Independence and domination in divisor graph and mod-difference graphs

Abstract

We initiate the study of domination and inverse domination in labeled graphs. In this paper, we determined the cardinality of maximal independent and minimum variant dominating (total dominating/independent dominating/co-independent dominating) sets and their inverse in divisor graph and in two new labeling definitions called 0-mod-difference and 1-mod-difference graphs.

Introduction

Consider $G(E,V)$ be a finite, undirected and simple graph. The independence number of $G$ denoted by $\beta \left(G\right)$ is the maximum cardinality over all independent sets. The domination number of G denoted by $\gamma (G$) is the minimum cardinality over all dominating sets. The inverse domination number of G denoted by ${\gamma }^{-1}\left(G\right)$ is the minimum cardinality over all inverse dominating sets.

We consider a finite undirected and simple graph $G(E,V)$ with a set $V(G)$ of vertices and a set $E(G)$ of edges.

A subgraph $H$ of a graph $G$ is said to be induced (or full) subgraph if, for any pair of vertices $x$ and y of $H$, $xy$ is an edge of $H$ if and only if $xy$ is an edge of $G$. If $H$ is an induced subgraph of $G$ and S is a set of its vertices then $H$ is said to be an induced subgraph by $S$ and denoted by $G\left[S\right].$

A set $I\subseteq G$ is an independent set or stable set in graph $G$ if no two of its vertices are adjacent. An independence number of $G$ denoted by $\beta \left(G\right)$ is the maximum cardinality over all independent sets.

A set $D \subseteq V(G)$ is a dominating set in $G$ if $N (v)\cap D\ne \varnothing$; for every vertex $v\in V(G)-D.$ the domination number of $G$, denoted by $\gamma ({G}$) $,$ is a minimum cardinality over all dominating sets in $G$.

A dominating set $D \subseteq V(G)$ is an independent dominating set in $G$ if $D$ is an independent set in $G$.The independence domination number of $G$, denoted by ${\gamma }_{{i}}({G})$, is a minimum cardinality of independent dominating sets in $G$.

A dominating set $D \subseteq V(G)$ is a total dominating set in G if $N (v)\cap D\ne \varnothing$; for every vertex $v\in {v}\left({G}\right)$. This means that $G\left[D\right]$ has no isolated vertex. A minimum cardinality over all total dominating sets in $G$ is the total domination number of $G$ and is denoted by ${\gamma }_{{t}} \left({G}\right)$ [10].

A dominating set $D \subseteq V(G)$ is a connected dominating set in G, if $G[D]$ is connected. The connected domination number of $G$, denoted by ${\gamma }_{{c}}({G}),$ is a minimum cardinality over all connected dominating sets in $G$ [8].

A dominating set $D \subseteq V(G)$ is a co-independent dominating set in $G$ if the complement of $D$ is an independent set. The co-independence domination number of $G$, denoted by ${\gamma }_{\mathrm{coi}}\left(G\right),$ is a minimum cardinality over all co-independent dominating sets of $G$ [10].

Let $D \subseteq V(G)$ be a minimum dominating (independent dominating/total dominating/connected dominating/co-independent dominating) set in graph $G$. If $V-D$ contains a dominating (an independence dominating/total dominating/connected dominating/co-independence dominating) set $ID$ of $G$, where $ID$ is called an inverse dominating (an independent dominating/total dominating/connected dominating/co-independent dominating) set with respect to $D$. The inverse domination (an independence domination/total domination/connected domination/co-independence domination) number of $G$, denoted by ( ${\gamma }^{-1}\left(G\right), {{\gamma }_{i}}^{-1}\left(G\right) ,{ {\gamma }_{t}}^{-1}\left(G\right) ,{ {{\gamma }_{c}}^{-1}\left(G\right) and {\gamma }_{coi}}^{-1}\left(G\right)$) is the minimum cardinality over all inverse dominating (an independent dominating/total dominating/connected dominating/co-independent dominating) sets of $G$ [6].

A graph labeling is an assignment of integers to the vertices or edges, or both, subject to certain conditions. Santhosh and Singh [7] call a graph $G(V,E)$ with vertex set $V$ and edge set $E$ a divisor graph if $V$ is labeled by a set of integers and for each edge $uv\in E$ either the label assigned to $u$ divides the label assigned to $v$ or vice versa. We studied the notion “divisor graph” in the sense that its vertices can be labeled with distinct integers $1, 2 ,\dots ,\left|V\right|$ such that for each edge $uv\in E$ either the label assigned to $u$ divides the label assigned to $v$ or vice versa. Also, we introduce two new definitions labeling called 0-mod-difference and 1-mod-difference.

There are more than 75 models of domination listed in the appendix of Haynes [5]. For more details about parameters of domination number, we refer to [2, 3]. In this paper, we study different formulas of cardinality of independence and domination (total domination, independence domination, co-independence domination) in divisor, 0-mod-difference and 1-mod-difference graph. The inverse domination (total domination, independence domination, co-independence domination) number of divisor (0-mod-difference/1-mod-difference graph) graph also determined.

Any notion or definition of graph labeling which is not found here could be found in [1].

Some new methods

In the following sections, we will study three new methods. The following are new notions.

Definition 1.1. [9]

Let $G(V,E)$ be a simple graph of order $n$ and $f:V\to \{1, 2 ,\dots ,n\}$ be a bijection. For each edge $uv$, if either $f(u)\backslash f(v)$ ( $f(u)$ divides $f(v)$) or $f(v)\backslash f(u)$ ( $f(v)$ divides $f(u)$) then $f$ is called a divisor labeling and $G$ is called a divisor graph. A graph which is not divisor is called a non-divisor graph.

Definition 1.2.

Let $G(V,E)$ be a simple graph of order $n$ and $f:V\to \{1, 2 ,\dots ,n \}$ be a bijection. A graph $G(V,E)$ with vertex set $V$ is said to be 0-mod-difference if for each edge $uv\in E, \left|f(u)-f(v)\right|\equiv 0 (mod m )$ where $2\le m\le \lfloor\frac{n}{2}\rfloor$. A graph which is not $0$-mod-difference is called a non-0-mod-difference graph [11].

Definition 1.3.

Let $G(V,E)$ be a simple graph of order $n$ and $f:V\to \{1, 2 ,\dots ,n \}$ be a bijection. A graph $G(V,E)$ with vertex set $V$ is said to be 1-mod-difference if for each edge $uv\in E, \left|f(u)-f(v)\right|\equiv 1 (mod m )$ where $2\le m\le \lfloor\frac{n}{2}\rfloor$. A graph which is not a $1$-mod-difference is called a non-1-mod-difference graph.

Definition 1.4.

A maximal divisor /0-mod-difference/1-mod-difference graph of $n$ vertices is a divisor/0-mod-difference/1-mod-difference graph such that adding any new edge yields a non-divisor (0-mod-difference/1-mod-difference) graph. Figure 1 gives a maximal divisor graph of order 10.

Fig. 1

Maximal divisor graph of order 10

Definition 1.5.

[4] Let $x$ be a nonnegative real number. The Gauss ̓ s function $\pi (x)$ is defined to be the number of primes not exceeding $x$. $i.e$, $\pi \left(x\right)=|\{p:\;p\;is\;prime,\; 2\le p\le x\}|$.

Note 1.6.

In all definitions in this article, we define the labeling function by:

$$f\left({v}_{i}\right)=i,i=1,\dots ,n$$

Divisor graph

Theorem 2.1.

If $G$ is a maximal divisor graph then,

• (i) $\beta (G)=\lceil\frac{n}{2}\rceil$.

• (ii) $\gamma (G)={\gamma }_{{i}}(G)={\gamma }_{{c}}(G)=1$

• (iii) ${\gamma }_{{t}}\left(G\right)=2$

• (iv) ${\gamma }_{\mathrm{coi}}\left(G\right)=\lfloor\frac{{n}}{2}\rfloor;n>3$

Proof

• (i) Consider $I=\left\{v\in G:f\left(v\right)>\lfloor\frac{n}{2}\rfloor\right\}$. Then, $I$ is an independence set, since for each vertex $v\in I$, the vertex of label $2f(v)$ does not belong to $G$ (see Fig. 1), therefore $\beta \left(G\right)\ge \left|I\right|=\lceil\frac{n}{2}\rceil$. If we assume that there is a set ${A}$ such that $\left|{A}\right|>\left|{i}\right|$ then ${A}$ must contain at least two adjacent vertices, since each vertex $v$ which $f(v)\le \lfloor\frac{{n}}{2}\rfloor$ is adjacent to a vertex of label $2f(v)$. Thus, $\beta \left(G\right)=\lceil\frac{n}{2}\rceil.$

• (ii) It is obvious, since the vertex of label one is adjacent to all vertices of $G$.

• (iii) Let ${{D}}_{1}=\{{v}_{1} ,{v}_{2}\}$. ${{D}}_{1}$ is a dominating set in $G$ with no isolated vertex and it is clear that it is the minimum total dominating set so ${\gamma }_{{t}}\left({G}\right)=2.$

• (iv) Consider ${{D}}_{2}=\left\{v\in G:f(v) \le \lfloor\frac{{n}}{2}\rfloor\right\}$. ${{D}}_{2}$ contains a vertex of label one therefore it is the dominating set in ${G}$ and ${v}-{{D}}_{2}$ is an independent set by (i). Thus, ${\gamma }_{\mathrm{coi}}\left({G}\right)\le \lfloor\frac{{n}}{2}\rfloor$. If we assume that there is a set ${c}$ such that $\left|{c}\right|<\left|{{D}}_{2}\right|$ then ${c}$ may be a dominating set, but ${v}-{c}$ cannot be an independent set by (i). Thus, ${\gamma }_{\mathrm{coi}}\left({G}\right)=\lfloor\frac{{n}}{2}\rfloor$.

Note 2.2.

• (1) If $G$ is a divisor graph and $\beta \left(G\right)>\lceil\frac{n}{2}\rceil$, $\gamma \left(G\right)>1$ or ${\gamma }_{i}\left(G\right)>1$ then $G$ is not a maximal divisor graph.

• (2) If $G$ is a divisor graph and ${\gamma }_{{c}}\left({G}\right)>1$ or there is no connected dominating set in $G$ then $G$ is not a maximal divisor graph.

• (3) If $G$ is a divisor graph and ${\gamma }_{{t}}\left(G\right)>2$ or there is no total dominating set in $G$ then $G$ is not a maximal divisor graph.

• (4) If $\beta (G)<\lceil\frac{{n}}{2}\rceil$ then $G$ is a non-divisor graph.

Theorem 2.3.

If $G$ is a maximal divisor graph then

• (i) ${\gamma }^{-1}\left(G\right)={{\gamma }_{i}}^{-1}\left(G\right)=\pi \left(n\right)$

• (ii) $G$ has no inverse total (connected/co-independence) dominating set.

Proof

• (i) Consider $\mathrm{ID}=\left\{{v}_{i}\in G;f\left({v}_{i}\right)=p, p\le n ,where p is a prime number\right\}.$ $\mathrm{ID}$ is a dominating set in ${G}$ and $\mathrm{ID}\subseteq {v}-{D}$ where ${D}$ is a minimum dominating set ( $D=\left\{{v}_{1}\right\}$) in ${G}.$ Therefore $\left|\mathrm{ID}\right|\ge {\gamma }^{-1}\left({G}\right)$ ( see Fig. 1). If we assume that there is a set ${A}$ such that $\left|{A}\right|<\left|\mathrm{ID}\right|$ then there is at least a vertex of prime label which is not belonging to the set ${A}$, so it cannot dominate this vertex. Therefore, $, \left|\mathrm{ID}\right|={\gamma }^{-1}\left({G}\right).$ Since $\mathrm{ID}$ is an independence set then ${\gamma }^{-1}\left(G\right)={{\gamma }_{i}}^{-1}\left(G\right)=\pi \left(n\right)$.

• (ii) $G$ has no inverse total (connected) dominating set, since there is an isolated vertex in $G[V-D]$ where D is a total (connected) dominating set, and there is no inverse co-independence set in G since all co-independence sets in G contain adjacent vertices.

0-mod-difference graph

Theorem 3.1.

Maximal 0-mod-difference graph is partitioned into $m$ complete induced subgraphs.

Proof

Let ${{S}}_{{i}}=\left\{{{v}}_{{j}}\in {v};{ j}\equiv {i}\left(\mathrm{mod}\;m\right),{ i}=\mathrm{0,1},\dots ,{m}-1\right\}$. It is clear that ${{G}[{S}}_{{i}}],{ i}=\mathrm{0,1},\dots ,{m}-1$ are disjoint graphs and ${\cup }_{{i}=0}^{{m}-1}{{S}}_{{i}}={v}({G})$. For each ${{v}}_{{i}1}\;\mathrm{and}\;{{v}}_{{i}2}\in {{S}}_{{i}},{ i}=\mathrm{0,1},\dots ,{m}-1$ there is an edge ${{v}}_{{i}1}{{v}}_{{i}2}\in {E}({G})$ since $\left|f\left({v}_{{{i}}_{1}}\right)-f\left({v}_{{{i}}_{2}}\right)\right|\equiv 0(\mathrm{mod}\;m)$ so $G{[S}_{i}]$ is a complete induced subgraph $\forall i$.

Example 3.2.

Figure 2; $n=9 ;m=3$ and Fig. 3; $n=10 ,m=3$ are illustrate the previous theorem.

Fig. 2

Case $n=9;\,m=3$

Fig. 3

Case $n=10;\,m=3$

Theorem 3.3.

If G(n, q) is a 0-mod-difference graph and $n\equiv r ( mod m)$ then

$$q\le \frac{1}{2}\lfloor\frac{n}{m}\rfloor\left(m\lfloor\frac{n}{m}\rfloor-m+2r\right)$$

Proof

Since the maximal 0-mod-difference is partitioned into ${m}$ complete induced subgraphs (Theorem 3.1),so If ${n}\equiv {r }(\mathrm{mod}\;m)$, then there is ${r}$ complete induced subgraphs of order $\lfloor\frac{{n}}{{m}}\rfloor+1$ and the others of order $\lfloor\frac{{n}}{{m}}\rfloor$, since if ${n}\equiv 0 (\mathrm{mod}\;m)$, then $\mathrm{m divides n}$ without residue (see Fig. 2), so all complete subgraphs have the same order equal to ${n}/{m}$.

If ${n}\equiv 1 (\mathrm{mod}\;m)$, then $\mathrm{m divides n}$ with residue one which is the vertex ${{v}}_{{n}}$, and it is clear that ${{{v}}_{{n}}\in {S}}_{1}$, since ${n}\equiv 1 (\mathrm{mod}\;m)$, therefore ${{S}}_{1}$ is of order $\lfloor\frac{{n}}{{m}}\rfloor+1$ and the others are of order $\lfloor\frac{{n}}{{m}}\rfloor$ (se Fig. 3). If ${n}\equiv 2 (\mathrm{mod}\;m)$, then $\mathrm{m divides n}$ with residue two which are vertices ${{v}}_{{n}}$ and ${{v}}_{{n}-1}$, ${{{v}}_{{n}-1}\in {S}}_{1}$, since ${n}-1\equiv 1 (\mathrm{mod}\;m)$ and ${{{v}}_{{n}}\in {S}}_{2}$, since ${n}\equiv 2 (\mathrm{mod}\;m)$, therefore ${{S}}_{1}$ and ${{S}}_{2}$ are of order $\lfloor\frac{{n}}{{m}}\rfloor+1$ and the others are of order $\lfloor\frac{{n}}{{m}}\rfloor$(see Fig. 4) and so on. The number of edges of any complete graph ${\mathrm{K}}_{{t}}$ is $\frac{{t}({t}-1)}{2}$, then the maximal number of edges in G is ${r}\frac{\left(\lfloor\frac{{n}}{{m}}\rfloor+1\right)\left(\lfloor\frac{{n}}{{m}}\rfloor\right)}{2}+ \left({m}-{r}\right)\frac{\lfloor\frac{{n}}{{m}}\rfloor\left(\lfloor\frac{{n}}{{m}}\rfloor-1\right)}{2}=\frac{1}{2}\lfloor\frac{{n}}{{m}}\rfloor\left({m}\lfloor\frac{{n}}{{m}}\rfloor-{m}+2{r}\right).$ Thus $\mathrm{q}\le \frac{1}{2}\lfloor\frac{{n}}{{m}}\rfloor\left({m}\lfloor\frac{{n}}{{m}}\rfloor-{m}+2{r}\right).$

Fig. 4

Case $n=11;\,m=4$

Theorem 3.4.

If ${G}$ is a maximal 0-m-mod-difference graph, then

• (i) $\beta \left({G}\right)={m}$

• (ii) $\gamma \left({G}\right)={\gamma }_{{i}}\left({G}\right)={m}$

• (iii) ${\gamma }_{\mathrm{coi}}\left({G}\right)={n}-{m}$

• (iv) ${\gamma }_{{t}}\left({G}\right)=2{m}$

• (v) G has no connected dominating set.

Proof

• (i) Let ${{D}}_{1}=\left\{{{v}}_{{i}0}; {{v}}_{{i}0}\mathrm{ only one vertex belong to }{{S}}_{{i}},{ i}=\mathrm{0,1},\dots ,{m}-1\right\}$. By Theorem 3.1${{S}}_{{i}},{ i}=\mathrm{0,1},\dots ,{m}-1$ are the vertices of complete subgraphs. It is clear that $\beta \left({G}[{{S}}_{{i}}]\right)=1$ therefore $\left|{{D}}_{1}\right|\le\beta \left({G}\right)$. If we assume that there is a set A such that $\left|{A}\right|>\left|{{D}}_{1}\right|$ then A contains at least two vertices belonging to the same set from ${{S}}_{{i}}$, since the graph of this set is a complete induced subgraph then these vertices are not independent (see Figs. 2, 3, 4). Thus, $\beta \left({G}\right)=\left|{{D}}_{1}\right|={m}.$

• (ii) By the same manner in (i) $\left\{{{v}}_{{i}0}\right\}$ is a dominating set in $[{{S}}_{{i}}]$ $\forall {i}$, so $\gamma \left({G}[{{S}}_{{i}}]\right)=1$. Therefore, $\left|{{D}}_{1}\right|\ge\gamma \left({G}\right)$. If we assume that there is a set B such that $\left|\mathrm{B}\right|<\left|{{D}}_{1}\right|$ then $\mathrm{B}$ does not contain at least one vertex belong to a set from ${ S}_{i}, i=\mathrm{0,1},\dots ,m-1$, so B cannot dominate the vertices of this set, since every ${G}\left[{{S}}_{{i}}\right]$ is a complete induced subgraph. Thus, $\gamma \left({G}\right)=\left|{{D}}_{1}\right|={m}$. Since ${{D}}_{1}$ is an independent set then $\gamma \left(G\right)={\gamma }_{i}\left(G\right)=m$.

• (iii) Let ${{D}}_{2}=\left\{\forall {{v}}_{{j}}\in {{S}}_{{i}}\mathrm{ except one vertex},{ i}=\mathrm{0,1},\dots ,{m}-1\right\}$. So ${{D}}_{2}$ is a dominating set and ${v}-{{D}}_{2}$ is an independent set therefore $\left|{{D}}_{2}\right|\ge {\gamma }_{\mathrm{coi}}\left({G}\right)$. If we assume that there is a set ${{c}}_{2}$ such that $\left|{{c}}_{2}\right|<\left|{{D}}_{2}\right|$ then ${v}\left({G}\right)-{{c}}_{2}$ cannot be an independence set since it contains at least two vertices in the same set from ${{S}}_{{i}},{ i}=\mathrm{0,1},\dots ,{m}-1$ then these vertices are adjacent. Thus, ${\gamma }_{\mathrm{coi}}\left({G}\right)=\left|{{D}}_{2}\right|={n}-{m}$.

• (iv) Let ${{D}}_{3}=\left\{\mathrm{only two vertices from }{{S}}_{{i}} ,{ i}=\mathrm{0,1},\dots ,{m}-1\right\}$. ${{D}}_{3}$ is a total dominating set since it is a dominating set and has no isolated vertex so $\left|{{D}}_{3}\right|\ge {\gamma }_{{t}}\left({G}\right)$. If we assume that there is a set ${{c}}_{3}$ such that $\left|{{c}}_{3}\right|<\left|{{D}}_{3}\right|$.Then ${{c}}_{3}$ contains at least one isolated vertex from one set from ${{S}}_{{i}} ,{ i}=\mathrm{0,1},\dots ,{m}-1$ or it has no any vertex from at least one set from ${\mathrm{ S}}_{{i}} ,{ i}=\mathrm{0,1},\dots ,{m}-1$. So ${{c}}_{3}$ is not a total dominating set in $G$. Thus, ${\gamma }_{{t}}\left({G}\right)=\left|{{D}}_{2}\right|=2{m}$.

• (v) $G$ has no connected dominating set since G is a disconnected graph by Theorem 3.1.

Theorem 3.5.

If ${G}$ is a maximal 0-mod-difference graph, then

• (i) ${\gamma }^{-1}\left(G\right)={{\gamma }_{i}}^{-1}\left(G\right)=m$

• (ii) ${{\gamma }_{coi}}^{-1}\left(G\right)={m}$ if and only if ${n}=2{m}$.

• (iii) ${{\gamma }_{t}}^{-1}\left(G\right)=2m$ if and only if $\lfloor\frac{n}{m} \rfloor\ge 4$.

Proof

• (i) Let ${\mathrm{ID}}_{1}=\left\{{{v}}_{{j}}\in {{S}}_{{i}}; {{v}}_{{j}}\in {v}-{{D}}_{1} ,{ i}=\mathrm{0,1},\dots ,{m}-1\right\}$ where ${{D}}_{1}$ is a dominating set in $G$ (Theorem 3.4). Similar to manner in Theorem 3.4 (i) ${\mathrm{ID}}_{1}$ is a minimum dominating set in $G$, so ${\gamma }^{-1}\left(G\right)=\left|{ID}_{1}\right|=m$. And since ${\mathrm{ID}}_{1}$ is an independent set in G, then

  $${\gamma }^{-1}\left(G\right)={{\gamma }_{i}}^{-1}\left(G\right)=m$$

• (ii) If ${{\gamma }_{\mathrm{coi}}}^{-1}\left({G}\right)={m}$ then there is a minimum co-independent inverse set in G ( ${\mathrm{ID}}_{2}$) such that ${\mathrm{ID}}_{2}\subseteq {v}- {{D}}_{2}$ where ${{D}}_{2}$ is a minimum co-independent dominating set in $G$ (Theorem 3.4) then ${ID}_{2}\cap {S}_{i}=1 \forall i$. Now if ${{D}}_{2}\cap {{S}}_{{i}}>1$ for some i then $V-{ID}_{2 }$ contains at least two vertices belonging to ${{S}}_{{i}}$ and these sets are complete subgraphs therefore $V-{ID}_{2}$ is not an independent set. Thus, ${{D}}_{2}\cap {{S}}_{{i}}=1$ implies that ${{S}}_{{i}}$ contain only two vertices $\forall {i}$ then ${n}=2{m}$ (see Fig. 5).

  Fig. 5

  

  $n=2m;\,m=5$

  Conversely If ${n}=2{m}$ then $\left|{{S}}_{{i}}\right|=2 \forall {i}$ (see Fig. 5), then ${{\gamma }_{\mathrm{coi}}}^{-1}\left({G}\right)={m}$.

• (iii) If $\lfloor\frac{{n}}{{m}}\rfloor\ge 4$ that means $\left|{{S}}_{{i}}\right|\ge 4$. Let ${\mathrm{H}}_{{i}}=\left\{{{v}}_{{i}1}, {{v}}_{{i}2}\right\}$ where ${{v}}_{{i}1}\;\mathrm{and}\;{{v}}_{{i}2}$ are any two vertices belong to ${{S}}_{{i}}$ and ${\mathrm{H}}_{{i}}\subseteq {v}-{{D}}_{3}$ where ${{D}}_{3}{ i}$ s the minimum total dominating set in $G$ (Theorem 3.4) (see Fig. 6). Consider ${\mathrm{ID}}_{3}=\cup \left\{{\mathrm{H}}_{{i}} ,{ i}=\mathrm{0,1},\dots ,{m}-1 \right\}$ as same manner in Theorem 3.4 (iv) ${\mathrm{ID}}_{3}$ is the minimum total dominating set in $G$ so ${{\gamma }_{{t}}}^{-1}\left({G}\right)=2{m}$.

  Fig. 6

  

  $m=3$

Conversely If ${{\gamma }_{{t}}}^{-1}\left({G}\right)=2{m}$ then there is a minimum dominating set in $G$ which contain at least two vertices in ${{S}}_{{i}}$ and belonging to ${v}-{{D}}_{3}$ where ${{D}}_{3}$ is a total dominating set in G (see Fig. 6; $m=3$), so $\left|{{S}}_{{i}}\right|\ge 4 \forall {i}$ then $\lfloor\frac{{n}}{{m}}\rfloor\ge 4$.

1-mod-difference graph

Lemma 4.1.

If ${G}$ is a 1-mod-difference graph, then $\Delta \left(v\right)\le \lfloor\frac{n}{m}\rfloor+1$

Proof

Let ${v}_{j } \in G$ there are two cases as follows:

• (i) (i) If $f({v}_{j })\le m$ and $j\equiv i (mod m)$, then ${v}_{j}$ joins with all vertices of labels which are congruent to $i+1$ (mod m) and with the vertex ${v}_{j-1}$ congruent to $i-1 (mod m)$, except the vertex ${v}_{1}$, since ${v}_{0}$ does not exist and ${v}_{m}$ which are joined with the vertex ${v}_{m-1}$ and all vertices of labeled in class $\left[1\right] \, \mathrm{except}\left\{1\right\}$. So the maximum number of vertices can be joined with vertex ${v}_{j}$ in this case is less than or equal to $\lfloor\frac{n}{m}\rfloor+1$.

• (ii) If $f\left({v}_{j }\right)\ge m+1$ and $j\equiv i (mod m)$, then ${v}_{j}$ would join with

  • (1) All labeled vertices ${v}_{r}$ which are congruent to $i-1$ (mod m) and $f\left({v}_{j }\right)>f\left({v}_{r}\right)$, the maximum number of these vertices is less than or equal $\lceil\frac{j}{m}\rceil$.

  • (2) All labeled vertices $f\left({v}_{w}\right)$ congruent to $i+1$ (mod m) and $f\left({v}_{j }\right)<f\left({v}_{w}\right)$, the maximum number of these vertices is $\lceil\frac{{n}- j}{m}\rceil$.

By 1 and 2, it is clear that $\mathrm{deg}\left({v}_{j }\right)\le \lceil\frac{ j}{m}\rceil+\lceil\frac{{n}- j}{m}\rceil\le \lfloor\frac{n}{m}\rfloor+1$

Theorem 4.2.

If ${G}$ is a maximal 1-mod-difference graph, then $\gamma \left({G}\right)={m}.$

Proof

Let $D=\left\{{v}_{i }, i=1, 2 ,\dots m\right\}$, $\forall {v}_{i }\in D$, ${v}_{i}$ is adjacent to all vertices of labeling that belong to class [i + 1]= $\left\{{v}_{j }:f\left({v}_{j }\right)\equiv i+1(mod m)\right\}$ (see Figs. 7, 8, 9). So $D$ is a dominating set since $V={\cup }_{i=0}^{m-1}\left[i\right]$. Now if there is a set of cardinal equal to $m-1$ then the number of vertices can be dominated by $m-1$ vertices is $\left(m-1\right)\left(\lfloor\frac{n}{m}\rfloor+1\right)-(m-2)<n$, by Lemma 4.1, since $m-2$ is the minimum number of common edges when we have $m-1$ successive vertices. Thus, $D$ is the minimum dominating set in $G$.

Fig. 7

$|D|=m=11$

Fig. 8

$|D|=m=15$

Fig. 9

$|D|=m=16$

Theorem 4.3.

If ${G}$ is a maximal 1-mod-difference graph, then ${\gamma }^{-1}\left(G\right)=m.$

Proof

Consider $ID=\left\{{v}_{i }, i=m+1,m+2,\dots ,2m\right\}$, any m successive vertices constitute minimum dominating set, since these vertices are adjacent to all classes in $G$, and it is clear that $ID\subseteq V-D$, where D is the minimum dominating set in $G$ (Theorem 4.2). Thus, ${\gamma }^{-1}\left(G\right)=m.$

Corollary 4.4.

If ${G}$ is a maximal 1-mod-difference graph, then

$${\gamma }_{{c}}\left({G}\right)={\gamma }_{{t}}\left({G}\right)={\gamma }_{{c}}^{-1}\left({G}\right)={\gamma }_{{t}}^{-1}\left({G}\right)={m}$$

Proof

It is clear, since the set $D$ in Theorem 4.2 and $ID$ in Theorem 4.3 are connected set.

Theorem 4.5.

If ${G}$ is a maximal 1-mod-difference graph where $m=2$, then ${\gamma }_{{i}}\left({G}\right)=\lfloor\frac{{n}}{2}\rfloor$.

Proof

To get an independent set ${S}$, we cannot take any set that contains vertices of odd and even labels together. Since if ${{v}}_{\mathrm{i }, }{{v}}_{\mathrm{j }}\in {{D}}_{1}$ such that ${{v}}_{{i}}$ is odd labels and ${{v}}_{{j}}$ is even labels, then $\left|{f}\left({{v}}_{\mathrm{j }}\right)-{f}\left({{v}}_{\mathrm{i }}\right)\right|\equiv 1 (\mathrm{ mod }2)$, these vertices are adjacent. Thus, ${S}$ is not an independent set. Then, S contains either vertices of odd labels or even labels. The cardinal of all vertices of even labels is less than or equal to the cardinal of odd labels, then let ${{D}}_{1}$ be the set of all vertices of even labels, ${{D}}_{1}$ is a dominating set, since if we take any vertex of ${{D}}_{1}$, this vertex is adjacent to all vertices of odd labels and it is an independent, since $\forall {{v}}_{\mathrm{j }},{{v}}_{\mathrm{i }}\in {{D}}_{1 ,}\left|{f}\left({{v}}_{\mathrm{j }}\right)-{f}\left({{v}}_{\mathrm{i }}\right)\right|\equiv 0 (\mathrm{ mod }2)$, thus ${\gamma }_{{i}}\left({G}\right)\le \left|{{D}}_{1}\right|=\lfloor\frac{{n}}{2}\rfloor$ (see Figs. 10, 11). Now if there is a set ${{ A}={D}}_{1}-\left\{{{v}}_{{r}}\right\}$, where $\left\{{{v}}_{{r }}\right\}\in {{D}}_{1 ,}$ then A cannot dominate the vertex ${{v}}_{{r}}$, therefore A cannot be a dominating set in G. Thus, ${\gamma }_{{i}}\left({G}\right)=\lfloor\frac{{n}}{2}\rfloor.$

Fig. 10

$|D_1|=n=11,m=2$

Fig. 11

$|D_1|=n=10,m=2$

Theorem 4.6.

If ${G}$ is a maximal 1-mod-difference graph where ${m}=2$, then ${\gamma }_{{i}}^{-1}\left({G}\right)=\lceil\frac{{n}}{2}\rceil$.

Proof

Consider ${ID}_{1}=\left\{\forall {v}_{j }; {v}_{j } is an odd vertex\right\}$, as the same manner in the previous theorem ${ID}_{1}$ is the minimum independent dominating set in $V-{D}_{1}$, where ${D}_{1}$ is an independent dominating set in $G$ (Theorem 4.5),(see Figs. 10, 11). Thus, ${\gamma }_{{i}}^{-1}\left({G}\right)=\lceil\frac{{n}}{2}\rceil$.

Corollary 4.7.

If ${G}$ is a maximal 1-mod-difference graph where ${m}=2$, then ${\gamma }_{\mathrm{coi}}\left({G}\right)=\lfloor\frac{{n}}{2}\rfloor$ and ${\gamma }_{\mathrm{coi}}^{-1}\left({G}\right)=\lceil\frac{{n}}{2}\rceil$.

Proof

We showed that the set ${D}_{1}$ in Theorem 4.5 is the dominating set in $G$ and $V-{D}_{1}=I{D}_{1}$ is an independent by Theorem 4.6, if we assume there is a set $A\subseteq V$ with cardinal less than ${D}_{1}$, so $A$ may be still dominating set but $V-A$ is not an independent set since it contains vertices of odd and even labels) (see Figs. 8, 9). Thus, ${\gamma }_{\mathrm{coi}}\left({G}\right)=\lfloor\frac{{n}}{2}\rfloor$. As the same manner with alternate two sets ${D}_{1} and I{D}_{1}$, we get ${\gamma }_{\mathrm{coi}}^{-1}\left({G}\right)=\lceil\frac{{n}}{2}\rceil$.

Theorem 4.8.

If ${G}$ is a maximal 1-mod-difference graph where ${m}=3$, then.

$${\gamma }_{i}\left(G\right)=\left\{\begin{array}{c}\lfloor\frac{n}{3}\rfloor , if n\equiv 0 \left(mod 3\right)\\ \lfloor\frac{n}{3}\rfloor+1 , if n\equiv 1, 2 \left(mod 3\right)\end{array}\right\}$$

Proof

Consider $S=\left\{{v}_{i };f\left({v}_{i }\right)\in \left[1\right]-\left\{1\right\}\right\}$ and $D=\left\{{v}_{2 }\right\}\cup S$. The vertex ${v}_{2}$ is adjacent to vertex ${v}_{1}$ and all vertices which there labels belong to class 0 ([0]) and the vertex ${v}_{4 }\in S$ is adjacent to all vertices which there labels belong to class 2 ([2]) except $\left\{2\right\}$, and S covers to all vertices of labeled in $\left[1\right]-\left\{1\right\}$. Thus, $D$ is the dominating set in $G$ and it is an independent, since $\forall {v}_{i }, {v}_{j }\in S$, $\left|f\left({v}_{i }\right)-f\left({v}_{2 }\right)\right|\equiv 2 (mod 3)$ and $\left|f\left({v}_{i }\right)-f\left({v}_{j }\right)\right|\equiv 0 (mod 3)$ (see Figs. 7, 8). Thus, ${\gamma }_{i}\left(G\right)\le \left|{D}\right|=\left\{\begin{array}{c}\lfloor\frac{n}{3}\rfloor , if n\equiv 0 \left(mod 3\right)\\ \lfloor\frac{n}{3}\rfloor+1 , if n\equiv 1, 2 \left(mod 3\right)\end{array}\right\}$. If there is an independent set $A\subseteq V$ with $\left|A\right|$< $\left|D\right|$, then $A$ is not a dominating set. Thus, we get the result.

Theorem 4.9.

If ${G}$ is a maximal 1-mod-difference graph where ${m}=3$, then

$${\gamma }_{{i}}^{-1}\left({G}\right)=\lfloor\frac{{n}}{3}\rfloor+1$$

Proof

Consider $S=\left\{{v}_{i };f\left({v}_{i }\right)\in \left[3\right]\right\}$ and $ID=\left\{{v}_{1 }\right\}\cup S$, it is obvious that $ID\subseteq V-D$, where D is the minimum independent dominating set in G (Corollary 4.7). The vertex ${v}_{1}$ is adjacent to all vertices which their labels belong to class 2 ([2]); the vertex ${v}_{3 }\in S$ is adjacent to all vertices which their labels belong to class 1 ([1]) except $\left\{1\right\}$. S covers all vertices which their labels belong to class 3 $\left[0\right]$. Thus, $ID$ is the dominating set in $G$ and it is an independent, since $\forall {v}_{i }\in S$, $\left|f\left({v}_{i }\right)-{f\left({v}_{1 }\right)}\right|\equiv 2 (mod 3)$ and $\left|f\left({v}_{i }\right)-f\left({v}_{j }\right)\right|\equiv 0 (mod 3)$. Thus, ${\gamma }_{{i}}^{-1}\left({G}\right)\le \left|\mathrm{ID}\right|=\lfloor\frac{{n}}{3}\rfloor+1$ (see Figs. 7, 8). If there is an independent set $A\subseteq V-D$ with $\left|A\right|$< $\left|D\right|$, then $A$ is not a dominating set. Thus, ${\gamma }_{{i}}^{-1}\left({G}\right)=\lfloor\frac{{n}}{3}\rfloor+1.$

Corollary 4.10.

If ${G}$ is a maximal 1-mod-difference graph where ${m}=3$, then

$${\gamma }_{\mathrm{coi}}\left({G}\right)\le \left\{\begin{array}{c}n-\lfloor\frac{n}{3}\rfloor , if n\equiv 0 \left(mod 3\right)\\ n-\lfloor\frac{n}{3}\rfloor-1 , if n\equiv 1, 2 \left(mod 3\right)\end{array}\right\}$$

Proof

Consider $M=V-D$, where $D$ is the set is in Theorem 4.8, it is clear that $M$ is the dominating set and D is an independent set. Thus,

$${\gamma }_{\mathrm{coi}}\left({G}\right)\le \left|\mathrm{M}\right|= \left\{\begin{array}{c}n-\lfloor\frac{n}{3}\rfloor , if n\equiv 0 \left(mod 3\right)\\ n-\lfloor\frac{n}{3}\rfloor-1 , if n\equiv 1, 2 \left(mod 3\right)\end{array}\right\}$$

Theorem 4.11.

If ${G}$ is a maximal 1-mod-difference graph, then

$$\beta \left( G \right) = \frac{n}{2},\;{\text{if}}\;m\;is\;{\text{even}}.$$

Proof

Consider $I=\left\{{v}_{i }\in G; {v}_{i } is an odd labeled vertex\right\}$ $\forall {v}_{j },{v}_{k }\in I, \left|f\left({v}_{j }\right)-f\left({v}_{k }\right)\right|\equiv w (mod m)$ where w is 0 or even number less than m, then I is an independent set. Now if we add any vertex ${v}_{h }\in V-I$ to the set I, then ${v}_{h}$ is an even labeled vertex, then ${v}_{h-1}$ is an odd labeled vertex, so ${v}_{h-1 }\in I$. Therefore, $\left|f\left({v}_{h }\right)-f\left({v}_{h-1 }\right)\right|\equiv 1 (mod\left(m\right))$ that means $I\cup \left\{{v}_{h}\right\}$ is not an independent set in $G.$ Thus $\beta \left(G\right)=\lceil\frac{n}{2}\rceil$.

Example 4.12.

The maximal 1-mod-difference graphs of order 11 and 15 where $m=3$, as shown in Figs. 7, 8, ${D}_{1}=\left\{{v}_{2 },{v}_{4 },{v}_{7 },{v}_{10}\right\}$, is the minimum independent dominating set in ${G}_{1}$, and $I{D}_{1}=\left\{{v}_{1 },{v}_{3 },{v}_{6 },{v}_{9}\right\}$, is the minimum independent sets in $V\left({G}_{1}\right)-D$, so ${\gamma }_{i}\left({G}_{1}\right)={\gamma }_{{i}}^{-1}\left({G}_{1}\right)=4$,, ${D}_{2}=\left\{{v}_{2 },{v}_{4 },{v}_{7 },{v}_{10},{v}_{13}\right\}$, is the minimum independent dominating set in ${G}_{2}$, and $I{D}_{1}=\left\{{v}_{1 },{v}_{3 },{v}_{6 },{v}_{9 },{v}_{12 },{v}_{15}\right\}$, is the minimum independent sets in $V\left({G}_{2}\right)-D$, so ${\gamma }_{i}\left({G}_{2}\right)=5 and {\gamma }_{{i}}^{-1}\left({G}_{2}\right)=6$.

Conclusion and discussion

In this work, we obtain the necessary condition(s) for a graph to be a maximal divisor graph and for a graph to be 0-mod-difference graph, also for a graph to be maximal o-mod-difference graph and finally for a graph to be maximal 1-mod-differnce graph.

These results will lead us to discuss in the future work to the independence and domination in multi-rooted graph.