$ \Biggl\vert \bigcap ^{\lambda }_{i=1}R^{D}_{n-k^{\prime }-1} \bigl(\{v_{i}\}\bigr) \Biggr\vert \geq \mu . $$

Therefore, $k(D,\lambda ,\mu ) \leq n-\lceil \frac{ d-\mu +1}{\lambda } \rceil$.

Consider the digraph $L_{n,d}$. Let

$$ t=n-\biggl\lceil \frac{ d-\mu +1}{\lambda }\biggr\rceil \quad \text{and}\quad k^{*}= \biggl\lceil \frac{ d-\mu +1}{\lambda }\biggr\rceil . $$

Then we consider the following two subcases.

Subcase 1. If $\lambda \leq d+1$.

When $\lambda =1$, then $t=n-d+\mu -1$. Noting that

$$ R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\{n-d+1,n-d+2,\ldots , n-d+ \mu -1\}, $$

then

$$ \bigl\vert R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr) \bigr\vert = \mu -1. $$

When $\lambda =2$, as

$$ R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\bigl\{ n-d+1,\ldots ,n-k^{*}\bigr\} $$

and

$$ R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-k^{*}-\mu +2\bigr\} \bigr)=\bigl\{ n-k^{*}-\mu +2,\ldots ,n,1, \ldots , n-2k^{*}-\mu +1\bigr\} , $$

we have

$$ \bigl\vert R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)\cap R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-k^{*}-\mu +2\bigr\} \bigr) \bigr\vert = \bigl\vert \bigl\{ n-k^{*}-\mu +2,\ldots ,n-1 \bigr\} \bigr\vert =\mu -1. $$

When $\lambda =3$, if $n-2k^{*}-\mu +1< n-d+1$, then $k^{*}=1$, $t=n-1$. Let

$$ X=\bigl\{ 1,n-k^{*}-\mu +2,u\bigr\} \subseteq V(L_{n,d}), $$

where $u\in V(L_{n,d})\setminus \{1, n-k^{*}-\mu +2\}$. Then $|X|=3$. Since

$$ R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\bigl\{ n-d+1,\ldots ,n-k^{*}\bigr\} $$

and

$$ R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-k^{*}-\mu +2\bigr\} \bigr)=\bigl\{ n-k^{*}-\mu +2,\ldots ,n,1, \ldots , n-2k^{*}-\mu +1\bigr\} , $$

we have

$$ R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)\cap R^{L_{n,d}}_{t-1} \bigl(\bigl\{ n-k^{*}-\mu +2\bigr\} \bigr)=\bigl\{ n-k^{*}- \mu +2,\ldots ,n-k^{*}\bigr\} . $$

Thus,

$$ \biggl\vert \bigcap _{x\in X}R^{L_{n,d}}_{t-1}\bigl(\{x\} \bigr) \biggr\vert \leq \bigl\vert R^{L_{n,d}}_{t-1}\bigl(\{1 \}\bigr) \cap R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-k^{*}- \mu +2\bigr\} \bigr) \bigr\vert =\mu -1. $$

If $n-2k^{*}-\mu +1\geq n-d+1$, then $n-2k^{*}-\mu +1< n-d+k^{*}+1< n-k^{*}-\mu +2$. Let

$$ X=\bigl\{ 1,n-k^{*}-\mu +2,n-d+k^{*}\bigr\} \subseteq V(L_{n,d}), $$

where $u\in V(L_{n,d})\setminus \{1, n-k^{*}-\mu +2\}$. Then $|X|=3$. Since

$$\begin{aligned}& R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\bigl\{ n-d+1,\ldots , n-k^{*}\bigr\} , \\& R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-d+k^{*}+1\bigr\} \bigr)=\bigl\{ n-d+k^{*}+1,\ldots , n,1,\ldots ,n-d \bigr\} \end{aligned}$$

and

$$ R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-k^{*}-\mu +2\bigr\} \bigr)=\bigl\{ n-k^{*}-\mu +2,\ldots , n,1, \ldots ,n-2k^{*}-\mu +1\bigr\} , $$

we have

$$ \bigcap _{x\in X}R^{L_{n,d}}_{t-1}\bigl(\{x\}\bigr)=\bigl\{ n-k^{*}-\mu +2,\ldots ,n-k^{*} \bigr\} , $$

which implies

$$ \biggl\vert \bigcap _{x\in X}R^{L_{n,d}}_{t-1}\bigl(\{x\} \bigr) \biggr\vert =\mu -1. $$

When $\lambda =4$, if $n-2k^{*}-\mu +1< n-d+1$, then $k^{*}=1$ and $t=n-1$. Let

$$ X=\{1,n-\mu +1,u_{1},u_{2}\}\subseteq V(L_{n,d}), $$

where $u_{1},u_{2}\in V(L_{n,d})\setminus \{1,n-\mu +1\}$. Then $|X|=4$. Since

$$ R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\{n-d+1,\ldots ,n-\mu +1, \ldots n-1\} $$

and

$$ R^{L_{n,d}}_{t-1}\bigl(\{n-\mu +1\}\bigr)=\{n-\mu +1,\ldots ,n,1,\ldots ,n-\mu -1 \}, $$

we have

$$ \bigl\vert R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)\cap R^{L_{n,d}}_{t-1}\bigl(\{n-\mu +1\}\bigr) \bigr\vert = \bigl\vert \{n- \mu +1,\ldots ,n-1\} \bigr\vert =\mu -1. $$

This implies

$$ |\bigcap _{x\in X}R^{L_{n,d}}_{t-1}\bigl(\{x\}\bigr)\leq \mu -1. $$

If $n-d+k^{*}+1>n-2k^{*}-\mu +1\geq n-d+1$, letting

$$ X=\bigl\{ 1,n-k^{*}-\mu +2,n-d+k^{*}+1,w\bigr\} \subseteq V(L_{n,d}), $$

where $w \in V(L_{n,d})\setminus \{1,n-k^{*}-\mu +2,n-d+k^{*}+1\}$, we have $|X|=4$. Since

$$\begin{aligned}& R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\bigl\{ n-d+1,\ldots , n-k^{*}\bigr\} , \\& R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-d+k^{*}+1\bigr\} \bigr)=\bigl\{ n-d+k^{*}+1,\ldots ,n,1,\ldots , n-d \bigr\} , \\& R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-\mu -k^{*}+2\bigr\} \bigr)=\bigl\{ n-\mu -k^{*}+2,\ldots ,n,1, \ldots , n-2k^{*}-\mu +1\bigr\} , \end{aligned}$$

we have

$$ R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)\cap R^{L_{n,d}}_{t-1} \bigl(\bigl\{ n-\mu -k^{*}+2\bigr\} \bigr) \cap R^{L_{n,d}}_{t-1} \bigl(\bigl\{ n-d+k^{*}+1\bigr\} \bigr)=\bigl\{ n-k^{*}-\mu +2,\ldots , n-k^{*} \bigr\} . $$

Thus,

$$ \biggl\vert \bigcap _{x\in X}R^{L_{n,d}}_{t-1}\bigl(\{x\} \bigr) \biggr\vert \leq \bigl\vert R^{L_{n,d}}_{t-1}\bigl(\{1 \}\bigr) \cap R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-\mu -k^{*}+2\bigr\} \bigr)\cap R^{L_{n,d}}_{t-1}\bigl( \bigl\{ n-d+k^{*}+1 \bigr\} \bigr) \bigr\vert \leq \mu -1. $$

If $n-d+k^{*}+1\leq n-2k^{*}-\mu +1$, letting

$$ X=\bigl\{ 1,n-d+k^{*}+1,n-k^{*}-\mu +2,n-2k^{*}-\mu +2\bigr\} \subseteq V(L_{n,d}), $$

we have $|X|=4$. Since

$$\begin{aligned}& R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\bigl\{ n-d+1,\ldots , n-k^{*}\bigr\} , \\& R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-d+k^{*}+1\bigr\} \bigr)=\bigl\{ n-d+k^{*}+1,\ldots ,n,1,\ldots , n-d \bigr\} , \\& R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-k^{*}-\mu +2\bigr\} \bigr)=\bigl\{ n-\mu -k^{*}+2,\ldots ,n,1, \ldots , n-2k^{*}-\mu +1\bigr\} , \\& R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-2k^{*}-\mu +2\bigr\} \bigr)=\bigl\{ n-2k^{*}-\mu +2,\ldots ,n,1, \ldots , n-3k^{*}-\mu +1\bigr\} , \end{aligned}$$

we have

$$ \biggl\vert \bigcap _{x\in X}R^{L_{n,d}}_{t-1}\bigl(\{x\} \bigr) \biggr\vert = \bigl\vert \bigl\{ n-k^{*}-\mu +2,n-k^{*}- \mu +3,\ldots , n-k^{*}\bigr\} \bigr\vert =\mu -1. $$

When $\lambda \geq 5$, if $n-2k^{*}-\mu +1< n-d+1$, letting

$$ X=\bigl\{ 1,n-k^{*}-\mu +2,v_{1},\ldots ,v_{\lambda -2}\bigr\} \subseteq V(L_{n,d}), $$

where $v_{i} \in V(L_{n,d})\setminus \{1,n-k^{*}-\mu +2\}$ for $i=1,2,\ldots ,\lambda -2$, then $|X|=\lambda$. Since

$$ R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\bigl\{ n-d+1,\ldots , n-k^{*}\bigr\} $$

and

$$ R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-k^{*}-\mu +2\bigr\} \bigr)=\bigl\{ n-\mu -k^{*}+2,\ldots ,n,1, \ldots , n-2k^{*}-\mu +1\bigr\} , $$

we have

$$ ( \bigcap _{x\in X}R^{L_{n,d}}_{t-1}\bigl(\{x\}\bigr) \subseteq \bigl( R^{L_{n,d}}_{t-1}\bigl( \{1\}\bigr)\cap R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-\mu -k^{*}+2\bigr\} \bigr)\bigr)=\bigl\{ n-k^{*}-\mu +2, \ldots , n-k^{*} \bigr\} , $$

which implies that

$$ \biggl\vert \bigcap _{x\in X}R^{L_{n,d}}_{t-1}\bigl(\{x\} \bigr) \biggr\vert \leq \mu -1. $$

If $n-d+k^{*}+1>n-2k^{*}-\mu +1\geq n-d+1$, letting

$$ X=\bigl\{ 1,n-d+k^{*}+1,n-k^{*}-\mu +2,v_{1},\ldots ,v_{\lambda -3}\bigr\} \subseteq V(L_{n,d}), $$

where $v_{i}\in V(L_{n,d})\setminus \{1,n-d+k^{*}+1,n-k^{*}-\mu +2\}$, for $i=1,2,\ldots ,\lambda -3$, then $|X|=\lambda$. As

$$\begin{aligned}& R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\bigl\{ n-d+1,\ldots , n-k^{*}\bigr\} , \\& R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-d+k^{*}+1\bigr\} \bigr)=\bigl\{ n-d+k^{*}+1,\ldots ,n,1,\ldots , n-d \bigr\} , \end{aligned}$$

and

$$ R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-k^{*}-\mu +2\bigr\} \bigr)=\bigl\{ n-\mu -k^{*}+2,\ldots ,n,1, \ldots , n-2k^{*}-\mu +1\bigr\} , $$

we have

$$ R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)\cap R^{L_{n,d}}_{t-1} \bigl(\bigl\{ n-\mu -k^{*}+2\bigr\} \bigr) \cap R^{L_{n,d}}_{t-1} \bigl(\bigl\{ n-d+k^{*}+1\bigr\} \bigr)=\bigl\{ n-k^{*}-\mu +2,\ldots , n-k^{*} \bigr\} . $$

Thus,

$$ \biggl\vert \bigcap _{x\in X}R^{L_{n,d}}_{t-1}\bigl(\{x\} \bigr) \biggr\vert \leq \bigl\vert R^{L_{n,d}}_{t-1}\bigl(\{1 \}\bigr)\cap R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-\mu -k^{*}+2\bigr\} \bigr)\cap R^{L_{n,d}}_{t-1}\bigl( \bigl\{ n-d+k^{*}+1\bigr\} \bigr) \bigr\vert =\mu -1. $$

If $n-(r+1)k^{*}-\mu +1< n-d+k^{*}+1\leq n-rk^{*}-\mu +1$ and $2\leq r\leq \lambda -3$, letting

$$ Y=\bigl\{ 1,n-d+k^{*}+1,n-k^{*}-\mu +2,n-2k^{*}-\mu +2,\ldots ,n-rk^{*}- \mu +2\bigr\} \subseteq V(L_{n,d}), $$

and

$$ X=\bigl(Y \cup \{v_{1},\ldots ,v_{\lambda -r-2}\}\bigr)\subseteq V(L_{n,d}), $$

where $v_{i} \in V(L_{n,d})\setminus \{1,n-d+k^{*}+1,n-k^{*}-\mu +2,n-2k^{*}- \mu +2,\ldots ,n-rk^{*}-\mu +2\}$ for $i=1,2,\ldots ,\lambda -r-2$, then $|X|=\lambda$. Since

$$\begin{aligned}& R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\bigl\{ n-d+1,\ldots , n-k^{*}\bigr\} , \\& R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-d+k^{*}+1\bigr\} \bigr)=\bigl\{ n-d+k^{*}+1,\ldots ,n,1,\ldots , n-d \bigr\} , \end{aligned}$$

and for $1\leq i\leq r$,

$$ R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-\mu -ik^{*}+2\bigr\} \bigr)=\bigl\{ n-\mu -ik^{*}+2,\ldots ,n,1, \ldots , n-(i+1)k^{*}-\mu +1\bigr\} , $$

we have

$$ \bigcap _{x\in Y}R^{L_{n,d}}_{t-1}\bigl(\{x\}\bigr)=\bigl\{ n-k^{*}-\mu +2,\ldots , n-k^{*} \bigr\} . $$

Thus,

$$ \biggl\vert \bigcap _{x\in Y}R^{L_{n,d}}_{t-1}\bigl(\{x\} \bigr) \biggr\vert \leq \biggl\vert \bigcap _{x\in X}R^{L_{n,d}}_{t-1} \bigl( \{x\}\bigr) \biggr\vert =\mu -1. $$

If $n-d+k^{*}+1\leq n-(\lambda -2)k^{*}-\mu +1$, letting

$$ X=\bigl\{ 1,n-d+k^{*}+1,n-k^{*}-\mu +2,n-2k^{*}-\mu +2,\ldots ,n-(\lambda -2)k^{*}- \mu +2 \bigr\} \subseteq V(L_{n,d}), $$

we have $|X|=\lambda$. Since

$$\begin{aligned}& R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\bigl\{ n-d+1,\ldots , n-k^{*}\bigr\} , \\& R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-d+k^{*}+1\bigr\} \bigr)=\bigl\{ n-d+k^{*}+1,\ldots ,n,1,\ldots , n-d \bigr\} , \end{aligned}$$

for $1\leq i\leq \lambda -2$

$$ R^{L_{n,d}}_{t-1}\bigl(\bigl\{ n-\mu -ik^{*}+2\bigr\} \bigr)=\bigl\{ n-\mu -ik^{*}+2,\ldots ,n,1, \ldots , n-(i+1)k^{*}-\mu +1\bigr\} , $$

and $n-d+k^{*}+1>n-(\lambda -1)k^{*}-\mu +1$, we have

$$ \bigcap _{x\in Y}R^{L_{n,d}}_{t-1}\bigl(\{x\}\bigr)=\bigl\{ n-k^{*}-\mu +2,\ldots , n-k^{*} \bigr\} . $$

Therefore,

$$ \biggl\vert \bigcap _{y\in Y}R^{L_{n,d}}_{t-1}\bigl(\{y\} \bigr) \biggr\vert \leq \biggl\vert \bigcap _{x\in X}R^{L_{n,d}}_{t-1} \bigl( \{x\}\bigr) \biggr\vert =\mu -1. $$

From the above, we have $k(L_{n,d},\lambda ,\mu )\geq n-\lceil \frac{ d-\mu +1}{\lambda }\rceil$, it follows that $k(L_{n,d},\lambda ,\mu )= n-\lceil \frac{ d-\mu +1}{\lambda } \rceil$.

Subcase 2. If $\lambda > d+1$.

If $\lambda > d+1$, then $t=n-1$. It is easy to see that

$$\begin{aligned}& R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\{n-d+1,n-d+2,\ldots ,n-1\}, \\& R^{L_{n,d}}_{t-1}\bigl(\{2\}\bigr)=\{n-d+1,n-d+2,\ldots ,n\}, \end{aligned}$$

for $3\leq i\leq \lambda -d$,

$$ R^{L_{n,d}}_{t-1}\bigl(\{\lambda -d\}\bigr)=\{n-d+1,\ldots ,n,1,\ldots , \lambda -d-2\} $$

and for $i=n-d+1,\ldots ,n$,

$$ R^{L_{n,d}}_{t-1}\bigl(\{i\}\bigr)=\{i,\ldots ,n,1,\ldots ,i-2 \}. $$

Let

$$ X_{1}=\{1,\ldots ,\lambda -d,n-d+1\}, \qquad X_{2}=\{n-d+2, \ldots ,n \}\quad \text{and}\quad X=X_{1}\cup X_{2}. $$

Then $|X|=\lambda$. Since

$$ \bigcap _{x\in X_{1}}R^{L_{n,d}}_{t-1}\bigl(\{x\}\bigr)= \{n-d+1,n-d+2,\ldots , n-1 \} $$

and

$$ \bigcap _{x\in X_{2}}R^{L_{n,d}}_{t-1}\bigl(\{x\}\bigr)=\{n,1, \ldots , n-d-1\}, $$

we have $\bigcap _{x\in X}R^{L_{n,d}}_{t-1}(\{x\})=\upphi$. Therefore,

$$ k(L_{n,d},\lambda ,\mu )\geq n-\biggl\lceil \frac{ d-\mu +1}{\lambda } \biggr\rceil , $$

it follows that

$$ k(L_{n,d},\lambda ,\mu )= n-\biggl\lceil \frac{ d-\mu +1}{\lambda } \biggr\rceil . $$

Case 2. If $\mu > d+1$.

Let $X \subseteq V(D)$ be a vertex set of any λ vertices. Set $X=\{v_{1},v_{2},\ldots , v_{\lambda }\}$. For any vertex $v_{i} \in X$, since $D\in P_{n}(d)\subseteq P_{n}$, by Lemma 3.2,

$$ \Biggl\vert \bigcup ^{n-1}_{t=0}R^{D}_{t} \bigl(\{v_{i}\}\bigr) \Biggr\vert \geq n-1+1=n, $$

where $i=1,2,\ldots ,\lambda$. Therefore, each loop vertex $u_{i}\in \bigcup ^{n-1}_{t=0}R^{D}_{t}(\{v_{i}\})$, where $i=1,2,\ldots ,d$. Then

$$ \{u_{1},u_{2},\ldots ,u_{d}\}\subseteq \Biggl(\bigcap ^{\lambda }_{i=1}R^{D}_{n-1} \bigl( \{v_{i}\}\bigr)\Biggr). $$

Since $u_{1},u_{2},\ldots ,u_{d}$ are loop vertices, there are at least $\mu -d$ vertices $w_{1},w_{2},\ldots ,w_{\mu -d}$ and $w_{i}\notin \{u_{1},u_{2},\ldots ,u_{d}\}$ such that

$$ \{w_{1},w_{2},\ldots ,w_{\mu -d}\}\subseteq R^{D}_{\mu -d}\bigl(\{u_{1},u_{2}, \ldots ,u_{d}\}\bigr), $$

where $i=1,2,\ldots ,\mu -d$. It follows that

$$ \Biggl\vert \bigcap ^{\lambda }_{i=1}R^{D}_{n+\mu -d+1} \bigl(\{v_{i}\}\bigr) \Biggr\vert \geq d+\mu -d= \mu . $$

We thus arrive at

$$ k(D,\lambda ,\mu )\leq n+\mu -d-1. $$

Next we consider the digraph $L_{n,d}$. Let $X\subseteq L_{n,d}$ be a vertex set of λ vertices and set $X=\{1,2,\ldots ,\lambda \}$. Let $t=n+\mu -d-1$. Since for $i=2,\ldots ,n-d$,

$$ R^{L_{n,d}}_{t-1}\bigl(\{1\}\bigr)=\{n-d+1,\ldots ,n, 1,\ldots ,\mu -d-1\} \subseteq R^{L_{n,d}}_{t-1}\bigl(\{i\}\bigr), $$

and for $j=n-d+1,\ldots ,n$,

$$ R^{L_{n,d}}_{t-1}\bigl(\{j\}\bigr)=\{1,\ldots ,n\}, $$

we have

$$ \bigcap _{x\in X}R^{L_{n,d}}_{t-1}\bigl(\{x\}\bigr)= \{n-d+1,\ldots ,n,1,\ldots , \mu -d-1\}, $$

which implies that

$$ \biggl\vert \bigcap _{x\in X}R^{L_{n,d}}_{t-1}\bigl(\{x\} \bigr) \biggr\vert =\mu -1. $$

Therefore, $k(L_{n,d},\lambda ,\mu )\geq n+\mu -d-1$. It follows that

$$ k(L_{n,d},\lambda ,\mu )= n+\mu -d-1. $$

Combining the proofs of Cases 1 and 2, the theorem follows as expected. □

Theorem 3.4

Let $D\in P_{n}$ with girth s. Then

$$ k(D,\lambda ,\mu ) \leq \textstyle\begin{cases} n-s+(n-1-\lfloor \frac{ n-\mu }{\lambda }\rfloor )s,&\lambda \leq s, \\ n-s+(n-1-\lfloor \frac{ n-\mu }{s}\rfloor )s,&\lambda >s. \end{cases} $$

Proof

Let $C_{s}$ be a directed cycle of length s in $D^{(s)}$. Consider the digraph $D^{(s)}$. Choose any r vertices $w_{1},w_{2},\ldots ,w_{r}$ of $C_{s}$. Let $\frac{n-\mu }{r}=k+\frac{{b}}{r}$ where $0\leq b< r-1$. In $D^{(s)}$, since $w_{i}$ is a loop vertex, $|R^{D^{s}}_{n-k-1}\{w_{i}\}|\geq n-k=n-\frac{n-\mu }{r}+\frac{{b}}{r}$ where $i=1,2,\ldots ,r$. Therefore

$$ \Biggl\vert \bigcap ^{r}_{i=1}R^{D^{s}}_{n-k-1} \{w_{i}\} \Biggr\vert \geq r\biggl(n-\frac{n-\mu }{r}+ \frac{{b} }{r}\biggr)-(r-1)n=\mu +b\geq \mu . $$

It follows that

$$ \Biggl\vert \bigcap ^{r}_{i=1}R^{D}_{(n-k-1)s} \{w_{i}\} \Biggr\vert \geq \mu . $$

For any λ vertices $v_{1},v_{2},\ldots ,v_{\lambda }\in V(D)$, there is a walk of length $n-s$ from $v_{i}$ to a vertex $u_{i}$ of $C_{s}$ where $i=1,2,\ldots ,\lambda$. If $\lambda \leq s$, then $|u_{1},u_{2},\ldots ,u_{\lambda }|\leq \lambda$ and if $\lambda > s$, then $|u_{1},u_{2},\ldots ,u_{\lambda }|\leq s$. Hence,

$$ k(D,\lambda ,\mu ) \leq \textstyle\begin{cases} n-s+(n-1-\lfloor \frac{ n-\mu }{\lambda }\rfloor )s,&\lambda \leq s, \\ n-s+(n-1-\lfloor \frac{ n-\mu }{s}\rfloor )s,&\lambda >s. \end{cases} $$

 □

4 Conclusions

In this paper, we studied μ-scrambling indices of primitive digraphs and gave some bounds for the λth lower and upper μ-scrambling indices of primitive digraphs with d loops. However, the digraphs attaining the sharp upper bounds are not determined completely. For a general given primitive digraph, its μ-scrambling indices are not given. It would be nice to settle these problems in further research.