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\begin{document}

\title{On the outer-independent double Italian domination number}
\author{{\normalsize Noor A'lawiah Abd Aziz, Hailiza Kamarulhaili}\\
{\normalsize School of Mathematical Sciences, Universiti Sains Malaysia}\\
{\normalsize 11800 USM Penang, Malaysia}\\
{\small E-mail: nooralawiah@gmail.com, hailiza@usm.my}\\
\\
{\normalsize Farzaneh Azvin and Nader Jafari Rad}\\
{\normalsize Department of Mathematics, Shahed University}\\
{\normalsize Tehran, Iran}\\
{\normalsize E-mails: azvin.f4@gmail.com, n.jafarirad@gmail.com}}
\date{}
\maketitle

\begin{abstract}
An outer-independent Italian dominating function (OIIDF) on a graph $G$ is a function $f:V(G)\longrightarrow \{0,1,2\}$ such that every vertex $v\in V(G)$ with $f(v)=0$ has at least two neighbors assigned $1$ under $f$ or one neighbor $w$ with $f(w)=2$, and the set $ \{u\in V(G)|f(u)=0\}$ is independent.
An outer-independent double Italian dominating function (OIDIDF)
on a graph $G$ is a function
$f:V(G)\longrightarrow \{0,1,2,3\}$ such that if $f(v)\in\{
0,1\}$ for a vertex $v\in V(G)$, then $\sum_{u\in N[v]}f(u)\geq3$
and the set $ \{u\in V(G)|f(u)=0\}$ is independent. The weight of
an OIIDF (respectively, OIDIDF) $f$ is the value $w(f)=\sum_{v\in V(G)}f(v)$. The
minimum weight of an OIIDF (respectively, OIDIDF) on a graph $G$ is called the
outer-independent Italian (respectively, outer-independent double Italian) domination number
of $G$. We characterize all trees $T$ with outer-independent double Italian domination number twice the outer-independent Italian domination number. We also present lower bounds on the outer-independent double Italian domination number of a connected graph $G$ in
terms of the order, minimum and maximum degrees.
\end{abstract}

\textbf{Keywords:} (Italian, double Italian, outer-independent Italian) domination, outer-independent double Italian domination.\\
\textbf{MSC: 05C69}


\section{Introduction}
We consider simple connected graphs $G$ with vertex
set $V=V(G)$ and edge set $E=E(G)$. The \textit{order} of $G$ is
$n(G)=|V|$. The \textit{open neighborhood} of a vertex $v$ is the
set $ N(v)=\{u\in V(G)\mid uv\in E\}$ and its \textit{closed
neighborhood} is the set $ N[v]=N(v)\cup \{v\}$. The
\textit{degree} of vertex $ v\in V $ is $\deg(v)=d(v)=|N(v)| $.
The \textit{maximum degree} and \textit{minimum degree} among the vertices of $G$ are
denoted by $ \Delta = \Delta (G) $ and $\delta =\delta (G)$,
respectively. A \textit{leaf} is a vertex of degree one, and its
neighbor is called a \textit{support vertex}. A \textit{strong
support vertex} is a support vertex adjacent to more than one
leaf. For a subset $D$ of vertices of $G$, we denote by $G[D]$ the subgraph of $G$ induced by $D$. A subset $D$ of
$V(G)$ is a \textit{dominating set} in $G$ if $\bigcup_{v\in D
}N[v]=V(G)$. A set $I$ of
vertices of $G$ is called \textit{independent} if no pair of vertices of $I$ are
adjacent. If $H$ is a subgraph of a graph $G$ and $f$ is  a function defined on $V(G)$, then we denote by $f\mid_H $ the restriction of $f$ on $V(H)$. For other notations and terminology not given here we refer to \cite{hhs}.

Cockayne et al. \cite{cdhh} introduced the concept of \textit{Roman domination} in graphs, and since then many generalizations and related variations have been considered by researchers, see \cite{cjsv,cjsv1,cjsv2,cjsv3,cjsv4}. A variation of Roman domination, namely, double Roman domination is introduced by Beeler et al. \cite{bhh}. A generalization of double Roman domination, namely Italian
domination, is introduced by Chellali et al. in \cite{chhm} and further studied in
\cite{hk} and \cite{km}. An \textit{Italian dominating function} (IDF) on a graph $G$ is a function $f:V(G)\longrightarrow \{0,1,2\}$ such that every vertex $v\in V(G)$ with $f(v)=0$ has at least two neighbors assigned $1$ under $f$ or one neighbor $w$ with $f(w)=2$. The \textit{weight} of an IDF is the value $w(f)=\sum_{u\in V(G)}f(u)$. The minimum weight of an IDF on a graph $G$ is called the \textit{Italian domination number} and denoted by $\gamma_I(G)$. For an IDF $f$ on $G$, let $V_i^f=\{v\in V(G): f(v)=i\}$, for $i=0,1,2$. We can equivalently write $f=(V_0^f,V_1^f,V_2^f)$ (or simply $ f=(V_0,V_1,V_2))$.

Fan et al. \cite{fmsss} initiated the study of outer independent Italian domination in graphs. An \textit{outer independent Italian dominating function} (OIIDF) on a graph $G$ is an IDF on $G$ that $V_0^f$ is an independent set. Mojdeh and Volkmann
\cite{mv} considered an extension of Italian domination as
follows. For a graph $G$, a \textit{double Italian dominating
function} (DIDF) is a function $f:V\longrightarrow \{0,1,2,3\}$
having the property for which every vertex $u\in V$, if $f(u)\in
\{0,1\}$, then $f(N[u])\geq 3$. The weight of a DIDF $f$ is the
sum $w(f)= f(V ) = \sum_{v\in V}f(v)$, and the minimum weight of
a DIDF in a graph $G$ is the \textit{double Italian domination
number}, denoted by $\gamma_{\{R3\}}(G)$, or $\gamma_{dI}(G)$ for convenience. Abdollahzadeh Ahangar et al. \cite{acsh} studied outer independent double Roman domination. An \textit{outer independent double Roman dominating function} (OIDRDF) on $G$ is an DRDF of $G$ for which $V_0^f$ is an independent set.

Azvin et al. \cite{ajv} (see also Volkmann \cite{v}) considered those double Italian dominating functions $f$ such that $ \{v\in V(G)\mid f(v)=0\}$ is an
independent set. An \textit{outer independent double Italian dominating function} (OIDIDF) of graph $G$ is a DIDF $f:V(G)\longrightarrow \{0,1,2,3\}$ for which  $V_0^f$  is independent. The \textit{outer independent double Italian domination number} denoted by $\gamma_{oidI}(G)$ is the minimum weight of an OIDIDF on $G$. An OIDIDF on graph $G$ with weight of $\gamma_{oidI}(G)$ is denoted by $\gamma_{oidI}(G)$-function. Benatallah \cite{b} studied outer independent double Italian domination number and presented the following bounds on the outer independent double Italian domination number.

\begin{theorem}[\cite{b}]\label{pr1}
For every graph $G$, $ \gamma_{oiI}(G)\leq \gamma_{oidI}(G)\leq 2\gamma_{oiI}(G)$ and these bounds are sharp.
\end{theorem}

In this paper we continue the study of the outer independent double Italian domination number of a graph. We characterize all trees $T$ achieving equality equality in both bounds given in Theorem \ref{pr1}. We also prove two lower bounds on the outer independent double Italian domination number of a graph.

\section{Trees $T$ with $\gamma_{oidI}(T)=2\gamma_{oiI}(T)$ or  $\gamma_{oidI}(T)=\gamma_{oiI}(T)+1$}

We aim to characterize trees achieving the equality in each of the lower or upper bounds given in Theorem \ref{pr1}. We begin with the following result.

\begin{observation}
If $\gamma_{oiI}(G)= \gamma_{oidI}(G)$ for a graph $G$, then  $\delta(G)\geq 2$.
\end{observation}

\begin{proof}
Suppose that $\gamma_{oiI}(T)= \gamma_{oidI}(T).$ Let $f=(V_0,V_1,V_2,V_3)$ be a $\gamma_{oidI}(T)$-function. If $|V_3|\neq 0$ then re-assigning $2$ to any vertex of $V_3$ and $1$ to any vertex of $V_2$ produces an OIIDF of $G$ of weight less than $\gamma_{oiI}(T)$, a contradiction. Thus $|V_3|= 0$. Similarly, $|V_2|= 0$.
Thus each vertex of $V_0$ is adjacent to at least three vertices of $V_1$ and each vertex of $V_1$ is adjacent to at least two vertices of $V_1$. Consequently, $\delta(G)\geq 2$.
\end{proof}

As the above observation shows, no tree achieves equality for the lower bound of Theorem \ref{pr1}. We will improve the lower bound of Theorem \ref{pr1} for trees.

\begin{proposition}
For any tree $T$ of order $n\geq 2$, $\gamma_{oidI}(T)\geq \gamma_{oiI}(T)+1$, with equality holds if and only if $T$ is a star.
\end{proposition}

\begin{proof}
Let $f=(V_0,V_1,V_2,V_3)$ be a $\gamma_{oidI}(T)$-function. If $|V_3|\neq 0$ and $|V_2|\neq 0$ then re-assigning $2$ to any vertex of $V_3$ and $1$ to any vertex of $V_2$ produces an OIIDF of $T$ of weight $\gamma_{oidI}(T)-2$, implying that $\gamma_{oidI}(T)\geq \gamma_{oiI}(T)+2$. Similarly, $\gamma_{oidI}(T)\geq \gamma_{oiI}(T)+2$ if $|V_3|\geq 2$ or $|V_2|\geq 2$. Thus we may assume that  $(|V_3|, |V_2|)\in \{(0,1),(1,0)\}$. If $(|V_3|, |V_2|)=(0,1)$, then  re-assigning $1$ to the vertex of $V_2$ produces an OIIDF of $G$ of weight $\gamma_{oidI}(T)-1$, implying that $\gamma_{oidI}(T)\geq \gamma_{oiI}(T)+1$, and if $(|V_3|, |V_2|)=(1,0)$, then  re-assigning $2$ to the vertex of $V_3$ produces an OIIDF of $G$ of weight $\gamma_{oidI}(T)-1$, implying that $\gamma_{oidI}(T)\geq \gamma_{oiI}(T)+1$.

Now assume that $\gamma_{oidI}(T)= \gamma_{oiI}(T)+1$. Let $f=(V_0,V_1,V_2,V_3)$ be a $\gamma_{oidI}(T)$-function. Following the above argument, we obtain that $(|V_3|, |V_2|)\in \{(0,1),(1,0)\}$. Suppose that $(|V_3|, |V_2|)=(0,1)$. Let $V_2=\{y\}$. Since $f$ is a $\gamma_{oidI}(T)$-function and $T$ has at least two leaves, for any leaf $x\neq y$ we have $f(x)=1$ and $x\in N(y)$. Since any vertex $x\neq y$ lies on a path from $y$ to a leaf of $T$, we deduce that any vertex $x\neq y$ is a leaf. Consequently, $T$ is a star centered at $y$. If $\deg(y)\geq 2$, then re-assigning $3$ to $y$ and $0$ to any leaf produces an OIDIDF of $T$ of weight less than $\gamma_{oidI}(T)$, a contradiction. Thus $T$ is star of order $2$. Next assume that $|V_3|=1$ and $|V_2|=0$. Let $V_3=\{y\}$. Clearly, for any leaf $x$, $f(x)=0$, and $x$ is adjacent to $y$, since  $f$ is a $\gamma_{oidI}(T)$-function. Consequently, $T$ is a star centered at $y$. The converse is obvious.
\end{proof}

We next aim to characterize trees achieving the equality in the upper bound of Theorem \ref{pr1}.
The following lemma has an important role in the rest of this section.

\begin{lemma}\label{lemn}
If $T$ is a tree for which $ \gamma_{oidI}(T)=2\gamma_{oiI}(T)$ and $f=(V_0,V_1,V_2)$ is a $\gamma_{oiI}(T)$-function, then:\\
(1) $V_2=\emptyset $ and $V_1$ is an independent set.\\
(2) $T$ does not have any strong support vertex.\\
(3) All leaves of $T$ belong to the same partite set of $T$, namely $X_{L(T)}$.\\
(4) The distance between any pair of vertices in $X_{L(T)}$ is even.\\
(5) There exists a $\gamma_{oidI}(T)$-function $g$ for which $ V_1^g\cup V_3^g=\emptyset$.
\end{lemma}

\begin{proof}
Assume that $\gamma_{oidI}(T)=2\gamma_{oiI}(T)$. Let $f=(V_0,V_1,V_2)$ be a $\gamma_{oiI}(T)$-function.

(1) As it is shown in the proof of Theorem \ref{pr1}, re-assigning $2$ to every vertex of $V_1$ and $3$ to every vertex of $V_2$ produces an OIDIDF for $T$ implying that
 $\gamma_{oidI}(T)\leq 2|V_1|+3|V_2|\leq 2|V_1|+4|V_2|=2\gamma_{oiI}(T)=\gamma_{oidI}(T)$ which implies that $V_2=\emptyset$. If there are two adjacent vertices $u$ and $v$ in $V_1$, then the function $g$ defined by $g(u)=1$ and $g(x)=2f(x)$ for every $x\in V(T)-\{u\}$ is an OIDIDF, and so $\gamma_{oidI}(T)\leqslant w(g)<2w(f)=2\gamma_{oiI}(T)$, a contradiction. Thus $V_1$ is an independent set.

(2) By (1), $V_2=\emptyset $ and both $V_0$ and $V_1$ are independent sets. Assume that $T$ has a strong support vertex $u$. Then $f(u)=0$ and $f(x)=1$ for every leaf neighbor of $u$. Then the function $g$ defined by $g(u)=3$, $g(t)=0$ for every leaf-neighbor of $u$, and $g(x)=2f(x) $ for any other vertex is an OIDIDF. Let $A$ be the set of all leaf-neighbors of $u$. Then $|A|\geqslant 2$, since $u$ is a strong support vertex. Thus $$\gamma_{oidI}(T)\leqslant w(g)=3+2(|V_1-A|)\leqslant 3+2|V_1|-4\leqslant 2|V_1|-1< 2|V_1|=2\gamma_{oiI}(T),$$ a contradiction. Consequently, $T$ does not have any strong support vertex.

(3) Since $V_2=\emptyset $ and both $V_0$ and $V_1$ are independent sets, it follows that $V_0$ and $V_1$ are partite sets of $T$. Since $f(x)=1$ for every leaf $x$, we find that all leaves belong to $V_1$.

(4) Let $X_{L(T)}$ be the partite set of $T$ containing all leaves. Let $x,y$ be two vertices in $X_{L(T)}$, and let $P:x_0x_1....x_ky$ be the shortest path between $x$ and $y$, where $x_0=x$. Then clearly $f(x)=f(y)=f(x_{2i})=1$ for $i\leq \lfloor \frac{k}{2} \rfloor $, and $f(u)=0$ for any other vertex of $P$. Since $V_0$ is an independent set, we conclude that $k$ is odd. Consequently, $d(x,y)$ is even.\\
\quad(5) In view of the proof of (1), we have $\gamma_{oidI}(T)=2\gamma_{oiI}(T)=2|V_1|$. It is now easy to see that $g=(V_0^g,V_1^g,V_2^g,V_3^g)=(V_0,\emptyset , V_1,\emptyset)$ is a $\gamma_{oidI}(T)$-function for which $V_1^g\cup V_3^g=\emptyset$.
\end{proof}

According to Lemma \ref{lemn}, for every tree $T$ with $\gamma_{oidI}(T)=2\gamma_{oiI}(T)$, we denote by $X_{L(T)}$ the partite set containing all leaves of $T$. Now we define the family $\mathcal{T}$ of trees as follows. Let $\mathcal{T}$  be the family of trees $T$ that can be obtained from a sequence $T_1,\dots,T_i$ $(i\geq 1)$ of trees such that $T_1$ is a path $P_5$ and if $i\geq 2$, then $T_{i+1}$ can be obtained, recursively, from $T_i$ by the following operation.\\

\textbf{Operation $\mathcal{O}$:}
Let $u\in X_{L(T_i)}$. Then $T_{i+1}$ is obtained from $T_i$ by joining $u$ to a leaf of a path $P_2$.

\begin{lemma}\label{a}
If $\gamma_{oidI}(T_i)=2\gamma_{oiI}(T_i)$ and $T_{i+1}$ is obtained from $T_i$ by Operation $\mathcal{O}$, then $\gamma_{oidI}(T_{i+1})=2\gamma_{oiI}(T_{i+1})$.
\end{lemma}
\begin{proof}
Let $v\in X_{L(T_i)}$ and suppose $T_{i+1}$ is obtained from $T_i$ by joining $v$ to the leaf $y$ of a $P_2$-path $P_2:yz$ according to Operation $\mathcal{O}$. We first show that $\gamma_{oiI}(T_{i+1})=\gamma_{oiI}(T_i)+1$. Let $f=(V_0,V_1,V_2)$ be a $\gamma_{oiI}(T_i)$-function. According to Lemma \ref{lemn} (1), $V_2=\emptyset$ and $V_1$ is an independent set. If $u$ is a leaf in $X_{L(T_i)}$, than $f(u)=1$. Since $d(u,v)$ is even, we find that $f(v)=1$. Then assigning $0$ to $y$ and $1$ to $z$ produces an OIIDF for $T_{i+1}$ implying that $\gamma_{oiI}(T_{i+1})\leq w(f)+1=\gamma_{oiI}(T_i)+1$. On the other hand, let $g$ be a $ \gamma_{oiI}(T_{i+1})$-function. Clearly, $g(v)+ g(z)+g(y)\geqslant 2$. We may assume that $g(z)=1$, $g(y)=0$ and $g(v)\geqslant 1$.  Therefore $\gamma_{oiI}(T_i)\leq w(g\mid_{T_i})\leqslant w(g)-1=\gamma_{oiI}(T_{i+1})-1$. Consequently, $\gamma_{oiI}(T_{i+1})=\gamma_{oiI}(T_i)+1.$

We next prove that $\gamma_{oidI}(T_{i+1})=2\gamma_{oiI}(T_{i+1})$. First we note that according to Proposition \ref{pr1}, $\gamma_{oidI}(T_{i+1})\leqslant 2\gamma_{oiI}(T_{i+1})$. Conversely let $f^*$ be a $\gamma_{oidI}(T_{i+1})$-function. Clearly $f^*(y)+f^*(z)\in \{2,3\}$. If $f^*(y)+f^*(z)=2$, then $f^*(y)=0$ and $f^*(z)=2$, and so $f^*|_{T_i}$ is an OIDIDF on $T_i$ and so $\gamma_{oidI}(T_i)\leqslant w(f^*)-2 =\gamma_{oidI}(T_{i+1})-2$. This implies that $$\gamma_{oidI}(T_{i+1})\geqslant \gamma_{oidI}(T_i)+2=2\gamma_{oiI}(T_i)+2=2(\gamma_{oiI}(T_i)+1)=2\gamma_{oiI}(T_{i+1})$$ as desired. Thus we assume that $f^*(y)+f^*(z)=3$. Without loss of generality, we may assume that $f^*(y)=3$ and $f^*(z)=0$. Then $f^*(v)<2$, since otherwise we can replace $f^*(y)$ by $0$ and $f^*(z)$ by $2$ to obtain an OIDIDF on $T_{i+1}$ with weight less than $w(f^*)$ which is a contradiction. If $f^*(v)=1$, then we define a function $g^*$ by $g^*(v)=g^*(z)=2, g^*(y)=0$ and $g^*(t)=f^*(t)$ otherwise. Then $w(g^*)=w(f^*)$ and $g^*\mid_{T_i}$ is an OIDIDF on $T_i$. Then $\gamma_{oidI}(T_i)\leqslant w(g^*\mid_{T_i})=w(g^*)-2$ and so $\gamma_{oidI}(T_{i+1})\geqslant \gamma_{oidI}(T_i)+2=2\gamma_{oiI}(T_i)+2=2(\gamma_{oiI}(T_i)+1)=2\gamma_{oiI}(T_{i+1}).$
Thus assume that $f^*(v)=0$. Note that $f^*(t)\geqslant 1 $ for $t\in N(v)$. If $ |N(v)-\{y\}|\geqslant 3$, then $f^*\mid_{T_i}$ is an OIDIDF on $T_i$ and so  $\gamma_{oidI}(T_i)\leqslant w(f^*\mid_{T_i})=w(f^*)-3< w(f^*)-2=\gamma_{oidI}(T_{i+1})$ and we obtain the result as before. If $|N(v)-\{y\}|=2$, then we may assume that $f^*(y)=1$ and $f^*(z)=2$ and define the function $f'$ by $f'(v)=1, f'(z)=2 $ and $f'(y)=0$. Then $f'\mid_{T_i}$ is an OIDIDF on $T_i$ and, as before, we find that $\gamma_{oidI}(T_{i+1})\geqslant2\gamma_{oiI}(T_{i+1}).$
Therefore assume that $|N(v)-\{y\}|=1$. Let $N(v)-\{y\}=\{w\}$. If $f^*(w)\geqslant 2 $, then $f'\mid_{T_i}$, where $f'$ is described above, is an OIDIDF on $T_i$, and as before the result follows.
Thus assume that $f^*(w)=1$. Then $h\mid_{T_i}$, where $h$ is defined by $h(v)=2$ and $h(t)=f^*(t)$ otherwise, is an OIDIDF on $T_i$.
Let $N(w)=\{v,w_1,w_2\dots w_k\}$, where $k\geq 1$. Let $g_1$ be a $\gamma_{oidI}(T)$-function on $T_i$ satsfying Lemma \ref{lemn} (5). Then $w(g_1)= 2\gamma_{oiI}(T_i)$, $ g_1(w)=0$ and $g_1(v)=g_1(w_i)=2$ for $i=1,2\dots, k $.
Assume that there is an intejer $j\in \{1,2\dots,k\}$ such that $w(h\mid_{T_{w_j}})\geq w(g_1\mid_{T_{w_j}})$. Then we change $h(x)$ by $g_1(x)$ for each $x\in V(T_{w_j})$ and define the function $h^*$ by $ h^*(w)=2 , h^*(w_j)=  h^*(v)=1 , h^*(x)=g_1(x)$ for $x\in V(T_{w_j})-\{w_j\}$ and $h^*(t)=h(t)$ otherwise. Since $d(v,w_j)=2$ and $v\in X_{L(T_i)}$, $h^*$ is an OIDIDF on $T_i$ and so $\gamma_{oidI}(T_i)\leqslant w(h^*)\leqslant w(h)-1=w(f^*)-2$. Therefore $\gamma_{oidI}(T_{i+1})\geqslant \gamma_{oidI}(T_i)+2 $, as desired.
Then we assume that $w(h\mid_{T_{w_i}})\leqslant w(g_1\mid_{T_{w_i}})-1$, for every $i=1,...,k$. We now have the following. $$\gamma_{oidI}(T_i)\leqslant h(v)+h(w)+\sum_{i=1}^{k}h(T_{w_i})\leqslant 2+1+ \sum_{i=1}^{k}g_1(T_{w_i})-k=2\gamma_{oiI}(T_i)+1-k. $$ If $k\geqslant 2$, then $\gamma_{oidI}(T_i)< 2\gamma_{oiI}(T_i)$, a contradiction. Thus $k=1$. Then $\deg(w)=2$. Observe that $f^*(w_1)=2$, since $f^*$ is an OIDIDF on $T_{i+1}$. Let $f_1$ be a function defined by $f_1(w)=f_1(y)=0$, $f_1(v)=f_1(z)=2$ and $f_1(t)=f^*(t)$ otherwise. Then $w(f_1)=w(f^*)$ and $f_1\mid_{T_i}$ is an OIDIDF on $T_i$. Then $ \gamma_{oidI}(T_i)\leqslant w(f_1\mid_{T_i})=\gamma_{oidI}(T_{i+1})-2$ and so $$\gamma_{oidI}(T_{i+1})\geqslant \gamma_{oidI}(T_i)+2=2\gamma_{oiI}(T_i)+2=2(\gamma_{oiI}(T_i)+1)=2\gamma_{oiI}(T_{i+1})$$ as desired.
\end{proof}

We are now ready to characterize trees with outer independent double Italian domination number twice the outer independent Italian domination number.

\begin{theorem}
For a tree $T$ of order $n\geq 5$, $\gamma_{oidI}(T)=2\gamma_{oiI}(T)$ if and only if $T\in \mathcal{T}$.
\end{theorem}
\begin{proof}
First we show that for every tree $T$ of the family $\mathcal{T} $, $\gamma_{oidI}(T)=2\gamma_{oiI}(T)$. We proceed by induction on the order $n\geqslant5$ of a tree $T\in \mathcal{T}$. Clearly $ 6=\gamma_{oidI}(P_5)=2\gamma_{oiI}(P_5)=2(3)$. This establishes the base step. For the inductive hypothesis, assume that for every tree $T'$ of order $n'$, $(5\leqslant n'< n)$ that belongs to the family $\mathcal{T}$, we have $\gamma_{oidI}(T')=2\gamma_{oiI}(T')$. Let $T$ be a tree of order $n$ that belongs to the family $\mathcal{T}$. Then $T$ is obtained from a sequence $T_1,\dots,T_i$ $(i\geq 1)$ of trees such that $T_1$ is a path $P_5$ and if $i\geq 2$, then $T_{i+1}$ can be obtained, recursively, from $T_i$ by Operation $\mathcal{O}$. Then by the inductive hypothesis and Lemma \ref{a} we obtain that $\gamma_{oidI}(T)=2\gamma_{oiI}(T)$.

Conversely, assume that $T$ is a tree of order $n\geqslant5$ with $\gamma_{oidI}(T)=2\gamma_{oiI}(T)$. We show that $T\in\mathcal{T}$. We proceed by induction on the order $n\geqslant5$. We first note that if $diam(T)=2$, then $T$ is a star and clearly $\gamma_{oidI}(T)=3$, $\gamma_{oiI}(T)=2$ and $\gamma_{oidI}(T)\neq 2\gamma_{oiI}(T)$. If $ diam(T)=3$, then $T$ is a double star in which $ \gamma_{oidI}(T)\in \{5,6\}$, $\gamma_{oiI}(T)\in \{3,4\}$ and $\gamma_{oidI}(T)\neq 2\gamma_{oiI}(T)$. Therefore we assume that $diam(T)\geqslant 4$ and so $n\geqslant 5$. If $n=5$, then $T\cong P_5\in \mathcal{T}.$ This establishes the base step of the induction. For the induction hypothesis assume that every tree $T'$ of order $n'$ $(5\leqslant n'<n)$ with $\gamma_{oidI}(T')=2\gamma_{oiI}(T')$ belongs to the family $\mathcal{T}$. Let $T$ be a tree of order $n$ with $\gamma_{oidI}(T)=2\gamma_{oiI}(T)$. We show $T\in \mathcal{T}$. We consider a diametrical path $x_1x_2\dots x_dx_{d+1}$ in $T$ and root $T$ at $x_1$. Let $f=(V_0,V_1,V_2)$ be a $\gamma_{oiI}(T)$-function. By Lemma \ref{lemn} (parts (1) and (2)), $\deg(x_d)=2$, $V_2=\emptyset $, $V_1$ and $V_0 $ are independent sets. Clearly, $f(x_{d+1})=f(x_{d-1})=1$ and $f(x_d)=0$. Thus $X_{L(T)}=\{x:f(x)=1\}$. Note that $x_{d-1}$ does not have any leaf neighbor, since $V_1$ is an independent set.

Let $T'=T-\{x_d,x_{d+1}\}$. We first show that $\gamma_{oiI}( T)=\gamma_{oiI}(T')+1$. Observe that  $f\mid_{T'}$ is an OIIDF on $T'$. Thus $\gamma_{oiI}(T')\leqslant \gamma_{oiI}(T)-1$. Assume that $f'=(V_0',V_1',V_2')$ is a $\gamma_{oiI}(T')$-function. If $f'(x_{d-1})\geqslant 1 $, then the function $g$ defined by $g(x_d)=0, g(x_{d+1})=1$ and $g(t)=f'(t)$ otherwise, is an  OIIDF on $T$ and so $\gamma_{oiI}(T)\leqslant \gamma_{oiI}(T')+1$. Thus we assume that $f'(x_{d-1})=0$. We consider the following cases.

Case 1. $\deg(x_{d-1})\geqslant 3$. Since $x_{d-1}$ does not have any leaf neighbor, we may assume that $x_{d-1}$ has a child $u\neq x_d$ which is a support vertex. By  Lemma \ref{lemn} $\deg(u)=2$. Let $w$ be the child of $u$. Then $f'(u)+f'(w)=2$. Then we replace $f'(x_{d-1})$ and $f'(w)$ by $1$ and $f'(u)$ by $0$ to obtain a $\gamma_{oiI}(T')$-function $f''$ with $f''(x_{d-1})\geq 1$. Then, as before, the function $g$ defined by $g(x_d)=0, g(x_{d+1})=1$ and $g(t)=f''(t)$ otherwise, is an  OIIDF on $T$ and so $\gamma_{oiI}(T)\leqslant \gamma_{oiI}(T')+1$, as desired.

Case 2. $\deg(x_{d-1})=2$. Observe that $f'(x_{d-2})=2$. Suppose that $x_{d-2}$ has a leaf neighbor $v$. Then $f(x_{d-2})=0$ and $f(v)=1$. Also $x_{d-3}$ doesn't have a leaf neighbor, since $d(x_{d-3},x_{d+1})$ is even. Now we define the function $f_1$ by $f_1(x_{d-2})=3$, $f_1(v)=0$, $f_1(x_{d-1})=f_1(x_{d-3})=1$ and $f_1(t)=2f(t)$ otherwise. Then $f_1$ is an OIDIDF on $T$ and so $\gamma_{oidI}(T)\leqslant w(f_1)=2w(f)-1<2\gamma_{oiI}(T)$, a contradiction. Thus $x_{d-2}$ does not have a leaf neighbor, and so every child of $x_{d-2}$ with value $0$ (if any), is adjacent to at least one other vertex with value at least $1$. Then we change $f'(x_{d-2})$ and $f'(x_{d-1})$ to $1$ to obtain a $\gamma_{oiI}(T')$-function $f''$ with $f''(x_{d-1})\geq 1$. Then the function $g$ defined by $g(x_d)=0, g(x_{d+1})=1$ and $g(t)=f''(t)$ otherwise, is an  OIIDF on $T$ and so $\gamma_{oiI}(T)\leqslant \gamma_{oiI}(T')+1$, as desired.

We conclude that  $\gamma_{oiI}( T)=\gamma_{oiI}(T')+1$. Thus $ \gamma_{oidI}(T)=2\gamma_{oiI}(T)=2(\gamma_{oiI}(T')+1).$ Then $\gamma_{oidI}(T)-2=2\gamma_{oiI}(T')$.

We next show that $\gamma_{oidI}(T')=\gamma_{oidI}(T)-2 $. Assume that $g$ is a $\gamma_{oidI}(T')$-function. If $ g(x_{d-1})\geqslant 1 $, then we set $ g(x_d)=0 $ and $ g(x_{d+1})=2 $ to extend $g$ to an OIDIDF on $T$. Therefore $\gamma_{oidI}(T)\leqslant \gamma_{oidI}(T')+2 $.
Assume that $ g(x_{d-1})=0 $. If $\deg_T(x_{d-1})=2$, then $ g(x_{d-2})=3 $ and $g(x_{d-3})+g(x_{d-4})\geqslant 1 $. Note that $diam(T')\geqslant4$, since $\gamma_{oidI}(T')=2\gamma_{oiI}(T')$. Since $\deg(x_{d-2})=2$, we may change $ g(x_{d-2})$ to $2$, and set $g(x_{d-1})=1$, $ g(x_d)=0 $ and $g(x_{d+1})=2$ to extend $g$ to an OIDIDF on $T$ and as before find that $\gamma_{oidI}(T)\leqslant \gamma_{oidI}(T')+2 $. Thus assume that  $\deg_T(x_{d-1})\geqslant3$. Since $x_{d-1}$ does not have a leaf neighbor, it has a child $y$ of degree $2$. Let $z$ be the child of $y$. Then $g(y)+g(z)=3$. Now we change $g(x_{d-1})$ to $1$, $g(y)$ to $0$ and $g(z)$ to $2$, and as  before obtain that $\gamma_{oidI}(T)\leqslant \gamma_{oidI}(T')+2 $. We conclude that $\gamma_{oidI}(T)\leqslant\gamma_{oidI}(T')+2$.
Next we show that $\gamma_{oidI}(T)\geqslant\gamma_{oidI}(T')+2$. Assume that $h$ is a $\gamma_{oidI}(T)$-function. If $h(x_{d-1})\geqslant 2$, then $h(x_{d+1})+h(x_d)=2$ and $h\mid_{V(T')}$ is an OIDIDF on $T'$, implying that $\gamma_{oidI}(T')\leqslant\gamma_{oidI}(T)-2$. Assume that $h(x_{d-1})=1$.
If $\sum_{(v\in N[x_{d-1}]-{x_d})}h(v)\geqslant 3$, then  we may assume that $ h(x_d)=0$ and $h(x_{d+1})=2$, in which the result follows as before. Thus assume that $\sum_{(v\in N[x_{d-1}]-{x_d})}h(v)< 3$. Then $h(x_d)+h(x_{d+1})=3$ and we may assume that $h(x_{d+1})=h(x_{d-1})=2$ and $g(x_d)=0 $ in which the result follows as before. We now assume that $h(x_{d-1})=0$. If $\deg(x_{d-1})\geqslant 3$, then $x_{d-1}$ has a child $y$ of degree two. Assume that $z$ is the child of $y$. Since $h(x_{d-1})=0$, we have $h(t)\geqslant 1$ for every vertex $t\in N(x_{d-1})$. Furthermore, $h(x_{d+1})+h(x_d)=h(y)+h(z)=3$. Let $h'$ be a function defined by $h'(x_{d-1})=h'(y)=1$, $h'(x_d)=0$, $h'(z)=h'(x_d)=2 $ and $h'(t)=h(t)$ otherwise. Then $h'\mid_{T'}$ is an OIDIDF on $T'$ and so $\gamma_{oidI}(T')\leqslant w(h')-2=w(h)-2$ as required. Thus assume that $\deg(x_{d-1})=2$. If $h(x_{d-2})\geqslant 2$, then we change $h(x_{d-1})$ to $1$, $h(x_d)$ to $0$ and $h(x_{d+1})$ to $2$, and as before we find that $\gamma_{oidI}(T')\leqslant\gamma_{oidI}(T)-2$. It remains to assume that $h(x_{d-2})=1$. Then $h(x_{d-3})\geqslant 2$, since $\deg(x_{d-2})=2$. Now we change $h(x_{d+1})$ and $h(x_{d-1})$ to $2$ and $h(x_d)$ and $h(x_{d-2})$ to $0$ to obtain $h'$. Then $h'\mid_{T'}$ is an OIDIDF on $T'$ and the result is obtained as before. Hence, $\gamma_{oidI}(T')=\gamma_{oidI}(T)-2 =2\gamma_{oiI}(T') $. We deduce $T'\in \mathcal{T}$ by the induction hypothesis. Since $f(x_{d-1})=1$, we have $x_{d-1}\in X_{L(T)}$ and so $T$ is formed by Operation $\mathcal{O}$ on $T'$.
\end{proof}

\section{New lower bounds}
In this section we prove two new lower bound on the the outer independent dou-
ble Italian domination number.

\begin{theorem}\label{m1}
For every graph $G$ of order $n$ with minimum degree $\delta=\delta(G)>0$ and maximum degree $\Delta=\Delta(G)$,  $\gamma_{oidI}(G)\geqslant \lfloor\frac{n\delta}{\delta+\Delta}\rfloor +1$ and this bound is sharp.
\end{theorem}
\begin{proof}
Let $G$ be a graph of order $n$ with minimum degree $\delta>0$, and $f=(V_0,V_1,V_2,V_3)$ be a $\gamma_{oidI}(G)-$function. Let $m$ be the number of edges having one end point in $V_0$ and the other end point in $V_1\cup V_2\cup V_3$. We then have $m \geq \delta | V_0|$, since $V_0$ is independent and  $f(N(u))\geq 3$ for each vertex $u\in V_0$. On the other hnd each vertex in $V_1\cup V_2\cup V_3$ has at most $\Delta$ neighbors in $V_0$. Thus, $ m\leq \Delta(|V_1|+|V_2|+|V_3|)$, and so $\delta(n-|V_1|-|V_2|-|V_3|)\leqslant \Delta (|V_1|+|V_2|+|V_3|)$ which implies that $ \delta n\leqslant (\Delta + \delta )(|V_1|+|V_2|+|V_3|) $. Therefore, $$ \frac{\delta n}{\Delta + \delta}\leqslant |V_1|+|V_2|+|V_3|=\gamma_{oidI}(G)-|V_2|-2|V_3|.$$ If $V_2\cup V_3\neq \emptyset$, then $\gamma_{oidI}(G)\geqslant \frac{n\delta}{\Delta+\delta}+|V_2|+2|V_3| \geqslant \frac{n\delta}{\Delta+\delta}+1 $, as desired. Thus assume that $ V_2\cup V_3 =\emptyset$. Then each vertex in $ V_1$ is adjacent to at least two vertices in $V_1$ and each vertex in $ V_0$ is adjacent to at least three vertices in $V_1$. Then if $|V_0|>0$, then $\Delta\geq 3$ while if $|V_0|=0$ then $\Delta\geq 2$. Furthermore, counting the number of edges having one end point in $V_0$ and the other end point in $V_1$ implies that $ \delta |V_0|\leqslant (\Delta -2 )|V_1|$. Since $ n=|V_0|+|V_1|$, we find that $ \delta (n-|V_1|)\leqslant (\Delta-2)|V_1|$. This implies that $\gamma_{oidI}(G)=|V_1| \geq \frac{\delta n}{\Delta +\delta -2}> \frac{\delta n}{\Delta +\delta}$. Since $\gamma_{oidI}(G)$ is an integer we deduce that $\gamma_{oidI}(G)\geqslant \lfloor\frac{n\delta}{\delta+\Delta}\rfloor +1 $. To see the sharpness, for each integer $n\geq 3$, let $G_n$ be a graph obtained from a cycle $C_n$ by adding $2n$ new vertices and joining each new vertex to all vertices of $C_n$. Note that $|V(G_n)|=3n$, $\delta(G_n)=n$, $\Delta(G_n)=2n+2$ and $\gamma_{oidI}(G_n)=n=\lfloor\frac{3n^2}{3n+2}\rfloor +1$.
\end{proof}

We now introduce a family of graphs as follows. Let $\mathcal{G}$ be the class of all graphs $G$ such that $G\in \mathcal{G}$ if and only if $G$ is obtained from a graph $H$ of order at least three and minimum degree at least two by adding at least $|V(H)|-\delta(H)$ new vertices and joining each new vertex to all vertices of $H$.

\begin{proposition}
If $G$ is a graph of order $n\geq 1$ with minimum degree $\delta$, $\gamma_{oidI}(G)\geq \delta $, with equality holds if and only if $G\in \mathcal{G}$.
\end{proposition}

\begin{proof}
The result is obvious if $\delta\leq 2$. Thus assume that $\delta\geq 3$. Let $ f=(V_0,V_1,V_2,V_3) $  be a $ \gamma_{oidI}(G) $-function. If $V_0=\emptyset $, then $  \gamma_{oidI}(G)\geq n \geq \delta+1$, since $\delta \leq n-1$. Thus assume that $V_0 \neq \emptyset $. Let $ x\in V_0 $. Therefore $ N(x)\subseteq V_1\cup V_2 \cup V_3 $, since $ V_0 $ is an independent set. Also $ \left |N(x) \right |\geq \delta $. Note that $$\gamma_{oidI}(G)=|V_1|+2|V_2|+3|V_3|= (\sum_{i=1}^{3}|V_i|)+|V_2|+2|V_3|.$$ If $(V_2\cup V_3 ) \neq \emptyset $, then $\gamma_{oidI}(G) \geq \left (\sum_{i=1}^{3}|V_i|\right )+1\geq \deg(x) +1\geq \delta +1$. Thus assume that $ V_2\cup V_3 =\emptyset $. Then $ \gamma_{oidI}(G)= |V_1|$. Since $ V_0 $ is an independent set, we find that $\gamma_{oidI}(G)=|V_1|\geq \delta $.

We next prove the equality part. Assume that $\gamma_{oidI}(G)= \delta$. Clearly $\gamma_{oidI}(G)\geq 2$. It is easy to see that $\gamma_{oidI}(G)=2$ if and only if $G=K_1$. Since $K_1$ has minimum degree $0$, from $\gamma_{oidI}(G)= \delta$ we obtain that $\delta\geq 3$. Let $ f=(V_0,V_1,V_2,V_3) $  be a $ \gamma_{oidI}(G) $-function. Following the above argument for the first part of the proof, we obtain that $V_2\cup V_3 =\emptyset$ and $\gamma_{oidI}(G)=|V_1|= \delta$. Let $H=G[V_1]$. Since $V_0$ is independent and $|V_1|= \delta$, every vertex of $V_0$ is adjacent to all vertices of $H$. Since any vertex of $V_1$ is adjacent to at least two other vertices of $V_1$, we obtain that $H$ has minimum degree at least $2$. Suppose that $|V_0|<|V(H)|-\delta(H)$. Let $v\in V_1$ be a vertex with $\deg_H(v)=\delta(H)$. Then $$\deg_G(v)=\deg_H(v)+|V_0|<|V(H)|=\delta(G).$$ This is a contradiction. Thus  $|V_0|\geq |V(H)|-\delta(H)$. We conclude that $G\in \mathcal{G}$.

Conversely assume that $G\in \mathcal{G}$. Thus $G$ is obtained from a graph $H$ of order $m\geq 3$ with $V(H)=\{y_1,...,y_m\}$ and $\delta(H)\geq 2$ by adding $t\geq m-\delta(H)$ new vertices $x_1,...,x_t$, and adding edges $x_iy_j$, for $i=1,...,t$ and $j=1,....,m$. Note that $\deg_G(x_i)=m$ for $i=1,...,t$, and $\deg_G(y_j)=\deg_H(y_j)+t\geq \delta(H)+t\geq m$ for $j=1,....,m$. Thus $\delta(G)=m$. Then $\gamma_{oidI}(G)\geq m$ by the first part of the proposition. Now let $g$ be a function on $G$ defined by $g(y_j)=1$ for each $j=1,...,m$ and $g(x_i)=0$ for each $x_i$, $i=1,...,t$. Then $g$ is an OIDIDF for $G$, implying that $\gamma_{oidI}(G)\leq m$. Consequently, $\gamma_{oidI}(G)= m=\delta(G)$.
\end{proof}


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