#### Abstract

We proved that is total product cordial. We also give sufficient conditions for the graph to admit (or not admit) a product cordial labeling.

#### 1. Introduction

Let denote a simple and finite connected graph with vertex set and edge set . Suppose and denote a vertex and an edge labeling of a graph, respectively. Let and denote the number of vertices and edges labeled with . In 2004, Sundaram et al. [1] introduced the notion of product cordial labelings.

*Definition 1. *Let be a vertex labeling of a graph that induces an edge labelings such that . One says is a product cordial labeling if and . A graph is called a product cordial graph if it admits a product cordial labeling.

Sundaram et al. [1] proved that many graphs are product cordial: trees; unicyclic graphs of odd order; triangular snakes; dragons; helms; ; if and only if is odd; ; ; ; ; ; ; if and only if is odd; if and only if . Kwong et al. [2] discussed product cordial index sets of 2 regular graphs. Kwong et al. [3] discussed product cordial index sets of cylinders.

In 2006, Sundaram et al. [4] introduced the notion of total product cordial labelings.

*Definition 2. *Let be a vertex labeling of a graph that induces edge labelings such that . We say is a total product cordial labeling if . A graph is called a total product cordial graph if it admits a total product cordial labeling.

Sundaram et al. [4, 5] also proved that graphs are total product cordial: every product cordial graph of even order or odd order and even size; trees; all cycles except ; ; with edges appended at each vertex; fans; wheels; helms. In [6], Ramanjaneyulu et al. proved that a family of planar graphs for which each face is a 4-cycle admit a total product cordial labeling.

In this paper, we determine the product cordiality and total product cordiality of the th power of the path , denoted by , which is defined as follows [7].

*Definition 3. *Let denote a path of length . The graph is obtained from by adding edges that join all vertices and whose distance is . The graph is illustrated in Figure 1. has edges.

In [5], the authors prove that all cycles except are total product cordial. Interested readers may refer to [8] for more results on product cordial labeling and total product cordial labeling.

*Definition 4. *The degree of a vertex , denoted by (or simply ), is the number of edges incident to .

#### 2. Total Product Cordial Labeling

When , is the cycle and is total product cordial except [4, 5]. Hence, we discuss the total product cordial labelings of for . We will let for .

Lemma 5. *For , is a total product cordial graph.*

*Proof. *Observe that , for and for . Also . Suppose when is even (or when is odd).*Case **1 (** is even)*. A total product cordial labeling gives . Define for and for such that .

If , we get . So is a total product cordial labeling.

If , we consider the following cases.

Suppose . If , we change the label of to 1 and the label of to 0. If , we change the label of to 1 and the label of to 0. In both possibilities, we get , and is a total product cordial labeling.

Suppose . If , we change the label of to 1 and the label of to 0. If , we change the label of to 1 and the label of to 0. In both possibilities above, we get , and is a total product cordial labeling.*Case **2 (** is odd)*. A total product cordial labeling gives . Define for and for such that .

If (or ), then (or ), and is a total product cordial labeling.

If . We change the label of to 1 so that reduces by 1 or by 2. We then get , and is a total product cordial labeling.

The proof is thus complete.

Lemma 6. *For , is a total product cordial graph.*

*Proof. *Observe that for and for . Note that if , then if and only if .*Case **1 (** is even)*. A total product cordial labeling gives . Define for and for such that .

If , we get ; then is a total product cordial labeling.

If , then we consider the following cases.(1)If , we change the label of to 0 and the label of to ; is increased by ; then ; then is a total product cordial labeling.(2)If , consider the following.(a)If . We change the label of to , because vertex is adjacent to vertex , so we define , for ; is increased by ; then ; then is a total product cordial labeling.(b)If , use the procedure in the proof of Lemma 5; we can obtain a total product cordial labeling.

If , we consider the following cases.(1)If , we change the label of to 0; is increased by ; then . Hence, is a total product cordial labeling.(2)If , consider the following.(a)If , we change the label of to 0; then .(b)If , use the procedure in the proof of Lemma 5, and we can obtain a total product cordial labeling.*Case **2 (** is odd)*. A total product cordial labeling gives . Define for for and for such that .

If (or ), then (or ), and is a total product cordial labeling.

If , we change the label of to 1 so that reduces by 1 or by 2. We then get , and is a total product cordial labeling.

The proof is thus complete.

Overall, we have the following.

Theorem 7. * is a total product cordial graph.*

#### 3. Product Cordial Labeling

When , is cycle and is product cordial if and only if ; and if and only if is even is product cordial [1]. Hence, we discuss the product cordiality of for .

Lemma 8. *In , if for , then the number of these vertices has the same parity with , and these vertices induce a path in the subgraph .*

*Proof. *Suppose for ; then . This implies that . So, the number of these vertices is which has the same parity with and induce a path in the subgraph .

The proof is thus complete.

Lemma 9. *In a product cordial labeling of , if is attained, then one has that all the 0-vertices induce a path or are a disjoint union of two paths in the subgraph .*

*Proof. *Note that if a vertex is labeled 0, then all the incident edges must be labeled 0. Suppose has been attained. Let ; it is clear that the minimum number of 0-edges in the path is or if and only if the 0-vertices induced a path or are a disjoint union of two paths in the subgraph .

The proof is thus complete.

Lemma 10. *When , is not product cordial.*

*Proof. *From Lemma 9 we know if is product cordial, when is got, then we get only when all the 0-vertices induce a path or are a disjoint union of two paths in the subgraph . Now, we prove that is not product cordial when . Because for , we consider the following two cases.(1) is odd; suppose is product cordial; then ; begin with (or ), and we choose vertices and define their labels with successively such that is minimum; , but , , a contradiction. Hence, is not a product cordial labeling.(2) is even; suppose is product cordial, and then ; similarly, we can get , but , contradiction. Hence, is not a product cordial labeling.

The proof is thus complete.

When , in , there exist vertices such that and . Let the number of such vertices be , and then .

Lemma 11. *When , is not product cordial in the follow cases.*(1)* is odd:(a) is even satisfying(i), , ;(ii), ;(iii), ;(iv), , .(b) is odd satisfying(i), , ; , ;(ii), ;(iii), ;(iv), , .*(2)

*is even:(a)*

*is even satisfying(i)*(b)*, , ;*(ii)*, ;*(iii)*, .**is odd:(i)**, , ;*(ii)*, ;*(iii)*, .**Proof. *In Figure 2, the hollow circles denote the vertices that ; the solid circles denote the vertices that . In any product cordial labeling, if , we first label vertices consecutively. Otherwise, we first label vertices and before labeling the remaining vertices according to the order given by the numeral . The value is then attained.*Case **1 (** is odd)**Subcase **1.1 (** is even)*. We consider three cases.(1)Consider . Since , we have , , and we get . We label all the vertices with by ; then . If , we get so that . Since is even, we must have , . Consequently, is not product cordial. Hence, (i) holds.(2)Consider , and then . We consider the following four subcases.(2.1). We let the labels of vertices be , and then . If , then . Hence, when , is not product cordial.(2.2). We let the labels of vertices be , and then . If , then . Hence, when , is not product cordial.(2.3). We let the labels of vertices be , and then . If , then . Hence, when , is not product cordial.(2.4). We let the labels of vertices be , and then . If , then . Hence, when , is not product cordial.

Hence, (ii) and (iii) hold.(3)Consider . If , then we let the labels of vertices be , and ; suppose , then we have , according to the parity of , have , because and , and have , . If , then let the labels of vertices be ; , and suppose , , contradiction.

Hence, (iv) holds.*Subcase **1.2 (** is odd)*. We consider three cases.(1)Consider . Since , we have , ; hence, we get . We let the labels of vertices be , and then . If , then . Hence, . Since is even, hence, only , ; , satisfy conditions that is odd, , , and is odd. Hence, (i) holds.(2)Consider . We have , and we consider the following four subcases.(2.1). We let the labels of vertices be ; then . If , then . Hence, when , is not product cordial.(2.2). We let the labels of vertices be ; then . If , then . Hence, when , is not product cordial.(2.3). We let the labels of vertices be ; then . If , then . Hence, when , is not product cordial.(2.4). We let the labels of vertices be ; then . If , then . Hence, when , is not product cordial.

Hence, (ii) and (iii) hold.(3)Consider . If , then we let vertex labels be , and ; suppose , then we have , according to the odd-even parity of , have , because and , and have , and therefore, , ; if , then let the labels of vertices be , and ; suppose , , contradiction.

Hence, (iv) holds.*Case **2 (** is even)**Subcase **2.1 (** is even)*(1)Consider . Since , we have , and hence have . We let vertices be labeled by ; then . If , , then . Since is even, hence, only , satisfy conditions that is odd, , , and is even. Hence, (i) holds.(2)Consider . We have , and we consider the following four cases.(2.1). We let the labels of vertices be , and then . If , then . Hence, when , is not product cordial.(2.2). We let the labels of vertices be , and then . If , then . Hence, when , is not product cordial.(2.3). We let the labels of vertices be , and then . If , then . Hence, when , is not product cordial.(2.4). We let the labels of vertices be , and then . If , then . Hence, when , is not product cordial.

Hence, (ii) and (iii) hold.(3)Consider . If , we let vertices be labeled by , and ; suppose , , according to the odd-even parity of , contradiction; if , let vertices be labeled by ; ; suppose , , contradiction.

Hence, (iv) holds.*Subcase **2.2 (** is odd)*(1)Consider . Since , then , , and hence . We let vertices be labeled by ; then . If , , so , because is odd; hence, only , satisfy conditions that is odd, , , and is odd. Hence, (i) holds.(2)Consider . We have , and we consider the following four cases.(2.1). We let vertices be labeled by ; then . If , then . Hence, when , is not product cordial.(2.2). We let vertices be labeled by ; then . If , then . Hence, when , is not product cordial.(2.3). We let vertices be labeled by ; then . If , then . Hence, when , is not product cordial.(2.4). We let vertices be labeled by ; then . If , then . Hence, when , is not product cordial.

Hence, (ii) and (iii) hold.(3)Consider . If , we let vertices be labeled by ; ; suppose , and then , according to the parity of , contradiction; if , we let vertices be labeled by ; then , and suppose , , contradiction. Hence, (iv) holds.

The proof is thus complete.

Lemma 12. *When and do not belong to the cases of Lemma 11, is product cordial.*

*Proof. *We prove the conclusion is right when is odd and is even, and the proofs of the other cases are similar to the proof of the facts that is odd and is even. Suppose do not belong to the conditions of Lemma 11; let vertices be labeled by , and then . If , then the conclusion is right. So, we discuss the conclusion when . First, we define for right now; . From Lemma 8, we know that exist for with ; suppose the number of these vertices is and satisfy . We consider the following three cases.*Case **1*. Consider , and then ; let the labels of vertices that be and the labels of vertices with be 1; then , exchanging the labels of ; then is increased by , if , exchanging the labels of ; then is increased by , after that, because when for . Hence, after the above some exchanging step, .*Case **2*. Consider ; then , and suppose , then let the labels of vertices with be , according to the sequence illustrated by Figure 2, the vertices with labeled with , and other vertices labeled with ; then . First exchanging the labels of with , the rules are(1);(2), for , , for .

After each exchange, is increased by , if some exchanging is completed; , and then exchange stops; when the above exchanges all are completed, , and then according to the exchanges of Case 1, we can get .*Case **3*. Consider , if ; let the vertices (except ) with be labeled by and other vertices labeled by ; then , and according to exchanges of Case 1, we can get ; if , let , for , and according to the exchanges of Case 1, we can get .

The proof is thus complete.

Overall, we can get the following.

Theorem 13. * is total product cordial when do not belong to the cases of Lemmas 10 and 11.*

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The authors would like to thank the anonymous referees very much for valuable suggestions, corrections, and comments, which result in a great improvement of the original paper. The work is supported by NSFC Grant no. 11371109.