Identifying codes and locating-dominating sets on paths and cycles

TL;DR

This paper characterizes $r$-identifying codes and 2-locating-dominating sets in paths and cycles, providing complete solutions for these graph classes and parameters.

Contribution

It offers the first complete characterization of $r$-identifying codes in paths and odd cycles, and of 2-locating-dominating sets in cycles.

Findings

Complete results for $r$-identifying codes in paths and odd cycles.

Complete results for 2-locating-dominating sets in cycles.

New characterizations and bounds for these graph parameters.

Abstract

Let $G=(V,E)$ be a graph and let $r\ge 1$ be an integer. For a set $D \subseteq V$, define $N_r[x] = \{y \in V: d(x, y) \leq r\}$ and $D_r(x) = N_r[x] \cap D$, where $d(x,y)$ denotes the number of edges in any shortest path between $x$ and $y$. $D$ is known as an $r$-identifying code ($r$-locating-dominating set, respectively), if for all vertices $x\in V$ ($x \in V\backslash D$, respectively), $D_r(x)$ are all nonempty and different. In this paper, we provide complete results for $r$-identifying codes in paths and odd cycles; we also give complete results for 2-locating-dominating sets in cycles.

Equations

$$i\vee i+(2r+1)\vee i+2(2r+1)\vee \cdot \cdot \cdot \vee i+g_i(2r+1)\vee h_i,$$

$$i\vee i+q\vee i+2q\vee \cdot \cdot \cdot \vee i+g_{i}q\vee h_i,$$

$$i\vee i+(2r+1)\vee i+2(2r+1)\vee \cdots \vee i+g_i(2r+1),$$

Full text

Identifying codes and locating-dominating sets on paths and cycles

††thanks: Supported in part by National Natural Science Foundation of China (Nos. 60673048 and 10871166) and Shanghai Leading Academic Discipline Project (No. B407).

**Chunxia Chen **     Changhong Lu111Correspond author. E-mail: chlu@math.ecnu.edu.cn 

Department of Mathematics

East China Normal University, Shanghai, 200241, China

**Zhengke Miao

**Department of Mathematics

Xuzhou Normal University, Xuzhou, 221116, China

Abstract Let $G=(V,E)$ be a graph and let $r\geq 1$ be an integer. For a set $D\subseteq V$, define $N_{r}[x]=\{y\in V:d(x,y)\leq r\}$ and $D_{r}(x)=N_{r}[x]\cap D$, where $d(x,y)$ denotes the number of edges in any shortest path between $x$ and $y$. $D$ is known as an $r$-identifying code  ($r$-locating-dominating set, respectively), if for all vertices $x\in V$ ($x\in V\backslash D$, respectively), $D_{r}(x)$ are all nonempty and different. In this paper, we provide complete results for $r$-identifying codes in paths and odd cycles; we also give complete results for $2$-locating-dominating sets in cycles.

Keywords: $r$-identifying codes; $r$-locating-dominating sets; cycles; paths.

1 Introduction

We investigate the well-known identifying codes problem which originated, for instance, from fault diagnosis in multiprocessor systems. The purpose of fault diagnosis is to test the multiprocessor system and locate faulty processors. A multiprocessor system can be modeled as an undirected graph $G=(V,E)$, where $V$ is the set of processors, $E$ is the set of links in the system. Specific detectors are executed on certain selected processors to carry out diagnosis. The selection of these processors is done by generating the code $D$ that allows for unique identification of faulty processors. Every processor corresponding to a codeword vertex tests itself and the processors that are in some areas. Hence, an optimal code(minimum number of codewords) minimizes the amount of overhead required to implement fault diagnosis.

More precisely, let $G=(V,E)$ be an undirected graph and let $r\geq 1$ be an integer. Assume that $D$ is a subset of $V$ at which we place detectors. We define $N_{r}[x]=\{y\in V:d(x,y)\leq r\}$ and $D_{r}(x)=N_{r}[x]\cap D$, where $d(x,y)$ denotes the number of edges in any shortest path between $x$ and $y$. In this sense, $D_{r}[x]$ is the set of all detectors that can detect an attack at $x$. We say that $D$ is an $r$-identifying code ($r$-IC) in $G$ if $D_{r}(x)\neq\emptyset$ for every vertex $x\in V$ and $D_{r}(x)\neq D_{r}(y)$ whenever $x\neq y$. In an $r$-IC, the set of detectors activated by an attack provides a unique signature that allows us to determine where the attack took place. We denote the minimum cardinality of an $r$-identifying code $D$ of $G$ by $M_{r}^{I}(G)$. Note that not all graphs admit an $r$-identifying code. A necessary and sufficient condition to admit an $r$-identifying code is that for any pair of distinct vertices $x$ and $y$ we have $N_{r}[x]\neq N_{r}[y]$ ([1]).

We define a closely related concept as follows. If for all vertices $x\in V\backslash D$, $D_{r}(x)$ are all not empty and different, then we say $D$ is an $r$-locating-dominating set or $r$-LD set for short. The smallest $d$ such that there is an $r$-LD set of size $d$ is denoted by $M_{r}^{LD}(G)$. The concept was introduces (for $r=1$) by Slater [20], motivated by nuclear power plant safety. It can be used for fault detection in distributed system. We also note that $r$-LD sets always exist, since the entire vertex set of a graph is an $r$-LD set.

Identifying codes were introduced in [16], locating-dominating sets in [12]. The literature about $r$-identifying codes and $r$-locating-dominating sets has become quite extensive. There are now numerous papers dealing with identifying codes and local-dominating sets (see for instance [17] for an up-to-date bibliography). The problems of finding optimal $r$-ICs or $r$-LDs in a graph are NP-complete (see [6, 10, 11, 12]). On the other hand, many special graphs have been investigated (see for instance, [2, 3, 4, 5, 7, 9, 14, 15, 19]). In this paper we are interested in studying $r$-IC for cycles and paths and and $2$-LD sets for cycles. This subject was already investigated in [1, 13, 18, 20, 21]. Let $P_{n}$ ($C_{n}$, respectively) be a path (cycle, respectively) of $n$ vertices. For $r=1$, the exact values of $M_{1}^{I}(P_{n})$ and $M_{1}^{I}(C_{n})$ for even cycles was given by [1]; Gravier et al. [13] gave the exact values of $M_{1}^{I}(C_{n})$ for odd cycles. Its analogue for $1$-LD sets was given by Slater [20]. For $r=2$, complete solution for $M_{2}^{I}(C_{n})$ and $M_{2}^{I}(P_{n})$ was provided in [18]. Bertand et al. [1] provided complete results about $M_{2}^{LD}(P_{n})$ and gave the exact value of $M_{2}^{LD}(C_{n})$ for $n=6k$. For all $r\geq 1$, complete results about $r$-ICs for even cycles are provided in [1]; Partial results about $r$-ICs for paths and odd cycles can be found in [1, 13, 21].

The structure of this paper is the following. Motivated by the method in [18], in Section $2$ we give all values of $M_{r}^{I}(C_{n})$ for odd cycle $C_{n}$. In Section $3$ we provide complete results for $r$-identifying codes in paths. In Section $4$ we find the values of $M_{2}^{LD}(C_{n})$.

2 $r$-identifying codes for odd cycle $C_{n}$

In the following, we assume that the vertices of $C_{n}$ have labeled consecutively as $x_{1},x_{2},\cdots,x_{n}$. When we are dealing with a cycle, we also use addition and subtraction modulo $n$, so that, for example, $x_{5n+2}$ means $x_{2}$. The case $n$ even is solved in [1], and it is obvious that $M_{r}^{I}(C_{n})$ is undefined for $n\leq 2r+1$. Hence, we assume that $n\geq 2r+3$.

Lemma 1

Suppose graph $G$ has maximum degree $2$, $y_{1},y_{2},\cdots,y_{2r+2}$ is a path in $G$, and $D$ is an $r$-IC for $G$. Then it is impossible to have $y_{1}\not\in D$ and $y_{2r+2}\not\in D$.

**Proof ** If $y_{1}\not\in D$ and $y_{2r+2}\not\in D$, then $D_{r}(y_{r+1})=D_{r}(y_{r+2})$. $\Box$

Lemma 2

*If $n\geq 4r+2$, $D$ is an $r$-IC for $C_{n}$ if and only if

(1) $x_{i}\in D$ or $x_{i+2r+1}\in D$ for all $i\in\{1,\cdots,n\}$;

(2) there are no $2r+1$ consecutive vertices none of which is in $D$.*

**Proof ** ($\Rightarrow$:) As $D_{r}(x)\neq\emptyset$ for all $x\in V$, condition (2) is necessary. Necessity of condition (1) follows from Lemma 1.

($\Leftarrow$:) Condition $2$ implies that $D_{r}(x)\neq\emptyset$ for each $x\in V$. Now we show that $D_{r}(x_{i})\neq D_{r}(x_{j})$ whenever $i\neq j$. Without loss of generality, we can assume that (A) $i<j$ and the distance from $x_{i}$ to $x_{j}$ is no larger in a clockwise direction around the cycle than in a counterclockwise direction. $x_{i-r}\neq x_{i+r+1}$ since $n\geq 4r+2$ (In fact, $n\geq 2r+2$ is sufficient). By condition (1), either $x_{i-r}\in D$ or $x_{i+r+1}\in D$. If $i+1\leq j\leq i+2r+1$, this implies that $D_{r}(x_{i})\neq D_{r}(x_{j})$. If $j>i+2r+1$, apply condition (2) and (A), we know that $\{x_{i-r},x_{i-r+1},\cdots,x_{i},\cdots,x_{i+r}\}\cap D\neq\emptyset$, and hence $D_{r}(x_{i})\neq D_{r}(x_{j})$. $\Box$

Theorem 3

*For the cycle $C_{2k+1}$, let $k=(2r+1)p+q$ with $p\geq 1$ and $q\in\{0,1,\cdots,2r\}$.

(a)  $M_{r}^{I}(C_{2k+1})={\rm gcd}(2r+1,2k+1)\lceil\frac{2k+1}{2{\rm gcd}(2r+1,2k+1)}\rceil$ if ${\rm gcd}(2r+1,2k+1)\neq 1$;

(b) $M_{r}^{I}(C_{2k+1})=k+2$ if ${\rm gcd}(2r+1,2k+1)=1$ and $q=0$, $2r$;

(c) $M_{r}^{I}(C_{2k+1})=k+1$ if otherwise.*

**Proof ** By Lemma 2, we have the constraint $x_{i}\in D$ or $x_{i+2r+1}\in D$ for $i=1,\cdots,2k+1$, then there are $2k+1$ such constraints and each $x_{i}$ is a term in exactly two of them. Thus, if $D$ has $d$ vertices, at most $2d$ such constrains are satisfied. It follows that $D$ must have at least $k+1$ vertices, i.e., $|D|\geq k+1$.

For notational convenience, we abbreviate $x_{i}$ by $i$ and $x_{i}\in D$ or $x_{j}\in D$ by $i\vee j$ in the constraints. Choose $i\in\{1,\cdots,2r+1\}$, consider the following stream of constraints, which we call stream $i$:

[TABLE]

where $i+g_{i}(2r+1)\leq 2k+1<i+(g_{i}+1)(2r+1)\equiv h_{i}~{}({\rm mod}(2k+1))$ and $h_{i}\in\{1,2,\cdots,2r+1\}$.

Suppose $k=(2r+1)p+q$, where $q\in\{0,1,\cdots,2r\}$. It is easy to know that $h_{1}=1$ if and only if $k=(2r+1)p+r$. Then stream $1$ leads into stream $h_{1}$, which leads into stream $h_{h_{1}}$, and so on, end with last $h_{i}=1$. For example, $k=10$ and $r=2$, stream $1$: $1\vee 6\vee 11\vee 16\vee 21\vee 5$ which leads into stream $5$: $5\vee 10\vee 15\vee 20\vee 4$, it leads into stream $4$, and so on, end with stream $1$. When ${\rm gcd}(2r+1,2k+1)=1$, putting the stream together in this order gives us all constraints, and we call it full constraint stream. When ${\rm gcd}(2r+1,2k+1)\neq 1$, it gives us a full stream, we call full stream $1$, which contains $\frac{2k+1}{{\rm gcd}(2r+1,2k+1)}$ constraints and $\frac{2k+1}{{\rm gcd}(2r+1,2k+1)}+1$ vertices (where two $1s$), similarly, we can get full stream $2$, full stream $3$, $\cdots$, full stream ${\rm gcd}(2r+1,2k+1)$. For example, if $k=10$ and $r=4$, ${\rm gcd}(2r+1,2k+1)=3$, full stream $3$ is given by $3\vee 12\vee 21\vee 9\vee 18\vee 6\vee 15\vee 3$.

Now we consider the case ${\rm gcd}(2r+1,2k+1)\neq 1$. In this case, there are ${\rm gcd}(2r+1,n)$ full streams and each full stream has $\frac{2k+1}{{\rm gcd}(2r+1,2k+1)}$ constraints, thus we need at least $\lceil\frac{2k+1}{2{\rm gcd}(2r+1,2k+1)}\rceil$ vertices from each to be put into $D$, which follows that $|D|\geq{\rm gcd}(2r+1,2k+1)\lceil\frac{2k+1}{2{\rm gcd}(2r+1,2k+1)}\rceil$. Note that $\frac{2k+1}{{\rm gcd}(2r+1,2k+1)}$ is odd, hence each full stream has even vertices. Let us denote $f_{i}^{j}$ the $j$-th vertex of full stream $i$ for $1\leq i\leq{\rm gcd}(2r+1,2k+1)$ and $1\leq j\leq\frac{2k+1}{{\rm gcd}(2r+1,2k+1)}+1$. Let $D=\{f_{i}^{j}:$ $1\leq i\leq{\rm gcd}(2r+1,2k+1)$, $1\leq j\leq\frac{2k+1}{{\rm gcd}(2r+1,2k+1)}+1$, $i+j$ is odd $\}$. It is easy to check that $D$ satisfies the conditions in Lemma 2, and hence $D$ is an $r$-IC of $C_{2k+1}$ with ${\rm gcd}(2r+1,2k+1)\lceil\frac{2k+1}{2{\rm gcd}(2r+1,2k+1)}\rceil$ vertices. The proof of (a) is complete. In the following, we assume that ${\rm gcd}(2r+1,2k+1)=1$.

We first consider the case $q=2r$. In this case, the full constraint stream consists of stream $1$, stream $2$, $\cdots$, stream $2r+1$ in turn. Suppose $D$ is an $r$-IC with $|D|=k+1$. Since the full constraint stream has $2k+1$ constraints and each vertex is in exactly two constraints, then if we can satisfy all the constraints with $k+1$ vertices of $D$, there must be exactly one constraint where both vertices are in $D$, and all other constraints have exactly one of their vertices in $D$. Without loss of generality, we assume that $1\in D$ and $1+(2r+1)\in D$. Then, the rest of the membership of $D$ is forced upon us:

$\bullet$ from stream $1$: use vertices $1$ and $1+z(2r+1)$, $z$ is odd;

$\bullet$ from stream $i$: use vertices $i+z(2r+1)$, $z$ is odd, for $i=2,3,\cdots,2r+1$.

This satisfies condition (1) of Lemma 2 and uses $k+1$ vertices. However, when $p\geq 1$, condition (2) of Lemma 2 is violated, since then $1+2(2r+1),2+2(2r+1),\cdots,(2r+1)+2(2r+1)$ are not in $D$. So, we conclude that $M_{r}^{I}(C_{2k+1})\geq k+2$. Now we construct an $r$-IC with $k+2$ vertices as follows:

$\bullet$ stream $i$: select vertices $i+z(2r+1)$, $z$ is odd, $i\neq r+1$;

$\bullet$ stream $r+1$: select vertices $r+1+z(2r+1)$, $z$ is even;

$\bullet$ add vertices $1$ and $r+1+(2p+1)(2r+1)$.

We now turn to the case $q=0$. In this case, $k=(2r+1)p$ and $p\geq 1$. The full constraint stream consists of stream $2r+1$, stream $2r$, $\cdots$, stream $1$ in turn. Similar to the case $q=2r$, we can prove that $M_{r}^{I}(C_{2k+1})\geq k+2$. We construct an $r$-IC $D$ with $k+2$ vertices as follows:

$\bullet$ stream $i$: select vertices $i+z(2r+1)$, $z$ is odd, $i=1,2r+1$;

$\bullet$ stream $j$: select vertices $j+z(2r+1)$, $z$ is even, $j=2,3,\cdots,2r$.

$\bullet$ add vertices $1$ and $1+2p(2r+1)$.

We now turn to the case $1\leq q\leq r-1$. In this case, we just need to construct an $r$-IC $D$ with $|D|=k+1$. As discussion above, if we select $1$ and $1+(2r+1)$ into $D$, then the other vertices of $D$ are fixed. Now we prove that $D$ satisfies two conditions of Lemma 2, and hence $D$ is an $r$-IC. Obviously, it satisfies condition (1) of Lemma 2. We just need to show it satisfies condition (2) of Lemma 2, i.e. there are no $2r+1$ consecutive vertices none of which is in $D$. Suppose to the contrary that $i+g(2r+1),i+1+g(2r+1),\cdots,2r+1+g(2r+1),1+(g+1)(2r+1),\cdots,i-1+(g+1)(2r+1)\not\in D$ for some $i\in\{1,2,\cdots,2r+1\}$ and $g\in\{0,1,\cdots,2p\}$. If $i=1$, then $1+g(2r+1),2+g(2r+1),\cdots,2r+1+g(2r+1)\not\in D$. Since $1,1+(2r+1)$ are selected into $D$, then $1+z(2r+1)\in D$ for all odd $z\leq 2p$. So $g$ must be even. On the other hand, stream $1:1\vee 1+(2r+1)\vee\cdots\vee 1+2p(2r+1)\vee 2r-2q+1$, leads into stream $2r-2q+1$, and $2r-2q+1+z(2r+1)\in D$ for all even $z\leq 2p$, it contradicts that $g$ is even and $2r-2q+1+g(2r+1)\not\in D$. So, $i>1$. Since $1+(g+1)(2r+1)\not\in D$, hence $g$ is odd.

As $2r-2q+1+z(2r+1)\in D$ for all even $z\leq 2p$, we know that $2r-2q+1\geq i$. If $2r-2q+1\leq 2q+1$, then stream $2r-2q+1$ leads into stream $4r-4q+1$ and $4r-4q+1+z(2r+1)\in D$ for all odd $z\leq 2p$, thus, $4r-4q+1<i$. It contradicts that $2r-2q+1\geq i$. So, $2r-2q+1>2q+1$. Then stream $2r-2q+1$ leads into stream $2r-4q$ and $2r-4q+z(2r+1)\in D$ for all even $z\leq 2p$. Therefore, $2r-4q\geq i$. Similarly, we can show that $2r-4q>2q+1$ and stream $2r-4q$ leads into stream $2r-6q-1$. Let $m_{0}$ be the minimum integer such that $2+2r-m_{0}(2q+1)\leq max\{i,2q+1\}$. We have that: stream $1\Rightarrow$ stream $2r+2-(2q+1)\Rightarrow$ stream $2r+2-2(2q+1)\Rightarrow\cdots\Rightarrow$ stream $2r+2-(m_{0}-1)(2q+1)\Rightarrow$ stream $2r+2-m_{0}(2q+1)$. By the selection of $m_{0}$, we know that $2r+2-m(2q+1)+z(2r+1)\in D$ for all $1\leq m\leq m_{0}$ and all even $z\leq 2p$. Hence, $2r+2-m_{0}(2q+1)\geq i$ and it implies that $2r+2-m_{0}(2q+1)\leq 2q+1$. Therefore, stream $2r+2-m_{0}(2q+1)$ leads into stream $4r-2q+2-m_{0}(2q+1)$ and $4r-2q+2-m_{0}(2q+1)+z(2r+1)\in D$ for all odd $z\leq 2p$. So, we know that $4r-2q+2-m_{0}(2q+1)<i$. It contradicts that $2r+2-m_{0}(2q+1)\geq i$. So, there are no $2r+1$ consecutive vertices none of which is in $D$, and hence $D$ is an $r$-IC. Therefore, $M_{r}^{I}(C_{2k+1})=k+1$ in this case.

Note that ${\rm gcd}(2k+1,2r+1)=1$ implies that $q\neq r$. At last, we consider the case $r+1\leq q\leq 2r-1$. In this case, $2k+1=(2p+1)(2r+1)+2(q-r)$. We just need to construct an $r$-IC $D$ with $|D|=k+1$. As discussion above, if we select $2r+1$ and $2r+1+(2r+1)$ into $D$, then the other vertices of $D$ are fixed. Obviously, $D$ satisfies condition (1) of Lemma 2. We just need to show that there are no $2r+1$ consecutive vertices none of which is in $D$. Suppose to the contrary that $i+g(2r+1),i+1+g(2r+1),\cdots,2r+1+g(2r+1),1+(g+1)(2r+1),\cdots,i-1+(g+1)(2r+1)\not\in D$ for some $i\in\{1,2,\cdots,2r+1\}$ and $g\in\{0,1,\cdots,2p+1\}$. Since $2r+1$ and $2r+1+(2r+1)$ are selected into $D$, $2r+1+z(2r+1)\in D$ for all odd $z\leq 2p$. Hence, $g$ is even.

If $i=1$, then $1+g(2r+1),2+g(2r+1),\cdots,2r+1+g(2r+1)\not\in D$. Consider stream $2r+1:2r+1\vee 2r+1+(2r+1)\vee\cdots\vee 2r+1+2p(2r+1)\vee 4r-2q+1$, which leads into stream $4r-2q+1$, and $4r-2q+1+z(2r+1)\in D$ for all even $z\leq 2p+1$, it contradicts that $g$ is even and $4r-2q+1+g(2r+1)\not\in D$. Hence $i>1$.

As $4r-2q+1+z(2r+1)\in D$ for all even $z\leq 2p+1$, we know that $4r-2q+1<i$. If $4r-2q+1>2(q-r)$, then stream $4r-2q+1$ leads into stream $6r-4q+1$ and $6r-4q+1+z(2r+1)\in D$ for all odd $z\leq 2p+1$, thus, $6r-4q+1\geq i$. It contradicts that $4r-2q+1<i$. So, $4r-2q+1\leq 2(q-r)$, then stream $4r-2q+1$ leads into stream $8r-4q+2$ and $8r-4q+2+z(2r+1)\in D$ for all even $z\leq 2p+1$. Therefore, $8r-4q+2<i$. Similarly, we can show that $8r-4q+2\leq 2(q-r)$ and stream $8r-4q+2$ leads into stream $12r-6q+3$. The following proof is similar to that of the case $1\leq q\leq r-1$. The detail is left to readers. $\Box$

Lemma 4

If $3r+2\leq n\leq 4r+1$, $D$ is a $r$-IC for $C_{n}$ if and only if $x_{i}\in D$ or $x_{i+2r+1}\in D$ for all $i\in\{1,\cdots,n\}$.

**Proof ** Necessity follows from Lemma 1. We shall now observe sufficiency. Assume that $D_{r}(x_{i})=\emptyset$ for some $x_{i}\in V$. Then, none of vertices in $\{x_{i-r},x_{i-r+1},\cdots,x_{i+r}\}$ is in $D$. As $x_{i}\in D$ or $x_{i+2r+1}\in D$ for all $i\in\{1,\cdots,n\}$, it follows that these vertices $x_{i+r+1},x_{i+r+2},\cdots,x_{i+3r+2}$ are all contained in $D$. This contradicts that $n\leq 4r+1$. Hence, $D_{r}(x_{i})\neq\emptyset$ for all $x_{i}\in C_{n}$. The proof of $D_{r}(x_{i})\neq D_{r}(x_{j})$ ( $i\neq j$) is same as that in Lemma 2, and hence omitted in here. $\Box$

Using Lemma 4, we can prove the following theorem, which has been shown by Gravier et al. (cf. Theorem 7 in [13]). The proof is similar to partial proof of Theorem 3, and omitted in here.

Theorem 5

(cf. Theorem 7 in [13]) $M_{r}^{I}(C_{2k+1})={\rm gcd}(2r+1,2k+1)\lceil\frac{2k+1}{2{\rm gcd}(2r+1,2k+1)}\rceil$ for $3r+2\leq 2k+1\leq 4r+1$.

Next, we discuss the value of $M_{r}^{I}(C_{n})$ for $2r+5\leq n\leq 3r+1$, which is the remainder case in [1, 13, 21]. A lemma is given firstly as follows.

Lemma 6

*If $2r+5\leq n\leq 3r+1$, let $n=2r+1+q$ ($4\leq q\leq r$), $D$ is an $r$-IC for $C_{n}$ if and only if

(1)  $x_{i}\in D$ or $x_{i+q}\in D$ for all $i\in\{1,\cdots,n\}$;

(2)  there is at most one set $\{x_{i+1},x_{i+2},\cdots,x_{i+q}\}$ such that none of which is in $D$.*

**Proof ** ($\Rightarrow$:) Suppose to the contrary that $x_{i}\not\in D$ and $x_{i+q}\not\in D$. Since $N_{r}[x_{i-r-1}]=V\backslash\{x_{i},x_{i+1},\cdots,x_{i+q-1}\}$ and $N_{r}[x_{i-r}]=V\backslash\{x_{i+1},x_{i+2},\cdots,x_{i+q}\}$, this leads to the equality $D_{r}(x_{i-r-1})=D_{r}(x_{i-r})$, a contradiction. If there exist two distinct sets $\{x_{i+1},x_{i+2},\cdots,x_{i+q}\}$ and $\{x_{j+1},x_{i+2},\cdots,x_{j+q}\}$ such that none of which is in $D$, then $D_{r}(x_{i-r})=D_{r}(x_{j-r})$, which follows from $N_{r}[x_{i-r}]=V\backslash\{x_{i+1},x_{i+2},\cdots,x_{i+q}\}$ and $N_{r}[x_{j-r}]=V\backslash\{x_{j+1},x_{j+2},\cdots,x_{j+q}\}$.

($\Leftarrow$:) For every vertex $x_{i}\in V$, $N_{r}[x_{i}]=\{x_{i-r},x_{i-r+1},\cdots,x_{i+r}\}$. As $q\leq r$, both $x_{i}$ and $x_{i+q}$ are in $N_{r}[x_{i}]$, and by condition (1), we can conclude that $D_{r}(x_{i})\neq\emptyset$. For distinct vertices $x_{i}$ and $x_{j}$, without loss of generality, we assume that $i<j$ and the distance from $x_{i}$ to $x_{j}$ in a clockwise direction around the cycle is no larger than in a counterclockwise direction. If $i+1\leq j\leq i+q$, we have $x_{i+r+1}\in N_{r}[x_{j}]\backslash N_{r}[x_{i}]$ and $x_{i+r+q+1}\in N_{r}[x_{i}]\backslash N_{r}[x_{j}]$. Hence, by condition (1), $D_{r}(x_{i})\neq D_{r}(x_{j})$. Assume that $j>i+q$. Let $A=\{x_{i+r+1},x_{i+r+2},\cdots,x_{i+r+q}\}$ and $B=\{x_{j+r+1},x_{j+r+2},\cdots,x_{j+r+q}\}$. Then $N_{r}[x_{i}]=V\backslash A$ and $N_{r}[x_{j}]=V\backslash B$. Since $j>i+q$ and the distance from $x_{i}$ to $x_{j}$ in a clockwise direction around the cycle is no larger than in a counterclockwise direction, it implies that $A\cap B=\emptyset$. By condition (2), either $A\cap D\neq\emptyset$ or $B\cap D\neq\emptyset$ holds. Without loss of generality, we assume that $A\cap D\neq\emptyset$ and $x_{i+r+t}\in D$ for some $t\in\{1,2,\cdots,q\}$. Then $x_{i+r+t}\not\in D_{r}(x_{i})$, but $x_{i+r+t}\in D_{r}(x_{j})$ as $A\cap B=\emptyset$. Hence, we have that $D_{r}(x_{i})\neq D_{r}(x_{j})$. $\Box$

Theorem 7

*For the cycle $C_{2k+1}$ with $2r+5\leq 2k+1\leq 3r+1$, let $2k+1=2r+1+q=lq+m$, where $l\geq 3$ is an integer and $m\in\{0,1,\cdots,q-1\}$, then

(1)  $M_{r}^{I}(C_{2k+1})=k+2$ if $l$ is odd, $m=q-1$, $2k+1\geq 5q$ or $l$ is even, $m=1$;

(2)  $M_{r}^{I}(C_{2k+1})={\rm gcd}(q,2k+1)\lceil\frac{2k+1}{2gcd(q,2k+1)}\rceil$ if otherwise.*

**Proof ** Since $2k+1=2r+1+q=lq+m$, then $q$ is even, and hence $m$ is odd. Let $D$ be an $r$-IC for $C_{2k+1}$, by Lemma 6, it must satisfy $2k+1$ constraints: $x_{i}\in D$ or $x_{i+q}\in D$ for $i=1,2,\cdots,2k+1$. Similarly, we abbreviate $x_{i}$ by $i$ and $x_{i}\in D$ or $x_{j}\in D$ by $i\vee j$ in the constraints. For $i\in\{1,2,\cdots,q\}$, we define stream $i$ as follows:

[TABLE]

where $i+g_{i}q\leq 2k+1<i+(g_{i}+1)q\equiv h_{i}~{}({\rm mod}(2k+1))$ and $h_{i}\in\{1,2,\cdots,q\}$.

Then stream $1$ leads into stream $h_{1}$, which leads into stream $h_{h_{1}}$, and so on, end with last $h_{i}=1$. When $gcd(q,2k+1)\neq 1$, it gives us a full stream, denoted by full stream $1$, which contains $\frac{2k+1}{gcd(q,2k+1)}$ constrains and $\frac{2k+1}{{\rm gcd}(q,2k+1)}+1$ vertices (where two $1$s). Similarly, we can get full stream $2$, $\cdots$, full stream $gcd(q,2k+1)$. There are ${\rm gcd}(q,2k+1)$ full streams and each full stream contains $\frac{2k+1}{{\rm gcd}(q,2k+1)}$ constrains and $\frac{2k+1}{{\rm gcd}(q,2k+1)}+1$ vertices (where two $i$s for $1\leq i\leq{\rm gcd}(q,2k+1)$). To satisfy all constrains, we need at least $\frac{2k+1}{2{\rm gcd}(q,2k+1)}$ vertices from each full stream to be put into $D$. This follows that $|D|\geq{\rm gcd}(q,2k+1)\lceil\frac{2k+1}{2{\rm gcd}(q,2k+1)}\rceil$.

Let us denote $f^{j}_{i}$ the $j$-th vertex of full stream $i$ for $1\leq i\leq{\rm gcd}(q,2k+1)$ and $1\leq j\leq\frac{2k+1}{{\rm gcd}(q,2k+1)}+1$. Since $2k+1$ is odd, thus each full stream has even vertices. Let $D=\{f^{j}_{i}:$ $1\leq i\leq{\rm gcd}(q,2k+1)$, $1\leq j\leq\frac{2k+1}{{\rm gcd}(q,2k+1)}+1$, $i+j$ is odd $\}$. It is easy to see that $D$ satisfies conditions of Lemma 6. Hence, $D$ is an $r$-IC with $|D|={\rm gcd}(q,2k+1)\lceil\frac{2k+1}{2{\rm gcd}(q,2k+1)}\rceil$.

We now turn to the case ${\rm gcd}(q,2k+1)=1$. ${\rm gcd}(q,2k+1)=1$ implies that there is only one full stream, which contains $2k+1$ constraints. We discuss it as the following cases.

Case 1. $m=q-1$, $2k+1=lq+q-1$.

In this case, the full stream consists of stream $1$, stream $2$, $\cdots$, stream $q$ in turn. Suppose that $D$ is an $r$-IC for $C_{2k+1}$ with $|D|=k+1$. Since there are $2k+1$ constraints, there must be exactly one constraint where both vertices are in $D$, and all other constraints have exactly one of their vertices in $D$. Without loss of generality, we take $1$ and $1+q$ in $D$, then the rest of the membership of $D$ is forced upon us. If $l$ is even, the membership of $D$ is just the following vertices:

$\bullet$ from stream $i$: use vertices $i+zq$, where the parity of $i$ and $z$ is the same, for $i=1,2,\cdots,q$.

It is easy to check that there are no $q$ consecutive vertices none of which is in $D$. By Lemma 6, $D$ is an $r$-IC with $|D|=k+1$.

If $l$ is odd, the membership of $D$ is just the following vertices:

$\bullet$ from stream $i$: use vertices $i+zq$, $z$ is odd, $i=1,2,3,\cdots,q$.

If $l$ is odd and $2k+1<5q$, i.e., $2k+1=4q-1$, then $D$ satisfies conditions of lemma 6. Hence, $D$ is also an $r$-IC with $|D|=k+1$. However, when $l$ is odd and $2k+1\geq 5q$, condition (2) of Lemma 6 is violated, there exist two sets $\{1+2q,2+2q,\cdots,q+2q\}$ and $\{1+4q,2+4q,\cdots,q+4q\}$ such that none of which is in $D$. Then $D$ is not an $r$-IC. So, we conclude that $M_{r}^{I}(C_{2k+1})\geq k+2$. Now we construct an $r$ - IC with $k+2$ vertices as follows:

$\bullet$ from stream $i:$ use vertices $i+zq$, $z$ is odd and $i\neq\frac{q}{2}+1$;

$\bullet$ from stream $\frac{q}{2}+1:$ use vertices $\frac{q}{2}+1+zq$, $z$ is even;

$\bullet$ add the vertex $\frac{q}{2}+1+lq$.

Case 2. $m=1,2k+1=lq+1$.

In this case, the full stream consists of stream $q$, stream $q-1$, $\cdots$, stream $1$ in turn. We can also prove that $M_{r}^{I}(C_{2k+1})=k+1$ if $l$ is odd and $M_{r}^{I}(C_{2k+1})\geq k+2$ if $l$ is even. The proof is analogous with case 1 and is omitted in here. If $l$ is even, we can construct an $r$-IC with $k+2$ vertices as follows:

$\bullet$ from stream $i:$ use vertices $i+zq$, $z$ is odd and $i\neq\frac{q}{2}+1$;

$\bullet$ from stream $\frac{q}{2}+1:$ use vertices $\frac{q}{2}+1+zq$, $z$ is even.

Case 3. $1<m<q-1,2k+1=lq+m$

Let $D$ denote an $r$-IC for $C_{2k+1}$ with $k+1$ vertices. When $l$ is odd, without loss of generality, we take $q$ and $2q$ in $D$, the rest of the membership of $D$ is forced upon us, and condition (1) of Lemma 6 holds. Next, we prove that there are no $q$ consecutive vertices none of which is in $D$. i.e., $D$ satisfies a stronger property than condition (2) of Lemma 6. Suppose to the contrary that $i+pq,i+1+pq,\cdots,q+pq,1+(p+1)q,2+(p+1)q,\cdots,(i-1)+(p+1)q\not\in D$ for some $i\in\{1,2,\cdots,q\}$ and $p\in\{0,1,\cdots l\}$. By the selection of $D$, we know that $q\in D$ and $q+zq\in D$ for all odd $z\leq l$. So, $p$ is an even positive integer. If $i=1$, it implies that $1+pq,2+pq,\cdots,q+pq\not\in D$. Stream $q:q\vee 2q\vee\cdots\vee lq\vee q-m$, leads into stream $q-m$, and $q-m+zq\in D$ for all even $z\leq l$, it contradicts that $p$ is even and $q-m+pq\not\in D$. Hence, $i>1$.

Since $q-m+zq\in D$ for all even $z\leq l$, then $q-m\leq i-1$. If $q-m>m$, then stream $q-m$ leads into stream $q-2m$, and $q-2m+zq\in D$ for all odd $z\leq l$. Thus, $q-2m\geq i$. It contradicts that $q-m\leq i-1$. Hence, $q-m\leq m$. Then stream $q-m$ leads into stream $2q-2m$, and $2q-2m+zq\in D$ for all even $z\leq l$. Therefore, $2q-2m\leq i-1$. Similarly, we have $2q-2m\leq m$ and stream $2q-2m$ leads into stream $3q-3m$. Let $t_{0}$ be the minimum integer such that $t_{0}(q-m)>max\{i-1,m\}$. We have that: stream $q\Rightarrow$ stream $q-m\Rightarrow$ stream $2(q-m))\Rightarrow$ $\cdots\Rightarrow$ stream $(t_{0}-1)(q-m)\Rightarrow$ stream $t_{0}(q-m)$. By the selection of $t_{0}$, we know that $t(q-m)+zq\in D$ for all $1\leq t\leq t_{0}$ and all even $z\leq l$. Hence, $t_{0}(q-m)\leq i-1$ and it implies that $t_{0}(q-m)>m$. Therefore, stream $t_{0}(q-m)$ leads into stream $t_{0}(q-m)-m$ and $t_{0}(q-m)-m+zq\in D$ for all odd $z\leq l$. So, we know that $t_{0}(q-m)-m\geq i$. It contradicts that $t_{0}(q-m)\leq i-1$. So, there are no $q$ consecutive vertices none of which is in $D$, and hence $D$ is an $r$-IC with $k+1$ vertices.

When $l$ is even, without loss of generality, we take $1$ and $1+q$ in $D$, the rest of the membership of $D$ is forced upon us. The remainder proof is analogous and is omitted in here. $\hfill\Box$

The value of $M_{r}^{I}(C_{2r+3})$ was first obtained in [13]. We present them here for completeness.

Theorem 8

(cf Theorem 5 in [13])  $M_{r}^{I}(C_{2r+3})=\lfloor\frac{4r+6}{3}\rfloor$ for all $r\geq 1$.

**Proof ** Let $D$ be an $r$-IC of $C_{2r+3}$. For all $i$ we have $D_{r}^{I}(x_{i-r-1})=D\backslash\{x_{i},x_{i+1}\}$. Thus, the two following assertions are true:

(a) there is at most a pair $\{x_{i},x_{i}+1\}$ such that $x_{i}\not\in D$ and $x_{i+1}\not\in D$;

(b) there is no pair $\{x_{i},x_{i+2}\}$ such that $x_{i}\not\in D$ and $x_{i+2}\not\in D$. (For otherwise, $D_{r}^{I}(x_{i-r-1})=D_{r}^{I}(x_{i-r})$).

Let us then partition the vertices of $C_{2r+3}$ into blocks of three consecutive vertices, plus possibly one block consisting in one or two vertices. If there is a pair $\{x_{i},x_{i+1}\}$ such that $x_{i}\not\in D$ and $x_{i+1}\not\in D$, then we can partition the vertices such that $x_{i}$ and $X_{i+1}$ are not in the same block. By (a) we know that there is at most one such pair. By (b), any three-element block of the partition contains at least one vertex in $D$. This leads to the inequality $M_{r}^{I}(C_{2r+3})\geq\lfloor\frac{4r+6}{3}\rfloor$.

Now we construct an $r$-IC of $C_{2r+3}$ to attaint this bound.

$\bullet$ $2r=0({\rm mod}\ \ 3)$, $D=\{x_{i}|i=1({\rm mod}\ \ 3)$ or $2({\rm mod}\ \ 3)\}$;

$\bullet$ $2r=1({\rm mod}\ \ 3)$, $D=\{x_{i}|1\leq i\leq 2r+2,i=1({\rm mod}\ \ 3)$ or $2({\rm mod}\ \ 3)\}$;

$\bullet$ $2r=2({\rm mod}\ \ 3)$, $D=\{x_{i}|1\leq i\leq 2r+1,i=1({\rm mod}\ \ 3)$ or $2({\rm mod}\ \ 3)\}\cup\{x_{2r+3}\}$. $\hfill\Box$

3 $r$-identifying codes for path $P_{n}$

We turn now to the path $P_{n}$. We assume that the vertices of $P_{n}$ have labeled consecutively as $x_{1},x_{2},\cdots,x_{n}$. First it is easy to know that $M_{r}^{I}(P_{n})$ is undefined if and only if $n\leq 2r$. In the following, we assume that $n\geq 2r+1$.

Lemma 9

If $D$ is an $r$-IC for $P_{n}$, then $x_{r+2},x_{r+3},\cdots,x_{2r+1}\in D$ and $x_{n-r-1},x_{n-r-2},\cdots,x_{n-2r}\in D$.

**Proof ** For $i=1,2,\cdots,r$, $D_{r}(x_{i})\neq D_{r}(x_{i+1})$ implies that $x_{i+r+1}\in D$, and $D_{r}(x_{n-i})\neq D_{r}(x_{n-i+1})$ implies that $x_{n-r-i}\in D$. $\hfill\Box$

Lemma 10

*If $n\geq 2r+1$, $D$ is an $r$-IC for $P_{n}$ if and only if the following conditions holds:

(1) there are no $2r+2$ consecutive vertices with the first and last not in $D$;

(2) there are no $2r+1$ consecutive vertices none of which is in $D$;

(3) $\{x_{r+2},x_{r+3},\cdots,x_{2r+1}\}\subseteq D$ and $\{x_{n-r-1},x_{n-r-2},\cdots,x_{n-2r}\}\subseteq D$.

(4) $\{x_{1},x_{2},\cdots,x_{r+1}\}\cap D\neq\emptyset$ and $\{x_{n},x_{n-1},\cdots,x_{n-r}\}\cap D\neq\emptyset$.*

**Proof ** ($\Rightarrow:$) Necessity of (1) follows from Lemma 1, and necessity of (2) follows from $D_{r}(x)\neq\emptyset$ for every vertex $x\in V$. Necessity of (3) follows from Lemma 9, and necessity of (4) follows from $D_{r}(x_{1})\neq\emptyset$ and $D_{r}(x_{n})\neq\emptyset$.

($\Leftarrow:$) By conditions (2), (3) and (4), $D_{r}(x)\neq\emptyset$ for every vertex $x\in V$. Consider $x_{i}$ and $x_{j}$, without loss of generality, we assume that $i<j$. If $i+1\leq j\leq i+2r+1$ and $i>r$, by condition (1), either $x_{i-r}\in D$ or $x_{i+r+1}\in D$ holds, and hence $D_{r}(x_{i})\neq D_{r}(x_{j})$. If $i+1\leq j\leq i+2r+1$ and $i\leq r$, by condition (3), we have $x_{i+r+1}\in D$, and hence $D_{r}(x_{i})\neq D_{r}(x_{j})$. If $j>i+2r+1$ and $i>r$, by condition (2), $\{x_{i-r},x_{i-r+1},\cdots,x_{i+r}\}\cap D\neq\emptyset$, so $D_{r}(x_{i})\neq D_{r}(x_{j})$. If $j>i+2r+1$ and $i\leq r$, by condition (4), $\{x_{1},x_{2},\cdots,x_{r+1}\}\cap D\neq\emptyset$, so $D_{r}(x_{i})\neq D_{r}(x_{j})$. $\hfill\Box$

Lemma 10 allows us to proceed for a path much as we did with cycle. Constraint streams are again the focus of our argument. Similarly, we use $i$ as an abbreviation for vertex $x_{i}$ and we modify the definition of constraint stream $i$ to omit the last term $h_{i}$. i.e., we define stream $i$ as follows:

[TABLE]

where $i+g_{i}(2r+1)\leq n$ and $1\leq i\leq 2r+1$.

The following theorem gives all results for $M_{r}^{I}(P_{n})$.

Theorem 11

*Let $n=(2r+1)p+q$, $p\geq 1$, $q\in\{0,1,2,\cdots,2r\}$.

(1) If $q=0$, then $M_{r}^{I}(P_{n})=\frac{(2r+1)p}{2}+1$ if $p$ is even; $M_{r}^{I}(P_{n})=\frac{(2r+1)(p-1)}{2}+2r$ if $p$ is odd.

(2) If $1\leq q\leq r+1$, then $M_{r}^{I}(P_{n})=\frac{(2r+1)p}{2}+q$ if $p$ is even; $M_{r}^{I}(P_{n})=\frac{(2r+1)(p-1)}{2}+2r+1$ if $p$ is odd.

(3) If $r+2\leq q\leq 2r$, then $M_{r}^{I}(P_{n})=\frac{(2r+1)p}{2}+q-1$ if $p$ is even; $M_{r}^{I}(P_{n})=\frac{(2r+1)(p-1)}{2}+2r+1$ if $p$ is odd.*

**Proof ** Let $D$ be an $r$-IC for $P_{n}$. We first discuss the case $q=0$.

(1) If $q=0$, then $r+2,r+3,\cdots,2r+1,1+(p-1)(2r+1),2+(p-1)(2r+1),\cdots,r+(p-1)(2r+1)\in D$, which follows from condition (3) of Lemma 10. For $i\in\{1,2,\cdots,2r+1\}$, the constraint stream $i$ is given as follows: $i\vee i+(2r+1)\vee\cdots\vee i+(p-1)(2r+1)$. To satisfy condition (4) of Lemma 10, there are four possible cases:

(1A) $r+1\in D$ and $r+1+(p-1)(2r+1)\in D$;

(1B) $i\in D$ for some $i\in\{1,2,\cdots,r\}$ and $r+1+(p-1)(2r+1)\in D$;

(1C) $r+1\in D$ and $j+(p-1)(2r+1)\in D$ for some $j\in\{r+2,r+3,\cdots,2r+1\}$;

(1D) $i\in D$ for some $i\in\{1,2,\cdots,r\}$ and $j+(p-1)(2r+1)\in D$ for some $j\in\{r+2,r+3,\cdots,2r+1\}$.

First consider the case (1A). For each stream $i$ ($i\in\{1,2,\cdots,r\}$), we have already taken $i+(p-1)(2r+1)$ into $D$, satisfying the last constrain, and there are $p-2$ remaining constraints. So, we need take at least $\lceil\frac{p-2}{2}\rceil$ vertices from each stream $i$ ($i\in\{1,2,\cdots,r\}$) into $D$ to satisfy the remaining constraint. Turn to stream $r+1$, since $r+1$ and $r+1+(p-1)(2r+1)$ are already put in $D$, satisfying the first and last constraints in stream $r+1$, so, we need at least $\lceil\frac{p-3}{2}\rceil$ to satisfy the remaining constraints. Similarly, it requires at least $\lceil\frac{p-2}{2}\rceil$ vertices to satisfy the remaining constraints in each stream $i$ for $i=r+2,r+3,\cdots,2r+1$. Hence, we need at least $2r+2+\lceil\frac{p-3}{2}\rceil+2r\lceil\frac{p-2}{2}\rceil$ vertices in all.

Now we consider the case (1B). For stream $i$, we have already taken $i$ and $i+(p-1)(2r+1)$ into $D$, satisfying the first and the last constrains, then we need at least $\lceil\frac{p-3}{2}\rceil$ vertices to satisfy the remaining constraints. For each of the other streams, we need at least $\lceil\frac{p-2}{2}\rceil$ vertices to satisfy the remaining constraints. Thus, we need at least $2r+2+\lceil\frac{p-3}{2}\rceil+2r\lceil\frac{p-2}{2}\rceil$ vertices.

We now turn to the case (1C). For stream $j$, we have already taken $j$ and $j+(p-1)(2r+1)$ into $D$, satisfying the first and the last constrains, then we need at least $\lceil\frac{p-3}{2}\rceil$ vertices to satisfy the remaining constraints. For each of the other streams, we need at least $\lceil\frac{p-2}{2}\rceil$ vertices to satisfy the remaining constraints. Thus, we need at least $2r+2+\lceil\frac{p-3}{2}\rceil+2r\lceil\frac{p-2}{2}\rceil$ vertices.

At last we consider the case (1D). For stream $i$, we have already taken $i$ and $i+(p-1)(2r+1)$ into $D$, satisfying the first and the last constrains, then we need at least $\lceil\frac{p-3}{2}\rceil$ vertices to satisfy the remaining constraints. For stream $j$, similarly, we need at least $\lceil\frac{p-3}{2}\rceil$ vertices to satisfy the remaining constraints. For stream $r+1$, we need at leat $\lceil\frac{p-1}{2}\rceil$ vertices to satisfy its constrains. For each of the other streams, we need at least $\lceil\frac{p-2}{2}\rceil$ vertices to satisfy the remaining constraints. Hence, we need at least $2r+2+2\lceil\frac{p-3}{2}\rceil+(2r-2)\lceil\frac{p-2}{2}\rceil+\lceil\frac{p-1}{2}\rceil$ vertices.

Finally, comparing the required minimum number of $D$ in all four cases, we see that when $p$ is even, the minimum is $\frac{(2r+1)p}{2}+1$, which is achieved in both cases (1A), (1B) and (1C) and when $p$ is odd, the minimum is $\frac{(2r+1)(p-1)}{2}+2r$, which is achieved in case (1D).

Next, we construct an $r$-IC, achieving the bound as follows:

When $p$ is even,

$\bullet$ from stream $i$: use vertices $i+z(2r+1)$, where $z$ is even, for $i=1,r+2,r+3,\cdots,2r+1$;

$\bullet$ from stream $j$: use vertices $j+z(2r+1)$, where $z$ is odd, for $j=2,3,\cdots,r+1$;

$\bullet$ add the vertex $1+(p-1)(2r+1)$.

When $p$ is odd,

$\bullet$ from stream $i$: use vertices $i+z(2r+1)$, where $z$ is even, for $i=1,r+2,r+3,\cdots,2r+1$;

$\bullet$ from stream $j$: use vertices $j+z(2r+1)$, where $z$ is odd, for $j=2,3,\cdots,r+1$;

$\bullet$ add vertices $i+(p-1)(2r+1)$ for $i=2,3,\cdots,r$.

(2) If $1\leq q\leq r+1$, then $r+2,r+3,\cdots,2r+1,q+1+(p-1)(2r+1),q+2+(p-1)(2r+1),\cdots,q+r+(p-1)(2r+1)\in D$, which follows from condition (3) of Lemma 10. For $i\in\{1,2,\cdots,q\}$, the constraint stream $i$ is given as follows: $i\vee i+(2r+1)\vee\cdots\vee i+p(2r+1)$. For $i\in\{q+1,\cdots,2r+1\}$, the constraint stream $i$ is given as follows: $i\vee i+(2r+1)\vee\cdots\vee i+(p-1)(2r+1)$. To satisfy condition (4) of Lemma 10, there are four possible cases:

(2A) $i\in D$ for some $i\in\{1,\cdots,q\}$ and $j+p(2r+1)\in D$ for some $j\in\{1,\cdots,q\}$;

(2B) $i\in D$ for some $i\in\{1,\cdots,q\}$ and $j+(p-1)(2r+1)\in D$ for some $j\in\{q+r+1,\cdots,2r+1\}$;

(2C) $i\in D$ for some $i\in\{q+1,\cdots,r+1\}$ and $j+p(2r+1)\in D$ for some $j\in\{1,\cdots,q\}$;

(2D) $i\in D$ for some $i\in\{q+1,\cdots,r+1\}$ and $j+(p-1)(2r+1)\in D$ for some $j\in\{q+r+1,\cdots,2r+1\}$.

First consider the case (2A). We first discuss the situation $i\neq j$. For stream $i$, we have already taken $i$ into $D$, satisfying the first constrain in stream $i$, and hence we need take at least $\lceil\frac{p-1}{2}\rceil$ vertices from stream $i$ to satisfy remaining constraints. For stream $j$, we have already taken $j+p(2r+1)$ into $D$, satisfying the last constrain in stream $j$, and hence we need take at least $\lceil\frac{p-1}{2}\rceil$ vertices from stream $j$ to satisfy remaining constraints. For each stream $t$ with $t\in\{1,\cdots,q\}\backslash\{i,j\}$, we need take at least $\lceil\frac{p}{2}\rceil$ vertices into $D$ to satisfy its constraint. For each stream $t$ with $t\in\{q+1,\cdots,r+1\}$, we have already taken the vertex $t+(p-1)(2r+1)$ into $D$, satisfying the last constrain in stream $t$, and there are $p-2$ remaining constraints. Hence, we need take at least $\lceil\frac{p-2}{2}\rceil$ vertices from stream $t$ to satisfy remaining constraints. For each stream $t$ with $t\in\{r+2,\cdots,r+q\}$, we have already taken $t$ and $t+(p-1)(2r+1)$ into $D$, satisfying the first and the last constrains in stream $t$, and there are $p-3$ remaining constraints. Hence, we need take at least $\lceil\frac{p-3}{2}\rceil$ vertices from stream $t$ to satisfy remaining constraints. For each stream $t$ with $t\in\{r+q+1,\cdots,2r+1\}$, we have already taken $t$ into $D$, satisfying the first in stream $t$. Hence, we need take at least $\lceil\frac{p-2}{2}\rceil$ vertices to satisfy remaining constraints. Therefore, we need at least $2r+2+(q-2)\lceil\frac{p}{2}\rceil+2\lceil\frac{p-1}{2}\rceil+(2r-2q+2)\lceil\frac{p-2}{2}\rceil+(q-1)\lceil\frac{p-3}{2}\rceil$ vertices in all.

We now discuss the situation $i=j$. For stream $i$, we have already taken $i$ and $i+p(2r+1)$ into $D$, satisfying the first and the last constrains in stream $i$, and hence we need take at least $\lceil\frac{p-2}{2}\rceil$ vertices from stream $i$ to satisfy remaining constraints. For each stream $t$ with $t\in\{1,\cdots,q\}\backslash\{i\}$, we need take at least $\lceil\frac{p}{2}\rceil$ vertices into $D$ to satisfy its constraint. For each stream $t$ with $t\in\{q+1,\cdots,2r+1\}$, the discussion is same as above. Therefore, we need at least $2r+2+(q-1)\lceil\frac{p}{2}\rceil+(2r-2q+3)\lceil\frac{p-2}{2}\rceil+(q-1)\lceil\frac{p-3}{2}\rceil$ vertices in all.

We now consider the case (2B). Similarly, we need at least $\lceil\frac{p-1}{2}\rceil$ vertices to satisfy the remaining constrains in stream $i$. For each stream $t$ with $t\in\{1,\cdots,q\}\backslash\{i\}$, we need at lest $\lceil\frac{p}{2}\rceil$ vertices to satisfy the remaining constrains. For each stream $t$ with $t\in\{q+1,\cdots,r+1\}$, we need at lest $\lceil\frac{p-2}{2}\rceil$ vertices to satisfy the remaining constrains. For each stream $t$ with $t\in\{r+2,\cdots,r+q\}$, we need at least $\lceil\frac{p-3}{2}\rceil$ vertices to satisfy the remaining constrains. For stream $j$, we need at least $\lceil\frac{p-3}{2}\rceil$ vertices to satisfy the remaining constrains. For each stream $t$ with $t\in\{q+r+1,\cdots,2r+1\}\backslash\{j\}$, we need at least $\lceil\frac{p-2}{2}\rceil$ vertices to satisfy the remaining constrains. Therefore, we need at least $2r+2+(q-1)\lceil\frac{p}{2}\rceil+\lceil\frac{p-1}{2}\rceil+(2r-2q+1)\lceil\frac{p-2}{2}\rceil+q\lceil\frac{p-3}{2}\rceil$ vertices in all.

We now turn to the case (2C). We need at least $\lceil\frac{p-1}{2}\rceil$ vertices to satisfy the remaining constrains in stream $j$. For each stream $t$ with $t\in\{1,\cdots,q\}\backslash\{j\}$, we need at least $\lceil\frac{p}{2}\rceil$ vertices to satisfy its constrains. For stream $i$, we need at least $\lceil\frac{p-3}{2}\rceil$ vertices to satisfy the remaining constrains. For each stream $t$ with $t\in\{q+1,\cdots,r+1\}\backslash\{i\}$, we need at least $\lceil\frac{p-2}{2}\rceil$ vertices to satisfy the remaining constrains. For each stream $t$ with $t\in\{r+2,\cdots,r+q\}$, we need at least $\lceil\frac{p-3}{2}\rceil$ vertices to satisfy the remaining constrains. For each stream $t$ with $t\in\{r+q+1,\cdots,2r+1\}$, we need at least $\lceil\frac{p-2}{2}\rceil$ vertices to satisfy the remaining constrains. Therefore, we need at least $2r+2+(q-1)\lceil\frac{p}{2}\rceil+\lceil\frac{p-1}{2}\rceil+(2r-2q+1)\lceil\frac{p-2}{2}\rceil+q\lceil\frac{p-3}{2}\rceil$ vertices in all.

At last we consider the case (2D). For each stream $t$ with $t\in\{1,\cdots,q\}$, we need at least $\lceil\frac{p}{2}\rceil$ vertices to satisfy its constrains. For stream $i$, we need at least $\lceil\frac{p-3}{2}\rceil$ vertices to satisfy its remaining constrains. For each stream $t$ with $t\in\{q+1,\cdots,r+1\}\backslash\{i\}$, we need at least $\lceil\frac{p-2}{2}\rceil$ vertices to satisfy its remaining constrains. For each stream $t$ with $t\in\{r+2,\cdots,r+q\}$, we need at least $\lceil\frac{p-3}{2}\rceil$ vertices to satisfy the remaining constrains. For stream $j$, we need at least $\lceil\frac{p-3}{2}\rceil$ vertices to satisfy its remaining constrains. For each stream $t$ with $t\in\{r+q+1,\cdots,2r+1\}\backslash\{j\}$, we need at least $\lceil\frac{p-2}{2}\rceil$ vertices to satisfy the remaining constrains. Therefore, we need at least $2r+2+q\lceil\frac{p}{2}\rceil+(2r-2q)\lceil\frac{p-2}{2}\rceil+(q+1)\lceil\frac{p-3}{2}\rceil$ vertices in all.

Finally, comparing the required minimum number of $D$ in all four cases, we see that when $p$ is even, the minimum is $\frac{(2r+1)p}{2}+q$, which is achieved in case (2A) under the situation $i=j$, and when $p$ is odd, the minimum is $\frac{(2r+1)(p-1)}{2}+2r+1$, which is achieved in cases (2C),(2D) and case (2A) under the situation $i\neq j$.

Next, we construct an $r$-IC, achieving the bound as follows:

When $p$ is even,

$\bullet$ from stream $i$: use vertices $i+z(2r+1)$, where $z$ is even, for $i=1,r+2,r+3,\cdots,2r+1$;

$\bullet$ from stream $j$: use vertices $j+z(2r+1)$, where $z$ is odd, for $j=2,3,\cdots,r+1$;

$\bullet$ add vertices $i+(p-1)(2r+1)$ for $i=r+2,r+3,\cdots,r+q$.

When $p$ is odd,

$\bullet$ from stream $i$: use vertices $i+z(2r+1)$, where $z$ is odd, for $i=1,2,\cdots,q$;

$\bullet$ from stream $j$: use vertices $j+z(2r+1)$, where $z$ is even, for $j=q+1,q+2,\cdots,2r+1$.

(3) The proof of (3) is analogous. We simply include the instruction for how to achieve an optimal set $D$ in this case.

When $p$ is even,

$\bullet$ from stream $i$: use vertices $i+z(2r+1)$, where $z$ is even, for $i=1,r+2,r+3,\cdots,2r+1$;

$\bullet$ from stream $j$: use vertices $j+z(2r+1)$, where $z$ is odd, for $j=2,3,\cdots,r+1$;

$\bullet$ add vertices $i+(p-1)(2r+1)$ for $i=q+1,q+2,\cdots,2r+1$ and $j+p(2r+1)$ for $j=2,3,\cdots,q-r-1$.

When $p$ is odd,

$\bullet$ from stream $i$: use vertices $i+z(2r+1)$, where $z$ is odd, for $i=1,2,\cdots,r$;

$\bullet$ from stream $j$: use vertices $j+z(2r+1)$, where $z$ is even, for $j=r+1,r+3,\cdots,2r+1$.

4 $2$-locating dominating sets for cycle $C_{n}$

Let $A$ and $B$ be two sets. Define $A\triangle B$ as $(A-B)\cup(B-A)$. For three vertices $x,u,v$, if $x\in D_{r}(u)\triangle D_{r}(v)$, then we say $\{u,v\}$ are $r$-separated by $x$, or $x$ $r$-separates $\{u,v\}$. Let $D$ be an $r$-LD for $C_{n}$. Two different vertices $x$ and $y$ not in $D$ are $D$-consecutive if ether $\{x+1,\cdots,y-1\}\subseteq D$ or $\{y+1,\cdots,x-1\}\subseteq D$ holds. Note that a pair of consecutive vertices $\{x,x+1\}$ not in $D$ are also $D$-consecutive.

Lemma 12

*([1]) Let $r\geq 1$ be an integer. Suppose $D$ is an $r$-LD for $C_{n}$. For every vertex $x$ in $D$, we have

(i) $x$ can $r$-separate at most two pairs of consecutive vertices;

(ii) $x$ can $r$-separate at most two pairs of $D$-consecutive vertices.*

**Proof ** (i) For every vertex $x$ in $D$, it $r$-separates at most two pairs of consecutive vertices $\{x-r-1,x-r\}$ and $\{x+r,x+r+1\}$.

(ii) Let $l$ and $l^{\prime}$ be integers such that $0<l\leq r$ and $l^{\prime}>r$. $x$ can at most $r$-separate the following two types of $D$-consecutive vertices: $(x\pm l,x+l^{\prime})$ and $(x\pm l,x-l^{\prime})$. $\hfill\Box$

Lemma 13

*([1])  For $r\geq 2,~{}n\geq 1$, $M_{2}^{LD}(C_{n})\geq\lceil\frac{n}{3}\rceil$.

**Proof ** Let $D$ be an $r$-LD of $C_{n}$. By Lemma 12, and since there are $n-|D|$ pairs of $D$-consecutive vertices, we have $2|D|\geq n-|D|$. $\hfill\Box$

In here, we focus on $r=2$. Our main results is the following theorem.

Theorem 14

*Let $C_{n}$ be a cycle with vertex set $\{x_{1},\cdots x_{n}\}$.

(1) $M_{2}^{LD}(C_{n})=n$ if $n\leq 5$;

(2) $M_{2}^{LD}(C_{n})=\lceil\frac{n}{3}\rceil+1$ if $n=6$ or $n=6k+3$ ($k\geq 1$);

(3) $M_{2}^{LD}(C_{n})=\lceil\frac{n}{3}\rceil$ if otherwise.*

**Proof ** When $n\leq 5$, the distance of any two vertices in $C_{n}$ is no more than $2$. Hence, $M_{2}^{LD}(C_{n})=n$. As a set with size two has only three nonempty subsets, we know that $M_{2}^{LD}(C_{6})\geq 3$. It is easy to know that $D=\{x_{1},x_{3},x_{5}\}$ is a $2$-LD of $C_{6}$. Therefore, $M_{2}^{LD}(C_{6})=3$. In the following, we assume that $n\geq 7$.

$M_{2}^{LD}(C_{n})\geq\lceil\frac{n}{3}\rceil$ holds by Lemma 13, next we construct a $2$-LD achieving the lower bound in the following cases:

$\bullet$ $n=6k$, $D=\{x_{i}|i=6p+4,p\geq 0\}\cup\{x_{i}|i=6q,q\geq 1\}$;

$\bullet$ $n=6k+1$ or $6k+2$, $D=\{x_{i}|i=6p+4,p\geq 0\}\cup\{x_{i}|i=6q,q\geq 1\}\cup\{x_{n}\}$;

$\bullet$ $n=6k+4$, $D=\{x_{i}|i=6p+4,p\geq 0\}\cup\{x_{i}|i=6q,q\geq 1\}\cup\{x_{n-2}\}$;

$\bullet$ $n=6k+5$ and $n>11$, $D=\{x_{i}|i=6p+2,0\leq p\leq k-2\}\cup\{x_{i}|i=6q,1\leq q\leq k-1\}\cup\{x_{n-8},x_{n-7},x_{n-2},x_{n-1}\}$;

$\bullet$ $n=11$, $D=\{x_{1},x_{2},x_{5},x_{9}\}$.

Now we turn to the case $n=6k+3$. By Lemma 13, we have known that $M_{2}^{LD}(C_{n})\geq 2k+1$. We first show that $M_{2}^{LD}(C_{n})\geq 2k+2$. Suppose to the contrary that $D$ is a $2$-LD for $C_{n}$ with $2k+1$ vertices. Then there are $4k+2$ pairs of $D$-consecutive vertices, and hence every vertex in $D$ $2$-separates exactly two pairs of $D$-consecutive vertices, and they are disjoint. We have the following claims.

Claim 1: $D$ contains at most two consecutive vertices in $C_{n}$.

Proof of Claim 1: Since each vertex in $D$ $2$-separates two pairs of $D$-consecutive vertices, it follows that $D$ contains at most four consecutive vertices in $C_{n}$. Suppose that $D$ contains four consecutive vertices in $C_{n}$, without loss of generality, we assume that $\{x_{1},x_{2},x_{3},x_{4}\}\subseteq D$. Then both $x_{1}$ and $x_{4}$ $2$-separate a pair of $D$-consecutive vertices $\{x_{n},x_{5}\}$, a contradiction. If $D$ contains three consecutive vertices in $C_{n}$, without loss of generality, we assume that $\{x_{1},x_{2},x_{3}\}\subseteq D$, then both $x_{1}$ and $x_{3}$ $2$-separate a pair of $D$-consecutive vertices $\{x_{n},x_{4}\}$, a contradiction. $\Box$ Assume that $D=\{x_{i_{1}},x_{i_{2}},\cdots,x_{i_{2k+1}}\}$ with $i_{1}<i_{2}<\cdots<i_{2k+1}$.

Claim 2: $|i_{j}-i_{j+1}|=2$ or $4$ for all $j\in\{1,\cdots,2k+1\}$.

Proof of Claim 2: Since $D_{2}(x)\neq\emptyset$ for any $x\not\in D$, it is easy to know that $|i_{j}-i_{j+1}|\leq 5$. If $|i_{j}-i_{j+1}|=5$ for some $j\in\{1,\cdots,2k+1\}$, then $x_{i_{j}},x_{i_{j+1}}$ both $2$-separate the pair of consecutive vertices $(x_{i_{j}+2},x_{i_{j}+3})$, a contradiction.

Suppose that $|i_{j}-i_{j+1}|=1$ for some $j\in\{1,\cdots,2k+1\}$, without loss of generality, we assume that $x_{1}\in D$ and $x_{2}\in D$. By Claim 1, we know that $x_{3}\not\in D$ and $x_{n}\not\in D$. If $x_{4}\in D$, then either $\{x_{3},x_{5}\}$ or $\{x_{3},x_{6}\}$ is a pair of $D$-consecutive vertices. So, $x_{1}$ and $x_{2}$ $2$-separate the same pair of $D$-consecutive vertices, a contradiction. Thus  $x_{4}\not\in D$. Similarly, $x_{n-1}\not\in D$. If $x_{n-2}$ and $x_{5}$ are both in $D$, then they both $2$-separate the pair of $D$-consecutive vertices $\{x_{n},x_{3}\}$, a contradiction. Without loss of generality, we take $x_{5}\not\in D$. $x_{6}\in D$ implies that the pair of $D$-consecutive vertices $\{x_{3},x_{4}\}$ are $2$-separated by both $x_{1}$ and $x_{6}$. It is a contradiction. $x_{7}\in D$ implies that the pair of $D$-consecutive vertices $\{x_{4},x_{5}\}$ are $2$-separated by both $x_{2}$ and $x_{7}$. It is a contradiction. Hence, $x_{6}$ and $x_{7}$ are both not in $D$. Thus, $D_{2}(x_{5})=\emptyset$, a contradiction. Therefore, $|i_{j}-i_{j+1}|\neq 1$.

Suppose that $|i_{j}-i_{j+1}|=3$ for some $j\in\{1,\cdots,2k+1\}$, without loss of generality, we assume that $x_{1}\in D$ and $x_{4}\in D$. Then $x_{n}\in D$ or $x_{5}\in D$, which follows from the pair of $D$-consecutive vertices $\{x_{2},x_{3}\}$ require to be $2$-separated, however, it contradicts with $|i_{j}-i_{j+1}|\geq 2$.   $\Box$

Since $C_{n}$ contains $6k+3$ vertices and there are $2k+1$ vertices in $D$, thus by Claim 2, there must exist some $j\in\{1,\cdots,2k+1\}$ such that $|i_{j}-i_{j+1}|=|i_{j}-i_{j-1}|$. However, if $|i_{j}-i_{j+1}|=|i_{j}-i_{j-1}|=2$, then $x_{i_{j-1}},x_{i_{j+1}}$ both $2$-separate $\{x_{i_{j}-1},x_{i_{j}+1}\}$; if $|i_{j}-i_{j+1}|=|i_{j}-i_{j-1}|=4$, then there is no vertex in $D$ $2$-separate $\{x_{i_{j}-1},x_{i_{j}+1}\}$. Therefore, $M_{2}^{LD}(C_{n})\geq 2k+2$.

Now, we construct a $2$-LD for $C_{n}$ with $2k+2$ vertices as follows: $D=\{x_{i}|i=6p+1$ or $6p+3,0\leq p\leq k-1\}\cup\{x_{n-1},x_{n-2}\}$. $\hfill\Box$

5 Conclusion

The main purpose of this paper is to give the exact value of $M_{r}^{I}(G)$ for paths and odd cycles for arbitrary positive integer $r$, and of $M_{2}^{LD}(C_{n})$. It would be of interest to extend the latter to $r$-LDs for $r>2$.