Broadcasts on Paths and Cycles

TL;DR

This paper determines various broadcast numbers for all paths and cycles in graphs, providing answers to previously open questions in the study of broadcast functions related to domination and independence.

Contribution

It explicitly calculates broadcast numbers for paths and cycles, advancing understanding of broadcast functions in graph theory.

Findings

Broadcast numbers for all paths and cycles are now explicitly known.

The results resolve an open question from prior research.

The study enhances the theoretical framework of broadcast functions in graphs.

Abstract

A broadcast on a graph $G=(V,E)$ is a function $f: V\longrightarrow \{0,\ldots,\operatorname{diam}(G)\}$ such that $f(v)\leq e\_G(v)$ for every vertex $v\in V$, where$\operatorname{diam}(G)$ denotes the diameter of $G$ and $e\_G(v)$ the eccentricity of $v$ in $G$. The cost of such a broadcast is then the value $\sum\_{v\in V}f(v)$.Various types of broadcast functions on graphs have been considered in the literature, in relation with domination, irredundence, independenceor packing, leading to the introduction of several broadcast numbers on graphs.In this paper, we determine these broadcast numbers for all paths and cycles, thus answering a questionraised in [D.~Ahmadi, G.H.~Fricke, C.~Schroeder, S.T.~Hedetniemi and R.C.~Laskar, Broadcast irredundance in graphs. {\it Congr. Numer.} 224 (2015), 17--31].

Tables

Table 1. Table 1: Broadcast parameters of paths and cycles

	${\gamma }_b=i{r}_b$	$i_b$	$p_b$	${\Gamma }_b=I{R}_b$	${\beta }_b$	$P_b$
$P_n$	$\lceil \frac{n}{3}\rceil$	$\begin{array}{cc}\lceil \frac{2n}{5}\rceil ,& \\ n\ne 3& \end{array}$		$\begin{array}{cc}n-1,& \\ n\ge 2& \end{array}$	$\begin{array}{cc}2n-4,& \\ n\ge 3& \end{array}$	$n-1$
	Th. 3.12	Th. 3.7	Th. 3.16	Th. 3.1	[20]	Th. 3.14
$C_n$	$\lceil \frac{n}{3}\rceil$	$\begin{array}{cc}\lceil \frac{2n}{5}\rceil ,& \\ n\ne 3& \end{array}$		$\begin{array}{cc}2\left(\lfloor \frac{n}{2}\rfloor -1\right),& \\ n\ge 4& \end{array}$	$\begin{array}{cc}n-2,& \\ n\ge 3& \end{array}$	$\lfloor \frac{n}{2}\rfloor$
	Th. 3.13	Th. 3.8	Th. 3.17	Th. 3.3, 3.4	Th. 3.5	Th. 3.14

Equations

$$N_{f}(v)=\big{\{}u:d_{G}(u,v)\leq f(v)\big{\}},$$

$$N_f(V_f^{+})={\cup }_{v\in V_f^{+}}{N}_f(v).$$

$$H_{f}(u)=\big{\{}v\in V^{+}_{f}:d_{G}(u,v)\leq f(v)\big{\}}.$$

$$PN_{f}(v)=\big{\{}u\in V:H_{f}(u)=\{v\}\big{\}},$$

$$PB_{f}(v)=\left\{\begin{array}[]{ll}\{v\}&\text{if $f(v)=1$ and $PN_{f}(v)=\{v\}$},\\[8.61108pt] \big{\{}u\in PN_{f}(v):d_{G}(u,v)=f(v)\big{\}}&\text{otherwise}.\end{array}\right.$$

$$\sigma (f)=\sum _{v\in V_f^{+}}f(v).$$

$$i{r}_b(G)\le {\gamma }_b(G)\le \gamma (G)\le \Gamma (G)\le {\Gamma }_b(G)\le I{R}_b(G).$$

$${\gamma }_b(G)\le i_b(G)\le {\beta }_b(G)\ge i(G)\ge \gamma (G)\ge {\gamma }_b(G).$$

$$|V(P_{n})|\geq t^{\prime}+\sum_{j=t^{\prime}+1}^{t}\big{(}d_{P_{n}}(x_{i_{j}},x_{i_{j}}^{p})+1\big{)}\geq\sum_{j=1}^{t^{\prime}}f(x_{i_{j}})+\sum_{j=t^{\prime}+1}^{t}\big{(}f(x_{i_{j}})+1\big{)}=I\!R_{b}(P_{n})+t-t^{\prime},$$

$$f(x_{i})=\left\{\begin{array}[]{cl}\left\lfloor\frac{n}{2}\right\rfloor-1&\text{if }i\in\big{\{}\left\lfloor\frac{n}{2}\right\rfloor,\left\lceil\frac{n}{2}\right\rceil+1\big{\}},\\[4.30554pt] 0&\text{otherwise}.\end{array}\right.$$

$$\sigma(f)=\left\{\begin{array}[]{ll}n-2&\text{if }n\text{ is even,}\\ n-3&\text{if }n\text{ is odd.}\end{array}\right.$$

$${\Gamma }_b(C_n)=\sum _{x_i\in V_g^{+}}g(x_i)=\sum _{x_i\in V_g^{+}}|Q_{x_i}|=|{\cup }_{x_i\in V_g^{+}}{Q}_{x_i}|\le n-|V_g^{+}|=n-t.$$

$$f(x_{i})=\left\{\begin{array}[]{cl}\left\lfloor\frac{n}{2}\right\rfloor-1&\text{if }i\in\big{\{}0,\left\lfloor\frac{n}{2}\right\rfloor\big{\}},\\[4.30554pt] 0&\text{otherwise},\end{array}\right.$$

$$\sigma(g)=g(x_{i})=e_{C_{n}}(x_{i})=\left\lfloor\frac{n}{2}\right\rfloor\leq 2\Big{(}\left\lfloor\frac{n}{2}\right\rfloor-1\Big{)},$$

$$\beta_{b}(C_{n})=\sigma(g)=\sum_{x_{i_{j}}\in V^{+}_{g}}g(x_{i_{j}})\leq\sum_{x_{i_{j}}\in V^{+}_{g}}(d_{C_{n}}(x_{i_{j}},x_{i_{j+1}})-1)=n-|V^{+}_{g}|\leq n-3\leq 2\Big{(}\left\lfloor\frac{n}{2}\right\rfloor-1\Big{)},$$

$$f_0(P_n)=(01010{)}^q,f_1(P_n)=(01010{)}^{q}1,f_2(P_n)=(01010{)}^{q}10,$$

$$f_3(P_n)=(01010{)}^{q}101,\text{mbox}and{f}_4(P_n)=(01010{)}^{q}0101.$$

$$n=d_{P_n}(x_1,x_{i_1})+d_{P_n}(x_{i_1},x_{i_2})+d_{P_n}(x_{i_2},x_n)+1\le 3f(x_{i_1})+2,$$

$$\sigma (g)-\sigma (f)=\lceil \frac{i_2}{2}\rceil -f(x_{i_1})-f(x_{i_2}).$$

$$i_2=d_{P_n}(x_1,x_{i_1})+d_{P_n}(x_{i_1},x_{i_2})+1\le 2f(x_{i_1})+2,$$

$$\sigma (g)-\sigma (f)\le f(x_{i_1})+1-f(x_{i_1})-f(x_{i_2})=1-f(x_{i_2}).$$

$$0^{i_1-1}f(x_{i_1})0^{i_2-i_1-1}f(x_{i_2})0^{i_3-i_2-f(x_{i_3})-2}$$

$$\sigma (g)-\sigma (f)=\lceil \frac{i_3-f(x_{i_3})-2}{2}\rceil -f(x_{i_1})-f(x_{i_2}).$$

$$i_2=d_{P_n}(x_1,x_{i_1})+d_{P_n}(x_{i_1},x_{i_2})+1\le 2f(x_{i_1})+2,$$

$$\sigma (g)-\sigma (f)\le f(x_{i_1})+\lceil \frac{f(x_{i_2})+1}{2}\rceil -f(x_{i_1})-f(x_{i_2})=\lceil \frac{1-f(x_{i_2})}{2}\rceil .$$

$$f(x_{i_{j-1}})0^{i_j-i_{j-1}-1}f(x_{i_j})0^{i_{j+1}-i_j-1}f(x_{i_{j+1}})$$

$$\sigma (f^{\prime })-\sigma (f)=\lfloor \frac{d}{2}\rfloor -f(x_{i_j})-f(x_{i_{j+1}}).$$

$$f(x_{i_j})+1\le d_{P_n}(x_{i_{j-1}},x_{i_j})\le f(x_{i_j})+2.$$

$$d=d_{P_n}(x_{i_{j-1}},x_{i_j})+d_{P_n}(x_{i_j},x_{i_{j+1}})\le f(x_{i_j})+1+f(x_{i_j})+f(x_{i_{j+1}})+1=2f(x_{i_j})+f(x_{i_{j+1}})+2,$$

$$\sigma (f^{\prime })-\sigma (f)\le f(x_{i_j})+\lfloor \frac{f(x_{i_{j+1}})}{2}\rfloor +1-f(x_{i_j})-f(x_{i_{j+1}})=\lfloor \frac{2-f(x_{i_{j+1}})}{2}\rfloor \le 0.$$

Full text

In this paper, we determine these broadcast numbers for all paths and cycles, thus answering a question raised in [D. Ahmadi, G.H. Fricke, C. Schroeder, S.T. Hedetniemi and R.C. Laskar, Broadcast irredundance in graphs. Congr. Numer. 224 (2015), 17–31].

Keywords: Broadcast; Dominating broadcast; Irredundant broadcast; Independent broadcast; Packing broadcast; Path; Cycle.

MSC 2010: 05C12, 05C69.

1 Introduction

Let $G=(V,E)$ be a graph of order $n=|V|$ and size $m=|E|$. The open neighborhood of a vertex $v\in V$ is the set $N_{G}(v)=\{u:uv\in E\}$ of vertices adjacent to $v$. Each vertex $u\in N_{G}(v)$ is a neighbor of $v$ in $G$. The closed neighborhood of $v$ is the set $N_{G}[v]=N_{G}(v)\cup\{v\}$. The open neighborhood of a set $S\subseteq V$ of vertices is $N_{G}(S)=\cup_{v\in S}N_{G}(v)$, while the closed neighborhood of $S$ is the set $N_{G}[S]=N_{G}(S)\cup S$. The degree of a vertex $v$ in $G$, denoted $\deg_{G}(v)$, is the size of the open neighborhood of $v$.

A $(u,v)$-geodesic in a graph $G$ is a shortest path joining $u$ and $v$. We denote by $d_{G}(u,v)$ the distance between the vertices $u$ and $v$ in $G$, that is, the length of a $(u,v)$-geodesic in $G$. The eccentricity $e_{G}(v)$ of a vertex $v$ in $G$ is the maximum distance from $v$ to any other vertex of $G$. The radius ${\rm rad}(G)$ and the diameter ${\rm diam}(G)$ of a graph $G$ are the minimum and the maximum eccentricity among the vertices of $G$, respectively.

A function $f:V\longrightarrow\{0,\ldots,\operatorname{diam}(G)\}$ is a broadcast on a graph $G=(V,E)$ if $f(v)\leq e_{G}(v)$ for every vertex $v\in V$. The value $f(v)$ is called the $f$-value of $v$. An $f$-broadcast vertex (or an $f$-dominating vertex) is a vertex $v$ for which $f(v)>0$. The set of all $f$-broadcast vertices is denoted $V^{+}_{f}(G)$. If $v\in V^{+}_{f}$ is an $f$-broadcast vertex, $u\in V$ and $d_{G}(u,v)\leq f(v)$, then the vertex $u$ hears a broadcast from $v$ and $v$ broadcasts to (or $f$-dominates) $u$. Note that, in particular, each vertex $v\in V^{+}_{f}$ hears a broadcast from itself and $f$-dominates itself.

The $f$-broadcast neighborhood of a vertex $v\in V^{+}_{f}$ is the set of vertices that hear $v$, that is

[TABLE]

and the broadcast neighborhood of $f$ is the set

[TABLE]

The set of $f$-broadcast vertices that a vertex $u\in V$ can hear is the set

[TABLE]

For a vertex $v\in V^{+}_{f}$, the private $f$-neighborhood of $v$ is the set of vertices that hear only $v$, that is

[TABLE]

and every vertex $u\in PN_{f}(v)$ is a private $f$-neighbor of $v$. Moreover, the private $f$-border of $v$ is either the set of private $f$-neighbors of $v$ that are at distance $f(v)$ from $v$, or the singleton $\{v\}$ if $f(v)=1$ and $PN_{f}(v)=\{v\}$, that is

[TABLE]

Every vertex in $PB_{f}(v)$ is a bordering private $f$-neighbor of $v$. In particular, if $f(v)=1$ and $PN_{f}(v)=\{v\}$, then $v$ is its own bordering private $f$-neighbor.

The cost of a broadcast $f$ on a graph $G$ is

[TABLE]

A broadcast $f$ on $G$ of some type is minimal (resp. maximal) if there does not exist any broadcast $g\neq f$ on $G$ of the same type such that $g(u)\leq f(u)$ (resp. $g(u)\geq f(u)$) for all $u\in V$. Several types of broadcasts have been defined in the literature, in relation with domination, irredundence, independence or packing, leading to the introduction of several broadcast numbers on graphs, corresponding to the minimum or maximum possible cost of a maximal or minimal broadcast of the corresponding type, respectively. For any such parameter, say $q(G)$, a broadcast $f$ on $G$ of the corresponding type with $\sigma(f)=q(G)$ will be simply called a $q$-broadcast. We will also say that such a broadcast is optimal.

We now introduce the various types of broadcasts we will consider in this paper.

Dominating broadcasts.

A broadcast $f$ on $G$ is a dominating broadcast if every vertex in $V-V^{+}_{f}$ is $f$-dominated by some vertex in $V^{+}_{f}$ or, equivalently, if for every vertex $v\in V$, $|H_{f}(v)|\geq 1$. The broadcast domination number $\gamma_{b}(G)$ of $G$ is the minimum cost of a dominating broadcast on $G$. The upper broadcast domination number $\Gamma_{b}(G)$ of $G$ is the maximum cost of a minimal dominating broadcast on $G$. If $f$ is a minimal dominating broadcast on $G$ such that $f(v)=1$ for each $v\in V^{+}_{f}$, then $V^{+}_{f}$ is a minimal dominating set in $G$, and the minimum (resp. maximum) cost of such a broadcast is the domination number $\gamma(G)$ (resp. the upper domination number $\Gamma(G)$) of $G$.

Irredundant broadcasts.

A broadcast $f$ on $G$ is an irredundant broadcast if $PB_{f}(v)\neq\emptyset$ for every vertex $v\in V^{+}_{f}$. Stated equivalently, a broadcast $f$ is irredundant if the following two conditions are satisfied : (i) for every $f$-broadcast vertex $v$ with $f(v)\geq 2$, there exists a vertex $u$ such that $H_{f}(u)=\{v\}$ and $d_{G}(u,v)=f(v)$, and (ii) for every $f$-broadcast vertex $v$ with $f(v)=1$, there exists a vertex $u\in N_{G}[v]$ such that $H_{f}(u)=\{v\}$ (note that, in this case, we can have $u=v$). The upper broadcast irredundance number $I\!R_{b}(G)$ of $G$ is the maximum cost of an irredundant broadcast on $G$. The broadcast irredundance number $ir_{b}(G)$ of $G$ is the minimum cost of a maximal irredundant broadcast on $G$. If $f$ is a maximal irredundant broadcast on $G$ such that $f(v)=1$ for each $v\in V^{+}_{f}$, then $V^{+}_{f}$ is a maximal irredundant set in $G$, and the minimum (resp. the maximum) cost of such a broadcast is the irredundance number $ir(G)$ (resp. the upper irredundance number $I\!R(G)$) of $G$.

Independent broadcasts.

A broadcast $f$ is an independent broadcast if no broadcast vertex $f$-dominates any other broadcast vertex or, equivalently, if for every $v\in V^{+}_{f}$, $|H_{f}(v)|=1$. The broadcast independence number $\beta_{b}(G)$ of $G$ is the maximum cost of an independent broadcast on $G$. The lower broadcast independence number $i_{b}(G)$ of $G$ is the minimum cost of a maximal independent broadcast on $G$. If $f$ is a maximal independent broadcast such that $f(v)=1$ for each $v\in V^{+}_{f}$, then $V^{+}_{f}$ is a maximal independent set in $G$, and the maximum (resp. minimum) cost of such a broadcast is the vertex independence number $\beta_{0}(G)$ (resp. the independent domination number $i(G)$) of $G$.

Packing broadcasts.

A broadcast $f$ is a packing broadcast if every vertex hears at most one broadcast, that is, for every vertex $v\in V$, $|H_{f}(v)|\leq 1$. The broadcast packing number $P_{b}(G)$ of $G$ is the maximum cost of a packing broadcast on $G$. The lower broadcast packing number $p_{b}(G)$ of $G$ is the minimum cost of a maximal packing broadcast on $G$. If $f$ is a maximal packing broadcast such that $f(v)=1$ for each $v\in V^{+}_{f}$, then $V^{+}_{f}$ is a maximal packing set in $G$, and the maximum (resp. the minimum) cost of such a broadcast is the packing number $P(G)$ (resp. the lower packing number $p(G)$) of $G$.

These four different types of broadcasts are illustrated in Figure 1 (broadcast vertices are drawn as black vertices, non-broadcast dominated vertices as gray vertices, and non dominated vertices as white vertices): $f_{1}$ is a dominating broadcast, $f_{2}$ is an irredundant broadcast (the $f_{2}$-broadcast vertices $x_{2}$, $x_{3}$, $x_{7}$ and $x_{8}$ all have a bordering private $f_{2}$-neighbor, namely $x_{1}$, $x_{4}$, $x_{6}$ and $x_{9}$, respectively), $f_{3}$ is an independent broadcast, and $f_{4}$ is a packing broadcast. Moreover, observe the following:

• •

$f_{1}$ is a minimal dominating broadcast and also a maximal irredundant broadcast. However, $f_{1}$ is neither an independent broadcast (the $f_{1}$-broadcast vertices $x_{2}$ and $x_{3}$ are both $f_{1}$-dominated twice), nor a packing broadcast ($x_{2}$ and $x_{3}$ both hear two $f_{1}$-broadcast vertices).

• •

$f_{2}$ is a maximal irredundant broadcast, but is neither a dominating broadcast ($x_{5}$ is not $f_{2}$-dominated), nor an independent broadcast (the $f_{2}$-broadcast vertices $x_{2}$, $x_{3}$, $x_{7}$ and $x_{8}$ are $f_{2}$-dominated twice), nor a packing broadcast ($x_{2}$, $x_{3}$, $x_{7}$ and $x_{8}$ all hear two $f_{2}$-broadcast vertices).

• •

$f_{3}$ is a maximal independent broadcast and a dominating broadcast, but is neither an irredundant broadcast (the $f_{3}$-broadcast vertex $x_{8}$ has no bordering private $f_{3}$-neighbor), nor a packing broadcast (vertices $x_{2}$, $x_{4}$ and $x_{6}$ are $f_{3}$-dominated twice).

• •

$f_{4}$ is a maximal packing broadcast, and also an irredundant broadcast and an independent broadcast. However, $f_{4}$ is neither a dominating broadcast ($x_{9}$ is not $f_{4}$-dominated), nor a maximal independent broadcast (we can increase the cost of $f_{4}$ by setting $f_{4}(x_{1})=f_{4}(x_{4})=f_{4}(x_{7})=2$), nor a maximal irredundant broadcast (we can increase the cost of $f_{4}$ by setting $f_{4}(x_{7})=2$, so that $x_{4}$ has still a bordering private $f_{4}$-neighbor, namely $x_{3}$, and $x_{9}$ is now the bordering private $f_{4}$-neighbor of $x_{7}$).

Directly from the definitions of these four types of broadcasts, we get the following observations.

Observation 1.1**.**

**

• (1)

Every maximal independent broadcast is a dominating broadcast, and thus $\gamma_{b}(G)\leq i_{b}(G)$ for every graph $G$.

• (2)

Every packing broadcast is an independent broadcast, and thus $P_{b}(G)\leq\beta_{b}(G)$ for every graph $G$.

• (3)

Every packing broadcast is an irredundant broadcast, and thus $P_{b}(G)\leq I\!R_{b}(G)$ for every graph $G$.

• (4)

Every dominating maximal irredundant broadcast is a minimal dominating broadcast.

Broadcast domination was introduced by Erwin [20] in his Ph.D. thesis, in which he discussed several types of broadcast parameters and the relationships between them. Many of these results appeared later in [19]. Since then, several papers have been published on various aspects of broadcasts in graphs, including the algorithmic complexity [7, 25, 26], the determination of the broadcast domination number for several classes of graphs [10, 12, 18, 24, 32, 33, 34], and a characterization of the classes of trees for which the broadcast domination number equals the radius [27] or equals the domination number [17, 28, 31]. The upper broadcast domination number is studied in [1, 11, 19, 21, 22, 29], the broadcast irredundance number is studied in [1, 29], and the broadcast independence number is studied in [2, 3, 4, 8, 9, 13]. Broadcast domination and multipacking are considered in [5, 6, 15, 16, 23, 30].

In this paper, we determine all the above defined numbers for paths and cycles. Ahmadi et al. observed in [1] that very little is known concerning these parameters. We first recall some preliminary results in Section 2, and prove our main results in Section 3. These results are summarized in Table 1. They confirm the conjectures given in [1] for $\gamma_{b}(P_{n})$, $\gamma_{b}(C_{n})$ and $\Gamma_{b}(P_{n})$, but disprove all other conjectures.

2 Preliminary results

The characterization of minimal dominating broadcasts was first given by Erwin in [21], and then restated in terms of private borders111In their paper, Mynhardt and Roux used a slightly different definition of the set $PB_{f}(v)$ when $f(v)=1$ and $N_{f}(v)\neq\{v\}$, by including the vertex $v$ in $PB_{f}(v)$. Moreover, they called the set $PB_{f}(v)$ the private $f$-boundary of $v$. We here use the term private $f$-border to avoid confusion between these two definitions. However, it is easy to check that the private $f$-boundary of $v$ is empty if and only if the private $f$-border of $v$ is empty, so that Proposition 2.1 is still valid in our setting. by Mynhardt and Roux in [29].

Proposition 2.1** (Erwin [21], restated in [29]).**

A dominating broadcast $f$ is a minimal dominating broadcast if and only if $PB_{f}(v)\neq\emptyset$ for each $v\in V^{+}_{f}$.

Dunbar et al. proved in [19] the following bound on the upper broadcast domination number of graphs.

Theorem 2.2** (Dunbar et al. [19]).**

For every graph $G$ with size $m$, $\Gamma_{b}(G)\leq m$. Moreover, $\Gamma_{b}(G)=m$ if and only if $G$ is a nontrivial star or path.

This upper bound was later improved in [11].

Theorem 2.3** (Bouchemakh and Fergani [11]).**

If $G$ is a graph of order $n$ with minimum degree $\delta(G)$, then $\Gamma_{b}(G)\leq n-\delta(G)$, and this bound is sharp.

From Proposition 2.1 and the definition of a maximal irredundant broadcast, one gets the following result.

Corollary 2.4** (Ahmadi et al. [1]).**

Every minimal dominating broadcast is a maximal irredundant broadcast.

Since the characteristic function of a minimal dominating set in a graph is a minimal dominating broadcast, Corollary 2.4 implies the following chain of inequalities.

Corollary 2.5** (Ahmadi et al. [1]).**

For every graph $G$,

[TABLE]

Moreover, Dunbar et al. [19] proved the following.

Proposition 2.6** (Dunbar et al. [19]).**

For every graph $G$,

[TABLE]

However, $\beta_{b}(G)$ and $\Gamma_{b}(G)$ are in general incomparable.

It is worth pointing out that the difference $I\!R_{b}(G)-\Gamma_{b}(G)$ can be arbitrarily large. Indeed, Mynhardt and Roux [29] constructed a family of graphs $\{G_{r}\}_{r\geq 3}$, where each $G_{r}$ is obtained by joining two copies of $K_{r+1}$ by $r$ independent edges, and proved the relation $\Gamma_{b}(G_{r})=3\leq I\!R_{b}(G_{r})=r$ for every $r\geq 3$. Nevertheless, we may have $I\!R_{b}(G)=\Gamma_{b}(G)$, as we will prove in Subsection 3.1 when $G$ is a path or a cycle.

Dunbar et al. observed in [19] that, for any graph $G$, neither $P(G)$ nor $p(G)$ is comparable with $p_{b}(G)$, while we have $p(G)\leq P(G)\leq P_{b}(G)$ and $p_{b}(G)\leq{\mbox{ rad}}(G)\leq{\mbox{ diam}}(G)\leq P_{b}(G)\leq\beta_{b}(G)$. For paths and cycles, we will prove in Section 3 that the lower bound ${\rm diam}(G)$ for $P_{b}(G)$ is achieved, while the difference between ${\rm rad}(G)$ and $p_{b}(G)$ can be arbitrarily large.

3 Broadcast numbers of paths and cycles

As mentioned by Ahmadi et al. in [1], it is quite surprising that the values of several broadcast parameters have not been determined yet for paths or cycles. Moreover, in the same paper, they conjecture the values of these parameters. In this section, we will determine the exact values of these parameters, which in some cases, but not all, correspond to their conjecture.

Throughout this section, we will denote by $P_{n}=x_{1}x_{2}\dots x_{n}$, $n\geq 2$, the path of order $n$, and by $C_{n}=x_{0}x_{1}\dots x_{n-1}$, $n\geq 3$, the cycle of order $n$. Moreover, we assume throughout this section that subscripts of vertices of $C_{n}$ are taken modulo $n$, and that the vertices $x_{1},\dots,x_{n}$ of $P_{n}$ are “ordered” from left to right, so that by the leftmost (resp. the rightmost) vertex in $P_{n}$ satisfying any property, we mean the vertex with minimum (resp. maximum) subscript satisfying this property.

3.1 Upper broadcast domination number and upper broadcast irredundance number

We first consider the case of paths.

Theorem 3.1**.**

For every integer $n\geq 2$, $\Gamma_{b}(P_{n})=I\!R_{b}(P_{n})={\rm diam}(P_{n})=n-1$.

Proof.

Theorem 2.2 directly gives $\Gamma_{b}(P_{n})=n-1$. By Corollary 2.5, we have $n-1=\Gamma_{b}(P_{n})\leq I\!R_{b}(P_{n})$. We now prove the opposite inequality. Let $f$ be an irredundant broadcast on $P_{n}$, and let $V^{+}_{f}=\{x_{i_{1}},\dots,x_{i_{t}}\}$, $i_{1}<\cdots<i_{t}$, $t\geq 1$. From the definition of an irredundant broadcast, we get that for every vertex $x_{i_{j}}\in V^{+}_{f}$ with $f(x_{i_{j}})\geq 2$, there exists a vertex $x_{i_{j}}^{p}$ such that $H_{f}(x_{i_{j}}^{p})=\{x_{i_{j}}\}$ and $d_{P_{n}}(x_{i_{j}},x_{i_{j}}^{p})=f(x_{i_{j}})$ and, for every vertex $x_{i_{j}}\in V^{+}_{f}$ with $f(x_{i_{j}})=1$, there exists a vertex $x_{i_{j}}^{p}\in N_{P_{n}}[x_{i_{j}}]$ such that $H_{f}(x_{i_{j}}^{p})=\{x_{i_{j}}\}$.

Let $t^{\prime}$, $0\leq t^{\prime}\leq t$, denote the number of $f$-broadcast vertices $x_{i_{j}}$ that are their own bordering private $f$-neighbor, that is, such that $x_{i_{j}}^{p}=x_{i_{j}}$ (which implies $f(x_{i_{j}})=1$), and suppose that these vertices are $\{x_{i_{1}},\dots,x_{i_{t^{\prime}}}\}$. We thus have

[TABLE]

which gives $I\!R_{b}(P_{n})\leq n-t+t^{\prime}$. If $t^{\prime}<t$, then $I\!R_{b}(P_{n})\leq n-1$ and we are done. Otherwise, every $f$-broadcast vertex is its own bordering private $f$-neighbor, which implies that $V^{+}_{f}$ is either the set of all vertices with even subscript, or the set of all vertices with odd subscript. In both cases, we get $\sigma(f)=I\!R_{b}(P_{n})\leq\left\lceil\frac{n}{2}\right\rceil\leq n-1$, as required. ∎

We now consider the case of cycles. For that, we will use the following lemma.

Lemma 3.2**.**

Let $f$ be an $I\!R_{b}$-broadcast on $C_{n}$. If $H_{f}(x_{i})=\emptyset$ for some vertex $x_{i}$, then $H_{f}(x_{i-1})\neq\emptyset$ and $H_{f}(x_{i+1})\neq\emptyset$.

Proof.

Assume to the contrary that we have $H_{f}(x_{i-1})=\emptyset$, so that $x_{i-1}$ is not $f$-dominated, which implies in particular $f(x_{i-2})=0$. Therefore, there exists an $f$-broadcast vertex $x_{j}$, $j<i-2$, which $f$-dominates $x_{i-2}$, for otherwise we could set $f(x_{i-1})=1$, contradicting the optimality of $f$.

We then necessarily have $d_{C_{n}}(x_{j},x_{i-2})=f(x_{j})$. But, in that case, the function $g$ obtained from $f$ by setting $g(x_{j})=0$ and $g(x_{j+1})=f(x_{j})+1$ would be an irredundant broadcast on $C_{n}$ with cost $\sigma(g)=\sigma(f)+1>\sigma(f)$, again a contradiction.

It follows that we have $H_{f}(x_{i-1})\neq\emptyset$ and, by symmetry, that we also have $H_{f}(x_{i+1})\neq\emptyset$. ∎

We can now prove the following result.

Theorem 3.3**.**

For every integer $n\geq 3$, $I\!R_{b}(C_{n})=\Gamma_{b}(C_{n})$.

Proof.

By Corollary 2.5, we only need to prove the inequality $I\!R_{b}(C_{n})\leq\Gamma_{b}(C_{n})$. For this, it is enough to construct, from any non-dominating $I\!R_{b}$-broadcast on $C_{n}$, a dominating irredundant broadcast (which is then a minimal dominating broadcast, by Observation 1.1(4)) with the same cost $I\!R_{b}(C_{n})$.

Let $f$ be a non-dominating $I\!R_{b}$-broadcast on $C_{n}$. Then, there exists $i\in\{0,\dots,n-1\}$ such that $H_{f}(x_{i})=\emptyset$. By Lemma 3.2, we have $H_{f}(x_{i-1})\neq\emptyset$ and $H_{f}(x_{i+1})\neq\emptyset$. Since $x_{i}$ is not $f$-dominated, we know that $x_{i-1}$ is $f$-dominated by a unique $f$-broadcast vertex, say $x_{j}$, $j<i-1$, such that $f(x_{j})=d_{C_{n}}(x_{j},x_{i-1})$, which implies $|PB_{f}(x_{j})|\geq 1$. We claim that we have $|PB_{f}(x_{j})|=1$. Indeed, if $|PB_{f}(x_{j})|=2$ , then we could set $f(x_{j+1})=f(x_{j})$, contradicting the optimality of $f$.

Now, observe that the function $g$ obtained from $f$ by setting $g(x_{j})=0$ and $g(x_{j+1})=f(x_{j})$ is an irredundant broadcast with $\sigma(g)=\sigma(f)$, such that $x_{i}$ is $g$-dominated, and all vertices that were $f$-dominated remain $g$-dominated (see Figure 2) . Repeating the same transformation for each vertex which is not dominated, we eventually produce a minimal dominating broadcast on $C_{n}$ with cost $I\!R_{b}(C_{n})$. ∎

We obviously have $\Gamma_{b}(C_{3})=1$. For $n\geq 4$, the value of $\Gamma_{b}(C_{n})$ is given by the following result.

Theorem 3.4**.**

For every integer $n\geq 4$, \Gamma_{b}(C_{n})=2\big{(}\left\lfloor\frac{n}{2}\right\rfloor-1\big{)}.

Proof.

From Theorem 2.3, we directly get $\Gamma_{b}(C_{n})\leq n-\delta(C_{n})=n-2$. Let now $f$ be the function defined by

[TABLE]

Clearly, $f$ is a minimal dominating broadcast on $C_{n}$ with cost

[TABLE]

Therefore, $\sigma(f)\leq\Gamma_{b}(C_{n})$. Combining this inequality with the previous one, we already infer that we have $\Gamma_{b}(C_{n})=n-2$ if $n$ is even, and $n-3\leq\Gamma_{b}(C_{n})\leq n-2$ if $n$ is odd.

It remains to discuss the case $n$ odd. For $n=5$, it is not difficult to check that we have $\Gamma_{b}(C_{5})=2$. Suppose $n\geq 7$ and let $g$ be any $\Gamma_{b}$-broadcast on $C_{n}$ such that $V^{+}_{g}=\{x_{i_{1}},\dots,x_{i_{t}}\}$, $i_{1}<\cdots<i_{t}$, $t\geq 1$. We first prove the following claim.

Claim A**.**

$t=2$.

Proof.

If $t=1$, then there is a unique $g$-broadcast vertex $x_{i}\in V^{+}_{g}$, and thus $\Gamma_{b}(C_{n})=g(x_{i})=e_{C_{n}}(x_{i})=\frac{n-1}{2}<n-3$, a contradiction. Hence, $t\geq 2$. We know by Proposition 2.1 that each $g$-broadcast vertex $x_{i}\in V^{+}_{g}$ has a bordering private $g$-neighbor, say $x_{i}^{p}$, with possibly $x_{i}^{p}=x_{i}$ (and, in that case, $g(x_{i})=1$). Let $Q_{x_{i}}$ be the set of edges defined as follows:

• •

if $g(x_{i})\geq 2$, then $Q_{x_{i}}$ is the set of edges of the unique $(x_{i},x_{i}^{p})$-geodesic,

• •

if $g(x_{i})=1$ and $x_{i}^{p}\in\{x_{i-1},x_{i+1}\}$ is a bordering private $g$-neighbor of $x_{i}$, then $Q_{x_{i}}$ is the singleton $\{x_{i}x_{i}^{p}\}$,

• •

if $g(x_{i})=1$ and $x_{i}$ is its own bordering private $g$-neighbor, then $Q_{x_{i}}$ is the singleton $\{x_{i}x_{i+1}\}$.

Clearly, $g(x_{i})=|Q_{x_{i}}|$ for every $x_{i}\in V^{+}_{g}$, and $Q_{x_{i}}\cap Q_{x_{j}}=\emptyset$ for every $x_{i},x_{j}\in V^{+}_{g}$.

We now claim that for every $Q_{x_{i}}=\{x_{i}x_{i+1},\dots,x_{i+t-1}x_{i+t}\}$ (resp. $Q_{x_{i}}=\{x_{i-t}x_{i-t+1},\dots,x_{i-1}x_{i}\}$), $t\geq 1$, the edge $x_{i+t}x_{i+t+1}$ (resp. $x_{i}x_{i+1}$) that “follows” $Q_{x_{i}}$ does not belong to any $Q_{x_{i^{\prime}}}$, with $x_{i^{\prime}}\in V^{+}_{g}$. Indeed, this directly follows from the following observations.

• $(a)$

Every $g$-broadcast vertex $x_{i}$ belongs to exactly one such path, namely $Q_{x_{i}}$.

• $(b)$

Every bordering private $g$-neighbor belongs to at most one such path.

• $(c)$

An end-vertex $x_{j}$ of such a path $Q$ is neither a $g$-broadcast vertex nor a bordering private $g$-neighbor if and only if $Q=Q_{x_{j-1}}$, $x_{j-1}$ is a $g$-broadcast vertex, $g(x_{j-1})=1$ and $x_{j-1}$ is its own bordering private $g$-neighbor.

Hence, we have

[TABLE]

If $t>3$, then $\Gamma_{b}(C_{n})<n-3$, a contradiction. Assume now that $t=3$ and let $V^{+}_{g}=\{x_{a},x_{b},x_{c}\}$, with $a<b<c$. Since $\Gamma_{b}(C_{n})=n-3$, any two of the paths $Q_{x_{a}}$, $Q_{x_{b}}$ and $Q_{x_{c}}$ are separated by exactly one edge.

Suppose first that one of these $g$-broadcast vertices, say $x_{c}$, is such that $g(x_{c})=1$ and $x_{c}$ is its own private $g$-neighbor, so that $Q_{x_{c}}=x_{c}x_{c+1}$. In that case, $x_{c-1}$ is neither a $g$-broadcast vertex nor a bordering private $g$-neighbor, which implies, by observation $(c)$ above, that $x_{b}=x_{c-2}$, $g(x_{b})=1$ and $x_{b}$ is its own bordering private $g$-neighbor. Using the same argument, we get that $x_{a}=x_{b-2}$, $g(x_{a})=1$, $x_{a}$ is its own bordering private $g$-neighbor, and $x_{c}=x_{a-2}$, leading to $n=6$, a contradiction since we assumed $n\geq 7$.

Hence, each end-vertex of any of the paths $Q_{x_{a}}$, $Q_{x_{b}}$ and $Q_{x_{c}}$ is either a $g$-broadcast vertex or a bordering private $g$-neighbor of the $g$-broadcast vertex belonging to the same path. But since we have three paths, one of $x_{a}$, $x_{b}$ or $x_{c}$ must be adjacent to a bordering private $g$-neighbor of another $g$-broadcast vertex, a contradiction. ∎

By Claim A, we can assume $V^{+}_{g}=\{x_{a},x_{b}\}$, which implies $|PB_{f}(x_{a})|=|PB_{f}(x_{b})|$. If $|PB_{f}(x_{a})|=|PB_{f}(x_{b})|=2$, then $n=2f(x_{a})+1+2f(x_{b})+1$, a contradiction since $n$ is odd. We thus have $|PB_{f}(x_{a})|=|PB_{f}(x_{b})|=1$, and the bordering private $g$-neighbors $x_{a}^{p}$ of $x_{a}$, and $x_{b}^{p}$ of $x_{b}$ are adjacent. Moreover, if we assume, without loss of generality, that $x_{a}x_{a+1}\dots x_{b}$ is a $(x_{a},x_{b})$-geodesic, then $b-a\leq 2$ for otherwise the function $h$ obtained from $g$ by setting $h(x_{a})=0$, $h(x_{b})=0$, $h(x_{a+1})=g(x_{a})+1$ and $h(x_{b-1})=g(x_{b})+1$, would be a minimal dominating broadcast with cost $\Gamma_{b}(h)=\Gamma_{b}(g)+2$, contradicting the optimality of $g$. Hence, we have $d_{C_{n}}(x_{a},x_{b})\leq 2$. If $x_{a}$ and $x_{b}$ are joined by an edge, we necessarily have $g(x_{a})=g(x_{b})$, implying that $n$ is even, contrary to our assumption. We thus have $d_{C_{n}}(x_{a},x_{b})=2$, and thus $g(x_{a})=g(x_{b})=\frac{n-3}{2}$, which gives $\sigma(g)=n-3$. ∎

3.2 Broadcast independence number and lower broadcast independence number

Dunbar et al. [19] noted that the upper broadcast domination number $\Gamma_{b}(G)$ and the broadcast independence number $\beta_{b}(G)$ of a graph $G$ are in general incomparable. Erwin gave in [20] the exact value of the broadcast independence number of paths. He proved that for every integer $n\geq 3$, $\beta_{b}(P_{n})=2(n-2)$, so that, by Theorem 3.1, $\beta_{b}(P_{n})>\Gamma_{b}(P_{n})$ for every $n>3$.

Our next result proves that the equality $\beta_{b}(C_{n})=\Gamma_{b}(C_{n})$ holds for every cycle $C_{n}$, $n\geq 3$ (recall Theorem 3.4).

Theorem 3.5**.**

For every integer $n\geq 3$, \beta_{b}(C_{n})=2\big{(}\left\lfloor\frac{n}{2}\right\rfloor-1\big{)}.

Proof.

It is easy to check that we have $\beta_{b}(C_{3})=1$ and $\beta_{b}(C_{4})=2$. Assume thus $n\geq 5$. Clearly, the function $f$ defined by

[TABLE]

is an independent broadcast on $C_{n}$ with cost $\sigma(f)=2(\lfloor\frac{n}{2}\rfloor-1)$, which implies $\beta_{b}(C_{n})\geq 2(\lfloor\frac{n}{2}\rfloor-1)$.

We now prove the opposite inequality. For this, let $g$ be any $\beta_{b}$-broadcast on $C_{n}$. If $|V^{+}_{g}|=1$, say $V^{+}_{g}=\{x_{i}\}$, then

[TABLE]

and, if $|V^{+}_{g}|=2$, then the two vertices of $V^{+}_{g}$ are antipodal, which gives $\sigma(g)=2(\lfloor\frac{n}{2}\rfloor-1)$.

Assume now that we have $|V^{+}_{g}|\geq 3$, and let $V^{+}_{g}=\{x_{i_{0}},\dots,x_{i_{k-1}}\}$, $k\geq 3$. Since $g$ is an independent broadcast, we have $d_{C_{n}}(x_{i_{j}},x_{i_{j+1}})\geq g(x_{i_{j}})+1$ for every $j$, $0\leq j\leq k-1$ (subscripts of $i$ are taken modulo $k$). We thus get

[TABLE]

as required. ∎

We now determine the value of the lower broadcast independence number of paths and cycles. For that, we will use the following lemma.

Lemma 3.6**.**

If $f$ is an $i_{b}$-broadcast on $P_{n}$, $n\geq 3$, with $V^{+}_{f}=\{x_{i_{1}},\dots,x_{i_{t}}\}$, $i_{1}<\cdots<i_{t}$, $t\geq 2$, then we have

$f(x_{i_{1}})\geq f(x_{i_{2}})$* and $f(x_{i_{t}})\geq f(x_{i_{t-1}})$,* 2. 2.

$d_{P_{n}}(x_{1},x_{i_{1}})\leq f(x_{i_{1}})$* and $d_{P_{n}}(x_{i_{t}},x_{n})\leq f(x_{i_{t}})$,* 3. 3.

for every $j$, $1\leq j\leq t-1$, $\max\{f(x_{i_{j}}),f(x_{i_{j+1}})\}+1\leq d_{P_{n}}(x_{i_{j}},x_{i_{j+1}})\leq f(x_{i_{j}})+f(x_{i_{j+1}})+1$, 4. 4.

$d_{P_{n}}(x_{i_{1}},x_{i_{2}})=f(x_{i_{1}})+1$* and $d_{P_{n}}(x_{i_{t-1}},x_{i_{t}})=f(x_{i_{t}})+1$.*

Proof.

If Item 1 is not satisfied, then we can increase by 1 the value of $f(x_{1})$ or $f(x_{n})$, contradicting the maximality of $f$.

If Item 2 is not satisfied, then we can set $f(x_{1})=1$, or $f(x_{n})=1$, contradicting the maximality of $f$.

For Item 3, the inequality $\max\{f(x_{i_{j}}),f(x_{i_{j+1}})\}+1\leq d_{P_{n}}(x_{i_{j}},x_{i_{j+1}})$ directly follows from the definition of an independent broadcast. Finally, if $d_{P_{n}}(x_{i_{j}},x_{i_{j+1}})>f(x_{i_{j}})+f(x_{i_{j+1}})+1$, then we can set $f(x_{i_{j}+f(x_{i_{j}})+1})=1$, contradicting the maximality of $f$.

Consider now Item 4. By items 1 and 3, we have $f(x_{i_{1}})+1\leq d_{P_{n}}(x_{i_{1}},x_{i_{2}})$ and $f(x_{i_{t}})+1\leq d_{P_{n}}(x_{i_{t-1}},x_{i_{t}})$. If any of these inequalities is strict, then we can increase by 1 the $f$-value of the involved $f$-broadcast vertex, again contradicting the maximality of $f$. ∎

In the rest of the paper, for convenience, we will often define a broadcast function $f$ on the path $P_{n}$ (resp. on the cycle $C_{n}$) by the word $f(x_{1})\dots f(x_{n})$ (resp. $f(x_{0})\dots f(x_{n-1})$), using standard notation from Formal Language Theory. In particular, recall that when we write $(a_{1}\dots a_{k})^{q}$ for some integer $q\geq 0$, we mean that the sequence $a_{1}\dots a_{k}$ is repeated exactly $q$ times (in particular, if $q=0$, then $(a_{1}\dots a_{k})^{q}$ is the empty word $\varepsilon$).

Theorem 3.7**.**

For every integer $n\geq 2$, $i_{b}(P_{n})=\left\lceil\frac{2n}{5}\right\rceil.$

Proof.

Let $n=5q+r$, with $q\geq 0$, $0\leq r\leq 4$ and $5q+r\geq 2$. It is easy to check that the functions $10$, $010$ and $0101$ are $i_{b}$-broadcasts on $P_{n}$ with cost $\left\lceil\frac{2n}{5}\right\rceil$ when $n=2$, $3$ and $4$, respectively.

Assume now $n\geq 5$. According to the value of $r$, we define the broadcasts $f_{r}$ on $P_{n}$ for each $r$, $0\leq r\leq 4$, as follows:

[TABLE]

[TABLE]

It is not difficult to check that each $f_{r}$, $0\leq r\leq 4$, is a maximal independent broadcast on $C_{5q+r}$, for each $q\geq 1$, with cost $\sigma(f_{r})=\left\lceil\frac{2n}{5}\right\rceil$, which gives $i_{b}(P_{n})\leq\lceil\frac{2n}{5}\rceil$.

Let us now consider the opposite inequality. Let $f$ be an $i_{b}$-broadcast on $P_{n}$. If $|V^{+}_{f}|=1$, then we have $i_{b}(P_{n})=rad(P_{n})=\left\lfloor\frac{n}{2}\right\rfloor$, which implies $n\in S=\{1,\ldots,9,11,13\}$, since otherwise we would have $i_{b}(P_{n})=\left\lfloor\frac{n}{2}\right\rfloor>\left\lceil\frac{2n}{5}\right\rceil$, contradicting the inequality we have established before. Observe also that we have $\left\lfloor\frac{n}{2}\right\rfloor=\left\lceil\frac{2n}{5}\right\rceil$ for every $n\in S$, so that we are done.

The remaining case we have to consider is thus $n\notin S$, which implies $|V^{+}_{f}|\geq 2$. Let $V^{+}_{f}=\{x_{i_{1}},\dots,x_{i_{t}}\}$, $i_{1}<\cdots<i_{t}$, with $t\geq 2$. The two following claims will prove that we can always choose $f$ such that $f(x_{i_{j}})=1$ for every $f$-broadcast vertex $v_{i_{j}}\in V^{+}_{f}$.

Claim B**.**

There exists an $i_{b}$-broadcast $g$ on $P_{n}$ such that $g(x_{i_{1}})=g(x_{i_{2}})=g(x_{i_{t-1}})=g(x_{i_{t}})=1$.

Proof.

We first prove that there exists an $i_{b}$-broadcast $g_{0}$ on $P_{n}$ such that $g_{0}(x_{i_{1}})=g_{0}(x_{i_{2}})=1$. If $f(x_{i_{1}})=f(x_{i_{2}})=1$, then we set $g_{0}:=f$ and we are done. So, suppose that we have $f(x_{i_{1}})+f(x_{i_{2}})\geq 3$.

If $|V^{+}_{f}|=2$, then, by Lemma 3.6(4), we have $d_{P_{n}}(x_{i_{1}},x_{i_{2}})=f(x_{i_{1}})+1=f(x_{i_{2}})+1$. Using Lemma 3.6(2) and (3), we then get

[TABLE]

and thus $\sigma(f)=2f(x_{i_{1}})\geq\frac{2(n-2)}{3}$. Now, recall that the above defined maximal independent broadcast $f_{r}$, with $r=n\mod 5$, is such that $\sigma(f_{r})=\left\lceil\frac{2n}{5}\right\rceil$. Since $n\geq 10$, this contradicts the optimality of $f$.

We thus have $|V^{+}_{f}|\geq 3$. We consider two cases, depending on the value of $d_{P_{n}}(x_{i_{2}},x_{i_{3}})$.

$f(x_{i_{3}})+1\leq d_{P_{n}}(x_{i_{2}},x_{i_{3}})\leq f(x_{i_{3}})+2$.

Let $g$ be the mapping obtained from $f$ by replacing the $f$-values $0^{i_{1}-1}f(x_{i_{1}})0^{i_{2}-i_{1}-1}f(x_{i_{2}})$ of $x_{1}\dots x_{i_{1}}\dots x_{i_{2}}$ by $(01)^{\frac{i_{2}}{2}}$ if $i_{2}$ is even, or by $1(01)^{\frac{i_{2}-1}{2}}$ if $i_{2}$ is odd (see Figure 3). In both cases, we have $g(x_{i_{2}})=1$, which implies that $g$ is a maximal independent broadcast on $P_{n}$. We then have

[TABLE]

By Lemma 3.6(2) and (4), we have

[TABLE]

and thus

[TABLE]

The optimality of $f$ then implies $f(x_{i_{2}})=1$, so that we have $\sigma(g)=\sigma(f)$ and we can set $g_{0}:=g$.

$d_{P_{n}}(x_{i_{2}},x_{i_{3}})\geq f(x_{i_{3}})+3$.

Let $g$ be the mapping obtained from $f$ by replacing the $f$-values

[TABLE]

of $x_{1}\dots x_{i_{1}}\dots x_{i_{2}}\dots x_{i_{3}-f(x_{i_{3}})-2}$ by $(01)^{\frac{i_{3}-f(x_{i_{3}})-2}{2}}$ if $i_{3}-f(x_{i_{3}})-2$ is even, or by $1(01)^{\frac{i_{3}-f(x_{i_{3}})-3}{2}}$ if $i_{3}-f(x_{i_{3}})-2$ is odd (see Figure 4).

In both cases, all vertices are $g$-dominated and $g$ is clearly a maximal independent broadcast on $P_{n}$. We then have

[TABLE]

By Lemma 3.6(2) and (4), we have

[TABLE]

and thus $i_{3}-f(x_{i_{3}})-2\leq 2f(x_{i_{1}})+f(x_{i_{2}})+1$. We then get

[TABLE]

The optimality of $f$ then implies $f(x_{i_{2}})\leq 2$, so that we have $\sigma(g)=\sigma(f)$ and we can set $g_{0}:=g$.

Observe now that in both of the above cases we have $g_{0}(x_{i_{j}})=f(x_{i_{j}})$ for every $j$, $3\leq j\leq t$. Therefore, using symmetry and starting from $g_{0}$ instead of $f$, we can similarly construct an $i_{b}$-broadcast $g$ on $P_{n}$ with $g(x_{i_{1}})=g(x_{i_{2}})=g(x_{i_{t-1}})=g(x_{i_{t}})=1$, as required. ∎

Claim C**.**

There exists an $i_{b}$-broadcast $g$ on $P_{n}$ such that $g(x_{i})=1$ for every vertex $x_{i}\in V_{g}^{+}$.

Proof.

By Claim B, we can suppose that $f(x_{i_{1}})=f(x_{i_{2}})=f(x_{i_{t-1}})=f(x_{i_{t}})=1$. If $f(x_{i})=1$ for every vertex $x_{i}\in V^{+}_{f}$, there is nothing to prove. Suppose thus that this is not the case, which implies $t\geq 5$, and let $x_{i_{j}}$, $3\leq j\leq t-2$, be the leftmost $f$-broadcast vertex for which $f(x_{i_{j}})\geq 2$.

We will prove that we can always construct an $i_{b}$-broadcast $f^{\prime}$ on $P_{n}$ such that the number of broadcast vertices with $f^{\prime}$-value at least 2 is strictly less than the number of broadcast vertices with $f$-value at least 2.

We consider two cases, depending on the value of $d_{P_{n}}(x_{i_{j+1}},x_{i_{j+2}})$.

$f(x_{i_{j+2}})+1\leq d_{P_{n}}(x_{i_{j+1}},x_{i_{j+2}})\leq f(x_{i_{j+2}})+2$.

Let $d=i_{j+1}-i_{j-1}=d_{P_{n}}(x_{i_{j-1}},x_{i_{j+1}})=d_{P_{n}}(x_{i_{j-1}},x_{i_{j}})+d_{P_{n}}(x_{i_{j}},x_{i_{j+1}})$, and $f^{\prime}$ be the mapping obtained from $f$ by replacing the $f$-values

[TABLE]

of $x_{i_{j-1}}\dots x_{i_{j}}\dots x_{i_{j+1}}$ by $1(01)^{\frac{d}{2}}$ if $d$ is even, or by $10(01)^{\frac{d-1}{2}}$ if $d$ is odd (see Figure 5).

Observe that we have $f^{\prime}(x_{i_{j-1}})=1$ and $f^{\prime}(x_{i_{j+1}})=1$ in both cases, which implies that $f^{\prime}$ is a maximal independent broadcast on $P_{n}$. Moreover, in both cases, we have

[TABLE]

Since $f(x_{i_{j-1}})=1<f(x_{i_{j}})$, Lemma 3.6(3) gives

[TABLE]

We thus consider two subcases, depending on the value of $d_{P_{n}}(x_{i_{j-1}},x_{i_{j}})$.

1. (a)

$d_{P_{n}}(x_{i_{j-1}},x_{i_{j}})=f(x_{i_{j}})+1$.

By Lemma 3.6(3), we have

[TABLE]

and thus

[TABLE]

The optimality of $f$ then implies the optimality of $f^{\prime}$, and the number of broadcast vertices with $f^{\prime}$-value at least 2 is strictly less than the number of broadcast vertices with $f$-value at least 2, as required. 2. (b)

$d_{P_{n}}(x_{i_{j-1}},x_{i_{j}})=f(x_{i_{j}})+2$.

Since $f$ is maximal, we necessarily have $d_{P_{n}}(x_{i_{j}},x_{i_{j+1}})=f(x_{i_{j}})+1$, since otherwise we could increase $f(x_{i_{j}})$ by 1. This gives

[TABLE]

and thus

[TABLE]

Again, the optimality of $f$ then implies the optimality of $f^{\prime}$, and the number of broadcast vertices with $f^{\prime}$-value at least 2 is strictly less than the number of broadcast vertices with $f$-value at least 2, as required.

$d_{P_{n}}(x_{i_{j+1}},x_{i_{j+2}})\geq f(x_{i_{j+2}})+3$.

Let $d^{\prime}=d_{P_{n}}(x_{i_{j+2}-f(x_{i_{j+2}})-2},x_{i_{j-1}+1})=i_{j+2}-f(x_{i_{j+2}})-2-i_{j-1}-1=i_{j+2}-i_{j-1}-f(x_{i_{j+2}})-3$, and $f^{\prime}$ be the mapping obtained from $f$ by replacing the $f$-values

[TABLE]

of $x_{i_{j-1}}\dots x_{i_{j}}\dots x_{i_{j+1}}\dots x_{i_{j+2}-f(x_{i_{j+2}})-2}$ by $10(01)^{\frac{d^{\prime}}{2}}$ if $d^{\prime}$ is even, or by $1(01)^{\frac{d^{\prime}+1}{2}}$ if $d^{\prime}$ is odd (see Figure 6).

Since $i_{j+2}-i_{j+1}\leq f(x_{i_{j+1}})+f(x_{i_{j+2}})+1$ and $i_{j+1}-i_{j-1}\leq 2f(x_{i_{j}})+f(x_{i_{j+1}})+2$, we get $i_{j+2}-i_{j-1}\leq 2\ f(x_{i_{j}})+2f(x_{i_{j+1}})+f(x_{i_{j+2}})+3$ and thus $d^{\prime}\leq 2f(x_{i_{j}})+2f(x_{i_{j+1}})$.

Hence, the mapping $f^{\prime}$ is a maximal independent broadcast on $P_{n}$ and we have

[TABLE]

The optimality of $f$ then implies the optimality of $f^{\prime}$, and the number of broadcast vertices with $f^{\prime}$-value at least 2 is strictly less than the number of broadcast vertices with $f$-value at least 2, as required.

In each case, we were able to construct an $i_{b}$-broadcast $f^{\prime}$ such that the number of broadcast vertices with $f^{\prime}$-value at least 2 is strictly less than the number of broadcast vertices with $f$-value at least 2, as required. Repeating this construction until no such broadcast vertex exists, we eventually get an $i_{b}$-broadcast $g$ such that $g(x_{i_{j}})=1$ for every $g$-broadcast vertex $x_{i_{j}}$. This completes the proof of Claim C. ∎

By Claim C, we can thus now assume that the $i_{b}$-broadcast $f$ is such that $f(x_{i_{j}})=1$ for every $f$-broadcast vertex $x_{i_{j}}$. It remains to prove that, for every $n\geq 5$, $\sigma(f)=\left\lceil\frac{2n}{5}\right\rceil$.

For that, we first prove the following claim. Let $w_{f}$ denote the word on the alphabet $\{0,1\}$ defined by $w_{f}=f(x_{1})\dots f(x_{n})$.

Claim D**.**

There exists an $i_{b}$-broadcast $f$ on $P_{n}$, $n\geq 5$, such that $f(x_{i_{j}})=1$ for every $f$-broadcast vertex $x_{i_{j}}$, that satisfies the following properties:

1. (1)

$w_{f}$* does not contain the factor $000$, and* 2. (2)

$w_{f}$* does not contain the factor $1001001$.* 3. (3)

$0101$* is a prefix of $w_{f}$,* 4. (4)

either $101$ or $1010$ is a suffix of $w_{f}$.

Proof.

If $f(x_{i-1})f(x_{i})f(x_{i+1})=000$, then we can set $f(x_{i})=1$, contradicting the maximality of $f$, which proves Item (1). Similarly, if $f(x_{i-3})\dots f(x_{i})\dots f(x_{i+3})=1001001$, then we can set $f(x_{i})=2$, again contradicting the maximality of $f$, which proves Item (2).

Now, observe that we can have neither $f(x_{1})f(x_{2})=00$, since otherwise we could set $f(x_{1})=1$, neither $f(x_{1})\dots f(x_{4})=0100$, since otherwise we could set $f(x_{2})=2$, nor $f(x_{1})\dots f(x_{4})=1001$, since otherwise we could set $f(x_{1})=2$, contradicting in each case the maximality of $f$. Therefore, either $0101$ or $1010$ is a prefix of $w_{f}$. In the former case we are done, so let us assume that $1010$ is a prefix of $w_{f}$. Suppose that $w_{f}$ contains the factor $100$ and consider its first occurrence, that is, suppose that $(10)^{k}100$ is a prefix of $w_{f}$ for some $k\geq 1$. In that case, we can replace the $f$-values $(10)^{k}100$ of $x_{1}\dots x_{2k+3}$ by $0(10)^{k}10$ and we are done. If $w_{f}$ does not contain the factor $100$, then we necessarily have either $w_{f}=(10)^{\frac{n}{2}}$ or $w_{f}=(10)^{\frac{n-1}{2}}1$, which gives $\sigma(f)=\left\lceil\frac{n}{2}\right\rceil>\left\lceil\frac{2n}{5}\right\rceil$, a contradiction. This proves Item (3).

By symmetry, using the same argument as for the previous item, we get that none of $00$, $0010$, or $1001$ can be a suffix of $w_{f}$. Hence, either $1010$ or $0101$ is a suffix of $w_{f}$, which proves Item (4). ∎

By Item (3) of Claim D, we let $w^{\prime}_{f}$ be the word defined by $w_{f}=0101w^{\prime}_{f}$. We will now “split” $w^{\prime}_{f}$ in factors (or blocks) $w_{1},\dots,w_{q}$, $q\geq 0$ ($q=0$ meaning that no such block appeared), each of length 2 or 5, inductively defined as follows.

• •

If $01$ is a prefix of $w^{\prime}_{f}$, then $w_{1}=01$,

• •

If $00101$ is a prefix of $w^{\prime}_{f}$, then $w_{1}=00101$,

• •

If $w^{\prime}_{f}=w_{1}\dots w_{k-1}w^{\prime\prime}$ and $01$ is a prefix of $w^{\prime\prime}$, then $w_{k}=01$,

• •

If $w^{\prime}_{f}=w_{1}\dots w_{k-1}w^{\prime\prime}$ and $00101$ is a prefix of $w^{\prime\prime}$, then $w_{k}=00101$.

We then have either $w_{f}=0101w^{\prime\prime}$ and $q=0$, or $w_{f}=0101w_{1}\dots w_{q}w^{\prime\prime}$, for some $q\geq 1$, with $w^{\prime\prime}$ being either empty or 0 (observe that $w^{\prime\prime}$ cannot start with a 1, and recall that 00 cannot be a suffix of $w_{f}$), and, if $q>0$, then $w_{i}\in\{01,00101\}$ for every $i$, $1\leq i\leq q$.

Observe that if $w_{i}=01$ and $w_{i+1}=00101$ for some $i$, $1\leq i\leq q-1$, then the mapping $g$ defined by $w_{g}=0101w_{1}\dots w_{i-1}.00101.01.w_{i+2}\dots w^{\prime\prime}$ is still an $i_{b}$-broadcast on $P_{n}$. Therefore, $f$ can be chosen in such a way that there exists some $q_{0}\leq q$ such that $w_{i}=00101$ if and only if $i\leq q_{0}$.

We now claim that $f$ can be chosen in such a way that we have at most two blocks equal to 01, that is, $q-q_{0}\leq 2$. Indeed, if we add at least three such blocks, then either $1010101$ or $10101010$ is a suffix of $w_{f}$. In the former case, $1010101$ could be replaced by $1001010$, contradicting the optimality of $f$. In the latter case, we may replace $10101010$ by $10010101$, so that $q-q_{0}=2$.

Finally, we get that the structure of $w_{f}$ (recall that $n\geq 5$) is either

[TABLE]

with ${q_{0}}\geq 1$, $w_{i}=00101$ for every $i$, $1\leq i\leq{q_{0}}$, and $w^{\prime}\in\{\varepsilon,0,01,010,0101,01010\}$. It is now routine to check that $\sigma(f)=\left\lceil\frac{2n}{5}\right\rceil$ in each case, which gives $i_{b}(P_{n})=\left\lceil\frac{2n}{5}\right\rceil$ for every $n\geq 3$.

This completes the proof. ∎

Using Theorem 3.7, we can also prove a similar result for cycles.

Theorem 3.8**.**

For every integer $n\geq 3$, $i_{b}(C_{n})=\left\lceil\frac{2n}{5}\right\rceil.$

Proof.

Observe first that 010 and 0101 are $i_{b}$-broadcasts on $C_{n}$ with cost $\left\lceil\frac{2n}{5}\right\rceil$, when $n=3,4$, respectively, and that the five functions $f_{0},\dots,f_{4}$, defined in the proof of Theorem 3.7, are also $i_{b}$-broadcasts on $C_{n}$, $n\geq 5$, with cost $\left\lceil\frac{2n}{5}\right\rceil$. We thus have $i_{b}(C_{n})=\left\lceil\frac{2n}{5}\right\rceil$ for $n=3,4$, and $i_{b}(C_{n})\leq\left\lceil\frac{2n}{5}\right\rceil$ for every $n\geq 5$.

We now prove the opposite inequality when $n\geq 5$. For that, let $f$ be any $i_{b}$-broadcast on $C_{n}$, $n\geq 5$. Suppose first that $|V^{+}_{f}|=1$, which implies $i_{b}(C_{n})={\rm diam}(C_{n})=\left\lfloor\frac{n}{2}\right\rfloor$. As observed in the proof of Theorem 3.7, according to the inequality we established before, this situation can only happen if $n\in S=\{1,\dots,9,11,13\}$.

Suppose now that $n\notin S$, which implies $|V^{+}_{f}|\geq 2$ and, in particular, $n\geq 10$. We now claim that we necessarily have $|V^{+}_{f}|\geq 3$. Indeed, suppose to the contrary that $|V^{+}_{f}|=2$, and let $V^{+}_{f}=\{x_{i_{1}},x_{i_{2}}\}$. Denote by $Q_{1}=x_{i_{1}}x_{i_{1}+1}\dots x_{i_{2}}$, and $Q_{2}=x_{i_{2}}x_{i_{2}+1}\dots x_{i_{1}}$ the two paths joining $x_{i_{1}}$ and $x_{i_{2}}$. Since $f$ is maximal, we can assume, without loss of generality, that $|Q_{1}|=d_{C_{n}}(x_{i_{1}},x_{i_{2}})=f(x_{i_{1}})+1=f(x_{i_{2}})+1$, which gives $f(x_{i_{1}})=f(x_{i_{2}})$. Since Item 3 of Lemma 3.6 also holds for cycles, we have $|Q_{2}|\leq f(x_{i_{1}})+f(x_{i_{2}})+1=2f(x_{i_{1}})+1$, which gives

[TABLE]

We then get $f(x_{i_{1}})\geq\left\lceil\frac{n-2}{3}\right\rceil$, and thus

[TABLE]

a contradiction with the inequality we established before since $2\left\lceil\frac{n-2}{3}\right\rceil>\left\lceil\frac{2n}{5}\right\rceil$ when $n\geq 10$.

We can thus assume $|V^{+}_{f}|\geq 3$, and let $V^{+}_{f}=\{x_{i_{0}},\dots,x_{i_{t-1}}\}$, $t\geq 3$. We first claim that there exists some $j$, $0\leq j\leq t-1$, such that $d_{C_{n}}(x_{i_{j}},x_{i_{j+1}})=f(x_{i_{j}})+f(x_{i_{j+1}})+1$ (subscripts are taken modulo $t$). Indeed, if this is not the case, we get

[TABLE]

which gives $i_{b}(C_{n})\geq\left\lceil\frac{n}{2}\right\rceil$, in contradiction with the inequality $i_{b}(C_{n})\leq\left\lceil\frac{2n}{5}\right\rceil$ we established before, since $n\notin S$.

We can thus suppose, without loss of generality, that $d_{C_{n}}(x_{i_{1}},x_{i_{2}})=f(x_{i_{1}})+f(x_{i_{2}})+1$, which implies that $d_{C_{n}}(x_{i_{2}},x_{i_{3}})=f(x_{i_{2}})+1$ (we may have $x_{i_{3}}=x_{i_{0}}$) and, similarly, that $d_{C_{n}}(x_{i_{0}},x_{i_{1}})=f(x_{i_{1}})+1$. To avoid confusion, let us denote $P_{n}=y_{0}y_{1}\dots y_{n-1}$, with $y_{f(x_{i_{2}})}=x_{i_{2}}$, and let $g$ be the function defined by $g(y_{j})=f(x_{j-f(x_{i_{2}})+i_{2}})$ for every $j$, $0\leq j\leq n-1$ (subscripts are taken modulo $n$). Observe that both $y_{0}$ and $y_{n-1}$ are $g$-dominated, since all vertices lying between $x_{i_{1}}$ and $x_{i_{2}}$ were $f$-dominated in $C_{n}$. Moreover, we cannot increase the $g$-value of $y_{f(x_{i_{2}})}$ (the leftmost $g$-broadcast vertex in $P_{n}$) since we had $d_{C_{n}}(x_{i_{2}},x_{i_{3}})=f(x_{i_{2}})+1$, neither the value of $y_{n-f(x_{i_{1}})-1}$ (the rightmost $g$-broadcast vertex in $P_{n}$) since we had $d_{C_{n}}(x_{i_{0}},x_{i_{1}})=f(x_{i_{1}})+1$.

Since $f$ was a maximal independent broadcast on $C_{n}$, we thus get that $g$ is a maximal independent broadcast on $P_{n}$, which gives $i_{b}(P_{n})\leq i_{b}(C_{n})$, and thus $i_{b}(C_{n})\geq\left\lceil\frac{2n}{5}\right\rceil$, as required. ∎

3.3 Broadcast irredundance number and broadcast domination number

Erwin proved in [21] that $\gamma_{b}(P_{n})=\gamma(P_{n})=\lceil n/3\rceil$. Knowing the value of $\gamma_{b}(P_{n})$, we can infer the value of $\gamma_{b}(C_{n})$. Indeed, Brešar and Špacapan proved in [14] that, for every connected graph $G$, there is a spanning tree $T$ of $G$ such that $\gamma_{b}(G)=\gamma_{b}(T)$. Since spanning trees of the cycle $C_{n}$ are all isomorphic to the path $P_{n}$, we get the following result.

Proposition 3.9**.**

For every integer $n\geq 3$, $\gamma_{b}(C_{n})=\gamma_{b}(P_{n})=\lceil\frac{n}{3}\rceil$.

We now consider the broadcast irredundance number of paths. For that, we first prove the two following lemmas.

Lemma 3.10**.**

Let $f$ be a maximal irredundant broadcast on $P_{n}$. If $H_{f}(x_{i})=\emptyset$ for some vertex $x_{i}$, then $N_{P_{n}}(x_{i})\cap N_{f}(V^{+}_{f})\neq\emptyset$.

Proof.

Assume to the contrary that we have $N_{P_{n}}(x_{i})\cap N_{f}(V^{+}_{f})=\emptyset$. In that case, we could set $f(x_{i})=1$, contradicting the maximality of $f$. ∎

Lemma 3.11**.**

For every integer $n\geq 3$, the following statements hold.

If $f$ is a maximal irredundant broadcast on $P_{n}$, then $H_{f}(x_{2})\neq\emptyset$ and $H_{f}(x_{n-1})\neq\emptyset$. 2. 2.

There exists an $ir_{b}$-broadcast $f$ on $P_{n}$ such that $H_{f}(x_{1})\neq\emptyset$ and $H_{f}(x_{n})\neq\emptyset$.

Proof.

We prove the two statements separately.

If $H_{f}(x_{2})=\emptyset$, then $x_{2}$ is not $f$-dominated, and consequently $x_{1}$ is also not $f$-dominated. We then have $N_{P_{n}}(x_{1})\cap N_{f}(V^{+}_{f})=\emptyset$, in contradiction with Lemma 3.10. The case $H_{f}(x_{n-1})=\emptyset$ is similar. 2. 2.

Let $g$ be an $ir_{b}$-broadcast on $P_{n}$. If $H_{g}(x_{1})\neq\emptyset$ and $H_{g}(x_{n})\neq\emptyset$, then we let $f:=g$ and we are done.

Suppose that we have $H_{g}(x_{1})=\emptyset$. By the previous item, we know that $x_{2}$ is $g$-dominated by some vertex $x_{i}$, $i>2$, such that $f(x_{i})=d_{P_{n}}(x_{2},x_{i})$. We then necessarily have $|PB_{g}(x_{i})|=1$ if $g(x_{i})\geq 2$, and $x_{i}\notin PN_{g}(x_{i})$ if $g(x_{i})=1$, since otherwise we could set $g(x_{1})=1$, contradicting the optimality of $g$.

Now, observe that the function $h$ obtained from $g$ by setting $h(x_{i})=0$ and $h(x_{i-1})=g(x_{i})$ is a maximal irredundant broadcast on $P_{n}$, with cost $\sigma(h)=\sigma(g)=ir_{b}(P_{n})$, that satisfies $H_{h}(x_{1})\neq\emptyset$.

If $H_{h}(x_{n})\neq\emptyset$, then we let $f:=h$ and we are done. Otherwise, using the same reasoning (by symmetry), we claim that can produce an $ir_{b}$-broadcast $f$ on $P_{n}$ such that $H_{f}(x_{1})\neq\emptyset$ and $H_{f}(x_{n})\neq\emptyset$. Observe first that we cannot have $|V_{g}^{+}|=2$ if $H_{g}(x_{1})=\emptyset$ and $H_{g}(x_{n})=\emptyset$, since in that case we could increase by 1 the $g$-values of $x_{i_{1}}$ and $x_{i_{2}}$, contradicting the maximality of $g$. Therefore, if $g$ was such that $H_{g}(x_{1})=\emptyset$ and $H_{g}(x_{n})=\emptyset$, then the optimality of $g$ implies $|V_{g}^{+}|\geq 3$, so that the modification of $h$ does not affect $h(x_{i})$, and thus $x_{1}$ is still $f$-dominated.

This concludes the proof. ∎

We are now able to determine the value of the broadcast irredundance number and of the broadcast domination number of paths.

Theorem 3.12**.**

For every integer $n\geq 2$, $ir_{b}(P_{n})=\gamma_{b}(P_{n})=\left\lceil\frac{n}{3}\right\rceil.$

Proof.

By Corollary 2.5 and Proposition 3.9, we only need to prove that $\gamma_{b}(P_{n})\leq ir_{b}(P_{n})$. For this, it is enough to construct, from any non-dominating $ir_{b}$-broadcast, a dominating $ir_{b}$-broadcast.

Let $f$ be an $ir_{b}$-broadcast on $P_{n}$. By Lemma 3.11, we can assume that $x_{1}$, $x_{2}$, $x_{n-1}$ and $x_{n}$ are $f$-dominated. If $f$ is dominating, then we are done. Thus suppose that $f$ is non-dominating, and let $x_{i}$, $3\leq i\leq n-2$, be the leftmost non-dominated vertex. We will prove that there exists a maximal irredundant broadcast $g$ on $P_{n}$, with $\sigma(g)=\sigma(f)=ir_{b}(P_{n})$, such that the number of vertices that are not $g$-dominated is strictly less than the number of vertices that are not $f$-dominated.

Let $x_{j}$, $j\leq i-2$, denote the $f$-broadcast vertex that dominates $x_{i-1}$. Since $x_{i}$ is not $f$-dominated, we necessarily have $x_{i-1}\in PB(x_{j})$. Observe that we have $|PB(x_{j})|=1$, that is, the bordering private $f$-neighbor of $x_{j}$ is $x_{i-1}$, since otherwise we could set $f(x_{i-1})=1$, contradicting the maximality of $f$. Note also that $x_{1}$ is not $f$-dominated by $x_{j}$, that is, $x_{j}$ is not the leftmost $f$-broadcast vertex, since otherwise we could increase the $f$-value of $x_{j}$ by 1, again contradicting the maximality of $f$. Let then $x_{j^{\prime}}$, $j^{\prime}<j$, denote the closest $f$-broadcast vertex to the left of $x_{j}$, and $x_{j^{\prime p}}$, $j^{\prime p}<j^{\prime}$, denote the bordering private $f$-neighbor of $x_{j^{\prime}}$.

Since $x_{j^{\prime p}}$ is the bordering private $f$-neighbor of $x_{j^{\prime}}$, we necessarily have $d_{P_{n}}(x_{j^{\prime p}},x_{j})\geq f(x_{j})+1$. Moreover, we necessarily have $d_{P_{n}}(x_{j^{\prime p}},x_{j})=f(x_{j})+1$, since otherwise we could increase the value of $f(x_{j})$ by 1 ($x_{i}$ becoming the bordering private $f$-neighbor of $x_{j}$), contradicting the maximality of $f$. Hence, we have $d_{P_{n}}(x_{j^{\prime p}},x_{j})=f(x_{j})+1$, and thus

[TABLE]

Let now $g$ be the function obtained from $f$ by setting

• •

$g(x_{j})=0$,

• •

$g(x_{j^{\prime p}+1})=1$ and $g(x_{j^{\prime}})=0$ if $x_{j^{\prime}}\neq x_{j^{\prime p}+1}$, and

• •

$g(x_{j+1})=f(x_{j})$ if $x_{i+1}$ is $f$-dominated, or $g(x_{j+2})=f(x_{j})$ if $x_{i+1}$ is not $f$-dominated (see Figure 7).

Observe that $x_{j^{\prime p}}$ is a bordering private $g$-neighbor of $x_{j^{\prime p}+1}$, and that either $x_{i}$ is a bordering private $g$-neighbor of $x_{j+1}$ (if $x_{i+1}$ is $f$-dominated), or $x_{i+1}$ is a bordering private $g$-neighbor of $x_{j+2}$ (if $x_{i+1}$ is not $f$-dominated). Moreover, all vertices $x_{1},\dots,x_{i}$ are $g$-dominated. Hence, $g$ is a maximal irredundant broadcast with cost

[TABLE]

The optimality of $f$ then imply $f(x_{j^{\prime}})=1$, so that $g$ is an $ir_{b}$-broadcast on $P_{n}$ such that the number of vertices that are not $g$-dominated is strictly less than the number of vertices that are not $f$-dominated.

By repeating this modification while they remain non-dominated vertices, we eventually get a dominating $ir_{b}$-broadcast on $P_{n}$, which concludes the proof. ∎

Using Theorem 3.12, we can prove a similar result for cycles.

Theorem 3.13**.**

For every integer $n\geq 3$, $ir_{b}(C_{n})=\gamma_{b}(C_{n})=\left\lceil\frac{n}{3}\right\rceil$.

Proof.

We already know by Proposition 3.9 that $\gamma_{b}(C_{n})=\left\lceil\frac{n}{3}\right\rceil$, so that we only need to prove that $ir_{b}(C_{n})=\left\lceil\frac{n}{3}\right\rceil$. By Corollary 2.5, we only need to prove that $ir_{b}(C_{n})\geq\gamma_{b}(C_{n})$ or, by Proposition 3.9 and Theorem 3.12, that $ir_{b}(C_{n})\geq ir_{b}(P_{n})$.

Observe that $010$, $0200$ and $00200$ are dominating $ir_{b}$-broadcasts for $C_{3}$, $C_{4}$ and $C_{5}$, respectively, with cost $\left\lceil\frac{n}{3}\right\rceil$. It thus remains to consider the case $n\geq 6$.

Let $f$ be an $ir_{b}$-broadcast on $C_{n}$, $n\geq 6$. If $f$ is dominating, then we are done. Thus suppose that $f$ is non-dominating. We claim that $|V^{+}_{f}|\geq 2$. Indeed, if $|V^{+}_{f}|=1$, then the maximality of $f$ implies $\sigma(f)=\left\lfloor\frac{n}{2}\right\rfloor$, which gives $ir_{b}(C_{n})=\left\lfloor\frac{n}{2}\right\rfloor>\left\lceil\frac{n}{3}\right\rceil=\gamma_{b}(C_{n})$, in contradiction with Corollary 2.5 since $n\geq 6$.

We thus have $|V^{+}_{f}|\geq 2$. Since $f$ is maximal, we cannot have three consecutive vertices that are not $f$-dominated. Moreover, since $f$ is non-dominating, we can assume, without loss of generality, that $x_{0}$ is not $f$-dominated, and that $x_{1}$ is $f$-dominated by some $f$-broadcast vertex $x_{i_{1}}$, $i_{1}>1$. Note that $x_{n-1}$ may be $f$-dominated or not.

To avoid confusion, let $P_{n}=y_{0}y_{1}\dots y_{n-1}$. Let then $g$ be the mapping defined by $g(y_{i})=f(x_{i})$ for every $i$, $0\leq i\leq n-1$. As in $C_{n}$, $x_{0}$ is not $g$-dominated, $x_{1}$ is $g$-dominated by $x_{i_{1}}$, and $g(x_{i_{1}})$ cannot be increased since otherwise $f(x_{i_{1}})$ could be increased, contradicting the maximality of $f$. Similarly, the $g$-value of the rightmost $g$-broadcast vertex in $P_{n}$ cannot be increased, since otherwise its $f$-value could be increased (since $x_{0}$ is not $f$-dominated, while $x_{n-1}$ may be $f$-dominated or not), again contradicting the maximality of $f$. we can apply the same reasoning if $x_{n-1}$ is not $g$-dominated, which implies that $x_{n-1}$ is not $f$-dominated. Hence, since $f$ is an $ir_{b}$-broadcast on $C_{n}$, we get that $g$ is also a maximal irredundant broadcast on $P_{n}$, which gives $ir_{b}(C_{n})=\sigma(f)=\sigma(g)\geq ir_{b}(P_{n})$ as required. ∎

3.4 Packing broadcast number and lower packing broadcast number

We first determine the broadcast packing number of paths and cycles.

Theorem 3.14**.**

For every $n\geq 2$, $P_{b}(P_{n})={\rm diam}(P_{n})=n-1$, and, for every $n\geq 3$, $P_{b}(C_{n})={\rm diam}(C_{n})=\left\lfloor\frac{n}{2}\right\rfloor$.

Proof.

We first consider the case of the path $P_{n}$, $n\geq 2$. Observe first that the function $f$ defined by $f(x_{1})=n-1$ and $f(x_{i})=0$ for every $i$, $2\leq i\leq n$ is a maximal broadcast packing with cost $n-1={\rm diam}(P_{n})$, which gives ${\rm diam}(P_{n})\leq P_{b}(P_{n})$. The opposite inequality directly follows from Observation 1.1(3) and Theorem 3.1.

Let us now consider the case of the cycle $C_{n}$, $n\geq 3$. Observe first that the function $f$ defined by $f(x_{0})=\left\lfloor\frac{n}{2}\right\rfloor$ and $f(x_{i})=0$ for every $i$, $1\leq i\leq n-1$ is a maximal broadcast packing with cost $\left\lfloor\frac{n}{2}\right\rfloor={\rm diam}(C_{n})$, which gives ${\rm diam}(C_{n})\leq P_{b}(C_{n})$.

Again, to establish the opposite inequality, it suffices to prove that, for every $P_{b}$-broadcast $f$ on $C_{n}$, $|V_{f}^{+}|=1$. Similarly as above, if we suppose that $f$ is a $P_{b}$-broadcast on $C_{n}$ with $|V_{f}^{+}|\geq 2$, we get

[TABLE]

which gives

[TABLE]

and thus

[TABLE]

again a contradiction. ∎

In order to determine the values of $p_{b}(P_{n})$, $n\geq 1$, we first prove the following lemma.

Lemma 3.15**.**

For every integer $n\geq 2$, there exists a $p_{b}$-broadcast $f$ on $P_{n}$ such that $f(x_{i})=1$ for every $f$-broadcast vertex $x_{i}$.

Proof.

Observe first that $10$, $010$, $1001$ and $10010$ define $p_{b}$-broadcasts on $P_{n}$, $n=2,3,4,5$, respectively, that satisfy the statement of the lemma. Suppose thus $n\geq 6$ and let $g$ be any $p_{b}$-broadcast on $P_{n}$. If $g(x_{i})=1$ for every $g$-broadcast vertex $x_{i}$, then we set $f:=g$ and we are done.

Otherwise, let $V_{g}^{+}=\{x_{i_{1}},\dots,x_{i_{t}}\}$, $i_{1}<\cdots<i_{t}$, $t\geq 1$. We first claim that we necessarily have $t\geq 2$. Indeed, if $t=1$, we get

[TABLE]

while the function $g^{\prime}$ defined by $g^{\prime}(x_{i})=1$, $1\leq i\leq n$, if and only if $i\equiv 1\pmod{3}$ is a maximal packing broadcast with cost $\sigma(g^{\prime})=\left\lfloor\frac{n}{3}\right\rfloor<\left\lfloor\frac{n}{2}\right\rfloor,$ a contradiction.

We thus have $|V_{g}^{+}|\geq 2$. Let $x_{i_{j}}$, $1\leq j\leq t$, be a $g$-broadcast vertex with minimum subscript such that $g(x_{i_{j}})\geq 2$. We will consider three cases, depending on the value of $i_{j}$. In each case, we will prove either that the case cannot occur, or that we can produce a $p_{b}$-broadcast $g^{\prime}$ on $P_{n}$, with $\sigma(g^{\prime})=\sigma(g)$, such that the subscript of the leftmost $g^{\prime}$-broadcast vertex with $g^{\prime}$-value at least $2$, if any, is strictly greater than the subscript of the leftmost $g$-broadcast vertex with $g$-value at least $2$.

$i_{j}=i_{1}$ or $i_{j}=i_{t}$.

Assume $i_{j}=i_{1}$, the case $i_{j}=i_{t}$ being similar, by symmetry. We first claim that we have $g(x_{i_{1}})\in\{i_{1}-2,i_{1}-1\}$. Indeed, if $g(x_{i_{1}})\geq i_{1}$, then the function $h$ obtained from $g$ by setting $h(x_{i_{1}})=0$ and $h(x_{i_{1}+1})=g(x_{i_{1}})-1$ is clearly a maximal packing broadcast with cost $\sigma(h)=\sigma(g)-1$, contradicting the optimality of $g$. Now, if $g(x_{i_{1}})\leq i_{1}-3$, then we could set $g(x_{1})=1$, contradicting the maximality of $g$. We thus have two cases to consider.

1. (a)

$g(x_{i_{1}})=i_{1}-1$, and thus $i_{1}\geq 3$.

Let $g_{1}$ be the function obtained from $g$ by modifying the $g$-values of $x_{1},\dots,x_{2g(x_{i_{1}})+1}$ as follows:

• •

$g_{1}(x_{1}\dots x_{2g(x_{i_{1}})+1})=(010)^{\alpha}$, if $2g(x_{i_{1}})+1\equiv 0\pmod{3}$,

• •

$g_{1}(x_{1}\dots x_{2g(x_{i_{1}})+1})=0(010)^{\alpha}$, if $2g(x_{i_{1}})+1\equiv 1\pmod{3}$,

• •

$g_{1}(x_{1}\dots x_{2g(x_{i_{1}})+1})=10(010)^{\alpha}$, if $2g(x_{i_{1}})+1\equiv 2\pmod{3}$,

where $\alpha=\left\lfloor\frac{2g(x_{i_{1}})+1}{3}\right\rfloor$ (see Figure 8(a)). It is then not difficult to check that $g_{1}$ is a maximal packing broadcast such that

[TABLE]

which gives $\sigma(g_{1})-\sigma(g)<0$ whenever $g(x_{i_{1}})\neq 2$. Consequently, $g(x_{i_{1}})=g(x_{3})=2$ and we can then define $g^{\prime}$ from $g$ by setting $g^{\prime}(x_{1}\dots x_{5})=10010$. 2. (b)

$g(x_{i_{1}})=i_{1}-2$, and thus $i_{1}\geq 4$.

Similarly to the previous case, let $g_{2}$ be the function obtained from $g$ by modifying the $g$-values of $x_{1},\dots,x_{2g(x_{i_{1}})+2}$ as follows:

• •

$g_{2}(x_{1}\dots x_{2g(x_{i_{1}})+2})=(010)^{\alpha}$, if $2g(x_{i_{1}})+2\equiv 0\pmod{3}$,

• •

$g_{2}(x_{1}\dots x_{2g(x_{i_{1}})+2})=0(010)^{\alpha}$, if $2g(x_{i_{1}})+2\equiv 1\pmod{3}$,

• •

$g_{2}(x_{1}\dots x_{2g(x_{i_{1}})+2})=10(010)^{\alpha}$, if $2g(x_{i_{1}})+2\equiv 2\pmod{3}$,

where $\alpha=\left\lfloor\frac{2g(x_{i_{1}})+2}{3}\right\rfloor$ (see Figure 8(b)) . Again, it is not difficult to check that $g_{2}$ is a maximal packing broadcast such that

[TABLE]

which gives $\sigma(g_{2})-\sigma(g)<0$ whenever $g(x_{i_{1}})\neq 2$ and $g(x_{i_{1}})\neq 3$. Consequently, either $g(x_{i_{1}})=g(x_{4})=2$ and we can then define $g^{\prime}$ from $g$ by setting $g^{\prime}(x_{1}\dots x_{6})=010010$, or $g(x_{i_{1}})=g(x_{5})=3$ and we can then define $g^{\prime}$ from $g$ by setting $g^{\prime}(x_{1}\dots x_{8})=10010010$. 2. 2.

$i_{j}\in\{i_{2},\dots,i_{t-1}\}$.

Since $g$ is a maximal packing broadcast, we necessarily have

[TABLE]

Moreover, we also have either

[TABLE]

We consider four subcases, depending on the value of $2g(x_{i_{j}})+1\mod 3$, and on the number $p$ of vertices lying between $x_{i_{j-1}}$ and $x_{i_{j}}$, or between $x_{i_{j}}$ and $x_{i_{j+1}}$, that are not $g$-dominated. Note that since $g$ is a maximal packing broadcast, we have either (i) $p=0$, or (ii) $p=1$ and either $x_{i_{j}-g(x_{i_{j}})-1}$ or $x_{i_{j}+g(x_{i_{j}})+1}$ is not $g$-dominated, or (iii) $p=2$ and either $x_{i_{j}-g(x_{i_{j}})-2}$ and $x_{i_{j}-g(x_{i_{j}})-1}$, or $x_{i_{j}+g(x_{i_{j}})+1}$ and $x_{i_{j}+g(x_{i_{j}})+2}$, are not $g$-dominated.

1. (a)

$2g(x_{i_{j}})+1\equiv 0\pmod{3}$.

Let $g_{0}$ be the function obtained from $g$ by setting $g_{0}(x_{i_{j}-g(x_{i_{j}})}\dots x_{i_{j}+g(x_{i_{j}})})=(010)^{\alpha}$, where $\alpha=\frac{2g(x_{i_{j}})+1}{3}$ (see Figure 9(a)). Since $g$ is maximal, $g_{0}$ is also maximal and we have

[TABLE]

which contradicts the optimality of $g$. 2. (b)

$2g(x_{i_{j}})+1\equiv 1\pmod{3}$ and $p=2$, or $2g(x_{i_{j}})+1\equiv 2\pmod{3}$ and $1\leq p\leq 2$.

Suppose that $2g(x_{i_{j}})+1\equiv 1\pmod{3}$, and $x_{i_{j}-g(x_{i_{j}})-2}$, $x_{i_{j}-g(x_{i_{j}})-1}$ are not $g$-dominated, or that $2g(x_{i_{j}})+1\equiv 2\pmod{3}$, and $x_{i_{j}-g(x_{i_{j}})-1}$ is not $g$-dominated. (The other cases are similar.)

Let $g^{\prime}$ be the function obtained from $g$ by setting

• •

$g^{\prime}(x_{i_{j}-g(x_{i_{j}})-2}\dots x_{i_{j}+g(x_{i_{j}})})=(010)^{\alpha}$ if $2g(x_{i_{j}})+1\equiv 1\pmod{3}$,

• •

$g^{\prime}(x_{i_{j}-g(x_{i_{j}})-1}\dots x_{i_{j}+g(x_{i_{j}})})=(010)^{\alpha}$ if $2g(x_{i_{j}})+1\equiv 2\pmod{3}$,

where $\alpha=\left\lfloor\frac{2g(x_{i_{j}})+p+1}{3}\right\rfloor$ (see Figure 9(b)). Since $g$ is maximal, $g^{\prime}$ is also maximal and we have

[TABLE]

The optimality of $g$ then implies either $p=1$ and $g(x_{i_{j}})=2$, or $p=2$ and $g(x_{i_{j}})=3$. In each case, $g^{\prime}$ is also optimal and the subscript of the leftmost $g^{\prime}$-broadcast vertex with $g^{\prime}$-value at least $2$, if any, is strictly greater than the subscript of the leftmost $g$-broadcast vertex with $g$-value at least $2$, as required. 3. (c)

$2g(x_{i_{j}})+1\equiv 1\pmod{3}$ and $0\leq p\leq 1$.

In that case, $x_{i_{j}-g(x_{i_{j}})-2}$ and $x_{i_{j}+g(x_{i_{j}})+2}$ are $g$-dominated, and at most one vertex among $x_{i_{j}-g(x_{i_{j}})-1}$ and $x_{i_{j}+g(x_{i_{j}})+1}$ is not $g$-dominated. Let $g^{\prime}$ be the function obtained from $g$ (see Figure 9(c)) by setting $g^{\prime}(x_{i_{j}-g(x_{i_{j}})}\dots x_{i_{j}+g(x_{i_{j}})})=(010)^{\alpha}0$, where $\alpha=\frac{2g(x_{i_{j}})}{3}$, and

• •

$g^{\prime}(x_{i_{j+1}})=0$, $g^{\prime}(x_{i_{j+1}-1})=g(x_{i_{j+1}})+1$, if $x_{i_{j}+g(x_{i_{j}})+1}$ is not $g$-dominated,

• •

$g^{\prime}(x_{i_{j+1}})=g(x_{i_{j+1}})+1$, if $i_{j}=i_{t-1}$ or $x_{i_{j}+g(x_{i_{j}})+1}$ is $g$-dominated, and $x_{i_{j+1}+g(x_{i_{j+1}})+1}$ is not $g$-dominated,

Note that we necessarily have one of the above cases, since otherwise $g^{\prime}$ would be a maximal packing broadcast with $\sigma(g^{\prime})<\sigma(g)$, contradicting the optimality of $g$. Since $g$ is maximal, $g^{\prime}$ is also maximal and we have

[TABLE]

(Recall that since $2g(x_{i_{j}})+1\equiv 1\pmod{3}$, we have $g(x_{i_{j}})\geq 3$.) The optimality of $g$ then implies $g(x_{i_{j}})=3$. We then get that $g^{\prime}$ is also optimal and the subscript of the leftmost $g^{\prime}$-broadcast vertex with $g^{\prime}$-value at least $2$, if any, is strictly greater than the subscript of the leftmost $g$-broadcast vertex with $g$-value at least $2$, as required. 4. (d)

$2g(x_{i_{j}})+1\equiv 2\pmod{3}$ and $p=0$.

Let $g^{\prime}$ be the function obtained from $g$ by setting $g^{\prime}(x_{i_{j}-g(x_{i_{j}})}\dots x_{i_{j}+g(x_{i_{j}})})=(010)^{\alpha}00$, where $\alpha=\frac{2g(x_{i_{j}})-1}{3}$, $g^{\prime}(x_{i_{j+1}})=0$, and $g^{\prime}(x_{i_{j+1}-1})=g(x_{i_{j+1}})+1$ (see Figure 9(d)). Since $g$ is maximal, $g^{\prime}$ is also maximal and we have

[TABLE]

The optimality of $g$ then implies $g(x_{i_{j}})=2$. Therefore, $g^{\prime}$ is also optimal and the subscript of the leftmost $g^{\prime}$-broadcast vertex with $g^{\prime}$-value at least $2$, if any, is strictly greater than the subscript of the leftmost $g$-broadcast vertex with $g$-value at least $2$, as required.

Repeating the same transformation for each vertex with $g$-value at least 2, we eventually produce a $p_{b}$-broadcast $g^{\prime}$ on $P_{n}$ all of whose broadcast vertices have $g^{\prime}$-value 1, as claimed in the statement of the lemma. This concludes the proof. ∎

We are now able to determine the lower broadcast packing number of paths.

Theorem 3.16**.**

For every integer $n\geq 2$,

[TABLE]

Proof.

Observe first that $10$, $010$, $1001$, $01001$, $001001$, $0010010$ and $00100100$ define optimal $p_{b}$-broadcasts on $P_{n}$ for $n=2,\dots,8$, respectively, whose costs are the values claimed by the theorem.

Suppose now $n\geq 9$ and let $n=8q+r$, with $q\geq 1$ and $0\leq r\leq 7$. According to the value of $r$, we define the broadcasts $f_{r}$, $0\leq r\leq 7$, on $P_{n}$ as follows:

[TABLE]

[TABLE]

[TABLE]

It is not difficult to check that each $f_{r}$ is indeed a maximal packing broadcast on $P_{8q+r}$, $8q+r\geq 9$, with cost

[TABLE]

which gives $p_{b}(P_{n})\leq\sigma(f_{r})$, that is, $p_{b}(P_{n})$ is not greater than the value claimed by the theorem.

We now prove the opposite inequality. By Lemma 3.15, we know that there exists a $p_{b}$-broadcast all of whose broadcast vertices have broadcast value 1. Let $g$ be such a broadcast. For every $k$, $0\leq k\leq q-1$, let

[TABLE]

From the definition of a packing broadcast, we get that the distance between any two consecutive $g$-broadcast vertices (with $g$-value $1$) is $3$, $4$ or $5$, which implies $2\leq\sigma_{k}\leq 3$ for every $k$, $0\leq k\leq q-1$. Observe also that $x_{2}$ and $x_{n-1}$ must be $g$-dominated, since otherwise we could set $g(x_{1})=1$ or $g(x_{n})=1$, contradicting the maximality of $g$, and that $d_{P_{n}}(x_{i_{1}},x_{i_{2}})=d_{P_{n}}(x_{i_{t-1}},x_{i_{t}})=3$, since otherwise we could increase the value of $g(x_{i_{1}})$ or $g(x_{i_{t}})$, contradicting the maximality of $g$.

We now consider three cases, depending on the value of $r$, and prove in each case that the cost of $g$ is the value claimed in the statement of the theorem.

$r=0$.

In that case, we have

[TABLE] 2. 2.

$r\in\{1,2,3\}$.

If $\sigma_{k}=3$ for some $k$, $0\leq k\leq q-1$, then

[TABLE]

and we are done.

Suppose now that $\sigma_{k}=2$ for every $k$, $0\leq k\leq q-1$, which implies $p_{b}(P_{n})\geq 2\left\lfloor\frac{n}{8}\right\rfloor$. Suppose, contrary to the statement of the theorem, that $p_{b}(P_{n})=2\left\lfloor\frac{n}{8}\right\rfloor$. This implies $g(x_{n-r+1})=\cdots=g(x_{n})=0$. If $r=3$, then we have a contradiction since $x_{n-1}$ must be $g$-dominated.

We thus have $r\in\{1,2\}$. Since $x_{n-1}$ must be $g$-dominated and $d_{P_{n}}(x_{i_{t-1}},x_{i_{t}})=3$, we necessarily have $\sigma_{q-1}\in\{00001001,00010010\}$ if $r=1$, and $\sigma_{q-1}=00001001$ if $r=2$. Since $g$ is maximal, we cannot have three consecutive vertices that are not $g$-dominated, which gives $\sigma_{j}\in\{00001001,00010010\}$ if $r=1$, and $\sigma_{j}=00001001$ if $r=2$, for every $j$, $0\leq j\leq q-1$, a contradiction since, in each case, this would imply that $x_{2}$ is not $g$-dominated.

Hence we have $p_{b}(P_{n})=2\left\lfloor\frac{n}{8}\right\rfloor+1$, as required. 3. 3.

$r\in\{4,5,6,7\}$.

Note first that we necessarily have $g(x_{n-r+1})+\cdots+g(x_{n})\geq 1$. Hence, if $\sigma_{k}=3$ for some $k$, $0\leq k\leq q-1$, then

[TABLE]

and we are done.

Suppose now that $\sigma_{k}=2$ for every $k$, $0\leq k\leq q-1$, and, contrary to the statement of the theorem, that $p_{b}(P_{n})\leq 2\left\lfloor\frac{n}{8}\right\rfloor+1$, which implies $g(x_{n-r+1})+\cdots+g(x_{n})=1$ since $x_{n-1}$ must be $g$-dominated. If $r=6$ or $r=7$, then we have a contradiction since we must have $d_{P_{n}}(x_{i_{t-1}},x_{i_{t}})=3$.

We thus have $r\in\{4,5\}$. Again, since $x_{n-1}$ must be $g$-dominated and $d_{P_{n}}(x_{i_{t-1}},x_{i_{t}})=3$, we necessarily have

[TABLE]

if $r=4$, and

[TABLE]

if $r=5$. Now, since $g$ is maximal, every $g$-broadcast vertex must be at distance $3$ from another $g$-broadcast vertex, and thus we get

[TABLE]

for every $j$, $0\leq j\leq q-1$. We then get a contradiction since $x_{2}$ must be $g$-dominated and we must have $d(x_{i_{1}},x_{i_{2}})=3$.

Hence, in this case also, we have $p_{b}(P_{n})=2\left\lfloor\frac{n}{8}\right\rfloor+2$, as required.

This completes the proof. ∎

Using Theorem 3.16, we can also prove a similar result for cycles.

Theorem 3.17**.**

For every integer $n\geq 3$,

[TABLE]

Proof.

Observe first that the broadcasts $f_{0},\dots,f_{7}$, defined in the proof of Theorem 3.16, are also maximal packing broadcast on $C_{n}$, with $n\geq 3$, $n=8q+r$ and $0\leq r\leq 7$, which gives $p_{b}(C_{n})\leq\sigma(f_{r})$, that is, $p_{b}(C_{n})$ is not greater than the value claimed by the theorem.

We now prove the opposite inequality. Observe first that $010$, $2000$, $20000$ and $001001$ are optimal solutions for $C_{3}$, $C_{4}$, $C_{5}$ and $C_{6}$, respectively, and that their cost is exactly the value claimed by the theorem. We can thus assume $n\geq 7$. Let $f$ be a $p_{b}$-broadcast on $C_{n}$, $n\geq 7$.

We first claim that we necessarily have $|V^{+}_{f}|\geq 2$. Indeed, suppose to the contrary that $|V^{+}_{f}|=1$ and that $V^{+}_{f}=\{x_{0}\}$, without loss of generality. Since $f$ is maximal, we necessarily have $\sigma(f)=f(x_{0})=\left\lfloor\frac{n-1}{2}\right\rfloor$, and thus $p_{b}(C_{n})=\left\lfloor\frac{n-1}{2}\right\rfloor$, which contradicts the inequality we previously established when $n\geq 7$.

We thus have $|V^{+}_{f}|\geq 2$. Let $V^{+}_{f}=\{x_{i_{0}},\dots,x_{i_{t-1}}\}$, $t\geq 2$. Since $f$ is maximal, we necessarily have

[TABLE]

for every $j$, $0\leq j\leq t-1$. Moreover, if $d_{C_{n}}(x_{i_{j}},x_{i_{j+1}})\geq f(x_{i_{j}})+f(x_{i_{j+1}})+2$, then we necessarily have $d_{C_{n}}(x_{i_{j+1}},x_{i_{j+2}})=f(x_{i_{j+1}})+f(x_{i_{j+2}})+1$ (we may have $i_{j}=i_{j+2}$), since otherwise we could increase the value of $f(x_{i_{j+1}})$ by 1.

We consider the two following cases.

If all vertices in $C_{n}$ are $f$-dominated, that is, $d_{C_{n}}(x_{i_{j}},x_{i_{j+1}})=f(x_{i_{j}})+f(x_{i_{j+1}})+1$ for every $j$, $0\leq j\leq t-1$, then, to avoid confusion, we let $P_{n}=y_{0}y_{1}\dots y_{n-1}$ with $y_{0}=x_{{i_{0}}+f(x_{i_{0}})+1}$. Let now $g$ be the function defined by $g(y_{j})=f(x_{j+i_{0}+f(x_{i_{0}})+1})$ for every $j$, $0\leq j\leq n-1$. Clearly, both $y_{0}$ and $y_{n-1}$ are $g$-dominated. Since $f$ was a maximal packing broadcast on $C_{n}$, we get that $g$ is also a maximal packing broadcast on $P_{n}$, which gives $p_{b}(P_{n})\leq p_{b}(C_{n})$. 2. 2.

If $C_{n}$ contains at least one vertex which is not $f$-dominated, then we can assume, without loss of generality, that $x_{i_{1}+f(x_{i_{1}})+1}$ is not $f$-dominated, which implies $d_{C_{n}}(x_{i_{0}},x_{i_{1}})=f(x_{i_{0}})+f(x_{i_{1}})+1$ and $d_{C_{n}}(x_{i_{2}},x_{i_{3}})=f(x_{i_{2}})+f(x_{i_{3}})+1$ (we may have $x_{i_{0}}=x_{i_{2}}$ and $x_{i_{1}}=x_{i_{3}}$). Again, to avoid confusion, we let $P_{n}=y_{0}y_{1}\dots y_{n-1}$ with $y_{0}=x_{{i_{1}}+f(x_{i_{1}})+2}$. Let now $g$ be the function defined by $g(y_{j})=f(x_{j+i_{1}+f(x_{i_{1}})+2})$ for every $j$, $0\leq j\leq n-1$. Since $d_{C_{n}}(x_{i_{0}},x_{i_{1}})=f(x_{i_{0}})+f(x_{i_{1}})+1$ and $d_{C_{n}}(x_{i_{2}},x_{i_{3}})=f(x_{i_{2}})+f(x_{i_{3}})+1$, the values of both the leftmost and the rightmost $g$-broadcast vertex in $P_{n}$ cannot be increased. Since $f$ was a maximal packing broadcast on $C_{n}$, we thus get that $g$ is also a maximal packing broadcast on $P_{n}$, which gives $p_{b}(P_{n})\leq p_{b}(C_{n})$.

We thus have $p_{b}(P_{n})\leq p_{b}(C_{n})$ in all cases and the result follows. ∎