JDM Journal of Discrete Mathematics 2090-9845 2090-9837 Hindawi Publishing Corporation 10.1155/2015/512696 512696 Research Article Product Cordial and Total Product Cordial Labelings of P n + 1 m Gao Zhen-Bin 1 Sun Guang-Yi 1 Zhang Yuan-Ning 1 Meng Yu 1 Lau Gee-Choon 2 Das Kinkar Ch. 1 College of Science Harbin Engineering University Harbin 150001 China hrbeu.edu.cn 2 Faculty of Computer and Mathematical Sciences Universiti Teknologi MARA, Segamat Campus 85000 Johor Malaysia uitm.edu.my 2015 512015 2015 23 07 2014 30 09 2014 15 12 2014 5 1 2015 2015 Copyright © 2015 Zhen-Bin Gao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We proved that P n + 1 m is total product cordial. We also give sufficient conditions for the graph to admit (or not admit) a product cordial labeling.

1. Introduction

Let G ( V , E ) denote a simple and finite connected graph G with vertex set V and edge set E . Suppose f and f * denote a vertex and an edge labeling of a graph, respectively. Let v f ( i ) and e f * ( i ) denote the number of vertices and edges labeled with i { 0,1 } . In 2004, Sundaram et al.  introduced the notion of product cordial labelings.

Definition 1.

Let f : V { 0,1 } be a vertex labeling of a graph G that induces an edge labelings f * : E { 0,1 } such that f * ( u v ) = f ( u ) f ( v ) . One says f is a product cordial labeling if | v f ( 0 ) - v f ( 1 ) | 1 and | e f * ( 0 ) - e f * ( 1 ) | 1 . A graph G is called a product cordial graph if it admits a product cordial labeling.

Sundaram et al.  proved that many graphs are product cordial: trees; unicyclic graphs of odd order; triangular snakes; dragons; helms; P m P n ; C m P n if and only if m + n is odd; P m K 1 , n ; W m F n ; K 1 , m K 1 , n ; W m K 1 , n ; W m P n ; W m C n ; P n 2 if and only if n is odd; C n if and only if n < 4 . Kwong et al.  discussed product cordial index sets of 2 regular graphs. Kwong et al.  discussed product cordial index sets of cylinders.

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

Definition 2.

Let f : V { 0,1 } be a vertex labeling of a graph G that induces edge labelings f * : E { 0,1 } such that f * ( u v ) = f ( u ) f ( v ) . We say f is a total product cordial labeling if | ( v f ( 0 ) + e f * ( 0 ) ) - ( v f ( 1 ) + e f * ( 1 ) ) | 1 . A graph G 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 C 4 ; K n , 2 n - 1 ; C n with m edges appended at each vertex; fans; wheels; helms. In , 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 m th power of the path P n + 1 , denoted by P n + 1 m , which is defined as follows .

Definition 3.

Let P n + 1 denote a path of length n . The graph P n + 1 m is obtained from P n + 1 by adding edges that join all vertices u and v whose distance is m . The graph P n + 1 m is illustrated in Figure 1. P n + 1 m has ( 2 n - m + 1 ) edges.

Graph P n + 1 m .

In , the authors prove that all cycles except C 4 are total product cordial. Interested readers may refer to  for more results on product cordial labeling and total product cordial labeling.

Definition 4.

The degree of a vertex u , denoted by d ( u ) (or simply d ), is the number of edges incident to u .

2. Total Product Cordial Labeling

When m = n , P n + 1 n is the cycle C n + 1 and is total product cordial except C 4 [4, 5]. Hence, we discuss the total product cordial labelings of P n + 1 m for 2 m < n . We will let g ( i ) = v f ( i ) + e f * ( i ) for i = 0,1 .

Lemma 5.

For 2 m [ n + 1 / 2 ] , P n + 1 m is a total product cordial graph.

Proof.

Observe that d ( u 1 ) = d ( u n + 1 ) = 2 , d ( u i ) = 3 for i { 2,3 , , m , n - m + 2 , n - m + 3 , , n } and d ( u i ) = 4 for i { m + 1 , m + 2 , , n - m + 1 } . Also | V | + | E | = 3 n - m + 2 . Suppose n - m / 2 + n + 1 = 3 k when n - m is even (or n - m + 1 / 2 + n + 1 = 3 k when n - m is odd).

Case 1 ( n - m is even). A total product cordial labeling gives g ( 0 ) = g ( 1 ) = 3 k . Define f ( u i ) = 0 for i { 1,2 , , k } and f ( u i ) = 1 for i { k + 1 , k + 2 , , n + 1 } such that 3 k - 2 g ( 0 ) 3 k .

If 3 k = g ( 0 ) , we get g ( 0 ) = g ( 1 ) = 3 k . So f is a total product cordial labeling.

If 3 k - 1 g ( 0 ) , we consider the following cases.

Suppose 3 k = g ( 0 ) + 1 . If k m , we change the label of u 1 to 1 and the label of u n + 1 to 0. If k m + 1 , we change the label of u k to 1 and the label of u k + 1 to 0. In both possibilities, we get g ( 0 ) = g ( 1 ) , and f is a total product cordial labeling.

Suppose 3 k = g ( 0 ) + 2 . If k m , we change the label of u 1 to 1 and the label of u n to 0. If k m + 1 , we change the label of u 1 to 1 and the label of u n + 1 to 0. In both possibilities above, we get g ( 0 ) = g ( 1 ) , and f is a total product cordial labeling.

Case 2 ( n - m is odd). A total product cordial labeling gives 3 k - 1 g ( 0 ) g ( 1 ) = 3 k . Define f ( u i ) = 0 for i { 1,2 , , k } and f ( u i ) = 1 for i { k + 1 , k + 2 , , n + 1 } such that 3 k - 1 g ( 0 ) 3 k + 1 .

If 3 k - 1 = g ( 0 ) (or 3 k = g ( 0 ) ), then 3 k = g ( 1 ) (or 3 k - 1 = g ( 1 ) ), and f is a total product cordial labeling.

If 3 k + 1 = g ( 0 ) . We change the label of u 1 to 1 so that g ( 0 ) reduces by 1 or by 2. We then get 3 k - 1 g ( 0 ) 3 k , and f is a total product cordial labeling.

The proof is thus complete.

Lemma 6.

For [ n + 1 / 2 ] < m n - 1 , P n + 1 m is a total product cordial graph.

Proof.

Observe that d ( u i ) = 2 for i { 1 , n - m + 2 , n - m + 3 , , m , n + 1 } and d ( u i ) = 3 for i { 2,3 , , n - m + 1 , m + 1 , m + 2 , , n } . Note that if n + 1 = 2 m , then d ( u i ) = 2 if and only if i = 1 , n + 1 .

Case 1 ( n - m is even). A total product cordial labeling gives g ( 0 ) = g ( 1 ) = 3 k . Define f ( u i ) = 0 for i { 1,2 , , k } and f ( u i ) = 1 for i { k + 1 , k + 2 , , n + 1 } such that 3 k - 2 g ( 0 ) 3 k .

If 3 k = g ( 0 ) , we get g ( 0 ) = g ( 1 ) ; then f is a total product cordial labeling.

If 3 k - 1 = g ( 0 ) , then we consider the following cases.

If d ( u k ) = 2 , we change the label of u k + 1 to 0 and the label of u k to 1 ; g ( 0 ) is increased by 1 ; then g ( 0 ) = 3 k = g ( 1 ) ; then f is a total product cordial labeling.

If d ( u k ) = 3 , consider the following.

If d ( u k + 1 ) = 2 . We change the label of u k to 1 , because vertex u k is adjacent to vertex u n + 1 , so we define f ( u n + 1 ) = 0 , f ( u i ) = 1 for k + 1 i n ; g ( 0 ) is increased by 1 ; then g ( 0 ) = 3 k = g ( 1 ) ; then f is a total product cordial labeling.

If d ( u k + 1 ) = 3 , use the procedure in the proof of Lemma 5; we can obtain a total product cordial labeling.

If 3 k - 1 = g ( 0 ) + 1 , we consider the following cases.

If d ( u k ) = 2 , we change the label of u k + 1 to 0; g ( 0 ) is increased by 2 ; then g ( 0 ) = 3 k = g ( 1 ) . Hence, f is a total product cordial labeling.

If d ( u k ) = 3 , consider the following.

If d ( u k + 1 ) = 2 , we change the label of u k + 1 to 0; then g ( 0 ) = 3 k = g ( 1 ) .

If d ( u k + 1 ) = 3 , use the procedure in the proof of Lemma 5, and we can obtain a total product cordial labeling.

Case 2 ( n - m is odd). A total product cordial labeling gives 3 k - 1 g ( 0 ) g ( 1 ) = 3 k . Define f ( u i ) = 0 for for i { 1,2 , , k } and f ( u i ) = 1 for i { k + 1 , k + 2 , , n + 1 } such that 3 k - 1 g ( 0 ) 3 k + 1 .

If 3 k - 1 = g ( 0 ) (or 3 k = g ( 0 ) ), then 3 k = g ( 1 ) (or 3 k - 1 = g ( 1 ) ), and f is a total product cordial labeling.

If 3 k + 1 = g ( 0 ) , we change the label of u 1 to 1 so that g ( 0 ) reduces by 1 or by 2. We then get 3 k - 1 g ( 0 ) 3 k , and f is a total product cordial labeling.

The proof is thus complete.

Overall, we have the following.

Theorem 7.

P n + 1 m ( 2 m n ) is a total product cordial graph.

3. Product Cordial Labeling

When m = n , P n + 1 n is cycle C n + 1 and is product cordial if and only if n < 4 ; and P n + 1 2 if and only if n is even is product cordial . Hence, we discuss the product cordiality of P n + 1 m for 3 m < n .

Lemma 8.

In P n + 1 m , if d ( u i ) = 2 for i 1 , n + 1 , then the number of these vertices has the same parity with n + 1 , and these vertices induce a path P 2 m - n - 1 in the subgraph P n + 1 .

Proof.

Suppose d ( u i ) = 2 for i 1 , n + 1 ; then m > [ n + 1 / 2 ] . This implies that n - m + 2 i m . So, the number of these vertices is 2 m - n - 1 which has the same parity with n + 1 and induce a path P 2 m - n - 1 in the subgraph P n + 1 .

The proof is thus complete.

Lemma 9.

In a product cordial labeling of P n + 1 m , if min { e f * ( 0 ) } is attained, then one has that all the 0-vertices induce a path or are a disjoint union of two paths in the subgraph P n + 1 .

Proof.

Note that if a vertex is labeled 0, then all the incident edges must be labeled 0. Suppose min { e f * ( 0 ) } has been attained. Let v f ( 0 ) = k ; it is clear that the minimum number of 0-edges in the path P n + 1 is k or k + 1 if and only if the 0-vertices induced a path or are a disjoint union of two paths in the subgraph P n + 1 .

The proof is thus complete.

Lemma 10.

When 2 < m [ n + 1 / 2 ] , P n + 1 m is not product cordial.

Proof.

From Lemma 9 we know if P n + 1 m is product cordial, when v f ( 0 ) is got, then we get min { e f * ( 0 ) } only when all the 0-vertices induce a path or are a disjoint union of two paths in the subgraph P n + 1 . Now, we prove that P n + 1 m is not product cordial when 2 < m [ n + 1 / 2 ] . Because d ( u i ) > 2 for i 1 , n + 1 , we consider the following two cases.

n is odd; suppose P n + 1 m is product cordial; then v f ( 0 ) = n + 1 / 2 ; begin with u 1 (or u n + 1 ), and we choose n + 1 / 2 vertices and define their labels with 0 successively such that e f * ( 0 ) is minimum; min { e f * ( 0 ) } = 2 + 2 + + 2 = n + 1 , but max { e f * ( 1 ) } < n - 1 , e f * ( 0 ) e f * ( 1 ) , a contradiction. Hence, f is not a product cordial labeling.

n is even; suppose P n + 1 m is product cordial, and then min { v f ( 0 ) } = n / 2 ; similarly, we can get min { e f * ( 0 ) } = n , but max { e f * ( 1 ) } < n - 1 , contradiction. Hence, f is not a product cordial labeling.

The proof is thus complete.

When [ n + 1 / 2 ] < m < n , in P n + 1 m , there exist vertices u i such that i 1 , n + 1 and d ( u i ) = 2 . Let the number of such vertices be l - 2 , and then l = 2 m - n + 1 > 2 .

Lemma 11.

When [ n + 1 / 2 ] < m < n , P n + 1 m is not product cordial in the follow cases.

n is odd:

m is even satisfying

l = n + 1 / 2 , n = 11 , m = 8 ;

n + 1 / 2 - l 0,2 ( m o d 4 ) ( n + 1 / 2 - l > 0 ) , n + 1 / 2 < m < min { n + 5 / 2 , 3 n - 1 / 4 } ;

n + 1 / 2 - l 1,3 ( m o d 4 ) ( n + 1 / 2 - l > 0 ) , n + 1 / 2 < m < min { n + 7 / 2 , 3 n - 1 / 4 } ;

l - n + 1 / 2 > 0 , n = 5 , m = 4 .

m is odd satisfying

l = n + 1 / 2 , n = 7 , m = 5 ; n = 15 , m = 11 ;

n + 1 / 2 - l 0,2 ( m o d 4 ) ( n + 1 / 2 - l > 0 ) , n + 1 / 2 < m < min { n + 7 / 2 , 3 n - 1 / 4 } ;

n + 1 / 2 - l 1,3 ( m o d 4 ) ( n + 1 / 2 - l > 0 ) , n + 1 / 2 < m < min { n + 9 / 2 , 3 n - 1 / 4 } ;

l - n + 1 / 2 > 0 , n = 9 , m = 7 .

n is even:

m is even satisfying

l = n / 2 , n = 6 , m = 4 ;

n / 2 - l 0,2 ( m o d 4 ) ( n + 1 / 2 - l > 0 ) , n / 2 < m < min { n + 2 / 2 , 3 n - 2 / 4 } ;

n / 2 - l 1,3 ( m o d 4 ) ( n + 1 / 2 - l > 0 ) , n / 2 < m < min { n + 4 / 2 , 3 n - 2 / 4 } .

m is odd:

l = n / 2 , n = 10 , m = 7 ;

n / 2 - l 0,2 ( m o d 4 ) ( n + 1 / 2 - l > 0 ) , n / 2 < m < min { n + 4 / 2 , 3 n - 2 / 4 } ;

n / 2 - l 1,3 ( m o d 4 ) ( n + 1 / 2 - l > 0 ) , n / 2 < m < min { n + 6 / 2 , 3 n - 2 / 4 } .

Proof.

In Figure 2, the hollow circles denote the vertices that d = 2 ; the solid circles denote the vertices that d = 3 . In any product cordial labeling, if v ( 0 ) l - 2 , we first label vertices u i , n - m + 2 i m consecutively. Otherwise, we first label vertices u 1 and u n + 1 before labeling the remaining vertices according to the order given by the numeral 1,2 , 3 , . The value min { e f * ( 0 ) } is then attained.

Case 1 ( n is odd)

Subcase 1.1 ( m is even). We consider three cases.

Consider l = n + 1 / 2 . Since l = 2 m - n + 1 , we have m = 3 n - 1 / 4 , m > n + 1 / 2 , and we get n > 3 . We label all the n + 1 / 2 vertices with d = 2 by 0 ; then min { e f * ( 0 ) } = l + 3 . If l + 3 > n - m / 2 + 1 , we get m > n - 5 so that 3 < n < 19 . Since l is even, we must have n = 11 , m = 8 . Consequently, P 12 8 is not product cordial. Hence, (i) holds.

Consider n + 1 / 2 - l = k > 0 , and then m < 3 n - 1 / 4 . We consider the following four subcases.

k 0 ( mod 4 ) . We let the labels of n + 1 / 2 vertices be 0 , and then min { e f * ( 0 ) } = l + 2 + 6 × k / 4 . If l + 2 + 6 × k / 4 > n - m / 2 + 1 , then m < n + 5 / 2 . Hence, when n + 1 / 2 < m < min { n + 5 / 2 , 3 n - 1 / 4 } , P n + 1 m is not product cordial.

k 1 ( mod 4 ) . We let the labels of n + 1 / 2 vertices be 0 , and then min { e f * ( 0 ) } = l + 4 + 6 × k - 1 / 4 . If l + 4 + 6 × k - 1 / 4 > n - m / 2 + 1 , then m < n + 7 / 2 . Hence, when n + 1 / 2 < m < min { n + 7 / 2 , 3 n - 1 / 4 } , P n + 1 m is not product cordial.

k 2 ( mod 4 ) . We let the labels of n + 1 / 2 vertices be 0 , and then min { e f * ( 0 ) } = l + 5 + 6 × k - 2 / 4 . If l + 5 + 6 × k - 2 / 4 > n - m / 2 + 1 , then m < n + 5 / 2 . Hence, when n + 1 / 2 < m < min { n + 5 / 2 , 3 n - 1 / 4 } , P n + 1 m is not product cordial.

k 3 ( mod 4 ) . We let the labels of n + 1 / 2 vertices be 0 , and then min { e f * ( 0 ) } = l + 7 + 6 × k - 3 / 4 . If l + 7 + 6 × k - 3 / 4 > n - m / 2 + 1 , then m < n + 7 / 2 . Hence, when n + 1 / 2 < m < min { n + 7 / 2 , 3 n - 1 / 4 } , P n + 1 m is not product cordial.

Hence, (ii) and (iii) hold.

Consider n + 1 / 2 - l < 0 . If l - n + 1 / 2 = 1 , then we let the labels of n + 1 / 2 vertices be 0 , and min { e f * ( 0 ) } = n + 1 / 2 + 2 ; suppose n + 1 / 2 + 2 > n - m - 2 / 2 , then we have m > n - 3 , according to the parity of n , m , have m = n - 1 , because l - n + 1 / 2 = 1 and l = 2 m - n + 1 , and have n = 5 , m = 4 . If l - n + 1 / 2 > 1 , then let the labels of n + 1 / 2 vertices be 0 ; min { e f * ( 0 ) } = n + 1 / 2 + 1 , and suppose n + 1 / 2 + 2 > n - m - 1 / 2 , m > n - 1 , contradiction.

Hence, (iv) holds.

Subcase 1.2 ( m is odd). We consider three cases.

Consider l = n + 1 / 2 . Since l = 2 m - n + 1 , we have m = 3 n - 1 / 4 , m > n + 1 / 2 ; hence, we get n > 3 . We let the labels of n + 1 / 2 vertices be 0 , and then min { e f * ( 0 ) } = l + 3 . If l + 3 > n - m - 1 / 2 , then m > n - 6 . Hence, 3 < n < 23 . Since l is even, hence, only n = 7 , m = 5 ; n = 15 , m = 11 satisfy conditions that n is odd, 3 < n < 13 , m = 3 n - 1 / 4 , and m is odd. Hence, (i) holds.

Consider n + 1 / 2 - l = k > 0 . We have m < 3 n - 1 / 4 , and we consider the following four subcases.

k 0 ( mod 4 ) . We let the labels of n + 1 / 2 vertices be 0 ; then min { e f * ( 0 ) } = l + 2 + 6 × k / 4 . If l + 2 + 6 × k / 4 > n - m - 1 / 2 , then m < n + 7 / 2 . Hence, when n + 1 / 2 < m < min { n + 7 / 2 , 3 n - 1 / 4 } , P n + 1 m is not product cordial.

k 1 ( mod 4 ) . We let the labels of n + 1 / 2 vertices be 0 ; then min { e f * ( 0 ) } = l + 4 + 6 × k - 1 / 4 . If l + 4 + 6 × k - 1 / 4 > n - m - 1 / 2 , then m < n + 9 / 2 . Hence, when n + 1 / 2 < m < min { n + 9 / 2 , 3 n - 1 / 4 } , P n + 1 m is not product cordial.

k 2 ( mod 4 ) . We let the labels of n + 1 / 2 vertices be 0 ; then min { e f * ( 0 ) } = l + 5 + 6 × k - 2 / 4 . If l + 5 + 6 × k - 2 / 4 > n - m - 1 / 2 , then m < n + 7 / 2 . Hence, when n + 1 / 2 < m < min { n + 7 / 2 , 3 n - 1 / 4 } , P n + 1 m is not product cordial.

k 3 ( mod 4 ) . We let the labels of n + 1 / 2 vertices be 0 ; then min { e f * ( 0 ) } = l + 7 + 6 × k - 3 / 4 . If l + 7 + 6 × k - 3 / 4 > n - m - 1 / 2 , then m < n + 9 / 2 . Hence, when n + 1 / 2 < m < min { n + 9 / 2 , 3 n - 1 / 4 } , P n + 1 m is not product cordial.

Hence, (ii) and (iii) hold.

Consider n + 1 / 2 - l < 0 . If l - n + 1 / 2 = 1 , then we let n + 1 / 2 vertex labels be 0 , and min { e f * ( 0 ) } = n + 1 / 2 + 2 ; suppose n + 1 / 2 + 2 > n - m - 1 / 2 , then we have m > n - 4 , according to the odd-even parity of n , m , have m = n - 2 , because l - n + 1 / 2 = 1 and l = 2 m - n + 1 , and have m = 3 n + 1 / 4 , and therefore, n = 9 , m = 7 ; if l - n + 1 / 2 > 1 , then let the labels of n + 1 / 2 vertices be 0 , and min { e f * ( 0 ) } = n + 1 / 2 + 1 ; suppose n + 1 / 2 + 1 > n - m - 1 / 2 , m > n - 2 , contradiction.

Hence, (iv) holds.

Case 2 ( n is even)

Subcase 2.1 ( m is even)

Consider l = n / 2 . Since l = 2 m - n + 1 , we have m = 3 n - 2 / 4 , m > n / 2 and hence have n > 4 . We let n / 2 vertices be labeled by 0 ; then min { e f * ( 0 ) } = l + 3 . If l + 3 > n - m / 2 + 1 , m > n - 4 , then 4 < n < 14 . Since l is even, hence, only n = 6 , m = 4 satisfy conditions that n is odd, 4 < n < 14 , m = 3 n - 2 / 4 , and m is even. Hence, (i) holds.

Consider n / 2 - l = k > 0 . We have m < 3 n - 2 / 4 , and we consider the following four cases.

k 0 ( mod 4 ) . We let the labels of n / 2 vertices be 0 , and then min { e f * ( 0 ) } = l + 2 + 6 × k / 4 . If l + 2 + 6 × k / 4 > n - m / 2 + 1 , then m < n + 2 / 2 . Hence, when n / 2 < m < min { n + 2 / 2 , 3 n - 2 / 4 } , P n + 1 m is not product cordial.

k 1 ( mod 4 ) . We let the labels of n / 2 vertices be 0 , and then min { e f * ( 0 ) } = l + 4 + 6 × k - 1 / 4 . If l + 4 + 6 × k - 1 / 4 > n - m / 2 + 1 , then m < n + 4 / 2 . Hence, when n / 2 < m < min { n + 4 / 2 , 3 n - 2 / 4 } , P n + 1 m is not product cordial.

k 2 ( mod 4 ) . We let the labels of n / 2 vertices be 0 , and then min { e f * ( 0 ) } = l + 5 + 6 × k - 2 / 4 . If l + 5 + 6 × k - 2 / 4 > n - m / 2 + 1 , then m < n + 2 / 2 . Hence, when n / 2 < m < min { n + 2 / 2 , 3 n - 2 / 4 } , P n + 1 m is not product cordial.

k 3 ( mod 4 ) . We let the labels of n / 2 vertices be 0 , and then min { e f * ( 0 ) } = l + 7 + 6 × k - 3 / 4 . If l + 7 + 6 × k - 3 / 4 > n - m / 2 + 1 , then m < n + 4 / 2 . Hence, when n / 2 < m < min { n + 4 / 2 , 3 n - 2 / 4 } , P n + 1 m is not product cordial.

Hence, (ii) and (iii) hold.

Consider n / 2 - l < 0 . If l - n / 2 = 1 , we let n / 2 vertices be labeled by 0 , and min { e f * ( 0 ) } = n / 2 + 2 ; suppose n / 2 + 2 > n - m - 2 / 2 , m > n - 2 , according to the odd-even parity of n , m , contradiction; if l - n + 1 / 2 > 1 , let n / 2 vertices be labeled by 0 ; min { e f * ( 0 ) } = n / 2 + 1 ; suppose n / 2 + 1 > n - m - 2 / 2 , m > n , contradiction.

Hence, (iv) holds.

Subcase 2.2 ( m is odd)

Consider l = n / 2 . Since l = 2 m - n + 1 , then m = 3 n - 2 / 4 , m > n / 2 , and hence n > 4 . We let n / 2 vertices be labeled by 0 ; then min { e f * ( 0 ) } = l + 3 . If l + 3 > n - m - 1 / 2 , m > n - 5 , so 4 < n < 18 , because l is odd; hence, only n = 10 , m = 7 satisfy conditions that n is odd, 3 < n < 13 , m = 3 n - 1 / 4 , and m is odd. Hence, (i) holds.

Consider n / 2 - l = k > 0 . We have m < 3 n - 2 / 4 , and we consider the following four cases.

k 0 ( mod 4 ) . We let n / 2 vertices be labeled by 0 ; then min { e f * ( 0 ) } = l + 2 + 6 × k / 4 . If l + 2 + 6 × k / 4 > n - m - 1 / 2 , then m < n + 4 / 2 . Hence, when n / 2 < m < min { n + 4 / 2 , 3 n - 2 / 4 } , P n + 1 m is not product cordial.

k 1 ( mod 4 ) . We let n / 2 vertices be labeled by 0 ; then min { e f * ( 0 ) } = l + 4 + 6 × k - 1 / 4 . If l + 4 + 6 × k - 1 / 4 > n - m - 1 / 2 , then m < n + 6 / 2 . Hence, when n / 2 < m < min { n + 6 / 2 , 3 n - 2 / 4 } , P n + 1 m is not product cordial.

k 2 ( mod 4 ) . We let n / 2 vertices be labeled by 0 ; then min { e f * ( 0 ) } = l + 5 + 6 × k - 2 / 4 . If l + 5 + 6 × k - 2 / 4 > n - m - 1 / 2 , then m < n + 4 / 2 . Hence, when n / 2 < m < min { n + 4 / 2 , 3 n - 2 / 4 } , P n + 1 m is not product cordial.

k 3 ( mod 4 ) . We let n / 2 vertices be labeled by 0 ; then min { e f * ( 0 ) } = l + 7 + 6 × k - 3 / 4 . If l + 7 + 6 × k - 3 / 4 > n - m - 1 / 2 , then m < n + 6 / 2 . Hence, when n / 2 < m < min { n + 6 / 2 , 3 n - 2 / 4 } , P n + 1 m is not product cordial.

Hence, (ii) and (iii) hold.

Consider n / 2 - l < 0 . If l - n / 2 = 1 , we let n / 2 vertices be labeled by 0 ; min { e f * ( 0 ) } = n / 2 + 2 ; suppose n / 2 + 2 > n - m - 1 / 2 , and then m > n - 3 , according to the parity of n , m , contradiction; if l - n / 2 > 1 , we let n / 2 vertices be labeled by 0 ; then min { e f * ( 0 ) } = n / 2 + 1 , and suppose n / 2 + 1 > n - m - 1 / 2 , m > n - 1 , contradiction. Hence, (iv) holds.

The proof is thus complete.

1, 2, 3, and so forth denote the order of vertices with d = 3 being labeled 0.

Lemma 12.

When [ n + 1 / 2 ] < m < n and n , m do not belong to the cases of Lemma 11, P n + 1 m is product cordial.

Proof.

We prove the conclusion is right when n is odd and m is even, and the proofs of the other cases are similar to the proof of the facts that n is odd and m is even. Suppose n , m do not belong to the conditions of Lemma 11; let n + 1 / 2 vertices be labeled by 0 , and then min { e f * ( 0 ) } n - m - 2 / 2 . If min { e f * ( 0 ) } = n - m - 2 / 2 , then the conclusion is right. So, we discuss the conclusion when min { e f * ( 0 ) } < n - m - 2 / 2 . First, we define f ( u i ) = 0 for 1 i n + 1 / 2 right now; e f * ( 0 ) > n - m - 2 / 2 . From Lemma 8, we know that u i exist for i 1 , n + 1 with d = 2 ; suppose the number of these vertices is l - 2 and satisfy l = 2 m - n + 1 > 2 . We consider the following three cases.

Case 1. Consider l = n + 1 / 2 , and then m = 3 n - 1 / 4 ; let the labels of vertices that d = 2 be 0 and the labels of vertices with d = 3 be 1; then e f * ( 0 ) = n + 1 / 2 + 3 < n - m - 2 / 2 , exchanging the labels of u m , u 2 ; then e f * ( 0 ) is increased by 1 , if e f * ( 0 ) < n - m - 2 / 2 , exchanging the labels of u m - 1 , u 3 ; then e f * ( 0 ) is increased by 1 , after that, because e f * ( 0 ) > n - m - 2 / 2 when f ( u i ) = 0 for 1 i n + 1 / 2 . Hence, after the above some exchanging step, e f * ( 0 ) = n - m - 2 / 2 .

Case 2. Consider n + 1 / 2 - l > 0 ; then m < 3 n - 1 / 4 , and suppose n + 1 / 2 - l = k , then let the labels of vertices with d = 2 be 0 , according to the sequence illustrated by Figure 2, the k vertices with d = 3 labeled with 0 , and other vertices labeled with 1 ; then e f * ( 0 ) < n - m - 2 / 2 . First exchanging the labels of u i , u j with d = 3 , the rules are

i < n + 1 / 2 < j ;

f ( u i ) = f ( u i + 1 ) = 1 , f ( u s ) = 0 for s < i , f ( u j - 1 ) = f ( u j ) = 0 , f ( u t ) = 1 for j < t < n + 1 .

After each exchange, e f * ( 0 ) is increased by 1 , if some exchanging is completed; e f * ( 0 ) = n - m - 2 / 2 , and then exchange stops; when the above exchanges all are completed, e f * ( 0 ) < n - m - 2 / 2 , and then according to the exchanges of Case 1, we can get e f * ( 0 ) = n - m - 2 / 2 .

Case 3. Consider n + 1 / 2 - l < 0 , if l = n + 1 / 2 + 1 ; let the vertices (except u n + 1 ) with d = 2 be labeled by 0 and other vertices labeled by 1 ; then e f * ( 0 ) < n - m - 2 / 2 , and according to exchanges of Case 1, we can get e f * ( 0 ) = n - m - 2 / 2 ; if l > n + 1 / 2 + 1 , let f ( u 1 ) = 0 , f ( u i ) = 0 for i { m - n + 1 / 2 + 2 , m - n + 1 / 2 + 3 , , m } , and according to the exchanges of Case 1, we can get e f * ( 0 ) = n - m - 2 / 2 .

The proof is thus complete.

Overall, we can get the following.

Theorem 13.

P n + 1 m is total product cordial when n , m 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.

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