Total Domination on Tree Operators

Abstract

Let G be a graph with vertex set V and edge set E, a set $D\subseteq V$ is a total dominating set if every vertex $v\in V$ has at least one neighbor in D. The minimum cardinality among all total dominating sets is called the total domination number, and it is denoted by $\gamma _{t}(G)$. Given an arbitrary tree graph T, we consider some operators acting on this graph; $\texttt {S}(T),\texttt {R}(T),\texttt {Q}(T)$ and $\texttt {T}(T)$, and we give bounds of the total domination number of these new graphs using other parameters in the graph T. We also give the exact value of the total domination number in some of them.

1 Introduction

Krausz introduced in [8] the concept of graph operators. These operators have applications in studies of graph dynamics (see [5, 10]), and topological indices and its applications in chemistry (see [3, 11, 15]). Many large graphs can be obtained by applying graph operators on smaller ones, thus some of their properties are strongly related. Motivated by the above works and by [12, 13], where the total domination number is studied for some graph operators acting on general graphs, we study here the total domination number of several graph operators acting on trees. Other parameters have also been studied in this type of graphs (see [1, 2, 9]).

We begin by stating some notation and terminology. We denote by $G= (V,E)$ a simple graph of order $n=|V|$ and size $m=|E|$. Given a vertex $v\in V$ we denote $N(v)=\{u \in V : uv\in E\}$ and $N[v]=N(v)\cup \{v\}$. The degree of a vertex $v \in V$ will be denoted by $\delta (v)=|N(v)|$, and $\delta$ and $\Delta$ represent the minimum and maximum degree of the graph, respectively. The distance between two vertices u and v in a graph is the number of edges in a shortest path connecting them, and it is denoted by d(u, v). The diameter of a graph G is $diam(G)=\max \{d(u,v):u,v\in V\}$. For a non-empty subset $D \subseteq V$ we denote by $G-D$ the graph whose set of vertices is $V{\setminus } D$ and set of edges is $\{uv\in E:u,v\in V{\setminus } D\}$. A set $D\subseteq V$ is a dominating set if every vertex $v\in V{\setminus } D$ has, at least, a neighbor in D. The domination number $\gamma (G)$ is the minimum cardinality among all dominating sets. A set $D\subseteq V$ is a total dominating set if every vertex $v\in V$ has, at least, a neighbor in D. The total domination number is the minimum cardinality among all total dominating sets, and it is denoted by $\gamma _{t}(G)$. The total domination number has been wisely studied in [6].

The graph $\texttt {S}(G)$ is the graph obtained from G by replacing each of its edge by a path of length two (or equivalently, by inserting an additional vertex into each edge of G). The graph $\texttt {R}(G)$ is defined as the graph obtained from G by adding a new vertex corresponding to each edge of G and by joining each new vertex to the end vertices of the edge corresponding to it. The graph $\texttt {Q}(G)$ is the graph obtained from G by inserting a new vertex into each edge of G and by joining every pair of these new vertices which lie on adjacent edges of G. Finally, the total graph $\texttt {T}(G)$ of a graph G is a graph whose vertex set is $V\cup E$, with two vertices of $\texttt {T}(G)$ being adjacent if and only if the corresponding elements of G are adjacent or incident (see Fig. 1). In this work we give bounds and some exact values for the total domination number in these graph operators acting on trees.

Fig. 1

Graph G and the corresponding graph operators $\texttt {S}(G)$, $\texttt {R}(G)$, $\texttt {Q}(G)$ and $\texttt {T}(G)$

2 Total Domination Number in the Subdivision Graph $\texttt {S}(T)$

We will use the following notation. If $G=(V,E)$, where $V=\{v_1,\ldots ,v_n\}$, then $\texttt {S}(G)=(V^{\prime },E^{\prime })$, such that $V^{\prime }=V\cup \{v^{i,j}:v_iv_j\in E,i<j\}$ and $E^{\prime }=\{v_iv^{i,j},v_jv^{i,j}:v_iv_j\in E,i<j\}$. To avoid writing all the time $i<j$ we will consider $v^{i,j}=v^{j,i}$.

Next, we give the exact value for the total domination number of $\texttt {S}(T)$. For that, we will need the following result, which was proved in [12].

Proposition 1

Let G be a graph with order n. Then, $\gamma _t(\texttt {S} (G))\ge n$.

Now, we will show that any tree attains the lower bound given in Proposition 1.

Theorem 2

If T is any tree with order $n\ge 2$, then

$$\begin{aligned} \gamma _t(\texttt {S} (T))= n. \end{aligned}$$

Proof

It can be checked that, for any tree T with order $n\le 4$ we have $\gamma _t(\texttt {S}(T))= n$. We suppose that the result is true for any tree with order smaller than n and let T be a tree with order n. Let $v_1,v_{d+1}\in V(T)$ such that $d(v_1,v_{d+1})=diam(T)=d$ and let $v_1,v_2,\ldots ,v_{d+1}$ be the shortest path from $v_1$ to $v_{d+1}$.

Case 1. We suppose that $\delta (v_d)\ge 3$ and let $v_{d+2}\in N(v_d){\setminus }\{v_{d+1}\}$. Since $v_{d+1}$ is a leaf in $\texttt {S}(T-\{v_{d+2}\})$, for any minimum total dominating set $D^{\prime }$ in $\texttt {S}(T-\{v_{d+2}\})$, we have $v^{d,d+1}\in D^{\prime }$. If $v_{d}\notin D^{\prime }$, then $v_{d+1}\in D^{\prime }$ and $D^{\prime \prime }=(D{\setminus }\{v_{d+1}\})\cup \{v_d\}$ is also a minimum total dominating set in $\texttt {S}(T-\{v_{d+2}\})$. Therefore, we suppose that $D^{\prime }$ is a minimum total dominating set in $\texttt {S}(T-\{v_{d+2}\})$ containing the vertices $v_d$ and $v^{d,d+1}$. In such a case, $D=D^{\prime }\cup \{v^{d,d+2}\}$ is a total dominating set in $\texttt {S}(T)$, so

$$\begin{aligned} \gamma _t(\texttt {S}(T))\le |D|=|D^{\prime }|+1=\gamma (\texttt {S}(T-\{v_{d+2}\}))=n-1+1=n, \end{aligned}$$

and, by Proposition 1, we have the result.

Case 2. We suppose that $\delta (v_d)=2$. If $D^{\prime }$ is a minimum total dominating set in $\texttt {S}(T-\{v_d,v_{d+1}\})$, then $D=D^{\prime }\cup \{v_d,v^{d,d+1}\}$ is a total dominating set in $\texttt {S}(T)$, thus

$$\begin{aligned} \gamma _t(\texttt {S}(T))\le |D|=|D^{\prime }|+2= \gamma _t(\texttt {S}(T-\{v_d,v_{d+1}\}))+2= n-2+2=n, \end{aligned}$$

and, by Proposition 1, we have the result. $\square$

3 Total Domination Number of $\texttt {R}(T)$

We will use the following notation. If $G=(V,E)$, where $V=\{v_1,\ldots ,v_n\}$, then $\texttt {R}(G)=(V^{\prime },E^{\prime })$, where $V^{\prime }=V\cup \{v^{i,j}:v_iv_j\in E\}$ and $E^{\prime }=E\cup \{v_iv^{i,j},v_jv^{i,j}:v_iv_j\in E\}$ (we will consider $v^{i,j}=v^{j,i}$).

A total dominating set D of a graph G is called a total co-independent dominating set if the set of vertices $V {\setminus } D$ is a non-empty independent set. The minimum cardinality of any total co-independent dominating set is the total co-independent domination number, and it is denoted by $\gamma _{t,coi}(G)$ (see [4, 14]).

Proposition 3

[12] For any graph G with order $n\ge 3$ we have

$$\begin{aligned} \gamma _t(\texttt {R} (G))=\gamma _{t,coi}(G). \end{aligned}$$

It was shown in [14] that, for any path $P_n$ with n vertices, $\gamma _{t,coi}(P_n)=\left\lfloor \frac{2n}{3}\right\rfloor$, so we have that $\gamma _t(\texttt {R}(P_n))=\left\lfloor \frac{2n}{3}\right\rfloor$. It is also known (see [4]) that $\gamma _{t,coi}(G)\le n-1$, and $\gamma _{t,coi}(G)\le n-2$ when $n\ge 4$ and $\delta =1$. We improve these upper bounds for trees. A leaf in a graph $G=(V,E)$ is a vertex $v\in V$ such that $\delta (v)= 1$. A support vertex in G is a non-leaf vertex adjacent to a leaf. A hair in G is an edge $uv\in E$ such that $\delta (u)=2$ and $\delta (v)=1$.

Theorem 4

For any tree T with order n we have

$$\begin{aligned} 2\le \gamma _t(\texttt {R} (T))\le \frac{2n}{3}. \end{aligned}$$

Proof

The lower bound is obvious. Let us prove that $\frac{2n}{3}$ is an upper bound by induction. It can be checked that the inequality is true for any tree T with $n\le 6$. By Proposition 3, we suppose that $\gamma _{t,coi}(T^{\prime })\le \frac{2n^{\prime }}{3}$ for any tree $T^{\prime }$ with order $n^{\prime }<n$. If there exists a support vertex $v\in V(T)$ with two leaves $v_1,v_2$ and we take a minimum total co-independent dominating set $D^{\prime }$ in $T-\{v_1\}$, since $v\in D^{\prime }$, $D^{\prime }$ is also a total co-independent dominating set in T, then

$$\begin{aligned} \gamma _{t,coi}(T)\le D^{\prime }\le \frac{2(n-1)}{3}< \frac{2n}{3}. \end{aligned}$$

Therefore, we assume that every support vertex has only one leaf. Let $v_1,v_{d+1} \in V(T)$ such that $d(v_1,v_{d+1})=diam(T)=d$ and let $v_1,v_2,\ldots ,v_{d+1}$ be the shortest path from $v_1$ to $v_{d+1}$. If $\delta (v_{d-1})\ge 3$, since every vertex $u\in N(v_{d-1}){\setminus }\{v_{d-2},v_d\}$ is a leaf or a support vertex with only one leaf, we can take a minimum total co-independent dominating set $D^{\prime }$ in $T-\{v_d,v_{d+1}\}$ such that $v_{d-1}\in D^{\prime }$, then $D=D^{\prime }\cup \{v_d\}$ is a total co-independent dominating set in T, and

$$\begin{aligned} \gamma _{t,coi}(T)\le D=D^{\prime }+1\le \frac{2(n-2)}{3}+1<\frac{2n}{3}. \end{aligned}$$

If $\delta (v_{d-1})=2$ and we take a minimum total co-independent dominating set $D^{\prime }$ in $T-\{v_{d-1},v_d,v_{d+1}\}$, we have that $D=D^{\prime }\cup \{v_{d-1},v_d\}$ is a total co-independent dominating set in T, and

$$\begin{aligned} \gamma _{t,coi}(T)\le D=D^{\prime }+2\le \frac{2(n-3)}{3}+2= \frac{2n}{3}. \end{aligned}$$

$\square$

Theorem 5

For any tree T with order $n\ge 4$ and l leaves we have

$$\begin{aligned} \frac{2(n-l+1)}{3}\le \gamma _t(\texttt {R} (T))\le \frac{7n-3l}{9}. \end{aligned}$$

Proof

Both inequalities can be checked for any tree T with $4\le n\le 6$. Firstly, let us prove that $\frac{7n-3l}{9}$ is an upper bound by induction on the number of vertices. By Proposition 3, we suppose that $\gamma _{t,coi}(T^{\prime })\le \frac{7n^{\prime }-3l^{\prime }}{9}$ for any tree $T^{\prime }$ with order $n^{\prime }<n$. If there exists a support vertex $v\in V(T)$ with two leaves $v_1,v_2$ and we take a minimum total co-independent dominating set $D^{\prime }$ in $T-\{v_1\}$, since $v\in D^{\prime }$, $D^{\prime }$ is also a total co-independent dominating set in T, then

$$\begin{aligned} \gamma _{t,coi}(T)\le D^{\prime }\le \frac{7(n-1)-3(l-1)}{9}=\frac{7n-3l-4}{9}< \frac{7n-3l}{9}. \end{aligned}$$

Therefore, we assume that every support vertex has only one leaf. Let $v_1,v_{d+1} \in V(T)$ such that $d(v_1,v_{d+1})=diam(T)=d$ and let $v_1,v_2,\ldots ,v_{d+1}$ be the shortest path from $v_1$ to $v_{d+1}$. If $\delta (v_{d-1})\ge 3$, we take a minimum total co-independent dominating set $D^{\prime }$ in $T-\{v_d,v_{d+1}\}$ such that $v_{d-1}\in D^{\prime }$, then $D=D^{\prime }\cup \{v_d\}$ is a total co-independent dominating set in T, and

$$\begin{aligned} \gamma _{t,coi}(T)\le D=D^{\prime }+1\le \frac{7(n-2)-3(l-1)}{9}+1< \frac{7n-3l}{9}. \end{aligned}$$

If $\delta (v_{d-1})=2$ and we take a minimum total co-independent dominating set $D^{\prime }$ in $T-\{v_{d-1},v_d,v_{d+1}\}$, we have that $D=D^{\prime }\cup \{v_{d-1},v_d\}$ is a total co-independent dominating set in T, and

$$\begin{aligned} \gamma _{t,coi}(T)\le D=D^{\prime }+2\le \frac{7(n-3)-3(l-1)}{9}+2= \frac{7n-3l}{9}. \end{aligned}$$

Now, let us prove that $\frac{2(n-l+1)}{3}$ is a lower bound also by induction on n. We suppose that $\gamma _{t,coi}(T^{\prime })\ge \frac{2(n^{\prime }-l^{\prime }+1)}{3}$ for any tree $T^{\prime }$ with order $n^{\prime }<n$ and $l^{\prime }$ leaves. If there exists a support vertex $v\in V(T)$ with two leaves $v_1,v_2$ and we take a minimum total co-independent dominating set D in T such that $v_1\notin D$, then D is also a total co-independent dominating set in $T-\{v_1\}$, then

$$\begin{aligned} \gamma _{t,coi}(T)= D\ge \gamma _{t,coi}(T-\{v_1\})\ge \frac{2((n-1)-(l-1)+1)}{3}=\frac{2(n-l+1)}{3}. \end{aligned}$$

Therefore, we assume that every support vertex has only one leaf. Let $v_1, v_{d+1}\in V(T)$ such that $d(v_1,v_{d+1})=diam(T)=d$ and let $v_1,v_2,\ldots ,v_{d+1}$ be the shortest path from $v_1$ to $v_{d+1}$. If $v_{d-1}$ is adjacent to a leaf $w_1$, we can take a minimum total co-independent dominating set D in T such that $v_{d-1},v_d\in D$, consequently, D is a total co-independent dominating set in $T-\{w_1\}$ and

$$\begin{aligned} \gamma _{t,coi}(T)= D\ge \gamma _{t,coi}(T-\{w_1\})\ge \frac{2((n-1)-(l-1)+1)}{3}=\frac{2(n-l+1)}{3}. \end{aligned}$$

If $v_{d-1}$ is adjacent to a hair $u_1u_2$ different from $v_dv_{d+1}$, then we can take a minimum total co-independent dominating set D in T such that $v_{d-1},v_d,u_1\in D$, consequently, $D{\setminus }\{u_1\}$ is a total co-independent dominating set in $T^{\prime }=T-\{u_1,u_2\}$ and

$$\begin{aligned} \gamma _{t,coi}(T)= D \ge \gamma _{t,coi}(T^{\prime })+1\ge \frac{2((n-2)-(l-1)+1)}{3}+1>\frac{2(n-l+1)}{3}. \end{aligned}$$

Therefore, it remains to study the case when $\delta (v_{d-1})=2$. Let D be a minimum total co-independent dominating set D in T. If $v_{d-2}\notin D$, then $D{\setminus }\{v_{d-1},v_d\}$ is a total co-independent dominating set in $T^{\prime }=T-\{v_{d-1},v_d, v_{d+1}\}$, consequently, since $v_{d-2}$ could be a leaf in $T^{\prime }$,

$$\begin{aligned} \gamma _{t,coi}(T)= D \ge \gamma _{t,coi}(T^{\prime })+2\ge \frac{2((n-3)-l)+1)}{3}+2=\frac{2(n-l+1)}{3}. \end{aligned}$$

Finally, if $v_{d-2}\in D$, since $(D{\setminus }\{v_{d+1}\})\cup \{v_{d-1}$ is a total co-independent dominating set in T, we can assume that $v_{d-1},v_d\in D$ and $v_{d+1}\notin D$. Then, $D{\setminus }\{v_d\}$ is a total co-independent dominating set in $T^{\prime }=T-\{v_{d+1}\}$, consequently,

$$\begin{aligned} \gamma _{t,coi}(T)= D \ge \gamma _{t,coi}(T^{\prime })+1\ge \frac{2((n-1)-l+1)}{3}+1>\frac{2(n-l+1)}{3}. \end{aligned}$$

$\square$

If we consider a tree T with vertices $V=\{v,w_1,w_2,z_1,z_2,z_3,\ldots ,$ $z_{3r-2},z_{3r-1},$ $z_{3r}\}$, such that $\delta (w_2)=1$, $N(w_1)=\{w_2,v\}$, $N(z_{3i-2})=\{v,z_{3i-1}\}$, $N(z_{3i-1})=\{z_{3i-2},z_{3i}\}$ and $\delta (z_{3i})=1$, for every $i\in \{1,2,\ldots ,r\}$, then we have that $n=3(r+1)$, $l=r+1$ and $\gamma _{t,coi}(T)=2(r+1)=\frac{7n-3l}{9}$ (see Fig. 2).

Fig. 2

Minimum total co-independent dominating set in T

The lower bound in Theorem 5 is attained in any path with $n=3k+1$ vertices.

Corollary 6

For any tree T with order $n\ge 4$ and l leaves we have

$$\begin{aligned} \max \left\{ 2,\left\lceil \frac{2(n-l+1)}{3}\right\rceil \right\} \le \gamma _t(\texttt {R} (T))\le \min \left\{ \left\lfloor \frac{2n}{3}\right\rfloor ,\left\lfloor \frac{7n-3l}{9}\right\rfloor \right\} . \end{aligned}$$

4 Total Domination Number of $\texttt {Q}(G)$

We will use the following notation. If $G=(V,E)$, where $V=\{v_1,\ldots ,v_n\}$, then $\texttt {Q}(G)=(V^{\prime },E^{\prime })$, where $V^{\prime }=V\cup \{v^{i,j}:v_iv_j\in E\}$ and $E^{\prime }=\{v_iv^{i,j},v_jv^{i,j}:v_iv_j\in E\}\cup \{v^{i,j}v^{j,k}:v_iv_j,v_jv_k\in E\}$ (we will consider $v^{i,j}=v^{j,i}$).

Given a graph G, we denote by $p_3(G)$ the maximum number of independent paths with three vertices we can get in G, that is, without vertices in common. The following result was proved in [12].

Theorem 7

Let G be a graph with order n. Then,

$$\begin{aligned} \gamma _t(\texttt {Q} (G))=n-p_3(G). \end{aligned}$$

Now, let us see a lower and upper bound for this parameter when the graph is a tree. For any vertex $v\in V$ we denote by L(v) the number of leaves adjacent to v, by H(v) the number of hairs adjacent to v and $lh(v)=L(v)+2H(v)$. We denote by $A_G=\{v\in V: lh(v)\ge 3\}$ and $c_G=\sum _{v\in A_G}(lh(v)-2)$. Then, we have the following theorem for trees.

Theorem 8

Let T be a tree with order n. Then,

$$\begin{aligned} \frac{2n+c_T}{3}\le \gamma _{t}(\texttt {Q} (T))\le \frac{4n-2+c_T}{5}. \end{aligned}$$

Proof

Since $\gamma _{t}(\texttt {Q}(T))=n-p_3(T)$, it is enough to prove that $\frac{n+2-c_T}{5}\le p_3(T)\le \frac{n-c_T}{3}$. Let T be a graph with order n, $t=p_3(T)$ and $M_1,M_2,\ldots ,M_t\subseteq V(T)$ the corresponding sets containing the three vertices of any independent path in T. Let $v\in V(T)$ be a vertex such that $lh(v)\ge 3$, and let $w_1,\ldots ,w_r$ and $z_1z_2,z_3z_4,\ldots ,z_{2s-1}z_{2s}$ be the corresponding leaves and hairs adjacent to v. Since $lh(v)\ge 3$, there exists $i\in \{1,2,\ldots ,t\}$ such that $v\in M_i$, then, at most, two vertices $u,u^{\prime }\in \{w_1,\ldots ,w_r,z_1,z_2,z_3,\ldots ,z_{2s}\}$ belong to $M_i$, therefore, the subtree $T^{\prime }$ induced by $V(T){\setminus }(\{w_1,\ldots ,w_r,z_1,z_2,z_3,\ldots ,z_{2s}\})\cup \{u,u^{\prime }\}$ satisfies $p_3(T^{\prime })=p_3(T)$ and $n(T^{\prime })\le n-(lh(v)-2)$. If $k=|A_T|$ and we do the same with every vertex in $A_T$ we obtain a subtree $T^{(k)}$ such that $p_3(T^{(k)})=p_3(T)$ and $n(T^{(k)})\le n-c_T$. Since $p_3(T^{(k)})\le \frac{n(T^{(k)})}{3}$, we get the upper bound. Moreover, since $p_3(T^{(k)})=p_3(T)$, to prove the lower bound, we can do it only for trees such that $c_T=0$ or, equivalently, $A_T=\emptyset$.

Let us prove by induction that, for any tree T with order n and $A_T=\emptyset$, we have $p_3(T)\ge \frac{n+2}{5}$. Note that, if T has $n\le 5$ vertices, then $A_T\ne \emptyset$. The result can be checked for any tree with order $n=6$, so we suppose that it is true for any tree with order $n^{\prime }<n$ and we consider a tree T with order n and $A_T=\emptyset$. Let $v_1,v_{d+1}\in V(T)$ such that $d(v_1,v_{d+1})=diam(T)=d$ and let $v_1,v_2,\ldots ,v_{d+1}$ be the shortest path from $v_1$ to $v_{d+1}$, so $1\le L(v_d)\le 2$, and we study several cases.

Case 1. We suppose that $L(v_d)=2$. If $N(v_d)=\{v_{d-1},v_{d+1},w\}$, $T_3=T-\{v_d,v_{d+1},w\}$ and $A_{T_3}=\emptyset$, by induction we have

$$\begin{aligned} p_3(T)\ge p_3(T_3)+1\ge \frac{(n-3)+2}{5}+1=\frac{n+4}{5}>\frac{n+2}{5}. \end{aligned}$$

Therefore, we suppose that $A_{T_3}\ne \emptyset$. This situation is possible only when $\delta (v_{d-1})=2$ or $\delta (v_{d-1})=3$ and $v_{d-1}$ is adjacent to a leaf.

Case 1.1. We suppose that $\delta (v_{d-1})=2$. Since $A_{T_3}\ne \emptyset$, we have two new cases.

Case 1.1.1. We suppose that that $lh(v_{d-2})=2$. In such a case, if $T_4=T-\{v_{d-1},v_d,v_{d+1},w\}$, we have that $A_{T_4}=\emptyset$ and, by induction

$$\begin{aligned} p_3(T)\ge p_3(T_4)+1\ge \frac{(n-4)+2}{5}+1=\frac{n+3}{5}>\frac{n+2}{5}. \end{aligned}$$

Case 1.1.2. We suppose that $\delta (v_{d-2})=2$ and $1\le lh(d_{v-3})\le 2$. In such a case, if we consider $T_5=T-\{v_{d-2},v_{d-1},v_d,v_{d+1},w\}$, by induction we have

$$\begin{aligned} p_3(T)\ge p_3(T_5)+1\ge \frac{(n-5)+2}{5}+1=\frac{n+2}{5}. \end{aligned}$$

Case 1.2. We suppose that $\delta (v_{d-1})=3$ and $v_{d-1}$ is adjacent to a leaf $w^{\prime }$. Since $A_{T_3}\ne \emptyset$, then $1\le lh(d_{v-2})\le 2$. If $lh(d_{v-2})= 2$ and we consider $T_5=T-\{v_{d-1},w^{\prime },v_d,v_{d+1},w\}$, then we have that $A_{T_5}=\emptyset$ and, by induction, that

$$\begin{aligned} p_3(T)\ge p_3(T_5)+1\ge \frac{(n-5)+2}{5}+1=\frac{n+2}{5}. \end{aligned}$$

If $lh(d_{v-2})= 1$ and we consider $T_4=T-\{w^{\prime },v_d,v_{d+1},w\}$, then we have that $A_{T_4}=\emptyset$, $p_3(T)\ge p_3(T_4)+1$ and, by induction, that

$$\begin{aligned} p_3(T)\ge p_3(T_4)+1\ge \frac{(n-4)+2}{5}+1=\frac{n+3}{5}>\frac{n+2}{5}. \end{aligned}$$

Case 2. We suppose that $L(v_d)=1$. Since $lh(v_{d-1})=2$, then, by Case 1, we can assume that $\delta (v_{d-1})=2$. If we consider $T_3=T-\{v_{d-1},v_d,v_{d+1}\}$ and $A_{T_3}=\emptyset$, by induction we have

$$\begin{aligned} p_3(T)\ge p_3(T_3)+1\ge \frac{(n-3)+2}{5}+1=\frac{n+4}{5}>\frac{n+2}{5}. \end{aligned}$$

Therefore, we suppose that $A_{T_3}\ne \emptyset$. This situation is possible only when $\delta (v_{d-2})=2$ or $\delta (v_{d-2})=3$ and $v_{d-2}$ is adjacent to a leaf. Hence, we can repeat the cases above with $v_{d-2}$ instead of $v_{d-1}$. $\square$

There exist infinite families of trees attaining the bounds given in Theorem 8. Let $T_k$ be the tree containing k paths of three vertices $\{v_1^i,v_2^i,v_3^i\}$, for $1\le i\le k$, as subgraphs, and the edges $\{v_2^1v_2^2,v_2^2v_2^3,\ldots ,v_2^{k-1}v_2^k\}$. It can be checked that $A_{T_k}=\emptyset$, $c_{T_k}=0$ and $p_3(T_k)=k=\frac{3k}{3}=\frac{n-c_T}{3}$ (see Fig. 3 on the left). Let $T^{\prime }_k$ be the tree containing k paths of five vertices $\{v_1^i,v_2^i,v_3^i,v_4^i,v_5^i\}$, for $1\le i\le k$, as subgraphs, and the edges $\{v_3^1v_1^2,v_3^2v_1^3,\ldots ,v_3^{k-1}v_1^k\}$. It can be checked that $A_{T^{\prime }_k}=\{v_3^1\}$, $lh(v_3^1)=4$, $c_{T^{\prime }_k}=2$ and $p_3(T^{\prime }_k)=k=\frac{5k+2-2}{5}=\frac{n+2-c_T}{5}$ (see Fig. 3 on the right).

Fig. 3

Two families of graphs attaining the bounds given in Theorem 8

5 Total Domination Number of $\texttt {T}(G)$

The total graph $\texttt {T}(G)$ of a graph G is a graph whose vertex set is $V\cup E$, with two vertices of $\texttt {T}(G)$ being adjacent if and only if the corresponding elements of G are adjacent or incident. Specifically, if $G=(V,E)$, where $V=\{v_1,\ldots ,v_n\}$, then $\texttt {T}(G)=(V^{\prime },E^{\prime })$, where $V^{\prime }=V\cup \{v^{i,j}:v_iv_j\in E\}$ and $E^{\prime }=E\cup \{v_iv^{i,j},v_jv^{i,j}:v_iv_j\in E\}\cup \{v^{i,j}v^{j,k}:v_iv_j,v_jv_k\in E\}$ (we will consider $v^{i,j}=v^{j,i}$).

Lemma 9

For any graph $G=(V,E)$, where $V=\{v_1,\ldots ,v_n\}$, there exists a minimum total dominating set D in $\texttt {T} (G)$ such that no vertex in $D\cap \{v_1,\ldots ,v_n\}$ is adjacent to a vertex in $D\cap \{v^{i,j}:v_iv_j\in E\}$.

Proof

Let D be a minimum total dominating set in $\texttt {T} (G)$ with the minimum number of vertices in $\{v^{i,j}:v_iv_j\in E\}$. If $v_i$ and $v^{i,j}$ belong to D, then we take $D^{\prime }=(D{\setminus }\{v^{i,j}\})\cup \{v_j\}$. The vertex $v^{i,j}$ is adjacent to $v_i,v_j$ and some vertices $v^{r,k}$ and, if $v^{i,j}$ is adjacent to $v^{r,k}$, it is because the edges $v_iv_j$ and $v_rv_k$ share a vertex, then $v_i$ or $v_j$ must be adjacent to $v^{r,k}$. Consequently, $D^{\prime }$ is a total dominating set in $\texttt {T}(G)$ with less vertices in $\{v^{i,j}:v_iv_j\in E\}$ than D, a contradiction. $\square$

Proposition 10

In any path $P_n$ with n vertices we have

$$\begin{aligned} \gamma _t(\texttt {T} (P_n))=\left\{ \begin{array}{cl} \left\lceil \frac{4n}{7}\right\rceil -1 &{} \quad { \mathrm {if}} \quad n\equiv 4\,(\mathrm{{mod}} \,\,7) \\ \left\lceil \frac{4n}{7}\right\rceil &{} \quad \mathrm{{otherwise}.} \\ \end{array}\right. \end{aligned}$$

Proof

We denote

$$\begin{aligned} f(n)=\left\{ \begin{array}{cl} \left\lceil \frac{4n}{7}\right\rceil -1 &{} \quad \text{ if } \quad n\equiv 4\,(\text {mod} \,\,7) \\ \left\lceil \frac{4n}{7}\right\rceil &{} \quad \text{ otherwise. } \\ \end{array}\right. \end{aligned}$$

It can be easily checked that $\gamma _t(\texttt {T}(P_n))=f(n)$ for every $n\le 10$ (see Figs. 4, 5).

Fig. 4

Minimum total dominating set in $\texttt {T}(P_3),\texttt {T}(P_4),{} \texttt {T}(P_5)$ and $\texttt {T}(P_{6})$

Fig. 5

Minimum total dominating set in $\texttt {T}(P_{7}),\texttt {T}(P_{8}),{} \texttt {T}(P_{9})$ and $\texttt {T}(P_{10})$

Since the choice at the right end of the $P_{10}$ is equal to the choice in $P_3$ we can generalize this process to obtain a total dominating set in $\texttt {T}(P_n)$ with cardinality f(n), so $\gamma _t(\texttt {T}(P_n))\le f(n)$.

Now, let us prove that $\gamma _t(\texttt {T}(P_n))\ge f(n)$ by induction on n. We suppose that the inequality is true for any $n^{\prime }<n$ and we consider the path $P_n$ with vertices $v_1,v_2,\ldots ,v_n$ ($n\ge 11$) and a minimum total dominating set D in $\texttt {T}(P_n)$ given by Lemma 9. We consider the paths $P_{n-7}$ and $P_7$ with vertices $v_1,v_2,\ldots ,v_{n-7}$ and $v_{n-6},v_{n-5},\ldots ,v_n$, respectively, then we denote the sets $D_1=D\cap V(\texttt {T}(P_{n-7}))$ and $D_2=D\cap V(\texttt {T}(P_7))$. We distinguish several cases.

Case 1. If $v_{n-6}\in D$, it is clear that $D_2\ge 5$ and, since $D_1\cup \{v_{n-8}\}$ is a total dominating set in $\texttt {T}(P_{n-7})$, by induction, we have

$$\begin{aligned} D=D_1+D_2\ge D_1 +1+4\ge f(n-7)+4=f(n). \end{aligned}$$

Case 2. If $v_{n-6}\notin D$ and $v^{n-7,n-6}\in D$, then we have that $D_2 \ge 4$ and, since $D_1\cup \{v_{n-8}\}$ is a total dominating set in $\texttt {T}(P_{n-7})$, by induction, we have

$$\begin{aligned} D = D_1 +1+ D_2 \ge D_1 +1+4\ge f(n-7)+4=f(n). \end{aligned}$$

Case 3. If $v_{n-6},v^{n-7,n-6}\notin D$, then we have that $D_2 \ge 4$ and, since $D_1$ is a total dominating set in $\texttt {T}(P_{n-7})$, by induction, we have

$$\begin{aligned} D = D_1 + D_2 \ge D_1 +4\ge f(n-7)+4=f(n). \end{aligned}$$

$\square$

The following result was proved in [13], but for the sake of completeness and clarity, we include the proof here.

Theorem 11

For any graph G with order n we have

$$\begin{aligned} \gamma _t(\texttt {T} (G))\ge \gamma _t(G). \end{aligned}$$

Proof

We take a minimum total dominating set D in $\texttt {T}(G)$ given by Lemma 9. Let us see that we can move the vertices in $D\cap \{v^{i,j}:v_iv_j\in E(G)\}$ to vertices in V(G) obtaining a total dominating set in G. We consider two adjacent vertices $v^{i,j},v^{j,k}\in D$. If there exist two vertices $v_r,v_t\in N(v_i){\setminus }\{v_j\}$ such that $v^{r,i},v^{t,i}\in D$, then we take $D^{\prime }=(D{\setminus }\{v^{i,j},v^{j,k}\})\cup \{v_j,v_k\}$, which is also a total dominating set in $\texttt {T}(G)$. If there exists only one vertex $v_r{\in }N(v_i){\setminus }\{v_j\}$ such that $v^{r,i}{\in }D$, we consider two cases. Firstly, if any vertex $v_k\in N(v_r){\setminus }\{v_i\}$ satisfies that $v^{k,r}\notin D$, then we take $D^{\prime }=(D{\setminus }\{v^{i,j},v^{j,k},v^{r,k}\})\cup \{v_i,v_j,v_k\}$. Let us observe that $D^{\prime }$ might not be a total dominating set in $\texttt {T}(G)$, but it dominates any vertex in V(G). Secondly, if there exists a vertex $v_k\in N(v_r){\setminus }\{v_i\}$ such that $v^{k,r}\in D$, then we take $D^{\prime }=(D{\setminus }\{v^{i,j},v^{j,k}\})\cup \{v_j,v_k\}$, which is also a total dominating set in $\texttt {T}(G)$. If we do this with any adjacent vertices in $D\cap \{v^{i,j}:v_iv_j\in E(G)\}$ we obtain a set $D^{\prime }\subseteq V(G)$, which is a total dominating set in G satisfying $D^{\prime } = D =\gamma _t(\texttt {T}(G))$. Therefore, $\gamma _t(G)\le \gamma _t(\texttt {T}(G))$. $\square$

By Theorem 11 we have $\gamma _t(G)\le \gamma _t(\texttt {T}(G))\le \gamma _t(\texttt {R}(G))=\gamma _{t,coi}(G)$. Therefore, if $\gamma _t(G)=\gamma _{t,coi}(G)$, then $\gamma _t(\texttt {T}(G))= \gamma _t(\texttt {R}(G))$.

Proposition 12

If T is a tree with diameter $d\le 6$, then $\gamma _t(T)=\gamma _{t,coi}(T)$.

Proof

By absurdum, we suppose that $\gamma _t(T)<\gamma _{t,coi}(T)$, it means, for any minimum total dominating set D in T, there exists an edge uv such that $u,v\in V{\setminus } D$. We take the minimum total dominating set D in T such that the number of edges in $V{\setminus } D$ is the smallest one. If $u,v\in V{\setminus } D$ are two adjacent vertices, since T does not contain cycles, there exist $v_1,v_2,u_1,u_2\in D$ such that $v_2v_1,v_1v,uu_1,u_1u_2\in E(T)$. If $v_2$ (or $u_2$) is a leaf, then $D^{\prime }=(D{\setminus }\{v_2\})\cup \{v\}$ is also a minimum total dominating set, but the number of edges in $V{\setminus } D^{\prime }$ is smaller the number of edges in $V{\setminus } D$, a contradiction. Consequently, $v_2$ and $u_2$ are not leaves and $d\ge 7$, which is also a contradiction. $\square$

Corollary 13

If T is a tree with diameter $d\le 6$, then $\gamma _t(\texttt {T} (T))= \gamma _t(\texttt {R} (T))$.

We saw that $\gamma _t(\texttt {R}(T))\le \min \left\{ \frac{2n}{3},\frac{7n-3l}{9} \right\}$ for any tree with order n and l leaves, and there exist some trees T (see Fig. 6) such that $\gamma _t(\texttt {T}(T))=\gamma _t(\texttt {R}(T))=\frac{2n}{3}=\frac{7n-3l}{9}$. Therefore, we cannot give a smaller upper bound for $\gamma _t(\texttt {T}(T))$ using these two parameters.

Fig. 6

Minimum total co-independent dominating set in two trees

It is known (see [6]) that, for any tree T with order n and l leaves, $\gamma _t(T)\ge \frac{n-l+2}{2}$, so this is a direct lower bound for $\gamma _t(\texttt {T}(T))$. We present in the next theorem a bigger lower bound when T is different from a star.

Theorem 14

For any tree T with order n and l leaves we have

$$\begin{aligned} \gamma _t(\texttt {T} (T))\ge \frac{4(n-l)+6}{7}. \end{aligned}$$

Proof

Let us prove the result by induction. The inequality can be checked for any tree T with $n\le 5$. We suppose that $\gamma _t(\texttt {T}(T^{\prime }))\ge \frac{4(n^{\prime }-l^{\prime })+6}{7}$ for any tree $T^{\prime }$ with order $n^{\prime }<n$ and $l^{\prime }$ leaves, and we take a tree T with order n and l leaves. If there exists a support vertex $v_1\in V(T)$ with $\delta (v_1)\ge 3$ and $v_2$ is one of its leaves, since $N[v_2,v^{1,2}]\subseteq N[v_1]$, we can take a minimum total dominating set D in $\texttt {T}(T)$ such that $v_2,v^{1,2}\notin D$. If we denote $T^{\prime }=T-\{v_2\}$, then D is also a total dominating set in $\texttt {T}(T^{\prime })$, then

$$\begin{aligned} \gamma _t(\texttt {T}(T))= D \ge \gamma _t(\texttt {T}(T^{\prime }))\ge \frac{4((n-1)-(l-1))+6}{7}=\frac{4(n-l)+6}{7}. \end{aligned}$$

Therefore, we assume that every support vertex has degree 2. Let $v_1,v_{d+1}\in V(T)$ such that $d(v_1,v_{d+1})=diam(T)=d$ and let $v_1,v_2,\ldots ,v_{d+1}$ be the shortest path from $v_1$ to $v_{d+1}$. If $v_{d-1}$ is adjacent to a hair $u_1u_2$ different from $v_dv_{d+1}$, then we can take a minimum total dominating set D in $\texttt {T}(T)$ such that $v_{d-1},v_d,u_1\in D$ and $T^{\prime }=T-\{u_1,u_2\}$, consequently, $D{\setminus }\{u_1\}$ is a total dominating set in $T(T^{\prime })$ and

$$\begin{aligned} \gamma _t(\texttt {T}(T))= D \ge \gamma _t(\texttt {T}(T^{\prime }))+1\ge \frac{4((n-2)-(l-1))+6}{7}+1>\frac{4(n-l)+6}{7}. \end{aligned}$$

Therefore, it remains to study the case when $\delta (v_{d-1})=2=\delta (v_d)$ (for any path with length equal to d joining two vertices u and v such that $d(u,v)=d$).

Case 1. There exists a hair $w_1w_2$ such that $v_{d-2}w_1\in E(T)$. In such a case, we can choose a minimum total dominating set D in $\texttt {T}(T)$ containing $v_{d-2}$ and $w_1$ and, if $T^{\prime }=T-\{v_{d-1},v_d,v_{d+1}\}$, $D{\setminus }\{v_{d-1},v_d\}$ is a total dominating in $\texttt {T}(T^{\prime })$, then

$$\begin{aligned} \gamma _t(\texttt {T}(T))= D \ge \gamma _t(\texttt {T}(T^{\prime }))+2\ge \frac{4((n-3)-(l-1))+6}{7}+2>\frac{4(n-l)+6}{7}. \end{aligned}$$

Case 2. There exists a path with three consecutive vertices $w_1,w_2,w_3$ such that $v_{d-2}w_1\in E(T)$. In such a case, we can choose a minimum total dominating set D in $\texttt {T}(T)$ containing $v_{d-1},v_d,w_1$ and $w_2$ and, if $T^{\prime }=T-\{w_1,w_2,w_3\}$, $D{\setminus }\{w_1,w_2\}$ is a total dominating in $\texttt {T}(T^{\prime })$, then

$$\begin{aligned} \gamma _t(\texttt {T}(T))= D \ge \gamma _t(\texttt {T}(T^{\prime }))+2\ge \frac{4((n-3)-(l-1))+6}{7}+2>\frac{4(n-l)+6}{7}. \end{aligned}$$

Case 3. We suppose that $\delta (v_{d-2})=2$. In this situation, we can choose D such that $v_{d-1},v_d\in D$ and $v_{d-2}\notin D$.

Case 3.1. If $\delta (v_{d-3})\ge 3$, we can take $T^{\prime }=T-\{v_{d-2},v_{d-1},v_d,v_{d+1}\}$, to have that $D{\setminus }\{v_{d-1},v_d\}$ is a total dominating in $\texttt {T}(T^{\prime })$ and

$$\begin{aligned} \gamma _t(\texttt {T}(T))= D \ge \gamma _t(\texttt {T}(T^{\prime }))+2\ge \frac{4((n-4)-(l-1))+6}{7}+2>\frac{4(n-l)+6}{7}. \end{aligned}$$

Case 3.2. We suppose that $\delta (v_{d-3})=2$. If $v_{d-3}\in D$, we can take $T^{\prime }=T-\{v_{d-1},v_d,v_{d+1}\}$, to have that $D{\setminus }\{v_{d-1},v_d\}$ is a total dominating in $\texttt {T}(T^{\prime })$ and

$$\begin{aligned} \gamma _t(\texttt {T}(T))= D \ge \gamma _t(\texttt {T}(T^{\prime }))+2\ge \frac{4((n-3)-l)+6}{7}+2>\frac{4(n-l)+6}{7}. \end{aligned}$$

Thus, we suppose that $v^{d-4,d-3}\in D$ and, by Lemma 9, that $v_{d-4},v_{d-3}\notin D$. Consequently, $v_{d-4}$ does not have any hair adjacent to it.

Case 3.2.1. We suppose that $\delta (v_{d-4})\ge 3$. If there exists $v_i$ adjacent to $v_{d-4}$ such that $v^{d-4,i}\notin D$, then we denote $e=v_{d-4}v_i$ and we consider the two trees in $T-e$, $T_1$ containing $v_{d-4}$ and $T_2$ containing $v_i$. Note that $D\cap V(\texttt {T}(T_1))$ is a total dominating set in $\texttt {T}(T_1)$ and $D\cap V(\texttt {T}(T_2))$ is a total dominating set in $\texttt {T}(T_2)$, then, by induction, we have

$$\begin{aligned} \gamma _t(\texttt {T}(T))= & {} D \ge \gamma _t(\texttt {T}(T_1))+\gamma _t(\texttt {T}(T_2))\\\ge & {} \frac{4(n_1-l_1)+6}{7}+\frac{4(n_2-l_2)+6}{7}=\frac{4((n_1+n_2)-(l_1+l_2))+12}{7}\\= & {} \frac{4((n_1+n_2)-(l_1+l_2-1))+8}{7}\ge \frac{4(n-l)+8}{7} >\frac{4(n-l)+6}{7}. \end{aligned}$$

If $v^{d-4,i}\in D$ for every vertex $v_i$ adjacent to $v_{d-4}$, then $D^{\prime }= D{\setminus }\{v^{d-4,d-3},v_{d-1},$ $v_d\}$ is a total dominating set in $\texttt {T}(T^{\prime })$, where $T^{\prime }=T-\{v_{d-3},v_{d-2},v_{d-1},v_d,v_{d+1}\}$, consequently,

$$\begin{aligned} \gamma _t(\texttt {T}(T))= D \ge \gamma _t(\texttt {T}(T^{\prime }))+3\ge \frac{4((n-5)-(l-1))+6}{7}+3>\frac{4(n-l)+6}{7}. \end{aligned}$$

Case 3.2.2. We suppose that $\delta (v_{d-4})=2$. In such a case, $v^{d-5,d-4}\in D$ and $v_{d-5}\notin D$. So we know that $v_{d-5}$ does not have any hair adjacent to it.

Case 3.2.2.1. We suppose that $\delta (v_{d-5})\ge 3$. If there exists $v_i$ adjacent to $v_{d-5}$ such that $v^{d-5,i}\notin D$, we do the same we did in Case 3.2.1, with $v_{d-5}$ instead of $v_{d-4}$. If $v^{d-5,i}\in D$ for every vertex $v_i$ adjacent to $v_{d-5}$, then $D^{\prime }=D{\setminus }\{v^{d-5,d-4},v^{d-4,d-3},v_{d-1},v_d\}$ is a total dominating set in $\texttt {T}(T^{\prime })$, where $T^{\prime }=T-\{v_{d-4},v_{d-3},v_{d-2},v_{d-1},v_d,v_{d+1}\}$, consequently,

$$\begin{aligned} \gamma _t(\texttt {T}(T))= D \ge \gamma _t(\texttt {T}(T^{\prime }))+4\ge \frac{4((n-6)-(l-1))+6}{7}+4>\frac{4(n-l)+6}{7}. \end{aligned}$$

Case 3.2.2.2. We suppose that $\delta (v_{d-5})=2$. If $v^{d-6,d-5}\notin D$, then $D^{\prime }=D{\setminus }\{v^{d-5,d-4},v^{d-4,d-3},v_{d-1},v_d\}$ is a total dominating set in $\texttt {T}(T^{\prime })$, where $T^{\prime }=T-\{v_{d-5},v_{d-4},$ $v_{d-3},v_{d-2},v_{d-1},v_d,v_{d+1}\}$, consequently,

$$\begin{aligned} \gamma _t(\texttt {T}(T))= D \ge \gamma _t(\texttt {T}(T^{\prime }))+4\ge \frac{4((n-7)-l)+6}{7}+4=\frac{4(n-l)+6}{7}. \end{aligned}$$

If $v^{d-6,d-5}\in D$ we distinguish two new cases.

Case 3.2.2.2.1. We suppose that $\delta (v_{d-6})\ge 3$. If there exists $v_i$ adjacent to $v_{d-6}$ such that $v^{d-6,i}\notin D$, we do the same we did in Case 3.2.1, with $v_{d-6}$ instead of $v_{d-4}$. If $v^{d-6,i}\in D$ for every vertex $v_i$ adjacent to $v_{d-6}$, then $D^{\prime }=D{\setminus }\{v^{d-6,d-5},v^{d-5,d-4},v^{d-4,d-3},v_{d-1},v_d\}$ is a total dominating set in $\texttt {T}(T^{\prime })$, where $T^{\prime }=T-\{v_{d-5},v_{d-4},v_{d-3},v_{d-2},v_{d-1},v_d,v_{d+1}\}$, consequently,

$$\begin{aligned} \gamma _t(\texttt {T}(T))= D \ge \gamma _t(\texttt {T}(T^{\prime }))+5\ge \frac{4((n-7)-(l-1))+6}{7}+5>\frac{4(n-l)+6}{7}. \end{aligned}$$

Case 3.2.2.2.2. We suppose that $\delta (v_{d-6})=2$. In this situation, if we take $D^{\prime }=(D{\setminus }\{v^{d-6,d-5},v^{d-5,d-4},v^{d-4,d-3},v_{d-1},v_d\})\cup \{v_{d-7}\}$ and $T^{\prime }=T-\{v_{d-5},v_{d-4},$ $v_{d-3},v_{d-2},v_{d-1},v_d,v_{d+1}\}$, we have that $D^{\prime }$ is a total dominating set in $\texttt {T}(T^{\prime })$ and

$$\begin{aligned} \gamma _t(\texttt {T}(T))= & {} D \ge D^{\prime } +4\ge \gamma _t(\texttt {T}(T^{\prime }))+4 \ge \frac{4((n-7)-l)+6}{7}+4\\= & {} \frac{4(n-l)+6}{7}. \end{aligned}$$

$\square$

In order to characterize all the trees attaining the lower bound given in Theorem 14, we consider the following family ${\mathcal {F}}$ of trees, which we define recursively. For any $k\ge 0$ we consider the path with $7k+4$ vertices in ${\mathcal {F}}$ and we construct new graphs in the family in two ways.

• (i) For any $k\ge 0$, the path $P_{7k+4}$ with $7k+4$ vertices, belongs to ${\mathcal {F}}$.

• (ii) If $T^{\prime }\in {\mathcal {F}}$ such that $diam(T^{\prime })=7k+3$, let $v_1,v_{7k+3}\in V(T^{\prime })$ such that $d(v_1,v_{7k+4})=7k+3$ and let $v_1,v_2,\ldots ,v_{7k+4}$ be the shortest path from $v_1$ and $v_{7k+4}$. If we add a new leaf to any vertex in $\{v_2,v_3,v_9,v_{10},\ldots ,$ $v_{7k-5},v_{7k-4},v_{7k+2},v_{7k+3}\}$, then this new graph belongs to ${\mathcal {F}}$.

Theorem 15

If T is a tree with order n and l leaves, then $\gamma _t(\texttt {T} (T))=\frac{4(n-l)+6}{7}$ if and only if $T\in {\mathcal {F}}$.

Proof

If $T\in {\mathcal {F}}$ has diameter $diam(T^{\prime })=7k+3=d(v_1,v_{7k+4})$ and $v_1,v_2,\ldots ,v_{7k+4}$ is the shortest path from $v_1$ to $v_{7k+4}$, then, looking at the proof of Proposition 10, we know that $D=\{v_2,v_3,v^{5,6},v^{6,7}v_9,v_{10},\ldots ,v_{7k-5}, v_{7k-4},$ $v^{7k-2,7k-1},v^{7k-1,7k},v_{7k+2},v_{7k+3}\}$ is a total dominating set in $\texttt {T}(T)$, then

$$\begin{aligned} \frac{4(n-l)+6}{7}=\left\lceil \frac{4n}{7}\right\rceil -1=4k+2= D \ge \gamma _t(\texttt {T}(T))\ge \frac{4(n-l)+6}{7}, \end{aligned}$$

which means that D is a minimum total dominating set in $\texttt {T}(T)$.

Now, we will prove that any tree T satisfying $\gamma _t(\texttt {T}(T))=\frac{4(n-l)+6}{7}$ belongs to the family ${\mathcal {F}}$. By absurdum, we suppose that there exists a tree T such that $\gamma _t(\texttt {T}(T))=\frac{4(n-l)+6}{7}$ and $T\notin {\mathcal {F}}$. We take the tree T satisfying it with the minimum number of vertices. Let $v_1,v_{d+1}\in V(T)$ such that $d(v_1,v_{d+1})={\text {diam}}(T)=d$ and let $v_1,v_2,\ldots ,v_{d+1}$ be the shortest path from $v_1$ to $v_{d+1}$. If $v_d$ or $v_{d-1}$ is adjacent to a leaf $v\ne v_{d+1}$, looking at the beginning of the proof of Theorem 14, we know that $\gamma _t(\texttt {T}(T-\{v\}))=\frac{4((n-1)-(l-1))+6}{7}$, thus, $T-\{v\}\in {\mathcal {F}}$. But, in such a case, there exists k such that $d=7k+3$ and $v_{d-1}=v_{7k+2}$ and $v_d=v_{7k+3}$, so $T\in {\mathcal {F}}$, a contradiction. Finally, since $\gamma _t(\texttt {T}(T))=\frac{4(n-l)+6}{7}$, following the cases in the proof of Theorem 14, we have that $\delta (v_i)=2$ for every $i\in \{d-6,d-5,\ldots ,d\}$ and, if $T^{\prime }=T-\{v_{d-5},v_{d-4},v_{d-3},v_{d-2},v_{d-1},v_d,v_{d+1}\}$, then $\gamma _t(\texttt {T}(T^{\prime }))=\frac{4((n-7)-l)+6}{7}$. Therefore, $T^{\prime }\in {\mathcal {F}}$ and, consequently, $T\in {\mathcal {F}}$, a contradiction. $\square$