Strong Equality Between the 2-Rainbow Domination and Independent 2-Rainbow Domination Numbers in Trees

Abstract

A 2-rainbow dominating function (2RDF) on a graph $G=(V, E)$ is a function f from the vertex set V to the set of all subsets of the set $\{1,2\}$ such that for any vertex $v\in V$ with $f(v)=\emptyset$ the condition $\bigcup _{u\in N(v)}f(u)=\{1,2\}$ is fulfilled. A 2RDF f is independent (I2RDF) if no two vertices assigned nonempty sets are adjacent. The weight of a 2RDF f is the value $\omega (f)=\sum _{v\in V}|f (v)|$. The 2-rainbow domination number $\gamma _{r2}(G)$ (respectively, the independent 2-rainbow domination number $i_{r2}(G)$) is the minimum weight of a 2RDF (respectively, I2RDF) on G. We say that $\gamma _{r2}(G)$ is strongly equal to $i_{r2}(G)$ and denote by $\gamma _{r2}(G)\equiv i_{r2}(G)$, if every 2RDF on G of minimum weight is an I2RDF. In this paper, we provide a constructive characterization of trees T with $\gamma _{r2}(T)\equiv i_{r2}(T)$.

1 Introduction

Let G be a simple graph with vertex set $V=V(G)$ and edge set $E=E(G)$. For every vertex $v\in V$, the open neighborhood N(v) is the set $\{u\in V\mid uv\in E\}$ and the closed neighborhood of v is the set $N[v] = N(v) \cup \{v\}$. The degree of a vertex $v\in V$ is $\deg _G(v)=\deg (v)=|N(v)|$. If $A\subseteq V(G)$, then G[A] is the subgraph induced by A. A vertex of degree one is called a leaf, and its neighbor is called a support vertex. If v is a support vertex, then $L_v$ will denote the set of all leaves adjacent to v. A support vertex v is called strong support vertex if $|L_v|>1$. For $r,s\ge 1$, a double star S(r, s) is a tree with exactly two vertices that are not leaves, with one adjacent to r leaves and the other to s leaves. For a vertex v in a rooted tree T, let C(v) denote the set of children of v, D(v) denote the set of descendants of v and $D[v]=D(v)\cup \{v\}$, and the depth of v, $\mathrm{depth}(v)$, is the largest distance from v to a vertex in D(v). The maximal subtree at v is the subtree of T induced by $D(v) \cup \{v\}$ and is denoted by $T_v$. For terminology and notation on graph theory not given here, the reader is referred to [11, 13].

For a positive integer k, a k-rainbow dominating function (kRDF) of a graph G is a function f from the vertex set V(G) to the set of all subsets of the set $\{1,2,\ldots ,k\}$ such that for any vertex $v\in V(G)$ with $f(v)=\emptyset$ the condition $\bigcup _{u\in N(v)}f(u)=\{1,2,\ldots ,k\}$ is fulfilled. The weight of a kRDF f is the value $\omega (f)=\sum _{v\in V}|f (v)|$. The k-rainbow domination number of a graph G, denoted by $\gamma _{rk}(G)$, is the minimum weight of a kRDF of G. A $\gamma _{rk}(G)$-function is a k-rainbow dominating function of G with weight $\gamma _{rk}(G)$. Note that $\gamma _{r1}(G)$ is the classical domination number $\gamma (G)$. The k-rainbow domination number was introduced by Bre$\check{\mathrm{s}}$ar, Henning and Rall [3] and has been studied by several authors (see, e.g., [4–7, 12]). To study other domination parameters, we refer the readers to [1, 2, 14, 15].

A k-rainbow dominating function f is called an independent k-rainbow dominating function (abbreviated IkRDF) on G if the set $V(G)-\{v\in V\mid f(v)=\emptyset \}$ is independent. The independent k-rainbow domination number, denoted by $i_{rk}(G)$, is the minimum weight of an IkRDF on G. An independent k-rainbow dominating function f is called an $i_{rk}(G)$-function if $\omega (f)=i_{rk}(G)$. Since each independent k-rainbow dominating function is a k-rainbow dominating function, we have $\gamma _{rk}(G)\le i_{rk}(G)$.

Clearly if $\gamma _{rk}(G)= i_{rk}(G)$, then every $i_{rk}(G)$-function is also a $\gamma _{rk}(G)$-function. However, not every $\gamma _{rk}(G)$-function is an $i_{rk}(G)$-function, even when $\gamma _{rk}(G)= i_{rk}(G)$. For example, the double star $S(k,k+1)$ has two $\gamma _{rk}(S(k,k+1))$-function, but only one of them is an $i_{rk}(S(k,k+1))$-function. We say that $\gamma _{rk}(G)$ and $i_{rk}(G)$ are strongly equal and denote by $\gamma _{rk}(G)\equiv i_{rk}(G)$, if every $\gamma _{rk}(G)$-function is an $i_{rk}(G)$- function.

Haynes and Slater in [10] were the first to introduce strong equality between two parameters. Also in [8, 9], Haynes, Henning and Slater gave constructive characterizations of trees with strong equality between some domination parameters.

Our purpose in this paper is to present a constructive characterizations of trees T with $\gamma _{r2}(T)\equiv i_{r2}(T)$.

We make use of the following result in this paper.

Proposition A

[5] Let G be a connected graph. If there is a path $v_3v_2v_1$ in G with $\deg (v_2)=2$ and $\deg (v_1)=1$, then G has a $\gamma _{r2}(G)$-function f such that $f(v_1)=\{1\}$ and $2\in f(v_3)$.

Corollary 1

Let T be a tree with $\gamma _{r2}(T)\equiv i_{r2}(T)$. If there is a path $v_3v_2v_1$ in T with $\deg (v_2)=2$ and $\deg (v_1)=1$ such that $v_3$ is a support vertex, then T has a $\gamma _{r2}(T)$-function f such that $f(v_3)=\{1,2\}$, $|f(v_1)|=1$ and $|f(x)|=0$ for every $x\in L_{v_3}\cup \{v_2\}$.

Observation 2

Let T be a tree and let z be a strong support vertex of T. Then

• (a) T has a $\gamma _{r2}(T)$-function such that $f(z)=\{1,2\}$.

• (b) $\gamma _{r2}(T)\not \equiv i_{r2}(T)$ if and only if T has a $\gamma _{r2}(T)$-function that is not independent and $f(z)=\{1,2\}$.

Proof

(a) The proof is immediate.

(b) Let $\gamma _{r2}(T)\not \equiv i_{r2}(T)$. Then T has a $\gamma _{r2}(T)$-function that is not independent. If $f(z)=\{1,2\}$, then we are done. If $|f(z)|=1$, then $|f(x)|=1$ for each $x\in L_z$ and the function $g:V(G)\rightarrow {\mathcal {P}}(\{1,2\})$ defined by $g(z)=\{1,2\}, g(x)=\emptyset$ for $x\in L_z$ and $g(u)=f(u)$ otherwise, is a 2RDF of T of weight less than $\omega (f)$ which is a contradiction. Let $f(z)=\emptyset$. Then clearly the function $g:V(G)\rightarrow {\mathcal {P}}(\{1,2\})$ defined by $g(z)=\{1,2\}, g(x)=\emptyset$ for $x\in L_z$ and $g(u)=f(u)$ otherwise, is a $\gamma _{r2}(T)$-function with the desired property. $\square$

2 Characterizations of Trees with $\gamma _{r2}(T)\equiv i_{r2}(T)$

Let ${\mathcal {F}}_1$ be the family of trees that can be obtained from $k\ge 1$ disjoint stars $K_{1,2}$ by adding either a new vertex v or a path uv and joining the centers of stars to v. Also let ${\mathcal {F}}_2$ be the family including $P_5$ and all trees obtained from $k\ge 2$ disjoint $P_3$ by adding either a new vertex v or a path uv and joining v to a leaf of each $P_3$. If T belongs to ${\mathcal {F}}_1 \cup {\mathcal {F}}_2-\{P_5\}$, then we call the vertex v, the special vertex of T and if $T=P_5$, then its support vertices are special vertices of T. Note that if $T\in {\mathcal {F}}_1 \cup {\mathcal {F}}_2$, then $\gamma _{r2}(T)\equiv i_{r2}(T)$.

Now we provide a constructive characterization of trees T with $\gamma _{r2}(T)\equiv i_{r2}(T)$. For this purpose, we define a family of trees as follows: Let ${\mathcal {F}}$ be the family of trees such that: ${\mathcal {F}}$ contains star $K_{1,2}$ and if T is a tree in ${\mathcal {F}}$, then the tree $T'$ obtained from T by the following seven operations which extend the tree T by attaching a tree to a vertex $y\in V (T)$, called an attacher, is also a tree in ${\mathcal {F}}$.

• Operation ${\mathcal {O}}_1$: If z is a strong support vertex of $T\in {{\mathcal {F}}}$, then $\mathcal {O}_1$ adds a new vertex x and an edge xz.

• Operation ${\mathcal {O}}_2$: If z is a vertex of $T\in {{\mathcal {F}}}$, then ${\mathcal {O}}_2$ adds a new tree $T_1\in {\mathcal {F}}_1$ with special vertex x and an edge xz provided that if x is a support vertex, then $\gamma _{r2}(T-z)\ge \gamma _{r2}(T)$.

• Operation ${\mathcal {O}}_3$: If z is a strong support vertex of $T\in {{\mathcal {F}}}$, then ${\mathcal {O}}_3$ adds a path zxy.

• Operation ${\mathcal {O}}_4$: If z is a vertex of $T\in {{\mathcal {F}}}$ which is adjacent to a support vertex of degree 2, then ${\mathcal {O}}_4$ adds a path zxy.

• Operation ${\mathcal {O}}_5$: If z is a vertex of $T\in {\mathcal {F}}$ which is adjacent to a strong support vertex, then ${\mathcal {O}}_5$ adds a path zxyw.

• Operation ${\mathcal {O}}_6$: If z is a vertex of $T\in {\mathcal {F}}$, then ${\mathcal {O}}_6$ adds a new tree $T_2\in {\mathcal {F}}_2$ with special vertex x and an edge xz provided that if x is a support vertex, then $\gamma _{r2}(T-z)\ge \gamma _{r2}(T)$.

• Operation ${\mathcal {O}}_7$: If z is a vertex of $T\in {{\mathcal {F}}}$ such that every $\gamma _{r2}(T)$-function assigns $\emptyset$ to z, then ${\mathcal {O}}_7$ adds the double star S(1, 2) and an edge zx where x is a leaf of S(1, 2) whose support vertex has degree 3.

Observation 3

The family ${{\mathcal {F}}}$ contains all graphs in $\{K_{1,t}\mid t\ge 2\}\cup {{\mathcal {F}}}_1\cup {{\mathcal {F}}}_2$.

Proof

Starting from $K_{1,2}\in {{\mathcal {F}}}$ and by applying $t-2$ times Operation ${{\mathcal {O}}}_1$, we obtain the star $K_{1,t}$ and hence ${{\mathcal {F}}}$ contains all stars. Furthermore, starting from $K_{1,2}$ and by applying Operation ${{\mathcal {O}}}_4$, we obtain that ${{\mathcal {F}}}$ contains $P_5$.

Now let $T\in {{\mathcal {F}}}_1$. If $|V(T)|=4$, then $T=K_{1,3}$ and immediately $T\in {{\mathcal {F}}}$. If $|V(T)|=5$, then T can be obtained from $K_{1,2}$ by applying Operation ${{\mathcal {O}}}_3$. If $|V(T)|\ge 6$, then T can be obtained from $K_{1,2}$ by applying Operation ${{\mathcal {O}}}_2$. Thus ${{\mathcal {F}}}$ contains all graphs in ${{\mathcal {F}}}_1$.

Finally, let $T\in {{\mathcal {F}}}_2-\{P_5\}$. If $|V(T)|=7$, then $T=P_7$ and T can be obtained from $P_5$ by applying Operation $\mathcal O_4$ twice and so $T\in {{\mathcal {F}}}$. If $|V(T)|\ge 9$, then T can be obtained from $K_{1,2}$ by applying Operation ${{\mathcal {O}}}_6$. Thus ${{\mathcal {F}}}$ contains all graphs in ${{\mathcal {F}}}_2$. $\square$

Lemma 4

Let T be a tree with $\gamma _{r2}(T)\equiv i_{r2}(T)$ and let $T'$ be the tree obtained from T by Operation ${\mathcal {O}}_1$. Then $\gamma _{r2}(T')\equiv i_{r2}(T')$.

Proof

Assume z is a strong support vertex of T and let x be a new vertex that is attached to z by applying Operation ${\mathcal {O}}_1$. By Observation 2(a), T has a $\gamma _{r2}$-function f that assigns $\{1,2\}$ to z. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is an I2RDF of T. Now we can extend f to an I2RDF of $T'$ by assigning $\emptyset$ to x, implying that $\gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)\le \gamma _{r2}(T)$. On the other hand, by Observation 2(a), there is a $\gamma _{r2}(T')$-function g which assigns $\{1,2\}$ to z, and clearly the function g, restricted to T, is a 2RDF of T of weight $\gamma _{r2}(T')$, implying that $\gamma _{r2}(T)\le \gamma _{r2}(T')$. Hence $\gamma _{r2}(T')=i_{r2}(T')$.

It will now be shown that $\gamma _{r2}(T')\equiv i_{r2}(T')$. Suppose h is a $\gamma _{r2}(T')$-function that is not independent. Since $|L_{z}|\ge 3$, we must have $f(z)=\{1,2\}$. Then the function h, restricted to T, is a $\gamma _{r2}(T)$-function that is not independent which leads to a contradiction. Thus $\gamma _{r2}(T')\equiv i_{r2}(T)$. $\square$

Lemma 5

Let T be a tree with $\gamma _{r2}(T)\equiv i_{r2}(T)$ and let $T'$ be a tree obtained from T by Operation ${\mathcal {O}}_2$. Then $\gamma _{r2}(T')\equiv i_{r2}(T')$.

Proof

Let $T_1\in {\mathcal {F}}_1$ be the tree which is attached by Operation ${\mathcal {O}}_2$ to T by the edge xz for obtaining the tree $T'$, where $z\in V(T)$ is the attacher vertex, and let $x_{1}, x_{2},\ldots ,x_{k}\in V(T_1)$ be the strong support vertices of $T_1$. Assume x is the special vertex of $T_1$. If x is a support vertex then let y be the leaf that is adjacent to x. Let t be a variant defined by $t=1$ if x is a support vertex, and $t=0$ otherwise. Every $i_{r2}(T)$-function can be extended to an I2RDF on $T'$ by assigning $\{1,2\}$ to $x_i$, $i=1,2,\ldots ,k$, $\emptyset$ to u for $u\in \cup _{i=1}^kN(x_i)$, and $\{1\}$ to y if x is a support vertex. This implies that $i_{r2}(T')\le i_{r2}(T)+2k+t$. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, we deduce that

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le \gamma _{r2}(T)+2k+t=i_{r2}(T)+2k+t. \end{aligned}$$

(1)

Now we show that $\gamma _{r2}(T')= \gamma _{r2}(T)+2k+t$. Let f be a $\gamma _{r2}(T')$-function. It is easy to see that $\sum _{u \in N[x_i]-\{x\}}|f(u)|\ge 2$, for $i=1,2,\ldots ,k$, and $|f(x)|+|f(y)|\ge 1$, if $t=1$. Then $\sum _{u\in V(T_1)}|f(u)|\ge 2k+t$. If $|f(x)|=0$ then $f|_{V(T)}$ is a 2RDF on T, and so $\sum _{u\in V(T)}|f(u)|\ge \gamma _{r2}(T)$. By adding two recent inequalities, we obtain $\gamma _{r2}(T')= \sum _{u\in V(T')}|f(u)|\ge \gamma _{r2}(T)+2k+t.$ Assume that $|f(x)|\ge 1$. Clearly if $t=1$ then $|f(x)|+|f(y)|\ge 2$. Thus $\sum _{u\in V(T_1)}|f(u)|\ge 2k+t+1$. If $|f(z)|\ne 0$ then $f|_{V(T)}$ is a 2RDF on T, and if $|f(z)|= 0$ then the function $f_1$ defined on V(T) by $f_1(z)=\{1\}$ and $f_1(u)=f(u)$ if $u\in V(T)-\{z\}$ is a 2RDF for T. It follows that $\gamma _{r2}(T')\ge \gamma _{r2}(T)+2k+t.$ Hence, we deduce that

$$\begin{aligned} \gamma _{r2}(T')= \gamma _{r2}(T)+2k+t. \end{aligned}$$

(2)

By (1) and (2), we have

$$\begin{aligned} i_{r2}(T')= i_{r2}(T)+2k+t=\gamma _{r2}(T)+2k+t=\gamma _{r2}(T'). \end{aligned}$$

It will now be shown that $\gamma _{r2}(T')\equiv i_{r2}(T')$. Assume h is a $\gamma _{r2}(T')$-function that is not independent. We may assume that h assigns $\{1,2\}$ to each support vertex adjacent to x. If $|h(x)|=0$ then clearly $h|_{V(T)}$ is a $\gamma _{r2}(T)$-function that is not independent, a contradiction with the assumption $\gamma _{r2}(T)\equiv i_{r2}(T)$. Thus $|h(x)|\ge 1$. Then $|h(z)|=0$ and $\sum _{v\in V(T_1)}|h(v)|\ge 2k+1+t$. If $|h(x)|=1$, then $\sum _{w\in N_{T}(z)}|h(w)|\ge 1$ and the function $g:V(T)\rightarrow {{\mathcal {P}}}(\{1,2\})$ defined by $g(z)=\{1\}$ and $g(u)=h(u)$ for $u\in V(T)-\{z\}$ is a $\gamma _{r2}(T)$-function that is not independent, contradicting $\gamma _{r2}(T)\equiv i_{r2}(T)$. Thus $|h(x)|=2$. Then x is a support vertex. Now

$$\begin{aligned} \gamma _{r2}(T-z)\le \sum _{u\in V(T-z)}|h(u)| = \gamma _{r2}(T^{\prime })-2k-1-t< \gamma _{r2}(T). \end{aligned}$$

This is a contradiction with the assumption $\gamma _{r2}(T-z)\ge \gamma _{r2}(T)$. Therefore, $\gamma _{r2}(T^{\prime })\equiv i_{r2}(T^{\prime })$ and the proof is complete. $\square$

Lemma 6

If T is a tree with $\gamma _{r2}(T)\equiv i_{r2}(T)$ and $T'$ is a tree obtained from T by Operation ${\mathcal {O}}_3$, then $\gamma _{r2}(T')\equiv i_{r2}(T')$.

Proof

Let $z\in V(T)$ be a strong support vertex and let zxy be the path added by Operation ${\mathcal {O}}_3$ to obtain $T'$. Let f be a $\gamma _{r2}(T)$-function such that $f(z)=\{1,2\}$ [Observation 2(a)]. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is an I2RDF of T. We can extend f to an I2RDF on $T'$ by assigning $\emptyset$ to x and $\{1\}$ to y, and thus

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)+1= \gamma _{r2}(T)+1. \end{aligned}$$

(3)

Let now $f_1$ be a $\gamma _{r2}(T')$-function. We can assume $f_1(z)=\{1,2\}$ by Observation 2(a). Since $f_1$ is a $\gamma _{r2}(T')$-function, we must have $|f_1(x)|=0$ and $|f_1(y)|=1$. Then $f_1|_{V(T)}$ is a 2RDF on T, and so

$$\begin{aligned} \gamma _{r2}(T)\le \gamma _{r2}(T')-1. \end{aligned}$$

(4)

It follows from (3) and (4) that $\gamma _{r2}(T')= i_{r2}(T')= \gamma _{r2}(T)+1=i_{r2}(T)+1.$

Finally, we shall show that $\gamma _{r2}(T')\equiv i_{r2}(T')$. Assume h is a $\gamma _{r2}(T')$-function that it is not independent. First let $|h(x)|\ge 1$. Then $|h(x)|+|h(y)|=2$. If $|h(z)|\ne 0$ then replace h(x) by $\emptyset$ and h(y) by $\{1\}$ or $\{2\}$ to obtain a 2RDF for $T'$ of weight less than $\gamma _{r2}(T')$, a contradiction. Thus $|h(z)|=0$. Then clearly $|h(u)|=1$ for any leaf u adjacent to z and the function $h_1:V(T')\rightarrow {{\mathcal {P}}}(\{1,2\})$ defined by $h_1(y)=\{1\}, h_1(z)=\{1,2\}, h_1(u)=\emptyset$ for $u\in L_z\cup \{x\}$ and $h_1(w)=h(w)$ otherwise, is a 2RDF for $T'$ of weight less than $\gamma _{r2}(T')$, a contradiction. Now let $|h(x)|=0$. Then clearly $|h(y)|=1$ (else we could make a change to be in the previous case $|h(x)|\ge 1$), and $h|_{V(T)}$ is a $\gamma _{r2}(T)$-function which is not independent, a contradiction. Hence, $\gamma _{r2}(T')\equiv i_{r2}(T')$. This completes the proof. $\square$

Lemma 7

If T is a tree with $\gamma _{r2}(T)\equiv i_{r2}(T)$ and $T'$ is a tree obtained from T by Operation ${\mathcal {O}}_4$, then $\gamma _{r2}(T')\equiv i_{r2}(T')$.

Proof

Let $z\in V(T)$ be a vertex which is adjacent to a support vertex of degree 2 such as w, and let Operation ${\mathcal {O}}_4$ add the path zxy to T.

First let $\deg _T(z)\ge 2$. Let $w'$ be the leaf adjacent to w. Assume f is a $\gamma _{r2}(T)$-function such that $2\in f(z)$ (Proposition A). Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is an $i_{r2}(T)$-function. Now f can be extended to an I2RDF on $T'$ by assigning $\emptyset$ to x and $\{1\}$ to y. Thus

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)+1=\gamma _{r2}(T)+1. \end{aligned}$$

(5)

On the other hand, if $f_1$ is a $\gamma _{r2}(T')$-function, then we may assume that $2\in f_1(z)$ by Proposition A. Clearly $|f_1(x)|+|f_1(y)|\ge 1$ and $f_1|_{V(T)}$ is a 2RDF on T of weight at most $\gamma _{r2}(T')-1$, implying that $\gamma _{r2}(T')\ge \gamma _{r2}(T)+1$. It follows from (5) and the recent inequality that $\gamma _{r2}(T')= i_{r2}(T')= i_{r2}(T)+1=\gamma _{r2}(T)+1.$

It will now be shown that $\gamma _{r2}(T')\equiv i_{r2}(T')$. Suppose h is a $\gamma _{r2}(T')$-function which it is not independent. If $|h(z)|> 0$ then we must have $|h(x)|=0$ and $|h(y)|=1$, and so $h|_{V(T)}$ is a $\gamma _{r2}(T)$-function which is not independent, a contradiction. Let $|h(z)|=0$. Then obviously $|h(x)|+|h(y)|=|h(w)|+|h(w')|=2$. Then the function $g:V(T')\rightarrow {{\mathcal {P}}}(\{1,2\})$ defined by $g(x)=g(w)=\emptyset$, $g(y)=g(w')=\{1\}, g(z)=\{2\}$ and $g(u)=f(u)$ for $u\in V(T')-\{x,y,w,w',z\}$, is a 2RDF of $T'$ of weight less than $\gamma _{r2}(T')$, a contradiction. Thus $\gamma _{r2}(T')\equiv i_{r2}(T')$.

Now let $\deg _T(z)=1$, i.e., z is a leaf.

Assume f is a $\gamma _{r2}(T)$-function. By Proposition A, we may assume that $f(z)=\{1\}$. Note that f is an $i_{r2}(T)$-function because $\gamma _{r2}(T)\equiv i_{r2}(T)$. Then f can be extended to an I2RDF on $T'$ by assigning $\emptyset$ to x and $\{2\}$ to y. This implies that

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)+1=\gamma _{r2}(T)+1. \end{aligned}$$

(6)

On the other hand, if $f_1$ is a $\gamma _{r2}(T')$-function, then by Proposition A we may assume $f_1(y)=\{1\}$ and $2\in f_1(z)$. Then $f_1|_{V(T)}$ is a 2RDF of T of weight at most $\gamma _{r2}(T')-1$, implying that $\gamma _{r2}(T')\ge \gamma _{r2}(T)+1$. It follows from the last inequality and (6) that $\gamma _{r2}(T')= i_{r2}(T')= \gamma _{r2}(T)+1=i_{r2}(T)+1$.

Next we show that $\gamma _{r2}(T')\equiv i_{r2}(T')$. Assume h is a $\gamma _{r2}(T')$-function that it is not independent. If $|h(z)|> 0$ then we may assume that $|h(x)|=0$ and $|h(y)|=1$, and so $h|_{V(T)}$ is a $\gamma _{r2}(T)$-function which is not independent, a contradiction. Let $h(z)=\emptyset$. Then $|h(x)|+|h(y)|\ge 2$. If $|h(w)|=0$ then $|h(x)|=2$ and $|h(y)|=0$, and the function $h_1:V(T)\rightarrow {{\mathcal {P}}}(\{1,2\})$ defined by $h_1(z)=\{1\}$ and $h_1(u)=h(u)$ if $u\in V(T)-\{z\}$ is a $\gamma _{r2}(T)$-function which is not independent, a contradiction. If $|h(w)|\ge 1$ then it follows from $|h(x)|+|h(y)|\ge 2$ that the function $h_1:V(T)\rightarrow {{\mathcal {P}}}(\{1,2\})$ defined above, is a $\gamma _{r2}(T)$-function which is not independent, a contradiction. Hence $\gamma _{r2}(T')\equiv i_{r2}(T')$. $\square$

Lemma 8

If T is a tree with $\gamma _{r2}(T)\equiv i_{r2}(T)$ and $T'$ is a tree obtained from T by Operation ${\mathcal {O}}_5$, then $\gamma _{r2}(T')\equiv i_{r2}(T')$.

Proof

Let $z\in V(T)$ be a vertex that has a strong support vertex u in its neighborhood and let Operation ${\mathcal {O}}_5$ add the path zxyw to T for obtaining $T'$. Any 2RDF of T can be extended to a 2RDF for $T'$ by assigning $\{1,2\}$ to y, and $\emptyset$ to x and w. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, we deduce that

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)+2=\gamma _{r2}(T)+2. \end{aligned}$$

(7)

Let f be a $\gamma _{r2}(T')$-function. We may assume $f(w)=\{1\}$, $f(y)=\emptyset$ and $2\in f(x)$, by Proposition A. Also we may assume that $|f(u)|=2$, since u is a strong support vertex. Then $f|_{V(T)}$ is a 2RDF on T of weight at most $\gamma _{r2}(T')-2$, and so $\gamma _{r2}(T)\le \gamma _{r2}(T')-2.$ It follows from (7) that

$$\begin{aligned} \gamma _{r2}(T')= i_{r2}(T')= i_{r2}(T)+2=\gamma _{r2}(T)+2. \end{aligned}$$

To show that $\gamma _{r2}(T')\equiv i_{r2}(T')$, suppose h is a $\gamma _{r2}(T')$-function that it is not independent. Since u is a strong support vertex, we may assume $|h(u)|=2$. Then clearly $h(z)=\emptyset$ and $|h(x)|+|h(y)|+|h(w)|=2$, and so $h|_{V(T)}$ is a $\gamma _{r2}(T)$-function which is not independent, a contradiction. Hence $\gamma _{r2}(T')\equiv i_{r2}(T')$ and the proof is completed. $\square$

The proof of next lemma is similar to the proof of Lemma 5, and therefore omitted.

Lemma 9

If T is a tree with $\gamma _{r2}(T)\equiv i_{r2}(T)$, and $T'$ is a tree obtained from T by Operation ${\mathcal {O}}_6$, then $\gamma _{r2}(T)\equiv i_{r2}(T)$.

Lemma 10

If T is a tree with $\gamma _{r2}(T)\equiv i_{r2}(T)$ and $T'$ is a tree obtained from T by Operation ${\mathcal {O}}_7$, then $\gamma _{r2}(T')\equiv i_{r2}(T')$.

Proof

Let z be a vertex of T such that every $\gamma _{r2}(T)$-function assign $\emptyset$ to it, and let x be a leaf of double star S(1, 2) whose support vertex has degree 3. Assume that Operation ${\mathcal {O}}_7$ adds the double star S(1, 2) and the edge xz to obtain $T'$ from T. Let $V(S(1, 2))=\{x,v, v_0, u, u_0\}$ where $N(v)=\{x,u,v_0\}$ and $u\in N(u_0)$. Any 2RDF of T can be extended to a 2RDF on $T'$ by assigning $\emptyset$ to x, u and $v_0$, $\{1, 2\}$ to v and $\{1\}$ to $u_0$. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, we deduce that

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)+3=\gamma _{r2}(T)+3. \end{aligned}$$

(8)

Let f be a $\gamma _{r2}(T')$-function such that $f(u_0)=\{1\}$ and $2\in f(v)$ by Observation A. Clearly $|f(v)|+|f(u_0)|+|f(u)|+|f(v_0)|\ge 3$. We may assume that $|f(x)|=0$, otherwise we replace f(x) by $\emptyset$ and f(z) by $f(z)\cup f(x)$. Then $f|_{V(T)}$ is a 2RDF of T, implying that $\gamma _{r2}(T)\le \gamma _{r2}(T')-3.$ By (8), we have $\gamma _{r2}(T')= i_{r2}(T')= \gamma _{r2}(T)+3= i_{r2}(T)+3$.

It will now be shown that $\gamma _{r2}(T')\equiv i_{r2}(T')$. Suppose h is a $\gamma _{r2}(T')$-function which is not independent. Clearly $\sum _{y\in V(S(1, 2))}|h(y)|\ge 3$. If $|h(z)|>0$, then $h|_{V(T)}$ is a $\gamma _{r2}(T)$-function assigning nonempty set to z which leads to a contradiction. Thus $|h(z)|=0$. If $\sum _{y\in V(S(1, 2))}|h(y)|\ge 4$, then we change the values of h on $V(S(1,2))\cup \{z\}$ to $h(z)=h(u_0)=\{1\}$, $h(v)=\{1,2\}$, and $h(x)=h(u)=h(v_0)=\emptyset$, and then the new function plays the role of h which has been considered earlier. Thus we assume that $\sum _{y\in V(S(1, 2))}|h(y)|=3$. Then clearly $|h(x)|=0$, and $h|_{V(T)}$ is a $\gamma _{r2}(T)$-function which is not independent, a contradiction. Hence $\gamma _{r2}(T')\equiv i_{r2}(T')$. $\square$

Theorem 11

Each tree T in family ${\mathcal {F}}\cup \{K_1\}$ satisfies $\gamma _{r2}(T)\equiv i_{r2}(T)$.

Proof

If $T=K_1$, then clearly $\gamma _{r2}(T)\equiv i_{r2}(T)$. Let $T\in {{\mathcal {F}}}$. Then T is obtained from a star $K_{1,2}$ by successive operations ${{{\mathcal {T}}}}^1,\ldots , {{{\mathcal {T}}}}^m$, where ${{{\mathcal {T}}}}^i\in \{{{{\mathcal {O}}}}_1,\ldots , {{{\mathcal {O}}}}_7\}$ if $m\ge 1$ and $T=K_{1,2}$ if $m=0$. The proof is by induction on m. If $m=0$, then clearly $\gamma _{r2}(K_{1,2})\equiv i_{r2}(K_{1,2})$. Let $m\ge 1$ and that the statement holds for all trees which are obtained from $K_{1,2}$ by applying $m-1$ operations in $\{{{{\mathcal {O}}}}_1,\ldots ,{{{\mathcal {O}}}}_7\}$. It follows from Lemmas 4, ..., 10 that $\gamma _{r2}(T)\equiv i_{r2}(T)$. $\square$

Observation 12

If S(p, q) is a double star with $q\ge p\ge 1$ and $\gamma _{r2}(S(p,q))\equiv i_{r2}(S(p,q))$, then $p=1$ and $q\ge 2$.

Theorem 13

Let T be a tree of order n. If $\gamma _{r2}(T)\equiv i_{r2}(T)$, then $T\in {\mathcal {F}}\cup \{K_1\}$.

Proof

The proof is by induction on n. If $n=1$ then $T=K_1$. Let the statement holds for all trees of order less than n and let T be a tree of order n with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Since $\gamma _{r2}(P_2)\not \equiv i_{r2}(P_2)$, we may assume that $n\ge 3$. If $\mathrm{diam}(T)=2$ then T is a star and by Observation 3, $T\in {{\mathcal {F}}}$. If $\mathrm{diam}(T)=3$, then T is a double star S(p, q) with $q\ge p\ge 1$. By Observation 12, we have $p=1$ and $q\ge 2$. Then T can be obtained from $K_{1,q}$ by Operation ${\mathcal {O}}_3$ and so $T\in {{\mathcal {F}}}$. Therefore, we may assume that $\mathrm{diam}(T)\ge 4$.

Let $v_1v_2\ldots v_k\;(k\ge 5)$ be a diametral path in T such that $|L_{v_2}|$ is as large as possible and root T at $v_k$. Also suppose among paths with this property we choose a path such that $|L_{v_3}|$ is as large as possible.

Assume first that $\deg (v_2)\ge 4$. Let f be a $\gamma _{r2}(T)$-function. Then clearly $f(v_2)=\{1,2\}$ and so f is a 2RDF of $T-v_1$. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is also an I2RDF of $T-v_1$, implying that $\gamma _{r2}(T)=i_{r2}(T)\ge i_{r2}(T-v_1)\ge \gamma _{r2}(T-v_1)$. On the other hand, by Observation 2(a), $T-v_1$ has a $\gamma _{r2}$-function g that assigns $\{1,2\}$ to $v_2$. Then g can be extended to a $\gamma _{r2}(T)$-function by assigning $\emptyset$ to $v_1$ that yields $\gamma _{r2}(T)\le \gamma _{r2}(T-v_1)$. Hence $\gamma _{r2}(T)=i_{r2}(T)= i_{r2}(T-v_1)=\gamma _{r2}(T-v_1).$

We show that $\gamma _{r2}(T-v_1)\equiv i_{r2}(T-v_1)$. Suppose that there is a $\gamma _{r2}(T-v_1)$-function g that is not independent. Since g is a $\gamma _{r2}(T-v_1)$-function, we must have $|g(v_2)|+\sum _{u\in L_{v_2}-\{v_1\}}|g(u)|=2$. Now the function $h:V(T-v_1)\rightarrow {{\mathcal {P}}}(\{1,2\})$ defined by $h(v_2)=\{1,2\}, h(u)=\emptyset$ for $u\in L_{v_2}-\{v_1\}$ and $h(x)=g(x)$ otherwise, is a 2RDF of $T-v_1$ which in not independent. It is clear that h can be extended to a $\gamma _{r2}(T)$-function which is not independent by assigning $\emptyset$ to $v_1$. This leads to a contradiction with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Thus $\gamma _{r2}(T-v_1)\equiv i_{r2}(T-v_1)$. It follows from the inductive hypothesis that $T-v_1\in {{\mathcal {F}}}$. Now it is clear that T can be obtained from $T-v_1\in {\mathcal {F}}$ by applying Operation ${\mathcal {O}}_1$.

Assume next that $\deg (v_2)=3$. Let $u\in L_{v_2}-\{v_1\}$. We claim that $v_3$ is not a strong support vertex. Assume to the contrary that $v_3$ is a strong support vertex. By Observation 2(a), T has a $\gamma _{r2}(T)$-function f such that $f(v_3)=\{1,2\}$. Clearly $|f(v_2)|+|f(v_1)|+|f(u)|= 2$. Now the function $g:V(T)\rightarrow {{\mathcal {P}}}(\{1,2\})$ defined by $g(v_2)=\{1,2\}, g(v_1)=g(u)=\emptyset$ and $g(x)=f(x)$ for $x\in V(T)-\{u,v_1,v_2\}$ is clearly a $\gamma _{r2}(T)$-function that is not independent, a contradiction with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Thus $v_3$ is not a strong support vertex. Using Proposition A and an argument similar to that described above, we deduce that $v_3$ is not adjacent to a support vertex of degree 2. By the choice of the diametral path, we deduce that any child of $v_3$ is a leaf or a support vertex of degree 3 and at most one of them is leaf. This implies that $T_{v_3}\in \mathcal {F}_1$. Let $T'=T-T_{v_3}$.

We claim that if $v_3$ is a support vertex, then $\gamma _{r2}(T'-v_4)\ge \gamma _{r2}(T')$. Let $v_3$ be a support vertex and let to the contrary that $\gamma _{r2}(T'-v_4)< \gamma _{r2}(T')$. Assume h is a $\gamma _{r2}(T'-v_4)$-function and define $g: V(T)\rightarrow {{\mathcal {P}}}(\{1,2\})$ by $g(x)=h(x)$ for $x\in V(T')-\{v_4\}$, $g(x)=\{1,2\}$ for $x\in N[v_3]-(L_{v_3}\cup \{v_4\})$ and $g(x)=\emptyset$ otherwise. Obviously g is a $\gamma _{r2}(T)$-function that is not independent, a contradiction with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Thus $\gamma _{r2}(T'-v_4)\ge \gamma _{r2}(T')$ when $v_3$ is a support vertex.

It will now be shown that $\gamma _{r2}(T')\equiv i_{r2}(T')$. First we show that $\gamma _{r2}(T')= i_{r2}(T')$. Since every $\gamma _{r2}(T')$-function can be extended to a 2RDF on T by assigning $\{1, 2\}$ to the strong support vertices in $N_{T_{v_3}}(v_3)$, $\{1\}$ to the leaf adjacent to $v_3$, if any, and $\emptyset$ to the other vertices in $T_{v_3}$, we deduce that

$$\begin{aligned} i_{r2}(T)=\gamma _{r2}(T)\le \gamma _{r2}(T')+2k+t\le i_{r2}(T')+2k+t \end{aligned}$$

(9)

where k is the number of strong support vertices adjacent to $v_3$ in $T_{v_3}$ and t is the number of leaf adjacent to $v_3$. On the other hand, let f be a $\gamma _{r2}(T)$-function. By Observation 2(a), we may assume that f assigns $\{1, 2\}$ to the strong support vertices in $T_{v_3}$. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is an I2RDF. Then f assigns $\emptyset$ to $v_3$ and $\{1\}$ or $\{2\}$ to the leaf adjacent to $v_3$, if any, and $f|_{V(T')}$ is an I2RDF on $T'$ with weight $i_{r2}(T)-2k-t$. Thus $i_{r2}(T')\le i_{r2}(T)-2k-t$. It follows from (9) that $i_{r2}(T)=\gamma _{r2}(T)=\gamma _{r2}(T')+2k+t= i_{r2}(T')+2k+t$ and hence $\gamma _{r2}(T')=i_{r2}(T')$.

Now we show that this equality is strong. Suppose h is a $\gamma _{r2}(T')$-function that it is not independent. We can extend h to a 2RDF on T by assigning $\{1,2\}$ to every strong support vertex of $T_{v_3}$ and $\{1\}$ to the leaf adjacent to $v_3$, if any, and $\emptyset$ to the other vertices in $T_{v_3}$, to obtain a $\gamma _{r2}(T)$-function which is not independent, a contradiction with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Therefore $\gamma _{r2}(T')\equiv i_{r2}(T')$. It follows from the induction hypothesis that $T'\in {\mathcal {F}}$. Then T can be obtained from $T'$ by applying Operation ${\mathcal {O}}_2$ and hence $T\in {\mathcal {F}}$.

We thus assume that $\deg (v_2)=2$. Furthermore, we may assume that every child of $v_3$ that is a support vertex has degree two. We now consider the following three cases on $|L_{v_3}|$.

Case 1 $|L_{v_3}|\ge 2$.

Let $T'=T-\{v_1,v_2\}$. We show that $\gamma _{r2}(T')\equiv i_{r2}(T')$. Suppose f is a $\gamma _{r2}(T)$-function that assigns $\{1, 2\}$ to $v_3$ [Observation 2(a)]. Clearly $|f(v_1)|+|f(v_2)|=1$. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is an $i_{r2}(T)$-function. Hence $|f(v_2)|=0$ and $f|_{V(T')}$ is an I2RDF on $T'$ implying that

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)-1=\gamma _{r2}(T)-1. \end{aligned}$$

(10)

Now let g be a $\gamma _{r2}(T')$-function that assigns $\{1,2\}$ to $v_3$ [Observation 2(a)]. Then g can be extended to a 2RDF on T by assigning $\emptyset$ to $v_2$ and $\{1\}$ to $v_1$. This yields $\gamma _{r2}(T)\le \gamma _{r2}(T')+1$, By (10), we have $\gamma _{r2}(T')= i_{r2}(T')$. To show that this equality is strong, assume h is a $\gamma _{r2}(T')$-function that it is not independent. We may assume $h(v_3)=\{1,2\}$. Now one can extend h to a $\gamma _{r2}(T)$-function which is not independent, by assigning $\emptyset$ to $v_2$ and $\{1\}$ to $v_1$, a contradiction with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Thus $\gamma _{r2}(T')\equiv i_{r2}(T')$. By induction hypothesis, $T'\in {\mathcal {F}}$ and so T can be obtain from $T'$ by Operation ${\mathcal {O}}_3$.

Case 2 $|L_{v_3}|=0$.

Then any child of $v_3$ is a support vertex of degree 2. We consider two subcases.

Subcase 2.1   $\deg (v_3)\ge 3$.

Let $z_2$ be a child of $v_3$ different from $v_2$, and let $z_1$ be the leaf adjacent to $z_2$. Suppose $T'=T-\{v_1,v_2\}$. We show that $\gamma _{r2}(T')\equiv i_{r2}(T')$. Let f be a $\gamma _{r2}(T)$-function. We may assume $2\in f(v_3)$ by Proposition A. Clearly $|f(v_1)|+|f(v_2)|=1$. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is a $i_{r2}(T)$-function. Clearly $f|_{V(T')}$ is an I2RDF on $T'$, implying that

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)-1=\gamma _{r2}(T)-1. \end{aligned}$$

(11)

On the other hand, by Proposition A, $T'$ has a $\gamma _{r2}(T')$-function g such that $2\in g(v_3)$. Then we can extend g on T by assigning $\emptyset$ to $v_2$ and $\{1\}$ to $v_1$, to obtain a 2RDF of weight $\gamma _{r2}(T')+1$. Thus $\gamma _{r2}(T')\ge \gamma _{r2}(T)-1$. It follows from (11) that $\gamma _{r2}(T')= i_{r2}(T')$.

To show that this equality is strong, assume h is a $\gamma _{r2}(T')$-function that it is not independent. First let $|h(v_3)|>0$. Assume without loss of generality that $2\in h(v_3)$. Then the function $h':V(T)\rightarrow {{\mathcal {P}}}(\{1,2\})$ defined by $h'(v_1)=\{1\}, h'(v_2)=\emptyset$ and $h'(x)=h(x)$ for $x\in V(T)-\{v_1,v_2\}$ is a $\gamma _{r2}(T)$-function that is not independent, a contradiction. Let now $|h(v_3)|=0$. Then $|h(z_2)|+|h(z_1)|=2$. If $\cup _{x\in N(v_3)-\{z_2\}}h(x)\ne \emptyset$, then we define $g:V(T)\rightarrow {{\mathcal {P}}}(\{1,2\})$ by $g(v_3)=\{1\}, g(z_2)=g(v_2)=\emptyset , g(z_1)=g(v_1)=\{2\}$ and $g(x)=h(x)$ otherwise, to produce a $\gamma _{r2}(T)$-function that is not independent, a contradiction. Let $\cup _{x\in N(v_3)-\{z_2\}}h(x)=\emptyset$. Then to rainbowly dominate $v_3$, we must have $h(z_2)=\{1,2\}$ and $|h(z_1)|=0$. Then the function $h_1:V(T')\rightarrow \mathcal {{\mathcal {P}}}(\{1,2\})$ defined by $h_1(v_3)=\{1\}, h_1(z_2)=\emptyset , h_1(z_1)=\{2\}$, and $h_1(x)=h(x)$ otherwise, is a $\gamma _{r2}(T')$-function that is not independent and $|h_1(v_3)|>0$. This leads to a contradiction as above. Thus $\gamma _{r2}(T')\equiv i_{r2}(T')$ and by inductive hypothesis we have $T'\in {\mathcal {F}}$. Now T can be obtained from $T'$ by Operation ${\mathcal {O}}_4$.

Subcase 2.2   $\deg (v_3)=2$.

First let $\deg (v_4)=2$. Let $T'=T-\{v_1,v_2\}$. We show that $\gamma _{r2}(T')\equiv i_{r2}(T')$. Let f be a $\gamma _{r2}(T)$-function such that $f(v_1)=\{1\}$ and $2\in f(v_3)$ (Proposition A). This implies that $|f(v_2)|=0$. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is an $i_{r2}(T)$-function. Obviously the function f, restricted to $T'$, is an I2RDF on $T'$, implying that

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)-1=\gamma _{r2}(T)-1. \end{aligned}$$

(12)

Now let g be a $\gamma _{r2}(T')$-function such that $g(v_3)=\{1\}$ by Proposition A. We can extend g to a $\gamma _{r2}(T)$-function by assigning $\emptyset$ to $v_2$ and $\{2\}$ to $v_1$. This implies that $\gamma _{r2}(T)\le \gamma _{r2}(T')+1$, and by (12) we obtain $\gamma _{r2}(T')= i_{r2}(T')$.

Now we show that this equality is strong. Assume h is a $\gamma _{r2}(T')$-function that is not independent. If $|h(v_3)|>0$, then we can extend h to a $\gamma _{r2}(T)$-function that is not independent by assigning $\emptyset$ to $v_2$ and $\{1\}$ to $v_1$ if $2\in h(v_3)$ and $\{2\}$ to $v_1$ if $1\in h(v_3)$, a contradiction with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Let $|h(v_3)|=0$. Then to rainbowly dominate $v_3$, we must have $h(v_4)=\{1,2\}$. Since h is a $\gamma _{r2}(T')$-function and $\deg (v_4)=2$, we must have $|h(v_5)|=0$. Then the function $h_1:V(T)\rightarrow {{\mathcal {P}}}(\{1,2\})$ defined by $h_1(v_5)=h_1(v_1)=\{1\}, h_1(v_3)=\{2\}, h_1(v_2)=h_1(v_4)=\emptyset$ and $h_1(x)=h(x)$ otherwise, is a $\gamma _{r2}(T)$-function which is not independent, a contradiction with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Hence $\gamma _{r2}(T')\equiv i_{r2}(T')$ and by inductive hypothesis, $T'\in {\mathcal {F}}$. Now T can be obtained from $T'$ by Operation ${\mathcal {O}}_4$.

Next let $\deg (v_4)\ge 3$. By Proposition A, T has a $\gamma _{r2}$-function f such that $f(v_1)=\{1\}$, $|f(v_2)|=0$ and $2\in f(v_3)$. Also suppose among $\gamma _{r2}(T)$-functions with this property we choose a $\gamma _{r2}(T)$-function such that $|f(v_4)|$ is as large as possible. If $|f(v_3)|=2$, then the function $g_1:V(T)\rightarrow {{\mathcal {P}}}(\{1,2\})$ defined by $g_1(v_1)=\{1\}, g_1(v_2)=\emptyset , g_1(v_3)=\{2\}, g_1(v_4)=\{1\}$ and $g_1(x)=f(x)$ for $x\in V(T)-\{v_1,v_2,v_3,v_4\}$ is a $\gamma _{r2}(T)$-function that is not independent, a contradiction. Therefore $|f(v_3)|=1$. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is an I2RDF of T and hence $f(v_4)=\emptyset$. This implies that neither $v_4$ is a strong support vertex nor $v_4$ has a support vertex of degree 2 in its neighborhood. If there is a path $v_4y_3y_2y_1$ in $T_4$ where $y_3\ne v_3$ and $\deg (y_1)=1$, then by the choice of diametral path $v_1\ldots v_k$, we have $|L_{v_2}|\ge |L_{y_2}|$ and $|L_{v_3}|\ge |L_{y_3}|$ that implies $\deg (y_2)=2$ and $|L_{y_3}|=0$. Hence, if there is a leaf at distance three from $v_4$ in $T_{v_4}$, then it plays the same role of $v_1$. Thus we may assume that each component of $T_{v_4}-v_4$ is isomorphic to $P_3$, $K_{1, t}, (t\ge 2)$ or a single vertex, where $v_4$ is adjacent to a leaf of each $P_3$, the center of $K_{1, t}$, or the single vertex, respectively.

Assume first that one of the components of $T_{v_4}-v_4$ is $K_{1, t}, (t\ge 2)$. That is, $v_4$ has a strong support vertex such as z in its neighborhood. Let $T'=T-\{v_1,v_2,v_3\}$ and let f be a $\gamma _{r2}(T)$-function. By Observation 2(a), we may assume $f(z)=\{1,2\}$. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is a $i_{r2}(T)$-function and hence $|f(v_4)|=0$. Then clearly $|f(v_1)|+|f(v_2)|+|f(v_3)|=2$ and $f|_{V(T')}$ is an I2RDF on $T'$, implying that

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)-2=\gamma _{r2}(T)-2. \end{aligned}$$

(13)

On the other hand, let $f_1$ be a $\gamma _{r2}(T')$-function such that $f_1(z)=\{1,2\}$ [Observation 2(a)]. We can extend $f_1$ to a 2RDF on T with weight $\gamma _{r2}(T')+2$ by assigning $\{2\}$, $\emptyset$ and $\{1\}$ to $v_3$, $v_2$ and $v_1$, respectively. Hence $\gamma _{r2}(T)\le \gamma _{r2}(T')+2$ and by (13), we have $\gamma _{r2}(T')= i_{r2}(T')$.

If there exists a $\gamma _{r2}(T')$-function h that is not independent, then as above we can extend h to a $\gamma _{r2}(T)$-function that is not independent, a contradiction with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Thus $\gamma _{r2}(T')\equiv i_{r2}(T')$. It follows from inductive hypothesis that $T'\in {\mathcal {F}}$ and so T can be obtained from $T'$ by Operation ${\mathcal {O}}_5$.

Now suppose that $v_4$ has no child which is a strong support vertex. We claim that $|L_{v_4}|\le 1$. Let to the contrary that $|L_{v_4}|\ge 2$. By Proposition A, T has a $\gamma _{r2}$-function f that $f(v_1)=\{1\}$ and $2\in f(v_3)$. Since $|L_{v_4}|\ge 2$, we may assume $f(v_4)=\{1,2\}$ which contradicts the assumption $\gamma _{r2}(T)\equiv i_{r2}(T)$. Hence $|L_{v_4}|\le 1$. Since $\deg (v_4)\ge 3$, we deduce that $T_{v_4}\in {{\mathcal {F}}}_2$. Let $T'=T-T_{v_4}$ and let g be a $\gamma _{r2}(T)$-function with $g(v_1)=\{1\}$ and $2\in g(v_3)$. By assumption, g is an I2RDF of T and hence $g(v_4)=\emptyset$. Then $g|_{V(T')}$ is an I2RDF of $T'$, implying that

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)-2\deg (v_4)+2-t=\gamma _{r2}(T)-2\deg (v_4)+2-t, \end{aligned}$$

(14)

where t is number of leaves adjacent to $v_4$.

On the other hand, each $\gamma _{r2}(T')$-function f can be extended to a 2RDF of T by assigning $\{2\}$ to $v_3$, $\{1\}$ to $v_1$, each vertex of $N(v_4){\setminus } (L_{v_4}\cup \{v_5,v_3\})$ and the leaf adjacent to $v_4$, if any, $\{2\}$ to every vertex in $T_{v_4}$ at distance 3 from $v_4$ except $v_1$, and $\emptyset$ to the other vertices of $T_{v_4}$. It follows that $\gamma _{r2}(T')\ge \gamma _{r2}(T)-2\deg (v_4)+2-t$. By (14) we obtain $\gamma _{r2}(T')=i_{r2}(T')$.

If h is a $\gamma _{r2}(T')$-function that is not independent, then we can easily extend h to a $\gamma _{r2}(T)$-function that is not independent, a contradiction with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Thus $\gamma _{r2}(T')\equiv i_{r2}(T')$. By inductive hypothesis, we have $T'\in {\mathcal {F}}$. It can be easily seen that $\gamma _{r2}(T'-v_5)\ge \gamma _{r2}(T')$ if $v_4$ is a support vertex. Now T can be obtained from $T'$ by Operation ${\mathcal {O}}_6$.

Case 3 $|L_{v_3}|=1$.

Let w be the leaf adjacent to $v_3$. We consider the following subcases.

Subcase 3.1   $\deg (v_3)>3$.

Then $v_3$ has a child $z_2\ne v_2$ that is a support vertex of degree 2. Let $z_1$ be the leaf adjacent to $z_2$. Set $T'=T-\{v_1,v_2\}$. We show that $\gamma _{r2}(T')\equiv i_{r2}(T')$. Assume that f is a $\gamma _{r2}(T)$-function. We may assume that $f(v_1)=\{1\}$ and $2\in f(v_3)$ by Proposition A. Clearly $|f(v_2)|=0$. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is an I2RDF of T. Now $f|_{V(T')}$ is an I2RDF of $T'$ of weight $\gamma _{r2}(T)-1$ which implies that

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)-1=\gamma _{r2}(T)-1. \end{aligned}$$

(15)

On the other hand, if $f_1$ is a $\gamma _{r2}(T')$-function, then we may assume that $2\in f_1(v_3)$ by Proposition A, and so $f_1$ can be extended to a 2RDF of T of weight $\gamma _{r2}(T')+1$ by assigning $\emptyset$ to $v_2$ and $\{1\}$ to $v_1$, implying that $\gamma _{r2}(T)\le \gamma _{r2}(T')+1$. By (15) we obtain $\gamma _{r2}(T')=i_{r2}(T')$.

To show that this equality is strong, suppose h is a $\gamma _{r2}(T')$-function which is not independent. We may assume $|h(v_3)|>0$, for otherwise we must have $|h(w)|=1$ and $|h(z_2)|+|h(z_1)|=2$ and the function $g:V(T')\rightarrow \mathcal P(\{1,2\})$ by $g(v_3)=\{1\}$, $g(z_2)=\emptyset , g(z_1)=g(w)=\{2\}$ and $g(x)=h(x)$ otherwise, is a $\gamma _{r2}(T')$-function with the desired property. Then we can easily extend h to a $\gamma _{r2}(T)$-function that is not independent, a contradiction with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Thus $\gamma _{r2}(T')\equiv i_{r2}(T')$ and by inductive hypothesis, $T'\in {\mathcal {F}}$. Now T can be obtained from $T'$ by Operation ${\mathcal {O}}_4$.

Subcase 3.2   $\deg (v_3)=3$.

First let $\deg (v_4)\ge 3$. Let f be a $\gamma _{r2}(T)$-function. By Corollary 1, we may assume $f(v_3)=\{1, 2\}$. Since $\gamma _{r2}(T)\equiv i_{r2}(T)$, f is an I2RDF of T. Then $|f(v_4)|=0$ and $|f(v_1)|=1$. If $1\in \cup _{x\in N(v_4)-\{v_3\}} f(x)$ (the case $2\in \cup _{x\in N(v_4)-\{v_3\}} f(x)$ is similar), then the function $f_1:V(T)\rightarrow {{\mathcal {P}}}(\{1,2\})$ defined by $f_1(v_1)=f_1(w)=\{1\}, f_1(v_3)=\{2\}, f_1(v_2)=\emptyset$ and $f_1(x)=f(x)$ otherwise, is a $\gamma _{r2}(T)$-function which is not independent, a contradiction with $\gamma _{r2}(T)\equiv i_{r2}(T)$. Thus $|\cup _{x\in N[v_4]-\{v_3\}} f(x)|=0$. This implies that $v_4$ has no child with depth 0 or 1. Assume that $v_4$ has a child z with depth 2. Then any leaf of $T_z$ at distance two from z plays the same role of $v_1$, and thus by the previous arguments, we may assume that $T_z\simeq T_{v_3}$ and as above we can define a $\gamma _{r2}(T)$-function g such that $g(z)=g(v_3)=\{1,2\}$ which leads to a contradiction. Thus $\deg (v_4)=2$. Suppose $T'=T-T_{v_4}$. We show that $i_{r2}(T')\equiv \gamma _{r2}(T')$. Let f be a $\gamma _{r2}(T)$-function that assigns $\{1, 2\}$ to $v_3$ and $\emptyset$ to $v_4$, according to Corollary 1. Note that f is also an I2RDF of T because $i_{r2}(T)\equiv \gamma _{r2}(T)$. Then $f|_{V(T')}$ is an I2RDF on $T'$, implying that

$$\begin{aligned} \gamma _{r2}(T')\le i_{r2}(T')\le i_{r2}(T)-3=\gamma _{r2}(T)-3. \end{aligned}$$

(16)

On the other hand, every $\gamma _{r2}(T')$-function can be extended to a 2RDF of T by assigning $\{1\}$ to $v_1$, $\emptyset$ to $v_2,v_4,w$ and $\{1, 2\}$ to $v_3$, and thus $\gamma _{r2}(T)\le \gamma _{r2}(T')+3$. It follows from (16) that $\gamma _{r2}(T')= i_{r2}(T')$.

If there is a $\gamma _{r2}(T')$-function g that is not independent then as above, we can extend it to a $\gamma _{r2}(T)$-function that is not independent, a contradiction. Thus $\gamma _{r2}(T')\equiv i_{r2}(T')$. By the inductive hypothesis, $T'\in {\mathcal {F}}$ and T can be obtained from $T'$ by Operation ${\mathcal {O}}_7$ and the proof is completed. $\square$

Now we are ready to state the main theorem of this paper.

Theorem 14

Let T be a tree. Then $i_{r2}(T)\equiv \gamma _{r2}(T)$ if and only if $T\in {{\mathcal {F}}}\cup \{K_1\}$.