Proper Total Domination in Generalized Fans and Wheels

Abstract

A set S of vertices in a graph G is a total dominating set if every vertex of G is adjacent to (totally dominated by) some vertex of S. The number of vertices in S that totally dominate a vertex v of G is denoted by ${\sigma }_S\left(v\right)$. A total dominating set S is a $pt$-dominating set (or $pt$-dominating set) if ${\sigma }_S\left(u\right)\ne {\sigma }_S\left(v\right)$ for every two adjacent vertices u and v of G. The $pt$-domination number ${\gamma }_{pt}\left(G\right)$ of a graph G is the minimum cardinality of a $pt$-dominating set in G. Two well-known classes of highly symmetric graphs are investigated here, namely fans and wheels. For $n\ge 2$, the fan $F_n$ is the join $P_n\vee K_1$ of the path $P_n$ of order n and the complete graph $K_1$ of order 1, while for $n\ge 3$, the wheel $W_n$ is the join $C_n\vee K_1$ of the cycle $C_n$ of order n and $K_1$. For a positive integer t, the general fan $F_{n,t}$ is the graph $P_n\vee {\overline{K}}_t$ and the general wheel $W_{n,t}$ is the graph $C_n\vee {\overline{K}}_t$ where ${\overline{K}}_t$ is the empty graph of order t. All fans, wheels, general fans and general wheels are determined that possess $pt$-dominating sets. Furthermore, $pt$-domination numbers of all these graphs are determined as well.

1. Introduction

A vertex u in a graph G is said to totally dominate a vertex v if u is adjacent to v. A set S of vertices in G is a total dominating set if every vertex of G is totally dominated by at least one vertex of S. Consequently, a graph G has a total dominating set if and only if G has no isolated vertices. The minimum cardinality of a total dominating set in G is the total domination number ${\gamma }_t\left(G\right)$ of G. A total dominating set S with $\left|S\right|={\gamma }_t\left(G\right)$ is called a ${\gamma }_t$-set of G. This concept was introduced by Cockayne, Dawes and Hedetniemi [1] in 1980. More information on total domination in graphs (as well as some special classes of graphs) can be found in [2,3,4,5,6,7,8], for example.

For a total dominating set S of a graph G, the number of vertices of S that totally dominate a vertex v of G is denoted by ${\sigma }_S\left(v\right)$. While there exists no graph G with a total dominating set S such that ${\sigma }_S\left(u\right)\ne {\sigma }_S\left(v\right)$ for every two distinct vertices u and v of G, such sets exist for every two adjacent vertices u and v of G. The concept of total domination has been studied extensively with many interesting results and applications (see [5,6], for example). A 2013 book by Henning and Yeo [6] deals exclusively with total domination in graphs with many real life applications. Another area of research in graph theory is the study of graph coloring; its origin is the famous four-color problem. Today, graph coloring is one of the most popular areas of graph theory because of its numerous fascinating problems and applications to modern society. In a proper coloring of a graph, adjacent vertices are required to be colored differently. The minimum number of colors in a proper coloring of a graph G is its chromatic number $\chi \left(G\right)$. Combining total domination and proper graph coloring gave rise to proper total domination in graphs.

A total dominating set S in a graph G is a proper total dominating set (or a $pt$-dominating set) if ${\sigma }_S\left(u\right)\ne {\sigma }_S\left(v\right)$ for every two adjacent vertices u and v of G. This concept was introduced by Chartrand and studied in [9,10,11,12]. This concept has a close connection to the topics studied in [13,14,15,16] and the methodological and applied themes in proper total dominations align closely with many recent research topics in other areas of science (see [17,18], for example).

In general, it is challenging to determine whether a given graph possesses a $pt$-dominating set. While several well-known classes of graphs have been shown to possess a $pt$-dominating set, for many classes this remains unknown (see [9,11,12]). Hence, the following observations are useful. A graph G is locally irregular if $\mathrm{deg}u\ne \mathrm{deg}v$ for every two adjacent vertices u and v of G (see [19]). The vertex set $S=V\left(G\right)$ of a locally irregular graph G is a total dominating set of G since ${\sigma }_S\left(v\right)=\mathrm{deg}v$ for every vertex v of G. In fact, locally irregular graphs are the only graphs G for which $V\left(G\right)$ is a $pt$-dominating set.

Observation 1

([11]). If S is a $pt$-dominating set of a connected graph G, then the subgraph $G\left[S\right]$ induced by S is locally irregular without isolated vertices. Thus, $\left|S\right|\ge 3$ while $\left|S\right|=3$ if and only if $G\left[S\right]=P_3$. Furthermore, the vertex set of a graph G is a $pt$-dominating set of G if and only if G is locally irregular.

The set $N\left(v\right)$ of vertices adjacent to a vertex v in a graph G is the neighborhood of v and the set $N\left[v\right]=\left\{v\right\}\cup N\left(v\right)$ is the closed neighborhood of v. Two vertices u and v in a graph G are called twins if $N\left(u\right)=N\left(v\right)$. If u and v are adjacent vertices in a graph G such that $N\left[u\right]=N\left[v\right]$, then u and v are adjacent twins.

Observation 2

([11]). If u and v are adjacent twins of a connected graph G with $pt$-dominating set S, then exactly one of u and v belongs to S. Consequently, if G contains a complete subgraph $K_t$ of order $t\ge 3$ such that every two vertices of $K_t$ are twins of G, then G has no $pt$-dominating set. In particular, every complete graph of order 3 or more has no $pt$-dominating set.

The minimum cardinality of a $pt$-dominating set in a graph G is the proper total domination number ($pt$ -domination number) ${\gamma }_{pt}\left(G\right)$ of G. A $pt$-dominating set S with $\left|S\right|={\gamma }_{pt}\left(G\right)$ is called a ${\gamma }_{pt}$-set of G. The $pt$-domination numbers have been determined for several well-known classes of graphs possessing a $pt$-dominating set, including paths, cycles, trees, complete multipartite graphs, and Cartesian products $G\square K_2$ of a graph G and $K_2$ (see [9,11,12]). Since proper total domination is more restrictive than total domination, it follows that ${\gamma }_t\left(G\right)\le {\gamma }_{pt}\left(G\right)$ for every graph G with a $pt$-dominating set. Relationships between the classic total domination number and proper total domination number were investigated in [9]. All pairs $a,b$ of positive integers were determined for which there is a graph G with a $pt$-dominating set such that ${\gamma }_t\left(G\right)=a$ and ${\gamma }_{pt}\left(G\right)=b$. For each graph G with a $pt$-dominating set S, the numbers in the set ${\Sigma }_S=\{{\sigma }_S\left(v\right):v\in V\left(G\right)\}$ give rise to a proper coloring of G. The proper total chromatic number ${\chi }_{pt}\left(G\right)$ of G is the minimum value of $|{\Sigma }_S|$ among all $pt$-dominating sets S of G. Hence, $\chi \left(G\right)\le {\chi }_{pt}\left(G\right)$ for every graph G with a $pt$-dominating set. Furthermore, it was shown in [10] that for every pair $a,b$ of positive integers with $2\le a\le b$, there is a graph G with $\chi \left(G\right)=a$ and ${\chi }_{pt}\left(G\right)=b$. In general, it is challenging to determine the exact value of the $pt$-domination number of a given graph having a $pt$-dominating set. Among those graphs whose $pt$-domination numbers have been determined are paths and cycles.

Theorem 1

([9]). For an integer $n\ge 3$, let $P_n$ be a path of order n and $C_n$ a cycle of order n.

(a)

The path $P_n$ has a $pt$-dominating set if and only if $n\equiv 3\left(\mathrm{mod}4\right)$. Furthermore, $P_n$ has a unique $pt$-dominating set and ${\gamma }_{pt}\left(P_n\right)={\displaystyle \frac{3(n+1)}{4}}$ when $n\equiv 3\left(\mathrm{mod}4\right)$.

(b)

The cycle $C_n$ has a $pt$-dominating set if and only if $n\equiv 0\left(\mathrm{mod}4\right)$. Furthermore, $C_n$ has a unique $pt$-dominating set and ${\gamma }_{pt}\left(C_n\right)={\displaystyle \frac{3n}{4}}$ when $n\ge 4$ and $n\equiv 0\left(\mathrm{mod}4\right)$.

Results on paths and cycles have led to a study of classes of highly symmetric connected graphs related to paths and cycles. One such class consists of Cartesian products $G\square H$ where G and H are either paths or cycles in [9,12]. In particular, the grid $P_n\square P_m$ has a $pt$-dominating set for all integers $n,m$ with $n\ge m\ge 2$ (see [9]). On the other hand, the prism $C_n\square K_2$ possesses a $pt$-dominating set if and only if $n\ge 4$ is even. While the cylindrical graph $C_n\square P_m$ possesses a $pt$-dominating set for every even integer $n\ge 4$ and every integer $m\ge 3$, many cylindrical graphs $C_n\square P_m$ have no $pt$-dominating set for odd integers $n\ge 3$. For example, $C_3\square P_m$ possesses a $pt$-dominating set if and only if $m=3$ (see [12]). For odd integers $n\ge 5$, it is not known whether there is a characterization of cylindrical graphs $C_n\square P_m$ possessing a $pt$-dominating set. While a cyclical graph $C_n\square C_m$ has a $pt$-dominating set if at least one of n and m is even, the graph $C_3\square C_m$ does not possess a $pt$-dominating set for each odd integer $m\ge 3$. It is not known which cyclical graphs $C_n\square C_m$ possess a $pt$-dominating set for odd integers $n,m\ge 5$. Furthermore, for those grids, prims, cylindrical graphs, and cyclical graphs possessing a $pt$-dominating set, the exact values of their $pt$-domination numbers are not known in general. All of these lead us to consider two other well-known classes of highly symmetric graphs closely related to paths and cycles, namely, fans, wheels, and generalized fans and wheels.

For two vertex-disjoint graphs F and H, the graph $F\vee H$ is the join of F and H, that is, $V(F\vee H)=V\left(F\right)\cup V\left(H\right)$ and $E(F\vee H)=E\left(F\right)\cup E\left(H\right)\cup \{uv:u\in V(F),v\in V(H\left)\right\}$. The graph $F_n=P_n\vee K_1$ of order $n+1\ge 3$ is referred to as a fan and the graph $W_n=C_n\vee K_1$ of order $n+1\ge 4$ is a wheel. More generally, for each positive integer t, the graph $F_{n,t}=P_n\vee {\overline{K}}_t$ is referred to as a generalized fan and the graph $W_{n,t}=C_n\vee {\overline{K}}_t$ is a generalized wheel, where ${\overline{K}}_t$ is the empty graph of order t. Here, we determine all generalized fans and all generalized wheels that possess a $pt$-dominating set as well as their $pt$-domination numbers. In Section 2, all fans possessing a $pt$-dominating set and their $pt$-domination numbers are determined. In Section 3, all wheels possessing a $pt$-dominating set and their $pt$-domination numbers are determined. We then extend results on fans and wheels to generalized fans and generalized wheels in Section 4. We refer to the books [5,20] for notation and terminology not presented here.

2. Proper Total Domination in Fans

Since the fan $F_2=P_2\vee K_1=K_3$ has no $pt$-dominating set by Observation 2, we are only concerned with fans $F_n$ with $n\ge 3$. The following observation is useful.

Observation 3.

Let G and H be connected graphs such that $G\subseteq H$. If G does not have a $pt$-dominating set and H has a $pt$-dominating set, then every $pt$-dominating set of H contains at least one vertex not belonging to G.

Proof. 

Assume, to the contrary, that H has a $pt$-dominating set T such that $T\subseteq V\left(G\right)$. Let u and v be adjacent vertices of G. Since T is a $pt$-dominating set of H, it follows that ${\sigma }_T\left(u\right)\ne {\sigma }_T\left(v\right)$, so T is a $pt$-dominating set of G, which is impossible. □

The following result describes those fans possessing a $pt$-dominating set.

Proposition 1.

For an integer $n\ge 3$, the fan $F_n$ has a $pt$-dominating set if and only if $n\ne 4$.

Proof. 

In the fan $F_n=P_n\vee K_1$, let $P_n=(v_1,v_2,\dots ,v_n)$ and $V\left(K_1\right)=\left\{v\right\}$. For $n=3,5,6$, the fan $F_n$ has a $pt$-dominating set. For example, $V\left(F_3\right)-\left\{v_2\right\}$ is a $pt$-dominating set of $F_3$, $V\left(F_5\right)-\left\{v_4\right\}$ is a $pt$-dominating set of $F_5$, and $V\left(F_6\right)-\{v_4,v_6\}$ is a $pt$-dominating set of $F_6$. (This is shown in Figure 1).

Next, we show that $F_4$ does not have a $pt$-dominating set. Assume, to the contrary, that $F_4$ has a $pt$-dominating set S. Since $F_4$ is not locally irregular, it follows that $V\left(F_4\right)$ is not a $pt$-dominating set by Observation 1 and thus $\left|S\right|\in \{3,4\}$. Furthermore, $F_4\left[S\right]$ is locally irregular by Observation 1 again. Since $P_4$ has no $pt$-dominating set, it follows that $v\in S$ by Observation 3.

⋆

First, suppose that $\left|S\right|=4$. Since $v\in S$ and the star $K_{1,3}$ of order 4 is the only connected locally irregular graph of order 4, it follows that $F_4\left[S\right]=K_{1,3}$, whose center is v. However, $F_4$ does not contain such an induced subgraph $K_{1,3}$, which is a contradiction.

⋆

Next, suppose that $\left|S\right|=3$. Since $v\in S$ and $P_3$ is the only connected locally irregular graph of order 3, it follows that $F_4\left[S\right]=P_3=(v_i,v,v_j)$ where $j\ge i+2$. We may assume that either (1) $F_4\left[S\right]=(v_2,v,v_4)$ or (2) $F_4\left[S\right]=(v_1,v,v_4)$. If (1) occurs, then ${\sigma }_S\left(v_1\right)={\sigma }_S\left(v\right)=2$, while if (2) occurs, then ${\sigma }_S\left(v_2\right)={\sigma }_S\left(v_3\right)=2$. A contradiction is produced in either case.

For $n\ge 7$, let $S=\left\{v\right\}\cup \{v_i:i \mathrm{i}\mathrm{s} \mathrm{o}\mathrm{d}\mathrm{d} \mathrm{a}\mathrm{n}\mathrm{d} 1\le i\le n\}$. If $n\ge 7$ is odd, then ${\sigma }_S\left(v_i\right)=1$ for odd integers i with $1\le i\le n$, ${\sigma }_S\left(v_i\right)=3$ for even integers i with $2\le i\le n-1$, and ${\sigma }_S\left(v\right)\ge (n+1)/2\ge 4$. If $n\ge 8$ is even, then ${\sigma }_S\left(v_i\right)=1$ for odd integers i with $1\le i\le n-1$, ${\sigma }_S\left(v_i\right)=3$ for even integers i with $2\le i\le n-2$, ${\sigma }_S\left(v_n\right)=2$, and ${\sigma }_S\left(v\right)\ge n/2\ge 4$. In either case, S is a $pt$-dominating set of $F_n$. □

In order to determine the exact value of ${\gamma }_{pt}\left(F_n\right)$ for all those integers n for which $F_n$ has a $pt$-dominating set, we first present some preliminary information. In the fan $F_n=P_n\vee K_1$, let $P_n=(v_1,v_2,\dots ,v_n)$ and $V\left(K_1\right)=\left\{v\right\}$. For a $pt$-dominating set S of $F_n$, a block B of $P_n$ with respect to S is a maximal set of consecutive vertices of $P_n$ (or a maximal subpath of $P_n$) such that either all vertices of B belong to S or no vertices of B belong to S. The path $P_n=(v_1,v_2,\dots ,v_n)$ can therefore be expressed as one of the following four sequences, depending on the locations of the initial block and terminal block in the sequence:

$$(B_1,B_1^{\prime },B_2,B_2^{\prime },\dots ,B_k,B_k^{\prime }), (B_1,B_1^{\prime },B_2,B_2^{\prime },\dots ,B_k,B_k^{\prime },B_{k+1}),$$

$$(B_1^{\prime },B_1,B_2^{\prime },B_2,\dots ,B_k^{\prime },B_k), \mathrm{and} (B_1^{\prime },B_1,B_2^{\prime },B_2,\dots ,B_k^{\prime },B_k,B_{k+1}^{\prime })$$

such that $B_i\subseteq S$ and $B_i^{\prime }\cap S=\varnothing$ for $1\le i\le k$ or $1\le i\le k+1$. The first block in a sequence is the initial block of the sequence and the last block in a sequence is the terminal block. First, we verify the following claims.

Claim 1:

If B is a block with $B\subseteq S$, then $\left|B\right|\in \{1,3\}$.

Claim 2:

Let $B^{\prime }$ be a block with $B^{\prime }\cap S=\varnothing$. If $B^{\prime }$ is the initial or terminal block, then $\left|B^{\prime }\right|\in \{1,2\}$ and if $B^{\prime }$ is an interior block, then $\left|B^{\prime }\right|\in \{1,3\}$.

Claim 3:

If $B_1^{\prime }$ is the initial block, then $\left|B_1\right|=1$. Similarly, if $B_i^{\prime }$ is the terminal block where $i\in \{k,k+1\}$, then $\left|B_i\right|=1$.

⋆

If Claim 1 is not true, then there is a block B with $B\subseteq S$ such that $\left|B\right|=2$ or $\left|B\right|\ge 4$. First, suppose that $\left|B\right|=2$ and $B=(v_i,v_{i+1})$ where $1\le i\le n-1$. If $v\in S$, then $v_i$ is dominated only by $v_{i+1}$ and v in S and $v_{i+1}$ is dominated only by $v_i$ and v in S. Thus, ${\sigma }_S\left(v_i\right)={\sigma }_S\left(v_{i+1}\right)=2$, which is impossible. If $v\notin S$, then $v_i$ is dominated only by $v_{i+1}$ in S and $v_{i+1}$ is dominated only by $v_i$ in S. Thus, ${\sigma }_S\left(v_i\right)={\sigma }_S\left(v_{i+1}\right)=1$, which is impossible. Next, suppose that $\left|B\right|=4$ and $B=(v_i,v_{i+1},v_{i+2},\dots ,v_{i+\ell })$ for some integer $\ell \ge 3$. If $v\in S$, then $v_{i+1}$ is dominated only by $v_i,v_{i+2}$, and v in S and $v_{i+2}$ is dominated only by $v_{i+1},v_{i+3}$ and v in S. Thus, ${\sigma }_S\left(v_{i+1}\right)={\sigma }_S\left(v_{i+2}\right)=3$, which is impossible. If $v\notin S$, then $v_{i+1}$ is dominated only by $v_i$ and $v_{i+2}$ in S and $v_{i+2}$ is dominated only by $v_{i+1}$ and $v_{i+3}$ in S. Hence, ${\sigma }_S\left(v_{i+1}\right)={\sigma }_S\left(v_{i+2}\right)=2$, which is impossible. This verifies Claim 1.

⋆

Assume, to the contrary, that Claim 2 is not true. First, suppose that $B^{\prime }$ is the initial block or the terminal block with $\left|B^{\prime }\right|\ge 3$. We may assume that $B^{\prime }=(v_1,v_2,\dots ,v_{\ell })$ for some integer $\ell \ge 3$. Since $v_1,v_2,v_3\notin S$, it follows that $v_1$ and $v_2$ are dominated only by the vertex v. This implies that $v\in S$ and ${\sigma }_S\left(v_1\right)={\sigma }_S\left(v_2\right)=1$, which is impossible. Next, suppose that $B^{\prime }$ is an interior block with $\left|B^{\prime }\right|\notin \{1,3\}$. Then either $\left|B^{\prime }\right|=2$ or $\left|B^{\prime }\right|\ge 4$. For $\left|B^{\prime }\right|=2$, let $B^{\prime }=(v_i,v_{i+1})$ where $2\le i\le n-2$ and $v_{i-1},v_{i+2}\in S$. If $v\in S$, then $v_i$ is dominated only by $v_{i-1}$ and v in S and $v_{i+1}$ is dominated only by $v_{i+2}$ and v in S. Thus, ${\sigma }_S\left(v_i\right)={\sigma }_S\left(v_{i+1}\right)=2$, which is impossible. If $v\notin S$, then $v_i$ is dominated only by $v_{i-1}$ in S and $v_{i+1}$ is dominated only by $v_{i+2}$ in S. Hence, ${\sigma }_S\left(v_i\right)={\sigma }_S\left(v_{i+1}\right)=1$, which is impossible. For $\left|B^{\prime }\right|\ge 4$, let $B^{\prime }=(v_{i+1},v_{i+2},\dots ,v_{i+\ell })$ where $i\ge 2$, $\ell \ge 4$, and $i+\ell \le n-1$. Then $v_{i+2}$ and $v_{i+3}$ are dominated only by v. This implies that $v\in S$ and ${\sigma }_S\left(v_{i+2}\right)={\sigma }_S\left(v_{i+3}\right)=1$, which is also impossible. This verifies Claim 2.

⋆

If Claim 3 is not true, then $\left|B_1\right|\ne 1$ and thus $\left|B_1\right|=3$ by Claim 1. Since $\left|B_1^{\prime }\right|\in \{1,2\}$ by Claim 2, it follows that $B_1=(v_j,v_{j+1},v_{j+2})$ where $j\in \{2,3\}$ and $v_{j-1}\in B_1^{\prime }$. This implies that ${\sigma }_S\left(v_{j-1}\right)={\sigma }_S\left(v_j\right)=2$ if $v\in S$ and ${\sigma }_S\left(v_{j-1}\right)={\sigma }_S\left(v_j\right)=1$ if $v\notin S$, which is impossible. Similarly, if $B_i^{\prime }$ is the terminal block where $i\in \{k,k+1\}$, then $\left|B_i\right|=1$. Thus, Claim 3 is true.

We are now prepared to determine the value of ${\gamma }_{pt}\left(F_n\right)$ for all those integers n for which $F_n$ has a $pt$-dominating set.

Theorem 2.

For an integer $n\ge 3$ with $n\ne 4$, ${\gamma }_{pt}\left(F_3\right)=3$, ${\gamma }_{pt}\left(F_n\right)=5$ for $5\le n\le 8$, ${\gamma }_{pt}\left(F_9\right)=4$, and for $n\ge 10$,

$${\gamma }_{pt}\left(F_n\right)=\left\{\begin{array}{cc}p+1\hfill & if n\in \{4p,4p+1\} for an integer p\ge 3\hfill \\ p+2\hfill & if n\in \{4p+2,4p+3\} for an integer p\ge 2.\hfill \end{array}\right.$$

Proof. 

In the fan $F_n=P_n\vee K_1$, let $P_n=(v_1,v_2,\dots ,v_n)$ and $V\left(K_1\right)=\left\{v\right\}$. First, suppose that $3\le n\le 8$ and $n\ne 4$. For $n=3$, let $S=V\left(F_3\right)-\left\{v_2\right\}$; for $n=5$, let $S=V\left(F_n\right)-\left\{v_4\right\}$; for $n=6$, let $S=V\left(F_n\right)-\{v_4,v_6\}$; and for $n=7,8$, let $S=\left\{v\right\}\cup \{v_i:i \mathrm{is} \mathrm{odd} \mathrm{and} 1\le i\le n\}$. In each case, S is a $pt$-dominating set of $F_n$. Since ${\gamma }_{pt}\left(F_3\right)=3$, it remains to show that ${\gamma }_{pt}\left(F_n\right)\ge 5$ for $5\le n\le 8$. Assume, to the contrary, that ${\gamma }_{pt}\left(F_n\right)\le 4$ for some integer n with $5\le n\le 8$. Let S be a ${\gamma }_{pt}$-set of $F_n$. Then $\left|S\right|\in \{3,4\}$ and $F_n\left[S\right]$ is a locally irregular graph (without isolated vertices) in $F_n$. There are two graphs of order 3 without isolated vertices, namely $P_3$ and $K_3$, while $K_3$ is not locally irregular. Thus, $P_3$ is the only locally irregular graph without isolated vertices of order 3. Hence, if $\left|S\right|=3$, then $F_n\left[S\right]=P_3$. There are seven graphs of order 4 without isolated vertices, namely, $K_{1,3}, K_{1,3}+e, P_4, C_4, C_4 + e, K_4, 2K_2$, while only $K_{1,3}$ is locally irregular. Hence, if $\left|S\right|=4$, then $F_n\left[S\right]=K_{1,3}$.

⋆

First, suppose that $\left|S\right|=3$ and $F_n\left[S\right]=P_3=(x,y,z)$. If $v\notin S$, then either there exists $x^{\prime }\in V\left(P_n\right)-S$ that is adjacent to x such that ${\sigma }_S\left(x\right)={\sigma }_S\left(x^{\prime }\right)=1$ or there exists $z^{\prime }\in V\left(P_n\right)-S$ that is adjacent to z such that ${\sigma }_S\left(z\right)={\sigma }_S\left(z^{\prime }\right)=1$, which is a contradiction. Thus, we may assume that $v\in S$. Then $P_3=(v_i,v,v_j)$ where $i<j$ and $j\ge i+2$ (since v is adjacent to every vertex of $P_n$). If $i\ge 2$, then ${\sigma }_S\left(v_{i-1}\right)={\sigma }_S\left(v\right)=2$; if $j\le n-1$, then ${\sigma }_S\left(v_{j+1}\right)={\sigma }_S\left(v\right)=2$; and if $i=1$ and $j=n$, then ${\sigma }_S\left(v_{i+1}\right)={\sigma }_S\left(v\right)=2$. In each case, a contradiction is produced.

⋆

Next, suppose that $\left|S\right|=4$ and $F_n\left[S\right]=K_{1,3}$. Since v is adjacent to every vertex of $P_n$, it is impossible to have an induced star of order 4 or more in $F_n$ whose center is not v. Thus, v is the center of $K_{1,3}$. Let $S=\{v,v_i,v_j,v_k\}$ where $1\le i<j<k$. Then $j\ge i+2$ and $k\ge j+2$. If $j=i+2$, then ${\sigma }_S\left(v_{i+1}\right)={\sigma }_S\left(v\right)=3$. Thus, $j\ge i+4\ge 5$ by Claim 2. Similarly, $k\ge j+4\ge 9$. However then, $n\ge 9$, which is impossible.

Thus, ${\gamma }_{pt}\left(F_n\right)=5$ for $5\le n\le 8$.

Next, suppose that $n=9$. Since $S=\{v,v_1,v_5,v_9\}$ is a $pt$-dominating set of $F_9$, it follows that ${\gamma }_{pt}\left(F_9\right)\le 4$ (see Figure 2). Next, we show that ${\gamma }_{pt}\left(F_9\right)\ne 3$. Assume, to the contrary, that ${\gamma }_{pt}\left(F_9\right)=3$. Let S be a ${\gamma }_{pt}$-set of $F_9$. Since $F_9\left[S\right]$ is locally irregular, $F_9\left[S\right]=P_3$. Thus, either (i) $F_9\left[S\right]=(v_i,v_{i+1},v_{i+2})$ for some integer i with $1\le i\le 7$ or (ii) $F_9\left[S\right]=(v_i,v,v_j)$ where $1\le i<j\le 9$ and $j\ge i+2$. In either case, we may assume that $1\le i\le 4$. If (i) occurs, then $v_9$ is not dominated by any vertex of S, a contradiction. If (ii) occurs, then ${\sigma }_S\left(v_{j+1}\right)={\sigma }_S\left(v\right)=2$ if $j\ne 9$ and ${\sigma }_S\left(v_{j-1}\right)={\sigma }_S\left(v\right)=2$ if $j=9$, also a contradiction. Thus, ${\gamma }_{pt}\left(F_9\right)=4$.

For an integer $n\ge 10$, let $n=4p+r$ where $r\in \{0,1,2,3\}$. We consider two cases.

Case 1. $n=4p$ or $n=4p+1$ for some integer $p\ge 3$. In this case, we show that ${\gamma }_{pt}\left(F_n\right)=p+1$. Let $S_1=\left\{v\right\}\cup \{v_i:i\equiv 3\left(\mathrm{mod}4\right) for 1\le i\le n\}$. Then $\left|S_1\right|=p+1$. Since ${\sigma }_{S_1}\left(v_i\right)=1$ if i is odd, ${\sigma }_{S_1}\left(v_i\right)=2$ if i is even, and ${\sigma }_{S_1}\left(v\right)=p\ge 3$, it follows that $S_1$ is a $pt$-dominating set of $F_n$, so ${\gamma }_{pt}\left(F_n\right)\le p+1$. (See Figure 3 for $n=13$ and a $pt$-dominating set in $F_{12}$ can be obtained by deleting the vertex $v_{13}$.) It remains to show that ${\gamma }_{pt}\left(F_n\right)\ge p+1$. Assume, to the contrary, that there is an integer $n\in \{4p, 4p+1\}$ for some integer $p\ge 3$ such that $F_n$ has a $pt$-dominating set S for which $\left|S\right|\le p$. By Claim 2, if $B^{\prime }$ is a block of consecutive vertices on $P_n$ such that $\left|B^{\prime }\right|\cap S=\varnothing$, then $\left|B^{\prime }\right|\le 3$. This implies that for every four consecutive vertices on $P_n$, at least one of these vertices belongs to S. Thus, $\left|S\right|=p$ and $S\subseteq V\left(P_n\right)$. Further, there is a block $B^{\prime }$ on $P_n$ with $B^{\prime }\cap S=\varnothing$ that is a 3-path $(v_j,v_{j+1},v_{j+2})$ for some integer j with $2\le j\le n-2$. However, in this case, $v_{j+1}$ is not dominated by any vertex in S, which is a contradiction. Notice that the set $S_1$ is not a $pt$-dominating set when $n=4p+1$ where $p=2$, since if $n=9$, then $S_1$ contains two vertices of $P_9$ and thus ${\sigma }_{S_1}\left(v\right)={\sigma }_{S_1}\left(v_2\right)=2$. Therefore, $p\ge 3$ when $n=4p+1\ge 13$ is necessary, as illustrated in Figure 3.

Case 2. $n=4p+2$ or $n=4p+3$ for an integer $p\ge 2$. In this case, we show that ${\gamma }_{pt}\left(F_n\right)=p+2$. Let $S_1=\left\{v\right\}\cup \{v_i:i\equiv 1\left(\mathrm{mod}4\right) for 1\le i\le n\}$. Then $\left|S_1\right|=p+2$. Since ${\sigma }_{S_1}\left(v_i\right)=1$ if i is odd, ${\sigma }_{S_1}\left(v_i\right)=2$ if i is even, and ${\sigma }_{S_1}\left(v\right)=p\ge 3$, it follows that $S_1$ is a $pt$-dominating set of $F_n$ and thus ${\gamma }_{pt}\left(F_n\right)\le p+2$. (See Figure 4 for $n=11$ and a $pt$-dominating set in $P_{10}\vee K_1$ can be obtained by deleting the vertex $v_{11}$.) It remains to show that ${\gamma }_{pt}\left(F_n\right)\ge p+2$. Assume, to the contrary, that that there is an integer $n\in \{4p+2,4p+3\}$ for some integer $p\ge 2$ such that $F_n$ has a $pt$-dominating set S for which $\left|S\right|\le p+1$. By Claims 1 and 2, for every four consecutive vertices in $P_n$, at least one of these vertices belongs to S. Thus, $\left|S\right|=p+1$ and $S\subseteq V\left(P_n\right)$. Further, there is a block $B^{\prime }$ in $P_n$ with $B^{\prime }\cap S=\varnothing$ that is a 3-path $(v_j,v_{j+1},v_{j+2})$ for some integer j with $2\le j\le n-2$. However then, $v_{j+1}$ is not dominated by any vertex in S, which is a contradiction. □

3. Proper Total Domination in Wheels

Since the wheel $W_3=K_4$ has no $pt$-dominating set, we assume that $n\ge 4$. First, we present some preliminary information. For the wheel $W_n=C_n\vee K_1$, let $C_n=(v_1,v_2,\dots ,v_n,v_1)$ and $V\left(K_1\right)=\left\{v\right\}$. For a $pt$-dominating set S of $W_n$, a block B of $C_n$ with respect to S is a maximal set of consecutive vertices (or a maximal subpath) of $C_n$ if either all vertices of B belong to S or no vertices of B belong to S. The cycle $C_n$ can therefore be expressed as a sequence $(B_1,B_1^{\prime },B_2,B_2^{\prime }\dots ,B_k,B_k^{\prime })$ of $2k$ blocks for some positive integer k such that $B_i\subseteq S$ and $B_i^{\prime }\cap S=\varnothing$ for $1\le i\le k$. We verify the following claim.

Claim: 

For each integer i with $1\le i\le k$, $\left|B_i\right|\in \{1,3\}$ and $\left|B_i^{\prime }\right|\in \{1,3\}$.

Suppose that this claim is false. We may assume, without loss of generality, that either $\left|B_1\right|\notin \{1,3\}$ or $\left|B_1^{\prime }\right|\notin \{1,3\}$. Let $B_1=(v_1,v_2,v_3,\dots ,v_p)$ and $B_1^{\prime }=(v_{p+1},v_{p+2},\dots ,v_{p+q})$ where either $p\notin \{1,3\}$ or $q\notin \{1,3\}$. If $p\ge 4$, then ${\sigma }_S\left(v_2\right)={\sigma }_S\left(v_3\right)$. If $p=2$, then ${\sigma }_S\left(v_1\right)={\sigma }_S\left(v_2\right)$. Thus, $p=2$ and $p\ge 4$ are impossible. Therefore, $\left|B_i\right|\in \{1,3\}$ for $1\le i\le k$. If $q\ge 4$, then ${\sigma }_S\left(v_{p+2}\right)={\sigma }_S\left(v_{p+3}\right)$; while if $q=2$, then ${\sigma }_S\left(v_{p+1}\right)={\sigma }_S\left(v_{p+2}\right)$. Thus, $q=2$ and $q\ge 4$ are impossible. Consequently, $\left|B_i^{\prime }\right|\in \{1,3\}$ for $1\le i\le k$. This verifies the claim.

Proposition 2.

For an integer $n\ge 4$, the wheel $W_n$ has a $pt$-dominating set if and only if n is even.

Proof. 

For the wheel $W_n$, let $C_n=(v_1,v_2,\dots ,v_n,v_1)$ and $V\left(K_1\right)=\left\{v\right\}$. If $n=4$, then $\{v_1,v_2,v_3\}$ is a $pt$-dominating set of $W_4$, while if $n=6$, then $\{v_1,v_2,v_3,v_5\}$ is a $pt$-dominating set of $W_6$. If $n\ge 8$, let $S=\left\{v\right\}\cup \{v_i:i is odd and 1\le i\le n-1\}$. See Figure 5. Since ${\sigma }_S\left(v_i\right)=1$ for odd integers i with $1\le i\le n-1$, ${\sigma }_S\left(v_i\right)=3$ for even integers i with $2\le i\le n$, and ${\sigma }_S\left(v\right)\ge n/2\ge 4$, it follows that S is a $pt$-dominating set of $W_n$.

There remains a need to verify the converse. Assume, to the contrary, that there is an odd integer $n\ge 5$ such that $W_n$ has a $pt$-dominating set S. Then the cycle $C_n$ in $W_n$ is then expressed as a sequence $(B_1,B_1^{\prime },B_2,B_2^{\prime }\dots ,B_k,B_k^{\prime })$ of $2k$ blocks for some positive integer k such that $B_i\subseteq S$ and $B_i^{\prime }\cap S=\varnothing$ for $1\le i\le k$. By the claim, $n={\sum }_{i=1}^k\left(\right|B_i|+|B_i^{\prime }\left|\right)$ is even, which is a contradiction. □

We now determine the value of ${\gamma }_{pt}\left(W_n\right)$ for every even integer $n\ge 4$.

Theorem 3.

For an even integer $n\ge 4$, ${\gamma }_{pt}\left(W_4\right)=3$, ${\gamma }_{pt}\left(W_6\right)=4$, ${\gamma }_{pt}\left(W_8\right)={\gamma }_{pt}\left(W_{10}\right)=5$, and for $n\ge 12$,

$${\gamma }_{pt}\left(W_n\right)=\left\{\begin{array}{cc}p+1& if n = 4p \mathit{for} \mathit{some} \mathit{integer} p \ge 3\hfill \\ p+2& if n = 4p + 2 \mathit{for} \mathit{some} \mathit{integer} p\ge 3.\hfill \end{array}\right.$$

Proof. 

For the wheel $W_n$ where $n\ge 4$ is an even integer, let $C_n=(v_1,v_2,\dots ,v_n,v_1)$ and $V\left(K_1\right)=\left\{v\right\}$. Since ${\gamma }_{pt}\left(W_4\right)=3$ where $\{v_1,v_2,v_3\}$ is a ${\gamma }_{pt}$-set of $W_4$, we may assume that $n\ge 6$. First, suppose that $n\in \{6,8,10\}$.

⋆

For $n=6$, the set $\{v_1,v_2,v_3,v_5\}$ is a $pt$-dominating set of $W_6$ and thus ${\gamma }_{pt}\left(W_6\right)\le 4$. Assume, to the contrary, that $W_6$ has a ${\gamma }_{pt}$-set S with $\left|S\right|=3$. Then $W_n\left[S\right]=P_3=(x,y,z)$ is locally irregular by Observation 1. If $v\notin S$, then S is not a total dominating set of $W_6$, a contradiction. If $v\in S$, then we may assume that $P_3=(v_1,v,v_j)$ where $j\in \{3,4\}$. However then, ${\sigma }_S\left(v\right)={\sigma }_S\left(v_6\right)=2$, a contradiction. Thus, ${\gamma }_{pt}\left(W_6\right)\ge 4$, so ${\gamma }_{pt}\left(W_6\right)=4$

⋆

For $n=8$, the set $\{v,v_1,v_3,v_5,v_7\}$ is a $pt$-dominating set of $W_8$ and thus ${\gamma }_{pt}\left(W_8\right)\le 5$. Assume, to the contrary, that $W_8$ has a ${\gamma }_{pt}$-set S with $\left|S\right|\le 4$. Then $\left|S\right|\in \{3,4\}$ and $W_n\left[S\right]$ is locally irregular by Observation 1. If $\left|S\right|=3$, then $W_n\left[S\right]=P_3=(x,y,z)$. If $v\notin S$, then S is not a total dominating set of $W_8$, a contradiction. If $v\in S$, then we may assume that $P_3=(v_1,v,v_j)$ where $3\le j\le 5$. However then, ${\sigma }_S\left(v\right)={\sigma }_S\left(v_8\right)=2$. If $\left|S\right|=4$, then $W_n\left[S\right]=K_{1,3}$ is a star centered at v. We may assume that $S=\{v,v_1,v_i,v_j\}$ where $i\ge 3$ and $j-i\ge 2$. If $i=3$, then ${\sigma }_S\left(v\right)={\sigma }_S\left(v_2\right)=3$, a contradiction. By the claim, $i=5$. Similarly, $j-i=4$ and thus $j=i+4=9$, which is impossible since $n=8$. Thus, ${\gamma }_{pt}\left(W_8\right)\ge 5$ and thus ${\gamma }_{pt}\left(W_8\right)=5$.

⋆

For $n=10$, the set $\{v,v_1,v_5,v_7,v_9\}$ is a $pt$-dominating set of $W_{10}$ and thus ${\gamma }_{pt}\left(W_{10}\right)\le 5$. Assume, to the contrary, that $W_{10}$ has a ${\gamma }_{pt}$-set S with $\left|S\right|\le 4$. Then $\left|S\right|\in \{3,4\}$ and $W_n\left[S\right]$ is locally irregular by Observation 1. If $\left|S\right|=3$, then $W_n\left[S\right]=P_3=(x,y,z)$. If $v\notin S$, then S is not a total dominating set of $W_{10}$, a contradiction. If $v\in S$, then we may assume that $P_3=(v_1,v,v_j)$ where $3\le j\le 6$. However, ${\sigma }_S\left(v\right)={\sigma }_S\left(v_{10}\right)=2$. If $\left|S\right|=4$, then $W_n\left[S\right]=K_{1,3}$ is a star centered at v. We may assume that $S=\{v,v_1,v_i,v_j\}$ where $i-1\ge 2$ and $j-i\ge 2$. If $i=3$, then ${\sigma }_S\left(v\right)={\sigma }_S\left(v_2\right)=3$, a contradiction. By the claim, $i=5$ and $j=i+4=9$. However, ${\sigma }_S\left(v\right)={\sigma }_S\left(v_{10}\right)=3$, a contradiction. Thus, ${\gamma }_{pt}\left(W_{10}\right)\ge 5$, so ${\gamma }_{pt}\left(W_{10}\right)=5$.

Next, suppose that $n\ge 12$. Thus, either $n=4p$ or $n=4p+2$ for some integer $p\ge 3$. We consider these two cases.

Case 1. $n=4p$ for some integer $p\ge 3$. We show that ${\gamma }_{pt}\left(W_n\right)=p+1$. Let

$$S_1=\left\{v\right\}\cup \{v_i:i\equiv 1\left(\mathrm{mod}4\right) \mathrm{for} 1\le i\le n-3\}.$$

Then $\left|S_1\right|=p+1$. Since ${\sigma }_{S_1}\left(v_i\right)=1$ if i is odd, ${\sigma }_{S_1}\left(v_i\right)=2$ if i is even, and ${\sigma }_{S_1}\left(v\right)=p\ge 3$, it follows that $S_1$ is a $pt$-dominating set of $W_n$, so ${\gamma }_{pt}\left(W_n\right)\le p+1$.

It remains to be shown that ${\gamma }_{pt}\left(W_n\right)\ge p+1$. Assume, to the contrary, that there is an integer $n=4p$ for some integer $p\ge 3$ such that $W_n$ has a $pt$-dominating set S for which $\left|S\right|\le p$. By the claim, for every four consecutive vertices on $C_n$, at least one of these vertices belongs to S. Thus, $\left|S\right|=p$ and $S\subseteq V\left(C_n\right)$. Further, there is a block $B^{\prime }=(x,y,z)$ on $C_n$ such that $B^{\prime }\cap S=\varnothing$. However, in this case, y is not dominated by any vertex in S, which is a contradiction. Therefore, ${\gamma }_{pt}\left(W_n\right)\ge p+1$ and thus ${\gamma }_{pt}\left(W_n\right)=p+1$.

Case 2. $n=4p+2$ for some integer $p\ge 3$. We show that ${\gamma }_{pt}\left(W_n\right)=p+2$. Let

$$S_2=\left\{v\right\}\cup \{v_i:i\equiv 1\left(\mathrm{mod}4\right) \mathrm{for} 1\le i\le n-1\}.$$

Then $\left|S_2\right|=p+2$. Since ${\sigma }_{S_2}\left(v_i\right)=1$ if i is odd, ${\sigma }_{S_2}\left(v_i\right)=2$ if i is even and $i\ne n$, ${\sigma }_{S_2}\left(v_n\right)=3$, and ${\sigma }_{S_2}\left(v\right)=p+1\ge 4$, it follows that $S_2$ is a $pt$-dominating set of $F_n$ and thus ${\gamma }_{pt}\left(W_n\right)\le p+2$.

It remains to show that ${\gamma }_{pt}\left(W_n\right)\ge p+2$. Assume, to the contrary, that there is an integer $n=4p+2$ for some integer $p\ge 3$ such that $W_n$ has a $pt$-dominating set S for which $\left|S\right|\le p+1$. By the claim, for every four consecutive vertices on $C_n$, at least one of these vertices belongs to S. Thus, $\left|S\right|=p+1$ and $S\subseteq V\left(C_n\right)$. Further, there is a block $B^{\prime }=(x,y,z)$ such that $B^{\prime }\cap S=\varnothing$. However then, y is not dominated by any vertex in S, which is a contradiction. Therefore, ${\gamma }_{pt}\left(W_n\right)\ge p+2$, so ${\gamma }_{pt}\left(W_n\right)=p+2$. □

4. Proper Total Domination in Generalized Fans and Wheels

In this section, we extend the results obtained in Section 2 and Section 3 to more generalized fans $F_{n,t}=P_n\vee {\overline{K}}_t$ and wheels $W_{n,t}=C_n\vee {\overline{K}}_t$ for every positive integer t, where ${\overline{K}}_t$ is the empty graph of order t. For a $pt$-dominating set S in a graph G, let ${\Sigma }_S\left(G\right)=\{{\sigma }_S\left(v\right):v\in V\left(G\right)\}.$ First, we establish a connection between a $pt$-dominating set in the graph $G\vee {\overline{K}}_1$ and a $pt$-dominating set in the graph $G\vee {\overline{K}}_t$ for an integer $t\ge 2$.

Theorem 4.

For a graph G, let $H_t=G\vee {\overline{K}}_t$ for a positive integer t.

(a)

If $H_1$ has a $pt$-dominating set S, then S is also a $pt$-dominating set of $H_t$ for each integer $t\ge 2$ and thus ${\gamma }_{pt}\left(H_t\right)\le {\gamma }_{pt}\left(H_1\right)$.

(b)

If $H_t$ has a $pt$-dominating set T for some integer $t\ge 2$ such that $\left|T\right|-1\notin {\Sigma }_T\left(G\right)$, then $H_1$ has a $pt$-dominating set.

Proof. 

Let $V\left({\overline{K}}_t\right)=\{w_1,w_2,\dots ,w_t\}$. If $t=1$, then $V\left({\overline{K}}_1\right)=\left\{w_1\right\}$ in $H_1=G\vee K_1$.

First, we verify (a). Suppose that S is a $pt$-dominating set in $H_1$. We show that S is a $pt$-dominating set of $H_t$ for each integer $t\ge 2$. Since a vertex in $V\left({\overline{K}}_t\right)$ is dominated only by a vertex in G, it follows that $p=|S\cap V(G\left)\right|\ge 1$ and ${\sigma }_S\left(w_i\right)=p$ for $1\le i\le t$. Let u and v be adjacent vertices of $H_t$. Then at least one of u and v belongs to G, say $u\in V\left(G\right)$. If $v\in V\left(G\right)$, then ${\sigma }_S\left(u\right)\ne {\sigma }_S\left(v\right)$ since S is a $pt$-dominating set of $H_1$. If $v\in V\left({\overline{K}}_t\right)$, then ${\sigma }_S\left(u\right)\ne {\sigma }_S\left(w_1\right)={\sigma }_S\left(v\right)$, again, since S is a $pt$-dominating set of $H_1$. Thus, S is a $pt$-dominating set of $H_t$. This is illustrated in Figure 6 for $G=P_6$ and $t=2$, so $H_1=F_6$ and $H_2=F_{6,2}$. Furthermore, if S is a ${\gamma }_{pt}$-set of $H_1$, then S is a $pt$-dominating set of $H_t$ for each integer $t\ge 2$ and thus ${\gamma }_{pt}\left(H_t\right)\le \left|S\right|={\gamma }_{pt}\left(H_1\right)$.

Next, we verify (b). Suppose that T is a $pt$-dominating set of $H_t=G\vee {\overline{K}}_t$ for some integer $t\ge 2$ such that $\left|T\right|-1\notin {\Sigma }_T\left(G\right)$. We show that $H_1$ has a $pt$-dominating set.

⋆

First, suppose that $T\subseteq V\left(G\right)$. Then ${\sigma }_S\left(w_i\right)=\left|T\right|$ for $1\le i\le t$. Let u and v be adjacent vertices of $H_1$. We may assume that $u\in V\left(G\right)$. If $v\in V\left(G\right)$, then ${\sigma }_T\left(u\right)\ne {\sigma }_T\left(v\right)$ since T is a $pt$-dominating set of $H_t$. If $v=w_1$, then ${\sigma }_T\left(u\right)\ne {\sigma }_T\left(w_1\right)=\left|T\right|$, again since T is a $pt$-dominating set of $H_t$. Thus, T is a $pt$-dominating set of $H_1$.

⋆

Next, suppose that $T⊈V\left(G\right)$. Thus, T contains at least one vertex in ${\overline{K}}_t$. Let $T=S\cup S^{\prime }$ where $S\subseteq V\left(G\right)$ and $S^{\prime }\subseteq V\left({\overline{K}}_t\right)$. Then $S\ne \varnothing$ and $S^{\prime }\ne \varnothing$. We may assume that $w_1\in S^{\prime }$. Let $p=\left|S\right|\ge 1$, $q=\left|S^{\prime }\right|\ge 1$, and $\left|T\right|=p+q$. Then ${\sigma }_T\left(w_i\right)=p$ for $1\le i\le t$. Next, let $T^{\prime }=S\cup \left\{w_1\right\}$. We show that $T^{\prime }$ is a $pt$-dominating set of $H_1$. Let u and v be adjacent vertices of $H_1$. We may assume that $u\in V\left(G\right)$. If $q=1$, then $T^{\prime }=T$ and thus ${\sigma }_{T^{\prime }}\left(u\right)\ne {\sigma }_{T^{\prime }}\left(v\right)$. Thus, we may assume that $q\ge 2$. If $v\in V\left(G\right)$, then ${\sigma }_{T^{\prime }}\left(u\right)={\sigma }_T\left(u\right)-(q-1)\ne {\sigma }_T\left(v\right)-(q-1)={\sigma }_{T^{\prime }}\left(v\right)$ since T is a $pt$-dominating set of $H_t$. If $v=w_1$, then ${\sigma }_{T^{\prime }}\left(u\right)={\sigma }_T\left(u\right)-(q-1)$ and ${\sigma }_{T^{\prime }}\left(v\right)={\sigma }_T\left(v\right)=p$. Since ${\sigma }_T\left(u\right)\ne \left|T\right|-1=p+q-1$, it follows that ${\sigma }_{T^{\prime }}\left(u\right)\ne {\sigma }_{T^{\prime }}\left(v\right)$. Thus, $T^{\prime }$ is a $pt$-dominating set of $H_1$. This is illustrated in Figure 7 for $G=P_4$ and $t=2$ and thus $H_1=F_4$ and $H_2=F_{4,2}$.

□

Figure 7. A $pt$-dominating set in $F_{4,2}$.

Figure 7. A $pt$-dominating set in $F_{4,2}$.

The condition that ${\sigma }_T\left(x\right)\ne \left|T\right|-1$ for every vertex x of G in Theorem 4 is necessary. For example, consider the generalized fan $F_{4,2}=P_4\vee {\overline{K}}_2$ where $P_4=(v_1,v_2,v_3,v_4)$ and $V\left({\overline{K}}_2\right)=\{w_1,w_2\}$ and let $T=\{v_2,w_1,w_2\}$. Since ${\sigma }_T\left(v_1\right)={\sigma }_T\left(v_3\right)=3$, ${\sigma }_T\left(v_2\right)={\sigma }_S\left(v_4\right)=2=\left|T\right|-1$, and ${\sigma }_T\left(w_1\right)={\sigma }_T\left(w_2\right)=1$, it follows that T is a $pt$-dominating set of $F_{4,2}$. By Proposition 1, the graph $F_4$ does not have a $pt$-dominating set.

The following are consequences of Propositions 1 and 2 and Theorem 4.

Corollary 1.

For positive integers n and t where $n\ge 3$,

(a)

The generalized fan $F_{n,t}$ has a $pt$-dominating set if and only if $(n,t)\ne (4,1)$;

(b)

The generalized wheel $W_{n,t}$ has a $pt$-dominating set if and only if n is even.

Proof. 

First, we verify (a). By Proposition 1, the graph $F_4=F_{4,1}$ does not have a $pt$-dominating set. For the converse, assume that $(n,t)\ne (4,1)$. For the graph $F_{4,t}$ where $t\ge 2$, let $P_4=(v_1,v_2,v_3,v_4)$ and let $w_1$ and $w_2$ be two vertices in ${\overline{K}}_t$. Then $\{v_2,w_1,w_2\}$ is a $pt$-dominating set of $P_4\vee {\overline{K}}_t$. By Proposition 1 and Theorem 4, the graph $F_{n,t}$ has a $pt$-dominating set for integers $n\ge 3$ and $t\ge 1$ such that $(n,t)\ne (4,1)$.

Next, we verify (b). An argument similar to that in the proof of Proposition 2 shows that if $n\ge 3$ is odd and $t\ge 2$, then $W_{n,t}$ does not have a $pt$-dominating set. For the converse, if n is even and $t\ge 1$, then $W_{n,t}$ has a $pt$-dominating set by Proposition 2 and Theorem 4. □

Next, we determine the $pt$-domination numbers of those generalized fans $F_{n,t}$ and generalized wheels $W_{n,t}$ possessing a $pt$-dominating set for each integer $t\ge 2$. In order to show this, we first present a lemma.

Lemma 1.

For a graph G, let $H=G\vee {\overline{K}}_t$ for a positive integer t and let T be a $pt$-dominating set of H. If $T\cap V\left(G\right)$ is a total dominating set of G, then $T\cap V\left(G\right)$ is a $pt$-dominating set of G. In particular, if $T\subseteq V\left(G\right)$, then T is a $pt$-dominating set of G.

Proof. 

Let $S=T\cap V\left(G\right)$ and let $|T-S|=p\ge 0$. Since each vertex $x\in V\left(G\right)$ is adjacent to every vertex in ${\overline{K}}_t$, it follows that ${\sigma }_T\left(x\right)={\sigma }_S\left(x\right)+p$. Furthermore, since S is a total dominating set of G, it follows that ${\sigma }_S\left(x\right)={\sigma }_T\left(x\right)-p\ge 1$ for every vertex x of G. Let u and v be adjacent vertices of G. Since T is a $pt$-dominating set of H, it follows that ${\sigma }_T\left(u\right)\ne {\sigma }_T\left(v\right)$. Thus, ${\sigma }_S\left(u\right)={\sigma }_T\left(u\right)-p\ne {\sigma }_T\left(v\right)-p={\sigma }_S\left(v\right)$. Therefore, S is a $pt$-dominating set of G.

In particular, if $T\subseteq V\left(G\right)$, then $T\cap V\left(G\right)=T$. Since T is a $pt$-dominating set of H, every vertex of G is totally dominated by a vertex of T and thus T is a total dominating set of G. Consequently, T is a $pt$-dominating set of G. □

The condition that $T\cap V\left(G\right)$ is a total dominating set of G in Lemma 1 is necessary. For example, consider the fan $F_{10}$ where $V\left(K_1\right)=\left\{v\right\}$ and $P_{10}=(v_1,v_2,\dots ,v_{10})$. Let $T=\{v,v_1,v_5,v_9\}$ in $F_{10}$ (see Figure 8). Since ${\sigma }_T\left(v\right)=3$, ${\sigma }_T\left(v_i\right)=1$ if i is odd and $1\le i\le 9$, while ${\sigma }_T\left(v_i\right)=2$ if i is even and $2\le i\le 10$, it follows that T is a $pt$-dominating set of $F_{10}$. However, $T\cap V\left(P_{10}\right)$ is not a total dominating set of $P_{10}$, so it is not a $pt$-dominating set of $P_{10}$. In fact, $P_{10}$ has no $pt$-dominating set by Theorem 1.

Theorem 5.

For integers $n\ge 3$ and $t\ge 2$,

$${\gamma }_{pt}\left(F_{n,t}\right)=\left\{\begin{array}{cc}3& if 3\le n\le 5\hfill \\ {\gamma }_{pt}\left(F_n\right)& if n\ge 6.\hfill \end{array}\right.$$

Proof. 

For the generalized fan $F_{n,t}=P_n\vee {\overline{K}}_t$, let $P_n=(v_1,v_2,\dots ,v_n)$ and $V\left({\overline{K}}_t\right)=\{w_1,w_2,\dots ,w_t\}$ where $n\ge 3$ and $t\ge 2$. First, we show that if $3\le n\le 5$, then ${\gamma }_{pt}\left(F_{n,t}\right)=3$. By Observation 1, it suffices to show that $F_{n,t}$ has a three-element $pt$-dominating set.

⋆

For $n=3$, let $S=\{v_1,v_3,w_1\}$. Since ${\sigma }_S\left(v_1\right)={\sigma }_S\left(v_3\right)=1$, ${\sigma }_S\left(v_2\right)=3$, and ${\sigma }_S\left(w_j\right)=2$ for $1\le j\le t$, it follows that S is a $pt$-dominating set of $F_{3,t}$.

⋆

For $n=4$, let $S=\{v_2,w_1,w_2\}$. Since ${\sigma }_S\left(v_i\right)=3$ for $i\in \{1,3\}$, ${\sigma }_S\left(v_i\right)=2$ for $i\in \{2,4\}$, and ${\sigma }_S\left(w_j\right)=1$ for $1\le j\le t$, it follows that S is a $pt$-dominating set of $F_{4,t}$.

⋆

For $n=5$, let $S=\{v_3,w_1,w_2\}$. Since ${\sigma }_S\left(v_i\right)=2$ for $i\in \{1,3,5\}$, ${\sigma }_S\left(v_i\right)=3$ for $i\in \{2,4\}$, and ${\sigma }_S\left(w_j\right)=1$ for $1\le j\le t$, it follows that S is a $pt$-dominating set of $F_{5,t}$.

Thus, ${\gamma }_{pt}\left(F_{n,t}\right)=3$ if $3\le n\le 5$ and $t\ge 2$.

Next, we show that if $n\ge 6$, then ${\gamma }_{pt}\left(F_{n,t}\right)={\gamma }_{pt}\left(F_n\right)$ for each integer $t\ge 2$. Since ${\gamma }_{pt}\left(F_{n,t}\right)\le {\gamma }_{pt}\left(F_n\right)$ by Theorem 4, it remains to show that ${\gamma }_{pt}\left(F_{n,t}\right)\ge {\gamma }_{pt}\left(F_n\right)$. Assume, to the contrary, that there exist integers $n\ge 6$ and $t\ge 2$ such that ${\gamma }_{pt}\left(F_{n,t}\right)<{\gamma }_{pt}\left(F_n\right)$. Let T be a ${\gamma }_{pt}$-set of $F_{n,t}$. We consider two cases, according to whether $T\subseteq V\left(P_n\right)$ or $T⊈V\left(P_n\right)$.

Case 1. $T\subseteq V\left(P_n\right)$. It follows by Lemma 1 that T is a $pt$-dominating set of $P_n$. Since $P_n$ has a $pt$-dominating set if and only if $n\equiv 3\left(\mathrm{mod}4\right)$ by Theorem 1, there is a contradiction if $n\not\equiv 3\left(\mathrm{mod}4\right)$. On the other hand, if $n\equiv 3\left(\mathrm{mod}4\right)$, then $P_n$ has a unique $pt$-dominating set of cardinality ${\displaystyle \frac{3(n+1)}{4}}$, so $\left|T\right|={\displaystyle \frac{3(n+1)}{4}}={\gamma }_{pt}\left(P_n\right)$ by Theorem 1. However, then, ${\displaystyle \frac{3(n+1)}{4}}=\left|T\right|={\gamma }_{pt}\left(F_{n,t}\right)<{\gamma }_{pt}\left(F_n\right)={\displaystyle \frac{n+5}{4}}$ by Theorem 2, which is impossible. Therefore, Case 1 cannot occur.

Case 2. $T⊈V\left(P_n\right)$. Let $T=S\cup S^{\prime }$ where $S\subseteq V\left(P_n\right)$ and $S^{\prime }\subseteq V\left({\overline{K}}_t\right)$. We may assume, without loss of generality, that $w_1\in S^{\prime }$. Then $S\ne \varnothing$ and $S^{\prime }\ne \varnothing$. Thus, $p=\left|S\right|\ge 1$ and $q=|S^{\prime }|\ge 1$. Notice that ${\sigma }_T\left(v_i\right)={\sigma }_S\left(v_i\right)+q$ for $1\le i\le n$ and ${\sigma }_T\left(w_j\right)=p$ for $1\le j\le t$. Let $T^{\prime }=S\cup \left\{w_1\right\}$. We show that $T^{\prime }$ is a $pt$-dominating set of $F_n$. Let x and y be adjacent vertices of $F_n$. Then at least one of x and y belongs to $P_n$. First, suppose that $x,y\in V\left(P_n\right)$, say $x=v_a$ and $y=v_b$ where $1\le a<b\le n$. Since ${\sigma }_T\left(v_a\right)\ne {\sigma }_T\left(v_b\right)$ and ${\sigma }_T\left(v_i\right)={\sigma }_S\left(v_i\right)+q$ for $1\le i\le n$, it follows that ${\sigma }_S\left(v_a\right)\ne {\sigma }_S\left(v_b\right)$. Hence, ${\sigma }_{T^{\prime }}\left(v_a\right)={\sigma }_S\left(v_a\right)+1\ne {\sigma }_S\left(v_b\right)+1={\sigma }_{T^{\prime }}\left(v_b\right)$. Next, suppose that exactly one of x and y belongs to $P_n$, say, $x\in V\left(P_n\right)$ and $y=w_1$. Then ${\sigma }_{T^{\prime }}\left(x\right)={\sigma }_S\left(x\right)+1$ and ${\sigma }_T\left(y\right)=p$. Since ${\sigma }_S\left(x\right)\in \{1,2\}$, it follows that ${\sigma }_{T^{\prime }}\left(x\right)\le 3$. Therefore, if $p\ge 4$, then ${\sigma }_{T^{\prime }}\left(x\right)\ne {\sigma }_{T^{\prime }}\left(y\right)$. There remains for consideration the situation where $p\le 3$. For this purpose, the block structure of the generalized fans needs to be investigated.

A block B of $P_n$ with respect to S is a maximal set of consecutive vertices of $P_n$ (or a maximal subpath of $P_n$) such that either all vertices of B belong to S or no vertices of B belong to S. The path $P_n=(v_1,v_2,\dots ,v_n)$ can therefore be expressed as one of the following four sequences, depending on the locations of the initial block and terminal block in the sequence

$$(B_1,B_1^{\prime },B_2,B_2^{\prime },\dots ,B_k,B_k^{\prime }), (B_1,B_1^{\prime },B_2,B_2^{\prime },\dots ,B_k,B_k^{\prime },B_{k+1}),$$

$$(B_1^{\prime },B_1,B_2^{\prime },B_2,\dots ,B_k^{\prime },B_k), \mathrm{and} (B_1^{\prime },B_1,B_2^{\prime },B_2,\dots ,B_k^{\prime },B_k,B_{k+1}^{\prime })$$

such that $B_i\subseteq S$ and $B_i^{\prime }\cap S=\varnothing$ for $1\le i\le k$ or $1\le i\le k+1$. The following claims can be verified by arguments similar to that used in the case of the fans.

Claim 1:

If B is a block with $B\subseteq S$, then $\left|B\right|\in \{1,3\}.$

Claim 2:

Let $B^{\prime }$ be a block with $B^{\prime }\cap S=\varnothing$. If $B^{\prime }$ is the initial or terminal block, then $|B^{\prime }|\in \{1,2\}$, while if $B^{\prime }$ is an interior block, then $|B^{\prime }|\in \{1,3\}.$

Claim 3:

If $B_1^{\prime }$ is the initial block, then $\left|B_1\right|=1$. Similarly, if $B_i^{\prime }$ is the terminal block where $i\in \{k,k+1\}$, then $\left|B_i\right|=1.$

First, we show that if $n\ge 14$, then $\left|S\right|=p\ge 4$. Assume, to the contrary, that $\left|S\right|\le 3$. By Claims 1 and 2, for every four consecutive vertices on $P_n$, at least one vertex is in S, so $\left|S\right|\ge 3$. Thus, $\left|S\right|=3$ and $\left|B_i\right|=1$ for every block $B_i$ in S. If $v_1\in S$, then $\left|S\right|\ge 4$ since S contains at least three vertices in $\{v_2,v_3,\dots ,v_{13}\}$, a contradiction. Thus, we may assume that $v_1\notin S$. If $v_2\in S$, then $\left|S\right|\ge 4$ since S contains at least three vertices in $\{v_3,v_4,\dots ,v_{14}\}$, a contradiction. Hence, $v_2\notin S$ and thus $v_3\in S$ by Claim 3. This forces $v_7\in S$, $v_{11}\in S$, and $v_i\notin S$ for $i\ge 12$. Since $n\ge 14$, the terminal block is not in S and contains at least three vertices, which contradicts Claim 2. Therefore, if $n\ge 14$, then $\left|S\right|=p\ge 4$.

Thus, we may assume that $6\le n\le 13$. By Theorem 2, if $6\le n\le 8$, then ${\gamma }_{pt}\left(F_n\right)=5$, while if $9\le n\le 13$, then ${\gamma }_{pt}\left(F_n\right)=4$. We consider these two possibilities.

Subcase $2.1$. $6\le n\le 8$. We show that ${\gamma }_{pt}\left(F_{n,t}\right)=5$ for each integer $t\ge 2$. Assume, to the contrary, that ${\gamma }_{pt}\left(F_{n,t}\right)\le 4$ for some integer n with $6\le n\le 8$ and an integer $t\ge 2$. Let T be a ${\gamma }_{pt}$-set of $F_{n,t}$. Then $\left|T\right|\in \{3,4\}$ and $F_{n,t}\left[T\right]$ is a locally irregular graph (without isolated vertices) in $F_{n,t}$. If $\left|T\right|=3$, then $F_{n,t}\left[T\right]=P_3$, while if $\left|T\right|=4$, then $F_{n,t}\left[T\right]=K_{1,3}$.

⋆

First, suppose that $\left|T\right|=3$ and $F_{n,t}\left[T\right]=P_3=(x,y,z)$. If $T\cap V\left({\overline{K}}_t\right)=\varnothing$, then since $n\ge 6$, there is a vertex of $P_n$ that is not totally dominated by any vertex in T, which is a contradiction. If $T\cap V\left({\overline{K}}_t\right)\ne \varnothing$, then we may assume that (i) $P_3=(v_i,w_1,v_j)$ where $j\ge i+2$ or (ii) $P_3=(w_1,v_i,w_2)$. If (i) occurs, then ${\sigma }_T\left(w_1\right)={\sigma }_T\left(v_{i-1}\right)=2$ (when $i\ge 2$), ${\sigma }_T\left(w_1\right)={\sigma }_T\left(v_{j+1}\right)=2$ (when $j\le n-1$), and ${\sigma }_T\left(w_1\right)={\sigma }_T\left(v_2\right)=2$ (when $i=1$ and $j=n$), which is impossible. If (ii) occurs, then we may assume that $P_3=(w_1,v_i,w_2)$ and $1\le i\le n/2$. Since $n\ge 6$, it follows that ${\sigma }_T\left(v_{n-1}\right)={\sigma }_T\left(v_n\right)=2$, which is impossible.

⋆

Next, suppose that $\left|T\right|=4$ and $F_{n,t}\left[T\right]=K_{1,3}$. Thus, either (i) the center of $K_{1,3}$ is a vertex in $V\left({\overline{K}}_t\right)$ or (ii) the center of $K_{1,3}$ is a vertex of $P_n$ and $t\ge 3$. If (i) occurs, we may assume that $w_1$ is the center of $K_{1,3}$. Let $T=\{w_1,v_i,v_j,v_k\}$ where $1\le i<j<k$. Then $j\ge i+2$ and $k\ge j+2$. By Claim 2, it follows that $j-i\in \{2,4\}$ and $k-j\in \{2,4\}$. If $j-i=2$, then ${\sigma }_T\left(v_{i+1}\right)={\sigma }_T\left(w_1\right)=3$, which is impossible. Thus, $j-i=4$. Similarly, $k-j=4$. This implies that $n\ge 9$, which is also impossible. If (ii) occurs, we may assume that $V\left(K_{1,3}\right)=\{v_i,w_1,w_2,w_3\}$ and $1\le i\le n/2$. Since $n\ge 6$, it follows that ${\sigma }_T\left(v_{n-1}\right)={\sigma }_T\left(v_n\right)=3$, which too is impossible.

Thus, ${\gamma }_{pt}\left(F_{n,t}\right)=5$ for $6\le n\le 8$ and $t\ge 2$.

Subcase $2.2$. $9\le n\le 13$. We show that ${\gamma }_{pt}\left(F_{n,t}\right)=4$ for each integer $t\ge 2$. Assume, to the contrary, that ${\gamma }_{pt}\left(F_{n,t}\right)=3$ for some integer n with $9\le n\le 13$ and an integer $t\ge 2$. Let T be a ${\gamma }_{pt}$-set of $F_{n,t}$. Then $F_{n,t}\left[T\right]=P_3=(x,y,z)$, an argument similar to that in Subcase $2.1$ when $\left|T\right|=3$ shows that a contradiction is produced. Thus, ${\gamma }_{pt}\left(F_{n,t}\right)=4$ for $9\le n\le 13$ and $t\ge 2$. □

We saw in Theorem 4 that ${\gamma }_{pt}\left(H_t\right)\le {\gamma }_{pt}\left(H_1\right)$ for every integer $t\ge 2$. By Theorem 5, ${\gamma }_{pt}\left(F_{5,t}\right)=3$ for every integer $t\ge 2$. Since ${\gamma }_{pt}\left(F_5\right)=5$ by Theorem 2, it follows that ${\gamma }_{pt}\left(F_{5,t}\right)<{\gamma }_{pt}\left(F_5\right)$ and hence strict inequality is possible.

Theorem 6.

For integers n and $t\ge 2$ where $n\ge 4$ is even,

$${\gamma }_{pt}\left(W_{n,t}\right)={\gamma }_{pt}\left(W_n\right).$$

Proof. 

For the generalized wheel $W_{n,t}=C_n\vee {\overline{K}}_t$ where $n\ge 4$ is even and $t\ge 2$, let $C_n=(v_1,v_2,\dots ,v_n,v_1)$ and $V\left({\overline{K}}_t\right)=\{w_1,w_2,\dots ,w_t\}$. By Theorem 4, ${\gamma }_{pt}\left(W_{n,t}\right)\le {\gamma }_{pt}\left(W_n\right)$ for every integer $t\ge 2$. It remains to show that ${\gamma }_{pt}\left(W_{n,t}\right)\ge {\gamma }_{pt}\left(W_n\right)$. Since ${\gamma }_{pt}\left(W_4\right)=3$, it follows by Observation 1 that ${\gamma }_{pt}\left(W_{4,t}\right)=3$ for every integer $t\ge 2$. Thus, we may assume that n is even and $n\ge 6$. Assume, to the contrary, that ${\gamma }_{pt}\left(W_{n,t}\right)<{\gamma }_{pt}\left(W_n\right)$ for some integers n and $t\ge 2$ where $n\ge 6$ is even. Let T be a ${\gamma }_{pt}$-set of $W_{n,t}$. We consider two cases, according to whether $T\subseteq V\left(C_n\right)$ or $T⊈V\left(C_n\right)$.

Case 1. $T\subseteq V\left(C_n\right)$. By Lemma 1, T is a $pt$-dominating set of $C_n$. Since $C_n$ has a unique $pt$-dominating set of cardinality ${\displaystyle \frac{3n}{4}}$, it follows that $\left|T\right|={\displaystyle \frac{3n}{4}}={\gamma }_{pt}\left(C_n\right)$ by Theorem 1. However, in this case, ${\displaystyle \frac{3n}{4}}=\left|T\right|={\gamma }_{pt}\left(W_{n,t}\right)<{\gamma }_{pt}\left(W_n\right)\le {\displaystyle \frac{n+10}{4}}$ by Theorem 3. This is impossible since $n\ge 6$. Therefore, Case 1 cannot occur.

Case 2. $T⊈V\left(C_n\right)$. Let $T=S\cup S^{\prime }$ where $S\subseteq V\left(C_n\right)$ and $S^{\prime }\subseteq V\left({\overline{K}}_t\right)$. We may assume, without loss of generality, that $w_1\in S^{\prime }$. Then $S\ne \varnothing$ and $S^{\prime }\ne \varnothing$. Thus, $p=\left|S\right|\ge 1$ and $q=|S^{\prime }|\ge 1$. Furthermore, ${\sigma }_T\left(v_i\right)={\sigma }_S\left(v_i\right)+q$ for $1\le i\le n$ and ${\sigma }_T\left(w_j\right)=p$ for $1\le j\le t$. Let $T^{\prime }=S\cup \left\{w_1\right\}$. We show that $T^{\prime }$ is a $pt$-dominating set of $W_n$. Let x and y be adjacent vertices of $W_n$. Then at least one of x and y belongs to $C_n$. First, suppose that $x,y\in V\left(C_n\right)$, say $x=v_a$ and $y=v_b$ where $1\le a<b\le n$. Since ${\sigma }_T\left(v_a\right)\ne {\sigma }_T\left(v_b\right)$ and ${\sigma }_T\left(v_i\right)={\sigma }_S\left(v_i\right)+q$ for $1\le i\le n$, it follows that ${\sigma }_S\left(v_a\right)\ne {\sigma }_S\left(v_b\right)$. Hence, ${\sigma }_{T^{\prime }}\left(v_a\right)={\sigma }_S\left(v_a\right)+1\ne {\sigma }_S\left(v_b\right)+1={\sigma }_{T^{\prime }}\left(v_b\right)$. Next, suppose that exactly one of x and y belongs to $P_n$, say $x\in V\left(P_n\right)$ and $y=w_1$. Then ${\sigma }_{T^{\prime }}\left(x\right)={\sigma }_S\left(x\right)+1$ and ${\sigma }_T\left(y\right)=p$. Since ${\sigma }_S\left(x\right)\in \{1,2\}$, it follows that ${\sigma }_{T^{\prime }}\left(x\right)\le 3$. Therefore, if $p\ge 4$, then ${\sigma }_{T^{\prime }}\left(x\right)\ne {\sigma }_{T^{\prime }}\left(y\right)$. There remains a need to consider a situation when $p\le 3$. Once again, the block structure of the generalized wheels needs to be investigated.

A block B of $C_n$ with respect to S is a maximal set of consecutive vertices of $C_n$ (or a maximal subpath of $C_n$) if either all vertices of B belong to S or no vertices of B belong to S. The cycle $C_n$ can therefore be expressed as a sequence $(B_1,B_1^{\prime },B_2,B_2^{\prime },\dots ,B_k,B_k^{\prime })$ of $2k$ blocks for some positive integer k such that $B_i\subseteq S$ and $B_i^{\prime }\cap S=\varnothing$ where $1\le i\le k$. The following claim can be verified by the argument used in the case of $W_n$.

Claim: 

For each integer i with $1\le i\le k$, $\left|B_i\right|\in \{1,3\}$ and $\left|B_i^{\prime }\right|\in \{1,3\}$.

First, we show that if n is even and $n\ge 14$, then $\left|S\right|=p\ge 4$. Assume, to the contrary, that $\left|S\right|\le 3$. By the claim, for every four consecutive vertices on $C_n$, there is at least one vertex in S and thus $\left|S\right|\ge 3$. Thus, $\left|S\right|=3$ and $\left|B_i\right|=1$ for every block $B_i$ in S. We may assume that $v_1\in S$. Since S contains at least three vertices in $\{v_2,v_4,\dots ,v_{13}\}$, it follows that $\left|S\right|\ge 4$, which is a contradiction. Therefore, if $n\ge 14$, then $\left|S\right|=p\ge 4$.

Thus, we may assume that n is even and $6\le n\le 12$. By Theorem 3, if $n=6,12$, then ${\gamma }_{pt}\left(W_n\right)=4$, while if $n=8,10$, then ${\gamma }_{pt}\left(W_n\right)=5$. We consider these two subcases.

Subcase $2.1$. $n=6,12$. We show that ${\gamma }_{pt}\left(W_{n,t}\right)=4$ for every integer $t\ge 2$. Assume, to the contrary, that ${\gamma }_{pt}\left(W_{n,t}\right)=3$ for some even integer $n\in \{6,12\}$ and an integer $t\ge 2$. Let T be a ${\gamma }_{pt}$-set of $W_{n,t}$. Since $W_{n,t}\left[T\right]$ is locally irregular, $W_{n,t}\left[T\right]=P_3$. If $T\cap V\left({\overline{K}}_t\right)=\varnothing$, then we may assume that $P_3=(v_1,v_2,v_3)$. However, $v_5$ is not dominated by any vertex of T, which is impossible. If $T\cap V\left({\overline{K}}_t\right)\ne \varnothing$, then we may assume that (i) $P_3=(v_1,w_1,v_j)$ where $3\le j\le 1+n/2$ or (ii) $P_3=(w_1,v_1,w_2)$. If (i) occurs, then ${\sigma }_T\left(w_1\right)={\sigma }_T\left(v_n\right)=2$ since $n\ge 6$, while if (ii) occurs, then ${\sigma }_T\left(v_3\right)={\sigma }_T\left(v_4\right)=2$. In either case, a contradiction is produced. Thus ${\gamma }_{pt}\left(W_{n,t}\right)=4$ if $n\in \{6,12\}$ and $t\ge 2$.

Subcase $2.2$. $n=8,10$. We show that ${\gamma }_{pt}\left(W_{n,t}\right)=5$ for every integer $t\ge 2$. Assume, to the contrary, that ${\gamma }_{pt}\left(W_{n,t}\right)\le 4$ for some integer $n\in \{8,10\}$ and an integer $t\ge 2$. Let T be a ${\gamma }_{pt}$-set of $W_{n,t}$. Then $\left|T\right|\in \{3,4\}$. The argument used in Subcase $2.1$ shows that $\left|T\right|\ne 3$. Hence, $\left|T\right|=4$ and $W_{n,t}\left[T\right]=K_{1,3}$. Thus, either (i) the center of $K_{1,3}$ is a vertex in $V\left({\overline{K}}_t\right)$ or (ii) the center of $K_{1,3}$ is a vertex of $C_n$ and $t\ge 3$. If (i) occurs, we may assume that $w_1$ is the center of $K_{1,3}$ where $T=\{w_1,v_1,v_i,v_j\}$ for $3\le i<j$. By Claim 2, it follows that $i\in \{3,5\}$ and $j-i\in \{2,4\}$. If $i=3$, then ${\sigma }_T\left(v_2\right)={\sigma }_T\left(w_1\right)=3$, which is impossible. Thus, $i=5$ and $j-i=4$ or $j=9$, which is impossible if $n=8$. If $n=10$, then ${\sigma }_T\left(v_{10}\right)={\sigma }_T\left(w_1\right)=3$, which is impossible. If (ii) occurs, we may assume that $V\left(K_{1,3}\right)=\{v_i,w_1,w_2,w_3\}$ and $1\le i\le n/2$. However, then ${\sigma }_T\left(v_{n-2}\right)={\sigma }_T\left(v_{n-1}\right)=3$, which too is impossible. □

5. Conclusions

In conclusion, we summarize the results obtained and state some open problems for further study.

(1)

Results on fans $F_n$ and wheels $W_n$ where $n\ge 3$.

⋆

The fan $F_n$ has a $pt$-dominating set if and only if $n\ne 4$. Furthermore, if $n\ge 3$ is an integer with $n\ne 4$, then ${\gamma }_{pt}\left(F_3\right)=3$, ${\gamma }_{pt}\left(F_n\right)=5$ for $5\le n\le 8$, ${\gamma }_{pt}\left(F_9\right)=4$, and for $n\ge 10$,

$${\gamma }_{pt}\left(F_n\right)=\left\{\begin{array}{cc}p+1\hfill & \mathrm{if} n\in \{4p,4p+1\} \mathrm{for} \mathrm{an} \mathrm{integer} p\ge 3\hfill \\ p+2\hfill & \mathrm{if} n\in \{4p+2,4p+3\} \mathrm{for} \mathrm{an} \mathrm{integer} p\ge 2.\hfill \end{array}\right.$$

⋆

The wheel $W_n$ has a $pt$-dominating set if and only if n is even. Furthermore, if $n\ge 4$ is an even integer, then ${\gamma }_{pt}\left(W_4\right)=3$, ${\gamma }_{pt}\left(W_6\right)=4$, ${\gamma }_{pt}\left(W_8\right)={\gamma }_{pt}\left(W_{10}\right)=5$, and for $n\ge 12$,

$${\gamma }_{pt}\left(W_n\right)=\left\{\begin{array}{cc}p+1& \mathrm{if} n=4p \mathrm{for} \mathrm{some} \mathrm{integer}p\ge 3\hfill \\ p+2& \mathrm{if} n=4p+2 \mathrm{for} \mathrm{some} \mathrm{integer}p\ge 3.\hfill \end{array}\right.$$

(2)

Results on generalized fans $F_{n,t}$ and wheels $W_{n,t}$ where $n\ge 3$ and $t\ge 2$.

⋆

The generalized fan $F_{n,t}$ has a $pt$-dominating set. Furthermore, if $3\le n\le 5$, then ${\gamma }_{pt}\left(F_{n,t}\right)=3$ and if $n\ge 6$, then ${\gamma }_{pt}\left(F_{n,t}\right)={\gamma }_{pt}\left(F_n\right)$.

⋆

The generalized wheel $W_{n,t}$ has a $pt$-dominating set if and only if n is even. Furthermore, if $n\ge 4$ is even, then ${\gamma }_{pt}\left(W_{n,t}\right)={\gamma }_{pt}\left(W_n\right)$.

A natural question here is to extend results obtained on generalized fans and wheels to more general classes of graphs. As mentioned earlier, it is challenging to determine whether a general graph possesses a $pt$-dominating set and to determine its $pt$-domination number. For example, while Theorem 1 provides a complete answer for all paths and cycles with a $pt$-dominating set as well as their $pt$-domination numbers, this is not the case for the Cartesian products of paths and cycles. On the other hand, with the aid of results on paths and cycles as well as the structure of $pt$-dominating sets in paths and cycles, a complete answer is obtained here for fans, wheels, and generalized fans and wheels. This observation suggests investigating the joins $T\vee {\overline{K}}_t$ of some well-known classes of trees T that are closely related to paths, and the joins $G\vee {\overline{K}}_t$ of some well-known classes of graphs G that are closely related to cycles. For example, for starlike trees (a subdivision of a star each of whose branches at the central vertex is a path), a characterization was established in [11] for every starlike tree possessing a $pt$-dominating set as well as the structures of $pt$-dominating sets. Another class of well-known trees is the caterpillars (a tree with the removal of all end-vertices results in a path), which have been investigated in [11]. A well-known class of graphs closely related to cycles is unicyclic graphs (connected graphs with exactly one cycle). Information on the join $G\vee K_t$ of a general graph G and ${\overline{K}}_t$ described in Theorem 4 and Lemma 1 might provide a potential approach to the investigation of $pt$-dominating sets in these graphs. Therefore, we close with the following problems.

Problem 1.

For a positive integer t, determine which graphs $G\vee {\overline{K}}_t$ possess a $pt$-dominating set, where G is a starlike tree, a caterpillar, or a unicyclic graph. For those graphs $G\vee {\overline{K}}_t$ possessing a $pt$-dominating set, determine its $pt$-domination number.

Of course, there is the more general problem of investigating proper total domination in the joins of any two graphs.